A Level Mathematics for OCR A Student Book 2 (Year 2) with Cambridge Elevate Edition (2 Years) (AS/A Level Mathematics for OCR) [Student ed.] 1316644677, 9781316644676, 9781316644362

Citation preview

Brighter Thinking

A Level Mathematics for OCR A Student Book 2 (Year 2)

Vesna Kadelburg, Ben Woolley, Paul Fannon and Stephen Ward

Contents Introduction How to use this resource Working with the large data set 1 Proof and mathematical communication Section 1: A reminder of methods of proof Section 2: Proof by contradiction Section 3: Criticising proofs 2 Functions Section 1: Mappings and functions Section 2: Domain and range Section 3: Composite functions Section 4: Inverse functions 3 Further transformations of graphs Section 1: Combined transformations Section 2: Modulus function Section 3: Modulus equations and inequalities 4 Sequences and series Section 1: General sequences Section 2: General series and sigma notation Section 3: Arithmetic sequences Section 4: Arithmetic series Section 5: Geometric sequences Section 6: Geometric series Section 7: Infinite geometric series Section 8: Using sequences and series to solve problems 5 Rational functions and partial fractions Section 1: Review of the factor theorem Section 2: Simplifying rational expressions Section 3: Partial fractions with distinct factors Section 4: Partial fractions with a repeated factor 6 General binomial expansion Section 1: General binomial expansion Section 2: Binomial expansions of compound expressions Focus on … Proof 1 Focus on … Problem solving 1 Focus on … Modelling 1 Cross-topic review exercise 1 7 Radian measure Section 1: Introducing radian measure Section 2: Inverse trigonometric functions and solving trigonometric equations Section 3: Modelling with trigonometric functions Section 4: Arcs and sectors Section 5: Triangles and circles Section 6: Small angle approximations 8 Further trigonometry

Section 1: Compound angle identities Section 2: Double angle identities Section 3: Expressions of the form Section 4: Reciprocal trigonometric functions 9 Calculus of exponential and trigonometric functions Section 1: Differentiation Section 2: Integration 10 Further differentiation Section 1: The chain rule Section 2: The product rule Section 3: The quotient rule Section 4: Implicit differentiation Section 5: Differentiating inverse functions 11 Further integration techniques Section 1: Reversing standard derivatives Section 2: Integration by substitution Section 3: Integration by parts Section 4: Using trigonometric identities in integration Section 5: Integrating rational functions 12 Further applications of calculus Section 1: Properties of curves Section 2: Parametric equations Section 3: Related rates of change Section 4: More complicated areas 13 Differential equations Section 1: Introduction to differential equations Section 2: Separable differential equations Section 3: Modelling with differential equations 14 Numerical solution of equations Section 1: Locating roots of a function Section 2: The Newton–Raphson method Section 3: Limitations of the Newton–Raphson method Section 4: Fixed-point iteration Section 5: Limitations of fixed-point iteration; alternative rearrangements 15 Numerical integration Section 1: Integration as the limit of a sum Section 2: The trapezium rule Focus on … Proof 2 Focus on … Problem solving 2 Focus on … Modelling 2 Cross-topic review exercise 2 16 Conditional probability Section 1: Set notation and Venn diagrams Section 2: Two-way tables Section 3: Tree diagrams Section 4: Modelling with probability 17 The normal distribution Section 1: Introduction to normal probabilities Section 2: Inverse normal distribution

Section 3: Finding unknown or Section 4: Modelling with the normal distribution 18 Further hypothesis testing Section 1: Distribution of the sample mean Section 2: Hypothesis tests for a mean Section 3: Hypothesis tests for correlation coefficients Focus on … Proof 3 Focus on … Problem solving 3 Focus on … Modelling 3 Cross-topic review exercise 3 19 Applications of vectors Section 1: Describing motion in two dimensions Section 2: Constant acceleration equations Section 3: Calculus with vectors Section 4: Vectors in three dimensions Section 5: Solving geometrical problems 20 Projectiles Section 1: Modelling projectile motion Section 2: The trajectory of a projectile 21 Forces in context Section 1: Resolving forces Section 2: Coefficient of friction Section 3: Motion on a slope Section 4: Further equilibrium problems 22 Moments Section 1: The turning effect of a force Section 2: Equilibrium Section 3: Non-uniform rods Section 4: Further equilibrium problems Focus on … Proof 4 Focus on … Problem solving 4 Focus on … Modelling 4 Cross-topic review exercise 4 Paper 1 practice questions Paper 2 practice questions Paper 3 practice questions Formulae Statistics Answers Gateway to A Level revision sheets Gateway to A Level revision sheet answers Worked solutions for chapter exercises 1 Proof and mathematical communication 2 Functions 3 Further transformations of graphs 4 Sequences and series 5 Rational functions and partial fractions 6 General binomial expansion

7 Radian measure 8 Further trigonometry 9 Calculus of exponential and trigonometric functions 10 Further differentiation 11 Further integration techniques 12 Further applications of calculus 13 Differential equations 14 Numerical solution of equations 15 Numerical integration 16 Conditional probability 17 The normal distribution 18 Further hypothesis testing 19 Applications of vectors 20 Projectiles 21 Forces in context 22 Moments Worked solutions for cross-topic review exercises Cross-topic review exercise 1 Cross-topic review exercise 2 Cross-topic review exercise 3 Cross-topic review exercise 4 Support and Extension sheet answers Support sheet answers Extension sheet answers Working with the large data set Working with the large data set answers Acknowledgements Copyright

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

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

Modelling Maths is valued because it helps us solve real-world problems. However, maths describes ideal situations and the real world is messy! Modelling is about deciding on the important features needed to describe the essence of a situation and turning that into a mathematical form, then using it to make predictions, compare to reality and possibly improve the model. In many situations the technical maths is actually the easy part – especially with modern technology. Deciding which features of reality to include or ignore and anticipating the consequences of these decisions is the hard part. Yet 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 resource 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 take on board these ideas (and do plenty of practice!) you will find maths examinations a much more approachable and possibly even enjoyable experience. However, always remember that the results of what you write down in a few hours by yourself in silence under exam conditions is not the only measure you should consider when judging your mathematical ability – it is only one variable in a much more complicated mathematical model!

How to use this resource Throughout this 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: review proof by deduction, proof by exhaustion and disproof by counter example learn a new method of proof called proof by contradiction practise criticising proofs.

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

Before you start… Student Book 1, Chapter 1

You should be able to use logical connectors.

1 Insert either and B:

Student Book 1, Chapter 1

You should be able to use disproof by counter example.

2 Disprove the statement ‘apart from 1 there are no other integers that can be written as both and

,

or

in the places marked A

’.

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.

Explore

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

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

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

CROSS-TOPIC REVIEW EXERCISE

Questions covering topics from across the preceding chapters, testing your ability to apply what you have learnt. 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 Paper 1, Paper 2, and Paper 3 practice questions, short answers 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.

Worksheet A support sheet for each chapter contains further worked examples and exercises on the most common question types. Extension sheets provide further challenge for the more ambitious.

Gateway to A Level

GCSE transition material that provides a summary of facts and methods you need to know before you start a new topic, with worked examples and practice questions. 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 below. 1

A sequence is defined by .

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

. Use the principle of mathematical induction to prove that

. 8 9

Prove that Prove using induction that

10 Prove that

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

Working with the large data set As part of your course you are expected to work with the large data set covering different methods of transport and age distributions in different parts of the country and in different years. This large data set is an opportunity to explore statistics in real life. We shall be using the large data set to guide you through the four following themes. You will not have to work with the full data set in the final examination, but familiarity with it will help you as many examination questions will be set in the context of this data set.

Practical difficulties with data Unlike most textbook or examination problems, the real world is messy. Often there are difficulties with being overwhelmed by too much data, or perhaps there are errors, missing items or labels which are ambiguous. For example, how do you deal with the fact that in 2001 Cornwall was made up of separate districts that were later combined into a single unitary authority, if you want to compare areas over time? If you are grouping data for a histogram, how big a difference does it make where you choose to put the class boundaries?

Using technology Modern statistics is heavily based on familiarity with technology. We will be encouraging you to use spreadsheets and graphing packages, looking at the common tools available to help simplify calculations and present data effectively. One important technique we can employ with modern technology is simulation. We will try to gain a better understanding of hypothesis testing by using the data set to simulate the effect of sampling on making inferences about the population.

Thinking critically about statistics Why might someone want to use a pie chart rather than a histogram? Whenever statistics are calculated or data sets are represented graphically, some information is lost and some information is highlighted. An important part of modern statistics is to ask critical questions about the way evidence provided by statistics is used to support arguments. One important part of this is the idea of validating statistics. For example, with the information presented it is not clear which category or categories a person would be included in if they travel to work by bicycle on some days and take the bus on others. We will look at ways in which we can interrogate the data to try to understand it more.

Statistical problem solving Technology can often do calculations for us. However the art of modern statistics is deciding what calculations to do on what data. One of the big difficulties is that we rarely have exactly the data we want, so we have to make indirect inferences from the data we have. For example, you will probably not see newspaper headlines saying ‘the correlation coefficient between median age and percentage of people cycling to work is ’, but you might see something saying ‘Pensioners promote pedalling!’ Deciding on an appropriate statistical technique to determine whether older people are more likely to use a bicycle and then interpreting results is the type of thing which is hard to examine but very valuable in real-world statistics. There are lots of decisions to be made. Should you use the total number of cyclists in an area? Or the percentage of people who cycle? Or the percentage of people who travel to work who cycle? We shall see how the answer to your main question depends on decisions like these.

1 Proof and mathematical communication In this chapter you will: review proof by deduction, proof by exhaustion and disproof by counter example learn a new method of proof called proof by contradiction practise criticising proofs.

Before you start… Student Book

You should be able to use

1, Chapter 1

logical connectors.

1 Insert either

,

or

in the places marked and

:

A B Student Book

You should be able to use

1, Chapter 1

disproof by counter example.

Student Book 1, Chapter 1

You should be able to use proof by deduction.

Student Book 1, Chapter 1

You should be able to use proof by exhaustion.

2 Disprove the statement ‘apart from there are no other integers that can be written as both

and

’.

3 Prove that the sum of any two odd numbers is always even. 4 Use proof by exhaustion to prove that

is a prime

number.

Developing proof One of the purposes of this chapter is to act as revision of the material from Student Book 1. It draws on all chapters from that book but, in particular, it builds on the fundamental ideas of proof from Chapter 1. This chapter introduces a new and very powerful method of proof that mathematicians often rely on: proof by contradiction.

Section 1: A reminder of methods of proof In Student Book 1 you met proof by deduction, proof by exhaustion and disproof by counter example. The following questions show how these methods can be used in topics from throughout Student Book 1. EXERCISE 1A 1

Use proof by exhaustion to prove that

2

Prove by exhaustion that all square numbers end in

3

The velocity of a particle after time is given by

is a prime number. or . . Prove that the particle never returns to its

original position. 4

a Prove from first principles that if

then

b Use a counter example to show that if

. , then it is not necessarily true that

5

Use a counter example to show that the following statement is not true:

6

Prove that

7

A set of data has mean , mode and median . Consider the following statement:

.

. .

Prove this statement or use a counter example to disprove it. 8

The diagram shows triangle

, where

lies on the circle with centre

is the midpoint of

A M 2θ B

O

a Use the cosine rule to prove that b Show that

.

c Hence, prove that 9

In quadrilateral the midpoint of

.

, is the origin and are the position vectors of points , , and . is , is the midpoint of , is the midpoint of and is the midpoint of .

a Show that b Hence, prove that c If

is a parallelogram.

is a rectangle, what can be said about the quadrilateral

10 a Use a counter example to disprove the statement

? .

.

b Prove that if c If

.

, does it follow that

11 a Show that b Hence, prove that 12 Prove algebraically that if take is .



, then

? , where is a constant to be determined.

is prime if and only if

.

, then the sum of the probabilities of the different values can

Section 2: Proof by contradiction Proof by contradiction starts from the opposite of the statement you are trying to prove, and shows that this results in an impossible conclusion. WORKED EXAMPLE 1.1

Use proof by contradiction to prove that there are an infinite number of prime numbers. Assume that there is a largest prime number, .

Proof by contradiction always starts by assuming the opposite of what you want to prove.

Construct another number, , that is the product of all the prime numbers up to and including .

Now set about trying to find a larger prime than .

Consider . This is greater than a number divisible by all the primes up to and including , so it cannot be divisible by



any of the primes up to and including . Therefore, larger than .

is either itself prime, or is divisible by primes



Either way, you have shown that there must be a prime larger than . This contradicts the premise that there is a largest prime number.

Here, the contradiction to the original assumption (that there is a largest prime) occurs.

Therefore, there are an infinite number of prime numbers.



Did you know? A variation on the previous proof can be found in Euclid’s masterpiece The Elements, a textbook written in approximately but still in use in many schools in the first half of the twentieth century!

WORKED EXAMPLE 1.2

Prove that Assume that

is irrational. You may use the fact that if is even, so is . ,

where and are integers with no common factors.

To use proof by contradiction, you start from the opposite statement: that is rational. Th is means that it can be written as a fraction. Every fraction can be cancelled down to its lowest terms.

Squaring both sides gives:



This means that is even so must also be even.

Using the fact given.

You can then write that



, for some integer , so

. Substituting this into the equation marked

gives:



This means that is even, so must be even.

Using the given fact again.

But you have shown that both and are even, so they

Here, the contradiction (to the fact that and share no common factors) arises.

share a factor of . This contradicts the original assertion (that and are integers with no common factors), so it must be incorrect. Therefore,

is irrational and cannot be written as .

EXERCISE 1B 1

Prove that if

2

Prove that

3

Prove that there are infinitely many even numbers.

4

Prove that the sum of a rational and irrational number is irrational.

5

Prove that if is even with

6

Prove that

7

Prove that

is even, then is also even. is irrational.

integers, then at least one of them is even.

is irrational. is irrational.

Worksheet See Support sheet 1 for an example of the same type as Question 7 and further practice questions on proof by contradiction. 8

Suppose that is a composite integer. Prove that has a prime factor less than or equal to

9

Prove that if any

different dates are chosen, some must be within the same month.

10 Prove that the value of 11 a Show that if

.

is never if and are whole numbers.

is a solution to the equation

, then

.

b Explain why there is no solution to this equation if is odd or is odd. c Prove that there are no rational solutions to 12 Prove that if a triangle has sides

and such that

. , then it is a right-angled triangle.



Section 3: Criticising proofs In Student Book 1 you were introduced to the notation used in logic: means that statements and are equivalent. means if is true, then so is . means if is true, then so is . When checking a proof (including solving equations, which is a type of proof!) you have probably looked out for errors in things like arithmetic or algebra. You now need to also look out for errors in logic. WORKED EXAMPLE 1.3

Yas was solving the equation

. Find the errors in her working.

1    2 3 4 On line , the symbol should be if is negative, line could be correct but line is not possible.

This is an error in logic.

In line the symbol should be

This is an error in logic.

, then

if

.

In line the positive square root of should be .

This is an arithmetic error.

Because one of the implications goes only one way, the final solutions might not work in the original equation. They should be checked.

This is an error in logic. Even if Yas had not made the arithmetic error, she would still need to state that is not a valid solution, because a negative number cannot be substituted into the original equation.

EXERCISE 1C 1

Lambert was asked to solve the equation

. Here is his working:

1 2 3 4 5 or a By checking his working, find the correct solution. b In which line of working is his mistake? 2

Craig was asked to solve the equation 1   2 a Show that

. Here is his solution:

is also a solution to the original equation.

b What logical symbol should Craig have used in the second line?

3

Freja was asked to solve the equation

. Here is her working:

1   2 3 4 5 6 7 a By checking her working, find the correct solution. b In which line of working is her mistake? 4

Jamie was asked to solve

. Here is her working:

1 2 3 4 5 6 7 In which line of working did Jamie make a mistake? 5

Andrew was asked to prove the following statement: The function

has a minimum at

. His working is shown below.

1 2 3 4 5 6 7 So it is a minimum. Describe the errors in this proof. 6

Criticise the following proof of the statement: If

is a factor of

, then the other factor is

If

is a factor of

, then you can write:

.

1 2 3 Comparing coefficients of 4 Therefore, the remaining factor is 7

.

Find the error in the following proof that 1 Assume that

, where and are integers with no common factors.

2 Squaring both sides gives 3 So



is irrational.

.

.

4 This means that is even, so must also be even.

5 6 7 8

You can then write that , so . Substituting this into the equation marked gives So . This means that is even, so must be even.

.

9 But you have shown that both and are even, so they share a factor of . 10 This contradicts the original assertion, so it must be incorrect. This means that

cannot be

written as .

Checklist of learning and understanding You should be able to apply counter examples, proof by exhaustion and proof by deduction to material from Student Book 1. Proof by contradiction is a method of proof that works by showing that assuming the opposite of the required statement leads to an impossible situation. When criticising proofs, look out for flaws in logic as well as mistakes in algebra or arithmetic.

Mixed practice 1 1

Prove that

2

Prove that

3

Prove that there are infinitely many square numbers.

4

Find the error(s) in the following working to solve

is always even. is irrational.

for

:

1 2 3 4 5 5

Choosing from options to , which symbol should be used to replace? in the working below? A B C D 1 2

6

Prove that

7

is irrational.

is a parallelogram with at the origin, and , , are the position vectors of points , , and Prove that

is the midpoint of

and is the point on

such that

is

.

is a straight line.

8

Prove that if and are whole numbers, then

9

a Prove that if is a polynomial of finite order with integer coefficients and is an integer, then is an integer.

is either odd or a multiple of .

b Use a counter example to show that the following statement is not always correct: If is a polynomial, where have integer coefficients.

is an integer whenever is an integer, then

10 Consider the following working to solve 1 2 3 4

:

must

5 6 7 a In which lines are there mistakes? b Rewrite the solution correctly, making appropriate use of logical connectors. 11 Fermat said that if is prime then counter example to this statement.

is prime. Find the smallest value of that provides a

12 The proof below is trying to demonstrate that there are an arbitrary number of consecutive composite (non-prime) numbers. A Consider for B

.

is divisible by .

C So the numbers

are not prime.

D This is a list of consecutive composite numbers. Which is the first line to contain an error? 13 Prove that if

and are integers such that

, then either or is even.

14 a By considering a right-angled triangle, prove that if is an acute angle, then . b Hence, prove that 15 Prove that for between

is a rational number. and

,

.

Worksheet See Extension sheet 1 to complete the details of a couple of famous proofs.

2 Functions In this chapter you will learn: about the difference between mappings and functions about one–one and many–one functions about the domain and range of a function how to find composite functions how to find the inverse of a function.

Before you start… Student Book 1, Chapter 1

You should be able to interpret function notation.

1 Given that

, evaluate:

a b

Student Book 1, Chapter 1

You should be able to use set notation and interval notation.

2 Write the following sets using the interval notation. a b

Student Book 1, Chapter 3

You should be able to complete the square.

3 a Express

in the form .

b Hence, state the coordinates of the turning point of . Student Book 1, Chapter 3

You should be able to solve quadratic inequalities.

4 Solve the inequality

.

Student Book 1, Chapter 7

You should be able to rearrange exponential and log expressions.

5 Make the subject of the following. a b

Student Book 1, Chapter 13

You should be able to establish where a function is increasing/decreasing.

6 Find the range of values for which is an increasing function.

Why study functions? Doubling, adding , finding the largest prime factor – these are all instructions that can be applied to numbers to produce a numerical result. This idea comes up a lot in mathematics. The formal study of it leads to the concept of a function.

Functions can be used whenever you need to express how one quantity changes with another, whether it is how the strength of the gravitational force varies with distance, or how the amount of paint needed depends on the area of a wall. In this chapter we will focus on developing the theory of functions. You have already seen many examples of modelling with linear, quadratic and exponential functions in Student Book 1. In Chapter 7 of this book you will meet further models using trigonometric functions.

Section 1: Mappings and functions A mapping takes numbers from a given set (inputs) and assigns to each of them one or more output values, using a mapping rule. For example: maps to , to , and to . maps to and , to and , and to . factor of maps to , to and , and to , , and .

Did you know? There is a more general way to describe a relationship between two sets, called a relation, where the output does not need to be written explicitly in terms of the input. For example, a relation could be given by ; you already know that the graph of this relation is a circle. Although you will not study relations in much detail, you will learn how to find gradients of some curves defined by relations (see Chapter 10, Section 4). We can represent a mapping using the arrow notation, as shown above, or we can give a name to the output value and write, for example, . We can also use a mapping diagram:

Another useful representation is on a graph, where the input values are shown on the horizontal axis and the output values on the vertical axis. You should remember that, to fully define a mapping, you need to state the set of the input values as well as the mapping rule. This set of all possible input values is called the domain of the mapping. For example, the mapping is different from the mapping , as can be seen from their graphs: y

-2

y

6

6

3

3

-1 O

1

2

x

-2

-1 O

-3

-3

-6

-6

1

2

x

Fast forward You will learn more about domains in Section 2. A mapping rule can assign more than one output to each input. A special type of mapping, where each input value has only one output, is called a function. Of the three mappings given previously, only is a function.

When a mapping is given by its graph, the easiest way to decide whether or not it is a function is to carry out a vertical line test:

y

y

8 7 6 5

O

x

4 3 2 1

O function

x

0

0

1

2

not a function

3

4

5

6

not a function

Key point 2.1 A mapping is a function if every input value maps to a single output value. Vertical line test: if a mapping is a function, any vertical line will meet its graph at most once. Functions are often named with letters, such as or . For example, the function notation . Then , and the image of .

. You can also use ; we say that is

Having decided that a mapping is a function, you can ask whether each output comes from just one input. To check this, you can apply the horizontal line test:

Key point 2.2 A function is: one–one if every value corresponds to only one value. many–one if there is at least one value that comes from more than one value. Horizontal line test: if a function is one–one, any horizontal line will meet the graph at most once. A mapping where a single input corresponds to more than one output (so is not a function) is called one–

7

8

many.

WORKED EXAMPLE 2.1

Which of the graphs below could represent functions? For those that could be functions, classify them as one–one or many–one.

y

a

x

O

y

b

O

x

y

c

O

x

Draw several vertical lines and see how many times they cross the graph.

y

a

x

O

Any vertical line meets the graph at most once; therefore it could be a function. y

Draw several horizontal lines and see how many times they cross the graph.

x

O

At least one of the horizontal lines meets the graph at more than one point; therefore it could be a many–one function.

y

b

O

Draw several vertical lines and see how many times they cross the graph.

x

At least one of the vertical lines meets the graph more than once; therefore it is not a function. Draw several vertical lines and see how many times they cross the graph. Remember that a vertical line through an open circle does not count as an intersection.

Tip An open circle on a graph means that that point is not a part of the graph, and a closed circle means that it is.

y

c

x

O

Any vertical line meets the graph at most once; therefore it could be a function.

y

Draw several horizontal lines and see how many times they cross the graph.

x

O

Any horizontal line meets the graph at most once; therefore it could be a one– one function. When you are given a rule for a function, rather than its graph, it can be more difficult to decide whether the function is one–one. If a function is not one–one, you need to find only one example of an output value with two corresponding input values. To prove that a function is one–one, you need to rearrange the equation and show that each comes from only one . You should remember that whether a function is one–one depends on its domain, as well as the function rule. WORKED EXAMPLE 2.2

a The function is defined for all real numbers and has the rule one function. b The function also has the rule function. a Two inputs have the same output, hence is not a one– one function. b Let

. Then:

but its domain is

. Show that is not a one–

. Show that is a one–one

Look for two input values that have the same output. In this case, you know that squaring a number gives the same result as squaring its negative.

You need to show that each possible comes from only one . One way to do this is to find in terms of .

However,

so .

This equation appears to give two values for each value. However, the domain of is , so must be positive.

Each value comes from just one value, so is a one–one function.

EXERCISE 2A 1

Which of these graphs could not represent functions? For each one that could represent a function, state if it is one–one or many–one.

y

a

x

O

y

b

x

O

y

c

O

x

y

d

x

O

y

e

x

O

y

f

O 2

x

Determine whether each of the following functions is one–one. a

i ii

b

i ii

c

i ii

Section 2: Domain and range The previous section mentioned that, in order to fully define a function, you must specify both the set of allowed inputs and the rule that tells you what to do with each input.

Key point 2.3 The set of allowed input values is called the domain of the function.

Tip Remember that stands for the set of all real numbers and for the set of whole numbers.

WORKED EXAMPLE 2.3

Sketch the graph of

over the domain:

a b a

Sketch the graph over the domain

.

Discard the part of the graph that is outside the required domain. Since the end point is not included you must label it with an open circle.

Use the same original graph as in part a, but this time include only integer values.

b

If no domain is explicitly mentioned, you can assume that the domain is the largest possible set of real numbers for which the values are defined. You may wonder why you would ever need any other domain. There are two main reasons. 1 You may be modelling a physical situation where the variables can take only particular values; for example, if the variable is age of humans, you would not want it to be negative or much beyond

.

2 The mathematical operation you are using may not be able to handle certain types of numbers. For example, if you are looking for the largest prime factor of a number, you would normally only be looking at positive integers. When working with real numbers, the four most common reasons to restrict the domain are: You cannot divide by zero. You cannot take a square root of a negative number. You cannot take the logarithm of a non-positive number. You cannot find the tan or cot of certain angles (for example,

or

).

Tip Unless told otherwise, you can assume that the domain is a subset of real numbers. For example, in fact means and .

WORKED EXAMPLE 2.4

Find the largest possible domain of a function with the rule You need

.

You can only take a logarithm of a positive number.

.

So the largest possible domain is

.

WORKED EXAMPLE 2.5

A function, , is defined on the domain Find the value of .

and given by the rule

Look for any values of for which

is not defined.

You may remember where the asymptotes of the tan graph are, or you may need to use the definition of tan and look for division by zero.

Not defined when . This is when so . Hence, .

.

,

Tip Remember that

is just an alternative notation for

.

WORKED EXAMPLE 2.6

What is the largest possible domain of a function with the rule

? Write

your answer using interval notation. There will be division by zero when

.

Look for division by zero.

There will be a square root of a negative number when .

Look for a square root of a negative number.

The largest possible domain is:

Decide what can therefore be allowed into the function.

Hence,

.

and

This describes two intervals: from to infinity, excluding .

to and from

As well as knowing the set of inputs for a function (the domain), it is useful to identify the set of possible outputs.

Tip Be aware that the range will depend upon the domain.

Key point 2.4 The set of all possible outputs of a function is called the range.

Tip Graph-plotting software or graphical calculators can be useful when investigating the domain and range of functions. The easiest way of finding the range is to sketch the graph. WORKED EXAMPLE 2.7

A function, f, is given by the rule

. Find its range if the domain is:

a b Sketch the graph

.

y

a

y = f(x)

Range

3 Domain -3

-2

-1 O

1

2

3

x

Use the graph to state which values can occur.

Range: y

b

Sketch the graph

.

y = f(x)

Range 7

3 Domain -3

-2

-1 O

Range:

1

2

3

x

Use the graph to state which values can occur.

EXERCISE 2B In this exercise, unless otherwise stated, is a real number.

Rewind Exponential and logarithm functions were covered in Student Book 1, Chapter 7. 1

Find the largest possible domain, and the corresponding range, of the functions with the following rules. a

i ii

b

i ii

2

Find the largest possible domain of the functions with the following rules. a

i ii

b

i ii

c

i

ii d

i ii

e

i ii

f

i ii

3

Find the range of the following functions. a

i ii

b

i ii

c

i ii

d

i ii

4

Find the largest possible domain and the corresponding range of the functions with the following rules. a

i ii

b

i ii

c

i ii

d

i ii

Rewind Some of these questions require completing the square and the solution of quadratic inequalities, covered in Student Book 1, Chapter 3. 5

a Write

in the form

.

b Hence, state the range of the function

.

6

Find the largest possible domain and the corresponding range of the function with the rule .

7

The function has the rule

8

. Find the largest possible domain of this function.

Find the largest possible domain of the function with the rule

.

9

a Sketch the graph of

.

b Hence, find the largest possible domain of the function with the rule f: 10 Find the largest possible domain of the function with the rule 11

Find the largest set of real values of such that the function with the rule

. . takes real

values. 12 a State the largest possible domain of the function with the rule i ii b Find

in each of the two cases.

if:

Section 3: Composite functions After applying a function to a number it is possible to apply another function to the image. The resulting rule is called a composite function.

Tip Whichever notation is being used, remember the correct order: the function nearest to acts first!

Key point 2.5 Applying the function to and then the function to the result is written:

It can be useful to refer to

as the inner function and

as the outer function.

WORKED EXAMPLE 2.8

If

and

, find:

a b c a

Evaluate

and then apply to the result.

Note that you don’t need to work out the general expression for . b

Replace in

with the expression for

.

c

Replace in

with the expression for

.

Tip Be careful:

and

are not the same function.

WORK IT OUT 2.1 Two functions are defined for all real numbers by

and

. Find

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1

Solution 2

.

Solution 3

You can also compose a function with itself.

Key point 2.6 can also be written as

.

WORKED EXAMPLE 2.9

Given that

, find and simplify an expression for means do

.

, then do to the result.

Multiply top and bottom of the fraction by

to simplify the denominator.

For the composite function to exist, the range of must lie entirely within the domain of , otherwise you would be trying to put values into that it cannot take.

g

f

In the diagram above, the large blue rectangle represents the domain of . Its image (which is the range of ) is represented by the smaller blue rectangle. The larger red rectangle represents the domain of . WORKED EXAMPLE 2.10

The functions and are defined by



a Explain why the composite function

is not defined.

b Find the largest possible domain for which a

You need

.

But, for example, , and this is not in the domain of .

is defined. In this case, state the range of

.

Check whether the output from is within the domain of . It is enough to find one counter example to show that not true for all .

is

You need

b

This is a quadratic inequality; the solution consists of two separate intervals.

or The range of

.

is

With the domain above, takes all real values , so can be any non-negative real number.

.

A more complex problem is to recover one of the original functions when you have a composite function. A good way to do this is to use a substitution. WORKED EXAMPLE 2.11

If

, find the expression for

in the form

.

Substitute for the inner function. Rearrange to get Replace all instances of .

Write the answer in terms of .

EXERCISE 2C In this exercise, functions are defined with their largest possible real domain, unless specified otherwise. 1

Given that a

and

, find:

and

, find:

i ii

b

i ii

2

Given that a

i ii

b

i ii

3

a Given that

and

, find:

i ii b Given that

and

, find:

i ii 4

Given that a b

and

, find:

5

Find a

, given the following conditions.

i ii

b

i ii

c

i ii

d

i ii

6

Given that

and

7

Given that

and

8

The function is defined by

, solve the equation , solve the equation . Find an expression for

. . in terms of in each of the following

cases. a b 9

Let and be two functions. Given that .

and

, find an expression for

Section 4: Inverse functions Functions transform an input into an output, but sometimes you want to reverse this process: to be able to say which input produced a given output. When this is possible, it is done by finding the inverse function, usually labelled . For example, if , then is a number that, when put into , produces output you are looking for a number such that . Hence, . 4

. In other words,

f(4) = 12

f - 1(12) = 4

12

Tip Make sure you don’t get confused about this notation. With numbers, the superscript

denotes reciprocal; for example,

With functions,

,

.

denotes the inverse function of .

Finding the inverse function To find the inverse function you must rearrange the formula to find the input

in terms of the output

Tip Finding the domain of the inverse function can be difficult; this will be discussed in the next subsection.

Key point 2.7 To find the expression for inverse function

, given an expression for

:

1 Start with . 2 Rearrange to get (the input) in terms of (the output). 3 Give by replacing every instance of with .

WORKED EXAMPLE 2.12

The function is defined for Start with

by the rule

. Find an expression for

.

Rearrange to make the subject. Taking to the power of both sides removes the logarithm.

Write the resulting function in terms of .

WORK IT OUT 2.2 Find the rule for the inverse function of

,

.

.

.

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1

Solution 2

Solution 3

The relationship between and Once you know how to find inverse functions, there are a couple of very important facts you need to know about them. When you are finding the inverse function you switch the inputs and the outputs, so on the graph you switch the and -axes:

Key point 2.8 The graph of

is a reflection of the graph of

in the line

.

When you do and undo a function, you get back to where you started.

Key point 2.9

WORKED EXAMPLE 2.13

The graph of

is shown below. Sketch the graphs of:

a b

y y = h(x)

x

-1 O

y

a

The graph of

y = h(x)

of

is a reflection in the line .

y = x y = h- 1(x)

-1 O

x -1

Simplify

to

.

y

b

y = x

x

O

The fact that the graphs of and are reflections of each other gives you a very useful trick to solve some equations that would otherwise involve complicated (or impossible) algebra. WORKED EXAMPLE 2.14

In this question you must show detailed reasoning. This diagram shows a part of the graph of function

,

.

y

O

x

Tip ‘You must show detailed reasoning’ means that you must solve equations algebraically rather than, for example, using intersections of the graph on your calculator. a On the same axes, sketch the graph of b Solve the equation

.

. The graph of the line .

is the reflection of the graph of

in

y

a

x

O

b

Finding the equation for the inverse function would involve solving a cubic equation, and you don’t know how to do that. Luckily, you can see from the graph that the graphs of and intersect on the line . This means that solving the equation is equivalent to solving the equation .

The reflection in the line coordinates).

swaps the domain and the range of a function (because it swaps and

Key point 2.10 The domain of The range of

is the same as the range of is the same as the domain of

. .

WORKED EXAMPLE 2.15

The function is defined by a Find an expression for

for

.

and state its domain and range.

b State the range of . Set

a

.

Make the subject.

Replace with . The domain of . The range of .

is is

b The range of is . EXERCISE 2D

In the expression for The range of question.

, the denominator cannot be zero.

is the same as the domain of , which is given in the

The range of is the same as the domain of found.

, which you have just

EXERCISE 2D 1

Find an expression for a

if:

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

g

i ii

h

i ii

2

Sketch the inverses of the following functions.

y

a

4 3 2 1 - 4 - 3 - 2 - 1O -1

1 2 3 4

x

-2 -3 -4 y

b

8 7 6 5 4 3 2 1 - 4 - 3 - 2 - 1O

1 2 3 4

x

y

c

4 3 2 1 - 4 - 3 - 2 - 1O -1

1 2 3 4

x

-2 -3 -4

y

d

4 3 2 1 - 4 - 3 - 2 - 1O -1

1 2 3 4

x

-2 -3 -4 3

Each of the following functions is defined on the largest possible real domain such that the inverse function exists. Find the expression for a

in each case. State the domain and range for both and

i ii

b

i ii

c

i ii

d

i ii

4

Below is a table giving selected values of the one–one function

a Evaluate

.

b Evaluate 5

.

In this question you must show detailed reasoning. The function is defined by Evaluate

.

.

.

.

6

Given that

:

a Find the inverse function b State the domain and range of

.

Worksheet See Support sheet 2 for a further example of finding inverse functions and their domains, and for more practice questions. 7

Given functions

8

Let and be two functions such that Let . Prove that

9

The diagram shows the graph of

and

, find the function

.

is defined, and suppose that both . . The lines

and

and

are the asymptotes of the graph.

y 10 5 - 15 - 10

-5 O -5

5

10

15

x

- 10

a Copy the graph and, on the same axes, sketch the graph of b State the domain and range of

.

.

c Write down the solutions to the equation 10 The functions and are defined by

. and

.

a Calculate b Show that

.

11 The function is defined for 12 Let a Find Let b Find

, for

by

Find an expression for

.

. for

.

, giving your answer in the form

Find the domain and range of

, where

.

13 In this question you must show detailed reasoning. The function is defined by The graph of

is shown in the diagram.

for

.

exist.

.

y 10 5

- 10

-5

O

5

10

x

-5 - 10 a On the same set of axes, sketch the graph of b Solve the equation

.

When does the inverse function exist? All functions have inverse mappings, but these inverse mappings are not necessarily themselves functions. Since an inverse function is a reflection in the line , for the result to pass the vertical line test the original function must pass the horizontal line test. But, as you saw in Section 1, this means it must be a one–one function.

Fast forward You will apply this idea in Chapter 7 when you define inverses of trigonometric functions.

Key point 2.11 Only one–one functions have inverse functions. This leads to one of the most important uses of domains. By restricting the domain you can turn any function into a one–one function, which allows you to find its inverse function. WORKED EXAMPLE 2.16

a Find the largest value of such that the function b For this value of , find

y

a

O

3

is one–one.

and state its range. Sketch the graph of

y = (x - 3)2 x

,

.

y

O

Eliminate the points towards the right of the graph which cause the horizontal line test to fail.

3

y = (x - 3)2 x Decide which section remains.

b

Follow standard procedure for finding inverse functions.

Since

Use the fact that

,

Write The range of

is

to decide which root to take.

.

The range of

is the domain of .

.

WORK IT OUT 2.3 What is the inverse function of

?

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1

Solution 2

Solution 3 It doesn’t exist.

It should be clear from the horizontal line test that if a function, either increases or decreases throughout

its domain, then it is one–one. As soon as there is a turning point, the function is no longer one–one (and therefore has no inverse). WORKED EXAMPLE 2.17

In this question you must show detailed reasoning. If

, for

, prove that has an inverse function.

If the function is either increasing or decreasing, then it will have an inverse, so this is a good thing to check first. Complete the square. Note that this is a common way of showing that a function is non-negative. for all

.

∴ is an increasing function for all

Although , the gradient is never negative. Since has no turning points or asymptotes, this means that it is increasing.

.

Hence, is one–one and so has an inverse function. There are two things to notice in problems about one–one functions. First of all, in Worked Example 2.17, the fact that the gradient is allowed you to conclude that the function is always increasing. This may not be true if its graph has asymptotes or other breaks, as illustrated by the two diagrams below. The graphs have positive gradients but are not always increasing.

Secondly, although a function increasing (or decreasing) implies that it is one–one, the converse is not true: a one–one function can have both increasing and decreasing sections. The diagram below shows the graph of the function

You can use the horizontal line test to show that this function is one–one, and therefore has an inverse function.

y 2 1 O

1

2

x

Most functions you will meet in this course will not require you to consider these issues, but it is useful to be aware of them. EXERCISE 2E

EXERCISE 2E 1

In the following situations, find the value of that gives the largest possible domain such that the inverse function exists. For this domain, find the inverse function. a b c d

2

For each function shown in the diagrams below, determine a possible domain of the given form for which the inverse function exists. a Domain: y 10 5

- 10

-5

O

5

10

5

10

5

10

x

-5

b Domain: y 10 5

- 10

-5

O

x

-5

c Domain: y 10 5

- 10

-5

O -5 - 10

3

Given that a Find

for

:

.

b Explain why

does not exist.

The domain of is changed to c Find the largest possible value of .

so that

now exists.

x

4

A function is called self-inverse if so that

5

for all in the domain. Find the value of the constant

is a self-inverse function.

A function is defined by which

, where is a constant. Find the set of values of for

exists for all .

Checklist of learning and understanding A mapping takes input values from a given set and maps each one of them to one or more output values. A mapping is a function if every value maps to a single value. The output value corresponding to the input is called the image of . The set of allowed input values of a mapping is called the domain. The set of all possible outputs of a mapping is called the range. A function is: one–one if every value corresponds to only one value. many–one if at least one value that come from more than one value. The vertical and horizontal line tests can be applied to graphs of mappings. Vertical line test: if a mapping is a function, any vertical line will meet its graph at most once. If a mapping fails a vertical line test, it is called one–many and it is not a function. Horizontal line test: if a function is one–one, any horizontal line will meet the graph at most once. The composite function formed by applying to and then to the result is written as: The inverse,

, of a function is such that:

To find the expression for the inverse function Start with

, given an expression for

:

.

Rearrange to get (the input) in terms of (the output). Obtain

by replacing every instance of with .

Only one–one functions have inverse functions. The graph of The domain of The range of

is a reflection of the graph of is the same as the range of . is the same as the domain of .

in the line

.

Mixed practice 2 In this exercise, if a domain of a function is not stated, you may assume that it is all real numbers. 1

Find the inverse of the following functions. a b

2

,

The diagram shows three graphs.

y

B

A is part of the graph of

A

.

B is part of the graph of

C

.

C is the reflection of graph B in line A.

x

O

Write down: a the equation of C in the form b the coordinates of the point where C cuts the -axis. 3

Let , and , where both functions are defined on their largest possible real domain. Solve the equation .

4

Let and

and

a Evaluate

.

.

b Find and simplify an expression for

.

c State the geometric relationship between the graphs of d

i Find an expression for ii Find the range of iii Find the domain of

.

. . .

e Explain why the equation 5

and

has no solutions.

Functions and are defined for all real values of by and

.

Evaluate, i

,

ii © OCR, GCE Mathematics, Paper 4723, January 2008

6

The function is given by a Write

, for

in the form

.

.

b Find the inverse function

, stating its domain.

Tip Whenever you find a function, you should always state its domain. 7

Let

,

and

.

a Find and simplify: i ii the range of iii iv v b Explain why c

does not exist.

i Find the form of

.

ii State the geometric relationship between the graphs of iii State the domain of

.

.

iv State the range of 8

and

.

In this question you must show detailed reasoning. The functions and are defined over the domain of all real numbers by . a Write

in the form

b Hence, sketch the graph of of the turning point. c State the range of

.

, labelling all axes intercepts and the coordinates

and

.

d Hence, or otherwise, find the range of 9

and

.

The functions and are defined for all real values of by and

,

where and are non-zero constants. i

Find the range of .

ii Explain why the function has no inverse. iii Given that iv Given further that

for all values of , show that

.

for all values of , find the set of possible values of . © OCR, GCE Mathematics, Paper 4723, June 2010

10 Let a Write

. in the form

.

b Hence, or otherwise, find the range of when its domain is all real numbers. c Another function is now defined with the same rule as , but with the largest possible domain of the form , for which it has an inverse function. Find an expression for . 11 a Show that if

, then

.

b A function satisfies the identity

.

By replacing all instances of with

find another identity satisfied by

c By solving these two identities simultaneously, express

.

in terms of .

12 In this question you must show detailed reasoning. The functions and is defined for

are given by and , except for the interval

. The function .

a Calculate the value of and of . b Find the range of

.

13 An odd function is any function a Show that

that satisfies

.

is an odd function.

b What type of symmetry must the graph of any odd function have? c Given any function

, show that

An even function is any function that satisfies d Show that

is an odd function. .

is an even function.

e What type of symmetry must the graph of any even function have? f

Given any function

, show that

is an even function.

g Hence, or otherwise, show that any function can be written as the sum of an even function and an odd function.

Worksheet See Extension sheet 2 for a selection of more challenging problems.

3 Further transformations of graphs In this chapter you will learn how to: draw a graph after two (or more) transformations find the equation of a graph after a combination of transformations sketch graphs of functions involving the modulus (absolute value) use modulus graphs to solve equations and inequalities.

Before you start… Student Book 1, Chapter 5

You should be able to recognise a graph transformation from the equation.

1 The graph of the diagram.

is shown in

y 4 3 2 1 - 2 - 1O

1 2 3 4 5

x

Sketch the graph of: a b Student Book 1, Chapter 5

You should be able to change the equation of a graph to achieve a given transformation.

2 A graph has equation Find the equation of the graph after:

.

a a translation of units in the positive direction b a horizontal stretch with scale factor . Student Book 1, Chapter 1

You should be able to use interval notation to express solution of inequalities.

3 In this question, write the solution using interval notation. a Solve the inequality

.

b Solve the system of inequalities and .

Combining transformations In this chapter you’ll take the transformations you met in Student Book 1, Chapter 5, and combine them to

produce a sequence of transformations of the original graph. You’ll also meet the modulus function, which is used in many contexts where a quantity needs to be positive. For example, the total distance travelled by a particle must be the sum of positive quantities even though it may change direction. You’ve also seen this idea used to find the area between a curve and the -axis when the enclosed region is below the axis.

Section 1: Combined transformations In this section we look at what happens when you apply two transformations to a graph. An important question to consider is: does the order in which the two transformations are done affect the outcome? To investigate this question, let us first consider transformations of a single point. y 6 5 4 3 2 1 - 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6

The point

1 2 3 4 5 6

x

is translated units up and then reflected in the -axis. The new point is

.

y 6 5 4 3 2 1 - 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6

The point

1 2 3 4 5 6

x

is reflected in the -axis first and then translated units up. The new point is

.

y 6 5 4 3 2 1 -3

-2

-1 O

The point

1

2

3

x

is translated units up and then reflected in the -axis. The new point is

.

y 6 5 4 3 2 1 -3

-2

-1 O

1

The point

2

x

3

is reflected in the -axis first and then translated units up. The new point is

.

These examples suggest the following rules for combining transformations.

Explore In advanced mathematics, algebra is much more than using letters to represent numbers. Unknowns can include transformations, as well as many other things. As demonstrated in this section, the rules for transformations are different from the rules for numbers, but there are certain similarities, too. The study of this more general form of algebra includes group theory, which has many applications, from particle physics to painting polyhedra.

Key point 3.1 When two vertical transformations or two horizontal transformations are combined, changing the order may affect the outcome. When one vertical and one horizontal transformation are combined, the outcome does not depend on the order.

Combining one vertical and one horizontal transformation WORKED EXAMPLE 3.1

The diagram shows the graph of the function

.

y 4 3 2 1 -2 -1 O

1

2

3

4

5

x

On separate diagrams, draw the graphs of:

Both these questions involve one horizontal and one vertical transformation.

a

Since the order doesn’t matter, you can carry out the transformation in brackets first.

b a

: translate units to the left.

is replaced by

, so the graph is translated to the left.

y 4 3 2 1 - 5 - 4 - 3 - 2 - 1O

1

x

2

: vertical stretch with scale factor . y 7 6 5 4 3 2 1 - 5 - 4 - 3 - 2 - 1O

1

x

2

: horizontal stretch with

b

is replaced by , so the graph is stretched horizontally.

scale factor y 4 3 2 1 -3 -2 -1O

1

2

3

4

5

6

7

8

9

x

Making the -coordinate negative results in a reflection in the -axis.

: reflection in the -axis y -3 -2 -1O -1

1

2

3

4

5

6

7

8

9

x

-2 -3 -4

Combining two vertical transformations To transform the graph of

you first multiply

into the graph of:

by and then add . The flow chart shows the order of operations and the

corresponding transformations: Stretch factor p f(x)

Translate 0 c p f(x) p f(x) + c

Multiply function by p

( )

Add c to function

Notice that this follows the normal order of operations: multiplication is done before addition. WORKED EXAMPLE 3.2

A graph has equation

. Find the equation of the graph after the following transformations.

a a vertical stretch with scale factor followed by a translation with vector b a translation with vector

followed by a vertical stretch with scale factor .

a Vertical stretch: becomes .

Vertical stretch: multiply the function by .

Vertical translation: add to the whole expression.

Vertical translation: becomes .

You could perform the two transformations in one go:

The final equation is .

b Vertical translation: becomes

The new equation is

This time, add to the expression first… . … then multiply the whole expression by .

Vertical stretch: becomes The final equation is

. .

Again, we can write this in one go:

Combining two horizontal transformations If we combine two horizontal transformations, we can transform the graph of

into the graph of

We can achieve this by first replacing with and then replacing all occurrences of by chart shows the resulting order of transformations.

. The flow

Stretch Translate factor 1 -d q 0 f(x) f(x + d) f(qx + d)

( )

Replace x with x + d

Replace x with qx

Notice that the transformations are in the ‘wrong’ order: the translation, which corresponds to addition, is done first. WORKED EXAMPLE 3.3

The following graph shows a b

. Sketch the graph of:

y

a

y = f(x) x

O

a This is a combination of two horizontal transformations, so you deal with the addition first.

There are two transformations: 1 is replaced by 2 is replaced by

. .

Replace with

1 is replaced by horizontal translation

.

Change

. to

y

y = f(x) + 1 x

O

2 is replaced by

horizontal

Replace with

.

Change

stretch, scale factor .

to

.

y

The red graph shows the final answer.

y = f(x + 1) y = f(2x + 1) x

O

b The two transformations are: 1 is replaced by

.

2 is replaced by

.

1 is replaced by

stretch,

scale factor .

The presence of brackets means that now we deal with the multiplication first.

Replace with Change

. to

.

y

y = f(2x) x

O

Replace with

2 is replaced by horizontal translation

.

y

Change

. to

.

The green graph shows the final answer. Note that this is a different answer from the one in part a.

y = f(2(x + 1)) x

O

As illustrated in Worked Example 3.3, if you want to perform a horizontal stretch before a translation you need to use brackets correctly in the equation. WORKED EXAMPLE 3.4

The graph of is transformed using a horizontal stretch with scale factor and then translated units to the right. Find the equation of the resulting graph. 1 Replace by :

A horizontal stretch with scale factor is achieved by replacing with .

is changed to

.

2 Replace by

:

A horizontal translation by units is achieved by replacing with

is changed to . The final equation is .

Notice that if we had performed the translation before the stretch, the resulting equation would have been

.

WORK IT OUT 3.1 Describe the sequence of two transformations that transform the graph of of

.

to the graph

.

Which of the following solutions is correct? Identify any mistakes in the other two. Solution 1 1 Add to ; this is a horizontal translation units to the left. 2 Replace by

; this is a horizontal stretch with scale factor .

Solution 2 1 Replace by

; this is a horizontal stretch with scale factor .

2 Add to ; this is a horizontal translation units to the left. Solution 3 1 Add to ; this is a horizontal translation units to the left. 2 Replace by

; this is a horizontal stretch with scale factor .

EXERCISE 3A 1

The following graphs show

and

.

y

f(x)

y

6 5 4 3 2 1

- 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6

6 5 4 3 2 1 1 2 3 4 5 6

Sketch the graphs of: a

i ii

b

i ii

c

i ii

d

i ii

e

i

x

- 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 (- 2.5, - 4) -5 -6

(2, 5) g(x)

1 2 3 4 5 6

x

ii 2

Given that

, express each of the following functions as

sequence of transformations mapping the graph of a

and, hence, describe the

to the graph of the given function.

i ii

b

i ii

3

Given that

, give the function

that represents the graph of

after the following

transformations. a

b

c

i Translation

followed by a vertical stretch of scale factor .

ii Translation

followed by a vertical stretch of scale factor .

i Vertical stretch of scale factor followed by a translation ii Vertical stretch of scale factor followed by a translation

.

i Reflection in the horizontal axis followed by a translation

.

ii Reflection in the horizontal axis followed by a translation d

.

.

i Reflection in the horizontal axis, followed by a vertical stretch of scale factor , followed by a translation

.

ii Reflection in the horizontal axis, followed by a translation

, followed by a vertical stretch

of, scale factor . 4

Given that , express each of the following functions as and, hence, describe the transformation mapping the graph of to the graph of the given function. a

i ii

b

i ii

5

Given that

, give the function

that represents the graph of

transformations. a

i Translation ii Translation

b

followed by a horizontal stretch of scale factor . followed by a horizontal stretch of scale factor .

i Horizontal stretch of scale factor followed by a translation ii Horizontal stretch of scale factor followed by a translation

c

i Translation

followed by a reflection in the -axis.

. .

after the following

ii Reflection in the vertical axis followed by a translation 6

Find the resulting equation after the graph of transformations.

.

is transformed using each sequence of

a A vertical translation units up, then a vertical stretch with scale factor . b A vertical stretch with scale factor followed by a vertical translation units up. c A horizontal stretch with scale factor , then a horizontal translation units to the left. d A horizontal translation units to the left followed by a horizontal stretch with scale factor . 7

a The graph of

is transformed using a horizontal stretch with scale factor followed by a

vertical stretch with scale factor . Find the equation of the resulting graph. Can you explain why this is the case? b The graph of is transformed using a horizontal translation units to the left followed by a vertical stretch with scale factor . The equation of the resulting graph is again . Find the value of . c The graph of is translated units up. What transformation (other than a translation units down) will return the graph to its original position? Use technology to sketch the graphs and see why this is the case. 8

The diagram shows the graph of

.

y

f(x)

6 5 4 3 2 1

- 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6

1 2 3 4 5 6

x

On separate axes, sketch the graphs of: a b 9

Sketch the following graphs. a b c In each case, indicate clearly the positions of the vertical asymptote and the -intercept.

10 The graph of

is translated units to the right and then reflected in the -axis. Find the

equation of the resulting graph, in the form 11 The graph of

.

is transformed by the following sequence:

translation by reflection in horizontal stretch with scale factor .

The resulting graph has equation 12 The graph of

Find the values of and .

is transformed by the following sequence:

reflection in translation by horizontal stretch with scale factor . The resulting graph has equation 13 Given that transforming

, give in simplest terms the formula for by the following sequence of transformations:

vertical stretch, scale factor translation by horizontal stretch, scale factor .



. Find the values of , and . , which is obtained by

Section 2: Modulus function The modulus (or absolute value) is a function that leaves non-negative numbers alone but reverses the sign of negative numbers. We use to denote the modulus of number ; for example, and . Note that . You can define the modulus function by this equation:

Key point 3.2

Did you know? You can think of the modulus function as giving the distance of a number from zero on the number line. You applied a similar idea to vectors in Student Book 1. But the distance between two objects is not always easy to define. For example, what length gives the ‘distance’ between two points on the surface of the Earth? This question leads to the idea of a metric. You may want to find out about the Minkowski Metric for finding the distance between two points in space–time in the theory of relativity. The graph of is shown. It will be useful to call the red branch the reflected branch (as it is the reflection of a part of the graph in the -axis) and the blue branch the unreflected branch. y

The domain of is all real numbers, whereas the range is all positive numbers and zero.

y = |x | O

x



You can combine this with the rules for transforming graphs to sketch some other functions involving the modulus. WORKED EXAMPLE 3.5

Sketch the graphs of: a b In each case, find the -intercept of the graph. a

-intercept: has been replaced by the right.

, so the graph is shifted units to

The -intercept is when

.

y

y = |x - 3| 3

O

b

x

3

-intercept: y

has been replaced by the left.

, so the graph is shifted units to

Subtracting shifts the graph units down. x

O

y = |x + 2|- 5 -3 (- 2, - 5)

You can also think about applying the modulus function to the graph of the function ‘inside’ the modulus. Any parts of the ‘original’ graph that are below the -axis will have their values changed from negative to positive; so those parts of the graph will be reflected in the -axis. WORKED EXAMPLE 3.6

Sketch the graph of

, indicating the intercepts with the coordinate axes.

y

Start by drawing the graph of This crosses the -axis at

5

. and the -axis at

.

y = 5 - 2x

O

2.5

x

y

The part of the graph to the right of the -intercept has negative values; taking the modulus will make those positive, so this part of the graph needs to be reflected in the -axis.

y = |5 - 2x|

5

O

2.5

x

Key point 3.3 To sketch the graph of below the -axis. In particular, the graph of .

, start with the graph of

and reflect any parts that are

has a V-shape with the vertex at

and -intercept

Using modulus notation in inequalities You often meet inequalities where the variable is between a number and its negative; for example, if then

. You can write this more concisely using modulus notation:

.

Fast forward You will use inequalities of this form with sequences (Chapter 4) and binomial expansion (Chapter 6). It is possible to extend this notation to write other inequalities that represent a single interval. For example, what does mean? If we replace by then, as in the previous example, means that . So , which can be rearranged into . There is a nice interpretation of this inequality on the number line: 3 1

2

3

3 4

5

6

7

8

9

x

The number is in the middle of the interval , units away from each end. So the inequality can be read as: ‘the distance between and is less than ’.

Key point 3.4 The modulus inequality

is equivalent to

.

Rewind You already use the modulus of a vector to represent the distance between two points.

Fast forward If you study Further Mathematics, in Student Book 1, you will extend the modulus notation to measure distances between points in the complex plane.

WORKED EXAMPLE 3.7

a Write the inequality b Write the interval a

in the form

.

using an inequality of the form

.

Use the result from Key point 3.4: So

. In this example,

and

.

Alternatively, you can think of on either side of .

as saying that is at most units

is the number in the middle of the interval.

b

is the distance from the middle of the interval to one end. So

.

The end points of the interval are included.

EXERCISE 3B 1

Sketch the following graphs, showing the axes intercepts.

a

i ii

b

i ii

c

i ii

d

i ii

2

Write the equation of each graph in the form a

y

i

4

O

x

4

y

ii

2 -2

b

x

O

y

i

4

-2

O

x

y

ii 6

-2

O

x

.

c

y

i

3

O

x

1

y

ii

5

O

3

x

2.5

Write the following inequalities in the form a

.

i ii

b

i ii

4

Write the following statements in the form a

.

i ii

b

i ii

c

i ii

5

Sketch the graphs of the following, showing the coordinates of the -intercept and the vertex. a b



6

Sketch the graph of

7

Sketch the graph of axis.

8

Sketch the graph of

, labelling the intercepts with the coordinate axes. . Give the coordinates of the points where the graph crosses the -

.

Section 3: Modulus equations and inequalities You can use graphs to solve equations and inequalities involving the modulus function.

Key point 3.5 When solving an equation involving a modulus function, sketch the graph. You need to use the graph to decide whether the intersection is on the reflected or the unreflected part of the graph. If it is on the unreflected part, you can rewrite the equation without the modulus sign. If it is on the reflected part, you need to replace the modulus sign by a minus sign. WORKED EXAMPLE 3.8

Solve the equation

.

y

Sketch the graphs of

and

.

There are two intersection points: y = |x - 1|

A

the blue graph intersects the reflected part of the red graph. the blue line intersects the unreflected part of the red graph.

y = x 2

B

x

O

You need to write a separate equation for each.

A

For the reflected part, replace the modulus sign with brackets and a minus sign.

B

For the unreflected part, just remove the modulus sign.

So the solution is: or You can also intersect two modulus graphs. WORKED EXAMPLE 3.9

Solve the equation

.

y

Sketch the graphs of y = |2x - 1|

and

.

There are two intersections: the unreflected blue line and the reflected red line.

B y = |x + 1|

the unreflected blue line and the unreflected red line.

A O

x

You need the reflected part of the red graph, so replace the modulus sign with brackets and a minus sign.

A

B

You need the unreflected parts for both graphs, so remove the modulus signs.

The solution is:

or

. There is an alternative method that can be used to solve an equation where both sides are inside a modulus. You then know that both sides of the equation are non-negative, so you can square both sides.

Key point 3.6

Tip Remember that squaring both sides is possible only if both sides are inside a modulus. For example, is not equivalent to , as could be negative. We illustrate this with the equation from Worked Example 3.9. See which method you prefer. WORKED EXAMPLE 3.10

Solve the equation

. Both sides are positive so can be squared.

Write as a quadratic equation with the

terms positive.

WORK IT OUT 3.2 Solve the equation

.

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1

or Solution:

Solution 2

or

or

Solution:

Solution 3

Solution:

or

To solve inequalities, first sketch the graphs and find their intersections. You can then decide on which parts of the graph the inequality is satisfied.

Rewind This is the same method as you use to solve quadratic inequalities – see Student Book 1, Chapter 3.

WORKED EXAMPLE 3.11

Solve

.

y

Sketch the graphs of inequality is satisfied. B

A

and

highlighting where the

y = x + 4

y = |2x - 3| x

O

A

Find the intersection points: is on the reflected part of the red line.

B

is on the unreflected part of the red line. Describe the highlighted region in terms of . Remember that you can use either the inequality notation or the interval notation to do this.

EXERCISE 3C 1

Solve the following equations. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

2

Solve the following inequalities. Write your answer using the stated notation. a Use interval notation. i ii b Use inequality notation. i ii c Use interval notation. i ii d Use inequality notation. i ii e Use interval notation. i ii f Use inequality notation. i ii

Worksheet See Support sheet 3 for a further example of modulus inequalities and for more practice questions. 3

4

a Solve the equation b Solve the inequality

.

a Solve the equation

.

b Solve the inequality 5

.

. Write your answer using the interval notation.

a Sketch the graph of b Hence, solve the inequality

6

Solve the inequality

7

Given that

8

Solve the equation

, find in terms of the solution of the inequality , where

.

.

Checklist of learning and understanding You can combine any two (or more) transformations of graphs: translations, stretches and reflections (both horizontal and vertical). The order in which transformations occur may affect the outcome. One horizontal and one vertical transformation can be done in either order. Changing the order of two horizontal or two vertical transformations may affect the outcome. For a function of the form

, the stretch is performed before the translation.

For a function of the form

, the translation is performed before the stretch.

The modulus function can be used to reflect the part of the graph below the -axis so that the whole graph is on or above it. The graph of

has a V-shape with the vertex at

To solve equations and inequalities involving the modulus function, always use graphs. You need to decide whether the solutions are on the reflected or unreflected part of the graph. You can solve some modulus equations by squaring both sides, using You can use modulus notation to express some inequalities. In particular, .

.

Mixed practice 3 1

The graph of

is shown.

y y = f(x)

O

(4, 0)

x

On separate diagrams, sketch the graphs of: a b 2

The graph of

is transformed by applying a translation with vector

, followed by

a vertical stretch with scale factor . Find the equation of the resulting graph in the form

3

a On the same set of axes, sketch the graphs of

and

.

b Hence, solve the inequality 4

a Describe two transformations that transform the graph of .

to the graph of

b Describe two transformations that transform the graph of .

to the graph of

c Hence, describe a sequence of transformations that transform the graph of to the graph of 5

The transformations R, S and T are defined as follows: R : reflection in the -axis S : stretch in the direction with scale factor T : translation in the positive direction by units. i The curve curve.

is transformed by R followed by T. Find the equation of the resulting

ii Find, in terms of S and T, a sequence of transformations that transforms the curve the curve

to

. You should make clear the order of the transformations. © OCR, GCE Mathematics, Paper 4723, June 2010

6

Find two transformations whose composition transforms the graph of of

7

.

Solve the inequality

.

to the graph

8

a Describe two transformations whose composition transforms the graph of graph of

to the

.

b Sketch the graph of

.

c Sketch the graph of

, marking clearly the positions of any asymptotes and

-intercepts. 9

a State the sequence of three transformations that transform the graph of graph of

. Hence, sketch the graph of

b Solve the equation

to the

.

.

c Write down the solution of the inequality

.

10 a Describe a transformation that transforms the graph of

to the graph of

.

b On the same diagram, sketch the graphs of: i ii Mark clearly any asymptotes and -intercepts on your sketches. c The graph of the function produce the graph of

has been translated and then reflected in the -axis to .

y y = h(x)

y = g(x)

(- 1, 7) (1, 3) x

O

(3, - 3) (1, - 7)

i State the translation vector. ii If

find constants , , , such that

.

y

11

x

O

The function is defined by i State the range of . ii Find the value of

.

for

. The graph of

is shown above.

iii Given that the equation the constant .

has two distinct roots, determine the possible values of

© OCR, GCE Mathematics, Paper 4723, January 2006

Worksheet See Extension sheet 3 for more challenging questions on modulus graphs and equations. 12 Solve the equation 13 Sketch the graph of

. .

4 Sequences and series In this chapter you will learn: how to determine the behaviour of some sequences how to use sigma notation for series about sequences with a constant difference between terms about finite series with a constant difference between terms about sequences with a constant ratio between terms about finite and infinite series with a constant ratio between terms how to apply sequences to real-life problems.

Before you start… GCSE

You should be able to find the formula for the term of a linear sequence.

1 Find the formula for the following sequences.

term of the

a b GCSE

You should be able to use term-to-term rules to generate sequences.

2 Find the second and third terms of the sequence defined by

Student Book 1, Chapter 3

You should be able to solve quadratic equations and inequalities.

3 Find the smallest positive integer that satisfies the inequality .

Student Book 1, Chapter 7

You should be able to solve exponential equations and inequalities.

4 Find the smallest integer value of such that .

GCSE

You should be able to solve linear simultaneous equations.

5 Solve:

Student Book 2, Chapter 3

You should be able to use modulus notation.

6 List all integers that satisfy

.

Modelling with sequences If you drop a ball, it will hit the ground and bounce but the new height that it reaches will be a little lower on each subsequent bounce. The heights the ball reaches after each bounce form a sequence. Although the idea of a sequence may just seem to be about abstract number patterns, it has a remarkable number of applications in the real world – from calculating mortgages to estimating harvests on farms.

h h1 h2 h3 h4 h5 t

Section 1: General sequences A sequence is a list of numbers in a specific order, and can be finite or infinite. A finite sequence has a finite number of terms. For example, the sequence (the number of days in each month) has terms. Such a sequence can be described by simply listing all terms, as we have done here. An infinite sequence continues forever, so it is not possible to list all the terms. Instead, we need to give a rule, either describing how to find each term, or how to get from one term to the next. For example, the sequence

can be described in two different ways: by a term-to-term rule: by a position-to-term rule (also called ‘the formula for the

term’):

Gateway to A Level For a reminder of how these two types of rules work, see Gateway to A Level Section X. There is also further practice at the beginning of Exercise 4A.

Tip is the standard notation for the nth term of a sequence. So in the sequence etc.



Did you know? Term-to-term rules can involve more than one previous term. You may already know the example of the famous Fibonacci Sequence, with . Based on a model Leonardo Fibonacci made for the breeding of rabbits, this has found many applications from patterns on pine cones to a proof of the infinity of prime numbers. There is also a beautiful link to the golden ratio,

.

Increasing, decreasing and periodic sequences Sequences can behave in many different ways but you need to be aware of three possibilities:

Key point 4.1 An increasing sequence is one where each term is larger than the previous one: for all . A decreasing sequence is one where each term is smaller than the previous one: for all . A periodic sequence is one where the terms start repeating after a while: for some number (the period of the sequence).

Tip When asked to comment on the behaviour of the sequence, you can use your calculator to generate terms, and investigate whether the sequence is increasing, decreasing or periodic.

WORKED EXAMPLE 4.1

Find the first five terms of each sequence, and describe the behaviour of the sequence. a

,

b c

,

a

Substitute each term into the formula to get the next term.

The sequence is increasing.

The terms are getting larger. Substitute

b

The sequence is decreasing. c

to calculate the terms.

The terms are getting smaller. Substitute each term into the formula to get the next term.

The sequence starts to repeat after two terms:

The sequence is periodic, with period .

Focus on … Focus on … Problem solving 1 looks further at the technique of using small cases to establish patterns in sequences.

The limit of a sequence In Worked Example 4.1, the sequence in part b decreases but the terms remain positive although they get closer and closer to zero. The sequence converges to zero. By contrast, the sequence in part a diverges – the terms increase without a limit.

You can use your calculator to investigate long-term behaviour of sequences by generating a large number of terms until you can see what is going on. But you can also find the limit of a convergent sequence by solving an equation.

Tip Use the table function if the sequence is given by the formula, or the ANS button if you have a term-to-term rule.

Key point 4.2 To find the limit, , of a convergent sequence defined by a term-to-term rule, set and solve for .

Fast forward You will see in Chapter 14 how you can use convergent sequences to solve some equations.

WORKED EXAMPLE 4.2

A convergent sequence is defined by

,

.

By setting up and solving an equation, find the limit of the sequence. Replace

Let

and

by .

Solve the equation for .

The limit is

.

Explore It is important to remember that the method in Worked Example 4.2 can be applied only if you know that a sequence converges. For example, trying to use it with , gives , but the sequence ( , , , , …) clearly does not converge. Find out about different methods for proving convergence of a sequence.

EXERCISE 4A 1

Write out the first five terms of the following sequences, defined by a term-to-term rule. a

i ii

b

i ii

c

ii d

,

i

i

,

ii 2

Write out the first four terms of the following sequences, each defined by an a

term formula.

i ii

b

i ii

c

i ii

d

i ii

3

Suggest a possible formula for the a

term of each sequence.

i ii

b

i ii

c

i ii

d

i ii

4

Using your calculator, determine the behaviour of these sequences. a

b

i

,

ii

, ,

i

,

ii 5

A sequence is defined by a Write down

and

.

b Show that 6

.

for a constant .

A sequence is defined by

, with

.

a Find the fifth term of the sequence. b Describe the long-term behaviour of the sequence. 7

A sequence is defined by Find the value of .

8

a A sequence is defined by

,

,

, where is a constant. The limit of the sequence is

, where is a constant. Show that

. b Given that

and that the sequence is increasing, find the value of .

Worksheet

.

See Extension sheet 4 for some more challenging questions on the long-term behaviour of sequences and series.

9

A sequence is defined by a Find

and

b State the value of

and

.

. .

10 A sequence is defined by a Find

,

,

.

.

b Determine, explaining your reasoning fully, whether the sequence converges. If it does, state the limit.

Section 2: General series and sigma notation When savings in a bank account earns interest each year, these interest payments form a sequence. While it may be useful to know how much interest is paid in each year, you may be even more interested to know how much will be paid altogether. This is one of many examples of a situation where you may want to add up the terms of a sequence. The sum of a sequence up to a certain point is called a series. Use the symbol to denote the sum of the first terms of a sequence, that is

Instead of writing using sigma notation:

, you will often see exactly the same thing in a shorter form

Key point 4.3 Sigma notation is a shorthand way to describe a series.

Tip Don’t be intimidated by this notation. If you are in doubt, try writing out the first few terms and the last term. There is nothing special about the letter here; any letter could be used but and are the most usual. You may also see the or being missed out above and below the sigma. WORKED EXAMPLE 4.3

Find the value of

. Put the starting value,

, into the expression to be summed, .

You’ve not reached the end value, so put in

.

You’ve still not reached the end value, so put in

.

You’ve reached the end value, so stop and evaluate.

WORKED EXAMPLE 4.4

Write the series

Describe each term of the series using a general term in the variable .

General term Starts at

in sigma notation.

.

Note the first value of .

Ends at

Note the final value of .

.

Summarise in sigma notation.

Notice that there is more than one possible answer to the question in Worked Example 4.4. For example, we could write the sum as

.

EXERCISE 4B 1

Evaluate the following expressions. a

i

ii b

i

ii c

i

ii 2

Write the following expressions in sigma notation. Be aware that there is more than one correct answer. a

i ii

b

i ii

c

i ii

3

A sequence is defined by ,

Find 4

.

A sequence is defined by ,

where is a constant. a Show that

.

b Given that

, find the value of .

5

Find the exact value of

6

a Find the exact value of

, giving your answer in the form

b Hence, find the exact value of

.

.

.

Section 3: Arithmetic sequences Arithmetic sequences have a common difference between each term: to get from one term to the next you add the common difference (which may be negative). For example:     (add to each term)     (add to each term)

Tip An arithmetic sequence is sometimes also called an arithmetic progression.

Gateway to A Level For practise at identifying arithmetic sequences and finding their Level Section Y.

term, see Gateway to A

In general, if the first term is and the common difference is , then you have: 1st term

2nd term

From this you can see how to form the

3rd term

4th term

5th term

term of an arithmetic sequence.

Key point 4.4 The

term of an arithmetic sequence with first term and common difference is

Tip The formula for the .

term involves

and not

. So, for example,

WORKED EXAMPLE 4.5

An arithmetic sequence has first term and common difference . Find the term with the value

.

You need to find when Use

with

. and

.

Solve the equation.

So

is the

term.

You will often need to set up simultaneous equations to find and . WORKED EXAMPLE 4.6

The fifth term of an arithmetic progression is and the eighth term is

.

and not

a Find the first term, , and common difference, . b Hence, find the

term. Use

with

But you know that

.

.

Repeat for the eighth term. :

Solve the simultaneous equations (1) and (2) using elimination (substitution would also work).

Now form the term formula. Note that you could tidy up by expanding the brackets and simplifying: . However, there is no need to, as you are not asked to state the formula at all. Substitute

.

Worksheet See Support sheet 4 for a further example of using simultaneous equations to find and .

WORKED EXAMPLE 4.7

The first three terms of an arithmetic sequence are

,

and

.

Find the possible values of . The difference between terms must be constant (i.e. ). Rearrange and solve the quadratic equation.

EXERCISE 4C

EXERCISE 4C 1

Find the formula for the a

b

c

term for the arithmetic sequence, given the following conditions.

i first term , common difference ii first term

, common difference

i first term

, common difference

ii first term

, common difference

i first term , second term ii first term , second term

d

i first term , second term ii first term

e

, second term

i third term , eighth term ii fifth term , eighth term

2

How many terms are there in the following sequences? a

i ii

b

i first term , common difference , last term ii first term , ninth term

3

, last term

An arithmetic sequence has and

as its first two terms, respectively.

a Write down, in terms of , an expression for the

term,

.

b Find the number of terms of the sequence that are less than term of an arithmetic sequence is

4

The

5

The

6

The height of the rungs in a ladder forms an arithmetic sequence. The third rung is above the ground and the tenth rung is above the ground. If the top rung is above the ground, how many rungs does the ladder have?

7

The first four terms of an arithmetic progression are , constants. Find and .

8

A book starts at page 1 and has a page number on every page.

term of an arithmetic sequence is

and the

.

and the

term is term is

,

. Find the value of the

term.

. Which term has the value

and

?

, where and are

a Show that the first eleven page numbers contain thirteen digits in total. b The total number of digits used for all the page numbers is

. How many pages are in the book?

Section 4: Arithmetic series When you add up terms of an arithmetic sequence you get an arithmetic series. Just as it was useful to have a formula for the term, so it is useful to have a formula that will find the sum of the first terms. There are two different versions of this.

Key point 4.5 For an arithmetic series with first term , common difference and first terms is

term , the sum of the

These formulae are given in your formula book.

Focus on … The first formula is proved in Focus on … Proof 1. The second formula follows immediately from the first, since the last term, , equals

.

PROOF 1

Starting from the formula

, prove that

.

Start from the formula given. You want to introduce , which is the

term:

Tip You will need the first formula when you know or want , and the second when you know or want .

WORKED EXAMPLE 4.8

a Find the sum of the first difference .

terms of an arithmetic progression with first term and common

b The first term of an arithmetic sequence is a

Use

b

Use

and the

term is

. Find the sum of all

terms.

, with

, with

,

and

.

You must be able to work backwards too, and find how many terms there are in a series when given the sum of the series. Remember that the number of terms can only be a positive integer. WORKED EXAMPLE 4.9

An arithmetic sequence has first term and common difference

. The sum of the first terms is

. Find the value of . Use

, with

.

Simplify and then solve the equation.

must be positive. Sometimes you have to interpret the question carefully to see that it is about an arithmetic series. WORKED EXAMPLE 4.10

Find the sum of all the multiples of between

This is an arithmetic series with and .

and

.

Write out the first few terms and the last term to see what is happening.

You need to know how many terms there are in this series, so set in the formula and solve for .

Now use

, with

,

,

.

EXERCISE 4D 1

Find the sum of the following arithmetic sequences. a

i ii

b

i ii

c

i ii

d

i ii

2

An arithmetic sequence has first term and common difference . How many terms are required to get a sum of each of the following? a

i ii

b

i ii

, for some positive even number , for some square number

3

For the arithmetic series

4

The sum of the first terms of a series is given by

, find the value of for which

.

, where

.

a Find the first three terms of the series. b Find an expression for the

term of the series.

5

The second term of an arithmetic sequence is . The sum of the first four terms is term, , and the common difference, , of the sequence.

6

The fourth term of an arithmetic sequence is . The sum of the first term, , and the common difference, , of the sequence.

7

The sum of the first three terms of an arithmetic progression is is . Find the first term, , and common difference, .

8

Prove that the sum of the first odd numbers is

9

Find the largest possible value of the sum of the arithmetic sequence

terms is

. Find the first

. Find the first

, and the sum of the first

terms

.

10 Find the least number of terms so that the sum of the series

is greater than

. 11 The sum of the first terms of an arithmetic sequence is

. Find the

term

.

12 A circular disc is cut into sectors whose areas are in an arithmetic sequence. The angle of the largest sector is twice the angle of the smallest sector. Find the size of the angle of the smallest sector. 13 a Find the sum of all multiples of between and

.

b Hence, find the sum of all integers between and

that are not divisible by .

14 The ratio of the fifth term to the twelfth term of a sequence in an arithmetic progression is term of this sequence is positive, and the product of the first term and the third term is sum of the first terms of this sequence. 15 a Find an expression for the sum of the first

terms of the series

. b Hence, find the exact value of for which

.

16 Find the sum of all three digit numbers that are multiples of

but not

.

. If each

, find the

Section 5: Geometric sequences A geometric sequence (or geometric progression) has a common ratio between each term: to get from one term to the next you multiply by the common ratio. For example:

    

In general, if the first term is and the common ratio is , then you have 1st term

2nd term

From this you can see how to form the

3rd term

4th term

5th term

term of a geometric sequence.

Key point 4.6 The

term of a geometric sequence with first term and common ratio is

Tip The formula for the

term involves

and not . So, for example,

and not

.

WORKED EXAMPLE 4.11

For the geometric sequence a Find a formula for the b Hence, find the

:

term.

term of the sequence.

a

If the common ratio isn’t obvious, divide the second term by the first term (or by , etc.) to find it. Now use Substitute

b

. .

As with arithmetic sequences, you will often need to set up simultaneous equations.

Worksheet See Support sheet 4 for a further example of using simultaneous equations to find and .

WORKED EXAMPLE 4.12

The term of a geometric sequence is common ratio. Use

. The

, with

But you know that

term is

. Find the possible values of the

. .

Repeat for the ninth term. gives: Solve simultaneously. Note that dividing the equations is a quick way of eliminating and solving for .

Tip Notice that the question asked for values rather than a value, so you should expect to give at least two answers.

WORKED EXAMPLE 4.13

Three consecutive terms of a geometric sequence with common ratio

are

.

a Find the possible values of . b For each value of , find the common ratio of the sequence. a

The ratio between terms must be constant (i.e. ). Rearrange and solve the quadratic.

b When

,

When

,

Find the ratio of the first two terms for each value of in turn.

When you are asked which term satisfies a particular condition, you normally need to use logarithms. WORKED EXAMPLE 4.14

A geometric sequence has first term term that is less than

and common ratio . What is the term number of the first

? Express the condition for the inequality. Use

with

and

term to be less than .

The unknown is in the power so solve using logarithms.

as an

is negative, so when you divide by it you must remember to change the inequality sign. is an integer, so we need the first integer greater than

.

Rewind See Student Book 1, Chapter 7, if you need a reminder of how to solve exponential equations.

EXERCISE 4E 1

Find an expression for the a

term of the following geometric sequences.

i ii

b

i ii

c

i ii

d

i ii

e

i ii

2

How many terms are there in the following geometric sequences? a

i ii

b

i ii

c

i ii

3

The second term of a geometric sequence is and the fifth term is a Find a formula for the

.

term of the sequence.

b Hence, find the tenth term. 4

The third term of a geometric progression is

and the fifth term is

.

a Find the possible values of the first term and the common ratio. b Hence, find the two possible values of the eighth term.

Tip ‘Progression’ is just another word for ‘sequence’. 5

The first three terms of a geometric sequence are ,

and

.

a Find the possible values of . b In each case, find the tenth term of the sequence. 6

The third term of a geometric sequence is ?

7

A geometric sequence has first term and common ratio a Show that

and the sixth term is

. Which term takes the value

. For a particular value of ,

.

b Hence, find the value of . 8

Which is the first term of the sequence

9

The difference between the fourth and the third terms of a geometric progression equals one-quarter of the second term. Find two possible values of the common ratio.

to be less than

10 The three terms , , are in arithmetic progression. The three terms progression. Find the value of and given that .

?

are in geometric

. Three 11 The sum of the first terms of an arithmetic sequence is given by the formula terms of this sequence, , and , are consecutive terms in a geometric sequence. Find .

.

Section 6: Geometric series When you add up the terms of a geometric sequence you get a geometric series. Just as for arithmetic series, there is a formula for the sum of the first terms of a geometric sequence.

Key point 4.7 The sum of the first terms of a geometric series with first term and common ratio is

or equivalently:

This formula is given in your formula book.

Focus on … This formula is proved in Focus on … Proof 1.

Tip In general, use the first of these formulae when the common ratio is less than and the second when the common ratio is greater than . This avoids having to work with negative numbers.

WORKED EXAMPLE 4.15

Find the exact value of the sum of the first six terms of the geometric sequence with first term and common difference . ,

and

. Use the first sum formula as

.

You will need to use logarithms if you are asked to find the term number or the number of terms. WORKED EXAMPLE 4.16

How many terms are needed for the sum of the geometric series ?

to exceed

You need , but you know and .

Use the second sum formula condition as an inequality.

and express the

The unknown is in the power, so use logarithms to solve the inequality.

But is a whole number, so are needed.

EXERCISE 4F

terms

EXERCISE 4F 1

Find the sums of the following geometric series. (There may be more than one possible answer!) a

i ii

b

i ii

c

d

2

i first term , common ratio

, last term

ii first term

, common ratio

i third term

, fifth term ,

terms

ii ninth term

,

, last term

term

, last term

Find the possible values of the common ratio if: a

i first term is

, sum of the first two terms is

ii first term is , sum of the first two terms is b

3

.

i first term is

, sum of the first three terms is

ii first term is

, sum of the first three terms is

The

term,

.

, of a geometric sequence is given by

.

a Find the common ratio . b Hence, or otherwise, find an expression for

the sum of the first terms of this sequence.

4

The first term of a geometric sequence is and the sum of the first three terms is common ratio.

5

The sum of the first two terms of a geometric sequence is

. Find the

and the sum of the first four terms is

. Find the first term and the common ratio. 6

The sum of the first four terms of a geometric sequence is and the sum of the first six terms is .

, the sum of the first five terms is

a Find the common ratio. b Find the sum of the first two terms. 7

a What is the sum of the first four terms of the geometric sequence with first term and common ratio ? b Factorise

into a linear factor and a polynomial of order .

,

Section 7: Infinite geometric series If you keep adding together terms of any arithmetic sequence, the sum increases (or decreases if it is negative) without limit. This can happen with geometric series too, but there is also a possibility that the sum will converge.

Tip Remember that converge means that the numbers are approaching a limit (without necessarily reaching it).

For example, consider adding the numbers

(this is a geometric sequence with common ratio

). Using your calculator you can see that, as you add more and more terms of the sequence, the sum gets closer and closer to . You can try some more sequences, with different common ratios; some have sums that grow without a limit, whereas others seem to converge.

Tip Here we are talking about the sum of the sequence converging to . The example sequence itself also converges, but its limit is . Looking at the formula for the sum of a geometric sequence,

you can see that the only part that is affected by making bigger is . When you raise most numbers to a large power the result gets bigger and bigger. The exception is when is a number between and ; in this case, gets smaller in magnitude as increases – in fact it tends towards zero. Thus, we have:

Key point 4.8 As increases the sum of a geometric series converges to This is called the sum to infinity of the series. If the series diverges (has an infinite sum). This formula is given in your formula book.

Tip The condition that

is just as important as the formula itself.

WORKED EXAMPLE 4.17

For the geometric series

, find:

a the sum of the first terms b the sum of the first

terms, correct to decimal places

c the sum to infinity. To find the common ratio, divide the second term by the first term. a

Use

. Be careful: is negative.

b

c

Use

. This is valid since

.

WORKED EXAMPLE 4.18

The sum to infinity of a geometric series is and the second term is Use Use

. Find the common ratio.

, with , with

. .

Solve simultaneously.

From (2):

Substituting into (1):

But since the sum to infinity exists,

Check that the series actually converges for the values of found.

Remember that some questions may focus on the condition for the sequence to converge, as well as the value that it converges to. WORKED EXAMPLE 4.19

The geometric series

converges. Find the range of possible values of

. Identify . Since the series converges:

Use the fact that the series converges.

Solve the inequality.

EXERCISE 4G 1

Find the value of the following infinite geometric series, or state that they are divergent. a

i

ii b

i ii

c

i ii

d

i ii

e

i ii

2

Find the values of that allow the following geometric series to converge. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

g

i ii

h

i ii

i

i ii

3

Find the sum to infinity of the geometric sequence

4

The first and fourth terms of a geometric series are

and

, respectively. Find:

a the sum of the first terms of the series b the sum to infinity of the series. 5

A geometric sequence has all positive terms. The sum of the first two terms is infinity is . Find the value of: a the common ratio

and the sum to

b the first term. 6

The sum to infinity of a geometric series is . The sum of the first four terms is and all the terms are positive. Find the difference between the sum to infinity and the sum of the first eight terms.

7

Consider the infinite geometric series a For what values of does the series converge? b Find the sum of the series if

.

8

The sum of an infinite geometric progression is first term.

9

An infinite geometric series is given by

, and the sum of the first three terms is

. Find the

.

a Find the values of for which the series has a finite sum. b When

, find the minimum number of terms needed to give a sum that is greater than

10 The common ratio of the terms in a geometric series is

.

a State the set of values of for which the sum to infinity of the series exists. b If the first term of the series is 11

, find the value of for which the sum to infinity is

is an infinite series. Evaluate where possible: a b

.

.

Section 8: Using sequences and series to solve problems In this section you will see how to model various situations using sequences and series. If the sequence in question is arithmetic or geometric, you can use the formulae from previous sections to calculate individual terms and sums of terms. If a sequence is not either of those two types, you can use your calculator to investigate its behaviour. When dealing with sequences and series problems, you usually need to: Identify whether it is a geometric or an arithmetic sequence (or neither). Identify whether it is asking for a term in the sequence or the sum of terms in the sequence. Translate the information given in the question into equations. WORKED EXAMPLE 4.20

A savings account pays annual compound interest, added at the end of each year. If £ is paid into the account at the start of the first year, how much will there be in the account at the start of the seventh year? For a increase, multiply by

Each year the balance of the account is increased by the same percentage, so this gives a geometric sequence. It is a good idea to write out the first few terms in questions like this, to make sure you know what is happening.

The balance of the account at the beginning of: Year Year Year This is a geometric sequence with .

and The balance at the start of the seventh year is

.

£

A common example of sequences and series is to calculations involving mortgages and other types of loans. In a typical example, the amount owed is increased by a fixed percentage, and a fixed amount is paid off at regular intervals (such as annually or monthly).

Tip A spreadsheet is a very useful tool to investigate loans and mortgages. You can also use an iterative function on your calculator to generate the sequence.

WORKED EXAMPLE 4.21

Sam borrows £ to buy a car. At the end of each year (including the first) she pays back £ the amount still owed if less than £ ). At the start of each year (starting from the second) interest is added.

(or

a Show that at the end of the third year, before the interest has been added, she will still owe £ b Let £

be the amount Sam will owe at the end of year . Find an expression for

c Hence, find how many years it will take for Sam to pay off the loan.

in terms of

. .

a End of first year:

The interest is not added until the start of the second year.

£

End of second year: £

End of third year: £

b

Each year, the amount owed is increased by decreased by £ .

c

Use your calculator to generate the terms of the sequence with the formula from part b.

and then

Show the first few and the last few terms.

Sam will pay off the loan at the end of the

When the terms become negative, the debt has been paid off.

year.

WORK IT OUT 4.1 A job advert isement states that the starting salary is £

with an annual increase of £

How much would you earn in total by the start of the fifth year? Which of the following is correct? Identify the mistake in the other two. Solution 1 Arithmetic sequence,

,

You need the sum of the first five terms:

£

Solution 2 Arithmetic sequence,

,

The fifth term is:

£

Solution 3 Arithmetic sequence,

,

The sum of the first four terms is:

£

EXERCISE 4H

.

EXERCISE 4H 1

Philippa invests £

for years, earning

compound interest each year.

a How much interest does she earn in the

year?

b What is the total amount of her investment at the end of years? 2

Lars starts a job on an annual salary of £ a What will be his

and is promised an annual increase of £

.

year’s salary?

b After how many complete years will Lars have earned a total of £ million? 3

A sum of £

is invested, earning

per annum compound interest.

a Write down an expression for the value of the investment after full years. b What will be the value of the investment at the end of years? c Given that the value of the investment will exceed £

after full years:

i Write an inequality to represent this information. ii Calculate the minimum value of . 4

A florist orders bunches of flowers. By the end of the day he has sold of his flowers. He orders more bunches for the next day. The same happens on each day afterwards: a Let be the number of bunches of flowers in the florist’s shop at the end of the down an expression for in terms of . b Describe the behaviour of the sequence long term?

5

A sum of £

day. Write

. How many bunches of flowers will be in the shop in the

is invested.

a If the interest is compounded annually at a rate of investment after . b If the interest is compounded monthly at a rate of

per year, find the total value, , of the

per month, find the minimum number of

months for the value of the investment to exceed . 6

A marathon is a

-mile race. In a training regimen for a marathon, a runner runs mile on her first

day of training and each day increases her distance by of a mile. a After how many days has she run for a total of b On which day does she first run over 7

miles?

Aaron and Blake each keep some money in a safe. Aaron deposits £ in the first month and then increases his deposits by £ each month. Blake deposits £ in the first month and then increases his deposits by Aaron?

8

miles?

each month. After how many months will Blake have more money in his safe than

A ball is dropped vertically from metres in the air. With each bounce, it bounces only up to a height of of its previous height. a What height does it reach after the fourth bounce? b How far has it travelled when it hits the ground for the ninth time? c Give one reason why this model is unlikely to be accurate after

9

bounces.

Miriam deposits £ into a bank account at the beginning of each year, starting in each year interest is added to the account.

. At the end of

a Show that at the beginning of

there is

in the account.

b Find an expression for the amount in the account at the beginning of the year. c When Miriam has a total of at least £ in her account at the beginning of a year, she will start looking to buy a house. In which year will this happen? 10 Each row of seats in a theatre has more seats than the row in front of it. There are front row and the designer wants the capacity to be at least .

seats in the

a How many rows are required? b Assuming the rows are equally spread, what percentage of people are seated in the front half of the theatre? 11 Dineth takes out a mortgage to buy a house. He borrows £ at an annual interest rate of interest is added at the end of each year, after which Dineth pays back £ . a Show that at the start at the third year, Dineth owes about £

. The

.

b Let be the amount Dineth owes at the start of year (so that Express in terms of .

and

).

c Hence find long it will take for Dineth to pay off his mortgage.

Checklist of learning and understanding A sequence can be defined by a formula for the term (a position-to-term rule) or by a rule that generates the next term of a sequence from the previous term or terms (a term-to-term rule). Behaviour of a sequence: an increasing sequence is one where each term is larger than the previous one: and a decreasing sequence is one where each term is smaller than the previous one: . a periodic sequence is one where the terms start repeating after a while: some number (the period of the sequence).

for

a convergent sequence is one where the terms approach a limit as increases, and a divergent sequence is one where they increase/decrease without a limit. A series is a sum of terms in a sequence and it can be described succinctly using sigma notation:

Arithmetic sequences have a constant difference, , between terms. If you know the first term

, the general term is:

If you know the first term

and last term

If you know the first term

and the common difference

, the sum of all terms in the sequence is:

Geometric sequences have a constant ratio, , between terms. If you know the first term

, the general term is:

The sum of the first terms is:

:

,

or equivalently

If

, the series converges and the sum to infinity is given by:

Mixed practice 4 1

The third term of an arithmetic sequence is and the sum of the first the first term and the common difference.

2

The fourth term of an arithmetic progression is

terms is

and the ninth term is

. Find

. Find the sum of

the first nine terms. 3

The third term of a geometric sequence is

and the sixth term is .

a Find the first term and the common ratio. b Find the sum to infinity. 4

The

5

The fifth term of an arithmetic sequence is three times the second term. Find the ratio:

6

A sequence, , has terms , for .

term of a geometric progression is

defined by

i Write down the values of ii Evaluate

. The sum to infinity is . Find the value of .

and

, and state what type of sequence is.

. © OCR, GCE Mathematics, Paper 4722, June 2005

7

Which is the first term of the sequence

8

The sum of the first three terms of a geometric progression is the common ratio, .

9

Find the exact value of

less than

? . The sum to infinity is

. Find

.

10 Ben builds a pyramid out of toy bricks. The top row contains one brick, the second row contains three bricks and each row after that contains two more bricks than the previous row. a How many bricks are in the b If a total of

row?

bricks are used, how many rows are there?

c In Ben’s largest ever pyramid, he noticed that the total number of bricks was four more than four times the number of bricks in the last row. What is the total number of bricks? 11 Kalinda is offered two investment plans, each requiring an initial investment of Plan offers a fixed return of Plan offers a return of

per year.

each year, reinvested in the plan.

Over what period of time is plan better than plan ? 12 The first three terms of a geometric sequence are a Find the two possible values of .

,

and

.

:

b Given that it exists, find the sum to infinity of the series. 13 Evaluate

.

14 Find the sum of all the integers between 15 A sequence is defined by a If

and

,

that are divisible by .

.

, find the range of possible values of the constant .

b Given that

, find the limit of the sequence as tends to infinity.

16 i John aims to pay a certain amount of money each month into a pension fund. He plans to pay £ in the first month, and then to increase the amount paid by £ each month; that is, paying £ in the second month, £ in the third month etc. If John continues making payments according to this plan for

months, calculate

a how much he will pay in the final month, b how much he will pay altogether over the whole period. months, ii Rachel also plans to pay money monthly into a pension fund over a period of starting with £ in the first month. Her monthly payments will form a geometric progression, and she will pay £ in the final month. Calculate how much Rachel will pay altogether over the whole period. © OCR, GCE Mathematics, Paper 4722, June 2006 17 A geometric progression has first term and common ratio , and the terms are all different. The first, second and fourth terms of the geometric progression form the first three terms of an arithmetic progression. i Show that

.

ii Given that the geometric progression converges, find the exact value of . iii Given also that the sum to infinity of this geometric progression is

, find the value of

the integer . © OCR, GCE Mathematics, Paper 4722, June 2010 18 A geometric sequence and an arithmetic sequence both start with a first term of . The third term of the arithmetic sequence is the same as the second term of the geometric sequence. The fourth term of the arithmetic sequence is the same as the third term of the geometric sequence. Find the possible values of the common difference of the arithmetic sequence. 19 The first, second and fourth terms of a geometric sequence form consecutive terms of an arithmetic sequence. Given that the sum to infinity of the geometric sequence exists, find the exact value of the common ratio. 20 Find an expression for the sum of the first

giving your answer in the form 21 A student writes

terms of the series

, where

.

on the first line of a page, then the next two integers

on the second

line of the page and then the next three integers pattern. a How many integers are on the

line?

b What is the last integer on the

line?

c What is the first integer on the

line?

d Show that the sum of all the integers on the

on the third line. He continues this

line is

.

e The sum of all the integers on the last line of the page is

. How many lines are on the

page? 22 Suresh has a mortgage of £ pays £ .

. At the end of each year

interest is added, then Suresh

a Show that at the end of the third year the amount owing (£ ) is:

b Find an expression for how much is owing at the end of the c After how many years will the mortgage be paid off?

year.

5 Rational functions and partial fractions In this chapter, you will learn how to: manipulate rational functions, including by polynomial division with remainders decompose rational functions into a sum of algebraic fractions when the denominator contains distinct linear factors decompose rational functions into a sum of algebraic fractions when the denominator contains repeated linear factors.

Before you start… GCSE

Factorise quadratic expressions.

1 Factorise

.

GCSE

Add algebraic fractions.

2 Simplify this expression into one fraction:

Student Book 1, Chapter 4

Carry out polynomial division.

3 Simplify

Student Book 1, Chapter 4

Use the factor theorem.

4 Find a linear factor of

.

.

What are rational functions? In Student Book 1 you studied polynomials. A rational function is a fraction where both the denominator and numerator are polynomials. These functions have a wide range of real–world applications, from economics to medicine. For example, the function

is often used to model the amount of anaesthetic

in a patient after a time, . In this chapter you will focus on applying algebraic techniques to change rational functions into forms that will allow more advanced manipulations later in the course. To do this you shall start by reviewing a very important theorem about polynomials – the factor theorem.

Focus on … Focus on … Modelling 1 compares models, using rational functions to exponential models.

Section 1: Review of the factor theorem In Student Book 1 you met the factor theorem as a way of checking if a polynomial had a particular factor.

Key point 5.1 If

, then

is a factor of

.

Once you have established that a polynomial has a particular factor, you can then factorise the whole polynomial. One major use of this is to solve equations. WORKED EXAMPLE 5.1

a Show that

is a factor of

.

b Hence, solve

. The factor theorem says that

a

is a factor if and only if

. You need to show enough working to convincingly demonstrate that By the factor theorem, factor of .

.

is a

b

You can use polynomial division to find the remaining factor.

You can factorise the quadratic expression. Writing the expression in the factorised form allows you to find where it equals zero.

So

When

, or

.

Tip Remember: if you do not like long division you can also write , then multiply out and compare coefficients.

as

Rewind You first met polynomial division in Student Book 1, Chapter 4.

The factor theorem also works in reverse, so if you know that

WORKED EXAMPLE 5.2

is a factor, then

.

Given that

is a factor of

find the value of . can be written as says that

so the factor theorem

.

EXERCISE 5A 1

Show that the following expressions have the given factor and, hence, factorise completely. a

b

i

has factor

ii

has factor has factor

i

has factor

ii 2

Show that

3 4

is a factor of

is a factor of a

. Find the value of .

is a factor of

b Hence, factorise 5

. Hence, factorise it completely.

a Show that

. Find the value of . .

is a factor of

b Hence, factorise 6

Show that

7

Prove that if

8

If

9

a

.

is a factor of

and

b Hence, solve

. Hence, solve

is a factor of

is a factor of

10 Show that

.

, then

.

.

, find an expression for in terms of .

are factors of

. Find the values of and .

. is a factor of

and, hence, solve

.

Section 2: Simplifying rational expressions You need to be able to simplify and manipulate rational functions using your knowledge of algebra. In particular, always look to factorise expressions and simplify. WORKED EXAMPLE 5.3

Simplify:

Always try to simplify first. You can do this by factorising and looking for common factors. Always look out for the difference of two squares! You can divide the top and bottom of the first fraction by the factor of . To divide, you flip the second fraction and multiply. Remember that you do not need a common denominator. There is a factor of in the denominator of the first fraction and the denominator of the second fraction. You can ‘cancel’ these when the fractions are being multiplied. You can use polynomial division to simplify rational functions when the degree of the numerator is at least the degree of the denominator. This turns the rational function into the sum of a polynomial and a simpler rational function, called the remainder term.

Tip In the previous example you used an identity symbol to emphasise that all the expressions are equivalent.

Key point 5.2 If

is a polynomial then

In the previous expression,

is also a polynomial, called the quotient, and is called the remainder.

Fast forward This is used in calculus (see Chapter 12), the binomial expansion (see Chapter 6) or partial fractions (see Section 5.3).

WORKED EXAMPLE 5.4

Write

in the form

. You can use polynomial division.

The remainder found in polynomial division forms the numerator of the rational expression.

WORK IT OUT 5.1 Simplify

.

Which of the following solutions is correct? Identify the mistakes in the other two. Solution 1

Solution 2

Solution 3 Cancelling

from top and bottom:

EXERCISE 5B 1

Simplify the following expressions. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

g

i ii

2

Simplify the following expressions. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

3

Simplify the following expressions. a

i ii

b

i ii

c

i ii

d

i ii

4

Write the following expressions in the form a

i

.

ii b

i ii

5

Simplify

6

Simplify

7

a Simplify

. .

.

b Hence, or otherwise, solve the equation 8

Solve the equation

9

Find the quotient and remainder when

.

. is divided by

10 Find the quotient and remainder when 11 Write 12

Show that

in the form

is divided by

and, hence, simplify

is a factor of

b Hence, simplify 14 The remainder when

.

.

is a factor of

13 a Show that

.

.

. .

is divided by

15 When the polynomial is divided by is divisible by .

is . Find the value of and then find the quotient.

, the quotient is the same as the remainder. Prove that

16 a A car travels at speed for , then at speed expression for the average speed during this journey.

for

. Find and simplify an

b If the average speed equals the arithmetic mean of the two speeds, show that either the speeds in the two sections of the journey are equal or the speed in the second section is twice that of the first section.

Section 3: Partial fractions with distinct factors You know that you can write a sum of two algebraic fractions as a single fraction:

However, there are situations when you need to reverse this process. The method for doing this is called partial fractions.

Fast forward You will see that you need to do this with binomial expansions in Chapter 6 and integration in Chapter 11. You can also use it to help sum some series if you study Further Mathematics, in Student Book 2. To do this you need to know the form the partial fractions will take:

Key point 5.3 decomposes into

.

decomposes into

.

Rewind Comparing coefficients was discussed in Student Book 1, Chapter 1. Once you know the form you are aiming for, you write it as an identity and multiply both sides by the denominator of the original fraction. You can then either compare coefficients or substitute convenient values to find the values of and . WORKED EXAMPLE 5.5

Write

in partial fractions. Write your expression in the form suggested in Key point 5.3. Multiply both sides by

When

:

When

to eliminate fractions.

You are working with an identity so choose a convenient value of to work with. When the coeffcient of is zero, so you can find out about . When about .

the coeffcient of is zero, so you can find out

So This method only works when the degree of the numerator is less than the degree of the denominator. If this is not the case, you need to use polynomial division to turn it into the required form. WORKED EXAMPLE 5.6

WORKED EXAMPLE 5.6

a Show that b Hence, write

can be written in the form

.

in partial fractions. You need to make a link between the top and the bottom.

a

You can then split the top into two fractions. This is the required form. You will first work with the fraction from part a. If you want to use partial fractions, you must factorise the denominator, using the difference of two squares.

b

You need to use the form suggested in Key point 5.3. Multiply by the denominator to eliminate fractions. When

:

When

:

When

Try to find values of that make lots of the brackets equal to zero.

:

Therefore

So

EXERCISE 5C 1

Write the following expressions in terms of partial fractions. a

i ii

b

i ii

c

i ii

2

Write the following expressions in terms of partial fractions. a

i

ii b

i ii

3

Given that

4

Express

5

Simplify

.

6

Decompose

into partial fractions.

7

Write

8

a Show that

, find the values of and . in terms of partial fractions.

in partial fractions.

b Hence, write 9

Write in partial fractions

. in terms of partial fractions. , where is a constant.

10 Write in partial fractions 11 a Show that

, where is a constant. can be written in the form

, where is a constant to be

determined. b Hence, write

in terms of partial fractions.

12 Check what happens if you try to write

in the form

.

Section 4: Partial fractions with a repeated factor If there is a repeated factor in the denominator, you need to use an alternative expression for the partial fractions.

Key point 5.4 decomposes into

.

WORKED EXAMPLE 5.7

Express

as partial fractions. Write out the expression using the form given in Key point 5.4. Multiply both sides by fractions.

to eliminate

You can choose convenient values for . When the coeffcient of and is zero, so you can find out about . When the coeffcient of and is zero, so you can find out about .

