A Level Further Mathematics for AQA: Student Book 2 (Year 2) 1316644316, 9781316644478, 9781316644317, 9781316644560, 9781316644577

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

Contents Introduction How to use this resource 1 Further complex numbers: powers and roots 1: De Moivre’s theorem 2: Complex exponents 3: Roots of complex numbers 4: Roots of unity 5: Further factorising 6: Geometry of complex numbers 2 Further complex numbers: trigonometry 1: Deriving multiple-angle formulae 2: Application to polynomial equations 3: Powers of trigonometric functions 4: Trigonometric series 3 Further transformations of the ellipse, hyperbola and parabola 1: Parametric and polar form 2: Rotations and enlargements 3: Combined transformations 4 Further graphs and inequalities 1: The graph of 2: The graph of 3: Modulus equations and inequalities 5 Further vectors 1: The vector product 2: Equation of a plane 3: Intersections between lines and planes 4: Angles between lines and planes 5: Distance between a point and a plane 6 Further matrices 1: Transposes and inverses 2: Row and column operations 3: Solving linear systems with three unknowns 4: Geometrical interpretation of -variable simultaneous equations 5: Eigenvalues and eigenvectors 6: Diagonalisation and applications Focus on … Proof 1 Focus on … Problem solving 1 Focus on … Modelling 1 Cross-topic review exercise 1 7 Further polar coordinates 1: Area enclosed by a curve 2: Area between two curves 8 Further hyperbolic functions 1: Domain and range of hyperbolic and inverse hyperbolic functions 2: Reciprocal hyperbolic functions 3: Using hyperbolic identities to solve equations

4: Differentiation 5: Integration 9 Further calculus 1: Differentiation of inverse trigonometric functions 2: Differentiation of inverse hyperbolic functions 3: Using inverse trigonometric and hyperbolic functions in integration 4: Using partial fractions in integration 5: Reduction formulae 6: Length of an arc 7: Area of a surface of revolution 10 Maclaurin series and limits 1: Maclaurin series 2: Limits 3: Improper integrals 11 Differential equations 1: Terminology of differential equations 2: The integrating factor 3: Homogeneous second order linear differential equations with constant coefficients 4: Inhomogeneous second order linear differential equations with constant coefficients 12 Applications of differential equations 1: Forming differential equations 2: Simple harmonic motion and Hooke’s law 3: Damping and damped oscillations 4: Coupled first order differential equations 13 Numerical methods 1: Mid-ordinate rule 2: Simpson’s rule 3: Euler’s method 4: Improved Euler’s method Focus on … Proof 2 Focus on … Problem solving 2 Focus on … Modelling 2 Cross-topic review exercise 2 Practice paper Formulae Answers to exercises Worked solutions for chapter exercises 1 Further complex numbers: powers and roots 2 Further complex numbers: trigonometry 3 Further transformations of the ellipse, hyperbola and parabola 4 Further graphs and inequalities 5 Further vectors 6 Further matrices 7 Further polar coordinates 8 Further hyperbolic functions 9 Further calculus 10 Maclaurin series and limits 11 Differential equations 12 Applications of differential equations

13 Numerical methods Worked solutions for cross-topic review exercises Cross-topic review exercise 1 Cross-topic review exercise 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. We hope, however, that anyone working through this book, as well as seeing how maths can be useful, 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 by 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 lots of 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 were developed by seeing problems being solved and reflecting on their 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 never need to integrate a rational function in the future, but we firmly believe that the problem-solving skills you will develop by trying it can be applied in 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 it is amazing how 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 above, 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 absorb 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 are 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 resource 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.

Chapter 1, Section 1

Chapter 1, 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 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.

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?

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

Common error Specific mistakes that are often made. These typically appear next to the point in the Worked example where the error could occur.

Tip Useful guidance, including ways of calculating or checking answers and using 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  

.

After every few 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 learned. Key terms are picked out in colour within chapters. You can hover over these terms to view their definitions, or find them in the Glossary tab. Towards the end of the resource you will find practice paper questions, short answers to all questions and worked solutions. Maths is all about making links, which is why throughout this resource 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.

1

A sequence is defined by

. Use the principle of mathematical induction to prove that

. 2

Show that

3

Show that

4

Prove by induction that

5

Prove by induction that

6

Prove that

7

Use the principle of mathematical induction to show that .

8 9

Prove that Prove using induction that

10 Prove that

Black – practice questions which 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. Purple – challenging questions that apply the concept of the current chapter across other areas of maths. Yellow – designed to encourage reflection and discussion.

indicates a question that requires a calculator

1 Further complex numbers: powers and roots 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… Further

You should

Mathematics Student Book 1,

know how to find the modulus and

Chapter 1

argument of a complex

1 Find the modulus and argument of

.

number. Further

You should be

Mathematics Student Book 1, Chapter 1

able to represent complex numbers on an Argand diagram.

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

Im 4 3 A

2 1

–4 –3 –2 –1 O –1

1

–2 –3 –4

B

2

3

4

Re

Further Mathematics Student

You should know how to work with

Book 1, Chapter 1

complex numbers in Cartesian form.

Further Mathematics Student Book 1, Chapter 1

You should be able to multiply and divide complex numbers in modulus– argument form.

Further Mathematics

You should be able to work

Student

with complex conjugates.

Book 1, Chapter 1

Further Mathematics

You should know how to

Student Book 1,

relate operations with

Chapter 1

complex numbers to

3 Given that

and

, evaluate:

a .

b

4

Given that

and

, find:

a .

b

Give the arguments in the range

.

5 Write down the complex conjugate of: a .

b

6 Let a

to

b

to

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

.

transformations 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 forms. Modulus– argument form is particularly well suited to multiplication and division. In this chapter you will see how you can use this to find powers and roots of complex numbers. This chapter also includes a definition of complex powers that can make calculations even simpler.

Rewind You met complex numbers in Further Mathematics Student Book 1, Chapter 1. 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. Because you can represent complex numbers as points on an Argand diagram, operations with complex numbers have a geometric interpretation. You can use this fact 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.

Fast forward You will learn more about links between complex numbers and trigonometry in Chapter 2.

Section 1: De Moivre’s theorem In Further Mathematics Student Book 1, Chapter 1, you learnt that you can write complex numbers in Cartesian form, , or in modulus–argument form, . 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 1.1 De Moivre’s theorem For a complex number, , with modulus and argument :

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

Rewind Proof by induction was covered in Further Mathematics Student Book 1, Chapter 12. 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 proof that extends 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.

Then for

Make a link between and . In this case use .

:

Group real and imaginary parts.

Use: and

Hence the result is true for

.

This is the result you are trying to prove, but with replaced by .

The result is true for , and if it is true for some then it is also true for . Therefore it is true for all integers

Remember to write the full conclusion.

by induction. You can use De Moivre’s theorem to evaluate powers of complex numbers. WORKED EXAMPLE 1.1 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:

WORK IT OUT 1.1 Evaluate Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

So, using De Moivre’s theorem

Solution 3

You can also prove that De Moivre’s theorem works for negative integer powers. WORKED EXAMPLE 1.2 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 and . Hence

and

.

This means that you can write . You need to prove that

b Using De Moivre’s theorem for positive powers:

.

Since you have already proved De Moivre’s theorem for positive powers, you can use the modulus and argument of found in part a.

Hence , as required.

WORKED EXAMPLE 1.3 Find the modulus and argument of

.

with

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 .

EXERCISE 1A

and the argument is

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

.

, so

EXERCISE 1A 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

and on the same Argand diagram.

a Given that

:

i write

and in modulus−argument form

ii represent

and on the same Argand diagram.

b For which natural numbers is 4

?

a Find the modulus and argument of

.

Hence, clearly showing your working, find:

5

b

in modulus−argument form.

c

in Cartesian form.

a Write

in the form

.

b Hence, clearly showing your working, find 6

7

in simplified Cartesian form.

Find the smallest positive integer value of for which

is real.

Find the smallest positive integer value of such that

is pure 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 1.2 Euler’s formula:

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. The Maclaurin series you know for , and provide a justification for Euler’s formula if you accept that the series for converges for imaginary as well as for real numbers.

Rewind See Further Mathematics Student Book 1, Chapter 11, for a reminder of the Maclaurin series. You will meet these again in Chapter 10. You can write a complex number with modulus and argument in exponential form as have two different ways of writing complex numbers with a given modulus and argument.

. You now

Key point 1.3

When working with complex numbers in exponential form you can use all the normal rules of indices. WORKED EXAMPLE 1.4 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 1.5 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 11.

Key point 1.4 The complex conjugate of

Tip

is

.

Note that if

, you also have

.

In Further Mathematics Student Book 1, Chapter 2, you used complex conjugates when solving polynomial equations. WORKED EXAMPLE 1.6 A cubic equation has real coefficients and two of its roots are and form . The roots are:

and

.

. Find the equation in the

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 1B You can use your calculator to perform operations with complex numbers in Cartesian, modulusargument 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

Let

and

. Show that Write each complex number in the form

a b 7

Write figures.

in the form

8

Write

in exact Cartesian form.

9

The equation values of and

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

has real coefficients, and two of its roots are and

10 A quartic equation has real coefficients and two of its roots are the form 11 Find 12 Find

in the form in the form

and

Find the

Find the equation in

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 Further Mathematics 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 Further Mathematics Student Book 1, Chapter 1, 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 1.7 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

Therefore,

Comparing the moduli:

Remember that, by definition, is a positive real number.

Comparing the arguments:

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

The solutions are:

Write down all three solutions in modulus-argument form.

If you plot the three solutions from Worked example 1.7 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. Im 3 z2

2 1

–3

–2

–1 O –1

z1 1

2

3

Re

z3 –2 –3

Key point 1.5 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

onto the argument does not change the

write different solutions in modulus-argument form. All solutions 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 1.7.

Key point 1.6 The solutions of

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

WORKED EXAMPLE 1.8 Draw an Argand diagram showing the solutions of the equation One solution is

.

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

Im 4 z3

3

z2

2 1

z4

–4 –3 –2 –1 O –1

z1 1

2

3

4

Re

–2 z5

–3

z6

–4 You can also find the equation with solutions that form a given regular polygon. WORKED EXAMPLE 1.9 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 solutions, so From the diagram:

.

Any equation of the form

has complex solutions.

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

is a solution. Hence

Remember that

.

EXERCISE 1C In this exercise you must clearly show your working. 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 on an Argand diagram.

and show them

a i ii b i ii 3

Solve the equation

4

a Find the modulus and the argument of b

for

. Give your answers in the form

Solve the equation

.

.

, giving your answers in the form

,

where and are integers. 5

Solve the equation

6

Find all complex solutions 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 solutions 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

10

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

a Find all the solutions 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 solutions are represented on an Argand diagram by points ¸ and , labelled clockwise with in the first quadrant. is the midpoint of and the corresponding complex number is .

b Find the modulus and argument of . c Write 12

in exact Cartesian form.

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

.

.

c Hence or otherwise solve the equation Cartesian form.

, giving any complex solution 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 solutions are called roots of unity. WORKED EXAMPLE 1.10 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 solutions.

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 1.7 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 1.8 You can write the th root of unity as: where

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 Further Mathematics Student Book 1, Chapter 1, 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 1.11 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 1.8.

b

This is a geometric series with first term and 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 1.9 If

are the th roots of unity, then

Rewind You learnt about sums and products of roots of polynomials in Further Mathematics Student Book 1, Chapter 2.

Tip You could also prove the result in Key point 1.9 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 1.9 with a specific value of to find some special values of trigonometric functions.

Fast forward You will learn more about the links between complex numbers and trigonometry in Chapter 2.

WORKED EXAMPLE 1.12 Let

.

a Show that

.

b Hence find the exact value of

.

a From the diagram: The five points form a regular pentagon.

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 1.9.

Pair up the terms with equal real parts.

b

Use the fact that

and .

Use

.

This is a quadratic equation in Take the positive root since

EXERCISE 1D 1

Write down, in the form

, all the solutions of each equation.

a i ii b i ii 2

For each equation from Question 1, write the solutions in exact Cartesian form.

3

a Write down, in the form

, the solutions of the equation

b Represent the solutions on an Argand diagram. 4

The diagram shows all the solutions of an equation

.

a Write down the value of . b Write down the value of

.

c Which statements are correct? Choose from these options.

.

. .

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 clearly show your working. a Find, in exact Cartesian form, all the complex solutions of the equation b Hence find the exact solutions of the equation

8

Multiply out and simplify

9

Let

, where

.

. .

.

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

.

.

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

is equivalent to

e Hence show that 10 Let positive argument.

for .

. be the solutions of the equation

, where is the one with the smallest

a Show these solutions on an Argand diagram. b Write in the form

:

i ii 11

.

a Show that 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 Further Mathematics Student Book 1, Chapter 2, 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 1.13 a Find all the complex solutions 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:

The solutions are:

b

You are looking for solutions, so add times.

three

Find the Cartesian form

The factors of equation

correspond to the roots of the 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 1E

.

EXERCISE 1E In this exercise you must clearly show your working. 1

a Find, in exponential form, all the complex solutions 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 solutions 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 solution 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 .

Rewind You already know, from Further Mathematics Student Book 1, Chapter 1, 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.

Key point 1.10 Multiplication by

corresponds to a rotation about the origin though angle and

an enlargement with scale factor .

WORKED EXAMPLE 1.14 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 that 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 1.10 is remarkably powerful in some situations that have nothing to do with complex numbers. WORKED EXAMPLE 1.15 An equilateral triangle has one vertex at the origin and another at coordinates of the third vertex.

. Find one possible set of

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

, corresponding to multiplication by

. You can obtain it by rotating .

Focus on… You can use several different approaches to solve the problem from Worked example 1.15. 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 Further Mathematics Student Book 1, Chapter 8.

EXERCISE 1F

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

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 . 6

Let a Represent

. and on an Argand diagram.

b Describe fully the transformation mapping to . 7

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. 8

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

a The line is the locus of

which satisfy

Point is the reflection of point in the line .

, and the point representing the

. Find the exact value of .

b Find the size of the angle

.

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

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 .

10

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

11

is rotated

anticlockwise about

.

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

Express the complex number

in terms of

and , respectively. is

.

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. 12 The diagram shows two equilateral triangles on an Argand diagram.

Find the complex number corresponding to the midpoint of form.

Checklist of learning and understanding

. Give your answer in exact Cartesian

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

onto the argument does not change the

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

, where

) are

.

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 angle anticlockwise about the origin.

Mixed practice 1 1

If

, find in exponential form.

Choose from these options. A B C D 2

a Find the modulus and argument of b Hence find

3

.

in exact Cartesian form.

a Write down, in the form

, all the solutions of the equation

.

b Show the solutions 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, centred at the origin, on an Argand diagram and one of its vertices is . Find an equation with solutions represented by the six vertices of the hexagon. 7

a Express

in the form

b Solve the equation .

, where

and

.

, giving your answers in the form

, where

and

[© AQA 2014] 8

If

, where is real and positive, find the exact value of .

Choose from these options. A B C D 9

10

Find the exact value of

a Express

in the form

, clearly showing your working.

.

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 If is a complex third root of unity and and are real numbers, prove that: a b

.

12 If

and

13 If 14

, find in its simplest form , prove that

a Express in the form

.

.

.

b Hence state the exact value of . 15 Let

.

a Write

and

in the form

b Explain why

.

c Show that

and

d Form a quadratic equation in 16 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 ,

and

.

17 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 . 18

a Express

in the form

b i Solve the equation and . ii The roots of the equation an Argand diagram.

, where

and

.

, giving your answers in the form

, where

are represented by the points

and on

Find the area of the triangle integer.

, giving your answer in the form

c By considering the roots of the equation

where is an

, show that

[© AQA 2013] 19 Point represents the complex number

on an Argand diagram. Point is rotated

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 20

.

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 and .

i Write down the complex number corresponding to point . ii Write down the number corresponding to point in modulus-argument form. iii Write down an expression for the length of iv Hence prove the

rule for the triangle

in terms of :

and .

2 Further complex numbers: 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 1, Section 1

Chapter 1, 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 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

Further Mathematics Student Book 1, Chapter 1

You should know how to divide complex numbers.

4 Find the real and imaginary

.

parts of

, where is

real. 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 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 2.1 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 2.1, you can obtain a similar expression for – see Question 1 in Exercise 2A.

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.

WORK IT OUT 2.1 Express in terms of . Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

So

Solution 2

Taking the imaginary part:

Solution 3

EXERCISE 2A

EXERCISE 2A 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

3

.

in terms of

a Expand

.

.

b Hence or otherwise express 4

in terms of

a Show that

a Find the values of

. .

, find the possible values of

a Find the real and imaginary parts of b Hence express

7

for

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 find the values of .

to

WORKED EXAMPLE 2.2 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

.

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

Hence

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 solution from part b:

is one of these four solutions. You can see from the cos 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 Further Mathematics Student Book 1, Chapter 2, for a reminder about roots of polynomials.

WORKED EXAMPLE 2.3 a Show that

.

b Show that the equation the value of .

can be written as

c Hence find the exact value of

a Write Then

and

.

Hence

where

.

Use De Moivre’s theorem.

Separate real and imaginary parts to find

Divide top and bottom by and use where

, and state

and

.

.

.

b

Rearrange the equation into the form from part a.

so c

Solve the cubic equation by solving

or

Although there are infinitely many values of , they only give three different values of periodic function).

Hence

.

(since tan is a

Use the result about the sum of the roots of a cubic polynomial:

EXERCISE 2B 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 .

and

.

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 2.1

You can further generalise this result.

Key point 2.2 If

, then

PROOF 2 Remember that .

Using De Moivre’s theorem for positive and negative integers:

and

Adding the two equations:

Subtracting the two equations:

You can use the results in Key point 2.2 to derive another class of trigonometric identities, expressing powers of trigonometric functions in terms of functions of multiple angles. For example: . WORKED EXAMPLE 2.4 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 2.2:

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 2.5 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 2.2. Divide by

c Using the result from part a:

EXERCISE 2C

.

Don’t forget to divide by the coefficient of .

EXERCISE 2C 1

Let

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

or

a i ii b i ii 2

Let

.

a Show that

.

b Hence show that found. 3

, where

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

5

c Find

.

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.

and are constants to be

.

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 ?

Did you know? 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 2.6 a Find an expression for

.

b Hence show that

a Geometric series with

.

.

This is a geometric series with common ratio . Use

for the sum of

the first terms. b

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 denominator. The imaginary part is:

in the

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 2.7

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 2.8 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

EXERCISE 2D 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 4

.

Use the geometric series Use the expansion of

to evaluate to show that

5

By considering

6

a Find an expression for the sum of the series

.

or otherwise, show that

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

, then In particular,

or

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 2 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

such that

.

b Hence find the exact value of 4

.

Show that

. Hence show that

equation 5

and

is a root of the

.

By considering the expansion of

, show that .

6

Show that

7

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. 8

a By considering

, find expressions for

b Show that

e By considering

is a root of the equation

is a factor of .

.

and hence find the exact solutions of the

, explain why

f Hence state the exact value of 9

.

.

c Hence show that d Show that equation

and

. .

a i Use De Moivre’s theorem to show that find a similar expression for .

and

ii Deduce that .

b Explain why

is a root of the equation

and write down the three other roots of this equation in trigonometrical form. c Deduce that . [© AQA 2011] 10

a Use De Moivre’s theorem to show that, if

b i Expand

, then

.

ii Show that

where

and are rational numbers.

c Hence solve the equation

for

, giving each solution in the form

.

d Show that

[© AQA 2012]

3 Further transformations of the ellipse, hyperbola and parabola In this chapter you will learn how to: recognise an ellipse, a hyperbola or a parabola from its parametric or polar equation find the equation of a curve after a combination of rotations, reflections and stretches identify a transformation, or a sequence of transformations, by considering the transformed equation.

Before you start… Further Mathematics Student Book 1, Chapter 3

Further Mathematics Student Book 1, Chapter 3

You should be able to recognise the equation and main features of an ellipse, hyperbola and parabola.

1 Name the curve described by

You should know how to find the equation of a curve after a translation, reflection of stretch.

2 Find the equation of the curve after:

the equation

and

find the equations of any asymptotes and the coordinates of any axis intercepts.

a a translation with vector

b a reflection in the line . A Level Mathematics Student Book 2, Chapter 3

You should be able to recognise a transformation, or combination of transformations, that map one function onto another.

3 Describe two transformations which map the graph of onto the graph of .

Conic sections Conic sections is a collective name for a group of curves that can be obtained as cross-sections of a cone. There are three distinct types: the ellipse (with a circle as a special case), hyperbola and parabola. In Further Mathematics Student Book 1, Chapter 3, you met Cartesian equations of these curves and also learnt how the equations are changed by translations, reflections and stretches. In this chapter you will add enlargements and rotations to this list of transformations. You will also see how you can apply techniques of parametric equations and polar coordinates to describe the curves in several different ways.

Section 1: Parametric and polar form Ellipse y b

–a

Hyperbola y b b —x —x y = y = – a a x2 y2 — — – = 1 a2 a2

x2 y2 — — + = 1 a2 a2

a

O

–b

x

–a

a

O



x



Parabola

y

y2 = 4ax

x

O

You can also describe these curves using parametric and polar equations.

Rewind You met parametric equations in A Level Mathematics Student Book 2, Chapter 12, and polar coordinates in Further Mathematics Student Book 1, Chapter 6.

WORKED EXAMPLE 3.1 Show that the curve described by the parametric equations is a parabola. You need to find a Cartesian equation. It looks like is related to . Hence the Cartesian equation is which is the equation of a parabola.

WORKED EXAMPLE 3.2

,

Recognise the standard equation of a parabola.

Find a Cartesian equation of the curve with polar equation

. Hence sketch the

curve. Using

The connection between polar and Cartesian coordinates is: and . It is often useful

:

to start by expressing any terms cancel.

Now use Divide by This is a hyperbola with -intercepts and asymptotes

.

in terms of and and see if

. to get on the right.

This is the equation of a hyperbola and

.

Transformations of curves You already know how to apply certain transformations to curves by changing the equation. Transformation translation with vector

Change to equation replace by

reflection in the -axis

replace by

reflection in the -axis

replace by

reflection in the line

swap and

reflection in the line

replace by

stretch horizontally with scale factor and vertically with scale factor

replace by

and by

and by

and by

with

WORKED EXAMPLE 3.3 Show that the curve with polar equation

is an ellipse and state the equations of its axes

of symmetry. Using

:

Start by expressing

in terms of and .

Note that can’t be zero, so you can divide by it. , so isolate on the left before squaring both sides. ( is, by definition, positive, so squaring both sides is fine.) You are looking for an equation of the ellipse, so complete the square for and make the RHS equal to .

This is an ellipse translated horizontally.

units

Recognise the translation with vector

.

Hence its axes of symmetry are and .

Did you know? The equation of an ellipse in the form shown in Worked example 3.3 is normally used to describe planetary orbits.

EXERCISE 3A

EXERCISE 3A 1

Show that the curve described by the parametric equations

2

Show that the curve with equations its vertex.

3

Find the Cartesian equation of the curve with parametric equations

is a parabola.

is a parabola and state the coordinates of

.

Hence sketch the curve. 4

Show that the curve with parametric equations equations of its asymptotes.

5

Sketch the curve with parametric equations

6

Show that the curve with parametric equations equations of its axes of symmetry.

is an ellipse and state the

7

Show that the curve with parametric equations coordinates of its vertex.

is a parabola. State the

8

is a hyperbola and find the

.

Find the Cartesian equation of the curve with polar equation

. What is the name

given to this curve? 9

Find the Cartesian equation of the polar curve

. What is the name given to this curve?

10 A curve is described by the polar equation

.

a Find the Cartesian equation of the curve. b Hence sketch the curve, indicating the position of the intercepts with the coordinate axes. 11 Find the Cartesian equation of the curve given by polar equation

. State the name given

to this curve and write down the equations of its axes of symmetry. 12 Curve

has equation

.

a Find the equation of the tangent to where and are integers. Curve

at the point

has equation

. Give your answer in the form .

b Describe the transformation that transforms c Hence find the equation of the tangent to

to

.

at the point

13 An ellipse is obtained from the circle

.

by a horizontal stretch with scale factor .

a Find the equation of the ellipse. b Write down the equation of the tangent to the circle at the point c Hence find the equation of the tangent to the ellipse at the point 14 The parabola

has equation

a Find the equations of the tangents to The parabola

. .

. that pass through the origin.

has parametric equations

b Describe the single transformation that maps c Hence find the equations of the tangents to

. to

.

that pass through the origin.

,

Section 2: Rotations and enlargements Key point 3.1 You can rotate a curve by changing its equation: Rotation anticlockwise

Change to equation replace by and by replace by

and by

replace by

and by

On this course, you only need to be able to use rotations through

and

.

WORKED EXAMPLE 3.4 The parabola with equation the resulting curve. A

is rotated

clockwise about the origin. Find the equation of

clockwise rotation is the same as a and by .

rotation anticlockwise. Replace by

An enlargement (with the centre at the origin) is a combination of a horizontal and a vertical stretch, both with the same scale factor.

Key point 3.2 Replacing by and by results in an enlargement with scale factor and centre at the origin.

WORKED EXAMPLE 3.5 Describe fully the transformation that transforms the curve

into the curve

. Relate the transformed equation to the original equation. Decide what and have been replaced by. This is an enlargement with centre at the origin and scale factor

EXERCISE 3B

.

and have been replaced by respectively.

and

,

EXERCISE 3B 1

Find the equation of each given curve after a rotations leave the curve unchanged?

and

rotation anticlockwise. Which of the

a Ellipse b Hyperbola c Parabola 2

Find the equation of each curve after it has been rotated

anticlockwise about the origin.

Which reflections result in the same curve? a Ellipse b Hyperbola c Parabola 3

Find the equation of the hyperbola

4

Find the equation of the parabola

5

Find the equation of the hyperbola

6

a Sketch the curve The curve

after it is rotated after it is rotated

b Find the equation of

.

with scale factor (and centre at the origin).

.

c Describe a horizontal stretch that transforms 7

anticlockwise about the origin.

after enlargement by scale factor .

with equation

is an enlargement of

anticlockwise about the origin.

back to

.

Describe fully the transformation that maps the hyperbola

onto the hyperbola

. 8

Describe fully a possible transformation that maps the curve Hence state the equations of the asymptotes of the curve

9

a Sketch the hyperbola with equation

onto the curve .

.

b Show that an enlargement with centre at the origin does not change the equations of the asymptotes. 10 Show that the curve with equation asymptotes.

is a hyperbola and find the equations of its

.

Section 3: Combined transformations You can combine any of the transformations from Sections 1 and 2. You know that sometimes the order of transformations matters – performing transformations in a different order may produce a different curve.

Rewind You met combined transformations in A Level Mathematics Student Book 2, Chapter 3.

WORKED EXAMPLE 3.6 Find the equation of the ellipse a

after each sequence of transformations.

Enlargement with scale factor followed by a translation with vector

b Translation with vector

a After the enlargement:

.

followed by an enlargement with scale factor .

The enlargement replaces by and by . Remember that both variables are squared.

After the translation:

b After the translation:

After the enlargement:

Replace by

and by

.

Now do the translation first.

For the enlargement, the and inside the brackets need to be replaced. You can see that this curve has the same shape as the one in part a but is in a different position.

You also need to be able to identify the sequence of transformations that maps one curve to another. You have already seen that sometimes there are several transformations giving the same result for a given curve (for example, because of the symmetry of the ellipse, a rotation gives the same result as a reflection in the line ). It is also possible that transformations can be carried out in a different order. Therefore there might be more than one correct answer to a question. WORKED EXAMPLE 3.7 The parabola stretch.

is transformed to the parabola

by a rotation followed by a horizontal

a Describe fully both transformations. b The rotation from part a can be replaced by a reflection and still result in the same curve. What is the reflection? c The same curve can also be obtained by a horizontal stretch followed by a rotation. Find the scale factor of the stretch.

a The rotation is the origin.

clockwise about

The rotation replaces by

The equation after the rotation is

).

The horizontal stretch with scale factor replaces by

. After the enlargement:

.

Hence . The enlargement has centre at the origin and scale factor

(and by or

Compare the last equation to

.

.

b The reflection is in the line

.

The reflection needs to replace by . (See the table in Key point 3.1)

c After the stretch:

and by or

The horizontal stretch with scale factor replaces by .

After the rotation:

Hence

The rotation still needs to be

, so the scale factor is

Compare this to

clockwise.

.

.

WORK IT OUT 3.1 Describe a sequence of three transformations that transforms the circle with equation

.

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

1

Translation with vector

2

Horizontal stretch with scale factor

3

Vertical stretch with scale factor .

; ;

Solution 2 1

Horizontal stretch with scale factor

2

Vertical stretch with scale factor ;

3

Translation with vector

.

Solution 3

1

Translation with vector

;

;

into the ellipse

2

Horizontal stretch with scale factor ;

3

Vertical stretch with scale factor .

EXERCISE 3C 1

Find the equation of each curve after the given sequence of transformations. a i

; rotation through

ii

ii

.

anticlockwise about the origin; translation with vector

.

; rotation through

b i

anticlockwise about the origin; translation with vector

; horizontal translation units to the right; rotation through ; vertical translation units up; rotation through

c

i

clockwise about the origin.

clockwise about the origin.

; vertical stretch with scale factor ; translation units in the positive -direction; horizontal stretch with scale factor .

ii

; horizontal stretch with scale factor ; vertical stretch with scale factor ; translation units in the negative -direction.

d i

; enlargement with scale factor ; translation units in the positive -direction; enlargement with scale factor .

ii

; translation units in the negative -direction; enlargement with scale factor ; translation units in the positive -direction.

2

Describe a possible sequence of transformations that transforms: a i

to

ii

to

b i

to

ii c

i

to

ii

to

d i ii 3

to

to to

.

Curve is obtained from the parabola by this sequence of transformations: translation units in the positive -direction; enlargement with scale factor . Find the equation of and find the coordinates of its vertex.

4

The parabola

is transformed by this sequence of transformations: translation units in the

negative -direction; enlargement with scale factor . The resulting curve has equation 5

The ellipse

. Find the value of .

is rotated

anticlockwise about the origin and then translated. The

equation of the resulting curve is

.

Find the translation vector. 6

The hyperbola

is rotated

clockwise about the origin and then translated. The

equation of the resulting curve is

.

Find the translation vector. 7

a Sketch the curve with equation

, labelling the axes intercepts.

b Find the equation of the curve after this sequence of transformations: translation units in the positive -direction; enlargement with scale factor ; rotation through

anticlockwise about

the origin. 8

Curve has equation

.

a State the name given to this curve and find the equations of its asymptotes. This sequence of transformations is applied to : translation units in the positive -direction; enlargement with scale factor ; rotation through The resulting curve is

.

b Find the equation of

.

c Find the equations of the asymptotes of 9

anticlockwise about the origin.

.

a Find the equations of the asymptotes of the hyperbola

.

b Describe a single transformation that transforms this hyperbola to 10

.

a Describe a combination of transformations that maps the circle equation

onto the ellipse with

.

b Hence find an expression for the area of the ellipse. 11 The curve has parametric equations

, for

.

a Find the Cartesian equation of . can be obtained from the parabola vertical stretch.

by a combination of a horizontal translation and a

b Describe fully each of the two transformations. 12 The hyperbola with equation

is translated units to the right and stretched vertically

with scale factor . Find the equation of the resulting curve and the equations of its asymptotes. 13

a Show that the curve with equation

can be obtained from the hyperbola

by a combination of a translation and a stretch. b Hence find the asymptotes of .

14 The ellipse

is translated and then rotated

equation of the resulting curve is

anticlockwise about the origin. The .

Find the translation vector. 15 The curve

is transformed into the curve

by a translation followed by

two stretches. Describe fully each of the three transformations.

Checklist of learning and understanding You need to be able to transform curves using any combination of these transformations: Transformation

Change to equation

translation with vector

replace by by

reflection in the -axis

replace by

reflection in the -axis

replace by

reflection in the line

swap and

reflection in the line

replace by

and by

stretch horizontally with scale factor and vertically with scale factor

replace by

and

enlargement with scale factor and the centre at the origin

replace by and by

rotation through

replace by and by

anticlockwise

and

by

rotation through

anticlockwise

replace by

and by

rotation through

anticlockwise

replace by

and by

Because of the symmetries of the ellipse, hyperbola and parabola, some of the listed transformations appear to have the same effect. You can identify a transformation, or a sequence of transformations, by considering how to get from one equation to another.

Mixed Practice 3 1

Find the scale factor of the enlargement that, when combined, transforms the curve into the curve . Choose from these options. A B C D

2

a Sketch the curve with equation

, showing the coordinates of any intercepts

with the axes. b The curve is rotated

anticlockwise about the origin and then stretched horizontally

with scale factor . Find the equation of the resulting curve. 3

The parabola

is enlarged with scale factor and then translated by vector

.

Find the equation of the resulting curve and the coordinates of its vertex. 4

The hyperbola

is enlarged with scale factor and then translated by vector

. Find the equation of the resulting curve, and the equations of its asymptotes. 5

The parabola the line

has equation

. The parabola

Find the coordinates of the points of intersection of 6

is obtained from

by a reflection in

followed by a horizontal stretch with scale factor .

a Find the coordinates of the points where the line

and

.

intersects the circle

b Describe fully two transformations that, when combined, map the circle the ellipse

.

c Hence find the points of intersection of the line

with the ellipse

. 7

The parabola

is translated by

origin. What is the equation of the resulting curve? Choose from these options. A B

and then rotated

clockwise about the

. onto

C D 8

Find the equation of the parabola rotated

after it has been translated by

and then

about the origin.

State the coordinates of the vertex of the transformed curve. 9

a Show that the line

is tangent to the circle

.

b Hence find the positive value of for which the line

is tangent to the ellipse

. 10

a

Find the exact values of such that the line

The ellipse

is translated by

is tangent to the ellipse

.

and then enlarged with scale factor .

b Find the equation of the transformed ellipse. c Find the exact value of such that the line 11

a The curve with Cartesian equation equation

is tangent to the transformed ellipse. is mapped onto the curve with polar

by a single geometrical transformation.

By writing the polar equation as a Cartesian equation in a suitable form, find the values of the constants and . b Hence describe the geometrical transformation referred to in part a. [© AQA 2016] 12 Find the equations of the asymptotes of the curve

.

Choose from these options. A

and

B C

and and

D 13 Curve

and is an ellipse with equation

anticlockwise about the origin.

. Curve

and

is obtained by rotating

intersect at four points, which form a square. Find,

in terms of and , the area of the square. 14 Curve has parametric equations of the hyperbola

with



. Curve is a part

.

a Find the Cartesian equation of . b Describe a possible sequence of two transformations that transforms into . c Hence sketch , stating the coordinates of its vertex and the equations of its asymptotes.

15 The curve has polar equation

.

a Find the Cartesian equation of . b Describe a sequence of two transformations that transforms into the circle c Hence find the exact area enclosed by .

.

4 Further graphs and inequalities In this chapter you will learn how to: draw graphs of draw graphs of for non-linear solve modulus equations and inequalities.

Before you start… Further Mathematics Student Book 1, Chapter 4

1 Sketch the graphs of:

You should be able to sketch graphs of rational functions.

a b

Further Mathematics Student Book 1, Chapter 5

You should be able to sketch graphs of hyperbolic functions.

.

2 Sketch the graphs of: a b

.

3 On the same axes,

A Level Mathematics Student Book 2, Chapter 3

You should be able to sketch modulus graphs for linear functions.

A Level Mathematics Student Book 2, Chapter 3

You should be able to solve modulus equations and inequalities for linear functions.

sketch the graphs of and . 4 Solve the inequality .

Sketching reciprocal and modulus graphs This chapter focuses on extending further the types of functions you can sketch. Building in particular on your knowledge of rational functions from Further Mathematics Student Book 1, Chapter 4, you will first look at techniques for drawing reciprocal graphs, i.e. graphs of

for some given function

. You

will then look at sketching general modulus functions rather than just the linear modulus functions you worked with in A Level Mathematics Student Book 2, Chapter 3. As in that chapter, you will be able to use these modulus graphs to help solve modulus equations and inequalities.

Section 1: The graph of A reciprocal transformation turns the graph of The simplest reciprocal transformation turns

into the graph of into

.

.

y

y y = x

1 y = — x

x

O

x

O

This example illustrates the behaviour of the graph of any function

.

Key point 4.1

The line The line asymptote.

is a vertical asymptote.

is a vertical

(tends to zero from above) (tends to zero from below)

has a horizontal asymptote .

has a horizontal asymptote

is a turning point.

.

is the opposite turning point, if

is

finite.

Tip If you are unsure about which side of an asymptote to sketch the graph, check a point on each side.

Using Key point 4.1 as a guide, it is possible to make a rough sketch of the reciprocal for any given graph. WORKED EXAMPLE 4.1 The diagram shows the graph of function

, with a root at and asymptotes

Sketch the corresponding graph of

y

and

.

.

x = b

a

O

x y = –c

y = f(x) k

Consider what happens when is large and negative.

As So

is a horizontal asymptote of

.

As So

is a horizontal asymptote of so

.

is a vertical asymptote of

Consider what happens when is large and positive. Consider what happens when

has a root.

so

Consider what happens when asymptote.

has a vertical

has no

Consider what happens when point.

has a turning

. is a vertical asymptote of . has no turning points so turning points.

y

Put this information together.

x = a

O 1 y = –— c

1 — k

b

x y = g(x)

EXERCISE 4A

1

By first sketching the graph of axes intercepts and asymptotes. a i ii

, sketch the graph of

, indicating the positions of any

b i ii c

i ii

d i ii e i ii f

i ii

g i ii 2

For each function

:

A find any axis intercepts and vertical asymptotes for B write down any axis intercepts and vertical asymptotes for a i ii

b i

for

ii 3

The graph of a i

is shown. Sketch the graph of y 2

x

O

ii

y

y = 5

–1 O –2

3

x

.

.

y

b i

3

(3, 2)

O

1

x

5

y

ii

x

O

c

y

i

(5, 3) –2

O

x

3

y

ii

(–2, 4)

–3

O

4

(2, –5)

x

y

d i

(0.5, 2)

O

x

1

(1.5, –2)

(–0.5, –2)

ii

y (6, 5)

O

x

2 (1, –1)

y

e i x = –1

4 — 3 O

1

x

x = 3 y

ii

y = x –1 –2

O

2

x

–4 x = –1

4

This is a graph of

. Sketch the graph of

, indicating clearly the positions of any

asymptotes and coordinates of any maximum and minimum points.

y

3

O 1

–2

x

3

(2, –3)

5

This is a graph of

.

y

(1, 1) O

x x = 2

On a separate diagram, sketch the graph of

6

is a function of the form Given that

where is non-zero, and

has a single vertical asymptote at

has a single vertical asymptote at 7

The graph of the functions

y y = f(x)

1.5

5

y = g(x)

x

are shown. Copy .

.

and a horizontal asymptote at

, find the values of

and

same axes, sketch the graph of

O

.

, while

and . and

and, on the

Section 2: The graph of The modulus function has the property that any negative values are changed to positive values of the same magnitude. Hence the graph of will be identical to that of wherever is positive, and will be the same as whenever is negative.

Key point 4.2 To draw axis.

, reflect any parts of

that occur below the -axis to appear above the -

WORKED EXAMPLE 4.2 Given that

, draw the graph of

y

.

Draw the graph of to see which parts are above and which parts are below the -axis. y = f(x)

O

x

2

y

Reflect the parts of the graph that occur below the -axis in the -axis. y = |f(x)|

O

2

x

EXERCISE 4B 1

For each graph of

, sketch a graph for

.

y

a i

(0.5, 2)

1

O

ii

x

y (6, 5)

O

2

x

(1, –1)

y

b i 1 x= – — 2

O

4 — 3 1

x

x = 3 y

ii

–2

O

x

2 –4

x = –1

2

These are the graphs of

and

.

y

y y = g(x) y = f(x)

–1

4 O

1

x

–2

O

x

1 2

Sketch: a i ii b i ii c

i ii

3

Sketch each graph, showing the axes intercepts and any asymptotes. a i ii b i ii c

i ii

d i ii e i ii 4

Sketch the graph of

5

Use graphs to find the number of real solutions of the equation

6

Use graphs to find the number of real roots of the equation

7

a Find the minimum value of

given that

. Label all axis intercepts. . .

.

Tip You could approach part a of Question 7 by differentiating, completing the square or by using the fact that the minimum occurs halfway between the roots. b For what values of does the equation 8

The diagram shows the graph of

.

have exactly two solutions?

y (1, 3)

y = f(x)

y = 2

1 –— 2

x

O

a Sketch the graph of

.

b State the set of values of for the which the equation 9

Solve the equation

.

10 Find the range of the function 11 This is a graph of

for

.

.

y (p, q)

y = f(x)

x

O

Sketch the graph of

.

12 The diagram shows the graph of the function a Sketch the graph of

.

b Solve the equation

.

y y = f(x) (p, f(p)) a

O

d c

b (q, f(q))

x

.

has exactly three real roots.

Section 3: Modulus equations and inequalities In A Level Mathematics Student Book 2, Chapter 3, you learnt how to solve modulus equations and inequalities involving linear functions, for example: or . You did this by sketching the graph and considering whether intersections occurred on the reflected part or the unreflected part of the graph. You can apply the same technique to equations and inequalities involving the modulus of other functions.

Tip An alternative method is to square both sides, but this only works if both sides of the equation (or inequality) are positive.

WORKED EXAMPLE 4.3 Solve the equation

. Sketch the graph to locate the points of intersection.

y y = |x2 – 7x + 10| A

y = 5 – x B

O

2

5

x

From the graph, one intersection is at .

The graph shows three points of intersection.

There are two others: For

Point is at

is on the unreflected part of the parabola.

.

For

Point is at

is on the reflected part of the parabola, so make the expression negative. .

Hence the solutions are . WORKED EXAMPLE 4.4 Solve the inequality

for

. Sketch the graph, remembering the asymptotes.

Points of intersection:

There are points of intersection on both the reflected part and the unreflected part of the graph.

The inequality holds when:

Describe the parts of the graph which are above the line , making sure you include the intersection points but exclude the asymptotes.

Sometimes it is possible to rearrange the equation or inequality before sketching graphs. Remember that you can multiply both sides of an inequality by an expression that is always positive. WORKED EXAMPLE 4.5 Solve the inequality

. Use

.

Since the denominator is positive, you can multiply without changing the direction of the inequality. Use y

You can now use the method from A Level Mathematics Student Book 2, Chapter 3: sketch the graph and find B

intersection points.

y = |2x + 1|

y = |3x – 9| A O

Point :

x

is on the unreflected part of the red graph and the reflected part of the blue graph.

Point :

is on the unreflected part of both graphs.

Hence

The red graph is above the blue graph between and .

WORK IT OUT 4.1 Solve . Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 Squaring both sides to remove the modulus:

Solution 2

y 10 y = 2 – x 8 6 4

y = |x + 1|

2 –6 –4 –2O –2

2 4 6 8

x

The intersection point is on the right-hand branch of the modulus function.

The inequality is satisfied to the right of this intersection, so the solution is Solution 3 If

, then the inequality becomes

which is a contradiction, so no solution here. If

, then the inequality becomes

which is also a contradiction, so no solution here either. Hence there is no solution.

EXERCISE 4C

.

EXERCISE 4C 1

Solve the equation

2

Solve the inequality

3

Solve the inequality

4

Solve the equation

.

5

Solve the inequality

.

6

Solve the inequality

7

Solve the inequality

.

8

Solve the inequality

.

9

Solve the equation

. . .

.

.

10 By sketching appropriate graphs or otherwise, solve 11 Solve the inequality

.

12 Solve the inequality

, giving your answer to three significant figures.

13 Solve the equation 14 If

for

.

, giving your answer in terms of . , what can you say about

?

Checklist of learning and understanding For a reciprocal transformation:

The line The line asymptote.

is a vertical asymptote.

is a vertical (tends to zero from above) (tends to zero from below) or

has a horizontal asymptote .

has a horizontal asymptote

.

is a turning point.

is the opposite turning point, if

is

finite.

To draw the -axis.

, reflect any parts of

that occur below the -axis so they appear above

You can use a graph to solve modulus equations and inequalities. When rearranging modulus equations and inequalities you can use

and

.

Mixed practice 4 1

Given that which

, find the complete set of possible values of the constant for has two solutions.

