A Level Further Mathematics for OCR A Pure Core Student Book 2 (Year 2) (AS/A Level Further Mathematics OCR) [Student ed.] 1316644391, 9781316644393, 9781316644249, 9781316644461, 9781316644454

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A Level Further Mathematics for OCR A Pure Core 2 Student Book (Year 2) Vesna Kadelburg, Ben Woolley, Stephen Ward, Paul Fannon

Contents Introduction How to use this resource 1 Series and induction Section 1: Review of proof by induction Section 2: Induction and series Section 3: Using standard series Section 4: The method of differences 2 Powers and roots of complex numbers Section 1: De Moivre’s theorem Section 2: Complex exponents Section 3: Roots of complex numbers Section 4: Roots of unity Section 5: Further factorising Section 6: Geometry of complex numbers 3 Complex numbers and trigonometry Section 1: Deriving multiple angle formulae Section 2: Application to polynomial equations Section 3: Powers of trigonometric functions Section 4: Trigonometric series 4 Lines and planes in space Section 1: Equation of a plane Section 2: Intersection between a line and a plane Section 3: Angles between lines and planes Section 4: Distances between points, lines and planes 5 Simultaneous equations and planes Section 1: Linear simultaneous equations Section 2: Intersections of planes Focus on … Proof 1 Focus on … Problem solving 1 Focus on … Modelling 1 Cross-topic review exercise 1 6 Hyperbolic functions Section 1: Defining hyperbolic functions Section 2: Inverse hyperbolic functions Section 3: Hyperbolic identities Section 4: Solving harder hyperbolic equations Section 5: Differentiation Section 6: Integration 7 Further calculus techniques Section 1: Differentiation of inverse trigonometric functions Section 2: Differentiation of inverse hyperbolic functions Section 3: Using inverse trigonometric and hyperbolic functions in integration Section 4: Using partial fractions in integration 8 Applications of calculus Section 1: Maclaurin series Section 2: Using standard Maclaurin series

Section 3: Improper integrals Section 4: Volumes of revolution Section 5: Mean value of a function 9 Polar coordinates Section 1: Curves in polar coordinates Section 2: Some features of polar curves Section 3: Changing between polar and Cartesian coordinates Section 4: Area enclosed by a polar curve Section 5: Area between two curves 10 Differential equations Section 1: Terminology of differential equations Section 2: The integrating factor method for first order equations Section 3: Homogeneous second order linear differential equations with constant coefficients Section 4: Non-homogeneous second order linear differential equations with constant coefficients 11 Applications of differential equations Section 1: Forming differential equations Section 2: Simple harmonic motion Section 3: Damping and damped oscillations Section 4: Linear systems Focus on … Proof 2 Focus on … Problem solving 2 Focus on … Modelling 2 Cross-topic review exercise 2 Practice paper 1 Practice paper 2 Formulae Answers Worked solutions for chapter exercises 1 Series and induction 2 Powers and roots of complex numbers 3 Complex numbers and trigonometry 4 Lines and planes in space 5 Simultaneous equations and planes 6 Hyperbolic functions 7 Further calculus techniques 8 Applications of calculus 9 Polar coordinates 10 Differential equations 11 Applications of differential equations Worked solutions for cross-topic review exercises Cross-topic review exercise 1 Cross-topic review exercise 2 Worked solutions for Pratice papers exercises Pratice papers 1 Pratice papers 2 Acknowledgements

Introduction You have probably been told that mathematics is very useful, yet it can often seem like a lot of techniques that just have to be learnt to answer examination questions. You are now getting to the point where you will start to see where some of these techniques can be applied in solving real problems. However, as well as seeing how maths can be useful, we hope that anyone working through this book will realise that it can also be incredibly frustrating, surprising and ultimately beautiful. The book is woven around three key themes from the new curriculum:

Proof Maths is valued because it trains you to think logically and communicate precisely. At a high level, maths is far less concerned about answers and more about the clear communication of ideas. It is not about being neat – although that might help! It is about creating a coherent argument that other people can easily follow but find difficult to refute. Have you ever tried looking at your own work? If you cannot follow it yourself it is unlikely anybody else will be able to understand it. In maths we communicate using a variety of means – feel free to use combinations of diagrams, words and algebra to aid your argument. And once you have attempted a proof, try presenting it to your peers. Look critically (but positively) at some other people’s attempts. It is only through having your own attempts evaluated and trying to find flaws in other proofs that you will develop sophisticated mathematical thinking. This is why we have included common errors in our ’work it out’ boxes – just in case your friends don’t make any mistakes!

Problem solving Maths is valued because it trains you to look at situations in unusual, creative ways, to persevere and to evaluate solutions along the way. We have been heavily influenced by a great mathematician and maths educator, George Polya, who believed that students were not just born with problem solving skills – these skills were developed by seeing problems being solved and reflecting on the solutions before trying similar problems. You may not realise it but good mathematicians spend most of their time being stuck. You need to spend some time on problems you can’t do, trying out different possibilities. If after a while you have not cracked it then look at the solution and try a similar problem. Don’t be disheartened if you cannot get it immediately – in fact, the longer you spend puzzling over a problem the more you will learn from the solution. You may, for example, never need to integrate a rational function in future, but we firmly believe that the problem solving skills you will develop by trying it can be applied to many other situations.

Modelling Maths is valued because it helps us solve real-world problems. However, maths describes ideal situations and the real world is messy! Modelling is about deciding on the important features needed to describe the essence of a situation and turning that into a mathematical form, then using it to make predictions, compare to reality and possibly improve the model. In many situations the technical maths is actually the easy part – especially with modern technology. Deciding which features of reality to include or ignore and anticipating the consequences of these decisions is the hard part. Yet some fairly drastic assumptions – such as pretending a car is a single point or that people’s votes are independent – can result in models that are surprisingly accurate. More than anything else, this book is about making links. Links between the different chapters, the topics covered and the themes just discussed, links to other subjects and links to the real world. We hope that you will grow to see maths as one great complex but beautiful web of interlinking ideas. Maths is about so much more than examinations, but we hope that if you take on board these ideas (and do plenty of practice!) you will find maths examinations a much more approachable and possibly even enjoyable experience. However, always remember that the results of what you write down in a few hours by yourself in silence under exam conditions is not the only measure you should consider when judging

your mathematical ability – it is only one variable in a much more complicated mathematical model!

How to use this resource Throughout this book you will notice particular features that are designed to aid your learning. This section provides a brief overview of these features. In this chapter you will learn how to: use De Moivre’s theorem to derive trigonometric identities find sums of some trigonometric series.

Learning objectives A short summary of the content that you will learn in each chapter.

WORKED EXAMPLE The left-hand side shows you how to set out your working. The right-hand side explains the more difficult steps and helps you understand why a particular method was chosen.

PROOF Step-by-step walkthroughs of standard proofs and methods of proof.

WORK IT OUT Can you identify the correct solution and find the mistakes in the two incorrect solutions?

Before you start… Chapter 2, Section 1

Chapter 2, Section 2

You should be able to use de Moivre’s theorem to raise a complex number to a power.

1 Find

You should be able to use the exponential form of a complex number.

2 a Write in exact Cartesian form.

in modulus–argument form.

b Write down the complex conjugate of .

Before you start Points you should know from your previous learning and questions to check that you’re ready to start the chapter.

Key point A summary of the most important methods, facts and formulae.

Explore Ideas for activities and investigations to extend your understanding of the topic.

Tip Useful guidance, including on ways of calculating or checking and use of technology. Each chapter ends with a Checklist of learning and understanding and a Mixed practice exercise, which includes past paper questions marked with the icon  

.

In between chapters, you will find extra sections that bring together topics in a more synoptic way.

Focus On … Unique sections relating to the preceding chapters that develop your skills in proof, problem solving and modelling.

CROSS-TOPIC REVIEW EXERCISE Questions covering topics from across the preceding chapters, testing your ability to apply what you have learnt. You will find practice questions towards the end of the book, as well as a glossary of key terms (picked out in colour within the chapters), and answers to all questions. Full worked solutions can be found on the Cambridge Elevate digital platform, along with a digital version of this Student Book. Maths is all about making links, which is why throughout this book you will find signposts emphasising connections between different topics, applications and suggestions for further research.

Rewind Reminders of where to find useful information from earlier in your study.

Fast forward Links to topics that you may cover in greater detail later in your study.

Focus on … Links to problem solving, modelling or proof exercises that relate to the topic currently being studied.

Did you know? Interesting or historical information and links with other subjects to improve your awareness about how mathematics contributes to society. Colour-coding of exercises The questions in the exercises are designed to provide careful progression, ranging from basic fluency to practice questions. They are uniquely colour-coded, as shown here.

2

For each equation from question 1, write the roots in exact Cartesian form.

5

Let

7

Solve the equation

8

Multiply out and simplify

and

. Show that

. , giving your answers in Cartesian form. , where

.

you found that two possible expressions were and 14 In the derivation of . Show that their sum is zero and hence explain why the expression chosen in Proof 3 is non-negative. 22 Point represents the complex number

on an Argand diagram. Point is rotated through

radians anticlockwise about the origin to point . Point is then translated by

to obtain

point .

Black – drill questions. Some of these come in several parts, each with subparts i and ii. You only need attempt subpart i at first; subpart ii is essentially the same question, which you can use for further practice if you got part i wrong, for homework, or when you revisit the exercise during revision. Green – practice questions at a basic level. Blue – practice questions at an intermediate level. Red – practice questions at an advanced level. Yellow – designed to encourage reflection and discussion. Purple – challenging questions that apply the concept of the current chapter across other areas of maths.

1 Series and induction In this chapter you will learn how to: use the principle of mathematical induction to prove results about sequences, series and differentiation use given results for the sums of integers, squares and cubes to find expressions for sums of other series use a technique called the method of differences to find an expression for the sum of terms of a series use the expression for the sum of the first terms to determine whether an infinite series converges and find its limit.

Before you start… Pure Core Student Book 1, Chapter 6

You should be able to use mathematical induction to prove results about matrices,

1 Prove that

.

divisibility and inequalities. GCSE

You should be able to use the th term formula to generate terms of a sequence.

2 A sequence is defined by . Find the first three terms.

GCSE

You should be able to simplify expressions by factorising.

3 Simplify .

A Level Mathematics Student Book 2, Chapter 4

You should be able to use sigma notation to write a series.

A Level Mathematics Student Book 2, Chapter 10

You should know how to differentiate using the product rule and chain rule.

5 Given that

A Level Mathematics Student Book 2, Chapter 5

You should know how to write an expression in partial fractions.

6 Write

4 Find

.

, find

.

in partial fractions.

Introduction In Pure Core Student Book 1, Chapter 6, you learnt about the method of proof by induction, which you can use to prove that observed patterns continue forever. The sorts of patterns you looked at included powers

of matrices (for example, prove that is divisible by for all

for all

), divisibility (for example, prove that

) and inequalities (for example, prove that

for

).

In this chapter you will revisit these ideas, including examples where you need to conjecture (guess) the pattern first, and then see how to extend them to other contexts, such as sums of series and differentiation.

Fast forward In Chapter 2 you will use induction to prove de Moivre’s theorem, a result about powers of complex numbers. Finding expressions for sums of series is one of the most difficult problems in mathematics as there is no general method that always works. For example, you know how to find the sum of the first terms of a geometric series:

. But what about a series such as

?

Fast forward In Chapter 8 you will learn about another type of series called the Maclaurin series. The method of mathematical induction is useful for proving that a conjectured formula for the sum of a series is correct, but it offers no help in finding what the formula might be. Sometimes you can guess the formula by looking at some examples, but most of the time the general expression is far from obvious. In this chapter you will meet the method of differences, which allows you to find the formula in some cases. You will also learn how to derive formulae for sums of more complicated series by combining results you have already derived.

Section 1: Review of proof by induction You can use induction to prove statements about a sequence or a pattern, where the statement holds for every natural number . The proof involves two steps: 1 Prove that the statement is true for some starting value of (usually, but not always, 2 Assuming that the statement is true for some , prove that it is also true for

).

.

Then the principle of mathematical induction states the statement is true for all values of . Sometimes you need to conjecture the pattern for yourself before using induction to prove it.

Rewind You will need to use the product rule for differentiation. This was covered in A Level Mathematics Student Book 2, Chapter 10.

WORKED EXAMPLE 1.1 Let

.

a Find

and

.

b Conjecture an expression for

and prove it by induction.

Use the product rule to differentiate.

a

The factorised form makes it easier to spot the pattern. b Conjecture:

Proof: When

:

Start by showing that the statement is true when .

So the statement is true for Assume it is true for

.

: Write down the statement with

When

,

Think about what you are trying to prove. Remember that you cannot use this result! You are working towards:

Relate

to

.

.

Use the result you have assumed for . Differentiate using the product rule. This the the required result for .

Hence the result is also true for The result is true for true for

, and if true for

it is also

Remember to write the conclusion.

.

Therefore, the result is true for all of mathematical induction.

by the principle

Sequences are often given by a term-to-term rule, but you might want to know a formula for the th term. You might be able to guess the formula by looking at the numbers and then you can use induction to prove that it works for all . For example, the term-to-term rule could be

describes a sequence whose first four terms are

. You might notice that these are all one less than a power of , so the formula for the th term . You can prove by induction that this formula indeed works for all .

WORKED EXAMPLE 1.2 A sequence is given by is

and

for

. Prove that the th term of the sequence

.

When



:

Show that the result is true for

.

So the formula works when Assume that the formula works when

:

Assume that the formula works for some . Think about what you are trying to prove. You are working towards: Use the result you assumed for

So, the formula also works when The formula works when

.

.

, and if it works for

some then it also works for . Hence, by the principle of mathematical induction, the formula works for all .

Remember to write the conclusion.

Sometimes each term in the sequence depends on more than one previous term. For example, the termto-term rule with produces this sequence:

and

and so on. You can still use proof by induction, but you need to show that the formula works for two starting values of . WORKED EXAMPLE 1.3 A sequence is given by the recurrence relation and Prove that the formula for the th term of the sequence is When

for

.

.

:

Check that the formula works for and .

So, the formula works for When

,

.

:

So, the formula works for Assume the formula works for

. and

:

Assume that the formula works for and

,

and prove that it works for . Think about the formula with and . When

,

Think about what you are trying to prove. You are working towards: Express and .

in terms of

Use the results for .

and

Look at what you are working towards; group the powers of and the powers of . So, the formula also works for

.

The formula works for and , and if it works for and then it also works for . Therefore, the formula works for all by the principle of mathematical induction.

EXERCISE 1A

Write a conclusion.

1

Given that

2

A sequence has first term and subsequent terms defined by

, prove by induction that

. . Prove by induction that

. 3

Given that

,

a find the first four terms of the sequence b conjecture a formula for the th term and prove it by induction. 4

Let

.

a Find

,

and

.

b Conjecture an expression for

and prove it by induction.

Focus on… Powers of matrices have many interesting applications. To explore one of them see Focus on … Modelling 1. 5

Let

.

a Find

for

.

b Which natural number do all

seem to be multiples of?

c Use mathematical induction to prove your conjecture from part b. 6

A sequence is given by the term-to-term rule the general term of the sequence is

7

Given that

8

Let a Find

and

. Prove that

.

, prove by induction that

.

. ,

and

.

b Conjecture an expression for 9

with

and prove it by induction.

a Suggest which natural number is a factor of all the numbers of the form

.

b Prove your claim by induction. 10 Given that

, use induction to prove that

11 Given that

, prove by induction that

.

.

12 Use mathematical induction to show that 13 Prove by induction that 14 Given that

. for

.

, use mathematical induction to prove that

15 The Fibonacci sequence is defined by of the Fibonacci sequence is given by

. for

. Show that the th term .

Explore Leonardo Fibonacci (c. 1170 - c. 1250) was an extremely influential mathematician in the Middle Ages, largely responsible for spreading the number system you use today. He also gave his name to the famous Fibonacci sequence. The formula in question shows the link between the Fibonacci sequence and the golden ratio, a quantity surprising places in mathematics.

which appears in many

Section 2: Induction and series A series is a sum of the terms of a sequence. If you add the terms up to a certain point you get a finite series, such as

. You can also try to form an infinite series, for example . Some infinite series, such as the geometric series given here, have a finite sum. In

this section you will only look at finite series; you will meet some infinite series in Section 4.

Rewind You met sequences, series and sigma notation in A Level Mathematics Student Book 2, Chapter 4. You can use sigma notation as a shorter way of writing a series. For example,

In the A Level Mathematics course you learnt how to find the general formula for the sum of the first terms of an arithmetic and a geometric series:

You also learnt about finite and infinite binomial series, for example:

Rewind Binomial series were covered in A Level Mathematics Student Book 1, Chapter 9, and A Level Mathematics Student Book 2, Chapter 6. In Chapter 8 of this book you will learn about Maclaurin series, which you can use to write a function as an infinite series, for example:

These are some examples of finite and infinite series where it is possible to find an exact expression for the sum. In general, finding an expression for the sum of the first terms of a series can be surprisingly difficult, if not impossible. For example, it is not possible to express a formula for the sum in terms of standard functions. In cases where you manage to guess the formula for the sum of a series, you can then try to prove it by induction. The inductive step relies on making the connection between the sum of the first terms and the sum of the first terms; this is done simply by adding the next term of the series.

Explore Although it is not possible to find a general expression for sum to infinity,

, it is possible to find its exact

. Find out what it is: the result may surprise you!

Key point 1.1 If

then

WORKED EXAMPLE 1.4 Prove by induction that for all For

.

:

Show that the statement is true for the starting value (in this case, ).

So, the result is true for

.

Assume that the result is true for

Let

:

State the assumption for

:

Consider using

and relate it to .

.

by

Substitute in the result for (assumed to be true). Combine this into one fraction and simplify. It is always a good idea to take out any common factors.

Show that this is in the required form by separating out in each place it occurs. So, the result is also true for The result is true for



.

, and if it is true for

it

is also true for . Therefore, the result is true for all , by induction.

EXERCISE 1B

Make sure you write a conclusion.

EXERCISE 1B 1

Prove by induction that, for all

2

Prove by induction that, for all integers

3

Using mathematical induction prove that, for all positive integers:

4

Prove by induction that, for all integers

5

Use mathematical induction to show that, for all integers

6

Prove by induction that, for all

7

Using mathematical induction prove that, for all integers

8

Prove by induction that, for all positive integers:

9

Prove using mathematical induction that, for all

:

:

:

:

:

10 Prove by induction that, for all integers

:

:

:

Section 3: Using standard series You will now look at finding expressions for the sums of series such as

by combining some

standard results. You can use the following formulae without proof (unless the question explicitly asks you to prove them).

Key point 1.2 Formulae for the sums of integers, squares and cubes:

The second and third formulae will be given in your formula book.

Rewind You proved the second and third formulae in Exercise 1B, Questions 2 and 3. The first formula in Key Point 1.2 is just a special case of an arithmetic series.

Explore The formula for the sum of the first integers,

, is the formula for the th term in the

sequence of triangular numbers:

Explore pictorial representations of the other two formulae from Key point 1.2. Before you use these results, notice how you can split up sums and take out constants. For example:

where

.

Key point 1.3 You can manipulate series in several ways.

where is a constant.

Tip Remember that a constant, , summed times is

and not just . For example,

and

not .

WORKED EXAMPLE 1.5 a Use the formula for b Hence find

to show that

.

.

You need to rearrange the expression into a form to which you can apply the standard formulae. Start

a

by splitting up the sum. Then take out of the first sum as a factor.

and Notice that it’s always a good idea to factorise first. In this case only factorises, but in more complicated examples this will avoid having to expand and then factorise a higher order polynomial later. You can only use the formula in part a if the sum starts from . Therefore, work out the sum of the first terms and subtract the sum of the first terms.

b

Now use the formula .

WORKED EXAMPLE 1.6 a Use the formulae for

,

and

to show that .

b Hence find an expression for

, simplifying your answer fully.

Expand the brackets.

with

and

a

Split up the series into separate sums. Take out constants. Substitute in the standard formulae.

Simplify the first two terms. Now factorise as many terms as possible. Note that this is much easier than expanding everything first.

Substitute part a.

b

for in the formula found in

Simplify and factorise, taking out a from the second bracket.

WORK IT OUT 1.1 Given that

, find an expression for

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

EXERCISE 1C

In this exercise, you can assume the formulae for 1

,

and

.

Evaluate each expression. a

i ii

b

i ii

2

Find a formula for each series, giving your answer in its simplest form. a

i ii

b

i ii

3

Show that

4

Show that

5

a Find an expression for

.

.

.

b Hence find the least value of such that

6

a Show that b Hence evaluate

.

. .

7

Show that

8

a Show that b Hence find a formula for

9

a Show that b Hence find, in the form

, where is an integer to be found.

. , fully simplifying your answer.

. , the exact value of

10 Show that the sum of the squares of the first odd numbers is given by an integer to be found.

. , where is

Section 4: The method of differences Whenever you are investigating a series, start by writing out a few terms to see if any patterns develop. For example, for the series

:

You can see that each term shares a common element with the next; in the first term, this element is positive and in the next it is negative. Therefore, when you complete the sum, these common elements will cancel out. This cancellation continues right through to the th term.

In fact, because

you have just shown that

which is the result for the sum of the first integers that you used in Section 3. This process for finding a formula for the sum of the first terms of a sequence is called the method of differences.

Key point 1.4 Method of differences If the general term of a series,

, can be written in the form

, then:

Tip The series won’t always take exactly this form, so write out several terms to see how the cancellation occurs. WORKED EXAMPLE 1.7

a Show that b Hence show that

. .

a

Use the binomial expansion to expand the cubed brackets.

Simplify to give the result required.

b

Sum both sides of the result from a. The RHS is a difference, so you expect cancellation. Write out the first few terms and the last couple of terms

.

Everything cancels except the terms shown. For the LHS, remember that

.

Make the expressions for the LHS and RHS equal. You now need to make

the subject.

Start by expanding the RHS and then simplify.

Factorise the RHS.

Finally, divide by .

Sometimes the cancellation occurs two terms apart. WORKED EXAMPLE 1.8

a Write

in partial fractions.

b Find an expression for

.

a

Each factor in the denominator corresponds to one partial fraction. Multiply through by the common denominator. Use suitable values of to make each bracket equal to zero. Sum both sides of the result in a.

b

Writing out several terms

shows

that the cancellations in the series occur two terms apart.

Continue the pattern of cancellation for the last few terms .

This leaves part of the first two terms and part of the last two. You could put this all over a common denominator and combine into one fraction but, as the question doesn’t specifically require this, there is no need to do anything else.

Rewind You learnt about partial fractions in A Level Mathematics Student Book 2, Chapter 5. As you keep adding more and more terms of a series, the sum could keep increasing without a limit, or it could approach a finite value. In the latter case you say that the series converges (is convergent) and you could try to find its sum to infinity. To do this, you can find an expression for the sum of the first terms and consider what this expression tends to when gets very large. WORKED EXAMPLE 1.9 Use the result from Worked example 1.8 to evaluate From Worked example 1.8:

.

Let

in this result. Then: As the denominator tends to to zero.

, these fractions tend

EXERCISE 1D 1

a Show that

.

b Hence show that 2

3

.

a Show that

.

b Hence show that

.

a Show that b

.

i Hence show that

.

ii Find

4

a Express b

.

Use the method of differences to show that

a Show that b Hence, find

7

.

. .

a Show that b Find an expression for c Hence show that

8

.

a Show that b

6

in partial fractions.

Use the method of differences to show that

c Find 5

.

Use the method of differences to find

. . . .

9

a Show that

.

b Use the method of differences to show that

where

and are constants to be found.

c Find 10 a Use the method of differences to find b Hence, prove that the series

. diverges.

Checklist of learning and understanding You can use proof by induction to prove results concerning matrices, divisibility, inequalities, sequences, series, powers and differentiation. When working with series, use

.

The formulae for the sums of integers, squares and cubes are:

You can manipulate series in these ways:

where is a constant. Method of differences If the general term of a series,

, can be written in the form

, then:

You might need to split an expression into partial fractions before using the method of differences. You can check whether an infinite series converges, and find its sum to infinity, by finding an expression for the first terms of the series and considering what happens to it as gets very large.

Mixed practice 1 1

Let

.

a Evaluate each .

for

and . Hence suggest a natural number which is a factor of

b Prove by induction that your conjecture from part a is correct for all 2

.

Prove by induction that, for . © OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, June 2010

3

Let

.

a Find

and

. Suggest an expression for

.

b Use mathematical induction to prove your conjecture from part a. 4

Find an expression for

5

Use the formulae for

6

Use the formulae for

7

a Show that

.

and

to show that

and

to find the value of

Find

.

.

b Hence show that

8

.

, where and are integers to be found.

, expressing your answer in a fully factorised form. © OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, June 2011

9

Use induction to prove that, for all integers

10 Prove by induction that, for all 11 a Show that b Prove that 12 a Show that

b Hence find the sum of

, . . .

.

13 a Show that

.

b Hence show that

.

14 a Show that

.

b Find

.

c Hence find

.

15 The sequence i Find

and

is defined by

.

.

ii Hence suggest a positive integer, other than , which divides exactly into every term of the sequence. iii By considering

, prove by induction that your suggestion in part ii is correct.

© OCR AS Level Mathematics, Unit 4725/01 Further Pure Mathematics 1, June 2014 16 i Show that

.

ii Hence show that

.

© OCR AS Level Mathematics, Unit 4725/01 Further Pure Mathematics 1, June 2013 17 Use mathematical induction to prove that

for all

18 Prove by induction that

for all

19 Find the smallest positive integer for which 20 Show that

for

and

. Prove that

.

. for all

.

.

21 Prove by induction that, for any positive integer : . 22 i Show that

.

ii Hence find an expression, in terms of , for

iii Hence find

.

.

© OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, January 2012

23 a Express

in partial fractions.

b Hence show that

where , and are

constants to be found. 24 Use the method of differences to show that and are coprime integers to be found.

, where ,

2 Powers and roots of complex numbers In this chapter you will learn how to: raise complex numbers to integer powers (de Moivre’s theorem) work with complex exponents find roots of complex numbers use roots of unity find quadratic factors of polynomials use a relationship between complex number multiplication and geometric transformations.

Before you start… Pure

You should

Core Student

know how to find the

Book 1, Chapter 4

modulus and argument of a complex number.

Pure

You should be

Core Student Book 1, Chapter 4

able to represent complex numbers on an Argand diagram.

1 Find the modulus and argument of

.

2 Write down the complex numbers corresponding to the points and .

Pure

You should

Core

know how to

Student Book 1,

work with complex

Chapter 2,

numbers in Cartesian form.

3 Given that

and

, evaluate:

a .

b

Section 2, Chapter 4 Pure

You should be

Core Student Book 1, Chapter

able to multiply and divide complex numbers in

4

4

modulus– argument form.

Pure Core Student Book 1,

You should be able to work with complex conjugates.

5 Write down the complex conjugate of:

Pure Core Student Book 1,

You should know how to relate operations with

6 Let

Chapter 4

complex numbers to transformations

Given that

and

, find:

a .

b

Give the arguments in the range

.

a .

b

Chapter 4

a

to

b

to

and be any complex number. Describe a geometrical transformation that maps:

.

on an Argand diagram.

Extending arithmetic with complex numbers You already know how to perform basic operations with complex numbers, both in Cartesian and in modulus–argument form. Modulus– argument form is particularly well suited to multiplication and division. In this chapter you will see how you can utilise this to find powers and roots of complex numbers. This chapter also includes a definition of complex powers which can make calculations even simpler.

Rewind You met complex numbers in Pure Core Student Book 1, Chapter 4. You will also meet roots of unity, which are the solutions of the equation . They have some useful algebraic and geometric properties. Some of the applications include finding exact values of trigonometric functions.

Fast forward You will learn more about links between complex numbers and trigonometry in Chapter 3.

Because you can represent complex numbers as points on an Argand diagram, operations with complex numbers have a geometric interpretation. You can use this to solve some problems that at first sight have nothing to do with complex numbers. This is just one example of the use of complex numbers to solve real-life problems.

Section 1: De Moivre’s theorem In Pure Core Student Book 1, Chapter 4, you learnt that you can write complex numbers in Cartesian form, , or in modulus–argument form, , or . You also learnt the rules for multiplying complex numbers in modulus–argument form: and You can apply this result to find powers of complex numbers. If a complex number has modulus and argument , then multiplying gives that has modulus and argument . Repeating this process, you can see that and In other words, when you raise a complex number to a power, you raise the modulus to the same power and multiply the argument by the power.

Key point 2.1 De Moivre’s theorem For a complex number, , with modulus and argument :

for every integer power .

Tip De Moivre’s theorem can also be written as

or

.

For positive integer powers, you can prove this result by induction.

Rewind For Proof 1 you will also need the compound angle formulae from A Level Mathematics Student Book 2, Chapter 8.

Focus on… See Focus on ... Proof 1 for a discussion of how to extend de Moivre’s theorem to all rational .

PROOF 1 When



:

Check that the result is true for so the result is true for

.

.

Assuming that the result is true for some :

Assume that the result is true for some and write down what that means. (Remember, you will need to use this later.)

Then for

Make a link between this case use

:

.

and

. In

Expand the brackets, remembering that . Group real and imaginary parts. Recognise the compound angle formulae:

and . This is the result you are trying to prove, but with replaced by . Hence the result is true for The result is true for

, and if it is true for

some then it is also true for is true for all



.

Remember to write the full conclusion.

. Therefore, it

by induction.

You can use de Moivre’s theorem to evaluate powers of complex numbers. WORKED EXAMPLE 2.1 Evaluate, without a calculator,

. First find the modulus and argument of each number.

By de Moivre’s theorem:

The argument needs to be between and : Dividing the moduli and subtracting the arguments:

Tip

.

If the power of with the fact that

had been smaller, you might have been able to use the binomial expansion , and . For example:

The usefulness of complex numbers is that the calculation does not get any longer or more difficult with larger powers. You can also prove that de Moivre’s theorem works for negative integer powers. WORKED EXAMPLE 2.2 Let

.

a Find the modulus and argument of . b Hence prove de Moivre’s theorem for negative integer powers. a Multiplying top and bottom by the complex conjugate:

Use

.

To find the modulus and argument, you need to write the number in this form. Remember that Hence

and

.

and

.

This means that you can write .

b Using de Moivre’s theorem for

Since you have already proved de Moivre’s theorem for

positive powers:

positive powers, you can use

with

the modulus and argument of found in part a.

Hence , as required.

WORKED EXAMPLE 2.3 Find the modulus and argument of Modulus and argument of

. :

The best way to find the modulus and argument is to sketch a diagram.

Applying de Moivre’s theorem for negative powers:

This is in the form so you can just read off the modulus and the argument. The modulus is is

EXERCISE 2A

.

and the argument

The argument needs to be between and need to take away .

, so you

EXERCISE 2A 1

Evaluate each expression, giving your answer in the form a

.

i ii

b

i ii

c

i

ii

2

Given that

:

a write , and in the form b represent 3

a Given that

and on the same Argand diagram. :

i write , and in modulus–argument form ii represent

on the same Argand diagram.

b For which natural numbers is

?

In questions 4 and 5 you must show detailed reasoning. 4

5

a Find the modulus and argument of

.

b Hence find

in modulus–argument form.

c Hence find

in Cartesian form.

a Write b Hence find

in the form

.

in simplified Cartesian form.

6

Find the smallest positive integer value of for which

7

Find the smallest positive integer value of such that

is real.

is purely imaginary.

Section 2: Complex exponents The rules for multiplying complex numbers in modulus–argument form look just like the rules of indices: Compare

with

You can extend the definition of powers to imaginary numbers so that all the rules of indices still apply.

Key point 2.2 Euler’s formula:

It is important to realise that Euler’s formula is a definition, and so it makes no sense to ask why it is true or how to prove it. You can, however, note that it seems to be a sensible definition, since it ensures that imaginary powers follow the same rules as real powers.

Fast forward With this definition of imaginary powers, all the usual properties of the exponential function still hold. For example, it turns out that the Maclaurin series, which you will meet in Chapter 8, can be extended to include imaginary powers. You can write a complex number with modulus and argument in exponential form as have four different ways of writing complex numbers with a given modulus and argument.

. You now

Key point 2.3

Did you know? Substituting into Euler’s formula and rearranging gives . This equation, called Euler’s identity, connects five important numbers from different areas of mathematics. It is often cited as ‘the most beautiful’ equation in mathematics. When working with complex numbers in exponential form you can use all the normal rules of indices. WORKED EXAMPLE 2.4 Given that

and

, find

in the form

.

You can do all the calculations in exponential form and then convert to Cartesian form at the end. Use rules of indices for the powers:

Now write in terms of trigonometric functions and evaluate.

You can combine Euler’s formula with rules of indices to raise any real number to any complex power. WORKED EXAMPLE 2.5 Find, correct to three significant figures, the value of: a b

.

a

Use rules of indices to separate the real and imaginary parts of the power. Use Euler’s formula for the imaginary power. Expand the brackets and give the answer to s.f. Remember that the arguments of the trigonometric functions are in radians.

b

You only know how to raise to a complex power, so express as a power of . Use rules of indices and then Euler’s formula. Note that and

The complex conjugate of a number is easy to find when written in exponential form. This is best seen on an Argand diagram, where taking the complex conjugate is represented by a reflection in the real axis. In this case it is best to take the argument between

and .

Fast forward You will use the exponential form of complex numbers when solving second order differential equations in Chapter 10.

Key point 2.4 The complex conjugate of

is

.

Tip Note that if

, you also have

.

In Pure Core Student Book 1, Chapter 5, you used complex conjugates when solving polynomial equations. WORKED EXAMPLE 2.6 A cubic equation has real coefficients and two of its roots are and form .

. Find the equation in the

The roots are: and

.

Complex roots occur in conjugate pairs, so you can write down the third root. Use the formulae for sums and products of roots to find the coefficients of the equation:

Remember that

, and that

Hence the equation is .

EXERCISE 2B You can use your calculator to perform operations with complex numbers in Cartesian, modulus–argument and exponential forms, as well as to convert from one form to another. Do the questions in this exercise without a calculator first, then use a calculator to check your answers. 1

Write each complex number in Cartesian form without using trigonometric functions. a

i ii

b

i ii

c

i ii

2

Write each complex number in the form

.

a

i ii

b

i ii

c

i ii

3

Write the answer to each calculation in the form a

.

i ii

b

i

ii

4

Represent each complex number on an Argand diagram. a

i ii

b

i ii

5

Let

and

6

In this question you must show detailed reasoning. Let

. Show that

and

.

. Write each complex number in the form

.

a b

.

7

Write figures.

8

Write

9

The equation values of , and .

in the form

, where and are real, giving your answer correct to three significant

in exact Cartesian form. has real coefficients, and two of its roots are and

10 A quartic equation has real coefficients and two of its roots are form

.

11 Find in the form 12 Find

in the form

. .

and

. Find the

. Find the equation in the

Section 3: Roots of complex numbers Now that you can use de Moivre’s theorem to find powers of complex numbers, it makes sense to ask whether you can also find roots. In Pure Core Student Book 1, you learnt how to find the two square roots of a complex number by writing and comparing real and imaginary parts. You also know that a polynomial equation of degree has complex roots. Just as a complex number has two square roots, it will have three cube roots, four fourth roots, and so on.

Rewind See Pure Core Student Book 1, Chapter 4, for an example of finding square roots of a complex number. You can’t always use the algebraic method to find all those roots. De Moivre’s theorem gives an alternative method. WORKED EXAMPLE 2.7 Solve the equation Let

. .

Then the equation is equivalent to

Use the modulus–argument form since raising to a power is easier in this form than in Cartesian form. Use de Moivre’s theorem and then compare the modulus and the argument of both sides. Find the modulus and argument of the RHS.

Therefore,

Comparing the moduli: Remember that, by definition, is a positive real number. Comparing the arguments:

If

then



Since adding to the argument returns to the same complex number, there are possible values for between and .

The roots are:

Write down all three roots in modulus–argument form.

If you plot the three roots from Worked example 2.7 on an Argand diagram, you will notice an interesting pattern. They all have the same modulus so they lie on a circle of radius . The arguments differ by

so

they are equally spaced around the circle. Therefore, the three points form an equilateral triangle.

Key point 2.5 To solve

:

write in modulus–argument form use de Moivre’s theorem to write compare moduli, remembering that they are always real compare arguments, remembering that adding number

to the argument does not change the

write different roots in modulus–argument form. All roots of the equation

will have the same modulus, and their arguments will differ by

. This

means that the Argand diagram will always show the pattern you noticed in Worked example 2.7.

Fast forward You will learn more about the connection between powers and rotations in Section 6.

Key point 2.6 The roots of

form a regular polygon with vertices on a circle centred at the origin.

WORKED EXAMPLE 2.8 Draw an Argand diagram showing the roots of the equation One root is

.

There are six roots, forming a regular hexagon. You only need to find one of them and then complete the diagram.

You can also find the equation whose roots form a given regular polygon. WORKED EXAMPLE 2.9 The diagram shows a regular pentagon inscribed in a circle on an Argand diagram. One of the vertices lies on the positive imaginary axis.

The five vertices of the pentagon correspond to the solutions of an equation of the form where is a complex number. Find the values of and . There are five roots, so From the diagram:

.

Any equation of the form

has complex roots.

Use the fact that one root is given in the question.

is a root. Hence

Remember that

.

EXERCISE 2C In this exercise you must show detailed reasoning. 1

Find all three cube roots of each number, giving your answers in the form a

i ii

b

i ii

c

i

.

,

ii 2

Find the fourth roots of each number. Give your answers in the form Argand diagram. a

and show them on an

i ii

b

i ii

3

Solve the equation

4

a Find the modulus and the argument of

for

. Give your answers in the form

b Solve the equation

.

.

, giving your answers in the form

,

where and are integers. 5

Solve the equation

6

Find all complex roots of the equation

, giving your answers in Cartesian form. , giving your answers in the form

, where and are integers. 7

a Write

in the form

.

b Hence solve the equation

, giving your answers in the form

.

c Show your solution on an Argand diagram. 8

The diagram shows a square with one vertex at . The complex numbers corresponding to the vertices of the square are solutions of an equation of the form , where and . Find the values of and .

9

a Solve the equation b Hence express

, giving your answers in Cartesian form. as a product of two real quadratic factors.

10 a Find all the roots of the equation

.

b Hence solve the equation 11 Consider the equation

. Give your answers in exact Cartesian form. .

a Solve the equation, giving your answers in the form

.

The roots are represented on an Argand diagram by points , and , labelled anticlockwise with in the first quadrant. is the midpoint of and the corresponding complex number is . b Find the modulus and argument of .

c Write

in exact Cartesian form.

12 a Find, in exponential form, the three roots of the equation b Expand

.

.

c Hence or otherwise solve the equation Cartesian form.

, giving any complex root in exact

Section 4: Roots of unity In Section 3 you learnt a method for finding all complex roots of a number. A special case of this is solving the equation . Its roots are called roots of unity. WORKED EXAMPLE 2.10 Find the fifth roots of unity, giving your answers in exponential form. Let the roots be

.

Write in exponential form and use de Moivre’s theorem.

Then:

has modulus and argument . Comparing the moduli:

Comparing the arguments:

Remember that there should be five roots.

The fifth roots of unity are:

As in Section 3, the five roots form a regular pentagon on the Argand diagram: The same procedure works for any power : there will be distinct roots, each with modulus , and with arguments differing by

. Remembering that one of the roots always equals , you can write down the

full set of roots.

Key point 2.7 The th roots of unity are: They form a regular -gon on an Argand diagram. Notice that all the arguments are multiples of

. But multiplying an argument by a number corresponds

to raising the complex number to the power of . Hence all the th roots of unity are powers of

. It is

usual to denote the roots

.

Key point 2.8 You can write the th root of unity as: .

You can use the fact that the roots form a regular polygon to deduce various relationships between them. For example, for

, you can use the symmetry of the pentagon to see that

and

.

One of the most useful results concerns the sum of all roots. You know from Pure Core Student Book 1, Chapter 4, that adding complex numbers corresponds to adding vectors on an Argand diagram. Since the points corresponding to the roots of unity are equally spaced around the circle, the sum of the corresponding vectors should be zero. You can also prove this result algebraically.

WORKED EXAMPLE 2.11 Let be a natural number, and let a Express

be the th roots of unity.

in terms of .

b Hence show that a

. Let

.

This is the result from Key point 2.8. This is a geometric series with first term and

b

common ratio . Note that so you can use the formula for the sum of the geometric series.

(since

)

is an th root of unity, which means that

Key point 2.9 If

are the th roots of unity, then .

Tip You could also prove the result in Key point 2.9 by using the result about the sum of the roots of a polynomial: these are the roots of the equation coefficient of which is .

, so their sum equals minus the

You can use the result in Key point 2.9 with a specific value of to find some special values of trigonometric functions.

Rewind You learnt about sums and products of roots of polynomials in Pure Core Student Book 1, Chapter 5.

Fast forward You will learn more about the links between complex numbers and trigonometry in Chapter 3.

WORKED EXAMPLE 2.12 Let

.

a Show that

.

b Hence find the exact value of

a

.

The five points form a regular pentagon.

From the diagram: and Hence

and

You are interested in the real parts.

.

Using the result taking the real part:

and

This is the result from Key point 2.9.

Pair up the terms with equal real parts.

b

Use the fact that

and .

Use the double angle formula to relate the two values.

This is a quadratic equation in Take the positive root since

EXERCISE 2D 1

Write down, in the form a

, all the roots of each equation.

i ii

b

i ii

2

For each equation from question 1, write the roots in exact Cartesian form.

3

a Write down, in the form

, the roots of the equation

b Represent the roots on an Argand diagram. 4

The diagram shows all the roots of an equation a Write down the value of . b Write down the value of c Which of these statements are correct? A B C D

.

.

.

. .

5

Let

.

a Express the seventh roots of unity in terms of . b Is there an integer such that

? Justify your answer.

c Write down the smallest positive integer such that d Write down an integer such that 6

.

Let

be the distinct sixth roots of unity.

a Show that

for

.

b Hence show that 7

.

.

In this question you must show detailed reasoning. a Find, in exact Cartesian form, all the complex roots of the equation b Hence find the exact roots of the equation

.

8

Multiply out and simplify

9

In this question you must show detailed reasoning. Let

.

, where

.

.

a Write, in terms of , the complex roots of the equation Consider the equation

.

.

b Find, in terms of , all roots of the equation. c Show that the roots can be written as d Show that

is equivalent to

e Hence show that

a Show these roots on an Argand diagram. :

i ii 11 a Show that

. .

.

10 Let be the roots of the equation positive argument.

b Write in the form

for

. .

, where is the solution with the smallest

Let b

.

i Show that

.

ii Hence deduce the value of c Show that

is a root of the equation

. .

Section 5: Further factorising In Pure Core Student Book 1, Chapter 5, you learnt that complex roots of a real polynomial come in conjugate pairs, and how you can use this fact to factorise a polynomial. You used the important result that, for any complex number ,

You can now combine this with your knowledge of roots of complex numbers to factorise expressions of the form . WORKED EXAMPLE 2.13 a Find all the complex roots of b Hence write a Let

, giving your answers in Cartesian form.

as a product of two real quadratic factors.

.

Then

Write in exponential form to find the roots, then turn answers into Cartesian form. The argument of

is .

Comparing the moduli: , so Comparing the arguments:

You are looking for four roots, so add

three times.

The roots are: Find the Cartesian form

b

The factors of correspond to the roots of the equation which you found in part a. To get real quadratic factors you need to pair up the factors corresponding to the conjugate roots. You can use the shortcut and .

EXERCISE 2E

EXERCISE 2E In this exercise you must show detailed reasoning. 1

a Find, in exponential form, all the complex roots of the equation

.

b Write your answers from part a in exact Cartesian form. c Hence express 2

By solving the equation

3

Show that

4

Let

as a product of two real quadratic factors. , express

as a product of four real quadratic factors. , where

.

a Write the roots of the equation

in terms of .

b Hence evaluate 5

.

a Show that b Solve the equation c Hence write

6

Let

.

. . as a product of four real quadratic factors.

.

a Write down the non-real roots of the equation b Show that c Hence show that

in terms of .

. is a root of the equation

.

Section 6: Geometry of complex numbers Multiplication of complex numbers has an interesting geometrical interpretation. On an Argand diagram, let be the point corresponding to the complex number , and let be the point corresponding to the complex number

Then

.

, and

. Hence the transformation that takes point to

point is a rotation through angle followed by an enlargement with scale factor .

Key point 2.10 Multiplication by corresponds to a rotation about the origin though angle and an enlargement with scale factor .

Rewind You already know, from Pure Core Student Book 1, Chapter 4, that adding a complex number corresponds to a translation with vector

, and that taking the complex conjugate

corresponds to a reflection in the real axis.

WORKED EXAMPLE 2.14 Points and on an Argand diagram represent complex numbers respectively.

and

a Find the modulus and argument of and . b Hence describe a combination of two transformations which maps to .

A diagram helps to find the modulus and the argument.

,

a

b Enlargement with scale factor and rotation through

about the

origin.

The angle of rotation is the difference between the arguments.

The result from Key point 2.10 is remarkably powerful in some situations that have nothing to do with complex numbers. WORKED EXAMPLE 2.15 An equilateral triangle has one vertex at the origin and another at

. Find one possible set of

coordinates of the third vertex. On an Argand diagram the point corresponds to the complex number

.

You can obtain the third vertex by rotation through anticlockwise about the origin and no enlargement. This corresponds to multiplication by the complex number with modulus and argument .

The complex number corresponding to the third vertex is

So the coordinates are

Tip There is another equilateral triangle with vertices clockwise through

and

. You can obtain it by rotating

, corresponding to multiplication by

.

Focus on… You can use several different approaches to solve the problem from Worked example 2.15. You could use coordinate geometry and trigonometry, or you could use a matrix to carry out the rotation. In Focus on ... Problem solving 1 you will explore different approaches to similar problems.

Rewind You studied rotation matrices in Pure Core Student Book 1, Chapter 3.

Division by

is the same as multiplication by

:

Geometrically, this represents a rotation through angle

.

Key point 2.11 Division by

corresponds to a rotation about the origin though angle

and an enlargement

with scale factor .

Tip If the angle is positive then multiplication corresponds to an anticlockwise rotation and division corresponds to a clockwise rotation. Since raising to a positive integer power is repeated multiplication, in an Argand diagram it corresponds to repeated rotation and enlargement.

Although you can solve Worked example 2.16 just by doing the algebra, thinking about rotations might help you visualise what’s going on. WORKED EXAMPLE 2.16 Find the smallest positive integer value of for which Raising

has argument If

is pure imaginary.

to a power corresponds to repeated rotation through

(combined with an enlargement with scale factor ).

.