There is no value of that can make the coeffcients of and zero, so you have to choose another simple alternative. In such a case, often makes the arithmetic simple. You can substitute the values of and found previously. So

EXERCISE 5D

EXERCISE 5D 1

Express the following in terms of partial fractions. a

i ii

b

i ii

c

i ii

2

Split

3

Express

4

Write

into partial fractions. as partial fractions. in terms of partial fractions.

Worksheet See Support sheet 5 for a further example of partial fractions with a repeated factor and for more practice questions. 5

a Use the factor theorem to show that b Hence, factorise c Write

6

a Write b Hence, write

7

Write

.

. in partial fractions. in the form

.

in partial fractions.

a Prove that if a function can be written as b Write

8

is a factor of

in the form

, it can also be written as

.

.

in terms of partial fractions.

Checklist of learning and understanding A rational function is a fraction where both the denominator and numerator are polynomials. In arithmetic they follow all the same rules as normal fractions. If the degree of the numerator is equal or greater than the degree of the denominator, you can use polynomial division to simplify the function:

Here,

is called the quotient and the remainder.

If the rational function has a numerator with a degree less than the denominator, it can be decomposed into partial fractions: decomposes into

.

decomposes into decomposes into

. .

Mixed practice 5 1

a Use polynomial division to simplify

.

b Hence, find the prime factors of 2

a Show that

.

is a factor of

b Factorise

.

.

c Solve

.

d Simplify

.

3

Write

in partial fractions.

4

Express

in partial fractions.

5

Write

in terms of partial fractions.

6

Write

in partial fractions.

7

Find the derivative of

.

Rewind Differentiation is covered in Student Book 1, Chapter 13. 8

Decompose

9

It is given that

into partial fractions.

for Express

.

in its simplest form. © OCR, GCE Mathematics, Paper 4724, January 2007

10 Express

in partial fractions. © OCR, GCE Mathematics, Paper 4724, June 2010

11 a Simplify b Hence, write

. in terms of partial fractions.

12 a Use the factor theorem to show that b Hence, factorise

is a factor of

.

c Solve

.

d Write

in partial fractions.

.

13 a Show that

is a factor of

b Solve

.

.

c Write

in partial fractions.

14 a Write

in the form

b Hence, write

.

in partial fractions.

15 Find the quotient and remainder when 16 a

is divided by

is divisible by

and by

. Find the values of and

. b Fully factorise c Solve

.

.

17 a Write

in terms of partial fractions.

b Hence, write

in terms of partial fractions.

18 a Use polynomial division to simplify b Hence, or otherwise, write 19 Write

.

in partial fractions.

in terms of partial fractions.

20

can be written as

21 The remainder when

. Find the value of , and . is divided by

22 The remainder when a polynomial, the value of

is

, is divided by

. Write in terms of . is the same as the quotient. Find

.

23 Given that

, prove that if

is linear, then

. 24 If two resistors with resistance resistance

and

are connected in parallel, the combined system has

. These are related by the equation

a Find and simplify an expression for b Hence, prove that

in terms of

. and

.

.

Worksheet See Extension sheet 5 for questions on continued fractions.

6 General binomial expansion In this chapter you will learn how to: expand , where is any rational power decide when a binomial expansion will converge use partial fractions to write expressions in the form required for the binomial expansion use binomial expansions to approximate functions.

Before you start… GCSE

You should be able to simplify expressions with exponents.

1 Simplify

Student Book 1, Chapter 9

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

2 Expand

Chapter 5

You should be able to write an expression in partial fractions.

3 Write

.

.

in the form .

Chapter 3

You should be able to write inequalities using the modulus function.

4 Write form

in the .

Extending the binomial theorem In Student Book 1 the binomial expansion was simply a quick way to expand brackets. You will see in this chapter that it can be extended to allow us to approximate many other functions by polynomials.

Section 1: General binomial expansion In Student Book 1 you met the binomial expansion when raising a sum to a positive integer power. For example: The binomial coefficients can be expanded:

Although difficult to prove, it turns out that this formula also works for any rational power .

Tip Remember that

is another way of writing

.

Key point 6.1 If

: , for any rational value of .

This formula will appear in your formula book. Notice that if is not a positive whole number, this is an infinite series.

Rewind The idea of a convergent series was covered in Chapter 4. You can always create the series on the right–hand side but it will only converge if . If the expression in the bracket on the left is different, for example , then the interval of convergence might be different, in this case

or

.

WORKED EXAMPLE 6.1

Find the first four terms in ascending powers of in the binomial expansion of

if

You need to rewrite the fraction in the form Susbstitute and presented in Key point 6.1.

. .

into the formula

The formula in Key point 6.1 only works if the first number in the bracket is . If it is something else, then you have to factorise the expression to turn the first number into . The interval over which the expansion converges might also then change.

Tip Saying that ‘the expansion is valid’ is the same as saying that it converges.

WORKED EXAMPLE 6.2

WORKED EXAMPLE 6.2

a Find the first three terms in ascending powers of in the binomial expansion of

.

b Find the values of for which this expansion is valid. You need to rewrite the square root as a power.

a

Take out a factor of inside the bracket to leave . Use the rules of exponents to separate out the numerical factor. Substitute

and

into the formula

presented in Key point 6.1.

Finally, multiply out the bracket. The expansion is valid when

b

used

, and you have

.

Multiplying through by . You could also write

.

You can generalise the method from Worked Example 6.2 to find the binomial expansion of any expression of the form .

Key point 6.2 The binomial expansion of

is valid for

.

Binomial expansions such as these produce infinite series, so you might wonder how they can be useful. If is small then very large powers of are extremely small, so they can be neglected. You can use the first few terms of the binomial expansion to approximate the original expression. WORKED EXAMPLE 6.3

Use the first two terms of the binomial expansion of places. a

to approximate

, to decimal

You need to rewrite the square root in the form

.

Take out a factor of inside the bracket to leave . Use the rules of exponents to separate out the numerical factor. Substitute

and

into the formula presented in

Key point 6.1. Multiply out the bracket. b You need such that

Find the value of that will give

.

.

Substitute

into

.

The terms after the second term will be so small that they won’t affect the first decimal places. (You can check that the next term is .)

Did you know? Calculators use a method similar to binomial expansion, called Taylor series, to find values of functions such as

or

.

You can also use the approximate binomial expression in solving equations, differentiation or integration. WORKED EXAMPLE 6.4

a Find the first three non-zero terms in the binomial expansion of

and state the range of values

for which the expansion is valid. b Use your expansion to find an approximate value of

.

You need to write the expression in the form and then take out a factor of .

a

Substitute

and

into Key Point 6.1.

Expand all brackets and simplify.

The expansion is valid when

You used

in place of

.

b

Replace the expression you want to integrate by its approximate expansion. Note that you can do this only when the expansion is valid within the limits of integration.

Fast Forward If you study Further Mathematics, in Pure Core Student Book 2, you will learn how to integrate this function exactly. The actual value of the integral in Worked Example 6.4 is .

EXERCISE 6A 1

Find the first three terms of the binomial expansion of each of the following expressions, stating the range of convergence.

a

i ii

b

i ii

c

i ii

d

i ii

2

Find the expansion of

3

Find the expansion of

4

a Find the first three terms in the expansion of

in ascending powers of , up to and including the term in

in ascending powers of , up to and including the term in

.

.

.

b Find the values of over which this expansion is valid.

Worksheet See Support sheet 6 for a further example of an expansion when the constant term isn’t , and for more practice questions.

5

a Use the binomial expansion to show that b State the range of values for which this expansion converges. c Deduce the first three terms of the binomial expansion of: i ii iii d Use the first three terms of the expansion to find an approximation for

6

a Find the first four terms of

, to decimal places.

in ascending powers of .

b State the range of values for which this expansion is valid. c Hence, approximate d Hence, approximate

, to decimal places. , to decimal places.

7

The cubic term in the expansion of

8

a Find the first three non-zero terms of the binomial expansion of

is

. Find the value of . and state the range of

values of for which the expansion is valid. b Use your expansion to find an approximate value of 9

.

a Find the first three non-zero terms, in ascending powers of , in the binomial expansion of Consider the definite integrals

and

.

b The expansion from part a can be used to find and approximate value of only one of those integrals. State, with a reason, which one it is and find its approximate value. 10 Given that the expansion of

is

11 Given that the expansion of

is

12 Assume that

, find the value of . , find the value of . .

By squaring both sides and comparing coefficients, find the values of , and . Does this agree with the binomial expansion of ?



Section 2: Binomial expansions of compound expressions Binomial expansions can be part of a larger expression. WORKED EXAMPLE 6.5

Find the quadratic term in the binomial expansion of

if

. Use the laws of indices to create expressions of the form . Expand both brackets, identifying for the first bracket and ,

for the second.

It is worth tidying up the expressions before you continue. You only need terms that result in . You can get this from the constant in the first bracket multiplied by the quadratic in the second, the two linear terms multiplied together or the quadratic in the first bracket multiplied by the constant in the second bracket.

Quadratic term:

Sometimes you have to use partial fractions to write an expression in a form where binomial expansion is possible.

Rewind Partial fractions were covered in Chapter 5.

WORKED EXAMPLE 6.6

a Find the first two terms in the expansion of

.

b State the range of values for which this expansion converges. You can write the expression using the standard partial fraction decomposition.

a

Substituting

So

:

.

Substituting

:

So

.

Therefore

The first term is

Write the first term in the required form and identify and .

The second term is

Write the second term in the required form and identify and .

Combine the two expansions together.

So b The first expansion converges if

so

Consider the range of convergence of each expansion separately.

.

The second expansion converges if

so

.

Therefore, both will converge if

.

You need both inequalities to be satisfied.

–1

–

1 1 – 2 3

1 1 3 2

0

1

WORK IT OUT 6.1 Find the first three non–zero terms, in ascending powers of , in the binomial expansion of . Which of the following solutions is correct? Identify the mistake in the other two. Solution 1 Set

. Then you need the expansion of

The expansion is

Substituting back for :

Solution 2

.

Solution 3

EXERCISE 6B

EXERCISE 6B 1

Find the first three terms, in ascending powers of , in the expansion of

2

Find the first three terms of the expansion of over which the expansion converges.

3

Find the first three terms in the expansion of

4

Find the first three terms in the expansion of

5

a Decompose

for

.

in ascending powers of , stating the range

. .

into partial fractions.

b Hence, find the first three terms in the binomial expansion of

.

c Write down the set of values for which this expansion is valid. 6

a Express

as partial fractions.

b Find the first three terms, in ascending powers of , in the binomial expansion of

.

c Find the values of for which this expansion converges. 7

Find the first three terms of the binomial expansion for

8

a Write

.

in partial fractions.

b Find the first three non-zero terms, in ascending powers of , in the binomial expansion of . Find the range of values of for which the expansion converges. c Use your expansion to find an approximate value of 9

.

a By first expressing it in partial fractions find the first two terms, in ascending powers of , in the expansion of

.

b Let be the largest integer for which the expansion from part a can be used to approximate the value of

. State the value of and find the approximate value of the integral

in this case. 10 a

Find the first two terms, in ascending powers of , in the expansion of

.

b Over what range of values is the expansion valid? c By substituting

find an approximation to

11 The first three non–zero terms of the expansion of 12 Is it possible to find a binomial expansion for

, to decimal places. are

?

Checklist of learning and understanding For any rational value of , the binomial expansion

. Find the value of .

. To find the binomial expansion of

, write it as

. This expansion is valid for

. A rational function may need to be written in partial fractions before using the binomial expansion. You can use the first few terms of a binomial expansion to approximate the expression for small values of .

Mixed practice 6 1

Find the first four terms, in ascending powers of , in the expansion of

, stating the

range of for which the expansion converges. 2

Find the first three non–zero terms, in ascending powers of , in the expansion of

,

stating the range of for which this is valid. 3

Find the first four terms in the expansion of

.

4

Find the first three non–zero terms in ascending order in the expansion of

5

i Expand

.

in ascending powers of , up to and including the term in

ii a Hence, or otherwise, expand the term in .

.

in ascending powers of , up to and including

b State the set of values of for which the expansion in part ii a is valid. © OCR, GCE Mathematics, Paper 4724, January 2010 6

a Find the first three terms of the binomial expansion of b Hence, find an approximation for

7

a Write

.

, to decimal places, showing your reasoning.

in partial fractions.

b Hence, find the first three terms in the expansion of

.

c Write down the set of values for which this expansion is valid. 8

Find the first three terms in the binomial expansion of

, stating the range of values

for which it is valid. 9

The first three terms in the binomial expansion of

are

. Find

the values of and . 10 i Show that ii By taking

, for , show that

.

. © OCR, GCE Mathematics, Paper 4724, June 2008

11 a

Find the first three non–zero terms of the binomial expansion of

b By setting

, find an approximation for

12 Given that the expansion of

14 a

, to decimal places.

is

13 Find the first three terms of the expansion of can be written in the form

.

, find the value of . . . Find the values of , and

, and state the set of values of for which this converges. b Show that

c

can be written in the form

can be written in the form

.

. Find the values of , and , and state

the set of values of for which this converges. d Use an appropriate expansion to approximate

to decimal places, showing your

reasoning. e Use an appropriate expansion to approximate

to significant figures, showing your

reasoning. 15 In special relativity the energy of an object with mass and speed is given by , where

is the speed of light.

a Find the first three non–zero terms of the binomial expansion in increasing powers of , stating the range over which it is valid. Let be the expansion containing two terms and let terms. b By what percentage is

bigger than

i

of the speed of light?

ii

of the speed of light?

c Prove that

be the expansion containing three

if is:

.

Worksheet See Extension sheet 6 for a look at Babylonian multiplication.

FOCUS ON … PROOF 1

Sums of series Arithmetic series You are going to demonstrate that, for an arithmetic series with first term and common difference , the sum of the first terms is

PROOF 2

Parts of this proof have been omitted. They are indicated by numbers in dark blue angled brackets. Work out what should be written there before turning to the answer. Write the first couple of terms of the series and the last couple of terms. Writing these terms in the opposite order has no effect on the outcome. Adding the two expressions gives identical terms. Collect like terms.

Did you know? Carl Friedrich Gauss (1777 – 1855) Carl Gauss was among the most eminent mathematicians of the 19th century. His many contributions to mathematics include great strides in number theory, statistics and physics. He was a child prodigy and there is a famous legend about a lesson where his teacher was hoping to keep him quiet by asking him to add together all the numbers from to . The teacher was somewhat disappointed when he replied with the correct answer within seconds. It is believed that he applied a procedure similar to the one used in this proof.

Geometric series You are going to demonstrate that for a geometric series with first term and common ratio the sum of

the first terms is

PROOF 3

Parts of this proof have been omitted. They are indicated by numbers in blue angled brackets. Work out what should be written there before turning to the answer. Write the first few terms and the last few terms of the sum.

By subtracting you can remove all the terms in common between the two series. Factorise both sides. Divide by

.

QUESTIONS 1

This proof does not work when when ?

2

This proof appeals to the very important mathematical idea of self-similarity – looking to get similar structures in two different ways so that things cancel out. Use this idea to evaluate:

. At which stage does it break down? What is the formula for

a b

Worksheet See Extension sheet 5 for more examples of continued fractions.



FOCUS ON … PROBLEM SOLVING 1

Trying small cases In this section you will look at solving problems about sequences, but the ideas apply in other contexts, too. In addition to thinking about trying small cases to spot patterns in sequences, you should also remember several other problem solving ideas. In particular: introduce letters to represent unknowns look for things that stay the same persevere. WORKED EXAMPLE

The terms of a sequence of positive integers

follow the rule

. Find

. Let

and

Then

.

First thoughts are that is a very large number and you should avoid simply working out all the numbers on the way up to it. Also you are not told what the first few terms are, so maybe it doesn’t matter. You’ll need to give them names to work with them though. Let’s keep working out some more cases and hope you see something useful! It’s worth simplifying these expressions to help anything useful stand out.

Persevere…

The algebra gets messy, but it works out quite nicely in the end!

This is nice – you’ve got back to followed by .

Importantly,

So

Since and and the sequence depends only on the previous two terms, it will continue to repeat every five terms.

QUESTIONS 1

The sequence of integers constant. Given that

satisfies

and

, where is a positive

, find the value of .

2

Find a formula for the sum of the first odd numbers.

3

An expression for the sum of the first terms of an arithmetic sequence is given by

. Find

an expression for the th term of the sequence. 4

Rebecca can walk upstairs one at a time or two at a time. For example, to go up stairs she might go , , , or , , . Her house has stairs with



steps. How many different ways can she go up the stairs?

FOCUS ON … MODELLING 1

Modelling with rational functions You are already familiar with exponential models, where the rate of change of a quantity is proportional to the quantity itself. In Student Book 1 you used them to model situations like chemical reactions, population growth and cooling of a liquid. Here are some graphs resulting from exponential models. y

y y = 6 - 6e- 0.5x y = 6e- 0.5x

x

O

x

O

But exponential equations are not the only ones that result in graphs of this shape. Here are the graphs of some rational functions:

y

y

y = 6x x + 1

y = 6 x + 1 x

O

x

O

Notice that for a simple rational function, the rate of change is proportional to the example, if

then

of its value. For

.

In this section you will meet an example of a model using a rational function and then learn a technique that allows you decide whether an exponential or a rational model is a better fit for given data. QUESTIONS 1

In biochemistry, the Michaelis–Menten model for the rate of enzymatic reaction relates the reaction rate, , to the concentration of the substrate, :

In this formula and are constants which depend on the substances involved in the reaction. a Given that and (in suitable units): i Find the rate of reaction when the concentration is ii



Find the concentration for which the rate of reaction is

iii Sketch the graph of against .

. .

b What does represent? In Student Book 1, Chapter 8, you learnt a method for determining the parameters in an exponential model by using logarithms to turn it into an equation of a straight line: If , then ln . You can use a similar idea for a rational function of the form like the Michaelis–Menten model. If then:

Hence, the graph of against is a straight line, with gradient and vertical axis intercept .

1 y

y

y = bx (x + c)

O

x

1 b

O 2

The data in the table can be modelled by the equation of the form

1 x

.

Draw the graph of against . Hence, determine the values of the constants and . The technique of transforming a graph into a straight line can also be used to decide whether a rational or an exponential function is a better model for a set of data. 3

Each graph shows a set of experimental data. Plot ln against and against . Hence, decide whether an exponential or a rational function is a better model for the data.

y

a 10

8

6

4

2

x

0 0

2

4

6

b Hint: consider

8

10

8

10

.

y 10

8

6

4

2

x

0 0



2

4

6

CROSS-TOPIC REVIEW EXERCISE 1 1

a Sketch the graph of

.

b Solve the equation

.

c Solve the inequality 2

.

The sum of the first terms of an arithmetic sequence is given by

. Find the

common difference of the sequence. 3

If

, sketch

for

,

.

4

Find an expression for the mean value of the first terms of a geometric series with first term and common ratio .

5

The function is defined by

, and the function g is defined by

where and are constants. Given that and the corresponding values of . 6

The polynomial a

i Find

,

, find the possible values of

is defined by

.

.

ii Show that b Simplify

is a factor of

.

, giving your answer in a fully factorised form.

y

7

x

O The function f is defined for all real values of by

The diagram shows the graph of i

Evaluate

.

.

ii Find the set of values of for which iii Find an expression for

.

iv State how the graphs of 8

and are related geometrically. © OCR, GCE Mathematics, Paper 4723, January 2010

a Find the range of the function

.

b Prove that there is no real value of such that terms of an arithmetic sequence. 9

The sequence

is defined by

a Find the exact value of

.

. .

,

and

are consecutive

b Find the exact value of

.

10 The functions and are defined by such that

and

. Find the value of

has equal roots.

11 For what values of is the series

convergent?

12 Functions and are defined by

and

a Find constants and such that

. .

b Using transformations, or otherwise, sketch the graph of

. Label the axes intercepts

and state the range of . c Find the domain and range of 13 It is given that a Express b

. .

in the form

, where and are integers.

i Find the first three terms of the binomial expansion of , and are rational numbers.

in the form

, where

ii Explain why the binomial expansion cannot be expected to give a good approximation to . 14 On its first trip between Malby and Grenlish, a steam train uses tonnes of coal. As the train does more trips, it becomes less efficient so that each subsequent trip uses more coal than the previous trip. i

Show that the amount of coal used on the fifth trip is figures.

tonnes, correct to significant

ii There are tonnes of coal available. An engineer wishes to calculate , the total number of trips possible. Show that satisfies the inequality iii Hence, by using logarithms, find the greatest number of trips possible. © OCR, GCE Mathematics, Paper 4722, January 2007 15 In the special theory of relativity, the energy of a particle with mass and speed is given by the formula

, where is the speed of light. Find the first three non-zero terms in

the binomial expansion, in ascending powers of , and comment on the result. 16

If

, find the exact value of .

17 The following table shows the values and gradient of

a Is

a one-to-one function?

b Evaluate

.

at various points.

c

The graph

is formed by translating the graph of

reflecting it in the -axis. Find

and then

.

18 a Find the coordinates of the image of reflection in the -axis. b

by a vector

after a reflection in the line

is a one-to-one function. The graph of

is rotated

followed by a

anticlockwise. Find the

equation of the resulting graph. 19 a If the polynomial

is divided by

the remainder is a linear function. Explain why

this statement can be written as b Find an expression for



in terms of

, where and

is a polynomial.

.

c Hence, show that the remainder is

when

d State the condition that must be satisfied if

is to be a factor of

is divided by .

.

7 Radian measure In this chapter you will learn: about different units for measuring angles, called radians how to calculate certain special values of trigonometric functions in radians how to use trigonometric functions in modelling real-life situations how to solve geometric problems involving circles that trigonometric functions can be approximated by polynomials. You will also revise solving trigonometric equations.

Before you start… Student Book 1, Chapter 10

You should be able to define trigonometric functions beyond acute angles, including exact values.

1 What is the exact value of

?

Student Book 1, Chapter 10

You should be able to solve trigonometric equations.

2 Solve

.

Chapter 3

You should be able to identify transformations of graphs.

3 The graph of is translated in the positive direction and stretched vertically with a scale factor of . Find the equation of the new graph.

Student Book 1, Chapter 11

You should be able to use the sine and cosine rules.

4 Find the smallest angle in a triangle with sides , and .

Chapter 6

You should be able to use the binomial expansion for negative and fractional powers.

5 Find the first three non-zero terms in the expansion of

for

, in ascending powers of .

What are radians? Measuring angles is related to measuring lengths around the perimeter of the circle. This observation leads to the introduction of a new unit for measuring angles, the radian, which will be a more useful unit of measurement in advanced mathematics.

Section 1: Introducing radian measure An angle measures the amount of rotation between two straight lines. You are already familiar with measuring angles in degrees, where a full turn measures . The measure of for a full turn may seem a little arbitrary. There are many other ways of measuring sizes of angles. In advanced mathematics, the most useful unit for measuring angles is the radian. This measure relates the size of the angle to the distance moved by a point around a circle. Consider a circle with centre and radius (this is called the unit circle), and two points, and , on its circumference. B 1

O

1

A

As the line rotates into position , point moves a distance equal to the length of the arc . The measure of the angle in radians is simply this arc length.

If point makes a full rotation around the circle, it will cover a distance equal to the length of the circumference of the circle. As the radius of the circle is , the length of the circumference is . Hence, a full turn measures . You can then deduce the sizes of other common angles in radians; for example, a right angle is one-quarter of a full turn, so it measures

radians. Although sizes of

common angles measured in radians are often expressed as fractions of , you can also use decimal approximations. Thus, a right angle measures approximately . WORKED EXAMPLE 7.1

a Convert

to radians.

b Convert

to degrees.

a

What fraction of a full turn is

?

Calculate the same fraction of

.

This is the exact answer. You can also find the decimal equivalent, to significant figures. What fraction of a full rotation is

b

Calculate the same fraction of

? .

Key point 7.1 To convert from degrees to radians, divide by and multiply by . To convert from radians to degrees, divide by and multiply by .

Did you know?

There are many different measures of angle. One historical attempt was gradians, which split a right angle into . It is conventional to draw the unit circle with the centre at the origin and measure angles anticlockwise from the positive -axis. WORKED EXAMPLE 7.2

Mark on the unit circle the points corresponding to the following angles, measured in radians. A B C D

C

radians is one-half of a full turn, so point represents half a turn around the circle.

D

π 3 A

is one-quarter of the full turn and the

sign represents clockwise

rotation; so point represents a quarter of a turn in the clockwise direction.

O

, so point represents a full turn followed by another B

quarter of a turn. , so point represents two full turns followed by another of a turn.

You can still find the values of trigonometric functions in radians, just as you could in degrees, but you’ll need to make sure you change your calculator to radian mode.

Gateway to A Level See Gateway to A Level Section Z for revision of working with exact values in degrees. However, as well as getting values of trigonometric functions from your calculator, you also need to recognise a few special angles for which you can find exact values. The method relies on properties of special right-angled triangles. WORKED EXAMPLE 7.3

Find the exact values of

,

and

.

If a right-angled triangle has a angle then the third angle is , so this is half of an equilateral triangle. You can choose any length for the side of the equilateral triangle. Let ; then

A

π 6

2

B

C

D

1

To find

, use Pythagoras’ theorem.

You can now use the definitions of triangles.

,

and tan in right-angled

The results for other values follow similarly and are summarised in the table.

Tip You should make sure you know how to change your calculator between degree and radian modes.

Key point 7.2

You can extend this table to include multiples of these angles and derive other properties, using symmetries of the trigonometric graphs. You therefore need to recognise -intercepts and turning points of these graphs in radians.

Key point 7.3 y y π ( , 1) 2 O

y = tan x

y

y=sinx 2π

π 3π ( , - 1) 2

WORKED EXAMPLE 7.4

x

(2π,1)

1

y=cosx O

π 2

( π, - 1)

3π 2

x

O

π 2

π

3π 2

x

WORKED EXAMPLE 7.4

Given that

, find the values of:

a b y

a

Label and

on the horizontal axis.

0.6 O

θ

x

π - θ π

y = sin x

The graph shows that

The graph has the same height at the two points.

.

y

b

Label and

on the horizontal axis.

0.6 π +θ O

θ

x

π

The graph shows that

.

WORK IT OUT 7.1 Given that

, find the value of

.

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1

Solution 2 From the graph,

Solution 3

So:

EXERCISE 7A

In this exercise, all angles are measured in radians unless stated otherwise. 1

Draw a unit circle for each part and mark the points corresponding to the following angles. a

i ii

b

i ii

c

i ii

d

i ii

2

Express the following angles in radians, giving your answer in terms of . a

i ii

b

i ii

c

i ii

d

i ii

3

Express the following angles in radians, correct to decimal places. a

i ii

b

i ii

c

i ii

d

i ii

4

Express the following angles in degrees. a

i ii

b

i ii

c

i ii

d

i

ii 5

Sketch the graph of: a

b

i

for

ii

for

i

for

ii

for

Solve questions 6 to 10 without using a calculator. 6

Given that

, find the value of:

a b c d 7

Given that

, find the value of:

a b c d 8

Given that

, find the value of:

a b c d 9

Find the exact values of: a

i ii

b

i ii

c

i ii

10 Evaluate the following, simplifying as far as possible. a

b c 11 Show that

.

12 Show that 13 Simplify

. .

14 Simplify the following expression.



Section 2: Inverse trigonometric functions and solving trigonometric equations When solving trigonometric equations in Student Book 1, Chapter 10, you used

to solve equations of the form . Based on the work in Chapter 2 of this book, you can see that this is an example of an inverse trigonometric function.

Rewind You met inverse functions in Chapter 2, Section 4. Not every function has an inverse function; we need the original function to be one–one. Looking at the graph of the sine function, it is clear that it is not one–one. So how can you define inverse sine? y

-π 2

y=sin x

x

O π 2

You need to consider the sine function on a restricted domain on which the function is one–one. From the graph on the previous page, a suitable domain is

. (There are other options but this is chosen

by convention.) It is now possible to define the inverse function. The graph of the inverse function is the reflection of the graph of the original function in the line . From the graph you can see the domain and range of the inverse function.

Tip You need to know the ranges of inverse trigonometric functions in both radians and degrees.

Key point 7.4 The inverse function of Its domain is

and its range

is

. (or

).

y π 2 1

-

f - 1(x) =arcsin(x) y=sin(x)

π 2 -1 O

1

π 2

x

-1 π 2

-

You can carry out similar analysis for cosine and tangent functions to identify the domains on which their inverse functions are defined. The results are as follows.

Tip As with the inverse sine function, it is possible to use a different restricted domain for when defining the inverse function (e.g. is another possibility). However, unless told otherwise, you should assume the standard restricted domain .

Key point 7.5 The inverse function of is Its domain is and its range (or

. ).

y π

π 2

f - 1(x) =arccos(x)

1 y =cos(x) -1

O

1

π 2

x

π

-1

Key point 7.6 The inverse function of

is

Its domain is and its range is

. (or

).

y y = tan (x)

-– π 2

f - 1(x) =arctan (x)

π 2 O

π 2 -– π 2

x

You should remember what happens when you compose a function with its inverse: this to inverse trigonometric functions, you get

. Applying

You need to be a little more careful when composing the other way around. For example, only when is in the restricted domain you used to define the arcsin function (so

).

For example,

WORK IT OUT 7.2 Evaluate

.

Which of the following solutions are valid? What errors have been made in the solution(s) that are not correct? Solution 1

Solution 2

Solution 3

This is undefined because

.

WORKED EXAMPLE 7.5

In this question you must show detailed reasoning. Solve the equation Evaluate the terms on the right-hand side.

Isolate the arcsine. Taking sine of both sides undoes the arcsine.

As you learnt in Student Book 1, Chapter 10, trigonometric equations can have more than one solution. Here we summarise how to find all the solutions when working in radians.

Key point 7.7 Equation

First solution,

Second solution,

Further solutions Add or subtract multiples of Add or subtract multiples of Add or subtract multiples of

WORKED EXAMPLE 7.6

Find the values of between

and

for which

Sketch the graph.

.

y

√2 2

x2 O

x1

x

x3

–π

2π

Note how many solutions.

Three solutions.

Use inverse cos to find the first solution. Use the symmetry of the graph to find the other solutions.

WORK IT OUT 7.3 Solve

for

.

Which of the following solutions are valid? What errors have been made in the solution(s) that are not correct? Solution 1

So

Solution 2

Adding would take you outside of the required region, so this is the only solution. Solution 3

are the only solutions in the required range..

EXERCISE 7B 1

Use your calculator to evaluate the following in radians, correct to significant figures. a

i ii

b

i ii

In questions 2 to 5 you must show detailed reasoning. 2

Find the exact value of the following, in radians. a

i ii

b

i ii

c

i ii

Tip Remember that ‘show detailed reasoning’ means you must use known values and algebra rather than solving with a calculator. 3

Find the exact value of: a

i ii

b

i ii

c

i ii

d

i ii

4

Solve the following equations. a b c

5

Find the values of between and a

for which:

i ii

b

i ii

c

i ii

d

i ii

6

Solve these equations in the given interval, giving your answers to significant figures. Do not use equation solver or graphs on the calculator. a

b

i

for

ii

for for

i

for

ii c

i

for for

ii d

i

for

ii

for

Worksheet See Support sheet 7 for a further example of solving equations in radians and for more practice questions. 7

Find the exact values of

8

Sketch the graph of

9

Find the exact solutions to

10 a If

for which for

. .

for

, show that

b Find all solutions to

. .

for

.

11 Find the exact solutions of the equation 12 Leo says that

.

for all . Prove by counter example that this statement is false.

13 Find the exact solutions of the equation 14 a Prove by a counter example that b Express

for

in terms of

.

for

.

c Solve the equation



.

Section 3: Modelling with trigonometric functions Trigonometric functions can be used to model real-life situations that show periodic behaviour; for example, the height of the tide, or the motion of a point on a Ferris wheel. To do this, you need to consider trigonometric functions with different periods and amplitudes. You can use your knowledge of transformations of graphs, from Chapter 3, to change the amplitude or the period of a function. Remember that the amplitude of an oscillating function is the vertical distance between the central line and the extreme values. This can be changed by applying a vertical stretch. For example, the function

has amplitude , while the period is unchanged.

y y=2sin x

2

1 -π

π y=sin x

O

2π

x

-–1 1 -–2 2

The period is the horizontal distance until the graph repeats. You can change it by applying a horizontal stretch to the graph. For example, the equation is of the form , so you need to apply a horizontal stretch with scale factor to the graph of

.

Tip is not the same as

.

y 1

-π

y=sinx y=sin2x

π O

2π

x

-–1 1

You can see that the amplitude of the function is still , but the period is halved to . The two types of transformation can be combined to change both the amplitude and the period of the function. The same transformations can also be applied to the graph of the cosine function.

Key point 7.8 The functions

and

have amplitude and period

WORKED EXAMPLE 7.7

a Sketch the graph of

for

.

b Write down the amplitude and the period of the function.

.

Vertical stretch with scale factor . Horizontal stretch with scale factor . y

You start with the graph of transformations to apply to it.

and think what

( )

x y=4cos 3

4

O

6π

x

-4

Hence,

and

As well as vertical and horizontal stretches, you can also apply translations to graphs. They will leave the period and the amplitude unchanged, but will change the positions of maximum and minimum points and the axes intercepts. WORKED EXAMPLE 7.8

a Sketch the graph of

for

.

b Find the coordinates of the maximum and the minimum points on the graph. a

y

y = sin(x - π) + 2 3

1

The equation is of the form

. This represents a

translation of units to the right and units up.

2

O

2π

x

y = sin(x)

b Maximum point:

The maximum and minimum points of

are

You need to apply the translation to these points. So the maximum point is . Minimum point:

So the minimum point is

The result of combining all four transformations can be summarised as follows.

Key point 7.9 The functions amplitude

and

have:

and

.

central value minimum value

and maximum value

period

You can use your knowledge of transformations of graphs to find an equation of a function when given its graph. WORKED EXAMPLE 7.9

The graph shown has the equation y

.

5π ( , 4) 4

π ( , 4) 4

x

O

3π ( , - 2) 4

Find the values of , and . is the amplitude, which is half the difference between the minimum and maximum values. is related to the period, which is the distance between the two consecutive maximum points. The formula is

.

represents the vertical translation of the graph. It is the value halfway between the minimum and the maximum values.

You can now use these ideas to create mathematical models of periodic motion, such as motion around a circle, oscillation of a particle attached to the end of a spring, water waves, or heights of tides. In practice, you would collect experimental data to sketch a graph and then use your knowledge of trigonometric functions to find its equation. You can then use the equation to do further calculations. WORKED EXAMPLE 7.10

The height of water in the harbour is metres at high tide, and metres at low tide The graph at right shows how the height of water changes with time over hours. a Find the equation for height (in metres) in terms of time (in hours) in the form b Find the first two times after the high tide when the height of water is

metres.

hours later. .

h 16

10

O

12

t

24

is the ‘central value’, halfway between the minimum and maximum values.

a

is the amplitude, which is half the distance between the minimum and maximum values. The period is

.

Write an equation that says that

So

Rearrange the equation into the form

b

The high tide is when answers with .

. .

, so you want the first two

Find two possible values of

.

To find : Solve for .

The height of the water will be and after the high tide.



WORKED EXAMPLE 7.11

A point moves with constant speed around a circle of radius , starting from the positive -axis and taking to complete one full rotation. Let be the height of the point above the -axis. Find an equation for in terms of time. Draw a diagram; you can see that between the radius and the -axis.

2 2

, where is the angle

You now need to find how depends on time. h

θ -2

-2

From the diagram:

2

You have information about the values of at the start and seconds later.

Let be the time, measured in seconds.

So

As the point moves with constant speed, is directly proportional to .

.

Hence,

.

EXERCISE 7C 1

State the amplitude and the period of the following functions, where is in radians. a b c d

2

Sketch the following graphs, giving coordinates of maximum and minimum points. a

b

c

d

3

i

for

ii

for

i

for

ii

for

i

for

ii

for

i

for

ii

for

The depth of water in a harbour varies during the day and is given by the equation , where is measured in metres and in hours after midnight. a Find the depth of the water at low and high tide. b At what times does high tide occur?

4

A small ball is attached to one end of an elastic spring, and the other end is fixed to the ceiling. The ball is pulled down and released, and starts to oscillate vertically. The graph shows how the length of the spring, , varies with time, seconds. The equation of the graph is . x(cm) 16

8

O

t (s)

a Write down the amplitude of the oscillation.

b How long does it take for the ball to perform five complete oscillations? c State one assumption that was made in this model.

Tip Remember that modelling assumptions are about aspects of the real situation which are not taken into account by the equation you are using. 5

The following graph has equation values of and .

for

. Find the

y 5

O

2π

x

-5

6

The following graph has equation

for

. Find the values of and .

y 2

O 110° 290° 470° 650°

x

-2

7

a On the same set of axes sketch the graphs

and

b Hence, state the number of solutions of the equation c Write down the number of solutions of the equation 8

for

.

for

.

for

The graph shows the height of water below the level of a walkway as a function of time. The equation of the graph is of the form . Find the values of , and . y 6

3

O

9

.

12

24

a Sketch the graph of

x

for

.

b Find the coordinates of the maximum and minimum points on the graph. c Write down the coordinates of the maximum and minimum points on the graph of for .

10 A point moves around a vertical circle of radius

, as shown in the diagram. It takes

seconds to

complete one revolution. y

5

h x

O

a The height of the point above the -axis is given by

, where is time, measured in

seconds. Find the values of and . b Find the time during the first revolution when the point is

below the -axis.

11 A ball is attached at its top and bottom to elastic strings, with each string attached at the far end and under tension. When the ball is pulled down and released, it starts moving up and down, so that the height of the ball above the ground is given by the equation in and is time in seconds.

, where is measured

a Find the least and greatest height of the ball above the ground. b Find the time required to complete one full oscillation. c Find the first time after the ball is released when it reaches the greatest height. 12 A Ferris wheel has radius

and the centre of the wheel is

above ground. The wheel takes

to complete a full rotation.

Seats are attached to the circumference of the wheel. Let radians be the angle between the radius connecting a seat to the centre of the wheel and the downward vertical, and let be the height of the seat

θ

above ground.

14m h

a State two modelling assumptions, relating to the seat, that have been made in the above diagram. b Find an expression for in terms of . c Initially, the seat is at the lowest point on the wheel. Assuming that the wheel rotates at constant speed, find an expression for in terms of , where is the time, measured in minutes. d Write down an expression for in terms of . For how long is the seat more than ground?



above

Section 4: Arcs and sectors The following diagram shows a circle with centre and radius , and points and on its circumference. The part of the circumference between points and is called an arc of a circle. B

l r

You can see that there are in fact two such parts; the shorter one is called the minor arc, and the longer one the major arc. The minor arc subtends an angle at the centre of the circle.

θ O

A

r

The ratio of the length of the arc to the circumference of the whole circle is the same as the ratio of angle to the angle measuring a full turn. If is the length of the arc, and angle is measured in radians, this means that

. Rearranging this equation gives the formula for the length of an arc.

Key point 7.10 The length of an arc is given by where is the radius of the circle and is the angle subtended at the centre, measured in radians.

WORKED EXAMPLE 7.12

Arc

of a circle with radius

subtends an angle of

at the centre, as shown on the diagram.

B 5cm O

0.6 rad 5cm

A

a Find the length of the minor arc

.

b Find the length of the major arc

.

a

You know the formula for the length of an arc.

b

The angle subtended by the major arc is equal to a full turn minus the smaller angle. Full turn is radians.

B

θ major sector

A sector is a part of a circle bounded by two radii and an arc. As with arcs, we distinguish between a minor sector and a major sector. minor sector

O

A

The ratio of the area of the sector to the area of the whole circle is the same as the ratio of to the angle measuring a full turn. If is the area of the sector, and is measured in radians, this means that

. Rearranging this equation gives the formula for the area of the sector.

Key point 7.11 The area of a sector of a circle is

where is the radius of the circle and is the angle subtended at the centre, measured in radians.

Tip You can only use the formulae for arc length and area of sector given in Key points 7.10 and 7.11 when the angle is in radians.

WORKED EXAMPLE 7.13

A sector of a circle has perimeter

and its angle at the centre

. Find the area of

the sector. You want to use the formula for the area, which is

. You need to find

. You are given the perimeter, and you know the formula for it. You can use this to find .

EXERCISE 7D 1

Calculate the length of the minor arc subtending an angle of radians at the centre of the circle of radius , when: a b

2

Points and lie on the circumference of a circle with centre and radius

. Angle

is

radians. Calculate the length of the major arc

, when:

a b 3

Points

and lie on the circumference of a circle with centre and radius . Calculate the area of the minor sector , when:

, and angle

a b 4

Points and lie on the circumference of a circle with centre and radius is

. Calculate the area of the major sector

The size of the angle

, when:

a b 5

Calculate the length of the minor arc

O

10 cm

6

10 cm

2.5 rad

A

in the diagram.

B

In the following diagram the radius of the circle is Calculate the size of the angle :

and the length of the minor arc

is

.

a in radians b in degrees.

B

7.5cm

8cm

A

8cm O

7

Points

and lie on the circumference of a circle with centre and radius

major arc

is

. Calculate the size of the smaller angle

. The length of the

.

8

Points and lie on the circumference of the circle with centre . The length of the minor arc and angle . Find the radius of the circle.

9

A circle has centre and radius area of the minor sector is

is

. Points and lie on the circumference of the circle so that the . Calculate the size of the angle .

10 Points and lie on the circumference of a circle with radius . The area of the major sector is . Find the size of the smaller angle , in degrees. 11 In the diagram, the length of the major arc

is

. Find the radius of the circle.

Y

O 2.1radians

X 12 A sector of a circle with angle

has area

. Find the radius of the circle.

13 The perimeter of the sector shown in the diagram is

. Find its area.

N

O

1.6 rad

M

14 The figure in the diagram shows an equilateral triangle with side , and three arcs of circles with centres at the vertices of the triangle. Calculate the perimeter of the figure. A

B

C

15 The figure shown in the diagram consists of a rectangle and an arc of a circle with centre at . Calculate the perimeter of the figure.

8cm

P

0.7 rad

5cm

8cm

16 A sector of a circle with angle

has area

. Find the radius of the circle.

17 The diagram shows a triangle and the segment of a circle. The centre of the circle is at point . Find the exact perimeter of the figure.

5cm

A

5cm

15°

18 A sector of a circle has perimeter circle.

and its angle at the centre

19 The diagram shows a sector of a circle with radius

. Find the radius of the

Find the area of the shaded region.

6 cm 3 cm 45°

20 A sector of a circle has perimeter circle.

and area

. Find the possible values of the radius of the

21 Points and lie on the circumference of a circle with centre and radius between the areas of the major sector and the minor sector is angle .

The difference . Find the size of the

22 A cone is made by rolling a piece of paper, as shown in the diagram.

12

θ 18

If the cone is to have height



and base diameter

, find the size of the angle marked .

Section 5: Triangles and circles In this section we look at two other important parts of circles: chords and segments. You will need to combine the results about arcs and sectors with the formulae for lengths and angles in triangles.

chord segment

Rewind See Student Book 1, Chapter 11 if you need a reminder of the cosine rule and the formula for the area of a triangle.

WORKED EXAMPLE 7.14

The diagram shows a sector of a circle of radius the shaded region shown, find:

and its angle at the centre

. For

a the perimeter b the area. A

r = 7cm

0.8 rad B

a For the arc:

The perimeter is made up of the arc The formula for the length of the arc is

For the chord: Cosine rule:

and the chord

.

.

The chord is the third side of the triangle . As you know two sides and the angle between them, you can use the cosine rule. Remember that the angle is in radians.

You can now find the perimeter. b For the sector:

If you subtract the area of triangle from the area of the whole sector, you are left with the area of the sector. Area of a sector

For the triangle:

Area of a triangle

You can now find the area of the segment.

You can follow the same method to derive general formulae for the length of a chord and area of a segment.

Key point 7.12 A

r

θ B

r

The length of a chord of a circle is given by and the area of the shaded segment is

where angle is measured in radians. The next example shows how you can solve more complex geometry problems by splitting up the figure into basic shapes, such as triangles and sectors. WORKED EXAMPLE 7.15

The diagram shows two equal circles of radius circumference of the other.

such that the centre of one circle is on the

P

C1

C2

Q

a Find the exact size of angle

, in radians.

b Calculate the exact area of the shaded region. P

C1

a The only thing you know is the radius of the circle, so draw all the lengths that are equal to the radius. C2

Q

The lengths and are all equal to the radius of the circle. Therefore, triangle is equilateral. b The shaded area is made up of two segments, each with an angle at the centre

.

P

C1

C2

Q

You can find the area of one segment using the formula. Remember to use the exact value of .

The shaded area consists of two equal segments.

EXERCISE 7E 1

Find the length of the chord a

in each of the following.

i B

1.2cm O

0.8rad A

ii

O 1.2rad A

b

3cm

B

i

O

A

2.8rad

11cm B

A

ii

2.6rad O

8.2cm

B

2

Find the perimeters of the minor segments from question 1.

3

Find the areas of the minor segments from question 1.

Worksheet See Extension sheet 7 for some more challenging questions of this type. 4

A circle has centre and radius . Chord subtends angle at the centre of the circle. Given that the area of the minor segment is , show that .

5

Two circles, with centres and , intersect at and . The radii of the circles are . P 6cm A

4cm

45°

θ

B

4cm

6cm Q

a Show that

.

b Find the size of angle

.

c Find the area of the shaded region.



and

, and

Section 6: Small angle approximations The following diagram shows the graphs of and , where is in radians. As you can see, near the origin, the two graphs are very close to each other. This means that for close to zero. y

x

O

PROOF 4

By considering the diagram, prove that, for small values of (measured in radians),

.

r

θ r

If the angle of the sector is radians, then its area is The area of the triangle is

For small values of , the sector and the triangle have approximately equal areas. So write down the expression for each.

.

When is small, the sector and the triangle have nearly the same area. Hence,

You can try to find a similar approximation for the cosine graph. y

O

x

For small values of , the graph of looks like a (negative) parabola. You can find its approximate equation by looking at the sector and the triangle again. PROOF 5

Prove that, for small values of ,

B r O

when is measured in radians.

appears in the cosine rule, so compare the length of the arc and the straight line .

θ r A

The length of the arc is . The length of the line is

When is small:

For small values of , the arc and the line have approximately equal lengths.

It is possible to derive a similar approximation for . All three results, called the small angle approximations, are summarised in Key point 7.13.

Key point 7.13 For small values of , measured in radians:

These results can be used to find approximate values of trigonometric functions.

Rewind This is similar to using the binomial expansion to find approximate values of powers, as you learnt to do in Student Book 1, Chapter 9.

WORKED EXAMPLE 7.16

a Use a small angle approximation to estimate the value of

.

b Find the percentage error in your estimate. a

Using

b

You need to use a calculator to find the actual value.

, as

is close to

Fast forward The ‘exact’ value of you get from the calculator is in fact also an approximation. It is obtained from a generalisation of small angle approximations, called the Maclaurin series, which

you will study if you take the Further Mathematics course, in Pure Core Student Book 2. Just as with the binomial expansion, you can replace by another function, such as

or

. You can also

multiply two expansions together. WORKED EXAMPLE 7.17

Assuming that is sufficiently small that terms in expression for:

and higher can be ignored, find an approximate

a b a

Replace by

b

Use the result from part a and

in

.

.

Expand the brackets, but only keep the terms up to

.

You need to be a little more careful when dividing two approximate expressions, you may need to turn the quotient into a product and then use the binomial expansion.

Rewind See Chapter 6 for a reminder of binomial expansion with negative powers.

WORKED EXAMPLE 7.18

Find an approximate expression for

, given that is small enough to neglect the terms in

and above. Replace

and

with their small angle approximations.

To complete the expansion, you need to rewrite the expression. Now do the binomial expansion. Remember that the bracket needs to be in the form .

Expand the bracket and ignore any terms in or higher.

EXERCISE 7F 1

Find the approximate value of: a

i ii

b

i

ii c

i ii

2

Find the small angle approximation for: a

i ii

b

i ii

c

i ii

3

Assuming is sufficiently small so that terms in and above can be ignored, find an approximate expression for each of the following. a

i ii

b

i ii

c

i ii

4

a Find a small angle approximation for

.

b Hence, find an approximate value of 5

a Given that is sufficiently small, write down an approximate expression for . b Use your expression with

6

. in the form

to find an approximate value of .

a Find an approximate expression for and higher can be ignored.

when is sufficiently small so that the terms in

b Find the percentage error when this approximation is used to estimate the value of: i ii 7

a For each of the three small angle approximations, find the largest value of for which the approximation error is less than . b Show that using the small angle approximation for and the identity gives the correct small angle approximation for . Why is it not appropriate to use in front of the square root?

8

Given that is close to zero, find an approximate expression for

, ignoring powers of

higher than . 9

Find an approximate expression for

when is close to zero.

10 Let be a small angle, measured in degrees. Use small angle approximations to find an approximate expression for and . 11 a Find an approximate expression for

for small values of , including terms up and

including . b Hence, find an approximate value of

.

12 Given that is close to zero, find an approximate expression for

13 Find an approximate expression for

.

when is sufficiently small to ignore the terms in or

higher.

Checklist of learning and understanding The radian is defined in terms of the distance travelled around the unit circle, so that a full turn radians. To convert from degrees to radians, divide by

and multiply by .

To convert from radians to degrees, divide by and multiply by

.

For some real numbers the three main trigonometric functions have exact values, which should be learnt:

The functions

and

have:

amplitude central value minimum value

and maximum value

period Properties of inverse trigonometric functions: Inverse function

Domain

Trigonometric equations can be solved in radians:

Range

Equation

First solution,

Second solution,

Further solutions

Add or subtract multiples of Add or subtract multiples of Add or subtract multiples of If is the radius of a circle and is the angle subtended at the centre, measured in radians, then the length of an

is

and the area of a sector is

For small , measured in radians:



.

Mixed practice 7 1

a The height of a wave, in metres, at a distance metres from a buoy is modelled by the function . State the amplitude of the wave. b Find the distance between consecutive peaks of the wave.

2

Solve the equation

3

In the diagram,

for

.

is a rectangle with sides

and

are circular arcs, with centre and angle C

.

is a straight line.

and

.

B

7cm

Q

2cm O

π 6

A

rad

P

a Write down the size of angle

.

b Find the area of the whole shape. c Find the perimeter of the whole shape. 4

A sector has perimeter

5

Use small angle approximations to estimate the value of

6

The diagram shows a circle with centre and radius The chord

and radius

subtends angle

P

. Find its area. . .

radians at the centre of the circle.

Q θ

r

r

O

Find: a the area of the shaded region b the perimeter of the shaded region. B

7

A

0.9rad D

C

The diagram shows a triangle, circle with centre and radius

, where angle .

is

.

is a sector of the

The area of the sector

i

is

ii The area of triangle

. Show that the length of

is twice the area of sector

iii Find the perimeter of the region

is

.

. Find the length of

.

. © OCR, GCE Mathematics, Paper 4722, June 2007

8

The diagram shows the graph of the function

. Find the values of and .

y (2, 5)

x

O

9

The shape of a small bridge can be modelled by the equation

, where is the

height of the bridge above water, and is the horizontal distance from one river bank, both measured in metres. a Find the width of the river. State one modelling assumption you used in your calculation. b A barge has height above the water level. Find the maximum possible width of the barge so it can pass under the bridge. c Another barge has width can pass under the bridge?

. What is the maximum possible height of the barge so it

10 A runner is jogging at a constant speed around a level circular track. His distance north of the centre of the track, in metres, is given by , where is measured in seconds. a How long does is take the runner to complete one lap? b What is the length of the track? c At what speed is the runner jogging? 11 Let

.

a State the period of the function. b Find the coordinates of the points where the graph of axis. c Hence, sketch the graph of

for

, for

, crosses the -

, showing the coordinates of the maximum

and minimum points. 12 Find an approximate expression for

, assuming is small enough to ignore terms in

and higher. 13 Two circles have equal radius and intersect at points and . The centres of the circles are and , and angle . a Explain why b Find the length

is also

.

in terms of .

c Find the area of the sector

.

d Find the area of the overlap of the two circles. 14 In this question you must show detailed reasoning. Find the exact values of

satisfying the equation

15 In the following diagram, is the centre of the circle and

. is the tangent to the circle at .

T

O

A

If 16 i

, and the circle has a radius of Sketch the graph of

, find the area of the shaded region.

for

.

On the same axes, sketch the graphs of

for

, indicating the

point of intersection with the -axis. ii Show that the equation

can be expressed in the form

Hence, solve the equation

for

.

© OCR, GCE Mathematics, Paper 4722, January 2012 [Abridged] 17 a Given that is small enough so that the terms in approximate expression for

and higher can be neglected, find an

.

b Hence, find an estimate for

.

18 Find the exact solutions to the equation

for

.

19 Two circular cogs are connected by a chain, as shown in the first diagram. The radii of the cogs are and , and the distance between their centres is . The second diagram shows the quadrilateral

. Line

is drawn parallel to

D O1

C

8 O1

25

3O

2

B

O2 θ

P

B

A

A

a Write down the size of b Explain why

in radians, giving a reason for your answer.

.

c Hence, find the length of

.

.

d Find the size of the angle marked , giving your answer in radians, correct to. e Calculate the length of the chain 20 a Write down an expression for b Show that c Hence, solve the equation

. .

. for

.

8 Further trigonometry In this chapter you will learn: how to find trigonometric functions of sums and differences of two angles (e.g. ) a useful method for working with sums of trigonometric functions (e.g. ) about some new trigonometric functions.

Before you start … Student Book 1, Chapter 10

You should be able to use the identities and .

1 Given that is an acute angle with



, find the exact value of: a b

Chapter 7 and

You should know and use graphs of

Student Book 1, Chapter 10

trigonometric functions, in degrees and radians.

Chapter 7 and Student Book 1, Chapter 10

You should be able to solve trigonometric equations in degrees and radians.

2 State the coordinates of the minimum point on the graph of for

,

.

3 Solve the following equations. a b

for for

.

Combining trigonometric functions Periodic phenomena, such as water waves, sound waves or motion around a circle, can be modelled using and functions. There are many situations where several functions need to be combined. For example, you can model the interference of two waves by adding the functions that describe their shapes. Some fairground rides, such as the ‘waltzer’, have groups of seats arranged in a circle, which rotates on top of a larger rotating platform. In this chapter you will learn how to simplify expressions involving some sums and products of trigonometric functions. This will enable you to, for example, sketch graphs such as

.

Section 1: Compound angle identities Working in radians, use your calculator to find: and and

Focus on … The proofs of the results from this section are given in Focus on … Proof 2. There seems to be no obvious connection between the values of, for example, fact, there are formulae, called compound angle identities, to express

, and in terms of the

. In

functions of the individual angles. The derivations are given in the Focus on … Proof 2, and the results are:

Tip Notice the signs in the cosine identities: in the identity for the sum you use the minus sign, and in the identity for the difference you use the plus sign.

Key point 2.1

This will appear in your formula book. In fact, you need to prove only the identities for and . They can then be used to prove the identities with

.

PROOF 6

Use the identity

to prove that .

Replace by

in the first identity:

The only difference between the two expressions on the left-hand side is the sign of , so see what happens when you replace with Now use the symmetries of the

as required.

and

graphs:

and

One of the simplest applications of these identities is to calculate exact values of trigonometric functions.

Rewind You met exact values of trigonometric functions in Chapter 7, Section 1.

WORKED EXAMPLE 8.1

Find the exact values of: a

b a

You only know the exact values of sine for a few angles: . You can write as a sum of two of those.

b

Look for a way to make

from the special angles.

Notice that the answers to parts a and b are the same. This is because .

Did you know? A sum of square roots can sometimes also be shown as a nested root. For example, an alternative representation for worked example 8.1 would be:

You can now use the sine and cosine compound angle identities to derive new ones. In particular:

Key point 1.1

WORKED EXAMPLE 8.2

Prove that If you can’t think of any relevant identity for always express it in terms of and .

, you can

You want to express this in terms of . Looking at the top of the fraction, if you divide by you will get in the first term, and if you divide by you will get in the second term. So divide top and bottom by .

EXERCISE 8A 1

Express the following in the form a b c d

, giving exact values of and .

2

Find the exact value of the following. a b c

3

Find the exact value of

for angles shown in these diagrams.

a

20cm

16cm

A B

5cm

b 5m

4m

6m

A

4

B

a Show that

.

b Simplify 5

a Express

. in terms of

b Given that

.

, find two possible values of

c Hence, solve the equation 6

for

. .

Write each of the following as a single trigonometric function. Hence, find the maximum value of each expression, and the smallest positive value of for which it occurs. a b

7

Assuming is small enough so that the terms in expression for

8

and higher can be ignored, find an approximate

in terms of non-negative powers of .

a Show that

.

b Hence, solve the equation 9

a Show that b Hence, solve the equation

for . for

.

.

Section 2: Double angle identities If you set

in the compound angle identities, you get the double angle identities.

Key point 8.3

Tip You can derive these identities from those given in the formula book; however, you will need them so often that it is a good idea to learn them. You can convert between the three different forms of the

identity by using

A useful application of these identities is finding exact values of half-angles. WORKED EXAMPLE 8.3

Using the exact value of

, show that

.

You know that formula to relate this to But you know that

, so you can use a double angle . .

You have to choose between the positive and negative square root here. , take the positive square root. Recognising the form of double angle formulae can be useful in solving trigonometric equations. WORKED EXAMPLE 8.4

Solve the equation Write the left-hand side in terms of only one trigonometric function. The LHS is similar to

Substitute Since

, which equals

.

Follow the standard procedure and look at the graph to find that there are four solutions in the given domain.

y 1

y = sin(A)

1 3

A

O A = - 2π

-1

If an equation contains both .

A = 2π

and

, you can use identities to turn it into an equation involving only

WORKED EXAMPLE 8.5

Showing detailed reasoning,find exact solutions of the equation

for

As before, you need to write the equation in terms of only one trigonometric function. (Note that and are not the same function!) The LHS involves a double angle, so choose the just . This is a quadratic in factorise.

identity involving

, so make one side equal to zero and try to

Solve each equation separately.

Remember to list all the solutions at the end. Although they are called the double angle identities, they can also be used with higher multiples.

Fast forward In Chapter 11 you will use double angle identities to integrate some trigonometric functions.

WORKED EXAMPLE 8.6

Find an expression for

in terms of:

a b a

and so use one of the cosine double angle identities. Since you want an expression involving only cosine, it has to be .

b From part a:

It is always a good idea to try using the answer from the previous part of the question. Use a double angle formula again to replace a.

in the answer in part

A combination of double angle and compound angle identities can be used to derive ‘triple-angle identities’. WORKED EXAMPLE 8.7

a Show that

.

b Solve the equation a

If you write , you can use a compound angle identity. You need an expression involving only single angles so you now use the double angle identities. Since the answer only involves sine, use You want only sine in the expression so replace with .

You need an equation in only one trigonometric function, which can be achieved by using the result in part a.

b

This is a cubic in , so rearrange to make the and then factorise.

The double angle formulae allow you to simplify some complex and awkward expressions involving functions and inverse functions. WORKED EXAMPLE 8.8

Let

. Find an expression in terms of for:

a b a

You know that in dealing with

but this doesn’t help you .

However, you do have a double angle here, so use the identity involving only cos (as you have no interest in introducing sin here). b

Use the sine double angle identity this time.

You can’t do anything with with .

but you can work

When you take the square root, there could be two possible answers.

EXERCISE 8B

1

a i ii b i ii c i ii

2

Given that Given that

, find the exact value of , find the exact value of

. .

Given that

and

, find the exact value of

.

Given that

and

, find the exact value of

.

Given that

and

, find the exact value of

.

Given that

and

, find the exact value of

.

Find the exact value of: a b c

3

Find the exact value of

4

Simplify using a double angle identity:

.

a b c d 5

Use double angle identities to solve the following equations. a b c d

6

Prove these identities: a b c

d 7

Find all the values of

8

Show that

.

9

a Given that

, find the possible values of

that satisfy the equation

b Solve the equation

.

, giving your answer in terms of

10 a Express

in terms of

.

b Express

in terms of

.

11 Express

.

in terms of:

a b 12 a Show that: i ii b Express 13 Given that

in terms of

.

and

, express

in terms of and .

14 Without using your calculator, find the exact value of: a b 15 a Show that b

Hence, show that

. for

.

Section 3: Expressions of the form In this section you look at a useful method for dealing with sums of trigonometric functions. Suppose you are trying to solve the equation . You know that you need to try to write everything in terms of one trigonometric function, so start by considering the identities met so far. You cannot just replace with a function of (or with a function of ), because the only identity you have linking these is and you have neither or in the equation. So instead look for an identity involving both and . From the list of compound angle identities there are several options. Try one with the understanding that if it does not work you will just try another.

Tip Notice that would work here as well as it also has

sign.

So consider the first one you met:

Now, compare this with the equation:

You can see that the LHS of the equation would be the same as the compound angle expression if you could find so that and . This is not possible, because and cannot be greater than . However, you can adjust the original identity by multiplying by a constant, :

Now you have:

which constitutes a pair of simultaneous equations in two unknowns ( and ). To find , you can use

:

(by convention

)

To find you can eliminate by dividing one equation by the other:

Tip There are other possible values of that satisfy

, but here we need to find only one such

value. So, you have shown that

can be written as

, which allows you to rewrite the

equation as:

This is now a type of equation that you know how to solve. Although this whole procedure looks rather long and complicated, with practice you will find that you can complete it quite quickly. In the next example, you will see that you can express the same function in terms of cosine rather than sine, following the same procedure, and then continue to solve the equation. WORKED EXAMPLE 8.9

a Write

in the form

b Solve

for

. .

a

You are told to use , so you first expand this using the compound angle identity. Comparing this with

we have

can now be found using

…

… and can be found using

.

∴ b

You can use the answer from the previous part.

You have met this type of equation before, so follow the standard procedure.

Let y 1

y = cos(A)

2 5

A

O

-1 A = - 0.664

A = 2π - 0.664

The procedure can be summarised as follows.

Key point 8.4 To write

in the form

or

Expand the brackets using compound angle identities. Equate coefficients of and to get equations for Use

: and

.

To get

, divide the

equation by the

equation.

The next example illustrates the method with a difference of two trigonometric functions. WORKED EXAMPLE 8.10

a Write

in the form

, where

b Hence, find the maximum value of occurs. c Sketch the graph of

and

.

and the smallest positive value of for which it

for

a

. You can now follow the standard procedure for this type of question. So, start by expanding

Comparing this with you have

.

∴ b

Since sine has a maximum value of , the maximum value of this function must be . This value occurs when For the smallest positive value of :

The maximum value of

c

occurs when

.

This is the graph of , translated units to the right and stretched vertically with scale factor . The -intercepts of to these values.

–intercepts: –intercept: Maximum points:

Subtract

find the other one.

Add and subtract to find the minimum points.

and

y (- 3.73, 7.21)

(2.55, 7.21)

0.98

- 2.16 - 5.3

4.12

O

-6 (- 0.59, - 7.21)

EXERCISE 8C

; add

You found one of the maximum points in part b.

and

Minimum points:

are

(5.70, - 7.21)

x

EXERCISE 8C 1

Express the following in the form

, where

and

:

a b 2

Express in the form

, where

and

, where

and

:

a b 3

Express in the form

:

a b 4

Express in the form

, where

and

:

a b 5

a Express

in the form

.

b Hence, give details of two successive transformations that transform the graph of graph of . 6

a Express

in the form

.

b Hence, find the range of the function 7

a Express

in the form

into the

. .

b Hence, find the smallest positive value of for which

.

Worksheet See Support sheet 8 for a further example of solving equations of the form and for more practice questions. 8

a Express

in the form

.

b Hence, find the coordinates of the minimum and maximum points on the graph of for . 9 10

Find, to 3 significant figures, all values of x in the interval By expressing .

in the form

for which

, solve the equation

. for

Section 4: Reciprocal trigonometric functions You already know that but having the notation for

, so you can do all the calculations you need with sine and cosine only, can simplify many expressions.

Similarly, since expressions of the form

often occur, it seems sensible to

have notations for these as well. This is part of the motive for introducing three additional trigonometric functions.

Key point 8.5 The secant function: The cosecant function: The cotangent function: for any integer Using the graphs of sine, cosine and tangent you can draw the graphs of For example, the graph of will have vertical asymptotes at the values of where (because you can’t divide by zero). As increases in magnitude, sec will decrease, and vice versa. When is positive, is also positive. Finally, when similar analysis can be applied to and as well.

–π

–π O 2

π 2

–π

–π

–π O 2

3π 2π 2

2π

π

O

x

x

y = sec x

y

–2π –3π 2

π

y = cosec x

y

–2π

has the same value. A

y = sec x

y

–2π –3π 2

,

π 2

π

3π 2π 2

x

The graphs show the domains and ranges of the reciprocal trigonometric functions.

.

Key point 8.6 Function

Domain

Range

In each domain, is an integer The following identities can be deduced from the familiar and

by dividing through by

, respectively.

Key point 8.7

Tip The reciprocal trigonometric functions follow the same conventions as the normal trigonometric functions, so

means

.

WORKED EXAMPLE 8.11

Solve the equation

. and therefore

As always, you require the equation to contain only one trigonometric function, so you use an identity to replace with a function of

You don’t know how to find inverse sec, so express Finally, you solve each equation separately.

WORKED EXAMPLE 8.12

Show that

in terms of

.

You have a formula for

You need .

, so start by introducing

in the answer so you change back from

Finally, you can simplify by multiplying top and bottom of the fraction by .

EXERCISE 8D In this exercise, all arguments of trigonometric functions are given in radians, unless specified otherwise. 1

Giving your answers to significant figures, find the value of: a i ii b i ii c i ii

2

Find the exact value of: a i ii b i ii c i ii d i ii

3

Find the values of

5m A

4

4m

and

6m

B

Solve the following equations for a i ii

in the following diagram.

, giving your answers to

.

b i ii c i ii d i ii 5

Find the exact solution of the following equations for a i ii b i ii c i ii d i ii

6

a i ii b i ii c i ii d i ii

Given that

, find the exact value of

.

Given that

, find the exact value of

.

Given that

find the exact value of

.

Given that

find the exact value of

.

Given that

, find the exact value of

.

Given that

, find the exact value of

Given that

, find the possible values of

.

Given that

, find the possible values of

.

7

Show that

8

Prove that

9

Prove that

.

.

10 a Given that

, show that

b Find the possible values of

11 Prove that

.

.

c Hence, solve the equation

12 Prove that

.

for .

.

13 Given that is small enough to neglect the terms in and above, find an approximate expression for in ascending powers of . 14 Find the inverse function of

in terms of the arccos function.

Checklist of learning and understanding Double angle identities and compound angle identities (Be careful to get the signs right!):

One particular application of these identities is to write or :

in the form

Expand the brackets using compound angle identities. Equate coefficients of

and

to get equations for

and

Find To get

, divide the

equation by the

equation.

Reciprocal trigonometric functions are defined by:

Two identities for reciprocal trigonometric functions can be derived from the Pythagorean identity:

Mixed practice 8 1

a Use the identity for

to prove that

b Find the exact solutions of the equation 2

a Write

in the form

for .

b Hence, find the exact value of 3

.

for which

The circle shown in the diagram has centre and

. .

is the diameter of the circle.

C θ

O

A

B

a Write down the lengths of

and

in terms of and .

b Write down an expression for the area of the triangle

.

c Write down an expression for the area of the triangle

.

d Hence, find the ratio of the two areas in the form

4

a Use the identity for b Write down the value of c Hence, find the exact value of

5

to show that

.

. .

A water wave has the profile shown in the graph, where represents the height of the wave, in metres, and is the horizontal distance, also in metres. y (m) 1.2 1.5 O

x (m)

a Given that the equation of the wave can be written as and . b

A second wave has the profile given by the equation amplitude and the period of the second wave.

, find the values of

. Write down the

When the two waves combine a new wave is formed, with the profile given by c Write the equation for in the form

for a suitable value of , where

. and

. d State the amplitude and the period of the combined wave. e Find the smallest positive value of for which the height of the combined wave is zero. f Find the first two positive values of for which the height of the combined wave is . 6

i

Sketch the graph of

ii

Solve the equation

for

.

for

, giving the roots correct to significant figures. © OCR, GCE Mathematics, Paper 4723, June 2007

iii Solve the equation figures. 7

for

a Use the identity for

, giving the roots correct to significant

to show that

.

b Hence, solve the equation 8

a Write

.

as a product of a linear and a quadratic factor.

b Show that

9

.

c Write down the exact value of

.

d Hence, find the exact value of

and

i

Express

ii

The expression

in the form

. , where

is defined by



a Determine a value of for which

and

.

. is not defined.

b Find the smallest positive value of satisfying



, giving your answer in an

exact form. © OCR, GCE Mathematics, Paper 4723, June 2010 10 a Express

in the form

.

b The function is defined by Using your answer to part , find i

the maximum value of

, giving your answer in the form

ii

the smallest value of for which this maximum occurs, giving your answer exactly, in terms of .

11 In this question, a Write down an expression for b Show that

. .

where

;

c Hence, find an expression for 12 a i ii

Expand

.

.

Hence, show that

.

iii State the value of for which the equality holds. A picture of height hangs on the wall. An observer stands at a distance from the wall with the lower end of the picture above their eye-level. The angle between the lines from the observer’s eyes to the top and bottom of the picture is , as shown in the diagram.

3m

2m

θ d m b Show that

.

c Use the answer to part a to show that

.

d Hence, find the value of that gives the largest possible value of e Find the largest possible value of angle .

Worksheet See Extension sheet 8 for a selection of more challenging problems.

.

9 Calculus of exponential and trigonometric functions In this chapter you will: learn how to differentiate , learn how to integrate , ,

,

,

and

and

review applications of differentiation to find tangents, normals and stationary points review applications of integration to find the equation of a curve and areas.

Before you start… Student Book 1, Chapters 2

You should be able to use rules of indices and logarithms.

1 Write

You should be able to differentiate

.

2

for

3

in the form

.

and 7 Student Book 1, Chapter 13 Student Book 1, Chapter 14

You should be able to integrate .

Chapters 7 and 8

You should be able to use compound angle formulae and small angle approximations.

Differentiate

.

Find the exact value of

.

4 Use small angle approximations to find the approximate value of .

Extending differentiation and integration You have already seen many situations that can be modelled using trigonometric, exponential and logarithm functions. For example, in Chapter 8 of Student Book 1 you studied exponential models for population growth, and in Chapters 7 and 8 of this book you saw an example of modelling water waves using sine and cosine functions. We are often interested in the rate of change of quantities in such models; for example, at what rate is the population increasing after ? Finding rates of change involves differentiation. Integration reverses this process; when given the rate of change you can find the equation for the original quantity. So far you have learnt how to differentiate and integrate functions of the form . In this section you will extend rules of differentiation and integration to include a wider variety of functions.

Section 1: Differentiation You already know from Student Book 1, Chapter 8 that the rate of growth of an exponential function is proportional to the value of the function.

Rewind You learnt about inverse functions and their graphs in Chapter 2, Section 4. In particular, for

the rate of growth, which is the same as the derivative, equals the value.

The derivative of the natural logarithm function is perhaps surprising, but you will see how it follows from the fact that is the inverse function of .

Key point 9.1 If

, then

If

.

, then

PROOF 7

Let point on the graph of have -coordinate . Let point be the point on the graph of , which is the reflection of in the line

.

The reflection swaps - and coordinates, so has coordinate .

y

a

y = ex

You know how to differentiate , and how the graphs of and are related. So you can use the gradient of the first graph to find the gradient of the second graph.

y = x

A y = ln x B

O

x

a

The gradient of

at is .

To see what happens to the gradient on reflection in , consider reflecting right-angled triangle.

You know that the gradient of

at equals the value.



This shows that, if a line with gradient is reflected in gradient of the reflected line is .

the

a = a gradient 1 1 a

A B

1

a gradient 1 a

So the gradient of the reflected line at is . The gradient of the graph at is . So the gradient of

is .

The rules for differentiating sums and constant multiples of expressions still apply. Sometimes an expression needs to be simplified before it can be differentiated. WORKED EXAMPLE 9.1

Differentiate You can differentiate only and so you need to simplify the expression using rules of indices and logarithms. is multiplied by a constant ( ). is a constant, so its derivative is . The derivative of

is

To find out how to differentiate and you need to remember differentiation from first principles (Student Book 1, Chapter 13). You also need to use compound angle formulae from Chapter 8, Section 1 and small angle approximations from Chapter 7, Section 6.

Tip Once you have derived the rules, as summarised in Key Point 9.2, on the next page you can use them without proof.

WORKED EXAMPLE 9.2

Use differentiation from first principles to prove that the derivative of measured in radians.

is

, where is

Use Use small angle approximations:

.

Let tend to . Differentiating

follows the same method. The result for

can also be proved in this way, as well

as by the quotient rule, which you will meet in Chapter 10, Section 3.

Key point 9.2 If

, then



If

, then



If

, then



. . .

Tip These formulae apply only if is in radians. The following two examples should help you remember two applications of differentiation: finding equations of tangents and normals, and finding stationary points. WORKED EXAMPLE 9.3

Find the equations of the tangent and the normal to the graph of the function point where . Give your answer in the form , where , , are integers. You need the gradient, which is When

:

.

To find the equation of a straight line, you also need coordinates of one point. The tangent passes through the point on the graph where coordinate is .

Tangent:

at the

. Its -

Put all the information into the equation of a line.

Normal: The gradient of the normal is

WORKED EXAMPLE 9.4

and it passes through the same point.

Find the coordinates of the stationary point on the graph of

and determine its nature.

You need to find where the gradient equals . Write

, so

.

The stationary point is

as

.

ln is only defined for

.

Use rules of logs to rewrite the -coordinate.

To determine the nature of the stationary point, you need to evaluate the second derivative at .

The stationary point is a minimum.

EXERCISE 9A 1

Differentiate the following. a i ii b i ii c i ii d i ii e i ii f

i ii

2

Differentiate after simplifying first: a i ii b i ii

c i ii d i ii e i ii f

i ii

3

Find the exact value of the gradient of the graph of

4

Find the exact value of the gradient of the graph

5

Find the value of where the gradient of

6

Find the value of where the gradient of

7

Find the rate of change of

8

Find the rate of change of

9

Find the equation of the tangent to the curve

that is parallel to

10 Find the equation of the tangent to the curve

that is parallel to

at the point when is

. is .

at the point

.

at the point

11 Given that

.

. . .

, find the values of for which

.

12 Find the equations of the tangent and the normal to the graph of

at

. Give

all the coefficients in an exact form. 13 Given that

solve the equation

14 Find and classify the stationary points on the curve 15 Show that the function

. in the interval

has a stationary point with -coordinate

16 Find the range of the function

,

. .

.

17 Find and classify the stationary points of: a b 18 The volume of water ( ), in millions of litres, in a tidal lake is modelled by the time, in days, after a hydroelectric plant is switched on.

, where is

a What is the smallest volume of the lake? b The hydroelectric plant produces an amount of electricity proportional to the rate of flow of water (measured as volume per unit time) through a tidal dam. Assuming all flow is through the dam, find the time, in the first , when the plant is producing maximum electricity.

Section 2: Integration You can reverse all the differentiation results from the previous section to integrate several new functions.

Key point 9.3

Note the modulus sign in

. ln( ) is not defined for

negative . However, you can see from the diagram that the area between the -axis and the negative part of the graph of is the same as the corresponding area between the -axis and the positive part of the graph. y y = 1 x

–b

–a O

a

b

x

Explore Is it possible to find the shaded area between the graph of integration gives

and the -axis? Using definite

. However, splitting the area in two doesn’t give

the same answer:

has no value, since

is undefined. It turns out that the

shaded area does not have a finite value.

y

–2 O

6

x

However, some graphs with asymptotes can still enclose a finite area; find out about improper integrals.

You can combine these facts with rules of integration you already know. Sometimes an expression needs to be rewritten in a different form before integrating. WORKED EXAMPLE 9.5

Find

. You don’t know how to integrate a quotient, but you can split the fraction and integrate each term separately.

WORK IT OUT 9.1 Which of the following statements is correct? Identify the mistake in the other three. Statement 1

Statement 2

Statement 3

Statement 4

Remember that integration is the reverse of differentiation, so if you know the derivative of a function you can use integration to find the function itself. WORKED EXAMPLE 9.6

A curve passes through the point

and its gradient is given by

of the curve. Integrate

to get .

Don’t forget the constant of integration. Using

. Find the equation

Use given values of and to find the constant of integration.

You can also use integration to find the area between the curve and the -axis. This involves evaluating definite integrals. It is always a good idea to draw a diagram to make sure you are finding the correct area.

Rewind You know from Student Book 1, Chapter 15 that when a curve is below the -axis the integral is negative.

WORKED EXAMPLE 9.7

Find the exact area enclosed between the -axis, the curve y

and the lines

and

.

Sketch the graph and identify the required area. y = sinx

O

π 3

x

Integrate and write in square brackets. Evaluate the integrated expression at the upper and lower limits, and subtract the lower from the upper.

EXERCISE 9B 1

Find the following integrals. a i ii b i ii c i ii d i

ii e i ii f

i ii

2

Find the following integrals. a i ii b i ii c i ii d i ii e i ii f

i ii

3

Find the exact value of these definite integrals. a i ii b i ii c i ii

d i ii e i ii f

i ii

g i ii 4

Find the exact value of the area enclosed by the curve . Give your answer in the form

5

Find the area enclosed by the curve

6

Find the exact value of

7

Find the exact value of

8

a Evaluate

, the -axis and the lines

and

. , the -axis, and the line

.

.

.

.

b State the value of the area between the graph of

, the -axis and the lines

and

. 9

Find the equation of the curve given that

and the curve passes through the point

. 10 Find

.

11 The derivative of the function

is

.

a Find an expression for all possible functions b If the curve

.

passes through the point

, find the equation of the curve.

Worksheet See Support sheet 9 for a further example of finding the equation of a curve and for more practice questions. 12 The diagram shows part of the curve

.

y

1

x

O

a Use your calculator to find the -intercept of the graph. b Find the shaded area enclosed between the curve, the -axis and the lines

and

.

Tip The equation cannot be solved exactly. You can use the solver function on your calculator. Note that the answer for the area is only an approximation. 13 The diagram shows a part of the curve with equation

.

y

O

6

x

a Show that the curve crosses the -axis at

and

.

b Find the exact value of the shaded area. Give your answer in the form

, where and are

rational numbers.

Fast forward Areas of regions that are partly above and partly below the -axis are discussed in more detail in Chapter 12, Section 4.

14 Show that the value of the integral

is independent of .

15 The gradient of the normal to a curve at any point is equal to the -coordinate at that point. If the curve passes through the point , find the equation of the curve in the form where is a rational function.

Checklist of learning and understanding You can now differentiate basic trigonometric, exponential and logarithm functions: ,

The results for

and

can be derived using differentiation from first principles.

You can also integrate some new functions: , You can combine these with the rules of differentiation and integration you already know, to find: Rates of change Equations of tangents and normals Stationary points Equation of a curve with a given gradient Area between a curve and the -axis.

Mixed practice 9 1

Find the equation of the tangent to the curve

2

If

3

Find the exact value of the gradient of the graph

and

, find

when

at the point where

.

.

.

4

Find the exact value of the integral

5

The diagram shows the curve with equation

. , which crosses the -axis at and

.

y

x

O

Find the exact value of the shaded area. 6

Find the indefinite integral

7

Find and classify the stationary points on the curve

8

Find and classify the stationary points on the curve

. in the interval for

.

. Give only

the -coordinates, and leave your answers in terms of . 9

Find the equation of the normal to the curve at the point the form , where , and are integers.

. Give your answer in

10 The population of bacteria ( ), in thousands, at a time , in hours, is modelled by . a i ii

b i ii c i ii

Find the initial population of bacteria. At what time does the number of bacteria reach on your calculator.) Find

million? (Hint: use a solver function

.

Find the time at which the bacteria are growing at a rate of million per hour. Find

and explain the physical significance of this quantity.

Find the minimum number of bacteria, justifying that it is a minimum.

11 The diagram shows the graphs of

and

. Find the exact value of the shaded

area.

y

x

O

12 Find

.

Worksheet See Extension sheet 9 for a selection of more challenging problems.

10 Further differentiation In this chapter you will learn: how to differentiate composite functions using the chain rule how to differentiate products and quotients of functions about implicit functions and their derivatives how to differentiate inverse functions.

Before you start… Student Book 1, Chapter 13 and Student Book 2, Chapter 9

You should be able to differentiate the following functions.

1 Differentiate: a b c d

Student Book 1, Chapter 14

You should be able to use differentiation to find the equations of tangents and normals, and stationary points.

2 A curve has equation . a Find the equations of the tangent and the normal at the point where . b Find the coordinates of the stationary point and show that it is a minimum point.

Student Book 1, Chapter 10

You should know basic trigonometric identities.

3 Simplify: a b

Chapter 8

You should know the definitions of reciprocal trigonometric functions.

4 Write in terms of a b

Student Book 1, Chapter 2 and Student Book 2,

You should be able to simplify expressions involving fractions and surds.

5 Simplify:

and

:

Chapter 5 a

b

GCSE and Student Book

You should be able to change the subject of a formula.

6 Make the subject of the formula:

1, Chapter 7

a b

Differentiating more complex functions In Chapter 9 you learnt how to differentiate a variety of functions. You can also differentiate sums, differences and constant multiples of those functions; for example, . How can you differentiate products and quotients of functions such as

or

?

You know from Student Book 1, Chapter 12 that you can’t just differentiate the separate components of a product (or quotient) and then multiply (or divide) the results, so you need a new rule for these situations. Before looking at products and quotients you will learn how to differentiate composite functions, such as or .You will also learn how to differentiate equations such as without making the subject, and apply this method to differentiate inverse functions.

Section 1: The chain rule If you were asked to differentiate

you could expand the brackets and differentiate term

by term. But in most situations this is either too difficult or not possible; for example, be ‘expanded’ and

cannot

would lead to a very long expression.

And what about functions such as or ? While you can differentiate have no rules so far telling you what to do when is replaced by or .

and

, you

These functions may seem quite different but they do have something in common – they are all composite functions:

, where

.

, where

 

, where

.

.

To differentiate a composite function you can think about how a change in affects the change in . If changes by this causes to change by , which, in turn, causes to change by . If you multiply the two rates of change you see that:

This leads to the rule for differentiating composite functions:

Key point 10.1 Chain rule: If , where

, then

Tip The chain rule also can be written using function notation:

WORKED EXAMPLE 10.1

Differentiate the following functions. a b c a

. These are all composite functions, so use the chain rule on each of them.

Write the answer in terms of . There is no need to expand the brackets. b

, where

and

In this example we use function notation.

.

Write the answer in terms of and rearrange into the conventional form. c

, where

.

Write the answer in terms of in the conventional form. Part b of the previous example illustrates a special case of the chain rule when the ‘inside’ function is of the form

:

Key point 10.2

For example:

and

It is useful to remember this shortcut as it occurs quite commonly; just look for an ‘inner function’ . In particular, you can apply this to the functions you learnt to differentiate in Chapter 9.

Key point 10.3

Tip Notice that the derivative of is the same as the derivative of . This is because , and is a constant which differentiates to zero. Sometimes you need to apply the chain rule more than once. WORKED EXAMPLE 10.2

Differentiate

.

Remember that , where

and

.

means

.

This is is a composite of three functions. Use the chain rule with three derivatives. For use the shortcut from Key point 10.3.

you can

Write everything in terms of and simplify.

You can now use the chain rule to differentiate reciprocal trigonometric functions.

Rewind For a reminder of how to differentiate

,

, and

see Chapter 9, Section 1.

WORKED EXAMPLE 10.3

Show that Express in terms of differentiate that. .

as you know how to

This is is a composite function … … so apply the chain rule.

You want the answer to contain

, which is

.

The proofs for the other two reciprocal trigonometric functions follow the same pattern, giving the results in Key point 10.4.

Key point 10.4

EXERCISE 10A 1

Use the chain rule to differentiate the following expressions with respect to . a

i ii

b

i ii

c

i

ii d

i ii

e

i ii

f

i ii

2

Use the chain rule to differentiate the following expressions with respect to . a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

g

i ii

h

i ii

3

Differentiate the following, using the short cut from Key point 10.2. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

f

i ii

4

Differentiate the following, using the chain rule twice. a

iii b

i ii

c

i ii

d

i ii

5

Find the equation of the tangent to the graph of

6

Find the equation of the normal to the curve

7

The function is defined by function have gradient ?

8

Find the coordinates of the stationary points on the curve with equation

9

A particle moves in a straight line, with displacement at time from given by At time

at the point where

.

at the point where for

.

. For what values of does the graph of this

. .

seconds, find:

a the velocity of the particle b the acceleration of the particle.

Rewind You met kinematics in Student Book 1, Chapter 19. 10 Find the coordinates of the stationary points on the graph of 11 Find the exact coordinates of stationary points on the curve

. for

12 Find the exact coordinates of the stationary point on the curve

. .

13 A population of bacteria increases exponentially so that the number of bacteria, , after minutes is given by Find the size of the population at the moment when its rate of increase is 197 bacteria per minute. 14 Given that a Find

: .

b Solve the equation

for

.

15 A non-uniform chain hangs from two posts. Its height, , above the ground satisfies the equation for . The left post is positioned at

, and the right post is positioned at

a State, with reasons, which post is taller. b Show that the minimum height occurs when

.

c Find the exact value of the minimum height of the chain.

Did you know?

.

Many people think that a chain fixed at both ends will hang as a parabola, but it can be proved that it hangs in the shape of the curve in question 15, called a catenary. The proof requires techniques from a mathematical area called differential geometry.

16 a Solve the equation

for

, giving your answers in terms of .

b Find the coordinates of the stationary points of the curve your answers correct to significant figures. c Hence, sketch the curve

for

.

for

, giving

Section 2: The product rule There is also a rule for differentiating products such as

or

.

Tip The product rule can also be written in function notation:

Key point 10.5 Product rule: If then

WORKED EXAMPLE 10.4

Differentiate: a b a Let

and

.

It doesn’t matter which function you call and which .

Then

and

So: Applying the product rule.

b Let

and

. Then:

is a product of two functions, so use the product rule.

So:

WORK IT OUT 10.1 Differentiate

.

Which of the following solutions is correct? Identify the mistakes in the other two.

Solution 1

Solution 2

Solution 3

When differentiating a more complicated product, you may need to use the chain rule as well as the product rule. When the function involves powers, the two terms in the product rule often have a common factor. WORKED EXAMPLE 10.5

Differentiate

and factorise your answer.

Let

. Then:

and

This is a product, so use the product rule.

is a composite function, so use the chain rule.

Now apply the product rule.

You are asked to factorise the answer, so instead of expanding the brackets look for common factors.

EXERCISE 10B

EXERCISE 10B 1

Differentiate the following using the product rule. a

i ii

b

i ii

c

i ii

d

i ii

2

Find a

and factorise your answer fully.

i ii

b

i ii

3

Differentiate

4

Given that

5

Find the -coordinates of the stationary points on the curve

6

Find the exact values of the -coordinates of the stationary points on the curve

7

Find the derivative of

8

a Given that b Hence, find

9

, giving your answer in the form , find

in the form

, where

. .

11 a Write

, find

.

.

, show that in the form

b Hence, or otherwise, find

.

, where and are constants to be found.

. .

c Find the exact coordinates of the stationary points of the curve 12 a If

.

with respect to .

Find the exact coordinates of the minimum point of the curve

10 Given that

is a polynomial.

.

and and are positive integers, find the -coordinate of the stationary point of the curve in the domain .

b Sketch the graph in the case when

and

.

c By considering the graph, or otherwise, identify a condition involving and/or to determine when this stationary point is a local minimum.

Section 3: The quotient rule If you need to differentiate a quotient such as

you could start by expressing it as

and use a combination of the chain rule and the product rule. However, there is a shortcut that allows you to differentiate quotients directly.

Key point 10.6 Quotient rule: If

then

WORKED EXAMPLE 10.6

Differentiate

, using the quotient rule, and simplify as far as possible. Use the quotient rule, making sure to get and the right way around.

You need the chain rule to differentiate .

Notice that you can cancel a factor of

Chapter 9 stated the result that the derivative of is with the derivatives of and , to prove this result.

. You can now use the quotient rule, together

WORKED EXAMPLE 10.7

Prove that

.

. You know how to differentiate . Use the quotient rule.

and

, so use

Notice that

.

WORK IT OUT 10.2 In each example, find a quicker method of differentiating the expression. How many of the solutions are correct? Identify any mistakes. Example 1 Differentiate

.

Solution 1 Using the product rule: ,

So,

Example 2 Differentiate

.

Solution 2

So use the product rule:

So,

Example 3 Differentiate

.

Solution 3 Using the quotient rule:

So,

Tip You need to use the quotient rule only when differentiating a quotient of two functions. If a function is multiplied or divided by a number, you can do it more simply. The quotient rule, just like the product rule, often leads to a long expression. If you are required to simplify it, you may need to work with fractions and roots, as in the following example.

Worksheet See Support sheet 10 for a further example of combining the chain rule with the product or quotient rule and some more practice questions.

WORKED EXAMPLE 10.8

Differentiate

, giving your answer in the form

Use the quotient rule.

, where

.

As you want a square root in the answer, turn the fractional powers back into roots. Remove ‘fractions within fractions’ by multiplying top and bottom by . Notice that

.

EXERCISE 10C 1

Differentiate the following using the quotient rule. a

i ii

b

i ii

c

i ii

d

i ii

2

Find the equation of the normal to the curve the form

at the point where

, giving your answer in

, where and are exact.

3

Find the coordinates of the stationary points on the graph of

4

The graph of

5

Find the exact coordinates of the stationary point on the curve

6

Find the range of values of for which the function

7

Given that

has gradient at the point

, show that

.

and

. Find the value of . and determine its nature. is increasing.

, stating clearly the value of the constants , and

. 8

Show that if the curve minimum stationary point at

has a maximum stationary point at as long as

.

then the curve

has a

Section 4: Implicit differentiation The functions you have differentiated so far have always been of the form . From time to time, you will come across equations of curves that are written differently. For example, the equation of the circle shown in the diagram is . Such equations are said to be implicit, where as those in the form are said to be explicit.

Tip Any equation that describes a relationship between and is called a relation. Mappings are a special type of relation (where the equation gives explicitly in terms of ) and functions are a special type of mapping (where each corresponds to just one ).

Explore Strictly speaking the equation does not define a function, but instead a relation. Explore the difference between functions and relations. The gradient of the tangent at any point on the circle is still given by

.

Rather than attempting to rearrange the equation to make it explicit, you can just differentiate term by term with respect to :

Some care needs to be taken when you come to will need the chain rule:

, as it is not expressed explicitly in terms of ; here, you

This will generally be the case when differentiating terms involving .

Key point 10.7 When differentiating implicitly, you need to use

WORKED EXAMPLE 10.9

Find an expression for

for the circle

.

Differentiate each term. Use the chain rule on the term involving . Remember that a constant

differentiates to zero.

You now need to rearrange the equation to find

Notice that the expression for

.

will often be in terms of both and .

Sometimes you may need to use the product rule as well as the chain rule in the process of implicit differentiation. WORKED EXAMPLE 10.10

Find an expression for

if Differentiate term by term, using the chain rule on all terms involving . sin is a product, so you need to use the product rule together with the chain rule on all terms involving . Group the terms involving

.

If you are interested only in the gradient at a particular point, or you are given the gradient and need to find the - and -coordinates, you can substitute the given value into the differentiated equation without rearranging it. WORKED EXAMPLE 10.11

For the curve with equation a Find the gradient at the point

: .

b Find the coordinates of the point where the gradient is . Differentiate each term with respect to . The term require the product rule. Use the chain rule on terms involving .

Group terms involving a When

and

b When

:

:

.

Substitute the numbers before rearranging.

Put in the given value of the gradient.

will

Substitute into

Using

:

the coordinates are and

This is a second equation relating and . You can solve it simultaneously with the original equation.

Remember to find both - and - coordinates.

.

The method used in part b can also be used to find stationary points. WORKED EXAMPLE 10.12

Find the coordinates of the turning points on the curve

.

Differentiate each term with respect to but be aware that the term will requirethe product rule. Use the chain rule on all terms involving .

For stationary points,

When

.

You know the value of

.

You have found a relationship between and at the stationary points, but to actually find the points you need to substitute back into the original equation.

:

is a stationary point. When

:

is a stationary point. One application of implicit differentiation is to differentiate exponential functions with base other than . WORKED EXAMPLE 10.13

Given that Let

. Then:

, show that Take ln of both sides.

You can now use implicit differentiation. Remember that ln is a constant.

Key point 10.8

Rewind This confirms that the gradient of any exponential function is proportional to the value, as you learnt in Student Book 1, Chapter 8.

EXERCISE 10D 1

Find the gradient of each curve at the given point. a

at

i

at

ii b i

at

ii

at at

c i ii

at

d i

at

ii 2

Find

at in terms of and .

a i ii b i ii c i ii d i ii 3

Find the coordinates of stationary points on the curves given by these implicit equations. a b

4

Find the exact value of the gradient at the given point. a

b

i

at

ii

at

i

when

when

ii c

d

i

when

ii

when

i

when

ii

when

5

Find the gradient of the curve

6

A curve has equation a Show that

at the point

.

. .

b Find the equation of the tangent to the curve at the point 7

A curve has equation

. Point has coordinates

. .

a Show that point lies on the curve. b Find the equation of the normal to the curve at . 8

Find the gradient of the curve with equation

9

Find the equation of the tangent to the curve with equation

at the point

at the point

10 Find the equation of the tangent to the curve with equation 11 A curve has implicit equation

. Find an expression for

at the point in terms of and .

12 Find the coordinates of the stationary point on the curve given by 13 The line is tangent to the curve , which has the equation

.

. when

and

.

a Find the equation of . b Show that meets again at the point with an -coordinate that satisfies the equation . c Find the coordinates of the point .

. .

Section 5: Differentiating inverse functions Another application of the chain rule is to differentiate inverse functions. If derivative of is

and the derivative of

is

then

. The

. But the chain rule says that

and so:

Key point The derivative of the inverse function is

You may remember that Chapter 9, Section 1 showed a graphical proof for the result for the derivative of . You can now prove it by using the fact that is the inverse function of . WORKED EXAMPLE 10.14

Use the derivative of to prove that the derivative of Let

Then

is .

You can’t use the result you are trying to prove, so cannot differentiate directly. Instead, you can use known properties of its inverse function, .

.

. Be careful – you are differentiating with respect to Use the inverse function rule here.

But

, so , as required.

You can use the derivative of the inverse function to solve some integration problems that you could not work out before. WORKED EXAMPLE 10.15

The deceleration of a parachute is proportional to the square of its velocity, so that When

the velocity is

. Find an expression for the velocity in terms of time. You can’t integrate respect to .

with respect to , but you can integrate it with

So you are trying to find the inverse function, in terms of . You can now integrate with respect to .

When

,

: Use initial conditions to find .

.

You need to express in terms of .

Fast Forward In Chapter 12 you will learn how to solve similar problems where the acceleration depends on both and .

EXERCISE 10E

EXERCISE 10E 1

If you prefer, you can differentiate inverse functions using implicit differentiation. For example, given that : a Express in terms of . b Use implicit differentiation to find

2

. Write your answer in terms of .

A population is increasing at a rate proportional to its current size, . The initial size of the population is and its initial rate of increase is . a Show that

.

b Solve the differential equation and, hence, find the size of the population when 3

A function is defined by a By considering

.

, prove that

has an inverse function.

b Find the gradient of the graph of 4

a Given that

at the point where

, write down an expression for

b Hence, prove that 5

.

.

.

.

A spherical balloon is being inflated at a constant rate of

. The radius at time seconds is

. a Find an equation for

6

in terms of .

b Given that initially the radius is

, find an expression for the radius in terms of time.

A function is defined by

.

a Prove that is an increasing function. b Show that the point c 7

Find the exact value of the gradient of the graph of

Given that a Find

. at the point

:

in terms of .

b Hence, find 8

lies on the graph of

Given that

in terms of . is an increasing function, prove that

is also an increasing function.

Checklist of learning and understanding The chain rule is used to differentiate composite functions:

The product rule is used to differentiate two functions multiplied together: If

then

.

The quotient rule is used to differentiate one function divided by another:

.

If

then

.

The derivatives of the reciprocal trigonometric functions are:

You need to know the following standard derivatives:

To differentiate an implicit equation: Differentiate each term. Use the chain rule for any term containing : Differentiation of inverse functions:

.

Mixed practice 10 1

Find

if:

a b c 2 3

Find the exact value of the gradient of the curve with equation

when

.

A particle moves in a straight line with velocity given by

.

Find the minimum value of the velocity in the particle’s motion. 4

A curve has equation

.

a Show that the point

lies on the curve.

b Find the gradient of the curve at . 5

The curve has equation

. The tangent to the curve at the point

the coordinate axes at points and . Find the area of the triangle 6

crosses

.

Differentiate each of the following with respect to . i ii © OCR, GCE Mathematics, Paper 4723, June 2007

7

The graph of

8

Find the exact coordinates of the stationary point on the curve with equation

9

a Find the value of so that

has a stationary point when

. Find the value of . .

.

b Hence, show that the derivative of

is

.

Rewind Compare this derivation with the one using implicit differentiation in Worked example 10.14. 10 Find the exact value of the gradient of the curve with equation 11 At time seconds, the displacement of a particle from is

at the point where .

a Find the value of when the particle is instantaneously at rest. b Find the particle’s maximum speed. 12 A curve has equation

.

a Write down the equation of the vertical asymptote of the curve. b Use differentiation to find the coordinates of stationary points on the curve.

.

c Determine the nature of the stationary points. d Sketch the graph of

.

13 a Prove that the derivative of

is

.

b Find the equation of the tangent to the graph of 14 A function is defined by a Find

for

and, hence, prove that

b Find the gradient of

i Find

.

.

has an inverse function.

at the point

15 The equation of a curve is

at the point where

.

.

in terms of and .

ii Show that the point iii The point

lies on the curve and that

, where

at this point.

, lies on the curve. Find the value of and the gradient of the

curve at this point. © OCR, GCE Mathematics, Paper 4724, January 2009 16 A curve is given by the implicit equation

.

a Find the coordinates of the stationary points on the curve. b Show that, at the stationary points, c Hence, determine the nature of the stationary points. 17 a Express b Let c

in terms of

.

. By first expressing in terms of , prove that

Find the equation of the normal to the curve

18 a Given that

, find

.

at the point where

. Give your answer in the form

.

, where and

are integers.

y

O

x 2

4

The diagram shows part of the curve

.

b Find the coordinates of the stationary point on the curve. c Find the shaded area enclosed by the curve, the -axis and the lines

and

.

Worksheet See Extension sheet 10 for questions on some properties of that use differentiation.

11 Further integration techniques In this chapter you will learn how to integrate using: known derivatives the chain rule in reverse a change of variable (substitution) the product rule in reverse (integration by parts) trigonometric identities the separation of a fraction into two fractions.

Before you start… Chapter 9

You should be able to differentiate and integrate polynomial, exponential and trigonometric functions.

1 Find: a b

2 Given that

, find:

a b

Chapter 10

You should be able to use the chain rule for differentiation.

3 Differentiate with respect to : a b

Chapter 8

You should know how to use double angle formulae.

4 Given that , find the possible values of

Chapter 5

You should know how to split an expression into partial fractions.

5 Write

in

partial fractions.

Integrating more complex functions Having extended the range of functions you can differentiate, you now need to do the same for integration. In some cases you will be able to use results from Chapter 11 directly, but in many others you will require new techniques. In this chapter you look at each of these in turn and then face the challenge of selecting the appropriate technique from the not inconsiderable list of options you have built up.



Section 1: Reversing standard derivatives You already know how to integrate many functions by reversing the corresponding differentiation results: when

In Chapter 10 you differentiated , and add the following results to your list:

and

. You can now reverse these standard derivatives too

Thinking about reversing the chain rule for differentiation allows you to go one step further and deal with integrals such as

.

You know that the answer must include

as sin is the integral of

If you work backwards and differentiate

, you get

Since you don’t want the in front of

, but this isn’t the final answer.

, you divide by (or multiply by ) to get:

WORKED EXAMPLE 11.1

Find

. integrates to

. But differentiating back, the chain rule will

also give a factor of (the derivative of

. Remove this by

multiplying by

WORKED EXAMPLE 11.2

If you differentiate derivative of

, the chain rule will give a factor of (the ). Remove this by multiplying by .

You may be noticing a pattern here: you always end up integrating the function and then dividing by the

coefficient of . This is indeed a general rule when the ‘inside’ function is of the form

.

Key point 11.1

where

is the integral of

.

Tip Note that this rule only applies when the ‘inside’ function is of the form It is worth remembering three particular examples of this rule:

Key point 11.2

WORKED EXAMPLE 11.3

Find

. Divide by the coefficient of .

WORK IT OUT 11.1 Find

.

Which of the following solutions are correct? (There may be more than one.) Identify the mistakes in those that are not correct. Solution 1

Solution 2

Solution 3

How can solutions 2 and 3 both be correct? The answer lies in the constant of integration. The answer in Solution 2 can be re-written as follows:

for an arbitrary constant, . But

is simply a number, so

written as

(where

is another arbitrary constant, so the answer in Solution 2 can be ).

WORKED EXAMPLE 11.4

Find

. Integrate

EXERCISE 11A 1

Find the following indefinite integrals. a

i ii

b

i ii

c

i ii

d

i ii

e

i

ii

to

and divide by the coefficient of

2

Find the following integrals. a

i ii

b

i ii

c

i ii

d

i ii

3

Find: a

i ii

b

i ii

c

i ii

d

i ii

4

Integrate the following. a b c d e f

5

Find the exact value of

6

Find the exact area enclosed by the graph of

7

Find the area enclosed by the -axis and the curve with equation

. , the -axis and the lines

and .

.

8

Given that is

and the area between the -axis, the lines , find the value of , correct to significant figures.

,

and the graph of

Section 2: Integration by substitution The shortcut for reversing the chain rule from the previous section (Key point 11.1) works only when the derivative of the ‘inside’ function is a constant. This is because a constant factor can ‘move through the integral sign’, as shown in the following example.