Choose from these options. A All real numbers B C D 2

or

The graph of

is shown.

y

f(x)

O

–5

x

5

(–3, –5)

(3, –5)

a Sketch the graph of

.

b State the coordinates of the maximum points of 3

Solve the inequality

4

A function is defined by

. and

Find all asymptotes of 5

The graph of

.

and sketch

for

. .

is shown. y f(x)

O

Sketch

4

x

. Indicate clearly the positions of any -intercepts and asymptotes.

6

a Sketch the graph of the curve meets the -axis.

, indicating the exact value of the -coordinate where

b i Solve the equation

.

ii Hence, or otherwise, solve the inequality

. [©AQA 2012]

7

a On separate diagrams, sketch: i the curve with equation ii the curve with equation

.

b i Solve the equation

.

ii Hence solve the inequality

. [©AQA 2011]

8

a Solve the equation

.

b Write down the solution of the inequality 9

Solve the inequality

.

10 Solve the inequality

.

11 Solve the equation 12

.

.

a Sketch the graphs of

and

.

b Hence state the number of solutions of the equation 13

a Sketch on the same set of axes the graphs of b Hence solve the inequality

and

a Explain why

, where is a positive real number?

.

b Solve the equation

.

16 The diagram shows part of the graph of y y = f(x) y = 2

2 –2

O

.

.

14 For which values of the real number is 15

.

x

.

On separate diagrams, sketch the graphs of: a b

.

17 The hyperbolic secant is defined by

for

.

a Sketch the graph of , indicating the coordinates of any stationary points and the equations of any asymptotes. b Find the exact solutions of the equation 18

a Given that

, find the exact value of .

b By sketching the graphs of inequality

.

and .

, or otherwise, solve the

5 Further vectors In this chapter you will learn how to: find a vector perpendicular to two given vectors (using the vector product) find the equation of a plane in several different forms find intersections between lines and planes calculate angles between lines and planes calculate the distance from a point to a plane.

Before you start… Further Mathematics Student

You should be able to find the

Book 1, Chapter 9

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.

Further Mathematics Student Book 1, Chapter 9

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

Further Mathematics Student Book 1, Chapter 9

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.

Further Mathematics Student Book 1, Chapter 9

You should know how to use the scalar product to find the distance from a point to a line and the distance between two skew lines.

4 Find the shortest distance between the skew lines

and

Another way of multiplying vectors You already know from Further Mathematics Student Book 1, Chapter 9, how to find the scalar or dot product of two vectors. In this chapter you will learn about a second way of multiplying together two vectors – the vector product or cross product. This provides a convenient way for finding a vector that is perpendicular to the two original vectors, and has many applications in Mechanics and Physics such as calculating the moment of a force and finding the effect of magnetic fields on charged particles.

Fast forward If you are studying the Mechanics option of Further Mathematics you will meet the cross product in Chapter 9 when looking at moments and couples of forces. As well as using the cross product to express the vector equation of a line in an alternative form, you will also use it to help you find the equation of a plane (a surface in three-dimensional space).

Section 1: The vector product You know that you can use the scalar product to check whether two lines are perpendicular, but you also need to be able to find a line that is perpendicular to a given line.

Rewind You saw in Further Mathematics Student Book 1, Chapter 9, that lines are perpendicular if . In two dimensions, if a line has gradient then a perpendicular line has gradient

and

. Furthermore,

there is only one perpendicular line passing through any given point.

In three dimensions, if you try to find a line through a given point and a perpendicular to a given line, you will find that there is more than one possible answer. In fact, there are infinitely many such lines.

However, if you are given two lines, then there is only one direction that is perpendicular to both of them. This direction is given by the vector product (or cross product).

Key point 5.1 The vector product of two vectors,

, is defined as

where is the angle between the vectors and , and is a unit vector perpendicular to and .

Fast forward Later in this section you will use the magnitude of the vector product:

Tip

.

It is important to use the symbol to denote the vector product, so that it is distinct from the scalar product. This means that the vector

is perpendicular to both and .

The definition in Key point 5.1 leads to a formula for finding the vector product from the components of the two vectors.

Key point 5.2 If

and

then:

Tip The formula for the vector product can also be written using the notation of the determinant of a matrix:

WORKED EXAMPLE 5.1

Find a vector perpendicular to both

and

The vector

will be perpendicular to both and .

Use the formula from Key point 5.2:

Did you know?

The vector product has many uses, both in pure mathematics and in its applications. For example, you can calculate the area of a triangle from the magnitude of the vector product. In Mechanics, the moment of a force is actually a vector, given by

. You might also

know the ‘right-hand rule’ for determining the direction of the magnetic force.

Tip One important example of perpendicular vectors are the base vectors; for example, you can check that

and

.

WORK IT OUT 5.1 Find a vector perpendicular to

and

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 A vector perpendicular is given by:

Solution 2 Let the perpendicular vector be

Then

so

and

so

Choose Then

So the vector is

.

Checking shows that the dot product is zero with both vectors. Solution 3

So

is a perpendicular vector.

WORKED EXAMPLE 5.2

Lines and have equations

a Calculate

and

.

b Hence find the equation of the line through the point

that is perpendicular to both and .

Use the formula from Key point 5.2:

a

The vector equation of a line is . The direction is the perpendicular vector you have found in part a.

b

The line passes through

.

The vector product has many properties similar to multiplication of real numbers but, like matrix multiplication, the vector product isn’t commutative.

Fast forward In Section 2 you will use the vector product to find the equation of a plane.

Key point 5.3 Properties of the vector product: ​• ​• ​• ​• Also, from the definition in Key point 5.2, you can see that

In fact, this is a specific case of the fact that the cross product of any two parallel vectors is the zero vector.

Key point 5.4 ​• If vectors and are parallel then ​• In particular,

.

WORKED EXAMPLE 5.3 Prove that

.

.

Expand exactly as you would with real numbers. and

and

.

Equation of a line In Further Mathematics Student Book 1, Chapter 9, you saw that the vector the line, , to a general point, , is parallel to the direction vector .

Using the vector product and the parallel vectors equation of a line.

from a point on

and , you now have an alternative vector

Key point 5.5 A line passing through the point with position vector and parallel to the vector has equation:

Tip Using the properties of the vector product, this can also be written as

.

WORKED EXAMPLE 5.4 Find a vector equation of the line that passes through the points your answer in the form .

and

The line is parallel to:

Find the direction of the line.

So an equation is:

Use

, giving

with either or .

Expand:

Evaluate the cross product:

The magnitude of the vector product The magnitude of the vector product is related to the area of the triangle determined by vectors and .

Key point 5.6 The area of a triangle with sides and is

.

Rewind You met the formula

for the area of a triangle in A Level Mathematics Student Book

1, Chapter 11.

WORKED EXAMPLE 5.5

Find the area of the triangle with vertices

You need to find the two vectors that define the sides of the triangle. Drawing a diagram will help you to see which vectors to use.

You need to calculate

first, and then find its magnitude.

EXERCISE 5A 1

In each case, calculate

and state the vector

a i

ii b i ii 2

In each case, calculate

and verify that it is perpendicular to both and

a i

ii b i ii 3

Find, in radians, the acute angle between the directions of vectors and given that: a

and

b

4

and

c

and

d

and

Find, in the form coordinates: a

and

b 5

, a vector equation of the line that passes through the points with

and

Find, in the form

, a vector equation of the line given by the equation:

a b 6

Calculate the area of the triangle with vertices: a i ii

and

b i

and

ii 7

and

and

Find a vector that is perpendicular to both

and

8

A line is perpendicular to vectors

and

and passes through the point

. Find the

vector equation of the line. 9

The points

and

are vertices of a parallelogram

.

a Find the coordinates of . b Calculate the area of the parallelogram. 10 A line passes through the point

and is perpendicular to the lines with equations

and

. Find the equation of the line in vector form.

11 Find the Cartesian equation of the line that passes through the point the lines with equations 12 Given that 13

 and 

.

.

a Explain why

.

b Evaluate

.

15 Find the Cartesian equation of the line through the point and

that is perpendicular to lines

.

16 Prove that for any two vectors and 17 Given that

18

, find the exact value of

.

b Simplify 14

.

and the angle between and is

a Prove that

and is perpendicular to

.

, show that

a Show that the lines

.

and

b Points and lie on and respectively, such that i Write down ii Show that

in terms of and . .

iii Find a second equation for and . iv Find the coordinates of and . v Hence find the shortest distance between and .

do not intersect. is perpendicular to both lines.

Section 2: Equation of a plane You are already used to describing positions of points by using unit vectors parallel to the - and -axes: for example, the position vector of the point is .

However, you can also use two directions other than those of and . The same point can be 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 scalars.

, where and are

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 that lie in the plane, as shown.

Key point 5.7 The vector equation of the plane containing point and parallel to the directions of vectors and is .

WORKED EXAMPLE 5.6 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 be , as they all lie in the plane. Vectors

Use

and

are parallel to the plane.

.

In Worked example 5.6, 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 5.7

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, and the two direction vectors are parallel to the plane.

WORKED EXAMPLE 5.8

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 5.9 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 5.6) 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.

Key point 5.8 gives a list of possible ways to describe a plane.

Key point 5.8 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. The normal vector is perpendicular to the line plane.

, so

. This means that

, which gives another form of an equation of the

Key point 5.9 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.

Key point 5.10 Writing

and expanding the scalar product gives the Cartesian equation of the plane in

the form

.

Tip The letter (capital ) is often used as the name for a plane.

WORKED EXAMPLE 5.10

Vector

is perpendicular to the plane that contains point

a Write an equation of in the form

.

.

b Find the Cartesian equation of the plane.

The equation of the plane is

a

The Cartesian equation involves

b

.

and (the

coordinates of ), which are the components of the position vector .

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 Section 1.

WORKED EXAMPLE 5.11 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

.

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

Hence the point lies in the plane.

You can also convert from a Cartesian to a vector equation by finding two vectors that are perpendicular to the normal. WORKED EXAMPLE 5.12 Find a vector equation of the plane with Cartesian equation Finding two vectors

.

You need two directions in the plane that are both perpendicular to the plane’s normal.

perpendicular to These vectors satisfy

.

This gives one equation, which means that you can select values for two of the variables and then find the third one.

The two direction vectors are and

.

For a point on the plane

: So

To find a point in the plane, pick its - and - coordinates and then find from the equation of the plane.

is a point in the plane. A vector equation of the plane is:

Now use

.

EXERCISE 5B 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

and .

a 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

.

and .

a i ii b i ii 9

Show that point lies in the 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 18

are perpendicular to . .

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 3: Intersections between lines and planes The methods introduced in this section require 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 might need to convert them first.

Intersection between a line and a plane 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 5.13 Find the intersection between the given line and plane, or show that they do not intersect.

a

and

b

and

a

Substitute the plane.





into the equation of

Now use this value of to find the coordinates. The intersection point is .

b



Impossible to find . The line and plane do not intersect.

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

It is impossible to find a value of for a point which satisfies both equations. This means that the line is parallel to the plane.

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

WORKED EXAMPLE 5.14

Show that the line

lies in 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

on the line also lies in the plane.

plane.

Intersection of two planes If two planes intersect, they do so along a straight line. As the line lies in both planes, it must be perpendicular to both normal vectors. You can therefore find the direction of the line.

Key point 5.11 The line of intersection of planes with normals

and

has direction

.

To find the equation of the intersection line you also need to find the position vector of one point on the line. WORKED EXAMPLE 5.15 Find the line of intersection of the planes with equations Direction vector:

and

.

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

You can change the magnitude of the direction vector.

You also need one point on the line; its satisfy both planes’ equations. Taking

:

coordinates will

There are three unknowns but only two equations, so you can pick a value for one of the unknowns and then solve for the other two.

These values give the coordinates of one point on the line.

You can now write down the equation of the line.

Intersection of three planes Two different planes are either parallel or intersect in a line. With three planes, there are more options. If two of the planes are parallel, the third plane can either intersect both of them (in two different lines) or be parallel to them.



Tip If two planes are parallel, their normals are multiples of each other. For example, the planes and are parallel. If two planes intersect in a line, there are three different options for the third plane.

The third plane intersects the line of intersection of the first two planes.

The third plane contains the line of intersection of the other two.

The third plane is parallel to the line of intersection of the other two planes.

The three planes intersect at a unique point.

The three planes intersect along a line and form a sheaf.

The three planes form a triangular prism.

You already know how to find the line of intersection of two planes. You can then try to find the intersection of this line with the third plane. WORKED EXAMPLE 5.16

The planes

and

and

are given by the equations:

intersect along a line .

a Find a vector equation of . b Find the point of intersection of all three planes. a Direction vector of :

The direction vector of the line is the cross product of the normals.

For a point on , set

:

The equation of is

b Intersection of and

To find one point on the line, you can pick a value for its -coordinate and then use the equations to find and .

Use

:

Then

.

Substitute , , from the equation of the line into the equation of the plane.

Substitute back into the equation of the line to find the intersection point.

So, the three planes intersect at the point .

Worked example 5.17 shows two possibilities when three planes do not intersect at a single point. It uses planes and from Worked example 5.16. WORKED EXAMPLE 5.17 The planes

and

are given by the equations:

.

and

intersect along a line with equation

a Show that, when

.

, the three planes form a triangular prism.

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

a When

:

Try to find the intersection between

No solution for , so and not intersect.

do

Hence the three planes form a triangular prism.

b The equation is satisfied for all when .

It is impossible to find such that

and .

.

You can also check that is parallel to . This is because the direction of is perpendicular to the normal vector of :

You now want to lie in the plane . This will happen when the equation for intersection is satisfied for all .

Fast forward In Chapter 6 you will learn that you can also use matrices to find the intersection of three planes and distinguish between different cases when the intersection is not a single point.

EXERCISE 5C 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 a vector equation of the line of intersection of each pair of planes. a i

and

ii

and

b i

and

ii

and

4

Find the point of intersection of the line

5

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

6

Find the coordinates of the point of intersection of the line

.

with the plane

. 7

a Calculate b Hence find the equation of the line of intersection of the planes

and

. 8

Let

and

a Calculate

.

.

b Hence find the equation of the line of intersection of the planes with equations . Give your answer in the form 9

The plane with equation

and

.

intersects the , , and axes at points

,

respectively. a Find the coordinates of

and .

b Find the area of the triangle

.

10 The line is the intersection of the planes with equations

and

.

a Find a vector equation of . A third plane has equation b By considering the intersection between this plane and , find the coordinates of the point of

intersection of all three planes. 11

a Find the line of intersection of the planes: and b Another plane is Show that the three planes form a triangular prism.

12 Three planes have equations:

. a Find an equation of the line of intersection of

and

.

b Find the value of for which the three planes form a sheaf. 13 Show that the intersection of the planes:

is a line and find its direction vector in the form

.

14 Three planes have equations:

a Find the Cartesian equation of the line of intersection of

and

.

b Find the value of for which the three planes do not intersect. c For this value of , describe the geometrical configuration of the three planes.

Section 4: 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 . Drawing a two-dimensional diagram of triangle makes it clearer what the angles are.

Key point 5.12 The angle between the line with direction vector and the plane with normal is where is the acute angle between and .

,

Rewind You can find the angle between two vectors by using the scalar product, which you met in Further Mathematics Student Book 1, Chapter 9.

WORKED EXAMPLE 5.18

Find the angle between line with equation

and the plane with equation

.

The angle between the line and the plane is

The angle between a line and a plane is – (angle between the lines’ 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 , so the two angles marked are equal.

Key point 5.13 The angle between two planes is equal to the angle between their normals.

WORKED EXAMPLE 5.19 Find the acute angle between 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 (1 d.p.).

EXERCISE 5D 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

and

.

b 3

and

Line has Cartesian equation

.

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

7

Show that the planes with equations other.

8

Line has Cartesian equation

.

and

are perpendicular to each

.

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 5 you will see another method to find the distance from a point to a plane.

and .

Section 5: Distance between a point and a plane Given a plane with equation and a point outside of the plane, the 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 .

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

.

Point is called the foot of the perpendicular from the point to the plane. WORKED EXAMPLE 5.20 Plane has Cartesian equation

.

a Find the equation of the line that is perpendicular to and passes through the point b Hence find the distance from

.

to .

The perpendicular line is the direction of the normal to the plane.

a

b Intersection with :

You need the distance from to the point where the perpendicular line intersects the plane.

The intersection point is

The distance is

Find the distance between point .

and the

You can use the same idea to find the reflection of a point in a plane. The reflection of in is the point such that is perpendicular to the plane and the distance of from is the same as the distance of

from . This means that

.

WORKED EXAMPLE 5.21 Find the coordinates of the reflection of the point Let be the foot of the perpendicular from to . Then has coordinates

.

These are the plane and the point from Worked example 5.20, so you have already found the

. Let

in the plane

coordinates of .

be the reflection of

.

You need to find a point

’ such that

.

Then The vector between two points equals the difference between their position vectors.

The coordinates of the reflection are .

EXERCISE 5E 1

Plane has equation .

. Line is perpendicular to and passes through the point

a Write down the direction vector of . b Find the coordinates of the point of intersection of and . c Hence find the exact distance of from . 2

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 . 3

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.

4

Plane has equation The line

. Point

is reflected in and the image is point .

intersects at .

a Write down a vector equation of the line b Find the coordinates of . c Hence find the coordinates of .

.

5

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 6

is the reflection of in . Find the coordinates of

Points a Find

,

and

.

lie in the plane .

.

b Hence find the area of the triangle

, correct to three 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 7

.

Two planes have equations: , and

.

a Show that

and

b Show that

passes through the origin.

are parallel.

c Write down the equation of the line through the origin which is perpendicular to d Hence find the distance between the planes 8

a Show that the planes

and

and

.

.

are parallel.

b Write down a vector equation of the line through the origin which is perpendicular to the two planes. c

i Find the coordinates of the foot of the perpendicular from the origin to

.

ii Find the coordinates of the foot of the perpendicular from the origin to

.

d Use your answers from part c to find the exact distance between the two planes.

Checklist of learning and understanding The vector product of two vectors,

, is defined as

where is the angle between the vectors and , and is a unit vector perpendicular to and . The vector product can be calculated as

The vector

is perpendicular to both and .

Properties of the vector product:

If vectors and are parallel then In particular,

.

.

The vector equation of a line passing through the point with position vector and parallel to the vector can also be written as The area of a triangle with sides and is

.

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 written in the scalar product form

. This can also be

, where

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

To find the distance between a point and a plane: find the equation of the line in the direction of the normal, passing through the point find where this line intersects the plane find the distance between this and the original point. If two planes intersect, the line of intersection has direction parallel to

.

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. To find the intersection of three planes, first find the line of intersection of two of the planes, and then intersect this line with the third plane.

Mixed practice 5 1

Find a vector perpendicular to

2

Given that

3

The vector

and

.

and

, find

.

is normal to a plane that passes through the point

.

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

lies on the plane.

The plane has equation

and the point has coordinates

.

a Write down the vector equation of the line through that 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 . 5

a Calculate

b Plane

.

has normal vector

and contains point

.

Find the Cartesian equation of the plane. c Plane

has equation

. Show that

contains point .

d Write down the vector equation of the line of intersection of the two planes.

e A third plane,

 , has equation

.

Find the coordinates of the point of intersection of all three planes. f Find the angle between 6

and

Plane has equation a Show that point

in degrees. 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 . 7

The fixed points and and the variable point have position vectors



and

respectively, relative to the origin , where is a scalar parameter.

a Find an equation of the line b Determine c

in the form

.

in terms of .

i Show that

is constant for all values of , and state the value of this constant.

ii Write down a geometrical conclusion that can be deduced from the answer to part c i. [©AQA 2010] 8

Two planes have equations and where is a non-zero constant. Given that the acute angle, , between the planes is such that

, find the value of . [©AQA 2013]

9

Line

has equation

and line

a Find

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. 10 The position vectors of points and are:

a Find the vector product

.

b Using your answer to part a, or otherwise, find the area of the parallelogram with two sides

and

11 The vectors

where

.

and are such that:

and

.

What can you say about

and ?

Choose from these options. A

is parallel to .

B

is perpendicular to .

C

is parallel to

D

is perpendicular to

. .

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 The planes

and

have Cartesian equations

respectively. a Find, in the form and .

, a vector equation for the line which is the intersection of

b i Determine whether meets , and use your answer to decide whether the system given by the equations of these three planes is consistent or inconsistent. ii Describe geometrically the arrangement of the three planes. c

i Find the coordinates of a common point of

and

.

ii Deduce a vector equation for the line of intersection of

and

. [© AQA 2012]

15 Consider the tetrahedron shown in the diagram and define vectors

and

.

a Write down an expression for the area of the base

in terms of vectors and only.

b

is the height of the tetrahedron,

and

. Express in terms of and .

c Use the results from parts a and b to prove that the volume of the tetrahedron is given by . d Find the volume of the tetrahedron with vertices

,

,

and

.

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 and

.

19 Two planes have equations

a Calculate

b Show that

and

.

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 20 The lines

and

which is parallel to both planes.

have equations

and

respectively. a Determine a vector, , which is perpendicular to both lines. b i The point on Show that:

and the point on

are such that

for some constant .

ii Find the position vectors of and . iii Deduce the shortest distance between

and

. [©AQA 2012]

21

a The plane

The point b The plane

is perpendicular to the line

lies on

c

. Find an equation for

contains the points

Show that an equation for

.

is

in the form

and

.

.

, where is an integer to be found.

Show that the acute angle, , between the planes

and

d The plane has the equation three planes , and have no point in common.

is such that

.

, where is an integer. The

i Show that one possible value of is and find the other possible value. ii State the geometrical configuration of the three planes for each of the two cases in part d i, giving reasons for your answers. [©AQA 2016]

6 Further matrices In this chapter you will learn how to: calculate determinants and inverses for

matrices

use matrices to solve simultaneous equations in three variables interpret geometrically instances where there is no unique solution to three simultaneous equations in three unknowns carry out row and column operations and understand how these affect the determinant of a matrix find eigenvalues and eigenvectors for and matrices use eigenvalues and eigenvectors to diagonalise a matrix.

Before you start… Further Mathematics Student Book 1, Chapter 7

You should know how to add, subtract and perform scalar multiplication with matrices and carry out multiplication of matrices.

1

and

Calculate: a b

Further Mathematics Student Book 1, Chapter 7

You should know how to calculate the determinant of a matrix.

Further Mathematics Student Book 1, Chapter 7

You should know how to calculate the inverse of a nonsingular matrix.

Further Mathematics Student Book 1, Chapter 7

You should know how to calculate the transpose of a matrix.

.

2 Calculate det

3 Find

.

.

4 Given that , find .

Further Mathematics Student Book 1, Chapter 7

You should be able to use matrices to solve a simultaneous equation in two unknowns.

5 Express the system of simultaneous equations

in the form

Find and hence find and .

Chapter 5

You should be able to express a plane in vector sum, vector

6 Points and have coordinates

normal or Cartesian equation form.

and respectively. Find the plane containing and , giving your answer in the form: a b c

Chapter 5

You should know how to find the equation of the line of

7 Find the vector equation of the line of

intersections of two planes.

A Level Mathematics Student Book

You should be able to use the

2, Chapter 5

factor theorem to find factors of polynomials.

.

intersections of the planes and 8 a Show that factor of . b Express

is a

as a

product of linear factors. Further Mathematics Student Book 1, Chapter 8

You should know how to find invariant lines for a linear transformation.

9 Transformation is given by matrix . Find any invariant lines under transformation and determine whether any are a line of invariant points.

Further uses of matrices In Further Mathematics Student Book 1, Chapters 7 and 8, you learnt about matrices and how to use them to solve equations and represent linear transformations. You can extend the ideas of determinants and inverses to matrices and then use them to solve systems of three linear equations. Such systems have many applications, including solving problems about planes, which you met in Chapter 5. The idea of invariant lines (lines that are mapped to themselves by a transformation) leads to the concept of eigenvectors and eigenvalues. You will learn how to use these to find a diagonalised form of a matrix, but they have many other uses beyond the scope of this course.

Focus on … Matrices have applications in many other situations. See Focus on… Modelling 1 for an application in biology.

Section 1: Transposes and inverses Transposes You met the idea of transposes in your previous study of matrices. The transpose of a matrix is formed by swapping its rows and columns. There are some rules associated with transposes that you might find useful.

Key point 6.1 The transpose of a matrix obeys these rules:

Determinant of a

matrix

There are several ways to calculate the determinant of a

matrix. Here you will learn the standard

method, which can be extended for calculating the determinant of a square matrix of any size.

Rewind You saw how to find the determinant of a Chapter 7.

matrix in Further Mathematics Student Book 1,

First you need a few concepts and terms that will make the calculation easier. For each element of the matrix, the submatrix is what remains when you delete that element’s row and column. So for matrix

, the submatrix of element

is the

matrix

you find by deleting row and column from the original matrix. You call the determinant of the submatrix of element the minor of element

, which

.

The cofactor for element is equal to its minor multiplied by It can help to envisage this power change as a checkerboard pattern of signs. The

sign change matrix

looks like this:

Finally, to find the determinant of a matrix, select any row or column and total the product of each element and its cofactor. It does not matter which row or column you choose, the value will be the same.

Key point 6.2 To find the determinant of a

matrix :

select a row or column for each element

in that row or column, calculate the cofactor

the determinant is the sum of the products

for the row or column selected.

You will most often expand along the first row, giving this rule:

However, if the first row doesn’t contain a zero but another row or column does, then choose the one that does to expand along. This will make the calculation easier.

Tip If the matrix has an unknown, expand along the row or column containing the unknown.

WORKED EXAMPLE 6.1

Find the determinant of matrix

.

Select a row or column, ideally containing at least one zero, to simplify the calculation. Expanding about the first row:

For each element in the top row, multiply by the sign change submatrix.

and then by the determinant of its

Tip You are permitted to use a calculator to find determinants (and inverses) of numerical matrices. You are expected to know how to calculate by hand as well, in the event of nonnumerical entries. You can always check your answer on a calculator by substituting one or more trial values for any unknowns and checking that your formula matches the numerical output from your calculator!

Rewind Remember from Further Mathematics Student Book 1, Chapter 7, that a matrix that has a zero determinant is said to be singular.

WORKED EXAMPLE 6.2

Find the value of for which matrix

Using the middle column:

is singular.

Select a row or column containing that unknown to minimise working. The centre column also contains a zero so is the easiest option. For each element in the centre column, multiply by the sign change

and then by the determinant of its submatrix.

For to be singular, det , so . As you learned in Further Mathematics Student Book 1, Chapter 8, the determinant of a gives the area scale factor of the linear transformation it describes. The determinant of a

matrix matrix

gives the volume scale factor of the transformation, where a negative determinant indicates that the orientation of the object is reversed by the transformation.

Key point 6.3 For a linear transformation in three dimensions defined by a

matrix :

for an object with volume , the image of under will have volume if det is negative, the transformation involves a reflection so that an object and its image have opposite orientations.

The inverse of a

matrix

Rewind You saw how to find the inverse of a Chapter 7.

matrix in Further Mathematics Student Book 1,

For the inverse, you now have to calculate cofactors for each of the nine elements, and assemble these into a cofactor matrix

For a non-singular matrix , the inverse the reciprocal of the determinant of .

equals the transpose of the cofactor matrix, multiplied by

Key point 6.4 For non-singular matrix , where is the matrix of cofactors of ,

WORKED EXAMPLE 6.3

Find the inverse of non-singular matrix

The matrix of cofactors:

.

For each cofactor, find the determinant of the relevant submatrix and apply the sign-change .

Write down

The determinant of equals any lead diagonal element of :

of

.

, so calculate the upper left element to find det .

WORKED EXAMPLE 6.4

is a non-singular matrix. Find the possible values for and give

in terms of

. The matrix of cofactors:

Rather than working out the determinant first to find the possible values of , you can move straight to finding the inverse in terms of and calculate the determinant in the course of the calculations. For each cofactor, find the determinant of the relevant submatrix and apply the sign-change .

Write down

The determinant of

equals any lead

diagonal element of

:

, so calculate the upper left element of For

So

is non-singular for

Tip

.

.

to find det

.

to be non-singular,

.

Even with a matrix containing an unknown, you can still use your calculator. Assign to the value and then calculate and . Since all the other elements are integers, multiples and powers of should be easy to read off from the display, as long as the coefficients are less than . However, you will be expected to answer any question of this form showing full working, so you can only use your calculator to check your working.

EXERCISE 6A 1

Write down the transpose of each matrix. a i

ii

b i

ii

2

Find the determinant of each matrix. a i

ii

b i

ii

c

i

ii

d i

ii

3

Find the inverse of each non-singular matrix in Question 2.

4

By considering

, demonstrate that:

a b

.

5

Find all possible values of for which

6

Find all possible solutions for

7

Prove that

8

Matrices and are given by

.

if and are

Given that both 9

is singular.

and

square matrices.

and

are singular, find the values of and .

Matrices and are given by a Find each of

and

b Without finding

10 Matrix

and

.

in terms of .

, determine all values of for which

is given by

a Find

.

is singular.

.

.

b A three-dimensional object with volume

is transformed by a linear transformation with

matrix . Write down the volume of the image of when: i ii

for some non-singular matrix where is a

iii

matrix with determinant

11 Transformation is defined by matrix

.

.

The image of point under transformation has coordinates

12 Transformation is given by matrix

a Given

where

, find

b Given that maps point to

. Find the coordinates of .

.

and .

, find the coordinates of .

13 Two invertible transformations and are defined by matrices

. Point has image Find

under both matrices.

and the coordinates of point .

and

14 An orthogonal matrix is defined as a matrix orthogonal matrices, then so is

.

satisfying

. Prove that if and are

Section 2: Row and column operations You have seen how you can use matrices to solve simultaneous equations. You know that swapping the two equations doesn’t change the solution – this is an example of a row operation.

Rewind See Further Mathematics Student Book 1, Chapter 7, for a reminder of solving simultaneous equations using matrices. There are three elementary row operations that you can perform on a matrix: swapping two rows multiplying a row by a non-zero constant adding a non-zero multiple of one row to another row. There are three equivalent types of column operations, and all the results shown here for row operations apply equally when considering columns instead of rows.

Key point 6.5 If matrix is obtained from matrix

by:

swapping two rows (or two columns), then multiplying one row (or one column) by constant , then adding a multiple of one row to another row (or a multiple of one column to another column), then . For example, if you start from the matrix

you can form a new matrix by adding

:

lots of the top row to the bottom row, getting:

Matrix will be much easier to work with because it has a row with two zeros. You can use the property of determinants that 6.5.

to prove the results in Key point

PROOF 3 Each of these three row operations is equivalent to multiplying the original matrix by a matrix , where is formed from the identity matrix by performing the desired operation. Left-multiplication by mimics a row operation and right-multiplication by mimics a column operation. Swapping two rows Swapping the first and second row of a matrix is the same as left-multiplying it by

.

As you know from Further Mathematics Student Book 1, Chapter 8, the matrix represents a reflection through

the plane , and so has determinant ; each of the other row swaps can be similarly represented by a reflection. So if is formed by swapping two rows of

,

for some reflection . Therefore, Multiplying a row by a non-zero constant Multiplying the third row of a matrix by a nonzero constant is the same as left-multiplying it by

As you know from Further Mathematics Student Book 1, Chapter 8, the matrix represents a stretch in the -

:

direction, and has determinant .

So if is formed by multiplying a row of by a non-zero constant , for some stretch . Therefore, Adding a non-zero multiple of one row to another row Adding lots of the third row to the second row of a matrix is the same as left-multiplying it by

The matrix

represents a

transformation that you have not encountered in three dimensions, a shear in the planes perpendicular to the -axis, and has determinant .

:

So if is formed by adding a multiple of one row of to another row, for some shear . Therefore, det

Using row and column operations to factorise determinants You can use these properties of row and column operations to deduce the results shown here; you can use the shorthand for the ith row and for the th column to keep the notation simple. If every element of a row (or column) has a common factor, the determinant of the matrix can be factorised by dividing through by that factor. For example:

A matrix with a row (or column) equal to a multiple of another row (or column) must be singular. For example:

because

so you could use the row operation

to

form a matrix with a row of zeros. WORKED EXAMPLE 6.5

Factorise completely

.

Rather than calculating the determinant immediately, use row and column operations to simplify the calculation. Factorise from

from

and from

.

Column operations:

Calculate the determinant, expanding through Factorise

from

and

from

.

.

Calculate the determinant. Rearrange for symmetry. It is important to notice that in Worked example 6.5, the repeated pattern through inevitably lead to a similarly consistent pattern cycle in the determinant.

and must

Sometimes symmetry in the matrix can help in finding the factorised form of a determinant. For example, if two rows are identical when , then subtracting these two rows would produce a row of zeros when . You could then use the factor theorem to deduce that is a factor of the determinant. WORKED EXAMPLE 6.6

a Show that

is a factor of

.

b Factorise completely into linear factors. a Replacing the top row with the difference of the two rows:

Since the top two rows are equal when , their difference is useful to consider. Factorising it highlights the factor of . From Key point 6.5 this operation does not change the value of . Use Key point 6.5 to take out to factor of

b

Use

.

Take out a factor from the middle row.

.

You can use

.

Take out a factor from the middle row.

Expand the determinant along the first column. Since the

determinant equals

WORK IT OUT 6.1 Find the determinant of

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 Using your calculator to find the determinant with numerical matrix:

Solution 2 Subtracting row 3 from row 1 and expanding along the first column:

Solution 3

Using the determinant to calculate a vector product You learned in Chapter 5, Section 1, how to calculate the vector product using the formula

If the vectors and are written as sums of the unit vectors , and , then you can also express the vector product using determinant notation.

Key point 6.6 If

and

, then

.

This result will be given in your formula book. Since the result in Key point 6.6 is a determinant, you can use column operations to simplify the determinant before calculating. WORKED EXAMPLE 6.7 Vectors and are given by Find

and

.

. Use

.

Expand along the first column.

EXERCISE 6B 1

Factorise each determinant completely. a i

ii

b i

ii

c

i

ii

2

Use

to find the vector product

a i

ii b i

,

for the given vectors.

ii 3

,

By using suitable row and column operations, show that each of these matrices is singular. a i

ii

b i

ii

4

Matrix is given by

.

a Show that is singular if b Calculate

5

Matrix

.

in terms of .

is given by

.

a Explain why the structure of the matrix shows that these must be factors of the determinant of i ii iii

.

b Find the determinant of

6

Matrix

, factorising your answer as far as possible.

and are given by

and

a Define the column operation which converts b By considering

.

to .

for some matrix , show that

.

7

Factorise completely the determinant

8

Three column vectors, , and are written as an array, to form matrix

a Show that det

=

.

:

.

b Use the result in part a to show that: i exchanging the first two columns of a matrix changes the sign of its determinant but not the absolute value of the determinant ii rotating the columns

does not change the determinant of a matrix

iii if two columns of a matrix are multiples of each other, then the matrix determinant is zero iv if the three columns are such that

for some non-zero values , and ,

:

then the matrix determinant is zero.

9

Express the determinant

10 Points

as the product of four linear factors.

, and have position vectors

and respectively, where

. a i Find the Cartesian equation of the plane containing does not lie in . ii By considering the cross product

and and show that the point

, find the exact area of triangle

.

iii Calculate the exact perpendicular distance between and plane . iv Use your answers to parts ii and iii to calculate the volume of tetrahedron

.

b Explain how your answer to part a iv relates to the value

Tip The volume of any straight-edged tapering solid equals

where is the base area and is

the perpendicular distance from base plane to vertex.

Did you know? Listing three vectors , and as the columns of a matrix and finding its determinant is a way of calculating the vector triple product with many applications in geometry and beyond. The vector triple product can be used to calculate the volumes of parallelepipeds and other solids in three-dimensional space.

Section 3: Solving linear systems with three unknowns In Further Mathematics Student Book 1, Chapter 7, you learnt how to recast a pair of simultaneous equations in two unknowns into a matrix problem, and use matrix inverses and multiplication to solve for the unknowns. Geometrically, the solution represents the intersection of two straight lines. If the matrix then the two lines must be parallel, and so have no unique intersection point.

is singular

You can use exactly the same approach to solve a set of three simultaneous equations in three unknowns.

Key point 6.7 For simultaneous equations in three unknowns

rewrite as If

where

is non-singular, the solution is

and , in the form

and

.

.

WORKED EXAMPLE 6.8 Using matrices, solve these linear simultaneous equations.

Recast the problem in matrix form as

Rearrange to

.

, find the inverse

and solve to find the unknowns.

Solution:

Remember to state the solution.

When is non-singular, a unique solution to and can be found. When is singular, there is no unique solution. You will look at the geometrical interpretations for these situations, and how to distinguish them, in Section 4. EXERCISE 6C 1

By expressing the set of simultaneous equations as a matrix problem solution in each case, or determine that there is no unique solution.

, find the unique

a i

ii

b i

ii

c

i

ii

2

Show that the system of equations

has no unique solution. 3

By calculating a matrix inverse, find the intersection between the planes and .

4

For what values of does the system of equations

have no unique solution?

5

Matrix

is given by

a Show that b Find c Use

6

is non-singular for every real value .

in terms of . to solve the equations

a Show that the system of equations has no unique solution for

.

b Find all values of for which there is no unique solution. c For

, find

and .

Section 4: Geometrical interpretation of 3-variable simultaneous equations Rewind In Chapter 5, Section 3, you learnt how to use vectors to find the line of intersection of two planes. Matrix and vector calculations offer complementary approaches to problems involving 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. As with lines, two planes can either intersect (planes intersect in a line) or be parallel or identical.

planes intersecting in a line

parallel planes

coincident planes

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

Consistent system: unique solution The matrix of coefficients is non-singular; therefore the three planes meet at a single point. Using the working described in Section 3, you can find the point.

three distinct planes intersecting in a point 2

Consistent system: infinitely many solutions The matrix of coefficients is singular. The three planes intersect either in a line or in a plane. a If all three equations describe the same plane, the solution is that plane.

coincident planes b If two of the equations describe the same plane, and the third plane is distinct and not parallel, the solution is a line.

two distinct planes intersecting in a line c If the three equations describe different planes, all of which meet at the same line (the planes form a sheaf), the solution is that line.

three distinct planes intersecting in a line 3

Inconsistent system: no solutions The matrix of coefficients is singular. The three planes coincide at no points. a All three planes are parallel and distinct.

three distinct parallel planes b Two of the equations describe the same plane and the third describes a distinct but parallel plane.

two distinct parallel planes c Two of the equations describe distinct, parallel planes and the third describes a plane which is not parallel to the other two.

two parallel planes and one non-parallel plane d 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 Cases 1 and 2 are said to arise from a consistent system of equations, because solving the equations (using algebraic elimination, for example) will reduce the three equations to: in case 1, a unique solution (such as ) in case 2, algebraic elimination will lead to an equation that is obviously always true, such as Case 3 arises from an Inconsistent system; attempting to solve the equations simultaneously will produce a contradiction, such as

.

After finding whether or not the solution is unique and whether the system is consistent, you can then consider how many of the planes are parallel to determine the geometric configuration of the system.

Key point 6.8 To determine the geometric configuration of a by system of equations: find the determinant of the matrix of coefficients if the matrix is singular, determine whether the system is consistent consider how many of the planes are parallel to each other.

WORKED EXAMPLE 6.9 For the system of simultaneous equations

where and are constants:

.

a show that there is a unique solution if

and find that solution when

b find the value of for which there is not a unique solution c for the value of found in part b, find the value of such that the system is consistent. Interpret your solution geometrically.

Recast the problem in matrix form as

a

Expanding about the first column: Show that to establish the existence of a unique solution. The matrix is non-singular for and therefore there will be a unique solution. For

,

: Rearrange to inverse

Solution:

,

,

find the

and solve to find the unknowns.

Remember to state the solution.

b There is no unique solution when the matrix determinant is .

Use the result for the determinant that you found in part a.

c When

Eliminate by using (1) and (2). Eliminate from (3) and (4). For a consistent solution,

, so

There are infinitely many solutions since the system is consistent. Since all three normals are different, the planes are all different so they must intersect in a line. They form a sheaf of planes.

WORKED EXAMPLE 6.10 For the system of simultaneous equations

The equation must reduce to system to be consistent.

for the

Being consistent but with no unique solution means that there are infinitely many solutions. You then only need to distinguish a plane of solutions from a line of solutions.

find the value of for which there is no unique solution, and interpret this situation geometrically. No unique solution when

.

There is no unique solution when When

:

Now use elimination to decide if the system is consistent.

Eliminate from (1) and (2). You could choose any variable to eliminate, but the numbers for look convenient. Eliminate from (1) and (3).

Eliminate from (4) and (5). Because the determinant was zero, this step should also eliminate . The system is therefore inconsistent.

This is since the last equation is clearly not true.

All three normals are in different directions, so no two planes are parallel. Therefore, they form a triangular prism.

You need to distinguish between the four possible geometric configurations of the inconsistent system. You can do this by considering how many planes are parallel.

Finding the solution set to a consistent system You already know that if the solution set to

is a single point, you can find its position as .

If the solution of the system is a line , at least two of the planes must be distinct (and not parallel). You can find the direction vector of by taking the cross product of the two distinct plane normals. To find a single point on line set one of , or equal to zero (pick one whose element in the direction vector is not zero) and solve for the other two in any two of the plane equations simultaneously.

Rewind You learnt how to find the line of intersection of two planes in Chapter 5, Section 3.

WORKED EXAMPLE 6.11 Three planes are described by the equations:

Given that the system is consistent, find the solution of this system of equations and interpret it geometrically.

Recast the problem in matrix form as Where . . , so there is not a single point of intersection.

You can find

No row of the matrix is a multiple of another, so there are no parallel planes. The three planes form a sheaf with a common intersecting line .

You now need to distinguish between a plane of solutions and a line of solutions. You can do this by considering the number of parallel planes from looking at the normal of each plane.

From the original matrix

To find the direction of the line, take the vector product of two rows of the matrix (equivalent to the vector product of the plane normals).

:

The direction vector of is

by using technology.

.

Setting in the first two equations of the system:

Fix ( or would also work since the direction vector has no zero elements).

, so So the point

lies on .

The planes intersect at line

Remember to state the solution.

.

EXERCISE 6D Questions 5 to 8 in this exercise also appear in Exercise 5C. Use this to compare the two methods of solving the equations. 1

For each set of simultaneous equations, determine whether they are consistent or inconsistent and interpret the geometrical relationship 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

A system of equations is given by a Show that the system has 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 equation system given by b Show that the system is inconsistent. c Interpret the system geometrically.

6

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. d For the value of found in part b, solve the system of equations. 7

a Show that the system of equations

is consistent. The three equations in part a represent three planes.

b Describe the geometrical configuration of the planes. c Find the solution of the system. 8

a Find the inverse of the matrix b Hence find, in terms of , the coordinates of the point of intersection of the planes and

9

.

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 system. 10 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 Describe the geometric configuration of the three planes represented by the three equations. d Find the solution of the system for the value of from part a and the larger of the two values of from part b.

Section 5: Eigenvalues and eigenvectors In Further Mathematics Student Book 1, Chapter 8, you learned how to find invariant lines for twodimensional linear transformations. For example, when finding an invariant line of the form

for the matrix

, you would set up an equation

.

You can turn this into simultaneous equations and solve to find and . In this case , . is said to be an eigenvalue of

, with associated eigenvector

,

or

.

The eigenvector does not have a unique form; any non-zero multiple is valid since it only serves to define the direction of the invariant line.

Key point 6.9 If

then is called an eigenvalue of

and is its associated eigenvector.

The only constraint on the eigenvector is that it cannot be a zero vector as this does not describe any direction. For any vector , the identity matrix , by its definition, has the property

.

Therefore,

(since if

were not singular, the equation could be solved to get

forms a polynomial in called the characteristic equation of matrix the eigenvalues of .

). and its roots are

To find the eigenvector or eigenvectors associated with each eigenvalue, substitute each value into the equation and solve the resulting equations. You will find that these equations do not have a unique solution, but you can set one (or sometimes more) of the components of equal to a convenient value to find a solution.

Tip This ability to set a component to a convenient value is equivalent to saying that the length of the eigenvector is not important.

WORKED EXAMPLE 6.12 Find the eigenvalues of  Characteristic equation:

and their associated eigenvectors. Solve the characteristic equation det eigenvalues.

to find the

For

:

Substitute into the equation

with

Write the matrix equation in a standard algebraic form. You will usually find that the two equations are equivalent. This means that there are infinitely many solutions.... When

then

, so

... but you only want one, so you are free to choose the one with

an eigenvector associated with For

.

is :

Substitute into the equation

with

.

Both rows give

Write the matrix equation in a standard algebraic form. Here you just get the same equation twice.