The question is therefore: how many rotations through

,

reach either or

which is not an

are needed to

?

integer. If

,

.

EXERCISE 2F 1

Points and represent complex numbers

and

on an Argand diagram.

a Find the modulus and argument of and . b Point is mapped to point by a combination of an enlargement and a rotation. Find the scale

factor of the enlargement and the angle of rotation. 2

Points and represent complex numbers a Show that

and

, respectively.

.

b Describe a single transformation that maps to . 3

The complex number corresponding to the point in the diagram is . The distance . Find, in surd form, the complex number corresponding to the point .

4

The diagram shows a square

, where has coordinates

.

Find the exact coordinates of and . 5

Point in the diagram corresponds to the complex number . The complex number equals

Write the complex numbers , and corresponding to the points , and in terms of and . 6

The diagram shows a right angled triangle

with angle

. The coordinates of are

.

.

a Find the exact length

.

b Using complex numbers, or otherwise, find the coordinates of . 7

Let

.

a Represent , and on an Argand diagram. b Describe fully the transformation mapping to . 8

The diagram shows line through the origin with gradient complex number .

a The line is the locus of

which satisfy

and the point representing the

. Find the exact value of .

Point is the reflection of point in the line . b Find the size of the angle

.

c Use complex numbers to find the exact coordinates of . 9

The diagram shows an equilateral triangle with its centre at the origin and one vertex

.

a Write down the complex number corresponding to the vertex . b Hence find the coordinates of the other two vertices. 10 The diagram shows a regular pentagon with one vertex at

Write down, in the form

.

, the complex numbers corresponding to the other four vertices.

11 a The point representing a complex number on an Argand diagram is reflected in the real axis and then rotated anticlockwise about the origin. Write down, in terms of , the complex number representing the resulting image. b If the rotation is applied before the reflection, show that the resulting image represents the complex number

.

12 a The point representing the complex number on an Argand diagram is rotated through angle about the point representing the complex number . The resulting point represents complex number . Explain why

.

b Find the exact coordinates of the image when the point the point .

is rotated

anticlockwise about

13 a On an Argand diagram, points , and represent complex numbers , and , respectively. is the image of after a rotation through angle , anticlockwise, about .

Express the complex number

in terms of

.

b The point is rotated anticlockwise about the origin. The image is then rotated anticlockwise about the point . Find the coordinates of the final image. 14 The diagram shows two equilateral triangles on an Argand diagram.

Find the complex number corresponding to the midpoint of

.

Give your answer in exact Cartesian form.

Checklist of learning and understanding De Moivre’s theorem:

for

.

Exponential form of a complex number: To solve

.

:

write in modulus–argument form and write compare moduli, remembering that they are always real compare arguments, remembering that adding number

to the argument does not change the

the roots form a regular polygon on an Argand diagram. The th roots of unity (solutions of You can write them as

, where

): and

.

. You can use roots of the equation to factorise the expression factors are found by combining each root with its complex conjugate:

. Real quadratic

Multiplication by corresponds to an enlargement with scale factor and a rotation through anticlockwise about the origin. Division by corresponds to rotation around the origin through angle

and an enlargement with scale factor .

Mixed practice 2 1

If

2

a Find the modulus and argument of

, write in exponential form.

b Hence find 3

.

in exact Cartesian form.

a Write down, in the form

, all the roots of the equation

.

b Show the roots on an Argand diagram. 4

Find

5

a Find the modulus and argument of

in the form

b Hence solve the equation 6

, where

.

. , giving your answers in the form

a Find the modulus and argument of

.

.

b A regular hexagon is inscribed in a circle on an Argand diagram, centred at the origin, and one of its vertices is . Find an equation whose roots are represented by the six vertices of the hexagon. 7

i ii

Express

in the form

, where

and

.

Hence find the smallest positive value of for which

is real and positive.

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2009 8

If

9

Find the exact value of

10 a Express

, where is real and positive, find the exact value of . , clearly showing your working.

in the form

.

b Hence show that

where is a real number to be found.

c Find one pair of possible values of positive integers and such that . 11 Use trigonometric identities to show that a b 12 If is a complex third root of unity and and are real numbers, prove that: a b

.

13 If

and

14 If

, find in its simplest form , prove that

15 a Express in the form

.

.

.

b Hence state the exact value of . 16 The complex numbers , and

are represented in an Argand diagram by the points ,

and respectively. i

Sketch the triangle

and show that it is equilateral.

ii Hence express

in polar form.

iii Hence find

, giving your answer in the form

where and are

rational numbers. © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2013 17 Let

.

a Write

and

in the form

b Explain why

.

c Show that

and

d Form a quadratic equation in 18 Let

.

. and hence show that

be the solutions of the equation

a Show that

.

.

.

b Find the value of i ii

.

c Hence find a cubic equation with integer coefficients and roots , 19 Point has coordinates

. Triangle

has

and

and angle

. .

a Write down the complex number corresponding to the point . b Find the two possible pairs of coordinates of . 20 Let and be points on an Argand diagram representing complex numbers and , respectively. The complex number represents the point obtained by translating using the vector and then rotating the image through angle anticlockwise about the origin. The complex number corresponds to the point obtained by first rotating anticlockwise through angle about the origin and then translating by vector

.

Show that the distance between the points represented by and is independent of .

21 i

Show that

.

ii Write down the seven roots of the equation in an Argand diagram. iii Hence express factors.

in the form

and show their positions

as the product of one real linear factor and three real quadratic

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2007 22 Point represents the complex number

on an Argand diagram. Point is rotated

through radians anticlockwise about the origin to point . Point is then translated by to obtain point . a Find, in Cartesian form, the complex number corresponding to . b Find the distance

.

23 a Points and on an Argand diagram correspond to complex numbers . Show that .

and

b The diagram shows a triangle with one vertex at the origin, one vertex at the point and one vertex at the point such that

i

and

.

Write down the complex number corresponding to point .

ii Write down the complex number corresponding to point in modulus–argument form. iii Write down an expression for the length of iv Hence prove the cosine rule for the triangle

in terms of , and . : .

3 Complex numbers and trigonometry In this chapter you will learn how to: use de Moivre’s theorem to derive trigonometric identities find sums of some trigonometric series.

Before you start… Chapter 2, Section 1

You should be able to use de Moivre’s theorem to raise a complex number to a power.

Chapter 2, Section 2

1 Find in modulus–argument form.

You should be able to use the exponential form of a complex number.

2 a Write in exact Cartesian form.

A Level Mathematics Student Book 1, Chapter 9

You should be able to use the binomial expansion for positive integer powers.

3 Expand and simplify .

Chapter 2, Section 2, Pure Core Student Book 1, Chapter 4

You should know how to divide complex numbers.

4 Find the real and imaginary

A Level Mathematics Student Book 2, Chapter 4

You should be able to use the formulae for the sum of a geometric series.

5 a Find an expression for the sum of the first terms of the geometric series

b Write down the complex conjugate of .

part of

.

b For which values of does the series in part a have a sum to infinity?

Using complex numbers to derive trigonometric identities The modulus–argument form of a complex number provides a link between complex numbers and trigonometry. This is a powerful tool for deriving new trigonometric identities. These trigonometric identities are one example of the use of complex numbers to establish facts about real numbers and functions. Other such applications include a formula for cubic equations, calculations involving alternating current, and analysing the motion of waves. The fact that complex numbers proved correct results in a concise way was a major factor in convincing mathematicians that they should be accepted.

Section 1: Deriving multiple angle formulae You can raise a complex number to a power in two different ways. You can either use the Cartesian form and multiply out the brackets, or you can write the complex number in modulus–argument form and use de Moivre’s theorem. Equating these two answers allows you to derive formulae for trigonometric ratios of multiple angles.

Rewind You have already met double angle formulae, such as

, in A Level

Mathematics Student Book 2, Chapter 8.

WORKED EXAMPLE 3.1 Derive a formula for Let

in terms of

.

Then

. Start with an expression for a complex number involving , and find in two different ways.

.

First using the binomial theorem:

. Now using de Moivre’s theorem: Equating real parts:

The two expressions for must have equal real parts and equal imaginary parts. You want the answer in terms of .

only, so use

Simplify the final expression.

Fast forward By equating imaginary parts of the two expressions in Worked example 3.1, you can obtain a similar expression for (see question 1 in Exercise 3A).

Did you know? These expressions for sines and cosines of multiple angles can also be derived through repeated application of compound angle identities. However, the calculations become increasingly long.

EXERCISE 3A

EXERCISE 3A 1

a Find the imaginary part of b Hence show that

.

2

Use the binomial expansion to find the real and imaginary parts of expression for in terms of .

3

a Expand

.

b Hence or otherwise express 4

in terms of

a Show that

. .

, find the possible values of

a Find the real and imaginary parts of b Hence express

7

for

a Find the values of , and such that b Given that

6

.

.

b Hence solve the equation 5

. Hence find an

in terms of

.

. .

a Show that

.

b Hence solve the equation: for 8

.

a Use the binomial expansion to find the real and imaginary parts of b Hence show that

.

.

c Assuming that is small enough that the terms in and higher can be ignored, find an approximate expression, in increasing powers of , for

.

Section 2: Application to polynomial equations In Section 1 you learnt how to express and as a polynomial in or . For example, . You can now use the roots of the polynomial and the solutions of to find the values of

.

WORKED EXAMPLE 3.2 a Find all the values of

for which

You are given that

.

b Write down the roots of the equation c Hence find the exact value of

.

in the form

, where

.

.

a

b Write Then

.

Hence

Making the substitution relates the equations from parts a and b.

The equation from part a has eight solutions but the equation from part b should only have four (since it is a degree polynomial). This is because, for example, .

c

You can actually solve the equation from part b exactly, as it is a quadratic in .

is the smallest positive one of the four solutions from part b:

is one of these four solutions. You can see from the graph that it is the smallest positive one of the four numbers.

Sometimes you can’t solve the polynomial equation, but you can still use the results about sums and products of roots to derive expressions involving combinations of trigonometric ratios.

Rewind See Pure Core Student Book 1, Chapter 5, for a reminder about roots of polynomials.

WORKED EXAMPLE 3.3 a Show that

.

b Show that the equation

can be written as

where

, and state

the value of . c Hence find the exact value of

a Write

and

.

.

Use de Moivre’s theorem.

Then

Hence

Separate real and imaginary parts to find

and

Divide top and bottom by and use where

.

.

.

b

Rearrange the equation into the form from part a.

(so

)

c

Solve the cubic equation by solving

or

Hence

Although there are infinitely many values of , they only give three different values of (since is a periodic function). Use the result about the sum of the roots of a cubic polynomial:

EXERCISE 3B

.

EXERCISE 3B 1

a Write down an expression for b Given that

in terms of

, find a quadratic equation in , where

c Hence find the exact value of 2

a Given that

, where

for

.

.

c Hence find the exact value of

.

You are given that

.

a Find the possible values of

for which

b Hence show that 4

.

.

, show that

b Solve the equation

3

.

.

.

a Show that

.

b Given that

and that

c Hence show that

, find the possible values of .

is a root of the equation

and find, in a similar form, the

other two roots. 5

You are given that a Show that b Let

6

. .

. Express

and

in terms of . Hence show that

You are given that

.

a Show that the equation b Hence find the exact value of

.

has roots .

.

Section 3: Powers of trigonometric functions Another important link with trigonometry comes from considering the exponential form of complex numbers:

and

By adding and subtracting these two equations you can establish two very useful identities.

Key point 3.1

You can further generalise this result.

Key point 3.2 If

, then

PROOF 2 Remember that

Using de Moivre’s theorem for positive and

and .

negative integers:

Adding the two equations:

Subtracting the two equations:

You can use these results to derive another class of trigonometric identities, expressing powers of trigonometric functions in terms of functions of multiple angles. For example: WORKED EXAMPLE 3.4 Show that Let

. .

Using the binomial expansion:

.

Simplify the fractions, taking care with negative signs. Group the terms to get expressions of the form . So

On both sides of the equation, use the result from Key point 3.2:

Trigonometric identities such as these are very useful when integrating powers of trigonometric functions.

Rewind In A Level Mathematics Student Book 2, Chapter 11, you used the identity to find

.

WORKED EXAMPLE 3.5 a Expand and simplify

.

b Show that

.

c Hence find

.

a Using the binomial expansion:

b Let

.

Group the terms on the right so that you can use the result from Key point 3.2. Divide by

c Using the result from part a:

EXERCISE 3C

.

Don’t forget to divide by the coefficient of .

EXERCISE 3C 1

Let a

. Express each of these as a sum of terms of the form

or

.

i ii

b

i ii

2

Let

.

a Show that

.

b Hence show that 3

where , and are constants to be found.

a Use the expansion of

, where

, to show that .

b Hence find the exact value of 4

.

A complex number is defined by a

i Show that

. .

ii Use de Moivre’s theorem to deduce that b

i Expand

.

.

ii Hence find integers , and such that . c Find 5

.

Let

.

a Show that b Expand

. and

.

c Hence show that 6

a Write down expressions for b Hence evaluate

. and

in terms of

.

, clearly showing your working.

Section 4: Trigonometric series In Section 1 you learnt about expressions for sine and cosine of multiple angles. What happens if you add several such expressions together? For example, is it possible to simplify a sum such as ?

Explore Sums like these come up when combining waves (interference). They are also used in Fourier series, which is a way of writing other functions in terms of sines and cosines. You can simplify certain sums of this type using the exponential form of complex numbers and the formula for the sum of geometric series. This is because is the imaginary part of , and the numbers form a geometric series.

Rewind You met geometric series in A Level Mathematics Student Book 2, Chapter 4.

WORKED EXAMPLE 3.6 a Find an expression for

.

b Hence show that

a Geometric series with

.

.

This is a geometric series with common ratio Use

b

.

for the sum of the first terms.

This is the imaginary part of the series from part a, with .

Multiply top and bottom by the complex conjugate of the denominator in order to separate real and imaginary parts.

Use The imaginary part is:

in the denominator.

Now the denominator is real, so you just need to take the imaginary part of the numerator. Use the double angle formula in the denominator: .

If the modulus of the common ratio is smaller than , a geometric series also has a sum to infinity. WORKED EXAMPLE 3.7 a Show that the geometric series

converges, and find an expression for its sum to

infinity. b Hence evaluate

.

a The geometric series has

The common ratio is

.

, hence it converges. Using

:

b

The required sum is the real part of the sum from part a.

Multiply top and bottom by the complex conjugate of the denominator to separate real and imaginary parts. Use

.

Now take the real part of the numerator, using . Another series you know how to sum is the binomial expansion. WORKED EXAMPLE 3.8 By considering the expansion of

, or otherwise, show that

Using the binomial expansion:

The required series is the imaginary part of this. Use the double angle formulae.

Hence

so

Therefore, the sum of the series equals

.

EXERCISE 3D 1

a Find an expression for the sum to infinity of the geometric series b Hence evaluate

2

.

.

a Show that the geometric series

converges and find an expression for its sum

to infinity. b Hence show that

.

3

Use the geometric series

4

Use the expansion of

5

By considering

to evaluate

.

to show that

.

or otherwise, show that .

6

a Find an expression for the sum of the series

.

b Hence prove that

.

c Find all the solutions to the equation

for

.

Checklist of learning and understanding By expanding

and comparing the real and imaginary parts to

you can derive expressions for

and

in terms of powers of

Considering these expressions as polynomials in values of trigonometric functions. If

, then

or

and

In particular,

and

and

.

you can find some exact

. .

You can use these expressions, together with the binomial expansion, to express powers of

and

in terms of

and

of multiples of .

By considering real and imaginary parts of geometric or binomial series involving derive expressions for sums of trigonometric series.

you can

Mixed practice 3 1

a Expand and simplify

.

b Hence find constants and such that 2

.

Use de Moivre’s theorem to show that

. Hence find the

largest and smallest values of 3

a By considering

.

, where

, find the values of constants , and

such that

.

b Hence find the exact value of 4

.

Show that

. Hence show that

is a root of the equation

. 5

By considering the expansion of

, show that .

6 7

Show that

.

i By expressing

in terms of

and

, show that

. ii Hence solve the equation

for

.

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2008 8

Let

.

a Show that

.

b Show that

.

c Consider the equation i

Show that the equation can be written as

.

ii Find all four complex roots of the original equation. 9

a By considering

, find expressions for

b Show that c Hence show that d Show that equation e By considering

and

.

. is a root of the equation is a factor of . , explain why

.

and hence find the exact solutions of the

.

f

Hence state the exact value of

.

10 i Use de Moivre’s theorem to prove that . ii Hence find the largest positive root of the equation , giving your answer in trigonometrical form. © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2007 11 Convergent infinite series and are defined by

i Show that ii Hence show that

. , and find a similar expression for .

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2010

4 Lines and planes in space In this chapter you will learn how to: find the equation of a plane in several different forms find intersections between lines and planes calculate angles between lines and planes calculate the distances between objects in three-dimensional space.

Before you start… Pure Core Student Book 1, Chapter 2

You should be able to find the vector and Cartesian equation of a line in three dimensions.

1 A line passes through the points and . a Find a vector equation of the line. b Write down a Cartesian equation of the line.

Pure Core Student Book 1, Chapter 2

You should be able to find the point of intersection of two lines.

2 Find the point of intersection of the line from question 1 and the line .

Pure Core Student Book 1, Chapter 2

You should know how to calculate the scalar product of two vectors and use it to calculate an angle between two lines.

3 Find the acute angle between the two lines from question 2.

Pure Core Student Book 1, Chapter 2

You should know how to calculate the vector product of two vectors and use it to find a vector perpendicular to two given vectors.

4 Find a vector perpendicular to both of the lines from question 2.

Introduction In Pure Core Student Book 1, Chapter 2, you learnt about equations of lines in three dimensions, and how to find intersections and angles between lines. You know that two lines might be skew (not intersecting but not parallel). In this chapter you will learn how to find the distance between two skew lines, as well as between two parallel lines.

You will also learn how to describe planes (flat surfaces) in three-dimensional space.

Section 1: Equation of a plane You are already used to describing positions of points in the and -axes: for example, the position vector of the point

plane using unit vectors parallel to the is .

However, you can also use two directions other than those of and . In the second diagram, the same point is reached from the origin by moving units in the direction of vector and units in the direction of vector . Hence its position vector is . In the same way, every point in the plane has a position vector of the form , where and are scalars.

Consider now a plane that does not pass through the origin. To reach a point in the plane starting from the origin, you can go to some other point in the plane first, and then move along two directions which lie in the plane, as shown.

Tip A plane is a flat surface in three-dimensional space. It extends indefinitely in all directions.

Key point 4.1 The vector equation of the plane containing a point with position vector and parallel to the directions of vectors and is .

WORKED EXAMPLE 4.1 Find a vector equation of the plane containing points

,

and

.

You need one point and two vectors parallel to the plane. Draw a diagram to see which vectors to use.

You can choose any of the three given points to find , as they all lie in the plane. Vectors

and

Use

are parallel to the plane.

.

In Worked example 4.1, the plane was determined by three points. Two points do not determine a plane: there is more than one plane containing the line determined by points and as shown in this diagram. You can pick out one of these planes by requiring that it also passes through a third point, point for example, which is not on the line , as illustrated here. This suggests that a plane can also be determined by a line and a point outside of that line.

  WORKED EXAMPLE 4.2

Find a vector equation of the plane containing the line

and the point

.

Point lies in the plane. The direction vector of the line is parallel to the plane.

You need another vector parallel to the plane. You can use any vector between two points in the plane. One point in the plane is For the second point, you can pick any point on the line, for example

.



Now use

.

You can also determine a plane by two intersecting lines whose two direction vectors are parallel to the plane.

WORKED EXAMPLE 4.3

Find a vector equation of the plane containing the lines

and

.

You can tell that the two lines intersect at the point , so you can take that as one point in the plane. The two lines’ direction vectors give two different directions in the plane. If you have four points, they don’t all necessarily lie in the same plane. WORKED EXAMPLE 4.4 Determine whether points Plane containing

and

lie in the same plane.

: The plan is to find the equation of the plane containing points , and (as in Worked example 4.1) and then check whether the point lies in that plane.

:

For to lie in the plane, you need values of and that make equal to the position vector of .

You can solve the first two equations, and then check whether the solutions satisfy the third equation.

does not lie in the same plane as , and .

There are no values of and that satisfy all three equations.

You can now summarise all possible ways to determine a plane.

Key point 4.2 A plane is uniquely determined by: three points, not on the same line, or a line and a point outside that line, or two intersecting lines.

Cartesian equation of a plane The vector equation of the plane can be a little difficult to work with, as it contains two parameters. It is also difficult to see whether two equations represent the same plane, because the two vectors parallel to the plane are not unique. Now you will look at the question: is there a way to describe the ‘direction’ of the plane using just one direction vector? The diagram shows a plane and a vector perpendicular to it. This vector is perpendicular to every line in the plane, and it is called the normal vector of the plane.

Suppose is a fixed point in the plane and let be any other point in the plane. The normal vector is perpendicular to the line equation of the plane.

, so

. This means that

, which gives another form of an

Key point 4.3 The scalar product equation of the plane is: where is the normal to the plane and is the position vector of a point in the plane. Remember that the position vector of a point is related to its coordinates. This means that you can use the scalar product equation to write the Cartesian equation of a plane.

Write

and

scalar product

The scalar product

is a constant, denoted in Key point 4.4, and the

can be expanded to get an expression in terms of , and .

Key point 4.4 The Cartesian equation of a plane can be written in the form

.

WORKED EXAMPLE 4.5

Vector

is perpendicular to the plane which contains point

a Write an equation for in the form

.

.

b Find the Cartesian equation of the plane.

The equation of the plane is

a

.

The Cartesian equation involves and (the coordinates of ), which are the components of the position vector .

b

Tip The letter (capital ) is often used as the name for a plane. You can convert from a vector to a Cartesian equation of the plane. This involves using the vector product to find the normal. The Cartesian equation is very convenient for checking whether a point lies in the plane: you just need to check that the coordinates of the point satisfy the equation.

Rewind You met the vector product in Pure Core Student Book 1, Chapter 2.

WORKED EXAMPLE 4.6 a Find the Cartesian equation of the plane with vector equation b Show that the point a

.

lies in the plane.

To find the Cartesian equation you need the normal vector and one point. Point

lies in the plane.

is perpendicular to all lines in the plane, so it is perpendicular to the direction vectors

and

.

The vector product of two vectors is perpendicular to both of them.

To get the Cartesian equation, write as

b Hence the point lies in the plane.

.

A point lies in the plane if its coordinates satisfy the Cartesian equation.

You can also convert from a Cartesian to a vector equation by finding two vectors that are perpendicular to the normal. WORKED EXAMPLE 4.7 Find a vector equation of the plane with Cartesian equation

.

Vector equation:

In the vector equation of the plane, is the position vector of one point in the plane and and are two direction vectors parallel to the plane.

Finding :

The coordinates of satisfy the Cartesian equation of the plane. You can choose any three numbers that satisfy this equation. For example, you can set find .

Hence a possible position vector is

and then



. Finding

and

Write

:

The two direction vectors parallel to the plane must be

. Then

perpendicular to the normal, which is

.

so: . When

and

:

As before, you can choose values for and and then find . To make the calculation simple, take

Write

and

(Notice that, in this case, you can’t set would make as well.)

because that

Repeat for

.

, but this time take

and

. Then:

Hence the two direction vectors are and

A possible vector equation of the plane is

.

Put all this together into the vector equation,

You should remember that the vector equation you found in Worked example 4.7 is not unique. You could have chosen any other point that satisfies the Cartesian equation, and there are infinitely many choices for pairs of direction vectors that are parallel to the normal. EXERCISE 4A 1

Write down the vector equation of the plane parallel to vectors and and containing point . a

i

ii b

i ii

2

Find a vector equation of the plane containing points a

and .

i ii

b

i ii

3

Find a vector equation of the plane containing line and point . a

i

ii

b

i

ii

4

A plane has normal vector and contains point . Find the equation of the plane in the form , and the Cartesian equation of the plane. a

i

ii

b

i

ii

5

Find a normal vector to the plane given by the vector equation: a

i

ii

b

i

ii

6

Find the equations of the planes from question 5 in the form

7

Find the Cartesian equations of the planes from question 5.

8

Find the Cartesian equation of the plane containing points a

.

and .

i ii

b

i ii

9

In each of the following show that point lies in the given plane . a

b c 10 A plane contains the point

. The vector

is perpendicular to the plane. Find the

Cartesian equation of the plane. 11 A plane contains points

and

.

Find the vector equation of the plane in the form 12 A plane contains points a Find

.

and

.

.

b Hence find the Cartesian equation of the plane. 13 A plane is determined by the points

and

a Find the equation of the plane in the form b Determine whether the point 14 a Calculate

.

.

lies in the same plane.

.

b Hence find the Cartesian equation of the plane with vector equation

.

15 a Calculate

.

b Two lines have equations and i Show that and intersect. ii Find the coordinates of the point of intersection. c Plane contains lines and . Find the Cartesian equation of . 16 Determine whether the points

and

17 A plane has Cartesian equation

lie in the same plane.

.

a Write down the normal vector, , of the plane. b Find the values of and such that the vectors

and

c Hence find a vector equation of the plane in the form

are perpendicular to . .

18 a Find the Cartesian equation of the plane with vector equation b Another plane has Cartesian equation form . 19 Find, in the form

.

. Find a vector equation of this plane in the

, a vector equation of the plane

.

Section 2: Intersection between a line and a plane The method introduced in this section requires equations of planes to be in Cartesian form and equations of lines to be in vector form. If in a question they are given in a different form, you will need to convert them first. The coordinates of the intersection point (if there is one) must satisfy both the equation of the line and the equation of the plane.

WORKED EXAMPLE 4.8 Find the intersection between the given line and plane, or show that they do not intersect.

a

and

b

and

a

Any point on the line has coordinates . The intersection point must also satisfy the equation of the plane, so substitute into the equation of the plane. Now use this value of to find the coordinates. The intersection point is .

b



Change the equation of the line into the vector form and then follow the same procedure as in part a.

Substitute the coordinates of a point on the line into the equation of the plane. Impossible to find . The line and plane do not intersect.

It is impossible to find a value of for a point that satisfies both equations. This means that the line and the plane have no common points.

Notice that if a line and a plane do not intersect, this means that they are parallel. For example, in part b of Worked example 4.8, the direction vector of the line is perpendicular to the normal of the plane:

; this shows that the line is parallel to the plane. By contrast, in part a, the direction of the line and the normal of the plane are not perpendicular:

.

It is possible for a line to lie entirely in a given plane. WORKED EXAMPLE 4.9

Show that the line

lies in the plane

.

A point on the line has coordinates . You need to show that every such point satisfies the equation of the plane. Every is a solution.

The equation is satisfied for all values of . This means that every point

So, the line lies in the plane.

on the line also lies in the plane.

EXERCISE 4B

EXERCISE 4B 1

Find the coordinates of the point of intersection of line and plane . a

i

,

ii

b

,

i

,

ii

2

,

Show that plane contains line . a

,

b

, 

c

,

d

,

3

Find the point of intersection of the line

4

A line has equation

and the plane

.

.

a Write down the direction vector of the line. b Find the coordinates of the point where the line intersects the plane with equation

5

Find the coordinates of the point of intersection of the line

with the plane

. 6

The plane with equation respectively. a Find the coordinates of

intersects the -, -, and -axes at points , and , and .

b Find the area of the triangle

.

.

Section 3: Angles between lines and planes The angle between a line and a plane is the smallest possible angle that makes with any of the lines in . In the diagram, this is the angle labelled between and the line , where is perpendicular to the plane (so that is in the direction of the normal). Drawing a two-dimensional diagram of triangle makes the angles clearer.

Key point 4.5 The angle between the line with direction vector and the plane with normal is

, where

is the acute angle between and .

Rewind You found the angle between two vectors using the scalar product in Pure Core Student Book 1, Chapter 2.

WORKED EXAMPLE 4.10

Find the angle between the line with equation

and the plane with equation

. Draw a diagram, labelling vectors and angles as in Key point 4.5.

The angle between the line and the plane is

The angle between a line and a plane is (angle between the line’s direction vector and the plane’s normal).

You can use a similar method to find the angle between two planes. Again, a diagram is helpful so you can see where the relevant angle is. The sum of angles in a quadrilateral is are equal.

, so the two angles marked

Key point 4.6 The angle between two planes is equal to the angle between their normals.

WORKED EXAMPLE 4.11 Find the acute angle between the planes with equations

and

.

You need to find the angle between the normals. The components of the normal vector are the coefficients in the Cartesian equation.

You need the acute angle. The angle between the planes is

.

EXERCISE 4C 1

Find the acute angle between line and plane , correct to the nearest a



i

ii

b

i

,

ii 2

Find the acute angle between each pair of planes. a b

and and

.

3

Line has Cartesian equation

.

a Write down the direction vector of . b Find the angle between and the plane with equation 4

Plane

has Cartesian equation

a Write down a normal vector of Plane

.

. .

has equation

.

b Find, correct to the nearest degree, the acute angle between 5

Line has equation

and

.

.

a Write down the direction vector of . b Find the acute angle that makes with the plane

6

A plane has vector equation

.

.

a Find the normal vector of the plane. b Find the angle that the plane makes with the line

7

Show that the planes with equations

.

and

are perpendicular to each

other. 8

Line has Cartesian equation

.

a Find the direction vector of . b Find the acute angle that makes with the plane 9

.

Plane has equation a Show that point

lies in the plane .

b Point has coordinates c Find the exact distance

. Find the exact value of the sine of the angle between .

d Hence find the exact distance of from .

Fast forward In Section 4 you will meet a formula for the distance from a point to a plane.

and .

Section 4: Distances between points, lines and planes Distance between a point and a plane Given a plane with equation and a point outside of the plane, the shortest distance from to the plane is equal to the distance , where the line is perpendicular to the plane. This means that the direction of is . (Point is called the foot of the perpendicular from the point to the plane.)

To find the distance

:

write down the vector equation of the line with direction through point find the intersection, , between the line and the plane calculate the distance . This procedure leads to the formula in Key point 4.7, which you can use unless the question explicitly asks you to carry out the three steps listed here (see questions 14 and 15 in Exercise 4D).

Key point 4.7 The shortest distance between the point with position vector and the plane with equation is given by

This will be given in your formula book.

WORKED EXAMPLE 4.12 Plane has Cartesian equation . has equation

with

. Find the shortest distance between and the point

To use the formula you need to identify the normal vector.

.

and The distance is

Now use the formula from Key point 4.7.

Distance between a point and a line Given a line with direction vector and a point not on the line, the shortest distance between the line is the distance from to the point on the line such that is perpendicular to .

and

Rewind In Pure Core Student Book 1, Focus on … Problem solving 1, you explored various methods for finding the shortest distance between a point and a line in three dimensions. For a reminder, see question 16 in Exercise 4D. In this course you only need to find this distance in two dimensions. In that case the equation of the line can be written as and you can use the formula in Key point 4.8.

Key point 4.8 The shortest distance between the point with coordinates

and the line with equation

is given by

This will be given in your formula book.

WORKED EXAMPLE 4.13 Line passes through the points

and

. Find the shortest distance between and the point

. Gradient of :

To use the formula you need to write the equation of in the form . First you need the gradient.

Equation of : Use

.

Now use the formula from Key point 4.8.

Distance between two skew lines Consider points possible when

and moving along two skew lines. The distance between them is the minimum is perpendicular to both lines. It may not be immediately obvious that such a position

of and always exists, but it does. In sketching a diagram of this situation, it is useful to envisage a cuboid, where one line runs along an upper edge, and the other runs along the diagonal of the base, as shown in the diagram. The shortest distance between the two lines is then the height of the cuboid.

You can find the position of the points and by using the fact that is perpendicular to both lines’ direction vector (see question 18 in Exercise 4D). However, in most questions you can use the following formula.

Key point 4.9 The shortest distance between two skew lines with equations given by

and

is

where This will be given in your formula book. You may need to use the equations of the lines to identify and and then calculate . WORKED EXAMPLE 4.14 Find the shortest distance between the lines with equations

and

. The direction vectors are:

The components of the direction vectors are given by the denominators in the Cartesian equation when written in the form

.

For the first line, the last numerator is rather than so the corresponding component of the direction vector is

, .

Use the vector product to calculate .

Now use the formula from Key point 4.9 with

and

.

Distance between two parallel lines Consider two parallel lines, both with a direction vector . You can measure the distance between them from any point on the first line: it is the distance , where is the point on the second line such that is perpendicular to . As you can see from the diagram, you can actually find the distance without finding the position vector of

. It equals

, where is any point on the second line and is the angle between

and .

Key point 4.10 The distance between parallel lines with equations

and

is given by

where

The formula in Key point 4.10 will not be given in the formula book, so it is a good idea to draw a diagram to make sure you get it right.

Fast forward You can also find the distance from point to the second line by using the fact that perpendicular to the direction vector. See question 17 in Exercise 4D for an example.

is

WORKED EXAMPLE 4.15 Show that the lines with equations

and

are parallel

and find the distance between them. The first line has direction vector

You need to identify the direction vectors of the two lines.

Second line:

For the second line you need to write each term in the form, .

has the direction vector

The two lines have parallel

The second direction vector is a multiple of the first direction

direction vectors so they are parallel.

vector. Draw a diagram to identify the required distance. You can see from the equation that the point lies on the first line and the point lies on the second line.

The distance is .



where is the angle between

EXERCISE 4D 1

Find the distance from the point a

to the plane .

i

ii b

i ii

2

Find the distance between the point a

and the line .

i ii

b

i ii

3

Find the distance between the lines and . a

b

i



ii



i



and



:

and



ii 4

and :

and



Show that the lines and are parallel and find the distance between them.

and

a

b

i

:

and

ii

:

and :

i

ii





and

:

and :



5

Find the shortest distance between the lines with equations .

6

Find the shortest distance between the point

7

Find the exact distance of the plane

8

Find the perpendicular distance between the point with coordinates through the points and .

9

Show that the lines with equations

and

and the plane with equation

.

from the origin.

and

and the line passing

do not intersect and

find the shortest distance between them. 10 Points the side

,

and

form a triangle. Find the height of the triangle corresponding to

and hence find the exact area of the triangle.

11 Show that the lines

and

are parallel and find the

distance between them. 12 Two planes have equations: , and

.

a Show that

and

are parallel.

b Show that

passes through the origin.

c Hence find the distance between the planes 13 a Show that the planes b Given that the point

and lies in

and

. are parallel.

find the value of .

c Use your answer from part b to find the exact distance between the two planes. 14 Plane has equation .

. Line is perpendicular to and passes through the point

a Find the equation of . b Find the coordinates of the point where intersects . c Find the shortest distance from to . 15 Plane has equation

. Line is perpendicular to and passes through the origin.

a Find the coordinates of the point of intersection of and . b Find the shortest distance of from the origin, giving your answer in exact form.

16 Line has equation

and point has coordinates

. Point lies on and

is perpendicular to . Find the coordinates of . Hence find the shortest distance of from . 17 Two lines are given by Cartesian equations:

a Show that and are parallel. b Show that the point

lies on .

c Find the coordinates of the point on such that

is perpendicular to the two lines.

d Hence find the distance between and , giving your answer to significant figures. 18

Two lines have vector equations

.

The point on and the point on are such that a Show that

is perpendicular to both lines.

.

b Find a second equation linking and . c Hence find the shortest distance between and , giving your answer as an exact value.

Checklist of learning and understanding The vector equation of a plane is

, where

and

are two vectors parallel

to the plane and is the position vector of one point in the plane. The Cartesian equation of a plane has the form in the scalar product form

, where

. This can also be written is the normal vector of the plane,

which is perpendicular to every line in the plane. To derive the Cartesian equation from a vector equation, use

.

The angle between two planes is the angle between their normals. The angle between a line and a plane is plane’s normal).

(angle between line’s direction vector and

To find the intersection between a line and a plane, express and substitute into the Cartesian equation of the plane.

for the line in terms of

The shortest distance between the point with position vector and the plane with equation is given by

.

The shortest distance between the point with coordinates is given by

.

The shortest distance between two skew lines with equations given by

and the line with equation

where

and

is

.

To find the distance between two parallel lines, find the perpendicular distance from a point

on one line to the other line. (A diagram will help you see how to find this distance).

Mixed practice 4 1

Find the exact perpendicular distance from the point

to the line with equation

. 2

Find the shortest distance of the point

3

The vector

from the plane

.

is normal to a plane which contains the point

.

a Find an equation for the plane. b Find if the point 4

lies on the plane.

a Calculate

b Plane

.

has normal vector

and contains point

. Find the Cartesian

equation of the plane. c Plane

has equation

d A third plane,

. Show that

, has equation

contains point .

. Find the angle between

and

in

degrees. 5

The plane passes through the points with coordinates i

Find a vector equation of in the form

,

and

.

.

ii Find a cartesian equation of . © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2013 6

Two skew lines have equations and i

.

Find the direction of the common perpendicular to the lines.

ii Find the shortest distance between the lines. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2009 7

Line has equation

a Find

and line has equation

.

.

b Find the coordinates of the point of intersection of the two lines. c Write down a vector perpendicular to the plane containing the two lines. d Hence find the Cartesian equation of the plane containing the two lines. 8

The plane has equation

and the point has coordinates

.

a Write down the vector equation of the line through which is perpendicular to . b Find the coordinates of the point of intersection of line and plane . c Hence find the shortest distance from point to plane . 9

Plane has equation a Show that point

and point has coordinates

.

lies in the plane .

b Find the vector equation of the line

.

c Write down the vector equation of the line through perpendicular to . d

is the foot of the perpendicular from to . Find the coordinates of .

e Find the exact distance of point from the plane . 10 Find the shortest distance between the skew lines with equations and 11 Consider the four points a Given that

.

,

,

and

.

is a parallelogram find the values of and .

b Find the distance between the lines c Hence find the area of

and

.

.

12 a Find the coordinates of the point of intersection of lines and

.

b Find a vector perpendicular to both lines. c Hence find the Cartesian equation of the plane containing and . 13 Point

lies on line which is perpendicular to plane

.

a Find the Cartesian equation of . b Find the point of intersection of the line and the plane . c Point is reflected in . Find the coordinates of the image of . d Point has coordinates

. Show that lies in .

e Find the distance between and . 14 A line has equation i

and a plane has equation

.

Find the point of intersection of and .

ii Find the equation of the plane which contains and is perpendicular to , giving your answer in the form . © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2009 15 A tetrahedron is such that the points , and are the plane i

is

is perpendicular to the base . The coordinates of , and respectively, and the equation of .

Find, in either order, the coordinates of and the length of

.

ii Find the acute angle between the planes

and

.

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2008 16

Line passes through point coordinates

and has direction vector

. Point has

.

Plane has normal vector , and contains the line and the point . a Write down a vector equation for . b Explain why

and are both perpendicular to .

c Hence find one possible vector . d Find the Cartesian equation of plane . 17 The plane

contains the line

. Find .

18 Find the Cartesian equation of the plane containing the lines

19 Two planes have equations

and

a Calculate

b Show that

.

.

and

are perpendicular.

c Show that the point

does not lie in either of the two planes.

d Find a vector equation of the line through

which is parallel to both planes.

20 Four points have coordinates a Show that

.

is perpendicular to both

and

.

b Write down the equation of the plane containing the points , and in the form . c Find the exact distance of point from plane . d Point 21 Points a Find

is the reflection of in . Find the coordinates of ,

and

.

lie in the plane .

.

b Find the area of the triangle

, correct to significant figures.

c Find the Cartesian equation of . Point has coordinates

.

d Find a vector equation of the line through perpendicular to the plane. e Find the intersection of this line with , and hence find the perpendicular distance of from .

f

Find the volume of the pyramid

.

22 A regular tetrahedron has vertices at the points . i

Obtain the equation of the face

in the form

(Answers which only verify the given equation will not receive full credit.) ii Give a geometrical reason why the equation of the face

can be expressed as

iii Hence find the cosine of the angle between two faces of the tetrahedron. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2010 23 With respect to the origin , the position vectors of the points , and are , and respectively. The mid points of the sides , and of the triangle are , and respectively. i

Show that

.

ii Verify that the point with position vector lines

,

and

, and deduce that the

intersect at .

iii Write down, in the form perpendicular to the plane iv It is now given that

lies on

, an equation of the line through which is . (It is not necessary to simplify the expression for .) and

. Find the perpendicular distance

from to the plane . © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012

5 Simultaneous equations and planes In this chapter you will learn how to: identify different geometrical configurations of two or three planes determine whether a set of simultaneous equations has a unique solution, no solutions or infinitely many solutions use simultaneous equations to determine the geometrical configuration of three planes.

Before you start… Pure Core Student Book 1, Chapter 1

Pure Core Student Book 1, Chapter 3

You should be able to find the determinant of a and a matrix and understand what is meant by a singular matrix.

1 Find the value of for which the matrix

You should be able to use matrices to solve simultaneous equations in two and three unknowns.

2 Express simultaneous equations

is singular.

in the form

Find and hence find and . Pure Core Student Book 1, Chapter 3

You should be able to determine when simultaneous equations do not have a unique solution.

3 Find the value of for which the simultaneous equations

do not have a unique solution. GCSE

You should be able to use elimination to solve simultaneous equations.

4 Use elimination to solve the simultaneous equations

Chapter 4

You should know how to find the Cartesian equation of a plane,

5 a Find the Cartesian equation of the plane

and to identify the normal

with the normal vector

vector of a plane with a given equation.

which contains the point . b Write down the normal vector of the plane with equation .

In Chapter 4 you learnt about equations of planes in three dimensions. A Cartesian equation of a plane is a linear equation with three unknowns. In Pure Core Student Book 1, Chapter 3, you learnt how to solve three simultaneous equations with three unknowns by using an inverse matrix, and also how to tell when the solution is not unique. In this chapter you will learn how to distinguish between different situations with non-unique solutions. You will then use simultaneous equations to find the intersection of three planes, and to determine the geometric configuration of the planes in the case when the intersection is not a single point.

Focus on … Matrices have applications in many other situations. You can explore one of them in Focus on … Modelling 1.

Section 1: Linear simultaneous equations In Pure Core Student Book 1, Chapter 3, you learnt how to recast a system of two or three simultaneous equations into a matrix problem, and to use an inverse matrix to solve it. This method only works when the corresponding matrix is non-singular, and it leads to a unique solution of the system (a single pair of values or a single triple of values).

Tip A non-singular matrix has an inverse and its determinant is not zero. If the matrix is singular (has a zero determinant) you need to use a different method, such as elimination, to determine whether there are any solutions. The three possible cases for two equations with two unknowns can be illustrated by the examples in this table. Unique solution

Infinitely many solutions

Non-singular matrix

No solutions

Singular matrix

Unique solution:

Infinitely many solutions: any

No solutions

with

Consistent system

Inconsistent system

Key point 5.1 For a system of simultaneous equations in matrix form, If det

the equations have a unique solution.

If det there is no unique solution. Use elimination to distinguish between two cases: Consistent equations: there are infinitely many solutions. Inconsistent equations: there are no solutions.

WORKED EXAMPLE 5.1 Consider simultaneous equations

a Find the value of for which the equations do not have a unique solution. b For this value of , find the value of for which the equations are consistent.

The solution is not unique when the determinant is zero.

a

b

Using

, try to solve the equations by elimination; eliminate

. For solutions:

For the system to be consistent, the RHS of the last equation must be zero.

You can apply the same method to a system of three equations with three unknowns. WORKED EXAMPLE 5.2 For the system of simultaneous equations

where and are constants: a find the value of for which there is not a unique solution b for the value of found in part a, find the value of such that the equations are consistent. a

Recast the problem in matrix form as . The determinant tells you whether there is a unique solution.

No unique solution when det b Using elimination:

.

Use elimination to determine when the equations are consistent. Eliminate from two equations. Eliminate from the final equation.

For a consistent solution, , so

EXERCISE 5A

Equation says possible if

, which is only .

1

For each pair of simultaneous equations, determine whether there is a unique solution, no solutions or infinitely many solutions. a

i ii

b

i ii

c

i ii

2

For each set of simultaneous equations determine whether or not there is a unique solution. Where there is a unique solution, use the inverse matrix to find it. Where there is no unique solution, determine whether the system is consistent or inconsistent. a

i

ii

b

i

ii

c

i

ii

3

Show that the system of equations

does not have a unique solution. 4

a For what values of does the system of equations

have no unique solution? b For each value of found in part a, determine whether the equations have no solutions or infinitely many solutions. 5

a Show that the system of equations

has no unique solution for

.

b Find the other values of for which there is no unique solution.

6

c For

, find

and .

d For

, determine whether the equations are consistent or inconsistent.

a Show that, for all values of , the simultaneous equations

do not have a unique solution. b Given that the equations have infinitely many solutions, show that 7

Find the value of and the value of for which the system of equations

has infinitely many solutions.

.

Section 2: Intersections of planes Simultaneous equations in two variables describe lines. If there is a unique solution, it represents the intersection of the lines, and if there is no unique solution this arises because the lines are parallel (no solution) or identical (infinitely many solutions). Equations in three variables describe planes. Two distinct planes can either intersect (planes intersect in a line) or be parallel.

planes intersecting in a line

parallel planes

Key point 5.2 Two planes are parallel if their normal vectors are multiples of each other.

WORKED EXAMPLE 5.3 Five planes have these equations:

a Identify pairs of parallel planes. b Which of the five plane(s) are identical to the plane a

b

and parallel. and parallel.

are

Parallel planes have normals in the same direction. This means that the LHS of the equations are multiples of each other.

are

and are parallel to .

First identify the plane(s) with the normals in the same direction. Check whether the RHS is also the same.

is identical to

.

With three distinct planes there are several possibilities; altogether, there are five different arrangements, but these fit into three cases: 1

Consistent system: unique solution The three planes meet at a single point.

three distinct planes intersecting at a point 2

Consistent system: infinitely many solutions The three planes intersect in a line (the planes form a sheaf).

three distinct planes intersecting in a line 3

Inconsistent system: no solutions There is no point common to all three planes. a

All three planes are parallel.

three distinct parallel planes b

Two of the planes are parallel.

two parallel planes and one non-parallel plane c

The planes enclose a triangular prism, so that each pair of planes intersects in a line, with the three distinct lines running parallel to each other.

three planes forming a triangular prism You can distinguish between the different cases by using the methods from Section 1.

Key point 5.3 To determine the geometrical arrangement of three planes described by a set of simultaneous equations, use If det

the three planes meet at a single point.

If det then: If the equations are consistent the planes meet in a line (form a sheaf). If the equations are inconsistent then either some of the planes are parallel, or the three planes form a prism.

Tip Remember that two planes are parallel if their normals are multiples of each other. If whole equations are multiples of each other then they describe the same plane.