This cannot be done with a variable:

is not the same as

. So you need a different rule

for integrating a product of two functions. In some cases this can be achieved by extending the principle of reversing the chain rule, leading to the method of integration by substitution.

Fast forward Another method for integrating products is integration by parts, which is the reverse of the product rule. You will meet it in Section 3.

Reversing the chain rule When using the chain rule to differentiate a composite function, you differentiate the outer function and multiply this by the derivative of the inner function; for example:

You can think of this as using a substitution

, and then

.

Tip When making a substitution of this type, only replace the inner function with Any other instances of will cancel out when changing to .

Now look at

. Since

. Thus the integral becomes You know how to integrate instead of

is a composite function, you can write it as

, where

.

But you need to be integrating with respect to , so you should have

. Those two are not the same thing but they are related because

, so

This can then be rearranged to make the subject so that it can be replaced in the integral: Substituting all of this into the integral gives

It follows from the chain rule that

.

Notice that having got an answer in terms of , it needed to be written back in terms of In practice, this method can be shortened by appreciating that gives

can be ‘split up’ just like a fraction, so

.

This is illustrated in the following examples. WORKED EXAMPLE 11.5

Find the following: a b a Let

.

Think of

as

; therefore, the inner function is

Then Make the substitution.

Write the answer in terms of . b Let

.

is a composite function with inner function

.

Then Make the substitution.

Write the answer in terms of .

When limits are given, you must ensure they are changed, too. In that case there is no need to change back to the original variable at the end. WORKED EXAMPLE 11.6

Evaluate

, giving your answer in the form

, where and are rational

numbers. Let

This is of the form something

Then

function is

.

, so the ‘inner’

Write the limits in terms of .

Limits:

Make the substitution. Simplify:

.

This particular case of substitution, where the top of the fraction is related to the derivative of the bottom, is definitely worth remembering:

Key point 11.3

You can use this result to integrate some trigonometric functions. WORKED EXAMPLE 11.7

Show that

. If form

then

. So you can write the integral in the

.

Using the result of Key point 11.3. You want the answer in terms of and use

, so remember that

.

EXERCISE 11B 1

Use either a suitable substitution or the reverse chain rule to find the following integrals.

Tip You don’t have to make a substitution if you can see that the expression is of the form ,

i.e. with both a function and its derivative present. You can just go

straight to the answer using the reverse chain rule:

a

i ii

b

i

.

ii c

i ii

d

i ii

e

i ii

f

i ii

g

i ii

h

i ii

i

i ii

2

Find

3

Find the exact value of

4

Show that

5

Find

6

Find

, showing all your working.

, where is an integer to be found. . .

General substitution In all the examples you have done so far, after making the substitution, the part of the integral that was still in terms of cancelled with a similar term coming from

. This will always happen when one part of

the expression to be integrated is an exact multiple of the derivative of the inner function. You can explain this by looking at the chain rule. In some cases you will not be able to cancel all terms in , so you will have to express in terms of to obtain a function of only. The full method of substitution will then be as follows:

Key point 11.4 1Select a substitution (if not already given). 2Differentiate the substitution and write 3Replace

in terms of

.

by the expression above, and replace any obvious occurrences of .

4Change the limits from to . 5Simplify as far as possible. 6If any terms with remain, write them in terms of . 7Do the new integral in terms of . 8Write the answer in terms of .

Tip If you are not told which substitution to use, look for a composite function and take function.

WORKED EXAMPLE 11.8

Find

, using the substitution

. Differentiate the substitution and write (step 2).

in terms of

Replace those parts that you already have expressions for, and simplify if possible (steps 3 to 5).

There is still an remaining, so replace it using (step 6). Now everything is in terms of so you can integrate (step 7). Remember that .

Write the answer in terms of , using 8).

(step

A substitution can be given as in terms of , rather than in terms of . WORKED EXAMPLE 11.9

Use the substitution

to evaluate

.

Differentiate the substitution. Note that now you are finding Find limits for . This involves solving an equation. Replace

by

and by

. Use the fact that

.

.

Simplify if possible. Now everything is in terms of so you can integrate. Remember the modulus sign with the ln.

EXERCISE 11C 1

Find the following integrals using the given substitutions. a

,

i

,

ii b

,

i

,

ii 2

Use the given substitutions to find these integrals. a

,

i

,

ii b

i ii

3

, ,

Find the following, using appropriate substitutions. a

i ii

b

i ii

c

i ii

4

Use the given substitutions to evaluate the following definite integrals. a

i ii

b

i ii

c

i ii

5

Use the substitution

6

a Show that

is a factor of

b Find



to find

7

Use the substitution

8

Show that

9

Using the substitution

. .

. to find

.

, where , and are integers to be found.

, find the exact value of

.

Section 3: Integration by parts You have already seen, in Section 2, cases in which you can integrate products of functions by the reverse chain rule or a substitution. But you still can’t do integrals such as

or

.

To integrate these you need to return to the product rule for differentiation:

Integrating with respect to we get:

Key point 11.5 The integration by parts formula is:

Tip This derivation shows that you can think of integration by parts as the ‘reverse product rule’. In practice, though, there is no need to write out the product rule every time; you can just use the integration by parts formula.

WORKED EXAMPLE 11.10

Find

. This is a product to which you can’t apply the reverse chain rule, so try integration by parts.

Apply the formula.

In Worked example 11.10, taking to be worked well because formula, the resulting integral was just

is a constant. So after applying the

. You could do this only because

integrate, so you could find . In some examples this is not possible. WORKED EXAMPLE 11.11

Find

.

is easy to

This is a product, so use integration by parts. You can’t take

because you don’t know how to integrate

. So try choosing them the other way round. Apply the formula. Always simplify before integrating.

You can summarise the strategy for choosing which function is and which is

as shown in Key point

11.6.

Key point 11.6 When using integration by parts for

, choose

Somewhat surprisingly, this strategy can even be used to integrate

in all cases except when

by itself.

WORKED EXAMPLE 11.12

Use integration by parts to find

. The trick is to write as a product of use integration by parts. As suggested in Key point 11.6, let

Apply the formula.

EXERCISE 11D

and so that you can

EXERCISE 11D 1

Find the following using integration by parts. a

i ii

b

i ii

c

i ii

d

i ii

2

Use integration by parts to evaluate

3

Find the exact value of

. .

Repeated integration by parts It may be necessary to use integration by parts more than once. WORKED EXAMPLE 11.13

Find the exact value of

. This is a product to which you can’t apply the reverse chain rule, so try integration by parts. Choose to be the polynomial. Apply the formula. Put in the limits on the away.

part straight

You have to integrate a product again, so use integration by parts for the second time. Choose to be the polynomial again. (Notice that if you were to now change and choose

, you

would end up back where you started!) Apply the formula again and use the limits.

Therefore,

Put both integrals together, making sure to keep track of negative signs by using brackets appropriately.

Tip When doing integration by parts twice, make sure you apply the negative sign in front of the integral to all of the second application of the parts formula.

EXERCISE 11E

EXERCISE 11E 1

Use integration by parts twice to find the following. a b c d

2

Use integration by parts to find. a b

3

Evaluate the following exactly. a b

4

When using the integration by parts formula, you start with

and find . Why do you not include a

constant of integration when you do this? Try a few examples, adding 5

Find

6

Evaluate

7

a Show that

.

.

b Hence, find

.

8

Use the substitution

9

Let

to find the exact value of

.

.

a Use integration by parts twice to show that b Hence, find

to and see what happens.

.

.

Section 4: Using trigonometric identities in integration Trigonometric identities are useful for integrating the squares of trigonometric functions. WORKED EXAMPLE 11.14

Find

. You can integrate

In Section 2 you integrated expressions such as cancels with the

, so use

using the substitution

. But if you try to integrate just

You could try rewriting it as

.

; the derivative

the same substitution doesn’t work.

, but you doesn’t know how to do this either.

Rewind Double angle identities were covered in Chapter 8, Section 2. The trick is to notice that appears in one of the versions of the double angle formulae for , and you know how to integrate . This leads to a method for integrating both

and

.

Key point 11.7 To integrate To integrate

use use

.

WORKED EXAMPLE 11.15

Find

. You can find an alternative expression for double angle identity.

Divide by the coefficient of when integrating

by using a

.

Sometimes trigonometric identities turn out to be useful in integrals that don’t appear to involve trigonometry. WORKED EXAMPLE 11.16

Use the substitution

to find

.



Make the substitution. Simplify each part separately first.

Use

.

Differentiate the substitution. Simplify fully before attempting to integrate.

Now we have:

Use the method from Key point 11.7:



. Remember to divide by when integrating

.

Find the limits for .

Limits:

Apply the limits to the indefinite integral.

In integrals of this type, you may need to use trigonometric identities again to write the answer in terms of at the end.

Tip The substitution in Worked example 11.16 is useful in other integrals involving expressions like .

WORKED EXAMPLE 11.17

a Show that b Use the substitution a

. to find

. There is no obvious identity linking the functions on the left to the ones on the right. So write everything in terms of and .

Differentiate the substitution and express terms of .

in

b

Replace those parts that you already have expressions for. There are no instances of remaining, so you can integrate. Notice that

.

Using the result from part , you now have a standard derivative (Section 1). To express the answer in terms of , you need to link to . Notice that, in choosing the substitution , we can choose to be between and . Then is non-negative, so we can take the positive square root here.

WORK IT OUT 11.2 Three students integrate

in three different ways.

How many of the solutions are correct? Identify any mistakes. Solution 1 Amara uses reverse chain rule with

.

so

Solution 2 Tim uses reverse chain rule with

so

Solution 3

.

Carlos uses the double angle formula

.

EXERCISE 11F 1

Simplify to get standard integrals, and then integrate the following. a b c d e

2

Use trigonometric identities before using a substitution (or reversing the chain rule) to integrate the following. a b c d e

3

Find the following. a

i ii

b

i ii

4

Find the exact value of the following. a

i ii

b

i ii

5

Use the given substitutions to find the following. a

b

i

,

ii

,

i

,

ii c

i

, ,

ii 6

Find

7

a Show that

, .

b Hence, find 8

. .

Given that

, where is positive, find the least value of that satisfies the

equation. 9

a Use the formula for b Hence, find

10 Use the substitution 11 Use the substitution

to show that

.

. to find

to find

12 a Show that b Hence, find the exact value of

.

. . .

13 The diagram shows a part of the circle with its centre at the origin and radius .

y

5

O

5

x

a Write the equation of the curve in the form

.

b Use integration to prove that the size of the shaded area is

.

Section 5: Integrating rational functions You can already use Key point 11.3 to do integrals such as

, where the derivative of the

denominator is a multiple of the numerator. This can be done with a substitution or directly with the reverse chain rule. If you change the numerator or the denominator just a little, this may not work any more. For example, using

in

gives

, so the terms do not cancel.

An alternative method is to split the fraction into two. You can either do this by splitting the numerator or by using partial fractions. You will look at partial fractions first.

Rewind See Chapter 5 for a reminder of partial fractions.

WORKED EXAMPLE 11.18

Use partial fractions to find

in the form

.

Solve for and using substitution or by comparing coefficients (as shown here).

Use

.

Remember that you can also use partial fractions when the denominator has a non-linear factor. WORKED EXAMPLE 11.19

Find the exact value of

. There are three fractions.

Remember the modulus signs when integrating . To integrate

, write it as

.

Before starting the partial fraction process, it is worth checking whether the fraction can be simplified. WORKED EXAMPLE 11.20

Find

. It is generally a good idea to check whether polynomials factorise.

You now have a standard integral. Remember to divide by the coefficient of . Improper fractions can be integrated after splitting them into a polynomial plus a proper fraction, either by polynomial division or by splitting a numerator so that one part cancels with the denominator.

Rewind See division of improper rational functions in Chapter 5.

WORKED EXAMPLE 11.21

Find: a b a

Polynomial division gives quotient

and remainder .

b

In this case, you can perform polynomial division informally by splitting the numerator in a convenient way.

EXERCISE 11G 1

State whether each of the following integrals can be done by using a substitution (or reversing the chain rule). For those that can, carry out the integration. a b c d

e f g 2

By first simplifying, find the following integrals. a b c d

3

Find the following integrals by splitting them into partial fractions. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

4

Find the following integrals by first performing polynomial division (or splitting the numerator). a

i ii

b

i ii

Worksheet See Support sheet 11 for further examples of integrating rational functions and for more practice questions.

5

Find the remaining integrals from question 1.

6

a Write

as a sum of partial fractions.

b Hence, find

, giving your answer in the form

7

Find the exact value of

8

Find the exact value of

9

a Split

.

.

.

into partial fractions.

b Given that

, find the value of .

10 Find the exact value of

. Give your answer in the form

, where and are

rational numbers.

Checklist of learning and understanding When using integration by substitution remember to: Differentiate the substitution and express

in terms of

.

Simplify the resulting expression. Find the limits for . Two special cases of integration by substitution should be remembered: , where

is the integral of

.

Integration by parts can be used for some products:

In integrals of the form

, take

unless

.

You may need to do integration by parts more than once. Some integrals can be simplified by using a trigonometric identity. The particularly useful ones to remember are: For

For

and

, use a rearrangement of the double angle formula:

, use the substitution

.

Some rational fractions can be integrated by splitting them into partial fractions. Look out for improper fractions where you may be able to split the numerator.

Mixed practice 11 1

Find the exact value of

2

Use integration by parts to find

3

Given that

4

Find the exact value of

5

Find the following integrals:

. .

calculate, to three significant figures, the value of .

.

a b 6

Use the substitution

to find the exact value of

.

© OCR, GCE Mathematics, Paper 4724, January 2010 7

Find

8

a Simplify

. .

b Hence find 9

Use the substitution

10 a Write b

to find in partial fractions.

Hence find, in the form

11 Find

.

, the exact value of

.

.

12 Using the substitution

, or otherwise, find

13 i Use integration by parts to find ii Hence find

.

.

. © OCR, GCE Mathematics, Paper 4724, January 2006

14 Find

15 Given that

.

, find the exact value of .

16 a Use a suitable substitution to find

.

b The ellipse shown in the diagram has equation

.

y 2

-3

O

3

x

-2 Use integration to prove that the area enclosed by the ellipse is 17 Let a Find

and

.

.

.

b Using the substitution c Hence, find

, find

.

.

18 A particle moves in a straight line with acceleration given by

When

, is at rest.

Find the distance travels in the first seconds of its motion.

Worksheet See Extension sheet 11 for a selection of more challenging problems.

12 Further applications of calculus In this chapter you will learn: how to use the second derivative to determine the shape of a curve about a new way of describing curves, using a parameter how to calculate rates of change of related quantities how to find the area between two curves, or between a curve and the -axis.

Before you start… Chapter 10

You should be able to find the first and second derivatives of various functions, including by using chain, product and quotient rules.

1 Differentiate: a b

Chapter 8

You should be able to use trigonometric identities

2 Solve the equation

to simplify expressions and solve equations. Student Book 1, Chapter 10

You should be able to use integration to find the area between a curve and the -axis.

for

.

3 Find the area between the axis and the graph of , between and .

Chapter 11

You should be able to integrate various functions, use substitution and integration by parts.

4 Integrate the following: a b c

More uses of differentiation and integration In the previous three chapters you learnt how to differentiate and integrate many different functions. You can now use these techniques to solve problems. Differentiation can be applied to analyse further properties of curves and to look at rate of change calculations that involve more than one variable. Integration is used to calculate more complex areas between curves. You will also learn about a new way of describing the equation of a curve, using a parameter, which has applications in mechanics and engineering; for example, to circular motion and projectile motion.

Worksheet See Extension sheet 12 for some uses of differentiation in economics.

Section 1: Properties of curves You already know that the first derivative tells you whether a function is increasing or decreasing. But an increasing function can have three different shapes: y

y

y B

A

C

x

O

x

O

x

O

Similarly, a decreasing function can decrease in three ways: y

y

y

E F

D O

x

O

x

x

O

The curves and curve upwards; they are called convex curves. The curves and , which curve downwards, are called concave curves. The two convex curves have something in common: their gradients are increasing (curve has negative gradient, which becomes less negative as we move to the right). So the rate of change of gradient is positive. But this rate of change is measured by the second derivative, so

for curves and .

For curves and , the second derivative is negative. Curves and have constant gradients, so their second derivatives equal zero.

Key point 12.1 A curve that curves upwards is called convex and has A curve that curves downwards is called concave and has

. .

WORKED EXAMPLE 12.1

Find the values of corresponding to the convex sections of the curve The curve is convex when

This is a quadratic inequality, so factorise and sketch the graph.

.

y

O

3

x

As you can see on the diagrams above, a point of inflection can be horizontal (so it is a stationary point, ie here as well) or non-horizontal (ie here). At a horizontal (stationary) point of inflection, and but you will see later on that the converse is not necessarily true; if and at a point then that point is not necessarily a horizontal point of inflection.

Key point 12.2 At a point of inflection,

and the curve changes from convex to concave or from concave

to convex.

Tip Remember that

is another notation for

, where

As you can see on the diagrams above, a point of inflection can be horizontal (so it is a stationary point) or non-horizontal. Notice that if you try to draw a tangent at a point of inflection, it will cross the curve at that point. WORKED EXAMPLE 12.2

The curve

has two points of inflection with The points of inflection satisfy

The points of inflection are and

. Find their coordinates. .

Remember to find the -coordinates as well.

.

Tip In Worked examples 12.2 and 12.3 you are told that the curve has two points of inflection, so you just need to find the two points with . You will see later that not every point with is a point of inflection, so you may need to do some additional checks.

WORKED EXAMPLE 12.3

WORKED EXAMPLE 12.3

The curve has two points of inflection. Find their coordinates and determine whether they are stationary or non-stationary. Hence, sketch the curve. Points of inflection have However not all points with are points of inflection.

and So

Stationary points have

.

.

is a stationary point of inflection.

and So is a non-stationary point of inflection. To sketch the curve, you also need the -intercepts and other stationary points.

To sketch the curve: -intercepts: Stationary points:

, so this is a maximum point. y

The curve has a point of inflection at the origin and a maximum point. The second point of inflection is between those two, where the curve changes from convex to concave.

(4.5, 136.6875) (3, 81)

O

You already know that is a point of inflection. Use the second derivative to find out about the other stationary point.

x

6

A stationary point with , which has has a minimum point there.

is not necessarily a point of inflection. For example, consider the graph of and . Although, at , both and are , the graph

Key point 12.3 A point where both and can be any of the three types of stationary point. To determine which one it is, you need to look at the gradient on either side of the point.

Explore Are there any other ways to identify points of inflection? For example, it is possible to use the third derivative,

.

WORKED EXAMPLE 12.4

Find the -coordinates of the stationary points on the graph of

and determine their

nature. Stationary points have

.

This equation can be solved by factorising. The second derivative tells us about the nature of the stationary points.

, so this is a minimum point.

Check the gradient on two sides of the point:

If the second derivative is positive, the stationary point is a minimum. The second derivative equals zero, so you can’t tell the nature of the stationary point. You need to check the gradient on either side.

The gradient changes from positive to negative, so the curve goes up then down.

So point.

is a maximum

This table summarises the possible shapes of a curve, depending on the signs of the first and second derivatives:

WORK IT OUT 12.1 Show that the function nature.

has a stationary point at the origin, and determine its

How many of the solutions are correct? Identify any mistakes. Solution 1

Solution 2

Solution 3

So this is a

Test the derivative on either side:

point of inflection.

Test the second derivative on either side:

The curve goes from increasing to increasing, so this is a point of inflection.

The curve changes from concave to convex, so this is a point of inflection.

EXERCISE 12A 1

Describe each section of the given curve using one or more of the words ‘increasing’, ‘decreasing’, ‘convex’, ‘concave’. a

i

to

ii

to

iii

to A

G

D C

E

B F

b

i

to

ii

to

iii

to E D

A C B

2

Mark the points

and on the given curves.

A Local maximum point B Stationary point of inflection C Non-stationary point of inflection

a

b

3

Mark one (or more if there is more than one) of each of the points

on the curve

that satisfy the given conditions.

A

and

B

and

C

and

D

and

E

and

4

Find the coordinates of the point of inflection on the curve

5

The curve

6

Show that all points of inflection on the curve

7

Find the set of values of corresponding to the convex sections of the curve

8

Find the coordinates of the points of inflection on the curve

.

has two points of inflection. Find their coordinates. lie on the -axis. .

for

. Justify

carefully that these are points of inflection. 9

Show that the graph of

has only one stationary point for

, and that this is a

point of inflection. 10 Show that the curve with equation nature. 11 The curve



has a stationary point at the origin, and determine its

has a stationary point of inflection. Show that

12 The curve

is concave for

. Find the value of .

13 Find the set of values of corresponding to the concave section of the curve 14 The graph shows

.

.

y

O

x

On a copy of the diagram: a Mark points corresponding to a local minimum of

with an .

b Mark points corresponding to a local maximum of

with a .

c Mark points corresponding to a point of inflection of



with a .

Section 2: Parametric equations You are by now very familiar with the idea of using equations to represent graphs and using differentiation to find their gradients. In Chapter 10, Section 4 you saw that some curves can’t be represented by equations of the form . y

P 5 θ O

x

For example, a circle with radius centred at the origin has an implicit equation . But there is another way to describe points on this circle, using the angle from the horizontal. Let be a point on the circle and let be the angle

makes with the -axis.

Then the coordinates of are

You have expressed both and in terms of a third variable, . These are called parametric equations and is called the parameter. You can check that these coordinates satisfy the original equation of the circle:

Tip The last equation, involving just and , is called a Cartesian equation. Each value of the parameter corresponds to a single point on the curve.

Tip You can use a calculator or graphing software to plot parametric equations.

WORKED EXAMPLE 12.5

A curve has parametric equations

and

.

a By filling in the table, mark the points corresponding to the given parameter values. Hence, sketch the curve. Point

b Find the parameter value(s) at the point: i ii a

Use the equations to find and for each .

Point

y

C –9

E

B D

x

O A

b i

The equation for is quadratic, so we can solve it.

When

When

There are two possible values of , so we need to check which one gives the correct value of .

,

,

So the value of is

.

ii

When

When

Although each parameter value gives a single point, there can be several parameter values giving the same point. Notice that also when , but there.

,

,

So the possible values of are

or

.

You can sometimes convert a parametric equation to a Cartesian equation by eliminating the parameter. Two of the most common ways of doing this are by substitution or by using a trigonometric identity. WORKED EXAMPLE 12.6

Find Cartesian equations of the following curves. a b a

for

.

for



.

Express from the equation and substitute it into the equation.

There is a trigonometric identity relating of .

b

and

, so write

The Cartesian equation does not need to be in the form equation containing just and is fine.

in terms

any

You can also write a Cartesian equation in parametric form. The simplest way to do this is to write . For example, the equation can be written in parametric form as , . But you could choose to be another function of . For example, you could take

and then

Explore Parametric equations can be used to describe many interesting curves that don’t have simple Cartesian equations. Find out about the cardioid and the cycloid.

WORKED EXAMPLE 12.7

A straight line, , has the Cartesian equation a Show that the point

lies on the line.

b Find the parametric equations for in the form

where and are constants to

be found, and the domain of is the set of real numbers. a So the point

lies on the line.

b

Check that the line.

and

satisfy the equation of

Substitute the given form of and into the equation of the line. Compare the coefficients of and the constant terms.

So the parametric equations are

In the example above, the parameter represents the distance along the line. For example, consider the points , which has , and , which has . The distance from to is , which is also the difference between the corresponding parameter values. Usually, the parameter does not have a meaning as such but is simply a construct.

Fast forward In Section 3 you will see other examples of parametric equations, when you study related rates of change.

Fast forward If you study Further Mathematics, in Student Book 2, you will see how to use parametric equations of this type to describe lines in three dimensions. In our first example of parametric equations of a circle, the parameter represented the angle between the radius and the -axis. In other contexts the parameter can represent time taken to travel between points; sometimes the parameter has no obvious meaning, but in all cases it determines where you are on the

curve. One important application of parametric equations is modelling motion in two dimensions.

Fast forward Kinematics in two dimensions and projectiles are covered in more detail in Chapters 19 and 20.

WORKED EXAMPLE 12.8

A ball is thrown upwards at an angle from the point coordinates of the ball are given by

. The -axis represents the ground level. The

where represents the time, in seconds. a Find the coordinates of the ball

after projection.

b Find the two times when the ball is

above ground.

c Show that the path of the ball is a parabola, and find its (approximate) Cartesian equation. a When

The coordinates are b

The height is measured by the -coordinate.

The two times are .

and

Eliminate from the first equation and substitute into the second.

c

EXERCISE 12B 1

By creating a table of values, sketch the curves given by the following parametric equations. a

for

i ,

ii b

for

i ii

c

i ii

Tip You can check your answers by using a calculator or graphing software. 2

Find the parameter value at the given point on the curve. a

i ii

point ; point

b

point

i

point

ii c

3

i

point

ii

point

)

Find a possible parametric equation of each curve. a

using

i

using

ii b

i

using

ii c

i ii

d

i ii

4



using using using in the form using

Find a Cartesian equation for each of these parametric curves. a

i ii

b

i ii

c

i ii

d

i ii

e

i ii

5

A ball is projected from point on the horizontal ground and moves in the vertical plane. The ball follows a parabolic path with parametric equations where is the time after the projection. The units on the - and -axes are metres, and time is measured in seconds. a Find the height of the ball above ground

after projection.

b Find the distance from of the point where the ball hits the ground. y (m)

O

6

x (m)

A straight line has parametric equations

.

a Find the -intercept of the line. b Find the distance between the points on the line with represent? c Find the Cartesian equation of the line in the form

and

What does the parameter

7

A model for infectious disease (such as a cold) consists of two variables: the number of infected individuals ( thousand) and the number of susceptible individuals ( thousand). The model proposes that, at time weeks after the start of the infection,

a Show that

is constant. Hence find the values of and when they are equal. Find also after

how many days this happens. b Find the values of and in the long term. 8

An economist models how the supply

and demand

for a good depend on the selling price

.

He uses the following equations:

a Find the value of when b Find an equation relating and . c The equilibrium price is the price of the good for which supply equals demand. Find the equilibrium price predicted by this model.

Differentiating parametric equations When a curve is given parametrically, we can find its gradient

using the chain rule:

Key point 12.4 Gradient of a parametric curve:

WORKED EXAMPLE 12.9

If a curve has parametric equations

, find the gradient of the curve:

a at the point where b at the point a

. The formula involves

and

.

b To find :

The formula for the gradient is in terms of , so we need to find the value of first.

at this point.

We need the value of that satisfies both equations. Use the equation from the previous part.

WORK IT OUT 12.2 Find the equation of the tangent to the curve with parametric equations point Which of the following solutions is correct? Identify the mistakes in the other two. Solution 1

Equation of the tangent:

Solution 2

When But When Tangent:

Solution 3

so

at the

When Then Tangent:

Curves given by parametric equations often have both horizontal and vertical tangents. You can find them by looking at the values of y

and

.

horizontal tangent

vertical tangent x

O

Key point 12.5 At a point where the tangent is parallel to the -axis,

.

At a point where the tangent is parallel to the -axis,

.

Tip Note that these statements are not reversible. So, for example,

does not necessarily

imply that the tangent is parallel to the -axis.

WORKED EXAMPLE 12.10

A curve has parametric equations

for

a Find the equation of the tangent parallel to the -axis. b Show that the curve has no tangents parallel to the -axis. If a tangent is parallel to the -axis, then

a

. This tangent is vertical, so it is of the form The equation of the tangent is Check that

at the point where

If a tangent is parallel to the -axis, then

b

. Hence, tangents.

, so there are no horizontal

The exponential function is always positive.

Parametric equations can be used to describe some interesting curves and to prove properties of their tangents and normals. In many of these proofs you need to work with the general value of the parameter. WORKED EXAMPLE 12.11

A curve has parametric equations

for

is a point on the curve.

y

B O

P x

A

a Find, in terms of , the equation of the tangent to the curve at . The tangent at a point, , meets the -axis at and the -axis at . b Prove that the length of

does not depend on the position of on the curve. The gradient of the tangent is

a

.

Use the chain rule to differentiate each function.

Equation of the tangent:

The equation of the tangent is

At ,

Set

:

and

.

to find axis intercepts.

Divide both sides by

.

You can factorise and use

At ,

:

The distance between is:

and

Now you have the coordinates of and , so you can find the distance between them.

This length does not depend on , so it is the same for every position of .

Did you know? The curve in this question is called the astroid. A ladder sliding down a wall is always tangent to a curve of this shape.

EXERCISE 12C

1

Find the expression for a

in terms of for the following parametric curves.

i ii

b

i ii

c

i ii

2

Find the gradient of each curve at the given point. (You need to find the parameter value first.) a

point

i

point

ii b

point

i

point

ii c

i

point

ii

point

3

A curve has parametric equations point where

4

A curve has parametric equations at the point where

5

The tangent to the curve with parametric equations coordinate axes at the points

Find the equation of the normal to the curve at the

Find the equation of the tangent to the curve

at the point

and . Find the exact area of triangle

crosses the

.

Worksheet See Support sheet 12 for a further example of finding tangents and normals with parametric equations and for more practice questions.

6

a Find the equation of the normal to the curve with equation

at the point

.

b Find the coordinates of the point where the normal crosses the curve again. 7

A parabola has parametric equations , . The normal to the parabola at the point where crosses the parabola again at the point . Find the coordinates of .

8

Prove that the curve with parametric equations



has no tangents parallel

to the -axis. 9

Let be the point on the curve

,

with coordinates

. The tangent to the curve at

meets the -axis at point and the -axis at point . Prove that 10 Point a Let

lies on the parabola with parametric equations be the point where the normal to the parabola at crosses the -axis. Find, in terms of

and , the coordinates of b

.

is the perpendicular from to the -axis. Prove that the distance

11 A predator-prey model describes the population sizes of two species. The number of snakes the number of rats , at time months, are modelled by the equations:

and

a Find all the times in the first two years when the number of snakes equals the number of rats. b Between which times in the first year are the numbers of both species increasing?

c Show that

. What does this quantity represent.

Integrating parametric equations When the equation of a curve is given in a parametric form, you can still use integration to find the area between the curve and the -axis.

Rewind This is very much like the idea used in integration by substitution in Chapter 11, Section 2 to change from to .

Key Point 12.6 The area under the curve

between points

and

is given by

where and are the parameter values at the points on the curve where y t1

a

and

t2

b

x

WORKED EXAMPLE 12.12

The diagram shows a part of the curve with parametric equations have -coordinates and .

Points and

y A

B 0

9

x

a Find the values of the parameter at and . b Find the area bounded by the curve, the -axis and the -axis. a At At

Notice that there are two possible values of that give one of them is consistent with

but only

Use the formula for parametric integration with the parameter values at and .

b

EXERCISE 12D 1

For each curve with the given parametric equations, find the shaded area. a

i y

t = 3

t = 0 x

ii y t = 0 t = –4 x

b

i y

O

2

6

x

ii y

O

c

2

8

x

i y

O

ii

x

y

x

O

2

a Find the coordinates of the points where the curve with parametric equations crosses the -axis. b Find the area enclosed by the curve and the -axis.

3

A curve is defined by the parametric equations y

x

O

a Find the exact values of at the points where

and

.

b Find the area bounded by the curve, the -axis and the line 4

Point

lies on the curve with parametric equations



. .

a Find the parameter value at . b The diagram shows a part of the curve and the normal at . Find the shaded area. y C

O

x

c Find the Cartesian equation of the curve.

Section 3: Related rates of change In the previous sections you learnt that you can use parametric equations to model situations where two variables both depend on a third one. Here we look at some more contexts where such situations occur, and how to calculate various rates of change in these situations. Consider, for example, inflating a balloon. Both the amount of gas in the balloon and the radius of the balloon vary with time, but they are also linked to each other. If we plotted a curve to show how changes with , then and would give parametric equations of that curve. We can consider three different rates of change: measures how quickly the amount (volume) of gas in the balloon is increasing. measures how quickly the radius is increasing. shows how the volume changes with radius.

You know from the previous section that

. But in this situation, you already know

(because

you know the formula linking the volume to the radius), and you can control the speed at which you inflate the balloon (this is

). You can then combine these two pieces of information to find the rate of change of

the radius. WORKED EXAMPLE 12.13

A spherical balloon is being inflated with air at a rate of increasing when the radius is

per minute. At what rate is the radius

? When solving problems in context you should always start by defining variables. Write the given rate of change and the required rate of change.

Relate these rates of change using the chain rule. So you need to find

.

Since the balloon is spherical

,

Use geometric context.

so Substitute into the chain rule.

So radius is increasing at about . Sometimes the chain rule needs to be combined with the product or the quotient rule. WORKED EXAMPLE 12.14

WORKED EXAMPLE 12.14

The length of a rectangle is increasing at a constant rate of per second, and the width is decreasing at a constant rate of per second. Find the rate of change of the area of the rectangle at the instant when the length is Let

be the length,

and

be the area.

Then

and

and the width is

.

Start by defining variables. The length and the width are both functions of time.

be the width

. Use the product rule to differentiate .

When

Use the given values of and .

and

The area is decreasing at the rate of

WORKED EXAMPLE 12.15

As a conical stalactite melts, the rate of decrease of height, , is per hour and the rate of decrease of the radius of the base, , is per hour. At what rate is the volume of the stalactite decreasing when the height is

and the base radius is

?

Write the given rates of change and the required rates of change. Remember that decrease means negative derivative.

Use geometric context to relate the variables. Differentiate both sides with respect to . This requires the product rule and the chain rule.

Put in given values.

The volume is decreasing at

.

EXERCISE 12E

1

In each of the following cases, find an expression for a

i ii

b

i ii

2

a

i Given that ii Given that

and and

, find , find

when when

in terms of .

b

c

i If

and

ii If

and

, find , find

i Given that

,

ii Given that 3

a

b

when

and

. , find the possible values of .

i Given that

find

when



ii Given that

find

when



i Given that

, find , find

i Given that ii Given that

4

.

, find the value of for which

ii Given that c

when

and and

. .

and

when



when



.

and

. .

and that when

, find

and that when

, find

A circular stain is spreading so that the radius is increasing at the constant rate of rate of increase of the area when the radius is

. . . Find the

.

5

The area of a square is increasing at the constant rate of side of the square when the length of the side is .

6

A spherical ball is inflated so that its radius increases at the rate of change of the volume when the radius is

7

A rectangle has length and width Both the length and the width are increasing at a constant rate of per second. Find the rate of increase of the area of the rectangle at the instant when and

8

The surface area of a closed cylinder is given by , where is the height and is the radius of the base. At the time when the surface area is increasing at the rate of , the radius is , the height is and is decreasing at the rate of . Find the rate of change of

. Find the rate of increase of the

per second. Find the rate of

radius at this time. 9

The radius of a cone is increasing at the rate of and its height is decreasing at the rate of . Find the rate of change of the volume of the cone at the instant when the radius is and the height is

.

10 A point is moving in the plane so that its coordinates are both functions of time, When the coordinates of the point are , the -coordinate is increasing at the rate of units per second and the -coordinate is increasing at the rate of units per second. At what rate is the distance of the point from the origin changing at this instant?

Section 4: More complicated areas In Student Book 1, Chapter 14, you learnt how to use integration to find the area between a curve and the -axis. In this section you will extend this technique to areas bounded by other lines and curves.

Area between two curves The area, , in the diagram is bounded by two curves with equations

and

y y = g(x)

A

O

a y = f(x)

b

x

The area can be found by taking the area bounded by by

and the -axis and subtracting the area bounded

and the -axis, that is

You can do the subtraction before integrating so that you have to integrate only one expression instead of two. This gives an alternative formula for the area:

Key point 12.7 The area enclosed by the curves

and

is given by

where and are the -coordinates of the intersection points of the two curves.

WORKED EXAMPLE 12.16

Find the exact area enclosed between the curves

and You first need to find the -coordinates of the intersections.

For intersection:

y

It may help to do a rough sketch to see the relative positions of the two curves.

y = x2 – 6x + 13 y = x + 15 2 A

O

1

11 2

x

Subtract the equation of the lower curve from the equation of the higher curve

before integrating.

Rewind Notice that in Student Book 1, Chapter 15 you found an area such as that in Worked example 12.16 by subtracting the area under the curve from the area of a trapezium. However, this new method is more direct. Subtracting the two equations before integrating is particularly useful when one of the curves is partly below the -axis. As long as is always above then the expression you are integrating, is always positive, so you do not have to worry about the signs of and themselves.

,

WORKED EXAMPLE 12.17

Find the area bounded by the curves y

and

Sketch the graph to see the relative position of two curves.

y = ex –3 –x

y = 5–7e O

Intersections:

x

Find the intersection points; start by multiplying through by .

This is a disguised quadratic that can be factorised.

Write down the integral representing the area. You can see from the graph that the ‘top’ curve is . Simplify before integrating.

Use

.

WORK IT OUT 12.3 Find the area enclosed between the line

and the curve

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1 Sketching the graph shows that part of the area is below the -axis, so you need to split it up into two parts.





y

B O

A

3

x

5

Solution 2 The intersection points are at

and

Solution 3 The intersection points are at

and

Area between a curve and the -axis How can you find the shaded area in the diagram alongside, bounded by the curve and the -axis? y

c A y = f(x)

d

x

O

You know how to find the area between the curve and the -axis, so one possible strategy is to draw vertical lines to create the region labelled . The required area is equal to the area of minus the orange rectangle plus the blue rectangle. y

c

d O

A1

y = f(x) x

Happily, there is a quicker way: you can treat as a function of , effectively reflecting the whole diagram in the line x

x = f(y)

A O

d

y

c

Key point 12.8 The area bounded by the curve , where

the -axis and the lines

and

is given by

is the expression for in terms of ; i.e.

Rewind You may have realised that this is related to inverse functions from Chapter 2.

WORKED EXAMPLE 12.8

The curve shown has equation

. Find the shaded area.

y y = 2 x - 1

O 1

10

x

Express in terms of .

When When

Find the limits on the -axis.

, ,

WORKED EXAMPLE 12.19

Find the exact area enclosed by the graph of y y= 1

.

Sketch the graph and identify the area required. The graph crosses the -axis at .

ln3 1

ln(x+3)

O

, the -axis and the line

x

The area between a curve and the -axis is given by , so we need to express in terms of … and then evaluate the definite integral.

Remember that

.

Notice that this value is negative; this is because the shaded area is to the left of the -axis.

So

EXERCISE 12F 1

Find the shaded areas below. You need to find the intersection points of the two curves first. a

y

i

y = (x - 1)2

y = 2x + 1

x

O y

ii

y = x + 1

y = 4x - x 2 - 1

x

O

b

y

i

y = x2 + 2x + 12

y = - x 2 - 4x + 12 O

x

y

ii

y = x 2 - 2x + 9

y = 4x - x 2 + 5 O

x

c

i

y y = x 2 - x

x

O y = 2x - x2

y

ii

y = x2 - 7x + 7

x

O

y = 3 - x - x2

d

y

i

y =

2 x

y = 5 - 2x

x

O y

ii

y =

5 x

y = 6 - x

x

O

2

Find the shaded areas in the following diagrams. a

y

i

y = x2 6

1

x

O y

ii

y = x3

3 1 O

x

b

y

i

y = x12

2 1 x

O y

ii

y = x

5

1 x

O

c

y

i

y = lnx 1 e

e

O

y

ii

y = O

3

x

1

6

3 x

x

A part of the graph of is shown below. The curve crosses the -axis at shaded region is enclosed by the curve, the -axis and the lines and

. The

y 3

O

1

2

x

y = x 2 - 4x + 3

Find the area of the shaded region. 4

Find the area enclosed between the graphs of

5

The diagram shows the graphs of

and

and

.

. Find the exact value of the shaded area.

y y = ex y = x2

O

6

2

x

Show that the area of the shaded region below is y y = x2

-1 O

7

2

x

The diagram below shows the curve y

.

y = x

2a

a

O

x

2

If the shaded area is

, find the value of .

8

Find the exact value of the area enclosed by the graph of

9

Find the exact value of the area between the graphs of

the line and

10 Find the total area enclosed between the graphs of

and

11 The area enclosed between the curve

is

12 The diagram below shows the graph of

and the line

and the -axis.

. . . Find the value of if

.

.

y y = x2

b

O

1

a

x

Prove that the area of the orange region is equal to twice the area of the blue region for all values of .

Checklist of learning and understanding A convex curve has

; a concave curve has

At a point of inflection,

.

and the curve changes from convex to concave, or vice versa.

Parametric equations are a way of describing a curve where both - and -coordinates are given in terms of a parameter (usually called or ). Each parameter value corresponds to a single point on the curve.

The gradient of a curve given in a parametric form is

.

The area between the -axis and a part of a curve with parametric equations given by

is

, where and are the parameter values at the end points.

The chain rule can be used to connect rates of change of two related variables. If depends on , and depends on , then

. You often need the geometric context of the

question to work out how depends on . The area enclosed between two curves with equations

and

is given by

, where and are -coordinates of the intersection points. The area between a curve, the -axis and the lines where

is the expression for in terms of .

and

is given by

,

Mixed practice 12 1 2

Find the coordinates of the point of inflection on the graph of A curve has parametric equations a Show that the point

lies on the curve, and find the value of at this point.

b Find the equation of the tangent to the curve at . 3

The diagram shows the graph of with a horizontal line drawn through its intercept. Find the exact value of the shaded area. y

x

O

4

a Find

.

b The graph of axis and the lines

is shown below. The shaded region is bounded by the curve, the and .

y y = x2 - 1

-1

O

1

3

x

Find the area of the shaded region. 5

y

O

x

The diagram shows the curve

and the line

i Find the -coordinates of the points of intersection of the curve and the line. ii Use integration to find the area of the shaded region bounded by the line and the curve. © OCR, GCE Mathematics, Paper 4722, June 2006 6

The diagram below shows the graph of

y

y = x

O

4

a

x

The shaded area is 7

. Find the value of .

a Find the coordinates of the stationary points on the graph of

and determine

their nature. b Prove that the graph has one non-horizontal point of inflection. 8

The curve shown in the diagram has parametric equations for A tangent to the curve is drawn at the point . Find the shaded area enclosed between the curve, the tangent and the -axis. y

x

O

9

a Solve the equation

for

b The diagram shows the curves area.

. and

. Find the exact value of the shaded

y

O

10 Consider the graph

x

of for

.

a Show that the -coordinates of the points of inflection satisfy b Use graphs to find the number of points of inflection on the graph. 11 The diagram shows a part of the graph of

for

.

y

b

1 O

x

a

1

The orange area is three times as large as the blue area. Find the value of . 12 Prove that the function



has a stationary point of inflection at the origin.

13 The diagram shows an isosceles right-angled triangle of side

. Point is moving along

the side towards point so that the area of the trapezium constant rate of

is decreasing at the

A E

D 100 h

B

Let

C

100

.

a Write down an expression for the area of the trapezium b Show that

in terms of .

.

Initially point is at vertex . c Given that

, find the value of .

14 The parametric equations of a curve are i Show that

.

.

ii Find the cartesian equation of the curve, giving your answer in a form not involving fractions. © OCR, GCE Mathematics, Paper 4724, June 2010 15 A particle, , moves in a straight line with velocity given by

for

a Find the times when is instantaneously at rest. b Find the total distance travelled in the first figures. 16 Show that the shaded area in the diagram below is .

of motion, giving your answer to

y y2 = x y = 2 - x

x

O

17 The ellipse shown in the diagram has parametric equations

, with

. y B (0, 2)

x A (5, 0)

O

a State the values of at the points marked and . b Find the shaded area and, hence, state the total area enclosed by the ellipse. 18 The graph below shows

.

y

O

x

On a sketch of this graph: a Mark points corresponding to a local minimum of

with an .

b Mark points corresponding to a local maximum of

with a .

c Mark points corresponding to a point of inflection of

with a .

19 a Show that b Find the coordinates of the points of intersection of the graphs .

and

c Find the area enclosed between these two graphs. d Show that the fraction of this area above the axis is independent of and state the value that this fraction takes.

13 Differential equations In this chapter you will learn: how to solve differential equations of the form how to write differential equations in a variety of contexts how to interpret a solution of a differential equation and decide whether it is realistic in the given context.

Before you start… Chapter 11

You should know how to integrate using partial fractions and simplify the answer using log rules.

1

Chapter 11

You should know how to use

2 Integrate:

integration by substitution and by parts.

Integrate and simplify

.

a

b

Chapter 11

You should know how to integrate using trigonometric identities.

3 Find

Chapter 12

You should know how to write equations involving related rates of change.

4 The rate of change of the radius, , of a sphere is . Find an expression for the rate of change of volume.

Student Book 1, Chapter 7

You should know how to rearrange expressions involving exponents and logarithms.

5 Given that terms of .

Student Book 1, Chapter 21

You should know how to draw force diagrams and find net force.

6 An object of weight falls under gravity. The magnitude of the air resistance is . Find the net force on the object.

.

, write in

Modelling using differential equations In Student Book 1, Chapter 19 you looked at problems that involved velocity as the rate of change of displacement, and acceleration as the rate of change of velocity. There are many processes in nature that can be modelled by equations involving the rate of change of some variable, such as population growth and cooling of bodies. In fact Newton’s well-known Second Law actually states that force is equal to the rate of change of momentum. To find the underlying variable from these rates of change involves solving differential equations.

In this chapter you look at forming differential equations with emphasis on real-world applications, and at a method for solving a particular type of differential equation.

Section 1: Introduction to differential equations To solve a differential equation means to go from an equation involving derivatives to one without. You have done this already for the case where the equation can be written in the form As an example, consider the differential equation needed is integration:

.

. To solve this differential equation all that is

. Because of the constant of integration, you find that there is

not just one solution to the differential equation; it could be any one of a family of solutions: y

y = x3 + 4 y = x3 + 2

4 2 O

y = x3 y = x3 - 2 x

-2

All the curves have the same gradient function, so they have the same shape. The solution is called the general solution to the differential equation. You may also be told that the curve passes through the point . This is called a boundary condition or initial condition and it allows you to narrow down the general solution to the particular solution, in this case . Sometimes it is necessary to rearrange the equation before integrating. WORKED EXAMPLE 13.1

Find the particular solution of the differential equation

with the boundary condition

when

.

You need to write the equation in the form Then integrate. Integration by substitution: Since the numerator is related to the derivative of the denominator, a substitution should work.

Make the substitution, and simplify as much as possible.

Remember

; this is the general solution.

Use the boundary condition to find .

When

So This value of gives the particular solution.

EXERCISE 13A

1

Find the general solution of the following differential equations. a i ii b i ii c i ii d i ii

2

Find the particular solution of the following differential equations. a i ii b i ii c i ii d i ii



, , ,

Section 2: Separable differential equations In the previous section you looked at differential equations where

depends just on . But there are

situations where the gradient depends on , or on both variables; for example,

.

You can’t solve this equation by direct integration as the right-hand side contains . However, if you divide through by , the equation becomes

. You can then integrate both sides of the equation with

respect to .

However,

so the equation above becomes

Tip On the final line we used

, where is a new arbitrary constant.

Just as with integration by substitution, though, you get the same results from just splitting up

as if it

were a fraction when you separate the terms and terms to different sides; this leads straight to

.

This method of solving differential equations is called separation of variables. If the equation can be written in the form

, then the procedure is as follows.

Key point 13.1 To solve a differential equation by separation of variables: Get all of the terms on one side and all of the terms on the other side by multiplication or division. Separate

as if it were a fraction.

Integrate both sides.

WORKED EXAMPLE 13.2

Solve the equation

given that

when

. Use multiplication or division to get the term onto the left-hand side. Separate

and integrate.

Since the difference of two constants is just another constant, you need on only one side.

 When

Use the initial condition to find .

:

Since at be written as .

 Therefore:

Using

, the curve will take positive values for , so

can

.

Sometimes the equation needs to be factorised first to get it into the correct form. WORKED EXAMPLE 13.3

Show that the general solution to the differential equation can be written as

.

You can use separation of variables if you can write the equation in the form . Separate variables: divide by

and multiply by

.

Then integrate. But since

:

Since is a constant, re-label it as .

WORK IT OUT 13.1 Find the general solution of the differential equation

.

Which of the following solutions is correct? Identify the mistake in the other two. Solution 1

Solution 2

Solution 3

If

depends just on , you can take a slight shortcut, using the fact that

.

Key point 13.2 If

then

.

Rewind is the derivative of the inverse function, which you met in Chapter 10, Section 5.

WORKED EXAMPLE 13.4

Newton’s law of cooling states that the rate of change of temperature of a cooling body is proportional to the difference between the body’s temperature and the surrounding temperature. A cup of coffee cools in the room where the air temperature is , satisfies the differential equation

. The temperature of the coffee,

where is the time, measured in minutes. The initial temperature of the coffee is

. Find the temperature of the coffee after

.

Integrate with respect to .

So:

When

,

When

:

Find using the initial conditions.

:

Substitute in the given value of before rearranging. Because of the context of the question, you are interested only in the solution in which decreases from . Hence, is positive, so you can remove the modulus signs.

EXERCISE 13B 1

Find the particular solutions of the following differential equations, giving your answer in the form , simplified as far as possible. a

i ii

b

i ii

2

Find the particular solutions of the following differential equations. You do not need to give the equation for explicitly. a

i ii

b

c

i

,

ii

,

ii 3

,

i ,

Find the general solution of the following differential equations, giving your answer in the form

simplified as far as possible. a

i ii

b

i ii

c

i ii

Worksheet See Support sheet 13 for a further example of solving a separable differential equation and for more practice questions.

4

Solve the differential equation form

5

, given that when

Give your answer in the

.

The function when

satisfies the differential equation

. When

. Find the value of

.

6

Given that

, where is a positive constant, show that

7

Find the general solution of the differential equation

. , giving your answer in the form

. 8

Given that constant to be found.



and that

when

, show that

, where is a

Section 3: Modelling with differential equations Now that you can solve various differential equations, we look at how such equations arise in a variety of contexts. There are many situations where you know (or it is reasonable to assume) what the rate of change of a quantity depends on. You can then write down and solve a differential equation to find out how the actual quantity behaves. In many applications, the rate of change is with respect to time. For example, in mechanics, the acceleration is the rate of change of velocity,

. But you also know that

, so the rate of change

of velocity depends on the force acting on the object. WORKED EXAMPLE 13.5

A skydiver of mass jumps out of an aeroplane with zero initial velocity. The air resistance is proportional to velocity and may be modelled as . Using , write and solve a differential equation to find an expression for the velocity of the skydiver in terms of time. Forces on the skydiver:

Always start by drawing a force diagram.

R = 0.8v

The weight acts downwards and the air resistance upwards.

60g

Write Newton’s second law equation, taking the positive direction to be downwards (because that’s the direction of motion).

Net force:

Since the left-hand side is in terms of , rewrite the equation in terms of Also use

. .

Integrate with respect to .

When

So:

,

:

Now use the initial condition.

You need to express in terms of .

If

then

.

Since you are interested in the solution for which initially, is positive so you can remove the modulus sign. Notice that in the penultimate line the modulus sign was removed. Thisis justified only if ; i.e. . Since initially , this is certainly true for some initial part of the motion. But you should ask whether eventually becomes larger than ; if it does then the equation needs to be changed. Looking at

the final solution,

is always positive, so is in fact always less than

. Therefore, according to our model, it is impossible for to breach the

and your solution is valid for all barrier.

Focus on… There are a number of examples of modelling real-life situations with differential equations in Focus on … Modelling 2.

Did you know? In fact,

decreases and tends to zero, so increases towards

without ever reaching it.

v

735

t

O

This is called terminal velocity. Of course, the object might hit the ground before getting close to terminal velocity. Often you have a model where some constants are unknown. You can find them using experimental or observational information. WORKED EXAMPLE 13.6

In a simple model of a population of bacteria, the growth rate is assumed to be proportional to the number of bacteria. a Let be the number of bacteria after increases to bacteria after

after .

. Initially there are

bacteria and this number

. Write and solve a differential equation to find the number of

b Comment on one limitation of this model. a

The rate of growth is

.

‘Proportional to ’ means you can write it as

for some constant

You have information about and but not about

, so you need

to solve the differential equation before you can put in the numbers. Since is positive, you don’t need modulus signs in . When

When

,

,

To find the constants and , use the two given conditions: first use initially (i.e. when ).

:

:

A second equation is needed to find , so next use the condition that when .

Use

.

Now put and back in and rearrange the equation to get in terms of . Check that you can follow all the steps!

So:

b This model predicts unlimited population growth, which is not realistic.

Exponential growth models often work initially, but need to be adapted for longer time periods.

Sometimes a problem has several variables and you need to use the geometric context and related rates of change to produce a single differential equation. WORKED EXAMPLE 13.7

A cylindrical tank with cross-sectional area

and height

is initially filled with water. The water

leaks out of the tank through a small hole at the bottom at the rate of height of water in the tank after . a Find an equation for

, where

is the

in terms of .

b Hence, find how long it takes for the tank to empty. The given rate of change is for the volume, so you need to relate volume to the height.

a

The rate of change is negative since the volume is decreasing.

You have

so

b

Solve the equation by integrating so

When

:

Initially the tank is full.

.

The tank is empty when

When

.

The tank is empty after .

EXERCISE 13C 1

Write differential equations to describe the following situations. You do not need to solve the equations. a

i A population increases at the rate equal to ii The mass of a substance

b

the size of the population

.

decreases at the rate equal to three times the current mass.

i The rate of change of velocity is directly proportional to the velocity and inversely proportional to the square root of time. ii The population size increases at a rate proportional to the square root of the population size and to the cube root of time.

c

i The area of a circular stain increases at a rate proportional to the square root of the radius. Find an equation for the rate of change of radius with respect to time. ii The volume of a sphere decreases at a constant rate of decrease of the radius.

2

. Find an equation for the rate of

In a simple model for a population of bacteria, the rate of growth is proportional to the size of the population. When the population size is , the rate of growth is . a Show that

.

Initially there were

.

b Solve the differential equation and predict the number of bacteria after 3

The mass of a radioactive substance is and it decays at the rate of a Show that

.

decays at a rate proportional to the mass. Initially the mass .

.

b Find the amount of time it takes for the mass to halve. 4

A particle of mass is moving with speed in a straight line on a horizontal table. A resistance force is applied to the particle in the direction of motion. The magnitude of the force is proportional to the square of the speed, so that . a Show that

.

b Find an expression for the velocity in terms of time and, hence, find how long it takes for the speed to decrease below . 5

The population of fish in a lake, thousand, can be modelled by the differential equation , where is the time, in years, since the fish were first introduced into the lake. Initially there were

fish.

a Show that the population initially increases and find when it starts to decrease. b Find the expression for in terms of .

c Hence, find the maximum population of fish in the lake. d What does this model predict about the size of the population in the long term? 6

In Economics, there is a model that shows how the demand for a commodity

depends on its price

. It states that the rate of change of with respect to is proportional to but inversely proportional to . a Explain how this model leads to the differential equation

,

where is a negative constant (called elasticity). b Find the general solution of this differential equation. c Sketch the graph of against , and describe how demand depends on price, in the following cases. i ii 7

Newton’s law of cooling states that the rate of change of temperature of a body is proportional to the difference in temperature between the body and its surroundings. A bottle of milk has a temperature of when it is initially taken out of the fridge. It is placed on the tablein the kitchen, where the room temperature is Initially, the milk is warming up at a rate of per minute. a Show that

, where

is the temperature of the milk and is the time, in minutes,

since the milk was taken out of the fridge. b Solve the differential equation and, hence, find how long it takes, to the nearest minute, for the temperature of the milk to reach the kitchen temperature, correct to the nearest degree. 8

A particle of mass is pulled through liquid using a light inextensible string. The tension in the string is . The resistance force is proportional to the velocity and equals . The particle starts from rest. a Find an expression for the velocity after time, . b Describe the velocity of the particle for large values of .

9

An inverted cone has base radius

and height

.

4cm

10cm h

The cone is filled with water at a constant rate of a Show that the height of water

.

satisfies the differential equation .

b Given that the cone is initially empty, find how long it takes to fill it.

10 A particle of mass is moving at a speed of when it enters a viscous liquid at a point, . Inside the liquid the resistance force is proportional to the velocity, and initially equal to . Apart from the resistance and the weight, no other forces are acting on the particle. a Show that the velocity of the particle satisfies the differential equation

.

b Find an expression for the velocity of the particle seconds after entering the liquid. c Find the displacement of the particle from point , and describe what happens to the displacement for large values of . 11 A population model for two competing species postulates that their numbers, and (in appropriate units), vary with time according to the equations: . a Find the range of values of and for which both variables are increasing. b Find an expression for

.

c Solve this differential equation to show that

for some constant .

Checklist of learning and understanding A differential equation is an equation for the derivative of a function. To solve a differential equation means to find an expression for the function itself. Some differential equations can be solved by separation of variables: Write the equation in the form

and integrate both sides.

Initial conditions can be used to find the constant of integration. Differential equations often describe the rate of change of a quantity; this is the derivative with respect totime. The rate of change is often proportional to one of the variables. You may need to use given information to find the constant of proportionality. Sometimes a problem involves more than one variable and you need to use related rates of change to write a differential equation.

Mixed practice 13 1

Solve the differential equation in the form

2

, given that

when

. Give your answer

.

Find the particular solution of the differential equation

such that

when

. 3

In a chemical reaction, the amount of the reactant

follows the differential equation

. Initially there was to be found. 4

of the reactant. Show that

, where and are constants

i Find the general solution of the differential equation

ii For the particular solution in which

when

, find the value of when

.

© OCR, GCE Mathematics, Paper 4724, January 2007 5

A population of fish initially contains fish and increases at the rate of be the number of fish after months.

fish per month. Let

In a simple model of population growth, the rate of increase is directly proportional to the population size. a Show that

.

b Solve the differential equation and find how long it takes for the number of fish to reach

c Comment on the long-term suitability of this model. An improved model takes into account seasonal variation:

d Given that initially there are months. 6

fish, find an expression for the size of the population after

A model for a relationship between the price of a commodity commodity

states that

and the demand for the

, where is the elasticity.

a Find the general solution to this differential equation. b Describe the relationship between the price and demand when c For most commodities, . 7

A cylindrical tank has radius

.

. Suggest, with a reason, what sort of commodity might have

and height

. Initially the tank is empty.

. The height is being filled with water so that

a Show that

and, hence, find an expression for the height of water in the

tank at time . b By writing your expression in the form

, or otherwise, determine whether the

tank will ever completely fill with water. 8

A ball of mass falls vertically downwards. When the velocity of the ball is resistance has magnitude . a Show that

the air

.

The ball falls from rest. b Show that

, stating the values of constants and .

c Hence, describe the motion of the ball. 9

A liquid is being heated in an oven maintained at a constant temperature of . It may be assumed that the rate of increase of the temperature of the liquid at any particular time, minutes, is proportional to

, where

is the temperature of the liquid at that time.

i Write down a differential equation connecting and . When the liquid was placed in the oven, its temperature was temperature had risen to .

and

later its

ii Find the temperature of the liquid, correct to the nearest degree, after another

.

© OCR, GCE Mathematics, Paper 4724, January 2009 10 Consider the following model of population growth: , where thousand is the population size at time months. a Suggest what the term b Given that initially

could represent. , solve the differential equation.

c Show that the solution may be written as

. Hence, describe what happens to

the population in the long term. 11 Variables and satisfy the differential equation Given that

when

, show that

.

12 A particle moves in a straight line. Its acceleration depends on the displacement as follows:

a Find an expression for

in terms of and .

Initially the particle is at the origin and its speed is

. The velocity of the particle remains

positive for b Show that

.

c Find expressions for the displacement and velocity in terms of time.

Worksheet See Extension sheet 13 for questions on some other types of differential equation.

14 Numerical solution of equations In this chapter you will learn: that some equations cannot be solved by algebraic rearrangement how to find an interval that contains a root of an equation, and how to check that a given solution is correct to a specified degree of accuracy (the sign change method) that you can approximate a part of the curve by a tangent, and use this to find an improved estimate for a solution (Newton–Raphson method) how to create a sequence that converges to a root of an equation (fixed-point iteration) how to identify situations in which the methods above fail to find a solution.

Before you start… Student Book 1, Chapters 4, 7 and 10

You should know how to rearrange equations involving polynomials, fractions, exponentials, logarithms and trigonometric functions.

1 Rearrange each equation into the required form. a b

into into

c

Chapter 10

You should know how to differentiate a variety of functions.

2 Differentiate the following. a b c

Chapter 4

You should know how to use the term-toterm rule to generate a sequence.

3 Find the first four terms of the sequence defined by .

What is a ‘numerical solution’? Throughout your studies of mathematics you have learnt to solve various types of equations: linear, quadratic, exponential, trigonometric … Some equations can be solved by algebraic rearrangement; for example:

Sometimes you can write solutions in an exact form, such as

or

Alternatively, you can use calculator buttons to find approximate solutions; for example:

Tip You can write the solution of an exact form as

in an exact form as

.

But there are some equations that can’t be solved by any combination of algebraic rearrangement and using calculator buttons. For example, how would you solve the equation ? However you try rearranging it, you can’t get it into the form . If you have a graphical calculator or graphing software, you can use them to find approximate solutions by drawing the graphs of and finding their intersection.

and

y 4

x ≈ 1.5

3 2 1 -4 -3 -2 -1O -1

1

2

3

4

x

-2 x ≈ - 1.5

-3 -4

Your calculator may have a ‘solver’ function that allows it to find approximate solutions without drawing the graph. (In the example of the equation it should give you solutions , and ; the solution is exact, but the other two are accurate to ) But how does it do that? Does it try lots of different values for until it finds one that works? Or is there some clever way to decide which values of to try? In this chapter you will learn several different methods for solving equations numerically. (This means that a solution is not found by algebraic manipulation, but by evaluating certain expressions to find increasingly accurate approximations.) You should be aware that these methods can give only approximate solutions, but you can choose the level of accuracy. You will also see that in some situations these methods fail to find an answer and will learn how to recognise such situations.

Section 1: Locating roots of a function The simplest numerical methods involve finding an interval in which the solution lies, and then improving accuracy by making this interval smaller. For example, try finding approximate solutions of the equation method works if you rearrange the equation into the form

. It is easier to see how the , so look at this equation:

Tip When equations involve trigonometric functions you should work in radians unless told otherwise. A sensible way to start is to try some values of , and then see if any of them give an answer close to zero. Here is a table of values:

From the table you can see that is a solution. But you can also see that the value of the expression changes sign between and : it is positive when but negative when . The graph is a continuous curve, so this means that it must equal zero somewhere between these two values.

Tip ‘Continuous’ is a technical term but, put simply, a function is continuous over an interval if you can draw its graph within that interval without taking your pen from the paper. You can see this if you draw the graph of negative, then the graph must cross the

. If the

changes from positive to

.

y x = 1

0.5

O

0.5

1.0

1.5

2.0

x

- 0.5 - 1.0 x = 2

You can try to locate this solution more accurately by looking for a smaller interval. For example:

Explore

Can you use a spreadsheet to find the solution, correct to decimal places? This tells you that the solution is between and . You can continue finding smaller and smaller intervals to find the solution to any degree of accuracy you want. This method uses the following result:

Key point 14.1 If

is a sufficiently well behaved and and are numbers such that

changes and , then

the equation has at least one root (solution) between and . However a lack of change of sign does not necessarily imply that there are no roots.

Tip This method finds only one solution. The previous equation

also has a negative

solution, between and . In general, with numerical methods, you can never be sure that you have found all the solutions.

Can the change of sign method fail? In Key point 14.1 we refer to ‘sufficiently well behaved’ functions. There are several situations in which the change of sign method cannot be used. If the graph of the function has a vertical asymptote, or another break in its graph, there may be a change of sign even if the graph does not cross the .

y

y

a O

b

a

x

O

x

b

It is also possible that the change of sign method does not detect a root when there is one (or more). This can happen if the graph is tangent to the , or if it crosses it several times in a small interval. y

y

aO

b

x

a

O

b

x

For the second graph above, the change of sign method may work if you picked a different interval. The problem is that, unless we know the approximate location of the roots already, there is no way of knowing how small an interval we need.

You should also note that a change of sign between and implies that there is at least one root between those two values; but it does not tell us whether there is more than one.

y

a O

b

x

WORKED EXAMPLE 14.1

Let

.

a Show that the equation b i ii

Evaluate

and

has a solution between

and

.

.

Explain why, in this case, the change of sign does not imply that between and .

has another solution

Show clearly which values of you used.

a

There is a change of sign, so there is a root between and .

Show clearly which values of you used.

b i

ii

Note that is sufficiently well-behaved in the domain .

has a vertical asymptote at , so the change of sign does not imply that the graph crosses the .

Think about the graph: if it has an asymptote then it does not necessarily cross the -axis. not defined for

so the graph will

have an asymptote there.

Checking the accuracy of a solution In practice, rather than knowing that a solution lies in a particular interval, you want to express it to a specified degree of accuracy, such as decimal place. Consider, as an example, the equation By trying some values of you can find that this equation has a solution around check that this is in fact correct to decimal place?

. But how can you

The solution, when rounded to decimal place, equals . But the numbers that round to and , so you should look for a change of sign between those two values.

are between

WORKED EXAMPLE 14.2

Show that the equation Let

.

has a root

.

, correct to decimal place. First write the equation in the form

… then look for a change of sign.

There is a change of sign, so there is a root between and . This root equals

, to d.p.

You are looking at the degree of accuracy of the value here, not the value. For the equation from Worked example , , which equals to decimal places. However, the root is not equal to to . Look at this table of values:

Focus on … Focus on … Problem Solving 2 compares a numerical and analytical solution to the same problem. The change of sign actually occurs between

and

so the root is

, correct to

.

EXERCISE 14A 1

Classify these equations as ‘can find exact solutions’ or ‘can’t rearrange algebraically’. a b c d e f g h

Explore There is, in fact, a formula for solving cubic equations, but you will not learn about it in this course. 2

Each of the following equations has a root between which the root lies. a b c d

and . In each case, find two integers between

3

For each of the following equations, show that there is a root in the given interval. , between and

a i

, between

ii

, between and

b i

, between

ii 4

and

For each equation, show that the given root is correct to the stated degree of accuracy. a i

,

ii

,

b i

,

ii

,

c i

,

ii

,

d i

,

ii 5

and

,

a Show that the equation

has a solution between and .

b Show that this solution equals 6

The equation

, correct to decimal place.

has two solutions.

a Show that one of the solutions equals

, correct to significant figures.

b The other solution lies between positive integers and 7

A function is defined by

.

a Show that the equation b i ii

8

Evaluate

and

.

a i

Sketch the graph of

b i ii

has no solutions. .

Alicia says that the change of sign implies that the equation . Explain why she is wrong.

Let

ii

. Find the value of .

for

.

State the number of solutions of the equation State the values of

and

has a root between and

between and .

.

George says, ‘There is no change of sign between and so the equation

has no

roots in this interval.’ Use your graph to explain why George’s reasoning is incorrect. iii Use the change of sign method (without referring to the graph) to show that the equation has two roots between and .

Section 2: The Newton–Raphson method The method of sign change allows you to show that a root of an equation lies in a certain interval, but how do you know which interval to try? In the first example, with , a table of integer values of helped you to find that there is a solution between and . You could then look at to locate the change of sign more accurately:

This shows that the root is between

and

. You can continue like this until you make the interval as

small as you like. If you wanted to locate the root to three or four decimal places, this could take quite a long time. In some applications, solutions of equations are required to even higher accuracy, and then this method becomes unfeasible, even with a fast computer. The numbers in the table above suggest that the root may be much closer to than to , so maybe the next search should not start from but from or , but how do you know which one? It would be good to have a method to tell you which number to try next. There are in fact many such methods. In this section you will meet the Newton–Raphson method, which uses the tangent to the graph of to suggest where to look for the root. The diagram below shows the section of the graph of for (in red). It crosses the between and . The diagram also shows the tangent to the graph at (in blue). Near this point, the tangent follows the graph closely, so it will cross the near the root of the equation.

1.3

1.4

1.5

1.6

1.7

x

Use a graph plotter to draw the graph and the tangent and find where the tangent crosses the will find that the value is

. You

You can now repeat the same procedure by drawing a tangent at this new value of ; this new tangent crosses the at You can check that this gives the correct root to three decimal places:

Repeating one more time gives the root correct to seven decimal places. This would have taken a very long time with the sign change method! So far you have used a graph plotter to draw tangents and find their . But is there a formula to calculate the of the tangent without having to rely on the graph? The answer is yes: you can find the equation of a tangent at any given point, and then use this equation to find where the tangent crosses the .

Key point 14.2 Newton–Raphson method:

xn + 1

xn

Given an approximate root

x

of the equation

, a better approximation is

Rewind You met equations of tangents in Student Book 1, Chapter 14.

PROOF 8

The tangent to the graph at where

has gradient

and it passes through the point

,

.

Therefore, its equation is

To find where this tangent crosses the

, you need to set

:

Repeating the Newton–Raphson procedure, starting with the new value of each time, gets you closer and closer to the root. WORKED EXAMPLE 14.3

a Show that the equation

has a root between and .

b Starting from , use three repetitions of the Newton–Raphson method to find a better approximation for the root. c Show that this approximation gives the root correct to three decimal places. a Let

.

Look for a change of sign between and .

There is a change of sign, so there is a root between and . b

To use the Newton–Raphson formula you need Call the next approximation Use

to find

.

.

.

Use c

to find

.

To show that the root equals to show that it is between and

., you need to .

There is a change of sign, so there is a root between and . Hence, the root equals

(

).

Repeating the Newton–Raphson calculation several times, as in Worked example 14.3, involves using the same formula with different numbers, whereby the number you need to put into the equation is the answer from the previous calculation. Most calculators have an ANS button that can be used to carry out such repetitive calculations. Here is how you could use it for Worked example 14.3, with the keys of a typical calculator.

Worksheet See Support sheet 14 for a further example of using the Newton–Raphson method and for more practice questions.

Start by entering the starting value of

(in this case 1.5) and press the

button.

Then type in

and press the repeatedly gives

button again. The answer is the value of

. Pressing the

button

and so on. This way you can quickly generate as many approximations to the root

as you like.

Tip You should familiarise yourself with the equivalent keys on your own calculator, as they may differ to this. You should also ensure that you show evidence of numerical reasoning, for example through writing down iterations.

EXERCISE 14B

EXERCISE 14B 1

Each equation below is to be approximately solved using Newton–Raphson method with the given starting value . In each case, use technology to sketch the graph, draw the tangent at and find the next approximation to the root. a i

,

ii

,

b i ii 2

, ,

For each equation, use the Newton–Raphson method with the given starting value to find the root correct to three significant figures. Use the change of sign method to show that your root is correct to three significant figures. a i ii

, , ,

b i ii

,

3

The equation has a root near . Using this value as the first approximation, use the Newton–Raphson method to find the next approximation to the root.

4

a Show that the equation

can be written as

b Given that the equation has a root near approximations. 5

The equation

has a root near

.

, use the Newton–Raphson method to find the next two

. Use the Newton–Raphson method to find the next two

approximations to the root. 6

a Show that the equation b Using

has a root between and .

as the starting value, use the Newton–Raphson method to find the next approximation

to the root. Give your answer correct to two decimal places. c Show that the approximation from part b is correct to one decimal place, but not to two decimal places. 7

a Show that the equation

has a root between and .

b Use the Newton–Raphson method to find this root correct to three decimal places, and show that the root you found is correct to three decimal places. 8

The equation has a root between and . Use the Newton–Raphson method to find this root correct to three significant figures, and show that the solution you found is correct to three significant figures.

Section 3: Limitations of the Newton–Raphson method The Newton–Raphson method can find a root to a high degree of accuracy very quickly, so it seems to be a very good method. Unfortunately, there are some situations when it doesn’t work. For example, try to find a root of the equation

.

The sign change check confirms that this equation has a root between and . The Newton–Raphson iteration formula is:

If you start from

you get

But you can’t divide by zero, so it’s impossible to find

could try changing the starting point slightly; for example, the root but then comes back again.

. You

. Try it! The sequence moves away from

Key point 14.3 The Newton–Raphson method doesn’t work if the starting value is a stationary point of . If the root is close to a stationary point, the sequence may initially (or permanently) move away from the root.

WORKED EXAMPLE 14.4

a Show that the equation b Explain why

has a root between and .

is not a suitable starting point for a Newton–Raphson iteration to find this root.

c Use the starting value

to find the root correct to three decimal places.

a Let Then:

There is a change of sign, so a root between and .

Use the sign change method.

has

b

The Newton–Raphson method fails when find to check.

, so

So there would be division by zero if the Newton–Raphson formula were used. c

Use the Newton–Raphson formula with

The sequence converges to .

Continue the sequence until you can see what the limit is. You are not being asked to write down all the iterations.

Explore Sketch the graph to see why the sequence initially moves farther away from the root before getting closer again. Another situation where the Newton–Raphson method can fail is if falls outside of the domain of the function. This can happen, for example, if the graph of the function has a vertical asymptote. WORKED EXAMPLE 14.5

WORKED EXAMPLE 14.5

a The equation from

has a root between

and . Explain why a Newton–Raphson iteration

fails to find this root.

b Starting from , carry out three iterations of the Newton–Raphson method, showing the values correct to three decimal places. In order to use the Newton–Raphson formula you first need to find using the quotient rule.

a

Use the formula with

But for

is not defined ned

to find

.

To find you would need to use the formula with has a vertical asymptote at and is not defined for is impossible.

. But so this

, so it is not

possible to find

.

b

You can use graphing software to see that the tangent at vertical asymptote.

crosses the

to the left of the

y 0.8 0.6 0.4 0.2 - 1.0 - 0.8 - 0.6 - 0.4 - 0.2O - 0.2

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4

x

- 0.4 - 0.6

Explore Try plotting tangents at different values of that needs to be less than about .)

to see when the iteration works. (You should find

EXERCISE 14C 1

For each equation below, carry out one iteration of the Newton–Raphson method starting with the given value of . In each case, sketch the graph (using technology) to explain why is not a better approximation to the root than . a i ii

, ,

b i

,

ii 2

,

Let

.

a Find the

of the stationary points of

b Show that the equation

has a root between and .

c The Newton–Raphson formula with improved approximation for the root. 3

.

is used to find

. Explain why

The diagram shows the curve with equation

may not be an

.

y

x

O

Find the of the stationary point of and, hence, explain, with an aid of a diagram, why a Newton–Raphson iteration with will not converge to the root of . 4

The diagram shows the curve with equation

.

y

x

O

a Find the coordinates of the turning points of The curve crosses the

at

,

.

and

, with

.

The Newton–Raphson method is to be used to find the roots of

, with

.

b To which root, if any, do the successive approximations converge when: i ? ii

?

c Write down the range of values of for which the Newton–Raphson iteration converges to . 5

Let

.

a Write down the values of between and The equation

for which

has a root between and .

is not defined.

, use the Newton–Raphson method to find . b Taking c Explain why the Newton–Raphson iteration cannot be continued to find a better approximation to the root. 6

The function

has three zeros (

) between

and .

a For each of the zeros, find two integers between which it lies. b Find the exact coordinates of the stationary points of

and sketch its graph.

c Find the coordinates of the point where the tangent to the graph at The Newton–Raphson method is to be used to find the root d Explain why the iteration with e Use an iteration with 7

Let

crosses the

.

.

does not converge to .

to find , correct to three significant figures. .

a Find the coordinates of the stationary point on the graph point. b Show that

and show that this is a maximum

has no points of inflection. Hence, sketch the graph of

The equation

.

has two roots.

c Show that one of the roots is between and and find two integers between which the other root lies. The Newton–Raphson iteration is to be used to find the roots. d Explain why the starting value e State the range of values of root. 8

cannot be used to find the smaller root.

for which the Newton–Raphson iteration converges to the larger

Each of the following equations has a root near the given starting value a Explain why starting with

. In each case:

does not lead to a better approximation to the root.

b Use technology to investigate for which starting values the Newton–Raphson method works. i ii iii

, , ,

iv 9

,

to find the root between and . to find the root between and . to find the root between and .

A function is defined by

.

a Show that if is any root of the equation The graph of

, then

.

is shown below. y

O

x

The Newton–Raphson formula with a positive value of b State the range of positive values of

for which

is used to find is closer than

.

to the positive root of

.

c Use a diagram to show that there is a positive value of such that approximations converge to the positive root of . 10 a Sketch the graphs of

and

for

, but subsequent

. Hence, show that the equation

has a solution, , in that interval. Consider the Newton–Raphson iteration with the starting value b Show that the iteration converges to when c There is a value such that the iteration with for

.

but not when

.

converges to . Sketch the graph of

. Show the value on your sketch and draw the tangent to the curve at

. d When , the Newton–Raphson iteration may not converge to . Describe two different cases that can arise.

Section 4: Fixed-point iteration You saw in the previous section that the Newton–Raphson method doesn’t always work. Luckily, there are alternative methods you can use in such situations. In this section you will learn about fixed-point iteration, which also involves creating a sequence that gets closer and closer to the root, but in a different way from Newton–Raphson. To use fixed-point iteration the equation needs to be rearranged into the form . Suppose you have a starting guess of . If you have found a solution of the equation. Otherwise we are looking for an improved guess. Since you want to equal works by looking at the graph.

, it makes sense to try

. You can see why this

The solution of the equation is the intersection of the graphs and . Starting from the point on the , you can find on the using the graph . To see whether is closer to the solution than , you need to find on the . You can do this by reflecting it in the line . y

y y = x

y = x

y = g(x)

y = g(x)

g(x1)

O

g(x1)

x1

x

O

x1

In the example shown in the previous diagram,

x

x2

is closer to the solution than

. If you now repeat the

same process you can hope to get closer and closer to the solution. In other words, you create the following sequence:

… You can write the general rule for the sequence as . This sequence may converge to a limit, meaning that the terms of the sequence get closer and closer to a certain number, . If this is the case, both and will get closer to , so the sequence equation becomes and so is a solution of the equation .

Rewind Sequence rules and convergence were covered in Chapter 4.

Did you know? The name ‘fixed-point iteration’ refers to the fact that the solution is a fixed point of the function . It is the value where the output equals the input. The procedure can be summarised as follows.

Key point 14.4 Fixed point iteration:

To solve an equation in the form Using a starting guess

:

, generate a sequence

.

If this sequence converges to a limit, then this limit is a solution of the equation.

y y = x y = g(x)

O

x1

x

x2 x3

WORKED EXAMPLE 14.6

The equation

has a solution between and .

a Use fixed-point iteration, with the starting value , to find the first five approximations to the solution. Give the values correct to five significant figures. b Show that this solution is correct to two decimal places. a Sequence:

b The solution is

Start by entering ‘ ’ into the calculator and use ln to generate subsequent values.

.

To check that the solution is correct to to rearrange the equation into the form for a sign change between and .

, we need and look

There is a sign change, so the solution is between and . It equals . In Worked example 14.6 the sequence increased towards the limit. But it is also possible for a sequence to oscillate above and below the limiting value. WORKED EXAMPLE 14.7

The equation has a root between and . The sequence used to find an approximation to this root.

, with

, is

a Draw a graph to illustrate the first three approximations. b Find the root correct to three decimal places. Draw the curve

and the line

Start with on the -axis. Find the corresponding point on the curve and reflect it in the line to find . Repeat to find

.

y

a

x1

O

x3

x

x2

b

Start with

. Use

to generate the sequence.

… The sequence converges to

Continue until the third decimal place stops changing.

Key point 14.5 The graphs showing fixed-point iteration are often called: staircase diagrams (if successive terms of an iterative sequence are all increasing or all decreasing); or cobweb diagrams (if successive terms of an iterative sequence oscillate either side of a value to which the sequence is converging).

EXERCISE 14D 1

Use fixed-point iteration with the given starting value to find the first five approximations to the roots of the following equations. Give your answers to three decimal places. a i

,

ii

,

b i

,

ii

,

c i

,

ii 2

Use fixed-point iteration with the given starting value to find an approximate solution, correct to two decimal places. a i ii b i ii

3

,

, , , ,

Each of the equations below is to be solved using fixed-point iteration. Draw the graphs and use technology to investigate the following. a b

c d i

Does the limit depend on the starting point?

ii

Does starting on different sides of the root give a different limit?

iii If there is more than one root, which one does the sequence converge to? Does it matter where you start? 4

The equation

has a root between and . Use fixed-point iteration to find this root,

correct to two decimal places. 5

Use the iterative formula

, with

, to find

. Give your answer correct to four

decimal places. This value of is an approximation to the root of the equation and suggest the value of the root, correct to two decimal places. 6

The equation

. Write down an expression for

has a root between and . Use fixed-point iteration to find this root

correct to three decimal places. Show the first three approximations, correct to five decimal places. 7

a Use the iterative formula

, with

, to find

and

, correct to four decimal

places. b The sequence and . 8

converges to the root of the equation

The diagram shows the graphs of and . The iteration of the equation . The starting value is shown on the diagram.

. Find the values of ,

converges to the root

y

B A

O

x

x1

The values of and are the corresponds to the point ? 9

a Show that the equation

has a root between and .

b Use fixed-point iteration with c Show that your value of

of the points marked and . Which of the two values

to find next three approximations to the root.

gives the root correct to two decimal places.

10 The diagram shows the graphs of

and

.

y

x

O

a Use the diagram to show that the iteration

converges to a root of the equation

. b Use the iteration from part a to find this root, correct to two decimal places. c By rewriting the equation in the form the percentage error in your approximation. 11 The diagram shows the graphs of

and

, find the exact value of the root. Hence, find

.

y

O

x

a Show on the diagram that the iteration b Prove that the iteration converges to

, with .

, converges.

Section 5: Limitations of fixed-point iteration; alternative rearrangements Fixed-point iteration seems a little easier to implement than the Newton–Raphson method, as there is no need to differentiate any of the expressions. It also works in some situations when the Newton–Raphson method doesn’t. However, there are other situations when fixed-point iteration doesn’t work because the sequence fails to converge to the root. For example, let us try solving the equation

for

by the iterative formula

Sketching graphs and trying some integer values of shows that there is a root between

y

. and

y = ex - 2 y = x

x

O

Starting the iteration at

gives the following sequence.

The sequence is clearly getting further away from the root; the sequence diverges. You can also try starting the sequence below the root, at ; the resulting sequence is:

Again, the sequence doesn’t seem to be approaching the root. You can see this on the graph by drawing the staircase diagram.

y

O

x

If you continue the second sequence it actually converges to the other root of the equation:

But is there any way you can use fixed-point iteration to find the first root?

.

The equation

can be written in a different way:

The sequence based on this rearrangement,

, starting at

, is:

Looking at the graph confirms that this sequence does indeed converge to the root between and . y

y = x y = ln(x + 2)

x

O

Interestingly, starting this iteration near the other root ( to one converging to the positive root. Starting at

Starting with we get is not defined for y

) leads either to a divergent sequence or gives the following sequence.

, and then the sequence cannot be continued further because .

y = x y = ln(x + 2)

O

x

In this investigation you have found the following.

Key point 14.6 Some rearrangements of the equation lead to convergent sequences, and others to divergent sequences. A divergent sequence does not find the required root. If an equation has more than one root, different rearrangements may converge to different roots.

WORKED EXAMPLE 14.8

a Show that the equation

has a root between and .

b Starting with , find the next four terms of the sequence behaviour of the sequence.

and describe the

c Illustrate the behaviour of the sequence using a cobweb diagram. d Show that the equation

can be written as

. Use this rearrangement to find an

approximate solution of the equation, correct to three decimal places. a Write

.

To use the change of sign method you need to rewrite the equation in the form .

There is a change of sign, so there is a root between and . b

The terms of the sequence are getting farther away from each other.

The sequence seems to diverge.

y

c

O d

1

x Make sure you show all the steps in a ‘show that’ type of question.

Use this new arrangement to form a sequence. It makes sense to start at again.

… You are not asked to show all the approximations, so just keep going until the third decimal place stops changing.

Is there any way you can tell, without actually trying it, whether a sequence from a particular rearrangement would converge? Looking at some staircase diagrams suggests that the answer has something to do with the gradient of the graph of at the point where it crosses the line .

y

y

g(x)

y = x y = x g(x) x

O

x

O

In the first diagram the gradient of the graph of is smaller than the gradient of and the sequence seems to converge. In the second diagram, where the sequence diverges, the graph of is steeper than the line . This suggests that the sequence converges when the gradient of is smaller than and diverges when it is greater than . It is less obvious what happens when the gradient of is negative, so the iteration produces a cobweb diagram. It turns out that it depends on whether the gradient of is smaller or greater than .

Both cases can be summarised as follows.

Key point 14.7 A fixed-point iteration converges if diverges if If

or

: near the root and near the root.

is sufficiently close to the root

it is impossible to tell whether the sequence will converge without trying it.

WORKED EXAMPLE 14.9

The equation

has a root near

. Two possible rearrangements of this equation are .

Determine which rearrangement will produce a sequence that converges to the root. Write the sequence as

When

The iteration converges if the root.

near

:

The root is near this point. near the root, so this iteration will not

, so evaluate the derivative at

converge. When

:

near the root, so this iteration will converge.

EXERCISE 14E 1

For each sequence below, find the first five terms, the behaviour of the sequence.

and the

term, and describe the

a b c d e f g 2

Each diagram below shows the graphs of happens to the iteration defined by

y

a

O

x

x1 y

b

O

x1

x

and .

, and the starting value

. Describe what

y

c

B

A O

x

x1 y

d

C B A O

x

x1 y

e

C B A O

x1

x

y

f

C B A O

x

x1 y

g

C B A O 3

x1

x

Find an alternative rearrangement for each of the following equations.

a

i ii

b

i ii

c

i ii

d

i ii

4

Find the missing constants to rearrange each of the following equations into the given equivalent form. a

i ii

b

i ii

c

i ii

d

i ii

5

The equations below each have more than one root. For each root, use technology to find a rearrangement that converges to it. a b c

6

The diagram below shows the graph of a function and the line . The equation has three roots, marked, , and . Which of the roots, if any, does the iteration converge to when: a b Justify your answer by showing several iterations on a sketch copy of the diagram.

y

p

α

Oβ

7

x

γ q

The iterative formula graphs of

is used to find an approximate solution of the equation

and

. The

are shown below.

y S R Q O b

a

x

c

P

Use the diagram to determine the behaviour of the sequence

when:

a b c 8

The Newton–Raphson iteration for solving point iteration for solving a Express

. This can also be considered as a fixed-

.

in terms of

.

Let be the solution of the equation, and assume that

for near .

b Find , and, hence, prove that the Newton–Raphson method always converges when the starting value is sufficiently close to . 9

a Show that the equation

has a root between

and

, and another one between and

. b Let

and define the sequence

with

and, hence, determine to which of the two roots the sequence 10 The function roots,

is defined by

. Solve the inequality converges.

. The equation

has two positive real

.

a Show that is between and

, and that is between

b The rearrangement

and

.

is used to find an approximate root of the equation

. Without carrying out the iteration, determine to which of the two roots the sequence converges.

c Show that an alternative rearrangement is

, and find the constants , and . Use this

rearrangement to find the root , correct to three decimal places. 11 a Find, in terms of , the roots of the equation Let , where . b Find

.

, where is the non-zero root of the equation

c The iterative formula equation

, with

.

, is used to find approximate roots of the

. Find the set of values of for which the iteration converges to the non-zero root.

Focus on … See Focus on … Problem solving 2 for some applications of this equation, and also for a comparison of numerical and analytical methods for analysing it.

Checklist of learning and understanding Some equations can be solved only by finding numerical approximations. Numerical methods are methods that tell you how to find an improved approximation. When an equation is written in the form between

and

by showing that

, you can show that it has a root (solution) and

have different signs.

The change of sign method works only for sufficiently well-behaved functions; it can fail if, for example, the graph of has a vertical asymptote, a break, or touches the . You can use the change of sign method to check the accuracy of the solution by ‘unrounding’ the number. The methods you met in this chapter involve creating a sequence that converges to the root of the equation. The first term of the sequence needs to be chosen from an interval that contains the root. The Newton–Raphson method works by approximating the curve by its tangent. The equation needs to be written in the form

. The resulting sequence is given by

The Newton–Raphson method fails to converge if the starting point is close to a stationary point of (where )

Q x1

x0

x

Fixed-point iteration requires the equation to be written in the form sequence is given by

. The iteration

This iteration can be represented graphically as a staircase or a cobweb diagram.

y

y y = x y = g(x)

O

x1

x2 x3

x

O

x1

x3

x2

x

Sometimes the sequence diverges, meaning that the terms get farther away from the required root. To get a convergent sequence we may need to rearrange the equation. If the equation has several roots, different rearrangements may converge to different roots. The iteration converges if

near the root and diverges if

.

Mixed practice 14 1

The graph of a

has a stationary point between and .

Show that the

of the stationary point satisfies

.

b Use fixed-point iteration with a suitable starting point to find the stationary point, correct to three decimal places. 2

The diagram shows a sector of a circle with radius shaded area equals .

of the

. The angle at the centre is . The

5 θ

a Show that

.

b Show that the equation given in part a has a root between and . c Use the Newton–Raphson method with a suitable starting point, to find the value of , correct to two decimal places. 3

a Sketch the graphs of and of solutions of the equation

on the same set of axes. State the number

b Show that the equation given in part a has a solution between and . c Use an iteration of the form decimal places. 4

to find this solution, correct to two

The diagram shows the graph of coordinate axes and the line .

. The shaded area is enclosed by the curve, the

y

O

a

x

a Given that the shaded area equals , show that

.

b Show that the equation given in part a has a root between and . c Use the Newton–Raphson method to find the value of , correct to

three decimal places.

5

The sequence defined by  

,   

converges to the number . i

Find the value of correct to decimal places, showing the result of each iteration.

ii

Find an equation of the form a root.

, where , and are integers, which has as

© OCR, GCE Mathematics, Paper 4723, January 2008 6

Let

for

a Show that the equation b Starting from

has a root between

and

.

, use the Newton–Raphson method to find

.

c Explain why this iteration cannot be continued to find the root. d Use the Newton–Raphson method with

to find the root, correct to three decimal

places. 7

A sector of a circle, with angle radians at the centre, is split into a segment and a triangle. as shown in the diagram. Given that the segment and the triangle have the same area:

θ

a Show that

.

b Use an iterative formula, with the starting value

, to find the value of , correct to

three decimal places. 8

A rectangle has two vertices on the , between and , and two vertices on the curve . Let the smaller of the of the vertices be . It is required to find the rectangle with a maximum possible area. a Show that, for this rectangle,

.

b Use the Newton–Raphson method, with the starting value , to find the value of , correct to four decimal places. Hence, find the maximum possible area of the rectangle. 9

Marek and Anjali each deposit £ into their respective bank accounts. Marek’s account earns simple interest per year, and Anjali’s account earns compound interest per year. Neither of them make withdrawals nor further deposits into their accounts. Using the Newton– Raphson method, find after how many whole years Anjali will first have more money than Marek.

y

10

y = x y = 2 ln(3x - 2)

O

α

The line

β

x

and the curve

meet where

and

, as shown in the

diagram. i

Use the iteration , with initial value correct to decimal places. Show all your working.

ii

With the help of a ‘staircase’ diagram, explain why this iteration will not converge to , whatever value of (other than ) is used.

iii

Show that the equation

, to find the value of

can be rewritten as

Newton–Raphson method with

. Use the

and

, to find correct to

decimal places. Show all your working. iv Given that root of

, explain why the Newton–Raphson method would not converge to a . © OCR, GCE Mathematics, Paper 4726, June 2010

11 A curve is defined by a Show that the

for of any stationary point on the curve satisfies the equation

One of the roots of this equation is between and . b By considering the derivative of converge to this root.

, prove that the iteration

c Find an alternative rearrangement of the equation of the stationary point on the curve

does not

and use it to find the between

and

. Give

your answer correct to three decimal places.

Worksheet See Extension sheet 14 for a look at Euler’s method for finding approximate solutions to differential equations.

15 Numerical integration In this chapter you will learn: why definite integration is connected to the area under a curve how to approximate integrals that can’t be found exactly how to establish whether these approximations are overestimates or underestimates.

Before you start… GCSE

You should be able to calculate the area of a trapezium.

1 Find the area of this shape.

6 4

1.5 Student Book 1, Chapter 15

You should know that a definite integral represents the area between the curve and the -axis.

2 Shade the area given by . y

1

O

Student Book 1, Chapter 19

You should know how to find distance from a velocity–time graph.

π

3 Find the total distance travelled in the first .

x

v (m s–1) 12 5 10 O

3

7

t (s)

–9

An approximation to definite integration Using definite integration to find the area between a curve and the -axis has many applications, for example in Mechanics (finding distance from a velocity-time graph); and Statistics, (where the area under the normal distribution–curve represents probability). Since you already know many different integration methods, you may think that in most cases the area under the curve can be found exactly. It turns out that in many real-world problems, it is actually not possible to integrate the function exactly (or the integration method is very complicated). In such cases you can find an approximation for the area using one of the methods from this chapter.

Section 1: Integration as the limit of a sum The simplest way to estimate the value of an area is to split it into rectangles. For example, an approximation to the area under the curve between and is needed because it turns out that it is impossible to find the exact value of the integral

.

y y = e- x2

O

0.2 0.4 0.6 0.8 1.0

x

Fast forward You will see in Chapter 17 that the function is used when finding probabilities from a standard normal distribution so it is important to be able to, find the area under its graph. In the previous diagram the required area has been split into five rectangles of equal width, . The height of each rectangle is the -coordinate of its top-right vertex; so the height of the first rectangle is the value of when

, which is

. You can make a table to show the heights of all five rectangles.

-coordinate

Height The total area of the five rectangles (each of width

) is

and this is an approximate value of the required area. You can see from the diagram that the actual area is a little bit larger; a lower bound for the area is . Unfortunately, this procedure doesn’t tell you anything about how good the approximation is. To assess this, it would also be useful to have an upper bound, so that you know the area lies between those two values. To find an upper bound use rectangles with the top edge above the curve. You can draw such rectangles using the left end point of each interval.

y

O

0.2 0.4 0.6 0.8 1.0

x

The heights of the five rectangles are: -coordinate

Height So their total area is

You can therefore say that the required area under the curve is certainly between

and

, and you

can write

Key point 15.1 You can find upper and lower bounds for the area under a curve by using rectangles that lie above and below the curve. The actual area lies between the lower and the upper bound.

WORKED EXAMPLE 15.1

The diagram shows a part of the graph of

.

y 1.2 1.0 0.8 0.6 0.4 0.2 O

0.2 0.4 0.6 0.8 1.0 1.2 1.4

x

Using six rectangles of equal width, find a lower bound for the value of y

.

From the graph you can see that for the lower bound (rectangles below the curve) you need to use -coordinates at the right end point of each interval.

1.2 1.0 0.8 0.6 0.4 0.2 O

0.2 0.4 0.6 0.8 1.0 1.2 1.4

x

The width of each rectangle is

So

WORKED EXAMPLE 15.2

Each rectangle has width equal to the -coordinate.

.

and height

A part of the curve with equation

is shown in the diagram.

y 4 2

-2

O

2

4

x

6

Use four rectangles of equal width to find an upper bound for

.

y From the graph we can see that for the upper bound we now need to use -coordinates at the right end point of each interval.

4 2

-2

O

2

4

6

x The width of each rectangle is

So

has an upper bound of

WORK IT OUT 15.1 Three students try to find an upper bound for the value of

, using six

rectangles. Which is the correct solution? Can you identify the errors made in the incorrect solutions?

Solution 1

Using

:

So

Solution 2

Using

:

.

So

Solution 3

Using

:

So

To get a more accurate approximation for the area you need to make the upper and lower bounds closer to each other. This can be done using more rectangles of smaller width. For the original example, estimating , you can use a spreadsheet to calculate areas using more and more rectangles. Here are some of the results.

Tip Always use the graph to decide which rectangles to use.

Number of rectangles

Width

Lower bound

Upper bound

You can use a graphical calculator or graphing software to check that the actual area is

.

Key point 15.2 As the number of rectangles increases, the upper and the lower bounds approach a limit, which is the actual value of a definite integral.

EXERCISE 15A 1

Use five rectangles to find upper and lower bounds for the value of each of the following integrals. Use technology to draw the graph first. a

i ii

b

i ii

c

i ii

2

Use a spreadsheet to find upper and lower bounds for each of the following integrals, using: i ii .

iii

In each case, find the difference between the upper and lower bounds. a b c d e How does the difference decrease when the number of rectangles doubles? 3

A part of the curve with equation

is shown in the diagram.

y 2 1 O –1

1

2

3

4

5

6

x

Using six rectangles of equal width, find upper and lower bounds for 4

The diagram shows the graph of

.

.

y 3 2 1 O a

1

2

3

4

5

x

Use five rectangles of equal width to find the upper and lower bounds for

b How could the difference between the upper and lower bounds be reduced? 5

a Sketch the graph of

for

.

b Use four rectangles of equal width to find an upper bound for c If 6

are used, will the upper bound increase or decrease?

The diagram shows a part of the graph

.

.

.

y

O

x

a State the exact coordinates of the maximum point on the curve. b Use four rectangles of equal width to find a lower bound for



.

Section 2: The trapezium rule You saw in Section 1 that you can approximate a definite integral using sums of areas of rectangles. However, you may need many rectangles to achieve high accuracy. You can improve this method by replacing each rectangle by a trapezium that connects the points on the curve corresponding to the end points of each interval (as shown in red in the diagram below). Each trapezium has an area somewhere between that of the upper and the lower rectangles. y y4

y3 y2 y1 y0 O

x0

x1

x2

x3

x4

x

The area of a trapezium is given by

, where and are the parallel sides and is the height

(perpendicular distance between the parallel sides). Looking at the first trapezium above, its height is equal to the width of the interval and the parallel sides have lengths equal to the -coordinates of the end points. If there are intervals, you can label the -coordinates coordinates . The areas of each trapezium are then:

and the corresponding -

Adding these together, each -coordinate appears twice, except the first and the last one. The following expression therefore gives an approximation for the total area under the curve. y y2

y1

y3

y0

h O

x0

h x1

h x2

x3

x

Key point 15.3 The trapezium rule using equal intervals with end-points

where

and

:

.

Tip It is a good idea to set out the and values in a table. The formula then says ‘ times ( first + last + twice the sum of the rest )’, where is the difference between the -coordinates.

WORKED EXAMPLE 15.3

Use the trapezium rule, with five equal intervals, to find an approximate value of

. Give

your answer correct to three decimal places. Divide the interval from to into five equal parts. The values start from and go up in steps of until they reach . Note that since there are five intervals, there should be six values. The values are calculated using . Since you want the answer correct to ., you should record the values to . and round at the end. Use the formula. You should show the numbers used in the calculation.

Tip You may have a TABLE function on your calculator that will produce the table of values. Alternatively, you can save the six numbers in memory to use in the trapezium rule calculation. If the expression for

is short, you can just type in the whole sum; e.g. .

Did you know? Your calculator may have a function for finding approximate values of definite integrals. It probably uses the trapezium rule. The actual value of the integral is , so our approximation is within The trapezium rule tends to be generally more efficient than using rectangles.

of the correct value.

In this case the approximation is a slight underestimate. Looking at the graph explains why this is the case: most of the trapezia lie below the curve.

y

0.0 0.2 0.4 0.6 0.8 1.0

x

Rewind You can see from the graph that the trapezium rule gives an overestimate when the function is convex. See Chapter 12, Section 1 for a definition of a convex function.

Key point 15.4 To determine whether the trapezium rule gives an underestimate or overestimate, you need to

look at the shape of the graph.

WORKED EXAMPLE 15.4

The diagram shows the graph of

.

y 1

O

π 16

a

π 3

3π 16

π 4

x

Use the trapezium rule, with four equal intervals, to estimate the value of

, giving your

answer correct to three significant figures. b Explain whether your answer is an overestimate or an underestimate. a

Divide the interval from to into four equal parts. The -coordinates start from and increase in increments of . The -coordinates are found using . Use the trapezium rule.

b Since the function is convex between

and

the trapezia are above the curve.

,

The explanation should refer to the shape of the graph.

y 1

O

π 16

π 3

3π 16

π 4

x

Hence, this approximation is an overestimate.

WORK IT OUT 15.2 Three students are asked to estimate the value of

using the trapezium rule, with five

strips. Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1

Solution 2

Solution 3

EXERCISE 15B 1

Use the trapezium rule, with the given number of intervals, to find the approximate value of each integral. Compare your answer to the upper and lower bounds found in question 1 of Exercise 15A. a

i ii

b

i

, , ,

ii c

i ii

2

, , ,

For each integral from question 1: a Use technology to find its value, correct to eight decimal places. b Use a spreadsheet to calculate trapezium rule approximations using

and

intervals.

c Find the percentage error in each estimate. How do the percentage errors decrease when you double the number of intervals? 3

For each of the following integrals, either find its exact value where possible, or an approximation using six trapezia.

a b c d e f 4

a Sketch the graph of b

.

Use the trapezium rule, with five strips, to estimate the value of

. Give your

answer to two decimal places. c Explain whether your answer is an overestimate or an underestimate. 5

a

Use the trapezium rule, with four intervals, to find an approximate value of

.

b Describe how you could obtain a more accurate approximation.

Worksheet See Support sheet 15 for a further example of using the trapezium rule and for more practice questions. 6

The diagram shows a part of the graph of

.

y 1.2 1.0 0.8 0.6 0.4 0.2 O

0.2 0.4 0.6 0.8 1.0 1.2 1.4

x

The graph crosses the -axis at the point where

.

a Find the exact value of . b Use the trapezium rule, with four intervals, to find an approximation for

.

c Is your approximation an overestimate or an underestimate? Explain your answer. 7

A particle moves in a straight line with velocity given by , where is measured in and in seconds. Use the trapezium rule, with six strips, to find the approximate distance travelled by the particle in the first .

8

The velocity, , of a particle moving in a straight line is given by the velocity-time graph for the particle.

. The diagram shows

v 1.5 1.0 0.5 O - 0.5

5 10 15 20 25 30 35 40

t

- 1.0 - 1.5

a The particle changes direction when

and

(with

). Find the exact values of and .

b Use the trapezium rule, with eight equal intervals, to estimate the total distance travelled by the particle during the first

.

