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Cambridge
International AS & A Level
Mathematics
Pure
Mathematics 2 & 3
Second edition
Sophie Goldie
Series editor: Roger Porkess
i
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Questions from the Cambridge International AS & A Level Mathematics papers are reproduced by
permission of Cambridge Assessment International Education. Unless otherwise acknowledged, the
questions, example answers, and comments that appear in this book were written by the authors.
Cambridge Assessment International Education bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.
®IGCSE is a registered trademark.
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Much of the material in this book was published originally as part of the MEI Structured Mathematics
series. It has been carefully adapted to support the Cambridge International AS & A Level Mathematics
syllabus. The original MEI author team for Pure Mathematics comprised Catherine Berry, Val Hanrahan,
Terry Heard, David Martin, Jean Matthews, Roger Porkess and Peter Secker.
© Roger Porkess and Sophie Goldie 2018
First published in 2012
This second edition published in 2018 by
Hodder Education, an Hachette UK company,
Carmelite House, 50 Victoria Embankment,
London EC4Y 0DZ
Impression number
5
Year 2022
2020
2021
4
3
2
2019
1
2018
All rights reserved. Apart from any use permitted under UK copyright law, no part of this publication
may be reproduced or transmitted in any form or by any means, electronic or mechanical, including
photocopying and recording, or held within any information storage and retrieval system, without
permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited.
Further details of such licences (for reprographic reproduction) may be obtained from the Copyright
Licensing Agency Limited, www.cla.co.uk.
Cover photo by Shutterstock/chatgunner
Illustrations by Pantek Media Maidstone, Kent and Integra Software Services Pvt. Ltd., Pondicherry, India
Typeset in Bembo Std 11/13 pt by Integra Software Services Pvt. Ltd., Pondicherry, India
Printed in Italy
A catalogue record for this title is available from the British Library
ISBN 9781510421738
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Contents
Note
Chapters 1–6 cover the Paper 2 (AS Level) syllabus content.
Chapters 1–11 cover the Paper 3 (A Level) syllabus content.
Introduction
How to use this book
The Cambridge International AS & A Level Mathematics
9709 syllabus
1 Algebra
1.1
1.2
1.3
Operations with polynomials
Solution of polynomial equations
The modulus function
2 Logarithms and exponentials
2.1
2.2
2.3
2.4
2.5
2.6
Exponential functions
Logarithms
Graphs of logarithms
Modelling curves
The natural logarithm function
The exponential function
vi
vii
x
1
2
7
17
24
25
28
33
37
45
48
3 Trigonometry
57
3.1
3.2
3.3
3.4
3.5
58
61
67
72
79
Reciprocal trigonometrical functions
Compound-angle formulae
Double-angle formulae
The forms r cos(θ ± α ), r sin(θ ± α )
The general solutions of trigonometrical equations
4 Differentiation
4.1
4.2
4.3
4.4
4.5
4.6
4.7
The product rule
The quotient rule
Differentiating natural logarithms and exponentials
Differentiating trigonometrical functions
Differentiating functions defined implicitly
Parametric equations
Parametric differentiation
83
84
86
90
96
101
108
112
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5 Integration
121
5.1
5.2
5.3
5.4
121
122
129
133
P2
Integrals involving the exponential function
Integrals involving the natural logarithm function
Integrals involving trigonometrical functions
Numerical integration
6 Numerical solution of equations
6.1
6.2
6.3
Interval estimation − change-of-sign methods
Fixed-point iteration
Problems with the fixed-point iteration method
7 Further algebra
7.1
7.2
7.3
7.4
The general binomial expansion
Review of algebraic fractions
Partial fractions
Using partial fractions with the binomial expansion
8 Further calculus
8.1
8.2
8.3
8.4
8.5
8.6
8.7
Differentiating tan–1 x
Integration by substitution
Integrals involving exponentials and natural logarithms
Integrals involving trigonometrical functions
The use of partial fractions in integration
Integration by parts
General integration
9 Differential equations
9.1
9.2
Forming differential equations from rates of change
Solving differential equations
10 Vectors
10.1 Vectors in two dimensions
10.2 Vectors in three dimensions
10.3 Vector calculations
10.4 The angle between two vectors
10.5 The vector equation of a line
10.6 The intersection of two lines
10.7 The angle between two lines
10.8 The perpendicular distance from a point to a line
144
146
151
156
162
163
172
174
180
184
184
186
191
194
199
203
212
217
218
223
234
235
236
240
250
258
265
272
273
iv
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11 Complex numbers
11.1 Extending the number system
11.2 Working with complex numbers
11.3 Sets of points in an Argand diagram
11.4 The modulus−argument form of complex numbers
11.5 Sets of points using the polar form
11.6 Working with complex numbers in polar form
11.7 Complex exponents
11.8 Complex numbers and equations
Answers
Index
280
281
282
294
297
304
307
310
313
319
355
v
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Introduction
This is one of a series of five books supporting the Cambridge International
AS & A Level Mathematics 9709 syllabus for examination from 2020. This
book follows on from Pure Mathematics 1. AS Level students should work
through the first six chapters, which cover the mathematics required for the
Paper 2 examination, and A Level students should work through all eleven
chapters for the Paper 3 examination. This part of the series also contains a
book for mechanics and two books for probability and statistics. The series
then continues with four more books supporting Cambridge International
AS & A Level Further Mathematics 9231.
These books are based on the highly successful series for the Mathematics
in Education and Industry (MEI) syllabus in the UK but they have been
redesigned and revised for Cambridge International students; where
appropriate, new material has been written and the exercises contain many
past Cambridge International examination questions. An overview of the
units making up the Cambridge International syllabus is given on the
following pages.
Throughout the series, the emphasis is on understanding the mathematics as
well as routine calculations. The various exercises provide plenty of scope for
practising basic techniques; they also contain many typical examination-style
questions.
The original MEI author team would like to thank Sophie Goldie who has
carried out the extensive task of presenting their work in a suitable form for
Cambridge International students and for her many original contributions.
They would also like to thank Cambridge Assessment International
Education for its detailed advice in preparing the books and for permission
to use many past examination questions.
Roger Porkess
Series editor
vi
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How to use this book
The structure of the book
This book has been endorsed by Cambridge Assessment International
Education. It is listed as an endorsed textbook for students taking the
Cambridge International AS & A Level Mathematics 9709 syllabus. The Pure
Mathematics 2 and 3 syllabus content is covered comprehensively and is
presented across eleven chapters, offering a structured route through the course.
If you are studying AS Mathematics, you should work through the first six
chapters, while for A Level you should work through all eleven chapters. This is
indicated on each chapter number by either showing P2 and P3, or P3 only.
The book is written on the assumption that you have covered and
understood the content of the Cambridge IGCSE® Mathematics 0580
(Extended curriculum) or Cambridge International O Level 4024/4029. Two
icons are used to indicate material that is not directly on the syllabus.
b Some of the early material is designed to provide an overlap and this
is designated background.You need to be familiar with the material
before you move on to develop it further.
e There are also places where the book goes beyond the requirements
of the syllabus to show how the ideas can be taken further or where
fundamental underpinning work is explored. Such work is marked as
extension.
Each chapter is broken down into several sections, with each section covering
a single topic. Topics are introduced through explanations, with key terms
picked out in red. These are reinforced with plentiful worked examples,
punctuated with commentary, to demonstrate methods and illustrate
application of the mathematics under discussion.
Regular exercises allow you to apply what you have learned.They offer a large
variety of practice and higher-order question types that map to the key concepts
of the Cambridge International syllabus. Look out for the following icons:
PS Problem-solving questions will help you to develop the ability
to analyse problems, recognise how to represent different situations
mathematically, identify and interpret relevant information, and select
appropriate methods.
M Modelling questions provide you with an introduction to the
important skill of mathematical modelling. In this, you take an everyday
or workplace situation, or one that arises in your other subjects, and
present it in a form that allows you to apply mathematics to it.
CP Communication and proof questions encourage you to become
a more fluent mathematician, giving you scope to communicate your
work with clear, logical arguments and to justify your results.
IGCSE® is a registered trademark.
9781510421738.indb 7
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Exercises also include questions you are likely to meet from real Cambridge
Assessment International Education past papers, so that you can become
familiar with the types of questions you will meet in formal assessments.
Answers to exercise questions, excluding long explanations and proofs,
are included in the back of the book, so you can check your work. It is
important, however, that you have a go at answering the questions before
looking up the answers if you are to understand the mathematics fully.
In addition to the exercises, a range of additional features are included to
enhance your learning.
ACTIVITY
Activities invite you to do some work for yourself, typically to introduce
you to ideas that are then going to be taken further. In some places,
activities are also used to follow up work that has just been covered.
PS
PROBLEM SOLVING
Mathematics provides you with the techniques to answer many standard
questions, but it also does much more than that: it helps you to develop
the capacity to analyse problems and to decide for yourself what methods
and techniques you will need to use. Questions and situations where this is
particularly relevant are highlighted as problem-solving tasks.
INVESTIGATION
In real life, it is often the case that as well as analysing a situation or
problem, you also need to carry out some investigative work. This allows you
to check whether your proposed approach is likely to be fruitful or to work
at all, and whether it can be extended. Such opportunities are marked as
investigations.
Other helpful features include the following.
? This symbol highlights points it will benefit you to discuss with
your teacher or fellow students, to encourage deeper exploration and
mathematical communication. If you are working on your own, there
are answers in the back of the book.
This is a warning sign. It is used where a common mistake,
misunderstanding or tricky point is being described to prevent you
from making the same error.
A variety of notes are included to offer advice or spark your interest:
Note
Notes expand on the topic under consideration and explore the deeper
lessons that emerge from what has just been done.
Historical note
Historical notes offer interesting background information about famous
mathematicians or results to engage you in this fascinating field.
viii
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Technology note
Although graphical calculators and computers are not permitted in the
examinations for this Cambridge International syllabus, we have included
Technology notes to indicate places where working with them can be helpful
for learning and for teaching.
Finally, each chapter ends with the key points covered, plus a list of the
learning outcomes that summarise what you have learned in a form that is
closely related to the syllabus.
Digital support
Comprehensive online support for this book, including further questions,
is available by subscription to MEI’s Integral® online teaching and learning
platform for AS & A Level Mathematics and Further Mathematics,
integralmaths.org. This online platform provides extensive, high-quality
resources, including printable materials, innovative interactive activities, and
formative and summative assessments. Our eTextbooks link seamlessly with
Integral, allowing you to move with ease between corresponding topics in
the eTextbooks and Integral.
Additional support
The Question & Workbooks provide additional practice for students. These
write-in workbooks are designed to be used throughout the course.
The Study and Revision Guides provide further practice for students as
they prepare for their examinations.
These supporting resources and MEI’s Integral® material have not been
through the Cambridge International endorsement process.
ix
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The Cambridge International
AS & A Level Mathematics
9709 syllabus
The syllabus content is assessed over six examination papers.
Paper 1: Pure Mathematics 1
Paper 4: Mechanics
• 1 hour 50 minutes
• 1 hour 15 minutes
• 60% of the AS Level; 30% of the • 40% of the AS Level; 20% of the
A Level
A Level
• Compulsory for AS and A Level • Offered as part of AS or A Level
Paper 2: Pure Mathematics 2
Paper 5: Probability & Statistics 1
• 1 hour 15 minutes
• 1 hour 15 minutes
• 40% of the AS Level
• 40% of the AS Level; 20% of the
A Level
• Offered only as part of AS Level;
not a route to A Level
• Compulsory for A Level
Paper 3: Pure Mathematics 3
Paper 6: Probability & Statistics 2
• 1 hour 50 minutes
• 1 hour 15 minutes
• 30% of the A Level
• 20% of the A Level
• Compulsory for A Level; not a
• Offered only as part of A Level;
route to AS Level
not a route to AS Level
The following diagram illustrates the permitted combinations for AS Level
and A Level.
AS Level
Mathematics
Paper 1 and Paper 2
Pure Mathematics only
A Level
Mathematics
(No progression to A Level)
Paper 1 and Paper 4
Pure Mathematics and Mechanics
Paper 1, 3, 4 and 5
Pure Mathematics, Mechanics and
Probability & Statistics
Paper 1 and Paper 5
Pure Mathematics and
Probability & Statistics
Paper 1, 3, 5 and 6
Pure Mathematics and
Probability & Statistics
x
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Prior knowledge
Knowledge of the content of the Cambridge IGCSE® Mathematics 0580
(Extended curriculum), or Cambridge O Level 4024/4029, is assumed.
Learners should be familiar with scientific notation for compound units,
e.g. 5 m s−1 for 5 metres per second.
In addition, learners should:
» be able to carry out simple manipulation of surds (e.g. expressing 12 as
6
2 3 and
as 3 2)
2
» know the shapes of graphs of the form y = kxn, where k is a constant and
n is an integer (positive or negative) or ± 1 .
2
For Paper 2: Pure Mathematics 2 and Paper 3: Pure Mathematics 3,
knowledge of the content of Paper 1: Pure Mathematics 1 is assumed, and
learners may be required to demonstrate such knowledge in answering
questions.
Command words
The table below includes command words used in the assessment for this
syllabus. The use of the command word will relate to the subject context.
Command word
What it means
Calculate
work out from given facts, figures or information
Describe
state the points of a topic / give characteristics and main
features
Determine
establish with certainty
Evaluate
judge or calculate the quality, importance, amount, or value
of something
Explain
set out purposes or reasons / make the relationships
between things evident / provide why and/or how and
support with relevant evidence
Identify
name/select/recognise
Justify
support a case with evidence/argument
Show (that)
provide structured evidence that leads to a given result
Sketch
make a simple freehand drawing showing the key features
State
express in clear terms
Verify
confirm a given statement/result is true
xi
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Key concepts
Key concepts are essential ideas that help students develop a deep
understanding of mathematics.
The key concepts are:
Problem solving
Mathematics is fundamentally problem solving and representing systems and
models in different ways. These include:
» Algebra: this is an essential tool which supports and expresses
mathematical reasoning and provides a means to generalise across a
number of contexts.
» Geometrical techniques: algebraic representations also describe a spatial
relationship, which gives us a new way to understand a situation.
» Calculus: this is a fundamental element which describes change in
dynamic situations and underlines the links between functions and graphs.
» Mechanical models: these explain and predict how particles and objects
move or remain stable under the influence of forces.
» Statistical methods: these are used to quantify and model aspects of the
world around us. Probability theory predicts how chance events might
proceed, and whether assumptions about chance are justified by evidence.
Communication
Mathematical proof and reasoning is expressed using algebra and notation so
that others can follow each line of reasoning and confirm its completeness
and accuracy. Mathematical notation is universal. Each solution is structured,
but proof and problem solving also invite creative and original thinking.
Mathematical modelling
Mathematical modelling can be applied to many different situations and
problems, leading to predictions and solutions. A variety of mathematical
content areas and techniques may be required to create the model. Once the
model has been created and applied, the results can be interpreted to give
predictions and information about the real world.
These key concepts are reinforced in the different question types included
in this book: Problem-solving, Communication and proof, and
Modelling.
xii
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P2
P3
1
Algebra
1 Algebra
No, it [1729]
is a very
interesting
number. It is
the smallest
number
expressible as
a sum of two
cubes in two
different ways.
Srinivasa
Ramanujan
(1887–1920)
A brilliant mathematician, Ramanujan was largely self-taught, being too
poor to afford a university education. He left India at the age of 26 to work
with G. H. Hardy in Cambridge on number theory, but fell ill in the English
climate and died six years later in 1920. On one occasion when Hardy
visited him in hospital, Ramanujan asked about the registration number of
the taxi he came in. Hardy replied that it was 1729, an uninteresting number;
Ramanujan’s instant response is quoted above.
?
❯ Find the two pairs of cubes referred to in the quote.
❯ How did you find them?
You will already have met quadratic expressions, like x2 − 5x + 6, and solved
quadratic equations, such as x2 − 5x + 6 = 0. Quadratic expressions have the
form ax2 + bx + c where x is a variable, a, b and c are constants and a is not
equal to zero. This work is covered in Pure Mathematics 1 Chapter 2.
An expression of the form ax 3 + bx 2 + cx + d, which includes a term in x 3, is
called a cubic in x. Examples of cubic expressions are
2x3 + 3x2 − 2x + 11,
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3y 3 − 1
and
4z3 − 2z.
1
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Similarly a quartic expression in x, like x 4 − 4x 3 + 6x 2 − 4x + 1, contains a
term in x 4; a quintic expression contains a term in x 5 and so on.
1
1 ALGEBRA
All these expressions are called polynomials. The order of a polynomial is
the highest power of the variable it contains. So a quadratic is a polynomial
of order 2, a cubic is a polynomial of order 3 and 3x 8 + 5x 4 + 6x is a
polynomial of order 8 (an octic).
1
Notice that a polynomial does not contain terms involving x , x , etc. Apart
from the constant term, all the others are multiples of x raised to a positive
integer power.
1.1 Operations with polynomials
b
Addition of polynomials
Polynomials are added by adding like terms; for example, you add the
coefficients of x 3 together (i.e. the numbers multiplying x 3 ), the coefficients
of x 2 together, the coefficients of x together and the numbers together.You
may find it easiest to set this out in columns.
Example 1.1
Add (5x 4 − 3x 3 − 2x) to (7x 4 + 5x 3 + 3x 2 − 2).
Solution
5x 4
−3x 3
−2x
4
+ (7x
+5x 3
+3x 2
−2)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
12x 4
+2x 3
+3x 2
−2x
−2
Note
This may alternatively be set out as follows:
(5x 4 − 3x 3 − 2x) + (7x 4 + 5x 3 + 3x 2 − 2)
= (5 + 7)x 4 + (−3 + 5)x 3 + 3x 2 − 2x − 2
= 12x 4 + 2x 3 + 3x 2 − 2x − 2
Subtraction of polynomials
Similarly polynomials are subtracted by subtracting like terms.
2
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Example 1.2
Simplify (5x 4 − 3x 3 − 2x) − (7x 4 + 5x 3 + 3x 2 − 2).
Solution
1
5x 4
−3x 3
−2x
+5x 3
+3x 2
−2)
− (7x 4
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
−2x 4
−8x 3
−3x 2
−2x
+2
1.1 Operations with polynomials
Be careful of the signs when subtracting.You may find it easier to change
the signs on the bottom line and then go on as if you were adding.
Note
This, too, may be set out alternatively, as follows:
(5x 4 − 3x 3 − 2x) − (7x 4 + 5x 3 + 3x 2 − 2)
= (5 − 7)x 4 + (−3 − 5)x 3 − 3x 2 − 2x + 2
= −2x 4 − 8x 3 − 3x 2 − 2x + 2
Multiplication of polynomials
When you multiply two polynomials, you multiply each term of the one by
each term of the other, and all the resulting terms are added. Remember that
when you multiply powers of x, you add the indices: x 5 × x 7 = x 12.
Example 1.3
Multiply (x 3 + 3x − 2) by (x 2 − 2x − 4).
Solution
Arranging this in columns, so that it looks like an arithmetical long
multiplication calculation you get:
+3x
−2
x3
×
x2
−2x
−4
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Multiply top line by x 2
x5
+3x 3 −2x 2
4
Multiply top line by −2x
−2x
−6x 2
+4x
3
−12x
+8
Multiply top line by −4
−4x
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Add
x 5 −2x 4
−x 3 −8x 2
−8x
+8
3
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Note
1
Alternatively:
(x3 + 3x − 2) × (x2 − 2x − 4)
= x3(x2 − 2x − 4) + 3x(x2 − 2x − 4) − 2(x2 − 2x − 4)
= x5 − 2x4 − 4x3 + 3x3 − 6x2 − 12x − 2x2 + 4x + 8
1 ALGEBRA
= x5 − 2x4 + (−4 + 3)x3 + (−6 − 2)x2 + (−12 + 4)x + 8
= x5 − 2x4 − x3 − 8x2 − 8x + 8
Division of polynomials
Division of polynomials is usually set out rather like arithmetical long division.
Example 1.4
Divide (2x3 − 3x2 + x − 6) by (x − 2).
Solution
If the dividend is missing a term, leave a blank
space. For example, write x 3 + 2x + 5 as
x 3 + _ + 2x + 5. Another way to write it is
x 3 + 0x 2 + 2x + 5.
Method 1
2x 2
x − 2 r 2x 3 − 3x 2 + x − 6
2x 3 − 4x 2
Found by dividing 2x 3 (the first term in
2x 3 − 3x 2 + x − 6) by x (the first term in x − 2).
2x 2(x − 2)
Now subtract 2x 3 − 4x 2 from 2x 3 − 3x 2, bring down the next term (i.e. x)
and repeat the method above:
2x 2 + x
3
x − 2 r 2x − 3x 2 + x − 6
2x 3 − 4x 2
x2 + x
x 2 − 2x
x2 ÷ x
x(x − 2)
Continuing gives:
This is the answer.
2x2 + x + 3
It is called the quotient.
3
x − 2 r 2x − 3x2 + x − 6
2x3 − 4x2
x2 + x
The final remainder of
x2 − 2x
zero means that
3x − 6
x − 2 divides exactly into
2x 3 − 3x 2 + x − 6.
3x − 6
0
Thus (2x 3 − 3x 2 + x − 6) ÷ (x − 2) = (2x 2 + x + 3).
4
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Method 2
Alternatively this may be set out as follows if you know that there is no
remainder.
The polynomial here
must be of order 2
because 2x 3 ÷ x will
give an x 2 term.
Let (2x 3 − 3x 2 + x − 6) ÷ (x − 2) ≡ ax 2 + bx + c
Multiplying both sides by (x − 2) gives
1
(2x3 − 3x2 + x − 6) ≡ (ax2 + bx + c)(x − 2)
2x3 − 3x2 + x − 6 ≡ ax3 + (b − 2a)x2 + (c − 2b)x − 2c
Comparing coefficients of x3
1.1 Operations with polynomials
The identity sign
is used here to
emphasise that this
is an identity and
true for all values
of x.
Multiplying out the expression on the right
2=a
Comparing coefficients of x2
⇒
−3 = b − 2a
−3 = b − 4
b=1
Comparing coefficients of x
⇒
1 = c − 2b
1=c−2
c=3
Checking the constant term
−6 = −2c (which agrees with c = 3).
So ax2 + bx + c is 2x2 + x + 3
i.e. (2x3 − 3x2 + x − 6) ÷ (x − 2) ≡ 2x2 + x + 3.
Method 3
With practice you may be able to do this method ‘by inspection’. The steps
in this would be as follows.
Needed to give the 2x3 term
3
2
2
when
multiplied by the x.
(2x − 3x + x − 6) = (x − 2)(2x + … + …)
This product gives −4x2. Only −3x2 is needed.
= (x − 2)(2x2 + x + …)
Introducing +x gives +x2 for this
product and so the x2 term is correct.
= (x − 2)(2x 2 + x + 3) This product gives −2x and +x is on the left-hand side.
This +3x product then gives the correct x term.
= (x − 2)(2x 2 + x + 3)
So (2x 3 − 3x 2 + x − 6) ÷ (x − 2) ≡ 2x 2 + x + 3.
Check that the constant
term (−6) is correct.
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Note
1
A quotient is the result of a division. So, in Example 1.4 the quotient is
2x 2 + x + 3.
1 ALGEBRA
Exercise 1A
b
1 State the orders of the following polynomials.
(i)
x3 + 3x2 − 4x
(ii) x12
(iii) 2 + 6x2 + 3x7 − 8x5
2 Add (x3 + x2 + 3x − 2) to (x3 − x2 − 3x − 2).
3 Add (x3 − x), (3x2 + 2x + 1) and (x4 + 3x3 + 3x2 + 3x).
4 Subtract (3x2 + 2x + 1) from (x3 + 5x2 + 7x + 8).
5 Subtract (x3 − 4x2 − 8x − 9) from (x3 − 5x2 + 7x + 9).
6 Subtract (x5 − x4 − 2x3 − 2x2 + 4x − 4) from (x5 + x4 − 2x3 − 2x2 + 4x + 4).
7 Multiply (x3 + 3x2 + 3x + 1) by (x + 1).
8 Multiply (x3 + 2x2 − x − 2) by (x − 2).
9 Multiply (x2 + 2x − 3) by (x2 − 2x − 3).
10 Multiply (x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x1 + 1) by (x − 1).
11 Simplify (x2 + 1)(x − 1) − (x2 − 1)(x − 1).
12 Simplify (x2 + 1)(x2 + 4) − (x2 − 1)(x2 − 4).
13 Simplify (x + 1)2 + (x + 3)2 − 2(x + 1)(x + 3).
14 Simplify (x2 + 1)(x + 3) − (x2 + 3)(x + 1).
15 Simplify (x2 − 2x + 1)2 − (x + 1)4.
16 Divide (x3 − 3x2 − x + 3) by (x − 1).
17 Find the quotient when (x3 + x2 − 6x) is divided by (x − 2).
18 Divide (2x3 − x2 − 5x + 10) by (x + 2).
19 Find the quotient when (x4 + x2 − 2) is divided by (x − 1).
20 Divide (2x3 − 10x2 + 3x − 15) by (x − 5).
21 Find the quotient when (x4 + 5x3 + 6x2 + 5x + 15) is divided by (x + 3).
22 Divide (2x4 + 5x3 + 4x2 + x) by (2x + 1).
23 Find the quotient when (4x4 + 4x3 − x2 + 7x − 4) is divided by (2x − 1).
24 Divide (2x4 + 2x3 + 5x2 + 2x + 3) by (x2 + 1).
25 Find the quotient when (x4 + 3x3 − 8x2 − 27x − 9) is divided by (x2 − 9).
26 Divide (x4 + x3 + 4x2 + 4x) by (x2 + x).
27 Find the quotient when (2x4 − 5x3 − 16x2 − 6x) is divided by (2x2 + 3x).
28 Divide (x4 + 3x3 + x2 − 2) by (x2 + x + 1).
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1.2 Solution of polynomial equations
You have already met the formula
1
2
x = −b ± b − 4ac
2a
for the solution of the quadratic equation ax2 + bx + c = 0.
1.2 Solution of polynomial equations
Unfortunately there is no such simple formula for the solution of a cubic
equation, or indeed for any higher power polynomial equation. So you have
to use one (or more) of three possible methods.
» Spotting one or more roots.
» Finding where the graph of the expression cuts the x-axis.
» A numerical method.
Example 1.5
Solve the equation 4x3 − 8x2 − x + 2 = 0.
Solution
Start by plotting the curve whose equation is y = 4x3 − 8x2 − x + 2. (You
may also find it helpful at this stage to display it on a graphical calculator or
computer.)
x
−1
0
1
2
3
y
−9
+2
−3
0
35
y
40
30
20
10
1
–1
0
2
3
x
–10
▲ Figure 1.1
Figure 1.1 shows that one root is x = 2 and that there are two others. One is
between x = −1 and x = 0 and the other is between x = 0 and x = 1.
➜
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1
Try x = − 21 .
Substituting x = − 21 in
y = 4x3 − 8x2 − x + 2 gives
y = 4 × (– 81) − 8 × 41 − (− 21 ) + 2
=0
1 ALGEBRA
So in fact the graph crosses the x-axis at x = − 21 and this is a root also.
Similarly, substituting x = + 21 in y = 4x 3 − 8x 2 − x + 2 gives
y = 4 × 81 − 8 × 41 − 21 + 2
=0
and so the third root is x = 21 .
The solution is x = − 21 , 1 or 2.
2
This example worked out nicely, but many equations do not have roots
which are whole numbers or simple fractions. In those cases you can find an
approximate answer by drawing a graph. To be more accurate, you will need
to use a numerical method, which will allow you to get progressively closer to
the answer, homing in on it. Such methods are covered in Chapter 6.
The factor theorem
The equation 4x 3 − 8x 2 − x + 2 = 0 has roots that are whole numbers or
fractions. This means that it could, in fact, have been factorised.
4x 3 − 8x 2 − x + 2 = (2x + 1)(2x − 1)(x − 2) = 0
Few polynomial equations can be factorised, but when one can, the solution
follows immediately.
Since (2x + 1)(2x − 1)(x − 2) = 0
it follows that either 2x + 1 = 0
⇒
x = – 21
or 2x − 1 = 0
⇒
x=2
x−2=0
⇒
x=2
or
1
and so x = − 21 , 1 or 2.
2
This illustrates an important result, known as the factor theorem, which
may be stated as follows.
Factor theorem: If (x − a) is a factor of the polynominal f (x), then f (a) = 0
and x = a is a root of the equation f (x) = 0. Conversely if f (a) = 0, then
(x − a) is a factor of f (x).
()
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()
So if (ax − b) is a factor of the polynomial f(x), then f ba = 0 and x = ba is a
root of the equation f(x) = 0. Conversely, if f ba = 0, then (ax − b) is a factor
of f(x).
02/02/18 1:11 PM
Example 1.6
Given that f(x) = x 3 − 6x 2 + 11x − 6:
(i)
find f(0), f(1), f(2), f(3) and f(4)
(ii)
factorise x 3 − 6x 2 + 11x − 6
1
(iii) solve the equation x 3 − 6x 2 + 11x − 6 = 0
(iv) sketch the curve whose equation is f(x) = x 3 − 6x 2 + 11x − 6.
1.2 Solution of polynomial equations
Solution
(i)
f(0) = 03 − 6 × 02 + 11 × 0 − 6 = −6
f(1) = 13 − 6 × 12 + 11 × 1 − 6 = 0
f(2) = 23 − 6 × 22 + 11 × 2 − 6 = 0
f(3) = 33 − 6 × 32 + 11 × 3 − 6 = 0
f(4) = 43 − 6 × 42 + 11 × 4 − 6 = 6
(ii) Since f(1), f(2) and f(3) all equal 0, it follows that (x − 1), (x − 2) and
(x − 3) are all factors. This tells you that
x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x − 3) × constant
By checking the coefficient of the term in x 3, you can see that the
constant must be 1, and so
x 3 − 6x 2 + 11x − 6 = (x − 1)(x − 2)(x − 3)
(iii) x = 1, 2 or 3
(iv)
f(x)
0
1
2
3
x
–6
▲ Figure 1.2
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In the previous example, all three factors came out of the working, but this will
not always happen. If not, it is often possible to find one factor (or more) by
‘spotting’ it, or by sketching the curve.You can then make the job of searching
for further factors much easier by dividing the polynomial by the factor(s) you
have found: you will then be dealing with a lower order polynomial.
1
1 ALGEBRA
Example 1.7
Given that f(x) = x3 − x2 − 3x + 2:
(i)
show that (x − 2) is a factor
(ii) solve the equation f(x) = 0.
Solution
To show that (x − 2) is a factor, it is necessary to show that f(2) = 0.
f(2) = 23 − 22 − 3 × 2 + 2
=8−4−6+2
=0
Therefore (x − 2) is a factor of x 3 − x 2 − 3x + 2.
(ii) Since (x − 2) is a factor you divide f(x) by (x − 2).
(i)
x2 + x − 1
x2 − 3x + 2
x3 − 2x2
x2 − 3x
x2 − 2x
−x + 2
−x + 2
0
So f(x) = 0 becomes (x − 2)(x 2 + x − 1) = 0,
⇒ either x − 2 = 0 or x 2 + x − 1 = 0.
Using the quadratic formula on x 2 + x − 1 = 0 gives
x − 2 r x3 −
−1 ± 1 − 4 × 1 × (−1)
2
= −1 ± 5
2
= −1.618 or 0.618 (to 3 d.p.)
x=
So the complete solution is x = −1.618, 0.618 or 2.
Note
You may prefer to use inspection or equate coefficients rather than long
division. You can write:
x 3 − x 2 − 3x + 2 = ( x − 2)(ax 2 + bx + c )
By inspection: a = 1 and c = −1
Equating coefficients of x2 or x gives b =1.
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Spotting a root of a polynomial equation
Most polynomial equations do not have integer (or fraction) solutions. It is
only a few special cases that work out nicely.
1
To check whether an integer root exists for any equation, look at the
constant term. Decide what whole numbers divide into it and test them.
Example 1.8
Solution
The constant term is −6 and this is divisible by −1, +1, −2, +2, −3, +3, −6 and
+6. So the only possible factors are (x ± 1), (x ± 2), (x ± 3) and (x ± 6). This
limits the search somewhat.
f(1) = −6
f(2) = −6
f(3) = 0
f(6) = 114
f(−1) = −12
f(−2) = −30
f(−3) = −66
f(−6) = −342
No;
No;
Yes;
No;
No;
No;
No;
No.
x = 3 is an integer root of the equation.
Example 1.9
1.2 Solution of polynomial equations
Spot an integer root of the equation x3 − 3x2 + 2x − 6 = 0.
Is there an integer root of the equation x3 − 3x2 + 2x − 5 = 0?
Solution
The only possible factors are (x ± 1) and (x ± 5).
f(1) = −5
No;
f(−1) = −11
f(5) = 55
No;
f(−5) = −215 No.
No;
There is no integer root.
Example 1.10
The polynomial, 3x 3 + ax 2 + b, where a and b are constants, is denoted by
p(x). It is given that (3x − 4) and (x + 2) are factors of p(x).
Find the values of a and b.
Solution
( x + 2) is a factor ⇒ p ( −2 ) = 0
⇒ 3 × ( −2 ) 3 + ( −2 ) 2 a + b = 0
⇒
−24 + 4a + b = 0
1
!
➜
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1
()
⇒ 3 × ( 43 ) + ( 43 ) a + b = 0
(3x − 4) is a factor ⇒ p 43 = 0
3
2
64 + 16 a + b = 0
9
9
⇒
2
!
1 ALGEBRA
1 − equation !
2 gives:
Equation !
−24 + 4a + b = 0
− 64
+ 16
a+b = 0
9
9
− 280
+ 20
a
9
9
=0
So 20 a = 280 ⇒ a = 14
9
9
1 gives b = −32.
Substituting a = 14 into equation !
The remainder theorem
Using the long division method, any polynomial can be divided by another
polynomial of lesser order, but sometimes there will be a remainder.
Look at (x 3 + 2x 2 − 3x − 7) ÷ (x − 2).
x2 + 4x + 5
x − 2 r x3 + 2x2 − 3x − 7
x3 − 2x2
4x2 − 3x
4x2 − 8x
5x − 7
5x − 10
3
The quotient is x2 + 4x + 5
and the remainder is 3.
You can write this as
x 3 + 2x 2 − 3x − 7 = (x − 2)(x 2 + 4x + 5) + 3
At this point it is convenient to call the polynomial x3 + 2x2 − 3x − 7 = f(x).
1
So f(x) = (x − 2)(x2 + 4x + 5) + 3.
!
1 gives f(2) = 3.
Substituting x = 2 into both sides of equation !
So f(2) is the remainder when f(x) is divided by (x − 2).
This result can be generalised to give the remainder theorem.
Remainder theorem: For a polynomial, f (x), f (a) is the remainder when f (x)
is divided by (x − a).
f (x) = (x − a)g(x) + f(a)
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Example 1.11
Find the remainder when 2x3 − 3x + 5 is divided by (x + 1).
1
Solution
The remainder is found by substituting x = −1 in 2x3 − 3x + 5.
2 × (−1)3 − 3 × (−1) + 5 = −2 + 3 + 5
=6
Example 1.12
1.2 Solution of polynomial equations
So the remainder is 6.
When 4x2 − 6x + a is divided by (2x − 1), the remainder is 2.
Find the value of a.
Solution
1
The remainder is found by substituting x = 2 into 4x2 − 6x + a.
( ) − 6 × 21 + a = 2
4 21
⇒
⇒
2
4 × 41 − 3 + a = 2
1 − 3+a = 2
⇒
a=4
A polynomial is divided by another of degree n.
❯ What can you say about the remainder?
?
When dividing by polynomials of order 2 or more, the remainder is usually
found most easily by actually doing the long division.
Example 1.13
Find the remainder when 2x 4 − 3x 3 + 4 is divided by (x 2 + 1).
Solution
x2 + 1 r 2x4 − 3x3
2x4
2x2 − 3x − 2
+4
+ 2x2
−3x3 − 2x2
−3x3
− 3x
− 2x2 + 3x + 4
−2
− 2x2
The remainder is 3x + 6.
3x + 6
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1 ALGEBRA
1
In a division such as the one in Example 1.13, it is important to keep a
separate column for each power of x and this means that sometimes it is
necessary to leave gaps, as in this example. In arithmetic, zeros are placed
in the gaps. For example, 2 thousand and 3 is written 2003.
Exercise 1B
1
2
3
4
5
6
PS
7
Given that f(x) = x3 + 2x2 − 9x − 18:
(i) find f(−3), f(−2), f(−1), f(0), f(1), f(2) and f(3)
(ii) factorise f(x)
(iii) solve the equation f(x) = 0
(iv) sketch the curve with the equation y = f(x).
The polynomial p(x) is given by p(x) = x 3 − 4x.
(i) Find the values of p(−3), p(−2), p(−1), p(0), p(1), p(2) and p(3).
(ii) Factorise p(x).
(iii) Solve the equation p(x) = 0.
(iv) Sketch the curve with the equation y = p(x).
You are given that f(x) = x 3 − 19x + 30.
(i) Calculate f(0) and f(3). Hence write down a factor of f(x).
(ii) Find p and q such that f(x) ! (x − 2)(x 2 + px + q).
(iii) Solve the equation x3 − 19x + 30 = 0.
(iv) Without further calculation draw a sketch of y = f(x).
(i) Show that (x − 3) is a factor of x3 − 5x2 − 2x + 24.
(ii) Solve the equation x 3 − 5x 2 − 2x + 24 = 0.
(iii) Sketch the curve with the equation y = x 3 − 5x 2 − 2x + 24.
(i) Show that x = 2 is a root of the equation x 4 − 5x 2 + 2x = 0 and
write down another integer root.
(ii) Find the other two roots of the equation x 4 − 5x 2 + 2x = 0.
(iii) Sketch the curve with the equation y = x 4 − 5x 2 + 2x.
(i) The polynomial p(x) = x 3 − 6x 2 + 9x + k has a factor (x − 4).
Find the value of k.
(ii) Find the other factors of the polynomial.
y
(iii) Sketch the curve with the
equation y = p(x).
The diagram shows the curve with
the equation y = (x + a)(x − b)2
c
where a and b are positive integers.
(i) Write down the values of a and
–2
–1
0
1
2
b, and also of c, given that the
curve crosses the y axis at (0, c).
x
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1
1.2 Solution of polynomial equations
Solve the equation (x + a)(x − b)2 = c using the values of a, b and c
you found in part (i).
8 The function f(x) is given by f(x) = x 4 − 3x 2 − 4 for real values of x.
(i) By treating f(x) as a quadratic in x 2, factorise it in the form
(x 2 + …)(x 2 + …).
(ii) Complete the factorisation as far as possible.
(iii) How many real roots has the equation f(x) = 0? What are they?
9
(i) Show that (x − 2) is not a factor of 2x 3 + 5x 2 − 7x − 3.
(ii) Find the quotient and the remainder when 2x 3 + 5x 2 − 7x − 3
is divided by (x − 2).
10 The equation f(x) = x 3 − 4x 2 + x + 6 = 0 has three integer roots.
(i) List the eight values of a for which it is sensible to check whether
f(a) = 0 and check each of them.
(ii) Solve f(x) = 0.
11 Factorise, as far as possible, the following expressions.
(i) x 3 − x 2 − 4x + 4 given that (x − 1) is a factor.
(ii) x 3 + 1 given that (x + 1) is a factor.
(iii) x 3 + x − 10 given that (x − 2) is a factor.
(iv) x 3 + x 2 + x + 6 given that (x + 2) is a factor.
12 (i) Show that neither x = 1 nor x = −1 is a root of x 4 − 2x 3 + 3x 2 − 8 = 0.
(ii) Find the quotient and the remainder when x 4 − 2x 3 + 3x 2 − 8 is
divided by
(a) (x − 1)
(b) (x + 1)
(c) (x 2 − 1).
(ii)
13 When 2x 3 + 3x 2 + kx − 6 is divided by (x + 1) the remainder is 7.
PS
14
PS
15
PS
16
17
Find the value of k.
When x 3 + px 2 + p 2x − 36 is divided by (x − 3) the remainder is 21.
Find a possible value of p.
When x 3 + ax 2 + bx + 8 is divided by (x − 3) the remainder is 2 and
when it is divided by (x + 1) the remainder is −2.
Find a and b and hence obtain the remainder on dividing by (x − 2).
When f(x) = 2x 3 + ax 2 + bx + 6 is divided by (x − 1) there is no
remainder and when f(x) is divided by (x + 1) the remainder is 10.
Find a and b and hence solve the equation f(x) = 0.
The polynomial f(x) is defined by
f(x) = 3x 3 + ax 2 + ax + a ,
where a is a constant.
(i) Given that (x + 2) is a factor of f (x), find the value of a.
(ii) When a has the value found in part (i), find the quotient when
f (x) is divided by (x + 2).
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q4 June 2011
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1
18 The polynomial 2x3 + 7x2 + ax + b, where a and b are constants, is
denoted by p(x). It is given that (x + 1) is a factor of p(x), and that
when p(x) is divided by (x + 2) the remainder is 5. Find the values of
a and b.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q4 June 2008
1 ALGEBRA
19 The polynomial 2x 3 − x 2 + ax − 6, where a is a constant, is denoted by
p(x). It is given that (x + 2) is a factor of p(x).
(i) Find the value of a.
(ii) When a has this value, factorise p(x) completely.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q2 November 2008
20 The polynomial x 3 + ax 2 + bx + 6, where a and b are constants, is
denoted by p(x). It is given that (x – 2) is a factor of p(x), and that when
p(x) is divided by (x – 1) the remainder is 4.
(i) Find the values of a and b.
(ii) When a and b have these values, find the other two linear factors
of p(x).
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q6 June 2009
21 The polynomial 4x3 + ax2 + bx − 2, where a and b are constants, is
denoted by p(x). It is given that (x + 1) and (x + 2) are factors of p(x).
(i) Find the values of a and b.
(ii) When a and b have these values, find the remainder when p(x) is
divided by (x2 + 1).
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q3 November 2014
22 (i)
(ii)
Find the quotient when 6x4 − x3 − 26x2 + 4x + 15 is divided by
(x2 − 4), and confirm that the remainder is 7.
Hence solve the equation 6x4 − x3 − 26x2 + 4x + 8 = 0.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q3 June 2014
23 The polynomials f(x) and g(x) are defined by f(x) = x3 + ax2 + b and
g(x) = x3 + bx2 − a, where a and b are constants. It is given that (x + 2)
is a factor of f(x). It is also given that, when g(x) is divided by (x + 1), the
remainder is −18.
(i) Find the values of a and b.
(ii) When a and b have these values, find the greatest possible value of
g(x) – f(x) as x varies.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q4 June 2015
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1.3 The modulus function
1
Look at the graph of y = f(x), where f(x) = x.
y
y = f(x) = x
1.3 The modulus function
x
O
▲ Figure 1.3
The function f(x) is positive when x is positive and negative when x is
negative.
Now look at the graph of y = g(x), where g(x) = | x |.
y
y = g(x) = |x|
O
x
▲ Figure 1.4
The function g(x) is called the modulus of x. g(x) always takes the positive
numerical value of x. For example, when x = −2, g(x) = 2, so g(x) is always
positive. The modulus is also called the magnitude of the quantity.
Another way of writing the modulus function g(x) is
g(x) = x
for x ! 0
g(x) = −x for x " 0.
?
❯ What is the value of g(3) and g(−3)?
❯ What is the value of | 3 + 3 |, | 3 − 3 |, | 3 | + | 3 | and | 3 | + | −3 |?
The graph of y = g(x) can be obtained from the graph of y = f(x) by
replacing values where f(x) is negative by the equivalent positive values.
This is the equivalent of reflecting that part of the line in the x-axis.
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1
Example 1.14
Sketch the graphs of the following on separate axes.
(i)
y=1−x
(ii) y = | 1 − x |
(iii) y = 2 + | 1 − x |
1 ALGEBRA
Solution
(i)
y = 1 − x is the straight line through (0, 1) and (1, 0).
y
1
y=1–x
O
1
x
▲ Figure 1.5
(ii) y = | 1 − x | is obtained by reflecting the part of the line for x # 1 in
the x-axis.
y
y = |1 – x|
1
O
x
1
▲ Figure 1.6
(iii) y = 2 + | 1 − x | is obtained from the previous graph by applying
⎛0 ⎞
the translation ⎜ ⎟ .
⎝2⎠
y
3
y = 2 + |1 – x|
(1, 2)
O
1
x
▲ Figure 1.7
18
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Inequalities involving the modulus sign
You will often meet inequalities involving the modulus sign.
?
1
Look back at the graph of y = | x | in Figure 1.4.
❯ How does this show that | x | " 2 is equivalent to −2 " x " 2?
Rule
Example
| x | = | −x |
| 3 | = | −3 |
|a − b| = |b − a|
| 8 − 5 | = | 5 − 8 | = +3
| x |2 = x2
| −3 |2 = (−3)2
|a | = |b|
⇔
a2 = b2
| −3 | = | 3 |
|x| $ a ⇔
−a $ x $ a
|x| $ 3
⇔
−3 $ x $ 3
|x| # a ⇔
x " −a or x # a
|x| # 3
⇔
x " −3 or x # 3
|x − 2| " 5
⇔ −3 " x " 7
|x − a| " b ⇔
Example 1.15
a−b"x"a+b
⇔
(−3)2 = 32
1.3 The modulus function
Here is a summary of some useful rules.
Solve the following.
(i)
|x + 3| $ 4
(ii) | 2x − 1 | # 9
(iii) 5 − | x − 2 | # 1
Solution
(i)
|x+3|$4
⇔
−4 $ x + 3 $ 4
⇔
−7 $ x $ 1
(ii) | 2x − 1 | # 9
⇔ 2x − 1 " −9 or 2x − 1 # 9
⇔ 2x " −8
or 2x # 10
⇔ x " −4
or x # 5
(iii) 5 − | x − 2 | # 1
⇔ 4#|x−2|
⇔ |x−2|"4
⇔ −4 " x − 2 " 4
⇔ −2 " x " 6
Note
The solution to part (ii) represents two separate intervals on the number line,
so cannot be written as a single inequality.
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1
Example 1.16
Express the inequality −2 " x " 6 in the form | x − a | " b, where a and b
are to be found.
Solution
|x−a|"b
⇔
−b " x − a " b
⇔
a−b"x"a+b
1 ALGEBRA
Comparing this with −2 " x " 6 gives
a − b = −2
a + b = 6.
Solving these simultaneously gives a = 2, b = 4, so | x − 2 | " 4.
Example 1.17
Solve 2x " | x − 3 |.
Solution
It helps to sketch a graph of y = 2x and y = | x − 3 |.
y
y = 2x
3
y = |x – 3|
O
c
3
x
▲ Figure 1.8
You can see that the graph of y = 2x is below y = | x − 3 | for x " c.
You can find the critical region by solving 2x " −(x − 3).
2x " −(x − 3)
2x " −x + 3
3x " 3
x"1
c is at the intersection
of the lines y = 2x
and y = −(x − 3).
20
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Example 1.18
(i)
Solve | 2x − 1 | = | x − 2 |.
(ii)
Solve | 2x − 1 | " | x − 2 |.
1
Solution
(i)
Sketching a graph of y = | 2x − 1 | and y = | x − 2 | shows that the
equation is true for two values of x.
1.3 The modulus function
y
y = |2x – 1|
2
y = |x – 2|
1
O
x
2
▲ Figure 1.9
You can find these values by solving | 2x − 1 | = | x − 2 |.
One method is to use the fact that | a | = | b | ⇔ a2 = b2.
| 2x − 1 | = | x − 2 |
Squaring:
(2x − 1)2 = (x − 2)2
Expanding:
4x2 − 4x + 1 = x2 − 4x + 4
Rearranging:
3x2 − 3 = 0
⇒
x2 − 1 = 0
Factorising:
(x − 1)(x + 1) = 0
So the solution is x = –1 or x = 1.
Exercise 1C
(ii)
When | 2x − 1 | " | x − 2 |, y = | 2x − 1 | (drawn in red) is below
y = | x − 2 | (drawn in blue) on the graph. So the solution to the
inequality is −1 " x " 1.
1
Solve the following equations.
(i) | x + 4 | = 5
(iii) | 3 − x | = 4
(v) | 2x + 1 | = 5
(vii) | 2x + 1 | = | x + 5 |
(ix) | 3x − 2 | = | 4 − x |
|x−3|=4
(iv) | 4x − 1 | = 7
(vi) | 8 − 2x | = 6
(viii) | 4x − 1 | = | 9 − x |
(ii)
21
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1
2
1 ALGEBRA
3
4
5
6
Solve the following inequalities.
(i) | x + 3 | " 5
(ii) | x − 2 | $ 2
(iii) | x − 5 | # 6
(iv) | x + 1 | % 2
(v) | 2x − 3 | " 7
(vi) | 3x − 2 | $ 4
Express each of the following inequalities in the form | x − a | " b,
where a and b are to be found.
(i) −1 " x " 3
(ii) 2 " x " 8
(iii) −2 " x " 4
(iv) −1 " x " 6
(v) 9.9 " x " 10.1
(vi) 0.5 " x " 7.5
Sketch each of the following graphs on a separate set of axes.
(i) y = | x + 2 |
(ii) y = | 2x − 3 |
(iii) y = | x + 2 | − 2
(iv) y = | x | + 1
(v) y = | 2x + 5 | − 4
(vi) y = 3 + | x − 2 |
Solve the following inequalities.
(i) | x + 3 | " | x − 4 |
(ii) | x − 5 | # | x − 2 |
(iii) | 2x − 1 | $ | 2x + 3 |
(iv) | 2x | $ | x + 3 |
(v) | 2x | # | x + 3 |
(vi) | 2x + 5 | % | x − 1 |
Solve the inequality | x | # | 3x − 2 |.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q1 June 2005
7
Given that a is a positive constant, solve the inequality | x − 3a | # | x − a |.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q1 November 2005
8
Solve the equation 3 x + 4 = 2x + 5 .
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q1 June 2011
9
Solve the equation x 3 − 14 = 13, showing all your working.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q1 June 2012
22
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KEY POINTS
1
2
4
5
6
1
1.3 The modulus function
3
A polynomial in x has terms in positive integer powers of x and may
also have a constant term.
The order of a polynomial in x is the highest power of x which
appears in the polynomial.
The factor theorem states that if (x − a) is a factor of a polynomial
f(x) then f(a) = 0 and x = a is a root of the equation f(x) = 0.
Conversely if f(a) = 0, then (x − a) is a factor of f(x).
The remainder theorem states that f(a) is the remainder when the
polynomial f(x) is divided by (x − a).
The modulus of x, written |x|, means the positive value of x.
The modulus function is
● | x | = x,
for x % 0
● | x | = −x,
for x " 0.
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■ divide a polynomial by a linear or quadratic expression and find the
quotient and remainder
■ understand and use the
■ factor theorem
■ remainder theorem
■ use the factor and remainder theorems to solve problems involving
polynomials including
■ finding unknown coefficients of a polynomial
■ solving a polynomial equation
■ solve equations and inequalities involving the modulus sign
■ know that
■ | x | = | −x |
■ |a − b| = |b − a|
■ | x |2 = x2
■ |x − a|$b ⇔ a − b $ x $ a + b
■ | x | # a ⇔ x " −a or x # a
■ sketch the graph of y = | ax + b |.
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P2
P3
2 LOGARITHMS AND EXPONENTIALS
2
We have got
so used to our
existing mental
software that
we see no fault
or limitation in
it. We cannot
see why it
should ever
need changing.
And yet, in
the last few
decades, it has
been shown
that different
thinking habits
can be very
much more
powerful.
Edward de Bono
(1933–)
Logarithms and
exponentials
?
During growth or reproduction in the human body, a cell divides into
M
two new cells roughly every 24 hours.
Assuming that this process takes exactly 1 day, and that none of the cells die off:
❯ Starting with one cell, how many cells will there be after
(i) 5 days
(ii) 10 days?
❯ Approximately how many days would it take to create one million
cells from a single cell?
To answer the question above, you need to use powers of 2.
An exponential function is a function which has the variable as the power,
such as 2x. (An alternative name for power is exponent.)
24
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ACTIVITY 2.1
2
Figure 2.1 shows the curves y = 2x and y = x2.
y
y = x2
B
A
O
1
2
3
4
5
6
x
▲ Figure 2.1
(i)
(a) Find the coordinates of the points A and B.
(b) For what values of x is 2x # x2?
If you have graphing software available to you, use it to draw
(a) y = 3x and y = x3
(b) y = 4x and y = x4.
(iii) Is it true that for all values of a
(a) y = ax and y = xa intersect when x = a
(b) for large enough values of x, ax > xa?
(ii)
2.1 Exponential functions
y = 2x
2.1 Exponential functions
All of the graphs of exponential functions seen in Activity 2.1 pass through
the point (0, 1) and have a gradient that is increasing. In fact this is true of all
exponential curves of the form y = kax where a > 1.
y
6
y = 3x
5
y = 2x
4
3
2
1
–4
–3
–2
–1
0
1
2
3
4
x
▲ Figure 2.2
The x-axis is a horizontal
asymptote.
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Exponential functions model many real-life situations, such as the growth of
cells in the question at the beginning of the chapter. Exponential functions
increase at an ever-increasing rate. In the case of exponential functions, this is
described as exponential growth.
2
It is also possible to have exponential functions for which the power is
negative, for example y = 2-x, y = 3-x, etc.
2 LOGARITHMS AND EXPONENTIALS
The effect of replacing x by −x for any function of x is to reflect the graph in
the y-axis.
y
6
5
y = 2–x
y = 2x
4
3
2
1
−5
−4
−3
−2
−1
0
1
2
3
▲ Figure 2.3
4
5 x
The x-axis is a
horizontal asymptote.
Notice that as x becomes very large, the value of 2–x becomes very small.
The graph of y = 2–x approaches the x-axis ever more slowly as x increases.
This is described as exponential decay.
The graphs of y = ax and y = a–x have a horizontal asymptote at y = 0
(the x-axis) and go through the point (0, 1).
?
❯ How will these features change if the graphs are
(i) translated vertically
(ii) translated horizontally
(iii) stretched horizontally
(iv) stretched vertically?
Example 2.1
26
9781510421738.indb 26
The cost, $C, of a machine t years after initial production is given by
C = 10 + 20 × 2–t.
(i)
What is the initial cost of the machine?
(ii)
What happens to the cost as t becomes very large?
(iii) Sketch the graph of C against t.
02/02/18 1:12 PM
Solution
(i)
2
When t = 0, C = 10 + 20 × 20 = 10 + 20 = 30.
Remember that a0 = 1.
So the initial cost is $30.
(ii)
(iii)
C
30
The line C = 10 is a
horizontal asymptote.
10
O
2.1 Exponential functions
As t becomes very large, 2–t becomes very small, and so the cost
approaches $10.
t
▲ Figure 2.4
Exercise 2A
1
2
M
3
M
4
Sketch each of the graphs below. Show the horizontal asymptote and the
coordinates of the point where the graph crosses the y-axis.
(i) y = 2x
(ii)
y = 2x + 1
(iii)
y = 2x − 1
Sketch each of the graphs below. Show the horizontal asymptote and the
coordinates of the point where the graph crosses the y-axis.
(i) y = 3–x
(ii)
y = 3–x + 2
(iii)
y = 3–x − 1
The growth in population P of a certain town after time t years can be
modelled by the equation P = 10 000 × 100.1t.
(i) State the initial population of the town and calculate the
population after 5 years.
(ii) Sketch the graph of P against t.
The height h m of a species of pine tree t years after planting is modelled
by the equation
h = 20 − 19 × 0.9t
What is the height of the trees when they are planted?
(ii) Calculate the height of the trees after 2 years.
(iii) Use trial and improvement to find the number of years for the
height to reach 10 metres.
(iv) What height does the model predict that the trees will eventually
reach?
(i)
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2 LOGARITHMS AND EXPONENTIALS
2
M
5
M
6
A flock of birds was caught in a hurricane and blown far out to sea.
Fortunately, they were able to land on a small and very remote island and
settle there. There was only a limited sustainable supply of suitable food
for them on the island.
Their population size, n birds, at a time t years after they arrived, can be
modelled by the equation n = 200 + 320 × 2−t.
(i) How many birds arrived on the island?
(ii) How many birds were there after 5 years?
(iii) Sketch a graph of n against t.
(iv) What is the long-term size of the population?
In music, the notes in the middle octave (eight consecutive notes) of
the standard scale, and their frequencies in hertz, are as follows. (The
frequencies have been rounded to the nearest whole numbers.)
A 220
B 247
C 262
D 294
E 330
F
349
G 392
In the next octave up, the notes are also called A to G and their
frequencies are exactly twice those given here; so, for example, the
frequency of A is 2 × 220 = 440 hertz.
The same pattern of multiplying by 2 continues for higher octaves.
Similarly, dividing by 2 gives the frequencies for the notes in lower octaves.
Find the frequency of B three octaves above the middle.
(ii) The lowest note on a standard piano has frequency 27.5 hertz.
What note is this and how many octaves is it below the middle?
(iii) Julia’s range of hearing goes from 75 up to 9000 hertz. How many
notes on the standard scale can she hear?
(i)
2.2 Logarithms
You can think of multiplication in two ways. Look, for example, at 81 × 243,
which is 34 × 35.You can work out the product using the numbers or you
can work it out by adding the powers of a common base − in this case base 3.
Multiplying the numbers:
Adding the powers of the base 3:
81 × 243 = 19 683
4 + 5 = 9 and 39 = 19 683
Another name for a power is a logarithm. Since 81 = 34, you can say that the
logarithm to the base 3 of 81 is 4. The word logarithm is often abbreviated to
log and the statement would be written log3 81 = 4. In general:
28
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y = ax ⇒
loga y = x
02/02/18 1:12 PM
Notice that since 34 = 81, 3log381 = 81. This is an example of a general result:
alogax = x
Example 2.2
(i)
Find the logarithm to the base 2 of each of these numbers.
64
(a)
(b)
1
2
(c)
1
(d)
2
Show that 2log264 = 64.
Solution
(i)
(ii)
(a)
64 = 26 and so log2 64 = 6
(b)
1
1
–1
2 = 2 and so log2 2 = −1
(c)
1 = 20 and so log2 1 = 0
(d)
2 = 2 2 and so log
1
2
2.2 Logarithms
(ii)
2
2 = 21
2log264 = 26 = 64 as required
Logarithms to the base 10
Any positive number can be expressed as a power of 10. Before the days
of calculators, logarithms to the base 10 were used extensively as an aid to
calculation. There is no need for that nowadays but the logarithm function
remains an important part of mathematics, particularly the natural logarithm
which you will meet later in this chapter. Base 10 logarithms continue to be
a standard feature on calculators, and occur in some specialised contexts: the
pH value of a liquid, for example, is a measure of its acidity or alkalinity and
is given by log10(1/the concentration of H+ ions).
Since 1000 = 103, log10 1000 = 3
Similarly
log10 100 = 2
log10 10 = 1
log10 1 = 0
log10 1 = log10 (10−1) = −1
(10 )
and so on.
1 = log (10−2) = −2
log10 (100
)
10
ACTIVITY 2.2
There are several everyday situations in which quantities are measured on
logarithmic scales.
What are the relationships between the following?
(i)
An earthquake of intensity 7 on the Richter scale and one of intensity 8.
(ii)
The frequency of the musical note middle C and that of the C
above it.
(iii) The intensity of an 85 dB noise level and one of 86 dB.
9781510421738.indb 29
29
02/02/18 1:12 PM
The laws of logarithms
2
The laws of logarithms follow from those for indices.
Multiplication
Writing xy = x × y in the form of powers (or logarithms) to the base a and
using the result that x = alogax gives
2 LOGARITHMS AND EXPONENTIALS
a loga xy = a loga x × a loga y
and so
a loga xy = a logax + logay.
Consequently logaxy = logax + logay.
Division
Similarly loga ⎛⎜ x ⎞⎟ = logax − loga y.
⎝y⎠
Power zero
Since a0 = 1, log a1 = 0.
However, it is more usual to state such laws without reference to the base of
the logarithms except where necessary, and this convention is adopted in the
key points at the end of this chapter. As well as the laws given here, others
may be derived from them, as follows.
Indices
Since
it follows that
and so
xn = x × x × x × … × x (n times)
log xn = log x + log x + log x + … + log x (n times),
log xn = n log x.
This result is also true for non-integer values of n and is particularly useful
because it allows you to solve equations in which the unknown quantity is
the power, as in the next example.
Example 2.3
Solve the equation 2n = 1000.
Solution
2n = 1000
Taking logarithms to the base 10 of both sides (since these can be found on a
calculator),
log10 (2n) = log10 1000
n log10 2 = log10 1000
log 10 1000
n = log 2 = 9.97 to 3 significant figures
10
Note
30
9781510421738.indb 30
Most calculators just have ‘log’ and not ‘log 10’ on their keys.
02/02/18 1:12 PM
Example 2.4
Solve the equation 32x + 2(3x) − 15 = 0.
Solution
Example 2.5
2.2 Logarithms
This is a quadratic equation in disguise.
Let y = 3x.
32x = (3x)2
So
y2 + 2y − 15 = 0
(3x)2 + 2(3x) − 15 = y2 + 2y − 15
⇒(y − 5)(y + 3) = 0
⇒y = 5 or y = −3
y = −3 ⇒ 3x = −3 which has no solutions.
y = 5 ⇒ 3x = 5
3x is always positive.
x
So log 3 = log 5
x log 3 = log 5
log 5
x=
= 1.46 (3 s.f.)
log 3
2
A geometric sequence begins 0.2, 1, 5, ... .
The kth term is the first term in the sequence that is greater than 500 000.
Find the value of k.
Solution
The kth term of a geometric sequence is given by ak = a × r k−1.
In this case a = 0.2 and r = 5, so:
0.2 × 5k–1 > 500 000
500 000
5k–1 >
0.2
k–1
5 > 2 500 000
Taking logarithms to the base 10 of both sides:
log10 5k−1 > log10 2 500 000
⇒ (k − 1)log10 5 > log10 2 500 000
⇒
k−1>
log 10 2 500 000
log 10 5
⇒
k − 1 > 9.15
⇒
k > 10.15
Since k is an integer, then k = 11.
So the 11th term is the first term greater than 500 000.
Check : 10th term = 0.2 × 510−1 = 390 625 (< 500 000) ✓
11th term = 0.2 × 511−1 = 1 953 125 (> 500 000) ✓
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2
Roots
A similar line of reasoning leads to the conclusion that:
log n x = n1 log x
The logic runs as follows:
2 LOGARITHMS AND EXPONENTIALS
{
Since n x × n x × n x × … × n x = x
n times
it follows that
and so
n log n x = log x
log n x = n1 log x
The logarithm of a number to its own base
Since 51 = 5, it follows that log5 5 = 1.
Clearly the same is true for any number, and in general,
loga a = 1
Reciprocals
Another useful result is that, for any base,
log ⎛⎜ 1y ⎞⎟ = −log y
⎝ ⎠
This is a direct consequence of the division law
loga ⎛⎜ x ⎞⎟ = loga x − loga y
⎝y⎠
with x set equal to 1:
⎛ ⎞
log ⎜ 1y ⎟ = log 1 − log y
⎝ ⎠
= 0 − log y
= − log y
⎛1⎞
If the number y is greater than 1, it follows that⎜ y ⎟lies between 0 and 1 and
⎝ ⎠
⎛
⎞
1
log ⎜ ⎟ is negative. So for any base (#1), the logarithm of a number between
⎝y ⎠
0 and 1 is negative.You saw an example of this on page 29: log10 1 = −1.
( 10 )
⎛ ⎞
The result log ⎜ 1y ⎟ = −log y is often useful in simplifying expressions involving
⎝ ⎠
logarithms.
ACTIVITY 2.3
Draw the graph of y = log2 x, taking values of x like 81, 41, 21, 1, 2, 4, 8, 16.
Use your graph to estimate the value of 2.
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2.3 Graphs of logarithms
Whatever the value, a, of the base (a #1), the graph of y = loga x has the same
general shape (shown in Figure 2.5).
y
2
y = loga x
1
1
a
2.3 Graphs of logarithms
O
x
▲ Figure 2.5
The graph has the following properties.
» The curve crosses the x-axis at (1, 0).
» The curve only exists for positive values of x.
» The line x = 0 is an asymptote and for values of x between 0 and 1 the
curve lies below the x-axis.
» There is no limit to the height of the curve for large values of x, but its
gradient progressively decreases.
» The curve passes through the point (a, 1).
?
❯ Each of the points above can be justified by work that you have
already covered. How?
The relationship y = loga x may be rewritten as x = ay, and so the graph of
x = ay is exactly the same as that of y = loga x. Interchanging x and y has the
effect of reflecting the graph in the line y = x, and changing the relationship
into y = ax, as shown in Figure 2.6.
y
y = ax
a
y=x
y = loga x
1
O
1
a
x
▲ Figure 2.6
33
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02/02/18 1:12 PM
The function y = a x, x ∈" is an exponential function. Notice that while
the domain of y = a x is all real numbers (x ∈"), the range is strictly the
positive real numbers. y = a x is the inverse of the logarithm function so the
domain of the logarithm function is strictly the positive real numbers and its
range is all real numbers. Remember that the effect of applying a function
followed by its inverse is to bring you back to where you started.
2
2 LOGARITHMS AND EXPONENTIALS
Thus loga (ax) = x and a(loga x) = x.
Exercise 2B
1
2
2x = 32 ⇔ x = log2 32
Write similar logarithmic equivalents of these equations. In each case
find also the value of x, using your knowledge of indices and not using
your calculator.
(i) 3x = 9
(ii)
4x = 64
1
1
(iii) 2x = 4
(iv)
5x = 5
(v) 7x = 1
(vi)
16x = 2
Write the equivalent of these equations in exponential form. Without
using your calculator, find also the value of y in each case.
(i) y = log3 9
(ii)
y = log5 125
(iii) y = log2 16
(iv)
y = log6 1
1
(v) y = log64 8
(vi)
y = log5 25
Write down the values of the following without using a calculator. Use your
calculator to check your answers for those questions which use base 10.
1
(i) log10 10 000
(ii) log10
( )
3
(iii) log10 10
(iv)
log3 81
(vi)
(v)
4
(ix) log4 2
(x)
( )
(ii)
log 6 − log 3
(iii) 2 log 6
(iv)
−log 7
1
2 log 9
(vi)
1
4 log 16 + log 2
1
log5 125
Write the following expressions in the form log x where x is a number.
log 5 + log 2
(vii) log 5 + 3 log 2 − log 10
9781510421738.indb 34
(viii) log 12 − 2 log 2 − log 9
16 + 2 log ( )
(x)
2 log 4 + log 9 − 2 log 144
Express the following in terms of log x.
1
(ix)
1
2 log
(i)
log x 2
(ii)
(iii) log x
(iv)
1
2
log x 5 − 2 log x
3
log x 2 + log 3 x
3 log x + log x 3
(vi)
log ( x )
Sketch each of the graphs below. Show the vertical asymptote and the
coordinates of the point where the graph crosses the x-axis.
(i) y = log10 x
(ii)
y = log2 (x + 1)
(iii) y = log2 (x − 2)
(v)
34
( )
1
log3 81
(viii) log3 4 3
(v)
6
10 000
(vii) log3 27
(i)
5
( )
log10 1
5
02/02/18 1:12 PM
7 Solve these inequalities.
2x " 128
(ii)
3x + 5 % 32
(iii) 4x + 6 % 70
(iv)
0.6x " 0.8
(vi)
0.5x + 0.2 $ 1
(i)
(v)
0.4x − 0.1 % 0.3
(vii) 2 $ 5x " 8
PS
M
(x)
2.3 Graphs of logarithms
PS
(viii) 1 $ 7x " 5
| 5x − 7 | " 4
8 Express the following as a single logarithm.
2 log10 x − log10 7
Hence solve
2 log10 x − log10 7 = log10 63.
9 Use logarithms to the base 10 to solve the following equations.
(i) 2x = 1 000 000
(ii) 2x = 0.001
(iii) 1.08x = 2
(iv) 1.1x = 100
(v) 0.99x = 0.000 001
(vi) 3(2x+1) = 20
(vii) 2(5x –2) = 30
(viii) 23x = 5(2x+1)
(x+3)
(x+4)
(ix) 7
=5
10 Solve the following equations.
(i) 22x + 3(2x) − 18 = 0
(ii) 52x + 5x − 2 = 0
(iii) 2(32x) − 7(3x) − 4 = 0
(iv) 62x − 5(6x) + 6 = 0
(v) 3(2x+1) + 5(3x) − 2 = 0
(vi) 2(72x) − 7(x+1) + 3 = 0
11 A geometric sequence has first term 5 and common ratio 7.
The kth term is 28 824 005.
Use logarithms to find the value of k.
12 Find how many terms there are in these geometric sequences.
(i) −1, 2, −4, 8, …, −16 777 216
(ii) 0.1, 0.3, 0.9, 2.7, …, 4 304 672.1
13 The strength of an earthquake is recorded on the Richter scale. This
provides a measure of the energy released. A formula for calculating
earthquake strength is
⎛ energy released in joules ⎞
⎟
earthquake strength = log10 ⎜
63000
⎝
⎠
(i) The energy released in an earthquake is estimated to be
2.5 × 1011 joules. What is its strength on the Richter scale?
(ii) An earthquake is 7.4 on the Richter scale. How much energy is
released in it?
(iii) An island had an earthquake measured at 4.2 on the Richter
scale. Some years later it had another earthquake and this one was
7.1. How many times more energy was released in the second
earthquake than in the first one?
(ix) | 2x − 4 | " 2
2
35
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2 LOGARITHMS AND EXPONENTIALS
2
M
14 The loudness of a sound is usually measured in decibels. A decibel is
1
10 of a bel, and a bel represents an increase in loudness by a factor of 10.
So a sound of 40 decibels is 10 times louder than one of 30 decibels;
similarly a sound of 100 decibels is 10 times louder than one of 90
decibels.
(i) Solve the equation x10 = 10.
(ii) Sound A is 35 decibels and sound B is 36 decibels. Show that (to
the nearest whole number) B is 26% louder than A.
(iii) How many decibels increase are equivalent to a doubling in
loudness? Give your answer to the nearest whole number.
(iv) Amir says ‘The percentage increase in loudness from 35 to
37 decibels is given by 37 − 35 × 100 = 5.7%’.
35
Explain Amir’s mistake.
15
Solve the inequality | y − 5 | < 1.
(ii) Hence solve the inequality | 3x − 5 | " 1, giving 3 significant
figures in your answer.
(i)
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q3 November 2007
16 Solve the equation 2|3x − 1| = 3x, giving your answers correct to 3
significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q2 November 2013
17 Using the substitution u = 3x, or otherwise, solve, correct to 3 significant
figures, the equation 3x = 2 + 3−x.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q4 June 2007
18 Solve the equation log10 (x + 9) = 2 + log10 x.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q1 June 2014
19 (i)
(ii)
Given that (x + 2) and (x + 3) are factors of 5x3 + ax2 + b,
find the values of the constants a and b.
When a and b have these values, factorise 5x3 + ax2 + b
completely, and hence solve the equation 53y + 1 + a ×52y + b = 0,
giving any answers correct to 3 significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q5 November 2014
20 The polynomial f(x) is defined by f(x) = 12x3 + 25x2 − 4x − 12.
(i)
(ii)
Show that f(−2) = 0 and factorise f(x) completely.
Given that 12 × 27y + 25 × 9y − 4 × 3y − 12 = 0,
state the value of 3y and hence find y correct to 3 significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q4 June 2011
36
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2.4 Modelling curves
When you obtain experimental data, you are often hoping to establish a
mathematical relationship between the variables in question. Should the data
fall on a straight line, you can do this easily because you know that a straight
line with gradient m and intercept c has equation y = mx + c.
Example 2.6
t
20
40
60
80
100
120
θ
16.3
20.4
24.2
28.5
32.0
36.3
100
120
t (seconds)
(i)
Plot a graph of θ against t.
(ii)
What is the relationship between θ and t ?
2.4 Modelling curves
In an experiment the temperature θ (in °C) was measured at different times t
(in seconds), in the early stages of a chemical reaction. The results are shown
in the table below.
2
Solution
(i)
40
35
30
25
20
15
10
0
20
40
60
80
▲ Figure 2.7
(ii)
Figure 2.7 shows that the points lie reasonably close to a straight line
and so it is possible to estimate its gradient and intercept.
Intercept: c = 12.3
36.3 − 16.3
Gradient: m = 120 − 20 = 0.2
In this case the equation is not y = mx + c but θ = mt + c, and so is given by
θ = 0.2t + 12.3
9781510421738.indb 37
37
02/02/18 1:12 PM
It is often the case, however, that your results do not end up lying on a
straight line but on a curve, so that this straightforward technique cannot be
applied. The appropriate use of logarithms can convert some curved graphs
into straight lines. This is the case if the relationship has one of two forms,
y = kxn or y = kax.
2 LOGARITHMS AND EXPONENTIALS
2
The techniques used in these two cases are illustrated in the following
examples. In theory, logarithms to any base may be used, but in practice
you would only use those available on your calculator: logarithms to the
base 10 and natural logarithms. The base of natural logarithms is a number,
2.718 28…, and is denoted by e. In the next section you will see how this
apparently unnatural number arises naturally; for the moment what is
important is that you can apply the techniques using base 10.
Relationships of the form y = kx n
Example 2.7
A water pipe is going to be laid between two points and an investigation is
carried out as to how, for a given pressure difference, the rate of flow R litres per
second varies with the diameter of the pipe d cm.The following data are collected.
d
1
2
3
5
10
R
0.02
0.32
1.62
12.53
199.80
It is suspected that the relationship between R and d may be of the form
R = kdn where k is a constant.
(i)
Explain how a graph of log d against log R tells you whether this is a
good model for the relationship.
(ii)
Make out a table of values of log10 d against log10 R and plot these on a graph.
(iii) If appropriate, use your graph to estimate the values of n and k.
Solution
(i)
If the relationship is of the form R = kdn, then taking logarithms gives
log R = log k + log dn
log R = n log d + log k.
or
This is in the form y = mx + c as n and log k are constants (so can replace
m and c) and log R and log d are variables (so can replace
y and x).
y
+
=
m x
log k
↔
n log d
↔
↔
=
↔
log R
+
c
So log R = n log d + log k is the equation of a straight line.
Consequently if the graph of log R against log d is a straight line, the
model R = kdn is appropriate for the relationship and n is given by the
38
9781510421738.indb 38
02/02/18 1:12 PM
gradient of the graph. The value of k is found from the intercept, log k,
of the graph with the vertical axis.
log10 k = intercept ⇒ k = 10intercept
(ii)
2
Working to 2 decimal places (you would find it hard to draw the
graph to greater accuracy) the logarithmic data are as follows.
0
0.30
0.48
0.70
1.00
log 10 R
−1.70
−0.49
0.21
1.10
2.30
2.4 Modelling curves
log 10 d
log10 R
3
2
1
0.4
0
0.2
0.6
0.8
1.0
1.2
log10 d
–1
–2
▲ Figure 2.8
(iii) In this case the graph in Figure 2.8 is indeed a straight line, with
gradient 4 and intercept −1.70, so n = 4 and k = 10–1.70 = 0.020 (to
2 significant figures).
The proposed equation linking R and d is a good model for their
relationship, and may be written as:
R = 0.02d 4
Exponential relationships
Example 2.8
The temperature in °C, θ, of a cup of coffee at time t minutes after it is made
is recorded as follows.
t
2
4
6
8
10
12
θ
81
70
61
52
45
38
(i)
Plot the graph of θ against t.
(ii)
Show how it is possible, by drawing a suitable graph, to test whether the
relationship between θ and t is of the form θ = kat, where k and a are
constants.
(iii) Carry out the procedure to find the values of a and of k.
➜
9781510421738.indb 39
39
02/02/18 1:12 PM
2
Solution
(i)
90
2 LOGARITHMS AND EXPONENTIALS
70
50
30
0
2
4
6
8
10
12
14
t (minutes)
▲ Figure 2.9
(ii)
If the relationship is of the form θ = kat, taking logarithms of both sides
gives
log θ = log k + log at
log θ = t log a + log k.
or
This is in the form y = mx + c as log a and log k are constants (so can
replace m and c) and log θ and t are variable (so can replace y and x).
=
y
m
x
+
log k
↔
log a t
↔
↔
=
↔
log θ
+
c
So log θ = t log a + log k is the equation of a straight line.