WORKED EXAMPLE 5.4

The simultaneous equations a Show that when

describe three planes.

the planes intersect at a single point and find its coordinates.

b Find the value of for which there is no unique solution, and describe the configuration of the three planes in that case.

Recast the problem in matrix form as . a Setting

: (from calculator)

Show that det to establish the existence of a unique solution.

The matrix is non-singular for and therefore there will be a unique solution. Find the solution by using the inverse matrix.

The intersection point is

. No unique solution when det

.

b

There is no unique solution when

.

Using elimination:

You now need to use elimination to determine whether the equations are consistent. Eliminate from two equations. Eliminate from the final equation. The equations are inconsistent, so the three

The resulting equation has no solutions.

planes do not intersect. No row of the matrix is a multiple of another, so there are no parallel planes.

Check whether any of the planes are parallel.

Therefore the three planes form a triangular prism. WORKED EXAMPLE 5.5 Planes

and

are given by the equations:

a Describe the geometric configuration of the three planes in the case when b In the case line.

.

find the value of and the value of for which the three planes intersect along a

a

so

and

It is a good idea to first check whether any of the planes are parallel, as that is easily seen from the equations.

are parallel.

is not parallel to them. There are two parallel planes, with a third plane intersecting them.

Describe the geometric configuration. A sketch might help you visualise it.

b

If the three planes intersect along a line then the equations don’t have any unique solutions. This means that the determinant is .

when

.

Using elimination:

You now need to use elimination to determine when the equations are consistent. Eliminate from two equations. Eliminate from the final equation.

Consistent equations, so

If the planes intersect along a line, the equations must be consistent.

EXERCISE 5B 1

For each set of simultaneous equations, determine whether they are consistent or inconsistent and interpret the geometrical configuration of the planes described by the equations. If there is a unique solution, find it. a

i

ii

b

i

ii

c

i

ii

d

i

ii

2

Consider a set of simultaneous equations a Show that the equations have a unique solution and find this solution. b The three equations represent planes. Describe the configuration of the three planes.

3

A system of equations is given by a Show that the system has a unique solution. b The three equations represent planes. Describe the configuration of the three planes.

4

Find the intersection of the planes

5

a Show that there is no unique solution to the simultaneous equations given by b Show that the equations are inconsistent. c Interpret this situation geometrically.

6

Describe the configuration of these three planes:

7

a Find the value of such that the planes b Now let

8

and

. Describe the configuration of the planes

and

are parallel. .

Consider this system of equations:

a Show that the system does not have a unique solution. b Find the value of for which the system is consistent. c The three equations represent planes. For the value of found in part b, describe the configuration of the three planes.

9

a Show that the system of equations

The three equations in part a represent three planes. b Describe the geometrical configuration of the planes.

is consistent.

10 a Find the inverse of the matrix

.

b Hence find, in terms of , the coordinates of the point of intersection of the planes and .

11 Consider the system of equations a Find the value of for which the system does not have a unique solution. b For the value of found in part a, determine whether the system is consistent, and describe the geometric configuration of the three planes represented by the equations. 12 a Find the value of for which the system of equations does not have a unique solution. b For the value of found in part a, find the two values of for which the system is consistent. c For the value of from part a and the values of from part b, describe the geometric configuration of the three planes represented by the three equations. 13 Find the values of and for which the intersection of the planes and

,

is a line.

Checklist of learning and understanding Three different planes could: intersect at a single point intersect along a line (form a sheaf) not intersect because the line of intersection of two of the planes is parallel to the third plane (form a triangular prism) not intersect because two of the planes are parallel or all three planes are parallel. Two planes are parallel if their normals are parallel. You can represent a system of three linear simultaneous equations in three unknowns in matrix form as where Each row of matrix

.

contains the coefficients of , and in the planes described.

If det planes.

there is a unique solution representing the point of intersection of the three

If det

then the three planes do not intersect at a single point and could be

inconsistent (no common intersection: parallel planes or a triangular prism) consistent (a line or a plane as the common intersection). When det If you get

, you can determine the geometrical interpretation using elimination. the three planes are identical or intersect along a line.

If you get then either at least two of the planes are parallel and distinct, or they form a triangular prism.

Mixed practice 5 1

The planes

and

are given by the equations:

Find the point of intersection of all three planes. 2

By using the determinant of an appropriate matrix, find the values of for which the simultaneous equations

do not have a unique solution. © OCR A Level Mathematics, Unit 4725 Further Pure Mathematics 1, June 2011 3

The planes

and

a Show that, when

are given by the equations:

, the three planes form a triangular prism.

b Find the value of for which the three planes intersect along a line. 4

Consider this system of equations, where and are real:

a Given that the system has no unique solution, find all possible values of . b When the system has a unique solution, find that solution in terms of when c Find such that the system is always consistent when 5

The matrix

is given by

i Find, in terms of , the determinant of ii Hence find the values of for which

.

.

.

. does not exist.

iii Determine whether the simultaneous equations

where is a non-zero constant, have a unique solution, no solution or an infinite number of solutions, justifying your answer.

© OCR A Level Mathematics, Unit 4725 Further Pure Mathematics 1, January 2011 6

The matrix is given by

.

i Find the determinant of in terms of . ii Three simultaneous equations are shown below.

For each of the following values of , determine whether or not there is a unique solution. If the solution is not unique, determine whether the equations are consistent or inconsistent. a b c © OCR A Level Mathematics, Unit 4725 Further Pure Mathematics 1, June 2012

FOCUS ON … PROOF 1

Extending the proof of de Moivre’s theorem De Moivre’s theorem states that

In Chapter 2, Section 1, you saw how to prove this result for integer values of . However, the result extends, with some careful consideration of conventions, to rational values.

Proving the result for rational numbers You have to be a little careful when raising a number to a rational power. For example, if you write as , then you can also write it as

. If you could apply de Moivre’s

theorem with a rational power , where and are integers with no common factors, then equal

would

.

Tip Notice that when the power is an integer (for example only one possible answer.

) then there is no problem: there is

This has different values, corresponding to . Multi-valued expressions are usually considered inconvenient, so you need to apply a convention that, when raising to a rational power, you choose to be the smallest positive value and . This is called the principal root.

Question 1

By considering

, prove that one possible value for and is

. You can assume de Moivre’s theorem for integer powers and that the normal rules for indices hold.

Not proving the result for irrational numbers It is tempting to think that if de Moivre’s theorem can be proved for all rational numbers, then it must hold for irrational numbers. However, this turns out to be difficult to define. For example: consider . You could write would have

. Then, if de Moivre’s theorem did extend to irrational numbers, you

. Unlike in the rational case there is no period to this expression. Each different value of therefore produces a different value for the expression. There are therefore infinitely many values (all lying on the circle with modulus ), which makes this expression very hard to work with.

Question 2

Use proof by contradiction to prove that if , then

.

You can assume that the period of the

function is

and that is an irrational number.

FOCUS ON … PROBLEM SOLVING 1

Using complex numbers to describe rotations You know two different ways to describe rotations in a plane. The first method is using matrices. You can find the image of a point with position vector

after a rotation through an angle about the origin by

multiplying it by the rotation matrix:

Rewind You met the rotation matrix in Pure Core Student Book 1, Chapter 3. The second method is using complex numbers. When a point corresponding to a complex number is rotated through an angle about the origin, you can find the complex number corresponding to the image point by multiplying by :

Rewind You met the idea of multiplication by

representing a rotation in Chapter 2, Section 6.

You can see that the two methods give the same coordinates for the image point. The advantage of the complex numbers method is that it results in a single equation, whereas the matrix method results in two equations (one for each component). Here you will compare the two methods when solving the following problem Three snails start at the vertices of an equilateral triangle. Each snail moves with the same constant speed towards an adjacent snail: towards , towards , and towards . Describe the path followed by each snail. After how long (if at all) do the snails meet?

First you need to specify the problem a little more precisely. Set up the coordinate axes so that the origin is at the centre of the equilateral triangle and let the initial position of each snail.

be at

. Let be the speed of

Because of the symmetry of the situation, the three snails will always form an equilateral triangle. You only need to find the path followed by . You can find the position of by rotating anticlockwise about the origin, and you can find the position of by rotating by the same amount. You will first approach the problem using position vectors and matrices.

Questions Let

be the position vector of

at time .

1

Use a rotation matrix to write down the position vector of

2

Explain why

3

Hence show that and satisfy the system of differential equations.

in terms of and .

for some constant . Show that

.

Although you will learn in Chapter 11 how to solve some systems of differential equations, these equations are non-linear so the methods from that chapter won’t work here. You will return to this system of equations later, but for now you will consider a different approach. Now, let be the complex number corresponding to the position of given by

.

at time . Then the position of

is

Questions 4

Show that

5

Write

(Remember that both and vary with time.) Show that: .

6

By equating real and imaginary parts, obtain this system of differential equations for and : .

7

Given that initially

8

The position of is then given by , by and by . At what time do the snails meet at the origin? What happens to the value of as approaches this value?

and

, show that

The curve described by the equations in question is called a logarithmic spiral. Although each snail travels a finite distance

, it performs an infinite number of rotations. This diagram shows

the paths of all three snails, and their positions when

and

.

In this problem, using complex numbers resulted in equations you could solve, while this was not the case when using position vectors and matrices. You should read the rest of this section after you have studied Chapter 9 on polar coordinates. In question 6 you derived two separate differential equations for and , the modulus and the argument of the complex number representing the position of . This suggests that you might also be able to solve this problem using polar coordinates, which are basically the same as the modulus and argument of a complex number.

Look again at the system of equations from question 3:

Let and be the polar coordinates of the point . Then the ‘problem’ term in the equations, is simply . You can rewrite the equations in terms of and .

Rewind Remember that

and

.

Questions 9

Show that

and obtain a similar expression for

10 Rewrite the system of equations in terms of and . Combine the two equations to show that .

Tip Notice that these are the same equations you derived in Question 6. 11 Obtain an expression for

and hence show that

The final equation in Question 11 is the polar equation of a logarithmic spiral, the path followed by each snail.

,

FOCUS ON … MODELLING 1

Leslie matrices Matrices are applied in many different real-life situations. Leslie matrices are a particular application to a biological population structured into different groups such as adults and juveniles. Imagine a group of rabbits. Each adult on average produces three juvenile rabbits each year. Each year of adult rabbits die and of juvenile rabbits die. Those juveniles who do not die become adults. The number of adult rabbits in year is denoted by

and the number of juvenile rabbits is denoted by

.

Questions 1

Explain why

2

Find an expression for

3

The equations found in questions 1 and 2 can be written in a matrix form as

Write down the matrix 4

. in terms of

.

.

An uninhabited island is populated with of adult rabbits in:

adult rabbits in year . Use technology to find the number

a year b year

.

5

By investigating the sequence formed, find the long-term growth rate of the population.

6

Find the long-term ratio of juveniles to adults.

7

The population of rabbits is infected with a disease which decreases the average number of juvenile rabbits produced per adult rabbit each year to . Find the smallest value of so that the population will not become extinct.

8

Describe the assumptions made in creating this model.

9

In an alternative model each adult rabbit produces exactly one juvenile each year and there is no death. If the population starts with one (presumably pregnant) adult, investigate the number of adult rabbits after years. Can you form a conjecture and prove it by induction?

Rewind For a reminder of proof by induction see Chapter 1.

Explore The situation in question 9 was described by Fibonacci, leading to his eponymous sequence.

CROSS-TOPIC REVIEW EXERCISE 1

1

a Show that

.

b Hence find the value of 2

a Express

correct to significant figures.

in the form

, where

b Solve the equation . 3

and

, giving your answers in the form

a Find the Cartesian equation of the plane . The plane

.

containing the points

has Cartesian equation

and

,

and

.

b Find, to significant figures, the acute angle between the planes 4

, where

Given that

and

.

find the values of the constants and .

© OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, January 2011 5

Find

giving your answer in a fully factorised form.

© OCR AS Level Mathematics, Unit 4725/01 Further Pure Mathematics 1, June 2013 6

i

Show that

ii Express as the product of four factors of the form , where . © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2012 7

i

Solve the equation where and

, giving the roots exactly in the form

,

.

ii Sketch an Argand diagram to show the lines from the origin to the point representing and from the origin to the points which represent the roots of the equation in part i. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 8

The line passes through the points and and the line passes through the points and . Find the shortest distance between and . © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2010

9

By using the determinant of an appropriate matrix, find the values of for which the simultaneous equations

do not have a unique solution for , and . © OCR AS Level Mathematics, Unit 4725/01 Further Pure Mathematics 1, January 2013 10 a

Given that and

are both real, with

b Find, in terms of , the argument of

.

, show that

.

c Hence show that 11 i

.

Show that

.

ii Hence find an expression, in terms of , for

iii

State, giving a brief reason, whether the series

converges.

© OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, June 2010 12 i

Show that

.

ii Hence find an expression, in terms of , for iii Given that

.

, find the value of .

© OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, June 2012 13 i

Show that

.

ii Hence find an expression, in terms of , for iii Find

.

, giving your answer in the form where and are integers.

© OCR AS Level Mathematics, Unit 4725/01 Further Pure Mathematics 1, June 2014 14 The sequence i

Find

and

is defined by , and show that

and

for

.

.

ii Hence suggest an expression for

.

iii Use induction to prove that your answer to part ii is correct. © OCR AS Level Mathematics, Unit 4725/01 Further Pure Mathematics 1, January 2013 15 The roots of the equation i

are denoted by , and

.

Sketch an Argand diagram to show these roots.

ii Show that iii Hence evaluate a b

. , .

iv Hence find a cubic equation, with integer coefficients, which has roots

and

.

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2008 16 i

Write down, in Cartesian form, the roots of the equation

.

ii Hence solve the equation , giving your answers in Cartesian form. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2010 17 The cube roots of are denoted by , and

, where the imaginary part of is positive.

i

Show that

In the diagram,

.

is an equilateral triangle, labelled anticlockwise. The points , and

represent the complex numbers , and respectively. ii State the geometrical effect of multiplication by and hence explain why . iii Hence show that . © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2011 18 i

Solve the equation

, giving your answers in polar form.

ii Hence, by considering the equation can be expressed in the form

, show that the roots of , stating the values of .

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2013 19 In an Argand diagram, the complex numbers , and and respectively. i

are represented by the points ,

Sketch a possible Argand diagram showing the triangle

. Show that the triangle is

isosceles and state the size of angle . The complex numbers and are represented by the points and respectively. The complex number is represented by the point , such that

and angle

.

ii Calculate the possible values of , giving your answers exactly in the form . © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2015 20 i

By expressing

and

in terms of

and

, prove that

. ii Hence show that all the roots of the equation are of the form

, where is any integer and is to be determined.

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 21 i

By expressing

in terms of

and

, show that .

ii Hence solve the equation for

.

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2014 22 The line has equations

and the point is

.

is the point

where the perpendicular from meets i

Find, in either order, the coordinates of

and the perpendicular distance from to .

ii Find the coordinates of the point on such that . © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2012 23 The plane has equation

and the line has equation

.

i

Express the equation of in the form

.

ii Find the point of intersection of and . iii The equation of may be expressed in the form

perpendicular to

, where is

Find .

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2012 24 The plane has equation and the line has equation . The line is parallel to and perpendicular to , and passes through the point with position vector in the form .

. Find the equation of , giving your answer

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 25 The lines and have equations

respectively. i

Find the shortest distance between the lines.

ii Find a Cartesian equation of the plane which contains and which is parallel to © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2013 26 a Integrate

with respect to .

b Show that, for

, the imaginary part of

c Hence find the exact value of

is

.

.

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 27 a Show that

.

b Hence find possible complex numbers for which

.

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 28 a If

, show that

.

b Find an expression for c Hence show that

.

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 29 a Find the smallest positive integer for which b Prove that

for all

.

.

30 a Prove using induction that for integer b Hence find the exact value of

. .

31 Prove, using induction, that for positive integer ,

32 i

Show that

ii Hence find an expression, in terms of , for

iii Show that © OCR AS Level Mathematics, Unit 4725 Further Pure Mathematics 1, January 2011 33 The integrals and are defined by

By considering

as a single integral, show that

and obtain a similar expression for . (You may assume that the standard result for constant, so that

remains true when is a complex

.)

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2008 34 i

Use de Moivre’s theorem to prove that

ii a By putting equation b Hence show that

in the identity in part i, show that

is a solution of the

iii Use the substitution

show that

where and are positive constants to be determined. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2009 35 i

Solve the equation

, for

.

ii By using de Moivre’s theorem, show that iii Hence find the exact value of

justifying your answer. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2010 36 i

Use de Moivre’s theorem to express

as a polynomial in

ii Hence prove that iii Use part ii to show that the only roots of the equation is an integer.

are

where

iv Show that only when © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2011 37 i

Use de Moivre’s theorem to prove that

ii Solve the equation iii Show that the roots of the equation may be expressed in the form stating the exact values of , where © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2012 38 Let i

a Show that, for

, where is an integer,

b State the value of for ii Hence show that, for

iii Hence show that

, where is an integer. , where is an integer,

is a root of

and find

another root in the interval © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2013

6 Hyperbolic functions In this chapter you will learn how to: define the hyperbolic functions , and draw the graphs of hyperbolic functions, showing their domains and ranges write the inverse hyperbolic functions in terms of logarithms define the reciprocal hyperbolic functions , and solve equations and prove identities involving hyperbolic functions differentiate hyperbolic functions.

Before you start… A Level Mathematics Student Book 2, Chapter 2

You should understand the terms domain and range of a function.

1 For the function state:

,

a the largest possible domain b the corresponding range.

A Level Mathematics Student Book 2, Chapter 3

You should be able to draw a graph after two (or more) transformations.

A Level Mathematics Student Book 2, Chapter 9

You should know how to differentiate and integrate the exponential function.

2 The graph of is shown. Sketch the graph of , giving the new coordinates of the three points labelled on the original graph.

3 Find: a b

A Level Mathematics Student Book 2, Chapter 9

You should know how to differentiate and integrate trigonometric

4 Find: a

.

functions.

b

A Level Mathematics Student Book

You should know how

5 Find

2, Chapter 10

to use the chain rule, product rule and

a

quotient rule for differentiation.

b

.

for these functions:

c

What are hyperbolic functions? Trigonometric functions are sometimes called circular functions. This is because of the definition that states: a point on the unit circle (with equation -axis, has coordinates .

) defining a radius at an angle to the positive

Related to the circle is a curve with equation , called a hyperbola. Points on this hyperbola have coordinates , although can no longer be interpreted as an angle.

Section 1: Defining hyperbolic functions Although the geometric definition of hyperbolic functions gives some helpful insight, a more useful definition is related to the number .

Key point 6.1

You can define

by analogy with the trigonometric definition of

.

Tip Cosh is pronounced as it reads, pronounced ‘tanch’ or ‘than’.

is pronounced either ‘sinch’ or ‘shine’ and

is

Key point 6.2

There are not many special values of these functions that you need to know, but, from Key points 6.1 and 6.2, you should be able to see that , and .

Did you know? The function is frequently used in physics, particularly in the context of special relativity and the study of entropy. As for trigonometric functions, you need to know the graphs of hyperbolic functions.

Key point 6.3 The graphs of

,



and

look like this.

Did you know?

You may think that the graph of

looks like a

parabola, but it is slightly flatter. It is called a catenary, which is the shape formed by a hanging chain.

You can establish the domains and ranges of the hyperbolic functions from their graphs.

Key point 6.4 The domains and ranges of the hyperbolic functions Function

,

Domain

and

are: Range

Rewind You met the domain and range of a function in A Level Mathematics Student Book 2, Chapter 2.

WORKED EXAMPLE 6.1 Given that

for

,

a sketch the graph of b state the range of . a

You need to apply two transformations to the graph of : Stretch by scale factor parallel to the -axis. Translation by

.

The horizontal asymptote at moves to and the one at is still at

b

, .

From the graph, the function is bounded by the asymptotes at and

.

Rewind You learnt about transformations of graphs in A Level Mathematics Student Book 1, Chapter 5.

EXERCISE 6A

EXERCISE 6A 1

For each hyperbolic function, sketch the graph of the corresponding range.

and state the largest possible domain and

a b c d e f 2

The diagram shows the graph of

, where and are integers.

Find the values of and . 3

The diagram shows the graph of

, where and are integers.

Find the values of and .

Rewind You saw in A Level Mathematics Student Book 2, Chapter 2, how to form the graphs of inverse functions from their original function by reflection in the line .

Section 2: Inverse hyperbolic functions The inverse functions of the hyperbolic functions are called

,

and

.

The graphs of these functions look like this.

You need to know the domain and range of each function.

Key point 6.5 The domains and ranges of the inverse hyperbolic functions Function

and

Domain

are:

Range

Tip and

are alternative notations for

and

.

WORKED EXAMPLE 6.2 Let

.

a State the largest possible domain of . b For the domain in part a, find the range of . a Domain:

The domain of

is

, so

.

It is always a good idea to sketch the graph when finding the range. You need to apply three transformations to the graph of : Stretch by scale factor parallel to the -axis. Reflection in the -axis. Translation by

.

b



Range:

You can use the inverse hyperbolic functions to solve simple equations involving hyperbolic functions. For example, if

then

, which you can evaluate, on a calculator, as

However, you can use the definition of

….

to derive a logarithmic form of this result. You can do this for

all three inverse hyperbolic functions.

Key point 6.6

These will be given in your formula book. These results can all be proved in the same way. The proof for

is given here.

PROOF 3 Prove that

.

Let

Let .

Then

Take

and then look to find an expression for of both sides.

Use the definition of

.

Rearrange into a disguised quadratic in .

Use the quadratic formula. So



Use the algebra of surds to simplify the expression.

Conventionally, you take the positive root, so this makes and .

But is a function so it can only take one value. But

WORKED EXAMPLE 6.3 Solve

, giving your answer in the form

for integers and .

Apply the inverse

function to find .

Use the logarithmic form

.

When you are trying to solve equations involving hyperbolic cosines, using the inverse function does not give all the solutions: it just gives the positive one. This is because the function is not one-to-one. As can be seen from the graph, there is a second, negative solution. (This is analogous to solving an equation like

, where taking the square root of both sides gives .).

, but there are in fact two solutions,

WORKED EXAMPLE 6.4 Given that

, express in the form

or

.

The expression is a disguised quadratic, so rearrange it to make one side zero and then factorise it. You could also have used the quadratic formula.

or But solutions.

so

has no

Use the inverse function to find . Remember that you need a plus or minus .

Use the logarithmic form

.



So or

Use the fact that

.

Simplify the second solution by rationalising the denominator to produce the required form.

So

or



You can generalise the method used at the end of Worked example 6.4 to write so the two solutions to for can always be written as

as .

EXERCISE 6B 1

For each inverse hyperbolic function, sketch the graph of domain and the corresponding range. a

and state the largest possible

i ii

b

i ii

c

i ii

2

Use your calculator to evaluate each expression where possible. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

3

Solve each equation, giving your answers to significant figures. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

4

Without using your calculator, find the exact value of each expression. a

i ii

,

b

i ii

c

i ii

d

i ii

e

i ii

5

Solve the equation

6

Find and simplify the exact value of

7

Find and simplify a rational expression for

8

The function is given by

. . . , where and are constants.

Find, in terms of , the set of values of the constant so that 9

Solve the equation

.

10 Solve the equation

.

11 Find and simplify an expression for 12 Prove that 13 Prove that 14 In the derivation of

has the largest domain possible.

. .

. you found that two possible expressions were

and

. Show that their sum is zero and hence explain why the expression chosen in Proof 3 is non-negative.

Section 3: Hyperbolic identities Just as there is the identity linking the trigonometric functions also an identity linking the hyperbolic functions and .

and

, there is

Key point 6.7

This will be given in your formula book. You can prove the identity in Key point 6.7 by using the definitions of

and

given in Key point 6.1.

Rewind The result in Key point 6.7 proves that described in the introduction to this chapter.

lies on the hyperbola

, as

WORKED EXAMPLE 6.5 Prove that

. Start from the definition of one of the hyperbolic functions. It doesn’t matter which one. It is squared in the expression so square it and simplify. Since

.

Repeat with the

term.

again. Combine the two terms and simplify.

You might be asked to prove other unfamiliar hyperbolic identities. To do this, always return to the definitions of the functions and follow a process similar to that in Worked examples 6.5 and 6.6. WORKED EXAMPLE 6.6 Prove that

. On the LHS use the definition of .

and replace each with

Then work from the RHS. Substitute the definitions of .

and

Multiply out the brackets, using the difference of two squares. Using the rules of indices.

EXERCISE 6C 1

Prove that

2

Simplify

3

Prove that

4

Prove that

.

5

Prove that

.

6

Prove that

. Hence prove that

7

Prove that

.

8

Use the binomial theorem to show that

9

a Explain why

. . .

. and

b Hence show that c Write 10 Given that

.

in terms of

.

show that

.

.

.

Section 4: Solving harder hyperbolic equations When you are solving equations involving hyperbolic functions you have several options: Rearrange to get a hyperbolic function that is equal to a constant and use inverse hyperbolic functions. Use the definition of hyperbolic functions to get an exponential function that is equal to a constant and use logarithms. Use an identity for hyperbolic functions to simplify the situation to one of the two preceding options. It is only with experience that you will develop an instinct about which method will be most efficient. WORKED EXAMPLE 6.7 Solve

. Use the definitions of

and

.



Tip When you are dealing with the sum or difference of two hyperbolic functions, it is often useful to use the exponential form.

WORKED EXAMPLE 6.8 Solve

, giving your answer in the form You could use the definitions of

. and

, but it is easier to use the

identity

Then use the logarithmic form of

.

WORKED EXAMPLE 6.9 Solve

, giving your answer in logarithmic form. The equation involves two types of function. You can use the identity to replace the term.

  Solve the resulting quadratic.

Use the logarithmic form of

.



or

WORK IT OUT 6.1 Solve

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 Dividing by :

or or or Solution 2 Dividing by

:

Solution 3

or or or

WORKED EXAMPLE 6.10 a Prove that b Solve the equation

a

. .

Since the RHS of the required identity contains it is easiest to start from the definition of terms of and . Write as a single fraction.

in

,

Use the identity from Key Point 6.7. b

From a,

.

This is a quadratic equation in

The range of possible.

Then use

EXERCISE 6D

is

.

, so

is not

.

EXERCISE 6D 1

Find the exact solution to

2

Solve

3

Solve

4

Solve

5

Solve the equation

6

Solve the equation

7

Find the exact solution to

8

Solve

9

Solve

.

, giving your answer in the form

.

, giving your answers correct to significant figures. , giving your answer in the form

, where is a rational number.

, giving your answer in the form

.

, giving exact answers. . , giving your answers in logarithmic form. , giving your answers in exact form.

10 Solve

, giving your answer in logarithmic form.

11 Solve

, giving your answer in logarithmic form.

12 Solve the equation

, giving your answer in the form

, where is a rational

number. 13 Using the identity in the form

, solve the equation

, giving your answers

.

14

a Show that

and

.

b Hence find the exact solutions to the simultaneous equations. 15 Find a sufficient condition on , and for 16 a Prove that

to have at least one solution.

.

b Hence solve the equation

, giving your answer in the form

17 a Show that, for any real number , . b Hence solve the equation giving your answers in the form

.

.

Section 5: Differentiation Key point 6.8

Only the final one of these will be given in your formula book.

Tip The first two of these formulae will not be given in your formula book. You can derive the results for

and

by returning to the definitions of these functions.

WORKED EXAMPLE 6.11 Show that

.

Use the definition of Differentiating:





You can show the result for

either from the definition again or by using

quotient rule. WORKED EXAMPLE 6.12 Use the derivatives of

and

to show that Use

Differentiating using the quotient rule:

Given that

, find

.

.



use

WORKED EXAMPLE 6.13

.

.

and the

Let

and

Use the product rule.

Then and

is a composite function so use the chain rule to differentiate.

Now apply the product rule formula.

EXERCISE 6E

EXERCISE 6E 1

Differentiate each function with respect to . a

i ii

b

i ii

c

i ii

2

Differentiate each function with respect to . a b

3

Find the exact coordinates of the turning point on the curve

4

Find the exact coordinates of the minimum point on the curve

5

Show that the equation of the tangent to the curve

6

Find the equation of the normal to the curve form .

7

a Find the exact values of the -coordinates of the turning points on the curve b Show that the maximum point has -coordinate

8

Find the coordinates of the stationary point on the curve

9

Show that the two points of inflection on the curve

. .

at

is

at

.

, giving your answer in the

.

. . have -coordinates

, stating the

value of . 10 a Find the exact value of the -coordinates of the stationary points on the curve with equation . b Prove that one of the stationary points from part a is a local maximum and that one is a local minimum point.

Section 6: Integration Key point 6.9

Tip These results will not be given in your formula book. As with differentiation, you can derive the results for the integrals of definitions of these functions.

and

by returning to the

WORKED EXAMPLE 6.14 a Show that

.

b Hence find the exact value of

.

a

Use the definition of

b

Use the result from part a together with the reverse chain rule.

Use the definition of that .

You can now find the integral of

in terms of , and the fact

either from the definition again or by using

applying the reverse chain rule. WORKED EXAMPLE 6.15 Show that

.

. Use

.

and



This is of the form

, so you can integrate it directly.

Tip Look out for integrals of the form

or

as you can integrate these

without need for a substitution, by reversing the chain rule. Often you will need to use a hyperbolic identity before integrating.

Rewind See A Level Mathematics Student Book 2, Chapter 11, for a reminder of integrating trigonometric functions using the reverse chain rule, trigonometric identities and integration by parts.

Tip When integrating hyperbolic functions, you can often use the same approach as with the corresponding trigonometric function.

WORKED EXAMPLE 6.16 Find

. Use the identity for

in terms of

.

Remember to divide by the coefficient of when integrating .

Sometimes it’s better to use the definition of the hyperbolic function, rather than the method you would have used with the corresponding trigonometric function. WORKED EXAMPLE 6.17 Find

. Use the definition of

.

Rewind If the integral in Worked example 6.17 had been

, you would have done integration

by parts twice and rearranged.

EXERCISE 6F 1

Find: a

i ii

b

i ii

c

i ii

2

Find the exact value of each integral. a

i ii

b

i ii

c

i ii

3

Use an appropriate hyperbolic identity to find each integral. a

i ii

b

i ii

c

i ii

4

Use integration by parts to find each integral. a

i ii

b

i

ii c

i ii

d

i ii

5

Use the definitions of a

and/or

to find each integral.

i ii

b

i ii

6

Find

.

7

By expressing

8

The diagram shows the region , which is bounded by the curve .

and

in terms of , evaluate

. , the -axis and the line

Show that the volume of the solid formed when the region is rotated through -axis is given by 9

radians about the

.

Find: a b

.

10 Find

.

11 The diagram shows the region bounded by curve .

, for

, the -axis and the line

Find the exact volume of the solid formed when the region is rotated through radians about the -axis. 12 Using the substitution

, show that

.

13 Show that 14 a Given that

, write down an expression for

b Use integration by parts to show that

Checklist of learning and understanding Definitions of hyperbolic functions:

Graphs of hyperbolic functions:

. .

Logarithmic form of inverse hyperbolic functions:

Domain and range of hyperbolic, inverse hyperbolic, and reciprocal hyperbolic functions: Function

Domain

Range

Identities: To prove other identities, return to the definitions of the functions involved. Derivatives of hyperbolic and reciprocal hyperbolic functions:

Integrals of hyperbolic functions:

Many hyperbolic integrals can be done by using the same method that you would use for the corresponding trigonometric integral.

Mixed practice 6 1

Given that

for

,

a sketch the graph of b state the range of . 2

Solve

3

Simplify

4

Solve

5

Solve

6

Find the exact solutions to

7

Solve the equation

8

Given

9

Show that the curve

, giving your answer in terms of . . , giving your answer in terms of logarithms. , giving your answer correct to significant figures. . , giving your answer in the form , find

. has no points of inflection.

10 Given that

, show that

11 By first expressing

.

.

and in terms of exponentials, solve the equation giving your answer in an exact logarithmic form.

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, January 2013 12 Find

.

13 Show that

.

14 Solve the equation

, giving your answers in exact form.

15 Solve 16 Prove that

, giving your answers in terms of natural logarithms. for all .

17 Find and simplify an expression for

.

18 Use the binomial theorem to show that 19 a Sketch the graph of b Given that that

.

. , use the definitions of

and

in terms of and

.

c i Show that the equation

can be written as .

ii Show that the equation

has only one solution for .

to show

Find this solution in the form 20 a Prove that

where is an integer.

.

b Hence solve the equation

, giving your answers in terms of natural

logarithms. 21 i Given that

, prove that

.

ii It is given that satisfies the equation for and to show that

. Use the logarithmic forms .

Hence, by squaring this equation, find the exact value of . © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, January 2012 22 Using the substitution

, find

.

23 The diagram shows the graphs of

and

.

Find the exact value of the area of the shaded region. 24 a Prove the identity b If

. and

find the value of .

c Hence find the exact real solution to without logarithms.

, giving your answer in a form

25 Solve these simultaneous equations, giving your answers in exact logarithmic form.

26 Using the logarithmic definition, prove that 27 Given that

, show that

28 Prove that

.

. .

7 Further calculus techniques In this chapter you will learn how to: differentiate inverse trigonometric and inverse hyperbolic functions reverse those results to find integrals of the form ,

,

and

use trigonometric and hyperbolic substitutions to find similar integrals integrate using partial fractions with a quadratic expression in the denominator.

Before you start… A Level Mathematics Student Book 2, Chapter 10

You should know how to differentiate functions defined implicitly.

1 Given that , find an expression for

A Level Mathematics Student Book 2, Chapter 11

You should be able to integrate using a substitution.

.

2 Use a suitable substitution to evaluate .

A Level Mathematics Student Book 2, Chapter 11

You should know how to integrate rational functions by splitting them into partial fractions.

3 Find .

Introduction In this chapter you will extend your range of integration methods and the variety of functions you can integrate. Differentiation of inverse trigonometric functions leads to rules for integrating functions of the form

and

and suggests that you can use trigonometric substitution to find other similar

integrals. Likewise, differentiation of inverse hyperbolic functions leads to rules for integrating and functions.

. You can use these results in combination with partial fractions to integrate many rational

Section 1: Differentiation of inverse trigonometric functions You already know how to differentiate use implicit differentiation.

,

and

. To differentiate their inverse functions you can

Rewind You met implicit differentiation in A Level Mathematics Student Book 2, Chapter 10.

WORKED EXAMPLE 7.1 Given that

, and that

, find

in terms of .

You know how to differentiate .

, so express in terms of

Differentiating each term with respect to : Remember the chain rule.

You want the answer in terms of , so you need to change to .

You should notice two important details in the derivation of the derivative of shown in Worked example 7.1. First, the function is defined for . However, you can see from the graph that the gradient at is not finite, so the condition is required for the derivative to exist. (You can also see that the expression for

is not defined when

Second, you used

to write

.) . When taking a square root, you need to

ask whether it should be positive or negative (or both). In this case, the range of the and in this range,

; this justifies taking the positive square root.

You can establish the results for the inverse

Key point 7.1

and

functions similarly.

function is

These will be given in your formula book. Notice that the derivative.

function is defined for all

, so there is no restriction on the domain of its

You can combine these results with other rules of differentiation. WORKED EXAMPLE 7.2 Differentiate: a b

and state the values for which the derivative is valid.

Tip Remember that

is alternative notation for

a Using the product rule:

.



Multiply by (the derivative of

b Using the chain rule:

) due to the chain rule.

Use

.

Multiply by

(the derivative of

).

The derivative is valid when and

EXERCISE 7A

The derivative of is only defined for the square root is only defined when

and .

EXERCISE 7A 1

Find a

for each function.

i ii

b

i ii

c

i ii

d

i ii

2

Find the exact value of the gradient of the graph of

at the point where

3

Find the exact value of the gradient of the graph of

at the point where

4

Differentiate

5

Find the derivative of

6

a Given that

, stating the range of values of for which your answer is valid. , show that

Given that

.

with respect to . , show that

and state the values of for which the

derivative is valid. 8

Given that

9

a Find

.

, simplifying your answer as far as possible.

b Hence differentiate 7

.

, find an expression for

.

.

b Hence find 10 Show that the graph of

. has no points of inflection.

Section 2: Differentiation of inverse hyperbolic functions You know how to differentiate hyperbolic functions, and so again you can use implicit differentiation to differentiate their inverse functions.

Rewind See Chapter 6, Section 5, for differentiation of hyperbolic functions.

WORKED EXAMPLE 7.3 Rewrite in terms of

.

Differentiating with respect to

, so take the reciprocal of both sides.

Use

You can establish the results for the inverse

.

and

functions similarly.

Key point 7.2

These will be given in your formula book.

WORKED EXAMPLE 7.4 Find the value of the -coordinate of the point on the curve parallel to the line

at which the tangent is

.

Differentiating using the chain rule:

Remember to multiply by the derivative of .

Simplify the denominator.

So has gradient , so set

The domain of

and solve for .

is

so take the positive

square root.

EXERCISE 7B 1

Differentiate each function with respect to . a

i ii

b

i ii

c

i ii

2

Given that

3

Given that

4

Given that

, find , find for

. .

, show that

where and are integers to be found. 5

Find the equation of the tangent to the curve answer in the form

6

The tangents at

at the point where

, giving your

. and

to the curve

coordinate of is

intersect at the point . Show that the .

7

Find the coordinates of the point of inflection on the curve

8

Show that

9

Prove that the -coordinate of the point on the curve , where is a value to be found.

satisfies

.

. at which the gradient is

is

Section 3: Using inverse trigonometric and hyperbolic functions in integration You can reverse the derivatives from Sections 1 and 2 to derive four more integration results:

Tip Notice that the results

and

included in this list, as the first is just the negative of partial fractions.

are not and the second can be done by

These results can be generalised slightly by making a linear substitution.

Key point 7.3

These will be given in your formula book.

Focus on … See Focus on … Problem solving 2 for an example of using one of these integrals. You also need to know how to derive these results, using trigonometric or hyperbolic substitutions. WORKED EXAMPLE 7.5 Use the substitution

to prove the result

when

Differentiate the substitution and express

Express the integrand in terms of .

. in terms of

Use

.

Since you are choosing the substitution, you can choose . For a given value of there are two possible values of . You can choose the that gives the positive value. Make the substitution and integrate.

Write the answer in terms of : .

You can derive the result the identity

similarly, using the substitution

and

.

Fast forward You will be asked to derive this result in Question 8 in Exercise 7C.

WORKED EXAMPLE 7.6 Use the substitution

to prove the result

.

Differentiate the substitution and express of .

in terms

Express the integrand in terms of .

Use

.

Since you are choosing the substitution, you can choose . Make the substitution and integrate.

Write the answer in terms of : .

You can derive the result the identity

similarly, using the substitution .

and

You can combine these results with algebraic manipulation to integrate an even wider variety of functions.

Fast forward You will be asked to derive this result Question 9 in Exercise 7C.

WORKED EXAMPLE 7.7 Find

. You can turn the integrand into the form ) by making a substitution

(with

.

Since the substitution is linear, you can simply divide by the coefficient of . If the denominator is not in the form it in this form.

or

, you might need to complete the square to write

WORKED EXAMPLE 7.8 Find

. The expression in the denominator is quadratic, so you should check whether you can complete the square to write it in the form

Hence



The integrand is of the form and

1

Find each indefinite integral. a

i ii

b

i ii

c

i ii

d

i

with

. Remember to divide by the

coefficient of when integrating.

EXERCISE 7C

.

ii e

i ii

f

i ii

2

Find each indefinite integral. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

3

By first completing the square, find: a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

4

Find the exact value of

.

5

Find the exact value of

.

6

Find the exact value of

.

7

Find: a b

8

a Use a trigonometric substitution to prove that b Hence evaluate

9

.

.

a Use a hyperbolic substitution to prove that b Hence evaluate

10 a Write

, giving your answer in terms of a natural logarithm. in the form

b Hence find 11 a Write

.

.

. in the form

.

b Hence find the exact value of

.

12 a Using a suitable substitution prove that, when . b Find

.

13 Use a suitable trigonometric substitution to show that .

,

14 Find

.

15 Show that

.

16 Find

.

17 Find

.

18 Find 19 a Write

. in the form

b Hence find

.

.

20 Use a suitable hyperbolic substitution to show that . 21 Use a suitable trigonometric substitution to show that . 22 a Given that

, express

and

in terms of .

b Use a suitable trigonometric substitution to show that .

Section 4: Using partial fractions in integration You have already used partial fractions to integrate rational expressions with linear and repeated linear factors in the denominator, such as

. You can now use the results from Section 3 to

extend the range of rational functions you can integrate to include those with denominators of the form .

Rewind You met partial fractions in A Level Mathematics Student Book 2, Chapter 5, and then used them in integration in Chapter 11. In general, when there is a quadratic factor in the denominator, there are three possibilities: The quadratic factorises into two different linear factors, fractions are

. The corresponding partial

.

The quadratic is a perfect square,

. The corresponding partial fractions are

.

The quadratic does not factorise (the quadratic factor is irreducible). For example, or . Then there is only one corresponding partial fraction, with a numerator of the form .

Key point 7.4 If

is a polynomial of order less than or equal to , then

WORKED EXAMPLE 7.9 a Express

in partial fractions.

b Hence find

.

Use the form from Key point 7.4.

a

Multiply through by the common denominator. Substitute in the values of which make some of the terms zero. Comparing coefficients of

:

Look at the coefficient of

to find .

Hence

b

Integrate each term separately before applying limits. You need to split the second integral into two in order to

apply standard results. Here you can use a substitution , or the reverse chain rule, as is half the derivative of Use

. with

. Hence

If there is a quadratic factor in the denominator, you first need to check whether it is irreducible or whether it can be factorised. If a quadratic factor is irreducible, you need to write it in completed square form before you can apply standard integration results. WORKED EXAMPLE 7.10 Given that

, find an expression for in terms of .

The denominator is

You need to split the function into partial fractions before integrating.

. Hence

Check whether the quadratic factors factorise. The second one has the discriminant so it is irreducible. Multiply through by the denominator. Substitute in suitable values of .

For the third integral, you need to complete the square and then use



EXERCISE 7D

EXERCISE 7D 1

Use partial fractions to find each integral. a

i ii

b

i ii

c

i ii

d

i ii

2

Use partial fractions to find the exact value of

3

a Write

in partial fractions.

b Given that 4

, and that

when

, find in terms of .

Use partial fractions to integrate: a b

5

.

Let a Write

. in partial fractions.

b Hence find the exact value of

6

Let

7

Use partial fractions to find

8

Show that

where

.

. Use partial fractions to evaluate

and are constants to be found.

Checklist of learning and understanding

.

.

Derivatives of inverse trigonometric functions:

Derivatives of inverse hyperbolic functions:

You can derive the corresponding integrals using a trigonometric substitution or a hyperbolic substitution or

or

You might need to write a quadratic expression in completed square form in order to apply one of the results shown. When splitting an expression into partial fractions, if the denominator has an irreducible quadratic factor

, then the corresponding partial fraction is

.

Mixed practice 7 1

Differentiate

2

Find

3

Differentiate

4

Find the -coordinates of the points on the curve

5

By first completing the square, find the exact value of

.

when

. . where the gradient is . .

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2015 6

It is given that i Express

.

in partial fractions.

ii Hence find

.

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2007 7

Given that

, find

8

Show that

satisfies

9

Find

. .

.

10 Find

.

In questions 11 and 12 you must show detailed reasoning.

Tip Remember that ‘show detailed reasoning’ means that you need to show full algebraic working, rather than using your calculator to evaluate indefinite integrals. 11 Given that

find the exact value of . © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2009 12 i Given that , find

in a simplified form.

ii Hence, or otherwise, find the exact value of

.

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2007 13 i Express

in partial fractions.

ii Show that © OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, January 2010 14

a Show that

.

b Hence find

.

15 Use the substitution

to show that

and state the value of the integer . 16

a Split

into partial fractions.

b Hence find

.

c Find the exact value of 17

.

a Show that b

Hence show that the area enclosed by the ellipse with equation

18 i Prove that

is

.

.

ii Hence, or otherwise, find iii By means of a suitable substitution, find

. .

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, January 2008

8 Applications of calculus In this chapter you will learn how to: find infinite series expansions (called Maclaurin series) of functions use given results to find the Maclaurin series of more complicated functions understand for which values of these series are valid find the value of definite integrals in certain cases where a limiting process is required (improper integrals) find the volume of a shape formed by rotating a curve around the -axis or the -axis find the mean value of a function.

Before you start… A Level Mathematics Student Book 2, Chapter 10

You should know how to differentiate functions using the chain rule.

1 For , find

A Level Mathematics Student Book 2, Chapter 9

You should know how to differentiate exponential, logarithmic and trigonometric functions.

.

2 For each function, find: i ii

.

a b c Chapter 6

You should know how to differentiate hyperbolic functions.

3 For find: a b

Chapter 6

You should know how to differentiate inverse hyperbolic functions.

.

4 For , find: a b

.

Chapter 7

You should know how to differentiate inverse trigonometric functions.

5 For find: a b

.

Introduction In the first part of this chapter you will see that, with certain restrictions, you can write many functions as infinite series in ascending positive integer powers of . Being able to take a number of terms of the infinite series as an approximation to a given function has many uses; for example, calculators and computers use these series to evaluate a function at a particular value or to produce an approximation for definite integrals of functions that can’t be integrated using standard functions. You will also look at definite integrals where either the integrand is not defined at one or more points in the range of integration or where the range of integration extends to infinity. These are known as improper integrals. Finally, you will see two further applications of calculus: finding volumes and finding the mean value of a function.

Section 1: Maclaurin series You know from A Level Mathematics Student Book 2, Chapter 6, that functions such as and can be written as infinite series using the binomial expansion. In general, for some function such a series will be of the form where

, etc. are real constants.

Differentiating this series several times:

Substituting

into

and each of its derivatives to find expressions for

Substituting these expressions for series formula for any function.

, etc. back into the expression for

, etc:

gives the Maclaurin

Did you know? Maclaurin series are named after the 18th-century mathematician Colin Maclaurin, who also developed some of Newton’s work on calculus, algebra and gravitation theory. You are assuming here that you can differentiate an infinite series term by term in the same way as a finite series. In fact, this is only possible for values of within the interval of convergence of the series.

Fast forward You will meet the interval of convergence for some standard functions in Section 2. You can use Key point 8.1 to find the first few terms of a Maclaurin series. You can sometimes spot a pattern in the derivatives, which enables you to also write down the general term,

Key point 8.1 The Maclaurin series of a function

is given by:

WORKED EXAMPLE 8.1 a Find the first three non-zero terms in the Maclaurin series of

.

b Write down a conjecture for the general term (you need not prove your conjecture). a

Find

.