Checklist of learning and understanding Some definite integrals cannot be evaluated exactly. In such cases it is possible to use rectangles to find upper and lower bounds. An upper bound for an area is a number that is larger than the area; a lower bound is a number that is smaller than the area. As the rectangles get smaller, the upper and lower bounds get closer to each other. The actual area is the limit of the sum of the rectangles. The trapezium rule is a way of estimating the area using trapezia. We need fewer trapezia than rectangles to achieve the same accuracy. Using equal intervals with end points

where

and

.

:

Mixed practice 15 1

The diagram shows a part of the graph of

.

y 2.5 2.0 1.5 1.0 0.5 O

5

10

15

x

20

Using five rectangles of equal width, find a lower bound for the value of

.

Give your answer correct to one decimal place. 2

The diagram shows a part of the curve with equation

.

y 15 10 5 - 100 O

100

200

300

x

a Use the trapezium rule, with five strips of equal width, to estimate the value of . b State, with a reason, whether your answer is an underestimate or an overestimate. c Explain how you could find a more accurate estimate. 3

Use six rectangles of equal width to find the values of and such that y

4

0.3 O

0.6

0.9

1.2

1.5

x

The diagram shows the curve with equation , for . The region bounded by the curve, the -axis and the line has area . The region is divided into five strips, each of width . i

By considering the set of rectangles indicated in the diagram, find an upper bound for . Give the answer correct to decimal places.

ii

By considering another set of five suitable rectangles, find a lower bound for . Give the answer correct to decimal places.

.

iii How could you reduce the difference between the upper and lower bounds for ? © OCR, GCE Mathematics, Paper 4726, June 2009 5

Use the trapezium rule, with

each of width , to estimate the value of

. © OCR, GCE Mathematics, Paper 4722, January 2008 6

The diagram shows a part of the curve with equation

. Use the trapezium

rule, with six strips of equal width, to estimate the area enclosed between the curve and the -axis. y 1.5 1.0 0.5 O

7

1

2

3

4

5

6

x

A particle moves in a straight line so that its velocity is given by the equation

.

y 1

O

1

2

3

4

x

-1 a Find the values of

when the velocity is zero.

b Use the trapezium rule, with strips of width , to find an approximate value of the distance travelled by the particle from 8

.

A part of the graph of

is shown in the diagram. Use four rectangles of equal

width to find a rational number such that

.

y 1.5 1.0 0.5 O

9

5

10

15

20

25

A curve has equation

x

.

a Sketch the curve, showing the coordinates of any stationary points. b

Use four rectangles of equal width to find a lower bound for

.

10 The diagram shows the velocity-time graph for a particle moving in a straight line.

v 10 5 O

10

20

30

40

t

The velocity of the particle is measured at following table.

intervals and the results given in the

Estimate the average speed of the particle during the

.

Worksheet See Extension sheet 15 for a look at how approximation methods were used in the development of calculus.

FOCUS ON … PROOF 2

Deriving the compound angle identities This proof will demonstrate that:

and

You have already seen trigonometric proofs that use right–angled triangles to prove results about more complicated figures. The same approach works here.

Rewind In Student Book 1, Focus on … Proof 2 used this strategy to prove the sine and cosine rules.

The sine compound angle formula You’ll prove the sine compound angle formula first:

.

PROOF 9

A x

B h

Create a triangle with angle by joining two right-angled triangles with angles and .

A + B y

x

y

From the left right-angled triangle:

Now express all the other lengths in terms of and .

Similarly, from the other right-angled triangle:

You can write in two different ways.

The area of the left–hand triangle is:

Use

on each of the triangles

individually. The area of the right–hand triangle is:

The area of the whole triangle is:

And then on the triangle as a whole.

Therefore:

Dividing by

QUESTIONS

.

QUESTIONS 1

Is it possible to draw two right-angled triangles with the same height for any pair of acute angles and ?

2

Does the identity still hold when the angles and are not acute? Can you prove it?

The cosine compound angle formula To derive the compound angle identities for

and

, you could use the same triangles

and the cosine rule. However, there is a simpler proof that uses the relationship between and 3 Write and use the compound angle identity for

to prove that

and

:

FOCUS ON … PROBLEM SOLVING 2

Choosing between analytical and numerical methods Many problems you have encountered in this course can be solved in more than one way. In real–life situations you are free to choose whatever method and approach you prefer. In making your decision you should think about the following. How difficult is the method? How efficient is it? Does it require lots of detailed or repeated calculations? How accurate is it? What level of accuracy do you actually need? Can it be easily adapted to solve other similar problems? For example, when solving an equation you have essentially two options: you can try to rearrange the equation using rules of algebra (an ‘analytical solution’), or you can use one of the iterative methods from Chapter 14 (a ‘numerical solution’). The former may not always be possible but, when it works, it gives you an exact solution (such as ). However, in many applications you need an answer to only a couple of decimal places, so you should consider whether the effort required to rearrange the equation is justified. On the other hand, your equation may have a parameter in it (for example,

) and you may

want to know how changing the value of the parameter affects the solution; in this case, finding the analytical solution once (in terms of the parameter) may be more efficient than repeating the numerical calculation lots of times. Here is a problem that can be investigated both analytically and numerically. You can try various approaches and decide for yourself which one suits you best.

The fishing lake problem The number of fish in a lake can be modelled by the equation

In this model,

is the number of fish in year . Each year, due to natural birth and death rates, the

population increases by a factor of . Meanwhile, number ( ) of fish are caught and removed.

die out due to lack of resources, and a constant

What is the maximum number of fish that can be removed each year without causing the population to die out? QUESTIONS 1

Use a spreadsheet to investigate how the population changes with the following parameter values. a

and: i ii iii

b

and: i ii iii

c What is the largest number of fish that can be caught each year without causing the population to die out? Does this depend on the initial size of the population?

2

Vary the values of and slightly. How does this affect the maximum possible value of ? You need to try many different values of the parameters to find the relationship between and . Is it possible to solve this problem analytically instead, to find an equation linking the three quantities? The sequence

is not one of the types you are familiar with; in fact, it is not

possible to find a general formula for . However, you are only interested in the long–term behaviour of the sequence: does it eventually decrease to zero or not? You know from Chapter 14 that, if a sequence 3

has a limit, then this limit is a solution of the equation

Consider the case when

and

a Solve the equation

.

. when:

i ii iii Compare the solution to what you observed in question . b Find the discriminant of the equation equation has a solution only when

in terms of . Hence, show that the .

c Use the quadratic formula to write the two solutions in terms of . Hence, show that, when , both solutions are positive. Could you tell, without doing the spreadsheet investigation, to which of the two solutions the sequence will converge to?

Rewind In Chapter 14, Section 5, you learnt that the iteration if 4

converges to a root of

near the root.

For the general case of the equation a Find the discriminant in terms of

and .

b Hence, show that the largest number of fish that can be removed without causing the population to die out is

.

5

Did you find the theoretical anal ysis or the spreadsheet investigation easier to follow? Which one do you think gives a more reliable answer? Which helps you understand the problem better?

6

When and , the formula we found previously says that the maximum number of fish that can be removed is . However, with and , the spreadsheet shows that there are still fish in the lake after 50 years. So is the additional accuracy you get from using the formula always required?

FOCUS ON … MODELLING 2

Translating information into equations The aim of a mathematical model is to describe a real–life situation using equations that can be solved and used to make predictions.

Tip In many examples, ‘rate of change’ means change in time; however, look out for examples where this is not the case! In this section you look at writing differential equations. These are equations involving the rate of change of a quantity. You need to remember the following. The rate of change of with respect to is ‘ is proportional to ’ means that

.

for some constant .

When writing differential equations, you usually have some information about particular values of the quantities involved. Sometimes you can use those to find constants in the equation (such as the in ), but sometimes you need to wait until you have solved the equation.

Rewind Solving some differential equations is covered in Chapter 13. In this section you will not need to solve any equations.

WORKED EXAMPLE 1

The speed of an object decreases at a rate proportional to the square root of its current speed. When the speed is it is decreasing at a rate of . Using for speed and for time, write an equation to represent this information. ‘Rate of change’ means the derivative with respect to time. The speed is decreasing, so you write ‘

’ to emphasise this.

You can use the given information to find . Remember that the rate of change is negative.

So the equation is:

Rewind You know that the rate of change of velocity is acceleration which is proportional to the force acting on an object. The model in Worked example 1 could therefore be used when there is a resistance force proportional to the square root of the speed, such as air resistance or drag when an object is moving through liquid.

WORKED EXAMPLE 2

In one possible model of population growth, the rate of growth depends on two factors: it is proportional to the current size of the population, and is also proportional to . (The last factor represents seasonal breeding patterns.) When the measurements began the population size was Using for the size of the population and for time, measured in months, write a differential

.

equation to represent this information. The two factors need to be multiplied together. Don’t forget to include the constant of proportionality. You only have information about when solved the equation.

, so you can’t find until you have

QUESTIONS Write a differential equation to represent each of the following situations. Where possible, find the values of any constants. 1

A population of a new town ( thousand) increases at a rate proportional to its size. Initially, the size of the population is

and it is increasing at the rate of

per year.

2

During the decay of a radioactive substance, the rate at which mass is lost is proportional to the mass present at that instant. Use for the mass of the substance, in grams, and for the time, in seconds. Initially there is of the substance and the mass is decreasing at the rate of .

3

In an electrical circuit, the voltage is decreasing at a rate proportional to the square of the present voltage. When the voltage is volts, it is decreasing at a rate of volts per second.

4

Newton’s law of cooling states that the rate at which a body cools is proportional to the difference between its temperature and the temperature of its surroundings. A cup of tea is initially at and is cooling at the rate of per minute in a room of temperature . Use for the temperature and for time (in minutes).

5

A metal rod, of length , has one of its ends heated. After a while the temperature remains constant in time. However, the temperature changes along the length of the rod, decreasing at a rate proportional to the distance from the hot end. The temperatures at the two ends are Use for temperature and for the distance from the hot end (measured in ).

6

and

.

The water pressure in the sea increases with depth. The pressure at depth is proportional to the density of the seawater ( ). The density also varies with depth, and is modelled by the equation . Write a differential equation for the rate of increase of pressure with depth.

7

A rumour spreads at a rate proportional to the square root of the number of people who have already heard it, and inversely proportional to the time it has been spreading. After 5 minutes, people have heard the rumour and it is spreading at the rate of people per minute. Write for the number of people who have heard the rumour and for the time, in minutes, since the rumour started. Write a differential equation to model this situation, and explain why the model needs to be modified for small values of .

8

A cylindrical tank has a base radius of . Water leaks out of the tank so that the rate at which the volume is decreasing is proportional to the height of the water remaining in the tank. Initially the height of water is and it is decreasing at the rate of rate of change of volume of water in the tank.

per minute. Find an equation for the

CROSS-TOPIC REVIEW EXERCISE 2

1

Find the equation of the curve that has gradient

and passes through the point (0,

4). 2

Find

3

a Given that b

. , find

Given that

.

is a solution of the differential equation

find the

possible values of . 4

The diagram below shows the graphs of area.

and

Find the value of the shaded

y y=cosx

y=sin x

x

O

5

The region in the diagram is enclosed between the graph of between

and

and the –axis

.

y

R O

2

5

x

a Find the shaded area between the curve and the -axis. b Hence, find the exact area of . 6

The diagram shows a part of the curve with parametric equations

y

O

x

.

a Find the values of at the points where the graph crosses the -axis. b Find the exact value of the shaded area. c Find the Cartesian equation of the curve. 7

A curve has equation i

Differentiate

, where is a non-zero constant.

, and show that

.

ii Given that the curve has exactly one stationary point, find the value of , and determine the exact coordinates of the stationary point. © OCR, GCE Mathematics, Paper 4723, January 2008 8

By taking natural logarithms of both sides, or otherwise, find

9

a Simplify

.

.

b Hence, or otherwise, find 10 a Given that b Find

given that

.

, express in terms of .

in terms of .

c Hence, show that

.

11 a Show that

.

b Find the coordinates of the point on the curve equal to 12 a Find

(

), where the gradient is

. in terms of and

b If

.

, find and simplify an expression for In

. c Hence, find the derivative of

.

13 Find the exact area enclosed between the graphs of

and

for

.

14 Triangle is made out of an elastic piece of string. Vertices and are being pulled apart so that the length of the base, , is increasing at the rate of and the height, , is decreasing at the rate of

. Initially,

for

a Find the rate at which the area of the triangle is changing when b Show that

and

.

, and find an expression for in terms of .

c Find an expression for in terms of . 15 A rectangle is drawn inside the region bounded by the curve in the diagram. The vertex has coordinates ( ).

y

B

O a

A(x, 0)

x

i Write down the coordinates of point .

and the -axis, as shown

ii Find an expression for the area of the rectangle in terms of . b

i Show that the stationary point of the area satisfies the equation. ii By sketching graphs, show that this equation has one root for

.

iii Use the second derivative to show that the stationary point is a maximum. c

i The equation for the stationary point can be written as suitable iterative formula, with correct to three decimal places.

. Use a

, to find the root of the equation

,

ii Hence, find the maximum possible area of the rectangle. 16 Consider the infinite geometric series

for



.

a Explain why the series converges. b Show that the sum of the series is

.

c Find the exact value of 17 Let

.

.

a Use the trapezium rule with four intervals of equal width to estimate the value if . b Using the first three non-zero terms of the binomial expansion of

, find another

approximation for . You are given that

.

c Use the substitution to calculate the exact value of . Hence determine whether method or method gives a better approximation. 18

i

Sketch the graph of

, where is a constant such that

. State the coordinates of

any points of intersection with the axes. has its -coordinate equal to ii The point on the curve coordinate of may be written as .

Show that the -

a Use the trapezium rule, with two strips each of width , to find an expression for the approximate value of . b Given that this approximate value is equal to

, find the value of .

© OCR, GCE Mathematics, Paper 4722, June 2009 19

i

Express

ii The expression

in the form

, where

and

.

is defined by

a Determine a value of for which

is not defined.

b Find the smallest positive value of satisfying

, giving your answer in an

exact form. © OCR, GCE Mathematics, Paper 4723, June 2010 20 Evaluate 21 a State an expression for

. .

b

Hence, or otherwise, show that

.

22 a Use the identity

to show that

b The diagram below shows part of the curve point . y

. . Write down the -coordinate of the

P

a

x

O

c Find the red-shaded area in terms of , writing your answer in a form without trigonometric functions. d By considering the blue–shaded area, find

for

.

23 A function is defined by a Find

.

b Show that the stationary points of

satisfy the equation

c Hence, show that the function has only one stationary point. 24 a Sketch the graph

.

b The tangent to this graph at the point ( , . c For what range of values of does

) passes through the origin. Find the value of have two solutions?

y

25

x

O

The function is defined for the domain

by

The diagram shows the curve with equation i

Find the range of .

ii The function is defined for the domain

by .

Given that is a one–to–one function, state the least possible value of . at which the gradient is . iii Show that there is no point on the curve © OCR, GCE Mathematics, Paper 4723, June 2008

16 Conditional probability In this chapter you will: use set notation to describe probabilities learn how to work with conditional probabilities in the context of Venn diagrams, two-way tables and tree diagrams learn a formula for conditional probability.

Before you start… Student Book 1, Chapter 17

You should understand the basic laws of probability, including the terms ‘mutually exclusive’ and ‘independent’.

1 In a certain village, having a car is mutually exclusive of having a motorbike. The probability of having a car is . The probability of having a motorbike is . What is the probability of having neither a car nor a motorbike?

GCSE

GCSE

You should be able to use tree diagrams to solve problems.

2 What is the probability of a mother having two children with the same gender, assuming they are born independently?

You should be able to find

3 a A bag contains ten red balls and five blue balls.

simple conditional probabilities.

One ball is taken from the bag and not replaced. A second ball is then taken. Given that the first ball is red, what is the probability that the second ball is also red? b In a group of children, have a pet. Out of those, have a dog. Find the probability that: i a child has a dog ii a child has a dog, given that they have a pet.

Student Book 1, Chapter 1

You should understand and be able to use set notation.

4 Write out the set

Student Book 1, Chapter 17

You should understand probability distributions, including the binomial distribution.

5 What is the probability of getting four heads when six fair coins are tossed?

What is conditional probability?



What is the probability that you will become a millionaire? You could just look at data for how many millionaires there are in the world, but this is not likely to give you a very reliable answer because it depends on lots of other factors. Where you were born, what your parents do and your attitude towards risk all change the probability. You may be glad to know that the fact that you are doing Maths A Level immediately increases your probability of becoming a millionaire! Information often changes probabilities. A probability that takes into account information is called a conditional probability. In reality nearly all probabilities are conditional: the probability of a patient having heart disease may change depending on their age; the probability of a defendant being guilty may change depending on their prior convictions; the probability of a football team winning may depend on the team they are playing. Most people have a very poor intuition for conditional probabilities. In this chapter you will visualise conditional probabilities in various ways and see how you can use them to solve problems.

Section 1: Set notation and Venn diagrams What is more likely when you roll a dice once: getting a prime number and an odd number? getting a prime number or an odd number? The first possibility is restrictive − we have to satisfy both conditions. The second opens up many more possibilities − we can satisfy either condition. So the second must be more likely.

Focus on … See Focus on … Problem solving 3 for the approach of using extreme values to evaluate possible solutions. These are examples of two of the most common ways of combining events: intersection (in normal language ‘and’) and union (in normal language ‘or’). These are given the following symbols.

Key point 16.1 is the intersection of and , meaning when both and happen. is the union of and , meaning when either happens, or happens, or both happen. is the complement of , meaning everything that could happen other than . You can use Venn diagrams to illustrate these concepts:

   

Explore If you have neither apples nor pears then you have no apples and no pears. In set notation this can be written as . This is one of De Morgan’s laws – a description of some of the algebraic rules obeyed by sets and, hence, probability. What other similar rules can you find?

Did you know? Why don’t mathematicians just use simple words? One of the reasons for this is the ambiguity of everyday language. If you say that you play rugby or hockey some people may think this means you do not play both.

There is a very important result that comes from looking at the Venn diagrams on the previous page:

Key point 16.2

Explore This is an example of a rule called the ‘inclusion–exclusion principle’, which can be extended to more complicated Venn diagrams. Investigate some applications of this rule. You can interpret this formula as saying ‘if you want to count the number of ways of getting or , count the number of ways of getting and add to that the number of ways of getting . However, you have then counted the number of ways of getting both and twice, so you need to compensate by subtracting that number’. If there is no possibility of and occurring at the same time, then mutually exclusive, and the formula reduces to

These events are

WORKED EXAMPLE 16.1

A chocolate is selected randomly from a box. The probability of it containing nuts is . The probability of it containing caramel is . The probability of it containing both nuts and caramel is . What is the probability of a randomly chosen chocolate containing either nuts or caramel or both? Use the formula.

Venn diagrams and conditional probability Conditional probability is the probability that an event happens given that we know that another event has already happened. For example, the probability that a randomly selected person owns a car might be . But if we know that this person has already passed the driving test, the probability that they own a car will be larger. Venn diagrams provide a good way of thinking about conditional probability. Remember that the probability that happens, given that has already happened, is denoted by .

The hatched region shows has happened. The probability of now happening depends upon the relative probability of the shaded region compared to the hatched region. This leads to an extremely useful formula for conditional probability.

Key point 16.3

where

is the probability of given that has happened.

WORKED EXAMPLE 16.2

WORKED EXAMPLE 16.2

The probability that a randomly chosen resident of a city in Japan is a millionaire is probability that a randomly chosen resident lives in a mansion is

. Only in

. The are

millionaires who live in mansions. What is the probability of a randomly chosen individual being a millionaire given that they live in a mansion? Write required probability in ‘given’ notation and apply the formula.

WORKED EXAMPLE 16.3

You are given that

,

, and

a Find b Hence, find a

A

B 1 - x 2

x

1 - x 4

It is often a good idea to start with labelling the intersection with an unknown and writing in all the remaining information in terms of the unknown.

5 12

Using the fact that all the probabilities together sum to . Make it clear that you know is

Use Key point 16.3.

WORKED EXAMPLE 16.4

In a class of students. students have a bicycle, have a mobile phone and have a laptop computer. have both a bike and a phone, have both a phone and a laptop, and have both a bike and a laptop. have none of these objects. a How many have a bike, a phone and a laptop? b What is the probability that a randomly chosen student from the class has all three of the items, given that they have at least two of them? a

B

P 11 - x x 6 - x

12 - x

2 L

Draw a Venn diagram showing three overlapping groups, and label the size of the central region as . Then work outwards. For example, the number who have a bicycle and a phone but not a laptop will be

B

Continue working outwards. For example, the total of all the bicycle regions must be , so the remaining section is which is

P 2 + x

11 - x

x - 2

There are two students outside

x 6 - x

and .

12 - x

x - 2

2 L

Use the fact that there are form an equation.

students in the class to

Therefore, three students have a bicycle, a phone and a laptop. b There are students who have at least two items and with three items. So

Use the Venn diagram to find the total number of students in overlapping regions.

.

Independence revisited You will recall from Student Book 1 (Chapter 17) that events and are independent if The formula in Key Point 16.3 provides another, more intuitive condition. If and are independent, then:

This simply says that, if and are independent, then the probability of happening given that has happened is the same as the overall probability of . In other words, having information about does not change our knowledge about probability of . This agrees with our intuitive understanding of ‘independence’.

Tip Notice that and can be swapped round, so if

then also

EXERCISE 16A 1

For each of the questions below write in mathematical notation the probability required. An expression rather than a number is required. a the probability that the outcome on a dice is prime and odd b the probability that a person is from either Senegal or Taiwan c the probability that a student studying A Levels is also studying French d the probability that a red playing card is a heart e the probability that a German person lives in Munich f the probability of someone wearing neither black nor white socks g the probability that a vegetable is a potato if it is not a cabbage

h the probability that a ball drawn is red given that the ball is either red or blue. 2

a

b

3

i If

and

find

ii If

and

find

i If

and

find

ii If

and

find

You may find that Venn diagrams are helpful in solving the questions below. a

i When a fruit pie is selected at random,

and

. Ten

per cent contain both apples and pears. Find of books are classed as fiction and are classed as 20th century. Half of the ii In a library books are 20th century fiction. What proportion of the books are either fiction or from the 20th century? b

i

of students in a school play either football or tennis. The probability of a randomly chosen students playing football is

and the probability that they play tennis is . What percentage of

students play both football and tennis? ii Two in students in a school study Spanish and in study French. Half of the school’s students study either French or Spanish. What fraction of students study both French and Spanish? c

i

of students in a school have a Facebook account and out of have a Twitter account. Onetwentieth of students have neither a Facebook nor a Twitter account. What percentage of students are on both Facebook and Twitter?

ii

of teams in a football league have French players and one-third have Italian players. have neither French nor Italian players. What percentage of teams have both French and Italian players?

d

i In a class of students, take French, take German and four take neither. What is the probability that a student who takes German also takes French? of like pizza and like lasagne. like neither pizza nor lasagne. Find the ii In a survey, probability that a participant likes lasagne if they like pizza.

4

Simplify the following expressions where possible. a b c d e f g h i j k l

5

a

i

and

Find

ii b

and

i

and

i

Find and

Find

and

ii d

Find

and

ii c

Find

Find

i Find

and and are mutually exclusive.

ii Find 6

Out of

students in a college,

and and are mutually exclusive. play football,

play badminton, and five play both sports.

a Draw a Venn diagram showing this information. b How many students play neither sport? c What is the probability that a randomly chosen student plays badminton? d If you know that the chosen student plays football, what is the probability that they also play badminton? 7

Out of students in a college, two subjects.

study Mathematics,

study Economics and

study neither of the

a Draw a Venn diagram to show this information. b How many students study both subjects? c A student tells you that she studies Mathematics. What is the probability that she studies both Mathematics and Economics? 8

a In a survey, of participants are in favour of a new primary school and are in favour of a new library. Half of all those surveyed would like both a new primary school and a new library. What percentage supported neither a new library nor a new primary school? b What proportion of those wanting a new primary school also wanted a new library?

9

Assume that

and

a By drawing a Venn diagram, or otherwise, find b Find 10 Given that

and

find:

a b 11 Events and satisfy a Find b Find 12 An integer is chosen at random from the first one thousand positive integers. Find the probability that the integer chosen is: a a multiple of b a multiple of both and c a multiple of , given that it is a multiple of . 13 Denise conducts a survey about food preferences in the college. She asks students which of the three meals (spaghetti bolognese, chilli con carne, and vegetable curry) they would eat. She finds out that,

of the

students: would eat spaghetti bolognese would eat vegetable curry would eat both the bolognese and the curry would eat both curry and chilli would eat both chilli and bolognese would eat all three meals. would not eat any of the three meals.

a Draw a Venn diagram showing this information. b How many students would eat only bolognese? c How many students would eat chilli? d What is the probability that a randomly selected student would eat only one of the three meals? e Given that a student would eat only one of the three meals, what is the probability that they would eat curry? f Find the probability that a randomly selected student would eat at least two of the three meals. 14 The probability that a person has dark hair is , the probability that they have blue eyes is the probability that they have both dark hair and blue eyes is

and

a Draw a Venn diagram showing this information. b Find the probability that a person has neither dark hair nor blue eyes. c Given that a person has dark hair, find the probability that they also have blue eyes. d Given that a person does not have dark hair, find the probability that they have blue eyes. e Are the characteristics of having dark hair and having blue eyes independent? Explain your answer. 15 The probability that it rains on any given day is probability that it is neither cold nor raining is

, and the probability that it is cold is

. The

.

a Find the probability that it is both cold and raining. b Draw a Venn diagram showing this information. c Given that it is raining, find the probability that it is not cold. d Given that it is not cold, find the probability that it is raining. e Are the events ‘it’s raining’ and ‘it’s cold’ independent? Explain your answer and show any supporting calculations. 16 If

and

17 If

and

18 If 19 a If

find

and

. , find

find the maximum and minimum values of

represents a probability, state the possible values that

b Express

in terms of

c By considering an expression for

and show that

can take.



Section 2: Two-way tables Another useful way of looking at conditional probability is to use a two-way table. This lists all the possible outcomes varying along two factors. WORKED EXAMPLE 16.5

These data show the arrival time of a random sample of posted 1st class or 2nd class.

Next day

letters, as well as whether they were

Later

1st class 2nd class On the basis of these data, find: a b c a

There are

letters in the ‘1st class’ and ‘Next day’ categories, out of

b

There are

‘Next day’ letters out of

‘1st class’ letters.

c

There are

‘1st class’ letters out of

‘Next day’ letters.

EXERCISE 16B 1

In each of the following two-way tables, find a

b

c

i



ii



i



ii



i



ii



letters.

2

The following two-way table describes the number of students in different year groups in a school.

a Complete the table. b Find the probability that a randomly selected student is a girl from Year 11. c Find the probability that a randomly selected girl is from Year 11. 3

The following two-way table describes the additions made to coffee in a drinks machine in one day.

a Find the probability of sugar being added. b Find the probability that sugar is added if milk is added. c Show that whether milk is added is independent of whether sugar is added. 4

The following table shows the number of returned shoes to three stores in one day.

a Find b Find c Find 5

The following table shows a general two-way table.

Find, in terms of

and :

a b c d e 6

The following table shows the results of the three top-performing countries in the 2016 Olympics.

A random result is chosen from amongst these a

results. Find:

b c d 7

A company makes three different sizes of T-shirt in three different colours. The following table shows the sales in a week.

a Find b Find c Find d e Find 8

The following table gives the probabilities of different midday temperatures in different air pressures in the United Kingdom based on long-term observations.

For a randomly chosen day, find: a b c 9

James investigates two identities: 1 2 Use a counterexample to show that one of these identities is incorrect.



Section 3: Tree diagrams A tree diagram is a useful way of illustrating situations where one outcome depends upon another. For example, the following diagram shows the experience of a restaurant trying to predict how many portions of chips it serves. Main course?

Chips? 7 8

Chips

1 8

No chips

0

Chips

1

No chips

1 5

Chips

4 5

No chips

Burger 2 5

1 2

Lasagne

1 10

Pie

The second, red probabilities all depend on the first, so they are conditional probabilities. For example,

You may already know that the probability of being on a branch is found by multiplying along the branch. For example, this means:

In general:

Key point 16.4

Tip This is actually just a rearrangement of Key point 16.3. If we wanted to find the overall probability of chips, we need to add different branches together:

WORKED EXAMPLE 16.6

Laila has a test at the end of each week for History. If she revises for a particular test, Laila estimates that there is an chance that she will pass. However, if she does not revise, she estimates only a chance of passing. The probability that Laila revises for any particular test is . What proportion of her History tests should Laila expect to pass? Decide which probability is not conditional. Start the tree diagram with

this event.

Revise?

The probability of passing the test is conditional on revision, so the revision branches have to come first.

R

3 4

1 4

R'

Revise?

Pass?

3 4

R

1 4

R'

80%

P

20%

P'

30%

P

70%

P'

Add the conditional event.

Identify which branches result in passing the test. Multiply to find the probability at the end of each branch.



Tip The questions in Worked examples 16.6 and 16.7 use the terms ‘chance’ and ‘proportion’. These are just other words for probability. Sometimes, you can use the information found from a tree diagram to find another conditional probability. WORKED EXAMPLE 16.7

If it is raining in the morning, there is a

chance that I will bring my umbrella. If it is not raining in

the morning, there is only a chance of me taking my umbrella. On any given morning the probability of rain is a What is the probability that I take my umbrella? b If you see me with my umbrella, what is the probability that it was raining that morning? a

First, draw a tree diagram.

Use the tree diagram to find relevant probabilities.

Since a conditional probability is required, use Key point 16.3.

b

In Worked example 16.7 we were given

and found

We can use Key point

16.4 to derive a formula for this. Key point 16.4 says that could also have written

but there is nothing special about and here. We

This leads to the following formula, which could be used to solve Worked example 16.7.

Tip Quite often in questions like this you use the answer to the first part in the second part.

Key point 16.5

Focus on … See Focus on … Proof 3 for the use of conditional probability and tree diagrams in analysing a claim of guilt in a legal case.

Explore This is related to a very important theorem in modern mathematics called Bayes’ theorem, named after the Reverend Thomas Bayes. It is a way of seeing how new information changes our knowledge about probabilities. Find out how this result may be used in hypothesis testing.

WORK IT OUT 16.1 The probability of an acorn landing more than from the original tree is . Of those that land more than from the tree germinate. Of those that land less than from the tree germinate. A seed germinates. What is the probability that it landed more than from the tree? Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1 We can work out P(germinate) from a tree diagram:

0.8

0.2

0.2

Germinate

0.8

Not germinate

> 2m

0.05

Germinate

0.95

Not germinate

< 2m

Solution 2

Solution 3 Using a Venn diagram:

Germinate

> 2m

x

0.8 - x

EXERCISE 16C 1

In the following questions, you may find that writing the information on a suitable tree diagram is helpful. a

b

c

2

i

and

ii

and

Find Find

i

and

Find

ii

and

Find

i

and

Find

ii

and

Find

A class contains six boys and eight girls. Two are picked at random. What is the probability that they are both boys?

3

A bag contains four red balls, three blue balls and two green balls. A ball is chosen at random from the bag and is not replaced. A second ball is chosen. Find the probability of choosing one green ball and one blue ball in any order.

4

and a Illustrate this information on a tree diagram. b Find c Find d Find

Worksheet See Support sheet 16 for a further example on conditional probability and tree diagrams, and for more practice questions.

5

Given that

and

, find:

a b 6

A factory has two machines making widgets. The older machine has larger capacity, so it makes of the widgets but are rejected by quality control. The newer machine has only a rejection rate. Find the probability that a randomly selected widget is rejected.

7

The school tennis league consists of players. Daniella has a chance of winning any game against a higher-ranked player, and a chance of winning any game against a lower-ranked player. If Daniella is currently in third place, find the probability that she wins her next game against a randomly selected opponent.

8

Box A contains six red balls and four green balls. Box B contains five red balls and three green balls. A standard fair cubic dice is thrown. If a ‘ ’ is obtained, a ball is selected from box A; otherwise a ball is selected from box B. a Calculate the probability that the ball selected is red. b Given that the ball selected is red, calculate the probability that it comes from box B.

9

Robert travels to work by train every weekday from Monday to Friday. The probability that he catches the weekday is

. train on Monday is . The probability that he catches the

. train on any other

. A weekday is chosen at random.

a Find the probability that he catches the b Given that he catches the Monday.

. train on that day.

train on that day, find the probability that the chosen day is

10 Bag contains six red cubes and ten blue cubes. Bag contains seven red cubes and three blue cubes. Two cubes are drawn at random, the first from bag and the second from bag . a Find the probability that the cubes are the same colour. b Given that the cubes selected are of different colours, find the probability that the red cube was selected from . 11 On any day in April there is a chance of rain in the morning. If it’s raining there is a chance I will remember my umbrella, but if it is not raining there is only a chance of remembering my umbrella.

a On a random day in April, what is the probability that I have my umbrella with me? b Given that I have my umbrella with me on a day in April, what is the probability that it was raining that morning? 12 A new blood test has been devised for early detection of a disease. Studies show that the probability that the blood test correctly identifies someone with this disease is , and the probability that the blood test correctly identifies someone without that disease is The incidence of this disease in the general population is The result of the blood test on a particular patient indicates that he has the disease. What is the probability that this patient has the disease? 13 You have two coins: one is a normal fair coin with heads on one side and tails on the other. The second coin has heads on both sides. You randomly pick a coin and flip it. The result comes up heads. What is the probability that you chose the fair coin? 14 There are discs in a bag. Some of them are black and the rest are white. Two are simultaneously selected at random. Given that the probability of selecting two discs of the same colour is equal to the probability of selecting two discs of different colour, how many black discs are there in the bag? 15 Prove that if and are independent, then so are



and

.

Section 4: Modelling with probability When answering probability questions, you often need to make various assumptions about the situation you are modelling; for example, that dice are fair, that each counter is equally likely to be picked, or that the probability of rain doesn’t change from one day to the next. In this section we look at some of the most common modelling assumptions and ask how justified they are, and how changing them would affect the result of our calculations. When you first met probability, you learnt to calculate it as the number of favourable outcomes divided by the number of all possible outcomes. For example, the probability of selecting a yellow ball from a bag containing five yellow and eight green balls is

. This calculation assumes that each ball is equally likely

to be picked. WORKED EXAMPLE 16.8

A bag contains replacement.

blue counters and nine yellow counters. Two counters are picked at random without

a Find the probability that both counters are blue. b Identify one assumption you have made in your calculation. Is this assumption justified? c If the blue counters were slightly larger than the yellow counters, how would that change the answer in part a? a

You can draw a tree diagram, or think in terms of conditional probabilities.

b We have assumed that each of the counters is equally likely to be picked. This assumption may be justified if all the counters are exactly the same shape and size. c The answer would be larger because larger counters would be more likely to be picked.

The probabilities

and

are based on the

assumption that each counter is equally likely to be picked.



Another common assumption in probability calculations is that probabilities don’t change over time, for example, if you have dice that are fair today, they will also be fair tomorrow. When considering the probability of two events happening simultaneously, or one after the other, it is also sometimes appropriate to assume that they are independent. WORKED EXAMPLE 16.9

a A simple weather model assumes that the probability that it rains on any given day is , independently of any other day. i

Find the probability that it rains on two consecutive days.

ii

Comment on the suitability of the two assumptions made in this model.

b In an improved model, the probability that it rains on the first day is still , but the probability that it

rains on the second day is if it rained on the first day and if it didn’t. i

Is the probability that it rains on two consecutive days larger or smaller than for the first model?

ii

The probability that it rains on the second day is still . Find the value of . The consecutive days are independent so you can just multiply the probability of rain for each.

a i

ii

The probability of rain will be different at different times of year; however, it won’t change so much from one day to the next, so the assumption that the probability is on both days

Is it reasonable to assume that the probability of rain is the same for each day?

is reasonable. If it rains on one day it is more likely that it will rain the next day, so the assumption of independence is probably not suitable. First day

b

Second day 1 2

1 5

Rain

4 5

No rain

No rain

q

Rain

1- q

This question seems a bit more complicated so draw a tree diagram.

Rain

1 2

Does rain on one day make it more or less likely that it will rain the next day?

No rain

i This is larger than predicted by the first model. ii

Multiply the probabilities along the two branches. There are two sets of branches corresponding to rain on the second day. This probability needs to equal .

Notice that in part b the probability of rain is the same on both days, but the two days are not independent:

, but

.

Focus on … See Focus on … Modelling 3 you will also consider modelling assumptions when using the normal distribution.

Rewind In Student Book 1, Chapter 17 you learnt that to use the binomial distribution, trials must be both independent and have constant probability of success. The Mixed practice at the end of this chapter contains questions about modelling with probability.

Checklist of learning and understanding

We can use set notation when describing probabilities: is the intersection of and , meaning when both and happen. is the union of and , meaning when either happens, or happens, or both happen. is the complement of , meaning everything that could happen other than . A Venn diagram leads to a formula relating the probabilities of the union and intersection: is the probability of happening if has happened. This can be visualised using Venn diagrams, two-way tables or tree diagrams. We can also use the formula:

In a tree diagram we often rearrange this formula to get: and

Mixed practice 16 1

A drawer contains six red socks, four black socks and eight white socks. Two socks are picked at random. a What is the probability that two socks of the same colour are drawn? b What assumptions have you made in calculating the probability?

2

Out of

flies studied in an experiment,

have the gene

have the gene

and

flies have neither gene. a Illustrate this information on a Venn diagram. b Find c Find 3

In an event,

and

a Illustrate this information on a tree diagram. b Find c Find d Find 4

The number of hours spent practising for a music examination is recorded by a teacher for each of her students. Hours practised

Male

Female

or fewer to or more a Find P(male |

or fewer hours practised).

b Two different students are randomly selected for further interviews. Find the probability that both are male. 5

The probability that a student plays badminton is . The probability that a student plays neither football nor badminton is , and the probability that a student plays both sports is . a Draw a Venn diagram showing this information. b Find the probability that a student plays badminton but not football. Given that a student plays football, the probability that they also play badminton is c Find the probability that a student plays both badminton and football. d Hence, complete your Venn diagram. What is the probability that a student plays only badminton? e Given that a student plays only one sport, what it the probability that they play badminton?

6

Only two international airlines fly daily into an airport. Pi Air has Air has

flights a day. Passengers flying with Air have a

flights a day and Lambda

probability of losing their

luggage, and passengers flying with Lambda Air have a probability of losing their luggage. Someone complains that their luggage has been lost. Find the probability that they travelled with Air. 7

A girl walks to school every day. If it is not raining, the probability that she is late is . If it is raining, the probability that she is late is . The probability that it rains on a particular day is . On a particular day the girl is late. a Find the probability that it was raining on that day. b Is the assumption of constant probability reasonable?

8

The table shows the number of male and female members of a vintage car club who own either a Jaguar or a Bentley. No member owns both makes of car.

Male

Female

Jaguar Bentley One member is chosen at random from these i

members.

Given that this member is male, find the probability that he owns a Jaguar.

Now two members are chosen at random from the

members. They are chosen one at a

time, without replacement. ii

Given that the first one of these members is female, find the probability that both own Jaguars. © OCR, GCE Mathematics, Paper 4732, January 2010

9

A game uses an unbiased dice with faces numbered to . The dice is thrown once. If it shows or or then this number is the final score. If it shows or or then the dice is thrown again and the final score is the sum of the numbers shown on the two throws. i

Find the probability that the final score is .

ii

Given that the dice is thrown only once, find the probability that the final score is .

iii Given that the dice is thrown twice, find the probability that the final score is . © OCR, GCE Mathematics, Paper 4732, January 2009 10 The probability that Thomas leaves his umbrella in any shop he visits is . After visiting two shops in succession, he finds he has left his umbrella in one of them. What is the probability that he left his umbrella in the second shop? 11 a A large bag of sweets contains eight red and twelve yellow sweets. Two sweets are chosen at random from the bag without replacement. i

Find the probability that two red sweets are chosen.

ii

What assumption is made in your calculation for part a i?

b A small bag contains red and yellow sweets. Two sweets are chosen without replacement from this bag. If the probability that two red sweets are chosen is

, show

that Ayesha has one large bag and two small bags of sweets. She selects a bag at random and

then picks two sweets without replacement. c Calculate the probability that two red sweets are chosen. d Given that two red sweets are chosen, find the probability that Ayesha selected from the large bag. 12 Two events, and , satisfy

.

a Find the possible values can take. b Find

in terms of .

13 The probability that it rains during a summer’s day in a certain town is . In this town, the probability that the daily maximum temperature exceeds is when it rains and when it does not rain. Given that the maximum daily temperature exceeded on a particular summer’s day, find the probability that it rained on that day. 14 Given that

and

, find

Worksheet See Extension sheet 16 for a selection of more challenging problems.

17 The normal distribution In this chapter you will learn: how to calculate probabilities for a normally distributed random variable that any normal distribution is related to the standard normal distribution how to calculate the value of the variable with a given cumulative probability how to find mean and standard deviation from information about probabilities how the normal distribution can be used as a model how the normal distribution can be used as an approximation to the binomial distribution.

Before you start… Student Book 1, Chapter 16

You should be able to interpret

1 What is the frequency density of a group of masses strictly between and ?

people with

histograms.

Chapter 16

You should be able to work with tree diagrams.

2 There is a chance of it raining. If it rains, there is a chance I am late. If it does not rain, there is a chance I am late. What is the probability that I am late?

Chapter 16

You should be able to calculate conditional probability.

3 Two dice are rolled: one red and one blue. If the total score is , what is the probability that the score on the red dice is ?

Student Book 1, Chapter 17

You should be able to use the binomial distribution.

4 The probability of rolling a on a biased dice is .

GCSE

You should be able to solve simultaneous equations.

5 Solve:

If the dice is rolled four times, find the probability of getting exactly one .

What is the normal distribution? In Student Book 1, Chapter 17 you met discrete random variables, which you could describe by listing all possible values and their probabilities. With a continuous variable, such as height or time, listing all values is impossible. You need a different way to describe how the probability is distributed across the possible values.

In Student Book 1, Chapter 16 you used histograms to represent continuous data. On a histogram, frequency is represented by the area of a bar, and the -axis shows frequency density. A similar idea can be used to describe the probability distribution of a continuous random variable. You can draw a curve such that the area under the curve represents probability. This curve is called the probability density function. There are many situations where a variable is very likely to be close to its average value, with values farther away from the average becoming increasingly unlikely. Many such situations can be modelled using the normal distribution. Natural measurements, such as heights of people or masses of animals, tend to follow a normal distribution. 2.0 1.5 1.0 0.5 0

90

140

190

Did you know? The normal distribution is not the only probability distribution that follows this ‘bell shape’. However, there are mathematical reasons why this particular distribution is a good model for many naturally occurring measurements. You will learn more about this if you study the Statistics option in Further Maths.

Section 1: Introduction to normal probabilities To specify a normal distribution fully you need to know its mean this distribution you use the notation .

and variance

. If a variable follows

X ~ N(µ, σ2)

µ – 2σ

µ – σ

µ

µ + σ

µ + 2σ

x

Tip Be careful with the notation:

is the variance, so

has standard deviation

.

The mean corresponds to the maximum point on the graph of the probability density function, and the standard deviation affects the width of the curve. Remember that the probability corresponds to the area under the graph. You know that the area under a curve can be found by integration. However, the expression for the normal distribution curve cannot be integrated exactly, so most calculators have a built-in function to find approximate probabilities. Different calculators may have slightly different ways of finding normal probabilities. Make sure you are familiar with your model. You usually need to enter the mean and standard deviation (not variance!). Some models can find the probability between any two values. For example, to find , where , you need to enter (in the correct order): , To find

,

,

you need to enter a negative number (e.g.

) for the lower bound.

Some models can find only cumulative probabilities; that is, probabilities of the form models, you would need to find as . You should remember that, since the normal distribution is continuous, .

. On those

is the same as

You may find it helpful to sketch a diagram and shade the area corresponding to the probability you are trying to find. X ~ N(5, 1.2)

5 6 P(X < 6)

X ~ N(20, 4)

x

18.5 20 21 P(18.5 < X < 21)

X ~ N(100, 64)

x

100 110 P(X > 110)

x

Tip With a continuous variable, and mean exactly the same thing, so you don’t have to worry about whether end points should be included. This is not the case for discrete variables! The diagram can also provide a useful check, as you can see whether to expect the probability to be

smaller or greater than

.

WORKED EXAMPLE 17.1

The height of people in a town can be modelled by a normal distribution with mean standard deviation

and

. Give the probability that a randomly selected resident:

a is less than

tall

b is between

and

tall.

is the ‘height of a town resident’ so

State the distribution used.

. a

X ~ N(170, 100)

165170 P(X < 165)

Sketch a normal distribution curve and shade the required area.

x

State the probability to be found and use the calculator.

(from calculator) b

X ~ N(170, 100)

170 180 190 P(180 < X < 190)

x

(from calculator) Note that the domain of the normal probability density function is all real numbers; the -axis is an asymptote to the graph. This means that, in theory, a normal variable could take any real value. However, most of the data lie within three standard deviations of the mean. It is useful to remember the following percentages.

Key point 17.1 Approximately of the data lie within three standard deviations of the mean. Approximately of data lie within two standard deviations of the mean. Approximately two-thirds of the data lie within one standard deviation of the mean. It turns out that the points of inflection of the normal distribution curve lie one standard deviation from the mean. This can be used to estimate the standard deviation.

Rewind You met points of inflection in Chapter 12.

PROOF 10

The curve corresponding to the normal distribution with mean zero and standard deviation is . Prove that the points of inflection of

occur at

.

To find a point of inflection we will use the fact that

Start by differentiating, using the chain rule.

The second derivative needs both the chain rule and the product rule.

At a point of inflection,

The best way to solve complicated equations is to factorise.

Factorising:

Since

EXERCISE 17A

so:

,

EXERCISE 17A 1

Shade the appropriate section of the normal distribution and find the following probabilities. a If

:

i ii b If

:

i ii c if

:

i ii d If has a normal distribution with mean

and standard deviation :

i ii e If has a normal distribution with mean

and standard deviation

:

i ii 2

A normal curve has points of inflection with -coordinates and deviation of this distribution.

3

The curve in diagram 1 represents a normal distribution. Copy the diagram and mark the approximate position of the points of inflection. Hence, estimate the mean and standard deviation of the normal

. Find the mean and standard

distribution. y

O

2 4 6 8 10 12 14 16 18 20 22 24 26

x

Diagram 1 y

O

2 4 6 8 10 12 14 16 18 20 22 24 26

x

Diagram 2 4

Estimate the mean and standard deviation of the normal distribution represented by the curve in diagram 2.

Using Z-scores and the standard normal distribution If a normally distributed random variable has mean , should a value of be considered unusually large? The answer depends on how spread out the variable is, and this is measured by its standard deviation. If the standard deviation is , then a value around will be quite common. However, if the standard deviation were , then

would indeed be very unusual.

It turns out that, for a normally distriuted random variable, the cumulative probability depends only on the number of standard deviations is away from the mean. This is called the -score.

Key point 17.2 For the Z-score measures the number of standard deviations away from the mean:

WORKED EXAMPLE 17.2

Given that

:

a How many standard deviations is b Find the value of that is

away from the mean?

standard deviations below the mean. The number of standard deviations away from the mean is measured by the -score.

a

is the variance, so take its square root to get the standard deviation. is

standard deviations

away from the mean. b

Values below the mean have a negative -score.

Starting with a random variable

you can create a new random variable, , which takes the

values equal to the -scores of the values of . It turns out that, whatever the original mean and standard deviation of , this new random variable always has normal distribution with mean and variance , called the standard normal distribution: .

Key point 17.3 If

and

then the probabilities of and are related by:

Fast Forward This is an extremely important property of the normal distribution and needs to be used in situations when the mean and standard deviation of are not known (see Section 3).

Did you know? Because of their importance, cumulative probabilities of the standard normal distribution are given special notation: Although you do not have to use this notation, you should understand what it means.

WORKED EXAMPLE 17.3

Let a

. Write the following in terms of probabilities of .

b a

We are given that

, so we can calculate .

b

We now have two x values, so find the corresponding value for each of them.

Questions on the normal distribution can be combined with other probability facts. In particular, watch out for questions that bring in conditional probability or the binomial distribution.

Rewind The binomial distribution was covered in Student Book 1, Chapter 17. Conditional probability was covered in Chapter 16.

WORKED EXAMPLE 17.4

The masses of fish caught by a trawler follow a normal distribution with mean deviation . A juvenile fish is classified as one having a mass less than .

and standard

a What is the probability that a randomly chosen fish is a juvenile? b If fish are caught, what is the probability that there are more than ten juveniles? State any assumptions you have to make, and comment on their validity. c Given that a fish is a juvenile, what is the probability that it has a mass more than a Let be ‘mass of a fish’ so

?

.

from calculator. b Let be the ‘number of juveniles’ so that

.

To use the binomial distribution we need to assume that the probability of catching a juvenile does not change. This may be valid if the

fish are caught in a short period of

time and in the same location; it may not be valid if they are caught over a period of several months. We also need

It is useful to write down the names of all the random variables to make it obvious which distribution you are using. Y has a binomial distribution, with the probability of ‘juvenile’ being the one found in part a. Use unrounded values in your calculations.

to assume that one fish being a juvenile is independent of other fish. This may not be valid, as fish of similar age may stay in a group. c

This is a question about conditional probability. It is useful to write the question in terms of probability statements involving the variables defined, then use the formula. The best way to find the intersection is to think about what it means: if is bigger than and less than , it is between and . It is very common in this type of question to use some of the probabilities already calculated.

EXERCISE 17B

1

Find the -score corresponding to the given value of . a

i ii

b

i ii

2

Given that a

, write the following in terms of probabilities of the standard normal variable.

i ii

b

i ii

c

i ii

3

It is found that the lifespan of a certain brand of laptop batteries follows normal distribution with mean hours and standard deviation . A particular battery has a lifespan of . a How many standard deviations below the mean is this? b What is the probability that a randomly chosen laptop battery has a lifespan shorter than this?

4

When Ali competes in long jump competitions, the lengths of his jumps are normally distributed with mean

and standard deviation

.

a What is the probability that Ali will record a jump between b Ali needs to jump

and

?

to qualify for the school team.

i What is the probability that he will qualify with a single jump? ii If he is allowed three jumps, what is the probability that he will qualify for the school team? c What assumptions did you have to make in your answer to b ii? Are these likely to be met in this situation? 5

Masses of a species of cat have a normal distribution with mean and variance the number in a sample of such cats that will have a mass above .

6

The

. Estimate

time of a group of athletes can be modelled by a normal distribution, with mean

seconds

and standard deviation seconds. a Find the probability that a randomly chosen athlete will run the

in under

seconds.

b Show that, if the binomial distribution can be used, the probability that all four athletes in a team run under

seconds is

.

c Miguel says that this means that the probability of the team breaking the school record of minutes and seconds is only . Give three reasons why this is likely to be incorrect. Do you think the real value will be greater or less than ? 7

If a

, find:

b

8

c



If

, find:

a

b 9

The masses of apples are normally distributed with mean mass Supermarkets classify apples as ‘medium’ if they are between

and standard deviation and .

.

a What proportion of apples are medium? b In a bag of ten apples, what is the probability that there are at least eight medium apples? 10 The wingspans of a species of pigeon are normally distributed with mean length deviation

and standard

. A pigeon is chosen at random and its wingspan measured.

a Find the probability that its length is greater than b Given that its length is greater than

.

, find the probability that its length is greater than

11 Grains of sand are believed to have a normal distribution with mean

and variance

a Find the probability that a randomly chosen grain of sand is larger than b The sand is passed through a filter that blocks grains wider than

. .

.

. The sand that passes

through the filter is examined. What is the probability that a randomly chosen grain of filtered sand is larger than ? 12 The amount of paracetamol per tablet is believed to be normally distributed with mean standard deviation

. A dose of less than

and

is ineffective in dealing with toothache. In a trial

of people suffering toothache, what is the probability that two or more of them have less than the effective dose? 13 A variable has a normal distribution with a mean that is seven times its standard deviation. What is the probability of the variable taking a value less than five times the standard deviation? 14 If

and

find

in terms of .

Section 2: Inverse normal distribution In Section 1 you saw how to find probabilities when you knew information about the variable. In real life it is often useful to work backwards from probabilities to estimate information about the data. This requires the inverse normal distribution.

Key point 17.4 For a given value of probability , the inverse normal distribution gives the value of such that

Note that many textbooks use the notation mentioned in the previous section to write inverse normal distribution: If , then

.

Fast Forward You will use inverse normal distribution to find critical values for a hypothesis tests in Chapter 18. To find inverse normal values on your calculator, you usually need to enter the value of , the mean and the standard deviation. Make sure you know the correct order to use on your calculator. WORKED EXAMPLE 17.5

The length of men’s feet is thought to be normally distributed with mean and variance . A shoe manufacturer wants only of men to be unable to find shoes large enough for them. How big should their largest shoe be? If is the continuous random variable ‘length of a man’s foot’ then .

Convert the question into mathematical terms.

You need to find the value of such that

Use the inverse normal distribution. You may have to convert into a probability of the form in order to use the calculator. So their largest shoe must fit a foot long.

EXERCISE 17C 1

a The random variable follows the normal distribution

. Find if:

i ii b The random variable follows the normal distribution

. Find if:

i ii c The random variable follows the normal distribution i

. Find if:

ii 2

IQ tests are designed to have a mean of to be in the top

3

Rabbits’ masses are normally distributed with an average mass of vet decides that the top

4

and a standard deviation of

and a variance of

and

.

a Calculate the probability that the machine dispenses less than b Find the value of if

. A

of rabbits are obese. What is the minimum mass for an obese rabbit?

The amount of coffee dispensed by a machine follows the normal distribution with mean standard deviation

5

. What IQ score is needed

of IQ scores?

of cups contain more than

of coffee.

of coffee.

The times taken for students to complete a test are normally distributed with a mean of a standard deviation of .

and

a Find the probability that a randomly chosen student completes the test in less than b

.

of students complete the test in less than minutes. Find the value of .

c A random sample of eight students has their completion time for the test recorded. Find the probability that exactly two of these students complete the test in less than

.

6

An old textbook says that the range of data can be estimated as six times the standard deviation. If the data are normally distributed, what percentage of the data are within this range?

7

The time taken to do a maths question can be modelled by a normal distribution with mean and standard deviation . The probability of two randomly chosen questions both taking longer than seconds is . Find the value of .

8

The concentration of salt in a cell, , can be modelled by a normal distribution with mean and standard deviation . Find the value of such that .

9

For a normal distribution, find the following ratios. a b

10 Evaluate , where standard normal distribution. 11 If

and

, for

is the inverse normal distribution function for the

, find

in terms of

.

12 Most calculators have a random number generator that generates random numbers from to . These random numbers are uniformly distributed, which means that the probability is evenly spread over all possible values. How can you use these to form random numbers drawn from a normal distribution?

Section 3: Finding unknown or One of the main applications of statistics is to determine parameters of the population when given information about the data. But how can you use the normal distribution calculations if the mean or the standard deviation is unknown? This is where the standard normal distribution comes in useful: you can replace all the values by their -scores, as they follow a known distribution, .

Tip This will involve solving equations, and sometimes simultaneous equations. As the numbers are usually not ‘nice’, you may want to use your calculator.

WORKED EXAMPLE 17.6

Random variable follows a normal distribution with standard deviation . An experiment estimates that . Estimate the mean of , correct to two significant figures. You may need the probability to be in the form If

.

Since you don‘t know , convert the probability into information about . Use inverse normal distribution to find . Relate to the given value.

WORKED EXAMPLE 17.7

The masses of gerbils are thought to be normally distributed. If of gerbils have a mass more than and have a mass less than , estimate the mean and the variance of the mass of a gerbil. If is ‘mass of a gerbil’, then

Convert the information into mathematical terms.

If you need all the probabilities to be in the form From (1):

Use the inverse normal distribution for and relate it to the given values.

From (2):

Solve the simultaneous equations.

(4)-(3) gives:

WORK IT OUT

17.1 If

and

, find the value of .

Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1 so, from the calculator, the corresponding -score is Therefore,

Solution 2

Therefore, So

Solution 3

EXERCISE 17D

.

EXERCISE 17D 1

a If

, find when:

i ii b If

, find when:

i ii 2

If a

b

3

, find and when: i

and

ii

and

i

and

ii

and

A manufacturer knows that his machines produce bolts whose diameters follow a normal distribution with standard deviation diameter greater than

. He takes a random sample of bolts and finds that . Find the mean diameter of the bolts.

of them have

4

The energy of an electron can be modelled by normal distribution with mean and standard deviation . If of electrons have an energy above , find the value of .

5

It is known that the heights of a certain plant follow a normal distribution. In a sample of are less than of the heights.

6

tall and

are more than

The time taken for a computer to start is modelled by a normal distribution. It is tested on of these it takes longer than seconds. On of the tests it takes less than Estimate the mean and standard deviation of the start-up times.

7

The actual voltage of a brand of

plants,

tall. Estimate the mean and the standard deviation

times and

seconds.

battery is thought to be normally distributed with standard

deviation and mean , where is the time, in hours, that the battery has been used. The batteries can no longer power a lamp when they drop below . In one batch of batteries only can power the lamp. Assuming that the model is correct, estimate how long the batteries have been used (assume that they were all used for the same length of time). 8

A scientist noticed that of temperature measurements were lower than the average. Assuming that the measurements follow a normal distribution, estimate the standard deviation.

9

The waiting time for a train is normally distributed with mean time is over . Find the probability that a person waits over

.

of the time the waiting on exactly two out of

three times they wait for the train. 10 The random variable models the temperature in an oven in . It follows a normal distribution, where the mean value is the temperature set on the oven. The probability of being within of the temperature set is . Find the probability of being within of the temperature set.

Section 4: Modelling with normal distribution Many real-world situations can be modelled using a normal distribution, such as the heights of people, masses of gerbils or error in experimental measurements. However, you should not just assume that a variable follows a normal distribution. There are various useful checks that you can use to help you.

Focus on … In Focus on … Modelling 3 you can investigate further examples of using the normal distribution as a model.

Key point 17.5 To decide whether a normal distribution is an appropriate model: When you look at a histogram you would expect the distribution to be approximately symmetrical with only one mode and no sharp cut-off. When you are looking at summary statistics you would expect nearly all of the data to fall within three standard deviations of the mean.

Fast forward If you study the Statistics option of Further Mathematics, you will find out about a statistical test called the chi-squared test, which allows us to decide more precisely whether data come from a normal distribution.

WORKED EXAMPLE 17.8

For each of the following histograms, explain why the normal distribution is not a good model for the data. a

1.4

Mass of mice

1.2 Number

1.0 0.8 0.6 0.4 0.2 0.0 90

b

190

Results in a test

3

Number

140 Mass (g)

2 1 0

0

50 Result (%)

100

a There appear to be two modes. b These data do not seem to be symmetrical and there is a sharp cut-off at the top end because it is

not possible to score over

.

WORKED EXAMPLE 17.9

In a sample, it is found that the mean time taken to complete a puzzle is with a standard deviation of . Explain why the normal distribution would not be an appropriate model to predict the time taken to complete this puzzle. There is a cut-off at zero seconds, which is

standard

deviations below the mean. The normal distribution would therefore predict that a significant number of people complete the puzzle in a negative amount of time, which is not possible.

With only the mean and standard deviation provided (no graph of the distribution), consider whether virtually all the data fall within standard deviations of the mean.

Many other statistical distributions can be approximated by the normal distribution. Historically, this was very important because calculations with the normal distribution were often much easier. For example, consider these bar charts for the binomial distribution.

0.3

X ~ B(100, 0.1)

0.20

X ~ B(100, 0.02) Probability

Probability

0.15

0.2 0.1 0

0

5 x

10

0.10 0.05 0

0

5

10 x

15

20

The one on the left has a sharp cut-off at and is not symmetrical, but the one on the right appears to have roughly the shape of a normal distribution. It turns out that it can indeed be approximated by a normal distribution, although the proof of this is beyond the scope of this book.

Key point 17.6 If with distribution with µ

and and

(where .

), then can be approximated by a normal

Explore Because the binomial distribution deals with discrete variables and the normal distribution deals with continuous variables, technically we should use something called a continuity correction. This says that in the binomial distribution is equivalent to in the normal distribution. You do not need to do this within this course, but you may want to find out about it and use it to check how good this approximation is.

WORKED EXAMPLE 17.10

A fair dice is rolled

. The random variable models the number of rolled.

a Explain why follows a binomial distribution. b The random variable can be approximated by a normal distribution. Find the mean and variance of this normal distribution. What properties of make this approximation valid? c Let be a random variable with the normal distribution from part b. Find d Find

. Explain why this is not the same as your answer to part c.

.

a There is a fixed number of rolls. The rolls are independent, with constant probability of rolling a ‘ ’. We can classify the outcomes into ‘ ’ or ‘not ’. b

and

, so

Recall the conditions for when using a binomial distribution is appropriate.

Use the fact that N( , ) is the approximate distribution.

and

. So the approximate normal distribution is . and , which are both larger than . This means that the binomial distribution will be reasonably symmetrical. c For

and

.

State the distribution of and use the calculator to find probability.

,

d For

Check whether

Use your a calculator to find the probability for the original binomial distribution.

,

and don’t have exactly the same distribution: is discrete and is continuous.

The normal distribution is only an approximation to the binomial, so the probabilities won’t be exactly the same.

In binomial hypothesis tests, when the value of is large the normal approximation is useful because the exact binomial probabilities can be difficult to find.

Rewind You met hypothesis testing using binomial distribution in Student Book 1, Chapter 18. Remember that the significance level gives the probability of incorrectly rejecting the null hypothesis.

Fast forward You will learn more about hypothesis testing using normal distribution in the next chapter.

WORKED EXAMPLE 17.11

A survey of a random sample of people in a large city is conducted in order to test a hypothesis about the proportion of people who cycle to work. In this sample, people cycle to work. The test conclude that there is evidence, at the significance level, that this proportion is less than . Use an appropriate normal distribution to estimate the maximum possible value of . This is a hypothesis test for the proportion of the binomial distribution. So start by stating the distribution and the hypotheses.

Let be the number of people in a sample of who cycle to work. Then

.

The hypotheses are:

We need such that: . Approximate normal distribution:

is the critical value for this test. The significance level is . We can use normal approximation because .

So Use inverse normal distribution to find the value of .

Using this normal distribution,

0.05

220 240 260 280 300 320 340 360 380

So

x

is the number of people who cycle to work, so needs to be a whole number.

.

EXERCISE 17E 1

For each of the following histograms, decide if the data could be modelled by a normal distribution. If it cannot, give a reason. a

2.0

Mass of mice

1.5 1.0 0.5 0.0 90

b

2.0

140

190

Time in a race

1.5 1.0 0.5 0.0 90

c

2.0

140

190

GDP of a country

1.5 1.0 0.5 0.0 90

190

Error in a measurement 70 60 50 40 30 20 10

d

–1.0

2

140

–0.5

0.5

1.0

The mean number of children in a family in the UK is

with a standard deviation of

. Use these

figures to explain why a normal distribution would not be a good model of the number of children in a family in the UK. 3

A psychology student asks people to estimate the value of an angle. Her results are summarised in the histogram below.

Estimate

24 20 16 12 8 4 0

40

50

60

a Copy and complete the frequency table below. Angle Frequency b Hence, estimate the mean and standard deviation of the data. c What features of the graph suggest that a normal distribution might be an appropriate model? d The student compares these data estimates to another group of people. If they follow the same normal distribution, how many would you expect to estimate over ? 4

A fair coin is tossed

times.

a Find the probability that there are more than i the binomial distribution

heads using:

ii the normal approximation. b Find the percentage error in using the normal distribution in this situation. 5

The number of people voting for a particular party in an election can be modelled by a binomial distribution. In the whole population of eligible voters, the probability of a person voting for this is . In a particular election, a random sample of voters actually vote. Use a normal approximation to the binomial to find the probability that this party wins a majority (more than of the vote) if: a b

6

Data collected over a long period of time indicate that of children contract a certain disease. Following a public awareness campaign, a doctor conducts a survey to find out whether this proportion has decreased. She uses a random sample of children and conducts a hypothesis test at the significance level. Use an appropriate normal distribution to find the approximate critical region for this test.

7

a Prove that if

and

, then

. Show, using a counterexample, that the reverse

is not true. b A binomial distribution 8

has a probability of

of being above

. Find the value of .

The random variable is approximated by the normal distribution . Prove using this approximation that if , then all values of within three standard deviations of the mean are positive.

Checklist of learning and understanding The normal distribution models many physical situations. It is described completely once we know its mean and its variance . Calculators can provide the probabilities of being in

any given range. The following are useful values to know: Approximately Approximately

of the data lie within three standard deviations of the mean. of data lie within two standard deviations of the mean.

Approximately two-thirds of the data lie within one standard deviation of the mean. The Z-score is the number of standard deviations above the mean that has a given cumulative probability. It is related to the original variable by the equation: . If we know probabilities relating to a variable with a normal distribution we can deduce information about the variable using the inverse normal distribution. We need to use the Z-score when the mean or the standard deviation are unknown. If with distribution with µ

and and

(where .

), then can be approximated by a normal

Mixed practice 17 1

a The test scores of a large group of students can be modelled by a normal distribution with mean and variance . Find the percentage of students with scores above . b What is the lowest score achieved by the top

2

of these students?

The masses of kittens can be modelled by a normal distribution with mean deviation . a Out of a group of b

and standard

kittens, how many would be expected to have a mass of less than

of kittens have a mass of more than

?

. Determine the value of .

3

The random variable is normally distributed with a mean of and standard deviation of By sketching a normal curve, or otherwise, find the value of such that

4

The masses, such that

5

Tennis balls are dropped from a standard height, and the height of the bounce, measured. is a random variable with the distribution . It is given that .

, of babies born at a certain hospital satisfy of the babies have a mass between and

. .

. Find the value of , where . , is

i

Find the value of

ii

tennis balls are selected at random. Use an appropriate approximation to find the probability that more than have © OCR, GCE Mathematics, Paper 4733, June 2010

6

The adult female of a breed of dog has average height

with variance

. If the

height follows a normal distribution, find the probability that in six independently selected dogs of this breed exactly four are above tall. 7

Heights of trees in a forest are distributed normally with mean . a Find the probability that a tree is more than b What is the probability that among tall?

8

and standard deviation

tall.

randomly selected trees at least two are more than

It is known that the scores on a test can be modelled by a normal distribution the scores are above and of the scores are below . a Show that µ

µ

.

of

.

b By writing another similar equation, find the mean and the standard deviation of the scores. 9

people are asked to estimate the size of an angle. give an estimate that is less than and give an estimate that is more than . Assuming that the data can be modelled by a normal distribution, estimate the mean and the standard deviation of the results.

10 The continuous random variable has the distribution i

µ

.

Each of the three following sets of probabilities is impossible. Give a reason in each case why the probabilities cannot both be correct. (You should not attempt to find µ and .)

a

and

b

and

c

and

ii Given that

and

, find the values of µ and . © OCR, GCE Mathematics, Paper 4733, January 2010

11

of students in a university are female. The discrete random variable , the number of female students in a group of size , is assumed to follow a binomial distribution. a Explain why, if is large, the binomial distribution can be approximated by the normal distribution, and state its parameters. b A dancing club contains

students. Assuming the binomial distribution is valid, use the

normal approximation to find the probability that more than

are female.

c Are the conditions for the binomial distribution met in this situation? Explain your answer. 12 The results of an examination have a mean of of .

, a median of

and a standard deviation

a Explain why a normal distribution could be a plausible model for these data. Grades are awarded in the following way. The top

get a distinction.

The next

get a merit.

The next

get a pass.

The remaining people get a fail. b Assuming a normal model, find the grade boundaries for this examination. 13 A company makes a large number of steel links for chains. They know that the force required to break any individual link is modelled by a normal distribution with mean . The company tests chains consisting of four links. If any link breaks, the chain will break. A force of is applied to all of the chains and break. a Estimate the probability of a single link breaking. b Hence, estimate the standard deviation in the breaking strength of the links. 14 a

of sand from the beach Playa Gauss falls through a sieve with gaps of

, but

passes through a sieve with gaps of . Assuming that the diameters of grains of sand are normally distributed, estimate the mean and standard deviation of the sand diameters. b



of sand from Playa Fermat falls through a sieve with gaps of . of this filtered sand passes through a sieve with gaps of . Assuming that the diameters of grains of sand are normally distributed, estimate the mean and standard deviation of the sand diameters.

18 Further hypothesis testing In this chapter you will learn: that the sample mean is a random variable how the sample mean is distributed how to test whether the mean of a normally distributed population is different from a predicted value how to test whether a set of bivariate data provides evidence for significant correlation.

Before you start… Student Book 1, Chapter 16

You should be able to interpret correlation coefficients.

1 Information on height, weight (mass), waist size and average time spent exercising per week is recorded from a random sample of adult males. Match the values of the product moment correlation coefficients with each of the sets of variables. A Height and weight B Height and time spent exercising C Waist measurement and time spent exercising 1 2 3

Student Book 1, Chapter 18

You should be able to conduct hypothesis tests using the binomial distribution.

2 A dice is rolled ten times and four are obtained. It is claimed that the dice is biased in favour of getting a . Test this claim at the level.

Chapter 17

You should be able to perform calculations using the normal distribution.

3

. Find: a b c

such that

Testing means and correlation coefficients A cereal packet claims that it weighs (has mass of) on average

. A sample of

packets contains a

mean of of cereal. Is this evidence that the company is systematically underfilling the packets? Intuition suggests probably not. You would not expect the mean of every sample to be exactly and the result seems to be reasonably close. You may wonder how far below

the mean would have to be

before there was significant evidence. To answer questions like this you can use a hypothesis test. In this chapter you will look at two different types of hypothesis tests. The first is a test to see if the mean of a sample is very different from a predicted value, such as in the previous example. To do this we first need to establish some theory about the distribution of sample means. In Student Book 1, Chapter 16 you saw that correlation coefficients could be used to describe the strength of correlation, but it was not clear how big the coefficient needed to be to have significant evidence of correlation. In Section 3 you will see how hypothesis tests can be used to decide if the sample correlation coefficient provides evidence for correlation in the population.

Section 1: Distribution of the sample mean If data follow a normal distribution then every observation is a random variable, meaning that an observation can take a different value each time. The histogram below shows samples taken from a distribution.

O

- 10

10

20

30

Instead of looking at a single value, you can look at a sample of independent observations and calculate a mean. This might take a different value every time you do it, so it is also a random variable and is given the symbol or simply . For the distribution above, you can take lots of samples of size and create a histogram of their mean, shown in dark blue in the diagram below.

O

- 10

10

20

30

Tip You might like to use technology to see if you can create a similar histogram. There are two important things to note about the sample means: They are clustered around the same mean as the original data. They are less spread out. It can be shown that the sample mean of independent observations from a normal distribution also follows a normal distribution, with parameters relating to the original distribution and .

Fast Forward You will find a proof for this if you study Further Mathematics, in Pure Core Student Book 1.

Key point 18.1 If the original distribution was

then

WORKED EXAMPLE 18.1

If

:

a Find b Find

. .

c Comment on your results in parts a and b. a

(from calculator)

To find probabilities for a given normal distribution you can go straight to the calculator. However, it can be useful to ‘sense check’ your result. The required region is within less than one standard deviation away from the mean, so the result must be less

than . You should always write down the distribution you are using if it is not given.

b So the standard deviation is .

Here the required region is within just over one standard deviation (from calculator) c The mean of

away from the mean, so the answer should be just over .

observations

is much more likely to be within unit of the true mean than a single observation is. You probably already had some intuition that calculating a mean of several observations provides a result that is likely to be closer to the true mean than a single observation. You can now have a mathematical justification for that intuition. EXERCISE 18A 1

Write down the distribution of the sample mean, given the original distribution.

Tip In this exercise we will use for the sample mean to emphasise the size of the sample. In later chapters, if the sample size is clearly given in the question, we will not include the subscript and simply write a i ii b i ii 2

, find

If

.

, find

If

, find

If

. .

, find

.

Find the following probabilities. a i ii b i ii

3

If

.

If

find

If

find

If

. .

, find

If

.

, find

.

is the energy (in eV) of beta particles emitted from a radioactive isotope. It is known that .

is the mean energy of

beta particles.

a Stating one necessary assumption, write down the distribution of b Find c Find 4

along with its parameters.

. .

The mass of a breed of dog is known to follow a normal distribution with a mean of

and a

standard deviation of . A random sample of four dogs is taken, and each dog’s mass is recorded. What is the probability that their mean mass is more than ? 5

The volume of apple juice in a carton follows a normal distribution with a mean of

and a

standard deviation of has a mean of less than 6

. A quality control process rejects a batch if a random sample of . Find the probability that a batch gets rejected.

Eggs are sold in boxes of six. The masses of eggs have a normal distribution with mean

cartons

and

variance of . What value of must be chosen if the average mass of eggs in a box must be more than in at least of boxes? 7

The lifetime of a light bulb,

, is modelled by

a mean lifetime of less than

.

of samples of

light bulbs have

. Find the value of .

8

The length of a species of fly follows a normal distribution with mean and standard deviation . of samples of flies have a mean of more than . Find the value of .

9

The diameter of an apple has mean and their mean diameter measured.

and standard deviation

a What is the probability that the mean diameter is between

. A sample of apples is chosen

and

if

?

b What is the smallest value of that must be chosen if the probability of the mean diameter being between and must be at least ? c By what factor must be increased (compared to your answer in part b) if the probability of being between and is required to double to ? 10 The mass of a student in a group is students find the probability that:

, where follows the

a The total mass of all the students is less than b The heaviest student has a mass less than

. .

distribution. In a sample of four

Section 2: Hypothesis tests for a mean One very common decision you have to make is whether a mean is different from a predicted value. For example, you may be told that the average is and want to see if students in a particular school have above-average . If you only have a sample from the school it is possible that the sample mean is above just through chance. You can conduct a hypothesis test to see if the difference above is big enough to be significant. You should take care with definitions: is the mean of the sample that you are using to make a decision about , the true mean of the population. In hypothesis testing, you assume that the conservative position – called the null hypothesis – is true and then see how likely you are to see something like the observed data. If the probability of seeing the observed data is very low, you reject the null hypothesis.

Rewind The terminology associated with hypothesis tests was introduced in Student Book 1, Chapter 18. To conduct a hypothesis test you need to make some assumptions about the underlying distribution. The test you will study requires the underlying distribution to be normal with known variance. To test the value of a population mean, , against a suggested value,

, at significance level :

1 Set up appropriate hypotheses, depending on the context, using one of the following.

Tip is called the ‘test statistic’. It is a value calculated from the sample that you use in the hypothesis. 2 Conduct the test. Do this by: either seeing if your observed mean falls into the acceptance region or the critical (rejection) region. To do this write down the distribution of (using Key point 18.1). Then find the regions at the ends of the distribution that have a total probability of (the critical region). Where these regions are depends on the alternative hypothesis:

y

y

H1 : µ > µ0

Acceptance region

α 2

α

x Critical region

H1 : µ ≠ µ0

H 1 : µ < µ 0

α µ0

y

µ0 Critical region

Acceptance region

x

α 2

µ0

x

Critical Acceptance Critical region region region

or by getting your calculator to find the -value of the observed mean. This is the probability of getting the observed value or more extreme (in the direction of the alternative hypothesis). 3 Reject the null hypothesis if the mean falls into the critical region or if the -value is less than the significance level. It is important that the conclusion is put in context and that it is not overly certain; you must show an

appreciation that you have found only evidence rather than stating a certain conclusion.

Testing using the critical region We will first look at the method using the critical region. WORKED EXAMPLE 18.2

It is believed that the healthy level of testosterone in blood is normally distributed with mean and standard deviation . Following a race a sprinter gives two samples with an average of

. Is this sufficiently high (at

significance) to suggest that the sprinter’s

testosterone level is above the mean? Define the variables.

= level of blood testosterone in a sprinter

State the hypotheses. This is a one-tailed test because the question is looking only for evidence of high testosterone. Under

State the distribution of

.

under

.

Find the critical region by sketching the normal curve and using the inverse normal distribution.

24

33.9

x

∴ Critical region:

Therefore, as falls in the rejection region, reject the null hypothesis.

On your calculator, you probably need to enter the mean, standard deviation, and . Some calculators can find the inverse normal value for the ‘right tail’ (using ) directly. Draw a conclusion.

There is evidence that the sprinter’s testosterone level is above average. Sometimes questions ask for you to find only the critical region without actually performing a test. WORKED EXAMPLE 18.3

The temperature of a water bath is normally distributed with a mean of

and a standard

deviation of . After the water bath is serviced, the temperature is measured on five independent occasions and a test is performed at the significance level to see if the temperature has changed from . It is assumed that the standard deviation is unchanged. What range of mean temperatures would provide sufficient evidence that the temperature has changed? = temperature of water bath

Define the variables.

State the hypotheses. This is a two-tailed test because the question does not specify that you are looking for evidence that the bath is too warm or too cold. State the distribution of

under

.

Use the inverse normal distribution to find the critical values of for the two-tailed region. The probability of being in each tail is half of the significance level, so it is .

2.5%

2.5% a

b

60

x

By symmetry about the mean,

. or

Write down the critical region.

Testing using the -value Many people find the -value method more straightforward.

Tip Many calculators can calculate the -value. You should check your manual to see how to do this with yours.

WORKED EXAMPLE 18.4

Standard light bulbs have a mean lifetime of

and a standard deviation of

. A low-

energy light bulb manufacturer claims that the lifetimes of their low-energy bulbs have the same standard deviations but that they last longer on average. A sample of low-energy light bulbs has a mean lifetime of . Test the manufacturer’s claim at the significance level. = lifetime of a bulb

Define the variables. State the hypotheses. State the test statistic and its distribution. Use the calculator to find the -value. Compare to the significance level and conclude.

Therefore, reject as there is evidence to support the manufacturer’s claim that the light bulbs last longer than on average.

WORK IT OUT 18.1 The wingspan of a species of butterfly is known to be normally distributed with mean and standard deviation . A scientist thinks she may have found some butterflies that belong to a different species with a different wingspan. The mean of a sample of six of these butterflies is . Test the scientist’s claim at the significance level, assuming the wingspans are still normally distributed with standard deviation . Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1

If

then (from calculator). , so accept that is, these butterflies do not belong to a different species.

Solution 2

If

then

(from calculator). The -value is twice this so reject . There is evidence for a different species. Solution 3

If the mean is

and the standard deviation is , then the critical region is

so the observed value lies in the critical region. Reject different species.



; that is, these butterflies are from a

EXERCISE 18B 1

Write null and alternative hypotheses for each of the following situations. a i ii b i ii c i ii

The average in a school over a long period of time has been . It is thought that changing the menu in the cafeteria might have an effect upon the average . It is claimed that the average size of photos created by a camera scientist believes that this figure is inaccurate. A consumer believes that steaks sold in portions of

is

. A computer

are, on average, underweight.

A careers adviser believes that the average extra amount earned by people with a degree is more than the figure he has been told at a seminar. The mean breaking tension of a brake cable ( ) does not normally exceed claims that it regularly does exceed this value. The average time taken to match a fingerprint ( ) is normally more than

. A new brand . A new

computer program claims to be able to do better. 2

In each of the following situations it is believed that of the following cases.

. Find the acceptance region in each

a i ii b i ii c i ii 3

In each of the following situations it is believed that

.

Find the -value of the observed sample mean. Hence, decide the result of the test if it is conducted at the significance level. a i ii b i ii c i ii d i ii 4

The average height of -olds in England is and the standard deviation is . Caroline believes that the students in her school are taller than average. To test her belief she measures the heights of a sample of students from her school. a State the hypotheses for Caroline’s test. You can assume that the heights follow the normal distribution and that the standard deviation of heights in Caroline’s school is the same as the standard deviation for the whole population. The students in Caroline’s school have mean height of b Test Caroline’s belief at the

5

.

level of significance.

All students in a large school are given a typing test and it is found that the times taken to type one page of text are normally distributed with mean and standard deviation . The students are given a month-long typing course and then a random sample of students is asked to take the typing test again. The mean time is now . Test at the significance level whether there is evidence that the time taken to type a page of text has decreased.

6

The national mean score in Mathematics GCSE is with a standard deviation of school the mean score of a sample of students is .

. In a particular

a State two assumptions that are needed to perform a hypothesis test to see if the mean is better in this school than the background population. b Assuming that these assumptions are met, test at the

significance level whether the school is

producing better results than the national average. 7

A farmer knows from experience that the average height of apple trees is with standard deviation . He buys a new orchard and wants to test whether the average height of apple trees is different. He assumes that the standard deviation of heights is still . a State the hypotheses he should use for his test. The farmer measures the heights of

trees and finds their mean.

b Find the critical region for the test at the c If the average height of the

trees is

level of significance. , state the conclusion of the hypothesis test.

Worksheet See Support sheet 18 for a further example of testing for the mean of a normal distribution and for more practice questions. 8

A doctor has a large number of patients starting a new diet in order to become slimmer. Before the diet the masses of the patients were normally distributed with mean and standard deviation

. The doctor assumes that the diet does not change the standard deviation of the masses. After the patients have been on the diet for a while, the doctor takes a sample of mean mass.

patients and finds their

a The doctor believes that the average mass of the patients has decreased following the diet. She wishes to test her belief at the level of significance. Find the critical region for this test. b State an additional assumption required in your answer to part a. c The mean average mass of the 9

patients after the diet is

. State the conclusion of the test.

The school canteen sells coffee in cups claiming to contain . It is known that the amount of coffee in a cup is normally distributed with standard deviation . Adam believes that, on average, the cups contain less coffee than claimed. He wishes to test his belief at the significance level. a Adam measures the amount of coffee in randomly chosen cups and finds the average to be . Can he conclude that the average amount of coffee in a cup is less than ? b Adam decides to collect a larger sample. He finds the average to be again, but this time this is sufficient evidence to conclude that the average amount of coffee in a cup is less than . What is the minimum sample size he must have used?

10 The null hypothesis is tested and a value alternative hypothesis is or ?

is observed. Will it have a higher -value if the

Section 3: Hypothesis tests for correlation coefficients You already know that values of the correlation coefficient close to or negative correlation, respectively.

represent strong positive or

However, values in between can be difficult to interpret, for instance, is evidence of significant correlation? The answer depends on the number of data points and how certain you want to be.

Rewind See Student Book 1, Chapter 16 for a reminder of the product moment correlation coefficient. You need to distinguish between two related values: the correlation coefficient of the sample, given the symbol and the correlation coefficient of the underlying population, given the symbol (the Greek letter ‘rho’). A hypothesis test uses to decide if there is evidence that is not zero. In a two-tailed test you are looking to see if there is correlation in either direction – positive or negative – so the alternative hypothesis is . In a one-tailed test you are looking for correlation in just one direction, so the alternative hypothesis would be either or . Finding the distribution of goes beyond the scope of this book, but the critical values can be found in tables that will be provided when needed. They are calculated assuming that both variables follow a normal distribution, and provide critical values for different numbers of data pairs .

Key point 18.2 If the modulus of is larger than the appropriate critical value, then reject the null hypothesis. Level of significance for one-tailed test Level of significance for two-tailed test

Explore The correlation coefficient used in this course is called the Pearson product- moment correlation coefficient. This is just one type of correlation coefficient. You could also use Spearman’s rank correlation coefficient or Kendall’s tau. They all have their own advantages and disadvantages.

WORKED EXAMPLE 18.5

The correlation coefficient between the mass and height of students in a sample of six students is . Test at the significance level whether height and mass of students from this school are positively correlated. Write down the hypotheses. You are looking only for positive correlation, so it is a one-tailed test we are performing. Level of significance for one-tailed test Level of significance for two-tailed test

Write down the critical value from the table. one-tailed and the level of significance is

, the test is .

The critical value from the table is .

Reject . There is evidence that mass and height are positively

State the conclusion in context.

correlated.

Did you know? You should remember from Student Book 1 that correlation does not imply causation. There are many examples of spurious correlations. For example, there is a very strong negative correlation between the number of pirates and global warming! Can you explain this?

WORKED EXAMPLE 18.6

Twenty students were asked to give the number of hours they spent watching television each week and their results in a reading test. The correlation coefficient for their results was . Test for evidence of correlation at the significance level. Write down the hypotheses. You are not told to look for correlation in any particular direction, so it is a two-tailed test. Level of significance for one-tailed test

Write down the critical value from the table. , level of significance

Level of significance for two-tailed test

The critical value is

.

Do not reject . There is not significant evidence for correlation between hours watching television and results in a reading test.

State the conclusion in context.

EXERCISE 18C A table of critical values of the correlation coefficient is given on the Statistics page. In the examination, a

relevant extract from the table will be given in each question that requires it. 1

Test each of the following sample correlation coefficients for positive correlation at where is the sample size. a i ii b i

significance,

, , ,

ii 2

Test each of the following sample correlation coefficients for correlation at

significance, where is

the sample size. a i ii b i ii 3

Information for students is used to investigate the hypothesis that there is a correlation between and results in a Maths test. a Write down the null and alternative hypotheses for this investigation. b Data are collected and the -value for the correlation coefficient is of the hypothesis test at the significance level?

4

. What is the conclusion

The average speed of cars is measured at six different checkpoints at varying distances from a junction. There is a belief that, in general, cars get faster as they are further from the junction. a Write down the null and alternative hypotheses for this investigation. b The -value of the observed data is found to be

5

. Test the data at the

significance level.

The amount spent by a government on unemployment support is expected to be negatively correlated with the amount spent on education. Data are collected across countries in 2016. a What is the population associated with this sample? b Write down appropriate null and alternative hypotheses. c The sample correlation coefficient is found to be test at

6

. What is the conclusion of the hypothesis

significance?

The correlation coefficient between the amount of water used in a town on temperature is .

summer days and the

a Jane thinks that there is a correlation between water usage and temperature on a summer day. Write down the null and alternative hypotheses Jane should use to test her suspicion. b Conduct the test at the

significance level.

c Karl says that if people use more water, the days will be warmer. Give two reasons why your hypothesis test does not support Karl’s statement. 7

The level of antibodies in blood is thought to go down as the dose of a medical drug is increased. a State appropriate null and alternative hypotheses to test this statement. b In a sample of patients the correlation coefficient between these two variables is found to be . Conduct an appropriate hypothesis test at the significance level. c What are the advantages of using a medical tests?

8

significance level rather than a

Data are collected on the height of a cake and the temperature.

significance level for

a When a hypothesis test is conducted to test for positive correlation, the -value is evidence of positive correlation at the significance level?

. Is this

b If the same data were instead used to test for correlation in either direction, what would be the value? Is there evidence of correlation at the significance level? c If the correlation coefficient increases, does this increase or decrease the -value found in part a? Justify your answer. 9

It is suspected that there is correlation between the height of a tree and the total surface area of its leaves. A random sample of trees is measured and the correlation coefficient is found to be . What is the smallest value of which makes this significant at significance?

10 Why do critical value tables for correlation coefficients start at

?

Checklist of learning and understanding The sample mean, If

, is a random variable.

, then

.

To test the value of a population mean, , against a suggested value,

:

Set up appropriate hypotheses, depending on the context. Either or or Then, use the distribution of and your calculator to either find the -value or set up the critical region for the given significance level, . If

(or if

is in the critical region), reject

.

To test whether there is correlation between two variables: Set up appropriate hypotheses depending on the context. Either or or Then look up in the tables the critical value for the given significance level and sample size. If the modulus of the sample correlation coefficient, , is greater than the critical value, then reject .

Mixed practice 18 1

A random sample of people have their height and mass measured. Which of the following correlation coefficients is the smallest that would suggest evidence of positive correlation between height and mass using a

significance level?

A B C D 2

The breaking load of steel wire is known to be normally distributed with mean and standard deviation . Find the probability that the mean breaking load of a sample of ten such wires is between and .

3

Data are collected on HIV rates and literacy rates of

countries in 2016.

a What is the population from which this sample is drawn? b A hypothesis test is conducted to see if there is correlation between HIV rates and literacy rates. Write down appropriate null and alternative hypotheses. c The -value for the observed correlation coefficient is hypothesis test at the 4

. What is the conclusion of the

significance level?

The mass of cakes produced by a bakery is known to be normally distributed with mean and standard deviation

. A new baker is employed.

a State appropriate null and alternative hypotheses to test if the mean mass of cakes has changed. b The mean mass of

cakes is found to be

the conclusion of the hypothesis test at the 5

What is the -value of these data? What is significance level?

The results of a group of students in a test are thought to follow an mean of a random sample of .

students is used to test the hypothesis

Which of the following is the critical region at the significant figures)?

distribution. The against

significance level (given to three

A B C D 6

A test is conducted to see if the time spent revising correlates with the results in a test. a State a distributional assumption that is required to use tables of critical values for the correlation coefficient. b A random sample of students is surveyed. The correlation coefficient for their responses is . Conduct a hypothesis test at the significance level.

7

The time taken for a full kettle to boil is known to follow a normal distribution with mean

seconds and standard deviation seconds. After cleaning the kettle it is boiled ten times to test if the time taken to boil has decreased. a Stating two necessary assumptions, find the critical region for this hypothesis test at the significance level. b The mean time is found to be

seconds. State the outcome of the hypothesis test.

c Why should there be a long delay between the ten observations for this test to be valid? 8

Over a long period the number of visitors per week to a stately home was known to have the distribution

. After higher car parking charges were introduced, a sample of four

randomly chosen weeks gave a mean number of visitors per week of . You should assume that the number of visitors per week is still normally distributed with variance . i

Test, at the

significance level, whether there is evidence that the mean number of

visitors per week has fallen. ii

Explain why it is necessary to assume that the distribution of the number of visitors per week (after the introduction of higher charges) is normal in order to carry out the test. © OCR, GCE Mathematics, Paper 4733, January 2008

9

Yukun wants to test the null hypothesis against the alternative hypothesis . He finds the mean sample of the data, , is . His calculator tells him that the value of his data is . Which of the following expressions defines the -value of his hypothesis test? A B C D

10 The distance an athlete jumps in a long jump is known to be normally distributed with mean and standard deviation . After a change to her technique, she looks at the average of jumps to see if her average distance has changed at the significance level. a Write down appropriate null and alternative hypotheses for this test. b If the jumps are still normally distributed with standard deviation acceptance region in terms of . c If the mean is found to be hypothesis being rejected.

, find the

, find the smallest value of that would result in the null

11 The viewing figures of a long-running television series is million. In the past it was known to follow a normal distribution with a standard deviation of million. A producer wants to know if a new presenter has changed the viewing figures across episodes. He conducts a hypothesis test, assuming the viewing figures are still normally distributed with standard deviation million. The acceptance region is found to be . a Deduce the null and alternative hypotheses. b Find the significance level of this hypothesis test, giving your answer as a percentage to the nearest whole number. 12 The continuous random variable has the distribution observations of is denoted by . It is given that

. The mean of . Find the value of .

© OCR, GCE Mathematics, Paper 4733, January 2009

Worksheet See Extension sheet 18 for some questions to make you think about appropriate significance levels in various situations.

FOCUS ON … PROOF 3

The prosecutor’s fallacy A man is accused of robbing a jeweller’s shop and stealing diamonds. The only evidence against him is a bag of diamonds found in his car during the police investigation. The prosecution argues that the probability that the bag is found in the car if the man is innocent is in and, hence, the probability of him being guilty is in (or ). You are going to investigate whether this is a valid argument.

QUESTIONS 1

Let denote the probability that the man is guilty of stealing the diamonds. Without taking into account any information (such as any findings of the police investigations), estimate the value of .

2

Denote by the event that the man is guilty, so

, and let be the event that the evidence

(in this case, the bag of diamonds) is found in his car. Then the prosecution’s statement says that . You also need to estimate the probability of finding the evidence if he is guilty; let’s assume this is quite likely, so set . a Complete the relevant tree diagram. E G

p

G'

b Using the tree diagram: i

Find

.

ii

Use the conditional probability formula to find

c Prove that, when argument?

,

and

in terms of .

. What does this tell you about the prosecutor’s

3

Why was it reasonable to assume that is very small? Think about what represents.

4

How do

5

Would it make sense to swap the branches on the tree diagram, so that the first ‘level’ has and

6

The prosecutor stated that ‘the probability that the bag is found in the car if the man is innocent is in ’. Write this event using conditional probability notation. Hence, write down and interpret the event that has the probability of in .

7

In law, the phrase ‘proof beyond reasonable doubt’ is used. What do you think this means? How does this compare to mathematical proof?

and

compare to each other if is larger (say,

)? ?

FOCUS ON … PROBLEM SOLVING 3

Using extreme values Probability can often be counter-intuitive, and people find it difficult to evaluate their solutions. One useful strategy can be to use extreme values to make the answer more obvious.

The Monty Hall problem The following game featured in a US television show. You are shown three closed doors and told that one door hides a car while the other two hide goats. You will win whatever is behind the door that you choose. The show host knows which door hides the car.

1

2

3

You are asked to choose a door. The host then opens one of the other two doors to reveal a goat. You are then given a choice: stick with the original door or switch to the third one. What should you do to maximise your probability of winning the car? Most people’s intuition is that it doesn’t matter – there are two closed doors and the car is equally likely to be behind either of them. But this doesn’t take into account the fact the host knows where the car is, so would never open that door.

QUESTIONS

QUESTIONS 1

Imagine that instead there are doors and you initially choose Door . The host opens other doors, leaving out Door . What would you do?

of the

The fact that the host knew which door to leave closed gives you additional information, so the probability is no longer ‘equally likely for each door’. 2

Suppose you play the game times. Call the door hiding the car Door . Your initial choice of the door is random. If you picked Door then the host can choose whether to open Door or Door ; assume he chooses randomly. If you initially picked Door or , then the host has no choice about which one to open. Fill in this table showing the possible outcomes, assuming the car is behind Door . You choose   Host opens

Use the table to find the probability that you win the car if you switch. You should find that it is not the intuitive probability discussed at the start of this example. 3

There are two children in the garden. One of them is a girl. What is the probability that both of them are girls? (You may assume that both genders are equally likely and that the genders of the two children are independent of each other so, for example, they are not identical twins.) a Which of the following arguments do you find most convincing? i Both genders are equally likely, and the genders of the two children are independent, so the probability that the second child is also a girl is . ii

The two genders are independent, so the probability of two girls is

iii The options for the two genders are

,

and

,

,

.

. Hence, the probability that both are girls

is . iv The options for the two genders are

and

. Hence, the probability of two girls is

. v You already know that one child is a girl, so there are fewer options to choose from. Hence, the probability of two girls is more than . b Now suppose there are ten children in the garden and nine of them are known to be girls. Do you think that the probability that all ten are girls is bigger or smaller than

?

c The table shows possible outcomes for two children. Fill in the probabilities and, hence, find the probability that both children are girls, given that one of them is a girl. First child Girl Boy Girl Second child Boy

FOCUS ON … MODELLING 3

When can you use the normal distribution? The normal distribution is commonly used as a model in many applications. However, it is important to be aware that this model isn’t always suitable. This section focuses on the properties of the normal distribution, which you should consider when deciding whether to use it as a model. WORKED EXAMPLE

The marks for a group of

students on a Statistics exam are summarised in the following and the standard deviation of the marks is

Frequency density

histogram. The mean mark is

200 180 160 140 120 100 80 60 40 20 0

0

.

10 20 30 40 50 60 70 80 90 Exam mark

a Estimate the number of students whose marks are more than two standard deviations away from the mean. b Hence, state, with a reason, whether a normal distribution could be used to model the distribution of the marks. a

On the histogram, there are no data values below or above . No students had marks more than standard deviations from the mean.

b For the normal distribution, around of the marks should be more than standard deviations from the mean. Hence, the normal distribution does not seem to be a good model for these marks.

For a normal distribution, of the data should be within two standard deviations of the mean.

QUESTIONS 1

The table summarises heights of a group of Height (

school children. )

Frequency

a Draw a histogram to represent the data (you may want to use technology to do this). b Hence, explain whether a normal distribution would be a suitable model for these heights. 2

a An old textbook says that the range of the data is about six times the standard deviation. For a normal distribution, what percentage of the values is contained in this range?

30

40 50 Length of throw (m)

60

The box plot alongside summarises the results of a discus throw competition (length, measured in metres). b Assuming the dataset follows a normal distribution, use the box plot to estimate its mean and standard deviation. Find the interquartile range for this normal distribution. c Hence, state, with a reason, whether a normal distribution is a suitable model for the lengths of the throws. 3

The mean, median and standard deviation for two sets of data are given below. For each set of data decide, based on this information, whether a normal distribution would be a suitable model.

4

a

,

,

b

,

,

In social science research, subjects are often asked to rank their opinions on a five-point scale, called the Likert scale (e.g. strongly disagree, disagree, no opinion, agree, strongly agree). For the purpose of statistical analysis, these responses are sometimes translated into numbers (e.g. ‘strongly disagree’ , ‘strongly agree’ ). Give a reason why data measured on a Likert scale should not be modelled using a normal distribution.

5

a A law firm has found that the average length of phone calls made by its employees is . Give a reason why a normal distribution may not be a suitable model for the length of phone calls. b It is often said that many naturally occurring measurements approximately follow a normal distribution. Discuss whether a normal distribution would be an appropriate model in the following situations. i

the weight of red squirrels in a forest

ii

the heights of all parents and children at a nursery school open day

iii the number of children in a family. The diagram shows three cumulative frequency curves. Which curves show: Cumulative frequency

6

100

b a normal distribution?

2

50

3 0

a a symmetrical distribution

1

0

10 20 30 40 50 60 70 80 90 100

x

CROSS-TOPIC REVIEW EXERCISE 3 1

Asher has a large bag of sweets, half of which are red. He rolls a fair six-sided dice once. If the dice shows or , he randomly picks two sweets from the bag. If the dice shows any other number, he randomly picks three sweets from the bag. Find the probability that Asher picks at least one red sweet.

2

Elsa is investigating whether there is any correlation between the average daily temperature and the daily amount of rainfall. a State suitable null and alternative hypotheses for her test. Elsa collects the data for a random sample of days and calculates that the correlation coefficient between the average temperature and the amount of rainfall is . b Conduct the hypothesis test at the context.

3

level of significance. State your conclusion in

It is known that the heights of a certain type of rose bush follow a normal distribution with mean and standard deviation . Larkin thinks that the roses in her garden have the same standard deviation of heights, but are taller on average. She measures the heights of rose bushes in her garden and finds that their mean height is . a State suitable hypotheses to test Larkin’s belief. b Showing your method clearly, test at the level of significance whether there is evidence that Larkin’s roses are taller than average.

4

The lifetime of a certain type of light bulb, hours, is modelled by the distribution It is given that .

.

a Find the value of . b Find the probability that, in a sample of more than hours.

randomly selected light bulbs, at least ten last

c By considering the range of values that contains nearly all values of , discuss whether the normal distribution is a reasonable model for the lifetime of the light bulbs. 5

Jenny and Omar are each allowed two attempts at a high jump. i

The probability that Jenny will succeed on her first attempt is . If she fails on her first attempt, the probability that she will succeed on her second attempt is . Calculate the probability that Jenny will succeed.

ii

The probability that Omar will succeed on his first attempt is . If he fails on his first attempt, the probability that he will succeed on his second attempt is also . The probability that he succeeds is . Find . © OCR, GCE Mathematics, Paper 4732, January 2011

6

The random variable has the distribution µ . One hundred observations of are taken. The results are summarised in the following table. Interval Frequency i

By considering

, write down an equation involving µ and

ii

Find a second equation involving µ and

.

. Hence calculate values for µ and .

iii Explain why your answers are only estimates. © OCR, GCE Mathematics, Paper 4733, June 2014

7

A dataset consists of four numbers: and . Find the value of for which the standard deviation of the data is the minimum possible.

8

Two events and are such that

,

and

. By use of a tree

diagram, or otherwise, find: a b c 9

.

Theo repeatedly rolls a fair dice until he gets a . a Show that the probability of him getting this on the third roll is b

.

is the probability of getting his first on the th roll. Find an expression for

in terms of

. c Prove algebraically that

.

10 A supermarket has a large stock of eggs.

of the stock are from a farm called Eggzact.

of the stock are brown eggs from Eggzact. An egg is chosen at random from the stock. Calculate the probability that i

this egg is brown, given that it is from Eggzact.

ii

this egg is from Eggzact and is not brown. © OCR, GCE Mathematics, Paper 4732, January 2008

11 The masses of bags of sugar are normally distributed with mean .

and standard deviation

a Find the probability that a randomly chosen bag of sugar weighs more than b Find the probability that in a box of .

.

bags there are at least two that weigh more than

c Darien picks up bags of sugar from a large crate at random. What is the probability that he has to pick up exactly bags before he finds one that weighs more than

?

12 A random variable, , has a normal distribution with mean and standard deviation . a Use your calculator to find

correct to five decimal places.

The exact value of the probability is given by

.

b Use the trapezium rule with six strips to estimate the value of this integral. Give your answer correct to five decimal places. c Find the percentage error in using the trapezium rule to estimate this probability. 13 The random variable has normal distribution distribution of .

. The line graph shows the probability

P(X = x) 0.06 0.04 0.02 0

50

60

70

80

90

100

110

120

x

The distribution of can be approximated by a normal distribution. a Use the graph to estimate the mean and standard deviation of the normal distribution.

Explain clearly how you arrived at your answer. b Hence, estimate the values of and . 14 A student is investigating whether there is any correlation between the amount of time spent revising and marks on a test. She uses information from a sample of six tests. In order to have a larger sample, she collects data from two of her friends. She finds that the correlation coefficient between hours spent revising and the percentage mark on the test is . She therefore suggests that there is negative correlation between the amount of time spent revising and the test marks. a State the null and alternative hypotheses for her test. b Test the hypotheses at the

level of significance and interpret the conclusion in context.

