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Cambridge International AS & A Level Mathematics Pure Mathematics 2 and 3 Second edition is available as a Whiteboard eTextbook which is an online interactive version of the printed textbook that enables teachers to: ● Display interactive pages to their class ● Add notes and highlight areas ● Add double-page spreads into lesson plans Additionally the Student eTextbook of Cambridge International AS & A Level Mathematics Pure Mathematics 2 and 3 Second edition is a downloadable version of the printed textbook that teachers can assign to students so they can: ● Download and view on any device or browser ● Add, edit and synchronise notes across two devices ● Access their personal copy on the move To find out more and sign up for free trials visit: www.hoddereducation.com/dynamiclearning Integral® A Level Mathematics online resources MEI’s Integral® A Level Mathematics online teaching and learning platform is available to enhance your use of this book. It can be used alongside our printed textbooks, and also links seamlessly with our eTextbooks, allowing you to move with ease between corresponding topics in the eTextbooks and Integral. Integral’s resources cover the revised Cambridge International AS & A Level Mathematics and Further Mathematics syllabuses, providing high-quality teaching and learning activities that include printable materials, innovative interactive activities, and formative and summative assessments. Integral is available by subscription. For details visit integralmaths.org. *MEI’s Integral® material has not been through the Cambridge International endorsement process. Cambridge International AS & A Level Mathematics Pure Mathematics 2 & 3 Second edition Sophie Goldie Series editor: Roger Porkess i 9781510421738.indb 1 02/02/18 1:11 PM Questions from the Cambridge International AS & A Level Mathematics papers are reproduced by permission of Cambridge Assessment International Education. 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The logging and manufacturing processes are expected to conform to the environmental regulations of the country of origin. Orders: please contact Bookpoint Ltd, 130 Park Drive, Milton Park, Abingdon, Oxon OX14 4SE. Telephone: (44) 01235 827720. Fax: (44) 01235 400401 Email: education@bookpoint.co.uk. Lines are open from 9 a.m. to 5 p.m., Monday to Saturday, with a 24-hour message answering service. You can also order through our website at www.hoddereducation.com. Much of the material in this book was published originally as part of the MEI Structured Mathematics series. It has been carefully adapted to support the Cambridge International AS & A Level Mathematics syllabus. The original MEI author team for Pure Mathematics comprised Catherine Berry, Val Hanrahan, Terry Heard, David Martin, Jean Matthews, Roger Porkess and Peter Secker. © Roger Porkess and Sophie Goldie 2018 First published in 2012 This second edition published in 2018 by Hodder Education, an Hachette UK company, Carmelite House, 50 Victoria Embankment, London EC4Y 0DZ Impression number 5 Year 2022 2020 2021 4 3 2 2019 1 2018 All rights reserved. Apart from any use permitted under UK copyright law, no part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or held within any information storage and retrieval system, without permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited. Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, www.cla.co.uk. Cover photo by Shutterstock/chatgunner Illustrations by Pantek Media Maidstone, Kent and Integra Software Services Pvt. Ltd., Pondicherry, India Typeset in Bembo Std 11/13 pt by Integra Software Services Pvt. Ltd., Pondicherry, India Printed in Italy A catalogue record for this title is available from the British Library ISBN 9781510421738 9781510421738.indb 2 02/02/18 1:11 PM Contents Note Chapters 1–6 cover the Paper 2 (AS Level) syllabus content. Chapters 1–11 cover the Paper 3 (A Level) syllabus content. Introduction How to use this book The Cambridge International AS & A Level Mathematics 9709 syllabus 1 Algebra 1.1 1.2 1.3 Operations with polynomials Solution of polynomial equations The modulus function 2 Logarithms and exponentials 2.1 2.2 2.3 2.4 2.5 2.6 Exponential functions Logarithms Graphs of logarithms Modelling curves The natural logarithm function The exponential function 3 Trigonometry 3.1 3.2 3.3 3.4 3.5 Reciprocal trigonometrical functions Compound-angle formulae Double-angle formulae The forms r cos(θ ± α ), r sin(θ ± α ) The general solutions of trigonometrical equations 4 Differentiation 4.1 4.2 4.3 4.4 4.5 4.6 4.7 The product rule The quotient rule Differentiating natural logarithms and exponentials Differentiating trigonometrical functions Differentiating functions defined implicitly Parametric equations Parametric differentiation vi vii x 1 2 7 17 24 25 28 33 37 45 48 57 58 61 67 72 79 83 84 86 90 96 101 108 112 iii 9781510421738.indb 3 02/02/18 1:11 PM 5 Integration P2 5.1 5.2 5.3 5.4 Integrals involving the exponential function Integrals involving the natural logarithm function Integrals involving trigonometrical functions Numerical integration 6 Numerical solution of equations 6.1 6.2 6.3 Interval estimation − change-of-sign methods Fixed-point iteration Problems with the fixed-point iteration method 7 Further algebra 7.1 7.2 7.3 7.4 The general binomial expansion Review of algebraic fractions Partial fractions Using partial fractions with the binomial expansion 8 Further calculus 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Differentiating tan–1 x Integration by substitution Integrals involving exponentials and natural logarithms Integrals involving trigonometrical functions The use of partial fractions in integration Integration by parts General integration 9 Differential equations 9.1 9.2 Forming differential equations from rates of change Solving differential equations 10 Vectors 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 Vectors in two dimensions Vectors in three dimensions Vector calculations The angle between two vectors The vector equation of a line The intersection of two lines The angle between two lines The perpendicular distance from a point to a line 121 121 122 129 133 144 146 151 156 162 163 172 174 180 184 184 186 191 194 199 203 212 217 218 223 234 235 236 240 250 258 265 272 273 iv 9781510421738.indb 4 02/02/18 1:11 PM 11 Complex numbers 11.1 Extending the number system 11.2 Working with complex numbers 11.3 Sets of points in an Argand diagram 11.4 The modulus−argument form of complex numbers 11.5 Sets of points using the polar form 11.6 Working with complex numbers in polar form 11.7 Complex exponents 11.8 Complex numbers and equations Answers Index 280 281 282 294 297 304 307 310 313 319 355 v 9781510421738.indb 5 02/02/18 1:11 PM Introduction This is one of a series of five books supporting the Cambridge International AS & A Level Mathematics 9709 syllabus for examination from 2020. This book follows on from Pure Mathematics 1. AS Level students should work through the first six chapters, which cover the mathematics required for the Paper 2 examination, and A Level students should work through all eleven chapters for the Paper 3 examination. This part of the series also contains a book for mechanics and two books for probability and statistics. The series then continues with four more books supporting Cambridge International AS & A Level Further Mathematics 9231. These books are based on the highly successful series for the Mathematics in Education and Industry (MEI) syllabus in the UK but they have been redesigned and revised for Cambridge International students; where appropriate, new material has been written and the exercises contain many past Cambridge International examination questions. An overview of the units making up the Cambridge International syllabus is given on the following pages. Throughout the series, the emphasis is on understanding the mathematics as well as routine calculations. The various exercises provide plenty of scope for practising basic techniques; they also contain many typical examination-style questions. The original MEI author team would like to thank Sophie Goldie who has carried out the extensive task of presenting their work in a suitable form for Cambridge International students and for her many original contributions. They would also like to thank Cambridge Assessment International Education for its detailed advice in preparing the books and for permission to use many past examination questions. Roger Porkess Series editor vi 9781510421738.indb 6 02/02/18 1:11 PM How to use this book The structure of the book This book has been endorsed by Cambridge Assessment International Education. It is listed as an endorsed textbook for students taking the Cambridge International AS & A Level Mathematics 9709 syllabus. The Pure Mathematics 2 and 3 syllabus content is covered comprehensively and is presented across eleven chapters, offering a structured route through the course. If you are studying AS Mathematics, you should work through the first six chapters, while for A Level you should work through all eleven chapters. This is indicated on each chapter number by either showing P2 and P3, or P3 only. The book is written on the assumption that you have covered and understood the content of the Cambridge IGCSE® Mathematics 0580 (Extended curriculum) or Cambridge International O Level 4024/4029. Two icons are used to indicate material that is not directly on the syllabus. b Some of the early material is designed to provide an overlap and this is designated background.You need to be familiar with the material before you move on to develop it further. e There are also places where the book goes beyond the requirements of the syllabus to show how the ideas can be taken further or where fundamental underpinning work is explored. Such work is marked as extension. Each chapter is broken down into several sections, with each section covering a single topic. Topics are introduced through explanations, with key terms picked out in red. These are reinforced with plentiful worked examples, punctuated with commentary, to demonstrate methods and illustrate application of the mathematics under discussion. Regular exercises allow you to apply what you have learned.They offer a large variety of practice and higher-order question types that map to the key concepts of the Cambridge International syllabus. Look out for the following icons: PS Problem-solving questions will help you to develop the ability to analyse problems, recognise how to represent different situations mathematically, identify and interpret relevant information, and select appropriate methods. M Modelling questions provide you with an introduction to the important skill of mathematical modelling. In this, you take an everyday or workplace situation, or one that arises in your other subjects, and present it in a form that allows you to apply mathematics to it. CP Communication and proof questions encourage you to become a more fluent mathematician, giving you scope to communicate your work with clear, logical arguments and to justify your results. IGCSE® is a registered trademark. 9781510421738.indb 7 vii 02/02/18 1:11 PM Exercises also include questions you are likely to meet from real Cambridge Assessment International Education past papers, so that you can become familiar with the types of questions you will meet in formal assessments. Answers to exercise questions, excluding long explanations and proofs, are included in the back of the book, so you can check your work. It is important, however, that you have a go at answering the questions before looking up the answers if you are to understand the mathematics fully. In addition to the exercises, a range of additional features are included to enhance your learning. ACTIVITY Activities invite you to do some work for yourself, typically to introduce you to ideas that are then going to be taken further. In some places, activities are also used to follow up work that has just been covered. PROBLEM SOLVING INVESTIGATION Mathematics provides you with the techniques to answer many standard questions, but it also does much more than that: it helps you to develop the capacity to analyse problems and to decide for yourself what methods and techniques you will need to use. Questions and situations where this is particularly relevant are highlighted as problem-solving tasks. PS In real life, it is often the case that as well as analysing a situation or problem, you also need to carry out some investigative work. This allows you to check whether your proposed approach is likely to be fruitful or to work at all, and whether it can be extended. Such opportunities are marked as investigations. Other helpful features include the following. ? This symbol highlights points it will benefit you to discuss with your teacher or fellow students, to encourage deeper exploration and mathematical communication. If you are working on your own, there are answers in the back of the book. This is a warning sign. It is used where a common mistake, misunderstanding or tricky point is being described to prevent you from making the same error. A variety of notes are included to offer advice or spark your interest: Note Notes expand on the topic under consideration and explore the deeper lessons that emerge from what has just been done. Historical note Historical notes offer interesting background information about famous mathematicians or results to engage you in this fascinating field. viii 9781510421738.indb 8 02/02/18 1:11 PM Technology note Although graphical calculators and computers are not permitted in the examinations for this Cambridge International syllabus, we have included Technology notes to indicate places where working with them can be helpful for learning and for teaching. Finally, each chapter ends with the key points covered, plus a list of the learning outcomes that summarise what you have learned in a form that is closely related to the syllabus. Digital support Comprehensive online support for this book, including further questions, is available by subscription to MEI’s Integral® online teaching and learning platform for AS & A Level Mathematics and Further Mathematics, integralmaths.org. This online platform provides extensive, high-quality resources, including printable materials, innovative interactive activities, and formative and summative assessments. Our eTextbooks link seamlessly with Integral, allowing you to move with ease between corresponding topics in the eTextbooks and Integral. Additional support The Question & Workbooks provide additional practice for students. These write-in workbooks are designed to be used throughout the course. The Study and Revision Guides provide further practice for students as they prepare for their examinations. These supporting resources and MEI’s Integral® material have not been through the Cambridge International endorsement process. ix 9781510421738.indb 9 02/02/18 1:11 PM The Cambridge International AS & A Level Mathematics 9709 syllabus The syllabus content is assessed over six examination papers. Paper 1: Pure Mathematics 1 • 1 hour 50 minutes • 60% of the AS Level; 30% of the A Level • Compulsory for AS and A Level Paper 2: Pure Mathematics 2 • 1 hour 15 minutes • 40% of the AS Level • Offered only as part of AS Level; not a route to A Level Paper 3: Pure Mathematics 3 • 1 hour 50 minutes • 30% of the A Level • Compulsory for A Level; not a route to AS Level Paper 4: Mechanics • 1 hour 15 minutes • 40% of the AS Level; 20% of the A Level • Offered as part of AS or A Level Paper 5: Probability & Statistics 1 • 1 hour 15 minutes • 40% of the AS Level; 20% of the A Level • Compulsory for A Level Paper 6: Probability & Statistics 2 • 1 hour 15 minutes • 20% of the A Level • Offered only as part of A Level; not a route to AS Level The following diagram illustrates the permitted combinations for AS Level and A Level. AS Level Mathematics Paper 1 and Paper 2 Pure Mathematics only A Level Mathematics (No progression to A Level) Paper 1 and Paper 4 Pure Mathematics and Mechanics Paper 1, 3, 4 and 5 Pure Mathematics, Mechanics and Probability & Statistics Paper 1 and Paper 5 Pure Mathematics and Probability & Statistics Paper 1, 3, 5 and 6 Pure Mathematics and Probability & Statistics x 9781510421738.indb 10 02/02/18 1:11 PM Prior knowledge Knowledge of the content of the Cambridge IGCSE® Mathematics 0580 (Extended curriculum), or Cambridge O Level 4024/4029, is assumed. Learners should be familiar with scientific notation for compound units, e.g. 5 m s−1 for 5 metres per second. In addition, learners should: » be able to carry out simple manipulation of surds (e.g. expressing 12 as 6 2 3 and as 3 2) 2 » know the shapes of graphs of the form y = kxn, where k is a constant and n is an integer (positive or negative) or ± 1 . 2 For Paper 2: Pure Mathematics 2 and Paper 3: Pure Mathematics 3, knowledge of the content of Paper 1: Pure Mathematics 1 is assumed, and learners may be required to demonstrate such knowledge in answering questions. Command words The table below includes command words used in the assessment for this syllabus. The use of the command word will relate to the subject context. Command word What it means Calculate work out from given facts, figures or information Describe state the points of a topic / give characteristics and main features Determine establish with certainty Evaluate judge or calculate the quality, importance, amount, or value of something Explain set out purposes or reasons / make the relationships between things evident / provide why and/or how and support with relevant evidence Identify name/select/recognise Justify support a case with evidence/argument Show (that) provide structured evidence that leads to a given result Sketch make a simple freehand drawing showing the key features State express in clear terms Verify confirm a given statement/result is true xi 9781510421738.indb 11 02/02/18 1:11 PM Key concepts Key concepts are essential ideas that help students develop a deep understanding of mathematics. The key concepts are: Problem solving Mathematics is fundamentally problem solving and representing systems and models in different ways. These include: » Algebra: this is an essential tool which supports and expresses mathematical reasoning and provides a means to generalise across a number of contexts. » Geometrical techniques: algebraic representations also describe a spatial relationship, which gives us a new way to understand a situation. » Calculus: this is a fundamental element which describes change in dynamic situations and underlines the links between functions and graphs. » Mechanical models: these explain and predict how particles and objects move or remain stable under the influence of forces. » Statistical methods: these are used to quantify and model aspects of the world around us. Probability theory predicts how chance events might proceed, and whether assumptions about chance are justified by evidence. Communication Mathematical proof and reasoning is expressed using algebra and notation so that others can follow each line of reasoning and confirm its completeness and accuracy. Mathematical notation is universal. Each solution is structured, but proof and problem solving also invite creative and original thinking. Mathematical modelling Mathematical modelling can be applied to many different situations and problems, leading to predictions and solutions. A variety of mathematical content areas and techniques may be required to create the model. Once the model has been created and applied, the results can be interpreted to give predictions and information about the real world. These key concepts are reinforced in the different question types included in this book: Problem-solving, Communication and proof, and Modelling. xii 9781510421738.indb 12 02/02/18 1:11 PM P2 P3 1 Algebra 1 Algebra No, it [1729] is a very interesting number. It is the smallest number expressible as a sum of two cubes in two different ways. Srinivasa Ramanujan (1887–1920) A brilliant mathematician, Ramanujan was largely self-taught, being too poor to afford a university education. He left India at the age of 26 to work with G. H. Hardy in Cambridge on number theory, but fell ill in the English climate and died six years later in 1920. On one occasion when Hardy visited him in hospital, Ramanujan asked about the registration number of the taxi he came in. Hardy replied that it was 1729, an uninteresting number; Ramanujan’s instant response is quoted above. ? ❯ Find the two pairs of cubes referred to in the quote. ❯ How did you find them? You will already have met quadratic expressions, like x2 − 5x + 6, and solved quadratic equations, such as x2 − 5x + 6 = 0. Quadratic expressions have the form ax2 + bx + c where x is a variable, a, b and c are constants and a is not equal to zero. This work is covered in Pure Mathematics 1 Chapter 2. An expression of the form ax 3 + bx 2 + cx + d, which includes a term in x 3, is called a cubic in x. Examples of cubic expressions are 2x3 + 3x2 − 2x + 11, 9781510421738.indb 1 3y 3 − 1 and 4z3 − 2z. 1 02/02/18 1:11 PM Similarly a quartic expression in x, like x 4 − 4x 3 + 6x 2 − 4x + 1, contains a term in x 4; a quintic expression contains a term in x 5 and so on. 1 1 ALGEBRA All these expressions are called polynomials. The order of a polynomial is the highest power of the variable it contains. So a quadratic is a polynomial of order 2, a cubic is a polynomial of order 3 and 3x 8 + 5x 4 + 6x is a polynomial of order 8 (an octic). 1 Notice that a polynomial does not contain terms involving x , x , etc. Apart from the constant term, all the others are multiples of x raised to a positive integer power. 1.1 Operations with polynomials b Addition of polynomials Polynomials are added by adding like terms; for example, you add the coefficients of x 3 together (i.e. the numbers multiplying x 3 ), the coefficients of x 2 together, the coefficients of x together and the numbers together.You may find it easiest to set this out in columns. Example 1.1 Add (5x 4 − 3x 3 − 2x) to (7x 4 + 5x 3 + 3x 2 − 2). Solution 5x 4 −3x 3 −2x 4 + (7x +5x 3 +3x 2 −2) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 12x 4 +2x 3 +3x 2 −2x −2 Note This may alternatively be set out as follows: (5x 4 − 3x 3 − 2x) + (7x 4 + 5x 3 + 3x 2 − 2) = (5 + 7)x 4 + (−3 + 5)x 3 + 3x 2 − 2x − 2 = 12x 4 + 2x 3 + 3x 2 − 2x − 2 Subtraction of polynomials Similarly polynomials are subtracted by subtracting like terms. 2 9781510421738.indb 2 02/02/18 1:11 PM Example 1.2 Simplify (5x 4 − 3x 3 − 2x) − (7x 4 + 5x 3 + 3x 2 − 2). Solution 1 5x 4 −3x 3 −2x +5x 3 +3x 2 −2) − (7x 4 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −2x 4 −8x 3 −3x 2 −2x +2 1.1 Operations with polynomials Be careful of the signs when subtracting.You may find it easier to change the signs on the bottom line and then go on as if you were adding. Note This, too, may be set out alternatively, as follows: (5x 4 − 3x 3 − 2x) − (7x 4 + 5x 3 + 3x 2 − 2) = (5 − 7)x 4 + (−3 − 5)x 3 − 3x 2 − 2x + 2 = −2x 4 − 8x 3 − 3x 2 − 2x + 2 Multiplication of polynomials When you multiply two polynomials, you multiply each term of the one by each term of the other, and all the resulting terms are added. Remember that when you multiply powers of x, you add the indices: x 5 × x 7 = x 12. Example 1.3 Multiply (x 3 + 3x − 2) by (x 2 − 2x − 4). Solution Arranging this in columns, so that it looks like an arithmetical long multiplication calculation you get: +3x −2 x3 × x2 −2x −4 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Multiply top line by x 2 x5 +3x 3 −2x 2 4 Multiply top line by −2x −2x −6x 2 +4x 3 −12x +8 Multiply top line by −4 −4x −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Add x 5 −2x 4 −x 3 −8x 2 −8x +8 3 9781510421738.indb 3 02/02/18 1:11 PM Note 1 Alternatively: (x3 + 3x − 2) × (x2 − 2x − 4) = x3(x2 − 2x − 4) + 3x(x2 − 2x − 4) − 2(x2 − 2x − 4) = x5 − 2x4 − 4x3 + 3x3 − 6x2 − 12x − 2x2 + 4x + 8 1 ALGEBRA = x5 − 2x4 + (−4 + 3)x3 + (−6 − 2)x2 + (−12 + 4)x + 8 = x5 − 2x4 − x3 − 8x2 − 8x + 8 Division of polynomials Division of polynomials is usually set out rather like arithmetical long division. Example 1.4 Divide (2x3 − 3x2 + x − 6) by (x − 2). Solution Method 1 2x 2 x−2r − 3x 2 + x − 6 3 2x − 4x 2 2x 3 If the dividend is missing a term, leave a blank space. For example, write x 3 + 2x + 5 as x 3 + _ + 2x + 5. Another way to write it is x 3 + 0x 2 + 2x + 5. Found by dividing 2x 3 (the first term in 2x 3 − 3x 2 + x − 6) by x (the first term in x − 2). 2x 2(x − 2) Now subtract 2x 3 − 4x 2 from 2x 3 − 3x 2, bring down the next term (i.e. x) and repeat the method above: 2x 2 + x 3 x − 2 r 2x − 3x 2 + x − 6 2x 3 − 4x 2 x2 + x x 2 − 2x x2 ÷ x x(x − 2) Continuing gives: This is the answer. 2x2 + x + 3 It is called the quotient. 3 x − 2 r 2x − 3x2 + x − 6 2x3 − 4x2 x2 + x The final remainder of x2 − 2x zero means that 3x − 6 x − 2 divides exactly into 2x 3 − 3x 2 + x − 6. 3x − 6 0 Thus (2x 3 − 3x 2 + x − 6) ÷ (x − 2) = (2x 2 + x + 3). 4 9781510421738.indb 4 02/02/18 1:11 PM Method 2 Alternatively this may be set out as follows if you know that there is no remainder. The polynomial here must be of order 2 because 2x 3 ÷ x will give an x 2 term. Let (2x 3 − 3x 2 + x − 6) ÷ (x − 2) ≡ ax 2 + bx + c Multiplying both sides by (x − 2) gives 1 (2x3 − 3x2 + x − 6) ≡ (ax2 + bx + c)(x − 2) 2x3 − 3x2 + x − 6 ≡ ax3 + (b − 2a)x2 + (c − 2b)x − 2c Comparing coefficients of x3 1.1 Operations with polynomials The identity sign is used here to emphasise that this is an identity and true for all values of x. Multiplying out the expression on the right 2=a Comparing coefficients of x2 ⇒ −3 = b − 2a −3 = b − 4 b=1 Comparing coefficients of x ⇒ 1 = c − 2b 1=c−2 c=3 Checking the constant term −6 = −2c (which agrees with c = 3). So ax2 + bx + c is 2x2 + x + 3 i.e. (2x3 − 3x2 + x − 6) ÷ (x − 2) ≡ 2x2 + x + 3. Method 3 With practice you may be able to do this method ‘by inspection’. The steps in this would be as follows. Needed to give the 2x3 term 3 2 2 when multiplied by the x. (2x − 3x + x − 6) = (x − 2)(2x + … + …) This product gives −4x2. Only −3x2 is needed. = (x − 2)(2x2 + x + …) Introducing +x gives +x2 for this product and so the x2 term is correct. = (x − 2)(2x 2 + x + 3) This product gives −2x and +x is on the left-hand side. This +3x product then gives the correct x term. = (x − 2)(2x 2 + x + 3) So (2x 3 − 3x 2 + x − 6) ÷ (x − 2) ≡ 2x 2 + x + 3. Check that the constant term (−6) is correct. 5 9781510421738.indb 5 02/02/18 1:11 PM Note 1 A quotient is the result of a division. So, in Example 1.4 the quotient is 2x 2 + x + 3. 1 ALGEBRA Exercise 1A b 1 State the orders of the following polynomials. (i) x3 + 3x2 − 4x (ii) x12 (iii) 2 + 6x2 + 3x7 − 8x5 2 Add (x3 + x2 + 3x − 2) to (x3 − x2 − 3x − 2). 3 Add (x3 − x), (3x2 + 2x + 1) and (x4 + 3x3 + 3x2 + 3x). 4 Subtract (3x2 + 2x + 1) from (x3 + 5x2 + 7x + 8). 5 Subtract (x3 − 4x2 − 8x − 9) from (x3 − 5x2 + 7x + 9). 6 Subtract (x5 − x4 − 2x3 − 2x2 + 4x − 4) from (x5 + x4 − 2x3 − 2x2 + 4x + 4). 7 Multiply (x3 + 3x2 + 3x + 1) by (x + 1). 8 Multiply (x3 + 2x2 − x − 2) by (x − 2). 9 Multiply (x2 + 2x − 3) by (x2 − 2x − 3). 10 Multiply (x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x1 + 1) by (x − 1). 11 Simplify (x2 + 1)(x − 1) − (x2 − 1)(x − 1). 12 Simplify (x2 + 1)(x2 + 4) − (x2 − 1)(x2 − 4). 13 Simplify (x + 1)2 + (x + 3)2 − 2(x + 1)(x + 3). 14 Simplify (x2 + 1)(x + 3) − (x2 + 3)(x + 1). 15 Simplify (x2 − 2x + 1)2 − (x + 1)4. 16 Divide (x3 − 3x2 − x + 3) by (x − 1). 17 Find the quotient when (x3 + x2 − 6x) is divided by (x − 2). 18 Divide (2x3 − x2 − 5x + 10) by (x + 2). 19 Find the quotient when (x4 + x2 − 2) is divided by (x − 1). 20 Divide (2x3 − 10x2 + 3x − 15) by (x − 5). 21 Find the quotient when (x4 + 5x3 + 6x2 + 5x + 15) is divided by (x + 3). 22 Divide (2x4 + 5x3 + 4x2 + x) by (2x + 1). 23 Find the quotient when (4x4 + 4x3 − x2 + 7x − 4) is divided by (2x − 1). 24 Divide (2x4 + 2x3 + 5x2 + 2x + 3) by (x2 + 1). 25 Find the quotient when (x4 + 3x3 − 8x2 − 27x − 9) is divided by (x2 − 9). 26 Divide (x4 + x3 + 4x2 + 4x) by (x2 + x). 27 Find the quotient when (2x4 − 5x3 − 16x2 − 6x) is divided by (2x2 + 3x). 28 Divide (x4 + 3x3 + x2 − 2) by (x2 + x + 1). 6 9781510421738.indb 6 02/02/18 1:11 PM 1.2 Solution of polynomial equations You have already met the formula 1 2 x = −b ± b − 4ac 2a for the solution of the quadratic equation ax2 + bx + c = 0. 1.2 Solution of polynomial equations Unfortunately there is no such simple formula for the solution of a cubic equation, or indeed for any higher power polynomial equation. So you have to use one (or more) of three possible methods. » Spotting one or more roots. » Finding where the graph of the expression cuts the x-axis. » A numerical method. Example 1.5 Solve the equation 4x3 − 8x2 − x + 2 = 0. Solution Start by plotting the curve whose equation is y = 4x3 − 8x2 − x + 2. (You may also find it helpful at this stage to display it on a graphical calculator or computer.) x −1 0 1 2 3 y −9 +2 −3 0 35 y 40 30 20 10 1 –1 0 2 3 x –10 ▲ Figure 1.1 Figure 1.1 shows that one root is x = 2 and that there are two others. One is between x = −1 and x = 0 and the other is between x = 0 and x = 1. ➜ 7 9781510421738.indb 7 02/02/18 1:11 PM 1 Try x = − 21 . Substituting x = − 21 in y = 4x3 − 8x2 − x + 2 gives y = 4 × (– 81) − 8 × 41 − (− 21 ) + 2 =0 1 ALGEBRA So in fact the graph crosses the x-axis at x = − 21 and this is a root also. Similarly, substituting x = + 21 in y = 4x 3 − 8x 2 − x + 2 gives y = 4 × 81 − 8 × 41 − 21 + 2 =0 and so the third root is x = 21 . The solution is x = − 21 , 1 or 2. 2 This example worked out nicely, but many equations do not have roots which are whole numbers or simple fractions. In those cases you can find an approximate answer by drawing a graph. To be more accurate, you will need to use a numerical method, which will allow you to get progressively closer to the answer, homing in on it. Such methods are covered in Chapter 6. The factor theorem The equation 4x 3 − 8x 2 − x + 2 = 0 has roots that are whole numbers or fractions. This means that it could, in fact, have been factorised. 4x 3 − 8x 2 − x + 2 = (2x + 1)(2x − 1)(x − 2) = 0 Few polynomial equations can be factorised, but when one can, the solution follows immediately. Since (2x + 1)(2x − 1)(x − 2) = 0 it follows that either 2x + 1 = 0 ⇒ x = – 21 or 2x − 1 = 0 ⇒ x= x−2=0 ⇒ x=2 or 1 2 and so x = − 21 , 1 or 2. 2 This illustrates an important result, known as the factor theorem, which may be stated as follows. Factor theorem: If (x − a) is a factor of the polynominal f (x), then f (a) = 0 and x = a is a root of the equation f (x) = 0. Conversely if f (a) = 0, then (x − a) is a factor of f (x). () 8 9781510421738.indb 8 () So if (ax − b) is a factor of the polynomial f(x), then f ba = 0 and x = ba is a root of the equation f(x) = 0. Conversely, if f ba = 0, then (ax − b) is a factor of f(x). 02/02/18 1:11 PM Example 1.6 Given that f(x) = x 3 − 6x 2 + 11x − 6: (i) 1 find f(0), f(1), f(2), f(3) and f(4) (ii) factorise x3 − 6x 2 + 11x − 6 (iii) solve the equation x 3 − 6x 2 + 11x − 6 = 0 (iv) sketch the curve whose equation is f(x) = x 3 − 6x 2 + 11x − 6. 1.2 Solution of polynomial equations Solution (i) f(0) = 03 − 6 × 02 + 11 × 0 − 6 = −6 f(1) = 13 − 6 × 12 + 11 × 1 − 6 = 0 f(2) = 23 − 6 × 22 + 11 × 2 − 6 = 0 f(3) = 33 − 6 × 32 + 11 × 3 − 6 = 0 f(4) = 43 − 6 × 42 + 11 × 4 − 6 = 6 (ii) Since f(1), f(2) and f(3) all equal 0, it follows that (x − 1), (x − 2) and (x − 3) are all factors. This tells you that x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x − 3) × constant By checking the coefficient of the term in x 3, you can see that the constant must be 1, and so x 3 − 6x 2 + 11x − 6 = (x − 1)(x − 2)(x − 3) (iii) x = 1, 2 or 3 (iv) f(x) 0 1 2 3 x –6 ▲ Figure 1.2 9 9781510421738.indb 9 02/02/18 1:11 PM In the previous example, all three factors came out of the working, but this will not always happen. If not, it is often possible to find one factor (or more) by ‘spotting’ it, or by sketching the curve.You can then make the job of searching for further factors much easier by dividing the polynomial by the factor(s) you have found: you will then be dealing with a lower order polynomial. 1 1 ALGEBRA Example 1.7 Given that f(x) = x3 − x2 − 3x + 2: (i) show that (x − 2) is a factor (ii) solve the equation f(x) = 0. Solution To show that (x − 2) is a factor, it is necessary to show that f(2) = 0. f(2) = 23 − 22 − 3 × 2 + 2 =8−4−6+2 =0 Therefore (x − 2) is a factor of x 3 − x 2 − 3x + 2. (ii) Since (x − 2) is a factor you divide f(x) by (x − 2). (i) x2 + x − 1 x − 2 r − x2 − 3x + 2 x3 − 2x2 x2 − 3x x2 − 2x −x + 2 −x + 2 0 So f(x) = 0 becomes (x − 2)(x 2 + x − 1) = 0, ⇒ either x − 2 = 0 or x 2 + x − 1 = 0. Using the quadratic formula on x 2 + x − 1 = 0 gives x3 −1 ± 1 − 4 × 1 × (−1) 2 = −1 ± 5 2 = −1.618 or 0.618 (to 3 d.p.) x= So the complete solution is x = −1.618, 0.618 or 2. Note You may prefer to use inspection or equate coefficients rather than long division. You can write: x 3 − x 2 − 3x + 2 = ( x − 2)(ax 2 + bx + c ) By inspection: a = 1 and c = −1 Equating coefficients of x2 or x gives b =1. 10 9781510421738.indb 10 02/02/18 1:11 PM Spotting a root of a polynomial equation Most polynomial equations do not have integer (or fraction) solutions. It is only a few special cases that work out nicely. 1 To check whether an integer root exists for any equation, look at the constant term. Decide what whole numbers divide into it and test them. Example 1.8 Solution The constant term is −6 and this is divisible by −1, +1, −2, +2, −3, +3, −6 and +6. So the only possible factors are (x ± 1), (x ± 2), (x ± 3) and (x ± 6). This limits the search somewhat. f(1) = −6 f(2) = −6 f(3) = 0 f(6) = 114 f(−1) = −12 f(−2) = −30 f(−3) = −66 f(−6) = −342 No; No; Yes; No; No; No; No; No. x = 3 is an integer root of the equation. Example 1.9 1.2 Solution of polynomial equations Spot an integer root of the equation x3 − 3x2 + 2x − 6 = 0. Is there an integer root of the equation x3 − 3x2 + 2x − 5 = 0? Solution The only possible factors are (x ± 1) and (x ± 5). f(1) = −5 No; f(−1) = −11 f(5) = 55 No; f(−5) = −215 No. No; There is no integer root. Example 1.10 The polynomial, 3x 3 + ax 2 + b, where a and b are constants, is denoted by p(x). It is given that (3x − 4) and (x + 2) are factors of p(x). Find the values of a and b. Solution ( x + 2) is a factor ⇒ p ( −2 ) = 0 ⇒ 3 × ( −2 ) 3 + ( −2 ) 2 a + b = 0 ⇒ −24 + 4a + b = 0 1 ! ➜ 11 9781510421738.indb 11 02/02/18 1:11 PM 1 ( 43 ) = 0 ⇒ 3 × ( 43 ) + ( 43 ) a + b = 0 (3x − 4) is a factor ⇒ p 3 2 64 9 ⇒ + 16 a + b = 0 2 ! 9 1 ALGEBRA 1 − equation ! 2 gives: Equation ! −24 + 4a + b = 0 − 64 + 16 a+b = 0 9 9 − 280 + 20 a 9 9 So 20 a 9 = 280 9 =0 ⇒ a = 14 1 gives b = −32. Substituting a = 14 into equation ! The remainder theorem Using the long division method, any polynomial can be divided by another polynomial of lesser order, but sometimes there will be a remainder. Look at (x 3 + 2x 2 − 3x − 7) ÷ (x − 2). x2 + 4x + 5 x − 2 r + 2x2 − 3x − 7 x3 − 2x2 x3 4x2 − 3x 4x2 − 8x 5x − 7 5x − 10 3 The quotient is x2 + 4x and the remainder is 3. +5 You can write this as x 3 + 2x 2 − 3x − 7 = (x − 2)(x 2 + 4x + 5) + 3 At this point it is convenient to call the polynomial x3 + 2x2 − 3x − 7 = f(x). 1 So f(x) = (x − 2)(x2 + 4x + 5) + 3. ! 1 gives f(2) = 3. Substituting x = 2 into both sides of equation ! So f(2) is the remainder when f(x) is divided by (x − 2). This result can be generalised to give the remainder theorem. Remainder theorem: For a polynomial, f (x), f (a) is the remainder when f (x) is divided by (x − a). f (x) = (x − a)g(x) + f(a) 12 9781510421738.indb 12 02/02/18 1:11 PM Example 1.11 Find the remainder when 2x3 − 3x + 5 is divided by (x + 1). 1 Solution The remainder is found by substituting x = −1 in 2x3 − 3x + 5. 2 × (−1)3 − 3 × (−1) + 5 = −2 + 3 + 5 =6 Example 1.12 1.2 Solution of polynomial equations So the remainder is 6. When 4x2 − 6x + a is divided by (2x − 1), the remainder is 2. Find the value of a. Solution The remainder is found by substituting x = 4 ⇒ ⇒ ( 21 ) 2 1 2 into 4x2 − 6x + a. − 6 × 21 + a = 2 4 × 41 − 3 + a = 2 1 − 3+a = 2 ⇒ a=4 A polynomial is divided by another of degree n. ❯ What can you say about the remainder? ? When dividing by polynomials of order 2 or more, the remainder is usually found most easily by actually doing the long division. Example 1.13 Find the remainder when 2x 4 − 3x 3 + 4 is divided by (x 2 + 1). Solution x2 +1r 2x4 2x4 − 3x3 2x2 − 3x − 2 +4 + 2x2 −3x3 − 2x2 −3x3 − 3x − 2x2 + 3x + 4 −2 − 2x2 The remainder is 3x + 6. 3x + 6 13 9781510421738.indb 13 02/02/18 1:11 PM 1 ALGEBRA 1 In a division such as the one in Example 1.13, it is important to keep a separate column for each power of x and this means that sometimes it is necessary to leave gaps, as in this example. In arithmetic, zeros are placed in the gaps. For example, 2 thousand and 3 is written 2003. Exercise 1B 1 2 3 4 5 6 PS 7 Given that f(x) = x3 + 2x2 − 9x − 18: (i) find f(−3), f(−2), f(−1), f(0), f(1), f(2) and f(3) (ii) factorise f(x) (iii) solve the equation f(x) = 0 (iv) sketch the curve with the equation y = f(x). The polynomial p(x) is given by p(x) = x 3 − 4x. (i) Find the values of p(−3), p(−2), p(−1), p(0), p(1), p(2) and p(3). (ii) Factorise p(x). (iii) Solve the equation p(x) = 0. (iv) Sketch the curve with the equation y = p(x). You are given that f(x) = x 3 − 19x + 30. (i) Calculate f(0) and f(3). Hence write down a factor of f(x). (ii) Find p and q such that f(x) ! (x − 2)(x 2 + px + q). (iii) Solve the equation x3 − 19x + 30 = 0. (iv) Without further calculation draw a sketch of y = f(x). (i) Show that (x − 3) is a factor of x3 − 5x2 − 2x + 24. (ii) Solve the equation x 3 − 5x 2 − 2x + 24 = 0. (iii) Sketch the curve with the equation y = x 3 − 5x 2 − 2x + 24. (i) Show that x = 2 is a root of the equation x 4 − 5x 2 + 2x = 0 and write down another integer root. (ii) Find the other two roots of the equation x 4 − 5x 2 + 2x = 0. (iii) Sketch the curve with the equation y = x 4 − 5x 2 + 2x. (i) The polynomial p(x) = x 3 − 6x 2 + 9x + k has a factor (x − 4). Find the value of k. (ii) Find the other factors of the polynomial. y (iii) Sketch the curve with the equation y = p(x). The diagram shows the curve with the equation y = (x + a)(x − b)2 c where a and b are positive integers. (i) Write down the values of a and –2 –1 0 1 2 b, and also of c, given that the curve crosses the y axis at (0, c). x 14 9781510421738.indb 14 02/02/18 1:11 PM 1 1.2 Solution of polynomial equations Solve the equation (x + a)(x − b)2 = c using the values of a, b and c you found in part (i). 8 The function f(x) is given by f(x) = x 4 − 3x 2 − 4 for real values of x. (i) By treating f(x) as a quadratic in x 2, factorise it in the form (x 2 + …)(x 2 + …). (ii) Complete the factorisation as far as possible. (iii) How many real roots has the equation f(x) = 0? What are they? 9 (i) Show that (x − 2) is not a factor of 2x 3 + 5x 2 − 7x − 3. (ii) Find the quotient and the remainder when 2x 3 + 5x 2 − 7x − 3 is divided by (x − 2). 10 The equation f(x) = x 3 − 4x 2 + x + 6 = 0 has three integer roots. (i) List the eight values of a for which it is sensible to check whether f(a) = 0 and check each of them. (ii) Solve f(x) = 0. 11 Factorise, as far as possible, the following expressions. (i) x 3 − x 2 − 4x + 4 given that (x − 1) is a factor. (ii) x 3 + 1 given that (x + 1) is a factor. (iii) x 3 + x − 10 given that (x − 2) is a factor. (iv) x 3 + x 2 + x + 6 given that (x + 2) is a factor. 12 (i) Show that neither x = 1 nor x = −1 is a root of x 4 − 2x 3 + 3x 2 − 8 = 0. (ii) Find the quotient and the remainder when x 4 − 2x 3 + 3x 2 − 8 is divided by (a) (x − 1) (b) (x + 1) (c) (x 2 − 1). (ii) 13 When 2x 3 + 3x 2 + kx − 6 is divided by (x + 1) the remainder is 7. PS 14 PS 15 PS 16 17 Find the value of k. When x 3 + px 2 + p 2x − 36 is divided by (x − 3) the remainder is 21. Find a possible value of p. When x 3 + ax 2 + bx + 8 is divided by (x − 3) the remainder is 2 and when it is divided by (x + 1) the remainder is −2. Find a and b and hence obtain the remainder on dividing by (x − 2). When f(x) = 2x 3 + ax 2 + bx + 6 is divided by (x − 1) there is no remainder and when f(x) is divided by (x + 1) the remainder is 10. Find a and b and hence solve the equation f(x) = 0. The polynomial f(x) is defined by f(x) = 3x 3 + ax 2 + ax + a , where a is a constant. (i) Given that (x + 2) is a factor of f (x), find the value of a. (ii) When a has the value found in part (i), find the quotient when f (x) is divided by (x + 2). Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 June 2011 15 9781510421738.indb 15 02/02/18 1:11 PM 1 18 The polynomial 2x3 + 7x2 + ax + b, where a and b are constants, is denoted by p(x). It is given that (x + 1) is a factor of p(x), and that when p(x) is divided by (x + 2) the remainder is 5. Find the values of a and b. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q4 June 2008 1 ALGEBRA 19 The polynomial 2x 3 − x 2 + ax − 6, where a is a constant, is denoted by p(x). It is given that (x + 2) is a factor of p(x). (i) Find the value of a. (ii) When a has this value, factorise p(x) completely. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q2 November 2008 20 The polynomial x 3 + ax 2 + bx + 6, where a and b are constants, is denoted by p(x). It is given that (x – 2) is a factor of p(x), and that when p(x) is divided by (x – 1) the remainder is 4. (i) Find the values of a and b. (ii) When a and b have these values, find the other two linear factors of p(x). Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 June 2009 21 The polynomial 4x3 + ax2 + bx − 2, where a and b are constants, is denoted by p(x). It is given that (x + 1) and (x + 2) are factors of p(x). (i) Find the values of a and b. (ii) When a and b have these values, find the remainder when p(x) is divided by (x2 + 1). Cambridge International AS & A Level Mathematics 9709 Paper 33 Q3 November 2014 22 (i) (ii) Find the quotient when 6x4 − x3 − 26x2 + 4x + 15 is divided by (x2 − 4), and confirm that the remainder is 7. Hence solve the equation 6x4 − x3 − 26x2 + 4x + 8 = 0. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2014 23 The polynomials f(x) and g(x) are defined by f(x) = x3 + ax2 + b and g(x) = x3 + bx2 − a, where a and b are constants. It is given that (x + 2) is a factor of f(x). It is also given that, when g(x) is divided by (x + 1), the remainder is −18. (i) Find the values of a and b. (ii) When a and b have these values, find the greatest possible value of g(x) – f(x) as x varies. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 June 2015 16 9781510421738.indb 16 02/02/18 1:11 PM 1.3 The modulus function 1 Look at the graph of y = f(x), where f(x) = x. y y = f(x) = x 1.3 The modulus function x O ▲ Figure 1.3 The function f(x) is positive when x is positive and negative when x is negative. Now look at the graph of y = g(x), where g(x) = | x |. y y = g(x) = |x| O x ▲ Figure 1.4 The function g(x) is called the modulus of x. g(x) always takes the positive numerical value of x. For example, when x = −2, g(x) = 2, so g(x) is always positive. The modulus is also called the magnitude of the quantity. Another way of writing the modulus function g(x) is g(x) = x for x ! 0 g(x) = −x for x " 0. ? ❯ What is the value of g(3) and g(−3)? ❯ What is the value of | 3 + 3 |, | 3 − 3 |, | 3 | + | 3 | and | 3 | + | −3 |? The graph of y = g(x) can be obtained from the graph of y = f(x) by replacing values where f(x) is negative by the equivalent positive values. This is the equivalent of reflecting that part of the line in the x-axis. 17 9781510421738.indb 17 02/02/18 1:11 PM 1 Example 1.14 Sketch the graphs of the following on separate axes. (i) y=1−x (ii) y = | 1 − x | (iii) y = 2 + | 1 − x | 1 ALGEBRA Solution (i) y = 1 − x is the straight line through (0, 1) and (1, 0). y y=1–x 1 O 1 x ▲ Figure 1.5 (ii) y = | 1 − x | is obtained by reflecting the part of the line for x # 1 in the x-axis. y y = |1 – x| 1 O x 1 ▲ Figure 1.6 (iii) y = 2 + | 1 − x | is obtained from the previous graph by applying ⎛0 ⎞ the translation ⎜ ⎟ . ⎝2⎠ y 3 y = 2 + |1 – x| (1, 2) O 1 x ▲ Figure 1.7 18 9781510421738.indb 18 02/02/18 1:11 PM Inequalities involving the modulus sign You will often meet inequalities involving the modulus sign. ? 1 Look back at the graph of y = | x | in Figure 1.4. ❯ How does this show that | x | " 2 is equivalent to −2 " x " 2? Rule Example | x | = | −x | | 3 | = | −3 | |a − b| = |b − a| | 8 − 5 | = | 5 − 8 | = +3 | x |2 = x2 | −3 |2 = (−3)2 |a | = |b| ⇔ a2 = b2 | −3 | = | 3 | |x| $ a ⇔ −a $ x $ a |x| $ 3 ⇔ −3 $ x $ 3 |x| # a ⇔ x " −a or x # a |x| # 3 ⇔ x " −3 or x # 3 |x − a| " b ⇔ Example 1.15 a−b"x"a+b ⇔ |x − 2| " 5 (−3)2 = 32 1.3 The modulus function Here is a summary of some useful rules. ⇔ −3 " x " 7 Solve the following. (i) |x + 3| $ 4 (ii) | 2x − 1 | # 9 (iii) 5 − | x − 2 | # 1 Solution (i) |x+3|$4 ⇔ −4 $ x + 3 $ 4 ⇔ −7 $ x $ 1 (ii) | 2x − 1 | # 9 ⇔ 2x − 1 " −9 or 2x − 1 # 9 ⇔ 2x " −8 or 2x # 10 ⇔ x " −4 or x # 5 (iii) 5 − | x − 2 | # 1 ⇔ 4#|x−2| ⇔ |x−2|"4 ⇔ −4 " x − 2 " 4 ⇔ −2 " x " 6 Note The solution to part (ii) represents two separate intervals on the number line, so cannot be written as a single inequality. 19 9781510421738.indb 19 02/02/18 1:11 PM 1 Example 1.16 Express the inequality −2 " x " 6 in the form | x − a | " b, where a and b are to be found. Solution |x−a|"b ⇔ −b " x − a " b ⇔ a−b"x"a+b 1 ALGEBRA Comparing this with −2 " x " 6 gives a − b = −2 a + b = 6. Solving these simultaneously gives a = 2, b = 4, so | x − 2 | " 4. Example 1.17 Solve 2x " | x − 3 |. Solution It helps to sketch a graph of y = 2x and y = | x − 3 |. y y = 2x 3 y = |x – 3| O c 3 x ▲ Figure 1.8 You can see that the graph of y = 2x is below y = | x − 3 | for x " c. You can find the critical region by solving 2x " −(x − 3). 2x " −(x − 3) 2x " −x + 3 3x " 3 x"1 c is at the intersection of the lines y = 2x and y = −(x − 3). 20 9781510421738.indb 20 02/02/18 1:11 PM Example 1.18 (i) Solve | 2x − 1 | = | x − 2 |. (ii) Solve | 2x − 1 | " | x − 2 |. 1 Solution (i) Sketching a graph of y = | 2x − 1 | and y = | x − 2 | shows that the equation is true for two values of x. 1.3 The modulus function y y = |2x – 1| 2 y = |x – 2| 1 O x 2 ▲ Figure 1.9 You can find these values by solving | 2x − 1 | = | x − 2 |. One method is to use the fact that | a | = | b | ⇔ a2 = b2. | 2x − 1 | = | x − 2 | Squaring: Expanding: Rearranging: (2x − 1)2 = (x − 2)2 4x2 − 4x + 1 = x2 − 4x + 4 3x2 − 3 = 0 ⇒ Factorising: x2 − 1 = 0 (x − 1)(x + 1) = 0 So the solution is x = –1 or x = 1. (ii) Exercise 1C 1 When | 2x − 1 | " | x − 2 |, y = | 2x − 1 | (drawn in red) is below y = | x − 2 | (drawn in blue) on the graph. So the solution to the inequality is −1 " x " 1. Solve the following equations. (i) | x + 4 | = 5 (iii) | 3 − x | = 4 (v) | 2x + 1 | = 5 (vii) | 2x + 1 | = | x + 5 | (ix) | 3x − 2 | = | 4 − x | |x−3|=4 (iv) | 4x − 1 | = 7 (vi) | 8 − 2x | = 6 (viii) | 4x − 1 | = | 9 − x | (ii) 21 9781510421738.indb 21 02/02/18 1:11 PM 1 2 1 ALGEBRA 3 4 5 6 Solve the following inequalities. (i) | x + 3 | " 5 (ii) | x − 2 | $ 2 (iii) | x − 5 | # 6 (iv) | x + 1 | % 2 (v) | 2x − 3 | " 7 (vi) | 3x − 2 | $ 4 Express each of the following inequalities in the form | x − a | " b, where a and b are to be found. (i) −1 " x " 3 (ii) 2 " x " 8 (iii) −2 " x " 4 (iv) −1 " x " 6 (v) 9.9 " x " 10.1 (vi) 0.5 " x " 7.5 Sketch each of the following graphs on a separate set of axes. (i) y = | x + 2 | (ii) y = | 2x − 3 | (iii) y = | x + 2 | − 2 (iv) y = | x | + 1 (v) y = | 2x + 5 | − 4 (vi) y = 3 + | x − 2 | Solve the following inequalities. (i) | x + 3 | " | x − 4 | (ii) | x − 5 | # | x − 2 | (iii) | 2x − 1 | $ | 2x + 3 | (iv) | 2x | $ | x + 3 | (v) | 2x | # | x + 3 | (vi) | 2x + 5 | % | x − 1 | Solve the inequality | x | # | 3x − 2 |. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q1 June 2005 7 Given that a is a positive constant, solve the inequality | x − 3a | # | x − a |. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q1 November 2005 8 Solve the equation 3 x + 4 = 2x + 5 . Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 June 2011 9 Solve the equation x 3 − 14 = 13, showing all your working. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 June 2012 22 9781510421738.indb 22 02/02/18 1:11 PM KEY POINTS 1 2 4 5 6 1 1.3 The modulus function 3 A polynomial in x has terms in positive integer powers of x and may also have a constant term. The order of a polynomial in x is the highest power of x which appears in the polynomial. The factor theorem states that if (x − a) is a factor of a polynomial f(x) then f(a) = 0 and x = a is a root of the equation f(x) = 0. Conversely if f(a) = 0, then (x − a) is a factor of f(x). The remainder theorem states that f(a) is the remainder when the polynomial f(x) is divided by (x − a). The modulus of x, written |x|, means the positive value of x. The modulus function is ● | x | = x, for x % 0 ● | x | = −x, for x " 0. LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ divide a polynomial by a linear or quadratic expression and find the quotient and remainder ■ understand and use the ■ factor theorem ■ remainder theorem ■ use the factor and remainder theorems to solve problems involving polynomials including ■ finding unknown coefficients of a polynomial ■ solving a polynomial equation ■ solve equations and inequalities involving the modulus sign ■ know that ■ | x | = | −x | ■ |a − b| = |b − a| ■ | x |2 = x2 ■ |x − a|$b ⇔ a − b $ x $ a + b ■ | x | # a ⇔ x " −a or x # a ■ sketch the graph of y = | ax + b |. 23 9781510421738.indb 23 02/02/18 1:11 PM P2 P3 2 LOGARITHMS AND EXPONENTIALS 2 We have got so used to our existing mental software that we see no fault or limitation in it. We cannot see why it should ever need changing. And yet, in the last few decades, it has been shown that different thinking habits can be very much more powerful. Edward de Bono (1933–) Logarithms and exponentials ? During growth or reproduction in the human body, a cell divides into M two new cells roughly every 24 hours. Assuming that this process takes exactly 1 day, and that none of the cells die off: ❯ Starting with one cell, how many cells will there be after (i) 5 days (ii) 10 days? ❯ Approximately how many days would it take to create one million cells from a single cell? To answer the question above, you need to use powers of 2. An exponential function is a function which has the variable as the power, such as 2x. (An alternative name for power is exponent.) 24 9781510421738.indb 24 02/02/18 1:12 PM ACTIVITY 2.1 2 Figure 2.1 shows the curves y = 2x and y = x2. y y = x2 B A O 1 2 3 4 5 6 x ▲ Figure 2.1 (i) (a) Find the coordinates of the points A and B. (b) For what values of x is 2x # x2? If you have graphing software available to you, use it to draw (a) y = 3x and y = x3 (b) y = 4x and y = x4. (iii) Is it true that for all values of a (a) y = ax and y = xa intersect when x = a (b) for large enough values of x, ax > xa? (ii) 2.1 Exponential functions y = 2x 2.1 Exponential functions All of the graphs of exponential functions seen in Activity 2.1 pass through the point (0, 1) and have a gradient that is increasing. In fact this is true of all exponential curves of the form y = kax where a > 1. y 6 y = 3x 5 y = 2x 4 3 2 1 –4 –3 –2 –1 0 1 2 3 4 x ▲ Figure 2.2 The x-axis is a horizontal asymptote. 25 9781510421738.indb 25 02/02/18 1:12 PM Exponential functions model many real-life situations, such as the growth of cells in the question at the beginning of the chapter. Exponential functions increase at an ever-increasing rate. In the case of exponential functions, this is described as exponential growth. 2 It is also possible to have exponential functions for which the power is negative, for example y = 2-x, y = 3-x, etc. 2 LOGARITHMS AND EXPONENTIALS The effect of replacing x by −x for any function of x is to reflect the graph in the y-axis. y 6 5 y = 2–x y = 2x 4 3 2 1 −5 −4 −3 −2 −1 0 1 2 3 ▲ Figure 2.3 4 5 x The x-axis is a horizontal asymptote. Notice that as x becomes very large, the value of 2–x becomes very small. The graph of y = 2–x approaches the x-axis ever more slowly as x increases. This is described as exponential decay. The graphs of y = ax and y = a–x have a horizontal asymptote at y = 0 (the x-axis) and go through the point (0, 1). ? ❯ How will these features change if the graphs are (i) translated vertically (ii) translated horizontally (iii) stretched horizontally (iv) stretched vertically? Example 2.1 26 9781510421738.indb 26 The cost, $C, of a machine t years after initial production is given by C = 10 + 20 × 2–t. (i) What is the initial cost of the machine? (ii) What happens to the cost as t becomes very large? (iii) Sketch the graph of C against t. 02/02/18 1:12 PM Solution (i) 2 When t = 0, C = 10 + 20 × 20 = 10 + 20 = 30. Remember that a0 = 1. So the initial cost is $30. (ii) (iii) C 30 The line C = 10 is a horizontal asymptote. 10 O 2.1 Exponential functions As t becomes very large, 2–t becomes very small, and so the cost approaches $10. t ▲ Figure 2.4 Exercise 2A 1 2 M 3 M 4 Sketch each of the graphs below. Show the horizontal asymptote and the coordinates of the point where the graph crosses the y-axis. (i) y = 2x (ii) y = 2x + 1 (iii) y = 2x − 1 Sketch each of the graphs below. Show the horizontal asymptote and the coordinates of the point where the graph crosses the y-axis. (i) y = 3–x (ii) y = 3–x + 2 (iii) y = 3–x − 1 The growth in population P of a certain town after time t years can be modelled by the equation P = 10 000 × 100.1t. (i) State the initial population of the town and calculate the population after 5 years. (ii) Sketch the graph of P against t. The height h m of a species of pine tree t years after planting is modelled by the equation h = 20 − 19 × 0.9t What is the height of the trees when they are planted? (ii) Calculate the height of the trees after 2 years. (iii) Use trial and improvement to find the number of years for the height to reach 10 metres. (iv) What height does the model predict that the trees will eventually reach? (i) 27 9781510421738.indb 27 02/02/18 1:12 PM 2 LOGARITHMS AND EXPONENTIALS 2 M 5 M 6 A flock of birds was caught in a hurricane and blown far out to sea. Fortunately, they were able to land on a small and very remote island and settle there. There was only a limited sustainable supply of suitable food for them on the island. Their population size, n birds, at a time t years after they arrived, can be modelled by the equation n = 200 + 320 × 2−t. (i) How many birds arrived on the island? (ii) How many birds were there after 5 years? (iii) Sketch a graph of n against t. (iv) What is the long-term size of the population? In music, the notes in the middle octave (eight consecutive notes) of the standard scale, and their frequencies in hertz, are as follows. (The frequencies have been rounded to the nearest whole numbers.) A 220 B 247 C 262 D 294 E 330 F 349 G 392 In the next octave up, the notes are also called A to G and their frequencies are exactly twice those given here; so, for example, the frequency of A is 2 × 220 = 440 hertz. The same pattern of multiplying by 2 continues for higher octaves. Similarly, dividing by 2 gives the frequencies for the notes in lower octaves. Find the frequency of B three octaves above the middle. (ii) The lowest note on a standard piano has frequency 27.5 hertz. What note is this and how many octaves is it below the middle? (iii) Julia’s range of hearing goes from 75 up to 9000 hertz. How many notes on the standard scale can she hear? (i) 2.2 Logarithms You can think of multiplication in two ways. Look, for example, at 81 × 243, which is 34 × 35.You can work out the product using the numbers or you can work it out by adding the powers of a common base − in this case base 3. Multiplying the numbers: Adding the powers of the base 3: 81 × 243 = 19 683 4 + 5 = 9 and 39 = 19 683 Another name for a power is a logarithm. Since 81 = 34, you can say that the logarithm to the base 3 of 81 is 4. The word logarithm is often abbreviated to log and the statement would be written log3 81 = 4. In general: 28 9781510421738.indb 28 y = ax ⇒ loga y = x 02/02/18 1:12 PM Notice that since 34 = 81, 3log381 = 81. This is an example of a general result: alogax = x Example 2.2 (i) Find the logarithm to the base 2 of each of these numbers. 64 (a) (b) 1 2 (c) 1 (d) 2 Show that 2log264 = 64. 2.2 Logarithms (ii) Solution (i) (a) 64 = 26 and so log2 64 = 6 (b) 1 2 (c) 1 = 20 and so log2 1 = 0 (d) (ii) 2 = 2–1 and so log2 21 = −1 1 2 = 2 2 and so log 2 2= 1 2 2log264 = 26 = 64 as required Logarithms to the base 10 Any positive number can be expressed as a power of 10. Before the days of calculators, logarithms to the base 10 were used extensively as an aid to calculation. There is no need for that nowadays but the logarithm function remains an important part of mathematics, particularly the natural logarithm which you will meet later in this chapter. Base 10 logarithms continue to be a standard feature on calculators, and occur in some specialised contexts: the pH value of a liquid, for example, is a measure of its acidity or alkalinity and is given by log10(1/the concentration of H+ ions). Since 1000 = 103, log10 1000 = 3 Similarly log10 100 = 2 log10 10 = 1 log10 1 = 0 log10 1 = log10 (10−1) = −1 (10 ) and so on. 1 = log (10−2) = −2 log10 (100 ) 10 ACTIVITY 2.2 There are several everyday situations in which quantities are measured on logarithmic scales. What are the relationships between the following? (i) An earthquake of intensity 7 on the Richter scale and one of intensity 8. (ii) The frequency of the musical note middle C and that of the C above it. (iii) The intensity of an 85 dB noise level and one of 86 dB. 9781510421738.indb 29 29 02/02/18 1:12 PM The laws of logarithms 2 The laws of logarithms follow from those for indices. Multiplication Writing xy = x × y in the form of powers (or logarithms) to the base a and using the result that x = alogax gives 2 LOGARITHMS AND EXPONENTIALS a loga xy = a loga x × a loga y and so a loga xy = a logax + logay. Consequently logaxy = logax + logay. Division Similarly loga ⎛⎜ x ⎞⎟ = logax − loga y. ⎝y⎠ Power zero Since a0 = 1, log a1 = 0. However, it is more usual to state such laws without reference to the base of the logarithms except where necessary, and this convention is adopted in the key points at the end of this chapter. As well as the laws given here, others may be derived from them, as follows. Indices Since it follows that and so xn = x × x × x × … × x (n times) log xn = log x + log x + log x + … + log x (n times), log xn = n log x. This result is also true for non-integer values of n and is particularly useful because it allows you to solve equations in which the unknown quantity is the power, as in the next example. Example 2.3 Solve the equation 2n = 1000. Solution 2n = 1000 Taking logarithms to the base 10 of both sides (since these can be found on a calculator), log10 (2n) = log10 1000 n log10 2 = log10 1000 log 10 1000 n = log 2 = 9.97 to 3 significant figures 10 Note 30 9781510421738.indb 30 Most calculators just have ‘log’ and not ‘log 10’ on their keys. 02/02/18 1:12 PM Example 2.4 Solve the equation 32x + 2(3x) − 15 = 0. Solution Example 2.5 2.2 Logarithms This is a quadratic equation in disguise. Let y = 3x. 32x = (3x)2 So y2 + 2y − 15 = 0 (3x)2 + 2(3x) − 15 = y2 + 2y − 15 ⇒(y − 5)(y + 3) = 0 ⇒y = 5 or y = −3 y = −3 ⇒ 3x = −3 which has no solutions. y = 5 ⇒ 3x = 5 3x is always positive. x So log 3 = log 5 x log 3 = log 5 log 5 x= = 1.46 (3 s.f.) log 3 2 A geometric sequence begins 0.2, 1, 5, ... . The kth term is the first term in the sequence that is greater than 500 000. Find the value of k. Solution The kth term of a geometric sequence is given by ak = a × r k−1. In this case a = 0.2 and r = 5, so: 0.2 × 5k–1 > 500 000 500 000 5k–1 > 0.2 k–1 5 > 2 500 000 Taking logarithms to the base 10 of both sides: log10 5k−1 > log10 2 500 000 ⇒ (k − 1)log10 5 > log10 2 500 000 ⇒ k−1> log 10 2 500 000 log 10 5 ⇒ k − 1 > 9.15 ⇒ k > 10.15 Since k is an integer, then k = 11. So the 11th term is the first term greater than 500 000. Check : 10th term = 0.2 × 510−1 = 390 625 (< 500 000) ✓ 11th term = 0.2 × 511−1 = 1 953 125 (> 500 000) ✓ 31 9781510421738.indb 31 02/02/18 1:12 PM 2 Roots A similar line of reasoning leads to the conclusion that: log n x = n1 log x The logic runs as follows: n x × n x × n x ×… × n x = x 2 LOGARITHMS AND EXPONENTIALS { Since n times it follows that and so x = log x log n x = n1 log x n log n The logarithm of a number to its own base Since 51 = 5, it follows that log5 5 = 1. Clearly the same is true for any number, and in general, loga a = 1 Reciprocals Another useful result is that, for any base, log ⎛⎜ 1y ⎞⎟ = −log y ⎝ ⎠ This is a direct consequence of the division law loga ⎛⎜ x ⎞⎟ = loga x − loga y ⎝y⎠ with x set equal to 1: ⎛ ⎞ log ⎜ 1y ⎟ = log 1 − log y ⎝ ⎠ = 0 − log y = − log y ⎛1⎞ If the number y is greater than 1, it follows that⎜ y ⎟lies between 0 and 1 and ⎝ ⎠ ⎛ ⎞ 1 log ⎜ ⎟ is negative. So for any base (#1), the logarithm of a number between ⎝y ⎠ 0 and 1 is negative.You saw an example of this on page 29: log10 1 = −1. ( 10 ) ⎛ ⎞ The result log ⎜ 1y ⎟ = −log y is often useful in simplifying expressions involving ⎝ ⎠ logarithms. ACTIVITY 2.3 Draw the graph of y = log2 x, taking values of x like 81, 41, 21, 1, 2, 4, 8, 16. Use your graph to estimate the value of 2. 32 9781510421738.indb 32 02/02/18 1:12 PM 2.3 Graphs of logarithms Whatever the value, a, of the base (a #1), the graph of y = loga x has the same general shape (shown in Figure 2.5). y 2 y = loga x 1 1 a 2.3 Graphs of logarithms O x ▲ Figure 2.5 The graph has the following properties. » The curve crosses the x-axis at (1, 0). » The curve only exists for positive values of x. » The line x = 0 is an asymptote and for values of x between 0 and 1 the curve lies below the x-axis. » There is no limit to the height of the curve for large values of x, but its gradient progressively decreases. » The curve passes through the point (a, 1). ? ❯ Each of the points above can be justified by work that you have already covered. How? The relationship y = loga x may be rewritten as x = ay, and so the graph of x = ay is exactly the same as that of y = loga x. Interchanging x and y has the effect of reflecting the graph in the line y = x, and changing the relationship into y = ax, as shown in Figure 2.6. y y = ax a y=x y = loga x 1 O 1 a x ▲ Figure 2.6 33 9781510421738.indb 33 02/02/18 1:12 PM The function y = a x, x ∈" is an exponential function. Notice that while the domain of y = a x is all real numbers (x ∈"), the range is strictly the positive real numbers. y = a x is the inverse of the logarithm function so the domain of the logarithm function is strictly the positive real numbers and its range is all real numbers. Remember that the effect of applying a function followed by its inverse is to bring you back to where you started. 