.

Then differentiate and evaluate each derivative at

.

Notice that you need to go as far as the fifth derivative to get three non-zero terms.



So

Substitute these values into the Maclaurin series formula:

b The general term is

The series will only contain terms with odd powers because even derivatives are and .



Because of the pattern of differentiating



alternate between and .

Not every function has a Maclaurin series. For example, for derivatives of derivatives at

at ). However, , do exist.

,

, the signs will

doesn’t exist (nor do any of the

does have a Maclaurin series as now

, and all the

WORKED EXAMPLE 8.2 a Find the Maclaurin series of

up to and including the term in

b Prove by induction that the general term is

a

.

.

Find

.

Then differentiate and evaluate each derivative at .

So

Substitute these values into the Maclaurin series formula:

b To prove:

To find the general term you need an expression for the th derivative, which you then divide by to get the coefficient of . You can conjecture the expression by looking at the first four derivatives you found in part a. Show that the expression is correct when

Hence the expression is correct for

.

.

Suppose that the expression is correct for

Now suppose that the expression is correct for some and show that it is still correct for .

the th derivative: Then the

th derivative is:

Differentiate

to get

The negative sign in whole expression.

.

changes the sign of the

Hence the expression is correct for the th derivative. The expression is correct for and, if it is correct for , then it is also

Write a conclusion.

correct for . It is therefore true for all , by the principle of mathematical induction.

One use of Maclaurin series is to approximate definite integrals of functions that can’t be integrated by standard methods.

Tip The question will make it clear whether you are required to find the Maclaurin series of a function from first principles or whether you can use one of the standard results in the formula book to find the series you need.

Fast forward In Section 2 you will see how to use the Maclaurin series of certain standard functions to find the series of more complicated functions.

WORKED EXAMPLE 8.3 a Use the Maclaurin series of .

to find the first three non-zero terms in the Maclaurin series of

b Hence find an approximate value for

a

b

EXERCISE 8A

, giving your answer to three decimal places.

Substitute

into the series for

Integrate the polynomial as usual.

.

1

Using Key point 8.1, find the first four non-zero terms of the Maclaurin series of these functions. Also conjecture the general term for each series (you need not prove your conjecture). a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

2

Find the Maclaurin series of

3

a Show that the first two non-zero terms in the Maclaurin series of

up to and including the term in

b Hence find, to three decimal places, an approximation to 4

It is given that a

.

are

.

.

.

i Find the first four derivatives of

.

ii Hence find the Maclaurin series of

, up to and including the term in

b Use your result from part a ii to find an approximation to

.

.

Give your answer in the form where and are integers. 5

a Find the Maclaurin series of

up to and including the term in

b Hence state the Maclaurin series of 6

It is given that

,

a

and

i Show that

.

up to and including the term in

.

.

ii Find the third and fourth derivatives of . b Hence find the Maclaurin series of c

up to and including the term in

Use your series from part b to find an approximation to

.

, giving your answer to

three decimal places. 7

The function is defined by a

i Find the first three derivatives of ii

,

.

.

Hence show that the Maclaurin series for

up to and including the

b Use the series from part a ii to find an approximate value for

a Given that

, use induction to show that

.

. Give your answer in the form ,

where and are integers. 8

term is

.

b Hence find the general term in the Maclaurin series for 9

a Given that

.

, prove by induction that

b Hence find the general term in the Maclaurin series for

.

10 a Find the first three non-zero terms in the Maclaurin series for b

i Let of

and , and

. State the relationship between

, the th derivative of

ii Hence show that 11 The diagram shows part of the graph of

, for any integer

b

, the th derivative

.

, where is a constant to be determined. .

Explain why neither of these can be Maclaurin series of the function a

.

:

Section 2: Using standard Maclaurin series The Maclaurin series of a few standard functions are given in your formula book. These can often be used, without needing to be derived, to find the series for more complicated functions.

Key point 8.2 Maclaurin series for some common functions and the values of for which they are valid: for all

for all for all

These will be given in your formula book.

Tip Don’t overlook the information on the values of for which these series are valid; this is a very important part of each result.

Rewind Note that the last result in Key point 8.2 is the binomial expansion, which is covered in A Level Mathematics Student Book 2, Chapter 6. Note also that replacing by in the series for gives the same series as adding , which agrees with Euler’s formula from Chapter 2, Section 2.

WORKED EXAMPLE 8.4 a Use the Maclaurin series for

to find the first four terms in the series for

.

b State the values of for which the series is valid.

Substitute

a

into the series for

:

Expand and simplify.

b Valid for all

.

Both

and

are valid for all

.

This process can be more complicated if it involves finding two separate Maclaurin series and then

combining them. WORKED EXAMPLE 8.5 a Use the Maclaurin series for

and that for to find the series for

as far as the term in

b State the values of for which your series is valid.

Start by replacing

a

with its

series. Now split this into a product of two terms. Then form the series for each of them. For

substitute

into

the series for . Expand term by term and simplify.

b Valid for all

.

The series for both and valid for all .

are

WORKED EXAMPLE 8.6 a Find the first three terms in the Maclaurin series for b Hence find the Maclaurin series up to the term in

.

for

.

c State the interval in which the expansion is valid.

a

You know the series expansion for so you need to write in this form. Start by factorising . Separate the , using

Then substitute of

into the series

:

Expand and simplify.

b

Again, you need everything in the form of . First, use the laws of logs.

.

For the first term, substitute the series for .

So:

into

You know the series expansion for the second term from part a. Now put both series together.

c Since

is valid when

•

is valid when This is when

•

:

Find the interval of validity separately for each function.

.

is valid when This is when

.

Therefore,

is valid .

For both to be valid, you need the smaller interval.

EXERCISE 8B 1

Find the first three non-zero terms and the general term of the Maclaurin series for each expression. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

2

By first manipulating it into an appropriate form, find the first three non-zero terms of the Maclaurin series for each expression. a

i ii

b

i ii

c

i ii

3

By combining Maclaurin series of different functions, find the series expansion as far as the term in for each expression. a

i ii

b

i ii

c

i ii

4

Find the Maclaurin series for

5

Find the Maclaurin series as far as the term in

6

Show that

7

a Find the first two non-zero terms of the Maclaurin series for b Hence find the Maclaurin series of

8

a By using the Maclaurin series for

and state the interval in which the series is valid. for

. . .

up to and including the term in

.

, find the series expansion for

b Hence find the first two non-zero terms of the expansion of

up to the term in .

c Use your result from b to find the first two non-zero terms of the series for 9

a Find the first four terms of the Maclaurin series for b Find the equation of the tangent to

10 a Find the Maclaurin series for

at

. .

.

, stating the interval in which the series is valid.

b Use the first three terms of this series to estimate the value of

, stating the value of used.

.

Section 3: Improper integrals Integrals where the range of integration extends to infinity You are by now very familiar with evaluating definite integrals where

.

In the examples you have encountered so far, the limits were often convenient, relatively small numbers such as . However, there is nothing to stop them from being very large numbers; this would make no difference to the method for evaluating the integral. If you continue along this line and let , you can still find a finite value for the integral in certain cases. In much the same way that you have seen that a sequence can either converge to a finite limit or diverge to infinity, so can an integral. Integrals of the form

are known as improper integrals.

To evaluate an improper integral, you need to replace the infinite limit by , find the value of the integral in terms of and then consider what happens when

.

Key point 8.3 The value of the improper integral

is ,

if this limit exists and is finite. If this limit is infinite you say that the improper integral diverges (does not have a value).

WORKED EXAMPLE 8.7 a Explain why b Evaluate

is an improper integral. .

a The integral is improper because the range of integration extends to infinity. b

Integrate as normal, but replace the upper limit with and take the limit as after you have completed the integration.

As

,

WORKED EXAMPLE 8.8 Explain why the improper integral

diverges.

. Therefore the integral converges.

Integrate as normal, but replace the upper limit with and consider the limit as after you have completed the integration.

When

,

tends

to infinity. Therefore the integral diverges. When evaluating improper integrals, you might need to use some more complicated limits. These will be given in each question. WORKED EXAMPLE 8.9 Evaluate

, showing clearly the limiting process used.

[You can use without proof the result

, where

.]

Integrate as normal, but replace the upper limit with and consider the limit as after you have completed the integration. Use integration by parts.

Now take the limit as Use the given result:

. , with

.

Integrals where the integrand is undefined at a point within the range of integration There is another type of improper integral, where the range of integration is finite but the integrand is not defined at a point within the range of integration (which could be at an end point or inside the range). Examples of such integrals are defined at

, which isn’t defined at

, and

, which isn’t

.

To evaluate the first of these integrals, you need to replace by as the lower limit, find the value of the integral in terms of and then consider the limit

Key point 8.4 If

and

is not defined at

, then

.

. If the limit is not finite, then the improper integral diverges (does not have a value).

WORKED EXAMPLE 8.10 Evaluate

, showing clearly the limiting process used.

[You may use without proof the result

, where

.]

In is not defined at Integrate as normal, but replace the lower limit with and consider the limit as after you have completed the integration. Use integration by parts. Remember that integrals with are an exception where you take .

Now take the limit as Use the given result:

. , with

.

If the point where the integrand is not defined is not an end point, you need to split the integral into two.

Key point 8.5 If

is undefined at

, then .

If either limit is not finite, then the improper integral diverges (does not have a value).

WORKED EXAMPLE 8.11 Find the exact value of

. The integrand is not defined at

, so you

need to split the integral in two. For each integral replace by , evaluate the integral and then find the limit when

.

Integrate as usual.

As

,

.

Repeat the same process for the second integral.

EXERCISE 8C 1

Determine which of these improper integrals converge. Evaluate the ones that do converge. a

b

c d 2

For what values of do each of these improper integrals converge? a b

3

Explain why

4

Evaluate

5

Evaluate

6

Evaluate the improper integral

is an improper integral and find its value. .

where is a constant.

. , giving your answer in the form

,

Section 4: Volumes of revolution In A Level Mathematics Student Book 1, Chapter 15, you saw that the area between a curve and the axis from

to

is given by

, as long as

. In this section, you will use a similar formula

to find the volume of a shape formed by rotating the curve about either the -axis or the -axis.

If a curve is rotated about the -axis or the -axis, the resulting shape is called a solid of revolution and the volume of that shape is referred to as the volume of revolution.

Key point 8.6 When the curve

between

and

of revolution is given by

.

When the curve

and

between

of revolution is given by

is rotated

about the -axis, the volume

is rotated

about the -axis, the volume

.

The proof of these results is very similar. The proof for rotation about the -axis is given in Proof 4.

PROOF 4 The solid can be split into small cylinders.

Draw an outline of a representative function to illustrate the argument.

The volume of each cylinder is .

The radius of each cylinder is the -coordinate and the height is .

The total volume is

You are starting at

approximately:

It is only approximate because the volume of revolution is not exactly the same as the total volume of the cylinders.

and stopping at

.

However, as you make the cylinders smaller the volume gets more and more accurate. The sum then becomes an integral. You can leave out of the integration and multiply by it at the end.

WORKED EXAMPLE 8.12 The graph of

,

, is rotated

about the -axis.

Find, in terms of , the volume of the solid generated. Use the formula:

.

Evaluate the definite integral.

To find the volume of revolution about the -axis you will often have to rearrange the equation of the curve to find in terms of .

Tip Remember that the limits of the integration need to be in terms of and not .

WORKED EXAMPLE 8.13 The part of the curve

between

and

is rotated

value of the volume of the solid generated. When

,

Find the limits in terms of .

about the -axis. Find the exact

When

, Express in terms of . Use the formula

, substituting in

.

You might also be asked to find a volume of revolution of an area between two curves.

From the diagram you can see that the volume formed when the region is rotated around the -axis is given by the volume of revolution of minus the volume of revolution of .

Key point 8.7 The volume of revolution of the region between curves

where the curve of

is above

and

and the curves intersect at

is:

and

.

Tip Make sure that you square each term within the brackets and do not make the mistake of squaring the whole expression inside the brackets: the formula is not

.

WORKED EXAMPLE 8.14 Find the volume formed when the region enclosed by about the -axis. For points of intersection:

and

is rotated through

First find the -coordinates of the points where the curves meet, by equating the RHS of both equations and solving. This will give you the limits of integration.

Sketch the graphs in the region concerned.

is above

.

Apply the formula

Use a calculator to evaluate the integral.

Tip Don’t forget that you can use your calculator to evaluate definite integrals. However, look out for the instruction to ‘show detailed reasoning’ which means that you must show full integration and evaluation.

Did you know? There are also formulae to find the surface area of a solid formed by rotating a region around an axis. Some particularly interesting examples arise if you allow one end of the region to tend to infinity; for example, rotating the region formed by the lines

,

and the -axis results

in a solid called Gabriel’s horn or Torricelli’s trumpet.

Areas and volumes can also be calculated using improper integrals, and it turns out that it is possible to have a solid of finite volume but infinite surface area! You can also use the formulae for the volume of revolution about the coordinate axes in the case where the curve is defined parametrically.

Key point 8.8 When the part of a curve with parametric equations , between points with parameter values and , is rotated about one of the coordinate axes, the resulting volume of revolution is for rotation about the -axis

for rotation about the -axis.

Rewind You learnt how to find the area defined by a parametric curve in A Level Mathematics Student Book 2, Chapter 12.

WORKED EXAMPLE 8.15 The curve shown in the diagram has parametric equations

The part of the curve between the points and exact value of the resulting volume of revolution. When

,

When but

.

, so

.

is rotated about the -axis. Find the

Find the values of corresponding to the end points. ;

. Use the formula

for the volume of revolution about the -

axis.

You can use your calculator to evaluate the integral (unless the question asks you to ‘show detailed reasoning’).

EXERCISE 8D In this exercise, whenever a question asks for an exact volume, you must show detailed reasoning. 1

The part of the curve for revolution formed in each case. a

b

i

;

ii

;

ii c

i ii

2

, ,

; ; ;

about the -axis. Find the exact volume of

,

;

i

is rotated

, , ,

Find the exact volume of revolution formed when each curve, for radians about the -axis. a

i

, is rotated through

ii b

i ii

c

i ii

3

The part of the curve for

is rotated

about the -axis.

Find the exact volume of revolution formed in each case. a

b

c

4

i

;

,

ii

;

,

i

;

,

ii

;

,

i

;

,

ii

;

,

The part of the curve

for

is rotated

about the -axis.

Find the exact volume of revolution formed in each case. a

;

i ii

b

c

5

i

, ;

;

,

,

ii

;

i

;

ii

;

,

The part of the curve for exact volume of revolution formed in each case. a

is rotated through

about the -axis. Find the

is rotated through

-axis. Find the exact

i ii

b

i ii

c

i ii

6

The part of the curve for volume of revolution formed in each case. a

i ii

b

i ii

c

i ii

7

The diagram shows the region, , bounded by the curve

, the -axis and the line

.

a Find the coordinates of the point where the curve crosses the -axis. This region is rotated

about the -axis.

b Find the exact volume of the solid generated.

8

The curve

, for

, is rotated through

about the -axis.

Find the volume of revolution generated, correct to 3 significant figures. 9

The part of the curve

between

and

is rotated through

radians about the -

axis. Find the exact volume of the solid generated. 10 The curve

, for

The resulting volume is

, is rotated through

about the -axis.

.

Find the value of . 11 The region enclosed by the curve

and the -axis is rotated

about the -axis.

Find an expression, in terms of , for the volume of revolution formed. 12 The part of the curve

between

and

axis. The volume of the resulting solid is

is rotated through

radians about the -

.

Find the exact value of . 13 a Find the coordinates of the points of intersection of curves

and

.

b Find the volume of revolution generated when the region between the curves is rotated through

and

about the -axis.

14 The region bounded by the curves

and

is rotated through

about the -axis.

Find the volume of the resulting solid. 15 a Find the coordinates of the points of intersection of the curves b The region between the curves

and

and

is rotated through

. about the -axis.

Find the volume of the solid generated. 16 By rotating the circle given by

around the -axis, prove that the volume of a sphere of radius is

.

17 The part of the curve with parametric equations , is rotated through

,

, between the points

about the -axis. Find the exact volume generated.

and

18 The diagram shows a part of the curve with parametric equations The section of the curve between points

and

.

is rotated a full turn about the -

axis. Find the exact volume of the resulting solid.

19 The diagram shows the curve with parametric equations

.

One of the loops of the curve is rotated 2 radians about the -axis. Find the exact value of the volume of revolution. 20 By choosing a suitable function to rotate around the -axis, prove that the volume of a circular cone with base radius and height is

.

21 Find the volume of revolution when the region enclosed by the graphs of rotated through about the line .

,

and

is

Section 5: Mean value of a function Suppose an object travels between time graph looks like this.

and

with a velocity given by

. Its velocity–time

Its average velocity can be found from:

Suppose, instead, the object has velocity given by

. Then you can compare the two velocity–time

graphs. The formula

would give the same mean velocity for the two graphs, which

can’t be correct because the red curve is underneath the blue line everywhere other than at the end points. You need a method of calculating the mean that takes into account the value of the function everywhere. One possibility is to use

.

You can then use the fact that total distance is the integral of velocity with respect to time.

For the blue line this gives:

For the red curve this gives:

This process can be generalised for any function.

Key point 8.9 The mean value of a function

between and is:

WORKED EXAMPLE 8.16 You are given that

. Find the mean value of

between and .

Use the formula for the mean value of a function: .

Notice that varies between and around seems reasonable.

EXERCISE 8E

, so a mean of

EXERCISE 8E 1

Find the mean value of each function between the given values of . a

b

c

2

i

for

ii

for

i

for

ii

for

i

for

ii

for

Find the mean value of each function over the domain given. a

b

i

for

ii

for

i

for

ii c

for

i ii

3

for

The velocity of a rocket is given by

where is time, in seconds, and is velocity, in metres

per second. Find the mean velocity in the first seconds. 4

The mean value of the function

for

is zero.

Find the value of . 5

for a

.

is the mean value of

b Given that

between and . Find an expression for

in terms of .

find an expression for in terms of .

6

Show that the mean value of

between and is inversely proportional to .

7

An alternating current has time period . The power dissipated by the current through a resistor is given by . Find the ratio of the mean power of one complete period to the maximum power.

8

The mean value of

between and is .

Prove that the mean value of 9

between and is

a Sketch the graph of

.

b Use the graph to explain why the mean value of and . c Hence prove that, if 10 If

is the mean value of

.

, for

between and is less than the mean of

. and

, then

Either prove this statement or disprove it using a counterexample.

.

Checklist of learning and understanding The Maclaurin series for a function

is given by

Maclaurin series for some common functions and the values of for which they are valid: for all

for all for all

Improper integrals are definite integrals where either: the range of integration is infinite or the integrand isn’t defined at every point in the range of integration. The value of the improper integral

is

if this limit exists and is finite. If

is not defined at

, then and

if the limits exist and are finite. If

is undefined at

, then

if the limits exist and are finite. The volume of a shape formed by rotating a curve about the -axis or the -axis is known as the volume of revolution. When the curve between volume of revolution is given by

and

is rotated

about the -axis, the

When the curve between volume of revolution is given by

and

is rotated

about the -axis, the

The volume of revolution of the region between curves

where the curve of

is above

and

and the curves intersect at

is:

and

.

When the part of a curve with parametric equations , between points with parameter values and , is rotated about one of the coordinate axes, the resulting volume of revolution is for rotation about the -axis

for rotation about the -axis The mean value of a function

between and is:

Mixed practice 8 1

a

Show that the first four terms in the Maclaurin series of are

b Use the series in part a to show that 2

.

.

a Find the first four derivatives of

.

b Hence find the first four non-zero terms of the Maclaurin series of

.

c Using this expansion, find the exact value of the infinite series 3

a Explain why

.

is an improper integral.

b Either find the value of the integral

or explain why it does not have a finite

value. 4

Given that Maclaurin series for

, find and . Hence show that the first term in the is , where the value of is to be found.

© OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, January 2012 5

The curve between and is rotated through solid has a volume of .

about the -axis. The resulting

Find the value of . 6

The curve

with

, is rotated through

about the -axis.

Find the volume of revolution generated, correct to significant figures. 7

For

, the mean value of is equal to the mean value of

.

Find the value of . 8

The mean value of

from to is .

Find the value of . 9

Find the set of values of for which the Maclaurin series of the function valid.

10 a Find the Maclaurin series of

up to and including the term in

b Hence show that 11 It is given that

is

.

. ,

a Find the first three derivatives of b Hence find the Maclaurin series of c Hence find

.

12 You are given that

.

. . up to and including the term in

.

i Find

and

.

ii Show that

and hence, or otherwise, find

iii Find a similar expression for

and hence, or otherwise, find

iv Find the Maclaurin series for

up to and including the term in

. . .

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, January 2013 13 The region bounded by the curve

and the -axis is rotated one full turn about the

-axis. Find, in terms of , the resulting volume of revolution. In questions 14 to 20 you must show detailed reasoning. 14 The diagram shows the curve

and the line

.

a Show that the two graphs intersect at (e, 1). The shaded region is rotated through

about the -axis.

b Find the exact value of the volume of revolution.

15 The region enclosed by the line .

and the line

is rotated through

about

Find the exact value of the resulting volume. 16 The diagram shows the curve coordinates

for

. The point on the curve has

. The shaded region is enclosed by the curve and the lines

and

. i Find the exact area of . ii Find the exact volume of the solid obtained when is rotated completely about the -axis.

© OCR A Level Mathematics, Unit 4723/01 Core Mathematics 3, June 2014

17 i Show that the derivative with respect to of

is

.

ii The diagram shows the curve with equation

. The point

curve. The shaded region is bounded by the curve and the lines

lies on the and

. Find the

exact volume of the solid produced when the shaded region is rotated completely about the -axis.

iii Hence find the volume of the solid produced when the region bounded by the curve and the lines , and is rotated completely about the -axis. © OCR A Level Mathematics, Unit 4723 Core Mathematics 3, June 2012 18 The region bounded by the curves and -axis. Find the resulting volume of revolution.

is rotated one full turn, about the

19 The shape of a rugby ball can be modelled as a solid obtained by rotating an ellipse about one of its axes of symmetry. An ellipse has parametric equations generated when the ellipse is rotated through

. Find the volume of the solid about the -axis.

20 The diagram shows a part of the curve with parametric equations The part of the curve between points

and

.

is rotated a full turn about the

-axis. Find the exact volume of the resulting solid.

21 a Using the series for , or otherwise, find the Maclaurin series of non-zero terms and the general term.

, stating the first four

b Hence find a series expansion of

.

c Hence show that

.

22 The part of the curve between axis. The volume of revolution formed is

and

is rotated

.

Find the value of . 23 Consider two curves with equations

and

.

a Find the coordinates of the points of intersection of the two curves. The region enclosed by the curves is rotated through

about the -axis.

b Write down an integral expression for the volume of the solid generated. c Evaluate the volume, giving your answer to the nearest integer. 24 The region enclosed by

and

is to be labelled .

a Draw a sketch showing . b Find the volume when is rotated through

about the -axis.

c Hence find the volume when is rotated through

about the -axis.

about the -

9 Polar coordinates In this chapter you will learn how to: use polar coordinates to represent curves establish various properties of those curves convert between polar and Cartesian equations of a curve find the area enclosed by a polar curve, or between two curves.

Before you start… A Level Mathematics Student Book 2, Chapter 7

You should be able to use radians.

1 a Express

radians in

degrees. b State the exact value of .

A Level Mathematics Student Book 1, Chapter 10

You should be familiar with graphs of trigonometric functions.

2 a Find the set of values of , between and which .

, for

b State the greatest possible value of .

A Level Mathematics Student Book 2, Chapter 11

You should be able to integrate trigonometric functions.

3 Evaluate: a

b

What are polar coordinates? You are familiar with describing positions of points in the plane by using Cartesian coordinates, which represent the distance of a point from the - and -axes. But you are also familiar with bearings, which determine a direction in terms of an angle from a fixed line. If you know that a point lies on a certain bearing, you can describe its exact position by also specifying the distance from the origin. For example, this diagram shows that is from on a bearing of .

Polar coordinates use a similar idea: positions of points are described in terms of a direction and the distance from the origin. They can be used to describe curves that cannot easily be represented in Cartesian coordinates. In this chapter, you will learn about equations of curves such as these.

    

    

Because the distance from the origin explicitly features as a variable, polar coordinates are often used to describe quantities that vary with distance from a point, such as the strength of a gravitational field.

Section 1: Curves in polar coordinates Polar coordinates describe the position of a point by specifying its distance from the origin (also called the pole) and the angle relative to a fixed line (called the initial line). By convention the initial line is drawn horizontally, in the direction of the positive -axis, and the angle is measured anticlockwise. You write polar coordinates as

, where is the distance from the pole and is the angle. For example,

point in the diagram has polar coordinates

, point

and point

.

Notice that polar coordinates of a point are not uniquely defined. For example, you could also say that the polar coordinates of are or . Conventionally, is taken to be between and (i.e. ). WORKED EXAMPLE 9.1 Points

and have polar coordinates

and

. Find:

a the length b the area of the triangle

, where is the pole.

a

Polar coordinates give information about lengths and angles.

In triangle

,

.

You can use the cosine rule in triangle

.

So:

b

Use

.

In Cartesian coordinates, an equation of a curve gives a relationship between the and -coordinate of any point on the curve. Similarly, a polar equation of a curve is a relationship between and that holds for any point on the curve. WORKED EXAMPLE 9.2 a Make a table of values for the curve with polar equation b Hence sketch the curve.

for

.

a

Use to calculator for various values of

b

Plot the points and join them up.

There might be values of for which is not defined. This happens when the expression for has a negative value; is a distance, so it must be non-negative. An effective way to identify such values is to sketch the graph of against .

Tip If you use a graphical calculator or graphing software to plot polar curves, you will find that some of them allow negative values of , showing parts of a curve where we claim the curve is not defined. In this course, we require that takes non-negative values.

WORKED EXAMPLE 9.3 Sketch the curve with equation

for

.

Finding disallowed values of :

The curve is not defined when to see where it is negative.

is not defined for: or

. Sketch the graph of

There should be no points in the sections

and

.

You can also see from the graph that increases for then decreases for between and

.

, then repeats the same values

,

EXERCISE 9A 1

Plot the points with the given polar coordinates. a

i ii

b

i ii

c

i ii

2

For points and with given polar coordinates, find the distance . a

b

c

3

i

,

ii

,

i

,

ii

,

i

,

ii

,

For each equation, make a table of values (for a

and the area of the triangle

) and sketch the curve.

i ii

b

i ii

c

i ii

4

Shade the region described by each inequality. They are given in polar coordinates.

a b c 5

A curve has polar equation

.

a State the values of for which the curve is not defined. b Hence sketch the curve. 6

A curve has polar equation

.

a Find the set of values of for which is not defined. b Show that the points and , with polar coordinates c Sketch the curve. d Find the exact length of

.

and

lie on the curve.

Section 2: Some features of polar curves When sketching curves in Cartesian coordinates you normally mark the axis intercepts, maximum and minimum points. For polar curves, there are similar features that you can deduce from the equation.

Minimum and maximum values of Since is a function of , you can use differentiation to find its minimum and maximum values.

Key point 9.1 The minimum and maximum values of occur where

.

WORKED EXAMPLE 9.4 A curve has polar equation

for

.

a Find the minimum and maximum values of , and the values of for which they occur. b Explain why there is a point on the curve corresponding to every value of . c Sketch the curve.

a

The maximum and minimum values of occur where

.

When

When

:

:

is a minimum at When : is maximum at When

so there . so there .

.

When

.

.

Hence the minimum value of occurs when . The maximum value of occurs when . b

When

is always positive for , so the curve exists for all .

The minimum value could occur at the end of the domain. You have already checked , so you just need to check .

For the curve to be defined, needs to be positive.

As increases from to , increases from to , which is just below ) and then decreases to

(when .

c

Polar equations often involve trigonometric functions. You might be able to use trigonometric graphs, rather than differentiation, to find the maximum and minimum values of . WORKED EXAMPLE 9.5 A curve has polar equation

for

.

a Find the largest and smallest values of . b Hence sketch the curve. a

Start by considering the minimum and maximum values of so

when

, and , when

and

:

,

.

The largest value is when The smallest value is when . b

Sketch the graph of

:

You can see that increases from to and decreases to again.

Tangents at the pole In Worked example 9.3 you saw a curve

that is only defined for certain values of . This

happens because we do not allow negative values of The value of

changes from positive to negative, or vice versa, when

. Each of

those values corresponds to a half-line, shown in red in the diagram. As the curve approaches each of the lines, gets closer to zero (so points on the curve get closer and closer to the pole). This means that each of the lines

is a tangent to the curve at the pole.

Key point 9.2 For a curve with polar equation on one side of the line.

, the line

is a tangent at the pole if

but

WORKED EXAMPLE 9.6 For the curve with polar equation curve.

, find the tangents at the pole and hence sketch the

Sketch the graph of to see which values of would produce negative values of .

The tangents at the pole are: .

The value of

passes through zero when and

.

Between the tangents where the curve is defined, the value of increases from to the maximum value of and then decreases back to .

EXERCISE 9B

EXERCISE 9B 1

For each curve, find the maximum and minimum possible value of , and the corresponding values of . Hence sketch the curve. (In all cases, .) a

i ii

b

i ii

2

Find the equations of the tangents at the pole for each curve. Hence sketch the curve. (In all cases, .) a

i ii

b

i ii

3

Consider the curve with polar equation

,

for

.

a Find the equations of the tangents at the pole. b State the set of values of for which the curve is not defined. c Hence sketch the curve. 4

a Sketch the curve with polar equation

,

.

b State the largest and smallest values of . 5

Consider the curve with polar equation

.

a Find the range of values of for which the curve exists. b Sketch the curve, labelling the tangents at the pole and indicating the points where has maximum value. 6

A curve has polar equation

.

a Find the largest and smallest value of and the values of at which they occur. b Hence sketch the graph. 7

a Sketch the graph of

for

.

b Sketch the curve with polar equation 8

.

a Find the smallest and largest values of

for

b Sketch the curve with the polar equation 9

Sketch the curve with equation

10 Sketch the curve with equation

, for for

for

.

.

. .

Section 3: Changing between polar and Cartesian coordinates You can use trigonometry to find the Cartesian coordinates of a point with given polar coordinates. Usually, the origin of the Cartesian coordinates is taken to be the pole, and the -axis to be the initial line.

Key point 9.3 A point with polar coordinates

has Cartesian coordinates

.

WORKED EXAMPLE 9.7 Points and have polar coordinates

and

.

a Show points and on the same diagram. b Find the Cartesian coordinates of and . a

The first coordinate is the distance from the origin and the second coordinate is the angle. .

b For :

Use

and

.

So the Cartesian coordinates of are . For :

So the Cartesian coordinates of are . To change from Cartesian to polar coordinates, consider the same diagram again. The value of is the distance from the origin, so . Remember that we require to be positive. You need to be a little careful when finding the angle. Since and , you can divide the two equations to get

. However, there are two values

need to consider the position of the point to decide which one is correct.

with the same value of

; you

Key point 9.4 For a point with Cartesian coordinates

the polar coordinates satisfy:

Rewind This should remind you of the modulus and argument of a complex number – see Pure Core Student Book 1, Chapter 4, and Chapter 2 of this book.

WORKED EXAMPLE 9.8 Find the polar coordinates of the points

and

.

Start by plotting the points to see which angle to use.

For :

Use

.

Find

, then add to get the two possible values

between and Hence the polar coordinates of are .

.

The angle for is smaller than .

For :

Hence the polar coordinates of are

The angle for is greater than .

.

You can now convert equations of curves between polar and Cartesian forms. WORKED EXAMPLE 9.9 Find the Cartesian equation of the curve with polar equation

.

Use

Hence:

and

Substitute for

.

in the equation of the curve.

Simplify if possible. In this case, multiply both sides by the ‘square root’ term.

WORKED EXAMPLE 9.10 Find the polar equation of the curve Use

. and

You also know that

. .

EXERCISE 9C 1

Each point is given in polar coordinates. Find the Cartesian coordinates. a

i ii

b

i ii

c

i ii

2

Each point is given in Cartesian coordinates. Find the polar coordinates. Take a

i ii

b

i ii

c

i ii

3

Find the polar equation of each curve. a

i ii

b

i ii

c

i ii

.

4

Find the Cartesian equation of each curve. a

i ii

b

i ii

c

i ii

5

A curve has polar equation the curve.

. Point , with polar coordinates

where

, lies on

a Find the value of . b Find Cartesian coordinates of . c Find the Cartesian equation of the curve. 6

Find the polar equation of the circle

7

Find the Cartesian equation of the curve with polar equation

8

A curve has polar equation

, giving your answer in the form

.

Show that the Cartesian equation of the curve can be written as 9

Find the Cartesian equation of the curve with polar equation: a b

.

.

.

.

Section 4: Area enclosed by a polar curve Finding the area bounded by a polar curve is similar to finding the area bounded by a Cartesian curve, except that rather than being the area between the curve, the -axis and vertical lines and , now it is the area between the curve, the pole and lines from the pole

and

.

Key point 9.5 The area enclosed between a polar curve and the half-lines

and

is

.

PROOF 5 Consider a curve

, where

and

. You can split the region into small sectors of angle and area .

The polar coordinates of the point are and the polar coordinates of the nearby point are .

The area of each sector is approximately the same as the area of a sector of a circle with angle and radius :

The area of a sector of a circle is

The total area is approximately:

Summing all these sectors between gives the approximate total area.

.

and

The approximation becomes more and more accurate as the angle gets smaller. In the limit as integral.

WORKED EXAMPLE 9.11

the sum becomes an

In this question you must show detailed reasoning. The diagram shows the curve with polar equation

.

Find the area enclosed by the curve. Use the formula

.

Expand the brackets. To integrate

, use the .

identity:

EXERCISE 9D 1

Find the area enclosed between these polar curves and half-lines. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

Focus on … Some areas can only be calculated exactly using polar coordinates. Focus on ... Proof 2 looks at one important example. 2

a Write down the polar equation of a circle of radius with centre at the pole.

3

b Using your answer to part a, show that the area of the circle is

.

Find the exact value of the area enclosed between the curve

, the initial line and the half-line

, clearly showing all your working. 4

The diagram shows the curve with polar equation

Show that the exact area enclosed by the curve is 5

.

.

The diagram shows the curve with polar equation

.

Find the exact area enclosed by the curve, clearly showing all your working. 6

a Sketch the curve with polar equation

.

b Show that the area enclosed between the lines of . 7

and

, where

a Sketch the curve with polar equation

, is independent

.

b Find the total area of the region enclosed by , clearly showing all your working. 8

The curve has polar equation

, for

and

.

a Sketch , giving the equations of any tangents at the pole. b Find, in terms of , the total area of the region enclosed by , clearly showing all your working. 9

The area of the region enclosed between the curve with polar equation and the half-line

is

, the initial line

.

Find the value of the positive constant . 10 The diagram shows the curve with polar equation Show that the area of the region enclosed by the curve is

. .

Section 5: Area between two curves To find the area enclosed between two polar curves, find the intersection points of the curves and calculate the part of the area bounded by each curve separately. WORKED EXAMPLE 9.12 Two curves have polar equations:

for

.

a Find the polar coordinates of the points of intersection of

and

.

b Find the exact value of the area of the finite region enclosed between a The curves intersect where

and

.

Equate the equations of the two curves.

Solve for . There are two values between and . From

: when

The points of intersection are

Find the corresponding values of .

. and

. b Sketching the curves:

It is a good idea to sketch the curves to see where the required region is.

The required region above the initial line is made up of two parts: one bounded by between

and

between

and

and one bounded by .

By symmetry about the initial line, the full area is double this. Expand and use

.

Tip In more complicated questions where the region is between two curves, remember that in polar coordinates you are finding the area of a sector bounded by the curve and two half-lines from the pole; not a region bounded by two vertical lines as in Cartesian coordinates.

EXERCISE 9E 1

The diagram shows the curve with polar equation both defined for

and the line with polar equation

.

The line intersects the curve at the point and the initial line at the point . a Find the polar coordinates of . b

i Find the exact area of the triangle

.

ii Hence show that the area of the shaded region is 2

The diagram shows the curve with polar equation

. .

,

Find, in terms of , the exact area of the shaded region . 3

The diagram shows the curves with polar equations

and

for

.

a Find the polar coordinates of the points of intersection of the two curves. b Find the exact area of the shaded region enclosed within both curves. 4

The diagram shows the curves with polar equations

and

.

a Find the polar coordinates of the points of intersection of the two curves. b Find the exact area enclosed inside both curves which is shaded on the diagram. 5

The diagram shows the curves .

and

with polar equations

and

a Find the polar coordinates of the points of intersection of the two curves. b Show that the area of the shaded region is 6

.

     On the same axes, sketch the curves with polar equations a

and

for

.

b

Show that the exact value of the area inside

but outside

is

.

Checklist of learning and understanding Polar coordinates describe the position of a point in terms of its distance from the pole and the angle measured anticlockwise from the initial line. The connection between polar and Cartesian coordinates is: , For a curve with equation given in polar coordinates: cannot be negative, so not all values of are possible there might be one or more tangents at the pole, given by the values of for which The area enclosed between a polar curve and the half-lines

and

is

. .

To find the area enclosed between two polar curves, find the intersection points of the curves and calculate the part of the area bounded by each curve separately.

Mixed practice 9 1

Find the greatest distance from the pole of any point on the curve

,

. 2

Find the polar equation of the curve

3

Points and have polar coordinates

. and

. Find:

a the distance b the area of the triangle

.

4

Sketch the curve with polar equation

5

Find the area of the region enclosed between the initial line and the curve with polar equation ,

6

.

.

a Sketch the curve with polar equation b

7

,

,

.

Show that the area bounded by the curve and the initial line is

a Sketch the curve with polar equation

.

,

b Find the exact area enclosed by the curve, clearly showing your working. 8

The Cartesian equation of a curve is

,

a Show that the polar equation of the curve is

.

.

b Find the equation of any tangents at the pole. c Hence sketch the curve for

.

d Show that the area enclosed by the curve is 9

Find a Cartesian equation for the curve

10 A curve has polar equation

. ,

,

. .

a Find the equations of the tangents at the pole. b Sketch the curve. c Find the Cartesian equation of the curve. 11 a Sketch the curve with polar equation

,

b Find the Cartesian coordinates of the point that is furthest away from the pole. 12 A curve has polar equation

,

.

a Find the equations of the tangents at the pole. b State the polar coordinates of the points at greatest distance from the pole. c Hence sketch the graph. 13 A curve is defined by the polar equation

for

.

a Sketch the curve. b Find the Cartesian coordinates of the point where the curve intersects the line 14 Sketch the curve with polar equation

,

.

. Indicate the equations of the

tangents at the pole, and give the polar coordinates of the point where the curve crosses the initial line. 15 The diagram shows the curve with polar equation

,

.

Find the exact value of the shaded area, clearly showing your working. 16 The diagram shows the curve with polar equation

a Show that the area of the region bounded by is The circle

,

.

.

intersects at the points and .

b Find the polar coordinates of and . c Find the area enclosed between the circle and . 17 The diagram shows the curve with polar equation

a i

Show that

,

.

ii Hence find the equations of the tangents at the pole. b Show that the area enclosed in the large loop is 18 The equation of a curve is i

.

Find the polar equation of this curve in the form

ii Sketch the curve.

.

.

.

iii The line

divides the region enclosed by the curve into two parts. Find the ratio

of the two areas. © OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2013 19

The diagram shows two curves, The polar equation of is curves,

and , which intersect at the pole and at the point . and the polar equation of is . For both

.

The value of at is . i

Show that

.

ii Show that the area of the region common to

and

, shaded in the diagram, is

. © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, January 2012

10 Differential equations In this chapter you will learn how to: understand and use the language associated with differential equations solve differential equations of the form solve differential equations of the form

use substitutions to turn differential equations into the required form.

Before you start… A Level Mathematics Student Book 2, Chapter 13

You should know how to solve separable differential equations.

1 Solve

A Level Mathematics Student Book 2, Chapter 10

You should know how to differentiate expressions, including using the product and chain rules.

2 Differentiate with respect to .

A Level Mathematics Student Book 2, Chapter 11

You should be able to use various methods to integrate expressions.

3 Integrate with respect to .

.

Introduction In many academic areas such as Physics and Economics it is important to describe situations in terms of rates of change. This produces differential equations. In this chapter you will extend the types of differential equations you can solve. You will then see in Chapter 11 how these methods can be applied in many real-life situations.

Rewind You met differential equations in A Level Mathematics Student Book 2, Chapter 13.

Did you know? If you read other books about differential equations you will see that the equations covered in this book are referred to as ordinary differential equations (ODEs). This is in contrast to another type of differential equation called partial differential equations (PDEs) which use a different type of differentiation.

Section 1: Terminology of differential equations Differential equations have an independent variable (which is the variable on the bottom of the derivatives) and at least one dependent variable (which is the variable on the top of the derivatives). For example, in the equation

the independent variable is and the dependent variable is .

There are many different types of differential equation. To decide which technique to use when solving differential equations you need to be able to categorise them. The order of a differential equation is the largest number of times the dependent variable is differentiated. For example:

is a third-order differential equation.

A linear differential equation is one in which the dependent variable ( in these examples) only appears to the power of (or not at all) in any expression. For example: linear differential equation, but any differential equation involving

,

is a or even

is non-linear.

A homogeneous differential equation is one where every term involves the dependent variable. For example:

is homogeneous, but

is a non-homogeneous differential

equation. Every non-homogeneous differential equation has a homogeneous differential equation associated with it, formed by removing all the terms not involving the dependent variable.

Tip In some books homogeneous is used to refer to a different property of differential equations. The definition given here is the only one relevant to this course.

WORKED EXAMPLE 10.1 Consider the differential equation

.

a State the order of the differential equation. b Is the equation linear? Explain your answer. c Explain why the equation is non-homogeneous, and write down the associated homogeneous equation. a The equation is first order.

The highest derivative of in the equation is

,

which is a first derivative. b The equation is linear because

and

are both linear terms.

The only occurrences of the dependent variable are and

, and they are both linear.

(Note that it doesn’t matter that

is multiplied

by .) c The equation is non-homogeneous because the term does not contain . The associated homogeneous equation is

Non-homogeneous equations include terms which do not contain the dependent variable. Remove all terms without to get a homogeneous equation.

To solve a differential equation, you need to find as a function of (if is the dependent variable and is the independent variable). When solving a differential equation, because the process is effectively integration, there will be arbitrary constants involved. The solution containing all the arbitrary constants is called the general solution.

Key point 10.1 The general solution to an th-order differential equation has arbitrary constants. To fix the arbitrary constants you either use initial conditions (values of , boundary conditions (values of ,

, etc. at one value of ) or

, etc. at several values of ). You need one piece of information for

each arbitrary constant, and then you normally use simultaneous equation techniques to find the values of each constant. When the constants in the general solution have the values required to fit the conditions, the result is called the particular solution. WORKED EXAMPLE 10.2 a Show that

is the general solution of the differential equation .

b Find the particular solution that satisfies

and

when

.



a

Then: Substitute and

into the given

differential equation.

Hence the given function is the general solution.

The given function satisfies the differential equation for all values of and . Notice that it also contains two arbitrary constants, as stated in Key point 10.1.

b When

:

Substitute the given values of , and into the general solution.

Solve the simultaneous equations (you can use a calculator). The particular solution is

Non-homogeneous linear equations In general, homogeneous equations are easier to solve than non-homogeneous ones. Fortunately, it turns out that, for linear equations, a general solution of a non-homogeneous equation can be found by using the general solution of the associated homogeneous equation, called the complementary function.

Tip The complementary function will contain arbitrary constants.

Fast forward In Section 4 you will learn how to ‘guess’ the form of a particular integral in some specific cases. Once you have the complementary function, all you need is one solution of the full non-homogeneous equation. This is called a particular integral. Combining this with the complementary function gives the general solution of the full non-homogeneous equation. For example, the differential equation homogeneous. You can check that

from Worked example 10.2 is non satisfies this equation, so it can be used as a particular

integral. The associated homogeneous equation is ). You can check that

(obtained by removing the term without

satisfies this homogeneous equation for all values of and ; hence

this is the complementary function. The general solution of the non-homogeneous equation is then , which is the sum of the particular integral and the complementary function.

Tip Particular integrals are not unique. For example,

and

can all be taken as particular integrals for our equation. You may wish to investigate how using different particular integrals affects the values of the constants and for the particular solution in part b of Example 10.2.

Key point 10.2 For a linear differential equation, the general solution is given by where is the complementary function and is a particular integral.

Tip Don’t confuse particular integrals with the particular solution. A particular integral is any solution of a differential equation. The particular solution needs to satisfy given initial conditions. The proof of the result in Key point 10.2 is shown in Proof 6. It generalises the observations made in the discussion above. You do not need to learn (or even understand) this proof, but it might help to explain the result. It also shows you an example of a type of proof often seen in advanced mathematics. Any linear differential equation can be written as For example: if Proof 6 uses the fact that

then

is

where is called a linear differential operator. and

.

. This is because the derivative of a sum equals the sum

of the derivatives.

PROOF 6 If is the complementary function and is the particular integral of a linear differential equation , then

will be a solution to the differential equation.

The complementary function, by:

is defined



is the solution of the associated homogeneous equation. The particular integral is any function, , satisfying



is a solution of the full equation. Then if



,

Use the given fact about linear differential operators. Use properties So also satisfies

and

.



.

WORKED EXAMPLE 10.3 The differential equation

is satisfied for

.

a Find the complementary function. b A particular integral has the form Find the values of and .

.

c Hence find the particular solution with initial condition a The associated homogenous differential equation is

when

.

The given equation is non-homogeneous. The complementary function is the solution of the associated homogeneous equation, obtained by removing any terms which do not contain .

You need to be able to recognise that this differential equation is separable.

Write as

in order to use rules of logarithms.

Remember that the complementary function should contain one arbitrary constant (since the equation is first order). b If



, then .

Substituting: So, Comparing coefficients of

Frequently when finding the particular integral you look at the coefficients on both sides of the equation.

Comparing coefficients of the constant term and using the fact that

So the particular integral is . c The general solution is

The general solution is the sum of the complementary function and the particular integral.

. Using the initial condition, substituting in when

:

To find the particular solution, use the initial condition to find the value of the constant .

Therefore the particular solution is

.