When then , so an eigenvector associated

You are free to choose the one with integers.

with

to keep all numbers as

is

Tip The lead diagonal of a square matrix runs from the top left element to the bottom right element. The sum of this lead diagonal is the sum of the eigenvalues, which provides a useful way of checking your answer.

Characteristic equations for

matrices

You can use exactly the same principle to find eigenvalues and eigenvectors for a

matrix.

WORKED EXAMPLE 6.13

Find the eigenvalues and eigenvectors of

.

Characteristic equation:

Find the characteristic equation roots are the eigenvalues.

, whose

Expanding about the first row:

If you expand the cubic fully you will need to use the Factor theorem to show that is a root. If possible, keep the cubic factorised to reduce working.

The eigenvalues are (repeated) and .

Use

For

Where all three rows are multiples of each other, a single plane is described. Pick any two non-parallel

:

to find the eigenvectors.

vectors in that plane as eigenvectors. All three rows give

.

Choose

,

, then

.

Choose

,

, then

.

To find these two vectors you are free to choose two values of the variables, then use the equation to find the third one. There are many possible choices, so there is no unique correct set.

So the eigenvectors associated with are For

.

:

When

so

and

, then:

You are free to choose a value of .

Notice that you did not have to use equation (3) - this is normal when solving for eigenvectors as it is just a

.

Substituting into (1) gives

.

combination of the other two equations so does not give additional information. In this example, .

Therefore an eigenvector associated with

is

.

Did you know? If an eigenvalue is a repeated solution of the characteristic equation with multiplicity then there can be up to eigenvectors associated with that eigenvalue. It is therefore possible for an by matrix to have fewer than eigenvectors. If this is the case, then it is usually associated with a geometric transformation called a shear. In finding the final eigenvector you might have wondered whether your choice of might fail. The answer is, sadly, yes – if in Worked example 6.13 you had chosen , you would have found and which is not an allowed eigenvector. Sometimes you can get equations with no solutions. In either case you need to look for another possible value. Sometimes, as in Worked example 6.14, you find that some values are constrained and only some are free to be chosen. WORKED EXAMPLE 6.14

The matrix

is given by

.

Find an eigenvector associated with an eigenvalue of . Put

This is equivalent to:

into the usual eigenvector equation.

Equation (2) seems very strange! Also, there is no mention of in any equation.

Equations (1) and (2) can be solved directly, so you are not free to choose or . So

and

There is no constraint on , so choose . So an eigenvector is

However, you are free to choose any convenient value for .

.

EXERCISE 6E 1

Find the eigenvalues and corresponding eigenvectors for each matrix. a i ii b i ii c

i ii

d i ii 2

Each matrix has only one eigenvalue. Determine the unknown constant and find the eigenvalue and associated eigenvector. a i ii b i

ii 3

Find the eigenvalues and eigenvectors of the matrix

4

Find the eigenvalues and eigenvectors of the matrix

5

Matrix is given by

.

.

.

Show that is an eigenvalue of and find the remaining eigenvalues and all associated eigenvectors of the matrix. 6

A transformation is given by

.

Given that has a line of invariant points, find all eigenvalues and associated eigenvectors of the matrix.

7

A matrix is given by

.

a Show that is an eigenvalue of . b Find all eigenvalues and associated eigenvectors of .

8

Matrix

. is the eigenplane associated with the eigenvalue of

.

a Find and . b Find the other eigenvalue and its associated eigenvector.

9

Transformation is described by matrix The sum of two of the eigenvalues of

.

is .

Find all the eigenvalues and associated eigenvectors of

.

10 Transformation is described by matrix

.

a Determine the characteristic equation of real value

and show that

has eigenvalues and , for some

, where is to be determined.

b i Find the eigenvectors associated with the eigenvalues. ii Show that all eigenvectors associated with associated with .

are perpendicular to all eigenvectors

is the composition of an enlargement with scale factor and a second transformation . c Use your answer to part b to describe the effect of transformation .

11 Matrix is given by associated eigenvectors.

and has two eigenvalues. Find the eigenvalues and

Section 6: Diagonalisation and applications Suppose matrix has linearly independent eigenvectors with associated eigenvalues . You can now use your understanding of eigenvectors and eigenvalues from Section 5 to diagonalise matrix ; that is, you can express in the form for some matrix where is a diagonal matrix. A diagonal matrix has zero for every element outside the lead diagonal.

Tip Remember, the lead diagonal of a square matrix runs from the top left element to the bottom right element:

.

To illustrate this, you can use the matrix

from Worked example 6.13, for which the

eigenvalues and eigenvectors were:

associated with

, associated with

and

.

Construct , with columns equal to the eigenvectors:

.

Remember that because the eigenvectors are linearly independent, you know from Section 2 that must be non-singular. Now since by definition

, it follows that

where is the diagonal matrix whose lead diagonal consists of the eigenvalues, listed in the same order as their associated eigenvectors in . But now, since

and is non-singular, right-multiplication by

gives

which is the diagonalised form required.

Key point 6.10 If

matrix

has eigenvectors

with associated eigenvalues

, then you

can write

in diagonalised form

where

and is a diagonal matrix with

as the lead diagonal elements.

WORKED EXAMPLE 6.15 The matrix

has eigenvalues

and Write

with associated eigenvectors

.

in the form

Let

and

, where is a diagonal matrix.

, the matrix with the eigenvectors as columns and

let

, the diagonal matrix with eigenvalues as elements of

Set

and

the diagonal matrix of eigenvalues.

the lead diagonal. Then Use

The method of diagonalisation relies on being non-singular: that is, there must be three non-parallel eigenvectors. As you saw in Section 5, this will be the case when the characteristic equation has no repeated roots, but might not be true otherwise. An matrix with fewer than eigenvectors cannot be diagonalised. The diagonalised form has several uses, but the most immediate application lies in calculating powers of a matrix. If

, then:

and because is a diagonal matrix, you can write down

without needing any matrix multiplication.

Key point 6.11 If matrix

can be written as

for some matrices and , then:

for any positive integer .

Tip You can easily show that Key point 6.11 is also true for negative integer values as long as non-singular (equivalently, as long as all eigenvalues are non-zero). It is not valid for noninteger values of .

WORKED EXAMPLE 6.16

is

It is known that the transformation given by

,

a Express

and

with associated eigenvectors

in the form

b Hence find

has eigenvalues

,

,

.

, where is a diagonal matrix.

for odd integer .

a Let

columns and let

, the matrix with the eigenvectors as

Set and the diagonal matrix of eigenvalues.

, the diagonal matrix with

eigenvalues as elements of the lead diagonal. Then Use

Calculate , using your calculator.

b

Find by raising each eigenvalue to the power . is odd so

Check that your answer is consistent with the original matrix .

EXERCISE 6F 1

For each matrix, if the matrix can be given in the form and . a i ii b i

where is a diagonal matrix, give

ii

Tip There is not a unique diagonalised form. For example, reordering the eigenvalues and eigenvectors will produce different matrices and . 2

Each of these matrices has an eigenvalue . For each matrix, if it can be written in diagonalised form where is a diagonal matrix, find and a i

ii

b i

ii

3

Express

4

Explain why

5

Matrix

in the form

cannot be written in the form

has eigenvectors

a Find two possible matrices b Find the product 6

, where is a diagonal matrix.

and and

, where is a diagonal matrix.

and eigenvalues and

.

.

and comment on your answer.

Matrix is given by

.

a Find a matrix and a diagonal matrix such that

.

b Find for integer when: i is odd ii is even. 7

Transformation is described by matrix a If

where

.

for some positive , calculate .

b Find the two eigenvectors of and their associated eigenvalues and show that the two eigenvectors are perpendicular. c

is the composition of an enlargement with scale factor and another transformation . Describe .

d By diagonalising

8

a Express b Hence find

, or otherwise, find

.

in the form for odd integer , in terms of .

, where is a diagonal matrix.

9

a Describe the transformation given by matrix

.

b Show that is an eigenvalue of and find the other eigenvalue and the complex eigenvector associated with each. c Use your answer to part to write in the form d Hence show that

for some diagonal matrix .

.

Checklist of learning and understanding The transpose of a matrix obeys these rules:

The determinant of a

matrix :

(expanding along the first row) is calculated as

the same value is obtained by expanding along any row or column represents the volume scale factor of linear transformation . The inverse of a non-singular

matrix is given by

where is the matrix of cofactors of . The vector product

can be calculated using matrix determinants:

Three simple row operations affect the determinant in predictable ways. Matrix is obtained from matrix

by swapping two rows:

Matrix is obtained from matrix

by multiplying one row by constant :

Matrix is obtained from matrix

by adding a multiple of one row to another row:

Equivalent column operations follow the same pattern. 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 intersection of the three

If det

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

inconsistent (no common intersection) consistent (line or plane as the common intersection). When , you can determine the geometrical interpretation by considering the consistency of the system and the number of parallel planes. An eigenvalue and associated eigenvector of a matrix The eigenvalues of an

matrix

have the property that

are the roots of the polynomial

Each eigenvalue has an associated eigenvector. Repeated eigenvalues can have multiple eigenvectors. If an matrix has non-parallel eigenvectors diagonalised form as where If

, then you can express

and is a diagonal matrix with lead diagonal elements , then

for integer .

in

Mixed practice 6 1

Find the sum of the eigenvalues of the matrix

.

Choose from these options. A B C D 2

Find the value of that results in the system of equations shown having a non-unique solution.

Choose from these options. A B C D 3

Find the eigenvalues and associated eigenvectors of

4

is an eigenvector of matrix

.

.

a Find the value of . b Find the eigenvalue associated with

and find the remaining eigenvalue and

associated eigenvector. 5

Let

.

a Use a row operation to show that

is a factor of .

b Hence, or otherwise, express as a product of linear factors. [©AQA 2011] 6

Matrix

is singular.

Using row operations, or otherwise, show that there is only one real value .

is a factor of

and hence show

7

Let a Determine

. .

b Show that det 

for all real values of .

c Find the value of for which the matrix

is singular. [©AQA 2012]

8

a Find the values of for which the system of equations

does not have a unique solution. b For the integer value of found in part a, find the relationship between

and such that

this system of equations is consistent. [©AQA 2011] 9

Let

a Show that

.

is a factor of .

b Factorise as completely as possible. [©AQA 2012] 10

a Determine the two values of for which the system of equations

does not have a unique solution. b Show that this system of equations is consistent for one of these values of , but is inconsistent for the other. (You are not required to find any solutions to this system of equations.) [©AQA 2010] 11 The matrices and are defined in terms of a real parameter by

and

.

a Find, in terms of , the matrix is a scalar multiple of the b For this value of , deduce

and deduce that there exists a value of such that identity matrix . . [©AQA 2010]

12 Factorise fully the determinant

.

[©AQA 2010] 13 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 14 The

.

.

matrices and satisfy

, where

and is constant. a Show that b Find c Find

is non-singular. in terms of .

. [©AQA 2013]

15 Let

.

a Use a row operation to show that

is a factor of .

b Factorise as fully as possible. [©AQA 2016]

16 The matrix is given by

.

a Given that

, find the value of and the value of .

b Given that

, prove that .

c Given that

, find the value of and the value of .

d Hence, or otherwise, find the solution of the system of equations

giving your answer in terms of . [©AQA 2013] 17

a By direct expansion, or otherwise, show that the value of

is

independent of . b i State, with a reason, whether the equations

are consistent or

inconsistent. ii The three equations given in part b i are the Cartesian equations of three planes. State the geometrical configuration of these three planes.

18 The matrix

where is a constant.

a Show that b Obtain c Use

is non-singular for all values of . in terms of .

to solve the equations

giving your solution in terms of . [©AQA 2014]

19 The matrix

, where is a constant, has three distinct eigenvalues

and

. a Given that

, find the value of .

b Given that

, find the values of

c The transformation has matrix

and

.

. Find a vector equation for any one of the invariant

lines of [©AQA 2016] 20 The transformation is represented by the matrix

where

and

with diagonalised form

.

a i State the eigenvalues, and corresponding eigenvectors, of

.

ii Find a Cartesian equation for the line of invariant points of . b

Write down

, and hence find the matrix

in the form

where

and are

integers. c By finding the element in the first row, first column position of

is a multiple of

, prove that

for all positive integers . [©AQA 2010]

FOCUS ON … PROOF 1

Extending the proof of De Moivre’s theorem De Moivre’s theorem states that . In Chapter 1, Section 1, you saw how to prove this result for integer values of . However, the result extends, with a little bit of careful consideration of conventions, to rational values.

Proving the result for rational values 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 with and integers with no common factors, then

would equal

. 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.

Tip Notice that when the power is an integer (for example:

) then there is no problem – there is

only one possible answer.

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 . As at the start of this section, you could write extend to irrational numbers, you would have

. Then, if De Moivre’s theorem did

. 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

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 involves 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:

y

(x, y) θ x

O

Rewind You met the rotation matrix in Further Mathematics Student Book 1, Chapter 8. The second method involves 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 :

Im zeiθ

z θ O

Re

Rewind You met the idea of multiplication by

representing a rotation in Chapter 1, Section 6.

You can see that the two methods give the same coordinates of 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). In this section you will compare the two methods when solving this problem: Three snails start at the vertices of an equilateral triangle. Each snail moves with a 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?

S2

v v S1

v

S3

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

be at

. Let be the speed of

each snail. 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

anticlockwise about the origin, and you can find the position of

by rotating

by rotating

through

through the same angle.

You will first approach the problem by using position vectors and matrices. S2

120° O

S1

S3

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 this system of differential equations.

in terms of and .

for some constant . Show that

.

Although you will learn in Chapter 12 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 the system

of equations in Question 3 later, but for now you will consider a different approach. QUESTIONS Now, let be the complex number corresponding to the position of given by

at time . Then the position of

is

.

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

and

, show that

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?

The curve described by the equations in Question 7 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. In Question 6, you derived two separate differential equations for and , which are 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 by 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

.

y (x, y) r θ O

x

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 . 11 Obtain an expression for

and hence show that

Tip Notice that these are the same equations you derived in Question 6.

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 have 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 juvenile rabbits each year. Each year 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 .

of

and the number of juvenile rabbits is denoted by

QUESTIONS 1

Explain why

2

Find an expression for

3

in terms of

.

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

4

.

.

An uninhabited island is populated with number of adult rabbits in:

adult rabbits in year . Use technology to find the

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

Find the eigenvalues and eigenvectors of

8

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.

9

Describe the assumptions made in creating this model.

. What do you notice?

10 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.

CROSS-TOPIC REVIEW EXERCISE 1

1

a Show that

.

b Hence find the value of 2

a Express

correct to three significant figures.

in the form

, where

b Solve the equation . 3

and

.

, giving your answers in the form

The hyperbola has equation

, where

and

.

Describe a sequence of two transformations that map onto the curve with equation . 4

The diagram shows the graph of

.

y

y = 3

y = f(x) O

x

2

On separate axes, sketch the graphs of: a b 5

.

a Sketch the graph of

.

b Solve the inequality 6

.

a On the same axes, sketch the graphs of: i ii

.

b Solve the inequality 7

Points

.

and have coordinates

a Find

,

and

respectively.

.

b Hence find the vector equation of the line through that is perpendicular to 8

a Find the Cartesian equation of the plane . The plane

containing the points

has Cartesian equation

a

Given that and

are both real, show that with

b Find, in terms of , the argument of c Hence show that

,

and

.

b Find, to three significant figures, the acute angle between the planes 9

and

. .

,

and .

.

.

10 Find the smallest positive integer values of and for which . [© AQA 2012] 11 Let

and

a Find

.

b Find the value of , given that 12

.

is parallel to .

a Expand and simplify, as far as possible,

where and are vectors.

b Given that and are perpendicular, deduce that

where

is an integer. [© AQA 2013] 13

a Determine the two values of the integer for which the system of equations

does not have a unique solution. b For the positive value of found in part a, determine whether the system is consistent or inconsistent, and interpret this result geometrically. [© AQA 2012] 14 The linear transformation is represented by the matrix a Find the determinant of

.

in terms of .

b The transformation is the composition of two transformations. The first transformation, , is an enlargement centre and scale factor , and the second is a transformation . When is applied to a solid shape of volume , the volume of the image is . i

Find the value of and hence state the scale factor of the enlargement.

ii Give a full geometrical description of

. [© AQA 2016]

15 The matrix a i

, where is a real constant.

Show that there is a value of for which

where is a rational number to be determined and is the ii Deduce the inverse matrix, b i

identity matrix.

, of for this value of .

Find det in terms of .

ii In the case when is singular, find the integer value of and show that there are no other possible real values of . iii Find the value of for which

is a real eigenvalue of . [© AQA 2012]

16 Let a Show that is a repeated eigenvalue of , and find the other eigenvalue of . b For each eigenvalue of , find a full set of eigenvectors. c The matrix represents the transformation . Describe the geometrical significance of the eigenvectors of in relation to . [© AQA 2011] 17

a Integrate

with respect to .

b Show that for

, the imaginary part of

is

.

c Hence find the exact value of 18

a Show that

.

b Hence find possible complex numbers for which 19

a If

and

.

, show that

.

b Let , where and are scalars. Show that is perpendicular to the line of intersection of the planes and for all values of and . 20 The matrix

is given by

where and are constants and

a Find the eigenvalues of

in terms of and .

b Find corresponding eigenvectors of

.

c Write down a matrix and a diagonal matrix such that d Show that

.

.

.

e Given that

and

and

as

, find matrix . [© AQA 2014]

21 The linear transformation is represented by the matrix a In the case when i

.

is a non-singular matrix:

find the possible values of

ii find

in terms of .

b In the case when , the matrix the volume of the image. c In the case when .

is applied to a solid shape of volume

. Find

, verify that the image of every point under lies in the plane

d Find the value of for which has a line of invariant points and obtain the Cartesian equations of this line. [© AQA 2015] 22

a i

Use De Moivre’s theorem to show that if

ii Write down a similar expression for b i

Expand

ii Hence show that

in terms of .

, then .

.

where and are integers. c

Hence, by means of the substitution

, find the exact value of

. [© AQA 2014]

23 The plane transformation is a rotation through radians anticlockwise about , and maps points onto image points such that

where

and

.

a Write down the inverse of the matrix and

 and hence show that

.

b The curve has equation . The image of under  is the curve with equation i Use the results of part a to show that

.

and express and in terms of and . ii Given that is an acute angle, find the values of and for which this case express the equation of in the form

and hence in

. iii Hence explain why is a hyperbola. [© AQA 2012] 24 The plane transformation has matrix points

and maps points

onto image

such that .

a i

Find

.

ii Hence express each of and in terms of and . b Give a full geometrical description of . c i

Show that the curve with equation

is an ellipse.

ii Deduce that the image of the curve under has equation iii Explain why the curve with equation

.

is an ellipse. [© AQA 2009]

7 Further polar coordinates In this chapter you will learn how to: find the area enclosed by a polar curve find the area enclosed between polar curves.

Before you start… Further Mathematics Student Book 1, Chapter 6

You should be able to sketch polar curves.

1 Sketch the curve with polar equation .

Further Mathematics Student Book 1, Chapter 6

You should be able to convert between Cartesian and polar equations of a curve.

2 Find the polar equation of the curve with Cartesian

You should be able to integrate trigonometric functions.

3 Evaluate:

A Level Mathematics Student Book 2, Chapter 11

equation

.

a b

Integration in polar coordinates In Further Mathematics Student Book 1, Chapter 6, you were introduced to polar coordinates. You saw how to sketch polar curves and establish certain characteristics such as tangents at the pole. In this chapter you will combine your knowledge of polar curves with your knowledge of integration to find areas inside and between polar curves.

Section 1: Area enclosed by a 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 7.1 The area enclosed between a polar curve and the half-lines

and

is

.

θ = β r = f(θ) A

O

θ = α

0, 2π

Focus on… Focus on... Proof 2 extends the idea of integration in polar coordinates to deal with otherwise impossible integrals.

PROOF 4 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 the sum becomes an integral.

WORKED EXAMPLE 7.1 The diagram shows the curve with polar equation

.

r = 3 – 2sin θ O

0, 2π

Find the area enclosed by the curve. Use the formula Expand the brackets. To integrate

, use the

EXERCISE 7A 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

identity:

ii

Focus on… Some areas can only be calculated exactly using polar coordinates. Focus on ... Proof 2 looks at one important example. 2

3

  a    Write down the polar equation of a circle of radius with centre at the pole. 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

.

r = 2 + cos θ

0, 2π

O

Show that the exact area enclosed by the curve is 5

.

The diagram shows the curve with polar equation

.

r = 5 + 2sin θ

O

0, 2π

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

a Sketch the curve with polar equation

, where

, 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 line and the half line

is

Find the value of the positive constant .

.

, the initial

10 The diagram shows the curve with polar equation

.

r = a sin θ sin 2θ

O

0, 2π

Show that the area of the region enclosed by the curve is

.

Section 2: 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 7.2 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

Find the corresponding values of .

.

The points of intersection are

and

. b Sketching the curves: r = 3 + cos θ

O

It is a good idea to sketch the curves to see where the required region is.

r = 7 cos θ

0, 2π

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

.

Common error 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 7B 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 intersects 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

. .

,

π 2

r = a(1 + cos θ) R

0, 2π

O

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. π 2 r = 2a sin 2θ

r = a O

4

0, 2π

The diagram shows the curves with polar equations

and

.

1 r = — 2 0, 2π

O

r = 1 – sin θ

a Find the polar coordinates of the points of intersection of the two curves. b Find the exact area of the shaded region enclosed inside both curves. 5

The diagram shows the curves .

and

with polar equations

and

r = 6 – 6 cos θ

O

0, 2π r = 2 + 2 cos θ

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

.

  a    On the same axes, sketch the curves with polar equations b

Show that the exact value of the area inside

and

but outside

for is

. .

Checklist of learning and understanding 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 7 1

Find the area of the region enclosed between the initial line and the curve with polar equation

.

Choose from these options. A B C D 2

  a    Sketch the curve with polar equation b

3

,

.

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. 4

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 5

.

The diagram shows a sketch of a curve , the pole and the initial line.

0, 2π

O

The polar equation of is

.

Show that the area of the shaded region, bounded by the curve and the initial line, is . [© AQA 2012] 6

The diagram shows the curve with polar equation

.

Find the exact value of the shaded area, clearly showing your working. 7

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 . 8

The diagram shows a sketch of a curve .

The polar equation of the curve is Show that the area of the region bounded by is 9

The diagram shows the curve with polar equation

. .

r = cos θ + cos 3θ

0, 2π

O

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

.

10 The diagram shows a sketch of a curve.

The polar equation of the curve is

.

The point is the point of the curve at which

.

The perpendicular from to the initial line meets the initial line at the point . a i Find the exact value of when ii

.

Show that the polar equation of the line

iii Find the area of triangle b i Using the substitution

is

in the form

. , where is an integer.

, or otherwise, find

, where

ii Find the area of the shaded region bounded by the line Give your answer in the form

, where

and the arc

. of the curve.

and are constants. [© AQA 2013]

8 Further hyperbolic functions In this chapter you will learn how to: define the reciprocal hyperbolic functions

,

and

draw the graphs of reciprocal hyperbolic functions know the domain and range of hyperbolic, inverse hyperbolic and reciprocal hyperbolic functions and work with transformations of their graphs write the inverse reciprocal hyperbolic functions in terms of logarithms use further hyperbolic identities to solve equations differentiate and integrate hyperbolic and reciprocal hyperbolic functions.

Before you start… Further Mathematics Student Book 1, Chapter 5

You should be able to use the definitions of and .

1 Use the definitions of and to show that

You should be able to draw the graphs of and .

2 a On the same axes, sketch the graphs of and .

Further Mathematics Student Book 1, Chapter 5

You should be able to draw the graphs of and .

3 On the same axes, sketch the graphs of and .

Further Mathematics Student Book 1, Chapter 5

You should know how to use the logarithmic form of and .

4 Give exact values for:

Further Mathematics Student Book 1, Chapter 5

.

b Hence state the number of solutions to the equation .

a b

Further Mathematics Student Book 1, Chapter 5

You should know how to use the identity .

.

5 Solve the equation , giving your answers in terms of natural logarithms.

A Level Mathematics Student Book

You should know how to

2, Chapter 2

understand the terms domain and range of a function.

6 For the function , state: a the largest possible domain b the corresponding range.

A Level Mathematics Student Book

You should be able to draw a

2, Chapter 3

graph after two (or more) transformations.

7 The graph of

is

shown. y y = f(x)

O

(0, 0)

(–2, –8)

x

(2, –8)

Sketch the graph of , giving the new coordinates of the three labelled points on the original graph. A Level Mathematics Student Book 2, Chapter 9

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

8 Find: a b

A Level Mathematics Student Book 2, Chapter 9

You should know how to differentiate and integrate trigonometric functions.

9 Find: a b

A Level Mathematics Student Book 2, Chapter 10

You should know how to use the chain rule, product rule and quotient rule for differentiation.

You should know how to integrate using trigonometric identities, the reverse chain

.

10 Find for these functions. a b c

A Level Mathematics Student Book 2, Chapter 11

.

11 Find: a

rule and integration by parts.

b c

More techniques with hyperbolic functions This chapter extends many of the ideas you met in Further Mathematics Student Book 1, Chapter 5, and introduces differentiation and integration of hyperbolic functions. You will already be familiar with many of the ideas and methods in this chapter from those you have used with the corresponding trigonometric functions.

Section 1: Domain and range of hyperbolic and inverse hyperbolic functions In Further Mathematics Student Book 1, Chapter 5, you met the graphs of the hyperbolic functions and

and the graphs of the inverse hyperbolic functions

,

and

.

Rewind You met domain and range in A Level Mathematics Student Book 2, Chapter 2. You now need to know the domain and range of these functions and be able to work with transformations of their graphs.

y

y

1 x

O

y = sinh x

x

O

y = cosh x





y

1

x

O

–1 y = tanh x

Key point 8.1 The domains and ranges of the hyperbolic functions table. Function

Domain

,

and

Range

are given in the

O

O

= arsinh

–1

1

= arcosh



O

1

= artanh



Key point 8.2 The domains and ranges of the inverse hyperbolic functions

and

are

given in the table. Function

Domain

Range

Tip and

are alternative notations for

and

.

WORKED EXAMPLE 8.1 Given that

for

,

a sketch the graph of b state the range of a

.

y

You need to apply two transformations to the graph of : • stretch by scale factor parallel to the -axis

5

• translation by

2

The horizontal asymptote at asymptote at moves to

f(x) = 3 tanh x + 2

O

. stays there, but the .

x

–1

b

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

WORKED EXAMPLE 8.2

and

.

Let

.

a State the largest possible domain of

.

b For the domain in part a, find the range of a Domain:

The domain of

y

b

.

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

(3, 1) O

x

( )

x –1 — f(x) = 1 – cosh 3

• reflection in the -axis • translation by

.

Range:

EXERCISE 8A 1

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

and state the largest possible domain and

a i ii b i ii c

i ii

2

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

and state the largest possible

a i ii b i ii c

i ii

3

The diagram shows the graph of

, where , and are integers.

y

y = a cosh(x + b) – 5

x

O (2, –1) Find the values of and . 4

The diagram shows the graph of

, where , and are integers.

y

y = 3

–0.5

x y = a tanh (2x + b)

y = –3

Find the values of and . 5

A function is given by

, where and are constants.

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

has the largest domain possible.

Section 2: Reciprocal hyperbolic functions The reciprocal hyperbolic functions are defined in the same way as the reciprocal trigonometric functions you met in A Level Mathematics Student Book 2, Chapter 8.

Key point 8.3 The reciprocal hyperbolic functions are:

Key point 8.4 The graph of y (0, 1)

O

x

Domain: Range:

Key point 8.5 The graph of y

O

Domain: Range:

Key point 8.6 The graph of

x

y

y = 1 x

O

Domain: Range:

y = –1

or

Just as for inverse hyperbolic functions, you can express inverse reciprocal hyperbolic functions in terms of natural logarithms. WORKED EXAMPLE 8.3 a Show that

.

b State the domain of

.

a

Use

together with the definition of

Complete the square (or use the quadratic formula).

Solve for .

If

, then since

: Taking the positive root guarantees that for positive . Take natural log of both sides.

.

b

The expression negative. Need

inside the square root must be non-

But the entire expression must be positive as ln is only defined for positive values.

Domain:

EXERCISE 8B 1

For each reciprocal hyperbolic function, sketch the graph of

and state the largest possible

domain and the corresponding range. a i ii b i ii c

i ii

2

a On the same axes, sketch the graphs of

and

.

b Hence state the number of solutions to the equation 3

a On the same axes, sketch the graphs of

. and

.

b Hence state the number of solutions to the equation 4

Prove that

5

a On the same axes, sketch the graphs of b For the function

. and

, state:

i the largest possible domain ii the corresponding range. c Prove that 6

.

a Prove that for . b Hence prove that

.

for all

.

.

Section 3: Using hyperbolic identities to solve equations In Further Mathematics Student Book 1, Chapter 5, you met the identities

and

. You now need to be familiar with some other identities and use these to solve hyperbolic equations.

Key point 8.7

These will be given in your formula book.

WORKED EXAMPLE 8.4 Prove that

. Use the definitions of

and

.

Expand and simplify.

Key point 8.8

WORKED EXAMPLE 8.5 a Prove that b Solve the equation are rational numbers.

a

. , giving your answer in the form

It is easiest to start with definition of . Create a common denominator.

where and

. Use the

b

Use

to get a quadratic in .

Factorise and solve.

Then use

EXERCISE 8C

.

EXERCISE 8C 1

Prove each identity using the definitions of

, and

a b c 2

Solve the equation

3

Solve the equation

4

Solve the equation

5

Solve the equation

6

Solve the equation

7

Solve the equation

, giving your answers to three significant figures. , giving your answers to three significant figures. , giving your answers in the form

.

, giving exact answers. , giving your answer in the form

.

, giving your answer in the form

, where is a rational

number. 8

Solve the equation rational numbers.

, giving your answer in the form

9

Solve the equation

, giving your answers in the form

10 Prove that 11

a Prove that

. , giving your answer in the form

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

.

.

b Hence solve the equation 12

, where and are

.

.

Section 4: Differentiation Key point 8.9

Only the final one of these will be given in your formula book. You can derive the results for the derivatives of functions.

and

by returning to the definitions of these

WORKED EXAMPLE 8.6 Show that

. Use the definition of

.

Differentiating:

You can show the result for

either from the definition again or by using

and the

quotient rule. WORKED EXAMPLE 8.7 Use the derivatives of

and

to show that

.

Use

Differentiating using the quotient rule:

Use

WORKED EXAMPLE 8.8 Given that Let

and

, find

. Use the product rule.

Then And

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

Now apply the product rule formula.

Key point 8.10

You can show each of the results in Key point 8.10 in a similar way. WORKED EXAMPLE 8.9 Prove that Express

in terms of

.

Differentiating, using the chain rule:

Rearrange into the form required.

WORKED EXAMPLE 8.10 Given that

, find

. Make sure you are thinking of

as

. Using the chain rule:

Remember to multiply by the derivative of as well.

EXERCISE 8D

EXERCISE 8D 1

Differentiate each function with respect to a i ii b i ii c

i ii

d i ii e i ii f

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

. .

at

is

. 6

Find the equation of the normal to the curve the form

7

, giving your answer in

.

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 value of .

10

at

.

. . have -coordinates

, stating the

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 5: Integration Key point 8.11

Only the final one of these will be given in your formula book. As with differentiation, you can derive the results for the integrals of

and

by returning to the

definitions of these functions. WORKED EXAMPLE 8.11 Show that

. Use the definition of

You can show the result for the integral of

.

either from the definition again or by using

and applying the reverse chain rule. WORKED EXAMPLE 8.12 Show that

. Use

.

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.

Rewind See A Level Mathematics Student Book 2, Chapter 11, for a reminder about integrating trigonometric functions, using the reverse chain rule, trigonometric identities and integration by parts. In Section 4, you differentiated and . You can now reverse these standard derivatives and add them to the list of functions you can integrate:

Often you will need to use a hyperbolic identity before integrating.

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

WORKED EXAMPLE 8.13 Find

. Expand the integrand. Use

Both terms are now standard integrals.

WORKED EXAMPLE 8.14 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 8.15 Find Use the definition of

Rewind

.

If the integral in Worked example 8.15 had been

, you would have done integration

by parts twice and rearranged.

EXERCISE 8E 1

Find: a i ii b i ii c

i ii

2

.

Find: a i ii b i ii c

i ii

d i ii 3

.

Use an appropriate hyperbolic identity to find each integral. a i ii b i ii c

i ii

4

Use a substitution or the reverse chain rule to find each integral.

a i ii b i ii c

i ii

5

Use integration by parts to find each integral. a i ii b i ii c

i ii

d i ii 6

Use the definitions of

and/or

to find each integral.

a i ii b i ii 7

Find

8

By expressing

9

a Find

. and

b Hence, or otherwise, find

in terms of , evaluate

.

. .

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

, the -axis and the line

y

y = sinh x

R x = ln 3 O

x

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

radians about the

.

11 Find: a b

.

12 Find

.

13 The diagram shows the region bounded by curve

, the -axis and the line

.

Find the exact volume of the solid formed when the region is rotated through -axis. 14

a Using the substitution

, show that

b Hence find

.

15 Using the substitution

16 Show that

, show that

.

.

radians about the

Checklist of learning and understanding Domain and range of hyperbolic and inverse hyperbolic functions: Function

Domain

Range

Definitions of reciprocal hyperbolic functions:

Graphs of reciprocal hyperbolic functions:



Domain and range of reciprocal hyperbolic functions: Function

Domain

Range



Identities:

Derivatives of hyperbolic, reciprocal hyperbolic and inverse hyperbolic functions:

Integrals of hyperbolic functions:

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

Mixed practice 8 1

Given

, find

.

Choose from these options. A B C D 2

Given that

,

a sketch the graph of b state the range of 3

Solve the equation

4

Show that the curve

5

Given

6

Given that

7

Find

8

Show that

9

Solve the equation rational number.

. , giving your answer in the form has no points of inflection.

, show that

. , show that

.

.

. , giving your answer in the form

10 Solve the equation 11 Given that

, show that

, find

. .

13 The diagram shows the graphs of

and

y

R y = sinh x + 7

O

where is a

, giving your answers in terms of natural logarithms. , where

12 Using the substitution

y = 5 cosh x

.

x

.

Find the exact value of the area of the shaded region. 14

a Sketch the graph of b Given that

. , use the definitions of

and

in terms of and

to show

that . c

i Show that the equation

can be written as . ii Show that the equation

has only one solution for . Find this solution in the form

, where is an integer. [©AQA 2009]

15

a i Sketch the graphs of

and

.

ii Use your graphs to explain why the equation

where is a constant, has exactly one solution. b A curve has equation . Show that has only one stationary point and show that its -coordinate is an integer. [©AQA 2013] 16

a Prove that the curve

has exactly one stationary point. b Given that the coordinates of this stationary point are

, show that

. [©AQA 2011]

17 Prove that 18

.

a Using the definition of

, show that

b Hence solve the equation 19

a Given that

. , giving your answer in the form

, use the definition of

in terms of and

.

to show that

. b A curve has equation

.

i Show that the curve has a single stationary point and find its -coordinate, giving your

answer in the form

, where is a rational number.

ii The curve lies entirely above the -axis. The region bounded by the curve, the coordinate axes and the line has area . Show that

where , and are integers. [©AQA 2016]

9 Further Calculus 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 substitutions to find similar integrals integrate, using partial fractions with a quadratic expression in the denominator derive and use reduction formulae for integrals find the length of an arc and the area of a surface of revolution.

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

A Level Mathematics Student Book 2, Chapter 11

You should know how to differentiate functions defined implicitly.

You should be able to integrate by using a substitution.

1 Given that , find an expression for

.

2 Use a suitable substitution to evaluate .

A Level Mathematics Student Book 2, Chapter 11

You should be able to use integration by parts.

A Level Mathematics Student Book 2, Chapter 11

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

3 Find

.

4 Find

Some new integration techniques and applications In this chapter you will extend the range of integration methods you can use and the range of functions you can integrate. Differentiation of inverse trigonometric functions leads to the 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

integrate many rational functions.

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

.

You will also see how to extend the idea of repeated integration by parts to derive reduction formulae. These are recursive formulae that allow you to calculate more complex integrals from simpler ones. So far you have used integration to find areas bounded by plane curves and volumes of revolution. You will now learn about formulae to calculate the length of an arc of a curve and the area of a surface of revolution.

Section 1: Differentiation of inverse trigonometric functions Rewind You met implicit differentiation in A Level Mathematics Student Book 2, Chapter 10. You already know how to differentiate

,

and

.

To differentiate their inverse functions you can use implicit differentiation. WORKED EXAMPLE 9.1 Given that

, and that

, find

in terms of .

You know how to differentiate sin, 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 .

y π — (1, ) 2 y = sin–1x x

O

π — (–1, – ) 2

You should notice two important details in the derivation of the derivative of shown in Worked example 9.1. 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 cos and tan functions similarly.

Key point 9.1

function is

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

function is defined for all

, so there is no restriction on the domain of its

derivative. You can combine these results with other rules of differentiation.

Tip Remember that

is alternative notation for

.

WORKED EXAMPLE 9.2 Differentiate: a b

and state the values for which the derivative is valid.

a Using the product rule:

Multiply by (the derivative of rule.

Use b Using the chain rule:

The derivative is valid when

EXERCISE 9A

Multiply by

The derivative of

) due to the chain

.

(the derivative of

is only defined for

).

.

EXERCISE 9A 1

Find

for each function.

a 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

.

, simplifying your answer as far as possible. , stating the range of values of for which your answer is valid. , show that

b Hence differentiate 7

.

Given that

.

with respect to . , show that

and state the values of for which the

derivative is valid. 8

Given that

9

a Find

, show that

, where and are integers to be found.

.

b Hence find 10 Show that the graph of

. has no points of inflection.

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

Rewind See Chapter 8, Section 4, for differentiation of hyperbolic functions.

WORKED EXAMPLE 9.3 Given that

, find

in terms of Rewrite in terms of

.

Differentiating with respect to :

, so take the reciprocal of both sides.

Use

.

You can establish the results for the inverse cosh and tanh functions in a similar way.

Key point 9.2

These will be given in your formula book.

WORKED EXAMPLE 9.4 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 square root.

EXERCISE 9B

and solve for .

is

, so take the positive

EXERCISE 9B 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

. and

to the curve

intersect at the point .

Show that the -coordinate of is

.

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 constant to be found.

10

satisfies

a Prove that

.

. at which the gradient is

is

.

b The tangent to the curve curve at

, giving your

at

crosses the -axis at the point . The normal to the

crosses the -axis at the point .

Find the distance

, giving your answer in the form

, where is a rational number.

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:

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

Tip Notice that the results

and

as the first is just the negative of

are not useful

and the second can be done by partial fractions.

Key point 9.3

These will be given in your formula book. You also need to know how to derive these results, using trigonometric or hyperbolic substitutions.

Focus on … See Focus on … Problem solving 2 for an example of using one of these integrals.

WORKED EXAMPLE 9.5 Use the substitution

to prove the result

when

Differentiate the substitution and express

. in terms of

Express the integrand 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 9C.

WORKED EXAMPLE 9.6 Use the substitution

to prove the result

.

Differentiate the substitution and express terms of .

in

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 .

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 in Exercise 9C, Question 9.

WORKED EXAMPLE 9.7

and

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 9.8 Find

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

Hence

The integrand is of the form and

. Remember to divide by the

coefficient of when integrating.

EXERCISE 9C 1

Find each indefinite integral. a i ii b i ii c

i ii

d i ii e i ii f

i

with

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

.

Showing all your working clearly, find the exact value of

5

Show that

6

Show that

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

.

, giving your answer in terms of a natural logarithm.

a Write

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 22

Use a suitable hyperbolic substitution to show that a Given that

, express

and

in terms of .

b Use a suitable trigonometric substitution to show that .

.

Section 4: Using partial fractions in integration Rewind You met partial fractions in A Level Mathematics Student Book 2, Chapter 5, and then used them in integration in Chapter 11. 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 . 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 9.4 If

is a polynomial of order less than or equal to , then .

WORKED EXAMPLE 9.9 a Express

in partial fractions.

b Hence find

.

Use the result from Key point 9.4.

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 factor is irreducible, you need to write it in completed square form before you can apply standard integration results. WORKED EXAMPLE 9.10 Given that The denominator is

, find an expression for in terms of . You need to split the function into partial fractions before integrating. 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 9D

EXERCISE 9D 1

Use partial fractions to find each integral. a i ii b i ii c

i ii

d i ii 2

Showing all your working clearly, 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

.

.

Showing all your working clearly, use partial fractions to evaluate 7

Use partial fractions to find

8 Show that where

and are constants to be found.

.

.

Section 5: Reduction formulae In A Level Mathematics Student Book 2, Chapter 11, you saw examples where you had to use integration by parts more than once. For example:

If you now wanted to find

, you would need to start by using integration by parts again:

However, rather than continuing with integration by parts, you can now use the answer you have already found:

You could now use this answer to find

, and so on.

For any positive integer , if you write

, then:

 

The last equation, which relates to

, is an example of a reduction formula. Since you know that

, you can use this formula to find for any

.

Rewind This process of building up from to any should remind you of proof by induction, which you met in Further Mathematics Student Book 1, Chapter 12.

WORKED EXAMPLE 9.11 Let a Derive the reduction formula b Hence find an expression for

. .

a Using integration by parts once:

Take the constants out of the integral and

relate to

Start with and use the reduction formula until you get to .

b

Sometimes you need to use integration by parts more than once to get a reduction formula. Also, the formula does not need to relate to

. In Worked example 9.12, the formula relates to

. This

example also illustrates how you can apply reduction formulae to evaluate definite integrals. WORKED EXAMPLE 9.12 Let a Express in terms of

.

b Hence find the exact values of and .

a

Use integration by parts with , so that

and

.

The integral is not directly related to use integration by parts again with and

so

so

.

The remaining integral is now a multiple of . b

To get you need to start from .

To get to you need to start from , which you need to find using integration by parts.

Reduction formulae do not always come from integration by parts. Another common way of deriving them is by using trigonometric identities. WORKED EXAMPLE 9.13 a Show that

.

b Hence evaluate

.

a

The identity relating

b Let

.

Then

You can find chain rule, as

and

is

.

using the reverse is the derivative of

Since has been expressed in terms of need to start from to build up to .

In more complicated examples, after applying integration by parts, both right-hand side of the equations.

.

, you

and could appear on the

WORKED EXAMPLE 9.14 Find a reduction formula for

. You can use integration by parts with because this can be

integrated to give

.

The integral on the right is not , but you can make an expression with in the denominator by writing . The expression on the right contains both and . Add to both sides.

EXERCISE 9E 1

Use integration by parts to derive each reduction formula. a i ii

, where , where

b i

, where

ii c

, where

i

, where

ii d i

, where

ii

, where

e i ii 2

, where

, where , where

Use the reduction formulae from Question 1 to find the exact value of each integral. a i ii b i ii c

i ii

d i

ii e i ii

3

Given that

:

a show that b find the exact value of . 4

a Find a reduction formula for b Hence evaluate

5

. .

a Use integration by parts to find b Given that

.

, show that

.

c Hence find the exact value of 6

Let

.

.

a Derive the reduction formula b Hence find 7

Let

.

in the form

, where

.

a Show that

.

b Hence find the exact value of 8

is a polynomial.

Given that

.

, where

:

a show that b find c use induction to show that 9

Let

.

a By writing

as

.

and integrating by parts, show that .

b Hence find the exact value of

10

a Given that

.

, show that

b Hence find the exact value of .

.

11 Let

.

a Show that

.

b Hence show that 12 Let

.

.

a Show that

.

b Hence find the exact value of c Prove by induction that

. .

13 Use a reduction formula to find the exact value of

.

Section 6: Length of an arc You have already met the idea of a definite integral as a limit of a sum and using this to find the area under a curve. You can adapt this idea to calculate the length of an arc of a curve.

Rewind See A Level Mathematics Student Book 1, Chapter 15, for a reminder of definite integration as the limit of the sum of areas of rectangles. Consider a small section of a 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

and so

Integrating this expression between two -values gives the length of the arc of the curve between two points.

(x + Δx, y + Δy)

Δs

Δy

(x, y) Δx

Key point 9.5 The length of an arc of a curve between the point

and

is

. This will be given in your formula book.

Focus on … The formula for the length of a curve is used in Focus on … Problem solving 2 to find the equation of a hanging chain.

WORKED EXAMPLE 9.15 Find the length of the arc of the curve with equation .

between the points

and

Use

.

