Edexcel A Level Further Maths Core Pure Book 2

11 - 19 PROGRESSION Edexcel A level Further Mathematics Core Pure Mathematics Book 2 Series Editor: Harry Smith • Fully updated to match the 2017 specifications, with more of a focus on problem-solving and modelling. • FREE additional online content to support your independent learning, including full worked solutions for every question in the book (SolutionBank) and GeoGebra interactives. • Includes access to an online digital edition (valid for 3 years once activated). • Includes worked examples with guidance, lots of exam-style questions, practice papers, and plenty of mixed and review exercises. Pearson Edexcel AS and A level Mathematics books Year 1/AS Year 2 Options For more information visit: www.pearsonschools.co.uk/edalevelmaths2017 www.pearsonschools.co.uk myorders@pearson.com 2 Edexcel A level Further Mathematics Core Pure Mathematics Book 2 line digital ion Compulsory on it ed Pearson Edexcel AS and A level Further Mathematics books Edexcel A level Further Mathematics Core Pure Maths Book 2 This book can be used alongside Book 1 to cover all the content needed for the compulsory Edexcel A level Core Pure Mathematics exams. ludes a inc n Pearson’s market-leading books are the most trusted resources for Edexcel AS and A level Further Mathematics. 11 – 19 PROGRESSION Edexcel A level Further Mathematics Core Pure Mathematics Book 2 Series Editor: Harry Smith Authors: Greg Attwood, Jack Barraclough, Ian Bettison, Lee Cope, Alistair Macpherson, Bronwen Moran, Johnny Nicholson, Laurence Pateman, Joe Petran, Keith Pledger, Harry Smith, Geoff Staley, Dave Wilkins F01_ALEVEL_CP2_83343_PRE_i-vi.indd 1 26/02/2018 16:26 Published by Pearson Education Limited, 80 Strand, London WC2R 0RL. www.pearsonschoolsandfecolleges.co.uk Copies of official specifications for all Pearson qualifications may be found on the website: qualifications.pearson.com Text © Pearson Education Limited 2018 Edited by Tech-Set Ltd, Gateshead Typeset by Tech-Set Ltd, Gateshead Original illustrations © Pearson Education Limited 2018 Cover illustration Marcus@kja-artists The rights of Greg Attwood, Jack Barraclough, Ian Bettison, Lee Cope, Alistair Macpherson, Bronwen Moran, Johnny Nicholson, Laurence Pateman, Joe Petran, Keith Pledger, Harry Smith, Geoff Staley, Dave Wilkins to be identified as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. First published 2018 21 20 19 18 10 9 8 7 6 5 4 3 2 1 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN 978 1 292 18334 3 (Print) ISBN 978 1 292 20727 8 (PDF) Copyright notice All rights reserved. No part of this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, Barnards Inn 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk). Applications for the copyright owner’s written permission should be addressed to the publisher. Printed in the UK by Bell & Bain Ltd, Glasgow Acknowledgements The authors and publisher would like to thank the following for their kind permission to reproduce their photographs: (Key: b-bottom; c-centre; l-left; r-right; t-top) 123RF: 170, 196r, Cobalt 77, 93r, Alamy Stock Photo: NASA Photo 52, 93cr, Getty Images: SteveDF 1, 93 l, SCIENCE PHOTO LIBRARY: Andrew Brookes, National Physical Laboratory 31, 93cl, Shutterstock: Tyler Olson 100, 196l, Spacedrone808 119, 196cl, Tatiana Shepeleva 147, 196cr. All other images © Pearson Education A note from the publisher In order to ensure that this resource offers high-quality support for the associated Pearson qualification, it has been through a review process by the awarding body. This process confirms that this resource fully covers the teaching and learning content of the specification or part of a specification at which it is aimed. 