Percentage score in test

The scatter graph of her data is shown below. 100 90 80 70 60 50 40 30 20 10 0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Time spent revising (hours)

c Suggest one possible interpretation of the data and the result of the hypothesis test. 15 The continuous random variable has the distribution µ . The mean of a random sample of observations of is denoted by . It is given that and . Obtain a formula for in terms of . Two students are discussing this question. Aidan says, ‘If you were told another probability, for instance , you could work out the value of .’ Binya says, ‘No, the value of is fixed by the information you know already.’ ii

State which of Aidan and Binya is right. If you think that Aidan is right, calculate the value of given that . If you think that Binya is right, calculate the value of . © OCR, GCE Mathematics, Paper 4733, January 2013

16 Daniel and Paolo play a game with a biased coin. They take it in turns to toss the coin. There is a probability of of the coin showing a head and a probability of of it showing a tail. If the coin shows a head the player who tossed the coin wins the game. If the coin shows a tail, the other player has the next toss. Daniel plays first and the game continues until there is a winner. a Write down the probability that Daniel wins on his first toss. b Calculate the probability that Paolo wins on his first toss. c Calculate the probability that Daniel wins on his second toss. d Show that the probability of Daniel winning is . e State the probability of Paolo winning. f They play the game with a different biased coin and find that the probability of Daniel

winning is five times the probability of Paolo winning. Find the probability of this coin showing a head. 17 In a large typesetting company the time taken for a typesetter to type one page is normally distributed with mean minutes and standard deviation minutes. A new training scheme is introduced and, after all the typesetters have completed the training, the times taken by a random sample of typesetters are recorded. The mean time for the sample is minutes. You may assume that the population standard deviation is unchanged. a Test, at the

significance level, whether the mean time to type a page has decreased.

b It is required to redesign the test so that the probability of incorrectly rejecting the null hypothesis is less than needed.

when the sample mean is

. Find the smallest sample size

18 A normal distribution can be represented by a curve shown in the diagram.

y

x

O The equation of the curve is

.

a Find the -coordinates of the points of inflection of the curve. Random variable follows this normal distribution. b Use the trapezium rule with four strips to estimate

. Give your answer to

three significant figures. c Is it possible to tell, without doing any further calculations, whether the answer in part b is an over-estimate or an underestimate? Explain your answer.

19 Applications of vectors In this chapter you will learn how to: use displacement, velocity and acceleration vectors to describe motion in two dimensions use some of the constant acceleration formulae with vectors use calculus to relate displacement, velocity and acceleration vectors in two dimensions when acceleration varies with time represent vectors in three dimensions using the base vectors i, j and k use vectors to solve geometrical problems in three dimensions.

Before you start… Student Book 1, Chapter 12

You should be able to link displacement vectors to coordinates and perform operations with vectors.

1 Consider the points

,

. Let and in column vector form:

and . Write

a b c d Student Book 1, Chapter 12

You should be able to find the magnitude and direction of a vector.

2 Find the magnitude and direction of the

Student Book 1, Chapter 19

You should understand the concepts of displacement and distance; instantaneous and average velocity and speed; acceleration.

3 In the diagram below, positive displacement is measured to the right.

vector

A

.

B 120m

C 180m

A particle takes seconds to travel from to and another seconds to travel from to . Find: a the average velocity b the average speed for the whole journey. Student Book 1, Chapter 20

You should be able to use constant acceleration formulae in one dimension.

4 A particle accelerates uniformly from to while covering the distance of in a straight line. Find the acceleration.

Student Book 1, Chapter 19

You should be able to use calculus to work with displacement, velocity and acceleration in one dimension.

5 A particle moves in a straight line with velocity . Find: a the acceleration when b an expression for the displacement from the starting position.

Chapter 12

You should know how to work with curves defined parametrically.

6 Find the Cartesian equation of the curve with parametric equations and .

Why do we need to use vectors to describe motion? In Student Book 1, Chapters 19 and 20, you studied motion in a straight line. You saw how displacement, velocity and acceleration are related through differentiation and integration:

In the special case when the acceleration is constant, you can use the constant acceleration equations:

But the world is three-dimensional, and objects don’t always move in a straight line. You need to be able to describe positions and motion in a plane (such as a car moving around a race track) or in space (e.g. flight paths of aeroplanes). This requires the use of vectors to describe displacement, velocity and acceleration.

Rewind Vectors were introduced in Student Book 1, Chapter 12. Remember that the vector



can also be written as

.

Section 1: Describing motion in two dimensions When a particle moves in two dimensions, the displacement, velocity and acceleration are vectors. The distance and speed are still scalars. WORKED EXAMPLE 19.1

Points , and have position vectors

and

, where the distance is measured in

metres. A particle travels along the straight lines between the points. It takes seconds to travel from to and then another seconds to travel from to . Find: a the average velocity and average speed from to b the final displacement of the particle from c the average velocity for the whole journey d the average speed for the whole journey.

y 6 5

B

4 3 C

2 1

-3 -2 -1O -1 -2

1

2

3 4 A

x

a The displacement is the difference between the position vectors.

b

The displacement from to is the difference between their position vectors. It is highlighted blue in the diagram.

c

d You need to add the distance from to to the distance from to . This is highlighted red in the diagram. The distance is the modulus of the vector between two points.

You have already found that the total time is

.

Notice that the average speed is not the magnitude of the average velocity vector. This is because the particle changes direction during the motion. You can calculate average acceleration by considering the change in velocity.

Tip You should include units with the final answer, even when it is a vector.

WORKED EXAMPLE 19.2

A particle moves in a plane. It passes point with velocity

and passes point

seconds later with velocity . Find the magnitude and direction of the average acceleration of the particle between and .

You need to find the acceleration vector first, then find its magnitude. The magnitude is:

The magnitude of a vector

is

.

The direction is represented by the angle measured anticlockwise from the horizontal (the direction of vector ). It is always a good idea to draw a diagram to see which angle you are looking for.

The direction is horizontal.

from the

Acceleration can cause a change in the direction of the velocity as well as its magnitude (speed). This means that the object will not necessarily move in a straight line. The velocity vector gives the direction of motion.

Rewind The displacement vector gives the parametric equations of the object’s path, with the parameter being time. See Chapter 12, Section 3 for a reminder.

If you know how the displacement vector varies with time, you can sometimes find the Cartesian equation of the object’s path.

Tip The path an object follows is also called a trajectory.

WORKED EXAMPLE 19.3

A particle moves in a plane. At time the particle is at a point and its displacement from the origin, , is given by the position vector

. Prove that the particle moves along a parabola;

find its Cartesian equation and sketch it. The coordinates of are: ,

The two components of the position vector give the - and coordinates of the particle’s position. We want to find in terms of , so substitute from the first equation into the second.

Hence, the path of the particle is a parabola.

y

Note that, since with .

, the object covers only a part of the parabola

(2, 1) x

O

. We can also look at two particles moving in a plane and ask questions about the distance between them, and whether they ever meet.

Rewind In Worked example 19.4, part b is an example of proof by contradiction. See Chapter 1 for a reminder.

WORKED EXAMPLE 19.4

Two particles, and , move in the plane. starts from the origin and moves with constant velocity . a Write down the position vector of in terms of . Particle starts from the point with position vector

m and moves with constant velocity

. b Prove that and never meet. c Find the minimum distance between the two particles. a

The position vector of is its displacement from the origin, and displacement equals velocity times time.

b Position vector of is:

The displacement of from its initial position is . So its position vector equals the initial position plus this displacement.

The particles meet when

We want to show that and are never in the same place at the same time.

:

So try to find the value of when the two displacements are equal and show that this is impossible. If two vectors are equal, then both components have to be equal. So we need a value of that works in both equations.

From the first equation:

Check in the second equation:

The second equation is NOT satisfied. There is no for which never meet.

, so the particles

c At time :

There is no value of that makes the two position vectors equal. The distance between the particles is the magnitude of the displacement between them, which is found by subtracting the two position vectors.

The distance between and is:

This expression has a minimum value when its square has a minimum value; so look at to avoid having to work with the square root. Let Then

. when

The minimum value of is:

.

You could complete the square to find the minimum value, but the numbers aren’t nice so differentiate instead. Don’t forget that this is the minimum value for .

Hence, the minimum distance

EXERCISE 19A 1

Points , and have position vectors , and metres. Find the average velocity when a particle travels: a

, where distance is measured in

i from to in seconds ii from to in seconds

b

i from to in seconds ii from to in seconds

c

i from to in seconds and then from to in seconds ii from to in seconds and then from to in seconds.

2

Find the average acceleration vector, and the magnitude of average acceleration in each of the following cases.

a

b

i The velocity changes from

to

ii The velocity changes from

to

in

in 8 seconds.

i A particle accelerates from rest to

in seconds.

ii A particle accelerates from rest to 3

seconds.

in

Three points have coordinates

and

seconds.

.

a A particle travels in a straight line from to in seconds. Find its average velocity and average speed. b Another particle travels in a straight line from to in seconds and then in a straight line from to in seconds. Find its average velocity and average speed. 4

A particle moves in the plane so that its displacement from the origin (measured in metres) at time seconds is given by the vector

.

a Find the particle’s distance from the origin when

.

b Find the Cartesian equation of the particle’s trajectory. 5

a An object’s velocity changes from of its average acceleration.

to

in seconds. Find the magnitude

b The object then moves for another 10 seconds with average acceleration direction of motion at the end of the 6

. Find its

seconds.

A particle travels in a straight line from point with coordinates to point with coordinates . The journey takes seconds and the distance is measured in metres. a Find the average speed of the particle. The particle then takes a further seconds to travel in a straight line to point with coordinates . b Find the displacement from to . c Find the average velocity of the particle for the whole journey. d Find the average speed for the whole journey from to . Explain why this is not equal to the magnitude of the average velocity.

7

Two particles, and , move in a plane. has constant velocity displacement from the origin is

and its initial

m. starts from the origin and moves with constant velocity

Show that the two particles meet and find the position vector of the meeting point. 8

A particle moves in a plane so that its displacement from the origin at time . a Find the distance of the particle from the origin when

is given by the vector

.

b Sketch the trajectory of the particle. 9

An object moves with a constant velocity .

. Its initial displacement from the origin is

a Find the Cartesian equation of the particle’s trajectory. b Find the minimum distance of the particle from the origin. 10 A particle moves in a plane so that its displacement from the origin at time seconds is

. Find the maximum distance of the particle from the origin.



Section 2: Constant acceleration equations When a particle moves with constant acceleration, we can use formulae analogous to those for onedimensional motion.

Rewind See Student Book 1, Chapter 20 for a reminder of the constant acceleration formulae.

Fast forward Notice that the list in Key point 19.1 does not contain the vector version of the formula . If you study the Mechanics option of Further Mathematics you will meet a way of multiplying vectors (called the scalar product) that enables us to extend this formula to two dimensions as well.

Key point 19.1 Constant acceleration formulae in two dimensions:

WORKED EXAMPLE 19.5

A particle starts with initial velocity and moves with constant acceleration. After 5 seconds its velocity is . Find its displacement and its distance from the initial position at this time. For displacement:

You know , and and want to find , so use the third formula from Key point 19.1.

For distance:

Distance is the magnitude of the displacement.

If a particle does not move in a straight line, its distance from the starting point (which is measured in a straight line) is not the same as distance travelled (which is along a curve). B

A

Explore Find out how you can use calculus to calculate the length of a curve. You need to be a little careful when solving equations with vectors. If you are comparing two sides of a vector equation, both components need to be equal. WORKED EXAMPLE 19.6

A particle moves with constant acceleration velocity is

. It is initially at the origin and its initial

. Find the time when the particle is at the point with position vector You have , and and want to find , so use the second formula from Key point 19.1.

First component:

The first component of the vectors gives a quadratic equation for . This will give us two possible values.

Second component:

We need to check for which of these values of the second component equals 4.

When When

: :

The particle is at

when

seconds.

WORK IT OUT 19.1 A particle moves with constant acceleration, starting from the origin. Its position vector at time t is given by . How many times does the particle pass through the origin during the subsequent motion? Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1 You need the displacement to be zero. When : or . So the particle passes the origin once, when

.

Solution 2 You need both components of the displacement to equal zero. When When

:

or . :

or

.

So the particle passes the origin twice, when

and

.

Solution 3 You need both components of the displacement to be zero. When : or . When

:

or

.

So the particle does not pass the origin again.

EXERCISE 19B

EXERCISE 19B 1

In each of the following questions a particle moves with constant acceleration. Time is measured in seconds and displacement in metres. a

i ii

b

i ii

c

d

.

. Find when . Find when . Find .

ii

. Find .

i

.

. Find when

i

ii 2

. Find when

. .

. Find . . Find .

An object moves with constant acceleration

and initial velocity

. Find its

velocity and displacement from the initial position after 7 seconds. 3

A particle moves with constant acceleration

. It is initially at the origin and its velocity is

a Find the distance of the particle from the origin after 3 seconds. b Find the direction of motion of the particle at this time. 4

A particle passes the origin with velocity

and moves with constant acceleration.

a Given that seconds later its velocity is

, find the acceleration.

b Find the time when the particle’s displacement from the origin is 5

An object moves with constant acceleration. When displacement from the initial position is

.

it has velocity

. When

its

. Find the magnitude of the acceleration.

6

A particle moves with constant acceleration. Its initial velocity is . 8 seconds later, its displacement from the initial position is . Find its direction of motion at this time.

7

An object moves with constant acceleration and initial velocity . When its displacement from the initial position is , its velocity is . Find the magnitude of the acceleration.

8

A particle moves with constant acceleration

. Given that its initial velocity is

find the time when its displacement from the initial position is 9

.

A particle starts with initial velocity and moves with constant acceleration that the speed of the particle increases with time.

. Prove

10 A particle moves with constant acceleration . When the particle is at rest, at the point with the position vector . Find the shortest distance of the particle from the origin during the subsequent motion.



Section 3: Calculus with vectors When the acceleration is not constant you need to use differentiation and integration to find expressions for displacement and velocity. In Student Book 1, Chapter 1 you learnt how to do that for motion in one dimension. The same principles apply to two-dimensional motion: differentiating the displacement equation gives the velocity equation, and differentiating the velocity equation gives the acceleration equation. The only difference is that those quantities are now represented by vectors.

Key point 19.2 To differentiate or integrate a vector, differentiate or integrate each component separately. In particular, If

, then



Tip Some people use a dot to denote differentiation with respect to time; for example, .

and

WORKED EXAMPLE 19.7

A particle moves in two dimensions. Its displacement from the starting point, measured in metres, varies with time (measured in seconds) as:

Find the speed of the particle when

.

The speed is the magnitude of the velocity vector. To find the velocity, differentiate the displacement equation with respect to . :

When using integration with vectors, the constant of integration will also be a vector. WORKED EXAMPLE 19.8

A particle moves with acceleration

. Its initial velocity is

.

Find the expression for the velocity at time . To find the velocity, integrate the acceleration vector.

When

,

:

Use the initial velocity to find .

The constant can be included within the existing vector.

So

Remember that, for two vectors to be equal, both components need to beequal. WORKED EXAMPLE 19.9

A particle starts from point with velocity

. Its acceleration is given by

. Show that the particle never returns to . Let be the displacement from at time .

You need to find an expression for the displacement from and show that it never equals for . First integrate to find .

when

Use the initial velocity to find .

, so

Now integrate the velocity to find the displacement. The initial displacement is , so .

You now need to check if there is any value of (other than ) when . You need to check both components of the vector.

When

is the starting position. When

:

Hence, the displacement is never for , so the particle does not return to the starting point.

Forces in two dimensions Newton’s second law still applies:

, where and are vectors and is a scalar.

WORKED EXAMPLE 19.10

A particle of mass the particle is

moves under the action of a constant force, N. When and when

the velocity of

seconds its displacement from the initial position is

. Find the vector . You are going to find the acceleration and then use

.

,

,

Use

,

Since the force is constant the acceleration is also constant, so we can use the constant acceleration equations.

:

Now use

Newton’s second law

.

means that the particle accelerates in the direction of the net force.

WORKED EXAMPLE 19.11

A particle moves under the action of two constant forces, . The particle’s acceleration is

and . Find the value of .

The resultant force should be in the same direction as the acceleration. If this is in the direction of , then

This means that the -component is zero.

If the net force changes with time you need to use integration to find an expression for the velocity. Remember that the direction of the velocity vector shows the direction of motion, whereas its magnitude gives the speed; both of these may change with time. WORKED EXAMPLE 19.12

A particle of mass starts from rest and moves under the action of the force Find the speed and direction of motion of the particle after seconds.

N.

You need to find the velocity, which we can do by integrating the acceleration. Use

first.

Remember to find . When

You can now use the given value of to find the velocity vector.

For the speed:

Speed is the magnitude of the velocity vector.

For the direction:

The direction of motion is given by the velocity vector; its direction is determined by the angle it makes with the vector .

The direction of motion is above the horizontal.

WORK IT OUT 19.2 A particle starts from rest and moves with acceleration

. Find an expression for

the velocity of the particle. Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1

.

Initially at rest means that the speed is zero. When

:

Solution 2

Initially at rest:

Solution 3

Initially at rest, so

.

EXERCISE 19C 1

For the particles moving with the given displacements, find expressions for the velocity and acceleration vectors. Also find the speed when . a

i ii

b

i ii

2

For a particle moving with the accelerations given below, find expressions for the velocity and displacement vectors. The initial displacement is zero, and the initial velocity is given in each question. Also find the distance from the starting point when a

i ii

b

i ii

c

, ,

i ii

3

An object of mass moves under the action of the force speed of the object at time seconds, in the following cases. a

. The object is initially at rest. Find the

i ii

b

i ii

4

A particle moves in a plane with the displacement from the starting point given by a Find an expression for the velocity of the particle at time . b Find the speed of the particle when

5

.

A particle moves in a plane, starting from rest. Its acceleration varies according to the equation

a Find an expression for the velocity of the particle at time .

.

b Find the displacement from the initial position after seconds. 6

The velocity, measured in

of a particle moving in a plane is given by .

a Find the initial speed of the particle. b Find the magnitude of the acceleration when

.

c Find an expression for the displacement from the initial position after seconds. 7

A particle of mass particle is

8

A particle starts from rest and moves with acceleration from the initial position after seconds.

9

The velocity of a particle moving in a plane is given by

is subjected to a constant force . The initial velocity of the . Find the velocity of the particle after seconds. . Find its distance

. Show that the particle

never returns to its initial position. 10 A particle is acted upon by two forces,

and

particle moves in the horizontal plane with acceleration

. The

. Find the value of .

11 A particle of mass

moves in the horizontal plane under the action of three forces, and . The particle moves with constant acceleration . Find the values of and .

12 For a particle moving in two dimensions, the displacement vector from the starting point is given by . a The components of the displacement vector give parametric equations of the trajectory of the particle, curve,

. Use parametric differentiation to find the gradient of the tangent to this .

b Find the velocity vector when

. What do you notice?

13 A particle moves with acceleration

. Its initial velocity is

. a Show that the speed of the particle is constant. b By considering the and components of the displacement vector, show that the particle moves in a circle. 14 A particle moves in the plane, from the initial position

m. Its velocity,

, at time s is given

by the equation: Find the time when the particle is closest to the origin, and find this minimum distance.



Section 4: Vectors in three dimensions In the previous sections you have learnt how to use vector equations to represent motion in two dimensions. But you live in a three-dimensional world, so you need to extend vector methods to be able to describe positions and various types of motion around you.

Did you know? Although our space is three-dimensional, it turns out that many situations can be modelled as motion in two dimensions. For example, it is possible to prove that the orbit of a planet lies in a plane, so two-dimensional vectors are sufficient to describe it.

To represent positions and displacements in three-dimensional space, you need three base vectors, all perpendicular to each other. They are conventionally called .

You can also show the components in a column vector:

.

B

k A

j i AB = 3i + 2j + 4k

Each point in a three-dimensional space can be represented by a position vector, which equals its displacement from the origin. The displacement from one point to another is the difference between their position vectors. WORKED EXAMPLE 19.13

Points and have coordinates a the position vectors of and

and

, respectively. Write as column vectors:

b the displacement vector

. The components of the position vectors are the coordinates of the point.

a

b

The formula for the magnitude of a three-dimensional vector is analogous to the two-dimensional vector.

Key point 19.3 The magnitude (modulus) of a vector

is

.

The distance between points with position vectors and is

.

Tip It is useful to remember that this formula comes from Pythagoras’ theorem.

WORKED EXAMPLE 19.14

Points and have position vectors

and

. Find the exact distance

.

The distance is the magnitude of the displacement vector, so you need to find first.

Now use the formula for the magnitude. Remember that you can use vector addition and subtraction to combine displacements. WORKED EXAMPLE 19.15

The diagram shows points

such that

. P Q i + j - 3k

- 2j + 5k M

3i - 2j + 6k N

Write the following vectors in component form. a b c

,

and

a

You can get from

b

Going from to

c

You can get from to via previous parts.

to via .

is the reverse of going from

, using the answers from

EXERCISE 19D 1

Write the following vectors in column vector notation (in three dimensions). a

i ii

b

i ii

2

Let

a

,

and

. Find the following vectors.

i ii

b

i ii

c

i ii

d

i ii

3

Let a

,

and

. Find the following vectors.

i ii

b

i ii

c

i ii

4

Find the magnitude of the following vectors in three dimensions.

a

b c d

to .

5

Find the distance between the following pairs of points in three dimensions. a

i

and

ii

and

b

and

i

and

ii 6

Find the distance between the points with the given position vectors. a

and

b

and

c

and and

d 7

Given that a

such that:

is the zero vector

b c d 8

Given that

and

9

Given that .

,

, find vector such that

and

.

, find the value of the scalar such that

10 Find the possible values of the constant such that the vector

has magnitude

.

Worksheet See Support sheet 19 for a further example of three-dimensional vectors and for more practice questions.

11 Let

and

. Find the possible values of such that

12 Points and are such that the possible values of such that

and

.

, where is the origin. Find

.

13 Points and have position vectors and . Find the value of for which the distance is the minimum possible and find this minimum distance.



Section 5: Solving geometrical problems This chapter concludes by reviewing how vector methods can be used to solve geometrical problems. You have already used these results: The position vector of the midpoint of line segment

is

If vectors and are parallel, then there is a scalar such that The unit vector in the same direction as is

. .

.

Rewind These results were introduced in Student Book 1, Chapter 12.

WORKED EXAMPLE 19.16

Points

and have position vectors

is the midpoint of

. Point

.

a Find the position vector of . b Show that

is a parallelogram. C

D

Draw a diagram to help you see what is going on.

E

A

B

a

The position vector of the midpoint is the average of the two position vectors.

b

In a parallelogram, opposite sides are equal and parallel, which means that the vectors corresponding to those sides are equal. So you need to show that .

So

is a parallelogram.

Tip Vector diagrams don’t have to be accurate or to scale to be useful. A two-dimensional sketch of a situation is often enough to show you what’s going on.

WORKED EXAMPLE 19.17

Given vectors

,

and

:

a Find the values of and such that is parallel to . b Find the value of scalar such that a Write Then:

is parallel to the vector

for some scalar .

.

When two vectors are parallel you can write

.

If two vectors are equal, then all their components are equal.

We can write vector

b

use Parallel to

in terms of and then .

:

You can find from the first equation, but you need to check that all three equations are satisfied.

WORKED EXAMPLE 19.18

a Find the unit vector in the same direction as

.

b Find a vector of magnitude parallel to . a

To find a unit vector, you need to divide the given vector by its magnitude (as the resulting vector will then have length 1).

b Let be parallel to and . Then

To get a vector of magnitude you need to multiply the unit vector by .

.

Tip Note that part b has two possible answers, as could be in the opposite direction. To get the second answer we would take the scalar to be instead of . You can use vectors to divide a line segment according to a givenratio. WORKED EXAMPLE 19.19

Points and have position vectors and . Find, in terms of and , the position vector of the point on such that . A

Always start by drawing a diagram. Since the question is about position vectors, include the origin. M

O B

You can get from to via either or . (You may want to confirm that both give you the same answer!) means that

.

EXERCISE 19E

1

a

i Find a unit vector parallel to ii Find a unit vector parallel to

b

. .

i Find the unit vector in the same direction as ii Find the unit vector in the same direction as

.

2

Points and have position vectors

a Write

and

as a column vector.

b Find the position vector of the midpoint of the line segment 3

Points , and have position vectors of point such that

4

Given that

vector

.

,

. and

. Find the position vector

is a parallelogram.

and

, find the value of the scalar such that

is parallel to the

.

5

Given that to vector .

6

Points and have position vectors

, find the value of the scalar such that

and

. Point lies on

is parallel

so that

. Find the position vector of . 7

Points and have position vectors a Find the position vector of the midpoint b Point lies on the line

8

9

Given that parallel to vector

such that

. of

. . Find the coordinates of .

, find the values of scalars and such that .

a Find a vector of magnitude parallel to

.

b Find a vector of magnitude in the same direction as 10 Points and have position vectors and . Point position vector of in terms of and .

lies on

. and

. Express the

11 In the following diagram, is the origin and points and have position vectors and are points on and extended such that , and A

P

Q

B

O

R

Prove that: a b

is

is a straight line is the midpoint of

.

and .

Checklist of learning and understanding Constant acceleration formulae in two dimensions:

To differentiate or integrate a vector, differentiate or integrate each component separately. Vectors in three dimensions can be expressed in terms of base vectors using components. The magnitude of a vector can be calculated using the components of the vector: The distance between the points with position vectors and is given by The unit vector in the same direction as is



.

Mixed practice 19 1

A particle of mass

moves with constant acceleration

.

a Find the magnitude of the net force acting on the particle. When

the particle has velocity

.

b Find the speed and the direction of motion of the particle 2

Points and have position vectors Find the exact distance

3

Points

and

later.

. is the midpoint of

.

and have position vectors

and

a Find the position vector of the point such that b Prove that

.

.

is a parallelogram.

is a rhombus.

4

A particle moves in the plane so that its position vector at time . Find the speed of the particle when

5

Points

and have position vectors .

a Prove that the triangle

is . and

is isosceles.

b Find the position vector of point such that the four points form a rhombus. 6

A particle of mass starts from rest and moves under the action of a constant force . Find how long it takes to reach the speed of .

7

A helicopter is initially hovering above the helipad. It sets off with constant acceleration , where the unit vectors and are directed east and north, respectively. The helicopter is modelled as a particle moving in two dimensions. a Find the bearing on which the helicopter is travelling. b Find the time at which the helicopter is

from its initial position.

c Explain in everyday language the meaning of the modelling assumption that the helicopter moves in two dimensions. 8

Points and have position vectors and . is the point on the line segment such that . Find the exact distance of from the origin.

9

A particle of mass

moves in the plane under the action of the force . The particle is initially at rest at the origin. Find the direction of motion of the particle after .

10 In this question, vectors and point due east and north, respectively. A port is located at the origin. One ship starts from the port and moves with velocity . a Write down the position vector at time hours. At the same time, a second ship starts .

north of the port and moves with velocity

b Write down the position vector of the second ship at time hours. c Show that, after half an hour, the distance between the two ships is

.

d Show that the ships meet, and find the time when this happens. e How long after the meeting are the ships 11 At time

apart?

two aircraft have position vectors

velocity

. The first moves with constant

and the second with constant velocity

.

a Write down the position vector of the first aircraft at time . Let be the distance between the two aircraft at time . b Find an expression for

in terms of . Hence, show that the two aircraft will not collide.

c Find the minimum distance between the two aircraft. 12 A position vector of a particle at time

is given by

.

a Find the Cartesian equation of the particle’s trajectory. b Find the maximum speed of the particle, and its position vector at the times when it has this maximum speed. 13 A particle of mass

moves on a horizontal surface under the action of the net force .

The particle is initially at the origin and has velocity are directed east and north, respectively.

. The unit vectors and

Find the distance of the particle from the origin at the time when it is travelling in the northerly direction.

Worksheet See Extension sheet 19 for questions on modelling rotation with vectors.

20 Projectiles In this chapter you will learn how to: model projectile motion in two dimensions find the maximum height and the range of a projectile find the Cartesian equation of the trajectory of a projectile.

Before you start… Student Book 1, Chapter 12

You should be able to find the

1 Find the magnitude and direction of the vector

magnitude and direction of a vector.

Student Book 1, Chapter 20

You should be able to use constant acceleration formulae in one dimension.

2 A particle accelerates uniformly from while covering a distance of straight line. Find the acceleration.

Chapter 19

You should be able to use constant acceleration formulae in two dimensions.

3 A particle initially has velocity and accelerates at . Find its velocity after 3 seconds.

Chapter 8

You should be able to use trigonometric identities and solve trigonometric equations.

4 Express

in terms of

to in a

.

5 Solve the equation .

for

Motion in two dimensions In Student Book 1, Chapter 20 you used the constant acceleration formulae to analyse the motion of a particle that was projected in one dimension, either horizontally or vertically. In this chapter we extend this idea to look at projectiles moving in a two-dimensional vertical plane. This extension enables us to model the motion of, for example, a bullet, a golf ball or water from a fountain.

u θ

Section 1: Modelling projectile motion Consider an object projected vertically upwards at an angle. If you model the object as a particle and ignore air resistance, then the only force acting on it is gravity. If we also assume that is constant, then the acceleration is constant with magnitude and is directed downwards. There is no acceleration horizontally as there is no force acting in this direction.

Fast Forward You will see in Section 2 how to prove that the path of a projectile is a parabola.

Rewind You met the particle model in Student Book 1, Chapter 19. It assumes that you can ignore an object’s size and any internal motion, such as spin. In Student Book 1 you also explored the assumption of constant , and effects of air resistance.

Key point 20.1 In projectile motion the acceleration of the particle is





where the vector (or the -axis) is directed vertically upwards.

Rewind You met the constant acceleration formulae with two-dimensional vectors in Chapter 19, Section 2. This means that the motion can be described using the constant acceleration equations in two dimensions. WORKED EXAMPLE 20.1

A particle is projected from point with velocity a the speed and direction of motion of the particle after

. Taking to be

, find:

seconds

b the distance of the particle from at this time. You first need to find the velocity vector after Use .

a

Speed is the magnitude of velocity. 7.2 θ v

3.04

seconds.

The direction of motion is the direction of the velocity vector. The direction of motion is the horizontal.

below

Draw a diagram to make sure you are finding the correct angle. Now we want the position vector.

b

Use

Distance from the origin:

.

Distance is the magnitude of displacement.

( d.p.) The initial velocity may be specified by giving the speed and the angle of projection.

Key point 20.2 If a particle is projected with an initial speed at an angle above the horizontal, then the components of the initial velocity are: u

u sinθ θ

u cosθ

horizontally: vertically: or as a vector:

Tip Since there is no acceleration horizontally, the horizontal component of velocity is constant throughout the motion, and so the horizontal displacement is . When solving problems involving projectiles, it is often useful to consider the horizontal and vertical motion separately rather than using vectors. This means using the one-dimensional constant acceleration formulae in each direction separately. We will use and to denote the horizontal and vertical components of the displacement vector. WORKED EXAMPLE 20.2

A particle is projected from ground level with speed

at an angle of

above the horizontal.

Using , find the height of the particle above horizontal ground level at the time when its horizontal displacement from the starting point is . Horizontally:

You need to find the time first, using the horizontal displacement equation.

Use a constant acceleration formula for the vertical direction.

Vertically:

If the particle is projected from a point above ground level, then the vertical displacement can be negative, corresponding to the particle falling below the starting point. WORKED EXAMPLE 20.3

A stone is thrown from the top of a cliff that is above sea level. The initial velocity has magnitude and is directed at above the horizontal. 8m s–1 20° 30m

Find the speed with which the stone hits the sea. Horizontally:

The horizontal component of the velocity is constant.

Vertically:

Note that point.

Use

as the stone finishes

below its starting

to find the vertical component of the velocity.

Now find the magnitude of the velocity.

We are often interested in the maximum height a projectile reaches, or in how far horizontally from the starting point it lands.

Key point 20.3 A projectile is at its maximum height when

.

For a particle projected from ground level, set travelled before it returns to ground level).

vy = 0

to find the range (the horizontal distance

vx

Greatest height y = 0 Range

Tip In practical problems, the ground may not be horizontal or the projectile might hit something before returning to ground level.

WORKED EXAMPLE 20.4

A particle is projected from ground level with speed in the direction a Show that the maximum height the particle reaches is b Given that

and

above the horizontal.

.

, find the range of the projectile.

c How will the answer to part b change if air resistance is included in the model? a Vertically:

The maximum height is reached when Use

b Vertically:

to find the vertical displacement at this point.

The projectile lands when lands using

.

. First find the time the particle

.

corresponds to the starting position, so you want the other value of . You can now find the horizontal distance at this time.

c The range will be smaller.

Air resistance would cause deceleration in the horizontal direction, so the projectile would travel less far.

In all of the previous examples, the particle was projected upwards at an angle. But the same equations still apply if the particle is projected downwards. WORKED EXAMPLE 20.5

A small ball is thrown from a window below the horizontal.

above ground with the initial velocity

directed at

30° 6m s–1 12m

Find how long it takes to reach the ground.

6 sin 30°

i

6 cos 30° 30° 6

j

The vertical component of the initial velocity is downwards. In this case, it is easier to take the positive direction (the direction of vector ) to be downwards.

The displacement is measured downwards from the points of projection, so the ground is at . The acceleration is downwards, so is also positive.

Vertically:

Use

. In this case, the vertical displacement s is

denoted by . Solve the quadratic for .

It takes

must be positive.

seconds.

EXERCISE 20A In this exercise use 1

unless instructed otherwise.

A particle is projected from a point, . Find its velocity vector after seconds if its initial velocity is: a

i ii

b

i ii

c

d

i

at

above the horizontal

ii

at

above the horizontal

i ii

at at

below the horizontal below the horizontal.

2

A particle is projected from the origin. Find its position vector after seconds if its initial velocity is as in question 1.

3

A small stone is projected from ground level with speed

at an angle of

above the

horizontal. a Find its height above the ground after

seconds.

b What is its speed at this time? 4

A particle is projected with initial velocity

. The unit vectors and are directed to the

right and vertically upwards. Find: a the time it takes the particle to reach its maximum height b the magnitude and direction of its velocity 5

seconds after projection.

A ball is projected from a point on a horizontal plane with speed . The ball returns to the plane at point . Find:

at an angle of elevation of

a the greatest height above the plane reached by the ball b the distance 6

.

A particle is projected with speed at an angle above the horizontal. The greatest height reached above the point of projection is . Find, to the nearest degree, the value of .

7

A particle is projected from a point, , with speed Find the length of time the particle is more than

8

at an angle of elevation of

.

above the horizontal level of .

A ball is hit from a point, , above the ground with speed The ball lands at the point , as shown in the diagram.

at an angle of elevation of

28m s–1 45°

P

1.5m Q

Find: a the time taken for the ball to travel from to b the distance c the speed with which the ball hits the ground.

Worksheet See Support sheet 20 for an example of finding the angle and speed of projection and for more practice questions. 9

A ball is projected with speed at an angle of above the horizontal. A high wall is located from the point of projection. Determine whether the ball will clear the wall. 12m s–1 1.6m

30° 6m

10 A ball is projected horizontally with speed

from the top of a

tall building.

a Find the distance from the foot of the building and the point where the ball hits the ground. A second ball is projected with speed

from the foot of the building.

b Find the possible angles of projection so that the second ball hits the ground at the same place as the first ball. 11 A golfer is aiming to land a ball on the green. The front of the green is green is long, as shown in the diagram. The ground is horizontal.

from his position and the

.

V m s- 1 α 190m

10m

a If the golfer strikes the ball with speed

at an angle above the horizontal, show that the

horizontal distance travelled by the ball when it lands is b If

.

, find the range of values of that will result in the golfer landing the ball on the green.

12 A film director wants to include the ‘human fired out of a cannon’ stunt in his film. For ethical reasons, he uses a scale model of a cannon, which is one-tenth of the real size, with a toy fired out of it. a What is the magnitude of the force of gravity that would be observed by someone watching the film? b Does the film need to speed up or be slowed down to correct this?



Section 2: The trajectory of a projectile So far we have only looked at how the displacement and velocity of a projectile change with time. But we can also find a relationship between the horizontal and vertical displacements. This leads to an equation describing the path, or trajectory, of the projectile.

Rewind This is an example of parametric equations, which you met in Chapter 12. You can apply the method of eliminating to find the Cartesian equation of the trajectory of a particle moving in two dimensions.

Key point 20.4 To find an equation for the trajectory of a projectile: Make the subject of the

equation.

Substitute this expression for into

.

WORKED EXAMPLE 20.6

A particle is projected from ground level with speed that

at an angle above the horizontal, such

. Let and be the horizontal and vertical displacements from the point of projection,

with measured upwards, and use

.

Find an expression for in terms of . 13m

5m

θ 12m

First you need to find the components of the velocity.

Horizontally:

Make the subject of the horizontal equation.

Vertically:

And substitute into the vertical equation.

As you can see, the trajectory of a projectile is a parabola. It is important to remember that the equations we have used include gravitational acceleration, but no other force. In particular, this model of a projectile assumes no air resistance. If the air resistance is included the trajectory is no longer a parabola.

without air resistance with air resistance

You may need to find the angle of projection in order for the particle to pass through a specific point. There will often be two possible values. WORKED EXAMPLE 20.7

A particle is projected from a point, , on horizontal ground, with speed at an angle above the horizontal. The particle passes through the point , which is at a horizontal distance of from and a height of

above the ground.

a Show that b Find the two possible values of . a Horizontally:

Make the subject of the horizontal equation.

Vertically:

And substitute into the vertical equation.

and

b

.

Factorise and solve for . Note that there are two possible trajectories that pass through :

y Q

P

x

WORK IT OUT 20.1 A ball is thrown from a point, , at ground level with speed

at an angle

horizontal. It lands at a point horizontally from on a platform that is The platform starts from and is long, as shown in the diagram.

above the

above the ground.

4m 15m s- 1 P

4m

θ 10m

The trajectory of the ball is given by

Taking

, find the angle .

Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1 ,

,

and

.

as this is the first point of contact with the platform. Solution 2

,

,

and

.

as the particle needs to be on the way down. Solution 3 ,

,

and

.

EXERCISE 20B In this exercise use 1

unless instructed otherwise.

A particle is projected horizontally with speed from the top of a tall cliff. Let the origin of an coordinate system be located at the top of the cliff, with the -axis pointing vertically downwards and the -axis horizontal in the direction of projection. Find the Cartesian equation of the trajectory of the particle.

2

A rugby ball is kicked from ground level, with speed , at an angle of elevation of towards a crossbar, which is above ground and away horizontally from the person kicking the ball.

ym 45°

xm

a If the ball hits the crossbar, show that

.

b If the crossbar is high and the kick is to be taken from the posts, find the minimum velocity with which the ball must be kicked to clear the crossbar.

c State two modelling assumptions you needed to make. 3

a Show that the equation of the trajectory of a particle projected at speed elevation is:

b An archer fires an arrow from a point elevation of . The target is diameter of the target is .

at an angle of

above the ground with speed

away and the centre of the target is

at an angle of above the ground. The

44m s–1 7° 1.5m

1.3m 50m

Determine whether the arrow hits the target. 4

A particle is projected from the origin with speed the point with position vector .

at an angle of elevation . It passes through

a Show that b The particle subsequently passes through the point with position vector

.

i Show that ii Find . 5

A particle is projected from a point on horizontal ground with speed at an angle of elevation . The maximum height reached by the particle is and the particle hits the ground from the point of projection. Find and .

6

A basketball player shoots at the hoop. The hoop is from the ground and the basketball player stands horizontally from the hoop. The ball is released from above the ground at an angle of above the horizontal.

10m s–1 α 3m

2m 6m

The initial speed of the ball is

. Assume that the ball passes through the hoop.

a Show that

.

b Hence, find the angle at which the ball was released, justifying your answer.

Checklist of learning and understanding In projectile motion the acceleration of the particle is vertically upwards. The modelling assumptions for projectile motion are: the projectile is modelled as a particle

, where the -axis points

air resistance can be ignored the value of remains constant. If a particle is projected with speed at an angle above the horizontal, then the components of the initial velocity are: horizontally: vertically: A projectile is at its maximum height when For a particle projected from ground level, set distance travelled).

. to find the range (the maximum horizontal

To find an equation for the trajectory of a projectile: Make the subject of the Substitute this expression for into

equation. .

Mixed practice 20 In this exercise use 1

unless instructed otherwise.

A particle is projected horizontally with a speed of from a point above horizontal ground. The particle moves freely under gravity. Calculate the speed and direction of motion of the particle at the instant it hits the ground. © OCR, GCE Mathematics, Paper 4729, June 2010

2

A particle of mass is projected horizontally off the edge of a cliff and lands in the sea below at a distance from the base of the cliff. A second particle of mass is projected horizontally from the same point at the same speed and lands in the sea at a distance from the base of the cliff. Which one of the following statements is true? A B C D It depends on the height of the cliff.

3

A ball is thrown horizontally at

from the top of a

tall building. Find:

a the time it takes for the ball to hit the ground below b the distance from the bottom of the building to the point where the ball hits the ground. 4

A projectile is launched from the point with velocity the point and is travelling with velocity

. After seconds it is at . Find:

a the value of b the distance 5

.

A ball is thrown from a window above the ground with velocity . It is caught by a child above the ground. The child stands from the point vertically below the window at ground level. (9i + 8j) m s- 1

5m 1m xm

Using

, find:

a the value of b the speed of the ball as it is caught. 6

A particle is projected from a point, , on a horizontal plane with speed above the horizontal. It lands on the plane at the point .

at an angle

a

i Show that ii Hence, deduce that for fixed , the maximum range is achieved when

b If

.

find, to the nearest degree, the two possible angles at which the particle was

projected. 7

Kim stands on a bridge that passes over a river, which is below. He throws a stone with speed directly at a small stationary rock, which is from the base of the bridge, as shown in the diagram.

15m s- 1 6m

8m How far away from the rock does the stone land in the river? 8

A particle is projected from with velocity time seconds. (8i + 11j) m s- 1

. It passes through the point at

P

O

Q

Given that

:

a Find the value of . At point the ball has the same speed as at . b Find the time taken for the ball to travel from to . 9

30m s- 1 P

40m s- 1

V1 250m θ1 O

A

A particle is projected with speed at an angle of elevation from a point on horizontal ground. When is vertically above a point on the ground its height is and its velocity components are diagram).

horizontally and

i Show that

, correct to significant figures.

and

vertically upwards (see

At the instant when is vertically above , a second particle is projected from with speed at an angle of elevation . and hit the ground at the same time and at the same place. ii Calculate the total time of flight of and the total time of flight of .

iii Calculate the range of the particles and, hence, calculate

and .

© OCR, GCE Mathematics, Paper 4729, January 2010 10 A particle is projected from a point with speed at an angle of elevation above the horizontal and it moves freely under gravity. The horizontal and upward vertical displacements of the particle from at any subsequent time, seconds, are and , respectively. i Express and in terms of and and, hence, show that

v θ O

A (h, - h)

The particle subsequently passes through the point with coordinates the diagram. It is given that and .

, as shown in

ii Calculate . iii Calculate the direction of motion of the particle at . iv Calculate the speed of the particle at . © OCR, GCE Mathematics, Paper 4729, January 2009 11 Take to be

in this question.

A particle is projected from a point shown in the diagram.

4m s- 1

above ground level with velocity

, as

u

3m s- 1 30m

Find the distance from the point of projection at the instant the particle is moving in a direction perpendicular to its initial velocity. Give your answer to an appropriate degree of accuracy. 12 A particle is projected horizontally with speed from the top of a vertical cliff. At the same instant a particle is projected from the bottom of the cliff, with speed at angle above the horizontal. and move in the same vertical plane. The height of the cliff is and the ground at the bottom of the cliff is horizontal. i Given that the particles hit the ground simultaneously, find the value of and find also the distance between the points of impact with the ground.

ii Given instead that the particles collide, find the value of , and determine whether is rising or falling immediately before the collision. © OCR, GCE Mathematics, Paper 4729, January 2012

Worksheet See Extension sheet 20 for a selection of more challenging problems.

21 Forces in context In this chapter you will learn: how to resolve forces in a given direction in order to calculate the resultant force about a model for friction how to determine the acceleration of a particle moving on an inclined plane.

Before you start… Student Book 1, Chapter 18

You should be able to add vectors and find magnitudes.

1 Three horizontal forces act on a particle. In newtons, the forces are , and . Calculate the magnitude of the resultant force and its angle from the direction .

Student Book 1, Chapter 18

You should be able to find horizontal and vertical components of a vector.

2 A force of acts at an angle of to the horizontal. Find the horizontal component of the force.

Student Book 1, Chapter 17

You should be able to solve problems involving motion with constant acceleration.

3 A force of

acts on a particle with mass

.

If the particle is initially at rest, after how many seconds will its displacement be equal to ?

Improving our physical model In Student Book 1, Chapter 18 you learnt to add and subtract forces in vector form and to determine the angle and magnitude of the resultant force. This chapter deals with more complex situations involving strings and planes in different orientations where the forces are not perpendicular. To solve these problems you will need to combine your knowledge of forces, vectors and trigonometry. In Student Book 1 you met problems where there is a constant resistance force acting on an object. Here we shall improve our model of friction to enable us to model more real-world situations.

Section 1: Resolving forces When considering forces acting on a moving body, it can be useful to split each force into components in two perpendicular directions. This process is called resolving the force. For example, consider a force acting in the from the direction of .

plane, with magnitude

at an angle of

anticlockwise

10N

10sin 50° 50° 10cos50° We can find the components of this force in the directions of and : The component in the direction is The component in the direction is

Rewind In Student Book 1, Chapter 12, you learnt how to find the components of a vector; in Chapter 21 you learnt how to add forces given in component form. In some situations it will be useful to resolve in directions other than and . For example, the force might be acting on a particle that lies on a slope making an angle of with the horizontal. In that case we can resolve the force in the directions parallel and perpendicular to the slope: 10 N in 30 10 s

20°

°

30° 0° os 3 10 c

The component up the slope is

.

The component perpendicular to the slope is

.

Once all forces have been resolved you can find the resultant force. WORKED EXAMPLE 21.1

Two forces act on a particle in the horizontal plane. Force has magnitude and acts at angle to the positive -axis. Force has magnitude and acts in the direction of the negativ axis. Find the magnitude and direction of the resultant force. Always draw a diagram.

F1 30°

F2

6N 6 sin 30°

30° 6 cos 30°

5N

Form a right-angled triangle with the force as hypotenuse and the sides parallel to the and directions. In this case, since the force is in the direction, we need to resolve only force . Calculate horizontal and vertical components of to the horizontal direction.

and add

The arrow notation indicates which component we are calculating; for example, indicates the component in the positive direction.

3N

Draw another triangle showing the resultant force. This helps find its magnitude and direction. Note that direction.

means that the force is acting in the negative

θ

0.196 N Magnitude: Direction:

is

, so the direction from the positive -axis.

Once forces have been found you may have to work with them using Newton’s second law

.

Rewind You met Newton’s second law in Student Book 1, Chapter 18, Section 1.

WORKED EXAMPLE 21.2

A particle with mass

moves on a horizontal (

, acted upon by two forces,

a

and

) plane with constant acceleration , where

, in the plane.

has magnitude

and acts at angle

has magnitude

and acts at acute angle clockwise from the positive -axis.

Show that

anticlockwise from the positive -axis.

.

b Calculate the exact value of a.

3N

a

a

Always draw a diagram.

30° θ 5N Resolve in the direction and set equal to zero (no acceleration, so equilibrium in this direction). Rearrange to find

.

Use to convert into . Since the angle is acute we only need to take the positive root as for acute angles.

b

Resolve in the direction and use Newton’s second law.

When an object is in contact with a surface there is a contact force between them. If the contact is smooth this contact force (also called the normal reaction force) is perpendicular to the surface.

Fast Forward In Section 2 you will learn how to find the contact force in the presence of friction.

WORKED EXAMPLE 21.3

A box of mass moves on a smooth horizontal surface under the action of a constant force of magnitude acting at an angle of above the horizontal. a Find the magnitude of the normal contact force between the box and the surface. b The box’s speed increases from a

65 N R

a

35°

to

. How far does the box travel in that time?

Draw a diagram showing all the forces. Don’t forget the weight and the normal contact force. The force needs to be resolved in horizontal and vertical directions.

18 g

There is no acceleration in the vertical direction. b

Use Newton’s second law in the horizontal direction to find the acceleration. Use

to find the distance travelled.

When an object is in equilibrium, the resolved part of the net force in any direction is zero.

Fast Forward You will meet more complex equilibrium problems in Section 4.

Fast Forward

In Section 4 you will see that you can also solve the following problem using a triangle of forces.

WORKED EXAMPLE 21.4

A particle of mass angles of and

is attached to the ceiling by two light inextensible strings. The strings make with the horizontal, as shown in the diagram.

30°

35°

4 kg Given that the particle is in equilibrium, find the magnitude of the tension in each string.

T1

35°

30°

Resolve the tension forces in horizontal and vertical directions…

T2

4 kg Since the particle is in equilibrium the sum of the horizontal components is zero… … and the sum of the vertical components is zero. These are two simultaneous equations for and .The best way to solve them is to substitute from the first equation into the second.

From the first equation:

Substitute into the second equation: It is best to enter the expression in the large bracket into your calculator in one go, then divide by it.

Use the ‘unrounded’ value of

in this calculation.

The next example illustrates (in part ) how to resolve forces in a direction that is not horizontal or vertical.

Fast Forward Resolving in the direction of motion will also be used with inclined planes in Section 3.

WORKED EXAMPLE 21.5

A toy helicopter of mass

has its rotors set so that the force provided by the engine is given by

, where and can be varied using a remote control, but the magnitude of is always The unit vector is horizontal and is vertical. Take . Find the acceleration of the helicopter in the following two cases. a the helicopter is set to fly horizontally b the helicopter starts from rest and accelerates at

above the horizontal.

Always draw a diagram. You could define to be the angle the

.

force makes with the vertical or the horizontal. Here it is convenient to choose the vertical.

a

There is no vertical acceleration.

Use Newton’s second law in the horizontal direction.

b Resolve in the direction of motion. The force acts at angle above :

The resultant force acts above the horizontal, so choose this direction to resolve in. The net force in the perpendicular direction is zero. (You could solve this problem by resolving horizontally and vertically. However, resolving in the direction of motion gives simpler equations.)

Perpendicular to movement:

In direction of movement:

EXERCISE 21A 1

In each of the following systems, a particle on a horizontal surface is acted on by two forces, and . Find the and components of the resultant force, where represents due north and represents due east. a

b

2

i

,

has magnitude

ii

,

has magnitude

and acts at bearing and acts at bearing

. .

i

has magnitude

and acts at bearing

,

has magnitude

and acts at bearing

.

ii

has magnitude

and acts at bearing

,

has magnitude

and acts at bearing

.

A particle of mass the horizontal.

hangs at rest attached to two light inextensible strings at angles and from

θ1

θ2

m

For each system, find the tension in each string. a

b

3

i

,

ii

,

, ,

i

,

,

ii

,

,

Find the resultant force acting on the object in the diagram below, giving your answer in the form

12N 5N

j i

30°

30N

4

In the following diagram, the particle is in equilibrium.

150N θ F

100N a Find the angle . b Find the force . 5

A mass of is being held in equilibrium, in the vertical plane, by a string at an angle to the vertical with tension and a horizontal force of . You may take .

Tip Remember that if a particle is in equilibrium, the resultant force is zero. a Show that

.

b Find the value of . c Find the value of . 6

Three friends are pulling a

load across a smooth horizontal surface, using horizontal ropes. Alf

pulls with a force of . Bert pulls with a force and is positioned at an angle clockwise from Alf. Charlie pulls with force and is positioned at an angle clockwise from Bert. Modelling the load as a particle, and given that the load begins to move exactly towards Bert, find and determine the initial acceleration of the load.

7

Two forces of magnitudes

and

act at a point, .

a Given that the two forces are perpendicular to each other, find: i the angle between the resultant and the force ii the magnitude of the resultant force. b If it is given instead that the resultant of the two forces acts in a direction perpendicular to the force,find: 8N

RN

P

15N

i the angle between the resultant and the

force

ii the magnitude of the resultant. 8

A particle with mass hangs at rest suspended by two light inextensible strings. One string is at an angle of to the horizontal and the other is at an angle of to the horizontal.

30°

45°

Find the tension in the two strings. 9

A small, smooth ring, , of weight is threaded on a light, inextensible, taut string. The ends of the string are attached to fixed points and at the same horizontal level. A horizontal force of magnitude is applied to . In the equilibrium position the angle is a right angle, and the portion of the string attached to makes an angle with the horizontal. A

B θ 4N R 5N

a Explain why the magnitude of the tension is the same in each part of the string. b Find and . 10 A light, smooth ring, , is threaded on a light, inextensible string. One end of the string is attached to a fixed point, . The other end of the string is threaded through a fixed smooth ring, directly below , and attached to a particle of mass .

. The angle

equals

A force of magnitude

is applied to ring at an angle with the horizontal.

a Given that the string is taut, find the exact value of required to maintain the system in equilibrium with angle b Given that

and

and the system is in equilibrium, find the relationship between and .

c If ring actually has mass with angle

.

and

, find the new force, , required to hold the system in equilibrium , and determine the new value of .

Section 2: Coefficient of friction In Student Book 1 you met friction as a constant force opposing motion. However, the magnitude of the frictional force depends on: the force pushing the object into the surface the roughness of the object and the surface any external forces applied to the object. Imagine a heavy box resting on a rough horizontal floor. You pull the box with a force . When is small, the box remains at rest and the frictional force exactly counter balances .

R Fr

P

mg If the pulling force is strong enough, friction will be overcome and the box will start to accelerate. There must therefore be a maximum value of friction, called limiting friction. When friction is at its limiting value, the object is on the point of moving and is said to be in limiting equilibrium. If the box is heavier it will take a larger force to move it; this suggests that the magnitude of the limiting friction depends on the magnitude of the normal contact force between the object and the surface. Similarly, a box is more difficult to move on a rough carpet than on a smooth floor.

Focus on … Focus on Problem solving 3 asks you to decide whether values of certain quantities (including frictional force and µ ) are reasonable in the given contexts.

Key point 21.1 The maximum or limiting value of friction,

, is given by: µ

where is the normal contact force between the object and the surface, and µ is the coefficient of friction.

Friction

The diagram shows how the frictional force depends on the external (pushing or pulling) force.

Fmax = µR

µR Driving force If the external force is not sufficiently strong to overcome the object remains at rest. In this case µ .

, the frictional force will just match it so that

Once the object is moving, the frictional force remains constant at

.

If more than one force is acting on an object, the frictional force will match the component of the resultant

force parallel to the surface, up to the limiting value.

Key point 21.2 If the sum of all the forces on a particle, excluding friction F, is smaller than

, then:

the particle will remain at rest will have an equal magnitude and opposite direction to the sum of the other forces.

Explore The coefficient of friction between the ground and a static object is in fact usually slightly higher

Friction

than the coefficient of friction between the ground and a moving object; the graph should look more like this.

kinetic

static

Pulling force Find out more about static and dynamic (kinetic) friction.

WORKED EXAMPLE 21.6

A block of mass lies at rest on a rough horizontal surface. Attached to opposite ends of the block are two light strings, each of which is under tension. The coefficient of friction, µ . Take .

T1 = 5N

10kg

T2 = 5N

a The tension in the left string is increased to Show that the block remains at rest.

.

b The tension in the right string is then increased to . Show that the block moves, and determine its acceleration.

R

a

10N

Always draw a diagram. If the block moves it will be to the left, so friction acts to the right.

5 N Fr 10g Since any movement will be horizontal, there is no vertical acceleration.

Maximum friction is µ .

Calculate limiting friction µ .

Horizontal forces without friction:

The frictional force can take

Friction will oppose the horizontal resultant without friction, up to a

maximum of limiting friction. Since the resultant without friction is less than limiting friction, the object will not move. Friction will just balance the resultant of the other forces.

the value 5 N, so the block will not move. b As before

.

If the block moves it will now be to the right, so friction acts to the left.

a R 10N

30N

Fr 10g

The resultant of the horizontal forces without friction exceeds the limiting friction, so friction takes it limiting value.

Horizontal forces without friction: The frictional force takes the limiting value

. Use

in the horizontal direction.

Notice that the frictional force acts in different directions in part a and part b of Worked example 21.6 because it must always be in the opposite direction to that in which the object would move in the absence of friction.

Tip In some questions you may be initially uncertain in which direction the friction acts. If you calculate a negative friction force in your answer, you have probably drawn the friction in the wrong direction. Change your diagram and adjust your equations.

Focus on … See Focus on … Proof 4 for proofs of formulae for the minimum force required to move a particle on a rough surface, in various situations.

WORKED EXAMPLE 21.7

Two blocks, and , lie at rest on a rough horizontal surface, with block on top of block . The coefficient of friction between the blocks is , the mass of block is and the mass of block is . The coefficient of friction between block and the surface is . A horizontal light, inextensible string is attached to block , and the tension in the string is . A

5kg

B

8kg

T

a Show that the system remains at rest. The tension in the string is increased to

and block begins to slide on block .

b What is the acceleration of each block? a Treating the blocks as a single object

Since you are interested in the motion of the whole system, you can treat the two blocks as a single object.

of mass between

, let frictional force

and the surface be

a

RA+B

A+B

13kg

T = 5N

Consider horizontal and vertical components separately. Remember to include the normal reaction force in your diagram.

FBS

13g There is no movement in the vertical direction, so the net force is zero.

Friction will oppose the tension up to a maximum of limiting friction.

Limiting friction: The tension is less than the limiting frictional force, so and the system remains at rest. b

exceeds the limiting friction so the blocks will move, under the action of force . Let the frictional force between the blocks be .

If is sliding on , itself must be sliding on the floor. (There will be different frictional force between the two blocks, and between block and the floor.)

Since the two blocks may move with different accelerations, consider forces for each block separately, starting with block .

For the upper block, : aA

RA 5kg

FAB

5g

Since the block is moving, the magnitude of the frictional force is .

For block :

Now look at the lower block, considering vertical components first. According to Newton’s third law, the normal reaction force acts downwards on block .

aB

RA RA + B

FAB 8kg

FBS

T = 100 N

8g

Substituting for

from (1):

The frictional force between block B and the surface is . Substituting

The frictional force acting on at the contact with equals from the calculation above (but acts to the left) — this follows from Newton’s third law.

from (2):

Did you know?

Wheel turning

Road pushing

Although you may normally think of friction as a force that resists motion, it in fact enables motion in some cases. For example, you can walk only because of the friction between your feet and the ground (which is why it is so difficult to walk on ice). Friction also enables a car to move forward: as the wheels rotate, the friction force prevents slipping, and this in turn produces a forward force on the car. It may be necessary to resolve all the forces before the direction of the frictional force can be calculated. WORKED EXAMPLE 21.8

A particle of mass

rests on a rough horizontal surface and the coefficient of friction between the

particle and the surface is

. The unit vectors and are both in the horizontal plane.

Two horizontal forces act on the particle: a The frictional force b A third force

and

.

. Find and is applied. Find the new frictional force in the form

a

Find the sum of all the given forces. Sum of all the forces without friction is:

Normal reaction equals weight since there is no vertical movement:

Compare to the maximal friction.

.

Limiting friction µ The particle will not move and the friction will counter the force.

b

The sum of the two forces is less than maximal friction, so the particles will be in equilibrium.

Find the sum of the three given forces. The magnitude is:

Limiting friction

Compare to the magnitude of the maximal friction (as found in part a).

The friction force has magnitude in the direction of . Hence, the friction force is:

The magnitude of the friction force is opposite to the that of the

and its direction is force (which has

magnitude ). To change the magnitude without changing the direction, divide by and multiply by .

The magnitude of the contact force When an object is in contact with a rough surface, two forces act on it due to this contact: the normal reaction force (also called the normal contact force) the frictional force. The resultant of these two forces is the total contact force between the object and the surface. Because the two forces are perpendicular to each other, it is straightforward to find the magnitude of the resultant.

Key point 21.3 The contact force between two surfaces has two components: the normal contact force ( ) and the frictional force ( ). The magnitude of the contact force is.

WORKED EXAMPLE 21.9

A box of mass the table is horizontal.

rests on a rough horizontal table. The coefficient of friction between the box and

. A force of magnitude

is applied to the box, at an angle of

above the

Given that the box is on the point of moving: a Find the value of . b Find the magnitude of the contact force between the box and the table. Draw a diagram showing all the forces. You need to include the

weight, the normal contact force and the frictional force, as well as the force .

a

Resolve horizontally and vertically. As the box is not moving, the net force in each direction is zero. As the box is on the point of moving, the friction has its limiting value. Since you want to find , substitute from the second equation into the first. Collect the terms containing and factorise. Take

b

The contact force is resultant of the normal contact force and the frictional force.

Use the second equation from part a to find . Use

µ

to find the frictional force.

The magnitude of the contact force is

EXERCISE 21B 1

In each of the following problems, a block of mass is pulled along a rough horizontal surface by a light horizontal rope with tension newtons. The acceleration is . The coefficient of friction is µ . Use

.

a m s–2 T N

m kg

a Find µ when: , i ii

,

, ,

b Find if: ,

i ii 2

,

, µ , µ

A full skip of mass

has a coefficient of friction of

required to move the skip? Use 3

A block of mass applied to it.

with the road. What horizontal force is

.

lies in limiting equilibrium on a horizontal surface, with a horizontal force of

a Find the coefficient of friction between the block and the surface. b Find the magnitude of the contact force between the block and the table. 4

A child pulls a toy box of mass

across a rough floor, using a light string tied to the box at one

end. The tension in the string is and the string remains at an angle of to the horizontal. If the coefficient of friction between the box and the floor is , what is the acceleration of the box? 5

Two particles, and , are connected by a light inextensible string.

A P

8N

B

5N Particle , which weighs

, is placed on a rough horizontal

surface. The connecting string runs horizontally to a pulley, , at the end of the surface, then vertically downward to , which weighs

. The pulley is light and

free to rotate. a The system is in limiting equilibrium. Calculate the coefficient of friction between and the surface. b A smooth ring with weight

is threaded on to the string so that it lies on particle . Calculate the

downward acceleration of . 6

Two particles, and are connected by a taut, light, inextensible string and lie at rest on a horizontal plane, with one particle at point and the other at point . weighs and weighs . The coefficient of friction between either particle and the surface near point is µ and the coefficient of friction between either particle and the surface near point is µ .

Rewind Connected particles, including pegs and pulleys, were covered in Student Book 1, Chapter 22. When is at , a horizontal force of

is applied to acting in direction,

limiting equilibrium. When is at , a horizontal force of

and the system is in

is applied to acting in direction,

and the system is once again in limiting equilibrium. Calculate 7

and

A car of weight

. travelling at

skids to a halt in

, taking seconds.

a Assuming a constant braking horizontal force, find the deceleration of the car. b Assume that friction is the only horizontal force acting on the car and that coefficient of friction between the car and the road. 8

. Find the

A particle of mass rests on a rough horizontal table. The coefficient of friction between the particle and the table is . A light inextensible string, inclined at an angle above the horizontal, is attached to the particle. The tension in the string is

.

a Show that the particle remains at rest. b Find the magnitude of the contact force between the particle and the table. 9

A particle of mass lies on a rough horizontal surface. The coefficient of friction between the particle and the surface is µ . A horizontal force acts on the particle, which is in limiting equilibrium. Show that the magnitude of the contact force between the particle and the surface is.

10 A particle of mass

lies on a rough horizontal surface, with the coefficient of friction between

surface and particle equal to

.

T N

1kg

θ

A light inextensible string is attached to the particle and tension is applied. The string is at an angle above the horizontal, where

.

a Given the system is in limiting equilibrium, show that the tension in the string satisfies the equation

for some value and find .

b The string will break if . Find the range of values for θ for which the particle will be caused to move by tension in the intact string. c Find the maximum possible acceleration for the particle. 11 In the model for friction used in this section, which of the following factors affect the frictional force of a surface acting on an object? A the contact surface area between the surface and object B the speed of the object C the acceleration of the object D lubrication between the surface and the object

Section 3: Motion on a slope If an object is on a slope (sometimes called an inclined plane) it is often convenient to resolve forces parallel or perpendicular to the slope. We use the same technique as for all resolving problems: drawing forces as the hypotenuse of a right-angled triangle with sides parallel and perpendicular to the slope. WORKED EXAMPLE 21.10

A block of mass

slides down a smooth slope inclined at

to the horizontal. Find the

acceleration of the block. a

Always draw a diagram. The only force acting on the block is its weight.

20°

Form a right-angled triangle with the force as hypotenuse and the sides parallel and perpendicular to the slope. The angle at the top of the triangle is . (See text after this example for a general explanation.)

2g cos 20°

20° 2g

2g sin 20°



Use

in the direction down the slope.

Weight comes up so often in inclined plane problems that it is useful to remember these general results:

Let be the angle that the slope makes with the horizontal. The green angle equals the blue angle is .

and, therefore,

Key point 21.4 The components of the weight of an object on a slope inclined at an angle to the horizontal are: down the slope: perpendicular to the slope:

Tip A good way of remembering this is to think about what happens when is zero. Then there is no component of weight parallel to the plane, and a component perpendicular to the plane.

WORKED EXAMPLE 21.11

A block with mass lies on a smooth slope inclined at equilibrium by a string with tension parallel to the slope.

to the horizontal, and is held in

Calculate and the normal reaction force, . The system is in equilibrium, so the components of the resultant force parallel and perpendicular to the slope will be zero. (In fact, resolving in

R

any direction will give a zero net force.)

T 15°

15°

3.5g Perpendicular to the slope: Resolving perpendicular to the slope gives an equation for . Parallel to the slope: Resolving parallel to the slope gives an equation for .

Some problems about motion on a slope also involve friction. As before, identify the direction of movement in the absence of friction, and assign the frictional force to act in the opposite direction. WORKED EXAMPLE 21.12

A rough inclined plane makes angle with the horizontal. A particle, , of mass lies at rest on the slope. a Given that is on the point of slipping down the slope find, in terms of , the coefficient of friction between and the slope. b The particle is now attached to a light inextensible string. The string passes over a light pulley at the top of the slope. A particle, is attached to the other end of the string and hangs freely under gravity. The particle is on the point of moving up the slope. Show that

R

a

Fr = mR

Draw a diagram: The forces acting on the particle are its weight and friction. Without friction, the particle would move down the slope. Hence, the friction acts upslope.

θ θ

mg Perpendicular to the slope:

Resolve perpendicular to the slope to find .

Down the slope:

The particle is in equilibrium, so the component of the resultant force parallel to the slope is zero.

But the particle is on the point of

Since the particle is on the point of slipping, the friction is limiting.

slipping, so

µ

R

b

The forces acting on now are its weight, friction, the tension in the string and the normal contact force.

T T

P Fr = mR

Q

θ

Since is on the point of moving up the slope, friction acts down the slope. The forces acting on are its weight and tension.

θ Mg mg

Perpendicular to the slope:

Consider, the contact normal forces on first.

Down the slope:

The particle is in limiting equilibrium again, so net force is zero.

µ

Use µ

µ

Use

µ

and the

from part a. .

Now consider forces on , which is also in equilibrium.

Forces on :

WORKED EXAMPLE 21.13

A block, , with mass

lies on a rough surface inclined at

friction between block and surface is

to the horizontal. The coefficient of

.

A light, inextensible string attached to the block passes over a smooth peg at the highest point of the surface (so that the length between block and peg is parallel to the slope) and is attached at its other end to a particle of mass , which hangs freely.

B 5 kg 3kg

15°

The system is initially held at rest and then released. You may assume that the block does not reach the peg or the bottom of the slope, and that the particle does not reach the peg or the floor. a Calculate the direction and magnitude of the frictional force between and the surface. b After the block has travelled metre, the string breaks. Calculate the total distance it travels in the subsequent 10 seconds. R

Always start by drawing a force diagram. You don’t know which way the friction acts, so you will first find the direction the system would move if there was no friction.

T T a

15° 15° 3g 5g

Assume that the block is moving down the plane. If you get negative acceleration that will tell you that it is in fact moving up.

For the block:

For the particle:

Since you are interested in the direction of , add the two equations to eliminate . Negative acceleration means that the block actually moves up the slope.

Limiting friction:

You have found that, if there was no friction, the block would move up the slope. With friction, the block will move only if the sum of forces without friction is larger than the limiting friction.

Block

Resolve perpendicular to the slope to find .

Net force on block without friction:

Hence, the block moves up the slope; the frictional force acts down the force and has magnitude . b There are two separate phases of motion: Phase 1: Moving up the slope with the string attached. Phase 2: Moving up the slope without string, slowing down. Phase 3: Possibly sliding back down the slope (or being kept still by

Compare the limiting friction forces without friction .

to the sum of all the other

The friction is not strong enough to prevent motion up the slope. You now know that the block initially accelerates up the slope. When the string breaks, the block will continue to move up for a while, slow down and stop. It then may start moving back down again, depending on whether the friction is strong enough to counteract gravity.

friction). You need to know the speed of the block when the string breaks, so that you can find how far it travels before stopping. For this, you need to find the acceleration.

Phase 1: R T T a

15° 15° 3g 5g

Particle:

Use ‘ ’ for both particle and block, in the direction of motion for each one.

Block:

You found in part that the magnitude of the frictional force is .

You now need to know the speed of the block after it has travelled , using constant acceleration formula ’.

When the string breaks, the forces on the block are the component of gravity and friction, both with the same

magnitude and direction as in equation

.

Use a constant acceleration formula again to find the distance travelled before stopping.

You also need to know whether the block takes less than seconds to stop. Phase 3: Movement after coming to rest

R

Fr

Once the block has stopped, it may or may not start to slip down the slope. You need to find out whether friction is strong enough to overcome the component of gravity pulling it down the slope. You already know that the limiting friction is The only movement after the string breaks is

. .

15°

15° 5g

Total force acting down the slope is . The force down the slope is less than maximal friction, so the block will remain at rest. Total distance travelled after the string breaks is therefore equal to

EXERCISE 21C Unless otherwise indicated, use 1

in this exercise.

A particle of mass is released from rest on a slope inclined at an angle to the horizontal, with coefficient of friction between particle and slope being µ . For each of the following cases, determine the force of friction acting on the particle. a

b

c

2

i

µ

ii

µ

i

µ

ii

µ

i

µ

ii

µ

A particle of mass

and a particle of mass

are connected by a light, inextensible string.

Particle lies on a slope inclined at an angle to the horizontal. The string passes from parallel to the line of greatest slope, and runs over a light pulley (which can rotate freely) at the top of the slope, then descends vertically to .

P m kg M kg

Q

θ

The coefficient of friction between and the slope is µ . For each of the following systems, find: A the force of friction acting on and its direction B the acceleration of when the system is released from rest. a

b

c

3

i

µ

ii

µ

i

µ

ii

µ

i

µ

ii

µ

A particle of mass is projected with velocity plane inclined at to the horizontal. Using

up the line of steepest slope of a long smooth calculate:

a the normal contact force between the plane and the particle b the total distance travelled in the first seconds. 4

A child of mass

is sliding down a slide at an angle of

to the horizontal.

a Find the normal contact force between the slide and the child. b Find the acceleration if: i the slide is smooth ii there is a coefficient of friction of 5

A block of mass

between the child and the slide.

lies at rest on a rough board, with µ

. If the board is slowly raised at one end,

beyond what angle will the block begin to slide? 6

A particle is projected upwards along a line of greatest slope from the foot of a surface inclined at to the horizontal. The initial speed of is and the coefficient of friction is . The particle comes to instantaneous rest before it reaches the top of the inclined surface. a Calculate the distance moves before coming to rest. b Calculate the time takes before coming to rest. c Find the time taken for to return to its initial position from its highest point.

7

A block of mass lies on a smooth plane inclined at to the horizontal. One end of a light, inextensible string is attached to B, and it runs from upslope parallel to the line of greatest slope on the plane to a smooth peg. The other end of the string is attached to a particle, , of mass hangs vertically below the peg exactly metre from the floor.

Worksheet See Support sheet 21 for a further example on limiting equilibrium on a rough inclined plane, and for more practice questions.

that

The system is released from rest. a Find the acceleration of up the slope. When hits the floor, the string breaks. b Assuming the initial distance between the block and the peg is sufficiently great that the block will not reach the peg, find the total distance travelled by the block when it instantaneously passes through its initial position. 8

A block of mass lies on a flat, rough surface. The coefficient of friction between the block and the surface is µ . The surface is initially horizontal, and its inclination is gradually increased. When the inclination of the surface exceeds

to the horizontal, the block begins to move.

a Find µ . b The inclination of the surface is increased further to contact force between the block and the surface. 9

. Find, in terms of , the magnitude of the

Two small blocks, and , connected by a light inextensible string, lie with above on the line of greatest slope of a rough plane inclined at to the horizontal. Block has mass and block has mass the slope.

. The connecting string is metre long, and block begins metres above the foot of

1m

B

A 2m

30° The coefficient of friction between each block and the surface is The system is released from rest at time

.

; when block reaches the foot of the slope, its motion

stops immediately. Find the time at which collides with . 10 A particle with mass lies on a rough plane inclined at to the horizontal. A light, inextensible string connects to , runs parallel with the line of greatest slope of the plane to a smooth peg, then vertically downwards through a smooth, free ring , with mass upwards to attach to a fixed point .

, and then vertically

S P R

30°

2g

4.5g The coefficient of friction between and the plane is

.

a Let be the acceleration of the ring when the system is released form rest. By considering the distance moved by each object, explain why the acceleration of is

.

b By considering forces on , find an equation linking and with friction . c Find the direction and magnitude of the frictional force.

d Determine whether will remain stationary, move up the slope or move down the slope when the system is released from rest. 11 A toy with mass

is placed on a rough surface inclined at

to the horizontal. Propulsion of the

toy is achieved by directing two small fans, one on either side of the toy, set at an angle to provide a driving force for movement. The total force produced by the fans is and is directed to make the toy move up the slope.

10N θ

30°

0.5kg

The coefficient of friction between the toy and the surface is

.

a Find the acceleration of the toy up the slope if the fans are set to blow horizontally. b Find the acceleration of the toy up the slope if the fans are set to blow parallel to the slope. c The fans are set so that the direction of the driving force is above the slope (i.e. at above the horizontal, where . Find the value of that maximises the acceleration. 12 A block of mass

lies on a surface inclined at

to the horizontal. A light, inextensible string is

attached at one end to , and runs from up the slope parallel to the line of greatest slope on the plane to a smooth peg . The string passes over the peg, through a smooth ring of mass , and is attached to a wall at

.

120°

W

B

R

45° Given that the angle

equals

and the system is in equilibrium, determine the possible values

for µ , the coefficient of friction between block and the surface. 13 A particle is projected up a rough slope inclined at an angle to the horizontal. The coefficient of friction between the particle and the slope is µ . The acceleration on the way up is twice the acceleration as it travels down. Prove that µ



Section 4: Further equilibrium problems Throughout this chapter, and in Student Book 1, you solved problems about particles in equilibrium using the fact that the resultant force is zero. You usually resolved forces either horizontally and vertically, or parallel and perpendicular to a slope. However, you could look at the sum of the components of all the forces in any direction.

Key point 21.5 A particle is in equilibrium if and only if the sum of the components of all the forces in any given direction is zero. It is often helpful to look at the components in a direction perpendicular to one of the unknown forces. You may need to calculate some additional angles. WORKED EXAMPLE 21.14

A particle of weight is attached to two light inextensible strings, which are also attached to a horizontal ceiling. The strings are perpendicular to each other and make angles of and with the ceiling.

35°

55°

80 N Find the tension in each string. 35°

Look at the components in the direction of each string.

55°

T1

T2

α

80 cos α

Since the strings are perpendicular, only the weight needs to be resolved. Adding a perpendicular line allows you to find the angle between the weight and one of the strings.

α 80 sin α 80 N

Since the particle is in equilibrium, the net force in the direction of each string is zero.

You could also solve the previous problem by resolving horizontally and vertically; you would then need to solve two simultaneous equations for and . There is an alternative way to solve equilibrium problems, without resolving any forces. Remember that vectors can be added by drawing arrows, joining the ‘tail’ of one vector to the ‘head’ of another. If the vectors add up to zero, this will result in a closed polygon.

d c a

a b

a + b + c = 0

c b

a + b + c + d = 0

Key point 21.6 If a particle is in equilibrium, the forces acting on it form a closed polygon.

WORKED EXAMPLE 21.15

Three forces act on a particle in the horizontal plane, as shown in the diagram. Given that the particle is in equilibrium, find, to the nearest degree, the angles between the two forces.

force and each of the other

8 N

7 N

10 N Label the angles you want to find.

‘Move’ the vectors to form a triangle.

Relate the angles you want to find to the ones in the triangle. You can now use the cosine rule to find the angles in the triangle.



You can now find the required angles.

EXERCISE 21D 1

In the following diagrams, the particle is in equilibrium. Find the forces and angles marked with italic letters. a

i

F θ

10 N

8N F

ii

8N

θ 5N

b

F2

i

130° 80°

F1

12 N ii

15N

13N θ2

θ1 26N

c

10N

i

110° 120° F2

F1

9N

ii

θ1 18N

θ2 12N

2

A particle of weight is suspended by two light inextensible strings, which make angles of with the horizontal ceiling. By considering components of the forces in the directions of the strings, find the tension in each string.

and

3

A particle is in equilibrium under the action of three horizontal forces, each making an angle of with the other two. Prove that the three forces all have equal magnitudes.

4

A box with mass rests in limiting equilibrium on a rough inclined plane. The magnitude of the frictional force between the box and the plane is . a Use a triangle of forces to find the angle that the plane makes with the horizontal. b Find the coefficient of friction between the box and the plane.

5

Three forces, of magnitudes and , act on a particle in equilibrium. All three forces act in the same horizontal plane. Find the angles that the force makes with the other two.

6

A particle of mass hangs in equilibrium suspended by two light inextensible strings. The strings make angles and with the upward vertical, as shown in the diagram. The tensions in the strings are and . Find, to the nearest degree, the values of and .

α

β

2.4g

Checklist of learning and understanding To resolve a force in a given direction, draw a right-angled triangle with the force as the hypotenuse and the other sides of the triangle parallel and perpendicular to the direction of interest. When calculating motion on a slope, resolve forces parallel and perpendicular to the slope rather than vertically and horizontally. In particular, the components of weight acting on a slope inclined at an angle to the horizontal are: down the plane: perpendicular to the plane: The contact force between an object and a surface has two components: the normal contact force, perpendicular to the surface the frictional force, parallel to the surface. The maximum or limiting value of friction, between an object and a surface is given by µ where is the normal reaction force between the object and the surface, and µ is

the coefficient of friction. If an object is stationary and the sum of all the other forces parallel to the surface, excluding friction, is smaller than , the object will remain at rest and the friction force, will equal the sum of the other forces. If the sum of all the forces parallel to the surface excluding friction, is larger than µ object will move and

, the

Mixed practice 21 Unless otherwise indicated use 1

A canal boat of mass has a rope with tension

throughout this exercise. is being pulled along a straight canal by two horses. Each horse acting at an angle to the canal.

a Assuming the resistance from the water is much smaller than the tension in the ropes, find the acceleration of the boat, to three significant figures. b How will your answer change if the assumption about the resistance due to the water is wrong? 2

A block with mass is projected up a rough slope, which makes a horizontal. The coefficient of friction is .

angle with the

a Find the magnitude of the initial acceleration of the block. b Find the magnitude of the contact force between the block and the slope. 3

The diagram below shows two forces acting on a particle.

8N 60° 10N a Find the component of the resultant force in the direction of the

force.

b The direction of the force is allowed to vary. Find the maximum and minimum value of the magnitude of the resultant force. 4

A block of mass slides freely from rest down a smooth slope inclined at horizontal. What is the acceleration of the block down the slope?

5

A particle is projected with an initial speed of comes to rest from its starting point.

to the

across a rough horizontal surface and

a Calculate the coefficient of friction between the particle and the surface. b A second, identical particle is projected across the same surface and comes to rest from its starting point. Determine its initial speed. 6

A small ball, , of mass grams is suspended from point by a light inextensible string. The string is displaced from the vertical (see diagram) and the ball is held in equilibrium by a force of . The string makes a angle with the downward vertical.

O θ° B 5N a Find the tension in the string.

b Find the value of . 7

A small box of mass is placed on a rough slope inclined at an angle of horizontal. It is released from rest and slides down the slope.

to the

a Draw a diagram showing the forces acting on the box. The slope is

long. The box takes

seconds to reach the bottom of the slope.

b Find the acceleration of the box. c Find the coefficient of friction between the box and the slope. d State an assumption that you have made about the forces acting on the box. 8

In this question take

.

A box of mass is held at rest on a plane inclined at an angle of box is then released and slides down the plane.

to the horizontal. The

a A simple model assumes that the only forces acting on the box are its weight and the normal reaction from the plane. Show that, according to this simple model, the acceleration of the box would be correct to three significant figures. b In fact, the box moves down the plane with constant acceleration and travels seconds. Using this information, find the acceleration of the box.

metres in

c Explain why the answer to part is less than the answer to part 9

Three horizontal forces, of magnitudes , and , act on a particle in the directions shown in the diagram. The angles and are such that , and

i



Show that the component, in the -direction, of the resultant of the three forces is zero.

ii Find the magnitude of the resultant of the three forces. iii State the direction of the resultant of the three forces. © OCR, GCE Mathematics, Paper 4728, January 2007 and acts along a 10 Two horizontal forces act at the point . One force has magnitude bearing of The other force has magnitude and acts along a bearing of (see diagram).

12N

14N

30°

O

i

Show that the resultant of the two forces has magnitude figures.

, correct to significant

ii Find the bearing of the line of action of the resultant. © OCR, GCE Mathematics, Paper 4728, June 2008 11

B

P 30°

The diagram shows a small block , of mass , and a particle , of mass , which are attached to the ends of a light inextensible string. The string is taut and passes over a small smooth pulley. is held at rest on a horizontal surface, and lies on a smooth plane inclined at to the horizontal. When is released from rest it accelerates at towards the pulley. i

By considering the motion of , show that the tension in the string is

.

ii Calculate the coefficient of friction between and the horizontal surface. © OCR, GCE Mathematics, Paper 4728, June 2009 is held at rest on a rough horizontal surface. The coefficient of friction 12 A block of mass between the block and the surface is . A light inextensible string, which passes over a smooth peg, is attached at one end to the block and at the other end to a particle of mass . The system is released from rest.

8kg

5kg

a Find the magnitude of the frictional force acting on the block. b Find the acceleration of the block. c Find the tension in the string. is pulled from rest up a smooth slope inclined at to the horizontal 13 A block with mass by a string with tension , maintained at an angle of to the horizontal. a After seconds, the block has moved

along the slope. Calculate .

b The block is allowed to come to rest again, then the tension is increased so that the block is

about to lift off the slope. Calculate the minimum tension needed to achieve this. , is projected up a line of greatest slope of a rough plane inclined at 14 A particle , of mass to the horizontal. The initial velocity is and the particle comes to instantaneous rest on the plane after travelling metres. a Calculate the frictional force acting on , and the coefficient of friction between and the plane. b Determine the acceleration of down the plane subsequent to the instant of rest. is projected with velocity 15 A particle of mass plane inclined at to the horizontal.

up the line of steepest slope of a

a If the plane is smooth, determine the velocity of the particle after

seconds, assuming it

does not reach the end of the plane. b If the plane is rough, with coefficient of friction between particle and plane, calculate the maximum vertical height above its starting point that the particle will reach. 16 A particle, , of mass is released from rest down the line of steepest slope of a rough plane inclined at an angle to the horizontal, where . The coefficient of friction between the particle and the plane is

.

The plane is now inclined at an angle to the horizontal and another particle, , of mass is released down it. The coefficient of friction is again . The time taken for to travel

is the same as the time taken for to travel

.

Show that is pulled in a straight line on a rough horizontal surface by a constant 17 A block of mass horizontal force of magnitude newtons. Assume that there is no air resistance acting on the block. a The acceleration of the block is . Find the coefficient of friction between the block and the surface. Give your answer correct to two significant figures. b Explain how and why your answer to part a will change if you assume that air resistance does act on the block. 18 Q P

60°

Particles and , of masses and respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley and the sections of the string not in contact with the pulley are vertical. rests in limiting equilibrium on a plane inclined at to the horizontal (see diagram). i a Calculate the components, parallel and perpendicular to the plane, of the contact force exerted by the plane on . b Find the coefficient of friction between and the plane.

is held stationary and a particle of mass released from rest.

is attached to . With the string taut, is

ii Calculate the tension of the string and the acceleration of the particles. © OCR, GCE Mathematics, Paper 4728, January 2010 A

19

B θ 5N R 7N

A small smooth ring of weight is threaded on a light inextensible string. The ends of the string are attached to fixed points and at the same horizontal level. A horizontal force of magnitude is applied to . The string is taut. In the equilibrium position the angle is a right angle, and the portion of the string attached to makes an angle with the horizontal (see diagram). i

Explain why the tension

is the same in each part of the string.

ii By resolving horizontally and vertically for the forces acting on , form two simultaneous equations in and . iii Hence, find and . © OCR, GCE Mathematics, Paper 4728, June 2011 20 A block of weight the block.

is at rest on a horizontal floor. A force of magnitude

i The block is in limiting equilibrium when the

is applied to

force is applied horizontally. Show that

the coefficient of friction is . ii

4.9N 30°

When the force of is applied at an angle of above the horizontal, as shown in the diagram, the block moves across the floor. Calculate a the vertical component of the contact force between the floor and the block, and the magnitude of the frictional force, b the acceleration of the block. iii Calculate the magnitude of the frictional force acting on the block when the at an angle of to the upward vertical, justifying your answer fully.

force acts

© OCR, GCE Mathematics, Paper 4728, January 2008 21 A block of weight is projected down a line of greatest slope of a plane inclined at an angle of to the horizontal. travels down the plane at constant speed. i a Find the components perpendicular and parallel to the plane of the contact force

between and the plane. b Hence, show that the coefficient of friction is ii

TN

, correct to significant figures.

B

45°

20°

is in limiting equilibrium when acted on by a force of directed towards the plane at an angle of to a line of the greatest slope (see diagram). Given that the frictional force on acts down the plane, find . © OCR, GCE Mathematics, Paper 4728, June 2009 22 Two right-angled triangular prisms of equal height, with angles of greatest slope and respectively, are positioned as shown, with a smooth peg, , between the two highest points. P A

B

g

3k g

2k

30°

45°

Block , with mass , is placed on the slope and block , with mass , is placed on the slope. The two blocks are connected by a light, inextensible string, which runs parallel to the line of greatest slope of each prism and passes over the smooth peg. The coefficient of friction between block and the slope surface is µ , and the coefficient of friction between block and the slope surface is µ. At time

block is projected down the

slope with speed

a Calculate the acceleration of block in terms of µ . It is determined that µ

.

b Calculate whether the blocks will return to their original positions and, if so, the time at which this will occur. 23 A particle of mass inclined at an angle of

moves upwards along a line of greatest slope of a rough plane to the horizontal. reaches its highest point and then moves back

down the plane. The coefficient of friction between and the plane is i Show that the magnitude of the frictional force acting on is figures.

. , correct to significant

ii Find the acceleration of when it is moving a up the plane, b down the plane. iii When is moving up the plane, it passes through a point with speed

.

a Find the length of time before reaches its highest point. b Find the total length of time for to travel from the point to its highest point and back

to . © OCR, GCE Mathematics, Paper 4728, January 2007 0.6m s- 1

24

P A

0.6m s- 1 Q

M

30° B

and are points at the upper and lower ends, respectively, of a line of greatest slope on a plane inclined at to the horizontal. is the midpoint of . Two particles, and , joined by a taut, light inextensible string, are placed on the plane at and respectively. The particles are simultaneously projected with speed down the line of greatest slope (see diagram). The particles move down the plane with acceleration . At the instant after projection, is at i

Find the distance

and is at . The particle subsequently remains at rest at . .

The plane is rough between and

, but smooth between

and .

ii Calculate the speed of when it reaches . has mass

and has mass

.

iii By considering the motion of , calculate the tension in the string while both particles are moving down the plane. iv Calculate the coefficient of friction between and the plane between and

.

© OCR, GCE Mathematics, Paper 4728, June 2014

Worksheet See Extension sheet 21 for a selection of more challenging problems.

22 Moments In this chapter you will learn: how to find the turning effect of a force about uniform rods and laminas how to find the centre of mass of a non-uniform rod about rotational equilibrium.

Before you start… Student Book 1, Chapter 21

You should be able to recognise types of force acting on a particle.

1 A particle is pulled across a smooth horizontal table by a string that is parallel to the table. Draw a diagram and label all the forces acting on the particle.

Student Book 1, Chapter 21

You should understand when a particle is in equilibrium.

2 Three forces act on a particle, as shown. 5N F

12N

The particle is in equilibrium. Find the magnitude of .

Modelling rotating systems Until now you have used the particle model to analyse forces and motion. However, there are situations in which this isn’t appropriate.

F

Consider, for example, closing a door. A force is applied to push the door closed, but when the door moves it doesn’t do so in a straight line. Instead it rotates about the hinge. In this chapter you will find out how to model situations like this.

Section 1: The turning effect of a force To describe the motion of a rotating object you need to specify the axis of rotation. In the example of a door, this is the vertical line passing through the hinges. In this chapter you will consider only situations where the object can be modelled as one- or two-dimensional, and the axis of rotation is perpendicular to the plane in which the object lies. In that case, you can talk about the object rotating about a point in the plane. From your experience of closing doors you probably know that it is easier to push if your hand is further away from the hinge. This is because the moment (the turning effect of a force) depends upon both the force applied and the distance away from the pivot point.

Key point 22.1 The moment of a force about a point is: where is the perpendicular distance of the line of action of the force from . F

P d

The units of moment are newton metres (

).

The moment will either cause clockwise or anticlockwise rotation about a point. P

anticlockwise movement

F

F

clockwise movement

P

If the line of action of the force acts through , then the moment about will be zero (as the perpendicular distance from is ). So in this case there is no rotational effect from the force. Many situations involving forces that cause rotation can be modelled using two basic shapes: a uniform lamina, which is a two-dimensional object (for example, in the shape of a rectangle, triangle or circle). We might use this to model objects such as a door or a book. a uniform rod, which has just one dimension. We might use this to model a see-saw, a snooker cue or a plank. In both cases ‘uniform’ means that the object has the same density throughout. The key fact you need to know is where the centre of mass is for both of these shapes.

Key point 22.2 The centre of mass is the point at which the object’s weight acts. For a uniform rod, this is at the midpoint. For a uniform rectangular lamina, this is at the intersection of its diagonals.

mg

mg

Fast forward Although this might sound obvious, determining the centre of mass of more complex shapes can be quite difficult. You will see how to do this if you study the Mechanics option of Further Mathematics.

WORKED EXAMPLE 22.1

A uniform rod of length 5m

P

has weight

. Point is at one end of the rod and point is

from .

Q

Find the moment of its weight about the point: a b a P

The perpendicular distance from to the centre of mass is 3m

2m

Q

.

The weight acts at the mid point. It would cause the rod to rotate clockwise about .

5N

b P

The perpendicular distance from to the centre of mass is 3m

2m

.

Q

5N

The centre of mass would cause the rod to rotate anticlockwise about .

WORKED EXAMPLE 22.2

A uniform rectangular lamina, measuring 0.6m

0.4m

by

, has centre and weight

12N

The lamina is free to rotate in a vertical plane about .

P

Find the moment about of: a the weight b the

force. The weight acts at the centre.

.

0.6m C

0.4m P

12N

d = 0.3m 8N

a

The perpendicular distance from to the line of action of the weight is . The rotation will be clockwise. 12N



d = 0.4m P

b

The perpendicular distance from to the line of action of the force is . The rotation will be clockwise.

When several forces act on a body they combine to give an overall resultant force; this is also true of moments.

Key point 22.3 To find the resultant moment about a point, find the sum of the clockwise and anticlockwise moments separately. The resultant moment will be the difference between the two sums (in the direction of the larger).

Focus on … Focus on … Modelling 4 applies moments to the context of levers.

WORKED EXAMPLE 22.3

A uniform rod of length diagram. 7N

and weight

is acted on by a

and

force, as shown in the

3N 2m

2m P

10N

Find the resultant moment about the point . Find the moment of each force in turn. The

force is

from .

Find the sum of the moments that act in the same direction. The sum of the clockwise moments is greater.

EXERCISE 22A Use 1

unless otherwise stated. Find the moment, about the point , of the weight of each of the following uniform rods. a

1m

i

P

3N

0.5m

ii P

4N

b

i

1m

5m

P 2N ii

4m

1.5m P

8N

c

i

2.5m P

8N ii

2

3m P

Find the moment, about the point , of the weight of each of the following uniform laminas.

a

i P

1.5m

1m

10N

4m

ii

P

5m

7N b

5m

i

7m

P 6N

8m

ii

10m

P 0.75N

c

2m

i

2m P

5m

12.5N 6m

ii

P

7m

15N 3

A a

long uniform rod weighs

. Find the net moment about the point in the following situations.

4m

i P

P

ii 1m

b

P

i 2m

4N

ii P 2m

c

i 10N P 1m 30N

100N

ii

50N P

1m

4

A

by

1.5m

uniform rectangular lamina weighs

.

Find the net moment about the point in the following situations, given that the lamina is in a vertical plane. a

6m

i

4m P 6m

ii

P

4m

b

50N

i

P 2m 50N

ii

P 2m c

i 100N

P 50N

ii

P

d

i 200N

P

200N

ii 100N

P 100N e

100N

i 50N

1m P

1m 1m

200N

100N

ii

2m

50N P

1m

50N

5

A steering wheel is modelled as a ring of diameter

. The driver applies two clockwise forces of

tangentially at diametrically opposite sides of the steering wheel. Find the net moment about the centre of the wheel. 6

Find the resultant moment, including the direction, about the point in the following diagram. 50N

100N

200N 3m P

3m

2m 50N

7

A diving board is modelled by a uniform plank of length

and mass

. A diver of mass

stands at one end of the plank. a Find the total moment around the opposite end of the board. b Explain how you used the fact that the plank is rigid in your calculation in part a. 8

The diagram below shows a regular hexagon. Find the resultant moment, including the direction, about the centre of the hexagon.

5N 5N

5N

5N

5N

20cm

5N

9

The diagram below shows a square of side with four forces of equal magnitudes acting at the corners. Prove that the net moment is the same about any point on the interior of the square.

F

F F

F 10 A badminton racquet is modelled as a uniform rod, of length

and mass

, connected to a

square lamina of side length and mass . The racquet is held horizontally at a point, , from the end of the rod. Find the net moment of the racquet about the point .



Section 2: Equilibrium To maintain an object in equilibrium, there need to be additional forces acting on it to counterbalance its weight. Some common situations you will meet in this chapter are: smooth supports providing a normal reaction force

RA

RB

A

B mg

or light strings providing a tension.

TA

TB

mg For an object to be at rest in equilibrium, you now need to add the condition that there is no rotation.

Fast forward In Section 4 you will also see examples of a rod leaning against a wall.

Rewind Remember from Chapter 21 that ‘smooth’ means there is no friction (so here the force at the support will be perpendicular to the surface of the object), and ‘light’ means the string has no mass.

Key point 22.4 If an object is in equilibrium, there is zero resultant force and zero resultant moment about any point. Notice that the resultant moment will, in general, be different about different points. However, if the object is in equilibrium, the resultant moment will be zero about any point.

Since you have a choice of which point to take moments about, it is a good idea to choose a point where at least one force acts. The moment of that force will be zero, making the calculation simpler. WORKED EXAMPLE 22.4

A plank of length and mass rests in equilibrium on two identical chairs. Chair is placed from one end of the plank, and chair is placed at the other end, as shown in the diagram.

1m

A

B

The chairs can provide a reaction force of modelled as a uniform rod and

RA 1m

, determine if either chair will break.

RB 3m

before breaking. Assuming that the plank can be

Since the plank is a uniform rod the centre of mass is from either end, so from .

4m

A

B mg

Taking moments around :

As the plank is in equilibrium, Notice that taking moments about either or allows you to ignore one of the unknown forces. You could have chosen either point to start with. , so doesn’t break.

So chair will not break. Vertical forces:

This exceeds the breaking force, so chair will break.

WORKED EXAMPLE 22.5



, so does break.

A shop sign is formed from a rectangular plastic sheet, , of weight . It is held in equilibrium by a vertical wire at and horizontal wires at and , as shown in the diagram. T1 A

T2

B

D

T3

C

If the sign can be modelled as a uniform lamina with Take moments

and

, find

,

and

.

Moments about eliminates two of the unknown forces.

about : Since it is in equilibrium, . The centre of mass is to the right of . Horizontal forces:

Vertical forces:

WORK IT OUT 22.1 A uniform rectangular lamina, , of width and height , has weight and is attached to a fixed bolt at point , about which it can rotate freely. A horizontal string is attached midway along under a tension of . A downward force of is applied at a point, , which lies on , with so that the lamina hangs in equilibrium with horizontal. Find the value of . 30 cm

D

20 cm

C

130 N x

A

B 20 N

Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1 Taking moments about : Clockwise moments: Anticlockwise moments:

Solution 2 Taking moments about the centre of mass: Clockwise moments:

Anticlockwise moments:

Solution 3 Taking moments about : Clockwise moments:

Anticlockwise moments:

EXERCISE 22B Use 1

unless otherwise stated. In each of the following diagrams, a uniform rod of length equilibrium by two vertical wires. Find the unknown values. a

40N

i

y N

4m x m 10N

ii

3m x m

y N

and weight

is being held in

b

i

x N

y N

1m

2m x N

ii

y N

2m 2

3m

The following uniform rectangular lamina have a weight of vertical plane. Find the unknown values. a

T1 N

i

10m

T2 N 6m

T3 N T1 N

ii 5m

T2 N

2.5m T3 N

b

T1 N

i 4m

T2 N 3m

T3 N

T1 N

ii

1m

T2 N

5m

T3 N

and are hanging in equilibrium in a

c

100N

i

4m

100N x m

100 N

100N

ii

8m

50N x m

50N 3

Two children sit on a see-saw formed from a uniform rod of length , balanced in the middle. One child, of mass , sits on one end. How far from the other end should the other child, of mass

,

sit so that the see-saw is balanced in a horizontal position? 4

A door is wide. A perpendicular force of is applied from the hinge but a wedge at the end of the door opposite the hinge is keeping it shut. Find the frictional force acting through the wedge.

5

In a simplified model of a crane, the arm

is modelled as a uniform rod of length

and mass

. The arm is attached to the main body of the crane from . A counterweight of

is suspended

2m

15000kg P 5m

from . A weight of

can be moved along the arm to keep the crane in equilibrium.

Q 5000kg

a Find at what distance from the counterweight should be attached. b Explain why the counterweight would not need to be placed precisely at the position found in part a. 6

A uniform plank, of length

and mass

the end of the plank and the other support. 7

A uniform beam of length painter of weight wires.

8

and mass

is standing

, rests in a horizontal equilibrium on two supports, one at

from the other end. Find the reaction force supplied by each

is suspended horizontally by wires at either end. A

from one end of the beam. Find the tension in each of the

A pole vaulter holds a pole in a horizontal position, with one hand on the end and the other The length of the pole is and its weight is . a Find, in terms of , the vertical forces exerted by his hands. b State one additional assumption you have made in part a.

away.

9

A spade is modelled as a uniform rod, of mass

and length

, attached to a uniform square

lamina, of side and mass . A gardener holds the spade horizontally with hands from the end of the rod. Find the vertical forces exerted by the gardener’s hands.

and

10 A model for the elbow joint models the bicep muscle connecting to the horizontal forearm by a vertical tendon

from the elbow joint. A mass is held in the hand

4cm

from the elbow joint.

30cm

If the maximum tension that can be exerted by the tendon before injury occurs is maximum mass that can be held in this way. 11 A shop sign, weight

, is modelled as a uniform rectangular lamina of width

, height

, find the

and

. It is resting on a rough support at with a horizontal wire connected at .

50m

A

B

40m x

X

D

C

a Find, in terms of where necessary: i the tension in the wire ii the normal reaction at iii the friction force at b How would your answer to part a i be different if a light rod rather than a wire were attached at ? 12 A door is modelled as a rectangular lamina of weight

, with height

and width

in equilibrium by two hinges at and .

1.2m 50cm F1

X 2m

F2 Y

F3 50cm

If the force at is entirely horizontal, find the magnitude of the forces at and .

, supported

13 In the film The Italian Job, a coach is balancing on the edge of a cliff with gold bullion at one end and a group of people at the other. Model the coach as a uniform rod of length and mass , with of gold at the end overhanging the cliff and of people at the other end. How much of the coach can overhang the cliff before it falls?



Section 3: Non-uniform rods When the mass of a rod is not evenly distributed throughout its length you say that the rod is said to be non-uniform. This could happen, for example, if the rod varies in thickness, or if it is made out of several different materials. The centre of mass of a non-uniform rod is not necessarily at its mid-point. You may be told the position of the centre of mass or may be able to determine it experimentally by measuring forces acting on the rod when it is in equilibrium. WORKED EXAMPLE 22.6

A non-uniform rod of length and mass hangs in equilibrium, supported by two light inextensible vertical strings of equal lengths attached to its ends. The tension in the first string is . Find the distance of the centre of mass from the end attached to the second string.

T

88.2

The weight acts at a distance from the right-hand end. (This is the position of the centre of mass.)

x 12g Taking moments about the righthand end:

Since the plank is in equilibrium, . Since you don’t know the force in the second string, take moments about that end.

The centre of mass is the right-hand end.

from



If you already know the position of the centre of mass, you can use it as before to solve problems about a rod in equilibrium. WORKED EXAMPLE 22.7

A non-uniform rod , of mass and length . It rests in equilibrium on two supports located placed from the support closer to end .

, has its centre of mass at a point from end from each end. A particle of mass is

m 1m

2.5 A 0.3

B 0.3 10g

Given that the reaction forces in the two supports are equal, find the value of .

R

R

0.3

1

A

0.3 B

2.2

2.2 10g

mg

Forces on the plank:

Since the rod is in equilibrium, the net force is zero…

Moments about the support closer to end :

EXERCISE 22C

…and the net moment is zero. You can choose to take moments either about one of the supports or about the centre of mass.

EXERCISE 22C In this exercise, take 1

.

A non-uniform plank , of length and mass , rests in equilibrium on two supports, one at each end. The centre of mass of the plank is located from end . Find the reaction force in each support.

2

A non-uniform plank of length rests in equilibrium on two supports, located and from each end. The reaction forces in the two supports are equal. Find the position of the centre of mass of the plank.