2 2 LOGARITHMS AND EXPONENTIALS Thus loga (ax) = x and a(loga x) = x. Exercise 2B 1 2 2x = 32 ⇔ x = log2 32 Write similar logarithmic equivalents of these equations. In each case find also the value of x, using your knowledge of indices and not using your calculator. (i) 3x = 9 (ii) 4x = 64 1 1 (iii) 2x = 4 (iv) 5x = 5 (v) 7x = 1 (vi) 16x = 2 Write the equivalent of these equations in exponential form. Without using your calculator, find also the value of y in each case. (i) y = log3 9 (ii) y = log5 125 (iii) y = log2 16 (iv) y = log6 1 1 (v) y = log64 8 (vi) y = log5 25 Write down the values of the following without using a calculator. Use your calculator to check your answers for those questions which use base 10. 1 (i) log10 10 000 (ii) log10 ( ) 3 (iii) log10 10 (v) 4 log3 81 ( 811 ) (x) ( ) (ii) log 6 − log 3 (iv) −log 7 (vi) 1 4 log 1 log5 125 Write the following expressions in the form log x where x is a number. log 5 + log 2 1 2 log 9 (vii) log 5 + 3 log 2 − log 10 16 + log 2 (viii) log 12 − 2 log 2 − log 9 16 + 2 log ( ) (x) 2 log 4 + log 9 − 2 log 144 Express the following in terms of log x. 1 (ix) 1 2 log (i) log x 2 (ii) (iii) log x (iv) 1 2 log x 5 − 2 log x 3 log x 2 + log 3 x 3 log x + log x 3 (vi) log ( x ) Sketch each of the graphs below. Show the vertical asymptote and the coordinates of the point where the graph crosses the x-axis. (i) y = log10 x (ii) y = log2 (x + 1) (iii) y = log2 (x − 2) (v) 9781510421738.indb 34 log3 (ix) log4 2 (v) 34 (vi) (viii) log3 4 3 (iii) 2 log 6 6 10 000 log10 1 (vii) log3 27 (i) 5 ( ) (iv) 5 02/02/18 1:12 PM 7 Solve these inequalities. (i) 2x " 128 (iii) 4x + 6 % 70 (v) 0.4x − 0.1 % 0.3 (vii) 2 $ 5x " 8 10 PS 11 PS 12 M 13 (iv) 0.6x " 0.8 (vi) 0.5x + 0.2 $ 1 2 (viii) 1 $ 7x " 5 (x) 2.3 Graphs of logarithms 9 3x + 5 % 32 | 5x − 7 | " 4 Express the following as a single logarithm. 2 log10 x − log10 7 Hence solve 2 log10 x − log10 7 = log10 63. Use logarithms to the base 10 to solve the following equations. (i) 2x = 1 000 000 (ii) 2x = 0.001 (iii) 1.08x = 2 (iv) 1.1x = 100 (v) 0.99x = 0.000 001 (vi) 3(2x+1) = 20 (vii) 2(5x –2) = 30 (viii) 23x = 5(2x+1) (x+3) (x+4) (ix) 7 =5 Solve the following equations. (i) 22x + 3(2x) − 18 = 0 (ii) 52x + 5x − 2 = 0 (iii) 2(32x) − 7(3x) − 4 = 0 (iv) 62x − 5(6x) + 6 = 0 (v) 3(2x+1) + 5(3x) − 2 = 0 (vi) 2(72x) − 7(x+1) + 3 = 0 A geometric sequence has first term 5 and common ratio 7. The kth term is 28 824 005. Use logarithms to find the value of k. Find how many terms there are in these geometric sequences. (i) −1, 2, −4, 8, …, −16 777 216 (ii) 0.1, 0.3, 0.9, 2.7, …, 4 304 672.1 The strength of an earthquake is recorded on the Richter scale. This provides a measure of the energy released. A formula for calculating earthquake strength is ⎛ energy released in joules ⎞ ⎟ earthquake strength = log10 ⎜ 63000 ⎝ ⎠ (i) The energy released in an earthquake is estimated to be 2.5 × 1011 joules. What is its strength on the Richter scale? (ii) An earthquake is 7.4 on the Richter scale. How much energy is released in it? (iii) An island had an earthquake measured at 4.2 on the Richter scale. Some years later it had another earthquake and this one was 7.1. How many times more energy was released in the second earthquake than in the first one? (ix) | 2x − 4 | " 2 8 (ii) 35 9781510421738.indb 35 02/02/18 1:12 PM 2 LOGARITHMS AND EXPONENTIALS 2 M 14 The loudness of a sound is usually measured in decibels. A decibel is 1 10 of a bel, and a bel represents an increase in loudness by a factor of 10. So a sound of 40 decibels is 10 times louder than one of 30 decibels; similarly a sound of 100 decibels is 10 times louder than one of 90 decibels. (i) Solve the equation x10 = 10. (ii) Sound A is 35 decibels and sound B is 36 decibels. Show that (to the nearest whole number) B is 26% louder than A. (iii) How many decibels increase are equivalent to a doubling in loudness? Give your answer to the nearest whole number. (iv) Amir says ‘The percentage increase in loudness from 35 to 37 decibels is given by 37 − 35 × 100 = 5.7%’. 35 Explain Amir’s mistake. 15 Solve the inequality | y − 5 | < 1. (ii) Hence solve the inequality | 3x − 5 | " 1, giving 3 significant figures in your answer. (i) Cambridge International AS & A Level Mathematics 9709 Paper 2 Q3 November 2007 16 Solve the equation 2|3x − 1| = 3x, giving your answers correct to 3 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q2 November 2013 17 Using the substitution u = 3x, or otherwise, solve, correct to 3 significant figures, the equation 3x = 2 + 3−x. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q4 June 2007 18 Solve the equation log10 (x + 9) = 2 + log10 x. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q1 June 2014 19 (i) (ii) Given that (x + 2) and (x + 3) are factors of 5x3 + ax2 + b, find the values of the constants a and b. When a and b have these values, factorise 5x3 + ax2 + b completely, and hence solve the equation 53y + 1 + a ×52y + b = 0, giving any answers correct to 3 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 November 2014 20 The polynomial f(x) is defined by f(x) = 12x3 + 25x2 − 4x − 12. (i) (ii) Show that f(−2) = 0 and factorise f(x) completely. Given that 12 × 27y + 25 × 9y − 4 × 3y − 12 = 0, state the value of 3y and hence find y correct to 3 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q4 June 2011 36 9781510421738.indb 36 02/02/18 1:12 PM 2.4 Modelling curves When you obtain experimental data, you are often hoping to establish a mathematical relationship between the variables in question. Should the data fall on a straight line, you can do this easily because you know that a straight line with gradient m and intercept c has equation y = mx + c. Example 2.6 t 20 40 60 80 100 120 θ 16.3 20.4 24.2 28.5 32.0 36.3 100 120 t (seconds) (i) Plot a graph of θ against t. (ii) What is the relationship between θ and t ? 2.4 Modelling curves In an experiment the temperature θ (in °C) was measured at different times t (in seconds), in the early stages of a chemical reaction. The results are shown in the table below. 2 Solution (i) 40 35 30 25 20 15 10 0 20 40 60 80 ▲ Figure 2.7 (ii) Figure 2.7 shows that the points lie reasonably close to a straight line and so it is possible to estimate its gradient and intercept. Intercept: c = 12.3 36.3 − 16.3 Gradient: m = 120 − 20 = 0.2 In this case the equation is not y = mx + c but θ = mt + c, and so is given by θ = 0.2t + 12.3 9781510421738.indb 37 37 02/02/18 1:12 PM It is often the case, however, that your results do not end up lying on a straight line but on a curve, so that this straightforward technique cannot be applied. The appropriate use of logarithms can convert some curved graphs into straight lines. This is the case if the relationship has one of two forms, y = kxn or y = kax. 2 LOGARITHMS AND EXPONENTIALS 2 The techniques used in these two cases are illustrated in the following examples. In theory, logarithms to any base may be used, but in practice you would only use those available on your calculator: logarithms to the base 10 and natural logarithms. The base of natural logarithms is a number, 2.718 28…, and is denoted by e. In the next section you will see how this apparently unnatural number arises naturally; for the moment what is important is that you can apply the techniques using base 10. Relationships of the form y = kx n Example 2.7 A water pipe is going to be laid between two points and an investigation is carried out as to how, for a given pressure difference, the rate of flow R litres per second varies with the diameter of the pipe d cm.The following data are collected. d 1 2 3 5 10 R 0.02 0.32 1.62 12.53 199.80 It is suspected that the relationship between R and d may be of the form R = kdn where k is a constant. (i) Explain how a graph of log d against log R tells you whether this is a good model for the relationship. (ii) Make out a table of values of log10 d against log10 R and plot these on a graph. (iii) If appropriate, use your graph to estimate the values of n and k. Solution (i) If the relationship is of the form R = kdn, then taking logarithms gives log R = log k + log dn log R = n log d + log k. or This is in the form y = mx + c as n and log k are constants (so can replace m and c) and log R and log d are variables (so can replace y and x). y + = m x log k ↔ n log d ↔ ↔ = ↔ log R + c So log R = n log d + log k is the equation of a straight line. Consequently if the graph of log R against log d is a straight line, the model R = kdn is appropriate for the relationship and n is given by the 38 9781510421738.indb 38 02/02/18 1:12 PM gradient of the graph. The value of k is found from the intercept, log k, of the graph with the vertical axis. log10 k = intercept ⇒ k = 10intercept (ii) 2 Working to 2 decimal places (you would find it hard to draw the graph to greater accuracy) the logarithmic data are as follows. 0 0.30 0.48 0.70 1.00 log 10 R −1.70 −0.49 0.21 1.10 2.30 2.4 Modelling curves log 10 d log10 R 3 2 1 0.4 0 0.2 0.6 0.8 1.0 1.2 log10 d –1 –2 ▲ Figure 2.8 (iii) In this case the graph in Figure 2.8 is indeed a straight line, with gradient 4 and intercept −1.70, so n = 4 and k = 10–1.70 = 0.020 (to 2 significant figures). The proposed equation linking R and d is a good model for their relationship, and may be written as: R = 0.02d 4 Exponential relationships Example 2.8 The temperature in °C, θ, of a cup of coffee at time t minutes after it is made is recorded as follows. t 2 4 6 8 10 12 θ 81 70 61 52 45 38 (i) Plot the graph of θ against t. (ii) Show how it is possible, by drawing a suitable graph, to test whether the relationship between θ and t is of the form θ = kat, where k and a are constants. (iii) Carry out the procedure to find the values of a and of k. ➜ 9781510421738.indb 39 39 02/02/18 1:12 PM 2 Solution (i) 90 2 LOGARITHMS AND EXPONENTIALS 70 50 30 0 2 4 6 8 10 12 14 t (minutes) ▲ Figure 2.9 (ii) If the relationship is of the form θ = kat, taking logarithms of both sides gives log θ = log k + log at log θ = t log a + log k. or This is in the form y = mx + c as log a and log k are constants (so can replace m and c) and log θ and t are variable (so can replace y and x). = y m x + log k ↔ log a t ↔ ↔ = ↔ log θ + c So log θ = t log a + log k is the equation of a straight line. Consequently if the graph of log θ against t is a straight line, the model θ = kat is appropriate for the relationship, and log a is given by the gradient of the graph. The value of a is therefore found as a = 10gradient. Similarly, the value of k is found from the intercept, log10 k, of the line with the vertical axis: k = 10intercept. (iii) The table gives values of log10 θ for the given values of t. t 2 4 6 8 10 12 log10 θ 1.908 1.845 1.785 1.716 1.653 1.580 40 9781510421738.indb 40 02/02/18 1:12 PM The graph of log 10 θ against t is as shown in Figure 2.10. log10 θ 2.0 2 1.974 1.9 1.8 2.4 Modelling curves 1.7 1.6 1.5 0 2 4 6 8 10 12 14 t ▲ Figure 2.10 The graph is indeed a straight line so the proposed model is appropriate. The gradient is −0.033 and so a = 10–0.033 = 0.927. The intercept is 1.974 and so k = 101.974 = 94.2. The relationship between θ and t is given by: θ = 94.2 × 0.927t Note Because the base of the exponential function, 0.927, is less than 1, the function’s value decreases rather than increases with t. Exercise 2C M 1 The planet Saturn has many moons. The table below gives the mean radius of orbit and the time taken to complete one orbit for five of the best-known of them. Tethys Dione Rhea Titan Iapetus Radius R (× 105 km) 2.9 3.8 5.3 12.2 35.6 Period T (days) 1.9 2.7 4.5 15.9 79.3 Moon It is believed that the relationship between R and T is of the form R = kT n. (i) How can this be tested by plotting log R against log T? (ii) Make out a table of values of log R and log T and draw the graph. (iii) Use your graph to estimate the values of k and n. 9781510421738.indb 41 41 02/02/18 1:12 PM 2 2 LOGARITHMS AND EXPONENTIALS M 2 In 1980 a Voyager spacecraft photographed several previously unknown moons of Saturn. One of these, named 1980 S.27, has a mean orbital radius of 1.4 × 10 5 km. (iv) Estimate how many days it takes this moon to orbit Saturn. The table below shows the area, A cm2, occupied by a patch of mould at time t days since measurements were started. t 0 1 2 3 4 5 A 0.9 1.3 1.8 2.5 3.5 5.2 It is believed that A may be modelled by a relationship of the form A = kb t . (i) Show that the model may be written as log A = t log b + log k. (ii) What graph must be plotted to test this model? (iii) Plot the graph and use it to estimate the values of b and k. (iv) (a) Estimate the time when the area of mould was 2 cm 2. Estimate the area of the mould after 3.5 days. (v) How is this sort of growth pattern described? The inhabitants of an island are worried about the rate of deforestation taking place. A research worker uses records over the last 200 years to estimate the number of trees at different dates. It is suggested that the number of trees, N, has been decreasing exponentially with the number of years, t, since 1930, so that N may be modelled by the equation N = kat where k and a are constants. (i) Show that the model may be written as log N = t log a + log k. The diagram shows the graph of log N against t. (b) M 3 log N 6.6 6.4 6.2 6 5.8 0 (ii) 20 40 60 80 100 t Estimate the values of k and a. What is the significance of k? 42 9781510421738.indb 42 02/02/18 1:12 PM M 4 The time after a train leaves a station is recorded in minutes as t and the distance that it has travelled in metres as s. It is suggested that the relationship between s and t is of the form s = ktn where k and n are constants. (i) Show that the graph of log s against log t produces a straight line. The diagram shows the graph of log s against log t. 2 log s 2.4 Modelling curves 4 3 2 1 –0.4 –0.2 0 0.2 0.6 log t 0.4 Estimate the values of k and n. (iii) Estimate how far the train travelled in its first 100 seconds. (iv) Explain why you would be wrong to use your results to estimate the distance the train has travelled after 10 minutes. The variables t and A satisfy the equation A = kb t, where b and k are constants. (i) Show that the graph of log A against t produces a straight line. The graph of log A against t passes through the points (0, 0.2) and (4, 0.75). (ii) 5 log A (4, 0.75) (0, 0.2) O (ii) t Find the values of b and k. 43 9781510421738.indb 43 02/02/18 1:12 PM 2 LOGARITHMS AND EXPONENTIALS 2 CP M 6 7 An experimenter takes observations of a quantity y for various values of a variable x. He wishes to test whether these observations conform to a formula y = A × xB and, if so, to find the values of the constants A and B. Take logarithms of both sides of the formula. Use the result to explain what he should do, what will happen if there is no relationship, and if there is one, how to find A and B. Carry this out accurately on graph paper for the observations in the table, and record clearly the resulting formula if there is one. (i) x 4 7 10 13 20 y 3 3.97 4.74 5.41 6.71 After the introduction of a vaccine, the numbers of new cases of a virus infection in successive weeks are given in the table below. Week number, t No. of new cases, y M 8 1 2 3 4 5 240 150 95 58 38 Plot the points (t, y) on graph paper and join them with a smooth curve. (ii) Decide on a suitable model to relate t and y, and plot an appropriate graph to find this model. (You may wish to use a spreadsheet or graphing software.) (iii) Find the value of y when t = 15, and explain what you find. A local newspaper wrote an article about the number of residents using a new online shopping site based in the town. Based on their own surveys, they gave the following estimates. Time in weeks, t Number of people, P 2 3 4 5 6 4600 5000 5300 5500 5700 Investigate possible models for the number of people, P, using the site at time t weeks after the site began, and determine which one matches the data most closely. (ii) Use your model to estimate how many people signed on in the first week that the site was open. (iii) Discuss the long-term validity of the model. The variables x and y satisfy the relation 3y = 4x + 2. (i) By taking logarithms, show that the graph of y against x is a straight line. Find the exact value of the gradient of this line. (ii) Calculate the x coordinate of the point of intersection of this line with the line y = 2x, giving your answer correct to 2 decimal places. (i) 9 44 9781510421738.indb 44 Cambridge International AS & A Level Mathematics 9709 Paper 2 Q2 June 2007 02/02/18 1:12 PM 2.5 The natural logarithm function The shaded region in Figure 2.11 is bounded by the x-axis, the lines x = 1 and 31 1 x = 3, and the curve y = x .The area of this region may be represented by x dx. 1 ∫ 2 y O 1 3 x ▲ Figure 2.11 ? ❯ Explain why you cannot apply the rule kx n +1 kxn dx = n + 1 + c to this integral. ∫ 2.5 The natural logarithm function 1 y= x However, the area in the diagram clearly has a definite value, and so we need to find ways to express and calculate it. ACTIVITY 2.4 Estimate, using numerical integration (for example by dividing the area up into a number of strips), the areas represented by these integrals. 31 21 6 1 dx dx dx (i) (ii) (iii) x x 1 1 1x What relationship can you see between your answers? ∫ ∫ ∫ a 1 The area under the curve y = x between x = 1 and x = a, that is x1 dx, 1 depends on the value a. For every value of a (greater than 1) there is a definite value of the area. Consequently, the area is a function of a. ∫ To investigate this function you need to give it a name, say L, so that L(a) is the area from 1 to a and L(x) is the area from 1 to x.Then look at the properties of L(x) to see if its behaviour is like that of any other function with which you are familiar. The investigation you have just done should have suggested to you that 31 2 6 dx + x1 dx = x1 dx. x 1 1 1 ∫ ∫ This can now be written as L(3) + L(2) = L(6). This suggests a possible law, that L(a) + L(b) = L(ab). 9781510421738.indb 45 ∫ 45 02/02/18 1:12 PM At this stage this is just a conjecture, based on one particular example. To prove it, you need to take the general case and this is done in the activity below. (At first reading you may prefer to leave the activity, accepting that the result can be proved.) 2 ACTIVITY 2.5 2 LOGARITHMS AND EXPONENTIALS CP Prove that L(a) + L(b) = L(ab), by following the steps below. (i) Explain, with the aid of a diagram, why ab L(a) + 1 dx = L(ab). a x (ii) Now call x = az, so that dx can be replaced by a dz. Show that ∫ ab 1 ∫a x dx = Explain why ∫ b 1 dz. 1z b ∫1 z1 dz = L(b). Notice that the limits of the left-hand integral, ab and a, are values for x but those for the right-hand integral, b and 1, are values for z. So, to find the new limits for the right-hand integral, you should find z when x = a (the lower limit) and when x = ab (the upper limit). Remember az = x. (iii) Use the results from parts (i) and (ii) to show that L(a) + L(b) = L(ab). What function has this property? For all logarithms log(a) + log(b) = log(ab). Could it be that this is a logarithmic function? ACTIVITY 2.6 Satisfy yourself that the function has the following properties of logarithms. (i) L(1) = 0 (ii) L(a) − L(b) = L (iii) L(an) = nL(a) ( ba ) The base of the logarithm function L(x) Having accepted that L(x) is indeed a logarithmic function (for x > 0), the remaining problem is to find the base of the logarithm. By convention this is denoted by the letter e. A further property of logarithms is that for any base p logp p = 1 ( p # 1). 46 9781510421738.indb 46 So to find the base e, you need to find the point such that the area L(e) under the graph is 1. See Figure 2.12. 02/02/18 1:12 PM y O 2 1 e x You have already estimated the value of L(2) to be about 0.7 and that of L(3) to be about 1.1 so the value of e is between 2 and 3. ACTIVITY 2.7 You will need a calculator with an area-finding facility, or other suitable technology, to do this. If you do not have this, read on. e Use the fact that 1 dx = 1 to find the value of e, knowing that it lies 1x between 2 and 3, to 2 decimal places. ∫ 2.5 The natural logarithm function ▲ Figure 2.12 The value of e is given to 9 decimal places in the key points on page 55. Like π, e is a number which occurs naturally within mathematics. It is irrational: when written as a decimal, it never terminates and has no recurring pattern. The function L(x) is thus the logarithm of x to the base e, loge x. This is often called the natural logarithm of x, and written as ln x. Values of x between 0 and 1 So far it has been assumed that the domain of the function ln x is the real numbers greater than 1 (x ∈ ", x > 1). However, the domain of ln x also includes values of x between 0 and 1. As an example of a value of x between 0 and 1, look at ln 21 . a = ln a − ln b Since ln b () ⇒ ln ( 21 ) = ln 1 − ln 2 = −ln 2 (since ln 1 = 0) In the same way, you can show that for any value of x between 0 and 1, the value of ln x is negative. When the value of x is very close to zero, the value of ln x is a large negative number. ln 1 = −ln 1000 = −6.9 (1000 ) ln ( 1 1000 000 ) = −ln 1 000 000 = −13.8 So as x → 0, ln x → −∞ (for positive values of x). 47 9781510421738.indb 47 02/02/18 1:12 PM 2 The graph of the natural logarithm function The graph of the natural logarithm function (shown in Figure 2.13) has the characteristic shape of all logarithmic functions and like other such functions it is only defined for x # 0. The value of ln x increases without limit, but ever more slowly: it has been described as ‘the slowest way to get to infinity’. 2 LOGARITHMS AND EXPONENTIALS y O y = ln x 1 x ▲ Figure 2.13 Historical note Logarithms were discovered independently by John Napier (1550–1617), who lived at Merchiston Castle in Edinburgh, and Jolst Bürgi (1552–1632) from Switzerland. It is generally believed that Napier had the idea first, and so he is credited with their discovery. Natural logarithms are also called Naperian logarithms but there is no basis for this since Napier’s logarithms were definitely not the same as natural logarithms. Napier was deeply involved in the political and religious events of his day and mathematics and science were little more than hobbies for him. He was a man of remarkable ingenuity and imagination and also drew plans for war chariots that look very like modern tanks, and for submarines. 2.6 The exponential function Making x the subject of y = ln x, using the theory of logarithms you obtain x = ey. y y = ex y=x Interchanging x and y, which has the effect of reflecting the graph in the line y = x, gives the exponential function y = ex. The graphs of the natural logarithm function and its inverse are shown in Figure 2.14. y = ln x O x You saw in Pure Mathematics 1 ▲ Figure 2.14 Chapter 5 that reflecting in the line y = x gives an inverse function, so it follows that ex and ln x are each the inverse of the other. Notice that e ln x = x, using the definition of logarithms, and ln(ex ) = x ln e = x. 48 9781510421738.indb 48 02/02/18 1:12 PM Although the function ex is called the exponential function, in fact any function of the form a x is exponential. Figure 2.15 shows several exponential curves. y y = 3x y = ex 2 y = 2x y = 1.5x 2.6 The exponential function y = 1x = 1 y = 0.5x x O ▲ Figure 2.15 The exponential function y = ex increases at an ever-increasing rate. This is described as exponential growth. By contrast, the graph of y = e –x, shown in Figure 2.16, approaches the x-axis ever more slowly as x increases. This is called exponential decay. y y = e –x 1 O x ▲ Figure 2.16 You will meet e x and ln x again later in this book. In Chapter 4 you learn how to differentiate these functions and in Chapter 5 you learn how to integrate them. In this secion you focus on practical applications which require you to use the In key on your calculator. Example 2.9 The number, N, of insects in a colony is given by N = 2000 e0.1t where t is the number of days after observations have begun. (i) Sketch the graph of N against t. (ii) What is the population of the colony after 20 days? (iii) How long does it take the colony to reach a population of 10 000? ➜ 49 9781510421738.indb 49 02/02/18 1:12 PM Solution 2 (i) N N = 2000 e0.1t 2000 When t = 0, N = 2000e0 = 2000 2 LOGARITHMS AND EXPONENTIALS O t ▲ Figure 2.17 (ii) When t = 20, N = 2000 e0.1 × 20 = 14 778 The population is 14 778 insects. (iii) When N = 10 000, 10 000 = 2000 e0.1t 5 = e0.1t Taking natural logarithms of both sides, ln 5 = ln(e0.1t ) Remember ln 5 = 0.1t ln(ex) = x. and so t = 10 ln 5 t = 16.09… It takes just over 16 days for the population to reach 10 000. Example 2.10 The radioactive mass, M grams in a lump of material is given by M = 25e–0.0012t where t is the time in seconds since the first observation. (i) Sketch the graph of M against t. (ii) What is the initial size of the mass? (iii) What is the mass after 1 hour? (iv) The half-life of a radioactive substance is the time it takes to decay to half of its mass. What is the half-life of this material? Solution (i) M 25 O t ▲ Figure 2.18 (ii) When t = 0, M = 25e0 M = 25 The initial mass is 25 g. 50 9781510421738.indb 50 02/02/18 1:12 PM t = 3600 M = 25e–0.0012×3600 M = 0.3324... (iii) After 1 hour, 2 The mass after 1 hour is 0.33 g (to 2 decimal places). (iv) The initial mass is 25 g, so after one half-life, M= 1 2 × 25 = 12.5 g 2.6 The exponential function At this point the value of t is given by 12.5 = 25e–0.0012t ⇒ 0.5 = e–0.0012t Taking logarithms of both sides: ln 0.5 = ln e –0.0012t ln 0.5 = −0.0012t ln 0.5 ⇒ t= –0.0012 t = 577.6 (to 1 decimal place). The half-life is 577.6 seconds. (This is just under 10 minutes, so the substance is highly radioactive.) Example 2.11 Make p the subject of ln(p) − ln(1 − p) = t. Solution ⎛ p ⎞ ln ⎜ ⎟ =t ⎝1 – p ⎠ Writing both sides as powers of e gives e ln (1–p p ) = et Using log a − () log b = log a b Remember eln x =x p t 1– p = e p = et(1 − p) p = et − pet p + pet = et p(1 + et) = et et p = 1 + et 51 9781510421738.indb 51 02/02/18 1:12 PM 2 Example 2.12 Solve these equations. (i) ln (x − 4) = ln x − 4 (ii) e2x + ex = 6 Solution (i) ln (x − 4) = ln x − 4 2 LOGARITHMS AND EXPONENTIALS ⇒ x − 4 = eln x−4 x − 4 = elnx e−4 x − 4 = x e−4 Rearrange to get all the x terms on one side: x − x e−4 = 4 x(l −e−4) = 4 4 x = 1 − e −4 So x = 4.07 (to 3 s.f.) (ii) e2x + ex = 6 is a quadratic equation in ex. Substituting u = ex: u2 + u = 6 +u−6=0 So Factorising: (u − 2)(u + 3) = 0 So u = 2 or u = −3. Since u = ex then ex = 2 or ex = −3. ex = −3 has no solution. ex = 2 ⇒ x = ln 2 So x = 0.693 (to 3 s.f.) u2 Exercise 2D 1 2 3 4 5 M 6 Make x the subject of ln x − ln x0 = kt. Make t the subject of s = s0e–kt. Make p the subject of ln p = −0.02t. Make x the subject of y − 5 = (y0 − 5)ex. Solve these equations. (i) ln(3 − x) = 4 + ln x (ii) ln(x + 5) = 5 + ln x (iii) ln(2 − x) = 2 + ln x 4 (iv) ex = x e (v) e2x − 8ex + 16 = 0 (vi) e2x + ex = 12 A colony of humans settle on a previously uninhabited planet. After t years, their population, P, is given by P = 100e0.05t. (i) Sketch the graph of P against t. (ii) How many settlers land on the planet initially? 52 9781510421738.indb 52 02/02/18 1:12 PM (iii) What is the population after 50 years? (iv) How long does it take the population to reach 1 million? M 7 The height h metres of a species of pine tree t years after planting is ln y 5 4 2 2.6 The exponential function modelled by the equation h = 20 − 19 × 0.9t. (i) What is the height of the trees when they are planted? (ii) Calculate the height of the trees after 2 years, and the time taken for the height to reach 10 metres. The relationship between the market value $y of the timber from the tree and the height h metres of the tree is modelled by the equation y = ahb, where a and b are constants. The diagram shows the graph of ln y plotted against ln h. 3 2 1 0 0.5 1 1.5 2 2.5 ln h –1 –2 (iii) Use the graph to calculate the values of a and b. (iv) Calculate how long it takes to grow trees worth $100. 8 It is given that ln(y + 5) − ln y = 2 ln x. Express y in terms of x, in a form not involving logarithms. Cambridge International AS & A Level Mathematics 9709 Paper 22 Q2 November 2009 9 Given that (1.25)x = (2.5)y, use logarithms to find the value of to 3 significant figures. x y correct Cambridge International AS & A Level Mathematics 9709 Paper 2 Q1 June 2009 10 Solve, correct to 3 significant figures, the equation ex + e2x = e3x. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q2 June 2008 53 9781510421738.indb 53 02/02/18 1:12 PM 2 11 The variables x and y satisfy the equation y = A(b–x), where A and b are constants. The graph of ln y against x is a straight line passing through the points (0, 1.3) and (1.6, 0.9), as shown in the diagram. ln y (0, 1.3) 2 LOGARITHMS AND EXPONENTIALS (1.6, 0.9) O x Find the values of A and b, correct to 2 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q3 November 2008 12 Solve the equation ln(2 + e−x) = 2, giving your answer correct to 2 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q1 June 2009 13 The variables x and y satisfy the equation y = Kx m , where K and m are constants. The graph of ln y against ln x is a straight line passing through the points (0, 2.0) and (6, 10.2), as shown in the diagram. ln y (6, 10.2) (0, 2.0) ln x O Find the values of K and m, correct to 2 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2011 14 ln y (5, 2.92) (2, 1.60) O x The variables x and y satisfy the equation p(x–1) y = Ae , where A and p are constants. 54 9781510421738.indb 54 02/02/18 1:12 PM The graph of ln y against x is a straight line passing through the points (2, 1.60) and (5, 2.92), as shown in the diagram. Find the values of A and p correct to 2 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q2 June 2015 1 2 3 4 5 6 7 8 A function of the form ax is described as exponential. y = loga x ⇔ ay = x. Logarithms to any base Multiplication: log xy = log x + log y ⎛ ⎞ Division: log ⎜ xy ⎟ = log x − log y ⎝ ⎠ Logarithm of 1: log 1 = 0 Powers: log xn = n log x ⎛ ⎞ Reciprocals: log ⎜ 1 ⎟ = −log y ⎝y ⎠ 1 Roots: log n x = n log x Logarithm to its own base: loga a = 1 Logarithms may be used to discover the relationship between the variables in two types of situation. n ● y = kx ⇔ log y = log k + n log x Plot log y against log x: this relationship gives a straight line where n is the gradient and log k is the intercept. x ● y = ka ⇔ log y = log k + x log a Plot log y against x: this relationship gives a straight line where log a is the gradient and log k is the intercept. logex is called the natural logarithm of x and denoted by ln x. e = 2.718 281 828 4… is the base of natural logarithms. ex and ln x are inverse functions: elnx = x and ln(ex) = x. 1 x dx = ln| x | + c 2.6 The exponential function KEY POINTS 2 ∫ 55 9781510421738.indb 55 02/02/18 1:12 PM 2 LEARNING OUTCOMES Now that you have finished this chapter, you should be able to 2 LOGARITHMS AND EXPONENTIALS ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ sketch graphs of exponential functions ■ showing where the graph crosses the y-axis ■ showing the horizontal asymptote x ■ know the difference in the shape of y = a when a < 1 and a > 1 use exponential models for real-life situations understand that the logarithm function is the inverse of the exponential function use the laws of logarithms ■ to rewrite combinations of logs as the log of a single expression ■ to split up a single logarithm into a combination of logarithms sketch the graphs of logarithmic functions use logarithms to solve equations and inequalities with an unknown power use the number e and the exponential function in context sketch a transformation of the graph of y = ex understand and use the natural logarithm function rewrite log statements as exponential statements and vice versa sketch transformations of the graph of y = ln x model curves by plotting ln y against ln x for curves of the form y = kxn using the gradient and intercept of the new graph to find the values of k and n model curves by plotting ln y against x for curves of the form y = kax using the gradient and intercept of the new graph to find the values of k and a. 56 9781510421738.indb 56 02/02/18 1:12 PM P2 P3 3 Trigonometry 3 Trigonometry Music, when soft voices die, vibrates in the memory Percy Bysshe Shelley (1792−1822) Many waves can be modelled as a sine curve. Estimate the amplitude in metres of the wave in the picture above (see Figure 3.1). y a y = a sin bx Amplitude O –a π b 2π b 3π b x Wavelength ▲ Figure 3.1 ? ❯ Use your estimates to suggest values of a and b which would make y = a sin bx a suitable model for the curve. ❯ Do you think a sine curve is a good model for the wave? 57 9781510421738.indb 57 02/02/18 1:12 PM 3.1 Reciprocal trigonometrical functions 3 As well as the three main trigonometrical functions, there are three more which are commonly used. These are their reciprocals − cosecant (cosec), secant (sec) and cotangent (cot), defined by 1 cos θ sec θ = 1 ; cot θ = tan θ = sin θ . cosec θ = 1 ; cos θ sin θ Each of these is undefined for certain values of θ. For example, cosec θ is undefined for θ = 0°, 180°, 360°, … since sin θ is zero for these values of θ. 3 TRIGONOMETRY ( ) Figure 3.2 shows the graphs of these functions. Notice how all three of the functions have asymptotes at intervals of 180°. Each of the graphs shows one of the main trigonometrical functions as a red line and the related reciprocal function as a blue line. y y 1 1 –360º –180º 0 –1 180º 360º x y = sin x –360º –180º y = cos x 0 –1 y = cosec x 180º 360º x y = sec x y y = tan x 1 –360º –180º 0 –1 180º 360º x y = cotx ▲ Figure 3.2 58 9781510421738.indb 58 02/02/18 1:12 PM Using the definitions of the reciprocal functions two alternative trigonometrical forms of Pythagoras’ theorem can be obtained. (i) 3 sin2 θ + cos2 θ ≡ 1 2 2 Dividing both sides by cos2 θ : sin 2 θ + cos 2 θ ≡ 12 cos θ cos θ cos θ ⇒ tan2 θ + 1 ≡ sec2 θ. This identity is sometimes used in mechanics. sin2 θ + cos2 θ ≡ 1 2 2 Dividing both sides by sin2 θ : sin 2 θ + cos2 θ ≡ 12 sin θ sin θ sin θ ⇒ 1 + cot2 θ ≡ cosec2 θ. Questions concerning reciprocal functions are usually most easily solved by considering the related function, as in the following examples. Example 3.1 Find cosec 120° leaving your answer in surd form. Solution Example 3.2 1 sin120° 3 =1÷ 2 = 2 3 cosec 120° = 3.1 Reciprocal trigonometrical functions (ii) Find values of θ in the interval 0° $ θ $ 360° for which sec2 θ = 4 + 2 tan θ. Solution First you need to obtain an equation containing only one trigonometrical function. sec2 θ = 4 + 2 tan θ ⇒ tan2 θ + 1 = 4 + 2 tan θ ⇒ tan2 θ − 2 tan θ − 3 = 0 ⇒ (tan θ − 3)(tan θ + 1) = 0 ⇒ tan θ = 3 or tan θ = −1 tan θ = 3 ⇒ θ = 71.6° (calculator) or θ = 71.6° + 180° = 251.6° (see Figure 3.3, overleaf ) tan θ = −1 ⇒ θ = −45° (not in the required range) or θ = −45° + 180° = 135° or θ = 135° + 180° = 315° (see Figure 3.3) ➜ 59 9781510421738.indb 59 02/02/18 1:12 PM y 3 3 y = tan θ 0 180º 360º 3 TRIGONOMETRY θ –1 ▲ Figure 3.3 The values of θ are 71.6°, 135°, 251.6°, 315°. Exercise 3A 1 2 3 4 5 Solve the following equations for 0° $ x $ 360°. (i) cosec x = 1 (ii) sec x = 2 (iii) cot x = 4 (iv) sec x = −3 (v) cot x = −1 (vi) cosec x = −2 Find the following giving your answers as fractions or in surd form. You should not need your calculator. (i) cot 135° (ii) sec 150° (iii) cosec 240° (iv) sec 210° (v) cot 270° (vi) cosec 225° In triangle ABC, angle A = 90° and sec B = 2. (i) Find the angles B and C. (ii) Find tan B. (iii) Show that 1 + tan2 B = sec2 B. In triangle LMN, angle M = 90° and cot N = 1. (i) Find the angles L and N. (ii) Find sec L, cosec L and tan L. (iii) Show that 1 + tan2 L = sec2 L. Malini is 1.5 m tall. At 8 pm one evening her shadow is 6 m long. Given that the angle of elevation of the sun at that moment is α (i) show that cot α = 4 (ii) find α. 60 9781510421738.indb 60 02/02/18 1:12 PM CP 6 3.2 Compound-angle formulae The photograph at the start of this chapter shows just one of the countless examples of waves and oscillations that are part of the world around us. 3 3.2 Compound-angle formulae 7 For what values of α, where 0° $ α $ 360°, are sec α, cosec α and cot α all positive? (ii) Are there any values of α for which sec α, cosec α and cot α are all negative? Explain your answer. (iii) Are there any values of α for which sec α, cosec α and cot α are all equal? Explain your answer. Solve the following equations for 0° $ x $ 360°. (i) cos x = sec x (ii) cosec x = sec x (iii) 2 sin x = 3 cot x (iv) cosec2 x + cot2 x = 2 (v) 3 sec2 x − 10 tan x = 0 (vi) 1 + cot2 x = 2 tan2 x (i) Because such phenomena are modelled by trigonometrical (and especially sine and cosine) functions, trigonometry has an importance in mathematics far beyond its origins in right-angled triangles. ACTIVITY 3.1 Find an acute angle θ so that sin(θ + 60°) = cos(θ − 60°). Tip: Try drawing graphs and searching for a numerical solution. You should be able to find the solution using either of these methods, but replacing 60° by, for example, 35° would make both of these methods rather tedious. In this chapter you will meet some formulae which help you to solve such equations more efficiently. It is tempting to think that sin(θ + 60°) should equal sin θ + sin 60°, but this is not so, as you can see by substituting a numerical value of θ. For example, putting θ = 30° gives sin(θ + 60°) = 1, but sin θ + sin 60° ≈ 1.366. To find an expression for sin(θ + 60°), you would use the compound-angle formula sin(θ + φ) = sin θ cos φ + cos θ sin φ. This is proved on the next page in the case when θ and φ are acute angles. It is, however, true for all values of the angles. It is an identity. 61 9781510421738.indb 61 02/02/18 1:12 PM 3 ? PS ❯ As you work through this proof make a list of all the results you are assuming. C φ θ a b 3 TRIGONOMETRY h A B D ▲ Figure 3.4 Using the trigonometrical formula for the area of a triangle (see Figure 3.4): area ABC = area ADC + area DBC 1 2 ab sin(θ + φ) = 21 bh sin θ + 21 ah sin φ h = a cos φ from #DBC ⇒ h = b cos θ from #ADC ab sin(θ + φ) = ab sin θ cos φ + ab cos θ sin φ which gives sin(θ + φ) = sin θ cos φ + cos θ sin φ 1 ! This is the first of the compound-angle formulae (or expansions), and it can be used to prove several more. These are true for all values of θ and φ. 1 gives Replacing φ by −φ in ! sin(θ − φ) = sin θ cos(−φ) + cos θ sin(−φ) cos(−φ) = cos φ sin(−φ) = −sin φ ⇒ sin(θ − φ) = sin θ cos φ − cos θ sin φ 2 ! ACTIVITY 3.2 Derive the rest of these formulae. (i) To find an expansion for cos(θ − φ) replace θ by (90° − θ ) in the expansion of sin(θ + φ). Tip: sin(90° − θ ) = cos θ and cos(90° − θ ) = sin θ To find an expansion for cos(θ + φ) replace φ by (−φ) in the expansion of cos(θ − φ). sin (θ + φ ) (iii) To find an expansion for tan(θ + φ), write tan(θ + φ) = . cos (θ + φ ) (ii) 62 9781510421738.indb 62 02/02/18 1:12 PM Tip: After using the expansions of sin(θ + φ) and cos(θ + φ), divide the numerator and the denominator of the resulting fraction by cos θ cos φ to give an expansion in terms of tan θ and tan φ. 3 (iv) To find an expansion for tan(θ − φ) in terms of tan θ and tan φ, replace φ by (−φ) in the expansion of tan(θ + φ). ? PS The four results obtained in Activity 3.2, together with the two previous results, form the set of compound-angle formulae. sin(θ + φ) = sin θ cos φ + cos θ sin φ sin(θ − φ) = sin θ cos φ − cos θ sin φ cos(θ + φ) = cos θ cos φ − sin θ sin φ cos(θ − φ) = cos θ cos φ + sin θ sin φ tan θ + tan φ 1 − tan θ tan φ tan θ – tan φ tan(θ − φ) = 1 + tan θ tan φ tan(θ + φ) = 3.2 Compound-angle formulae ❯ Are your results valid for all values of θ and φ? ❯ Test your results with θ = 60°, φ = 30°. (θ + φ) ≠ 90°, 270°, ... (θ − φ) ≠ 90°, 270°, ... You are now in a position to solve the earlier problem more easily. To find an acute angle θ such that sin(θ + 60°) = cos(θ − 60°), you expand each side using the compound-angle formulae. sin(θ + 60°) = sin θ cos 60° + cos θ sin 60° = 1 sin θ + 3 cos θ 2 2 cos(θ − 60°) = cos θ cos 60° + sin θ sin 60° = 1 cos θ + 3 sin θ 2 2 1 ! 2 ! 2 1 and ! From ! 1 sin θ + 3 cos θ = 1 cos θ + 3 sin θ 2 2 2 2 sin θ + 3 cos θ = cos θ + 3 sin θ Collect like terms: ⇒ ( 3 – 1) cos θ = ( 3 – 1) sin θ cos θ = sin θ Divide by cos θ : 1 = tan θ θ = 45° This gives an equation in one trigonometrical ratio. Since an acute angle was required, this is the only root. 63 9781510421738.indb 63 02/02/18 1:12 PM Uses of the compound-angle formulae 3 3 TRIGONOMETRY You have already seen compound-angle formulae used in solving a trigonometrical equation and this is quite a common application of them. However, their significance goes well beyond that since they form the basis for a number of important techniques. Those covered in this book are as follows. » The derivation of double-angle formulae The derivation and uses of these are covered on pages 67 to 70. » The addition of different sine and cosine functions This is covered on pages 72 to 75. It is included here because the basic wave form is a sine curve. It has many applications, for example in applied mathematics, physics and chemistry. » Calculus of trigonometrical functions This is covered in Chapters 4 and 5 and also in Chapter 8. Proofs of the results depend on using either the compound-angle formulae or the factor formulae which are derived from them. You will see from this that the compound-angle formulae are important in the development of the subject. Some people learn them by heart, others think it is safer to look them up when they are needed. Whichever policy you adopt, you should understand these formulae and recognise their form. Without that you will be unable to do the next example, which uses one of them in reverse. Example 3.3 Simplify cos θ cos 3θ − sin θ sin 3θ. Solution The formula which has the same pattern of cos cos − sin sin is cos(θ + φ) = cos θ cos φ − sin θ sin φ Using this, and replacing φ by 3θ, gives cos θ cos 3θ − sin θ sin 3θ = cos(θ + 3θ ) = cos 4θ Exercise 3B 1 2 Use the compound-angle formulae to write the following as surds. (i) sin 75° = sin(45° + 30°) (ii) cos 135° = cos(90° + 45°) (iii) tan 15° = tan(45° − 30°) (iv) tan 75° = tan(45° + 30°) Expand each of the following expressions. (i) sin(θ + 45°) (ii) cos(θ − 30°) (iii) sin(60° − θ ) (iv) cos(2θ + 45°) (v) tan(θ + 45°) (vi) tan(θ − 45°) 64 9781510421738.indb 64 02/02/18 1:12 PM 3 4 PS 5 (ii) PS 6 ( ) π π 2 cos (θ – ) = cos (θ + ) 3 2 3 3.2 Compound-angle formulae PS Simplify each of the following expressions. (i) sin 2θ cos φ − cos 2θ sin θ (ii) cos φ cos 7φ − sin φ sin 7φ (iii) sin 120° cos 60° + cos 120° sin 60° (iv) cos θ cos θ − sin θ sin θ Solve the following equations for values of θ in the range 0° $ θ $ 180°. (i) cos(60° + θ ) = sin θ (ii) sin(45° − θ ) = cos θ (iii) tan(45° + θ ) = tan(45° − θ ) (iv) 2 sin θ = 3 cos(θ − 60°) (v) sin θ = cos(θ + 120°) Solve the following equations for values of θ in the range 0 $ θ $ π. (When the range is given in radians, the solutions should be in radians, using multiples of π where appropriate.) π (i) sin θ + 4 = cos θ The diagram shows three points A (−4, 1), B (0, 4) and C (6, −2) joined to form a triangle. The angles α and β and the point P are also shown in the diagram. y B (0, 4) α A (−4, 1) β P x O C (6, −2) (i) Show that sin α = 4 5 and write down the value of cos α. Find the values of sin β and cos β. 7 (iii) Show that sin (α + β ) = 5 2 (iv) Show that tan (α + β ) = −7 and comment on the significance of the negative value. (ii) 65 9781510421738.indb 65 02/02/18 1:12 PM 3 TRIGONOMETRY 3 7 The angle α lies between 0° and 90° and is such that 2 tan2 α + sec2 α = 5 − 4 tan α. (i) Show that 3 tan2 α + 4 tan α − 4 = 0 and hence find the exact value of tan α. (ii) It is given that the angle β is such that cot(α + β ) = 6. Without using a calculator, find the exact value of cot β. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2014 8 (i) (ii) Show that the equation tan(45° + x) − tan x = 2 can be written in the form tan2 x + 2 tan x − 1 = 0. Hence solve the equation tan(45° + x) − tan x = 2, giving all solutions in the interval 0° $ x $ 180°. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q5 November 2007 9 The angles α and β lie in the interval 0° " x " 180°, and are such that tan α = 2 tan β and tan(α + β ) = 3. Find the possible values of α and β. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q4 November 2009 10 (i) (ii) Show that the equation tan(30° + θ ) = 2 tan(60° − θ ) can be written in the form tan2 θ + (6 3) tan θ − 5 = 0. Hence, or otherwise, solve the equation tan(30° + θ ) = 2 tan(60° − θ ), for 0° $ θ $ 180°. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q4 June 2008 11 The angles A and B are such that sin(A + 45°) = (2 2)cos A and 4sec2 B + 5 = 12 tan B. Without using a calculator, find the exact value of tan(A − B). Cambridge International AS & A Level Mathematics 9709 Paper 33 Q6 November 2015 66 9781510421738.indb 66 02/02/18 1:12 PM 3.3 Double-angle formulae As you work through these proofs, think how you can check the results. ❯ Is a check the same as a proof? PS sin(θ + φ) = sin θ cos φ + cos θ sin φ When φ = θ, this becomes sin(θ + θ ) = sin θ cos θ + cos θ sin θ giving (ii) sin 2θ = 2 sin θ cos θ cos(θ + φ ) = cos θ cos φ − sin θ sin φ 3.3 Double-angle formulae Substituting φ = θ in the relevant compound-angle formulae leads immediately to expressions for sin 2θ, cos 2θ and tan 2θ, as follows. (i) 3 ? When φ = θ, this becomes cos(θ + θ ) = cos θ cos θ − sin θ sin θ giving cos 2θ = cos2 θ − sin2 θ Using the Pythagorean identity cos2 θ + sin2 θ = 1, two other forms for cos 2θ can be obtained. cos 2θ = (1 − sin2 θ ) − sin2 θ ⇒ cos 2θ = 1 − 2 sin2 θ cos 2θ = cos2 θ − (1 − cos2 θ ) ⇒ cos 2θ = 2 cos2 θ − 1 These alternative forms are often more useful since they contain only one trigonometrical function. (iii) tan(θ + φ) = tan θ + tan φ 1 – tan θ tan φ (θ + φ) ≠ 90°, 270°, ... When φ = θ, this becomes tan(θ + θ ) = giving tan θ + tan θ 1 – tan θ tan θ tan 2θ = 2 tan θ2 1 – tan θ θ ≠ 45°, 135°, ... Uses of the double-angle formulae e In modelling situations You will meet situations, such as that on the next page, where using a double-angle formula not only allows you to write an expression more neatly but also thereby allows you to interpret its meaning more clearly. 67 9781510421738.indb 67 02/02/18 1:12 PM 3 height 3 TRIGONOMETRY u α R O ground horizontal distance ▲ Figure 3.5 When an object is projected, such as a golf ball being hit as in Figure 3.5, with speed u at an angle α to the horizontal over level ground, the horizontal distance it travels before striking the ground, called its range, R, is given by the product of the horizontal component of the velocity u cos α and its time α of flight 2u sin g . 2 R = 2u singα cos α Using the double-angle formula sin 2α = 2 sin α cos α allows this to be written as 2 2α . R = u sin g Since the maximum value of sin 2α is 1, it follows that the greatest value of 2 the range R is ug and that this occurs when 2α = 90° and so α = 45°. Thus an angle of projection of 45° will give the maximum range of the projectile over level ground. (This assumes that air resistance may be ignored.) In this example, the double-angle formula enabled the expression for R to be written tidily. However, it did more than that because it made it possible to find the maximum value of R by inspection and without using calculus. In calculus The double-angle formulae allow a number of functions to be integrated and you will meet some of these later (see page 130). The formulae for cos 2θ are particularly useful in this respect since cos 2θ = 1 − 2 sin2 θ ⇒ sin2 θ = 21(1 − cos 2θ ) cos 2θ = 2 cos2 θ − 1 ⇒ cos2 θ = 21(1 + cos 2θ ) and and these identities allow you to integrate sin2 θ and cos2 θ. 68 9781510421738.indb 68 02/02/18 1:12 PM In solving equations You will sometimes need to solve equations involving both single and double angles as shown by the next two examples. Example 3.4 3 Solve the equation sin 2θ = sin θ for 0° $ θ $ 360°. Solution ⇒ 2 sin θ cos θ = sin θ ⇒ 2 sin θ cos θ − sin θ = 0 ⇒ sin θ (2 cos θ − 1) = 0 ⇒ sin θ = 0 or cos θ = 1 2 The principal value is the one which comes from your calculator. 3.3 Double-angle formulae sin 2θ = sin θ Be careful here: don’t cancel sin θ or some roots will be lost. sin θ = 0 ⇒ θ = 0° (principal value) or 180° or 360° (see Figure 3.6) y 1 y = sin θ O 180° 360° θ –1 ▲ Figure 3.6 cos θ = 1 2 ⇒ θ = 60° (principal value) or 300° (see Figure 3.7) y 1 y = cos θ 1 2 O 60° 300° 360° θ –1 ▲ Figure 3.7 The full set of roots for 0° $ θ $ 360° is θ = 0°, 60°, 180°, 300°, 360°. 69 9781510421738.indb 69 02/02/18 1:12 PM 3 When an equation contains cos 2θ, you will save time if you take care to choose the most suitable expansion. Example 3.5 Solve 2 + cos 2θ = sin θ for 0 $ θ $ 2π. 3 TRIGONOMETRY Solution This is the most suitable expansion since the right-hand side contains sin θ. Using cos 2θ = 1 − 2 sin2 θ gives 2 + (1 − 2 sin2 θ ) = sin θ ⇒ 2 sin2 θ + sin θ − 3 = 0 ⇒ (2 sin θ + 3)(sin θ − 1) = 0 ⇒ sin θ = − 23 (not valid since −1 $ sin θ $ 1) or sin θ = 1 Figure 3.8 shows that the principal value θ = π 2 is the only root for 0 $ θ $ 2π. y 1 O Notice that the request for 0 $ θ $ 2π, i.e. in radians, is an invitation to give the answer in radians. y = sin θ π 2 π 2π θ ▲ Figure 3.8 Exercise 3C 1 2 3 PS 4 5 Solve the following equations for 0° $ θ $ 360°. (i) 2 sin 2θ = cos θ (ii) tan 2θ = 4 tan θ (iii) cos 2θ + sin θ = 0 (iv) tan θ tan 2θ = 1 (v) 2 cos 2θ = 1 + cos θ Solve the following equations for −π $ θ $ π. (i) sin 2θ = 2 sin θ (ii) tan 2θ = 2 tan θ (iii) cos 2θ − cos θ = 0 (iv) 1 + cos 2θ = 2 sin2 θ (v) sin 4θ = cos 2θ Tip: Write the expression in part (v) as an equation in 2θ. By first writing sin 3θ as sin(2θ + θ ), express sin 3θ in terms of sin θ. Hence solve the equation sin 3θ = sin θ for 0 $ θ $ 2π. Solve cos 3θ = 1 − 3 cos θ for 0° $ θ $ 360°. 1 + cos 2θ Simplify . sin 2θ 70 9781510421738.indb 70 02/02/18 1:12 PM CP 6 (i) Sketch on the same axes the graphs of y = cos 2x and y = sin x for 0 $ x $ 2π. (ii) Show that these curves meet at points whose x coordinates are solutions of the equation 2 sin2 x + sin x − 1 = 0. 3 (iii) Solve this equation to find the values of x in terms of π for CP 7 PS 8 CP 9 Show that 1 – tan 2 θ = cos 2θ. 1 + tan θ (i) Show that tan π + θ tan π – θ = 1. 4 4 (ii) Given that tan 26.6° = 0.5, solve tan θ = 2 without using your calculator. Give θ to 1 decimal place, where 0° " θ " 90°. (i) Sketch on the same axes the graphs of y = cos 2x and y = 3 sin x − 1 for 0 $ x $ 2π. 2 ( ) ( ) 3.3 Double-angle formulae 0 $ x $ 2π. Show that these curves meet at points whose x coordinates are solutions of the equation 2 sin2 x + 3 sin x − 2 = 0. (iii) Solve this equation to find the values of x in terms of π for 0 $ x $ 2π. 10 (i) Show that cos(θ − 60°) + cos(θ + 60°) ≡ cos θ . cos(2x − 60°) + cos(2x + 60°) (ii) Given that = 3 , find the exact value cos( x − 60°) + cos( x + 60°) of cos x. (ii) Cambridge International AS & A Level Mathematics 9709 Paper 33 Q4 November 2014 11 Solve the equation tan 2x = 5cot x , for 0° < x < 180°. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q3 June 2013 12 (i) (ii) Prove the identity cosec 2θ + cot 2θ ≡ cot θ. Hence solve the equation cosec 2θ + cot 2θ = 2, for 0° $ θ $ 360°. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q3 June 2009 13 It is given that cos a = 53 , where 0° $ a $ 90°. Showing your working and without using a calculator to evaluate a, (i) find the exact value of sin(a − 30)°, (ii) find the exact value of tan 2a, and hence find the exact value of tan 3a. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q3 June 2010 71 9781510421738.indb 71 02/02/18 1:12 PM 3 3.4 The forms r cos(θ ± α), r sin(θ ± α) Another modification of the compound-angle formulae allows you to simplify expressions such as 4 sin θ + 3 cos θ and hence solve equations of the form a sin θ + b cos θ = c. 3 TRIGONOMETRY To find a single expression for 4 sin θ + 3 cos θ, you match it to the expression r sin(θ + α) = r (sin θ cos α + cos θ sin α). This is because the expansion of r sin(θ + α) has sin θ in the first term, cos θ in the second term and a plus sign in between them. It is then possible to choose appropriate values of r and α. 4 sin θ + 3 cos θ ≡ r(sin θ cos α + cos θ sin α) Coefficients of sin θ : Coefficients of cos θ : 4 = r cos α 3 = r sin α. Looking at the right-angled triangle in Figure 3.9 gives the values for r and α. r r = 32 + 4 2 =5 The sides, 4 and 3, come from the expression 4 sin θ + 3 cos θ. 3 α 4 ▲ Figure 3.9 In this triangle, the hypotenuse is 4 2 + 3 2 = 5, which corresponds to r in the expression above. The angle α is given by sin α = 3 5 and cos α = 4 5 ⇒ α = 36.9°. So the expression becomes 4 sin θ + 3 cos θ = 5 sin(θ + 36.9°). The steps involved in this procedure can be generalised to write a sin θ + b cos θ = r sin(θ + α) where b a =b cos α = =a a2 + b2 r a2 + b2 r The same expression may also be written as a cosine function. In this case, rewrite 4 sin θ + 3 cos θ as 3 cos θ + 4 sin θ and notice that: r = a2 + b2 72 9781510421738.indb 72 sin α = (i) The expansion of cos(θ − β ) starts with cos θ … just like the expression 3 cos θ + 4 sin θ. (ii) The expansion of cos(θ − β ) has + in the middle, just like the expression 3 cos θ + 4 sin θ. 02/02/18 1:12 PM The expansion of r cos(θ − β ) is given by 3 r cos(θ − β ) = r (cos θ cos β + sin θ sin β ). To compare this with 3 cos θ + 4 sin θ, look at the triangle in Figure 3.10, in which r= 32 + 4 2 = 5 cos β = 53 sin β = 45 r β = 53.1°. 3.4 The forms r cos(θ ± α), r sin(θ ± α) r = 32 + 4 2 =5 ⇒ 4 β 3 ▲ Figure 3.10 This means that you can write 3 cos θ + 4 sin θ in the form r cos(θ − β ) = 5 cos(θ − 53.1°). The procedure used here can be generalised to give the result a cos θ + b sin θ = r cos(θ − α) where r= a cos α = r a2 + b2 sin α = br . Note The value of r will always be positive, but cos α and sin α may be positive or negative, depending on the values of a and b. In all cases, it is possible to find an angle α for which −180° " α " 180°. You can derive alternative expressions of this type based on other compoundangle formulae if you wish α to be an acute angle, as is done in the next example. Example 3.6 (i) (ii) Express 3 sin θ − cos θ in the form r sin(θ − α), where r # 0 and π 0 " α " 2. State the maximum and minimum values of 3 sin θ − cos θ. (iii) Sketch the graph of y = (iv) Solve the equation 3 sin θ − cos θ for 0 $ θ $ 2π. 3 sin θ − cos θ = 1 for 0 $ θ $ 2π. ➜ 73 9781510421738.indb 73 02/02/18 1:12 PM 3 Solution (i) r sin(θ − α ) = r(sin θ cos α − cos θ sin α ) = (r cos α )sin θ − (r sin α )cos θ Comparing this with 3 sin θ − cos θ, the two expressions are identical if 3 TRIGONOMETRY r cos α = 3 r sin α = 1. and From the triangle in Figure 3.11 r = 1+ 3 = 2 and tan α = 1 3 π ⇒ α=6 π 3 sin θ − cos θ = 2 sin(θ − 6 ). The sine function oscillates between 1 and −1, π so 2 sin(θ − 6 ) oscillates between 2 and −2. so (ii) r Maximum value = 2. 1 α 3 Minimum value = −2. (iii) To sketch the curve y = 2 sin(θ − π 6 ), notice that ▲ Figure 3.11 » it is a sine curve » its y values go from −2 to 2 π 7π 13π » it crosses the horizontal axis where θ = 6 , 6 , 6 ,… The curve is shown in Figure 3.12. y 2 O π 6 π 2 π 7π 6 3π 2 2π 13π 6 5π 2 θ –2 ▲ Figure 3.12 3 sin θ − cos θ = 1 is equivalent to π 2 sin(θ − 6 ) = 1 π ⇒ sin(θ − 6 ) = 21 π Let x = (θ − 6 ) and solve sin x = 21. π Solving sin x = 21 gives x = 6 (principal value) π or x = π − 6 = 5π (from the graph in Figure 3.13) 6 π π π 5π + π = π. θ= + = θ= or giving 6 6 3 6 6 (iv) The equation 74 9781510421738.indb 74 02/02/18 1:12 PM 3 y y = sin x π O π 6 π– π 6 x π The roots in 0 $ θ $ 2π are θ = 3 and π. Always check (for example by reference to a sketch graph) that the number of roots you have found is consistent with the number you are expecting. When solving equations of the form sin(θ − α ) = c by considering sin x = c, it is sometimes necessary to go outside the range specified for θ since, for example, 0 $ θ $ 2π is the same as −α $ x $ 2π − α. 3.4 The forms r cos(θ ± α), r sin(θ ± α) ▲ Figure 3.13 Using these forms There are many situations which produce expressions which can be tidied up using these forms. They are also particularly useful for solving equations involving both the sine and cosine of the same angle. The fact that a cos θ + b sin θ can be written as r cos(θ − α ) is an illustration of the fact that any two waves of the same frequency, whatever their amplitudes, can be added together to give a single combined wave, also of the same frequency. Exercise 3D 1 Express each of the following in the form r cos(θ − α ), where r # 0 and 0° " α " 90°. (i) cos θ + sin θ (ii) 20 cos θ + 21 sin θ (iii) cos θ + 3 sin θ (iv) 5 cos θ + 2 sin θ 2 Express each of the following in the form r cos(θ + α ), where r # 0 and 0 " α " π. 2 (i) cos θ − sin θ (ii) 3 cos θ − sin θ Express each of the following in the form r sin(θ + α ), where r # 0 and 0° " α " 90°. (i) sin θ + 2 cos θ (ii) 2 sin θ + 5 cos θ 3 75 9781510421738.indb 75 02/02/18 1:12 PM 4 3 3 TRIGONOMETRY 5 6 7 8 9 PS Express each of the following in the form r sin(θ − α ), where r # 0 and 0 " α " π. 2 (i) sin θ − cos θ (ii) 7 sin θ − 2 cos θ Express each of the following in the form r cos(θ − α ), where r # 0 and −180° " α " 180°. (i) cos θ − 3 sin θ (ii) 2 2 cos θ − 2 2 sin θ (iii) sin θ + 3 cos θ (iv) 5 sin θ + 12 cos θ (v) sin θ − 3 cos θ (vi) 2 sin θ − 2 cos θ (i) Express 5 cos θ − 12 sin θ in the form r cos(θ + α ), where r # 0 and 0° " α " 90°. (ii) State the maximum and minimum values of 5 cos θ − 12 sin θ. (iii) Sketch the graph of y = 5 cos θ − 12 sin θ for 0° $ θ $ 360°. (iv) Solve the equation 5 cos θ − 12 sin θ = 4 for 0° $ θ $ 360°. (i) Express 3 sin θ − 3 cos θ in the form r sin(θ − α ), where r # 0 and 0 " α " π. 2 (ii) State the maximum and minimum values of 3 sin θ − 3 cos θ and the smallest positive values of θ for which they occur. (iii) Sketch the graph of y = 3 sin θ − 3 cos θ for 0 $ θ $ 2π. (iv) Solve the equation 3 sin θ − 3 cos θ = 3 for 0 $ θ $ 2π. (i) Express 2 sin 2θ + 3 cos 2θ in the form r sin(2θ + α), where r # 0 and 0° < α < 90°. (ii) State the maximum and minimum values of 2 sin 2θ + 3 cos 2θ and the smallest positive values of θ for which they occur. (iii) Sketch the graph of y = 2 sin 2θ + 3 cos 2θ for 0° $ θ $ 360°. (iv) Solve the equation 2 sin 2θ + 3 cos 2θ = 1 for 0° $ θ $ 360°. (i) Express cos θ + 2 sin θ in the form r cos(θ − α ), where r # 0 and 0° < α < 90°. (ii) State the maximum and minimum values of cos θ + 2 sin θ and the smallest positive values of θ for which they occur. (iii) Sketch the graph of y = cos θ + 2 sin θ for 0° $ θ $ 360°. (iv) State the maximum and minimum values of 1 3 + cos θ + 2 sin θ and the smallest positive values of θ for which they occur. 10 The diagram opposite shows a table stuck in a corridor. The table is 120 cm long and 80 cm wide, and the width of the corridor is 130 cm. (i) Show that 12 sin θ + 8 cos θ = 13. (ii) Hence find the angle θ. (There are two answers.) 76 9781510421738.indb 76 02/02/18 1:12 PM 80 cm 3 130 cm θ 11 (i) (ii) Express 3 cos x + 4 sin x in the form R cos(x − α), where R # 0 and 0° " α " 90°, stating the exact value of R and giving the value of α correct to 2 decimal places. Hence solve the equation 3 cos x + 4 sin x = 4.5, giving all solutions in the interval 0° " x " 360°. Cambridge International AS & A Level Mathematics 9709 Paper 22 Q6 November 2009 Express 5 cos θ − sin θ in the form R cos(θ + α ), where R # 0 and 0° " α " 90°, giving the exact value of R and the value of α correct to 2 decimal places. (ii) Hence solve the equation 5 cos θ − sin θ = 4, giving all solutions in the interval 0° $ θ $ 360°. 12 (i) 3.4 The forms r cos(θ ± α), r sin(θ ± α) 120 cm Cambridge International AS & A Level Mathematics 9709 Paper 2 Q5 June 2008 Express 7 cos θ + 24 sin θ in the form R cos(θ − α ), where R # 0 and 0° " α " 90°, giving the exact value of R and the value of α correct to 2 decimal places. (ii) Hence solve the equation 7 cos θ + 24 sin θ = 15, giving all solutions in the interval 0° $ θ $ 360°. 13 (i) Cambridge International AS & A Level Mathematics 9709 Paper 3 Q4 June 2006 14 By expressing 8 sin θ − 6 cos θ in the form R sin(θ − α), solve the equation 8 sin θ − 6 cos θ = 7, for 0° $ θ $ 360°. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q5 November 2005 77 9781510421738.indb 77 02/02/18 1:12 PM 3 INVESTIGATION Currents The simplest alternating current is one which varies with time t according to I = A sin 2πft, 3 TRIGONOMETRY where f is the frequency and A is the maximum value. The frequency of the public AC supply is 50 hertz (cycles per second). Investigate what happens when two alternating currents A1 sin 2πft and A2 sin(2πft + α) with the same frequency f but a phase difference of α are added together. The previous exercises have each concentrated on just one of the many trigonometrical techniques which you will need to apply confidently. The following exercise requires you to identify which technique is the correct one. Exercise 3E 1 Simplify the following. (i) 2 sin 3θ cos 3θ (ii) (iii) cos2 3θ + sin2 3θ (iv) 1 − 2 sin2 sin(θ − α)cos α + cos(θ − α)sin α sin 2θ (vii) 2 sin θ Express (i) (cos x − sin x)2 in terms of sin 2x (ii) cos4 x − sin4 x in terms of cos 2x (iii) 2 cos2 x − 3 sin2 x in terms of cos 2x. Prove that 1 − cos 2θ 2 (i) 1 + cos 2θ ≡ tan θ (vi) 3 sin θ cos θ (v) 2 CP 3 cos2 3θ − sin2 3θ (θ2 ) (viii) cos 2θ − 2 cos2 θ cosec 2θ + cot 2θ ≡ cot θ 4t(1 − t 2 ) (iii) tan 4θ ≡ where t = tan θ. 1 − 6t 2 + t 4 Solve the following equations. (i) sin(θ + 40°) = 0.7 0° $ θ $ 360° 2 (ii) 3 cos θ + 5 sin θ − 1 = 0 0° $ θ $ 360° π (iii) 2 cos(θ − ) = 1 −π $ θ $ π 6 (iv) cos(45° − θ ) = 2 sin(30° + θ ) −180° $ θ $ 180° (v) cos 2θ + 3 sin θ = 2 0 $ θ $ 2π (vi) cos θ + 3 sin θ = 2 0° $ θ $ 360° (vii) tan2 θ − 3 tan θ − 4 = 0 0° $ θ $ 180° (ii) 4 78 9781510421738.indb 78 02/02/18 1:12 PM e 3.5 The general solutions of trigonometrical equations 3 The equation tan θ = 1 has infinitely many roots: …, −315°, −135°, 45°, 225°, 405°, … (in degrees) 5π , 9π , … (in radians). 4 4 Only one of these roots, namely 45° or π , is denoted by the function tan−1 1. 4 This is the value which your calculator will give you. It is the principal value. The principal value for any inverse trigonometrical function is unique and lies within a specified range: − π " tan−1 x " π 2 2 π π − $ sin−1 x $ 2 2 0 $ cos−1 x $ π. It is possible to deduce all other roots from the principal value and this is shown below. To solve the equation tan θ = c, notice how all possible values of θ occur at intervals of 180° or π radians (see Figure 3.14). So the general solution is θ = tan−1 c + nπ n∈ 3.5 The general solutions of trigonometrical equations …, − 7π , − 3π , − π , 4 4 4 (in radians). y y = tan θ c –90° – π 2 ▲ Figure 3.14 90° O π 2 270° 3π 2 θ tan−1 c is the principal value. 