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If you spot an error, please do contact us at resourcescorrections@pearson.com so we can make sure it is corrected. ii F01_ALEVEL_CP2_83343_PRE_i-vi.indd 2 27/02/2018 11:57 Contents Contents Overarching themes iv Extra online content vi 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 2 2.1 2.2 2.3 2.4 3 3.1 3.2 3.3 3.4 3.5 4 4.1 4.2 4.3 4.4 Complex numbers 1 Exponential form of complex numbers 2 Multiplying and dividing complex numbers5 De Moivre’s theorem 8 Trigonometric identities 11 Sums of series 16 nth roots of a complex number 20 Solving geometric problems 25 Mixed exercise 1 27 Series31 The method of differences 32 Higher derivatives 38 Maclaurin series 40 Series expansions of compound functions44 Mixed exercise 2 48 Methods in calculus52 Improper integrals 53 The mean value of a function 58 Differentiating inverse trigonometric functions62 Integrating with inverse trigonometric functions65 Integrating using partial fractions 69 Mixed exercise 3 74 Volumes of revolution77 Volumes of revolution around the x-axis78 Volumes of revolution around the y-axis81 Volumes of revolution of parametrically 83 defined curves Modelling with volumes of revolution 87 Mixed exercise 4 89 Review exercise 1 93 5 5.1 5.2 5.3 5.4 Polar coordinates100 Polar coordinates and equations 101 Sketching curves 104 Area enclosed by a polar curve 109 Tangents to polar curves 113 Mixed exercise 5 116 6 6.1 6.2 6.3 6.4 6.5 Hyperbolic functions119 Introduction to hyperbolic functions 120 Inverse hyperbolic functions 123 Identities and equations 125 Differentiating hyperbolic functions 130 Integrating hyperbolic functions 135 Mixed exercise 6 142 7 7.1 7.2 Methods in differential equations147 First-order differential equations 148 Second-order homogeneous differential equations153 Second-order non-homogeneous 157 differential equations Using boundary conditions 162 Mixed exercise 7 165 7.3 7.4 8 8.1 8.2 8.3 8.4 Modelling with differential equations170 Modelling with first-order differential equations171 Simple harmonic motion 175 Damped and forced harmonic motion 180 Coupled first-order simultaneous differential equations 186 Mixed exercise 8 191 Review exercise 2196 Exam-style practice: Paper 1 209 Exam-style practice: Paper 2211 Answers213 Index256 iii F01_ALEVEL_CP2_83343_PRE_i-vi.indd 3 26/02/2018 16:26 Overarching themes Overarching themes The following three overarching themes have been fully integrated throughout the Pearson Edexcel AS and A level Mathematics series, so they can be applied alongside your learning and practice. 1. Mathematical argument, language and proof • Rigorous and consistent approach throughout • Notation boxes explain key mathematical language and symbols • Dedicated sections on mathematical proof explain key principles and strategies • Opportunities to critique arguments and justify methods 2. Mathematical problem solving • Hundreds of problem-solving questions, fully integrated into the main exercises • Problem-solving boxes provide tips and strategies • Structured and unstructured questions to build confidence • Challenge boxes provide extra stretch 3. Mathematical modelling The Mathematical Problem-solving cycle specify the problem interpret results collect information process and represent information • Dedicated modelling sections in relevant topics provide plenty of practice where you need it • Examples and exercises include qualitative questions that allow you to interpret answers in the context of the model • Dedicated chapter in Statistics & Mechanics Year 1/AS explains the principles of modelling in mechanics Finding your way around the book Each chapter starts with a list of objectives The Prior knowledge check helps make sure you are ready to start the