Use partial fractions. This is an improper fraction because the highest power in both the numerator and the denominator is .

You can also use the formula for the length of an arc when the equation of the curve is given parametrically, using

.

Key point 9.6 The length of an arc of a curve between points with parameter values and is given by

This will be given in your formula book.

WORKED EXAMPLE 9.16 A circle with radius has parametric equations the circle between the points with

and

Use

,

. Find the length of the arc of

.

.

EXERCISE 9F 1

Find and simplify an integral expression for the length of each arc and use technology to evaluate the integral. a i

from

to

.

ii

from

to

.

b i

c

between

ii

between

i

from

.

and to

. .

ii

from

to

.

d i

from

to

.

ii 2

and

from

to

.

Without using a calculator, find the exact value of the length of each arc. a i

between

ii

and

between

and

b i

between

ii

between

c

i

. .

and and

between

ii

.

and

between

d i

. .

and

between

. and

between 3

.

and

Find the length of the graph of the curve with equation

 between the points

and

. 4

Find the length of the arc of

5

A curve has parametric equations

6

between

and

.

,

. Find the length of the arc of the

curve between the points with parameter values

and

.

A curve has parametric equations

. The arc of the curve between

and

has length . a Show that

.

b Hence find the length of the arc. 7

A curve has parametric equations

for

. Find the length of the

curve. 8

A curve has parametric equations between and . a Show that

. is the length of the arc of the curve

.

b The Cartesian equation of the curve is and 9

. By considering the values of corresponding to

, find the exact length of the arc of the curve

Consider the curve with parametric equations

between

and

.

a Find the length of the arc of the curve between the points with

and

.

b Find the Cartesian equation of the curve. Sketch the portion of the curve for 10 A curve is defined parametrically by the points with and is . a Show that

.

.

. The length of the arc of the curve between

, stating the value of the constant .

b Find the exact value of . 11 A curve is defined by with parameter and is a Show that b Show that

. The length of the arc of the curve between the points .

. .

12 Let denote the length of the arc of the curve a Show that b

between

and

.

.

Use the substitution

to show that

and state the value of the

integer . 13 A curve has equation curve between

and

. Use the substitution is

to show that the length of the arc of the

where and are constants to be found.

Section 7: Area of a surface of revolution In Further Mathematics Student Book 1, Chapter 10, you learned how to find volumes of solids of revolution, which are formed by rotating a region about the - or the -axis. It is also possible to calculate their surface area. You only need to be able to use the formulae for surfaces generated by rotation about the -axis, and there is both a Cartesian and a parametric version.

Key point 9.7 The area of a surface of revolution rotated about the -axis:

This will be given in your formula book. Note that the formula in Key point 9.7 gives the curved surface area, not including the base(s) of the resulting solid.  WORKED EXAMPLE 9.17 Find the curved surface area of the cone with base radius 5 and height 12.

y

The curved surface area of the cone is generated by rotating the line segment about the -axis. You can see from the diagram that the gradient of this line segment is

5

equation is

O

12

so its

.

x

Use

The -values go from to

You might recognise the formula area of the cone. In this case,

for the curved surface and

WORKED EXAMPLE 9.18 A semi-circle with radius is given by parametric equations for . The semi-circle is rotated radians about the -axis. Show that the surface area of the resulting sphere is . Use

.

.

.

EXERCISE 9G

EXERCISE 9G 1

The given curve is rotated radians about the -axis. Write down and simplify the integral to evaluate the surface of revolution. Use technology to evaluate the integral. a i ii

from from

to

.

to

.

b i

from

to

.

ii

from

to

.

c

i ii

d i ii

from

to

.

from

to

from

to

from

. .

to

.

2

The arc of the curve between and is rotated Find the exact value of the area of the surface of revolution.

3

The arc of the curve between and exact area of the resulting surface of revolution.

4

Find the area of the surface of revolution formed when the arc of the curve

is rotated

radians about the -axis.

radians about the -axis. Find the

for

is

rotated through a full turn about the -axis. 5

A frustum is formed by rotating the line segment connecting points Find the surface area of the frustum, including the two bases.

6

A curve is defined parametrically by .

The arc of the curve between

and

is rotated through a full turn about the -axis.

a Show that the exact area of the resulting surface of revolution is to be found.

where is a rational number

An ellipse has parametric equations . The arc of the ellipse between and is rotated fully about the -axis. Let be the area of the resulting surface of revolution. a Show that b

8

about the -axis.

.

a Show that

7

and

.

Use a suitable trigonometric substitution to show that

The arc of the curve

between

and

is rotated

. radians about the -axis. The

resulting surface of revolution has area . a Show that

.

b Use an appropriate hyperbolic substitution to find the exact value of .

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

.

A reduction formula relates an integral to a similar but simpler one, by reducing the power of one of the expressions. You can derive a reduction formula by using integration by parts or a trigonometric or algebraic identity. Length of an arc of a curve in Cartesian and parametric form:

Surface area when a curve is rotated about the -axis:

Mixed Practice 9 1

Differentiate

.

Choose from these options. A B C D 2

Find

3

Differentiate

4

Find the -coordinates of the points on the curve

5

A curve is defined parametrically by

when

. .

a Show that

where the gradient is .

.

b The arc of from

to

is rotated through

radians about the -axis.

Find the area of the surface generated, giving your answer in the form where and are integers.

,

[© AQA 2015] 6

Given that

, find

7

Show that

satisfies

8

Find

9

An ellipse has parametric equations

. .

. .

a Show that the perimeter of the ellipse is given by . b Show that this perimeter is equal to the length of one complete wave of the curve . 10 Find 11

.

Using integration by parts, or otherwise, show that

.

12 The function , where

, has domain

and has inverse function

, where

. a Show that

.

b Hence show that

. [© AQA 2012]

13 A curve is defined parametrically by The arc length between the points where a Show that b Hence show that

and

on the curve is .

, stating the value of the constant . . [© AQA 2013]

14

a Show that b Hence find

15

a Express

.

. in partial fractions.

b Hence find

.

c Show that 16

a

.

Using an appropriate substitution, show that

.

b Hence show that 17

a Show that b

18

.

Hence show that the area enclosed by the ellipse with equation

a Show that b Let

is

.

. . Show that

c Hence show that 19 A curve has Cartesian equation

for all

.

. .

a Show that b The points and on the curve have -coordinates and respectively. Find the arc length , giving your answer in the form , where and are rational numbers.

[© AQA 2012] 20

between the points where and is rotated a The arc of the curve through radians about the -axis. Show that the area of the curved surface formed is given by . b By means of the substitution

, show that . [© AQA 2011]

10 Maclaurin series and limits In this chapter you will learn how to: find Maclaurin series for functions without using any standard results use Maclaurin series to approximate particular values of functions or definite integrals find limits of functions in certain circumstances, using Maclaurin series or using an alternative method called l’Hôpital’s rule find the value of definite integrals in certain cases where a limiting process is required (improper integrals).

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 8

You should know how to differentiate hyperbolic functions.

3 For find: a b

Chapter 8

You should know how to differentiate inverse hyperbolic functions.

.

4 For , find: a b

Chapter 9

You should know how to differentiate

5 For

.

inverse trigonometric functions. find: a b

.

Analysis of functions In Further Mathematics Student Book 1, Chapter 11, you were given the Maclaurin series for some common functions. In the first part of this chapter you will see where these results come from and use this method to find Maclaurin series for other functions. In the second part of the chapter you will use both Maclaurin series and an alternative rule for evaluating some awkward limits of functions. Finally, you will look at definite integrals where either the integrand is not defined at point(s) in the range of integration, or where the range of integration extends to infinity. These are known as improper integrals.

Did you know? Analysis is a branch of mathematics that deals with infinite series, limits and calculus among other ideas. It forms a significant component of most mathematics degrees.

Section 1: Maclaurin series You know from Further Mathematics Student Book 1, Chapter 11, that many functions can be written as infinite series. In general, for some function such a series will be of the form where

etc. are real constants.

Differentiating this series several times:

Did you know? 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.

Rewind You met the interval of convergence in Further Mathematics Student Book 1, Chapter 11. 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

Key point 10.1 The Maclaurin series of a function

is given by:

You can use Key point 10.1 to find the first few terms of the Maclaurin series. You can sometimes spot a pattern in the derivatives, which enables you also to write down the general term,

.

Tip Different but equivalent forms of the general term can be given, depending on whether starts from or .

WORKED EXAMPLE 10.1 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

b The general term is

for

Substitute these values into the Maclaurin series formula:

ensures that the sign alternates. only odd terms are present.

ensures that

Note that you could also give the general term as, for example,

for

Not every function has a Maclaurin series. For example, for , doesn’t exist (nor do any of the derivatives of at ). However, does have a Maclaurin series, as now and all the derivatives at do exist. WORKED EXAMPLE 10.2 a Find the Maclaurin series of

up to and including the term in

b Prove by induction that the general term is

a

.

, for

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 the th derivative:

Now suppose that the expression is correct for some and show that it is still correct for .

Then the

st derivative is:

Differentiate

to get

.

Hence the expression is correct for the st derivative. The expression is correct for

and, if

it is correct for then it is also correct for . It is therefore true for all , by induction.

Write a conclusion.

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.

Rewind In Further Mathematics Student Book 1, Chapter 11, you saw how to use the Maclaurin series of certain standard functions to find the series of more complicated functions.

WORKED EXAMPLE 10.3 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

, giving your answer to three decimal places.

Substitute

a

into the series for

.

Integrate the polynomial as usual.

b

EXERCISE 10A 1

Using Key point 10.1, find the first four non-zero terms of the Maclaurin series of these functions. Also conjecture the general term (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

By differentiating an appropriate number of times, find the Maclaurin series of and including the term in

3

4

.

a Show that the first two non-zero terms in the Maclaurin series of

are

b Hence find, correct to three decimal places, an approximation to

.

It is given that

up to

.

.

a 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

up to and including the term in

,

a i Show that

.

.

and

ii Find the third and fourth derivatives of b Hence find the Maclauin series of c

.

. up to and including the term in

Use your series from part b to find an approximation to

.

, giving your answer

correct to three decimal places. 7

A function is defined as

,

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

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

.

. Give your answer in the form ,

where and are integers. 8

term is

.

10

a Find the first three non-zero terms in the Maclaurin series for b i Let and State the relationship between of , for any integer . ii Hence show that

.

. , the th derivative of

, and

, the th derivative

, where is a constant to be determined.

11 The diagram shows part of the graph of

.

y

1 O

y = f(x) x

Explain why neither of these can be the Maclaurin series of the function a b

:

Section 2: Limits You are already familiar with the idea of taking limits, from the process of differentiation from first principles:

.

In that case you took the limit as

but found that, to avoid getting , you first needed to cancel

throughout the expression. You can apply a similar idea to any rational function where the limit is of the form

.

Rewind You met differentiation from first principles in A Level Mathematics Student Book 1, Chapter 12.

WORKED EXAMPLE 10.4 Find

. Just allowing

straight away would give

, so instead

start by dividing through the expression by the highest power of , which is . Simplify and then let

As

,

.

and

Using Maclaurin series to find limits If you don’t have a rational function, then you can’t directly use the method in Worked example 10.4 when faced with limits such as and

. However, if you can find the Maclaurin series of any non-polynomial

elements of the function, then you can use the method in Worked example 10.4 and again divide through by the highest power of . WORKED EXAMPLE 10.5 Using the Maclaurin series for

and

, evaluate

.

Use

Simplify and then cancel expression.

and

throughout the

Now let

.

You need to be able to recognise two important limits that often occur.

Key point 10.2

,

Fast forward You will use both of these limits in Section 3. You can prove the first result in Key point 10.2 using Maclaurin series.

Fast forward The second result in Key point 10.2 is proved in Worked example 10.9 later in this section.

WORKED EXAMPLE 10.6 Prove that

. Use the Maclaurin series for .

Divide through by . All the terms in the denominator from onwards will tend to , so the fraction tends to .

L’Hôpital’s rule An alternative to using Macaurin series to evaluate limits of the form or rule.

Key point 10.3

is given by l’Hôpital’s

L’Hôpital’s rule Given functions

and

such that either or

,

then

provided that

exists.

Common error It is important not to try applying l’Hôpital’s rule to a quotient that is not of the form or

. It

is only valid in these cases.

WORKED EXAMPLE 10.7 Find

. Check the limits of the numerator and denominator of the function.

and

Using l’Hôpital’s rule: Since the limit of the function is of the form , use l’Hôpital’s rule.

This limit exists, so by l’Hôpital’s rule this is the limit of the original function.

Sometimes it may be necessary to use l’Hôpital’s rule more than once. WORKED EXAMPLE 10.8 Evaluate

.

and

Check the limits of the numerator and denominator of the function.

Using l’Hôpital’s rule: Since the limit of the function is of the form , use l’Hôpital’s rule.

Since

and

, the limit of the rational

function is again of the form ... Using l’Hôpital’s rule again: ...so use l’Hôpital’s rule again.

WORK IT OUT 10.1 Find

.

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

Solution 2 Using l’Hôpital’s rule:

Solution 3 Using l’Hôpital’s rule twice:

You can also use l’Hôpital’s rule to find limits of the form these expressions into a quotient that is of the form or

or .

You can use this idea to prove the second result in Key point 10.2. WORKED EXAMPLE 10.9 Prove that

, for

.

. First you need to rearrange

This is a limit of the form becomes of the form

, but rewriting

as

, the limit

.

Using l’Hôpital’s rule: Apply l’Hôpital’s rule: and Tidy up the expression and then let

.

EXERCISE 10B 1

Find the limit of each function, where the limit exists. a i   ii b i   ii c

i   ii

d i   ii 2

Use standard results to form the Maclaurin series of each function, and use this to find the limit in each case. a b c d e f

3

Use l’Hôpital’s rule to find each limit. a

b c d

e f 4

a Write down the expansion of

in ascending powers of up to and including the term in

b Hence evaluate 5

.

.

In each case find the limit if it exists. a b c

6

Use l’Hôpital’s rule to find

7

Find

8

Show that

9

Show that

10

.

. .

.

a By using the Maclaurin series for series expansion of

.

b Hence find 11 Find 12

, or otherwise, find the first three non-zero terms of the

. .

a State the function for which

.

b By integrating an appropriate series, find the Maclaurin series of term in

.

c Hence evaluate d Use your series for

. from part b to estimate .

up to and including the

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 10.4 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 10.10 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

,

. Therefore the integral converges.

WORKED EXAMPLE 10.11 Explain why the improper integral

diverges. Integrate as normal, but replace the upper limit with and

consider the limit as integration.

When

,

after you have completed the

tends to

infinity. Therefore the integral diverges. When evaluating improper integrals, you might need to use some of the limits you met in Section 2. WORKED EXAMPLE 10.12 Evaluate

, showing clearly the limiting process used. Integrate as normal, but replace the upper limit with and consider the limit as completed the integration.

after you have

Use integration by parts.

Now take the limit as . Use the result from Key point 10.2:

, 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 some point(s) 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 10.5 If

is not defined at

, then

and . If the limit is not finite, then the improper integral diverges (does not have a value). WORKED EXAMPLE 10.13

Evaluate

, showing clearly the limiting process used. 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 ln are an exception where you take .

Now take the limit as . Use the result from Key point 10.2: , with .

If the point where the integrand is not defined is not an end point, you need to split the integral into two parts.

Key point 10.6 If

is undefined at

, then .

If either limit is not finite, then the improper integral diverges (does not have a value).

WORKED EXAMPLE 10.14 Find the exact value of

. The integrand is not defined at need to split the integral in two.

, so you

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 10C 1

Determine which of these improper integrals converge. Evaluate those 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. . .

showing the limiting process used and giving your answer in the form

, where is a

constant. [©AQA 2013]

Checklist of learning and understanding The Maclaurin series for a function

is given by

Two common applications of Maclaurin series are: to find limits to approximate integrals.

L’Hôpital’s rule can often be used to find limits of the form or Given functions

and

:

such that either

then

provided that

exists.

Two useful limits that arise often are: , where , where

.

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

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

if the limits exist and are finite.

, then

is

Mixed practice 10 1

Which is an example of an improper integral? Choose from these options. A

B

C

D 2

a

Show that the first four terms in the Maclaurin series of are

b Use the series in part a to show that 3

Use the Maclaurin series for

4

a Find the first four derivatives of

.

.

and

to find .

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 5

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. [© AQA 2015] 6

a Find the Maclaurin series of

up to and including the term in

b Hence show that 7

Find

8

It is given that

.

, clearly showing your working. ,

a Find the first three derivatives of b Hence find the Maclaurin series of c Hence find 9

.

. . up to and including the term in

.

.

a Write down the expansion of in ascending powers of up to and including the term in . Give your answer in its simplest form. b i Given that

, find

, and

.

(You may leave your expression for

unsimplified)

ii Hence, by using Maclaurin’s theorem, show that the first three non-zero terms in the expansion, in ascending powers of , of are . c Find . [© AQA 2010] 10 Find

.

11 Using l’Hôpital’s rule, show that 12

.

a Using the series for , or otherwise, find the Maclaurin series of non-zero terms and the general term. b Hence find a series expansion of

.

c Hence show that 13

.

a It is given that i Show that ii Find

, stating the first four

. .

.

b Hence use Maclaurin’s theorem to show that the first three non-zero terms in the expansion, in ascending powers of , of c Write down the expansion of to and including the term in

are

.

, where is a constant, in ascending powers of up .

d i Find the value of for which

exists.

ii Hence find the value of

when takes the value found in part d i. [© AQA 2013]

14

a It is given that i Show that ii Find

. .

.

b i Hence use Maclaurin’s theorem to show that the first three non-zero terms in the expansion, in ascending powers of , of

are

.

ii Write down the first three non-zero terms in the expansion, in ascending powers of , of .

c Hence find the first three non-zero terms in the expansion, in ascending powers of , of . [© AQA 2014]

11 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.

A Level Mathematics Student Book 2, Chapter 10

You should know how to differentiate expressions, including using the product and chain rules.

A Level Mathematics Student Book 2, Chapter 11

You should be able to integrate complicated expressions.

1

Solve

2 Differentiate with respect to .

3 Integrate with respect to .

Rates of change In many academic areas such as physics and economics it is easier 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 12 how these methods can be applied in many reallife situations.

Rewind You met differential equations in A Level Mathematics Student Book 2, Chapter 13.

Section 1: Terminology of differential equations Differential equations usually 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 independent variable is and the dependent variable is .

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. 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 ( in these examples) is differentiated. For example:

is a third order differential

equation. A linear differential equation is one in which the dependent variable (or any of its derivatives) only appears to the power of or in any term. For example:

is a linear

differential equation, but any differential equation involving

or even a

is non-linear.

A homogeneous differential equation is one in which 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. 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 11.1 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. Any solution to a differential equation is called a particular integral. The solution to the associated homogeneous differential equation is called the complementary function – and this is usually where the

arbitrary constants are found. If the differential equation is linear, these are combined to give the general solution of the differential equation.

Key point 11.2 For a linear differential equation, the general solution is given by

where is the complementary function and is the particular integral. is a differential equation where is called a linear differential operator. For example: if , then is

and

In Proof , you can use the fact that for a linear differential operator

.

PROOF 5 If is the complementary function and is the particular integral, then

will be a

solution to the differential equation. The complementary function, , is defined by:

The particular integral is any function, , satisfying

Then if Use the given fact about linear differential operators. Use properties (1) and (2). So also satisfies

.

WORKED EXAMPLE 11.1 The differential equation

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 a The associated homogenous differential equation is

when

.

The complementary function is the solution of the associated homogeneous equation. The differential equation is separable.

b

If

, then

Substituting:

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

Use Key point 11.1. The general solution is the sum of the complementary function and the particular integral.

Using the initial condition, substituting in

when

:

Therefore the particular solution is

EXERCISE 11A 1

Write an example of each 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. 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 i

;

when

ii

;

b i

;

ii c

i

when when ;

;

ii d i

when when ;

;

ii 4

when

when

and ;

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

.

a Find the values of the constants and for which integral of the differential equation

b Hence find the solution of this differential equation, given that

is a particular

when

. [©AQA 2014]

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 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, 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

, you have the left-hand side in the required

form. From here you can proceed exactly as before:

The only remaining question is to decide on the function You need

This function

is known as the integrating factor.

to make this work.

Key point 11.3 Given a first order linear differential equation

multiply through by the integrating factor:

The solution will be:

Tip When calculating

you do not actually need to include the ‘

’. It turns out that it does

not actually matter whether there is a constant as it would cancel later in the process.

WORKED EXAMPLE 11.2 Solve the differential equation for given that

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 not to miss the sign on

So

Now multiply through by LHS is of the form

making sure .

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 11.3 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 in turns the given differential equation into:

which is a linear differential equation. The integrating factor is

.

Use Key point 11.3. Use Key point 11.3 again.

Therefore

Notice that the Therefore

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 11B 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

Find the particular solution of the linear differential equation

for which

. 3

Find the general solution to the differential equation

.

when

4 5

Find the general solution of the differential equation

.

Find the particular solution of the linear differential equation through the point

that passes

.

6

Given that

7

Prove that when finding

and that

when

, find in terms of .

in Key point 11.3, it does not matter whether or not the

constant of integration is included. 8

9

Find the general solution to the differential equation 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

when

.

, or otherwise, solve the equation:

given that when

. Give your answer in the form

.

b Use another substitution to find the general solution to the equation . 11 A differential equation is given by . a Show that the substitution

transforms this differential equation into . b Hence find the general solution of the differential equation

giving your answer in the form

. [©AQA 2011]

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 11.4 The auxiliary equation to

is

Solving the auxiliary equation gives you important information about the solution of the differential equation, as set out in Proofs 6, 7 and 8. However, in most instances you will be able to just quote the results set out in Key point 11.5. If the auxiliary equation has real, distinct roots, .

and

, then the solution to the differential equation is

PROOF 6 If

and

then

, , where

and

are the roots of the associated auxiliary equation.

One possible solution to the differential equation could occur if you had a function of which the derivative and second derivative are proportional to the original function, allowing everything to ‘cancel’ and result in zero. is one example of a function that has this property (although you will see later that there are others).

If Then

and . Substituting this into the differential equation:

Take out a factor of Since If , then it will have two real solutions. Call these and . Therefore two possible solutions to the

.

can never be zero. This is the auxiliary equation.

differential equation are and

. This was proved in Proof 5.

will also be a solution, since any linear combination of solutions is also a solution. This is a solution with two arbitrary constants, therefore it is the

Use Key point 11.1 to justify that your ‘guess’ gives the complete solution.

general solution. If the roots of the auxiliary equation are complex, you could still write the solution as However, you can use the work from Chapter 1, Section 2, to rewrite this in terms of trigonometric functions. If the solutions are

, then you can write the solution as .

PROOF 7 Given that

with .

and

, then can be written in the form

Substituting in the given information:

Take out a factor of

.

Rewrite the complex exponential into modulusargument form. Separate out the cosine and sine terms. with

and

If the roots of the auxiliary equation are equal, then you need to try another possible complementary function .

PROOF 8 Given that

and

, then

is a possible solution for a suitably

chosen . If then

and . Substituting into the differential equation:

Divide through by

(which can never be

.

) and tidy up.

Comparing coefficients:

: Substituting

into the second equation:

This is the given condition. So 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 be able to just write down the solution without going through Proofs 6, 7 and 8.

Key point 11.5 Solution to auxiliary equation Two distinct roots,

General solution to differential equation

and

Repeated root, Complex roots,

WORKED EXAMPLE 11.4 Solve the differential equation given that

and

when

.

The auxiliary equation is . This has roots and , so the general solution to the differential equation is .

Since the roots are real and distinct you can write the solution to the differential equation in exponential form using Key point 11.5. 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

.

You can use technology to solve these types of simultaneous equation. .

EXERCISE 11C 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. 4

a Find the general solution of the differential equation b Find the particular solution that satisfies

5

a

and

a

and

a

and

, find

.

and

. . .

.

Find the general solution of the differential equation

b Find the particular solution that satisfies 9

.

a Find the general solution of the differential equation b Given that

8

.

Find the general solution of the differential equation

b Find the particular solution that satisfies 7

when

Find the general solution of the differential equation

b Find the particular solution that satisfies 6

.

and

, writing your answer in terms of .

Find the general solution of the differential equation

.

10 Find the general solution of the differential equation 11 By trying the function

.

, find the general solution of 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: Inhomogeneous 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 11.2. You first of all solve the associated homogeneous equation to find a complementary function (using Key point 11.5) and then find a particular integral. The form of the particular integral will depend on .

Key point 11.6 Trial function

Polynomial

General polynomial of the same order



WORKED EXAMPLE 11.5 Find the general solution of 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 .

is

Differentiating twice:

. Substituting into the left-hand side of the differential equation and comparing to the exponential part of the right-hand side:

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.

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 this into the left-hand side of the differential equation and comparing to the linear part of the right-hand side: Compare coefficients to find and . Comparing the coefficient of :

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 11.2.

.

Sometimes the trial function given in Key point 11.6 already appears as a part of the complementary function. In that case, you need to modify the particular integral.

Key point 11.7 If your trial function is already part of the complementary function, try multiplying the trial function by .

WORKED EXAMPLE 11.6 Find the general solution of the differential equation . The associated homogenous equation is

First solve the associated homogenous equation to find the complementary function.

. This has auxiliary equation

Therefore, the complementary function is . The trial function associated with .

is

Although the right-hand side of the equation involves only , you

need to include both the trial function.

and

in

This already appears the complementary function, so try .

You need to adjust the trial function according to Key point 11.7.

Differentiating twice:

You need to use the product rule to differentiate.

. Substituting into the left-hand side of the differential equation:

Notice that the terms containing and all cancel. This will always happen in the situation described in Key point 11.7.

The particular integral is

.

The general solution is

The general solution is the sum of the complementary function and the particular integral, from Key point 11.2.

.

EXERCISE 11D

1

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

For the differential equation

:

a find the complementary function b hence find the general solution c find the particular solution for which 5

For the differential equation a the general solution

and

when , find:

.

b the particular solution for which 6

and

For the differential equation

when

.

, find:

a the general solution b the particular solution for which 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 for which

8

and

when

.

Find the solution of the differential equation

given that

when

and

when

. [©AQA 2016]

9

A differential equation is given by

.

a Show that the substitution where is a function of , transforms this differential equation into . b Hence find the general solution of the differential equation

giving your answer in the form

. [©AQA 2013]

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 the particular integral.

, where is the

The complementary function is the solution of the associated homogeneous equation. The particular integral is any solution of the differential equation. Given a first order linear differential equation integrating factor

, multiply through by the

. The solution will be

.

The auxiliary equation to the homogeneous differential equation . The solution to the auxiliary equation gives the general solution of the homogeneous equation:

is

Solution to auxiliary equation Two distinct roots,

General solution to differential equation

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 11 1

What is the complementary function of the differential equation

?

Choose from these options. A B C D 2

What is the integrating factor for the differential equation

?

Choose from these options. A B C D 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 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 for which 6

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. 7

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 for which 8

and

when

.

Solve the differential equation

given that

as

and that

when

. [©AQA 2014]

9

Find the general solution of the differential equation . [©AQA 2010]

10

a Find the general solution of b Show that the substitution into c

11

for

.

transforms the differential equation

.

Hence find the general solution of the differential equation

a Find the general solution of the differential equation b Find the particular solution if

12 It is given that, for

for

.

.

.

, satisfies the differential equation

a Show that the substitution

transforms this differential equation into

. b Hence find the general solution of the differential equation

giving your answer in the form

. [©AQA 2012]

13

a By using an integrating factor, find the general solution of the differential equation

giving your answer in the form

.

b Show that the substitution

transforms the differential equation

into

c Hence, given that

, find the general solution of the differential equation . [©AQA 2014]

14

a Express

in the form

b Use the substitution

given that

and

, where and are integers.

to solve the differential equation

when

. Give your answer in the form

. [©AQA 2016]

12 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 use Hooke’s law model damped oscillations using second order differential equations and interpret their solution solve coupled first order differential equations.

Before you start… Chapter 11

You should know how to solve second order

1 Find the general solution to .

differential equations. A Level Mathematics Student Book 1, Chapter 18

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 when .

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 differential equations, and in this chapter you will look at some common situations modelled by differential equations and how you can use the methods from Chapter 11 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.

Rewind See A Level Mathematics Student Book 1, Chapter 18, for a reminder of using , and A Level Mathematics Student Book 2, Chapter 13, for its use in setting up 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 in which differential equations arise, but this time you will often need to use methods from Chapters 9 and 11 to solve them. You will also look at the types of assumptions that are made when writing these differential equations. WORKED EXAMPLE 12.1 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 proportional to newtons, where is the initial speed of the car. Find the time taken for the car to come to rest. a

The resistance force is

for some constant .

mk(u 2 + v2)

Using

and

The resistance force is negative as the car is moving in the opposite direction to that in which this force acts.

:

Separating the variables:



This is a standard arctan integral.

Use the initial condition to find .

When

:

EXERCISE 12A

The car will come to rest when

1

A stone, of mass , falls vertically downwards under gravity through still water. At time , the stone has speed and it experiences a resistance force of magnitude , where is a constant. a Show that

.

b The initial speed of the stone is . Find an expression for at time . [©AQA 2016] 2

A car, of mass

, is moving along a straight horizontal road. At time seconds, the car has

speed . As the car moves, it experiences a resistance force of magnitude No other horizontal force acts on the car. a Show that

.

b The initial speed of the car is Show that

newtons.

.

. [©AQA 2011]

3

The current in a circuit with resistance

, capacitance and inductance is modelled by: .

Sketch the solution in each situation. a b c 4

The rate of immigration into a country is modelled as being 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 model.

million, find the long-term population predicted by the

c The model is refined by adding the term 5

. Suggest what this term represents.

A chicken 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 it is 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 the 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 and Hooke’s law In Chapter 11, Section 3, you saw that some second order 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 12.1 The differential equation for simple harmonic motion is

where is a constant with units

.

Tip The constant is sometimes referred to as the angular frequency. In the equation in Key point 12.1, represents the displacement of an object and is the time. 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 12.2 The general solution to the simple harmonic motion differential equation is . You need to know some pieces of terminology that are useful in describing these solutions. The average position around which the object oscillates is called the central line. The maximum distance from the central line is called the amplitude. The motion repeats itself after time, , which is called the period. If initially the object is: on the central line, then the solution will be at the maximum displacement from the central line, then the solution will be

.

In both of these cases the amplitude is given by . Since

and

repeat when gets to

, one full period, , occurs when

.

Key point 12.3 The period, , of a particle moving with simple harmonic oscillation is . The object has its maximum speed as it is going through the central line, and it is instantaneously at rest when it reaches the maximum displacement.

Key point 12.4 The relationship between velocity and displacement for a particle moving with simple harmonic oscillation is .

PROOF 9 Prove that If

.

when

,

then

.

Since you are only looking for a relationship between and , choose to start the time when the object moves through the central line. Use Use

. .

Group the terms together to make a link with the expression for displacement. One common context for simple harmonic motion is situations with springs. You can use a standard model from physics for the force from a spring. This says that the tension, , in the spring is proportional to the extension, , from the spring’s natural length. The tension is always directed back towards the equilibrium position.

Key point 12.5 Hooke’s law:

The value is sometimes referred to as the spring constant or the stiffness of the spring.

Fast forward If you are taking the Mechanics option of Further Mathematics, you will meet Hooke’s Law again in Chapter 5 of the Further Mechanics Student Book.

WORKED EXAMPLE 12.2 A spring of natural length and spring constant mass is attached to the other end of the spring. Use

is attached to a hook in the ceiling. A

, giving your final answers to an appropriate degree of accuracy.

a Show that the equilibrium length of the spring is

.

b Show that if the spring is displaced from the equilibrium it will undergo simple harmonic motion and find the time period of oscillations about this equilibrium. c Given that the spring is stretched an additional

and then released, find its maximum speed.

Draw a diagram to help visualise the situation. Only tension and weight are acting on the mass. These must balance.

a

T

0.5 g When the spring is in equilibrium:

Substituting

and

So the extension is

:



and the

equilibrium length is

.

b

This is Newton’s second law vertically. The acceleration is . If is positive when below the equilibrium position, then you need to use down as positive, so the resultant force is weight – tension.

T

y

0.5 g If is the extension below the equilibrium position, then:

So,

, where

.

Rearrange into the standard form for simple harmonic motion.

This is the equation for simple harmonic motion. Then

So

Use

s.f.)

from Key point 12.3.

c If the spring is stretched an additional , then the amplitude is

It is important to be consistent with units – in this case working only in metres.

. .

This is maximised when the maximum speed is

Use

from Key point 12.4.

, so .

So, maximum speed



s.f.)

EXERCISE 12B In this exercise, unless otherwise instructed, use appropriate degree of accuracy. 1

, giving your final answers to an

State the amplitude and the period of the simple harmonic motion described by each equation. Also state whether the particle is at rest or passing through the equilibrium point when . a i ii b i ii c

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 i Amplitude

, period seconds; at rest when

.

ii Amplitude

, period seconds; at rest when

.

b i Amplitude

, period

ii Amplitude

, period

c

i Amplitude ii Amplitude

3

, period , period

seconds; in equilibrium when

.

seconds; 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.

Tip is an alternative notation for

a i ii b i

.

ii c

i ii

d i ii 4

A particle performs simple harmonic motion with amplitude . The particle passes through the equilibrium position when

and angular frequency rad .

a Find the distance of the particle from the equilibrium position when

.

b Find the maximum speed of the particle. 5

A small ball is attached to one end of an elastic spring. When rest

the ball is released from

from the equilibrium position and performs simple harmonic motion with period

.

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 and angular frequency . 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 the acceleration of the ball at this time.

7

A particle performs simple harmonic motion with amplitude through the equilibrium point is .

. Its speed as it passes

a Find the period 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 passes through the equilibrium position when after .

. The particle

with positive displacement immediately

a Find the displacement of the particle when b Find the first time when the particle is

and period

.

. 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 . When its displacement from the equilibrium position is

the speed of the particle is

.

a Find the period 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

is at . a Write down the amplitude and the period of the simple harmonic motion.

the particle

b Write down an expression for in terms of . c Point is between and , and through . d The mass of the particle is when it passes through . 11 A particle of mass

. Find the first time when the particle passes

. Find the magnitude of the force acting on the particle

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 and then released. When the extension of the spring is the elastic force in the spring is where is a constant. All other forces on the particle can be ignored.

,

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. 12 A cart of mass is moving in a straight line with a speed of that is attached to a fixed wall by a light spring.

when it hits a buffer

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 acing on the cart at that point. c Find the time taken to reach the point of maximum compression. 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 the 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 end of the spring and hangs vertically below .

A

B x

is attached to the other

When the extension of the spring is metres, the magnitude of the tension in the spring is newtons. 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 of the resultant force on the particle is

, the magnitude

.

c Hence show that the particle performs simple harmonic motion and find the period of the motion. 15 A particle , of mass

, is attached to one end of a light elastic string, and the other end of

the string is attached to a fixed point a The particle hangs in equilibrium at a point , vertically below , where the extension of the string is metres. Calculate the stiffness of the string. b The particle is pulled vertically downwards from the point by a distance of

metres,

and released from rest. The displacement of from at time seconds after being released is metres. i Given that the string does not become slack during the subsequent motion, show that where is a constant to be determined. ii Hence deduce that the motion of is simple harmonic. iii Show that the period of this motion is

seconds, correct to three significant figures.

iv Calculate the maximum speed of during its motion. [©AQA 2012] 16 A particle moves along a straight line between the points and with simple harmonic motion. The point is the midpoint of . At time seconds, the particle is metres from and moving with speed

.

The motion of the particle satisfies the equation a The particle completes one oscillation in Show that

.

seconds.

.

b When the particle passes through the point , as shown in the diagram,

and

.

–1

2 m s A

O

C

B

i Show that the amplitude of the motion is

metres, correct to three significant figures.

ii When the particle first passes through , it is heading away from towards . Find the time that it takes to move from to and back to , giving your answer to two significant figures. iii Find the maximum speed of the particle during the oscillations. [©AQA 2010] 17 A particle, of mass , is attached to two identical springs. The other ends of the springs are attached to fixed points, and , which are metres apart on a smooth horizontal surface. Each spring has a natural length

and stiffness

The particle is released from rest at a distance of

. metres from and moves on the line

.

The midpoint of is . At time seconds after release, the displacement of the particle from is metres, where the direction from to is taken to be positive.

x

A

C

B

a Show that the resultant force on the particle, at time , is

newtons.

b Hence show that the particle moves with simple harmonic motion. c State the period of this motion. d Find the speed of the particle when it is

metres from .

e Write down an expression for in terms of . [©AQA 2013] 18 A spring has natural length

and stiffness

. A particle of mass

is attached to

one end of the spring. The other end of the spring is attached to a peg which moves up and down between two points, and . The midpoint of is . The point is metres above , and is metres below . At time seconds, the displacement of the peg from the point is metres, where the downward direction is taken as positive. The displacement of the particle from at time is metres, as shown in the diagram.

A O

0.1 sin 4t

B

x

Assume that there is no air resistance. a Show that . b At time

, the particle is at rest with

. Find an expression for at time . [©AQA 2014]

Section 3: Damping and damped oscillations When bodies 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 assume that the drag force, , is proportional to the speed, acting in the opposite direction.

Key point 12.6 The drag force is given by

If you add this to the standard equation for simple harmonic motion, the differential equation becomes:

where

is a positive constant. This is called damped simple harmonic motion.

As you saw in Chapter 11, the solutions to this differential equation depend upon how many solutions there are to the auxiliary equation, and each is given a different name.

Key point 12.7 x

x

x

t

O

t

O

t

O

Heavy damping

Critical damping

where and are roots of the auxiliary equation.

Light damping

where

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 12.3 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 that produces critical damping. As the time period is , the value of is given by

Rearrange the formula in Key point 12.3 to express in terms of .

.

Therefore, in air the differential equation is





The force is therefore given by

In oil, there must be the additional drag force:

Now

.



Since Critical damping occurs when

Only the positive solution to the equation is required, since from the context you need .

EXERCISE 12C 1

Each differential equation describes damped harmonic motion. In each case determine whether the damping is light, heavy or critical. a i ii b i ii c

i ii

d i ii 2

A particle performs damped harmonic motion 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 newtons, where

newtons. The resistance force acting on the particle has magnitude is the speed of the particle.

a Show that the equation of motion of the particle is

.

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 and the resistance force on has magnitude newtons.

newtons

a Write down a differential equation which 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 . When

and

.

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 , where is the speed of the particle. a

Show that the equation of motion for the particle is

When

the particle is at rest,

.

from .

b Find 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

In this question use accuracy.

, giving your final answers to an appropriate degree of

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 When the extension of the string is metres the magnitude of the tension in the string is

.

newtons. a

hangs in equilibrium at the point . Find the extension of the string at this point.

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 newtons. 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 whether the type of damping is light, heavy or critical. At

and

.

b Find an expression for in terms of .

c Show that the maximum extension of the spring is approximately 10 In this question use accuracy.

.

, giving your final answers to an appropriate degree of

A bungee jumper has mass . He uses an elastic rope of natural length and stiffness . He falls vertically and, when the rope becomes taut for the first time, he is travelling at . Assume that, once the rope is taut, the bungee jumper experiences an air resistance force that has magnitude newtons, where is his speed. At time seconds after the rope has become taut, the extension of the rope is metres. a Show that . b Find in terms of . c Find the value of when the bungee jumper first comes instantaneously to rest. [©AQA 2013, adapted] 11 A light spring lies at rest and unstretched on a smooth horizontal surface. A particle, , of mass is attached to the end of the spring. The spring has stiffness , where is a positive constant. The end of the spring is set in motion and moves with constant acceleration of magnitude in the direction , as shown in the diagram. The particle is consequently forced into motion. A

B

P f

At time after the motion begins, is moving with speed and experiences a resistant force of magnitude . The extension of the spring is and the displacement of from its initial position in the direction is . a Show that . b Hence show that has velocity

c Hence show that . d Given that

, and that a particular integral for this differential equation is

find an expression for , in terms of and . [©AQA 2010]

Section 4: Coupled first order differential equations There are many situations in which two variables are linked by differential equations. If both of these differential equations are linear and first order, then you can eliminate one of the variables to form a second order differential equation. WORKED EXAMPLE 12.4 In a population of foxes ( thousands) and rabbits ( thousands) the foxes have a birth rate of and a death rate of

. 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 three 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 the expression for

Rearranging (1):

from (2):





Substituting this in:

.

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 into 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, so the population of foxes tends towards , but the population of rabbits grows without limit.

EXERCISE 12D 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 Form a second order differential equation for . When

,

and

.

.

b Find expressions for and in terms of . 6

Three identical cylindrical cans, each with cross-sectional area are placed vertically one above another. A hole is drilled in the base of each of the top two cans so that water can flow from the top can to the middle one and from the middle to the bottom one.

For each of the top two cans, when the height of water in the can is of the can is . Initially, the height of water in the top can is

, the rate of flow of water out and the middle can is

empty. Let be the height of water in the top can and the height of water in the middle can at time . The time taken for water to fall between cans 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 can never empties. c Find the maximum height of water in the second can. 7

A system contains sharks and fishes . The sharks have a birth rate given by and a death rate given by . The fishes 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… See Focus on... Modelling 2 for an improved version of the model in Question 7.

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 . Time period of the solution: Speed, , is given by: Hooke’s law: The drag force is given by:

.

If there is a drag force there can be light damping, heavy damping 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 12 1

Find the period of the oscillations of a particle modelled by: . Choose from these options. A B C D

2

Find the value of which would result in critical damping in the system modelled by: . Choose from these options. A B C D

3

A particle of mass is acted on by a force measured in seconds.

newtons, where is the time

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

.

. Find the magnitude of the maximum force acting on the ball

A particle of mass moves in a straight line so that, when the displacement of the particle from the origin is metres, the force acting on the particle is directed towards the origin and has magnitude newtons. a

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 and .

. , where

metres from the origin. Find the value of the constants

d Hence find the maximum speed of the particle. 6

A particle moves along a straight line with simple harmonic motion. The point is the midpoint of . When the displacement of the particle relative to is metres, its speed is . When

,

and when

,

.

a Show that the amplitude of the motion is

metres.

b Find the maximum speed of the particle during the motion. [©AQA 2011] 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 metres the force acting on the particle is

newtons.

a Given that the mass of the ball is

, show that the equation of motion of the ball is

. b Show that and .

satisfies the equation and find the value of the constants

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 these 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

and are initially at rest. moves under the

action of a force newtons, modelled by , where is the time measured in seconds. moves under the action of a force newtons, modelled by , where metres is the displacement from the initial position. Which particle travels further in the first 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 displacement of the particle from .

, where is the

a By writing

as

, show that

.

b Given that the particle takes 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 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 , the magnitude of the tension in the spring is given by . a

Show that the equation of motion of the ball can be written as

b By writing

and integrating, show that

. . Hence 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 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 the tension in the spring has magnitude . a Show that b By writing

. as

, show that

.

c Hence 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? c In an improved model there is also a resistance force on the cart of magnitude where is the speed of the cart.

,

Find an expression for is 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 of snakes and badgers at time years are modelled by these differential equations: , Initially there are

badgers and

animals on the plain after 3 months.

snakes on the plain. Find the total number of

17 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 is .

apart on a

x

A

C

The particle is released from rest

B

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 .

18 A railway truck, of mass , is travelling in a straight line along a horizontal track. At time , the truck strikes one end of a buffer which is fixed at its other end. The buffer may be modelled as a light spring of natural length and stiffness constant. At time , the compression of the buffer is .

, where is a positive

a In a simple model of the motion, the only force affecting the truck during this motion is the thrust from the buffer. i Show that, while the truck is in contact with the buffer, the truck performs simple harmonic motion. ii Find, in terms of , the period of this motion. b In a more realistic model, the motion of the truck is affected by a resistance force of magnitude , where is the speed of the truck and is a positive constant. i Show that, while the buffer is being compressed, satisfies the equation . ii At time

, the truck is travelling with speed . Given that

, find in terms of

, and . iii By means of a sketch, or otherwise, explain whether the type of damping is light, critical or heavy. [©AQA 2011] 19 A uniform metal bar, of mass , is held at rest in a horizontal position. The ends of the bar are attached to identical light elastic strings, which each have stiffness . The strings are also attached to fixed points that are directly above the ends of the bar. A damping device is also connected to the bar. The bar is released from rest with the strings vertical and at their natural length. As the bar falls, it remains horizontal and the damping device exerts an upward force of magnitude on the centre of the bar, where is a constant, in appropriate units, and is the speed of the bar. The motion of the bar is critically damped.