79 9781510421738.indb 79 02/02/18 1:12 PM The cosine graph (see Figure 3.15) has the y-axis as a line of symmetry. Notice how the values ±cos−1 c generate all the other roots at intervals of 360° or 2π. So the general solution is 3 θ = ±cos−1 c + 2nπ n∈ (in radians). y 3 TRIGONOMETRY y = cos θ c –450° –270° – 5π 2 –90° 90° –π O 2 – 3π 2 −cos–1 c ▲ Figure 3.15 270° 3π 2 π 2 450° 5π 2 630° 7π 2 θ cos–1 c is the principal value. Now look at the sine graph (see Figure 3.16). As for the cosine graph, there are two roots located symmetrically. The line of symmetry for the sine graph is θ = π , which generates all the other possible roots. This gives rise to the 2 slightly more complicated expressions π π θ = ± ( − sin−1 c) + 2nπ 2 2 π θ = (2n + 21 )π ± ( − sin−1 c) 2 or n∈ . You may, however, find it easier to remember these as two separate formulae: θ = 2nπ + sin−1 c θ = (2n + 1)π − sin−1 c. or y y = sin θ c –540° –3π –360° –2π ▲ Figure 3.16 –180° –π 180° O sin−1 c is the principal value. π 360° 540° 2π 3π θ (180° − sin−1 c) or (π − sin−1 c) ACTIVITY 3.3 Show that the general solution of the equation sin θ = c may also be written θ = nπ + (−1)n sin−1 c. 80 9781510421738.indb 80 02/02/18 1:12 PM KEY POINTS 1 2 3 1 1 cosec θ = sin θ ; cot θ = tan θ tan2 θ + 1 = sec2 θ; 1 + cot2 θ = cosec2 θ Compound-angle formulae ● sin(θ + φ) = sin θ cos φ + cos θ sin φ ● sin(θ − φ) = sin θ cos φ − cos θ sin φ ● cos(θ + φ) = cos θ cos φ − sin θ sin φ ● cos(θ − φ) = cos θ cos φ + sin θ sin φ tan θ + tan φ ● tan(θ + φ) = (θ + φ) ≠ 90°, 270°, ... 1 − tan θ tan φ tan θ − tan φ ● tan(θ − φ) = (θ − φ) ≠ 90°, 270°, ... 1 + tan θ tan φ Double-angle and related formulae ● sin 2θ = 2 sin θ cos θ ● cos 2θ = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1 2 tan θ ● tan 2θ = 1 − tan 2 θ θ ≠ 45°, 135°, ... ● 1 cos2 θ = 2(1 + cos 2θ ) The r, α formulae ● a sin θ + b cos θ = r sin(θ + α ) ● a sin θ − b cos θ = r sin(θ − α ) ● a cos θ + b sin θ = r cos(θ − α ) ● a cos θ − b sin θ = r cos(θ + α ) ● 5 1 sin2 θ = 2(1 − cos 2θ ) } where r = a 2 + b 2 a cos α = r sin α = br 3 3.5 The general solutions of trigonometrical equations 4 1 sec θ = cos θ ; 81 9781510421738.indb 81 02/02/18 1:12 PM 3 TRIGONOMETRY 3 LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ understand and use ■ the sec, cosec and cot functions ■ the relationships between the graphs of the sin, cos, tan, cosec, sec and cot functions ■ use the identities ■ sin2 θ + cos2 θ ≡ 1 ■ tan2 θ + 1 ≡sec2 θ ■ cot2 θ + 1 ≡cosec2 θ ■ solve equations involving sec, cosec and cot ■ understand and use the identities for sin (A ± B ), cos (A ± B ), tan (A ± B ) ■ know and use the identities for sin 2A, cos 2A, tan 2A ■ write a cos θ + b sin θ in the equivalent forms r sin (θ ± α ) and r cos (θ ± α ) ■ use for finding maximum and minimum values ■ use for solving equations ■ use trigonometrical identities ■ in solving equations ■ in proofs ■ to solve problems. 82 9781510421738.indb 82 02/02/18 1:12 PM P2 P3 4 Differentiation 4 Differentiation Nothing takes place in the world whose meaning is not that of some maximum or some minimum. Leonhard Euler (1707–1783) Displacement Many physical systems, such as a simple pendulum or swing or a mass on an elastic spring, can be modelled as having displacement–time graphs which have a sine wave shape. O Time ▲ Figure 4.1 To be able to perform calculations involving velocity and acceleration for these systems, you need to be able to differentiate the sine function. ? ❯ Think of some other situations in which it would be useful to be able to differentiate functions other than polynomials. 83 9781510421738.indb 83 02/02/18 1:12 PM 4 DIFFERENTIATION 4 4.1 The product rule Figure 4.2 shows a sketch of the curve of y = 20x(x − 1)6. y x ▲ Figure 4.2 If you wanted to find the gradient function, dy , for the curve, you could dx expand the right-hand side then differentiate it term by term – a long and cumbersome process! There are other functions like this, made up of the product of two or more simpler functions, which are not just time-consuming to expand – they are impossible to expand. One such function is 1 y = ( x − 1) 2 ( x + 1)6 (for x > 1). Clearly you need a technique for differentiating functions that are products of simpler ones, and a suitable notation with which to express it. The most commonly used notation involves writing y = uv, dy where the variables u and v are both functions of x. Using this notation, is dx given by dy dv du =u +v . dx dx dx This is called the product rule and it is derived from first principles in the next section. The product rule from first principles A small increase δx in x leads to corresponding small increases δu, δv and δy in u, v and y. And so y + δy = (u + δu)(v + δv) = uv + v δu + u δv + δu δv. Since y = uv, the increase in y is given by δy = v δu + u δv + δu δv. 84 9781510421738.indb 84 02/02/18 1:12 PM Example 4.1 Given that y = (2x + 3)(x 2 − 5), find dy using the product rule. dx Solution 4 4.1 The product rule Dividing both sides by δx, δy = v δu + u δv + δu δv . δx δx δx δx In the limit, as δx → 0, so do δu, δv and δy, and dy δy δu δv . → du , → dv and → δx δ x δ x dx dx dx The expression becomes dy = v du + u dv . dx dx dx Notice that since δu → 0, the last term on the right-hand side has disappeared. y = (2x + 3)(x 2 − 5) Let u = 2x + 3 and v = x 2 − 5. du dv Then = 2 and = 2x. dx dx dy Using the product rule, = v du + u dv dx dx dx = (x 2 − 5) × 2 + (2x + 3) × 2x = 2(x 2 − 5 + 2x 2 + 3x) = 2(3x 2 + 3x − 5) Note In this case you could have multiplied out the expression for y. y = 2x 3 + 3x 2 − 10x − 15 dy = 6x 2 + 6x − 10 dx = 2(3x 2 + 3x − 5) Example 4.2 Differentiate y = 20x(x − 1)6. Figure 4.2 shows the graph of this function. Solution Let u = 20x and v = (x − 1)6. du Then = 20 and dv = 6(x − 1)5 (using the chain rule). 20(x − 1)5 is a dx dx common factor. dy = v du + u dv Using the product rule, dx dx dx = (x − 1)6 × 20 + 20x × 6(x − 1)5 = 20(x − 1)5 × (x − 1) + 20(x − 1)5 × 6x = 20(x − 1)5[(x − 1) + 6x] = 20(x − 1)5(7x − 1) 85 9781510421738.indb 85 02/02/18 1:12 PM 4 The factorised result is the most useful form for the solution, as it allows you to find stationary points easily.You should always try to factorise your answer as much as possible. Once you have used the product rule, look for factors straight away and do not be tempted to multiply out. 4 DIFFERENTIATION 4.2 The quotient rule In the last section, you met a technique for differentiating the product of two functions. In this section you will see how to differentiate a function which is the quotient of two simpler functions. As before, you start by identifying the simpler functions. For example, the function y = 3x + 1 (for x ≠ 2) x−2 u can be written as y = v where u = 3x + 1 and v = x − 2. Using this notation, dy is given by dx v du − u dv dy dx dx = dx v2 This is called the quotient rule and it is derived from first principles in the next section. The quotient rule from first principles A small increase, δx in x results in corresponding small increases δu, δv and δy in u, v and y. The new value of y is given by y + δy = u + δu v + δv u and since y = , you can rearrange this to obtain an expression for δy in v terms of u and v. u + δu − u δy = v + δv v v(u + δu ) − u(v + δv ) = v(v + δv ) uv + v δu − uv − u δv = v(v + δv ) v δu − u δv = v(v + δv ) Dividing both sides by δx gives To divide the right-hand δu δv side by δx you only divide the v −u δy δx δx numerator by δx. = δx v(v + δv ) 86 9781510421738.indb 86 In the limit as δx → 0, this is written in the form you met above. dv du dy v dx − u dx = dx v2 02/02/18 1:12 PM ACTIVITY 4.1 Verify that the quotient rule gives Example 4.3 Given that y = dy correctly when u = x 10 and v = x 7. dx dy 3x + 1 , find using the quotient rule. x−2 dx Letting u = 3x + 1 and v = x − 2 gives du dv dx = 3 and dx = 1. du dv dy v dx − u dx Using the quotient rule, = dx v2 ( x − 2) × 3 − (3x + 1) × 1 = ( x − 2) 2 = 3x − 6 − 3x2 − 1 ( x − 2) = 4.2 The quotient rule Solution Example 4.4 4 −7 ( x − 2) 2 2 Given that y = x + 1 , find dy using the quotient rule. 3x − 1 dx Solution Letting u = x 2 + 1 and v = 3x − 1 gives du dv = 2x and = 3. dx dx du dv dy v dx − u dx Using the quotient rule, = dx v2 (3x − 1) × 2x − ( x 2 + 1) × 3 = (3x − 1) 2 2 2 = 6x − 2x − 3x2 − 3 (3x − 1) 2 = 3x − 2x −2 3 (3x − 1) Exercise 4A 9781510421738.indb 87 1 Differentiate the following using the product rule or the quotient rule. (i) y = (x 2 − 1)(x 3 + 3) (ii) y = x 5(3x 2 + 4x − 7) 2x (iii) y = x 2(2x + 1)4 (iv) y = 3 x −1 x3 (v) y = 2 (vi) y = (2x + 1)2(3x 2 − 4) x +1 x−2 2x − 3 (vii) y = (viii) y = 2 ( x + 3) 2 2x + 1 (ix) y = ( x + 1) x − 1 87 02/02/18 1:12 PM 4 DIFFERENTIATION 4 CP 2 3 PS 4 PS 5 The diagram shows the graph of y = x . x −1 dy (i) Find . dx y (ii) Find the gradient of the curve at (0, 0), and the equation of the tangent at (0, 0). (iii) Find the gradient of 1 the curve at (2, 2), and the equation of the tangent at (2, 2). O 1 x (iv) What can you deduce about the two tangents? Given that y = (x + 1)(x − 2)2 (i) find dy dx (ii) find any stationary points and determine their nature (iii) sketch the curve. x−3 Given that y = x−4 (i) find dy dx (ii) find the equation of the tangent to the curve at the point (6, 1.5) (iii) find the equation of the normal to the curve at the point (5, 2) (iv) use your answer from part (i) to deduce that the curve has no stationary points, and sketch the graph. 2x The diagram shows the graph of y = , which is undefined for x −1 x < 0 and x = 1. P is a minimum point. y P O x 88 9781510421738.indb 88 02/02/18 1:12 PM (b) the normal found in part (iv), call it R. (vi) Show that the area of the triangle PQR is 441 . 8 6 2 The diagram shows the graph of y = x − 2x − 5 . 2x + 3 4 4.2 The quotient rule Find dy . dx (ii) Find the gradient of the curve at (9, 9), and show that the equation of the normal at (9, 9) is y = −4x + 45. (iii) Find the coordinates of P and verify that it is a minimum point. (iv) Write down the equation of the tangent and the normal to the curve at P. (v) Write down the point of intersection of the normal found in part (ii) and (a) the tangent found in part (iv), call it Q (i) y –1.5 O x Find dy . dx (ii) Use your answer from part (i) to find any stationary points of the curve. (iii) Classify each of the stationary points and use calculus to justify your answer. x2 A curve has the equation y = . 2x + 1 Find the coordinates of the stationary points on the curve and identify their nature. (i) 7 89 9781510421738.indb 89 02/02/18 1:12 PM 4 DIFFERENTIATION 4 PS 8 The diagram shows part of the graph with the equation y = x 9 − 2x 2 . It crosses the x axis at (a, 0). (i) Find the value of a, giving your answer as a multiple of 2. y Show that the result of O (ii) (a, 0) x differentiating 9 − 2x 2 is −2x . 9 − 2x 2 Hence show that if y = x 9 − 2x 2 then 2 dy = 9 − 4x . dx 9 − 2x 2 (iii) Find the x coordinate of the maximum point on the graph of y = x 9 − 2x 2 . Write down the gradient of the curve at the origin. What can you say about the gradient at the point (a, 0)? 4.3 Differentiating natural logarithms and exponentials 1 In Chapter 2 you learnt that the integral of x is ln x. It follows, therefore, that the differential of ln x is 1. x So y = ln x ⇒ dy = 1 dx x The differential of the inverse function, y = e x, may be found by interchanging y and x. dx 1 x = ln y ⇒ dy = y dy 1 ⇒ dx = dx = y = e x . dy Therefore d e x = e x . dx The differential of ex is itself ex. This may at first seem rather surprising. ? The function f(x) (x ! ") is a polynomial in x of order n. So f(x) = anx n + an−1x n−1 + ... + a1x + a0 CP where an, an−1, ..., a0 are all constants and at least an is not zero. d f(x) cannot equal f(x)? ❯ How can you prove that dx 90 9781510421738.indb 90 02/02/18 1:12 PM Since the differential of ex is ex, it follows that the integral of ex is also ex. ∫ ex dx = ex + c. This may be summarised as in the following table. Differentiation Integration dy dx y ⎯→ y dx y ⎯→ ∫ 1 ⎯→ ln x + c x ex ⎯→ ex ex ⎯→ ex + c Differentiate y = e5x. Solution Make the substitution u = 5x to give y = eu. dy = eu = e 5x and du = 5. Now dx du d u d y d y By the chain rule, = × dx du dx = e5x × 5 4.3 Differentiating natural logarithms and exponentials ln x ⎯→ 1 x These results allow you to extend very considerably the range of functions which you are able to differentiate and integrate. Example 4.5 4 = 5e5x This result can be generalised as follows. y = eax ⇒ dy = ae ax dx where a is any constant. This is an important standard result, and you would normally use it automatically, without recourse to the chain rule. Example 4.6 Differentiate y = 42x . e Solution ⇒ 4 y = e 2 x = 4e −2 x dy = 4 × (−2e −2 x ) dx = −8e−2x 91 9781510421738.indb 91 02/02/18 1:12 PM 4 Example 4.7 Differentiate y = 3e(x2+1). Solution Let u = x 2 + 1, then y = 3eu. ⇒ 4 DIFFERENTIATION By the chain rule, dy = 3e u = 3e ( x 2 +1) du and du dx = 2x dy dy du = × dx du dx = 3e(x2+1) × 2x = 6x e (x2+1) Example 4.8 Differentiate the following. y = 2ln x (i) (ii) y = ln(3x) Solution (i) (ii) dy = 2× 1 x dx 2 = x Let u = 3x, then y = ln u dy 1 = = 1 and du u 3x By the chain rule, dy dy du = × dx du dx = 1 ×3 3x =1 x ⇒ du = 3 dx Note An alternative solution to part (ii) is y = ln(3x) = ln 3 + ln x ⇒ dy = 0 + 1 = 1. x x dx ❯ The gradient function found in part (ii) above for y = ln(3x) is the same as that for y = ln x. What does this tell you about the shapes of the two curves? For what values of x is it valid? ? 92 9781510421738.indb 92 02/02/18 1:12 PM Example 4.9 Differentiate the following. (i) y = ln(x4) (ii) 4 y = ln(x2 + 1) Solution (i) By the properties of logarithms (ii) Let u = x 2 + 1, then y = ln u dy 1 = = 1 and du u x 2 + 1 dy dy du By the chain rule, dx = du × dx 1 = x 2 + 1 × 2x 2x = x2 + 1 ⇒ du = 2x dx If you need to differentiate expressions similar to those in the examples above, follow exactly the same steps. The results can be generalised as follows. dy a = dx x dy 1 = y = ln(ax) ⇒ dx x dy f '( x ) = y = ln(f(x)) ⇒ dx f( x ) y = a ln x ⇒ Example 4.10 Differentiate y = dy = ae x dx dy = ae ax y = eax ⇒ dx dy = f '( x )e f(x ) y = ef(x) ⇒ dx y = aex ⇒ 4.3 Differentiating natural logarithms and exponentials ⇒ y = ln(x 4) = 4 ln(x) dy 4 = dx x ln x . x Solution u where u = ln x and v = x v du = 1 and dv = 1. ⇒ dx x dx du dv dy v dx − u dx By the quotient rule, = dx v2 1− x × x 1 × ln x = x2 1 − ln x = x2 Here y is of the form 9781510421738.indb 93 93 02/02/18 1:12 PM 4 Exercise 4B 1 Differentiate the following. (i) y = 3 ln x (ii) 2 (iii) y = ln(x ) (iv) 1 (v) y = ln (vi) x (vii) y = x 2 ln(4x) (viii) ( ) 4 DIFFERENTIATION (ix) y = ln x 2 − 1 (x) 2 Differentiate the following. (i) y = 3ex (iii) y = ex y = x e4x x (vii) y = x e (v) M 2 y = ln(4x) y = ln(x 2 + 1) y = x ln x y = ln x + 1 x ln x y= 2 x ( ) (ii) y = e 2x (iv) y = e(x +1) (vi) y = 2x 3e −x y = (e2x + 1)3 (viii) 2 3 Knowing how much rain has fallen in a river basin, hydrologists are often able to give forecasts of what will happen to a river level over the next few hours. In one case it is predicted that the height h, in metres, of a river above its normal level during the next 3 hours will be 0.12e0.9t, where t is the time elapsed, in hours, after the prediction. dh (i) Find dt , the rate at which the river is rising. (ii) At what rate will the river be rising after 0, 1, 2 and 3 hours? 4 The graph of y = xex is shown below. y O x P (i) (ii) CP 5 dy d 2y Find and 2 . dx dx Find the coordinates of the minimum point P. The graph of f(x) = x ln(x 2) is shown below. x O x 94 9781510421738.indb 94 02/02/18 1:12 PM 6 7 y O x x You are given that the curve y = e x has one stationary point. (i) Find the exact coordinates of this point. (ii) Determine whether this point is a maximum or a minimum point. ln x 9 Given that f(x) = ,x > 0 x (i) write down the coordinates of the point where the graph of y = f(x) crosses the x-axis (ii) find the exact coordinates of the stationary point of the curve y = f(x) (iii) sketch the curve of y = f(x). − 1x 10 Find the exact coordinates of the point on the curve y = xe 2 at d 2y which 2 = 0. dx 8 4 4.3 Differentiating natural logarithms and exponentials Describe, giving a reason, any symmetries of the graph. (ii) Find f ′(x) and f ″(x). (iii) Find the coordinates of any stationary points. ex Given that y = x dy (i) find dx (ii) find the coordinates of any stationary points on the curve (iii) sketch the curve. The graph of f(x) = x ln x is shown below. Find the exact coordinates of the stationary point. (i) Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 November 2008 11 It is given that the curve y = (x − 2)ex has one stationary point. (i) (ii) Find the exact coordinates of this point. Determine whether this point is a maximum or a minimum point. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 June 2008 95 9781510421738.indb 95 02/02/18 1:12 PM 4 12 The equation of a curve is y = x3e–x. (i) (ii) Show that the curve has a stationary point where x = 3. Find the equation of the tangent to the curve at the point where x = 1. 4 DIFFERENTIATION Cambridge International AS & A Level Mathematics 9709 Paper 22 Q5 June 2010 4.4 Differentiating trigonometrical functions ACTIVITY 4.2 Figure 4.3 shows the graph of y = sin x, with x measured in radians, together with the graph of y = x. You are going to sketch the graph of the gradient function for the graph of y = sin x. y=x y 1 y = sin x –π 2 –2π – 3π 2 3π 2 –π 0 π 2 π 2π x –1 ▲ Figure 4.3 Draw a horizontal axis for the angles, marked from −2π to 2π, and a vertical axis for the gradient, marked from −1 to 1, as shown in Figure 4.4. dy dx 1 –2π –π 0 π 2π x –1 ▲ Figure 4.4 First, look for the angles for which the gradient of y = sin x is zero. Mark zeros at these angles on your gradient graph. 96 9781510421738.indb 96 02/02/18 1:12 PM Decide which parts of y = sin x have a positive gradient and which have a negative gradient. This will tell you whether your gradient graph should be above or below the x-axis at any point. 4 Look at the part of the graph of y = sin x near x = 0 and compare it with the graph of y = x. What do you think the gradient of y = sin x is at this point? Mark this point on your gradient graph. Also mark on any other points with plus or minus the same gradient. 4.4 Differentiating trigonometrical functions Now, by considering whether the gradient of y = sin x is increasing or decreasing at any particular point, sketch in the rest of the gradient graph. The gradient graph that you have drawn should look like a familiar graph. What graph do you think it is? Sketch the graph of y = cos x, with x measured in radians, and use it as above to obtain a sketch of the graph of the gradient function of y = cos x. ❯ Is y = x still a tangent of y = sin x if x is measured in degrees? ? Activity 4.2 showed you that the graph of the gradient function of y = sin x resembled the graph of y = cos x. You will also have found that the graph of the gradient function of y = cos x looks like the graph of y = sin x reflected in the x-axis to become y = −sin x. ? ❯ Both of these results are in fact true but the work above does not amount to a proof. Explain why. CP Summary of results d (sin x) = cos x dx d (cos x) = −sin x dx Remember that these results are only valid when the angle is measured in radians, so when you are using any of the derivatives of trigonometrical functions you need to work in radians. ACTIVITY 4.3 sin x By writing tan x = cos x , use the quotient rule to show that d (tan x ) = 2 sec x where x is measured in radians. dx You can use the three results met so far to differentiate a variety of functions involving trigonometrical functions, by using the chain rule, product rule or quotient rule, as in the following examples. 9781510421738.indb 97 97 02/02/18 1:12 PM Differentiate y = cos 2x. Solution As cos 2x is a function of a function, you may use the chain rule. du Let u = 2x ⇒ dx = 2 dy = −sin u y = cos u ⇒ du dy dy du = × dx du dx 4 DIFFERENTIATION 4 Example 4.11 = −sin u × 2 = −2 sin 2x With practice it should be possible to do this in your head, without needing to write down the substitution. This result may be generalised. y = cos kx ⇒ dy = −k sin kx dx y = sin kx ⇒ dy = k cos kx dx y = tan kx ⇒ dy = k sec2 kx dx Similarly and Example 4.12 Differentiate y = x 2 sin x. Solution x 2 sin x is of the form uv, so the product rule can be used with u = x 2 and v = sin x. du = 2 x dv = cos x dx dx Using the product rule dy du dv =v +u dx dx dx dy ⇒ = 2x sin x + x 2 cos x dx 98 9781510421738.indb 98 02/02/18 1:12 PM Example 4.13 Differentiate y = etanx. 4 Solution = eu sec2 x = etanx sec2 x Example 4.14 Differentiate y = 1 + sin x . cos x Solution u 1 + sin x is of the form v so the quotient rule can be used, with cos x u = 1 + sin x and du ⇒ dx = cos x and The quotient rule is du dv dy v dx − u dx = dx v2 4.4 Differentiating trigonometrical functions etanx is a function of a function, so the chain rule may be used. du = 2 Let u = tan x ⇒ sec x dx dy y = eu ⇒ = eu du Using the chain rule dy dy du = × dx du dx v = cos x dv = −sin x dx Substituting for u and v and their derivatives gives dy (cos x )(cos x ) − (1 + sin x )(− sin x ) = dx (cos x ) 2 2 2 = cos x + sin2x + sin x cos x + x 1 sin = using sin2 x + cos2 x = 1 cos 2 x = (sec2 x)(1 + sin x) 99 9781510421738.indb 99 02/02/18 1:12 PM 4 Exercise 4C 1 2 4 DIFFERENTIATION 3 4 5 Differentiate each of the following. (i) 2 cos x + sin x (ii) tan x + 5 (iii) sin x − cos x Use the product rule to differentiate each of the following. (i) x tan x (ii) sin x cos x (iii) e x sin x Use the quotient rule to differentiate each of the following. ex x + cos x sin x (i) (ii) (iii) cos x x sin x Use the chain rule to differentiate each of the following. (i) tan(x 2 + 1) (ii) sin 2x (iii) ln(sin x) Use an appropriate method to differentiate each of the following. ex tan x (iii) sin 4x 2 sin x (iv) ecos 2x (v) (vi) ln(tan x) 1 + cos x (i) Differentiate y = x cos x. (ii) Find the gradient of the curve y = x cos x at the point where x = π. (iii) Find the equation of the tangent to the curve y = x cos x at the point where x = π. (iv) Find the equation of the normal to the curve y = x cos x at the point where x = π. The diagram shows the part of the curve y = 21 tan 2x for 0 $ x $ 1 π. (i) 6 7 cos x (ii) 2 y O 1π 4 1π 2 x Find the x coordinates of the points on this part of the curve at which the gradient is 4. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 November 2011 8 The equation of a curve is y = 6 sin x − 2 cos 2x. Find the equation of the tangent to the curve at the point ( π,2). 1 6 Give the answer in the form y = mx + c, where the values of m and c are correct to 3 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2015 100 9781510421738.indb 100 02/02/18 1:13 PM 9 Find the gradient of each of the following curves at the point for which x = 0. (i) y = 3 sin x + tan 2x 6 (ii) y = 1 + e 2x Cambridge International AS & A Level Mathematics 9709 Paper 21 Q2 June 2014 4 10 The equation of a curve is y = x + 2 cos x. Find the x coordinates of the Cambridge International AS & A Level Mathematics 9709 Paper 2 Q3 June 2006 11 A curve has equation y = 2 − tan x . 1 + tan x Find the equation of the tangent to the curve at the point for which x = 41 π, giving the answer in the form y = mx + c, where c is correct to 3 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q3 November 2015 12 The curve with equation y = e–x sin x has one stationary point for which 0 $ x $ π. (i) Find the x coordinate of this point. (ii) Determine whether this point is a maximum or a minimum point. 4.5 Differentiating functions defined implicitly stationary points of the curve for 0 $ x $ 2π, and determine the nature of each of these stationary points. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q4 November 2007 ex 1 −1π cos x , for 2 " x "− 2 π , has one stationary point. Find the x coordinate of this point. 13 The curve y = Cambridge International AS & A Level Mathematics 9709 Paper 3 Q3 November 2008 4.5 Differentiating functions defined implicitly All the functions you have differentiated so far have been of the form y = f(x). However, many functions cannot be arranged in this way at all, for example x 3 + y 3 = xy, and others can look clumsy when you try to make y the subject. An example of this is the semicircle x 2 + y 2 = 4, y % 0, illustrated in Figure 4.5. y 2 P 2 P O –2 ▲ Figure 4.5 O 2 x y x By Pythagoras’ theorem, x2 + y2 = 22 101 9781510421738.indb 101 02/02/18 1:13 PM Because of Pythagoras’ theorem, the curve is much more easily recognised in this 4 form than in the equivalent y = 4 − x 2. When a function is specified by an equation connecting x and y that does not have y as the subject, it is called an implicit function. dy dy du = × and the product rule d (uv ) = u dv + v du are dx du dx dx dx dx used extensively to help in the differentiation of implicit functions. 4 DIFFERENTIATION The chain rule Example 4.15 Differentiate each of the following with respect to x. (i) y2 xy (ii) (iii) 3x2y3 (iv) sin y Solution (i) (ii) (iii) d 2 dy d 2 dx (y ) = dy ( y ) × dx (c dy = 2y dx d (xy) = x dy + y dx dx ( ( (product rule) Example 4.16 ) ) d d d (3x2y3) = 3 x 2 ( y 3 ) + y 3 ( x 2 ) dx dx dx dy + y 3 × 2x = 3 x 2 × 3y 2 dx ( = 3xy 2 3x (iv) (chain rule) dy + 2y dx d d dy dx (sin y) = dy (sin y) × dx dy = (cos y) dx ) (product rule) (chain rule) (chain rule) The equation of a curve is given by y3 + xy = 2. dy (i) Find an expression for in terms of x and y. dx (ii) Hence find the gradient of the curve at (1, 1) and the equation of the tangent to the curve at that point. 102 9781510421738.indb 102 02/02/18 1:13 PM Solution (i) ⇒ = 1, y = 1 into dy . the expression for dx Substitute x Using y − y1 = m(x − x1) the equation of the tangent is (y − 1) = − 41 (x − 1) ⇒ x + 4y − 5 = 0 Figure 4.6 shows the graph of the curve with the equation y 3 + xy = 2. ? y O 4.5 Differentiating functions defined implicitly (ii) 4 y3 + xy = 2 dy dy ⇒ 3y 2 + (x + y) = 0 dx dx dy ⇒ (3y 2 + x) = −y dx −y dy ⇒ = dx 3y 2 + x dy 1 At (1, 1), = −4 dx x ▲ Figure 4.6 ❯ Why is this not a function? Stationary points dy As before, these occur where dx = 0. dy = 0 will not usually give values of x directly, but will give a dx relationship between x and y. This needs to be solved simultaneously with the equation of the curve to find the coordinates. Putting 103 9781510421738.indb 103 02/02/18 1:13 PM 4 Example 4.17 (i) Differentiate x 3 + y 3 = 3xy with respect to x. (ii) Hence find the coordinates of any stationary points. Solution 4 DIFFERENTIATION (i) (ii) d ( x 3 ) + d ( y 3 ) = d (3xy ) dx dx dx ⎛ dy ⎞ d y = 3 ⎜x + y⎟ ⇒ 3x2 + 3y2 dx ⎝ dx ⎠ dy At stationary points, dx = 0 ⇒ 3x 2 = 3y ⇒ Notice how it is not necessary to find an expression for x2 = y dy unless dx you are asked to. To find the coordinates of the stationary points, solve x2 = y x 3 + y 3 = 3xy } simultaneously. Substituting for y gives x 3 + (x 2)3 = 3x(x 2) ⇒ x 3 + x 6 = 3x 3 ⇒ x 6 = 2x 3 ⇒ x 3(x 3 – 2) = 0 ⇒ x = 0 or x = 3 2 y = x2 so the stationary points are (0, 0) and ( 3 2, 3 4 ). The stationary points are A and B in Figure 4.7. y 2 –2 A B 2 x –2 ▲ Figure 4.7 104 9781510421738.indb 104 02/02/18 1:13 PM Types of stationary points As with explicit functions, the nature of a stationary point can be determined by d 2y considering the sign of either side of the stationary point. dx 2 Example 4.18 4 The curve with equation sin x + sin y = 1 for 0 $ x $ π, 0 $ y $ π is shown in Figure 4.8. π π O x ▲ Figure 4.8 (i) (ii) Differentiate the equation of the curve with respect to x and hence find the coordinates of any stationary points. Show that the points π , π , π , 5π , 5π , π and 5π , 5π all lie on 6 6 6 6 6 6 6 6 the curve. ( )( )( ) ( ) 4.5 Differentiating functions defined implicitly y Find the gradient at each of these points. What can you conclude about the natures of the stationary points? Solution (i) sin x + sin y = 1 dy ⇒ cos x + (cos y) = 0 dx dy cos x = − cos y ⇒ dx dy = 0 ⇒ cos x = 0 At any stationary point dx ⇒ x = π (only solution in range) 2 Substitute in sin x + sin y = 1. π When x = 2, sin x = 1 ⇒ sin y = 0 ⇒ y = 0 or y = π (2 ) (2 ) ⇒ stationary points at π , 0 and π , π . ➜ 105 9781510421738.indb 105 02/02/18 1:13 PM 4 (ii) sin π6 = 21 , sin 5π = 1 6 2 So, for each of the four given points, sin x + sin y = 21 + 21 = 1. Therefore they all lie on the curve. dy cos x The gradient of the curve is given by dx = − cos y 4 DIFFERENTIATION cos π = 3 , cos 5π = − 3 6 2 6 2 ( ) dy = − dx At π , π , 6 6 ( ) dy = − dx ( ) dy = − dx At π , 5π , 6 6 At 5π , π , 6 6 ( ) At 5π , 5π , 6 6 dy = − dx y 3 2 = −1 3 2 3 2 = 1 − 23 3 − 2 = 1 3 2 3 − 2 = −1 3 − 2 y π π 5π 6 π 6 O π 6 π 2 5π 6 x O π 6 π 2 5π 6 x ▲ Figure 4.9 These results show that π , 0 is a minimum π , π is a maximum 2 2 These points are confirmed by considering the sketch in Figure 4.8 on page 105. ( ) Exercise 4D 1 ( ) Differentiate each of the following with respect to x. (i) y4 (ii) x2 + y3 − 5 (iii) xy + x + y (iv) cos y (v) e(y+2) (vi) xy3 (vii) 2x2y5 (viii) x + ln y − 3 (ix) xey − cos y (x) x2 ln y (xi) x esin y (xii) x tan y − y tan x 106 9781510421738.indb 106 02/02/18 1:13 PM 2 3 4 5 6 PS 7 8 ( ) 4 4.5 Differentiating functions defined implicitly PS Find the gradient of the curve xy3 = 5 ln y at the point (0, 1). Find the gradient of the curve esin x + ecos y = e + 1 at the point π , π . 2 2 (i) Find the gradient of the curve x2 + 3xy + y2 = x + 3y at the point (2, −1). (ii) Hence find the equation of the tangent to the curve at this point. Find the coordinates of all the stationary points on the curve x2 + y2 + xy = 3. A curve has the equation (x − 6)(y + 4) = 2. dy (i) Find an expression for in terms of x and y. dx (ii) Find the equation of the normal to the curve at the point (7, −2). (iii) Find the coordinates of the point where the normal meets the curve again. b (iv) By rewriting the equation in the form y − a = , identify any x−c asymptotes and sketch the curve. A curve has the equation y = x x for x # 0. (i) Take logarithms to base e of both sides of the equation. (ii) Differentiate the resulting equation with respect to x. (iii) Find the coordinates of the stationary point, giving your answer to 3 decimal places. (iv) Sketch the curve for x # 0. The equation of a curve is 3x2 + 2xy + y2 = 6. It is given that there are two points on the curve where the tangent is parallel to the x-axis. (i) Show by differentiation that, at these points, y = −3x. (ii) Hence find the coordinates of the two points. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q5 June 2006 9 The curve with equation 6e 2 x + ke y + e 2 y = c , where k and c are constants, passes through the point P with coordinates (ln 3, ln 2). (i) Show that 58 + 2k = c. (ii) Given also that the gradient of the curve at P is −6, find the values of k and c. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q5 June 2011 10 The equation of a curve is x2 + y2 − 4xy + 3 = 0. (i) (ii) dy 2y − x Show that dx = y − 2x. Find the coordinates of each of the points on the curve where the tangent is parallel to the x-axis. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 June 2008 107 9781510421738.indb 107 02/02/18 1:13 PM 4 11 The equation of a curve is x3 – x2y – y3 = 3. (i) (ii) dy Find dx in terms of x and y. Find the equation of the tangent to the curve at the point (2, 1), giving your answer in the form ax + by + c = 0. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q3 November 2009 4 DIFFERENTIATION 12 The equation of a curve is xy(x + y) = 2a3, where a is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q6 June 2008 4.6 Parametric equations When you go on a ride like the one in the picture, your body follows a very unnatural path and this gives rise to sensations which you may find exhilarating or frightening. You are accustomed to expressing curves as mathematical equations. How would you do so in a case like this? Figure 4.10 shows a simplified version of such a ride. (a) At the start (b) Some time later AP has in total turned through angle 3θ. O 108 9781510421738.indb 108 4m A 2m P O P 2θ A θ θ ▲ Figure 4.10 02/02/18 1:13 PM The passenger’s chair is on the end of a rod AP of length 2 m which is rotating about A. The rod OA is 4 m long and is itself rotating about O. The gearing of the mechanism ensures that the rod AP rotates twice as fast relative to OA as the rod OA does. This is illustrated by the angles marked on Figure 4.10(b), at a time when OA has rotated through an angle θ. At this time, the coordinates of the point P, taking O as the origin, are given by 4 y 2 x = 4 cos θ + 2 cos 3θ 2 sin 3 θ 4 y = 4 sin θ + 2 sin 3θ (see Figure 4.11). 3θ A 4 sin θ θ 4 cos θ O 2 cos 3 θ x ▲ Figure 4.11 These two equations are called parametric equations of the curve. They do not give the relationship between x and y directly in the form y = f(x) but use a third variable, θ, to do so. This third variable is called the parameter. 4.6 Parametric equations P To plot the curve, you need to substitute values of θ and find the corresponding values of x and y. θ = 0° Thus ⇒ x=4+2=6 y=0+0=0 Point (6, 0) θ = 30° ⇒ x = 4 × 0.866 + 0 = 3.464 y = 4 × 0.5 + 2 × 1 = 4 Point (3.46, 4) and so on. Joining points found in this way reveals the curve to have the shape shown in Figure 4.12. y 4 θ = 150° θ = 30° θ = 60°, 120° 2 θ = 90° θ = 180° –6 –4 0 –2 2 θ = 270° θ = 0° 4 6 x –2 θ = 240°, 300° θ = 210° –4 θ = 330° ▲ Figure 4.12 ❯ At what points of the curve would you feel the greatest sensations? ? 109 9781510421738.indb 109 02/02/18 1:13 PM b Graphs from parametric equations 4 DIFFERENTIATION 4 Parametric equations are very useful in situations such as this, where an otherwise complicated equation may be expressed reasonably simply in terms of a parameter. Indeed, there are some curves which can be given by parametric equations but cannot be written as cartesian equations (in terms of x and y only). The next example is based on a simpler curve. Make sure that you can follow the solution completely before going on to the rest of the chapter. Example 4.19 36 . t2 Find the coordinates of the points corresponding to t = 1, 2, 3, −1, −2 and −3. A curve has the parametric equations x = 2t, y = (i) (ii) Plot the points you have found and join them to give the curve. (iii) Explain what happens as t → 0. Solution (i) t –3 –2 –1 1 2 3 x –6 –4 –2 2 4 6 y 4 9 36 36 9 4 The points required are (−6, 4), (−4, 9), (−2, 36), (2, 36), (4, 9) and (6, 4). (ii) The curve is shown in Figure 4.13. y 40 t = –1 t=1 30 20 10 t = –2 t=2 t = –3 t=3 –6 –4 –2 0 2 4 6 x ▲ Figure 4.13 (iii) As t → 0, x → 0 and y → ∞. The y-axis is an asymptote for the curve. 110 9781510421738.indb 110 02/02/18 1:13 PM Example 4.20 A curve has the parametric equations x = t 2, y = t 3 − t. (i) Find the coordinates of the points corresponding to values of t from −2 to +2 at half-unit intervals. (ii) Sketch the curve for −2 $ t $ 2. 4 (iii) Are there any values of x for which the curve is undefined? (i) (ii) t –2 –1.5 –1 x 4 2.25 1 y –6 –1.875 0 –0.5 0 0.5 1 1.5 2 0.25 0 0.25 1 2.25 4 0.375 0 –0.375 0 1.875 6 y 4.6 Parametric equations Solution 6 4 2 0 1 2 3 4 x –2 –4 –6 ▲ Figure 4.14 (iii) The curve in Figure 4.14 is undefined for x " 0. Technology note Graphical calculators can be used to sketch parametric curves but, as with cartesian curves, you need to be careful when choosing the range. Finding the equation by eliminating the parameter For some pairs of parametric equations, it is possible to eliminate the parameter and obtain the cartesian equation for the curve. This is usually done by making the parameter the subject of one of the equations, and substituting this expression into the other. 111 9781510421738.indb 111 02/02/18 1:13 PM 4 Example 4.21 t Eliminate t from the equations x = t3 − 2t2, y = 2. Solution t y= 2 ⇒ t = 2y. Substituting this in the equation x = t 3 − 2t 2 gives 4 DIFFERENTIATION x = (2y)3 − 2(2y)2 or x = 8y 3 − 8y 2. 4.7 Parametric differentiation To differentiate a function which is defined in terms of a parameter t , you need to use the chain rule: Since dy dy dt = × . dx dt dx dt = 1 dx dx dt it follows that dy dy dt = dx dx dt provided that dx ≠ 0. dt Example 4.22 A curve has the parametric equations x = t2, y = 2t. dy (i) Find in terms of the parameter t. dx (ii) Find the equation of the tangent to the curve at the general point (t2, 2t). (iii) Find the equation of the tangent at the point where t = 3. (iv) Eliminate the parameter, and hence sketch the curve and the tangent at the point where t = 3. Solution (i) x = t2 ⇒ y = 2t ⇒ dx = 2t dt dy =2 dt dy dy dt = = 2 =1 dx dx 2t t dt 112 9781510421738.indb 112 02/02/18 1:13 PM (ii) Using y − y1 = m(x − x1) and taking the point (x1, y1) as (t2, 2t), the equation of the tangent at the point (t2, 2t) is 1 y − 2t = t (x − t2) ⇒ ty − 2t2 = x − t2 ⇒ x − ty + t2 = 0 4 This equation still contains the parameter, and is called the equation of the tangent at the general point. 4.7 Parametric differentiation (iii) Substituting t = 3 into this equation gives the equation of the tangent at the point where t = 3. The tangent is x − 3y + 9 = 0. (iv) Eliminating t from x = t2, y = 2t gives 2 ⎛y⎞ or y2 = 4x. x=⎜ ⎟ ⎝2⎠ This is a parabola with the x-axis as its line of symmetry. The point where t = 3 has coordinates (9, 6). The tangent x − 3y + 9 = 0 crosses the axes at (0, 3) and (−9, 0). The curve is shown in Figure 4.15. y (9, 6) 6 3 –9 0 9 x –6 ▲ Figure 4.15 Example 4.23 A curve has parametric equations x = 4 cos θ, y = 3 sin θ. (i) Find dy at the point with parameter θ. dx (ii) Find the equation of the normal at the general point (4 cos θ, 3 sin θ ). π (iii) Find the equation of the normal at the point where θ = . 4 π (iv) Find the coordinates of the point where θ = . 4 (v) 9781510421738.indb 113 Show the curve and the normal on a sketch. ➜ 113 02/02/18 1:13 PM 4 Solution (i) x = 4 cos θ ⇒ 4 DIFFERENTIATION y = 3 sin θ (ii) ⇒ dx = −4 sin θ dθ dy = 3 cos θ dθ dy dy dθ = = 3cos θ dx dx −4 sin θ dθ = − 3cos θ 4 sin θ The tangent and normal are perpendicular, so the gradient of the normal is m1m2 = –1 for − 1 which is + 4 sin θ . perpendicular lines. dy 3cos θ dx Using y − y1 = m(x − x1) and taking the point (x1, y1) as (4 cos θ, 3 sin θ ), the equation of the normal at the point (4 cos θ, 3 sin θ ) is 4 sin θ y − 3 sin θ = 3cos θ (x − 4 cos θ ) 3y cos θ − 9 sin θ cos θ = 4x sin θ − 16 sin θ cos θ ⇒ ⇒ 4x sin θ − 3y cos θ − 7 sin θ cos θ = 0 1 1 π (iii) When θ = , cos θ = and sin θ = , so the equation of the 4 2 2 normal is 4x × ⇒ 1 1 1 1 − 3y × −7× × =0 2 2 2 2 4 2x − 3 2y − 7 = 0 ⇒ 4x − 3y − 4.95 = 0 (to 2 decimal places) (iv) The coordinates of the point where θ = π are 4 1 1 ⎞ ⎛ 4 cos π ,3 sin π = ⎜ 4 × ,3 × 4 4 2 2 ⎟⎠ ⎝ ≈ (2.83, 2.12) ( ) y (v) This curve is an ellipse. 3 (2.83, 2.12) –4 O 4 x –3 ▲ Figure 4.16 114 9781510421738.indb 114 02/02/18 1:13 PM Stationary points Example 4.24 Find the stationary points of the curve with parametric equations x = 2t + 1, y = 3t − t 3, and distinguish between them. Solution dx =2 dt dy ⇒ = 3 − 3t2 y = 3t − t3 dt dy 2 dy 3(1 − t 2 ) dt = = 3 − 3t = 2 2 dx dx dt dy Stationary points occur when = 0: dx ⇒ t2 = 1 ⇒ t=1 or x = 2t + 1 ⇒ At t = 1: x = 3, y = 2 At t = 0.9: x = 2.8 (to the left); At t = −1: x = −1, y = 2 At t = −1.1: x = −1.2 (to the left); 4 4.7 Parametric differentiation When the equation of a curve is given parametrically, the easiest way to dy distinguish between stationary points is usually to consider the sign of dx . If you use this method, you must be careful to ensure that you take points which are to the left and right of the stationary point, i.e. have x coordinates smaller and larger than those at the stationary point. These will not necessarily be points whose parameters are smaller and larger than those at the stationary point. t = −1 dy = 0.285 (positive) dx dy = −0.315 (negative) At t = 1.1: x = 3.2 (to the right); dx There is a maximum at (3, 2). dy = −0.315 (negative) dx dy = 0.285 (positive) At t = −0.9: x = −0.8 (to the right); dx There is a minimum at (−1, −2). e An alternative method d 2y dy Alternatively, to find dx 2 when is expressed in terms of a parameter dx requires a further use of the chain rule: d 2y d ⎛ dy ⎞ d ⎛ dy ⎞ dt = . ⎜ ⎟= ⎜ ⎟ × 2 dx dx ⎝ dx ⎠ dt ⎝ dx ⎠ dx 115 9781510421738.indb 115 02/02/18 1:13 PM 4 DIFFERENTIATION 4 Exercise 4E 1 2 3 4 116 9781510421738.indb 116 PS 5 PS 6 For each of the following curves, find dy in terms of the parameter. dx (i) x = 3t 2 (ii) x = θ − cos θ y = 2t 3 y = θ + sin θ 1 (iii) x = t + t (iv) x = 3 cos θ y = 2 sin θ 1 y=t−t (v) x = (t + 1)2 (vi) x = θ sin θ + cos θ 2 y = (t − 1) y = θ cos θ − sin θ t 2t (vii) x = e + 1 (viii) x = 1 + t y = et t y= 1−t A curve has the parametric equations x = tanθ, y = tan 2θ. Find dy π (i) the value of when θ = 6 dx π (ii) the equation of the tangent to the curve at the point where θ = 6 π (iii) the equation of the normal to the curve at the point where θ = . 6 1 A curve has the parametric equations x = t2, y = 1 − for t # 0. Find 2t (i) the coordinates of the point P where the curve cuts the x-axis (ii) the gradient of the curve at this point (iii) the equation of the tangent to the curve at P (iv) the coordinates of the point where the tangent cuts the y-axis. A curve has parametric equations x = at2, y = 2at, where a is constant. Find (i) the equation of the tangent to the curve at the point with parameter t (ii) the equation of the normal to the curve at the point with parameter t (iii) the coordinates of the points where the normal cuts the x- and y-axes. A curve has parametric equations x = cos θ, y = cos 2θ. dy (i) Show that = 4 cos θ. dx dy d 2y (ii) By writing in terms of x , show that 2 − 4 = 0. dx dx b The parametric equations of a curve are x = at , y = , where a and b are t constant. Find in terms of a, b and t dy (i) dx b (ii) the equation of the tangent to the curve at the general point (at, ) t (iii) the coordinates of the points X and Y where the tangent cuts the x- and y-axes. (iv) Show that the area of triangle OXY is constant, where O is the origin. 02/02/18 1:13 PM PS 7 The diagram shows a sketch of the curve given parametrically in terms of t by the equations x = 4t and y = 2t2 where t takes positive and negative values. y PS 8 9 x P is the point on the curve with parameter t. (i) Show that the gradient at P is t. (ii) Find and simplify the equation of the tangent at P. The tangents at two points Q (with parameter t1) and R (with parameter t2) meet at S. (iii) Find the coordinates of S. (iv) In the case when t1 + t2 = 2 show that S lies on a straight line. Give the equation of the line. A particle P moves in a plane so that at time t its coordinates are given by x = 4 cos t, y = 3 sin t. Find dy (i) in terms of t dx (ii) the equation of the tangent to its path at time t (iii) the values of t for which the particle is travelling parallel to the line x + y = 0. The parametric equations of a curve are x = e3t, y = t2et + 3. dy t(t + 2) (i) Show that . = dx 3e 2t (ii) Show that the tangent to the curve at the point (1, 3) is parallel to the x-axis. (iii) Find the exact coordinates of the other point on the curve at which the tangent is parallel to the x-axis. 4.7 Parametric differentiation O 4 Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2011 10 The parametric equations of a curve are x = e −t cos t, y = e −t sin t . Show that dy = tan(t − 1 π). 4 dx Cambridge International AS & A Level Mathematics 9709 Paper 31 Q4 November 2013 117 9781510421738.indb 117 02/02/18 1:13 PM 4 11 The parametric equations of a curve are x = 3t + ln(t − 1), (i) 4 DIFFERENTIATION (ii) y = t2 + 1, for t # 1. dy in terms of t. dx Find the coordinates of the only point on the curve at which the gradient of the curve is equal to 1. Express Cambridge International AS & A Level Mathematics 9709 Paper 2 Q3 June 2007 12 The parametric equations of a curve are y = 3 – 2 cos 2θ, dy where − 21 π " θ "− 21 π . Express in terms of θ, simplifying your dx answer as far as possible. x = 4 sinθ, Cambridge International AS & A Level Mathematics 9709 Paper 2 Q4 June 2009 13 The parametric equations of a curve are y = et + e−t. x = 1 − e−t, (i) (ii) dy = e 2t − 1. dx Hence find the exact value of t at the point on the curve at which the gradient is 2. Show that Cambridge International AS & A Level Mathematics 9709 Paper 22 Q4 November 2009 14 The parametric equations of a curve are x = 2θ + sin 2θ, dy Show that = tan θ. dx y = 1 − cos 2θ. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q3 June 2006 15 The parametric equations of a curve are x = a cos3 t, y = a sin3 t, where a is a positive constant and 0 " t "− 21 π. dy (i) Express in terms of t. dx (ii) Show that the equation of the tangent to the curve at the point with parameter t is x sin t + y cos t = a sin t cos t. (iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY is always equal to a. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q6 June 2009 118 9781510421738.indb 118 02/02/18 1:13 PM KEY POINTS 1 2 4 5 6 7 8 ( ) 4 4.7 Parametric differentiation 3 dy = knxn−1 where k and n are real constants. dx dy dy du = × . Chain rule: dx du dx dy du dv =v +u . Product rule (for y = uv): dx dx dx du dv dy v dx − u dx u = Quotient rule for y = : . v dx v2 dy = 1 dx d x dy d (ln x ) = 1 x dx d (e x ) = e x dx d (sin kx ) = k cos kx dx d (cos kx ) = −k sin kx dx d (tan kx ) = k sec 2 kx dx y = kxn ⇒ 9 An implicit function is one connecting x and y where y is not the subject. When you differentiate an implicit function: dy ● differentiating y2 with respect to x gives 2 y dx dy ● differentiating 4x3y2 with respect to x gives 12x2 ×y2 + 4x3 ×2 y dx ● the derivative of any constant is 0. 10 In parametric equations the relationship between two variables is expressed by writing both of them in terms of a third variable or parameter. 11 To draw a graph from parametric equations, plot the points on the curve given by different values of the parameter. dy dy dt = provided that dx ≠ 0. 12 dt dx dx dt 119 9781510421738.indb 119 02/02/18 1:13 PM 4 LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ 4 DIFFERENTIATION ■ ■ ■ ■ differentiate ekx and ln x and related sums, differences and constant multiples differentiate sin kx, cos kx and tan kx where x is measured in radians use the chain rule, product rule and quotient rule to differentiate functions involving the functions above differentiate functions defined implicitly or parametrically find ■ stationary points ■ equations of tangents for functions defined implicitly or parametrically. 120 9781510421738.indb 120 02/02/18 1:13 PM P2 P3 5 Integration 5.1 Integrals involving the exponential function The moving power of mathematics invention is not reasoning, but imagination. Augustus de Morgan (1806−71) ? ❯ How could you estimate the numbers of birds in this picture? 5.1 Integrals involving the exponential function Since you know that d (eax+b) = aeax+b, dx you can see that ∫ eax+b dx = 1a eax+b + c. This increases the number of functions which you are able to integrate, as in the following example. 121 9781510421738.indb 121 02/02/18 1:13 PM 5 Example 5.1 Find the following integrals. ∫ e 2x–3 dx (i) (ii) 5 ∫1 6e 3x dx Solution ∫ e2x–3 dx = 2 e2x–3 + c 1 (i) 5 INTEGRATION 5 6e 3 x ∫ 1 6e3x dx = ⎡⎢⎣ 3 ⎤⎦⎥1 = [ 2e 3 x ]15 5 (ii) = 2(e15 − e3) = 6.54 × 106 (to 3 s.f.) 5.2 Integrals involving the natural logarithm function You have already seen that 1 ∫ x dx = ln x + c. There are many other integrals that can be reduced to this form. Example 5.2 Evaluate 5 1 ∫ 2 2x dx. Solution 1 2 5 1 ∫ 2 x dx = 21 [ ln x ] 25 = 21 (ln 5 − ln 2) = 0.458 (to 3 s.f.) In this example the 21 was taken outside the integral, allowing the standard result for x1 to be used. Since y = ln(ax + b) ⇒ So dy a = dx ax + b a ∫ ax + b dx = ln(ax + b ) + c and 122 9781510421738.indb 122 1 c means ‘an arbitrary constant’ and so does not necessarily have the same value from one equation to another. 1 ∫ ax + b dx = a ln(ax + b ) + c 02/02/18 1:13 PM Example 5.3 Find ∫ 2 0 5 1 dx . 5x + 3 Solution 2 ∫ 0 5x1+ 3 dx = ⎡⎣ 51 ln(5x + 3)⎤⎦ 0 2 = 0.293 5.2 Integrals involving the natural logarithm function = 51 ln13 − 51 ln 3 (to 3 s.f.) Extending the domain for logarithmic integrals 1 The use of ∫ x dx = ln x + c has so far been restricted to cases where x # 0, since logarithms are undefined for negative numbers. Look, however, at the area between −b and −a on the left-hand branch of the curve y = x1 in Figure 5.1.You can see that it is a real area, and that it must be possible to evaluate it. y –b B –a A O a b x ▲ Figure 5.1 ACTIVITY 5.1 (i) What can you say about the areas of the two shaded regions? (ii) Try to prove your answer to part (i) before reading on. CP 123 9781510421738.indb 123 02/02/18 1:13 PM Proof 5 −a 1 Let A = ∫ −b x dx. Now write the integral in terms of a new variable, u, where u = −x. 5 INTEGRATION This gives new limits: x = −b ⇒ u = b x = −a ⇒ u = a. du = −1 ⇒ dx = −du. dx So the integral becomes a A = ∫ 1 (−du) b −u a = ∫ 1 du bu [ = ln a − ln b] = −[ ln b − ln a] = −area B So the area has the same size as that obtained if no notice is taken of the fact that the limits a and b have minus signs. However it has the opposite sign, as you would expect because the area is below the axis. Consequently the restriction that x > 0 may be dropped, and the integral is written 1 x dx = ln | x | + c. ∫ f '( x ) Similarly, f ( x ) dx = ln | f(x) | + c. ∫ Example 5.4 Find the value of ∫ 7 5 1 dx. 4− x Solution To make the top line into the differential of the bottom line, you write the integral in one of two ways. −∫ 7 −1 dx = −3ln |4 − x |47 5 54 − x = −3(ln |−3|) − (ln |−1|)4 = −3ln 3 − ln 14 = −1.10 (to 3 s.f.) −∫ 7 5x 1 dx = −3ln |x − 4|47 5 −4 = −3ln 3 − ln 14 = −1.10 (to 3 s.f.) 124 9781510421738.indb 124 02/02/18 1:13 PM y Consequently in the integral q 1 x dx both the limits p and q must O ∫ 5 x p have the same sign, either + or −. The integral is invalid otherwise. ▲ Figure 5.2 ? p (x ) The equation of a curve is y = 1 where p1(x) and p2(x) are p 2 (x ) polynominals. ❯ How can you tell from the equation whether the curve has a discontinuity? ❯ How can you prove that y = x 2 − 2x + 3 has no discontinuities? Exercise 5A 1 Find the following indefinite integrals. (i) 2 3 ∫ x3 dx ∫ 4x1 dx (ii) ∫ x −1 5 dx (iii) (iv) PS 5.2 Integrals involving the natural logarithm function Since the curve y = 1 is not x defined at the discontinuity at x = 0 (see Figure 5.2), it is not possible to integrate across this point. ∫ 2x1− 9 dx Find the following indefinite integrals. (i) ∫ e3x dx (ii) ∫ e−4x dx (iv) ∫ e105x dx (v) ∫ e e 2+x 4 dx (iii) x ∫ e − 3 dx 3x Find the following definite integrals. Where appropriate give your answers to 3 significant figures. 4 (i) ∫ 0 4e2x dx (iii) ∫ –1 (ex + e–x) dx 1 3 (ii) ∫ 1 2x4+ 1 dx (iv) ∫ −2 e3x −2 dx 1 125 9781510421738.indb 125 02/02/18 1:13 PM 4 5 The graph of y = x + 4 is shown below. x y P –4 5 INTEGRATION O 5 x Q Find the coordinates of the minimum point, P, and the maximum point, Q. (ii) Find the area of each shaded region. The diagram illustrates the graph of y = ex. The point A has coordinates (ln 5, 0), B has coordinates (ln 5, 5) and C has coordinates (0, 5). (i) CP 5 y C B (ln5, 5) E O (i) (ii) A x Find the area of the region OABE enclosed by the curve y = ex, the x-axis, the y-axis and the line AB. Hence find the area of the shaded region EBC. The graph of y = ex is transformed into the graph of y = ln x. Describe this transformation geometrically. 126 9781510421738.indb 126 02/02/18 1:13 PM (iii) Using your answers to parts (i) and (ii), or otherwise, show that 5 ∫ 1 ln x dx = 5 ln 5 − 4. (iv) Deduce the values of 5 (a) ∫ 1 ln(x3) dx (b) ∫ 1 ln(3x) dx. 5 Differentiate ln(2x + 3). (ii) Hence, or otherwise, show that 3 1 ∫−1 2x + 3 dx = ln 3. (iii) Find the quotient and remainder when 4x 2 + 8x is divided by 2x + 3. (iv) Hence show that (i) 4 x 2 + 8x dx = 12 − 3 ln 3. ∫−1 2x + 3 3 Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 June 2006 7 dy = e2x − 2e−x. The point (0, 1) lies on the curve. dx Find the equation of the curve. The curve has one stationary point. Find the x coordinate of this point and determine whether it is a maximum or a minimum point. A curve is such that (i) (ii) 5.2 Integrals involving the natural logarithm function 6 5 Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 November 2005 8 (i) (ii) Find the equation of the tangent to the curve y = ln(3x − 2) at the point where x = 1. (a) Find the value of the constant A such that 6x ≡ 2 + A . 3x − 2 3x − 2 6 6x dx = 8 + 8 ln 2. (b) Hence show that ∫ 3 2 3x − 2 Cambridge International AS & A Level Mathematics 9709 Paper 2 Q8 June 2009 9 Find the exact value of the constant k for which ∫ k 1 1 dx = 1 . 2x − 1 Cambridge International AS & A Level Mathematics 9709 Paper 3 Q1 November 2007 127 9781510421738.indb 127 02/02/18 1:13 PM e 5 INVESTIGATIONS A series for ex The exponential function can be written as the infinite series e x = a0 + a 1x + a 2x 2 + a 3x 3 + a4x 4 + … (for x ! ") where a 0, a 1, a 2, … are numbers. 5 INTEGRATION You can find the value of a 0 by substituting the value zero for x. Since e 0 = 1, it follows that 1 = a0 + 0 + 0 + 0 + … , and so a0 = 1. You can now write: ex = 1 + a 1x + a 2x 2 +a 3 x 3+ a 4 x 4 + … . Now differentiate both sides: ex = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 + … , and substitute x = 0 again: 1 = a 1 + 0 + 0 + 0 + … , and so a 1 = 1 also. Now differentiate a second time, and again substitute x = 0. This time you find a 2. Continue this procedure until you can see the pattern in the values of a 0, a 1, a 2, a 3, … . When you have the series for ex, substitute x = 1. The left-hand side is e 1 or e, and so by adding the terms on the right-hand side you obtain the value of e. You will find that the terms become small quite quickly, so you will not need to use very many to obtain the value of e correct to several decimal places. If you are also studying statistics you will meet this series expansion of ex in connection with the Poisson distribution. Compound interest You win $100 000 in a prize draw and are offered two investment options. A You are paid 100% interest at the end of 10 years, or B You are paid 10% compound interest year by year for 10 years. Under which scheme are you better off? final money $200 000 Clearly in scheme A, the ratio R = −−−−−−−−−− is −−−−−−−− = 2. original money $100 000 What is the value of the ratio R in scheme B? Suppose that you asked for the interest to be paid in 20 half-yearly instalments of 5% each (scheme C). What would be the value of R in this case? Continue this process, investigating what happens to the ratio R when the interest is paid at increasingly frequent intervals. Is there a limit to R as the time interval between interest payments tends to zero? 128 9781510421738.indb 128 02/02/18 1:13 PM 5.3 Integrals involving trigonometrical functions Since d (sin(ax + b )) = a cos(ax + b ) dx it follows that ∫ cos(ax + b )dx = 1a sin(ax + b ) + c ∫ a cos(ax + b) dx = sin(ax + b) + c d (cos(ax + b )) = −a sin(ax + b ) dx it also follows that ∫ sin(ax + b )dx = − 1a cos(ax + b ) + c Example 5.5 d (tan(ax + b )) = a sec 2 (ax + b ) dx and so ∫ sec 2 (ax + b )dx = 1a tan(ax + b ) + c Find ∫ sec 2 x dx (i) ∫ sin 2x dx (ii) (iii) ∫ cos(3x − π)dx. Solution ∫ sec 2 x dx = tan x + c (ii) ∫ sin 2x dx = − 21 cos 2x + c (iii) ∫ cos(3x − π) dx = 1 sin(3x − π) + c 3 (i) Example 5.6 Find the exact value of ∫ π 3 0 (sin 2x − cos 4 x )dx. Solution ∫ π 3 0 5.3 Integrals involving trigonometrical functions ∫ −a sin(ax + b) dx = cos(ax + b) + c Similarly, since Also 5 π (sin 2x − cos4 x )dx = ⎡⎣− 21 cos 2x − 41 sin 4 x ⎤⎦ 3 0 =0 = ⎡⎢− 1 cos 2π − 1 sin 4π ⎤⎥ − ⎡⎢− 1 cos 0 − 1 sin 0 ⎤⎥ ⎣ 2 4 ⎦ 3 4 3⎦ ⎣ 2 ⎡ ⎛ ⎞ = ⎢− 21 × − 21 − 41 × ⎜− 3 ⎟ ⎝ 2 ⎠ ⎣ ( ) = 41 + 3+ 8 ⎤ 1 ⎥ − ⎡⎣− 2 × 1⎤⎦ ⎦ 1 2 3+ 3 8 4 = 6+ 3 8 = 129 9781510421738.indb 129 02/02/18 1:13 PM Using trigonometrical identities in integration 5 Sometimes, when it is not immediately obvious how to integrate a function involving trigonometrical functions, it may help to rewrite the function using one of the trigonometrical identities. 5 INTEGRATION Example 5.7 Find ∫ sin2 x dx. Solution Use the identity cos 2x = 1 − 2 sin2 x. (Remember that this is just one of the three expressions for cos 2x.) This identity may be rewritten as sin2 x = 21 (1 − cos 2x). By putting sin2x in this form, you will be able to perform the integration. ∫ sin2 x dx ∫ = 21 (1 − cos 2x) dx = 1 2 (x − 21 sin 2x) + c 1 = 21 x − 4 sin 2x + c You can integrate cos2 x in the same way, by using cos2 x = 21 (cos 2x + 1). Other even powers of sin x or cos x can also be integrated in a similar way, but you have to use the identity twice or more. Example 5.8 Find ∫ cos4 x dx. Solution First express cos4 x as (cos2 x)2: cos4 x = [ 2 (cos 2x + 1)] 1 2 1 = 4 (cos2 2x + 2 cos 2x + 1) Next, apply the same identity to cos2 2x : cos2 2x = 21 (cos 4x + 1) Hence cos4 x = 41 ( 2 cos 4x + 1 1 2 + 2 cos 2x + 1) = 41 ( 21 cos 4x + 2 cos 2x + 23 ) = 81 cos 4x + 21 cos 2x + 83 130 9781510421738.indb 130 02/02/18 1:13 PM This can now be integrated. ∫ cos4 x dx = ∫ ( 81 cos 4x + 21 cos 2x + 83 ) dx = Exercise 5B 1 Integrate the following with respect to x. (i) sin x − 2 cos x (ii) 3 cos x + 2 sin x 2 (iv) 4 sec x (v) sin(2x + 1) 2 (vii) 6 sec 2x (viii) 3 sec2 3x − sin 2x Find the exact value of the following. (i) (iii) (v) (vii) PS 4 ∫ ∫ ∫ ∫ ∫ π sin x dx (ii) ∫ cos x dx π (iv) ∫ 3 0 π 3 6 5π 6 0 cos 3x dx ( ) cos 2x + π dx 2 0 π π ∫ ∫ (vi) (viii) π 4 0 π (vi) cos(5x − π) (ix) 4 sec2 x − cos 2x sec 2 x dx 2π 3 0 (iii) 5 sin x + 4 cos x sin 2x dx 6 π 8 sec 2 2x dx 4 (sec 2 x + cos 4 x )dx π 0 (cos x + sin 2x )dx (i) Show that sin x cos x = 21 sin 2x. π (ii) Hence find the exact value of 3 sin x cos x dx . 0 Use a suitable trigonometrical identity to help you find these. (ix) 3 + 41 sin 2x + 83 x + c 6 0 5 5.3 Integrals involving trigonometrical functions 2 1 32 sin 4x ∫ π (i) ∫ cos x dx ∫ (a) (b) ∫ sin x dx ∫ sin x dx Express ( 3 ) cos x + sin x in the form R cos( x − α ), where R > 0 (a) cos 2 xdx (b) 2 2 0 π (ii) 5 (i) (ii) 3 2 2 0 and 0 < α < 21 π , giving the exact values of R and α. 1π 1 1 Hence show that 12 2 dx = 4 3 . π 6 3 cos x + sin x ∫ (( ) ) Cambridge International AS & A Level Mathematics 9709 Paper 33 Q4 June 2013 6 (i) (ii) Prove that tan θ + cot θ ≡ 2 . sin 2θ Hence 1 1 (a) find the exact value of tan 8 π + cot 8 π 1π 2 6 dθ . (b) evaluate θ + tan cot θ 0 ∫ Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2014 131 9781510421738.indb 131 02/02/18 1:13 PM 5 7 (i) (ii) Express cos2 x in terms of cos 2x. Hence show that ∫ 1π 3 cos 2 x d x = 61 π + 1 3 . 8 0 (iii) By using an appropriate trigonometrical identity, deduce the exact 5 INTEGRATION value of ∫ 8 (i) 1π 3 0 sin 2 x d x. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 June 2007 Prove the identity (cos x + 3 sin x)2 ≡ 5 − 4 cos 2x + 3 sin 2x. (ii) Using the identity, or otherwise, find the exact value of ∫ 9 (i) (ii) (ii) 11 (i) (ii) 0 (cos x + 3 sin x ) 2 d x. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 November 2007 ∫ 1π Show that 4 cos 2x d x = 21. 0 By using an appropriate trigonometrical identity, find the exact value of ∫ 10 (i) 1π 4 1 π 3 1π 6 3 tan 2 x d x. Cambridge International AS & A Level Mathematics 9709 Paper 22 Q4 June 2010 ∫ 4e (3 + e )dx. Show that ∫ (3 + 2 tan θ )dθ = 21 (8 + π). Find x 2x 1π 4 − 1π 4 2 Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2011 Prove the identity cos 4θ + 4 cos 2θ ≡ 8 cos 4 θ − 3. Hence (a) solve the equation cos 4θ + 4 cos 2θ = 1 for − 21 π & θ & 21 π (b) find the exact value of ∫ 1π 4 0 cos 4 θ dθ . Cambridge International AS & A Level Mathematics 9709 Paper 31 Q9 June 2011 132 9781510421738.indb 132 02/02/18 1:13 PM 12 (i) By first expanding cos (2x + x), show that cos 3x ≡ 4 cos3x − 3 cos x. (ii) Hence show that ∫ 1π 6 0 (2 cos 3 x − cos x )dx = 5 5. 12 Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 November 2011 (ii) Find ∫ 4 cos 2 ( 21 θ ) dθ . Find the exact value of ∫ 6 −1 1 dx . 2x + 3 Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 November 2014 5.4 Numerical integration 5.4 Numerical integration 13 (i) Note Section 5.4, including Exercise 5C, is about understanding and using the trapezium rule to estimate the value of a definite integral. This is required knowledge only for Paper 2: Pure Mathematics 2 (AS Level). If you are studying for the A Level, although you may find it interesting to extend your knowledge, it is not required for Paper 3: Pure Mathematics 3. There are times when you need to find the area under a graph but cannot do this by the integration methods you have met so far. » The function may be one that cannot be integrated algebraically. (There are many such functions.) » The function may be one that can be integrated algebraically but which requires a technique with which you are unfamiliar. » It may be that you do not know the function in algebraic form, but just have a set of points (perhaps derived from an experiment). In these circumstances you can always find an approximate answer using a numerical method, but you must: » have a clear picture in your mind of the graph of the function, and how your method estimates the area beneath it » understand that a numerical answer without any estimate of its accuracy, or error bounds, is valueless. The trapezium rule In this chapter just one numerical method of integration is introduced, namely the trapezium rule. As an illustration of the rule, it is used to find the area under the curve y = 5x − x 2 for values of x between 0 and 4. 133 9781510421738.indb 133 02/02/18 1:13 PM 5 It is in fact possible to integrate this function algebraically, but not using the techniques that you have met so far. Note 5 INTEGRATION You should not use a numerical method when an algebraic (sometimes called analytic) technique is available to you. Numerical methods should be used only when other methods fail. Figure 5.3 shows the area approximated by two trapezia of equal width. y 3 y = 5x – x 2 2 1 A O 1 B 2 3 4 5 x ▲ Figure 5.3 Remember the formula for the area of a trapezium, Area = 21 h(a + b), where a and b are the lengths of the parallel sides and h is the distance between them. In the cases of the trapezia A and B, the parallel sides are vertical. The left-hand side of trapezium A has zero height, and so the trapezium is also a triangle. When x=0 ⇒ y= 0 =0 When x=2 ⇒ y= 6 = 2.4495 (to 4 d.p.) When x = 4, ⇒ y= 4 =2 A (0) 2.4495 (2) 2.4495 (2) B 2 (4) ▲ Figure 5.4 The area of trapezium A = 1 2 × 2 × (0 + 2.4495) = 2.4495 The area of trapezium B = 1 2 × 2 × (2.4495 + 2) = 4.4495 Total 6.8990 134 9781510421738.indb 134 02/02/18 1:13 PM For greater accuracy you can use four trapezia, P, Q, R and S, each of width 1 unit as in Figure 5.5. The area is estimated in just the same way. y 5 3 2 P Q 2 O R 6 1 S 6 2 2 3 4 5 x ▲ Figure 5.5 These figures are given to 4 decimal places but the calculation has been done to more places on a calculator. Trapezium P: 1 2 × 1 × (0 + 2) = 1.0000 Trapezium Q: 1 2 × 1 × (2 + 2.4495) = 2.2247 Trapezium R: 1 2 × 1 × (2.4495 + 2.4495) = 2.4495 Trapezium S: 1 2 × 1 × (2.4495 + 2) 5.4 Numerical integration 1 = 2.2247 Total 7.8990 Accuracy In this example, the first two estimates are 6.8989… and 7.8989… .You can see from Figure 5.5 that the trapezia all lie underneath the curve, and so in this case the trapezium rule estimate of 7.8989… must be too small.You cannot, however, say by how much. To find that out you will need to take progressively more strips to find the value to which the estimate converges. Using 8 strips gives an estimate of 8.2407…, and 16 strips gives 8.3578… . The first figure, 8, looks reasonably certain but it is still not clear whether the second is 3, 4 or even 5.You need to take even more strips to be able to decide. In this example, the convergence is unusually slow because of the high curvature of the curve. Technology note You can use a graph-drawing program with the capability to calculate areas using trapezia. Calculate the area using progressively more strips and observe the convergence. For example, 32 strips: 8.398 50 strips: 8.409 100 strips: 8.416 1000 strips: 8.420 135 9781510421738.indb 135 02/02/18 1:13 PM 5 ? It is possible to find this area without using calculus at all. ❯ How can this be done? How close is the 16-strip estimate? The procedure 5 INTEGRATION In the previous example, the answer of 7.8990 from four strips came from adding the areas of the four trapezia P, Q, R and S: 1 2 1 2 × 1 × (0 + 2) + + 1 2 × 1 × (2 + 2.4495) + 1 2 × 1 × (2.4495 + 2.4495) × 1 × (2.4495 + 2) These are the heights of the intermediate vertical lines. and this can be written as 1 2 ×1 ×[0 + 2 ×(2 + 2.4495 + 2.4495) + 2] These are the heights of the ends of the whole area: 0 and 2. This is the strip width: 1. This is a useful way to remember the formula. This is often stated in words as Area " 1 2 × strip width ×[ends + twice middles] or in symbols, for n strips of width h A" 1 2 × h ×[y0 + yn + 2(y1 + y2 + … + yn − 1)]. This is called the trapezium rule for width h (see Figure 5.6). y y = f(x) y1 y0 yn y2 yn–1 yn–2 a b h h h x h ▲ Figure 5.6 136 9781510421738.indb 136 02/02/18 1:13 PM ? ❯ Look at the three graphs in Figure 5.7, and in each case state whether the trapezium rule would underestimate or overestimate the area, or whether you cannot tell. y y y x x (ii) (iii) ▲ Figure 5.7 Exercise 5C M This exercise covers the trapezium rule, which is required knowledge only for Paper 2: Pure Mathematics 2 (AS Level). It is not required for Paper 3: Pure Mathematics 3 (A Level). 