chapter Access an online digital edition using the code at the front of the book. The real world applications of the maths you are about to learn are highlighted at the start of the chapter with links to relevant questions in the chapter iv F01_ALEVEL_CP2_83343_PRE_i-vi.indd 4 26/02/2018 16:26 Overarching themes Exercise questions are carefully graded so they increase in difficulty and gradually bring you up to exam standard Exercises are packed with examstyle questions to ensure you are ready for the exams Challenge boxes give you a chance to tackle some more difficult questions Exam-style questions E are ﬂagged with E Problem-solving questions are ﬂagged with P Each section begins with explanation and key learning points Each chapter ends with a Mixed exercise and a Summary of key points Step-by-step worked examples focus on the key types of questions you’ll need to tackle Problem-solving boxes provide hints, tips and strategies, and Watch out boxes highlight areas where students often lose marks in their exams Every few chapters a Review exercise helps you consolidate your learning with lots of exam-style questions Two A level practice papers at the back of the book help you prepare for the real thing. v F01_ALEVEL_CP2_83343_PRE_i-vi.indd 5 26/02/2018 16:26 Extra online content Extra online content Whenever you see an Online box, it means that there is extra online content available to support you. Differentiation 12A SolutionBank SolutionBank provides a full worked solution for every question in the book. Online Full worked solutions are available in SolutionBank. Download all the solutions as a PDF or quickly find the solution you need online 1 a Examples of estimates of gradients: Gradient of tangent at x = −1 is y2 − y1 3 −1 = x2 − x1 (−1) − (−0.5) = −4 Gradient of tangent at x = 0 is y2 − y1 1 − (−1) = x2 − x1 (−0.5) − (0.5) = −2 Gradient of tangent at x = 1 is y2 − y1 (−1) − (−1) = x2 − x1 2−0 =0 Gradient of tangent at x = 2 is y2 − y1 (−1) − 1 = x2 − x1 1.5 − 2.5 =2 Gradient of tangent at x = 3 is y2 − y1 3 −1 = =4 x2 − x1 3 − 2.5 x-coordinate Estimate for gradient of curve −1 0 1 2 3 −4 −2 0 2 4 2 c i Gradient of AD = y2 − y1 x2 − x1 0.8 − 0.19 = 0.6 − 0.9 = −1.21 (3 s.f.) ii Gradient of AC = y2 − y1 x2 − x1 0.8 − 0.6 = 0.6 − 0.8 = −1 iii Gradient of AB = y2 − y1 x2 − x1 0.8 − 0.51 = 0.6 − 0.7 = − 0.859 (3 s.f.) d As the points move closer to A, the gradient tends to − 0.75. 3 a i Gradient= 16 − 9= 7= 7 4−3 1 12.25 − 9 3.25 = = 6.5 3.5 − 3 0.5 b The gradient of the curve at the point where x = p is 2p − 2. ii Gradient = c Gradient of tangent at x = 1.5 is y2 − y1 (−1.7) − 0.3 = x2 − x1 0.5 − 2.5 =1 2p − 2 = 2(1.5) − 2 = 1 iii Gradient = 9.61 − 9 0.61 = = 6.1 3.1 − 3 0.1 iv Gradient = 9.0601 − 9 0.0601 = = 6.01 3.01 − 3 0.01 v Gradient = ( 3 + h )2 − 9 (3 + h) − 3 2 a Substituting x = 0.6 into = y 1 − x2 : y = 1 − 0.6 2 = 0.64 = 0.8 , therefore the point A (0.6, 0.8) lies on the curve. 6h + h 2 h h (6 + h) = h =6+h = b Gradient of tangent at x = 0.6 is y2 − y1 1.1 − 0.8 = x2 − x1 0.2 − 0.6 = − 0.75 © Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 1 Use of technology Explore topics in more detail, visualise problems and consolidate your understanding using pre-made GeoGebra activities. Online Find the point of intersection graphically using technology. GeoGebra-powered interactives Interact with the maths you are learning using GeoGebra's easy-to-use tools Access all the extra online content for free at: www.pearsonschools.co.uk/cp2maths You can also access the extra online content by scanning this QR code: vi F01_ALEVEL_CP2_83343_PRE_i-vi.indd 6 26/02/2018 16:26 1 Complex numbers Objectives After completing this chapter you should be able to: ● Express a complex number in exponential form → pages 2–5 ● Multiply and divide complex numbers in exponential form ● Understand de Moivre’s theorem → pages 5–8 → pages 8–11 ● Use de Moivre’s theorem to derive trigonometric identities → pages 11–15 ● Use de Moivre’s theorem to find sums of series → pages 16–19 ● Use complex roots of unity to solve geometric problems → pages 25–27 ● Know how to solve completely equations of the form zn − a − ib = 0, → pages 20–25 giving special attention to cases where a = 1 and b = 0 Prior knowledge check 1 __ π π z = 4 + 4i√ 3  and w = 2(cos  __  + i sin  __ ). 