EXERCISE 10A 1

Write an example of each type of differential equation. a

i Linear ii Non-linear

b

i Second order ii Third order

c

i Homogeneous ii Non-homogeneous

2

Classify each differential equation, stating its order, whether or not it is linear, and whether it is homogeneous or non-homogeneous. a

i ii

b

i ii

c

i ii

d

i ii

3

Given these solutions to differential equations and the initial or boundary conditions, find the particular solution to each differential equation. a

b

;

i

when

ii

;

i

;

when ;

ii c

;

i

;

i

when

;

when

when and

;

ii 4

when

;

ii d

when

;

when

when and

The differential equation

when is defined for

.

a Find the complementary function. b A particular integral has the form

. Find the values of and .

c Hence find the particular solution with initial condition 5

The differential equation

when

is defined for

.

a Find the complementary function. b A particular integral exists of the form

. Find the values of and .

c Hence find the general solution to the differential equation

defined for

. 6

The differential equation

is defined for

.

a Show that this differential equation has a particular integral of the form

, stating the

values of and . b Solve the associated homogenous differential equation. c Explain why the general solution cannot be written as the sum of the complementary function and the particular integral. d By considering the expression

, or otherwise, find the general solution of the differential

equation. 7

The differential equation a Find the complementary function.

is defined for all .

b Find a particular integral. c Hence find the general solution.

Section 2: The integrating factor method for first order equations You already have the necessary tools to solve a differential equation such as:

because the left-hand side is of a convenient form. Notice that

is the derivative of

and that

is the

derivative of , which means you have an expression that has resulted from the differentiation of a product using the product rule. Therefore you can write the equation equivalently as

Now, integrating both sides and rearranging:

When faced with a differential equation like this where you cannot separate the variables, it will not often be the case that the left-hand side is quite so convenient. However, this method does suggest a way forward in such cases. Consider, for example, the equation throughout by . This is

. This can be rewritten as

by dividing

.

Consider, in general, a similar linear first order differential equation:

where

and

are just functions of . Note that if there is a function in front of

, you can divide

through the equation by that function to get it in this form. To make the left-hand side the derivative of a product as before, you can multiply through the equation by a function

:

and then notice that if is chosen such that required form. From here you can proceed exactly as before:

The only remaining question is to decide on the function You need

you have the left-hand side in the

to make this work.

This function

is known as the integrating factor.

Tip When calculating

you do not need to include the

. It turns out that it does not

matter whether there is a constant, as it would cancel later in the process.

Key point 10.3 Given a first order linear differential equation

multiply through by the integrating factor: then the general solution will be: .

WORKED EXAMPLE 10.4 Solve the differential equation

where

when

. You can’t write the LHS as the derivative of a product: always check for this first. (Note that if the LHS had been

, i.e.

without the , you could have written it as .) Therefore, start by dividing through by to get the equation in the correct form for applying the integrating factor.

Find the integrating factor sure not to miss the sign on

So

Now multiply through by LHS is of the form

, making .

and check that the

You can now integrate both sides. Since



:

Finally, you need to find the constant and rearrange into the form .

You can transform some differential equations into the required form by using a substitution. In an examination, you would generally be given the required substitution. WORKED EXAMPLE 10.5 Show that the substitution

transforms the differential equation

into a linear differential equation. Hence find the general solution of the given differential equation. If

then

.



Substituting this turns the given differential equation into:

which is a linear differential equation. The integrating factor is

Use Key point 10.3.

. Use Key point 10.3 again. Notice that the Therefore

EXERCISE 10B

or

is in the brackets.

You need to give the solution in terms of and . If you can easily rewrite it to make the subject, this is the conventional thing to do.

EXERCISE 10B 1

Use an integrating factor to find the general solution to each linear differential equation. a

i ii

b

i ii

c

i ii

2 3

Find the particular solution of the linear differential equation

which has

Find the general solution to the differential equation

.

4

Find the general solution of the differential equation

5

Find the particular solution of the linear differential equation point

when

.

. that passes through the

.

6

Given that

7

Prove that when finding

and that

when

, find in terms of .

in Key point 10.3, it does not matter whether or not the constant

of integration is included. 8

Find the general solution to the differential equation

9

a Use the substitution

to transform the equation

into a linear differential equation in and . b Solve the resulting equation, writing in terms of . c Find the particular solution to the original equation that has 10 a Using the substitution

given that when

,

when

.

, or otherwise, solve the equation:

. Give your answer in the form

b Use another substitution to find the general solution to the equation

. .

Section 3: Homogeneous second order linear differential equations with constant coefficients A differential equation of the form

is called a homogeneous second order linear differential equation with constant coefficients. To find the solution to this type of differential equation, you need to create an auxiliary equation.

Key point 10.4 The auxiliary equation of the differential equation

is .

Tip How the auxiliary equation arises is shown in Proof 7. Solving the auxiliary equation gives you important information about the solution of the differential equation, as set out in Proofs 7, 8 and 9. However, in most instances you will be able to just quote the results, which will be summarised in Key point 10.5. If the auxiliary equation has real, distinct roots, .

and

, then the solution to the differential equation is

PROOF 7 If

and

then

, where

If

and

are the roots of the equation.

One possible solution to the differential equation could occur if you had a function whose derivative and second derivative are proportional to the original function, allowing everything to ‘cancel’ and result in zero. is one example of a function which has this property (although you will see later that there are others).

Then

and . Substituting these into the differential equation:



Taking out a factor of Since If then the auxiliary equation will



.

can never be zero. This is the auxiliary equation.

have two real solutions. Call these

and

.

Therefore two possible solutions to the differential equation are and

. will

This can be proved in a similar way to Proof 6.

also be a solution, since any linear combination of solutions is also a solution. This is a solution with two arbitrary constants,

Use Key point 10.1 to justify that your ‘guess’ gives the complete solution.

therefore it is the general solution. If the roots of the auxiliary equation are complex, you could still write the solution as However, you can then rewrite this in terms of trigonometric functions.

Rewind You learnt in Chapter 2, Section 2, that

.

PROOF 8 If

with

and

, then can be written in the form

.

Substituting in the given information:

Take out a factor of

.

Rewrite the complex exponential into polar form. Separate out the cosine and sine terms. with



and

If the auxiliary equation has equal roots,

, the general solution becomes

, which contains only one arbitrary constant. According to Key point 10.1, you need to find another complementary function. Proof 9 shows you how to do this.

PROOF 9 If If then

and

, then

is a possible solution for a suitably chosen .

.

and . Substituting into the differential equation:

Divide through by ) and tidy up.

Comparing coefficients:

Checking

in the second equation:

You are given that

So

(which can never be

.



is a solution to the differential

equation in this case. To get to the general solution a linear combination of the two possible values is required, leading to . When solving differential equations normally you will just be able to write down the solution without going through the Proofs 7, 8 and 9.

Key point 10.5 Solution to auxiliary equation Two distinct roots,

General solution to differential equation

and

Repeated root, Complex roots,

WORKED EXAMPLE 10.6 Solve the differential equation given that

and

when

The auxiliary equation is

. .



This has roots and , so the general solution to the differential

Since the roots are real and distinct you can write the solution to the differential equation in exponential form

equation is

using Key point 10.5.

.

You need to differentiate the expression to make use of the initial conditions.

Using the initial conditions, when

Solving gives

and

So the particular solution is .

EXERCISE 10C

:



.

You can use technology to solve these types of simultaneous equations.

EXERCISE 10C 1

a

Write the auxiliary equation associated with the differential equation

.

b Hence find the general solution of the differential equation. 2

a

Write the auxiliary equation associated with the differential equation

.

b Hence find the general solution of the differential equation. 3

a Write the auxiliary equation associated with the differential equation

.

b Hence find the general solution of the differential equation.

Tip Remember that is another notation for

4

a Find the general solution to the differential equation b Find the particular solution that satisfies

5

and

when

.

a

.

and

.

Find the general solution to the differential equation

b Find the particular solution that satisfies 7

.

a Find the general solution to the differential equation b Find the particular solution that satisfies

6

.

and

. .

a Find the general solution to the differential equation b Given that

and

, find

.

.

Tip Remember that means 8

a

.

Find the general solution to the differential equation

b Find the particular solution that satisfies 9

and

. , writing your answer in terms of .

Find the general solution to the differential equation

.

10 Find the general solution to the differential equation 11 By using the trial function

.

, find the general solution to the differential equation

. 12 a

Use the substitution

to turn the differential equation

into a

second order differential equation with constant coefficients involving and b Solve the differential equation to find as a function of . c Hence solve the original differential equation given that when

,

and

.

Section 4: Non-homogeneous second order linear differential equations with constant coefficients The second order differential equations you need to solve can all be written in the form . To solve these differential equations you use the method in Key point 10.2. You first of all solve the associated homogeneous equation to find a complementary function (using Key point 10.5) and then find a particular integral. The form of the particular integral will depend on . Key point 10.6 gives trial functions for some common situations. This trial function needs to be substituted into the differential equation to find the unknown constants.

Key point 10.6 Trial function

Polynomial

General polynomial of the same order

Tip You need to learn these forms of trial functions. If a different trial function is needed, it will be given in the question.

WORKED EXAMPLE 10.7 Find the general solution to the differential equation . The associated homogenous equation is

First solve the associated homogenous equation to find the complementary function.

. This has auxiliary equation

which has roots

and

.

Therefore, the complementary function is . The trial function associated with . Differentiating twice:

is

You can find the particular integral in two stages. Notice that the coefficient of in the power of the trial function mirrors the original function.

. Substituting into the left-hand side of the differential equation and comparing to the exponential part of the right-hand side:

The trial function associated with

is

Notice that although the expression on the right

.

only involves a term in , you need the general linear expression in the trial function.

Differentiating twice:

. Substituting these into the left-hand side of



the differential equation and comparing to the linear part of the right-hand side:

Comparing the coefficient of :

Compare coefficients to find and .

Comparing the constant term:

You can use the fact that

The particular integral is

to solve for .

.

The general solution is .

The general solution is the sum of the complementary function and the particular integral, from Key point 10.2.

Sometimes the trial function given in Key point 10.6 already appears as a part of the complementary function. In that case, you need to modify the particular integral.

Key point 10.7 If your trial function is already part of the complementary function, try multiplying the trial function by .

WORKED EXAMPLE 10.8 Find the general solution to the differential equation . The associated homogenous equation is

.

This has auxiliary equation

First solve the associated homogenous equation to find the complementary function.

which has roots and

.

Therefore, the complementary function is .

From Key point 10.5.

The trial function associated with is .

Although the right-hand side of the equation involves only , you need to include both and in the trial function.

This already appears in the complementary function, so try .

You need to adjust the trial function according to Key point 10.7.

Differentiating twice:

You need to use the product rule to differentiate.

Substituting into the left-hand side of

Notice that the terms containing

the differential equation:

cancel. This will always happen in the situation described in Key point 10.7.

The particular integral is

.

The general solution is the sum of the complementary function and the particular integral, from Key point 10.2.

EXERCISE 10D

For the differential equation

:

a find the complementary function b find the particular integral c hence write down the general solution. 2

For the differential equation

:

a find the complementary function b find the particular integral c hence write down the general solution. 3

For the differential equation

:

a find the complementary function b find the particular integral c hence find the general solution. 4

all

.

The general solution is

1

and

For the differential equation a find the complementary function b hence find the general solution

:

c find the particular solution given that 5

and

when

For the differential equation

.

, find:

a the general solution b the particular solution given that 6

and

For the differential equation

when

.

, find:

a the general solution b the particular solution given that 7

and

when

For the differential equation

.

:

a find the complementary function b show that there is a particular integral of the form c hence find the general solution d find the particular solution given that 8

The function

and

when

.

satisfies the differential equation

with boundary conditions

. Given that there is a particular integral of the form solution of the differential equation. 9

, find the particular

Find the general solution of the differential equation

10 For the differential equation

, find

a the general solution b the particular solution which satisfies

and

when

.

Checklist of learning and understanding The general solution to an th-order differential equation has arbitrary constants. For a linear differential equation, the general solution is given by complementary function and is a particular integral.

, where is the

The complementary function is the solution of the associated homogeneous equation, obtained from the original equation by removing the terms that do not contain . Given a first order linear differential equation integrating factor

. The solution will be

, multiply through by the .

The auxiliary equation to the homogeneous differential equation . The solution to the auxiliary equation gives the form of the general solution of the homogeneous equation: Solution to auxiliary equation Two distinct roots,

and

General solution to differential equation

is

Two distinct roots,

and

Repeated root, Complex roots, The form of the particular integral for the homogeneous differential equation depends on

:

Trial function

Polynomial

General polynomial of the same order

If the trial function is already part of the complementary function, multiply the trial function by .

Mixed practice 10 1

Find the complementary function of the differential equation

2

What is the integrating factor for the differential equation

3

a Solve the quadratic equation b

4

. ?

.

Hence write down the general solution of the differential equation

.

a Show that the integrating factor for the differential equation

is

.

b Hence find the general solution of the differential equation. 5

The variables and satisfy the differential equation . i Find the complementary function. ii Hence, or otherwise, find the general solution. iii Find the approximate range of values of when is large and positive. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2011

6

The differential equation

is defined for all .

a By considering the associated homogeneous differential equation, find the complementary function. b Show that a function of the form

forms a particular integral, and find the value of .

c Hence write down the general solution of the differential equation. d Find the particular solution with initial conditions 7

a Find the value of the constant for which equation

and

.

is a particular integral of the differential

.

b Hence find the general solution of the differential equation. 8

The differential equation

is defined for all .

a By considering the associated homogeneous differential equation, find the complementary function. b Show that a function of the form and .

forms a particular integral, and find the values of

c Hence write down the general solution of the differential equation. d Find the particular solution with initial conditions 9

Find the particular solution of the differential equation

and

when

.

for which

when

, giving in terms of .

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2015 10 Solve the differential equation

subject to the conditions

when

.

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2014 11 Solve the differential equation

for in terms of , given that

when

. © OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2013 12 The variables and satisfy the differential equation . i Find the complementary function of the differential equation. ii Given that there is a particular integral of the form iii Find the solution of the equation for which

, find the constant .

and

when

.

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2012 13 i Find the general solution of the differential equation

ii Find the particular solution for which

and

when

.

iii Write down the function to which approximates when is large and positive. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, January 2011 14 a Find the general solution of b Show that the substitution into c

for

.

transforms the differential equation

.

Hence find the general solution of the differential equation

for

15 a Find the general solution to the differential equation b Find the particular solution with

.

.

16 Find the general solution of the differential equation

.

.

11 Applications of differential equations In this chapter you will learn how to: use differential equations in modelling, in kinematics and in other contexts solve the equation for simple harmonic motion and relate the solution to the motion model damped oscillations using second order differential equations and interpret their solution solve coupled first order differential equations and use them to model situations with two dependent variables.

Before you start… Chapter 10

You should know how to solve second order differential

1 Find the general solution to

equations. A Level Mathematics Student Book 1, Chapter 21

You should be able to use Newton’s second law.

2 A falling object of mass is subjected to a constant air resistance of . Find the acceleration of the object if .

Real world modelling In reality, nearly everything of interest – be it the effect of a medicine or the price of a share – changes over time. The tool that mathematicians use to model these situations is a differential equation. In this chapter you will look at some common situations modelled by differential equations and how you can use the methods from Chapter 10 to solve them and interpret their solutions in context.

Section 1: Forming differential equations When modelling real-life situations, it is often the case that the description can be interpreted in terms of differential equations. You have already met many examples of setting up differential equations in A Level Mathematics Student Book 2, Chapter 13. In this section you will see further types of situations where differential equations arise, but this time you will often need to use methods from Chapter 10 to solve them. You will also look at the type of assumptions which are made when writing these differential equations.

Rewind See A Level Mathematics Student Book 1, Chapter 21, for a reminder of using

, and A

Level Mathematics Student Book 2, Chapter 13, for its use in setting up differential equations.

WORKED EXAMPLE 11.1 A car, of mass , is moving along a straight horizontal road. At time seconds, the car has speed . The only force acting is a resistance, which is modelled as being proportional to newtons. Initially the car is moving with speed

and deceleration

.

Find the time taken for the car to come to rest. The resistance force is

Using

and

for some constant .

: The resistance force is negative as the car is moving in the opposite direction to which this force acts.

Initially:

Use the initial conditions of . Separating the variables and integrating:

This is a standard arctan integral. When Use the initial condition again to find .

to find the value

When

: The car will come to rest when

.

Tip You can shortcut having to find and then setting by using definite instead of indefinite integration. After separating variables, integrate with limits and for , and and for :

Then will be the time taken for the car to stop.

EXERCISE 11A 1

A stone of mass falls vertically downwards under gravity. At time , the stone has speed , and it experiences air resistance of magnitude a Find an expression for

, where is a constant.

in terms of , and .

b The initial speed of the stone is . Find an expression for at time . 2

A car of mass

is moving along a straight horizontal road. At time seconds, the car has speed

. The magnitude of the resistance force, in newtons, is modelled by force acts on the car. The initial speed of the car is . a Show that, according to this model,

. No other horizontal

.

b A student performs an experiment to measure the speed of the car. She finds that the speed of the car after half a second is and the speed after two seconds is . Comment on the suitability of the model. 3

The current ( ) at time in a circuit with resistance (2 ), capacitance ( ) and inductance ( ) is modelled by the differential equation:

Solve to find as a function of and sketch the solution in each situation. a b c 4

The rate of immigration into a country is modelled as exponentially decreasing. The initial rate is per year. One year later the rate is per year. a Write a differential equation for the population ( ), assuming that changes in the population are due only to immigration. b Given that the initial population is

million, find the long-term population predicted by the model.

c The model is refined by adding the term represents.

to the rate of change. Suggest what this term

5

A chicken is to be cooked and is placed into an oven. The temperature of the oven, , follows the rule , where is the time in minutes after the chicken is put into the oven. The rate of increase of the temperature of the chicken ( ) is modelled as proportional to the difference between the chicken’s temperature and the oven’s temperature. a Write a differential equation for the temperature of the chicken. b If the temperature of the chicken is originally

and increasing at a rate of

, find the

particular solution of the differential equation. c Find an estimate of the chicken’s temperature after appropriate degree of accuracy.

minutes, giving your answer to an

d Describe one way in which the model is a simplification of the chicken’s temperature. 6

A school has students. The rate of spread of a rumour in a school is thought to be proportional to both the number of students who know the rumour ( ) and the number who do not know the rumour. a Write this information as a differential equation. b Find the number of students who know the rumour when the rumour is spreading fastest. c Write down two assumptions that are being made in this situation.

7

A bacterium is modelled as a sphere. According to one biological model the volume of the bacterium ( ) follows this differential equation:

a Explain the biological significance of the b By using the substitution

term.

, solve the differential equation given that initially

.

c Sketch the solution and hence find the long-term volume of the bacterium.

Did you know? The model in question 7 is called Von Bertalanffy growth. It is very important in mathematical biology.

Section 2: Simple harmonic motion In Chapter 10, you saw that some differential equations have solutions involving sines and cosines. These describe oscillating behaviour. The differential equation which has pure sinusoidal behaviour is called simple harmonic motion. It occurs in a surprisingly wide range of physical situations.

Tip Simple harmonic motion is often abbreviated to SHM.

Key point 11.1 The differential equation for simple harmonic motion is

Tip Using the dot notation to represent differentiation with respect to time, you can also write this equation as . In the equation in Key point 11.1, represents the displacement of an object from its equilibrium position (the position where the acceleration of the object is zero). To solve this differential equation you find the auxiliary equation:

This has solutions

, which lead to the general solution to the differential equation.

Key point 11.2 The general solution to the simple harmonic motion differential equation is

Rewind The general form of solutions for second order differential equations was given in Key point 10.5. There is some terminology that is useful in describing these solutions: The average position around which the object oscillates (corresponding to ) is the equilibrium position. The maximum distance from the equilibrium position is called the amplitude. The motion repeats itself after time, , which is called the period. The value is called the angular frequency. If initially the object is: at the equilibrium position, then the solution will be at the maximum displacement from the equilibrium position, then the solution will be In both of these cases the amplitude is given by .

.

You know from A Level Mathematics Student Book 2, Chapter 8 that in the form

can also be written

.

Key point 11.3 The general solution to the simple harmonic motion can also be written as

.

In the equation in Key point 11.3, the amplitude is and is called the phase shift. sin gives the initial displacement from the equilibrium position. Since

and

repeat when gets to

, one full period, , occurs when

.

Key point 11.4 The period, , of a particle moving with simple harmonic motion is . The object has its maximum speed as it is going through the equilibrium position, and it is instantaneously at rest when it reaches the maximum displacement, .

Key point 11.5 The relationship between velocity and displacement for a particle moving with simple harmonic motion is

PROOF 10 Prove that If then

when

. ,

Since you are only looking for a relationship between and , choose to start the time when the object moves through the equilibrium position.

. Use the fact that Use

from kinematics. .

Group the terms together to make a link with the expression for displacement.

A common context for simple harmonic motion is the situation with springs. One of the forces acting on the object is the tension, , in the spring. The tension is always directed back towards the equilibrium position. There is a standard model in physics for the magnitude of the tension, but in this course you will always be told the form to use.

Explore The tension in an elastic spring can be modelled using Hooke’s Law. You will learn about it if you study the Further Mechanics option.

If the only force acting on the object is the tension from the spring (for example, when the object is moving on a smooth horizontal table) then in the equilibrium position the spring is neither extended nor compressed (it is at its natural length). However, if there are additional forces, this spring might be extended or compressed at the equilibrium position.

WORKED EXAMPLE 11.2 A spring of natural length is attached to a hook in the ceiling. A particle of mass is attached to the other end of the spring. When the extension of the spring from its natural length is , the tension in the spring has magnitude . Use

, giving your final answers to an appropriate degree of accuracy.

a Show that, in the equilibrium position, the length of the spring is

.

b Show that, if the spring is displaced from the equilibrium, the particle will perform simple harmonic motion and find the time period of oscillations about this equilibrium. c The spring is stretched

from the equilibrium position and then released. Find the maximum

speed of the particle. a

Draw a diagram to help visualise the situation. Only tension and weight are acting on the mass. These must balance in equilibrium.

When the spring is in equilibrium:

To keep consistent units you need to use metres for . So the extension is equilibrium length is

and the .

b

If is the extension below the equilibrium position then: This is Newton’s second law vertically. The acceleration is . The total extension of the spring is magnitude of the tension is

so the .

If is positive when below the equilibrium position, then you need to use down as positive, so the resultant force is weight tension. Rearrange into the standard form for simple harmonic motion. So,

, where

.

This is the equation for simple harmonic motion. Then

Use (

from Key point 11.4.

)

c If the spring is stretched an additional , then the amplitude is . . This is maximised when the maximum speed is

It is important to be consistent with units, in this case working only in metres. Use

from Key point 11.5.

, so (

)

EXERCISE 11B 1

State the amplitude and the period of the simple harmonic motion described by each equation. Also state whether the particle is at rest, passing through the equilibrium position, or neither, when . a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

2

For each description of simple harmonic motion write an equation for in terms of (where is in metres and is in seconds). a

b

c

i Amplitude

, period seconds; at rest when

.

ii Amplitude

, period seconds; at rest when

.

i Amplitude

, period

ii Amplitude

, period

i Amplitude ii Amplitude

3

, period , period

seconds; in equilibrium when

.

seconds; at in equilibrium when seconds; in equilibrium when seconds; at rest when

. .

.

Each differential equation models a particle performing simple harmonic motion. Find the period of the motion. a

i ii

b

i ii

c

i ii

d

i ii

4

A particle performs simple harmonic motion with amplitude particle passes through the equilibrium position when .

and angular frequency

a Find the distance of the particle from the equilibrium position when

. The

.

b Find the maximum speed of the particle. 5

A small ball is attached to one end of an elastic spring. When the ball is released from rest from the equilibrium position and performs simple harmonic motion with angular frequency . a Find the displacement of the ball from the equilibrium position after seconds. b Find the time when the ball first passes through the equilibrium position, and the speed of the ball at this time.

6

A small ball attached to the end of a spring performs simple harmonic motion with amplitude angular frequency

and

.

a Find the maximum speed of the ball. b The ball is at rest when . Find the speed of the ball seconds later. Find also the magnitude of acceleration of the ball at this time. 7

A particle performs simple harmonic motion with amplitude equilibrium position is

. Its speed as it passes through the

.

a Find the angular frequency of the simple harmonic motion. b Find the speed of the particle when its displacement from the equilibrium position is 8

.

A particle performs simple harmonic motion with amplitude and angular frequency . The particle passes through the equilibrium position when with positive displacement immediately after . a Find the displacement of the particle when b Find the time when the particle is first

from the equilibrium position.

c Find the speed of the particle at that point. Is it moving towards or away from the equilibrium position? 9

A particle is attached to one end of an elastic spring. It is displaced from its equilibrium position and performs simple harmonic motion with amplitude position is the speed of the particle is

. When its displacement from the equilibrium .

a Find the angular frequency of the simple harmonic motion. b Hence find the distance from the equilibrium position when the speed of the particle is

.

10 A particle moves in a straight line between points and which are apart. The midpoint of is and the displacement of the particle from at time seconds is metres. The motion of the particle is described by the equation

. When

the particle is at

. a Write down the amplitude and the period of the simple harmonic motion. b Write down an expression for in terms of . c Point is between and , and

. Find the time when the particle first passes through

. d The mass of the particle is

. Find the magnitude of the force acting on the particle when it

passes through . 11 A cart of mass

is moving in a straight line with a speed of

when it hits a buffer which is

attached to a fixed wall by a light spring. At time seconds after the impact the compression of the spring is metres and the force in the spring is given by

newtons. Any other forces acting on the cart can be ignored.

a Show that the cart performs simple harmonic motion as long as it remains in contact with the buffer. b Find the maximum compression of the spring and the magnitude of the force acting on the cart at that point. c Find the time taken to reach the point of maximum compression. 12 A particle of mass

is attached to one end of a light spring and rests on a smooth horizontal table.

The string is horizontal and its other end is attached to a fixed wall. The particle is displaced away from the wall so that the extension of the spring is released. When the extension of the spring is the elastic force in the spring is a constant. All other forces on the particle can be ignored.

and then , where is

a Show that the particle performs simple harmonic motion and find, in terms of , the period of the motion. b Find the maximum speed and the maximum acceleration of the particle. c Find the extension of the spring at the moment when the speed of the particle equals half of its maximum speed. 13 A particle of mass

rests on a smooth horizontal table. The particle is attached to two light

springs and the other ends of the springs are attached to fixed points and , which are apart. The natural length of each spring is and the magnitude of the tension in each spring is given by , where is the extension of the spring. The particle is released from rest

from , the midpoint of

.

At time the displacement of the particle from is . a Find the magnitude of the resultant force on the particle at time . b Hence show that the particle performs simple harmonic motion. c Find an expression for in terms of . 14 A light spring is attached to a fixed point . A particle of mass

is attached to the other end of

the spring and hangs vertically below . When the extension of the spring is metres, the magnitude of the tension in the spring is a The particle hangs in equilibrium at point . Find the extension of the spring. The particle is displaced

downward from the equilibrium position.

b Write down the extension of the spring. Hence show that, as long as the resultant force on the particle is .

, the magnitude of

c Hence show that the particle performs simple harmonic motion and find the period of the motion.

.

Section 3: Damping and damped oscillations When objects are moving they are usually subjected to resistive forces such as air resistance or drag in water. There are several ways in which you can model this situation. One common model is to say that the drag force, , is proportional to the speed, acting in the opposite direction.

Key point 11.6 The drag force on an object moving with speed is given by where is a constant. If you add this to the standard equation for simple harmonic motion, the differential equation becomes:

where

is a positive constant. Damped oscillations result, or damped simple harmonic motion.

As you saw in Chapter 10, the solutions to this differential equation depend upon how many solutions there are to the auxiliary equation, , and each case is given a different name.

Key point 11.7

Overdamping

Critical damping

Underdamping

Did you know? In physical situations, such as the suspension of a car, critical damping is often desirable as it minimises vibrations without too much jerkiness.

WORKED EXAMPLE 11.3 A bob of mass is connected to a spring. In air the bob is found to follow simple harmonic motion with period seconds. The bob is then placed into oil where there is a drag force of magnitude . Find the value of which produces critical damping. If the period is , then the value of is given by

.

Therefore, in air the differential equation is:

Rearrange the formula in Key point 11.4 to express in terms of .

The force is given by:

In oil, there must be the additional drag force:

Divide through by . Since Critical damping occurs when

.

Critical damping occurs when the auxiliary equation has a repeated root. Only the positive solution to the equation is required, since from the context you need .

EXERCISE 11C 1

Each differential equation describes damped harmonic motion. In each case determine whether the damping is critical, underdamping or overdamping. a

i ii

b

i ii

c

i ii

d

i ii

Tip Remember that is an alternative notation for 2

, and is an alternative notation for

.

A particle performs damped oscillations described by the differential equation

.

Given that the damping is critical, find the value of . 3

A particle of mass is attached to one end of a light spring and rests on a horizontal table, with the spring horizontal. When the extension of the spring is metres the tension in the spring has magnitude . The resistance force acting on the particle has magnitude speed of the particle. a Show that the equation of motion of the particle is

, where

is the

.

b Given that the motion of the particle is critically damped, find the exact value of . c Name the type of damping that occurs when 4

A particle of mass

.

is attached to one end of a spring. When the displacement of from its

equilibrium position is metres, the magnitude of the tension in the spring is resistance force on has magnitude .

and the

a Write down a differential equation that models the motion of the particle. b Given that the motion of the particle is critically damped, express in terms of and . 5

A particle is attached to one end of a spring and moves under the action of a tension and a resistance force. The motion of the particle is described by the differential equation ,

and

, when

.

a Find an expression for in terms of . b Name the type of damping that occurs and sketch the graph of as a function of . 6

A particle of mass is attached to one end of a spring and moves in a straight line on a horizontal table. When the displacement of the particle from a fixed point is the tension in the spring has magnitude . The resistance force acting on the particle has magnitude 1.4 the speed of the particle. a

Show that the equation of motion for the particle is

When

the particle is at rest,

, where

is

.

from .

b Find an equation for in terms of . c Show that the particle never reaches . d Name the type of damping that occurs in this case. 7

A particle of mass moves in a straight line under the action of two forces. When the particle’s displacement from a fixed point is there is a force towards of magnitude as well as a resistance force of magnitude (where is the speed of the particle). When the particle is at rest

from .

a Show that the motion of the particle is described by the equation

.

b Given that the motion of the particle is critically damped, find the value of . c In this case, find an expression for in terms of . 8

A particle of mass

is suspended by a light elastic string, and the other end of the string is

attached to a fixed point vertically above . The natural length of the string is extension of the string is , the magnitude of the tension in the string is a

hangs in equilibrium at a point . Taking point.

. When the .

, find the extension of the string at this

is held at rest with the string at its natural length, and then released. When the speed of is the resistance force acting on has magnitude . b Show that the subsequent motion of can be modelled by the differential equation . c Name the type of damping that occurs in this case and find an expression for in terms of . d According to this model, what will the length of the string be in the long term? e Find the speed of when it passes through for the first time. 9

A particle is attached to one end of a light spring and performs damped oscillations described by the differential equation , where is the extension of the spring beyond the equilibrium position.

It is given that

.

a Determine the type of damping that occurs. At

,

and

.

b Find an expression for in terms of . c Show that the maximum extension of the spring is approximately

.

Section 4: Linear systems There are many situations where two variables are linked by coupled differential equations; for example, each variable might change with time, but the rate of change might depend of the value of the other variable. If both of these differential equations are linear and first order, then you can eliminate one of the variables to form a second order linear differential equation.

Rewind You learnt how to solve linear second order differential equations in Chapter 10, Sections 3 and 4.

WORKED EXAMPLE 11.4 In a population of foxes ( thousands) and rabbits ( thousands), the foxes have a birth rate and a death rate . The rabbits have a birth rate of and a death rate of . a Write this information in the form of a pair of differential equations. b Rewrite these differential equations as a second order differential equation for . c Solve this second order differential equation given that initially

and

.

d Hence find the solution for , given that the initial population of rabbits is five thousand. e What is the long-term population of foxes and rabbits?

The rate of change of the fox population will be (birth rate – death rate), and likewise for the rabbit population.

a

b Differentiating (1) with respect to :

Substituting in

from (2):

Rearranging (1):

Substituting this in:

So c Auxiliary equation is:

When

,

When

,

so so

. .

So the solution is

.

d Substituting the solution from part c into (2):

This is a first order linear differential equation, so you can write it in an appropriate form to use integrating factors.

The integrating factor is

When

,

so

, so:

.

e As gets very large gets very small, but gets very large. The population of foxes tends towards , but the population of rabbits grows without limit.

EXERCISE 11D 1

Write each pair of differential equations as a single second order equation for . Hence find the general solution for and in terms of . a

i ii

b

,

i ii

c

, ,

i ii

,

2

Find the general solution of

3

Find the general solution for and in terms of for this system of differential equations:

4

The variables and satisfy the differential equations

,

. When

,

and

.

Find expressions for and in terms of .

.

5

Consider the system of differential equations . a Find a second order differential equation for . When

,

and

.

b Find expressions for and in terms of . 6

Three identical cylindrical buckets, each with cross-sectional area , are placed vertically above each other. A hole is drilled in the base of each of the top two buckets so that water can flow from the top bucket to the middle one and from the middle one to the bottom one. For each of the top two buckets, when the height of water in the bucket is , the rate of flow of water out of the bucket is . Initially, the height of water in the top bucket is and the middle bucket is empty. Let be the height of water in the top bucket and be the height of water in the middle bucket at time . The time taken for water to fall between buckets can be ignored. a Show that

and write a differential equation for in terms of .

b Find an expression for in terms of and show that this model predicts that the second bucket never empties. c Find the maximum height of water in the second bucket.

7

A system contains sharks ( thousand) and fish ( million). The sharks have a birth rate given by and a death rate given by . The fish have a birth rate given by and a death rate given by . a Write this information in the form of a pair of differential equations. b Rewrite these differential equations as a second order differential equation for . c Solve this second order differential equation given that initially

and

.

d By writing the solution for in the form , where , find the time of the first peak in the shark population. Find the equivalent time at which the fish population first peaks.

e Describe the long-term behaviour of the two populations.

Focus on … You can find out how this model can be improved in Focus on ... Modelling 2. 8

A predator-prey system is of the form

,

Prove that the system will only oscillate if

. .

Checklist of learning and understanding The differential equation for simple harmonic motion is The general solution to the simple harmonic motion differential equation is , which can also be written as . Period of the solution: Speed, , is given by: Drag force is given by

.

If there is a drag force there can be underdamping, overdamping or critical damping depending on the number of solutions to the auxiliary equation. You can rewrite coupled pairs of linear first order differential equations as a second order differential equation in one variable.

Mixed practice 11 1

Find the period of the oscillations of a particle whose motion is modelled by the differential equation .

2

Find the value of which would result in critical damping in the system modelled by .

3

A particle of mass

is acted on by a force

newtons, where is the time

measured in seconds. a Write down a differential equation satisfied by the displacement, metres, of the particle from its initial position. b Given that the particle is initially at rest, find its displacement after seconds. 4

A ball is attached to one end of an elastic string and performs simple harmonic motion with amplitude and angular frequency . a Find the maximum speed of the ball. b Find the speed of the ball when its displacement from the equilibrium position is c The mass of the ball is during the motion.

5

A particle of mass from the origin is magnitude . a

.

. Find the magnitude of the maximum force acting on the ball

moves in a straight line so that, when the displacement of the particle , the force acting on the particle is directed towards the origin and has

Show that the displacement of the particle satisfies the differential equation

b Verify that, for some value of which you should state, and are constants, satisfies this differential equation. c The particle is initially at rest .

. , where

from the origin. Find the value of the constants and

d Hence find the maximum speed of the particle. 6

A particle of mass moves on a smooth horizontal plane. Initially it is projected with velocity from a fixed point towards another fixed point . At time after projection, is from and is moving with velocity , with the direction being positive. A force of acts on in the direction parallel to . i

Find an expression for in terms of .

ii Find the time when the velocity of is next

.

iii Find the times when subsequently passes through . iv Find the distance travels in the third second of its motion. © OCR A Level Mathematics, Unit 4730/01 Mechanics 3, June 2013 7

One end of a light elastic spring is attached to a fixed wall and a small ball is attached to the

other end. The ball rests on a smooth horizontal table. At

the ball is given the velocity of

away from the equilibrium position. When the

displacement of the ball from the equilibrium position is . a Given that the mass of the ball is

the force acting on the particle is

, show that the equation of motion of the ball is

. b Show that

satisfies the equation and find the value of the constants

and . c Find the time when the particle first returns to the equilibrium position. 8

The spread of a disease through a population is modelled using the following differential equations:

where is the number of uninfected individuals and is the number of infected individuals in the population at time months. Initially there are uninfected individuals and infected individual. According to this model, how long will it take for half the population to become infected? 9

Solve this system of differential equations:

given that

and

.

10 Two particles, and , each have a mass of action of a force newtons, modelled by seconds. moves under the action of a force

and are initially at rest. moves under the , where is the time measured in newtons, modelled by , where

metres is the displacement from the initial position. Which particle travels further in the first five seconds? 11 A particle moves with simple harmonic motion in a straight line between points and , which are

apart. The midpoint of

is .

The motion of the particle satisfies the differential equation

, where is the

displacement of the particle from . a Show that

satisfies the differential equation. Hence show that .

b Given that the particle takes four seconds to travel from to , find the value of . c Given that the mass of the particle is particle.

grams, find the maximum force acting on the

12 One end of a light spring is attached to a fixed wall. A ball of mass is attached to the other end of the spring and rests on a smooth horizontal table. The ball is displaced from the equilibrium position and then released. When the extension of the spring is magnitude of the tension in the spring is given by .

, the

a

Show that the equation of motion of the ball can be written as

.

b Find the maximum speed of the ball. 13 One end of a light spring is attached to a fixed wall. A particle of mass is attached to the other end of the spring and rests on a smooth horizontal table with the spring horizontal. is displaced from its equilibrium position and released from rest. When the displacement of from equilibrium is m the tension in the spring has magnitude . a Show that

.

b Show that and . Hence show that

satisfies the differential equation and find the values of .

c Find the maximum speed of the particle. 14 A cart of mass is attached to one end of a horizontal spring. When the extension of the spring is the tension in the spring is . Initially the cart is displaced from its equilibrium position along the axis of the spring. It is held at rest and then released. In a simple model the only force acting on the cart is the tension in the spring. a Find an expression for in terms of time. b How long does it take for the cart to reach the equilibrium position for the first time? In an improved model there is also a resistance force on the cart of magnitude is the speed of the cart.

, where

c Find an expression for in terms of for the second model. d Which model predicts the cart reaching the equilibrium position later? 15 Find the general solution of this system of differential equations:

16 Snakes and badgers are in competition for resources on a plain. There are no other types of animals on this plain. The populations snakes ( ) and badgers ( ) at time months are modelled by these differential equations:

Initially there are badgers and on the plain after three months.

snakes on the plain. Find the total number of animals

17 A particle starts from rest from a point and moves in a straight line with simple harmonic motion about a point . At time seconds after the motion starts the displacement of from is towards . The particle is next at rest when having travelled a distance of . i

Find the maximum velocity of .

ii Find the value of and the velocity of when

.

, at which ’s speed is the same as when iii Find the other values of , for Find also the corresponding values of .

.

© OCR A Level Mathematics, Unit 4730/01 Mechanics 3, June 2015

18

is a fixed point on a horizontal plane. A particle of mass is released from rest at and moves in a straight line on the plane. At time after release the only horizontal force acting on has magnitude

and . The force acts in the direction of ’s motion. ’s velocity at time i

Find an expression for in terms of , valid for greater when than it is when .

ii Sketch the ( , ) graph for

is

.

, and hence show that is three times

.

© OCR A Level Mathematics, Unit 4730 Mechanics 3, June 2010 19 A particle of mass is attached to two identical springs, each of natural length . The magnitude of the tension in each spring is , where is the extension of the spring. The other ends of the springs are attached to points and , which are smooth horizontal surface. The midpoint of

The particle is released from rest

apart on a

is .

from .

a Show that, when the displacement of the particle from is , the magnitude of the force acting on the particle is . b Hence show that the particle performs simple harmonic motion, and find the period of the motion. c Find the speed of the particle when it is

from .

20 In a strongman competition the competitors pull a truck (initially at rest) for winner is the person who pulls the truck furthest.

seconds. The

The truck has a mass of and is subject to a constant resistance force of . Brawny Bill initially pulls the truck with a force of , but by the end of the seconds he is pulling it with a force of . a State one assumption needed to model this force as a linear function of time. Comment on the appropriateness of this assumption. b Given the assumption from part a, write down a differential equation satisfied by the displacement, , of the truck from its initial position. c Solve your differential equation and hence find the displacement of the truck at the end of the seconds. d

Muscly Mike’s force, newtons, is modified as Determine who wins the competition.

at time seconds.

FOCUS ON … PROOF 2

Elements of area and Gaussian integrals You have seen that you can find areas under a curve using an integral which you thought of as summing up lots of little rectangles. In more advanced work it is useful to sum up lots of little elements of area instead and do a double sum over all coordinates. You can write this as:

where, in Cartesian coordinates,

.

Tip This Focus on ... section extends significantly beyond the scope of the specification, but it will be of interest to anyone wanting to go on to study Mathematics, Physics, Chemistry, Engineering or Theoretical Economics. The double integrals become

If you are looking for the area between a curve and the -axis, then the limits on are from to the area is:

so

which is the formula you are used to using. In Chapter 9 you found that the area between two half-lines in polar coordinates is given by

You can derive this using a similar method to the one shown for Cartesian coordinates. In the diagram the shaded area is approximately a rectangle with one dimension (using the formula for arc length in radians) and the other .

The area element in polar coordinates is therefore:

Question 1

Prove that the area bounded by the lines

,

and

is given by the formula

. These area elements have some lovely consequences, including allowing you to evaluate otherwise impossible integrals. Consider the integral

.

You cannot find the indefinite integral of using standard functions, however over this range you can evaluate the integral exactly. The in the integral is just a dummy variable. You could also write:

Rewind This is an example of an improper integral which you met in Chapter 8. Multiplying the two expressions together:

It turns out that you can combine these two integrals into one double integral:

But is just an element of area, so you could rewrite it as . You can recast the whole expression in terms of polar coordinates, noting that and that the limits represent the whole plane:

Questions 2

Complete the proof to evaluate .

3

Hence evaluate

where is a constant.

Rewind The integral in question 3 is of vital importance in working with the normal distribution, which you met in A Level Mathematics Student Book 2, Chapter 17.

FOCUS ON … PROBLEM SOLVING 2

Finding the shape of a hanging chain Consider this problem: A uniform chain is suspended from two fixed points at the same height and hangs under its own weight. Find the shape of the chain. The first step is to express the question in a mathematical form. If you set up the coordinate axes so that the two end points have the same -coordinate, then you can describe the shape of the chain by a function . The task is thens to find an expression for . It is clear that the shape of the chain will be symmetrical, with the lowest point halfway between the end points. Note that the position of the -axis is irrelevant, since the shape of the chain does not change if the end points are moved vertically. Next you need to introduce some parameters: what could the exact shape of the chain depend on? It seems reasonable to consider these factors: the mass of the chain the length of the chain the distance between the end points

.

As already noted, the height of the end points does not affect the shape of the chain. The shape of the chain is determined by the forces acing on it. As well as the mass of the chain, there is a tension force acting along the chain. At each point the tension acts along the tangent to the chain. So, if you can determine the direction of the tension at each point, you will know the gradient of the tangent, which is

.

Knowing the gradient will enable you to find the equation for in terms of . Consider the part of the chain between the lowest point and another point with a variable -coordinate. The forces acting on this part of the chain are shown in the diagram ( is the mass of this part of the chain). The force is fixed (it is the force from the left half of the chain on the right half of the chain), but changes with .

Resolving forces horizontally and vertically gives:

and therefore

. But, since the force is directed along the tangent to the curve,

the gradient of the curve at that point. Hence

equals

.

Rewind For a reminder of resolving forces, see A Level Mathematics Student Book 2, Chapter 21. In the expression for the gradient, and

are constants, but (the mass of this part of the chain)

depends on the -coordinate. If you can express in terms of , you can then integrate

to obtain your

required equation for the shape of the chain. Since the chain is uniform, the mass of a part of the chain is proportional to the length of that part. The whole chain has length and mass

, so

where is the length of the section of the chain

between -coordinates and . To find an expression for the length of the curve in terms of , consider a small section of the curve between points with coordinates and ), and denote the length of this small section The small section of the curve is close to a straight line, so

As

and

You now have the equation Differentiating this gives

, and so

, since , and so

.

.

You can now proceed to solve this differential equation.

Questions 1

Make a substitution

and show that .

Rewind You met integrals of this type in Chapter 7, Section 3. 2

Explain why the constant of integration is zero. Hence show that

3

Hence find an expression for in terms of . Explain why the constant of integration can be taken to be zero.

.

In the expression you found in Question 3, is a constant and and are fixed properties of the chain. However, you don’t yet know what is; you defined it as the magnitude of the tension acting at the lowest point of the chain. You should also notice that you have not yet used the condition that the end points of the chain are a distance apart. It seems reasonable that the tension in the chain will depend on how far apart the end points are.

Questions 4

Show that the length of the curve

between points with coordinates

and

is

.

Tip You have seen that the length of the curve satisfies

. Integrating this

expression, you find that

5

Use the fact that the total length of the chain is , and that the end points are at to show that

6

and

,

.

Use technology to show that this equation has a solution for

whenever

. Explain why this

condition always holds in this problem. In summary, you have found that a chain suspended freely from two fixed points hangs in the shape of a curve,

, where is a constant depending on the mass and the length of the chain and

the distance between the end points.

Did you know? The

curve is called a catenary, meaning ‘relating to a chain’.

FOCUS ON … MODELLING 2

The Lotka-Volterra model and phase planes During World War I the marine biologist Umberto D’Ancona noticed something puzzling about fish in the Adriatic Sea. Although they were being fished less (and so their natural death rate decreased), the numbers of small fish were actually decreasing while the numbers of predator fish were increasing. His father-in-law, Vito Volterra, applied the work of Alfred Lotka to try to explain this observation. Consider a population of a species of fish ( million) and sharks ( thousand). The natural net birth rate of the fish (i.e. the birth rate minus the death rate) is proportional to the number of fish, with constant of proportionality . There is also a death rate due to predation which is proportional to both the number of fish and the number of sharks, with constant of proportionality . This means that: . A similar differential equation governs the population of sharks:

where the term represents the growth in the shark population due to their predation on the fish and the term is the natural net death rate of the sharks.

Question 1

Describe some modelling assumptions that have been made in creating this model.

When analysing systems like this, it is often the case that solving the differential equation is less important than finding fixed points of the system (values of the population where there is no change in the population i.e. places where

and

.)

Questions 2

Find all fixed points of the differential equations in the Lotka-Volterra model. Which correspond to the biological equilibrium values if the populations do not go extinct?