3

A non-uniform plank of weight and length is placed on a support at its midpoint. The plank is held in equilibrium by a downward force of magnitude acting on one end. Find the distance of the centre of mass of the plank from its midpoint.

4

A non-uniform rod is held in a horizontal position by two light inextensible strings attached to its ends. The centre of mass of the rod is located from one end, and the tensions in the strings are and . Find the possible lengths of the rod.

5

A non-uniform rod

, of mass

and length

. It rests in equilibrium on two supports located placed from the support closer to end .

, has its centre of mass at a point from each end. A particle of mass

from end is

m 0.6

2.5 A 0.3

B 0.3 10g

Given that the reaction forces in the two supports are equal, find the value of . 6

A non-uniform plank of mass and length rests on two supports, and , located from its ends. The centre of mass of the plank is from support . A particle of mass is placed on the plank at the distance from the centre of mass, towards support , so that the normal reaction forces in the two supports are equal. Find the distance .

7

A snooker cue is formed by connecting end-to-end two uniform rods of length . One has mass and the other has mass . The cue rests in equilibrium when supported at a point a distance from the exterior end of the

8

section. Find the value of .

A non-uniform rod has length . The rod is placed on a support at its midpoint. It is in equilibrium when a particle of weight is placed at one end and a particle of weight on the other end. The particles are then removed and the rod is then suspended by two light inextensible strings attached to its ends. When the rod is horizontal, the tension in one of the strings is . Find the two possible values for the weight of the rod.



Section 4: Further equilibrium problems Remember that at the start of this chapter the moment about a point was defined as the force times the perpendicular distance from the point to the force. In all the examples with rods so far, the forces have been acting perpendicular to the rod so that the perpendicular distance was measured along the rod.

P Moment = Wd

d W

Now look at a rod that is held at an angle to the horizontal. To calculate the moment of its weight about an end point you need to use the perpendicular distance from that point to the vertical line, as shown in the diagram.

Moment = Wd

P d W

When a rod is in equilibrium, you can still use the rule that the net force is zero and the net moment about any point is zero. However, it is now possible to write three equations, because you can consider horizontal and vertical components separately.

Key point 22.5 If an object is in equilibrium, then: The sum of horizontal components of all the forces equals zero. The sum of vertical components of all the forces equals zero. The sum of clockwise moments equals the sum of anticlockwise moments about any point.

WORKED EXAMPLE 22.8

A uniform rod of length and weight is held in equilibrium by four light inextensible strings, as shown in the diagram. Two of the strings are horizontal and two are vertical. The rod makes a angle with the horizontal. The tension in the vertical string to the left of the diagram is . Find the tensions in the two horizontal strings.

20N 2l 30°

Label all four tensions and the weight on the diagram. Since the rod is in equilibrium…

T3 T2 l

20N

l

T1

30° A

50N

…the net horizontal force is zero, …the net vertical force is zero, 30N T2 l 2l sin 30°

l

30°

l cos 30°

…and the net clockwise moment equals the net anticlockwise moment. Take moments about the lower end of the rod (marked ). You need perpendicular distances from to the line of the forces marked , and .

50N

2l cos 30°

When the rod is placed on the ground or against a wall, you may also need to include friction in your force diagram. WORKED EXAMPLE 22.9

A ladder can be modelled as a uniform rod of length and mass . The ladder is placed on rough horizontal ground and against a smooth vertical wall. The coefficient of friction between the ladder and the ground is and the ladder is on the point of slipping. Find the angle the ladder makes with the horizontal. Remember to include the normal reaction forces from both the ground and the wall. The friction force acts only at the ground contact point (since the wall is smooth). Since the ladder is on the point of slipping, the friction is limiting (so equals ) and directed towards the wall.

The net vertical force is zero. The net horizontal force is zero. The clockwise moments equal the anti-clockwise moments about any point. We chose the point of contact with the ground since this eliminates two forces.

EXERCISE 22D

EXERCISE 22D 1

Find the moment of each force about the point , stating whether it is clockwise or anticlockwise. a

i

2m 8N

20° P ii P

30° 12 N

5 m

b

i P

40° 4 m

30 N ii

P 40°

3 m

50 N c

20 N

i P

ii

P

30°

8 m

5 m 40° 20 N

d

30 N

i

4 m P

25°

60 N

ii

5 m 20°

P 2

A uniform rod of length

and weight

is freely hinged at one end. The other end is attached to a

light inextensible string. The rod is held in equilibrium with the string horizontal.

30° 5 m

Find the tension in the string required to keep the rod at an angle of 3

to the horizontal.

A uniform rod of length and weight is freely hinged at end . End is attached to a light inextensible string. The rod is in equilibrium with the string horizontal. The tension in the string is

.

A

θ

2l

4 N B Find the angle that the rod makes with the vertical. 4

A uniform ladder of length and mass rests against a smooth vertical wall. The ladder makes a angle with the rough horizontal ground and is in limiting equilibrium. Find the coefficient of friction between the ladder and the ground.

5

A uniform ladder of length

rests on rough horizontal ground against a rough vertical wall. The

coefficient of friction between the ladder and the ground is , and the coefficient of friction between the ladder and the wall is . The equilibrium is limiting at both contact points. Find the angle the ladder makes with the horizontal. 6

A ladder of mass rests on rough horizontal ground against a smooth vertical wall, making a angle with the wall. The coefficient of friction between the ladder and the ground is . The ladder is modelled as a uniform rod of length . A man of mass climbs up the ladder. How far can he climb before the ladder slips?

Checklist of learning and understanding The moment of a force about an axis is:

where is the perpendicular distance of the line of action of the force from the axis. In two dimensions, the distance is measured from the point where the axis of rotation passes through the plane of the body. The centre of mass is the point at which the object’s weight acts. For a uniform rod, this is at the midpoint. For a uniform rectangular or other regular lamina, this is at its point of symmetry. For a non-uniform rod, you can find its position by considering moments. To find the resultant moment about a point, find the sum of the clockwise and anticlockwise moments separately. The resultant moment will be the difference between the two sums (in the direction of the larger). If an object is in equilibrium, there is zero resultant force and zero resultant moment about any point.

Mixed practice 22 Use 1

unless otherwise stated. What is the resultant moment about point on the uniform rectangular lamina of weight shown in the diagram? 5N 4m 2m 4N

O 10N

2

A rigid uniform rod of length weight sits

and weight

rests on two supports, as shown below. A

from support A. Find the value of the reaction force at .

2m

3m

A

B 30N

3

The diagram below shows a uniform rod of length and weight suspended from two wires. The tensions in the wires are and and , respectively. Find the value of .

20N

lying in equilibrium and are set at distance

30N

x m 2x m 4

A non-uniform plank of length is placed on two supports, one at each end. The reaction forces in the supports are and . Find the distance of the centre of mass from the midpoint of the plank.

5

A ladder of length and weight rests on rough horizontal ground and against a smooth vertical wall. The ladder is modelled as a uniform rod, and makes an angle of with the horizontal. The coefficient of friction between the ladder and the ground is . A boy of weight stands at the top of the ladder. Given that the ladder is on the point of slipping, find the value of .

6

A uniform rod of mass and length rests with on rough horizontal ground. The rod makes an angle of with the horizontal and is supported by a fixed smooth peg, . The distance is (see diagram).

B

P 1.6 m 60° A

Tip The hinge is fixed in space and applies a horizontal force and a vertical force to the rod. i

Calculate the magnitude of the force exerted by the peg on the rod.

ii Find the least value of the coefficient of friction between the rod and the ground needed to maintain equilibrium. © OCR, GCE Mathematics, Paper 4729, January 2012 7

An oar is modelled as a uniform rod, , of length and mass with an additional mass of attached to end . The oar is hung by a single wire. How far from must this be attached if the oar is to hang horizontally?

8

A uniform plank, , of mass and length rests on supports from and from . A mass, is placed in a position to equalise the reaction forces on the two supports. Find the distance of the mass from .

9

A uniform plank, , of mass and length hangs from two vertical ropes attached to and . When a particle, , of weight is attached to , the plank rests horizontally in equilibrium.

P

A

C

5m

B

If the tension in the rope at is three times the tension at , find: a the tension at b the distance

.

10 A uniform plank, , of length and weight smooth supports, as shown in the diagram.

rests horizontally in equilibrium on two

a Find the reaction at . A child of weight stands on the plank at . The plank remains in equilibrium. The reactions on the plank at and are now equal. b Find the distance

.

11 A non-uniform plank has length shown in the diagram.

and weight

. It rests horizontally in equilibrium, as

1m A

C

B

6m A particle of weight

is placed on the plank at . The plank remains in equilibrium and the

reaction at is

. The centre of mass of the plank is a distance from .

a Show that

.

The particle is now removed from and placed at . The rod remains in equilibrium and the reaction at is now .

0.5m A

C

1m

D

B

4m b Find

and .

Worksheet See Support sheet 22 for an example of non-uniform rods in equilibrium and for more practice questions. 12 A uniform plank, , of length and weight rests horizontally in equilibrium on two smooth supports, and , as shown in the diagram.

1.5m A

C

D

2.5m

B

6m A particle of weight is attached to a point on the plank equilibrium and the reactions at and are now equal. a Show that

from . The plank remains in

.

b Hence, find the range of possible values of . 13 A rectangular lamina is formed by connecting two square uniform laminas of side masses and , where .

2m

2m

m

M

and

2m

x m The rectangular lamina balances on a point a distance from the line joining the two square laminas, as shown in the diagram below. Find an expression for in terms of and .

Worksheet See Extension sheet 22 for some questions involving laminas formed of a rectangle and a triangle. 14 A uniform rod of length and weight is being pushed over a fixed smooth peg ( ) and a fixed support ( ), with a coefficient of friction . and are apart. is the length of the rod overhanging .

x m P

S

Find, as a function of , the force required for the rod to move at a constant speed. 15 In this question take

.

A car with its contents is modelled by a uniform rectangular lamina (having length and height only since width will not be relevant) of mass and length . The wheels are located from the front of the car and from the rear of the car. Assume that air resistance is negligible. a By taking moments about the rear wheel, show that the normal reaction of the ground on the front wheel is . b The car has front wheel drive. Given that the coefficient of friction between the front wheel and the road is , and assuming that the car has sufficient power, find the maximum acceleration of the car. c What modelling assumptions have been made in part b? 16 A uniform rod , of weight attached at . The tension, equilibrium.

and length , is freely hinged at with a vertical string , in the string is sufficient to maintain the rod in a horizontal

a Find the value of . b Find the magnitude and direction of the force provided by the hinge.

FOCUS ON … PROOF 4

Overcoming friction A block of weight rests on rough horizontal ground. The coefficient of friction between the block and the ground is µ . You are going to derive the formulae for the minimum force required to move the block in various situations.

QUESTIONS

A horizontal force of magnitude acts on the block F

W 1

Find, in terms of

and µ , the minimum value of required to move the block.

The force acts at a fixed angle above the horizontal F θ

W 2

Prove that the minimum value of is

µ

.

Both and can vary 3

a Find the maximum value of µ for . Hence, find the minimum magnitude of the force required to move the box. Prove that this minimum magnitude is always less than . (In other words, however large µ is, it is always possible to move the box using a force smaller than its weight.) b Prove that the angle at which the minimal force needs to act is

µ.

The force pushes the block at an angle below the horizontal F θ

W 4

a Prove that

µ

µ

.

b By considering the graph of µ for , or otherwise, prove that the required magnitude of the force is minimal when the force is horizontal, and state this magnitude.

FOCUS ON … PROBLEM SOLVING 4

Checking for reasonableness When solving problems in real contexts answers are rarely ‘nice’ numbers. It can therefore be difficult to tell whether your answer is correct. However, there are still some checks you can do, such as confirming that your calculation gives correct units and making sure that the answer isn’t completely unreasonable. Below are proposed answers to some Mechanics problems. Decide which ones are obviously wrong. 1 A car’s deceleration is

.

2 The tension in the cable supporting the lift is 3 The frictional force acting on the box is

.

.

4 The stone was dropped from a height of 5 The road is inclined at 6 The mass of the car is

to the horizontal. .

7 The coefficient of friction between the box and the floor is 8 An athlete runs

at an average speed of

.

9 The maximum height reached by the projectile is 10 The two trains will meet after

.

.

.

11 The coefficient of friction between the skates and the ice is 12 The car accelerates at

.

13 The coefficient of friction between the box and the ice is 14 The ball takes

.

to fall from the tenth floor.

15 The journey from London to Manchester takes

.

.

FOCUS ON … MODELLING 4

Modelling with moments It is alleged that Archimedes once said: Give me a place to stand and I will move the Earth.

It is theoretically possible to move any weight using a long enough lever. To move the Earth, Archimedes would need to be standing on a different planet and he would also need a support for the lever.

QUESTIONS 1

Suppose the Earth is modelled as a particle of mass , resting on one end of the lever, and that Archimedes is a particle of mass , sitting on the other end. The lever rests on a support from the Earth. How long would the other side of the lever need to be?

2

Other than modelling the Earth and Archimedes as particles, what other assumption did you make in your calculation?

3

Is modelling the Earth as a particle realistic? What could you change in your model to make the particle assumption reasonable? (The radius of the Earth is .) How does this change your answer to question 1?

4

Using the modelling assumptions from question 1, suppose Archimedes moves his end of the lever for a year, at an average speed of . By how much would he move the Earth?

Did you know? The Ancient Greek mathematician Archimedes is perhaps best known for exclaiming ‘Eureka!’ on realising that a body submerged in water displaces its own volume of water. This idea is now commonly known as Archimedes’ principle. However, his work extended well beyond this discovery and his work on levers. He designed the screw pump for raising water, compound pulleys and siege machines, and also developed the fundamentals of integral calculus before Newton and Leibniz eventually formalised it.

CROSS-TOPIC REVIEW EXERCISE 4 1

A uniform plank of mass

and length

rests on two supports, and . The support

is from one end of the plank and support is from the other end. The plank is horizontal. A box of mass is placed on the plank, from .

0.8m

1.2m

0.3m 7kg

Find, in terms of , the force acting on the plank at each support. 2

Points and have position vectors a Find the vector

and

.

.

b Find the exact distance between the two points. 3

A particle of mass by:

moves on a horizontal plane. Its position,

, at time seconds is given

a Find an expression for the velocity of the particle at time . b Find the speed and the direction of motion of the particle when c Find, when : i the magnitude of the acceleration ii the magnitude of the resultant force on the particle. 4

Points

, and have coordinates

,

a Find the coordinates of the point so that b Show that

, respectively.

is a parallelogram.

is a rhombus.

c Find the coordinates of the point on the line 5

and

such that

A non-uniform plank of mass and length rests on a support at its midpoint. A particle of mass rests on the plank, to the left of the support. Given that the plank is in equilibrium, find: a the position of its centre of mass b the magnitude of the force acting on the plank at the support. 2.5N

6 2.4N

θ

FN

A particle rests on a smooth horizontal surface. Three horizontal forces of magnitudes , and act on the particle on the bearings and , respectively (see diagram). The particle is in equilibrium. i a Find and . The

force suddenly ceases to act on the particle, which has mass

.

b Find the magnitude and direction of the acceleration of the particle. © OCR, GCE Mathematics, Paper 4728, June 2014

7

The position vectors of two particles, and , at time , are given by and . Prove that the distance between the two particles is constant.

8

Three points have coordinates that is a straight line.

9

Two small blocks, each of mass

,

and

. Find the values of and so

, are connected by a light inextensible string passing over

a smooth pulley. One block rests on a rough horizontal table and the other hangs freely with the string vertical. The coefficient of friction between the block and the table is .

3kg

3kg

The first block is initially

from the edge of the table when the system is released from

rest. a Find, in terms of , the acceleration of the system. b How long does it take for the first block to reach the edge of the table? c Find the magnitude and direction of the force acting on the pulley from the string. d How did you use the modelling assumption that: i the pulley is smooth ii the string is inextensible? 10 A uniform rectangular lamina has sides of length and and mass grams. The lamina is freely hinged at point and is held in equilibrium, with the longer side horizontal, by means of a light inextensible string attached to point . A

68cm

B

42cm

a Find the tension in the string. b Find the magnitude and direction of the force acting on the lamina at . 11 A non-uniform beam

, of length

and mass

, has its centre of mass at the point of

the beam where . The beam is freely suspended from its end and is held in a horizontal position by means of a wire attached to the end . The wire makes an angle of with the vertical and the tension is (see diagram).

i

Calculate .

ii Calculate the magnitude and the direction of the force acting on the beam at . © OCR, GCE Mathematics, Paper 4729, June 2010 12 A particle is projected with speed

from the top of a smooth inclined plane of length

. After its projection moves downwards along a line of greatest slope with acceleration . At the instant after projection has moved half way down the plane. reaches the foot of he plane after the instant of projection. Form two simultaneous equations in and , and hence calculate the speed of projection

i

of and the length of the plane. ii Find the inclination of the plane to the horizontal. iii Given that the contact force exerted on by the plane has magnitude mass of .

, calculate the

© OCR, GCE Mathematics, Paper 4728, June 2013 13

B 4m

θ A

A uniform ladder

, of weight

and length

, rests with its end on rough horizontal

ground and its end against a smooth vertical wall. The ladder is inclined at an angle to the horizontal, where

(see diagram). A small object of weight

ladder at a point , which is

is placed on the

from . The coefficient of friction between the ladder and the

ground is µ and the system is in limiting equilibrium. i

Show that µ

.

A small object of weight is placed on the ladder at its midpoint, and the object of weight is placed on the ladder at its lowest point . ii Given that the system is in equilibrium, find the set of possible values of . © OCR, GCE Mathematics, Paper 4729, June 2015 14 A golfer hits a ball from a point on horizontal ground with a velocity of at an angle above the horizontal. The horizontal range of the ball is metres and the time of flight is seconds. a

i Express in terms of and, hence, show that

The golfer hits the ball so that it lands

.

from .

ii Calculate the two possible values of . © OCR, GCE Mathematics, Paper 4729, June 2008 15 A particle of mass moves on a smooth horizontal plane. The unit vectors and are directed east and north, respectively. The particle moves under the action of a horizontal force, . a The particle is initially moving with velocity time . b At time

. Find the velocity of the particle at

, the particle is moving due south.

i Show that

.

ii Hence, find the speed of the particle when it is moving due south. 16 The acceleration, , of a particle moving in a straight line is given by , where is the velocity of the particle and is a positive constant. When the velocity of the particle is . Find an expression for in terms of , and . 17 A particle moves in a horizontal plane with acceleration . The particle is initially at the origin and has velocity . Prove that the particle moves in a

straight line. 18 A netball player takes a shot at the hoop, above ground. The player stands from the foot of the post and releases a ball at a height of , as shown in the diagram. The ball is released with speed at an angle above horizontal.

V α

3m

1.5m 5m a Given that the ball passes through the hoop, show that b

i Find the maximum value of

.

ii Hence, find the minimum speed with which the ball can be released and still pass through the hoop.

PAPER 1 PRACTICE QUESTIONS 2 hours, 100 marks 1

The graph of

is translated two units to the right. What is the equation of the resulting

graph? Give your answer in the form [2 marks] 2

A sequence is defined by a Find the first four terms of the sequence. Hence, describe the behaviour of the sequence. [2 marks] b Find the value of

. [1 mark]

c Find

. [3 marks]

3

In this question you must show detailed reasoning. Find the exact value of [3 marks]

4

In this question you must show detailed reasoning. Solve the equation Give your answer in terms of e. [4 marks]

5

Find an approximate expression for

when is small enough to neglect the terms in

and above. [3 marks] 6

The triangle in the diagram has area

. The angle is obtuse.

A 6cm

θ

B Find the length of

11cm

C , correct to one decimal place. [6 marks]

7

Find the equation of the normal to the curve at the point where the form where and are integers.

Give your answer in [6 marks]

8

The sector shown in the diagram has perimeter .

θ r cm Find, in terms of , the largest possible area of the sector. [6 marks] 9

a Express

in the form

where and are integers. [3 marks]

b Hence, or otherwise, find the binomial expansion of

up to and including the term in

. [6 marks] c Find the range of values of for which the binomial expansion of

is valid. [1 mark]

10 a Three consecutive terms in an arithmetic sequence are b Prove that there is no value of for which

Find the possible values of . [5 marks]

are consecutive terms of a geometric

sequence. [3 marks] 11 Use a substitution to find

in the form

, where

are rational numbers. [7 marks]

12 The diagram shows a part of the graph with equation point are

2π

3π

6 6.5 7 7.5 8 8.5 9 9.5 10 a

The coordinates of the maximum

x

Use the trapezium rule with four strips to find an approximate value of State whether your answer is an overestimate or an underestimate. [4 marks]

b

Use four rectangles of equal width to find an upper bound for [4 marks]

c Write down values of and such that . How can the difference between and be reduced? [2 marks]

13 A curve is defined by the implicit equation a Find an expression for

in terms of and [5 marks]

b Hence, prove that the curve has no tangents parallel to the -axis. [3 marks] 14 Consider the equation

where is measured in radians.

a By means of a sketch, show that this equation has only one solution. [3 marks] b Show that this solution lies between and [2 marks] c Show that the equation can be rearranged into the form to be found.

where and are constants [2 marks]

d Hence, use a suitable iterative formula to find an approximate solution to the equation correct to three decimal places. [3 marks] 15 The polynomial

is defined by

a Use the Factor theorem to show that

. is a factor of

. [2 marks]

b Hence, express

as a product of three linear factors. [2 marks]

c

i Show that the equation

can be written as

, where

ii Hence, find all solutions of the equation in the interval

giving your solutions to the nearest degree. [8 marks]

PAPER 2 PRACTICE QUESTIONS 2 hours, 100 marks

Section A 1

The diagram shows the graph of the function . The graph crosses the -axis at the points and . The area labelled equals and the area labelled equals .

,

y

R O

1

3

6

x

S

What is the value of

? [1 mark]

2

The largest possible domain of the function

is

. Find the values

of and . [2 marks] 3

The first term of an arithmetic sequence is terms.

and the 15th term is

. Find the sum of the first [3 marks]

4

The polynomial gives the remainder when divided by factor of . Find the values of and .

, and

is a [5 marks]

5

a Write

in the form

. [3 marks]

b A function is defined by is an increasing function. i Prove that

for

. [2 marks]

ii Does

have an inverse function? Explain your answer. [2 marks]

iii Find the range of values of for which

is convex. [3 marks]

6

a Express

in the form

, where

. Give the value of correct to

three decimal places. [3 marks] b Hence, find the minimum value of Give your answer in the form

. [3 marks]

7

The diagram shows a part of the graph with equation

.

One of the roots of the equation

lies between and .

y 2 1

A

O -1

1

-2

2

3

x

B

a The Newton–Raphson method is used to find an approximation to the root at . , find the second approximation, . Give your answer i Taking the first approximation to be to three significant figures. ii Illustrate the relationship between

and

on a copy of the diagram. [5 marks]

b The point is the minimum point on the curve. i Find the exact -coordinate of . ii Explain why an iteration starting at would not converge to a root of the equation . [4 marks] 8

A student is investigating the number of friends people have on a large social networking site. He collects some data on the percentage ( ) of people who have friends and plots the graph shown.

P 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0

0 20 40 60 80 100120140160

n

The student proposes two possible models for the relationship between and . Model 1:

  Model 2:

To check which model is a better fit, he plots the graph of against , where The graph is approximately a straight line with equation

and

.

.

a Is model 1 or model 2 a better fit for the data? Explain your answer. [2 marks] b Find the values of and . [3 marks] 9

A curve is given by parametric equations

,

axis at point , and is a maximum point on the curve.

, for

. The curve crosses the -

p B

–1

A

O

1

n

a Find the exact coordinates of . [4 marks] b

i Find the values of at the points and . ii Find the shaded area. [6 marks]

Section B 10 A large school wants to investigate the reasons why students are late in the morning. a The Deputy Principal selects ten classes at random, and randomly selects four pupils from each class to take part in a survey. What is the name of this sampling procedure? [1 mark] The Deputy Principal thinks that there may be a correlation between how far a pupil lives from school and the number of times in a month they are late. In the sample of students, the correlation coefficient is . b A test is carried out using the following hypotheses:

Show that, at the

significance level,

is not rejected. [2 marks]

c Decide, based on the test, and explaining your reasoning, whether each of the following statements is correct. i There is no correlation between the distance a student lives from school and the number of times they are late. [2 marks] ii The number of times a student is late is independent of the distance they live from school. [2 marks] 11 A company carried out a ‘Travel to Work’ survey among its employees. The two-way table below shows the number of people in the survey using different modes of transport, split by gender. Male

Female

Walk Bus Cycle Drive a Find P(male walk). [2 marks]

b Find P(male walk). [1 mark] c Find P(male | walk). [2 marks] d Two employees who walk to work are chosen without replacement. Find the probability that they are both male. [2 marks] e Use appropriate figures from this table to provide a counterexample to the identity [2 marks] 12 There are

girls at a school.

of them describe themselves as having blonde hair.

a Use the binomial distribution to find the probability that a randomly selected group of contains exactly girls with blonde hair.

girls [2 marks]

b A club contains

girls of whom

have blonde hair. Use the binomial distribution to test at

significance whether this is more than the expected number of girls with blonde hair. State your null and alternative hypotheses and your -value. [4 marks] c Give two reasons why the binomial distribution may not be an appropriate model for the number of girls with blonde hair in this club. [2 marks] 13 The distribution of the number of books, , borrowed by members of a library follows the following distribution.

a Find the value of . [1 mark] b Find the probability that a randomly chosen member has borrowed at least one book. [1 mark] c The librarian proposes the following model for the probability of a book being overdue. If a member has borrowed books, the probability that they have an overdue book is Find the probability that a randomly chosen member has an overdue book. [3 marks] 14 Two events, and , satisfy: a Find

. [3 marks]

b Show that and are not independent [2 marks] 15 Alex takes the bus to work every day. He keeps a record of the length of time it takes him to travel to work, over the course of several years. The mean is minutes and standard deviation minutes. The lowest value is minutes and the highest value is minutes. a Give two reasons why the numerical information given supports a normal model for the data. [2 marks] b Assuming that a normal model holds, find the probability that Alex takes longer than travel to work.

minutes to [1 mark]

c Stating a necessary assumption, find the probability that Alex takes more than to work on at least three days in a 5-day week.

minutes to travel [4 marks]

d After a change to the bus route, Alex times how long it takes him to travel to work every day during a -day week. The total time for the five days is minutes. Test at the significance level if his average time to travel to work has changed, assuming that it still comes from a normal distribution with standard deviation

minutes. State your null and alternative hypotheses. [4 marks]

e Ming takes a different bus to work. It takes her more than minutes on of days and less than minutes on of days. Assuming a normal distribution, estimate the value of the mean and variance of Ming’s travel times. [4 marks]

PAPER 3 PRACTICE QUESTIONS 2 hours, 100 marks

Section A 1

Let be the smallest positive solution of the equation of , the next two positive solutions.

, where is in radians. Find, in terms [2 marks]

2

The diagram shows the graphs of

and

.

y y = x2 A

B

O

y = 6 - x x

a Find the coordinates of points and . [2 marks] b Shade the region determined by the inequalities

and

. [1 mark]

c State the smallest integer value of which satisfies both inequalities. [1 mark] 3

Points and have position vectors

and

a Find the position vector of the midpoint of

.

. [2 marks]

b Point has position vector

It is given that

. Find the value of . [3 marks]

4

Find the exact coordinates of the point of inflection on the graph of

. [5 marks]

5

Prove by contradiction that

is an irrational number. [4 marks]

6

a Show that the equation can be written in the form State the values of the constants , and .

. [3 marks]

b Hence, solve the equation nearest degree.

for

, giving your answers to the [4 marks]

7

Find

, giving your answer in the form

. [5 marks]

8

The circle shown in the diagram has equation

.

y B P

O

x

A

a Show that the point

lies on the circle. [2 marks]

b The line

is tangent to the circle at . Find the area shaded in the diagram.

0.5m

0.5m

L

R A

B [5 marks]

9

The diagram shows the curves with equations

y

y = 3ex - 3 B

and

.

y = 13 - 5e- x

A

O

x

a Write down the coordinates of point . [1 mark] b The curves intersect at the point . i Show that the -coordinate of satisfies the equation

.

ii Hence, find the exact coordinates of . [4 marks] c Find the exact value of the shaded area. [6 marks]

Section B 10 A cyclist moves along a straight horizontal road. The velocity–time graph of her journey is shown in the diagram.

v 10 5 0 -5

8 12

27

35

t

- 10 - 15 a Calculate the acceleration of the cyclist during the first seconds. [2 marks] b Show that the cyclist returns to the starting point after

seconds. [3 marks]

c State the average velocity of the cyclist for the whole journey. [1 mark] 11 A non-uniform plank of length and mass rests on two supports, and . The supports are located from the ends of the plank, marked and on the diagram. Given that the reaction force acting on the plank at the support is

:

a Find the reaction force acting on the plank at . [1 mark] b Find the distance of the centre of mass of the plank from the end marked . [3 marks] The plank is now placed against a smooth vertical wall, with end on the ground. The ground is horizontal and the coefficient of friction between the plank and the ground is . c Given that the plank is on the point of slipping, find the angle it makes with the horizontal. Give your answer correct to the nearest degree. [5 marks] 12 Four forces act on a particle in a horizontal plane, as shown in the diagram. The unit vectors and are directed east and north, respectively.

F2 = (4j) N F3 = (5i + j) N

F1 = (- 6i) N F4 a The particle is in equilibrium. Find the force

in the form

. [2 marks]

b The force

is suddenly removed. Given that the mass of the particle is

, find:

i the magnitude of the acceleration of the particle ii the angle the acceleration vector makes with the direction of vector . [5 marks] 13 A particle moves in a straight line with constant acceleration is . At time seconds, the velocity of the particle is initial position is .

. The initial velocity of the particle and its displacement from the

v V U t

T

O

By considering the velocity–time graph shown above: a Write down an equation relating , , and . b Hence, show that

. [4 marks]

14 A child kicks a ball from ground level at an angle above the horizontal. The speed of projection is . The ball is modelled as a particle and air resistance can be ignored. a At time seconds the ball’s displacement from the point of projection is

. Show that:

[4 marks] b The angle of projection is . The ball passes over a of projection. Find the minimum possible value of

tall fence that is

from the point [3 marks]

c How will the answer change if your model includes air resistance? [1 mark] 15 Two particles, and , are connected by a light inextensible string, which passes over a smooth pulley. Particle has mass and lies on a rough plane. The plane is inclined at to the horizontal, and the coefficient of friction between and the plane is µ . Particle has mass and hangs freely below the pulley. The system is in equilibrium.

m kg P

Q 5.2kg

35°

a Show that [5 marks] b Obtain another similar inequality for . [2 marks] c In the case when equilibrium.

, find the range of values of for which the system remains in [2 marks]

16 A particle of mass moves in a straight line. At time seconds the velocity of the particle is and the force acting on the particle is , where . When the velocity of the particle is . Find the velocity of the particle after

.

[7 marks]

FORMULAE

The following formulae will be given on the A Level assessment papers. Arithmetic series

Geometric series

for Binomial series

where

Differentiation

Quotient rule: Differentiation from first principles

Integration

Integration by parts Small angle approximations where is measured in radians. Trigonometric identities

Numerical methods Trapezium rule: The Newton–Raphson iteration for solving

:

Probability

or Standard deviation or The binomial distribution If

then

Mean of is

; Variance of is

).

Hypothesis test for the mean of a normal distribution If

then

and

Percentage points of the normal distribution If has a normal distribution with mean and variance then, for each value of , the table gives the value of such that

Kinematics Motion in a straight line

Motion in twostraight line

STATISTICS

Critical values for the product-moment correlation coefficient, . 1-Tailed test

½

½

2-Tailed test

–

–

–

–

–

–

–

–



Answers Chapter 1 Before you start… 1

;

2 e.g. 3 Proof 4 Proof

Exercise 1A 1 Proof 2 Proof 3 Proof 4 a Proof b e.g. 5 e.g. 6 Proof 7 e.g. , ,

,

,

8 a Proof b Proof c Proof 9 a Proof b Proof c Its diagonals are perpendicular (so it must be a kite). 10 a e.g.

,

b Proof c Only when 11 a Proof, b Proof 12 Proof

Exercise 1B 1 Proof 2 Proof 3 Proof 4 Proof

.

5 Proof 6 Proof 7 Proof 8 Proof 9 Proof 10 Proof 11 a Proof b Proof c Proof 12 Proof

Exercise 1C 1 a b Line 2 a Proof b 3 a b Line 4 5 Step : should say

or

; Step : differentiating a specific value rather than the function; Step :

does not imply minimum. 6 If

the suggested factorisation is not necessarily true. Not all coefficients were compared at line .

7 Line does not follow from line .

Mixed practice 1 1 Proof 2 Proof 3 Proof 4 Line does not necessarily follow from line . Line is not equivalent to line . 5 C 6 Proof 7 Proof 8 Proof 9 a Proof b Proof 10 a

: Not equivalent. : Missing

: Final solution may not be a solution to the original equation.

b

or Check: only 11 12 C 13 Proof 14 a Proof b Proof 15 Proof

works.

Chapter 2 Before you start… 1 a b 2 a b 3 a b 4

or

5 a b 6

Exercise 2A 1 a Function: many–one b Mapping c Mapping d Function: one–one e Function: one–one f Function: one–one 2 a i No ii No b i Yes ii Yes c i Yes ii Yes

Exercise 2B 1 a i Domain: ; Range: ii Domain: ; Range: b i Domain:

; Range:

ii Domain:

; Range:

2 a i ii b i

or

ii c i ii d i ii e i

or

ii f i ii 3 a i ii b i ii c i

,

ii

,

d i

or

ii 4 a i

,

ii

,

b i

,

ii

,

c i

or

ii

,

d i

or

ii

or

5 a b 6

,

7 8

,

,

,

, ,

y

9 a

-2

b 10 11 12 a i ii Ø b

Work it out 2.1

Exercise 2C 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii 4 a b 5 a i ii b i

O

3 2

x

ii c i ii d i ii 6

,

7 8 a b 9

Work it out 2.2

Exercise 2D 1 a i ii b i ii c i ii d i ii e i ii f i ii g i ii h i ii

y

2 a 4 3 2 1

- 4 - 3 - 2 - 1O -1

1

2

3

4

7

8

x

-2 -3 -4 y

b 4 3 2 1

O -1

1

2

3

4

5

6

x

-2 -3 -4 y

c 4 3 2 1

-4 -3 -2 -1O -1

1

2

3

4

1

2

3

4

x

-2 -3 -4

y

d 4 3 2 1

- 4 - 3 - 2 - 1O -1 -2 -3 -4

3

a

i ii

b

i ii

x

c

i ii

d

i ii

4 a b 5 6 a b Domain:

; range:

7 8 Proof y

9 a 10 5

- 10

-5

O

5

10

x

-5 - 10

b Domain: c

; range:

, , (Hint: Use the graph.)

10 a b Proof 11 12 a b

; domain

; range

y

13 a 10 5

- 10

-5

O -5 - 10

5

10

x

b

Work it out 2.3

Exercise 2E 1 a

;

b

;

c

;

d

;

2 a b c 3 a b

has a turning point, so is not one-to-one.

c 4 5

Mixed practice 2 1 a

,

b 2 a b 3 4 a b c Reflection in the line

.

d i ii iii e

or

but domain of

therefore there are no solutions. 5 i ii 6 a b

is limited to

.

and

are not in the domain,

7 a i ii iii iv v b

can be , which is not in the domain of .

c i ii Reflection in the line

.

iii iv 8 a

y

b

9 (- 2, 5) O c Range of f is

x

; range of g is

d 9 i ii Not one-to-one. iii Proof iv 10 a b c 11 a Proof b c 12 a

,

b 13 a Proof b Rotation

around the origin.

.

c Proof d Proof e Reflection in the -axis. f Proof g Proof

Chapter 3 Before you start… y

1 a 4 3 2 1

- 4 - 3 - 2 - 1O

1

2

3

3

4

5

x

y

b

1 - 2 - 1O -1

1

2

x

-2 -3 -4 2 a b 3 a b

Work it out 3.1

Exercise 3A y

1 a i 6 5 4 3 2 1

- 6 - 5 - 4 - 3 - 2 - 1O -1

1

2

3

4

5

-2 -3

y

ii

(2, 5.5) (- 5, 3) y = 3 (- 2.5, 1) O

x

6

x

y

b i 6 5 4 3 2 1

- 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6

ii

1 2 3 4 5 6

x

y (- 2.5, 10)

y = 2 (- 5, 2)

x

O

(2, - 8) y

c i 12 10 8 6 4 2

- 6 - 5 - 4 - 3 - 2 - 1O -2

1

2

3

4

5

-4 -6 -8 - 10 - 12

ii

y (- 2.5, 2.5)

(- 5, 0.5)

y = 0.5 x

O

(2, - 2)

6

x

y

d i 8 6 4 2

- 10 - 8 - 6 - 4 - 2 O -2

2

4

6

8 10 12

x

-4

ii

(- 0.5, 5)

y

(- 4, 0)

x

O

(- 2.75, - 4) y

e i 6 5 4 3 2 1

- 24 - 20 - 16 - 12 - 8 - 4 O -1

4

8 12 16 20 24 28 32

x

-2 -3

y

ii

(7, 5)

(- 7, 0) O

x

(- 2, - 4)

2 a i

ii b i

ii 3 a i ii b i ii

. A vertical stretch of and translation

. A stretch of relative to . A stretch of relative to

. A stretch of relative to

(in either order).

and translation , reflection in

, reflection in

. and translation

and translation

.

.

c i ii d i ii 4 a i

: translation axis then translation

ii

then reflection in -axis. Alternatively, reflection in -

.

: translation axis then translation

b i

.

: translation relative to

ii

then stretch of relative to

then translation

: translation relative to

then reflection in -axis. Alternatively, reflection in -

. Alternatively, stretch of

.

then stretch of relative to

then translation

5 a i ii b i ii c i ii 6 a b c d 7 a b c Horizontal stretch with scale factor .

.

. Alternatively, stretch of

y

8 a

6 5 y = f(2x) - 3 4 3 2 1 - 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6

1 2 3 4 5 6

y

b 8 7 6 5 4 3 2 1

y = 1 - 3f(x)

- 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1O -1 -2 -3 -4 -5 -6 -7 -8

1 2 3 4 5 6 7 8

y

9 a

y = lnx O

b

x

y x = - 2

- 1O

1

x

y = 3ln (x + 2)

x

x

y

c

x =

O

1 2

x

1

y = ln(2x - 1)

10 11 12 13

Exercise 3B y

1 a i

3

y = |x - 3|

O

x

3

y

ii 5 y = |x + 5|

x

O

-5

y

b i

y = |3x + 5|

5

- 5 3

O

x

y

ii

y = |2x - 1|

1 O

x

1 2

y

c i

4

y = |4 - 2x|

O

x

2

y

ii

y = |- 3x + 2|

2

O

x

2 3

y

d i

y = |x| + 1

1 x

O

y

ii

y = |x| - 2

-2

O -2

2 a i

2

x

ii b i ii c i ii 3 a i ii b i ii 4 a i ii b i ii c i ii y

5 a

y = |x - 2|

2

O

x

2

y

b

y = |x - 2| + 3

5

(2, 3)

x

O y

6 y = |2x + 3|

3

–1.5 O

x

y

7

3

-3

y = 5 - |x - 2|

O

x

7

y

8

y = x|x|

O

Work it out 3.2

Exercise 3C 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

x

ii d i ii

or

e i ii f i ii 3 a b 4 a b

or y

5 a

y = |3x - 7|

7

x

O

b No solutions 6 7

or



8

Mixed practice 3 1 a

y 3 f(x - 2)

(2, 0) O

(6, 0)

x

y

b y = 3 - f(2x)

(0, 3)

x

O

2 y

3 a 5

y = |2x - 1|

4 3

y = x

2 1 -3 -2 -1O -1

1

2

3

4

x

-2

b 4 a Translation by b Translation by c Translation by

and vertical stretch with scale factor .

and translation by

.

and vertical stretch with scale factor .

5 a b T, S, S 6 Translation by 7

or

and vertical stretch with scale factor .



8 a Vertical stretch with scale factor ; horizontal stretch with scale factor . b

y

( )

y = 3ln x 2

O

(2, 0)

x

c x = - 2

y

( )

y = 3ln x + 1 2

x

O (0, 0)

9 a Vertical stretch with scale factor ; reflection in the -axis; translation units up. y

5

y = 5 - 3|x|

x

O

b c 10 a Translation by

.

y

b y = - 2

y = ln(x + 2)

y = 3 - ln(x + 2) -1

c i ii 11 a b c 12

O

x

13

y y = |x| + x

O

x

Chapter 4 Before you start… 1 a b 2

;

3 4 5 6

, ,

,

, , , ,

Exercise 4A 1 a i ii b i ii

,

,

,

,

,

,

, , ,

, ,

,

,

,

,

,

, ,

,

c i ii d i ii 2 a i

, , , ,

b i ii c i

,

, , ,

, ,

ii

,

, , , ,

,

,

, ,

, ,

, ,

,

, ,

,

, , , , ,

ii d i ii 3 a i ii b i ii c i ii d i

, ,

, ,

,

ii 4 a i Increases, converges to ii Decreases, diverges b i Periodic, period ii Periodic, period 5 a

,

b Proof 6 a b Converges to (and oscillates) 7 8 a Proof b 9 a

,

,

, ,

b 10 a

,

b It doesn’t converge; the terms increase without limit.

Exercise 4B 1 a i ii b i ii c i ii 2 a i

ii

b i

ii

c i

ii

3 4 a Proof b 5 6 a b

Exercise 4C 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii 3 a b 4 5 6 7

,

8 a Proof b

pages

Exercise 4D 1 a i ii b i ii c i

ii d i ii 2 a i ii b i ii 3 4 a

, ,

b 5

,

6

,

7

,

8 Proof 9 10 11 12 13 a b 14 15 a b 16

Exercise 4E 1 a i ii b i

ii c i ii d i

ii



e i ii 2 a i ii b i ii c i ii 3 a b 4 a

,

b 5 a

,

b

,

6 7 a Proof b 8 9 10

,

11

Exercise 4F 1 a i ii b i ii c i ii d i

or

ii

or

2 a i ii b i

,

ii

,

3 a b 4 5

, ,

6 a b 7 a b

Exercise 4G 1 a i ii b i ii c i Divergent ii Divergent d i ii e i Divergent ii 2 a i ii b i ii c i ii d i ii e i

;

ii f i ii g i ii h i ii i i ii 3

4 a

b 5 a b 6 7 a b 8 9 a b 10 a b 11 a b undefined

Work it out 4.1

Exercise 4H 1 a £ b £ 2 a £

b

years

3 a b £ c i ii

years

4 a b Decreases; tends to 5 a b 6 a

in the long term.

£

months days

b Day 7

months

8 a b c For smaller bounces, motions will be damped by physical effects such as air resistance so that the object will come to rest 9 a Proof b c Year 10 a b 11 a Proof b c

years

Mixed practice 4 1

,

2 3 a

,

b 4 5 6 i ii 7

, , ; arithmetic

8 9 10 a b c 11 12 a

,

b 13 14 15 a b 16 i a b ii 17 i Proof ii iii 18 19 20 21 a b c d Proof e 22 a Proof b c

years

Chapter 5 Before you start… 1 2 3 4

Exercise 5A 1 a i ii b i ii 2 3 4 a b 5 a Proof b 6 Proof;

,

7 Proof 8 9 a

,

b 10

, ,

Work it out 5.1

Exercise 5B 1 a i ii b i ii c i ii

or

d i ii e i ii f i ii g i ii 2 a i ii b i ii c i ii d i ii e i ii 3 a i ii b i ii c i ii d i ii

4 a i

ii

b i ii 5 6 7 a b

or

8 9 Quotient: ; remainder: 10 Quotient: 11 12 13 a Proof b 14

;

15 Proof 16 a b Proof

Exercise 5C 1 a i ii b i ii c i ii 2 a i ii b i ii 3

,

; remainder:

4 5 6 7 8 a Proof b 9 10 11 a b 12 Proof

Exercise 5D 1 a i

ii b i ii c i

ii

2 3 4 5 a Proof b c 6 a b

7 a Proof b

8

Mixed practice 5 1 a b

,

,

2 a Proof b c

, ,

d 3 4 5 6 7 8 9 10 11 a b 12 a Proof b c

or

d 13 a Proof b c

,

14 a b 15 Quotient: 16 a

; remainder:

,

b c

,

,

17 a b 18 a b 19 20

,

21 22 23 Proof 24 a b Proof

,

,

Chapter 6 Before you start… 1 2 3 4

Exercise 6A 1 a i ii b i ii c i ii d i ii 2 3 4 a b 5 a Proof b c i ii iii d 6 a b c d

7 8 a

for

b 9 a b

(The expansion is only valid for

).

10 11 12

Work it out 6.1

Exercise 6B 1 2 3 4 5 a b c 6 a

b c 7

Tip There are several ways to do this – you could treat it as , where

8

a

b c 9

a

.

or

or

b 10 a b c 11 12 No, as the expression is not defined for small values of .

Mixed practice 6 1 2 3 4 5 i ii a b 6 a b 7 a

b c 8 9

or

10 i Proof ii Proof 11 a b 12 13 14 a b Proof c

d e 15 a b i ii c Proof

Focus on … Proof 1 Arithmetic series proof 1 1 2 3 4 5 6 Divide by .

Geometric series proof 2 1 Multiply through by . 2 3 4 5

Questions 1

, so can’t divide by it on last line.

2 a b

Focus on … Problem solving 1 1 2 3 4

Focus on … Modelling 1 1 a i ii

v

iii 0.6 0.5 0.4 0.3 0.2 0.1

0 40 0 0 60 0 0 80 0 10 00 0 12 00 0 14 00 0 16 00 0 18 00 0 20 00 00 0

S

20

O

b The maximum possible rate (asymptote, can’t be reached). 2

,

3 a Rational function b Exponential function

Cross-topic review exercise 1 y

1 a

9

O

x

3

b

or

c

or

2

y

3

1

x

O

–1 4 5

,

6 a i ii Proof b

, or

,

7 i ii iii iv Reflection in the line

.

8 a b Proof 9 a b 10 11 12 a

,

b Range: y

2 - 1O

1

x

-3

c Domain: 13 a

or

; range:

and

,

b i ii The expansion is valid only for

.

14 i Proof ii Proof iii 15

. When is small, this is

16 17 a No b c 18 a b 19 a Proof

plus the kinetic energy.

b c Proof d

,

Chapter 7 Before you start… 1 2

,

,

,

3 4 5

Work it out 7.1

Exercise 7A 1 a i and ii

y π 3 π 4

x

O

b i and ii

y 3π 4

O

4π 3 c i and ii

x

y

x

O

-

π 3

-

π 6

d i and ii

y

O

2 a i ii b i ii c i ii d i ii 3 a i ii b i ii c i ii d i ii 4 a i

x - 2π, - 4π

ii b i ii c i ii d i ii

y

5 a i

-π 2 O

π 2

x

y

ii

-π

2π O

π

x

y

b i

π -

π 2

O

π 2

3π 2

x

y

ii

–π

2π O

6 a b c d 7 a b c d 8 a b c d 9 a i ii b i ii c i ii 10 a b c 11 Proof 12 Proof 13

π

x

14

Work it out 7.2

Work it out 7.3

Exercise 7B 1 a i ii b i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii d i ii 4 a b c 5 a i ii b i ii

c i ii d i ii 6 a i ii

,

,

,

,

,

,

b i ii

,

,

,

,

,

,

c i

,

ii

,

d i

,

,

ii 7 y

8 5 4 3 2 1

- 1O -1

1

-2 -3 -4 -5

9 10 a Proof b

,

11 12 e.g. 13 14 a e.g. b c

Exercise 7C

x

1 a

;

b

;

c

;

d

;

2 a i

y





π ( 3, 2)

x

O

4π, - 2) ( 3 y

ii

(2π, 3)

(0, 3)

x

O

(π, - 3)

b i

y π ( , 1) 4

3π (- , 1) 4

x

O

π (- , - 1) 4

3π, - 1) ( 4

ii

y

2π ( , 1) 3

(0, 1)

x

O

π ( , - 1) 3

(π, - 1)

y

c i

O

π

x

y

ii

O

x

π 6

y

d i

(0, 1)

(2π, 1)

(4π, 1)

x

O

(π, - 5)

(3π, - 5)

y

ii

π ( , 3) 2

x

O π (- , - 1) 2

3 a

,

b 4 a b c There is no loss of energy, for example due to air resistance, so that the amplitude of oscillation remains constant. 5

,

6

, y

7 a

y = 2 cosx

2

y = 1+ sin2x

O

2π

x

–2

b c 8

;

;

y

9 a 2 1

O

60°120° 180° 240° 300° 360°

1 -2

b

,

x

c

,

10 a b 11 a

,

b c 12 a The size of the seat is ignored, and it is attached exactly on the circumference of the wheel. b c d

Exercise 7D 1 a b 2 a b 3 a b 4 a b 5 6 a b 7 8 9 10 11 12 13 14 15 16 17

;

18 19 20

or

21 22

radians

Exercise 7E 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii 4 Proof 5 a Proof b c

Exercise 7F 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii

3 a i ii b i ii c i ii 4 a b 5 a b 6 a b i ii 7 a

;

b Proof 8 9 10 11 a b 12

13

Mixed practice 7 1 a b 2

,

3 a b c 4 5

;

6 a b 7 i Proof ii iii 8 9 a

; The bridge comes down to water level exactly at the edge of the bank.

b c 10 a b c 11 a b c Proof 12 13 a The quadrilateral SATB has four equal sides (all ) and a right angle, so it is a square. b c d 14 15

y

16 i

3

O

ii 17 a b

,

x

18 19 a

, right angle between a tangent and a radius.

b c d e 20 a b Proof c

is a rectangle because there are right angles at and , and

is parallel to

.

Chapter 8 Before you start… 1 a b 2 3 a b

Exercise 8A 1 a b c d 2 a b c 3 a b 4 a Proof b 5 a b c 6 a b 7 8 a Proof b 9 a Proof b

Exercise 8B 1 a i ii b i ii c i ii 2 a b c 3 4 a b c d 5 a b c d 6 a Proof b Proof c Proof d Proof 7

,

8 Proof 9 a b 10 a b 11 a

b 12 a i Proof ii Proof b 13 14 a b 15 a Proof b Proof

Exercise 8C 1 a b 2 a b 3 a b 4 a b 5 a b Vertical stretch with scale factor 6 a b 7 a b 8 a b 9 a 10

Exercise 8D 1 a i ii b i ii

; translation

units to the left

c i ii 2 a i ii b i ii c i ii d i ii 3 4 a i ii b i ii c i ii d i ii 5 a i ii b i ii c i ii d i ii 6 a i ii b i ii

c i ii d i ii 7 Proof 8 Proof 9 Proof 10 a Proof b c 11 Proof 12 Proof 13 14

Mixed practice 8 1 a Proof b 2 a b 3 a b c d 4 a Proof b c 5 a b c

d Amplitude e f y

6 i

O

ii iii 7 a Proof b 8 a b Proof c d 9 i ii a b 10 a b i ii 11 a b Proof c 12 a i ii Proof iii b Proof c Proof d e

x

Chapter 9 Before you start… 1 2 3 4

Exercise 9A 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 4 5 6 7 8 9 10 11 12 Tangent:

;

Normal: 13 14

local maximum; local minimum

15 Proof 16 17 a

local maximum

b

local minimum

18 a

million litres

b

Work it out 9.1

Exercise 9B 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 g i ii

4 5 6 7 8 a b 9 10 11 a b 12 13 a Proof b 14 Proof 15

Mixed practice 9 1 2 3 4 5 6 7

local maximum;

8 9 10 a i ii b i ii

local minimum

c i

(Shows that the rate of growth increases exponentially.)

ii 11 12

Tip Use an identity.

(so a minimum point),

Chapter 10 Before you start… 1 a b c d 2 a b 3 a b 4 a b 5 a b 6 a b

Exercise 10A 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 g i ii h i

ii 3 a i ii b i ii c i ii d i ii e i ii f i ii 4 a i ii b i ii

c i ii d i ii 5 6 7 8 9 a



b



10 11 12 13

or

14 a b 15 a Left post b Proof c 16 a b y

c

O

Work it out 10.1

Exercise 10B 1 a i

x

ii b i ii c i ii d i ii 2 a i ii b i ii 3 4 5 6 7 8 a b 9 10 11 a b c 12 a y

b

O

c

is odd.

Work it out 10.2

a

b

x

Exercise 10C 1 a i

ii

b i

ii c i

ii d i ii 2 3 4 5 6 7 8 Proof

Exercise 10D 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 a b 4 a i ii b i ii c i ii d i ii 5 6 a Proof b 7 a Proof b 8 9 10 11 12 13 a b Proof c

Exercise 10E 1 a

b 2 a Proof b 3 a Proof b 4 a b Proof 5 a

b 6 a Proof b Proof c 7 a b 8 Proof

Mixed practice 10 1 a b

c 2 3



4 a Proof b 5 6 i ii 7 8 9 a b Proof

10 11 a

seconds

b 12 a b c

local min;

d

y

O

local max

1 2

x (1, - 1)

13 a Proof b 14 a

. Proof

b 15 i ii Proof iii 16 a b Proof c

local max;

17 a b Proof c 18 a b c

local min

Chapter 11 Before you start… 1 a b 2 a b 3 a b 4 5

Work it out 11.1

Exercise 11A 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii c i ii

d i ii 3 a i ii b i ii c i ii d i ii 4 a b c d e f 5 6 7 8

Exercise 11B 1 a i ii b i ii c i ii d i ii e i

ii f i ii g i ii h i ii i i ii 2 3 4 ln

, so

5 6

Exercise 11C 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii c i

.

ii 4 a i ii b i ii c i ii 5 6 a Proof b 7 8

, so

9

Exercise 11D 1 a i ii b i ii c i ii d i ii 2 3

Exercise 11E 1 a b c d 2 a

b 3 a b 4 Proof 5 6 7 a Proof b 8 9 a Proof b

Work it out 11.2 Answer = They are all correct, with different values for :

Exercise 11F 1 a b c d e 2 a b c d e 3 a i ii b i

ii 4 a i ii b i ii 5 a i ii b i ii c i ii 6 7 a Proof b 8 9 a Proof b 10 11 12 a Proof b 13 a b Proof

Exercise 11G 1 a b c No d e No f g

2 a b c d 3 a i ii b i ii c i ii d i ii e i ii 4 a i ii b i ii 5 c d 6 a b 7 8 9 a b 10

Mixed practice 11 1

2 3 4 5 a b 6 7 8 a b 9 10 a b 11 12 13 i ii 14 15 16 a b Proof 17 a b c 18

Chapter 12 Before you start… 1 a Proof b Proof 2 Proof 3 Proof 4 a Proof b Proof c Proof

Work it out 12.1 .

Exercise 12A 1 a i Decreasing, convex ii Concave iii Increasing, convex b i Increasing ii Increasing, concave iii Increasing, convex

A

2 a

C

B

A

b

C B

C

D

3

E

C E A

B

4 5 6 Proof 7 8 9 Proof 10 Minimum point 11 Proof 12 13

y

14

A B

C O

x

C

C

Exercise 12B y

1 a i

x

O

y

ii

O

x

y

b i

x

O

y

ii

x

O

y

c i

x

O

y

ii

O

2 a i ii b i

x

ii c i ii 3 a i ii b i ii c i ii d i ii 4 a i ii b i ii c i ii d i ii e i ii 5 a b 6 a b

; the distance along the line.

c 7 a Proof;

weeks

b 8 a b c

Work it out 12.2

Exercise 12C

1 a i ii b i ii c i ii 2 a i ii b i ii c i ii 3 4 5 6 a b 7 8 Proof 9 Proof 10 a b Proof 11 a b between

and

months

c Proof; the rate of change of the number of rats as the number of snakes increases.

Exercise 12D 1 a i ii b i ii c i

ii 2 a b 3 a b 4 a b c

Exercise 12E 1 a i ii b i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii 4 5 6 7 8 9 10

Work it out 12.3

Exercise 12F 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii 3 4 5 6 Proof 7 8 9 10 11 12 Proof

Mixed practice 12 1



2 a b 3 4 a

b 5 i ii 6 7 a

minimum;

maximum

b Proof 8 9 a b 10 a Proof b 11 12 Proof 13 a b Proof c 14 i Proof ii 15 a b 16 Proof 17 a b

y

18

C B

C

AO

B

19 Proof 20

and (

)

C

A

x

21 22

Chapter 13 Before you start… 1 2 a b 3 4 5 6

downwards

Exercise 13A 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

Work it out 13.1

Exercise 13B 1 a i

ii b i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii 4 5 6 Proof 7 8

Exercise 13C 1 a i ii b i ii c i ii 2 a Proof b 3 a Proof b

4 a Proof b 5 a When

; decreases from

.

b c d It will decay to zero. 6 a

is the rate of change. ‘Proportional to and inversely proportional to ’ means that so change

. Here is negative because demand decreases as price increases, so the rate of is negative.

b c i

ii

7 a Proof b

minutes

8 a b The velocity approaches

.

9 a Proof b

seconds

10 a Proof b c 11 a b c Proof

Mixed practice 13

; it approaches

.

,

1 2 3 4 i ii 5 a Proof b

months

c Not suitable, as it predicts indefinite growth.

d 6 a b Demand is inversely proportional to the root of the price. c Luxury goods; the demand increases with price. 7 a b It will never fill, as

for all .

8 a Proof b c The velocity increases, but tends to

as increases.

9 i ii 10 a Decrease in size due to, for example, competition for food. b c Increases with the limit of . 11 Proof 12 a b Proof c

Chapter 14 Before you start… 1 a b c 2 a b c 3

Exercise 14A 1 a Exact solution b Can’t rearrange c Exact solution d Exact solution e Can’t rearrange f Exact solution g Can’t rearrange h Can’t rearrange 2 a

and

b

and

c

and

d

and

3 a i Proof ii Proof b i Proof ii Proof 4 a i Proof ii Proof b i Proof ii Proof c i Proof ii Proof d i Proof

ii Proof 5 a, b Proof 6 a Proof b 7 a Proof b i ii

is not continuous (has an asymptote) at y

8 a i

O

0.5

1

ii Two b i ii

changes sign twice.

iii Use, for example,

Exercise 14B 1 a i ii b i ii 2 a i ii b i ii 3 4 a Proof b 5 6 a Proof b c Proof 7 a Proof b 8

1.5

x

.

Exercise 14C 1 a i

close to stationary point

ii

close to stationary point

b i

; on the other side of an asymptote

ii

; outside the domain

2 a b Proof c It may be close to a stationary point. 3

; the tangent crosses the -axis further away from the root, so subsequent values of

y

O

x

3.5

4 a b i ii Doesn’t converge c 5 a b c 6 a

on the far side of a discontinuity and a turning point. and

and

and

b

y 5

O

c d It will converge to .

x

increase.

e 7 a

y

b

(1, 1.5)

x

O c

and

d

, which is outside of the domain of

e 8 a i Proof ii Proof iii Proof iv Proof b i ii iii iv



9 a Proof b

,



y

c

x2

x3 x4 O x1

x0

x

.

10 a

b Proof y

c

k O

x

d Doesn’t converge, or converges to another root (in a different interval).

Exercise 14D 1 a i ii b i ii c i ii 2 a i ii b i ii 3 a

,

b

c

d

a i No ii No iii N/A b i No ii No iii N/A c i Yes ii Yes iii Greater root d i Yes ii Yes iii Smaller root

4 5 6 7 a b 8 9 a Proof b c 10 a Proof b c 11 a

y

O

x

2

b Proof

Exercise 14E 1 a Diverges (increases without a limit) b Diverges (decreases without a limit) c Converges to

(and oscillates)

d Converges to

(decreases)

e Oscillates between and . f Converges (increases to g Undefined after 2 a Converges b Converges c Converges to d Converges to e Converges to f Converges to

.

.)

g Diverges 3 a i ii b i ii c i ii d i ii 4 a i



ii b i

,

ii c i ii





d i ii 5 a Proof b Proof c Proof 6 a b 7 a Converges to b Converges to c Converges to 8 a

b

, so

9 a Proof b 10 a Proof b c

; the positive root.

11 a

and

b c

Mixed practice 14 1 a Proof b 2 a Proof b Proof c 3 a Two solutions;

y

O

x

b Proof c 4 a Proof b Proof c 5 i ii 6 a Proof b c

is on the other side of the asymptote.

d 7 a Proof b 8 a Proof b 9 10 i ii Proof

iii iv 11 a Proof b Proof c

Chapter 15 Before you start… 1

y

2

1

O

1

2

x

π

3

Work it out 15.1

Exercise 15A LB: 0.386; UB: 1.39

y

1 a i 1

O

1

2

3

4

5

LB: 0.910; UB: 1.27

y

ii 1

O

1

x

2

LB: 1.29; UB: 1.68

y

b i 1

O

1

x

2

y

ii

LB: 3.61; UB: 3.79

5 4 3 2 1 0

x

3

4

x

LB: 2.74; UB: 3.99

y

c i 4 3 2 1 O

1

LB: 0.264; UB: 0.576

y

ii

x

2

2 1

O

1

x

2 a i ii iii b i ii iii c i ii iii d i ii iii e In general, doubling the number of rectangles from between the upper and lower bounds. 3 LB:

UB:

4 a LB:

UB:

b By using more rectangles.

y

5 a

5 4 3 2 1 O b

π 2

x

to

or from

to

halves the difference

c Decrease 6 a b

Work it out 15.2

Exercise 15B 1 a i ii b i ii c i ii 2 a,b See table in Worked solutions. c Usually error decreases by a factor of to . 3 a Approximation b c d Approximation e f Exact integral y

4 a 4

x = 1

3 2 1 O -1

1

2

3

4

x

-2 -3 -4

b c Concave curve: underestimate 5 a b Use more trapezia; exact integration is possible. 6 a b

c Concave curve: underestimate 7 8 a b

Mixed practice 15 1 2 a b Concave curve: underestimate c By using more intervals. 3 4 i ii iii Use more rectangles. 5 6 7 a

and

b 8 y

9 a

(0, 1) O

x

b 10

Focus on … Proof 2 Questions 1 Yes, by varying the base (or hypotenuse) accordingly. 2 Yes 3 Proof

Focus on … Problem solving 2

1 a i Tends to

.

ii Stays at

.

iii Dies out after

years.

b i Same as in part a. ii Proof iii Proof c Proof 2 Increases with , decreases with . 3 a i ii iii No solutions b c 4 a b Proof 5 Investigation 6 Investigation

Focus on … Modelling 2 1 2 3 4 5 6 7

this model predicts infinite initial rate of growth.

8

Cross-topic review exercise 2 1 2 3 a

b 4 5 a b 6 a b c 7 i Proof ii 8 9 a b 10 a b c Proof 11 a Proof b 12 a b

c

13 14 a b c 15 a i ii b i Proof ii Proof iii Proof c i ii 16 a

.

b Proof c 17 a b c

trapezium rule is better. y

18 i

(0, 4) O

ii Proof iii iv 19 i ii a b 20 21 a b Proof 22 a Proof b c d 23 a b Proof c Proof,



x

y

24 a

O

b e c 25 i ii iii Proof

1

x

Chapter 16 Before you start… 1 2 3 a b i ii 4 5

Exercise 16A 1 a b c d e f g h 2 a i ii b i ii 3 a i ii All b i ii c i ii d i ii 4 a

b c d e f g h i j k l 5 a i ii b i ii c i ii d i ii 6 a

F

B 29

5

13 98

b c d 7 a

M

E 26

32

15 72

b

c 8 a b 9 a b 10 a b 11 a b 12 a b c 13 a

B

C 23

0

32

12 8

12 48

10 V

b c d e f 14 a

Dark hair

0.5

Blue eyes

0.2

0.2 0.1

b c d

e 15 a b

Cold

0.3

Raining

0.3

0.15 0.25

c d e 16 17 18

to

19 a b c Proof

Exercise 16B 1 a i ii b i ii

or

c i ii 2 a

Year 9 Girls Boys Total

b a b 3 a b c Proof 4 a

Year 10

Year 11

Total

b c 5 a b c d e 6 a b c d 7 a b c d e 8 a b c 9 For example, playing cards in a deck with

Work it out 16.1

Exercise 16C 1 a i ii b i ii c i ii

and

.

2 3 4 a

0.3

0.7

0.6

B

0.4

B'

0.8

B

0.2

B'

A

A'

b c d 5 a b 6 7 8 a b 9 a b 10 a b 11 a b 12 13 14

or

15 Proof

Mixed practice 16 1 2 a

28

ig9

24 − x

xar3

x

52 − x

b c 3 a 0.2

0.8

0.4

A

0.6

A'

0.5

A

0.5

A'

X

X'

b c d 4 a b 5 a

B

0.3 - x

F

x

0.2 0.5

b c d

B

0.1

F

0.2

0.2 0.5

e 6 7 8 i ii 9 i ii

iii 10 11 a b Proof c d 12 a b 13 14

Chapter 17 Before you start… 1 2 3 4 5

Exercise 17A 1 a i ii b i ii c i ii d i ii e i ii 2 Mean: , Standard deviation: 3 Mean:

, standard deviation:

4 Mean:

, standard deviation:

Exercise 17B 1 a i ii b i ii 2 a i ii b i ii c i ii 3 a b 4 a

b i ii c Assumes that they are independent, but Ali might be getting tired. 5 6 a b Proof c Best athletes likely to be chosen for the team; not independent as will depend on the weather / track / how well other athletes are doing; all four going under condition. 7 a b c 8 a b 9 a b 10 a b 11 a b 12 13 14

Exercise 17C 1 a i ii b i ii c i ii 2 3 4 a b 5 a b

seconds is a sufficient but not necessary

c 6 7 8

Tip Did you try sketching a graph? What is the probability of being below 9 a b 10 11 12 Investigation

Work it out 17.1

Exercise 17D 1 a i ii b i ii 2 a i ii b i ii 3 4 5 6 7 8 9 10

Exercise 17E 1 a Yes b No, not symmetrical. c No, two modes.

?

d No, not a characteristic bell curve. 2 Three standard deviations below the mean would be impossible results. 3 a

;

b Mean:

, standard deviation:

c Symmetric, bell-shaped curve d About

students

4 a i ii b 5 a b 6 7 a Proof b 8 Proof

Mixed practice 17 1 a b 2 a b 3 4 5 i ii accept

,

or

6 7 a b 8 a Proof b

,

9 10 i a Probabilities don’t add up to . b

must be the smaller one.

c ii

, but this should be the smaller one. ,

children, so the normal model would predict

11 a It is roughly symmetric. Mean , Variance . b c No, probably not independent. 12 a Within standard deviations of the mean are scores from

to

, which are all possible. The

mean is about the same as the median, suggesting a symmetrical distribution. b Distinction: 13 a b 14 a b

, Merit:

, Pass:

Chapter 18 Before you start… 1 2

.

3 a b c

Exercise 18A 1 a i ii b i ii 2 a i ii b i ii 3 a

, assuming independence of emissions.

b c 4 5 6 7 8 9 a b c 10 a b

Work it out 18.1

Exercise 18B 1 a i ii

b i ii c i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii d i ii

, reject

.

, do not reject , reject

.

, reject

.

.

, do not reject , do not reject

. .

, do not reject , do not reject

. .

4 a b

, do not reject

5

, reject

.

.

6 a The scores of students in the school follow a normal distribution. The standard deviation is still b

, do not reject

.

7 a b c Reject

.

8 a b The weights follow a normal distribution. c Reject

.

9 a b 10 Higher

if

, as you have the probability from both tails.

Exercise 18C 1 a i Significant evidence ii Significant evidence b i No significant evidence

.

ii No significant evidence 2 a i No significant evidence ii No significant evidence b i Significant evidence ii Significant evidence 3 a b Reject

.

4 a b Do not reject

.

5 a All countries in 2016. b c Reject

.

6 a b Reject

.

c No, correlation does not imply causation and the test was not for positive correlation. 7 a b Reject

.

c Less likely to get a false positive, which would result in a useless drug being used. 8 a Significant evidence b

no significant evidence

c Decrease 9 10 When

the correlation coefficient is always or

.

Mixed practice 18 1 2 3 a All countries in 2016. b c Do not reject

.

4 a b

; do not reject

.

5 6 a Both variables must be normally distributed. b Reject

.

7 a Still normally distributed. Still standard deviation seconds.

.

b Reject

.

c They would not be independent – the water needs to cool down to the original temperature. 8

reject

; Sufficient evidence that the mean number of visitor has fallen.

9

Tip Many people think that values are a direct measure of how likely sure you do not fall into this trap.

is to be true. Make

10 a b c 11 a b 12

Focus on… Proof 3 1 Very small (

)

2 a p

1 - p

0.99

E

0.01

E'

0.0001

E

0.9999

E'

G

G'

b i ii

;

.

c The argument is flawed. 3

is the probability of an arbitrary person being guilty.

4 The relative likelihood is reversed. 5 No, because he is either guilty or not, irrespective of whether evidence has been found. 6 7 In legal matters, some alternative explanation may still have non-zero probability, so some doubt remains. Mathematical proof can be absolute.

Focus on … Problem solving 3 1 Investigation 2









You choose



Host opens



.

3 a Proof b Larger, because we already know that there are girls. c









Second child

First child girl

boy

girl

¼

¼

boy

¼

¼

.

Focus on … Modelling 3 3.5 Frequency density

1 a

3 2.5 2 1.5 1 0.5 0 120

130

140 150 Height (cm)

160

170

b No, not symmetrical 2 a b c It could be a suitable model (the IQR of the data is and the box plot looks symmetrical.) 3 a Yes b No 4 It is discrete, but the normal distribution is continuous. 5 a It is likely that the distribution would not be symmetrical, as the minimum possible length of a phone call is (cannot be less than ), but the maximum can be more than b i Yes ii No; bimodal iii No; discrete 6 a Curves and b Curve

Cross-topic review exercise 3 1

minutes.

2 a b Do not reject

; there is no significant evidence of correlation between temperature and rainfall.

3 a

.

b There is evidence that they’re taller 4 a b c Not reasonable; predicts nearly all bulbs will last more than

hours.

5 i ii 6 i ii iii Frequencies are estimated from a sample. 7 8 a b c 9 a Proof b c Proof 10 i ii 11 a b c 12 a b c 13 a b

(a range of answers is acceptable)

14 a b Reject

. Significant evidence of negative correlation; as the amount of time increases the test

marks decrease. c The two sections of the scatter graph represent two different students; for each student there is positive correlation, but one student got better marks despite spending less time on revision.

15 i ii Binya is right; 16 a b c d Proof e f 17 a Reject

, sufficient evidence that the mean time has decreased

b 18 a

.

b c No, because some of the trapezia are above and some below the curve.

Chapter 19 Before you start… 1 a

b

c

d 2

from horizontal

3 a b 4 5 a b 6

Exercise 19A 1 a i ii b i ii c i ii 2 a i ii b i ii 3 a b 4 a b 5 a b 6 a b c

from horizontal

d

; the particle changes direction during the motion

7 8 a

y

b

(0, 9)

x

O

9 a b 10

Work it out 19.1

Exercise 19B 1 a i ii b i ii c i ii d i ii 2 3 a b

from horizontal

4 a b 5 6 7

from horizontal

8 9 Proof 10

Work it out 19.2

Exercise 19C 1 a i ii b i

ii

2 a i

ii

b i

ii

c i

ii

3 a i ii b i ii

4 a b 5 a

b 6 a b c 7 8 9 Proof 10 11 12 a b 13 a Proof b Proof 14

,

Exercise 19D 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 c i ii 4 a b c d 5 a i ii b i ii 6 a b c d 7 a b c d

8

9 10 11 12 13

;

Exercise 19E 1 a i

ii

b i

ii

2 a

b

3

4 5

6

7 a b

8

9 a

b

10 11 a Proof b Proof

Mixed practice 19 1 a b

from the horizontal

2 3 a b Proof 4 5 a Proof b 6 7 a b c It stays at a constant height 8 9

below the horizontal

10 a

b c Proof d e 11 a b c

, or

12 a b

when

13

Chapter 20 Before you start… 1

at

from vector

2 3 4 5

Exercise 20A 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 a b 4 a b 5 a b 6 7 8 a b

below the horizontal.

c 9 Yes; height is

at that point.

10 a b 11 a Proof b 12 a Ten-times that of normal. b Slowed down.

Work it out 20.1

Exercise 20B 1 2 a Proof b c No air resistance; rugby ball is a particle (we ignore its size). 3 a Proof b Yes 4 a Proof b i Proof ii 5 6 a Proof b

or

it must be falling when it goes through the hoop.

Mixed practice 20 1

below the horizontal

2 3 a b 4 a b 5 a b 6 a i, ii Proof b 7 8 a

b 9 i Proof ii iii 10 i Proof ii iii

below horizontal

iv 11 12 i ii

, falling

Chapter 21 Before you start… 1

; magnitude

angle

from direction

2 3

Exercise 21A 1 a i ii b i ii 2 a i ii b i ii 3 4 a b 5 a Proof b c 6 7 a i ii b i ii 8 9 a They are two parts of the same string. b 10 a b

Â

c

Exercise 21B 1 a i ii b i

ii 2 3 a b 4 5 a b 6 7 a b 8 a Proof b 9 Proof 10 a b c 11 and only.

Exercise 21C 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii 3 a b 4 a b i ii

up the slope down the slope up the slope;

up the slope

up the slope; up the slope; up the slope;

down the slope

5 6 a b c 7 a b 8 a b 9 10 a Proof b c Up the slope with magnitude d Down the slope, with acceleration 11 a b c 12 13 Proof.

Exercise 21D 1 a i ii b i ii c i ii 2 3 Proof 4 a b 5

and

6

Mixed practice 21 1 a b Acceleration would be less 2 a

b 3 a b 4 5 a b 6 a b 7 a

b c d No air resistance force acting; no other forces acting on the box; no turning effect (due to forces). 8 a Proof b c The acceleration is reduced because of air resistance or the fact that there is friction. 9 i Proof ii iii To the right 10 i Proof ii 11 i Proof ii 12 a b c 13 a b 14 a b 15 a b

directed down the slope.

16 Proof 17 a b Less friction, so a smaller coefficient of friction. 18 i a b ii 19 i Because the ring is smooth. ii iii 20 i Proof ii a b iii 21 i a b Proof ii 22 a

(in direction of moving upslope)

b Yes; 23 i Proof ii a b iii a b 24 i ii iii iv

seconds after the start of movement.

Chapter 22 Before you start…

R

1

T

mg 2

Exercise 22A 1 a i ii b i ii

anticlockwise clockwise clockwise anticlockwise

c i ii 2 a i ii b i ii

clockwise anticlockwise anticlockwise clockwise

c i ii 3 a i ii

clockwise clockwise

b i ii c i ii 4 a i ii b i ii c i ii d i

clockwise anticlockwise anticlockwise clockwise anticlockwise clockwise clockwise clockwise clockwise anticlockwise

ii

anticlockwise

e i

anticlockwise

ii

clockwise

5

clockwise

6

clockwise

7 a b Since there is no bending, the diver is

from the other end.

8 9 Proof

Tip Write a general point

relative to the bottom left corner of the square.

10

Work it out 22.1

Exercise 22B 1 a i ii b i ii 2 a i ii b i ii c i ii 3 4 5 a b The crane attachment would be able to handle some net moment. 6 7 8 a

and

b That the pole is straight. 9

10 11 a b c d

could be greater than

12 At



13

Exercise 22C 1

,

2

from each support

3 4

or

5 6 7 8

and

Exercise 22D 1

a i ii b i ii c i ii d i ii

2 3 4 5 6

Mixed practice 22 1 2 3

.

4 5 6

i ii

7 8 9

a b

10 a b 11 a Proof b 12 a Proof b 13 14 15 a Proof b c There are no resistive forces other than the friction. 16 a b

Focus on… Proof 4 1 2 Proof 3

a b Proof

4

a Proof b Proof

Focus on … Problem solving 4 The answers that are clearly wrong are:

Focus on … Modelling 4 1 2 The gravitational acceleration on Archimedes’ planet is also

.

3 We need the support to be further away from the point where the Earth is resting on the lever. If this distance is increased to , the length of the other end would need to be . 4

Cross-topic review exercise 4 1 2

a b

3

a b

,

above horizontal

c i ii 4

a b Proof c

5

a b

6

i ii

, bearing

7 Proof 8 9

a b c d i The tension is the same on both sides. ii The two blocks have the same speed and acceleration.

10 i ii 11 i ii 12 i ii iii 13 i Proof ii 14 i

,

ii 15 a b i Proof ii

16 17 Proof 18 a Proof b i ii

Paper 1 practice questions 1 2

a b c

3 4 5 6 7 8 9

a

,

b c

.

10 a b Proof 11 12 a b c 13 a b Proof

,

, use more rectangles and trapezia.

14 a Proof b Proof c d 15 a Proof b c i Proof ii

Paper 2 practice questions 1 2 3 4 5

a b i Proof ii Yes; it is increasing and therefore one-to-one. iii

6

a b

7

a i ii

y

O

x2 x1

x

b i ii The tangent does not cross the -axis. 8

a Model 2; taking logs gives b

9

a b i ii

which is of the form

10 a Cluster sampling b c i No – we can only say that this sample does not provide significant evidence of correlation. ii No – there may be a non-linear relationship. 11 a b c d e 12 a b

-value

Do not reject

: this is not significantly above the

expected number. c May not be independent; club not representative of whole school. 13 a b c 14 a b Proof 15 a Mean is around middle of range. Range is about standard deviations. b c

assuming independence.

d

-value time.

e

Paper 3 practice questions 1 2

a b

y y = x2

y = 6 - x O

x

Do not reject

No significant change in the mean travel

c 3

a b

4 5 Proof 6

a b

7 8

a Proof b

9

a b i Proof ii c

10 a b Proof c 11 a b c 12 a b i ii 13 a b Proof 14 a Proof b c

would need to be larger.

15 a Proof b c 16

Gateway to A level – GCSE revision X. Sequences A sequence is a list of numbers or terms, for example, The first term is usually denoted by

, the second term by

and so on.

You should be familiar with two ways of defining a sequence: by a position-to-term rule (also called ‘the formula for the th term’), for example, by a term-to-term rule, for example, .

Worked example 1 a Find the tenth term of the sequence defined by

.

b Find the second and third terms of the sequence defined by Solution

.

Comments

a

Substitute

b

Use .

into the formula.

with

to link

to

Now substitute in as given in the definition of the sequence.

Use …

with

, to link

to

…and substitute in

A particular type of sequence you have met is a linear or arithmetic sequence. This is one in which there is a common difference between terms. For example: is an arithmetic sequence with common difference is an arithmetic sequence with common difference The th term formula of an arithmetic sequence is always of the form difference.

. , where is the common

You may have used the method shown in Worked example 2 to find the th term formula.

Worked example 2 Find a formula for the th term of: a

b Solution

Comments

a Common difference is

where

so

so

b Common difference is

so

.

Taking , you can see that you need to subtract from to get .

so

where

Taking add to

.

, you can see that you need to to get

.

Tip At A Level you will use a slightly different method to find the th term formula.

EXERCISE X

EXERCISE X 1 Find the first five terms of each sequence defined by a position to term rule. a b c d e f g h 2 Find

and

for each sequence defined by a term-to-term rule.

a b c d e f 3 Find the th term formula for each arithmetic sequence. a b c d e f



Gateway to A level – GCSE revision Y. Adding algebraic fractions To add algebraic fractions, always find the lowest common denominator first.

Worked example 1 Write

as a single fraction.

Solution

Comments You need to find a common denominator before you can add fractions. One that will always work is the product of the two denominators – in this case . You can create equivalent fractions with this denominator by multiplying top and bottom of the first fraction by and multiplying top and bottom of the second fraction by .

Once you have a common denominator you can add fractions by just adding the numerators.



Worked example 2 Write

as a single fraction.

Solution

Comments To find the common denominator, you need to factorise both denominators first.

In this case, multiplying the two denominators together gives . However, you can find a lower common denominator by using

.

Remember that the minus sign applies to both and in the second fraction!

Worked example 3

Solve Solution

Comments Add together the fractions on the LHS in the normal way: here the common denominator is

.

Multiply through by the common denominator to remove the fraction.

Rearrange and solve the quadratic equation.

or

Tip In Worked example 3, instead of adding the fractions you could multiply through the entire equation by at the beginning.

EXERCISE Y

EXERCISE Y 1 Write each expression as a single fraction. a i ii b i ii c i ii d i ii e i ii f i ii 2 Solve these equations. a b c d



Gateway to A level – GCSE revision Z. Exact values of trigonometric functions You can find the exact values of sin, cos and tan of that contain these angles.

,

and

by constructing right-angled triangles



Worked example 1 Find the exact values of Solution

,

and

.

Comments

B

Draw a right-angled triangle that has a 45°

You can choose any length for the sides of the triangle. Let .

1

45° A

angle (which makes it

isosceles).

1

C

Use Pythagoras’ theorem to find

.

Now use the definitions of sin, cos and tan. Rationalise the denominator of

The results for other values are summarised in this table.

sin

cos

tan

You need to be able to use these values in more complicated expressions.

Not defined

Worked example 2 Show that

.

Solution

Comments Use

and

Rationalise the denominator of

.

.

Create a common denominator and add the fractions.

EXERCISE Z Do not use your calculator in this exercise. 1 Evaluate each expression, simplifying as far as possible. a b c d 2 Show that: a b c d



.

Gateway to A level – GCSE revision answers X. Sequences 1

a b c d e f g h

2

a b c d e f

3

a b c d e f



Gateway to A level – GCSE revision answers Y. Adding algebraic fractions 1

a i ii b i ii c i ii d i ii e i ii f i

ii 2

a b c d



Gateway to A level – GCSE revision answers Z. Exact values of trigonometric functions 1

a b c d

2

 Proof.

Worked solutions 1 Proof and mathematical communication Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 1A 1 You need to show that

is not a multiple of any prime number smaller than it. In fact, because factors

of a number come in pairs, it is only necessary to check prime numbers up to

So

:

is not a multiple of any prime number smaller than it. Hence it is a prime number.

2 Every integer can be uniquely written as where is an integer and is a single digit positive integer. Then the square of an integer can be expressed as:

The final digit of must then be the final digit of expression which will lie in the units position of the value.

, since this is the only element of the above

Final digit of

Thus the final digit of a square integer must be

or .

3 by definition of displacement, so

So has only one solution, at from its original position. 4

a Let

. Then

.

, and therefore the particle never again has zero displacement

b If

then it is still the case that

, so the implication does not hold in the opposite

direction. 5 When

it is again the case that

(as it is for

for any integer ).

6 7 The statement is false. For example, consider: , mode 8

, median

. So

,

,

,

but

,

,

,

,

,

,

. Then mean

, contrary to the given statement.

a

so that By the cosine rule, in triangle

b In right-angled triangle

:

:

.

c

y

9

C

R B

S

Q

c

b a

O

A

P x

a b So and are parallel and of equal length, which is a sufficient condition to show that parallelogram.

is a

c If

is a rectangle then

is perpendicular to

.

So is parallel to . We also know that is parallel to that is, the diagonals of must be perpendicular.

. Hence

is perpendicular to

,

10 a For example:

b Taking each side as an exponent of :

, since

, so division by

is valid here without restriction.

c The steps are valid in both directions as long as any arguments of the logarithm lie within its domain. However, if (for example)

and

, then although

, the statement

is problematic, because we require that a logarithm only takes a positive value as argument.

Tip If you study further mathematics, you will find out that the logarithm of a negative number can be calculated as a complex number. In this example, and

so we still have

. However, even if we allow that the arguments can be negative, we must still require that they are non-zero, so cannot calculate

remains a problem. Although

at the origin, we

as any finite value.

11 a b

is prime if it has no factors other than itself and . From a we know that only be prime if

and .

Then

or

.

If

then

which is not prime.

If

then

which is prime.

are factors.

must be greater than , so

can

If

then

Thus 12 If

which is not prime.

is prime if and only if

.

then

for integer such that

So the sum of the probabilities is

.

.

Consider now the expansion of the binomial expression The binomial theorem gives that this equals

. .

Rewriting as , this is the same expression as the sum of the probabilities:

EXERCISE 1B 1 Suppose that is odd. Then

So

for some integer .

is odd, and therefore:

For even , suppose that is odd. By the above, that would mean that the initial statement, meaning that cannot be odd. Therefore, if

must be odd, contradicting

is even, must also be even.

2 Suppose that

, where and are integers with no common factors. Squaring both sides gives:

         (1) This means that is a multiple of so must also be a multiple of . Therefore

, so

.

Substituting this into equation (1) gives:

This means that is a multiple of , so must be a multiple of . But we have shown that both and are multiples of , so they share a factor of . This contradicts the original assertion, so it must be incorrect. Therefore

cannot be written as , so is not rational.

3 Suppose that there is a finite number of even numbers. Then this finite list can then be ordered so that , and the largest even number is . But

would also be even and clearly greater than

, so is not in the list.

(The same argument could be made using

.)

Therefore there are more than even numbers. This contradicts the initial proposition, and therefore there are infinitely many even numbers.

Tip Be careful to ensure that the contradiction applies directly to the initial assumptions. In the above proof we assume a finite number of even numbers and produce a contradiction showing that there must be more. We could alternatively assume a largest even number and show that there must be a larger one. Either approach is fine, but they must not be mixed. 4 Suppose that there exists a rational number ( and coprime integers) and an irrational number such that their sum

is rational. Then

for some coprime and .

Tip Remember that two integers are coprime if their highest common factor is . But then

, which is rational, since (

) and are integers.

This contradicts the initial supposition, and therefore there is no example of a rational and an irrational whose sum is rational. This proves the original proposition that the sum of a rational and irrational number is always irrational. 5 Suppose that and are both odd and is even. odd means that

for some integer .

odd means that

for some integer .

Then

.

So is one more than an even number and is therefore odd, which contradicts the initial supposition. This proves the original proposition that if is even for integers and , at least one of and must also be even. 6 Suppose that

where and are coprime integers.

Cubing both sides gives:

   (1) This means that is a multiple of , so must also be a multiple of , and hence .

, so that

Substituting this into equation (1) gives:

This means that is a multiple of , so must be a multiple of . But we have shown that both and are multiples of , so they share a factor of . This contradicts the original assertion, so it must be incorrect.

Therefore

cannot be written as , so is not rational.

7 Suppose that

where and are coprime integers,

.

Taking each side as a power of gives Because is an integer, each side of the equation can be raised to the power :     (1) No integer power of three is a multiple of , so equation (1) can only be true if This contradicts the original assertion, and therefore 8

.

cannot be written as ; so is not rational.

is a composite integer, so has at least one prime factor where Suppose has no prime factors less than or equal to

.

.

Therefore Since is a factor of not equal to , there is an integer Then

such that

.

, so

If is prime, this contradicts the supposition that no prime factor is less than or equal to If is not prime, then it must have a prime factor

.

.

But if is a factor of then it must also be a factor of . Since

, this again contradicts the supposition that no prime factor is less than

.

This proves the original proposition, that a composite integer must have at least one prime factor less than or equal to . 9 Suppose that at most are within each month. Then at most

dates are chosen, which

conflicts with the initial information. This proves the original proposition, that in at least one month there must be at least dates chosen. 10 Suppose

for some integers and .

Then

.

Then is a multiple of and so must be a multiple of . That is,

for some integer .

Substituting:

. The left-hand side is a multiple of but the right-hand side is not. This proves the original proposition that there are no integers and such that 11 a Substituting

into the cubic:

.

Multiplying through by :

b If and are both odd then each term on the left-hand side must be odd (products of odd values are themselves odd). So

and

for some integers

and .

Substituting: The left-hand side is odd but the right-hand side is zero, which is even. This proves the original proposition that there can be no solution with and both odd. c If one of and is odd and the other even then two of the terms on the left-hand side must be even and the third must be odd. Again, the left-hand side ( inconsistent with the right-hand side equalling zero.

) must be odd, but this is

The only possibility then is that both and are even, but this means that and have a common multiple and are therefore not coprime. If there is a rational solution to the cubic then that rational solution must have a simplest form, where numerator and denominator are coprime. The above working shows that they cannot be coprime, which contradicts the assertion. This proves that there are no rational solutions to 12 Suppose that a triangle with sides

.

and has

.

By the cosine rule, the angle is such that

.

Comparing it to the initial proposition, we see that We reject Therefore

or

, since then the shape is not a triangle, but a line.       (1)

Since is an internal angle of a planar triangle, The only solution to equation (1) in that interval is

.

Tip Using trigonometry can feel like a circular argument, since many trigonometric laws proofs refer to some properties of right-angled triangles, such as this one. In such a question, you should feel free to use standard results unless specifically told otherwise.

EXERCISE 1C 1

a b In line the double implication is incorrect. From line , the correct equivalent line would be , and this explains the introduced solution .

2

a If

then both

b The correct symbol is solution.

and

so

is a solution to the original equation.

(if), which shows that

is a solution but not necessarily the only

3

a b In line the implication sign is incorrect, and should be an ‘only if’ . When multiplying both sides of the equation by in line , a false solution is introduced into the work. In fact, noticing that the numerator and denominator have a common factor

in line , a faster

solution would show 3. 4.



5.



4 In line the implication sign is incorrect, and should be an ‘only if’

. When combining two logs using

the rule log , the rule assumes that both and are known to be positive. Since we have no such information here, a forward implication sign should be used, or (as in this case) a false solution may be introduced. 5 From line to line he takes the positive square root but omits the negative root. Line should read , or the statement ‘so is a stationary point’ (it is acceptable to mention only since that is the focus of the question, but there should not be a suggestion that it is unique). Line is an error in understanding; when taking the derivative, the function expression should be used, not the derivative of a value (which is always zero). Having supposedly found that the second derivative is zero, Andrew makes the incorrect assertion in line than this shows that the point is a minimum on the curve. Line onwards should read 5.  6.  7. At

,

so

is a minimum.

6 The working is correct but incomplete; it is always necessary to compare all coefficients to ensure that all criteria are found and are consistent. Only the

term is evaluated.

To complete the solution:

From the coefficient of

, substituting

gives

It would therefore also be accurate to give , and there are infinitely more possibilities. However, this is not really important; the question asserts that is a factor, and does not require that the answer clarifies the relationship between and necessary for this to be the case, though a diligent student might feel it wise to give this information in an examination: For

to be a factor of

.

Then the cubic factorises as 7 Step 8 is false:

MIXED PRACTICE 1

shows that is even, not .

.

MIXED PRACTICE 1 1 If is even then this product is even. If is odd then

is even, so the product is even.

Either way, the product is even. 2 Suppose that

where and are integers with no common factors.

Squaring both sides gives:

(1)

This means that is a multiple of , so must also be a multiple of . That is,

, so

.

Substituting this into equation (1) gives:

This means that is a multiple of , so must be a multiple of . But we have shown that both and are multiples of , so they share a factor of . This contradicts the original assertion, so it must be incorrect. Therefore

cannot be written as , so is not rational.

3 Suppose that there are only a finite number of square numbers. Let the total number of squares be . Then the squares can be listed in increasing order

, so that

is the largest square

number. But

is a square number, which must be greater than

So there is a square larger than

, since

, which means that there are more than squares.

This contradicts the initial proposition, and it is therefore proved that there must be infinitely many square numbers. 4 Moving from line to line , both sides are multiplied by

.

Whenever two sides of an equation are multiplied by a function , the roots of may be introduced as additional solutions, if not already solutions of the original problem, and the implication sign has to be one way (or a caveat introduced). Step 2 should either read 2a.



or (better) 2b.



,

Then from step 2 to step 3 contains the more critical error; both sides are divided by . Whenever two sides of an equation are divided by a common factor , the roots of should be noted as possible solutions of the problem to be carried forward in the working.

Step 3 should read (continuing from 2b) 3.

(

or

),

Tip Always check that your working does not introduce false solutions. In fact, in this case, multiplying by does not introduce any, but this does not change the fact that the working should have noted the possibility and excluded it. 5 If

then

(both sides equal

). However, if

could also be . Hence the correct symbol is 6 Suppose that

, it does not follow that has to be : it

.

where and are coprime integers,

. Taking each side as a power of

gives Because is an integer, each side of the equation can be raised to the power

(1)

No integer power of five is a multiple of , so equation (1) can only be true if

.

This contradicts the original assertion. Therefore 7

cannot be written as

, so is not rational.

y B P C

Q b

c

A a x

O

If

is a parallelogram then

so that

Since the displacement vector from to is a multiple of the displacement vector from to , it follows that

and are collinear.

8 If is even then multiple of : for some integer .

is also even, so this is a product of two even numbers and must be a

If odd:

is odd then

is also odd, so this is a product of two odd numbers and must also be

for some integer .

Tip Sometimes rearranging the equation can reduce the number of relevant cases, thereby saving working. This question could be checked by working through an exhaustive list of the four combinations of and being even and odd, but the rearrangement reduces the number of relevant cases to only . 9

a

Tip When you are trying to make a coherent proof of something which seems obvious, carefully break down your argument and ensure that you do not use any circular reasoning.

Let

be a polynomial of order given by

with integer coefficients . Note that:

(1) The product of two integers is an integer ( is closed under multiplication). (2) Consequently, an integer to an integer power is also an integer. (3) The sum of two integers is an integer ( is closed under addition). Then for an integer , By (2), By (1),

is an integer for each value . is an integer for each value .

By repeated application of (3), the finite sum b

is an integer.

is a polynomial with non-integer coefficients which nonetheless always gives an integer result for an integer value .

10 a Line has an incorrect implication sign (or lacks an additional condition). It should read either 2a.



or, more usefully when working out the solution, 2b.



and

(The condition ‘and

’ can be then added to each line up to .)

Line has incorrectly found the completed square form with a sign error in the bracket. 5a.



Line is missing the negative root from line . Following through from 6a.



and

(with the sign correction):

The final line should then resolve the multiple statements in line 7a.



2.



3.



4.



5.



6.



7.



:

b

11 The problem is to find a prime number such that When

, which is prime.

When

, which is prime.

When

, which is prime.

When

, which is not prime.

is not prime.

Hence the required value of is . 12 Line is incorrect. Statement correctly implies that is divisible by ; however, this does not imply that is a prime number (as every prime number is divisible by ). 13 Let

for integers

and .

Suppose that and are both odd. Then

and

for some integers and . .

Clearly this is both even and not a multiple of . Now consider : If is odd then is odd. But

has been shown to be even.

If is even then for some so that not to be a multiple of . No integer exists which can equal

and is a multiple of . But

has been shown

, contradicting the initial statement.

This proves that the initial supposition must be incorrect and that at least one of or must be odd. 14 a For right-angled triangle

with a right angle at :

and But Hence b Then:

, so .

So the product of all the 15

from the four lines above equals , which is rational.

Tip There are often many ways of proving a trigonometric result, using a variety of trigonometric facts. Below are three simple options for this question, but there are many more. Method 1: Squaring and using known features

Between

and

,

(1) 

is negative, and in consequence:

; and

(2)  From

is positive and

. and

,

between

Square rooting, and using

:

and between

and

You will be able to understand methods and after completing Chapters 8 and 10. Method 2: Reducing to a single trigonometric function

Between

and

,

.

By the above working, between

and

,

.

Method 3: Calculus Let At

and at

Taking the derivative of

,

. :

has a stationary point where

In the interval

.

, there is a single solution to this, at

, which corresponds to

. Since

is continuous in the interval, and has a single stationary point greater than the two interval

end points, that stationary point represents a maximum and therefore minimum of the two end points. Hence

in the interval.

is never less than the



Worked solutions 2 Functions Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 2B 5

a

b 6 Domain: Require the argument of the logarithm to be positive:

Range: Range of logarithm is the whole real line . 7 Domain: Require the argument of the root to be non-negative.

8 Cannot have division by zero:

.

Square root only takes non-negative values. Require Domain of 9

.

is

a This is a negative quadratic with roots at

and

.

y

–2

O

x

3 2

b The function ‘square root’ requires a non-negative argument, and therefore requires . From the graph, 10

. But ln takes only positive values. Therefore: or

11 Cannot have division by zero: therefore

.

Because square root only takes non-negative values, either . and Domain of

or

and

and

is therefore

12 a Because square root only takes non-negative values, Also, because ln takes only positive values, i



Domain is

ii



Function has no domain.

b

EXERCISE 2C

.

.

.

or

and

EXERCISE 2C 6

7

8 a b 9 a Domain of

: Require non-negative value for

b Over the domain

takes all values except

.

and so the range of is

.

10 a Range of is

.

The range of lies within the domain of , so is a valid composition. Range of is

.

Values lie within the range of but not within the domain of , so is not a valid composition for the full domain of . b Require the range of to be limited to 11 By observation,

EXERCISE 2D 4

a

.

so restrict the domain to

.

b Since

, then (because is stated as being one-to-one and so

is well-defined)

5

6

a

b Range of is Domain of is

so domain of so range of

is

.

is

7

8 Suppose that

, defined over the range of .

That is, for any element in the range of , in the domain of , . Then, by definition, 9

. Similar working shows that for any element

.

a The graph of the inverse function is the original graph reflected through the line

.

Tip Always be alert for axes with different scales; the scale is the same in this question, so is the usual line.

y 10 5

–10

O

–5

5

10

x

–5 –10

b The domain of one):

is the range of

c Since is an increasing function, .

. The range of

is the domain of (since is one-to-

will have solutions on the line

. The solutions are

Tip Remember that the solutions to will lie on the line if the functions are always increasing. This is NOT the case for a decreasing function; for example, the function is self-inverse, with every value in the domain , but only for does .

being a solution to

10 a

b

11

(Selecting negative root since can only take non-positive values)

has an asymptote at Hence the domain of

(as

) and is always decreasing, so range of is

is

Tip The domain is an essential part of the definition of a function, so if you are asked to find a function, always specify its domain as part of your answer for anything other than the entire real line.

12 a

The range of is , so the domain of

is also .

b 13 a A graph of an inverse function is the original graph reflected through the line

.

y 10 5

–10

O

–5

5

10

x

–5 –10

b Since

is an increasing function, all solutions of

also lie on the line algebraically:

EXERCISE 2E

. From the graph, this appears to be at

, intersections of the two lines, will , which can be shown

EXERCISE 2E 3

a b The function has a turning point at for which .

and is continuous. Therefore there are values and with

The function is not one-to-one across its domain, so c For domain

is not defined.

, the function is strictly decreasing .

and so the function is one-to-one over the

4

so

.

For to be self-inverse,

.

Comparing these, it is evident that

.

(Alternatively, multiply out and rearrange to find so

for all values of , from which

.)

5 For the inverse to be defined, must one-to-one and so there can be at most one stationary point in the cubic.

Tip A single stationary point on a cubic (as for example in

) will be a horizontal inflection,

which does not prevent it being one-to-one. However, if there are two stationary points, they must be a local maximum and a local minimum, and the function will not be one-to-one.

For there to be at most one real root for this quadratic, the discriminant

Positive quadratic is less than zero between the roots

MIXED PRACTICE 2

.

1

a

The domain of is

, so range of is

. Hence domain of

is

.

b

The range of is 2

a The reflection of

, so domain of in the line

is

gives the graph of

b Where cuts the -axis is the reflection in so cuts the -axis at 3

(domain Then

4

.

, range

),

(domain , range

is the same as the domain of

as

.

of where cuts the -axis. cuts the -axis at

with the same domain and range as

The range of

so is

).

.

so the composite function

is well-defined

with domain , range

a b

for

.

c The graphs of a function and its inverse are reflections of each other through

.

d i

(Since

, we choose the positive root.)

ii The domain of is iii The range of is

so the range of

is

so the domain of

. is

.

e

But

so there are no solutions to this equation, since the roots are and

.

5

a

, so

b When 6

. , so

and

. Hence

.

a b

(Select the positive root as the domain of is

The domain of 7

is the range of . Since

, the domain of

is

.)

.

a i ii Range of is . iii iv v b The value

is in the range of but not in the domain of , so

is not defined.

c i

Note that

is self-inverse.

ii The graphs of a function and its inverse are reflections of each other through iii

is self-inverse so the domain and range of be the same value set. The domain of is

iv The range of 8

is

.

a b Vertex at

, positive quadratic, -intercept at y

9

(–2, 5) x = –2 O

x

and the domain and range of .

. must all

c Range of d 9

is

. Range of

is

. Range of on a domain

is

. The range of

is

.

a The easiest way to find the range of a quadratic function is by completing the square:

Since

is always

, the range of is

.

b The graph of is a parabola, so it is not one-to-one. Hence f has no inverse function. c First obtain an expression for

:

, so Hence Comparing the coefficient of

, so

or

.

Comparing the constant term: When

, so

When

. But the question says that is non-zero, so this is not possible.

, which is fine. Hence

d Now

.

. Then: for all . for all .

This means that the graph does not cross the -axis (it is a negative quadratic), so the discriminant must be negative: . Rearranging this gives

.

10 a b c

11 a

b

   (1) Substituting with :     (2)

c Equation (1)

equation (2) gives:

. Then

12 a The domain of

is

, so restrict the domain of g so that the range excludes values less than .

The excluded interval has

. Then

Positive quadratic is less than zero between the roots

b The range of for domain

is

The range of for domain

is

So the range of the composite is 13 a So

for any value of . Hence by definition,

b Rotational symmetry of c Let

is an odd function.

about the origin.

.

Then

. Hence by the definition given,

d

is an odd function.

. So modulus is an even function.

e An even function has reflective symmetry through the line f Let

(the -axis).

.

Then

. Hence, by the definition given above,

is an even function.

g Since a multiple of an odd function must be odd and a multiple of an even function must be even, as defined above must be odd and must be even. But

.

Any function is therefore the sum of an even function and an odd function.



Worked solutions 3 Further transformations of graphs Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 3A 6

a Vertical translation by :

becomes

:

Vertical stretch by sf :

becomes

:

b Vertical stretch by sf :

becomes

:

Vertical translation by :

becomes

. . .

:

.

c Horizontal stretch by sf : Replace with :

7

.