1 CP 2 3 5.4 Numerical integration x (i) 5 The speed v in m s−1 of a train at time t seconds is given in the following table. t 0 10 20 30 40 50 60 v 0 5.0 6.7 8.2 9.5 10.6 11.6 The distance that the train has travelled is given by the area under the graph of the speed (vertical axis) against time (horizontal axis). (i) Estimate the distance the train travels in this 1-minute period. (ii) Give two reasons why your method cannot give a very accurate answer. 1 1 dx is known to equal π . The definite integral ∫ 4 0 1+ x2 (i) Using the trapezium rule for four strips, find an approximation for π. (ii) Repeat your calculation with 10 and 20 strips to obtain closer estimates. (iii) If you did not know the value of π, what value would you give it with confidence on the basis of your estimates in parts (i) and (ii)? 1.6 1 + x 2 dx. The trapezium rule is used to estimate the value of I = ∫ 0 (i) Draw the graph of y = 1 + x 2 for 0 $ x $ 1.6. (ii) Use strip widths of 0.8, 0.4, 0.2 and 0.1 to find approximations to the value of the integral. (iii) State the value of the integral to as many decimal places as you can justify. 137 9781510421738.indb 137 02/02/18 1:13 PM 5 INTEGRATION 5 4 5 6 1 The trapezium rule is used to estimate the value of ∫ sin x dx. 0 (i) Draw the graph of y = sin x for 0 $ x $ 1. (ii) Use 1, 2, 4, 8 and 16 strips to find approximations to the value of the integral. (iii) State the value of the integral to as many decimal places as you can justify. 1 4 dx . The trapezium rule is used to estimate the value of + x2 1 0 4 (i) Draw the graph of y = for 0 $ x $ 1. 1+ x2 (ii) Use strip widths of 1, 0.5, 0.25 and 0.125 to find approximations to the value of the integral. (iii) State the value of the integral to as many decimal places as you can justify. A student uses the trapezium rule to estimate the value of ∫ 2 ∫ 0(2 − cos 2πx) dx. (i) (ii) 7 (iii) Use integration to find the exact value of this integral. Use the trapezium rule with four intervals to find an approximation to ∫ 5 1 8 Find approximations to the value of the integral by applying the trapezium rule using strip widths of (a) 2 (b) 1 (c) 0.5 (d) 0.25. Sketch the graph of y = 2 − cos 2πx for 0 $ x $ 2. On copies of your graph shade the areas you have found in parts (i)(a) to (d). | 2x − 8 | dx. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 November 2014 ln x for 0 " x $ 4. x The curve cuts the x-axis at A and its maximum point is M. The diagram shows the part of the curve y = y M O A 4 x 138 9781510421738.indb 138 02/02/18 1:13 PM Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 June 2005 9 The diagram shows the part of the curve y = ex cos x for 0 $ x $ 21 π. The curve meets the y-axis at the point A. The point M is a maximum point. y 5 5.4 Numerical integration Write down the coordinates of A. (ii) Show that the x coordinate of M is e, and write down the y coordinate of M in terms of e. (iii) Use the trapezium rule with three intervals to estimate the value of 4 ∫ 1 lnxx dx , correct to 2 decimal places. (iv) State, with a reason, whether the trapezium rule gives an underestimate or an overestimate of the true value of the integral in part (iii). (i) M A O 1 π 2 x Write down the coordinates of A. (ii) Find the x coordinate of M. (iii) Use the trapezium rule with three intervals to estimate the value of (i) ∫ 1π 2 0 e x cos x d x, giving your answer correct to 2 decimal places. (iv) State, with a reason, whether the trapezium rule gives an underestimate or an overestimate of the true value of the integral in part (iii). Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 June 2007 139 9781510421738.indb 139 02/02/18 1:13 PM 5 10 The diagram shows the curve y = x 2 e–x and its maximum point M. y 5 INTEGRATION M x O Find the x coordinate of M. (ii) Show that the tangent to the curve at the point where x = 1 passes through the origin. (iii) Use the trapezium rule, with two intervals, to estimate the value of (i) ∫ 3 1 x 2e − x d x , giving your answer correct to 2 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q8 November 2007 11 (i) (ii) 16 6 dx = ln125 . Show that 6 2x − 7 Use the trapezium rule with four intervals to find an approximation to ∫ ∫ 17 1 log 10 x dx, giving your answer correct to 3 significant figures. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2014 140 9781510421738.indb 140 02/02/18 1:13 PM KEY POINTS 1 2 ∫ kx dx = kxn+ 1 + c ∫ e dx = e + c ∫ e dx = 1a e + c 1 ∫ x dx = ln | x | + c ∫ ax1+ b dx = 1a ln | ax + b | + c 1 ∫ cos(ax + b)dx = a sin(ax + b) + c 1 ∫ sin(ax + b)dx = − a cos(ax + b) + c ∫ sec (ax + b)dx = 1a tan(ax + b) + c n +1 n x x ax + b 4 ax + b 5.4 Numerical integration 3 5 2 5 (Paper 2: Pure Mathematics 2 only) You can use the trapezium rule, with n strips of width h, to find an approximate value for a definite integral as A ≈ h ⎣⎡y 0 + 2( y 1 + y 2 + ... + y n − 1 ) + y n ⎦⎤ 2 In words this is Area ≈ 21 × strip width × [ends + twice middles]. LEARNING OUTCOMES Now you have finished this chapter, you should be able to ■ extend the idea that integration is the reverse of differentiation to integrate ■ e ax +b 1 ■ ax + b ■ ■ ■ ■ 9781510421738.indb 141 ■ sin(ax + b ) ■ cos(ax + b ) ■ sec 2 (ax + b ) and related sums, differences and constant multiples use integration in cases where the process is the reverse of the chain rule use trigonometrical identities in carrying out integration (Paper 2: Pure Mathematics 2 only) understand and use the trapezium rule to estimate the value of a definite integral (Paper 2: Pure Mathematics 2 only) use a sketch graph to determine whether the trapezium rule gives an overestimate or an underestimate. 141 02/02/18 1:13 PM 5 PROBLEM SOLVING PS Numerical integration Vesna is carrying out numerical integration, approximating the area under a curve by rectangles. She makes the following statements. • Using rectangles drawn from the left-hand ends of the intervals is a 5 INTEGRATION poor method. Using the right-hand ends is no better. They always give you extreme estimates for the area under the curve. You can see it in my diagrams (Figures 5.8 and 5.9). y y 2 2 1 1 O 1 2 3 4 5 x ▲ Figure 5.8 Low estimate in this case O 1 2 3 4 x 5 ▲ Figure 5.9 High estimate in this case • Taking the average of the left-hand and right-hand rectangles is better. The answer is always the same as you get from the trapezium rule. • An even better method is to draw the rectangles at the midpoints of the intervals as shown in Figure 5.10. For any number of intervals that will be more accurate. y 2 1 O 1 2 3 4 5 x ▲ Figure 5.10 Best estimate Investigate whether Vesna’s statements are true. 1 Problem specification and analysis Vesna has made three statements. Look at them carefully and decide how you are going to proceed. • Which of them are little more than common sense and which need real work? • How you are going to investigate those that need work? - What technology will you use? - What examples will you choose to work with? - How many different examples do you expect to use? - To what level of accuracy do you expect to work? 142 9781510421738.indb 142 02/02/18 1:13 PM • How you are going to report the outcomes? - Will you be happy just to say ‘true’ or ‘false’, or do you expect to make statements such as ‘it is usually true but there are exceptions such as ...’? - How much explanation do you expect to give? It will be helpful, at least in some cases, to choose functions which you know how to integrate, allowing you to know the answer to which a numerical method should be converging. However, the whole point of using a numerical method is to find an answer when an analytical method is not available to you. It may be that there is one that you don’t know or it may be that one just does not exist. So you should also use at least one example where you will only know the answer (to your chosen level of accuracy) when you have completed your work. 5 Problem solving 2 Information collection You may be able to make some comments on Vesna’s statements just by thinking about them but you will also need to carry out some investigations of your own. This will require the use of technology. 5 Do not be content to work with just one or two types of functions. Try a variety of functions but always start with a sketch of the curve of the function.You can of course use graphing software to obtain this. 3 Processing and representation The previous stage will probably result in you having a lot of information. Now you need to sort through it and to organise it in a systematic way that allows you to comment on Vesna’s three statements. Where you can explain your results using algebra, then you should do so. In other cases you may present them as experimental outcomes. 4 Interpretation The method illustrated in Figure 5.10 is called the midpoint rule. Much of your work on this task will have been focused on midpoint rule, and you need to comment on whether this is a good method for numerical integration. In order to do so you will need to explain what you mean by ‘good’ and what the desirable features are in such a method. There are other methods of numerical integration and you may choose to conclude by saying something about them. 143 9781510421738.indb 143 02/02/18 1:13 PM P2 P3 6 NUMERICAL SOLUTION OF EQUATIONS 6 Numerical solution of equations It is the true nature of mankind to learn from his mistakes. Fred Hoyle (1915–2001) A golfer doesn’t often hit a ball into the hole at the first attempt! Instead, he or she will try to hit the ball as close to the hole as possible. After that, successive attempts will usually be closer and closer to the hole, until the ball finally lands in the hole. Think of some other situations where you need to make a rough approximation for your first attempt, and then gradually improve your attempts. ❯ Which of the following equations can be solved algebraically, and which cannot? For each equation find a solution, accurate or approximate. (i) x2 − 4x + 3 = 0 (ii) x2 + 10x + 8 = 0 ? ❯ (iii) x5 − 5x + 3 = 0 (v) (iv) x3 − x = 0 ex = 4x 144 9781510421738.indb 144 02/02/18 1:13 PM You probably realised that the equations x5 − 5x + 3 = 0 and ex = 4x cannot be solved algebraically.You may have decided to draw their graphs, either manually or using a graphical calculator or computer package, as in Figure 6.1. y y 6 f(x) = 4x (2.15, 8.6) (–1, 7) f(x) = ex (0.357, 1.43) x O (1, –1) O x ▲ Figure 6.1 The graphs show you that » x5 − 5x + 3 = 0 has three roots, lying in the intervals [−2, −1], [0, 1] and [1, 2]. » ex = 4x has two roots, lying in the intervals [0, 1] and [2, 3]. 6 Numerical solution of equations f(x) = x5 – 5x + 3 Note An interval written as [a, b] means the interval between a and b, including a and b. This notation is used in this chapter. If a and b are not included, the interval is written (a, b). You may also elsewhere meet the notation ]a, b[, indicating that a and b are not included. The problem now is how to find the roots to any required degree of accuracy, and as efficiently as possible. In many real problems, equations are obtained for which solutions using algebraic or analytic methods are not possible, but for which you nonetheless want to know the answers. In this chapter you will be introduced to numerical methods for solving such equations. In applying these methods, keep the following points in mind. » Only use numerical methods when algebraic ones are not available. If you can solve an equation algebraically (e.g. a quadratic equation), that is the right method to use. » Before starting to use a calculator or computer program, always start by drawing a sketch graph of the function whose equation you are trying to solve. This will show you how many roots the equation has and their approximate positions. It will also warn you of possible difficulties with particular methods. When using a graphical calculator or computer package ensure that the range of values of x is sufficiently large to, hopefully, find all the roots. 145 9781510421738.indb 145 02/02/18 1:13 PM » Always give a statement about the accuracy of an answer (e.g. to 5 decimal places, or ± 0.000 005). An answer obtained by a numerical method is worthless without this; the fact that at some point your calculator display reads, say, 1.676 470 588 2 does not mean that all these figures are valid. » Your statement about the accuracy must be obtained from within the numerical method itself. Usually you find a sequence of estimates of everincreasing accuracy. » Remember that the most suitable method for one equation may not be that for another. 6 NUMERICAL SOLUTION OF EQUATIONS 6 6.1 Interval estimation − change-of-sign methods Assume that you are looking for the roots of the equation f(x) = 0. This means that you want the values of x for which the graph of y = f(x) crosses the x-axis. As the curve crosses the x-axis, f(x) changes sign, so provided that f(x) is a continuous function (its graph has no asymptotes or other breaks in it), once you have located an interval in which f(x) changes sign, you know that that interval must contain a root. In both of the graphs in Figure 6.2, there is a root lying between a and b. y y a O b b x O a x ▲ Figure 6.2 Example 6.1 Show that the equation ex − x = x2 + 2 has a root between x = 2 and x = 3. Solution There are two methods you can use to show there is a root in the interval [a, b]. The left-hand side, LHS, and the right-hand side, RHS. Method 1 Substitute x = a into each side of the equation and show that LHS < RHS for x = a. Or vice versa: Then show LHS > RHS for x = b. LHS < RHS Hence there exists a value of x in the when x = b. interval [a, b] where the LHS = RHS, which is the root of the equation. Or vice versa: LHS > RHS when x = a. 146 9781510421738.indb 146 02/02/18 1:13 PM When x = 2, e x − x = e 2 − 2 = 5.389... 6 and x 2 + 2 = 2 2 + 2 = 6. So when x = 2, e x − x < x 2 + 2 . When x = 3, e x − x = e 3 − 3 = 17.085... and x 2 + 2 = 3 2 + 2 = 11. So when x = 3, e x − x > x 2 + 2 . ex − x = x 2 + 2. Method 2 Rearrange the equation so it is equal to zero. Then show there is a change of sign in the interval [a, b]. ex − x = x 2 + 2 ⇒ ex − x 2 − x − 2 = 0 When x = 2: e 2 − 2 2 − 2 − 2 = −0.6109... When x = 3: e 3 − 3 2 − 3 − 2 = 6.0855... There is a change of sign in the interval [2, 3], so there must be a root between x = 2 and x = 3. You have seen that x5 − 5x + 3 = 0 has roots in the intervals [−2, −1], [0, 1] and [1, 2]. There are several ways of homing in on such roots systematically. Two of these are now described, using the search for the root in the interval [0, 1] as an example. 6.1 Interval estimation − change-of-sign methods Hence there must be a value of x between x = 2 and x = 3 where Decimal search In this method you first take increments in x of size 0.1 within the interval [0, 1], working out the value of f(x) = x5 − 5x + 3 for each one.You do this until you find a change of sign. x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 f (x) 3.00 2.50 2.00 1.50 1.01 0.53 0.08 −0.33 147 9781510421738.indb 147 02/02/18 1:13 PM 6 There is a sign change, and therefore a root, in the interval [0.6, 0.7] since the function is continuous. Having narrowed down the interval, you can now continue with increments of 0.01 within the interval [0.6, 0.7]. x 0.60 0.61 0.62 f (x) 0.08 0.03 −0.01 6 NUMERICAL SOLUTION OF EQUATIONS This shows that the root lies in the interval [0.61, 0.62]. Alternative ways of expressing this information are that the root can be taken as 0.615 with a maximum error of ± 0.005, or the root is 0.6 (to 1 decimal place). This process can be continued by considering x = 0.611, x = 0.612, … to obtain the root to any required number of decimal places. ❯ How many steps of decimal search would be necessary to find each of the values 0.012, 0.385 and 0.989, using x = 0 as a starting point? ? When you use this procedure on a computer or calculator you should be aware that the machine is working in base 2, and that the conversion of many simple numbers from base 10 to base 2 introduces small rounding errors. This can lead to simple roots such as 2.7 being missed and only being found as 2.699 999. Interval bisection This method is similar to the decimal search, but instead of dividing each interval into ten parts and looking for a sign change, in this case the interval is divided into two parts − it is bisected. Looking as before for the root in the interval [0, 1], you start by taking the midpoint of the interval, 0.5. f(0.5) = 0.53, so f(0.5) # 0. Since f(1) " 0, the root is in [0.5, 1]. Now take the midpoint of this second interval, 0.75. f(0.75) = −0.51, so f(0.75) " 0. Since f(0.5) # 0, the root is in [0.5, 0.75]. The midpoint of this further reduced interval is 0.625. f(0.625) = −0.03, so the root is in the interval [0.5, 0.625]. The method continues in this manner until any required degree of accuracy is obtained. However, the interval bisection method is quite slow to converge to the root, and is cumbersome when performed manually. ACTIVITY 6.1 Investigate how many steps of this method you need to achieve an accuracy of 1, 2, 3 and n decimal places, having started with an interval of length 1. 148 9781510421738.indb 148 02/02/18 1:13 PM Error (or solution) bounds 6 Change-of-sign methods have the great advantage that they automatically provide bounds (the two ends of the interval) within which a root lies, so the maximum possible error in a result is known. Knowing that a root lies in the interval [0.61, 0.62] means that you can take the root as 0.615 with a maximum error of ± 0.005. Problems with change-of-sign methods 6.1 Interval estimation − change-of-sign methods There are a number of situations which can cause problems for change-ofsign methods if they are applied blindly, for example by entering the equation into a computer program without prior thought. In all cases you can avoid problems by first drawing a sketch graph, provided that you know what dangers to look out for. The curve touches the x -axis In this case there is no change of sign, so change-of-sign methods are doomed to failure (see Figure 6.3). f(x) x O ▲ Figure 6.3 There are several roots close together Where there are several roots close together, it is easy to miss a pair of them. The equation f(x) = x3 − 1.9x2 + 1.11x − 0.189 = 0 has roots at 0.3, 0.7 and 0.9. A sketch of the curve of f(x) is shown in Figure 6.4. In this case f(0) " 0 and f(1) # 0, so you know there is a root between 0 and 1. A decimal search would show that f(0.3) = 0, so that 0.3 is a root.You would be unlikely to search further in this interval. f(x) O 0.3 0.7 0.9 x ▲ Figure 6.4 Interval bisection gives f(0.5) # 0, so you would search the interval [0, 0.5] and eventually arrive at the root 0.3, unaware of the existence of those at 0.7 and 0.9. 149 9781510421738.indb 149 02/02/18 1:13 PM There is a discontinuity in f (x ) 6 1 The curve y = x − 2.7 has a discontinuity at x = 2.7, as shown by the asymptote in Figure 6.5. 6 NUMERICAL SOLUTION OF EQUATIONS y O 2.7 x ▲ Figure 6.5 1 The equation x − 2.7 = 0 has no root, but all change-of-sign methods will converge on a false root at x = 2.7. None of these problems will arise if you start by drawing a sketch graph. Technology note It is important that you understand how each method works and are able, if necessary, to perform the calculations using only a scientific calculator. However, these repeated operations lend themselves to the use of a spreadsheet or a programmable calculator. Many packages, such as Autograph, will both perform the methods and illustrate them graphically. Exercise 6A 1 2 3 4 Show that the equation x3 + 3x − 5 = 0 has no turning (stationary) points. (ii) Show with the aid of a sketch that the equation can have only one root, and that this root must be positive. (iii) Find the root, correct to 3 decimal places. (i) How many roots has the equation ex − 3x = 0? (ii) Find an interval of unit length containing each of the roots. (iii) Find each root correct to 2 decimal places. (i) Sketch y = 2x and y = x + 2 on the same axes. (ii) Use your sketch to deduce the number of roots of the equation 2x = x + 2. (iii) Find each root, correct to 3 decimal places if appropriate. Find all the roots of x3 − 3x + 1 = 0, giving your answers correct to 2 decimal places. (i) 150 9781510421738.indb 150 02/02/18 1:13 PM CP 5 (i) (ii) PS 6 (i) Find the roots of x5 − 5x + 3 = 0 in the intervals [−2, −1] and [1, 2], correct to 2 decimal places, using (a) decimal search (b) interval bisection. Comment on the ease and efficiency with which the roots are approached by each method. Use a systematic search for a change of sign, starting with x = −2, to locate intervals of unit length containing each of the three roots of 6 x3 − 4x2 − 3x + 8 = 0. 7 PS 8 6.2 Fixed-point iteration Sketch the graph of f(x) = x3 − 4x2 − 3x + 8. (iii) Use the method of interval bisection to obtain each of the roots correct to 2 decimal places. (iv) Use your last intervals in part (iii) to give each of the roots in the form a ± (0.5)n where a and n are to be determined. The diagram shows a sketch of the graph of f(x) = ex − x3 f(x) f(x) = ex – x3 without scales. (i) Use a systematic search for a change of sign to locate x O intervals of unit length containing each of the roots. (ii) Use a change-of-sign method to find each of the roots correct to 3 decimal places. For each of the equations below (a) sketch the curve (b) write down any roots (c) investigate what happens when you use a change-of-sign method with a starting interval of [−0.3, 0.7]. x2 1 x (iii) y = (i) y = (ii) y = 2 2 x x +1 x +1 (ii) 6.2 Fixed-point iteration In fixed-point iteration you find a single value or point as your estimate for the value of x, rather than establishing an interval within which it must lie.This involves an iterative process, a method of generating a sequence of numbers by continued repetition of the same procedure. If the numbers obtained in this manner approach some limiting value, then they are said to converge to this value. 151 9781510421738.indb 151 02/02/18 1:13 PM 6 NUMERICAL SOLUTION OF EQUATIONS 6 INVESTIGATION Iterations Notice what happens in each of the following cases, and try to find some explanation for it. (i) Set your calculator to the radian mode, enter zero if not automatically displayed and press the cosine key repeatedly. (ii) Enter any positive number into your calculator and press the square root key repeatedly. Try this for both large and small numbers. (iii) Enter any positive number into your calculator and press the sequence + 1 = √ = repeatedly. Write down the number which appears each time you press = . The sequence generated appears to converge. You may recognise the number to which it appears to converge: it is called the Golden Ratio. Rearranging the equation f(x ) = 0 into the form x = F(x ) The first step, with an equation f(x) = 0, is to rearrange it into the form x = F(x). Any value of x for which x = F(x) is a root of the original equation, as shown in Figure 6.6. When f(x) = x2 − x − 2, f(x) = 0 is the same as x = x2 − 2. y –1 y = f(x) = x2 – x – 2 O x 2 y=x y y = F(x) = x2 – 2 –1 O 2 x ▲ Figure 6.6 152 9781510421738.indb 152 02/02/18 1:13 PM The equation x5 − 5x + 3 = 0 which you met earlier can be rewritten in a number of ways. One of these is 5x = x5 + 3, giving 5 x = F(x) = x + 3 . 5 Figure 6.7 shows the graphs of y = x and y = F(x) in this case. y= y –1 O 1 y=x 6.2 Fixed-point iteration –2 x5 + 3 5 6 x 2 ▲ Figure 6.7 This provides the basis for the iterative formula 5 xn + 1 = xn + 3 . 5 Taking x = 1 as a starting point to find the root in the interval [0, 1], successive approximations are: x1 = 1, x2 = 0.8, x3 = 0.6655, x4 = 0.6261, x5 = 0.6192, x6 = 0.6182, x7 = 0.6181, x8 = 0.6180, x9 = 0.6180. In this case the iteration has converged quite rapidly to the root for which you were looking. ? Another way of arranging − 5x + 3 = 0 is x = 5x − 3. ❯ What other possible rearrangements can you find? ❯ How many are there altogether? x5 5 The iteration process is easiest to understand if you consider the graph. Rewriting the equation f(x) = 0 in the form x = F(x) means that instead of looking for points where the graph of y = f(x) crosses the x-axis, you are now finding the points of intersection of the curve y = F(x) and the line y = x. 153 9781510421738.indb 153 02/02/18 1:13 PM 6 NUMERICAL SOLUTION OF EQUATIONS 6 What you do What it looks like on the graph Choose a value, x1, of x Take a starting point on the x-axis Find the corresponding value of F(x1) Move vertically to the curve y = F(x) Take this value F(x1) as the new value of x, i.e. x2 = F(x1) Move horizontally to the line y = x Find the value of F(x2) and so on Move vertically to the curve y y=x 1 y = F(x) O x4 x3 1 x1 x2 x ▲ Figure 6.8 The effect of several repeats of this procedure is shown in Figure 6.8. The successive steps look like a staircase approaching the root: this type of diagram is called a staircase diagram. In other examples, a cobweb diagram may be produced, as shown in Figure 6.9. y y=x y = F(x) O x1 x3 x5 x4 x2 x ▲ Figure 6.9 154 9781510421738.indb 154 02/02/18 1:13 PM Successive approximations to the root are found by using the formula 6 xn +1 = F(xn). This is an example of an iterative formula. If the resulting values of xn approach some limit, a, then a = F(a), and so a is a fixed point of the iteration. It is also a root of the original equation f(x) = 0. Note 6.2 Fixed-point iteration In the staircase diagram, the values of xn approach the root from one side, but in a cobweb diagram they oscillate about the root. From Figures 6.8 and 6.9 it is clear that the error (the difference between a and xn) is decreasing in both diagrams. Accuracy of the method of rearranging the equation Iterative procedures give you a sequence of point estimates. A staircase diagram, for example, might give the following. 1, 0.8, 0.6655, 0.6261, 0.6192 What can you say at this stage? Looking at the pattern of convergence it seems as though the root lies between 0.61 and 0.62, but you cannot be absolutely certain from the available evidence. To be certain you must look for a change of sign. f(0.61) = +0.034… ❯ f(0.62) = −0.0083… Explain why you are certain that your judgement is now correct. ? CP Note Estimates from a cobweb diagram oscillate above and below the root and so naturally provide you with bounds. Using different arrangements of the equation In the calculations the full calculator values of xn were used, but only the first 4 decimal places have been written down. So far only one possible arrangement of the equation x5 − 5x + 3 = 0 has been used. What happens when you use a different arrangement, for example x = 5 5x − 3 , which leads to the iterative formula xn+1 = 5 5x n − 3? The resulting sequence of approximations is: x1 = 1, x5 = 1.2679..., x9 = 1.2755..., x2 = 1.1486..., x6 = 1.2727..., x10 = 1.2756..., x3 = 1.2236..., x7 = 1.2745..., x11 = 1.2756..., x4 = 1.2554..., x8 = 1.2752..., x12 = 1.2756.... 155 9781510421738.indb 155 02/02/18 1:13 PM 6 The process has clearly converged, but in this case not to the root for which you were looking: you have identified the root in the interval [1, 2]. If instead you had taken x1 = 0 as your starting point and applied the second formula, you would have obtained a sequence converging to the value −1.6180, the root in the interval [−2, −1]. 6 NUMERICAL SOLUTION OF EQUATIONS ❯ ? If a numerical method finds a root of an equation, but not the one you were looking for, is it a failure of the method? 6.3 Problems with the fixed-point iteration method You have already seen that sometimes a particular rearrangement will result in the iterations diverging, or converging to the wrong root. What is happening when this happens? The equation xe x + x 2 − 1 = 0 has a solution of x = 0.48 correct to 2 decimal 1 − xe x . places. The equation can be rearranged into the form x = x xn However, using the iterative formula xn +1 = 1 − xn e with a starting point xn of x1 = 0.5 produces x 2 = 0.35, x 3 = 1.43, x 4 = −3.46, …, with repeated iterations giving values further from the solution. y 4 x y = 1 – xe x 3 2 y =x 1 0 -1 x1 0.5 x2 1 x3 1.5 x ▲ Figure 6.10 The cobweb diagram in Figure 6.10 shows the positions of x1, x2 and x3 for the above iterative formula. 1 − xe x ❯ What feature of the curve y = causes the iteration to diverge? x ? A particular rearrangement of the equation f ( x ) = 0 into the form x = g ( x ) will give an iteration formula that converges to a root α of the equation if the gradient of the curve y = g ( x ) is not too steep near x = α. 156 9781510421738.indb 156 02/02/18 1:13 PM ACTIVITY 6.2 x5 + 3 Try using the iterative formula xn+1 = n to find the roots in the 5 intervals [−2, −1] and [1, 2]. In both cases use each end point of the interval as a starting point. What happens? x5 + 3 . Explain what you find by referring to a sketch of the curve y = 5 1 (i) (ii) Show that the equation x3 − x − 2 = 0 has a root between 1 and 2. The equation is rearranged into the form x = F(x), where F(x) = 3 x + 2. Use the iterative formula suggested by this rearrangement to find the value of the root to 3 decimal places. CP 2 Show that the equation x 4 + x − e x = 0 can be rearranged into each of the following forms: (i) x = e x − x 4 (ii) x = ln ( x 4 + x ) (iii) x = 4 ex − x ex − x x2 Show that the equation e−x − x + 2 = 0 has a root in the interval [2, 3]. The equation is rearranged into the form x = e−x + 2. Use the iterative formula suggested by this rearrangement to find the value of the root to 3 decimal places. (iv) x = CP 3 (i) (ii) CP 4 5 6 6.3 Problems with the fixed-point iteration method Exercise 6B 6 Show that the equation ex + x − 6 = 0 has a root in the interval [1, 2]. (ii) Show that this equation may be written in the form x = ln(6 − x). (iii) Use an iterative formula based on the equation x = ln(6 − x) to calculate the root correct to 3 decimal places. (i) Sketch the curves y = ex and y = x2 + 2 on the same graph. (ii) Use your sketch to explain why the equation ex − x2 − 2 = 0 has only one root. (iii) Rearrange this equation in the form x = F(x). (iv) Use an iterative formula based on the equation found in part (iii) to calculate the root correct to 3 decimal places (i) Show that x2 = ln(x + 1) for x = 0 and for one other value of x. (ii) Use the method of fixed-point iteration to find the second value to 3 decimal places. (i) 157 9781510421738.indb 157 02/02/18 1:13 PM 6 7 (i) (ii) 8 (i) Sketch the graphs of y = x and y = cos x on the same axes, for 0 $ x $ π. 2 Find the solution of the equation x = cos x to 5 decimal places. Given that a ∫ (3e 1x 2 0 satisfies the equation a = 2 ln 16 − a . 6 ⎛ 16 − a n ⎞ Use the iterative formula a n +1 = 2 ln ⎜ with a1 = 2 to find ⎝ 6 ⎟⎠ the value of a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 6 NUMERICAL SOLUTION OF EQUATIONS ( (ii) + 1)dx = 10 , show that the positive constant a ) Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2015 9 (i) The sequence of values given by the iterative formula 2x n 4 + 2, xn +1 = xn 3 with initial value x1 = 2, converges to α. (ii) Use this iterative formula to determine α correct to 2 decimal places, giving the result of each iteration to 4 decimal places. (iii) State an equation that is satisfied by α and hence find the exact value of α. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q2 November 2007 10 (i) By sketching a suitable pair of graphs, show that the equation cos x = 2 − 2x, where x is in radians, has only one root for 0 $ x $ 21 π. (ii) Verify by calculation that this root lies between 0.5 and 1. (iii) Show that, if a sequence of values given by the iterative formula xn +1 = 1 − 21 cos xn converges, then it converges to the root of the equation in part (i). (iv) Use this iterative formula, with initial value x1 = 0.6, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 November 2008 158 9781510421738.indb 158 02/02/18 1:13 PM 11 The diagram shows the curve y = x 2 cos x, for 0 $ x $ 21 π, and its maximum point M. y 6 M 1 π 2 x Show by differentiation that the x coordinate of M satisfies the equation tan x = 2 . x (ii) Verify by calculation that this equation has a root (in radians) between 1 and 1.2. 2 (iii) Use the iterative formula xn +1 = tan−1 x to determine this root (i) ( ) n correct to 2 decimal places. Give the result of each iteration to 4 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 22 Q7 November 2009 12 The diagram shows the curve y = x e2x and its minimum point M. y O 6.3 Problems with the fixed-point iteration method O x M Find the exact coordinates of M. (ii) Show that the curve intersects the line y = 20 at the point whose x coordinate is the root of the equation x = 21 ln 20 . x (iii) Use the iterative formula 20 xn+1 = 21 ln x , (i) ( ) ( ) n with initial value x1 = 1.3, to calculate the root correct to 2 decimal places, giving the result of each iteration to 4 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 June 2009 9781510421738.indb 159 159 02/02/18 1:13 PM 6 13 (i) By sketching a suitable pair of graphs, show that the equation ln x = 2 − x 2 has only one root. (ii) Verify by calculation that this root lies between x = 1.3 and x = 1.4. (iii) Show that, if a sequence of values given by the iterative formula 6 NUMERICAL SOLUTION OF EQUATIONS xn+1 = √(2 − ln xn) converges, then it converges to the root of the equation in part (i). (iv) Use the iterative formula xn +1 = √(2 − ln xn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 22 Q6 June 2010 14 The equation x3 − 8x − 13 = 0 has one real root. (i) (ii) Find the two consecutive integers between which this root lies. Use the iterative formula xn +1 = (8xn + 13) 1 3 to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q2 November 2009 15 The equation x3 − 2x − 2 = 0 has one real root. Show by calculation that this root lies between x = 1 and x = 2. (ii) Prove that, if a sequence of values given by the iterative formula 2x 3 + 2 xn +1 = n2 3xn – 2 converges, then it converges to this root. (iii) Use this iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. (i) Cambridge International AS & A Level Mathematics 9709 Paper 3 Q4 June 2009 16 A curve has parametric equations 1 x= , y = (t + 2 ) . ( 2t + 1)2 The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1. 1 (i) (ii) 160 9781510421738.indb 160 1 Show that p = ( p + 2)6 − 2 . Use an iterative process based on the equation in part (i) to find the value of p correct to 3 decimal places. Use a starting value of 0.7, show the result of each iteration to 5 decimal places. Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2012 02/02/18 1:13 PM KEY POINTS 1 2 3 LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ ■ ■ ■ ■ find an interval in which the root of an equation lies, using change-of-sign methods know circumstances under which change-of-sign methods fail rearrange an equation in the form f(x) = 0 into the form x = F( xn ) carry out a fixed-point iteration of an equation in the form xn +1 = F( xn ), to find the root of an equation to a given degree of accuracy understand that a particular rearrangement of f(x) = 0 may produce an iterative formula that fails to converge. 6 6.3 Problems with the fixed-point iteration method 4 When f(x) is a continuous function, if f(a) and f(b) have opposite signs, there will be at least one root of f(x) = 0 in the interval [a, b]. When an interval [a, b] containing a root has been found, this interval may be reduced systematically by decimal search or interval bisection. Fixed-point iteration may be used to solve an equation f(x) = 0. You can sometimes find a root by rearranging the equation f(x) = 0 into the form x = F(x) and using the iteration xn+1 = F(xn ). Successive iterations of xn+1 = F(xn ) may fail to converge if the gradient of F(x) is not very steep for values of x close to the root. 161 9781510421738.indb 161 02/02/18 1:13 PM P3 7 FURTHER ALGEBRA 7 Further algebra At the age of twenty-one he wrote a treatise upon the Binomial Theorem. ... On the strength of it, he won the Mathematical Chair at one of our smaller Universities. Sherlock Holmes on Professor Moriarty in ‘The Final Problem’ by Sir Arthur Conan Doyle (1859– 1930) ? ❯ How would you find 101 correct to 3 decimal places, without using a calculator? Many people are able to develop a very high degree of skill in mental arithmetic, particularly those whose work calls for quick reckoning. There are also those who have quite exceptional innate skills. Shakuntala Devi, pictured above, is famous for her mathematical speed. On one occasion she found the 23rd root of a 201-digit number in her head, beating a computer by 12 seconds. On another occasion she multiplied 7 686 369 774 870 by 2 465 099 745 779 in just 28 seconds. While most mathematicians do not have Shakuntala Devi’s high level of talent with numbers, they do acquire a sense of when something looks right or wrong. This often involves finding approximate values of numbers, such as 101, using methods that are based on series expansions, and these are the subject of the first part of this chapter. 162 9781510421738.indb 162 02/02/18 1:13 PM INVESTIGATION Square roots Using your calculator, write down the values of 1.02, 1.04, 1.06, …, giving your answers correct to 2 decimal places. What do you notice? Use your results to complete the following, giving the value of the constant k. 7 1 1.02 = (1 + 0.02) 2 ≈ 1 + 0.02k 1 1.04 = (1 + 0.04) 2 ≈ 1 + 0.04k 7.1 The general binomial expansion In Pure Mathematics 1 Chapter 4 you met the binomial expansion in the form ⎛ n⎞ ⎛ n⎞ ⎛ n⎞ ⎛ n⎞ (1 + x )n = 1 + ⎜ 1⎟ x + ⎜ 2⎟ x 2 + ⎜ 3⎟ x 3 + … + ⎜ r ⎟ x r + … ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ which holds when n is any positive integer (or zero), that is n ! $. 7.1 The general binomial expansion What is the largest value of x such that 1 + x ≈ 1 + kx is true for the same value of k? This may also be written as n(n − 1) 2 n (n − 1)(n − 2) 3 x + x +… 2! 3! n(n − 1)(n − 2)…(n − r + 1) r + x +… r! (1 + x )n = 1 + nx + This is a short way of writing ‘n is a natural number’. A natural number is any positive integer or zero. which, being the same expansion as above, also holds when n ! $. The general binomial theorem states that this second form, that is (1 + x )n = 1 + nx + + n(n − 1) 2 n(n − 1)(n − 2) 3 x + x +… 2! 3! n(n − 1(n − 2)…(n − r + 1) r x +… r! is true when n is any real number, but there are two important differences to note when n " $. This is a short way of writing ‘n is not a » The series is infinite (or non-terminating). » The expansion of (1 + x)n is valid only if |x| " 1. natural number’. Proving this result is beyond the scope of an A-level course but you can assume that it is true. Consider now the coefficients in the binomial expansion: 1, n, n(n − 1) , 2! n(n − 1(n − 2) , 3! n(n − 1)(n − 2)(n − 3) , 4! … 163 9781510421738.indb 163 02/02/18 1:13 PM 7 When n = 0, we get 1 0 0 0 0 … n=1 1 1 0 0 0 … ditto n=2 1 2 1 0 0 … ditto n=3 1 3 3 1 0 … ditto n=4 1 4 6 4 1 … ditto (infinitely many zeros) 7 FURTHER ALGEBRA so that, for example (1 + x)2 = 1 + 2x + x2 + 0x3 + 0x4 + 0x5 + … (1 + x)3 = 1 + 3x + 3x2 + x3 + 0x4 + 0x5 + … (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 + 0x5 + … Of course, it is usual to discard all the zeros and write these binomial coefficients in the familiar form of Pascal’s triangle: 1 1 1 1 1 2 1 3 1 3 4 6 1 4 1 and the expansions as (1 + x)2 = 1 + 2x + x2 (1 + x)3 = 1 + 3x + 3x2 + x3 (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 However, for other values of n (where n " $) there are no zeros in the row of binomial coefficients and so we obtain an infinite sequence of non-zero terms. For example: (−3)(−4) (−3)(−4)(−5) (−3)(−4)(−5)(−6) n = −3 gives 1 –3 … 2! 3! 4! −10 … that is 1 −3 6 15 n= 1 2 1 1 2 that is 1 1 2 gives ( 21 )(− 21 ) ( 21) (− 21 )(− 23 ) ( 21 )(− 21 )(− 23 )(− 52 ) 2! − 81 3! 1 16 4! 5 − 128 … … so that (1 + x )−3 = 1 − 3x + 6x 2 − 10 x 3 + 15x 4 + … and 1 1 x3 − 5 x4 + … (1 + x ) 2 = 1 + 21 x − 81 x 2 + 16 128 But remember: these two expansions are valid only if | x | " l. 1 2 ❯ Show that the expansion of (1 + x ) is not valid when x = 8. ? CP 164 9781510421738.indb 164 02/02/18 1:13 PM These examples confirm that there will be an infinite sequence of non-zero coefficients when n " $. In the investigation at the beginning of this chapter you showed that 1+ x ≈ 1+ 7 1 x 2 is a good approximation for small values of x. Notice that these are the first 1 two terms of the binomial expansion for n = 2 . If you include the third term, the approximation is Take y = 1 + 21 x, y = 1 + 21 x − 81 x 2 and y = 1 + x . They are shown in the graph in Figure 7.1 for values of x between −1 and 1. y 1.5 y = 1 + 12 x y = 1 + 12 x – 18 x2 1.0 7.1 The general binomial expansion 1 + x ≈ 1 + 21 x − 81 x 2. 0.5 y= 1+x –1.0 –0.5 0 0.5 1.0 x ▲ Figure 7.1 ACTIVITY 7.1 For n = 21 the first three terms of the binomial expansion are 1 + 21 x − 81 x 2 . Use your calculator to verify the approximate result 1 + x ≈ 1 + 21 x − 81 x 2 for ‘small’ values of x. What values of x can be considered as ‘small’ if you want the result to be correct to 2 decimal places? Now take n = −3. Using the coefficients found earlier suggests the approximate result (1 + x)−3 ≈ 1 − 3x + 6x2. Comment on values of x for which this approximation is correct to 2 decimal places. When | x | " 1, the magnitudes of x2, x3, x4, x5, … form a decreasing geometric sequence. In this case, the binomial expansion converges ( just as a geometric progression converges for −1 " r " 1, where r is the common ratio) and has a sum to infinity. 9781510421738.indb 165 165 02/02/18 1:13 PM ACTIVITY 7.2 7 Compare the geometric progression 1 − x + x2 − x3 + … with the series obtained by putting n = −1 in the binomial expansion.What do you notice? 7 FURTHER ALGEBRA To summarise: when n is not a positive integer or zero, the binomial expansion of (1 + x)n becomes an infinite series, and is only valid when some restriction is placed on the values of x. The binomial theorem states that for any value of n: (1 + x )n = 1 + nx + n(n − 1) 2 n(n − 1)(n − 2) 3 x + x +… 2! 3! where » if n ! $, x may take any value; » if n " $, | x | " 1. Note The full statement is the binomial theorem, and the right-hand side is referred to as the binomial expansion. Example 7.1 Expand (1 − x)−2 as a series of ascending powers of x up to and including the term in x 3, stating the set of values of x for which the expansion is valid. Solution n(n − 1) 2 n(n − 1)(n − 2) 3 x + x +… 2! 3! Replacing n by −2, and x by (−x) gives (1 + x )n = 1 + nx + −2 (1 + (−x )) = 1 + (−2)(−x ) + (−2)(−3) (−2)(−3)(−4) ( −x ) 2 + ( −x ) 3 + … 2! 3! when | – x | < 1 It is important to put brackets round the term −x, since, for example, (−x)2 is not the same as −x2. which leads to (1 − x)−2 ≈ 1 + 2x + 3x 2 + 4 x 3 when | x | < 1. 166 9781510421738.indb 166 02/02/18 1:13 PM Note In Example 7.1 the coefficients of the powers of x form a recognisable sequence, and it would be possible to write down a general term in the expansion. The coefficient is always one more than the power, so the r th term would be rxr −1. Using sigma notation, the infinite series could be written as 7 ∞ ∑ rx r −1 r =1 1 Find a quadratic approximation for and state for which values of t 1 + 2t the expansion is valid. Solution 1 1 −1 2 = 1 = (1 + 2t ) 1 + 2t (1 + 2t ) 2 The binomial theorem states that n(n – 1) 2 n(n − 1)(n − 2) 3 (a + x)n = 1 + nx + x + x +… 2! 3! Replacing n by − 1 and x by 2t gives Remember to put brackets round the term 2t, since (2t)2 is not the same as 2t2. 2 (1 + 2t ) −1 2 = 1+ ( ) −1 2 − 1 )(− 3 ) ( 2 (2t ) + 2 (2t ) 1 2! ⇒ (1 + 2t )− 2 ≈ 1 − t + 3 t 2 2 INVESTIGATION 7.1 The general binomial expansion Example 7.2 2 + … when | 2t | < 1 when | t | < 1 2 Finding coefficients Example 7.1 showed how using the binomial expansion for (1 − x)−2 gave a sequence of coefficients of powers of x which was easily recognisable, so that the particular binomial expansion could be written using sigma notation. Investigate whether a recognisable pattern is formed by the coefficients in the expansions of (1 − x)n for any other negative integers n. The equivalent binomial expansion of (a + x)n when n is not a positive integer is rather unwieldy. It is easier to start by taking a outside the brackets: n (a + x)n = a n 1 + x a The first entry inside the bracket is now 1 and so the first few terms of the expansion are ⎡ ⎤ n(n − 1) x 2 n(n − 1)(n − 2) x 3 (a + x)n = a n ⎢1 + n x + + + …⎥ a a a 2! 3! ⎣ ⎦ x for < 1. a ( ) () 9781510421738.indb 167 () () 167 02/02/18 1:14 PM Note 7 Since the bracket is raised to the power n, any quantity you take out must be raised to the power n too, as in the following example. 7 FURTHER ALGEBRA Example 7.3 Expand (2 + x)−3 as a series of ascending powers of x up to and including the term in x 2, stating the values of x for which the expansion is valid. Solution (2 + x ) –3 = = 1 (2 + x ) 3 1 ( ) (1 + x2 ) 23 1 + x 2 = 81 3 Notice that this is the –3 x . same as 2−3 1 + 2 ( ) –3 Take the binomial expansion n(n − 1) 2 n(n − 1)(n − 2) 3 x + x +… 2! 3! x and replace n by −3 and x by to give 2 –3 2 1 1+ x 1 ⎡1 + ( −3) x + ( −3)( −4) x + …⎤ = when x < 1 ⎢ ⎥⎦ 8 8⎣ 2 2 2! 2 2 (1 + x )n = 1 + nx + ( ) () ≈ 1 8 − 3 x 16 + 3 2 x 16 () when | x | < 2. ? ❯ The chapter began by asking how you would find 101 to 3 decimal places without using a calculator. How would you find it now? Example 7.4 (2 + x ) Find a quadratic approximation for , stating the values of x for which (1 – x 2 ) the expansion is valid. Solution (2 + x ) = (2 + x )(1 − x 2 )−1 (1 – x 2 ) Take the binomial expansion (1 + x )n = 1 + nx + n(n − 1) 2 n(n − 1)(n − 2) 3 x + x +… 2! 3! 168 9781510421738.indb 168 02/02/18 1:14 PM Replace n by −1 and x by (−x 2) to give (−1)(−2)(−x 2 ) 2 −1 +… when | −x 2 | < 1 (1 + (−x 2 )) = 1 + (−1)(−x 2 ) + 2! (1 − x2)−1 = 1 + x2 + … when | x2 | " 1, i.e. when | x | " 1. 7 Multiply both sides by (2 + x) to obtain (2 + x)(1 − x 2)−1: (2 + x)(1 − x 2)−1 ≈ (2 + x)(1 + x 2) ≈ 2 + x + 2x2 when | x | " 1. Sometimes two or more binomial expansions may be used together. If these impose different restrictions on the values of x, you need to decide which is the strictest. Example 7.5 Find a and b such that 7.1 The general binomial expansion The term in x3 has been omitted because the question asked for a quadratic approximation. 1 ≈ a + bx (1 − 2x )(1 + 3x ) and state the values of x for which the expansions you use are valid. Solution 1 −1 −1 (1 − 2x )(1 + 3x ) = (1 − 2x) (1 + 3x) Using the binomial expansion: and ⇒ (1 − 2x)−1 ≈ 1 + (−1)(−2x) for | −2x | " 1 (1 + 3x)−1 ≈ 1 + (−1)(3x) for | 3x | " 1 (1 − 2x)−1(1 + 3x)−1 ≈ (1 + 2x)(1 − 3x) ≈ 1−x (ignoring higher powers of x) giving a = 1 and b = −1. For the result to be valid, both | 2x | " 1 and | 3x | " 1 need to be satisfied. and | 2x | " 1 ⇒ − 21 " x " 21 | 3x | " 1 ⇒ − 13 " x " 13 Both of these restrictions are satisfied if − 13 " x " 13. This is the stricter restriction. 169 9781510421738.indb 169 02/02/18 1:14 PM Note 7 The binomial expansion may also be used when the first term is the variable. For example: ( ) 7 FURTHER ALGEBRA (x + 2)−1 may be written as (2 + x)−1 = 2−1 1 + x 2 and (2x − 1)−3 = [(−1)(1 − 2x)]−3 = (−1)−3(1 − 2x)−3 = −(1 − 2x)−3 −1 ? ❯ What happens when you try to rearrange x − 1 so that the binomial expansion can be used? Exercise 7A 1 2 3 Expand each of the following as a series of ascending powers of x up to and including the term in x3, stating the set of values of x for which the expansion is valid. (i) (1 + x)−3 (ii) (1 + 2x)−3 (iii) (1 − 2x)−3 Expand each of the following as a series of ascending powers of x up to and including the term in x2, stating the set of values of x for which the expansion is valid. 1 1 −1 −1 (i) (1 + x) 2 (ii) (1 + x) 2 (iii) (1 + x) 4 (iv) (1 + x) 4 Expand each of the following as a series of ascending powers of x up to and including the term in x3, stating the set of values of x for which the expansion is valid. −2 2x −2 2x −2 1− 1+ 1+ x (i) (ii) (iii) 3 3 3 For each of the expressions below (a) write down the first three non-zero terms in their expansion as a series of ascending powers of x (b) state the values of x for which the expansion is valid (c) substitute x = 0.1 in both the function and its expansion and calculate the percentage error, where absolute error percentage error = × 100%. true value ( 4 ) ( ) ( ) (1 + x)−2 (ii) (1 + 2x)−1 (iii) 1 − x2 (i) Write down the expansion of (1 − x)3. (ii) Find the first three terms in the expansion of (1 + x)−4 in ascending powers of x. For what values of x is this expansion valid? (1 − x ) 3 (iii) When the expansion is valid, can be written as (1 + x )4 1 + ax + bx2 + higher powers of x. (i) 5 Find the values of a and b. 170 9781510421738.indb 170 02/02/18 1:14 PM CP ( ) 1 Show that ( ) 3 x + cx 2. 8 The expansion of (a + bx)−3 may be approximated by 81 + 16 (i) Find the values of the constants a, b and c. (ii) 9 (i) (ii) For what range of values of x is the expansion valid? 1 Find a quadratic function that approximates to for 3 (1 − 3x ) 2 values of x close to zero. For what values of x is the approximation valid? 10 Expand (2 + 3x)−2 in ascending powers of x, up to and including the term in x 2, simplifying the coefficients. 7 7.1 The general binomial expansion 1 = 1 1 + x − 2. 4 4+ x 2 (ii) Write down the first three terms in the binomial expansion of − 21 in ascending powers of x, stating the range of values of x 1+ x 4 for which this expansion is valid. 2(1 − x ) (iii) Find the first three terms in the expansion of in ascending 4+x powers of x, for small values of x. (3 − x )(1 + x ) 7 Find a quadratic approximation for and state the values of (4 − x ) x for which this is a valid approximation. 6 (i) Cambridge International AS & A Level Mathematics 9709 Paper 3 Q1 June 2007 11 Expand (1 + x) (1 − 2x ) in ascending powers of x, up to and including the term in x2, simplifying the coefficients. 2 Cambridge International AS & A Level Mathematics 9709 Paper 3 Q2 November 2008 12 When (1 + 2x)(1 + ax) 3 , where a is a constant, is expanded in ascending powers of x, the coefficient of the term in x is zero. (i) Find the value of a. (ii) When a has this value, find the term in x3 in the expansion of 2 (1 + 2x)(1 + ax) 3 , simplifying the coefficients. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q5 June 2009 (1 − 6x ) in ascending powers of x up to and including the term in x3, simplifying the coefficients. 13 Expand 3 Cambridge International AS & A Level Mathematics 9709 Paper 31 Q1 June 2011 171 9781510421738.indb 171 02/02/18 1:14 PM b 7 FURTHER ALGEBRA 7 7.2 Review of algebraic fractions f (x ) is an algebraic fraction g(x ) or rational function. It may also be called a rational expression. There are many occasions in mathematics when a problem reduces to the manipulation of algebraic fractions, and the rules for this are exactly the same as those for numerical fractions. If f(x) and g(x) are polynomials, the expression Simplifying fractions To simplify a fraction, you look for a factor common to both the numerator (top line) and the denominator (bottom line) and cancel by it. For example, in arithmetic 15 = 5 × 3 = 3 20 5 × 4 4 and in algebra 6a = 2 × 3 × a = 2 9a 2 3 × 3 × a × a 3a Notice how you must factorise both the numerator and denominator before cancelling, since it is only possible to cancel by a common factor. In some cases this involves putting brackets in. 2a + 4 = 2(a + 2) = 2 a 2 − 4 (a + 2)(a – 2) (a − 2) Multiplying and dividing fractions Multiplying fractions involves cancelling any factors common to the numerator and denominator. For example: 10a × 9ab = 2 × 5 × a × 3 × 3 × a × b = 6a 2 5×5 5b 3b 2 25 3 × b × b As with simplifying, it is often necessary to factorise any algebraic expressions first. a 2 + 3a + 2 × 12 = (a + 1)(a + 2) × 3 × 4 a +1 (a + 1) 9 3×3 (a + 2) 4 = × 1 3 4(a + 2) = 3 Remember that when one fraction is divided by another, you change ÷to × and invert the fraction which follows the ÷symbol. For example: ( x + 1) 12 ÷ 4 = 12 × 4 x 2 − 1 x + 1 ( x + 1)( x − 1) = 3 ( x − 1) 172 9781510421738.indb 172 02/02/18 1:14 PM Adding and subtracting fractions To add or subtract two fractions they must be replaced by equivalent fractions, both of which have the same denominator. 7 For example: 2 3 + 1 4 = 8 12 3 + 12 = 11 12 2x + x = 8x + 3x = 11x 3 4 12 12 12 and 2 + 1 = 8 + 3 = 11 3x 4 x 12x 12x 12x Notice how you only need 12x here, not 12x2. You must take particular care when the subtraction of fractions introduces a sign change. For example: 4 x − 3 − 2x + 1 = 2(4 x − 3) − 3(2x + 1) 6 4 12 = 8x − 6 − 6x − 3 12 = 2x − 9 12 7.2 Review of algebraic fractions Similarly, in algebra: Notice how in addition and subtraction, the new denominator is the lowest common multiple of the original denominators. When two denominators have no common factor, their product gives the new denominator. For example: 2 + 3 = 2( y − 2) + 3( y + 3) y+3 y−2 ( y + 3)( y − 2) = 2 y − 4 + 3y + 9 ( y + 3)( y − 2) = 5y + 5 ( y + 3)( y − 2) = 5( y + 1) ( y + 3)( y − 2) It may be necessary to factorise denominators in order to identify common factors, as shown here. 2b − 3 = 2b − 3 a 2 − b 2 a + b (a + b )(a − b ) (a + b ) 2b − 3(a − b ) (a + b )(a − b ) = 5b − 3a (a + b )(a − b ) = (a + b) is a common factor. 173 9781510421738.indb 173 02/02/18 1:14 PM 7 FURTHER ALGEBRA 7 Exercise 7B Simplify the expressions in questions 1 to 10. 1 6a × a b 9b 2 2 5xy ÷ 15xy 2 3 3 x2 − 9 x − 9x + 18 4 5x − 1 × x 2 + 6 x + 9 x + 3 5x 2 + 4 x − 1 5 4 x 2 − 25 4 x 2 + 20x + 25 6 a 2 + a − 12 × 3 5 4a − 12 7 4 x 2 − 9 ÷ 2x − 3 x 2 + 2x + 1 x 2 + x 8 2p + 4 ÷ ( p 2 − 4) 5 9 a2 − b2 2a + ab − b 2 2 2 10 x 2 + 8x + 16 × x 2 + 2x − 3 x 2 + 6x + 9 x 2 + 4x In questions 11 to 24 write each of the expressions as a single fraction in its simplest form. x − ( x + 1) 11 1 + 1 12 3 4 4 x 5x 13 a + 1 a +1 a −1 14 2 + 3 x−3 x−2 15 x − 1 x −4 x+2 16 p2 p2 − 2 p −1 p +1 17 2 − a a + 1 a2 + 1 18 2y − 4 ( y + 2) 2 y + 4 19 x+ 1 x +1 20 2 − 3 b 2 + 2b + 1 b + 1 21 2 3 + 3( x − 1) 2( x + 1) 22 6 + 2x 5( x + 2) ( x + 2) 2 23 2 − a−2 a + 2 2a 2 + a − 6 24 1 +1+ 1 x−2 x x+2 2 2 7.3 Partial fractions Sometimes, it is easier to deal with two or three simple separate fractions than it is to handle one more complicated one. For example: 1 (1 + 2x )(1 + x ) may be written as 2 – 1 . (1 + 2x ) (1 + x ) 174 9781510421738.indb 174 02/02/18 1:14 PM partial fractions. When finding partial fractions you must always assume the most general numerator possible, and the method for doing this is illustrated in the following examples. 7 7.3 Partial fractions 2 1 – 1 you can then do is written as (1 + 2x ) (1 + x ) (1 + 2x )(1 + x ) binomial expansions on the two fractions, and so find an expansion for the original fraction. » When integrating, it is easier to work with a number of simple fractions than a combined one. For example, the only analytic method for 1 2 involves first writing it as integrating − 1 . (1 + 2x )(1 + x ) (1 + 2x ) (1 + x ) You will meet this application in Chapter 8. 1 This process of taking an expression such as and writing it in (1 + 2x )(1 + x ) 2 1 the form (1 + 2x ) − (1 + x ) is called expressing the algebraic fraction in » When Type 1: Denominators of the form (ax + b)(cx + d)(ex + f ) Example 7.6 Express 4+x as a sum of partial fractions. (1 + x )(2 – x ) Solution Assume 4+x ≡ A + B (1 + x )(2 – x ) 1 + x 2 – x Remember: a linear denominator ⇒ a constant numerator if the fraction is to be a proper fraction. Multiplying both sides by (1 + x)(2 − x) gives 4 + x ≡ A(2 − x) + B(1 + x). 1 ! This is an identity; it is true for all values of x. There are two possible ways in which you can find the constants A and B. You can either 1 (two values are needed to give two » substitute any two values of x in ! equations to solve for the two unknowns A and B); or » equate the constant terms to give one equation (this is the same as putting x = 0) and the coefficients of x to give another. Sometimes one method is easier than the other, and in practice you will often want to use a combination of the two. Method 1: Substitution Although you can substitute any two values of x, the easiest to use are x = 2 and x = −1, since each makes the value of one bracket zero in the identity. ➜ 175 9781510421738.indb 175 02/02/18 1:14 PM 4 + x ≡ A(2 − x) + B(1 + x) 7 x=2 ⇒ 4 + 2 = A(2 − 2) + B(1 + 2) 6 = 3B x = −1 ⇒ ⇒ B=2 4 − 1 = A(2 + 1) + B(1 − 1) 3 = 3A ⇒ A=1 Substituting these values for A and B gives 7 FURTHER ALGEBRA 4+x 1 2 ≡ + (1 + x )(2 − x ) 1 + x 2 − x Method 2: Equating coefficients In this method, you write the right-hand side of 4 + x ≡ A(2 − x) + B(1 + x) as a polynomial in x, and then compare the coefficients of the various terms. 4 + x ≡ 2A − Ax + B + Bx 4 + 1x ≡ (2A + B) + (−A + B)x Equating the constant terms: 4 = 2A + B Equating the coefficients of x: 1 = −A + B These are simultaneous equations in A and B. Solving these simultaneous equations gives A = 1 and B = 2 as before. ? ❯ In each of these methods the identity (≡) was later replaced by equality (=). Why was this done? In some cases it is necessary to factorise the denominator before finding the partial fractions. Example 7.7 Express x(5x + 7) as a sum of partial fractions. (2x + 1)( x 2 − 1) Solution x(5x + 7) x(5x + 7) ≡ (2x + 1)( x 2 − 1) (2x + 1)( x + 1)( x − 1) Start by factorising the denominator fully, replacing (x 2 − 1) with (x + 1)(x − 1). There are three factors in the denominator, so write x(5x + 7) ≡ A + B + C (2x + 1)( x + 1)( x − 1) 2x + 1 x + 1 x − 1 Multiplying both sides by (2x + 1)( x + 1)( x − 1) gives x(5x + 7) ≡ A( x + 1)( x − 1) + B(2x + 1)( x − 1) + C (2x + 1)( x + 1) 176 9781510421738.indb 176 02/02/18 1:14 PM Substituting x = 1 gives: 12 = 6C ⇒ Notice how a combination of the two methods is used. C=2 Substituting x = −1 gives: −2 = 2B ⇒ B = −1 Equating coefficients of x2 gives: 5 = A + 2B + 2C As B = −1 and C = 2: 5=A−2+4 A=3 x(5x + 7) ≡ 3 − 1 + 2 Hence (2x + 1)( x + 1)( x − 1) 2x + 1 x + 1 x − 1 In the next example the orders of the numerator (top line) and the denominator (bottom line) are the same. 7.3 Partial fractions ⇒ Example 7.8 7 2 Express 6 − x 2 as a sum of partial fractions. 4−x Solution Start by dividing the numerator by the denominator. In this case the quotient is 1 and the remainder is 2. So Now find 6 − x2 2 = 1+ 4 − x2 4 − x2 You can also use this method when the order of the numerator is greater than that of the denominator. 2 . 4 − x2 2 2 A B ≡ ≡ + 4 − x 2 (2 + x )(2 − x ) 2 + x 2 − x Multiplying both sides by (2 + x)(2 − x) gives 2 ≡ A(2 − x) + B(2 + x) 2 ≡ (2A + 2B) + x(B − A) 2 = 2A + 2B Equating constant terms: A+B=1 so 0 = B − A, so B = A Equating coefficients of x: 1 gives Substituting in ! 1 ! A=B= 1 2 Using these values 1 1 1 1 2 + ≡ 2 + 2 ≡ (2 + x )(2 − x ) 2 + x 2 − x 2(2 + x ) 2(2 − x ) So 6 − x2 ≡ 1+ 1 1 + 2(2 + x ) 2(2 − x ) 4 − x2 177 9781510421738.indb 177 02/02/18 1:14 PM 7 FURTHER ALGEBRA 7 Exercise 7C Write the expressions in questions 1 to 15 as a sum of partial fractions. 1 5 ( x − 2)( x + 3) 2 1 x( x + 1) 3 6 ( x − 1)( x − 4) 4 x+5 ( x − 1)( x + 2) 5 3x (2x − 1)( x + 1) 6 4 x 2 − 2x 7 2 ( x − 1)(3x − 1) 8 x −1 x 2 − 3x − 4 9 x+2 2x 2 − x 10 7 2x 2 + x − 6 11 2x − 1 2x 2 + 3x − 20 12 2x + 5 18x 2 − 8 6x 2 + 22x + 18 13 ( x + 1)( x + 2)( x + 3) 4 x 2 − 25x − 3 14 (2 x + 1)( x − 1)( x − 3) 15 5x 2 + 13x + 10 (2x + 3)( x 2 − 4) Type 2: Denominators of the form (ax + b)(cx 2 + d) Example 7.9 Express 2x + 3 as a sum of partial fractions. ( x − 1)( x 2 + 4) Solution You need to assume a numerator of order 1 for the partial fraction with a denominator of x 2 + 4, which is of order 2. Bx + C is the most 2x + 3 ≡ A + Bx + C general numerator of ( x − 1)( x 2 + 4) x − 1 x 2 + 4 order 1. Multiplying both sides by (x − 1)(x 2 + 4) gives 2x + 3 ≡ A(x2 + 4) + (Bx + C )(x − 1) x=1 ⇒ 5 = 5A ⇒ 1 ! A=1 The other two unknowns, B and C, are most easily found by equating 1 may be rewritten as coefficients. Identity ! 2x + 3 ≡ (A + B)x 2 + (−B + C )x + (4A − C) Equating coefficients of x2: 0=A+B ⇒ B = −1 Equating constant terms: 3 = 4A − C ⇒ C=1 This gives 2x + 3 ≡ 1 + 1− x ( x − 1)( x 2 + 4) x − 1 x 2 + 4 178 9781510421738.indb 178 02/02/18 1:14 PM Type 3: Denominators of the form (ax + b)(cx + d )2 The factor (cx + d)2 is of order 2, so it would have an order 1 numerator in the partial fractions. However, in the case of a repeated factor there is a simpler form. 4x + 5 Consider (2x + 1) 2 7 2(2x + 1) + 3 (2x + 1) 2 This can be written as 2(2x + 1) 3 + (2x + 1) 2 (2x + 1) 2 ≡ 2 3 + (2x + 1) (2x + 1) 2 7.3 Partial fractions ≡ Note In this form, both the numerators are constant. px + q In a similar way, any fraction of the form can be written as (cx + d ) 2 A + B (cx + d ) (cx + d ) 2 When expressing an algebraic fraction in partial fractions, you are aiming to find the simplest partial fractions possible, so you would want the form where the numerators are constant. Example 7.10 Express x +1 as a sum of partial fractions. ( x − 1)( x − 2) 2 Solution Let x +1 ≡ A + B + C ( x − 1)( x − 2) 2 ( x − 1) ( x − 2) ( x − 2) 2 Notice that you only need (x − 2)2 here and not (x − 2)3. Multiplying both sides by (x − 1)(x − 2)2 gives x + 1 ≡ A(x − 2)2 + B(x − 1)(x − 2) + C(x − 1) x = 1 (so that x − 1 = 0) ⇒ 2 = A(−1)2 x = 2 (so that x − 2 = 0) ⇒ 3=C ⇒ 0=A+B Equating coefficients of x2: ⇒ A=2 ⇒ B = −2 This gives x +1 3 ≡ 2 − 2 + ( x − 1)( x − 2) 2 x − 1 x − 2 ( x − 2) 2 179 9781510421738.indb 179 02/02/18 1:14 PM 7 Example 7.11 2 Express 52x − 3 as a sum of partial fractions. x ( x + 1) Solution Let 5x 2 − 3 ≡ A + B + C x 2 ( x + 1) x x 2 x + 1 Multiplying both sides by x 2(x + 1) gives 7 FURTHER ALGEBRA 5x2 − 3 ≡ Ax(x + 1) + B(x + 1) + Cx2 x=0 x = −1 ⇒ ⇒ −3 = B +2 = C Equating coefficients of x2: +5 = A + C 5x 2 − 3 3 3 2 This gives x 2 ( x + 1) ≡ x − x 2 + x + 1 ⇒ A=3 7.4 Using partial fractions with the binomial expansion One of the most common reasons for writing an expression in partial fractions is to enable binomial expansions to be applied, as in the following example. Example 7.12 2x + 7 in partial fractions and hence find the first three terms ( x − 1)( x + 2) of its binomial expansion, stating the values of x for which this is valid. Express Solution 2x + 7 ≡ A + B ( x − 1)( x + 2) ( x − 1) ( x + 2) Multiplying both sides by (x − 1)(x + 2) gives 2x + 7 ≡ A(x + 2) + B(x − 1) x=1 ⇒ 9 = 3A ⇒ A=3 x = −2 ⇒ 3 = −3B ⇒ B = −1 3 1 2x + 7 This gives ( x − 1)( x + 2) ≡ ( x − 1) − ( x + 2) In order to obtain the binomial expansion, each bracket must be of the form (1 ± …), giving 2x + 7 1 ≡ −3 − ( x − 1)( x + 2) (1 − x ) 2 1+ x 2 ≡ −3(1 − x )−1 180 9781510421738.indb 180 ( ) − (1 + x ) 2 1 2 −1 1 ! 