3

When trawler fishing is reduced, the net birth rate of the fish will increase and the net death rate of the sharks will decrease. Use the Lokta-Volterra model to explain this.

A common way to visualise these systems of equations is to use a phase plane. These plot the ‘flow’ of the system at each value of and . You can find phase plane plotters online. For , the phase plane for Lotka-Volterra is shown in the diagram.

Questions 4

Sketch the curve in the phase plane above, corresponding to the initial values this curve to estimate the maximum fish population.

5

Hence sketch the behaviour of against and of against for these initial conditions.

6

The effect of competition amongst the fish can be included in the model by adding another term in

and

. Use

to the original differential equation:

Assuming that all parameters are positive, explain why this adaptation introduces a competition effect into the differential equations. 7

By using online technology, investigate this system with introduction of competition changed the behaviour of the system?

and

. How has the

CROSS-TOPIC REVIEW EXERCISE 2 1

The curve has polar equation

, where

and

.

a Sketch , clearly stating the range of values of for which it is defined. b Find the total area enclosed by . 2

Solve the equation

giving your answer in the form

3

Solve the equation logarithms.

4

The Cartesian equation of a circle is

.

giving your answer in terms of natural

Using the origin as the pole and the positive -axis as the initial line, find the polar equation of this circle, giving your answer in the form 5

The shaded region in the diagram is bounded by the curve with equation axis and lines

the -

and

Calculate the volume of revolution when the shaded region is rotated

6

Express

7

By first completing the square in the denominator, find the exact value of

about the -axis.

in terms of exponentials and hence, by using the substitution

, find

© OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, January 2012 8

The diagram shows part of the curve

. The shaded region is bounded by the

curve and the lines , and . Find the exact volume of the solid produced when this shaded region is rotated completely about the -axis. © OCR A Level Mathematics, Unit 4723 Core Mathematics 3, January 2012

9

Given that the first three terms of the Maclaurin series for

are identical to the

first three terms of the binomial series for , find the values of the constants and . (You may use appropriate results given in the List of Formulae (MFl).) © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2010 10 a By using the substitution

, show that

The part of the curve is rotated through

between

and

, where is a positive constant,

about the -axis. The volume of the solid formed is

.

b Find the exact value of . 11 Show that 12 a Show that b Given that 13 i

find the exact value of

.

Use the definitions of hyperbolic function in terms of exponentials to prove that .

ii Solve the equation giving your answer in logarithmic form. © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, January 2011 14 i

Show that

ii Given that

where is a constant, show that

© OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2010 15 i

Prove that, if

, then

ii Find the Maclaurin series for

, up to and including the term in

.

iii Use the result of part ii and the Maclaurin series for to find the Maclaurin series for up to and including the term in . © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2011 16 It is given that i

Show that

ii Hence find the Maclaurin series for

up to and including the term in

.

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2013 17 The equation of a curve in polar coordinates is i

for

Sketch the curve.

ii Find the area of the region enclosed by this curve. iii By expressing

in terms of

, show that a Cartesian equation for the curve is

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2015 18 i

Find the general solution of the differential equation

ii Show that, when is large and positive, the solution approximates to a linear function, and state its equation. © OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2010 19

Find the solution of the differential equation

for which

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, June 2013 20 Find the solution of the differential equation

for which

when

Give your answer in the form

© OCR A Level Mathematics, Unit 4727 Further Pure Mathematics 3, June 2012 21 At time a particle , of mass , is away from a point on a smooth horizontal plane and is moving away from with speed . The only horizontal force acting on has magnitude N, where is the distance , and acts away from . i

Show that the speed of ,

is given by

ii Find an expression for in terms of . © OCR A Level Mathematics, Unit 4730/01 Mechanics 3, January 2013 22 A particle starts from rest at a point and moves in a straight line with simple harmonic motion. At time after the motion starts, displacement from a point on the line is towards . The particle returns to for the first time when The maximum speed of is and occurs when passes through . i

Find the distance

.

ii Find the values of and the velocity of when iii Find the number of occasions in the interval at which that when and find the corresponding values of and .

speed is the same as

© OCR A Level Mathematics, Unit 4730 Mechanics 3, January 2012 23 A particle of mass is projected horizontally with speed from a fixed point on a smooth horizontal surface and moves in a straight line on the surface. The only horizontal force acting on has magnitude , where is the velocity of at time after it is projected from . This force is directed towards . i

Find an expression for in terms of . The particle passes through a point with speed

ii Find the average speed of for its motion between and . © OCR A Level Mathematics, Unit 4730 Mechanics 3, June 2011 24 Find the set of values of for which

has at least one solution. 25 The mean value of the function

between and is

Find the value of . 26 a Find, up to the term in

, the Maclaurin series for

b Find the set of values for which the expansion is valid. c By evaluating the series in part a at an appropriate value of , find a rational approximation to . 27 Prove that if

, then

28 a Use the substitution

b Hence find

. to show that

.

29 Evaluate the improper integral

, clearly showing the limiting process

used. 30 i

Using the definition of

ii Use the substitution

in terms of and

, show that

to find, in terms of

, the real root of the equation

© OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2010 31 i

Use the substitution

to find

giving your answer in the form

ii Hence calculate the exact area of the region between the curve and the lines

and

the -axis

(see diagram).

iii What can you say about the solid of revolution obtained when the region defined in part ii is rotated completely about the -axis? Justify your answer. © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2011 32 A curve has polar equation i

for

.

Sketch the curve, indicating the line of symmetry and stating the polar coordinates of the point on the curve which is furthest away from the pole.

ii Calculate the area enclosed by the curve. iii Find the Cartesian equation of the tangent to the curve at . iv Show that a Cartesian equation of the curve is © OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, January 2013 33 The differential equation i

In the case

is to be solved, where is a constant.

by using a particular integral of the form

find the

general solution. ii Describe briefly the behaviour of when iii In the case

explain whether would exhibit the same behaviour as in part ii when

© OCR A Level Mathematics, Unit 4727/01 Further Pure Mathematics 3, January 2013 34

Particles and are each moving with simple harmonic motion along the same straight line. motion has centre , period and amplitude ; motion has centre , period

and amplitude

. The points

and

are

apart. The displacements of

and from their centres of oscillation at time are denoted by and respectively. The diagram shows the positions of the particles at time , when and i

State expressions for

and

in terms of , which are valid until the particles collide.

The particles collide when

correct to significant figures.

ii Find the distance travelled by

and

before the collision takes place.

iii Find the velocities of and immediately before the collision, and state whether the particles are travelling in the same direction or in opposite directions. © OCR A Level Mathematics, Unit 4730 Mechanics 3, June 2010 35 i

Given that

show that

ii Show that iii Solve the equation

giving your answer in logarithmic form.

© OCR A Level Mathematics, Unit 4726/01 Further Pure Mathematics 2, June 2014 36 A function is defined by i

When

show that the value of

, for for which

is

ii

The diagram shows the graph of and state the range of values that

Sketch the graph of can take for

for

© OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2012 37 a Find b Show that

. .

The area is bounded by the curve with equation as shown.

, the -axis and the line

c i

Find the area in terms of .

ii Hence state

,

.

38

The diagram shows the curve with equation i

between the points

and

Find the area of the region bounded by the curve, the -axis and the line . Hence find the area of the region bounded by the curve and the lines and , where is the origin.

ii Show that the curve between and can be expressed in polar coordinates as , where iii Deduce from parts i and ii that © OCR A Level Mathematics, Unit 4726 Further Pure Mathematics 2, June 2010

PRACTICE PAPER 1 1 hour 30 minutes, 75 marks 1

Let

,

and

.

Find the value of such that is perpendicular to and . 2

Use the definitions of

and

[3 marks]

to prove that .

3

In this question you must show detailed reasoning. Evaluate

4

and are roots of the equation

The region bounded by the curve rotated through

, find the value of

, the -axis and the lines

and

. [4 marks] is

about the -axis.

Show that the volume generated is 6

[3 marks]

, explaining clearly why the integral converges.

In this question you must show detailed reasoning. Given that

5

[3 marks]

, where

a Write down the th roots of unity in the form

and are integers to be found.

.

[5 marks] [3 marks]

b

The diagram shows a regular seven-sided polygon with the centre at the origin and one vertex at , whose vertices represent the roots of the equation , with and . Find the values of and . [2 marks] 7

Find the general solution of the system of differential equations: [5 marks]

8

a Write b Hence find

in partial fractions. .

[4 marks] [3 marks]

9

Plane contains the points

,

and

.

a Find the Cartesian equation of . b Point has coordinates to the nearest .

[5 marks]

. Find the angle between and the line

10 A curve has polar equation

, giving your answer [4 marks]

.

a Find its Cartesian equation in the form The curve intersects the line

. [4 marks]

.

b Find the value of at the points of intersection. 11 Consider the matrix

[3 marks]

.

a Show that is non-singular for all values of .

[2 marks]

b Find

[5 marks]

in terms of .

c Hence solve the simultaneous equations [3 marks]

12 Given that

when

, solve the differential equation ,

Give your answer in the form 13 a Prove by induction that, for

. .

[6 marks]

, .

b Hence find the exact value of

.

[6 marks] [2 marks]

PRACTICE PAPER 2 1 hour 30 minutes, 75 marks 1

In this question you must show detailed reasoning. Let

.

a Find in the form

.

[2 marks]

b Find the modulus and argument of .

2

[3 marks]

A transformation is described by the matrix

a The image of point is the point

.

. Find the coordinates of .

[2 marks]

b Describe the transformation fully.

[2 marks]

3

Find the Cartesian equation of the curve with polar equation

4

Use the formulae for

5

In this question you must show detailed reasoning.

and

.

to show that

.

By first completing the square, find the exact value of 6

[2 marks]

[6 marks]

.

a Show that the Maclaurin series of the function

[5 marks]

up to the term in

is [6 marks]

b A student claims that this series is valid for all is wrong. 7

Matrix is given by a Find

,

and

. Show by means of a counterexample that he [2 marks]

. .

b Conjecture an expression for

[2 marks] for

.

[1 mark]

c Use mathematical induction to prove your conjecture. 8

[5 marks]

Show that the lines with Cartesian equations

and

are

skew and find the shortest distance between them. 9

[7 marks]

Show that the mean value of the function

between and is

and are constants to be found. 10 a If

[7 marks]

, show that

b Show that c Hence solve 11 Three planes have equations

, where

[1 mark]

. for

[4 marks]

. .

[3 marks] ,

,

a Show that, for all values of , the planes do not intersect at a unique point.

. [2 marks]

b Find the value of for which the intersection of the three planes is a line.

[4 marks]

12 A particle of mass is attached to one end of a light horizontal spring. The other end of the spring is attached to a fixed point. The magnitude of the tension in the spring is given by at time seconds and is a constant.

, where is the extension in the spring

The particle experiences a resistance to motion of magnitude

, where is the speed of the

particle at time seconds. a Show that b Given that when , and i find in terms of and

[3 marks]

. :

ii state whether the damping is underdamping, overdamping or critical.

[5 marks] [1 mark]

FORMULAE

Learners will be given the following formulae in the Formulae Booklet in each assessment. Pure Mathematics Arithmetic series

Geometric series

for Binomial series

where

Series

Maclaurin series

for all

for all

for all

Matrix transformations Reflection in the line Rotations through about the coordinate axes. The direction of positive rotation is taken to be anticlockwise when looking towards the origin from the positive side of the axis of rotation.

Differentiation

Quotient rule Differentiation from first principles

Integration

Integration by parts The mean value of

on the interval

Area of sector enclosed by polar curve is

is

or

or

Numerical methods Trapezium rule:

where

The Newton-Raphson iteration for solving Complex numbers Circles: Half-lines: Lines: De Moivre’s theorem: Roots of unity: The roots of

are given by

Vectors and 3D coordinate geometry Cartesian equation of the line through the point with position vector

in direction

is Cartesian equation of a plane is

Vector product:

The distance between skew lines is

, where and are position vectors of points on each

line and is a mutual perpendicular to both lines The distance between a point and a line is

, where the coordinates of the point are

and the equation of the line is given by The distance between a point and a plane is the equation of the plane is given by

, where is the position vector of the point and

Small angle approximations where is small and measured in radians Trigonometric identities

Hyperbolic functions

Simple harmonic motion

Answers Chapter 1 Before you start… 1 Proof 2 3 4 5 6 Exercise 1A 1, 2 Proof 3 a ; Proof

b 4 a

; Proof

b 5 a b c Proof 6, 7 Proof 8 a b

; Proof

9 a b Proof 10 –15 Proof Exercise 1B 1–10 Proof Work it out 1.1 The correct answer is Solution 2. Exercise 1C 1 a i ii b i ii 2 a i ii b i ii 3, 4 Proof

5 a b 6 a Proof b 7 Proof; 8 a Proof b 9 a Proof b 10 Proof; Exercise 1D 1, 2 Proof 3 a Proof b i Proof ii 4 a b Proof c 5 Proof 6 a Proof b 7 a Proof b c Proof 8 9 a Proof b Proof; c 10 a b Proof Mixed practice 1 1 a b Proof 2 Proof 3 a b Proof 4 5 Proof 6 7 a Proof b Proof; 8 9–11 Proof

,

,

12 a Proof b 13 a Proof b Proof 14 a Proof b c 15 i ii iii Proof 16–18 Proof 19 ; Proof , 21 Proof 20 22 i

Proof

ii iii 23 a b Proof; 24 Proof;

Chapter 2 Before you start… ;

1

radians

2 3 a b 4 a b 5 a b 6 a Reflection in the real axis. b Translation by

.

Exercise 2A 1 a i ii b i ii c

i ii

2 a

;

;

b

3 a i

;

,

ii

b 4 a b c 5 a b 6 7 Exercise 2B 1 a i ii b i ii c

i ii

2 a i ii b i ii c

i ii

3 a i ii b i ii

4

5 Proof 6 a b 7 8 9 10 11 12 Exercise 2C 1 a i ii b i ii c

i ii

2 a i

ii

;

;

b i

;

ii

;

3 4 a

;

b 5

,

6

, ,

7 a b

,

,

c

8 9 a b 10 a b 11 a

,

b

,

;

c 12 a b c Exercise 2D 1 a i ii b i

ii 2 a i ii b i ii 3 a

,

,

, ,

,

,

,

,

,

,

,

b

4 a b c 5 a b No. Consider c d 6 a, b Proof 7 a

,

b 8 9 a b

for

c–e Proof 10 a

b i ii 11 a Proof b i Proof ii c Proof Exercise 2E 1 a b c

, or an Argand diagram.

2 3 Proof; 4 a b 5 a Proof b c

6 a b, c Proof Exercise 2F 1 a b 2 a Proof b Rotation

radians

about the origin.

3 4 5 6 a b 7 a

b Rotation through

radians about the origin.

8 a b c 9 a b 10

,

11 a b Proof 12 a Proof b 13 a b

,

,

14 Mixed practice 2 1 2 a b 3 a

for

b

4 5 a

;

b

for

6 a b 7

i ii

8 9 10 a b Proof; ,

c

 11, 12 Proof 13 14 Proof 15 a b 16 i

Proof

ii iii 17 a b, c Proof d 18 a Proof b i ii c 19 a b 20 Proof; it equals

.

Proof

21 i ii

;

iii

22 a b 23 a Proof b i ii iii iv Proof

Chapter 3 Before you start… 1 2 a b 3 4 5 a b Exercise 3A 1 a b Proof 2 3 a b 4 a Proof b 5 a b 6 a Real: b 7 a Proof b 8 a b Proof c Exercise 3B 1 a b c 2 a Proof b c 3 a b Proof 4 a Proof b

;

c Proof; 5 a Proof ;

b

; proof

6 a Proof b Exercise 3C 1 a i ii b i ii 2 a Proof b Proof; 3 a Proof b 4 a i, ii Proof b i ii c 5 a Proof

b c Proof 6 a b Exercise 3D 1 a

or

b 2 a Proof; b Proof 3  4, 5 Proof 6 a b Proof c Mixed practice 3 1 a

or

2 b Proof; 3 a b  4–6 Proof 7

i

Proof

ii   8 a, b Proof i Proof

c

ii 9 a

b, c Proof d Proof; e Proof f 10 i

Proof

ii 11 i ii

Proof Proof;

Chapter 4 Before you start… 1 a b 2 3 4 Exercise 4A 1 a i

ii b i ii 2 a i

ii

b i

ii

3 a i

ii

b i

ii

4 a i

ii

b i

ii

5 a i

ii

b i

ii

6 a i

ii

b i

ii 7 a i ii b i ii 8 a i ii b i ii 9 a–c Proof 10 11

12 a b 13 a b No

14 a b 15 a b i Proof ii c 16 No 17 a

b c 18 a b For example:

19 For example: Exercise 4B 1 a i ii b i ii 2 a–d Proof 3 4 a b 5 6 a b Exercise 4C 1 a i ii b i ii 2 a b

3 a b 4 a b 5 a b 6 a b 7 Proof 8 a b 9 a Proof b c d Exercise 4D 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii 4 a i Proof; ii Proof; b i Proof; ii Proof;

5 6 7 8 9 Proof; 10 11 Proof; 12 a, b Proof c 13 a proof b c 14 a b c 15 a b 16

;

17 a, b Proof c d 18 a proof b c Mixed practice 4 1 2 3 a b 4 a b c Proof d 5

i ii

6

i ii

7 a b c d 8 a b c 9 a Proof b c d e 10 11 a b c 12 a b c 13 a b c d Proof e 14 i ii 15 i ii 16 a b Proof c d 17

18 19 a

b, c Proof d 20 a Proof b c d 21 a b c d e

;

f 22 i Proof ii Symmetry in the plane iii 23 i, ii Proof iii iv

.

Chapter 5 Before you start… 1 2

;

;

3 4 5 a b Exercise 5A 1 a i Unique ii Unique b i No solutions ii Infinitely many c

i Infinitely many ii No solutions

2 a i Unique;

,

,

ii Unique; b i Unique;

,

,

ii Not unique; inconsistent c

i Not unique; consistent

ii Not unique; consistent 3 Proof 4 a b No solutions for both 5 a Proof b c d Inconsistent 6 a, b Proof 7 Exercise 5B 1 a i Consistent; unique solution: ii Consistent; unique solution: b i Inconsistent; two parallel planes with a single intersecting plane. ii Consistent; line intersection of three distinct planes (sheaf) c

i Consistent; unique solution;

,

,

ii Inconsistent; two parallel planes with a single intersecting plane d i Consistent; unique solution; ii Consistent; unique solution: 2 a Proof (

);

,

,

b The planes intersect at a single point. 3 a Proof b The planes intersect at a single point. 4 5 a, b Proof c Triangular prism and are parallel, 6

intersects them.

7 a b Triangular prism 8 a Proof b c Intersect along a line (sheaf) 9 a Proof b Intersect along a line (sheaf) 10 a

b 11 a b Inconsistent; triangular prism 12 a b c Intersect in a line (form a sheaf) 13 Mixed practice 5 1 2 3 a Proof b 4 a b c 5

i ii iii Infinitely many

6

i ii a Unique b Not unique, inconsistent c Not unique, consistent

Focus on … 1 Focus on … Proof 1 1, 2 Proof Focus on … Problem solving 1

1 2 The velocity is parallel to

; proof

   3–7 Proof 8 9 Proof;

10

; proof

11

; proof

Focus on … Modelling 1 1 Proof 2 3 4 a b 5 6 7 8 For example: all adults are the same; there is no reference to gender; the average of might not give a good prediction with small numbers; no randomness; there are no limiting factors such as the size of the island; there are no direct effects of predators. 9 Investigation Cross-topic review exercise 1 1 a Proof b 2 a b 3 a b 4 5 6

i Proof ii

7

i

and

ii

8 9 10 a Proof b c Proof 11 i Proof ii iii Does not converge; as 12 i Proof ii iii 13 i Proof ii iii 14 i ii iii Proof 15 i

ii Proof iii a b iv

,

16 i ii 17 i Proof ii Rotation of a point in the Argand plane by iii Proof 18 i ii Proof; 19 i

ii 20 i Proof ii Proof; 21 i Proof ii 22 i ii 23 i ii iii

24 25 i ii 26 a b Proof c 27 a Proof b 28 a Proof b

or

about the origin; proof

c Proof 29 a b Proof 30 a Proof b 31 Proof 32 a Proof b c Proof 33 Proof; 34  i, ii Proof iii  Proof; 35 i ii Proof iii 36 i ii–iv Proof 37 i Proof ii iii 38 i a Proof b ii Proof iii Proof;

Chapter 6 Before you start… 1 a b 2

3 a b 4 a b 5 a b c Exercise 6A 1 a

Domain:

; range:

Domain:

; range:

b

c

Domain:

; range:

Domain:

; range:

Domain:

; range:

Domain:

; range:

d

e

f

2 3 Exercise 6B

1 a i

Domain:

; range:

Domain:

; range:

ii

b i

Domain:

; range:

ii

Domain:

; range:

c

i

Domain:

; range:

ii

Domain: 2 a i ii b i ii c

i ii

d i ii e i Not possible ii Not possible 3 a i ii b i ii c

i ii

d i No solution ii No solution e i ii 4 a i ii b i

; range:

ii c

i ii

d i Doesn‘t exist ii Doesn‘t exist e i ii or

5 6 7 8 9 10 11

 12–14 Proof Exercise 6C 1 Proof 2 3–8 Proof 9 a, b Proof  c 10 Proof Work it out 6.1 Solution 3 is correct. Exercise 6D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 a Proof b 15

or

16 a Proof b 17 a Proof b Exercise 6E 1 a i ii b i ii c

i ii

2 a b 3 4 5 Proof 6 7 a b Proof 8 9 Proof; 10 a b Proof Exercise 6F 1 a i ii b i ii c

i ii

2 a i ii b i ii

c

i ii

3 a i ii b i ii c

i ii

4 a i ii b i ii c

i ii

d i ii 5 a i ii b i ii 6 7 8 Proof 9 a b 10 11  12, 13 Proof 14 a b Proof Mixed practice 6

1 a

b 2 3 4 5 6 7 8  9, 10 Proof 11 12 13 Proof 14 or 15 16 Proof 17 18 Proof 19 a

b Proof c i

Proof

ii Proof; 20 a Proof b 21 i Proof ii Proof; 22

23 24 a Proof b c 25 26 –28 Proof

or

Chapter 7 Before you start… 1 2 3 Exercise 7A 1 a i

ii b i ii c i

ii d i

ii 2 3 4 5 6 a Proof b 7 Proof; 8 9 a b 10 Proof Exercise 7B 1 a i

ii

b i ii c i ii 2 3 4 Proof; 5 6 Proof 7 8 Proof 9 Proof; Exercise 7C 1 a i ii b i ii c i ii d i ii e i ii f i

ii 2 a i ii

b i ii c i ii d i ii e i ii f i ii 3 a i ii b i ii c i ii d i ii e i ii f i ii 4 5 6 7 a b 8 a Proof b

9 a Proof b 10 a b 11 a b 12 a Proof b 13 Proof 14 15 Proof 16 17 18 19 a b  20, 21 Proof 22 a b Proof Exercise 7D 1 a i ii b i ii c i ii d i ii 2 3 a b 4 a b 5 a

b 6 7 8 Proof; Mixed practice 7 1 2 3 4 5 6

i ii

7 8 Proof 9 10 11 12 i ii 13 i ii Proof 14 a Proof b 15 Proof; 16 a b c 17 a, b Proof 18 i Proof ii iii

Chapter 8 Before you start… 1 2 a i ii b i ii c i ii 3 a b 4 a b 5 a b Exercise 8A 1 a i ii b i

ii c i ii d i ii e i

ii f i

ii

2 3 a Proof b 4 a i

; ;

ii b 5 a b 6 a i

Proof

ii

b c ;

7 a i

;

ii Proof b 8 a Proof b 9 a Proof b 10 a b i ii Proof; 11 a Not equal to

.

is an increasing function at

b

Exercise 8B 1 a i ii b i ii

, but the first derivative of the series is negative at

.

c i

ii d i

ii

e i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii ;

4 5 6 Proof 7 a b 8 a b c 9 a b

10 a

; convergence for

b

;

Exercise 8C 1 a b c Integral diverges. d 2 a b 3 Proof; 4 5 6 Exercise 8D 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii 4 a i

ii b i ii c i ii 5 a i ii b i ii c i ii 6 a i ii b i ii c i ii 7 a b 8 9 10 11 12 13 a b 14 15 a

,

b 16 Proof 17 18 19 20 Proof; use

.

21 Exercise 8E 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii 3 4 5 a b 6 Proof 7 8 Proof 9 a

b Curve is concave up. c Proof 10 Not true. For example: Mixed practice 8 1 a, b Proof 2 a

between

and .

b c 3 a The integrand is not defined at

.

does not converge to a finite value.

b 4 5 6 7 8 9 10 a b Proof ,

11 a b c 12 i ii Proof; iii iv 13 14 a Proof b 15 16 i ii 17 i

Proof

ii iii 18 19 20 21 a b c Proof

,

22 23 a b c 24 a

b c

and

Chapter 9 Before you start… 1 a b 2 a b 3 a b Exercise 9A 1 a-c

Distance

; area

ii Distance

; area

Distance

; area

ii Distance

; area

Distance

; area

ii Distance

; area

2 a i

b i

c i

3 a i

ii

b i

ii

c i

ii

4 a

b

c

5 a b

6 a b Proof

and

c

d Exercise 9B 1 a i

Maximum

ii Maximum

b i

Maxima minima

; minimum

; minimum

and and

;

ii Maxima minima

2 a i

ii

and and

;

b i

ii

3 a b

and

c

4 a

b 5 a b

,

6 a Largest

,

,

when

; maximum at

; smallest

and

when

.

b

7 a

b

8 a Maximum is b

; minimum is

9

10

Exercise 9C 1 a i

ii b i ii c i ii 2 a i ii b i ii c i ii 3 a i ii b i

ii c i ii 4 a i ii b i ii c i ii 5 a b c 6 7 8 Proof 9 a b Exercise 9D 1 a i ii b i ii c i ii d i ii e i ii 2 a b Proof 3 4 Proof 5

6 a

b Proof 7 a

b 8 a

Tangents: b 9 10 Proof Exercise 9E 1 a b i

ii Proof 2 3 a

and

b 4 a

,

b 5 a b Proof. 6 a

b Proof Mixed practice 9 1 2 3 a b 4

5

6 a

b Proof 7 a

b 8 a Proof b c

d Proof 9 10 a

and

b

c 11 a

b 12 a b c

13 a

b

and

14

15 16 a Proof b

and

c 17 a i

Proof

ii b Proof 18 i ii

iii 19 i, ii Proof

Chapter 10 Before you start… 1 2 3 Exercise 10A 1 a i

For example:

ii For example: b i

For example:

ii For example: c i

For example:

ii For example: 2 a i

Second order linear non-homogeneous

ii Second order linear non-homogeneous b i

Second order non-linear homogeneous

ii First order non-linear non-homogeneous c i

Second order non-linear non-homogeneous

ii Third order linear homogeneous d i

Second order linear homogeneous

ii First order non-linear non-homogeneous 3 a i ii b i ii c i ii d i ii 4 a b c 5 a b

c 6 a Proof; b c The equation is not linear. d 7 a b c Exercise 10B 1 a i ii b i ii c i ii 2 3 4 5 6 7 Proof 8 9 a b c 10 a b Exercise 10C 1 a b 2 a b 3 a b 4 a b 5 a

b 6 a b 7 a b 8 a b 9 10 11 12 a b c Exercise 10D 1 a b c 2 a b c 3 a b c 4 a b c 5 a b 6 a b 7 a b Proof c d 8 9 10 a b Mixed practice 10 1

2 3 a b 4 a Proof b 5

i ii

iii For large , the function oscillates approximately between 6 a b Proof; c d 7 a b 8 a b Proof; c d 9 10 11 12 i ii iii 13 i ii iii 14 a b Proof c 15 a b 16

and

.

Chapter 11 Before you start… 1 2 Exercise 11A 1 a b 2 a Proof b The model seems suitable initially, but is not accurate for later times. 3 a

b

c

4 a b c A natural net birth rate of the population 5 a

b c d Different parts of the chicken are likely to have different temperatures. 6 a b c For example: interest in the rumour remains constant; the number of students who know the rumour is modelled as a continuous variable; there are no people outside the school spreading the rumour. 7 a This is proportional to the surface area of the bacterium. Larger surface areas make the bacterium more efficient in taking up nutrients so it will grow faster. b c

( )

Exercise 11B 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii c i ii

; at rest ; equilibrium ; equilibrium ; at rest ; equilibrium ; at rest ; neither ; neither ; neither ; neither

3 a i ii b i ii c i ii d i ii 4 a b 5 a b 6 a b 7 a b 8 a b ; away

c 9 a b 10 a b c d 11 a Proof b c 12 a Proof; b c 13 a b Proof c 14 a b

; proof

c Proof; Exercise 11C 1 a i

Underdamping

ii Overdamping b i

Critical

ii Underdamping c i

Underdamping

ii Critical d i

Critical

ii Underdamping 2 3 a Proof b c Overdamping 4 a b 5 a b Underdamping

6 a Proof b c Proof d Overdamping 7 a Proof b c 8 a b Proof c Underdamping;

d e 9 a Underdamping b c Proof Exercise 11D

1 a i

;

ii

;

;

b i

,

;

ii

,

c i

;

,

;

ii

, 2 3 4 5 a b 6 a Proof; b

; proof

c 7 a

b c d

; shark peak at

, fish peak at

time units), and with a phase delay of e The populations oscillate with the same period ( units between fish population peaking and shark population peaking. 8 Proof Mixed practice 11 1 2 3 a b 4 a b c 5 a Proof b Proof;

time

c d 6 i ii iii iv 7 a Proof b Proof; c

seconds months

8 9 10

11 a Proof b c 12 a Proof b

.

13 a Proof b

; proof

c 14 a b c d The second model (after

). ,

15 16 17 i ii iii

18 i

; proof

ii

19 a Proof b Proof; c 20 a The force decreases at a constant rate. In reality the force will vary over each stride; he might be more motivated towards the end.

b c

;

but you don’t need to find this. A sketch of both forces makes it clear d Bill (Mike’s distance is that Mike is never pulling harder.)

Focus on … 2 Focus on … proof 2 1 Proof 2 3 Focus on … problem solving 2 1 Proof 2 The tangent to the chain at

is horizontal; proof

moving the chain vertically does not change its shape.

3 4 Proof 5 Proof

6 Investigation (using graphing software); the distance between the end-points ( than the length of the chain ( ).

) has to be smaller

Focus on … modelling 2 1 For example: all fish and sharks are treated as equivalent so that the effects of age or disease average out over the population; there is no randomness, which might be acceptable if the populations are large enough; there are no external populations (i.e. no other predators or sources of food); there is no seasonality so the birth rate stays constant over time. 2

or

; the second solution is the biologically relevant one.

3 The equilibrium value of the fish goes down when decreases. The equilibrium value of the sharks goes up when increases. 4

Maximum fish population is about

million.

5

term will 6 When is small the term will be relatively more important. When is large the dominate, meaning that when the population is too large there is a net death, as would be expected with internal competition. 7

The systems tends to an equilibrium at Cross-topic review exercise 2 1 a

.

b 2 or

3 4 5 6 7 8 9

10 a Proof b 11 Proof 12 a Proof b 13 i Proof ii 14 i, ii Proof 15 i Proof ii iii 16 i Proof ii 17 i

ii iii Proof 18 i

ii Proof; 19 20 21 i Proof ii 22 i ii iii

, velocity and

and

23 i ii 24 25 26 a b c 27 Proof 28 a Proof b 29 30 i Proof ii 31 i ii iii The volume of revolution is infinite. 32 i

ii iii iv Proof

and

33 i oscillates increasingly widely, with amplitude proportional to .

ii

iii oscillates with a stable amplitude throughout all values of , and does not grow without limit. 34 i ii The period of

is

. The particles are travelling in opposite directions.

iii 35 i, ii Proof iii   36 i Proof ii

37 a b Proof c i ii 38 i ii, iii Proof

Practice paper 1 1 2 Proof proof 3 4 5 Proof;

,

6 a

b 7 8 a b 9 a

,

so the particle has not completed a full cycle.

,

b 10 a b 11 a Proof b c 12 13 a Proof b

Practice paper 2 1 a b 2 a b Rotation through

anticlockwise about the origin

3 4 Proof 5 6 a Proof b For example, 7 a

.

b c Proof 8 Proof; 9 Proof; 10 a, b Proof b 11 a Proof b 12 a Proof b i ii Overdamping

Worked solutions 1 Series and induction Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 1A

1 Proposition: Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers.

2 Proposition: Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 3

a

b For the first four terms, it appears that Proposition: Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 4

a

b For the first four terms, it appears that Proposition: Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 5

a





is a multiple of in all the cases above.

b

c Proposition:

is a multiple of for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

for some integer

Working towards:

for some integer

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers.

6

and

Proposition: Base cases: so the proposition is true for so the proposition is true for Inductive case: Assume that the proposition is true for some positive integers and So

and

Working towards:

So if the proposition is true for

and

then it is also true for

Conclusion: The proposition is true for .

and

and if true for

and

then it is also true for

By the principle of mathematical induction, the proposition is true for all positive integers.

7

and

Proposition: Base cases: so the proposition is true for so the proposition is true for Inductive case: Assume that the proposition is true for some positive integers and So

and

Working towards:

So if the proposition is true for

and

then it is also true for

Conclusion: The proposition is true for .

and

and if true for

and

then it is also true for

By the principle of mathematical induction, the proposition is true for all positive integers. 8

a

b For the first four terms, it appears that

Proposition:

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer

So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 9

a Let Then

,

If there is to be a natural number which is a factor of must be . b Proposition:

for all integer values then that factor

is a multiple of for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

for some integer

Working towards:

for some integer

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 10 Proposition:

Base case:

for integer

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer

So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers,

.

11 Proposition:

Base case:

for integer

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers,

.

12 Let Proposition: Base case:

for integer so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers,

.

13 Let Proposition: Base case:

for so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

By the principle of mathematical induction, the proposition is true for all integers all integer .

Tip

. , and hence for

Although the question asked for a proof for , since it is clear that the proposition is true for , this makes a far simpler base case to start from and avoids the labour of differentiating twice with the product rule! The proof given above encompasses a proof of the required, more limited, statement. While in general it will be preferable (and safer) to start your base case at the point required in a question, if you are confident that you can prove the statement more broadly and with less work, you may do so. 14 Let Proposition:

for integer

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers, 15 Proposition:

.

for positive integers

Base cases: so the proposition is true for

so the proposition is true for Inductive case: Assume that the proposition is true for some positive integers So

and

and

.

Working towards:

So if the proposition is true for

and

then it is also true for

Conclusion: The proposition is true for .

and

and if true for

and

then it is also true for

By the principle of mathematical induction, the proposition is true for all positive integers.

EXERCISE 1B 1

Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 2

Proposition:

for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 3

Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 4

Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 5

Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 6

Proposition:

for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 7

Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 8

Proposition:

for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 9

Tip This induction could either use

(the difference of two sums) or

single sum whose terms are given as a function of ). Either is acceptable.

Proposition:

Base case:

for all positive integers

so the proposition is true for

(a

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 10 Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer

Tip Since the dummy variable in the proposition is , we need a different letter to use for the induction! Don’t make the mistake of using the same letter for two different contexts.

So

Working towards:

So if the proposition is true for Conclusion:

then it is also true for

The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers.

EXERCISE 1C 3

Standard formulae:

and

4

Standard formulae:

and

5

a Standard formulae:

and

b

Positive quadratic is greater than zero outside the roots.

Roots at

or

So the least integer for which 6

a Standard formulae:

is

,

and

b

7

Standard formulae:

So 8

and

.

a Standard formulae:

b Replacing with

and

in the formula found in part a:

9

a Standard formulae:

b Since

10 Standard formulae:

and

:

,

and

If is the sum of the squares of the first odd numbers then

So

EXERCISE 1D 1

a

b

2

a

b

3

a

b i

ii

4

a

for some constants and Multiplying through by the denominator on the LHS:

Substituting values:

so so Then

b

c As So 5

a b

6

a b

,

As

,

So 7

a

b

c

So as

,

for some constants , and .

8

Multiplying through by the denominator on the LHS:

Substituting values of : so so so So Using this partial fraction form:

9

a

b

so c

10 a Using this rearrangement:

using the method of differences. b

so the series does not converge.

MIXED PRACTICE 1 1

a

is a multiple of in all the cases above. b Proposition:

is a multiple of for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

for some integer

Working towards:

So if the proposition is true for

for some integer

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 2

Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 3

a So



and

It appears that b Proposition:

for natural numbers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 4

Formula for the sum of the first integers:

5

Formulae for the sum of the first integers and the first squares: and

6

Formulae for the sum of the first squares and the first cubes: and

7

a b So



8

Formulae for the sum of the first squares:

9

Proposition:

for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 10 Proposition:

Base case:

for all positive integers

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers.

11 a Using compound angle identity Then b Proposition:

for natural numbers

Base case:

by part a so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 12 a

b

13 a

since the term for

is zero

b

14 a b

c

so as

,

15 i ,

,

ii All of these terms are multiples of . iii

Proposition:

is divisible by for all positive integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

for some integer

Working towards:

for some integer

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 16 i

ii

17 Proposition:

for all integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers greater than 1. 18 Proposition:

for all integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 19

So is the smallest positive integer for which Proposition:

for all integers

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers greater than

or equal to . 20 Proof by induction on , independent of the value : Proposition:

for all positive integers and real values

Base case:

so the proposition is true for

for all values

Inductive case: Assume that the proposition is true for some integer So

for all real values

Working towards:

for all real values

Expanding the product on the right side:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers.

Tip Take great care to establish clearly which variable you are applying the induction to; the proof either has to show that the result is true for all values of the other variable (as here) or split into cases and show that it works the same in each case (for example, if the sign of had changed the logic of the proof, you would have to show for each of positive and negative ).

21 Proposition:

for all integers

Tip The large Pi notation takes the product of terms, in the same way that the large Sigma notation indicates the sum of terms. This notation is used in this solution to keep the working easy to read, but would not be required in an examination; any clear notation, including use of … to indicate a continuation of a pattern, is acceptable.

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

.

By the principle of mathematical induction, the proposition is true for all positive integers. 22 i ii

iii

as

so

Then for some constants and

23 a

Multiplying through by the denominator on the LHS:

Substituting so so

Then b

So for some constants , and

24

Multiplying through by the denominator on the LHS:

Substituting so so so Then

So

,

and

.

Worked solutions 2 Powers and roots of complex numbers Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 2A 2

a Using de Moivre’s theorem:

b

3

a i Using de Moivre’s theorem:

ii

b Since 4

a

, it follows that

for all integers and so

for

.

(not adding because

)

b From part a: Then using de Moivre’s theorem:

c 5

a (adding because So b Using de Moivre’s theorem:

6

Using de Moivre’s theorem:

Require

, so

for some positive integer .

The least such is 7

Using de Moivre’s theorem:

Require

The least such is

EXERCISE 2B 5

So

6

a

, so

for some positive integer

)

b

7 8 9

Complex roots of a polynomial with real coefficients occur in conjugate pairs. The three roots are therefore ,

,

and

and

Tip Instead of expanding the factorised form, consider using the formulae you know for the coefficients: sum of roots,

sum or root pair products and

product of roots.

10 Complex roots of a polynomial with real coefficients occur in conjugate pairs. The four roots are therefore

The polynomial is 11 12

EXERCISE 2C 3

If One solution is

, then

and

.

The other solutions will differ in argument by

4

:

a (not adding because b If

)

, then

One solution is The other solutions will differ in argument by : ,

and

5 (not adding because If

)

, then

One solution is The other solutions will differ in argument by

:

,

Using double angle formula:

So

Similarly, In exact form, Then

Tip This nested surd form is equivalent to a surd sum form, so you could also give these as ,

.

Try squaring the real part and show that the two forms are equivalent.

6

If

, then

One solution is The other solutions will differ in argument by :

7

a (not adding because

)

So b If

, then

One solution is The other solutions will differ by factors of ,

:

and

c

8

The point shown is complex number The vertices form a square, so are the fourth roots of unity of . So

9

and

a If

, then

One solution is The arguments of other solutions will differ by :

.

b Then

can be factorised as

Multiplying the conjugate pair factors:

10 a If

, then

One solution is The arguments of other solutions will differ by

:

so

b Then

.

Rearranging gives Substituting the three values for :

Similarly (first rationalising the denominator and then ensuring a real denominator):

11 a

and So If One solution is

, then

The arguments of other solutions will differ by

:

is an equilateral triangle with centre at the origin.

b

The distance from the origin to any of the vertices is Then triangle

is isosceles, with altitude

Angle

,

. .

and so

.

Since is the midpoint of and which are equidistant from the origin, its argument is the mean of their arguments. Therefore c So, using de Moivre’s theorem: . 12 a If

, then

One solution is The exponents of other solutions will differ by

:

and b Using binomial expansion:

c Comparing with part b: and so Then using part a: or In Cartesian form:

or

or

EXERCISE 2D 3

a

so

,

,

,

and

b

4

a b For the roots of unity (where

), the sum of the roots is zero.

Therefore, and

c So

is TRUE is TRUE is TRUE so is FALSE 5

a The seventh roots of unity are

.

b No; on an Argand diagram, the points form a regular heptagon. Since there are an odd number of sides, the line passing through the origin and point does not pass through a vertex of the heptagon on the other side, it bisects a side instead. Therefore is not one of the heptagon vertices, so for any integer .

c

so Smallest positive integer

d Since the roots of unity form a regular polygon which is symmetrical about the real axis:

So

Smallest positive integer 6

a Let the roots be Then Comparing arguments: , so Then

and

is a geometric series with common ratio

b

7

a Let the roots be Using de Moivre’s theorem:

Considering real and imaginary parts: , so So b Let

so that

.

By part a: Rearranging: Substituting the solutions for : When

, is not defined.

When

,

8

, so

, and

Then The fifth roots of unity are

and

where

.

9

a b Rearranging: From part a: the solutions are However,

for

must be excluded, since

has no solution.

Rearranging: and so c

(excluding

)

for

d Expanding

using the binomial theorem:

e But from part c, the four roots to this quartic are Since

and

, these could alternatively be written as

and

Then the quartic must factorise as for some constant . Multiplying the conjugate factors and then expanding fully:

Comparing coefficients of and between this and the final line of part d:

. .

10 a

b i So

and

Then ii

and from part i

11 a Using compound and double angle formulae:

b i

is the primary seventh root of unity, so that

.

Then using the formula for the sum of a geometric sequence:

ii Using de Moivre’s theorem: and because

, it follows that

Then, taking the real part only of part i:

Hence c With

, the conclusion to part b ii can be rewritten as

.

Applying part a and the double angle formula for cosine:

So

is a root of the cubic

for

EXERCISE 2E 1

a Let

.

Then So the primary root is

and further roots differ in argument by .

Hence the roots are: b can be factorised as

c

Expanding the conjugate factor pairs:

2

Let

.

Then So the primary root is

and further roots differ in argument by .

Hence the roots are: which can be expressed in Cartesian form as

Then

can be factorised as

Expanding the conjugate factor pairs:

3

Solutions to

have the form

That is, abbreviating with

and

Therefore, the quintic can be factorised as:

for :

4

a The solutions to b Then

are

can be factorised into linear complex factors:

Therefore, substituting

:

So 5

a Solving

using the quadratic formula: for

.

In Euler form, these roots are Factorising using these complex roots:

b If

, then

so the primary solution has

and subsequent solutions differ in

argument by . So c In part b, you set The equivalent solution set for

will have

and The complete factorisation of

6

a The roots of

is therefore

are

The non-real roots are therefore b The roots of unity sum to zero, so Note that

so that

so that

Then c Using compound and double angle formulae:

, of which only the first is real. .

Then taking

, the result in part b can be written as:

Substituting the result for

So

, and using the double angle formula:

is a root of the cubic

for

EXERCISE 2F 1

a

and and

b The scale of the enlargement is ( 2

and the angle of rotation is

radians

). and

a b

and The transformation from to is a rotation through

3

A rotation through

radians (

).

on the Argand plane is equivalent to multiplication by

So corresponds to the complex number . 4

is a rotation of about the origin through

and an enlargement scale factor

equivalent to multiplication of the corresponding complex number by So corresponds to the complex number is produced by a rotation of about the origin through corresponding complex number by

5

.

and has coordinates

.

, equivalent to multiplication of the

.

So corresponds to the complex number

,

and has coordinates

All the points have the same modulus, since they lie on the same circle about the origin.

.

is at 6

, is at

and is at

.

a Using trigonometry:

b

is achieved by rotating through and enlarging by

, which is equivalent to multiplication of

the corresponding complex number by

So the coordinates of are 7

a

b So a rotation through 8

a The line has equation

about the origin maps to . so a point on the line has argument

b Therefore the angle between

and line is

Since is the reflection of in the line, angle c

. .

can therefore also be obtained from by a rotation through about the origin, equivalent to multiplication by

So has coordinates 9

a b Rotation through

on the Argand diagram is equivalent to multiplication by

corresponds to

so has coordinates

corresponds to

so has coordinates

10 Fifth roots of unity: ,

,

,

.

11 a Reflection of in the real axis produces Rotation through

about the origin is equivalent to multiplication by , so the resultant point is

.

b Reversing the order of transformations: Rotation through produces and the reflection then produces

.

12 a Rotation through angle about the origin on an Argand diagram is equivalent to multiplying by Rotating about point instead can be implemented by translating by at the origin), then rotating and then translating back by . So is translated to . That is,

, then rotated to become

(so that the centre is now

and then translated again to

and so

b In the terms given in part a: ,

so

,

So has coordinates 13 a

is the complex number equivalent to vector

and

is equivalent to

A rotation through (of a vector about its origin point) is equivalent to multiplication by b A rotation through

Then, using part a, with centre of rotation rotation

so that

, so

about the origin is equivalent to multiplication by

So after the first rotation, the point corresponding to

.

has image

, initial point

, the final image is given by where:

and angle of

14 On an Argand diagram, a rotation through about the origin is equivalent to multiplication by So corresponds to The midpoint of

and corresponds to

corresponds to the arithmetic mean of these complex values:

MIXED PRACTICE 2 so

1 Then 2

a (adding because b Then

3

a If

)

, and so , then

for

b

4

(subtracting integer multiples of 5

a

to ensure the argument is in the required interval)

(not adding because b If

)

, then

The primary solution is

with further solutions having

arguments differing by . So 6

for

a (not adding because

)

b A regular hexagon centred at the origin will be described by an equation of the form for some . If

is a solution, then

So the equation is 7

i

So ii Then

so

.