At time after the bar has been released, the displacement of the bar below its initial position is . a Show that

.

b Find an expression for in terms of and . c Find the value of as tends to infinity. d Find the maximum speed of the bar. [©AQA 2016] 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 with a force of , but by the end of the seconds he is pulling 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 that the assumption from part a is satisfied, 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 models his force, newtons, at time seconds as Determine who wins.

.

13 Numerical methods In this chapter you will learn how to: approximate definite integrals, using the mid-ordinate rule approximate definite integrals, using Simpson’s rule find approximate numerical solutions to differential equations, using Euler’s method and the improved Euler method.

Before you start… A Level Mathematics Student Book 2, Chapter 15

You should know how to use the trapezium rule.

1 Use the trapezium rule with four strips to estimate correct to three significant figures.

A Level Mathematics Student Book 2, Chapter 4

You should know how to use an iterative formula.

2 If

A Level Mathematics Student Book 2, Chapter 13

You should be able to recognise and solve a separable differential equation.

3 Find the particular solution to the differential equation

You should know how to use the method of integrating factors for solving differential equations.

4 Find the particular solution to the differential equation

You should be able to use implicit differentiation.

5 Given that

Chapter 11

A Level Mathematics Student Book 2, Chapter 10

, and , find , correct to four decimal places.

, where

when

.

, where when

.

,

use implicit differentiation to find an expression for in terms of and .

Approximating integrals In integral problems you have met so far, you have been aiming either to calculate the area enclosed by

the curve of a function over a given interval (definite integration) or to find the equation of the curve of which the derivative is that function (indefinite integration).

Rewind See A Level Mathematics Student Book 2, Chapter 15, for a reminder of upper and lower bounds for definite integrals. In A Level Mathematics Student Book 2, Chapter 15, you met the trapezium rule as a means of approximating the area under a curve where the function is either not explicitly known or cannot be integrated. You also learned methods for giving upper and lower bounds on the area, using rectangular strips anchored to the curve at either their upper left or upper right corners. In both cases, you can generally improve the approximation by taking increasing numbers of strips. In the pre-computer era, doubling the number of strips would at least double the amount of calculation effort, and beyond a certain level the improvement in accuracy would not justify the additional effort involved. So alternative, more efficient methods were devised, which would improve accuracy with less additional effort; two such methods are the mid-ordinate rule and Simpson’s rule.

Section 1: Mid-ordinate rule The simplest way to approximate an area under a curve is to split it into vertical strips, each of which is approximated as being rectangular. Rather than taking the end points of the intervals to form the rectangles, as you did when finding the upper and lower bounds, the mid-ordinate rule uses the midpoint of each interval to fit the rectangles. With an expected error approximately half that of the trapezium rule, this method also uses fewer calculations and is therefore preferable where the number of calculations is a relevant cost. y

O

1 4

—

1 2

—

3 4

—

1

x

Key point 13.1 The mid-ordinate rule using equal-sized intervals with end-points given by

where

is

.

This will be given in your formula book.

Tip As with other numerical methods, it can be useful to lay out all the values in a table when showing your working.

WORKED EXAMPLE 13.1 Using the mid-ordinate rule with four strips, approximate

to three decimal places.

Divide the interval from to into four strips. The -values start at and go up to in increments of . The mid-values therefore start at in increments of . Calculate the -values, using

and also rise

.

Since you want a final result accurate to three decimal places, you should either hold values in your calculator or record them to at least four decimal places of accuracy and round at the end of

the calculation. Using the formula:

So

You should show the values used in the calculation.

( d.p.).

Tip Remember that your calculator might have a TABLE function that will produce the table of values for you. You already know that the trapezium rule will overestimate the area for a curve with a positive second derivative and underestimate the area for a curve with a negative second derivative. To see the equivalent criteria for the mid-ordinate rule, consider a single strip. The area of the rectangular strip will be the same as the area of any trapezium with the same width and base with slanting edge that passes through the same midpoint. In particular, it will be the same as the trapezium that is tangent to the curve at the midpoint.

x  

x

It is clear that for a convex curve like this (having a positive second derivative throughout the interval considered), the mid-ordinate rule will give an underestimate. Similarly, for a concave curve (having a negative second derivative), the mid-ordinate rule will give an overestimate. Again, it should be clear from the graph that the error will be in the opposite direction to the error from the trapezium rule in the same interval.

Rewind For a reminder of convex and concave, see A Level Mathematics Student Book 2, Chapter 12. For a reminder of how this affects the trapezium rule, see A Level Student Book 2, Chapter 15.

Key point 13.2 The mid-ordinate rule: underestimates the area when the curve is convex overestimates the area when the curve is concave.

EXERCISE 13A 1

For each of these curves, use the mid-ordinate rule to estimate the area beneath the curve for the given interval to three decimal places with: i four strips

ii six strips. a b c

 for  for  for

d 2

 for

Use the mid-ordinate rule, with: i four intervals ii five intervals to find the approximate value of each of these integrals, giving your answers to three significant figures. a b c d

3

For each integral in Question 2: i use technology to find the value correct to eight decimal places ii use a spreadsheet to calculate mid-ordinate approximations, using

and

intervals

iii find the percentage errors in each approximation and assess how the error changes as the number of intervals is doubled. 4

a Sketch the graph of b

.

Use the mid-ordinate rule with five strips to estimate the value of

, giving your

answer to two decimal places. c Explain whether your answer is an overestimate or an underestimate. 5

a

Use the mid-ordinate rule with four intervals to find an approximate value for

to three

decimal places. b Describe how you could use the mid-ordinate rule to obtain a more accurate approximation. 6

The diagram shows a part of the graph of is the first positive root of the equation

. .

y

y = cos x3

O

a Find the exact value of .

0.2 0.4 0.6 0.8

1

1.2

x

b Use the mid-ordinate rule with four intervals to find an approximation to four decimal places for . c Is your approximation an overestimate or an underestimate? Explain your answer. 7

The velocity

of a particle moving in a straight line is given by

. The diagram

shows the velocity–time graph for the particle.

a The particle changes direction at

and

. Find the values and .

b Use the mid-ordinate rule, with eight intervals, to estimate the total distance travelled by the particle during the first seconds of movement.

Section 2: Simpson’s rule When applying the trapezium rule or the mid-ordinate rule, you are using a straight line to approximate the curve for each strip; the trapezium rule uses the chord that connects the two end-points of the curve in the strip interval and the mid-ordinate rule uses a rectangle at the midpoint of the strip.

x 

x

For a more accurate approximation of the curve in the strips, it is possible to fit a quadratic instead of a straight line. While a straight line is defined by any two points on it, a quadratic is defined by any three points on the curve. Conventionally, the area to be approximated is divided into an even number of strips, and a quadratic curve fitted to each pair of strips so that it passes through the three points defined on the curve by the strip boundaries.

p – d

p

p + d

x

PROOF 10 A quadratic curve passing through points

and

will have area

over that interval. The general equation for a quadratic . Substituting the values of the coordinates: Substitute the quadratic.

The area under the curve between is:

and



and

into

Use equations

and

.

Using the fact in Proof 10, you can approximate the area beneath a curve of several pairs of strips.

y1

y0

y2

y3

y4 y5 y6

x0 The first pair has area

x1

x2

x3

x4

x5

x6

x

.

The second pair has area

.

The third pair has area

.

Adding these up: the end points and are seen once; the odd-numbered values and , which are the heights at the centre of each pair of strips, are each seen once with a multiple of ; the even numbered values and at the junctions between pairs are each seen twice with a multiple of . This can be generalised to give Simpson’s rule.

Key point 13.3 Simpson’s rule using an even number of equal-sized intervals with end points is given by:

where

.

This will be given in your formula book.

Tip As with other numerical methods, it can be useful to lay out all the values in a table when showing your working. WORKED EXAMPLE 13.2

Using Simpson’s rule with four strips, approximate

to three decimal places.

Divide the interval from to into strips. Multiple

The -values start at and go up to in increments of . Calculate the -values, using

.

Since you want a final result accurate to three decimal places, you should either hold values in your calculator or record them to at least four decimal places of accuracy and round at the end of the calculation.

sum Using the formula:

You should show the values used in the calculation.

Area

.

So

( s.f).

Since the true value of the integral in Worked example 13.2 is

, the

Simpson’s rule approximation shown is an improvement on the mid-ordinate rule approximation of from Worked example 13.1, and in general you can expect Simpson’s rule, for the same number of strips, to give a better approximation than any of the linear methods.

Did you know? Since the error for the mid-ordinate rule is approximately half that of the trapezium rule, and in the opposite direction (one will overestimate and the other will underestimate, depending on whether the curve is convex or concave), taking a weighted average of the two in an attempt to cancel out the bulk of the error might seem a sensible step. If

represents the approximation

by the mid-ordinate rule and the approximation by the trapezium rule, then should be an improvement on both, and is an alternative representation of Simpson’s rule.

EXERCISE 13B 1

For each of these curves, use Simpson’s rule to estimate the area beneath the curve for the given interval to three decimal places with: i four strips ii six strips. a b c

  for   for   for

d 2

  for

Use Simpson’s rule, with: i five ordinates ii seven ordinates to find the approximate value of each of these integrals, giving your answers to four significant figures.

a b c d

Common error A question might specify the number of strips, intervals or ordinates when using Simpson’s rule. Remember that the number of ordinates means the number of -values, so five ordinates means , the boundary values for four strips. 3

For each integral in Question 2: i use technology to find the value, correct to six decimal places ii use a spreadsheet to calculate Simpson’s rule approximations, using

and

intervals

iii find the percentage errors in each approximation and assess how the error changes as the number of intervals is doubled. 4

a Sketch the graph of b

for

.

Use Simpson’s rule with six strips to estimate the value of

, giving your answer

to three decimal places. 5

a

Use Simpson’s rule with five ordinates to find an approximate value for

.

b Describe how you could use Simpson’s rule to obtain a more accurate approximation. 6

The diagram shows a part of the graph of

.

and are, respectively, the first and second positive roots of the equation

.

a Find the exact values of and . b

Use Simpson’s rule with six intervals to find an approximation for

to three

decimal places. 7

a Write down the exact area of the semicircle with curved edge defined by b i Explain why, when estimating the value of

.

, using Simpson’s rule with seven

ordinates, it is preferable to use ordinates from to than from

to .

ii

Use Simpson’s rule with seven ordinates to find the approximate value of your answer correct to three decimal places.

c Hence find an approximation to the value of .

, giving

Section 3: Euler’s method In Sections 1 and 2, you met two new methods for approximating the definite integral for a function given solely in terms of .

Rewind See Chapter 11 for a reminder of how to solve linear differential equations. However, in many mathematical models, a differential equation is formed where

is a function of both

and . You already know how to approach some such problems if they are separable or can be rearranged as linear differential equations, but there are many differential equations that cannot be solved analytically. If you know a point on the required solution curve, then you can use the principles encountered in Chapter 10, Section 1, on Maclaurin series: The Maclaurin series of a function is If you apply a translation to this expansion such that arbitrary value rather than about .

, you obtain the expansion about an

The simplest approximation to track the curve uses the first two terms of this expansion: .

Tip Notice that this formula is a rearrangement of

; in the limit as

you

know this as the first principles definition of the derivative.

Did you know? This generalisation of Maclaurin series is called a Taylor series after the English mathematician Brook Taylor who formalised this idea. The Scottish mathematician Colin Maclaurin then made use of the particular case where . Geometrically, this means that if you know a point on the curve and the gradient at that point, you can estimate the position of another point on the curve by extending the tangent.

y = 2 + x 4

x

For values of far from the known point, the approximation gets increasingly unreliable. To keep the

approximated path close to the curve, you can recalculate the gradient according to the formula for the derivative. As in all the approximations encountered so far, the smaller the value of , the more accurate you can expect the approximation to be. The diagram shows the curve along with two approximations that both begin at the point on the curve with . You can see that the approximation error is approximately halved by halving the distance .

y = 2 + x 4

h = 0.5

h = 1 x

You can take this principle a step further for the general case where the derivative is a function of both and . To track the series of tangents you use the general formula for Euler’s method.

Did you know? A plot of tangents at regular points in the plane is called a slope field diagram and can be used to sketch curves. Try plotting slope field diagrams for Worked example 13.3 and for part a of Question 1 in Exercise 13C and use them to sketch curves.

Key point 13.4 Euler’s method: for

and known point

,

, where

.

This will be given in your formula book. Notice that in the approximation to the curve

shown, each line section was parallel to the

tangent to the curve vertically above it (because the gradient is a function solely in ). This will not be the case when the gradient is a function of both and . WORKED EXAMPLE 13.3 The function

satisfies the differential equation

Use the Euler formula your answer to three decimal places. , Using the Euler formula:

, with

where

and

, to obtain an approximation to

. , giving

State the start value and the interval . At each step, apply the Euler formula , using the previous value of and as and .

You should recognise that the differential equation in Worked example 13.3 is solvable using the method of integrating factors, with solution .

Plotting the solution from Worked example 13.3 against the actual curve shows the approximation accumulating errors with each step.

y 3.2

y = 4ex–1 – x – 1

2.8

2.4

2

1

1.1

1.2

1.3

x

Rewind See Chapter 11, Section 2, for the method of solving a first order linear differential equation using the integrating factor.

EXERCISE 13C

EXERCISE 13C 1

Use Euler’s method with step length differential equations. a i

,

b i

c

i ii

2

, , , ,

For the differential equation

with

to find an approximate value of 3

for these

,

ii

ii

to find an approximate value of

. Give your answer correct to three decimal places.

Consider the differential equation a Use Euler’s method with

, use Euler’s method with step length

with the boundary condition to find an approximate value of

.

.

b Solve the differential equation. c

i Find the percentage error in your approximation from part a. ii How can this error be decreased?

4

The function

satisfies the differential equation

a i Use Euler’s method with step length

with

.

to find an approximate value of

.

ii How can your approximation be made more accurate? b Solve the differential equation and hence find the actual value of

.

c Sketch the graph of your solution and use it to explain why your approximation from part a is smaller than the actual value of . 5

Consider the differential equation a Find an expression for

where

and

when

.

in terms of .

b Use Euler’s method with step length

to find the approximate value of when

.

Section 4: Improved Euler’s method As seen in Section 3, when making the second approximation using Euler’s method, if the curve continues convex or concave then you will compound any error from the first approximation. If at

is the known point on the curve, you obtain point over the interval distance , so that and

, using Euler’s method, taking the tangent .

Just using the original Euler method, you would obtain the second approximation point by taking a tangent at point over interval . If the curve continues concave or convex, this tangent at will be at a distance from the curve.

P2

P1

P0

However, translating the tangent at so that it passes through leads to an amended approximation when iterating.

will give a better estimate

; this

Q2

P2

P1

P0

Key point 13.5 The improved Euler method: for

and known point , where

, .

This will be given in your formula book.

WORKED EXAMPLE 13.4 It is given that

satisfies the differential equation

, where

and

. a Use the Euler formula b Use the improved Euler formula

, with

, to obtain an approximation to

.

with your answer to part a to obtain an

approximation to a

, giving your answer to three significant figures.

,

State the start value and the interval .

Using the Euler formula:

Apply the Euler formula

b

.

Calculate the gradient at the approximated point

.

Apply the improved Euler formula

.

EXERCISE 13D 1

In each part of this question, function

and known value

satisfies the differential equation .

Use the Euler formula to find a i

,

ii

,

b i

,

c

ii

,

i

,

ii

to find

and then use the improved Euler formula

, giving each answer correct to three significant figures.

,

d i

,

ii 2

for the given

,

It is given that

satisfies the differential equation

, where

and

. a Use the Euler formula

, with

, to obtain an approximation to

.

b Use the improved Euler formula with your answer to part a, to obtain an approximation to , giving your answer to three significant figures. 3

It is given that

satisfies the differential equation

, where

and

. a Use the Euler formula

, with

, to obtain an approximation to

.

b Use the Euler formula with followed by the improved Euler formula to obtain an alternative approximation to , giving your answer to three decimal places. c Without further calculation, state, with reasoning, which approximation you consider likely to be more accurate. 4

It is given that

satisfies the differential equation

a Use the Euler formula terms of . b The improved Euler formula in part a to obtain an approximation

, with

, where

and

, to obtain an approximation to

. , in

is used in conjunction with the Euler formula . Find .

5

It is given that

satisfies the differential equation

, where

and

. a Use the Euler formula

, with

, to obtain an approximation to

.

b Use the improved Euler formula , with your answer to part a, to obtain an approximation to , giving your answer to four decimal places. c

i Using the substitution in the form

, or otherwise, solve the differential equation, giving your answer .

ii Find the percentage error in the approximation from part b.

Checklist of learning and understanding The mid-ordinate rule and Simpson’s rule are more sophisticated methods of approximating a definite integral than the trapezium rule. The mid-ordinate rule using equal-sized intervals with end points given by

is

, where

.

The mid-ordinate rule: underestimates the area when the curve is convex overestimates the area when the curve is concave. Simpson’s rule using an even number of equal-sized intervals with end points is given by , where

.

Euler’s method and associated variations are methods of approximating the path of a curve from a formula for the first derivative. Euler’s method: for

and known point , where

Improved Euler method: for

. .

and known point , where

. .

Mixed practice 13 1

Simpson’s rule with nine ordinates is used to find an estimate for

.

What is the width of each interval? Choose from these options. A B C D 2

Use the mid-ordinate rule with four strips to estimate

, giving your answer

correct to three decimal places. 3

Use the mid-ordinate rule with five strips to estimate , giving your answer correct to four decimal places.

4

The function

.

Use Simpson’s rule with seven ordinates to find the approximate value of

, giving

your answer correct to three decimal places. 5

Use Simpson’s rule with seven ordinates (six strips) to estimate

giving your answer correct to three decimal places. 6

The function where

satisfies the differential equation and

.

Use the Euler formula with giving your answer to three decimal places. 7

The function where

satisfies the differential equation and

,

, to obtain an approximation to

,

,

.

a Use the Euler formula . b Use the improved Euler formula significant figures.

, with

, to find the approximate value of

to find

, correct to three

8

The function

satisfies the differential equation

, where

and

a Use the Euler formula,

with

, to obtain an approximation to

, giving your answer to four decimal

places. b Use the improved Euler formula

with

, to obtain an approximation to

, giving your answer to four decimal

places. [©AQA 2010] 9

The function

satisfies the differential equation

, where

and

.

a Use the Euler formula , with places.

, to obtain an approximation to

, giving your answer to four decimal

b Use the improved Euler formula , with your answer to part a, to obtain an approximation to three decimal places.

, giving your answer to

[©AQA 2010] 10

The mid-ordinate rule with equal intervals is used to find an estimate for for

,

, where

and

What will happen to the value obtained by the mid-ordinate rule if now used? Choose from these options. A Increase B Decrease C Stay the same D It depends on the particular function .

equal intervals are

11

a

Use the mid-ordinate rule with six strips to estimate

, correct to three

decimal places. b Using the substitution places.

, calculate the true value of the integral to three decimal

c Calculate the percentage error of the approximation. d How might the approximation with the mid-ordinate rule be improved? 12 A class is asked to use Simpson’s rule with five ordinates (four strips) to estimate . a Most of the students used . Show that the approximate value the students calculated is , correct to three decimal places. Some of the students answered the question using

instead.

b Explain their reasoning and, without further working, suggest whether they will get a more or less accurate estimate. 13

Simpson’s rule, with two pairs of strips, is used to approximate

.

a Show that the value obtained by Simpson’s rule is exactly accurate. b Without further calculation, explain why this result can be generalised to the approximation of any integral of a cubic equation

using Simpson’s rule with two pairs of strips. 14 The function

satisfies the differential equation

a Use the Euler formula

to estimate

, where

and

to four decimal places, using

a step size: i ii

.

b Which of your answers to part a do you expect to be more accurate? Justify your answer. 15

a

Use the mid-ordinate rule with four strips to find an estimate for

,

giving your answer to three decimal places. b Find the exact value of the gradient of the curve curve where .

at the point on the

[©AQA 2015] 16

a Use Simpson’s rule with seven ordinates (six strips) to find an approximation to , giving your answer to three significant figures.

b Hence find an approximation to

. [©AQA 2016, adapted]

17 Use Simpson’srule, with five ordinates (four strips), to calculate an estimate for . Give your answer to four significant figures. [©AQA 2014] 18

The mid-ordinate rule with two strips is used to estimate

.

What is the percentage error in the estimate? 19 The function

satisfies the differential equation

, where

and

, where is a constant. a The Euler formula approximation

is used with a step size

to give the

. Find .

b Using the method of integrating factors, or otherwise, find the particular solution to the differential equation. c Calculate the percentage error in the approximation 20

a It is given that

.

satisfies the differential equation ,

where

and

.

Use the Euler formula , with a step size , to obtain an approximation to , giving your answer to three decimal places. b It is given that

and

when

satisfies the differential equation

.

i Use implicit differentiation to find

, giving your answer in terms of and .

ii Hence find the first three non-zero terms in the expansion, in ascending powers of , of . Give your answer in an exact form. iii Use your answer to part b ii to obtain an approximation to three decimal places.

, giving your answer to

[©AQA 2016, adapted]

FOCUS ON … PROOF 2

Elements of area and Gaussian integrals 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. 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,

.

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. y

dy dx O

x

In Chapter 7 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:

dr

r dθ 0, 2π

O 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

.

Rewind This is an example of an improper integral, which you met in Chapter 10. You cannot find the indefinite integral of , using standard functions, however you can evaluate this definite integral exactly. The in the integral is just a dummy variable. You could also write

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

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 21.

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 question then becomes to find an expression for . y D

D

x

O

It is clear that the shape of the chain will be symmetrical, with the lowest point half-way 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, but changes with . y T

T0

mg O

x

x

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

. y T T sin θ θ T cos θ

T0

mg O

x

x

Rewind For a reminder about resolving forces, see A Level Mathematics Student Book 2, Chapter 18. 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 . The length of a curve is given by

and so

.

Rewind You met the expression for the length of the curve in Chapter 9, Section 6. Substituting this into

results in an equation involving both a derivative and an integral. You

don’t know how to solve such equations. However, differentiating it gives

You can now proceed to solve this differential equation.

QUESTIONS

QUESTIONS 1

Make a substitution

and show that .

Rewind You met integrals of this type in Chapter 9, Section 3. 2 3

Explain why the constant of integration is zero. Hence show that

.

Hence find an expression for in terms of . Explain why the constant of integration can be taken to be zero.

Did you know? The

curve is called a catenary, meaning ‘relating to a chain’.

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

. 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 the equation in Question 5 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 catenary,

, where is a constant depending on the mass and the length of the chain and the

distance between the end points.

FOCUS ON … MODELLING 2

The Lotka-Volterra model and phase planes During World War 1 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. These differential equations comprise the Lotka-Volterra model. 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 corresponds 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. Assuming that , the phase plane for Lotka-Voltera is shown in the diagram.

S

F

O

QUESTIONS 4

If the original value is and , sketch the trajectory of the system over time. Hence 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 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 the system in Question 6 with How has the introduction of competition changed the behaviour of the system?

and

.

CROSS-TOPIC REVIEW EXERCISE 2 1

The curve has polar equation

, where

and

.

a Sketch , including the equations of any tangents at the pole. b Find the total area enclosed by . 2

a Sketch the graph of

and state the equations of its asymptotes.

b Use the definitions of

and

in terms of and

to show that

. c Solve the equation logarithms.

, giving your answers in terms of natural [©AQA 2015]

3

Use l’HÔpital’s rule to evaluate each limit. a b

4

a Find, in terms of and , the value of the integral

.

b Show that only one of the following improper integrals has a finite value, and find that value: i ii

. [©AQA 2011]

5

The length of the arc of the curve

between

and

is .

Find the exact positive value of . 6

A particle, of mass particle has speed magnitude

, is moving along a straight horizontal line. At time seconds, the . As the particle moves, it experiences a resistance force of

. No other horizontal force acts on the particle.

The initial speed of the particle is

.

a Show that

b Find the value of when the particle comes to rest. [©AQA 2013] 7

a Write down the expansion of in .

in ascending powers of up to and including the term

b Find . [©AQA 2010]

8

The diagram shows a sketch of the graph of

.

a The area of the shaded region, bounded by the curve, the -axis and the lines , is given by

and

.

Use the mid-ordinate rule with five strips to find an estimate for this area. Give your answer to three significant figures. b With the aid of a diagram, explain whether the mid-ordinate rule applied in part a gives an estimate which is smaller than or greater than the area of the shaded region. [©AQA 2013] 9

It is given that the function

satisfies the differential equation

where

and . a Use the Euler formula with , giving your answer to three decimal places.

to obtain an approximation to

b Use your answer to part a and the improved Euler formula , with , to obtain an approximation to , giving your answer to three decimal places. [©AQA 2013] 10 The polar equation of a curve

a i

is

Find the Cartesian equation of

ii Deduce that centre.

.

is a circle and find its radius and the Cartesian coordinates of its

b The diagram shows the curve

with polar equation

r = 4 + sin θ

0, 2π

O

i

Find the area of the region that is bounded by

ii Prove that the curves

and

.

do not intersect.

iii Find the area of the region that is outside

but inside

. [©AQA 2010]

11 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 . 12 Given that a show that, for b hence evaluate

, giving your answer in terms of .

13 A curve is defined parametrically by curve between and is . a Show that

. The length of the arc of the

.

b Find the exact value of . 14 a Explain why

is an improper integral.

b Evaluate

, showing the limiting process used. [©AQA 2015]

15 a Show that the substitution

into

transforms the integral

.

b Hence show that . [©AQA 2010] 16 a Express

in the form

, where

and are integers.

b By means of a suitable substitution, or otherwise, find the exact value of

. [©AQA 2012] 17 a Given that

, show that

where is an integer. b Hence show that

where and are rational numbers. [©AQA 2015] 18 a Given that

, find

and

(You may leave your expression for

.

unsimplified.)

b Hence, using Maclaurin’s theorem, find the first two non-zero terms in the expansion, in ascending powers of , of . c Find . [©AQA 2011] 19 Solve the differential equation

given that

and

when

. Give your answer in the form

. [©AQA 2012]

20 a Given that

and is a function of , show that:

i ii

.

b Hence show that the substitution

transforms the differential equation

into

c Hence find the general solution of the differential equation

giving your answer in the form

. [©AQA 2010]

21 A particle moves with simple harmonic motion on a line between two points and , which are metres apart. The maximum speed of the particle is . The particle passes

through a point that is

metres from .

a Find the period of the motion. b Find the speed of the particle when it is at . c Given that the particle is at rest at at time of the particle from at time seconds.

, find an expression for the displacement

d Find the time that it takes for the particle to move from to . [©AQA 2015] 22 The diagram shows a sketch of a curve and a circle. The polar equation of the curve is . The circle, whose polar equation is

, intersects the curve at the points

and , as shown in the diagram.

a Find the polar coordinates of and the polar coordinates of . b A straight line, drawn from the point through the pole , intersects the curve again at the point . i

Find the polar coordinates of .

ii Find, in surd form, the length of

.

iii Hence, or otherwise, explain why the line

is a tangent to the circle

c Find the area of the shaded region which lies inside the circle . Give your answer in the form

.

but outside the curve

, where and are integers. [©AQA 2013]

23 The hyperbola has parametric equations

.

a Show that the equation of a normal to is

.

The normal to the point on intersects the -axis at the point . b Show that as varies, the locus of the mid-point of 24 Prove that if

, then

25 a Use the substitution

is part of another hyperbola.

. to show that .

b Hence find 26 a Find

. .

b Show that

.

The area is bounded by the curve with equation shown.

, the -axis and the line

as

y y = sin2 x p A x

O c i

Find the area in terms of .

ii Hence state

.

27 Evaluate the improper integral

, clearly showing the limiting process

used. 28 A curve has equation has length .

for all

a Write down an expression for It is given that, for all

. The arc from the point

to the point

.

is proportional to .

b Show that the only curve with this property is a straight line. 29 A differential equation is given by

a Show that the substitution

tranforms this differential equation into

b Hence find the general solution of the differential equation

[© AQA 2015] 30 A particle, , of mass , moves in a straight horizontal line. At time seconds, the displacement of from a fixed point on the line is metres, and is moving with velocity . Throughout the motion, two horizontal forces act on : a force of magnitude newtons directed towards , and a resistance force of magnitude newtons, where and are positive constants. a Show that b In one case, i

. When

and

Show that

ii Show that passes through when

. .

c In a different case, i

.

Find a general solution for at time seconds.

ii Hence state the type of damping which occurs. [©AQA 2012] 31 A curve is given parametrically by the equations

a Express

in terms of

.

b The arc of from i

to

is rotated through

radians about the -axis.

Show that , the area of the curved surface generated, is given by

ii Find the exact value of . [©AQA 2010] 32 a i

Show that

where is an integer. ii Hence show that

where and are rational numbers. b The arc of the curve with equation is rotated through i

between the points where

and

radians about the -axis.

Show that the area of the curved surface formed is given by .

ii Use the substitution

to find the exact value of . [© AQA 2013]

PRACTICE PAPER 2 hours, 100 marks 1

Find the value of for which the matrix

is singular.

Choose from these options. A B C D [1 mark] 2

Find the Cartesian equation of the curve with polar equation

.

Choose from these options. A B C D [1 mark] 3

Let

and

Find the value of such that is perpendicular to and . [3 marks] 4

Use the definitions of

and

to prove that . [3 marks]

5

Given that

when

, solve the differential equation .

Give your answer in the form

. [6 marks]

6

Show that the mean value of the function

between and is

, where

and are constants to be found. [7 marks] 7

The function is defined by a Show that

. , where and are constants in terms of .

b Hence state a transformation that maps the graph of

onto the graph of

.

c

i Sketch the graph of ii Find, in terms of , the range of values of for which the equation

has exactly two

solutions. [10 marks] 8

a If

, show that

.

b Show that

, for integer .

c Hence solve

. [8 marks]

9

Given that

,

a show that, for

.

b Hence evaluate

. [8 marks]

10 Three planes have equations

.

a Show that for all values of , the planes do not intersect at a unique point. b Find the value of for which the intersection of the three planes is a line, and find the vector equation of this line. [9 marks]

11 Given the matrix a i show that is an eigenvalue of

and find the other two eigenvalues

ii find eigenvectors corresponding to these three eigenvalues. The transformation has matrix

.

b Describe the geometrical significance of the eigenvectors of

in relation to . [14 marks]

12 a i Find the first four derivatives of ii Hence find the Maclaurin series for

. up to and including the term in

.

b Use your result to part a ii and the series for to find the Maclaurin series for including the term in c Evaluate

, up to and

. , clearly showing your working. [12 marks]

13 a Prove by induction that, for

, .

b Hence find the exact value of

. [8 marks]

14 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 i find in terms of and

. and

:

ii state whether the damping is light, heavy or critical. [10 marks]

FORMULAE

Further Pure Mathematics Differentiation

Integration

or or

Complex numbers

The roots of

are given by

for

Matrix transformations Anticlockwise rotation through about :

Reflection in the line

:

The matrices for rotations (in three dimensions) through an angle about one of the axes are:

for the -axis

for the -axis

for the -axis

Summations

Maclaurin’s series

for all

for all

for all

Vectors The resolved part of in the direction of is

The vector product

If is the point with position vector

, then

the straight line through with direction vector (Cartesian form) or

has equation (vector product form)

the plane through and parallel to and has vector equation

Area of a sector (polar coordinates)

Hyperbolic functions

Conics Ellipse

Parabola

Hyperbola

Standard form

Parametric form Asymptotes

none

Further numerical integration The mid-ordinate rule: where Simpson’s rule: where

and is even

Numerical solution of differential equations For

and small , recurrence relations are: Euler’s method:

For Euler’s method: Improved Euler method:

Arc length (Cartesian coordinates)

(parametric form)

Surface area of revolution (Cartesian coordinates)

none



(parametric form)

Pure mathematics Binomial series

where

Arithmetic series

Geometic series

  for  

Trigonometry: small angles For small angle ,

Trigonometric identities

Differentiation

Differentiation from first principles

Integration

( constant;

where relevant)

Numerical solution of equations The Newton-Raphson iteration for solving

Numerical integration The trapezium rule:

where

Answers to exercises 1 Further complex numbers: powers and roots 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

.

WORK IT OUT 1.1 Solution 1 is correct.

EXERCISE 1A 1 a

i ii

b

i ii

c

i ii

2 a

Im

b

z4

3 1 z

z2 z

–1

O

–1 3 a

i

1

Re

Im

ii

z, z4

O

–1

z2 b 4 a b c 5 a b 6 7

EXERCISE 1B 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

z31

Re

Im

4

5 4

bi

3 2 aii 1

ai

–5 –4 –3 –2 –1 O –1 –2 –3

1

2

3

4

5

Re

bii

–4 –5

5 Proof. 6 a b 7

( s.f.)

8 9 10 11 12

EXERCISE 1C 1 a

i ii

b

i ii

c

i ii

2 a

i

Im 2 z2

z1 1

–2

–1

O

1

2

–1 z3

z4 –2

Re

ii

Im 4 z2

3 2

z1

1 –4 –3 –2 –1 O –1 z3 –2 –3

1

2

3

4

Re

z4

–4 b

i

Im z2 4 3 2 1 –4 –3 –2 –1 O –1 z3

z1 1

2

3

4

Re

–2 –3 –4

z4

ii

Im z2

1 z1

–1

O

1

z3 –1 3 4 a b

z4

Re

5 6

7 a b

Im

c

z2 2 1 z1 –2 z3

–1

O

1

–1 –2 8 9 a b 10 a b 11 a b c 12 a b c

EXERCISE 1D 1 a

i ii

b

i

ii 2 a

i ii

b

i ii

z4

2

Re

3 a

Im

b

1

w1

w2

Re 1w0

O

–1 w3

–1

w4

4 a b c 5 a

(or

),

b No. Consider

, or an Argand diagram.

c d 6 a, b Proof. 7 a b 8 9 a ( ), b

for

c–e Proof.

Im

10 a

ω2

ω –1 3

i ii

11 a Proof. b

i Proof. ii

c Proof.

ω1

ω Re 1

O

ω4 b

1

–1

ω5

EXERCISE 1E 1 a b c 2 3 Proof; 4 a b 5 a Proof. b c 6 a

(or

)

b, c Proof.

EXERCISE 1F 1 a

( s.f.)

b

( s.f.)

2 a Proof. b Rotation through

radians

about the origin.

3 4 5 a b

lm

6 a

z

2

1

z

z3 –1

O

1

Re

–1 b Rotation through 7 a b 8 a b c 9 a b Proof. 10 a Proof.

radians

( s.f.) about the origin.

b 11 a b 12

MIXED PRACTICE 1 1 A 2 a



b 3 a

for

Im

b

1

z1

z2

Re 1z0

O

–1 z3

–1

z4

4 5 a b

for

6 a b 7 a b

for

8 C 9 10 a b Proof;

.

c For example, 11 a, b Proof. 12 13 Proof. 14 a b 15 a b–d Proof. 16 a Proof. b

i

ii c 17 Proof; it equals 18 a b

i

for

ii c Proof. 19 a b

( s.f.)

20 a Proof. b

i ii iii iv Proof.

.

2 Further complex numbers:   trigonometry BEFORE YOU START 1 2 a b 3 4 5 a b

WORK IT OUT 2.1 Solution 2 is correct.

EXERCISE 2A 1 a b Proof. 2 3 a b 4 a Proof. b 5 a b 6 a

;

b 7 a Proof. b 8 a

b Proof. c

EXERCISE 2B 1 a b

;

c 2 a Proof. b c 3 a b Proof. 4 a Proof. b c Proof; 5 a Proof. b

;

6 a Proof. b

EXERCISE 2C 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 2D 1 a

or, alternatively,

; proof.

b 2 a Proof;

or, alternatively,

b Proof. 3 4 Proof. 5 Proof. 6 a b Proof. c

MIXED PRACTICE 2 1 a b 2 Proof; 3 a b 4 Proof. 5 Proof. 6 Proof. 7 a, b Proof. c

i Proof. ii

8 a

b, c Proof. d Proof; e Proof. f 9 a

i Proof; ii Proof.

b Proof; c Proof. 10 a Proof. b

i ii Proof;

c d Proof.

.

3 Further transformations of the ellipse, hyperbola and parabola BEFORE YOU START 1 Hyperbola; asymptotes

axis intercepts

.

2 a b 3 Reflection in the -axis, horizontal stretch with scale factor .

EXERCISE 3A 1 Proof; 2 Proof; 3

y 5

x 2 y2 — + = 1 — 16 25

O

–4

4

x

–5 4 Proof;

y

5

y = –x

y = x x2 – y2 = 25

–5

O

5

6 Proof; 7 Proof; 8 9 10 a

(rectangular) hyperbola. ellipse.

x

b

11

ellipse;

.

12 a b Translation with vector

.

c 13 a b c 14 a b Horizontal stretch with scale factor . c

EXERCISE 3B 1

unchanged under rotations

a

b

c

–

2 a

reflections in

.

b

reflections in

.

c 3 4 5

reflection in

.

y

6 a

y2 = 5x

O

x

b c Horizontal stretch with scale factor . 7 Enlargement, centre at the origin, scale factor 8 Rotation through 9 a

b Proof. 10 Proof;

WORK IT OUT 3.1 Solution 3 is correct.

EXERCISE 3C 1 a

i ii

b

i ii

c

i ii

d

i

.

(in either direction) about the origin or reflection in

.

ii 2 a

i Translation with vector ii Translation with vector

b

ii

.

Enlargement with scale factor

i Translation with vector

(Or reflection in

(Or reflection in i Reflection in

and translation with vector

and reflection in the line

followed by translation

ii Translation with vector

d

and enlargement with scale factor .

i Horizontal stretch with scale factor , vertical stretch with scale factor and translation with vector

c

and enlargement with scale factor .

or

and reflection in the line

followed by translation

.

.

rotation instead of the reflection.) .

.)

, vertical stretch with scale factor and translation units in the positive -

direction. ii Reflection in direction.

, vertical stretch with scale factor and translation unit in the positive -

3 4 5 6 7 a

b 8 a Hyperbola; b

(or

)

c (or, for example:

)

9 a b Enlargement with scale factor , centre origin. 10 a Stretch scale factor in the -direction and scale factor in the -direction. b 11 a b Translation unit to the left; vertical stretch with scale factor

.

12 13 a Proof. b 14 15 Translation with vector

horizontal stretch with scale factor

vertical stretch with scale factor

.

MIXED PRACTICE 3 1 C 2 a

b 3 4

and

5

and

6 a

and

.

b Translation units in the positive -direction, vertical stretch with scale factor . c

and

.

7 A 8

(or

9 a Proof. b 10 a

);

b c 11 a b Translation

.

12 B 13 14 a b For example, reflection in

translation

.

c

vertex

asymptotes

and

.

15 a b Translation units to the right; vertical stretch with scale factor c

.

4 Further graphs and inequalities BEFORE YOU START 1 a

b

2 a

y y = sinh (x + 2)

–2 O

x

y

b

x

O y = cosh x – 4 –3 y

3

y = |2x – 3| y = |x + 1|

3 1 –1 O

3 — 2

x

4

EXERCISE 4A 1 a

i

vertical asymptote:

; horizontal asymptote:

; axis intercept

.

y

ii

y = 0

x

O –0.2

x = 5 vertical asymptote: b

; horizontal asymptote:

; axis intercept

.

y

i

x = –1

1

y = 0 O

vertical asymptote:

x

; horizontal asymptote:

; axis intercept

.

y

ii

x = –3

1 — 3

y = 0 O

vertical asymptote:

x

; horizontal asymptote:

; axis intercept

.

c

y

i

y = 0

x

O x = 0

vertical asymptote:

; horizontal asymptote:

; no axis intercepts.

y

ii

1 y = 0

x

O x = 1

vertical asymptote: d

; horizontal asymptote:

; axis intercept

.

y

i

y = 0

x

O –0.25

x = –2 vertical asymptotes:

x = 2 ; horizontal asymptote:

; axis intercept

.

ii

vertical asymptotes: e

; horizontal asymptote:

y

i

y = 1 0.5 y = 0

x

O

horizontal asymptotes:

; axis intercept

y

ii

e–2 y = 0

O

horizontal asymptote:

x

; axis intercept

.

.

; axis intercept

.

f

y

i

x

O

x = e–2 vertical asymptote:

; axis intercept

.

y

ii

1 — ln3 –3

x

O

x = –2 vertical asymptote: g

i

; axis intercepts

,

.

y

x = nπ π O –2π–— 3π –π –— 2 2

vertical asymptotes:

π — 2

3π 2π π — 2

x

for integer ; no axis intercepts.

y

ii

π O –2π–— 3π –π –— 2 2

π — 2

3π 2π π — 2

x

x = nπ

vertical asymptotes: 2 a

for integer ; axis intercepts

i

ii

b

i

ii

3 a

i

y

2

O

x

.

ii

b

i

y

1 —) (3, 2

1 — 3 O ii

x = 5

x = 1

x

y

O

x

c

y

i

(5, —13)

O

x = –2

x

x = 3

ii

d

i

y

(0.5, 0.5) x

O (–0.5, –0.5)

(1.5, –0.5)

x = 1

ii

y

(6, 0.2) x

O (1, –1) x = 2 e

y

i

x = 1

0.75 –1 O

x

3

y

ii

x = 2

x = –2

–0.25 –1 O

x

4

y

5

O

x

(1, –1)

(0, –2)

(2, –2)

6 7

y y = f(x)

y =

y = g(x)

O

EXERCISE 4B

1.5

5

f(x) g(x)

x

1 a

y

i

(0.5, 2)

O

ii

1

x

y (6, 5) (1, 1)

O

b

i

2

x

y

ii

4

–2 O

x

2

x = –1 2 a

y

i

–1 O

x

1

y

ii

4 –2

O

1

2

x

b

y

i

(–1, 4) (1, 4)

x

O y

ii

6

(–2, –2)

c

O

x (1, –2) (2, –2)

y

i

(–1, 4)

(1, 4)

O

x

y

ii

6

–2

3 a

O

1

x

2

y

i

6 –2

O

1

3

x

y

ii

45

–3

O

3

5

x

b

y

i

10

O

–2

5

x

y

ii

O c

1

1

2

x

y

i

y = 2

O

ln 2

x

y

ii

y = 3

d

x

O

–ln 3

y

i

x

O

y

ii

–1.317

O

1.317

x

e

y

i

1.5

O

y = 1 x

3

x = –2 y

ii

y = 1

1.667

x

O

–5

x = 3 y

4

–ab a O b

5 Three. 6 Four. 7 a b

or

x

8 a

y

(1, 3)

y = |f(x)| y = 2

x

1O –— 2

b 9 10

y

11

(p, 2q) y = f(x) + |f(x)|

O 12 a

b

WORK IT OUT 4.1

x

Solution 2 is correct.

EXERCISE 4C 1 2 3 4

( s.f.)

5

for integer and

6 7 8 9 10

,

11

or

12

( s.f.)

13 14

MIXED PRACTICE 4 1 D

y

2 a

x = –5

x = 5

x

O 1 — f(x)

b 3 4

;

y

y = 2 – x

y = x – 2

g(x)

x

4 x = 2

O

y

5

x = 4 1 — f(x) x

O

y

6 a

x = e

y = |4 ln (x – e)|

O b

i ii

1 + e or or

x

7 a

y

i

y = |3x + 3|

3 x

–1 O y

ii

y = |x2 – 1|

1

–1 b

i ii

8 a b 9 10 11

or or

O

1

x

y

12 a

y = |ex – 1|

y = |ln (x – 1)| x

O x = 1 b One. 13 a

b 14 15 a Proof. b

y

16 a

x = –2

1 y = — f(x)

1 y = — 2 O

x

y

b

y = xf(x)

–2

y = 2x

x

O

17 a Asymptote

y

(0, 1) O b 18 a b

y = sech x x

5 Further vectors BEFORE YOU START 1 a

b 2 3 4

WORK IT OUT 5.1 Solution 3 is correct.

EXERCISE 5A 1 a

i

ii b

2 a

i

;

ii

;

i

ii b

i ii

3 a b c d 4 a

b 5 a b 6 a

i

ii b

i ii

7 8 9 a b

( s.f.)