Consequently if the graph of log θ against t is a straight line, the model
θ = kat is appropriate for the relationship, and log a is given by the
gradient of the graph. The value of a is therefore found as
a = 10gradient. Similarly, the value of k is found from the intercept,
log10 k, of the line with the vertical axis: k = 10intercept.
(iii) The table gives values of log10 θ for the given values of t.
t
2
4
6
8
10
12
log10 θ
1.908
1.845
1.785
1.716
1.653
1.580
40
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02/02/18 1:12 PM
The graph of log 10 θ against t is as shown in Figure 2.10.
log10 θ
2.0
2
1.974
1.9
1.8
2.4 Modelling curves
1.7
1.6
1.5
0
2
4
6
8
10
12
14
t
▲ Figure 2.10
The graph is indeed a straight line so the proposed model is appropriate.
The gradient is −0.033 and so a = 10–0.033 = 0.927.
The intercept is 1.974 and so k = 101.974 = 94.2.
The relationship between θ and t is given by:
θ = 94.2 × 0.927t
Note
Because the base of the exponential function, 0.927, is less than 1, the
function’s value decreases rather than increases with t.
Exercise 2C
M
1
The planet Saturn has many moons. The table below gives the mean
radius of orbit and the time taken to complete one orbit for five of the
best-known of them.
Moon
Tethys
Dione
Rhea
Titan
Iapetus
Radius R (× 105 km)
2.9
3.8
5.3
12.2
35.6
Period T (days)
1.9
2.7
4.5
15.9
79.3
It is believed that the relationship between R and T is of the form R = kT n.
(i) How can this be tested by plotting log R against log T?
(ii)
Make out a table of values of log R and log T and draw the graph.
(iii) Use your graph to estimate the values of k and n.
9781510421738.indb 41
41
02/02/18 1:12 PM
2
2 LOGARITHMS AND EXPONENTIALS
M
2
In 1980 a Voyager spacecraft photographed several previously unknown
moons of Saturn. One of these, named 1980 S.27, has a mean orbital
radius of 1.4 × 10 5 km.
(iv) Estimate how many days it takes this moon to orbit Saturn.
The table below shows the area, A cm2, occupied by a patch of mould at
time t days since measurements were started.
t
0
1
2
3
4
5
A
0.9
1.3
1.8
2.5
3.5
5.2
It is believed that A may be modelled by a relationship of the form A = kb t .
(i) Show that the model may be written as log A = t log b + log k.
(ii) What graph must be plotted to test this model?
(iii) Plot the graph and use it to estimate the values of b and k.
(iv) (a) Estimate the time when the area of mould was 2 cm 2.
Estimate the area of the mould after 3.5 days.
(v) How is this sort of growth pattern described?
The inhabitants of an island are worried about the rate of deforestation
taking place. A research worker uses records over the last 200 years to
estimate the number of trees at different dates.
It is suggested that the number of trees, N, has been decreasing
exponentially with the number of years, t, since 1930, so that N may be
modelled by the equation
N = kat
where k and a are constants.
(i) Show that the model may be written as log N = t log a + log k.
The diagram shows the graph of log N against t.
(b)
M
3
log N
6.6
6.4
6.2
6
5.8
0
(ii)
20
40
60
80
100 t
Estimate the values of k and a.
What is the significance of k?
42
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02/02/18 1:12 PM
M
4
The time after a train leaves a station is recorded in minutes as t and
the distance that it has travelled in metres as s. It is suggested that the
relationship between s and t is of the form s = ktn where k and n are
constants.
(i) Show that the graph of log s against log t produces a straight line.
The diagram shows the graph of log s against log t.
2
log s
2.4 Modelling curves
4
3
2
1
–0.4
–0.2
0
0.2
0.6 log t
0.4
Estimate the values of k and n.
(iii) Estimate how far the train travelled in its first 100 seconds.
(iv) Explain why you would be wrong to use your results to estimate
the distance the train has travelled after 10 minutes.
The variables t and A satisfy the equation A = kb t, where b and k are
constants.
(i) Show that the graph of log A against t produces a straight line.
The graph of log A against t passes through the points (0, 0.2) and (4, 0.75).
(ii)
5
log A
(4, 0.75)
(0, 0.2)
O
(ii)
t
Find the values of b and k.
43
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02/02/18 1:12 PM
2 LOGARITHMS AND EXPONENTIALS
2
CP
M
M
6
7
8
An experimenter takes observations of a quantity y for various values of
a variable x. He wishes to test whether these observations conform to a
formula y = A × xB and, if so, to find the values of the constants A and B.
Take logarithms of both sides of the formula. Use the result to explain
what he should do, what will happen if there is no relationship, and if
there is one, how to find A and B.
Carry this out accurately on graph paper for the observations in the
table, and record clearly the resulting formula if there is one.
(i)
x
4
7
10
13
20
y
3
3.97
4.74
5.41
6.71
After the introduction of a vaccine, the numbers of new cases of a
virus infection in successive weeks are given in the table below.
Week number, t
1
2
3
4
5
No. of new cases, y
240
150
95
58
38
Plot the points (t, y) on graph paper and join them with a smooth curve.
(ii) Decide on a suitable model to relate t and y, and plot an
appropriate graph to find this model. (You may wish to use a
spreadsheet or graphing software.)
(iii) Find the value of y when t = 15, and explain what you find.
A local newspaper wrote an article about the number of residents using a
new online shopping site based in the town. Based on their own surveys,
they gave the following estimates.
Time in weeks, t
Number of people, P
2
3
4
5
6
4600
5000
5300
5500
5700
Investigate possible models for the number of people, P, using the
site at time t weeks after the site began, and determine which one
matches the data most closely.
(ii) Use your model to estimate how many people signed on in the first
week that the site was open.
(iii) Discuss the long-term validity of the model.
The variables x and y satisfy the relation 3y = 4x + 2.
(i) By taking logarithms, show that the graph of y against x is a straight
line. Find the exact value of the gradient of this line.
(ii) Calculate the x coordinate of the point of intersection of this line
with the line y = 2x, giving your answer correct to 2 decimal places.
(i)
9
44
9781510421738.indb 44
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q2 June 2007
02/02/18 1:12 PM
2.5 The natural logarithm function
The shaded region in Figure 2.11 is bounded by the x-axis, the lines x = 1 and
31
1
x = 3, and the curve y = x .The area of this region may be represented by x dx.
1
∫
2
y
O
1
3
x
▲ Figure 2.11
?
❯ Explain why you cannot apply the rule
kx n +1
kxn dx = n + 1 + c
to this integral.
∫
2.5 The natural logarithm function
1
y= x
However, the area in the diagram clearly has a definite value, and so we need
to find ways to express and calculate it.
ACTIVITY 2.4
Estimate, using numerical integration (for example by dividing the area up
into a number of strips), the areas represented by these integrals.
31
21
6
1 dx
dx
dx
(i)
(ii)
(iii)
x
x
1
1
1x
What relationship can you see between your answers?
∫
∫
∫
1
The area under the curve y = x between x = 1 and x = a, that is
a
∫1 x1 dx,
depends on the value a. For every value of a (greater than 1) there is a
definite value of the area. Consequently, the area is a function of a.
To investigate this function you need to give it a name, say L, so that L(a) is the area
from 1 to a and L(x) is the area from 1 to x.Then look at the properties of L(x) to
see if its behaviour is like that of any other function with which you are familiar.
The investigation you have just done should have suggested to you that
31
2
6
dx + x1 dx = x1 dx.
x
1
1
1
∫
∫
This can now be written as
L(3) + L(2) = L(6).
This suggests a possible law, that
L(a) + L(b) = L(ab).
9781510421738.indb 45
∫
45
02/02/18 1:12 PM
At this stage this is just a conjecture, based on one particular example. To
prove it, you need to take the general case and this is done in the activity
below. (At first reading you may prefer to leave the activity, accepting that the
result can be proved.)
2
ACTIVITY 2.5
2 LOGARITHMS AND EXPONENTIALS
CP
Prove that L(a) + L(b) = L(ab), by following the steps below.
(i)
Explain, with the aid of a diagram, why
ab
L(a) + 1 dx = L(ab).
a x
(ii) Now call x = az, so that dx can be replaced by a dz. Show that
∫
ab 1
∫a
x dx =
Explain why
∫
b
1 dz.
1z
b
∫1 z1 dz = L(b).
Notice that the limits of the left-hand
integral, ab and a, are values for x
but those for the right-hand integral,
b and 1, are values for z. So, to find
the new limits for the right-hand
integral, you should find z when x = a
(the lower limit) and when x = ab (the
upper limit). Remember az = x.
(iii) Use the results from parts (i) and (ii) to show that
L(a) + L(b) = L(ab).
What function has this property? For all logarithms
log(a) + log(b) = log(ab).
Could it be that this is a logarithmic function?
ACTIVITY 2.6
Satisfy yourself that the function has the following properties of
logarithms.
(i)
L(1) = 0
(ii)
L(a) − L(b) = L
(iii) L(an) = nL(a)
( ba )
The base of the logarithm function L(x)
Having accepted that L(x) is indeed a logarithmic function (for x > 0), the
remaining problem is to find the base of the logarithm. By convention this is
denoted by the letter e. A further property of logarithms is that for any base p
logp p = 1 ( p # 1).
46
9781510421738.indb 46
So to find the base e, you need to find the point such that the area L(e) under
the graph is 1. See Figure 2.12.
02/02/18 1:12 PM
y
O
2
1
e
x
You have already estimated the value of L(2) to be about 0.7 and that of L(3)
to be about 1.1 so the value of e is between 2 and 3.
ACTIVITY 2.7
You will need a calculator with an area-finding facility, or other suitable
technology, to do this. If you do not have this, read on.
e
Use the fact that 1 dx = 1 to find the value of e, knowing that it lies
1x
between 2 and 3, to 2 decimal places.
∫
2.5 The natural logarithm function
▲ Figure 2.12
The value of e is given to 9 decimal places in the key points on page 55. Like
π, e is a number which occurs naturally within mathematics. It is irrational:
when written as a decimal, it never terminates and has no recurring pattern.
The function L(x) is thus the logarithm of x to the base e, loge x. This is often
called the natural logarithm of x, and written as ln x.
Values of x between 0 and 1
So far it has been assumed that the domain of the function ln x is the real
numbers greater than 1 (x ∈ ", x > 1). However, the domain of ln x also
includes values of x between 0 and 1. As an example of a value of x between
0 and 1, look at ln 21 .
a
= ln a − ln b
Since
ln
b
()
⇒
(1)
ln 2 = ln 1 − ln 2 = −ln 2 (since ln 1 = 0)
In the same way, you can show that for any value of x between 0 and 1, the
value of ln x is negative.
When the value of x is very close to zero, the value of ln x is a large negative
number.
ln 1 = −ln 1000 = −6.9
(1000 )
(
)
ln 10001000 = −ln 1 000 000 = −13.8
So as x → 0, ln x → −∞ (for positive values of x).
47
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02/02/18 1:12 PM
2
The graph of the natural logarithm function
The graph of the natural logarithm function (shown in Figure 2.13) has the
characteristic shape of all logarithmic functions and like other such functions
it is only defined for x # 0. The value of ln x increases without limit, but ever
more slowly: it has been described as ‘the slowest way to get to infinity’.
2 LOGARITHMS AND EXPONENTIALS
y
O
y = ln x
1
x
▲ Figure 2.13
Historical note
Logarithms were discovered independently by John Napier (1550–1617), who
lived at Merchiston Castle in Edinburgh, and Jolst Bürgi (1552–1632) from
Switzerland. It is generally believed that Napier had the idea first, and so he
is credited with their discovery. Natural logarithms are also called Naperian
logarithms but there is no basis for this since Napier’s logarithms were
definitely not the same as natural logarithms. Napier was deeply involved
in the political and religious events of his day and mathematics and science
were little more than hobbies for him. He was a man of remarkable ingenuity
and imagination and also drew plans for war chariots that look very like
modern tanks, and for submarines.
2.6 The exponential function
Making x the subject of y = ln x, using
the theory of logarithms you obtain
x = ey.
y
y = ex
y=x
Interchanging x and y, which has the
effect of reflecting the graph in the line
y = x, gives the exponential function
y = ex.
The graphs of the natural logarithm
function and its inverse are shown in
Figure 2.14.
y = ln x
O
x
You saw in Pure Mathematics 1
▲ Figure 2.14
Chapter 5 that reflecting in the line
y = x gives an inverse function, so it follows that ex and ln x are each the
inverse of the other.
Notice that e ln x = x, using the definition of logarithms, and ln(ex ) = x ln e = x.
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Although the function ex is called the exponential function, in fact any
function of the form a x is exponential. Figure 2.15 shows several exponential
curves.
y
y = 3x
y = ex
2
y = 2x
y = 1.5x
2.6 The exponential function
y = 1x = 1
y = 0.5x
x
O
▲ Figure 2.15
The exponential function y = ex increases at an ever-increasing rate. This is
described as exponential growth.
By contrast, the graph of y = e –x, shown in Figure 2.16, approaches the x-axis
ever more slowly as x increases. This is called exponential decay.
y
y = e –x
1
O
x
▲ Figure 2.16
You will meet e x and ln x again later in this book. In Chapter 4 you learn
how to differentiate these functions and in Chapter 5 you learn how to
integrate them. In this secion you focus on practical applications which
require you to use the In key on your calculator.
Example 2.9
The number, N, of insects in a colony is given by N = 2000 e0.1t where t is
the number of days after observations have begun.
(i)
Sketch the graph of N against t.
(ii)
What is the population of the colony after 20 days?
(iii) How long does it take the colony to reach a population of 10 000?
➜
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Solution
2
(i)
N
N = 2000 e0.1t
2000
When t = 0, N = 2000e0 = 2000
2 LOGARITHMS AND EXPONENTIALS
O
t
▲ Figure 2.17
(ii)
When t = 20,
N = 2000 e0.1 × 20 = 14 778
The population is 14 778 insects.
(iii) When N = 10 000,
10 000 = 2000 e0.1t
5 = e0.1t
Taking natural logarithms of both sides,
ln 5 = ln(e0.1t )
Remember
ln 5 = 0.1t
ln(ex) = x.
and so
t = 10 ln 5
t = 16.09…
It takes just over 16 days for the population to reach 10 000.
Example 2.10
The radioactive mass, M grams in a lump of material is given by M = 25e–0.0012t
where t is the time in seconds since the first observation.
(i)
Sketch the graph of M against t.
(ii)
What is the initial size of the mass?
(iii) What is the mass after 1 hour?
(iv) The half-life of a radioactive substance is the time it takes to decay to
half of its mass. What is the half-life of this material?
Solution
(i)
M
25
O
t
▲ Figure 2.18
(ii)
When t = 0,
M = 25e0
M = 25
The initial mass is 25 g.
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t = 3600
M = 25e–0.0012×3600
M = 0.3324...
(iii) After 1 hour,
2
The mass after 1 hour is 0.33 g (to 2 decimal places).
(iv) The initial mass is 25 g, so after one half-life,
1
M = 2 × 25 = 12.5 g
2.6 The exponential function
At this point the value of t is given by
12.5 = 25e–0.0012t
⇒
0.5 = e–0.0012t
Taking logarithms of both sides:
ln 0.5 = ln e –0.0012t
ln 0.5 = −0.0012t
ln 0.5
⇒
t=
–0.0012
t = 577.6 (to 1 decimal place).
The half-life is 577.6 seconds. (This is just under 10 minutes, so the
substance is highly radioactive.)
Example 2.11
Make p the subject of ln(p) − ln(1 − p) = t.
Solution
⎛ p ⎞
ln ⎜
⎟ =t
⎝1 – p ⎠
Writing both sides as powers of e gives
e (1– p ) = et
p
t
1– p = e
p = et(1 − p)
ln
p
Using log a − log b = log
(ba )
Remember eln x = x
p = et − pet
p + pet = et
p(1 + et) = et
et
p = 1 + et
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2
Example 2.12
Solve these equations.
(i) ln (x − 4) = ln x − 4
(ii)
e2x + ex = 6
Solution
(i)
ln (x − 4) = ln x − 4
2 LOGARITHMS AND EXPONENTIALS
⇒ x − 4 = eln x−4
x − 4 = elnx e−4
x − 4 = x e−4
Rearrange to get all the x terms on one side:
x − x e−4 = 4
x(l −e−4) = 4
4
x = 1 − e −4
So x = 4.07 (to 3 s.f.)
(ii)
e2x + ex = 6 is a quadratic equation in ex.
Substituting u = ex:
u2 + u = 6
u2 + u − 6 = 0
So
Factorising: (u − 2)(u + 3) = 0
So u = 2 or u = −3.
Since u = ex then ex = 2 or ex = −3.
ex = −3 has no solution.
ex = 2 ⇒ x = ln 2
So
x = 0.693 (to 3 s.f.)
Exercise 2D
1
2
3
4
5
M
6
Make x the subject of ln x − ln x0 = kt.
Make t the subject of s = s0e–kt.
Make p the subject of ln p = −0.02t.
Make x the subject of y − 5 = (y0 − 5)ex.
Solve these equations.
(i) ln(3 − x) = 4 + ln x
(ii) ln(x + 5) = 5 + ln x
(iii) ln(2 − x) = 2 + ln x
4
(iv) ex = x
e
(v) e2x − 8ex + 16 = 0
(vi) e2x + ex = 12
A colony of humans settle on a previously uninhabited planet. After
t years, their population, P, is given by P = 100e0.05t.
(i) Sketch the graph of P against t.
(ii) How many settlers land on the planet initially?
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(iii) What is the population after 50 years?
(iv) How long does it take the population to reach 1 million?
M
7 The height h metres of a species of pine tree t years after planting is
ln y
5
4
2
2.6 The exponential function
modelled by the equation h = 20 − 19 × 0.9t.
(i) What is the height of the trees when they are planted?
(ii) Calculate the height of the trees after 2 years, and the time taken
for the height to reach 10 metres.
The relationship between the market value $y of the timber from the
tree and the height h metres of the tree is modelled by the equation
y = ahb, where a and b are constants.
The diagram shows the graph of ln y plotted against ln h.
3
2
1
0
0.5
1
1.5
2
2.5
ln h
–1
–2
(iii) Use the graph to calculate the values of a and b.
(iv) Calculate how long it takes to grow trees worth $100.
8 It is given that ln(y + 5) − ln y = 2 ln x. Express y in terms of x, in a form
not involving logarithms.
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q2 November 2009
9 Given that (1.25)x = (2.5)y, use logarithms to find the value of
to 3 significant figures.
x
y correct
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q1 June 2009
10 Solve, correct to 3 significant figures, the equation
ex + e2x = e3x.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q2 June 2008
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2
11 The variables x and y satisfy the equation y = A(b–x), where A and b are
constants. The graph of ln y against x is a straight line passing through the
points (0, 1.3) and (1.6, 0.9), as shown in the diagram.
ln y
(0, 1.3)
2 LOGARITHMS AND EXPONENTIALS
(1.6, 0.9)
O
x
Find the values of A and b, correct to 2 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q3 November 2008
12 Solve the equation ln(2 + e−x) = 2, giving your answer correct to
2 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q1 June 2009
13 The variables x and y satisfy the equation y = Kx m , where K and m are
constants. The graph of ln y against ln x is a straight line passing through
the points (0, 2.0) and (6, 10.2), as shown in the diagram.
ln y
(6, 10.2)
(0, 2.0)
ln x
O
Find the values of K and m, correct to 2 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q3 June 2011
14
ln y
(5, 2.92)
(2, 1.60)
O
x
The variables x and y satisfy the equation
p(x–1)
y = Ae
, where A and p are constants.
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The graph of ln y against x is a straight line passing through the points
(2, 1.60) and (5, 2.92), as shown in the diagram.
Find the values of A and p correct to 2 significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q2 June 2015
1
2
3
4
5
6
7
8
A function of the form ax is described as exponential.
y = loga x ⇔ ay = x.
Logarithms to any base
Multiplication:
log xy = log x + log y
⎛ ⎞
Division:
log ⎜ xy ⎟ = log x − log y
⎝ ⎠
Logarithm of 1:
log 1 = 0
Powers:
log xn = n log x
⎛ ⎞
Reciprocals:
log ⎜ 1 ⎟ = −log y
⎝y ⎠
1
Roots:
log n x = n log x
Logarithm to its own base: loga a = 1
Logarithms may be used to discover the relationship between the
variables in two types of situation.
n
● y = kx ⇔ log y = log k + n log x
Plot log y against log x: this relationship gives a straight line where n is
the gradient and log k is the intercept.
x
● y = ka ⇔ log y = log k + x log a
Plot log y against x: this relationship gives a straight line where log a is
the gradient and log k is the intercept.
logex is called the natural logarithm of x and denoted by ln x.
e = 2.718 281 828 4… is the base of natural logarithms.
ex and ln x are inverse functions: elnx = x and ln(ex) = x.
1
x dx = ln| x | + c
2.6 The exponential function
KEY POINTS
2
∫
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2
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
2 LOGARITHMS AND EXPONENTIALS
■
■
■
■
■
■
■
■
■
■
■
■
■
sketch graphs of exponential functions
■ showing where the graph crosses the y-axis
■ showing the horizontal asymptote
x
■ know the difference in the shape of y = a when a < 1 and a > 1
use exponential models for real-life situations
understand that the logarithm function is the inverse of the exponential
function
use the laws of logarithms
■ to rewrite combinations of logs as the log of a single expression
■ to split up a single logarithm into a combination of logarithms
sketch the graphs of logarithmic functions
use logarithms to solve equations and inequalities with an unknown
power
use the number e and the exponential function in context
sketch a transformation of the graph of y = ex
understand and use the natural logarithm function
rewrite log statements as exponential statements and vice versa
sketch transformations of the graph of y = ln x
model curves by plotting ln y against ln x for curves of the form y = kxn
using the gradient and intercept of the new graph to find the values of
k and n
model curves by plotting ln y against x for curves of the form y = kax
using the gradient and intercept of the new graph to find the values of
k and a.
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P2
P3
3
Trigonometry
3 Trigonometry
Music, when
soft voices
die, vibrates
in the memory
Percy Bysshe
Shelley
(1792−1822)
Many waves can be modelled as a sine curve. Estimate the amplitude in
metres of the wave in the picture above (see Figure 3.1).
y
a
y = a sin bx
Amplitude
O
–a
π
b
2π
b
3π
b
x
Wavelength
▲ Figure 3.1
?
❯ Use your estimates to suggest values of a and b which would make
y = a sin bx a suitable model for the curve.
❯ Do you think a sine curve is a good model for the wave?
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3.1 Reciprocal trigonometrical functions
3
As well as the three main trigonometrical functions, there are three more
which are commonly used. These are their reciprocals − cosecant (cosec),
secant (sec) and cotangent (cot), defined by
1
cos θ
sec θ = 1 ;
cot θ = tan θ = sin θ .
cosec θ = 1 ;
cos θ
sin θ
Each of these is undefined for certain values of θ. For example, cosec θ is
undefined for θ = 0°, 180°, 360°, … since sin θ is zero for these values of θ.
3 TRIGONOMETRY
(
)
Figure 3.2 shows the graphs of these functions. Notice how all three of the
functions have asymptotes at intervals of 180°. Each of the graphs shows one
of the main trigonometrical functions as a red line and the related reciprocal
function as a blue line.
y
y
1
1
–360º
–180º
0
–1
180º
360º x
y = sin x
–360º
–180º
y = cos x
0
–1
y = cosec x
180º
360º x
y = sec x
y
y = tan x
1
–360º
–180º
0
–1
180º
360º
x
y = cotx
▲ Figure 3.2
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Using the definitions of the reciprocal functions two alternative
trigonometrical forms of Pythagoras’ theorem can be obtained.
(i)
3
sin2 θ + cos2 θ ≡ 1
2
2
Dividing both sides by cos2 θ : sin 2 θ + cos 2 θ ≡ 12
cos θ cos θ cos θ
⇒
tan2 θ + 1 ≡ sec2 θ.
This identity is sometimes used in mechanics.
sin2 θ + cos2 θ ≡ 1
2
2
Dividing both sides by sin2 θ : sin 2 θ + cos2 θ ≡ 12
sin θ sin θ sin θ
⇒ 1 + cot2 θ ≡ cosec2 θ.
Questions concerning reciprocal functions are usually most easily solved by
considering the related function, as in the following examples.
Example 3.1
Find cosec 120° leaving your answer in surd form.
Solution
Example 3.2
1
sin120°
3
=1÷ 2
= 2
3
cosec 120° =
3.1 Reciprocal trigonometrical functions
(ii)
Find values of θ in the interval 0° $ θ $ 360° for which sec2 θ = 4 + 2 tan θ.
Solution
First you need to obtain an equation containing only one trigonometrical
function.
sec2 θ = 4 + 2 tan θ
⇒
tan2 θ + 1 = 4 + 2 tan θ
⇒
tan2 θ − 2 tan θ − 3 = 0
⇒
(tan θ − 3)(tan θ + 1) = 0
⇒
tan θ = 3 or tan θ = −1
tan θ = 3
⇒ θ = 71.6°
(calculator)
or θ = 71.6° + 180° = 251.6° (see Figure 3.3, overleaf )
tan θ = −1
⇒ θ = −45° (not in the required range)
or θ = −45° + 180° = 135°
or θ = 135° + 180° = 315°
(see Figure 3.3)
➜
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y
3
3
y = tan θ
0
180º
360º
3 TRIGONOMETRY
θ
–1
▲ Figure 3.3
The values of θ are 71.6°, 135°, 251.6°, 315°.
Exercise 3A
1
2
3
4
5
Solve the following equations for 0° $ x $ 360°.
(i) cosec x = 1
(ii) sec x = 2
(iii) cot x = 4
(iv) sec x = −3
(v) cot x = −1
(vi) cosec x = −2
Find the following giving your answers as fractions or in surd form.
You should not need your calculator.
(i) cot 135°
(ii) sec 150°
(iii) cosec 240°
(iv) sec 210°
(v) cot 270°
(vi) cosec 225°
In triangle ABC, angle A = 90° and sec B = 2.
(i) Find the angles B and C.
(ii) Find tan B.
(iii) Show that 1 + tan2 B = sec2 B.
In triangle LMN, angle M = 90° and cot N = 1.
(i) Find the angles L and N.
(ii) Find sec L, cosec L and tan L.
(iii) Show that 1 + tan2 L = sec2 L.
Malini is 1.5 m tall.
At 8 pm one evening her shadow is 6 m long.
Given that the angle of elevation of the sun at that moment is α
(i) show that cot α = 4
(ii) find α.
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CP
6
3.2 Compound-angle formulae
The photograph at the start of this chapter shows just one of the countless
examples of waves and oscillations that are part of the world around us.
3
3.2 Compound-angle formulae
7
For what values of α, where 0° $ α $ 360°, are sec α, cosec α and
cot α all positive?
(ii) Are there any values of α for which sec α, cosec α and cot α are all
negative?
Explain your answer.
(iii) Are there any values of α for which sec α, cosec α and cot α are all
equal?
Explain your answer.
Solve the following equations for 0° $ x $ 360°.
(i) cos x = sec x
(ii) cosec x = sec x
(iii) 2 sin x = 3 cot x
(iv) cosec2 x + cot2 x = 2
(v) 3 sec2 x − 10 tan x = 0
(vi) 1 + cot2 x = 2 tan2 x
(i)
Because such phenomena are modelled by trigonometrical (and especially
sine and cosine) functions, trigonometry has an importance in mathematics
far beyond its origins in right-angled triangles.
ACTIVITY 3.1
Find an acute angle θ so that sin(θ + 60°) = cos(θ − 60°).
Tip: Try drawing graphs and searching for a numerical solution.
You should be able to find the solution using either of these methods,
but replacing 60° by, for example, 35° would make both of these methods
rather tedious. In this chapter you will meet some formulae which help
you to solve such equations more efficiently.
It is tempting to think that sin(θ + 60°) should equal sin θ + sin 60°, but this
is not so, as you can see by substituting a numerical value of θ. For example,
putting θ = 30° gives sin(θ + 60°) = 1, but sin θ + sin 60° ≈ 1.366.
To find an expression for sin(θ + 60°), you would use the compound-angle
formula
sin(θ + φ) = sin θ cos φ + cos θ sin φ.
This is proved on the next page in the case when θ and φ are acute angles.
It is, however, true for all values of the angles. It is an identity.
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3
?
PS
❯ As you work through this proof make a list of all the results you are
assuming.
C
φ
θ
a
b
3 TRIGONOMETRY
h
A
B
D
▲ Figure 3.4
Using the trigonometrical formula for the area of a triangle (see Figure 3.4):
area ABC = area ADC + area DBC
1
1
1
2 ab sin(θ + φ) = 2 bh sin θ + 2 ah sin φ
h = a cos φ
from #DBC
⇒
h = b cos θ
from #ADC
ab sin(θ + φ) = ab sin θ cos φ + ab cos θ sin φ
which gives
sin(θ + φ) = sin θ cos φ + cos θ sin φ
1
!
This is the first of the compound-angle formulae (or expansions), and it can
be used to prove several more. These are true for all values of θ and φ.
1 gives
Replacing φ by −φ in !
sin(θ − φ) = sin θ cos(−φ) + cos θ sin(−φ)
cos(−φ) = cos φ
sin(−φ) = −sin φ
⇒ sin(θ − φ) = sin θ cos φ − cos θ sin φ
2
!
ACTIVITY 3.2
Derive the rest of these formulae.
(i)
To find an expansion for cos(θ − φ) replace θ by (90° − θ ) in the
expansion of sin(θ + φ).
Tip: sin(90° − θ ) = cos θ and cos(90° − θ ) = sin θ
To find an expansion for cos(θ + φ) replace φ by (−φ) in the
expansion of cos(θ − φ).
sin (θ + φ )
(iii) To find an expansion for tan(θ + φ), write tan(θ + φ) =
.
cos (θ + φ )
(ii)
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Tip: After using the expansions of sin(θ + φ) and cos(θ + φ),
divide the numerator and the denominator of the resulting fraction
by cos θ cos φ to give an expansion in terms of tan θ and tan φ.
3
(iv) To find an expansion for tan(θ − φ) in terms of tan θ and tan φ,
replace φ by (−φ) in the expansion of tan(θ + φ).
?
PS
The four results obtained in Activity 3.2, together with the two previous
results, form the set of compound-angle formulae.
sin(θ + φ) = sin θ cos φ + cos θ sin φ
sin(θ − φ) = sin θ cos φ − cos θ sin φ
cos(θ + φ) = cos θ cos φ − sin θ sin φ
cos(θ − φ) = cos θ cos φ + sin θ sin φ
tan θ + tan φ
1 − tan θ tan φ
tan θ – tan φ
tan(θ − φ) =
1 + tan θ tan φ
tan(θ + φ) =
3.2 Compound-angle formulae
❯ Are your results valid for all values of θ and φ?
❯ Test your results with θ = 60°, φ = 30°.
(θ + φ) ≠ 90°, 270°, ...
(θ − φ) ≠ 90°, 270°, ...
You are now in a position to solve the earlier problem more easily. To find
an acute angle θ such that sin(θ + 60°) = cos(θ − 60°), you expand each side
using the compound-angle formulae.
sin(θ + 60°) = sin θ cos 60° + cos θ sin 60°
= 1 sin θ + 3 cos θ
2
2
cos(θ − 60°) = cos θ cos 60° + sin θ sin 60°
= 1 cos θ + 3 sin θ
2
2
1
!
2
!
2
1 and !
From !
1 sin θ + 3 cos θ = 1 cos θ + 3 sin θ
2
2
2
2
sin θ + 3 cos θ = cos θ + 3 sin θ
Collect like terms:
⇒
( 3 – 1) cos θ = ( 3 – 1) sin θ
cos θ = sin θ
Divide by cos θ :
1 = tan θ
θ = 45°
This gives an equation in
one trigonometrical ratio.
Since an acute angle was required, this is the only root.
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Uses of the compound-angle formulae
3
3 TRIGONOMETRY
You have already seen compound-angle formulae used in solving a
trigonometrical equation and this is quite a common application of them.
However, their significance goes well beyond that since they form the basis
for a number of important techniques. Those covered in this book are as
follows.
» The derivation of double-angle formulae
The derivation and uses of these are covered on pages 67 to 70.
» The addition of different sine and cosine functions
This is covered on pages 72 to 75. It is included here because the basic
wave form is a sine curve. It has many applications, for example in applied
mathematics, physics and chemistry.
» Calculus of trigonometrical functions
This is covered in Chapters 4 and 5 and also in Chapter 8. Proofs of the
results depend on using either the compound-angle formulae or the
factor formulae which are derived from them.
You will see from this that the compound-angle formulae are important in
the development of the subject. Some people learn them by heart, others
think it is safer to look them up when they are needed. Whichever policy
you adopt, you should understand these formulae and recognise their form.
Without that you will be unable to do the next example, which uses one of
them in reverse.
Example 3.3
Simplify cos θ cos 3θ − sin θ sin 3θ.
Solution
The formula which has the same pattern of cos cos − sin sin is
cos(θ + φ) = cos θ cos φ − sin θ sin φ
Using this, and replacing φ by 3θ, gives
cos θ cos 3θ − sin θ sin 3θ = cos(θ + 3θ )
= cos 4θ
Exercise 3B
1
2
Use the compound-angle formulae to write the following as surds.
(i) sin 75° = sin(45° + 30°)
(ii) cos 135° = cos(90° + 45°)
(iii) tan 15° = tan(45° − 30°)
(iv) tan 75° = tan(45° + 30°)
Expand each of the following expressions.
(i) sin(θ + 45°)
(ii) cos(θ − 30°)
(iii) sin(60° − θ )
(iv) cos(2θ + 45°)
(v) tan(θ + 45°)
(vi) tan(θ − 45°)
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3
4
PS
5
PS
( )
π
π
(ii) 2 cos (θ – ) = cos (θ + )
3
2
6
3
3.2 Compound-angle formulae
PS
Simplify each of the following expressions.
(i) sin 2θ cos φ − cos 2θ sin θ
(ii) cos φ cos 7φ − sin φ sin 7φ
(iii) sin 120° cos 60° + cos 120° sin 60°
(iv) cos θ cos θ − sin θ sin θ
Solve the following equations for values of θ in the range 0° $ θ $ 180°.
(i) cos(60° + θ ) = sin θ
(ii) sin(45° − θ ) = cos θ
(iii) tan(45° + θ ) = tan(45° − θ )
(iv) 2 sin θ = 3 cos(θ − 60°)
(v) sin θ = cos(θ + 120°)
Solve the following equations for values of θ in the range 0 $ θ $ π.
(When the range is given in radians, the solutions should be in radians,
using multiples of π where appropriate.)
π
(i) sin θ + 4 = cos θ
The diagram shows three points A (−4, 1), B (0, 4) and C (6, −2) joined
to form a triangle. The angles α and β and the point P are also shown in
the diagram.
y
B (0, 4)
α
A (−4, 1)
β
P
x
O
C (6, −2)
(i)
Show that sin α = 45 and write down the value of cos α.
Find the values of sin β and cos β.
7
(iii) Show that sin (α + β ) =
5 2
(iv) Show that tan (α + β ) = −7 and comment on the significance of the
negative value.
(ii)
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3 TRIGONOMETRY
3
7
The angle α lies between 0° and 90° and is such that
2 tan2 α + sec2 α = 5 − 4 tan α.
(i) Show that 3 tan2 α + 4 tan α − 4 = 0 and hence find the exact value
of tan α.
(ii) It is given that the angle β is such that cot(α + β ) = 6. Without
using a calculator, find the exact value of cot β.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q7 November 2014
8
(i)
(ii)
Show that the equation
tan(45° + x) − tan x = 2
can be written in the form
tan2 x + 2 tan x − 1 = 0.
Hence solve the equation
tan(45° + x) − tan x = 2,
giving all solutions in the interval 0° $ x $ 180°.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q5 November 2007
9
The angles α and β lie in the interval 0° " x " 180°, and are such that
tan α = 2 tan β and tan(α + β ) = 3.
Find the possible values of α and β.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q4 November 2009
10 (i)
(ii)
Show that the equation tan(30° + θ ) = 2 tan(60° − θ ) can be
written in the form
tan2 θ + (6 3) tan θ − 5 = 0.
Hence, or otherwise, solve the equation
tan(30° + θ ) = 2 tan(60° − θ ),
for 0° $ θ $ 180°.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q4 June 2008
11 The angles A and B are such that
sin(A + 45°) = (2 2)cos A and
4sec2 B + 5 = 12 tan B.
Without using a calculator, find the exact value of tan(A − B).
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q6 November 2015
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3.3 Double-angle formulae
As you work through these proofs, think how you can check the results.
❯ Is a check the same as a proof?
PS
sin(θ + φ) = sin θ cos φ + cos θ sin φ
When φ = θ, this becomes
sin(θ + θ ) = sin θ cos θ + cos θ sin θ
giving
(ii)
sin 2θ = 2 sin θ cos θ
cos(θ + φ ) = cos θ cos φ − sin θ sin φ
3.3 Double-angle formulae
Substituting φ = θ in the relevant compound-angle formulae leads
immediately to expressions for sin 2θ, cos 2θ and tan 2θ, as follows.
(i)
3
?
When φ = θ, this becomes
cos(θ + θ ) = cos θ cos θ − sin θ sin θ
giving
cos 2θ = cos2 θ − sin2 θ
Using the Pythagorean identity cos2 θ + sin2 θ = 1, two other forms for
cos 2θ can be obtained.
cos 2θ = (1 − sin2 θ ) − sin2 θ
⇒
cos 2θ = 1 − 2 sin2 θ
cos 2θ = cos2 θ − (1 − cos2 θ )
⇒
cos 2θ = 2 cos2 θ − 1
These alternative forms are often more useful since they contain only
one trigonometrical function.
(iii) tan(θ + φ) = tan θ + tan φ
1 – tan θ tan φ
(θ + φ) ≠ 90°, 270°, ...
When φ = θ, this becomes
tan(θ + θ ) =
giving
tan θ + tan θ
1 – tan θ tan θ
tan 2θ = 2 tan θ2
1 – tan θ
θ ≠ 45°, 135°, ...
Uses of the double-angle formulae
e
In modelling situations
You will meet situations, such as that on the next page, where using a
double-angle formula not only allows you to write an expression more neatly
but also thereby allows you to interpret its meaning more clearly.
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3
height
3 TRIGONOMETRY
u
α
R
O
ground
horizontal distance
▲ Figure 3.5
When an object is projected, such as a golf ball being hit as in Figure 3.5,
with speed u at an angle α to the horizontal over level ground, the horizontal
distance it travels before striking the ground, called its range, R, is given by
the product of the horizontal component of the velocity u cos α and its time
α
of flight 2u sin
g .
2
R = 2u singα cos α
Using the double-angle formula sin 2α = 2 sin α cos α allows this to be
written as
2
2α .
R = u sin
g
Since the maximum value of sin 2α is 1, it follows that the greatest value of
2
the range R is ug and that this occurs when 2α = 90° and so α = 45°. Thus
an angle of projection of 45° will give the maximum range of the projectile
over level ground. (This assumes that air resistance may be ignored.)
In this example, the double-angle formula enabled the expression for R to be
written tidily. However, it did more than that because it made it possible to
find the maximum value of R by inspection and without using calculus.
In calculus
The double-angle formulae allow a number of functions to be integrated and
you will meet some of these later (see page 130).
The formulae for cos 2θ are particularly useful in this respect since
cos 2θ = 1 − 2 sin2 θ
⇒
sin2 θ = 21(1 − cos 2θ )
cos 2θ = 2 cos2 θ − 1
⇒
cos2 θ = 21(1 + cos 2θ )
and
and these identities allow you to integrate sin2 θ and cos2 θ.
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In solving equations
You will sometimes need to solve equations involving both single and double
angles as shown by the next two examples.
Example 3.4
3
Solve the equation sin 2θ = sin θ for 0° $ θ $ 360°.
Solution
⇒
2 sin θ cos θ = sin θ
⇒
2 sin θ cos θ − sin θ = 0
⇒
sin θ (2 cos θ − 1) = 0
⇒
sin θ = 0 or cos θ = 21
The principal value is
the one which comes
from your calculator.
3.3 Double-angle formulae
sin 2θ = sin θ
Be careful here:
don’t cancel sin θ
or some roots
will be lost.
sin θ = 0 ⇒ θ = 0° (principal value) or 180° or 360° (see Figure 3.6)
y
1
y = sin θ
O
180°
360°
θ
–1
▲ Figure 3.6
cos θ = 21 ⇒ θ = 60° (principal value) or 300° (see Figure 3.7)
y
1
y = cos θ
1
2
O
60°
300°
360°
θ
–1
▲ Figure 3.7
The full set of roots for 0° $ θ $ 360° is θ = 0°, 60°, 180°, 300°, 360°.
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3
When an equation contains cos 2θ, you will save time if you take care to
choose the most suitable expansion.
Example 3.5
Solve 2 + cos 2θ = sin θ for 0 $ θ $ 2π.
3 TRIGONOMETRY
Solution
This is the most
suitable expansion
since the right-hand
side contains sin θ.
Using cos 2θ = 1 − 2 sin2 θ gives
2 + (1 − 2 sin2 θ ) = sin θ
⇒
2 sin2 θ + sin θ − 3 = 0
⇒
(2 sin θ + 3)(sin θ − 1) = 0
⇒ sin θ = − 23 (not valid since −1 $ sin θ $ 1)
or sin θ = 1
Figure 3.8 shows that the principal value θ = π
2
is the only root for 0 $ θ $ 2π.
y
1
O
Notice that the request
for 0 $ θ $ 2π, i.e. in
radians, is an invitation
to give the answer in
radians.
y = sin θ
π
2
π
2π
θ
▲ Figure 3.8
Exercise 3C
1
2
3
PS
4
5
Solve the following equations for 0° $ θ $ 360°.
(i) 2 sin 2θ = cos θ
(ii) tan 2θ = 4 tan θ
(iii) cos 2θ + sin θ = 0
(iv) tan θ tan 2θ = 1
(v) 2 cos 2θ = 1 + cos θ
Solve the following equations for −π $ θ $ π.
(i) sin 2θ = 2 sin θ
(ii) tan 2θ = 2 tan θ
(iii) cos 2θ − cos θ = 0
(iv) 1 + cos 2θ = 2 sin2 θ
(v) sin 4θ = cos 2θ
Tip: Write the expression in part (v) as an equation in 2θ.
By first writing sin 3θ as sin(2θ + θ ), express sin 3θ in terms of sin θ.
Hence solve the equation sin 3θ = sin θ for 0 $ θ $ 2π.
Solve cos 3θ = 1 − 3 cos θ for 0° $ θ $ 360°.
1 + cos 2θ
Simplify
.
sin 2θ
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CP
6
(i)
Sketch on the same axes the graphs of
y = cos 2x and y = sin x for 0 $ x $ 2π.
(ii)
Show that these curves meet at points whose x coordinates are
solutions of the equation
2 sin2 x + sin x − 1 = 0.
3
(iii) Solve this equation to find the values of x in terms of π for
CP
7
PS
8
CP
9
Show that 1 – tan 2 θ = cos 2θ.
1 + tan θ
(i) Show that tan π + θ tan π – θ = 1.
4
4
(ii) Given that tan 26.6° = 0.5, solve tan θ = 2 without using your
calculator. Give θ to 1 decimal place, where 0° " θ " 90°.
(i) Sketch on the same axes the graphs of
y = cos 2x and y = 3 sin x − 1 for 0 $ x $ 2π.
2
( ) (
)
3.3 Double-angle formulae
0 $ x $ 2π.
Show that these curves meet at points whose x coordinates are
solutions of the equation 2 sin2 x + 3 sin x − 2 = 0.
(iii) Solve this equation to find the values of x in terms of π for
0 $ x $ 2π.
10 (i) Show that cos(θ − 60°) + cos(θ + 60°) ≡ cos θ .
cos(2x − 60°) + cos(2x + 60°)
(ii) Given that
= 3 , find the exact value
cos( x − 60°) + cos( x + 60°)
of cos x.
(ii)
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q4 November 2014
11 Solve the equation tan 2x = 5cot x , for 0° < x < 180°.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q3 June 2013
12 (i)
(ii)
Prove the identity cosec 2θ + cot 2θ ≡ cot θ.
Hence solve the equation cosec 2θ + cot 2θ = 2, for 0° $ θ $ 360°.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q3 June 2009
13
It is given that cos a = 53 , where 0° $ a $ 90°. Showing your working
and without using a calculator to evaluate a,
(i) find the exact value of sin(a − 30)°,
(ii) find the exact value of tan 2a, and hence find the exact value of
tan 3a.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q3 June 2010
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3
3.4 The forms r cos(θ ± α), r sin(θ ± α)
Another modification of the compound-angle formulae allows you to simplify
expressions such as 4 sin θ + 3 cos θ and hence solve equations of the form
a sin θ + b cos θ = c.
3 TRIGONOMETRY
To find a single expression for 4 sin θ + 3 cos θ, you match it to the expression
r sin(θ + α) = r (sin θ cos α + cos θ sin α).
This is because the expansion of r sin(θ + α) has sin θ in the first term, cos θ
in the second term and a plus sign in between them. It is then possible to
choose appropriate values of r and α.
4 sin θ + 3 cos θ ≡ r(sin θ cos α + cos θ sin α)
Coefficients of sin θ :
Coefficients of cos θ :
4 = r cos α
3 = r sin α.
Looking at the right-angled triangle in Figure 3.9 gives the values for r and α.
r
r = 32 + 4 2
=5
The sides, 4 and 3, come
from the expression
4 sin θ + 3 cos θ.
3
α
4
▲ Figure 3.9
In this triangle, the hypotenuse is 4 2 + 3 2 = 5, which corresponds to r in the
expression above.
The angle α is given by
3
sin α = 5
and
cos α = 45
⇒
α = 36.9°.
So the expression becomes
4 sin θ + 3 cos θ = 5 sin(θ + 36.9°).
The steps involved in this procedure can be generalised to write
a sin θ + b cos θ = r sin(θ + α)
where
b
a
=b
cos α =
=a
a2 + b2 r
a2 + b2 r
The same expression may also be written as a cosine function. In this case,
rewrite 4 sin θ + 3 cos θ as 3 cos θ + 4 sin θ and notice that:
r = a2 + b2
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sin α =
(i)
The expansion of cos(θ − β ) starts with cos θ … just like the expression
3 cos θ + 4 sin θ.
(ii)
The expansion of cos(θ − β ) has + in the middle, just like the expression
3 cos θ + 4 sin θ.
02/02/18 1:12 PM
The expansion of r cos(θ − β ) is given by
3
r cos(θ − β ) = r (cos θ cos β + sin θ sin β ).
To compare this with 3 cos θ + 4 sin θ, look at the triangle in Figure 3.10,
in which
r=
32 + 4 2 = 5
cos β = 53
sin β = 45
r
β = 53.1°.
3.4 The forms r cos(θ ± α), r sin(θ ± α)
r = 32 + 4 2
=5
⇒
4
β
3
▲ Figure 3.10
This means that you can write 3 cos θ + 4 sin θ in the form
r cos(θ − β ) = 5 cos(θ − 53.1°).
The procedure used here can be generalised to give the result
a cos θ + b sin θ = r cos(θ − α)
where
r=
a
cos α = r
a2 + b2
sin α = br .
Note
The value of r will always be positive, but cos α and sin α may be positive or
negative, depending on the values of a and b. In all cases, it is possible to find
an angle α for which −180° " α " 180°.
You can derive alternative expressions of this type based on other compoundangle formulae if you wish α to be an acute angle, as is done in the next
example.
Example 3.6
(i)
(ii)
Express 3 sin θ − cos θ in the form r sin(θ − α), where r # 0 and
π
0 " α " 2.
State the maximum and minimum values of 3 sin θ − cos θ.
(iii) Sketch the graph of y =
(iv) Solve the equation
3 sin θ − cos θ for 0 $ θ $ 2π.
3 sin θ − cos θ = 1 for 0 $ θ $ 2π.
➜
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3
Solution
(i)
r sin(θ − α ) = r(sin θ cos α − cos θ sin α )
= (r cos α )sin θ − (r sin α )cos θ
Comparing this with 3 sin θ − cos θ, the two expressions are identical if
3 TRIGONOMETRY
r cos α = 3
r sin α = 1.
and
From the triangle in Figure 3.11
r = 1+ 3 = 2
tan α =
π
3 sin θ − cos θ = 2 sin(θ − 6 ).
so
(ii)
and
1
3
π
⇒ α=6
The sine function oscillates between 1 and −1,
π
so 2 sin(θ − 6 ) oscillates between 2 and −2.
r
Maximum value = 2.
1
α
3
Minimum value = −2.
(iii) To sketch the curve y = 2 sin(θ −
π
6 ), notice that
▲ Figure 3.11
» it is a sine curve
» its y values go from −2 to 2
π 7π 13π
» it crosses the horizontal axis where θ = 6 , 6 , 6 ,…
The curve is shown in Figure 3.12.
y
2
O
π
6
π
2
π 7π
6
3π
2
2π 13π
6
5π
2
θ
–2
▲ Figure 3.12
3 sin θ − cos θ = 1 is equivalent to
π
2 sin(θ − 6 ) = 1
π
⇒
sin(θ − 6 ) = 21
π
Let x = (θ − 6 ) and solve sin x = 21.
π
Solving sin x = 21 gives x = 6 (principal value)
π
or
x = π − 6 = 5π (from the graph in Figure 3.13)
6
π π π
5π + π = π.
θ= + =
θ=
or
giving
6 6 3
6
6
(iv) The equation
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3
y
y = sin x
π
O
π
6
π– π
6
x
π
The roots in 0 $ θ $ 2π are θ = 3 and π.
Always check (for example by reference to a sketch graph) that the
number of roots you have found is consistent with the number you
are expecting. When solving equations of the form sin(θ − α ) = c by
considering sin x = c, it is sometimes necessary to go outside the range
specified for θ since, for example, 0 $ θ $ 2π is the same as
−α $ x $ 2π − α.
3.4 The forms r cos(θ ± α), r sin(θ ± α)
▲ Figure 3.13
Using these forms
There are many situations which produce expressions which can be tidied
up using these forms. They are also particularly useful for solving equations
involving both the sine and cosine of the same angle.
The fact that a cos θ + b sin θ can be written as r cos(θ − α ) is an illustration of
the fact that any two waves of the same frequency, whatever their amplitudes,
can be added together to give a single combined wave, also of the same
frequency.
Exercise 3D
1
Express each of the following in the form r cos(θ − α ), where r # 0 and
0° " α " 90°.
(i) cos θ + sin θ
(ii) 20 cos θ + 21 sin θ
(iii) cos θ + 3 sin θ
(iv)
5 cos θ + 2 sin θ
2
Express each of the following in the form r cos(θ + α ), where r # 0 and
0 " α " π.
2
(i) cos θ − sin θ
(ii)
3 cos θ − sin θ
Express each of the following in the form r sin(θ + α ), where r # 0 and
0° " α " 90°.
(i) sin θ + 2 cos θ
(ii) 2 sin θ + 5 cos θ
3
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4
3
3 TRIGONOMETRY
5
6
7
8
9
PS
Express each of the following in the form r sin(θ − α ), where r # 0 and
0 " α " π.
2
(i) sin θ − cos θ
(ii)
7 sin θ − 2 cos θ
Express each of the following in the form r cos(θ − α ), where r # 0 and
−180° " α " 180°.
(i) cos θ − 3 sin θ
(ii) 2 2 cos θ − 2 2 sin θ
(iii) sin θ + 3 cos θ
(iv) 5 sin θ + 12 cos θ
(v) sin θ − 3 cos θ
(vi)
2 sin θ − 2 cos θ
(i) Express 5 cos θ − 12 sin θ in the form r cos(θ + α ), where r # 0 and
0° " α " 90°.
(ii) State the maximum and minimum values of 5 cos θ − 12 sin θ.
(iii) Sketch the graph of y = 5 cos θ − 12 sin θ for 0° $ θ $ 360°.
(iv) Solve the equation 5 cos θ − 12 sin θ = 4 for 0° $ θ $ 360°.
(i) Express 3 sin θ − 3 cos θ in the form r sin(θ − α ), where r # 0 and
0 " α " π.
2
(ii) State the maximum and minimum values of 3 sin θ − 3 cos θ and
the smallest positive values of θ for which they occur.
(iii) Sketch the graph of y = 3 sin θ − 3 cos θ for 0 $ θ $ 2π.
(iv) Solve the equation 3 sin θ − 3 cos θ = 3 for 0 $ θ $ 2π.
(i) Express 2 sin 2θ + 3 cos 2θ in the form r sin(2θ + α), where r # 0
and 0° < α < 90°.
(ii) State the maximum and minimum values of 2 sin 2θ + 3 cos 2θ and
the smallest positive values of θ for which they occur.
(iii) Sketch the graph of y = 2 sin 2θ + 3 cos 2θ for 0° $ θ $ 360°.
(iv) Solve the equation 2 sin 2θ + 3 cos 2θ = 1 for 0° $ θ $ 360°.
(i) Express cos θ + 2 sin θ in the form r cos(θ − α ), where r # 0 and
0° < α < 90°.
(ii) State the maximum and minimum values of cos θ + 2 sin θ and the
smallest positive values of θ for which they occur.
(iii) Sketch the graph of y = cos θ + 2 sin θ for 0° $ θ $ 360°.
(iv) State the maximum and minimum values of
1
3 + cos θ + 2 sin θ
and the smallest positive values of θ for which they occur.
10 The diagram opposite shows a table stuck in a corridor. The table is
120 cm long and 80 cm wide, and the width of the corridor is 130 cm.
(i) Show that 12 sin θ + 8 cos θ = 13.
(ii) Hence find the angle θ. (There are two answers.)
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80 cm
3
130 cm
θ
11 (i)
(ii)
Express 3 cos x + 4 sin x in the form R cos(x − α), where R # 0 and
0° " α " 90°, stating the exact value of R and giving the value of
α correct to 2 decimal places.
Hence solve the equation
3 cos x + 4 sin x = 4.5,
giving all solutions in the interval 0° " x " 360°.
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q6 November 2009
Express 5 cos θ − sin θ in the form R cos(θ + α ), where R # 0 and
0° " α " 90°, giving the exact value of R and the value of α
correct to 2 decimal places.
(ii) Hence solve the equation
5 cos θ − sin θ = 4,
giving all solutions in the interval 0° $ θ $ 360°.
12 (i)
3.4 The forms r cos(θ ± α), r sin(θ ± α)
120 cm
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q5 June 2008
Express 7 cos θ + 24 sin θ in the form R cos(θ − α ), where R # 0
and 0° " α " 90°, giving the exact value of R and the value of α
correct to 2 decimal places.
(ii) Hence solve the equation
7 cos θ + 24 sin θ = 15,
giving all solutions in the interval 0° $ θ $ 360°.
13 (i)
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q4 June 2006
14 By expressing 8 sin θ − 6 cos θ in the form R sin(θ − α), solve the equation
8 sin θ − 6 cos θ = 7,
for 0° $ θ $ 360°.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q5 November 2005
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3
INVESTIGATION
Currents
The simplest alternating current
is one which varies with time t
according to
I = A sin 2πft,
3 TRIGONOMETRY
where f is the frequency and A is the
maximum value. The frequency of
the public AC supply is 50 hertz
(cycles per second).
Investigate what happens when
two alternating currents
A1 sin 2πft and A2 sin(2πft + α) with
the same frequency f but a phase
difference of α are added together.
The previous exercises have each concentrated on just one of the many
trigonometrical techniques which you will need to apply confidently. The
following exercise requires you to identify which technique is the correct one.
Exercise 3E
1
Simplify the following.
(i) 2 sin 3θ cos 3θ
(ii)
(iii) cos2 3θ + sin2 3θ
(iv) 1 − 2 sin2
sin(θ − α)cos α + cos(θ − α)sin α
sin 2θ
(vii)
2 sin θ
Express
(i) (cos x − sin x)2 in terms of sin 2x
(ii) cos4 x − sin4 x in terms of cos 2x
(iii) 2 cos2 x − 3 sin2 x in terms of cos 2x.
Prove that
1 − cos 2θ
2
(i)
1 + cos 2θ ≡ tan θ
(vi) 3 sin θ cos θ
(v)
2
CP
3
cos2 3θ − sin2 3θ
(θ2 )
(viii) cos 2θ − 2 cos2 θ
cosec 2θ + cot 2θ ≡ cot θ
4t(1 − t 2 )
(iii) tan 4θ ≡
where t = tan θ.
1 − 6t 2 + t 4
Solve the following equations.
(i) sin(θ + 40°) = 0.7
0° $ θ $ 360°
2
(ii) 3 cos θ + 5 sin θ − 1 = 0
0° $ θ $ 360°
π
(iii) 2 cos(θ − ) = 1
−π $ θ $ π
6
(iv) cos(45° − θ ) = 2 sin(30° + θ )
−180° $ θ $ 180°
(v) cos 2θ + 3 sin θ = 2
0 $ θ $ 2π
(vi) cos θ + 3 sin θ = 2
0° $ θ $ 360°
(vii) tan2 θ − 3 tan θ − 4 = 0
0° $ θ $ 180°
(ii)
4
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e
3.5 The general solutions of
trigonometrical equations
3
The equation tan θ = 1 has infinitely many roots:
…, −315°, −135°, 45°, 225°, 405°, … (in degrees)
5π , 9π , …
(in radians).
4
4
Only one of these roots, namely 45° or π , is denoted by the function tan−1 1.
4
This is the value which your calculator will give you. It is the principal
value.
The principal value for any inverse trigonometrical function is unique and
lies within a specified range:
− π " tan−1 x " π
2
2
π
π
− $ sin−1 x $
2
2
0 $ cos−1 x $ π.
It is possible to deduce all other roots from the principal value and this is
shown below.
To solve the equation tan θ = c, notice how all possible values of θ occur at
intervals of 180° or π radians (see Figure 3.14). So the general solution is
θ = tan−1 c + nπ
n∈_
3.5 The general solutions of trigonometrical equations
…, − 7π , − 3π , − π ,
4
4
4
(in radians).
y
y = tan θ
c
–90°
– π
2
▲ Figure 3.14
90°
O
π
2
270°
3π
2
θ
tan−1 c is the
principal value.
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The cosine graph (see Figure 3.15) has the y-axis as a line of symmetry.
Notice how the values ±cos−1 c generate all the other roots at intervals of
360° or 2π. So the general solution is
3
θ = ±cos−1 c + 2nπ
n∈_
(in radians).
y
3 TRIGONOMETRY
y = cos θ
c
–450°
–270°
– 5π
2
–90°
90°
–π O
2
– 3π
2
−cos–1 c
▲ Figure 3.15
270°
3π
2
π
2
450°
5π
2
630°
7π
2
θ
cos–1 c is the
principal value.
Now look at the sine graph (see Figure 3.16). As for the cosine graph, there
are two roots located symmetrically. The line of symmetry for the sine graph
is θ = π , which generates all the other possible roots. This gives rise to the
2
slightly more complicated expressions
π π
θ = ± ( − sin−1 c) + 2nπ
2 2
π
n ∈ _.
2
You may, however, find it easier to remember these as two separate formulae:
θ = (2n + 21 )π ± ( − sin−1 c)
or
θ = 2nπ + sin−1 c
θ = (2n + 1)π − sin−1 c.
or
y
y = sin θ
c
–540°
–3π
–360°
–2π
▲ Figure 3.16
–180°
–π
180°
O
sin−1 c is the
principal value.
π
360°
540°
2π
3π
θ
(180° − sin−1 c)
or (π − sin−1 c)
ACTIVITY 3.3
Show that the general solution of the equation sin θ = c may also be written
θ = nπ + (−1)n sin−1 c.
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KEY POINTS
1
1
sec θ = cos θ ;
2
tan2 θ + 1 = sec2 θ;
3
Compound-angle formulae
● sin(θ + φ) = sin θ cos φ + cos θ sin φ
● sin(θ − φ) = sin θ cos φ − cos θ sin φ
● cos(θ + φ) = cos θ cos φ − sin θ sin φ
● cos(θ − φ) = cos θ cos φ + sin θ sin φ
tan θ + tan φ
● tan(θ + φ) =
(θ + φ) ≠ 90°, 270°, ...
1 − tan θ tan φ
tan θ − tan φ
● tan(θ − φ) =
(θ − φ) ≠ 90°, 270°, ...
1 + tan θ tan φ
Double-angle and related formulae
● sin 2θ = 2 sin θ cos θ
● cos 2θ = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1
2 tan θ
● tan 2θ =
1 − tan 2 θ θ ≠ 45°, 135°, ...
●
3
1 + cot2 θ = cosec2 θ
1
sin2 θ = 2(1 − cos 2θ )
1
cos2 θ = 2(1 + cos 2θ )
The r, α formulae
● a sin θ + b cos θ = r sin(θ + α )
● a sin θ − b cos θ = r sin(θ − α )
● a cos θ + b sin θ = r cos(θ − α )
● a cos θ − b sin θ = r cos(θ + α )
●
5
1
cot θ = tan θ
}
where r = a 2 + b 2
a
cos α = r
sin α = br
3.5 The general solutions of trigonometrical equations
4
1
cosec θ = sin θ ;
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3 TRIGONOMETRY
3
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■ understand and use
■ the sec, cosec and cot functions
■ the relationships between the graphs of the sin, cos, tan, cosec, sec
and cot functions
■ use the identities
■ sin2 θ + cos2 θ ≡ 1
■ tan2 θ + 1 ≡sec2 θ
■ cot2 θ + 1 ≡cosec2 θ
■ solve equations involving sec, cosec and cot
■ understand and use the identities for sin (A ± B ), cos (A ± B ), tan (A ± B )
■ know and use the identities for sin 2A, cos 2A, tan 2A
■ write a cos θ + b sin θ in the equivalent forms r sin (θ ± α ) and r cos (θ ± α )
■ use for finding maximum and minimum values
■ use for solving equations
■ use trigonometrical identities
■ in solving equations
■ in proofs
■ to solve problems.
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P2
P3
4
Differentiation
4 Differentiation
Nothing takes
place in the
world whose
meaning is not
that of some
maximum or
some
minimum.
Leonhard Euler
(1707–1783)
Displacement
Many physical systems, such as a simple pendulum or swing or a mass on an
elastic spring, can be modelled as having displacement–time graphs which
have a sine wave shape.
O
Time
▲ Figure 4.1
To be able to perform calculations involving velocity and acceleration for
these systems, you need to be able to differentiate the sine function.
?
❯ Think of some other situations in which it would be useful to be able
to differentiate functions other than polynomials.
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4 DIFFERENTIATION
4
4.1 The product rule
Figure 4.2 shows a sketch of the curve of y = 20x(x − 1)6.
y
x
▲ Figure 4.2
If you wanted to find the gradient function, dy , for the curve, you could
dx
expand the right-hand side then differentiate it term by term – a long and
cumbersome process!
There are other functions like this, made up of the product of two or more
simpler functions, which are not just time-consuming to expand – they are
impossible to expand. One such function is
1
y = ( x − 1) 2 ( x + 1)6
(for x > 1).
Clearly you need a technique for differentiating functions that are products
of simpler ones, and a suitable notation with which to express it.
The most commonly used notation involves writing
y = uv,
dy
where the variables u and v are both functions of x. Using this notation,
is
dx
given by
dy
dv
du
=u +v .
dx
dx
dx
This is called the product rule and it is derived from first principles in the
next section.
The product rule from first principles
A small increase δx in x leads to corresponding small increases δu, δv and δy
in u, v and y. And so
y + δy = (u + δu)(v + δv)
= uv + v δu + u δv + δu δv.
Since y = uv, the increase in y is given by
δy = v δu + u δv + δu δv.
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Example 4.1
Given that y = (2x + 3)(x 2 − 5), find
dy
using the product rule.
dx
Solution
4
4.1 The product rule
Dividing both sides by δx,
δy
= v δu + u δv + δu δv .
δx
δx
δx
δx
In the limit, as δx → 0, so do δu, δv and δy, and
dy
δy
δu
δv
.
→ du ,
→ dv and
→
δx
δ
x
δ
x
dx
dx
dx
The expression becomes
dy
= v du + u dv .
dx
dx
dx
Notice that since δu → 0, the last term on the right-hand side has disappeared.
y = (2x + 3)(x 2 − 5)
Let u = 2x + 3 and v = x 2 − 5.
du
dv
Then
= 2 and
= 2x.
dx
dx
dy
Using the product rule,
= v du + u dv
dx
dx
dx
= (x 2 − 5) × 2 + (2x + 3) × 2x
= 2(x 2 − 5 + 2x 2 + 3x)
= 2(3x 2 + 3x − 5)
Note
In this case you could have multiplied out the expression for y.
y = 2x 3 + 3x 2 − 10x − 15
dy
= 6x 2 + 6x − 10
dx
= 2(3x 2 + 3x − 5)
Example 4.2
Differentiate y = 20x(x − 1)6.
Figure 4.2 shows the graph of this function.
Solution
Let u = 20x and v = (x − 1)6.
du
Then
= 20 and dv = 6(x − 1)5 (using the chain rule). 20(x − 1)5 is a
dx
dx
common factor.
dy
= v du + u dv
Using the product rule,
dx
dx
dx
= (x − 1)6 × 20 + 20x × 6(x − 1)5
= 20(x − 1)5 × (x − 1) + 20(x − 1)5 × 6x
= 20(x − 1)5[(x − 1) + 6x]
= 20(x − 1)5(7x − 1)
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4
The factorised result is the most useful form for the solution, as it allows you
to find stationary points easily.You should always try to factorise your answer
as much as possible. Once you have used the product rule, look for factors
straight away and do not be tempted to multiply out.
4 DIFFERENTIATION
4.2 The quotient rule
In the last section, you met a technique for differentiating the product of two
functions. In this section you will see how to differentiate a function which is
the quotient of two simpler functions.
As before, you start by identifying the simpler functions. For example, the
function
y = 3x + 1 (for x ≠ 2)
x−2
u
can be written as y = v where u = 3x + 1 and v = x − 2. Using this notation,
dy is given by
dx
v du − u dv
dy
dx
dx
=
dx
v2
This is called the quotient rule and it is derived from first principles in the
next section.
The quotient rule from first principles
A small increase, δx in x results in corresponding small increases δu, δv and δy
in u, v and y. The new value of y is given by
y + δy = u + δu
v + δv
u
and since y = , you can rearrange this to obtain an expression for δy in
v
terms of u and v.
u + δu − u
δy =
v + δv v
v(u + δu ) − u(v + δv )
=
v(v + δv )
uv + v δu − uv − u δv
=
v(v + δv )
v δu − u δv
=
v(v + δv )
Dividing both sides by δx gives
To divide the right-hand
δu
δv
side by δx you only divide the
v −u
δy
δx
δx
numerator by δx.
=
δx
v(v + δv )
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In the limit as δx → 0, this is written in the form you met above.
dv
du
dy v dx − u dx
=
dx
v2
02/02/18 1:12 PM
ACTIVITY 4.1
Verify that the quotient rule gives
Example 4.3
Given that y =
dy
correctly when u = x 10 and v = x 7.
dx
dy
3x + 1
, find
using the quotient rule.
x−2
dx
Letting u = 3x + 1 and v = x − 2 gives
du
dv
dx = 3 and dx = 1.
du
dv
dy v dx − u dx
Using the quotient rule,
=
dx
v2
( x − 2) × 3 − (3x + 1) × 1
=
( x − 2) 2
= 3x − 6 − 3x2 − 1
( x − 2)
=
4.2 The quotient rule
Solution
Example 4.4
4
−7
( x − 2) 2
2
Given that y = x + 1 , find dy using the quotient rule.
3x − 1
dx
Solution
Letting u = x 2 + 1 and v = 3x − 1 gives
du
dv
= 2x and
= 3.
dx
dx
du
dv
dy v dx − u dx
Using the quotient rule,
=
dx
v2
(3x − 1) × 2x − ( x 2 + 1) × 3
=
(3x − 1) 2
2
2
= 6x − 2x − 3x2 − 3
(3x − 1)
2
= 3x − 2x −2 3
(3x − 1)
Exercise 4A
9781510421738.indb 87
1
Differentiate the following using the product rule or the quotient rule.
(i) y = (x 2 − 1)(x 3 + 3)
(ii) y = x 5(3x 2 + 4x − 7)
2x
(iii) y = x 2(2x + 1)4
(iv) y =
3
x
−1
x3
(v) y = 2
(vi) y = (2x + 1)2(3x 2 − 4)
x +1
x−2
2x − 3
(vii) y =
(viii) y =
2
(
x
+ 3) 2
2x + 1
(ix) y = ( x + 1) x − 1
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02/02/18 1:12 PM
4 DIFFERENTIATION
4
CP
2
3
PS
4
PS
5
The diagram shows the graph of y = x .
x −1
dy
(i) Find
.
dx
y
(ii) Find the gradient of
the curve at (0, 0),
and the equation of
the tangent at (0, 0).
(iii) Find the gradient of
1
the curve at (2, 2),
and the equation of
the tangent at (2, 2).
O
1
x
(iv) What can you
deduce about the
two tangents?
Given that y = (x + 1)(x − 2)2
(i) find dy
dx
(ii) find any stationary points and determine their nature
(iii) sketch the curve.
x−3
Given that y =
x−4
(i) find dy
dx
(ii) find the equation of the tangent to the curve at the point (6, 1.5)
(iii) find the equation of the normal to the curve at the point (5, 2)
(iv) use your answer from part (i) to deduce that the curve has no
stationary points, and sketch the graph.
2x
The diagram shows the graph of y =
, which is undefined for
x −1
x < 0 and x = 1. P is a minimum point.
y
P
O
x
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(b)
the normal found in part (iv), call it R.
(vi) Show that the area of the triangle PQR is 441 .
8
6
2
The diagram shows the graph of y = x − 2x − 5 .
2x + 3
4
4.2 The quotient rule
Find dy .
dx
(ii) Find the gradient of the curve at (9, 9), and show that the equation
of the normal at (9, 9) is y = −4x + 45.
(iii) Find the coordinates of P and verify that it is a minimum point.
(iv) Write down the equation of the tangent and the normal to the
curve at P.
(v) Write down the point of intersection of the normal found in
part (ii) and
(a) the tangent found in part (iv), call it Q
(i)
y
–1.5
O
x
Find dy .
dx
(ii) Use your answer from part (i) to find any stationary points of the
curve.
(iii) Classify each of the stationary points and use calculus to justify your
answer.
x2
A curve has the equation y =
.
2x + 1
Find the coordinates of the stationary points on the curve and identify
their nature.
(i)
7
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4 DIFFERENTIATION
4
PS
8
The diagram shows part of the
graph with the equation
y = x 9 − 2x 2 .
It crosses the x axis at (a, 0).
(i) Find the value of a, giving
your answer as a multiple of
2.
y
Show that the result of
O
(ii)
(a, 0)
x
differentiating 9 − 2x 2 is
−2x
.
9 − 2x 2
Hence show that if y = x 9 − 2x 2 then
2
dy
= 9 − 4x .
dx
9 − 2x 2
(iii) Find the x coordinate of the maximum point on the graph of
y = x 9 − 2x 2 .
Write down the gradient of the curve at the origin.
What can you say about the gradient at the point (a, 0)?
4.3 Differentiating natural logarithms
and exponentials
1
In Chapter 2 you learnt that the integral of x is ln x. It follows, therefore, that
the differential of ln x is 1.
x
So y = ln x ⇒ dy = 1
dx x
The differential of the inverse function, y = e x, may be found by
interchanging y and x.
dx 1
x = ln y ⇒ dy = y
dy
1
⇒ dx = dx = y = e x .
Therefore d e x = e x .
dx
The differential of ex is itself ex. This may at first seem rather surprising.
dy
?
The function f(x) (x ! ") is a polynomial in x of order n.
So
f(x) = anx n + an−1x n−1 + ... + a1x + a0
CP
where an, an−1, ..., a0 are all constants and at least an is not zero.
d
f(x) cannot equal f(x)?
❯ How can you prove that
dx
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Since the differential of ex is ex, it follows that the integral of ex is also ex.
∫ ex dx = ex + c.
This may be summarised as in the following table.
Differentiation
Integration
dy
dx
y ⎯→ y dx
y ⎯→
∫
1 ⎯→ ln x + c
x
ex ⎯→ ex
ex ⎯→ ex + c
Differentiate y = e5x.
Solution
Make the substitution u = 5x to give y = eu.
dy = eu = e 5x and du = 5.
Now
dx
du
d
u
d
y
d
y
By the chain rule,
=
×
dx du dx
= e5x × 5
4.3 Differentiating natural logarithms and exponentials
ln x ⎯→ 1
x
These results allow you to extend very considerably the range of functions
which you are able to differentiate and integrate.
Example 4.5
4
= 5e5x
This result can be generalised as follows.
y = eax
⇒
dy
= ae ax
dx
where a is any constant.
This is an important standard result, and you would normally use it
automatically, without recourse to the chain rule.
Example 4.6
Differentiate y = 42x .
e
Solution
4
y = e 2 x = 4e −2 x
⇒
dy
= 4 × (−2e −2 x )
dx
= −8e−2x
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4
Example 4.7
Differentiate y = 3e(x2+1).
Solution
Let u = x 2 + 1, then y = 3eu.
⇒
4 DIFFERENTIATION
By the chain rule,
dy
= 3e u = 3e ( x 2 +1)
du
and
du
dx = 2x
dy dy du
=
×
dx du dx
= 3e(x2+1) × 2x
= 6x e (x2+1)
Example 4.8
Differentiate the following.
y = 2ln x
(i)
(ii)
y = ln(3x)
Solution
(i)
(ii)
dy
= 2× 1
x
dx
2
=
x
Let u = 3x, then y = ln u
dy 1
= = 1 and
du u 3x
By the chain rule,
dy dy du
=
×
dx du dx
= 1 ×3
3x
=1
x
⇒
du = 3
dx
Note
An alternative solution to part (ii) is
y = ln(3x) = ln 3 + ln x
⇒
dy
= 0 + 1 = 1.
x x
dx
❯ The gradient function found in part (ii) above for y = ln(3x) is the
same as that for y = ln x. What does this tell you about the shapes of
the two curves? For what values of x is it valid?
?
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Example 4.9
Differentiate the following.
(i)
y = ln(x4)
(ii)
4
y = ln(x2 + 1)
Solution
(i)
By the properties of logarithms
(ii)
Let u = x 2 + 1, then y = ln u
dy 1
= = 1
and
du u x 2 + 1
dy dy du
By the chain rule, dx = du × dx
1
= x 2 + 1 × 2x
2x
= x2 + 1
⇒
du = 2x
dx
If you need to differentiate expressions similar to those in the examples
above, follow exactly the same steps. The results can be generalised as follows.
dy a
=
dx x
dy 1
=
y = ln(ax) ⇒
dx x
dy f '( x )
=
y = ln(f(x)) ⇒
dx f( x )
y = a ln x ⇒
Example 4.10
Differentiate y =
dy
= ae x
dx
dy
= ae ax
y = eax ⇒
dx
dy
= f '( x )e f(x )
y = ef(x) ⇒
dx
y = aex ⇒
4.3 Differentiating natural logarithms and exponentials
⇒
y = ln(x 4) = 4 ln(x)
dy 4
=
dx x
ln x
.
x
Solution
u
where u = ln x and v = x
v
du = 1 and dv = 1.
⇒
dx x
dx
du
dv
dy v dx − u dx
By the quotient rule,
=
dx
v2
1−
x × x 1 × ln x
=
x2
1 − ln x
=
x2
Here y is of the form
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4
Exercise 4B
1
Differentiate the following.
(i) y = 3 ln x
(ii)
y = ln(4x)
2
(iii) y = ln(x )
(iv)
y = ln(x 2 + 1)
1
(v) y = ln
(vi)
y = x ln x
x
(vii) y = x 2 ln(4x)
(viii) y = ln x + 1
x
ln
x
(ix) y = ln x 2 − 1 (x)
y= 2
x
Differentiate the following.
4 DIFFERENTIATION
( )
2
y = 3ex
(ii)
y = e 2x
(iii) y = ex
(iv)
y = e(x +1)
(vi)
y = 2x 3e −x
y = (e2x + 1)3
(i)
2
y = x e4x
x
(vii) y = x
e
(v)
M
( )
(viii)
2
3
Knowing how much rain has fallen in a river basin, hydrologists are
often able to give forecasts of what will happen to a river level over the
next few hours. In one case it is predicted that the height h, in metres, of
a river above its normal level during the next 3 hours will be 0.12e0.9t,
where t is the time elapsed, in hours, after the prediction.
dh
(i) Find dt , the rate at which the river is rising.
(ii) At what rate will the river be rising after 0, 1, 2 and 3 hours?