Horizontal translation by

: Replace with

:

d Horizontal translation by

: Replace with

:

Horizontal stretch by sf

: Replace with :

.

a Horizontal stretch by sf : Replace with : Vertical stretch by sf :

becomes

:

Because the function is of a power of , a stretch along one axis is equivalent to a stretch along the other axis. For any , a horizontal stretch with scale factor will be reversed by a vertical stretch with scale factor . b Horizontal translation by Vertical stretch by sf :

: Replace with becomes

:

:

Hence c Vertical translation by : To replace with

becomes

:

requires a horizontal stretch with scale factor .

It is a defining quality of the exponential curve that a horizontal translation is equivalent to a vertical stretch. A logarithm, being the inverse function to an exponential, has its graph and consequent properties reflected through ; for a logarithm, a horizontal stretch is equivalent to a vertical translation. 8 Graph connects a

to

to

to

: Translation

and horizontal stretch, factor relative to the -axis

New graph connects

y 6 5 4

y = f(2x) – 3

3 2 1 –6 –5 –4 –3 –2 –1 O –1

1

2

3

4

5

6

x

–2 –3 –4 –5 –6

b Vertical stretch scale factor with reflection through the -axis followed by a translation New graph connects

to y 8 7 6

y = 1 – 3f(x)

5 4 3 2 1

–8 –7 –6 –5 –4 –3 –2 –1 O –1

1

2

3

4

5

5

–2 –3 –4 –5 –6 –7 –8

9

a

: Vertical asymptote.

, intercept

y

y = lnx O

1

x

7

8

x

.

: Translation

b

Vertical asymptote y

intercept

y = 3ln (x + 2)

x = –2

x

–1 O

: Translation

c

and vertical stretch, factor relative to the -axis.

Vertical asymptote

and horizontal stretch, factor relative to the -axis.

intercept

y x =

O

1 2

x

1

y = ln(2x – 1)

10 Translation

: Replace with

Function becomes Reflection through the -axis: Multiply function by Function becomes 11 Translation by

Reflection through

: Replace with

, add to function.

: Replace with

Horizontal stretch, factor : Replace with

Comparing coefficients:

12 Reflection in the line

Translation by

: Replace with

: Replace with

, add to function.

Horizontal stretch, factor : Replace with .

Comparing coefficients: : : : 13 Vertical stretch, factor : Multiply function by .

Translation by

: Replace with

, add to function.

Horizontal stretch, factor : Replace with .

EXERCISE 3B 5

a Graph of

crosses -axis at

.

y

y = |x – 2|

2

O

x

2

b Graph in a translated units upwards.

6 Graph of

crosses the -axis at

y y = |2x + 3|

3

–1.5 O

7

x

.

y

3

y = 5 – |x – 2|

O

–3

x

7

8 y

y = x|x|

x

O

EXERCISE 3C Each problem in this exercise can be solved either by an algebraic method (starting by squaring both sides) or by drawing graphs. The graphical method may be better if the question involves an inequality. Both methods are shown for the first question, then the graphical method alone for subsequent inequality questions. 3 Method 1: Graphical approach a

y = |4x + 1|

y

x1 O

The line

y = x + 3

x2

x

crosses the modulus graph once on each side.

At

:

At

:



b From the graph,

for

Method 2: Algebraic approach a Squaring both sides:

b Testing an intermediate value: When

, the inequality

required lies between the two boundary values. Hence 4

for

a

y

y = |2x – 5|

y = |x + 2|

x2

O x1

x

intersects both arms of At

:



At

:



b From the graph, the required region lies outside the interval for

5

a

is true, so the region

y

y = | 3x – 7|

7

x

O

b There are no intersections; there is no solution to y

y = |3x – 7|

y = 1 – x

x

O

6

y

y = |3x + 1|

O

The graph of is 7

y = 2x

x

lies above the graph of

for all values of , so the solution set for

y

y = |k – 2x|

y = |k – x |

k x2

O

x

k 2

Both intersections lie in the left arm of

, which is

: :





From the graph, the inequality

has solution set

8 Squaring:

That is, for

, the solution is

.

MIXED PRACTICE 3 1

a After translation and

b

and vertical stretch, factor , points

and

on original will shift to

.

to After horizontal stretch, factor , reflection through the -axis followed by translation and

on original will shift to

and

.

, points

y y = 3 – f(2x)

(0, 3) O

x

2 Original curve: Translation

: Replace with

.

Vertical stretch, factor : Multiply function by .

3

a

y

y = |2x – 1|

O

x1

y = x

x2

x

b One intersection on each arm of the modulus function. At

:



At

:



From the graph, the inequality

is satisfied between these two intersection values. Hence

. 4

a Transform Replace with

to

. : Translation

.

Multiply function by : Vertical stretch, factor . b Transform

to

.

Replace with

Add

: Translation

.

to function: Translation

c Transform

.

to

This is achieved by (b) followed by (a): Translation 5

a

followed by a vertical stretch, factor .

changes to replaces by

, so the equation after transformation is .

So the resulting equation is b

is replaced by

.

.

, then by and then by again. This corresponds to the sequence of

transformations T, S, S. 6 Transform

to

Replace with

.

: Translation

.

Multiply function by : Vertical stretch, factor .

7

y

y = |2x – 1|

y = |x – 6|

x1

O

x2

x

Both intersections lie on the left arm of At

:

At

:

From the graph, the inequality is satisfied outside the interval 8

a

. Hence

or

.

is replaced by and is multiplied by . This corresponds to a horizontal stretch with scale factor and a vertical stretch with scale factor .

b Apply the two transformations from part a to the graph of

.

y

( )

y = 3ln x 2

O

x

(2, 0)

, so this is the function from b with replaced by

c

, which corresponds to a

horizontal translation units to the left. Hence the vertical asymptote is through the origin.

and the curve passes

y

x = –2

( )

y = 3ln x + 1 2

O

9

a Transform

x

(0, 0)

to

Multiply function by : Vertical stretch, factor . Multiply function by

: Reflection through -axis.

Add to function: Translation

.

Note that the stretch and reflection can be taken in either order but the specified translation must occur last. Alternative correct answers include: Vertical stretch, factor , translation

then reflection through -axis.

Reflection through -axis, translation

then vertical stretch, factor .

Translation

then reflection through -axis and vertical stretch, factor (in either order).

y

5

x

O

y = 5 – 3| x |

b

y y = |2x – 1|

x1 O

x

x2 y = 5 – 3 | x|

At

:

At

:

c From the graph, the inequality holds in the interval enclosed by these two intersection values. Hence

10 a To transform b

to

has asymptote intercept

, replace with and intercept

. It also has an intercept at

: Translation

. So

.

has asymptote

and

.

takes the previous graph and reflects it through the -axis then translates it by . This graph will have the same asymptote will have axis intercepts at

and

and will pass through .

, descending. It

y y = –2 y = ln(x + 2)

–1

y = 3 – ln(x + 2) x

O

c i The image of the first (left-most) turning point must be the left-most turning point on the image graph . If the only transformation prior to the reflection in the -axis is a translation, the translation takes

to

and so is a translation

.

ii Translation

: Replace with

. Then

Reflection through -axis: Multiply function by

. So

Hence 11 a Since

, the range is

.

b

.

c To see the number of solutions of the equation, sketch the graph of

.

y 3

2

1

O

5

10

15

20

x

Looking at the dashed horizontal lines, we can see that the line . 12

or

13

will cross the curve twice when

y y = |x| + x

O



x

Worked solutions 4 Sequences and series Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 4A 5 a



b

, so

.

6 a

b Continuing the sequence on the calculator, it is clear that the sequence approaches oscillating (consecutive terms are above and below the limit). 7

. If the sequence has a limit , then we can write

:

8 a

b

or But

9

gives a sequence which decreases from

.

as

,

a

The sequence cycles through these values, so b 10 a b Continuing the sequence: It appears that the sequence increases without a limit.

EXERCISE 4B

.

EXERCISE 4B 3

4

a

b

5

6

a

b The sum of every six consecutive terms in the sequence is zero.

EXERCISE 4C 3

a so

b Let

be the last term less than or equal to

Then is the greatest integer such that so There are 4

and so

terms less than

.

5

Let

.

Then

6

Let

.

Then

7

Let be the common difference between consecutive terms. Then (1) (2) (3) Also, Substituting

8

and

:

a First pages are numbered with single digits for a total of digits. The numbered with two digits. So the total for the first pages is digits. b First pages: pages at digit per page: Total Pages Pages

pages at digits per page: Total pages at digits per page: Total

th and

th pages are each

There are

three-digit pages.

Therefore there are in total

pages.

EXERCISE 4D 3

is an integer, so reject the first factor; 4 a

b

5

(1) (2)

6

(1) (2)

7 (1) (2)

.

8 The odd numbers form an arithmetic sequence with

.

9 For the largest possible sum, require all the positive values in the sequence (and no negative values). Since

, it follows that

, so the largest possible sum is given by:

10

:

Roots of

are

Least positive such that

and is therefore

.

11 , so

. Hence

The th term is

. .

12

(1) (2) Substituting

into

The smallest sector is

:

.

13 a The multiples of form an arithmetic sequence with and . The greatest multiple not greater than is , and is the nd term in the sequence. Hence:

b The sum of integers between and arithmetic series with ):

inclusive is the .

th triangle number (or

The difference between the two sums is the sum of all positive integers up to multiples of :

14 Rearranging:





for the

which are not

Substituting:

All terms in the sequence are positive so

.

So 15 a

The multiple

is an arithmetic series with

The sum of the logarithms is therefore

and

.

.

b 16 This question can be approached in two different ways: by considering the sum of two simpler series or the difference of two simpler series. Method 1: Sum of two series

This can be considered as the sum of two series: is the sum of values with form is the sum of values with form Series with Series with

Method 2: Difference of two series Let be the sum of all three-digit multiples of Let

be the sum of all three digit multiples of Series with Series with

EXERCISE 4E 3

a

(1)

(2)

so

.

Hence the th term b 4

a

(1) (2)

b 5

a

(1) (2) (3)

Substituting into

b 6

(1) (2)



Then So 7

a

The negative sign indicates that is an even number, so

b



;

8 The sequence is a geometric sequence with first term

and common ratio

.

Require

Tip Remember that log

The first term less than

when

so dividing through causes a reversal in the inequality.

is

9

Since

, we can cancel:

Solving the quadratic equation gives 10

Tip In sequences and series, letters

and normally have standard meanings, so take extra

care in questions which use these letters in other ways and do not use any letter for more than one meaning. is an arithmetic progression so: and

(1)

is a geometric progression so: and Substituting

(2) into



or gives So

which contradicts the requirement

.

The common difference of the arithmetic sequence is

But





The geometric sequence is So

(reject



.

since this clearly does not lie in the (increasing) arithmetic

sequence).

So





EXERCISE 4F 3 a Comparing to the general form for a term from a geometric sequence

,

b

4

Using your GDC, there are two solutions: 5

or

(1) (2) The expression for

can be written as

Hence we have: Solving this equation gives Substituting into (a) gives 6

a

(1) (2) (3)

(4)

.

.



(5)

b First term: Second term: 7

a This sum can be shown in two different ways: either by writing out all the terms, or by using the formula a geometric sequence. Hence:

b

appears in the expression for the sum of the first six terms of a geometric series with common ratio :

Rearranging this gives:

EXERCISE 4G

Tip If you study If you are told that the sum to infinity has a finite value, you can assume that without further information. If you are not told that the series converges, you should always explicitly show that as part of your answer when evaluating or using . 3

(1) (2)

4 The first step is to find the common ratio. (1) (2)



a The expression for the sum of the first terms is:

b

5

a

(1) (2)

Each term of the series is positive so reject the negative solution, and conclude that

.

b 6

(1) (2)

Each term of the series is positive so reject the negative solution and conclude that So

. Then

7

, so a The convergence criterion is: That is, b If

and

..

. Therefore

.

and

8

,

. (1) (2)



9 Comparing features:

and

a Using the convergence criterion:

.

.

b



, so



and

Require least integer such that

. .

Therefore at least terms are required to reach a sum greater than

.

Tip You may also be able to use a TABLE function on your calculator to find the required value of . 10 a Using the convergence criterion: b If





,



So





.



11 The series can be written as and common ratio a When

b When

This is a geometric series with first term

. Its sum to infinity is

,

, which exists when

.

.

the series diverges (so the sum is infinite).

EXERCISE 4H 1

a Let

represent the balance at the start of year .

Tip Be careful to define the terms you use; in finance questions it will often be critical whether you consider as the value at the start of year or the end of year . If you are defining your own variables, always state this clearly at the start of the question. Geometric sequence:

The interest for the sixth year is £ b The balance after six years is the balance at the start of the seventh year,

The balance after six years £ 2

a Let

be Lars’ salary in the th year.

Arithmetic sequence:

.

In the twentieth year his salary will be £

.

b Require So

The roots of this positive quadratic are after 3 Let

and

. So he will have earned more than

years.

be the balance at start of year . Then

.

£

a After full years the balance is the same as at the start of year b The balance at the end of years: £

£

£

.

.

c i ii

So the balance will exceed 4

after

whole years.

a b

and The number sold exceeds the restocking rate of

so the number of bunches will decrease.

In the long term, the number of bunches would stabilise: Long-term number of bunches f is such that So

from which

In the long term, there will be a stable 5

a The balance at the start of year is £ £

bunches in the shop. .

£

£

million

b The balance at the end of month is £

.

Require £

It takes 6 Let

months, equivalent to

years and months for the investment value to exceed

be the number of miles run on day . Then the relevant arithmetic sequence is

a Require

.

The roots of this positive quadratic are exceeds miles. b Require

and

So it is after

days that the total distance

.





. So it is on the

7 At the end of the th month, let

nd day he runs more than

miles.

and be the amounts paid in by Aaron and Blake, respectively.

is an arithmetic progression with

. His balance is the sum

is a geometric progression with

Require the least integer such that

. His balance is the sum

a Let So

.

.

Tables on a calculator will show that Blake will have more money than Aaron after 8

.

months.

be the greatest height in metres reached after the ball has bounced times. and forms a geometric sequence with ratio

b Let the distance travelled until the th bounce be

Then

and

So c The small distances of the bounces predicted by the model after bounces would be overtaken by the inaccuracies in the model; the distance of the initial drop and the percentage decrease are only given to one significant figure, and any error would be compounded by the model at each bounce. Also other effects of ball and surface irregularities, air resistance etc would have a greater impact at smaller bounces. A model such as this would predict that the ball would bounce for eternity, but small motions are damped by other physical effects so that bouncing objects do come to rest. 9 Let the account balance at the beginning of year be given by £

. Then

and a

b The pattern shows that

c Require

So after 10 Let

is the sum of a geometric sequence with

:

years of saving, Samantha will have accumulated at least £

.

be the number of seats in row . Then the relevant arithmetic sequence is

a Require

.

The roots of this positive quadratic are excess of seats.

and

b

The percentage of seats in the first rows is 11 a Let the amount owed at the start of the th year be Then

and from

. So

rows are required for there to be in

b c

Tip Take extra care when a sum is from zero instead of one. You could approach this by direct calculation – using a calculator or spreadsheet to find the first for which would be negative – or by finding a formula for in terms of and then solving Both are given below as some questions may require an algebraic approach.

for least .

Method 1: Direct iteration By iteration of the formula:

and

Therefore Dineth will pay off his mortgage by the end of the

th year.

Method 2: Rearrangement to formula

Generalising to

:

Recognising the summation as the value of a geometric series:

Solving for

, which would give the number of years that the Dineth begins the year owing

money:

so So Dineth owes money for the final time at the start of the mortgage completely over the course of years.

MIXED PRACTICE 4 1

(1) (2)

2

(1)

th year, and therefore pays of the

(2)

3

a

(1) (2)

b





so the sum to infinity converges to a finite value.

4

so

So



If

then



5

6







a

This is an arithmetic sequence (since the terms increase by each time). b From part a,

and

. Hence the sum of the

7 Geometric sequence Hence Require least integer such that

.

terms is:

So the smallest integer satisfying this condition is 8

.

(1) (2)



9

Tip Take extra care when a sum is from zero instead of one.

Infinite geometric series:

10 Let

be the number of bricks in row , where row is the top row.

This is an arithmetic sequence

.

a b





c

Therefore 11 Let

(reject the solution

, in context). So the total number of bricks

be the amount after years in plan and

Using tables on the calculator it can be seen 12 a

(1) (2)

the amount after years in plan .

is larger until

.

.

(3) So



b If



or

then

Since If

and

, the series does not converge as then

Since

.

and

. .

, the series converges as

:

.

13 This is the sum of two infinite geometric series:

The first series

where

,

.

and

has

. The sum to infinity

.

The second series is

where

,

and

has

. The sum to infinity

. The total value is therefore 14 The least multiple of greater

Let

is

and the greatest less than or equal to

be an arithmetic sequence with

Then the sum of the multiples of between

, and

and

is

.

.

is

15 a

A positive quadratic is less than zero between the roots. The required range is therefore b Continuing the sequence on the calculator suggests that the limit is setting

. This can be checked by

.

Rearranging: Given

,

16 a This scheme can be represented by an arithmetic sequence with first term difference . i

and common

£

ii b

, so £

17 a In an arithmetic progression, Dividing by :

. Therefore

.

.

b To solve this equation, note that

The solutions are

and

is a solution so

is a factor:

.

Since the terms of the sequence are all different, . Therefore the only possible solution is

. Since the geometric progression converges, .

c

This equals

when

18 Let the geometric sequence be common difference .

. and have common ratio , and the arithmetic sequence have

(1)

(2)

(constant sequence) or 19 Let the geometric sequence be

and have common ratio , and the arithmetic sequence have

common difference and first term .

(1)



(or

(2)

which produces the constant zero sequence, for which the common ratio is not

defined) To factorise this cubic, note that if be met, so is a factor.

then the two sequences are constant and the conditions would

The quadratic term has roots

.

If the sum to infinity converges to a finite value then

so only the root

meets the

criteria. 20 Let the sequence be identified as

, where

Comparing this to the general form for a term of an arithmetic sequence to be an arithmetic sequence with

21 a b

,

this is seen

. Then

integers The final integer on the th line is the th triangle number:

c The first integer on the th line must

.

less than the final integer:

Alternatively, it is one more than the final integer on the line d Arithmetic sequence of consecutive values from ,

, which gives to

:

.

e Using a table on the calculator,

.

22 a Consider the mortgage as held in one account

and the payments into a separate account

The mortgage account then just rises at its interest rate: is a geometric sequence with

,

At the end of three years the mortgage account stands at The payments account works as previous payments and then a new £

£

. .

£

, since each year interest is added to the payment is made. is the geometric series

. At the end of three years, the payments account stands at So, after three years, the outstanding mortgage balance is Balance at 3 years

£

b Continuing the pattern: Balance at years £ £

c Require mortgage balance at years less than zero.

So the mortgage is paid off after



years.

.

£

.

.

Worked solutions 5 Rational functions and partial fractions Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 5A

Tip When proving factors for a polynomial expression, it is often useful to define the expression as first, so that the factor theorem can more easily be expressed, citing . 2 Let Then By the factor theorem, if So

then

is a factor of

.

for some integers , and .

Comparing coefficients:











: Consistent with

3 Let By the factor theorem, if

is a factor of

then

.

4

a By the factor theorem, if

is a factor of

then

.

b So

for some integers , and .

Tip Remember never to re-use an unknown constant in a different context within the same question. There is already a value from the first part, so a different letter for the quadratic

coefficient of the other factor must be used here. Comparing coefficients:













5





is consistent.

a Let

By the factor theorem, if b

then

is a factor of

for some integers , and . Comparing coefficients:















is consistent with

.

6 By the factor theorem, if

then

So

is a factor of

.

for some integers , and .

Comparing coefficients:













is consistent with

has solutions 7 Let If

or

.

. is a factor of

8 Let

,



then by factor theorem,

.

.

If

9

is a factor of

, then by the factor theorem,

a By the factor theorem,



.

.

(1)



(2)

Substituting into (1): b So

for some integers , and .

Expanding:

Comparing coefficients:











Consistent with

,

Consistent with the above.

.

for 10

so by the factor theorem, so by the factor theorem, So

is a factor of

.

for some integers , and . Comparing coefficients:















Consistent with Consistent with

for

EXERCISE 5B 5

6

7

a b or or

8



is a factor of is a factor of

. .

9

Tip There are a number of ways of solving this type of question. Shown below are the longdivision method, a gradual fraction rearrangement method equivalent to long division, and two approaches using a predicted general solution where unknowns are found either by substitution or by comparing coefficients. Make sure you understand how each works and choose the one you prefer. Method 1: Long division

Quotient is , remainder is

.

Method 2: Converting improper rational function to polynomial plus proper rational function

Quotient is , remainder is

.

Method 3: General solution specified by substitution and

are both of degree so

Substituting





Substituting



Substituting

into



for some constants and . (1) (2)



Quotient is , remainder is

.

Method 4: General solution specified by comparing coefficients and

are both of degree so

Comparing coefficients:







Quotient is , remainder is

.

10 Using method from the solution to question :

for some constants and .

Quotient:

, remainder

11 Using method from the solution to question :

12 Let Then By the factor theorem, if So

then

is a factor of

.

for some integers , and .

Comparing coefficients:









is consistent with

.

Then 13 a Let

.

Then By the factor theorem, if b So

then

for some integers , and .

Comparing coefficients:

is a factor of









is consistent with

.

.

Then 14

is a quadratic and

is linear so

for some integers ,

and . The remainder is known to be . Comparing coefficients:





Therefore the quotient 15

.

where is the quotient function and is the remainder function. Note that the remainder must be a lower order than so is a constant term in this case. If

then

So

is divisible by

.

16 a Total distance: Total time:

hours

Mean speed over journey: b

So either

(constant speed throughout the journey) or

(the speed in the second section is

double the speed in the first section).

EXERCISE 5C 3 Multiply through by RHS denominator:

Tip As detailed in the chapter, from here you can either compare coefficients or substitute convenient values. Be careful if you choose the values method: if your general form is wrong you will still get a solution which seems to work. Be thorough and compare all the coefficients.

Doing so will alert you to an error. Method 1: Substitute

Tip Try to pick either ‘easy’ values such as , , for simple simultaneous equations or deliberately select values of which eliminate one or other of the unknowns.

Substitute



Substitute



so

.

so

.

Method 2: Compare coefficients Comparing coefficients:







(1)

Substituting

: so

.

Then from (1):

.

4

for some constants and . Multiplying through by the LHS denominator:

Substituting



Substituting

so

so

. .

Hence 5

for some constants , and . Multiplying through by the LHS denominator:

Substituting



so

Substituting



so

Substituting



so

Hence 6

for some constants , and . Multiplying through by the LHS denominator:

Substituting



so

Substituting



Substituting



so so

Hence 7

Partial fractions: Let for some constants , , and . Multiplying through by the denominator on the LHS:

Substituting values to find the constants: Therefore 8

a b

for some constants and . Multiplying through by the LHS denominator:

Substituting



Substituting



so

Hence 9

for some constants and . Multiplying through by the LHS denominator:

Substituting



Substituting



so so

Hence 10

for some constants and . Multiplying through by the LHS denominator:

Substituting



Substituting

so

so

Hence 11 a b

for some constants and . Multiplying through by the LHS denominator:

Substituting



Substituting



so so

Hence 12

Tip Make sure that you can see how this question highlights the dangers of the substitution method for completing partial fractions. The question is answered first with substitution and then with the safer method of comparing variables.

Suppose

for some constants and .

Multiplying through by the LHS denominator:

Substituting



Substituting



so

The above working appears to show that

.

However, testing a third substitution shows this is not valid. For example, with that

, the above asserts

, which is clearly false.

The problem is that the proposed form

is not valid.

If we compare coefficients, it is immediately apparent that the proposed form is incorrect. Multiplying through by the LHS denominator:

Comparing coefficients:

No values of or will make this true, so the proposed form is invalid. In general, for any rational function, first rewrite (as in ) as the sum of a polynomial and a proper rational function, where the numerator has a lower degree than the denominator:

If the proper rational function has a denominator which factorises then it can be split into partial fractions. Within the scope of this syllabus, all such factors will be linear (though if there are repeated factors a modified approach is needed, as seen in Section 4 of this chapter). for some constants and . Multiplying through by the LHS denominator:

Substituting



Substituting



so

Hence

EXERCISE 5D 2

for some constants , and . Multiplying through by the LHS denominator:

Comparing coefficients:



,

,

Hence 3

for some constants , and . Multiplying through by the LHS denominator:

Comparing coefficients:

;



;





,

,

Hence 4

for some constants , and . Multiplying through by the LHS denominator:

Comparing coefficients:

(1)

(2)

(3)

Substituting (1)



Hence 5

a Let

.

Then By the factor theorem, if b

then

is a factor of

.

for some constants , and . Comparing coefficients:









is consistent with the above. Hence

.

c

for some constants , and . Multiplying through by the LHS denominator:

Comparing coefficients:

(1)

(2)



(3)

Substituting



Hence 6

a

is an improper rational function with numerator degree and denominator degree , so can be rewritten as the sum of a degree polynomial and a proper rational function with numerator of lower degree than the denominator. Suppose

for some constants , and .

Multiplying through by the LHS denominator:

Comparing coefficients:





is consistent with

Since all results are consistent, b

can be written as

for some constants , and . Multiplying through by the LHS denominator:

Comparing coefficients:







So 7

a

b

for some constants , and .

.

Multiplying through by the LHS denominator:

Comparing coefficients:

(1)

(2)



(3)

Hence

8

for some constants , and . Multiplying through by the LHS denominator:

Comparing coefficients:

(1)



(2) (3)

Hence

Tip Note that even if , this still holds true; in some other circumstances you might need to specify because repeated factors in the denominator must be treated differently. For example, consider the following problem: Write

in partial fractions

For distinct and , the solution is

where For

. This solution fails if

, because then and are infinite.

, the solution is

Remember that if you are finding partial fractions with an unknown value in the problem, set conditions or take separate cases to ensure that the constants you find cannot have a zero denominator.

MIXED PRACTICE 5

MIXED PRACTICE 5

1

a

So b Let Then That is, The prime factors of 2

a Let

are ,

and

.

. Then

By the factor theorem, if b So

then

is a factor of

.

for some integers , and .

Expanding and comparing coefficients: is consistent with the above

c

when

d

3

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)

(2)

Hence 4

for some constants , and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1) (2) (3)

Hence 5

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

Hence 6

for some constants , and . Multiplying through by the denominator of the LHS:

Comparing coefficients:



,

,

Hence 7 Always check whether a fractional expression can be simplified first. To do this, try to factorise using as one factor.

Tip You can check that this is a factor by using factor theorem, with

Hence

so the derivative is

8

.

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)

Substitute

into

(2) :

, Hence 9 Factorise both the denominator and the numerator to see if there are any common factors:

10 There is a repeated factor in the denominator, so the required form is

Multiplying through by the common denominator:

When

, so

When When

so

Hence, 11 a

b

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)

(2)

Hence 12 a Let Then

.

By the factor theorem, if

then

b So

is a factor of

for some integers , and .

Expanding and comparing coefficients: is consistent with the above. Hence

c

for

or .

d for some constants , and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

.



(1)

(2)



(3)



,

(4)

,

Hence 13 a

By the factor theorem, if

then

b So

is a factor of

.

for some constants , and .

Comparing coefficients:







is consistent with the above. Hence for

or

c

for some constants , , and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)



(2)

(3)

Hence

(4)

14 a

for some constants , , and . Multiplying through by the denominator of the LHS:

Comparing coefficients:











Hence b

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)

(2)

Hence

15

Hence the quotient is

and the remainder is

.

16 a By the factor theorem, if

is divisible by

and

then



(1)

Substituting (1):

b So

for some integer , and .

Expanding:

Comparing coefficients:







is consistent with the above. is consistent with the above Hence c

when

17 a

or

.

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)

Hence

(2)

b

18 a

Quotient , remainder

b

So let

for some constants and .

Multiplying through by the denominator of the LHS:

Comparing coefficients:

(1)

(2)

Hence 19 If

then

Otherwise,

.

for some constants , and .

Multiplying through by the denominator of the LHS:

Comparing coefficients:

Hence

.

Note that this general form simplifies to when

.

20 Multiplying through by

Comparing coefficients:



21 Expanding and comparing coefficients:

22 Given

So

then

.

23 Multiplying through by the denominator of the RHS:

Comparing coefficients: If Coefficient of 24 a b

is linear then the coefficient of . so

on RHS is zero.



Worked solutions 6 General binomial expansion Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 6A

Tip As with the standard binomial expansion, it is always wise to write out an expansion in its most general format first, bracketing elements where appropriate. Only after you have the essential structure written down clearly should you perform any multiplication, division or application of indices needed to calculate coefficients. This is something you should bear in mind throughout this exercise. 2

3

4

a

b Require

5

a

or, equivalently,

.

b Require

or, equivalently,

c i Substituting

ii Substituting

.

for in :

for in

:

iii

d Let

in the expansion from , which is within the convergence interval.

Then 6

.

a

b Require

or, equivalently,

.

c Let

in the expansion in which is within the convergence interval .

Then

.

d Let

in the expansion in

7 The cubic term has the form So 8

a

Valid for

b

which is within the convergence interval . Then

.

9

a

b The expansion is valid for

. Hence only I can be approximated using the expansion.

10

Comparing coefficients:

11

Comparing coefficients:

12 Then

Comparing coefficients: is consistent with the first coefficient of the binomial is consistent with the second coefficient

.

.

is consistent with the third coefficient

.

EXERCISE 6B 1

2

For convergence, require that

or, equivalently,

.

3

4

5

a

for some constants and . Multiplying through by the denominator on the LHS:

Substituting Substituting Hence

b

c Require

and

; the more stringent requirement is

or, equivalently,

, so

this is the convergence criterion for the whole expansion. 6

a

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence b

Tip It is perfectly reasonable to find the three expansions for the three rational functions in and sum them, and this is the intention in the question. Try amalgamating the second and third first; you will find that this requires slightly less work, because their denominators consist of the same linear factor to different powers. Recombining the second and third fractions:

Then

c Require

and

; the more stringent requirement is

, so this is the convergence

criterion for the whole expansion. 7

Tip This question can be approached either using partial fractions and adding two expansions or by direct multiplication of the two expansions. Both methods are shown below; in this case, direct multiplication is not particularly arduous but there are many circumstances where the small amount of preparatory work establishing the partial fractions is preferable to the multiplication, particularly if more terms are required. Method 1: Direct multiplication

Method 2: Partial fractions for some constants and . Multiplying through by the denominator on the LHS:

Substituting Substituting

8

a

Equating coefficients:

So, b Re-write to:

Expand:

as

and as

as

and as

So, the expansion converges when c

9

a

Equating coefficients:

So, Re-write to:

Expand:

b

So,

,

and

.

10 a

b The first expansion is valid for

, the second is valid for

condition. The whole expansion is therefore valid only for

, which is a more restrictive , equivalent to

.

c Substituting

So 11

Comparing coefficients:

Hence

12

is not real for values

so even some cunning rearrangement does not allow convergence.

Tip If you study further mathematics, you may be interested to consider the complex rearrangement available: for any complex z such that

. This is indeed convergent .

MIXED PRACTICE 6 1

The expansion is valid for

, equivalent to

2

The expansion is valid for

, equivalent to

.

3

4

5

a b i

so replacing with

ii The expansion is valid for

so

in answer :

.

6

a

b This expansion is valid for so substitute

7

a

, equivalent to

.

, which is within the convergence interval.

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence

.

b

c The first expansion is valid for

, equivalent to

.

The second expansion is valid for

, equivalent to

.

The sum is valid where both these are valid; the more restrictive condition is 8

.

for some constants and . Multiplying through by the denominator on the LHS:

Substituting





Substituting

The first expansion converges as long as

and the second expansion converges as long as

. The latter is the stronger condition so for the sum to converge, require that equivalent to

.

9

Comparing coefficients:

Substituting

Substituting:

:

, which is

10 a

Expansion valid for

, equivalent to

.

Tip An alternative approach is to multiply binomial expansions for and , but you will find that this approach minimises the work (and opportunity for errors) in writing out the expansions. b Substituting

11 a

:

The product of the two expansions is

b Substitute

:

So

. A d.p. approximation is

.

12

Comparing coefficients:

Substituting

into



so

.

. Then from 13

Tip Be careful here! When the coefficient of

turns out to be , it becomes necessary to extend

the expansion to to get three terms. At that point you have to go back and continue the expansion structure to include the fourth term. If you got as your third term you probably neglected to check this. 14 a

The expansion is valid for b

.

c

The sum converges for

, equivalent to

d Using the expansion from with

e Using the expansion from with

15 a

b

i

So

is

greater than

.

so

ii

So

is

greater than

.

c The kth term in the expansion has the form

.

That is, every term is positive, so the sequence increases the more terms are included in any approximation: .



7 Radian measure Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 7A 11 12 13 14

So

.

EXERCISE 7B 7 Primary solution: Secondary solution: Periodic solutions lie outside the required interval. Hence

. y

8

5

O

–1

–5

1

x

9 Let

Then

.

Primary solution: Periodic solutions:

,

,

10 a

This is a quadratic in

Since

so, using the quadratic formula:

has range

Hence

reject the negative root. .

b Primary solution: Secondary solution: 11 or

,

Tip Always remember that if you divide both sides of an equation by an expression (here or ) you must also allow for the possibility that the expression equals zero. If you multiply both sides by an expression (here ), you may introduce a false solution where that expression equals zero.

or 12 Since has the real numbers as its domain and arccos has only the interval value outside that interval will be a counter example. Let Then 13 Let

. Then

as its range, any

Primary solution: Secondary solution: Periodic solutions:

,

Further periodic solutions lie outside the interval for .

14 a For example:

gives

,

and

Clearly in this case,

.

Tip Note that the false idea being disproved by counterexample here is that division ‘passes through’ the inversion of a function; i.e. that for a function ; this is generally not the case.

b Let Then So c

EXERCISE 7C 3 a

has range

so has range

.

Low tide has depth metres and high tide has depth b High tide occurs when

.

So high tide occurs at hours after midnight 4

a Amplitude b

. seconds.

metres.

each day.

it should follow that

Time for complete oscillations is

seconds.

c There is no loss of energy, for example, due to air resistance, so that the amplitude of oscillation remains constant. 5 Amplitude

. , so

6 Amplitude

.

.

Period is

7

, so the maximum is one quarter

before

.

y

a

y = 2 cosx

2

y = 1+ sin2x

O

2π

x

–2

b From the graph, there are intersections in this interval. c The two functions each have a whole number of periods in the interval shown. In an interval four times as large there will be four times as many intersections: solutions to the equation

8 Centre of oscillation is at Amplitude is

9

.

.

a Amplitude , roots at

and

, min at

and max at

y 2 1

O

60° 120° 180° 240° 300° 360°

x

1 –2

b Min

, max

c Graph translated by 10

. Min

, max

a

Amplitude equals the radius: b If the particle starts at and moves upwards initially, it will pass the lowest point three quarters of the way through its first rotation, after seconds. 11 a Cosine ranges from Least height is

to so has range

.

above ground and greatest height is

b Period equals



above ground.

.

c The ball will move from lowest to highest position in half a period:

.

12 a The size of the seat is ignored, and it is attached exactly on the circumference of the wheel. b Vertical distance below the centre is

.

So vertical distance above the ground is c One rotation

.

takes minutes.

At

is directly proportional to .

d Substituting into a:

Require

:

when So

for

EXERCISE 7D 5 6

a b

or

, i.e.

or

, a duration of minutes or

seconds.

7 Angle in the major sector So the minor angle

Ĉ

8 9 10 Angle in the major sector

.

So the minor angle

.

11 12 13 So

.

Then

.

14 Perimeter is composed of three arcs, each with radius So

.

15 16

17 So

.

18 So

.

19

. . So

.

20 Perimeter Area



(1) (2)

Substituting Multiplying both sides by and rearranging:

and angle

.

21 Let the minor angle be , so the major angle is Minor area:



Major area:

.

(1)

(2)

(2) – (1):

22 On the cone, slant height base

, which is the radius of the sector. The perimeter of the

, which is the arc length of the sector.

EXERCISE 7E 4

.

So 5

a By the cosine rule in triangle

:

b By the cosine rule in triangle

,

So

.

c The shaded area is the sum of two segments: radius , angle

and radius , angle

. .

EXERCISE 7F 4

a

5

b Let equal

. Then using a:

a

for small

b From the above, So

6

a So Ignoring terms in

and higher, the approximation is

.

b

7

i

Percentage error equals

ii

Percentage error equals

a For small

.

Require that

, for which the solution is

, so the upper bound is

. For small

.

Require that bound is

, for which the solution is

, so the upper

.

For small

.

Require that

, for which the solution is

, so the upper bound is

. b Approximating

with :

But the binomial expansion of this expression is (to the first term containing a power of , since is small):

So, approximating for small For small values of inappropriate to include

.

is a positive value, lying either in the first or fourth quadrant, so it is .

8

for small angles . Using a binomial expansion for the second bracketed expression:

9

for small angles . Using a binomial expansion for the second bracketed expression:

10

. So

and

.

11 a Expanding using a binomial approximation:

b

Note that

is already quite large enough to make the approximation fairly inaccurate. See Q7 for

a boundary on the approximation having a

or lower error.

12 Expanding the second bracket using a binomial approximation:

13 Expanding the second bracket using a binomial approximation:

MIXED PRACTICE 7 1

a Amplitude is

.

b 2

Primary solution:

, alternative primary solution:

Secondary solution:

, alternative secondary solution

Hence solutions are 3

.

a Using angles on a straight line:

.

b

c

4

5 6

a

b 7

a

so

.

.

b So

.

c Using the cosine rule to calculate

:

8 Maximum at

and passes through so

Amplitude

, so the period is .

.

.

9 a Bridge has zero height at either end:

. The bridge width is therefore

. The

bridge comes down to water level exactly at the edge of the bank. b Require that a rectangle

by

fits under the curve.

when Let

.

. Then

Primary solution:

.

Secondary solution: So

. .

The width between these is

metres, which is the maximum width of the barge.

c Centring the barge under the bridge, it extends

either side of the midpoint at

Height of the bridge at the ends of the barge cross-section: 10 a One lap takes a single period. b Circle radius is

.

, so its circumference is

c 11 a b Let

. Then

.

.

.

, so solutions are

.

Solutions for are Hence coordinates are

.

c y

(7π/12, 3)

(19π/12 ,3)

x

O

(π/12, −3)

(13π/12, −3)

12 Using a binomial expansion and ignoring terms in

13 a

and higher:

is a rhombus since each side is a radius in length. But if square, so each internal angle is .

b

, a quarter circle.

d The overlap consists of two segments, each subtending square :

. Then

Primary solution Periodic solutions

, secondary solution and

So by Pythagoras,

. .

Solutions for are because

, or (equivalently) two sectors less the

.

Solutions for are

15

must be a

, the diagonal of a square with side .

c Sector

14 Let

then

is a tangent to the circle. .

Hence the area of the triangle

is

.

The shaded area equals the area of triangle

less the sector area. .

y

16 a

3

x

O

b

Multiplying both sides by

:

Hence

as required.

By quadratic formula: .

so Primary solution:

.

Secondary solution:

.

Periodic solutions lie outside the required interval. Hence solutions:

or

.

17 a Using a binomial expansion and ignoring terms in

and higher:

b

18



(ordered:

19 a

, since

).

is tangent to the circle.

b By the same reasoning, c

and hence

is a rectangle, so

.

, So

d

e

20 a

Tip Remember that the reverse is not true: . b Let

so that

.

Then

.

So

.

c So Using a and b:

Since

, use only positive root.

Hence

.

only if lies in the range of arcsin



Worked solutions 8 Further trigonometry Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 8A 4

a

b

5

a

b

c Primary solutions:

, but these are not in the

required interval. Periodic solutions: 6

a This function has a maximum value , occurring at at which there is a maximum is

, so the smallest positive value of

.

b This function has a maximum value , occurring at at which there is a maximum is . 7

, so the smallest positive value of

for small values of . Hence 8

a

b Let

.

Then from



is the only solution in the given interval. 9

a

b By So or

(for which there are no real solutions)

so

EXERCISE 8B 7 Double angle formula: Then

so

8 Double angle formula: Then 9

a

So

. .

and

b Then For

, solutions are

10 a

b

Using the second line of

Then Dividing numerator and denominator by

:

11 a b

Tip Note that it is at least as fast in this case to start again as to convert the answer from using , because the first line of working can be reused.

12 a i ii

b

13 or

Hence

(no solutions in the given interval)

.

14 a

and

b

15 a Double angle formula gives Let

.

.

Then

.

So

.

b

Restrict to lie within the domain of arccos , so

.

EXERCISE 8C 5

a Require

, so

Require

b

and

Translation

Multiply function by

, so

.

: Vertical stretch, scale factor

.

a Require Require

b 7

is in the first quadrant and

to Replace with

6

.

, so and

. is in the first quadrant and

has range

a Require

so

.

, so

.

Require

and

is in the first quadrant and

, so

.

b so Hence smallest positive value of is 8

.

a Require

, so

Require

and

b For

. is in the first quadrant and

:

Maximum occurs when the argument is a multiple of Minimum occurs when the argument is 9 Let

for some and

and so

so then

Primary solution:

so

Secondary solution: Solutions are

so or

10 So Let

, so

Primary solution: Secondary solution:



. .

Then

If

, so

.

Periodic solutions: Hence solutions are:

EXERCISE 8D 7

8

9

10 a But Substituting:

b or c or Hence solutions are 11

12 Divide numerator and denominator by

13 Using the first few terms of a binomial expansion to approximate this:

14 Let

. Then

But

.

so

.

Therefore

, that is, the inverse to the function

is the arccosine of the reciprocal

of .

MIXED PRACTICE 8 1

a Setting

b Let



Hence solutions are 2

.

a b By similar working,

.

If the two expressions are equal then

So

, which occurs at

.

for integer .

Within the given interval, solutions are 3

a Diameter Therefore b

c

and angle and

is a right-angled triangle so

. Then by the sine rule for area:

. (angle in a semicircle). .

d So 4

a Setting

:

b c Let Then by

so that

. Call

.



So This quadratic has roots Since

must be positive.

Therefore, 5

a Amplitude

, so

b Amplitude is

,

c

and

so lies in the first quadrant and

.

Hence

d Amplitude is e Require

, the period remains . . for some integer ; so least positive occurs when

f Require

.

.

Let

with

so that

Primary solution:

.

Secondary solution:

.

So the first two solutions are

.

Hence 6

.

y

a

x

O

b

so Primary solution:

.

Secondary solution: c

.

so Primary solution:

.

Periodic solution: 7

.

.

a Setting

b

Using the formula in and the double angle formula for

Dividing through by

: or

Require solutions with

.

Primary solutions:

.

Periodic solution:

(further periodic solutions are outside required interval).

Hence solutions: 8



a Let By inspection,

so by the factor theorem,

So

for some integers

is a factor

.

and .

Expanding and comparing coefficients:

is consistent with the above. Hence b

Tip This was shown in Exercise 8B, question , but that question had already derived an expression for in terms of . There is no such preliminary work available here. In theory it is possible to follow the same path: establish and , take the ratio and manipulate. But below is the alternative, direct approach using compound and double angle formulae for

.

Compound angle formula:

.

Double angle formula:

.

Multiplying numerator and denominator by

c d Suppose

for

. Since this is a single period, there should be three values

which satisfy the equation. Then Let

or

and so

or

.

. Then using the formula from , for any of the value

,

Rearranging gives

, for which the solutions, from are

As noted, within the period of there are three values for which . By the above working, the tangent of each of these must correspond to one of the roots of the cubic . Since

9

is an increasing function in the interval

, it follows that

a Let Then

from which

and

b

c i The denominator will equal zero when For example, when

for any integer

so

ii

so solutions are of the form Lowest positive solution for so 10 a

and

so lies in the first quadrant and

so

b i

Maximum value for occurs with minimum positive denominator (denominator is always nonzero). So require

, at which:

ii

, so Hence



11 a

for .

b Let

.

Then So

for

c Using the double angle formula: , for 12 a i ii The LHS, being a squared expression, is never negative for real so:

iii The equality will hold when the initial expression

, when

b Let the angle of inclination to the base of the picture be ; then the top of the picture is

, and

.

. The angle of inclination to

.

Using the compound angle formula

with

and

Multiplying numerator and denominator by

c From So d From

,

. .

, equality occurs for

.

e It is clear from context that

is an increasing function within this domain.

Consequently the greatest value of corresponds with the greatest value of When

.

.

:



Worked solutions 9 Calculus of exponential and trigonometric functions Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 9A 3 4 5

6 Require

.

Checking the answers for validity: Reject Hence

as not within the domain of .

.

Tip Always check for validity of solutions in any question containing ln or a square root, since the working can give rise to solution values outside the domain of the original function. 7 8

9 Require gradient to be 3.

Point on curve is Tangent has equation

, gradient is . .

Hence

.

10 Require gradient to be .

If

then

.

Point on curve is

, gradient is .

Tangent has equation

11

.

,

12

Gradient of tangent is , gradient of normal is

Tangent has equation

.

Normal has equation

13

Require

Hence 14

.

.

.

Stationary points where

.

Stationary points are

and

.

is a local maximum. is a local minimum. 15 Stationary points occur where

When

.

,

.

Hence stationary point is

.

16 Stationary points occur where Stationary point is

, so the point is a minimum. Range of is

.

17 a

Stationary point is

Hence

.

is a local maximum.

b

Stationary point is

.

Hence 18 a

is a local minimum.

has range

so the minimum volume is

million litres.

b Since all water flow is through the dam, water flow will equal the change in lake volume.

Assume that the hydroelectric dam can generate electricity when flow is in either direction, so that there is generation at all times except when flow is zero. Maximum flow occurs when

.

Hence maximum flow in the first days occurs at

EXERCISE 9B 4

5

6

7 8

a b Area is

(below the -axis).

9 Curve passes through

Hence 10 11 a

.

.

Tip An equivalent solution is where the unknown is restricted to a positive value. But unless the question requires it, or doing so simplifies the appearance of the equation, there is no need to rewrite logarithm solutions in this way. b Curve passes through

Hence

.

.

12 Curve crosses -axis when From calculator, this is at

. .

13 a

The curve has roots at

and

.

b

14 The value of the integral is indeed independent of . 15 Normal gradient equals

tangent gradient equals

.

Require that

.

Hence

.

MIXED PRACTICE 9 1

Equation of the tangent with gradient

through

is:

2

3 4

5

6 7

at stationary points. so Since Hence

, a positive indicates maximum and a negative indicates minimum. is the local maximum and

is the local minimum.

Tip An alternative way of solving this problem is to express as a single trigonometric function to find the stationary points.) 8 Stationary points occur where

.

Using the chain rule:

local minimum local maximum local maximum

local minimum 9

Gradient of normal at Equation of normal is

is

. .

10 a i

, so the initial population is

bacteria.

ii From calculator, b i

. So the population reaches

million after

hours.

ii

Hence the rate of growth is million per hour after

hours.

c i This shows that the rate of growth increases exponentially. ii The minimum number of bacteria will occur either at a boundary point or when

The stationary point at

has a lower bacterial population than the initial condition.

so this is a minimum point on the curve. Hence, according to the model, the minimum number of bacteria is 11 Intersection of the graphs when

Enclosed area consists of two parts:   Below

for

Area   Below

for

Area Hence 12



.

.

.

.

Worked solutions 10 Further differentiation Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 10A 5

Tangent at

is given by

.

6

Normal at

has gradient

and is given by

.

7 Require

But the domain of 8

is

, so the only solution is

.

The gradient is zero at a stationary point.

Hence stationary points have coordinates 9

a

.

so at

b

so at

10 The gradient is zero at a stationary point.

Hence stationary points have coordinates 11 Stationary points occur where

.

since Hence stationary points are at

and

.

12

The gradient is zero at a stationary point.

Hence stationary point has coordinates

.

13 Require The calculation gives

, so the population is either

14 a b since 15 a

Since

, the post at

(the left post) is taller.

or

.

b Stationary point occurs when

.

Classifying the point:

The stationary point is therefore a minimum. Minimum height is:

c

16 a Using the trigonometric identity for

:

b Stationary point occurs when Using the identity for

Quadratic in

.

:

. Using the quadratic formula:

or Hence stationary points are

or ,

,

,

.

c Knowing the roots and the stationary points for the oscillating function means that it can be sketched (very roughly). Additional points can be used as well: , etc.

y

x

O

EXERCISE 10B 3 Let

. Then

4 Let

. Then

Let

.

5 Let

. Then

Stationary points occur when

6 Let

. Then

Stationary points occur when

7 Let

.

.

.

Then

and

.

By chain rule, By product rule,

8

a Let

. Then

b

9 Let

. Then

Stationary points occur when

.

Hence stationary point lies at 10 Let

.

. Then

Hence 11 a b By chain rule, By product rule,

c Stationary points occur where since

[See question 8a]

.

Hence stationary point is at

.

12 a By product rule,

Stationary points occur where

The solutions

requires

In the interval

.

and the solution

requires

, the only stationary point is

.

.

Tip You do not have to show this in your answer, but you should satisfy yourself that this must lie in the given interval:

b

: Double root at

.

: Triple root at

.

Positive polynomial of order 5. Stationary point is at Turning point is from to .

y

O

a

b

x

c The curve will finish with a positive gradient for any ,

.

If the second root is repeated an odd number of times (i.e. is odd), the curve will be negative in the interval , and the stationary point will be minimum.

If is even, the curve will be positive in

and the stationary point will be a maximum.

So the stationary point is maximum if (and only if) is even.

EXERCISE 10C

2 Let

. Then

Normal at

has gradient

.

Normal equation is given as:

3 Let

.

. Then

.

Stationary points occur where

.

Hence stationary points are at

and

4 Let

Require

5 Let

.

. Then

.

.

. Then

Stationary points occur where

.

.

Let

. Then

Hence stationary point at 6 Let

is a local maximum.

. Then

Require

. (since denominator is always non-negative).

Checking for validity in the function, For

,

7 Let

8

is increasing.

. Then

Hence

is not in the domain.

.

.

has a local maximum at Let

. Then

So

for small

.

.

; there is a stationary point at

Also,

and

so

is a local minimum.

Tip Although it might seem better to use second derivative analysis to determine the nature of the stationary point, there are good reasons not to do this; it is possible for a local maximum to have a zero second derivative, such as in the curve , so any proof would require contingencies for this circumstance, and in any case calculating the second derivative of requires multiple uses of chain rule, product rule and quotient rule and is unnecessarily complicated.

EXERCISE 10D 5 Using implicit differentiation:

Hence at 6

.

a Using implicit differentiation:

b At Equation of tangent is:

7

a When

,

, so lies on the curve.

b Using implicit differentiation:

Gradient of normal at

is

Hence equation of normal is:

8 Using implicit differentiation:

Substitute

So at the point

, the gradient is .

9 Using implicit differentiation:

Substitute

Equation of tangent is:

10

Substitute

At point

, the gradient is infinite; that is, the tangent is vertical.

Hence the tangent at

is given as

11

12

Stationary point occurs when Substituting into the curve equation:

Hence stationary point is at 13 a

.

.

At

so is tangent at

The gradient at

is

.

.

Hence has equation

.

b To find the intersection of and , substitute

c One (repeated) root must be

into

since is tangent at this point.

Comparing coefficients gives intersects again when Substituting into the equation for Point of intersection is

EXERCISE 10E 1

a b Using implicit differentiation:

(i.e. on the lower part of curve ).

:

2

where

a

when N = 2500, so

b Separating variables: so so so Then 3

a

.

for all , so there are no turning points. The cubic is therefore one-to-one and so must have an inverse.

b When

,

, by examination of the function.

The gradient of the inverse at Let

is the reciprocal of the gradient at

.

Then 4

.

a Let Then

.

. so

.

b

5

so

a Then

. so

b Integrating with respect to :

When

,

Then 6

so

a So f is an increasing function. b

, so this point is on the graph.

c At the point

Therefore, at the point

on the graph of

, the gradient is

on the graph of

, the gradient is the reciprocal of this:

7

a b

8

is an increasing function if

for all values of in the domain. Assuming

is defined, we

know that its gradient at is the reciprocal of the gradient of f at . Since the reciprocal of a positive value is also positive, wherever the gradient of is defined, it is positive, so is also an increasing function.

Tip If the requirement was that

should be a strictly increasing function (that is, the gradient is

greater than zero and never equal to zero), then the inverse exists (f must be one-to-one) and there is no need to be concerned about the possibility of a stationary point on f. However, considering the example of , it is clear that this restriction is not needed for the result to stand.

MIXED PRACTICE 10 1

a b By chain rule:

c By product rule:

2

3 Differentiating to find the values of at which reaches a stationary value: so

Also check the boundary value:

or

So the minimum velocity for 4

a Substituting

,

occurs at

and is

into the LHS:

So point does lie on the curve. b Using implicit differentiation:

When

,

:

.

5

When

, gradient equals

Tangent at

has gradient

.

so has equation:

Axis intercepts of this tangent line are Area of the right-angled triangle 6

a By the product rule:

is

.

7

.

8 Using implicit differentiation:

At a stationary point,

.

.

b By the chain rule:

Hence

and

.

Substituting into the curve equation:

Hence stationary point is 9

.

a b

10

So for

11 a

, the only time at which

so stationary values for the velocity occur at

b .

Rejecting the negative value, and also checking the boundary value at and So the maximum speed is 12 a Vertical asymptote where denominator equals zero:

.

b Using the quotient rule:

c

is

Stationary point is where

.

Hence stationary points are

and

.

Tip Second derivative analysis is possible to establish the nature of the stationary points but the calculation is arduous. Easier, in this case, to assess the sign changes of the first derivative. Always remember if doing this to include consideration of any vertical asymptotes, as behaviour may change either side.

\ From the table,

is a local minimum and

is a local maximum.

d

13 a Let Using the chain rule:

b When

,

.

Equation of the tangent at

is therefore

.

14 a

For

so is a strictly increasing function. is therefore one-to-one and has an inverse.

b At

has gradient 4. Therefore at

has gradient

.

15 a Using implicit differentiation: so

b Substituting

and

into the curve equation:

and

so

does lie on the curve.

Substituting into the numerator of the derivative equation: so the derivative equals zero at this point. c When At

,

so

,

16 a Using implicit differentiation:

. Given

, the only solution is

.

Stationary points are when Therefore

.

. Substituting into the original equation:

Hence stationary points are at

and

.

b Using implicit differentiation:

At stationary points,

so this reduces to

.

c

At At

,

so ,

is a local maximum.

so

is a local minimum.

17 a Starting from the identity

and dividing through by

b Implicit differentiation gives

Hence c Use the chain rule and the above result. If

then

The gradient when

18 a

. is .

The gradient of the normal at

is

.

The equation of the normal at

is therefore:

gives

Hence

.

b

Stationary point is when c

, so at

.

Tip You could approach this problem directly and without reference to the above work, using a substitution . However, part a offers a faster route which avoids integration and substitution, as shown here. Either approach would be acceptable.

Let

If

. Then

and

then

Since the integral of

.

is

: .



Worked solutions 11 Further integration techniques Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 11A

5 6 Graph lies above the -axis throughout.

Tip Remember to be alert for graphs which dip below the -axis in case you need to split the integral into multiple regions to find total area enclosed.

7 Hence curve intersects the -axis at

8 If

then

Hence

EXERCISE 11B

.

.

and

.

EXERCISE 11B 2 Let

.

Then

so

Let

. . Then

3 Let

.

Then

so

.

Let

4 Let

.

.

Then

so

Let

5 Let

. .

.

Then

so

Let

.

. Then

6 Let

.

Then

EXERCISE 11C

Tip

so

.

It can often be helpful in questions of this sort to assign a letter to equal the integral (typically is used): this gives an algebraic ‘handle’ which makes referring to the integral easier in the later working. 5 Let

.

Then

so

.

Let

6

.

a Let

. so by the factor theorem,

is a factor of

.

b Let

.

Then

so

.

Let

.

Tip You should aim to recognise this sort of situation and shortcut the working. Any time you integrate a function of the form

, the result will be ln

, and it is

acceptable to show no intermediate steps of working (though it may be wise to briefly justify this step). 7 Let Then

. so

.

Let

8 Let Then

.

. so

.

Let

.

Hence 9 Let Then

. . so

.

Let

.

EXERCISE 11D

2 Let

.

Integration by parts: Set

3 Let

so that

and

.

.

Integration by parts: Set

EXERCISE 11E

and

and

so that

and

.

4 When using the integration by parts formula, if you include an unknown constant when integrating , this is what happens:

As you can see, the constant is cancelled out in the final answer, leaving only the standard unknown constant (here using ). 5 Let

.

Integration by parts: Set

6

and

so that

and

.

.

Tip Remember that it is usual to differentiate the polynomial and integrate the other function but, when dealing with a logarithm, it is always the logarithm which must be differentiated to simplify the integrand in future steps of the integration.

Integration by parts: Set

7

a Let Then

and

so that

and

.

. so

b Let Integration by parts: Set

. and

so that

and

.

Then 8 Let

.

Then

so

.

Substituting bounds: When

and when

.

Tip In this question, substituting for the bounds is faster than keeping the bounds in terms of and substituting back, but either approach will of course yield the same answer.

Let

. Then

Integration by parts: Set

9

. and

so

and

.

a Integration by parts: Set

So

and

so that

and

.

Now set

and

so that

and

.

Tip Remember that as soon as you eliminate the last of the integration signs you must allow for an unknown constant. b Adding to both sides of this equation and halving:

Tip

but both are unknown constants. It is best to change the constant rather than use in both lines; you may see or used for this in other texts, to represent a constant which, while still unknown, is different from the previous version of the unknown constant.

EXERCISE 11F 6 Using the identity

7

:

a b

Tip You should not need to undertake a substitution technique to integrate either of these terms, and should be able to move directly to the answer with reference to a formula booklet; but the full method is given below.

Let

.

Let

.

Then

so

Let

, and

.

Let

.

Then

So

so

and

.

Tip Because tan and sec are closely interlinked in calculus problems, you will often find that sec is preferred to cos in such results; this answer could also be given as 8

Require This has infinitely many solutions for , since the curve times for the primary curve (though

intersects the line

is not a solution of the original problem) and once for each

other periodic repeat of the curve.

y

O

x

3

The least, positive solution is 9

.

a Let

:

Then

.

b Let Then

10 Let Then

so

. so

three

.

11 Let

Then

.

so

.

12 a b

Let

.

Then

so

.

13 a Full circle has equation .

. This region has and positive, so rearranging for

b Let Then

. so

Bounds of integral: When

. ,

and when

,

EXERCISE 11G 6

a

for some and .

Tip Note that the indefinite integral will be valid but any definite integral whose interval includes either or both of the roots of the denominator and would be illdefined. Multiplying through by the denominator on the LHS:

Comparing coefficients:

Tip Remember that you can compare coefficients or substitute trial values of ; the advantage of the coefficients method is that if you have made an error in your assertion of the partial fraction form (for example, missed out a term) then you will be unable to consistently solve the resultant coefficient equations, but substitution – unless you try further values to check – will not give you this safeguard.

Hence b

7

for some and .

Tip Note that if the roots of the denominator lie outside the integral interval (as they do here, with ), the function is defined throughout the interval of integration

Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence

, and then

8

9

a

for some and . The roots of the denominator and throughout the interval of integration.

lie outside the integral interval, so the function is defined

Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence b

10

for some

and .

The roots of the denominator and lie outside the integral interval, so the function is defined throughout the interval of integration. Multiplying through by the denominator on the LHS:

Comparing coefficients:

gives Then

.

gives

Substituting into

. :

Hence

, and then

MIXED PRACTICE 11

1 . 2 Let

.

Integration by parts: Set

and

so that

Then

and

.

.

3

However, the function

has a discontinuity at

, and any integral with bound at

infinite, so the negative solution is not valid for the original problem. Hence

.

is

4 5

a Function to be integrated has the form So b

Tip If you want to show more working you can perform a substitution function to a power, this is sufficient.

but for a linear

6 Let so when

and when

.

so So

.

7 Let

.

Integration by parts: Set

and

so that

and

.

Then 8

Tip In general, simplifying a function of this sort would involve reducing as much as possible the number of instances of . However, this actually makes the subsequent calculations slightly harder. Method 1 is more thorough, method 2 has slightly easier integration. Method 1: a Multiply the numerator and denominator by

and then factorise. or

b Let Let So

. Then . Then

. so

Using partial fractions,

for some constants and .

Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence

Method 2: a

b Let

. Then

Let

.

. Then

so

.

Notice that the two answers arrived at by Methods 1 and 2 differ only by a constant, which can be accounted for within the logarithm. 9 Let Then Let

. so

. . Then

10 a

for some constants and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence

.

b

11 Let Substitution: Let

.

so Then

Tip If you are tempted to try integration by parts, you may run into difficulties… Integration by parts: : Set

and

so that

and

Unfortunately, with on both sides of the equation (rather than some multiple on the RHS), this does not yield a tidy wraparound integration as was seen in earlier exercises such as Exercise 11E Q9. 12 Let

. Then

so

.

Let

13 a Let

.

Integration by parts: Set

Then b

and

so that

.

and

.

So 14

Tip Although it would be a perfectly valid approach, using the standard partial fractions approach is unnecessarily long-winded if the denominator is just a repeated factor.

So

15

for some constants and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

Hence

so For the function to be defined over the integral interval, logarithms to have real values). That being the case, the modulus signs.

16 a Let Let Then

. (This is also necessary for the individual and

are both positive, so we can drop

. . so

Limits of the integration: When

. and when

.

b The internal area of the ellipse is twice the area of the upper half. Rearranging the equation

for non-negative:

so Therefore the area of the upper half of the ellipse is given by From a,

.

.

Therefore the total area of the ellipse is

.

17 a b Let

.

Then

so

.

c 18 When

, the particle is at rest so

Since

can be written as simply

Using

.

:

Displacement equals zero at

Since

and therefore

for all

, by definition, so

, the displacement at

, from which

equals the total distance travelled.



Worked solutions 12 Further applications of calculus Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 12A 4

and At a point of inflection, So

passes through .

and the point of inflection is

5

and At a point of inflection, So

6

passes through .

. The points of inflection are

and

.

and At a point of inflection, But

so

7

passes through .

at all points of inflection. and

Curve is convex where i.e.

.

or

8

and At a point of inflection, So

or

passes through .

; since

, at both of these points the second derivative changes sign so

they are points of inflection. so at points of inflection, Hence the points of inflection are 9 Using the chain rule,

. and

.

.

In the interval given the derivative only equals zero at

So at

, so this is the only stationary point.

.

At a point of inflection,

passes through .

Since is an odd function (so that

for all ), it follows that the first derivative must be

even and, if non-zero, the second derivative must be odd. Therefore

is also a point of inflection.

Tip As an alternative you could try values either side of such as

to show that the second

derivative passes from negative to positive through . But remember that it is not sufficient to show that the second derivative is zero. 10 Using the product rule,

.

So the derivative is zero at

(and

so this is indeed at the origin).

The second derivative is also zero at the origin, since the original function has derivative analysis a small distance from the origin:

Slope

\

–

as a factor. Using

/

So the origin is a local minimum of the curve. 11 For the cubic to have a stationary point of inflection, it must have a single stationary point (a cubic with two stationary points must have a maximum and a minimum). So An alternative solution is as follows.

At a point of inflection,

passes through zero, so

If this is to be a stationary point,

12

and Curve is convex where

.

. Then

.

Curve is convex for so 13

and Curve is concave where

Since

is positive for all , require that

.

Positive quadratic is less than zero between the roots. Roots of

are

.

Solution is y

14

A

C O

B

C

x

C

a Local minimum:

with

and

for small positive .

b Local maximum:

with

and

for small positive .

c Inflection points of are stationary points of

EXERCISE 12B

which are not horizontal inflections.

EXERCISE 12B 5

a b When At

6

a

(launch) or

(hitting the ground).

,

.

when

.

When

so the -intercept is

b When

and

.

When

and

.

Distance between these points is

.

.

is in fact equal to the distance travelled along the path.

Tip Note that this specific example does not of course prove the generality, but we can easily do that by finding the distance between the point at and When

and

When

and

Distance between these points is

c Substituting

: The Cartesian equation is

Hence

.

EXERCISE 12C 3

and

so

When

. and

The normal at

has gradient

. .

So the equation of the normal at

4

and When

has equation

so and

.

.)

Hence the tangent at 5 At

is so

and

.

.

so

So the tangent at

has gradient

.

Hence the equation of the tangent is

When

, so

When

, so is

The area of triangle 6

is

.

. .

is

a At

so

and

. .

so

So the normal has gradient . The normal at

has equation

.

b Substituting the parametric equations into the normal equation:

at the point point 7

so the point where the normal meets the curve again is at .

and

so

Gradient of the normal is At

.

, the point of the curve is

so the normal has equation

.

Substituting the parametric form:

The source point of the normal is at So is

so is at

.

8 Tangent parallel to the -axis occurs when

.

for all values of so there is no point at which the tangent is parallel to the -axis.

which is

9

and At

so

, the gradient is

The tangent has equation

At

so

At

,so

.

Hence

Hence 10 a

. at . and

so

At , the gradient of the curve is

so the gradient of the normal is

.

The equation of the normal is

When b

is

so so

is

is

.

.

11 a

b

for an increasing function i.e. Re-writing in the form

So, i.e. This is true for i.e.

and and

for an increasing function i.e. Re-writing in the form

So, i.e. This is true for i.e. Within the first year both these functions are increasing where these regions overlap, i.e. between and months. c

This represents the rate of change of the number of rats as the number of snakes increases.

EXERCISE 12D 2

a

where

or

When

. When

. .

Hence coordinates of the points where the curve crosses the -axis are

and

.

b

This is below the -axis, so required area is 3

a

square units.

when when

b When

4

and when

a

: . Checking, this gives the desired values for and . b For the normal at : , Then

so .

So the normal at has gradient

and hence has equation

.

Substituting

: This line crosses the -axis at

Call

and

The curve has root where

c

and

.

. at point

.

so

is the Cartesian equation of the curve.

Tip If you don’t see this immediately, you can still find the solution by the more arduous method of substitution. so

from which

Substituting into the parametric equation for :

So If you substitute both and , the working is even more unpleasant, requiring that you rationalise the denominator and cancel terms to get the same result.

EXERCISE 12E 4 Let

radius

;

Then

and

.

so

Hence when 5 Let

, the rate of change of the area is

side length

;

. . Then

and

.

so

Hence when

, the rate of change of the side length is

6 Let

.

. Then and



So when

and

, the rate of change of the volume is

7 Let

. . Then

and So when

and

, the rate of change of area of the rectangle is

8 Let

.

. Then and

When Hence when

, so and

. , the rate of change of the radius is

9 Let

. Then so

.

. and

When

10

and

:

and

At

: units per second

EXERCISE 12F 3

4 Intersections of the two curves occur when

The line lies above the quadratic so the difference equation is Let the enclosed area

. Then

5 Let the enclosed area

. Then

6 The oblique line bounding the shaded area passes through points . Hence its equation is

.

and

. Its

. Let the shaded area be . Then

Tip An alternative approach would be to integrate between and and subtract that from the area of the trapezium. Provided that your working is clearly laid out, any valid method will be acceptable.

7

so

So

.

8

so

9 Intersections occur where

.

So Intersections at

and

has no discontinuities in the interval The line overlies the reciprocal in the interval

. . so the difference function is

.

y

10

x

O

Intersections occur where So

.

which has solutions at

The difference equation is

and

. .

11 Intersections occur where

, at

or

.

y

12

y = x2

b

O

a

1

x

Let the blue region have area and the red region have area . Note that

.

Then The area of the enclosing rectangle with vertices

and

is clearly

The small white rectangle is in fact the unit square, distorted by the scale of the graph. So

, independently of the value of .

MIXED PRACTICE 12

1

, so At a point of inflection,

2

a When

and

so

and the coordinates of the point of inflection are

so lies on the curve at

.

.

b

,

so

At

so the equation of the tangent at is

3 The line has equation about

.

, and meets the curve again at

, since the quadratic is symmetrical

.

The difference function between the line and the curve is be . Then

4

. Let the shaded area

a b The difference between the areas of the two shaded regions is known to be from part a so only one needs to be calculated to find their sum.

5

a At intersections, and

, so

.

b Difference function is

6

. Hence intersections are at

. Let the enclosed area be . Then

so At At

. . .

So 7

a Stationary points occur where Hence stationary points are

so

so

or

and

is a local maximum.

. .

so classify by determining gradient either side:

\ So

–

/

is a local minimum.

b At a point of inflection,

passes through zero.

Hence

or

It is already established that there is a turning point at

.

The curve changes from convex to concave or vice versa at horizontal point of inflection. 8

. So at

, there is a non-

Tip There is more than one way to answer this question. You may choose whether to integrate this using the method of parameters or to find the Cartesian equation and integrate directly. You may also choose either to calculate the trapezium area under the tangent and deduct the area under the curve or calculate the triangle area above the tangent and subtract this from the area between the curve and the -axis. Each of these methods is shown below. Method 1: Parameters and area between the curve and the -axis At

, so ,

.

, so

The gradient of the tangent at

. is

.

The equation of the tangent is

.

The triangle above the tangent has area

. Let the shaded area be . Then

Method 2: Cartesian equation and area between the curve and the -axis

The shaded area is the area of the trapezium curve between

and

. Let the shaded area be . Then

less the area under the

9

a or Hence

or or

b In the interval

. ,

is greater than

so the difference function is

. Let the

shaded area be . Then

10 a

Points of inflection occur when

passes through zero.

Hence y

b

O

y =tan2x

y = x−1 x

From the diagram, there are values of in the interval

for which

, and since in

none of the cases are the two curves tangent, each must represent a point of inflection. 11 The upper right vertex of the white rectangle enclosed by the coloured zones is rectangle has vertices at the origin, and and hence has area

so the inner .

The upper right vertex of the coloured zones is so the outer rectangle containing the areas of interest has vertices at the origin, and and hence has area . Let the blue area be and the red area . Then

By subtraction,

So if 12

then

.

so the curve does pass through the origin. so

. There is therefore a stationary point at the origin. so

long as

. There is therefore a point of inflection at the origin, as

. so

. Therefore there is a point of inflection

at the origin. 13 a The area of the trapezium is the difference in the two triangle areas:

b The derivative of the area with respect to time is

.

So c

Tip To solve equation (1) either use the method of separating variables or take the reciprocal and integrate with respect to .

Method 1 Separating variables for (*) gives:

At So Doubling and rearranging:

So

and given cannot exceed

Method 2: , so Integrating with respect to :

When

so

; so

.

(positive root since cannot exceed

14 a

so

Then

for all .

b 15 a

)

so

, or equivalently

.

is instantaneously at rest when at

b Let

for integer . Within the interval

,

so

and

so

and

, the two values are

Integration by parts: Let Then Let

,

Integration by parts:

So

16 The curves are

and

Intersections occur when

. . or

.

The line is to the right of the quadratic so the difference equation is

Tip

.

and

You could alternatively reflect the entire graph in the line integration with respect to . 17 a At :

so

At :

so

and solve the more routine

b The shaded area is a quarter of the ellipse.

So the total ellipse area is

.

18 a Local minimum:

and gradient on graph of

is positive.

b Local maximum:

and gradient on graph of

is negative.

c Inflection points: Stationary points of

which are not horizontal inflections.

y

C B

A O C

B

A

x

C

19 a

b Intersections occur when

.

So or Hence intersections are at c The difference function is

d

or or

.

So the ratio of the enclosed area lying above the -axis is



Worked solutions 13 Differential equations Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 13B 4 Separating variables:

Substituting

so

:

and the curve uses the positive branch.

Hence

.

5 Separating variables:

So

where

Substituting

. :

and the curve uses the positive branch: Hence when

.

.

6 Separating variables:

Hence

where

.

7 Separating variables:

The left integral can be approached using the method of partial fractions:

Multiplying through:

Substituting Substituting

Hence

.

8 Separating variables:

Substituting

:

However, So Hence

.

EXERCISE 13C 2

a Rate of population growth is proportional to population size.

When

Hence b

. When

Hence when 3

, the model predicts

a Rate of decay is proportional to mass.

When Hence

.

b When Half-life is such that

4

a

b Separating variables:

When Hence Require

So after 5

:

seconds, the speed has decreased below

a For

The population decreases after b Separating variables:

At

years.

.

Hence

.

c Completing the square in the exponent:

So the maximum population is d As 6

, that is,

fish.

: In the long term, the fish population drops towards zero, according to the model.

a Rate of change of with respect to

is proportional to

. So

for some constant of

proportionality . Therefore

.

is expected to be negative because demand would normally decrease as price increases, so the rate of change

is negative.

b Separating variables: where Hence c i

. : Demand is inversely proportional to price.

Q

P

O

ii

: Demand is constant, independent of price. Q

A

O

7

a The milk remains no greater than

P

, so the temperature difference is

Rate of temperature change proportional to the temperature difference:

so Hence

.

b Separating variables

Since temperature measurement is to the nearest degree, need to find when

Hence, to the nearest minute, it takes 8

minutes for the milk to reach room temperature.

a Resistance force Acceleration

Separating variables:

Tip Note that it is possible to tell from the differential equation that cannot exceed if it starts from less than , so it is not necessary to use a modulus sign in the logarithm of .

Hence

.

b As

so for large values of , the velocity approaches

.

4cm

9

10cm h

a Cone volume is

. In this cone, the filled volume will always be a cone with same ratio of

height and radius as the large cone: So

.

.

The fill rate is

by implicit differentiation.

Hence

.

b Separating variables:

When

.

Therefore it takes approximately

seconds to fill the cone.

10 a Resistance force Acceleration Hence

.

b Separating variables:

Tip Since the only force in the system resists motion, the velocity will never be negative; therefore it is not necessary to use a modulus sign in the logarithm of .

Hence

.

c

As

MIXED PRACTICE 13 1 Separating variables:

so for large values of , the displacement from approaches

.

where

Hence

.

2 Separating variables:

Hence

or

.

3 Separating variables:

where

Hence 4

, and

a Separating variables: , so

b

so Then

5

, so

a Growth rate proportional to population:

Hence b

.

Tip Watch out for instances where, instead of separating variables, the derivative can be manipulated, because there is only one variable in the expression for the rate.

Rearranging: When

Hence it takes

:

months (just under years) to reach

fish in the population, according to the

model. c The model has the population growing exponentially without limit; at some stage, limitations on space, food resources, etc. will constrain the growth, and the model will become increasingly inaccurate. d Separating variables:

Hence 6

.

a Separating variables:

Note that no modulus signs are needed as negative price or demand is not considered. Hence

where

.

b

: Demand is inversely proportional to the root of the price; if the price quadruples, the demand halves.

c

means that as the price rises, demand also rises. This is rarely the case for anything but luxury items, where rarity and prestige become relevant in the purchase decision.

7 Volume of a cylinder of radius a By implicit differentiation,

Separating variables:

and height .

is

.

So

Hence

.

b As 8

but there is no finite time at which

, so the tank will never completely fill.

a Resultant forces:

Hence

.

b Separating variables:

where

Hence

, and

.

c The velocity increases towards 9

a



b

with time. (1)

Tip There are two ways to tackle this question. The ‘quick method’ is available because the time increments are the same (

and

given,

required); but in general you

are expected to integrate and find the function for as a function of time, then substitute values to determine the unknown constant and finally calculate the value at , as shown in the ‘standard method’. Quick method: At

, the liquid is at

At

, the liquid is at

. so over minutes the temperature difference between liquid

and oven has decreased to

its original value.

Exponential behaviour indicates that the change factor will be consistent throughout the model for any fixed time period, so the temperature difference will decrease by the same factor over the next minutes, to

(to the nearest degree). The temperature of the liquid will then be .

Standard method: Separating variables from equation (1): so

where So

; since the temperature begins below

and therefore

At

so

At

so

, it will never exceed

.

. from which

Then at

.

.

10 a The term causes a slowing in growth with the square of the population; the square of the population will be proportional to the number of interactions within the population, and so may reflect competition for resources or space. b Separating variables:

Using the method of partial fractions on the right side:

Substituting

:

Substituting

:

where

Hence c As

. ; the population increases towards

as time progresses.

Tip This is an example of a logistic growth model, which is used in many population models for simple growth with an externally imposed population limit.

11 Separating variables:

.

12 a By the chain rule, b Separating variables:

Hence c Separating variables:

Substituting into the expression for :



Worked solutions 14 Numerical solution of equations Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 14A 1

a Factorisable quadratic – exact solutions available. b Can't rearrange algebraically – in exponent and base position. c Can rearrange using trigonometric identities – exact solutions available. d Can rearrange using logarithms – exact solutions available. e Can't rearrange algebraically – in argument of trigonometric function and outside as well. f Can rearrange using exponential – exact solutions available. g Can't rearrange algebraically – in argument of two dissimilar functions. h Can't rearrange algebraically – cubic-solving equation not covered in this course.

5

a For

:

Cubic equation is continuous so

must have a root between and .

b

Change of sign indicates there must be a root of correct to d.p. 6 Let

between

and

, which would equal

.

a

Change of sign indicates there must be a root of to s.f.

between

and

, which would equal

y

b

O

1

2

3

4

x

( )

x x2 ln – + 2 3 4

From the diagram there is a root between and , so 7

.

a For a rational function to equal zero, the numerator must equal zero. for all so

cannot equal zero so has no roots.

b i ii The change of sign principle cannot be applied where there is a discontinuity in the interval. Since is a vertical asymptote, the change of sign from to does not imply a root. 8

y

a i

O

0.5

1

1.5

x

ii b i ii A graph can cross the axis and cross back. In fact, for a continuous graph, George's comment could be modified to say that there must be an even number of roots in the interval – in this case, two. so by the change of sign principle there must be a root between and and another

iii

between and

EXERCISE 14B 3

4

a Multiplying through by :

b

5

6

a

The only discontinuity is at

so the change of sign principle shows that there is a root between

and . b

c

There must be a root between

and

The change of sign does not lie between

, which equals

and

to d.p.

so the root we are seeking does not equal

to d.p.

Tip Be careful of your phrasing here – see Ex 14A Q7b(ii) for why without a graph you should not declare that there is no root in the interval because of a lack of sign-change. 7

a

By the change of sign principle, since and .

is continuous in this interval, there must be a root between

b

!

Test solution:

By the change of sign principle there is a root between

and

, which equals

to s.f.

8

This sequence of values is heading away from the known solution; a better start value is needed.

Test the solution:

By the change of sign principle there is a root between

EXERCISE 14C 2 a

b

at any stationary points.

and

, which equals

to s.f.

By the change of sign principle, there is a root between and . c

is close to the stationary point and so may be expected to have a tangent close to horizontal, which will intersect the -axis a long distance from the interval. Note that you are not expected to actually use the value, simply to predict the result. If you choose to check the suggestion, you will find

is on the other side of the stationary point at , so is both a lot further from the point and also unlikely to return to the desired root. 3 There is a stationary point at , so any value will have a tangent with negative gradient, so that the iterative sequence increases, rather than decreasing towards the root between and .

y

O 4

x

3.5

a Turning points are at b i Start values less than

or

and

converge to the first root

ii Start values at a turning point c Start values greater than 5

a

. .

or do not converge.

converge to the third root

.

is not defined at (due to the logarithm) or odd multiples of (due to the tan function), so the required values are

b

c The value of is on the far side of a discontinuity of the graph (also on the far side of a turning point) and so continuing the iteration will not be expected to return to the root value close to 6

a So turning points are at

and .

b

: Stationary point at : Stationary point at y 5

x

O

c

Hence tangent at

crosses the -axis at

.

d Because is close to a stationary point, its tangent is close to horizontal. As seen, results in below the stationary point at and so would be expected to converge to instead of . e

Test the solution:

By the change of sign principle, there is a root between

and

, which equals

to s.f.

7 a Hence there is a stationary point at

(reject negative point as outside the domain of which is

due to the logarithm).

, so

is a maximum.

b An inflection occurs where there can be no points of inflection.

. But

has no solutions since

. So

y (1, 1.5)

x

O

c As seen on the graph, there is a root between and . Analytically, because

as

, the change of sign principle also indicates that there must be a root between and .

Hence there must also be a root between and . d

For , the next value from the Newton-Raphson iteration is below the lower limit of the function domain, so cannot be used. e For 8

, the iteration will converge to the greater root.

i a Stationary points at

and

is likely too close to the stationary point, so will lead to

outside the interval

which is the locality of the root. b Limits for then

for which there is convergence:

, where if

then

and

.

From calculator,

.

ii a Stationary point at

.

is likely too close to the stationary point, so will lead to b Require

such that

. Limits for

outside the domain, with

for which there is convergence:

iii a Stationary points at odd multiples of , curve has discontinuities at multiples of leads to

, which is in a different region of the curve.

b This equation can be solved algebraically. There is no need for an iterative method!

and

Limits for

for which there is convergence:

.

iv a Stationary points at multiples of , curve has discontinuities at odd multiples of leads to b Limits for 9

above the discontinuity at

for which there is convergence:

.

a Using the quotient rule:

So b Both

when and , and so

, but at both such points,

are positive for greater than the positive root (at will be closer to .

There is a discontinuity at For

closer to than

(where

where

) and for

Therefore the positive values of . Solution: c

as well.

,

) so

when

. the ratio

must be negative, which is true for

(where for which

is closer to than

). are restricted to

and

y

10 a

x

O

b

Sequence converges to a root in the given interval for

but not for

c The requirement on is that the tangent at must meet the -axis at , so that for all values between and , the tangent meets the axis at a point to the left of and continues on a convergent sequence.

From calculator, the approximate solution to this is

.

y

k O

x

d As seen with , if , the next term may lie on the far side of a discontinuity and the sequence may not converge at all, or may converge to a different root.

EXERCISE 14D 3

a

i There is only one intersection point; all staircase diagrams will lead to it, irrespective of starting point. ii The same result comes from either side of the root. iii N/A b

i There is only one intersection point; all staircase diagrams will lead to it, irrespective of starting point. ii The same result comes from either side of the root. iii N/A c

i There are two intersection points, and with

. The greater root will be approached by

any staircase diagram with a starting point greater than . ii A starting term less than yields no solution (the iterated value decreases until it drops below

,

where the function is not defined. iii This iteration will never converge to root . d

i There are two intersection points, and with

. The lesser root will be approached by

any staircase diagram with a starting point less than . ii A starting term greater than yields no solution (the iterated value increases forever). iii This iteration will never converge to root . 4

Test the solution: For

:

By the change of sign principle, there is a root between

and

, which equals

to d.p.

5

The iteration is converging to a root of 6

. To d.p., the root appears to be equal to

.

Test the solution: Let

.

By the change of sign principle, there is a root between 7

, which equals

to d.p.

a

Hence

.

b The sequence converges to a root such that

Hence

find

.

.

8 The iteration pattern involves starting at

9

and

. This will clearly occur at , with

, rising to the curve, moving horizontally to the line at .

a has a discontinuity at

.

By the change of sign principle, there is a root between and . b

c

to d.p. Check solution:

to

By the change of sign principle, there is a root between

and

, which equals

y

10 a

x

O

b

To d.p. the root is c At the root,

. .

The upper root, towards which the iteration converged, is Percentage error is 11 a

y

O

2

x

.

to d.p.

Sequence converges towards b At the root,

(approximately).

. , and clearly the iteration above converges to the positive

root.

EXERCISE 14E 5

y

a

y = x y = 2 + 3ln( x − 2)

a

O

Two roots and y

O

x

b

. For starting values greater than , iteration will converge to .

( )

y = 2 + e (x − 2) 3

a

b

y = x

x

For starting values less than , the iteration will converge to .

y

b

y = x

y = 1 + 3arctan( x − 2)

a

O

x

b

Two roots and

. For starting values greater than , iteration will converge to .

y

( )

y = 2 + tan (x − 1) 3

O

a

y = x

x

b

For starting values less than , the iteration will converge to . Note that this second arrangement of the equation has infinitely many solutions, but because the problem posed was in terms of the function arctan, the domain for solutions is restricted to values which can equal , which is the interval c

or

.

can be rearranged in many ways. Two possibilities are: for the iteration for the iteration y (x3 – 3x2 + 3) y = 5

a O

b

c

y = x

x

, or

Three roots

and

. For starting values between and , iteration will converge to .

y

y = x

a

x

c

b

O

( 1 )

(3x2 + 5x – 3) 3 y = 5

For starting values greater than , the iteration will converge to . For starting values between approximately and , the iteration will converge to (iterations with starting values much less than will jump into the upper iteration towards . 6 Iteration moves vertically to the curve

and horizontally to

.

and horizontally to

.

y

α

p

O

γ

β

x

q

a b 7 Iteration moves vertically to the curve y S

R a

b Q

O

P

a Converges to . b Converges to .

c

x

c Converges to . 8

a b Using the Quotient Rule:

since 9

has a stationary point at the root.

a Let

By the change of sign principle, there is a root between

and

.

By the change of sign principle, there is a root between and .

b







is always positive, and since for (and can be shown to be less than 1 for ) convergence is to the upper (positive) root. 10 a

⁴

³

y

O

x

By the change of sign principle, there is a root between and

By the change of sign principle, there is a root between

and

.

.

b

In the vicinity of the upper root , c

is consistently below , so the iteration will converge to .

³

⁴

⁴ ⁴ ⁴

⁴

Hence

.

Test the solution:

By the change of sign principle, there is a root between 11 a







Hence

.

b



Hence c For convergence to , require

MIXED PRACTICE 14 1

a



and

, which equals

to d.p.

At stationary point,

so

b Use iteration

. .

Test solution:

By the change of sign principle, there is a stationary point between to d.p. 2

and

, which equals

a Segment area is Hence

.

b Let

By the change of sign principle, there is a root between and . c

Test solution:

By the change of sign principle, there is a stationary point between to d.p. 3

y

a

a O

b

x

and

, which equals

It can be seen from the diagram that there are two solutions

with

.

b Let

By the change of sign principle, there is a stationary point between and . c





Use iteration

;

Test solution:

By the change of sign principle, there is a stationary point between to d.p. 4

and

, which equals

a Let the shaded area be . Then

Hence

.

b Let

By the change of sign principle, there is a stationary point between and . c





Test the solution:

By the change of sign principle, there is a stationary point between to d.p.

and

, which equals

5

a Use

as an iteration to solve

Solutions will be roots of

Test the solution

.

.

:

By the change of sign principle, there is a stationary point between to d.p.

and

, which equals

b At convergent value, so 6

, from which

, a Discontinuity at

, which is not in the domain.

Tip Remember, always find any discontinuities first, then you can be confident in using the change of sign rule.

By the change of sign principle, there is a root between

and

.

b

c

which is the far side of the discontinuity at ; the method fails because the iteration will not converge to the desired root, but will grow without limit, since and for and tends towards as tends to infinity, so each term will be greater than the previous one, increasingly so.

d

By the change of sign principle, there is a stationary point between equals 7

and

, which

to d.p.

a using the sine rule for triangle area. Taking the difference, If the segment and triangle have the same area then

so

b Let Then Using the Newton-Raphson method:

and By the change of sign principle, the root lies between

and

so has value

8

a The rectangle has width When this is maximal,

Rearranging: b Let

and height so

so has area



correct to s.f.

Then Using the Newton-Raphson method:

and By the change of sign principle, the root lies between

and

so has value

correct

to d.p. 9 Let

and

be the amounts Marek and Anjali have in their accounts respectively at the start of

the th year. and Intersecting the two to find when Anjali′s balance overtakes Marek′s balance: Let Then

(using

Using the Newton-Raphson method:

) , starting with

:

and By the change of sign principle, the root lies between

and

will have a greater balance than Marek. Therefore it takes 10 a Use

as an iteration to solve

, so at the start of the

th year, Anjali

whole years of saving to accomplish this. . Solutions will be roots of

.

Test the solution

:

By the change of sign principle, there is a stationary point between to d.p.

and

, which equals

y

b

y = x y = 2 ln(3x − 2)

O

α

For For c

x

β

, the sequence diverges (rapidly reach and for

, the sequence converges to .

so

Test the solution

, beyond the domain of the curve.

, from which

and so

.

:

By the change of sign principle, there is a stationary point between

and

, which equals

to d.p. d For

so the Newton Raphson method would fail to produce a second term in the

sequence, since division by zero gives no defined value. 11 a

y 5 4 3 2 1 O

At stationary point,

x

b Let

. Then

Between

and

criterion is that

.

,

so the iteration will not converge (the convergence

in the vicinity of the root).

c Normally, taking the inverse relationship would result in a root. rearranges to The iteration (it converges to

. converges to a root… but not the one required

). This is because the range of arctan is

.

To get the required value, the formula has to be amended:

This rearranges to Use the iteration

. .

Test the solution: For

By the change of sign principle, there is a stationary point between to d.p.



and

, which equals

Worked solutions 15 Numerical integration Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 15A 2 The table shows that in every case, doubling the number of rectangles from halves the difference between the upper and lower bounds. Number of rectangles Integral

Bound Lower

a

Upper Difference Lower

b

Upper Difference Lower

c

Upper Difference Lower

d

Upper Difference

3 LB

UB

TOTAL Hence lower bound 4

; upper bound

a LB

TOTAL

UB

.

to

or from

to

Hence lower bound

; upper bound

.

b To reduce the gap between the bounds, divide the area into more strips and approximate with more rectangles. 5

y

a

5 4 3 2 1 O

π 2

x

b LB

UB

TOTAL Hence upper bound c If

.

rectangles were used, the bounds would move closer to the true value, so the upper bound

would decrease. 6

a The maximum point will occur at

.

b Two possible interpretations may be made here. For the lower bound, rectangles which lie beneath the curve are required; since the curve is symmetrical, it would be sensible to fit four rectangles beneath the first half of the curve (the rising portion, so that the upper left corners of the rectangles lie on the curve) and double this. Alternatively, four rectangles could be fitted beneath the entire curve, with the first two attached at upper left to the curve and the second two attached at the upper right. Below is shown the first of these, which gives the more accurate estimate. Note that the interval is divided into five, not four, because the left-most rectangle attaches to the curve at and has zero area, so is ignored.

y

x

O

LB

TOTAL Lower bound on the total area under the curve is twice this total

.

EXERCISE 15B 2 a and b Number of intervals Integral a i

a ii

b i

b ii

c i

c ii

Exact Error

Error

Error

Error

Error

Error

iii The percentage errors shrink rapidly, particularly (as in a i) when they begin very high, due to a rapidly changing gradient. In most cases, doubling strips appears to shrink the error by a factor of between and , though this is not an absolute rule.

3

a Use trapezium rule approximation with

.

TOTAL TOTAL Hence approximate area b Substitute

. Then

and

.

c d Use trapezium rule approximation with

TOTAL TOTAL Hence approximate area

.

e

f

4

a There is a vertical asymptote at

.

.

y 4

x = 1

3 2 1 O –1

1

2

3

x

4

–2 –3 –4 b Trapezium rule approximation:

.

TOTAL TOTAL Hence approximate

.

c The curve is concave so the trapezium rule will give an underestimate; all the trapezia will lie beneath the curve, with the oblique edges being chords under the curve. 5

a Trapezium rule approximation:

.

TOTAL TOTAL Hence approximate area

.

b A more accurate approximation could be found by increasing the number of trapezia in the interval.

Tip An exact value could be obtained by using a substitution and integration by parts: Let

so

Integration by parts:

and

.

so

6

a

when

so

.

b Trapezium rule approximation:

.

TOTAL TOTAL Hence approximate area

.

c The curve is concave so the trapezium rule will give an underestimate; all the trapezia will lie beneath the curve, with the oblique edges being chords under the curve. 7 Trapezium rule approximation for

:

.

TOTAL TOTAL Hence approximate area

.

So the distance travelled in the first seconds is approximately 8

.

a The changes of direction occur at the roots of the function (as changes sign). Hence

and

.

b Total distance travelled equals total enclosed area. Require Approximate the integral for

. with two strips:

.

TOTAL TOTAL Approximate area

.

Approximate the integral for

with six strips:

.

TOTAL TOTAL Approximate area

.

So total enclosed area (distance travelled) is approximately

MIXED PRACTICE 15 1 LB

TOTAL Hence approximate area 2

a

TOTAL TOTAL

.

.

Hence approximate area

.

b The curve is concave so the trapezium rule will give an underestimate; all the trapezia will lie beneath the curve, with the oblique edges being chords under the curve. c A more accurate estimate could be obtained by using more intervals.

Tip The function can be integrated exactly, using integration by parts: Take

so

3 LB

UB

TOTAL Hence lower bound 4

; upper bound

.

a LB

UB

TOTAL Hence upper bound for area is b Lower bound for area is

. .

c Dividing the interval into more strips would decrease the errors in both bounds, reducing the difference between them. 5 Trapezium rule approximation for

:

.

TOTAL TOTAL Hence approximate area

.

6 Roots are where . Approximate

with

.

TOTAL TOTAL Hence approximate area 7

a Roots of

. :

at

and

at

and

The product has roots at

b Approximate

and

.

with .

TOTAL

TOTAL Hence approximate area

.

Tip This area can be calculated exactly, using the identity , to find the area

8 Upper bound for

: use

.

.

UB

The upper bound for the integral is given by where:

Hence 9

.

a Stationary points where

.

so the only stationary point is at

, coordinate

.

The curve exists for all values of and is an even function so will be symmetrical about

.

y

(0, 1) x

O

b Since the curve is symmetrical, we should use the four rectangles on only one side of the -axis, and double the result. Using : LB

TOTAL So lower bound for half the integral is

; lower bound for the entire integral is

.

10 Using the trapezium rule to estimate the area under the curve, which will give total distance:

TOTAL TOTAL Hence approximate total distance covered is .



, and an estimate of average speed is

Worked solutions 16 Conditional probability Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 16A

Tip Each of the questions in this exercise could be answered using any of a variety of methods: abstract algebra, Venn diagram illustration, tables or tree diagrams, for example. A range of methods is given below but you should be able to see that all ultimately use the same logic and working. Note also that where a Venn diagram is used to illustrate an answer, preliminary calculations for populating the diagram are given in the worked solution, but need not usually be shown in an examination. 6

a

A

B 29

5

13 98

b The number playing neither equal c d 7

a

M

E 26

32

15 72

b From diagram: c

8

a

Then

P

L 10%

50%

35% 5%

b 9

a Taking

, then

A

B 0.6 – x

x

0.5 – x 0.1

Since the total in the diagram must equal :

b 10

Then

A

B 0.1

0.1

0.5 0.3

a b

11

B

A 0.3

0.2

0.3 0.2

a

b 12 a Since

, there are

multiples of .

Hence b LCM Since

, there are

multiples of

.

Hence c 13 a Labelling Bolognese as , chilli as and vegetable curry as :

B

C 23

0

32

12 8

12 48

10 V

b From diagram: c d

e f

14 a

Dark hair

0.5

Blue eyes

0.2

0.2 0.1

b From diagram: c d e

, so blue eyes and dark hair are not independent.

15 a

b

Cold

0.3

Raining

0.3

0.15 0.25

c d e So the two events are not independent. 16

and

Then

17 Let

. Then

and

.

Then So 18 The maximum value of

is

, when is included within .

is

, when

A B 0.4

0.4

0.2

The minimum value of

A

.

B 0.6

0.2

0.2 0

Tip Be careful! It is an easy mistake to say that the smallest the intersection could be is zero; they cannot be mutually exclusive because .

Then Hence

.

19 a Any probability must be non-negative and no greater than . So b

c .

EXERCISE 16B

.

EXERCISE 16B 2

a Ensuring that all row totals and column totals are achieved (including the grand total): Year 9

Year 10

Year 11

Total

Girls Boys Total b

c

3 Extending the table in the question: Milk

No Milk

Total

Sugar No Sugar Total a

b

c

so milk and sugar are independent; knowing the presence or absence of milk gives no information as to whether sugar is present.

4 Expanding the table: Store A

Store B

Returned Unreturned TOTAL a

b

c

5 Expanding the table: TOTAL

TOTAL a

Store C

TOTAL

b c d

e 6 Expanding the table: Gold

Silver

Bronze

TOTAL

USA GB China TOTAL a b c d 7 Expanding the table: S

M

L

TOTAL

White Black Green TOTAL a b c

d

e 8 Expanding the table: TOTAL

TOTAL a i

ii iii 9

is incorrect. For example, assume James rolls a fair be ‘even number’ and be ‘prime number’.

–sided die with sides numbered to

, and let

TOTAL

TOTAL

But On the other hand, is true. If you know is the case then either or

must be true so

.

EXERCISE 16C

Tip In all the questions in this exercise a tree, populated with relevant values, is sufficient preliminary working. The tree diagrams have been given in these answers. Note that the full algebraic working, such as would be needed for an answer if a tree diagram were not drawn, is also given in the answers below. 2 First

6 14

8 14

3

Second 5 13

B

8 13

G

B

G

Tip In a question like this, there is no need to draw a ‘full’ tree, since only a few results are of interest. Thus, after Blue in the first draw, all that is of interest is whether the second ball is Green or not; Similarly, only

branches are sufficient – there is no need for three branches . are needed after and a first draw of need not be further detailed.

First

Second 2 8

G

6 8

G9

3 8

B

5 8

B9

B 3 9 2 9

G

4 9 R

4

a

0.6

B

0.4

B9 P(A ∩ B9) = 0.3 × 0.4 = 0.12

0.8

B

0.2

B9 P(A9 ∩ B9) = 0.7 × 0.2 = 0.14

2 9

Y

7 9

1 × 7 = 7 Y9 P(X ∩ Y9) = 3 9 27

1 3

Y

2 3

2 × 2 = 12 Y9 P(X9 ∩ Y9) = 3 3 27

P(A ∩ B) = 0.3 × 0.6 = 0.18

A

0.3

0.7

P(A9 ∩ B) = 0.7 × 0.8 = 0.56

A9

b c d 5 1 3

2 3

X

X9

2 × 1 = 6 P(X9 ∩ Y) = 3 3 27

a b

Alternatively:

Tip Normally this alternative of calculating the complement to the union would be the faster calculation, but in this question we can with equal ease harness the answer to part a.

Machine

6

Quality 6%

60%

40%

R P(R ∩ M1) = 60% × 6% = 3.6%

M1 R9

3%

R P(R ∩ M1) = 40% × 3% = 1.2%

M2 R9

7

Opponent

Result 30%

2 × 30% = 6 Win P(High ∩ Win) = 11 110

High

2 11

Lose

70%

9 11

9 × 70% = 63 Win P(Low ∩ Win) = 11 110

Low Lose

8

Bag

Ball 6 10

1 6

5 6

R

6 = 1 1 × P(A ∩ R) = 6 10 10

A R9 5 8

R

5 × 5 = 25 P(B ∩ R) = 6 8 48

B R9

a

b

9

Catch

Day

2 3 1 5

4 5

2 = 1 1 × C P(M ∩ C) = 3 10 5

M C9

90% M9

4 × 9 = 36 C P(M9 ∩ C) = 5 10 50 C9

a

b

10

Bag 2

Bag 1

3 8

5 8

7 10

R

7 = 21 3 × P(R, R) = 8 10 80

3 10

B

3 = 9 3 × P(R, B) = 8 10 80

7 10

R

5 × 7 =35 P(B, R) = 8 10 80

3 10

B

5 × 3 =15 P(B, B) = 8 10 80

R

B

a b

11

Umbrella?

Rain?

4 5 2 3

1 3

4 = 8 2 × C P(R ∩ C) = 5 15 3

R C9 2 5 R9

2 1 × = 2 C P(R9 ∩ C) = 3 5 15 C9

a

b

12 Let be the event that a patient has the disease, and be the event that the patient tested positive to the disease.

Disease state

0.0003

0.9997

Test positive 0.95

× 0.95 = 0.000285 C P(D ∩ T) = 0.0003

0.05

C9

0.01

× 0.01 = 0.09997 C P(D9 ∩ T) = 0.9997

0.99

C9

D

D9

Tip This is a very real problem in medical tests; for a condition which is very rare, a test regime has to be incredibly accurate to be reliable. Something to watch out for is situations where, as shown here, despite a seemingly very predictive test, the probability of a positive test result being a false positive is much more likely than a true positive, because the incidence of the disease .

is so much lower than the probability of an incorrect test on a healthy patient

13 Let be the event of picking the fair coin, and be the event of flipping heads.

Coin

Flip result H P(F ∩ H) = 0.5 × 0.5 = 0.25

0.5

H9

F

0.5

0.5

14

0.5

F9

H P(F9∩ H) = 0.5 × 1 = 0.5

1

Second

First

B n 36

36 – n 36

B 36 – n 35 n 35

W

n × 36 – n = n(36 – n) P(B, W) = 36 35 35 × 36

B

36 – n × n = n(36 – n) P(W, B) = 35 36 35 × 36

W W

Let the number of black counters be . .

Hence 15

or

.

and are independent if and only if

This, by definition, means that

and

are independent.

MIXED PRACTICE 16 1

a

b It is assumed that each sock is equally likely to be selected. This could be justified if all the socks are of the same size and made out of the same material. 2

a If is the number with both genes, then

b c 3

a 0.2

0.8

0.4

A

0.6

A'

0.5

A

0.5

A'

X

X'

b c d 4 Adding totals to the table:

Hours practised

Total

Total a b 5

a

B

F

0.3 – x

x

0.2 0.5

b c

d

B

F

0.1

0.2

0.2 0.5

e 6 40 65

25 65

1 10

Lose

9 10

Lose9

1 4

Lose

3 4

Lose9

40 × 1 = 4 P(Π ∩ Lose) = 65 10 65

Π

Λ

25 × 1 = 25 P(Λ ∩ Late) = 65 4 260

7

a 1 4

3 4

2 3

Late

1 3

Late9

1 5

Late

4 5

Late9

1 × 2=1 P(Rain ∩ Late) = 4 3 6

Rain

Rain9

3 × 1 = 3 P(Rain9 ∩ Late) = 4 5 20

b No, it probably varies with the seasons. 8

a

b

9 Let the score be and the individual rolls be listed as events

and

.

a

b

c

10 Let mean that the man left the umbrella in the first shop, that he left it in the second shop and , represent his keeping the umbrella in each case. Clearly terminates the events and so needs no branches from it in the probability tree, so the three possible event paths are

or

,

or

,

.

First shop

Second shop

1 5

P(L1) = 1 5

L

4 5

1 5

L

4 5

4 × 4 L9 P(K1 , K2) = 5 5

K

4× 1 P(K1 , L2) = 5 5

11 a i ii That each sweet is equally likely to be chosen. For example, if yellow sweets were larger, the answer would be smaller. b

Hence

, as reject

as not valid in context.

c

d

12 a

If

, then

.