10 11 12 13 a Proof. b 14 a Proof. b 15 16 Proof. 17 Proof. 18 a Proof. b

i ii Proof. iii iv v

EXERCISE 5B 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 For example,

12 a b 13 a b No. 14 a b 15 a b

i Proof. ii

c 16 No. 17 a

b c For example, 18 a b For example,

19 For example,

EXERCISE 5C 1 a

i ii

b

i ii

2 a–d Proof. 3 a

i

ii

b

i

ii 4 5 a b 6 7 a

b 8 a b 9 a b 10 a For example,

b 11 a For example, b Proof. 12 a For example, b 13 Proof; 14 a

b c They form a triangular prism.

EXERCISE 5D 1 a

i ii

b

i ii

2 a b 3 a b

( s.f.)

4 a b

(to the nearest degree)

5 a b

( s.f.)

6 a b

( s.f.)

7 Proof. 8 a b

( s.f.)

9 a Proof. b c d

EXERCISE 5E 1 a b c 2 a b c

3 a b 4 a b c 5 a Proof. b c d 6 a b c d e f 7 a, b Proof. c

d 8 a Proof. b c

i ii

d

MIXED PRACTICE 5 1 For example, 2 3 a b 4 a b

c 5 a b c Proof. d For example, e f 6 a Proof. b

c d e 7 a

b c

i Proof; ii a, b and c aren't in the same plane.

8 9 a b c d 10 a b 11 C 12 a b c

13 a b c d Proof. e 14 a b

i They do not meet; inconsistent. ii Triangular prism.

c

i For example, ii

15 a b c Proof. d 16 a b Proof. c d 17 18 19 a b Proof. c Proof. d

20 a b

i Proof. ii iii

21 a b Proof;

c Proof. d

i Proof; ii

and

are parallel (

not parallel).

: planes form a triangular prism.

6 Further matrices BEFORE YOU START 1 a b 2 3

4 5

;

;

6 a

b c 7 8 a Proof. b 9

. Neither is a line of invariant points.

EXERCISE 6A 1 a

i ii

b

i

ii 2 a

i ii

b

i ii

c

i ii

d

i ii

3 a

i

ii

b

i ii Singular matrix.

c

i

ii

d

i

ii 4

 Proof.

5 6

or

7 Proof. 8 9 a b

or

10 a b

i ii iii

11 12 a b 13 14 Proof.

WORK IT OUT 6.1 Solution 2 is correct.

EXERCISE 6B 1 a

i ii

b

i

ii c

i ii

2 a

i

ii b

i ii

3 a

i ii

b

or or

i ii

4 a Proof. b 5 a i–iii Proof. b 6 a b Proof. 7 8

 Proof.

9 10 a

i

; proof.

ii iii iv b The volume of tetrahedron vectors , and .

equals one sixth of the determinant of the matrix with column

where is the base area and is the perpendicular height.

EXERCISE 6C 1 a

i ii

b

i

ii c

i ii

2 Proof. 3 4 5 a Proof. b c 6 a Proof. b c

EXERCISE 6D 1 a

b

i Consistent; unique solution:

.

ii Consistent; unique solution:

.

i Consistent; line intersection of two distinct planes. ii Consistent; line intersection of three distinct planes (sheaf).

c

i Inconsistent; prism. 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

 Proof. c Triangular prism.

6 a Proof. b c Intersect along a line (sheaf). d 7 a Proof. b Intersect along a line (sheaf). c 8 a

. .

b 9 a b Inconsistent; triangular prism. 10 a b c Intersect in a line (sheaf). d

EXERCISE 6E i and

1 a

b

and

ii and

and

i and

and

ii and c

and

i and ii and

d

i and ii

2 a

and

and

and

i ii

b

i ii

3

and

4

and

and and

5 Proof; and

6

and

,

, and

and

7 a Proof. b 8 a

and

, and

,

and

b

9

and

, and

10 a b

,

and

; proof; i For example,

and

;

ii Proof. c Reflection in the plane 11 For example,

.

and

;

EXERCISE 6F 1 a

i ii

b

i Not diagonalisable. ii Not diagonalisable.

2 a

i

ii

b

i

ii

3 4 Only one eigenvector 5 a

.

and

b

. All vectors in the plane can be expressed as eigenvectors has eigenvalue

through

6 a b

i ii

7 a b

and

and

proof.

.

, and each of the

c Reflection in the line

.

d

8 a

b 9 a Rotation about the origin through b Proof;

and

and

c

MIXED PRACTICE 6 1 C 2 A 3

. .

d Proof.

and

and

4 a b

;

5 a Proof. b 6 Proof. 7 a b Proof. c 8 a b 9 a Proof. b 10 a b Proof. 11 a

b 12 13 a b

;

anticlockwise.

c 14 a Proof. b

c 15 a Proof. b 16 a b Proof. c d 17 a Proof. b

i Consistent. ii Planes intersect at a single point

.

18 a Proof. b c 19 a b c 20 a

for i

and

or and

ii b c Proof.

FOCUS ON … PROOF 1 1 Proof. 2 Proof.

FOCUS ON … PROBLEM SOLVING 1 1

2 The velocity is parallel to 3 Proof. 4 Proof. 5 Proof. 6 Proof.

; proof.

.

7 Proof. 8 9 Proof;

10

; proof.

; proof.

11

FOCUS ON … MODELLING 1 1 Proof. 2 3 4 a b 5 6 7

with

and

with

.

The larger eigenvalue is the growth rate and the ratio of the components of the eigenvector is the ratio of juveniles to adults. 8 9 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. 10 This is the situation described by Fibonacci leading to his eponymous sequence.

CROSS-TOPIC REVIEW EXERCISE 1 1 a Proof. b

.

2 a b 3 Translation by

stretch parallel to -axis with scale factor

y

4 a

x = 2

1 y = — f(x)

1 y = — 3 O

x

.

y

b

y = |f(x)|

y = 3 O

x

2

y

5 a

y = |arsinh x| x

O

b y

6 a

y = |x2 – x – 12| y = |3x – 7|

12

7

O

–3

b 7 a

b 8 a b

( s.f.)

9 a Proof. b c Proof. 10 11 a

b 12 a b Proof;

.

7 — 3

4

x

13 a b Inconsistent; form a triangular prism. 14 a i

; scale factor .

ii Rotation about the -axis through 15 a

i Proof;

.

.

ii

b

i ii

; proof.

iii 16 a Proof; . b

c

and, for example,

is an enlargement in the plane

factor . The line

b Proof. c 18 a Proof. b 19 a, b Proof. 20 a b c

,

d Proof. e i ii b c Proof. d

;

(or equivalently

) with scale

, which is normal to the plane, is a line of invariant points.

17 a

21 a

.

22 a

i Proof. ii

b

i ii Proof;

c 23 a b

; proof. i Proof;

,

ii iii Since 24 a

.

; is a hyperbola that is just a rotation of

must also be a hyperbola.

i ii

,

b Enlargement, scale factor

; Rotation through

c i, ii Proof. iiiIt’s an enlarged rotation of , so it is still an ellipse.

.

7 Further polar coordinates BEFORE YOU START 1

r = 2 – 2cos θ 0, 2π

O

2 3 a b

EXERCISE 7A 1 a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

2 a b Proof. 3 4 Proof. 5 1 —

6 a

r = θ – 2 O

b Proof.

0, 2π

π 2

7 a

π

O

r = 3 sin 2θ

0, 2π

3π 2

b π 4

8 a 3π 4

r = a cos 2θ O

7π 4

5π 4

Tangents: b 9 10 Proof.

EXERCISE 7B 1 a b

i ii Proof.

2 3 a

and

b 4 a b 5 a b Proof.

0, 2π

,

6 a r = 2 0, 2π

O r = 3 – 2 cos θ

b Proof.

MIXED PRACTICE 7 1 B r = θ

2 a

0, 2π

O

b Proof. 3 a

r = 5 – 4 cos θ 0, 2π

O

b 4 a Proof. b

or

c 2π

3

r = 1 + 2 cos θ

O

4π 3 d Proof. 5 Proof. 6 7 a Proof. b

and

c 8 Proof. 9 a

i Proof.

0, 2π

ii b Proof. 10 a

i ii Proof. iii

b

i ii

8 Further hyperbolic functions BEFORE YOU START 1 Proof.

y

2 a

y = cosh x

y = 1 + tanh x x

O

b

solutions.

y

3

y = sinh x y = x

x

O

y = arsinh x

4 a b 5 6 a b

y

7

(–1, 11)

(1, 11)

(0, 3) O

y = 3 – f(2x) 8 a b 9 a

x

b 10 a b c 11 a b c

EXERCISE 8A 1 a

y

i

3

( )

x f(x) = 3 – sinh — 2

O

Domain:

range: y

ii

O

Domain: b

i

x

1

f(x) = 2 sinh (x – 1) x

range:

f(x) = cosh (2x + 3)

y

(–1.5, 1) x

O Domain:

range:

.

y

ii

(0, 3)

f(x) = 4 – cosh x

x

O

Domain: c

range:

.

y

i

y = 7 f(x) = 5 + 2 tanhx 5 y = 3 x

O

Domain:

range:

.

y

ii

y = 4

x

O f(x) = 4 tanh (– x) y = –4 Domain: 2 a

range: y

i

.

f(x) = 3 sinh –1x + 2 2 O

x

Domain:

range:

y

ii

2 x

O

f(x) = 2 + sinh–1(–x)

Domain: b

i

range:

y

f(x) = 1 + cosh–12x

(0.5, 1) x

O

Domain: ii

range:

.

y

( )

x f(x) = 2cosh–1 — 3

O

Domain: c

x

(3, 0)

range:

.

y

i

f(x) = 2 tanh –1(x + 1) + 3 x = –2

x

O

x = 0 Domain:

range:

y

ii

f(x) = 3 tanh –1(x – 2) + 1

x

O

x = 1

x = 3

Domain:

range:

3 4 5

EXERCISE 8B 1 a

y

i

y = 3 f(x) = 3 – 2 sechx

(0, 1) O

Domain:

x

range:

.

y

ii

(–2, 0) x O f(x) = sech(x + 2) – 1 y = –1

Domain: b

range:

y

i

y = 2 O

f(x) = cosech (–x) + 2 x

.

Domain:

range:

.

y

ii

f(x) = cosech( 2 x – 3)

x

O

3 x = — 2

Domain: c

range:

.

y

i

f(x) = 3 coth x – 4

y = –1 y = –7

x

O

Domain:

range:

or

.

y

ii

f(x) = 1 – 2 coth x

y = 3 y = –1

Domain:

x

O

range:

or

.

y

2 a

x = –1 y = 2 sech x y = coth (x + 1)

y = 1

x

O

b

y = –1

solutions.

y

3 a

y = cosech x + 1

y = 2

y = 2 – sech x O

b

(0, 1)

y = 1

x

solution.

4 Proof. y

5 a

y = arcothx

x = –1

y = x y = coth x

y = 1

x

O

y = –1

x = 1

b

i ii

c Proof. 6 a, b Proof.

EXERCISE 8C 1 a–c Proof. 2 3 4

(3 s.f.) (3 s.f.)

5 6 7 8 9 10 Proof. 11 a Proof. b 12 a Proof. b

EXERCISE 8D 1 a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

2 a b 3 4 5 Proof. 6 7 a b Proof. 8 9 Proof; 10 a b Proof.

EXERCISE 8E

.

1 a

i ii

b

i ii

c

i ii

2 a

i ii

b

i ii

c

i ii

d

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

d

i ii

6 a

i ii

b

i ii

7 8 9 a b 10 Proof. 11 a b 12 13 14 a Proof. b 15 Proof. 16 Proof.

MIXED PRACTICE 8 1 B

y

2 a

y = 4

f(x) = 3 tanh2 x + 1 (0, 1) O

b 3 4 Proof. 5 Proof. 6 Proof. 7 8 Proof. 9 10 11 Proof.

x

12 13

y

14 a

1 y = tanhx x

O

–1 b Proof. c

i Proof. ii Proof;

15 a

i

.

y

y = sinh x x

O

y

y = cosh x 1 O

ii Proof. b Proof. 16 a, b Proof. 17 Proof. 18 a Proof. b 19 a Proof. b

i Proof;

x

ii Proof;

.

9 Further calculus BEFORE YOU START 1 2 3 4

EXERCISE 9A 1 a

i ii

b

i ii

c

i ii

d

i ii

2 3 4 5 6 a Proof. b 7 Proof; 8 Proof; 9 a b 10 Proof.

EXERCISE 9B 1 a

i ii

,

.

b

i ii

c

i ii

2 3 4 Proof;

.

5 6 Proof. 7 8 Proof. 9 Proof; 10 a Proof. b

EXERCISE 9c 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 Proof. 6 Proof. 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 Proof. 21 Proof. 22 a b Proof.

EXERCISE 9D 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;

.

EXERCISE 9E 1 a–e Proof. 2 a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

3 a Proof. b 4 a b 5 a b Proof. c 6 a Proof. b 7 a Proof. b 8 a Proof. b c Proof. 9 a Proof. b 10 a Proof. b 11 a, b Proof. 12 a Proof. b c Proof. 13

EXERCISE 9F

1 a

i ii

b

i ii

c

i ii

d

i ii

2 a

i ii

b

i ii

c

i ii

d

i ii

3 4 5 6 a Proof. b 7 8 a Proof. b 9 a b

;

y

1 1 —) + y =— (x – 2 4 2

O

10 a Proof;

2

1

1 — 2

.

b 11 a, b Proof. 12 a Proof. b Proof;

.

13 Proof;

.

EXERCISE 9G 1 a

i ii

b

i ii

c

i ii

d

i

ii 2 3 4 5 6 a Proof. b Proof;

.

7 a, b Proof. 8 a Proof. b

MIXED PRACTICE 9 1 D

x

2 3 4 5 a Proof. b 6 7 Proof. 8 9 a, b Proof. 10 11 Proof. 12 a, b Proof. 13 a Proof; b Proof. 14 a Proof. b 15 a b c Proof. 16 a, b Proof. 17 a, b Proof. 18 a–c Proof. 19 a Proof. b 20 a, b Proof.

.

10 Maclaurin series and limits BEFORE YOU START 1 2 a

i ii

b

i ii

c

i ii

3 a b 4 a b 5 a b

EXERCISE 10A 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

( d.p.) i

; ;

ii b 5 a b 6 a

i Proof. ii

b c 7 a

( d.p.) ;

i

;

ii Proof. b 8 a Proof. b 9 a Proof. b 10 a b

i ii Proof;

11 a

.

but in the diagram the -intercept is not at

b

is an increasing function at

WORK IT OUT 10.1 Solution is correct.

EXERCISE 10B 1 a

i ii

b

i ii

c i, ii Limit doesn’t exist.

.

, but the first derivative of the series is negative at

.

d

i ii

2 a b c d e f 3 a b c d e f 4 a b 5 a b Limit doesn’t exist. c Limit doesn’t exist. 6 7 8 Proof. 9 Proof. 10 a b 11 12 a b c d

EXERCISE 10C 1 a b c Integral diverges. d

2 a b 3 Proof; . 4 5 6

MIXED PRACTICE 10 1 D 2 a, b Proof. 3 4 a

;

; ;

b c 5 a Integrand isn’t defined at

.

b Integral doesn’t have a finite value. 6 a b Proof. 7 ;

8 a

;

b c 9 a b

i ii Proof.

c 10 11 Proof. 12 a b c Proof. 13 a

i Proof. ii

b Proof. c

;

; 

d

i ii

14 a

i Proof. ii

b

i Proof. ii

c

11 Differential equations BEFORE YOU START 1 2 3

EXERCISE 11A 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

b 7 a b c

EXERCISE 11B 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 11 a Proof. b

EXERCISE 11C 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 11D 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 a Proof. b

MIXED PRACTICE 11 1 A

2 B 3 a b 4 a Proof. b 5 a b Proof;

.

c d 6 a b 7 a b Proof; c d 8 9 10 a b Proof. c 11 a b 12 a Proof. b 13 a b Proof. c 14 a b

.

12 Applications of differential equations BEFORE YOU START 1 2

EXERCISE 12A 1 a Proof. b 2 a, b Proof. 3 a

b

I

I = e–kt (Asinbt + B cosbt)

O

t R k = –— L CL – R2C2 – b =— CL

c

4 a b c A natural net birth rate of the population. 5 a b

c

( s.f.)

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 external people spreading the rumour, students mix freely and do not get grouped. 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 V

c

8 t —

–3 3 V = (2 – e )

1

t

O

As

EXERCISE 12B 1 a

i

; at rest.

ii

; equilibrium.

b a b c a b 2 a

i ii

b

i ii

c

i ii

3 a

i ii

b

i ii

; equilibrium. ; at rest. ; equilibrium. ; at rest.

c

i ii

d

i ii

4 a b 5 a b 6 a b 7 a b 8 a b c

; away.

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;

.

15 a b

i Proof; ii, iii Proof. iv

16 a Proof. b

i Proof. ii

( s.f.)

.

iii 17 a, b Proof. c d e 18 a Proof. b

EXERCISE 12C 1 a

i Light. ii Heavy.

b

i Critical. ii Light.

c

i Heavy. ii Critical.

d

i Critical. ii Light.

2 3 a Proof. b c Heavy. 4 a b 5 a b Light. x x = 0.9e–2t x = 0.9e–2t sin 3t t

O

x = –0.9e–2t

6 a Proof. b c Proof. d Heavy damping. 7 a Proof. b c 8 a

b Proof. c Light; d e

( s.f.)

9 a Light. b c Proof. 10 a Proof. b c 11 a–c Proof. d

EXERCISE 12D 1 a

i

;

ii

b

i

;

; ,

ii

; ,

c

i

; ,

ii

; ,

2 3 4 5 a b 6 a Proof; b c

. ; proof.

7 a

;

b c

,

d

; first shark peak at

, first fish peak at

.

e The populations oscillate with the same period ( time units), and with a phase delay of units of time between fish population peaking and shark population peaking. 8 Proof.

MIXED PRACTICE 12 1 C 2 D 3 a b 4 a b c 5 a Proof. b Proof;

.

c d 6 a Proof. b 7 a Proof. b Proof; c 8

.

seconds months ( s.f.)

9 10 Particle

.

11 a Proof. b c 12 a Proof. b Proof;

.

13 a, b Proof. c 14 a b

seconds

c d The second model (after 15 16

seconds). ,

17 a Proof. b Proof; . c 18 a

i Proof. ii

b

i Proof. ii iii

x

t

O

Heavy damping. 19 a Proof. b c d 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

;

d Bill (Mike’s distance is but you don’t need to find this. A sketch of both forces makes it clear that Mike is never pulling harder.)

13 Numerical methods BEFORE YOU START 1

( s.f.)

2

( d.p.)

3 4 5

EXERCISE 13A 1 Answers are given to three decimal places. a

i ii

b

i ii

c

i ii

d

i ii

2 Answers are given to three significant figures. a

i ii

b

i ii

c

i ii

d

i ii

3 i, ii

Exact

Approx.

% error

Approx.

% error

a b c d

Approx.

% error

Approx.

%error

a b c d iiiIn most cases, doubling the number of intervals reduces the error by between a half and three quarters. 4 a

y x = 1

O

b

2

y = 4 ln (x – 1)

3

4

5

x

( d.p.)

c Overestimate because the curve is concave. 5 a

( d.p.)

b Use more intervals. 6 a b

( d.p.)

c Overestimate because the curve is concave. 7 a b

( d.p.)

EXERCISE 13B 1 Answers are given to three decimal places. a

i ii

b

i ii

c

i ii

d

i ii

2 Answers are given to four significant figures. a

i ii

b

i ii

c

i

ii d

i ii

3 i,ii

Exact

Approx.

% error

Approx.

% error

a b c d

Approx.

% error

Approx.

% error

a b c d iiiIn most cases, doubling the number of intervals reduces the error by a factor of to y 4 a

y = 2 ln (x2 + 1)

O b

1

2

( d.p.)

5 a

( d.p.)

b Use more intervals. 6 a b

( d.p.)

7 a b

i Proof. ii

c

( d.p.) ( d.p.)

EXERCISE 13C

3

4

x

(or more).

1 a

i ii

b

i ii

c

i ii

2

( d.p.)

3 a

( d.p.)

b c

i ii Use smaller .

4 a

i

( d.p.)

ii Use smaller . b

;

( d.p.)

c

The tangent is always below the curve (curve is convex). 5 a b

( d.p.)

EXERCISE 13D 1 Answers are given to three significant figures. a

i ii

b

i ii

c

i ii

d

i ii

2 a

( s.f.)

b

( s.f.)

3 a

( s.f.)

b

( d.p.)

c Part b: It is a convex curve, so the improved Euler formula is more accurate than the Euler method. 4 a b 5 a

( s.f.)

b c

( d.p.) i ii

MIXED PRACTICE 13 1 D 2

( d.p.)

3

( d.p.)

4

( d.p.)

5

( d.p.)

6

( d.p.)

7 a b

( s.f.)

8 a

( d.p.)

b

( d.p.)

9 a

( d.p.)

b

( d.p.)

10 B 11 a

( d.p.)

b

( d.p.)

c d Use more strips. 12 a Proof. b Curve is symmetrical about -axis so estimate for and double. More accurate; effectively using strips for the whole interval instead of . 13 a Proof. b

for quadratic and, by part a, the Simpson’s rule estimate of the first two integral values will be exact and Simpson’s rule will fit a curve exactly to any quadratic so the third term is also exact. 14 a

i

( d.p.)

ii b A smaller step will lead to a more accurate answer. 15 a

( d.p.)

b 16 a

( s.f.)

b 17

( s.f.)

18 19 a b c 20 a b

( d.p.) i

ii iii

( d.p.)

FOCUS ON … PROOF 2 1 Proof. 2 3

FOCUS ON … PROBLEM SOLVING 2 1 Proof. 2 The tangent to the chain at 3

is horizontal; proof.

moving the chain vertically does not change its shape.

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 OK 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 goes down. The equilibrium value of the sharks goes up when increases.

S

4 3 2.75 2.5 2.25 2 1.75 1.5 1.25 1 0.75 0.5 0.25

O

0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3

Maximum fish population is about

F

million.

F

5 1.75

0.5

O

t

S

1.75

1 0.5

O

6 When is small the

t

term will be relatively more important. When is large the

term will

dominate, meaning that when the population is too large there is a net death, as would be expected with internal competition.

S

7 3 2.75 2.5 2.25 2 1.75 1.5 1.25 1 0.75 0.5 0.25

O

0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3

The systems tends to an equilibrium at

CROSS-TOPIC REVIEW EXERCISE 2 π 2

1 a

r 2 = a sin 4θ

3π 4

π 4

π

0, 2π

O

5π 4

7π 4 3π 2

b 2 a Asymptotes

y

b

y = 1

1 y = tanh x x

O y = –1 –1 c Proof. d 3 a b

F .

4 a b

i Not finite. ii Finite; .

5 6 a Proof. b

seconds

7 a b 8 a

( s.f.)

b Smaller. 9 a

( d.p.)

b

( d.p.)

10 a

i ii Centre

b

, radius

.

i ii Proof. iii

11 a Proof. b 12 a Proof. b 13 a Proof. b 14 a Interval of integration is infinite. b 15 a, b Proof. 16 a b 17 a Proof; b Proof; 18 a

b c 19 20 a i, ii Proof. b Proof.

. .

c 21 a b

( s.f.)

c d

( s.f.)

22 a b

; i ii iii Angle

(since Pythagoras theorem holds) so

c 23 a, b Proof. 24 Proof. 25 a Proof. b 26 a b Proof. c

i ii

27 28 a b Proof. 29 a Proof. b

30 a Proof. b i, ii Proof. c

i ii Critical.

31 a b

i Proof. ii

32 a

i Proof;

.

ii Proof; b

i Proof. ii

PRACTICE PAPER 1 D

.

is a tangent.

2 C 3 4 Proof. 5 6 Proof;

.

7 a Proof;

.

b Translation by c

.

y

i

|

|

1 — y = a + |x| – a

y = a

1 a – — a O

1 – a — a x = –a

x 1 a – — a x = a

ii 8 a Proof. b Proof. c 9 a Proof. b 10 a Proof. b 11 a

i Proof;

.

ii

b

is a line of invariant points.

and 12 a

i

are invariant lines.

ii b c 13 a Proof. b 14 a Proof. b

i ii Heavy.

Worked solutions 1 Further complex numbers: powers and roots Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 1A 2

a Using De Moivre’s theorem:

Im

b

3 1 z

z4

z2 z

O

–1

1

Re

–1

3

a

i Using De Moivre’s theorem:

Im

ii z, z4

O

–1

z31

Re

z2

b Since 4

, it follows that

for all integers and so

for

a (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

The smallest such is 7

for some positive integer .

Using De Moivre’s theorem:

Require

, so

The smallest such is

EXERCISE 1B 5

So

6

a

for some positive integer

.

b

( s.f.)

7 8 9

Complex roots of a polynomial with real coefficients occur in conjugate pairs. The three roots are therefore ,

and

.

Tip Instead of expanding the factorised form, consider using the formulae you know for the coefficients: sum of roots, sum of root pair of products and product of roots. 10 Complex roots of a polynomial with real coefficients occur in conjugate pairs. The four roots are therefore

and

.

The polynomial is

.

11 12

EXERCISE 1C 3

If

, then

One solution is 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 ,

:

,

Im

c

z2 2 1 z1 –2 z3

–1

O

1

2

Re

–1 –2

8

z4

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

b

:

so Then

.

Rearranging: 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 ,

(or, equivalently,

:

)

b Using the binomial expansion:

c Comparing with part b: and so Then using part a: or In Cartesian form: or So

or

EXERCISE 1D 3

a

so

,

,

,

and

Im

b

1

w1

w2

Re 1w0

O

–1 w3

–1

4

w4

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.

: 5

so is False.

a The seventh roots of unity are

,

.

b No; on an Argand diagram, the points form a regular heptagon. Since there is 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 . Im z2 z1 z3 z0 Re

O z4

–z1 z6 z5

so

c

Smallest positive integer

.

d Since the roots of unity form a regular polygon that 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

.

Using part a: Rearranging: Substituting the solutions for : When

, is not defined.

When

,

When

,

8

, so Then

, and

9

a The fifth roots of unity are

and

where

.

b Rearranging: From part a: the solutions are However,

for

.

must be excluded, since

has no solution.

Rearranging: and so c

(excluding

for

d Using the binomial theorem to expand

:

e But from part c, the four roots to this quartic are Since

and

, these could also 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:

. .

Im

10 a

1

ω2

ω –1 3

ω1

ω Re 1

O

ω4

ω5

–1

b i

so

and

Then and

ii

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 1E 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

for

and

:

,

,

,

Therefore, the quintic can be factorised as:

4

a The solutions to b Then

are

can be factorised into linear complex factors:

,

.

Therefore, substituting

:

So 5

a Using the quadratic formula to solve

:

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

, of which only the first is real.

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:

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 1F 1

a

and

( s.f.)

and

( s.f.)

b The scale factor of the enlargement is radians ( 2

and the angle of rotation is

) ( s.f.). and

a b

and

( s.f.)

The transformation from to is a rotation through the origin. 3

A rotation through

radians (

) ( s.f.) about

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 with scale factor

equivalent to multiplication of the corresponding complex number by So corresponds to the complex number

corresponding complex number by

.

, equivalent to multiplication of the

.

So corresponds to the complex number 5

.

and has coordinates

is produced by a rotation of about the origin through

,

and has coordinates

.

a Using trigonometry:

b

is achieved by rotating through and enlarging by the corresponding complex number by

.

, which is equivalent to multiplication of

So the coordinates of are 6

.

lm

a

z2

1

z

z3 –1

O

1

Re

–1

( s.f.)

b So a rotation through 7

( s.f.) about the origin maps to .

a b Rotation through

on the Argand diagram is equivalent to multiplication by

corresponds to

.

so has coordinates

corresponds to

.

so has coordinates .

8

a The line has equation

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 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

.

10 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

, then rotated to become

(so that the centre is now

and then translated again to

.

. That is,

and so

.

b In the terms given in part a: ,

so

,

So has coordinates 11 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

about the origin is equivalent to multiplication by

So after the first rotation, the point corresponding to Then, using part a, with centre of rotation rotation

, so

so that

.

has image

.

, initial point

and angle of

, the final image is given by where:

So the final image has coordinates (

,

).

12 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 1

.

so

1 Then 2

(Answer A)

a (adding because b Then

3

a If

, and so , then

for

Im

b

1

z1

z2

–1

Re 1z0

O

z3 –1

z4

4 (subtracting integer multiples of 5

to ensure the argument is in the required interval)

a

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 some . If

is a solution, then

So the equation is 7

for

.

.

a b If

, then

The primary solution is therefore So

for

, with further solutions at multiples of

.

(requiring the argument to be given within the interval)

is equivalent to the statement

8

. , so

, so the positive root is the one required. (Answer C) 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 If is a complex third root of unity then

.

Then But since

, it follows that

b From part a:

12

for some Then 13

, so

.

.

14 a b Then 15 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

c

and since Then

, it follows that

(that is,

)

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

16 a If is a complex third root of unity then

.

Then But since b

, it follows that

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

.

17 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

.

18 a (adding because So b

i If

, then

.

The primary solution is therefore Then

, with further solutions differing by powers of

.

or

ii The points form an equilateral triangle centred at the origin. Dividing the triangle into three equal isosceles triangles similar to

But

So c

and angle

:

, so

. is one of the complex third roots of unity, so

The sum of the three roots of the cubic In Cartesian form, the three roots are:

. are

.

Then, since

as shown earlier, it follows that

Substituting:

.

as required.

19 a Rotation through about the origin on the Argand diagram is equivalent to multiplication by

So point corresponds to b

.

Then point corresponds to

20 a The vectors

and

.

, so

Then b

Im

B b

O

θ

a

A

Re

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 2 Further complex numbers: trigonometry Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 2A 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

,

,

.

b

The solutions are 6

.

a Using the binomial theorem:

So and b Also, using De Moivre’s theorem:

Equating real and imaginary parts:

Dividing numerator and denominator on the RHS by

:

7

a Using the binomial theorem:

So and Also, using De Moivre’s theorem:

Equating real and imaginary parts:

Dividing numerator and denominator on the RHS by

b Rearranging, the sextic given is equivalent to

So

8

for

:

, using the formula in part a. Then

(further solutions lie outside the required interval)

a Using the binomial theorem:

So and b Using De Moivre’s theorem:

Equating imaginary parts:

c Using the approximation

and disregarding powers of and higher:

EXERCISE 2B 1

a b

, so

c Solutions to

:

or

, so

or

Solutions to the quadratic equation in part a:

Of the two solutions

or

, only

gives

.

So 2

, so

a b Solutions to So

or

from which

:

for integer

within the interval

.

c Solutions to the quadratic in part a: Since 3

,

a Dividing both sides by

Solutions to

. :

for which , so

Within the interval b Since

: , but

, the solutions are

.

, it follows that the quartic factorises as:

where

and

the domain

.

Comparing coefficients: , so

, both of which are positive, since cosine has a positive range in

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:

So:

on the RHS:

.

,

and

.

Then if

:

Expanding: Therefore 6

a Writing

as required. :

So when

,

Solutions to

or :

so

The solutions to which . Since

. 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

, simplifying:

So

EXERCISE 2C 2

, so using De Moivre’s theorem:

a

and Then

.

b Using the binomial theorem:

, 3

, ,

a So

.

But also, using the binomial expansion and collecting terms for the formulae shown:

Therefore: So

.

b

4

a

so

i Therefore

.

ii Using De Moivre’s theorem:

and Then b

.

i Using the binomial theorem:

ii From part a ii: From part b i:

Equating the two expressions:

So ,

,

c Using

5

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:

So

6

a b

Then

EXERCISE 2D 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:

is a geometric series with common ratio

3

Since

, the series converges to:

Substituting

4

:

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 starting 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 2 1

a Using the binomial theorem:

b Also, using De Moivre’s theorem:

Equating the imaginary parts:

, 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:

,

,

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

or

for integer value . .

Therefore, the five distinct roots of the quintic are:

5

and

, so

for values of for

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

a If

, then

Hence

, and therefore

b Using De Moivre’s theorem:

Substituting

:

Then 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 the double angle formula for cosine:

So

or and using or

8

, the four values for are:

.

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

, rearranging:

must be a solution to the this cubic.

d Let Since factor of

, by the factor theorem

Tip Or observe that in the working of part c it is shown that Factorising the cubic: Since

is a factor of

, then one root of

The other two root values are e In the interval Since f 9

a

Therefore

, the curve , it follows that .

i Using De Moivre’s theorem:

is

.

. is strictly increasing. .

is a solution.

is a

Expanding, using the binomial theorem:

Equating the real parts of these two forms:

Equating the imaginary parts similarly:

ii Taking the ratio of these two results and dividing numerator and denominator by

b Taking

and writing

:

in the formula in part a ii: (1)

But since

, this gives

So when

,

The other three roots of the equation must also have the form where is a solution to (excluding , which is accounted for in the factor in the numerator of the formula in (1)). has solutions

for integer .

The four distinct non-zero values for

are therefore:

. c Since

, these roots can be written as

or

where

and

. The product of the roots equals the ratio of the constant coefficient to the lead coefficient. So

.

Then

, so

(since both

and

have positive values)

Tip Take care to observe explicitly that you know the sign of the results, or the square root step is not rigorous, since in the absence of other information it would allow the possibility that . 10 a De Moivre’s theorem states that Then

and

for integer . .

Taking the sum of these: b

.

i Using the binomial theorem:

ii Pairing up terms and applying the formula from part a:

But also, Equating these:

So

,

,

.

c From part b:

, so the given condition means that

. The solutions to this are: , so Solutions in the interval d Using the formula from part b:

are

.

, so

Worked solutions 3 Further transformations of the ellipse, hyperbola and parabola Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 3A , so

1

for

.

The parameterisation shows that the equation has the form of a parabola. 2

and

, so

.

This is a parabola with vertex where , so

3

, so the vertex is at

.

.

y 5

–4

O

x 2 y2 — + = 1 — 16 25

4

x

–5

, so

4

This hyperbola has asymptotes

. , alternately given as

.

5 So

6

, which is a rectangular hyperbola passing through

so

.

with asymptotes

.

This is the ellipse symmetry

after a translation

and

, so has centre

and therefore axes of

. , so

7

which is the parabola

after a translation

. The vertex is therefore at and

8 Then

,

, so that

.

, which is the rectangular hyperbola and

9

,

Then

, so that

.

, which is the ellipse and

10 a

,

Then

.

, so that

, so that

.

. .

(1) Completing the square:

So

(2)

In standard format: b This is the hyperbola

after a translation

The centre is therefore at

.

and the curve has asymptotes

.

Axes intercepts: from

: when

,

from

: when

,

and

11 Then

, so

, , so that

or .

, so that

. .

(1)

Completing the square: (2) In standard format, dividing through by This is the ellipse

:

after a translation

, so has axes of symmetry

and

.

12 a Using implicit differentiation: so At

, the gradient is

.

Then the tangent equation is

.

Rearranging to the required form: b

has equation (after completing the square):

This is curve

after a translation

.

c The same translation maps point after the translation:

to

, so the tangent line at

is

Rearranging: 13 a Horizontal stretch scale factor 5: Replace by . The ellipse has equation b At

.

, the tangent to the circle has gradient

(perpendicular to the radius from the origin to

). The tangent equation is c The horizontal stretch maps the stretch is applied:

. to

, so the tangent at

is the previous tangent after

Rearranging:

14 a Using implicit differentiation: , so

.

Then the general tangent at point . Rearranging:

has tangent with gradient and therefore equation

For this to pass through the origin, require the constant term to equal zero: , so

and so

.

Then the tangents have gradients b Parabola

is given by

Then

,

and so have equations

.

.

, which, to compare with the equation for

, can be given as

. This is the original curve after a horizontal stretch with scale factor .

Tip There are several rearrangements available here. Alternatives might include arranging the equation for as , which could be interpreted as a vertical stretch with scale factor combined with a translation

. But, for part c to be aided by

the answer to part b, you need to find a transformation that comprises exclusively stretches, not translations (because you still want tangents passing through the origin). c Applying the same stretch to the tangents found in part a, the new tangents passing through the origin are: .

EXERCISE 3B 3

Rotation

anticlockwise about the origin: replace with and with

New equation: 4

Rotation

anticlockwise about the origin: replace with and with

New equation: Rearranging: 5

Enlargement scale factor : replace with and replace with . New equation:

6

a

y y2 = 5x

O

x

.

b Enlargement scale factor : replace with and replace with . Equation of c From

to

: :

replace with , which is the result of a horizontal stretch with scale factor . 7

Dividing the second equation through by

to facilitate comparison:

becomes has been replaced by

and has been replaced by

centred at the origin with scale factor has been replaced with

8

, which is the effect of an enlargement

.

and vice versa.

This could arise from a rotation about the origin through through the line . The original curve has asymptotes 9

(in either direction) or a reflection

, so the new curve has asymptotes

.

a

b The curve has asymptotes

, so

Enlargement with scale factor : replace with and replace with . Applying this transformation to the asymptote equations: is equivalent to the original asymptote equations.

Tip Alternatively, observe that any line through the origin is invariant to an enlargement centred at the origin. 10 General equation of a hyperbola in standard orientation:

So

is a hyperbola with asymptotes

Applying a reflection through

: exchange and in the equation.

This produces the given curve equation, which must (applying the same transformation) have asymptotes

.

EXERCISE 3C 3

Initial parabola Translation

has vertex at the origin. :

replace with

.

New equation:

has vertex at

.

Enlargement scale factor : replace with and with . has equation:

, which has vertex at

Rearranging: 4

, which has vertex at

Translation

:

replace with New equation: Enlargement scale factor : replace with

and with

New equation: Rearranging: Comparing with

5

:

Rotation through

:

replace with and with New equation: Translation

:

replace with

and with

New equation: Comparing with translation is 6

Rotation through

: . : replace with

and with

. .

New equation: Translation

: replace with

and with

.

New equation: Comparing with translation is 7

: .

a

b Translation

: replace with

.

New equation: Enlargement scale factor : replace with

and with

.

New equation: Rotation through

: replace with and with

.

New equation: 8

a

is a hyperbola, with asymptotes

b Translation

: replace with

, alternatively written as .

New equation: Enlargement scale factor : replace with

and with

New equation: Rotation through New equation: c Asymptotes of Rearranging:

:

: replace with and with

.

.

.

9

a The curve

, alternatively written as

The curve asymptotes.

, has asymptotes

and

is this same curve after an enlargement with scale factor

The curve

is then obtained by a translation

. , and has the same

, so has asymptotes

and

. b A further enlargement with scale factor , centre origin: replace with

and with

New equation: Rearranging to the required curve:

10 a Replace with

and with

.

Horizontal stretch with scale factor and vertical stretch with scale factor . b The unit circle has area so the ellipse has area

.

, so

11 a

Curve has equation b From

. .

, which can be written as

:

Replacing with

(vertical stretch, scale factor

Replacing with

(translation

12 Translation

: replace with

):

):

, which is curve .

.

New equation: Vertical stretch scale factor : replace with . New equation: Asymptotes are

.

Rearranging: 13 a Start from

, which can be rearranged to

.

Vertical stretch with scale factor : replace with . New equation: Translation

: replace with

.

New equation: Expanding: b The original asymptotes are

, which can also be expressed as

.

After the vertical stretch: After the translation: 14

Reversing the transformations from the new curve and rotating

through

clockwise: replace with and with Midway equation: From

to

replace with

:

and with

15 Translation by vector

. This is a translation

: replace with

and with

.

.

New equation: Horizontal stretch with scale factor : replace with

.

Vertical stretch with scale factor : replace with .

New equation: Rearranging to match the format of the given equation:

Comparing with the given equation: ,

,

and

, so

.

So the transformations are a translation vertical stretch with scale factor .

MIXED PRACTICE 3 1

Enlargement with scale factor : replace with and with . New equation: Rearranging: Comparing:

Tip

so

(Answer C)

then a horizontal stretch with scale factor and a

Although is also a solution to , an enlargement with negative scale factor is usually interpreted as a combined transformation of reflection through both axes (or rotation of ) together with an enlargement. 2

a

b Rotation through

: replace with and with

.

New equation: Horizontal stretch with scale factor : replace with

.

New equation: 3

Enlargement with scale factor : replace with and with . New equation: Rearranging: Translation

: replace with

and with

.

New equation: The vertex of the original curve is at the origin; following the same transformations, the vertex remains at the origin after the enlargement, and moves to under the translation. 4

Enlargement with scale factor : replace with and with . New equation: Rearranging: Translation

: replace with

and with

.

New equation: The asymptotes of the original curve are

, which are unaffected by the enlargement.

Under the translation, the asymptote equations become equations and . 5

Reflection through

: replace with and with .

New equation: Horizontal stretch with scale factor : replace with .

which simplify to the line

New equation: Rearranging: The intersection of the two curves can be found by substituting one into the other: so Then 6

,

or

or

a Substituting

,

.

, so the intersections are at

and

.

into the circle equation:

so

.

The intersections are at

and

.

b Rearranging the original circle equation:

Vertical stretch with scale factor : replace with

.

New equation: Translation

: replace with

New equation:

.

as required.

The transformations are a vertical stretch with scale factor and a translation

.

gives the line so the intersection c Applying the same transformations to the line points will be the image points of the answers to part a under the transformation. is mapped to

and then to

is mapped to

and then to

The new intersection points are at 7

Translation

.

: replace with

.

and

.

and with

.

New equation: Rotation through

: replace with

and with .

New equation: Rearranging: 8

(Answer A)

Original curve has equation Translation

: replace with

New equation: Rotation through New equation:

and vertex

.

and with

.

, with vertex at : replace with

and with

, with vertex at

.

9

a Form a quadratic for the -coordinates of the intersection between line and circle. Substituting

:

The only intersection is when

, so the line is tangent at

Tip Alternatively, show that the discriminant of the quadratic is zero, so that there is only one solution. b

Tip Either solve the question from first principles, exactly as for part a, or use transformations. Both methods are given here. Method 1: Direct solution Form a quadratic for the -coordinates of the intersection between line and ellipse. Substituting

:

For the line to be tangent, require a single root to this equation, so the discriminant equals zero:

so

.

The question calls for the positive value, so

.

Method 2: Transformation Transforming the circle replace with translation replace with

to the ellipse

:

: ; :

horizontal stretch with scale factor

.

Also, possible reflection through -axis (replacing with Applying the same transformations to the tangent in part a: After the translation,

becomes

.

).

The stretch transforms this to

, so

.

The reflection, which is possible but not necessary, transforms this to the negative value

which gives

, rejected by the question.

Tip The question and topic clearly are looking for you to use the second method, but it is far less reliable; any number of transformations could have been selected to shift the circle to the ellipse, but only a few will transform the initial tangent to a tangent of the desired form. You could spend a long time trying to get the suitable combination of transformations, so the direct method is preferable unless you are specifically instructed to do otherwise. 10 a Using substitution

to find points of intersection:

For the line to be tangent there must be a single point of intersection, so the discriminant of this quadratic is zero:

so b Translation

: replace with

and with

.

New equation: Enlargement with scale factor : replace with

and with

.

New equation: Rearranging: c Neither an enlargement nor a translation will change the gradient of a line. So under the transformation, the tangent with gradient image also of gradient .

will be transformed to a tangent to the

After the translation: After the enlargement: Rearranging: the new tangents are and

11 a Then

,

so , so that

.

Rearranging:

so

Completing the square: In standard format: If a single simple transformation maps the ellipse

to this curve, the transformation

must be a translation, and so the denominators are the same in the two equations. , b Replace with

: translation

.

12 Putting the equation into standard format:

The asymptotes are

.

Rearranging: , so the asymptotes are

and

. (Answer B)

Tip Notice that an alternative approach is to read off the centre-point of the hyperbola as , which must be the point of intersection of the two asymptotes. Substituting into each of the proposed line equations eliminates all options except the correct one.

13 Rotating So

through

has equation

anticlockwise: replace with and with

.

.

Since the new ellipse could also have been obtained by reflecting either through line (exchanging and ) or through line (replace with and with ), the four intersection points of and must all lie on these lines. Substituting

into curve , so

: , and therefore

It follows that the side length of the square is therefore

.

and then the area of the square is

.

Tip The question can be solved by using substitution to find the coordinates of a point of intersection rather than the given geometric argument, but the working is unnecessary. A thorough, well-reasoned argument is an acceptable alternative. 14 a So

b Starting from

:

Replace with and with : reflection in the line

.

New equation: Replace with

and with

: translation

.

New equation:

Tip There are very many valid alternatives. For example, any have the same effect as the reflection in the line

rotation will – in this case –

. A translation

followed by the

reflection in would also be valid; using rotations instead of the reflection would require a different translation first. c

is a hyperbola with asymptotes

, passing through vertex

After the reflection, the asymptotes are

( is specified for

and the vertex is

After the translation, the asymptotes are

, which rearrange to

. The vertex is

.

and

15 a

,

, so that

.