02/02/18 1:14 PM The two binomial expansions are and for | x | " 1 (1 − x)−1 = 1 + (−1)(−x) + (−1)(−2) (−x)2 + … 2! ≈ 1 + x + x2 −1 ( −1)( −2) x 2 1+ x = 1 + ( −1) x + + for x " 1 2 2 2! 2 2 2 ≈ 1− x + x 2 4 ( ) () 7 () 2x + 7 ≈ −3(1 + x + x 2 ) − ( x − 1)( x + 2) = − 72 − 11 x− 4 1 2 (1 − x2 + x4 ) 2 25 2 x 8 The expansion is valid when | x | " 1 and P x P " 1. The stricter of these is 2 | x | " 1. ACTIVITY 7.3 Find a binomial expansion for the function 1 f(x) = (1 + 2x )(1 − x ) and state the values of x for which it is valid (i) by writing it as (1 + 2x)−1(1 − x)−1 (ii) by writing it as [1 + (x − 2x2)]−1 and treating (x − 2x2) as one term (iii) by first expressing f(x) as a sum of partial fractions. 7.4 Using partial fractions with the binomial expansion 1 gives Substituting these in ! Decide which method you find simplest for the following cases. (a) When a linear approximation for f(x) is required. (b) When a quadratic approximation for f(x) is required. (c) When the coefficient of xn is required. Exercise 7D 1 Express each of the following fractions as a sum of partial fractions. 5 − 2x 4 4 + 2x (i) (ii) (iii) 2 x − ( 1) 2 ( x + 2) (1 − 3x )(1 − x ) 2 (2x − 1)( x + 1) (iv) 2x + 1 ( x − 2)( x 2 + 4) (v) 2x 2 + x + 4 (2x 2 − 3)( x + 2) (vi) x2 − 1 x (2x + 1) (vii) x2 + 3 x(3x 2 − 1) (viii) 2x 2 + x + 2 (2x 2 + 1)( x + 1) (ix) 4x 2 − 3 x(2x − 1) 2 2 181 9781510421738.indb 181 02/02/18 1:14 PM 7 2 Find the first three terms in ascending powers of x in the binomial expansion of the following fractions. 4 + 2x 4 (i) (ii) (2 x − 1)( x 2 + 1) (1 − 3x )(1 − x ) 2 2x + 1 5 − 2x (iv) ( x − 2)( x 2 + 4) ( x − 1) 2 ( x + 2) 6x − 8 (i) Write 2 in the form Ax2 + B + C . ( x + 1)( x + 1) x +1 x +1 (ii) Hence find the first three terms in the binomial expansion of 6x − 8 , stating the values of x for which the expansion is valid. ( x 2 + 1)( x + 1) (iii) 7 FURTHER ALGEBRA 3 4 (i) 5 Find a cubic approximation for 6 (i) Express (ii) 10 in (2 − x )(1 + x 2 ) ascending powers of x, up to and including the term in x3, simplifying the coefficients. Write down the first three terms in the binomial expansion of 1 in ascending powers of x. (1 + 2x )(1 + x ) (ii) For what values of x is this expansion valid? 2 , stating the range of ( x + 1)( x 2 + 1) values of x for which the expansion is valid. 10 in partial fractions. (2 − x )(1 + x 2 ) Hence, given that | x | " 1, obtain the expansion of Cambridge International AS & A Level Mathematics 9709 Paper 3 Q9 June 2006 7 (i) (ii) 3x 2 + x in partial fractions. ( x + 2)( x 2 + 1) 3x 2 + x Hence obtain the expansion of in ascending powers ( x + 2)( x 2 + 1) of x, up to and including the term in x3. Express Cambridge International AS & A Level Mathematics 9709 Paper 3 Q9 November 2005 8 (i) (ii) 2 − x + 8x 2 in partial fractions. (1 − x )(1 + 2x )(2 + x ) 2 − x + 8x 2 Hence obtain the expansion of in ascending (1 − x )(1 + 2x )(2 + x ) powers of x, up to and including the term in x2. Express Cambridge International AS & A Level Mathematics 9709 Paper 3 Q9 November 2007 9 182 9781510421738.indb 182 2 Let f( x ) = 2x − 7x2 − 1 . ( x − 2)( x + 3) 2x 2 − 7x − 1 (i) Express f( x ) = in partial fractions. ( x − 2)( x 2 + 3) (ii) Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x2. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q7 November 2013 02/02/18 1:14 PM KEY POINTS 1 ( ) 3 When multiplying algebraic fractions, you can only cancel when the same factor occurs in both the numerator and the denominator. 4 When adding or subtracting algebraic fractions, you first need to find a common denominator. 5 The easiest way to solve any equation involving fractions is usually to multiply both sides by a quantity which will eliminate the fractions. 6 A proper algebraic fraction with a denominator which factorises can be decomposed into a sum of proper partial fractions. 7 The following forms of partial fraction should be used. px 2 + qx + r A B C ≡ + + (ax + b )(cx + d )(ex + f ) ax + b cx + d ex + f px 2 + qx + r A Bx + C ≡ + (ax + b )(cx 2 + d ) ax + b cx 2 + d px 2 + qx + r A B C ≡ + + (ax + b )(cx + d )2 ax + b cx + d (cx + d )2 7 7.4 Using partial fractions with the binomial expansion 2 The general binomial expansion for n ! " is n(n − 1) 2 n(n − 1)(n − 2) 3 x + x +…. (1 + x)n = 1 + nx + 2! 3! In the special case when n ! $, the series expansion is finite and valid for all x. When n " $, the series expansion is non-terminating (infinite) and valid only if | x | " 1. n When n " $, (a + x)n should be written as a n 1 + x before a obtaining the binomial expansion. LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ ■ ■ ■ find and use the binomial expansion (1 + x)n, where n is a rational number and | x | " 1 adapt the standard binomial expansion for cases in which the constant term is not 1 know the condition for a binomial expansion to be valid express algebraic fractions as partial fractions when the fraction has a denominator of the form ■ (ax + b)(cx + d )(ex + f ) ■ (ax + b)(cx2 + d ) ■ (ax + b)(cx + d )2. 183 9781510421738.indb 183 02/02/18 1:14 PM P3 8 FURTHER CALCULUS 8 Further calculus The mathematical process has a reality and virtue in itself, and once discovered it constitutes a new and independent factor. Winston Churchill (1876–1965) ? ❯ How would you estimate the volume of water in a sea wave? In this chapter you will learn how to integrate more complicated functions. For example, you might model a sea wave as a sine wave, so finding the volume of water in the wave might involve integrating a sine function. 8.1 Differentiating tan−1x You can use implicit differentiation to differentiate inverse trigonometrical functions. Example 8.1 Find dy in terms of x when y = tan−1x. dx Solution y = tan−1x 184 9781510421738.indb 184 ⇒ tan y = tan (tan−1x) ⇒ tan y = x A function and its inverse cancel out. 1 ! 02/02/18 1:14 PM d (tan y ) = d ( x ) dx dx d (tan y ) dy = d ( x ) dx dx dy Use implicit differentiation. 8 The derivative of tan y is sec2 y. ⇒ dy =1 dx dy = 1 dx sec 2 y dy 1 Using sec2 y ≡tan2 y + 1 gives dx = 1 + tan 2 y dy 1 into ! 2 gives Substituting ! = 1 dx 1 + x 2 Make dy the subject. dx 2 ! Remember: tan y = x 8.1 Differentiating tan−1x ⇒ sec 2 y You can also differentiate inverse functions by using the fact that dy 1 = dx dx dy −1 Given y = tan x then x = tan y dy dx 1 1 = sec 2 y ⇒ = dy dx sec 2 y = 1 + x 2 , as before. e You can use the same method to differentiate y = sin−1kx and y = cos−1 kx to obtain: dy k y = sin−1kx ⇒ = dx 1 −k 2 x 2 and y = cos−1kx ⇒ Exercise 8A 1 dy −k = dx 1 −k 2 x 2 dy in terms of x for each of the following. dx (i) y = tan −1 2x (ii) y = tan −1 3x x (iii) y = tan −1 (iv) y = tan −1 x 2 3 Find () 9781510421738.indb 185 CP 2 CP 3 CP 4 CP 5 () k2 . k x2 + 1 Given f ( x ) = 1 tan −1 x where k is a constant, prove that f '( x ) = 2 1 2 . x +k k k ⎛ ⎞ Prove that the equation of the tangent to the curve y = tan − 1 ⎜ x ⎟ at the ⎝ 3⎠ point where x = 3 is 12y = 3x + 4π − 3 3. Given f ( x ) = k tan −1(kx ) where k is a constant, prove that f '( x ) = () 2 Prove that the equation of the normal to the curve y = tan −1( 2x ) at the 2 point where x = 2 is y + 2x = 4 + π . 4 185 02/02/18 1:14 PM e 6 8 8 FURTHER CALCULUS PS 7 dy in terms of x for each of the following. dx y = cos −1x (i) (ii) y = sin −1x Find the equations of the tangent and the normal to the curve y = cos −1 2x at the point 1 , π . 4 3 Find ( ) 8.2 Integration by substitution Figure 8.1 shows the graph of y = y x. 2 y= x 1 0 1 2 3 4 x ▲ Figure 8.1 ? ❯ How does it allow you to find the shaded area in the graph in Figure 8.2? y 2 y= x+1 1 0 1 2 3 4 x ▲ Figure 8.2 The graph of y = Figure 8.3. x − 1 is shown in y y= x–1 The shaded area is given by ∫ 5 1 x − 1d x = ∫ 5 1 1 ( x – 1) 2 d x. O 1 5 x ▲ Figure 8.3 186 9781510421738.indb 186 02/02/18 1:14 PM You may remember how to investigate this by inspection. However, you can also transform the integral into a simpler one by using the substitution u = x − 1 to get b ∫ u d u. 1 2 a 8 When you make this substitution it means that you are now integrating with respect to a new variable, namely u. The limits of the integral, and the ‘dx’, must be written in terms of u. x=1 The new limits are given by ⇒ u=1−1=0 The integral now becomes: ∫ 4 ⎡ 23 ⎤ u u du = ⎢ 3 ⎥ u =0 ⎢⎣ 2 ⎥⎦0 u =4 8.2 Integration by substitution and x=5 ⇒ u = 5 − 1 = 4. d u Since u = x − 1, = 1. dx Even though du is not a fraction, it is usual to treat it as one in this situation dx (see the warning at the bottom of the page), and to write the next step as ‘du = dx’. 1 2 4 ⎡ 23 ⎤ 2u = ⎢⎣ 3 ⎥⎦ 0 = 5 13 This method of integration is known as integration by substitution. It is a very powerful method which allows you to integrate many more functions. Since you are changing the variable from x to u, the method is also referred to as integration by change of variable. The last example included the statement ‘du = dx’. Some mathematicians are reluctant to write such statements on the grounds that du and dx may only be used in the form du , i.e. as a gradient. This is not in fact dx true; there is a well-defined branch of mathematics which justifies such statements but it is well beyond the scope of this book. In the meantime it may help you to think of it as shorthand for ‘in the limit as δx → 0, δu → 1, and so δu = δx ’. δx Example 8.2 Evaluate ∫ 3 1 ( x + 1) 3 dx by making a suitable substitution. ➜ 187 9781510421738.indb 187 02/02/18 1:14 PM Solution 8 y = (x + 1)3 y Let u = x + 1. x=1 ⇒ u=1+1=2 x=3 ⇒ u=3+1=4 Converting the limits: Converting dx to du: 8 FURTHER CALCULUS du = 1 ⇒ du = dx. dx ∫ 3 ( x + 1) 3 dx = 1 ∫ 4 2 u 3 du 4 4 = ⎡⎢u ⎤⎥ ⎣ 4 ⎦2 4 4 =4 − 2 4 4 = 60 1 O 1 3 x ▲ Figure 8.4 ? ❯ Can integration by substitution be described as the reverse of the chain rule? Example 8.3 Evaluate 4 ∫3 2x (x 2 − 4) 1 2 dx by making a suitable substitution. Solution Notice that 2x is the derivative of the expression in the brackets, x2 − 4, and so u = x2 − 4 is a natural substitution to try. du This gives = 2x ⇒ du = 2x dx dx Converting the limits: x = 3 ⇒ u=9−4=5 x=4 ⇒ u = 16 − 4 = 12 So the integral becomes: 4 1 2 12 ∫3 (x2 − 4) 2x dx = ∫5 1 u 2 du 12 ⎡ 23 ⎤ = ⎢ 2u ⎥ ⎣ 3 ⎦5 = 20.3 (to 3 significant figures) Note In the last example there were two expressions multiplied together; the second expression is raised to a power. The two expressions are in this case related, since the first expression, 2x, is the derivative of the expression in brackets, x 2 − 4. It was this relationship that made the integration possible. 188 9781510421738.indb 188 02/02/18 1:14 PM Example 8.4 ∫ Find x(x2 + 2)3 dx by making an appropriate substitution. 8 Solution Since this is an indefinite integral there are no limits to change, and the final answer will be a function of x. ∫ You only have x dx in the integral, not 2x dx. ∫ = ∫ u3 × 21 du u4 +c 8 ( x 2 + 2)4 +c = 8 = 8.2 Integration by substitution Let u = x2 + 2, then: du = 2x ⇒ 1 du = x dx 2 dx So x(x2 + 2)3 dx = (x2 + 2)3x dx Always remember, when finding an indefinite integral by substitution, to substitute back at the end. The original integral was in terms of x, so your final answer must be too. Example 8.5 ∫ By making a suitable substitution, find x x − 2 dx. Solution This question is not of the same type as the previous ones since x is not the derivative of (x − 2). However, by making the substitution u = x − 2 you can still make the integral into one you can do. Let u = x − 2, then: du = 1 ⇒ du = dx dx There is also an x in the integral so you need to write down an expression for x in terms of u. Since u = x − 2 it follows that x = u + 2. 1 In the original integral you can now replace x − 2 by u 2 , dx by du and x by u + 2. ∫x ∫ = ∫(u 1 x − 2 dx = (u + 2)u 2 du 3 2 5 1 + 2u 2 )du 3 = 25 u 2 + 43 u 2 + c 3 2 Replacing u by x − 2 and tidying up gives 15 (3x + 4)(x − 2) 2 + c. 189 9781510421738.indb 189 02/02/18 1:14 PM ACTIVITY 8.1 8 Complete the algebraic steps involved in tidying up the answer to Example 8.5. 8 FURTHER CALCULUS Exercise 8B 1 Find the following indefinite integrals by making the suggested substitution. Remember to give your final answer in terms of x. (i) 3x2(x3 + 1)7 dx, u = x3 + 1 (ii) 2x(x2 + 1)5 dx, u = x2 + 1 (iii) (v) 2 ∫1 5 ∫1 x x − 1 dx Find the area of the shaded region for each of the following graphs. + 1) O 5 y y= O 1 PS (ii) y = 6x(x2 + 1)3 y –1 4 2 ∫−1 (i) PS ∫ (iv) ∫ x 2x − 5 dx, u = 2x2 − 5 x dx, u = x + 9 (vi) ∫ x+9 Evaluate each of the following definite integrals by using a suitable substitution. Give your answer to 3 significant figures where appropriate. 5 2 (i) x 2(x 3 + 1)2 dx (ii) 2x (x − 3)5 dx (iii) 3 ∫ ∫ 3x2(x3 − 2)4 dx, u = x3 − 2 ∫ x 2x + 1 dx, u = 2x + 1 1 2 x (x – 1)3 4 x x dy 1 and passes through the = dx 2x + 3 point 0, 2 3 . Find the equation of the curve. The diagram shows the curve y = x + 1 and the line y = 2. Find the area of the shaded region. A curve has a gradient function ( ) y 2 y = x+ 1 1 O 6 190 9781510421738.indb 190 Evaluate ∫ 2 1 3 x x 2 ( x 3 − 4) d x using the substitution u = x 3 − 4 . 02/02/18 1:14 PM 7 ∫ ∫ 8 ∫ 8.3 Integrals involving exponentials and natural logarithms In Chapter 5 you met integrals involving logarithms and exponentials. That work is extended here using integration by substitution. Example 8.6 By making a suitable substitution, find 4 ∫0 2x e x dx. 2 Solution 4 4 ∫0 2x ex dx = ∫0 ex 2x dx 2 2 Since 2x is the derivative of x2, let u = x2. du = 2x ⇒ du = 2x dx dx The new limits are given by x=0 ⇒ and x=4 ⇒ The integral can now be written as 16 ∫0 Example 8.7 Evaluate 5 ∫1 x eu du [ ] = eu 16 0 = e16 − e0 = 8.89 × 106 (to 3 significant figures) 2x dx. +3 2 Solution In this case, substitute u = x2 + 3, so that du dx = 2x ⇒ du = 2x dx The new limits are given by x=1 ⇒ u=4 and x = 5 ⇒ u = 28 9781510421738.indb 191 u=0 u = 16 8 8.3 Integrals involving exponentials and natural logarithms Integrate with respect to x. 1 4 + 3 (a) (b) 6x(1 + x2) 2 3 x x 4 (1 + x )3 (ii) Show that the substitution x = u2 transforms 1 dx into b x 3 an integral of the form k(1 + u) du. a State the values of k, a and b. Evaluate this integral. ln x Find the integral (1 + ln x )dx using the substitution u = 1 + ln x . x (i) y y= O ▲ Figure 8.5 1 2x x2 + 3 5 x ➜ 191 02/02/18 1:14 PM 5 ∫1 x 8 2x dx = +3 2 28 1 ∫4 u du [ ] 28 = ln u 4 = ln 28 − ln 4 = 1.95 (to 3 significant figures) ∫ ff('(xx)) dx, where f(x) = x2 + 3. In such cases the substitution u = f(x) transforms the integral into ∫ 1 du. The answer is then u 8 FURTHER CALCULUS The last example is of the form ln u + c or ln(f(x)) + c (assuming that u = f(x) is positive). This result may be stated as the working rule below. If you obtain the top line when you differentiate the bottom line, the integral is the natural logarithm of the bottom line. So, ∫ ff('(xx)) dx = ln | f(x) | + c. Example 8.8 Evaluate 5x 4 + 2 x 5 + x 2 + 4 dx. 2 ∫1 x Solution You can work this out by substituting u = x5 + x2 + 4 but, since differentiating the bottom line gives the top line, you could apply the rule above and just write: 2 2 5x 4 + 2 x 5 2 5 + x 2 + 4 dx = ln(x + x + 4) 1 = ln 40 − ln 6 [ ∫1 x ] = 1.90 (to 2 significant figures) In the next example some adjustment is needed to get the top line into the required form. Example 8.9 Evaluate 1 ∫0 x x 5 dx. +7 6 Solution The differential of x6 + 7 is 6x5, so the integral is rewritten as Integrating this gives Exercise 8C 1 1 6 [ ] 1 ∫ ln(x6 + 7) 0 or 0.022 (to 2 significant figures). Find the following indefinite integrals. 2x + 3 2x dx (i) (ii) 2 + 9 x − 1 dx 2 3 x x +1 ∫ 1 1 6x 5 dx. 6 0 x6 + 7 ∫ (iii) ∫12x 2 ex dx 3 192 9781510421738.indb 192 02/02/18 1:14 PM 2 Find the following definite integrals. Where appropriate give your answers to 3 significant figures. 3 6 x−3 (i) 2x e−x 2 dx (ii) 2 − 6 x + 9 dx x 2 4 The sketch shows the graph y of y = x ex2. (i) Find the area of region A. (ii) Find the area of region B. –1 (iii) Hence write down the total O A area of the shaded region. ∫ 3 B 2 x x+2 is shown below. x 2 + 4x + 3 The graph of y = y –5 –4 –3 CP 5 –2 –1 0 1 2 x Find the area of each shaded region. A curve has the equation y = (x + 3)e−x. dy (i) Find . dx x + 2 dx . (ii) Hence find ex (iii) Find the x and y coordinates of the stationary point S on the curve. d 2y (iv) Calculate at the point S. dx 2 What does its value indicate about the stationary point? 2 + ln u du into 2 + x dx. (v) Show that the substitution u = ex converts ex u2 e 2 + ln u (vi) Hence evaluate du. u2 1 ex for values of x between (i) Sketch the curve with equation y = x e +1 0 and 2. (ii) Find the area of the region enclosed by this curve, the axes and the line x = 2. e 2t dt. (iii) Find the value of 2 1 t +1 (iv) Compare your answers to parts (ii) and (iii). Explain this result. 8.3 Integrals involving exponentials and natural logarithms 4 8 ∫ ∫ PS ∫ ∫ 6 ∫ ∫ 193 9781510421738.indb 193 02/02/18 1:14 PM 7 8 FURTHER CALCULUS 8 x and its maximum point x2 + 1 M. The shaded region R is bounded by the curve and by the lines y = 0 and x = p. y (i) Calculate the x coordinate of M. M (ii) Find the area of R in terms of p. The diagram shows part of the curve y = (iii) Hence calculate the value R O of p for which the area of R is 1, giving your answer correct to 3 significant figures. 8 (i) (ii) ∫ 4 x Cambridge International AS & A Level Mathematics 9709 Paper 3 Q9 June 2005 1 dx. x(4 − x ) Use the substitution u = x to show that I = Hence show that I = 21 ln 3. Let I = p 1 ∫ 2 1 2 du. u(4 − u ) Cambridge International AS & A Level Mathematics 9709 Paper 3 Q7 June 2007 8.4 Integrals involving trigonometrical functions In Chapter 5 you met integrals involving trigonometrical functions. That work is extended here using integration by substitution. Example 8.10 ∫ Find 2x cos(x2 + 1) dx. Solution Make the substitution u = x 2 + 1. Then differentiate. du = 2x ⇒ 2x dx = du dx 2x cos(x 2 + 1) dx = cos u du ∫ ∫ = sin u + c = sin(x 2 + 1) + c Notice that the last example involves two expressions multiplied together, namely 2x and cos(x2 + 1). These two expressions are related by the fact that 2x is the derivative of x2 + 1. Because of this relationship, the substitution u = x2 + 1 may be used to perform the integration.You can apply this method to other integrals involving trigonometrical functions, as in the next example. 194 9781510421738.indb 194 02/02/18 1:14 PM Example 8.11 Find π 2 ∫ 0 cos x sin2 x dx. Remember that sin2 x means the same as (sin x)2. Solution 8 This integral is the product of two expressions, cos x and (sin x)2. Differentiating: du du = cos x dx. dx = cos x ⇒ The limits of integration need to be changed as well: Therefore x=0 ⇒ u=0 x = π2 ⇒ u=1 π 2 1 ∫ 0 cos x sin2 x dx = ∫0 u 2 du 1 3 = ⎡⎢u ⎤⎥ ⎣ 3 ⎦0 = Example 8.12 1 3 ∫ 8.4 Integrals involving trigonometrical functions Now (sin x)2 is a function of sin x, and cos x is the derivative of sin x, so you should use the substitution u = sin x. Find cos3 x dx. Solution First write cos3 x = cos x cos2 x. Now remember that cos2 x + sin2 x = 1 ⇒ cos2 x = 1 − sin2 x. This gives cos3 x = cos x (1 − sin2 x) = cos x − cos x sin2 x The first part of this expression, cos x, is easily integrated to give sin x. The second part is more complicated, but you can see that it is of a type that you have met already, as it is a product of two expressions, one of which is a function of sin x and the other of which is the derivative of sin x. This can be integrated either by making the substitution u = sin x or simply in your head (by inspection). ∫ cos3 x dx = ∫(cos x − cos x sin2 x) dx = sin x − 13 sin3 x + c 195 9781510421738.indb 195 02/02/18 1:14 PM 8 Example 8.13 Find (i) ∫ cot x dx ∫ (ii) π 3 π 6 tan x dx Solution (i) cos x Rewrite cot x as sin x . cos x cot x dx = sin x .dx Now you can use the substitution u = sin x. 8 FURTHER CALCULUS ∫ ∫ du dx = cos x ⇒ du = cos x dx cos x . 1 1 sin x dx = sin x × cos x dx = u du = ln | u | + c = ln | sin x | + c cos x You may have noticed that the integral sin x .dx is in the form f '( x ) dx = ln | f(x) | + c, f (x ) and so you could have written the answer down directly. ∫ ∫ ∫ ∫ The ‘top line’ is the derivative of the ‘bottom line’. ∫ (ii) ∫ π 3 π 6 tan x dx = ∫ π 3 π 6 sin x cos x dx Adjusting the numerator to make it the derivative of the denominator gives: ∫ π 3 π 6 sin x cos x dx = − [ ∫ π 3 π 6 – sin x cos x dx = −ln | cos x | ] π 3 π 6 = ⎡− ln 1 ⎤ − ⎡− ln 3 ⎤ 2 ⎦ ⎢⎣ ⎣ 2 ⎥⎦ = −ln 1 + ln 3 2 2 = ln 3 1 = 2 ln 3 cos π = 1 and 3 2 cos π = 3 6 2 Use the laws of logs: ⎛ ⎞ ln 3 – ln 21 = ln ⎜ 3 ÷ 21 ⎟ 2 ⎝ 2 ⎠ Use the laws of logs: 1 ln 3 = ln3 2 = 21 ln3 Note You may find that as you gain practice in this type of integration you become able to work out the integral without writing down the substitution. However, if you are unsure, it is best to write down the whole process. 196 9781510421738.indb 196 02/02/18 1:14 PM Question 3 of Exercise 8A on page 185 asked you to prove that () f ( x ) = 1 tan −1 x k k when where k is a constant, 1 . x2 + k2 Using this result and the fact that integration is the reverse of differentiation you can write k dx = tan −1 x + c . 1 dx = 1 tan −1 x + c and k k k x2 + k2 x2 + k2 8 f '( x ) = Example 8.14 () ∫ Find (i) ∫x 2 1 + 4 dx (ii) Solution (i) (ii) ∫ () 1 dx = 1 tan −1 x + c 2 2 +4 1 dx = 1 1 dx 5 x2 + 3 5x 2 + 3 5 1 1 = dx ∫x 2 ∫ 5 ∫x 2 + ( ) 3 5 ∫ 5x 1 dx . +3 2 Start by rewriting the integral in the form 2 ∫x 2 1 dx . + k2 k 2 = 3 so k = 3 5 5 = 1 × 1 tan − 1 ⎛⎜ x ⎞⎟ + c 5 3 3 ⎝ 5⎠ 5 = 1× 5 3 = 1 tan − 1 15 Exercise 8D 1 ( x) + c ( x) + c 5 tan − 1 5 3 5 3 1 × 5 3 5 = 5 5 3 = 1 5 3 = 1 15 Integrate the following by using the substitution given, or otherwise. (i) cos 3x u = 3x (ii) sin(1 − x) u=1−x 3 (iii) sin x cos x u = cos x sin x u = 2 − cos x (iv) 2 − cos x (v) tan x u = cos x write tan x as sin x cos x (vi) sin 2x (l + cos 2x)2 u = 1 + cos 2x Use a suitable substitution to integrate the following. ( 2 8.4 Integrals involving trigonometrical functions () ∫ ) 2x sin(x 2) (ii) cos x e sin x tan x cos x (iii) cos 2 x (iv) sin 2 x Evaluate the following definite integrals by using suitable substitutions. π π 4 π 2 (i) cos 2x − 2 dx (ii) cos x sin3x dx 0 0 (i) 3 ∫ 9781510421738.indb 197 ( ) ∫ 197 02/02/18 1:14 PM 8 ∫0 (iii) x sin(x 2) dx π 4 (iv) Find the following. 1 dx (i) x2 + 1 8 FURTHER CALCULUS ∫ 2x (iv) 6 ∫0 e tan x cos 2 x dx 1 2 ∫ 5 π 4 ∫ 0 cos x (1 + tan x) dx (v) 4 π 1 2 +1 dx (ii) ∫x 1 dx + 16 (iii) ∫ x 2+ 9 dx (v) ∫ 3x 1+ 4 dx (vi) ∫ 2x 2 2 2 10 dx . x 2 + 25 y The diagram shows the curve y = sin2 2x cos x for 0 $ x $ 21 π, M and its maximum point M. (i) Find the x coordinate of M. O (ii) Using the substitution u = sin x, find by integration the area of the shaded region bounded by the curve and the x-axis. Find ∫ 1 dx +5 2 5 3 0 1 2 x π Cambridge International AS & A Level Mathematics 9709 Paper 33 Q9 June 2013 7 The diagram shows the curve y = e 2 sin x cos x for 0 & x & 21 π , and its maximum point M. (i) Using the substitution u = sin x, find the exact value of the area of the shaded region bounded by the curve and the axes. (ii) Find the x coordinate of M, giving your answer correct to 3 decimal places. y M Cambridge International AS & A Level Mathematics 9709 Paper 33 Q9 June 2014 8 (i) (ii) Prove that cot θ + tan θ ≡ 2cosec2θ . Hence show that ∫ 1π 3 1π 6 O 1π 2 x cosec 2θ dθ = 21 ln 3 . Cambridge International AS & A Level Mathematics 9709 Paper 31 Q5 November 2013 9 (i) Use the substitution x = sin2θ to show that (1 −x x ) dx = ∫ (ii) 9781510421738.indb 198 2 θ dθ Hence find the exact value of ∫ 198 ∫ 2 sin 1 4 0 (1 −x x ) dx. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q6 November 2005 02/02/18 1:14 PM ∫ 10 (a) Find (4 + tan 2 2x )dx . (b) Find the exact value of ∫ 1π 2 1π 4 8 sin( x + 61 π) . dx sin x Cambridge International AS & A Level Mathematics 9709 Paper 31 Q5 June 2015 11 The diagram shows part of the curve y = sin 3 2x cos 3 2x . The shaded 8.5 The use of partial fractions in integration region shown is bounded by the curve and the x-axis and its exact area is denoted by A. y x O (i) (ii) Use the substitution u = sin 2x in a suitable integral to find the value of A. kπ sin 3 2x cos 3 2x d x = 40 A, find the value of the Given that 0 constant k. ∫ Cambridge International AS & A Level Mathematics 9709 Paper 33 Q7 November 2012 8.5 The use of partial fractions in integration ❯ Why is it not possible to use any of the integration techniques you 2 have learnt so far to find x 2 − 1 dx? ? ∫ Partial fractions reminder In Chapter 7 you met partial fractions. Here is a reminder of the work you did there. Since x2 − 1 can be factorised to give (x + 1)(x − 1), you can write the expression to be integrated as partial fractions. 2 = A + B x2 − 1 x − 1 x + 1 2 ≡ A(x + 1) + B(x − 1) 9781510421738.indb 199 This is true for all values of x. It is an identity and to emphasise this point we use the identity symbol ≡. Let x = 1 2 = 2A ⇒ A=1 Let x = −1 2 = −2B ⇒ B = −1 199 02/02/18 1:14 PM Substituting these values for A and B gives 2 = 1 − 1 . x2 − 1 x − 1 x + 1 The integral then becomes 8 2 dx = 1 dx − 1 dx. x −1 x +1 −1 Now the two integrals on the right can be recognised as logarithms. ∫x ∫ 2 ∫ 2 dx = ln ⎢x − 1 ⎢− ln ⎢x + 1 ⎢+ c −1 = ln x − 1 + c x +1 Here you worked with the simplest type of partial fraction, in which there are two different linear factors in the denominator. This type will always result in two fractions, both of which can be integrated to give logarithmic expressions. Now look at the other types of partial fraction. 8 FURTHER CALCULUS ∫x 2 ⎪ ⎪ A repeated factor in the denominator Example 8.15 x+4 Find (2x − 1)( x + 1) 2 dx. ∫ Solution First write the expression as partial fractions: x+4 A + B + C = (2x − 1)( x + 1) 2 (2x − 1) ( x + 1) ( x + 1) 2 x + 4 ≡ A(x + 1)2 + B(2x − 1)(x + 1) + C(2x − 1). where Let x = −1 3 = −3C ⎛3⎞ 2 ⎜ ⎟ ⎝2⎠ Let x = 21 9 2 Let x = 0 4 =A−B−C =A ⇒ C = −1 ⇒ 9 2 ⇒ B = A − C − 4 = 2 + 1 − 4 = −1 = 49 A ⇒ A=2 Substituting these values for A, B and C gives x+4 2 1 = − 1 − (2x − 1)( x + 1) 2 (2x − 1) ( x + 1) ( x + 1) 2 Now that the expression is in partial fractions, each part can be integrated separately. x+4 2 dx − 1 dx − 1 dx dx = (2x − 1)( x + 1) 2 ( 1) 2 x + ( 2x − 1) ( x + 1) ∫ ∫ ∫ ∫ The first two integrals give logarithmic expressions as you saw above. The third, however, is of the form u −2 and therefore can be integrated by using the substitution u = x + 1, or by inspection (i.e. in your head). x+4 ∫ (2x − 1)(x + 1) 2 dx = ln | 2x − 1 | − ln | x + 1 | + 1 +c x +1 = ln 2x − 1 + 1 + c x +1 x +1 ⎪ ⎪ 200 9781510421738.indb 200 02/02/18 1:14 PM A quadratic factor in the denominator Example 8.16 Find ∫ (x 2 8 x−2 dx. + 2)( x + 1) Solution where 8.5 The use of partial fractions in integration First write the expression as partial fractions: x−2 C Ax + B ( x 2 + 2)( x + 1) = ( x 2 + 2) + ( x + 1) x − 2 ≡ (Ax + B)(x + 1) + C(x2 + 2) Rearranging gives x − 2 ≡ (A + C )x2 + (A + B)x + (B + 2C ) Equating coefficients: x2 ⇒ A+C=0 x ⇒ A+B=1 constant terms ⇒ B + 2C = −2 Solving these gives A = 1, B = 0, C = −1. Hence x−2 x 1 ( x 2 + 2)( x + 1) = ( x 2 + 2) − ( x + 1) x−2 x 1 ( x 2 + 2)( x + 1) dx = ( x 2 + 2) dx − ( x + 1) dx 2x 1 = 21 dx − ( x + 1) dx ( x 2 + 2) 1 2 2 – 2 ln | x + 2 | = ln x + 2 = 21 ln | x 2 + 2 | − ln | x + 1 | + c 2 Notice that (x + 2) is 2 positive for all values of x. = ln x + 2 + c x +1 ∫ ∫ ∫ ∫ ⎪ ∫ ⎪ e Note If B had not been zero, you would have had an expression of the form integrate. This can be split into Ax + B . x2 + 2 x2 + 2 Ax + B to x2 + 2 The first part of this can be integrated as in Example 8.16, but to integrate the second part you would need to use the standard result: ∫ (x 2 () 1 dx = 1 tan−1 x + c. + k2) k k which you met on page 197. 201 9781510421738.indb 201 02/02/18 1:14 PM 8 FURTHER CALCULUS 8 Exercise 8E 1 2 Express the fractions in each of the following integrals as partial fractions, and hence perform the integration. 7x − 2 1 (i) dx (ii) dx ( x − 1) 2 (2x + 3) (1 − x )(3x − 2) x +1 3x + 3 (iii) dx (iv) ( x − 1)(2x + 1) dx ( x 2 + 1)( x − 1) 1 1 dx (v) (vi) ( x + 1)( x + 3) dx x 2 (1 − x ) 5x + 1 2x − 4 (vii) dx (viii) ( x + 2)( 2x + 1) 2 dx ( x 2 + 4)( x + 2) ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ Express in partial fractions 3x + 4 f(x) = 2 ( x + 4)( x − 3) and hence find PS 3 (i) 2 ∫0 f(x) dx. (a) 3 Express (1 + x )(1 − 2x ) in partial fractions. (b) Hence find 0.1 ∫0 (ii) (a) (b) 3 dx (1 + x )(1 − 2x ) giving your answer to 5 decimal places. Find the first three terms in the binomial expansion of 3(1 + x)−1(1 − 2x)−1. Use the first three terms of this expansion to find an approximation for 0.1 3 dx. 0 (1 + x )(1 − 2 x ) What is the percentage error in your answer to part (b)? ∫ (c) 4 (i) (ii) Find the values of the constants A, B, C and D such that 2x 3 – 1 ≡ A + B + C + D . x 2 (2x – 1) x x 2 2x – 1 Hence show that 2 2x 3 – 1 3 1 16 2 (2 x – 1) dx = 2 + 2 ln 27 . x 1 ( ) ∫ 5 Cambridge International AS & A Level Mathematics 9709 Paper 32 Q10 June 2010 x 2 + 3x + 3 . Let f ( x ) ≡ ( x + 1)( x + 3) (i) Express f(x) in partial fractions. (ii) Hence show that ∫ 3 0 f ( x )d x = 3 − 21 ln 2. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q7 June 2008 202 9781510421738.indb 202 02/02/18 1:14 PM 6 2 Let f ( x ) = 12 + 8x − x 2 . (2 − x )(4 + x ) (i) (ii) 8 A + Bx + C 2 − x 4 + x2 1 . Show that 0 f ( x )d x = ln ( 25 2) Express f(x) in the form ∫ Cambridge International AS & A Level Mathematics 9709 Paper 31 Q8 November 2011 (i) (ii) Show that (x + 1) is a factor of 4 x 3 − x 2 − 11x − 6. 4 x 2 + 9 x − 1 dx . Find 4 x 3 − x 2 − 11x − 6 ∫ Cambridge International AS & A Level Mathematics 9709 Paper 33 Q7 November 2015 8 By first using the substitution u = e x , show that ln 4 e 2x dx = ln ( 85 ) . 0 e 2 x + 3e x + 2 ∫ 8.6 Integration by parts 7 Cambridge International AS & A Level Mathematics 9709 Paper 33 Q10 November 2014 8.6 Integration by parts There are still many integrations which you cannot yet do. In fact, many functions cannot be integrated at all, although virtually all functions can be differentiated. However, some functions can be integrated by techniques which you have not yet met. Integration by parts is one of those techniques. Example 8.17 ∫ Find x cos x dx. Solution The expression to be integrated is clearly a product of two simpler expressions, x and cos x, so your first thought may be to look for a substitution to enable you to perform the integration. However, there are some expressions which are products but which cannot be integrated by substitution. This is one of them.You need a new technique to integrate such expressions. Take the expression x sin x and differentiate it, using the product rule. d dx (x sin x) = x cos x + sin x Now integrate both sides. This has the effect of ‘undoing’ the differentiation, so ∫ ∫ x sin x = x cos x dx + sin x dx ➜ 203 9781510421738.indb 203 02/02/18 1:14 PM 8 Rearranging this gives ∫ x cos x dx = x sin x − ∫ sin x dx = x sin x − (−cos x) + c = x sin x + cos x + c 8 FURTHER CALCULUS This has enabled you to find the integral of x cos x. The work in this example can be generalised into the method of integration by parts. Before coming on to that, do the following activity. ACTIVITY 8.2 For each of the following (a) differentiate using the product rule (b) rearrange your expression to find an expression for the given integral I (c) use this expression to find the given integral. (i) y = x cos x I = x sin x dx (ii) ∫ I = ∫ 2x e2x dx y = xe2x The work in Activity 8.2 has enabled you to work out some integrals which you could not previously have done, but you needed to be given the expressions to be differentiated first. Effectively you were given the answers. ? Look at the expressions you found in part (b) of Activity 8.2. ❯ Can you see any way of working out these expressions without starting by differentiating a given product? The general result for integration by parts The method just investigated can be generalised. Look back at Example 8.17. Use u to stand for the function x, and v to stand for the function sin x. Using the product rule to differentiate the function uv, d (uv) = v du + u dv . dx dx dx Integrating gives uv = v du dx + u dv dx. dx dx Rearranging gives u dv dx = uv − v du dx. dx dx This is the formula you use when you need to integrate by parts. ∫ ∫ 204 9781510421738.indb 204 ∫ ∫ 02/02/18 1:14 PM ∫ So now you can find x cos x dx. Put u=x ⇒ du = 1 dx dv = cos x ⇒ and v = sin x dx Substituting in 8 8.6 Integration by parts In order to use it, you have to split the function you want to integrate into two simpler functions. In Example 8.17 you split x cos x into the two functions x and cos x. One of these functions will be called u and the other dv , to fit dx the left-hand side of the expression.You will need to decide which will be which. Two considerations will help you. du on the right-hand side of the expression, u should » As you want to use dx be a function which becomes a simpler function after differentiation. So in this case, u will be the function x. » As you need v to work out the right-hand side of the expression, it must be possible to integrate the function dv to obtain v. In this case, dv will be dx dx the function cos x. ∫u ddxv dx = uv − ∫v ddxu dx gives ∫x cos x dx = x sin x − ∫1 ×sin x dx = x sin x − (−cos x) + c = x sin x + cos x + c Example 8.18 ∫ Find 2x ex dx. Solution First split 2x ex into the two simpler expressions, 2x and ex. Both can be integrated easily but, as 2x becomes a simpler expression after differentiation and ex does not, take u to be 2x. du = 2 u = 2x ⇒ dx dv x ⇒ v = ex dx = e Substituting in dv gives du ∫u dx dx = uv − ∫v dx dx ∫2x ex dx = 2x ex − ∫2ex dx = 2x ex − 2ex + c In some cases, the choices of u and v may be less obvious. 205 9781510421738.indb 205 02/02/18 1:14 PM 8 Example 8.19 ∫ Find x ln x dx. Solution It might seem at first that u should be taken as x, because it becomes a simpler expression after differentiation. 8 FURTHER CALCULUS du =1 u=x ⇒ dx dv dx = ln x Now you need to integrate ln x to obtain v. Although it is possible to integrate ln x, it has to be done by parts, as you will see in the next example. The wrong choice has been made for u and v, resulting in a more complicated integral. So instead, let u = ln x. u = ln x du = 1 dx x ⇒ dv dx = x Substituting in v = 21 x 2 ⇒ dv du ∫u dx dx = uv − ∫v dx dx gives 1 x2 ∫x ln x dx = 2 x 2 ln x − ∫ 2 x 1 dx 1 = 21 x 2 ln x − 2 x dx ∫ = Example 8.20 1 2 2 x ln x − 41 x 2 + c ∫ Find ln x dx. Solution You need to start by writing ln x as 1 ln x and then use integration by parts. As in the last example, let u = ln x. u = ln x dv dx = 1 Substituting in dv ⇒ du = 1 dx x ⇒ v=x du ∫u dx dx = uv − ∫v dx dx 206 9781510421738.indb 206 02/02/18 1:15 PM gives 8 1 ∫ 1 ln x dx = x ln x − ∫ x × x dx = x ln x − ∫ 1 dx = x ln x − x + c Sometimes it is necessary to use integration by parts twice or more to complete the integration successfully. Example 8.21 ∫ Find x 2 sin x dx. Solution First split x 2 sin x into two: x 2 and sin x. As x 2 becomes a simpler expression after differentiation, take u to be x 2. u = x2 dv dx = sin x Substituting in ⇒ gives du dx = 2x ⇒ dv 8.6 Integration by parts Using integration by parts twice v = −cos x du ∫ u dx dx = uv − ∫ v dx dx ∫ x 2 sin x dx = −x 2 cos x − ∫ −2x cos x dx = −x 2 cos x + ∫ 2x cos x dx 1 ! Now the integral of 2x cos x cannot be found without using integration by parts again. It has to be split into the expressions 2x and cos x and, as 2x becomes a simpler expression after differentiation, take u to be 2x. u = 2x dv dx = cos x Substituting in dv gives ⇒ ⇒ du dx = 2 v = sin x du ∫ u dx dx = uv − ∫ v dx dx ∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x − (−2 cos x) + c = 2x sin x + 2 cos x + c 1 So in ! ∫ x 2 sin x dx = −x 2 cos x + 2x sin x + 2 cos x + c. 207 9781510421738.indb 207 02/02/18 1:15 PM The technique of integration by parts is usually used when the two functions are of different types: polynomials, trigonometrical functions, exponentials, logarithms. There are, however, some exceptions, as in questions 3 and 4 of Exercise 8F. 8 Integration by parts is a very important technique which is needed in many other branches of mathematics. For example, integrals of the form x f(x) dx are used in statistics to find the mean of a probability density function, and in mechanics to find the centre of mass of a shape. Integrals of the form x2 f(x) dx are used in statistics to find variance and in mechanics to find moments of inertia. 8 FURTHER CALCULUS ∫ ∫ Exercise 8F 1 For each of these integrals (a) write down the expression to be taken as u and the dv expression to be taken as dx (b) use the formula for integration by parts to complete the integration. x (i) x e dx (ii) x cos 3x dx (iii) (v) 2 3 4 5 6 7 PS 8 ∫ ∫ (2x + 1)cos x dx ∫ x e−x dx ∫ ∫ x e−2x dx ∫ x sin 2x dx (iv) (vi) Use integraton by parts to integrate (i) x 3 ln x (ii) 3x e 3x (iii) 2x cos 2x (iv) x 2 ln 2x Find x 1 + x dx (i) by using integration by parts (ii) by using the substitution u = 1 + x. Find 2x(x − 2)4 dx (i) by using integration by parts (ii) by using the substitution u = x − 2. (i) By writing ln x as the product of ln x and 1, use integration by parts to find ln x dx. (ii) Use the same method to find ln 3x dx. (iii) Write down ln px dx where p # 0. Find x 2 ex dx. ∫ ∫ ∫ ∫ ∫ ∫ Find ∫ (2 − x)2 cos x dx. Find ∫ x tan−1x dx. Definite integration by parts When you use the method of integration by parts on a definite integral, it is important to remember that the term uv on the right-hand side of the expression has already been integrated and so should be written in square brackets with the limits indicated. 208 9781510421738.indb 208 b b u dv dx = uv − v du dx a dx a dx a ∫ b [ ] ∫ 02/02/18 1:15 PM Example 8.22 Evaluate 2 ∫ 0 x e x dx. 8 Solution du dx = 1 v = ex u=x ⇒ dv x and dx = e ⇒ Substituting in Put b 8.6 Integration by parts b ∫a u dx dx = [uv]a − ∫av dx dx dv b du gives 2 ∫ 0x ex dx [ ] ∫ e dx = [x e ] − [e ] 2 2 0 0 = x ex − x 2 x x 0 2 0 = (2e 2 − 0) − (e2 − e0) = 2e2 − e2 + 1 = e2 + 1 Example 8.23 Find the area of the region between the curve y = x cos x and the x-axis, between x = 0 and x = π . 2 Solution y Figure 8.6 shows the region whose area is to be found. 1 To find the required area, you need to integrate the function O x cos x between the limits 0 and π . 2 You therefore need to work out –1 π 2 ∫ 0 x cos x dx. Put u=x dv dx = cos x Substituting in and dv x ▲ Figure 8.6 ⇒ du dx = 1 ⇒ v = sin x ∫a u dx dx = [uv] a − ∫a v dx dx b π 2 b b du ➜ 209 9781510421738.indb 209 02/02/18 1:15 PM gives 8 π 2 π 2 π 2 ∫0 x cos x dx = [x sin x]0 − ∫0 sin x dx π 2 π 2 = [x sin x] 0 − [−cos x]0 π 2 8 FURTHER CALCULUS = [x sin x + cos x]0 = ( π2 + 0) − (0 + 1) = ( ) π 2 −1 So the required area is π − 1 square units. 2 Exercise 8G 1 Evaluate these definite integrals. π 1 (i) x e3x dx (ii) (x − 1)cos x dx (iii) (v) 2 3 4 5 6 PS 7 ∫0 2 ∫0 (x + 1)e x dx ∫ 0 x sin 2x dx π 2 (iv) (vi) ∫0 2 ∫1 ln 2x dx 4 ∫1 x 2 lnx dx Find the coordinates of the points where the graph of y = (2 − x)e−x cuts the x- and y-axes. (ii) Hence sketch the graph of y = (2 − x)e−x. (iii) Use integration by parts to find the area of the region between the x-axis, the y-axis and the graph y = (2 − x)e−x. (i) Sketch the graph of y = x sin x from x = 0 to x = π and shade the region between the curve and the x-axis. (ii) Find the area of this region using integration by parts. Find the area of the region between the x-axis, the line x = 5 and the graph y = ln x. Find the area of the region between the x-axis and the graph y = x cos x π from x = 0 to x = 2 . Find the area of the region between the negative x-axis and the graph y = x x +1 (i) using integration by parts (ii) using the substitution u = x + 1. Find (i) (i) (ii) ∫ ∫ 0 1 2 0 3 6x tan − 1 x dx x tan − 1(2x )dx . 210 9781510421738.indb 210 02/02/18 1:15 PM 8 Find the exact value of ∫ 4 1 ln x d x.. x Cambridge International AS & A Level Mathematics 9709 Paper 31 Q3 November 2013 9 ∫ a 8 1 The constant a is such that 0 x e 2 x dx = 6. (i) Show that a satisfies the equation 1 x = 2 + e – 2x . Cambridge International AS & A Level Mathematics 9709 Paper 3 Q9 November 2008 8.6 Integration by parts By sketching a suitable pair of graphs, show that this equation has only one root. (iii) Verify by calculation that this root lies between 2 and 2.5. (iv) Use an iterative formula based on the equation in part (i) to calculate the value of a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. (ii) 1 10 The diagram shows the curve y = e – 2 x √(1 + 2x) and its maximum point M. The shaded region between the curve and the axes is denoted by R. y M R O (i) (ii) x Find the x coordinate of M. Find by integration the volume of the solid obtained when R is rotated completely about the x-axis. Give your answer in terms of π and e. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q9 June 2008 11 The expression f(x) is defined by f(x) = 3x e −2 x . (i) Find the exact value of f ' (− 21 ). (ii) Find the exact value of ∫ 0 −1 2 f ( x )dx. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q5 November 2012 211 9781510421738.indb 211 02/02/18 1:15 PM 1 8 12 The diagram shows the curve y = x 2 ln x. The shaded region between the curve, the x-axis and the line x = e is denoted by R. y R e 8 FURTHER CALCULUS O (i) (ii) x Find the equation of the tangent to the curve at the point where x = 1, giving your answer in the form y = mx + c. Find by integration the volume of the solid obtained when the region R is rotated completely about the x-axis. Give your answer in terms of π and e. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q9 June 2012 8.7 General integration You now know several techniques for integration which can be used to integrate a wide variety of functions. One of the difficulties which you may now experience when faced with an integration is deciding which technique is appropriate! This section gives you some guidelines on this, as well as revising all the work on integration that you have done so far. ❯ Look at the integrals below and try to decide which technique you would use and, in the case of a substitution, which expression you would write as u. Do not attempt actually to carry out the integrations. Make a note of your decisions − you will return to these integrals later. x−5 x +1 (i) (ii) x 2 + 2 x − 3 dx x 2 + 2x − 3 dx ∫ (iii) x e dx ∫ 2x + cos x (v) ∫ x + sin x dx x 2 ? ∫ (iv) x e dx ∫ (vi) cos x sin2 x dx ∫ x2 212 9781510421738.indb 212 02/02/18 1:15 PM Choosing an appropriate method of integration You have now met the following standard integrals. 8 ∫f(x) dx f(x) (n ≠ −1) x n+1 n +1 1 x (x ≠ 0) ln|x | ex (x ! ") ex sin x (x ! ") −cos x cos x (x ! ") sin x If you are asked to integrate any of these standard functions, you may simply write down the answer. 8.7 General integration xn For other integrations, the following table may help. Type of expression to be integrated Examples Method of integration Simple variations of any of the standard functions cos(2x + 1) e3x Substitution may be used, but it should be possible to do these by inspection. Product of two expressions of the form f ′(x)g[f(x)] d Note that f ′(x) means [f(x)] dx 2x ex x2(x3 + 1)6 Substitution u = f(x) Other products, particularly when one expression is a small positive integer power of x or a polynomial in x x ex x2 sin x Integration by parts 2 x f ′ (x ) Quotients of the form f( x ) or expressions which can easily be converted to this form sinx cosx Polynomial quotients which may be split into partial fractions x +1 x( x − 1) 2 x +1 Substitution u = f(x) or by inspection: k ln | f(x) | + c, where k is known Split into partial fractions and integrate term by term x−4 x2 − x − 2 Odd powers of sin x or cos x cos3 x Use cos2 x + sin2 x = 1 and write in form f ′(x)g[f(x)] Even powers of sin x or cos x sin2 x cos4 x Use the double-angle formulae to transform the expression before integrating. Expressions in the form x2 1 + k2 x2 1 1 or + 25 3x 2 + 4 Use standard result: x 1 1 +c dx = tan -1 x2 + k2 k k ∫ () 213 9781510421738.indb 213 02/02/18 1:15 PM It is impossible to give an exhaustive list of possible types of integration, but the tables on the previous page cover the most common situations that you will meet. 8 ACTIVITY 8.3 8 FURTHER CALCULUS Now look back at the integrals in the discussion point on page 212 and the decisions you made about which method of integration should be used for each one. Now find these integrals. x−5 x +1 (i) dx (ii) dx x 2 + 2x − 3 x 2 + 2x − 3 (iii) (v) Exercise 8H 1 Note Often when a question requires integration by substitution then you will be given a suitable substitution, however in this exercise you need to decide on the best substitution to use yourself. ∫ ∫x ex dx 2x + cos x ∫ x + sin x dx (vi) 2 Choose an appropriate method and integrate the following. You may find it helpful to discuss in class first which method to use. 2x + 1 (i) cos(3x − 1) dx (ii) dx ( x 2 + x − 1) 2 (iii) e1−x dx (iv) cos 2x dx ∫ ∫ (v) ∫ln 2x dx (vii) ∫ 2x − 3 dx (ix) x 3 ln x dx ∫ (xi) (x + 1)ex + 2x dx ∫ (xiii) x 2 sin 2x dx ∫ (vi) (viii) (x) (xii) (iv) (vii) (x) 3 (xiv) ∫ ∫ x ∫ (x − 1) dx 4x − 1 ∫ (x − 1) (x + 2) dx 5 ∫ 2x − 7x + 3 dx sin x − cos x ∫ sin x + cos x dx ∫sin3 2x dx 2 3 2 2 Evaluate the following definite integrals. (i) 24 ∫8 π 2 (ii) dx 3x − 8 ∫0 ∫ 2 lnx x dx sin3x dx 2 ∫ 1 0 24 (ii) ∫8 (v) 2 dx 3x − 8 ∫1 ∫ x cos 3x dx π 3 (viii) 1 0 x 2 ln x dx (iii) 24 ∫8 (vi) ∫ (ix) ∫ 9x dx 3x − 8 1 x 2 dx 1+ x3 π 4 sin x dx 0 cos 4 x 0 x dx e 2x Let f ( x ) = (i) 9781510421738.indb 214 2 (iv) 2 2 214 ∫ ∫x e x dx ∫cos x sin2 x dx 6 + 6x . (2 − x )(2 + x 2 ) A + Bx + C . − x 2 + x2 2 1 Show that f ( x ) d x = 3 ln 3. Express f(x) in the form ∫ −1 Cambridge International AS & A Level Mathematics 9709 Paper 33 Q8 June 2014 02/02/18 1:15 PM The diagram shows the curve y = x 2 e 2 − x and its maximum point M. 4 y M O x ∫ (ii) 0 Cambridge International AS & A Level Mathematics 9709 Paper 31 Q9 June 2015 The integral I is defined by I = 5 ∫ 2 0 4t 3 ln(t 2 + 1)dt. Use the substitution x = t 2 + 1 to show that (i) I = ∫ 5 1 8.7 General integration Show that the x coordinate of M is 2. 2 Find the exact value of x 2 e 2 − x dx . (i) 8 (2x − 2) ln x d x. Hence find the exact value of I. (ii) Cambridge International AS & A Level Mathematics 9709 Paper 31 Q7 June 2011 KEY POINTS 1 You can differentiate tan−1 x by using either implicit differentiation, or dy ● = 1 dx dx dy kx n +1 2 kxn dx = n + 1 + c where k and n are constants but n ≠ −1. 3 Substitution is often used to change a non-standard integral into a standard one. ● ∫ 4 ∫ ex dx = ex + c ∫ e dx = 1a e ax + b 5 ax + b +c 1 ∫ x dx = ln | x | + c ∫ ax1+ b dx = 1a ln | ax + b | + c 6 ∫ f ′(x ) f(x ) dx = ln | f(x) | + c 215 9781510421738.indb 215 02/02/18 1:15 PM 8 7 ∫ cos(ax + b) dx = 1a sin(ax + b) + c 1 ∫ sin(ax + b) dx = − a cos(ax + b) + c ∫ sec (ax + b) dx = 1a tan(ax + b) + c 8 FURTHER CALCULUS 2 8 ∫x 2 () 1 dx 1 tan − 1 x = + c k k + k2 9 Using partial fractions often makes it possible to use logarithms to integrate the quotient of two polynomials. 10 Some products may be integrated by parts using the formulae u dv dx = uv − v du dx dx dx b dv b b u dx = uv u − v du dx x d a a a dx ∫ ∫ ∫ [ ] ∫ LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ differentiate tan −1 x ■ extend the idea of ‘reverse differentiation’ to include the integration of 1 x2 + a2 ■ use integration by substitution in cases where the process is the reverse of the chain rule (either by inspection or by writing down the working for the substitution) ■ use integration by substitution in other cases, finding a suitable substitution for both definite and indefinite integrals k f ′( x ) ■ use dx = k lln | f(x) |++ c f( x ) ■ use partial fractions in integration ■ use trigonometrical relationships in integration ■ use the method of integration by parts including when more than one application of the method may be required to integrate ln x. ∫ 216 9781510421738.indb 216 02/02/18 1:15 PM P3 9 Differential equations 9 Differential equations The greater our knowledge increases, the more our ignorance unfolds. John F. Kennedy (1917–1963) Temperature T ❯ How long do you have to wait for a typical cup of coffee to be drinkable? ❯ How long does it take to go cold? ❯ What do the words ‘drinkable’ and ‘cold’ mean in this context? T0 O Time t ? M Newton’s law of cooling states that the rate of change of the temperature of an object is proportional to the difference between the object’s temperature and the temperature of its surroundings. dT = k (T − T 0 ) , where This leads to the equation dt T is the temperature of the object at time t, T 0 is the temperature of the surroundings, and k is a constant of proportionality. ▲ Figure 9.1 217 9781510421738.indb 217 02/02/18 1:15 PM 9 DIFFERENTIAL EQUATIONS 9 This is an example of a differential equation. To be able to predict the temperature of the object at different times, you need to solve the differential equation. In this chapter, you will learn how to solve problems like this involving rates of change. 9.1 Forming differential equations from rates of change If you are given sufficient information about the rate of change of a quantity, such as temperature or velocity, you can work out a differential equation to model the situation, like the one given earlier for Newton’s law of cooling. It is important to look carefully at the wording of the problem which you are studying in order to write an equivalent mathematical statement. For example, if the altitude of an aircraft is being considered, the phrase ‘the rate of change of height’ might be used. This actually means ‘the rate of change of height with respect to time’ and could be written as dh . However, you dt might be more interested in how the height of the aircraft changes according to the horizontal distance it has travelled. In this case, you would talk about ‘the rate of change of height with respect to horizontal distance’ and could write dh this as , where x is the horizontal distance travelled. dx Some of the situations you meet in this chapter involve motion along a straight line, and so you will need to know the meanings of the associated terms. The position of an object (+5 in Figure 9.2) is its distance from the origin O in the direction you have chosen to define as being positive. O –1 0 1 2 3 4 5 6 s ▲ Figure 9.2 The rate of change of position of the object with respect to time is its velocity, and this can take positive or negative values according to whether the object is moving away from the origin or towards it. v = ds dt The rate of change of an object’s velocity with respect to time is called its acceleration, a. a = dv dt Velocity and acceleration are vector quantities but in one-dimensional motion there is no choice in direction, only in sense (i.e. whether positive or negative). Consequently, as you may already have noticed, the conventional bold type for vectors is not used in this chapter. 218 9781510421738.indb 218 02/02/18 1:15 PM Example 9.1 An object is moving through a liquid so that the rate at which its velocity decreases is proportional to its velocity at any given instant. When it enters the liquid, it has a velocity of 5 m s–1 and the velocity is decreasing at a rate of 1 m s–2. Find the differential equation to model this situation. 9 Solution − dv ∝ v or dv = −kv dt dt where k is a positive constant. When the object enters the liquid its velocity is 5 m s–1, so v = 5, and the velocity is decreasing at the rate of 1 m s–2, so dv = −1 dt Putting this information into the equation dv = −kv gives dt –1 = –k × 5 ⇒ k = 51. So the situation is modelled by the differential equation dv = − v 5 dt Example 9.2 9.1 Forming differential equations from rates of change The rate of change of velocity means the rate of change of velocity with respect to time and so can be written as dv . As it is decreasing, the rate of dt change must be negative, so A model is proposed for the temperature gradient within a star, in which the temperature decreases with respect to the distance from the centre of the star at a rate which is inversely proportional to the square of the distance from the centre. Express this model as a differential equation. Solution In this example the rate of change of temperature is not with respect to time but with respect to distance. If θ represents the temperature at a point in the star and r the distance from the centre of the star, the rate of change of temperature with respect to distance may be written as – dθ , so dr − dθ ∝ 12 or dθ = − k2 dr dr r r where k is a positive constant. Note This model must break down near the centre of the star, otherwise it would be infinitely hot there. 219 9781510421738.indb 219 02/02/18 1:15 PM 9 Example 9.3 The area A of a square is increasing at a rate proportional to the length of its side s. The constant of proportionality is k. Find an expression for ds . dt s 9 DIFFERENTIAL EQUATIONS s A s s % Figure 9.3 Solution The rate of increase of A with respect to time may be written as dA. dt As this is proportional to s, it may be written as dA = ks dt where k is a positive constant. dA. You can use the chain rule to write down an expression for ds in terms of dt dt ds = ds × dA dt dA dt ds You now need an expression for dt . Because A is a square A = s2 ⇒ ⇒ dA = 2s ds ds 1 = dA 2 s Substituting the expressions for ds and dA into the expression for ds dt dA dt ds 1 ⇒ dt = 2 s × ks ds = 1 k ⇒ dt 2 Exercise 9A 1 The differential equation M dv = 5v 2 dt models the motion of a particle, where v is the velocity of the particle in dv m s–1 and t is the time in seconds. Explain the meaning of dt and what the differential equation tells you about the motion of the particle. 220 9781510421738.indb 220 02/02/18 1:15 PM 2 A spark from a firework is moving in a straight line at a speed which is 3 5 6 7 8 9 10 11 9 9.1 Forming differential equations from rates of change 4 inversely proportional to the square of the distance which the spark has travelled from the firework. Find an expression for the speed (i.e. the rate of change of distance travelled) of the spark. The rate at which a sunflower increases in height is proportional to the natural logarithm of the difference between its final height H and its height h at a particular time. Find a differential equation to model this situation. In a chemical reaction in which substance A is converted into substance B, the rate of increase of the mass of substance B is inversely proportional to the mass of substance B present. Find a differential equation to model this situation. After a major advertising campaign, an engineering company finds that its profits are increasing at a rate proportional to the square root of the profits at any given time. Find an expression to model this situation. The coefficient of restitution e of a squash ball increases with respect to the ball’s temperature θ at a rate proportional to the temperature, for typical playing temperatures. (The coefficient of restitution is a measure of how elastic, or bouncy, the ball is. Its value lies between zero and one, zero meaning that the ball is not at all elastic and one meaning that it is perfectly elastic.) Find a differential equation to model this situation. A cup of coffee cools at a rate proportional to the temperature of the coffee above that of the surrounding air. Initially, the coffee is at a temperature of 95°C and is cooling at a rate of 0.5°C s–1. The surrounding air is at 15°C. Find a differential equation to model this situation. The rate of increase of bacteria is modelled as being proportional to the number of bacteria at any time during their initial growth phase. When the bacteria number 2 × 106 they are increasing at a rate of 105 per day. Find a differential equation to model this situation. The acceleration (i.e. the rate of change of velocity) of a moving object under a particular force is inversely proportional to the square root of its velocity. When the speed is 4 m s–1 the acceleration is 2 m s–2. Find a differential equation to model this situation. The radius of a circular patch of oil is increasing at a rate inversely proportional to its area A. Find an expression for dA . dt A poker, 80 cm long, has one end in a fire. The temperature of the poker decreases with respect to the distance from that end at a rate proportional to that distance. Halfway along the poker, the temperature is decreasing at a rate of 10°C cm–1. Find a differential equation to model this situation. 221 9781510421738.indb 221 02/02/18 1:15 PM 12 A spherical balloon is allowed to deflate. The rate at which air is leaving 9 DIFFERENTIAL EQUATIONS 9 the balloon is proportional to the volume V of air left in the balloon. When the radius of the balloon is 15 cm, air is leaving at a rate of 8 cm3 s–1. Find an expression for dV . dt 13 A tank is shaped as a cuboid with a square base of side 10 cm. Water runs out through a hole in the base at a rate proportional to the square root of the height, h cm, of water in the tank. At the same time, water is pumped into the tank at a constant rate of 2 cm3 s–1. Find an expression for dh . dt INVESTIGATION Under pressure Figure 9.4 shows the isobars (lines of equal pressure) on a weather map featuring a storm. The wind direction is almost parallel to the isobars and its speed is proportional to the pressure gradient. 20 ° 30° 70° 60° ° 10 70° Scale of nautical miles 100 300 500 700 0° 60° 50° 200 400 600 40° 10° ▲ Figure 9.4 Draw a line from the point H to the point L on a copy of this diagram. This runs approximately perpendicular to the isobars. It is suggested that along this line the pressure gradient (and so the wind speed) may be modelled by the differential equation dp = –a sin bx dx Suggest values for a and b, and comment on the suitability of this model. 222 9781510421738.indb 222 02/02/18 1:15 PM 9.2 Solving differential equations 9 The general solution of a differential equation Finding an expression for f(x) from a differential equation involving derivatives of f(x) is called solving the equation. Example 9.4 Solve the differential equation 9.2 Solving differential equations Some differential equations may be solved simply by integration. dy = 3x2 − 2. dx Solution Integrating gives ∫ y = (3x 2 − 2) dx y = x 3 − 2x + c Notice that when you solve a differential equation, you get not just one solution, but a whole family of solutions, as c can take any value. This is called the general solution of the differential equation. The family of solutions for the differential equation in the example above would be translations in the y direction of the curve y = x3 – 2x. Graphs of members of the family of curves can be found in Figure 9.5 on page 226. The method of separation of variables It is not difficult to solve a differential equation like the one in Example 9.4, because the right-hand side is a function of x only. So long as the function can be integrated, the equation can be solved. dy = xy. Now look at the differential equation dx This cannot be solved directly by integration, because the right-hand side is a function of both x and y. However, as you will see in the next example, you can solve this and similar differential equations where the right-hand side consists of a function of x and a function of y multiplied together. Example 9.5 Find, for y # 0, the general solution of the differential equation dy = xy. dx Solution The equation may be rewritten as 1 dy =x y dx so that the right-hand side is now a function of x only. ➜ 223 9781510421738.indb 223 02/02/18 1:15 PM 9 Integrating both sides with respect to x gives 1 dy ∫ y dx d x = ∫ x d x dy As dx dx can be written as dy 9 DIFFERENTIAL EQUATIONS 1 ∫ y dy = ∫ x dx Both sides may now be integrated separately. ln | y | = 21 x 2 + c Since you have been told y # 0, you may drop the modulus symbol. In this case, |y| = y. ? ❯ Explain why there is no need to put a constant of integration on both sides of the equation. You now need to rearrange the solution above to give y in terms of x . Making both sides powers of e gives 1 elny = e 2 x ⇒ ⇒ 2 + cc y=e 1 2 +c 2x c y=e 1x 2 c c 2 e Notice that the right-hand 1 x 2 +c c side is e 2 x 1x and not e 2 x + ec. 2 c This expression can be simplified by replacing ec with a new constant A. 1 y = Ae 2 x 2 Note Usually the first part of this process is carried out in just one step. dy = xy dx can immediately be rewritten as ∫ 1y d y = ∫ x d x This method is called separation of variables. It can be helpful to do this dy by thinking of the differential equation as though were a fraction and dx trying to rearrange the equation to obtain all the x terms on one side and all the y terms on the other. Then just insert an integration sign on each side. Remember that dy and dx must both end up in the numerator (top line). 224 9781510421738.indb 224 02/02/18 1:15 PM Example 9.6 Find the general solution of the differential equation dy = e−y. dx Solution 9 Separating the variables gives −y ⇒ y ∫ dx dy = ∫ dx dy = The right-hand side can be thought of as integrating 1 with respect to x. ey = x + c Taking logarithms of both sides gives y = ln⎪x + c ⎪ 9.2 Solving differential equations ∫ e1 ∫e ln⎪x + c ⎪ is not the same as ln ⎪x ⎪ + c. Exercise 9B 1 2 Solve the following differential equations by integration. dy dy 2 (i) (ii) dx = x dx = cos x dy dy = x (iii) = ex (iv) dx dx Find the general solutions of the following differential equations by separating the variables. dy dy x 2 (i) = xy 2 (ii) = y dx dx dy dy (iii) =y (iv) = e x −y dx dx dy dy y =x y (v) (vi) = dx dx x dy x( y 2 + 1) dy = (vii) = y 2 cos x (viii) dx y( x 2 + 1) dx dy x ln x dy = 2 (ix) = x ey (x) dx y dx The particular solution of a differential equation You have already seen that a differential equation has an infinite number of different solutions corresponding to different values of the constant of dy integration. In Example 9.4, you found that = 3x 2 − 2 had a general d x solution of y = x 3 − 2x + c. 225 9781510421738.indb 225 02/02/18 1:15 PM Figure 9.5 shows the curves of the solutions corresponding to some different values of c. 9 y 6 y = x3 – 2x + 2 (c = 2) 9 DIFFERENTIAL EQUATIONS y = x3 – 2x (c = 0) y = x3 – 2x – 1 (c = –1) 0 –3 –2 –1 1 2 3 x –6 % Figure 9.5 If you are given some more information, you can find out which of the possible solutions is the one that matches the situation in question. For example, you might be told that when x = 1, y = 0. This tells you that the correct solution is the one with the curve that passes through the point (1, 0). You can use this information to find out the value of c for this particular solution by substituting the values x = 1 and y = 0 into the general solution. ⇒ y = x 3 − 2x + c 0=1−2+c c=1 So the solution in this case is y = x 3 − 2x + 1. This is called the particular solution. Example 9.7 (i) (ii) dy = y2. dx Find the particular solution for which y = 1 when x = 0. Find the general solution of the differential equation Solution (i) Separating the variables gives ∫ y1 dy = ∫ d x 2 1 − =x+c y 1 The general solution is y = − x + c . Figure 9.6 shows a set of solution curves. 