4
The graph of y = xex is shown below.
y
O
x
P
(i)
(ii)
CP
5
dy
d 2y
Find
and 2 .
dx
dx
Find the coordinates of the minimum point P.
The graph of f(x) = x ln(x 2) is shown below.
x
O
x
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6
7
y
O
x
x
You are given that the curve y = e x has one stationary point.
(i) Find the exact coordinates of this point.
(ii) Determine whether this point is a maximum or a minimum point.
ln x
9 Given that f(x) =
,x > 0
x
(i) write down the coordinates of the point where the graph of
y = f(x) crosses the x-axis
(ii) find the exact coordinates of the stationary point of the curve
y = f(x)
(iii) sketch the curve of y = f(x).
− 1x
10 Find the exact coordinates of the point on the curve y = xe 2 at
d 2y
which 2 = 0.
dx
8
4
4.3 Differentiating natural logarithms and exponentials
Describe, giving a reason, any symmetries of the graph.
(ii) Find f ′(x) and f ″(x).
(iii) Find the coordinates of any stationary points.
ex
Given that y = x
dy
(i) find
dx
(ii) find the coordinates of any stationary points on the curve
(iii) sketch the curve.
The graph of f(x) = x ln x is shown below.
Find the exact coordinates of the stationary point.
(i)
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q6 November 2008
11 It is given that the curve y = (x − 2)ex has one stationary point.
(i)
(ii)
Find the exact coordinates of this point.
Determine whether this point is a maximum or a minimum point.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q6 June 2008
95
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4
12 The equation of a curve is y = x3e–x.
(i)
(ii)
Show that the curve has a stationary point where x = 3.
Find the equation of the tangent to the curve at the point where x = 1.
4 DIFFERENTIATION
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q5 June 2010
4.4 Differentiating trigonometrical
functions
ACTIVITY 4.2
Figure 4.3 shows the graph of y = sin x, with x measured in radians,
together with the graph of y = x. You are going to sketch the graph of the
gradient function for the graph of y = sin x.
y=x
y
1
y = sin x
–π
2
–2π
– 3π
2
3π
2
–π
0
π
2
π
2π
x
–1
▲ Figure 4.3
Draw a horizontal axis for the angles, marked from −2π to 2π, and a
vertical axis for the gradient, marked from −1 to 1, as shown in Figure 4.4.
dy
dx
1
–2π
–π
0
π
2π
x
–1
▲ Figure 4.4
First, look for the angles for which the gradient of y = sin x is zero.
Mark zeros at these angles on your gradient graph.
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Decide which parts of y = sin x have a positive gradient and which have a
negative gradient. This will tell you whether your gradient graph should
be above or below the x-axis at any point.
4
Look at the part of the graph of y = sin x near x = 0 and compare it with
the graph of y = x. What do you think the gradient of y = sin x is at this
point? Mark this point on your gradient graph. Also mark on any other
points with plus or minus the same gradient.
4.4 Differentiating trigonometrical functions
Now, by considering whether the gradient of y = sin x is increasing or
decreasing at any particular point, sketch in the rest of the gradient graph.
The gradient graph that you have drawn should look like a familiar graph.
What graph do you think it is?
Sketch the graph of y = cos x, with x measured in radians, and use it as
above to obtain a sketch of the graph of the gradient function of y = cos x.
❯ Is y = x still a tangent of y = sin x if x is measured in degrees?
?
Activity 4.2 showed you that the graph of the gradient function of y = sin x
resembled the graph of y = cos x. You will also have found that the graph of
the gradient function of y = cos x looks like the graph of y = sin x reflected in
the x-axis to become y = −sin x.
?
❯ Both of these results are in fact true but the work above does not
amount to a proof. Explain why.
CP
Summary of results
d (sin x) = cos x
dx
d
(cos x) = −sin x
dx
Remember that these results are only valid when the angle is measured in
radians, so when you are using any of the derivatives of trigonometrical
functions you need to work in radians.
ACTIVITY 4.3
sin x
By writing tan x = cos x , use the quotient rule to show that
d (tan x ) = 2
sec x where x is measured in radians.
dx
You can use the three results met so far to differentiate a variety of functions
involving trigonometrical functions, by using the chain rule, product rule or
quotient rule, as in the following examples.
9781510421738.indb 97
97
02/02/18 1:12 PM
Differentiate y = cos 2x.
Solution
As cos 2x is a function of a function, you may use the chain rule.
du
Let
u = 2x
⇒ dx = 2
dy
= −sin u
y = cos u ⇒
du
dy dy du
=
×
dx du dx
4 DIFFERENTIATION
4
Example 4.11
= −sin u × 2
= −2 sin 2x
With practice it should be possible to do this in your head, without needing
to write down the substitution.
This result may be generalised.
y = cos kx
⇒
dy
= −k sin kx
dx
y = sin kx
⇒
dy
= k cos kx
dx
y = tan kx
⇒
dy
= k sec2 kx
dx
Similarly
and
Example 4.12
Differentiate y = x 2 sin x.
Solution
x 2 sin x is of the form uv, so the product rule can be used with u = x 2 and
v = sin x.
du = 2 x
dv = cos x
dx
dx
Using the product rule
dy
du
dv
=v
+u
dx
dx
dx
dy
⇒
= 2x sin x + x 2 cos x
dx
98
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Example 4.13
Differentiate y = etanx.
4
Solution
= eu sec2 x
= etanx sec2 x
Example 4.14
Differentiate y =
1 + sin x
.
cos x
Solution
u
1 + sin x
is of the form v so the quotient rule can be used, with
cos x
u = 1 + sin x and
du
⇒ dx = cos x
and
The quotient rule is
du
dv
dy v dx − u dx
=
dx
v2
4.4 Differentiating trigonometrical functions
etanx is a function of a function, so the chain rule may be used.
du = 2
Let u = tan x ⇒
sec x
dx
dy
y = eu
⇒
= eu
du
Using the chain rule
dy dy du
=
×
dx du dx
v = cos x
dv
= −sin x
dx
Substituting for u and v and their derivatives gives
dy (cos x )(cos x ) − (1 + sin x )(− sin x )
=
dx
(cos x ) 2
2
2
= cos x + sin2x + sin x
cos x
+
x
1
sin
=
using sin2 x + cos2 x = 1
cos 2 x
= (sec2 x)(1 + sin x)
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4
Exercise 4C
1
2
4 DIFFERENTIATION
3
4
5
Differentiate each of the following.
(i) 2 cos x + sin x (ii)
tan x + 5
(iii)
sin x − cos x
Use the product rule to differentiate each of the following.
(i) x tan x
(ii)
sin x cos x
(iii)
e x sin x
Use the quotient rule to differentiate each of the following.
ex
x + cos x
sin x
(i)
(ii)
(iii)
cos
x
x
sin x
Use the chain rule to differentiate each of the following.
(i) tan(x 2 + 1)
(ii)
sin 2x
(iii)
ln(sin x)
Use an appropriate method to differentiate each of the following.
ex tan x
(iii)
sin 4x 2
sin x
(iv) ecos 2x
(v)
(vi)
ln(tan x)
1 + cos x
(i) Differentiate y = x cos x.
(ii) Find the gradient of the curve y = x cos x at the point where x = π.
(iii) Find the equation of the tangent to the curve y = x cos x at the
point where x = π.
(iv) Find the equation of the normal to the curve y = x cos x at the
point where x = π.
The diagram shows the part of the curve y = 21 tan 2x for 0 $ x $ 1 π.
(i)
6
7
cos x
(ii)
2
y
O
1π
4
1π
2
x
Find the x coordinates of the points on this part of the curve at which
the gradient is 4.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q3 November 2011
8
The equation of a curve is y = 6 sin x − 2 cos 2x.
Find the equation of the tangent to the curve at the point
( π,2).
1
6
Give the answer in the form y = mx + c, where the values of m and c are
correct to 3 significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q3 June 2015
100
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9
Find the gradient of each of the following curves at the point for which x = 0.
(i) y = 3 sin x + tan 2x
6
(ii) y =
1 + e 2x
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q2 June 2014
4
10 The equation of a curve is y = x + 2 cos x. Find the x coordinates of the
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q3 June 2006
11 A curve has equation
y = 2 − tan x .
1 + tan x
Find the equation of the tangent to the curve at the point for which
x = 41 π, giving the answer in the form y = mx + c, where c is correct to
3 significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q3 November 2015
12 The curve with equation y = e–x sin x has one stationary point for which
0 $ x $ π.
(i) Find the x coordinate of this point.
(ii) Determine whether this point is a maximum or a minimum point.
4.5 Differentiating functions defined implicitly
stationary points of the curve for 0 $ x $ 2π, and determine the nature
of each of these stationary points.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q4 November 2007
ex
1
−1π
cos x , for 2 " x "− 2 π , has one stationary point. Find
the x coordinate of this point.
13 The curve y =
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q3 November 2008
4.5 Differentiating functions defined
implicitly
All the functions you have differentiated so far have been of the form y = f(x).
However, many functions cannot be arranged in this way at all, for example
x 3 + y 3 = xy, and others can look clumsy when you try to make y the subject.
An example of this is the semicircle x 2 + y 2 = 4, y % 0, illustrated in Figure 4.5.
y
2
P
2
P
O
–2
▲ Figure 4.5
O
2
x
y
x
By Pythagoras’ theorem,
x2 + y2 = 22
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Because of Pythagoras’ theorem, the curve is much more easily recognised in this
4
form than in the equivalent y = 4 − x 2.
When a function is specified by an equation connecting x and y that does
not have y as the subject, it is called an implicit function.
dy dy du
=
×
and the product rule d (uv ) = u dv + v du are
dx du dx
dx
dx
dx
used extensively to help in the differentiation of implicit functions.
4 DIFFERENTIATION
The chain rule
Example 4.15
Differentiate each of the following with respect to x.
(i) y2
xy
(ii)
(iii)
3x2y3
(iv)
sin y
Solution
(i)
(ii)
(iii)
d 2
dy
d 2
dx (y ) = dy ( y ) × dx (c
dy
= 2y dx
d (xy) = x dy + y
dx
dx
(
(
(chain rule)
(product rule)
)
)
d
d
d
(3x2y3) = 3 x 2 ( y 3 ) + y 3 ( x 2 )
dx
dx
dx
dy
+ y 3 × 2x
= 3 x 2 × 3y 2
dx
( ddxy + 2y)
(product rule)
(chain rule)
= 3xy 2 3x
(iv)
Example 4.16
d
d
dy
dx (sin y) = dy (sin y) × dx
dy
= (cos y)
dx
(chain rule)
The equation of a curve is given by y3 + xy = 2.
dy
(i) Find an expression for
in terms of x and y.
dx
(ii) Hence find the gradient of the curve at (1, 1) and the equation of the
tangent to the curve at that point.
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Solution
(i)
⇒
Substitute x = 1, y = 1 into
the expression for
dy .
dx
Using y − y1 = m(x − x1) the equation of the tangent is
(y − 1) = − 41 (x − 1)
⇒
x + 4y − 5 = 0
Figure 4.6 shows the graph of the curve with the equation y 3 + xy = 2.
?
y
O
4.5 Differentiating functions defined implicitly
(ii)
4
y3 + xy = 2
dy
dy
⇒ 3y 2 + (x
+ y) = 0
dx
dx
dy
⇒
(3y 2 + x) = −y
dx
−y
dy
⇒
=
dx 3y 2 + x
dy
1
At (1, 1),
= −4
dx
x
▲ Figure 4.6
❯ Why is this not a function?
Stationary points
dy
As before, these occur where dx = 0.
dy
= 0 will not usually give values of x directly, but will give a
dx
relationship between x and y. This needs to be solved simultaneously with the
equation of the curve to find the coordinates.
Putting
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4
Example 4.17
(i)
Differentiate x 3 + y 3 = 3xy with respect to x.
(ii)
Hence find the coordinates of any stationary points.
Solution
4 DIFFERENTIATION
(i)
(ii)
d ( x 3 ) + d ( y 3 ) = d (3xy )
dx
dx
dx
⎛ dy
⎞
d
y
= 3 ⎜x
+ y⎟
⇒ 3x2 + 3y2
dx
⎝ dx
⎠
dy
At stationary points, dx = 0
⇒ 3x 2 = 3y
⇒
Notice how it is not
necessary to find an
expression for
x2 = y
dy
unless
dx
you are asked to.
To find the coordinates of the stationary points, solve
x2 = y
x 3 + y 3 = 3xy
}
simultaneously.
Substituting for y gives
x 3 + (x 2)3 = 3x(x 2)
⇒
x 3 + x 6 = 3x 3
⇒
x 6 = 2x 3
⇒
x 3(x 3 – 2) = 0
⇒
x = 0 or x = 3 2
y = x2 so the stationary points are (0, 0) and ( 3 2, 3 4 ) .
The stationary points are A and B in Figure 4.7.
y
2
–2
A
B
2
x
–2
▲ Figure 4.7
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Types of stationary points
As with explicit functions, the nature of a stationary point can be determined by
d 2y
considering the sign of
either side of the stationary point.
dx 2
Example 4.18
4
The curve with equation sin x + sin y = 1 for 0 $ x $ π, 0 $ y $ π is shown
in Figure 4.8.
π
π
O
x
▲ Figure 4.8
(i)
(ii)
Differentiate the equation of the curve with respect to x and hence find
the coordinates of any stationary points.
Show that the points π , π , π , 5π , 5π , π and 5π , 5π all lie on
6 6
6 6
6 6
6 6
the curve.
( )(
)(
) (
)
4.5 Differentiating functions defined implicitly
y
Find the gradient at each of these points.
What can you conclude about the natures of the stationary points?
Solution
(i)
sin x + sin y = 1
dy
⇒ cos x + (cos y) = 0
dx
dy
cos x
= − cos y
⇒
dx
dy
= 0 ⇒ cos x = 0
At any stationary point
dx
⇒ x = π (only solution in range)
2
Substitute in sin x + sin y = 1.
π
When x = 2, sin x = 1 ⇒ sin y = 0
⇒ y = 0 or y = π
(2 )
(2 )
⇒ stationary points at π , 0 and π , π .
➜
105
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4
(ii)
sin π6 = 21 , sin 5π = 1
6
2
So, for each of the four given points, sin x + sin y = 21 + 21 = 1.
Therefore they all lie on the curve.
dy
cos x
The gradient of the curve is given by dx = − cos y
4 DIFFERENTIATION
cos π = 3 , cos 5π = − 3
6
2
6
2
( )
3
At π , π ,
6 6
dy
= − 2 = −1
dx
3
(
)
dy
= − 2 = 1
dx
− 3
(
)
dy
=
dx
2
At π , 5π ,
6 6
At 5π , π ,
6 6
(
)
At 5π , 5π ,
6 6
3
dy
=
dx
y
2
3
−
− 2 = 1
3
2
3
− 2
−
= −1
3
− 2
y
π
π
5π
6
π
6
O
π
6
π
2
5π
6
x
O
π
6
π
2
5π
6
x
▲ Figure 4.9
These results show that
π , 0 is a minimum
π
, π is a maximum
2
2
These points are confirmed by considering the sketch in Figure 4.8 on
page 105.
( )
Exercise 4D
1
( )
Differentiate each of the following with respect to x.
(i) y4
(ii)
x2 + y3 − 5
(iii)
xy + x + y
(iv) cos y
(v)
e(y+2)
(vi)
xy3
(vii) 2x2y5
(viii) x + ln y − 3
(ix)
xey − cos y
(x) x2 ln y
(xi)
x esin y
(xii)
x tan y − y tan x
106
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2
3
4
5
6
PS
7
8
( )
4
4.5 Differentiating functions defined implicitly
PS
Find the gradient of the curve xy3 = 5 ln y at the point (0, 1).
Find the gradient of the curve esin x + ecos y = e + 1 at the point π , π .
2 2
(i) Find the gradient of the curve x2 + 3xy + y2 = x + 3y at the point
(2, −1).
(ii) Hence find the equation of the tangent to the curve at this point.
Find the coordinates of all the stationary points on the curve
x2 + y2 + xy = 3.
A curve has the equation (x − 6)(y + 4) = 2.
dy
(i) Find an expression for
in terms of x and y.
dx
(ii) Find the equation of the normal to the curve at the point (7, −2).
(iii) Find the coordinates of the point where the normal meets the
curve again.
b
(iv) By rewriting the equation in the form y − a =
, identify any
x−c
asymptotes and sketch the curve.
A curve has the equation y = x x for x # 0.
(i) Take logarithms to base e of both sides of the equation.
(ii) Differentiate the resulting equation with respect to x.
(iii) Find the coordinates of the stationary point, giving your answer to
3 decimal places.
(iv) Sketch the curve for x # 0.
The equation of a curve is 3x2 + 2xy + y2 = 6. It is given that there are
two points on the curve where the tangent is parallel to the x-axis.
(i) Show by differentiation that, at these points, y = −3x.
(ii) Hence find the coordinates of the two points.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q5 June 2006
9
The curve with equation 6e 2 x + ke y + e 2 y = c ,
where k and c are constants, passes through the point P with coordinates
(ln 3, ln 2).
(i) Show that 58 + 2k = c.
(ii) Given also that the gradient of the curve at P is −6, find the values
of k and c.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q5 June 2011
10 The equation of a curve is x2 + y2 − 4xy + 3 = 0.
(i)
(ii)
dy 2y − x
Show that dx = y − 2x.
Find the coordinates of each of the points on the curve where the
tangent is parallel to the x-axis.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q7 June 2008
107
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4
11 The equation of a curve is x3 – x2y – y3 = 3.
(i)
(ii)
dy
Find dx in terms of x and y.
Find the equation of the tangent to the curve at the point (2, 1),
giving your answer in the form ax + by + c = 0.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q3 November 2009
4 DIFFERENTIATION
12 The equation of a curve is xy(x + y) = 2a3, where a is a non-zero
constant. Show that there is only one point on the curve at which the
tangent is parallel to the x-axis, and find the coordinates of this point.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q6 June 2008
4.6 Parametric equations
When you go on a ride like the one in the picture, your body follows a very
unnatural path and this gives rise to sensations which you may find
exhilarating or frightening.
You are accustomed to expressing curves as mathematical equations. How
would you do so in a case like this?
Figure 4.10 shows a simplified version of such a ride.
(a) At the start
(b) Some time later
AP has in total
turned through
angle 3θ.
O
108
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4m
A
2m
P
O
P
2θ
A
θ
θ
▲ Figure 4.10
02/02/18 1:13 PM
The passenger’s chair is on the end of a rod AP of length 2 m which is
rotating about A. The rod OA is 4 m long and is itself rotating about O. The
gearing of the mechanism ensures that the rod AP rotates twice as fast relative
to OA as the rod OA does. This is illustrated by the angles marked on
Figure 4.10(b), at a time when OA has rotated through an angle θ.
At this time, the coordinates
of the point P, taking O as the
origin, are given by
4
y
2
x = 4 cos θ + 2 cos 3θ
2 sin 3 θ
4
y = 4 sin θ + 2 sin 3θ
(see Figure 4.11).
3θ
A
4 sin θ
θ
4 cos θ
O
2 cos 3 θ
x
▲ Figure 4.11
These two equations are called parametric equations of the curve. They
do not give the relationship between x and y directly in the form y = f(x) but
use a third variable, θ, to do so. This third variable is called the parameter.
4.6 Parametric equations
P
To plot the curve, you need to substitute values of θ and find the
corresponding values of x and y.
θ = 0°
Thus
⇒ x=4+2=6
y=0+0=0
Point (6, 0)
θ = 30° ⇒ x = 4 × 0.866 + 0 = 3.464
y = 4 × 0.5 + 2 × 1 = 4
Point (3.46, 4)
and so on.
Joining points found in this way reveals the curve to have the shape shown in
Figure 4.12.
y
4
θ = 150°
θ = 30°
θ = 60°, 120°
2
θ = 90°
θ = 180°
–6
–4
0
–2
2
θ = 270°
θ = 0°
4
6
x
–2
θ = 240°, 300°
θ = 210°
–4
θ = 330°
▲ Figure 4.12
❯ At what points of the curve would you feel the greatest sensations?
?
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b
Graphs from parametric equations
4 DIFFERENTIATION
4
Parametric equations are very useful in situations such as this, where an
otherwise complicated equation may be expressed reasonably simply in
terms of a parameter. Indeed, there are some curves which can be given by
parametric equations but cannot be written as cartesian equations (in terms of
x and y only).
The next example is based on a simpler curve. Make sure that you can follow
the solution completely before going on to the rest of the chapter.
Example 4.19
36
.
t2
Find the coordinates of the points corresponding to t = 1, 2, 3, −1, −2
and −3.
A curve has the parametric equations x = 2t, y =
(i)
(ii)
Plot the points you have found and join them to give the curve.
(iii) Explain what happens as t → 0.
Solution
(i)
t
–3
–2
–1
1
2
3
x
–6
–4
–2
2
4
6
y
4
9
36
36
9
4
The points required are (−6, 4), (−4, 9), (−2, 36), (2, 36), (4, 9) and (6, 4).
(ii)
The curve is shown in Figure 4.13.
y
40
t = –1
t=1
30
20
10
t = –2
t=2
t = –3
t=3
–6
–4
–2
0
2
4
6
x
▲ Figure 4.13
(iii) As t → 0, x → 0 and y → ∞. The y-axis is an asymptote for the curve.
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Example 4.20
A curve has the parametric equations x = t 2, y = t 3 − t.
(i)
Find the coordinates of the points corresponding to values of t from
−2 to +2 at half-unit intervals.
(ii)
Sketch the curve for −2 $ t $ 2.
4
(iii) Are there any values of x for which the curve is undefined?
(i)
(ii)
t
–2
–1.5
–1
–0.5
0
0.5
1
1.5
2
x
4
2.25
1
0.25
0
0.25
1
2.25
4
y
–6
–1.875
0
0.375
0 –0.375
0
1.875
6
y
4.6 Parametric equations
Solution
6
4
2
0
1
2
3
4
x
–2
–4
–6
▲ Figure 4.14
(iii) The curve in Figure 4.14 is undefined for x " 0.
Technology note
Graphical calculators can be used to sketch parametric curves but, as with
cartesian curves, you need to be careful when choosing the range.
Finding the equation by eliminating the parameter
For some pairs of parametric equations, it is possible to eliminate the
parameter and obtain the cartesian equation for the curve. This is usually
done by making the parameter the subject of one of the equations, and
substituting this expression into the other.
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4
Example 4.21
t
Eliminate t from the equations x = t3 − 2t2, y = 2.
Solution
t
y= 2
⇒
t = 2y.
Substituting this in the equation x = t 3 − 2t 2 gives
4 DIFFERENTIATION
x = (2y)3 − 2(2y)2 or
x = 8y 3 − 8y 2.
4.7 Parametric differentiation
To differentiate a function which is defined in terms of a parameter t , you need
to use the chain rule:
Since
dy dy dt
=
× .
dx dt dx
dt = 1
dx dx
dt
it follows that
dy
dy dt
=
dx dx
dt
provided that dx ≠ 0.
dt
Example 4.22
A curve has the parametric equations x = t2, y = 2t.
dy
(i) Find
in terms of the parameter t.
dx
(ii) Find the equation of the tangent to the curve at the general point
(t2, 2t).
(iii) Find the equation of the tangent at the point where t = 3.
(iv) Eliminate the parameter, and hence sketch the curve and the tangent at
the point where t = 3.
Solution
(i)
x = t2
⇒
y = 2t
⇒
dx
= 2t
dt
dy
=2
dt
dy
dy dt
=
= 2 =1
dx dx 2t t
dt
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(ii)
Using y − y1 = m(x − x1) and taking the point (x1, y1) as (t2, 2t), the
equation of the tangent at the point (t2, 2t) is
1
y − 2t = t (x − t2)
⇒
ty − 2t2 = x − t2
⇒
x − ty + t2 = 0
4
This equation still contains
the parameter, and is called
the equation of the tangent
at the general point.
4.7 Parametric differentiation
(iii) Substituting t = 3 into this equation gives the equation of the tangent
at the point where t = 3.
The tangent is x − 3y + 9 = 0.
(iv) Eliminating t from x = t2, y = 2t gives
2
⎛y⎞
or
y2 = 4x.
x=⎜ ⎟
⎝2⎠
This is a parabola with the x-axis as its line of symmetry.
The point where t = 3 has coordinates (9, 6).
The tangent x − 3y + 9 = 0 crosses the axes at (0, 3) and (−9, 0).
The curve is shown in Figure 4.15.
y
(9, 6)
6
3
–9
0
9
x
–6
▲ Figure 4.15
Example 4.23
A curve has parametric equations x = 4 cos θ, y = 3 sin θ.
(i) Find dy at the point with parameter θ.
dx
(ii) Find the equation of the normal at the general point (4 cos θ, 3 sin θ ).
π
(iii) Find the equation of the normal at the point where θ = .
4
π
(iv) Find the coordinates of the point where θ = .
4
(v)
9781510421738.indb 113
Show the curve and the normal on a sketch.
➜
113
02/02/18 1:13 PM
4
Solution
(i)
x = 4 cos θ ⇒
4 DIFFERENTIATION
y = 3 sin θ
(ii)
⇒
dx = −4 sin θ
dθ
dy = 3 cos θ
dθ
dy
dy dθ
=
= 3cos θ
dx dx −4 sin θ
dθ
= − 3cos θ
4 sin θ
The tangent and normal are perpendicular, so the gradient of the
normal is
m1m2 = –1 for
− 1 which is + 4 sin θ .
perpendicular lines.
dy
3cos θ
dx
Using y − y1 = m(x − x1) and taking the point (x1, y1) as (4 cos θ, 3 sin θ ),
the equation of the normal at the point (4 cos θ, 3 sin θ ) is
4 sin θ
y − 3 sin θ = 3cos θ (x − 4 cos θ )
⇒
3y cos θ − 9 sin θ cos θ = 4x sin θ − 16 sin θ cos θ
⇒
4x sin θ − 3y cos θ − 7 sin θ cos θ = 0
1
1
π
(iii) When θ = , cos θ =
and sin θ =
, so the equation of the
4
2
2
normal is
4x ×
⇒
1
1
1
1
− 3y ×
−7×
×
=0
2
2
2
2
4 2x − 3 2y − 7 = 0
⇒
4x − 3y − 4.95 = 0 (to 2 decimal places)
(iv) The coordinates of the point where θ = π are
4
1
1 ⎞
⎛
4 cos π ,3 sin π = ⎜ 4 ×
,3 ×
4
4
2
2 ⎟⎠
⎝
≈ (2.83, 2.12)
(
)
y
(v)
This curve is
an ellipse.
3
(2.83, 2.12)
–4
O
4
x
–3
▲ Figure 4.16
114
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Stationary points
Example 4.24
Find the stationary points of the curve with parametric equations x = 2t + 1,
y = 3t − t 3, and distinguish between them.
Solution
dx
=2
dt
dy
⇒
= 3 − 3t2
y = 3t − t3
dt
dy
2
dy
3(1 − t 2 )
dt
=
= 3 − 3t =
2
2
dx dx
dt
dy
Stationary points occur when
= 0:
dx
⇒
t2 = 1
⇒
t=1
or
x = 2t + 1
⇒
At t = 1:
x = 3, y = 2
At t = 0.9:
x = 2.8 (to the left);
At t = −1:
x = −1, y = 2
At t = −1.1:
x = −1.2 (to the left);
4
4.7 Parametric differentiation
When the equation of a curve is given parametrically, the easiest way to
dy
distinguish between stationary points is usually to consider the sign of dx . If you
use this method, you must be careful to ensure that you take points which are to
the left and right of the stationary point, i.e. have x coordinates smaller and
larger than those at the stationary point. These will not necessarily be points
whose parameters are smaller and larger than those at the stationary point.
t = −1
dy
= 0.285 (positive)
dx
dy
= −0.315 (negative)
At t = 1.1:
x = 3.2 (to the right);
dx
There is a maximum at (3, 2).
dy
= −0.315 (negative)
dx
dy
= 0.285 (positive)
At t = −0.9:
x = −0.8 (to the right);
dx
There is a minimum at (−1, −2).
e
An alternative method
d 2y
dy
Alternatively, to find dx 2 when
is expressed in terms of a parameter
dx
requires a further use of the chain rule:
d 2y
d ⎛ dy ⎞ d ⎛ dy ⎞ dt
=
.
⎜ ⎟= ⎜ ⎟ ×
2
dx
dx ⎝ dx ⎠ dt ⎝ dx ⎠ dx
115
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4 DIFFERENTIATION
4
Exercise 4E
1
2
3
4
116
9781510421738.indb 116
PS
5
PS
6
For each of the following curves, find dy in terms of the parameter.
dx
(i) x = 3t 2
(ii)
x = θ − cos θ
y = 2t 3
y = θ + sin θ
1
(iii) x = t + t
(iv)
x = 3 cos θ
y = 2 sin θ
1
y=t−t
(v) x = (t + 1)2
(vi)
x = θ sin θ + cos θ
2
y = (t − 1)
y = θ cos θ − sin θ
t
2t
(vii) x = e + 1
(viii) x =
1
+
t
y = et
t
y=
1−t
A curve has the parametric equations x = tanθ, y = tan 2θ. Find
dy
π
(i) the value of
when θ = 6
dx
π
(ii) the equation of the tangent to the curve at the point where θ =
6
π
(iii) the equation of the normal to the curve at the point where θ = .
6
1
A curve has the parametric equations x = t2, y = 1 − for t # 0. Find
2t
(i) the coordinates of the point P where the curve cuts the x-axis
(ii) the gradient of the curve at this point
(iii) the equation of the tangent to the curve at P
(iv) the coordinates of the point where the tangent cuts the y-axis.
A curve has parametric equations x = at2, y = 2at, where a is constant.
Find
(i) the equation of the tangent to the curve at the point with
parameter t
(ii) the equation of the normal to the curve at the point with
parameter t
(iii) the coordinates of the points where the normal cuts the x- and
y-axes.
A curve has parametric equations x = cos θ, y = cos 2θ.
dy
(i) Show that
= 4 cos θ.
dx
dy
d 2y
(ii) By writing
in terms of x , show that 2 − 4 = 0.
dx
dx
b
The parametric equations of a curve are x = at , y = , where a and b are
t
constant. Find in terms of a, b and t
dy
(i)
dx
b
(ii) the equation of the tangent to the curve at the general point (at, )
t
(iii) the coordinates of the points X and Y where the tangent cuts the
x- and y-axes.
(iv) Show that the area of triangle OXY is constant, where O is the origin.
02/02/18 1:13 PM
PS
7
The diagram shows a sketch of the curve given parametrically in terms
of t by the equations x = 4t and y = 2t2 where t takes positive and
negative values.
y
PS
8
9
x
P is the point on the curve with parameter t.
(i) Show that the gradient at P is t.
(ii) Find and simplify the equation of the tangent at P.
The tangents at two points Q (with parameter t1) and R (with
parameter t2) meet at S.
(iii) Find the coordinates of S.
(iv) In the case when t1 + t2 = 2 show that S lies on a straight line.
Give the equation of the line.
A particle P moves in a plane so that at time t its coordinates are given by
x = 4 cos t, y = 3 sin t. Find
dy
(i)
in terms of t
dx
(ii) the equation of the tangent to its path at time t
(iii) the values of t for which the particle is travelling parallel to the line
x + y = 0.
The parametric equations of a curve are x = e3t, y = t2et + 3.
dy t(t + 2)
(i) Show that
.
=
dx
3e 2t
(ii) Show that the tangent to the curve at the point (1, 3) is parallel to
the x-axis.
(iii) Find the exact coordinates of the other point on the curve at which
the tangent is parallel to the x-axis.
4.7 Parametric differentiation
O
4
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q7 November 2011
10 The parametric equations of a curve are x = e −t cos t, y = e −t sin t .
Show that
dy
= tan(t − 1 π).
4
dx
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q4 November 2013
117
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4
11 The parametric equations of a curve are
x = 3t + ln(t − 1),
(i)
4 DIFFERENTIATION
(ii)
y = t2 + 1,
for t # 1.
dy
in terms of t.
dx
Find the coordinates of the only point on the curve at which the
gradient of the curve is equal to 1.
Express
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q3 June 2007
12 The parametric equations of a curve are
y = 3 – 2 cos 2θ,
dy
where − 21 π " θ "− 21 π . Express
in terms of θ, simplifying your
dx
answer as far as possible.
x = 4 sinθ,
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q4 June 2009
13 The parametric equations of a curve are
y = et + e−t.
x = 1 − e−t,
(i)
(ii)
dy
= e 2t − 1.
dx
Hence find the exact value of t at the point on the curve at which
the gradient is 2.
Show that
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q4 November 2009
14 The parametric equations of a curve are
x = 2θ + sin 2θ,
dy
Show that
= tan θ.
dx
y = 1 − cos 2θ.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q3 June 2006
15 The parametric equations of a curve are
x = a cos3 t,
y = a sin3 t,
where a is a positive constant and 0 " t "− 21 π.
dy
(i) Express
in terms of t.
dx
(ii) Show that the equation of the tangent to the curve at the point
with parameter t is
x sin t + y cos t = a sin t cos t.
(iii) Hence show that, if this tangent meets the x-axis at X and the
y-axis at Y, then the length of XY is always equal to a.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q6 June 2009
118
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KEY POINTS
1
2
4
5
6
7
8
(
)
4
4.7 Parametric differentiation
3
dy
= knxn−1 where k and n are real constants.
dx
dy dy du
=
× .
Chain rule:
dx du dx
dy
du
dv
=v
+u .
Product rule (for y = uv):
dx
dx
dx
du
dv
dy v dx − u dx
u
=
Quotient rule for y = :
.
v dx
v2
dy
= 1
dx d x
dy
d (ln x ) = 1
x
dx
d (e x ) = e x
dx
d (sin kx ) = k cos kx
dx
d (cos kx ) = −k sin kx
dx
d (tan kx ) = k sec 2 kx
dx
y = kxn ⇒
9
An implicit function is one connecting x and y where y is not the
subject. When you differentiate an implicit function:
dy
● differentiating y2 with respect to x gives 2 y
dx
dy
● differentiating 4x3y2 with respect to x gives 12x2 ×y2 + 4x3 ×2 y
dx
● the derivative of any constant is 0.
10 In parametric equations the relationship between two variables is
expressed by writing both of them in terms of a third variable or
parameter.
11 To draw a graph from parametric equations, plot the points on the
curve given by different values of the parameter.
dy
dy dt
=
provided that dx ≠ 0.
12
dt
dx dx
dt
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4
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■
4 DIFFERENTIATION
■
■
■
■
differentiate ekx and ln x and related sums, differences and constant
multiples
differentiate sin kx, cos kx and tan kx where x is measured in radians
use the chain rule, product rule and quotient rule to differentiate
functions involving the functions above
differentiate functions defined implicitly or parametrically
find
■ stationary points
■ equations of tangents
for functions defined implicitly or parametrically.
120
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P2
P3
5
Integration
5.1 Integrals involving the exponential function
The moving
power of
mathematics
invention is not
reasoning, but
imagination.
Augustus de
Morgan
(1806−71)
?
❯ How could you estimate the numbers of birds in this picture?
5.1 Integrals involving the exponential
function
Since you know that
d (eax+b) = aeax+b,
dx
you can see that
∫ eax+b dx = 1a eax+b + c.
This increases the number of functions which you are able to integrate, as in
the following example.
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5
Example 5.1
Find the following integrals.
∫ e 2x–3 dx
(i)
(ii)
5
∫1 6e 3x dx
Solution
∫ e2x–3 dx = 2 e2x–3 + c
1
(i)
5 INTEGRATION
5
6e 3 x
∫ 1 6e3x dx = ⎡⎣⎢ 3 ⎤⎦⎥1
= [ 2e 3 x ]15
5
(ii)
= 2(e15 − e3)
= 6.54 × 106
(to 3 s.f.)
5.2 Integrals involving the natural
logarithm function
You have already seen that
1
∫ x dx = ln x + c.
There are many other integrals that can be reduced to this form.
Example 5.2
Evaluate
5
1
∫ 2 2x dx.
Solution
1
2
5
1
∫ 2 x dx = 21 [ ln x ] 25
= 21 (ln 5 − ln 2)
= 0.458
(to 3 s.f.)
In this example the 21 was taken outside the integral, allowing the standard
result for x1 to be used.
Since
y = ln(ax + b) ⇒
So
dy
a
=
dx ax + b
a
∫ ax + b dx = ln(ax + b ) + c
and
122
9781510421738.indb 122
1
c means ‘an arbitrary constant’
and so does not necessarily
have the same value from one
equation to another.
1
∫ ax + b dx = a ln(ax + b ) + c
02/02/18 1:13 PM
Example 5.3
Find ∫
2
0
5
1 dx .
5x + 3
Solution
2
∫ 0 5x1+ 3 dx = ⎡⎣ 51 ln(5x + 3)⎤⎦ 0
2
= 0.293
5.2 Integrals involving the natural logarithm function
= 51 ln13 − 51 ln 3
(to 3 s.f.)
Extending the domain for logarithmic integrals
1
The use of ∫ x dx = ln x + c has so far been restricted to cases where x # 0,
since logarithms are undefined for negative numbers.
Look, however, at the area between −b and −a on the left-hand branch of the
curve y = x1 in Figure 5.1.You can see that it is a real area, and that it must be
possible to evaluate it.
y
–b
B
–a
A
O
a
b
x
▲ Figure 5.1
ACTIVITY 5.1
(i)
What can you say about the areas of the two shaded regions?
(ii)
Try to prove your answer to part (i) before reading on.
CP
123
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Proof
5
−a 1
Let A = ∫ −b x dx.
Now write the integral in terms of a new variable, u, where u = −x.
5 INTEGRATION
This gives new limits: x = −b ⇒ u = b
x = −a ⇒ u = a.
du = −1 ⇒ dx = −du.
dx
So the integral becomes
a
A = ∫ 1 (−du)
b −u
a
= ∫ 1 du
bu
[
= ln a − ln b]
= −[ ln b − ln a]
= −area B
So the area has the same size as that obtained if no notice is taken of the fact
that the limits a and b have minus signs. However it has the opposite sign, as
you would expect because the area is below the axis.
Consequently the restriction that x > 0 may be dropped, and the integral is
written
1
x dx = ln | x | + c.
∫
f '( x )
Similarly, f ( x ) dx = ln | f(x) | + c.
∫
Example 5.4
Find the value of
7
1
∫ 4 − x dx.
5
Solution
To make the top line into the differential of the bottom line, you write the
integral in one of two ways.
−∫
7
−1 dx = −3ln |4 − x |47
5
54 − x
= −3(ln |−3|) − (ln |−1|)4
= −3ln 3 − ln 14
= −1.10 (to 3 s.f.)
−∫
7
1 dx = −3ln |x − 4|47
5
5x −4
= −3ln 3 − ln 14
= −1.10 (to 3 s.f.)
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02/02/18 1:13 PM
y
Consequently in the integral
q
1
x dx both the limits p and q must
O
∫
5
x
p
have the same sign, either + or −.
The integral is invalid otherwise.
▲ Figure 5.2
?
p (x )
The equation of a curve is y = 1
where p1(x) and p2(x) are
p 2 (x )
polynominals.
❯ How can you tell from the equation whether the curve has a
discontinuity?
❯ How can you prove that y = x 2 − 2x + 3 has no discontinuities?
Exercise 5A
1
Find the following indefinite integrals.
(i)
2
3
∫ x3 dx
∫ 4x1 dx
(ii)
∫ x −1 5 dx
(iii)
(iv)
PS
5.2 Integrals involving the natural logarithm function
Since the curve y = 1 is not
x
defined at the discontinuity at x = 0
(see Figure 5.2), it is not possible to
integrate across this point.
∫ 2x1− 9 dx
Find the following indefinite integrals.
(i)
∫ e3x dx
(ii)
∫ e−4x dx
(iv)
∫ e105x dx
(v)
∫ e e 2+x 4 dx
(iii)
x
∫ e − 3 dx
3x
Find the following definite integrals.
Where appropriate give your answers to 3 significant figures.
4
(i)
∫ 0 4e2x dx
(iii)
∫ –1 (ex + e–x) dx
1
3
(ii)
∫ 1 2x4+ 1 dx
(iv)
∫ −2 e3x −2 dx
1
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02/02/18 1:13 PM
4
5
The graph of y = x + 4 is shown below.
x
y
P
–4
5 INTEGRATION
O
5
x
Q
Find the coordinates of the minimum point, P, and the maximum
point, Q.
(ii) Find the area of each shaded region.
The diagram illustrates the graph of y = ex. The point A has coordinates
(ln 5, 0), B has coordinates (ln 5, 5) and C has coordinates (0, 5).
(i)
CP
5
y
C
B (ln5, 5)
E
O
(i)
(ii)
A
x
Find the area of the region OABE enclosed by the curve y = ex, the
x-axis, the y-axis and the line AB. Hence find the area of the shaded
region EBC.
The graph of y = ex is transformed into the graph of y = ln x.
Describe this transformation geometrically.
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02/02/18 1:13 PM
(iii) Using your answers to parts (i) and (ii), or otherwise, show that
5
∫ 1 ln x dx = 5 ln 5 − 4.
(iv) Deduce the values of
5
(a)
∫ 1 ln(x3) dx
(b)
∫ 1 ln(3x) dx.
5
Differentiate ln(2x + 3).
(ii) Hence, or otherwise, show that
3
1
∫−1 2x + 3 dx = ln 3.
(iii) Find the quotient and remainder when 4x 2 + 8x is divided
by 2x + 3.
(iv) Hence show that
(i)
4 x 2 + 8x dx = 12 − 3 ln 3.
∫−1 2x + 3
3
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q7 June 2006
7
dy
= e2x − 2e−x. The point (0, 1) lies on the curve.
dx
Find the equation of the curve.
The curve has one stationary point. Find the x coordinate of this
point and determine whether it is a maximum or a minimum
point.
A curve is such that
(i)
(ii)
5.2 Integrals involving the natural logarithm function
6
5
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q6 November 2005
8
(i)
(ii)
Find the equation of the tangent to the curve y = ln(3x − 2) at the
point where x = 1.
(a) Find the value of the constant A such that
6x ≡ 2 + A .
3x − 2
3x − 2
6 6x
dx = 8 + 8 ln 2.
(b) Hence show that ∫
3
2 3x − 2
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q8 June 2009
9
Find the exact value of the constant k for which ∫
k
1
1 2x − 1
dx = 1.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q1 November 2007
127
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02/02/18 1:13 PM
e
5
INVESTIGATIONS
A series for ex
The exponential function can be written as the infinite series
e x = a0 + a 1x + a 2x 2 + a 3x 3 + a4x 4 + …
(for x ! ")
where a 0, a 1, a 2, … are numbers.
5 INTEGRATION
You can find the value of a 0 by substituting the value zero for x.
Since e 0 = 1, it follows that 1 = a0 + 0 + 0 + 0 + … , and so a0 = 1.
You can now write: ex = 1 + a 1x + a 2x 2 +a 3 x 3+ a 4 x 4 + … .
Now differentiate both sides: ex = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 + … ,
and substitute x = 0 again: 1 = a 1 + 0 + 0 + 0 + … , and so a 1 = 1 also.
Now differentiate a second time, and again substitute x = 0. This time
you find a 2. Continue this procedure until you can see the pattern in the
values of a 0, a 1, a 2, a 3, … .
When you have the series for ex, substitute x = 1. The left-hand side is
e 1 or e, and so by adding the terms on the right-hand side you obtain the
value of e. You will find that the terms become small quite quickly, so
you will not need to use very many to obtain the value of e correct to
several decimal places.
If you are also studying statistics you will meet this series expansion of
ex in connection with the Poisson distribution.
Compound interest
You win $100 000 in a prize draw and are offered two investment
options.
A You are paid 100% interest at the end of 10 years, or
B You are paid 10% compound interest year by year for 10 years.
Under which scheme are you better off?
final money
$200 000
Clearly in scheme A, the ratio R = −−−−−−−−−− is −−−−−−−− = 2.
original money $100 000
What is the value of the ratio R in scheme B?
Suppose that you asked for the interest to be paid in 20 half-yearly
instalments of 5% each (scheme C). What would be the value of R in
this case?
Continue this process, investigating what happens to the ratio R when the
interest is paid at increasingly frequent intervals.
Is there a limit to R as the time interval between interest payments tends
to zero?
128
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5.3 Integrals involving
trigonometrical functions
Since
d (sin(ax + b )) = a cos(ax + b )
dx
it follows that
∫ cos(ax + b )dx = 1a sin(ax + b ) + c
Example 5.5
Also
d (tan(ax + b )) = a sec 2 (ax + b )
dx
and so
∫ sec 2 (ax + b )dx = 1a tan(ax + b ) + c
∫ a cos(ax + b) dx
= sin(ax + b) + c
∫ −a sin(ax + b) dx
= cos(ax + b) + c
Find
∫ sec 2 x dx
(i)
∫ sin 2x dx
(ii)
(iii)
∫ cos(3x − π)dx.
Solution
∫ sec 2 x dx = tan x + c
(ii) ∫ sin 2x dx = − 21 cos 2x + c
(iii) ∫ cos(3x − π) dx = 1 sin(3x − π) + c
3
(i)
Example 5.6
Find the exact value of
5.3 Integrals involving trigonometrical functions
d (cos(ax + b )) = −a sin(ax + b )
dx
it also follows that ∫ sin(ax + b )dx = − 1a cos(ax + b ) + c
Similarly, since
5
π
3
∫ (sin 2x − cos 4x )dx.
0
Solution
π
3
π
3
0
0
∫ (sin 2x − cos4x )dx = ⎡⎣− 21 cos 2x − 41 sin 4x⎤⎦
=0
= ⎡⎢− 1 cos 2π − 1 sin 4π ⎤⎥ − ⎡⎢− 1 cos 0 − 1 sin 0 ⎤⎥
⎣ 2
4
⎦
3 4
3⎦ ⎣ 2
⎡
⎛
⎞⎤
= ⎢− 21 × − 21 − 41 × ⎜− 3 ⎟ ⎥ − ⎡⎣− 21 × 1⎤⎦
⎝ 2 ⎠⎦
⎣
( )
= 41 +
3+ 1
8 2
3+ 3
8 4
= 6+ 3
8
=
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Using trigonometrical identities in integration
5
Sometimes, when it is not immediately obvious how to integrate a function
involving trigonometrical functions, it may help to rewrite the function using
one of the trigonometrical identities.
5 INTEGRATION
Example 5.7
Find ∫ sin2 x dx.
Solution
Use the identity
cos 2x = 1 − 2 sin2 x.
(Remember that this is just one of the three expressions for cos 2x.)
This identity may be rewritten as
sin2 x = 21 (1 − cos 2x).
By putting sin2x in this form, you will be able to perform the integration.
∫ sin2 x dx = 21 ∫ (1 − cos 2x) dx
= 21 (x − 21 sin 2x) + c
1
= 21 x − 4 sin 2x + c
You can integrate cos2 x in the same way, by using cos2 x = 21 (cos 2x + 1).
Other even powers of sin x or cos x can also be integrated in a similar way,
but you have to use the identity twice or more.
Example 5.8
Find ∫ cos4 x dx.
Solution
First express cos4 x as (cos2 x)2:
cos4 x = [ 2 (cos 2x + 1)]
1
2
1
= 4 (cos2 2x + 2 cos 2x + 1)
Next, apply the same identity to cos2 2x :
cos2 2x = 21 (cos 4x + 1)
Hence
cos4 x = 41 ( 2 cos 4x + 2 + 2 cos 2x + 1)
1
1
= 41 ( 21 cos 4x + 2 cos 2x + 23 )
= 81 cos 4x + 21 cos 2x + 83
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This can now be integrated.
∫ cos4 x dx = ∫ ( 81 cos 4x + 21 cos 2x + 83 ) dx
1
sin 4x + 41 sin 2x + 83 x + c
= 32
Exercise 5B
1
(vi) cos(5x − π)
(ix) 4 sec2 x − cos 2x
π
π
(i)
(iii) 5 sin x + 4 cos x
∫ sin x dx
3
∫ sec x dx
4
(ii)
0
2
0
π
3
2π
3
π
0
(iv)
∫ sin 2x dx
∫ cos x dx
(v)
(vi)
∫ sec 2x dx
∫ cos 3x dx
π
(vii) ∫ cos ( 2x + ) dx
(viii) ∫ (sec x + cos 4 x )dx
2
(ix) ∫ (cos x + sin 2x )dx
1
(iii)
6
5π
6
π
6
π
8
0
π
π
0
0
2
4
2
π
6
3
PS
4
0
5.3 Integrals involving trigonometrical functions
2
Integrate the following with respect to x.
(i) sin x − 2 cos x (ii)
3 cos x + 2 sin x
2
(iv) 4 sec x
(v)
sin(2x + 1)
2
(vii) 6 sec 2x
(viii) 3 sec2 3x − sin 2x
Find the exact value of the following.
5
Show that sin x cos x = 2 sin 2x. π
(ii) Hence find the exact value of 3 sin x cos x dx .
0
Use a suitable trigonometrical identity to help you find these.
(i)
∫
π
∫ cos x dx
∫
(ii) (a)
(b)
∫ sin x dx
∫ sin x dx
(i) Express ( 3 ) cos x + sin x in the form R cos( x − α ), where R > 0
(i)
(a)
cos 2 xdx
(b)
2
2
0
π
3
2
2
0
5
(ii)
and 0 < α < 21 π , giving the exact values of R and α.
1π
1
1
Hence show that 12
2 dx = 4 3 .
π
6
3 cos x + sin x
∫
(( )
)
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q4 June 2013
6
(i)
(ii)
Prove that tan θ + cot θ ≡ 2 .
sin 2θ
Hence
1
1
(a) find the exact value of tan 8 π + cot 8 π
1π
2
6
dθ .
(b) evaluate
θ
+
tan
cot θ
0
∫
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q5 June 2014
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5
7
(i)
(ii)
Express cos2 x in terms of cos 2x.
Hence show that
1π
3
∫ cos x d x = 61 π + 81 3 .
2
0
(iii) By using an appropriate trigonometrical identity, deduce the exact
5 INTEGRATION
value of
1π
3
∫ sin x d x.
2
0
8
(i)
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q6 June 2007
Prove the identity
(cos x + 3 sin x)2 ≡ 5 − 4 cos 2x + 3 sin 2x.
(ii)
Using the identity, or otherwise, find the exact value of
1π
4
∫ (cos x + 3 sin x ) d x.
2
0
9
(i)
(ii)
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q7 November 2007
∫
1π
Show that 4 cos 2x d x = 21.
0
By using an appropriate trigonometrical identity, find the exact
value of
1
π
3
1π
6
∫ 3 tan x d x.
2
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q4 June 2010
∫ 4e (3 + e )dx.
(ii) Show that ∫ (3 + 2 tan θ )dθ = 1 (8 + π).
2
10 (i)
Find
x
2x
1π
4
− 1π
4
11 (i)
(ii)
2
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q6 June 2011
Prove the identity cos 4θ + 4 cos 2θ ≡ 8 cos 4 θ − 3.
Hence
(a) solve the equation cos 4θ + 4 cos 2θ = 1 for − 21 π & θ & 21 π
(b)
find the exact value of
1π
4
∫ cos θ dθ .
4
0
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q9 June 2011
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12 (i)
By first expanding cos (2x + x), show that
cos 3x ≡ 4 cos3x − 3 cos x.
(ii)
Hence show that
1π
6
5
∫ (2 cos x − cos x )dx = 125 .
3
0
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q8 November 2011
(ii)
Find
∫ 4 cos ( θ ) dθ .
2
1
2
Find the exact value of
∫ 2x1+ 3 dx .
6
−1
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q3 November 2014
5.4 Numerical integration
5.4 Numerical integration
13 (i)
Note
Section 5.4, including Exercise 5C, is about understanding and using the
trapezium rule to estimate the value of a definite integral. This is required
knowledge only for Paper 2: Pure Mathematics 2 (AS Level). If you are
studying for the A Level, although you may find it interesting to extend your
knowledge, it is not required for Paper 3: Pure Mathematics 3.
There are times when you need to find the area under a graph but cannot do
this by the integration methods you have met so far.
» The function may be one that cannot be integrated algebraically. (There
are many such functions.)
» The function may be one that can be integrated algebraically but which
requires a technique with which you are unfamiliar.
» It may be that you do not know the function in algebraic form, but just
have a set of points (perhaps derived from an experiment).
In these circumstances you can always find an approximate answer using a
numerical method, but you must:
» have a clear picture in your mind of the graph of the function, and how
your method estimates the area beneath it
» understand that a numerical answer without any estimate of its accuracy,
or error bounds, is valueless.
The trapezium rule
In this chapter just one numerical method of integration is introduced,
namely the trapezium rule. As an illustration of the rule, it is used to find
the area under the curve y = 5x − x 2 for values of x between 0 and 4.
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5
It is in fact possible to integrate this function algebraically, but not using the
techniques that you have met so far.
Note
5 INTEGRATION
You should not use a numerical method when an algebraic (sometimes called
analytic) technique is available to you. Numerical methods should be used
only when other methods fail.
Figure 5.3 shows the area approximated by two trapezia of equal width.
y
3
y = 5x – x 2
2
1
A
O
1
B
2
3
4
5
x
▲ Figure 5.3
Remember the formula for the area of a trapezium, Area = 21 h(a + b),
where a and b are the lengths of the parallel sides and h is the distance
between them.
In the cases of the trapezia A and B, the parallel sides are vertical. The
left-hand side of trapezium A has zero height, and so the trapezium is also
a triangle.
When
x=0 ⇒
y=
0
=0
When
x=2
⇒
y=
6
= 2.4495 (to 4 d.p.)
When
x = 4, ⇒
y=
4
=2
A
(0)
2.4495
(2)
2.4495
(2)
B
2
(4)
▲ Figure 5.4
The area of trapezium A = 21 × 2 × (0 + 2.4495) =
2.4495
The area of trapezium B = 21 × 2 × (2.4495 + 2) =
4.4495
Total
6.8990
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For greater accuracy you can use four trapezia, P, Q, R and S, each of width
1 unit as in Figure 5.5. The area is estimated in just the same way.
y
5
3
2
P
Q
2
O
R
6
1
S
6
2
2
3
4
5
x
▲ Figure 5.5
These figures
are given to
4 decimal
places but the
calculation has
been done to
more places on
a calculator.
Trapezium P:
1
× 1 × (0 + 2)
2
= 1.0000
Trapezium Q:
1
× 1 × (2 + 2.4495)
2
= 2.2247
Trapezium R:
1
× 1 × (2.4495 + 2.4495)
2
= 2.4495
Trapezium S:
1
× 1 × (2.4495 + 2)
2
= 2.2247
Total
5.4 Numerical integration
1
7.8990
Accuracy
In this example, the first two estimates are 6.8989… and 7.8989… .You can
see from Figure 5.5 that the trapezia all lie underneath the curve, and so in
this case the trapezium rule estimate of 7.8989… must be too small.You
cannot, however, say by how much. To find that out you will need to take
progressively more strips to find the value to which the estimate converges.
Using 8 strips gives an estimate of 8.2407…, and 16 strips gives 8.3578… .
The first figure, 8, looks reasonably certain but it is still not clear whether
the second is 3, 4 or even 5.You need to take even more strips to be able to
decide. In this example, the convergence is unusually slow because of the high
curvature of the curve.
Technology note
You can use a graph-drawing program with the capability to calculate areas
using trapezia. Calculate the area using progressively more strips and
observe the convergence.
For example,
32 strips:
8.398
50 strips:
8.409
100 strips:
8.416
1000 strips:
8.420
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5
?
It is possible to find this area without using calculus at all.
❯ How can this be done? How close is the 16-strip estimate?
The procedure
5 INTEGRATION
In the previous example, the answer of 7.8990 from four strips came from
adding the areas of the four trapezia P, Q, R and S:
1
1
1
2 × 1 × (0 + 2) + 2 × 1 × (2 + 2.4495) + 2 × 1 × (2.4495 + 2.4495)
1
These are the heights
+ 2 × 1 × (2.4495 + 2)
of the intermediate
vertical lines.
and this can be written as
1
×1 ×[0 + 2 ×(2 + 2.4495 + 2.4495) + 2]
2
These are the heights of the
ends of the whole area: 0 and 2.
This is the
strip width: 1.
This is a useful
way to remember
the formula.
This is often stated in words as
Area " 21 × strip width ×[ends + twice middles]
or in symbols, for n strips of width h
A " 21 × h ×[y0 + yn + 2(y1 + y2 + … + yn − 1)].
This is called the trapezium rule for width h (see Figure 5.6).
y
y = f(x)
y1
y0
yn
y2
yn–1
yn–2
a
b
h
h
h
x
h
▲ Figure 5.6
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?
❯ Look at the three graphs in Figure 5.7, and in each case state whether
the trapezium rule would underestimate or overestimate the area, or
whether you cannot tell.
y
y
y
x
x
(ii)
(iii)
▲ Figure 5.7
Exercise 5C
M
This exercise covers the trapezium rule, which is required knowledge only
for Paper 2: Pure Mathematics 2 (AS Level). It is not required for Paper 3:
Pure Mathematics 3 (A Level).
1
CP
2
3
5.4 Numerical integration
x
(i)
5
The speed v in m s−1 of a train at time t seconds is given in the following
table.
t
0
10
20
30
40
50
60
v
0
5.0
6.7
8.2
9.5
10.6
11.6
The distance that the train has travelled is given by the area under the
graph of the speed (vertical axis) against time (horizontal axis).
(i) Estimate the distance the train travels in this 1-minute period.
(ii) Give two reasons why your method cannot give a very accurate
answer.
1
1 dx is known to equal π .
The definite integral ∫
4
0 1+ x2
(i) Using the trapezium rule for four strips, find an approximation for π.
(ii) Repeat your calculation with 10 and 20 strips to obtain closer
estimates.
(iii) If you did not know the value of π, what value would you give it
with confidence on the basis of your estimates in parts (i) and (ii)?
1.6
1 + x 2 dx.
The trapezium rule is used to estimate the value of I = ∫
0
(i) Draw the graph of y = 1 + x 2 for 0 $ x $ 1.6.
(ii) Use strip widths of 0.8, 0.4, 0.2 and 0.1 to find approximations to
the value of the integral.
(iii) State the value of the integral to as many decimal places as you can
justify.
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5 INTEGRATION
5
4
5
6
1
The trapezium rule is used to estimate the value of ∫ sin x dx.
0
(i) Draw the graph of y = sin x for 0 $ x $ 1.
(ii) Use 1, 2, 4, 8 and 16 strips to find approximations to the value of
the integral.
(iii) State the value of the integral to as many decimal places as you can
justify.
1
4 dx .
The trapezium rule is used to estimate the value of
+
x2
1
0
4
(i) Draw the graph of y =
for 0 $ x $ 1.
1+ x2
(ii) Use strip widths of 1, 0.5, 0.25 and 0.125 to find approximations to
the value of the integral.
(iii) State the value of the integral to as many decimal places as you can
justify.
A student uses the trapezium rule to estimate the value of
∫
2
∫ 0(2 − cos 2πx) dx.
(i)
(ii)
7
Find approximations to the value of the integral by applying the
trapezium rule using strip widths of (a) 2 (b) 1 (c) 0.5 (d) 0.25.
Sketch the graph of y = 2 − cos 2πx for 0 $ x $ 2. On copies of
your graph shade the areas you have found in parts (i)(a) to (d).
(iii) Use integration to find the exact value of this integral.
Use the trapezium rule with four intervals to find an approximation to
5
∫ | 2x − 8 | dx.
1
8
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q1 November 2014
ln x
for 0 " x $ 4.
x
The curve cuts the x-axis at A and its maximum point is M.
The diagram shows the part of the curve y =
y
M
O
A
4
x
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Cambridge International AS & A Level Mathematics
9709 Paper 2 Q6 June 2005
9
The diagram shows the part of the curve y = ex cos x for 0 $ x $ 21 π.
The curve meets the y-axis at the point A. The point M is a maximum
point.
y
5
5.4 Numerical integration
Write down the coordinates of A.
(ii) Show that the x coordinate of M is e, and write down the y
coordinate of M in terms of e.
(iii) Use the trapezium rule with three intervals to estimate the value of
4
∫ 1 lnxx dx ,
correct to 2 decimal places.
(iv) State, with a reason, whether the trapezium rule gives an
underestimate or an overestimate of the true value of the integral in
part (iii).
(i)
M
A
O
1
π
2
x
Write down the coordinates of A.
(ii) Find the x coordinate of M.
(iii) Use the trapezium rule with three intervals to estimate the value of
(i)
1π
2
∫ e cos x d x,
x
0
giving your answer correct to 2 decimal places.
(iv) State, with a reason, whether the trapezium rule gives an
underestimate or an overestimate of the true value of the integral in
part (iii).
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q7 June 2007
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5
10 The diagram shows the curve y = x 2 e–x and its maximum point M.
y
5 INTEGRATION
M
x
O
Find the x coordinate of M.
(ii) Show that the tangent to the curve at the point where x = 1 passes
through the origin.
(iii) Use the trapezium rule, with two intervals, to estimate the value of
(i)
3
∫ x e dx ,
2 −x
1
giving your answer correct to 2 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q8 November 2007
11 (i)
(ii)
16
6 dx = ln125 .
Show that
6 2x − 7
Use the trapezium rule with four intervals to find an approximation
to
∫
17
∫ log x dx,
1
10
giving your answer correct to 3 significant figures.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q6 June 2014
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KEY POINTS
∫ kx dx = kxn+ 1 + c
2
∫ e dx = e + c
∫ e dx = 1a e + c
1
3
∫ x dx = ln | x | + c
∫ ax1+ b dx = 1a ln | ax + b | + c
1
4
∫ cos(ax + b)dx = a sin(ax + b) + c
1
∫ sin(ax + b)dx = − a cos(ax + b) + c
∫ sec (ax + b)dx = 1a tan(ax + b) + c
1
n +1
n
x
5
x
ax + b
ax + b
5.4 Numerical integration
2
5
(Paper 2: Pure Mathematics 2 only) You can use the trapezium
rule, with n strips of width h, to find an approximate value for a
definite integral as
A ≈ h ⎣⎡y 0 + 2( y 1 + y 2 + ... + y n − 1 ) + y n ⎦⎤
2
In words this is
Area ≈ 21 × strip width × [ends + twice middles].
LEARNING OUTCOMES
Now you have finished this chapter, you should be able to
■ extend the idea that integration is the reverse of differentiation to integrate
■ e ax +b
1
■
ax + b
■
■
■
■
9781510421738.indb 141
■
sin(ax + b )
■
cos(ax + b )
■
sec 2 (ax + b )
and related sums, differences and constant multiples
use integration in cases where the process is the reverse of the chain rule
use trigonometrical identities in carrying out integration
(Paper 2: Pure Mathematics 2 only) understand and use the
trapezium rule to estimate the value of a definite integral
(Paper 2: Pure Mathematics 2 only) use a sketch graph to determine
whether the trapezium rule gives an overestimate or an underestimate.
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5
PROBLEM SOLVING
PS
Numerical integration
Vesna is carrying out numerical integration, approximating the area under
a curve by rectangles. She makes the following statements.
• Using rectangles drawn from the left-hand ends of the intervals is a
5 INTEGRATION
poor method. Using the right-hand ends is no better. They always give
you extreme estimates for the area under the curve. You can see it in my
diagrams (Figures 5.8 and 5.9).
y
y
2
2
1
1
O
1
2
3
4
5
x
▲ Figure 5.8 Low estimate in this case
O
1
2
3
4
x
5
▲ Figure 5.9 High estimate in this case
• Taking the average of the left-hand and right-hand rectangles is better.
The answer is always the same as you get from the trapezium rule.
• An even better method is to draw the rectangles at the midpoints of the
intervals as shown in Figure 5.10. For any number of intervals that will be
more accurate.
y
2
1
O
1
2
3
4
5
x
▲ Figure 5.10 Best estimate
Investigate whether Vesna’s statements are true.
1 Problem specification and analysis
Vesna has made three statements. Look at them carefully and decide how you
are going to proceed.
• Which of them are little more than common sense and which need real
work?
• How you are going to investigate those that need work?
- What technology will you use?
- What examples will you choose to work with?
- How many different examples do you expect to use?
- To what level of accuracy do you expect to work?
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• How you are going to report the outcomes?
- Will you be happy just to say ‘true’ or ‘false’, or do you expect to make
statements such as ‘it is usually true but there are exceptions such as ...’?
- How much explanation do you expect to give?
It will be helpful, at least in some cases, to choose functions which you know
how to integrate, allowing you to know the answer to which a numerical
method should be converging.
However, the whole point of using a numerical method is to find an answer
when an analytical method is not available to you. It may be that there is one
that you don’t know or it may be that one just does not exist. So you should
also use at least one example where you will only know the answer (to your
chosen level of accuracy) when you have completed your work.
5 Problem solving
2 Information collection
You may be able to make some comments on Vesna’s statements just by
thinking about them but you will also need to carry out some investigations
of your own. This will require the use of technology.
5
Do not be content to work with just one or two types of functions. Try
a variety of functions but always start with a sketch of the curve of the
function.You can of course use graphing software to obtain this.
3 Processing and representation
The previous stage will probably result in you having a lot of information.
Now you need to sort through it and to organise it in a systematic way that
allows you to comment on Vesna’s three statements.
Where you can explain your results using algebra, then you should do so. In
other cases you may present them as experimental outcomes.
4 Interpretation
The method illustrated in Figure 5.10 is called the midpoint rule. Much
of your work on this task will have been focused on midpoint rule, and
you need to comment on whether this is a good method for numerical
integration. In order to do so you will need to explain what you mean by
‘good’ and what the desirable features are in such a method.
There are other methods of numerical integration and you may choose to
conclude by saying something about them.
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P2
P3
6 NUMERICAL SOLUTION OF EQUATIONS
6
Numerical solution of
equations
It is the true
nature of
mankind to
learn from his
mistakes.
Fred Hoyle
(1915–2001)
A golfer doesn’t often hit a ball into the hole at the first attempt! Instead,
he or she will try to hit the ball as close to the hole as possible. After that,
successive attempts will usually be closer and closer to the hole, until the
ball finally lands in the hole.
Think of some other situations where you need to make a rough
approximation for your first attempt, and then gradually improve
your attempts.
❯ Which of the following equations can be solved algebraically,
and which cannot? For each equation find a solution, accurate or
approximate.
(i) x2 − 4x + 3 = 0
(ii) x2 + 10x + 8 = 0
?
❯
(iii) x5 − 5x + 3 = 0
(v)
(iv)
x3 − x = 0
ex = 4x
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You probably realised that the equations x5 − 5x + 3 = 0 and ex = 4x cannot be
solved algebraically.You may have decided to draw their graphs, either manually
or using a graphical calculator or computer package, as in Figure 6.1.
y
y
6
f(x) = 4x
(2.15, 8.6)
(–1, 7)
f(x) = ex
(0.357, 1.43)
x
O
(1, –1)
O
x
▲ Figure 6.1
The graphs show you that
» x5 − 5x + 3 = 0 has three roots, lying in the intervals [−2, −1], [0, 1] and
[1, 2].
» ex = 4x has two roots, lying in the intervals [0, 1] and [2, 3].
6 Numerical solution of equations
f(x) = x5 – 5x + 3
Note
An interval written as [a, b] means the interval between a and b, including a
and b. This notation is used in this chapter. If a and b are not included, the
interval is written (a, b). You may also elsewhere meet the notation ]a, b[,
indicating that a and b are not included.
The problem now is how to find the roots to any required degree of
accuracy, and as efficiently as possible.
In many real problems, equations are obtained for which solutions using
algebraic or analytic methods are not possible, but for which you nonetheless
want to know the answers. In this chapter you will be introduced to
numerical methods for solving such equations. In applying these methods,
keep the following points in mind.
» Only use numerical methods when algebraic ones are not available. If you
can solve an equation algebraically (e.g. a quadratic equation), that is the
right method to use.
» Before starting to use a calculator or computer program, always start by
drawing a sketch graph of the function whose equation you are trying to solve.
This will show you how many roots the equation has and their approximate
positions. It will also warn you of possible difficulties with particular methods.
When using a graphical calculator or computer package ensure that the range
of values of x is sufficiently large to, hopefully, find all the roots.
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» Always give a statement about the accuracy of an answer (e.g. to 5 decimal
places, or ± 0.000 005). An answer obtained by a numerical method is
worthless without this; the fact that at some point your calculator display
reads, say, 1.676 470 588 2 does not mean that all these figures are valid.
» Your statement about the accuracy must be obtained from within the
numerical method itself. Usually you find a sequence of estimates of everincreasing accuracy.
» Remember that the most suitable method for one equation may not be
that for another.
6 NUMERICAL SOLUTION OF EQUATIONS
6
6.1 Interval estimation −
change-of-sign methods
Assume that you are looking for the roots of the equation f(x) = 0. This
means that you want the values of x for which the graph of y = f(x) crosses
the x-axis. As the curve crosses the x-axis, f(x) changes sign, so provided that
f(x) is a continuous function (its graph has no asymptotes or other breaks in
it), once you have located an interval in which f(x) changes sign, you know
that that interval must contain a root. In both of the graphs in Figure 6.2,
there is a root lying between a and b.
y
y
a
O
b
b
x
O
a
x
▲ Figure 6.2
Example 6.1
Show that the equation ex − x = x2 + 2 has a root between x = 2 and x = 3.
Solution
There are two methods you can use to show there is a root in the interval [a, b].
The left-hand
side, LHS, and
the right-hand
side, RHS.
Method 1
Substitute x = a into each side of the equation and show that LHS < RHS
for x = a.
Or vice versa:
Then show LHS > RHS for x = b.
LHS < RHS
Hence there exists a value of x in the when x = b.
interval [a, b] where the LHS = RHS,
which is the root of the equation.
Or vice versa:
LHS > RHS when
x = a.
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When x = 2, e x − x = e 2 − 2 = 5.389...
6
and x 2 + 2 = 2 2 + 2 = 6.
So when x = 2, e x − x < x 2 + 2 .
When x = 3, e x − x = e 3 − 3 = 17.085...
and x 2 + 2 = 3 2 + 2 = 11.
So when x = 3, e x − x > x 2 + 2 .
ex − x = x 2 + 2.
Method 2
Rearrange the equation so it is equal to zero. Then show there is a change of
sign in the interval [a, b].
ex − x = x 2 + 2 ⇒ ex − x 2 − x − 2 = 0
When x = 2: e 2 − 2 2 − 2 − 2 = −0.6109...
When x = 3: e 3 − 3 2 − 3 − 2 = 6.0855...
There is a change of sign in the interval [2, 3], so there must be a root
between x = 2 and x = 3.
You have seen that x5 − 5x + 3 = 0 has roots in the intervals [−2, −1], [0, 1]
and [1, 2]. There are several ways of homing in on such roots systematically.
Two of these are now described, using the search for the root in the interval
[0, 1] as an example.
6.1 Interval estimation − change-of-sign methods
Hence there must be a value of x between x = 2 and x = 3 where
Decimal search
In this method you first take increments in x of size 0.1 within the interval
[0, 1], working out the value of f(x) = x5 − 5x + 3 for each one.You do this
until you find a change of sign.
x
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
f (x)
3.00
2.50
2.00
1.50
1.01
0.53
0.08
−0.33
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6
There is a sign change, and therefore a root, in the interval [0.6, 0.7] since
the function is continuous. Having narrowed down the interval, you can now
continue with increments of 0.01 within the interval [0.6, 0.7].
x
0.60
0.61
0.62
f (x)
0.08
0.03
−0.01
6 NUMERICAL SOLUTION OF EQUATIONS
This shows that the root lies in the interval [0.61, 0.62].
Alternative ways of expressing this information are that the root can be taken as
0.615 with a maximum error of ± 0.005, or the root is 0.6 (to 1 decimal place).
This process can be continued by considering x = 0.611, x = 0.612, … to
obtain the root to any required number of decimal places.
❯
How many steps of decimal search would be necessary to find each of
the values 0.012, 0.385 and 0.989, using x = 0 as a starting point?
?
When you use this procedure on a computer or calculator you should be aware
that the machine is working in base 2, and that the conversion of many simple
numbers from base 10 to base 2 introduces small rounding errors. This can lead
to simple roots such as 2.7 being missed and only being found as 2.699 999.
Interval bisection
This method is similar to the decimal search, but instead of dividing each
interval into ten parts and looking for a sign change, in this case the interval
is divided into two parts − it is bisected.
Looking as before for the root in the interval [0, 1], you start by taking the
midpoint of the interval, 0.5.
f(0.5) = 0.53, so f(0.5) # 0. Since f(1) " 0, the root is in [0.5, 1].
Now take the midpoint of this second interval, 0.75.
f(0.75) = −0.51, so f(0.75) " 0. Since f(0.5) # 0, the root is in [0.5, 0.75].
The midpoint of this further reduced interval is 0.625.
f(0.625) = −0.03, so the root is in the interval [0.5, 0.625].
The method continues in this manner until any required degree of accuracy
is obtained. However, the interval bisection method is quite slow to converge
to the root, and is cumbersome when performed manually.
ACTIVITY 6.1
Investigate how many steps of this method you need to achieve an accuracy
of 1, 2, 3 and n decimal places, having started with an interval of length 1.
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Error (or solution) bounds
6
Change-of-sign methods have the great advantage that they automatically
provide bounds (the two ends of the interval) within which a root lies, so
the maximum possible error in a result is known. Knowing that a root lies
in the interval [0.61, 0.62] means that you can take the root as 0.615 with a
maximum error of ± 0.005.
Problems with change-of-sign methods
6.1 Interval estimation − change-of-sign methods
There are a number of situations which can cause problems for change-ofsign methods if they are applied blindly, for example by entering the equation
into a computer program without prior thought. In all cases you can avoid
problems by first drawing a sketch graph, provided that you know what
dangers to look out for.
The curve touches the x -axis
In this case there is no change of sign, so change-of-sign methods are
doomed to failure (see Figure 6.3).
f(x)
x
O
▲ Figure 6.3
There are several roots close together
Where there are several roots close together, it is easy to miss a pair of them.
The equation
f(x) = x3 − 1.9x2 + 1.11x − 0.189 = 0
has roots at 0.3, 0.7 and 0.9. A sketch of the
curve of f(x) is shown in Figure 6.4.
In this case f(0) " 0 and f(1) # 0, so you
know there is a root between 0 and 1.
A decimal search would show that f(0.3) = 0,
so that 0.3 is a root.You would be unlikely
to search further in this interval.
f(x)
O
0.3
0.7
0.9
x
▲ Figure 6.4
Interval bisection gives f(0.5) # 0, so
you would search the interval [0, 0.5] and eventually arrive at the root 0.3,
unaware of the existence of those at 0.7 and 0.9.
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There is a discontinuity in f (x )
6
1
The curve y = x − 2.7 has a discontinuity at x = 2.7, as shown by the
asymptote in Figure 6.5.
6 NUMERICAL SOLUTION OF EQUATIONS
y
O
2.7
x
▲ Figure 6.5
1
The equation x − 2.7 = 0 has no root, but all change-of-sign methods will
converge on a false root at x = 2.7.
None of these problems will arise if you start by drawing a sketch graph.
Technology note
It is important that you understand how each method works and are
able, if necessary, to perform the calculations using only a scientific
calculator. However, these repeated operations lend themselves to the use
of a spreadsheet or a programmable calculator. Many packages, such as
Autograph, will both perform the methods and illustrate them graphically.
Exercise 6A
1
2
3
4
Show that the equation x3 + 3x − 5 = 0 has no turning (stationary)
points.
(ii) Show with the aid of a sketch that the equation can have only one
root, and that this root must be positive.
(iii) Find the root, correct to 3 decimal places.
(i) How many roots has the equation ex − 3x = 0?
(ii) Find an interval of unit length containing each of the roots.
(iii) Find each root correct to 2 decimal places.
(i)
Sketch y = 2x and y = x + 2 on the same axes.
(ii) Use your sketch to deduce the number of roots of the equation
2x = x + 2.
(iii) Find each root, correct to 3 decimal places if appropriate.
Find all the roots of x3 − 3x + 1 = 0, giving your answers correct to
2 decimal places.
(i)
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CP
5
(i)
(ii)
PS
6
(i)
Find the roots of x5 − 5x + 3 = 0 in the intervals [−2, −1] and
[1, 2], correct to 2 decimal places, using
(a) decimal search
(b) interval bisection.
Comment on the ease and efficiency with which the roots are
approached by each method.
Use a systematic search for a change of sign, starting with x = −2, to
locate intervals of unit length containing each of the three roots of
6
x3 − 4x2 − 3x + 8 = 0.