For the value to be real and positive, the argument is required to be an integer multiple of The smallest positive such is

.

is equivalent to the statement

8

, so , so the positive root is the one required.

and

9 So

.

Then

10 a b Then, using de Moivre’s theorem:

.

So c

and Require that the arguments differ by an integer multiple of

:

so is a multiple of

. A pair of such values is

.

11 a

Also, since

and

,

b Using part a,

12 a If is a complex third root of unity then Then But since

, it follows that

b From part a:

13

for some Then , so

14 Then

.

15 a b Then 16 i

and angle Two equal sides with a ii Then

angle between shows that the triangle is equilateral.

is represented by vector

, which has argument

and length .

So iii 17 a b The sum of the roots of unity equals zero (geometrically, the mean position of the vertices lying on a regular polygon centred about the origin must be the origin). Algebraic proof: so

But since

, it follows that and since

c Then

(that is,

, it follows that )

Therefore: So

and

d Applying part c to the result in part b, and writing

:

Using the double angle formula:

Solution using the quadratic formula: Selecting the positive root (because cosine has positive range for

18 a If is a complex third root of unity then

):

.

Then But since

, it follows that

b i ii Using part b i:

c The cubic is where is the negative sum of the roots: and is the sum of the paired products: and is the negative product of the roots: Then the cubic is represents

19 a

b Multiplying by

rotates the point by

about the origin and enlarges it by a

factor of . Then can represent

:

or 20 Translation is equivalent to the addition of complex numbers. Rotation through angle about the origin is equivalent to multiplication by So

and

Then the distance between the points:

which is independent of .

Tip

.

Show that this value can also be written as

.

and

21 i So ii Let If

then

so

for ,

,

,

iii Alternatively writing

,

,

,

,

and

, and using part a:

factorises into seven linear factors which can be paired up:

22 a Rotation through about the origin on the Argand diagram is equivalent to multiplication by

So point corresponds to b

Then point corresponds to

23 a The vectors

and

, so

Then b

i The number corresponding to point is . ii The number corresponding to point is

.

iii Using part a:

iv  From part b iii: .

Worked solutions 3 Complex numbers and trigonometry Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 3A 1

a

b Also, using de Moivre’s theorem:

Equating these two expressions:

2 So and Using de Moivre’s theorem:

Equating imaginary parts: 3

a Using the binomial theorem:

b Using de Moivre’s theorem: . Equating imaginary parts:

4

a Using the binomial theorem:

Using de Moivre’s theorem: . Equating real parts:

b If

, then

Factorising:

Since the range of

is

, the expression in parentheses cannot equal zero (for real

values of ) so the only solutions occur for 5

:

a Using the binomial theorem:

Using de Moivre’s theorem:

Equating imaginary parts:

So

,

and

b

The solutions are 6

a Using the binomial theorem:

So

b Also, using de Moivre’s theorem:

Equating real and imaginary parts:

and

7

a Using the binomial theorem:

So and Also, using de Moivre’s theorem:

Equating real and imaginary parts:

b Rearranging, the sextic given is equivalent to

, using the formula in part a.

Then So

8

for

a Using the binomial theorem:

So and b Using de Moivre’s theorem:

Equating imaginary parts:

(further solutions lie outside the required interval)

c Using the approximation

and disregarding powers of four and higher:

EXERCISE 3B 1

a b

, so

c Solutions to

:

or

, so

or

Solutions to the quadratic equation in part a:

Of the two solutions

or

So 2

gives

. , so

a b Solutions to So

, only

or

from which

:

for integer

within the interval

c Solutions to the quadratic in part a: Since 3

,

a Solutions to

for which

:

, so Within the interval b Since

, the solutions are , it follows that the quartic factorises as

where

and

, both of which are positive, since cosine has a positive range

in the domain Comparing coefficients: , so 4

a Using the binomial expansion:

So Using de Moivre’s theorem:

Equating imaginary parts:

b Solutions for So

:

or

or

.

Within the given interval, the solutions are or c If

, then by part a: , so the roots of

all equal

for one of the values of given

in part b. Since 5

, the three distinct root values are

a Using the compound angle formula:

So if

, then

b Using the binomial expansion:

Using de Moivre’s theorem:

Equating real and imaginary parts: and So (dividing through by By equivalent reasoning:

)

,

and

Then if

:

Expanding: Therefore 6

a Writing

as required. :

So when

,

Solutions to

or :

The solutions to which Since

so

. must be the six distinct values of

, there are distinct values for

for solutions for

:

b The product of the roots will be equal to the ratio of the constant term to the lead coefficient of the sextic, therefore:

Since

, this can be simplified to:

So

EXERCISE 3C 2

, so using de Moivre’s theorem:

a

and Then b

, 3

and ,

a So

But also, using the binomial expansion and collecting terms for the formulae shown:

Therefore: So b

4

so

a i Therefore

ii Using de Moivre’s theorem:

Then b i Using the binomial theorem:

ii From part a ii: From part b i:

Equating the two expressions:

So , c Using

5

and :

a Using de Moivre’s theorem:

and Then and b Using the binomial theorem:

c From part a: and From part b and part a:

6

a

b

Then

EXERCISE 3D 1

a Geometric series with common ratio , so the series converges to

b For the sum to infinity for

2

in part a:

a Geometric series with common ratio

.

, so the series converges to

b Using the result from part a:

3

is a geometric series with common ratio

Since

Substituting 4

, the series converges to

:

Using the binomial theorem:

so

and first term

.

So Therefore:

5

Using the binomial theorem:

, so

So Therefore:

6

a

is a geometric series with start term

So b

, so Since

:

and common ratio

.

c Substituting

into the result in part b:

so Solutions are

for integer , so

.

Solutions in the required interval are:

MIXED PRACTICE 3 1

a Using the binomial theorem:

b Also, using de Moivre’s theorem:

Equating the imaginary parts:

and 2

Using the binomial theorem:

Using de Moivre’s theorem:

Equating real parts:

Then 3

a

, which has range , so using de Moivre’s theorem:

and

and therefore

Using the binomial theorem:

But also, Equating the two expressions:

,

and

b

4

Using de Moivre’s theorem: , so Using the binomial theorem:

So Equating the two expressions:

Then

is equivalent to

Solutions to the quintic which

must therefore take the form

.

Solutions: or So

for integer value

or

Therefore, the five distinct roots of the quintic are: .

for values of for

and

5

, so

Therefore: Using the binomial theorem:

So Equating the two expressions: 6

Using the binomial theorem:

Since

, it follows that

Alternatively:

So

, and therefore

Equating the two expressions:

7

i

ii From part i, So

So

For 8

, solutions are

a If

.

, then

Hence

, and therefore

b Using de Moivre’s theorem:

Substituting

gives

and substituting

:

Then c

i Since

is clearly not a root to the equation, dividing through by and collecting pairs:

Using the result from part b:

ii Using double angle formula for cosine:

So

or and using

, the four values for are:

or 9

a Using de Moivre’s theorem:

Expanding using the binomial theorem:

Equating real and imaginary parts of these two forms:

b Taking the ratio of these two results and dividing numerator and denominator by

c For

, writing

and using the formula in part b: .

But since

So d Let

, rearranging:

must be a solution to this cubic.

:

Since factor of

, by the factor theorem

is a

Tip Or observe that in the working of part c it is shown that

is a solution.

Factorising the cubic: The other two root values are: e In the interval Since f

, the curve

is strictly increasing.

, it follows that

Therefore

10 i De Moivre’s theorem gives that Then using the binomial expansion:

ii Let

. Using part i,

, so

.

The six roots of the sextic are the six distinct solutions Solutions:

where

for integer ;

Since the graph of

decreases in the interval

, the largest positive root of is therefore

. 11 i and

This is the sum of a geometric series with constant ratio Then

and first term

ii

Worked solutions 4 Lines and planes in space Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 4A

10

, so the Cartesian equation of the plane is

11

, so

, so

The equation of the plane is

Tip Remember that there is no unique correct answer. In fact, you could simplify this further, since any multiple of either direction vector can be offset against the other or the position vector , so you could also give this as the three points you can quickly see that they all have all lie.

12 a

,

, so

b The normal to the plane is

and

The Cartesian equation of the plane is therefore 13 a

,

, so

, so the normal equation of the plane is

b

, so does not lie in the plane.

. In fact, if you look at so this is the plane in which they

14 a b The answer to part a is a vector perpendicular to both directions in the plane and so is the normal to the plane. , so the Cartesian equation of the plane is

15 a b i Assuming and intersect:

Solving: and Then ii Substituting

which is true, so the system of simultaneous equations is consistent. , the point of intersection is

c From part a, the normal vector to the plane is

. .

, so the Cartesian equation of the plane is

16 Vectors from the first point to the other three are

,

and

.

The normal to the plane containing the first three points is

But

, so the third vector is not perpendicular to this normal, and therefore the

fourth point does not lie in the plane containing the first three points.

Tip This is equivalent to showing that the tetrahedron with the four points as vertices has a nonzero volume; see Chapter 6 for more about the vector triple product .

17 a The normal vector

b

, so

for the vectors to be perpendicular.

, so

for the vectors to be perpendicular.

c Then two direction vectors of the plane (taking a multiple of for integer elements) are and

.

From the Cartesian equation, it is clear that point

lies in the plane.

The equation of the plane can be given as 18 a The normal to the plane can be found by taking the vector product of the two direction vectors: , so normal vector

.

, so the Cartesian equation of the plane is

b The normal to the plane has vector

.

Two vectors perpendicular to this (by inspection) are

From the Cartesian equation it is clear that point

and

.

lies in the plane.

The plane can be described by the vector equation

19 The normal to the plane has vector

.

Two vectors perpendicular to this (by inspection) are

From the Cartesian equation it is clear that point The plane can be described by the vector equation

EXERCISE 4B

and

lies in the plane.

.

EXERCISE 4B 3

Substituting the line parameters

,

and

into the plane equation:

so The point of intersection is therefore 4

.

a The direction vector of the line is

.

b The line equation can be given as

.

Substituting into the plane equation: , so

and therefore

.

The intersection point is 5

Substituting the parameterised form of the line equation:

,

and

into the plane

so The point of intersection is therefore 6

a

.

-axis:

; substituting into plane equation gives

-axis:

; substituting into plane equation: , so is

-axis: , so is

, so is

.

.

; substituting into plane equation: .

b

EXERCISE 4C

3

a

Tip Always remember to make sure that the numerators have as the coefficient of , and before reading off the denominators to find the direction vector.

b Normal vector

The angle between the normal and the line is given by

So the angle between the line and the plane is 4

a Normal vector

b Normal vector

The angle between the planes is the same as the angle between their normals.

5

a

b Normal vector

The angle between the normal and the line is given by

So the angle between the line and the plane is 6

a The normal vector can be found from the cross product of the two direction vectors.

b The direction vector of the line

.

The angle between the normal and the line is given by

So the angle between the line and the plane is 7

The two normal vectors are

and

.

But

The normals are perpendicular so the planes must also be perpendicular. 8

a Line can be parameterised as and so the direction vector is

b Normal vector

.

so in vector form is .

The angle between the normal and the line is given by

So the angle between the line and the plane is 9

a Substituting

into the plane equation: is true, so lies in .

b

and the plane normal

If the angle between .

.

and is then the angle between

and the plane normal will be

c d If is the point on such that angle

is

:

Then the distance from to the plane is

EXERCISE 4D 5

Tip If you know the formula for the distance between skew lines, you can quote it and then apply it. Alternatively, use a standard method which will produce an equivalent calculation. Let on the first line (with parameter ) and on the second line (with parameter ) be the closest points so that is the shortest distance between the lines. Then

vectors

must be perpendicular to the two direction

and

.

Simplifying: so

and

Then

6

so the shortest distance

Let be the point on the plane such that

is perpendicular to the plane for

Line

; substituting into the plane equation

has equation

so has parameter

. gives

.

Then the shortest distance from the point to the plane is

7

The plane normal is

A plane with equation

lies at distance

from the origin.

So in this case, the distance to the origin is

Tip More generally, two parallel planes

and

have a separation distance

; the distance of a plane to the origin is the same as the distance to the parallel plane passing through the origin, for which 8

Using the formula for the distance from a point

The line through

and

has gradient

Using the formula, the distance from

. to a line

:

so has equation

to the line is

Tip Here is an alternative method, which would work equally well for a three-dimensional problem. Consider the area of the triangle formed by the three points in space.

,

and

This can be calculated as

It can also be calculated as distance from to

.

where is the perpendicular

Equating these,

.

Tip

9

Either use a longhand method such as shown for Question 5, or the standard formula. The shortest distance between two skew lines with equations

and

is given by

where

The lines are

and

So

and therefore

Since this is not a zero distance, the two lines do not intersect. 10 Using the formula for the distance from a point

The line through

and

has gradient

Using the formula, the distance from

to a line

:

so has equation

to the line is

so using the formula for area of a triangle,

11

Rewriting both lines in the form

so that direction vector

can be

read directly: First line is

so has direction vector

Second line is

with the same direction vector, so the two lines are

parallel. Distance separating lines where

and .

is given by

So

So

and therefore

12 a Reinterpreting the Cartesian equations into the normal product form :

and

, so

and therefore the planes

are parallel. b Since the constant term in the second equation is zero, the origin must lie in the plane. c

The perpendicular distance between two parallel planes

For

, the equations are

So the distance between

and

is

.

.

is

is

is

and

13 a The normal to plane

normal to

and

so the normal to

is

and

. Since the two normals are parallel, the two planes must also be

parallel. b Substituting into the plane equation: c There must be a point in Then has coordinates so .

so

that lies along the plane normal from point for some . Substituting into the equation for

Then the separation between the planes is 14 a The normal vector of is

.

.

, which is also the direction vector of .

Equation of : b To find the intersection of this line with , substituting the parameterised values into the plane equation: , so

, and therefore the point of intersection is at

.

:

c The shortest distance from to is

15 a The normal vector of is

, which is also the direction vector of .

Equation of :

To find the intersection of this line with , substitute the parameterised values into the plane equation: , so

, and therefore the point of intersection is at

.

b The shortest distance from the origin to is

Tip You can calculate this result immediately if you know either of the following facts: the shortest distance from the origin to a plane

the perpendicular distance between two parallel planes

16

equals

and

is

.

for some value

so

Since

is perpendicular to , require

Then , so has coordinates

.

Then the shortest distance from to is

17 a The lines are given by

so both have direction vector

and

.

passes through point and passes through point and since is not a multiple of , the two lines have the same direction but do not intersect so are parallel but not the

same. b Substituting

,

and

into the equation for :

so satisfies the equation and does lie on . c

on has coordinates

so

.

Require that So so Therefore, has coordinates

d

18 a

so

has position vector

and has position vector

.

Then

Since

is perpendicular to :

b Since

is also perpendicular to :

c

: So

so and since

is perpendicular to both lines, distance

between the lines.

MIXED PRACTICE 4 The formula for the distance from a point

to a line

is:

is the shortest distance

1

Using the formula, the distance from

2

to the line

The normal to the plane is

.

Let be the foot of the perpendicular from Then has position vector

Substituting,

to the plane. . Since lies in the plane,

so

.

.

Then the distance from the point 3



to the plane is

a

, so the plane is given by

b Substituting

into the answer to part a:

, so

4

a

, so

b

has equation

and

c

which in Cartesian form is

, so also lies on plane

equation d The angle between the planes equals the angle between their normals.

5

i For the points and

,

and

:

, which has

Then a vector equation for the plane is

ii

Then the normal to the plane containing the three points is given by

so the Cartesian equation of the plane is

6

i The two direction vectors are

and

.

respectively.

The common perpendicular is given by ii The shortest distance between two skew lines with equations by

and

is given

where

The lines are

and

and

Then

7

a b Taking the components separately to form a set of simultaneous equations:

Substituting: So the consistent solution is

and the intersection point is

.

c Part a gives a vector perpendicular to the two direction lines; the plane normal is d

lies on line and so lies in the required plane. , so the equation of the plane is

In Cartesian form:

.

8

a The plane normal

is the direction vector of the line.

The line equation is b Substituting the parameterised form , so

into the plane equation:

and therefore the intersection is at

.

c The shortest distance to the plane is therefore

9

a Plane equation is

where the plane normal

Since

b

.

, the point

.

does lie in the plane.

, so the line equation is

.

c d To find the intersection of the line in part c with plane , substitute the parameterised form into the plane equation: , so

, and the foot of the perpendicular is

e The distance from to the plane is

.

.

10 The shortest distance between two skew lines with equations

and

is given by

where

and

so

Then

11 a

so

b Distance separating lines where

so

and and .

is given by

,

.

Then

so

c The area of a parallelogram is the product of base and perpendicular distance:

12 a Parameterising

.

Substituting this into This is consistent for

, so the intersection point is

b The direction vectors are

and

.

.

is a vector perpendicular to both lines.

c

, so the plane equation is

In Cartesian form: 13 a The normal vector of the plane,

, is the direction vector of the line.

The line has equation b Substituting the parameterised form , so

and the intersection point is

into the plane equation: .

and the line intersects the plane at c On the line, is at is at , with coordinates . , so

d

so the image of in the plane

does lie in the plane

e The shortest distance to the line will lie within the plane, from to the answer to part b:

14 i Parameterised form of line is

.

Substituting into the plane equation:

So

and therefore

and the point of intersection is

ii The new plane contains the direction vector of , given by

is given by

.

. , and the normal of the plane

Then the normal to the new plane is given by contains the point

and the plane

which lies on . so the new plane has Cartesian equation

15 i The normal to the plane

is

.

, so lies both in the plane

and on the line

. Substituting the second equation into the first: so

Then has coordinates

, and therefore

and

.

.

ii The angle between the planes is equal to the angle between the plane normals. Plane

contains vectors

and

.

so the normal to plane

is

.

Then the angle between the normal is where

So the angle between the planes is

16 a b Points and both lie in the plane, and so the plane normal must be perpendicular to .

The whole of the line lies in the plane, so similarly the direction must be perpendicular to . c Calculating

to find the plane normal:

d The plane is given by In Cartesian form: 17 Putting the line in standard form: If the line is to lie within the plane, the direction vector of the line must be perpendicular to the plane.

, so

;

Tip Note also that the point

lies on the line and since

, this point lies

in the plane and so the line does indeed lie within the plane rather than running parallel to it. 18 Writing the line equations in standard form: and

.

The two direction vectors are

and

, so the normal to the plane is

.

A point on the first line is

.

The plane equation is

, or in Cartesian form:

19 a

b The two plane normals are

and

.

The angle between the planes equals the angle between the normals and since that the planes are perpendicular. and

c

, so

does not lie in either

or

, it follows

.

d From part a: a line parallel to both planes must have direction vector

The line has equation

20 a

,

and

, so b From part a:

.

is perpendicular to both

and

.

is normal to the plane containing , and . , so has equation

c Since

is perpendicular to the plane containing , it follows that

is the shortest distance from

to the plane.

d

, so

21 a

,

has coordinates

.

, so

b

c Part a gives a vector normal to the plane,

.

, so the equation of the plane is

In Cartesian form, this is d The line through with direction vector has equation e Substituting the parameterised form c: , so

and the intersection is at

into the answer to part

.

The distance from to the plane is f

The volume of a linearly tapering solid with base area and perpendicular height is

, so the

volume of the tetrahedron (triangular-based pyramid) is

Tip Very often in questions of this sort (as here), roots will recombine at the end of the question; always use the exact values in your future calculations to avoid truncation errors.

and

22 i

so the normal to the plane is

.

so the equation of the plane is

In Cartesian form, this is

.

, as required.

, and is the reflection of in the plane ii Both and lie in the plane plane will be the reflection of plane through the plane

; therefore the

Reflecting the equation of a shape described by a Cartesian equation through substituting for throughout. From part i, the equation for plane Therefore, the equation for plane

is

is achieved by

.

is

iii The angle between two planes equals the angle between their normals.

and

so , the angle between the normals, is given by

Since this value is negative, it represents the cosine of the obtuse angle between the planes; the acute angle

has

.

23 i

ii Then line When

is given by , this gives the point with vector position

By the symmetry of the equations,

is given by

. and

is given by

and in each case, setting the parameter to equal gives the point . Since lies on all three (distinct) lines, it is the common intersection of the lines. iii The line passes through so

.

The line is perpendicular to the plane containing vectors direction . iv The plane is given by

, so the point with position vector

from the origin to the plane. Then the distance from the origin to the plane is

and

so has a

is the foot of the normal

Worked solutions 5 Simultaneous equations and planes Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 5A

3

where

Expanding along the first row:

is singular so there is no unique solution.

Tip Alternatively, observe that linearly independent, hence

4

a

(or is non-singular.

) so the rows (columns) are not

where

There is no unique solution when the determinant is zero, which occurs when b For

:

so the system is inconsistent and has no solution. For

:

so the system is inconsistent and has no solution.

or

.

5

The system of simultaneous equations can be expressed as: where a When

so

and there is no unique solution.

b Expanding about the first row:

From part a,

is a root of

, by the factor theorem.

Factorising:

is non-singular, and so the system has no unique solution, for c When

or

.

,

so

So d When

,

so the system is inconsistent and there is no solution. 6

a

The system does not consist of three independent equations so there is no unique solution. b From part a, for the equations to be consistent,

.

7

For infinitely many solutions, require that the system is consistent with elimination resulting in an equation So

and

EXERCISE 5B

EXERCISE 5B 2

a Writing the system of equations in matrix form: where

and

Expanding about the first row:

The matrix is non-singular so there is a unique solution given by Using your calculator:

.

, so the solution is

.

b The three planes intersect at a single point. 3

a Writing the system of equations in matrix form: where

and

Expanding about the first row:

The matrix is non-singular so there is a unique solution given by

.

b The three planes intersect at a single point. 4

Writing the system of equations in matrix form: where

and

Expanding about the first row:

The matrix is non-singular so there is a unique solution given by Using your calculator: 5

, so the planes intersect at the point

a Writing the system of equations in matrix form: where

.

and

Expanding about the first row:

The matrix is singular so there is no unique solution.

.

b Using row operations on the extended array

The third row shows that the system is inconsistent; any solution would require so there can be no solution to all three equations simultaneously.

,

c The system is inconsistent, but none of the planes represented by the equations are parallel (no row of the matrix is a multiple of another). The system therefore describes a triangular prism. 6

Equation shows that planes and are parallel and distinct (same normal but with inconsistent equations), while plane is not parallel to them, and intersects both of them. 7

a For the planes to be parallel, their normals must also be parallel. and

So

, from which

b

No two of the planes are parallel and since they are inconsistent, they must form a triangular prism. 8

a Writing the system of equations in matrix form: where

and

Expanding about the first row:

The matrix is singular so there is no unique solution. b Using row operations on the extended array:

For the system to be consistent, the third row must consist entirely of zero entries, so therefore .

and

c No row of the matrix is a multiple of another so no two of the planes are parallel. For the three planes to be consistent, not parallel and yet have no unique solution, they must intersect along a common line and form a sheaf. 9

a Since the constant term is zero in each case, the three equations all have common solution and so are consistent. b Writing the system of equations in matrix form: where

and

Expanding about the first row:

So the matrix is singular and there is no unique solution to the simultaneous equations. Since no row of the matrix is a multiple of another, no two of the planes are parallel. Three non-parallel planes in a consistent system form a sheaf, with a common line of intersection. 10 a Using your calculator: b The system of planes can be written as

where

and

.

Then the solution to the system is given by

So the point of intersection of the planes has coordinates

.

11 a Writing the system of equations in matrix form: where

and

Expanding about the third row:

For a system with no unique solution, require

, so

.

b Using row operations on the extended array:

When

, the system is inconsistent.

No row of the matrix is a multiple of another, so no two of the planes are parallel. Three non-parallel planes in an inconsistent system form a triangular prism. 12 a Writing the system of equations in matrix form: where

and

Expanding about the third row:

For a system with no unique solution, require

, so

.

b Using row operations on the extended array:

For

:

The system is consistent if

or c No row of the matrix is a multiple of another so no two planes are parallel. A consistent system of three non-parallel planes with a non-unique solution must intersect in a common line and form a sheaf arrangement. 13 Use elimination; if there is a line intersection then the system of equations will be consistent and elimination will produce .

So

,

MIXED PRACTICE 5 1

The system of simultaneous equations can be expressed as where

Then

(from calculator)

so the point of intersection is

2

The system of simultaneous equations can be expressed as where Then

. When

is singular, the system will not have a unique solution. So

3

Elimination shows that there is no unique solution to the three equations, so there cannot be a single point of intersection. a No two of the planes are parallel: plane normals are

When b When 4

,

and

the system is inconsistent, so the three planes form a prism. the system is consistent so the three planes form a sheaf, intersecting along a line.

a Writing the system of equations in matrix form: where

and

Expanding about the second row:

.

If there is no unique solution then for b If

is singular.

.

is not singular, then

.

So the unique solution is

So the values are

,

,

.

c The system is always consistent when there is a unique solution. When

:

Using row operations on the extended array

When , the first row of the matrix side has only zero elements, so the system will only be consistent if .

So when

5

, the system is always consistent.

i Expanding about the first row:

ii

is singular (has zero determinant and so has no inverse defined) when

iii The system can be expressed as

From part ii, there is no unique solution because

where

.

is singular for that value of .

so the system is consistent for all values of . This system of equations has infinitely many solutions. 6

i Expanding about the first row:

ii The system can be expressed as a When

, is non-singular so there is a unique solution.

b When

,

so there is not a unique solution.

so the system is inconsistent. c When

,

Very clearly,

or .

so there is not a unique solution.

is a solution, so the equations are consistent.

Worked solutions 6 Hyperbolic functions Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 6A 2

is transformed to

, with vertex

mapped to

. Replace with :

stretch parallel to the -axis with scale factor Vertex at

is moved to

Replace with Vertex at

and with

: translation

is moved to

Comparing: 3

.

.

,

is transformed to asymptotes

mapped to

, with origin mapped to

and horizontal

.

Replace with : stretch parallel to the -axis with scale factor Centre point at origin is unchanged; asymptotes ) Replace with

Centre point at

; asymptotes are unchanged.

: stretch parallel to the -axis with scale factor is mapped to

End graph has centre point at End graph has asymptotes at

EXERCISE 6B 5 or 6

(reflected vertically if

: translation

Centre point at origin is mapped to Replace with

are mapped to

; asymptotes unchanged. , so

.

(and the graph is reflected vertically), so

.

7

8

The domain of

is

The domain of

, so the domain of

is

is

, so the domain of

. is

For a domain with maximal interval width of , require that the lower end of the domain of is within the domain of . Then

9

10

or Reject

as it is outside the range of

11 12 Let

so that

.

Quadratic in :

Tip Select the positive root since you require that Taking the natural logarithm of both sides: Hence 13 Let

, as required. so that

.

.

.

so

Taking natural logarithm of both sides: Hence

, as required.

14

Tip This is a proof that is an even function, where, for any given value of the range, the corresponding values in the domain come in pairs. The range of

is

For

and

Therefore,

and so

, so the domain of

is

.

. .

This first possible expression is shown to be non-negative (and so the other is non-positive, since they must sum to zero).

EXERCISE 6C 1 2

3

Tip Some identities are easier to prove if you start with the RHS and work towards the LHS.

4

Tip Here the result from Key point 6.7 has been used to simplify the numerator.

5

Tip The identities established in question 3 and Worked example 6.6 have been used. Dividing numerator and denominator by 6 Therefore 7

8

9

so

a Similarly,

so

Adding these together:

b

10 If

EXERCISE 6D

, then

gives

1

Using the definitions of

So

and

:

and

:

and

:

and

:

and therefore

2 So or

3 So 4

or

Using the definitions of

5

so

6

or

or 7

Using the definitions of

So Multiplying through by

:

Reject negative solution. So 8

Using the definitions of

Multiplying through by and factorising:

So, 9

Using the identity

:

or 10 so Reject negative root. So Therefore,

Tip , solutions can always be given in the form 11 So Therefore 12

Multiplying by and factorising:

.

(reject) or

13

(reject

since

14 a

for real )

so

Therefore,

and

, as required.

b Then So Expanding:

Multiplying through by and rearranging:

Solutions:

and

or

and

.

Tip Since the equations are symmetrical, it shouldn’t surprise you that the answers are also symmetrical. 15 Multiplying through by

and rearranging:

Quadratic in has at least one real root if the discriminant

:

So

Tip Since the quadratic can be expressed as , where both and must be positive, the symmetry line of the quadratic will always have a positive value for , so that, if there is at least one root, there must be at least one positive root and there is indeed a real solution for .

16 a

using part a for

b

Factorising, noting that if

the equation is valid, so

(The quadratic expression has discriminant

is a factor:

so has no real solutions.)

and

17 a So b

EXERCISE 6E

Tip Note that if

for

then

, and you may use

either form (with the inside or outside the logarithm) in your answer. Proof:

2

a b at turning point.

3 So

or

Turning point at

(reject because , at the point

for real ) .

4

at the turning points.

When

,

, so

and

Minimum is at

.

, so

5

At

,

and

The equation of the tangent at

is

. Multiplying through by

and rearranging:

, so

6 At

,

and

The gradient of the normal at this point is therefore

and so the equation of the normal is

Rearranging:

7

a

, so At the turning points,

, so

Then b The graph of

looks like:

(rejecting the negative root)

so the second derivative is negative for occurs at At this point:

so

8 Stationary point occurs when Multiplying through by

and rearranging:

, so At this point:

The stationary point is at

9

, so

.

and

Points of inflection occur where so , so

10 a

, so Stationary points occur when

, so

, so

and therefore the maximum on the graph of

b Since

and

are both odd functions and

is an even function, it follows that

is

an odd function. Therefore Consequently, for the stationary points with

, the second derivative will be

positive at one (indicating a local minimum) and negative at the other (indicating a local maximum).

Tip You could go so far as to calculate values, or sketch the graph of or

; however, since

the question does not ask which is the maximum and which the minimum, the reasoning shown is perfectly sufficient.

EXERCISE 6F 6 and

7 So

8

9

Using the formula for the volume of rotation about the -axis,

a Using the fact that

and

:

:

b

10 Using integration by parts:

11

Using the formula for the volume of rotation about the -axis, so when

:

,

Rotating the area below the line around the axis produces a cylinder with volume The volume to be excluded from the desired volume of rotation is therefore:

So Then the volume to exclude So the required volume is 12 Let When

, so ,

. and when

,

.

13 Using the chain rule:

Then using integration by parts:

14 a b Using integration by parts:

MIXED PRACTICE 6 1

and

a

b , so 2 so

3

4 So Therefore

Tip If you calculate the solution without using the logarithm formula for arcosh, you might get the result as your final answer instead. Is this a problem? Evaluate the two answers on your calculator and explain the result. 5

So

6

(reject So 7

, so

as it is outside the range of

)

8

9

for all , so there can be no point of inflection, at which the second derivative would equal zero. 10

11 Collecting terms, rearranging and multiplying through by

This is a quadratic in ; Only the positive root produces a real solution for

12

13 Let When

, so ,

and when

,

:

14 Rearranging and factorising: or

(rejecting the negative root as it is outside the range of

or 15 Using the identity

:

So since

16 17 Let So Then

18

19 a

b

So Therefore c

i Dividing by

:

for all .

)

So the equation becomes So

, as required.

ii Factorising: So

(reject because the range of

is

)

or Then, using the formula found in b:

20 a

So Dividing both sides of this identity by

gives

b

or

:

:

21 i

so Multiplying through by and rearranging gives Using the quadratic formula: Only the positive root gives a positive value for , so Therefore

ii Similarly,

.

, which is a quadratic in .

Substituting into

:

Rearranging: Squaring:

Squaring:

(Because the initial equation contained 22 Let

, so

23 Intersections where

or or

24 a

So

:

, the negative root is not a valid solution.)

b If

then

c

25

so

From (4):

.

Substituting into (3): , so , so Solutions are 26 The logarithmic formula for Therefore:

27

or is

Tip You could also answer this question using a hyperbolic identity from the A Level Further Mathematics course: Using

and taking

Rearranging:

and

28 Using

:

of both sides of the initial equation:

Worked solutions 7 Further calculus techniques Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 7A so by the chain rule,

2

Then at

3

,

so by the product rule and chain rule,

Then at

,

4

so by the chain rule,

5

so by the chain rule, domain Here,

6



has

so its derivative has domain so if

then

and so the domain is

so

a

Differentiating with respect to : But

(standard identity) so so by the chain rule,

b

so by the chain rule,

7

so its derivative has domain



Here,

so the domain is

8 9

. Using chain rule: a b Then

. Using product rule and chain rule:

so if

has domain

then

and

10 Inflection points occur only when the second derivative is zero. . Using chain rule:

Then using the product rule:

The domain of the derivative is , for which the denominator is always positive; the numerator has minimum value . Therefore the value of the expression is always positive. It follows that there can be no points of inflection.

EXERCISE 7B 2

Using the chain rule:

3

Using product rule and chain rule:

4 Using the chain rule:

So

and so

5

Then

and

So the tangent line is given by equation Rearranging: 6

so

and

,

and

Then the tangent at the origin has equation

and the tangent at

has equation

For the intersection of the two lines, substitute the first tangent equation into the second:

so

7

Then Point of inflection occurs when

so

so the point of inflection is at 8 Using the chain rule:

Using implicit differentiation:

Multiplying through by

:

9 Using the chain rule:

If

then

Solving the quadratic: Then So

EXERCISE 7C

or

(reject)

EXERCISE 7C 4

5

6

7

a b

8

a Let Then

so that

and

so

Substituting into the integral:

b 9

a Let

b Using part a,

10 a

so

and

. The substitution is valid for

b

11 a b

12 a Let

so that

and

The substitution is valid for Then

.

so

Substituting: b

. so

13 Let

so that

and

so Limits: when Then

14

15

.

, and when

,

.

16

Tip Remember you can split elements of a fraction numerator. Look for that part of the numerator which is the derivative of the internal function of the denominator to use the reverse chain rule.

17

18

Tip The modulus signs are not needed in this case because is always positive, but there is no harm in using them in your answer. If the internal function could take negative values, the answer would be incorrect without modulus signs. 19 a b

20 Let

so that

and

Substituting:

21

Tip This question can be approached directly, using substitution, or with a little forethought, you

can set up the answer as in Exercise 7A Question 9, differentiating the non-trigonometric expression on the right side of the equation. Both methods are offered below. Method 1: Direct approach Let Then

so that

and

.

so

Substituting:

Method 2: Rearranging first Let

. Using product rule and chain rule:

Then

Let Then

so that so

and

.

22 a

b Let

so that

Then

and

so

Substituting:

EXERCISE 7D 2

for some constants , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

so

3

a

, then

and

for some constants , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

so

, then

and

b

so

4

a

for some constants , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

so

, then

b

and

for some constants , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

so

5

a

, then

and

for some constants , , and

Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

so ,

so

Then b

6

for some constants , , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

Then So,

, so

,

, so

,

and

7

for some constants , , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

,

,

8

and

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

So

,

and

,

.

and

MIXED PRACTICE 7 1 Using the chain rule:

.

2 Using the product rule:

3 Using the chain rule:

so by the chain rule,

4

If

then

5

6

i

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so Then ii

7 Using the chain rule:

, so

and

Then using the product rule:

8 Using the chain rule:

So Then using implicit differentiation:

Multiplying through by

9

10

11

12 i

ii

and rearranging:

13 i

for some constants , , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so , so , so

, and

Then ii

14 a

b

15 Let Limits: when

so ,

and and when

,

16 a

for some constants , and Multiplying through by the denominator on the LHS:

Comparing coefficients:

so

, then

So b

c

17 a Let

Let

and

so

and

and

Integration by parts:

Rearranging:

Tip It would also be acceptable to show that the derivative of the expression on the right gives the function being integrated on the left. b If

then the upper half of the ellipse is given by

, for

. Using part a, the area of the upper half of the ellipse is therefore

Then the total area of the ellipse is double this: 18 i Let

.

so

Then Taking the reciprocal of both sides: ii Let

iii Let

so

so

so

and

.

Worked solutions 8 Applications of calculus Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 8A

so

2

so

so

Maclaurin series:

3

so

a

so so so

Maclaurin series:

b Then 4

a i

so so so so so

ii Maclaurin series:

b

5

a

so

so so so so so so Maclaurin series:

Tip Alternatively, use the expansion for

up to

b 6

a i so using the chain rule: Then the quotient rule gives

ii

b Then

,

Maclaurin series:

,

,

and

and square it.

c

7

a i

so using the chain rule:

ii Then

,

,

and

Maclaurin series:

b

for

.

Substituting: 8

a Proposition:

for

Base case:

so the proposition is true for

Inductive case: Assume the proposition is true for

so

Working towards:

Conclusion: The proposition is true for

and if true for

then it is also true for

Hence, by the principle of mathematical induction, the proposition is true for all integers b From part a, Maclaurin series:

Then

General term 9

a Proposition:

Base case:

for

so the proposition is true for

. .

Inductive case: Assume the proposition is true for

so

Working towards:

Conclusion: The proposition is true for

and if true for

then it is also true for

Hence, by the principle of mathematical induction, the proposition is true for all integers b From part a, Maclaurin series:

Then

, so the general term of the series is

10 a

so

so

so

so

so

so

Maclaurin series:

b i Since

and

, it follows that, for

(as found by repeated differentiation) So

,

.

ii Taking the sum of the two series, all terms cancel except the constant term, which is

Tip You may have noticed that this question uses somewhat circular arguments; indeed, as soon as we know that

, integration immediately gives that

, and evaluation at

shows

.

In many exam questions requiring 'proof', you have to start with some basic assumptions; unless explicitly advised otherwise, you can use any information in your formula book as a starting point. In this question, you can assume the results for

and

. In

another question you might be required to prove those results, in which case you would need to assume that in a substitution, and in yet another question you might have to prove that for all . 11 a

For the Maclaurin series

,

-ntercept in the diagram is not at b

.

For the Maclaurin series

,

this is clearly not the case in the diagram, since the function has a positive gradient at

EXERCISE 8B 4

The expansion is valid for 5

6

but the

so

but .

7

Tip

a

You could calculate the terms directly from the definition of the Maclaurin series using derivatives evaluated at zero, or compound the two series for and . Both are shown here. Method 1: From derivatives Let

Maclaurin series: So Method 2: Using given series and

b

8

a

Then

b c Assuming Maclaurin series can be differentiated term by term to find the expansion of a derivative function:

9

a

Maclaurin series: So b

and So the equation of the tangent at

is

10 a

The two expansions of and so the combined expansion is valid for b Let

are valid for .

so that

Tip Alternatively, and more accurately, use

so that

and

respectively,

This is more accurate because although the end result has to be doubled, it uses a smaller value for , so that the discarded terms are less significant to the end estimate.

has

the advantage that the first three terms of the approximation each produce terminating decimals, so are easier to evaluate without technology.

EXERCISE 8C 3

The function

is not defined at

, one of the integral end points.

The integral converges, so this is an improper integral. 4

5

6

Tip It is tempting to just take the limit through the logarithm function:

In this particular case, this approach reaches the correct solution, but taking a limit through a function is not generally a reliable method, and you should avoid it. In questions concerning limits, you should restrict yourself to methods shown in the Student Book. In this case, you can use the Maclaurin expansion for to obtain a series of negative powers of , to which you can apply the limit directly. Remove factors from the logarithms so that each is a logarithm of the form only of for some polynomial . Then either use the Maclaurin expansion for repeatedly or restate as logarithms so that you can apply the limit. Continuing the calculation from (*):

EXERCISE 8D 7

a When

,

so has coordinates

b

8

When

,

and when

9

10 So

, from which

11 Roots of the curve are at

,

12

So

and therefore and

13 a

intersect where

so

so

Intersection points are

or

and

b In the interval

so

and

14

Intersection where Intersections are at

:

so

and

, and in the interval

15 a Intersection where Squaring both sides:

or Intersections are b The two curves are

and which is equivalent to

and which is equivalent to

16 Only need the upper semicircle: , from

to

.

17 The point

is described by

and

is described by

18 The point

is described by

and

is described by

19 The point of intersection with the positive horizontal occurs at at

and the point at the origin occurs

.

20 Taking the apex of the cone at the origin, require a straight line graph such that when the function to be used is

21 The intersection of

and

is at

.

The volume described is equivalent to the volume of revolution when the region beneath between and is rotated around the -axis.

EXERCISE 8E

3 4

Mean value of

between and is

So

. Given

, the only solution is

so

5

a Mean value of

between and is

b

so

6

Mean value of

between and is

7

Maximum power is Mean power over a single period is

So the ratio 8

Let

is . The mean value of

Then the mean value of

9

, where

between and is

between and is

a

, so b The curve is concave up (the gradient is negative and the second derivative positive) for all the curve between and lies entirely below the chord connecting and and in consequence, the average value of the function is less than the -oordinate of the midpoint of that chord. c For

, the mean value of the function between and is

However, by the argument in part b,

So Multiplying through by

:

then the assertion will be true, but if the 10 If the function is increasing for the entire interval function has a local maximum or minimum value which lies outside the interval , then the assertion may not be true. For a simple counter example, choose a non-constant function with a minimum, and select end values and so that the minimum lies within the interval, but with very little of the graph lying in the box formed by the lines , , and . For example, throughout

for , . Then and and very clearly the mean value will be negative.

but the function is negative

Tip In fact, it is easy to show what the criterion is for the mean value lying within Divide the curve up into the intervals where it lies below the line , the intervals where it lies above the line and the intervals where it lies within Then if the total area enclosed by the curve above the box is and the total area enclosed by the curve below the box is , the area below the curve within the box is and the area above the curve within the box is , the criterion for the mean value lying within is that and . In a single inequality:

.

MIXED PRACTICE 8

1

so

a

so Maclaurin series:

. b Substituting

2

:

so

a

so

so so so b Maclaurin series:

c Substituting 3

a

gives

is defined for So

.

is an improper integral because the function being integrated is not defined at the

lower end of the integral interval. b

The limit does not converge, so the integral does not have a finite value. so

4 Then

and ,

and

.

Maclaurin series gives So for

, the first term in the Maclaurin series is

, so

5 6

7 and If

, then

.

8

9

The expansion of converges for all values of The expansion of So the expansion of

converges for is valid for so

10 a

so

so so so

Maclaurin series:

, so for

.

.

b Substituting

into the result for part a:

So

Tip Note that substituting

into the result for part a gives an alternative approximation:

While equally valid, this is not the result required … Inspection of the denominator should clue you in to which value of is required. In fact,

and

, while

, so the

substitution gives a better approximation. Can you explain why you could anticipate that, given the series expansion? 11 a

so

so

so

so b Maclaurin series:

c

12 i

Using the product rule:

So ii Using the product rule:

From part i:

iii Using implicit differentiation on the result from part ii:

Then iv Maclaurin series: So 13 The curve has roots

at and

Taking

By symmetry, if

then the volume will have the same positive value, so in general, the volume of

revolution is and

14 a So

and

. Then

lies on both curves.

b The shaded area can be split into two; the area to the left of for of for .

and the area to the left

Rearranging the two curves to give in terms of :

and

15 Translating the curve down by unit so that the volume is rotated around the -axis: , which is bounded by the -axis between and Then the volume of revolution is

16 i The region is bounded by the curve

and the line

ii Rotating the region about the -axis:

17 i Using the product rule:

ii Rearranging the equation of the curve to write in terms of : gives

.

iii The rectangle bounded by the lines and height

,

,

and

forms a cylinder with radius

when rotated about the -axis. This cylinder has volume

, so the volume

produced when the non-shaded portion of that rectangle is rotated about the -axis is

18 Intersections of the two curves occur when The region to be rotated is bounded by for .

so and the -axis for

19 Consider the part of the curve above the horizontal for The volume of revolution is given by At

,

and at

,



so

.

and by and

20 The volume of revolution is given by At

,

and at

,

Using integration by parts:

21 a So b

Tip Line of the working assumes that it is valid to say that an infinite series can be integrated term by term; in this and many other cases, the result is true. However, if you study this branch of mathematics at university you will learn that convergence properties should be assessed before making such assumptions. c Substituting

into the integral:

Then

22

so the positive solution is 23 a Let

and

.

Intersections occur when : , so the intersection points are b In the interval

,

so .

and

Solutions are

and

, so the volume of revolution is given by

c 24 a

Intersection points of

and

are at

and

.

b c The region is symmetrical about

, so the volume of revolution is the same about either axis;

the volume of revolution about the -axis is therefore also

.

Worked solutions 9 Polar coordinates Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 9A 4

a

is the locus of points no more than units from the pole, which is the interior of the circle, centred at the origin, with radius (including the circumference).

b

is the locus of points between and units (inclusive) from the pole, which is the interior of an annulus, centred at the origin, with inner radius and outer radius . Both circumferences are included in the locus.

c

is the locus of points in the wedge of space between the (non-included) line at angle (this is the positive

5

) anticlockwise to the (included) line at angle (this is the negative -axis).

a Question requires values of for which for b

and

: so the curve is not defined for these values of .

Tip Sketch using a table of values or by previous experience, using known properties of period and symmetry to minimise the work needed. You can convert the locus formula using a table of values:

Then evaluating further values using periodic property

6

a

is not defined for

b When

,

and symmetry property

; that is, for

and when

,

so both and lie on the curve. c Using a table of values:

Tip Can you convert the locus formula into a parametric Cartesian equation and prove that it is a circle? d

:

EXERCISE 9B 3

a Tangent occurs when

:

for b The curve is restricted to

:

In the intervals

and

, the function would give

so the curve is not defined for these values of . c

Tip Sketch using a table of values or by previous experience, using known properties of period and symmetry to minimise the work needed. Using a table of values:

Then evaluating further values using periodic property symmetry property

4

a

b at

, , and

:

and

5

Maxima at

and

a Require that

so

; minima at

for

and

so

Then the curve exists for b Tangents at the pole occur when

:

so Then

.

Maximum at

and

So maximum occurs at 6

has range

a

and

.

so the maximum value of is at

, and the minimum value is at

. b

Tip Sketch using a table of values or by previous experience, using known properties of period and symmetry to minimise the work needed. Using a table of values:

Then evaluating further values using symmetry property

:

7

a

b

8

a at is the maximum value of . is the minimum value of .

b

9

10

EXERCISE 9C 5

a

b c

6

or Since =

gives

7

8

9

a

b

EXERCISE 9D 2

a

, the first solution is included in the second, so the equation is:

b The area enclosed by a curve using polar coordinates is So the area enclosed by the circle

is

3

4

5

6

a

b

The area is constant, independent of the value of . 7

a  

.

b

8

a  

for

Tangents at the pole occur when So tangents occur when Solutions:

. .

, so

b Due to the symmetry of this curve, the total area for this curve is double the area of one ‘petal’.