226 9781510421738.indb 226 02/02/18 1:15 PM y = – 1 (c = 0) x y y = – 1 (c = 1) x+1 3 y=– 9 1 x – 1 (c = –1) 2 –4 –3 –2 –1 0 1 2 9.2 Solving differential equations 1 x –1 –2 –3 ▲ Figure 9.6 (ii) Example 9.8 When x = 0, y = 1, which gives 1 1 = −c ⇒ c = −1. So the particular solution is 1 1 y = − x − 1 or y = 1 − x This is the blue curve illustrated in Figure 9.6. The acceleration of an object is inversely proportional to its velocity at any given time and the direction of motion is taken to be positive. When the velocity is 1 m s−1, the acceleration is 3 m s−2. (i) Find a differential equation to model this situation. (ii) Find the particular solution to this differential equation for which the initial velocity is 2 m s−1. (iii) In this case, how long does the object take to reach a velocity of 8 m s−1? Solution (i) (ii) dv = k dt v dv 3 dv When v = 1, dt = 3 so k = 3, which gives dt = v . Separating the variables: ∫ v d v = ∫ 3 dt 1v 2 2 = 3t + c ➜ 227 9781510421738.indb 227 02/02/18 1:15 PM When t = 0, v = 2 so c = 2, which gives 9 1 2 2v = 3t + 2 v 2 = 6t + 4 Since the direction of motion is positive 9 DIFFERENTIAL EQUATIONS v= 6t + 4 (iii) When v = 8: 64 = 6t + 4 60 = 6t ⇒ t = 10 The object takes 10 seconds to reach a velocity of 8 m s−1. The graph of the particular solution is shown in Figure 9.7. v v = 6t + 4 2 O The remainder of the curve for t " 0 and v " 2 is not shown as it is not relevant to the situation. t % Figure 9.7 Sometimes you will be asked to verify the solution of a differential equation. In that case you are expected to do two things: » substitute the solution in the differential equation and show that it works » show that the solution fits the conditions you have been given. Example 9.9 Show that sin y = x is a solution of the differential equation dy 1 = dx 1 − x2 given that y = 0 when x = 0. Solution ⇒ ⇒ sin y = x dy cos y =1 dx dy = 1 dx cos y 228 9781510421738.indb 228 02/02/18 1:15 PM Substituting into the differential equation LHS: RHS: 1 cos y 1 1 = = 1 1− x2 1 − sin 2 y cos y dy 1 : = dx 1 − x2 Substituting x = 0 into the solution sin y = x gives sin y = 0 and this is satisfied by y = 0. So the solution also fits the particular conditions. Exercise 9C M 1 2 Find the particular solution of each of the following differential equations. dy (i) = x 2 − 1 y = 2 when x = 3 dx dy (ii) = x2y y = 1 when x = 0 dx dy (iii) = x e–y y = 0 when x = 0 dx dy (iv) = y2 y = 1 when x = 1 dx dy (v) = x(y + 1) y = 0 when x = 1 dx dy (vi) = y 2 sin x y = 1 when x = 0 dx A cold liquid at temperature θ °C, where θ < 20, is standing in a warm room. The temperature of the liquid obeys the differential equation dθ dt = 2(20 − θ ) where the time t is measured in hours. 9.2 Solving differential equations So the solution fits the differential equation. 9 Find the general solution of this differential equation. (ii) Find the particular solution for which θ = 5 when t = 0. (iii) In this case, how long does the liquid take to reach a temperature of 18°C? A population of rabbits increases so that the number of rabbits N (in hundreds), after t years is modelled by the differential equation dN = N. dt (i) Find the general solution for N in terms of t. (ii) Find the particular solution for which N = 10 when t = 0. (iii) What will happen to the number of rabbits when t becomes very large? Why is this not a realistic model for an actual population of rabbits? (i) M 3 229 9781510421738.indb 229 02/02/18 1:15 PM 9 4 9 DIFFERENTIAL EQUATIONS 5 ( ) An object is moving so that its velocity v = ds is inversely proportional dt to its displacement s from a fixed point. If its velocity is 1 m s−1 when its displacement is 2 m, find a differential equation to model the situation. Find the general solution of your differential equation. Given that y = 1 when x = 0, solve the differential equation dy = 4 x(3y 2 + 10y + 3), dx obtaining an expression for y in terms of x. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q7 June 2015 6 The temperature of a quantity of liquid at time t is θ. The liquid is cooling in an atmosphere whose temperature is constant and equal to A. The rate of decrease of θ is proportional to the temperature differerence (θ − A). Thus θ and t satisfy the differential equation dθ = −k(θ − A), dt where k is a positive constant. (i) Find, in any form, the solution of this differential equation, given that θ = 4A when t = 0. (ii) Given also that θ = 3A when t = 1, show that k = ln 23. (iii) Find θ in terms of A when t = 2, expressing your answer in its simplest form. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q9 November 2009 7 The variables x and t are related by the differential equation e 2t dx = cos 2 x, dt where t % 0. When t = 0, x = 0. (i) Solve the differential equation, obtaining an expression for x in terms of t. (ii) State what happens to the value of x when t becomes very large. (iii) Explain why x increases as t increases. Cambridge International AS & A Level Mathematics 9709 Paper 32 Q7 June 2010 8 An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time t hours after filling begins, the volume of liquid is V m3 and the depth of liquid is h m. It is given that V = 43 h 3. 230 9781510421738.indb 230 02/02/18 1:15 PM 9 dh = 5 − 1 . dt h 2 20 2000 20h 2 ≡ −20 + (ii) Verify that . (10 − h )(10 + h ) 100 − h 2 (iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h. 9.2 Solving differential equations The liquid is poured in at a rate of 20 m3 per hour, but owing to leakage, liquid is lost at a rate proportional to h2. When h = 1, dh = 4.95.. dt (i) Show that h satisfies the differential equation Cambridge International AS & A Level Mathematics 9709 Paper 3 Q8 November 2008 9 Naturalists are managing a wildlife reserve to increase the number of plants of a rare species. The number of plants at time t years is denoted by N, where N is treated as a continuous variable. (i) It is given that the rate of increase of N with respect to t is proportional to ( N − 150). Write down a differential equation relating N, t and a constant of proportionality. (ii) Initially, when t = 0, the number of plants was 650. It was noted that, at a time when there were 900 plants, the number of plants was increasing at a rate of 60 per year. Express N in terms of t. (iii) The naturalists had a target of increasing the number of plants from 650 to 2000 within 15 years. Will this target be met? Cambridge International AS & A Level Mathematics 9709 Paper 33 Q10 November 2015 10 The number of birds of a certain species in a forested region is recorded over several years. At time t years, the number of birds is N, where N is treated as a continuous variable. The variation in the number of birds is modelled by dN = N (1800 − N ) . 3600 dt It is given that N = 300 when t = 0. (i) Find an expression for N in terms of t. (ii) According to the model, how many birds will there be after a long time? Cambridge International AS & A Level Mathematics 9709 Paper 31 Q10 June 2011 9781510421738.indb 231 231 02/02/18 1:15 PM 9 11 The variables x and θ satisfy the differential equation 2cos 2 θ dx = (2x + 1), dθ and x = 0 when θ = 41 π. Solve the differential equation and obtain an expression for x in terms of θ. 9 DIFFERENTIAL EQUATIONS Cambridge International AS & A Level Mathematics 9709 Paper 33 Q5 June 2014 12 The variables x and y are related by the differential equation (i) (ii) dy 1 21 = xy sin ( 13 x ) . dx 5 Find the general solution, giving y in terms of x. Given that y = 100 when x = 0, find the value of y when x = 25. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q8 November 2014 13 A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°, as shown in the diagram. At time t = 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to h, where h is the depth of water at time t. The tank becomes empty when t = 60. 60° h C Show that h and t satisfy a differential equation of the form dh = − Ah − 23 , dt where A is a positive constant. (ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H. 1 (iii) Find the time at which the depth reaches H . 2 [The volume V of a cone of vertical height h and base radius r is (i) given by V = 13 πr 2h .] Cambridge International AS & A Level Mathematics 9709 Paper 31 Q10 November 2013 232 9781510421738.indb 232 02/02/18 1:15 PM INVESTIGATION Cooling Investigate the coffee cooling problem introduced on page 217.You will need to make some assumptions about the initial temperature of the coffee and the temperature of the room. Would it be better to allow the coffee to cool first before adding the milk? KEY POINTS 1 2 3 4 5 A differential equation is an equation involving derivatives such as dy d 2y and x2 d dx A first-order differential equation involves a first derivative only. Some first-order differential equations may be solved by separating the variables. A general solution is one in which the constant of integration is left in the solution, and a particular solution is one in which additional information is used to calculate the constant of integration. A general solution may be represented by a family of curves, a particular solution by a particular member of that family. 9.2 Solving differential equations What difference would it make if you were to add some cold milk to the coffee and then leave it to cool? 9 LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ formulate first-order differential equations using information about rates of change ■ solve a first-order differential equation using separation of variables ■ find the general solution of a first-order differential equation ■ find the particular solution of a first-order differential equation ■ solve problems using differential equations and interpret the solution. 233 9781510421738.indb 233 02/02/18 1:15 PM P3 10 VECTORS 10 Vectors We drove into the future looking into a rear view mirror. Herbert Marshall McLuhan (1911–1980) ? ❯ What information do you need to decide how closely the aircraft which left these vapour trails passed to each other? A quantity which has both size and direction is called a vector. The velocity of an aircraft through the sky is an example of a vector, having size (e.g. 600 mph) and direction (on a course of 254°). By contrast the mass of the aircraft (100 tonnes) is completely described by its size and no direction is associated with it; such a quantity is called a scalar. Vectors are used extensively in mechanics to represent quantities such as force, velocity and momentum, and in geometry to represent displacements. They are an essential tool in three-dimensional coordinate geometry and it is this application of vectors which is the subject of this chapter. However, before coming on to this, you need to be familiar with the associated vocabulary and notation, in two and three dimensions. 234 9781510421738.indb 234 02/02/18 1:15 PM 10.1 Vectors in two dimensions Terminology The vector in Figure 10.1 has magnitude 5, direction +30°. This is written (5, 30°) and said to be in magnitude–direction form or in polar form. The general form of a vector written in this way is (r, θ ) where r is its magnitude and θ its direction. 5 + 10.1 Vectors in two dimensions In two dimensions, it is common to represent a vector by a drawing of a straight line with an arrowhead. The length represents the size, or magnitude, of the vector and the direction is indicated by the line and the arrowhead. Direction is usually given as the angle the vector makes with the positive x-axis, with the anticlockwise direction taken to be positive. 10 30° ▲ Figure 10.1 Note In the special case when the vector is representing real travel, as in the case of the velocity of an aircraft, the direction may be described by a compass bearing with the angle measured from north, clockwise. However, this is not done in this chapter, where directions are all taken to be measured anticlockwise from the positive x direction. An alternative way of describing a vector is in terms of components in given directions. The vector in Figure 10.2 is 4 units in the x direction, and 4 2 in the y direction, and this is denoted by ⎛ ⎞ . ⎝ 2⎠ )) 4 or 4i + 2j 2 2 4 ▲ Figure 10.2 j This may also be written as 4i + 2j, where i is a vector of magnitude 1, a unit vector, in the x direction and j is a unit vector in the y direction (Figure 10.3). i ▲ Figure 10.3 In a book, a vector may be printed in bold, for example p or OP, or as a line between two points with an arrow above it to indicate its direction, 235 9781510421738.indb 235 02/02/18 1:15 PM ⎯ → such as OP. When you write a vector by hand, it is usual to underline it, for ⎯ → example, p or OP, or to put an arrow above it, as in O P. 10 VECTORS 10 The magnitude of a vector is its length or size. The magnitude of a vector is also called its modulus and denoted by the symbols ||. You can use Pythagoras’ theorem to find the magnitude. Another convention for writing the magnitude of a vector is to use the same letter, but in italics and not bold type; so the magnitude of a may be written a. Example 10.1 Find the magnitude and direction of the vector a = 4i + 2j. Solution a 2 θ 4 ▲ Figure 10.4 The magnitude of a is given by the length a in Figure 10.4. a = 42 + 22 using Pythagoras’ theorem = 4.47 (to 3 significant figures) The direction is given by the angle θ. tan θ = 2 4 = 0.5 θ = 26.6° (to 3 significant figures) The vector a is (4.47, 26.6°). 10.2 Vectors in three dimensions 236 9781510421738.indb 236 02/02/18 1:15 PM Points In three dimensions, a point has three coordinates, usually called x, y and z. z This point is (3, 4, 1). 2 –1 –2 –1 2 1 O 1 2 3 4 y P –1 3 x ▲ Figure 10.5 The axes are conventionally arranged as shown in Figure 10.5, where the point P is (3, 4, 1). Even on correctly drawn three-dimensional grids, it is often hard to see the relationship between the points, lines and planes, so it is seldom worth your while trying to plot points accurately. 10.2 Vectors in three dimensions 1 –3 10 The unit vectors i, j and k are used to describe vectors in three dimensions. Equal vectors The statement that two vectors a and b are equal means two things. » The direction of a is the same as the direction of b. » The magnitude of a is the same as the magnitude of b. If the vectors are given in component form, each component of a equals the corresponding component of b. Position vectors Saying the vector a is given by 3i + 4j + k tells you the components of the vector, or equivalently its magnitude and direction. It does not tell you where the vector is situated; indeed it could be anywhere. All of the lines in Figure 10.6 represent the vector a. a a a a k i ▲ Figure 10.6 9781510421738.indb 237 j 237 02/02/18 1:15 PM There is, however, one special case which is an exception to the rule, that of a vector which starts at the origin. This is called a position vector. Thus ⎛ 3⎞ the line joining the origin to the point P(3, 4, 1) is the position vector ⎜ 4 ⎟ ⎜⎝ 1 ⎟⎠ or 3i + 4j + k. 10 Another way of expressing this is to say that the point P(3, 4, 1) has the 10 VECTORS ⎛ 3⎞ position vector ⎜ 4 ⎟ . ⎜⎝ 1 ⎟⎠ Displacement vectors Vectors can also be used to represent displacement. For example, start at A and walk 5 km east and then 12 km north to point B can be represented ⎛ 5⎞ by the vector AB = ⎜ ⎟ . Note that magnitude of displacement is not the ⎝ 12 ⎠ same as distance – the distance travelled is 5 km + 12 km = 17 km; whereas the magnitude of displacement is 5 2 + 12 2 = 13 km. Example 10.2 Points L, M and N have coordinates (4, 3), (−2, −1) and (2, 2). (i) Write down, in component form, the position vector of L and the vector ⎯→ MN. (ii) What do your answers to part (i) tell you about the lines OL and MN? Solution (i) ⎯ → 4 The position vector of L is OL = ⎛ ⎞ . 3⎠ ⎝ ⎯→ 4 The vector MN is also ⎛ ⎞ (see Figure 10.7). ⎝ 3⎠ y 4 L 3 N 2 1 –2 M 238 9781510421738.indb 238 (ii) –1 O 1 2 3 4 x –1 ▲ Figure 10.7 ⎯ → ⎯→ Since OL = MN, lines OL and MN are parallel and equal in length. 02/02/18 1:15 PM Note 10 A line joining two points, like MN in Figure 10.7, is often called a line segment, meaning that it is just that particular part of the infinite straight line that passes through those two points. ⎯→ The length of a vector In two dimensions, the use of Pythagoras’ theorem leads to the result that a vector a1i + a2 j has length | a | given by | a | = a12 + a 22 . ? CP ❯ Show that the length of the three-dimensional vector a1i + a2 j + a3k is given by |a|= Example 10.3 10.2 Vectors in three dimensions The vector MN is an example of a displacement vector. Its length represents the magnitude of the displacement when you move from M to N. a12 + a 22 + a 32 . ⎛ 2⎞ Find the magnitude of the vector a = ⎜⎜ −5 ⎟⎟ . ⎝ 3⎠ Solution |a| = 22 + (−5)2 + 32 = 4 + 25 + 9 = 38 = 6.16 (to 2 d.p.) Exercise 10A 1 Express the following vectors in component form. (i) (ii) y 3 3 a 2 2 1 1 –2 –1 0 –1 y 1 2 3 4 x –2 –1 0 –1 b 1 2 3 x 239 9781510421738.indb 239 02/02/18 1:15 PM (iii) 10 (iv) y 3 3 c 2 10 VECTORS 2 1 1 2 −i + 2j ⎛ 1⎞ ⎜ –2⎟ ⎜⎝ 3⎟⎠ 5 0 x (v) 1 2 3 4 x 3i − 4j ⎛ 4⎞ (ii) ⎜ 0⎟ (iii) 2i + 4j + 2k ⎛ 6⎞ ⎜ –2⎟ ⎜⎝ −3⎟⎠ (vi) i − 2k ⎜⎝ −2⎠⎟ (iv) i + j − 3k PS 4 Find the magnitude of these vectors. (i) 4 3 Draw diagrams to show each of these vectors and find the magnitude and direction. ⎛ 3⎞ ⎛ –4 ⎞ (i) 2i + 3j (ii) (iii) ⎝ –2⎠ ⎝ –4 ⎠ (iv) 3 d 2 1 0 y (v) Write, in component form, the vectors represented by the line segments joining the following points. (i) (2, 3) to (4, 1) (ii) (4, 0) to (6, 0) (iii) (0, 0) to (0, −4) (iv) (0, −4) to (0, 0) (v) (0, 0, 0) to (0, 0, 5) (vi) (0, 0, 0) to (−1, −2, 3) (vii) (−1, −2 , 3) to (0, 0, 0) (viii) (0, 2, 0) to (4, 0, 4) (ix) (1, 2, 3) to (3, 2, 1) (x) (4, −5, 0) to (−4, 5, 1) The points A, B and C have coordinates (2, 3), (0, 4) and (−2, 1). (i) Write down the position vectors of A and C. (ii) Write down the vectors of the line segments joining AB and CB. (iii) What do your answers to parts (i) and (ii) tell you about (a) AB and OC (b) CB and OA? (iv) Describe the quadrilateral OABC. 10.3 Vector calculations Multiplying a vector by a scalar 240 9781510421738.indb 240 When a vector is multiplied by a number (a scalar) its length is altered but its direction remains the same. 02/02/18 1:15 PM The vector 2a in Figure 10.8 is twice as long as the vector a but in the same direction. a 10 2a When the vector is in component form, each component is multiplied by the number. For example: 2 × (3i − 5j + k) = 6i − 10j + 2k ⎛ 3⎞ ⎛ 6 ⎞ 2 × ⎜ –5⎟ = ⎜ –10 ⎟ . ⎜ ⎟ ⎜ ⎟ ⎝ 1⎠ ⎝ 2 ⎠ 10.3 Vector calculations ▲ Figure 10.8 The negative of a vector In Figure 10.9 the vector −a has the same length as the vector a but the opposite direction. a –a ▲ Figure 10.9 When a is given in component form, the components of −a are the same as those for a but with their signs reversed. So ⎛ 23⎞ ⎛ –23⎞ – ⎜ 0⎟ = ⎜ 0⎟ ⎜ ⎟ ⎜ ⎟ ⎝ –11⎠ ⎝ +11⎠ Adding vectors When vectors are given in component form, they can be added component by component. This process can be seen geometrically by drawing them on graph paper, as in Example 10.4 on the next page. 241 9781510421738.indb 241 02/02/18 1:15 PM 10 Example 10.4 Add the vectors 2i − 3j and 3i + 5j. Solution 10 VECTORS 2i − 3j + 3i + 5j = 5i + 2j 5i + 2j 2i 3i + 5j 5j –3j 2i – 3j 3i ▲ Figure 10.10 The sum of two (or more) vectors is called the resultant and is usually indicated by being marked with two arrowheads. Adding vectors is like adding the legs of a journey to find its overall outcome (see Figure 10.11). resultant leg 1 leg 3 leg 2 ▲ Figure 10.11 When vectors are given in magnitude−direction form, you can find their resultant by making a scale drawing, as in Figure 10.11. If, however, you need to calculate their resultant, it is usually easiest to convert the vectors into component form, add component by component, and then convert the answer back to magnitude−direction form. Subtracting vectors Subtracting one vector from another is the same as adding the negative of the vector. 242 9781510421738.indb 242 02/02/18 1:15 PM Example 10.5 Two vectors a and b are given by 10 a = 2i + 3j b = −i + 2j. (i) Find a − b. (ii) Draw diagrams showing a, b, a − b. (i) 10.3 Vector calculations Solution a − b = (2i + 3j) − (−i + 2j) = 3i + j (ii) b –b a a j a + (–b) = a – b i ▲ Figure 10.12 When you find the vector represented by the line segment joining two points, you are in effect subtracting their position vectors. If, for example, P is the point (2, 1) and Q is the ⎯ → 1 point (3, 5), PQ is ⎛ 4 ⎞ , as ⎝ ⎠ Figure 10.13 shows. You find this by saying ⎯ → ⎯ → ⎯ → PQ = PO + OQ = −p + q. In this case, this gives ⎯ → 2 3 1 PQ = – ⎛ 1 ⎞ + ⎛ 5⎞ = ⎛ 4 ⎞ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ as expected. This is an important result: y 6 Q(3, 5) 5 4 )) 1 4 3 2 1 0 P(2, 1) 1 2 3 4 5 x ▲ Figure 10.13 ⎯ → PQ = q − p where p and q are the position vectors of P and Q. 243 9781510421738.indb 243 02/02/18 1:15 PM Geometrical figures 10 It is often useful to be able to express lines in a geometrical figure in terms of given vectors. ACTIVITY 10.1 10 VECTORS The diagram shows a cuboid OABCDEFG. P, Q, R, S and T are the midpoints of the edges they lie on. The origin is at O and the axes lie along OA, OC and OD, as shown in Figure 10.14. ⎛ 6⎞ ⎯→ ⎛ 0⎞ ⎯→ ⎛ 0 ⎞ OA = ⎜ 0⎟ , OC = ⎜ 5⎟ , OD = ⎜ 0 ⎟ ⎜⎝ 0⎟⎠ ⎜⎝ 0⎠⎟ ⎜⎝ 4 ⎟⎠ ⎯ → S G F T R D E z B C y O Q x A P ▲ Figure 10.14 (i) (ii) Example 10.6 Name the points with the following coordinates. (a) (6, 5, 4) (b) (0, 5, 0) (d) (0, 2.5, 4) (e) (3, 5, 4) (c) (6, 2.5, 0) Use the letters in the diagram to give displacements which are equal to the following vectors. Give all possible answers; some of them have more than one. ⎛ 6⎞ ⎛ 6⎞ ⎛ 0⎞ ⎛ −6⎞ ⎛ −3 ⎞ (a) ⎜ 5 ⎟ (b) ⎜ 0 ⎟ (c) ⎜ 5 ⎟ (d) ⎜ −5⎟ (e) ⎜ 2.5⎟ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ Figure 10.15 shows a hexagonal prism. G H r B q C I p A D J F 244 9781510421738.indb 244 E ▲ Figure 10.15 02/02/18 1:15 PM ⎯ → ⎯ → The hexagonal cross-section is regular and consequently A D = 2BC. ⎯ → ⎯ → 10 ⎯ → AB = p, BC = q and BG = r. Express the following in terms of p, q and r. ⎯ → ⎯ → ⎯ → → (i) AC (ii) AD (iii) HI (iv) (v) EF (vi) BE (vii) AH (viii) FI ⎯ → (i) (ii) ⎯ → ⎯→ q B ⎯ → ⎯ → AC = AB + BC =p+q ⎯ → p C p+q ⎯ → AD = 2BC = 2q ⎯ → ⎯→ A ⎯ → (iii) H I = CD ⎯ → ⎯ → C ⎯ → Since AC + CD = AD ⎯ → 10.3 Vector calculations Solution ⎯ → IJ p+q p + q + CD = 2q ⎯ → CD = q − p ⎯ → So → HI = q − p ⎯ → (iv) I J = DE ⎯ → = −AB = −p D 2q B ⎯ → (v) A C ⎯ → E F = −BC = −q ⎯ → D ⎯ → ⎯ → ⎯ → (vi) BE = BC + CD + DE = q + (q − p) + −p = 2q − 2p ⎯ → ⎯ → ⎯ → ⎯→ Notice that BE = 2CD. ⎯→ ⎯ → (vii) AH = AB + BC + CH E ▲ Figure 10.16 ⎯ → ⎯ → CH = B G =p+q+r → ⎯ → → → (viii) FI = FE + EJ + J I ⎯→ ⎯ → → ⎯ ⎯ → → ⎯ → F E = B C, E J = B G, J I = AB =q+r+p Unit vectors A unit vector is a vector with a magnitude of 1, like i and j. To find the unit vector in the same direction as a given vector, divide that vector by its magnitude. 245 9781510421738.indb 245 02/02/18 1:15 PM Thus the vector 3i + 5j (in Figure 10.17) has magnitude 32 + 52 = 34 , and 3 5 so the vector j is a unit vector. It has magnitude 1. i+ 34 34 The unit vector in the direction of vector a is written as â and read as ‘a hat’. 10 y 10 VECTORS 5j 4j 3j 3i + 5j This is the unit vector 3 i + 34 2j 5 j 34 j O 2i i 3i 4i x ▲ Figure 10.17 Example 10.7 Relative to an origin O, the position vectors of the points A, B and C are given by ⎛ 0⎞ ⎛ −2⎞ ⎯ → ⎛ −2 ⎞ ⎯ → ⎯ → OA = ⎜ 3⎟ , OB = ⎜ 1⎟ and OC = ⎜ 3⎟ . ⎜⎝ −3⎠⎟ ⎝⎜ −2⎟⎠ ⎝⎜ 1⎟⎠ ⎯ → (i) Find the unit vector in the direction AB. (ii) Find the perimeter of triangle ABC. Solution ⎯ → ⎯ → ⎯ → For convenience call OA = a, OB = b and OC = c. ⎛ 0⎞ ⎛ −2⎞ ⎛ 2⎞ ⎯ → (i) AB = b − a = ⎜ 1⎟ − ⎜ 3⎟ = ⎜ −2⎟ ⎜⎝ −3⎟⎠ ⎜⎝ −2⎟⎠ ⎜⎝ −1⎟⎠ ⎯ → ⎯ → To find the unit vector in the direction AB, you need to divide AB by its magnitude. | A⎯→B | = = 22 + (−2)2 + (−1)2 This is the 9 ⎯ → magnitude of A B. = 3 2 ⎛ 2⎞ ⎛ 3 ⎞ So the unit vector in the direction AB is 13 ⎜ −2⎟ = ⎜ − 23 ⎟ ⎜ −1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ − 3⎠ ⎯ → 246 9781510421738.indb 246 02/02/18 1:15 PM (ii) The perimeter of the triangle is given by | AB | + | AC | + | B C |. ⎯ → ⎯ → ⎯ → 10 ⎛ −2⎞ ⎛ −2⎞ ⎛ 0⎞ ⎯ → AC = c − a = ⎜ 3⎟ − ⎜ 3⎟ = ⎜ 0⎟ ⎜⎝ 1⎟⎠ ⎜⎝ −2⎟⎠ ⎜⎝ 3⎟⎠ | A⎯→C | = ⇒ 02 + 02 + 32 10.3 Vector calculations =3 ⎛ −2⎞ ⎛ 0⎞ ⎛ −2⎞ ⎯ → BC = c − b = ⎜ 3⎟ − ⎜ 1⎟ = ⎜ 2⎟ ⎜⎝ 1⎟⎠ ⎜⎝ −3⎟⎠ ⎜⎝ 4 ⎟⎠ | B⎯→C | = ⇒ ( −2) 2 + 2 2 + 4 2 = 24 Perimeter of ABC = | AB | + | AC | + | B C | ⎯ → ⎯ → =3+3+ ⎯ → 24 = 10.9 Example 10.8 A Figure 10.18 shows triangle AOB. C is a point on AB and divides it in the ratio 2 : 3. C a ⎯→ Find OC in terms of the vectors a and b. B b O ▲ Figure 10.18 Solution ⎯→ When you divide AB in the ratio 2 : 3 then ⎯→ 2 ⎯→ 2 AC is 2 + 3 = OC = OA + 5 AB ⎯→ ⎯→ OA = a and AB = b − a ⎯→ OC = a + 2 ( b − a ) 2 5 of AB and CB is 3 = 3 of AB. 2+3 5 5 = a + 2b− 2a 5 = 3a 5 + 5 2b 5 The above example made use of the ratio theorem. ❯ Prove that when C divides AB in the ratio s : t then ⎯→ OC = ? CP t a+ s b s+t s+t 247 9781510421738.indb 247 02/02/18 1:15 PM 1 Simplify the following. ⎛ 2⎞ + ⎛ 4 ⎞ (i) ⎝ 3⎠ ⎝ 5 ⎠ ⎛ 3⎞ + ⎛ –3 ⎞ (iii) ⎝ 4 ⎠ ⎝ –4 ⎠ (v) 10 VECTORS 10 Exercise 10B 2 (ii) ⎛ 2⎞ + ⎛ –1⎞ ⎝ –1⎠ ⎝ 2⎠ 2 1 (iv) 3 ⎛ ⎞ + 2 ⎛ ⎞ ⎝1⎠ ⎝ –2⎠ 6(3i − 2j) − 9(2i − j) The vectors p, q and r are given by p = 3i + 2j + k q = 2i + 2j + 2k r = −3i − j − 2k. Find, in component form, the following vectors. (i) p + q + r (ii) p − q (iv) 3(p − q) + 2(p + r) (v) 4p − 3q + 2r CP 3 ⎯ → (iii) p+r ⎯ → In the diagram, PQRS is a parallelogram and PQ = a, PS = b. (i) Write, in terms of a and b, Q the following vectors. ⎯ → ⎯ → (a) QR (b) PR (c) ⎯ → R a QS The midpoint of PR is M. Find ⎯ → ⎯ → (a) PM (b) QM. P (iii) Explain why this shows you that the diagonals of a parallelogram bisect each other. In the diagram, ABCD is a kite. AC and BD meet at M. ⎯ → ⎯ → AB = i + j and AD = i − 2j A (i) Use the facts that the diagonals j of a kite meet at right angles and that M is the midpoint of i AC to find, in terms of i and j, ⎯ → ⎯ → (a) AM (b) AC (ii) CP 4 (c) (ii) CP 248 9781510421738.indb 248 5 ⎯ → BC (d) ⎯ → CD. S b B M C D Verify that | A B | = | BC | and | AD | = | CD |. ⎯ → ⎯ → ⎯ → ⎯ → A In the diagram, ABC is a triangle. L, M and N are the midpoints of the sides BC, CA and AB. ⎯ → ⎯ → A B = p and AC = q. M N (i) Find, in terms of p and q, ⎯ → ⎯→ ⎯→ ⎯ → BC, MN, L M and LN. (ii) Explain how your results B L from part (i) show you that the sides of triangle LMN are parallel to those of triangle ABC, and half their lengths. C 02/02/18 1:15 PM 6 7 ⎛ −2⎞ ⎛ 4⎞ (iv) ⎜ 4 ⎟ (v) 5i – 3j + 2k (vi) ⎜ 0⎟ ⎜⎝ −3⎟⎠ ⎜⎝ 0⎟⎠ Relative to an origin O, the position vectors of the points A, B and C are given by ⎛ −2⎞ ⎯ → ⎛ 2⎞ ⎯ → ⎯ → OA = ⎜ 1⎟ , OB = ⎜ 4 ⎟ and OC = ⎜⎝ 3⎟⎠ ⎜⎝ 3⎟⎠ 9 ⎛ −1⎞ ⎜ 2⎟ . ⎜⎝ 1⎟⎠ Find the perimeter of triangle ABC. Relative to an origin O, the position vectors of the points P and Q are given by ⎯ → CP 10 10.3 Vector calculations 8 Find unit vectors in the same directions as the following vectors. 2 –2 (i) ⎛ ⎞ (ii) 3i + 4j (iii) ⎛ ⎞ (iv) 5i − 12j 3 ⎝ ⎠ ⎝ –2⎠ Find unit vectors in the same direction as the following vectors. ⎛ 1⎞ (i) ⎜ 2⎟ (ii) 2i – 2j + k (iii) 3i – 4k ⎜⎝ 3⎟⎠ ⎯ → O P = 3i + j + 4k and OQ = i + xj − 2k. Find the values of x for which the magnitude of PQ is 7. ⎯→ ⎯→ ⎯→ 10 In the cuboid, OA = p, OE = q , OG = r . B (i) Express the following vectors in terms of p, q and r. ⎯→ ⎯→ A (a) GF (b) CF ⎯→ (c) OB (e) OC ⎯→ M ⎯→ C D (d) OD p E q F The point M divides O r G AD in the ratio 3 : 2. ⎯→ Find OM in terms of p, q and r. (iii) Use vectors to prove that OC and BG bisect each other. Relative to an origin O, the position vectors of the points A, B, C and D are given by (ii) CP 11 ⎛ 6⎞ ⎛ 5⎞ ⎛ 8⎞ ⎛ 3⎞ ⎯→ ⎯→ ⎯→ ⎟ ⎜ 5 , OC = ⎜ 2 ⎟ , and OD = ⎜ −2 ⎟ OA = ⎜ 1 ⎟ , OB = ⎜ −1 ⎟ ⎟ ⎜ ⎜ 7⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎠ ⎝ ⎝ 13 ⎠ ⎯→ Use vectors to prove that ABCD is a parallelogram. 249 9781510421738.indb 249 02/02/18 1:15 PM 10 12 Relative to an origin O, the position vectors of the points A and B are given by ⎛ 4⎞ OA = ⎜ 1⎟ ⎜⎝ −2⎟⎠ ⎯ → ⎛ 3⎞ OB = ⎜ 2⎟ . ⎜⎝ –4 ⎟⎠ ⎯ → and ⎯ → ⎯ → Given that C is the point such that AC = 2AB, find the unit vector ⎯ → in the direction of OC. ⎛ 1⎞ ⎯ → The position vector of the point D is given by OD = ⎜ 4 ⎟ , where k is a ⎜⎝ k ⎟⎠ 10 VECTORS (i) ⎯ → ⎯ → ⎯ → constant, and it is given that OD = mOA + nOB, where m and n are constants. (ii) Find the values of m, n and k. Cambridge International AS & A Level Mathematics 9709 Paper 1 Q9 June 2007 10.4 The angle between two vectors ? CP ❯ As you work through the proof in this section, make a list of all the results that you are assuming. To find the angle θ between the two vectors ⎯ → OA = a = a1i + a2 j and y B (b1, b2) ⎯ → OB = b = b1i + b2 j A start by applying the cosine rule to triangle OAB in Figure 10.19. cos θ = (a1, a2) OA2 + OB2 – AB2 2OA × OB In this, OA, OB and AB are the ⎯ → ⎯ → lengths of the vectors OA, OB and ⎯ → AB, and so OA = | a | = ⎯ → a21 + a22 a b θ O x ▲ Figure 10.19 and OB = | b | = b21 + b22 . The vector AB = b − a = (b1i + b2 j) − (a1i + a2 j) = (b1 − a1)i + (b2 − a2)j and so its length is given by AB = | b − a | = (b1 – a1)2 + (b2 – a2)2. 250 9781510421738.indb 250 02/02/18 1:15 PM Substituting for OA, OB and AB in the cosine rule gives cos θ = = 2 a21 + a22 × b21 + b22 a21 + a22 + b21 + b22 – (b21 – 2a1b1 + a21 + b22 – 2a2b2 + a22 2 |a || b| cos θ = a b + a2 b2 2a1b1 + 2a2b2 = 11 | a || b | 2 | a || b | The expression on the top line, a1b1 + a2b2, is called the scalar product (or dot product) of the vectors a and b and is written a .b. Thus a.b cos θ = . |a || b| This result is usually written in the form a .b = | a | | b | cos θ. The next example shows you how to use it to find the angle between two vectors given numerically. ) 10.4 The angle between two vectors This simplifies to Example 10.9 10 (a12 + a 22) + (b21 + b22) – [(b1 – a1)2 + (b2 – a2)2] 5 3 Find the angle between the vectors ⎛ 4 ⎞ and ⎛ –12⎞ . ⎝ ⎠ ⎝ ⎠ Solution Let 3 a = ⎛ 4⎞ ⎝ ⎠ ⇒ |a|= 32 + 42 = 5 and 5 b = ⎛ –12 ⎞ ⎝ ⎠ ⇒ |b|= 52 + (–12)2 = 13. The scalar product ⎛3⎞ ⎛ 5 ⎞ ⎜ ⎟ ⋅⎜ ⎟ ⎝4 ⎠ ⎝ –12 ⎠ = 3 × 5 + 4 × (−12) = 15 − 48 = −33 Substituting in a .b = | a | | b | cos θ gives ⇒ −33 = 5 × 13 × cos θ cos θ = –33 65 θ = 120.5° Perpendicular vectors Since cos 90° = 0, it follows that if vectors a and b are perpendicular then a .b = 0. Conversely, if the scalar product of two non-zero vectors is zero, they are perpendicular. 9781510421738.indb 251 251 02/02/18 1:15 PM 10 Example 10.10 2 6 Show that the vectors a = ⎛ 4 ⎞ and b = ⎛ –3 ⎞ are perpendicular. ⎝ ⎠ ⎝ ⎠ Solution 10 VECTORS The scalar product of the vectors is ⎛2⎞ ⎛ 6⎞ a.b = ⎜ ⎟ . ⎜ ⎟ ⎝4 ⎠ ⎝ –3 ⎠ = 2 × 6 + 4 × (−3) = 12 − 12 = 0 Therefore the vectors are perpendicular. Further points concerning the scalar product » You will notice that the scalar product of two vectors is an ordinary number. It has size but no direction and so is a scalar, rather than a vector. It is for this reason that it is called the scalar product. There is another way of multiplying vectors that gives a vector as the answer; it is called the vector product. This is beyond the scope of this book. » The scalar product is calculated in the same way for three-dimensional vectors. For example: ⎛ 2⎞ ⎛ 5⎞ ⎜ 3⎟ . ⎜ 6⎟ = 2 × 5 + 3 × 6 + 4 × 7 = 56 ⎜⎝ 4 ⎟⎠ ⎜⎝ 7⎟⎠ In general ⎛ a1 ⎞ ⎛ b1 ⎞ ⎜ a2 ⎟ . ⎜ b2 ⎟ = a1b1 + a2b2 + a3b3 ⎜⎝ a ⎟⎠ ⎜⎝ b ⎟⎠ 3 3 » The scalar product of two vectors is commutative. It has the same value whichever of them is on the left-hand side or right-hand side. Thus a .b = b .a, as in the following example. ⎛ 2⎞ . ⎛ 6⎞ = 2 × 6 + 3 × 7 = 33 ⎝ 3⎠ ⎝ 7⎠ ⎛ 6⎞ . ⎛ 2⎞ = 6 × 2 + 7 × 3 = 33. ⎝ 7⎠ ⎝ 3⎠ ❯ How would you prove this result? ? CP 252 9781510421738.indb 252 02/02/18 1:15 PM The angle between two vectors in three dimensions 10 The angle θ between the vectors a = a1i + a2 j and b = b1i + b2 j in two dimensions is given by a1b1 + a2b2 a.b = cos θ = | || b| a a12 + a22 × b12 + b22 ❯ Show that the angle between the three-dimensional vectors a = a1i + a2 j + a3k ? CP b = b1i + b2 j + b3k and is also given by cos θ = a.b | a || b | but that the scalar product a .b is now a .b = a1b1 + a2b2 + a3b3. 10.4 The angle between two vectors where a .b is the scalar product of a and b. This result was proved by using the cosine rule on page 250. When working in two dimensions you found the angle between two lines by using the scalar product. As you have just proved, this method can be extended into three dimensions, and its use is shown in the following example. Example 10.11 The points P, Q and R are (1, 0, −1), (2, 4, 1) and (3, 5, 6). Find ∠QPR. Solution ⎯ → ⎯ → The angle between PQ and PR is given by θ in cos θ = ⎯→ ⎯→ ⎯→ ⎯→ PQ . PR |PQ||PR| In this ⎛ 2 ⎞ ⎛ 1⎞ ⎛ 1⎞ ⎯→ PQ = ⎜4 ⎟ – ⎜ 0 ⎟ = ⎜4 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1⎠ ⎝ –1⎠ ⎝ 2 ⎠ Similarly ⎛ 3⎞ ⎛ 1⎞ ⎛ 2 ⎞ PR == ⎜5 ⎟ – ⎜ 0 ⎟ = ⎜5 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝6 ⎠ ⎝ –1⎠ ⎝7 ⎠ ⎯ → ⎯ → | PQ | = ⎯ → | PR | = 12 + 42 + 22 = 21 22 + 52 + 72 = 78 ➜ 253 9781510421738.indb 253 02/02/18 1:15 PM Therefore 10 (3, 5, 6) ⎛ 1⎞ ⎛ 2 ⎞ PQ. . PR = ⎜4 ⎟ . ⎜ 5⎟ ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝7⎠ R ⎯ → ⎯ → )) 10 VECTORS = 1 ×2 + 4 ×5 + 2 ×7 = 36 2 5 7 Substituting gives 36 21 × 78 ⇒ θ = 27.2° cos θ = (2, 4, 1) )) 1 4 2 θ P Q (1, 0, –1) ▲ Figure 10.20 You must be careful to find the correct angle. To find ∠QPR ⎯ → ⎯ → (see Figure 10.21), you need the scalar product PQ . PR, as in the ⎯→ ⎯ → example above. If you take QP . PR, you will obtain ∠Q´PR, which is (180° − ∠QPR). R Q θ P Q' ▲ Figure 10.21 Exercise 10C 1 2 Find the angles between these vectors. (i) 2i + 3j and 4i + j (ii) 2i − j and i + 2j (iii) ⎛ –1⎞ ⎝ –1⎠ (v) ⎛2⎞ ⎜ ⎟ ⎝3⎠ and and ⎛ –1⎞ ⎝ –2⎠ ⎛ –6 ⎞ ⎜ ⎟ ⎝ 4⎠ (iv) 4i + j and i + j (vi) ⎛ 3⎞ ⎝ –1⎠ and ⎛ –6⎞ ⎝ 2⎠ The points A, B and C have coordinates (3, 2), (6, 3) and (5, 6), respectively. ⎯ → ⎯ → (i) Write down the vectors A B and BC. (ii) Show that the angle ABC is 90°. | ⎯→ | | ⎯→ | (iii) Show that A B = BC . (iv) The figure ABCD is a square. 254 9781510421738.indb 254 Find the coordinates of the point D. 02/02/18 1:15 PM 3 Find the angles between these pairs of vectors. (i) ⎛ 2⎞ ⎜1⎟ ⎜⎝ 3⎟⎠ and ⎛ 2⎞ ⎜ –1⎟ ⎜⎝ 4 ⎟⎠ ⎛ 1⎞ ⎜ –1⎟ ⎜⎝ 0⎟⎠ (ii) and 10 ⎛ 3⎞ ⎜1⎟ ⎜⎝ 5⎟⎠ (iii) 3i + 2j − 2k and −4i − j + 3k 4 The diagram shows a room with rectangular walls, floor and ceiling. A string has been stretched in a straight line between the corners A and G. G F z string E (0, 0, 3) D y C (0, 4, 0) B spider x A (5, 0, 0) O (0, 0, 0) 10.4 The angle between two vectors PS The corner O is taken as the origin. A is (5, 0, 0), C is (0, 4, 0) and D is (0, 0, 3), where the lengths are in metres. A spider walks up the string, starting at A. (i) Write down the coordinates of G. (ii) Find the vector A G and the distance the spider walks along the string from A to G. ⎯ → (iii) Find the angle of elevation of the spider’s journey along the string. 5 In the diagram, OABCDEFG is a cube in which each side has length 6. ⎯ → ⎯ → ⎯ → Unit vectors i, j and k are parallel to OA, OC and OD respectively. The ⎯ → ⎯ → point P is such that A P = 13 A B and the point Q is the midpoint of DF. F G Q D E k B C j O P A i ⎯ → ⎯ → (i) Express each of the vectors OQ and PQ in terms of i, j and k. (ii) Find the angle OQP. Cambridge International AS & A Level Mathematics 9709 Paper 12 Q6 November 2009 9781510421738.indb 255 255 02/02/18 1:15 PM 10 6 Relative to an origin O, the position vectors of points A and B are 2i + j + 2k and 3i − 2j + pk respectively. (i) Find the value of p for which OA and OB are perpendicular. (ii) In the case where p = 6, use a scalar product to find angle AOB, correct to the nearest degree. 10 VECTORS ⎯ → (iii) Express the vector A B in terms of p and hence find the values of p for which the length of AB is 3.5 units. Cambridge International AS & A Level Mathematics 9709 Paper 1 Q10 June 2008 7 Relative to an origin O, the position vectors of the points A and B are given by ⎯ → OA = 2i − 8j + 4k ⎯ → and ⎯ → OB = 7i + 2j − k. ⎯ → (i) Find the value of OA . OB and hence state whether angle AOB is acute, obtuse or a right angle. (ii) The point X is such that A X = 5 A B. Find the unit vector in the direction of OX. ⎯ → → 2⎯ Cambridge International AS & A Level Mathematics 9709 Paper 1 Q6 June 2009 8 Relative to an origin O, the position vectors of the points A and B are given by ⎯ → OA = 2i + 3j − k and ⎯ → OB = 4i − 3j + 2k. (i) Use a scalar product to find angle AOB, correct to the nearest degree. (ii) Find the unit vector in the direction of A B. ⎯ → ⎯ → (iii) The point C is such that OC = 6j + pk, where p is a constant. ⎯ → ⎯ → Given that the lengths of A B and A C are equal, find the possible values of p. Cambridge International AS & A Level Mathematics 9709 Paper 1 Q11 June 2005 9 Relative to an origin O, the position vectors of the points P and Q are given by ⎯ → ⎛ −2 ⎞ OP = ⎜ 3⎟ ⎜⎝ 1⎟⎠ and ⎛ 2⎞ ⎯ → OQ = ⎜ 1 ⎟ , ⎜⎝ q ⎟⎠ where q is a constant. (i) In the case where q = 3, use a scalar product to show that cos POQ = 71 . (ii) Find the values of q for which the length of PQ is 6 units. ⎯ → Cambridge International AS & A Level Mathematics 9709 Paper 1 Q4 November 2005 256 9781510421738.indb 256 02/02/18 1:15 PM 10 The diagram shows a semicircular prism with a horizontal rectangular base ABCD. The vertical ends AED and BFC are semicircles of radius 6 cm. The length of the prism is 20 cm. The midpoint of AD is the origin O, the midpoint of BC is M and the midpoint of DC is N. The points E and F are the highest points of the semicircular ends of the prism. The point P lies on EF such that EP = 8 cm. 10 F E B M C k A N j 6 cm O 20 cm i D Unit vectors i, j and k are parallel to OD, OM and OE respectively. ⎯ → ⎯ → (i) Express each of the vectors PA and PN in terms of i, j and k. (ii) Use a scalar product to calculate angle APN. Cambridge International AS & A Level Mathematics 9709 Paper 1 Q4 November 2008 10.4 The angle between two vectors P 8 cm 11 The diagram shows the roof of a house. The base of the roof, OABC, is rectangular and horizontal with OA = CB = 14 m and OC = AB = 8 m. The top of the roof DE is 5 m above the base and DE = 6 m. The sloping edges OD, CD, AE and BE are all equal in length. 6m D E B 8m C A k j 14 m O i Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. ⎯→ (i) Express the vector OD in terms of i, j and k, and find its magnitude. (ii) Use a scalar product to find angle DOB. Cambridge International AS & A Level Mathematics 9709 Paper 1 Q8 June 2006 257 9781510421738.indb 257 02/02/18 1:15 PM 10 12 The diagram shows a cube OABCDEFG in which the length of each ⎯→ ⎯→ ⎯→ side is 4 units. The unit vectors i, j and k are parallel to OA, OC and OD respectively. The midpoints of OA and DG are P and Q respectively and R is the centre of the square face ABFE. F G 10 VECTORS Q D E R B C k j O i P A ⎯ → ⎯ → (i) Express each of the vectors PR and PQ in terms of i, j and k. (ii) Use a scalar product to find angle QPR. (iii) Find the perimeter of triangle PQR, giving your answer correct to 1 decimal place. Cambridge International AS & A Level Mathematics 9709 Paper 1 Q10 November 2007 10.5 The vector equation of a line The vector joining two points In Figure 10.22, start by looking at two points A(2, −1) and B(4, 3); that is the ⎯→ ⎛ 2 ⎞ ⎯→ ⎛ 4 ⎞ points with position vectors OA = ⎜ ⎟ and OB = ⎜ ⎟ , alternatively 2i − j ⎝ 3⎠ ⎝ −1⎠ and 4i + 3j. y 3 B(4, 3) N 2 1 –1 O –1 M 1 2 3 4 5 x A(2, –1) –2 258 9781510421738.indb 258 ▲ Figure 10.22 02/02/18 1:15 PM ⎯ → The vector joining A to B is AB and this is given by ⎯ → ⎯ → 10 ⎯→ AB = AO + OB ⎯→ ⎯→ = −OA + OB ⎯→ ⎯→ = OB − OA 10.5 The vector equation of a line ⎛ 4 ⎞ ⎛ 2⎞ ⎛ 2⎞ =⎜ ⎟− ⎜ ⎟ =⎜ ⎟ ⎝ 3⎠ ⎝ –1⎠ ⎝ 4 ⎠ ⎯ → ⎛ 2⎞ Since AB = ⎜ ⎟ , then it follows that the length of AB is given by ⎝ 4⎠ ⎯ → ⏐AB⏐ = 2 2 + 4 2 = 20 You can find the position vectors of points along AB as follows. ⎯→ The midpoint, M, has position vector OM, given by ⎯→ ⎯→ ⎯ → OM = OA + 21AB ⎛ 2⎞ ⎛2⎞ = ⎜ ⎟ + 21 ⎜ ⎟ ⎝ –1⎠ ⎝4 ⎠ ⎛ 3⎞ =⎜ ⎟ ⎝ 1⎠ In the same way, the position vector of the point N, three-quarters of the distance from A to B, is given by ⎯→ ⎛ 2⎞ ⎛ 2⎞ ON = ⎜ ⎟ + 43 ⎜ ⎟ ⎝ –1⎠ ⎝ 4⎠ ⎛ 3 1⎞ = ⎜ 2⎟ ⎝2 ⎠ and it is possible to find the position vector of any other point of subdivision of the line AB in the same way. ⎯ → ⎯→ ⎯ → A point P has position vector OP = OA + λAB where λ is a fraction. ❯ Show that this can be expressed as ⎯ → ⎯→ ? CP ⎯ → OP = (1 − λ)OA + λOB. 259 9781510421738.indb 259 02/02/18 1:15 PM 10 VECTORS 10 The vector equation of a line It is now a small step to go from finding the position vector of any point on the line AB to finding the vector form of the equation of the line AB. To take this step, you will find it helpful to carry out the following activity. ACTIVITY 10.2 The position vectors of a set of points are given by λ is the Greek ⎛ 2⎞ ⎛ 2⎞ r = ⎜ ⎟ + λ⎜ ⎟ letter ‘lamda’. –1 4 ⎝ ⎠ ⎝ ⎠ where λ is a parameter which may take any value. Show that λ = 2 corresponds to the point with position ⎛ 6⎞ vector ⎜ ⎟ . ⎝ 7⎠ (ii) Find the position vectors of points corresponding to values of λ of −2, −1, 0, 21 , 43 , 1, 3. (i) (iii) Mark all your points on a sheet of squared paper and show that when they are joined up they give the line AB in Figure 10.23. (iv) State what values of λ correspond to the points A, B, M and N. (v) What can you say about the position of the point if (a) 0 < λ < 1? (b) λ > 1? (c) λ < 0? This activity should have convinced you that ⎛ 2⎞ ⎛ 2⎞ r =⎜ ⎟ +λ⎜ ⎟ ⎝ –1⎠ ⎝ 4⎠ The number λ is called a parameter and it can take any value. Of course, you can use other letters for the parameter such as µ, s and t. is the equation of the line passing through (2, −1) and (4, 3), written in vector form. 260 9781510421738.indb 260 02/02/18 1:15 PM You may find it helpful to think of this in these terms. ⎛ 2⎞ r=⎜ ⎟ +λ ⎝ –1⎠ 2 Move to the point A with ( ) 2 . position vector −1 ⎛ 2⎞ ⎜⎝ 4 ⎟⎠ ( ) 2 (i.e. 3 Move λ steps of 4 ⎯ → in the direction AB). λ need 10 not be a whole number and may be negative. 3 B(4, 3) 2 1 Start at the origin. 1 –1 O 1 2 –1 3 4 5 x 10.5 The vector equation of a line and then y A(2, –1) –2 ▲ Figure 10.23 You should also have noticed that when: λ= 0 the point corresponds to the point A λ= 1 the point corresponds to the point B 0"λ"1 the point lies between A and B λ#1 the point lies beyond B λ"0 the point lies beyond A. The vector form of the equation is not unique; there are many (in fact infinitely many) different ways in which the equation of any particular line may be expressed. There are two reasons for this: direction and location. Direction ⎛ 2⎞ The direction of the line in the example is ⎜ ⎟ . That means that for every ⎝ 4⎠ 2 units along (in the i direction), the line goes up 4 units (in the j direction). This is equivalent to stating that for every 1 unit along, the line goes up 2 units, corresponding to the equation ⎛ 2⎞ ⎛1 ⎞ r= ⎜ ⎟ + λ⎜ ⎟. ⎝ –1⎠ ⎝2⎠ 261 9781510421738.indb 261 02/02/18 1:15 PM 10 VECTORS 10 The only difference is that the two equations have different values of λ for ⎛ 4⎞ particular points. In the first equation, point B, with position vector ⎜ ⎟ , ⎝ 3⎠ corresponds to a value of λ of 1. In the second equation, the value of λ for B is 2. ⎛1⎞ ⎛ 3⎞ ⎛1⎞ ⎛ 2⎞ The direction ⎜ ⎟ is the same as ⎜ ⎟ , or as any multiple of ⎜ ⎟ such as ⎜ ⎟ , ⎝ 2⎠ ⎝ 6⎠ ⎝ 2⎠ ⎝ 4⎠ ⎛ –5⎞ or ⎛ 100.5⎞ . Any of these could be used in the vector equation of the line. ⎜⎝ –10⎠⎟ ⎜⎝ 201 ⎟⎠ Location In the equation ⎛ 2⎞ ⎛ 2⎞ r=⎜ ⎟ + λ⎜ ⎟ ⎝ –1 ⎠ ⎝ 4⎠ ⎛ 2⎞ ⎜⎝ –1⎟⎠ is the position vector of the point A on the line, and represents the point at which the line was joined. However, this could have been any other point on the line, such as M(3, 1), B(4, 3), etc. Consequently ⎛ 3⎞ ⎛ 2⎞ r = ⎜ 1 ⎟ + λ ⎜ 4⎟ ⎝ ⎠ ⎝ ⎠ and ⎛ 4⎞ ⎛ 2⎞ r = ⎜ 3⎟ + λ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ are also equations of the same line, and there are infinitely many other possibilities, one corresponding to each point on the line. Notes 1 It is usual to refer to any valid vector form of the equation as the vector equation of the line even though it is not unique. 2 It is often a good idea to give the direction vector in its simplest integer form: ⎛2⎞ ⎛1 ⎞ for example, replacing ⎜ ⎟ with ⎜ ⎟ . ⎜4 ⎟ ⎝ ⎠ ⎜2⎟ ⎝ ⎠ The general vector form of the equation of a line If A and B are points with position a and b, then the equation ⎯→ ⎯ → r = OA + λ AB 262 9781510421738.indb 262 may be written as r = a + λ(b − a) which implies r = (1 − λ)a + λb. This is the general vector form of the equation of the line joining two points. 02/02/18 1:16 PM ACTIVITY 10.3 Plot the following lines on the same sheet of squared paper. When you have done so, explain why certain among them are the same as each other, others are parallel to each other, and others are in different directions. ⎛ 2⎞ ⎛1 ⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ –1⎠ ⎝2⎠ ⎛0 ⎞ ⎝2⎠ ⎛1 ⎞ ⎝2⎠ (iii) r = ⎜ ⎟ + λ ⎜ ⎟ (v) (ii) ⎛ 2⎞ ⎛ –1⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ –1⎠ ⎝ 2⎠ (iv) 1 3 r = ⎛⎜ ⎞⎟ + λ ⎛⎜ ⎞⎟ ⎝ –3⎠ ⎝ 6⎠ ⎛ 4⎞ ⎛ 1⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ 3⎠ ⎝ –2⎠ The same methods can be used to find the vector equation of a line in three dimensions, as shown in the next example. Example 10.12 10.5 The vector equation of a line (i) 10 The coordinates of A and B are (–2, 4, 1) and (2, 1, 3) respectively. (i) (ii) (iii) Find the vector equation of the line AB. Does the point P(6, –2, 7) lie on the line AB? The point N lies on the line AB. ⎯→ ⎯→ Given that 3⏐AN⏐=⏐NB⏐ find the coordinates of N. Solution (i) ⎛ 2⎞ ⎯→ ⎛ −2 ⎞ ⎯ → a = OA = ⎜ 4 ⎟ and b = OB = ⎜ 1⎟ ⎝ 3⎠ ⎝ 1⎠ ⎛ 2 ⎞ ⎛− 2 ⎞ ⎛ 4 ⎞ ⎯ → AB = b − a = ⎜⎜ 1 ⎟⎟ − ⎜⎜ 4 ⎟⎟ = ⎜⎜− 3⎟⎟ ⎝ 3 ⎠ ⎝ 1⎠ ⎝ 2 ⎠ The vector equation of a line can be written as ⎯→ ⎯ → r = OA + λAB ⎛−2 ⎞ ⎛ 4⎞ ⇒ r = ⎜ 4 ⎟ + λ ⎜−3⎟ ⎜ 1⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ (ii) There are other ways of writing this equation, for example ⎛2⎞ ⎛ 4⎞ r = ⎜ 1⎟ + λ ⎜− 3⎟ or ⎝ 3⎠ ⎝ 2⎠ ⎛− 6 ⎞ ⎛ 4⎞ r = ⎜ 7 ⎟ + λ ⎜− 3⎟ ⎝ − 1⎠ ⎝ 2⎠ but they are all equivalent to each other. If P lies on the line AB then for some value of λ ⎛ 4⎞ ⎛ x ⎞ ⎛ 6 ⎞ ⎛− 2 ⎞ ⎜ y ⎟ = ⎜− 2 ⎟ = ⎜ 4 ⎟ + λ ⎜− 3⎟ ⎜ z ⎟ ⎜ 7 ⎟ ⎜ 1⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Find the value of λ for the x-coordinate. x : 6 = −2 + 4λ ⇒ λ=2 ➜ 263 9781510421738.indb 263 02/02/18 1:16 PM Then check whether this value of λ gives a y-coordinate of −2 and a z-coordinate of 7. 10 y : −2 = 4 − 3 × 2 z : 7 ≠ 1 + 2 ×2 So the point P(6, –2, 7) does not lie on the line. 10 VECTORS (iii) ⎯→ ⎯→ Since 3⏐AN⏐=⏐NB⏐, N must lie the value of λ is 41 . ⎯→ ⎯→ 1 4 of the way along the line AB so → 1 ⎯ ON = OA + 4 AB ⎛−2 ⎞ ⎛ 4 ⎞ ⎛ −1⎞ ⎜ ⎟ 1 ⎜ ⎟ ⎜3 1 ⎟ ON = 4 + −3 = 4 ⎜ ⎟ 4 ⎜ ⎟ ⎜ 1⎟ ⎝ 1⎠ ⎝ 2⎠ ⎝ 1 2 ⎠ ⎯→ So the coordinates of N are (–1, 3.25, 1.5). Exercise 10D 1 For each of these pairs of points, A and B, write down: ⎯ → (a) the vector AB ⎯ → (b) ⏐AB⏐ (c) the position vector of the midpoint of AB. (i) A is (2, 3), B is (4, 11). (ii) A is (4, 3), B is (0, 0). (iii) A is (−2, −1), B is (4, 7). (iv) A is (−3, 4), B is (3, −4). (v) 2 A is (−10, −8), B is (−5, 4). Find the equation of each of these lines in vector form. (i) Joining (2, 1) to (4, 5) (ii) Joining (3, 5) to (0, 8) (iii) Joining (−6, −6) to (4, 4) (iv) Through (5, 3) in the same direction as i + j Through (2, 1) parallel to 6i + 3j ⎛ –1⎞ (vi) Through (0, 0) parallel to ⎜⎝ 4 ⎟⎠ (vii) Joining (0, 0) to (−2, 8) (v) (viii) Joining (3, −12) to (−1, 4) 264 9781510421738.indb 264 02/02/18 1:16 PM 3 Find the equation of each of these lines in vector form. (i) ⎛ 3⎞ Through (2, 4, −1) in the direction ⎜ 6⎟ ⎝ 4⎠ ⎛ 1⎞ Through (1, 0, −1) in the direction ⎜ 0⎟ ⎝ 0⎠ (iii) Through (1, 0, 4) and (6, 3, −2) 10 (ii) (v) 4 Through (1, 2, 3) and (−2, −4, −6) Determine whether the given point P lies on the line in each of the following cases. (i) ⎛ 1⎞ ⎛ 2⎞ P(5, 1, 4) and the line r = ⎜ 3⎟ + λ ⎜− 1⎟ ⎜4 ⎟ ⎜ 0⎟ ⎝ ⎠ ⎝ ⎠ (ii) ⎛ 2⎞ ⎛ 1⎞ P(−1, 5, 1) and the line r = ⎜⎜ 3⎟⎟ + λ ⎜⎜− 2 ⎟⎟ ⎝4 ⎠ ⎝ 3⎠ ⎛ 1⎞ ⎛−2 ⎞ ⎝−2 ⎠ ⎝ 5⎠ 10.6 The intersection of two lines (iv) Through (0, 0, 1) and (2, 1, 4) (iii) P(−5, 3, 12) and the line r = ⎜⎜ 0 ⎟⎟ + λ ⎜⎜ 1⎟⎟ ⎛ 1⎞ ⎛ 4⎞ ⎝0⎠ ⎝− 2 ⎠ (iv) P(9, 0, −6) and the line r = ⎜⎜ 2 ⎟⎟ + λ ⎜⎜ − 1⎟⎟ (v) PS 5 ⎛ 1⎞ ⎛2⎞ P(−9, −2, −17) and the line r = ⎜⎜ 3⎟⎟ + λ ⎜⎜ 1⎟⎟ ⎝−2 ⎠ ⎝ 3⎠ The coordinates of three points are A(−1, −2, 1), B( −3, 4, −5) and C(0, −2, 4). (i) Find a vector equation of the line AB. (ii) Find the coordinates of the midpoint M of AB. (iii) The point N lies on BC. ⎯ → ⎯→ Given that 2⏐BN⏐=⏐NC⏐, find the equation of the line MN. 10.6 The intersection of two lines Hold a pen and a pencil to represent two distinct straight lines as follows: » hold them to represent parallel lines; » hold them to represent intersecting lines; » hold them to represent lines which are not parallel and which do not intersect (even if you extend them). 9781510421738.indb 265 265 02/02/18 1:16 PM In three-dimensional space two or more straight lines which are not parallel and which do not meet are known as skew lines. In a plane two distinct lines are either parallel or intersecting, but in three dimensions there are three possibilities: the lines may be parallel, or intersecting, or skew. The next example illustrates a method of finding whether two lines meet, and, if they do meet, the coordinates of the point of intersection. 10 VECTORS 10 Example 10.13 Find the position vector of the point where the following lines intersect. ⎛6 ⎞ ⎛ 1⎞ ⎛2⎞ ⎛1 ⎞ r = ⎜ ⎟ + λ ⎜ ⎟ and r = ⎜ ⎟ + µ ⎜ ⎟ ⎝1 ⎠ ⎝ –3⎠ ⎝3⎠ ⎝2⎠ Note here that different letters are used for the parameters in the two equations to avoid confusion. Solution When the lines intersect, the position vector is the same for each of them. ⎛x ⎞ ⎛2⎞ ⎛1 ⎞ ⎛6 ⎞ ⎛ 1⎞ r = ⎜ ⎟ = ⎜ ⎟ + λ⎜ ⎟ = ⎜ ⎟ + µ⎜ ⎟ ⎝y ⎠ ⎝3⎠ ⎝ 2 ⎠ ⎝1 ⎠ ⎝− 3⎠ This gives two simultaneous equations for λ and µ. x: 2 + λ = 6 + µ ⇒ λ−µ=4 y : 3 + 2λ = 1 − 3µ ⇒ 2λ + 3µ = −2 Solving these gives λ = 2 and µ = −2. Substituting in either equation gives ⎛ 4⎞ r=⎜ ⎟ ⎝ 7⎠ which is the position vector of the point of intersection. Example 10.14 Find the coordinates of the point of intersection of the lines joining A(1, 6) to B(4, 0), and C(1, 1) to D(5, 3). y A(1, 6) 6 3 D(5, 3) 1 C(1, 1) B(4, 0) O 1 4 x ▲ Figure 10.24 266 9781510421738.indb 266 02/02/18 1:16 PM Solution 10 ⎛ 4 ⎞ ⎛ 1 ⎞ ⎛ 3⎞ ⎯ → AB = ⎜ 0 ⎟ − ⎜ 6⎟ = ⎜ –6⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ and so the vector equation of line AB is ⎯→ ⎯ → r = OA + λAB 10.6 The intersection of two lines ⎛1 ⎞ ⎛ 3⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝6 ⎠ ⎝ –6 ⎠ ⎯→ ⎛ 5 ⎞ ⎛ 1⎞ ⎛ 4⎞ CD = ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ ⎝ 3⎠ ⎝ 1⎠ ⎝ 2⎠ and so the vector equation of line CD is ⎯→ ⎯→ r = OC + µ CD ⎛1⎞ ⎛4 ⎞ r = ⎜ ⎟ + µ⎜ ⎟ ⎝1⎠ ⎝2⎠ The intersection of these lines is at ⎛1 ⎞ ⎛ 3⎞ ⎛1⎞ ⎛4 ⎞ r = ⎜ ⎟ + λ⎜ ⎟ = ⎜ ⎟ + µ⎜ ⎟ ⎝6 ⎠ ⎝− 6 ⎠ ⎝1⎠ ⎝2⎠ x : 1 + 3λ = 1 + 4µ ⇒ 3λ − 4µ = 0 1 ! y : 6 − 6λ = 1 + 2µ ⇒ 6λ + 2µ = 5 2 ! 1 and ! 2 simultaneously: Solve ! 1 : ! 2 × 2: ! Add: 3λ − 4µ = 0 12λ + 4µ = 10 15λ = 10 ⇒ λ= Substitute λ = 2 3 2 3 in the equation for AB: ⎛ 1⎞ ⎛ 3⎞ ⇒ r = ⎜ ⎟ + 23 ⎜ ⎟ ⎝ 6⎠ ⎝ –6⎠ ⎛ 3⎞ ⇒ r = ⎜ ⎟ . The point of intersection has coordinates (3, 2). ⎝ 2⎠ Note Alternatively, you could have found µ = 1 and substituted in the equation 2 for CD. 267 9781510421738.indb 267 02/02/18 1:16 PM Example 10.15 10 VECTORS 10 In three dimensions, lines may be parallel, they may intersect or they may be skew. Determine whether each pair of lines are parallel, intersect or are skew. (i) ⎛ 1⎞ ⎛−3⎞ ⎛ 1⎞ ⎛ 6⎞ r = ⎜⎜−2 ⎟⎟ + λ ⎜⎜ 2 ⎟⎟ and r = ⎜⎜ 3⎟⎟ + µ ⎜⎜−4 ⎟⎟ ⎝ 1⎠ ⎝ 1⎠ ⎝−2 ⎠ ⎝ −2 ⎠ (ii) ⎛ 1⎞ ⎛ 2⎞ ⎛ 4⎞ ⎛−1⎞ r = ⎜⎜ 2 ⎟⎟ + λ ⎜⎜−3⎟⎟ and r = ⎜−2 ⎟ + µ ⎜ 2 ⎟ ⎜ −5⎟ ⎜ 1⎟ ⎝−1⎠ ⎝ 4⎠ ⎝ ⎠ ⎝ ⎠ Solution (i) ⎛ −3⎞ ⎛ 6⎞ The vectors −2 ⎜ 2⎟ and ⎜ −4 ⎟ are in the same direction as ⎝ 1⎠ ⎝ −2⎠ ⎛ 6⎞ ⎛ −3⎞ ⎜ −4 ⎟ = −2 ⎜ 2⎟ ⎝ −2⎠ ⎝ 1⎠ Note the lines are different as one line passes through (1, –2, 1) and the other through (1, 3, –2). So the lines are parallel. (ii) These lines are not parallel, so either they intersect or they are skew. If the two lines intersect then there is a point (x, y, z) that lies on both lines. ⎛ 2⎞ ⎛ x ⎞ ⎛ 1⎞ ⎜ y ⎟ = ⎜ 2 ⎟ + λ ⎜− 3⎟ ⎜ z ⎟ ⎜− 1⎟ ⎜ 4⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ and ⎛x ⎞ ⎛ 4 ⎞ ⎛− 1⎞ ⎜ y ⎟ = ⎜− 2 ⎟ + µ ⎜ 2 ⎟ ⎜ z ⎟ ⎜ − 5⎟ ⎜ 1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ This gives three simultaneous equations for λ and µ. x : 1 + 2λ = 4 − µ ⇒ 2λ + µ = 3 1 ! y : 2 − 3λ = −2 + 2µ ⇒ 3λ + 2µ = 4 2 ! z : −1 + 4λ = −5 + µ ⇒ 4λ − µ = −4 3 ! Now solve any two of the three equations above simultaneously. 1 and ! 2 : Using ! { 23 λ+µ=3 λ + 2µ = 4 } ⇒ {34 λ + 2µ = 6 λ + 2µ = 4 } ⇒ λ = 2, µ = −1 3 If these solutions satisfy the previously unused equation (equation ! here) then the lines meet, and you can substitute the value of λ (or 1,! 2 and ! 3 to find the coordinates of the point of µ) into equations ! intersection. 3 then the lines are skew. If these solutions do not satisfy equation ! 3 4λ − µ = −4 ! 268 9781510421738.indb 268 02/02/18 1:16 PM When λ = 2 and µ = −1 4λ − µ = 9 ≠ −4 As there are no values for λ and µ that satisfy all three equations, the lines do not meet and so are skew; you have already seen that they are not parallel. If the equation of the second line was ⎛ x ⎞ ⎛ 4⎞ ⎛ − 1⎞ ⎜ y ⎟ = ⎜ − 2⎟ + µ ⎜ 2⎟ ⎜⎝ z ⎟⎠ ⎜⎝ 8⎟⎠ ⎜⎝ 1 ⎟⎠ then the values of λ = 2 and µ = –1 would produce the same point for both lines: ⎛ x ⎞ ⎛ 1⎞ ⎛ 2 ⎞ ⎛ 5⎞ ⎜ y ⎟ = ⎜ 2 ⎟ + 2 ⎜ − 3⎟ = ⎜ − 4 ⎟ ⎜⎝ z ⎟⎠ ⎜⎝ − 1 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 7⎟⎠ 10.6 The intersection of two lines Note 10 ⎛ x ⎞ ⎛ 4⎞ ⎛ − 1 ⎞ ⎛ 5⎞ ⎜ y ⎟ = ⎜ − 2⎟ − 1⎜ 2⎟ = ⎜ − 4 ⎟ ⎜⎝ z ⎟⎠ ⎜⎝ 8⎟⎠ ⎜⎝ 1 ⎟⎠ ⎜⎝ 7⎟⎠ . and So the lines would intersect at (5, –4, 7). Exercise 10E 1 Find the position vector of the point of intersection of each of these pairs of lines. (i) ⎛2⎞ ⎛1 ⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝1 ⎠ ⎝0 ⎠ and ⎛ 3⎞ ⎛1⎞ r = ⎜ ⎟ + µ⎜ ⎟ ⎝1⎠ ⎝0 ⎠ (ii) ⎛ 2⎞ ⎛ 1⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ –1⎠ ⎝2⎠ and ⎛1⎞ r = µ⎜ ⎟ ⎝1⎠ and ⎛ 0⎞ ⎛ 1⎞ r = ⎜ ⎟ + µ⎜ ⎟ ⎝2⎠ ⎝ –7 ⎠ and ⎛1 ⎞ ⎛ 2⎞ r = ⎜ ⎟ + µ⎜ ⎟ ⎝ 3⎠ ⎝ –1⎠ and ⎛5⎞ ⎛ 1⎞ r = ⎜ ⎟ + µ⎜ ⎟ ⎝2⎠ ⎝1 ⎠ ⎛0 ⎞ ⎝5 ⎠ ⎛ –2 ⎞ ⎟ ⎝ –2 ⎠ (iii) r = ⎜ ⎟ + λ ⎜ ⎛−2 ⎞ ⎛−1⎞ ⎟ + λ⎜ ⎟ ⎝ 3⎠ ⎝ –3 ⎠ (iv) r = ⎜ (v) ⎛2⎞ ⎛ 1⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ –1⎠ ⎝7 ⎠ 269 9781510421738.indb 269 02/02/18 1:16 PM 2 10 VECTORS 10 Decide whether each of these pairs of lines intersect, are parallel or are skew. If the lines intersect, find the coordinates of the point of intersection. ⎛ 1⎞ ⎛ 1⎞ ⎛9⎞ ⎛ 2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3⎟ + = + r = − 6 2 and r 7 λ µ (i) ⎜ −1⎟ ⎜ 3⎟ ⎜2⎟ ⎜−1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 1⎞ ⎛ 6⎞ r = ⎜⎜−6 ⎟⎟ + λ ⎜⎜−9 ⎟⎟ ⎝ 0⎠ ⎝ −3⎠ and ⎛−5⎞ ⎛ 2⎞ r = ⎜⎜ 3⎟⎟ + µ ⎜⎜−3⎟⎟ ⎝ 0⎠ ⎝ −1⎠ ⎛ 6⎞ ⎛ 1⎞ ⎜ ⎟ ⎜−2 ⎟ + r = − λ 4 (iii) ⎜ 2⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ and ⎛ 1⎞ ⎛ 1⎞ r = ⎜⎜ 4 ⎟⎟ + µ ⎜⎜−1⎟⎟ ⎝−17 ⎠ ⎝ 2⎠ (ii) ⎛2⎞ ⎛−1⎞ ⎜ ⎟ ⎜0⎟ + r = 2 λ (iv) ⎜ 4⎟ ⎜ 3⎟ ⎝ ⎠ ⎝ ⎠ and ⎛ 2⎞ ⎛ 4⎞ r = ⎜⎜ 5⎟⎟ + µ ⎜⎜−3⎟⎟ ⎝−1⎠ ⎝ 2⎠ ⎛ 9⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ 2⎟ + r = λ 3 (vi) ⎜−4 ⎟ ⎜−3⎟ ⎝ ⎠ ⎝ ⎠ and ⎛ 1⎞ ⎛ 1⎞ r = ⎜⎜−4 ⎟⎟ + µ ⎜⎜−1⎟⎟ ⎝ 5⎠ ⎝ 2⎠ ⎛2⎞ ⎛ 1⎞ ⎝ 1⎠ ⎝− 2 ⎠ (vii) r = ⎜⎜ 3⎟⎟ + λ ⎜⎜ 1⎟⎟ 3 ⎛ 5⎞ ⎛−4 ⎞ r = ⎜⎜ 4 ⎟⎟ + µ ⎜⎜−2 ⎟⎟ ⎝ 6⎠ ⎝ 1⎠ ⎛ 0⎞ ⎛ 5⎞ r = ⎜⎜−1⎟⎟ + λ ⎜⎜ 3⎟⎟ ⎝ 4⎠ ⎝−3⎠ (v) PS and and ⎛ − 1⎞ ⎛ 1⎞ r = ⎜⎜− 3⎟⎟ + µ ⎜⎜ 3⎟⎟ ⎝ − 1⎠ ⎝2⎠ In this question the origin is taken to be at a harbour and the unit vectors i and j to have lengths of 1 km in the directions E and N. A cargo vessel leaves the harbour and its position vector t hours later is given by r1 = 12t i + 16t j. A fishing boat is trawling nearby and its position at time t is given by r2 = (10 − 3t)i + (8 + 4t)j. (i) How far apart are the two boats when the cargo vessel leaves harbour? (ii) How fast is each boat travelling? (iii) What happens to the boats? 270 9781510421738.indb 270 02/02/18 1:16 PM 4 The points A(1, 0), B(7, 2) and C(13, 7) are the vertices of a triangle. The midpoints of the sides BC, CA and AB are L, M and N. (i) Write down the position vectors of L, M and N. (ii) 10 Find the vector equations of the lines AL, BM and CN. (iii) Find the intersections of these pairs of lines. AL and BM (b) BM and CN (iv) What do you notice? 5 ⎛ −4 ⎞ ⎛ 2⎞ ⎛ 4⎞ ⎛ 2⎞ The line r = ⎜⎜ 4 ⎟⎟ + q ⎜⎜−10 ⎟⎟ meets r = ⎜⎜ −15⎟⎟ + s ⎜⎜−3⎟⎟ at A and meets ⎝−12 ⎠ ⎝ 11⎠ ⎝−16 ⎠ ⎝−5⎠ ⎛ −1⎞ ⎛ 1⎞ r = ⎜⎜−29 ⎟⎟ + t ⎜⎜ 1⎟⎟ at B. ⎝ −3⎠ ⎝8 ⎠ Find the coordinates of A and the length of AB. PS 6 7 10.6 The intersection of two lines (a) To support a tree damaged in a gale a tree surgeon attaches wire guys to four of the branches (see the diagram). He joins (2, 0, 3) to (−1, 2, 6) and (0, 3, 5) to (−2, −2, 4). Do the guys, assumed straight, meet? ⎛ −7⎞ ⎛ 4⎞ ⎛ 3⎞ ⎛ 2⎞ Show that the three lines r = ⎜ 24 ⎟ + q ⎜ −7⎟ , r = ⎜ −10⎟ + s ⎜ 2⎟ and ⎝ −4 ⎠ ⎝ 4⎠ ⎝ 15⎠ ⎝ −1⎠ ⎛ 8⎞ ⎛ −3⎞ r = ⎜ 6⎟ + t ⎜ −3⎟ form a triangle and find the lengths of its sides. ⎝ 2⎠ ⎝ 6⎠ 271 9781510421738.indb 271 02/02/18 1:16 PM 10.7 The angle between two lines 10 Earlier in this chapter, you learnt that the angle, θ, between two 10 VECTORS ⎛ a1 ⎞ ⎛ b1 ⎞ vectors a = ⎜ a 2 ⎟ and b = ⎜ b 2 ⎟ can be found ⎝ a3 ⎠ ⎝ b3 ⎠ using the formula a.b cos θ = |a||b| where a . b is the scalar product and a . b = a1b1 + a2b2 + a3b3. b a θ ▲ Figure 10.25 The angle between two lines is the same as the angle between their direction vectors. Example 10.16 Verify that the lines ⎛ −9 ⎞ ⎛ 2 ⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ 2 ⎟ + λ ⎜ − 2 ⎟ and r = ⎜ 2 ⎟ + µ ⎜ − 3 ⎟ are perpendicular. ⎜3⎟ ⎜3⎟ ⎜ 4 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Solution When two lines are perpendicular, the angle between their direction vectors is 90°. Since cos 90° = 0 then a . b = 0. So if the scalar product of two non-zero vector lines is zero then the lines are perpendicular. ⎛−9 ⎞ ⎛ 2⎞ The direction vectors of these two lines are ⎜⎜−2 ⎟⎟ and ⎜⎜−3⎟⎟ . ⎝ 4⎠ ⎝ 3⎠ ⎛ −9⎞ ⎛ 2⎞ ⎜ −2⎟ . ⎜ −3⎟ = ( −9) × 2 + ( −2) × ( −3) + 4 × 3 ⎝ 4 ⎠ ⎝ 3⎠ = ( −18) + 6 + 12 =0 Therefore, the two lines are perpendicular. 272 9781510421738.indb 272 02/02/18 1:16 PM Even if two lines do not meet, it is still possible to specify the angle between them. The lines l and m shown in Figure 10.26 do not meet; they are skew. 10 l θ m ▲ Figure 10.26 The angle between them is that between their directions; it is shown in Figure 10.26 as the angle θ between the lines l and m', where m' is a translation of the line m to a position where it does intersect the line l. Example 10.17 Find the angle between the skew lines ⎛1⎞ ⎛ 2⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜0 ⎟ + λ ⎜− 1 ⎟ ⎜− 1 ⎟ ⎜4 ⎟ ⎝ ⎠ ⎝ ⎠ and ⎛ 2 ⎜ r = ⎜ −1 ⎜ 3 ⎝ ⎛3⎞ ⎞ ⎜ ⎟ ⎟ ⎟⎟ + µ ⎜⎜ 0 ⎟⎟ . ⎠ ⎝1⎠ Solution The angle between the lines is the angle between their direction vectors 10.8 The perpendicular distance from a point to a line m' ⎛ 3⎞ ⎛ 2⎞ ⎜ –1⎟ and ⎜ 0⎟ . ⎜ ⎟ ⎜ ⎟ ⎜⎝ –1⎟⎠ ⎜⎝ 1 ⎟⎠ Using cos θ = cos θ = a.b |a||b| 2 × 3 + (–1) × 0 + (–1) × 1 2 2 + ( −1) 2 + ( −1) 2 × 3 2 + 0 2 + 12 5 × 6 10 θ = 49.8° cos θ = ⇒ 10.8 The perpendicular distance from a point to a line The scalar product is also useful when determining the distance between a point and a line. 273 9781510421738.indb 273 02/02/18 1:16 PM 10 Example 10.18 Find the shortest distance from point P(11, −5, −3) to the line l with equation ⎛1⎞ ⎛− 3 ⎞ r = ⎜5⎟ + λ⎜ 1 ⎟ . ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝ 4⎠ 10 VECTORS Solution ⎯→ The shortest distance from P to the line l is ⏐NP⏐ where N is a point on the line l and PN is perpendicular to the line l. P N l ▲ Figure 10.27 ⎯→ You need to find the coordinates of N and then you can find ⏐NP⏐. N lies on the line l. Let the value of λ at N be t. So, relative to the origin O ⎛ 1⎞ ⎛ − 3⎞ ⎛ 1 − 3t ⎞ ⎯→ ON = ⎜ 5⎟ + t ⎜ 1⎟ = ⎜ 5 + t ⎟ ⎝ 0⎠ ⎝ 4 ⎠ ⎝ 4t ⎠ and ⎯ → ⎯ → ⎯→ NP = OP − ON ⎛ 11⎞ ⎛ 1 − 3t ⎞ = ⎜ −5⎟ − ⎜ 5 + t ⎟ ⎝ −3⎠ ⎝ 4t ⎠ ⎛ 10 + 3t ⎞ = ⎜ −10 − t ⎟ ⎝ −3 − 4t ⎠ When two vectors are perpendicular, their scalar product is 0. ⎯ → As NP is perpendicular to the line l, ⎯ → ⎛ −3⎞ NP . ⎜ 1⎟ = 0 ⎝ 4⎠ ⎛ 10 + 3t ⎞ ⎯ → ⎛ −3⎞ NP . ⎜ 1⎟ = ⎜⎜ −10 − t ⎟⎟ ⎝ 4 ⎠ ⎝−3 − 4t ⎠ The direction of the line l • ⎛−3⎞ ⎜ 1⎟ ⎜ 4⎟ ⎝ ⎠ = (10 + 3t) × (−3) + (−10 − t) × 1 + (−3 − 4t) × 4 = −30 − 9t − 10 − t − 12 − 16t = −52 − 26t 274 9781510421738.indb 274 02/02/18 1:16 PM The scalar product is 0, so −52 − 26t = 0 ⎯→ ⇒ ⎯ → 10 t = −2 Substituting t = −2 into ON and NP gives ⎛ 10 + 3 × ( −2) ⎞ ⎛ 4 ⎞ ⎯ → NP = ⎜ −10 − ( −2) ⎟ = ⎜ −8⎟ ⎝ −3 − 4 × ( −2)⎠ ⎝ 5⎠ and ⎯→ ⏐NP⏐ = So 4 2 + ( −8) 2 + 5 2 = 105 = 10.25 units Exercise 10F ⎛ 1⎞ Remember i = ⎜ 0⎟ , j = ⎝ 0⎠ ⎛ 0⎞ ⎜ 1⎟ and k = ⎝ 0⎠ ⎛ 0⎞ ⎜ 0⎟ . ⎝ 1⎠ In questions 1 to 5, find the angle between each pair of lines. 1 ⎛ 2⎞ ⎛ 1⎞ r = ⎜ 1⎟ + s ⎜ 4 ⎟ ⎝ 3⎠ ⎝ 0⎠ and ⎛ 6⎞ ⎛ 2⎞ r = ⎜ 10⎟ + t ⎜ 1⎟ ⎝ 4⎠ ⎝ 1⎠ 2 ⎛ 4⎞ r = s ⎜ 1⎟ ⎝ 4⎠ and ⎛ 7 ⎞ ⎛ 1⎞ r = ⎜⎜ 0 ⎟⎟ + t ⎜⎜ 2 ⎟⎟ ⎝−3⎠ ⎝−1⎠ 3 ⎛ 4⎞ ⎛ 3⎞ r = ⎜ 2⎟ + s ⎜ 7⎟ ⎝ −1⎠ ⎝ −4 ⎠ and ⎛ 5⎞ ⎛ 2⎞ r = ⎜ 1⎟ + t ⎜ 8 ⎟ ⎝ 0⎠ ⎝ − 5⎠ 4 r = 2i + 3j + 4k + s(i + j − k) and r = t(i − k) 5 r = i − 2j − k + s(2i + 3j + 2k) and r = 2i + j + tk 6 For each point P and line l find (a) the coordinates of the point N on the line such that PN is perpendicular to the line (b) (i) 10.8 The perpendicular distance from a point to a line ⎛ 1⎞ ⎛ − 3⎞ ⎛ 7 ⎞ ⎯→ ON = ⎜ 5⎟ − 2 ⎜ 1⎟ = ⎜ 3⎟ ⎝ 0⎠ ⎝ 4 ⎠ ⎝ − 8⎠ the distance PN. P(–2, 11, 5) and ⎛ 0⎞ ⎛ −1⎞ r = ⎜ 2⎟ + t ⎜ 2⎟ ⎝ −3⎠ ⎝ 5⎠ 275 9781510421738.indb 275 02/02/18 1:16 PM 10 P(7, –1, 6) and ⎛ 2⎞ ⎛ 1⎞ r = ⎜ 1⎟ + t ⎜ − 2 ⎟ ⎝ 3⎠ ⎝ 4⎠ (iii) P(8, 4, –1) and ⎛ 1⎞ ⎛ −1⎞ r = ⎜ 5⎟ + t ⎜ −2⎟ ⎝ −3⎠ ⎝ 0⎠ (ii) 10 VECTORS 7 Find the perpendicular distance of the point P(–7, –2, 13) to the line ⎛ 1⎞ ⎛ 1⎞ r = ⎜⎜ 2 ⎟⎟ + λ ⎜⎜ 3⎟⎟ . ⎝ 5⎠ ⎝− 4 ⎠ M 8 Find the distance of the point C(0, 6, 0) to the line joining the points A(–4, 2, –3) and B(–2, 0, 1). 9 The diagram shows an extension to a house. Its base and walls are rectangular and the end of its roof, EPF, is sloping, as illustrated. Q (2, 5, 4) H G (4, 5, 3) (2, 1, 4) P E (0, 0, 3) (0, 0, 0) F C O B (4, 5, 0) A (i) Write down the coordinates of A and F. (ii) Find, using vector methods, the angles FPQ and EPF. The owner decorates the room with two streamers which are pulled taut. One goes from O to G, the other from A to H. She says that they touch each other and that they are perpendicular to each other. (iii) Is she right? 276 9781510421738.indb 276 02/02/18 1:16 PM 10 The points A and B have position vectors, relative to the origin O, given by ⎯→ OA = i + 2j + 3k and ⎯→ OB = 2i + j + 3k. The line l has vector equation 10 r = (1 − 2t)i + (5 + t)j + (2 − t)k. Show that l does not intersect the line passing through A and B. (ii) The point P lies on l and is such that angle PAB is equal to 60°. Given that the position vector of P is (1 − 2t)i + (5 + t)j + (2 − t)k, show that 3t 2 + 7t + 2 = 0. Hence find the only possible position vector of P. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q10 June 2008 11 With respect to the origin O, the position vectors of two points A and B ⎯→ ⎯→ are given by OA = i + 2 j + 2k and OB = 3i + 4 j. The point P lies on ⎯→ ⎯→ the line through A and B, and AP = λ AB . ⎯→ (i) Show that OP = (1 + 2λ )i + (2 + 2 λ ) j + (2 − 2λ )k. (ii) By equating expressions for cos AOP and cos BOP in terms of λ , find the value of λ for which OP bisects the angle AOB. (iii) When λ has this value, verify that AP : PB = OA : OB. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q7 November 2011 10.8 The perpendicular distance from a point to a line (i) 12 The equations of two straight lines are r = i + 4 j − 2k + λ( i + 3k ) and r = ai + 2 j − 2k + µ( i + 2 j + 3ak ), where a is a constant. (i) Show that the lines intersect for all values of a. (ii) Given that the point of intersection is at a distance of 9 units from the origin, find the possible values of a. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q7 November 2014 13 The straight line l 1 passes through the points (0, 1, 5) and (2, −2, 1). The straight line l 2 has equation r = 7i + j + k + µ( i + 2 j + 5k ) . (i) Show that the lines l 1 and l 2 are skew. (ii) Find the acute angle between the direction of the line l 2 and the direction of the x-axis. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q6 June 2015 277 9781510421738.indb 277 02/02/18 1:16 PM 10 KEY POINTS 1 2 10 VECTORS 3 4 5 6 A vector quantity has magnitude and direction. A scalar quantity has magnitude only. ⎯→ Vectors are typeset in bold, a or OA, or in the form⎯→ OA. They are handwritten either in the underlined form a, or as OA. The length (or modulus or magnitude) of the vector a is written as a or as |a|. Unit vectors in the x, y and z directions are denoted by i, j and k, respectively. A vector may be specified in ● magnitude−direction form: (r, θ ) (in two dimensions) ● () x component form: x i + y j or y (in two dimensions) ⎛ x⎞ x i + y j + z k or ⎜ y ⎟ (in three dimensions). ⎝ z⎠ 7 8 9 ⎯ → The position vector OP of a point P is the vector joining the origin to P. ⎯ → The vector AB is b − a, where a and b are the position vectors of A and B. The angle between two vectors, a and b, is given by θ in cos θ = a⋅b |a||b| where a . b = a1b1 + a2b2 (in two dimensions) = a1b1 + a2b2 + a3b3 (in three dimensions). ⎯ → 10 The position vector OP of a point P is the vector joining the origin to P. ⎯ → 11 The vector AB is b – a, where a and b are the position vectors of A and B. 12 The vector r often denotes the position vector of a general point. 13 The vector equation of the line through A with direction vector u is given by r = a + λu. 14 The vector equation of the line through points A and B is given by ⎯ → ⎯ → r = OA + λ AB = a + λ(b − a) = (1 − λ)a + λb. 278 9781510421738.indb 278 02/02/18 1:16 PM 15 The vector equation of the line through (a1, a2, a3) in the direction ⎛ u1 ⎞ ⎜ u 2 ⎟ is r = ⎜u ⎟ ⎝ 3⎠ ⎛a1 ⎞ ⎛u 1 ⎞ ⎜ a ⎟ + λ ⎜u ⎟ . ⎜ 2⎟ ⎜ 2⎟ ⎝a 3 ⎠ ⎝u 3 ⎠ Now you have finished this chapter, you should be able to ■ understand the terms vector and scalar ■ understand vectors in two and three dimensions, and express them ■ using i, j and k vectors ■ using column vectors ⎯→ ■ using OA or a notation ■ understand equal vectors, position vectors and displacement vectors ■ understand the link between the coordinates of a point and its position vector ■ multiply a vector by a scalar ■ add and subtract vectors ■ find a unit vector in the direction of a given vector ■ understand that vectors are parallel when one is a scalar multiple of the other ■ find the vector equation of a line ■ calculate the scalar product between two vectors and use it to find the angle between ■ two vectors ■ two lines ■ determine whether two lines are ■ parallel ■ intersect ■ skew ■ use vectors in geometry problems. 10.8 The perpendicular distance from a point to a line LEARNING OUTCOMES 10 279 9781510421738.indb 279 02/02/18 1:16 PM P3 11 COMPLEX NUMBERS 11 Complex numbers …that wonder of analysis, that portent of the ideal world, that amphibian between being and not-being, which we call the imaginary root of negative unity. Gottfried Wilhelm Leibniz (1646–1716) Real numbers Rational numbers Integers Natural numbers ▲ Figure 11.1 What is the meaning of each of the terms shown in Figure 11.1? ❯ Suggest two numbers that could be placed in each part of the diagram. ? ❯ 280 9781510421738.indb 280 02/02/18 1:16 PM 11.1 Extending the number system 11 The number system we use today has taken thousands of years to develop. To classify the different types of numbers used in mathematics the following letter symbols are used: Natural numbers & Integers ' Rational numbers ' Irrational numbers " Real numbers ? ❯ Why is there no set shown on the diagram for irrational numbers? You may have noticed that some of these sets of numbers fit within the other sets. This can be seen in Figure 11.1. 11.1 Extending the number system $ ACTIVITY 11.1 On a copy of Figure 11.1 write the following numbers in the correct positions. 7 5 –13 227 109 – 5 3.1415 π 0.33 . 0.3 What are complex numbers? ACTIVITY 11.2 Solve each of these equations and decide which set of numbers the roots belong to in each case. x+7=9 (ii) 7x = 9 (iii) x2 = 9 (iv) x + 10 = 9 (v) x2 + 7x = 0 (i) Now think about the equation x2 + 9 = 0. You could rewrite it as x2 = –9. However, since the square of every real number is positive or zero, there is no real number with a square of –9.This is an example of a quadratic equation which, up to now, you would have classified as having no real roots. 9781510421738.indb 281 Writing this quadratic equation as x2 + 0x + 9 = 0 and calculating the discriminant for this quadratic gives b2 – 4ac = –36 which is less than zero. 281 02/02/18 1:16 PM The existence of such equations was recognised for hundreds of years, in the same way that Greek mathematicians had accepted that x + 10 = 9 had no solution; the concept of a negative number had yet to be developed. The number system has expanded as mathematicians increased the range of mathematical problems they wanted to tackle. 11 11 COMPLEX NUMBERS You can solve the equation x2 + 9 = 0 by extending the number system to include a new number, i (sometimes written as j). This has the property that i2 = −1 and it follows the usual laws of algebra. i is called an imaginary number. The square root of any negative number can be expressed in terms of i. For example, the solution of the equation x2 = −9 is x = ± −9. This can be written as ± 9 × −1 which simplifies to ±3i. 11.2 Working with complex numbers Faced with the problem of wanting the square root of a negative number, we make the following Bold Hypothesis. The real number system can be extended by including a new number, denoted by i, which combines with itself and the real numbers according to the usual laws of algebra, but which has the additional property that i2 = −1. The original notation for i was ι, the Greek letter iota. The letter j is also commonly used instead of i. The first thing to note is that we do not need further symbols for other square roots. For example, since −196 = 196 × (−1) = 142 × i2, we see that −196 has two square roots, ±14i. The following example uses this idea to solve a quadratic equation with no real roots. Example 11.1 Solve the equation z2 − 6z + 58 = 0, and check the roots. We use the letter z for the variable here because we want to keep x and y to stand for real numbers. Solution Using the quadratic formula: z= = 6 ± 6 2 − 4 × 58 2 6 ± −196 2 6 ± 14i 2 = 3 ± 7i = 282 9781510421738.indb 282 02/02/18 1:16 PM To check: 11 z = 3 + 7i ⇒ z2 − 6z + 58 = (3 + 7i)2 − 6(3 + 7i) + 58 = 9 + 42i + 49i2 − 18 − 42i + 58 Notice that here 0 means 0 + 0i. = 9 + 42i − 49 − 18 − 42i + 58 =0 i2 = −1 Check the other root, z = 3 − 7i. Notice in particular that the imaginary part is real! A number z of the form x + iy, where x and y are real, is called a complex number. x is called the real part of the complex number, denoted by Re(z), and y is called the imaginary part, denoted by Im(z). So if, for example, z = 3 − 7i then Re(z) = 3 and Im(z) = −7. In Example 11.1 you did some simple calculations with complex numbers. The general methods for addition, subtraction and multiplication are similarly straightforward. » Addition: add the real parts and add the imaginary parts. 11.2 Working with complex numbers ACTIVITY 11.3 (x + iy) + (u + iv) = (x + u) + i(y + v) » Subtraction: subtract the real parts and subtract the imaginary parts. (x + iy) − (u + iv) = (x − u) + i(y − v) » Multiplication: multiply out the brackets in the usual way and simplify, remembering that i2 = −1. (x + iy)(u + iv) = xu + ixv + iyu + i2yv = (xu − yv) + i(xv + yu) Division of complex numbers is dealt with later in the chapter. What are the values of i3, i4, i5? ❯ Explain how you would work out the value of in for any positive integer value of n. ? ❯ Complex conjugates The complex number x − iy is called the complex conjugate, or just the conjugate, of x + iy. Simarly x + iy is the complex conjugate of x − iy. x + iy and x − iy are a conjugate pair. The complex conjugate of z is denoted by z*. If a polynomial equation, such as a quadratic, has real coefficients, then any complex roots will be conjugate pairs. This is the case in Example 11.1. If, however, the coefficients are not all real, this is no longer the case. 283 9781510421738.indb 283 02/02/18 1:16 PM You can solve quadratic equations with complex coefficients in the same way as an ordinary quadratic, either by completing the square or by using the quadratic formula. This is shown in the next example. 11 Example 11.2 Solve z2 − 4iz − 13 = 0. 11 COMPLEX NUMBERS Solution Substitute a = 1, b = −4i and c = −13 into the quadratic formula. 2 z = −b ± b − 4ac 2a 4i ± (−4i) 2 − 4 × 1 × (−13) = 2 = 4i ± −16 + 52 2 4i 36 ± = 2 = 4i ± 6 2 = 2i ± 3 So the roots are 3 + 2i and –3 + 2i. ACTIVITY 11.4 (i) Let z = 3 + 5i and w = 1 − 2i. Find the following. (a) z + z* (b) w + w* (c) zz* (d) ww* What do you notice about your answers? (ii) Let z = x + iy. Show that z + z* and zz* are real for any values of x and y. Equality of complex numbers Two complex numbers z = x + i y and w = u + i v are equal if both x = u and y = v. If y ≠ u or y ≠ v, or both, then z and w are not equal. You may feel that this is obvious, but it is interesting to compare this situation with the equality of rational numbers. 284 9781510421738.indb 284 02/02/18 1:16 PM The rational numbers x and u are equal if x = u and y = v. y v x u ❯ Is it possible for the rational numbers y and v to be equal if u ≠ x and v ≠ y? Example 11.3 The complex numbers z1 and z2 are given by : z1 = (3 − a) + (2b − 4)i and z2 = (7b − 4) + (3a − 2)i Given that z1 and z2 are equal, find the values of a and b. (ii) Check your answer by substituting your values for a and b into the expressions above. (i) Solution (i) Equating real and imaginary parts leads to two equations. (3 − a) + (2b − 4)i = (7 − 2b) + (3a − 2)i Equating real parts: 11 11.2 Working with complex numbers For two complex numbers to be equal, the real parts must be equal and the imaginary parts must be equal. Using this result is described as equating real and imaginary parts, as shown in the following example. ? 3 − a = 7b − 4 Equating imaginary parts: 2b − 4 = 3a − 2 Simplifying the equations. 7b + a = 7 2b − 3a = 2 Solving simultaneously gives b = 1 and a = 0. (ii) Substituting a = 0 and b = 1 gives z1 = 3 − 2i and z2 = 3 − 2i so z1 and z2 are indeed equal. Exercise 11A 1 Express the following in the form x + iy. (i) (8 + 6i) + (6 + 4i) (ii) (iii) (2 + 7i) − (5 + 3i) (iv) (v) 3(4 + 6i) + 9(1 − 2i) (vi) (vii) (9 + 2i)(1 + 3i) (viii) 2 (ix) (7 + 3i) (x) (xi) (1 + 2i)(3 − 4i)(5 + 6i) (xii) (9 − 3i) + (−4 + 5i) (5 − i) − (6 − 2i) 3i(7 − 4i) (4 − i)(3 + 2i) (8 + 6i)(8 − 6i) (3 + 2i)3 285 9781510421738.indb 285 02/02/18 1:16 PM 2 11 COMPLEX NUMBERS 11 3 4 CP 5 CP 6 CP 7 Solve each of the following equations, and check the roots in each case. (i) z2 + 2z + 2 = 0 (ii) z2 − 2z + 5 = 0 (iii) z2 − 4z + 13 = 0 (iv) z2 + 6z + 34 = 0 (v) 4z2 − 4z + 17 = 0 (vi) z2 + 4z + 6 = 0 Solve each of the following equations. (i) z2 − 4iz − 4 = 0 (ii) z2 − 2iz + 15 = 0 (iii) z2 − 2iz − 2 = 0 (iv) z2 + 6iz − 13 = 0 (v) z2 + 8iz − 17 = 0 (vi) z2 + iz + 6 = 0 Given that z = 2 + 3i and w = 6 − 4i, find the following. (i) Re(z) (ii) Im(w) (iii) z* (iv) w* (v) z* + w* (vi) z* − w* (vii) Im(z + z*) (viii) Re(w − w*) (ix) zz* − ww* (x) (z3)* (xi) (z*)3 (xii) zw* − z*w Let z = x + iy. Show that (z*)* = z. Let z1 = x1 + iy1 and z2 = x2 + iy2. Show that (z1 + z2)* = z1* + z2*. Given that the complex numbers z1 = a2 + (3 + 2b)i z2 = (5a − 4) +b2i are equal, find the possible values of a and b. Hence list the possible pairs of complex numbers z1 and z2. 286 9781510421738.indb 286 02/02/18 1:16 PM Division of complex numbers You can use the method of equating real and imaginary parts when you divide by a complex number. The following example illustrates this. Example 11.4 Find real numbers p and q such that p + iq = 1 . 3 + 5i You need to find real numbers p and q such that (p + iq)(3 + 5i) = 1. Expanding gives 3p − 5q + i(5p + 3q) = 1. Equating real and imaginary parts gives 3p − 5q = 1 Imaginary: 5p + 3q = 0 These simultaneous equations give p = 3 34 5 , q = − 34 and so 11.2 Working with complex numbers Solution Real: 11 1 3 5 = − i 3 + 5i 34 34 ACTIVITY 11.5 x − iy 1 1 = By writing x + iy = p + iq, show that . x + iy x 2 + y 2 This result shows that there is an easier way to find the reciprocal of a complex number. First, notice that (x + iy)(x − iy) = x 2 − i2y 2 = x2 + y2 which is real. So to find the reciprocal of a complex number you multiply numerator and denominator by the complex conjugate of the denominator. 287 9781510421738.indb 287 02/02/18 1:16 PM 1 Find the real and imaginary parts of 5 + 2i . Solution Multiply numerator and denominator by 5 − 2i. 5 − 2i 1 = 5 + 2i (5 + 2i)(5 – 2i) 5 − 2i = 25 + 4 11 COMPLEX NUMBERS 11 Example 11.5 = so the real part is 5 29 5 – 2i is the conjugate of the denominator, 5 + 2i. 5 − 2i 29 2 and the imaginary part is − 29 . Note You may have noticed that this process is very similar to the process of rationalising a denominator. To make the denominator of 1 rational 3+ 2 you have to multiply the numerator and denominator by 3 − 2. Similarly, division of complex numbers is carried out by multiplying both numerator and denominator by the conjugate of the denominator, as in the next example. Example 11.6 Express 9 − 4i as a complex number in the form x + iy. 2 + 3i Solution 9 − 4i = 9 − 4i × 2 − 3i 2 + 3i 2 + 3i 2 − 3i 2 = 18 − 27i2 − 8i2+ 12i 2 +3 − 6 35i = 13 6 35 = 13 − 13 i The square root of a complex number The next example shows you how to find the square root of a complex number. 288 9781510421738.indb 288 02/02/18 1:16 PM Example 11.7 Find the two square roots of 8 + 6i. 11 Solution Let (x + iy)2 = 8 + 6i ⇒ x 2 + 2ixy – y 2 = 8 + 6i +i2y2 = –y2 Real: x2 – y2 = 8 1 ! 2xy = 6 2 ! Imaginary: 11.2 Working with complex numbers Equating the real and imaginary parts gives: 2 gives Rearranging ! y=3 x 3 1 Substituting ! into ! gives x 2 − 92 = 8 x x 4 − 9 = 8x 2 3 ! x 4 − 8x 2 − 9 = 0 ( x 2 − 9)( x 2 + 1) = 0 ⇒ x 2 = −1 which has no real roots This is a quadratic in x2. Remember that x and y are both real numbers. or x 2 = 9 ⇒ x = ±3. When x = 3, y = 1 When x = –3, y = –1 So the square roots of 8 + 6i are 3 + i and –3 – i. ? 1 1 1 ❯ What are the values of , 2 and 3 ? i i i ❯ Explain how you would work out the value of 1 for any positive in integer value of n. Exercise 11B 1 Express these complex numbers in the form x + iy. 1 1 5i (i) (ii) (iii) 3+ i 6−i 6 − 2i (iv) 7 + 5i 6 − 2i (v) 3 + 2i 1+ i (vi) 47 − 23i 6+i (vii) 2 − 3i 3 + 2i (viii) 5 − 3i 4 + 3i (ix) 6+i 2 − 5i (x) 12 − 8i ( 2 + 2i) 2 289 9781510421738.indb 289 02/02/18 1:16 PM 2 11 Find real numbers a and b with a > 0 such that (a + ib)2 = 21 + 20i (ii) (a + ib)2 = −40 − 42i (iii) (a + ib)2 = −5 − 12i (iv) (a + ib)2 = −9 + 40i (vi) (a + ib)2 = i. (i) (v) 11 COMPLEX NUMBERS 3 4 5 6 CP 7 PS 8 (a + ib)2 = 1 − 1.875i Find real numbers a and b such that a + b 3 + i 1 + 2i = 1 − i. Solve these equations. (i) (1 + i)z = 3 + i (ii) (3 − 4i)(z − 1) = 10 − 5i (iii) (2 + i)(z − 7 + 3i) = 15 − 10i (iv) (3 + 5i)(z + 2 − 5i) = 6 + 3i Find all the complex numbers z for which z 2 = 2z*. For z = x + iy, find 1 + 1 in terms of x and y. z zz* Show that z + zz* (i) Re(z) = 2 z − zz*. (ii) Im(z) = 2i (i) Expand and simplify (a + ib)3. (ii) Deduce that if (a + ib)3 is real then either b = 0 or b 2 = 3a 2. (iii) Hence find all the complex numbers z for which z 3 = 1. PS 9 (i) Expand and simplify (z − α )(z − β ). Deduce that the quadratic equation with roots α and β is z 2 − (α + β )z + αβ = 0, that is: z 2 − (sum of roots)z + product of roots = 0. (ii) Using the result from part (i), find quadratic equations in the form az 2 + bz + c = 0 with the following roots. 7 + 4i, 7 − 4i (b) 5i , − 5i 3 3 (c) −2 + 8 i , −2 − 8 i (a) (d) 2 + i, 3 + 2i 10 Find the two square roots of each of these. (i) –9 (iii) –16 + 30i (v) 21 – 20i (ii) 3 + 4i (iv) –7 – 24i (vi) –5 – 12i 290 9781510421738.indb 290 02/02/18 1:16 PM Representing complex numbers geometrically Since each complex number x + iy can be defined by the ordered pair of real numbers (x, y), it is natural to represent x + iy by the point with cartesian coordinates (x, y). 4 2 + 3i 2 –4 2 + 3i is represented by (2, 3) −5 − 4i is represented by (−5, −4) 2i is represented by (0, 2) 7 is represented by (7, 0). –2 O 2i 2 4 6 8 Re –2 –5 – 4i –4 ▲ Figure 11.2 All real numbers are represented by points on the x-axis, which is therefore called the real axis. Pure imaginary numbers (of the form 0 + iy) give points on the y-axis, which is called the imaginary axis. It is useful to label these Re and Im respectively.This geometrical illustration of complex numbers is called the complex plane or the Argand diagram after Jean-Robert Argand (1768−1822), a self-taught Swiss book-keeper who published an account of it in 1806. 11.2 Working with complex numbers For example, in Figure 11.2, 11 Im ACTIVITY 11.6 Copy Figure 11.2. For each of the four given points z mark also the point −z. Describe the geometrical transformation which maps the point representing z to the point representing −z. (ii) For each of the points z mark the point z*, the complex conjugate of z. Describe the geometrical transformation which maps the point representing z to the point representing z*. (i) You will have seen in this activity that the points representing z and −z have half-turn symmetry about the origin, and that the points representing z and z* are reflections of each other in the real axis. ❯ How would you describe points that are reflections of each other in the imaginary axis? ? Representing the sum and difference of complex numbers Several mathematicians before Argand had used the complex plane representation. In particular, a Norwegian surveyor, Caspar Wessel (1745−1818), wrote a paper in 1797 (largely ignored until it was republished in French a century later) in which the complex number x + iy is represented by the ⎛x ⎞ position vector ⎜ y ⎟ , as shown in Figure 11.3. ⎝ ⎠ 9781510421738.indb 291 291 02/02/18 1:16 PM Im 11 z = x + iy 11 COMPLEX NUMBERS O Re ▲ Figure 11.3 The advantage of this is that the addition of complex numbers can then be shown by the addition of the corresponding vectors. ⎛ x 1 + x 2⎞ ⎜⎝ y + y ⎟⎠ 1 2 ⎛ x 1⎞ ⎛ x 2⎞ ⎜⎝ y ⎟⎠ + ⎜⎝ y ⎟⎠ = 1 2 In an Argand diagram the position vectors representing z1 and z2 form two sides of a parallelogram, the diagonal of which is the vector z1 + z2 (see Figure 11.4). Im z2 z1 + z2 z1 O Re ▲ Figure 11.4 You can also represent z by any other directed line segment with ⎛ x⎞ Im components ⎜ ⎟ , not anchored at the origin as a ⎝ y⎠ position vector. Then addition can be shown as a z1 + z2 triangle of vectors (see Figure 11.5). z2 z1 O Re ▲ Figure 11.5 If you draw the other diagonal of the parallelogram, and let it represent the complex number w (see Figure 11.6), then Im w = z 1 – z2 z2 z 2 + w = z 1 ⇒ w = z 1 − z 2. w z1 O Re ▲ Figure 11.6 292 9781510421738.indb 292 02/02/18 1:16 PM This gives a useful illustration of subtraction: the complex number z 1 − z 2 is represented by the vector from the point representing z 2 to the point representing z 1, as shown in Figure 11.7. Notice the order of the points: the vector z 1 − z 2 starts at the point z 2 and goes to the point z 1. 11 Im z2 z1 O Re ▲ Figure 11.7 ACTIVITY 11.7 Draw a diagram to illustrate z2 − z1. (ii) Draw a diagram to illustrate that z1 − z2 = z1 + (−z2). Show that z1 + (−z2) gives the same vector, z1 − z2 as before, but represented by a line segment in a different place. (i) 11.2 Working with complex numbers z1 – z2 The modulus of a complex number Figure 11.8 shows the point representing z = x + iy on an Argand diagram. Im x + iy y O x Re ▲ Figure 11.8 Using Pythagoras’ theorem, you can see that the distance of this point from the origin is x 2 + y 2 . This distance is called the modulus of z, and is denoted by | z |. So for the complex number z = x + iy, | z | = x2 + y2 . If z is real, z = x say, then | z | = x 2 , which is the absolute value of x, i.e. | x |. So the use of the modulus sign with complex numbers fits with its previous meaning for real numbers. 293 9781510421738.indb 293 02/02/18 1:16 PM 11 Exercise 11C 1 Represent each of the following complex numbers on a single Argand diagram, and find the modulus of each complex number. (i) 3 + 2i (iv) −2 11 COMPLEX NUMBERS 4i (iii) −5 + i (viii) −6 − 5i (vi) 4 − 3i Given that z = 2 − 4i, represent the following by points on a single Argand diagram. 2 (ii) −z (iii) z* (iv) −z* (v) iz (vi) −iz (vii) iz* (viii) (iz)* (i) PS (ii) z 3 Given that z = 10 + 5i and w = 1 + 2i, represent the following complex numbers on an Argand diagram. (i) z (ii) w (iii) z + w (iv) z − w (v) w−z 4 Given that z = 3 + 4i and w = 5 − 12i, find the following. |z| (ii) z (v) (iv) w What do you notice? (i) | | |w| w z | | (iii) | zw | CP 5 Let z = 1 + i. (i) Find zn and | zn | for n = −1, 0, 1, 2, 3, 4, 5. (ii) Plot each of the points zn from part (i) on a single Argand diagram. Join each point to its predecessor and to the origin. (iii) What do you notice? CP 6 Give a geometrical proof that (−z)* = −(z*). 11.3 Sets of points in an Argand diagram In the last section, you saw that | z | is the distance of the point representing z from the origin in the Argand diagram. ❯ ? What do you think that | z 2 − z 1 | represents? If z 1 = x 1 + iy1 and z 2 = x 2 + iy2 , then z 2 − z 1 = x 2 − x 1 + i(y2 − y1). So | z 2 − z 1 | = ( x 2 − x 1 )2 + ( y 2 − y1 )2 . 294 9781510421738.indb 294 02/02/18 1:16 PM Figure 11.9 shows an Argand diagram with the points representing the complex numbers z 1 = x 1 + iy1 and z 2 = x 2 + iy2 marked. Im 11 x2 + iy2 y2 – y1 x2 – x1 O Re ▲ Figure 11.9 Using Pythagoras’ theorem, you can see that the distance between z 1 and z 2 is given by ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . So | z 2 − z 1 | is the distance between the points z 1 and z 2. This is the key to solving many questions about sets of points in an Argand diagram, as in the following examples. Example 11.8 11.3 Sets of points in an Argand diagram x1 + iy1 Draw an Argand diagram showing the set of points z for which | z − 3 − 4i | = 5. Solution | z − 3 − 4i | can be written as | z − (3 + 4i) |, and this is the distance from the point 3 + 4i to the point z. This equals 5 if the point z lies on the circle with centre 3 + 4i and radius 5 (see Figure 11.10). Im 3 + 4i O Re ▲ Figure 11.10 295 9781510421738.indb 295 02/02/18 1:16 PM 11 COMPLEX NUMBERS 11 ❯ ? How would you show the sets of points for which (i) | z − 3 − 4i | $ 5 (ii) | z − 3 − 4i | " 5 (iii) | z − 3 − 4i | % 5? Example 11.9 Draw an Argand diagram showing the set of points z for which | z − 3 − 4i | $ | z + 1 − 2i |. Solution The condition can be written as | z − (3 + 4i) | $ | z − (−1 + 2i) |. | z − (3 + 4i) | is the distance of point z from the point 3 + 4i, point A in Figure 11.11, and | z − (−1 + 2i) | is the distance of point z from the point −1 + 2i, point B in Figure 11.11. Im A 3 + 4i B –1 + 2i O Re ▲ Figure 11.11 These distances are equal if z is on the perpendicular bisector of AB. So the given condition holds if z is on this bisector or in the half plane on the side of it containing A, shown shaded in Figure 11.11. ❯ How would you show the sets of points for which ? (i) | z − 3 − 4i | = | z + 1 − 2i | (ii) | z − 3 − 4i | " | z + 1 − 2i | (iii) | z − 3 − 4i | # | z + 1 − 2i |? 296 9781510421738.indb 296 02/02/18 1:16 PM Exercise 11D 1 For each of parts (i) to (viii), draw an Argand diagram showing the set of points z for which the given condition is true. |z |=2 (ii) |z − 4| $ 3 (iii) | z − 5i | = 6 (iv) | z + 3 − 4i | " 5 (vi) | z + 2 + 4i | = 0 (i) (v) |6− i − z| % 2 CP 2 Draw an Argand diagram showing the set of points z for which | z − 12 + 5i | $ 7. Use the diagram to prove that, for these z, 6 $ | z | $ 20. PS 3 (i) (ii) PS 4 PS 5 On an Argand diagram, show the region R for which | z − 5 + 4i | $ 3. Find the greatest and least values of | z + 3 − 2i | in the region R. By using an Argand diagram see if it is possible to find values of z for which | z − 2 + i | % 10 and | z + 4 + 2i | $ 2 simultaneously. For each of parts (i) to (iv), draw an Argand diagram showing the set of points z for which the given condition is true. (i) | z | = | z − 4 | (ii) | z | % | z − 2i | (iii) | z + 1 − i | = | z − 1 + i | (iv) | z + 5 + 7i | $ | z − 2 − 6i | 11.4 The modulus−argument form of complex numbers 11.4 The modulus−argument form of complex numbers (viii) Re(z) = −2 (vii) 2 $ | z − 1 + i | $ 3 11 The position of the point z in an Argand diagram can be described by means of the length of the line connecting this point to the origin, and the angle which this line makes with the positive real axis (see Figure 11.12). Im z When describing complex numbers, it is usual to give the angle θ in radians. r θ O Re ▲ Figure 11.12 The distance r is of course | z |, the modulus of z as defined on page 293. The angle θ is slightly more complicated: it is measured anticlockwise from the positive real axis, normally in radians. However, it is not uniquely 297 9781510421738.indb 297 02/02/18 1:16 PM 11 defined since adding any multiple of 2π to θ gives the same direction. To avoid confusion, it is usual to choose that value of θ for which −π " θ $ π. This is called the principal argument of z, denoted by arg z. Then every complex number except zero has a unique principal argument. The argument of zero is undefined. For example, with reference to Figure 11.13, 11 COMPLEX NUMBERS arg(−4) = π Im –3 + 3i π arg(−2i) = − 2 –4 arg(1.5) = 0 –2 0 1.5 Re –2i 3π arg(−3 + 3i) = 4 ▲ Figure 11.13 Remember that π radians = 180°. ❯ ? Without using your calculator, state the values of the following. (i) arg i (ii) arg(−4 − 4i) (iii) arg(2 − 2i) You can see from Figure 11.14 that x = r cos θ y = r sin θ y tan θ = x and the same relations hold in the other quadrants too. r= x2 + y2 Im r y θ O x Re ▲ Figure 11.14 Since x = r cos θ and y = r sin θ, we can write the complex number z = x + iy in the form z = r (cos θ + i sin θ ). This is called the modulus−argument or polar form. 298 9781510421738.indb 298 02/02/18 1:16 PM ACTIVITY 11.8 (i) Set your calculator to degrees and use it to find the following. (a) tan−1 1 (b) tan−1 2 (c) tan−1 100 (d) tan−1(−2) (e) tan−1(−50) (f) tan−1(−200) 11 What are the largest and smallest possible values, in degrees, of tan−1 x? Find tan−1 x for some different values of x. What are the largest and smallest possible values, in radians, of tan−1 x? If you know the modulus and argument of a complex number, it is easy to use the relations x = r cos θ and y = r sin θ to find the real and imaginary parts of the complex number. Similarly, if you know the real and imaginary parts, you can find the modulus and argument of the complex number using the relations r = x 2 + y 2 and y tan θ = , but you do have to be quite careful in finding the argument. It is x ⎛y⎞ tempting to say that θ = tan−1 ⎜ x ⎟ , but, as you saw in the last activity, this gives ⎝ ⎠ a value between − π and π , which is correct only if z is in the first or fourth 2 2 quadrants. For example, suppose that the point z1 = 2 − 3i has argument θ1, and the point z2 = −2 + 3i has argument θ2. It is true to say that tan θ1 = tan θ2 = − 23 . In the case of z1, which is in the fourth quadrant, θ1 is correctly given by 11.4 The modulus−argument form of complex numbers (ii) Now set your calculator to radians. (− 23) ≈ −0.98 rad (≈ −56°). However, in the case of z2, which is in the second quadrant, θ2 is given by (− 23 ) + π ≈ 2.16 rad (≈ 124°). These two tan−1 points are illustrated in Figure 11.15. Im z2 θ2 –2 0 θ1 2 Re –2 z1 ▲ Figure 11.15 Figure 11.16 shows the values of the argument in each quadrant. It is wise always to draw a sketch diagram when finding the argument of a complex number. 299 9781510421738.indb 299 02/02/18 1:16 PM 11 + π 2 arg + arg – 11 COMPLEX NUMBERS arg + π is positive y arg z = tan−1 x + π y arg z = tan−1 x arg z = tan−1 arg z = tan−1 arg – () y ( x) – π () y ( x) –π 2 ▲ Figure 11.16 ACTIVITY 11.9 Mark the points 1 + i, 1 − i, −1 + i, −1 − i on an Argand diagram. Find arg z for each of these, and check that your answers are consistent with Figure 11.16. Note The modulus−argument form of a complex number is also called the polar form, as the modulus of a complex number is its distance from the origin, sometimes called the pole. ACTIVITY 11.10 Most calculators can convert from (x, y) to (r, θ ) (called rectangular to polar, and often shown as R → P) and from (r, θ ) to (x, y) (polar to rectangular, P → R). Find out how to use these facilities on your calculator, and compare with other available types of calculator. Does your calculator always give the correct θ, or do you sometimes have to add or subtract π (or 180°)? A complex number in polar form must be given in the form z = r (cos θ + i sin θ ), not, for example, in the form z = r (cos θ − i sin θ ). The value of r must also be positive. So, for example, the complex number −2(cos θ + i sin θ ) is not in polar form. However, by using some of the relationships cos (π − α) = −cos α cos (α − π) = −cos α cos (−α) = cos α sin (π − α) = sin α sin (α − π) = −sin α sin (−α) = −sin α you can rewrite the complex number, for example −2(cos α + i sin α) = 2(−cos α − i sin α) = 2(cos (α − π) + i sin (α − π)). 300 9781510421738.indb 300 This is now written correctly in polar form. The modulus is 2 and the argument is α − π. 02/02/18 1:16 PM ❯ ? How would you rewrite the following in polar form? (i) −2(cos α − i sin α) (ii) 2(cos α − i sin α) 11 When you use the polar form of a complex number, remember to give the argument in radians, and to use a simple rational multiple of π where possible. 11.4 The modulus−argument form of complex numbers ACTIVITY 11.11 Copy and complete this table. Give your answers in terms of 2 or 3 where appropriate, rather than as decimals. You may find Figure 11.17 helpful. π π 4 π 6 3 2 tan 2 1 sin 1 cos 1 1 ▲ Figure 11.17 Example 11.10 Write the following complex numbers in polar form. (i) 4 + 3i (ii) −1 + i (iii) −1 − 3 i Solution (i) x = 4, y = 3 Modulus = 32 + 4 2 = 5 Since 4 + 3i lies in the first quadrant, the argument = tan−1 43 . 4 + 3i = 5(cos α + i sin α), where α = tan−1 (ii) 3 4 ≈ 0.644 radians x = −1, y = 1 Modulus = 12 + 12 = 2 Since −1 + i lies in the second quadrant, argument = tan−1(−1) + π π 3π = −4 +π = 4 −1 + i = 2 cos 3π + i sin 3π 4 4 ( ) ➜ 301 9781510421738.indb 301 02/02/18 1:16 PM 11 11 COMPLEX NUMBERS x = −1, y = − 3 (iii) Modulus = 1 + 3 = 2 Since −1 − 3i lies in the third quadrant, argument = tan−1 3 − π π 2π = 3−π =− 3 −1 − 3i = 2 cos − 2π + i sin − 2 π 3 3 ( ( ) Exercise 11E 1 Write down the values of the modulus and the principal argument of each of these complex numbers. cos 2.3 + i sin 2.3 (i) (ii) 8 cos π + i sin π 5 5 4 π π (iii) 4 cos − i sin (iv) −3(cos (−3) + i sin (−3)) 3 3 ( ( 2 ( )) ) ) For each complex number, find the modulus and principal argument, and hence write the complex number in polar form. Give the argument in radians, either as a simple rational multiple of π or correct to 3 decimal places. (i) 1 (iv) −4i (vii) 1 − 3i (ii) −2 (iii) 3i (v) 1+i (vi) (viii) 6 3 + 6i −5 − 5i (ix) 3 − 4i −12 + 5i (xi) 4 + 7i (xii) −58 − 93i Write each complex number with the given modulus and argument in the form x + iy, giving surds in your answer where appropriate. π π (i) | z | = 2, arg z = 2 (ii) | z | = 3, arg z = 3 5π π (iii) | z | = 7, arg z = 6 (iv) | z | = 1, arg z = − 4 2π (v) | z | = 5, arg z = − 3 (vi) | z | = 6, arg z = −2 Given that arg(5 + 2i) = α, find the principal argument of each of the following in terms of α. (i) −5 − 2i (ii) 5 − 2i (iii) −5 + 2i (iv) 2 + 5i (v) −2 + 5i (x) 3 4 302 9781510421738.indb 302 02/02/18 1:16 PM 5 The variable complex number z is given by z = 1 + cos 2θ + i sin 2θ, where θ takes all values in the interval − 21 π " θ " 21 π. (i) Show that the modulus of z is 2 cos θ and the argument of z is θ. 1 (ii) Prove that the real part of z is constant. 6 The variable complex number z is given by z = 2 cos θ + i(1 – 2 sin θ ), where θ takes all values in the interval −π " θ $ π. (i) (ii) Show that | z − i | = 2, for all values of θ. Hence sketch, in an Argand diagram, the locus of the point representing z. 1 Prove that the real part of z + 2 − i is constant for −π " θ " π. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q5 June 2008 7 4 − 3i . 1 − 2i (a) Express z in the form x + iy, where x and y are real. (b) Find the modulus and argument of z. Find the two square roots of the complex number 5 – 12i, giving your answers in the form x + iy, where x and y are real. The complex number z is given by z = (i) (ii) Cambridge International AS & A Level Mathematics 9709 Paper 3 Q8 November 2007 8 The complex number 2 + i is denoted by u. Its complex conjugate is denoted by u*. (i) Show, on a sketch of an Argand diagram with origin O, the points A, B and C representing the complex numbers u, u* and u + u* respectively. Describe in geometrical terms the relationship between the four points O, A, B and C. u (ii) Express ui in the form x + iy, where x and y are real. * u (iii) By considering the argument of ui , or otherwise, prove that * tan–1 43 = 2 tan–1 21 . () 9 11.4 The modulus−argument form of complex numbers Cambridge International AS & A Level Mathematics 9709 Paper 32 Q8 June 2010 11 () Cambridge International AS & A Level Mathematics 9709 Paper 3 Q7 June 2006 The complex number z is defined by z = a + ib, where a and b are real. The complex conjugate of z is denoted by z*. Show that | z|2 =zz* and that(z − ki)* = z* + ki , where k is real. In an Argand diagram a set of points representing complex numbers z is defined by the equation (i) | z − 10i| = 2| z − 4i|. 303 9781510421738.indb 303 02/02/18 1:16 PM 11 COMPLEX NUMBERS 11 (ii) Show, by squaring both sides, that zz* − 2iz* + 2iz − 12 = 0. Hence show that | z − 2i| = 4. (iii) Describe the set of points geometrically. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q7 June 2013 11.5 Sets of points using the polar form ? You already know that arg z gives the angle between the line connecting the point z with the origin and the real axis. ❯ What do you think arg (z2 − z1) represents? If z1 = x 1 + iy1 and z2 = x 2 + iy 2, then z 2 − z 1 = x 2 − x 1 + i(y2 − y1). y 2 − y1 arg (z 2 − z 1) = tan−1 x − x 2 1 Figure 11.18 shows an Argand diagram with the points representing the complex numbers z1 = x 1 + iy1 and z2 = x 2 + iy 2 marked. Im x2 + iy2 y2 – y1 x1 + iy1 x2 – x1 O Re ▲ Figure 11.18 The angle between the line joining z1 and z2 and a line parallel to the real axis is given by y 2 − y1 tan−1 x − x . 2 1 So arg (z1 − z2) is the angle between the line joining z1 and z2 and a line parallel to the real axis. 304 9781510421738.indb 304 02/02/18 1:16 PM Example 11.11 Draw Argand diagrams showing the sets of points z for which π (i) arg z = 4 π (ii) arg (z − i) = 4 π (iii) 0 $ arg (z − i) $ 4 . (i) π arg z = 4 π ⇔ the line joining the origin to the point z has direction 4 π ⇔ z lies on the half-line from the origin in the 4 direction, see Figure 11.19. Im O Re 11.5 Sets of points using the polar form Solution 11 ▲ Figure 11.19 (ii) (Note that the origin is not included, since arg 0 is undefined.) π arg (z − i) = 4 π ⇔ the line joining the point i to the point z has direction 4 π ⇔ z lies on the half-line from the point i in the 4 direction, see Figure 11.20. Im i O Re ▲ Figure 11.20 π (iii) 0 $ arg (z − i) $ 4 ⇔ the line joining the point i to the point z has direction between 0 π and 4 (inclusive) ⇔ z lies in the one-eighth plane shown in Figure 11.21. Im i O ▲ Figure 11.21 9781510421738.indb 305 Re 305 02/02/18 1:16 PM 11 COMPLEX NUMBERS 11 Exercise 11F 1 2 3 For each of parts (i) to (vi) draw an Argand diagram showing the set of points z for which the given condition is true. π (i) arg z = − 3 (ii) arg (z − 4i) = 0 π 3π (iii) arg (z + 3) % 2 (iv) arg (z + 1 + 2i) = 4 π π π (v) arg (z − 3 + i) $ − (vi) − $ arg (z + 5 − 3i) $ 3 6 4 Find the least and greatest possible values of arg z if | z − 8i | $ 4. 3 The complex number w is given by w = − 21 + i . 2 (i) Find the modulus and argument of w. (ii) The complex number z has modulus R and argument θ, where − 13 π " θ " − 13 π. State the modulus and argument of wz and the z modulus and argument of w . (iii) Hence explain why, in an Argand diagram, the points representing z z, wz and w are the vertices of an equilateral triangle. (iv) In an Argand diagram, the vertices of an equilateral triangle lie on a circle with centre at the origin. One of the vertices represents the complex number 4 + 2i. Find the complex numbers represented by the other two vertices. Give your answers in the form x + iy, where x and y are real and exact. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q10 November 2008 4 The complex number w is defined by w = 22 + 4i2 . (2 − i) (i) Without using a calculator, show that w = 2 + 4i. (ii) It is given that p is a real number such that 41 π & arg(w + p ) & 43 π. Find the set of possible values of p. (iii) The complex conjugate of w is denoted by w*. The complex numbers w and w* are represented in an Argand diagram by the points S and T respectively. Find, in the form | z − a | = k, the equation of the circle passing through S,T and the origin. Cambridge International AS & A Level Mathematics 9709 Paper 31 Q8 June 2015 5 Solve the equation z2 + (2√3)iz − 4 = 0, giving your answers in the form x + iy, where x and y are real. (ii) Sketch an Argand diagram showing the points representing the roots. (iii) Find the modulus and argument of each root. (iv) Show that the origin and the points representing the roots are the vertices of an equilateral triangle. (i) Cambridge International AS & A Level Mathematics 9709 Paper 3 Q7 June 2009 306 9781510421738.indb 306 02/02/18 1:16 PM 6 Cambridge International AS & A Level Mathematics 9709 Paper 32 Q7 November 2009 7 2 The complex number −1 + i is denoted by u. (i) Find the modulus and argument of u and u2. (ii) Sketch an Argand diagram showing the points representing the complex numbers u and u2. Shade the region whose points represent the complex numbers z which satisfy both the inequalities | z | " 2 and | z − u2 | " | z − u |. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q8 June 2007 8 The complex number u is defined by u = 6 − 3i . 1 + 2i (i) Showing all your working, find the modulus of u and show that the argument of u is − 21 π. 11 11.6 Working with complex numbers in polar form The complex numbers −2 + i and 3 + i are denoted by u and v respectively. (i) Find, in the form x + iy, the complex numbers (a) u + v, u (b) v , showing all your working. u (ii) State the argument of v . In an Argand diagram with origin O, the points A, B and C represent the complex numbers u, v and u + v respectively. 3 (iii) Prove that angle AOB = 4 π. (iv) State fully the geometrical relationship between the line segments OA and BC. For complex numbers z satisfying arg ( z − u ) = 41 π, find the least possible value of | z |. (iii) For complex numbers z satisfying | z − (1 + i) u | = 1, find the greatest possible value of | z |. (ii) Cambridge International AS & A Level Mathematics 9709 Paper 31 Q8 June 2011 11.6 Working with complex numbers in polar form The polar form quickly leads to an elegant geometrical interpretation of the multiplication of complex numbers. For if z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2) then z1z2 = r1r2(cos θ1 + i sin θ1)(cos θ2 + i sin θ2) = r1r2[cos θ1cos θ2 − sin θ1 sin θ2 + i (sin θ1 cos θ2 + cos θ1 sin θ2)]. Using the compound-angle formulae gives z1z2 = r1r2[cos (θ1 + θ2) + i sin (θ1 + θ2)]. 9781510421738.indb 307 307 02/02/18 1:16 PM 11 This is the complex number with modulus r1r2 and argument (θ1 + θ2), so we have the beautiful result that | z1z2 | = | z1 | | z2 | and 11 COMPLEX NUMBERS arg (z1z2) = arg z1 + arg z2 (±2π if necessary, to give the principal argument). So to multiply complex numbers in polar form you multiply their moduli and add their arguments. ACTIVITY 11.12 Using this interpretation, investigate (i) multiplication by i (ii) multiplication by –1. z The corresponding results for division are easily obtained by letting 1 = w. z2 Then z1 = wz2 so that | z1 | = | w | | z2 | and arg z1 = arg w + arg z2 (±2π if necessary). and | | z1 | z1 | = –––– z2 | z2 | z arg w = arg 1 z2 = arg z1 – arg z2 (±2π if necessary, to give the principal argument). Therefore | w | = So to divide complex numbers in polar form you divide their moduli and subtract their arguments. This gives the following simple geometrical interpretation of multiplication and division. To obtain the vector z1z2 enlarge the vector z1 by the scale factor | z2 | and rotate it through arg z2 anticlockwise about O (see Figure 11.22 (ii)). 1 z To obtain the vector z 1 enlarge the vector z1 by scale factor –––– and rotate | z 2 2| it clockwise through arg z2 about O (see Figure 11.22 (iii)). This combination of an enlargement followed by a rotation is called a spiral dilatation. In summary: and | z1z2 | = | z1 | | z2 | arg (z1z2) = arg z1 + arg z2 | z1 | | z1 | –––– = –––– | z2 | | z2 | ⎛z ⎞ arg ⎜ 1 ⎟ = arg z1 − arg z2 ⎝z2 ⎠ 308 9781510421738.indb 308 02/02/18 1:16 PM Im 9 arg (z1z2) = arg z1 + arg z2 8 7 Im 6 5 5 r1r2 4 z1 3 3 r1 2 2 r2 1 z2 1 2 3 4 Re (i) z1 and z2 O 3 arg z2 1 arg z2 4 z1 r1 2 1 arg z1 1 2 3 4 Re O (ii) Multiplying z1 by z2 z1 r1 arg z2 z1 r1 z2 1 r2 2 3 4 Re (iii) Dividing z1 by z2 ▲ Figure 11.22 ACTIVITY 11.13 Check this by accurate drawing and measurement for the case z1 = 2 + i, z2 = 3 + 4i. Then do the same with z1 and z2 interchanged. Example 11.12 Find (i) (ii) ( ) ( ) 6 (cos π + i sin π ) ÷ 2 (cos π + i sin π ) 2 2 4 4 6 cos π + i sin π × 2 cos π + i sin π 2 2 4 4 11.6 Working with complex numbers in polar form 4 11 Im 6 6 5 O ⎛z ⎞ arg ⎜ z 1 ⎟ = arg z1 − arg z2 ⎝ 2⎠ z1z2 Solution (i) Remember that r1(cos θ1 + i sin θ1) × r2(cos θ2 + i sin θ2) = r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2)) So to multiply complex numbers you 6 × 2 = 12; π + π = 3π 2 4 4 » multiply the moduli » add the arguments. ( ) ( ) ( 6 cos π + i sin π × 2 cos π + i sin π = 12 cos 3π + i sin 3π 2 2 4 4 4 4 (ii) To divide complex numbers you » divide the moduli » subtract the arguments. ( ) ( ) 6 ÷ 2 = 3; π − π = π 2 4 4 ) ( 6 cos π + i sin π ÷ 2 cos π + i sin π = 3 cos π + i sin π 2 2 4 4 4 4 ) 309 9781510421738.indb 309 02/02/18 1:16 PM 11 This leads to an alternative method of finding the square root of a complex number. Im Writing 8 + 6i in polar form gives r 11 COMPLEX NUMBERS r = 8 2 + 6 2 = 10 tan θ = 6 ⇒ θ = 0.6435 radians 8 So 8 + 6i = 10(cos 0.6435 + i sin 0.6435). 6 0.6435... 8 O Re Notice that if you add 2π to the argument you will ▲ Figure 11.23 end up with exactly the same complex number on the Argand diagram (as you have just rotated through one full turn). So 8 + 6i is also the same as 10[cos(0.6435 + 2π) + i sin(0.6435 + 2π)]. Let r(cos θ + i sin θ ) be the square root of 8 + 6i so that r (cos θ + i sin θ ) × r (cos θ + i sin θ ) = r 2 (cos 2θ + i sin 2θ ) = 8 + 6i ⇒ r(cos 2θ + i sin 2θ ) = 10(cos 0.6435 + i sin 0.6435) and 2 r(cos 2θ + i sin 2θ ) = 10[cos(0.6435 + 2π) + i sin(0.6435 + 2π)] ! 1 or ! 2 From ! r 2 = 10 ⇒ r = 10 1 From ! 2θ = 0.6435 ⇒ θ = 0.321 75 ⇒ θ = 0.321 75 + π 2 From ! 2θ = 0.6435 + 2π So one square root of 8 + 6i is 10 (cos 0.321 75 + i sin 0.321 75) = 3 + i The square root of the modulus of 8 + 6i. Half of the argument of 8 + 6i, plus π. and the other square root is 10 [cos(0.321 75 + π) + i sin(0.321 75 + π)] = −3 − i Compare this method with that used on page 288 to find the square roots of 8 + 6i. ❯ 1 ! Half of the argument of 8 + 6i plus π. What are the square roots of 10[cos(0.6435 + 2nπ) + i sin(0.6435 + 2nπ)], where n is an integer? ? 11.7 Complex exponents When multiplying complex numbers in polar form you add the arguments, and when multiplying powers of the same base you add the exponents. This suggests that there may be a link between the familiar expression cos θ + i sin θ and the seemingly remote territory of the exponential function. This was first noticed in 1714 by the young Englishman Roger Cotes two years before his death at the age of 28 (when Newton remarked ‘If Cotes had lived we might have known something’), and made widely known through an influential book published by Euler in 1748. 310 9781510421738.indb 310 02/02/18 1:16 PM Let z = cos θ + i sin θ. Since i behaves like any other constant in algebraic manipulation, to differentiate z with respect to θ you simply differentiate the real and imaginary parts separately. This gives dz dθ = –sin θ + i cos θ = i2 sin θ + i cos θ 11 = i(cos θ + i sin θ ) So z = cos θ + i sin θ is a solution of the differential equation dz = iz. dθ If i continues to behave like any other constant when it is used as an index, then the general solution of dz = iz is z = eiθ+c, where c is a constant, just as dθ dx x = ekt+c is the general solution of dt = kx. cos θ + i sin θ = eiθ+c. Therefore 11.7 Complex exponents = iz Putting θ = 0 gives cos θ + i sin θ = e0+c ⇒ 1 = ec ⇒c=0 and it follows that cos θ + i sin θ = eiθ. The problem with this is that you have no way of knowing how i behaves as an index. But this does not matter. Since no meaning has yet been given to ez when z is complex, the following definition can be made, suggested by this work with differential equations but not dependent on it: eiθ = cos θ + i sin θ. Note The particular case when θ = π gives eiπ = cos π + i sin π = –1, so that eiπ + 1 = 0. This remarkable statement, linking the five fundamental numbers 0, 1, i, e and π, the three fundamental operations of addition, multiplication and exponentiation, and the fundamental relation of equality, has been described as a ‘mathematical poem’. Example 11.13 Find (i) (ii) (a) 4e 5i × 3e 2i (b) 6e 9i ÷ 3e 2i Write these results as complex numbers in polar form. ➜ 311 9781510421738.indb 311 02/02/18 1:16 PM Solution 11 (i) 11 COMPLEX NUMBERS (ii) Exercise 11G 1 (a) 4e 5i × 3e 2i = 12e 7i 4 × 3 = 12; 5i + 2i = 7i (b) 6e 9i ÷ 3e 2i = 2e 7i 6 ÷ 3 = 2; 9i − 2i = 7i (a) 4(cos 5 + i sin 5) × 3(cos 2 + i sin 2) = 12(cos 7 + i sin 7) (b) 6(cos 9 + i sin 9) ÷ 3(cos 2 + i sin 2) = 2(cos 7 + i sin 7) Find the following. 8(cos 0.2 + i sin 0.2) × 4(cos 0.4 + i sin 0.4) (i) (ii) 8(cos 0.2 + i sin 0.2) ÷ 4(cos 0.4 + i sin 0.4) π π × 2 cos π + i sin π (iii) 6 cos + i sin 3 3 6 6 π π π ÷ 2 cos + i sin π (iv) 6 cos + i sin 6 6 3 3 π π (v) 12(cos π + i sin π) × 2 cos + i sin 4 4 π π (vi) 12(cos π + i sin π) ÷ 2 cos + i sin 4 4 Given that z = 2 cos π + i sin π and w = 3 cos π + i sin π , find the 3 3 4 4 following in polar form. w (i) wz (ii) (iii) z z w 1 (iv) (v) w 2 (vi) z 5 z (vii) w 3z 4 (viii) 5iz (ix) (1 + i)w ( ( 2 ( ) ( ) ( ( ( ) ) ) ) ) ( ) 1 Prove that, in general, arg z = –arg z, and deal with the exceptions. CP 3 CP 4 Given the points 1 and z on an Argand diagram, explain how to find the following points by geometrical construction. (i) 3z (ii) 2iz (iii) (3 + 2i)z (iv) z* (v) | z | (vi) z2 CP 5 Find the real and imaginary parts of Express –1 + i and 1 + 6 −1 + i . 1+ 3 i 3 i in polar form. Hence show that cos 5π = 3 − 1, and find an exact expression for 2 2 12 5π sin . 12 Express ez in the form x + iy where z is the given complex number. (i) –iπ (ii) iπ 4 (iii) 2 + 5iπ 6 (iv) 3 – 4i 312 9781510421738.indb 312 02/02/18 1:16 PM 7 Find the following. (i) (a) 2e 3i × 5e −2i (b) 8e 5i ÷ 2e 5i (c) 3e 7i × 2e i (d) 12e 5i ÷ 4e 4i (e) 3e 2i × e i (f) 8e 3i ÷ 2e 4i (ii) Write these results as complex numbers in polar form. 8 (i) It is given that (1 + 3i)w = 2 + 4i . Showing all necessary working, (ii) On a single Argand diagram sketch the loci | z | = 5 and | z − 5 | = | z |. Hence determine the complex numbers represented by points common to both loci, giving each answer in the form reiθ. Cambridge International AS & A Level Mathematics 9709 Paper 33 Q9 November 2015 9 The complex numbers w and z are defined by w = 5 + 3i and z = 4 + i. (i) (ii) Express iw in the form x + iy, showing all your working and giving z the exact values of x and y. Find wz and hence, by considering arguments, show that tan −1 ( 53 ) + tan −1 ( 41 ) = 41 π. 11.8 Complex numbers and equations prove that the exact value of w 2 is 2 and find arg(w2) correct to 3 significant figures. 11 Cambridge International AS & A Level Mathematics 9709 Paper 33 Q5 November 2014 11.8 Complex numbers and equations The reason for inventing complex numbers was to provide solutions for quadratic equations which have no real roots, i.e. to solve az2 + bz + c = 0 when the discriminant b2 − 4ac is negative. This is straightforward since if b2 − 4ac = −k2 (where k is real) then the formula for solving quadratic equations gives z = −b ± ik . These are the two complex roots of the 2a equation. Notice that when the coefficients of the quadratic equation are real, these roots are a pair of conjugate complex numbers. It would be natural to think that to solve cubic equations would require a further extension of the number system to give some sort of ‘super-complex’ numbers, with ever more extensions to deal with higher degree equations. But luckily things are much simpler. It turns out that all polynomial equations (even those with complex coefficients) can be solved by means of complex numbers. This was realised as early as 1629 by Albert Girard, who stated that an nth degree polynomial equation has precisely n roots, including complex roots and taking into account repeated roots. (For example, the fifth degree equation (z − 2)(z − 4)2(z 2 + 9) = 0 has five roots: 2, 4 (twice), 3i and −3i.) Many great mathematicians tried to prove this. The chief difficulty is to show that every polynomial equation must have at least one root: this is called the Fundamental Theorem of Algebra and was first proved by Carl Friedrich Gauss in 1799. 313 9781510421738.indb 313 02/02/18 1:16 PM The Fundamental Theorem, which is too difficult to prove here, is an example of an existence theorem: it tells us that a solution exists, but does not say what it is. To find the solution of a particular equation you may be able to use an exact method, such as the formula for the roots of a quadratic equation. (There are much more complicated formulae for solving cubic or quartic equations, but not in general for equations of degree five or more.) Alternatively, there are good approximate methods for finding roots to any required accuracy, and your calculator probably has this facility. 11 COMPLEX NUMBERS 11 ACTIVITY 11.14 Find out how to use your calculator to solve polynomial equations. You have already noted that the complex roots of a quadratic equation with real coefficients occur as a conjugate pair. The same is true of the complex roots of any polynomial equation with real coefficients. This is very useful in solving polynomial equations with complex roots, as shown in the following examples. Example 11.14 Given that 1 + 2i is a root of 4z 3 − 11z 2 + 26z − 15 = 0, find the other roots. Solution Since the coefficients are real, the conjugate 1 − 2i is also a root. Therefore [z − (1 + 2i)] and [z − (1 − 2i)] are both factors of 4z3 − 11z2 + 26z − 15 = 0. This means that (z − 1 − 2i)(z − 1 + 2i) is a factor of 4z3 − 11z2 + 26z − 15 = 0. (z − 1 − 2i)(z − 1 + 2i) = [(z − 1) − 2i][(z − 1) + 2i] = (z − 1)2 + 4 = z 2 − 2z + 5 By looking at the coefficient of z3 and the constant term, you can see that the remaining factor is 4z − 3. 4z 3 − 11z 2 + 26z − 15 = (z 2 − 2z + 5)(4z − 3) The third root is therefore Example 11.15 3 4 . Given that −2 + i is a root of the equation z 4 + az 3 + bz 2 + 10z + 25 = 0, find the values of a and b, and solve the equation. Solution z = −2 + i z2 = (−2 + i)2 = 4 − 4i + (i)2 = 4 − 4i − 1 = 3 − 4i 314 9781510421738.indb 314 02/02/18 1:16 PM −2a + 3b − 2 = 0 11a − 4b − 14 = 0 Solving these equations simultaneously gives a = 2, b = 2. The equation is z4 + 2z3 + 2z2 + 10z + 25 = 0. Since −2 + i is one root, −2 − i is another root. So (z + 2 − i)(z + 2 + i) = (z + 2)2 + 1 = z2 + 4z + 5 is a factor. Using polynomial division or by inspection z 4 + 2z 3 + 2z 2 + 10z + 25 = (z 2 + 4z + 5)(z 2 − 2z + 5). The other two roots are the solutions of the quadratic equation z2 − 2z + 5 = 0. Using the quadratic formula 11 11.8 Complex numbers and equations z3 = (−2 + i)z2 = (−2 + i)(3 − 4i) = −6 + 11i + 4 = −2 + 11i z4 = (−2 + i)z3 = (−2 + i)(−2 + 11i) = 4 − 24i − 11 = −7 − 24i Now substitute these into the equation. −7 − 24i + a(−2 + 11i) + b(3 − 4i) + 10(−2 + i) + 25 = 0 (−7 − 2a + 3b − 20 + 25) + (−24 + 11a − 4b + 10)i = 0 Equating real and imaginary parts gives z= 2± 4−4×5 2 = 2 ± −16 2 = 2 ± 4i 2 = 1 ± 2i The roots of the equation are −2 ± i and 1 ± 2i. Exercise 11H 1 4 − 5i is one root of a quadratic equation with real coefficients. Write down the second root of the equation and hence find the equation. 2 Check that 2 + i is a root of z3 − z2 − 7z + 15 = 0, and find the other roots. One root of z3 − 15z2 + 76z − 140 = 0 is an integer. Solve the equation. Given that 1 − i is a root of z3 + pz2 + qz + 12 = 0, find the real numbers p and q, and the other roots. One root of z4 − 10z3 + 42z2 − 82z + 65 = 0 is 3 + 2i. Solve the equation. The equation z4 − 8z3 + 20z2 − 72z + 99 = 0 has a pure imaginary root. Solve the equation. One root of z3 – 15z2 + 76z – 140 = 0 is an integer. Solve the equation and show all three roots on an Argand diagram. 3 4 5 6 7 9781510421738.indb 315 315 02/02/18 1:16 PM 8 The equation z3 − 2z2 − 6z + 27 = 0 has a real integer root in the 11 COMPLEX NUMBERS 11 PS range −6 $ z $ 0. (i) Find the real root of the equation. (ii) Hence, solve the equation and find the exact value of all three roots. 9 Given that 4 is a root of the equation z3 − z2 − 3z − k = 0, find the value of k and hence find the exact value of the other two roots of the equation. 10 The three roots of a cubic equation are shown on the Argand diagram below. Im 2 z2 ö2 1 z1 0 −1 1 2 Re −1 z3 −ö2 −2 CP Find the equation in polynomial form. 11 For each of these statements about polynomial equations with real coefficients, say whether the statement is TRUE or FALSE, and give an explanation. (i) A cubic equation can have three complex roots. (ii) Some equations of order 6 have no real roots. (iii) A cubic equation can have a single root repeated three times. (iv) A quartic equation can have a repeated complex root. 12 Given that z = −2 + i is a root of the equation z4 + az3 + bz2 + 10z + 25 = 0, find the values of a and b, and solve the equation. 13 The equation 2x 3 + x 2 + 25 = 0 has one real root and two complex roots. (i) Verify that 1 + 2i is one of the complex roots. (ii) Write down the other complex root of the equation. (iii) Sketch an Argand diagram showing the point representing the complex number 1 + 2i. Show on the same diagram the set of points representing the complex numbers z which satisfy | z | = | z − 1 − 2i |. Cambridge International AS & A Level Mathematics 9709 Paper 3 Q7 November 2005 316 9781510421738.indb 316 02/02/18 1:16 PM KEY POINTS 1 2 (x1 + iy1) ± (x2 + iy2) = (x1 ± x2) + i(y1 ± y2) 4 To multiply complex numbers, multiply out the brackets and simplify. (x1 + iy1)(x2 + iy2) = (x1x2 − y1y2) + i(x1y2 + x2y1) 5 To divide complex numbers, multiply top and bottom by the conjugate of the bottom. x 1 + iy 1 ( x 1x 2 + y 1y 2 ) + i(x 2 y 1 –x 1y 2 ) = x 2 + iy 2 x 22 + y 22 6 The complex number z can be represented geometrically as the point (x, y). This is known as an Argand diagram. 7 The modulus of z = x + iy is | z | = x 2 + y 2 . This is the distance of the point z from the origin. 8 The distance between the points z1 and z2 in an Argand diagram is | z2 − z2 |. 9 The principal argument of z, arg z, is the angle θ, −π < θ < π, between the line connecting the origin and the point z and the positive real axis. 10 The modulus−argument or polar form of z is z = r(cos θ + i sin θ ), where r = | z | and θ = arg z. 11 x = r cos θ y = r sin θ y 2 + y2 x r= tan θ = x 12 To multiply complex numbers in polar form, multiply the moduli and add the arguments. 11 11.8 Complex numbers and equations 3 Complex numbers can be written in the form z = x + iy with i2 = −1. x is called the real part, Re(z), and y is called the imaginary part, Im(z). The conjugate of z is z* = x − iy. To add or subtract complex numbers, add or subtract the real and imaginary parts separately. z1z2 = r1r2[cos(θ 1 + θ 2) + i sin(θ 1 + θ 2)] 13 To divide complex numbers in polar form, divide the moduli and subtract the arguments. z 1 r1 [cos(θ 1 − θ 2) + i sin(θ 1 − θ 2)] z 2 = r2 14 eiθ = cos θ + i sin θ, e–iθ = cos θ – i sin θ 15 A polynomial equation of degree n has n roots, taking into account complex roots and repeated roots. In the case of polynomial equations with real coefficients, complex roots always occur in conjugate pairs. 317 9781510421738.indb 317 02/02/18 1:16 PM 11 LEARNING OUTCOMES Now that you have finished this chapter, you should be able to ■ 11 COMPLEX NUMBERS ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ understand how complex numbers extend the number system solve quadratic equations with complex roots know what is meant by the terms real part, imaginary part and complex conjugate know that two complex numbers are equal when both their real and imaginary parts are equal add, subtract, multiply and divide complex numbers solve problems involving complex numbers by equating real and imaginary parts find the two square roots of a complex number represent a complex number on an Argand diagram represent addition and subtraction of two complex numbers on an Argand diagram find the modulus of a complex number find the principal argument of a complex number using radians express a complex number in modulus–argument form multiply and divide complex numbers in modulus–argument form represent multiplication and division of two complex numbers on an Argand diagram represent and interpret sets of complex numbers as loci on an Argand diagram ■ circles of the form | z – a | = r ■ half-lines of the form arg (z – a) = θ ■ lines of the form | z – a | = | z – b | represent and interpret regions defined by inequalities based on the above multiply and divide complex numbers expressed in polar form understand that complex roots of polynomial equations with real coefficients occur in conjugate pairs solve cubic and quartic equations with complex roots. 318 9781510421738.indb 318 02/02/18 1:16 PM Answers Answers The questions, with the exception of those from past question papers, and all example answers that appear in this book were written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication. Non-exact numerical answers should be given correct to three significant figures (or one decimal place for angles in degrees) unless a different level of accuracy is specified in the question.You should avoid rounding figures until reaching your final answer. Chapter 1 ? (Page 1) 13 + 123 = 93 + 103 = 1729 Ramanujan must have been referring to positive cubes only, as if you allow negative cubes then 6 3 + (−5) 3 = 4 3 + 3 3 = 91. 22 x3 23 2x3 + 24 2x2 + 2x + 3 25 x2 + 3x + 1 26 x2 +4 (i) 3 2 2x 3 3 x 4 + 4x 3 + 6x 2 + 4x + 1 4 x 3 + 2x 2 + 5x + 7 5 −x2 + 15x + 18 6 2x 4 + 8 7 x 4 + 4x 3 + 6x 2 + 4x + 1 8 x 4 − 5x 2 + 4 9 x 4 − 10x 2 + 9 (ii) 12 (iii) 7 −4 + 2x2 +x 3x2 + x+4 (i) 30, 0, (x − 3) (ii) p = 2, q = −15 (iii) −5, 2 or 3 (iv) y 30 –5 27 x 2 − 4x − 2 x 2 3 28 x2 + 2x − 2 Exercise 1A (Page 6) 1 3 21 x 3 + 2x 2 + 5 ? 