7
PS
8
6.2 Fixed-point iteration
Sketch the graph of f(x) = x3 − 4x2 − 3x + 8.
(iii) Use the method of interval bisection to obtain each of the roots
correct to 2 decimal places.
(iv) Use your last intervals in part (iii) to give each of the roots in the
form a ± (0.5)n where a and n are to be determined.
The diagram shows a sketch
of the graph of f(x) = ex − x3
f(x)
f(x) = ex – x3
without scales.
(i) Use a systematic search for
a change of sign to locate
x
O
intervals of unit length
containing each of the
roots.
(ii) Use a change-of-sign
method to find each of the
roots correct to
3 decimal places.
For each of the equations below
(a) sketch the curve
(b) write down any roots
(c) investigate what happens when you use a change-of-sign
method with a starting interval of [−0.3, 0.7].
x2
1
x
(iii)
y
=
(i) y =
(ii)
y
=
2
2
x
x +1
x +1
(ii)
6.2 Fixed-point iteration
In fixed-point iteration you find a single value or point as your estimate for
the value of x, rather than establishing an interval within which it must lie.This
involves an iterative process, a method of generating a sequence of numbers
by continued repetition of the same procedure. If the numbers obtained in this
manner approach some limiting value, then they are said to converge to this value.
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6 NUMERICAL SOLUTION OF EQUATIONS
6
INVESTIGATION
Iterations
Notice what happens in each of the following cases, and try to find some
explanation for it.
(i) Set your calculator to the radian mode, enter zero if not automatically
displayed and press the cosine key repeatedly.
(ii) Enter any positive number into your calculator and press the square root
key repeatedly. Try this for both large and small numbers.
(iii) Enter any positive number into your calculator and press the sequence
+ 1 = √
= repeatedly. Write down the number which appears each
time you press = . The sequence generated appears to converge. You
may recognise the number to which it appears to converge: it is called
the Golden Ratio.
Rearranging the equation f(x ) = 0 into the form
x = F(x )
The first step, with an equation f(x) = 0, is to rearrange it into the form
x = F(x). Any value of x for which x = F(x) is a root of the original equation,
as shown in Figure 6.6.
When f(x) = x2 − x − 2, f(x) = 0 is the same as x = x2 − 2.
y
–1
y = f(x) = x2 – x – 2
O
x
2
y=x
y
y = F(x) = x2 – 2
–1
O
2
x
▲ Figure 6.6
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The equation x5 − 5x + 3 = 0 which you met earlier can be rewritten in a
number of ways. One of these is 5x = x5 + 3, giving
5
x = F(x) = x + 3 .
5
Figure 6.7 shows the graphs of y = x and y = F(x) in this case.
y=
y
–1
O
1
2
y=x
6.2 Fixed-point iteration
–2
x5 + 3
5
6
x
▲ Figure 6.7
This provides the basis for the iterative formula
5
xn + 1 = xn + 3 .
5
Taking x = 1 as a starting point to find the root in the interval [0, 1],
successive approximations are:
x1 = 1, x2 = 0.8, x3 = 0.6655, x4 = 0.6261, x5 = 0.6192,
x6 = 0.6182, x7 = 0.6181, x8 = 0.6180, x9 = 0.6180.
In this case the iteration has converged quite rapidly to the root for which
you were looking.
?
Another way of arranging x5 − 5x + 3 = 0 is x = 5 5x − 3.
What other possible rearrangements can you find?
❯ How many are there altogether?
❯
The iteration process is easiest to understand if you consider the graph.
Rewriting the equation f(x) = 0 in the form x = F(x) means that instead of
looking for points where the graph of y = f(x) crosses the x-axis, you are now
finding the points of intersection of the curve y = F(x) and the line y = x.
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6 NUMERICAL SOLUTION OF EQUATIONS
6
What you do
What it looks like on the graph
Choose a value, x1, of x
Take a starting point on the x-axis
Find the corresponding value of F(x1)
Move vertically to the curve y = F(x)
Take this value F(x1) as the new value of
x, i.e. x2 = F(x1)
Move horizontally to the line y = x
Find the value of F(x2) and so on
Move vertically to the curve
y
y=x
1
y = F(x)
O
x4 x3
1
x1
x2
x
▲ Figure 6.8
The effect of several repeats of this procedure is shown in Figure 6.8. The
successive steps look like a staircase approaching the root: this type of diagram
is called a staircase diagram. In other examples, a cobweb diagram may
be produced, as shown in Figure 6.9.
y
y=x
y = F(x)
O
x1
x3 x5
x4 x2
x
▲ Figure 6.9
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Successive approximations to the root are found by using the formula
6
xn +1 = F(xn).
This is an example of an iterative formula. If the resulting values of
xn approach some limit, a, then a = F(a), and so a is a fixed point of the
iteration. It is also a root of the original equation f(x) = 0.
Note
6.2 Fixed-point iteration
In the staircase diagram, the values of xn approach the root from one side,
but in a cobweb diagram they oscillate about the root. From Figures 6.8 and
6.9 it is clear that the error (the difference between a and xn) is decreasing in
both diagrams.
Accuracy of the method of rearranging the equation
Iterative procedures give you a sequence of point estimates. A staircase
diagram, for example, might give the following.
1, 0.8, 0.6655, 0.6261, 0.6192
What can you say at this stage?
Looking at the pattern of convergence it seems as though the root lies
between 0.61 and 0.62, but you cannot be absolutely certain from the
available evidence.
To be certain you must look for a change of sign.
f(0.61) = +0.034…
❯
f(0.62) = −0.0083…
Explain why you are certain that your judgement is now correct.
?
CP
Note
Estimates from a cobweb diagram oscillate above and below the root and so
naturally provide you with bounds.
Using different arrangements of the equation
In the
calculations the
full calculator
values of xn
were used, but
only the first 4
decimal places
have been
written down.
So far only one possible arrangement of the equation x5 − 5x + 3 = 0 has
been used. What happens when you use a different arrangement, for example
x = 5 5x − 3 , which leads to the iterative formula
xn+1 = 5 5x n − 3?
The resulting sequence of approximations is:
x1 = 1,
x5 = 1.2679...,
x9 = 1.2755...,
x2 = 1.1486...,
x6 = 1.2727...,
x10 = 1.2756...,
x3 = 1.2236...,
x7 = 1.2745...,
x11 = 1.2756...,
x4 = 1.2554...,
x8 = 1.2752...,
x12 = 1.2756....
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6
The process has clearly converged, but in this case not to the root for which
you were looking: you have identified the root in the interval [1, 2]. If instead
you had taken x1 = 0 as your starting point and applied the second formula,
you would have obtained a sequence converging to the value −1.6180, the
root in the interval [−2, −1].
6 NUMERICAL SOLUTION OF EQUATIONS
❯
?
If a numerical method finds a root of an equation, but not the one
you were looking for, is it a failure of the method?
6.3 Problems with the fixed-point
iteration method
You have already seen that sometimes a particular rearrangement will result
in the iterations diverging, or converging to the wrong root.
What is happening when this happens?
The equation xe x + x 2 − 1 = 0 has a solution of x = 0.48 correct to 2 decimal
1 − xe x
.
places. The equation can be rearranged into the form x =
x
xn
However, using the iterative formula xn +1 = 1 − xn e with a starting point
xn
of x1 = 0.5 produces x 2 = 0.35, x 3 = 1.43, x 4 = −3.46, …, with repeated
iterations giving values further from the solution.
y
4
x
y = 1 – xe
x
3
2
y =x
1
0
-1
x1
0.5
x2
1
x3 1.5
x
▲ Figure 6.10
The cobweb diagram in Figure 6.10 shows the positions of x1, x2 and x3
for the above iterative formula.
1 − xe x
❯ What feature of the curve y =
causes the iteration to diverge?
x
?
A particular rearrangement of the equation f ( x ) = 0 into the form x = g ( x )
will give an iteration formula that converges to a root α of the equation if the
gradient of the curve y = g ( x ) is not too steep near x = α.
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ACTIVITY 6.2
x5 + 3
Try using the iterative formula xn+1 = n
to find the roots in the
5
intervals [−2, −1] and [1, 2].
In both cases use each end point of the interval as a starting point.
What happens?
x5 + 3
.
Explain what you find by referring to a sketch of the curve y =
5
1
(i)
(ii)
Show that the equation x3 − x − 2 = 0 has a root between 1 and 2.
The equation is rearranged into the form x = F(x), where
F(x) = 3 x + 2 .
Use the iterative formula suggested by this rearrangement to find
the value of the root to 3 decimal places.
CP
2
Show that the equation x 4 + x − e x = 0 can be rearranged into each of
the following forms:
(i) x = e x − x 4
(ii) x = ln ( x 4 + x )
4 x
(iii) x = e − x
ex − x
x2
Show that the equation e−x − x + 2 = 0 has a root in the interval
[2, 3].
The equation is rearranged into the form x = e−x + 2.
Use the iterative formula suggested by this rearrangement to find
the value of the root to 3 decimal places.
(iv) x =
CP
3
(i)
(ii)
CP
4
5
6
6.3 Problems with the fixed-point iteration method
Exercise 6B
6
Show that the equation ex + x − 6 = 0 has a root in the interval [1, 2].
(ii) Show that this equation may be written in the form x = ln(6 − x).
(iii) Use an iterative formula based on the equation x = ln(6 − x) to
calculate the root correct to 3 decimal places.
(i) Sketch the curves y = ex and y = x2 + 2 on the same graph.
(ii) Use your sketch to explain why the equation ex − x2 − 2 = 0 has
only one root.
(iii) Rearrange this equation in the form x = F(x).
(iv) Use an iterative formula based on the equation found in part (iii) to
calculate the root correct to 3 decimal places
(i) Show that x2 = ln(x + 1) for x = 0 and for one other value of x.
(ii) Use the method of fixed-point iteration to find the second value to
3 decimal places.
(i)
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6
7 (i)
(ii)
8 (i)
Sketch the graphs of y = x and y = cos x on the same axes, for
0 $ x $ π.
2
Find the solution of the equation x = cos x to 5 decimal places.
Given that
a
∫ (3e + 1)dx = 10 , show that the positive constant a
1x
2
0
satisfies the equation
a = 2 ln 16 − a .
6
⎛ 16 − a n ⎞
Use the iterative formula a n +1 = 2 ln ⎜
with a1 = 2 to find
⎝ 6 ⎟⎠
the value of a correct to 3 decimal places. Give the result of each
iteration to 5 decimal places.
6 NUMERICAL SOLUTION OF EQUATIONS
(
(ii)
)
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q5 June 2015
9 (i)
The sequence of values given by the iterative formula
2x n 4
+ 2,
xn +1 =
xn
3
with initial value x1 = 2, converges to α.
(ii) Use this iterative formula to determine α correct to 2 decimal
places, giving the result of each iteration to 4 decimal places.
(iii) State an equation that is satisfied by α and hence find the exact
value of α.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q2 November 2007
10 (i)
By sketching a suitable pair of graphs, show that the equation
cos x = 2 − 2x,
where x is in radians, has only one root for 0 $ x $ 21 π.
(ii) Verify by calculation that this root lies between 0.5 and 1.
(iii) Show that, if a sequence of values given by the iterative formula
xn +1 = 1 − 21 cos xn
converges, then it converges to the root of the equation in part (i).
(iv) Use this iterative formula, with initial value x1 = 0.6, to determine
this root correct to 2 decimal places. Give the result of each
iteration to 4 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q7 November 2008
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11 The diagram shows the curve y = x 2 cos x, for 0 $ x $ 21 π, and its
maximum point M.
y
6
M
1
π
2
x
Show by differentiation that the x coordinate of M satisfies the
equation
tan x = 2 .
x
(ii) Verify by calculation that this equation has a root (in radians)
between 1 and 1.2.
2
(iii) Use the iterative formula xn +1 = tan−1 x to determine this root
(i)
( )
n
correct to 2 decimal places. Give the result of each iteration to
4 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q7 November 2009
12 The diagram shows the curve y = x e2x and its minimum point M.
y
O
6.3 Problems with the fixed-point iteration method
O
x
M
Find the exact coordinates of M.
(ii) Show that the curve intersects the line y = 20 at the point whose
x coordinate is the root of the equation
x = 21 ln 20 .
x
(iii) Use the iterative formula
20
xn+1 = 21 ln x ,
(i)
( )
( )
n
with initial value x1 = 1.3, to calculate the root correct to 2 decimal
places, giving the result of each iteration to 4 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 2 Q7 June 2009
9781510421738.indb 159
159
02/02/18 1:13 PM
6
13 (i)
By sketching a suitable pair of graphs, show that the equation
ln x = 2 − x 2
has only one root.
(ii) Verify by calculation that this root lies between x = 1.3 and
x = 1.4.
(iii) Show that, if a sequence of values given by the iterative formula
6 NUMERICAL SOLUTION OF EQUATIONS
xn+1 = √(2 − ln xn)
converges, then it converges to the root of the equation in part (i).
(iv) Use the iterative formula xn +1 = √(2 − ln xn) to determine the root
correct to 2 decimal places. Give the result of each iteration to
4 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 22 Q6 June 2010
14 The equation x3 − 8x − 13 = 0 has one real root.
(i)
(ii)
Find the two consecutive integers between which this root lies.
Use the iterative formula
xn +1 = (8xn + 13)
1
3
to determine this root correct to 2 decimal places. Give the result
of each iteration to 4 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q2 November 2009
15 The equation x3 − 2x − 2 = 0 has one real root.
Show by calculation that this root lies between x = 1 and x = 2.
(ii) Prove that, if a sequence of values given by the iterative formula
2x 3 + 2
xn +1 = n2
3xn – 2
converges, then it converges to this root.
(iii) Use this iterative formula to calculate the root correct to 2 decimal
places. Give the result of each iteration to 4 decimal places.
(i)
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q4 June 2009
16
A curve has parametric equations
1
x=
, y = (t + 2 ) .
( 2t + 1)2
The point P on the curve has parameter p and it is given that the
gradient of the curve at P is −1.
1
160
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1
(i)
Show that p = ( p + 2)6 − 2 .
(ii)
Use an iterative process based on the equation in part (i) to find the
value of p correct to 3 decimal places. Use a starting value of 0.7,
show the result of each iteration to 5 decimal places.
Cambridge International AS & A Level Mathematics
9709 Paper 21 Q6 June 2012
02/02/18 1:13 PM
KEY POINTS
1
2
3
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■
■
■
■
■
find an interval in which the root of an equation lies, using
change-of-sign methods
know circumstances under which change-of-sign methods fail
rearrange an equation in the form f(x) = 0 into the form x = F( xn )
carry out a fixed-point iteration of an equation in the form
xn +1 = F( xn ), to find the root of an equation to a given degree of
accuracy
understand that a particular rearrangement of f(x) = 0 may produce an
iterative formula that fails to converge.
6
6.3 Problems with the fixed-point iteration method
4
When f(x) is a continuous function, if f(a) and f(b) have opposite signs,
there will be at least one root of f(x) = 0 in the interval [a, b].
When an interval [a, b] containing a root has been found, this interval
may be reduced systematically by decimal search or interval bisection.
Fixed-point iteration may be used to solve an equation f(x) = 0.
You can sometimes find a root by rearranging the equation f(x) = 0
into the form x = F(x) and using the iteration xn+1 = F(xn ).
Successive iterations of xn+1 = F(xn ) may fail to converge if the
gradient of F(x) is not very steep for values of x close to the root.
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P3
7 FURTHER ALGEBRA
7
Further algebra
At the age of
twenty-one
he wrote a
treatise upon
the Binomial
Theorem. ... On
the strength of
it, he won the
Mathematical
Chair at one
of our smaller
Universities.
Sherlock
Holmes on
Professor
Moriarty in
‘The Final
Problem’ by Sir
Arthur Conan
Doyle (1859–
1930)
?
❯
How would you find 101 correct to 3 decimal places, without
using a calculator?
Many people are able to develop a very high degree of skill in mental
arithmetic, particularly those whose work calls for quick reckoning.
There are also those who have quite exceptional innate skills. Shakuntala
Devi, pictured above, is famous for her mathematical speed. On one
occasion she found the 23rd root of a 201-digit number in her head,
beating a computer by 12 seconds. On another occasion she multiplied
7 686 369 774 870 by 2 465 099 745 779 in just 28 seconds.
While most mathematicians do not have Shakuntala Devi’s high level of
talent with numbers, they do acquire a sense of when something looks
right or wrong. This often involves finding approximate values of numbers,
such as 101, using methods that are based on series expansions, and these
are the subject of the first part of this chapter.
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INVESTIGATION
Square roots
Using your calculator, write down the values of 1.02, 1.04, 1.06, …,
giving your answers correct to 2 decimal places. What do you notice?
Use your results to complete the following, giving the value of the
constant k.
7
1
1.02 = (1 + 0.02) 2 ≈ 1 + 0.02k
1
1.04 = (1 + 0.04) 2 ≈ 1 + 0.04k
7.1 The general binomial expansion
In Pure Mathematics 1 Chapter 4 you met the binomial expansion in the form
⎛ n⎞
⎛ n⎞
⎛ n⎞
⎛ n⎞
(1 + x )n = 1 + ⎜ 1⎟ x + ⎜ 2⎟ x 2 + ⎜ 3⎟ x 3 + … + ⎜ r ⎟ x r + …
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
which holds when n is any positive integer (or zero), that is n ! $.
7.1 The general binomial expansion
What is the largest value of x such that 1 + x ≈ 1 + kx is true for the
same value of k?
This may also be written as
n(n − 1) 2 n (n − 1)(n − 2) 3
x +
x +…
2!
3!
n(n − 1)(n − 2)…(n − r + 1) r
+
x +…
r!
(1 + x )n = 1 + nx +
This is a short way of
writing ‘n is a natural
number’. A natural
number is any positive
integer or zero.
which, being the same expansion as above, also holds when n ! $.
The general binomial theorem states that this second form, that is
(1 + x )n = 1 + nx +
+
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x +…
2!
3!
n(n − 1(n − 2)…(n − r + 1) r
x +…
r!
is true when n is any real number, but there are two important differences
to note when n " $.
This is a short way
of writing ‘n is not a
» The series is infinite (or non-terminating).
» The expansion of (1 + x)n is valid only if |x| " 1.
natural number’.
Proving this result is beyond the scope of an A-level course but you can
assume that it is true.
Consider now the coefficients in the binomial expansion:
1,
n,
n(n − 1)
,
2!
n(n − 1(n − 2)
,
3!
n(n − 1)(n − 2)(n − 3)
,
4!
…
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7
When
n = 0, we get
1
0
0
0
0
…
n=1
1
1
0
0
0
…
ditto
n=2
1
2
1
0
0
…
ditto
n=3
1
3
3
1
0
…
ditto
n=4
1
4
6
4
1
…
ditto
(infinitely many zeros)
7 FURTHER ALGEBRA
so that, for example
(1 + x)2 = 1 + 2x + x2 + 0x3 + 0x4 + 0x5 + …
(1 + x)3 = 1 + 3x + 3x2 + x3 + 0x4 + 0x5 + …
(1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 + 0x5 + …
Of course, it is usual to discard all the zeros and write these binomial
coefficients in the familiar form of Pascal’s triangle:
1
1
1
1
1
2
1
3
1
3
4
6
1
4
1
and the expansions as
(1 + x)2 = 1 + 2x + x2
(1 + x)3 = 1 + 3x + 3x2 + x3
(1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4
However, for other values of n (where n " $) there are no zeros in the row
of binomial coefficients and so we obtain an infinite sequence of non-zero
terms. For example:
(−3)(−4) (−3)(−4)(−5) (−3)(−4)(−5)(−6)
n = −3 gives 1 –3
…
2!
3!
4!
−10
…
that is 1 −3
6
15
n = 21
1
1
2
that is 1
1
2
gives
( 21 )(− 21 ) ( 21) (− 21 )(− 23 ) ( 21 )(− 21 )(− 23 )(− 52 ) …
2!
− 81
3!
1
16
4!
5
− 128
…
so that (1 + x )−3 = 1 − 3x + 6x 2 − 10 x 3 + 15x 4 + …
and
1
1 x3 − 5 x4 + …
(1 + x ) 2 = 1 + 21 x − 81 x 2 + 16
128
But remember: these two expansions are valid only if | x | " l.
1
2
❯ Show that the expansion of (1 + x ) is not valid when x = 8.
?
CP
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These examples confirm that there will be an infinite sequence of non-zero
coefficients when n " $.
In the investigation at the beginning of this chapter you showed that
7
1 + x ≈ 1 + 21 x
is a good approximation for small values of x. Notice that these are the first
1
two terms of the binomial expansion for n = 2 . If you include the third term,
the approximation is
Take y = 1 + 21 x, y = 1 + 21 x − 81 x 2 and y = 1 + x .
They are shown in the graph in Figure 7.1 for values of x between −1 and 1.
y
1.5
y = 1 + 12 x
y = 1 + 12 x – 18 x2
1.0
7.1 The general binomial expansion
1 + x ≈ 1 + 21 x − 81 x 2.
0.5
y= 1+x
–1.0
–0.5
0
0.5
1.0
x
▲ Figure 7.1
ACTIVITY 7.1
For n = 21 the first three terms of the binomial expansion are 1 + 21 x − 81 x 2 .
Use your calculator to verify the approximate result
1 + x ≈ 1 + 21 x − 81 x 2
for ‘small’ values of x.
What values of x can be considered as ‘small’ if you want the result to be
correct to 2 decimal places?
Now take n = −3. Using the coefficients found earlier suggests the
approximate result
(1 + x)−3 ≈ 1 − 3x + 6x2.
Comment on values of x for which this approximation is correct to
2 decimal places.
When | x | " 1, the magnitudes of x2, x3, x4, x5, … form a decreasing
geometric sequence. In this case, the binomial expansion converges ( just as
a geometric progression converges for −1 " r " 1, where r is the common
ratio) and has a sum to infinity.
9781510421738.indb 165
165
02/02/18 1:13 PM
ACTIVITY 7.2
7
Compare the geometric progression 1 − x + x2 − x3 + … with the series
obtained by putting n = −1 in the binomial expansion.What do you notice?
7 FURTHER ALGEBRA
To summarise: when n is not a positive integer or zero, the binomial
expansion of (1 + x)n becomes an infinite series, and is only valid when some
restriction is placed on the values of x.
The binomial theorem states that for any value of n:
(1 + x )n = 1 + nx +
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x +…
2!
3!
where
» if n ! $, x may take any value;
» if n " $, | x | " 1.
Note
The full statement is the binomial theorem, and the right-hand side is
referred to as the binomial expansion.
Example 7.1
Expand (1 − x)−2 as a series of ascending powers of x up to and including the
term in x 3, stating the set of values of x for which the expansion is valid.
Solution
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x +…
2!
3!
Replacing n by −2, and x by (−x) gives
(1 + x )n = 1 + nx +
−2
(1 + (−x )) = 1 + (−2)(−x ) +
(−2)(−3)
(−2)(−3)(−4)
( −x ) 2 +
( −x ) 3 + …
2!
3!
when | – x | < 1
It is important to put brackets round the
term −x, since, for example, (−x)2 is not
the same as −x2.
which leads to
(1 − x)−2 ≈ 1 + 2x + 3x 2 + 4 x 3 when | x | < 1.
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Note
In Example 7.1 the coefficients of the powers of x form a recognisable
sequence, and it would be possible to write down a general term in the
expansion. The coefficient is always one more than the power, so the r th term
would be rxr −1. Using sigma notation, the infinite series could be written as
7
∞
∑ rx
r −1
r =1
1
Find a quadratic approximation for
and state for which values of t
1 + 2t
the expansion is valid.
Solution
1
1
−1
2
=
1 = (1 + 2t )
1 + 2t (1 + 2t ) 2
The binomial theorem states that
n(n – 1) 2 n(n − 1)(n − 2) 3
(a + x)n = 1 + nx +
x +
x +…
2!
3!
Replacing n by − 1 and x by 2t gives
Remember to put brackets
round the term 2t, since
(2t)2 is not the same as 2t2.
2
(1 + 2t )
−1
2
( )
= 1 + − 1 (2t ) +
2
1
⇒ (1 + 2t )− 2 ≈ 1 − t + 3 t 2
2
INVESTIGATION
7.1 The general binomial expansion
Example 7.2
(− 21 )(− 23 ) (2t ) + … when | 2t | < 1
2!
2
when | t | < 1
2
Finding coefficients
Example 7.1 showed how using the binomial expansion for (1 − x)−2 gave
a sequence of coefficients of powers of x which was easily recognisable, so
that the particular binomial expansion could be written using sigma notation.
Investigate whether a recognisable pattern is formed by the coefficients in
the expansions of (1 − x)n for any other negative integers n.
The equivalent binomial expansion of (a + x)n when n is not a positive
integer is rather unwieldy. It is easier to start by taking a outside the brackets:
n
(a + x)n = a n 1 + x
a
The first entry inside the bracket is now 1 and so the first few terms of the
expansion are
⎡
⎤
n(n − 1) x 2 n(n − 1)(n − 2) x 3
(a + x)n = a n ⎢1 + n x +
+
+ …⎥
a
a
a
2!
3!
⎣
⎦
x
for
< 1.
a
( )
()
9781510421738.indb 167
()
()
167
02/02/18 1:14 PM
Note
7
Since the bracket is raised to the power n, any quantity you take out must be
raised to the power n too, as in the following example.
7 FURTHER ALGEBRA
Example 7.3
Expand (2 + x)−3 as a series of ascending powers of x up to and including the
term in x 2, stating the values of x for which the expansion is valid.
Solution
(2 + x ) –3 =
1
(2 + x ) 3
=
1
( )
= (1 + x )
2
23 1 + x
2
3
Notice that this is the
–3
x
.
same as 2−3 1 +
2
( )
–3
1
8
Take the binomial expansion
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x +…
2!
3!
x
and replace n by −3 and x by to give
2
–3
2
1 1+ x
1 ⎡1 + ( −3) x + ( −3)( −4) x + …⎤
=
when x < 1
⎢
⎥⎦
8
8⎣
2
2
2!
2
2
(1 + x )n = 1 + nx +
( )
()
≈ 81 − 3 x + 3 x 2
16
16
()
when | x | < 2.
?
❯ The chapter began by asking how you would find 101 to 3 decimal
places without using a calculator. How would you find it now?
Example 7.4
(2 + x )
Find a quadratic approximation for
, stating the values of x for which
(1 – x 2 )
the expansion is valid.
Solution
(2 + x )
= (2 + x )(1 − x 2 )−1
(1 – x 2 )
Take the binomial expansion
(1 + x )n = 1 + nx +
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x +…
2!
3!
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Replace n by −1 and x by (−x 2) to give
(−1)(−2)(−x 2 ) 2
−1
+…
when | −x 2 | < 1
(1 + (−x 2 )) = 1 + (−1)(−x 2 ) +
2!
(1 − x2)−1 = 1 + x2 + …
when | x2 | " 1, i.e. when | x | " 1.
7
Multiply both sides by (2 + x) to obtain (2 + x)(1 − x 2)−1:
(2 + x)(1 − x 2)−1 ≈ (2 + x)(1 + x 2)
≈ 2 + x + 2x2
when | x | " 1.
Sometimes two or more binomial expansions may be used together. If these
impose different restrictions on the values of x, you need to decide which is
the strictest.
Example 7.5
Find a and b such that
7.1 The general binomial expansion
The term in x3 has been omitted
because the question asked for a
quadratic approximation.
1
≈ a + bx
(1 − 2x )(1 + 3x )
and state the values of x for which the expansions you use are valid.
Solution
1
−1
−1
(1 − 2x )(1 + 3x ) = (1 − 2x) (1 + 3x)
Using the binomial expansion:
and
⇒
(1 − 2x)−1 ≈ 1 + (−1)(−2x)
for | −2x | " 1
(1 + 3x)−1 ≈ 1 + (−1)(3x)
for | 3x | " 1
(1 − 2x)−1(1 + 3x)−1 ≈ (1 + 2x)(1 − 3x)
≈ 1−x
(ignoring higher powers of x)
giving a = 1 and b = −1.
For the result to be valid, both | 2x | " 1 and | 3x | " 1 need to be satisfied.
and
| 2x | " 1
⇒
− 21 " x " 21
| 3x | " 1
⇒
− 13 " x " 13
Both of these restrictions are satisfied if − 13 " x " 13. This is the stricter
restriction.
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Note
7
The binomial expansion may also be used when the first term is the variable.
For example:
( )
7 FURTHER ALGEBRA
(x + 2)−1 may be written as (2 + x)−1 = 2−1 1 + x
2
and
(2x − 1)−3 = [(−1)(1 − 2x)]−3
= (−1)−3(1 − 2x)−3
= −(1 − 2x)−3
−1
?
❯ What happens when you try to rearrange x − 1 so that the binomial
expansion can be used?
Exercise 7A
1
2
3
Expand each of the following as a series of ascending powers of x up to
and including the term in x3, stating the set of values of x for which the
expansion is valid.
(i) (1 + x)−3
(ii) (1 + 2x)−3
(iii) (1 − 2x)−3
Expand each of the following as a series of ascending powers of x up to
and including the term in x2, stating the set of values of x for which the
expansion is valid.
1
1
−1
−1
(i) (1 + x) 2
(ii) (1 + x) 2
(iii) (1 + x) 4
(iv) (1 + x) 4
Expand each of the following as a series of ascending powers of x up to
and including the term in x3, stating the set of values of x for which the
expansion is valid.
−2
2x −2
2x −2
1−
1+
1+ x
(i)
(ii)
(iii)
3
3
3
For each of the expressions below
(a) write down the first three non-zero terms in their expansion
as a series of ascending powers of x
(b) state the values of x for which the expansion is valid
(c)
substitute x = 0.1 in both the function and its expansion and
calculate the percentage error, where
absolute error
percentage error =
× 100%.
true value
(
4
)
(
)
(
)
(1 + x)−2
(ii) (1 + 2x)−1
(iii)
1 − x2
(i) Write down the expansion of (1 − x)3.
(ii) Find the first three terms in the expansion of (1 + x)−4 in ascending
powers of x. For what values of x is this expansion valid?
(1 − x ) 3
(iii) When the expansion is valid,
can be written as
(1 + x )4
1 + ax + bx2 + higher powers of x.
(i)
5
Find the values of a and b.
170
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CP
(
)
1
Show that
(
)
3 x + cx 2.
8 The expansion of (a + bx)−3 may be approximated by 81 + 16
Find the values of the constants a, b and c.
(ii) For what range of values of x is the expansion valid?
1
9 (i) Find a quadratic function that approximates to
for
3
(1
−
3x ) 2
values of x close to zero.
(i)
(ii)
For what values of x is the approximation valid?
10 Expand (2 + 3x)−2 in ascending powers of x, up to and including the
term in x 2, simplifying the coefficients.
7
7.1 The general binomial expansion
1 = 1 1 + x − 2.
4
4+ x 2
(ii) Write down the first three terms in the binomial expansion of
− 21
in ascending powers of x, stating the range of values of x
1+ x
4
for which this expansion is valid.
2(1 − x )
(iii) Find the first three terms in the expansion of
in ascending
4+x
powers of x, for small values of x.
(3 − x )(1 + x )
7 Find a quadratic approximation for
and state the values of
(4 − x )
x for which this is a valid approximation.
6 (i)
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q1 June 2007
11 Expand (1 + x) (1 − 2x ) in ascending powers of x, up to and including
the term in x2, simplifying the coefficients.
2
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q2 November 2008
12 When (1 + 2x)(1 + ax) 3 , where a is a constant, is expanded in ascending
powers of x, the coefficient of the term in x is zero.
(i) Find the value of a.
(ii) When a has this value, find the term in x3 in the expansion of
2
(1 + 2x)(1 + ax) 3 , simplifying the coefficients.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q5 June 2009
13 Expand 3 (1 − 6x ) in ascending powers of x up to and including the
term in x3, simplifying the coefficients.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q1 June 2011
171
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b
7 FURTHER ALGEBRA
7
7.2 Review of algebraic fractions
f (x )
is an algebraic fraction
g(x )
or rational function. It may also be called a rational expression.
There are many occasions in mathematics when a problem reduces to the
manipulation of algebraic fractions, and the rules for this are exactly the same
as those for numerical fractions.
If f(x) and g(x) are polynomials, the expression
Simplifying fractions
To simplify a fraction, you look for a factor common to both the numerator
(top line) and the denominator (bottom line) and cancel by it.
For example, in arithmetic
15 = 5 × 3 = 3
20 5 × 4 4
and in algebra
6a = 2 × 3 × a = 2
9a 2 3 × 3 × a × a 3a
Notice how you must factorise both the numerator and denominator before
cancelling, since it is only possible to cancel by a common factor. In some cases
this involves putting brackets in.
2a + 4 = 2(a + 2) = 2
a 2 − 4 (a + 2)(a – 2) (a − 2)
Multiplying and dividing fractions
Multiplying fractions involves cancelling any factors common to the
numerator and denominator. For example:
10a × 9ab = 2 × 5 × a × 3 × 3 × a × b = 6a 2
5×5
5b
3b 2 25 3 × b × b
As with simplifying, it is often necessary to factorise any algebraic expressions
first.
a 2 + 3a + 2 × 12 = (a + 1)(a + 2) × 3 × 4
a +1
(a + 1)
9
3×3
(a + 2) 4
=
×
1
3
4(a + 2)
=
3
Remember that when one fraction is divided by another, you change ÷to ×
and invert the fraction which follows the ÷symbol. For example:
( x + 1)
12 ÷ 4 =
12
×
4
x 2 − 1 x + 1 ( x + 1)( x − 1)
=
3
( x − 1)
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Adding and subtracting fractions
To add or subtract two fractions they must be replaced by equivalent
fractions, both of which have the same denominator.
7
For example:
2
8
3
11
+ 41 = 12
+ 12
= 12
3
2x + x = 8x + 3x = 11x
3 4 12 12 12
and
2 + 1 = 8 + 3 = 11
3x 4 x 12x 12x 12x
Notice how you only need
12x here, not 12x2.
You must take particular care when the subtraction of fractions introduces a
sign change. For example:
4 x − 3 − 2x + 1 = 2(4 x − 3) − 3(2x + 1)
6
4
12
= 8x − 6 − 6x − 3
12
= 2x − 9
12
7.2 Review of algebraic fractions
Similarly, in algebra:
Notice how in addition and subtraction, the new denominator is the lowest
common multiple of the original denominators. When two denominators
have no common factor, their product gives the new denominator. For
example:
2 + 3 = 2( y − 2) + 3( y + 3)
y+3 y−2
( y + 3)( y − 2)
=
2 y − 4 + 3y + 9
( y + 3)( y − 2)
=
5y + 5
( y + 3)( y − 2)
=
5( y + 1)
( y + 3)( y − 2)
It may be necessary to factorise denominators in order to identify common
factors, as shown here.
2b − 3 =
2b
− 3
a 2 − b 2 a + b (a + b )(a − b ) (a + b )
2b − 3(a − b )
(a + b )(a − b )
= 5b − 3a
(a + b )(a − b )
=
(a + b) is a
common factor.
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7 FURTHER ALGEBRA
7
Exercise 7B
Simplify the expressions in questions 1 to 10.
1
6a × a
b 9b 2
2
5xy
÷ 15xy 2
3
3
x2 − 9
x − 9x + 18
4
5x − 1 × x 2 + 6 x + 9
x + 3 5x 2 + 4 x − 1
5
4 x 2 − 25
4 x 2 + 20x + 25
6
a 2 + a − 12 × 3
5
4a − 12
7
4 x 2 − 9 ÷ 2x − 3
x 2 + 2x + 1 x 2 + x
8
2p + 4
÷ ( p 2 − 4)
5
9
a2 − b2
2a + ab − b 2
10
x 2 + 8x + 16 × x 2 + 2x − 3
x 2 + 6x + 9
x 2 + 4x
2
2
In questions 11 to 24 write each of the expressions as a single fraction in its
simplest form.
x − ( x + 1)
11 1 + 1
12
3
4
4 x 5x
13
a + 1
a +1 a −1
14
2 + 3
x−3 x−2
15
x − 1
x −4 x+2
16
p2
p2
− 2
p −1 p +1
17
2 − a
a + 1 a2 + 1
18
2y
− 4
( y + 2) 2 y + 4
19
x+
1
x +1
20
2
− 3
b 2 + 2b + 1 b + 1
21
2
3
+
3( x − 1) 2( x + 1)
22
6
+ 2x
5( x + 2) ( x + 2) 2
23
2 − a−2
a + 2 2a 2 + a − 6
24
1 +1+ 1
x−2 x x+2
2
2
7.3 Partial fractions
Sometimes, it is easier to deal with two or three simple separate fractions than
it is to handle one more complicated one.
For example:
1
(1 + 2x )(1 + x )
may be written as
2
– 1 .
(1 + 2x ) (1 + x )
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partial fractions.
When finding partial fractions you must always assume the most general
numerator possible, and the method for doing this is illustrated in the
following examples.
7
7.3 Partial fractions
2
1
– 1 you can then do
is written as
(1 + 2x ) (1 + x )
(1 + 2x )(1 + x )
binomial expansions on the two fractions, and so find an expansion for
the original fraction.
» When integrating, it is easier to work with a number of simple fractions
than a combined one. For example, the only analytic method for
1
2
involves first writing it as
integrating
− 1 .
(1 + 2x )(1 + x )
(1 + 2x ) (1 + x )
You will meet this application in Chapter 8.
1
This process of taking an expression such as
and writing it in
(1 + 2x )(1 + x )
2
1
the form (1 + 2x ) − (1 + x ) is called expressing the algebraic fraction in
» When
Type 1: Denominators of the form
(ax + b)(cx + d)(ex + f )
Example 7.6
Express
4+x
as a sum of partial fractions.
(1 + x )(2 – x )
Solution
Assume
4+x
≡ A + B
(1 + x )(2 – x ) 1 + x 2 – x
Remember: a linear
denominator ⇒ a
constant numerator if
the fraction is to be a
proper fraction.
Multiplying both sides by (1 + x)(2 − x) gives
4 + x ≡ A(2 − x) + B(1 + x).
1
!
This is an identity; it is true for all values of x.
There are two possible ways in which you can find the constants A and B.
You can either
1 (two values are needed to give two
» substitute any two values of x in !
equations to solve for the two unknowns A and B); or
» equate the constant terms to give one equation (this is the same as putting
x = 0) and the coefficients of x to give another.
Sometimes one method is easier than the other, and in practice you will
often want to use a combination of the two.
Method 1: Substitution
Although you can substitute any two values of x, the easiest to use are x = 2
and x = −1, since each makes the value of one bracket zero in the identity.
➜
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4 + x ≡ A(2 − x) + B(1 + x)
7
x=2
⇒
4 + 2 = A(2 − 2) + B(1 + 2)
6 = 3B
x = −1 ⇒
⇒
B=2
4 − 1 = A(2 + 1) + B(1 − 1)
3 = 3A
⇒
A=1
Substituting these values for A and B gives
7 FURTHER ALGEBRA
4+x
1
2
≡
+
(1 + x )(2 − x ) 1 + x 2 − x
Method 2: Equating coefficients
In this method, you write the right-hand side of
4 + x ≡ A(2 − x) + B(1 + x)
as a polynomial in x, and then compare the coefficients of the various terms.
4 + x ≡ 2A − Ax + B + Bx
4 + 1x ≡ (2A + B) + (−A + B)x
Equating the constant terms:
4 = 2A + B
Equating the coefficients of x:
1 = −A + B
These are simultaneous
equations in A and B.
Solving these simultaneous equations gives A = 1 and B = 2 as before.
?
❯ In each of these methods the identity (≡) was later replaced by
equality (=). Why was this done?
In some cases it is necessary to factorise the denominator before finding the
partial fractions.
Example 7.7
Express
x(5x + 7)
as a sum of partial fractions.
(2x + 1)( x 2 − 1)
Solution
x(5x + 7)
x(5x + 7)
≡
(2x + 1)( x 2 − 1) (2x + 1)( x + 1)( x − 1)
Start by factorising
the denominator fully,
replacing (x 2 − 1)
with (x + 1)(x − 1).
There are three factors in the denominator, so write
x(5x + 7)
≡ A + B + C
(2x + 1)( x + 1)( x − 1) 2x + 1 x + 1 x − 1
Multiplying both sides by (2x + 1)( x + 1)( x − 1) gives
x(5x + 7) ≡ A( x + 1)( x − 1) + B(2x + 1)( x − 1) + C (2x + 1)( x + 1)
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Substituting x = 1 gives: 12 = 6C
⇒
Notice how a
combination of the
two methods is used.
C=2
Substituting x = −1 gives: −2 = 2B
⇒
7
B = −1
Equating coefficients of x2 gives: 5 = A + 2B + 2C
As B = −1 and C = 2:
5=A−2+4
A=3
x(5x + 7)
≡ 3 − 1 + 2
Hence
(2x + 1)( x + 1)( x − 1) 2x + 1 x + 1 x − 1
In the next example the orders of the numerator (top line) and the
denominator (bottom line) are the same.
Example 7.8
7.3 Partial fractions
⇒
Express 6 − x 2 as a sum of partial fractions.
4−x
2
Solution
Start by dividing the numerator by the denominator.
In this case the quotient is 1 and the remainder is 2.
So
Now find
6 − x2
2
= 1+
4 − x2
4 − x2
You can also use
this method when
the order of the
numerator is
greater than that of
the denominator.
2 .
4 − x2
2
2
A
B
≡
≡
+
4 − x 2 (2 + x )(2 − x ) 2 + x 2 − x
Multiplying both sides by (2 + x)(2 − x) gives
2 ≡ A(2 − x) + B(2 + x)
2 ≡ (2A + 2B) + x(B − A)
2 = 2A + 2B
Equating constant terms:
A+B=1
so
1
!
Equating coefficients of x:
0 = B − A, so B = A
1 gives
Substituting in !
A=B=2
1
Using these values
1
1
1
1
2
+
≡ 2 + 2 ≡
(2 + x )(2 − x ) 2 + x 2 − x 2(2 + x ) 2(2 − x )
So
6 − x2 ≡ 1+
1
1
+
2(2 + x ) 2(2 − x )
4 − x2
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7 FURTHER ALGEBRA
7
Exercise 7C
Write the expressions in questions 1 to 15 as a sum of partial fractions.
1
5
( x − 2)( x + 3)
2
1
x( x + 1)
3
6
( x − 1)( x − 4)
4
x+5
( x − 1)( x + 2)
5
3x
(2x − 1)( x + 1)
6
4
x 2 − 2x
7
2
( x − 1)(3x − 1)
8
x −1
x 2 − 3x − 4
9
x+2
2x 2 − x
10
7
2x 2 + x − 6
11
2x − 1
2x 2 + 3x − 20
12
2x + 5
18x 2 − 8
6x 2 + 22x + 18
13 ( x + 1)( x + 2)( x + 3)
4 x 2 − 25x − 3
14 (2 x + 1)( x − 1)( x − 3) 15
5x 2 + 13x + 10
(2x + 3)( x 2 − 4)
Type 2: Denominators of the form (ax + b)(cx 2 + d)
Example 7.9
Express
2x + 3
as a sum of partial fractions.
( x − 1)( x 2 + 4)
Solution
You need to assume a numerator of order 1 for the partial fraction with a
denominator of x 2 + 4, which is of order 2.
Bx + C is the most
2x + 3
≡ A + Bx + C
general numerator of
( x − 1)( x 2 + 4) x − 1 x 2 + 4
order 1.
Multiplying both sides by (x − 1)(x 2 + 4) gives
2x + 3 ≡ A(x2 + 4) + (Bx + C )(x − 1)
x=1
⇒
5 = 5A
⇒
1
!
A=1
The other two unknowns, B and C, are most easily found by equating
1 may be rewritten as
coefficients. Identity !
2x + 3 ≡ (A + B)x 2 + (−B + C )x + (4A − C)
Equating coefficients of x2:
0=A+B
⇒
B = −1
Equating constant terms:
3 = 4A − C
⇒
C=1
This gives
2x + 3
≡ 1 + 1− x
( x − 1)( x 2 + 4) x − 1 x 2 + 4
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Type 3: Denominators of the form (ax + b)(cx + d )2
The factor (cx + d)2 is of order 2, so it would have an order 1 numerator
in the partial fractions. However, in the case of a repeated factor there is a
simpler form.
4x + 5
Consider
(2x + 1) 2
7
2(2x + 1) + 3
(2x + 1) 2
This can be written as
2(2x + 1)
3
+
(2x + 1) 2 (2x + 1) 2
≡
2
3
+
(2x + 1) (2x + 1) 2
7.3 Partial fractions
≡
Note
In this form, both the numerators are constant.
px + q
In a similar way, any fraction of the form
can be written as
(cx + d ) 2
A +
B
(cx + d ) (cx + d ) 2
When expressing an algebraic fraction in partial fractions, you are aiming to
find the simplest partial fractions possible, so you would want the form where
the numerators are constant.
Example 7.10
Express
x +1
as a sum of partial fractions.
( x − 1)( x − 2) 2
Solution
Let
x +1
≡ A + B + C
( x − 1)( x − 2) 2 ( x − 1) ( x − 2) ( x − 2) 2
Notice that you only
need (x − 2)2 here
and not (x − 2)3.
Multiplying both sides by (x − 1)(x − 2)2 gives
x + 1 ≡ A(x − 2)2 + B(x − 1)(x − 2) + C(x − 1)
x = 1 (so that x − 1 = 0)
⇒
2 = A(−1)2
x = 2 (so that x − 2 = 0)
⇒
3=C
⇒
0=A+B
Equating coefficients of x2:
⇒
A=2
⇒
B = −2
This gives
x +1
3
≡ 2 − 2 +
( x − 1)( x − 2) 2 x − 1 x − 2 ( x − 2) 2
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7
Example 7.11
2
Express 52x − 3 as a sum of partial fractions.
x ( x + 1)
Solution
Let
5x 2 − 3 ≡ A + B + C
x 2 ( x + 1) x x 2 x + 1
Multiplying both sides by x 2(x + 1) gives
7 FURTHER ALGEBRA
5x2 − 3 ≡ Ax(x + 1) + B(x + 1) + Cx2
x=0
x = −1
⇒
⇒
−3 = B
+2 = C
Equating coefficients of x2: +5 = A + C
5x 2 − 3 3 3
2
This gives x 2 ( x + 1) ≡ x − x 2 + x + 1
⇒
A=3
7.4 Using partial fractions with the
binomial expansion
One of the most common reasons for writing an expression in partial
fractions is to enable binomial expansions to be applied, as in the following
example.
Example 7.12
2x + 7
in partial fractions and hence find the first three terms
( x − 1)( x + 2)
of its binomial expansion, stating the values of x for which this is valid.
Express
Solution
2x + 7
≡ A + B
( x − 1)( x + 2) ( x − 1) ( x + 2)
Multiplying both sides by (x − 1)(x + 2) gives
2x + 7 ≡ A(x + 2) + B(x − 1)
x=1
⇒
9 = 3A
⇒
A=3
x = −2
⇒
3 = −3B
⇒
B = −1
3
1
2x + 7
This gives ( x − 1)( x + 2) ≡ ( x − 1) − ( x + 2)
In order to obtain the binomial expansion, each bracket must be of the form
(1 ± …), giving
2x + 7
1
≡ −3 −
( x − 1)( x + 2) (1 − x )
2 1+ x
2
( )
≡ −3(1 − x ) − (1 + x )
2
−1
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1
2
−1
1
!
02/02/18 1:14 PM
The two binomial expansions are
and
for | x | " 1
(1 − x)−1 = 1 + (−1)(−x) + (−1)(−2) (−x)2 + …
2!
≈ 1 + x + x2
−1
( −1)( −2) x 2
1+ x
= 1 + ( −1) x +
+#
for x " 1
2
2
2!
2
2
2
≈ 1− x + x
2 4
( )
()
7
()
(
2
2x + 7
≈ −3(1 + x + x 2 ) − 21 1 − x + x
( x − 1)( x + 2)
2 4
)
= − 72 − 11
x − 25
x2
4
8
The expansion is valid when | x | " 1 and P x P " 1. The stricter of these is
2
| x | " 1.
ACTIVITY 7.3
Find a binomial expansion for the function
1
f(x) = (1 + 2x )(1 − x )
and state the values of x for which it is valid
(i) by writing it as (1 + 2x)−1(1 − x)−1
(ii) by writing it as [1 + (x − 2x2)]−1 and treating (x − 2x2) as one term
(iii) by first expressing f(x) as a sum of partial fractions.
7.4 Using partial fractions with the binomial expansion
1 gives
Substituting these in !
Decide which method you find simplest for the following cases.
(a) When a linear approximation for f(x) is required.
(b) When a quadratic approximation for f(x) is required.
(c) When the coefficient of xn is required.
Exercise 7D
1
Express each of the following fractions as a sum of partial fractions.
5 − 2x
4
4 + 2x
(i)
(ii)
(iii)
2
x
−
(
1) 2 ( x + 2)
(1 − 3x )(1 − x ) 2
(2x − 1)( x + 1)
(iv)
2x + 1
( x − 2)( x 2 + 4)
(v)
2x 2 + x + 4
(2x 2 − 3)( x + 2)
(vi)
x2 − 1
x (2x + 1)
(vii)
x2 + 3
x(3x 2 − 1)
(viii)
2x 2 + x + 2
(2x 2 + 1)( x + 1)
(ix)
4x 2 − 3
x(2x − 1) 2
2
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7
2
Find the first three terms in ascending powers of x in the binomial
expansion of the following fractions.
4 + 2x
4
(i)
(ii) (2 x − 1)( x 2 + 1)
(1 − 3x )(1 − x ) 2
2x + 1
5 − 2x
(iv)
( x − 2)( x 2 + 4)
( x − 1) 2 ( x + 2)
6x − 8
(i) Write 2
in the form Ax2 + B + C .
( x + 1)( x + 1)
x +1 x +1
(ii) Hence find the first three terms in the binomial expansion of
6x − 8
, stating the values of x for which the expansion is valid.
( x 2 + 1)( x + 1)
(iii)
7 FURTHER ALGEBRA
3
4
(i)
5
Find a cubic approximation for
6
(i)
Express
(ii)
10
in
(2 − x )(1 + x 2 )
ascending powers of x, up to and including the term in x3,
simplifying the coefficients.
Write down the first three terms in the binomial expansion of
1
in ascending powers of x.
(1 + 2x )(1 + x )
(ii) For what values of x is this expansion valid?
2
, stating the range of
( x + 1)( x 2 + 1)
values of x for which the expansion is valid.
10
in partial fractions.
(2 − x )(1 + x 2 )
Hence, given that | x | " 1, obtain the expansion of
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q9 June 2006
7
(i)
(ii)
3x 2 + x
in partial fractions.
( x + 2)( x 2 + 1)
3x 2 + x
Hence obtain the expansion of
in ascending powers
( x + 2)( x 2 + 1)
of x, up to and including the term in x3.
Express
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q9 November 2005
8
(i)
(ii)
2 − x + 8x 2
in partial fractions.
(1 − x )(1 + 2x )(2 + x )
2 − x + 8x 2
Hence obtain the expansion of
in ascending
(1 − x )(1 + 2x )(2 + x )
powers of x, up to and including the term in x2.
Express
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q9 November 2007
9
182
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2
Let f( x ) = 2x − 7x2 − 1 .
( x − 2)( x + 3)
2x 2 − 7x − 1
(i) Express f( x ) =
in partial fractions.
( x − 2)( x 2 + 3)
(ii) Hence obtain the expansion of f(x) in ascending powers of x, up to
and including the term in x2.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q7 November 2013
02/02/18 1:14 PM
KEY POINTS
1
(
)
3
When multiplying algebraic fractions, you can only cancel when the
same factor occurs in both the numerator and the denominator.
4
When adding or subtracting algebraic fractions, you first need to find
a common denominator.
5
The easiest way to solve any equation involving fractions is usually to
multiply both sides by a quantity which will eliminate the fractions.
6
A proper algebraic fraction with a denominator which factorises can
be decomposed into a sum of proper partial fractions.
7
The following forms of partial fraction should be used.
px 2 + qx + r
A
B
C
≡
+
+
(ax + b )(cx + d )(ex + f ) ax + b cx + d ex + f
px 2 + qx + r
A
Bx + C
≡
+
(ax + b )(cx 2 + d ) ax + b cx 2 + d
7
7.4 Using partial fractions with the binomial expansion
2
The general binomial expansion for n ! " is
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x +….
(1 + x)n = 1 + nx +
2!
3!
In the special case when n ! $, the series expansion is finite and
valid for all x.
When n " $, the series expansion is non-terminating (infinite) and
valid only if | x | " 1.
n
When n " $, (a + x)n should be written as a n 1 + x before
a
obtaining the binomial expansion.
px 2 + qx + r
A
B
C
≡
+
+
(ax + b )(cx + d )2 ax + b cx + d (cx + d )2
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■
■
■
■
find and use the binomial expansion (1 + x)n, where n is a rational
number and | x | " 1
adapt the standard binomial expansion for cases in which the constant
term is not 1
know the condition for a binomial expansion to be valid
express algebraic fractions as partial fractions when the fraction has a
denominator of the form
■ (ax + b)(cx + d )(ex + f )
■ (ax + b)(cx2 + d )
■ (ax + b)(cx + d )2.
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P3
8 FURTHER CALCULUS
8
Further calculus
The mathematical
process has a
reality and virtue
in itself, and once
discovered it
constitutes a new
and independent
factor.
Winston
Churchill
(1876–1965)
?
❯ How would you estimate the volume of water in a sea wave?
In this chapter you will learn how to integrate more complicated functions.
For example, you might model a sea wave as a sine wave, so finding the
volume of water in the wave might involve integrating a sine function.
8.1 Differentiating tan−1x
You can use implicit differentiation to differentiate inverse trigonometrical
functions.
Example 8.1
Find
dy
in terms of x when y = tan−1x.
dx
Solution
y = tan−1x
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9781510421738.indb 184
⇒
tan y = tan (tan−1x)
⇒
tan y = x
A function and its
inverse cancel out.
1
!
02/02/18 1:14 PM
d (tan y ) = d ( x )
dx
dx
d (tan y ) dy = d ( x )
dx dx
dy
Use implicit
differentiation.
8
The derivative of tan y is sec2 y.
⇒
dy
=1
dx
dy
= 1
dx sec 2 y
dy
1
Using sec2 y ≡tan2 y + 1 gives dx = 1 + tan 2 y
1 into !
2 gives
Substituting !
dy
= 1
dx 1 + x 2
Make
dy
the subject.
dx
2
!
Remember: tan y = x
8.1 Differentiating tan−1x
⇒ sec 2 y
You can also differentiate inverse functions by using the fact that
dy
1
=
dx dx
dy
−1
Given y = tan x then x = tan y
dy
dx
1
1
= sec 2 y ⇒
=
dy
dx sec 2 y = 1 + x 2 , as before.
e
You can use the same method to differentiate y = sin−1kx and y = cos−1 kx to
obtain:
dy
k
y = sin−1kx ⇒
=
dx
1 −k 2 x 2
and y = cos−1kx ⇒
Exercise 8A
1
dy
−k
=
dx
1 −k 2 x 2
dy
in terms of x for each of the following.
dx
(i)
y = tan −1 2x
(ii) y = tan −1 3x
x
(iii) y = tan −1
(iv) y = tan −1 x
2
3
Find
()
9781510421738.indb 185
CP
2
CP
3
CP
4
CP
5
()
k2 .
k x2 + 1
Given f ( x ) = 1 tan −1 x where k is a constant, prove that f '( x ) = 2 1 2 .
x +k
k
k
⎛ ⎞
Prove that the equation of the tangent to the curve y = tan − 1 ⎜ x ⎟ at the
⎝ 3⎠
point where x = 3 is 12y = 3x + 4π − 3 3.
Given f ( x ) = k tan −1(kx ) where k is a constant, prove that f '( x ) =
()
2
Prove that the equation of the normal to the curve y = tan −1( 2x ) at the
2
point where x = 2 is y + 2x = 4 + π .
4
185
02/02/18 1:14 PM
e
6
8
8 FURTHER CALCULUS
PS
7
dy
in terms of x for each of the following.
dx
y = cos −1x
(i)
(ii) y = sin −1x
Find the equations of the tangent and the normal to the curve
y = cos −1 2x at the point 1 , π .
4 3
Find
( )
8.2 Integration by substitution
Figure 8.1 shows the graph of y =
y
x.
2
y= x
1
0
1
2
3
4
x
▲ Figure 8.1
?
❯ How does it allow you to find the shaded area in the graph in
Figure 8.2?
y
2
y= x+1
1
0
1
2
3
x − 1 is shown in
y
4
x
▲ Figure 8.2
The graph of y =
Figure 8.3.
y= x–1
The shaded area is given by
∫
5
1
x − 1d x =
5
∫ (x – 1) d x.
1
2
1
O
1
5
x
▲ Figure 8.3
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You may remember how to investigate this by inspection. However, you can
also transform the integral into a simpler one by using the substitution
u = x − 1 to get
b
∫ u d u.
1
2
a
8
When you make this substitution it means that you are now integrating with
respect to a new variable, namely u. The limits of the integral, and the ‘dx’,
must be written in terms of u.
x=1
The new limits are given by
⇒
u=1−1=0
The integral now becomes:
∫
4
⎡ 23 ⎤
u
u du = ⎢ 3 ⎥
u =0
⎢⎣ 2 ⎥⎦0
u =4
8.2 Integration by substitution
and
x=5 ⇒
u = 5 − 1 = 4.
d
u
Since u = x − 1,
= 1.
dx
Even though du is not a fraction, it is usual to treat it as one in this situation
dx
(see the warning at the bottom of the page), and to write the next step as
‘du = dx’.
1
2
4
⎡ 23 ⎤
2u
= ⎢⎣ 3 ⎥⎦
0
= 5 13
This method of integration is known as integration by substitution. It is
a very powerful method which allows you to integrate many more functions.
Since you are changing the variable from x to u, the method is also referred
to as integration by change of variable.
The last example included the statement ‘du = dx’. Some mathematicians
are reluctant to write such statements on the grounds that du and dx
may only be used in the form du , i.e. as a gradient. This is not in fact
dx
true; there is a well-defined branch of mathematics which justifies such
statements but it is well beyond the scope of this book. In the meantime
it may help you to think of it as shorthand for ‘in the limit as δx → 0,
δu → 1, and so δu = δx ’.
δx
Example 8.2
Evaluate
3
∫ (x + 1) dx by making a suitable substitution.
3
1
➜
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Solution
8
y = (x + 1)3
y
Let u = x + 1.
x=1 ⇒ u=1+1=2
x=3 ⇒ u=3+1=4
Converting the limits:
Converting dx to du:
8 FURTHER CALCULUS
du = 1 ⇒ du = dx.
dx
3
4
∫ (x + 1) dx = ∫ u du
3
1
3
2
4 4
= ⎡⎢u ⎤⎥
⎣ 4 ⎦2
4
4
=4 − 2
4
4
= 60
1
O
1
3
x
▲ Figure 8.4
?
❯ Can integration by substitution be described as the reverse of the
chain rule?
Example 8.3
Evaluate
4
1
2
∫3 2x (x 2 − 4) dx by making a suitable substitution.
Solution
Notice that 2x is the derivative of the expression in the brackets, x2 − 4, and
so u = x2 − 4 is a natural substitution to try.
du
This gives
= 2x ⇒ du = 2x dx
dx
Converting the limits: x = 3 ⇒
u=9−4=5
x=4
⇒
u = 16 − 4 = 12
So the integral becomes:
4
1
2
12
1
2
∫3 (x2 − 4) 2x dx = ∫5 u du
12
⎡ 23 ⎤
= ⎢ 2u ⎥
⎣ 3 ⎦5
= 20.3 (to 3 significant figures)
Note
In the last example there were two expressions multiplied together; the
second expression is raised to a power. The two expressions are in this case
related, since the first expression, 2x, is the derivative of the expression in
brackets, x 2 − 4. It was this relationship that made the integration possible.
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Example 8.4
∫
Find x(x2 + 2)3 dx by making an appropriate substitution.
8
Solution
Since this is an indefinite integral there are no limits to change, and the final
answer will be a function of x.
∫
You only have x dx
in the integral, not
2x dx.
∫
= ∫ u3 × 21 du
u4
+c
8
( x 2 + 2)4
+c
=
8
=
8.2 Integration by substitution
Let u = x2 + 2, then:
du
= 2x ⇒ 1 du = x dx
2
dx
So x(x2 + 2)3 dx = (x2 + 2)3x dx
Always remember, when finding an indefinite integral by substitution, to
substitute back at the end. The original integral was in terms of x, so your
final answer must be too.
Example 8.5
∫
By making a suitable substitution, find x x − 2 dx.
Solution
This question is not of the same type as the previous ones since x is not the
derivative of (x − 2). However, by making the substitution u = x − 2 you can
still make the integral into one you can do.
Let u = x − 2, then:
du
= 1 ⇒ du = dx
dx
There is also an x in the integral so you need to write down an expression
for x in terms of u. Since u = x − 2 it follows that x = u + 2.
1
In the original integral you can now replace x − 2 by u 2 , dx by du and
x by u + 2.
∫ x x − 2 dx = ∫(u + 2)u du
= ∫(u + 2u )du
1
2
3
2
1
2
5
3
= 25 u 2 + 43 u 2 + c
3
2
Replacing u by x − 2 and tidying up gives 15
(3x + 4)(x − 2) 2 + c.
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ACTIVITY 8.1
8
Complete the algebraic steps involved in tidying up the answer to
Example 8.5.
8 FURTHER CALCULUS
Exercise 8B
1
Find the following indefinite integrals by making the suggested
substitution. Remember to give your final answer in terms of x.
(i)
3x2(x3 + 1)7 dx, u = x3 + 1
(ii) 2x(x2 + 1)5 dx, u = x2 + 1
∫
(iii) ∫ 3x2(x3 − 2)4 dx, u = x3 − 2
(v) ∫ x 2x + 1 dx, u = 2x + 1
2
3
∫1
5
(iii) ∫ x x − 1 dx
1
∫−1
Find the area of the shaded region for each of the following graphs.
+ 1)
O
5
y
y=
O
1
PS
(ii)
y = 6x(x2 + 1)3
y
–1
4
2
Evaluate each of the following definite integrals by using a suitable
substitution. Give your answer to 3 significant figures where appropriate.
5
2
(i)
x 2(x 3 + 1)2 dx
(ii)
2x (x − 3)5 dx
(i)
PS
∫
(iv) ∫ x 2x − 5 dx, u = 2x2 − 5
x dx, u = x + 9
(vi) ∫
x+9
1
2
x
(x – 1)3
4
x
x
dy
1
and passes through the
=
dx
2x + 3
point 0, 2 3 . Find the equation of the curve.
The diagram shows the curve y = x + 1 and the line y = 2.
Find the area of the shaded region.
A curve has a gradient function
(
)
y
2
y = x+ 1
1
O
6
190
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Evaluate
2
3
x
∫ x (x − 4) d x using the substitution u = x − 4.
2
3
3
1
02/02/18 1:14 PM
7
∫
∫
8
∫
8.3 Integrals involving exponentials
and natural logarithms
In Chapter 5 you met integrals involving logarithms and exponentials. That
work is extended here using integration by substitution.
Example 8.6
By making a suitable substitution, find
4
∫0 2x e x dx.
2
Solution
4
4
∫0 2x ex dx = ∫0 ex 2x dx
2
2
Since 2x is the derivative of x2, let u = x2.
du
= 2x ⇒ du = 2x dx
dx
The new limits are given by
x=0
⇒
and
x=4
⇒
The integral can now be written as
16
∫0
Example 8.7
Evaluate
5
eu du
[ ]
16
= eu 0
= e16 − e0
= 8.89 × 106
(to 3 significant figures)
2x
∫1 x + 3 dx.
2
Solution
In this case, substitute
u = x2 + 3, so that
du
dx = 2x ⇒ du = 2x dx
The new limits are given by
x=1 ⇒ u=4
and x = 5 ⇒ u = 28
9781510421738.indb 191
u=0
u = 16
8
8.3 Integrals involving exponentials and natural logarithms
Integrate with respect to x.
1
4 + 3
(a)
(b) 6x(1 + x2) 2
3
x x
4 (1 +
x )3
(ii) Show that the substitution x = u2 transforms 1
dx into
b
x
3
an integral of the form k(1 + u) du.
a
State the values of k, a and b.
Evaluate this integral.
ln x
Find the integral
(1 + ln x )dx using the substitution u = 1 + ln x .
x
(i)
y
y=
O
▲ Figure 8.5
1
2x
x2 + 3
5
x
➜
191
02/02/18 1:14 PM
2x
5
28 1
∫1 x + 3 dx = ∫4 u du
8
2
[ ]
28
= ln u 4
= ln 28 − ln 4
= 1.95
(to 3 significant figures)
∫ ff('(xx)) dx, where f(x) = x2 + 3. In such cases the
substitution u = f(x) transforms the integral into ∫ 1 du. The answer is then
u
8 FURTHER CALCULUS
The last example is of the form
ln u + c or ln(f(x)) + c (assuming that u = f(x) is positive). This result may be
stated as the working rule below.
If you obtain the top line when you differentiate the bottom line, the
integral is the natural logarithm of the bottom line. So,
∫ ff('(xx)) dx = ln | f(x) | + c.
Example 8.8
Evaluate
2
5x 4 + 2 x
∫1 x + x + 4 dx.
5
2
Solution
You can work this out by substituting u = x5 + x2 + 4 but, since
differentiating the bottom line gives the top line, you could apply the rule
above and just write:
5x 4 + 2 x
∫1 x + x + 4 dx = [ln(x 5 + x 2 + 4)]1
2
5
2
2
= ln 40 − ln 6
= 1.90
(to 2 significant figures)
In the next example some adjustment is needed to get the top line into the
required form.
Example 8.9
Evaluate
1
x5
∫0 x + 7 dx.
6
Solution
The differential of x6 + 7 is 6x5, so the integral is rewritten as 61
[
]
1
1
6x 5
∫0 x + 7 dx.
6
Integrating this gives 61 ln(x6 + 7) 0 or 0.022 (to 2 significant figures).
Exercise 8C
1
Find the following indefinite integrals.
2x + 3
2x dx
(i)
(ii)
2 + 9 x − 1 dx
2
3
x
x +1
∫
∫
(iii)
∫12x 2 ex dx
3
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2
Find the following definite integrals.
Where appropriate give your answers to 3 significant figures.
3
6
x−3
(i)
2x e−x 2 dx
(ii)
2 − 6 x + 9 dx
x
2
4
The sketch shows the graph
y
of y = x ex2.
(i) Find the area of region A.
(ii) Find the area of region B.
–1
(iii) Hence write down the total
O
A
area of the shaded region.
∫
3
B
2
x
x+2
is shown below.
x 2 + 4x + 3
The graph of y =
y
–5
–4
–3
CP
5
–2
–1
0
1
2
x
Find the area of each shaded region.
A curve has the equation y = (x + 3)e−x.
dy
(i) Find
.
dx
x + 2 dx .
(ii) Hence find
ex
(iii) Find the x and y coordinates of the stationary point S on the curve.
d 2y
(iv) Calculate
at the point S.
dx 2
What does its value indicate about the stationary point?
2 + ln u du into 2 + x dx.
(v) Show that the substitution u = ex converts
ex
u2
e 2 + ln u
(vi) Hence evaluate
du.
u2
1
ex
for values of x between
(i) Sketch the curve with equation y = x
e +1
0 and 2.
(ii) Find the area of the region enclosed by this curve, the axes and the
line x = 2.
e 2t
dt.
(iii) Find the value of
2
1 t +1
(iv) Compare your answers to parts (ii) and (iii). Explain this result.
8.3 Integrals involving exponentials and natural logarithms
4
8
∫
∫
PS
∫
∫
6
∫
∫
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7
8 FURTHER CALCULUS
8
x and its maximum point
x2 + 1
M. The shaded region R is bounded by the curve and by the lines y = 0
and x = p.
y
(i) Calculate the x coordinate
of M.
M
(ii) Find the area of R in terms
of p.
The diagram shows part of the curve y =
(iii) Hence calculate the value
R
O
of p for which the area of
R is 1, giving your answer
correct to 3 significant figures.
8
Let I =
(i)
(ii)
4
∫ x(4 −1 x ) dx.
p
x
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q9 June 2005
1
Use the substitution u = x to show that I =
Hence show that I = 21 ln 3.
2
2
∫ u(4 − u) du.
1
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q7 June 2007
8.4 Integrals involving trigonometrical
functions
In Chapter 5 you met integrals involving trigonometrical functions. That
work is extended here using integration by substitution.
Example 8.10
∫
Find 2x cos(x2 + 1) dx.
Solution
Make the substitution u = x 2 + 1. Then differentiate.
du = 2x
⇒ 2x dx = du
dx
2x cos(x 2 + 1) dx = cos u du
∫
∫
= sin u + c
= sin(x 2 + 1) + c
Notice that the last example involves two expressions multiplied together,
namely 2x and cos(x2 + 1). These two expressions are related by the fact that
2x is the derivative of x2 + 1. Because of this relationship, the substitution
u = x2 + 1 may be used to perform the integration.You can apply this method
to other integrals involving trigonometrical functions, as in the next example.
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Example 8.11
Find
π
2
∫ 0 cos x sin2 x dx.
Remember that sin2 x means
the same as (sin x)2.
Solution
8
This integral is the product of two expressions, cos x and (sin x)2.
Differentiating:
du
du = cos x dx.
dx = cos x ⇒
The limits of integration need to be changed as well:
Therefore
x=0
⇒
u=0
x = π2
⇒
u=1
π
2
1
∫ 0 cos x sin2 x dx = ∫0 u 2 du
3 1
= ⎡⎢u ⎤⎥
⎣ 3 ⎦0
= 13
Example 8.12
∫
8.4 Integrals involving trigonometrical functions
Now (sin x)2 is a function of sin x, and cos x is the derivative of sin x, so you
should use the substitution u = sin x.
Find cos3 x dx.
Solution
First write cos3 x = cos x cos2 x.
Now remember that
cos2 x + sin2 x = 1
⇒
cos2 x = 1 − sin2 x.
This gives
cos3 x = cos x (1 − sin2 x)
= cos x − cos x sin2 x
The first part of this expression, cos x, is easily integrated to give sin x.
The second part is more complicated, but you can see that it is of a type that
you have met already, as it is a product of two expressions, one of which is a
function of sin x and the other of which is the derivative of sin x. This can be
integrated either by making the substitution u = sin x or simply in your head
(by inspection).
∫ cos3 x dx = ∫(cos x − cos x sin2 x) dx
= sin x − 13 sin3 x + c
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8
Example 8.13
Find
(i)
∫
cot x dx
(ii)
π
3
∫ tan x dx
π
6
Solution
(i)
cos x
Rewrite cot x as sin x .
cos x
8 FURTHER CALCULUS
∫ cot x dx = ∫ sin x .dx
Now you can use the substitution u = sin x.
du
dx = cos x ⇒ du = cos x dx
cos x .
1
1
sin x dx = sin x × cos x dx = u du
= ln | u | + c
= ln | sin x | + c
cos x
You may have noticed that the integral sin x .dx is in the form
f '( x )
dx = ln | f(x) | + c,
f (x )
and so you could have written the answer down directly.
∫
∫
∫
∫
The ‘top line’ is
the derivative of
the ‘bottom line’.
∫
(ii)
∫
π
3
π
6
tan x dx =
π
3
sin x
∫ cos x dx
π
6
Adjusting the numerator to make it the derivative of the denominator
gives:
π
3
π
sin x
3 – sin x
dx
dx = −
x
cos
π cos x
π
∫
6
∫
[
6
= −ln | cos x |
]
π
3
π
6
= ⎡− ln 1 ⎤ − ⎡− ln 3 ⎤
2 ⎦ ⎢⎣
⎣
2 ⎥⎦
= −ln 1 + ln 3
2
2
= ln 3
1
= 2 ln 3
cos π = 1 and
3 2
cos π = 3
6
2
Use the laws of logs:
⎛
⎞
ln 3 – ln 21 = ln ⎜ 3 ÷ 21 ⎟
2
⎝ 2
⎠
Use the laws of logs:
1
ln 3 = ln3 2 = 21 ln3
Note
You may find that as you gain practice in this type of integration you become
able to work out the integral without writing down the substitution. However,
if you are unsure, it is best to write down the whole process.
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Question 3 of Exercise 8A on page 185 asked you to prove that
when
()
f ( x ) = 1 tan −1 x
k
k
where k is a constant,
1 .
x2 + k2
Using this result and the fact that integration is the reverse of differentiation
you can write
k dx = tan −1 x + c .
1 dx = 1 tan −1 x + c and
k
k
k
x2 + k2
x2 + k2
8
f '( x ) =
Example 8.14
()
∫
Find
(i)
1
∫ x + 4 dx
2
(ii)
1
∫ 5x + 3 dx .
2
Solution
∫ x + 4 dx = 2 tan ( 2 ) + c
1
1
1
(ii)
∫ 5x + 3 dx = 5 ∫ x + 53 dx
1
=1∫
dx
5
(i)
2
1
1
x
−1
2
Start by rewriting the integral
in the form
2
x2 +
( )
3
5
2
1
∫ x + k dx .
2
2
k 2 = 3 so k = 3
5
5
= 1 × 1 tan − 1 ⎛⎜ x ⎞⎟ + c
5
3
3
⎝ 5⎠
5
( x) + c
= 1 tan ( x ) + c
15
= 1×
5
5 tan − 1
Exercise 8D
1
5
3
3
−1
5
3
1
× 5= 5 =
3
5
5 3
1
= 1
5 3
15
Integrate the following by using the substitution given, or otherwise.
(i) cos 3x
u = 3x
(ii) sin(1 − x)
u=1−x
3
(iii) sin x cos x
u = cos x
sin x
u = 2 − cos x
(iv)
2 − cos x
(v) tan x
u = cos x
write tan x as sin x
cos x
(vi) sin 2x (l + cos 2x)2
u = 1 + cos 2x
Use a suitable substitution to integrate the following.
(
2
8.4 Integrals involving trigonometrical functions
()
∫
)
2x sin(x 2)
(ii) cos x e sin x
tan x
cos x
(iii) cos 2 x
(iv) sin 2 x
Evaluate the following definite integrals by using suitable substitutions.
π
π
4
π
2
(i)
cos 2x − 2 dx
(ii)
cos x sin3x dx
0
0
(i)
3
∫
9781510421738.indb 197
(
)
∫
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02/02/18 1:14 PM
8
∫0
(iii)
x sin(x 2) dx
π
4
(iv)
Find the following.
1 dx
(i)
x2 + 1
8 FURTHER CALCULUS
1
∫ 2x + 1 dx
(iv)
6
∫0
e tan x
cos 2 x dx
1
2
∫
5
π
4
∫ 0 cos x (1 + tan x) dx
(v)
4
π
2
1
(ii)
∫ x + 16 dx
(iii)
∫ x 2+ 9 dx
(v)
∫ 3x 1+ 4 dx
(vi)
∫ 2x + 5 dx
2
2
2
1
2
10 dx .
x 2 + 25
y
The diagram shows the curve
y = sin2 2x cos x for 0 $ x $ 21 π,
M
and its maximum point M.
(i) Find the x coordinate of M.
O
(ii) Using the substitution u = sin x,
find by integration the area of the
shaded region bounded by the curve and the x-axis.
Find
∫
5 3
0
1
2
x
π
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q9 June 2013
7
The diagram shows the curve y = e 2 sin x cos x for
0 & x & 21 π , and its maximum point M.
(i) Using the substitution u = sin x, find the
exact value of the area of the shaded region
bounded by the curve and the axes.
(ii) Find the x coordinate of M, giving your
answer correct to 3 decimal places.
y
M
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q9 June 2014
8
(i)
(ii)
Prove that cot θ + tan θ ≡ 2cosec2θ .
Hence show that
1π
3
O
∫ cosec 2θ dθ = ln 3 .
1π
2
x
1
2
1π
6
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q5 November 2013
9
(i)
Use the substitution x = sin2θ to show that
∫ (1 −x x ) dx = ∫ 2 sin θ dθ
2
(ii)
Hence find the exact value of
∫
198
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1
4
0
(1 −x x ) dx.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q6 November 2005
02/02/18 1:14 PM
∫
10 (a)
Find (4 + tan 2 2x )dx .