Then Rearranging:

, so

Completing the square: In standard format: b From Replace with

to the circle : translation

: .

).

.

and

New equation: Replace with

: vertical stretch with scale factor

.

New equation: Curve is transformed to the circle by a translation

and a vertical stretch with scale factor

. c The circle has radius

so has area

The area before the stretch is therefore so the area of is

.

. , and the area is unaffected by the translation,

Worked solutions 4 Further graphs and inequalities Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 4A 4

Vertical asymptotes where

,

Maximum at minimum of

, point is

Minimum at maximum of

, point is

and . .

y x = 1 x = 3

(0, —13) (2, – —13)

O

x

x = –2

5

Reciprocal transformation followed by translation No vertical asymptotes as Maximum at Roots of and

.

.

. occur at the asymptotes of

and

, so the new curve passes through

. y

O (0, –2)

6

x

(1, –1) (2, –2)

has a single vertical asymptote at Therefore, the denominator must be

, so the quadratic denominator only has a root at .

.

has a horizontal asymptote at

, so the ratio of lead coefficients

has a single vertical asymptote at numerator of , has a single root at

, so the quadratic denominator of .

Therefore, the numerator must be

7

Roots at roots of

,

, which is the

.

.

Asymptotes at roots of As

.

.

from above.

y y = f(x)

y =

y = g(x)

O

1.5

5

f(x) g(x)

x

EXERCISE 4B 4

Positive quadratic with roots at and with the -axis. y

–ab a O b

5

x

Plotting the two graphs on the same axes:

, with the part of the curve with

reflected in

y y = |x2 – 5x|

y = |lnx|

O

1

x

5

Three intersection points. 6

Plotting the two graphs on the same axes: y y = |x2 – 1|

y = |sin x|

–1

O

x

1

Four intersection points. 7

a Completing the square: , so the graph has a minimum value of b

will have two solutions at , four solutions for and two solutions again for . No solutions for . So

8

a

.

or

, four solutions for

gives two solutions.

y

(1, 3)

y = |f(x)| y = 2

1O –— 2

b A horizontal line 9

x

will intersect the graph in part a three times for

The question is equivalent to solving

.

A positive quadratic is less than zero between the roots. Solution:

.

10 The range of

(for

So the range for

is

) is .

11 The required graph will be:

(The graphs will ‘cancel’ when y (p, 2q) y = f(x) + |f(x)|

O

12 a

b From the graph:

EXERCISE 4C 1

x

.

.)

Intersections are on the parts of the graphs representing so

, from which

.

2 y y = |x2 – x – 6| y = 3 – x

O

–2

x

3

Intersections are at

and one additional intersection with each part of the quadratic curve. , so the additional root on the outer curve is at

.

, so the additional root on the inner curve is at From the graph, the interval for which

is

.

.

3 y

y = |x2 – 2x – 8|

y = |x – 3|

–2 O

x

4

Intersections are at and one additional intersection with each part of the quadratic curve and the left wing of the linear modulus graph , so the additional root on the outer curve is at

.

, so the additional root on the inner curve is at From the graph, the intervals for which

are

,

4 Algebraic solution: restricting solutions to

, squaring both sides:

(Both solutions are valid as both are positive.) or

( s.f.)

and

. .

5 , so

or

at

.

or

at

for integer .

or

The inequality is true for all

for integer for integer and

.

6 for

, so

and therefore

.

for

, so

and therefore

.

Solutions: 7

or so y

y = |2x + 1|

y = |x – 3|

1O –— 2

3

x

Both intersections are with the left arm of the red graph:

.

and and From the graph, the region of interest lies between the two intersections, so the solution is . 8

, so

y y = |5x – 7|

y = |2x + 2|

–1O

7 — 5

x

Both intersections are with the right arm of the red graph:

.

and and From the graph, the region of interest lies outside the two intersections, so the solution is . 9

or or 10 The graph of

is the same as:

Simplifying:

Intersections in the first and third sections of the three-part graph.

or

Solutions are

or

.

11 So

or

.

The full domain of

is

gives

, so the solution is a subset of that domain. .

gives

.

So the solution is

or

.

12 So y y = cosh x – 3

y = 1 x

O y = –1

gives gives

, so

is an even function).

, so

Solution is As decimals:

(because

or or

. ( s.f.)

13

Tip This question can be solved either by using a graph and algebra by intervals or by direct algebraic calculation. Both methods are given here, and either would be acceptable in an examination. Method 1: Graphical identification of interval

y y = |x – 2q2| y = |x + q2|

–q2 O

x

2q2

Intersection of

and

occurs in the interval

.

Method 2: Algebraic solution using squaring to eliminate modulus signs

14 Squaring both sides:

Then

; that is,

.

MIXED PRACTICE 4 1

has minimum point Therefore,

has a maximum at

solutions for solutions for solutions for solutions for solutions for

, and so

. has:

.

(Answer D) 2

a The roots of

are asymptotes of

The asymptotes of

Minima

for

are roots of

become maxima

. no roots.

for

.

y x = –5

x = 5

x

O 1 — f(x)

b 3

Graphs of

and

y

y = |2x – 1|

O

y = x

1 — 2

x

One intersection with each arm of the modulus graph. and and From the graph, the region of interest lies between the two intersections, so the solution is 4 Vertical asymptotes at roots of the denominator:

Oblique asymptotes

and

Roots at roots of the numerator:

. and

.

.

.

y

y = 2 – x

y = x – 2

g(x)

5

x

4 x = 2

O

The roots of

are the asymptotes of the reciprocal:

as

.

, so the reciprocal tends to zero from below.

as for

and

, so the reciprocal tends to zero from above. and

, so these are the negative regions of the reciprocal.

y x = 4 1 — f(x) x

O

6

has vertical asymptote

a y x = e

y = |4 ln (x – e)|

O

b

x

1 + e

when

i So

;

or

. .

ii From the graph and part b i: for

or

and -axis intercept at

.

7

a

y

i

y = |3x + 3|

3 x

–1 O y

ii

y = |x2 – 1|

1

O

–1

b

x

1

i , so Solutions are

or

.

or .

ii From the diagrams: outside the intersection points, so for 8

a Graphs of

or

and

From the diagram, the intersections are between: and , so and , so Solutions:

or

b From the diagram, the required interval lies between the intersections:

9

,

.

Tip Remember to check when rearranging that you are not allowing a false solution; multiplying by an expression is fine but you must account for the possibility that the expression equals zero! Graphs of

and

y y = |2x – 4| y = |x + 1|

–1 O

x

2

Two intersections, both with , so , so Then

over the interval

So

for

or

.

.

with

10 Graphs of

and y

y = 3|x + 3|

y = |x – 1|

–3

O

x

1

Two intersections, both with , so , so Then So 11 Factorising:

for for

or

or .

.

So

or

Solutions:

or

12 a Graphs of

and

y y = |ex – 1|

y = |ln(x – 1)| x

O

x = 1

b There is a single solution to has roots at

13 a

, so

has vertical asymptotes

and

.

y

y = |x2 – 3x| 1 y = — x2 – 3x

x

O

x = 3

b Intersections are when , so has solution So

for where

14

or ,

and

( s.f.).

( s.f.).

.

Since both sides are positive, squaring gives an equivalent equation: Therefore, the requirement is simply that

; that is,

.

Tip Note that this result is not the same for equivalent solution for this situation. 15 a For real values of ,

. Use graphs or algebraic reasoning to find the

b Multiplying both sides by

(noting that

for any value ):

Using part a:

has a vertical asymptote at the root of

16 a

.

has a horizontal asymptote at the reciprocal of the horizontal asymptote of , approached from below as The graph of

.

appears fairly linear until close to the maximum; that line has equation

so for an equivalent domain, the reciprocal will closely approximate has a minimum at the value where

The -intercept of

is at

, so

.

has a maximum.

has -intercept

.

y

1 y = — f(x)

1 y = — 2 O

x

x = –2

b

has roots at

and at the root of

has a horizontal asymptote approached from above as .

for large , so

. has an oblique asymptote

The graph of appears fairly linear until close to the maximum; that line has equation so for an equivalent domain, the reciprocal will closely approximate .

,

y y = xf(x)

–2

y = 2x

x

O

17 a Vertical asymptotes at roots of

none

Roots at vertical asymptotes of

none

Minimum/maximum at maximum/minimum, respectively, of For



as

so

, so maximum is at

.

is an asymptote for the graph

y

(0, 1)

y = sech x x

O

b

, so

18 a

, so

b

From the graph, the intersection is when .

; by symmetry and from part a, this is at

There is no second intersection for a large positive value of because (from part a), the only solution of occurs for negative . So

for

Worked solutions 5 Further vectors Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 5A

7

, so the direction vector of the line,

8

9

The vector equation of the line is

.

a

.

, so has coordinates

.

b Parallelogram area equals

10 The direction vector of the line must be perpendicular to the two given direction vectors.

, so the new line has equation

11 The direction vector of line

is

The direction vector of line

is

.

.

.

The new line is perpendicular to both these lines;

12

The new line has equation

.

In Cartesian form, this is

.

where is the angle between vectors and .

So for all non-zero vectors .

13 a

The vector product is distributive across addition. for all vectors and .

b

14 a The scalar (dot) product of two perpendicular vectors is zero. The vector (cross) product of two non-zero vectors produces a vector perpendicular to both. So

is perpendicular to and therefore

.

b

15 Direction vectors of the two given lines are

, so

and

.

is the direction vector of the required line.

Then the Cartesian equation of the required line is

.

16 For vectors and , with least angle between them: and so

.

17 Then and 18 a If and intersect, then:

Substituting

into the other two equations:

Using equation

:

Using equation

:

Since these two conditions are incompatible, the two lines must not intersect. b

i ii If

is perpendicular to , then:

(4) iii If

is perpendicular to , then:

(5) iv Solving simultaneously for and : so So is at

and is at

, .

v From part b i:

so

By construction, since between them.

is perpendicular to both lines, this represents the shortest distance

EXERCISE 5B

, so the Cartesian equation of the plane is

10

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

. In fact, if you look at

the three points you can quickly see that they all have all lie.

12 a

,

, so

.

so this is the plane in which they

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.

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: from

:

Then from consistent. ii Substituting

and from :

: , which is true, so the system of simultaneous equations is

, 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

,

The normal to the plane containing the first three points is

and

.

.

.

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 containing the four points has a non-zero volume; see Chapter 6 for more about the vector triple product .

17 a The normal vector

b

.

, so

, so

for the vectors to be perpendicular.

for the vectors to be perpendicular.

c Then two direction vectors of the plane (taking a multiple of for integer elements) are

. From the Cartesian equation, it is clear that point

and

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

and

.

From the Cartesian equation it is clear that point

lies in the plane.

The plane can be described by the vector equation

.

EXERCISE 5C 4

Substituting the line parameters

,

and

into the plane equation:

so The point of intersection is therefore 5

.

a The direction vector of the line is

.

b The line equation can be given as

.

Substituting into the plane equation:

, so

The intersection point is 6

and therefore

.

Substituting the parameterised form of the line equation:

,

and

so The point of intersection is therefore 7

.

a b The direction of the line of intersection is perpendicular to both plane normals. For a point on the intersection line: setting

and solving:

, so

.

The line has equation

Taking

gives the simpler form

.

.

into the plane

8

a b The direction of the line of intersection is perpendicular to both plane normals. For a point on the intersection line:

setting

and solving:

So

.

The line has equation

Taking

.

gives the simpler form

In the desired form, the line has equation 9

a

-axis:

; substituting into the plane equation gives

-axis:

; substituting into the plane equation: , so is

-axis:

.

. , so is

.

; substituting into the plane equation:

, so is

.

b

10 a The direction of the line of intersection is perpendicular to both plane normals.

For a point on the intersection line: setting

and solving:

, so

.

.

The line has equation

.

b Substituting the parameterised form , so

into the plane equation:

.

Then the point of intersection is

.

11 a The direction of the line of intersection is perpendicular to both plane normals.

For a point on the intersection line: setting

and solving:

, so

.

The line has equation

.

b

So the direction of the line of intersection of and is perpendicular to the normal of so is the direction vector of each line of intersection between any two of the planes.

However,

, so the line itself does not lie in

, and

. Since none of the

normal vectors are parallel, no two of the planes are parallel; the three planes therefore form a triangular prism (rather than having a common line of intersection). 12 a The direction of the line of intersection is perpendicular to both plane normals.

For a point on the intersection line: setting

and solving:

, so

The line has equation

.

.

b If the three planes form a sheaf, then the line of intersection is common to all three planes. Substituting the answer from part a into the equation for

:

, so

.

13 The direction of the line of intersection is perpendicular to both plane normals.

The line of intersection of

and

has direction vector

.

It is clear that all three planes pass through the origin (all equations have a zero constant).

Since

, it follows that the line of intersection also lies in the

third plane, and so is a common line of intersection to all three planes.

14 a The direction of the line of intersection is perpendicular to both plane normals.

The line of intersection of

and

has direction vector

.

For a point on the intersection line: setting

and solving:

, so

.

The line has equation

.

In Cartesian form, this is

.

b For the planes not to intersect, the line of intersection of the first two planes must be parallel to (but not lie in) the third plane.

, so

Since

.

, the line does not lie within

for this value of .

c No two of the normal vectors are multiples of each other, so no two of the planes are parallel. Therefore the three planes form a triangular prism.

EXERCISE 5D 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 acute angle between the line and the plane is 4

( s.f.)

a Normal vector

b Normal vector

The angle between the planes is the same as the angle between their normals. (to the nearest degree)

5

a

b Normal vector

The angle between the normal and the line is given by . So the acute angle between the line and the plane is 6

( s.f.).

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 acute angle between the line and the plane is

7

The two normal vectors are

and

( s.f.).

.

But

The normals are perpendicular so the planes must also be perpendicular.

8

a The line can be parameterised as

so in vector form it is

and so the direction vector is

.

b Normal vector

The angle between the normal and the line is given by . So the acute angle between the line and the plane is 9

a Substituting

( s.f.).

into the plane equation: is true, so lies in . and the plane normal

b

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 5E

1

a The normal vector of is

b So has equation

, which is also the direction vector 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

c The distance from to is

.

.

2

a The normal vector of is

, which is also the direction vector of .

Equation of : b 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

c The shortest distance from to is

3

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 these facts: The shortest distance from the origin to a plane

equals

The perpendicular distance between two parallel planes

. and

is

.

4

a The normal vector

The equation of line b

is the direction vector of the line

is

is the intersection of line equation:

. and the plane . Substituting the answer to part a into the plane

, so

c Since

.

, has coordinates

.

, and therefore has coordinates

.

5

,

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 from to the plane.

, so

d

6

,

a

has coordinates

, so

is the shortest distance

.

.

b Area

( s.f.)

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

into the answer to part

and the intersection is at

The distance from to the plane is f

.

( s.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 subsequent calculations to avoid truncation errors. 7

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 d

The perpendicular distance between two parallel planes

For

, the equations are

So the distance between

8

a For

and

and

is

and

is

.

, the equations are

and

.

Since the two normals are the same, the planes must be parallel. b c Substituting the answer to part b into the plane equations shown in part a: i In ii In

, so , so

and so the foot of the perpendicular is and so the foot of the perpendicular is

d The distance between the planes is

.

MIXED PRACTICE 5

1

2

3

a

b Substituting

, so the plane is given by

into the answer to part a:

, so

.

. .

.

4

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

5

.

a

, so

b

has equation

and

c equation

, which in Cartesian form is

, so also lies on plane

, which has

.

d The line of intersection must be perpendicular to both normals.

From part a: direction vector

From part c: lies in both planes, so the line of intersection has equation

.

e Substituting the answer to part d into the plane equation:

, so f

6

The angle between the planes equals the angle between their normals.

a Plane equation is

Since

b

and the intersection point is

where the plane normal

, the point

, so the line equation is

.

does lie in the plane.

.

.

c d To find the intersection of the line in part c with plane , substituting the parameterised form into the plane equation: , so

, and the foot of the perpendicular is

e The distance from to the plane is

7

, so the line

a

.

.

is

b

c

, which is independent of .

i ii The general vector represents the points on the line

, which does not pass

through the origin. represents the vector perpendicular to both and ; since , it follows that does not run perpendicular to the normal of the plane containing and . That is, , and cannot lie in the same plane.

8

The normal to the first plane is

and the normal to the second plane is

The angle between the planes equals the angle between their normals, so .

(Reject 9

as the question states that is non-zero.)

a b Taking the components separately to form a set of simultaneous equations:

From

:

.

Substituting into: 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: 10 a b where neither nor

11

So is parallel to

is a zero vector.

. (Answer C)

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.

, so the plane equation is

c

.

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

and the line intersects the plane at c On the line, is at is at , with coordinates . d

, so

into the plane equation: . 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 a The direction vector of line is given by the vector product of the first two normals.

, so

To find a point on the line, setting

, so

and therefore

has equation b

i If intersects

and solving:

.

. , then substituting the parameterised version of the line :

But this reduces to , which is a contradiction. There is no value at which the line meets the plane, so they are parallel, and therefore the system of planes is inconsistent. ii Since the intersection of the first two planes runs parallel to the third plane, the three together form a triangular prism. c

i Setting

and solving:

, so

and therefore

So a point on the intersection is

.

.

ii The line of intersection of the second and third plane must have the same direction vector as , since the three intersection lines are the parallel edges of the triangular prism. Therefore the line of intersection of

and

can be given as

.

15 a Area b c Volume of a linearly tapering solid equals

where is the base area.

So the volume of the tetrahedron is But since is the angle between and the vector perpendicular to the base (and so perpendicular to and , which can be expressed as ), it follows that Therefore d Taking as

, the three side vectors can be given as

,

and

.

The volume is therefore

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

19 a

. , or in Cartesian form:

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

, it follows

or

.

d From part a: a line parallel to both planes must have direction vector

The line has equation

b

.

and

20 a

i If

.

.

, then

.

Writing these as a set of simultaneous equations and simplifying:

ii so Then from and from

: :

, so

Then has coordinates

.

and has coordinates

iii The shortest distance between the two lines is construction).

21 a The normal vector

An equation for

because

. is perpendicular to both lines (by

and the point with position vector

is therefore

b Two direction vectors in the plane are

lies in the plane.

.

and

.

The normal vector to the plane is perpendicular to both of these directions:

So

and the point with position vector

An equation for

lies in the plane.

can be given as

Taking the general position vector

, this can be written as

so

c The angle between the planes is equal to the angle between the plane normals.

d

i

Tip You will have a further method of solving this question using matrix determinants when you have studied further matrices in Chapter 6. The line of intersection between

and

has direction given by

.

, so that the direction of the intersection line is

.

A point on the intersection can be found by setting

, so

and then

So the line of intersection has equation

Intersecting this with

and substituting

in both equations:

.

.

into the plane equation:

If then there is no solution for , which indicates that the line cannot intersect with the plane.

Tip

Alternatively, note that if

then the normal vector of , which would indicate that

is is parallel to

, so there can

be no intersection of all three planes. The other value for which there is no solution for is ii If

.

, then

However, is given by and by so the second and third planes are (distinct and) parallel, so there can be no common intersection point.

If

, then

, so that

That is, the line of intersection between However, the line does not lie within

and

is parallel to

.

, so the three planes form a triangular prism.

Tip If the intersection line lay within so that the three planes formed a sheaf, then the equation for the intersection would have been always true and independent of , so that equation would have some value for which . In the given problem, of are , neither of which is a root of

as calculated in part d i, so the roots .

Worked solutions 6 Further matrices Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 6A 4

a

, so

and

So b

, so

and

So 5

Expanding about the first row:

Determinant is zero when 6

Expanding about the first column:

For det 7

.

Let

,

or . and

Then

.

, so

and

.

So 8

.

Expanding about the first row:

Expanding about the third column:

9

a Expanding about the first row:

Expanding about the second column:

b or 10 a Expanding about the first row:

b Determinant represents volume scale factor. i Volume of image is ii

. , so scale factor is still

.

Volume of image is

.

iii Volume of image is

.

11 Using your calculator to find the inverse:

has coordinates

.

12 a

From the second row: From the upper left cell:

, so

From the lower right cell:

, so

b

has coordinates 13 Let have coordinates

. .

From the third element: Taking the difference: But from the first element:

, so

.

( is non-singular, so

( is non-singular, so

.)

.)

Comparing:

Simplifying:

so Substituting this:

So

consistently.

Substituting

into

:

So

or

(false solution, introduced from multiplying through by .

Then

.

Point has position vector

has coordinates 14

is orthogonal, so Since

.

. and similarly

is also orthogonal.

EXERCISE 6B a If

.

and matrix multiplication is associative, it follows that:

Therefore

4

at the start of the calculations)

, then

;

the rows are not linearly independent, so

.

b Expanding about the first row:

5

a

i is a factor of each element in the second row, so can be taken out as a factor of the determinant. ii Subtracting the third row from the second does not affect the determinant, but the second row elements then each have as a factor, so can be taken out as a factor of the determinant. iii Subtracting the first row from the second does not affect the determinant, but the second row elements then each have as a factor, so can be taken out as a factor of the determinant.

b

6

a A multiple of times the second column is added to the first column: b

where

, so

7

8

a Expanding about the third column:

b

i Since

, from part a:

So swapping the first two columns of a matrix reverses the sign of its determinant. ii Expanding the determinant of

about the first column, by the same working as in part a:

From part a: So, determinant.

, so rotating the columns from

to

does not affect the

iii Using parts b ii and b i: swapping any two columns of a matrix results in the determinant changing sign. If any column of

is identically zero then rotate the columns until it is on the right.

By part a: If two columns of are non-zero multiples of each other then, again rotating until these two columns are left and centre, by part a: iv For matrix

, if

, then by part a:

9

10 a

i

, so the normal to plane is

.

The Cartesian equation for is

.

Then plane has equation

.

Since

, does not lie in the plane.

ii Area of triangle

equals

iii Plane is given by

.

.

The perpendicular distance from a point with position vector to a plane with normal passing through a point with position is given by:

So the perpendicular distance between and is

. iv

Volume of a tetrahedron with base area and perpendicular height is is

.

b The volume of tetrahedron vectors and .

equals one sixth of the determinant of the matrix with column

where is the base area and is the perpendicular height.

EXERCISE 6C

2

where

Expanding about the first row:

is singular so there is no unique solution.

Tip

so the volume of

Alternatively, observe that linearly independent, hence 3

(or is non-singular.

) so the rows (columns) are not

The system of simultaneous equations can be expressed as:

where

.

Using your calculator:

4

where

There is no unique solution when the determinant is zero, which occurs when 5

a Expanding about the first row:

Then

cannot equal zero for any real .

b

c Taking

, the system can be expressed as:

so

So 6

.

The system of simultaneous equations can be expressed as:

or

.

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

:

so

So

.

EXERCISE 6D 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:

or

.

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 Eliminating variables from the equations:

So the system is inconsistent, and 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

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 Eliminating variables from the equations:

So the system is only consistent if

.

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. d The common line has a direction perpendicular to all the plane normals.

, so

To find a point on the common line, set

, so

.

.

Then the common line has vector equation 7

.

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. c The common line has a direction perpendicular to all the plane normals.

, so

Since the line passes through the origin, it has vector equation

8

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 9

.

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 Eliminating variables from the equations:

So the system is inconsistent if

.

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. 10 a Writing the system of equations in matrix form:

where

and

.

Expanding about the third row: . For a system with no unique solution, require b Eliminating variables from the equations:

For

:

, so

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. d Taking the vector product of two of the plane normals to find the direction vector of the common line:

so

.

Taking

:

Setting

to find a point on the intersection line:

, so

and then

.

So the vector equation of the line of intersection is

.

EXERCISE 6E

Tip It can be very useful to note that since the characteristic equation has the eigenvalues as its roots, the sum of the eigenvalues must (by the way the characteristic equation is generated) equal the sum of the lead diagonals of the matrix. Also, the determinant always equals the product of the eigenvalues. For a matrix, you can use these two facts to write down the characteristic equation directly. 3

Product of eigenvalues Sum of eigenvalues = sum of lead diagonal elements Characteristic equation:

4

Eigenvalues are

,

Solving

to find associated eigenvectors

for

:

for

:

.

, so

, so

.

Product of eigenvalues Sum of eigenvalues = sum of lead diagonal elements Characteristic equation:

and

:

Eigenvalues are

,

Solving

to find associated eigenvectors

for

.

:

for

and

:

, so

:

, so

.

5 Characteristic equation: let

Expanding about the first row:

Then So is a root of

and therefore is an eigenvalue of .

By the factor theorem:

is therefore a factor of

.

Factorising further:

So the eigenvalues are Solving

for

,

and

.

to find associated eigenvectors

,

and

:

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

6

.

Characteristic equation: let

Expanding about the second row:

If there is a line of invariant points then one eigenvalue must equal , and it is clear that so is indeed an eigenvalue of .

and

Factorising further:

So the eigenvalues are Solving

(repeated value) and

to find associated eigenvectors

for

gives the conditions

associated with this eigenvalue:

eigenvalue:

a

and

:

and

, so there is only a single eigenvector

.

:

Substituting

7

,

:

Substituting

For

.

gives the conditions

and

, so the eigenvector associated with this

.

Characteristic equation: let

Expanding about the second row:

Then So is a root of

and therefore is an eigenvalue of .

By the factor theorem,

is therefore a factor of

.

b

So the eigenvalues are Solving

for

and

(repeated value).

to find associated eigenvectors

,

and

:

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

For

:

Substituting

gives the condition

with this eigenvalue:

8

.

so there is a single eigenvector associated

.

a

If is an eigenplane of associated with eigenvalue , then any two vectors within the plane will have the property .

Two such vectors are

and

, so

.

and

.

b Characteristic equation: let

.

Expanding about the second row:

So the repeated eigenvalue is

, associated with the eigenplane

The remaining eigenvalue is the final root of the characteristic equation: Solving

for

to find associated eigenvector

:

.

:

. .

and

Substituting

9

gives the conditions

and

, so

.

Tip Remember the key fact that the product of the eigenvalues (including repeated values) is equal to the matrix determinant and their sum equals the sum of the lead diagonal elements. If the three eigenvalues are

,

and

:

sum of lead diagonal elements

(1) (2)

Given

, then by

Then from If

:

:

, so

.

and

, then

and

,

and

The three eigenvalues are Solving

for

.

. .

to find associated eigenvectors

:

,

and

:

.

Taking the vector product of two distinct two rows of the matrix gives the eigenvector:

, so

For

.

:

Taking the vector product of two rows of the matrix gives the eigenvector:

, so

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

10 a

Characteristic equation: let

Expanding about the first row:

.

Tip If you do not spot a common factor during the working and continue to expand the cubic fully, substituting and using and then summing the resultant equations will give , at which point the cubic can be fully factorised. The eigenvalues are the roots of the characteristic equation, so b

i So the eigenvalues are Solving

(repeated value) and

to find associated eigenvectors

for

:

Substituting

. .

,

and

:

.

gives the equation

associated with the eigenplane

, so the eigenvalue

is

.

The associated eigenvectors are any two non-parallel vectors spanning the plane, such as and

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

ii The eigenplane

.

can be expressed as

; that is, eigenvector

to the eigenplane, and hence must be perpendicular to both

and

is normal

.

c The enlargement must have scale factor , leaving transformation with eigenvalues (so that the eigenplane consists of invariant points) and , with associated vector perpendicular to the plane. So represents a reflection in the eigenplane 11

.

Tip Calculating the characteristic equation and seeking to factorise it is clearly unpalatable, given the values involved. Using knowledge of the sum and product of the eigenvalues, together with the fact that there are only two eigenvalues, offers a faster route to the solution.

Finding the determinant: adding the third row to the first row and then subtracting the third column from the first, to simplify the calculation:

If the two distinct eigenvalues are

and

the sum of the three eigenvalues

: sum of lead diagonal elements

the product of the three eigenvalues

. (1) (2)

The solution (either by observation or by substituting and trying values in the resulting cubic) is , . Solving

to find associated eigenvectors

for

:

Substituting

,

and

:

.

gives the repeated equation

associated with the eigenplane

, so the eigenvalue

.

The associated eigenvectors are any two non-parallel vectors spanning the plane, such as and

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

.

EXERCISE 6F

Tip It can be very useful to note that since the characteristic equation has the eigenvalues as its roots, the sum of the eigenvalues must (by the way the characteristic equation is generated) equal the sum of the lead diagonals of the matrix. Also, the determinant always equals the product of the eigenvalues. For a matrix, you can use these two facts to write down the characteristic equation directly.

is

, sum of lead diagonal elements

3

Characteristic equation: Eigenvalues are

,

.

Solving

to find associated eigenvectors

for

:

, so

for

:

, so

Let

and

:

.

, the matrix with the eigenvectors as columns and let

, the diagonal

matrix with eigenvalues as elements of the lead diagonal. Then

so

.

, sum of lead diagonal elements

4

Characteristic equation: There is a single eigenvalue; Solving for

.

to find any associated eigenvectors: :

, so

.

The matrix has only one eigenvector and so cannot be diagonalised.

Tip This is a sufficient argument to answer the question. If you had to justify why having insufficient eigenvectors makes diagonalisation impossible, you could argue that if for a diagonal matrix then , and therefore each column of must be an eigenvector with corresponding eigenvalues on the lead diagonal of . If there are not enough eigenvectors, then cannot be constructed without repeated columns, in which case and so cannot be found. This argument applies to the diagonalisation of any square matrix. 5

a Matrix is formed with the two eigenvectors as its columns:

.

Then

If the pairing of eigenvalues and eigenvectors is

,

and

,

, then

If the corresponding eigenvalues are the other way round so that

,

, then

b Since the two eigenvectors span the whole of the plane, it follows that any position vector in the plane can be expressed as for some and .

(Each matrix leaves one eigenvector unchanged and stretches the other by factor is the same as stretching each eigenvector by .)

; applying both

Tip Alternatively,

6

, sum of lead diagonal elements

a

Characteristic equation: Eigenvalues are

,

Solving

to find associated eigenvectors

for

for

Let

:

. and

:

, so

:

, so

.

, the matrix with the eigenvectors as columns and let

matrix with corresponding eigenvalues as elements of the lead diagonal. Then b

so

.

, so

i If is odd, then

, so

, the diagonal

ii If is even, then 7

, so

.

, so

a b Sum of lead diagonal elements Characteristic equation: Solving for

, so eigenvalues

and

to find associated eigenvectors :

for

and

:

, so

:

, so

.

, so the eigenvectors are perpendicular. c

has eigenvalues and

associated with two perpendicular vectors; this describes a reflection in

the line of invariant points, given by eigenvector is a reflection in the line

.

.

d Two methods are shown; the first is purely algebraic and the second uses geometric understanding. Method 1: Diagonalisation where

is the matrix with the eigenvectors as columns and

is the diagonal matrix with the corresponding eigenvalues as lead diagonal. Then

Method 2: Geometric interpretation Transformation is a reflection combined with an enlargement scale factor . So applying twice is equivalent to just enlarging with scale factor . That is,

.

Therefore: , where is the enlargement with scale factor

and is the reflection.

8

a

Characteristic equation: let

Expanding about the first row:

Roots of the characteristic equation are eigenvalues of Solving

for

to find associated eigenvectors

:

: and

,

and

.

:

.

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

For

.

:

Taking the vector product of two rows of the matrix gives the eigenvector:

, so

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

Let

.

, the matrix with the eigenvectors as columns and let

diagonal matrix with corresponding eigenvalues as elements of the lead diagonal. Then

, so

b

For odd :

Using your calculator:

.

, the

Tip Don’t miss the opportunity to check that substituting produces the original matrix as a means of spotting arithmetic errors. Also remember the trick of using your calculator to find the result directly, by substituting a recognisable dummy value (for example ) instead of the variable value ( in this case) and letting technology take the strain!

9

for

a

, so represents an anticlockwise rotation about the

origin through , sum of lead diagonal elements

b

Characteristic equation:

has roots

which are the

eigenvalues of . Solving

c

to find associated eigenvectors

for

:

for

:

Let

and

:

, so

, so

.

, the matrix with the eigenvectors as columns and let

, the

diagonal matrix with corresponding eigenvalues as elements of the lead diagonal. Then d

, so

.

, so

.

But

.

So

.

(As expected, performing a

rotation twelve times results in a return to the start position!)

MIXED PRACTICE 6 1

The sum of eigenvectors equals the sum of the lead diagonal elements: . (Answer C)

2

The system has a non-unique solution if the matrix

has

and the system is

consistent. Taking a factor of from the third row and then expanding about the first row:

So for

, the system has a non-unique solution. (Answer A) , sum of lead diagonal elements

3

Characteristic equation:

4

Eigenvalues are

,

.

Solving

to find associated eigenvectors

for

:

, so

for

:

, so

a If

is an eigenvector of

and

.

for an eigenvalue , then

, so

:

and

.

.

b From part a: which is the product of the eigenvalues, so , so 5

is the eigenvector associated with

.

a Adding the second row to the third row (which does not affect the value of the determinant):

b Subtracting the first column from each of the others and then calculating:

6

So

is a factor of

.

Expanding about the first row:

But So

because is singular. or

which has roots

. Therefore

is the only real value for

which is singular. 7

, so

a

and therefore b

Then Since this value is the sum of two non-positive values (for real ) it must always be non-positive. c Factorising the determinant:

So 8

for

(since

for real ).

a Writing the system of equations in matrix form:

where

Expanding about the first row:

and

.

If there is no unique solution then for b When

or

, the system becomes:

Then on the left side: 9

is singular.

, so the system is consistent if

.

a

Adding the second column to the third (which does not affect the value of ):

So

is a factor of .

b Adding the second column to the first, factorising and then subtracting the first column from the third factorising and expanding about the third column:

10 a Writing the system of equations in matrix form:

where

and

.

Expanding about the third column:

If there is no unique solution then for

is singular.

or

b Eliminating variables from the equations:

Since

when

is the case when

is singular, the third row can only be consistent if

.

11 a

If , then all the elements other than those on the lead diagonal have value zero, and all elements on the lead diagonal have value , so when . b So when

,

.

12

13 a Writing the system of equations in matrix form:

, which

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

:

eliminating variables from the equations:

When

So when

, the system will only be consistent if

, the system is always consistent.

.

So

.

14 a Expanding about the third column:

b

c

15 a

b

16 a

, so

, b

, so

and therefore

c

So

and

.

d The system can be given by the matrix equation

where

.

The solution is given by

So

,

.

and

.

17 a

So

b

, independent of the value of .

i The equations can be given as

for

.

Since the matrix is known to be non-singular, by part a, this system has a unique solution given by

and so the system is consistent.

ii The planes intersect at a single point.

, so the point of intersection is 18 a Expanding about the second column:

The matrix is always non-singular, since

b

for all values of .

.

c The system of equations can be expressed as

This rearranges to

So

,

.

and

.

, so for the corresponding eigenvector

19 a

If

.

,

.

, then

From

:

Then from So from

: :

, from which

.

b Characteristic equation: let

Expanding about the first column:

Since

is a known root of

, by the factor theorem

is a factor of

.

So The remaining two eigenvalues are therefore c The invariant lines are given by

and

.

where is one of the eigenvectors of

.

From the working in part a: eigenvector 20 a

i Reading from the diagonalised form: eigenvalue

eigenvalue

is associated with eiegenvector

is associated with eigenvector

.

ii The second eigenvector spans the line of invariant points, so the equation of that line is b

, so

c

for some values , and . But since all elements of are integers, then all elements of being the sum of products of elements of . Therefore multiple of

must be integers as well, each

must be an integer for all positive integers and so for all positive integers .

must be a

Worked solutions 7 Further 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 black-coded practice questions. EXERCISE 7A 2

a b The area enclosed by a curve using polar coordinates is So the area enclosed by the circle

is

.

3

4

5

6

1 —

a

r = θ – 2 O

0, 2π

b

The area is constant, independent of the value of . 7

a

.

π 2

π

r = 3 sin 2θ

0, 2π

O

3π 2 b

8

for

a

π 4

3π 4

r = a cos 2θ 0, 2π

O

7π 4

5π 4

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 twice the area of one ‘petal’.

The total area is twice this: 9

When

,

Since

is clearly increasing (because the derivative is

originally being integrated, which is always positive for

, the function

), this is the only solution in that

interval; there is a second solution in the interval

, but this is not a valid solution, as the

integral used to calculate the area in this interval is negative. Therefore

.

10

EXERCISE 7B 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 area

.

Tip Or calculate the area directly using the polar integral

The shaded region is the area enclosed by the curve, less triangle

.

ii

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

, which is one sixth of a circle, with area for

and

symmetry).

4

a The two curves intersect where

, so

.

(which are equal, by

Solution:

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

different values of ). Solution:

or

Polar coordinates of the intersection points are

and

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.

r = 2 0, 2π

O r = 3 – 2 cosθ 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 7

1 2

(Answer B) a

forms a spiral from the pole:

r = θ 0, 2π

O

b 3

a

r = 5 – 4 cos θ

O

0, 2π

b

4

a

where

and

Then the Cartesian equation is equivalent to or more simply b A tangent at the pole occurs when , so

:

or

c

2π 3 r = 1 + 2 cos θ

O

0, 2π

4π 3 d Using the symmetry along the horizontal:

, so that

.

5

The shaded area is bound by the curve for

6

The curve touches the pole when

, so

.

:

or

So the shaded area is enclosed by the curve for

7

.

.

a

b The two curves intersect where

, so

So the intersection points are

and

π 2

c

r = 10 – 10cos θ r = 5 π

O

3π 2 The enclosed area is:

0, 2π

: .

or

.

bounded by the curve

for

bounded by the circle

for

(that is, two thirds of the circle, with area

).

8

Tangents at the pole occur when Solutions:

at

and

, so

. at

Therefore the area is enclosed by the curve over the interval

.

Tip This might be something you would assume, given the diagram. However, you should always show explicitly that you have established the extent of the interval enclosing an area to be found.

9

a

i

Tip 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:

. so

Solutions for

:

at

or

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

, which is double the area enclosed by the

by symmetry across the horizontal.

(1)

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

:

:

:

10 a

i

ii On line iii Triangle

,

, so the polar equation of the line is has base length

and perpendicular height

, so has area b

i Let

.

, so

Then ii The curve meets the pole when

, which occurs when

at

or in the area of

interest. The area is therefore enclosed by the curve over the interval

.

Worked solutions 8 Further 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 black-coded practice questions.

Tip Note that if for , then either form (with the inside or outside the logarithm) in your answer.

EXERCISE 8A

, and you can use

EXERCISE 8A 3

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: 4

.

.

,

is transformed to asymptotes

.

, with the origin mapped to

mapped to

and horizontal

.

Replace with : stretch parallel to the -axis with scale factor . Centre point at the origin is unchanged; asymptotes if ). Replace with

: translation

Centre point at

; asymptotes are unchanged.

: stretch parallel to the -axis with scale factor . is mapped to

; asymptotes unchanged.

End graph has centre point at End graph has asymptotes at 5

(reflected vertically

.

Centre point at origin is mapped to Replace with

are mapped to

The domain of

is

The domain of

is

, so

.

(and the graph is reflected vertically), so , so the domain of , so the domain of

is

.

. is

.

For a domain with maximal interval width of , require that the lower end of the domain of is within the domain of .

EXERCISE 8B

2

y

a

x = –1 y = 2 sech x y = coth (x + 1)

y = 1

x

O

y = –1

b The graphs intersect at two points so there are two solutions to 3

.

y

a

y = cosech x + 1

y = 2

y = 2 – sech x

O

y = 1

(0, 1)

x

b The graphs intersect at only one point so there is one solution to 4

5

y

a

y = arcoth x

x = –1

y = x y = coth x

y = 1

x

O

y = –1

x = 1

b

i

or

ii c Let Then

, so that

.

.

So

6

a Let

, so that

.

Then Rearranging: Multiplying through by : This is a quadratic in . Using the quadratic formula: Reject the negative root because

.

So

b If

as before, then

Then

. , proving the identity.

EXERCISE 8C 2

Using

So

Solutions: 3

:

or

or

( s.f.)

Using the double angle formulae:

or

( s.f.)

4

(Reject

since

for all

.)

So, 5 Using

:

or

or

6

so

7

Multiplying by and factorising:

(reject) or

8

(Reject

because

for real .)

9

(Reject

since

for real .)

10 Let

, so that

11 a Using the binomial theorem:

b Using the result from part a:

Factorising, noting that if

the equation is valid, so

(The quadratic expression has discriminant

so has no real solutions.)

and

12 a So

b Using the result from part a:

EXERCISE 8D 2

a b

3

at turning point. So

or

Turning point at 4

(Reject because , at the point

is a factor:

for real .) . at the turning points.

, so

When

,

and

Minimum is at 5

.

.

, so

At

,

and

The equation of the tangent at Multiplying through by

6

is

.

and rearranging:

, so 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

(rejecting the negative root).

Then b The graph of

looks like this: y

(0,1)

dy — = 2sech2 2x – 1 dx x

O y = –1

So the second derivative is negative for occurs at At this point:

and therefore the maximum on the graph of .

, 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 , so

10 a

Stationary points occur when

, so

.

, so b Since

and

are both odd functions and

an odd function. Therefore Consequently, for the stationary points with

is an even function, it follows that

is

. , 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 is the minimum, the reasoning shown is perfectly sufficient.

EXERCISE 8E 7 8

and So

9

a Using the chain rule:

b Using the identity

10

and the result from part a:

Using the formula for the volume of rotation about the -axis,

11 a Using the fact that

b

12 Using integration by parts:

and

:

:

13

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 14 a Let

, so

b

15 Let When

, so ,

. and when

,

.

16 Using the chain rule:

Then using integration by parts:

MIXED PRACTICE 8 1

Using the double angle formulae:

(Answer B) 2

y

a y = 4

f(x) = 3 tanh2 x + 1 (0, 1) O

x

b , so 3

, so

4

for all , so there can be no point of inflection, since at this point the second derivative would equal zero. 5

6

7 8

Let When

9

, so ,

and when

,

Multiplying both sides by

So

:

(rejecting the negative solution)

10

So

or

:

:

11

12 Let

, so

13 Intersections where

or or

:

y

14 a

1 y = tanh x x

O

–1

b Then

, so , so

c

i

ii Factorising the LHS of the equation from part i:

(reject the root

15 a

i

since the range of

y

y = sinh x O

x

is

for real .)

y

y = cosh x (0, 1) x

O

for real , so if

ii Since

is a

then

function, with range

, there is a single solution to this.

b

From part a ii: has a single solution, so there is only one stationary point on . The curve equation is

.

At the stationary point,

, so

.

16 a

Stationary point where

So

:

(Reject the negative root.)

b At the stationary point found in part a, So 17

and Using

:

:

.

18 a Using the binomial theorem:

b From part a: Multiplying through by

So

and rearranging:

(rejecting the negative root) (rejecting the negative square root since

for real )

19 a So

So

b

(rejecting the negative root since

i

Then

when

Using the formula in part a: At the stationary point, ii

since

for all real .

for all real )

So

Worked solutions 9 Further calculus Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 9A , so using the chain rule:

2

Then at

:

, so using the product rule and chain rule:

3

Then at

:

4

, so using the chain rule:

5

, so using the chain rule:

has domain Here, 6

, so if

, so its derivative has domain , then

.

and so the domain is

.

, so

a

Differentiating with respect to :

But b

7

(standard identity), so , so using the chain rule:

, so using the chain rule:

.

has domain Here,

, so if

, so its derivative has domain , then

.

and so the domain is

.

8 Using the chain rule: So 9

and

a Using the product rule and chain rule: b Then, using the result from part a:

10 Inflection points occur only when the second derivative is zero.

Using the 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 9B 2

Method 1: Direct differentiation using the chain rule Using the chain rule and differentiating directly:

Method 2: Rearranging and using implicit differentiation , so Using implicit differentiation:

So

3

Using the product rule and chain rule:

4 Using the chain rule:

So 5

and , so

Then

and

So the tangent line is given by the 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, substituting the first tangent equation into the second:

7

, so

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: or

. (reject)

Then So 10 a Let Then

so

Taking the reciprocal:

So

b

and

So the tangent at

has equation

.

Rearranging: Then has -coordinate

.

The normal at

has equation

.

Rearranging: Then has -coordinate

.

So distance

.

So

EXERCISE 9C

4

5

6

7

a b

8

a Let Then

, so that

and

, so

Substituting into the integral:

b 9

a Let

so

The substitution is valid for

and .