4 (Page 13) Its order will be less than n. (ii) −2, 3 or 4 y (iii) 24 Exercise 1B (Page 14) 1 –2 0, 0, −8, −18, −24, −20, 0 (ii) (x + 3)(x + 2)(x − 3) (iii) −3, −2 or 3 (iv) y 3 4 x (i) 5 (i) 0 (ii) –1 ± 2 (iii) y –3 10 x11 − 1 –2 O x 3 11 2x − 2 –2 2 x 12 10x 2 13 4 14 2x 2 − 2x 15 −8x 3 − 8x 16 x2 − 2x − 3 17 x2 + 3x 2 (i) −15, 0, 3, 0, −3, 0, 15 (ii) x(x + 2)(x − 2) (iii) −2, 0 or 2 y (iv) 6 y –2 O 2 −4 (x − 1)2 (iii) 18 2x2 − 5x + 5 19 x 3 + x 2 + 2x + 2 (i) (ii) 1 4 x x 20 2x 2 + 3 –4 9781510421738.indb 319 319 02/02/18 1:16 PM 7 Answers 8 (i) a = 2, b = 1, c = 2 (ii) 0, 3 or – 3 (i) (x2 (ii) (x + 2)(x − 2)(x2 + 1) (iii) Two real roots: −2 and 2 − 4)(x2 23 10 (i) ±6, ±3, ±2, ±1 (ii) −1, 2 or 3 (i) (x − 1)(x − 2) (x + 2) (ii) (x + − x + 1) (iv) (x + 2)(x2 − x + 3) (a) ? x3 − x2 + 2x + 2 remainder −6 (b) x3 − 3x2 + 6x − 6 remainder −2 (c) x2 1 (i) x = −9 or x = 1 3 2 (ix) x = −1 or x = 2 (i) 3 2 (–2, 2) (iv) y 1 O x (v) y −8 " x " 2 (ii) 0 $ x $ 4 1 (iii) x " −1 or x # 11 (iv) x $ −3 or x % 1 a = −13 (vi) − 23 $ x $ 2 a = −4, b = 1 3 (i) |x−1|"2 (ii) x − 13 (iii) | x − 1 | " 3 (i) 6x 2 − x − 2 remainder 7 (iv) | x − 2.5 | " 3.5 1, 2 2 3 –4 12 – 12 O x (v) −2 " x " 5 (ii) | x − 5 | " 3 9781510421738.indb 320 x O (viii) x = − 83 or x = 2 a = 11, b = 5 320 –4 (vii) x = −2 or x = 4 a=8 (ii) −2, 2, − y (vi) x = 1 or x = 7 (ii) (x − 3) and (x + 1) 21 (i) x 1.5 (v) x = −3 or x = 2 (ii) (x + 2)(2x + 1)(x − 3) 22 (iii) 3 18 a = 2, b = −3 (i) y O (iv) x = − 2 or x = 2 (ii) 3x 2 + 2x + 4 20 x O 3 (iii) x = −1 or x = 7 − 2x + 4 remainder −2x − 4 16 a = −1; b = −7; 1, −2 or (i) (ii) (Page 19) (ii) x = 7 or x = −1 15 a = −5; b = 4; 4 19 –2 Exercise 1C (Page 21) 14 2 or −5 (i) y 2 | x | " 2 and x % 0 ⇒0$x"2 | x | " 2 and x " 0 ⇒ −2 " x " 0 13 −12 17 (i) (Page 17) g(3) = 3, g(−3) = 3 | 3 + 3 | = 6, | 3 − 3 | = 0, | 3 | + | 3 | = 6, | 3 | + | −3 | = 6 ? (iii) (x − 2)(x2 + 2x + 5) 12 (ii) 4 − 1) (ii) 2x2 + 9x + 11 remainder 19 1)(x2 a = 5, b = −12 (ii) 7 9 11 (i) (–2 12 , –4) (vi) y 5 (v) | x − 10 | " 0.1 (vi) | x − 4 | " 3.5 (2, 3) O x 02/02/18 1:16 PM 5 (i) x " Exercise 2A (page 27) (ii) x 1 (iii) x 1 2 " 72 % − 21 (i) y (v) 6 7 x " −1 or x # 3 (vi) x $ −6 or x % − 43 1 1 2 O "x " 1 x > 2a (ii) x = 1, x = 9 x = 1, x = 3 1 O (Page 24) 32 (ii) 1024 (iii) Activity 2.1 (Page 25) 2 (a) A(2, 4), B(4, 16) (b) 0 < x < 2 and x > 4 (iii) (a) Yes (b) No (ii) If the graphs are translated horizontally by k units, the horizontal asymptote will be y = 0 and the graphs will both go through (k, 1). (iii) If the graphs are stretched horizontally, the horizontal asymptote will be y = 0 and the graphs will both go through (0, 1). (iv) If the graphs are stretched vertically by a scale factor k, the horizontal asymptote will be y = 0 and the graphs will both go through (0, k). x y (Page 26) If the graphs are translated vertically by k units, the horizontal asymptote will be y = k and the graphs will go through the point (0, k). y (i) (i) (i) x O −1 About 20 days ? y 2 Chapter 2 (i) x −9 5 8 ? Answers (iv) −1 $ x $ 3 1 O (ii) x y 3 2 O (iii) x y O x −1 321 9781510421738.indb 321 02/02/18 1:17 PM 3 Initial population = 10 000; population after 5 years = 31 623 (i) (ii) ? (Page 33) » a0 = 1, loga 1 = 0 » am = x # 0 (for a # 0) so loga (x) = m is defined P Answers only for x # 0 » Putting x = 1y in log ⎛⎜⎝ 1y ⎞⎟⎠ = − log y ⇒ log x = − log ⎛⎜⎝ 1y ⎞⎟⎠ : as x → 0, − log ⎛⎜⎝ x1 ⎞⎟⎠ → − ∞ 10 000 t O 4 (i) (ii) (iii) (iv) 1m 4.61 m Just over 6 years 20 m 5 (i) (ii) 520 210 (iii) » There is no limit to m in am = x and loga x = m; think, for example, of base 2, i.e. a = 2. Then x = 2y. When y = 1, 2, 3, 4, … then x = 2, 4, 8, 16, … . So increases in y are accompanied by ever larger increases in x and so a decreasing gradient. This is the case not just for a = 2 but for any value of a greater than 1. » loga a = 1 n Exercise 2B (Page 34) 520 1 (i) (ii) (iii) (iv) (v) (vi) x = log3 9 = 2 x = log4 64 = 3 x = log2 41 = −2 x = log5 51 = −1 x = log7 1 = 0 x = log16 2 = 41 2 (i) (ii) (iii) (v) 3y = 9 ⇔ y = 2 5y = 125 ⇔ y = 3 2y = 16 ⇔ y = 4 64y = 8 ⇔ y = 21 200 O t (iv) 200 birds 6 (i) 1976 Hz (ii) A, three octaves below the middle (iii) 48 (from E two octaves below the middle, up to C five octaves above the middle) Activity 2.2 (Page 29) (vi) 5y = 3 (i) 10 (ii) 21 (iii) 10 8.6/85 = 1.26 Activity 2.3 (Page 32) y 4 3 2 11 2 0 –1 –2 –3 (16, 4) (1, 0) (4, 2) 4 (2, 1) 2 4 6 8 10 12 14 16 x ( 12 , –1) ( 14 , –2) ( 18 , –3) 1 2 = 2 2 ≈ 1.4 ⇔ y = −2 −4 0 (i) 4 (iii) 21 (ii) (iv) (v) 4 (vii) 23 (vi) −4 (viii) 41 1 2 (x) −3 (i) log 10 (iii) log 36 (ii) (iv) log 2 log 1 (v) log 3 (vi) log 4 (vii) log 4 (viii) log 1 (ix) log 1 (x) (ix) (8, 3) 1 25 2 7 3 log 12 322 9781510421738.indb 322 02/02/18 1:17 PM 5 (i) 2 log x (iii) 1 2 (ii) 3 log x log x (iv) 11 log x 6 5 log x 2 (v) 6 log x 9 (i) y O x 1 10 (ii) y –1 O log10 x , x = 21 7 2 (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) x = 19.93 x = −9.97 x = 9.01 x = 48.3 x = 1375 x = 0.863 x = 1.38 x = −1.41 x = 1.78 (i) (ii) (iii) (iv) (v) (vi) x = 1.58 x=0 x = 1.38 x = 0.387 or 6.13 x = −1 x = −0.356 or 0.564 x 12 13 y O 7 (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) 2 3 x<7 x%3 x%3 x > 0.437 x&1 x % 0.322 0.431 & x < 1.29 0 & x < 0.827 1 < x < 2.58 0.68 < x < 1.49 x (ii) 3 y = 23 , y = −0.369 Exercise 2C (Page 41) Some of the questions in this exercise involve drawing a line of best fit by eye. Consequently your answers may reasonably vary a little from those given. 1 7 11 k = 9 (iii) 20 (i) ( x + 2)(4 x + 3)(3x − 2) 14 (i) (ii) (i) 6.6 (ii) 1.58 × 1012 joules (iii) 794 times more energy (i) x = 1.259 (3 d.p.) (iii) 3 decibels (iv) It should be 15 (i) (ii) 16 x = 0.631, x = −0.369 4<y<6 1.26 < x < 1.63 17 x = 0.802 x= If the relationship is of the form R = kT n, the graph of log R against log T will be a straight line. (ii) Values of log R: 5.46, 5.58, 5.72, 6.09, 6.55 Values of log T: 0.28, 0.43, 0.65, 1.20, 1.90 log R 6 25 17 10 3.7 − 10 3.5 × 100 = 58.5% 10 3.5 18 (i) Answers 6 (vi) 8 1 11 19 (i) a = 19,b = −36 (ii) y = 0.113 5 0 2.0 log T 1.0 (iii) k = 1.8 × 105, n = 0.690 (iv) 0.7 days 2 (ii) Plotting log A against t will test the model: if it is a straight line the model fits the data. (iii) b = 1.4, k = 0.89 log A 0.8 0.6 0.4 0.2 0 –0.2 1 2 3 4 5 t (iv) (a) 2.4 days (b) 3.0 cm2 (v) Exponential growth 323 9781510421738.indb 323 02/02/18 1:17 PM 3 Answers 4 k = 3.2 × 106, a = 0.98 The constant k is the original number of trees. (ii) (iii) The model predicts growth in local user numbers without limit, which is impossible. However, the growth becomes so slow that this may well not be a problem. (ii) k = 1100, n = 1.6 (iii) 2500 m (iv) The train would not continue to accelerate like this throughout its journey. After 10 minutes it would probably be travelling at constant speed, or possibly even slowing down. 5 (ii) b = 1.37, k = 1.58 6 log y = B log x + log A. He should plot log y against log x. If this gives a straight line, there is a relationship of the form y = AxB. If there is no such relationship, the points will not be in a straight line. The value of log A is given by the intercept on the log y axis. The value of B is the gradient of the line. log y 1.0 0.8 0.6 0.4 0.2 (ii) = x−1. This means that n = −1 and so n + 1 = 0.You cannot divide by zero. ∫ ∫ ∫ () ∫ e = 2.72 (2 d.p.) (ii) 0.693 Exercise 2D (Page 52) (iii) 1.792 2 1 6 1 1 ∫ 1 x dx + ∫ 1 x dx = ∫ 1 x dx 1 x = x0ekt 2 0 t = 1 ln ⎜⎝ s ⎟⎠ 3 Activity 2.5 (Page 46) 4 y y= ∫ Activity 2.7 (Page 47) 1.099 (i) ∫ ∫ Activity 2.4 (Page 45) 3 1 L(a) − L(b) = a1 dx − b 1 dx = a 1 dx 1x 1x bx Let x = bz a1 –a 1 dx = b z dz x 1 b =La b an 1 (iii) L(an) = dx 1 x Let x = zn then dx = nzn−1 dz. a1 a dx = 1n × nz n−1 dz 1x 1z a = n 1 dz 1z = n L(a) (ii) ∫ 1 x (i) 1 ∫ 1 x dx = 0 ∫ (Page 45) O 5 1 x 1 a ⎛s ⎞ k p = 25e−0.02t ⎛y–5⎞ x = ln ⎜ ⎟ ⎝ y0 – 5 ⎠ (i) x = 0.0540 (ii) x = 0.0339 (iv) x = 0.693 ab x = az ⇒ dx = adz Converting the limits: x=a⇒z=1 x = ab ⇒ z = b ab 1 dx = b 1 ×a dz 1 az a x b 1 = dz 1 z b 1 dz = b x1 dx = L(b) 1 z 1 ab 1 (iii) L(a) + x dx = L(ab) a ⇒ L(a) + L(b) = L(ab) x (v) (ii) log x y = 400 × 0.63t (iii) y = 0.4. The infection is under control. 8 (ii) 3.42 L(1) = (iii) x = 0.238 From the graph, A = 1.5, B = 0.5. The formula is therefore y = 1.5x0.5. 7 ? (i) (i) L(a) ab 1 a ∫ x dx 0. 2 0. 4 0. 6 0. 8 1. 0 1. 2 1. 4 0 9 log 4 log 3 Activity 2.6 (Page 46) (i) P = 4000 × t 0.2 (ii) 4000 ∫ ∫ ∫ ∫ ∫ ∫ x = 1.386 (vi) x = 1.099 6 (i) P 100 O (ii) t 100 (iii) 1218 (iv) 184 years 324 9781510421738.indb 324 02/02/18 1:17 PM 7 1m (ii) 4.61 m, 6.09 years (iii) a = e−2 = 0.135, b = 2.5 (iv) 11 years y = 25 x –1 (i) 0 $ α $ 90° (ii) No, for each of the second, third and fourth quadrants a different function is positive. ? (iii) No, the graphs of the three functions do not intersect at a single point. 9 4.11 10 x = 0.481 11 A = 3.67, b = 1.28 12 x = −1.68 13 K = 7.39, m = 1.37 14 p = 0.44, A = 3.2 7 (i) (ii) (iii) (iv) x = 0°, 180°, 360° x = 45°, 225° x = 60°, 300° x = 54.7, 125.3°, 234.7°, 305.3° Chapter 3 (v) x = 18.4°, 71.6°, 198.4°, 251.6° (Page 57) (vi) x = 45°, 135°, 225°, 315° ? Answers will vary. For example, a ≈ 1.6 to 1.8, b ≈ 3 ⇒ y = 1.5 sin 3x. A sine function is a reasonable model for a wave. Exercise 3A (Page 60) 1 2 (i) x = 90° (ii) (iii) (iv) (v) (vi) x = 60°, 300° x = 14.0°, 194.0° x = 109.5°, 250.5° x = 135°, 315° x = 210°, 330° (i) (iii) 3 (iv) (i) B = 60°, C = 30° (i) (ii) (vi) − 2 0 (ii) 5 (ii) –2 3 (v) (ii) 4 −1 –2 3 –2 3 3 L = 45°, N = 45° 2, 2 , 1 14.0° Area of a triangle = 21 × base × height. The definitions of sine and cosine in a right-angled triangle. Activity 3.2 (Page 62) (i) (ii) Activity 3.1 (Page 61) 1 0 y = sin( + 60 °) A y = cos( – 60 °) 180 ° sin(θ + φ) = sin θ cos φ + cos θ sin φ ⇒ sin[(90° − θ ) + φ] = sin(90° − θ )cos φ + cos(90° − θ )sin φ ⇒ sin[90° − (θ − φ)] = cos θ cos φ + sin θ sin φ ⇒ cos(θ − φ) = cos θ cos φ + sin θ sin φ ⇒ cos[θ − (−φ)] = cos θ cos(−φ) + sin θ sin(−φ) ⇒ cos(θ + φ) = cos θ cos φ − sin θ sin φ sin (θ + φ ) (iii) tan(θ + φ) = y = sin(θ + 60°) is obtained from ⎛ –60° ⎞ y = sin θ by a translation ⎜ ⎟. ⎝ 0 ⎠ y = cos(θ − 60°) is obtained from ⎛60° ⎞ y = sin θ by a translation ⎜ ⎟ . ⎝ 0⎠ y (Page 62) = Answers 8 6 (i) cos (θ + φ ) sin θ cos φ + cosθ sin φ cosθ cos φ – sin θ sin φ sin θ cos φ cosθ sin φ + cosθ cos φ cosθ cosφ = cosθ cosφ sin θ sin φ – cosθ cosφ cosθ cosφ = tan θ + tan φ 1 – tan θ tan φ (iv) tan[θ + (−φ)] = tan θ + tan ( –φ ) 1 – tan θ tan ( –φ ) ⇒ tan(θ − φ) 360 ° tan θ – tan φ = 1 + tan θ tan φ It appears that the θ coordinate of A is midway between the two maxima (30°, 1) and (60°, 1). Checking: θ = 45° ⇒ sin(θ + 60°) = 0.966 cos(θ − 60°) = 0.966. If 60° is replaced by 35°, using the trace function on a graphical calculator would enable the solutions to be found. ? (Page 63) No. In part (iii) you get tan 90° = 3+ 1 3 1– 3× 1 3 Neither tan 90° nor 1 is defined. 1–1 For the result to be valid you must exclude the case when θ + φ = 90° (or 270°, 450°, ...). Similarly in part (iv) you must exclude θ − φ = 90°, 270°, etc. 325 9781510421738.indb 325 02/02/18 1:17 PM Exercise 3B (Page 64) 1 (i) Answers (ii) (iii) (iv) 2 (i) 3 + 1 2 2 2 2 1 – 2 3–1 3 +1 3 +1 3–1 1 2 (sin θ + cos θ ) (ii) 21 ( 3 cos θ + sin θ ) (iii) 21( 3 cos θ − sin θ ) (iv) (v) (vi) 3 1 (cos 2θ 2 − sin 2θ) tan θ + 1 1 – tan θ tan θ – 1 1 + tan θ (i) sin θ (ii) cos 8φ (iii) 0 (iv) cos 2θ 4 tan α = (ii) cot β = − 20 9 8 (ii) x = 22.5°, 112.5° 9 α = 26.6° and β = 45° or α = 135° and β = 116.6° 10 (ii) 11 3 11 ? For sin 2θ and cos 2θ, substituting θ = 45° is helpful. You know that sin 45° = cos 45° = and that sin 90° = 1 and cos 90° = 0. For tan 2θ you cannot use θ = 45°. Take θ = 30° instead; tan 30° = and tan 60° = 3. θ = 157.5° 1 (i) (ii) 6 (i) θ= −1 8 9 (iii) π , 5π , 3π 6 6 2 (ii) θ = 63.4° (i) y 2 y = cos 2x π O 2π x –2 y = 3sinx – 1 –4 (iii) θ = 90°, 210°, 330° 11 40.2°, 139.8° θ = 30°, 150°, 210°, 330° 12 (ii) θ = 26.6°, 206.6° 13 (i) 1 10 (4 (ii) 44 tan 2a = − 24 7 ,tan 3a = − 117 (v) sin β = cos β = 2 θ = 14.5°, 90°, 165.5°, 270° 2 x 3π 2 10 (iv) cos α = 3 5 2 2 (iv) Negative sign means angle (α +β ) is greater than 90° but less than 180°. (ii) 1 3 π 2 π 5π (iii) x = 6 , 6 (ii) cos x = 41 (3 − 17) (ii) (iv) θ = 111.7° y = sin x y = cos 2x O −0.5 No, checking like this is not the same as proof. (ii) π 8 θ = 2.79 radians (i) 1 2 Exercise 3C (Page 70) (i) 6 0.5 (Page 67) (iii) θ = 0° or 180° 5 cot θ y 1 θ = 15° θ = 165° 5 3 θ = 24.7°, 95.3° (i) (v) 4 θ = –11π , –3π , –7π , –π , 12 4 12 4 π , π ,5π , 3π 12 4 12 4 3 sin θ − 4 sin3 θ, π 3π 5π 7π θ = 0, 4 , 4 , π, 4 , 4 , 2π θ = 51°, 309° (v) 2 3 (i) 7 θ = 0°, 35.3°, 144.7°, 180°, 215.3°, 324.7°, 360° θ = 0°, 138.6°, 221.4°, 360° (i) θ = –π, 0, π (ii) θ = –π, 0, π 2π (iii) θ = –2π , 0, 3 3 –π –3π (iv) θ = , , π , 3π 4 4 4 4 3 − 3) Exercise 3D (Page 75) 1 (i) 2 cos(θ − 45°) (ii) 29 cos(θ − 46.4°) (iii) 2 cos(θ − 60°) (iv) 3 cos(θ − 41.8°) 326 9781510421738.indb 326 02/02/18 1:17 PM 2 (i) (ii) 3 5 (iii) 3 sin(θ + 48.2°) (i) (ii) π ) 4 3 sin(θ − 0.49 rad) (i) 2 cos(θ − (−60°)) (ii) 4 cos(θ − (−45°)) (iv) θ = 53.8°, 159.9°, 233.8°, 339.9° 2 sin(θ − 9 (i) 3 cos(θ − 54.7°) (ii) Max 3 , θ = 54.7°; min − 3 , θ = 234.7° (iii) (ii) Max 13, min −13 (iii) 360° θ – 3 (vi) 2 cos(θ − 135°) 13 cos(θ + 67.7°) 180° O 2 cos(θ − 150°) (i) y 3 (iv) 13 cos(θ − 22.6°) 6 360° θ – 13 (iii) 2 cos(θ − 30°) (v) 180° O 5 sin(θ + 63.4°) (ii) y 13 Answers 4 (i) π 2 cos(θ + ) 4 π 2 cos(θ + 6 ) (iv) Max 3 –1 , θ = 234.7°; min y 3 1 ,θ 3+ 3 = 54.7° 10 (ii) θ = 30.6°, 82.0° 11 (i) 5 cos(x − 53.1°) (ii) x = 27.3°, 79.0° 13 5 112.6° O 360° θ 12 −13 (iv) θ = 4.7°, 220.5° 7 (i) (ii) (iii) π 2 3 sin(θ − 6 ) Max 2 3, θ = 2π ; 3 min −2 3 , θ = 5π 3 13 14 y –2 3 π (iv) θ = 3 , π 8 (i) 13 sin(2θ + 56.3°) (ii) Max 13, θ = 16.8°; min − 13, θ = 106.8° 26 cos(θ + 11.3°) (ii) θ = 27.02°, 310.4° (i) 25 cos(θ − 73.7°) (ii) θ = 20.6°, 126.9° θ = 81.3°, 172.4° Investigation (Page 78) 2 3 O – 3 (i) The total current is 2π θ I = A1 sinω t + A2 sin(ω t + α)(where ω = 2πf ). I = A1 sin ω t + A2 sin ω t cos α + A2 cos ω t sin α = (A1 + A2 cos α)sin ω t + (A2 sin α)cos ω t Let A1 + A2 cos α = P and A2 sin α = Q so I = P sin ω t + Q cos ω t = P 2 + Q 2 sin(ω t + ε) () Q . P This is a sine wave with the same frequency but a greater amplitude.The phase angle ε is between 0 and α. where ε = tan–1 327 9781510421738.indb 327 02/02/18 1:17 PM Exercise 3E (Page 78) 1 (i) sin 6θ (ii) cos 6θ Answers (iii) 1 (iv) cos θ 2 4 (v) sin θ (vi) 23 sin 2θ (vii) cos θ Exercise 4A (Page 87) (viii) −1 1 (i) x(5x 3 − 3x + 6) x4(21x2 + 24x − 35) y 1 O 5 (i) 1 − sin 2x (ii) (ii) cos 2x (iii) 1 2(5 (i) θ = 4.4°, 95.6° (iii) 2x(6x + 1)(2x + 1)3 2 (iv) − (3x – 1) 2 x 2 ( x 2 + 3) (v) ( x 2 + 1) 2 (vi) 2(2x + 1)(12x2 + 3x − 8) cos 2x − 1) θ = 199.5°, 340.5° –π π (iii) θ = 6 , 2 (iv) θ = −15.9°, 164.1° (v) θ = π, π, 5π 2 6 6 (ii) (vii) θ = 76.0°, 135° 2(1 + 6x – 2x 2 ) (2x 2 + 1) 2 7–x (viii) ( x + 3) 3 3x – 1 (ix) 2 x –1 1 (i) − ( x – 1) 2 2 (ii) Chapter 4 Many possible answers: e.g. amplitude of waves in physics, rate of radioactive decay, rate of population growth and so on. (ii) (0, 4), maximum; (2, 0), minimum 4 du dv = 10x9 and = 7x6 dx dx Using the quotient rule, gives –1 O 4 (i) (ii) 2 1 4)2 4y + x = 12 − Q( 37 4 , 8) (b) R(4, 29) 2( x + 1)( x + 2) (2x + 3) 2 (−1, −2); (−2, −3) 7 Maximum at (−1,−1) and minimum at (0, 0) 8 (i) 3 2 2 (iii) 3 ; 3; gradient 2 ? y y = uv where u = x10 and v = x7 (i) (a) =∞ (Page 90) d dx (f(x)) is a polynomial of order (n − 1) so it has no term in xn. (iii) Activity 4.1 (Page 87) 1 4 (v) ? 3x(x − 2) (ii) (iii) (−1, −2), minimum; (−2, −3), maximum −1; y = −x (i) x–2 ( x – 1) 2 (iv) Tangent: y = 8; normal: x = 4 (iv) The two tangents are parallel. 3 (i) (ii) (iii) −1; y = −x + 4 (Page 83) x 4 (iii) (4, 8) 6 (vii) (vi) θ = 20.8°, 122.3° ? (iii) y = x − 3 dy ≠ 0 for any value of x (iv) dx dv du dy v dx – u dx = dx v2 7 9 10 6 = x × 10x – x × 7x x 14 16 – 7 x 16 = 10x 14 = 3x2 x 10 dy u y = v = x 7 = x3 ⇒ = 3x2 x dx x (Page 92) y = ln (3x) is a translation of ⎛0 ⎞ y = ln(x) through ⎜ ⎟ . ⎝ 3⎠ The curves have the same shape. The gradient function is valid for x > 0. (x – 328 9781510421738.indb 328 02/02/18 1:17 PM Exercise 4B (Page 94) 1 6 2x x2 + 1 (v) − x1 (vi) 1 + ln x (iii) 1 x ( x + 1) x (ix) x2 – 1 1 – 2 ln x (x) x3 (ii) 2e2x 1 8 9 + 4x) (v) e4x(1 (vi) 2x2e−x(3 (vii) 1– x ex y ? Maximum (i) (1, 0) e, 1 e (e, 1e ? ( ) x (1, 0) 4 (i) 0.108e0.9t 10 (4, 4e−2) (ii) 0.108 m h−1; 0.266 m h−1; 0.653 m h−1; 1.61 m h−1 11 (i) (1, −e) (ii) Minimum (ii) ey − 2x + 1 = 0 (i) dy = (1 + x)ex; dx (ii) 5 (i) (ii) ( −1, − 1e ) Rotation symmetry, centre (0, 0) of order 2. f(x) is an odd function since f(−x) = −f(x). f '(x) = 2 + ln(x2); 2 f ((x) = x 1 –2π –π 0 2π x (Page 97) (Page 97) sin x y = tan x = cos x dy cos x (cos x ) – sin x ( – sin x ) dx = cos 2 x 2 2 1 = cos x +2 sin x= cos 2 x cos x = sec2x Exercise 4C (Page 100) 1 Activity 4.2 (Page 96) d 2y = (2 + x)ex dx 2 π dy dx Activity 4.3 (Page 97) (viii) 6e2x(e2x + 1)2 12 x This is a demonstration but ‘looking like’ is not the same as proof. − x) O 2π No.You can see this if you draw both graphs. (ii) (iii) 2 (iv) 2(x + 1)e(x +1) π –1 ( 1e , − 1e) (i) (1, 1e ) (ii) 2 (iii) 2xex 3 –π x 1 ( 3ex 7 0 –1 –2π (viii) − (i) –π –2π e O y = cos x 1 y (vii) x(1 + 2 ln(4x)) 2 y e x ( x – 1) x2 (1, e), minimum (i) (ii) (iv) (− 1e , 2e ) , maximum; ( 1e , − 2e ) ,minimum Answers 3 (i) x (ii) x1 (iii) x2 (iii) 0 2 –1 2π dy dx against x looks like the graph of cos x. When y = sin x the graph of −2 sin x + cos x (ii) sec2x (iii) cos x + sin x dy dx π (i) (i) x sec2x + tan x (ii) cos2 x − sin2 x = cos 2x (iii) ex(sin x + cos x) x 3 (i) (ii) x cos x – sin x x2 x e (cos x + sin x ) sec2x (iii) sin x(1 – sin x ) – cos x( x + cos x ) sin 2 x 329 9781510421738.indb 329 02/02/18 1:17 PM 4 (i) 2x sec2(x2 + 1) (iii) x Answers (ii) 5 2 cos 2x 1 (iii) tan x sin x (i) − 2 cos x (ii) ex(tan x (iii) 8x + (iv) (iv) −sin y dy dx d y (v) e(y+2) dx sec2x) (vi) y3 + 3xy2 cos 4x2 (vii) (iv) −2 sin 2x ecos 2x 1 (v) 1 + cos x 1 (vi) sin x cos x 6 (iii) (0.368, 0.692) dy dy +y+1+ d x dx (i) cos x − x sin x (ii) −1 4xy5 + 1 dy dx O dy 10x2y4 dx dy (viii) 1 + 1y dx (ix) xey (x) dy dy + ey + sin y dx dx x 2 dy + 2x ln y y dx (iv) y = x − 2π 1 8 (ii) (1, −3), (−1, 3) 9 (ii) k = 5, c = 68 10 (ii) (2, 1), (−2, −1) 11 (i) (ii) 12 dy (xi) esiny + x cosy esiny dx dy (xii) tan y + x sec 2 y dx (iii) y = −x y ? 3x 2 − 2xy x 2 + 3y 2 8x − 7y − 9 = 0 (a, −2a) (Page 109) 7 x = π,x = π 6 3 8 y = 8.66x − 2.53 2 1 5 Exercise 4E (Page 116) 9 (i) 5 3 0 1 (ii) −3 4 (i) (ii) 5 (1, −2) and (−1, 2) 6 (i) 10 − (tan x Maximum at x = 61 π, minimum at x = 5π 6 11 y = − 23 x + 1.68 (ii) 12 (i) 1 π 4 (iii) (ii) Maximum 13 ? (ii) 330 9781510421738.indb 330 t–1 t +1 (vi) −tan θ 1 (vii) 2e t (1 + t ) 2 (viii) (1 − t ) 2 (v) (2, − 4 ) 1 2 6 4y3 dy dx 2x + 3y2 dy dx x 3 7 (i) (ii) ln y = x ln x 1 dy = 1 + ln x y dx (i) 6 (ii) y = 6x − 3 (iii) 3x + 18y − 19 3 = 0 4 Exercise 4D (Page 106) (i) (iv) − 23cot θ y+4 6–x x − 2y − 11 = 0 O t (ii) 2 (Page 103) (i) 1 + cos θ 1 + sin θ t2 + 1 (iii) 2 t –1 0 y = −1 y The mapping is one-many. 1 At points where the rate of change of gradient is greatest. dy + y sec2x) dx (iv) Asymptotes x = 6, y = −4 − 41 π x (i) ( 41 , 0) (ii) 2 (iii) y = 2x − (iv) (0, − 21 ) 1 2 02/02/18 1:17 PM 4 x − ty + at2 = 0 (ii) tx + y = at3 + 2at (iii) (at2 + 2a, 0), (0, at3 x2 − 2x + 3 = (x − 1)2 + 2 so is defined for all values of x and is always greater than or equal to 2. + 2at) − b2 at (ii) at2y + bx = 2abt (iii) X(2at, 0),Y 0, 2b t (iv) Area = 2ab (i) ( ) 7 (ii) y = tx − 2t2 (iii) [2(t1 + t2), 2t1t2] (i) (ii) 2 9 (iii) (e −6 ,4e −2 11 (i) 2t(t − 1) 3t − 2 (6, 5) (ii) 12 3 (ii) 15 (i) (Page 121) Many possible answers. One method would be to divide the flock into squares, estimate the number of birds in one square and then multiply by the estimated number of squares in the flock. 4 5 6 7 8 Activity 5.1 (Page 123) (i) ? The areas of the two shaded regions are equal since y = x1 is an odd function. (Page 125) The polynomial p2(x) can take the value zero. 9781510421738.indb 331 106 instalments: R agrees with the value of e to 5 d.p. (iv) 1 2 ln (i) 1 3x 3e + c − 41 e−4x + c x −3e− 3 + c |2x − 9| + c Exercise 5B (Page 131) 1 9 (i) −cos x − 2 sin x + c (ii) 3 sin x − 2 cos x + c (iii) −5 cos x + 4 sin x + c (iv) 4 tan x + c (v) − 21 cos(2x + 1) + c 1 5 sin(5x (i) 2(e8 − 1) = 5960 (vi) (ii) ln 49 9 (vii) 3 tan 2x + c = 1.69 (ix) 4 tan x − 21 sin 2x + c (i) P(2, 4); Q(−2, −4) (ii) 8.77; 14.2 (to 3 s.f.) (i) 4; 5 ln 5 − 4 (ii) Ref lection in y = x 2 (i) y = 21 e 2 x + 2e − x − 23 (ii) Minimum when x = 0.231 (i) y = 3x − 3 (ii) (a) 4 (i) 1 2 (ii) 1 (iii) (iv) (a) 3(5 ln 5 − 4) (b) 4 ln 3 + 5 ln 5 − 4 2 (i) 2x + 3 (iii) Quotient = 2x + 1, remainder = −3 1 2 2 (e (iv) 3 4 (v) 1 3 (vi) (vii) 0 3 4 3 (ii) 3 8 4 (i) (a) (ii) ex 1 a 2! 3 = e = 2.718 281 83 (8 d.p.) 1 a 3! 4 = 1 4! 3 −1 2 (ix) Investigations (Page 128) a0 = 1 a1 = 1 a2 = 3 −1 2 (viii) 1 + 1) A series for – π) + c (viii) tan 3x + 21 cos 2x + c (iv) 0.906 Chapter 5 ? 104 instalments: R = 2.718 |x| + c (iii) 4.70 2 sin θ 13 1 4 ln 1000 instalments: R = 2.717 (iii) (iv) − 25x + c e (v) ex − 2e−2x + c + 3) ln3 2 –tan t (ii) (ii) 3 cos t − 4 sin t 3x cos t + 4y sin t = 12 (iii) t = 0.6435 + nπ 3 ln |x| + c (iii) ln |x − 5| + c (iv) x = 4 8 (i) Scheme B: R = 2.594 Scheme C: R = 2.653 Exercise 5A (Page 125) 1 Compound interest Answers 6 (i) 5 (i) 1x 2 (b) π 4 + 41 sin 2x + c (a) 1x 2 – 1 sin 2x + c (b) π− 3 6 8 4 2 cos( x − 61 π) 331 02/02/18 1:17 PM 6 Answers 7 8 (ii) (a) 2 2 (b) 3 (i) 1 2 (iii) 1 6π (ii) 1 4 (5π 1 O (ii) (i) 12e x + 4 e 3x + c 11 (ii) (a) θ = ±0.572 (b) 3 π 32 3 4 (i) (i) 2sinθ + 2θ + c (ii) 1 In15 2 3 2 y 1 1 (d) 0 (ii) (Page 136) ) 1.5 2 x 0.5 1 1.5 2 x y 0.458 658, 0.575 532, 0.618 518, 0.634 173, 0.639 825 2 1 0 5 (i) y (iii) 4 4 2 0 Underestimates – all trapezia below the curve (ii) Impossible to tell 7 27 8 (i) 1 x 6 (i) 458 m (ii) A curve is approximated by a straight line. The speeds in question are only correct to 2 s.f. y (i) 3.1349... 1 (ii) 3.1399..., 3.1411... 9 (a) 2, (b) 2, (c) 4, (d) 4 (i) (0, 1) (ii) 1 π 4 (iii) 1.77 (ii) (iv) Underestimate 3 10 2 0 (1, 0) (ii) 1e (iii) 0.89 (iv) Underestimate, as all trapezia are below the curve 3, 3.1, 3.131 176, 3.138 988 (iii) 3.14 (This actually converges to π.) (i) (iii) 3.14 1 3 1 x (Page 137) Exercise 5C (Page 137) 0.5 (iii) 0.64 (iii) Overestimates – all trapezia above the curve 2 (c) 2 x 1 y 0 Error from 16-strip estimate is about 0.7%. 1 0 + 41 Area required is half a major segment = 8.4197 units2. (ii) x 2.179 218, 2.145 242, 2.136 756, 2.134 635 (iii) 2.13 The curve is part of the circle centre 2 21 , 0 , radius 2 21 . (i) y 1 − 2) 10 ? (b) − 81 3 (ii) ( y 2 + 21 cos 2x 9 ? (i) 3 2 3−π 2 13 3 (i) 2 (iii) 0.95 11 1 2 (ii) 13.5 x 332 9781510421738.indb 332 02/02/18 1:17 PM Chapter 6 ? 2 (Page 144) 2 roots (ii) [0, 1]; [1, 2] (i) y = 2x 2 1 –2 (ii) O x 2 roots For 1 d.p., an interval length of " 0.05 is usually necessary, requiring 5 steps. However, it depends on the position of the end points of the interval. 4 –1.88, 0.35, 1.53 5 (i) –1.62 (ii) 1.28 (i) [−2, −1]; [1, 2]; [4, 5] 6 (i) [1, 2]; [4, 5] (ii) 1.857, 4.536 (i) (a) y O y (iii) 1.154 x x=0 (c) Failure to find root (ii) Converges to 1. (b) x < x for x > 1, x > x for x < 1 and 1 = 1 (iii) Converges to 1.6180 (to 4 d.p.) since this is the solution of x = x + 1 (i.e. the positive solution of x2 − x − 1 = 0). ? Exercise 6A (Page 150) O –5 (b) x Converges to 0.7391 (to 4 d.p.) since cos 0.7391 = 0.7391 (to 4 d.p.). x (iv) a = −1.511 718 75, n = 8 a = 1.244 384 766, n = 12 a = 4.262 695 313, n = 10 8 y (i) (iii) −1.51, 1.24, 4.26 The expected number of steps for 2 d.p., requiring an interval of length " 0.005, is 8 steps. Success Investigation (Page 152) O 7 (c) O (ii) In cases like this, 2 and 3 d.p. accuracy is obtained very quickly after 1 d.p. x=0 1 f(x) For example, the interval [0.25, 0.3125] obtained in 4 steps gives 0.3 (1 d.p.) but the interval [0.3125, 0.375] obtained in 4 steps is inconclusive. As are the interval [0.343 75, 0.375] obtained in 5 steps, the interval [0.343 75, 0.359 375] obtained in 6 steps, the interval [0.343 75, 0.351 562 5] obtained in 7 steps, etc. (b) (iii) (a) (iii) 2, –1.690 Activity 6.1 (Page 148) (ii) x y=x+2 Answers 3 y O y (Page 148) 0.012 takes 5 steps 0.385 takes 18 steps 0.989 takes 28 steps In general 0.abc takes (a + b + c + 2) steps. 1 (a) (iii) 0.62, 1.51 (i), (ii) and (iv) can be solved algebraically; (iii) and (v) cannot. ? (ii) (i) x No root (c) Convergence to a non-existent root (Page 153) Writing as gives x5 − 5x + 3 = 0 x5 − 4x + 3 = x g(x) = x5 − 4x + 3 Generalising this to x5 + (n − 5)x + 3 = nx x 5 + (n – 5 ) x + 3 gives g(x) = n and indicates that infinitely many rearrangements are possible. 333 9781510421738.indb 333 02/02/18 1:17 PM ? (Page 155) 5 (i) y Answers Bounds for the root have now been established. ? y = x2 + 2 (Page 156) ? (ii) 6 x0 = 2 gives divergence to +∞. Gradient is just greater than zero here. O 2 x 1 0.618 At this root −1 " gradient " 1 so the root is found. At this root gradient # 1 so the root is not found. 7 (ii) 0.747 (i) y (ii) 2.120 4 (iii) 1.503 2 –1 O 1 2 x (i) 3 and 4 (ii) 3.43 15 (iii) 1.77 16 (ii) 0.678 Chapter 7 ? y=x 1 y = cos x π 2 O x (Page 162) One possibility, using a pen and paper, would be to use trial and improvement; another is that there is a structured method, which is rather like long division. (ii) 0.739 09 Investigation (Page 163) 8 (ii) a = 1.732 9 (i) 1.01, 1.02, 1.03 1 + x ≈ 1 + 21 x or k = 21 0.20 10 (i) 2.29 (ii) x = 2x + 42 ; α = 3 12 3 x y ? 2 x ≈ 21 (1 + x) (Page 164) 1 (1 + x)2 = 3 but substituting x = 8 into the expansion gives successive approximations of 1, 5, −3, 29, −131, … and these are getting further from 3 rather than closer to it. y = 2 – 2x Exercise 6B (Page 157) 3 x O other root x = −1, gradient = 1. x = 1, gradient = 1. 14 y = ln(x + 1) At this root gradient # 1 so the root is not found. 1.521 y y = 2 – x2 –2 y = x2 y x0 = 1 gives convergence to 0.618 (ii) (i) (iv) x = 1.31 x0 = −1 gives convergence to 0.618 1 ) y = lnx (i) x0 = −2 gives divergence to −∞ –1 1 ,− 1 2 2e Only one point of intersection Activity 6.2 (Page 157) –2 (− (iv) 1.319 The iteration diverges because of the gradient of the curve; it is steeper than the line y = x. y (i) (iii) F(x) = ln(x2 + 2) is possible. (Page 156) Between x = −1 and x = 1, −1 " gradient " 1. 12 13 x O Often it is not so much a failure of the method as a failure to use an appropriate starting point or an appropriate rearrangement of the equation. (iii) 1.08 (iii) 1.35 2 1 y = ex 11 1 y = cos x Activity 7.1 (Page 165) O (iv) 0.58 1 π 2 x −0.19 " x " 0.60 −0.08 " x " 0.07 334 9781510421738.indb 334 02/02/18 1:17 PM Activity 7.2 (Page 166) For | x | " 1 the sum of the 1 geometric series is 1 + x which is the same as (1 + x)−1. 5 2 (iv) 1 – 41 x + 32 x Valid for –1 " x " 1 3 (1 − = 1 + 3x + + The coefficients of x are the triangular numbers. x)−3 ? 6x2 10x3 4 x3 1 − 23 x + 13 x 2 − 27 Valid for −3 " x " 3 (ii) 1 − 43 x + 43 x 2 − 32 x3 27 Valid for −1.5 " x " 1.5 … (iii) (Page 168) 101 = 100 × 1.01 4 (i) = 10 1.01 = 10(1 + 0.01) 1 2 = 10[1 + 21 (0.01) + ( 21) (– 21 ) (ii) (0.01)2 + …] 2! = 10.050 (3 d.p.) ? x – 1 is only defined for x # 1. A possible rearrangement is ( ) ( ) 1 2 x 1 – x1 = x 1 – x1 . 1 Since x # 1 ⇒ 0 " x " 1 the binomial expansion could be used but the resulting expansion would not be a series of positive powers of x. 5 (b) |x|"1 (c) 0.43% (a) 1 − 2x + 4x2 6 (b) | x | " 21 7 5 (b) |x|"1 (c) 0.000 006 3% 1 − 3x + 3x2 − x3 (ii) 1 – 4x + 10x2 Valid when | x | " 1 1− 1x + 8 3 2 x 128 (iii) 1 − 9 x + 8 19 x 2 128 1 − 3x + 6x2 − 10x3 Valid for –1 " x " 1 1 − 6x + 24x2 − 80x3 Valid for – 21 " x " 21 7 3 + 11 x − 5 x 2 , | x | " 4 4 16 64 8 (i) a = 2, b = −1, c = 16 (ii) |x|"2 (iii) 1 + 6x + 24x2 + 80x3 Valid for – 21 " x " 21 9 (i) 1 + 2x + 5x2 (ii) |x|" (i) (ii) 2 4 Valid when | x | " 4 Exercise 7A (Page 170) 1 1 − 2x + 3x2 (i) (ii) (i) (ii) 1 + 21 x – 81x2 Valid for –1 " x " 1 1x 2 3x2 8 + 1– Valid for –1 < x " 1 3 2 x (iii) 1 + 41 x – 32 Valid for –1 " x " 1 1 3 1 − 3 x + 27 x 2 4 4 16 11 1 − 23 x 2 (i) a = −3 1 − 2x − 4 x 2 − 40 x3 3 2a 2 3b 3 1 9y x+3 x–6 x+3 x +1 2x – 5 2x + 5 3(a + 4) 20 x(2x + 3) ( x + 1) 8 2 5( p – 2) 9 a–b 2a – b 10 ( x + 4)( x – 1) x( x + 3) 11 9 20x 12 x–3 12 13 a2 + 1 a2 – 1 14 5x – 13 ( x – 3)( x − 2) 15 2 ( x + 2)( x − 2) 16 2p2 ( p 2 – 1)( p 2 + 1) a2 – a + 2 (a + 1)(a 2 + 1) 3 10 12 2 (a) (iii) a = –7, b = 25 6 1 3 0.8% x2 x4 (iii) (a) 1 − 2 − 8 3 Exercise 7B (Page 174) 1 + 43 x + 43 x 2 + 32 x3 27 Valid for –1.5 " x " 1.5 (c) (Page 170) 13 − 10 x 3 17 Answers Investigation (Page 167) (i) (ii) 18 –2( y 2 + 4 y + 8) ( y + 2) 2 ( y + 4) 19 x2 + x + 1 x +1 20 – (3b + 1) (b + 1) 2 335 9781510421738.indb 335 02/02/18 1:17 PM Answers 21 13x – 5 6( x – 1)( x + 1) 22 4(3 – x ) 5( x + 2) 2 23 3a – 4 (a + 2)( 2a – 3) 24 ? 3x 2 –4 x( x – 2)( x + 2) (Page 176) The identity is true for all values of x. Once a particular value of x is substituted you have an equation. Equating constant terms is equivalent to substituting x = 0. Activity 7.3 (Page 181) 5 2(1 − x); | x | " 1 The binomial expansion is 1 − x + 3x2. The expansion is valid when | x | " 21 . Which method is preferred is a matter of personal preference for (a) and (b) but for (c) must be (iii). 6 (i) 2 + 2x + 4 (2 − x ) (1 + x 2 ) (ii) 15 3 2 5 + 52 x − 15 4 x − 8 x (i) 2 + x −1 (2 + x ) ( x 2 + 1) (ii) 1 x 2 (i) 1 + 2 − 4 (1 – x ) (1 + 2x ) (2 + x ) (ii) 2 1 − 2x + 17 2 x (i) −1 + 3x − 1 ( x − 2) x 2 + 3 (ii) 1 6 Exercise 7D (Page 181) 1 (ii) Exercise 7C (Page 178) 1 2 1 1 x – ( x + 1) 3 4 2 – 1 ( x – 1) ( x + 2) 5 1 + 1 ( x + 1) ( 2x – 1) 6 2 –2 ( x – 2) x 7 1 – 3 ( x – 1) ( 3x – 1) 8 3 + 2 5( x – 4 ) 5( x + 1) 9 10 11 5 –2 ( 2x – 1) x 2 – 1 ( 2x – 3) ( x + 2) 8 9 + 13( 2x – 5) 13( x + 4 ) 12 19 11 – 24( 3x – 2) 24( 3x + 2) 13 1 + 2 + 3 ( x + 1) ( x + 2) ( x + 3) 14 4 + 3 + 2 ( x − 1) ( 3 − x ) ( 2x + 1) 1 − 2 − 1 (2 + x ) (2 − x ) (2x + 3) 15 4 – 2x ( 2x – 1) ( x 2 + 1) 1 – 1 + 1 (iii) ( x – 1) 2 ( x – 1) ( x + 2) 1 – 1 ( x – 2) ( x + 3) 2 – 2 ( x – 4 ) ( x – 1) 9 2 – 3 – (i) ( 1 – 3x ) (1 – x ) (1 – x ) 2 (iv) 5 + 6 – 5x 8( x – 2) 8( x 2 + 4 ) (v) 5 – 2x + 2 ( 2 x 2 – 3) ( x + 2 ) Can be taken further using surds. 3 2– 1 – x x 2 ( 2x + 1) 10x – 3 (vii) ( 3x 2 – 1) x Can be taken further using surds. 1 + 1 (viii) ( 2x 2 + 1) ( x + 1) (vi) 2 3 8 9 ? 1 −4 − 10x − 16x2 (iii) 5 2 (iv) –1 – (i) 7 – 1 x2 (i) 1 − 3x + (ii) | x | " 21 (i) 2 4x 2 + 1 (ii) 3 9x 2 + 1 2 x2 + 4 y= 8 7x2 72 3 x2 + 9 − 1 (i) 1− x2 1 (ii) 1− x2 y = − 4x + π + 1 ; 3 3 3 8 7x − 1 − 7 x2 + 1 x + 1 − 8 + 14x − 6x 2 Valid for | x | " 1 4 (iv) (ii) 5 x 16 + 5 x + 17 x 2 + (Page 184) (iii) 4 + 20x + 72x2 8 8 Exercise 8A (Page 185) (i) 4 4 Answer in text following 3 8 4 – – ( 2x – 1) (2x – 1) 2 x 6 + 11 x + 33 x 2 + 5 x2 − 9 x3 Chapter 8 (ix) (ii) 4 7 ? 3x + π − 3 4 3 16 (Page 186) The area is the same as ∫ 4 1 x dx. 336 9781510421738.indb 336 02/02/18 1:18 PM ? 3 (Page 188) ∫ ∫ Activity 8.1 (Page 190) 2 (x 5 − 5 2) 2 4 (x 3 + − 3 2) 2 +c 3 2) 2 2 [3(x − 2) + (x − 15 3 2 (3x + 4)( x − 2) 2 + c = 15 = 2 3 (vi) (i) 222 000 (ii) 586 (ii) (iii) (iv) (v) 2 (i) 22 21 (ii) 19 4 y= 2x + 3 + 3 5 4 3 7 6 6 7 (i) 4 5 8 (1 + (ii) ln 3 (iv) (i) 1 2 (e − 1) 1 4 2 (e − 1) 1 (e + e4) − 2 3 6 2)e−x −(x + (ii) –(x + 3)e−x + c O 2 ln(e + 1) ≈ 1.434 2 2 +1 e ) ≈ 1.434 (iii) ln( 2 (iv) The same. The substitution ex = t2 transforms the integral in part (ii) into that in part (iii). (i) 1 (ii) (p 2 3 2 2 2(1 + x ) + c ) k = 2, a = 1, b = 2; 32.5 ln x ) 2 6 (ii) 1 3 ln |3x2 + 9x − 1| + c (ii) cos(1 − x) + c (iii) − 41 cos4 x (v) (i) tan −1 x + c 1 tan −1 x + c 4 4 2 tan −1 3 () ( x3 ) + c (iv) 2 tan −1 2 ( 2x ) + c (v) ⎛ 3 tan − 1 ⎜ 6 ⎝ 3x ⎞ + c 2 ⎟⎠ (vi) 1 tan −1 10 ( 25x ) + c 5 2π 3 6 (i) x = 0.685 (ii) 8 15 (i) 1 (e 2 2 (ii) 0.896 9 (ii) 1 π − 41 6 10 (a) 3x + 21 tan 2x + c (b) 1 8 (i) 1 24 (ii) k = 10 7 11 ( − 1) 3 3π + ln 4 ) +c +c (iv) ln | 2 − cos x | + c (2 ln x − 1) + c ln |x2 + 1| + c (i) 1 3 sin 3x ln 2 (iii) + 1 Exercise 8D (Page 197) 1 1 16 (v) (ii) x (iii) 2.53 (a) 8 x − 3 2 + c 2x 4 y 1 1 2 ln –1 + c sin x 1 (iv) e − 1 1 2 7 +c (iii) 1 1 = 27.7 (i) (i) (i) (ii) 0.490; 0.314 (ii) (i) 9781510421738.indb 337 1 2 2 tan x (iv) −e2; max. at x = −2 4 (vi) 3 − e Exercise 8C (Page 192) 1 (iii) (iii) (–2, e2) 1 (b) (ii) 0.018 (to 3 s.f.) (iii) 18.1 3 (i) (iii) 10] + c 1 3 8 8 (x + 1) + c 1 2 6 6 (x + 1) + c 1 3 5 5 (x − 2) + c 3 1 2 2 6 (2x − 5) + c 3 1 2 15(2x + 1) (3x − 1) + c 2 ( x + 9) 21 (x − 18) + c 3 (i) esin x + c (ii) Exercise 8B (Page 190) 1 (ii) Answers Yes: Using the chain rule dy dy du = × dx du dx Integrating both sides with respect to x, dy du y= ( )dx × du dx dy = ( )du du (iii) 4ex + c −ln | cos x | + c ? (Page 199) Substitution using u = x2 − 1 needs 2x in the numerator. Not a product. (vi) − 61 (cos 2x + 1)3 + c 2 (i) −cos(x2) + c 337 02/02/18 1:18 PM 1 (i) Answers (ii) ⎪ 1 + ln x – 1 + c ⎪2x + 3⎪ 1– x ln ⎪ 3x – 2 +c 1– x (iv) (vi) 1 2 ln ⎪ ⎪ ⎪ ⎪ ⎪ (vii) ln (viii) ln 2x + 1 + x+2 1 +c 2 ( 2x + 1) 3 (i) x2 (ii) (c) ? x2 + 4 + c x+2 (i) 5 (i) 6 (i) 7 (ii) 2⎞ 6 ⎟⎠ (a) 2 + 1 1 – 2x 1 + x (b) ln (118) = 0.318 45 (a) 3 + 3x + 9x2 + … 1 (i) (ii) 0.14% 1 3 − 2( x + 1) 2( x + 3) 3 4x + 2 − x 4 + x2 1+ (v) 338 9781510421738.indb 338 ∫ ∫ ∫ 2 (i) 3 2 1+ x ) 2 15 ( 4 1 15 (x 5 (i) x ln x − x + c (ii) x ln 3x − x + c (ii) (iii) (3x − 2) + c − 2)5(5x + 2) + c 6 x2ex − 2x ex + 2ex + c 7 (2 − x)2 sin x − 2(2 − x)cos x − 2 sin x + c 8 1 2 (( x 2 + 1)tan −1 x − x ) + c Exercise 8G (Page 210) 1 (a) u = x, dv = ex dx (b) xex − ex + c (a) u = x, dv = cos 3x dx (b) 13x sin 3x + (i) 2 3 9e (ii) −2 + 1 9 (iii) 2e2 (iv) 3 ln 2 − 1 (v) π 4 64 (vi) 3 ln 4 − 7 cos 3x + c 2 (i) (2, 0), (0, 2) (ii) y 2 y = (2 – x)e–x +c (a) u = x, dv = e−x dx (b) −xe−x − e−x + c 1 4 1 4 4 x ln x − 16 x + x e3x − 13e3x + c x sin 2x + 21 cos 2x − 91 x3 + c (iii) x ln px − x + c dx ⇒ 2x 1 2x 2x = xe − 2 e + c (vi) (a) u = x, dv = sin 2x dx 1 (b) − 2 x cos 2x + 1 sin 2x + c 4 d (a) dx (x cos x) = −x sin x + cos x 1 3 3 x ln 2x 3 1e−2x 4 ln 2 − x − 2 ln x + 1 + 2 ⇒ x cos x = −x sin x dx + cos x dx ⇒ x sin x dx = −x cos x + cos x dx (iv) dx (iii) (a) u = 2x + 1, dv = cos x dx (b) (2x + 1)sin x + 2 cos x + c (iv) (a) u = x, dv = e−2x dx (b) − 21x e−2x − A = 1, B = 2, C = 1, D = –3 ∫ ∫ − e2x 1 9 Activity 8.2 (Page 204) (b) ∫ e2x xe2x (Page 204) + 1 + 2 ln 4 x + 3 + c (i) dx Each of the integrals in Activity 8.2 is of the form ∫ x dv dx and is dx found by starting with the product xv. (b) 0.318 00 4 ∫ ∫ 2xe2x Exercise 8F (Page 208) x + 1 , ln ⎛⎜ +4 x – 3 ⎝ (c) ∫ = ⎪ ⎪ xx ++ 13⎪ + c ⇒ xe2x = 2xe2x dx + e2x dx ⇒ 2 1 x ln 1 – x − x + c – (b) ⎪ ⎪ ( x − 1) ln ⎪ 2x + 1 ⎪ + c (v) 2 (ii) x –1 +c x2 + 1 (iii) ln ∫ ⇒ x sin x dx = −x cos x + sin x + c d (a) dx (x e2x) = x ×2e2x + e2x (c) Exercise 8E (Page 202) 2 O x (iii) e–2 + 1 3 (i) y y = xsinx c +c π O (ii) x π 02/02/18 1:18 PM 4 5 ln 5 − 4 2 π −1 2 6 4 4 − 15 so area = 15 square units 7 (i) 4π − 3 3 (ii) 1 (π 16 8 4(ln 4 − 1) 9 (ii) y This integral can also be found using partial fractions, but using logarithms is quicker. 3 x (iv) 2.31 ? (i) 1 2 (ii) π ( 2 e − 3) (i) 6e (ii) − (i) y = x −1 (ii) 1 4 ln | x2 + sin x | + c. (vi) This is a product: sin2 x is a function of sin x, and cos x is the derivative of sin x, so you can use the substitution u = sin x. Using u = sin x = ln | x2 ∫ cos x sin2 x dx = ∫ u2 du = 13 u3 + c + 2x − 3 | + c (iii) This is a product of x and There is no relationship between one expression and the derivative of the other, so you cannot use substitution. As one of the expressions is x, you can use integration by parts. = 13 sin3 x + c ex. 3 4 π( e2 − 2) (Page 212) You will return to these integals in Activity 8.3. ∫x ex dx ex dx (iv) This is also a product, this 2 2 time of x and ex . ex is a function of x2, and 2x is the derivative of x2, so you can use the substitution u = x2. Activity 8.3 (Page 214) Using u = (i) ∫ ∫ This is a quotient. The derivative of the expression on the bottom is not related to the expression on the top, so you cannot use substitution. However, as the expression on the bottom can be factorised, you can write it as partial fractions. ∫ =x − = x ex − ex + c ex x2 9781510421738.indb 339 2x + cos x ∫ x 2 + sin x dx = ∫ x 2 + 2x – 3 dx = 21 ∫ 2 2x + 2 dx x + 2x – 3 1 2 In this case the numerator is the differential of the denominator and so the integral is the natural logarithm of the modulus of the denominator. Since d 2 (x + sin x) = 2x + cos x, dx then x +1 O 12 f ′(x ) dx = k ln | f(x) | + c. f(x ) ∫ – 1x 2 11 The derivative of the expression on the bottom line is 2x + 2, which is twice the expression on the top line. So the integral is of the form k y=x y=2+e 10 = 2 ln | x + 3 | − ln | x − 1 | + c (ii) − 2) (v) 2 x ex dx = = = 1 u 2e 1 u 2e + c 1 x2 2e + c du Answers 5 x–5 ∫ x + 2x – 3 dx 1 2 =∫ dx − ∫ ( x – 1) dx ( x + 3) Exercise 8H (Page 214) 1 (i) 1 3 sin(3x (ii) –1 +c ( x 2 + x – 1) − l) + c (iii) −e1−x + c (iv) 1 2 sin 2x (v) x ln 2x − x + c (vi) –1 +c 4( x 2 – 1) 2 (vii) 1 ( 2x − 3) 3 + 3 +c 2 c (viii) ln ⎪ x – 1⎪ + 1 + c x+2 x –1 1 4 − 16 x +c (ix) 1 4 4 x ln x (x) ln ⎪2xx––31⎪ + c (xi) 1 x2 + 2x 2e +c (xii) −ln (sin x + cos x) + c 339 02/02/18 1:18 PM 2 + 41 cos 2x + c 3 (xiv) 1 − 2 cos 2x 4 (i) 8 3 1 ln 4 3 (ii) + 61 cos3 2x + c 5 (iii) 48 + 8 ln 4 (iv) (v) (vi) 2 3 8 ln 2 − 79 3 2 2 −1 3 ) ( (vii) (ln 2) 2 2 9 6 de dθ = kθ 7 dθ = – (θ – 15) 160 dt 8 dN = N dt 20 9 dv = 4 v dt 1+ 3 4 4e 2 11 3 (i) 12 dV = – 2V 1125π dt 4 (ii) 3 + 3x 2 − x 2 + x2 2e 2 − 10 5 (ii) 15 ln 5 − 4 13 dh = (2 – k h ) 100 dt Answers depend on the size of the cup, the initial temperature of the coffee, whether the cup is insulated and whether milk is added, as well as personal preference. It takes around 5–10 minutes for a cup to become drinkable (around 60°C) and after about 25 minutes the coffee is probably too cool for most people (around 30°C). Exercise 9A (Page 220) 1 dv dt is the rate of change of velocity with respect to time, i.e. the acceleration. The differential equation tells you that the acceleration is proportional to the square of the velocity. 9781510421738.indb 340 can be rewritten as ln | y | = 12 x2 + (c2 − c1). dA = 2k π = k ′ A A dt dθ = – s 4 ds 2 y = 13 x3 + c (ii) y = sin x + c (i) (ii) 2 (x 2 + c ) y2 = 23x3 + c y=− (iii) y = Aex (iv) y = ln | ex + c | (v) H is about (70° N, 35°W) and L is about (62° N, 5°W) so they are separated by 30° in longitude at a mean latitude of 66°. Reference to the scale shows this to be about 900 nautical miles. y = Ax (vi) y = ( 41x2 + c)2 1 (vii) y = − ( sin x + c ) (viii) y2 = A(x2 + 1) − 1 Investigation (Page 222) (ix) y = −ln(c − 21 x2) (x) y3 = 23 x2 ln x − 43 x2 + c Exercise 9C (Page 229) 1 1035 1 (i) y = 3 x3 − x − 4 (ii) y = e3 1x 2 (iii) y = ln( 21 x2 + 1) 1 (iv) y = ( 2 – x ) 1 (v) y = e 2( x 2 −1) − 1 996 957 O (i) 3 (x) (Page 217) ln | y | + c1 = 21x2 + c2 (iv) y = 23 x 2 + c 1 (2 3 ? (Page 224) (iii) y = ex + c (ix) 2 − 1) ? 1 − 10 The model covers the main features of the situation. Exercise 9B (Page 225) (viii) Chapter 9 340 ds = k dt s 2 dh = k ln(H − h) dt dm = k dt m dP = k P dt 2 isobars Answers (xiii) − 21 x2 cos 2x + 21x sin 2x 900 nautical miles The mean level is 996 and the amplitude 39 so a model is πx p = 996 + 39 cos 900 dp –39π and = sin πx 900 dx 900 dp or = −a sin bx dx ( ) ( ) (vi) y = sec x 2 (i) θ = 20 − Ae−2t (ii) θ = 20 − 15e−2t (iii) t = 1.01 hours 3 (i) N = Aet (ii) N = 10et with a = 0.136 and b = 0.0035. 02/02/18 1:18 PM 4 ds 2 = ;s = dt s 5 16 x y = 3e 16−x 21 3−e 6 (i) 4t + c Using the assumptions in Exercise 9A, question 7, the rate of cooling is proportional to the temperature of the coffee above the surrounding air. The initial temperature is 95°C and the cooling rate is 0.5°C s−1. So − t 160 θ = 15 + 71e (iii) 7A 3 (i) tan −1 (ii) The value of x tends to tan −1 21 . ( 21 − 21 e −2t ) ( 21 − 21 e −2t ). (iii) 100 ln ⎛⎜ 10 + h ⎞⎟ − 20h ⎝ 10 − h ⎠ dN = k( N − 150) (i) dt 9 N = (ii) 500e 0.08t + 150 (iii) No; when t = 15 then N = 1810 or when N = 1500 then t = 16.4 10 1 2t (i) N = 1800e1 t 5 + e2 (ii) 1800 11 x = 81 (tan θ + 1) 2 − 12 (i) ( y= − a1 a2 Length = (ii) + P Chapter 10 a3 ? (Page 234) To find the distance between the vapour trails you need two pieces of information for each of them: either two points that it goes through, or else one point and its direction. All of these need to be in three dimensions. However, if you want to find the closest approach of the aircraft you also need to know, for each of them, the time at which it was at a given point on its trail and the speed at which it was travelling. (This answer assumes constant speeds and directions.) O OP = a 21 + a 22 + a 32 ⇒ Exercise 10A (Page 239) 1 (i) 3i + 2j (ii) 5i − 4j (iii) 3i (iv) −3i − j 2 (Page 239) +c ) For all question 2: j i (i) 2 a3 203 2 52 h 5 = 5 1 H 2t 150 ( ) P 5 + 25 H 2 ⎛ (iii) t = 60 ⎜⎜1 − h H ⎝ Q OP2 = OQ2 + QP2 = (a12 + a 22 ) + a 32 The vector a1i + a2 j + a3k is shown in the diagram. 9 sin 1 x 10 3 a 21 + a 22 The final temperature is lower if the milk is added at the end. ? 1 2 Q Now look at the triangle OQP. . z 3 x cos 1 x 10 3 (ii) y x Adding 10% milk at 5°C gives (iii) As 21 − 21 e −2t increases so does 8 O − t 160. θ = A(1 + 3e −kt ) tan −1 Start with the vector ⎯→ OQ = a1i + a2 j. θ = 15 + 80e 2 7 13 Investigation (Page 233) Answers (iii) N tends to ∞, which would never be realised because of the combined effects of food shortage, predators and human controls. 5 2 ⎞ ⎟⎟ ⎠ O a2 a1 x y ( 13, 56.3°) Q 341 9781510421738.indb 341 02/02/18 1:18 PM (ii) (v) 5k Exercise 10B (Page 248) (vi) −i − 2j + 3k (vii) i + 2j − 3k (viii) 4i − 2j + 4k ( 13, −33.7°) Answers 1 (ix) 2i − 2k (iii) (x) 5 (4 2, −135°) (iv) −8i + 10j + k A: 2i + 3j, C: −2i + j ⎯→ (ii) A B = −2i + j, ⎯→ C B = 2i + 3j ⎯→ ⎯→ (iii) (a) A B = ΟC ⎯→ ⎯→ (b) C B = ΟA (i) ( 5, 116.6°) 2 (ii) 3.74 (ii) 4.47 (iii) 4.90 (iv) 3.32 (v) 7 (vi) 2.24 4 (i) 2i − 2j (ii) 2i (iii) −4j (iv) 4j ? ⎛1⎞ ⎜ ⎟ ⎝1⎠ (iii) ⎛0 ⎞ ⎜ ⎟ ⎝0 ⎠ (v) ⎛ 8⎞ ⎜ ⎟ ⎝−1⎠ –3j (i) 2i + 3j + k (ii) i–k (iii) j – k (iv) 3i + 2j – 5k (a) F (b) C (v) (i) (ii) (iv) Activity 10.1 (Page 244) 3 ⎛6 ⎞ ⎜ ⎟ ⎝8 ⎠ (i) (iv) A parallelogram (5, −53.1°) (i) 3 (v) –6k (i) (a) b (c) Q (b) a + b (d) T (c) –a + b (e) S ⎯→ (a) Ο F ⎯→ ⎯→ (b) OΕ, CF ⎯→ → ⎯→ (c) OG, PS, AF ⎯→ (d) BD ⎯→ ⎯→ (e) QS, PT (ii) (b) 1 2 (a + b) 1 2 (–a + b) (iii) PQRS is any parallelogram and ⎯→ 4 (i) ⎯→ ⎯→ ⎯→ PM = 21 PR, Q M = 21QS (a) i (b) 2i (Page 247) (c) i − j !!" ⎯→ !⎯→ s ⎯→ OC AB ΟC == OA ΟA + s + t AB ⎯→ ⎯→ ΟA = a and A B = b − a OC = a + s ( b − a ) s+t = a+ s b− s a s+t s+t + s t s = a− a+ s b s+t s+t s+t t s = a+ b s+t s+t (a) 5 (ii) (d) −i − 2j ⎯→ ⎯→ | AB | = | Β C | = 2 , ⎯→ ⎯→ | A B | = | CB | = 5 (i) −p + q, 1p − 1q, 2 − 21p, − 21q (ii) 2 ⎯→ 1 ⎯→ NM = 2 BC, ⎯→ ⎯→ N L = 21 A C, ⎯→ ⎯→ ML = 21 A B 342 9781510421738.indb 342 02/02/18 1:18 PM 6 2 (ii) 3i 5 + (iii) ⎛ –1 ⎞ ⎜ 2⎟ ⎜ –1 ⎟ ⎜ ⎟ ⎝ 2⎠ (i) (iv) 7 (i) (ii) (iii) (iv) 4 5 j ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 12 ⎞ ⎟ 14 ⎟ 3 ⎟ (Page 253) 3 4 5 θ b a A B b – a = (b1 – a1)i + 2 k 38 6 x = 4 or x = −2 10 (i) (b) −p (a) q (c) q + p (d) r + p (e) p+q+r (ii) p + 53 r (i) ⎛ 2⎞ ⎜ ⎟ ⎜ 3⎟ ⎜−6 ⎟ ⎝ ⎠ m = −2, n = 3, k = −8 (ii) 76.2° (i) (0, 4, 3) (ii) ⎛ −5 ⎞ ⎜ ⎟,5 2 ⎜ 4⎟ ⎝ 3⎠ (iii) 25.1° ⎯→ (i) OQ = 3i + 3j + 6k, ⎯→ PQ = −3i + j + 6k (ii) 53.0° (i) −2 40° ⎯→ (iii) AB = i − 3j + (p − 2)k; p = 0.5 or p = 3.5 7 OB2 = b12 + b22 + b32 AB2 = (b1 − a1)2 + (b2 − a2)2 + (b3 − a3)2 2(a1b1 + a 2b 2 + a 3b 3 ) ⇒ cos θ = 2|a||b| 29.0° (ii) (b2 – a2)j + (b3 – a3)k 2 2 2 cos θ = OA + OB – AB 2 × OA × OB OA2 = a12 + a22 + a32 ⎛1 ⎞ ⎜ ⎟ ⎜0 ⎟ ⎜0 ⎟ ⎝ ⎠ (i) (iii) 162.0° O ⎞ 29 ⎟ ⎟ 4 ⎟ 29 ⎟ –3 ⎟ 29 ⎠ ⎛ 3⎞ ⎛− 1⎞ ⎜ ⎟, ⎜ ⎟ ⎝1 ⎠ ⎝ 3⎠ ⎯→ ⎯→ (ii) BA . BC = 0 ⎯→ ⎯→ (iii) | AB | = | BC | = 10 (i) (iv) (2, 5) Consider the triangle OAB with angle AOB = θ, as shown in the diagram. –2 9 9781510421738.indb 343 ? 14 ⎠ 11.74 (ii) 2 These are the same because ordinary multiplication is commutative. 14 ⎟ 2 ⎟ 90° (vi) 180° (Page 252) ⎛ b1 ⎞ ⎛ a 1 ⎞ ⎜ ⎟ . ⎜ ⎟ = b1a1 + b2a2 ⎝b 2 ⎠ ⎝ a 2 ⎠ 3 5 i− j+ 38 38 1 7 (v) ⎛ a 1 ⎞ ⎛ b1 ⎞ ⎜ ⎟ . ⎜ ⎟ = a1b1 + a2b2 ⎝ a 2 ⎠ ⎝b 2 ⎠ 8 12 ? – 13 j 1 (Page 250) The cosine rule Pythagoras’ theorem 2i − 2 j + 1 k 3 3 3 3i − 4k 5 5 (v) (vi) ? ⎟ 13 ⎠ 5 13 i ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ 13 ⎟ 3 ⎟ (i) 8 −6, obtuse (i) ⎛ 2⎞ ⎜ 3⎟ ⎜ – 2⎟ ⎜ 3⎟ ⎜ 1⎟ ⎝ 3⎠ 99° (ii) 1(2i 7 (ii) a.b = |a || b| − 6j + 3k) (iii) p = −7 or p = 5 Exercise 10C (Page 254) 9 (ii) q = 5 or q = −3 1 10 (i) ⎯→ P A = −6i − 8j − 6k, ⎯→ P N = 6i + 2j − 6k (ii) 99.1° (i) 42.3° (ii) 90° (iii) 18.4° (iv) 31.0° Answers ⎛ ⎜ ⎜ ⎜ ⎝ 343 02/02/18 1:18 PM 11 Answers 12 (i) 4i + 4j + 5k, 7.55 m (ii) 43.7° (or 0.763 radians) (i) ⎯→ P R = 2i + 2j + 2k, ⎯→ P Q = −2i + 2j + 4k (iii) is parallel to (i) since the direction vector is the same. (iv) is parallel to (ii) since ⎛ –1⎞ ⎛ 1⎞ ⎜ ⎟ = −⎜ ⎟ . ⎝ 2⎠ ⎝ –2 ⎠ (ii) 61.9° Exercise 10D (Page 264) (iii) 12.8 units ? 1 (i) (Page 259) (ii) (iii) ⎛ 8⎞ ⎜⎜ ⎟⎟ ⎝11⎠ (v) It lies between A and B. (v) 2 y 8 6 (i), (iv) (ii) 2 2 (iii) 4 (ii) 6 8 (v) (i) and (iv) are the same since putting λ = −1 in (i) gives 4 x ⎛1⎞ ⎛ 2⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ 2⎠ ⎝1⎠ ⎛ 3⎞ ⎛ –1⎞ r = ⎜ ⎟ + λ⎜ ⎟ ⎝ 5⎠ ⎝ 1⎠ ⎛ −6⎞ ⎛ 1⎞ (iii) r = ⎜ ⎟ + λ ⎜ ⎟ ⎝ –6 ⎠ ⎝ 1⎠ ⎛ 5⎞ ⎛ 1⎞ (iv) r = ⎜ ⎟ + λ ⎜ ⎟ ⎝ 3⎠ ⎝ 1⎠ (v) ⎛2⎞ r = λ⎜ ⎟ ⎝ 1⎠ ⎛ 1⎞ and ⎛ 1⎞ is parallel ⎜ ⎟ ⎜⎝ 2⎟⎠ ⎝ –3⎠ ⎛ –1⎞ (vi) r = λ ⎜ ⎟ ⎝ 4⎠ 3 to ⎛ ⎞ . ⎜⎝ 6⎟⎠ ⎛ –1⎞ (vii) r = λ ⎜ ⎟ ⎝ 4⎠ (v) ⎛ 1⎞ r = λ ⎜ 2⎟ ⎜ ⎟ ⎝ 3⎠ (i) Yes, λ = 2 (ii) Yes, λ = −1 (iii) No Note:These answers are not unique. (i) 4 (a) 5i + 12j (c) −7.5i − 2j Activity 10.3 (Page 263) 0 (a) 6i − 8j (b) 13 (c) It lies beyond A. (ii) ⎛ 1⎞ ⎛ 1⎞ ⎜ ⎟ r = 0 + λ ⎜ 0⎟ ⎜ ⎟ ⎜ ⎟ ⎝ –1⎠ ⎝ 0⎠ ⎛ 0⎞ ⎛ 2⎞ ⎜ ⎟ (iv) r = 0 + λ ⎜ 1⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1⎠ ⎝ 3⎠ (c) 0 (b) It lies beyond B. –2 (a) 6i + 8j (b) 10 1 3 2, 4 (i) ⎛ 2⎞ ⎛ 3⎞ ⎜ ⎟ r = 4 + λ ⎜ 6⎟ ⎜ ⎟ ⎜ ⎟ ⎝ –1⎠ ⎝ 4⎠ ⎛ 1⎞ ⎛ 5⎞ ⎜ ⎟ (iii) r = 0 + λ ⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4⎠ ⎝ –6⎠ (c) i + 3j (iv) Note:These answers are not unique. (a) −4i − 3j (b) 10 (iv) 0, 1, –2 68 (c) 2i + 1.5j (ii) ⎛−2 ⎞ ⎛ 0 ⎞ ⎛ 2 ⎞ ⎛ 3⎞ ⎜ ⎟ , ⎜ ⎟, ⎜ ⎟ , ⎜ ⎟, ⎝−9 ⎠ ⎝ –5 ⎠ ⎝ –1 ⎠ ⎝1 ⎠ (a) (b) (b) 5 Activity 10.2 (Page 260) ⎛4 ⎞ ⎜⎜ ⎟⎟ , ⎝3⎠ 2i + 8j 3 (c) 3i + 7j ⎯ → ⎯ → ⎯ → ⎯ → O P = O A + λ(O B − O A) ⎯ → ⎯ → = (1 − λ)O A + λO B ⎛3 1 ⎞ ⎜ 2 ⎟, ⎝2 ⎠ (a) ⎛ –1⎞ ⎛ 3⎞ (viii) r = ⎜ + λ⎜ ⎟ ⎟ ⎝ 4⎠ ⎝ –12⎠ (iv) No 5 (v) Yes, λ = −5 (i) ⎛ −1⎞ ⎛ −1⎞ r = ⎜ −2 ⎟ + λ ⎜ 3 ⎟ ⎜ −3 ⎟ ⎜⎝ 1⎟⎠ ⎝ ⎠ ⎛ −2 ⎞ ⎛ −1⎞ or r = ⎜ −2 ⎟ + λ ⎜ 6 ⎟ ⎜ −6 ⎟ ⎜⎝ 1⎟⎠ ⎝ ⎠ (ii) (–2, 1, –2) (iii) ⎛ 0⎞ ⎛ −2 ⎞ r = ⎜ 1⎟ + λ ⎜ 1⎟ ⎜ 0⎟ ⎜⎝ −2 ⎟⎠ ⎝ ⎠ 344 9781510421738.indb 344 02/02/18 1:18 PM Exercise 10E (Page 269) 5 (–2, –6, –1); 30 units ⎛ 5⎞ ⎜⎝ 5⎟⎠ 6 No 7 6 units, 9 units, 77 units 1 3 ⎛ 4⎞ ⎜⎝ 1⎟⎠ (ii) (iii) ⎛ 12⎞ ⎜⎝ 17⎟⎠ –5 (iv) ⎛⎜ ⎞⎟ ⎝ 6⎠ (v) ⎛ 6⎞ ⎜⎝ 3⎟⎠ (i) (ii) (iii) (iv) (v) (vi) (vii) Intersect at (3, –2, 5) Parallel Intersect at (3, 2, –13) Intersect at (1, 2, 7) Skew Intersect at (4, –7, 11) Skew (i) (ii) 12.8 km 20 km h−1, 5 km h−1 Exercise 10F (Page 275) 1 53.6° 2 81.8° 3 8.72° 4 35.3° 5 61.0° 6 (i) 4 (i) ⎯ → ⎛ 10 ⎞ OL = ⎜ ⎟ ; ⎝ 4.5⎠ ⎯ → 7 OM = ⎛⎜ ⎞⎟ ; 3.5 ⎝ ⎠ (ii) ⎯ → ⎛ 4⎞ ON = ⎜ ⎟ ⎝ 1⎠ ⎛ 1⎞ ⎛ 2⎞ AL: r = ⎜ ⎟ + λ ⎜ ⎟ ; ⎝ 0⎠ ⎝ 1⎠ ⎛ 7⎞ ⎛ 0⎞ BM: r = ⎜ ⎟ + µ ⎜ ⎟ ; ⎝ 2⎠ ⎝ 1⎠ ⎛ 13⎞ ⎛ 3⎞ CN: r = ⎜ ⎟ + ν ⎜ ⎟ ⎝ 7⎠ ⎝ 2⎠ (iii) (a) (7, 3) 29 units (a) (3, –1, 7) (b) (iii) 17 units (a) (2, 7, –3) (b) 7 units 7 2 10 units 8 35 units 9 (Page 281) Any real number is either rational or irrational. This means that all real numbers will either lie inside the set of rational numbers, or inside the set of real numbers but outside the set of rational numbers. Therefore no separate set is needed for irrational numbers. – The symbol ℚ is used for irrational numbers – numbers which cannot be expressed exactly as a fraction, such as π. Activity 11.1 (Page 281) (i) A(4, 0, 0), F(4, 0, 3) (ii) 114.1°, 109.5° Real Rational Integers (iii) They touch but are not perpendicular. 10 (ii) −13 227 109 3.1415 0.3 −√5 √5 π 3 8 (ii) λ= 12 (ii) a = −2, a = 3 13 (ii) 79.5° Chapter 11 ? Natural numbers 7 5i + 3j + 4k 11 (b) (7, 3) (iv) The lines AL, BM and CN are concurrent. (They are the medians of the triangle, and this result holds for the medians of any triangle.) (–2, 6, 7) (b) (iii) After 40 minutes there is a collision. ℕ Natural numbers – nonnegative whole numbers (although there is some debate amongst mathematicians as to whether zero should be included!) ? (a) (ii) ℤ Integers – positive or negative whole numbers, including zero Answers 2 (i) ℚ Rational numbers – numbers which can be expressed exactly as a fraction (Page 280) ℝ Real numbers – any number which is not complex Activity 11.2 (Page 281) (i) x = 2 Natural number (or integer) (ii) x= 9 7 Rational number (iii) x = ±3 Integers (iv) x = −1 Integer (v) x = 0, −7 Integers 345 9781510421738.indb 345 02/02/18 1:18 PM Answers Activity 11.3 (Page 283) (vii) 3 + 29i Activity 11.5 (Page 287) z = 3 − 7i ⇒ z2 − 6z + 58 = (3 − 7i)2 − 6(3 − 7i) + 58 = 9 − 42i + 49i2 − 18 + 42i + 58 = 9 − 42i − 49 − 18 + 42i + 58 =0 (viii) 14 + 5i (ix) 40 + 42i 1 = p + iq x + iy ⇒ (p + iq)(x + iy) = 1 (x) 100 ⇒ px + ipy + iqx + iqy2 = 1 (xi) 43 + 76i ⇒ (px − qy) + i(py + qx) = 1 ? (xii) −9 + 46i 2 (Page 283) = −i, i4 = 1, i5 = i All numbers of the form i3 » Activity 11.4 (Page 284) 1 2 ± 2i (vi) −2 ± 2 i 3 (i) 2i (ii ) 5i and –3i (a) 6 (b) 2 (c) (d) (iii) 1 + i and –1 + i 34 5 (iv) 2 – 3i and –2 – 3i They are all real. (v) z + z∗ = (x + iy) + (x − iy) = 2x zz∗ = (x + iy)(x − iy) = x2 − ixy + ixy − i2y2 –1 – 4i and 1 – 4i (vi) –3i and 2i 4 = x2 + y2 These are real for any real values of x and y. ? 1 ± 2i (v) are equal to i » are equal to −1 4n+3 » i are equal to −i. (ii) (ii) (iv) −3 ± 5i i4n+2 (i) −1 ± i (iii) 2 ± 3i » i4n are equal to 1 i4n+1 (i) (i) 2 (ii) −4 Solving simultaneously gives –y x p= 2 ,q = 2 x + y2 x + y2 x − iy 1 so x + iy = 2 x + y2 ? (Page 289) 1 = −i, 1 = −1, 1 = i i i2 i3 All numbers of the form » 14n are equal to 1 i » 41n+1 are equal to –i i » 4n1+ 2 are equal to –1 i » 4n1+ 3 are equal to i. i Exercise 11B (Page 289) (i) 3 10 – 1 10 i (iii) 2 − 3i (ii) 6 37 + 1 37 i (iv) 6 + 4i (iii) – 41 + 43 i (iv) 4 5 5 2 (v) (Page 285) px − qy = 1 and py + qx = 0 1 8+i 11 + 10 i (vi) −4 − 7i (v) (vii) 0 (vi) 7 − 5i (viii) 0 Exercise 11A (Page 285) (vii) −i (ix) −39 1 (x) (viii) Yes, for example 2 ≠ 4 and 3 ≠ 6. 2 = 4 , although 3 6 (i) 14 + 10i (ii) 5 + 2i (xi) −46 − 9i (xii) 52i (iii) −3 + 4i (iv) −1 + i (v) 21 (vi) 12 + 21i −46 − 9i 7 a = 1 or 4, b = −1 or 3 The possible complex numbers are 1 + 9i, 1 + i; 16 + 9i, 16 + i 2 – 21 i (ix) 11 25 7 29 (x) –1 – (i) a = 5, b = 2 (ii) a = 3, b = −7 – + 27 25 i 32 29 i 3 2i (iii) a = 2, b = −3 (iv) a = 4, b = 5 346 9781510421738.indb 346 02/02/18 1:18 PM a= 5 4 , b = − 43 (vi) a = 1 1 (v) 3 ,b = (i) 2 z2 a = 2, b = 2 (i) z=2−i (ii) z=3+i 5 0, 2, −1 ± 3 i 6 2x x2 + y2 8 (i) a3 − 3ab2 (v) O (ii) Re Im z1 O z1 –z2 + − b3)i 5 (i) (z − α)(z − β) = z2 − (α + β)z + αβ (ii) (a) z2 + 4z + 12 = 0 (d) z2 − (5 + 3i)z + 4 + 7i = 0 (i) 3i and –3i (ii) 2 + i and –2 – i (iii) 3 + 5i and –3 – 5i (v) (ii) (i) 13 (ii) (iii) 26 (iv) 2 (v) 61 (vi) 5 –z Im Re (iii) The half-squares formed are enlarged by 2 and rotated through π each time. 4 iz O –z* (i) z and −z∗ (or −z and z∗) are reflections of each other in the imaginary axis. 2 4 z* iz* Activity 11.6 (Page 291) (Page 291) 2 1 2 Im (iz)* –iz ? – 1 i, z5 = −4 − 4i, | z5 | = 4 2 –6 – 5i 3 1 2 z4 = −4, | z4 | = 4 Re 4 – 3i (vi) 2 – 3i and –2 + 3i Rotation through 180° about the origin (ii) Reflection in the real axis z−1 = z3 = −2 + 2i, | z3 | = 2 2 3 + 2i –2 O 5 – 2i and –5 + 2i 13 5 z1 = 1 + i, | z1 | = 4i (iv) 3 – 4i and –3 + 4i 5 13 z2 = 2i, | z2 | = 2 –5 + i 2 (iv) z 0 = 1, | z 0 | = 1 (b) 9z2 + 25 = 0 (c) 13 | z−1 | = Im z2 − 14z + 65 = 0 (i) Re Exercise 11C (Page 294) 1 (ii) w ,w = z z , z1 + (–z2) (3a2b 5 z zw = z w , z = , w w z1 1 1 (iii) z = 1, − 2 ± 2 3 i 10 (i) (iii) 65 z2 – z1 (iii) z = 11 − 10i (iv) z = –35 + 149i 34 9 Im 4 Answers 4 2 Activity 11.7 (Page 293) z Re 6 Points: (i) 10 + 5i (ii) 1 + 2i (iii) 11 + 7i (iv) 9 + 3i (v) −9 − 3i ? Half a turn about O followed by reflection in the x-axis is the same as reflection in the x-axis followed by half a turn about O. (Page 294) | z2 − z1 | is the distance between the points representing z1 and z2 in the Argand diagram. 347 9781510421738.indb 347 02/02/18 1:18 PM ? ? (Page 296) (i) Im Answers (i) (Page 296) Im 3 + 4i 3 + 4i –1 + 2i O (ii) O Re Re (ii) Im Im 3 + 4i 3 + 4i –1 + 2i O Re O (iii) Re (iii) Im Im 3 + 4i 3 + 4i –1 + 2i O O Re Re Exercise 11D (Page 297) 1 (i) Im 2 O Re 348 9781510421738.indb 348 02/02/18 1:18 PM (ii) (vii) Im Re 4 O Answers O Im Re 1–i (iii) Im (viii) Im 5i –2 O (iv) O Re Re 2 Im Im A –3 + 4i Re 12 – 5i B O (v) Re | z | is least at A and greatest at B. Im | 12 − 5i | = 144 + 25 = 13 At A, | z | = 13 − 7 = 6 At B, | z | = 13 + 7 = 20 Re O 6–i 3 (i) O 2 4 6 8 Re –2 (vi) Im –4 O R 5 – 4i –6 Re –8 Im (ii) –2 – 4i 4 7, 13 Not possible 349 9781510421738.indb 349 02/02/18 1:18 PM 5 (i) (e) −88.9° Im Answers (ii) 2 Activity 11.9 (Page 300) 4 Re arg(1 + i) = π , 4 arg(1 − i) = − π , 4 arg(−1 + i) = 3π , 4 arg(−1 − i) = − 3π 4 Im 2i i O (iii) Re ? (Page 301) (i) 2(cos (π − α) + i sin (π − α)) (ii) 2(cos (−α) + i sin (−α)) Activity 11.11 (Page 301) Im π 4 π 6 tan 1 1 3 3 sin 1 2 1 2 3 2 cos 1 2 3 2 1 2 –1 + i O (iv) Re 1–i Im 2 + 6i Re O (i) 1 π r = 8, θ = 5 (ii) r = 41 , θ = 2.3 π (iii) r = 4, θ = − 3 (iv) r = 3, θ = π – 3 2 (i) (Page 298) π 2 – 3π 4 (iii) – π 4 (ii) (a) (i) r = 1, θ = 0, z = 1(cos 0 + i sin 0) r = 2, θ = π, z = 2(cos π + i sin π) π (iii) r = 3, θ = 2 , z = 3(cos π + i sin π ) 2 2 (ii) Activity 11.8 (Page 299) (i) π 3 Exercise 11E (Page 302) –5 – 7i ? −89.7° −90° " tan−1 x " 90° π – π " tan−1 x " 2 2 (ii) O (f) 45° (c) 89.4° (b) 63.4° (d) −63.4° 350 9781510421738.indb 350 02/02/18 1:19 PM 6 (i) 3 2 1 3π (vi) r = 5 2 , θ = − 4 , z = 5 2 (cos (− 3π ) + i sin (− 3π )) 4 4 –3 7 (ix) r = 5, θ = −0.927, z = 5(cos(−0.927) + i sin(−0.927)) 8 Real part = (i) (a) 2 + i (b) r = α−π (ii) −α (i) Im 2 5 (ii) Real part = A 1 O C 1 –1 2 3 4 5 Re B –2 OACB is a rhombus. (ii) 53 + 45 i 9 ? (iii) The locus is a circle, centre 2i, radius 5. (Page 304) arg(z1 − z2) is the angle between the line joining z1 and z2 and a line parallel to the real axis. Exercise 11F (Page 306) 1 (iii) π − α (iv) π − α 2 (v) π + α 2 3 Re 3 (i) (i) 2 5 , θ = 0.464 −3 + 2i and 3 − 2i (xii) r = 12013, θ = −2.128, z = 12013 (cos (−2.128)+ i sin (−2.128) 4 1 –2 1 4 (ii) r = 13, θ = 2.747, z = 13(cos 2.747 + i sin 2.747) z = 2i (ii) z = 3 + 3 3 i 2 2 7 3 7 (iii) z = – + i 2 2 1 1 – i (iv) z = 2 2 (v) z = – 5 – 5 3 i 2 2 (vi) z = −2.497 − 5.456i O –1 (ii) (xi) r = 65 , θ = 1.052, z = 65 (cos 1.052 + i sin 1.052) 3 –2 –1 (vii) r = 2, θ = − π , 3 π z = 2(cos (− 3 ) + i sin (− π )) 3 π (viii) r = 12, θ = 6 , z = 12(cos π + i sin π ) 6 6 (x) Im 4 Answers (iv) r = 4, θ = − π , 2 z = 4(cos (− π ) + i sin (− π )) 2 2 π (v) r = 2 , θ = , 4 z = 2 (cos π + i sin π ) 4 4 (i) Im O –π 3 Re 1 2 351 9781510421738.indb 351 02/02/18 1:19 PM (ii) (iii) The three points are the same Im distance from the origin and separated by equal angles of 2π (i.e. 120°). 3 (iv) –(2 + 3) + (2 3 – 1)i –(2 – 3) – (2 3 + 1)i Answers 4i O Re 4 (iii) Im 5 (ii) −6 & p & 2 (iii) z−5 =5 (i) 1 − 3i, − 1 − 3i (ii) –3 O Re 1 –3 (iv) Im 2 –2 Im –1 O –1 1 2 3 Re –2 O (iii) 1 − 3i: r = 2, θ = − π 3 −1 − 3i: r = 2, 2π θ=− 3 (iv) The three points are the same distance from the origin and separated by equal angles of 2π (i.e. 120°). 3 Re –1 – 2i (v) Im 6 O 3–i (vi) –π 6 7 (i) (ii) π 3 –π 4 (a) 2 , θ = – 43 π u2: r = 2, θ = 21 π u: r = Im 4 3 O Re 1 + 2i (b) – 21 + 21 i (ii) 3π 4 (iv) OA = BC and OA and BC are parallel Re Im –5 + 3i (i) 2 u2 1 O 2 π , 2π 3 3 3 (i) (ii) 352 9781510421738.indb 352 r = 1, θ = 23 π wz: modulus = R, argument = θ + 23 π 2 z w : modulus = R, argument = θ – 3π –4 –3 –2 –1 –1 u –2 1 2 3 4 Re –3 –4 02/02/18 1:19 PM 8 (i) (ii) ( 3 3 at z 2 = 3 2 (ix) 3 2 cos 7π + i sin 7π 12 12 − 23 i 3 1 2 (iii) 3 + (ii) Half turn of vector z ( = two successive π 2 rotations: –1 = i × i ) 4 Enlarge from O ×2 and rotate + π 2 (iii) Complete the parallelogram 3z, 0, 2iz (iv) Reflect in the real axis (v) Exercise 11G (Page 312) 32(cos0.6 + i sin 0.6) (ii) 2(cos( −0.2) + i sin( −0.2)) (iii) 12 cos π2 + i sin π2 ( ) ( ) 5π 24 cos 5π 4 + i sin 4 (i) (ii) (iii) (iv) (v) (vi) ( ) 6 ) ( ) 7π 6 (cos 7π 12 + i sin 12 ) (cos 12π + i sin 12π ) ⎡ π π ⎤ ⎢⎣cos ( – 12 ) + i sin ( – 12 )⎥⎦ ⎡ ⎤ cos − π + i sin (− π ) ⎥ 4 ⎦ ⎣⎢ ( 4 ) 2π 9 (cos 2π 3 + i sin 3 ) ⎡ ⎤ 32 ⎢cos ( – 34π ) + i sin ( – 34π )⎥ ⎣ ⎦ (i) –1 1+ i (ii) 2 (iii) –1.209 + 0.698i (iv) –13.129 + 15.201i 7 (i) 3 2 2 3 (ii) (a) 10e i (b) 4 (c) 6e 8i (d) 3e i (e) 3e 3i (f) 4e − i (a) 2(cos 3 + i sin 3) × 5(cos( −2) + i sin(−2)) = 10(cos 1 + i sin 1) 1 2 (vii) 432(cos 0 + isin 0) ( π π (viii) 10 cos 34 + i sin 34 ) + 2 cos π + i sin π ; 3 1 3 3 2 2 3π (vi) 6 cos 3π 4 + i sin 4 2 3 − 1, 3 + 1; 4 4 3π 2 cos + i sin 3π , 4 4 5 (i) (v) Find where the circle with centre O through z meets the positive real axis (vi) Complete the similar triangles 0, 1, z and 0, z, z2 3 + i and −3 − i ( Enlarge from O ×3 (i) (ii) (Page 309) (iv) 3 cos π6 + i sin π6 ( z1 ) = arg z Answers Rotation of vector z through + π 2 ( if z = 0 then 1 does not exist z (iii) if z = real and negative then arg (i) 1 Exceptions (i) Activity 11.12 (Page 308) ? ) ) (b) 8(cos 5 + i sin 5) ÷ 2(cos 5 + i sin 5) =4 (c) 3(cos 7 + i sin 7) × 2(cos 1 + i sin 1) = 6(cos 8 + i sin 8) 353 9781510421738.indb 353 02/02/18 1:19 PM (d) 12(cos 5 + i sin 5) ÷ 4(cos 4 + i sin 4) = 3(cos 1 + i sin 1) 2 2 – i, –3 3 z = 7, 4 ± 2i 3(cos 2 + i sin 2) × (cos 1 + i sin 1) = 3(cos 3 + i sin 3) 4 p = 4, q = –10, other roots 1 + i, –6 5 z = 3 ± 2i, 2 ± i 8(cos 3 + i sin 3) ÷ 2(cos 4 + i sin 4) = 4(cos( −1) + i sin(−1)) 6 z = ±3i, 4 ± 5 7 7, 4 ± 2i 8 (i) z = –3 (ii) z = –3, 5 ± 11 i 2 2 Answers (e) (f) 8 (i) −16.3° (ii) Im 5 −5 O 2.5 5 Re 9 k = 36, other roots are − 5 ± 3 3 i 2 2 10 z3 − z – 6 = 0 11 (i) (ii) False (iii) True −5 Complex numbers common to both loci: 5e ± 3 π i 1 9 (i) 7 23 − 17 + 17 i (ii) wz = 17 + 17i Exercise 11H (Page 315) 1 True (iv) True 12 a = 2, b = 2, z = −2 ± i,1 ± 2i 13 (ii) (iii) 1 – 2i Im 2 O 1 Re 4 + 5i is the other root. The equation is z2 − 8z + 41 = 0 354 9781510421738.indb 354 02/02/18 1:19 PM Index A B binomial expansion general 163–71 use of partial fractions 180–2 binomial theorem 166 C change-of-sign methods 146–51 cobweb diagram 154, 155 common factors 173 complex conjugates 283–4, 313 complex exponents 310–13 complex numbers addition and subtraction 291–3 definition 283 equality 284–6 and equations 313–16 geometrical representation 291 modulus 293–4 modulus–argument (polar) form 297–304, 307–10 multiplication and division 287–8, 307–10 need for 281–2, 313 notation 282 real and imaginary parts 283 square root 288–90 complex plane 291 compound interest 128 compound-angle formulae 61–6, 72–8 convergence to a limiting value 151 cosecant (cosec) 58 cotangent (cot) 58 cubic expressions see polynomials curves, discontinuous 125 D decimal search 146–8 Devi, Shakuntala 162 9781510421738.indb 355 differential equations forming from rates of change 218–22 general solution 223 particular solution 225–32 verifying the solution 228–9 differentiation of exponentials 90–3 implicit 184 of implicitly defined functions 101–8 of inverse trigonometrical functions 184–816 of natural logarithms 90–3 parametric 112–18 product rule 84–6 quotient rule 86–7 of trigonometrical functions 96–101 direction, of a line 261–2 displacement vectors 238–9 displacement-time graphs 83 distance, from a point to a line 273–7 division see multiplication and division dot product 251–3 double-angle formulae 67–71 E e, base of natural logarithms 38, 46–7 equation of a line 37 vector form 258–65 equations and complex numbers 313–16 rearranging 152–6 error, bounds 149 estimates, accuracy 135 existence theorem 314 experimental data, curve fitting 37–41 exponential functions applications 39–41, 48–52 and complex numbers 310–13 definition 24, 34, 48–52 differentiation and integration 90–3, 121–2, 191–4 graphs 25–6 growth and decay 24, 26–7, 49 as infinite series 128 F factor theorem 8–10 fixed-point iteration 151–61 fractions addition and subtraction 173 multiplication and division 172 simplification 172 Fundamental Theorem of Algebra 313–14 G Index acceleration, definition 218 addition and subtraction of complex numbers 291–3 of fractions 173 of polynomials 2–3 of vectors 241–3 algebraic fractions 172–4 alternating current (AC) 78 angle between two lines 272–3 between two vectors 250–8 Argand diagrams 291–3, 294–7, 304–7 gradient of exponential curves 25 of a straight line 37 graphs displacement–time 83 of exponential functions 25–6 of logarithms 33 of the natural logarithm function 48 of parametric equations 110 of polynomial equations 7 of trigonometrical functions 58–9 growth and decay, exponential 24, 26–7, 49 I identities see trigonometrical identities identity symbol 199 imaginary number 282 implicit functions, differentiation 101–8 indices, laws 30 inequalities, involving the modulus sign 19–21 integrals involving exponentials 191–4 involving natural logarithms 191–4 standard 213 integration by change of variable 186–91 by parts 203–12 by substitution 186–91 choice of technique 212–14 of exponentials 90–3 involving the exponential function 121–2 involving the natural logarithm function 122–7 of natural logarithms 90–3 numerical 133–40 of trigonometrical functions 129–33, 194–9 use of partial fractions 199–203 intervals bisection 148 estimation 146–51 notation 145 inverse functions 34, 48 iteration 151, 153, 155 355 02/02/18 1:19 PM Index L line segment 239 lines angle between 272–3 direction 261–2 gradient 37 intersection 265–71 location on 262–3 perpendicular distance from 273–7 skew 266, 273 location, on a line 262–3 logarithm function 34, 46, 47 inverse 34 logarithmic integrals 123–4 logarithmic relationships 38–9 logarithms 28–33 see also natural logarithms lowest common multiple 173 M magnitude of a quantity 17 of a vector 236 modelling use of logarithms 37–41 use of trigonometric functions 57, 61, 83 of waves 57, 61, 75, 184 modulus of a complex number 293–4 of a vector 236 modulus function 17–21 multiplication and division of complex numbers 287–8, 307–10 of fractions 172 of logarithms 30 of polynomials 3–6 of a vector by a scalar 240–1 N Napier, John 48 natural logarithm function 45–8, 122–7 natural logarithms base 38 differentiation and integration 90–3, 191–4 natural numbers 163 negative number, square root 282 Newton’s law of cooling 217 numbers, types 280, 281–2 numerical methods accuracy 135, 144–6 integration 133–40, 142–3 for polynomial equations 8 P parameter, eliminating 111–12 parametric differentiation 112–18 parametric equations 108–12 partial fractions types 174–80 use in integration 199–203 using with the binomial expansion 180–2 Pascal’s triangle 164 point coordinates 237 perpendicular distance from a line 273–7 polynomial equations graphs 7 roots 11–12 solution 7–16 with complex numbers 313 polynomials addition and subtraction 2–3 definition 1–2 multiplication and division 3–6 order 2 position, definition 218 position vector 237–8, 262 principal argument, of a complex number 298 principal value 69, 79 product rule 84–6 spiral dilatation 309 square root of a complex number 288–90 of a negative number 282 staircase diagram 154, 155 stationary points of implicit functions 103–6 of a parametric curve 115 straight line see equation of a line, line segment; lines subtraction see addition and subtraction Q unit vector 235, 245–7 quadratic expressions 1 quadratic formula 7, 313 quotient 4, 6 quotient rule 86–7 R Ramanujan, Srinivasa 1 rate of change 218 rational functions see algebraic fractions real and imaginary axes 291 reciprocal trigonometrical functions 58–60 remainder theorem 12–13 resultant 242 roots of logarithms 32 number 75, 145 of polynomial equations 7–8, 11–12 of trigonometric equations 79–80 S scalar, definition 234 scalar product 251–3 secant (sec) 58 separation of variables 223–5 sets of points, on an Argand diagram 294–7, 304–7 sine function, differentiating 83 T trapezium rule 133–40 trial and improvement 144 trigonometrical equations, general solution 79–80 trigonometrical functions differentiation 96–101 graphs 58–9 integration 129–33, 194–9 inverse, differentiation 184–6 reciprocal 58–60 trigonometrical identities 59, 61, 67, 130–1 U V vector product 252 vectors addition and subtraction 241–3 angle between 250–8 calculations 240–50 components 235 definition 234 equal 237 of geometrical figures 244 joining two points 258–9 length 239 magnitude 236 magnitude—direction (polar) form 235 multiplication by a scalar 240–1 negative of 241 notation and terminology 218, 235–6 perpendicular 251–2 polar form 235 in three dimension 236–40, 252–3 in two dimensions 235–6 uses in mathematics 234 velocity, definition 218 W waves, modelling 57, 61, 75, 184 356 9781510421738.indb 356 02/02/18 1:19 PM Take mathematical understanding to the next level with this accessible series, written by experienced authors, examiners and teachers. The five Student’s Books fully cover the Cambridge International AS & A Level Mathematics syllabus (9709) for examination from 2020. Each is accompanied by a Workbook*, and Student and Whiteboard eTextbooks. This series is endorsed by Cambridge Assessment International Education to support the full syllabus for examination from 2020. Student’s Book Student eTextbook Whiteboard eTextbook Workbook Pure Mathematics 1 9781510421721 9781510420762 9781510420779 9781510421844 Pure Mathematics 2 and 3 9781510421738 9781510420854 9781510420878 9781510421851 Mechanics 9781510421745 9781510420953 9781510420977 9781510421837 Probability & Statistics 1 9781510421752 9781510421066 9781510421097 9781510421875 Probability & Statistics 2 9781510421776 9781510421158 9781510421165 9781510421882 * The Workbooks have not been through the Cambridge International endorsement process. 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