(b)
Find the exact value of
∫
1π
2
1π
4
8
sin( x + 61 π) .
dx
sin x
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q5 June 2015
11 The diagram shows part of the curve y = sin 3 2x cos 3 2x . The shaded
8.5 The use of partial fractions in integration
region shown is bounded by the curve and the x-axis and its exact area
is denoted by A.
y
x
O
(i)
(ii)
Use the substitution u = sin 2x in a suitable integral to find the
value of A.
kπ
sin 3 2x cos 3 2x d x = 40 A, find the value of the
Given that
0
constant k.
∫
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q7 November 2012
8.5 The use of partial fractions in
integration
❯ Why is it not possible to use any of the integration techniques you
2
have learnt so far to find x 2 − 1 dx?
?
∫
Partial fractions reminder
In Chapter 7 you met partial fractions. Here is a reminder of the work you
did there.
Since x2 − 1 can be factorised to give (x + 1)(x − 1), you can write the
expression to be integrated as partial fractions.
2 = A + B
x2 − 1 x − 1 x + 1
2 ≡ A(x + 1) + B(x − 1)
9781510421738.indb 199
This is true for all values of x. It is
an identity and to emphasise this
point we use the identity symbol ≡.
Let x = 1
2 = 2A
⇒
A=1
Let x = −1
2 = −2B
⇒
B = −1
199
02/02/18 1:14 PM
Substituting these values for A and B gives
2 = 1 − 1 .
x2 − 1 x − 1 x + 1
The integral then becomes
8
2
1
1
∫ x − 1 dx = ∫ x − 1 dx − ∫ x + 1 dx.
2
Now the two integrals on the right can be recognised as logarithms.
2
8 FURTHER CALCULUS
∫ x − 1 dx = ln ⎢x − 1 ⎢− ln ⎢x + 1 ⎢+ c
2
= ln x − 1 + c
x +1
Here you worked with the simplest type of partial fraction, in which there
are two different linear factors in the denominator. This type will always
result in two fractions, both of which can be integrated to give logarithmic
expressions. Now look at the other types of partial fraction.
⎪
⎪
A repeated factor in the denominator
Example 8.15
x+4
Find (2x − 1)( x + 1) 2 dx.
∫
Solution
First write the expression as partial fractions:
x+4
A + B + C
=
(2x − 1)( x + 1) 2 (2x − 1) ( x + 1) ( x + 1) 2
x + 4 ≡ A(x + 1)2 + B(2x − 1)(x + 1) + C(2x − 1).
where
Let x = −1
3 = −3C
⎛3⎞ 2
⎜ ⎟
⎝2⎠
Let x = 21
9
2
Let x = 0
4 =A−B−C
=A
⇒
C = −1
⇒
9
9
2 = 4A
⇒
B = A − C − 4 = 2 + 1 − 4 = −1
⇒
A=2
Substituting these values for A, B and C gives
x+4
2
1
=
− 1 −
(2x − 1)( x + 1) 2 (2x − 1) ( x + 1) ( x + 1) 2
Now that the expression is in partial fractions, each part can be integrated
separately.
x+4
2 dx −
1 dx −
1 dx
dx =
(2x − 1)( x + 1) 2
(
1) 2
x
+
( 2x − 1)
( x + 1)
∫
∫
∫
∫
The first two integrals give logarithmic expressions as you saw above. The
third, however, is of the form u −2 and therefore can be integrated by using
the substitution u = x + 1, or by inspection (i.e. in your head).
x+4
1
∫ (2x − 1)(x + 1) dx = ln | 2x − 1 | − ln | x + 1 | + x + 1 + c
2
= ln 2x − 1 + 1 + c
x +1
x +1
⎪
⎪
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A quadratic factor in the denominator
Example 8.16
Find
8
x−2
∫ (x + 2)(x + 1) dx.
2
Solution
where
8.5 The use of partial fractions in integration
First write the expression as partial fractions:
x−2
C
Ax + B
( x 2 + 2)( x + 1) = ( x 2 + 2) + ( x + 1)
x − 2 ≡ (Ax + B)(x + 1) + C(x2 + 2)
Rearranging gives
x − 2 ≡ (A + C )x2 + (A + B)x + (B + 2C )
Equating coefficients:
x2
⇒
A+C=0
x
⇒
A+B=1
constant terms
⇒
B + 2C = −2
Solving these gives A = 1, B = 0, C = −1.
Hence
x−2
x
1
( x 2 + 2)( x + 1) = ( x 2 + 2) − ( x + 1)
x−2
x
1
( x 2 + 2)( x + 1) dx = ( x 2 + 2) dx − ( x + 1) dx
2x
1
= 21
dx − ( x + 1) dx
( x 2 + 2)
1
2
2
–
2 ln | x + 2 | = ln x + 2
= 21 ln | x 2 + 2 | − ln | x + 1 | + c
2
Notice that (x + 2) is
2
positive for all values of x.
= ln x + 2 + c
x +1
∫
∫
∫
∫
⎪
∫
⎪
e
Note
If B had not been zero, you would have had an expression of the form
integrate. This can be split into
Ax + B .
x2 + 2 x2 + 2
Ax + B to
x2 + 2
The first part of this can be integrated as in Example 8.16, but to integrate the
second part you would need to use the standard result:
∫ (x + k ) dx = k tan−1 ( xk ) + c.
2
1
2
1
which you met on page 197.
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8 FURTHER CALCULUS
8
Exercise 8E
1
2
Express the fractions in each of the following integrals as partial fractions,
and hence perform the integration.
7x − 2
1
(i)
dx
(ii)
dx
( x − 1) 2 (2x + 3)
(1 − x )(3x − 2)
x +1
3x + 3
(iii)
dx
(iv)
( x − 1)(2x + 1) dx
( x 2 + 1)( x − 1)
1
1
dx
(v)
(vi)
( x + 1)( x + 3) dx
x 2 (1 − x )
5x + 1
2x − 4
(vii)
dx
(viii)
( x + 2)( 2x + 1) 2 dx
( x 2 + 4)( x + 2)
∫
∫
∫
∫
∫
∫
∫
∫
Express in partial fractions
3x + 4
f(x) = 2
( x + 4)( x − 3)
and hence find
PS
3
(i)
2
∫0 f(x) dx.
(a)
3
Express (1 + x )(1 − 2x ) in partial fractions.
(b)
Hence find
3
0.1
∫ 0 (1 + x )(1 − 2x ) dx
(ii)
(a)
(b)
giving your answer to 5 decimal places.
Find the first three terms in the binomial expansion of
3(1 + x)−1(1 − 2x)−1.
Use the first three terms of this expansion to find an
approximation for
0.1
3
dx.
0 (1 + x )(1 − 2 x )
What is the percentage error in your answer to part (b)?
∫
(c)
4
(i)
(ii)
Find the values of the constants A, B, C and D such that
2x 3 – 1 ≡ A + B + C + D .
x 2 (2x – 1)
x x 2 2x – 1
Hence show that
2 2x 3 – 1
3 1
16
2 (2 x – 1) dx = 2 + 2 ln 27 .
x
1
( )
∫
5
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q10 June 2010
(i)
x 2 + 3x + 3 .
( x + 1)( x + 3)
Express f(x) in partial fractions.
(ii)
Hence show that
Let f ( x ) ≡
3
∫ f (x )d x = 3 − 21 ln 2.
0
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q7 June 2008
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6
Let f ( x ) = 12 + 8x − x 2 .
(2 − x )(4 + x )
2
(i)
(ii)
8
A + Bx + C
2
− x 4 + x2
1
.
Show that 0 f ( x )d x = ln ( 25
2)
Express f(x) in the form
∫
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q8 November 2011
(i)
(ii)
Show that (x + 1) is a factor of 4 x 3 − x 2 − 11x − 6.
4 x 2 + 9 x − 1 dx .
Find
4 x 3 − x 2 − 11x − 6
∫
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q7 November 2015
8
By first using the substitution u = e x , show that
ln 4
e 2x
dx = ln ( 85 ) .
0 e 2 x + 3e x + 2
∫
8.6 Integration by parts
7
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q10 November 2014
8.6 Integration by parts
There are still many integrations which you cannot yet do. In fact, many
functions cannot be integrated at all, although virtually all functions can be
differentiated. However, some functions can be integrated by techniques
which you have not yet met. Integration by parts is one of those techniques.
Example 8.17
∫
Find x cos x dx.
Solution
The expression to be integrated is clearly a product of two simpler
expressions, x and cos x, so your first thought may be to look for a
substitution to enable you to perform the integration. However, there are
some expressions which are products but which cannot be integrated by
substitution. This is one of them.You need a new technique to integrate such
expressions.
Take the expression x sin x and differentiate it, using the product rule.
d
dx (x sin x) = x cos x + sin x
Now integrate both sides. This has the effect of ‘undoing’ the differentiation, so
∫
∫
x sin x = x cos x dx + sin x dx
➜
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8
Rearranging this gives
∫ x cos x dx = x sin x − ∫ sin x dx
= x sin x − (−cos x) + c
= x sin x + cos x + c
8 FURTHER CALCULUS
This has enabled you to find the integral of x cos x.
The work in this example can be generalised into the method of integration
by parts. Before coming on to that, do the following activity.
ACTIVITY 8.2
For each of the following
(a) differentiate using the product rule
(b) rearrange your expression to find an expression for the given
integral I
(c) use this expression to find the given integral.
(i) y = x cos x
I = x sin x dx
(ii)
∫
I = ∫ 2x e2x dx
y = xe2x
The work in Activity 8.2 has enabled you to work out some integrals
which you could not previously have done, but you needed to be given
the expressions to be differentiated first. Effectively you were given the
answers.
?
Look at the expressions you found in part (b) of Activity 8.2.
❯ Can you see any way of working out these expressions without
starting by differentiating a given product?
The general result for integration by parts
The method just investigated can be generalised.
Look back at Example 8.17. Use u to stand for the function x, and v to stand
for the function sin x.
Using the product rule to differentiate the function uv,
d (uv) = v du + u dv .
dx
dx
dx
Integrating gives
uv = v du dx + u dv dx.
dx
dx
Rearranging gives
u dv dx = uv − v du dx.
dx
dx
This is the formula you use when you need to integrate by parts.
∫
∫
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∫
∫
02/02/18 1:14 PM
∫
So now you can find x cos x dx.
Put
u=x
⇒ du = 1
dx
dv = cos x ⇒
and
v = sin x
dx
Substituting in
8
8.6 Integration by parts
In order to use it, you have to split the function you want to integrate into
two simpler functions. In Example 8.17 you split x cos x into the two functions
x and cos x. One of these functions will be called u and the other dv , to fit
dx
the left-hand side of the expression.You will need to decide which will be
which. Two considerations will help you.
du
on the right-hand side of the expression, u should
» As you want to use
dx
be a function which becomes a simpler function after differentiation. So
in this case, u will be the function x.
» As you need v to work out the right-hand side of the expression, it must be
possible to integrate the function dv to obtain v. In this case, dv will be
dx
dx
the function cos x.
∫u ddxv dx = uv − ∫v ddxu dx
gives
∫x cos x dx = x sin x − ∫1 ×sin x dx
= x sin x − (−cos x) + c
= x sin x + cos x + c
Example 8.18
∫
Find 2x ex dx.
Solution
First split 2x ex into the two simpler expressions, 2x and ex. Both can be
integrated easily but, as 2x becomes a simpler expression after differentiation
and ex does not, take u to be 2x.
du = 2
u = 2x ⇒
dx
dv
x
⇒
v = ex
dx = e
Substituting in
dv
gives
du
∫u dx dx = uv − ∫v dx dx
∫2x ex dx = 2x ex − ∫2ex dx
= 2x ex − 2ex + c
In some cases, the choices of u and v may be less obvious.
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8
Example 8.19
∫
Find x ln x dx.
Solution
It might seem at first that u should be taken as x, because it becomes a
simpler expression after differentiation.
8 FURTHER CALCULUS
du
=1
u=x
⇒
dx
dv
dx = ln x
Now you need to integrate ln x to obtain v. Although it is possible to
integrate ln x, it has to be done by parts, as you will see in the next
example. The wrong choice has been made for u and v, resulting in a more
complicated integral.
So instead, let u = ln x.
u = ln x
dv
dx = x
Substituting in
⇒
du = 1
dx x
⇒
v = 21 x 2
dv
du
∫u dx dx = uv − ∫v dx dx
gives
1 x2
∫x ln x dx = 2 x 2 ln x − ∫ 2 x dx
1
∫
1
= 21 x 2 ln x − 2 x dx
= 21 x 2 ln x − 41 x 2 + c
Example 8.20
∫
Find ln x dx.
Solution
You need to start by writing ln x as 1 ln x and then use integration by parts.
As in the last example, let u = ln x.
u = ln x
dv
dx = 1
Substituting in
dv
⇒
du = 1
dx x
⇒
v=x
du
∫u dx dx = uv − ∫v dx dx
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gives
8
1
∫ 1 ln x dx = x ln x − ∫ x × x dx
= x ln x − ∫ 1 dx
= x ln x − x + c
Sometimes it is necessary to use integration by parts twice or more to
complete the integration successfully.
Example 8.21
∫
Find x 2 sin x dx.
Solution
First split x 2 sin x into two: x 2 and sin x. As x 2 becomes a simpler expression
after differentiation, take u to be x 2.
u = x2
dv
dx = sin x
Substituting in
⇒
gives
du
dx = 2x
⇒
dv
8.6 Integration by parts
Using integration by parts twice
v = −cos x
du
∫ u dx dx = uv − ∫ v dx dx
∫ x 2 sin x dx = −x 2 cos x − ∫ −2x cos x dx
= −x 2 cos x + ∫ 2x cos x dx
1
!
Now the integral of 2x cos x cannot be found without using integration
by parts again. It has to be split into the expressions 2x and cos x and, as 2x
becomes a simpler expression after differentiation, take u to be 2x.
u = 2x
dv
dx = cos x
Substituting in
dv
gives
⇒
du
dx = 2
⇒
v = sin x
du
∫ u dx dx = uv − ∫ v dx dx
∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx
= 2x sin x − (−2 cos x) + c
= 2x sin x + 2 cos x + c
1
So in !
∫ x 2 sin x dx = −x 2 cos x + 2x sin x + 2 cos x + c.
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The technique of integration by parts is usually used when the two functions are
of different types: polynomials, trigonometrical functions, exponentials, logarithms.
There are, however, some exceptions, as in questions 3 and 4 of Exercise 8F.
8
Integration by parts is a very important technique which is needed in many
other branches of mathematics. For example, integrals of the form x f(x) dx
are used in statistics to find the mean of a probability density function, and in
mechanics to find the centre of mass of a shape. Integrals of the form x2 f(x) dx
are used in statistics to find variance and in mechanics to find moments of inertia.
8 FURTHER CALCULUS
∫
∫
Exercise 8F
1
2
3
4
5
6
7
PS
8
For each of these integrals
(a) write down the expression to be taken as u and the
dv
expression to be taken as dx
(b) use the formula for integration by parts to complete the
integration.
x
(i)
x e dx
(ii)
x cos 3x dx
∫
(iii) ∫ (2x + 1)cos x dx
(v) ∫ x e−x dx
∫
(iv) ∫ x e−2x dx
(vi) ∫ x sin 2x dx
Use integraton by parts to integrate
(i) x 3 ln x
(ii) 3x e 3x
(iii) 2x cos 2x
(iv) x 2 ln 2x
Find x 1 + x dx
(i) by using integration by parts
(ii) by using the substitution u = 1 + x.
Find 2x(x − 2)4 dx
(i) by using integration by parts
(ii) by using the substitution u = x − 2.
(i)
By writing ln x as the product of ln x and 1, use integration by
parts to find ln x dx.
(ii) Use the same method to find ln 3x dx.
(iii) Write down ln px dx where p # 0.
Find x 2 ex dx.
∫
∫
∫
∫
∫
∫
Find ∫ (2 − x)2 cos x dx.
Find ∫ x tan−1x dx.
Definite integration by parts
When you use the method of integration by parts on a definite integral, it
is important to remember that the term uv on the right-hand side of the
expression has already been integrated and so should be written in square
brackets with the limits indicated.
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b
b
u dv dx = uv − v du dx
a dx
a dx
a
∫
b
[ ] ∫
02/02/18 1:15 PM
Example 8.22
Evaluate
2
∫ 0 x e x dx.
8
Solution
du
dx = 1
v = ex
u=x ⇒
dv
x
and
dx = e ⇒
Substituting in
Put
b
8.6 Integration by parts
b
∫a u dx dx = [uv]a − ∫av dx dx
dv
b
du
gives
2
∫ 0x ex dx
[ ] ∫ e dx
= [x e ] − [e ]
2
2
0
0
= x ex −
x
2
x
x
0
2
0
= (2e 2 − 0) − (e2 − e0)
= 2e2 − e2 + 1
= e2 + 1
Example 8.23
Find the area of the region between the curve y = x cos x and the x-axis,
between x = 0 and x = π .
2
Solution
y
Figure 8.6 shows the region whose area
is to be found.
1
To find the required area, you
need to integrate the function
O
x cos x between the limits 0 and π .
2
You therefore need to work out
–1
π
2
∫ 0 x cos x dx.
Put
u=x
dv
dx = cos x
Substituting in
and
dv
x
▲ Figure 8.6
⇒
du
dx = 1
⇒
v = sin x
∫a u dx dx = [uv] a − ∫a v dx dx
b
π
2
b
b
du
➜
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gives
8
π
2
π
2
π
2
∫0 x cos x dx = [x sin x]0 − ∫0 sin x dx
π
2
π
2
= [x sin x] 0 − [−cos x]0
π
2
8 FURTHER CALCULUS
= [x sin x + cos x]0
=
=
( )
( π2 + 0) − (0 + 1)
π
2 −1
So the required area is π − 1 square units.
2
Exercise 8G
1
Evaluate these definite integrals.
π
1
(i)
x e3x dx
(ii)
(x − 1)cos x dx
∫0
2
(iii) ∫ (x + 1)e x dx
0
(v) ∫ x sin 2x dx
0
π
2
2
3
4
5
6
PS
7
∫0
2
(iv) ∫ ln 2x dx
1
4
(vi) ∫ x 2 lnx dx
1
Find the coordinates of the points where the graph of y = (2 − x)e−x
cuts the x- and y-axes.
(ii) Hence sketch the graph of y = (2 − x)e−x.
(iii) Use integration by parts to find the area of the region between the
x-axis, the y-axis and the graph y = (2 − x)e−x.
(i)
Sketch the graph of y = x sin x from x = 0 to x = π and shade the
region between the curve and the x-axis.
(ii) Find the area of this region using integration by parts.
Find the area of the region between the x-axis, the line x = 5 and the
graph y = ln x.
Find the area of the region between the x-axis and the graph y = x cos x
π
from x = 0 to x = 2 .
Find the area of the region between the negative x-axis and the graph
y = x x +1
(i) using integration by parts
(ii) using the substitution u = x + 1.
Find
(i)
(i)
(ii)
3
∫ 6x tan x dx
∫ x tan (2x )dx.
0
1
2
−1
−1
0
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8
Find the exact value of
4
∫ lnxx d x..
1
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q3 November 2013
9
∫
a
8
1
The constant a is such that 0 x e 2 x dx = 6.
(i) Show that a satisfies the equation
1
x = 2 + e – 2x .
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q9 November 2008
8.6 Integration by parts
By sketching a suitable pair of graphs, show that this equation has
only one root.
(iii) Verify by calculation that this root lies between 2 and 2.5.
(iv) Use an iterative formula based on the equation in part (i) to
calculate the value of a correct to 2 decimal places. Give the result
of each iteration to 4 decimal places.
(ii)
1
10 The diagram shows the curve y = e – 2 x √(1 + 2x) and its maximum point
M. The shaded region between the curve and the axes is denoted by R.
y
M
R
O
(i)
(ii)
x
Find the x coordinate of M.
Find by integration the volume of the solid obtained when R is
rotated completely about the x-axis. Give your answer in terms of π
and e.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q9 June 2008
11 The expression f(x) is defined by f(x) = 3x e −2 x .
(i)
Find the exact value of f ' (− 21 ).
(ii)
Find the exact value of
0
∫ f (x )dx.
−1
2
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q5 November 2012
211
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02/02/18 1:15 PM
1
8
12 The diagram shows the curve y = x 2 ln x. The shaded region between
the curve, the x-axis and the line x = e is denoted by R.
y
R
e
8 FURTHER CALCULUS
O
(i)
(ii)
x
Find the equation of the tangent to the curve at the point where
x = 1, giving your answer in the form y = mx + c.
Find by integration the volume of the solid obtained when the
region R is rotated completely about the x-axis. Give your answer
in terms of π and e.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q9 June 2012
8.7 General integration
You now know several techniques for integration which can be used to
integrate a wide variety of functions. One of the difficulties which you may
now experience when faced with an integration is deciding which technique
is appropriate! This section gives you some guidelines on this, as well as
revising all the work on integration that you have done so far.
❯ Look at the integrals below and try to decide which technique
you would use and, in the case of a substitution, which expression
you would write as u. Do not attempt actually to carry out the
integrations. Make a note of your decisions − you will return to these
integrals later.
x−5
x +1
(i)
(ii) x 2 + 2 x − 3 dx
x 2 + 2x − 3 dx
∫
(iii) x e dx
∫
2x + cos x
(v)
∫ x + sin x dx
x
2
?
∫
(iv) x e dx
∫
(vi) cos x sin2 x dx
∫
x2
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Choosing an appropriate method of integration
You have now met the following standard integrals.
8
∫f(x) dx
f(x)
(n ≠ −1)
x n+1
n +1
1
x
(x ≠ 0)
ln|x |
ex
(x ! ")
ex
sin x
(x ! ")
−cos x
cos x
(x ! ")
sin x
If you are asked to integrate any of these standard functions, you may simply
write down the answer.
8.7 General integration
xn
For other integrations, the following table may help.
Type of expression to be integrated
Examples
Method of integration
Simple variations of any of the standard
functions
cos(2x + 1)
e3x
Substitution may be used, but it
should be possible to do these by
inspection.
Product of two expressions of the form
f ′(x)g[f(x)]
d
Note that f ′(x) means
[f(x)]
dx
2x ex
x2(x3 + 1)6
Substitution u = f(x)
Other products, particularly when one
expression is a small positive integer
power of x or a polynomial in x
x ex
x2 sin x
Integration by parts
2
Substitution u = f(x) or by inspection:
k ln | f(x) | + c,
where k is known
x
f ′ (x )
Quotients of the form f( x )
or expressions which can easily
be converted to this form
sinx
cosx
Polynomial quotients which may be split
into partial fractions
x +1
x( x − 1)
2
x +1
Split into partial fractions and
integrate term by term
x−4
x2 − x − 2
Odd powers of sin x or cos x
cos3 x
Use cos2 x + sin2 x = 1 and write
in form f ′(x)g[f(x)]
Even powers of sin x or cos x
sin2 x
cos4 x
Use the double-angle formulae to
transform the expression before
integrating.
Expressions in the form
1
x2 + k2
1
x 2 + 25
or
1
3x 2 + 4
Use standard result:
x
1
1
+c
dx = tan -1
x2 + k2
k
k
∫
()
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It is impossible to give an exhaustive list of possible types of integration, but
the tables on the previous page cover the most common situations that you
will meet.
8
ACTIVITY 8.3
8 FURTHER CALCULUS
Now look back at the integrals in the discussion point on page 212 and
the decisions you made about which method of integration should be
used for each one. Now find these integrals.
x−5
x +1
(i)
dx
(ii)
dx
x 2 + 2x − 3
x 2 + 2x − 3
∫
(iii) x ex dx
∫
2x + cos x
(v)
∫ x + sin x dx
1
Note
Often when a
question requires
integration by
substitution then
you will be given a
suitable substitution,
however in this
exercise you need to
decide on the best
substitution to use
yourself.
(vi)
Choose an appropriate method and integrate the following.
You may find it helpful to discuss in class first which method to use.
2x + 1
(i)
cos(3x − 1) dx
(ii)
dx
( x 2 + x − 1) 2
(iii) e1−x dx
(iv)
cos 2x dx
∫
∫
(v)
∫ln 2x dx
(vii)
∫ 2x − 3 dx
(ix) x 3 ln x dx
∫
(xi) (x + 1)ex + 2x dx
∫
(xiii) x 2 sin 2x dx
∫
2
dx
3x − 8
24
∫8
π
2
∫0
2 ln x dx
(vii) ∫
x
(iv)
2
2
(ii)
dx
(iii)
x 2 ln x dx
(vi)
π
3
0
24
9x
∫ 8 3x − 8 dx
∫
∫
1
x 2 dx
1+ x3
π
4 sin x
dx
0 cos 4 x
0
1
∫ ex dx
0
2x
Let f ( x ) =
(i)
2
(v)
1
(x)
24
∫ 8 3x − 8
(ii)
∫1
(viii) ∫ x cos 3x dx (ix)
sin3x dx
2
3
3
Evaluate the following definite integrals.
(i)
9781510421738.indb 214
∫
∫
x
(vi)
∫ (x − 1) dx
4x − 1
(viii)
∫ (x − 1) (x + 2) dx
5
(x)
∫ 2x − 7x + 3 dx
sin x − cos x
(xii)
∫ sin x + cos x dx
(xiv) sin3 2x dx
∫
2
2
214
2
(iv)
2
Exercise 8H
∫
∫x e x dx
∫cos x sin2 x dx
6 + 6x
.
(2 − x )(2 + x 2 )
A + Bx + C .
− x 2 + x2
2
1
Show that
f ( x ) d x = 3 ln 3.
Express f(x) in the form
∫
−1
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q8 June 2014
02/02/18 1:15 PM
The diagram shows the curve y = x 2 e 2 − x and its maximum point M.
4
y
M
O
x
∫
(ii)
0
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q9 June 2015
The integral I is defined by I =
5
2
∫ 4t ln(t + 1)dt.
3
2
0
Use the substitution x = t 2 + 1 to show that
(i)
I =
5
∫ (2x − 2) ln x d x.
8.7 General integration
Show that the x coordinate of M is 2.
2
Find the exact value of
x 2 e 2 − x dx .
(i)
8
1
Hence find the exact value of I.
(ii)
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q7 June 2011
KEY POINTS
1 You can differentiate tan−1 x by using either
implicit differentiation, or
dy
●
= 1
dx dx
dy
kx n +1
2 kxn dx =
n + 1 + c where k and n are constants but n ≠ −1.
3 Substitution is often used to change a non-standard integral into a
standard one.
●
∫
4
∫ ex dx = ex + c
∫ e dx = 1a e
ax + b
5
ax + b
+c
1
∫ x dx = ln | x | + c
∫ ax1+ b dx = 1a ln | ax + b | + c
6
f ′(x )
∫ f(x ) dx = ln | f(x) | + c
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8
7
∫ cos(ax + b) dx = 1a sin(ax + b) + c
1
∫ sin(ax + b) dx = − a cos(ax + b) + c
∫ sec (ax + b) dx = 1a tan(ax + b) + c
8 FURTHER CALCULUS
2
8
∫ x + k dx = k tan ( k ) + c
2
1
2
1
−1
x
9 Using partial fractions often makes it possible to use logarithms to
integrate the quotient of two polynomials.
10 Some products may be integrated by parts using the formulae
u dv dx = uv − v du dx
dx
dx
b dv
b
b
u dx = uv
u − v du dx
x
d
a
a
a dx
∫
∫
∫
[ ] ∫
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■ differentiate tan −1 x
■ extend the idea of ‘reverse differentiation’ to include the integration of
1
x2 + a2
■ use integration by substitution in cases where the process is the reverse
of the chain rule (either by inspection or by writing down the working
for the substitution)
■ use integration by substitution in other cases, finding a suitable
substitution for both definite and indefinite integrals
k f ′( x )
■ use
dx = k lln | f(x) |++ c
f( x )
■ use partial fractions in integration
■ use trigonometrical relationships in integration
■ use the method of integration by parts including when more than one
application of the method may be required to integrate ln x.
∫
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P3
9
Differential equations
9 Differential equations
The greater
our knowledge
increases,
the more our
ignorance
unfolds.
John F. Kennedy
(1917–1963)
Temperature T
❯ How long do you have to wait for a typical cup of coffee to be
drinkable?
❯ How long does it take to go cold?
❯ What do the words ‘drinkable’ and ‘cold’ mean in this context?
T0
O
Time t
?
M
Newton’s law of cooling states that the rate
of change of the temperature of an object is
proportional to the difference between the object’s
temperature and the temperature of its surroundings.
dT
= k (T − T 0 ) , where
This leads to the equation
dt
T is the temperature of the object at time t, T 0 is
the temperature of the surroundings, and k is a
constant of proportionality.
▲ Figure 9.1
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9 DIFFERENTIAL EQUATIONS
9
This is an example of a differential equation. To be able to predict the
temperature of the object at different times, you need to solve the differential
equation. In this chapter, you will learn how to solve problems like this
involving rates of change.
9.1 Forming differential equations
from rates of change
If you are given sufficient information about the rate of change of a quantity,
such as temperature or velocity, you can work out a differential equation to
model the situation, like the one given earlier for Newton’s law of cooling.
It is important to look carefully at the wording of the problem which
you are studying in order to write an equivalent mathematical statement.
For example, if the altitude of an aircraft is being considered, the phrase
‘the rate of change of height’ might be used. This actually means ‘the rate of
change of height with respect to time’ and could be written as dh . However, you
dt
might be more interested in how the height of the aircraft changes according
to the horizontal distance it has travelled. In this case, you would talk about
‘the rate of change of height with respect to horizontal distance’ and could write
dh
this as , where x is the horizontal distance travelled.
dx
Some of the situations you meet in this chapter involve motion along a
straight line, and so you will need to know the meanings of the associated
terms.
The position of an object (+5 in Figure 9.2) is its distance from the origin O
in the direction you have chosen to define as being positive.
O
–1
0
1
2
3
4
5
6
s
▲ Figure 9.2
The rate of change of position of the object with respect to time is its
velocity, and this can take positive or negative values according to whether
the object is moving away from the origin or towards it.
v = ds
dt
The rate of change of an object’s velocity with respect to time is called its
acceleration, a.
a = dv
dt
Velocity and acceleration are vector quantities but in one-dimensional
motion there is no choice in direction, only in sense (i.e. whether positive
or negative). Consequently, as you may already have noticed, the conventional
bold type for vectors is not used in this chapter.
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Example 9.1
An object is moving through a liquid so that the rate at which its velocity
decreases is proportional to its velocity at any given instant. When it enters
the liquid, it has a velocity of 5 m s–1 and the velocity is decreasing at a rate of
1 m s–2. Find the differential equation to model this situation.
9
Solution
− dv ∝ v or dv = −kv
dt
dt
where k is a positive constant.
When the object enters the liquid its velocity is 5 m s–1, so v = 5, and the
velocity is decreasing at the rate of 1 m s–2, so
dv = −1
dt
Putting this information into the equation dv = −kv gives
dt
–1 = –k × 5
⇒
k = 51.
So the situation is modelled by the differential equation
dv = − v
5
dt
Example 9.2
9.1 Forming differential equations from rates of change
The rate of change of velocity means the rate of change of velocity with
respect to time and so can be written as dv . As it is decreasing, the rate of
dt
change must be negative, so
A model is proposed for the temperature gradient within a star, in which the
temperature decreases with respect to the distance from the centre of the star
at a rate which is inversely proportional to the square of the distance from the
centre. Express this model as a differential equation.
Solution
In this example the rate of change of temperature is not with respect to time
but with respect to distance. If θ represents the temperature at a point in
the star and r the distance from the centre of the star, the rate of change of
temperature with respect to distance may be written as – dθ , so
dr
− dθ ∝ 12 or dθ = − k2
dr
dr
r
r
where k is a positive constant.
Note
This model must break down near the centre of the star, otherwise it would
be infinitely hot there.
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9
Example 9.3
The area A of a square is increasing at a rate proportional to the length of its
side s. The constant of proportionality is k. Find an expression for ds .
dt
s
9 DIFFERENTIAL EQUATIONS
s
A
s
s
% Figure 9.3
Solution
The rate of increase of A with respect to time may be written as dA.
dt
As this is proportional to s, it may be written as
dA = ks
dt
where k is a positive constant.
dA.
You can use the chain rule to write down an expression for ds in terms of
dt
dt
ds = ds × dA
dt dA dt
ds
You now need an expression for dt . Because A is a square
A = s2
⇒
⇒
dA = 2s
ds
ds
1
=
dA 2 s
Substituting the expressions for ds and dA into the expression for ds
dt
dA
dt
ds
1
⇒
dt = 2 s × ks
ds = 1 k
⇒
dt 2
Exercise 9A
1 The differential equation
M
dv = 5v 2
dt
models the motion of a particle, where v is the velocity of the particle in
dv
m s–1 and t is the time in seconds. Explain the meaning of dt and what
the differential equation tells you about the motion of the particle.
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2 A spark from a firework is moving in a straight line at a speed which is
3
5
6
7
8
9
10
11
9
9.1 Forming differential equations from rates of change
4
inversely proportional to the square of the distance which the spark has
travelled from the firework. Find an expression for the speed (i.e. the rate
of change of distance travelled) of the spark.
The rate at which a sunflower increases in height is proportional to the
natural logarithm of the difference between its final height H and its
height h at a particular time. Find a differential equation to model this
situation.
In a chemical reaction in which substance A is converted into substance
B, the rate of increase of the mass of substance B is inversely proportional
to the mass of substance B present. Find a differential equation to model
this situation.
After a major advertising campaign, an engineering company finds that
its profits are increasing at a rate proportional to the square root of the
profits at any given time. Find an expression to model this situation.
The coefficient of restitution e of a squash ball increases with respect
to the ball’s temperature θ at a rate proportional to the temperature,
for typical playing temperatures. (The coefficient of restitution is a
measure of how elastic, or bouncy, the ball is. Its value lies between
zero and one, zero meaning that the ball is not at all elastic and one
meaning that it is perfectly elastic.) Find a differential equation to
model this situation.
A cup of coffee cools at a rate proportional to the temperature of
the coffee above that of the surrounding air. Initially, the coffee is
at a temperature of 95°C and is cooling at a rate of 0.5°C s–1. The
surrounding air is at 15°C. Find a differential equation to model this
situation.
The rate of increase of bacteria is modelled as being proportional to the
number of bacteria at any time during their initial growth phase.
When the bacteria number 2 × 106 they are increasing at a rate of
105 per day. Find a differential equation to model this situation.
The acceleration (i.e. the rate of change of velocity) of a moving object
under a particular force is inversely proportional to the square root of
its velocity. When the speed is 4 m s–1 the acceleration is 2 m s–2. Find a
differential equation to model this situation.
The radius of a circular patch of oil is increasing at a rate inversely
proportional to its area A. Find an expression for dA .
dt
A poker, 80 cm long, has one end in a fire. The temperature of the
poker decreases with respect to the distance from that end at a rate
proportional to that distance. Halfway along the poker, the temperature
is decreasing at a rate of 10°C cm–1. Find a differential equation to
model this situation.
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12 A spherical balloon is allowed to deflate. The rate at which air is leaving
9 DIFFERENTIAL EQUATIONS
9
the balloon is proportional to the volume V of air left in the balloon.
When the radius of the balloon is 15 cm, air is leaving at a rate of 8 cm3 s–1.
Find an expression for dV .
dt
13 A tank is shaped as a cuboid with a square base of side 10 cm. Water runs
out through a hole in the base at a rate proportional to the square root of
the height, h cm, of water in the tank. At the same time, water is pumped
into the tank at a constant rate of 2 cm3 s–1. Find an expression for dh .
dt
INVESTIGATION
Under pressure
Figure 9.4 shows the isobars (lines of equal pressure) on a weather map
featuring a storm. The wind direction is almost parallel to the isobars and its
speed is proportional to the pressure gradient.
20
°
30°
70°
60°
°
10
70°
Scale of nautical miles
100 300 500 700
0°
60°
50°
200
400
600
40°
10°
▲ Figure 9.4
Draw a line from the point H to the point L on a copy of this diagram. This
runs approximately perpendicular to the isobars. It is suggested that along this
line the pressure gradient (and so the wind speed) may be modelled by the
differential equation
dp
= –a sin bx
dx
Suggest values for a and b, and comment on the suitability of this model.
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9.2 Solving differential equations
9
The general solution of a differential equation
Finding an expression for f(x) from a differential equation involving
derivatives of f(x) is called solving the equation.
Example 9.4
Solve the differential equation
9.2 Solving differential equations
Some differential equations may be solved simply by integration.
dy
= 3x2 − 2.
dx
Solution
Integrating gives
∫
y = (3x 2 − 2) dx
y = x 3 − 2x + c
Notice that when you solve a differential equation, you get not just one
solution, but a whole family of solutions, as c can take any value. This is called
the general solution of the differential equation. The family of solutions
for the differential equation in the example above would be translations in
the y direction of the curve y = x3 – 2x. Graphs of members of the family of
curves can be found in Figure 9.5 on page 226.
The method of separation of variables
It is not difficult to solve a differential equation like the one in Example 9.4,
because the right-hand side is a function of x only. So long as the function
can be integrated, the equation can be solved.
dy
= xy.
Now look at the differential equation
dx
This cannot be solved directly by integration, because the right-hand side
is a function of both x and y. However, as you will see in the next example,
you can solve this and similar differential equations where the right-hand side
consists of a function of x and a function of y multiplied together.
Example 9.5
Find, for y # 0, the general solution of the differential equation
dy
= xy.
dx
Solution
The equation may be rewritten as
1 dy
=x
y dx
so that the right-hand side is now a function of x only.
➜
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9
Integrating both sides with respect to x gives
1 dy
∫ y dx d x = ∫ x d x
dy
As dx dx can be written as dy
9 DIFFERENTIAL EQUATIONS
1
∫ y dy = ∫ x dx
Both sides may now be integrated separately.
ln | y | = 21 x 2 + c
Since you have been
told y # 0, you may drop
the modulus symbol. In
this case, |y| = y.
?
❯ Explain why there is no need to put a constant of integration on
both sides of the equation.
You now need to rearrange the solution above to give y in terms of x .
Making both sides powers of e gives
1
elny = e 2 x + cc
⇒
⇒
2
y=e
1 2 +c
2x c
y=e
1x 2 c
c
2
Notice that the right-hand
1 x 2 +c
side is e 2 x
and not e
e
c
1x 2 c
2x +
ec.
This expression can be simplified by replacing ec with a new constant A.
1
y = Ae 2 x
2
Note
Usually the first part of this process is carried out in just one step.
dy
= xy
dx
can immediately be rewritten as
∫ 1y d y = ∫ x d x
This method is called separation of variables. It can be helpful to do this
dy
by thinking of the differential equation as though
were a fraction and
dx
trying to rearrange the equation to obtain all the x terms on one side and all
the y terms on the other. Then just insert an integration sign on each side.
Remember that dy and dx must both end up in the numerator (top line).
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Example 9.6
Find the general solution of the differential equation
dy
= e−y.
dx
Solution
9
Separating the variables gives
−y
⇒
y
The right-hand side can be thought of as integrating 1 with respect to x.
ey = x + c
Taking logarithms of both sides gives
y = ln⎪x + c ⎪
9.2 Solving differential equations
∫ e1 d y = ∫ d x
∫ e dy = ∫ dx
ln⎪x + c ⎪ is not the same as ln ⎪x ⎪ + c.
Exercise 9B
1
2
Solve the following differential equations by integration.
dy
dy
2
(i)
(ii)
dx = x
dx = cos x
dy
dy
= x
(iii)
= ex
(iv)
dx
dx
Find the general solutions of the following differential equations by
separating the variables.
dy
dy x 2
(i)
= xy 2
(ii)
=
y
dx
dx
dy
dy
(iii)
=y
(iv)
= e x −y
dx
dx
dy
dy y
=x y
(v)
(vi)
=
dx
dx x
dy x( y 2 + 1)
dy
=
(vii)
= y 2 cos x (viii)
dx y( x 2 + 1)
dx
dy x ln x
dy
= 2
(ix)
= x ey
(x)
dx
y
dx
The particular solution of a differential equation
You have already seen that a differential equation has an infinite number
of different solutions corresponding to different values of the constant of
dy
integration. In Example 9.4, you found that
= 3x 2 − 2 had a general
d
x
solution of y = x 3 − 2x + c.
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Figure 9.5 shows the curves of the solutions corresponding to some different
values of c.
9
y
6
y = x3 – 2x + 2 (c = 2)
9 DIFFERENTIAL EQUATIONS
y = x3 – 2x (c = 0)
y = x3 – 2x – 1 (c = –1)
0
–3
–2
–1
1
2
3
x
–6
% Figure 9.5
If you are given some more information, you can find out which of
the possible solutions is the one that matches the situation in question.
For example, you might be told that when x = 1, y = 0. This tells you that the
correct solution is the one with the curve that passes through the point (1, 0).
You can use this information to find out the value of c for this particular
solution by substituting the values x = 1 and y = 0 into the general solution.
⇒
y = x 3 − 2x + c
0=1−2+c
c=1
So the solution in this case is y = x 3 − 2x + 1.
This is called the particular solution.
Example 9.7
(i)
(ii)
dy
= y2.
dx
Find the particular solution for which y = 1 when x = 0.
Find the general solution of the differential equation
Solution
(i)
Separating the variables gives
∫ y1 dy = ∫ d x
2
1
− =x+c
y
1
The general solution is y = − x + c .
Figure 9.6 shows a set of solution curves.
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y = – 1 (c = 0)
x
y
y = – 1 (c = 1)
x+1
3
y=–
9
1
x – 1 (c = –1)
2
–4
–3
–2
–1
0
1
2
9.2 Solving differential equations
1
x
–1
–2
–3
▲ Figure 9.6
(ii)
Example 9.8
When x = 0, y = 1, which gives
1
1 = −c
⇒
c = −1.
So the particular solution is
1
1
y = − x − 1 or y = 1 − x
This is the blue curve illustrated in Figure 9.6.
The acceleration of an object is inversely proportional to its velocity at any
given time and the direction of motion is taken to be positive.
When the velocity is 1 m s−1, the acceleration is 3 m s−2.
(i)
Find a differential equation to model this situation.
(ii)
Find the particular solution to this differential equation for which the
initial velocity is 2 m s−1.
(iii) In this case, how long does the object take to reach a velocity of 8 m s−1?
Solution
(i)
(ii)
dv = k
dt v
dv 3
dv
When v = 1, dt = 3 so k = 3, which gives dt = v .
Separating the variables:
∫ v d v = ∫ 3 dt
1 v 2 = 3t + c
2
➜
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When t = 0, v = 2 so c = 2, which gives
9
1 2
2 v = 3t + 2
v 2 = 6t + 4
Since the direction of motion is positive
9 DIFFERENTIAL EQUATIONS
v=
6t + 4
(iii) When v = 8: 64 = 6t + 4
60 = 6t
⇒
t = 10
The object takes 10 seconds to reach a velocity of 8 m s−1.
The graph of the particular solution is shown in Figure 9.7.
v
v = 6t + 4
2
O
The remainder of the
curve for t " 0 and
v " 2 is not shown as
it is not relevant to the
situation.
t
% Figure 9.7
Sometimes you will be asked to verify the solution of a differential equation.
In that case you are expected to do two things:
» substitute the solution in the differential equation and show that
it works
» show that the solution fits the conditions you have been given.
Example 9.9
Show that sin y = x is a solution of the differential equation
dy
1
=
dx
1 − x2
given that y = 0 when x = 0.
Solution
⇒
⇒
sin y = x
dy
cos y
=1
dx
dy
= 1
dx cos y
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Substituting into the differential equation
LHS:
RHS:
1
cos y
1
1
=
= 1
1− x2
1 − sin 2 y cos y
dy
1 :
=
dx
1 − x2
Substituting x = 0 into the solution sin y = x gives sin y = 0 and this is
satisfied by y = 0.
So the solution also fits the particular conditions.
Exercise 9C
1
M
2
Find the particular solution of each of the following differential equations.
dy
(i)
= x 2 − 1 y = 2 when x = 3
dx
dy
(ii)
= x2y
y = 1 when x = 0
dx
dy
(iii)
= x e–y
y = 0 when x = 0
dx
dy
(iv)
= y2
y = 1 when x = 1
dx
dy
(v)
= x(y + 1) y = 0 when x = 1
dx
dy
(vi)
= y 2 sin x y = 1 when x = 0
dx
A cold liquid at temperature θ °C, where θ < 20, is standing in a warm
room. The temperature of the liquid obeys the differential equation
dθ
dt = 2(20 − θ )
where the time t is measured in hours.
9.2 Solving differential equations
So the solution fits the differential equation.
9
Find the general solution of this differential equation.
(ii) Find the particular solution for which θ = 5 when t = 0.
(iii) In this case, how long does the liquid take to reach a temperature
of 18°C?
A population of rabbits increases so that the number of rabbits N
(in hundreds), after t years is modelled by the differential equation
dN = N.
dt
(i) Find the general solution for N in terms of t.
(ii) Find the particular solution for which N = 10 when t = 0.
(iii) What will happen to the number of rabbits when t becomes very
large? Why is this not a realistic model for an actual population of
rabbits?
(i)
M
3
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9
4
9 DIFFERENTIAL EQUATIONS
5
( )
An object is moving so that its velocity v = ds is inversely proportional
dt
to its displacement s from a fixed point.
If its velocity is 1 m s−1 when its displacement is 2 m, find a differential
equation to model the situation.
Find the general solution of your differential equation.
Given that y = 1 when x = 0, solve the differential equation
dy
= 4 x(3y 2 + 10y + 3),
dx
obtaining an expression for y in terms of x.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q7 June 2015
6
The temperature of a quantity of liquid at time t is θ. The liquid is
cooling in an atmosphere whose temperature is constant and equal to A.
The rate of decrease of θ is proportional to the temperature differerence
(θ − A). Thus θ and t satisfy the differential equation
dθ = −k(θ − A),
dt
where k is a positive constant.
(i) Find, in any form, the solution of this differential equation, given
that θ = 4A when t = 0.
(ii) Given also that θ = 3A when t = 1, show that k = ln 23.
(iii) Find θ in terms of A when t = 2, expressing your answer in its
simplest form.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q9 November 2009
7
The variables x and t are related by the differential equation
e 2t dx = cos 2 x,
dt
where t % 0. When t = 0, x = 0.
(i) Solve the differential equation, obtaining an expression for x in
terms of t.
(ii) State what happens to the value of x when t becomes very large.
(iii) Explain why x increases as t increases.
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q7 June 2010
8
An underground storage tank is being filled with liquid as shown in the
diagram. Initially the tank is empty. At time t hours after filling begins,
the volume of liquid is V m3 and the depth of liquid is h m. It is given
that V = 43 h 3.
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9
dh = 5 − 1 .
dt h 2 20
2000
20h 2 ≡ −20 +
(ii) Verify that
.
(10 − h )(10 + h )
100 − h 2
(iii) Hence solve the differential equation in part (i), obtaining an
expression for t in terms of h.
9.2 Solving differential equations
The liquid is poured in at a rate of 20 m3 per hour, but owing to leakage,
liquid is lost at a rate proportional to h2. When h = 1, dh = 4.95..
dt
(i) Show that h satisfies the differential equation
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q8 November 2008
9
Naturalists are managing a wildlife reserve to increase the number of
plants of a rare species. The number of plants at time t years is denoted
by N, where N is treated as a continuous variable.
(i) It is given that the rate of increase of N with respect to t is
proportional to ( N − 150). Write down a differential equation
relating N, t and a constant of proportionality.
(ii) Initially, when t = 0, the number of plants was 650. It was noted
that, at a time when there were 900 plants, the number of plants
was increasing at a rate of 60 per year. Express N in terms of t.
(iii) The naturalists had a target of increasing the number of plants from
650 to 2000 within 15 years.
Will this target be met?
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q10 November 2015
10 The number of birds of a certain species in a forested region is recorded
over several years. At time t years, the number of birds is N, where N is
treated as a continuous variable. The variation in the number of birds is
modelled by
dN = N (1800 − N ) .
3600
dt
It is given that N = 300 when t = 0.
(i) Find an expression for N in terms of t.
(ii) According to the model, how many birds will there be after a long time?
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q10 June 2011
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231
02/02/18 1:15 PM
9
11 The variables x and θ satisfy the differential equation
2cos 2 θ dx = (2x + 1),
dθ
and x = 0 when θ = 41 π. Solve the differential equation and obtain an
expression for x in terms of θ.
9 DIFFERENTIAL EQUATIONS
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q5 June 2014
12 The variables x and y are related by the differential equation
(i)
(ii)
dy 1 21
= xy sin ( 13 x ) .
dx 5
Find the general solution, giving y in terms of x.
Given that y = 100 when x = 0, find the value of y when x = 25.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q8 November 2014
13
A tank containing water is in the form of a cone with vertex C. The axis
is vertical and the semi-vertical angle is 60°, as shown in the diagram. At
time t = 0, the tank is full and the depth of water is H. At this instant, a
tap at C is opened and water begins to flow out. The volume of water in
the tank decreases at a rate proportional to h, where h is the depth of
water at time t. The tank becomes empty when t = 60.
60° h
C
Show that h and t satisfy a differential equation of the form
dh = − Ah − 23 ,
dt
where A is a positive constant.
(ii) Solve the differential equation given in part (i) and obtain an
expression for t in terms of h and H.
1
(iii) Find the time at which the depth reaches H .
2
[The volume V of a cone of vertical height h and base radius r is
(i)
given by V = 13 πr 2h .]
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q10 November 2013
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INVESTIGATION
Cooling
Investigate the coffee cooling problem introduced on page 217.You will
need to make some assumptions about the initial temperature of the coffee
and the temperature of the room.
Would it be better to allow the coffee to cool first before adding the milk?
KEY POINTS
1
2
3
4
5
A differential equation is an equation involving derivatives such as
dy
d 2y
and
x2
d
dx
A first-order differential equation involves a first derivative only.
Some first-order differential equations may be solved by separating
the variables.
A general solution is one in which the constant of integration is left
in the solution, and a particular solution is one in which additional
information is used to calculate the constant of integration.
A general solution may be represented by a family of curves, a
particular solution by a particular member of that family.
9.2 Solving differential equations
What difference would it make if you were to add some cold milk to the
coffee and then leave it to cool?
9
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■ formulate first-order differential equations using information about
rates of change
■ solve a first-order differential equation using separation of variables
■ find the general solution of a first-order differential equation
■ find the particular solution of a first-order differential equation
■ solve problems using differential equations and interpret the solution.
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P3
10 VECTORS
10
Vectors
We drove into
the future
looking into
a rear view
mirror.
Herbert
Marshall
McLuhan
(1911–1980)
?
❯ What information do you need to decide how closely the aircraft
which left these vapour trails passed to each other?
A quantity which has both size and direction is called a vector. The
velocity of an aircraft through the sky is an example of a vector, having size
(e.g. 600 mph) and direction (on a course of 254°). By contrast the mass of
the aircraft (100 tonnes) is completely described by its size and no direction
is associated with it; such a quantity is called a scalar.
Vectors are used extensively in mechanics to represent quantities such as
force, velocity and momentum, and in geometry to represent displacements.
They are an essential tool in three-dimensional coordinate geometry and it
is this application of vectors which is the subject of this chapter. However,
before coming on to this, you need to be familiar with the associated
vocabulary and notation, in two and three dimensions.
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10.1 Vectors in two dimensions
Terminology
The vector in Figure 10.1 has magnitude 5, direction +30°. This is written
(5, 30°) and said to be in magnitude–direction form or in polar form.
The general form of a vector written in this way is (r, θ ) where r is its
magnitude and θ its direction.
5
+
10.1 Vectors in two dimensions
In two dimensions, it is common to represent a vector by a drawing of a
straight line with an arrowhead. The length represents the size, or magnitude,
of the vector and the direction is indicated by the line and the arrowhead.
Direction is usually given as the angle the vector makes with the positive
x-axis, with the anticlockwise direction taken to be positive.
10
30°
▲ Figure 10.1
Note
In the special case when the vector is representing real travel, as in the case
of the velocity of an aircraft, the direction may be described by a compass
bearing with the angle measured from north, clockwise. However, this is
not done in this chapter, where directions are all taken to be measured
anticlockwise from the positive x direction.
An alternative way of describing a vector is in terms of components in
given directions. The vector in Figure 10.2 is 4 units in the x direction, and
4
2 in the y direction, and this is denoted by ⎛ ⎞ .
⎝ 2⎠
))
4
or 4i + 2j
2
2
4
▲ Figure 10.2
j
This may also be written as 4i + 2j, where i is a vector of magnitude 1,
a unit vector, in the x direction and j is a unit vector in the y direction
(Figure 10.3).
i
▲ Figure 10.3
In a book, a vector may be printed in bold, for example p or OP, or as a
line between two points with an arrow above it to indicate its direction,
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⎯
→
such as OP. When you write a vector by hand, it is usual to underline it, for
⎯
→
example, p or OP, or to put an arrow above it, as in O P.
10 VECTORS
10
The magnitude of a vector is its length or size. The magnitude of a
vector is also called its modulus and denoted by the symbols ||. You can use
Pythagoras’ theorem to find the magnitude. Another convention for writing
the magnitude of a vector is to use the same letter, but in italics and not bold
type; so the magnitude of a may be written a.
Example 10.1
Find the magnitude and direction of the vector a = 4i + 2j.
Solution
a
2
θ
4
▲ Figure 10.4
The magnitude of a is given by the length a in Figure 10.4.
a =
42 + 22
= 4.47
using Pythagoras’ theorem
(to 3 significant figures)
The direction is given by the angle θ.
tan θ = 42 = 0.5
θ = 26.6°
(to 3 significant figures)
The vector a is (4.47, 26.6°).
10.2 Vectors in three dimensions
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Points
In three dimensions, a point has three coordinates, usually called x, y and z.
z
This point is
(3, 4, 1).
2
–1
–2
–1
2
1
O
1
2
3
4
y
P
–1
3
x
▲ Figure 10.5
The axes are conventionally arranged as shown in Figure 10.5, where the
point P is (3, 4, 1). Even on correctly drawn three-dimensional grids, it is
often hard to see the relationship between the points, lines and planes, so it is
seldom worth your while trying to plot points accurately.
10.2 Vectors in three dimensions
1
–3
10
The unit vectors i, j and k are used to describe vectors in three dimensions.
Equal vectors
The statement that two vectors a and b are equal means two things.
» The direction of a is the same as the direction of b.
» The magnitude of a is the same as the magnitude of b.
If the vectors are given in component form, each component of a equals the
corresponding component of b.
Position vectors
Saying the vector a is given by 3i + 4j + k tells you the components of the
vector, or equivalently its magnitude and direction. It does not tell you where
the vector is situated; indeed it could be anywhere.
All of the lines in Figure 10.6 represent the vector a.
a
a
a
a
k
i
▲ Figure 10.6
9781510421738.indb 237
j
237
02/02/18 1:15 PM
There is, however, one special case which is an exception to the rule, that of
a vector which starts at the origin. This is called a position vector. Thus
⎛ 3⎞
the line joining the origin to the point P(3, 4, 1) is the position vector ⎜ 4 ⎟
⎜⎝ 1 ⎟⎠
or 3i + 4j + k.
10
Another way of expressing this is to say that the point P(3, 4, 1) has the
10 VECTORS
⎛ 3⎞
position vector ⎜ 4 ⎟ .
⎜⎝ 1 ⎟⎠
Displacement vectors
Vectors can also be used to represent displacement. For example, start at
A and walk 5 km east and then 12 km north to point B can be represented
⎛ 5⎞
by the vector AB = ⎜ ⎟ . Note that magnitude of displacement is not the
⎝ 12 ⎠
same as distance – the distance travelled is 5 km + 12 km = 17 km; whereas the
magnitude of displacement is 5 2 + 12 2 = 13 km.
Example 10.2
Points L, M and N have coordinates (4, 3), (−2, −1) and (2, 2).
(i)
Write down, in component form, the position vector of L and the vector
⎯→
MN.
(ii)
What do your answers to part (i) tell you about the lines OL and MN?
Solution
(i)
⎯
→
4
The position vector of L is OL = ⎛ ⎞ .
3⎠
⎝
⎯→
4
The vector MN is also ⎛ ⎞ (see Figure 10.7).
⎝ 3⎠
y
4
L
3
N
2
1
–2
M
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(ii)
–1
O
1
2
3
4
x
–1
▲ Figure 10.7
⎯
→ ⎯→
Since OL = MN, lines OL and MN are parallel and equal in length.
02/02/18 1:15 PM
Note
10
A line joining two points, like MN in Figure 10.7, is often called a line
segment, meaning that it is just that particular part of the infinite straight
line that passes through those two points.
⎯→
The length of a vector
In two dimensions, the use of Pythagoras’ theorem leads to the result that a
vector a1i + a2 j has length | a | given by
| a | = a12 + a 22 .
?
CP
❯ Show that the length of the three-dimensional vector a1i + a2 j + a3k
is given by
|a|=
Example 10.3
10.2 Vectors in three dimensions
The vector MN is an example of a displacement vector. Its length represents
the magnitude of the displacement when you move from M to N.
a12 + a 22 + a 32 .
⎛ 2⎞
Find the magnitude of the vector a = ⎜⎜ −5 ⎟⎟ .
⎝ 3⎠
Solution
|a| =
22 + (−5)2 + 32
=
4 + 25 + 9
=
38
= 6.16 (to 2 d.p.)
Exercise 10A
1
Express the following vectors in component form.
(i)
(ii)
y
3
3
a
2
2
1
1
–2 –1 0
–1
y
1
2
3
4
x
–2 –1 0
–1
b
1
2
3
x
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(iii)
10
(iv)
y
3
3
c
2
10 VECTORS
2
1
1
2
−i + 2j
⎛ 1⎞
⎜ –2⎟
⎜⎝ 3⎟⎠
5
0
x
(v)
1
2
3
4
x
3i − 4j
⎛ 4⎞
(ii) ⎜ 0⎟
(iii)
2i + 4j + 2k
⎛ 6⎞
⎜ –2⎟
⎜⎝ −3⎟⎠
(vi)
i − 2k
⎜⎝ −2⎠⎟
(iv) i + j − 3k
PS
4
Find the magnitude of these vectors.
(i)
4
3
Draw diagrams to show each of these vectors and find the
magnitude and direction.
⎛ 3⎞
⎛ –4 ⎞
(i) 2i + 3j
(ii)
(iii)
⎝ –2⎠
⎝ –4 ⎠
(iv)
3
d
2
1
0
y
(v)
Write, in component form, the vectors represented by the line
segments joining the following points.
(i) (2, 3) to (4, 1)
(ii)
(4, 0) to (6, 0)
(iii) (0, 0) to (0, −4)
(iv) (0, −4) to (0, 0)
(v) (0, 0, 0) to (0, 0, 5)
(vi) (0, 0, 0) to (−1, −2, 3)
(vii) (−1, −2 , 3) to (0, 0, 0)
(viii) (0, 2, 0) to (4, 0, 4)
(ix) (1, 2, 3) to (3, 2, 1)
(x)
(4, −5, 0) to (−4, 5, 1)
The points A, B and C have coordinates (2, 3), (0, 4) and (−2, 1).
(i) Write down the position vectors of A and C.
(ii) Write down the vectors of the line segments joining AB
and CB.
(iii) What do your answers to parts (i) and (ii) tell you about
(a) AB and OC
(b)
CB and OA?
(iv) Describe the quadrilateral OABC.
10.3 Vector calculations
Multiplying a vector by a scalar
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When a vector is multiplied by a number (a scalar) its length is
altered but its direction remains the same.
02/02/18 1:15 PM
The vector 2a in Figure 10.8 is twice as long as the vector a but in the same
direction.
a
10
2a
When the vector is in component form, each component is multiplied by the
number. For example:
2 × (3i − 5j + k) = 6i − 10j + 2k
⎛ 3⎞ ⎛ 6 ⎞
2 × ⎜ –5⎟ = ⎜ –10 ⎟ .
⎜ ⎟ ⎜
⎟
⎝ 1⎠ ⎝ 2 ⎠
10.3 Vector calculations
▲ Figure 10.8
The negative of a vector
In Figure 10.9 the vector −a has the same length as the vector a but the
opposite direction.
a
–a
▲ Figure 10.9
When a is given in component form, the components of −a are the same as
those for a but with their signs reversed. So
⎛ 23⎞ ⎛ –23⎞
– ⎜ 0⎟ = ⎜ 0⎟
⎜ ⎟ ⎜
⎟
⎝ –11⎠ ⎝ +11⎠
Adding vectors
When vectors are given in component form, they can be added component
by component. This process can be seen geometrically by drawing them on
graph paper, as in Example 10.4 on the next page.
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10
Example 10.4
Add the vectors 2i − 3j and 3i + 5j.
Solution
10 VECTORS
2i − 3j + 3i + 5j = 5i + 2j
5i + 2j
2i
3i + 5j
5j
–3j
2i – 3j
3i
▲ Figure 10.10
The sum of two (or more) vectors is called the resultant and is usually
indicated by being marked with two arrowheads.
Adding vectors is like adding the legs of a journey to find its overall outcome
(see Figure 10.11).
resultant
leg 1
leg 3
leg 2
▲ Figure 10.11
When vectors are given in magnitude−direction form, you can find their
resultant by making a scale drawing, as in Figure 10.11. If, however, you
need to calculate their resultant, it is usually easiest to convert the vectors
into component form, add component by component, and then convert the
answer back to magnitude−direction form.
Subtracting vectors
Subtracting one vector from another is the same as adding the negative of the
vector.
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Example 10.5
Two vectors a and b are given by
10
a = 2i + 3j b = −i + 2j.
(i)
Find a − b.
(ii)
Draw diagrams showing a, b, a − b.
(i)
10.3 Vector calculations
Solution
a − b = (2i + 3j) − (−i + 2j)
= 3i + j
(ii)
b
–b
a
a
j
a + (–b) = a – b
i
▲ Figure 10.12
When you find the vector
represented by the line segment
joining two points, you are in
effect subtracting their position
vectors. If, for example,
P is the point (2, 1) and Q is the
⎯
→
1
point (3, 5), PQ is ⎛ 4 ⎞ , as
⎝ ⎠
Figure 10.13 shows.
You find this by saying
⎯
→
⎯
→
⎯
→
PQ = PO + OQ = −p + q.
In this case, this gives
⎯
→
2
3
1
PQ = – ⎛ 1 ⎞ + ⎛ 5⎞ = ⎛ 4 ⎞
⎝ ⎠ ⎝ ⎠
⎝ ⎠
as expected.
This is an important result:
y
6
Q(3, 5)
5
4
))
1
4
3
2
1
0
P(2, 1)
1
2
3
4
5
x
▲ Figure 10.13
⎯
→
PQ = q − p
where p and q are the position vectors of P and Q.
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Geometrical figures
10
It is often useful to be able to express lines in a geometrical figure in terms of
given vectors.
ACTIVITY 10.1
10 VECTORS
The diagram shows a cuboid OABCDEFG. P, Q, R, S and T are the
midpoints of the edges they lie on. The origin is at O and the axes lie
along OA, OC and OD, as shown in Figure 10.14.
⎛ 6⎞ ⎯→ ⎛ 0⎞ ⎯→ ⎛ 0 ⎞
OA = ⎜ 0⎟ , OC = ⎜ 5⎟ , OD = ⎜ 0 ⎟
⎜⎝ 0⎟⎠
⎜⎝ 0⎠⎟
⎜⎝ 4 ⎟⎠
⎯
→
S
G
F
T
R
D
E
z
B
C
y
O
Q
x
A
P
▲ Figure 10.14
(i)
(ii)
Example 10.6
Name the points with the following coordinates.
(a)
(6, 5, 4)
(b) (0, 5, 0)
(d)
(0, 2.5, 4)
(e) (3, 5, 4)
(c) (6, 2.5, 0)
Use the letters in the diagram to give displacements which are equal
to the following vectors. Give all possible answers; some of them have
more than one.
⎛ 6⎞
⎛ 6⎞
⎛ 0⎞
⎛ −6⎞
⎛ −3 ⎞
(a) ⎜ 5 ⎟
(b) ⎜ 0 ⎟
(c) ⎜ 5 ⎟
(d) ⎜ −5⎟
(e) ⎜ 2.5⎟
⎜⎝ 4 ⎟⎠
⎜⎝ 4 ⎟⎠
⎜⎝ 4 ⎟⎠
⎜⎝ 4 ⎟⎠
⎜⎝ 4 ⎟⎠
Figure 10.15 shows a hexagonal prism.
G
H
r
B
q
C
I
p
A
D
J
F
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E
▲ Figure 10.15
02/02/18 1:15 PM
⎯
→
⎯
→
The hexagonal cross-section is regular and consequently A D = 2BC.
⎯
→
⎯
→
10
⎯
→
AB = p, BC = q and BG = r. Express the following in terms of p, q and r.
⎯
→
⎯
→
⎯
→
→
(i)
AC
(ii)
AD
(iii)
HI
(iv)
(v)
EF
(vi)
BE
(vii)
AH
(viii) FI
⎯
→
(i)
(ii)
⎯
→
⎯→
⎯→
q
B
⎯
→
⎯
→
AC = AB + BC
=p+q
⎯
→
⎯
→
⎯
→
⎯
→
p
AD = 2BC = 2q
(iii) H I = CD
⎯
→
⎯
→
C
p+q
A
C
⎯
→
Since AC + CD = AD
⎯
→
10.3 Vector calculations
Solution
⎯
→
IJ
p+q
p + q + CD = 2q
⎯
→
CD = q − p
⎯
→
So
→
HI = q − p
⎯
→
(iv) I J = DE
⎯
→
= −AB
= −p
D
2q
B
⎯
→
(v)
A
C
⎯
→
E F = −BC
= −q
⎯
→
D
⎯
→
⎯
→
⎯
→
(vi) BE = BC + CD + DE
= q + (q − p) + −p
= 2q − 2p
⎯
→
⎯
→
⎯
→
⎯→
Notice that BE = 2CD.
⎯→
⎯
→
(vii) AH = AB + BC + CH
E
▲ Figure 10.16
⎯
→ ⎯
→
CH = B G
=p+q+r
→
⎯
→
→
→
(viii) FI = FE + EJ + J I
⎯→ ⎯
→ →
⎯
⎯
→ → ⎯
→
F E = B C, E J = B G, J I = AB
=q+r+p
Unit vectors
A unit vector is a vector with a magnitude of 1, like i and j. To find the
unit vector in the same direction as a given vector, divide that vector by its
magnitude.
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Thus the vector 3i + 5j (in Figure 10.17) has magnitude 32 + 52 = 34 , and
3
5
so the vector
j is a unit vector. It has magnitude 1.
i+
34
34
The unit vector in the direction of vector a is written as â and read as ‘a hat’.
10
y
10 VECTORS
5j
4j
3j
3i + 5j
This is the unit vector
3 i +
34
2j
5 j
34
j
O
2i
i
3i
4i
x
▲ Figure 10.17
Example 10.7
Relative to an origin O, the position vectors of the points A, B and C are
given by
⎛ 0⎞
⎛ −2⎞
⎯
→ ⎛ −2 ⎞ ⎯
→
⎯
→
OA = ⎜ 3⎟ , OB = ⎜ 1⎟ and OC = ⎜ 3⎟ .
⎜⎝ −3⎠⎟
⎝⎜ −2⎟⎠
⎝⎜ 1⎟⎠
⎯
→
(i) Find the unit vector in the direction AB.
(ii)
Find the perimeter of triangle ABC.
Solution
⎯
→
⎯
→
⎯
→
For convenience call OA = a, OB = b and OC = c.
⎛ 0⎞ ⎛ −2⎞
⎛ 2⎞
⎯
→
(i) AB = b − a = ⎜ 1⎟ − ⎜ 3⎟ = ⎜ −2⎟
⎜⎝ −3⎟⎠ ⎜⎝ −2⎟⎠
⎜⎝ −1⎟⎠
⎯
→
⎯
→
To find the unit vector in the direction AB, you need to divide AB by
its magnitude.
| A⎯→B | =
=
22 + (−2)2 + (−1)2
This is the
9
⎯
→
magnitude of A B.
= 3
2
⎛ 2⎞ ⎛ 3 ⎞
So the unit vector in the direction AB is 13 ⎜ −2⎟ = ⎜ − 23 ⎟
⎜ −1⎟ ⎜ 1 ⎟
⎝ ⎠ ⎝ − 3⎠
⎯
→
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9781510421738.indb 246
02/02/18 1:15 PM
(ii)
The perimeter of the triangle is given by | AB | + | AC | + | B C |.
⎯
→
⎯
→
⎯
→
10
⎛ −2⎞ ⎛ −2⎞
⎛ 0⎞
⎯
→
AC = c − a = ⎜ 3⎟ − ⎜ 3⎟ = ⎜ 0⎟
⎜⎝ 1⎟⎠ ⎜⎝ −2⎟⎠
⎜⎝ 3⎟⎠
⇒
| A⎯→C | =
02 + 02 + 32
10.3 Vector calculations
=3
⎛ −2⎞ ⎛ 0⎞
⎛ −2⎞
⎯
→
BC = c − b = ⎜ 3⎟ − ⎜ 1⎟ = ⎜ 2⎟
⎜⎝ 1⎟⎠ ⎜⎝ −3⎟⎠
⎜⎝ 4 ⎟⎠
⇒
| B⎯→C | =
( −2) 2 + 2 2 + 4 2
=
24
Perimeter of ABC = | AB | + | AC | + | B C |
⎯
→
⎯
→
=3+3+
24
⎯
→
= 10.9
Example 10.8
A
Figure 10.18 shows triangle AOB.
C is a point on AB and divides it in the
ratio 2 : 3.
C
a
⎯→
Find OC in terms of the vectors a and b.
B
b
O
▲ Figure 10.18
Solution
⎯→
When you divide AB in the ratio 2 : 3 then
⎯→
2 ⎯→
2
AC is 2 + 3 = 25 of AB and CB is
OC = OA + 5 AB
⎯→
⎯→
OA = a and AB = b − a
⎯→
OC = a + 2 ( b − a )
3 = 3 of AB.
2+3 5
5
= a + 2b− 2a
5
5
= 3a + 2b
5
5
The above example made use of the ratio theorem.
❯ Prove that when C divides AB in the ratio s : t then
⎯→
OC =
?
CP
t a+ s b
s+t
s+t
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1
Simplify the following.
⎛ 2⎞ + ⎛ 4 ⎞
(i)
⎝ 3⎠ ⎝ 5 ⎠
⎛ 3⎞ + ⎛ –3 ⎞
(iii)
⎝ 4 ⎠ ⎝ –4 ⎠
(v)
10 VECTORS
10
Exercise 10B
2
(ii)
⎛ 2⎞ + ⎛ –1⎞
⎝ –1⎠ ⎝ 2⎠
2
1
(iv) 3 ⎛ ⎞ + 2 ⎛ ⎞
⎝1⎠
⎝ –2⎠
6(3i − 2j) − 9(2i − j)
The vectors p, q and r are given by
p = 3i + 2j + k
q = 2i + 2j + 2k
r = −3i − j − 2k.
Find, in component form, the following vectors.
(i) p + q + r
(ii) p − q
(iv) 3(p − q) + 2(p + r)
(v) 4p − 3q + 2r
CP
3
⎯
→
(iii)
p+r
⎯
→
In the diagram, PQRS is a parallelogram and PQ = a, PS = b.
(i) Write, in terms of a and b,
Q
the following vectors.
⎯
→
⎯
→
(a) QR
(b)
PR
(c)
⎯
→
R
a
QS
The midpoint of PR is M. Find
⎯
→
⎯
→
(a) PM
(b) QM.
P
(iii) Explain why this shows you that the
diagonals of a parallelogram bisect each other.
In the diagram, ABCD is a kite.
AC and BD meet at M.
⎯
→
⎯
→
AB = i + j and AD = i − 2j
A
(i) Use the facts that the diagonals
j
of a kite meet at right angles
and that M is the midpoint of
i
AC to find, in terms of i and j,
⎯
→
⎯
→
(a) AM
(b)
AC
(ii)
CP
4
(c)
(ii)
CP
248
9781510421738.indb 248
5
⎯
→
BC
(d)
⎯
→
CD.
S
b
B
M
C
D
Verify that | A B | = | BC | and | AD | = | CD |.
⎯
→
⎯
→
⎯
→
⎯
→
A
In the diagram, ABC is a triangle.
L, M and N are the midpoints of
the sides BC, CA and AB.
⎯
→
⎯
→
A B = p and AC = q.
M
N
(i) Find, in terms of p and q,
⎯
→ ⎯→ ⎯→
⎯
→
BC, MN, L M and LN.
(ii) Explain how your results
B
L
from part (i) show you that
the sides of triangle LMN are parallel to those of triangle ABC,
and half their lengths.
C
02/02/18 1:15 PM
6
7
10
⎛ −2⎞
⎛ 4⎞
(iv) ⎜ 4 ⎟
(v) 5i – 3j + 2k (vi) ⎜ 0⎟
⎜⎝ −3⎟⎠
⎜⎝ 0⎟⎠
Relative to an origin O, the position vectors of the points A, B and C are
given by
⎛ −2⎞
⎛ −1⎞
⎯
→ ⎛ 2⎞ ⎯
→
⎯
→
OA = ⎜ 1⎟ , OB = ⎜ 4 ⎟ and OC = ⎜ 2⎟ .
⎜⎝ 1⎟⎠
⎜⎝ 3⎟⎠
⎜⎝ 3⎟⎠
9
Find the perimeter of triangle ABC.
Relative to an origin O, the position vectors of the points P and Q are
given by
⎯
→
CP
10.3 Vector calculations
8
Find unit vectors in the same directions as the following vectors.
2
–2
(i) ⎛ ⎞
(ii) 3i + 4j
(iii) ⎛ ⎞
(iv) 5i − 12j
3
⎝ ⎠
⎝ –2⎠
Find unit vectors in the same direction as the following vectors.
⎛ 1⎞
(i) ⎜ 2⎟
(ii) 2i – 2j + k (iii) 3i – 4k
⎜⎝ 3⎟⎠
⎯
→
O P = 3i + j + 4k and OQ = i + xj − 2k.
Find the values of x for which the magnitude of PQ is 7.
⎯→
⎯→
⎯→
10 In the cuboid, OA = p, OE = q , OG = r .
B
(i) Express the following vectors
in terms of p, q and r.
⎯→
⎯→
A
(a) GF
(b) CF
⎯→
(c)
OB
(e)
OC
⎯→
M
⎯→
C
D
(d) OD
p
E
q
F
The point M divides
O
r
G
AD in the ratio 3 : 2.
⎯→
Find OM in terms of p, q and r.
(iii) Use vectors to prove that OC and BG bisect each other.
Relative to an origin O, the position vectors of the points A, B, C and D
are given by
(ii)
CP
11
⎛ 6⎞
⎛ 5⎞
⎛ 8⎞
⎛ 3⎞
⎯→
⎯→
⎯→
⎟
⎜
5 , OC = ⎜ 2 ⎟ , and OD = ⎜ −2 ⎟
OA = ⎜ 1 ⎟ , OB =
⎜ −1 ⎟
⎟
⎜
⎜ 7⎟
⎜ 5⎟
⎝ ⎠
⎝ ⎠
⎠
⎝
⎝ 13 ⎠
⎯→
Use vectors to prove that ABCD is a parallelogram.
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10
12
Relative to an origin O, the position vectors of the points A and B are
given by
⎛ 4⎞
OA = ⎜ 1⎟
⎜⎝ −2⎟⎠
⎯
→
⎛ 3⎞
OB = ⎜ 2⎟ .
⎜⎝ –4 ⎟⎠
⎯
→
and
⎯
→
⎯
→
Given that C is the point such that AC = 2AB, find the unit vector
⎯
→
in the direction of OC.
⎛ 1⎞
⎯
→
The position vector of the point D is given by OD = ⎜ 4 ⎟ , where k is a
⎜⎝ k ⎟⎠
10 VECTORS
(i)
⎯
→
⎯
→
⎯
→
constant, and it is given that OD = mOA + nOB, where m and n are
constants.
(ii) Find the values of m, n and k.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q9 June 2007
10.4 The angle between two vectors
?
CP
❯ As you work through the proof in this section, make a list of all the
results that you are assuming.
To find the angle θ between the two vectors
⎯
→
OA = a = a1i + a2 j
and
y
B
(b1, b2)
⎯
→
OB = b = b1i + b2 j
A
start by applying the cosine rule to
triangle OAB in Figure 10.19.
(a1, a2)
OA2 + OB2 – AB2
2OA × OB
a
In this, OA, OB and AB are the
⎯
→ ⎯
→
lengths of the vectors OA, OB and
⎯
→
AB, and so
θ
cos θ =
OA = | a | =
⎯
→
a21 + a22
b
O
x
▲ Figure 10.19
and
OB = | b | =
b21 + b22 .