So

.

Then when b

for .

, with

when and are mutually exclusive,

13

Tip Note that the diagram is the clearest and fastest way to show working, but for completeness the full algebraic working is also given here. Hot / Cold

Rain

0.3 0.2

0.8

A

R C 0.6

A

R9 C

14



P(R ∩ A) = 0.2 × 0.3 = 0.06

P(R9 ∩ A) = 0.8 × 0.6 = 0.48

Worked solutions 17 The normal distribution Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 17A 2 Points of inflection are at

standard deviation from the mean. So the mean is the midpoint:

The standard deviation is 3

0 2 4 6 8 10 12 14 16 18 20 22

x

Mean is at the mode:

Standard deviation is distance from inflection to mean:

4 Mean is at the mode:

Standard deviation is distance from inflection to mean:

EXERCISE 17B 3 a b 4 a b i ii c It is assumed that his probability of success is identical on each attempt, and that the attempts are independent of each other. It might be speculated that he would improve with the practice or get increasingly determined (so his chance of failure would decrease) or that he might get tired or dispirited (so his chance of failure would increase). You are free to argue in whichever direction you think reasonable, having established what the assumptions are. 5

So the expected number in a sample of 6

is

.

a b Assuming the runners are selected at random from the group, the number of runners running under seconds in a squad of four is given by a binomial distribution Then c The real probability is likely to be greater than

because:

i To beat a total time of minutes seconds does not require all four runners to beat a second time, it requires their mean to beat seconds. If one or more runs the distance in well under seconds, they can still beat the record even if the rest run relatively slowly. ii The athletes chosen for the team are likely to be at the upper end of the distribution rather than randomly selected, so their team mean time should be less than seconds and their standard deviation also smaller. iii Unless the data relates to the exact conditions under which a race will be run (weather, track, competition) then the conditions encountered will cause the experienced mean to be different. 7 a b c

8 a

b

9 a

b Let be the number of medium apples in a bag of ten.

10 a b 11 a b

12

Let be the number of people in the sample of

who receive a less than effective dose. Then

13

14

and By symmetry about ,

for any . Then

EXERCISE 17C 2 Let be the score in an

test. Then

3 Let be the mass of a rabbit. Then

4 Let be the amount of coffee dispensed

.

.

. Then

.

a b

5 Let be the time taken on a test. Then

.

a b

c Let be the number of students in the sample who complete the test in less than

6

minutes. Then

Note that this assumes that the range is centred on the mean, which while not necessarily the case is a valid assumption. 7 Let be the time taken to do a maths question (seconds). Then





Hence

.

8 Let be the concentration

. Then

.

Distribution is symmetrical about so Then 9

.

. so

.

a Normal distribution is symmetrical about the mean, so median = mean. Hence

.

b , Then So

.

10 Let

, so that

Then But

, by the symmetry about as shown in the diagram. y

–a

O

a

x

So

So

.

11

12 If is a uniform continuous distribution across normal , take .

then

for

To transform to a

EXERCISE 17D 3 Let be the diameter of a bolt. Then

Hence

.

.

4 Let be the energy of an electron (eV). Then

Hence

.

.

5 Let be the height of a plant. Then

.

and



:

Substituting into

:

6 Let be the time taken to start a computer (seconds). Then and



:

Substituting into

Tip

:

.

Being astute, you will notice that this calculation puts the mean less than standard deviations above zero for a distribution which can only take positive values, and should therefore have misgivings as to the appropriateness of it being called ‘normal’. 7 Let be the voltage of a battery. Then

.

Estimate that the batteries have been used for

hours (or

8 Let be the measured temperature. Then



minutes)

.



Hence standard deviation is 9 Let be the time waited for a train (minutes). Then



.



Let be the number of times out of three that a person waits over

minutes.

Then

10 Let be the temperature in the oven Then

and the set temperature be

.

.

Using the symmetry of the distribution: So

So approximately



of the time, the oven temperature will be within

of the set temperature.

EXERCISE 17E 2 For a normal distribution to be appropriate, ideally three standard deviations either side of the mean should be possible values. However,

children, so the normal model would have a

substantial proportion of families predicted to have negative number of children. 3

a Reading from the graph: Angle Frequency

b Estimate of the mean: taking as the midpoint of each category:

Tip A parameter with a ‘hat’ on is the best estimate for that parameter. So could also be written as , since it is the best estimate for available. You might also write here as

,

the standard deviation of the sample. If you study statistics at a higher level you will also encounter

, the estimate of the population standard deviation based on a sample.

c The graph is symmetrical and bell-shaped. d Using these estimated parameters, estimate over

so

out of

students can be expected to

.

4 Let be the number of Heads in

throws.

a i

ii

b Note that, as mentioned in the chapter, this should properly be calculated using the ‘continuity correction’, whereby we would interpret the probability of more than Heads as , since the event is equivalent to when read from a continuous scale. which is much closer to the binomial being estimated, with an error of . However, the continuity correction is not required in the A Level syllabus. 5 Let be the number of people voting for the Orange party. Then .

can be approximated by

a b 6 Let be the number of children out of

contracting the disease. Then

, approximated

by

.

(one-tailed test) Use significance level Critical zone is

.

where

So

The critical region is therefore 7

a If

and

.

then summing these together:

That is, However, if for example b

and

, it is true that

but

is approximated by



The solution (The solution

gives

and sorelates to

gives

and relates to

introduced when squaring both sides of equation

, a false solution .)

The only valid solution to the problem is 8 Standard deviation If

then

since .

That is, the mean is greater than three standard deviations above zero.

MIXED PRACTICE 17 1

is a test score; a

So

of students score more than

in the test.

b By the symmetry of the distribution, the mean score of students. 2 Let be the mass of a kitten

. Then

is the boundary between the upper and lower

.

a So the expected number of kittens out of

with a mass less than

is

Tip Remember that an expectation, like a mean, need not be a realisable value. b

Hence

.

3 Require So

4 Since

5

a

, require that

and

So b Let be the number of balls in a sample of

randomly selected which have

Then Alternatively, use Normal approximation: Then

or, if using the continuity correction,

6 Let be the height of an adult female of the breed described. Then

.

Let be the number of dogs, in a random sample of females, which are more than

tall. Then

.

7 Let be the height of a tree in the forest

. Then

a b Let be the number of trees, in a random sample of

8 Let be a test score. Then

, which are more than

tall. Then

.

a

. Rearranging:



b

. Rearranging:



:

so

Substituting into

:

9 Let be an angle estimate (degrees). Then

.



:

so

Substituting into

:

10 i a Since for a continuous distribution, and should sum to . b

so

c

but as with

ii By symmetry, µ Then Hence

, the two probabilities

.

cannot be less than cannot be greater than

.

11 If is the number of female students in a group of size , then a The binomial can be approximated by the normal if

and

Since

, for

can be approximated by a normal b c The binomial distribution requires a constant probability for each ‘trial’, that is, the gender of each student attending must be independent of all others, which is probably not valid in this case. 12 a The range of scores included within standard deviations from the mean is

to

, all of which

are possible scores. The mean and median are very close, suggesting that the distribution is symmetrical.

Tip The distribution could still be far from normal – it could for example be bimodal, but it is not possible to tell any more from the statistics given. The point to bear in mind is that there are no strong reasons the distribution should not be normal. b Let be the score of a student in the examination. Then For distinction boundary , For merit boundary

so

,

so

For pass boundary ,

so

13 Let be the breaking force for a chain

. Then

a Let be the probability of one link breaking under

. force.

Then the probability of a -link chain breaking is

b



14 a Let be the diameter of a Playa Gauss grain of sand

. Then

.

so so : From

so

:

b Let be the diameter of a Playa Fermat grain of sand so

. Then

.

,



Then

and so :

From



:

so

Worked solutions 18 Further hypothesis testing Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions.

Tip In the worked solutions for this chapter, it is assumed that calculations on a stated normal distribution can be made accurately, either by calculator or by use of the formula , so the working often moves directly between probabilities and -values. Make sure you understand how this works. Refer to Chapter 17 for further detail.

EXERCISE 18A 3

a Assuming the energy values of particles are independent: b c

4 Let be the mass of a dog of the given breed

. Then

5 Let be the volume in a carton of apple juice

. Then

6 Let be the mass of an egg

.

Require that

. Then

.

Then

Hence µ must be at least 7

so

.

so

.

8 Let be the length of a fly

. Then

.

so

Hence

.

9 Let be the diameter of an apple

. Then

because this is a symmetrical interval about the mean.

a : b Require

so

Hence must choose at least c Require

Hence must choose

. so

times as many apples.

10 a b

Tip With a little care, you can usually adjust your working to consider the least or greatest value in a group. The calculations for second or third greatest/least are much more arduous. If the heaviest is less than independent weights,

then all of them must be. So, assuming the sample students have

EXERCISE 18B 4 Let be the height of an a

µ

-year-old in England

. Then

.

(one-tailed)

µ

,

b

,

, test at

significance level.

Do not reject : There is insufficient evidence to show that the class mean height is significantly greater than the general population of -year-olds. 5 Let be the time taken in the writing test

. Then

.

µ

(one-tailed)

µ

,

Reject

,

, test at

: There is evidence at the

significance level.

significance level to show that the mean typing time is

decreased after the typing course. 6 Let be a score in Mathematics GCSE. Then

.

a Assume that: the scores in the school follow a normal distribution the standard deviation in the school is also the students chosen were randomly selected from the cohort in the school. b

µ

(one-tailed)

µ

,

,

, test at

significance level.

Do not reject : There is insufficient evidence to show that the school mean is significantly greater than the national mean. 7 Let be the height of an apple tree a

. Then

.

µ

(two-tailed)

µ

b

,

, test at

The critical region will lie outside

significance level (

where is such that

Hence the critical region is c

.

lies within the critical region so reject : There is evidence that the mean height of trees in the new orchard differs significantly from .

8 Let be the weight of a patient. Then a

in each tail).

µ

.

(one-tailed)

µ

,

, test at

significance level.

Critical value is such that

.

Hence critical region is

.

b The test assumes that the post-diet weights also follow a normal distribution. c

is within the critical region of the test so reject at significance: There is significant evidence that the diet has reduced the mean weight of patients.

9 Let be the amount of coffee in a cup

. Then

.

µ

(one-tailed)

µ

Test at a

significance level. ,

,

Cannot reject

: There is insufficient evidence that the mean volume in a cup is less than

.

,

b

Require

Hence the minimum sample size is 10 The -value represents the probability of finding the observed value or ‘more extreme’. With a twotailed test, ‘more extreme’ means further away from the mean in either direction, whereas for a onetailed test it means further away only on one side. For

:

, the -value will be

For : greater.

, the -value will be

, which will be

EXERCISE 18C 3

a (two-tailed test) b

so reject in a maths test.

4

a

: There is evidence of a significant correlation between IQ and results

(one-tailed test) b 5

so do not reject : There is insufficient evidence to show that car speed is positively correlated with distance from the junction.

a The population being sampled is ‘countries in 2013’. b (one-tailed test) c

, test at

significance.

From the table, the critical region for Reject

at

is

significance: There is significant evidence of a negative correlation between

unemployment support and amount spent on education. 6

a (two-tailed test) b

,

, test at

significance.

From the table, the critical region for Reject at temperature.

is

significance: There is significant evidence of a correlation between water use and

c Correlation does not imply causation (and it would be hard to see how water use could drive daily temperature anyway – the reverse causal link would seem more probable!) Furthermore, the test only showed that a correlation exists, not the direction of that correlation. A one-tailed test with (using fresh data, to avoid data snooping) would be needed to find evidence that there is a positive correlation. 7

a (one-tailed test) b

,

, test at

significance.

From the table, the critical region for

is approximately

.

Reject at significance: There is significant evidence of a negative correlation between dose and antibody level. c A more strict significance level reduces the probability of a false positive and therefore helps prevent ineffective therapies being mistakenly classed as effective. Of course, there is always a pay-off. Reducing the false positives will potentially increase the probability of a false negative, which in context would mean discarding an effective therapy because it was not shown to be significantly effective in a drug trial. Where a drug is only therapeutic for a subset of patients this is particularly problematic – most medical conditions are multi-factorial; if drug will cure patients with a particular rare gene, but is ineffective in all other patients, then it may be discarded at an early stage of trials if this linkage is not detected, since it will not have a sufficiently significant benefit in general trials. 8

a

so this is significant evidence of positive correlation.

b The -value for a two-tailed test would be twice that for a one-tailed test. so there would not be significant evidence of correlation.

c If increases then it becomes less likely that this is a chance result from a population with -value therefore decreases.

The

9 (two-tailed test) From the value table,

is significant for a two-tailed

significance test for

10 With only two data points there is always a perfectly fitted straight line, with positive gradient or negative gradient . (Zero gradient has zero probability if the dependent variable is normally distributed.) Because of this, the value of cannot have any useful means of showing whether the two variables are correlated.

MIXED PRACTICE 18 1 (one-tailed test) Test at

significance

From the table, the critical value is For a positive correlation, require so is the answer to the question: it is the least of the values given which would result in rejecting . 2 Let be the breaking load of steel wire. Then

3

.

a The population is ‘countries in 2015’. b (two-tailed test) Test at

c

significance.

The critical region is Do not reject

does not lie within the critical region.

: There is insufficient evidence of correlation between HIV rates and literacy rates.

4 Let be the mass of cakes a

so

. Then

µ µ

(two-tailed test)

b is the -value for the data. so do not reject : There is insufficient evidence to conclude that the mean cake mass of the new baker differs from the previous mean. 5

(one-tailed test)

Critical value is such that So 6

, which is Solution D.

a For the Pearson Moment Correlation Coefficient critical values to be used, both variables (times spent revising and test results) must be normally distributed. b (two-tailed test)

Tip You might assume that revision would be be linked with improved test results, indicating a one-tailed test. However, the question phrasing is neutral and since it could equally be argued that weaker students would spend more time revising, you should be conservative and take a two-tailed test. Whatever you choose, be clear so that your working can be assessed on the basis of your interpretation. The critical region for Since

in a

, reject

significance two-tailed test is

: There is evidence of a significant correlation between time spent

revising and test results. 7 Let be the time taken for a full kettle to boil (seconds). Then

.

a Assume that after cleaning the distribution of times remains normal with the same standard deviation seconds. µ µ

(one-tailed test) , test at

significance level.

Critical region is

where

So the critical region of the test is b

so reject

.

: There is evidence that cleaning the kettle caused a significant reduction in

boiling time. c Since a single device is being used in a heating experiment, both it and any contents must return to room temperature between each trial. 8

a Let be the number of visitors in a week; then µ

µ

µ

, so , so there is evidence to reject

.

There is sufficient evidence that the mean number of visitors has fallen. b If was not normal then the probability.

would not (necessarily) be normal, so we would not be able to work out

Tip If you study the Statistics option of Further Mathematics you will learn that, when the sample is large, the distribution of is approximately normal even if is not. 9 The -value is the probability of the observed result or more extreme, calculated under the assumption that

is true. , which is Solution D.

10 Let be the distance the athlete jumps in a long jump

. Then

.

a (two-tailed test) b

, test at

significance level.

The acceptance region is

where is given by

So

So the acceptance region is c Require

.

to lie beyond the acceptance region.

so the least value of resulting in a rejection of

is .

11 a The null hypothesis mean must lie at the midpoint of the acceptance region.

(two-tailed test) b The acceptance region can be rewritten as

But

.

.

The rejection region to the right of the acceptance region represents region to the left of the acceptance region must also represent

.

, and therefore the rejection

To the nearest integer percentage, the test has significance level 12

so

so



.

.

Worked solutions 19 Applications of vectors Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 19A 3

a

For a straight path, average speed is the modulus of average velocity.

b Average velocity is

, since there is no overall displacement.

Tip You could also explicitly split the answer into components as the zero vector .

4

rather than use

a

so its distance from the origin at

is

. b

5

and

. Substituting

:

a

Magnitude of average acceleration is b End velocity is This direction is between

and

anticlockwise from , so the exact angle is given by

anticlockwise from .

6

a Distance

equals

b c

d Distance

equals

This is not equal to the modulus of the average velocity because points , and are not collinear. The average velocity only takes account of start and end points and , irrespective of the route taken. The average speed is calculated for total distance travelled and so must be calculated by summing the distance travelled in each leg of the journey. The average speed must be greater than equal to the average velocity, with equality only occurring if the path is a straight line for the entire journey. 7

The particles will meet if there is a solution

So

;

.

The particles meet at 8

so

a

Hence distance when b

, seconds after starting.

is

.

so Substituting into

:

Negative quadratic with vertex

and -intercept

. Roots at

y (1, 10)

(0, 9)

t = 3

t = 0 9.22 O

9

(1 + 2√2, 0)

x

a So

and

Eliminating b

is the trajectory.

Tip There are many methods to calculate the shortest distance from the origin to the straight line. A couple are given below, but you will already be familiar with other approaches. Method 1: Closest point Let be the closest point on the line. Then Trajectory

has gradient

lies at the intersection of Substituting:

is perpendicular to the trajectory.

so

has gradient and hence equation

and

:

,

Hence distance Method 2: Vector distance For a line with vector equation

So line has equation

, the closest distance to the origin is always

and hence

.

10 Distance from the origin at time is where

Since

has maximum value ,

has maximum value

So the maximum distance of the particle from the origin is

EXERCISE 19B 2

,

,

. Find and

.

. .

.

3

a

,

,

. Find

.

So the distance from its start point at time

is

. b The direction of movement is at angle above the horizontal. The vector is in the first quadrant so is between and . Hence 4

.

a

,

b

,

,

. Find .

,

:

. Find .

so

Check for consistency: :

is consistent.

:

is consistent.

So the particle's displacement from the origin is 5

,

,

seconds.

. Find .

Hence magnitude of the acceleration is 6

when

.

.

Since we are interested in displacement from starting point, set ,

,

.

. Find .

Angle of anticlockwise from is between

and

(second quadrant) so is given by

. 7

. Since we are interested in displacement from starting point, set ,

,

.

. Find .

is a consistent solution to this pair of simultaneous equations.

So magnitude of the acceleration is 8

.

. Since we are interested in displacement from starting point, set

.

,

So require :

Checks show that this is a valid solution for both

and

.

Hence the particle's displacement from its initial position is 9

10

, so

.

seconds after it starts moving.

. Therefore, as increases, speed increases.

.

Particle is initially at rest at

Trajectory has

, so

and

.

and

So the path is

, for

Tip Remember to limit the position to ; it may transpire that the closest position is actually the start point (though there may be a closer position on the extended path).

Hence shortest distance from this path to the origin is

.

EXERCISE 19C 4

a

b When 5

,

so the speed is

a

So

.

b

So

When 6

a

,

. so

Hence initial speed b

. so

Hence magnitude of the acceleration when

is

.

c

So 7

, so

.

Then

8

So

.

Then

So displacement from initial position is

At

.

,

So the distance from the start position is

.

9

Particle returns to its original position when

So

when

or

and

when

.

or

There is therefore no time after when both never returns to its original position. 10 The component must be zero so 11 Resolving components: 12

and

. and

at the same time; that is, the particle

a

so so

At

,

and

.

b

So at

,

.

As expected, this vector has gradient , since the calculations for are the same as the ones to find the gradient by parametric means.

which has gradient

calculates as

13 a

, and the parametric method for finding gradient also

.

so

The speed of the particle is

.

b Then

.

So Thus the path of the particle is a circle of radius , centred units to the right position. 14

of the starting

Hence

.

Distance to the origin is where

From this completed square form, minimum when .

is

when

. So minimum is

EXERCISE 19D

8

9

,

and

Require such that

is a consistent solution for each element of this vector equation.

10

Require that

, so

.

Tip Take great care when taking a common factor out of a modulus sign; if the factor is an expression rather than a value you must keep it in its own modulus and ultimately allow that factor to take either sign in calculations like this.

11

Require that

so

12

Require

, so

.

so

13

Hence minimum

occurs for

, when

.

EXERCISE 19E

2

a

b Midpoint

of

is given by

3 Require that

so

Then

is the position vector of .

4 Require

.



From the component, Substituting So

.

into the component gives

.

.

This is consistent for all three components in equation

Tip

.

In this case, since the component gives the same equation as the component there are only two different equations for the two unknowns and . In problems where there are fewer unknowns than equations, be careful to check that the solution is consistent with all the information available. 5 Require

.

Components of and equal zero for

.

6

7

a

b Either If

or is the midpoint of then has coordinates

So has coordinates

.

8



9

a Require

.

. .

So the two vectors of magnitude parallel to

b Require

are

.

.

So the two vectors of magnitude parallel to

are

.

10

. 11 a

,

Then

, and

.

Two vectors are scalar multiples of each other if they are parallel. Lines b From

and

are parallel and both pass through and so , and lie on a straight line.

,

, so is the midpoint of

.

MIXED PRACTICE 19 1

a The magnitude of the force is therefore:

b Constant acceleration:

,

,

. Find .

Speed Direction of motion is in the first quadrant so angle direction .

2

anticlockwise from

So 3

must be half this distance:

a Require that

. So

Hence

b

. .

.

Tip If you are familiar with geometric properties, and have studied scalar products in the Further Mathematics course, you could prove this by demonstrating that the diagonals are perpendicular. You are not limited by the syllabus in this course when writing your answer, if you know a rigorous method from other work.

So

is a parallelogram with equal length adjacent sides, so is a rhombus. Note that you do not

need to find all the side lengths; it is already established that is sufficient to show it is a rhombus.

is a parallelogram, so the above

4

Then

5

, so the speed is

.

a

So

So

So Hence

. , so

is isosceles.

b For the points to form a rhombus, must lie opposite (so that the two pre-existing rhombus sides shared with triangle

are equal length). Then require that

.

Then

6

.

so

.

The particle will move in a straight line so reduce the problem to a one-dimensional situation:

Constant acceleration problem:

7

,

,

. Find .

a Since the helicopter moves with constant acceleration from rest, the direction of movement will be equal to the direction of the acceleration. Bearing

Tip Remember that for a bearing, you must calculate the angle from the vertical (north) rather than the usual angle over the horizontal. The formula is therefore

.

b Since all movement is in a straight line, the problem can be reduced to a linear scenario. The magnitude of acceleration

.

Constant acceleration problem:

Hence

,

,

. Find .

seconds.

c The helicopter is flying horizontally (at a constant height).

8

Hence 9

. , so

Particle begins at rest: so

So

.

Then

.

Position is in the fourth quadrant so angle is between

and

anticlockwise from the horizontal.

Angle So the particle's movement is directed

below the horizontal.

10 a

b c Distance between the ships is where

Hence when

,

d From ,

. so

when

; the ships meet hours after the start of the movement.

e The ships start apart and move at constant velocity, so it will take a further hours after they meet to once again be apart. 11 a Let the two aircraft be and .

,

,

b

Hence

. So the aircraft are at all times at least

collide. c At

, the aircraft are at minimum separation:

.

apart so will not

12 a

, so This is an ellipse with horizontal semi-axis and vertical semi-axis .

b For speed : So the maximum value of occurs when When sin 13

,

so

, at which times

.

.

, so

Hence

Require such that

for some .

Taking the initial position as the origin, require

Then

.

.

Hence distance of the particle from the origin is



.

Worked solutions 20 Projectiles Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 20A 3

a

;

,

b When

,

So velocity has magnitude 4

;

.

,

a Maximum height occurs when

b When

,

.

.

Velocity has magnitude

and direction (that is,

5

a

;

below the horizontal).

,

Completing the square for :

So the greatest height achieved is

.

b From the symmetry of the completed square, the maximum height is achieved when

, so the particle returns to the plane after twice that time, at . Substituting into the formula for : . 6

;

,

The vertex of the quadratic equals .

Tip There are many ways to approach this problem. Remember that you can use any known properties of quadratics to give you a quick approach to an answer. You could alternatively find the value of at which the particle returns to the ground and then evaluate at the midpoint of the flight, since the vertex of a quadratic lies midway between the roots. Completing the square for :

So the maximum height attained is

.

Hence 7

;

Require

,

.

So This quadratic has roots

or

the particle is more than 8

;

a When

c

,

:

b When

.

.

,

, which has positive root ,

, so

. A positive quadratic is less than zero between the roots. Hence

above the horizontal level of for

.

9

;

When

,

,

so

.

So the ball will clear the wall by over 10 a

;

When

,

,

. ,

, so

The ball hits the ground

. from the foot of the building.

b For the second ball,

where is the angle at which the ball is hit.

Require that

gives

(ball is at ground level when it is hit) or

Substituting into

gives

Hence 11 a

. ;

If the initial velocity is at angle of inclination then

The ball lands when

:

(at the strike) or

.

Substituting into : At landing,

b

. Require

.

.

or Hence

or

.

12 a What would be observed would be the acceleration due to gravity, since we can't see the actual forces involved, only their effects. In both cases the acceleration due to gravity is clearly , but in the scale model it will appear to be ten times this so the gravitational force acting will appear to be ten times the normal force acting. Justification: In freefalling

from rest,

so

.

In the film, a scaled down object would fall but it would still be observed with speed so where is the acceleration due to gravity observed. So

.

b To address this, the film would need to be slowed down by a factor of appear to fall with an appropriate acceleration.

so that objects

Justification: To freefall 1 metre from rest,

so

.

In the film, a scaled-down object would fall To give the effect that this speed.

from rest in

is actually a full

EXERCISE 20B 1

;

,

Substituting

2

a

into the equation for :

;

Substituting

,

,

b To clear the bar, require that

when

.

.

, the film must be run about one-third normal









Hence c The equation models the ball as a particle, so neglects both the ball's dimensions and its propensity to spin. Spin, in conjunction with air resistance (also ignored in the model) could cause the path to deviate significantly both within and away from the vertical plane of the model. 3

a

;

,

a Substituting

:

b Relative to the height of the initial position of the arrow, when .

, to hit the target would require

Tip Another way to tackle this is to use the modified trajectory equation and look for

,

. Either way, make clear

what frame of reference you are using. and

.

Using the trajectory equation from : When

,

This is within the interval 4

;

required for the arrow to strike the target. ,

a Substituting

:

                                                      Trajectory passes through

:

So Hence b i Substituting

                                              ,

into

:

So

                                          :

Hence

.

ii Substituting these into

Hence 5

:

. ;

,

Substituting

into the equation for :

                          The maximum height occurs at half the distance to the landing,                          

Hence

.

Substituting

So

into

:

. ;

6

,

,

a Require that the trajectory passes through the position

From

:

Substituting into

. :

, so

Rearranging:

Multiplying through by

and using

Hence

:

.

b Solving this quadratic in

:

The ball was thrown at angle rising, at the point given.

, since to go through the hoop it must be dropping rather than

MIXED PRACTICE 20 1 Taking the origin of the system as the ground immediately below the point of projection: ,

,

The particle hits the ground when so at Then the speed is

, and the angle is

particle is travelling at angle 2 In both cases Therefore

, so

, so the

below the horizontal.

so the trajectory paths are identical. , represented by Solution C. This may seem counterintuitive but the difference you

would experience (if you were the one throwing the particles) would be the energy expended to give the particles their initial velocities. You would have to put more effort into throwing the heavier particle, but having done so it would follow the same path if the initial velocities were the same. 3 Taking the origin of the system as the point of projection: ,

,

a The ball hits the ground when

b When 4

,

. So the ball hits the ground

;

away from the foot of the building.

,

a If

then

b Taking the origin of the system as the point of projection: ,

At

,

,

Hence

.

5 Taking the origin of the system as the point of projection: ,

,

a Relative to the point of projection, the ball is caught at At the catch,

so

.

.

Then Hence

(taking the positive root).

b At the catch, Hence speed 6

so

. .

a i Taking the origin of the system as the point of projection:

,

When

,

,

When

or

.

,

so

ii Maximum value of b

, so

.

by varying will occur when

, at

.

. or

Hence

or

.

7 The initial velocity has

and is at angle

.

Taking the origin of the system as the point of projection: ,

,

The stone hits the water when

.

So At 8

(taking the positive root). ,

, so the stone falls

short of its target.

a Taking the origin of the system as the point of projection: ,

,

At ,

Hence

(moment of projection) or

.

b If the ball has the same speed at , then by the symmetry of the parabola, will be at the same elevation as and an equal distance away from the midpoint of the trajectory (both in distance and in time). Completing the square for :

The midpoint of the trajectory occurs at

, so the ball is at at

. 9

a

. Considering movement in the vertical direction: So

,

and

.

          

Considering movement in the horizontal direction: So

,

,

                                               :

Then

,

so

:

, so

.

b Let the total time of flight be

and

for and respectively.

so for vertical movement,

when

. Then

. so when is at , vertical speed is Then

.

so

.

.

c Horizontal speed is constant from launch until the particle returns to ground, so distance travelled is . But then for particle , So

.                                                  

Vertical displacement for : Hence

.

                                        :

Then

so

so

:

so

.

10 a Taking the origin of the system as the point of projection: ,

,

Substituting b Substituting

into ,

,

: and

:

.

Then c

so

or

.

so the particle passes through at

.

so at , the angle of motion is (that is,

below the horizontal)

d 11

Require that this be perpendicular to the initial velocity, so

.

So so at the starting point is

,

and therefore distance from .

Tip All parameters and initial values are given to s.f. so the answer should not be given to a greater accuracy than that, something to bear in mind for any calculation based on an initial set of figures. 12 a Taking the origin as the base of the cliff:

reaches ground level when Then

and

.

so

.

Horizontal distance travelled at the time of impact for

.

Horizontal distance travelled at the time of impact for Therefore, the particles hit the ground b At the point of collision,

apart.

so

Considering the horizontal position:

. , so

For , substituting

into the vertical position:

For , substituting

into the vertical position:

Equating these:

.

.

. . .

But For ,



so the collision is at

. so when

,

so the particle is falling.

Worked solutions 21 Forces in context Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 21A 3 Resolving forces in the unit directions:

The resultant force is 4

.

a Resolving forces in equilibrium:

b





5

θ T 5N

0.2g

a Forces in equilibrium. Resolving forces: Taking

:



b : c From 6

15°



:

. A 200N

40° B 150N

C kN



Resolving perpendicular to and parallel to therefore resultant force) is in the direction of .



7

Hence

.

a 8N

F

because all motion (and



15N

P Taking the

force as having direction and

as having direction , the sum of the forces is

. i The angle of from the direction is

.

ii The magnitude of is

.

b 15N

8N

RN

Let the

force be

, with magnitude

i The angle between and ii



is



8

30°

45° T2

T1

1g Resolving forces at theparticle:

So







.

.

Substituting in Equation



Hence, from Equation 9

:

:

.

a They are two parts of the same string threaded through a smooth ring. , so

b

.

Resolving forces at R: To solve these two equations for and , you can either use substitution to derive equations for and and then taking the ratio or sum of squares to find or , or you can use double angle formulae. Method 1: Substitution From

:

Substituting into

:

Then from :

:





Hence

.

Method 2: From

:

From

: :

Hence From

:



and

10 a F N

A α°

T θ° R

T

S T

2g

If

then with

,

The system is in equilibrium Resolving at the particle: Resolving at :



With

:

From

:

From

:

so

Hence

.

b From

:

From

:

Dividing these: Requiring Hence

restricts the angles so that .

c Resolving at with

,

:

still gives

,

From

:

So

Then

and

EXERCISE 21B 2

Resolving for the skip:



Limiting friction, so Require

.

, so

.

The horizontal force must exceed 3

for the skip to move.

R

a

6N

F

1.5g Resolving for the block:



Limiting friction, so Hence





b The normal reaction force equals the weight: So the total contact force is

, which has magnitude

a

4

R

28N 25°

F

3.5g

Resolving for the box:



Limiting friction, so

.

Hence 5

. R

a A

T

F T 8N

B 5N

Resolving for particle in equilibrium:



Resolving for particle in equilibrium and limiting friction:



At limiting friction, Hence





b Now the weight at is

. The force of friction is already maximal at

.

Resolving for :

Tip Be careful here! Because each particle. Resolving for :

you need to divide the weight by to find the mass of

:





Hence

.

6 For each particle the normal reaction force equals the weight, since all other forces are horizontal. When is at : P

Q

5

8

7N

Fp

A

B

FQ

Friction opposing movement of is

and friction opposing movement of is

Resolving horizontally for the whole system (so that tension forces can be ignored as they will cancel): When is at :

FQ

Q

P

8

5 Fp

A

6N

B

Friction opposing movement of is

and friction opposing movement of is

Resolving horizontally as before: :



Substituting into 7



:

.

a Constant acceleration:

,

,

. Find .

b Normal reaction from the road must equal the weight. Limiting friction: But also

.

So

. R

8

µ .

15 20°

F

5g

a Resolving for the particle: so

Therefore at limiting friction, This is greater than the horizontal pull from the string so the frictional force will exactly counter the horizontal pull and the particle will remain at rest. b

and the vertical reaction is The magnitude of the contact force between particle and table is

R

9

P

F

mg Limiting equilibrium so

and

µ

The contact force is therefore

, which has magnitude

R

10

T θ F 1g a Resolving forces in equilibrium:



Limiting friction: So

.

Let Then

. and

So

. and

. lies in the first quadrant (

. Substituting into equation

Hence

:

.

and

both positive) so

b Require

with the particle moving so require

Solutions to

are

and

at limiting equilibrium.

.

So c Set

with the particle moving.

Resolving forces vertically gives the same result:

Resolving forces horizontally:

Limiting friction:

Hence maximum occurs for

, at which

.

11 The frictional force equals the lesser of µ and the resultant driving force acting in the contact plane. a is not a feature of the model used since the object is assumed to be a particle and therefore has no dimension. If the surface area were taken into consideration it would affect µ . b plays no part in the model used, so doesn’t affect the frictional force. c relates directly to the resultant of driving and frictional forces, so does affect the frictional force. d will affect µ and so does affect the frictional force, albeit the model used doesn’t give us a way of quantifying this – a different value of µ would just be given for a lubricated surface.

EXERCISE 21C

a

3

R

30°

5g a Resolving perpendicular to the slope:

b Taking the positive direction of movement as upslope (so that acceleration will be negative) and then

resolving along the slope:

Constant acceleration:

,

Hence

,

. Find .

.

4 Model the child on the slide as a particle on a slope.

R

F

a

40°

30g a Resolving perpendicular to the slope:

b Resolving along the slope:

Limiting friction:

µ

i Smooth slide: µ ii µ R

5

F

θ 3g

Require that the system be at limiting equilibrium. Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ





So the angle at which limiting equilibrium occurs for is 6 Take the dominant direction of motion as being upslope throughout the question, so that acceleration is negative.

R a

F mg

45°

Moving upslope, and resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ

So

.

Constant acceleration:

,

,

. Find and .

a b c

R

F

a

45°

mg

Moving downslope,

as before.

Resolving along the slope:

Limiting friction:

µ

So Constant acceleration:

. ,

,

. Find .

a

7 R

T T P

35°

a

2g 1g

a Resolving forces for along the slope: Resolving forces for vertically:

b During the initial upslope motion: Constant acceleration problem:

,

,

. Find .

During the second part of the upslope motion, reduces to so that

Constant acceleration problem:

The total distance travelled upslope is is twice that, . 8

a

,

,

. Find .

, so the total travelled when it passes the initial position

F

R

42°

mg

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ

b At greater inclination, the block will move, acted upon by limiting friction.

Normal reaction force

and

The magnitude of the contact force is therefore

Tip The unit in this answer arises from the fact that force can be given either in newtons or in the equivalent unit (since force in newtons equals mass in kg multiplied by acceleration in ). Since we have been given the mass as and not , leaving the answer as simply would mean we were answering that the magnitude of the force was measured in , and giving the answer as is no better. The alternative of leaving the answer in the form may be preferable 9 Initial motion: Take the two particles as a single body with mass RB B FB

RA A FA 1.5g

30° 2.5g

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

and

After travelling two metres under constant acceleration: ,

,

, find .

New system only involves the upper block. and

are unchanged and there is now no tension.

So

.

To travel a further one metre under constant acceleration: ,

,

, find .

, so only external forces apply.

seconds (reject negative solution to the quadratic)

Tip Note that the acceleration on block is constant throughout the motion. If you are sufficiently confident in your understanding of the system to assert this, you can shorten the working considerably, solving the constant acceleration problem with total travelled by . 10

,

and

for the

Tip Take great care here – the acceleration of the ring will not be the same as the acceleration of the particle. a aP

RP

T

T

aR

R

F 30°

2g 4.5g

Resolving at :

For the ring, the vertical acceleration is given by Hence

.

.

b Resolving for the particle away from the slope:

Resolving for the particle up the slope, taking friction as acting downslope and acceleration as upslope:

Now it is important to note that the acceleration of the ring will be half that of the particle, because every two units of distance the particle moves up the slope will permit the ring to lower by only one unit of distance, since the string suspending it between the peg and the ceiling will lengthen by one unit on each side. Hence

.

c Using the equations from and to eliminate :

This shows that the overall force excluding friction is to move the particle downslope, so that in fact friction will act upslope. Correcting for this, and reorienting the accelerations:

aP

RP

T

T

R

F 30°

aR

2g 4.5g

The frictional force is acting up the slope to oppose a net force of Limiting friction is

.

.

Friction is therefore maximal (and the system will move) and friction acts up the slope with magnitude . d

will move downslope, with acceleration

.

R

11 a

10 F 30° 0.5g

Resolving perpendicular to the slope: Resolving up the slope:

Friction is opposing a force of

.

Maximal friction equals friction. Hence

. R

b

, so the hovercraft will move up slope under limiting

10

F 30° 0.5mg

Resolving perpendicular to the slope:

Resolving up the slope:

Friction is opposing a force of

.

Maximal friction equals

, so the hovercraft will move up slope under limiting

friction. Hence

. 10

c R θ°

F 30° 0.5mg

Resolving perpendicular to the slope:

Resolving up the slope:

Assuming limiting friction (which must apply since the object can move even with .

):

This needs to be differentiated, so terms should be adjusted to radians instead of degrees. Let where is measured in radians. Then differentiating and setting the derivative equal to zero to find the stationary point:

Hence 12

.

RB

T

W

P T

T

B

R

4g

45°

3g

Considering horizontal forces on , it is clear that the inclination of the string either side must be the same (or the ring would slide until this was the case). Resolving vertically at

Resolving perpendicular to the slope for

Resolving downslope:

Substituting for :

Friction cannot exceed µ

, so

R

13

aup F θ

mg R F

adown

θ

mg For the upward movement: Resolving along the slope:

For the downward movement:

In both cases, resolving perpendicular to the slope gives

There is movement, so the system moves with limiting friction: Cancelling and setting

:

µ

.

EXERCISE 21D 2 30°

60° T2

T1

18N

The strings are perpendicular. Resolving in the direction of

so

Resolving in the direction of

so

3 Using a vector polygon, the three horizontal forces form a triangle with internal angles all . Since this must be an equilateral triangle, all the side lengths, which are the vector magnitudes, must be equal. 4

R

a

120.7

ϑ 36g

Rearranging the forces into a vector polygon, with

:

R

36g φ

120.7

The normal reaction force is perpendicular to the frictional force angled triangle. Therefore

so the angle of inclination of the plane is

b By Pythagoras, Therefore the coefficient of friction,

so this is a right-

.

21N

5 17N ϑ1 ϑ2

30N

Reforming to make the vector polygon: 17N φ1

21N φ2

30N

where Using the cosine rule: so

so 6 Reforming the forces as a vector polygon: α 18 2.4g β

12

Using the cosine rule:

MIXED PRACTICE 21 1

a Force in the direction of motion is , so b Water resistance would reduce the resultant force in the direction of movement, so the acceleration would be less; if the water resistance increased with the velocity of the canal boat, the acceleration would decrease towards zero over time as the resistance gradually equilibrated with the resultant pull force.

2

R

a a

F 35° 2g

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ

b Normal reaction force

, .Frictional force

Magnitude of contact force is 3

a b The maximum would occur if the would occur if the

and

and

forces were in the same direction:

forces were in the opposite direction:

. The minimum

.

R

4 a

25° 5g

Resolving along the slope:

Tip Remember that in the absence of friction there is no need to resolve perpendicular to the slope as there is no need of the normal reaction force. Also, the mass of the block is not relevant, as the acceleration down a smooth slope at inclination is always just 5

a Resolving vertically:

Resolving horizontally:

Limiting friction:

µ

.

Constant acceleration:

Then

,

. Find .

.

b Constant acceleration:

6

,

,

,

. Find .

a Resolving forces on :

So



So

so

7

b From

:

a

R

so

F

20° 5g

b Constant acceleration problem:

,

,

. Find .

c Resolving perpendicular to the slope:

Resolving downslope, noting that at limiting friction,

µ :

µ

Hence

.

d It is assumed that the only resistive force acting on the box is friction with the slope (so no air resistance) and that the box can be modelled as a particle moving down the slope in a vertical plane (so no rotation of the box occurs).

R

8

F

a

40° 4g

a Assume resistive forces represented by are zero. Resolving down the slope:

b Constant acceleration problem:

,

,

. Find .

c The acceleration is reduced because of resistive forces such as friction and air resistance which reduce the resultant force driving the block downslope. 9

a Resolving in the -direction: Resultant force b Therefore the resultant must lie in the -direction. Resolving in the -direction:

c The resultant force is to the right (positive -direction). 10 Taking as the unit vector with bearing

and as the unit vector with bearing

:

Resultant a b

Resultant force acts in the northeast quadrant so the angle to The bearing of the resultant force is therefore 0.2

11 a

RB

F

B

T

RP T

0.2 P

3g

30° 0.8g

Resolving downslope for :

.

is

.

b Resolving vertically for so Limiting friction so

µ

Resolving horizontally for so Then

µ

, so

.

12 a Resolving vertically for the block:

Resolving vertically for the particle: Resolving horizontally for the block: The force against which friction is acting is Friction will therefore be the lesser of

.

and µ (limiting friction).

µ

So the particle will move with limiting friction

.

Tip Remember that you cannot necessarily assume that

µ . You must first determine the

resultant driving force against which the friction will be acting, since the friction cannot exceed that driving force. The question clearly implies that limiting friction will apply but you should not assume so unless it is explicitly stated that the system moves when released. b From

:

c From

:

13 a Constant acceleration problem:

Resolving along the slope:

b

is increased until

.

Resolving perpendicular to the slope:

. Find .

Substituting

:

14 Taking the dominant direction of movement as downslope throughout the question (so that there is positive acceleration but in the first part of the movement both initial velocity and displacement are negative): a

F 30°

1g

a Constant acceleration problem:

Acceleration is

. Find

directed downslope.

Resolving perpendicular to the slope:



Resolving parallel to the slope (downwards)



Maximal friction (friction is limiting since the particle is moving) Hence

µ

.

b In the downward movement: Resolving parallel to the slope (downwards)

Since maximal equals Hence

, there will be downward movement with limiting friction. .

15 Taking the dominant direction of movement as downslope throughout the question (so that there is positive acceleration but in the first part of the movement both initial velocity and displacement are negative): R a

F 20°

2.2g

Resolving perpendicular to the slope:



Resolving down the slope:

a Smooth plane:

Constant acceleration problem:

So after

. Find .

seconds, the velocity is

directed down the slope.

b Rough plane; the particle will move under limiting friction:

Constant acceleration problem:

. Find .

So the furthest point the particle reaches is

upslope from the original position. It then turns

around and moves back downslope. The high point is

more elevated than the starting point.

R

16

F

a m

ϕ

mg

In both cases

.

µ

When the plane is inclined at :

So,



(1)

When the plane is inclined at

So,



(2)

Since moves twice as far as in the same time with both starting from rest,

.

Hence, using (1) and (2),

17 a Resolving forces vertically: so The block is moving, so friction

µ

.

Resolving forces in the direction of movement: so Then

.

b With other resistive forces, less of the resistance would be attributable to friction so the coefficient of friction would be smaller (the coefficient is calculated as the ratio of the friction force divided by the normal reaction force; the friction force is reduced but the normal reaction is unchanged.) 18 a i

T

T F

Q

RP P

0.3g

0.4g 60° Resolving vertically for : so Resolving perpendicular to the slope for :



Resolving parallel to the slope for : so ii The system is at limiting equilibrium so Then

.

b With the increase in mass at , the normal force (and so the friction force) at will immediately decrease to zero.

T

T Q

a

P

a

0.5g

0.4g 60°

Resolving vertically for : Resolving vertically for : :

so

Substituting into

:

19 a The ring is smooth. b Resolving at c

:

so

:

so :

so

4.9N

F

14.7

Resolving forces: so Limiting equilibrium so so

and then from

R

20 a



µ

,

.

a

b

R

4.9N 30°

F

14.7

i Resolving forces: so Movement so friction is maximal:

µ

ii So

Tip Remember that

; it is a common error to use the weight instead of the mass in a

question which supplies the weight directly. c Resolving forces: so Then so the driving force opposed by friction is only The block will remain stationary (not in limiting equilibrium) and

.

R

21

B

F 20° 10

a i Resolving forces: so so ii The block is moving at zero acceleration so friction is maximal: So

.

µ

.

R

b TN

B

45° F 20° 10

Resolving forces: so Block is still in limiting equilibrium so

µ

.

So

22 a With block heavier and on the steeper slope, anticipate that the net force (and acceleration) will be in favour of moving downwards. With travelling downwards initially, friction will act to slow the movement. R F

30° 40g

For block : Resolving perpendicular to the slope:



Resolving up the slope:

Friction is limiting while the blocks move so

µ

For block Resolving perpendicular to the slope:



Resolving down the slope:

Friction is limiting while the blocks move so

µ

Substituting and adding

to

to eliminate :

b When the blocks come to rest, forces generating movement will remain the same but the forces opposing movement will swap direction. a RA

a T

T

RB

FB A

B

FA

45°

30°

2g

3g

From block



From block

Summing again to eliminate :

Maximum frictional forces are:

Total frictional force to oppose motion is

.

Therefore the blocks will accelerate in the direction opposite to the initial velocity and will pass again through their original positions. Initial movement:

Constant acceleration problem:

So the blocks come to instantaneous rest after

, find and .

seconds,

from their start positions.

Subsequent movement:

Constant acceleration problem:





, find .

seconds

Total time from the start:

seconds.

So the blocks are back in their initial positions

seconds after the start of movement.

23 a During the upward movement (upward acceleration being negative): a (0)

R

P F

40° 0.5g

Resolving perpendicular to the slope:



Particle is moving so friction is maximal: µ

b i

so So acceleration is

downslope

ii When the particle is moving downwards, the friction will oppose the downward force (which exceeds maximum friction so the particle will still move). so So acceleration is

downslope.

c i Constant acceleration problem, taking as the position at the start of the motion being calculated: . Find and .

So the time taken for the particle to reach the highest point after leaving is ii Second part of the motion: so

. Find . and so

The total time taken to return to from is therefore 24 a Constant acceleration problem:

. Find and .

.

.

The distance travelled is

so the distance

b From i, the velocity of when it reaches between

and

is

. . Acceleration on a smooth slope is

will have acceleration

Constant acceleration problem:

. . Find .

c The only forces acting on parallel to the slope are tension and Resolving upslope: so d For , resolving forces perpendicular to the slope: so Then resolving parallel to the slope:

So

.

Since the particle is moving, friction is maximal so Hence



.

µ .

.

so

Worked solutions 22 Moments Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions.

EXERCISE 22A 5 Clockwise moment: 6 Clockwise moment: Anticlockwise moment: Resultant moment: 7

clockwise

a Assuming the plank remains rigid:

b In asserting that the perpendicular distance between diver's weight and the pivot point is constant , and that the perpendicular distance between the weight acting at the board's centre of mass is a constant

we require that the board remains exactly horizontal and rigid.

8 Perpendicular distance from the force to the centre is

. Note that this is clear for the points top and

bottom, and since it is a regular hexagon must also be the case for the other side: don’t start doing trigonometry unnecessarily!

. 9 Let the bottom left corner of the square be the origin, and any point in the square be defined by its coordinate relative to that corner. Then for the point , the four forces have perpendicular distances and .

. That is, the clockwise moment is the same for every point, as the moment is independent of and . 10 Distance from centre of rod to is

and distance from centre of lamina to is .

EXERCISE 22B

.

EXERCISE 22B 3 Let the distance from the end the second child sits be

. Then, taking moments about the pivot point

of the seesaw:







4 Taking moments about the vertical line of the hinges: , where is the perpendicular frictional force acting through the wedge. Hence 5

.

a Let the counterweight be metres from . Then, taking moments about the point of support with the crane's upright:





So the counterweight should be

away from .

b The point of attachment with the crane upright should have tolerance for some net moment if the crane is to be useable in practice. 6 Let the reaction force at the end be and the reaction force total upward force must equal the weight of the plank:

from the other end be

. Then the

Taking moments about the end with the support: So



.

7 Let the tension in the wire nearer the painter be

and the tension in the further wire be

. The total

tension must equal to the total weight carried:

Taking moments about the end nearer the painter: So 8

.

a Let the downward force at the end of the pole be and the upward force at the other handhold be . The total upward force must equal the weight of the pole so

.

Taking moments about the end of the pole being held:







b The working in assumes that the pole is rigid and horizontal. 9 Let the vertical force nearer the handle be and the force nearer the blade be . Then the total must equal the total weight of the spade:

.

Taking moments about the point of :







So 10 Require equilibrium in moments. Taking moments at the elbow joint:



11 a Let the tension the wire be . Then, taking moments at :





b For equilibrium of forces, the vertical component of the force at (the normal reaction) equals

.

c For equilibrium of forces, the horizontal component of the force at (the friction) equals . d A rod can be under tension or compression but a wire can only be under tension; since , in a system with a wire must be positive so (the resting point is on the same side as the wire tether). In a system with a rod, can take any value between and

.

12 Let the horizontal component of the force at the upper hinge be (away from the door). Then taking moments at the upper hinge, with the lower hinge below it and the line of action of the weight away:

Vertical forces have equal magnitude: So magnitude of force at is Horizontal forces have equal magnitude:

So magnitude of force at is 13 Let the distance from the people end to the pivot point be metres. It is assumed that the coach will not slip! Clearly, for equilibrium, . Then the moment favouring falling is

.

The coach will be in exact equilibrium when this moment is zero:

So



metres of the coach overhangs the cliff.

EXERCISE 22C In this exercise, take

.

1 Let the reaction forces at and be

and

respectively.

Taking moments about :

so

Taking moments about :

so

2 Let the centre of mass be a distance from one of the supports, so that it is from the other. If the reaction force at each support is then, taking moments about the centre of mass: so So the centre of mass is at the

from each support.

Tip Generally, if a solution is intuitively obvious then you should be able to show it concisely by a few lines of working and using algebra can be faster than writing a verbal explanation. Here, you could fairly argue that knowing the reactions in the two supports are equal immediately shows that the centre of mass lies midway between them. Such an argument is acceptable but to be sure it is sufficiently rigorous can be challenging. 3 Let the centre of mass be

from the midpoint. Then, taking moments about the midpoint:

so So the centre of mass is therefore 4 Let the rod have length If the tension is .

from the midpoint of the plank.

. Then, taking moments about the centre of mass:

metres from the centre of mass then

If the tension is metres from the centre of mass then . So the possible lengths of the rod are

and

so

so

and the rod has length

and the rod has length

.

5 Let the centre of mass be at and the particle at , with supports at and . so 0.3 A

2.2

C

2.6 X

R

must be the remaining 0.6 0.3

Y

mg

10g

B

D

R

Resolving vertically: Moments at : Substituting and cancelling : So 6

.

0.3 A



C

2.2

x X

R

Mg

Resolving vertical forces,

y Y

10Mg

so

0.3 B

D

R

.

Taking moments at :

But

, so

Substituting:

.

so

7 Let the heavy rod be

and therefore

and the lighter rod be

Moment in favour of descending:



1

8

1 x

A

B

X

M 50

30 W

Taking moments about the support: so that TA

1

1

TB

x A

B

X

M

W

If

then, taking moments about :

If



then, taking moments about :



EXERCISE 22D 2 Taking moments about the hinged end of the rod:

Then 3 Taking moments about : So

, from which

.

.

. Let the balance point be : then

.

RA

4

RB 30g 70° FB

Resolving vertical forces in limiting equilibrium:                 (1) Resolving horizontal forces in limiting equilibrium:                 (2)

µ

Taking moments about : Dividing through by

and substituting using (1) and (2):

µ RA

5

RB mg θ FB

Resolving vertical forces in limiting equilibrium:               (1) Resolving horizontal forces in limiting equilibrium:                    (2) and

in limiting equilibrium.

Then

so from (1):

Taking moments about : Dividing through by

So

and multiplying by

.

:

RA

6 40°

x RB

70g 12g F

Assume the man has climbed up the ladder a distance so that the system is in limiting equilibrium. Then: Resolving vertical forces: Resolving horizontal forces:

and also

in limiting equilibrium.

So Taking moments about point :

So Then So the man can climb

before the ladder will slip.

MIXED PRACTICE 22 In this mixed practice exercise, use

unless otherwise stated.

1 Moment about :

2 Taking moments about :



3 Taking moments about the left end of the rod: 4







a Taking moments about :



b Resolving the system vertically shows that the vertical component of the force at is . There are no horizontal forces in the system. The force at is therefore 5 Let

upwards.

be the distance of the centre of mass from the midpoint in the direction of the

reaction.

Resolving vertically shows that the weight of the plank is Taking moments about the support with the

So the centre of mass is

.

reaction:

from the midpoint.

RA

6

400

RB

50 θ° F

Resolving vertically: Resolving horizontally: Limiting friction: Taking moments about Substituting values and dividing by

:

B

7 MP 0.4m AM 1.2m

RP RA

M

10g 60°

P

F

A

a Taking moments about :



b Resolving forces on the rod: so

    (1)

                 (2) At limiting equilibrium,

µ

. Substituting this and (1) into (2):

8 Let the distance from to the wire be . Then, taking moments about the point of connection of the wire:









9 Total load on the supports is is

. Therefore, if equal, the normal reaction at each support

.

Let the distance of the mass from be . Taking moments about the support nearer to :









10 a Let the tension in the rope at be given as so the tension in the rope at is . Then the total tension must equal the total weight of plank and particle, so . Then

and

b Let distance

, the tension at , equals

.

be given as . Then, taking moments at :









11 a Taking moments about , with reaction at given as

:



b Let distance be given as . The sum of the two reactions must equal the total weight of plank and child, . Therefore, if the two reactions are equal then . Taking moments about again:







12 a Taking moments about after the weight is added at :

Multiplying through by :



           (1)

b With the weight at , again take moments about :

(1)

              (2) (2):



So the weight is

and the distance is

.

13 a Let each of the normal forces at and be . Then, taking moments about :    (1)

The supports are holding the plank and weight so Substituting

.

into (1):





b Since must be finite (and given we have no information as to the strength of the plank), the above equation implies that . In fact will have a maximum value somewhat lower than significantly – and then break – for high values of beyond the scope of this syllabus.

, because the plank will bend

. But that is a topic for material sciences and is

14 Taking moments about the pivot point, requiring equilibrium:

So





Tip Always remember that you can check an answer for sense even when the question has abstract values. In this case, it is easy to check that if then the pivot point should be at the join line and if then the pivot point should be in the centre of the block , both of which can be seen immediately from the answer. 15 If the overhang is then the distance from to the centre of mass is also , since and are apart, for a rod of length . Let the normal force at be . Then, taking moments about :

So

for

If the rod is moving then friction at is limiting and frictional force For the rod to move at constant speed, the force moving the rod must exactly counter this frictional force, so the rod must be pushed with force . 16 a Let the normal reaction of the ground to the rear wheel be to the front wheel be

and the normal reaction of the ground

. Then, taking moments about the rear wheel:

b At maximum acceleration, friction will be limiting (the car would slip rather than be provided greater horizontal force).

so So the greatest acceleration available is

.

c There are no resistive forces other than the friction.



Cross topic review exercise 1 worked solutions Cross-topic review exercise 1 1

a y

9

O

x

3

b One solution on each arm of the graph:

c From the graph,

outside the solutions in :

2 By the formula for the sum of terms:

Comparing the coefficient in 3

so

and

is not defined.

y 1

O

x

–1

4 The mean value equals the sum of the terms divided by :

5 Comparing coefficients:

so

is consistent with Solution: 6

or

a i ii

. By the factor theorem, if

then

is a factor of

. b From ,

is also a factor of

So

.

for some and .

Comparing coefficients:

and

Then 7

a where

b

.

so

, from which

.

c Let Then





Then d The graphs of 8

and

are reflections of each other through the line

.

a Completing the square:

The range of

is

.

and for some (so that the three expressions given b If represent consecutive terms of an arithmetic sequence with difference ), then :

So But as shown in , is outside the range of

for real values of .

Therefore the three expressions cannot represent consecutive terms of a real arithmetic sequence. 9

a Geometric sequence with first term

, common ratio

Tip Always be alert to the fact that standard formulae must be adjusted if the sequence is numbered with the first term u0 instead of . You can easily check the validity of your formulae using

or

.

b 10 For equal roots, require discriminant

:



11 Geometric series with common ratio So



. A geometric series converges for



The expression inside the modulus sign is never less than disregarded. So

.

so the modulus sign can be

.

Quadratic has roots Positive quadratic is less than zero between the roots. Hence solution is





12 a Hence

.

b Vertical asymptote at

, horizontal asymptote at

. Axis intercepts

y

2 - 1O

x

1 -3

c Require domain such that

from graph, this is

The range of h for this restricted domain is So the range of

is

. .

,

13 a

for some constants and . Multiplying both sides by the denominator on the LHS:

Comparing coefficients:





and

Hence

.

b i Expansion converges for

Expansion converges for

so

The sum of these two expansions gives the expansion for

ii The first expansion is only convergent for for .

so this approximation is not likely to be accurate

14 Let the amount of coal (tonnes) used on trip be a Geometric sequence:

and

:

. Then

and

.

so

b Require

So c Then The train can complete

journeys from the

tonnes of coal.

15

For small values of (relative to ), this is equal to formula) plus the kinetic energy of the particle 16

(Einstein's famous Mass-Energy equivalence .

Tip Remember that the large Pi notation is for products, as the large Sigma notation is for sums.

17 a

ln





so the function is not one to one.

b c The derivative is unaffected by the vertical translation and has its sign reversed by the reflection. Hence 18 a

.

under reflection in

becomes

under reflection in

becomes

So the image of is

.

.

b Under a rotation anticlockwise, a point has image . As shown in part , a is equivalent to a reflection through followed by a reflection through . Reflection through

is the transformation from a function to its inverse.

Reflection through

Replace with

So the new graph has equation 19 a If a polynomial

. .

when divided by a polynomial

for some polynomial

has a remainder

then:

is the polynomial analogue of a 'remainder' when dividing integers.

Rearranging:

.

Substituting in this question for some constants and . So

is called

and the remainder

.

b Using the product rule:

c The remainder

d For the remainder to be equal to zero for all values of (so that for all Substituting Substituting

rotation

is a factor of

is linear so

Cross topic review exercise 2 worked solutions Cross-topic review exercise 2 1 When

so

So the equation of the curve is

.

2 3

a

so

and then

b So the solutions are

or .

Tip In fact it is certain that the solution to the differential equation is limited to all functions of the form because: All such equations are solutions to the equation A general solution to a second order differential equation (in its simplest form) can have at most two unknown constants. You will learn more about methods to solve second order differential equations if you study Further Mathematics A level. 4 The two graphs meet where under

5

a

for

in plus the area under

, at

. The shaded area is therefore the area for

.

so .

b Area of is the area of the outer rectangle

6

a b

, less the inner rectangle

when is a multiple of , so the roots shown in the graph are

or

and the shaded area.

.

Integration by parts: Set

and

so

and



c Eliminating : 7

a By product rule, Then using quotient rule:

b Single stationary point so the quadratic

has a single root so discriminant

so For

and since

, the quadratic simplifies to

so

8

Using implicit differentiation and the product rule:

Tip Alternatively, use the chain rule. 9

a

So b Set

. and

. Then:

10 a If

and range (domain

and range

) then: )

b

c So 11 a

and therefore so using the chain rule:

.

. and the stationary point has coordinates

b Require that

so

.

Quadratic in cos :

or

Reject the first root as outside the range of So the point has coordinates

. If

then

.

12 a

b

c

13 The graphs intersect where

, at

and

.

The difference function is

14 a Let

. From the question,

and

.

so (using implicit differentiation and product rule):

So when

and

,

b

c 15 a i By the symmetry of the sine graph, is

.

ii b i Let the area be . Then:

At a stationary value for area, Dividing through by

:

y

ii

y = 2tan x

y = π – 2x

O

x

iii At the stationary point, Using the result in b i,

because the point lies in

.

at the stationary point so

The negative second derivative indicates that the value is a local maximum. c i Try iteration

…

Test the solution: Let

By the change of sign principle, the solution is

to d.p.

Tip Not unexpectedly, this iteration is slow. For faster convergence, it would be better to use the Newton Raphson method: Let

so

The same sign test would be used to demonstrate the validity of this solution. ii For 16 a The common ratio , which for a geometric series to converge.

has value such that

b The first term

and

Using the double angle formula

c

17 a

b

c

. Limits:

with

:

, which is the criterion for

This equals 18 a

, so the trapezium rule gives a better approximation.

is a positive exponential curve with axis intersect

and asymptote

.

y

(0, 4) x

O

b

so

. Taking logarithms with base

c Use the trapezium rule to estimate

:

TOTAL TOTAL Hence approximate area is d

then

so

. .

19 a Then

and lies in the first quadrant so Hence

.

.

b i

is not defined for integer so for example

is not defined at

.

ii

So Require

. so

Primary solution:

, so by symmetry

is also a solution (and is smaller

within the required interval). So

is the smallest positive solution to the equation.

20 The expression is the binomial expansion of Using the identity

:

21 a This is the sum of the first

b

terms of a geometric series with

and

:

Tip Always There are many ways to get to this result but below are two tidy variations. Method 1: Let Then So From a:

Method 2: From a: The LHS is a finite order polynomial, so can be differentiated term by term:

Using the quotient rule on the LHS:

22 a But b

, so

has -coordinate so has -coordinate

c Let the red shaded area be . Then

.

d Let the blue shaded area be . Then can be found by subtracting the result in c from the rectangle area

.

But can also be found by finding the area to the left of the curve:

Equating these results and changing the dummy variable in the integration from to :

(

since must be positive and within the range of

)

23 a Hence

.

b Stationary points occur where

:

Using double angle formula for

Hence

and multiplying through by

:

.

c Factorising:

or

(reject as

)

. Within the interval where

:

24 a Asymptote

is always positive so the only stationary points are

.

and axis intercept

.

y

O

b

1

x

so the tangent at

has gradient

.

The equation of the tangent is Require that this passes through the origin. Substitute

:

c From the nature of the graph, any line through the origin with positive gradient less than

will

intersect the graph twice, while any line with negative gradient will intersect the graph once and any line with gradient greater than will not intersect the graph at all. For two intersections: 25 a

Using quotient rule:

Positive stationary point must therefore be at

, so the maximum is

.

The denominator has a greater order than the numerator and is always positive so for But

as

.

Therefore the range of is b For

.

to be one-one, the domain cannot include both sides of the turning point so the least is .

c If

then

.

So Discriminant is no solution to the equation

, so this quadratic has no real roots and therefore there .

.

Cross topic review exercise 3 worked solutions Cross-topic review exercise 3 1

2

a b Critical value for –tailed

test with

is

so

cannot be rejected; there is

insufficient evidence to show that there is a linear correlation between average temperature and rainfall. 3

Let be the height of a rose bush a

µ

µ

. Then

µ

so

.

(one-tailed test)

b

so reject

at

significance level; there is

evidence that Larkin's rose bushes have a population mean height significantly greater than 4 a and so b Let be the number of lightbulbs in a sample of

which last more than

hours. Then

So c Although the mean is many standard deviations above zero (so the model does not suffer from the problem of a significant proportion being predicted to have a negative life-expectancy), the fact that the mean is so far above zero is also a problem – in this model there is a vanishingly small probability of a bulb failing in early life, which is contrary to real life experience. The model predicts that nearly all bulbs will have a lifetime between and hours ( standard deviations above and below the mean). 5

a b , from which the only valid solution is

6

µ

.

a

from the sample. , so         (1)

b

, so         (2) :

, so

and from

.

c The probabilities are estimated from a sample and are therefore not exact. 7 Minimum standard deviation will use the same value as minimum variance.

From this completed square, variance is minimised by

.

B

8 A

B' B

A' B'

a b

c

9

a

b c

10 a b

is the infinite sum of a geometric series with

and

.

11 Let be the mass of a bag of sugar

. Then

a b Let be the number of bags in a box of with randomly selected bags). Then

which weigh more than

(assuming boxes are packed

c 12 a

is a standard normal, so is equivalent to .

b

Trapezium rule approximation:

TOTAL TOTAL × So approximate area = c 13 a The mode of the graph appears to be between mean. The graph becomes undetectable below

and

and above

, so

would be the best estimate of the

; if this interval is approximately six

standard deviations (about a midpoint which is , close to the estimate of the mean from the central symmetry) then the standard deviation would be . Alternatively, looking at the points of inflection in the graph (when made continuous by joining the tops of the bars to form a smooth curve), these would be at approximately and , at a distance of approximately from the estimate of the mean, again giving a standard deviation approximately . b Mean gives 14 a

is approximately so

and variance and then

is approximately

. The ratio of the two values

.

(one-tailed test)

b For data points, the critical value for significance is so an observed value is significant. Therefore reject and conclude that there is evidence of negative correlation: more revision is associated with worse test scores.

c Taking the data from two such dissimilar individuals and treating it as from a single source has resulted in a false result; from the scatter graphs it appears that a student with high attainment revised less on each occasion than a student with low attainment but (critically) for each student, longer revision time was associated with an improved score. The disparity between the students was substantially greater than the improvement due to revision time, so the amalgamated data hides this detail, which is clear in the graph. 15

µ

so

a

so

Hence

          so

Hence Then

            :

, so

.

b From the simultaneous equations it follows that µ

and it is known that that

. Any

probability of the distribution can therefore be calculated, and Binya is correct.

16 Use the notation that is the event of Daniel winning on his winning on his toss, with being a throw of heads so that

toss and is the event of Theo is a throw of tails.

a b c

d

e f Let be the probability of throwing Heads. Require that P(Daniel wins):           

so (reject denominator of the RHS in (1)).

as a false solution introduced when multiplying through by the

17

Let be the time taken to type a page (mins). Then a

µ

so

.

(one-tailed test)

µ

Observed mean

Therefore reject

µ

, test at

significance

: There is evidence to show that the mean is significantly lower than

minutes.

b Require sample size such that

Then

So require a sample size at least 18 a Inflections are at

and

.

.

b Trapezium rule approximation with

TOTAL TOTAL . c Given the inflection is at , which lies within the area being estimated, some of the trapezia will lie above and some below the curve. In consequence, without further working it is not possible to say whether will be an over- or under-estimate.

Cross topic review exercise 4 worked solutions Cross-topic review exercise 4 1 Let

be the midpoint of the plank (

RA

RB 0.3m

7g

2m M

X

A

to the right of the box) and be the position of the box.

B

12g

Resolving vertical forces: Taking moments about : So 2

and therefore

.

a b

3

a b Hence speed is

and the direction is

above the horizontal. c i

so So

ii 4

a Require

so has coordinates

.

b and

. Two adjacent sides of the

parallelogram are equal length so it is a rhombus.

c 5

or

a Suppose the centre of mass is of contact with the support:

so that has coordinates to the right of the support. Then, taking moments about the point

So

; the centre of mass is

b Total weight of the system is

6

to the right of the support.

, to be exactly equalled by the reaction at the support

a Resolving horizontally:

so

Resolving vertically:

so

b Resultant force must now be right (

to the right, so the acceleration will be

to the

bearing).

7

So

for all values of .

8

. For

to be a straight line,

so

and therefore

.

and

.

a

9

R

T

F

T a

3g

3g

a Resolving vertically for the hanging block:           (1) Resolving for the supported block: so

        (2)

             (3) (1) (3):

          (4)

Maximal friction is friction will equal

.

From (4):

which is less than the driving force

so

b Constant acceleration problem:

. Find .

so the blocks will move and

which gives

, so

c The two parts of the string each affect the pulley with force , so the resultant is at horizontal and of magnitude . From (1):

, so the resultant force on the pulley has magnitude

to the

.

d i There is no friction at the pulley so the tension is the same on both sides of the pulley. ii The acceleration of the suspended block downwards equals the acceleration of the supported block horizontally. 10 a Let the tension in the string be . Taking moments about in equilibrium: so b All other forces acting on the system are vertical, so in equilibrium the force at must also be vertical. If the force acting at is upwards then resolving vertical forces:

Using ,

(upwards). T

11 Ry

20° Rx

A

G B

5g

a Calculating moments about :

so

b Resolving vertical forces:             (1) Resolving horizontal forces:             (2) The magnitude of the reaction at is

and the angle is

to the left of vertical. 12 a

; when

and when

so (2)

(1):

so

and therefore

b Acceleration down a smooth plane at inclination is

.

So c Reaction force must equal the component of the weight acting perpendicular to the slope: , so

13

RB RA

4m C F

θ A 2W

W

a Resolving vertically:

so

Resolving horizontally: µ

so

because the system is at limiting equilibrium so

µ

.

Taking moments about Dividing by Then

µ

so

.

b The equations in a change to the following: Resolving vertically: Resolving horizontally:

(equality in limiting equilibrium)

Taking moments about : So so 14 a

where

and

When the -component returns to zero: Hence

.

Substituting this into the -component:

b

, so Then Substituting into

, so :

or

15 a

Hence

.

. .

b i For the particle to be moving south, require that

From (1):

so

and from (2), the negative root is

indicated, so and

ii

so

and the particle has speed

16 Separating variables:

so

So so Therefore when

,

.

17

so

so Then at all times,

and then , so the trajectory is along the straight line given by that equation.

18 a ,

,

Known point on the trajectory as the ball passes through the hoop:

So From the -component of this equation,

. Substituting into the -component:

.

Hence

          (1)

b i Let and Then

and lies in the first quadrant so so has maximum value

ii Equation (1) gives

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

1. Proof and mathematical communication

1–3 Proof.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

2. Functions

2 1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

3. Further transformations of graphs

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

4. Sequences and series

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

5. Rational functions and partial fractions

1

2 3

4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

6. General binomial expansion

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

7. Radian measure

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

8. Further trigonometry

1

a b

2

a b

3

a b



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

9. Calculus of exponential and trigonometric functions

1 2 3

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

10. Further differentiation

1 2 3

4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

11. Further integration techniques

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

12. Further applications of calculus

1 2 3 Proof.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

13. Differential equations

1 2 3

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

14. Numerical solutions of equations

1

a Proof. b

,

,

c Proof. 2

a Proof. b

,

c Proof. 3

a Proof. b c



is a stationary point.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

15. Numerical integration

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

16. Conditional probability

1

a b c

2

a b c

3

a b c

4

a b c



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

17. The normal distribution

1 2 3

a Proof. b µ

4 µ

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

18. Further hypothesis testing

insufficient insufficient mean plates 1 Insufficient evidence that the mean length has changed

.

2 Sufficient evidence that the lifetime is longer 3 Sufficient evidence of a decrease 4 Insufficient evidence that his mean is different

. . .

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

19. Applications of vectors

1

2

3 4

or

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

20. Projectiles

1 2

a b

3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

21. Forces in context

1 2 3 4

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

22. Moments

1 2 3 4

 

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

1. Proof and mathematical communication 1

a b Proof. Use the fact that

2

.

a Proof. is always divisible by . b Proof. Consider the numbers

3

.

a b Any digit other than . c Proof. Compare the th digits of and

.

d Proof (by contradiction: start by assuming that the numbers have been put into a list).

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

2. Functions 1

y

a 8 7 6 5 4 3 2 1

O

1

2

3

4

5

6

7

8

1

2

3

4

5

6

7

8

1

2

3

4

5

6

7

8

1

2

3

x

y

b 4 3 2 1

O

x

y

c 6 5 4 3 2 1

O

x

y

d 5 4 3 2 1

O

4

5

6

7

8

2 3 4

a b

for some for some

c d

for some

x



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

3. Further transformations of graphs 1

a

y

O

b 2 3 4 5 6

x

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

4. Sequences and series 1

a b It approaches . c

2 It approaches

.

3 It doesn’t approach a limit. 4 Investigation.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

5. Rational functions and partial fractions 1

a b c Proof.

2

a Proof. Notice that b

, which is the golden ratio!

c 3

a Proof. b Proof. c

4 It converges to .

.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

6. General binomial expansion 1

a b c d e

2

a b c

3 4 Proof. 5

a b c



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

7. Radian measure 1 2

a b Proof. c i ii

3 4

a b

5

a b c



because

lies on circles with centres and .

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

8. Further trigonometry 1 2 Proof. 3 Proof. 4

a Proof. b c Proof.

5

,

,

6 Proof. 7

a Proof. b c



,

,

,

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

9. Calculus of exponential and trigonometric functions 1 2 One if 3

or if

. Two if

. None otherwise.

a Proof. b Proof. c Proof.

4

a b

c

for odd , zero for even .

d

y 1

y = sin x –3

–2

–1 O

1

2

3

x

–1

e f Proof. Set

and use the first two terms.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

10. Further differentiation 1  a–c Proof. 2

a Proof. b Proof. Consider the second derivative.

3

a b c



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

11. Further integration techniques 1 2 3 Proof. 4

a Proof. b

5 6

, a Proof. b c



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

12. Further applications of calculus 1 About 2

a b

3

t-shirts; pair of shorts.

a b

4

chocolate bars.

computer games; DVDs.

a b

5

a

bikes

b

bikes

c



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

13. Differential equations 1

a b

2 3

; oscillation with frequency

.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

14. Numerical solutions of equations 1

a b Proof. c

2

a y

b 3.5 3 2.5 2 1.5 1 0.5

O



0.5 1 1.5 2 2.5 3

x

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

15. Numerical integration 1

a b Proof. c

d e Because 2

a Proof; b c d

3

a b

green; yellow

c Proof. d Proof. 4 Discussion. 5 Discussion. 6 Discussion.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

16. Conditional probability 1

a b c

2

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

17. The normal distribution This is an investigation. These answers indicate approximately what results you should get. 1 2 3 4 5 Investigation.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

18. Further hypothesis testing 1

a b Discussion.

2 Discussion. 3 Discussion. 4

a b c d Discussion.



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

19. Applications of vectors 1

a b Proof. c d e

2

a

is the length of the arc that has been in contact with the ground.

b c

y

O

d

2π

4π

6π

x

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

20. Projectiles 1 Proof. 2 3

a Proof. b

4 5 Skimming is faster.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

21. Forces in context 1

a b

2

a b

3 4 5



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

22. Moments 1

a Proof. b Discussion. Consider areas on each side of a median. c

2

a b c



Working with the large data set OCR have provided you with a data set looking at different methods for travelling to work and age structures in different local authorities across England. The data is taken from censuses in 2001 and 2011. You can find the large data set on the OCR website. This chapter is meant to be used to complement the statistical chapters within the main book.

Learning about the data set Every data set has its own peculiarities. These questions are designed to get you to think critically about exactly what is meant by some of the data given. QUESTIONS 1 a Which local authority had the highest percentage of their workforce travelling to work by bicycle in 2011? b What is the modal method of travel to work in 2011? c Which local authority had the highest proportion of the working population driving to work? d Which region showed the largest decrease in the proportion of people driving to work between 2001 and 2011? e Which categories might you include in ‘public transport’? f Which local authority had the highest proportion of people commuting by public transport in 2011? 2 ‘All usual residents’ of County Durham in 2011 has a value of

.

a What do you think the term ‘all usual residents’ means? b Find the total number of people from County Durham included in the 2011 travel to work survey. c Explain why your answer to part b is not . By exploring the age structure data, can you suggest which age groups are included in the travel to work survey? 3 a Suggest some methods of travel which might be included in ‘Other method of travel to work’. b Consider Birmingham in 2011. i ii

Does the sum of all the methods of travel equal the number of people in employment? What does this mean for people who have more than one method of travel to work?

4 a Estimate the number of people that can fit on a typical bus. b Use the figures from the large data set to estimate the number of buses required each morning in Blackpool in 2011. c Give two reasons why your answer is likely to be an underestimate. 5 The ‘car occupancy rate’ is defined as the sum of the number of car drivers and passengers, divided by the number of drivers. a Calculate the car occupancy rate in 2001. b Calculate the car occupancy rate in 2011. c What do these two numbers tells you about the changing use of cars between these two

periods? 6 a Estimate the number of people driving to work in Sheffield in a car with no passengers in 2011. b Explain why your answer is only an estimate. 7 Cornwall has no underground, trams or metro services, but some people in Cornwall commute to work using these methods. Explain why this might be the case. 8 a Find the percentage increase in the number of the people working from home between 2001 and 2011. b Find the percentage increase in the proportion of the working population who work from home. c The 2011 Census Analysis – Method of Travel to Work in England and Wales Report (Gower) states that: “between 2001 and 2011 the proportion of the working population that worked at or from home increased by percentage points”. Explain how they found this figure. (This document is available online from the Office of National Statistics: www.ons.gov.uk/ons/dcp171766_299766.pdf) d Why might each of the comparisons in parts a, b and c be useful?



A reminder of statistical diagrams You need to be able to interpret a range of diagrams including: pie charts bar charts histograms cumulative frequency charts box-and-whisker plots. You may find it helpful to explore how to create these (and other) diagrams using technology, and think about when each type of diagram is useful. The following questions might help you. QUESTIONS 9 The pie charts show some of the methods of transport to work used in Lewisham.

a Summarise the main changes in travel to work illustrated in the pie chart. b What changes in method of transport are not illustrated by the pie chart? 10 Draw a pie chart to show the proportion of employed people who travel to work by car in Wales in 2011. 11 The following three bar charts all display information on the number of people who cycle to work in different areas of England in 2011. Graph 1

Graph 2

Graph 3

a Which area has the most cyclists? Which graph illustrates this best? b Is cycling more popular in the North West or the East Midlands? Explain your answer. c Why might the second graph be considered misleading? 12 The histogram shows the proportions of people of working age who are employed across the different local authorities.

a Approximately how many of the local authorities have an employment rate between ?

and

b Using technology, create a histogram of the same data with class widths of size . What are the advantages and disadvantages of this plot compared to the histogram with class widths of size ? 13 This histogram shows the age distribution in County Durham in 2011.

It could also be represented using a bar chart:

a What is the advantage of the histogram compared to the bar chart? b The first group is described as ages to . Explain why this group is plotted over the region from to . c The modal group is defined as the group with the highest frequency density. i ii

Why is this a better definition than the group with the highest frequency? What is the modal age group for these data?

14 The following cumulative frequency diagram shows the population of each local authority in 2011.

a Use the graph to estimate the number of local authorities in 2011. b Use the graph to estimate the median size of a local authority. c Given that the biggest local authority is Birmingham and the smallest local authority is the Isles of Scilly, sketch a box-and-whisker plot for these data. 15 The box-and-whisker plots show ages of populations in 2001 in two different areas, labelled A and B.

a Which region has the larger interquartile range of ages? b Which region has the lower median age? c One of the areas is London and the other is Wales. By looking at the original data in the large data set, decide which box-and-whisker plot corresponds to which area. 16 The box-and-whisker plots show the percentage of people in 2001 who travelled to work by bus, minibus or coach and the percentage who travelled by train.

The data is also displayed as dot plots:

Which of these graphs corresponds to travelling by train? 17 The 2011 Census Analysis – Method of Travel to Work in England and Wales Report (Gower) shows the following bar chart:

a Give reasons why it is better to use: i ii

percentage of workers rather than number of workers a stacked bar chart rather than a parallel bar chart.

b Summarise the main points highlighted by this chart. c An environmental campaigner uses this chart to claim that London is less polluting than other areas of the country. State three further assumptions that the campaigner has made when making this claim. 18 The 2011 Census Analysis – Method of Travel to Work in England and Wales Report (Gower) also shows this bar chart of percentage point change in drivers and passengers in cars between 2001 and 2011.

a Why is a parallel bar chart better than a stacked bar chart to illustrate this information? b A politician claims that the London congestion charge has caused a decrease in the number of Londoners travelling by car. Does this graph support this claim? c Politicians are debating whether a campaign to increase car sharing on the way to work in London has been successful. Provide evidence to suggest that: i ii

the campaign failed the campaign succeeded.

19 Suggest a reason to choose each of the following diagrams to illustrate data. a pie chart b bar chart c cumulative frequency diagram d histogram e box-and-whisker plot



Standard deviation, interquartile range and other statistical calculations Spreadsheets can be used to find the standard deviation of data. A common syntax might look like this:

Be careful! The command shown in Excel gives the standard deviation of the values presented. There is also a command STDEV which actually gives an estimate of the population’s standard deviation based on the data presented being a sample. This idea comes up a lot in statistics and you will meet this more if you study the statistics option of Further Mathematics. You can also use spreadsheet commands to find the range and the interquartile range. A common syntax might look like this:

QUESTIONS

QUESTIONS 20 a i

Find the standard deviation of the number of people who take the train to work in the North East in 2011.

ii

Find the standard deviation of the number of people who take the train to work in the North West in 2011.

b Hence compare the distribution of the number of people who take the train in the two areas. 21 a i ii

Find the interquartile range for the number of people who travel to work on foot in the West Midlands in 2011. Find the interquartile range for number of people who travel to work on foot in the East Midlands in 2011.

b Hence compare the distribution of the two populations. 22 a i

Find the interquartile range for the number not in employment in the North East in 2011.

ii

Find the standard deviation of the number not in employment in the South West in 2011.

b Hence compare the distribution of the two populations. 23 a i ii

Find the standard deviation for the number of people who travel to work by driving a car or van in the entire country in 2011. Find the standard deviation for the number of people who travel to work by motorcycle, moped or scooter in the entire country in 2011.

b Asher says: ‘Since the standard deviation of the number of people who travel to work by driving a car or van is larger than the standard deviation of the number of people who travel to work by motorcycle, moped or scooter, there was greater variation in the number of people who travel to work by driving a car or van in this period.’

Do you agree? c Elsa suggest that a better measure to compare the variation is to use:

What are the advantages and disadvantages of this measure? 24 Daria and Leonie want to calculate the average percentage of the employable population not in employment in 2011. a Daria calculates the percentage not in employment in each local authority and finds the mean across all local authorities. i ii

Show that the percentage of people not in employment in County Durham in 2011 is approximately . What answer does she get for the average of people not in employment across England and Wales?

b Leonie finds the total of all the people in employment and not in employment across England and Wales and uses that figure to calculate the average percentage not in employment. What answer does she get? c Explain why your answers to parts a ii and b are different. d Suggest why your answers to part a could differ from publicly reported statistics into the percentage of unemployed people. 25 a i

Use an appropriate technological method to estimate the median age in County Durham in 2011. (Do not use the quoted value!)

ii

Stating an appropriate assumption, estimate the mean age in County Durham in 2011. (Do

not use the quoted value!) iii

Explain why these are slightly different from the values quoted.

b Estimate the standard deviation and interquartile range for the age in County Durham in 2011. c In 2001 the local authority equivalent to County Durham is split up into seven local authorities. Use the information sheet on the large data set to find the names of these seven local authorities. d Hence find the mean, median, standard deviation and interquartile range of the area equivalent to County Durham in 2001. e Compare the age distributions in County Durham from 2001 to 2011. f Has the change in County Durham reflected the change across the country? g A geographer thinks that there may be some economic migration involved in the changing demographics of County Durham. He looks at the changes in the population in each age group and plots the following bar chart.

Does this provide evidence for the geographer’s claim?



Outliers and cleaning data There are two standard definitions for deciding if something is an outlier. An outlier is any number more than

interquartile ranges away from the nearest quartile.

An outlier is anything more than two standard deviations from the mean. Once an outlier has been identified you must investigate and use your judgement to decide if it is an error (i.e. a clear mistake – such as putting a decimal point in the wrong place) or just an anomaly. If it is an anomaly then you can choose whether or not to include it, depending upon exactly what question you are trying to answer.

QUESTIONS

QUESTIONS 26 The following diagram shows the percentage of workers who walk to work in 2001.

a Using the first definition for an outlier, given above, show that there must be at least one outlier in this data. b Is the largest outlier likely to be an error? Explain your answer. c Using the original data set calculate the standard deviation and interquartile range of the percentage of workers who walk to work in 2001. d Remove the outliers and recalculate the standard deviation and interquartile range. e Use your calculations to explain whether it is better to calculate the standard deviation or interquartile range of data with outliers. 27 The following is an extract from the methodology used in the travel to work survey 2011: The method of travel used for the longest part, by distance, of the usual journey to work. This topic is only applicable to people who were in employment in the week before the census. This table prioritises workplace address information over method of travel to identify home-workers, to allow a direct comparison with data from the 2001 Census. For example, a person who has indicated their place of work as their home address and said that they travel to work by driving a car or van (eg visiting clients) appears in the category 'Work mainly at or from home', as was the case in 2001. (Method of travel to work (2001 specification) by NS-SeC by sex: www.nomisweb.co.uk/census/2011/dc7604ewla)

This is a method used to clean the data when someone has made a mistake in filling out the survey. Is this always appropriate? 28 a Explain why the mean number of people taking the underground, metro, light rail or tram across each local authority is not a useful statistic. b Theo is trying to calculate the average of the percentages of employed people who take the underground, metro, light rail or tram in 2001. He uses a spreadsheet to find the percentage in each local authority and then finds the mean of all of these figures. i ii



What problem occurs if you try to follow Theo’s procedure? By suitably editing the data, find an appropriate value for the average of the percentages of employed people who take the underground, metro, light rail or tram in 2001.

Scatter diagrams and correlation You can use a spreadsheet to find the correlation coefficient between two variables. One common syntax is shown here:

QUESTIONS 29 The 2001 Census gathered data on how people commuted to their workplace. This scatter graph shows, for each local authority surveyed, the number of people who drove a car or van to work against the number of people who said they were a passenger in a car or van.

a Which of the following is the correlation coefficient for these data? A B C D b How would you describe the relationship between these two variables? c There are three points that might be described as outliers. By looking at the original data identify the local authorities associated with these points. Should these results be discarded? d Would the correlation coefficient be higher or lower with the three outliers discarded? Recalculate the correlation coefficient with the three outlying points discarded.

e Conduct a test at the positive.

significance level to see if the underlying correlation coefficient is

f Suggest a third factor that could explain the relationship between these two variables. 30 a Use technology to find the correlation coefficient between the percentage of workers who walk to work and the percentage of people who cycle to work in London in 2001. b Test the above correlation coefficient at the underlying correlation.

significance level to decide if there is

c Use technology to plot a scatter diagram showing the relationship between the percentage of workers who walk to work and the percentage of workers who cycle to work in London in 2001. d Identify the outlier from the scatter graph. Is it reasonable to remove this point? e Repeat parts a and b without the outlying point. f Use technology to find the correlation coefficient between the percentage of workers who walk to work and the percentage who drive to work in a car or van in London in 2001. (Do not remove any outliers.) g Test the above correlation coefficient at negative correlation.

significance to decide if there is underlying

h Use technology to plot a scatter diagram showing the relationship between the percentage of workers who walk to work and the percentage who drive to work in a car or van in London in 2001. Comment on your answer to part f in this context. 31 The following scatter graph shows the percentage of people who go to work by train against the percentage travelling by underground, metro, tram or light rail in the East of England, the South East and London.

a Describe the pattern observed. b The graph is then changed to highlight which region each point comes from. Does this change your answer to part a?



Sampling and hypothesis testing Whenever we are doing real statistics we have to take into account the fact that our data is only a sample of the whole population we are interested in. We need to understand the sampling method to understand what population the sample represents. To make any decisions about the population we also need to take into account the fact that we might just have an unusual sample. The method for doing this is hypothesis testing. Many spreadsheets can take random samples from data. For example, you can use the data analysis ToolPak in Excel:



QUESTIONS

QUESTIONS 32 This data set comes from the 2001 and 2011 UK Censuses. These are sent out via post to every household in the country and answers refer to one week in the year. In 2011, of households responded. a Explain why this census is not a simple random sample of the population. b The fact that the survey asked about only one week is also a sample. What is the population that it is drawn from? What sampling method has been chosen? c Explain why the results of the census should not simply be multiplied by results for the entire population.

to estimate

d To try to correct for people not returning their census forms a more detailed survey was undertaken in areas. These areas were chosen at random, although it was ensured that the balance of regions and economic prosperity matched the country as a whole. Name the type of sampling method used. 33 It is thought that

of local authorities in 2011 had a total population of less than

a Use a two-tailed binomial test at from: i ii

.

significance to determine if this is the case using data

the South West all local authorities.

b Select random samples of size from the data. In how many of these random samples is there sufficient evidence to reject the hypothesis that of local authorities in 2011 had a total population of less than ? 34 In 2011 the mean age across local authorities is It is thought that the average age is increasing.

years with a standard deviation of

years.

a The average age in a mini-census of local authorities is . What is the probability of this value or greater being observed if the statistics from 2011 are unchanged? Is this sufficient evidence at the significance level to say that the average age is increasing? b Create random samples of size from the data for local authorities in 2011. How many of these samples would provide sufficient evidence of an increase in the mean age from using a significance level?



Statistical problem solving In real life we can use data to answer questions. However, we rarely have the perfect data to answer our questions so we have to make inferences from the data we have. For each question below, suggest a statistical method that could be used to answer or illustrate it, and describe the assumptions you are making. QUESTIONS 35 Are people in England and Wales getting healthier? 36 Is the increase in popularity of cycling reducing the use of cars purchased? 37 Are the transport preferences of Londoners very different to those of people in the rest of the country? 38 Do people drive more in 2011 than 2001? 39 Does prosperity affect the mode of transport people use?



Working with the large data set – answers 1 a Cambridge

.

b Driving by car or van. c South Staffordshire

.

d London – about percentage point decrease. e ‘Underground, metro, light rail, tram’, ‘bus, minibus or coach’ and ‘trains’ should definitely be included. It is perhaps subjective as to whether taxis might be considered ‘public transport’. (The official report did not do so.) f Using the definition in part e, Newham

.

2 a The people who regularly live in County Durham – for example, if someone was just staying there with relatives during the census they would not count. b c Only working age people are included. It seems that this is people aged from

to

.

3 a For example, helicopter, lorry, horse. b i The sum does equal all people in employment. This means that people with more than one method of travel are only counted once. ii They might have self-classified as ‘Other method of travel’ but the actual instruction on the survey was to put the method used for the longest part of the journey to work. 4 a About

.

b c Not every bus will be full to capacity. There may be non-workers using the bus. 5 a b c This might indicate that there is a decrease in the amount of ‘car sharing’ on the way to work, although the numbers are very similar so it might be down to normal statistical variation. 6 a An estimate might be to subtract the number of passengers from the number of drivers, getting . b There are several reasons this is an estimate. It assumes that each car has at most one passenger. Giving significant figures is hard to justify as on any given day there will be variation. Not everybody driving to work in Sheffield will live in Sheffield, so the figure might be larger owing to people commuting from nearby towns. 7 A resident of Cornwall might work in London during the week, possibly from a second home. 8 a

(from

b

(from

to to

c A direct subtraction of

) ) from

.

d A home office supplier might be interested in the actual number of people who might buy their products, so use the value in a. People interested in changing patterns of behaviour, such as economists, could use either b or c. The answer to part b emphasises the change relative to the starting proportion while part c emphasises the change relative to the whole population. Both are valid. 9 a The biggest change is that the proportion of car usage has decreased from 2001 to 2011. More

people travel to work by bus, minibus or coach, taxi. b The increase in population from 2001 to 2011. It also does not show the ‘other’ category, although this would be negligible. 10

11 a London has the greatest number of people who cycle to work. Best shown by graph 1. b Although more people cycle to work in the North West than in the East Midlands, a larger percentage cycle to work in the East Midlands. c The vertical axis starting at other areas. 12 a About

distorts the impression given of London compared to the

.

b There is perhaps too much detail here, which makes the overall picture harder to see. However, if fine detail is needed this might be better – particularly at the edges. 13 a A bar chart does not take into account the different widths of the groups. b Ages are rounded down so someone would be described as being up until the day that they are . c i The total frequency depends on the arbitrary decision about the width of each group. ii 14 a About b About c

15 a B b A c A = London, B = Wales 16 B 17 a i The number of workers would reflect the population of the areas more than the patterns of

travel choices. ii A stacked bar chart emphasises that the total proportion is constant. b Most areas follow a similar pattern with the majority of people travelling by car/van/taxi/motorcycle. London shows a significantly different pattern with many more people using public transport. c There are lots of possibilities – that the population of London is the same as the other areas; that public transport is less polluting per person than cars; that the average number of miles travelled to work is the same in each region; that the types of car used in each region are equally polluting; that transport is the main source of pollution; that commuting to work is the main type of journey (or is proportional to the total number of journeys) – many more! 18 a It allows the patterns in passengers and drivers to be compared to each other and amongst regions more easily. b This graph does provide some evidence that supports the claim in so far as the change in the pattern of driving to work is very different in London compared to other areas. However, you might have expected the number of passengers in cars to increase under the congestion charge scheme. There may be other factors (such as improvements in public transport, rising costs of fuel) which caused the decrease in driver numbers, so it is hard to determine causality from this graph. c i The percentage of passengers has decreased. ii The decrease in London has been less than in other areas. 19 a Shows proportions of each category. b Quickly shows a value associated with each category. c Mainly used to read off median/quartiles of grouped data. d Displays the frequencies of grouped continuous data. e Illustrates key stats (median, quartiles, range) and allows distributions to be quickly compared. 20 a i ii b At first sight there seems to be a greater variation in the number of train journeys in the North West. However, you might think that this variation is actually due to the different sizes of the local authorities in the two regions, so maybe it would be better to compare standard deviations in the proportions. 21 a i ii b It seems that there is a greater variation in the West Midlands, although this might be due to the different sizes of local authorities. 22 a i ii b We can’t compare

with standard deviation.

23 a i ii b It is a valid comparison, but it might simply be a reflection of the different numbers of people who tend to take these modes of transport rather than a greater underlying variability. c This allows quantities with different units to be compared, but it can cause a problem if the mean is very small – for example it is perfectly possible for the mean to be zero, which would make this undefined. 24 a i Use the formula

.

ii b c The answer to a ii does not take into account the different sizes of the local authorities. d People who are not employed includes students and perhaps people who are not looking for employment (e.g. people staying at home to look after children or people who have retired), who are not normally included in unemployment figures. 25 a i

(You might perhaps have to plot a cumulative frequency diagram if your spreadsheet does not do this automatically.)

ii Assuming that the maximum age is

, the mean is

.

iii The values quoted come from the original data. By using grouped data some information is lost. b

.

c Chester-le-Street, Derwentside, Durham, Easington, Sedgefield, Teesdale and Wear Valley. d Median:

, mean:

,

,

e The average age has increased and it has become more spread out. f That data for all of England and Wales: In 2001: median: . In 2011: median: , mean: , ,

, mean:

,

,

So County Durham is slightly older on average than the rest of the country, but the whole country has generally seen an increase in average age and slight increase in spread of age from 2001 to 2011. This reflects a generally ageing population with more older people. g It does provide some evidence. It seems that the biggest change in County Durham is a decrease in the 30–44 age group. This might be because working age people have migrated to other areas of the country. You have to be slightly careful here because the 30–44 age group in 2011 includes some people who were in the 20–24 and 25–29 age groups in 2001, so direct comparisons are tricky, but this graph is suggestive of emigration from the area. 26 a The is and the is about so anything above about so this is an outlier.

is an outlier. The maximum value is

b This is by some way the largest value in any region, so it is suspicious but it is not a clear error – it is a plausible value (i.e. not a misplaced negative sign or a decimal point). So we can treat it as an anomaly, but it is not necessarily an error. You may use your geographical knowledge to realise that ‘City of London’ is an atypical region. c d e The standard deviation is far more affected by the presence of the outlier, so the stable measure when there are outliers.

is a more

27 Someone might have a business based at their house, but they travel to clients at different addresses. 28 a Each local authority is a different size. b i The Isles of Scilly has a ‘ ’ instead of a value, resulting in an error when the percentage is calculated. ii 29 a b Strong linear correlation. c Leeds, Birmingham and County Durham. These are amongst the largest local authorities, so we can explain why these numbers should be large so they should not really be discarded. d It would be lower, since the three points are at the extreme of the line. The new value is

.

e There is significant evidence of correlation.

Tip This hypothesis test assumes that the underlying data follows a bivariate normal distribution, which has not really been established here – in fact, it looks positively skewed. f The size of the population in the local authority. 30 a b No significant evidence. c

d City of London – this is a tiny, atypical borough of London, so it might be reasonable to remove it. e New coefficient is

. There is significant evidence of correlation.

f correlation coefficient g Significant evidence of negative correlation. h

There is a general negative trend, but it is not clear if it is linear. There might actually be two lines going on, or perhaps some sort of curve. Therefore looking for one linear correlation coefficient might not be appropriate. 31 a No real pattern. b It seems that London is a very different area to the rest. It actually now looks more as if there are two distinct patterns within London (perhaps corresponding to inner and outer London). This data would have to be removed to get a better idea of the behaviour in the East of England and the South East. 32 a Not all households are equally likely to be included in the census, since those who decline are not represented.

b It is an opportunity sample of all weeks in the year. c Not all types of people are equally likely to have not responded. d Stratified sampling. 33 a i Evidence that this is not the case. ii Insufficient evidence that this is not the case. b This is a powerful method called simulation. The hypothesis should be rejected in about samples 34 a

, which is sufficient evidence.

b About of these samples should be extreme enough to say that the mean is above though it really is ).

(even

35 You might choose to look at the mean (or perhaps the median) age of all people in the country as a proxy variable for healthiness in 2001 and 2011 and compare them – but how big a difference would be big enough? You might look at the standard deviation to get a sense of whether the difference is significant. You could also compare a histogram or look at the percentage of people aged over 60. Of course, just because people are getting older doesn’t mean that they are more healthy. You might instead want to look at the percentage who cycle or walk to work to infer something about attitudes towards fitness. 36 You might look at the correlation between percentage of cyclists and motorists in each local authority, but this does nothing to establish anything about changes. You might instead plot percentage change in cyclists from 2001 to 2011 against percentage change in motorists. However, this would not tell us about causality. 37 You might plot a stacked bar chart showing the percentages of different methods of travelling to work in different regions, but the problem is that travelling to work is different from general travel. You might also argue that the preference is different from their main method, which is down to availability and cost. 38 You could look at the total number of people who drive in these two years, but the increase might be due to increasing population. You could look at the percentage of people who drive to work, but that does not tell you about the distances they travel or the time they take. 39 You could plot the percentage in transport category against the percentage of the population in work on a scatter graph and look for a trend, maybe using a correlation coefficient. However, the percentage of the population in work might be a poor proxy variable for prosperity as authorities with many students, babies or retired people might appear less prosperous.

Acknowledgements The authors and publishers acknowledge the following sources of copyright material and are grateful for the permissions granted. While every effort has been made, it has not always been possible to identify the sources of all the material used, or to trace all copyright holders. If any omissions are brought to our notice, we will be happy to include the appropriate acknowledgements on reprinting. Thanks to the following for permission to reproduce images: Cover image: huskyomega/Getty Images Back cover: Fabian Oefner www.fabianoefner.com dunlin/Getty Images; Hulton Archive/Stringer/Getty Images; litu92458/Getty Images; Photograph by Patrick Murphy/Getty Images; David Soanes Photography/Getty Images; Bettmann/Contributor/Getty Images; Esben_H/Getty Images; Stefan Swalander/EyeEm/Getty Images; Hulton Archive/Stringer/Getty Images; Thai Yuan Lim/EyeEm/Getty Images; Dave Porter Peterborough UK/Getty Images; Elli Thor Magnusson/Getty Images; Kalawin/Getty Images; vili45/Getty Images; Mark Tipple/Getty Images; Graiki/Getty Images; Vico Collective/Michael Shay/Getty Images; Rafe Swan/Getty Images; shank_ali/Getty Images; Hulton Archive/Stringer/Getty Images; Wikipedia; James Aiken/EyeEm/Getty Images; Jack Andersen/Getty Images; Bettmann/Contributor/Getty Images; Shoji Fujita/Getty Images; BSIP/UIG /Getty Images; Iryna Rieber/EyeEm/Getty Images; Ed Freeman/Getty Images; Yoshinori Kuwahara/Getty Images; Peter Beavis/Getty Images; Stock Montage/Getty Images.

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781316644300 (Paperback) www.cambridge.org/9781316644676 (Paperback with Cambridge Elevate edition) www.cambridge.org/9781316644362 (2 year licence) www.cambridge.org/9781316644379 (Site licence) © Cambridge University Press 2017 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2017 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Printed in Malaysia by Vivar Printing A catalogue record for this publication is available from the British Library ISBN 978-1-316-64430-0 Paperback ISBN 978-1-316-64467-6 Paperback with Cambridge Elevate edition ISBN 978-1-316-64436-2 (2 year licence) ISBN 978-1-316-64437-9 (Site licence)

Additional resources for this publication at www.cambridge.org/education Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. NOTICE TO TEACHERS IN THE UK

It is illegal to reproduce any part of this work in material form (including photocopying and electronic storage) except under the following circumstances: (i) where you are abiding by a licence granted to your school or institution by the Copyright Licensing Agency; (ii) where no such licence exists, or where you wish to exceed the terms of a licence, and you have gained the written permission of Cambridge University Press; (iii) where you are allowed to reproduce without permission under the provisions of Chapter 3 of the Copyright, Designs and Patents Act 1988, which covers, for

example, the reproduction of short passages within certain types of educational anthology and reproduction for the purposes of setting examination questions. This resource is endorsed by OCR for use with specification A Level Mathematics A (H240). In order to gain OCR endorsement, this resource has undergone an independent quality check. Any references to assessment and/or assessment preparation are the publisher’s interpretation of the specification requirements and are not endorsed by OCR. OCR recommends that a range of teaching and learning resources are used in preparing learners for assessment. OCR has not paid for the production of this resource, nor does OCR receive any royalties from its sale. For more information about the endorsement process, please visit the OCR website, www.ocr.org.uk.