The total area is double this: 9

When

,

Since

is clearly increasing (because the derivative is

originally being integrated, which is always positive for interval; there is a second solution in the interval

, the function

), this is the only solution in that , but this is not a valid solution, as it

arises from the integral calculating the area in this interval as negative. Therefore

.

10

EXERCISE 9E 1

a

lies on the curve, so at

.

also lies on the line

, so at ,

is at b i

. is a right-angled triangle with base

and height

, so the

Tip Or calculate the area directly using the polar integral:

ii The shaded region is the area enclosed by the curve less triangle

2

The curve meets the initial line at

and the vertical at

.

, so the right-angled triangle formed

by these end points and the origin has area . The total area enclosed by the curve is given by:

So the shaded area is this total enclosed area, less the triangle area.

3

a The two curves intersect where Solution:

or

, so

, so or

.

Polar coordinates of the intersection points are

and

.

b The shaded region comprises: the interior of the circle for the interior of the curve symmetry).

, which is one sixth of a circle, with area for

and

;

(which are equal, by

4

a The two curves intersect where Solution:

, so

or

Polar coordinates of the intersection points are

and

.

b The shaded region comprises: the interior area of the circle for

(that is, two-thirds of the circle, which has area

; the interior of the curve

for

.

For simplicity of calculations, using the symmetry of the shape through the vertical:

5

a The two curves intersect where

, so

(and also at the pole, for

and

.

different values of ) Solution: Polar coordinates of the intersection points are

b The two shaded regions are identical. The upper one comprises:

the interior of the larger curve for the interior of the smaller curve for

, .

Therefore the total shaded area is 6

a

.

: Circle with radius centred at the pole.

b The two curves intersect where

, so

or

.

The required area is the difference between the area enclosed by the curve for

and the

area enclosed by the circle for the same interval. Since the shapes are symmetrical about the horizontal, this is the same as double the difference between the areas for

.

MIXED PRACTICE 9 1

Minimum value of

is

so maximum is

2

Tip The solution 3

is included within this, at

a Cartesian coordinates are

, so you do not need to state it separately.

and

Distance b Area

4

(using sine rule for area):

Tip With no guidance, you may approach this question either by using a graphical calculator, or by a table of points which you plot and join, or by attempting to transform into a Cartesian equation. Be warned that it is unusual to have to transform a question into a simple Cartesian equation unless you are actively steered in that direction! Method 1: Table of values

Tip Use

and repeat the shape for the second period of the function, rotated

.

Method 2: Transform to Cartesian equation

Tip As perhaps expected, although you can transform into a Cartesian equation, it does not particularly help! In this case, a calculator or join-the-dots approach is best.

5 6

a

b 7

a  

forms a spiral from the pole:

b

8

where

a

and

Then the Cartesian equation is equivalent to or more simply b A tangent at the pole occurs when , so c

or

:

, so that

d Using the symmetry along the horizontal:

9

10 a

when

or

The tangents at the pole are b

c

and

11 a

b Maximum occurs when

, so has polar coordinate

Cartesian coordinates for this point are 12 a Tangents occur when for b

:

c

13 a

,

and

, which occurs at

Polar coordinates of the points are

b When

.

; solutions are

is maximal when

.

.

and

and .

.

.

The point is 14

Tangents at the pole when , ,

,

,

:

,

Curve crosses the initial line when 15 The curve touches the pole when

, i.e. at the point

  , so

:

So the shaded area is enclosed by the curve for

.

16 a

b The two curves intersect where So the intersection points are

, so and

: .

or

.

c

The enclosed area is: bounded by the curve bounded by the circle

for for

, (that is, two-thirds of the circle, with area

).

Tip 17 a i

Remember you can choose your method; de Moivre’s theorem proof or repeated use of compound angle formulae are both acceptable. For , the compound angle method is arguably briefer. Using double angle identities and the compound angle identity for cosine:

ii Tangents at the pole occur when Using part a i: Solutions for

so :

at

and

at

So the tangents at the pole (since the tangent lines are found on both sides of the pole, since is included for this function) are b The large loop is enclosed by the curve for curve for

by symmetry across the horizontal.

, which is double the area enclosed by the

Tip There are many ways you could approach this integral; one possible method is shown.

Let

.

Then (using integration by parts) for

:

So Since

Then from

:

:

18 i Substituting

,

so

:

so ii (The table shows

; use symmetry about the initial line)

crosses the axes at

iii

and

, which are both points on the curve.

The area above the line, shaded blue, is the area enclosed by the curve in the interval less the triangle with vertices at the origin and the two intersection points.

The total enclosed area is given by

Then

so the ratio of the areas is

19 i The curves intersect where So

or Then the intersection point must be at angle where ii The shaded area is bounded above by Therefore it is bounded by

But So

so .

for

and below by and by

. .

for

.

Worked solutions 10 Differential equations Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 10A 4

a Homogeneous equation is

Complementary function:

for unknown constant .

so

b

Substituting into the differential equation: (for all ) Comparing sides of the equation: Particular integral:

c General solution:

Substituting

: , so

Particular solution:

5

a Homogeneous equation is

Complementary function: for unknown constant .

, so

, so

b

Substituting into the differential equation: (for all ) Comparing sides of the equation:

and

Particular integral:

c General solution:

6

a Try

so

Substituting: So

so

and

so

Particular integral: b Homogeneous equation

Complementary function:

so

.

c For a linear differential equation,

but this equation is not linear: there is a

and a

term in the differential equation, so the sum of the complementary function and the particular integral will not be a valid solution. d Substitute

so

Then Substituting back to the original variable:

7

a Homogeneous equation is

Complementary function: for unknown constant b Try

, so

Substituting into the differential equation:

Particular integral:

c General solution:

EXERCISE 10B 2 Integrating factor:

General solution: Boundary condition: Particular solution: 3 Integrating factor:

General solution: 4 Integrating factor:

General solution: 5

, so

Integrating factor:

General solution: Boundary condition:

, so

Particular solution: 6 Integrating factor:

General solution: Boundary condition: Particular solution: 7

For an equation of the form including the constant of integration would make the integrating factor where Multiplying through by the integrating factor:

But at this point (or any line previously), the exponential of a constant can just be cancelled:

So inclusion of the constant in calculations is irrelevant and does not affect the end solution. 8

Integrating factor:

General solution: 9

a

for unknown constant

, so Substituting into

:

b Integrating factor:

General solution: c Converting back to the original variable: General solution:

Boundary condition:

, so

Particular solution: 10 a

, so Substituting into

:

Integrating factor:

General solution: Converting back to the original variable:

General solution:

for unknown constant

Boundary condition: so the curve follows the negative root and

, so

Particular solution: b Let

, so that

Substituting into

:

Integrating factor:

General solution: Converting back to the original variable: General solution:

Tip Notice that usually you would say that

, with modulus signs around

the argument of the logarithm; for

, the primary interval

over which the function

can be integrated coincides with the primary

interval over which , so the modulus signs can be dropped in this case. In any further working you would need to be checking that the function is restricted to domain .

EXERCISE 10C 1

a Auxiliary equation: b

, so General solution:

2

a Auxiliary equation:

or

, so

b General solution: 3

a Auxiliary equation: b

, so

(repeated root)

General solution: 4

a Auxiliary equation: , so

or

General solution: b So Initial conditions: , and so

,

Particular solution: 5

a Auxiliary equation: , so

(repeated root)

General solution: b So Initial conditions:

Particular solution: 6

a Auxiliary equation: , so General solution: b so Initial conditions:

Particular solution: 7

a

Auxiliary equation: , so

or

General solution: b Boundary conditions: , so

,

Particular solution: Then 8

a Auxiliary equation: , so

(repeated root)

General solution: b Boundary conditions:

Particular solution: 9 Auxiliary equation: By inspection,

, so by the factor theorem,

is a factor of

.

Factorising further:

So

or

General solution: 10 Auxiliary equation: So the roots are

, recognising the binomial coefficients. (triple root)

General solution: 11 Try

so

and

Substituting: , so

;

General solution:

Tip Notice that having found two possible solutions any multiple of those solutions are also valid,

because this is a homogeneous equation Therefore, two unknown constants can be incorporated into a solution, and since this is a second order differential equation, this solution is the general solution. 12 a

Let

, so

and

Then

, so

b Auxiliary equation: , so

or

General solution: c Converting back to the original variable: General solution: So

Initial conditions:

Substituting So

into

:

and so

Particular solution:

EXERCISE 10D 1

a Auxiliary equation: , so

or

Complementary function: b Try

, so

and

Substituting: , so Particular integral: c General solution: 2

a Auxiliary equation:

, so

or

Complementary function: b Try

, so

and

Substituting: Comparing coefficients:

, so

Particular integral: c General solution: 3

a Auxiliary equation: , so

or

Complementary function: b Try

, so

and

Substituting: Comparing coefficients: , so Particular integral: c General solution: 4

a Auxiliary equation:

, so

Complementary function: b Try

, so

Substituting:

and , so

Particular integral: General solution: c Initial conditions:

, so

Particular solution: 5

a Auxiliary equation: , so

(repeated root)

,

Complementary function: Try So

and

Substituting:

Comparing coefficients:

, so

Particular integral: General solution: b Initial conditions:

, so

,

Particular solution: 6

a Auxiliary equation: , so Complementary function: Try So

and

Substituting:

Comparing coefficients:

Particular integral: General solution: b Initial conditions: Particular solution: 7

a

, so

,

Auxiliary equation: , so

(repeated root)

Complementary function: b Try So

and

Substituting:

Comparing coefficients:

Particular integral: c General solution: d Initial conditions:

, so

Particular solution: where

8

Auxiliary equation: so

or

Complementary function: Try So

and

Substituting: Comparing coefficients:

so

Particular integral: General solution: Boundary conditions: so

,

Particular solution: 9 Auxiliary equation: so

or

,

Complementary function: Since

is a term of the complementary function, try

So

and

Substituting:

so

Particular integral: General solution: 10 a Auxiliary equation:

so

Complementary function: Since

is a term of the complementary function, try

So and Substituting: Comparing coefficients:

so

Particular integral: General solution: b So Initial conditions:

so

and

Particular solution:

MIXED PRACTICE 10 1 Auxiliary equation:

, so

Complementary function: 2

3

a b

, so

or

and

4

has the form

a

The integrating factor b

General solution: 5

i Auxiliary equation:

so

Complementary function: ii Particular integral: Try

so

Substituting: Comparing coefficients: so

and

General solution: iii As

, the exponential term becomes insignificant, so for large , the function is approximately .

For

,

and

Then So for large , the function oscillates approximately between 6

a Auxiliary equation: , so

or

Complementary function: b Try

, so

Substituting: Particular integral: c General solution: d

and , so

and

Initial conditions:

, so

,

Particular solution: 7

a Try

, so

and

Substituting:

, so

Particular integral: b Auxiliary equation:

, so

Complementary function: General solution: 8

a Auxiliary equation: , so Complementary function: b Try

, so

Substituting: Comparing coefficients: so Particular integral: c General solution: d

Initial conditions: then Particular solution: 9 Integrating factor:

and

General solution: Boundary condition:

so

Particular solution: 10 Auxiliary equation: so

or

Complementary function: Particular integral: Try Substituting:

so

and

so

General solution: Then

Initial conditions:

so

and

Particular solution: 11 Integrating factor:

Boundary condition:

so

Particular solution: 12 i Auxiliary equation: so

or

Complementary function: ii Particular integral: Try

so

and

Substituting:

so

.

General solution: iii Then

Initial conditions:

so

and

Particular solution: 13 i Auxiliary equation: so

or

Complementary function: Particular integral: Try

so

and

Substituting: Comparing coefficients:

so

and

General solution: ii Then

Initial conditions:

so

and

Particular solution: , the exponential part becomes insignificant, and so for large , the curve closely iii As approximates .

14 a Integrating factor:

b Let

, so

Substituting into

:

c Converting to the original variable: so

(where unknown constant

15 a Integrating factor:

General solution: b Boundary condition: Particular solution: 16 Auxiliary equation: so

or

Complementary function: Particular integral: Try Substituting:

General solution:

so

so

and

)

Worked solutions 11 Applications of differential equations Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 11A 1

a Using

and

, with the positive aligned downwards:

, so b Separating variables:

Tip When separating variables it is often preferable to leave multiples relating to rate constants on the same side as to reduce rearrangements after integration.

When

2

,

, so

a Force So so Integrating: Rearranging:

Initial condition:

so

Multiplying numerator and denominator by for integer values throughout:

b The model predicts

when

and

when

. So the model seems good for

small values of , but it doesn’t work for larger values of . 3 Auxiliary equation:

So a If

where , then

and

General solution:

b If

, then

, so where

and

where

General solution:

c If

, then

General solution:

and

where

and

4

for population and time (years)

a

Boundary conditions:

, so Then b Then

, so

Initial condition: So Then So as

,

c If the population change has a positive element proportional to the size of the population, this element is likely to represent a natural net birth rate (i.e. birth rate less death rate). 5

a

b Integrating factor:

(using integration by parts)

Initial conditions:

so Then

, so

Then c From part b: Given the accuracy level of the parameters used, a realistic estimate of the temperature would be ‘approximately ’. d Different parts of the chicken will have different temperatures (it is not uniform; the temperature has to penetrate through the chicken). 6

a b

as a function of is a quadratic with roots at point occurs when

and

, so by symmetry, the maximum

.

c The model is assuming that the spread of the rumour involves an informed person meeting an uninformed person, with random mixing and a fixed probability of the rumour being passed on. (In a random mixing situation, the number of meetings of the informed and uninformed would be proportional to the product of those populations, .) The underlying assumptions include: the probability of passing the rumour at such a meeting is constant (i.e. interest in the rumour is constant) mixing of students is random (this is VERY unlikely to be the case, since students typically form social groups or at the very least might socialise differently within classes/years than between them) mixing continues throughout the course of the model (again, this is not valid if students are in classes and therefore potentially segregated in smaller populations for periods of time) the spread of the rumour occurs solely between members of the population in question (no external interaction) the total number of students, and initial population with knowledge, are both large enough that the population can reasonably be modelled as a continuous variable throughout the course of the model. 7

a If is the volume of a sphere, then Then

and therefore is proportional to .

is proportional to , and so is proportional to the surface area of the bacterium.

Surface area could reasonably be considered a limiting factor when considering how readily the bacterial cell can take in nutrients for growth. b Let

, so

Separation of variables:

Converting to the original variable:

Initial condition:

so

c

As

,

EXERCISE 11B 4

a Let be displacement above the equilibrium position ( ). SHM with amplitude and angular frequency has general solution Initial condition:

, so

Particular solution: Then At

above the equilibrium position.

, so the maximum speed of the particle is

b 5

, the particle is

a Let be displacement above the equilibrium position ( ). SHM with amplitude and angular frequency has general solution So Initial conditions: Then At

, so

and

, so , the particle is

below the equilibrium position.

b The particle first passes through

when

, so

Maximum speed is

and is achieved when the ball passes through the equilibrium position.

The speed of the particle at this time is 6

a For SHM with amplitude and angular frequency , the maximum speed is

.

So in this instance, the maximum speed is b General solution: So Initial condition:

, so

(taking displacement as maximum at

).

Particular solution: Then

Then 7

8

a For SHM with amplitude and angular frequency , Substituting

,

b Substituting

,

and

:

and

:

a Let be displacement above the equilibrium position ( ), where ‘above’ represents the direction to the particle immediately after . General solution: Amplitude

; angular frequency

Initial condition:

.

, so

Particular solution: Then , so

b c

, so at

, and so ,

The particle has positive displacement and positive velocity, so is moving away from the equilibrium position. 9

a For SHM with amplitude and angular frequency , Substituting

b Now substituting

10 a The amplitude is

,

and

,

:

and

into the formula:

and the angular frequency

b Taking displacement at as positive c The displacement at is

,

so the period is

Solving

:

, so

d Acceleration

, so when

,

11 a Force So , and since for constant angular frequency describes simple harmonic motion. b At the moment of impact,

, the equation

and

Under SHM with amplitude and angular frequency , Substituting:

, so

, so maximal acceleration occurs at maximum compression At that compression, force c SHM with amplitude equation

, angular frequency

and initial position

has

Maximum compression occurs when , so

12 a The extension of the spring is (so that negative values of represent compression). The tension will be directed against the extension. , so

which describes SHM with angular frequency . The period is therefore

. b From the question, the amplitude is therefore and maximum acceleration is c When

,

and at , (and ), so , so the maximum speed is

, and so

and and the

.

Therefore at that time, 13 a Let be directed so that the initial displacement is positive, then The extension of the positive-side spring is

.

and the extension in the other is

.

Then the resultant of the tensions in the two springs (resolved in this positive direction) is given by , so the magnitude of the resultant force is b

, so

, which describes SHM with angular frequency

c SHM with amplitude At

,

and angular frequency

and

:

, so

14 a The particle is in equilibrium when the resultant force is zero. , so

and therefore

.

.

.

The equilibrium position is at b At displacement

extension.

below , the extension is

Then the resultant force in the downward direction is The magnitude of the resultant force is therefore c

, and so

.

which describes SHM with angular frequency

The period is therefore

.

.

EXERCISE 11C 2

Critical damping occurs when the auxiliary equation has a zero discriminant: Auxiliary equation: Discriminant

(Reject

because that would convert the differential equation to

, which does not

describe damped harmonic motion). 3

a Tension

and resistance

Resultant force

.

, so

Then b Critical damping occurs when the auxiliary equation has a zero discriminant: Auxiliary equation: Discriminant

c For greater values of , the discriminant is positive and the damping is then overdamping. 4

, so

a

b Critical damping occurs when the auxiliary equation has a zero discriminant: Auxiliary equation: Discriminant

5

a Auxiliary equation: , so General solution: So

Initial conditions: So

and

Particular solution: b This is underdamping.

6

a Tension

and resistance

Resultant force

.

, so

Then b Auxiliary equation: , so General solution: So Initial conditions: , so

and

Particular solution: c Solving for

So

:

which has no solution for

d This is overdamping. 7

a Attraction force Resultant force

and resistance

.

, so

Then b Critical damping occurs when the auxiliary equation has a zero discriminant: Auxiliary equation: Discriminant

c General solution:

So Initial conditions:

, so

and

Particular solution: 8

a At equilibrium, the weight of the particle is equal to the tension in the string. , so at equilibrium, extension b For displacement below the equilibrium, tension

Resultant force

, weight

and resistance

, so

Then c Auxiliary equation: Discriminant

, so this equation describes underdamping.

General solution: So Initial conditions:

, so

and

Particular solution: d As , , so the displacement reduces towards zero and the extension of the string therefore tends towards . e At ,

. The particle passes here when , so

, so

The first such time is at particle begins at

, and the velocity will be positive, since the .

, so 9

a Auxiliary equation: The discriminant of the auxiliary equation If

then

b Solutions to the auxiliary equation: General solution: So

, so the damping is underdamping.

Initial conditions:

, so

,

Particular solution: , so initially the spring is extending.

c

The maximum extension therefore occurs on the first occasion when the speed is zero.

So

when

So At

, the spring is at maximum extension.

When

:

So at this time,

EXERCISE 11D , so

2

Substituting

:

Then Auxiliary equation:

, so

Complementary function: Particular integral: General solution: So Paired system general solution: , so

3

Substituting

Then

:

Auxiliary equation:

, so

General solution: So Paired system general solution: , so

4

Substituting

:

Then Auxiliary equation:

, so

Complementary function: Particular integral: Substituting gives

, so

General solution: So Paired system general solution:

Initial conditions:

, so

Paired system particular solution: 5

, so

a

Substituting

and

:

Then b Auxiliary equation: , so General solution: So Paired system general solution:

Initial conditions: , so

and

Paired system particular solution:

.

,

.

6

a The rate of water loss from the top bucket is proportional to the height of water. Cross-sectional area is metres. , so

, so the volume of the top bucket

from which

where is measured in

.

Integrating factor:

General solution: Initial condition: Particular solution: For the second bucket, water flows in at the same rate that it flows out of the first bucket (ignoring transit time between buckets) and flows out under the same relationship to height as for the first bucket.

So b Integrating factor: as in part a.

General solution: Initial condition: Particular solution: So only has a solution at empties. c From part b: .

, and

for all

; that is, the second bucket never fully

, so maximum height occurs when

At that time, the height in the second bucket is 7

or

a b

, so Substituting:

and

So c Auxiliary equation:

, so

which is at

:

.

Complementary function: Particular integral:

; substituting:

, so

General solution: So , so Initial conditions:

so

,

Particular solutions: d For some

and

,

and So

and the first peak in shark population occurs when

, so

By similar calculations:

and So

and peak fish population first occurs at

time units), and with a phase delay of e The populations oscillate with the same period ( of time between the fish population peaking and the shark population peaking. and

8 Then

; substituting

and

:

So Auxiliary equation: Discriminant If the discriminant is positive, the general solution will have the form If the discriminant is zero, the general solution will have the form If the discriminant is negative, the general solution will have the form and only in this case will the system oscillate: so

Tip

units

Notice that the system could be written in matrix form as . The auxiliary equation for the second order differential equation is exactly the same as the characteristic equation of the matrix: . The three general solutions then relate to the transformation described by the matrix, which can be given as conditions on the discriminant of the characteristic equation: : The matrix has distinct real eigenvalues and is diagonalisable, with the result that for some diagonal matrix . This is equivalent to a solution to the system of a sum of two exponentials. : The matrix has repeated eigenvalues and cannot be diagonalised; it represents an enlarging skew on the horizontal plane. : The matrix has complex eigenvalues and represents a rotation around the origin. This is equivalent to an oscillating solution for the system; if and are plotted against each other, they will be seen to rotate about an equilibrium position, echoing the rotation about the origin described by the matrix.

MIXED PRACTICE 11 1

describes SHM with angular frequency and period So

2

describes SHM with period

Critical damping occurs when the auxiliary equation has a zero discriminant: Auxiliary equation: Discriminant

3

so

a Then

b Integrating: Initial condition:

so

Integrating: Initial condition:

so

Then 4

a SHM with amplitude equilibrium position So b

and angular frequency

and therefore maximum speed is

has displacement from the

When

, , so maximum acceleration occurs when

c

:

, so maximum force is 5

, so

a b Try Then

, and so

So

, as required.

satisfies the differential equation for

.

A second order differential equation has a general solution with two unknowns, so this is the general solution to the differential equation.

Tip Take care to address the question as it requires. If solving this without a ‘show that’, this working would be suitable: Auxiliary equation:

so

General solution: However, the question tells you to ‘verify’ that this form is valid, so you should instead work from the answer and demonstrate its validity rather than deriving it in the normal way. More care is needed to show reasoning in a ‘show that’ question; no credit will be awarded for the solution equation, since it is given in the question! c Initial conditions:

, so

and

d For SHM with amplitude and angular velocity , maximum speed is 6

so

i Then

Integrating: Initial condition:

ii Require Therefore

so again at

iii Integrating: Initial condition: So So passes through the origin at iv Since passes the origin at

and and

, the cubic path must have turning points between

and

and again between

and

, so will not turn again in the third second.

Then the distance travelled in the third second can be calculated as

7

a Force

, so

If

then

b If

, so , then

and so The general solution to a second order differential equation must have two unknown constants, so is the general solution to . Initial conditions:

, so

Particular solution: c Then

when

, so the particle first returns to the equilibrium position at

so

8

Initial condition: Then

; rearranging,

The total population in the model is

, so when

,

so

9

Substituting,

and

:

So Auxiliary equation:

so

General solution: So Then

Initial condition:

Particular solutions:

10

so Direct integration gives Initial condition:

so

and

months

.

Initial condition: Particular solution: So so

, which rearranges to

Auxiliary equation:

so

Complementary function: Particular integral: Try

; substituting gives

so

General solution: So Initial conditions: so

and

Then Both particles have increasing displacements for positive time, so after the first seconds, particle has travelled further than particle . 11 a

Separating variables:

, so At the ends of the movement and , , so

and

. Substituting into the formula:

Then b The period of oscillation is

(the time taken to travel from to and then back to again).

So , so maximum acceleration occurs when

c

At this moment, Force

.

12 a Force

, so

b Separating variables:

, so At the start of the motion,

and

. Substituting into the formula:

, so Then

, so

Maximum speed will occur at 13 a Force

, at which point

.

, so

b Separating variables:

, so At the start of the motion,

and

. Substituting into the formula:

Then reaches maximum value when

c 14 a Force

, so

Auxiliary equation:

, so

General solution: So Initial conditions: Particular solution: b Find the least positive for which , so c Force

, so

Auxiliary equation:

:

at which point

, so

, so General solution: So

Initial conditions:

, so

and

Particular solution: d Under the new model, So

when

, from which

The cart reaches the equilibrium position later under the second model.

Tip With adequate explanation, you could answer part d without explicit calculation; since the model includes an additional resistance element, the cart's acceleration will be lower in the initial phase of movement, so it will reach the equilibrium point later than predicted in the first model. 15

so

Then Auxiliary equation:

so

General solution: Then Since

16

,

Tip It is reasonable to answer this question by finding the particular solutions for and ; this method is given below the faster solution you may be able to spot! However, there is a problem with the fast approach, detailed below the longer answer. Quick method Let

be the total number of animals on the plain.

Then

so the total number of animals is always constant. and therefore

Standard method so

Auxiliary equation:

so

General solution: So

Initial conditions:

so

and

Particular solution: Then the total number of animals is Then at

, a constant value.

, the total number of animals is 4000.

Tip You should always check the component models for validity. It would appear that is always constant, but notice that for

years, the model for the number of badgers

predicts a negative number of badgers, so the model cannot be extended past this time. 17 i The particle starts at rest, so is at one extreme of the oscillation; it will next be at rest at the opposite extreme, so the amplitude Period

and angular frequency

Maximum velocity is ii Initial displacement is

so the motion is described by

Then iii Require that Then

and for for integer

and so

: : and : and

18 i

so

Then

Integrating:

Boundary conditions:

so

and

Then

So ii

19 a

so the transition between the two equations occurs at the maximum of the first (negative) cubic and the minimum of the second (positive) cubic.

is measured as positive towards , as shown in the diagram. At , each spring is in extension by

. So when the particle is displaced

in the direction of :

the tension in the spring attached at is and the tension in the spring attached at is

.

The resultant force acting upon the particle in the direction of positive is

So the magnitude of the force is b Force

, so

Then

.

This equation has the form

which describes SHM with angular frequency

.

The period of the motion is given by c The amplitude of motion is

Then when

and speed and displacement are related by the equation

,

20 a The proposed model assumes that the force decreases at a constant rate, but in reality the force would probably not follow a linear path. Bill is not on wheels with a continuous movement, but taking steps, so the amount of force he can exert will change depending on where he is within each stride, resulting in a series of pulls rather than a single long pull. The force he can apply will change as he tires, which will not necessarily occur in a linear manner. He may increase his effort towards the end as he tries to burn all his remaining energy (in a final sprint), so the final force may not be the lowest applied during the challenge. b If the pulling force follows a linear pattern

(for time seconds and

force Newtons) then: Boundary conditions: Substituting

into

gives

so

Then if metres is the displacement from the starting point, , noticing that

throughout (because

although after the resistance exceeds the force being applied, the truck will still be moving. The resistance force will slow the truck to a stop but will not cause it to reverse, due to the symmetry of the resultant force in the time interval .) c Integrating: Initial condition: Integrating: Initial condition: So

and therefore

d The force Mike applies follows a negative exponential curve with the same start and end points as the linear function of Bill’s force, so throughout the time period, Bill is pulling with a force no less than Mike, and greater than Mike for all . Therefore Bill must win the competition.

Worked solutions Cross-topic review exercise 1 1

so

a Therefore

( s.f.)

b 2

a b Let Then and

so

Then for in the required interval,

3

a Two vectors in the plane are

and

A vector equation of the plane is therefore

so the normal to the plane is

, which in Cartesian form is

b The normal to the second plane is

The angle between the planes is equal to the angle between the normals:

So the acute angle between the planes is 4

Formulae for the sum of the first integers and the first cubes: and

Then



Comparing coefficients:

,

Tip Notice that since you are given the formula, and not required to justify it in any way, you could more simply just substitute and and solve the resulting set of simultaneous equations: Setting :

so



(1)

Setting :



Substituting 5

into

:

,

.

Formulae for the sum of the first integers, the first squares and the first cubes: ,

6

so

(2)

and

and so

i So

ii Using the result from part i:

7

i Let

. By de Moivre’s theorem, and

So Then

and

so

The solutions with So

are ,

,

and

ii

8

Tip There are several reasonable approaches; if you find the vector perpendicular to both lines you can use the formula

. Alternatively, use a method such as given below.

Line has general equation

line has general equation

and

.

Let on line and on line be the closest points on the lines so that between the two lines. Then

is the shortest distance

must be perpendicular to both lines.





so

So

The shortest distance between the lines is 9

Expressing the system of simultaneous equations as a matrix equation: where

,

and

.

There will be no unique solution when Expanding about the second row:

So there is no unique solution when

or

10 a Using the binomial theorem: Since this is known to be real, Solving this quadratic in using the quadratic formula:

b Using de Moivre’s theorem, c But

because

is known to be real.

Then So

, and therefore

Of the five possible values of

for

:

is not defined, and the greatest positive value will occur for Therefore

.

Tip Take care not to ignore cases and explicitly assign the required case to the relevant angle using knowledge of the function. 11 i

ii

iii The limit of the sum does not converge to a finite value;

So as

,

12 i

ii

iii From part ii, as

,

so

So

Taking only the positive integer solution, 13 i

ii

iii

14 i

,

and

ii It appears that

iii Proposition: Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

By the principle of mathematical induction, the proposition is true for all positive integers. 15 i

ii If , and are distinct roots of a cubic Since , and

then their sum

are the distinct roots of

, it immediately follows that

iii a From part ii, Then b iv The new cubic Sum of roots: Sum of paired products: Product of roots:

has roots ,

and

using part iii b

. .

Then the new cubic is Multiplying through by for integer coefficients:

so

16 i

and therefore

or

so let

ii From part i,

or

Rearranging the substitution,

so the solutions are

: : : : is the sum of a geometric series.

17 i If

and

then

ii Multiplication by results in the rotation of a point in the Argand plane by

about the origin.

The centre of the triangle is at , representing complex number Translating the triangle so that the centre lies at the origin is equivalent to subtracting . Multiplying the result by would rotate one vertex to the next. Therefore

and

and

.

Using the above:

iii If

then

But by part i, So 18 i Let

.

Then So

for integer . , so

ii Let

or so

, since But if

.

is not defined for

so by part i,

and so

(excluding

)

then by the binomial expansion,

.

The four solutions to this equation are therefore 19 i Multiplication by So angle

for

or

is equivalent to rotation by about the origin.

is

Also, the modulus will be unaffected by rotation about the origin, so is therefore isosceles.

and the triangle

from each of the values involved translates to the origin, to , representing ii Subtracting value and to representing value , but does not change their relative positions. If then lies on the perpendicular bisector of , and so the origin is on the perpendicular bisector of and ; then forms an isosceles triangle. If angle

is then so is angle

.

By the same argument used in part i, it follows that (allowing that

That is,

or and so

20 i Then

Using these results:

and

may be clockwise or anticlockwise from

)

ii If

then by part i,

So

(solutions

or

(solutions

Overall general solution is

for integer ) for integer ). for integer . So

.

and so

21 i Then

Using this result and the binomial expansion:

ii Using the result from part i, So if

then or

Then

or

Solutions in the interval

are

or

22 i Require that

runs perpendicular to the direction of line where

lies on so has position vector

and so

Then

so

has coordinates

.

Then ii Require on But

so the distance such that so

Then

so is at

23 i The normal is perpendicular to both direction vectors of the plane: so

Then

so is given by

ii Substituting the line equation to find the intersection point: so

and the intersection point is at

iii The two direction vectors of the plane must be perpendicular to the normal. So is perpendicular to both and

.

so

Tip Take care here – you cannot assert that

(or any other specific multiple of this

vector); the question contains enough information to find the direction of but not its magnitude.

24 The normal vector to plane is

and the direction vector of is

The direction vector of is perpendicular to both and so

passes through the point with position vector 25 i  

Tip

so has equation

If you can remember the formula for the shortest distance between two lines you can quote and use it directly; otherwise, derive it or use a standard method as shown here. Let on and on be the closest points on the two lines so that between the lines.

But

must be perpendicular to both direction vectors:

so

so

so

and then

So

Then the shortest distance is ii The plane must have normal vector parallel to

so

The plane passes through the point with position vector

so the plane is given by

In Cartesian form: 26 a b

so

.

.

is the shortest distance

c

so

27 a Then b

so Multiplying through by

:

so

and therefore so

28 a b Then the infinite series c By de Moivre’s theorem,

Taking the imaginary part of the equation in part b:

29 a

So b Proposition: Base case:

for so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

By the principle of mathematical induction, the proposition is true for integers 30 a Proposition:

. .

for

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

Taking

and

, this shows that

Then So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

By the principle of mathematical induction, the proposition is true for integers b Let

and

Using part a:

.

. .

31 Proposition:

for

Tip Be careful when setting the start value; here the base case is that your base case matches the first term in the sequence.

not

. Always check

Tip Remember that the symbol

is used for taking the product of multiple terms much as

used for taking the sum of multiple terms, avoiding the need for an ellipsis (…) in the working.

Base case:

so the proposition is true for

Inductive case: Assume that the proposition is true for some integer So

Working towards:

So if the proposition is true for

then it is also true for

Conclusion: The proposition is true for

and if true for

then it is also true for

By the principle of mathematical induction, the proposition is true for integers 32 i

.

is

ii

iii By part ii, as

,

Then

33

Then

and

34 i Let

then by de Moivre’s theorem,

Then

and

Using the binomial theorem:

Then Dividing numerator and denominator by

ii a Let

and

. Then substituting into the result from part i:

Rearranging: b If

:

for then

So

, which by part i has solutions

where

and then

So the three roots of the cubic are the distinct values Since one of the roots is cubic.

,

, it follows by the factor theorem that

Factorising:

and must be a factor of the



So the values of the three distinct roots are Since

is an increasing function for

Then

, it follows that

, and only one of the three root values lies in this interval.

Therefore iii Substituting When

,

so and when

: ,

by part ii b

35 i

has solution

and so

In the required interval, ii If

then by de Moivre’s theorem,

Then Using the binomial expansion:

From part i, solutions to the equation and

include the values

and

, for which

respectively.

The sextic therefore must have factors

and

.

, so taking this factor out of the sextic:

iii Writing

, this can be expressed as

The cosines of the six values of found in part i are distinct (since ). and

are the roots of the quadratic factor

is decreasing in the interval

and so the cosines of the

remaining values must be the four roots of the quartic factor. For a general quartic equation Then since

,

,

, the product of the roots will equal . and

are the roots of

,

Tip Alternatively, calculate the values of the four roots of the quartic directly and find their product. 36 i Let

. By de Moivre’s theorem,

So Using binomial expansion:

ii Then

 

iii So

Observing this has a solution

Solutions are

or

, by the factor theorem

so the only real solutions are for

The only real solutions for are therefore

for integer .

iv Using the result from part ii again:

So

Solutions are

or

so the only real solutions have

37 i Let

. By de Moivre’s theorem,

So

and

Using the binomial theorem,

is a factor. Factorising:

, so

So Dividing numerator and denominator by

ii If

then

so

gives

for integer

.

Within the interval

, the solutions are

iii Multiplying the quartic by

Let

gives:

(this substitution is valid without restriction since the range of

Then by part i, this is equivalent to

Since

, with five distinct solutions for

, this corresponds to the root of the introduced linear factor

is all real values) :

, so the

remaining four distinct values must be the roots to the quartic. has solutions 38 i a

is the sum of the first ten terms of a geometric series with first term and constant ratio When

,

and so

Dividing numerator and denominator by

But

so

Then

b When ii Since

,

and so ,

:

.

iii Using part ii, if

Then

or

Simplifying: So Taking

then

so

for integer values or

or gives the solution

(rejecting

part i and does not give a valid solution for

Taking

gives the solutions

So another solution in the interval

since )

or is

is excluded from the identity in

Worked solutions Cross-topic review exercise 2 1

so every quadrant has the same shape.

a

no curve

The curve is not defined wherever

:

b Total enclosed area consists of four equal ‘petals’.

2

so

Then

3

Factorising:

or

4

Substituting

and

so that

:

5

6 Let

so

Using partial fractions,

, so the integral becomes:

Reversing the substitution

Multiply top and bottom by

:

:

7 Let

so

Limits: When

,

and when

,

8

9

Equating the first three terms of the two expansions:

Substituting

into

so

10 a

b

So

:

so

, from which

11

12 a b Method 1: Find and hence , so

for which the only real solution is

Then Method 2: Use identities

Dividing by

and using: so

so

So and

13 i

Using binomial expansion:

ii Using part i: If

then

But So

(rejecting the negative solution)

and

14 i Let

so that

Then Taking the reciprocal: ii Using the chain rule and part i:

So Using implicit differentiation and the product rule:

Multiplying through by

and simplifying:

So

as required.

15 i If

then

Implicit differentiation with respect to (taking the positive root because

is an increasing function in the primary domain)

Taking the reciprocal: so

ii

so

so

so

Maclaurin series:

iii

16 i Let

so

Then using implicit differentiation: .

But

So Then

ii

so

so

Maclaurin series:

17 i

Tip

You do not need to draw and label tangent or symmetry lines but since credit is often given for diagrams clearly showing these qualities, it is appropriate to do so. ii

iii Using compound angle formulae:

Substituting

and

so that

18 i Auxiliary equation: so Complementary function: Particular integral: Try

so

and

Substituting: Comparing variables:

so

and

Then General solution: , the complementary function tends towards zero, so the solution approximates to the ii As particular integral: . 19 Auxiliary equation: so Complementary function: Particular integral: Try Substituting: Then

so so

and

General solution: Then

Initial conditions:

so

and

Particular solution: 20 Integrating factor:

Let

and

so

and

Integration by parts: So General solution: Boundary condition:

so

Particular solution: 21 i Force So But using the chain rule, So Separating variables: so Initial conditions:

so

and the movement uses the positive root.

Tip When evaluating a variable according to a fixed value of a second variable, you can’t always use brackets unambiguously. Here, for example, we generally consider that the independent variable governing motion is time so would mean velocity at time , not velocity at position . To be clear about this, we use a vertical line notation, with a subscript indicating the condition. Then

ii Separating variables again:

so

Then Initial conditions:

so

and

22 i Maximum speed occurs at the equilibrium position in SHM, so is the equilibrium position, is displacement from the equilibrium and is the amplitude of the motion. If the angular frequency of the motion is then the period of the motion is Maximum speed is given by

so

so

ii The movement begins at maximum displacement so the motion is governed by the equation and therefore Then So at

and ,

and the velocity is

iii Require that So

and for integer so there are three possible values in the interval

At At At

, , ,

23 i Force So Separating variables:

Initial condition: ii Rearranging:

so so when

and so ,

From the equation it is clear that is always positive, so

.

.

Tip

24

There are several approaches to this problem; three are shown here. Method 1: Solving for and

Multiplying through by

produces a quadratic in :

Require values of for which there are real, positive solutions for

For real values, require that For positive values, require that

since

So require that Method 2: Finding the minimum algebraically

for some

Tip This is equivalent to the method you already know for writing as a single function or . With hyperbolic functions, you can write as when and as when , using the identities and You can check, and derive, these identities yourself by writing and . Comparing:

and

so

and

in terms of

and

So Then the minimum value of

is

, so

will have at least one solution for

.

.

Method 3: Finding the minimum by calculus Let Then Stationary point occurs when

so

and therefore

Since there is only one stationary point and be a global minimum.

So when

as

is

, so

will have at least one solution for

25

and rearranging:

or or From context,

is an invalid solution, so

26 a

b Valid for c Require Substituting

, the stationary point must

,

Then the minimum value of

Multiplying by

or

.

, equivalent to so

from which

into the expansion in part a:

.

27

so

28 a

so

and

b

for some constants , and .

29

Multiplying by the denominator on the LHS:

Comparing coefficients:

so

and then

and

So Therefore, where defined,

The function

is not defined at

, which is outside the interval of

integration.

30 i Using binomial expansion:

So ii Let

Converting to the original variable:

31 i Let

so

Converting to the original variable:

and

ii

iii

Like the integral in part ii, this is an improper integral since the function to be integrated is not defined at one of the integral limits (when ). Unlike the integral in part ii, this integral does not converge to a finite value:

But

does not converge as

The volume of revolution is infinite. 32 i

The maximum value of occurs at

; the point has polar coordinates

ii

iii By the symmetry of the graph, the tangent at has gradient The Cartesian coordinates of are

.

:

The equation of the tangent is therefore iv Substituting

and

so that

,

so

Squaring both sides:

as required.

33 i Auxiliary equation:

so

Complementary function: Particular integral: Then and Substituting: Comparing coefficients:

so

,

So General solution: ii As

,

so oscillates increasingly widely, with amplitude proportional to .

iii If

then

, so

; substitution gives

.

Then Then oscillates with a stable amplitude throughout all values of , and does not grow without limit as seen in part ii. 34 i Both particles begin at the maximum limit of their amplitude, so their motion is described by cosine functions with no phase shift. For angular frequency ,

so

and

and ii The period of

is

so at

, the particle has moved one complete period and the

start of the next. so the total distance travelled up to that time is one cycle travelled in the second period is . So the total distance travelled is

.

and

iii So when

,

and

The particles are travelling in opposite directions. 35 i

so Multiplying through by

and rearranging:

and the distance

Solving this quadratic in Then

. The range of

is

so pick the positive root:

ii Implicit differentiation gives

So Then so

iii 36 i

so using the chain rule,

Then

ii

when

.

is an odd function and so is also an odd function; therefore is an odd function and therefore , so the graph for is the same as the graph for , rotated about the origin.

From part i and using symmetry, the stationary points are at the function is 37 a

so the range of

Tip Choose your preferred method: integration by parts directly and with substitution are shown here. Method 1: Integration by parts

Integration by parts: Let

,

and

so

Method 2: Substitution and integration by parts Let

so

and

Substituting:

Converting to the original variable:

b Using double angle identities for cosine and sine:

c

i The curve meets the line

where

Then the rectangle bounded by the axes, the line .

and the line

Then

ii The shaded area can also be calculated directly as

Equating the two answers, and changing the variable to :

has area

38 i

Let be the point Then triangle has area ii Substituting

.

has area

so the region bounded by the curve and the lines

and

. and

into

:

Squaring:

At

, the angle to the origin is

and at

the angle to the origin is , so the

curve between and is given by the polar curve for

.

iii Using a double angle formula for cosine, the polar equation can be expressed as

Calculating the area bounded by the curve and the lines

Equating this with the solution from part i:

So

and

using polar integration:

PRACTICE PAPER 1 1

Require

(checked as true)

and So

and therefore [3 marks] and

2

[3 marks] 3

The function to be integrated is not defined at an integral with lower limit as .

, so the integral must be calculated as the limit of

[3 marks] 4

Sum of roots Sum of paired products of the roots

[4 marks] 5

So

and [5 marks]

6

a

[3 marks] b Regular seven-sided polygon: The vertex at

represents so [2 marks]

7

so Auxiliary equation:

so

so [5 marks] 8

a

for some constants , and Multiplying through by the denominator of the LHS:

Comparing coefficients:

so

,

and

Then [4 marks] b

[3 marks] 9

a

,

so

So the normal to the plane is

so the equation of the plane is

In Cartesian form,

.

. [5 marks]

b

The acute angle between

and the normal to the plane is where

So Then the angle between

and the plane is [4 marks]

10 a Substituting

and

:

so

[4 marks] b The line equation rearranges to Substituting into the equation for : so [3 marks] 11 a Expanding about the third row:

Since

, cannot be singular (have zero determinant) for any real value . [2 marks]

b

where is the matrix of cofactors. Since is non-singular,

exists for any real

value .

So

[5 marks] c Expressing the system of simultaneous equations in matrix form: so

Then

[3 marks] 12 Using the method of integrating factors for

:

Integrating factor Then When

,

So [6 marks] 13 a Proposition: for Base case: , so the proposition is true for Inductive case: Suppose the proposition is true for

for some

, so

Working towards:

Conclusion: The proposition is true for

and if it is true for integer

then it is also true for

Hence by the principle of mathematical induction, the proposition is true for all

.

. [6 marks]

b

[2 marks]

PRACTICE PAPER 2 1

a [2 marks] b (adding because

) [3 marks]

2

a

so has coordinates [2 marks] b

for

so represents a rotation by

(anticlockwise)

about the origin. [2 marks] where

3

and so

Then So [2 marks] 4

[5 marks]

5

Let Substituting:

so

. When

and when

[6 marks] 6

a

[6 marks] is valid for all but the expansion for b The expansion for when the expansion will not converge.

is valid for

so

is an example value for which the series will not converge. [2 marks] 7

a

[2 marks] b [1 mark] c Proposition:

for

Base case: Inductive case: Assume that Working towards: Using the assumption:

so the proposition is true for for some

Conclusion: The proposition is true for

and if true for

then it is also true for

By the principle of mathematical induction, the proposition is true for

.

. [5 marks]

8

The shortest distance between two skew lines with equations where

and

is given by

.

The two lines are

and

Then

and

Using the formula:

Since this shortest separation distance is non-zero, the two lines do not intersect in any point, so are skew. [7 marks] 9

Partial fractions:

for some constants

Multiplying through by the denominator on the LHS:

Comparing coefficients:

so

and therefore

and .

So [7 marks] 10 a Therefore if [1 mark] b

Tip There are several valid approaches you will have seen in the course. Here the method using induction is shown. Proof by induction: Proposition:

for integer

Base case: When Inductive case: Suppose

so the proposition is true for

.

for integer

Working towards:

Conclusion: The proposition is true for

and if it is true for

then it is true for

.

By the principle of mathematical induction, the proposition is true for all non-negative integers . Using part a: if

then

Therefore the proposition is true for all integers both positive and negative. Using this result:

[4 marks] c Dividing through by :

Using part b:

or If If

Tip Notice that an alternative approach is to observe, via the symmetry of the coefficients, that the polynomial can be expressed as sum of the binomial expansions . Factorising this swiftly leads to the same solution. [3 marks] 11 a Interpreting simultaneous equations as a matrix equation: Intersection is at position vector where

for matrix

.

Expanding about the second row:

So

is singular and there is no unique point of intersection of the three planes. [2 marks]

b Since

is singular, the rows of the matrix

and

are not linearly independent.

By observation: For the three plane equations to be consistent, require that the same relationship holds for the constant terms; that is, so . [4 marks] 12 a Total forces (both resisting extension): Then

so

Rearranging and cancelling :

[3 marks] b i Auxiliary equation: so General solution: Initial conditions:

or so

so

and then

Particular solution: [5 marks] ii A motion equation of the form

describes overdamping. [1 mark]

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