The vector AB = b − a = (b1i + b2 j) − (a1i + a2 j)
= (b1 − a1)i + (b2 − a2)j
and so its length is given by
AB = | b − a | =
(b1 – a1)2 + (b2 – a2)2.
250
9781510421738.indb 250
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Substituting for OA, OB and AB in the cosine rule gives
cos θ =
=
2 a21 + a22 ×
b21 + b22
a21 + a22 + b21 + b22 – (b21 – 2a1b1 + a21 + b22 – 2a2b2 + a22 )
2 |a || b|
cos θ =
a b + a2 b2
2a1b1 + 2a2b2
= 11
| a || b |
2 | a || b |
The expression on the top line, a1b1 + a2b2, is called the scalar product
(or dot product) of the vectors a and b and is written a .b. Thus
a.b
cos θ =
.
|a || b|
This result is usually written in the form
a .b = | a | | b | cos θ.
The next example shows you how to use it to find the angle between two
vectors given numerically.
10.4 The angle between two vectors
This simplifies to
Example 10.9
10
(a12 + a 22) + (b21 + b22) – [(b1 – a1)2 + (b2 – a2)2]
5
3
Find the angle between the vectors ⎛ 4 ⎞ and ⎛ –12⎞ .
⎝ ⎠
⎝
⎠
Solution
Let
3
a = ⎛ 4⎞
⎝ ⎠
⇒
|a|=
32 + 42 = 5
and
5
b = ⎛ –12 ⎞
⎝
⎠
⇒
|b|=
52 + (–12)2 = 13.
The scalar product
⎛3⎞ ⎛ 5 ⎞
⎜ ⎟ ⋅⎜
⎟
⎝4 ⎠ ⎝ –12 ⎠ = 3 × 5 + 4 × (−12)
= 15 − 48
= −33
Substituting in a .b = | a | | b | cos θ gives
⇒
−33 = 5 × 13 × cos θ
cos θ = –33
65
θ = 120.5°
Perpendicular vectors
Since cos 90° = 0, it follows that if vectors a and b are perpendicular then
a .b = 0.
Conversely, if the scalar product of two non-zero vectors is zero, they are
perpendicular.
9781510421738.indb 251
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02/02/18 1:15 PM
10
Example 10.10
2
6
Show that the vectors a = ⎛ 4 ⎞ and b = ⎛ –3 ⎞ are perpendicular.
⎝ ⎠
⎝ ⎠
Solution
10 VECTORS
The scalar product of the vectors is
⎛2⎞ ⎛ 6⎞
a.b = ⎜ ⎟ . ⎜ ⎟
⎝4 ⎠ ⎝ –3 ⎠
= 2 × 6 + 4 × (−3)
= 12 − 12 = 0
Therefore the vectors are perpendicular.
Further points concerning the scalar product
» You will notice that the scalar product of two vectors is an ordinary
number. It has size but no direction and so is a scalar, rather than a vector.
It is for this reason that it is called the scalar product. There is another way
of multiplying vectors that gives a vector as the answer; it is called the
vector product. This is beyond the scope of this book.
» The scalar product is calculated in the same way for three-dimensional
vectors. For example:
⎛ 2⎞ ⎛ 5⎞
⎜ 3⎟ . ⎜ 6⎟ = 2 × 5 + 3 × 6 + 4 × 7 = 56
⎜⎝ 4 ⎟⎠ ⎜⎝ 7⎟⎠
In general
⎛ a1 ⎞ ⎛ b1 ⎞
⎜ a2 ⎟ . ⎜ b2 ⎟ = a1b1 + a2b2 + a3b3
⎜⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
3
3
» The scalar product of two vectors is commutative. It has the same value
whichever of them is on the left-hand side or right-hand side. Thus
a .b = b .a, as in the following example.
⎛ 2⎞ . ⎛ 6⎞ = 2 × 6 + 3 × 7 = 33
⎝ 3⎠ ⎝ 7⎠
⎛ 6⎞ . ⎛ 2⎞ = 6 × 2 + 7 × 3 = 33.
⎝ 7⎠ ⎝ 3⎠
❯ How would you prove this result?
?
CP
252
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The angle between two vectors in three dimensions
10
The angle θ between the vectors a = a1i + a2 j and b = b1i + b2 j in two
dimensions is given by
a1b1 + a2b2
a.b
=
cos θ =
|
|| b|
a
a12 + a22 × b12 + b22
❯ Show that the angle between the three-dimensional vectors
a = a1i + a2 j + a3k
?
CP
b = b1i + b2 j + b3k
and
is also given by
cos θ =
a.b
| a || b |
but that the scalar product a .b is now
a .b = a1b1 + a2b2 + a3b3.
10.4 The angle between two vectors
where a .b is the scalar product of a and b. This result was proved by using
the cosine rule on page 250.
When working in two dimensions you found the angle between two lines by
using the scalar product. As you have just proved, this method can be extended
into three dimensions, and its use is shown in the following example.
Example 10.11
The points P, Q and R are (1, 0, −1), (2, 4, 1) and (3, 5, 6). Find ∠QPR.
Solution
⎯
→
⎯
→
The angle between PQ and PR is given by θ in
cos θ =
⎯→
⎯→
⎯→
⎯→
PQ . PR
|PQ||PR|
In this
⎛ 2 ⎞ ⎛ 1⎞ ⎛ 1⎞
⎯→
PQ = ⎜4 ⎟ – ⎜ 0 ⎟ = ⎜4 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ 1⎠ ⎝ –1⎠ ⎝ 2 ⎠
Similarly
⎛ 3⎞ ⎛ 1⎞ ⎛ 2 ⎞
PR == ⎜5 ⎟ – ⎜ 0 ⎟ = ⎜5 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝6 ⎠ ⎝ –1⎠ ⎝7 ⎠
⎯
→
⎯
→
| PQ | =
⎯
→
| PR | =
12 + 42 + 22 =
21
22 + 52 + 72 =
78
➜
253
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Therefore
10
(3, 5, 6)
⎛ 1⎞ ⎛ 2 ⎞
PQ. . PR = ⎜4 ⎟ . ⎜ 5⎟
⎜ ⎟ ⎜ ⎟
⎝2⎠ ⎝7⎠
R
⎯
→ ⎯
→
))
10 VECTORS
= 1 ×2 + 4 ×5 + 2 ×7
= 36
2
5
7
Substituting gives
36
21 × 78
⇒ θ = 27.2°
cos θ =
(2, 4, 1)
))
1
4
2
θ
P
Q
(1, 0, –1)
▲ Figure 10.20
You must be careful to find the correct angle. To find ∠QPR
⎯
→ ⎯
→
(see Figure 10.21), you need the scalar product PQ . PR, as in the
⎯→ ⎯
→
example above. If you take QP . PR, you will obtain ∠Q´PR, which is
(180° − ∠QPR).
R
Q
θ
P
Q'
▲ Figure 10.21
Exercise 10C
1
2
Find the angles between these vectors.
(i) 2i + 3j and 4i + j
(ii) 2i − j and i + 2j
(iii)
⎛ –1⎞
⎝ –1⎠
and
⎛ –1⎞
⎝ –2⎠
(v)
⎛2⎞
⎜ ⎟
⎝3⎠
and
⎛ –6 ⎞
⎜ ⎟
⎝ 4⎠
(iv) 4i + j and i + j
(vi)
⎛ 3⎞
⎝ –1⎠
and
⎛ –6⎞
⎝ 2⎠
The points A, B and C have coordinates (3, 2), (6, 3) and (5, 6),
respectively.
⎯
→
⎯
→
(i) Write down the vectors A B and BC.
(ii)
Show that the angle ABC is 90°.
| ⎯→ | | ⎯→ |
(iii) Show that A B = BC .
(iv) The figure ABCD is a square.
254
9781510421738.indb 254
Find the coordinates of the point D.
02/02/18 1:15 PM
3
Find the angles between these pairs of vectors.
(i)
⎛ 2⎞
⎜1⎟
⎜⎝ 3⎟⎠
and
⎛ 2⎞
⎜ –1⎟
⎜⎝ 4 ⎟⎠
⎛ 1⎞
⎜ –1⎟
⎜⎝ 0⎟⎠
(ii)
and
10
⎛ 3⎞
⎜1⎟
⎜⎝ 5⎟⎠
(iii) 3i + 2j − 2k and −4i − j + 3k
4
The diagram shows a room with rectangular walls, floor and ceiling.
A string has been stretched in a straight line between the corners A and G.
G
F
z
string
E
(0, 0, 3) D
y
C
(0, 4, 0)
B
spider
x
A
(5, 0, 0)
O (0, 0, 0)
10.4 The angle between two vectors
PS
The corner O is taken as the origin. A is (5, 0, 0), C is (0, 4, 0) and
D is (0, 0, 3), where the lengths are in metres.
A spider walks up the string, starting at A.
(i)
Write down the coordinates of G.
(ii)
Find the vector A G and the distance the spider walks along the
string from A to G.
⎯
→
(iii) Find the angle of elevation of the spider’s journey along the string.
5
In the diagram, OABCDEFG is a cube in which each side has length 6.
⎯
→ ⎯
→
⎯
→
Unit vectors i, j and k are parallel to OA, OC and OD respectively. The
⎯
→
⎯
→
point P is such that A P = 13 A B and the point Q is the midpoint of DF.
F
G
Q
D
E
k
B
C
j
O
P
A
i
⎯
→
⎯
→
(i)
Express each of the vectors OQ and PQ in terms of i, j and k.
(ii)
Find the angle OQP.
Cambridge International AS & A Level Mathematics
9709 Paper 12 Q6 November 2009
9781510421738.indb 255
255
02/02/18 1:15 PM
10
6
Relative to an origin O, the position vectors of points A and B are
2i + j + 2k and 3i − 2j + pk respectively.
(i)
Find the value of p for which OA and OB are perpendicular.
(ii)
In the case where p = 6, use a scalar product to find angle AOB,
correct to the nearest degree.
10 VECTORS
⎯
→
(iii) Express the vector A B in terms of p and hence find the values of p
for which the length of AB is 3.5 units.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q10 June 2008
7
Relative to an origin O, the position vectors of the points A and B are
given by
⎯
→
OA = 2i − 8j + 4k
⎯
→
and
⎯
→
OB = 7i + 2j − k.
⎯
→
(i)
Find the value of OA . OB and hence state whether angle AOB is
acute, obtuse or a right angle.
(ii)
The point X is such that A X = 5 A B. Find the unit vector in the
direction of OX.
⎯
→
→
2⎯
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q6 June 2009
8 Relative to an origin O, the position vectors of the points A and B are
given by
⎯
→
OA = 2i + 3j − k
and
⎯
→
OB = 4i − 3j + 2k.
(i)
Use a scalar product to find angle AOB, correct to the nearest
degree.
(ii)
Find the unit vector in the direction of A B.
⎯
→
⎯
→
(iii) The point C is such that OC = 6j + pk, where p is a constant.
⎯
→
⎯
→
Given that the lengths of A B and A C are equal, find the possible
values of p.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q11 June 2005
9 Relative to an origin O, the position vectors of the points P and Q are
given by
⎯
→ ⎛ −2 ⎞
OP = ⎜ 3⎟
⎜⎝ 1⎟⎠
and
⎛ 2⎞
⎯
→
OQ = ⎜ 1 ⎟ ,
⎜⎝ q ⎟⎠
where q is a constant.
(i)
In the case where q = 3, use a scalar product to show that
cos POQ = 71 .
(ii)
Find the values of q for which the length of PQ is 6 units.
⎯
→
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q4 November 2005
256
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02/02/18 1:15 PM
10 The diagram shows a semicircular prism with a horizontal rectangular
base ABCD. The vertical ends AED and BFC are semicircles of radius
6 cm. The length of the prism is 20 cm. The midpoint of AD is the origin
O, the midpoint of BC is M and the midpoint of DC is N. The points E
and F are the highest points of the semicircular ends of the prism. The
point P lies on EF such that EP = 8 cm.
10
F
E
B
M
C
k
A
N
j
6 cm
O
20 cm
i
D
Unit vectors i, j and k are parallel to OD, OM and OE respectively.
⎯
→
⎯
→
(i)
Express each of the vectors PA and PN in terms of i, j and k.
(ii)
Use a scalar product to calculate angle APN.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q4 November 2008
10.4 The angle between two vectors
P
8 cm
11 The diagram shows the roof of a house. The base of the roof, OABC, is
rectangular and horizontal with OA = CB = 14 m and OC = AB = 8 m.
The top of the roof DE is 5 m above the base and DE = 6 m. The sloping
edges OD, CD, AE and BE are all equal in length.
6m
D
E
B
8m
C
A
k
j
14 m
O
i
Unit vectors i and j are parallel to OA and OC respectively and the unit
vector k is vertically upwards.
⎯→
(i)
Express the vector OD in terms of i, j and k, and find its
magnitude.
(ii)
Use a scalar product to find angle DOB.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q8 June 2006
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10
12 The diagram shows a cube OABCDEFG in which the length of each
⎯→ ⎯→
⎯→
side is 4 units. The unit vectors i, j and k are parallel to OA, OC and OD
respectively. The midpoints of OA and DG are P and Q respectively and
R is the centre of the square face ABFE.
F
G
10 VECTORS
Q
D
E
R
B
C
k
j
O
i
P
A
⎯
→
⎯
→
(i)
Express each of the vectors PR and PQ in terms of i, j and k.
(ii)
Use a scalar product to find angle QPR.
(iii) Find the perimeter of triangle PQR, giving your answer correct to
1 decimal place.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q10 November 2007
10.5 The vector equation of a line
The vector joining two points
In Figure 10.22, start by looking at two points A(2, −1) and B(4, 3); that is the
⎯→ ⎛ 2 ⎞
⎯→ ⎛ 4 ⎞
points with position vectors OA = ⎜ ⎟ and OB = ⎜ ⎟ , alternatively 2i − j
⎝ 3⎠
⎝ −1⎠
and 4i + 3j.
y
3
B(4, 3)
N
2
1
–1
O
–1
M
1
2
3
4
5
x
A(2, –1)
–2
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▲ Figure 10.22
02/02/18 1:15 PM
⎯
→
The vector joining A to B is AB and this is given by
⎯
→
⎯
→
10
⎯→
AB = AO + OB
⎯→
⎯→
⎯→
⎯→
= −OA + OB
= OB − OA
10.5 The vector equation of a line
⎛ 4 ⎞ ⎛ 2⎞ ⎛ 2⎞
=⎜ ⎟− ⎜ ⎟ =⎜ ⎟
⎝ 3⎠ ⎝ –1⎠ ⎝ 4 ⎠
⎯
→ ⎛ 2⎞
Since AB = ⎜ ⎟ , then it follows that the length of AB is given by
⎝ 4⎠
⎯
→
⏐AB⏐ = 2 2 + 4 2
=
20
You can find the position vectors of points along AB as follows.
⎯→
The midpoint, M, has position vector OM, given by
⎯→
⎯→
⎯
→
OM = OA + 21AB
⎛ 2⎞
⎛2⎞
= ⎜ ⎟ + 21 ⎜ ⎟
⎝ –1⎠
⎝4 ⎠
⎛ 3⎞
=⎜ ⎟
⎝ 1⎠
In the same way, the position vector of the point N, three-quarters of the
distance from A to B, is given by
⎯→
⎛ 2⎞
⎛ 2⎞
ON = ⎜ ⎟ + 43 ⎜ ⎟
⎝ –1⎠
⎝ 4⎠
⎛ 3 1⎞
= ⎜ 2⎟
⎝2 ⎠
and it is possible to find the position vector of any other point of subdivision
of the line AB in the same way.
⎯
→
⎯→
⎯
→
A point P has position vector OP = OA + λAB where λ is a fraction.
❯ Show that this can be expressed as
⎯
→
⎯→
?
CP
⎯
→
OP = (1 − λ)OA + λOB.
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10 VECTORS
10
The vector equation of a line
It is now a small step to go from finding the position vector of any point on
the line AB to finding the vector form of the equation of the line AB. To take
this step, you will find it helpful to carry out the following activity.
ACTIVITY 10.2
The position vectors of a set of points are given by
λ is the Greek
⎛ 2⎞
⎛ 2⎞
r = ⎜ ⎟ + λ⎜ ⎟
letter ‘lamda’.
–1
4
⎝ ⎠
⎝ ⎠
where λ is a parameter which may take any value.
Show that λ = 2 corresponds to the point with position
⎛ 6⎞
vector ⎜ ⎟ .
⎝ 7⎠
(ii)
Find the position vectors of points corresponding to values
of λ of −2, −1, 0, 21 , 43 , 1, 3.
(i)
(iii)
Mark all your points on a sheet of squared paper and show that
when they are joined up they give the line AB in Figure 10.23.
(iv)
State what values of λ correspond to the points A, B, M and N.
(v)
What can you say about the position of the point if
(a) 0 < λ < 1?
(b) λ > 1?
(c) λ < 0?
This activity should have convinced you that
⎛ 2⎞
⎛ 2⎞
r =⎜ ⎟ +λ⎜ ⎟
⎝ –1⎠
⎝ 4⎠
The number λ is called a parameter
and it can take any value. Of course,
you can use other letters for the
parameter such as µ, s and t.
is the equation of the line passing through (2, −1) and (4, 3), written in vector
form.
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You may find it helpful to think of this in these terms.
⎛ 2⎞
⎛ 2⎞
r=⎜ ⎟ +λ⎜ ⎟
⎝ –1⎠
⎝ 4⎠
2 Move to the point A with
( )
2 .
position vector
−1
( )
2 (i.e.
3 Move λ steps of 4
⎯
→
in the direction AB). λ need
10
not be a whole number and
may be negative.
3
B(4, 3)
2
1 Start at the origin.
1
–1
O
1
2
–1
3
4
5
x
10.5 The vector equation of a line
and then
y
A(2, –1)
–2
▲ Figure 10.23
You should also have noticed that when:
λ= 0
the point corresponds to the point A
λ= 1
the point corresponds to the point B
0"λ"1
the point lies between A and B
λ#1
the point lies beyond B
λ"0
the point lies beyond A.
The vector form of the equation is not unique; there are many (in fact
infinitely many) different ways in which the equation of any particular line
may be expressed. There are two reasons for this: direction and location.
Direction
⎛ 2⎞
The direction of the line in the example is ⎜ ⎟ . That means that for every
⎝ 4⎠
2 units along (in the i direction), the line goes up 4 units (in the j direction).
This is equivalent to stating that for every 1 unit along, the line goes up 2
units, corresponding to the equation
⎛ 2⎞
⎛1 ⎞
r= ⎜ ⎟ + λ⎜ ⎟.
⎝ –1⎠
⎝2⎠
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10 VECTORS
10
The only difference is that the two equations have different values of λ for
⎛ 4⎞
particular points. In the first equation, point B, with position vector ⎜ ⎟ ,
⎝ 3⎠
corresponds to a value of λ of 1. In the second equation, the value of λ
for B is 2.
⎛1⎞
⎛ 3⎞
⎛1⎞
⎛ 2⎞
The direction ⎜ ⎟ is the same as ⎜ ⎟ , or as any multiple of ⎜ ⎟ such as ⎜ ⎟ ,
⎝ 2⎠
⎝ 6⎠
⎝ 2⎠
⎝ 4⎠
⎛ –5⎞ or ⎛ 100.5⎞ . Any of these could be used in the vector equation of the line.
⎜⎝ –10⎠⎟ ⎜⎝ 201 ⎟⎠
Location
In the equation
⎛ 2⎞
⎛ 2⎞
r=⎜ ⎟ + λ⎜ ⎟
⎝ –1 ⎠
⎝ 4⎠
⎛ 2⎞
⎜⎝ –1⎟⎠ is the position vector of the point A on the line, and represents the
point at which the line was joined. However, this could have been any other
point on the line, such as M(3, 1), B(4, 3), etc. Consequently
⎛ 3⎞
⎛ 2⎞
r = ⎜ 1 ⎟ + λ ⎜ 4⎟
⎝ ⎠
⎝ ⎠
and
⎛ 4⎞
⎛ 2⎞
r = ⎜ 3⎟ + λ ⎜ 4 ⎟
⎝ ⎠
⎝ ⎠
are also equations of the same line, and there are infinitely many other
possibilities, one corresponding to each point on the line.
Notes
1
It is usual to refer to any valid vector form of the equation as the vector
equation of the line even though it is not unique.
2
It is often a good idea to give the direction vector in its simplest integer
form:
⎛2⎞
⎛1 ⎞
for example, replacing ⎜ ⎟ with ⎜ ⎟ .
⎜4 ⎟
⎝ ⎠
⎜2⎟
⎝ ⎠
The general vector form of the equation of a line
If A and B are points with position a and b, then the equation
⎯→
⎯
→
r = OA + λ AB
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9781510421738.indb 262
may be written as
r = a + λ(b − a)
which implies
r = (1 − λ)a + λb.
This is the general vector form of the equation of the line joining two points.
02/02/18 1:16 PM
ACTIVITY 10.3
Plot the following lines on the same sheet of squared paper. When you
have done so, explain why certain among them are the same as each other,
others are parallel to each other, and others are in different directions.
⎛ 2⎞
⎛1 ⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ –1⎠
⎝2⎠
⎛0 ⎞
⎝2⎠
⎛1 ⎞
⎝2⎠
(iii) r = ⎜ ⎟ + λ ⎜ ⎟
(v)
(ii)
⎛ 2⎞
⎛ –1⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ –1⎠
⎝ 2⎠
(iv)
1
3
r = ⎛⎜ ⎞⎟ + λ ⎛⎜ ⎞⎟
⎝ –3⎠
⎝ 6⎠
⎛ 4⎞
⎛ 1⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ 3⎠
⎝ –2⎠
The same methods can be used to find the vector equation of a line in three
dimensions, as shown in the next example.
Example 10.12
10.5 The vector equation of a line
(i)
10
The coordinates of A and B are (–2, 4, 1) and (2, 1, 3) respectively.
(i)
(ii)
(iii)
Find the vector equation of the line AB.
Does the point P(6, –2, 7) lie on the line AB?
The point N lies on the line AB.
⎯→
⎯→
Given that 3⏐AN⏐=⏐NB⏐ find the coordinates of N.
Solution
(i)
⎛ 2⎞
⎯→ ⎛ −2 ⎞
⎯
→
a = OA = ⎜ 4 ⎟ and b = OB = ⎜ 1⎟
⎝ 3⎠
⎝ 1⎠
⎛ 2 ⎞ ⎛− 2 ⎞ ⎛ 4 ⎞
⎯
→
AB = b − a = ⎜⎜ 1 ⎟⎟ − ⎜⎜ 4 ⎟⎟ = ⎜⎜− 3⎟⎟
⎝ 3 ⎠ ⎝ 1⎠ ⎝ 2 ⎠
The vector equation of a line can
be written as
⎯→
⎯
→
r = OA + λAB
⎛−2 ⎞
⎛ 4⎞
⇒ r = ⎜ 4 ⎟ + λ ⎜−3⎟
⎜ 1⎟
⎜ 2⎟
⎝ ⎠
⎝ ⎠
(ii)
There are other ways of
writing this equation, for
example
⎛2⎞
⎛ 4⎞
r = ⎜ 1⎟ + λ ⎜− 3⎟ or
⎝ 3⎠
⎝ 2⎠
⎛− 6 ⎞
⎛ 4⎞
r = ⎜ 7 ⎟ + λ ⎜− 3⎟
⎝ − 1⎠
⎝ 2⎠
but they are all equivalent
to each other.
If P lies on the line AB then for some value of λ
⎛ 4⎞
⎛ x ⎞ ⎛ 6 ⎞ ⎛− 2 ⎞
⎜ y ⎟ = ⎜− 2 ⎟ = ⎜ 4 ⎟ + λ ⎜− 3⎟
⎜ z ⎟ ⎜ 7 ⎟ ⎜ 1⎟
⎜ 2⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎝ ⎠
Find the value of λ for the x-coordinate.
x : 6 = −2 + 4λ
⇒
λ=2
➜
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Then check whether this value of λ gives a y-coordinate of −2 and
a z-coordinate of 7.
10
y : −2 = 4 − 3 × 2
z : 7 ≠ 1 + 2 ×2
So the point P(6, –2, 7) does not lie on the line.
10 VECTORS
(iii)
⎯→
⎯→
Since 3⏐AN⏐=⏐NB⏐, N must lie 41 of the way along the line AB so
the value of λ is 41 .
⎯→
⎯→
→
1 ⎯
ON = OA + 4 AB
⎛−2 ⎞
⎛ 4 ⎞ ⎛ −1⎞
⎜
⎟
1 ⎜ ⎟ ⎜3 1 ⎟
ON = 4 + −3 = 4
⎜ ⎟ 4 ⎜ ⎟ ⎜ 1⎟
⎝ 1⎠
⎝ 2⎠ ⎝ 1 2 ⎠
⎯→
So the coordinates of N are (–1, 3.25, 1.5).
Exercise 10D
1
For each of these pairs of points, A and B, write down:
⎯
→
(a) the vector AB
⎯
→
(b) ⏐AB⏐
(c) the position vector of the midpoint of AB.
(i)
A is (2, 3), B is (4, 11).
(ii)
A is (4, 3), B is (0, 0).
(iii) A is (−2, −1), B is (4, 7).
(iv) A is (−3, 4), B is (3, −4).
(v)
2
A is (−10, −8), B is (−5, 4).
Find the equation of each of these lines in vector form.
(i)
Joining (2, 1) to (4, 5)
(ii)
Joining (3, 5) to (0, 8)
(iii)
Joining (−6, −6) to (4, 4)
(iv) Through (5, 3) in the same direction as i + j
Through (2, 1) parallel to 6i + 3j
⎛ –1⎞
(vi) Through (0, 0) parallel to
⎜⎝ 4 ⎟⎠
(vii) Joining (0, 0) to (−2, 8)
(v)
(viii) Joining (3, −12) to (−1, 4)
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3
Find the equation of each of these lines in vector form.
(i)
⎛ 3⎞
Through (2, 4, −1) in the direction ⎜ 6⎟
⎝ 4⎠
⎛ 1⎞
Through (1, 0, −1) in the direction ⎜ 0⎟
⎝ 0⎠
(iii) Through (1, 0, 4) and (6, 3, −2)
10
(ii)
(v)
4
Through (1, 2, 3) and (−2, −4, −6)
Determine whether the given point P lies on the line in each of the
following cases.
(i)
⎛ 1⎞
⎛ 2⎞
P(5, 1, 4) and the line r = ⎜ 3⎟ + λ ⎜− 1⎟
⎜4 ⎟
⎜ 0⎟
⎝ ⎠
⎝ ⎠
(ii)
⎛ 2⎞
⎛ 1⎞
P(−1, 5, 1) and the line r = ⎜⎜ 3⎟⎟ + λ ⎜⎜− 2 ⎟⎟
⎝4 ⎠
⎝ 3⎠
⎛ 1⎞
⎛−2 ⎞
⎝−2 ⎠
⎝ 5⎠
10.6 The intersection of two lines
(iv) Through (0, 0, 1) and (2, 1, 4)
(iii) P(−5, 3, 12) and the line r = ⎜⎜ 0 ⎟⎟ + λ ⎜⎜ 1⎟⎟
⎛ 1⎞
⎛ 4⎞
⎝0⎠
⎝− 2 ⎠
(iv) P(9, 0, −6) and the line r = ⎜⎜ 2 ⎟⎟ + λ ⎜⎜ − 1⎟⎟
(v)
PS
5
⎛ 1⎞
⎛2⎞
P(−9, −2, −17) and the line r = ⎜⎜ 3⎟⎟ + λ ⎜⎜ 1⎟⎟
⎝−2 ⎠
⎝ 3⎠
The coordinates of three points are A(−1, −2, 1), B( −3, 4, −5) and
C(0, −2, 4).
(i) Find a vector equation of the line AB.
(ii)
Find the coordinates of the midpoint M of AB.
(iii) The point N lies on BC.
⎯
→
⎯→
Given that 2⏐BN⏐=⏐NC⏐, find the equation of the line MN.
10.6 The intersection of two lines
Hold a pen and a pencil to represent two distinct straight lines as follows:
» hold them to represent parallel lines;
» hold them to represent intersecting lines;
» hold them to represent lines which are not parallel and which do not
intersect (even if you extend them).
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265
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In three-dimensional space two or more straight lines which are not parallel
and which do not meet are known as skew lines. In a plane two distinct
lines are either parallel or intersecting, but in three dimensions there are
three possibilities: the lines may be parallel, or intersecting, or skew. The next
example illustrates a method of finding whether two lines meet, and, if they
do meet, the coordinates of the point of intersection.
10 VECTORS
10
Example 10.13
Find the position vector of the point where the following lines intersect.
⎛6 ⎞
⎛ 1⎞
⎛2⎞
⎛1 ⎞
r = ⎜ ⎟ + λ ⎜ ⎟ and r = ⎜ ⎟ + µ ⎜ ⎟
⎝1 ⎠
⎝ –3⎠
⎝3⎠
⎝2⎠
Note here that different letters are used for the parameters in the two
equations to avoid confusion.
Solution
When the lines intersect, the position vector is the same for each of them.
⎛x ⎞ ⎛2⎞
⎛1 ⎞ ⎛6 ⎞
⎛ 1⎞
r = ⎜ ⎟ = ⎜ ⎟ + λ⎜ ⎟ = ⎜ ⎟ + µ⎜ ⎟
⎝y ⎠ ⎝3⎠
⎝ 2 ⎠ ⎝1 ⎠
⎝− 3⎠
This gives two simultaneous equations for λ and µ.
x: 2 + λ = 6 + µ
⇒ λ−µ=4
y : 3 + 2λ = 1 − 3µ ⇒ 2λ + 3µ = −2
Solving these gives λ = 2 and µ = −2. Substituting in either equation gives
⎛ 4⎞
r=⎜ ⎟
⎝ 7⎠
which is the position vector of the point of intersection.
Example 10.14
Find the coordinates of the point
of intersection of the lines joining
A(1, 6) to B(4, 0), and C(1, 1) to
D(5, 3).
y
A(1, 6)
6
3
D(5, 3)
1
C(1, 1)
B(4, 0)
O
1
4
x
▲ Figure 10.24
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Solution
10
⎛ 4 ⎞ ⎛ 1 ⎞ ⎛ 3⎞
⎯
→
AB = ⎜ 0 ⎟ − ⎜ 6⎟ = ⎜ –6⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
and so the vector equation of line AB is
⎯→
⎯
→
r = OA + λAB
10.6 The intersection of two lines
⎛1 ⎞
⎛ 3⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝6 ⎠
⎝ –6 ⎠
⎯→ ⎛ 5 ⎞ ⎛ 1⎞
⎛ 4⎞
CD = ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟
⎝ 3⎠ ⎝ 1⎠ ⎝ 2⎠
and so the vector equation of line CD is
⎯→
⎯→
r = OC + µ CD
⎛1⎞
⎛4 ⎞
r = ⎜ ⎟ + µ⎜ ⎟
⎝1⎠
⎝2⎠
The intersection of these lines is at
⎛1 ⎞
⎛ 3⎞ ⎛1⎞
⎛4 ⎞
r = ⎜ ⎟ + λ⎜ ⎟ = ⎜ ⎟ + µ⎜ ⎟
⎝6 ⎠
⎝− 6 ⎠ ⎝1⎠
⎝2⎠
x : 1 + 3λ = 1 + 4µ
⇒
3λ − 4µ = 0
1
!
y : 6 − 6λ = 1 + 2µ
⇒
6λ + 2µ = 5
2
!
1 and !
2 simultaneously:
Solve !
1 :
!
2 × 2:
!
Add:
3λ − 4µ = 0
12λ + 4µ = 10
15λ
= 10
⇒
λ = 23
Substitute λ = 23 in the equation for AB:
⎛ 1⎞
⎛ 3⎞
⇒ r = ⎜ ⎟ + 23 ⎜ ⎟
⎝ 6⎠
⎝ –6⎠
⎛ 3⎞
⇒ r = ⎜ ⎟ . The point of intersection has coordinates (3, 2).
⎝ 2⎠
Note
Alternatively, you could have found µ = 1 and substituted in the equation
2
for CD.
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Example 10.15
10 VECTORS
10
In three dimensions, lines may be parallel, they may intersect or they may be
skew.
Determine whether each pair of lines are parallel, intersect or are skew.
(i)
⎛ 1⎞
⎛−3⎞
⎛ 1⎞
⎛ 6⎞
r = ⎜⎜−2 ⎟⎟ + λ ⎜⎜ 2 ⎟⎟ and r = ⎜⎜ 3⎟⎟ + µ ⎜⎜−4 ⎟⎟
⎝ 1⎠
⎝ 1⎠
⎝−2 ⎠
⎝ −2 ⎠
(ii)
⎛ 1⎞
⎛ 2⎞
⎛ 4⎞
⎛−1⎞
r = ⎜⎜ 2 ⎟⎟ + λ ⎜⎜−3⎟⎟ and r = ⎜−2 ⎟ + µ ⎜ 2 ⎟
⎜ −5⎟
⎜ 1⎟
⎝−1⎠
⎝ 4⎠
⎝ ⎠
⎝ ⎠
Solution
(i)
⎛ −3⎞
⎛ 6⎞
The vectors
−2 ⎜ 2⎟ and ⎜ −4 ⎟ are in the same direction as
⎝ 1⎠
⎝ −2⎠
⎛ 6⎞
⎛ −3⎞
⎜ −4 ⎟ = −2 ⎜ 2⎟
⎝ −2⎠
⎝ 1⎠
Note the lines are different as one
line passes through (1, –2, 1) and
the other through (1, 3, –2).
So the lines are parallel.
(ii)
These lines are not parallel, so either they intersect or they are skew.
If the two lines intersect then there is a point (x, y, z) that lies on
both lines.
⎛ 2⎞
⎛ x ⎞ ⎛ 1⎞
⎜ y ⎟ = ⎜ 2 ⎟ + λ ⎜− 3⎟
⎜ z ⎟ ⎜− 1⎟
⎜ 4⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
and
⎛x ⎞ ⎛ 4 ⎞
⎛− 1⎞
⎜ y ⎟ = ⎜− 2 ⎟ + µ ⎜ 2 ⎟
⎜ z ⎟ ⎜ − 5⎟
⎜ 1⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
This gives three simultaneous equations for λ and µ.
x : 1 + 2λ = 4 − µ
⇒
2λ + µ = 3
1
!
y : 2 − 3λ = −2 + 2µ
⇒
3λ + 2µ = 4
2
!
z : −1 + 4λ = −5 + µ
⇒
4λ − µ = −4
3
!
Now solve any two of the three equations above simultaneously.
1 and !
2 :
Using !
{ 23 ++ 2 ==34 } ⇒ {34 ++ 22 == 64 } ⇒
λ
µ
λ
µ
λ
µ
λ
µ
λ = 2, µ = −1
3
If these solutions satisfy the previously unused equation (equation !
here) then the lines meet, and you can substitute the value of λ (or
1,!
2 and !
3 to find the coordinates of the point of
µ) into equations !
intersection.
3 then the lines are skew.
If these solutions do not satisfy equation !
3
4λ − µ = −4
!
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When λ = 2 and µ = −1
4λ − µ = 9 ≠ −4
As there are no values for λ and µ that satisfy all three equations, the
lines do not meet and so are skew; you have already seen that they are
not parallel.
If the equation of the second line was
⎛ x ⎞ ⎛ 4⎞
⎛ − 1⎞
⎜ y ⎟ = ⎜ − 2⎟ + µ ⎜ 2⎟
⎜⎝ z ⎟⎠ ⎜⎝ 8⎟⎠
⎜⎝ 1 ⎟⎠
then the values of λ = 2 and µ = –1 would produce the same point for both
lines:
⎛ x ⎞ ⎛ 1⎞
⎛ 2 ⎞ ⎛ 5⎞
⎜ y ⎟ = ⎜ 2 ⎟ + 2 ⎜ − 3⎟ = ⎜ − 4 ⎟
⎜⎝ z ⎟⎠ ⎜⎝ − 1 ⎟⎠
⎜⎝ 4 ⎟⎠ ⎜⎝ 7⎟⎠
10.6 The intersection of two lines
Note
10
⎛ x ⎞ ⎛ 4⎞
⎛ − 1 ⎞ ⎛ 5⎞
⎜ y ⎟ = ⎜ − 2⎟ − 1⎜ 2⎟ = ⎜ − 4 ⎟
⎜⎝ z ⎟⎠ ⎜⎝ 8⎟⎠
⎜⎝ 1 ⎟⎠ ⎜⎝ 7⎟⎠ .
and
So the lines would intersect at (5, –4, 7).
Exercise 10E
1
Find the position vector of the point of intersection of each of these
pairs of lines.
(i)
⎛2⎞
⎛1 ⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝1 ⎠
⎝0 ⎠
and
⎛ 3⎞
⎛1⎞
r = ⎜ ⎟ + µ⎜ ⎟
⎝1⎠
⎝0 ⎠
(ii)
⎛ 2⎞
⎛ 1⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ –1⎠
⎝2⎠
and
⎛1⎞
r = µ⎜ ⎟
⎝1⎠
⎛ –2 ⎞
⎟
⎝ –2 ⎠
and
⎛ 0⎞
⎛ 1⎞
r = ⎜ ⎟ + µ⎜ ⎟
⎝2⎠
⎝ –7 ⎠
⎛−2 ⎞
⎛−1⎞
⎟ + λ⎜ ⎟
⎝ 3⎠
⎝ –3 ⎠
and
⎛1 ⎞
⎛ 2⎞
r = ⎜ ⎟ + µ⎜ ⎟
⎝ 3⎠
⎝ –1⎠
⎛2⎞
⎛ 1⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ –1⎠
⎝7 ⎠
and
⎛5⎞
⎛ 1⎞
r = ⎜ ⎟ + µ⎜ ⎟
⎝2⎠
⎝1 ⎠
⎛0 ⎞
⎝5 ⎠
(iii) r = ⎜ ⎟ + λ ⎜
(iv) r = ⎜
(v)
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2
10 VECTORS
10
Decide whether each of these pairs of lines intersect, are parallel or
are skew. If the lines intersect, find the coordinates of the point of
intersection.
⎛ 1⎞
⎛ 1⎞
⎛9⎞
⎛ 2⎞
⎜
⎟
⎜
⎟
⎜
⎟
⎜ 3⎟
+
=
+
r
=
−
6
2
and
r
7
λ
µ
(i)
⎜ −1⎟
⎜ 3⎟
⎜2⎟
⎜−1⎟
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎛ 1⎞
⎛ 6⎞
r = ⎜⎜−6 ⎟⎟ + λ ⎜⎜−9 ⎟⎟
⎝ 0⎠
⎝ −3⎠
and
⎛−5⎞
⎛ 2⎞
r = ⎜⎜ 3⎟⎟ + µ ⎜⎜−3⎟⎟
⎝ 0⎠
⎝ −1⎠
⎛ 6⎞
⎛ 1⎞
⎜
⎟
⎜−2 ⎟
+
r
=
−
λ
4
(iii)
⎜ 2⎟
⎜ 5⎟
⎝ ⎠
⎝ ⎠
and
⎛ 1⎞
⎛ 1⎞
r = ⎜⎜ 4 ⎟⎟ + µ ⎜⎜−1⎟⎟
⎝−17 ⎠
⎝ 2⎠
⎛2⎞
⎛−1⎞
⎜
⎟
⎜0⎟
+
r
=
2
λ
(iv)
⎜ 4⎟
⎜ 3⎟
⎝ ⎠
⎝ ⎠
and
⎛ 5⎞
⎛−4 ⎞
r = ⎜⎜ 4 ⎟⎟ + µ ⎜⎜−2 ⎟⎟
⎝ 6⎠
⎝ 1⎠
⎛ 0⎞
⎛ 5⎞
r = ⎜⎜−1⎟⎟ + λ ⎜⎜ 3⎟⎟
⎝ 4⎠
⎝−3⎠
and
⎛ 2⎞
⎛ 4⎞
r = ⎜⎜ 5⎟⎟ + µ ⎜⎜−3⎟⎟
⎝−1⎠
⎝ 2⎠
⎛ 9⎞
⎛ 1⎞
⎜
⎟
⎜ 2⎟
+
r
=
λ
3
(vi)
⎜−4 ⎟
⎜−3⎟
⎝ ⎠
⎝ ⎠
and
⎛ 1⎞
⎛ 1⎞
r = ⎜⎜−4 ⎟⎟ + µ ⎜⎜−1⎟⎟
⎝ 5⎠
⎝ 2⎠
and
⎛ − 1⎞
⎛ 1⎞
r = ⎜⎜− 3⎟⎟ + µ ⎜⎜ 3⎟⎟
⎝ − 1⎠
⎝2⎠
(ii)
(v)
⎛2⎞
⎛ 1⎞
⎝ 1⎠
⎝− 2 ⎠
(vii) r = ⎜⎜ 3⎟⎟ + λ ⎜⎜ 1⎟⎟
PS
3
In this question the origin is taken to be at a harbour and the unit
vectors i and j to have lengths of 1 km in the directions E and N.
A cargo vessel leaves the harbour and its position vector t hours later is
given by
r1 = 12t i + 16t j.
A fishing boat is trawling nearby and its position at time t is given by
r2 = (10 − 3t)i + (8 + 4t)j.
(i)
How far apart are the two boats when the cargo vessel leaves
harbour?
(ii)
How fast is each boat travelling?
(iii) What happens to the boats?
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4
The points A(1, 0), B(7, 2) and C(13, 7) are the vertices of a triangle.
The midpoints of the sides BC, CA and AB are L, M and N.
(i) Write down the position vectors of L, M and N.
(ii)
10
Find the vector equations of the lines AL, BM and CN.
(iii) Find the intersections of these pairs of lines.
AL and BM
(b)
BM and CN
(iv) What do you notice?
5
⎛ −4 ⎞
⎛ 2⎞
⎛ 4⎞
⎛ 2⎞
The line r = ⎜⎜ 4 ⎟⎟ + q ⎜⎜−10 ⎟⎟ meets r = ⎜⎜ −15⎟⎟ + s ⎜⎜−3⎟⎟ at A and meets
⎝−12 ⎠
⎝ 11⎠
⎝−16 ⎠
⎝−5⎠
⎛ −1⎞ ⎛ 1⎞
r = ⎜⎜−29 ⎟⎟ + t ⎜⎜ 1⎟⎟ at B.
⎝ −3⎠ ⎝8 ⎠
Find the coordinates of A and the length of AB.
PS
6
7
10.6 The intersection of two lines
(a)
To support a tree damaged in a gale a tree surgeon attaches wire guys to
four of the branches (see the diagram). He joins (2, 0, 3) to (−1, 2, 6) and
(0, 3, 5) to (−2, −2, 4). Do the guys, assumed straight, meet?
⎛ −7⎞
⎛ 4⎞
⎛ 3⎞
⎛ 2⎞
Show that the three lines r = ⎜ 24 ⎟ + q ⎜ −7⎟ , r = ⎜ −10⎟ + s ⎜ 2⎟ and
⎝ −4 ⎠
⎝ 4⎠
⎝ 15⎠
⎝ −1⎠
⎛ 8⎞
⎛ −3⎞
r = ⎜ 6⎟ + t ⎜ −3⎟ form a triangle and find the lengths of its sides.
⎝ 2⎠
⎝ 6⎠
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10.7 The angle between two lines
10
Earlier in this chapter, you learnt that the angle, θ, between two
10 VECTORS
⎛ a1 ⎞
⎛ b1 ⎞
vectors a = ⎜ a 2 ⎟ and b = ⎜ b 2 ⎟ can be found
⎝ a3 ⎠
⎝ b3 ⎠
using the formula
a.b
cos θ =
|a||b|
where a . b is the scalar product
and a . b = a1b1 + a2b2 + a3b3.
b
a
θ
▲ Figure 10.25
The angle between two lines is the same as the angle between their direction
vectors.
Example 10.16
Verify that the lines
⎛ −9 ⎞
⎛ 2 ⎞
⎛1⎞
⎛1⎞
⎜
⎟
⎜
⎟
⎜ ⎟
⎜ ⎟
r = ⎜ 2 ⎟ + λ ⎜ − 2 ⎟ and r = ⎜ 2 ⎟ + µ ⎜ − 3 ⎟ are perpendicular.
⎜3⎟
⎜3⎟
⎜ 4 ⎟
⎜ 3 ⎟
⎝ ⎠
⎝ ⎠
⎝
⎠
⎝
⎠
Solution
When two lines are perpendicular, the angle between their direction
vectors is 90°.
Since cos 90° = 0 then a . b = 0.
So if the scalar product of two non-zero vector lines is zero then the lines are
perpendicular.
⎛−9 ⎞
⎛ 2⎞
The direction vectors of these two lines are ⎜⎜−2 ⎟⎟ and ⎜⎜−3⎟⎟ .
⎝ 4⎠
⎝ 3⎠
⎛ −9⎞ ⎛ 2⎞
⎜ −2⎟ . ⎜ −3⎟ = ( −9) × 2 + ( −2) × ( −3) + 4 × 3
⎝ 4 ⎠ ⎝ 3⎠
= ( −18) + 6 + 12
=0
Therefore, the two lines are perpendicular.
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Even if two lines do not meet, it is still possible to specify the angle between
them. The lines l and m shown in Figure 10.26 do not meet; they are skew.
10
l
θ
m
▲ Figure 10.26
The angle between them is that between their directions; it is shown
in Figure 10.26 as the angle θ between the lines l and m', where m' is a
translation of the line m to a position where it does intersect the line l.
Example 10.17
Find the angle between the skew lines
⎛1⎞
⎛ 2⎞
⎜ ⎟
⎜ ⎟
r = ⎜0 ⎟ + λ ⎜− 1 ⎟
⎜− 1 ⎟
⎜4 ⎟
⎝ ⎠
⎝ ⎠
and
⎛3⎞
⎛ 2⎞
⎜ ⎟
⎜ ⎟
r = ⎜ −1 ⎟ + µ ⎜ 0 ⎟ .
⎜ 3⎟
⎜1⎟
⎝ ⎠
⎝ ⎠
Solution
The angle between the lines is the angle between their direction vectors
10.8 The perpendicular distance from a point to a line
m'
⎛ 3⎞
⎛ 2⎞
⎜ –1⎟ and ⎜ 0⎟ .
⎜ ⎟
⎜ ⎟
⎜⎝ –1⎟⎠
⎜⎝ 1 ⎟⎠
Using cos θ =
cos θ =
a.b
|a||b|
2 × 3 + (–1) × 0 + (–1) × 1
2 2 + ( −1) 2 + ( −1) 2 × 3 2 + 0 2 + 12
5
×
6
10
θ = 49.8°
cos θ =
⇒
10.8 The perpendicular distance from
a point to a line
The scalar product is also useful when determining the distance between a
point and a line.
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10
Example 10.18
Find the shortest distance from point P(11, −5, −3) to the line l with equation
⎛1⎞
⎛− 3 ⎞
r = ⎜5⎟ + λ⎜ 1 ⎟ .
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝ 4⎠
10 VECTORS
Solution
⎯→
The shortest distance from P to the line l is ⏐NP⏐ where N is a point on the
line l and PN is perpendicular to the line l.
P
N
l
▲ Figure 10.27
⎯→
You need to find the coordinates of N and then you can find ⏐NP⏐.
N lies on the line l. Let the value of λ at N be t.
So, relative to the origin O
⎛ 1⎞
⎛ − 3⎞ ⎛ 1 − 3t ⎞
⎯→
ON = ⎜ 5⎟ + t ⎜ 1⎟ = ⎜ 5 + t ⎟
⎝ 0⎠
⎝ 4 ⎠ ⎝ 4t ⎠
and
⎯
→
⎯
→
⎯→
NP = OP − ON
⎛ 11⎞ ⎛ 1 − 3t ⎞
= ⎜ −5⎟ − ⎜ 5 + t ⎟
⎝ −3⎠ ⎝ 4t ⎠
⎛ 10 + 3t ⎞
= ⎜ −10 − t ⎟
⎝ −3 − 4t ⎠
When two vectors are
perpendicular, their
scalar product is 0.
⎯
→
As NP is perpendicular to the line l,
⎯
→ ⎛ −3⎞
NP . ⎜ 1⎟ = 0
⎝ 4⎠
The direction of
the line l
⎛ 10 + 3t ⎞ ⎛−3⎞
⎯
→ ⎛ −3⎞
NP . ⎜ 1⎟ = ⎜⎜ −10 − t ⎟⎟ ⎜⎜ 1⎟⎟
⎝ 4 ⎠ ⎝−3 − 4t ⎠ ⎝ 4 ⎠
•
= (10 + 3t) × (−3) + (−10 − t) × 1 + (−3 − 4t) × 4
= −30 − 9t − 10 − t − 12 − 16t
= −52 − 26t
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The scalar product is 0, so
−52 − 26t = 0
⎯→
⇒
⎯
→
10
t = −2
Substituting t = −2 into ON and NP gives
⎛ 10 + 3 × ( −2) ⎞ ⎛ 4 ⎞
⎯
→
NP = ⎜ −10 − ( −2) ⎟ = ⎜ −8⎟
⎝ −3 − 4 × ( −2)⎠ ⎝ 5⎠
and
⎯→
⏐NP⏐ =
So
4 2 + ( −8) 2 + 5 2
= 105
= 10.25 units
Exercise 10F
⎛ 0⎞
⎛ 0⎞
⎛ 1⎞
Remember i = ⎜ 0⎟ , j = ⎜ 1⎟ and k = ⎜ 0⎟ .
⎝ 0⎠
⎝ 1⎠
⎝ 0⎠
In questions 1 to 5, find the angle between each pair of lines.
1
⎛ 2⎞
⎛ 1⎞
r = ⎜ 1⎟ + s ⎜ 4 ⎟
⎝ 3⎠
⎝ 0⎠
and
⎛ 6⎞
⎛ 2⎞
r = ⎜ 10⎟ + t ⎜ 1⎟
⎝ 4⎠
⎝ 1⎠
2
⎛ 4⎞
r = s ⎜ 1⎟
⎝ 4⎠
and
⎛ 7 ⎞ ⎛ 1⎞
r = ⎜⎜ 0 ⎟⎟ + t ⎜⎜ 2 ⎟⎟
⎝−3⎠ ⎝−1⎠
3
⎛ 4⎞
⎛ 3⎞
r = ⎜ 2⎟ + s ⎜ 7⎟
⎝ −1⎠
⎝ −4 ⎠
and
⎛ 5⎞
⎛ 2⎞
r = ⎜ 1⎟ + t ⎜ 8 ⎟
⎝ 0⎠
⎝ − 5⎠
4
r = 2i + 3j + 4k + s(i + j − k)
and
r = t(i − k)
5
r = i − 2j − k + s(2i + 3j + 2k)
and
r = 2i + j + tk
6
For each point P and line l find
(a) the coordinates of the point N on the line such that PN is
perpendicular to the line
(b)
(i)
10.8 The perpendicular distance from a point to a line
⎛ 1⎞
⎛ − 3⎞ ⎛ 7 ⎞
⎯→
ON = ⎜ 5⎟ − 2 ⎜ 1⎟ = ⎜ 3⎟
⎝ 0⎠
⎝ 4 ⎠ ⎝ − 8⎠
the distance PN.
P(–2, 11, 5)
and
⎛ 0⎞
⎛ −1⎞
r = ⎜ 2⎟ + t ⎜ 2⎟
⎝ −3⎠
⎝ 5⎠
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10
P(7, –1, 6)
and
⎛ 2⎞
⎛ 1⎞
r = ⎜ 1⎟ + t ⎜ − 2 ⎟
⎝ 3⎠
⎝ 4⎠
(iii) P(8, 4, –1)
and
⎛ 1⎞
⎛ −1⎞
r = ⎜ 5⎟ + t ⎜ −2⎟
⎝ −3⎠
⎝ 0⎠
(ii)
10 VECTORS
7
Find the perpendicular distance of the point P(–7, –2, 13) to the line
⎛ 1⎞
⎛ 1⎞
r = ⎜⎜ 2 ⎟⎟ + λ ⎜⎜ 3⎟⎟ .
⎝ 5⎠
⎝− 4 ⎠
M
8
Find the distance of the point C(0, 6, 0) to the line joining the points
A(–4, 2, –3) and B(–2, 0, 1).
9
The diagram shows an extension to a house. Its base and walls are
rectangular and the end of its roof, EPF, is sloping, as illustrated.
Q (2, 5, 4)
H
G (4, 5, 3)
(2, 1, 4)
P
E
(0, 0, 3)
(0, 0, 0)
F
C
O
B (4, 5, 0)
A
(i)
Write down the coordinates of A and F.
(ii)
Find, using vector methods, the angles FPQ and EPF.
The owner decorates the room with two streamers which are pulled
taut. One goes from O to G, the other from A to H. She says that they
touch each other and that they are perpendicular to each other.
(iii) Is she right?
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10 The points A and B have position vectors, relative to the origin O, given by
⎯→
OA = i + 2j + 3k and
⎯→
OB = 2i + j + 3k.
The line l has vector equation
10
r = (1 − 2t)i + (5 + t)j + (2 − t)k.
Show that l does not intersect the line passing through A and B.
(ii)
The point P lies on l and is such that angle PAB is equal to 60°.
Given that the position vector of P is (1 − 2t)i + (5 + t)j + (2 − t)k,
show that 3t 2 + 7t + 2 = 0. Hence find the only possible position
vector of P.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q10 June 2008
11 With respect to the origin O, the position vectors of two points A and B
⎯→
⎯→
are given by OA = i + 2 j + 2k and OB = 3i + 4 j. The point P lies on
⎯→
⎯→
the line through A and B, and AP = λ AB .
⎯→
(i)
Show that OP = (1 + 2λ )i + (2 + 2 λ ) j + (2 − 2λ )k.
(ii)
By equating expressions for cos AOP and cos BOP in terms of λ ,
find the value of λ for which OP bisects the angle AOB.
(iii) When λ has this value, verify that AP : PB = OA : OB.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q7 November 2011
10.8 The perpendicular distance from a point to a line
(i)
12 The equations of two straight lines are
r = i + 4 j − 2k + λ( i + 3k ) and r = ai + 2 j − 2k + µ( i + 2 j + 3ak ),
where a is a constant.
(i)
Show that the lines intersect for all values of a.
(ii)
Given that the point of intersection is at a distance of 9 units from
the origin, find the possible values of a.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q7 November 2014
13 The straight line l 1 passes through the points (0, 1, 5) and (2, −2, 1). The
straight line l 2 has equation r = 7i + j + k + µ( i + 2 j + 5k ) .
(i)
Show that the lines l 1 and l 2 are skew.
(ii)
Find the acute angle between the direction of the line l 2 and the
direction of the x-axis.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q6 June 2015
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10
KEY POINTS
1
2
10 VECTORS
3
4
5
6
A vector quantity has magnitude and direction.
A scalar quantity has magnitude only.
⎯→
Vectors are typeset in bold, a or OA, or in the form⎯→
OA. They are
handwritten either in the underlined form a, or as OA.
The length (or modulus or magnitude) of the vector a is written as a
or as |a|.
Unit vectors in the x, y and z directions are denoted by i, j and k,
respectively.
A vector may be specified in
● magnitude−direction form: (r, θ ) (in two dimensions)
●
()
x
component form: x i + y j or y (in two dimensions)
⎛ x⎞
x i + y j + z k or ⎜ y ⎟ (in three dimensions).
⎝ z⎠
7
8
9
⎯
→
The position vector OP of a point P is the vector joining the origin
to P.
⎯
→
The vector AB is b − a, where a and b are the position vectors of A
and B.
The angle between two vectors, a and b, is given by θ in
cos θ =
a⋅b
|a||b|
where a . b = a1b1 + a2b2 (in two dimensions)
= a1b1 + a2b2 + a3b3 (in three dimensions).
⎯
→
10 The position vector OP of a point P is the vector joining the origin
to P.
⎯
→
11 The vector AB is b – a, where a and b are the position vectors of A
and B.
12 The vector r often denotes the position vector of a general point.
13 The vector equation of the line through A with direction vector u is
given by
r = a + λu.
14 The vector equation of the line through points A and B is given by
⎯
→
⎯
→
r = OA + λ AB
= a + λ(b − a)
= (1 − λ)a + λb.
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15 The vector equation of the line through (a1, a2, a3) in the direction
⎛a1 ⎞
⎛u 1 ⎞
⎛ u1 ⎞
⎜ u 2 ⎟ is r = ⎜a 2 ⎟ + λ ⎜u 2 ⎟ .
⎜ ⎟
⎜ ⎟
⎜u ⎟
⎝ 3⎠
⎝a 3 ⎠
⎝u 3 ⎠
Now you have finished this chapter, you should be able to
■ understand the terms vector and scalar
■ understand vectors in two and three dimensions, and express them
■ using i, j and k vectors
■ using column vectors
⎯→
■ using OA or a notation
■ understand equal vectors, position vectors and displacement vectors
■ understand the link between the coordinates of a point and its position
vector
■ multiply a vector by a scalar
■ add and subtract vectors
■ find a unit vector in the direction of a given vector
■ understand that vectors are parallel when one is a scalar multiple of the
other
■ find the vector equation of a line
■ calculate the scalar product between two vectors and use it to find the
angle between
■ two vectors
■ two lines
■ determine whether two lines are
■ parallel
■ intersect
■ skew
■ use vectors in geometry problems.
10.8 The perpendicular distance from a point to a line
LEARNING OUTCOMES
10
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P3
11 COMPLEX NUMBERS
11 Complex numbers
…that wonder
of analysis, that
portent of the
ideal world,
that amphibian
between being
and not-being,
which we call
the imaginary
root of negative
unity.
Gottfried
Wilhelm Leibniz
(1646–1716)
Real numbers
Rational numbers
Integers
Natural numbers
▲ Figure 11.1
What is the meaning of each of the terms shown in Figure 11.1?
❯ Suggest two numbers that could be placed in each part of the
diagram.
?
❯
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11.1 Extending the number system
11
The number system we use today has taken thousands of years to develop.
To classify the different types of numbers used in mathematics the following
letter symbols are used:
Natural numbers
&
Integers
'
Rational numbers
'
Irrational numbers
"
Real numbers
?
❯
Why is there no set shown on the diagram for irrational numbers?
You may have noticed that some of these sets of numbers fit within the other
sets. This can be seen in Figure 11.1.
11.1 Extending the number system
$
ACTIVITY 11.1
On a copy of Figure 11.1 write the following numbers in the correct
positions.
7
5
–13
227
109
– 5
3.1415
π
0.33
.
0.3
What are complex numbers?
ACTIVITY 11.2
Solve each of these equations and decide which set of numbers the roots
belong to in each case.
x+7=9
(ii) 7x = 9
(iii) x2 = 9
(iv) x + 10 = 9
(v) x2 + 7x = 0
(i)
Now think about the equation x2 + 9 = 0.
You could rewrite it as x2 = –9. However, since
the square of every real number is positive or zero,
there is no real number with a square of –9.This
is an example of a quadratic equation which, up
to now, you would have classified as having no
real roots.
9781510421738.indb 281
Writing this quadratic
equation as x2 + 0x + 9 = 0
and calculating the
discriminant for this
quadratic gives
b2 – 4ac = –36 which is
less than zero.
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The existence of such equations was recognised for hundreds of years, in
the same way that Greek mathematicians had accepted that x + 10 = 9 had
no solution; the concept of a negative number had yet to be developed.
The number system has expanded as mathematicians increased the range of
mathematical problems they wanted to tackle.
11
11 COMPLEX NUMBERS
You can solve the equation x2 + 9 = 0 by extending the number system to
include a new number, i (sometimes written as j). This has the property that
i2 = −1 and it follows the usual laws of algebra. i is called an imaginary
number.
The square root of any negative number can be expressed in terms of i. For
example, the solution of the equation x2 = −9 is x = ± −9. This can be
written as ± 9 × −1 which simplifies to ±3i.
11.2 Working with complex numbers
Faced with the problem of wanting the square root of a negative number, we
make the following Bold Hypothesis.
The real number system can be extended by including a new number, denoted by i,
which combines with itself and the real numbers according to the usual laws of algebra,
but which has the additional property that i2 = −1.
The original notation for i was ι, the Greek letter iota. The letter j is also
commonly used instead of i.
The first thing to note is that we do not need further symbols for other
square roots. For example, since −196 = 196 × (−1) = 142 × i2, we see that
−196 has two square roots, ±14i. The following example uses this idea to
solve a quadratic equation with no real roots.
Example 11.1
Solve the equation z2 − 6z + 58 = 0,
and check the roots.
We use the letter z for the variable
here because we want to keep x
and y to stand for real numbers.
Solution
Using the quadratic formula:
z=
=
6 ± 6 2 − 4 × 58
2
6 ± −196
2
6 ± 14i
2
= 3 ± 7i
=
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To check:
11
z = 3 + 7i ⇒ z2 − 6z + 58 = (3 + 7i)2 − 6(3 + 7i) + 58
= 9 + 42i + 49i2 − 18 − 42i + 58
Notice that here
0 means 0 + 0i.
= 9 + 42i − 49 − 18 − 42i + 58
=0
i2 = −1
Check the other root, z = 3 − 7i.
Notice in particular
that the imaginary
part is real!
A number z of the form x + iy, where x and y are real, is called a complex
number. x is called the real part of the complex number, denoted by Re(z),
and y is called the imaginary part, denoted by Im(z). So if, for example,
z = 3 − 7i then Re(z) = 3 and Im(z) = −7.
In Example 11.1 you did some simple calculations with complex numbers.
The general methods for addition, subtraction and multiplication are similarly
straightforward.
» Addition: add the real parts and add the imaginary parts.
11.2 Working with complex numbers
ACTIVITY 11.3
(x + iy) + (u + iv) = (x + u) + i(y + v)
» Subtraction: subtract the real parts and subtract the imaginary parts.
(x + iy) − (u + iv) = (x − u) + i(y − v)
» Multiplication: multiply out the brackets in the usual way and simplify,
remembering that i2 = −1.
(x + iy)(u + iv) = xu + ixv + iyu + i2yv
= (xu − yv) + i(xv + yu)
Division of complex numbers is dealt with later in the chapter.
What are the values of i3, i4, i5?
❯ Explain how you would work out the value of in for any positive
integer value of n.
?
❯
Complex conjugates
The complex number x − iy is called the complex conjugate, or just the
conjugate, of x + iy. Simarly x + iy is the complex conjugate of x − iy. x + iy
and x − iy are a conjugate pair. The complex conjugate of z is denoted by z*.
If a polynomial equation, such as a quadratic, has real coefficients, then any
complex roots will be conjugate pairs. This is the case in Example 11.1. If,
however, the coefficients are not all real, this is no longer the case.
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You can solve quadratic equations with complex coefficients in the same
way as an ordinary quadratic, either by completing the square or by using the
quadratic formula. This is shown in the next example.
11
Example 11.2
Solve z2 − 4iz − 13 = 0.
11 COMPLEX NUMBERS
Solution
Substitute a = 1, b = −4i and c = −13 into the quadratic formula.
2
z = −b ± b − 4ac
2a
4i ± (−4i) 2 − 4 × 1 × (−13)
=
2
= 4i ± −16 + 52
2
4i
36
±
=
2
= 4i ± 6
2
= 2i ± 3
So the roots are 3 + 2i and –3 + 2i.
ACTIVITY 11.4
(i)
Let z = 3 + 5i and w = 1 − 2i.
Find the following.
(a)
z + z*
(b)
w + w*
(c)
zz*
(d)
ww*
What do you notice about your answers?
(ii)
Let z = x + iy.
Show that z + z* and zz* are real for any values of x and y.
Equality of complex numbers
Two complex numbers z = x + i y and w = u + i v are equal if both x = u
and y = v. If y ≠ u or y ≠ v, or both, then z and w are not equal.
You may feel that this is obvious, but it is interesting to compare this situation
with the equality of rational numbers.
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The rational numbers x and u are equal if x = u and y = v.
y
v
x
u
❯ Is it possible for the rational numbers
y and v to be equal if u ≠ x
and v ≠ y?
Example 11.3
The complex numbers z1 and z2 are given by : z1 = (3 − a) + (2b − 4)i
and
z2 = (7b − 4) + (3a − 2)i
Given that z1 and z2 are equal, find the values of a and b.
(ii) Check your answer by substituting your values for a and b into the
expressions above.
(i)
Solution
(i)
(3 − a) + (2b − 4)i = (7 − 2b) + (3a − 2)i
Equating real parts:
3 − a = 7b − 4
11
11.2 Working with complex numbers
For two complex numbers to be equal, the real parts must be equal and the
imaginary parts must be equal. Using this result is described as equating real
and imaginary parts, as shown in the following example.
?
Equating real and
imaginary parts leads
to two equations.
Equating imaginary parts: 2b − 4 = 3a − 2
7b + a = 7
2b − 3a = 2
Simplifying the
equations.
Solving simultaneously gives b = 1 and a = 0.
(ii) Substituting a = 0 and b = 1 gives z1 = 3 − 2i and z2 = 3 − 2i
so z1 and z2 are indeed equal.
Exercise 11A
1
Express the following in the form x + iy.
(i) (8 + 6i) + (6 + 4i)
(ii) (9 − 3i) + (−4 + 5i)
(iii) (2 + 7i) − (5 + 3i)
(iv) (5 − i) − (6 − 2i)
(v) 3(4 + 6i) + 9(1 − 2i)
(vi) 3i(7 − 4i)
(vii) (9 + 2i)(1 + 3i)
(viii) (4 − i)(3 + 2i)
2
(ix) (7 + 3i)
(x) (8 + 6i)(8 − 6i)
(xi) (1 + 2i)(3 − 4i)(5 + 6i)
(xii) (3 + 2i)3
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2
11 COMPLEX NUMBERS
11
3
4
CP
5
CP
6
CP
7
Solve each of the following equations, and check the roots in each case.
(i) z2 + 2z + 2 = 0
(ii) z2 − 2z + 5 = 0
(iii) z2 − 4z + 13 = 0
(iv) z2 + 6z + 34 = 0
(v) 4z2 − 4z + 17 = 0
(vi) z2 + 4z + 6 = 0
Solve each of the following equations.
(i) z2 − 4iz − 4 = 0
(ii) z2 − 2iz + 15 = 0
(iii) z2 − 2iz − 2 = 0
(iv) z2 + 6iz − 13 = 0
(v) z2 + 8iz − 17 = 0
(vi) z2 + iz + 6 = 0
Given that z = 2 + 3i and w = 6 − 4i, find the following.
(i) Re(z)
(ii) Im(w)
(iii) z*
(iv) w*
(v) z* + w*
(vi) z* − w*
(vii) Im(z + z*)
(viii) Re(w − w*)
(ix) zz* − ww*
(x) (z3)*
(xi) (z*)3
(xii) zw* − z*w
Let z = x + iy.
Show that (z*)* = z.
Let z1 = x1 + iy1 and z2 = x2 + iy2.
Show that (z1 + z2)* = z1* + z2*.
Given that the complex numbers
z1 = a2 + (3 + 2b)i
z2 = (5a − 4) +b2i
are equal, find the possible values of a and b.
Hence list the possible pairs of complex numbers z1 and z2.
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Division of complex numbers
You can use the method of equating real and imaginary parts when you
divide by a complex number. The following example illustrates this.
Example 11.4
Find real numbers p and q such that p + iq =
1 .
3 + 5i
You need to find real numbers p and q such that
(p + iq)(3 + 5i) = 1.
Expanding gives
3p − 5q + i(5p + 3q) = 1.
Equating real and imaginary parts gives
3p − 5q = 1
Imaginary:
5p + 3q = 0
5
3
These simultaneous equations give p = 34
, q = − 34
and so
11.2 Working with complex numbers
Solution
Real:
11
1
3
5
=
−
i
3 + 5i 34 34
ACTIVITY 11.5
1
By writing x + iy = p + iq, show that
x − iy
1
=
.
x + iy x 2 + y 2
This result shows that there is an easier way to find the reciprocal of a
complex number. First, notice that
(x + iy)(x − iy) = x 2 − i2y 2
= x2 + y2
which is real.
So to find the reciprocal of a complex number you multiply numerator and
denominator by the complex conjugate of the denominator.
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1
Find the real and imaginary parts of 5 + 2i .
Solution
Multiply numerator and denominator by 5 − 2i.
5 − 2i
1 =
5 + 2i (5 + 2i)(5 – 2i)
5 − 2i
=
25 + 4
11 COMPLEX NUMBERS
11
Example 11.5
=
5 – 2i is the
conjugate of the
denominator, 5 + 2i.
5 − 2i
29
5
2
so the real part is 29
and the imaginary part is − 29
.
Note
You may have noticed that this process is very similar to the process of
rationalising a denominator. To make the denominator of
1
rational
3+ 2
you have to multiply the numerator and denominator by 3 −
2.
Similarly, division of complex numbers is carried out by multiplying both
numerator and denominator by the conjugate of the denominator, as in the
next example.
Example 11.6
Express
9 − 4i
as a complex number in the form x + iy.
2 + 3i
Solution
9 − 4i = 9 − 4i × 2 − 3i
2 + 3i 2 + 3i 2 − 3i
2
= 18 − 27i2 − 8i2+ 12i
2 +3
−
6
35i
=
13
6
35
= 13 − 13
i
The square root of a complex number
The next example shows you how to find the square root of a complex
number.
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Example 11.7
Find the two square roots of 8 + 6i.
11
Solution
Let (x + iy)2 = 8 + 6i
⇒
x 2 + 2ixy – y 2 = 8 + 6i
+i2y2 = –y2
Real:
x2 – y2 = 8
1
!
2xy = 6
2
!
Imaginary:
11.2 Working with complex numbers
Equating the real and imaginary parts gives:
2 gives
Rearranging !
y=3
x
3
1
Substituting ! into ! gives
x 2 − 92 = 8
x
x 4 − 9 = 8x 2
3
!
x 4 − 8x 2 − 9 = 0
( x 2 − 9)( x 2 + 1) = 0
⇒ x 2 = −1 which has no real roots
This is a
quadratic in x2.
Remember that
x and y are both
real numbers.
or x 2 = 9 ⇒ x = ±3.
When x = 3, y = 1
When x = –3, y = –1
So the square roots of 8 + 6i are 3 + i and –3 – i.
?
1 1
1
❯ What are the values of , 2 and 3 ?
i
i i
❯ Explain how you would work out the value of 1 for any positive
in
integer value of n.
Exercise 11B
1
Express these complex numbers in the form x + iy.
1
1
5i
(i)
(ii)
(iii)
3+ i
6−i
6 − 2i
(iv)
7 + 5i
6 − 2i
(v)
3 + 2i
1+ i
(vi)
47 − 23i
6+i
(vii)
2 − 3i
3 + 2i
(viii)
5 − 3i
4 + 3i
(ix)
6+i
2 − 5i
(x)
12 − 8i
( 2 + 2i) 2
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2
11
Find real numbers a and b with a > 0 such that
(a + ib)2 = 21 + 20i
(ii)
(a + ib)2 = −40 − 42i
(iii) (a + ib)2 = −5 − 12i
(iv)
(a + ib)2 = −9 + 40i
(vi)
(a + ib)2 = i.
(i)
(v)
11 COMPLEX NUMBERS
3
4
5
6
CP
7
PS
8
(a + ib)2 = 1 − 1.875i
Find real numbers a and b such that
a + b
3 + i 1 + 2i = 1 − i.
Solve these equations.
(i) (1 + i)z = 3 + i
(ii) (3 − 4i)(z − 1) = 10 − 5i
(iii) (2 + i)(z − 7 + 3i) = 15 − 10i
(iv) (3 + 5i)(z + 2 − 5i) = 6 + 3i
Find all the complex numbers z for which z 2 = 2z*.
For z = x + iy, find 1 + 1 in terms of x and y.
z zz*
Show that
z + zz*
(i) Re(z) =
2
z
−
zz*.
(ii) Im(z) =
2i
(i) Expand and simplify (a + ib)3.
(ii)
Deduce that if (a + ib)3 is real then either b = 0 or b 2 = 3a 2.
(iii) Hence find all the complex numbers z for which z 3 = 1.
PS
9
(i)
Expand and simplify (z − α )(z − β ).
Deduce that the quadratic equation with roots α and β is
z 2 − (α + β )z + αβ = 0,
that is:
z 2 − (sum of roots)z + product of roots = 0.
(ii)
Using the result from part (i), find quadratic equations in the form
az 2 + bz + c = 0 with the following roots.
7 + 4i, 7 − 4i
(b) 5i , − 5i
3
3
(c) −2 + 8 i , −2 − 8 i
(a)
(d)
2 + i, 3 + 2i
10 Find the two square roots of each of these.
(i)
–9
(iii) –16 + 30i
(v)
21 – 20i
(ii)
3 + 4i
(iv) –7 – 24i
(vi) –5 – 12i
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Representing complex numbers geometrically
Since each complex number x + iy
can be defined by the ordered pair
of real numbers (x, y), it is natural to
represent x + iy by the point with
cartesian coordinates (x, y).
4
2 + 3i
2
–4
2 + 3i is represented by (2, 3)
−5 − 4i is represented by (−5, −4)
2i
is represented by (0, 2)
7
is represented by (7, 0).
–2
O
2i
2
4
6
8 Re
–2
–5 – 4i
–4
▲ Figure 11.2
All real numbers are represented by points on the x-axis, which is therefore called
the real axis. Pure imaginary numbers (of the form 0 + iy) give points on the
y-axis, which is called the imaginary axis. It is useful to label these Re and
Im respectively.This geometrical illustration of complex numbers is called the
complex plane or the Argand diagram after Jean-Robert Argand (1768−1822),
a self-taught Swiss book-keeper who published an account of it in 1806.
11.2 Working with complex numbers
For example, in Figure 11.2,
11
Im
ACTIVITY 11.6
Copy Figure 11.2.
For each of the four given points z mark also the point −z.
Describe the geometrical transformation which maps the point
representing z to the point representing −z.
(ii) For each of the points z mark the point z*, the complex conjugate
of z. Describe the geometrical transformation which maps the point
representing z to the point representing z*.
(i)
You will have seen in this activity that the points representing z and −z have
half-turn symmetry about the origin, and that the points representing z and
z* are reflections of each other in the real axis.
❯
How would you describe points that are reflections of each other in
the imaginary axis?
?
Representing the sum and difference of complex
numbers
Several mathematicians before Argand had used the complex plane
representation. In particular, a Norwegian surveyor, Caspar Wessel (1745−1818),
wrote a paper in 1797 (largely ignored until it was republished in French
a century later) in which the complex number x + iy is represented by the
⎛x ⎞
position vector ⎜ y ⎟ , as shown in Figure 11.3.
⎝ ⎠
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Im
11
z = x + iy
11 COMPLEX NUMBERS
O
Re
▲ Figure 11.3
The advantage of this is that the addition of complex numbers can then be
shown by the addition of the corresponding vectors.
⎛ x 1 + x 2⎞
⎜⎝ y + y ⎟⎠
1
2
⎛ x 1⎞
⎛ x 2⎞
⎜⎝ y ⎟⎠ + ⎜⎝ y ⎟⎠ =
1
2
In an Argand diagram the position vectors representing z1 and z2 form two
sides of a parallelogram, the diagonal of which is the vector z1 + z2 (see
Figure 11.4).
Im
z2
z1 + z2
z1
O
Re
▲ Figure 11.4
You can also represent z by any other directed line segment with
⎛ x⎞
Im
components ⎜ ⎟ , not anchored at the origin as a
⎝ y⎠
position vector. Then addition can be shown as a
z1 + z2
triangle of vectors (see Figure 11.5).
z2
z1
O
Re
▲ Figure 11.5
If you draw the other diagonal of the parallelogram,
and let it represent the complex number w (see
Figure 11.6), then
Im
w = z 1 – z2
z2
z 2 + w = z 1 ⇒ w = z 1 − z 2.
w
z1
O
Re
▲ Figure 11.6
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This gives a useful illustration of subtraction: the complex number z 1 − z 2
is represented by the vector from the point representing z 2 to the point
representing z 1, as shown in Figure 11.7. Notice the order of the points: the
vector z 1 − z 2 starts at the point z 2 and goes to the point z 1.
11
Im
z2
z1
O
Re
▲ Figure 11.7
ACTIVITY 11.7
Draw a diagram to illustrate z2 − z1.
(ii) Draw a diagram to illustrate that z1 − z2 = z1 + (−z2).
Show that z1 + (−z2) gives the same vector, z1 − z2 as before, but
represented by a line segment in a different place.
(i)
11.2 Working with complex numbers
z1 – z2
The modulus of a complex number
Figure 11.8 shows the point representing z = x + iy on an Argand diagram.