.

b

10 a 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:

,

and when

,

.

14

15

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 You could approach this question directly, using substitution, or with a little forethought, you can set up the answer as in question 9 in Exercise 9A, differentiating the non-trigonometric expression on the right side of the equation. Both methods are shown. Method 1: Direct approach Let Then

, so that , so

Substituting:

Method 2: Rearranging first Let

.

Using the product rule and chain rule:

and

.

Then, from the previous working:

Let

, so that

Then

and

, so

22 a

b Let

, so that

Then

and

, so

Substituting:

EXERCISE 9D 2

for some constants , and .

.

Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

3

, then

a

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

and

b

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

5

, then

a

and

for some constants , , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

so so Substituting From (2): Then b

into (1): , so

, so

6

for some constants , , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so , so

,

and

Then

7

for some constants , , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

,

,

and

8

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so

So

,

,

and

and

EXERCISE 9E 3 Let

and

Integrating by parts:

, so that

and

.

a Then b Using the formula from part a:

4

a Let

and

, so that

and

Integrating by parts:

Then b Let

From part a:

Then:

Tip An alternative form of the answer: 5

a Let

and

, so that

and

.

.

Integrating by parts:

Then b Let

and

, so that

and

.

Integrating by parts:

Then c From part a: From part b:

6

a Let

and

, so that

and

.

Integrating by parts:

Then b From part a:

Tip You should be able to spot the formula for the coefficients by the manner in which they are iteratively constructed. For example, the constant coefficient of within is

.

7

a

b Let

From part a:

8

a Let

and

, so that

and

.

Integrating by parts:

Then

(requires that

b From part a:

c Proposition:

for

Base case: From the working in part b: So

and therefore the proposition is true for

Inductive step:

.

)

Assume that

for some integer

.

Working towards: Then, using part a and the assumption:

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 integers 9

a Let

and

, so that

Integrating by parts:

Rearranging: b Let Then From part a:

10 a

and

and

.

. .

b From part a:

11 a

Let

and

, so that

and

.

Integrating by parts:

Rearranging: b From part a:

12 a Let

and

Integrating by parts:

Rearranging: b From part a:

, so that

and

.

c Proposition:

for

Base case: From the working in part b: So the proposition is true for

.

Inductive step: Assume that

for some integer

.

Working towards: Then using part a and the assumption:

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 integers 13 Let Let

and

, so that

Integrating by parts:

For

:

Rearranging:

, so

and

.

. .

EXERCISE 9F 3

, so

4

, so

5

6

,

a

,

, so

, so

and

and

b 7

and

, so

and

8

a

,

b When

, so

,

and

and when

,

, so

(Reject negative root as Therefore

.

.)

and

, so

.

From part a:

9

a

Tip Although you could calculate the arc length with the parameterisation given, you might recognise that substituting from the start makes the calculations more straightforward and saves work later. , Substitute Limits: when

, so that ,

and when

and ,

. .

, so , so

, so

b

.

Rearranging:

, so

Completing the square: This is a circle, radius , centred at and

. The portion described for

, so is the upper right quarter of the circle circumference.

y

1 1 —) + y =— (x – 2 4 2

O

1

1 — 2

10 a

,

So

2

, so

x

and

.

b 11 a

,

b

So

, so

and

holds

12 a

, so

b Let

, so

Limits: when

So 13

,

.

, so

, so

Limits: when

So

EXERCISE 9G

3

and when

.

.

Substitute

2

,

and

, so

, so

,

and and when

,

.

4

5

, so

The equation of the line is

, for which

The two bases have radii and

, so have total area

The total surface area is therefore 6

a

, so the curved surface area is

so

.

. and

, so

Then b

So 7

a

, so

and

Substitute Limits: when

, so ,

, so

. and when

,

.

So, reversing limits and changing the sign of the integrand:

b Substitute

, so

and

.

Limits: when

8

a

,

and when

,

.

, so

Then Substitute

, so

Limits: when

,

and and when

,

.

.

So, reversing limits and changing the sign of the integrand:

b Substitute

, so

Limits: when

,

and and when

,

So

MIXED PRACTICE 9 1

Using the chain rule: (Answer D)

2 Then using the product rule:

Tip Alternatively, using implicit differentiation: , so

. where

.

Rearranging:

3 Using the chain rule:

, so by the chain rule:

4

If 5

, then

a

,

, so

and

Then b

Let

and

, so that

Integrating by parts:

, so using the chain rule:

6

Then 7 Using the chain rule:

and

.

So Then using implicit differentiation:

Multiplying through by

and rearranging:

8

9

a



, so

b For the curve

and

,

and the period is

.

Then length of a single complete wave is given by

Changing the variable: let Limit: when

,

, so

.

.

So the length of one complete wave of the curve in part a. 10

11 Using integration by parts: and

is equal to the perimeter of the ellipse

12 a Let

for

Then

, so

and

and also

for

.

, so

.

b Using the chain rule:

13 a

,

So

, so

and

.

b Substituting: , so that Limits: when

,

and and when

,

14 a Multiplying numerator and denominator by

b Using the result from part a:

. .

:

15 a

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

, so So b

c

16 a Let So,

b Using integration by parts:

, then

and

and

Using part a:

17 a Let

.

Using integration by parts:

So b

The equation of the ellipse can be rearranged to

and the area

enclosed by the ellipse will be twice the area between the upper curve and the axis.

Using part a:

18 a

b Using the result from part a:

Let

and

, so that

and

.

Integrating by parts:

This is valid for all non-zero . Rearranging:

But when

,

, so the formula is also valid for

values . c

which has a finite value.

Tip In fact, From part b, substituting Rearranging: 19 a

b

but this is not needed. :

and so is true for all real

So 20 a

, , so

b Substituting: , so that Limits: when

,

and and when

,

. .

Worked solutions 10 Maclaurin series and limits Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions.

Tip In these worked solutions, sometimes the notation ellipsis (…).

is used in a sequence instead of an

This means that the missing terms all have raised to a power or greater. For example, the Maclaurin series for is

but you could write this more precisely as because the missing terms consist of raised to power and

higher. For convergent series, where only a few terms are used to calculate an approximation, knowing the order of the subsequent terms can be useful in assessing the likely error and is particularly useful when determining the convergence of combinations of series.

EXERCISE 10A so

2

so so

Maclaurin series:

3

so

a

so so so Maclaurin series:

b Then 4

a

i

( d.p.) so so so so

so ii Maclaurin series:

b

5

a

so so so so so so so

Tip You should be able to see that for

Maclaurin series:

Tip Alternatively, use the expansion for

up to

b 6

a

i

, so using the chain rule:

Then using the quotient rule:

ii

and square it.

b Then

,

,

,

and

Maclaurin series:

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: Then, using the assumption followed by the product rule:

Conclusion: The proposition is true for .

and if it is true for

then it is also true for

Hence, by the principle of mathematical induction, the proposition is true for all integers

.

b Maclaurin series: Then:

The general term is 9

a Proposition:

for

Base case: So the proposition is true for

.

Inductive case: Assume the proposition is true for So

Working towards: Then, using the assumption:

Conclusion: The proposition is true for .

and if it is 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

. so

10 a

so so so so

so

Maclaurin series:

b

i Since

and

, it follows that

, and so

for

(as found by

repeated differentiation). So ii Taking the sum of the two series, all terms cancel except the constant term, which is . So

Tip You might have noticed that this question uses somewhat circular arguments; indeed, as soon as you 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 intercept in the diagram is not at

b

For the Maclaurin series

,

but the -

. ,

clearly not the case in the diagram, since the function has a positive gradient at

but this is .

EXERCISE 10B 4

a

b Cancelling through by in the numerator and denominator and then using l’Hôpital’s rule twice:

5

a b c

which is infinite, of the form , i.e. the limit doesn’t exist. is infinite, of the form , i.e. the limit doesn’t exist.

6

7

Using l’Hôpital’s rule and then rearranging and using l’Hôpital’s rule again:

8

Using the double angle formulae and then using l’Hôpital’s rule twice:

9

Using the Maclaurin expansion and then cancelling and multiplying the numerator and denominator by :

10 a Maclaurin series: Then

has series

b From part a:

11 Using l’Hôpital’s rule and then multiplying the numerator and denominator by then using l’Hôpital’s rule again:

12 a b

, so is the infinite sum for a geometric series with common ratio

(with

):

Therefore the integral of such a function gives:

So c Using the Maclaurin series:

d So

EXERCISE 10C 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 just to 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 use the Maclaurin expansion for repeatedly. In the limit bracket you now have just a polynomial of , and can apply the limit directly. Continuing the calculation:

MIXED PRACTICE 10 1

A is a proper integral: the function is defined except at interval. B is a proper integral: the function is defined except at interval. C is a proper integral: the function is defined for D is an improper integral: the function is defined for integral interval.

, so is defined throughout the integral

, so is defined throughout the integral

, so is defined throughout the integral interval. , so is not defined at the lower end of the

(Answer D)

Tip Ensure that you can show that the improper integral in D does converge.

For the final step of this calculation, recall from Key point 10.2 in the Student Book that . 2

a

so so Maclaurin series:

. b Substituting

:

3

4

a

so so so so so

b Maclaurin series:

c Substituting

5

a

into the result from part b:

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

This limit does not converge, so the integral does not have a finite value. 6

a

so so so so so Maclaurin series:

b Substituting

into the result from part a:

So

Tip Note that substituting

into the result from part a gives an alternative

approximation:

While equally valid, this is not the result required; inspection of the denominator should give a hint as 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?

7 Then, using l’Hôpital’s rule:

8

so

a

so

so so b Maclaurin series:

c

9

a

b

i

so so so

so ii Maclaurin series:

c

10 Using l’Hôpital’s rule:

11 Using l’Hôpital’s rule twice:

12 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.

Tip The general terms for parts a and b would also commonly be written as

and

, respectively. Make sure you can see that these are equivalent to the general terms stated in the worked solutions shown here. c Substituting

Then

into the integral and using integration by parts:

13 a

i

So ii

b Maclaurin series: ,

,

,

and

So the first three non-zero terms of the Maclaurin series are:

c

d

i

This limit can only converge if after cancellation.

, so that the powers of inside the limit are non-negative

ii Using the result from part d i:

14 a i

ii b

i Maclaurin series: ,

,

and

So the first three non-zero terms of the Maclaurin series are:

ii So, substituting

c

for in the expansion found in part i:

Worked solutions 11 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 black-coded practice questions. EXERCISE 11A

4 a Homogeneous equation is

Complementary function: b

.

for unknown constant .

, so Substituting into the differential equation: (for all ) Comparing sides of the equation:

, so

Particular integral:

c General solution: Substituting

: , so

Particular solution: 5 a Homogeneous equation is

Complementary function: b

, so

for unknown constant .

Substituting into the differential equation: (for all ) Comparing sides of the equation:

and

Particular integral: c General solution: 6 a Try

, so

Substituting:

So

and , so

and

Particular integral: b Homogeneous equation is

Complementary function:

.

for unknown constant .

General solution: , so Particular solution: 7 a Homogeneous equation is

Complementary function: b Try

.

for unknown constant

, so

Substituting into the differential equation: Particular integral: c General solution:

EXERCISE 11B

2 Integrating factor:

General solution: Boundary condition:

, so

Particular solution: 3 Dividing through by

to rearrange into the form

:

to rearrange into the form

:

Integrating factor:

General solution: 4 Integrating factor:

General solution: 5 Dividing through by

Integrating factor:

General solution: Boundary condition:

, so

Particular solution: 6

Dividing through by

to rearrange into the form

:

Integrating factor:

General solution: Boundary condition: Particular solution: 7 For an equation of the form integrating factor

including the constant of integration would make the 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: Boundary condition:

for unknown constant

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 over which the function

, the primary interval 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 check that the function is restricted to the domain .

11 a Let

, so that

Substituting into

:

Rearranging: So b

Integrating factor:

.

as required.

General solution: Converting back to the original variable:

General solution:

EXERCISE 11C

1 a Auxiliary equation: b

, so

or

General solution: 2 a Auxiliary equation: b

, so General solution:

3 a Auxiliary equation: b

, so

(repeated root)

General solution: 4 a Auxiliary equation: , so

or

General solution: b From part a: Initial conditions: , and so Particular solution:

,

5 a Auxiliary equation: , so

(repeated root)

General solution: b From part a: Initial conditions: From Particular solution: 6 a Auxiliary equation: , so General solution: b From part a:

Initial conditions: From 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:

,

From 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

and so

General solution:

Tip Notice that having found two possible solutions, because this is a homogeneous equation any multiples of those solutions are also valid; 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

Then

and

, so

b Auxiliary equation: , so

or

General solution: c Converting back to the original variable: General solution: So

.

.

Initial conditions:

Substituting So

into

:

and

Particular solution:

EXERCISE 11D

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: Particular integral: c General solution: 3 a Auxiliary equation: , so

or

, so

.

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

and

Substituting:

.

, so

Particular integral: General solution: c Initial conditions: So

,

Particular solution: 5 a Auxiliary equation: , so

(repeated root)

Complementary function: Try

so

Substituting:

Comparing coefficients:

, so

and

From From Particular integral: General solution: b From part a: Initial conditions: So

,

Particular solution: 6 a Auxiliary equation: , so Complementary function: Try

so

and

Substituting:

Comparing coefficients:

From

:

From

:

From

:

Particular integral: General solution: b From part a: Initial conditions: So

,

Particular solution: 7 a Auxiliary equation: , so

(repeated root)

Complementary function: b Try

, so

and

.

Substituting:

Comparing coefficients:

So Particular integral: c General solution: d From part c:

Initial conditions: So

,

Particular solution: 8 Auxiliary equation:

, so

Complementary function: Try

, so

and .

Substituting:

Comparing coefficients:

From

:

From

:

From

:

Particular integral: General solution:

Boundary conditions:

So

,

Particular solution:

9 a Let

, so

and

.

Substituting:

b Auxiliary equation:

, so

Complementary function: Try

, so

Substituting:

Comparing coefficients: So

,

Particular integral: General solution: Converting to the original variable:

MIXED PRACTICE 11 1

and

.

Auxiliary equation:

, so

Complementary function:

(Answer A)

2 Integrating factor: 3

(Answer B) , so

a

or

b 4

a

has the form Integrating factor:

b

General solution: 5

a Auxiliary equation: , so

or

Complementary function: b Try

, so

and

Substituting:

. , so

Particular integral: c General solution: d

Initial conditions:

, so

,

Particular solution: 6

a Try Substituting:

, so

and , so

.

Particular integral: b Auxiliary equation:

, so

Complementary function: General solution: 7

a Auxiliary equation: , so Complementary function: b Try

, so

and

.

Substituting: Comparing coefficients: From

:

From

:

Particular integral: c General solution: d

Initial conditions: From, From Particular solution: 8 Auxiliary equation: , so

or

Complementary function: Try

, so

Substituting:

Comparing coefficients:

and

.

So Particular integral: General solution:

Boundary conditions:

From From Particular solution: 9 Auxiliary equation:

, so

Complementary function: Try

so .

Substituting:

Comparing coefficients:

So

,

,

,

and

Particular integral: General solution: 10 a Dividing through by to rearrange into the form : Integrating factor:

and

b Let

, so

.

Substituting into

:

c Converting to the original variable: , so

(where unknown constant

11 a Dividing through by

to rearrange into the form

Integrating factor:

General solution: b Boundary condition: Particular solution: 12 a Let

, so

and

.

Rearranging the original differential equation:

Substituting:

b Auxiliary equation: , so

(repeated root)

Complementary function: Try

, so

Substituting:

Comparing coefficients: From From

and

.

:

)

Particular integral: General solution: Converting to the original variable:

13 a

Integrating factor:

b Let

then

, so

Substituting:

c From part a: Then since

:

Using the fact that

14 a

and rewriting:

for some constants and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

.

(1) (2) , so b Let

, so

.

Integrating factor:

Converting to the original variable: , so

Initial conditions: So

and

Particular solution:

Worked solutions 12 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 black-coded practice questions. EXERCISE 12A

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 a Using

, so

and

, with the positive aligned to the direction of travel:

, so b Separating variables:

When

,

, so

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

or where

and

.

4 a

for population and time (years).

Boundary conditions:

From From Then b From part a:

, 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:

From

so

Then from

, so

Then c From part b: Given the accuracy level of the parameters used, a realistic estimate of the temperature would not be more accurate than approximately . d Different parts of the chicken are likely to be at different temperatures (it is not uniform, the thermal energy has to travel through the chicken). 6 a b

as a function of is a quadratic with roots at maximum point occurs when

and

, so by symmetry, the

.

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 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 into 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

V

c

8 t —

–3 3 V = (2 – e )

1

t

O As

,

EXERCISE 12B 4 a Let be the displacement above the equilibrium position

.

SHM with amplitude and angular frequency has general solution Initial condition:

.

, so

Particular solution: Then At

, the particle is

b

above the equilibrium position.

, so the maximum speed of the particle is

5 a Let be the displacement above the equilibrium position SHM with amplitude and period has general solution frequency So

.

. . , where angular

Initial conditions: So

,

,

Then At

, so , the particle is

below the equilibrium position.

b The particle first passes through Maximum speed is

when

, so

.

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 a For SHM with amplitude and angular frequency , Substituting

,

and

.

:

Period b Substituting

,

and

:

8 Let be the displacement above the equilibrium position

.

General solution: Amplitude

and period

Initial condition:

so angular frequency

.

, 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

,

and

.

:

Period b Now substituting

10 a

The amplitude is

,

and

and the angular frequency

b Taking displacement at as positive c The displacement at is Solving

into the formula:

so the period is

,

.

.

.

:

, so

d Acceleration

, so when

,

11 a The extension of the spring is (so that negative values of represent compression). The tension will be directed against the extension. , so therefore

which describes SHM with angular frequency . The period is .

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,

.

12 a Force So , and since for constant angular frequency describes simple harmonic motion. b At the moment of impact,

and

.

Under SHM with amplitude and angular frequency , Substituting:

, the equation

.

, so

, so maximal acceleration, which occurs at maximum compression At that compression, force c SHM with amplitude equation:

.

. , angular frequency

Maximum compression occurs when

:

and initial position

has

, so

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 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

.

15 a At equilibrium, the tension (acting upwards) is equal to the weight:

By Hooke’s law: So stiffness b i Extension

, measuring positive as displacement below point .

Then So ii

for

.

describes SHM; in this case, angular frequency

iii Period iv

( s.f.) and

Then

, so so maximum speed is

.

.

16 a Period is

, so

b i SHM with angular frequency and amplitude :

When

,

, so

Then

( s.f.)

ii The question relates to time taken to pass between points so for simplicity of calculations, set at . Then At ,

so the times at which the particle passes this point satisfy

Then

.

(primary solution)

The secondary solution is So the time to pass from to and back to is iii Maximum speed is

( s.f.).

.

17 a At displacement , the extension of the spring attached at is the spring attached at is

and the extension of

Then the tension in the spring attached at is and the tension in the spring attached at is

.

The resultant force (in the direction of positive ) is therefore

b

, so Therefore

, which describes SHM for angular frequency

.

c The period of motion is d SHM with angular frequency

When e At

and amplitude

.

, ,

and

, so

18 a The extension of the spring at time is The spring has stiffness

, so

Then resultant forces on the particle are , so Then b Complementary function:

. .

Particular integral: try Then Substituting into the differential equation and comparing coefficients:

So

,

and

So General solution: So

Initial conditions:

So

and

Particular solution:

EXERCISE 12C 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 heavy.

4 a

, so

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 light damping. x x = 0.9e–2t x = 0.9e–2t sin 3t t

O

x = –0.9e–2t

6 Tension

and resistance

Resultant force

.

, so

.

Then b Auxiliary equation: , so

or

General solution: So Initial conditions: , so Particular solution: c Solving for

:

and

So

which has no solution for

.

d This is heavy damping. 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

, so

, weight

and resistance

.

Then c Auxiliary equation: Discriminant

, so this equation describes light damping.

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

( s.f.)

9 a Auxiliary equation: The discriminant of the auxiliary equation If

then

, so the damping is light.

b Solutions to the auxiliary equation: General solution: So Initial conditions: So

,

Particular solution: c

, so initially the spring is extending. The maximum extension therefore occurs on the first occasion when the speed is zero.

So

when

So At When

, the spring is at maximum extension. :

So at this time, 10 a Forces acting on the jumper after

(as long as the rope remains taut; i.e. for

):

tension weight resistance Then

, so

Rearranging:

b Auxiliary equation:

Complementary function: Particular integral: try

; substituting gives

, so

General solution: So Initial conditions: From Then from Particular solution: c Then So the jumper will be instantaneously at rest when Then

for

So

.

: seconds

11 a Displacement of is equal to the displacement of , less any extension of the spring. moves according to constant acceleration , so the displacement of is given by the constant acceleration equation acceleration

, so

. In this case, the initial velocity

.

Therefore b The velocity of is given by the derivative of the displacement:

c Taking the derivative again: The acceleration of is governed by Forces governing the movement of are:

.

and constant

tension

(acting in the direction of movement)

resistance

(acting against the movement)

Equating the two forms for :

Rearranging and cancelling :

d

, so Auxiliary equation: , so

or

Complementary function: Particular integral: General solution: So

Initial conditions:

, so

, and so

Particular solution:

EXERCISE 12D 2

, so Substituting

:

Then Auxiliary equation:

, so

Complementary function: Particular integral: General solution: So Paired system general solution:

3

, so Substituting

:

Then Auxiliary equation:

, so

General solution: So Paired system general solution: 4

, so Substituting

:

Then Auxiliary equation:

, so

Complementary function: Particular integral: Substituting:

, so

General solution: So Paired system general solution:

Initial conditions: So

,

Paired system particular solution: 5 a

, so Substituting

and

Then b Auxiliary equation: , so General solution: So

or

:

Paired system general solution:

Initial conditions: , so

and

Paired system particular solution: 6 a The rate of water loss from the top can is proportional to the height of water in the can. Cross-sectional area is metres. , so

, so the volume of the top can

from which

where is measured in

.

Integrating factor:

General solution: Initial condition: Particular solution: For the second can, water flows in at the same rate that it flows out of the first can (ignoring transit time between cans) and flows out under the same relationship to height as for the first can.

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

; that is, the second can never fully

, so maximum height occurs when

At that time, the height in the second can is 7 a

for all

, which is at

or

.

b

, so Substituting

and

:

So c Auxiliary equation:

, so

Complementary function: Particular integral:

; substituting:

, so

General solution: So , so Initial conditions: So

,

Particular solutions: d For some

and

:

and So so

and the first peak in shark population occurs when .

By similar calculations:

and So e

and peak fish population first occurs at

The populations will oscillate with the same period be out of phase by population.

8

time units. They will

time units, with the shark population peaking after the fish

and 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 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 about 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 12 describes SHM with angular frequency and period

1 So 2

describes SHM with

. (Answer C)

Critical damping occurs when the auxiliary equation has a zero discriminant. Auxiliary equation: Discriminant (Answer D)

3

.

a Force

, so

b Integrating: Initial condition:

, so

Integrating again: Initial condition:

, so

Then 4

a SHM with amplitude equilibrium position So

and angular frequency .

has displacement from the

and therefore maximum speed is

.

b When

, , so maximum acceleration occurs when

c

, so maximum force is 5

:

.

, so

a b Try

.

Then

, and so

as required.

A second-order differential equation has a general solution with two unknowns. So for

, the general solution is

Tip Take care to address the question as it requires you to ‘show that’. 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

a In SHM with amplitude and angular frequency :

When

,

and when

,

.

Substituting these values into the equation:

.

Rearranging: , so b Substituting into , so

: .

Maximum speed is 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 8

when

, so the particle first returns to the equilibrium position at

, so Initial condition: Then Rearranging: The total population in the model is

, so when

, so

9

Substituting

and

:

So Auxiliary equation: General solution: So Then

Initial conditions:

So

and

, so

,

( s.f.).

.

Particular solutions:

, so

10

Integrating directly: Initial condition:

Initial condition: Particular solution: So , so Rearranging: Auxiliary equation:

, so

Complementary function: Particular integral: try

; substituting:

, so

.

General solution: So Initial conditions: , so Then

and

, so

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 , Substituting into the formula:

and

.

, so

.

Then b The period of oscillation is

(the time taken to travel from to and then back to again).

So c

, so maximum acceleration occurs when

.

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:

, so

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

:

at which point

.

Auxiliary equation: , 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. , so

15

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 showing the faster solution you might be able to spot! However, there is a problem with the fast approach, detailed in the longer answer shown. 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

or

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

.

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 a

is measured as positive towards , as shown in the diagram. At , each spring is in extension by :

. So when the particle is displaced

the tension in the spring attached at is and the tension in the spring attached at is

.

in the direction of

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 18 a

. and speed and displacement are related by the equation:

,

i Force

.

.

, so , which describes SHM with angular frequency

.

ii Period equals b

i Force

, so

, so

ii

Auxiliary equation: , so

or

General solution: So Initial conditions: , so

and

Particular solution: iii The discriminant of the auxiliary equation is positive so that there are distinct real roots:

Therefore the model shows heavy damping.

x

O

t

19 a Each elastic string experiences an extension of , producing a tension

.

Resultant forces

, so

Auxiliary equation: In critical damping, the discriminant of the auxiliary equation equals zero. , so b

, so

(repeated root)

Complementary function: Particular integral: try Substituting:

, so

General solution: So Initial conditions:

So

and

Particular solution: c As d

, , so

Maximum speed when

, so

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: He 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 might increase his effort towards the end as he tries to burn all his remaining energy (in a final sprint), so the final force might not be the lowest applied during the challenge. If the pulling force follows a linear pattern, (for time seconds and force b Newtons). Then: boundary conditions: Substituting

into

:

, so

Then if metres is the displacement from the starting point: , noticing that

throughout. (This is

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

( s.f.).

d The force Bill applies follows a negative exponential curve with the same start and end points as the linear function of Mike’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 13 Numerical methods Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. EXERCISE 13A 3

i, ii

Exact

Approx.

Exact

Approx.

error

Approx.

error

a b c d

error

Approx.

error

a b c d iii In most cases, doubling the number of intervals reduces the error by between a half and three quarters. 4

a

y x = 1

O

b

2

y = 4 ln (x – 1)

3

4

;

x

5

;

,

Total

So

( d.p.)

c The curve is concave (has negative second derivative), so the mid-ordinate rule produces an overestimate. 5

a

;

;

,

Total

So

( d.p.)

b You could improve the accuracy of the estimate by using more intervals. 6

a If

, then ;

b

, so ;

,

Total

So

( d.p.)

c The curve is concave (has negative second derivative), so the mid-ordinate rule produces an overestimate. 7

a Direction changes when changes sign; that is, when Then

, so

.

,

b Distance travelled is equal to the area enclosed between the curve and the horizontal axis (irrespective of whether the area is above or below the axis). So distance Estimating each integral, using the mid-ordinate rule, use

so that the area above the axis is

estimated by two strips and the area below the axis by six strips, with all strips then having equal width. ;

;

,

Total

So

( d.p.)

EXERCISE 13B 3

i, ii

Exact a b

Approx.

error

Approx.

error

c d

Exact

Approx.

error

Approx.

error

a b c d iiiThe error reduces by a factor of to 4

a

(or more) when the number of intervals is doubled.

y

y = 2 ln (x2 + 1)

O

1

2

3

;

b

x

4

;

-value

, -value

Total

So 5

a

( d.p.) ;

;

,

Multiple

-value

-value

Multiple

Total

So

( d.p.)

b You could improve the accuracy of the estimate by using more intervals. 6

a

, so , ;

b

;

,

-value

-value

Total

So 7

a Roots are at

( d.p.) , so the radius of the semicircle is .

Area b

i Using the symmetry of the semicircle:

Multiple

Using six strips on the quarter circle and doubling the result can be expected to give a better estimate than six strips on the semicircle, since it splits the area into narrower strips with lower errors when fitting a quadratic to sets of three points. ;

ii

;

,

-value

-value

Multiple

Total

So

and therefore

c Then

( d.p.)

EXERCISE 13C 2 Euler’s method:

Using the Euler method with step ( d.p.) 3

a

,

:

( d.p.)

Euler’s method:

,

Using the Euler method with step

:

( d.p.) b Using integration by parts:

Boundary condition: Particular solution: c

i

ii Iterating with smaller values of can generally be expected to give a more accurate estimate. 4

a

i Euler’s method:

,

Using the Euler method with step

:

( d.p.) ii Iterating with smaller values of can generally be expected to give a more accurate estimate. b Separating variables: , so

Boundary condition: Particular solution: Then

( d.p.)

y

c

1 3 — 4

1 1 — 3 x3 — y = e 2

1 — 2 1 — 4 O

1 — 4

1 — 2

3 — 4

1

x

The curve is convex, so the tangent is always below the curve and also has a lower gradient than the curve at each approximation, increasing the error. 5

a

, so Boundary condition:

, so

So b Euler’s method:

,

Using the Euler method with step

:

( d.p.)

EXERCISE 13D 2

a Euler’s method:

Using the Euler method with step ( s.f.)

,

:

b Improved Euler formula: ( s.f.) 3

a Euler’s method:

,

Using the Euler method with step

:

( s.f.) b Euler’s method:

,

Using the Euler method with step

:

Improved Euler formula: ( d.p.) c The curve is consistently convex throughout the interval of interest so the improved Euler formula will be more accurate than the simple Euler method. 4

a Euler’s method:

,

b Improved Euler formula:

5

a Euler’s method:

,

( s.f.) b Improved Euler formula: ( d.p.) c

i , so Separable equation:

Boundary condition: when Then

,

so

, so , so

(selecting the positive root so that

when

Then ii Substituting

into the equation from part c i:

MIXED PRACTICE 13 1

Nine ordinates: strips across an interval of width will each have width (Answer D) ;

2

Total

So 3

( d.p.) ;

Total

.

)

So

( d.p.) ;

4

;

,

-value

-value

Multiple

Total

So

( d.p.)

5 -value

-value

Total

So 6

( d.p.) ,

Multiple

( d.p.) 7

a Euler’s method:

,

b Improved Euler formula: 8

( s.f.)

,

a

( d.p.) b

( d.p.) 9

,

a

( d.p.) ( d.p.)

b

10 The second derivative is negative over the interval, so the curve is concave, and therefore the midordinate rule will overestimate the integral. Since taking additional intervals should improve the accuracy of the estimate, increasing the number of intervals will decrease the error, bringing the estimate down towards the true integral value. So the value of the estimate will decrease. (Answer B) ;

11 a

Total

So b If

( d.p.) , then

, so

Partial fractions:

for some constants and .

Multiplying through by the denominator on the LHS:

Substituting Substituting

, so , so

Then

( d.p.)

c d The mid-ordinate approximation would be likely to improve when using more strips. 12 a -value

-value

Multiple

Total

So

( d.p.)

b The function being integrated is even (its curve is symmetrical about the -axis), so the area between and is the same as twice the area between and .

Using an approximation for the area between and and doubling the result is the same as taking strips for the interval , which gives a more accurate result than strips for the same interval. 13 a Using Simpson’s rule approximation with strips:

-value

-value

Multiple

Total

So the approximation is exactly equal to the true value. b Generalising to a cubic

where

is a quadratic:

By part a, the Simpson's rule estimate of the first two integral values will be exact. Since Simpson’s rule fits a quadratic curve, it will approximate any quadratic by that exact quadratic so Simpson’s rule (with any number of strips) will always be accurate for a quadratic. Therefore Simpson’s rule, with two strips, is always exact for the integral of any cubic. 14 a

,

i

( d.p.) ,

ii

b The answer to part a i will be more accurate; the curve gradient will always be increasing, so iterating with a smaller step size will lead to a more accurate approximation. 15 a

;

Total

So

( d.p.) so, using the product rule:

b

Then 16 a -value

-value

Multiple

Total

So

( s.f.)

b

17 -value

-value

Multiple

Total

So

( s.f.)

18 Mid-ordinate rule approximation: ;

Total

19 a

, so b

, so Integrating factor:

Using integration by parts:

Initial condition: Particular solution:

, so

c Using the result from part b: Percentage error: ,

20 a

( d.p.) b

i

ii Using the Maclaurin series:

, and from part b i:

So iii Substituting

into the equation from part b ii: ( d.p.)

Worked solutions Cross-topic review exercise 1 Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions. 1

so

a Therefore

( s.f.)

b 2

a b Let

.

Then and

, so

.

Then for in the required interval:

3

New curve is Rearranging: and so

. transformed to

replacing with replacing with 4

a Root

of

:

represents a translation represents a stretch parallel to the -axis with scale factor becomes vertical asymptote

Vertical asymptote Horizontal asymptote

of of

becomes root

of

.

of

.

becomes horizontal asymptote

of

.

.

y x = 2

1 y = — f(x)

1 y = — 3

x

O

y

b

y = |f(x)|

y = 3 O

5

x

2

y

a

y = |arsinh x|

x

O b Since

is an even function,

, so 6

or

a i, ii y y = |x2 – x – 12| y = |3x – 7|

12

7

–3

O

7 — 3

4

x

b Four intersections. From left to right: meets

can be rearranged to

, so meets , so meets meets Then 7

for

.

a b The line perpendicular to

8

or

and

a Two vectors in the plane are

will have direction vector

and

so the normal to the plane is:

A vector equation of the plane is therefore b The normal to the second plane is

, which in Cartesian form is .

The angle between the planes is equal to the angle between the normals:

So the acute angle between the planes is 9

a Using the binomial theorem:

Since this is known to be real:

Using the quadratic formula to solve this quadratic in :

b Using De Moivre’s theorem:

c But Then

because

is known to be real.

( s.f.)

.

So

, so

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.

10

So

for some integer .

The smallest integers and for which this is true are

,

11 a

b If

, then

Then

, and so

12 a Since

and

for all vectors and :

b If and are perpendicular, then

.

From part a:

So 13 a Writing the system of equations in matrix form,

and

.

There is no unique solution when

.

Expanding about the first row:

So there is no unique solution for

or

b Eliminating variables from the equations:

.

where

.

So 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. 14 a Expanding about the second row:

b

i The determinant gives the volume scale factor of the transformation. , so So the scale factor is .

ii

is given by matrix

for

This describes a rotation about the -axis through

.

Tip A common mistake is getting the direction (or equivalently, angle) of rotation wrong. Remember that you can’t just ‘drop the centre row and column out’ to assess the direction of rotation. When rotating about the -axis (i.e . in the normal 2-dimensional plane), goes to and to . When rotating about the -axis, goes to and to

.

When rotating about the -axis, goes to and to

.

So each of these is a rotation about the unchanged axis:

,

15 a

i

Then when So

,

,

ii If

b

, then

i Expanding about the first row:

, then every element of ii From part a i: if therefore it would have zero determinant. Since

,

would have the same value, and

, so when

,

.

, so The cubic from part b i must factorise as

, so

Then the only real value for for which

is

.

.

Tip You could find this value by trial and error or calculator search; however, you should always look to see if a previous part of a question gives you a more direct algebraic approach to solving a higher order polynomial.) iii If

is an eigenvalue of , then for some associated eigenvector ,

, so 16 a

Eigenvalues are the roots of the characteristic equation

.

Expanding about the first row:

So the roots are b Solving

For

and

(repeated root).

to find associated eigenvectors

:

,

and

:

:

Substituting

:

So the eigenvalue

is associated with the eigenplane

.

The associated eigenvectors are any two non-parallel vectors spanning the plane, such as and

For

.

:

Taking the vector product of two distinct rows of the matrix gives the eigenvector:

, so

.

(spanned by and ) is invariant under the transformation; is an c The plane enlargement in this plane with scale factor .

, associated with eigenvalue

invariant points

.

17 a , so

b c

, so

18 a Then b

, so Multiplying through by , so Therefore,

:

, is the normal to this plane and spans the line of

and so

, where

.

Tip you get and not just multiple of still satisfies :

. You can check that the addition of any .)

19 a

and

b The normal to the two planes are and respectively. The line of intersection of two planes must be perpendicular to both normals, so the direction vector of the intersection line must be parallel to . The intersection line has equation intersection line, where . Then line

for some position vector of a point on the

must be perpendicular to the intersection line, since because by definition, is perpendicular to both

and . 20 a Characteristic equation:

Eigenvalues are

,

b Solving

to find associated eigenvectors

For

:

For

:

c Let

and

:

, so

, so , the matrix with the eigenvectors as columns and let

diagonal matrix with eigenvalues as elements of the lead diagonal. Then

d so

, so

, the

e

If

and

Then as

, then

and

as

.

,

So 21 a

i Expanding about the first row:

So the matrix is non-singular for

.

ii The matrix of cofactors

Then

b When

,

reversed under

, so the volume scale factor of

and the orientation would be

.

The image of an object with volume c If

is

would have volume

.

, then

But

That is, the image of any point under transformation will lie on the plane Cartesian form is written

.

, which in

d For a line of invariant points, require at least one eigenvalue

.

That is, Expanding about the first column:

Setting

, the eigenvector associated with

will have

:

Taking the cross product of two distinct rows of the matrix gives :

so

The line spanned by this eigenvector is So for 22 a

,

.

has a line of invariant points

.

, so using De Moivre’s theorem:

i

Then So ii b

i ii From part a: and Therefore

c Let

, so that

Limits: when

,

and and when

,

.

23 a Then

So b

and

i Substituting the result from part a: becomes

where

,

and

ii Using

: (1)

or Require

, so

Then from (1):

. , so

, and therefore

,

Then and So Rearranging: iii 24 a

i

is a hyperbola, and is the image of under a rotation; therefore also describes a hyperbola. , so , so

ii , b

where

and

represents an enlargement (scale factor

. ) together with a rotation through

anticlockwise about the origin. c

i Rearranging:

, which conforms to the standard equation

oriented ellipse centred at the origin, with horizontal semi-radius radius

for an axis and vertical semi-

.

ii From part a ii: under transformation ,

and

Substituting:

iii As defined in part b, transformation produces an enlargement and a rotation, so cannot alter the shape of a curve; this new curve also therefore describes an ellipse.

Worked solutions Cross-topic review exercise 2 Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black-coded practice questions.

Tip In these worked solutions, sometimes the notation ellipsis .

is used in a sequence instead of an

This means that the missing terms all have raised to a power or greater. For example, the Maclaurin series for is

but you could write this more precisely as because the missing terms consist of raised to power 4 and

higher. For convergent series, where only a few terms are used to calculate an approximation, knowing the order of the subsequent terms can be useful in assessing the likely error and is particularly useful when determining the convergence of combinations of series. 1

a The graph will repeat in every quadrant, since

–

Tangents at the pole occur when for

for integer , so

Tangents at the pole:

.

–

–

π 2

r 2 = a sin 4θ

3π 4

π 4

π

0, 2π

O

5π 4

7π 4 3π 2

b Each of the four ‘petals’ has the same area.

2

a Asymptotes:

.

y

y = 1 y = tanh x x

O

y = –1

b

c Using part b:

or

So 3

or

a Using l’Hôpital’s rule:

b Using l’Hôpital’s rule twice:

4

a b

i

is not finite.

ii

5

So 6

a Using resultant force

 :

Separating variables:

So At

,

so

and therefore

Rearranging: and therefore b When

: and so

7

a

b

8

a

strips for the interval Mid-strip

: strip width Mid-strip

Total

So

( s.f.)

b In each of the intervals of the region considered, the curve is concave (has a positive second derivative, as seen by the persistently increasing gradient) so the mid-ordinate rule will always underestimate.

The area enclosed by the black outlined rectangle is the one calculated for a strip by the midordinate rule, and is the same as the area under the red tangent line (the two pink triangles must have equal area, as the black line and red line cross at the half-way mark of the strip).

But the red line is the tangent to the curve at the mid-ordinate mark, and so the area under the tangent for a concave curve must be less than the area under the curve, as seen. 9

,

a

( d.p.) ( d.p.)

b 10 a

with

i

,

so that

Then ii Rearranging the equation from part a i: Completing the squares: This is the equation of a circle, centre b

i

and radius

.

The area enclosed by a curve, using polar coordinates, is So the area bounded by the curve

.

is:

ii At an intersection of the two curves:

So where

and

Therefore So which this can be true.

for some angle , but since

there is no real value for

Therefore, there is no value at which the two curves intersect. iii The area of circle

with radius

is

Since, by part b i, lies entirely within difference in their areas:

11 a

, so

. , the area enclosed between the curves must be the

b

So

12 a Using integration by parts twice:

b Using the result from part a:

Then

13 a Arc length

, so

and

, so

Then b 14 a The interval of integration is infinite, so this is an improper interval. b Rewriting the integral as a limit and then using integration by parts:

15 a Dividing numerator and denominator of the integrand by :

b Using part a:

16 a

b Let

17 a

, so that

and

and

and then using

, so that

So b Using part a:

So 18 a

b Using the results from part a:

,

and

Maclaurin series: So c

19 Auxiliary equation:

, so

Complementary function: Particular integral: try General solution: Then

; then

, so

Initial conditions: So

,

Particular solution: 20 a

i Using the chain rule:

So ii Applying the chain rule again:

Then using the product rule:

b Substituting

into the differential equation and using part a:

Collecting like terms and dividing through by

:

c Using the inhomogeneous equation from part b: Auxiliary equation:

, so

or

Complementary function: Particular integral: try

so

and

Substituting: Comparing coefficients:

so

,

, so

Then General solution: Converting to original variable:

, so

and so the equation of motion is 21 a The amplitude of movement is some constants and measuring displacement from point . So

, and the maximum speed

for

Therefore

and so the period of the motion

b When the particle is at :

, so Therefore the speed is c Using the fact that at

So

( s.f.) ,

:

and therefore

Since

, simplifying the model:

d Using the result from part c:

, so

when

( s.f.)

22 a Intersection of curve

and circle

:

, so Solutions are

or

So the intersection polar coordinates are b

i

and

.

has angle So at So the point is

.

ii Angle So, using the cosine rule:

iii Then

, so angle is a right angle. Since is a radius of the circle and , it follows that must be tangent to the circle (since it meets a radius at the circumference at a right angle).

c The shaded region is the difference of the areas bounded by curve circle

for:

and (one third of a circle, enclosed area

The area is calculated as:

).

, so

23 a

and

Then

, so

, so the gradient of a normal to at a point with parameter is

The equation of a line through point

with gradient

has equation:

Rearranging: as required. b When

, the point described on is point .

The normal at intersects the -axis. Substituting

into the equation from part a with

, so Then the midpoint of

Then for all values of ,

and

lies on the curve

which follows the general format 24 Letting

is:

of a hyperbola.

:

Multiplying numerator and denominator on the RHS by

:

:

25 a

, so

and

b

Using the result from part a:

26 a Integrating by parts: Let

and

, so

b Using double angle identities:

c

i

, so From the graph,

and so when

,

.

The shaded area equals the area of the rectangle bounded by the axes, less the area under the curve for . For simplicity in working, set

so that

and therefore

and

,

.

ii Calculating the shaded area directly:

Equating the two answers, and changing the variable to :

27

for some constants , and . Multiplying by the denominator on the LHS:

Comparing coefficients: (1) (2) (3) , so

and then

and

So Therefore, where defined:

The function integration.

is not defined at

, which is outside the interval of

so

28 a b If

, then

Then

so

Then 29 a

and therefore , a straight line.

, and so

,

and

Using the chain rule:

The differential equation can be written as Applying the substitutions detailed:

Dividing through by :

b Auxiliary equation:

, so

Complementary function: Particular integral: try

Substituting into the differential equation:

Comparing coefficients:

, which is constant.

, so , so , so , so Then General solution: Converting to the original variable, using

:

30 a Resultant force Both forces resist displacement, so that when , each force is aligned towards positive displacement and when , each force is aligned towards negative displacement. So Cancelling and rearranging:

b

,

i

and

Auxiliary equation:

General solution: Then Initial conditions:

So

Particular solution: ii

passes through Dividing through by

when :

(since

)

c

i Auxiliary equation: (repeated root) General solution: ii An equation of the form roots) describes critical damping.

(resulting from an auxiliary equation with repeated

,

31 a Then

and

So b

i Surface of revolution

Then ii

32 a

i Using the chain rule and the product rule:

So ii Integrating both sides of the result from part a i with respect to :

So

,

b

i Surface of rotation

, so

,

and

Tip When completing a ‘show that ’ question in which you need to substitute values into a standard formula to reach a given result, provide all details of the substitution, explicitly; any marks available will only be awarded for complete clarity. For example, in the mark scheme for this part of the past paper question, marks were only awarded if a student had written that

and had shown that

within the working. ii Let

, so then

Limits: when

,

and when

,

Applying these substitutions and using the result from part a ii:

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