Im
x + iy
y
O
x
Re
▲ Figure 11.8
Using Pythagoras’ theorem, you can see that the distance of this point from the
origin is x 2 + y 2 . This distance is called the modulus of z, and is denoted by | z |.
So for the complex number z = x + iy, | z | =
x2 + y2 .
If z is real, z = x say, then | z | = x 2 , which is the absolute value of x, i.e. | x |.
So the use of the modulus sign with complex numbers fits with its previous
meaning for real numbers.
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11
Exercise 11C
1
Represent each of the following complex numbers on a single Argand
diagram, and find the modulus of each complex number.
(i)
3 + 2i
(iv) −2
11 COMPLEX NUMBERS
4i
(iii)
−5 + i
(viii)
−6 − 5i
(vi)
4 − 3i
Given that z = 2 − 4i, represent the following by points on a single
Argand diagram.
2
z
(ii)
−z
(iii)
z*
(iv) −z*
(v)
iz
(vi)
−iz
(vii) iz*
(viii)
(iz)*
(i)
PS
(ii)
3
Given that z = 10 + 5i and w = 1 + 2i, represent the following complex
numbers on an Argand diagram.
(i) z
(ii)
w
(iii) z + w
(iv) z − w
(v)
w−z
4
Given that z = 3 + 4i and w = 5 − 12i, find the following.
|z|
(ii)
z
(v)
(iv) w
What do you notice?
(i)
| |
|w|
w
z
| |
(iii)
| zw |
CP
5
Let z = 1 + i.
(i) Find zn and | zn | for n = −1, 0, 1, 2, 3, 4, 5.
(ii) Plot each of the points zn from part (i) on a single Argand diagram.
Join each point to its predecessor and to the origin.
(iii) What do you notice?
CP
6
Give a geometrical proof that (−z)* = −(z*).
11.3 Sets of points in an Argand
diagram
In the last section, you saw that | z | is the distance of the point
representing z from the origin in the Argand diagram.
❯
?
What do you think that | z 2 − z 1 | represents?
If z 1 = x 1 + iy1 and z 2 = x 2 + iy2 , then z 2 − z 1 = x 2 − x 1 + i(y2 − y1).
So | z 2 − z 1 | =
( x 2 − x 1 )2 + ( y 2 − y1 )2 .
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Figure 11.9 shows an Argand diagram with the points representing the
complex numbers z 1 = x 1 + iy1 and z 2 = x 2 + iy2 marked.
Im
11
x2 + iy2
y2 – y1
x2 – x1
O
Re
▲ Figure 11.9
Using Pythagoras’ theorem, you can see that the distance between z 1 and
z 2 is given by ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 .
So | z 2 − z 1 | is the distance between the points z 1 and z 2.
This is the key to solving many questions about sets of points in an Argand
diagram, as in the following examples.
Example 11.8
11.3 Sets of points in an Argand diagram
x1 + iy1
Draw an Argand diagram showing the set of points z for which
| z − 3 − 4i | = 5.
Solution
| z − 3 − 4i | can be written as | z − (3 + 4i) |, and this is the distance from the
point 3 + 4i to the point z.
This equals 5 if the point z lies on the circle with centre 3 + 4i and radius 5
(see Figure 11.10).
Im
3 + 4i
O
Re
▲ Figure 11.10
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11 COMPLEX NUMBERS
11
❯
?
How would you show the sets of points for which
(i) | z − 3 − 4i | $ 5
(ii) | z − 3 − 4i | " 5
(iii) | z − 3 − 4i | % 5?
Example 11.9
Draw an Argand diagram showing the set of points z for which
| z − 3 − 4i | $ | z + 1 − 2i |.
Solution
The condition can be written as | z − (3 + 4i) | $ | z − (−1 + 2i) |.
| z − (3 + 4i) | is the distance of point z from the point 3 + 4i, point A in
Figure 11.11, and | z − (−1 + 2i) | is the distance of point z from the point
−1 + 2i, point B in Figure 11.11.
Im
A 3 + 4i
B
–1 + 2i
O
Re
▲ Figure 11.11
These distances are equal if z is on the perpendicular bisector of AB.
So the given condition holds if z is on this bisector or in the half plane on
the side of it containing A, shown shaded in Figure 11.11.
❯
How would you show the sets of points for which
?
(i) | z − 3 − 4i | = | z + 1 − 2i |
(ii) | z − 3 − 4i | " | z + 1 − 2i |
(iii) | z − 3 − 4i | # | z + 1 − 2i |?
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Exercise 11D
1
For each of parts (i) to (viii), draw an Argand diagram showing the set of
points z for which the given condition is true.
|z |=2
(ii)
|z − 4| $ 3
(iii) | z − 5i | = 6
(iv)
| z + 3 − 4i | " 5
(vi)
| z + 2 + 4i | = 0
(i)
(v)
|6− i − z| % 2
CP
2
Draw an Argand diagram showing the set of points z for which
| z − 12 + 5i | $ 7. Use the diagram to prove that, for these z,
6 $ | z | $ 20.
PS
3
(i)
(ii)
PS
4
PS
5
On an Argand diagram, show the region R for which
| z − 5 + 4i | $ 3.
Find the greatest and least values of | z + 3 − 2i | in the region R.
By using an Argand diagram see if it is possible to find values of z for
which | z − 2 + i | % 10 and | z + 4 + 2i | $ 2 simultaneously.
For each of parts (i) to (iv), draw an Argand diagram showing the set of
points z for which the given condition is true.
(i) | z | = | z − 4 |
(ii) | z | % | z − 2i |
(iii) | z + 1 − i | = | z − 1 + i |
(iv) | z + 5 + 7i | $ | z − 2 − 6i |
11.4 The modulus−argument form of
complex numbers
11.4 The modulus−argument form of complex numbers
(viii) Re(z) = −2
(vii) 2 $ | z − 1 + i | $ 3
11
The position of the point z in an Argand diagram can be described by means
of the length of the line connecting this point to the origin, and the angle
which this line makes with the positive real axis (see Figure 11.12).
Im
z
When describing
complex numbers, it is
usual to give the angle
θ in radians.
r
θ
O
Re
▲ Figure 11.12
The distance r is of course | z |, the modulus of z as defined on page 293.
The angle θ is slightly more complicated: it is measured anticlockwise from
the positive real axis, normally in radians. However, it is not uniquely
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11
defined since adding any multiple of 2π to θ gives the same direction. To
avoid confusion, it is usual to choose that value of θ for which −π " θ $ π.
This is called the principal argument of z, denoted by arg z. Then
every complex number except zero has a unique principal argument. The
argument of zero is undefined.
For example, with reference to Figure 11.13,
11 COMPLEX NUMBERS
arg(−4) = π
Im
–3 + 3i
π
arg(−2i) = − 2
–4
arg(1.5) = 0
–2
0
1.5
Re
–2i
3π
arg(−3 + 3i) = 4
▲ Figure 11.13
Remember that
π radians = 180°.
❯
?
Without using your calculator, state the values of the following.
(i) arg i
(ii) arg(−4 − 4i)
(iii) arg(2 − 2i)
You can see from Figure 11.14 that
x = r cos θ
r=
y = r sin θ
y
tan θ = x
x2 + y2
and the same relations hold in the other quadrants too.
Im
r
y
θ
O
x
Re
▲ Figure 11.14
Since x = r cos θ and y = r sin θ, we can write the complex number z = x + iy in
the form
z = r (cos θ + i sin θ ).
This is called the modulus−argument or polar form.
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ACTIVITY 11.8
(i) Set your calculator to degrees and use it to find the following.
(a) tan−1 1
(b) tan−1 2
(c) tan−1 100
(d) tan−1(−2)
(e) tan−1(−50)
(f) tan−1(−200)
11
What are the largest and smallest possible values, in degrees, of tan−1 x?
Find tan−1 x for some different values of x.
What are the largest and smallest possible values, in radians, of tan−1 x?
If you know the modulus and argument of a complex number, it is easy to use
the relations x = r cos θ and y = r sin θ to find the real and imaginary parts of the
complex number.
Similarly, if you know the real and imaginary parts, you can find the modulus
and argument of the complex number using the relations r = x 2 + y 2 and
y
tan θ = , but you do have to be quite careful in finding the argument. It is
x
⎛y⎞
tempting to say that θ = tan−1 ⎜ x ⎟ , but, as you saw in the last activity, this gives
⎝ ⎠
a value between − π and π , which is correct only if z is in the first or fourth
2
2
quadrants.
For example, suppose that the point z1 = 2 − 3i has argument θ1, and the point
z2 = −2 + 3i has argument θ2. It is true to say that tan θ1 = tan θ2 = − 23 . In the
case of z1, which is in the fourth quadrant, θ1 is correctly given by
11.4 The modulus−argument form of complex numbers
(ii) Now set your calculator to radians.
tan−1 (− 23 ) ≈ −0.98 rad (≈ −56°). However, in the case of z2, which is in the
second quadrant, θ2 is given by (− 23 ) + π ≈ 2.16 rad (≈ 124°). These two
points are illustrated in Figure 11.15.
Im
z2
θ2
–2
0
θ1 2
Re
–2
z1
▲ Figure 11.15
Figure 11.16 shows the values of the argument in each quadrant. It is wise
always to draw a sketch diagram when finding the argument of a complex
number.
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11
+
π
2
arg +
arg +
π is positive
y
arg z = tan−1 x + π
()
y
arg z = tan ( x ) – π
−1
11 COMPLEX NUMBERS
arg –
arg –
y
arg z = tan−1 x
()
y
arg z = tan ( x )
−1
–π
2
▲ Figure 11.16
ACTIVITY 11.9
Mark the points 1 + i, 1 − i, −1 + i, −1 − i on an Argand diagram.
Find arg z for each of these, and check that your answers are consistent
with Figure 11.16.
Note
The modulus−argument form of a complex number is also called the polar
form, as the modulus of a complex number is its distance from the origin,
sometimes called the pole.
ACTIVITY 11.10
Most calculators can convert from (x, y) to (r, θ ) (called rectangular to polar,
and often shown as R → P) and from (r, θ ) to (x, y) (polar to rectangular,
P → R). Find out how to use these facilities on your calculator, and
compare with other available types of calculator.
Does your calculator always give the correct θ, or do you sometimes have
to add or subtract π (or 180°)?
A complex number in polar form must be given in the form z = r (cos θ + i sin θ ),
not, for example, in the form z = r (cos θ − i sin θ ). The value of r must also
be positive. So, for example, the complex number −2(cos θ + i sin θ ) is not in
polar form. However, by using some of the relationships
cos (π − α) = −cos α
cos (α − π) = −cos α
cos (−α) = cos α
sin (π − α) = sin α
sin (α − π) = −sin α
sin (−α) = −sin α
you can rewrite the complex number, for example
−2(cos α + i sin α) = 2(−cos α − i sin α)
= 2(cos (α − π) + i sin (α − π)).
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This is now written correctly in polar form. The modulus is 2 and the argument
is α − π.
02/02/18 1:16 PM
❯
?
How would you rewrite the following in polar form?
(i)
−2(cos α − i sin α)
(ii)
2(cos α − i sin α)
11
When you use the polar form of a complex number, remember to give the
argument in radians, and to use a simple rational multiple of π where possible.
11.4 The modulus−argument form of complex numbers
ACTIVITY 11.11
Copy and complete this table.
Give your answers in terms of 2 or 3 where appropriate, rather than as
decimals. You may find Figure 11.17 helpful.
π
π
4
π
6
3
2
tan
2
1
sin
1
cos
1
1
▲ Figure 11.17
Example 11.10
Write the following complex numbers in polar form.
(i)
4 + 3i
(ii)
−1 + i
(iii)
−1 − 3 i
Solution
(i)
x = 4, y = 3
Modulus =
32 + 4 2 = 5
Since 4 + 3i lies in the first quadrant, the argument = tan−1 43 .
4 + 3i = 5(cos α + i sin α), where α = tan−1 43 ≈ 0.644 radians
(ii)
x = −1, y = 1
Modulus = 12 + 12 = 2
Since −1 + i lies in the second quadrant,
argument = tan−1(−1) + π
π
3π
= −4 +π = 4
−1 + i = 2 cos 3π + i sin 3π
4
4
(
)
➜
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11
11 COMPLEX NUMBERS
x = −1, y = − 3
(iii)
Modulus = 1 + 3 = 2
Since −1 − 3i lies in the third quadrant,
argument = tan−1 3 − π
π
2π
= 3−π =− 3
−1 − 3i = 2 cos − 2π + i sin − 2 π
3
3
( ( )
Exercise 11E
1
Write down the values of the modulus and the principal argument of
each of these complex numbers.
cos 2.3 + i sin 2.3
(i)
(ii)
8 cos π + i sin π
5
5
4
π
π
(iii) 4 cos − i sin
(iv)
−3(cos (−3) + i sin (−3))
3
3
(
(
2
( ))
)
)
For each complex number, find the modulus and principal argument,
and hence write the complex number in polar form.
Give the argument in radians, either as a simple rational multiple of π or
correct to 3 decimal places.
1
(ii)
−2
(iii) 3i
(iv) −4i
(v)
1+i
(vi)
(i)
(vii) 1 −
3i
(viii) 6 3 + 6i
−5 − 5i
(ix) 3 − 4i
−12 + 5i
(xi) 4 + 7i
(xii) −58 − 93i
Write each complex number with the given modulus and argument in
the form x + iy, giving surds in your answer where appropriate.
π
π
(i) | z | = 2, arg z = 2
(ii) | z | = 3, arg z = 3
5π
π
(iii) | z | = 7, arg z = 6
(iv) | z | = 1, arg z = − 4
2π
(v) | z | = 5, arg z = − 3
(vi) | z | = 6, arg z = −2
Given that arg(5 + 2i) = α, find the principal argument of each of the
following in terms of α.
(i) −5 − 2i
(ii)
5 − 2i
(iii) −5 + 2i
(iv) 2 + 5i
(v)
−2 + 5i
(x)
3
4
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5
The variable complex number z is given by
z = 1 + cos 2θ + i sin 2θ,
where θ takes all values in the interval − 21 π " θ " 21 π.
(i) Show that the modulus of z is 2 cos θ and the argument of z is θ.
1
(ii) Prove that the real part of
z is constant.
6
The variable complex number z is given by
z = 2 cos θ + i(1 – 2 sin θ ),
where θ takes all values in the interval −π " θ $ π.
(i)
(ii)
Show that | z − i | = 2, for all values of θ. Hence sketch, in an
Argand diagram, the locus of the point representing z.
1
Prove that the real part of z + 2 − i is constant for −π " θ " π.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q5 June 2008
7
4 − 3i
.
1 − 2i
(a) Express z in the form x + iy, where x and y are real.
(b) Find the modulus and argument of z.
Find the two square roots of the complex number 5 – 12i, giving
your answers in the form x + iy, where x and y are real.
The complex number z is given by z =
(i)
(ii)
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q8 November 2007
8
The complex number 2 + i is denoted by u. Its complex conjugate is
denoted by u*.
(i) Show, on a sketch of an Argand diagram with origin O, the points
A, B and C representing the complex numbers u, u* and u + u*
respectively. Describe in geometrical terms the relationship between
the four points O, A, B and C.
u
(ii) Express ui in the form x + iy, where x and y are real.
*
u
(iii) By considering the argument of ui , or otherwise, prove that
*
tan–1 43 = 2 tan–1 21 .
()
9
11.4 The modulus−argument form of complex numbers
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q8 June 2010
11
()
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q7 June 2006
The complex number z is defined by z = a + ib, where a and b are real.
The complex conjugate of z is denoted by z*.
Show that | z|2 =zz* and that(z − ki)* = z* + ki , where k is real.
In an Argand diagram a set of points representing complex numbers z is
defined by the equation
(i)
| z − 10i| = 2| z − 4i|.
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11 COMPLEX NUMBERS
11
(ii)
Show, by squaring both sides, that
zz* − 2iz* + 2iz − 12 = 0.
Hence show that | z − 2i| = 4.
(iii) Describe the set of points geometrically.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q7 June 2013
11.5 Sets of points using the polar
form
?
You already know that arg z gives the angle between the line connecting
the point z with the origin and the real axis.
❯ What do you think arg (z2 − z1) represents?
If z1 = x 1 + iy1 and z2 = x 2 + iy 2, then z 2 − z 1 = x 2 − x 1 + i(y2 − y1).
y 2 − y1
arg (z 2 − z 1) = tan−1 x − x
2
1
Figure 11.18 shows an Argand diagram with the points representing the
complex numbers z1 = x 1 + iy1 and z2 = x 2 + iy 2 marked.
Im
x2 + iy2
y2 – y1
x1 + iy1
x2 – x1
O
Re
▲ Figure 11.18
The angle between the line joining z1 and z2 and a line parallel to the real
axis is given by
y 2 − y1
tan−1 x − x .
2
1
So arg (z1 − z2) is the angle between the line joining z1 and z2 and a line parallel
to the real axis.
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Example 11.11
Draw Argand diagrams showing the sets of points z for which
π
(i) arg z = 4
π
(ii) arg (z − i) = 4
π
(iii) 0 $ arg (z − i) $ 4 .
(i)
π
arg z = 4
π
the line joining the origin to the point z has direction 4
π
⇔ z lies on the half-line from the origin in the 4 direction, see
Figure 11.19.
⇔
Im
O
Re
11.5 Sets of points using the polar form
Solution
11
▲ Figure 11.19
(ii)
(Note that the origin is not included, since arg 0 is undefined.)
π
arg (z − i) = 4
π
⇔ the line joining the point i to the point z has direction 4
π
⇔ z lies on the half-line from the point i in the 4 direction, see
Figure 11.20.
Im
i
O
Re
▲ Figure 11.20
π
(iii) 0 $ arg (z − i) $ 4
⇔
the line joining the point i to the point z has direction between 0
π
and 4 (inclusive)
⇔ z lies in the one-eighth plane shown in Figure 11.21.
Im
i
O
▲ Figure 11.21
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Re
305
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11 COMPLEX NUMBERS
11
Exercise 11F
1
2
3
For each of parts (i) to (vi) draw an Argand diagram showing the set of
points z for which the given condition is true.
π
(i) arg z = − 3
(ii) arg (z − 4i) = 0
π
3π
(iii) arg (z + 3) % 2
(iv) arg (z + 1 + 2i) = 4
π
π
π
(v) arg (z − 3 + i) $ −
(vi) − $ arg (z + 5 − 3i) $ 3
6
4
Find the least and greatest possible values of arg z if | z − 8i | $ 4.
3
The complex number w is given by w = − 21 + i .
2
(i) Find the modulus and argument of w.
(ii) The complex number z has modulus R and argument θ, where
− 13 π " θ " − 13 π. State the modulus and argument of wz and the
z
modulus and argument of w .
(iii) Hence explain why, in an Argand diagram, the points representing
z
z, wz and w are the vertices of an equilateral triangle.
(iv) In an Argand diagram, the vertices of an equilateral triangle lie on
a circle with centre at the origin. One of the vertices represents the
complex number 4 + 2i. Find the complex numbers represented by
the other two vertices. Give your answers in the form x + iy, where x
and y are real and exact.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q10 November 2008
4
The complex number w is defined by w = 22 + 4i2 .
(2 − i)
(i) Without using a calculator, show that w = 2 + 4i.
(ii) It is given that p is a real number such that 41 π & arg(w + p ) & 43 π.
Find the set of possible values of p.
(iii) The complex conjugate of w is denoted by w*. The complex
numbers w and w* are represented in an Argand diagram by the
points S and T respectively. Find, in the form | z − a | = k, the
equation of the circle passing through S,T and the origin.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q8 June 2015
5
Solve the equation z2 + (2√3)iz − 4 = 0, giving your answers in the
form x + iy, where x and y are real.
(ii) Sketch an Argand diagram showing the points representing the
roots.
(iii) Find the modulus and argument of each root.
(iv) Show that the origin and the points representing the roots are the
vertices of an equilateral triangle.
(i)
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q7 June 2009
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6
Cambridge International AS & A Level Mathematics
9709 Paper 32 Q7 November 2009
7
2
The complex number −1 + i is denoted by u.
(i) Find the modulus and argument of u and u2.
(ii) Sketch an Argand diagram showing the points representing the
complex numbers u and u2. Shade the region whose points
represent the complex numbers z which satisfy both the
inequalities | z | " 2 and | z − u2 | " | z − u |.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q8 June 2007
8
The complex number u is defined by u = 6 − 3i .
1 + 2i
(i) Showing all your working, find the modulus of u and show that the
argument of u is − 21 π.
11
11.6 Working with complex numbers in polar form
The complex numbers −2 + i and 3 + i are denoted by u and v respectively.
(i) Find, in the form x + iy, the complex numbers
(a) u + v,
u
(b) v , showing all your working.
u
(ii) State the argument of v .
In an Argand diagram with origin O, the points A, B and C represent the
complex numbers u, v and u + v respectively.
3
(iii) Prove that angle AOB = 4 π.
(iv) State fully the geometrical relationship between the line segments
OA and BC.
For complex numbers z satisfying arg ( z − u ) = 41 π, find the least
possible value of | z |.
(iii) For complex numbers z satisfying | z − (1 + i) u | = 1, find the
greatest possible value of | z |.
(ii)
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q8 June 2011
11.6 Working with complex numbers
in polar form
The polar form quickly leads to an elegant geometrical interpretation of the
multiplication of complex numbers. For if
z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2)
then
z1z2 = r1r2(cos θ1 + i sin θ1)(cos θ2 + i sin θ2)
= r1r2[cos θ1cos θ2 − sin θ1 sin θ2 + i (sin θ1 cos θ2 + cos θ1 sin θ2)].
Using the compound-angle formulae gives
z1z2 = r1r2[cos (θ1 + θ2) + i sin (θ1 + θ2)].
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02/02/18 1:16 PM
11
This is the complex number with modulus r1r2 and argument (θ1 + θ2), so we
have the beautiful result that
| z1z2 | = | z1 | | z2 |
and
11 COMPLEX NUMBERS
arg (z1z2) = arg z1 + arg z2 (±2π if necessary, to give the
principal argument).
So to multiply complex numbers in polar form you multiply their moduli and
add their arguments.
ACTIVITY 11.12
Using this interpretation, investigate
(i)
multiplication by i
(ii)
multiplication by –1.
z
The corresponding results for division are easily obtained by letting 1 = w.
z2
Then z1 = wz2 so that
| z1 | = | w | | z2 | and arg z1 = arg w + arg z2 (±2π if necessary).
and
| |
z1
| z1 |
= ––––
z2
| z2 |
z
arg w = arg 1
z2
= arg z1 – arg z2 (±2π if necessary, to give the principal argument).
Therefore | w | =
So to divide complex numbers in polar form you divide their moduli and
subtract their arguments.
This gives the following simple geometrical interpretation of multiplication
and division.
To obtain the vector z1z2 enlarge the vector z1 by the scale factor | z2 | and
rotate it through arg z2 anticlockwise about O (see Figure 11.22 (ii)).
1
z
To obtain the vector z 1 enlarge the vector z1 by scale factor –––– and rotate
|
z
2
2|
it clockwise through arg z2 about O (see Figure 11.22 (iii)).
This combination of an enlargement followed by a rotation is called a spiral
dilatation.
In summary:
and
| z1z2 | = | z1 | | z2 |
arg (z1z2) = arg z1 + arg z2
| z1 | | z1 |
–––– = ––––
| z2 | | z2 |
⎛z ⎞
arg ⎜ 1 ⎟ = arg z1 − arg z2
⎝z2 ⎠
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Im
9
arg (z1z2) = arg z1 + arg z2
8
7
Im
6
5
5
r1r2
4
z1
3
3
r1
2
2
r2
1
z2
1
2
3
4 Re
(i) z1 and z2
O
3
arg z2
1
arg z2
4
z1
r1
2
1
arg z1
1
2
3
4 Re
O
(ii) Multiplying z1 by z2
z1
r1
arg z2
z1
r1 z2
1 r2 2 3
4 Re
(iii) Dividing z1 by z2
▲ Figure 11.22
ACTIVITY 11.13
Check this by accurate drawing and measurement for the case
z1 = 2 + i, z2 = 3 + 4i. Then do the same with z1 and z2 interchanged.
Example 11.12
Find
(
) (
)
(ii) 6 (cos π + i sin π ) ÷ 2 (cos π + i sin π )
2
2
4
4
(i)
6 cos π + i sin π × 2 cos π + i sin π
2
2
4
4
11.6 Working with complex numbers in polar form
4
11
Im
6
6
5
O
⎛z ⎞
arg ⎜ z 1 ⎟ = arg z1 − arg z2
⎝ 2⎠
z1z2
Solution
(i)
Remember that
r1(cos θ1 + i sin θ1) × r2(cos θ2 + i sin θ2) = r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2))
So to multiply complex numbers you
6 × 2 = 12; π + π = 3π
2 4
4
» multiply the moduli
» add the arguments.
(
) (
)
(
6 cos π + i sin π × 2 cos π + i sin π = 12 cos 3π + i sin 3π
2
2
4
4
4
4
(ii)
To divide complex numbers you
» divide the moduli
» subtract the arguments.
(
) (
)
6 ÷ 2 = 3; π − π = π
2 4 4
) (
6 cos π + i sin π ÷ 2 cos π + i sin π = 3 cos π + i sin π
2
2
4
4
4
4
)
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11
This leads to an alternative method of finding
the square root of a complex number.
Im
Writing 8 + 6i in polar form gives
r
11 COMPLEX NUMBERS
r = 8 2 + 6 2 = 10
tan θ = 6 ⇒ θ = 0.6435 radians
8
So 8 + 6i = 10(cos 0.6435 + i sin 0.6435).
6
0.6435...
8
O
Re
Notice that if you add 2π to the argument you will ▲ Figure 11.23
end up with exactly the same complex number on
the Argand diagram (as you have just rotated through one full turn).
So 8 + 6i is also the same as 10[cos(0.6435 + 2π) + i sin(0.6435 + 2π)].
Let r(cos θ + i sin θ ) be the square root of 8 + 6i so that
r (cos θ + i sin θ ) × r (cos θ + i sin θ ) = r 2 (cos 2θ + i sin 2θ ) = 8 + 6i
⇒
r(cos 2θ + i sin 2θ ) = 10(cos 0.6435 + i sin 0.6435)
and
2
r(cos 2θ + i sin 2θ ) = 10[cos(0.6435 + 2π) + i sin(0.6435 + 2π)] !
1 or !
2
From !
r 2 = 10
⇒
r = 10
1
From !
2θ = 0.6435
⇒
θ = 0.321 75
⇒
θ = 0.321 75 + π
2
From !
2θ = 0.6435 + 2π
So one square root of 8 + 6i is
10 (cos 0.321 75 + i sin 0.321 75) = 3 + i
The square
root of the
modulus of
8 + 6i.
Half of the argument
of 8 + 6i, plus π.
and the other square root is
10 [cos(0.321 75 + π) + i sin(0.321 75 + π)] = −3 − i
Compare this method with that used on page 288 to find
the square roots of 8 + 6i.
❯
1
!
Half of the
argument of
8 + 6i plus π.
What are the square roots of
10[cos(0.6435 + 2nπ) + i sin(0.6435 + 2nπ)], where n is an integer?
?
11.7 Complex exponents
When multiplying complex numbers in polar form you add the arguments,
and when multiplying powers of the same base you add the exponents. This
suggests that there may be a link between the familiar expression cos θ + i sin θ
and the seemingly remote territory of the exponential function. This was first
noticed in 1714 by the young Englishman Roger Cotes two years before his
death at the age of 28 (when Newton remarked ‘If Cotes had lived we might
have known something’), and made widely known through an influential
book published by Euler in 1748.
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Let z = cos θ + i sin θ. Since i behaves like any other constant in algebraic
manipulation, to differentiate z with respect to θ you simply differentiate the
real and imaginary parts separately. This gives
dz
dθ = –sin θ + i cos θ
= i2 sin θ + i cos θ
11
= i(cos θ + i sin θ )
So z = cos θ + i sin θ is a solution of the differential equation dz = iz.
dθ
If i continues to behave like any other constant when it is used as an index,
then the general solution of dz = iz is z = eiθ+c, where c is a constant, just as
dθ
dx
x = ekt+c is the general solution of dt = kx.
cos θ + i sin θ = eiθ+c.
Therefore
11.7 Complex exponents
= iz
Putting θ = 0 gives
cos θ + i sin θ = e0+c
⇒ 1 = ec
⇒c=0
and it follows that
cos θ + i sin θ = eiθ.
The problem with this is that you have no way of knowing how i behaves as
an index. But this does not matter. Since no meaning has yet been given to
ez when z is complex, the following definition can be made, suggested by this
work with differential equations but not dependent on it:
eiθ = cos θ + i sin θ.
Note
The particular case when θ = π gives eiπ = cos π + i sin π = –1, so that
eiπ + 1 = 0.
This remarkable statement, linking the five fundamental numbers 0, 1, i,
e and π, the three fundamental operations of addition, multiplication and
exponentiation, and the fundamental relation of equality, has been described
as a ‘mathematical poem’.
Example 11.13
Find
(i)
(ii)
(a)
4e 5i × 3e 2i
(b)
6e 9i ÷ 3e 2i
Write these results as complex numbers in polar form.
➜
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Solution
11
(i)
11 COMPLEX NUMBERS
(ii)
Exercise 11G
1
(a)
4e 5i × 3e 2i = 12e 7i
4 × 3 = 12; 5i + 2i = 7i
(b)
6e 9i ÷ 3e 2i = 2e 7i
6 ÷ 3 = 2; 9i − 2i = 7i
(a)
4(cos 5 + i sin 5) × 3(cos 2 + i sin 2) = 12(cos 7 + i sin 7)
(b)
6(cos 9 + i sin 9) ÷ 3(cos 2 + i sin 2) = 2(cos 7 + i sin 7)
Find the following.
8(cos 0.2 + i sin 0.2) × 4(cos 0.4 + i sin 0.4)
(i)
(ii) 8(cos 0.2 + i sin 0.2) ÷ 4(cos 0.4 + i sin 0.4)
π
π × 2 cos π + i sin π
(iii) 6 cos + i sin
3
3
6
6
π
π
π
÷ 2 cos + i sin π
(iv) 6 cos + i sin
6
6
3
3
π
π
(v) 12(cos π + i sin π) × 2 cos + i sin
4
4
π
π
(vi) 12(cos π + i sin π) ÷ 2 cos + i sin
4
4
Given that z = 2 cos π + i sin π and w = 3 cos π + i sin π , find the
3
3
4
4
following in polar form.
w
(i) wz
(ii)
(iii) z
z
w
1
(iv)
(v) w 2
(vi) z 5
z
(vii) w 3z 4
(viii) 5iz
(ix) (1 + i)w
(
(
2
(
) (
) (
(
(
)
)
)
)
)
(
)
1
Prove that, in general, arg z = –arg z, and deal with the exceptions.
CP
3
CP
4
Given the points 1 and z on an Argand diagram, explain how to find the
following points by geometrical construction.
(i) 3z
(ii) 2iz
(iii) (3 + 2i)z
(iv) z*
(v)
| z |
(vi) z2
CP
5
Find the real and imaginary parts of
Express –1 + i and 1 +
6
−1 + i
.
1+ 3 i
3 i in polar form.
Hence show that cos 5π = 3 − 1, and find an exact expression for
2 2
12
5π
sin
.
12
Express ez in the form x + iy where z is the given complex number.
(i)
–iπ
(ii)
iπ
4
(iii)
2 + 5iπ
6
(iv)
3 – 4i
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7
Find the following.
(i) (a) 2e 3i × 5e −2i
(b) 8e 5i ÷ 2e 5i (c) 3e 7i × 2e i
(d) 12e 5i ÷ 4e 4i
(e) 3e 2i × e i
(f) 8e 3i ÷ 2e 4i
(ii) Write these results as complex numbers in polar form.
8
(i) It is given that (1 + 3i)w = 2 + 4i . Showing all necessary working,
(ii)
On a single Argand diagram sketch the loci | z | = 5 and
| z − 5 | = | z |. Hence determine the complex numbers represented
by points common to both loci, giving each answer in the form reiθ.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q9 November 2015
9
The complex numbers w and z are defined by w = 5 + 3i and z = 4 + i.
(i)
(ii)
Express iw in the form x + iy, showing all your working and giving
z
the exact values of x and y.
Find wz and hence, by considering arguments, show that
tan −1 ( 53 ) + tan −1 ( 41 ) = 41 π.
11.8 Complex numbers and equations
prove that the exact value of w 2 is 2 and find arg(w2) correct to
3 significant figures.
11
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q5 November 2014
11.8 Complex numbers and equations
The reason for inventing complex numbers was to provide solutions for
quadratic equations which have no real roots, i.e. to solve az2 + bz + c = 0
when the discriminant b2 − 4ac is negative. This is straightforward since if
b2 − 4ac = −k2 (where k is real) then the formula for solving quadratic
equations gives z = −b ± ik . These are the two complex roots of the
2a
equation. Notice that when the coefficients of the quadratic equation are real,
these roots are a pair of conjugate complex numbers.
It would be natural to think that to solve cubic equations would require a
further extension of the number system to give some sort of ‘super-complex’
numbers, with ever more extensions to deal with higher degree equations.
But luckily things are much simpler. It turns out that all polynomial
equations (even those with complex coefficients) can be solved by means of
complex numbers. This was realised as early as 1629 by Albert Girard, who
stated that an nth degree polynomial equation has precisely n roots, including
complex roots and taking into account repeated roots. (For example, the fifth
degree equation (z − 2)(z − 4)2(z 2 + 9) = 0 has five roots: 2, 4 (twice), 3i and
−3i.) Many great mathematicians tried to prove this. The chief difficulty is
to show that every polynomial equation must have at least one root: this is
called the Fundamental Theorem of Algebra and was first proved by Carl
Friedrich Gauss in 1799.
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The Fundamental Theorem, which is too difficult to prove here, is an
example of an existence theorem: it tells us that a solution exists, but does
not say what it is. To find the solution of a particular equation you may be
able to use an exact method, such as the formula for the roots of a quadratic
equation. (There are much more complicated formulae for solving cubic or
quartic equations, but not in general for equations of degree five or more.)
Alternatively, there are good approximate methods for finding roots to any
required accuracy, and your calculator probably has this facility.
11 COMPLEX NUMBERS
11
ACTIVITY 11.14
Find out how to use your calculator to solve polynomial equations.
You have already noted that the complex roots of a quadratic equation with
real coefficients occur as a conjugate pair. The same is true of the complex roots
of any polynomial equation with real coefficients. This is very useful in solving
polynomial equations with complex roots, as shown in the following examples.
Example 11.14
Given that 1 + 2i is a root of 4z 3 − 11z 2 + 26z − 15 = 0, find the other roots.
Solution
Since the coefficients are real, the conjugate 1 − 2i is also a root.
Therefore [z − (1 + 2i)] and [z − (1 − 2i)] are both factors of
4z3 − 11z2 + 26z − 15 = 0.
This means that (z − 1 − 2i)(z − 1 + 2i) is a factor of
4z3 − 11z2 + 26z − 15 = 0.
(z − 1 − 2i)(z − 1 + 2i) = [(z − 1) − 2i][(z − 1) + 2i]
= (z − 1)2 + 4
= z 2 − 2z + 5
By looking at the coefficient of z3 and the constant term, you can see that
the remaining factor is 4z − 3.
4z 3 − 11z 2 + 26z − 15 = (z 2 − 2z + 5)(4z − 3)
The third root is therefore 43 .
Example 11.15
Given that −2 + i is a root of the equation z 4 + az 3 + bz 2 + 10z + 25 = 0,
find the values of a and b, and solve the equation.
Solution
z = −2 + i
z2 = (−2 + i)2 = 4 − 4i + (i)2 = 4 − 4i − 1 = 3 − 4i
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−2a + 3b − 2 = 0
11a − 4b − 14 = 0
Solving these equations simultaneously gives a = 2, b = 2.
The equation is z4 + 2z3 + 2z2 + 10z + 25 = 0.
Since −2 + i is one root, −2 − i is another root.
So (z + 2 − i)(z + 2 + i) = (z + 2)2 + 1
= z2 + 4z + 5
is a factor.
Using polynomial division or by inspection
z 4 + 2z 3 + 2z 2 + 10z + 25 = (z 2 + 4z + 5)(z 2 − 2z + 5).
The other two roots are the solutions of the quadratic equation
z2 − 2z + 5 = 0.
Using the quadratic formula
11
11.8 Complex numbers and equations
z3 = (−2 + i)z2 = (−2 + i)(3 − 4i) = −6 + 11i + 4 = −2 + 11i
z4 = (−2 + i)z3 = (−2 + i)(−2 + 11i) = 4 − 24i − 11 = −7 − 24i
Now substitute these into the equation.
−7 − 24i + a(−2 + 11i) + b(3 − 4i) + 10(−2 + i) + 25 = 0
(−7 − 2a + 3b − 20 + 25) + (−24 + 11a − 4b + 10)i = 0
Equating real and imaginary parts gives
z= 2± 4−4×5
2
= 2 ± −16
2
= 2 ± 4i
2
= 1 ± 2i
The roots of the equation are −2 ± i and 1 ± 2i.
Exercise 11H
1
4 − 5i is one root of a quadratic equation with real coefficients. Write
down the second root of the equation and hence find the equation.
2
Check that 2 + i is a root of z3 − z2 − 7z + 15 = 0, and find the other roots.
One root of z3 − 15z2 + 76z − 140 = 0 is an integer. Solve the equation.
Given that 1 − i is a root of z3 + pz2 + qz + 12 = 0, find the real
numbers p and q, and the other roots.
One root of z4 − 10z3 + 42z2 − 82z + 65 = 0 is 3 + 2i.
Solve the equation.
The equation z4 − 8z3 + 20z2 − 72z + 99 = 0 has a pure imaginary
root. Solve the equation.
One root of z3 – 15z2 + 76z – 140 = 0 is an integer.
Solve the equation and show all three roots on an Argand diagram.
3
4
5
6
7
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8 The equation z3 − 2z2 − 6z + 27 = 0 has a real integer root in the
11 COMPLEX NUMBERS
11
PS
range −6 $ z $ 0.
(i) Find the real root of the equation.
(ii) Hence, solve the equation and find the exact value of all three roots.
9 Given that 4 is a root of the equation z3 − z2 − 3z − k = 0, find the
value of k and hence find the exact value of the other two roots of the
equation.
10 The three roots of a cubic equation are shown on the Argand diagram
below.
Im
2
z2
ö2
1
z1
0
−1
1
2
Re
−1
z3
−ö2
−2
CP
Find the equation in polynomial form.
11 For each of these statements about polynomial equations with real
coefficients, say whether the statement is TRUE or FALSE, and give an
explanation.
(i) A cubic equation can have three complex roots.
(ii) Some equations of order 6 have no real roots.
(iii) A cubic equation can have a single root repeated three times.
(iv) A quartic equation can have a repeated complex root.
12 Given that z = −2 + i is a root of the equation
z4 + az3 + bz2 + 10z + 25 = 0, find the values of a and b, and solve
the equation.
13 The equation 2x 3 + x 2 + 25 = 0 has one real root and two complex
roots.
(i) Verify that 1 + 2i is one of the complex roots.
(ii) Write down the other complex root of the equation.
(iii) Sketch an Argand diagram showing the point representing the
complex number 1 + 2i. Show on the same diagram the set of
points representing the complex numbers z which satisfy
| z | = | z − 1 − 2i |.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q7 November 2005
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KEY POINTS
1
2
(x1 + iy1) ± (x2 + iy2) = (x1 ± x2) + i(y1 ± y2)
4
To multiply complex numbers, multiply out the brackets and simplify.
(x1 + iy1)(x2 + iy2) = (x1x2 − y1y2) + i(x1y2 + x2y1)
5
To divide complex numbers, multiply top and bottom by the
conjugate of the bottom.
x 1 + iy 1 ( x 1x 2 + y 1y 2 ) + i(x 2 y 1 –x 1y 2 )
=
x 2 + iy 2
x 22 + y 22
6
The complex number z can be represented geometrically as the point
(x, y). This is known as an Argand diagram.
7 The modulus of z = x + iy is | z | = x 2 + y 2 .
This is the distance of the point z from the origin.
8 The distance between the points z1 and z2 in an Argand diagram is
| z2 − z2 |.
9 The principal argument of z, arg z, is the angle θ, −π < θ < π,
between the line connecting the origin and the point z and the
positive real axis.
10 The modulus−argument or polar form of z is z = r(cos θ + i sin θ ),
where r = | z | and θ = arg z.
11 x = r cos θ
y = r sin θ
y
2 + y2
x
r=
tan θ =
x
12 To multiply complex numbers in polar form, multiply the moduli and
add the arguments.
11
11.8 Complex numbers and equations
3
Complex numbers can be written in the form z = x + iy with i2 = −1.
x is called the real part, Re(z), and y is called the imaginary part,
Im(z).
The conjugate of z is z* = x − iy.
To add or subtract complex numbers, add or subtract the real and
imaginary parts separately.
z1z2 = r1r2[cos(θ 1 + θ 2) + i sin(θ 1 + θ 2)]
13 To divide complex numbers in polar form, divide the moduli and
subtract the arguments.
z 1 r1
[cos(θ 1 − θ 2) + i sin(θ 1 − θ 2)]
z 2 = r2
14 eiθ = cos θ + i sin θ, e–iθ = cos θ – i sin θ
15 A polynomial equation of degree n has n roots, taking into account
complex roots and repeated roots. In the case of polynomial equations
with real coefficients, complex roots always occur in conjugate pairs.
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11
LEARNING OUTCOMES
Now that you have finished this chapter, you should be able to
■
11 COMPLEX NUMBERS
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
understand how complex numbers extend the number system
solve quadratic equations with complex roots
know what is meant by the terms real part, imaginary part and
complex conjugate
know that two complex numbers are equal when both their real and
imaginary parts are equal
add, subtract, multiply and divide complex numbers
solve problems involving complex numbers by equating real and
imaginary parts
find the two square roots of a complex number
represent a complex number on an Argand diagram
represent addition and subtraction of two complex numbers on an
Argand diagram
find the modulus of a complex number
find the principal argument of a complex number using radians
express a complex number in modulus–argument form
multiply and divide complex numbers in modulus–argument form
represent multiplication and division of two complex numbers on an
Argand diagram
represent and interpret sets of complex numbers as loci on an Argand
diagram
■ circles of the form | z – a | = r
■ half-lines of the form arg (z – a) = θ
■ lines of the form | z – a | = | z – b |
represent and interpret regions defined by inequalities based on the
above
multiply and divide complex numbers expressed in polar form
understand that complex roots of polynomial equations with real
coefficients occur in conjugate pairs
solve cubic and quartic equations with complex roots.
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Answers
Answers
The questions, with the exception of those from past question papers, and all example answers that appear in this
book were written by the authors. Cambridge Assessment International Education bears no responsibility for the
example answers to questions taken from its past question papers which are contained in this publication.
Non-exact numerical answers should be given correct to three significant figures (or one decimal place for angles
in degrees) unless a different level of accuracy is specified in the question.You should avoid rounding figures until
reaching your final answer.
Chapter 1
?
(Page 1)
13 + 123 = 93 + 103 = 1729
Ramanujan must have been
referring to positive cubes only, as
if you allow negative cubes then
6 3 + (−5) 3 = 4 3 + 3 3 = 91.
3
21 x 3 + 2x 2 + 5
22
x3 + 2x2 + x
23 2x3 + 3x2 + x + 4
24 2x2 + 2x + 3
25 x2 + 3x + 1
30
26 x2 + 4
–5
27 x 2 − 4x − 2
x
2 3
28 x2 + 2x − 2
Exercise 1A (Page 6)
?
4
(Page 13)
1
(i) 3
2
2x 3 − 4
Its order will be less than n.
3
x 4 + 4x 3 + 6x 2 + 4x + 1
4
Exercise 1B (Page 14)
x 3 + 2x 2 + 5x + 7
5
−x2 + 15x + 18
6
2x 4 + 8
7
x 4 + 4x 3 + 6x 2 + 4x + 1
8
x 4 − 5x 2 + 4
9
x 4 − 10x 2 + 9
(ii) 12 (iii) 7
(i) 30, 0, (x − 3)
(ii) p = 2, q = −15
(iii) −5, 2 or 3
(iv)
y
1
(ii) −2, 3 or 4
y
(iii)
24
–2
0, 0, −8, −18, −24,
−20, 0
(ii) (x + 3)(x + 2)(x − 3)
(iii) −3, −2 or 3
(iv)
y
3
4 x
(i)
5
(i) 0
(ii) –1 ± 2
(iii)
y
–3
10 x11 − 1
–2 O
x
3
11 2x − 2
–2
2
x
12 10x 2
13 4
14 2x 2 − 2x
15 −8x 3 − 8x
16 x2 − 2x − 3
17 x2 + 3x
2
(i) −15, 0, 3, 0, −3, 0, 15
(ii) x(x + 2)(x − 2)
(iii) −2, 0 or 2
y
(iv)
6
y
–2
O
2
−4
(x − 1)2
(iii)
18 2x2 − 5x + 5
19 x 3 + x 2 + 2x + 2
(i)
(ii)
1
4 x
x
20 2x 2 + 3
–4
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319
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7
Answers
8
(i)
a = 2, b = 1, c = 2
(ii)
0, 3 or – 3
(i)
(x2 − 4)(x2 − 1)
(ii)
(x + 2)(x − 2)(x2 + 1)
?
(iii)
Two real roots: −2 and 2
g(3) = 3, g(−3) = 3
| 3 + 3 | = 6, | 3 − 3 | = 0,
| 3 | + | 3 | = 6, | 3 | + | −3 | = 6
9
(ii) 2x2 + 9x + 11
remainder 19
10
(i)
±6, ±3, ±2, ±1
(ii)
−1, 2 or 3
(i)
(x − 1)(x − 2) (x + 2)
(ii)
(x + 1)(x2 − x + 1)
11
(iii) (x − 2)(x2 + 2x + 5)
(iv) (x + 2)(x2 − x + 3)
12 (ii)
(a)
x3 − x2 + 2x + 2
remainder −6
(b)
x3 − 3x2 + 6x − 6
remainder −2
(c)
x2 − 2x + 4
23
?
1
(i)
22
O
(iii)
x
1.5
y
–4
x
O
(–2, 2)
(iv)
y
1
(viii) x = − 83 or x = 2
(ix) x = −1 or x = 23
2
(i)
O
x
(v)
y
−8 " x " 2
(ii) 0 $ x $ 4
1
(iii) x " −1 or x # 11
(iv) x $ −3 or x % 1
–4 12
– 12
O
x
(v) −2 " x " 5
(vi) − 23 $ x $ 2
a = −4, b = 1
3
(i)
|x−1|"2
(ii)
x − 13
(iii) | x − 1 | " 3
(i)
6x 2 − x − 2
remainder 7
(iv) | x − 2.5 | " 3.5
(ii)
−2, 2, − 21 , 23
(vi) | x − 4 | " 3.5
9781510421738.indb 320
3
(vii) x = −2 or x = 4
(ii) | x − 5 | " 3
320
y
(vi) x = 1 or x = 7
a = 11, b = 5
21 (i)
(ii)
x
O
(v) x = −3 or x = 2
a = −13
(ii) (x − 3) and (x + 1)
–2
x = −9 or x = 1
3
(ii) (x + 2)(2x + 1)(x − 3)
20
(i)
(iv) x = − 2 or x = 2
18 a = 2, b = −3
(i)
2
(iii) x = −1 or x = 7
(ii) 3x 2 + 2x + 4
19
y
(Page 19)
(ii) x = 7 or x = −1
15 a = −5; b = 4; 4
a=8
(i)
Exercise 1C (Page 21)
14 2 or −5
(i)
4
(Page 17)
| x | " 2 and x % 0
⇒0$x"2
| x | " 2 and x " 0
⇒ −2 " x " 0
13 −12
17
a = 5, b = −12
(ii) 7
remainder
−2x − 4
16 a = −1; b = −7; 1, −2 or 23
(i)
(–2 12 , –4)
(vi)
y
5
(v) | x − 10 | " 0.1
(2, 3)
O
x
02/02/18 1:16 PM
5
(i)
x " 21
Exercise 2A (page 27)
(ii)
x " 72
1
(iii)
x % − 21
(i)
y
(v)
6
x " −1 or x # 3
(vi) x $ −6 or x % − 43
1
1
2 "x " 1
O
7
x > 2a
8
x = 1, x = − 9
5
9
x = 1, x = 3
32
(ii)
1024
1
O
(iii)
2
(a) A(2, 4), B(4, 16)
(b) 0 < x < 2 and x > 4
(iii) (a) Yes
(b) No
If the graphs are translated vertically by
k units, the horizontal asymptote will be
y = k and the graphs will go through the
point (0, k).
(ii)
If the graphs are translated horizontally
by k units, the horizontal asymptote will
be y = 0 and the graphs will both go
through (k, 1).
(iii) If the graphs are stretched horizontally, the
horizontal asymptote will be y = 0 and
the graphs will both go through (0, 1).
(iv) If the graphs are stretched vertically by a
scale factor k, the horizontal asymptote
will be y = 0 and the graphs will both go
through (0, k).
x
y
(Page 26)
(i)
y
(i)
(i)
?
x
O
−1
About 20 days
Activity 2.1 (Page 25)
y
2
(Page 24)
(i)
x
(ii)
Chapter 2
?
Answers
(iv) −1 $ x $ 3
1
O
(ii)
x
y
3
2
O
(iii)
x
y
O
x
−1
321
9781510421738.indb 321
02/02/18 1:17 PM
3
Initial population = 10 000;
population after 5 years = 31 623
(i)
?
(Page 33)
» a0 = 1, loga 1 = 0
» am = x # 0 (for a # 0) so loga (x) = m is defined
(ii) P
Answers
only for x # 0
» Putting x = 1y in log ⎛⎜⎝ 1y ⎞⎟⎠ = − log y
⇒ log x = − log ⎛⎜⎝ 1y ⎞⎟⎠ :
as x → 0,
− log ⎛⎜⎝ x1 ⎞⎟⎠ → − ∞
10 000
t
O
4
(i)
(ii)
(iii)
(iv)
1m
4.61 m
Just over 6 years
20 m
5
(i)
(ii)
520
210
» There is no limit to m in am = x and loga x = m;
think, for example, of base 2, i.e. a = 2. Then
x = 2y. When y = 1, 2, 3, 4, … then x = 2, 4,
8, 16, … . So increases in y are accompanied by
ever larger increases in x and so a decreasing
gradient. This is the case not just for a = 2 but
for any value of a greater than 1.
» loga a = 1
(iii) n
Exercise 2B (Page 34)
520
1
(i)
(ii)
(iii)
(iv)
(v)
(vi)
x = log3 9 = 2
x = log4 64 = 3
x = log2 41 = −2
x = log5 51 = −1
x = log7 1 = 0
x = log16 2 = 41
2
(i)
(ii)
(iii)
(v)
3y = 9 ⇔ y = 2
5y = 125 ⇔ y = 3
2y = 16 ⇔ y = 4
64y = 8 ⇔ y = 21
200
O
t
(iv) 200 birds
6
(i) 1976 Hz
(ii) A, three octaves below the middle
(iii) 48 (from E two octaves below the middle,
up to C five octaves above the middle)
Activity 2.2 (Page 29)
1
(vi) 5y = 25
⇔ y = −2
3
(i) 10
(ii) 21
(iii) 10 8.6/85 = 1.26
Activity 2.3 (Page 32)
y
4
3
2
11
2
0
–1
–2
–3
(16, 4)
(1, 0) (4, 2)
(2, 1)
2 4 6 8 10 12 14 16 x
( 12 , –1)
( 14 , –2)
( 18 , –3)
(ii)
(iv)
(v) 4
(vii) 23
(vi) −4
(viii) 41
1
2
(x)
−3
(i) log 10
(iii) log 36
(ii)
(iv)
log 2
log 1
(v) log 3
(vi)
log 4
(vii) log 4
(viii) log 1
(ix) log 1
(x)
(ix)
4
(8, 3)
1
2 = 2 2 ≈ 1.4
−4
0
(i) 4
(iii) 21
2
7
3
log 12
322
9781510421738.indb 322
02/02/18 1:17 PM
5
2 log x
(ii)
3 log x
(iii)
1
2 log x
(iv)
(v) 6 log x
(vi)
11 log x
6
5 log x
2
8
9
(i)
y
O
x
1
10
(ii)
y
–1
O
log10 x , x = 21
7
2
(i) x = 19.93
(ii) x = −9.97
(iii) x = 9.01
(iv) x = 48.3
(v) x = 1375
(vi) x = 0.863
(vii) x = 1.38
(viii) x = −1.41
(ix) x = 1.78
12
O
13
7
2
3
(i) x < 7
(ii) x % 3
(iii) x % 3
(iv) x > 0.437
(v) x & 1
(vi) x % 0.322
(vii) 0.431 & x < 1.29
(viii) 0 & x < 0.827
(ix) 1 < x < 2.58
(x) 0.68 < x < 1.49
x
Exercise 2C (Page 41)
Some of the questions in this
exercise involve drawing a line of
best fit by eye. Consequently your
answers may reasonably vary a
little from those given.
1
7
11 k = 9
y
(ii) 3 y = 23 , y = −0.369
x = 1.58
x=0
x = 1.38
x = 0.387 or 6.13
x = −1
x = −0.356 or 0.564
(i)
(ii)
(iii)
(iv)
(v)
(vi)
x
(iii)
20 (i) ( x + 2)(4 x + 3)(3x − 2)
14
(i)
(ii)
If the relationship is of
the form R = kT n, the
graph of log R against
log T will be a straight
line.
(ii)
Values of log R: 5.46,
5.58, 5.72, 6.09, 6.55
Values of log T: 0.28,
0.43, 0.65, 1.20, 1.90
log R
6
25
17
5
(i) 6.6
(ii) 1.58 × 1012 joules
(iii) 794 times more energy
0
(i) x = 1.259 (3 d.p.)
(iii) 3 decibels
(iv) It should be
2
10 3.7 − 10 3.5 × 100 = 58.5%
10 3.5
15
(i)
(ii)
16
x = 0.631, x = −0.369
4<y<6
1.26 < x < 1.63
17 x = 0.802
18
(i)
Answers
6
(i)
x= 1
11
19 (i) a = 19,b = −36
(ii) y = 0.113
2.0
log T
1.0
(iii) k = 1.8 × 105, n = 0.690
(iv) 0.7 days
(ii)
Plotting log A against
t will test the model: if
it is a straight line the
model fits the data.
(iii) b = 1.4, k = 0.89
log A
0.8
0.6
0.4
0.2
0
–0.2
1
2
3
4
5
t
(iv) (a) 2.4 days
(b) 3.0 cm2
(v)
Exponential growth
323
9781510421738.indb 323
02/02/18 1:17 PM
3
Answers
4
k = 3.2 × 106, a = 0.98
The constant k is the
original number of trees.
(ii)
(iii) The model predicts
growth in local user
numbers without limit,
which is impossible.
However, the growth
becomes so slow that
this may well not be a
problem.
(ii) k = 1100, n = 1.6
(iii) 2500 m
(iv) The train would not
continue to accelerate
like this throughout its
journey. After 10 minutes
it would probably be
travelling at constant
speed, or possibly even
slowing down.
5
(ii) b = 1.37, k = 1.58
6
log y = B log x + log A. He
should plot log y against
log x. If this gives a straight
line, there is a relationship
of the form y = AxB. If
there is no such relationship,
the points will not be in a
straight line.
The value of log A is given
by the intercept on the log y
axis. The value of B is the
gradient of the line.
log y
1.0
0.8
0.6
0.4
0.2
(ii)
and so n + 1 = 0.You cannot
divide by zero.
∫
∫
∫
∫
()
∫
Activity 2.7 (Page 47)
1.099
e = 2.72 (2 d.p.)
(ii) 0.693
Exercise 2D (Page 52)
(iii) 1.792
2
1
6
1
1
∫ 1 x dx + ∫ 1 x dx = ∫ 1 x dx
1
x = x0ekt
2
t = 1 ln ⎜⎝ s ⎟⎠
3
Activity 2.5 (Page 46)
4
y
(i)
∫
∫
Activity 2.4 (Page 45)
3
L(a) − L(b) =
a1
dx − b 1 dx = a 1 dx
1x
1x
bx
Let x = bz
a1
–a 1
dx = b z dz
x
1
b
=La
b
an 1
(iii) L(an) =
dx
1 x
Let x = zn then dx = nzn−1 dz.
a1
a
dx = 1n × nz n−1 dz
1x
1z
a
= n 1 dz
1z
= n L(a)
(ii)
∫
1
−1
x = x . This means that n = −1
(i)
11
∫ 1 x dx = 0
∫
(Page 45)
5
1
y= x
O
1
a
⎛ s0 ⎞
k
p = 25e−0.02t
⎛y–5⎞
x = ln ⎜
⎟
⎝ y0 – 5 ⎠
(i)
x = 0.0540
(ii)
x = 0.0339
(iv) x = 0.693
ab
x = az ⇒ dx = adz
Converting the limits:
x=a⇒z=1
x = ab ⇒ z = b
ab 1
dx = b 1 ×a dz
1 az
a x
b 1
=
dz
1 z
b 1
dz = b x1 dx = L(b)
1 z
1
ab
1
(iii) L(a) + x dx = L(ab)
a
⇒ L(a) + L(b) = L(ab)
x
(v)
(ii)
log x
y = 400 × 0.63t
(iii) y = 0.4. The infection
is under control.
8
(ii) 3.42
L(1) =
(iii) x = 0.238
From the graph, A = 1.5,
B = 0.5. The formula is
therefore y = 1.5x0.5.
7
?
(i)
(i)
L(a) ab 1
a ∫ x dx
0.
2
0.
4
0.
6
0.
8
1.
0
1.
2
1.
4
0
9
log 4
log 3
Activity 2.6 (Page 46)
(i)
P = 4000 × t 0.2
(ii)
4000
∫
∫
∫
∫
∫
∫
x = 1.386
(vi) x = 1.099
6
(i)
P
100
O
(ii)
t
100
(iii) 1218
(iv) 184 years
324
9781510421738.indb 324
02/02/18 1:17 PM
7
1m
(ii)
4.61 m, 6.09 years
(iii)
a = e−2 = 0.135, b = 2.5
(iv) 11 years
y = 25
x –1
(i)
0 $ α $ 90°
(ii)
No, for each of the
second, third and fourth
quadrants a different
function is positive.
?
(iii) No, the graphs of the
three functions do not
intersect at a single
point.
9
4.11
10
x = 0.481
11
A = 3.67, b = 1.28
12
x = −1.68
13
K = 7.39, m = 1.37
14
p = 0.44, A = 3.2
7
(i)
(ii)
(iii)
(iv)
x = 0°, 180°, 360°
x = 45°, 225°
x = 60°, 300°
x = 54.7, 125.3°,
234.7°, 305.3°
Chapter 3
(v)
x = 18.4°, 71.6°,
198.4°, 251.6°
(Page 57)
(vi)
x = 45°, 135°, 225°, 315°
?
Answers will vary.
For example, a ≈ 1.6 to 1.8,
b ≈ 3 ⇒ y = 1.5 sin 3x.
A sine function is a reasonable
model for a wave.
Exercise 3A (Page 60)
1
2
3
(i)
x = 90°
(ii)
(iii)
(iv)
(v)
(vi)
x = 60°, 300°
x = 14.0°, 194.0°
x = 109.5°, 250.5°
x = 135°, 315°
x = 210°, 330°
(i)
−1
(ii)
(iii)
–2
3
(iv)
–2
3
–2
3
(v)
0
(vi)
(i)
B = 60°, C = 30°
(ii)
4
5
− 2
3
(i)
L = 45°, N = 45°
(ii)
2, 2 , 1
(ii)
14.0°
Area of a triangle = 21 × base × height.
The definitions of sine and cosine
in a right-angled triangle.
Activity 3.2 (Page 62)
(i)
(ii)
Activity 3.1 (Page 61)
1
0
y = sin( + 60 °)
A
y = cos( – 60 °)
180 °
sin(θ + φ)
= sin θ cos φ + cos θ sin φ
⇒ sin[(90° − θ ) + φ]
= sin(90° − θ )cos φ +
cos(90° − θ )sin φ
⇒ sin[90° − (θ − φ)]
= cos θ cos φ + sin θ sin φ
⇒ cos(θ − φ)
= cos θ cos φ + sin θ sin φ
⇒ cos[θ − (−φ)]
= cos θ cos(−φ) + sin θ sin(−φ)
⇒ cos(θ + φ)
= cos θ cos φ − sin θ sin φ
(iii) tan(θ + φ) =
y = sin(θ + 60°) is obtained from
⎛ –60° ⎞
y = sin θ by a translation ⎜
⎟.
⎝ 0 ⎠
y = cos(θ − 60°) is obtained from
⎛60° ⎞
y = sin θ by a translation ⎜ ⎟ .
⎝ 0⎠
y
(Page 62)
=
Answers
8
6
(i)
sin (θ + φ )
cos (θ + φ )
sin θ cos φ + cosθ sin φ
cosθ cos φ – sin θ sin φ
sin θ cos φ cosθ sin φ
+
cosθ cos φ cosθ cosφ
=
cosθ cosφ sin θ sin φ
–
cosθ cosφ cosθ cosφ
=
tan θ + tan φ
1 – tan θ tan φ
(iv) tan[θ + (−φ)]
=
tan θ + tan ( –φ )
1 – tan θ tan ( –φ )
⇒ tan(θ − φ)
360 °
tan θ – tan φ
= 1 + tan θ tan φ
It appears that the θ coordinate
of A is midway between the two
maxima (30°, 1) and (60°, 1).
Checking:
θ = 45° ⇒ sin(θ + 60°) = 0.966
cos(θ − 60°) = 0.966.
If 60° is replaced by 35°, using the
trace function on a graphical
calculator would enable the
solutions to be found.
?
(Page 63)
No. In part (iii) you get
tan 90° =
3+ 1
3
1– 3× 1
3
Neither tan 90° nor 1 is defined.
1–1
For the result to be valid you must
exclude the case when θ + φ = 90°
(or 270°, 450°, ...).
Similarly in part (iv) you must
exclude θ − φ = 90°, 270°, etc.
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Exercise 3B (Page 64)
1
(i)
Answers
(ii)
(iii)
(iv)
2
(i)
3 + 1
2 2 2 2
1
–
2
3–1
3 +1
3 +1
3–1
1
2 (sin θ + cos θ )
(ii) 21 ( 3 cos θ + sin θ )
(iii) 21( 3 cos θ − sin θ )
(iv)
(v)
(vi)
3
1
(cos 2θ − sin 2θ)
2
tan θ + 1
1 – tan θ
tan θ – 1
1 + tan θ
(i)
sin θ
(ii)
cos 8φ
(iii) 0
(iv) cos 2θ
4
(i)
tan α = 23
(ii)
cot β = − 20
9
8
(ii)
x = 22.5°, 112.5°
9
α = 26.6° and β = 45° or
α = 135° and β = 116.6°
7
10
(ii)
11
3
11
?
For sin 2θ and cos 2θ, substituting
θ = 45° is helpful.
and that sin 90° = 1 and cos 90° = 0.
For tan 2θ you cannot use θ = 45°.
Take θ = 30° instead; tan 30° = 1
3
and tan 60° = 3.
θ = 157.5°
1
(i)
(ii)
(iv) θ = 111.7°
(ii)
6
(i)
θ=
2
2
(iv) Negative sign means
angle (α +β ) is greater
than 90° but less than
180°.
(ii)
θ = 0°, 35.3°, 144.7°,
180°, 215.3°, 324.7°,
360°
2 x
3π
2
−1
8
9
(iii)
π , 5π , 3π
6 6 2
(ii)
θ = 63.4°
(i)
y
2
y = cos 2x
π
O
2π
x
–2
y = 3sinx – 1
–4
π 5π
(iii) x = 6 , 6
cos x = 41 (3 − 17)
(ii)
(iii) θ = 90°, 210°, 330°
11
40.2°, 139.8°
θ = 30°, 150°, 210°,
330°
12
(ii)
θ = 26.6°, 206.6°
13
(i)
1
10 (4
(ii)
44
tan 2a = − 24
7 ,tan 3a = − 117
(v)
sin β = cos β =
2
θ = 14.5°, 90°, 165.5°,
270°
π
2
10
(iv)
cos α = 3
5
y = sin x y = cos 2x
O
−0.5
No, checking like this is not the
same as proof.
(ii)
π
8
θ = 2.79 radians
(i)
2
Exercise 3C (Page 70)
(i)
6
You know that
sin 45° = cos 45° = 1
θ = 15°
θ = 165°
cot θ
0.5
(i)
(v)
5
3
y
1
(Page 67)
(iii) θ = 0° or 180°
5
θ = 24.7°, 95.3°
4
θ = –11π , –3π , –7π , –π ,
12
4 12 4
π , π ,5π , 3π
12 4 12 4
3 sin θ − 4 sin3 θ,
π 3π 5π 7π
θ = 0, 4 , 4 , π, 4 , 4 , 2π
θ = 51°, 309°
(v)
θ = 0°, 138.6°, 221.4°,
360°
(i)
θ = –π, 0, π
(ii)
θ = –π, 0, π
2π
(iii) θ = –2π , 0,
3
3
–π
–3π
(iv) θ =
, , π , 3π
4 4 4 4
3 − 3)
Exercise 3D (Page 75)
1
(i)
2 cos(θ − 45°)
(ii)
29 cos(θ − 46.4°)
(iii) 2 cos(θ − 60°)
(iv) 3 cos(θ − 41.8°)
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2
(i)
(ii)
3
5
(iii)
3 sin(θ + 48.2°)
(i)
(ii)
π
)
4
3 sin(θ − 0.49 rad)
(i)
2 cos(θ − (−60°))
(ii)
4 cos(θ − (−45°))
(iv) θ = 53.8°, 159.9°, 233.8°, 339.9°
2 sin(θ −
9
(i)
3 cos(θ − 54.7°)
(ii)
Max 3 , θ = 54.7°;
min − 3 , θ = 234.7°
(iii)
(ii)
Max 13, min −13
(iii)
360°
θ
– 3
(vi) 2 cos(θ − 135°)
13 cos(θ + 67.7°)
180°
O
2 cos(θ − 150°)
(i)
y
3
(iv) 13 cos(θ − 22.6°)
6
360° θ
– 13
(iii) 2 cos(θ − 30°)
(v)
180°
O
5 sin(θ + 63.4°)
(ii)
y
13
Answers
4
(i)
π
2 cos(θ + )
4
π
2 cos(θ + 6 )
(iv)
y
Max 3 –1 , θ = 234.7°;
3
1
min
, θ = 54.7°
3+ 3
10
(ii)
θ = 30.6°, 82.0°
11
(i)
5 cos(x − 53.1°)
(ii)
x = 27.3°, 79.0°
(i)
26 cos(θ + 11.3°)
(ii)
θ = 27.02°, 310.4°
(i)
25 cos(θ − 73.7°)
(ii)
θ = 20.6°, 126.9°
13
5
112.6°
O
360°
θ
12
−13
(iv) θ = 4.7°, 220.5°
7
(i)
(ii)
(iii)
π
2 3 sin(θ − 6 )
Max 2 3, θ = 2π ;
3
min −2 3 , θ = 5π
3
13
14
y
Investigation (Page 78)
2 3
O
– 3
–2 3
π
(iv) θ = 3 , π
8
(i)
13 sin(2θ + 56.3°)
(ii)
Max 13, θ = 16.8°;
min − 13, θ = 106.8°
θ = 81.3°, 172.4°
The total current is
2π θ
I = A1 sinω t + A2 sin(ω t + α)(where ω = 2πf ).
I = A1 sin ω t + A2 sin ω t cos α + A2 cos ω t sin α
= (A1 + A2 cos α)sin ω t + (A2 sin α)cos ω t
Let A1 + A2 cos α = P and A2 sin α = Q
so I = P sin ω t + Q cos ω t
= P 2 + Q 2 sin(ω t + ε)
()
Q
.
P
This is a sine wave with the same frequency but a
greater amplitude.The phase angle ε is between 0 and α.
where ε = tan–1
327
9781510421738.indb 327
02/02/18 1:17 PM
Exercise 3E (Page 78)
1
(i)
sin 6θ
(ii)
cos 6θ
Answers
(iii) 1
(iv) cos θ
2
4
(v) sin θ
(vi) 23 sin 2θ
(vii) cos θ
Exercise 4A (Page 87)
(viii) −1
1
(i)
x(5x 3 − 3x + 6)
x4(21x2 + 24x − 35)
y
1
O
5
(i)
1 − sin 2x
(ii)
(ii)
cos 2x
(iii)
1
2(5 cos 2x − 1)
(i)
θ = 4.4°, 95.6°
(iii) 2x(6x + 1)(2x + 1)3
2
(iv) −
(3x – 1) 2
x 2 ( x 2 + 3)
(v)
( x 2 + 1) 2
(vi) 2(2x + 1)(12x2 + 3x − 8)
θ = 199.5°, 340.5°
–π π
(iii) θ = 6 , 2
(iv) θ = −15.9°, 164.1°
(v) θ = π, π, 5π
2 6 6
(ii)
(vii) θ = 76.0°, 135°
2(1 + 6x – 2x 2 )
(2x 2 + 1) 2
7–x
(viii)
( x + 3) 3
3x – 1
(ix)
2 x –1
1
(i) −
( x – 1) 2
2
(ii)
Chapter 4
Many possible answers: e.g.
amplitude of waves in physics,
rate of radioactive decay, rate of
population growth and so on.
(ii)
(0, 4), maximum;
(2, 0), minimum
gives
4
du
dv
= 10x9 and
= 7x6
dx
dx
Using the quotient rule,
–1 O
4
2
1
(i)
−
(ii)
4y + x = 12
Q( 37
4 , 8)
(b)
R(4, 29)
2( x + 1)( x + 2)
(2x + 3) 2
(−1, −2); (−2, −3)
7
Maximum at (−1,−1) and
minimum at (0, 0)
8
(i)
3 2
2
(iii)
3 ; 3; gradient = ∞
2
?
y
y = uv where u = x10 and v = x7
(i)
(a)
(Page 90)
d
dx (f(x)) is a polynomial of order
(n − 1) so it has no term in xn.
(iii)
Activity 4.1 (Page 87)
1
4
(v)
?
3x(x − 2)
(ii)
(iii) (−1, −2), minimum;
(−2, −3), maximum
−1; y = −x
(i)
x–2
( x – 1) 2
(iv) Tangent: y = 8;
normal: x = 4
(iv) The two tangents are
parallel.
3
(i)
(ii)
(iii) −1; y = −x + 4
(Page 83)
x
4
(iii) (4, 8)
6
(vii)
(vi) θ = 20.8°, 122.3°
?
(iii) y = x − 3
dy
≠ 0 for any value of x
(iv)
dx
dv
du
dy v dx – u dx
=
dx
v2
7
9
10
6
= x × 10x – x × 7x
x 14
16 – 7 x 16
= 10x 14
= 3x2
x
10
dy
u
y = v = x 7 = x3 ⇒
= 3x2
x
dx
x
(Page 92)
y = ln (3x) is a translation of
⎛0 ⎞
y = ln(x) through ⎜ ⎟ .
⎝ 3⎠
The curves have the same shape.
The gradient function is valid for
x > 0.
(x – 4)2
328
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02/02/18 1:17 PM
Exercise 4B (Page 94)
1
6
2x
x2 + 1
(v) − x1
(vi) 1 + ln x
(iii)
O
(ii)
2e2x
8
9
(v)
e4x(1 + 4x)
(vi)
2x2e−x(3 − x)
(vii)
1– x
ex
y
?
4
(e, 1e
( )
Activity 4.3 (Page 97)
(1, 0)
sin x
y = tan x = cos
x
x
(ii)
0.108 m h−1;
0.266 m h−1;
0.653 m h−1;
1.61 m h−1
11
(i)
(1, −e)
(ii)
Minimum
(ii)
ey − 2x + 1 = 0
5
(i)
(ii)
Rotation symmetry,
centre (0, 0) of order 2.
f(x) is an odd function
since f(−x) = −f(x).
f '(x) = 2 + ln(x2);
2
f ((x) = x
dy cos x (cos x ) – sin x ( – sin x )
dx =
cos 2 x
2
2
1
= cos x +2 sin x=
cos 2 x
cos x
= sec2x
Exercise 4C (Page 100)
1
Activity 4.2 (Page 96)
d 2y
= (2 + x)ex
dx 2
)
1
–2π
(Page 97)
This is a demonstration but
‘looking like’ is not the same
as proof.
(4, 4e−2)
(
(Page 97)
(1, 0)
e, 1
e
10
−1, − 1e
x
(i)
0.108e0.9t
(ii)
2π
0
?
(i)
(i)
π
dy
dx
No.You can see this if you draw
both graphs.
(viii) 6e2x(e2x + 1)2
dy
= (1 + x)ex;
dx
x
Maximum
O
12
2π
(ii)
(iii)
2
(iv) 2(x + 1)e(x +1)
π
–1
( 1e , − 1e)
(i) (1, 1e )
(ii)
2
(iii) 2xex
3
–π
x
1
(
3ex
7
0
1
–2π
1
x ( x + 1)
x
(ix)
x2 – 1
1 – 2 ln x
(x)
x3
y = cos x
–1
e
(viii) −
(i)
–π
–2π
1
y
(vii) x(1 + 2 ln(4x))
2
y
e x ( x – 1)
x2
(1, e), minimum
(i)
(ii)
(iv)
(− 1e , 2e ) , maximum;
( 1e , − 2e ) ,minimum
Answers
3
(i)
x
(ii) x1
(iii) x2
(iii)
–π
0
2
–1
2π
dy
dx
against x looks like the graph of
cos x.
When y = sin x the graph of
−2 sin x + cos x
(ii)
sec2x
(iii) cos x + sin x
dy
dx
π
(i)
(i)
x sec2x + tan x
(ii)
cos2 x − sin2 x = cos 2x
(iii) ex(sin x + cos x)
x
3
(i)
(ii)
x cos x – sin x
x2
x
e (cos x + sin x ) sec2x
(iii)
sin x(1 – sin x ) – cos x( x + cos x )
sin 2 x
329
9781510421738.indb 329
02/02/18 1:17 PM
4
(i)
2x sec2(x2 + 1)
(iii) x
Answers
(ii)
5
2 cos 2x
1
(iii)
tan x
sin x
(i) −
2 cos x
(ii)
ex(tan x + sec2x)
(iii)
8x cos 4x2
(i)
cos x − x sin x
(ii)
−1
(iv)
(iv) −sin y dy
dx
d
y
(v) e(y+2)
dx
(vi) y3 + 3xy2
(vii)
(iv) −2 sin 2x ecos 2x
1
(v)
1 + cos x
1
(vi) sin x cos x
6
(iii) (0.368, 0.692)
dy
dy
+y+1+
d
x
dx
1
dy
dx
O
dy
4xy5 + 10x2y4
dx
dy
(viii) 1 + 1y
dx
(ix) xey
(x)
y
dy
dy
+ ey + sin y
dx
dx
x 2 dy + 2x ln y
y dx
(iii) y = −x
dy
(xi) esiny + x cosy esiny dx
(iv) y = x − 2π
(xii) tan y + x sec 2 y
8
(ii)
(1, −3), (−1, 3)
9
(ii)
k = 5, c = 68
10
(ii)
(2, 1), (−2, −1)
11
(i)
(ii)
12
dy
dx
1
?
3x 2 − 2xy
x 2 + 3y 2
8x − 7y − 9 = 0
(a, −2a)
(Page 109)
7
x = π,x = π
6
3
8
y = 8.66x − 2.53
2
1
5
Exercise 4E (Page 116)
9
(i)
5
3
0
1
(ii)
−3
4
(i)
(ii)
5
(1, −2) and (−1, 2)
6
(i)
10
− (tan x
Maximum at x = 61 π,
minimum at x = 65 π
11
y = − 23 x + 1.68
(ii)
12
(i)
1
π
4
(iii)
(ii)
Maximum
13
y
2
6
(ii)
330
9781510421738.indb 330
4y3
dy
dx
2x + 3y2
dy
dx
x
7
(i)
(ii)
ln y = x ln x
1 dy
= 1 + ln x
y dx
(i)
6
(ii)
y = 6x − 3
(iii) 3x + 18y − 19 3 = 0
3
Exercise 4D (Page 106)
(i)
t–1
t +1
(vi) −tan θ
1
(vii) 2e t
(1 + t ) 2
(viii)
(1 − t ) 2
(v)
1
2
4
1
(iv) − 23cot θ
(2, − 4 )
The mapping is one-many.
t
(ii)
y+4
6–x
x − 2y − 11 = 0
O
(i)
1 + cos θ
1 + sin θ
t2 + 1
(iii) 2
t –1
0
y = −1
(Page 103)
?
At points where the rate of change
of gradient is greatest.
dy
+ y sec2x)
dx
(iv) Asymptotes x = 6,
y = −4
− 41 π
x
(i)
( 41 , 0)
(ii)
2
(iii) y = 2x − 21
(iv)
(0, − 21 )
02/02/18 1:17 PM
4
x − ty + at2 = 0
(ii)
tx + y = at3 + 2at
(iii)
(at2 + 2a, 0), (0, at3 + 2at)
− b2
at
(ii) at2y + bx = 2abt
(iii) X(2at, 0),Y 0, 2b
t
(iv) Area = 2ab
(i)
x2 − 2x + 3 = (x − 1)2 + 2 so is
defined for all values of x and is
always greater than or equal to 2.
( )
7
(ii)
y = tx − 2t2
(iii) [2(t1 + t2), 2t1t2]
(i)
(ii)
2
9
(iii)
(e −6 ,4e −2 + 3)
11
(i)
2t(t − 1)
3t − 2
(6, 5)
(ii)
12
13
(ii)
15
(i)
3
(Page 121)
Many possible answers. One method
would be to divide the flock into
squares, estimate the number
of birds in one square and then
multiply by the estimated number
of squares in the flock.
?
The areas of the two shaded
regions are equal since
y = x1 is an odd function.
(Page 125)
The polynomial p2(x) can take the
value zero.
9781510421738.indb 331
106 instalments: R agrees with the
value of e to 5 d.p.
(iv)
1
2 ln |2x − 9| + c
(i)
1 3x
3e + c
− 41 e−4x + c
x
−3e− 3 + c
Exercise 5B (Page 131)
1
4
5
6
7
8
9
(i)
−cos x − 2 sin x + c
(ii)
3 sin x − 2 cos x + c
(iii) −5 cos x + 4 sin x + c
(iv) 4 tan x + c
(v)
− 21 cos(2x + 1) + c
1
5 sin(5x – π) + c
(i)
2(e8 − 1) = 5960
(vi)
(ii)
ln 49
9 = 1.69
(vii) 3 tan 2x + c
(viii) tan 3x + 21 cos 2x + c
(iv) 0.906
Activity 5.1 (Page 123)
(i)
104 instalments: R = 2.718
(iii) 4.70
Chapter 5
?
1
4 ln |x| + c
1000 instalments: R = 2.717
(iii)
(iv) − 25x + c
e
(v) ex − 2e−2x + c
2 sin θ
ln3
2
–tan t
(ii)
(ii)
3 cos t
− 4 sin t
3x cos t + 4y sin t = 12
(iii) t = 0.6435 + nπ
3 ln |x| + c
(iii) ln |x − 5| + c
(iv) x = 4
8
(i)
Scheme B: R = 2.594
Scheme C: R = 2.653
Exercise 5A (Page 125)
1
Compound interest
Answers
6
(i)
(i)
P(2, 4); Q(−2, −4)
(ii)
8.77; 14.2 (to 3 s.f.)
(i)
4; 5 ln 5 − 4
(ii)
Ref lection in y = x
(ix) 4 tan x − 21 sin 2x + c
2
y = 21 e 2 x + 2e − x − 23
(ii)
Minimum when
x = 0.231
(i)
y = 3x − 3
(ii)
(a) 4
1
2
(ii)
1
(iii)
(iv) (a) 3(5 ln 5 − 4)
(b) 4 ln 3 + 5 ln 5 − 4
2
(i) 2x + 3
(iii) Quotient = 2x + 1,
remainder = −3
(i)
(i)
(iv)
3
4
(v)
1
3
(vi)
(vii) 0
(ix)
3
4
3
(ii)
3
8
4
(i)
(a)
(ii)
Investigations (Page 128)
A series for ex
1
3!
e = 2.718 281 83 (8 d.p.)
4!
3 −1
2
(viii) 1
1 2
2 (e + 1)
a0 = 1 a1 = 1 a2 = 2 ! a3 = 1 a4 = 1
3 −1
2
5
(i)
1 x + 1 sin 2x + c
4
2
(b)
π
4
(a)
1 x – 1 sin 2x + c
2
4
(b)
π− 3
6 8
2 cos( x − 61 π)
331
02/02/18 1:17 PM
6
Answers
7
8
(ii)
(a)
2 2
(b)
3
(i)
1
1
2 + 2 cos 2 x
(iii)
1
1
6π − 8
(ii)
1
4 (5π − 2)
(ii)
10
(i)
12e x + 4 e 3x + c
13
?
3
(a)
θ = ±0.572
(b)
3 π+ 1
32
4
(i)
2sinθ + 2θ + c
(ii)
1
In15
2
2
1
1
O
(ii)
4
(i)
(ii)
2
3
1
1
1
1.5
2
x
(ii)
0.5
1
1.5
2
x
3
1 x
0.458 658, 0.575 532,
0.618 518, 0.634 173,
0.639 825
2
1
(iii) 0.64
0
5
(i) y
(iii) 4
4
2
7
27
8
(i)
(Page 137)
0
Underestimates – all trapezia
below the curve
(ii)
Impossible to tell
1 x
6
(i)
458 m
(ii)
(ii)
A curve is approximated
by a straight line. The
speeds in question are
only correct to 2 s.f.
y
(i)
3.1349...
1
(ii)
3.1399..., 3.1411...
9
(a) 2, (b) 2, (c) 4, (d) 4
(i)
(0, 1)
(ii)
1
π
4
(iii) 1.77
(iv) Underestimate
3
10
2
0
(1, 0)
(ii) 1e
(iii) 0.89
(iv) Underestimate, as all
trapezia are below the
curve
3, 3.1, 3.131 176, 3.138
988
(iii) 3.14 (This actually
converges to π.)
(i)
(iii) 3.14
0.5
(d) y
)
Exercise 5C (Page 137)
2 x
1
y
2
0
(iii) Overestimates – all trapezia
above the curve
1
(c)
(i) y
Error from 16-strip estimate is
about 0.7%.
?
0
0
The curve is part of the circle
centre 2 21 , 0 , radius 2 21 .
Area required is half a major
segment = 8.4197 units2.
x
2.179 218, 2.145 242,
2.136 756, 2.134 635
(iii) 2.13
(Page 136)
(
(b) y
3
9
(ii)
(i) y
3
2 3−π
2
11
3
(i)
2
(iii) 0.95
11
1
2
(ii)
13.5
x
332
9781510421738.indb 332
02/02/18 1:17 PM
Chapter 6
?
2
(Page 144)
2 roots
(ii)
[0, 1]; [1, 2]
(i)
y = 2x
2
1
–2
(ii)
O
x
2 roots
For 1 d.p., an interval length of
" 0.05 is usually necessary,
requiring 5 steps. However, it
depends on the position of the
end points of the interval.
4
–1.88, 0.35, 1.53
5
(i)
–1.62
(ii)
1.28
(i)
[−2, −1]; [1, 2]; [4, 5]
6
(i)
[1, 2]; [4, 5]
(ii)
1.857, 4.536
(i)
(a)
y
O
y
(iii) 1.154
x
x=0
(c)
Failure to find root
(ii)
Converges to 1.
(b)
x < x for x > 1, x > x
for x < 1 and 1 = 1
(iii) Converges to 1.6180
(to 4 d.p.) since this is the
solution of x = x + 1
(i.e. the positive solution of
x2 − x − 1 = 0).
?
Exercise 6A (Page 150)
O
–5
(b)
x
Converges to 0.7391
(to 4 d.p.) since
cos 0.7391 = 0.7391 (to 4 d.p.).
x
(iv) a = −1.511 718 75, n = 8
a = 1.244 384 766, n = 12
a = 4.262 695 313, n = 10
8
y
(i)
(iii) −1.51, 1.24, 4.26
The expected number of steps for
2 d.p., requiring an interval of
length " 0.005, is 8 steps.
Success
Investigation (Page 152)
O
7
(c)
O
(ii)
In cases like this, 2 and 3 d.p.
accuracy is obtained very quickly
after 1 d.p.
x=0
1
f(x)
For example, the interval
[0.25, 0.3125] obtained in 4 steps
gives 0.3 (1 d.p.) but the interval
[0.3125, 0.375] obtained in 4
steps is inconclusive. As are the
interval [0.343 75, 0.375] obtained
in 5 steps, the interval [0.343 75,
0.359 375] obtained in 6 steps, the
interval [0.343 75, 0.351 562 5]
obtained in 7 steps, etc.
(b)
(iii) (a)
(iii) 2, –1.690
Activity 6.1 (Page 148)
(ii)
x
y=x+2
Answers
3
y
O
y
(Page 148)
0.012 takes 5 steps
0.385 takes 18 steps
0.989 takes 28 steps
In general 0.abc takes
(a + b + c + 2) steps.
1
(a)
(iii) 0.62, 1.51
(i), (ii) and (iv) can be solved
algebraically; (iii) and (v) cannot.
?
(ii)
(i)
x
No root
(c) Convergence to a
non-existent root
(Page 153)
Writing
as
gives
x5 − 5x + 3 = 0
x5 − 4x + 3 = x
g(x) = x5 − 4x + 3
Generalising this to
x5 + (n − 5)x + 3 = nx
x 5 + (n – 5 ) x + 3
gives
g(x) =
n
and indicates that infinitely many
rearrangements are possible.
333
9781510421738.indb 333
02/02/18 1:17 PM
?
(Page 155)
5
(i)
y
Answers
Bounds for the root have now
been established.
?
y = x2 + 2
(Page 156)
?
(ii)
6
x0 = 2 gives divergence to +∞.
Gradient is just
greater than
zero here.
O
2 x
1
0.618
At this root −1 "
gradient " 1 so
the root is found.
At this root gradient # 1
so the root is not found.
7
(ii)
0.747
(i)
y
(ii)
2.120
4
(iii) 1.503
2
–1
O
1
2
x
(i)
3 and 4
(ii)
3.43
15
(iii) 1.77
16
(ii)
0.678
Chapter 7
?
y=x
1
y = cos x
π
2
O
x
(Page 162)
One possibility, using a pen and
paper, would be to use trial and
improvement; another is that there
is a structured method, which is
rather like long division.
(ii)
0.739 09
Investigation (Page 163)
8
(ii)
a = 1.732
9
(i)
1.01, 1.02, 1.03
1 + x ≈ 1 + 21 x or
10
(i)
2.29
(ii) x = 2x + 42 ; α = 3 12
3 x
k = 21
0.20
x ≈ 21 (1 + x)
(Page 164)
y
?
2
(1 + x)2 = 3 but substituting x = 8
into the expansion gives successive approximations of 1, 5, −3,
29, −131, … and these are getting
further from 3 rather than closer
to it.
y = 2 – 2x
Exercise 6B (Page 157)
3
x
O other
root
x = −1,
gradient = 1.
x = 1,
gradient = 1.
14
y = ln(x + 1)
At this root gradient
# 1 so the root is
not found.
1.521
y
y = 2 – x2
–2
y = x2
y
x0 = 1 gives convergence to 0.618
(ii)
(i)
(iv) x = 1.31
x0 = −1 gives convergence to 0.618
1
1
2e
y = lnx
(i)
x0 = −2 gives divergence to −∞
–1
1
2
Only one point of
intersection
Activity 6.2 (Page 157)
–2
(− , − )
(iv) 1.319
The iteration diverges because
of the gradient of the curve; it is
steeper than the line y = x.
y
(i)
(iii) F(x) = ln(x2 + 2) is
possible.
(Page 156)
Between x = −1
and x = 1, −1 "
gradient " 1.
12
13
x
O
Often it is not so much a failure
of the method as a failure to use
an appropriate starting point or an
appropriate rearrangement of the
equation.
(iii) 1.08
(iii) 1.35
2
1
y = ex
11
1
y = cos x
1
Activity 7.1 (Page 165)
O
(iv) 0.58
1
π
2
x
−0.19 " x " 0.60
−0.08 " x " 0.07
334
9781510421738.indb 334
02/02/18 1:17 PM
Activity 7.2 (Page 166)
For | x | " 1 the sum of the
1
geometric series is 1 + x which is
the same as (1 + x)−1.
5 2
(iv) 1 – 41 x + 32
x
Valid for –1 " x " 1
3
4 x3
1 − 23 x + 13 x 2 − 27
Valid for −3 " x " 3
(ii)
1 − 43 x + 43 x 2 − 32
x3
27
Valid for
−1.5 " x " 1.5
(1 − x)−3 = 1 + 3x + 6x2 + 10x3 …
The coefficients of x are the
triangular numbers.
?
(iii)
(Page 168)
101 = 100 × 1.01
4
(i)
= 10 1.01
= 10(1 + 0.01)
1
2
= 10[1 + 21 (0.01)
+
(ii)
( 21) (– 21 ) (0.01)2 + …]
2!
= 10.050 (3 d.p.)
?
( )
( )
x 1 – x1 = x 1 – x1 .
1
Since x # 1 ⇒ 0 " x " 1
the binomial expansion could be
used but the resulting expansion
would not be a series of positive
powers of x.
5
1
4
(b)
|x|"1
(c)
0.43%
(a)
1 − 2x + 4x2
6
(b)
| x | " 21
7
5
2a 2
3b 3
1
9y
x+3
x–6
x+3
x +1
2x – 5
2x + 5
3(a + 4)
20
x(2x + 3)
( x + 1)
2
5( p – 2)
9
a–b
2a – b
10
( x + 4)( x – 1)
x( x + 3)
11
9
20x
12
x–3
12
13
a2 + 1
a2 – 1
(iii) 1 − 9 x + 19 x 2
14
5x – 13
( x – 3)( x − 2)
15
2
( x + 2)( x − 2)
16
2p2
( p 2 – 1)( p 2 + 1)
17
a2 – a + 2
(a + 1)(a 2 + 1)
(b)
|x|"1
(c)
0.000 006 3%
(i)
1 − 3x + 3x2 − x3
(ii)
1 – 4x + 10x2
Valid when | x | " 1
(ii)
1− 1 x + 3 x2
8
128
8
128
1 − 3x + 6x2 − 10x3
Valid for –1 " x " 1
1 − 6x + 24x2 − 80x3
Valid for – 21 " x " 21
7
3 + 11 x − 5 x 2 , | x | " 4
4 16
64
8
(i)
a = 2, b = −1, c = 16
(ii)
|x|"2
(iii) 1 + 6x + 24x2 + 80x3
Valid for – 21 " x " 21
9
(i)
1 + 2x + 5x2
(ii)
|x|"3
18
10
1 − 3 x + 27 x 2
4 4
16
–2( y 2 + 4 y + 8)
( y + 2) 2 ( y + 4)
19
11
1 − 23 x 2
x2 + x + 1
x +1
20
–
(i)
(ii)
2
1 − 2x + 3x2
Valid when | x | " 4
Exercise 7A (Page 170)
2
(a)
(iii) a = –7, b = 25
6
1 − 2x − 4 x 2 − 40
x3
3
8
x – 1 is only defined for x # 1.
A possible rearrangement is
1
2
1
3
0.8%
x2 x4
(iii) (a) 1 − 2 − 8
3
Exercise 7B (Page 174)
1 + 43 x + 43 x 2 + 32
x3
27
Valid for
–1.5 " x " 1.5
(c)
(Page 170)
13
− 10 x 3
(i)
1 + 21 x – 81x2
Valid for –1 " x " 1
(ii)
1 – 21 x + 83x2
Valid for –1 < x " 1
3 2
x
(iii) 1 + 41 x – 32
Valid for –1 " x " 1
12
(i)
3
1
a = −3
Answers
Investigation (Page 167)
(i)
(ii)
(3b + 1)
(b + 1) 2
335
9781510421738.indb 335
02/02/18 1:17 PM
Answers
21
13x – 5
6( x – 1)( x + 1)
22
4(3 – x )
5( x + 2) 2
23
3a – 4
(a + 2)( 2a – 3)
24
3x 2 – 4
?
x( x – 2)( x + 2)
(Page 176)
The identity is true for all values
of x. Once a particular value
of x is substituted you have an
equation. Equating constant terms
is equivalent to substituting x = 0.
Activity 7.3 (Page 181)
5
2(1 − x); | x | " 1
The binomial expansion is
1 − x + 3x2.
The expansion is valid when
| x | " 21 .
Which method is preferred is a
matter of personal preference for
(a) and (b) but for (c) must be
(iii).
6
(i)
2 + 2x + 4
(2 − x ) (1 + x 2 )
(ii)
15 3
2
5 + 52 x − 15
4 x − 8 x
(i)
2 + x −1
(2 + x ) ( x 2 + 1)
(ii)
1
x + 5 x2 − 9 x3
2
4
8
(i)
1 +
2
− 4
(1 – x ) (1 + 2x ) (2 + x )
(ii)
2
1 − 2x + 17
2 x
(i)
−1 + 3x − 1
( x − 2) x 2 + 3
(ii)
1 5
+ x + 17 x 2 + $
6 4
72
Exercise 7D (Page 181)
1
(ii)
Exercise 7C (Page 178)
1
2
1
1
x – ( x + 1)
3
4
2 – 1
( x – 1) ( x + 2)
5
1 + 1
( x + 1) ( 2x – 1)
6
2 –2
( x – 2) x
7
1 – 3
( x – 1) ( 3x – 1)
8
3 + 2
5( x – 4 ) 5( x + 1)
9
5 –2
( 2x – 1) x
10
2 – 1
( 2x – 3) ( x + 2)
8
9
+
13( 2x – 5) 13( x + 4 )
11
12
19
11
–
24( 3x – 2) 24( 3x + 2)
13
1 + 2 + 3
( x + 1) ( x + 2) ( x + 3)
14
4 + 3 +
2
( x − 1) ( 3 − x ) ( 2x + 1)
15
1 − 2 − 1
(2 + x ) (2 − x ) (2x + 3)
4
– 2x
( 2x – 1) ( x 2 + 1)
1
– 1 + 1
(iii)
( x – 1) 2 ( x – 1) ( x + 2)
1 – 1
( x – 2) ( x + 3)
2 – 2
( x – 4 ) ( x – 1)
9
2
– 3 –
(i) (
1 – 3x ) (1 – x ) (1 – x ) 2
(iv)
5
+ 6 – 5x
8( x – 2) 8( x 2 + 4 )
(v)
5 – 2x + 2
( 2 x 2 – 3) ( x + 2 )
Can be taken further
using surds.
3
2– 1 –
x x 2 ( 2x + 1)
10x – 3
(vii)
( 3x 2 – 1) x
Can be taken further
using surds.
1
+ 1
(viii)
( 2x 2 + 1) ( x + 1)
(vi)
2
3
8
9
Chapter 8
?
Exercise 8A (Page 185)
1
4 + 20x + 72x2
(ii)
−4 − 10x − 16x2
(iii)
5 11
+ x + 33 x 2
2
4
8
(iv)
– 1 – 5 x – 1 x2
(i)
(i)
1 − 3x + 7x2
(ii)
| x | " 21
(ii)
3
9x 2 + 1
2
x2 + 4
3
x2 + 9
− 1
(i)
1− x2
1
(ii)
1− x2
y = − 4x + π + 1 ;
3 3
3
7
y=
8
7x − 1 − 7
x2 + 1 x + 1
− 8 + 14x − 6x 2
Valid for | x | " 1
2
4x 2 + 1
(iv)
(i)
16
(i)
(iii)
3
8
4
–
–
( 2x – 1) (2x – 1) 2 x 6
8
(Page 184)
Answer in text following
(ix)
(ii)
4
7
?
3x + π − 3
4
3 16
(Page 186)
The area is the same as
∫
4
1
x dx.
336
9781510421738.indb 336
02/02/18 1:18 PM
?
(Page 188)
∫
∫
Activity 8.1 (Page 190)
3
5
4
2
( x − 2) 2 + ( x − 2) 2 + c
3
5
2
3
4
5
(vi)
(i)
222 000
(ii)
586
(iv)
(v)
2
6
(i)
22 21
(ii)
19
4
y=
2x + 3 + 3
5
4
3
7
6
6
7
(i)
8
7
(iv)
(i)
1
2 (e − 1)
1 4
2 (e − 1)
1 (e + e4) − 1 = 27.7
2
(a) 8 x − 3 2 + c
2x
3
2 2
2(1 + x ) + c
(i)
−(x + 2)e−x
(ii)
–(x + 3)e−x + c
1
(i)
ln |x2 + 1| + c
(ii)
1
2
3 ln |3x + 9x − 1| + c
(ii)
1
16
(iii) 1
(iv) e − 1
4
(v)
ln 2
(i)
tan −1 x + c
1
tan −1 x + c
4
4
(ii)
()
tan ( x ) + c
3
2
3
(iv)
2
tan −1
2
1
(v)
⎛
3
tan − 1 ⎜
6
⎝
1
2
(vi)
1
tan −1
10
(i) y
x
2
ln(e + 1) ≈ 1.434
2
2 +1
e
) ≈ 1.434
(iii) ln(
2
(iv) The same. The
substitution ex = t2
transforms the integral
in part (ii) into that in
part (iii).
(i) 1
1
2 ln
( p + 1)
−1
( 2x ) + c
3x ⎞ + c
2 ⎟⎠
( 25x ) + c
5
2π
3
6
(i)
x = 0.685
(ii)
8
15
(i)
1 (e 2 − 1)
2
(ii)
0.896
9
(ii)
1
π − 41
6
10
(a)
3x + 21 tan 2x + c
(b)
1
8
(i)
1
24
(ii)
k = 10
7
2
11
3
( 3π + ln 4 )
(i)
1
3 sin 3x + c
(ii)
cos(1 − x) + c
?
(iii)
− 41 cos4 x + c
Substitution using u = x2 − 1
needs 2x in the numerator. Not
a product.
(iv) ln | 2 − cos x | + c
(2 ln x − 1) + c
(i)
–1 + c
sin x
1
(iii)
Exercise 8D (Page 197)
k = 2, a = 1, b = 2;
32.5
6
3
0.490; 0.314
(iii) 2.53
(1 + ln x ) 2
9781510421738.indb 337
ln 3
(ii)
(v)
Exercise 8C (Page 192)
1
(ii)
(ii)
1
(b)
(ii)
1
2
2 tan x + c
O
(iii) 18.1
3
(iii)
(iv) −e2; max. at x = −2
4
(vi) 3 − e
Exercise 8B (Page 190)
(iii)
0.018
(iii) (–2, e2)
3
(ii)
(i)
(to 3 s.f.)
2
(3x + 4)( x − 2) 2 + c
= 15
1 3
8
8 (x + 1) + c
1 2
6
6 (x + 1) + c
1 3
5
5 (x − 2) + c
3
1
2
2
6 (2x − 5) + c
3
1
2
15(2x + 1) (3x − 1) + c
2 ( x + 9) 21 (x − 18) + c
3
esin x + c
(iii)
= 15
(i)
(ii)
(ii)
3
2
( x − 2) 2 [3(x − 2) + 10] + c
1
3
Answers
Yes: Using the chain rule
dy dy du
=
×
dx du dx
Integrating both sides with
respect to x,
dy du
y= (
)dx
×
du dx
dy
= ( )du
du
(iii) 4ex + c
−ln | cos x | + c
(Page 199)
(vi) − 61 (cos 2x + 1)3 + c
2
(i)
−cos(x2) + c
337
02/02/18 1:18 PM
Answers
1
⎪
1 + ln x – 1 + c
(ii)
⎪2x + 3⎪
1– x
(i)
ln
⎪
3x – 2
+c
1– x
(ii)
(b)
x –1
+c
x2 + 1
⎪ ⎪
( x − 1)
(iv) ln
⎪ 2x + 1 ⎪ + c
(iii) ln
1
x
ln 1 – x − x + c
(vi)
1
2 ln
⎪
(c)
⎪ xx ++ 13⎪ + c
?
⎪
⎪
⎪
⎪
(vii) ln
–
3
(i)
(ii)
⎝
⎠
(a)
2 + 1
1 – 2x 1 + x
(b)
ln (118) = 0.318 45
(a)
3 + 3x + 9x2 + …
1
4
(i)
5
(i)
6
(i)
7
(ii)
(Page 204)
(i)
(ii)
0.14%
A = 1, B = 2, C = 1,
D = –3
1
3
−
2( x + 1) 2( x + 3)
3
4x
+
2 − x 4 + x2
1+
(v)
ln 2 − x − 2 ln x + 1 + 2
∫
338
9781510421738.indb 338
∫
∫
∫
5
(i)
x ln x − x + c
(ii)
x ln 3x − x + c
(iii) x ln px − x + c
6
x2ex − 2x ex + 2ex + c
7
(2 − x)2 sin x − 2(2 − x)cos x
− 2 sin x + c
8
1
( x 2 + 1)tan −1 x − x
2
(
2
(i)
(ii)
(iii)
1 4
1 4
4 x ln x − 16 x + c
x e3x − 13e3x + c
x sin 2x + 21 cos 2x + c
)+ c
Exercise 8G (Page 210)
1
(i)
2 3 1
9e + 9
(ii)
−2
(iii) 2e2
(iv) 3 ln 2 − 1
(v) π
4
64
(vi) 3 ln 4 − 7
2
(i)
(2, 0), (0, 2)
(ii)
y
2
y = (2 – x)e–x
2
O
(a) u = x, dv = e−x
dx
(b) −xe−x − e−x + c
(vi) (a) u = x, dv = sin 2x
dx
1
(b) − 2 x cos 2x +
1 sin 2x + c
4
d
(a) dx (x cos x)
= −x sin x + cos x
⇒ x cos x
= −x sin x dx + cos x dx
⇒ x sin x dx
= −x cos x + cos x dx
1
5
15 (x − 2) (5x + 2) + c
1e−2x + c
4
Activity 8.2 (Page 204)
(b)
4
(a) u = x, dv = ex
dx
(b) xex − ex + c
(a) u = x, dv = cos 3x
dx
(b) 13x sin 3x +
(iii) (a) u = 2x + 1,
dv = cos x
dx
(b) (2x + 1)sin x +
2 cos x + c
(iv) (a) u = x, dv = e−2x
dx
(b) − 21x e−2x −
+ 1 + 2 ln 4 x + 3 + c
(i)
3
2
1+ x ) 2 (3x − 2) + c
15 (
1
cos 3x + c
9
(b) 0.318 00
(c)
∫
∫
⇒ 2x e2x dx
= xe2x − 21 e2x + c
1 3
1 3
3 x ln 2x − 9 x + c
3
e2x dx
Exercise 8F (Page 208)
1 , ln ⎛⎜ 2 ⎞⎟
6
x
+
x2 + 4 x – 3
∫
∫
(iv)
2xe2x dx
Each of the integrals in Activity 8.2
is of the form ∫ x dv dx and is
dx
found by starting with the
product xv.
(viii) ln 2x + 1 +
x+2
1
+c
2 ( 2x + 1)
2
∫
= xe2x −
⎪
x2 + 4 + c
x+2
⇒ xe2x
= 2xe2x dx + e2x dx
⇒
2
(v)
∫
⇒ x sin x dx
= −x cos x + sin x + c
d
(a) dx (x e2x) = x ×2e2x + e2x
(c)
Exercise 8E (Page 202)
x
(iii) e–2 + 1
3
(i)
y
y = xsinx
π
O
(ii)
x
π
02/02/18 1:18 PM
4
5 ln 5 − 4
2
π −1
2
6
4
4
− 15
so area = 15
square units
7
(i)
4π − 3 3
(ii)
1
(π − 2)
16
8
4(ln 4 − 1)
9
(ii)
y
= 2 ln | x + 3 | − ln | x − 1 | + c
(ii)
This integral can also be
found using partial fractions,
but using logarithms is
quicker.
3
?
x
(i)
1
2
(ii)
π ( 2 e − 3)
(i)
6e
(ii)
−
(i)
y = x −1
(ii)
1
2
4 π( e − 2)
(vi) This is a product: sin2 x is a
function of sin x, and cos x is
the derivative of sin x, so you
can use the substitution
u = sin x.
3
4
You will return to these integals in
Activity 8.3.
∫ cos x sin2 x dx = ∫ u2 du
= 13 u3 + c
= 13 sin3 x + c
This is a product of x and ex.
There is no relationship
between one expression and
the derivative of the other, so
you cannot use substitution.
As one of the expressions is
x, you can use integration
by parts.
∫
x ex dx = x ex −
∫
ex dx
= x ex − ex + c
(iv) This is also a product, this
2
2
time of x and ex . ex is a
function of x2, and 2x is the
derivative of x2, so you can
use the substitution u = x2.
Activity 8.3 (Page 214)
Using u = x2
(i)
∫
This is a quotient. The
derivative of the expression
on the bottom is not related
to the expression on the
top, so you cannot use
substitution. However, as the
expression on the bottom
can be factorised, you can
write it as partial fractions.
Using u = sin x
= 21 ln | x2 + 2x − 3 | + c
(iii)
(Page 212)
9781510421738.indb 339
ln | x2 + sin x | + c.
∫ x 2 + 2x – 3 dx
= 21 ∫ 2 2x + 2 dx
x + 2x – 3
(iv) 2.31
12
2x + cos x
∫ x 2 + sin x dx =
x +1
O
In this case the numerator
is the differential of the
denominator and so the
integral is the natural
logarithm of the modulus of
the denominator.
Since
d 2
(x + sin x) = 2x + cos x,
dx
then
f ′(x )
∫ f(x) dx = k ln | f(x) | + c.
– 1x
11
The derivative of the
expression on the bottom
line is 2x + 2, which is twice
the expression on the top
line. So the integral is of the
form
k
y=x
y=2+e 2
10
(v)
2
x ex dx =
∫
1 u
2 e du
= 21 eu + c
1 2
= 2 ex + c
Answers
5
x–5
∫ x + 2x – 3 dx
1
2
=∫
dx − ∫ ( x – 1) dx
( x + 3)
Exercise 8H (Page 214)
1
(i)
1
3 sin(3x − l) + c
(ii)
–1
+c
( x 2 + x – 1)
(iii) −e1−x + c
(iv)
1
2 sin 2x + c
(v)
x ln 2x − x + c
(vi)
–1
+c
4( x 2 – 1) 2
(vii)
1 ( 2x − 3) 3 + c
3
2
(viii) ln ⎪ x – 1⎪ + 1 + c
x+2
x –1
(ix)
1 4
1 4
4 x ln x − 16 x + c
(x)
ln ⎪2xx––31⎪ + c
(xi)
1 x2 + 2x
+c
2e
(xii) −ln (sin x + cos x) + c
339
02/02/18 1:18 PM
(xiii) − 21 x2 cos 2x + 21x sin 2x
2
+ 41 cos 2x + c
3
2
(i)
(ii)
8
3
1 ln 4
3
5
(iii) 48 + 8 ln 4
(iv)
(v)
(vi)
2
3
8
ln 2 − 79
3
2
2 −1
3
)
(
(vii) (ln 2) 2
(viii) − 2
9
6
de
dθ = kθ
7
dθ = – (θ – 15)
160
dt
8
dN = N
dt 20
9
dv = 4
v
dt
10
dA = 2k π = k ′
A
A
dt
dθ = – s
4
ds
3
(i)
12
dV = – 2V
1125π
dt
4
(ii)
3 + 3x
2 − x 2 + x2
2e 2 − 10
5
(ii)
15 ln 5 − 4
13
dh = (2 – k h )
100
dt
Exercise 9A (Page 220)
1
dv
dt is the rate of change of
velocity with respect to time,
i.e. the acceleration.
The differential equation tells
you that the acceleration is
proportional to the square
of the velocity.
9781510421738.indb 340
can be rewritten as
ln | y | = 12 x2 + (c2 − c1).
2
y = 13 x3 + c
(ii)
y = sin x + c
(i)
(ii)
2
(x 2 + c )
y2 = 23x3 + c
y=−
(iii) y = Aex
(iv) y = ln | ex + c |
(v)
H is about (70° N, 35°W) and L
is about (62° N, 5°W) so they are
separated by 30° in longitude at a
mean latitude of 66°. Reference
to the scale shows this to be about
900 nautical miles.
y = Ax
(vi) y = ( 41x2 + c)2
1
(vii) y = − (
sin x + c )
(viii) y2 = A(x2 + 1) − 1
Investigation (Page 222)
(ix) y = −ln(c − 21 x2)
(x)
y3 = 23 x2 ln x − 43 x2 + c
Exercise 9C (Page 229)
1
1035
1
(i)
y = 3 x3 − x − 4
(ii)
y = e3
1x 2
(iii) y = ln( 21 x2 + 1)
1
(iv) y = ( 2 – x )
1
(v) y = e 2( x 2 −1) − 1
996
957
O
(i)
3
11
Answers depend on the size of the
cup, the initial temperature of the
coffee, whether the cup is insulated
and whether milk is added, as well as
personal preference. It takes around
5–10 minutes for a cup to become
drinkable (around 60°C) and after
about 25 minutes the coffee is
probably too cool for most people
(around 30°C).
ln | y | + c1 = 21x2 + c2
(iv) y = 23 x 2 + c
1+ 3
4 4e 2
(Page 217)
(Page 224)
(iii) y = ex + c
(x)
?
?
1
1
(2
3
2 − 1)
The model covers the main features
of the situation.
Exercise 9B (Page 225)
(ix)
Chapter 9
340
4
isobars
Answers
1
(xiv) − 2 cos 2x + 61 cos3 2x + c
ds = k
dt s 2
dh = k ln(H − h)
dt
dm = k
dt m
dP = k P
dt
900
nautical miles
The mean level is 996 and the
amplitude 39 so a model is
πx
p = 996 + 39 cos 900
dp –39π
and
=
sin πx
900
dx 900
dp
or
= −a sin bx
dx
( )
( )
(vi) y = sec x
2
(i)
θ = 20 − Ae−2t
(ii)
θ = 20 − 15e−2t
(iii) t = 1.01 hours
3
(i)
N = Aet
(ii)
N = 10et
with a = 0.136 and b = 0.0035.
02/02/18 1:18 PM
4
ds 2
= ;s =
dt s
5
16 x
y = 3e 16−x 21
3−e
6
(i)
θ = A(1 + 3e −kt )
(iii)
7A
3
(i)
tan −1 21 − 21 e −2t
(ii)
The value of x tends to
tan −1 21 .
4t + c
Using the assumptions in
Exercise 9A, question 7, the rate
of cooling is proportional to the
temperature of the coffee above
the surrounding air. The initial
temperature is 95°C and the
cooling rate is 0.5°C s−1. So
− t
)
)
tan −1 21 − 21 e −2t .
(iii) 100 ln ⎛⎜ 10 + h ⎞⎟ − 20h
⎝ 10 − h ⎠
dN = k( N − 150)
(i)
dt
8
9
N = 500e 0.08t + 150
(ii)
(iii) No; when t = 15 then
N = 1810 or when
N = 1500 then t = 16.4
10
1
2t
(i)
N = 1800e1 t
5 + e2
(ii)
1800
x = 81 (tan θ + 1) 2 − 21
12
(i)
(
a1
a2
Length =
P
Chapter 10
a3
?
(Page 234)
To find the distance between the
vapour trails you need two pieces
of information for each of them:
either two points that it goes
through, or else one point and its
direction. All of these need to be in
three dimensions. However, if you
want to find the closest approach
of the aircraft you also need to
know, for each of them, the time at
which it was at a given point on its
trail and the speed at which it was
travelling. (This answer assumes
constant speeds and directions.)
O
OP = a 21 + a 22 + a 32
⇒
Exercise 10A (Page 239)
1
(i)
3i + 2j
(ii)
5i − 4j
(iii) 3i
(iv) −3i − j
2
(Page 239)
10
3
(ii)
203
5
(ii) 2 h 2 =
5
10
3
For all question 2:
j
i
(i)
2
a3
5
5
1
H 2t + 25 H 2
150
5⎞
⎛
2
(iii) t = 60 ⎜⎜1 − h ⎟⎟
H
⎝
⎠
Q
OP2 = OQ2 + QP2
= (a12 + a 22 ) + a 32
The vector a1i + a2 j + a3k is
shown in the diagram.
)
a 21 + a 22
Now look at the triangle OQP.
z
y = − 3 x cos 1 x + 9 sin 1 x + c
Q
The final temperature is lower if
the milk is added at the end.
?
11
y
x
θ = 15 + 71e 160.
(iii) As 21 − 21 e −2t increases
so does
(
O
− t
Adding 10% milk at 5°C gives
(
Start with the vector
⎯→
OQ = a1i + a2 j.
θ = 15 + 80e 160.
2
7
13
Investigation (Page 233)
Answers
(iii) N tends to ∞, which
would never be
realised because of
the combined effects
of food shortage,
predators and human
controls.
P
O
( )
a2
a1
x
y
( 13, 56.3°)
Q
341
9781510421738.indb 341
02/02/18 1:18 PM
(ii)
(v)
5k
Exercise 10B (Page 248)
(vi) −i − 2j + 3k
(vii) i + 2j − 3k
(viii) 4i − 2j + 4k
( 13, −33.7°)
Answers
1
(ix) 2i − 2k
(iii)
(x)
5
(4 2, −135°)
(iv)
−8i + 10j + k
A: 2i + 3j, C: −2i + j
⎯→
(ii) A B = −2i + j,
⎯→
C B = 2i + 3j
⎯→ ⎯→
(iii) (a) A B = ΟC
⎯→ ⎯→
(b) C B = ΟA
(i)
( 5, 116.6°)
2
(ii)
3.74
(ii)
4.47
(iii) 4.90
(iv) 3.32
(v)
7
(vi) 2.24
4
(i)
2i − 2j
(ii)
2i
(iii) −4j
(iv) 4j
?
⎛1⎞
⎜ ⎟
⎝1⎠
(iii)
⎛0 ⎞
⎜ ⎟
⎝0 ⎠
(v)
⎛ 8⎞
⎜ ⎟
⎝−1⎠
–3j
(i)
2i + 3j + k
(ii)
i–k
(iii) j – k
(iv) 3i + 2j – 5k
(a) F
(b) C
(v)
(i)
(ii)
(iv)
Activity 10.1 (Page 244)
3
⎛6 ⎞
⎜ ⎟
⎝8 ⎠
(i)
(iv) A parallelogram
(5, −53.1°)
(i)
3
(v)
–6k
(i)
(a) b
(c) Q
(b) a + b
(d) T
(c) –a + b
(e) S
⎯→
(a) Ο F
⎯→ ⎯→
(b) OΕ, CF
⎯→ → ⎯→
(c) OG, PS, AF
⎯→
(d) BD
⎯→ ⎯→
(e) QS, PT
(ii)
(b)
1
2 (a + b)
1
2 (–a + b)
(iii) PQRS is any
parallelogram and
⎯→
4
(i)
⎯→
⎯→
⎯→
PM = 21 PR, Q M = 21QS
(a) i
(b) 2i
(Page 247)
(c) i − j
!!"
⎯→ !⎯→
s ⎯→
OC
AB
ΟC == OA
ΟA + s + t AB
⎯→
⎯→
ΟA = a and A B = b − a
OC = a + s ( b − a )
s+t
= a+ s b− s a
s+t
s+t
+
s
t
s
=
a−
a+ s b
s+t
s+t
s+t
t
s
=
a+
b
s+t
s+t
(a)
5
(ii)
(d) −i − 2j
⎯→
⎯→
| AB | = | Β C | = 2 ,
⎯→
⎯→
| A B | = | CB | = 5
(i)
−p + q, 1p − 1q,
2
− 21p, − 21q
(ii)
2
⎯→ 1 ⎯→
NM = 2 BC,
⎯→
⎯→
N L = 21 A C,
⎯→
⎯→
ML = 21 A B
342
9781510421738.indb 342
02/02/18 1:18 PM
6
(i)
(iii)
(iv)
7
(i)
(ii)
(iii)
(iv)
(Page 250)
The cosine rule
Pythagoras’ theorem
?
⎛ –1 ⎞
⎜ 2⎟
⎜ –1 ⎟
⎜ ⎟
⎝ 2⎠
⎛ a 1 ⎞ ⎛ b1 ⎞
⎜ ⎟ . ⎜ ⎟ = a1b1 + a2b2
⎝ a 2 ⎠ ⎝b 2 ⎠
⎛ b1 ⎞ ⎛ a 1 ⎞
⎜ ⎟ . ⎜ ⎟ = b1a1 + b2a2
⎝b 2 ⎠ ⎝ a 2 ⎠
5
12
13 i – 13 j
(v)
2
?
(Page 253)
(iv) (2, 5)
3
4
5
θ
b
a
A
B
b – a = (b1 – a1)i +
(v)
3
5
i−
j+
38
38
(vi)
⎛1 ⎞
⎜ ⎟
⎜0 ⎟
⎜0 ⎟
⎝ ⎠
2
k
38
6
9
x = 4 or x = −2
10
(i)
(b) −p
(a)
q
(c)
q + p (d) r + p
(e)
p+q+r
AB2 = (b1 − a1)2 + (b2 − a2)2 +
(b3 − a3)2
2(a1b1 + a 2b 2 + a 3b 3 )
⇒ cos θ =
2|a||b|
(ii)
76.2°
(i)
(0, 4, 3)
(ii)
⎛ −5 ⎞
⎜ ⎟,5 2
⎜ 4⎟
⎝ 3⎠
(iii) 25.1°
⎯→
(i) OQ = 3i + 3j + 6k,
⎯→
PQ = −3i + j + 6k
(ii)
53.0°
(i)
−2
40°
⎯→
(iii) AB = i − 3j + (p − 2)k;
p = 0.5 or p = 3.5
7
OB2 = b12 + b22 + b32
11.74
29.0°
(ii)
(b2 – a2)j + (b3 – a3)k
2
2
2
cos θ = OA + OB – AB
2 × OA × OB
OA2 = a12 + a22 + a32
(i)
(iii) 162.0°
O
⎛ –2 ⎞
⎜ 29 ⎟
⎜ 4 ⎟
⎜
⎟
⎜ 29 ⎟
⎜ –3 ⎟
⎝ 29 ⎠
⎛ 3⎞ ⎛− 1⎞
⎜ ⎟, ⎜ ⎟
⎝1 ⎠ ⎝ 3⎠
⎯→ ⎯→
BA . BC = 0
⎯→
⎯→
(iii) | AB | = | BC | = 10
Consider the triangle OAB with
angle AOB = θ, as shown in the
diagram.
2i − 2 j + 1 k
3
3
3
3i − 4k
5
5
(i)
(ii)
These are the same because
ordinary multiplication is
commutative.
⎛ 1 ⎞
⎜ 14 ⎟
⎜ 2 ⎟
⎜ ⎟
⎜ 14 ⎟
⎜ 3 ⎟
⎝ 14 ⎠
90°
(vi) 180°
(Page 252)
3i + 4 j
5
5
8
12
?
(i)
8
−6, obtuse
(i)
⎛ 2⎞
⎜ 3⎟
⎜ – 2⎟
⎜ 3⎟
⎜ 1⎟
⎝ 3⎠
99°
(ii)
1(2i − 6j + 3k)
7
(ii)
a.b
=
|a || b|
(iii) p = −7 or p = 5
(ii)
p + 53 r
Exercise 10C (Page 254)
9
(ii)
q = 5 or q = −3
1
10
(i)
(i)
⎛ 2⎞
1 ⎜ 3⎟
7 ⎜ ⎟
⎜−6 ⎟
⎝ ⎠
⎯→
P A = −6i − 8j − 6k,
⎯→
P N = 6i + 2j − 6k
(ii)
99.1°
(ii)
9781510421738.indb 343
m = −2, n = 3, k = −8
(i)
42.3°
(ii)
90°
(iii) 18.4°
(iv) 31.0°
Answers
(ii)
⎛ 2 ⎞
⎜ 13 ⎟
⎜ 3 ⎟
⎜ ⎟
⎝ 13 ⎠
343
02/02/18 1:18 PM
11
Answers
12
(i)
4i + 4j + 5k, 7.55 m
(ii)
43.7° (or 0.763 radians)
(i)
⎯→
P R = 2i + 2j + 2k,
⎯→
P Q = −2i + 2j + 4k
(iii) is parallel to (i) since the
direction vector is the same.
(iv) is parallel to (ii) since
⎛ –1⎞
⎛ 1⎞
⎜ ⎟ = −⎜ ⎟ .
⎝ 2⎠
⎝ –2 ⎠
(ii)
61.9°
Exercise 10D (Page 264)
(iii) 12.8 units
?
1
(i)
(Page 259)
(ii)
(iii)
(v)
(b) It lies beyond B.
2
y
8
6
(i), (iv)
(ii)
2
2
4
(ii)
6
8
(v)
(i) and (iv) are the same since
putting λ = −1 in (i) gives
4
x
⎛1⎞
⎛ 2⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ 2⎠
⎝1⎠
⎛ 3⎞
⎛ –1⎞
r = ⎜ ⎟ + λ⎜ ⎟
⎝ 5⎠
⎝ 1⎠
⎛ −6⎞
⎛ 1⎞
(iii) r = ⎜ ⎟ + λ ⎜ ⎟
⎝ –6 ⎠
⎝ 1⎠
⎛ 5⎞
⎛ 1⎞
(iv) r = ⎜ ⎟ + λ ⎜ ⎟
⎝ 3⎠
⎝ 1⎠
(v)
⎛2⎞
r = λ⎜ ⎟
⎝ 1⎠
⎛ 1⎞ and ⎛ 1⎞ is parallel
⎜ ⎟
⎜⎝ 2⎟⎠
⎝ –3⎠
⎛ –1⎞
(vi) r = λ ⎜ ⎟
⎝ 4⎠
3
to ⎛ ⎞ .
⎜⎝ 6⎟⎠
⎛ –1⎞
(vii) r = λ ⎜ ⎟
⎝ 4⎠
(v)
⎛ 1⎞
r = λ ⎜ 2⎟
⎜ ⎟
⎝ 3⎠
(i)
Yes, λ = 2
(ii)
Yes, λ = −1
(iii) No
Note:These answers are not
unique.
(i)
4
(a) 5i + 12j
(c) −7.5i − 2j
Activity 10.3 (Page 263)
(iii)
(a) 6i − 8j
(b) 13
(c) It lies beyond A.
(ii)
⎛ 1⎞
⎛ 1⎞
⎜
⎟
r = 0 + λ ⎜ 0⎟
⎜ ⎟
⎜ ⎟
⎝ –1⎠
⎝ 0⎠
⎛ 0⎞
⎛ 2⎞
⎜
⎟
(iv) r = 0 + λ ⎜ 1⎟
⎜ ⎟
⎜ ⎟
⎝ 1⎠
⎝ 3⎠
(c) 0
It lies between A and B.
0
(a) 6i + 8j
(b) 10
(iv) 0, 1, 21 , 43
(i)
⎛ 2⎞
⎛ 3⎞
⎜
⎟
r = 4 + λ ⎜ 6⎟
⎜ ⎟
⎜ ⎟
⎝ –1⎠
⎝ 4⎠
⎛ 1⎞
⎛ 5⎞
⎜
⎟
(iii) r = 0 + λ ⎜ 3 ⎟
⎜ ⎟
⎜ ⎟
⎝ 4⎠
⎝ –6⎠
(c) i + 3j
(iv)
Note:These answers are not
unique.
(a) −4i − 3j
(b) 10
⎛ 3 1 ⎞ ⎛4 ⎞ ⎛ 8 ⎞
⎜ 2 ⎟ , ⎜⎜ ⎟⎟ , ⎜⎜ ⎟⎟
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝11⎠
–2
68
(c) 2i + 1.5j
(ii) ⎛−2 ⎞ ⎛ 0 ⎞ ⎛ 2 ⎞ ⎛ 3⎞
⎜ ⎟ , ⎜ ⎟, ⎜ ⎟ , ⎜ ⎟,
⎝−9 ⎠ ⎝ –5 ⎠ ⎝ –1 ⎠ ⎝1 ⎠
–2
(b)
(b) 5
Activity 10.2 (Page 260)
(a)
2i + 8j
3
(c) 3i + 7j
⎯
→ ⎯
→
⎯
→ ⎯
→
O P = O A + λ(O B − O A)
⎯
→
⎯
→
= (1 − λ)O A + λO B
(v)
(a)
⎛ –1⎞
⎛ 3⎞
(viii) r = ⎜
+ λ⎜ ⎟
⎟
⎝ 4⎠
⎝ –12⎠
(iv) No
5
(v)
Yes, λ = −5
(i)
⎛ −1⎞
⎛ −1⎞
r = ⎜ −2 ⎟ + λ ⎜ 3 ⎟
⎜ −3 ⎟
⎜⎝ 1⎟⎠
⎝ ⎠
⎛ −2 ⎞
⎛ −1⎞
or r = ⎜ −2 ⎟ + λ ⎜ 6 ⎟
⎜ −6 ⎟
⎜⎝ 1⎟⎠
⎝ ⎠
(ii)
(–2, 1, –2)
(iii)
⎛ 0⎞
⎛ −2 ⎞
r = ⎜ 1⎟ + λ ⎜ 1⎟
⎜ 0⎟
⎜⎝ −2 ⎟⎠
⎝ ⎠
344
9781510421738.indb 344
02/02/18 1:18 PM
Exercise 10E (Page 269)
5
(–2, –6, –1); 30 units
⎛ 5⎞
⎜⎝ 5⎟⎠
6
No
7
6 units, 9 units, 77 units
1
3
⎛ 4⎞
⎜⎝ 1⎟⎠
(ii)
(iii)
⎛ 12⎞
⎜⎝ 17⎟⎠
–5
(iv) ⎛⎜ ⎞⎟
⎝ 6⎠
(v)
⎛ 6⎞
⎜⎝ 3⎟⎠
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Intersect at (3, –2, 5)
Parallel
Intersect at (3, 2, –13)
Intersect at (1, 2, 7)
Skew
Intersect at (4, –7, 11)
Skew
(i)
(ii)
12.8 km
20 km h−1, 5 km h−1
Exercise 10F (Page 275)
(i)
⎯
→ ⎛ 10 ⎞
OL = ⎜ ⎟ ;
⎝ 4.5⎠
⎯
→
7
OM = ⎛⎜ ⎞⎟ ;
3.5
⎝ ⎠
(ii)
⎯
→ ⎛ 4⎞
ON = ⎜ ⎟
⎝ 1⎠
⎛ 1⎞
⎛ 2⎞
AL: r = ⎜ ⎟ + λ ⎜ ⎟ ;
⎝ 0⎠
⎝ 1⎠
⎛ 7⎞
⎛ 0⎞
BM: r = ⎜ ⎟ + µ ⎜ ⎟ ;
⎝ 2⎠
⎝ 1⎠
⎛ 13⎞
⎛ 3⎞
CN: r = ⎜ ⎟ + ν ⎜ ⎟
⎝ 7⎠
⎝ 2⎠
(iii)
(a) (7, 3)
53.6°
2
81.8°
3
8.72°
4
35.3°
?
5
61.0°
6
(i)
Any real number is either rational
or irrational. This means that all
real numbers will either lie inside
the set of rational numbers, or
inside the set of real numbers
but outside the set of rational
numbers. Therefore no separate set
is needed for irrational numbers.
–
The symbol ℚ is used for irrational
numbers – numbers which cannot
be expressed exactly as a fraction,
such as π.
(ii)
(iii)
(a)
(–2, 6, 7)
(b)
29 units
(a)
(3, –1, 7)
(b)
17 units
(a)
(2, 7, –3)
(b) 7 units
7
2 10 units
8
35 units
9
(i)
A(4, 0, 0), F(4, 0, 3)
(ii)
114.1°, 109.5°
(Page 281)
Activity 11.1 (Page 281)
Real
Rational
Integers
(iii) They touch but are not
perpendicular.
10
(ii)
5i + 3j + 4k
11
(ii)
λ = 83
12
(ii)
a = −2, a = 3
13
(ii)
79.5°
(b) (7, 3)
(iv) The lines AL, BM and
CN are concurrent.
(They are the medians
of the triangle, and
this result holds for
the medians of any
triangle.)
ℕ Natural numbers – nonnegative whole numbers (although
there is some debate amongst
mathematicians as to whether zero
should be included!)
1
(iii) After 40 minutes there
is a collision.
4
ℤ Integers – positive or negative
whole numbers, including zero
Chapter 11
?
(Page 280)
ℝ Real numbers – any number
which is not complex
Answers
2
(i)
ℚ Rational numbers – numbers
which can be expressed exactly as
a fraction
Natural
numbers
7
−13
227
109
3.1415
0.3
−√5
√5
π
Activity 11.2 (Page 281)
(i)
x = 2 Natural number
(or integer)
(ii)
x = 7 Rational number
9
(iii) x = ±3 Integers
(iv) x = −1 Integer
(v)
x = 0, −7 Integers
345
9781510421738.indb 345
02/02/18 1:18 PM
Answers
Activity 11.3 (Page 283)
(vii) 3 + 29i
Activity 11.5 (Page 287)
z = 3 − 7i
⇒ z2 − 6z + 58
= (3 − 7i)2 − 6(3 − 7i) + 58
= 9 − 42i + 49i2 − 18 + 42i + 58
= 9 − 42i − 49 − 18 + 42i + 58
=0
(viii) 14 + 5i
(ix) 40 + 42i
1 = p + iq
x + iy
⇒ (p + iq)(x + iy) = 1
(x)
100
⇒ px + ipy + iqx + iqy2 = 1
(xi) 43 + 76i
⇒ (px − qy) + i(py + qx) = 1
(xii) −9 + 46i
px − qy = 1 and py + qx = 0
All numbers of the form
(iii) 2 ± 3i
Solving simultaneously gives
–y
x
p= 2
,q = 2
x + y2
x + y2
x − iy
1
so x + iy = 2
x + y2
» i4n are equal to 1
(iv) −3 ± 5i
?
?
2
(Page 283)
i3 = −i, i4 = 1, i5 = i
»
»
» i4n+3 are equal to −i.
i4n+2 are equal to −1
Activity 11.4 (Page 284)
(ii)
1 ± 2i
1
2 ± 2i
(vi) −2 ± 2 i
3
(i)
2i
(ii ) 5i and –3i
(a)
6
(b)
2
(c)
(d)
(iii) 1 + i and –1 + i
34
5
(iv) 2 – 3i and –2 – 3i
They are all real.
(ii)
−1 ± i
(v)
i4n+1 are equal to i
(i)
(i)
(v)
z + z∗ = (x + iy) + (x − iy) = 2x
zz∗ = (x + iy)(x − iy)
= x2 − ixy + ixy − i2y2
–1 – 4i and 1 – 4i
(vi) –3i and 2i
4
(i)
2
(ii)
−4
(Page 289)
1 = −i, 1 = −1, 1 = i
i
i2
i3
All numbers of the form
» 14n are equal to 1
i
» 41n+1 are equal to –i
i
» 4n1+ 2 are equal to –1
i
» 4n1+ 3 are equal to i.
i
Exercise 11B (Page 289)
(i)
3
1
10 – 10 i
(iii) 2 − 3i
(ii)
6
1
37 + 37 i
(iv) 6 + 4i
(iii)
– 41 + 43 i
8+i
(iv)
(vi) −4 − 7i
(v)
(vii) 0
4 11
5 + 10 i
5 1
2 – 2i
(vi) 7 − 5i
(viii) 0
Exercise 11A (Page 285)
(vii) −i
(ix) −39
1
(x)
(viii)
= x2 + y2
These are real for any real
values of x and y.
?
(v)
(Page 285)
Yes, for example 23 = 46 , although
2 ≠ 4 and 3 ≠ 6.
(i)
14 + 10i
(ii)
5 + 2i
(v)
21
(vi) 12 + 21i
−46 − 9i
(xii) 52i
7
a = 1 or 4, b = −1 or 3
The possible complex
numbers are 1 + 9i, 1 + i;
16 + 9i, 16 + i
(x)
11 27
25 – 25 i
7
32
29 + 29 i
–1 – 23 i
(i)
a = 5, b = 2
(ii)
a = 3, b = −7
(ix)
(xi) −46 − 9i
(iii) −3 + 4i
(iv) −1 + i
1
2
(iii) a = 2, b = −3
(iv) a = 4, b = 5
346
9781510421738.indb 346
02/02/18 1:18 PM
(v)
a = 45 , b = − 43
(vi) a =
3
2
,b =
1
z2
(i)
z=2−i
(ii)
z=3+i
0, 2, −1 ± 3 i
6
2x
x2 + y2
(i)
a3 − 3ab2 + (3a2b − b3)i
1 1
(iii) z = 1, − 2 ± 2 3 i
(i)
(z − α)(z − β)
= z2 − (α + β)z + αβ
(ii)
(a)
10
(d)
z2 − (5 + 3i)z +
4 + 7i = 0
(i)
3i and –3i
(ii)
2 + i and –2 – i
(iii) 3 + 5i and –3 – 5i
O
(ii)
Re
O
z1
–z2
5
1
z−1 = 1 – 1 i,
2
2
z1 = 1 + i, | z1 | =
4i
2
z3 = −2 + 2i, | z3 | = 2 2
3 + 2i
–2 O
z4 = −4, | z4 | = 4
Re
4 – 3i
z5 = −4 − 4i, | z5 | = 4 2
(ii)
–6 – 5i
(i)
13
(ii)
(iii)
26
(iv) 2
(v)
61
(vi) 5
Im
4
Re
Im
–z
z*
iz*
(iii) The half-squares
formed are enlarged
by 2 and rotated
through π each time.
4
iz
O
–z*
(i)
3
(i)
z2 = 2i, | z2 | = 2
–5 + i
2
13
5
| z−1 | = 12
Im
Activity 11.6 (Page 291)
z and −z∗ (or −z and z∗) are
reflections of each other in the
imaginary axis.
5
13
z 0 = 1, | z 0 | = 1
(iz)*
–iz
(Page 291)
(iv)
Re
Exercise 11C (Page 294)
(vi) 2 – 3i and –2 + 3i
?
(iii) 65
w
,w =
z
z
z1 + (–z2)
5 – 2i and –5 + 2i
Rotation through 180° about
the origin
(ii) Reflection in the real axis
13
,
Im
z1
(iv) 3 – 4i and –3 + 4i
(v)
(ii)
z
zw = z w , z =
,
w
w
(b) 9z2 + 25 = 0
z2 + 4z + 12 = 0
5
z1
z2 − 14z + 65 = 0
(c)
(i)
(v)
z2 – z1
a = 2, b = 2
5
9
Im
2
(iii) z = 11 − 10i
(iv) z = –35 + 149i
34
8
(i)
4
Answers
4
1
Activity 11.7 (Page 293)
z
Re
6
Points:
(i)
10 + 5i
(ii)
1 + 2i
(iii) 11 + 7i
(iv) 9 + 3i
(v)
−9 − 3i
?
Half a turn about O
followed by reflection in
the x-axis is the same as
reflection in the x-axis
followed by half a turn
about O.
(Page 294)
| z2 − z1 | is the distance between
the points representing z1 and z2
in the Argand diagram.
347
9781510421738.indb 347
02/02/18 1:18 PM
?
?
(Page 296)
(i)
Im
Answers
(i)
(Page 296)
Im
3 + 4i
3 + 4i
–1 + 2i
O
(ii)
O
Re
Re
(ii)
Im
Im
3 + 4i
3 + 4i
–1 + 2i
O
Re
O
(iii)
Re
(iii)
Im
Im
3 + 4i
3 + 4i
–1 + 2i
O
O
Re
Re
Exercise 11D (Page 297)
1
(i)
Im
2
O
Re
348
9781510421738.indb 348
02/02/18 1:18 PM
(ii)
(vii)
Im
Re
4
O
Answers
O
Im
Re
1–i
(iii)
Im
(viii)
Im
5i
–2
O
(iv)
O
Re
Re
2
Im
Im
A
–3 + 4i
Re
12 – 5i
B
O
(v)
Re
| z | is least at A and greatest at B.
Im
| 12 − 5i | = 144 + 25 = 13
At A, | z | = 13 − 7 = 6
At B, | z | = 13 + 7 = 20
Re
O
6–i
3
(i)
O
2
4
6
8
Re
–2
(vi)
Im
–4
O
R
5 – 4i
–6
Re
–8
Im
(ii)
–2 – 4i
4
7, 13
Not possible
349
9781510421738.indb 349
02/02/18 1:18 PM
5
(i)
(e) −88.9°
Im
Answers
(ii)
2
Activity 11.9 (Page 300)
4
Re
arg(1 + i) = π ,
4
arg(1 − i) = − π ,
4
arg(−1 + i) = 3π ,
4
arg(−1 − i) = − 3π
4
Im
2i
i
O
(iii)
Re
?
(Page 301)
(i)
2(cos (π − α) + i sin (π − α))
(ii)
2(cos (−α) + i sin (−α))
Activity 11.11 (Page 301)
Im
π
4
π
6
π
3
tan
1
1
3
3
sin
1
2
1
2
3
2
cos
1
2
3
2
1
2
–1 + i
O
(iv)
−89.7°
−90° " tan−1 x " 90°
π
– π " tan−1 x " 2
2
(ii)
O
(f)
Re
1–i
Im
2 + 6i
Re
O
Exercise 11E (Page 302)
1
(i)
π
r = 8, θ = 5
r = 41 , θ = 2.3
π
(iii) r = 4, θ = − 3
(iv) r = 3, θ = π – 3
(ii)
–5 – 7i
?
(i)
(Page 298)
π
2
2
– 3π
4
(iii) – π
4
(ii)
(a)
r = 1, θ = 0,
z = 1(cos 0 + i sin 0)
r = 2, θ = π,
z = 2(cos π + i sin π)
π
(iii) r = 3, θ = 2 ,
z = 3(cos π + i sin π )
2
2
(ii)
Activity 11.8 (Page 299)
(i)
(i)
45°
(b)
63.4°
(c) 89.4°
(d)
−63.4°
350
9781510421738.indb 350
02/02/18 1:19 PM
6
(i)
Im
4
3
2
1
3π
(vi) r = 5 2 , θ = − 4 ,
–3
–2
(vii) r = 2, θ = − π ,
3
π
z = 2(cos (− 3 ) + i sin (− π ))
3
π
(viii) r = 12, θ = 6 ,
z = 12(cos π + i sin π )
6
6
7
(ix) r = 5, θ = −0.927,
z = 5(cos(−0.927) + i sin(−0.927))
8
(ii)
–2
1
Real part = 4
(i)
(a) 2 + i
(b) r =
−3 + 2i and 3 − 2i
(i)
Im
2
α−π
(ii)
−α
O
5
(ii)
C
1
–1
2
3
4
5 Re
B
–2
OACB is a rhombus.
(ii) 53 + 45 i
9
?
(iii) The locus is a circle, centre 2i, radius 5.
(Page 304)
arg(z1 − z2) is the angle between the line joining
z1 and z2 and a line parallel to the real axis.
Exercise 11F (Page 306)
1
(iii) π − α
(iv) π − α
2
(v) π + α
2
A
1
(i)
(i)
3 Re
3
(xii) r = 12013, θ = −2.128,
z = 12013 (cos (−2.128)+ i sin (−2.128)
4
2
5 , θ = 0.464
(ii)
r = 13, θ = 2.747,
z = 13(cos 2.747 + i sin 2.747)
z = 2i
(ii) z = 3 + 3 3 i
2
2
7
3
7
(iii) z = –
+ i
2
2
1
1
–
i
(iv) z =
2
2
(v) z = – 5 – 5 3 i
2
2
(vi) z = −2.497 − 5.456i
1
–1
(xi) r = 65 , θ = 1.052,
z = 65 (cos 1.052 + i sin 1.052)
3
O
–1
z = 5 2 (cos (− 3π ) + i sin (− 3π ))
4
4
(x)
Answers
(iv) r = 4, θ = − π ,
2
z = 4(cos (− π ) + i sin (− π ))
2
2
π
(v) r = 2 , θ = ,
4
z = 2 (cos π + i sin π )
4
4
(i)
Im
O
–π
3
Re
Real part = 1
2
351
9781510421738.indb 351
02/02/18 1:19 PM
(ii)
(iii) The three points are the same
Im
distance from the origin and separated by
equal angles of 2π (i.e. 120°).
3
(iv) –(2 + 3) + (2 3 – 1)i
–(2 – 3) – (2 3 + 1)i
Answers
4i
O
Re
4
(iii)
Im
5
(ii)
−6 & p & 2
(iii)
z−5 =5
(i)
1 − 3i, − 1 − 3i
(ii)
–3
O
Re
1
–3
(iv)
Im
2
–2
Im
–1 O
–1
1
2
3 Re
–2
O
(iii) 1 − 3i: r = 2, θ = − π
3
−1 − 3i: r = 2,
2π
θ=− 3
(iv) The three points are the same distance
from the origin and separated by equal
angles of 2π (i.e. 120°).
3
Re
–1 – 2i
(v)
Im
6
O
3–i
(vi)
–π
6
7
(i)
(ii)
π
3
–π
4
(a)
2 , θ = – 43 π
u2: r = 2, θ = 21 π
u: r =
Im
4
3
O
Re
1 + 2i
(b) – 21 + 21 i
(ii) 3π
4
(iv) OA = BC and OA and BC are parallel
Re
Im
–5 + 3i
(i)
2
u2
1
O
2
π , 2π
3 3
3
(i)
(ii)
352
9781510421738.indb 352
r = 1, θ = 23 π
wz: modulus = R, argument = θ + 23 π
2
z
w : modulus = R, argument = θ – 3π
–4 –3 –2 –1
–1
u
–2
1
2
3
4 Re
–3
–4
02/02/18 1:19 PM
8
(i)
3
(ii)
3 at z = 3 − 3 i
2 2
2
(
(ix) 3 2 cos 7π + i sin 7π
12
12
3
(iii) 3 + 1
Rotation of vector z through + π
2
(ii)
Half turn of vector z
( = two successive π
2
rotations: –1 = i × i )
4
Enlarge from O ×2 and rotate + π
2
(iii) Complete the parallelogram 3z, 0, 2iz
(iv) Reflect in the real axis
(v)
Exercise 11G (Page 312)
32(cos0.6 + i sin 0.6)
(ii)
2(cos( −0.2) + i sin( −0.2))
(iii) 12 cos π2 + i sin π2
(v)
(
)
(
)
5π
24 cos 5π
4 + i sin 4
(
)
6
)
(
)
7π
7π
(i) 6 (cos 12 + i sin 12 )
(ii)
(cos 12π + i sin 12π )
(iii) ⎡⎢cos ( – π ) + i sin ( – π )⎤⎥
12
12 ⎦
⎣
⎡
⎤
cos − π + i sin (− π ) ⎥
(iv)
4 ⎦
⎣⎢ ( 4 )
2π
(v) 9 (cos 2π
3 + i sin 3 )
(vi) 32 ⎡⎢cos ( – 3π ) + i sin ( – 3π )⎤⎥
4
4 ⎦
⎣
(i)
–1
1+ i
(ii)
2
(iii) –1.209 + 0.698i
(iv) –13.129 + 15.201i
7
(i)
3
2
2
3
(ii)
(a) 10e i
(b)
4
(c)
6e 8i
(d)
3e i
(e)
3e 3i
(f)
4e − i
(a)
2(cos 3 + i sin 3)
× 5(cos( −2) + i sin(−2))
= 10(cos 1 + i sin 1)
1
2
(vii) 432(cos 0 + isin 0)
(
π
π
(viii) 10 cos 34 + i sin 34
)
+
2 cos π + i sin π ; 3 1
3
3 2 2
3π
(vi) 6 cos 3π
4 + i sin 4
2
3 − 1, 3 + 1;
4
4
3π
2 cos + i sin 3π ,
4
4
5
(i)
(
Find where the circle with centre O
through z meets the positive real axis
(vi) Complete the similar
triangles 0, 1, z and 0, z, z2
3 + i and −3 − i
(iv) 3 cos π6 + i sin π6
Enlarge from O ×3
(i)
(ii)
(Page 309)
(
( z1 ) = arg z
Answers
(i)
1
if z = 0 then 1 does not exist
z
(iii) if z = real and negative then arg
Activity 11.12 (Page 308)
?
Exceptions
(i)
2
)
)
(b) 8(cos 5 + i sin 5)
÷ 2(cos 5 + i sin 5)
=4
(c)
3(cos 7 + i sin 7)
× 2(cos 1 + i sin 1)
= 6(cos 8 + i sin 8)
353
9781510421738.indb 353
02/02/18 1:19 PM
(d) 12(cos 5 + i sin 5)
÷ 4(cos 4 + i sin 4)
= 3(cos 1 + i sin 1)
2
2 – i, –3
3
z = 7, 4 ± 2i
3(cos 2 + i sin 2)
× (cos 1 + i sin 1)
= 3(cos 3 + i sin 3)
4
p = 4, q = –10, other roots 1 + i, –6
5
z = 3 ± 2i, 2 ± i
8(cos 3 + i sin 3)
÷ 2(cos 4 + i sin 4)
= 4(cos( −1) + i sin(−1))
6
z = ±3i, 4 ± 5
7
7, 4 ± 2i
8
(i)
z = –3
(ii)
z = –3, 5 ± 11 i
2
2
Answers
(e)
(f)
8
(i)
−16.3°
(ii)
Im
5
−5
O
2.5
5 Re
9
k = 36, other roots are − 5 ± 3 3 i
2
2
10
z3 − z – 6 = 0
11
(i)
(ii)
False
(iii) True
−5
Complex numbers common to both loci:
5e ± 3 π i
1
9
(i)
7
23
− 17
+ 17
i
(ii)
wz = 17 + 17i
Exercise 11H (Page 315)
1
True
(iv) True
12
a = 2, b = 2, z = −2 ± i,1 ± 2i
13
(ii)
(iii)
1 – 2i
Im
2
O
1
Re
4 + 5i is the other root.
The equation is z2 − 8z + 41 = 0
354
9781510421738.indb 354
02/02/18 1:19 PM
Index
A
B
binomial expansion
general 163–71
use of partial fractions 180–2
binomial theorem 166
C
change-of-sign methods 146–51
cobweb diagram 154, 155
common factors 173
complex conjugates 283–4, 313
complex exponents 310–13
complex numbers
addition and subtraction 291–3
definition 283
equality 284–6
and equations 313–16
geometrical representation 291
modulus 293–4
modulus–argument (polar) form
297–304, 307–10
multiplication and division
287–8, 307–10
need for 281–2, 313
notation 282
real and imaginary parts 283
square root 288–90
complex plane 291
compound interest 128
compound-angle formulae 61–6, 72–8
convergence to a limiting value 151
cosecant (cosec) 58
cotangent (cot) 58
cubic expressions see polynomials
curves, discontinuous 125
D
decimal search 146–8
Devi, Shakuntala 162
9781510421738.indb 355
differential equations
forming from rates of change
218–22
general solution 223
particular solution 225–32
verifying the solution 228–9
differentiation
of exponentials 90–3
implicit 184
of implicitly defined functions
101–8
of inverse trigonometrical
functions 184–816
of natural logarithms 90–3
parametric 112–18
product rule 84–6
quotient rule 86–7
of trigonometrical functions
96–101
direction, of a line 261–2
displacement vectors 238–9
displacement-time graphs 83
distance, from a point to a line 273–7
division see multiplication and division
dot product 251–3
double-angle formulae 67–71
E
e, base of natural logarithms 38, 46–7
equation of a line 37
vector form 258–65
equations
and complex numbers 313–16
rearranging 152–6
error, bounds 149
estimates, accuracy 135
existence theorem 314
experimental data, curve fitting 37–41
exponential functions
applications 39–41, 48–52
and complex numbers 310–13
definition 24, 34, 48–52
differentiation and integration
90–3, 121–2, 191–4
graphs 25–6
growth and decay 24, 26–7, 49
as infinite series 128
F
factor theorem 8–10
fixed-point iteration 151–61
fractions
addition and subtraction 173
multiplication and division 172
simplification 172
Fundamental Theorem of Algebra
313–14
G
Index
acceleration, definition 218
addition and subtraction
of complex numbers 291–3
of fractions 173
of polynomials 2–3
of vectors 241–3
algebraic fractions 172–4
alternating current (AC) 78
angle
between two lines 272–3
between two vectors 250–8
Argand diagrams 291–3, 294–7, 304–7
gradient
of exponential curves 25
of a straight line 37
graphs
displacement–time 83
of exponential functions 25–6
of logarithms 33
of the natural logarithm function
48
of parametric equations 110
of polynomial equations 7
of trigonometrical functions 58–9
growth and decay, exponential 24,
26–7, 49
I
identities see trigonometrical identities
identity symbol 199
imaginary number 282
implicit functions, differentiation
101–8
indices, laws 30
inequalities, involving the modulus
sign 19–21
integrals
involving exponentials 191–4
involving natural logarithms
191–4
standard 213
integration
by change of variable 186–91
by parts 203–12
by substitution 186–91
choice of technique 212–14
of exponentials 90–3
involving the exponential function
121–2
involving the natural logarithm
function 122–7
of natural logarithms 90–3
numerical 133–40
of trigonometrical functions
129–33, 194–9
use of partial fractions 199–203
intervals
bisection 148
estimation 146–51
notation 145
inverse functions 34, 48
iteration 151, 153, 155
355
02/02/18 1:19 PM
Index
L
line segment 239
lines
angle between 272–3
direction 261–2
gradient 37
intersection 265–71
location on 262–3
perpendicular distance from 273–7
skew 266, 273
location, on a line 262–3
logarithm function 34, 46, 47
inverse 34
logarithmic integrals 123–4
logarithmic relationships 38–9
logarithms 28–33
see also natural logarithms
lowest common multiple 173
M
magnitude
of a quantity 17
of a vector 236
modelling
use of logarithms 37–41
use of trigonometric functions
57, 61, 83
of waves 57, 61, 75, 184
modulus
of a complex number 293–4
of a vector 236
modulus function 17–21
multiplication and division
of complex numbers 287–8,
307–10
of fractions 172
of logarithms 30
of polynomials 3–6
of a vector by a scalar 240–1
N
Napier, John 48
natural logarithm function 45–8, 122–7
natural logarithms
base 38
differentiation and integration
90–3, 191–4
natural numbers 163
negative number, square root 282
Newton’s law of cooling 217
numbers, types 280, 281–2
numerical methods
accuracy 135, 144–6
integration 133–40, 142–3
for polynomial equations 8
P
parameter, eliminating 111–12
parametric differentiation 112–18
parametric equations 108–12
partial fractions
types 174–80
use in integration 199–203
using with the binomial expansion
180–2
Pascal’s triangle 164
point
coordinates 237
perpendicular distance from a line
273–7
polynomial equations
graphs 7
roots 11–12
solution 7–16
with complex numbers 313
polynomials
addition and subtraction 2–3
definition 1–2
multiplication and division 3–6
order 2
position, definition 218
position vector 237–8, 262
principal argument, of a complex
number 298
principal value 69, 79
product rule 84–6
spiral dilatation 309
square root
of a complex number 288–90
of a negative number 282
staircase diagram 154, 155
stationary points
of implicit functions 103–6
of a parametric curve 115
straight line see equation of a line, line
segment; lines
subtraction see addition and
subtraction
Q
unit vector 235, 245–7
quadratic expressions 1
quadratic formula 7, 313
quotient 4, 6
quotient rule 86–7
R
Ramanujan, Srinivasa 1
rate of change 218
rational functions see algebraic
fractions
real and imaginary axes 291
reciprocal trigonometrical functions
58–60
remainder theorem 12–13
resultant 242
roots
of logarithms 32
number 75, 145
of polynomial equations 7–8,
11–12
of trigonometric equations 79–80
S
scalar, definition 234
scalar product 251–3
secant (sec) 58
separation of variables 223–5
sets of points, on an Argand diagram
294–7, 304–7
sine function, differentiating 83
T
trapezium rule 133–40
trial and improvement 144
trigonometrical equations, general
solution 79–80
trigonometrical functions
differentiation 96–101
graphs 58–9
integration 129–33, 194–9
inverse, differentiation 184–6
reciprocal 58–60
trigonometrical identities 59, 61, 67,
130–1
U
V
vector product 252
vectors
addition and subtraction 241–3
angle between 250–8
calculations 240–50
components 235
definition 234
equal 237
of geometrical figures 244
joining two points 258–9
length 239
magnitude 236
magnitude—direction (polar)
form 235
multiplication by a scalar 240–1
negative of 241
notation and terminology 218,
235–6
perpendicular 251–2
polar form 235
in three dimension 236–40,
252–3
in two dimensions 235–6
uses in mathematics 234
velocity, definition 218
W
waves, modelling 57, 61, 75, 184
356
9781510421738.indb 356
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