6 6 Find: a |z| b arg(z) z __     e w z f arg(__    ) w | | 2 The relationships between complex numbers and trigonometric functions allow electrical engineers to analyse oscillations of voltage and current in electrical circuits more easily. c |zw| d arg(zw) ← Book 1, Chapter 2 f(z) = z4 + 4z3 + 9z2 + 4z + 8 Given that z = i is a root of f(z) = 0, show all the roots of f(z) = 0 on an Argand ← Book 1, Chapters 1, 2 diagram. 3 Use the binomial expansion to find the n4 term in the expansion of (2 + n)9. ← Pure Year 1, Chapter 8 1 M01_ALEVEL_CP2_83343_U01_001-030.indd 1 26/02/2018 09:17 Chapter 1 1.1 Exponential form of complex numbers You can use the modulus−argument form of a complex number to express it in the exponential form: z = reiθ. Links The modulus−argument form of a complex number is z = r(cos θ + i sin θ), where r = |z| and θ = arg z. You can write cos θ and sin θ as infinite series of powers of θ: θ 2 θ 4 __ θ6 (−1)r θ 2r     + __     −     + … + _______     + … cos θ = 1 − __ 2! 4! 6! (2r)! ← Book 1, Section 2.3 (1) θ 3 θ 5 __ θ7 (−1)r θ 2r + 1     + __     −     + … +  _________  + … (2) sin θ = θ − __ 3! 5! 7! (2r + 1)! You can also write ex, x ∈ ℝ, as a series expansion in powers of x. Links These are the Maclaurin series expansions of sin θ, cos θ and ex . 2 x3  +  __ x4  +  __ x5  + … +  __ xr  + …  x   +  __ ex = 1 + x + __ r! 2! 3! 4! 5! → Chapter 2 You can use this expansion to define the exponential function for complex powers, by replacing x with a complex number. In particular, if you replace x with the imaginary number iθ, you get (iθ )2 (iθ )3 ____ (iθ )4 (iθ )5 ____ (iθ )6     + ____ eiθ = 1 + iθ + ____     +     + ____     +     + … 2! 3! 4! 5! 6! 2 2 3 3 4 4 5 5 6 6   i θ   + ____   i θ   + ____   i θ   + ____   i θ   + … = 1 + iθ + ____   i θ   + ____ 2! 3! 4! 5! 6! 2 3 4 5 6  θ   − ___   iθ   + __  θ   + …  θ   + ___   iθ   − __ = 1 + iθ − __ 2! 3! 4! 5! 6! ( ) ( ) 2 4 6 3 5  θ   + __  θ   + … + i θ − __  θ   + __   θ   − __   θ   − … = 1 − __ 2! 4! 6! 3! 5! By comparing this series expansion with (1) and (2), you can write eiθ as eiθ = cos θ + i sin θ This formula is known as Euler’s relation. It is important for you to remember this result. ■ You can use Euler’s relation, eiθ = cos θ + i sin θ, to write a complex number z in exponential form: z=r where r = |z| and θ = arg z. eiθ Note Substituting θ = πinto Euler’s relation yields Euler’s identity: eiπ + 1 = 0 This equation links the five fundamental constants 0, 1, π , e and i, and is considered an example of mathematical beauty. 2 M01_ALEVEL_CP2_83343_U01_001-030.indd 2 26/02/2018 09:17 Complex numbers Example 1 Express the following in the form r eiθ, where −π , θ < π. __ π π π π a z=√ 2  cos ___     + i sin ___ b z = 5 cos __     − i sin __          8 8 10 10 ( ( ) ( ) ) __ π π a z = √ 2 cos __ 10 + i sin __ 10 __ π So r = √ 2 and θ = __ 10 Compare with r(cos θ + i sin θ). πi __ __ Therefore, z = √ 2 e10 ( z = reiθ ) π π b z = 5 cos __− i sin __ 8 8 ( ( ) ( )) Problem-solving π π z = 5 cos − __ + i sin − __ 8 8 Use cos (−θ) = cos θ and sin (−θ) = −sin θ. π So r = 5 and θ = − __ 8 Compare with r(cos θ + i sin θ). πi −  __ Therefore, z = 5e 8 Example z = reiθ. 2 Express z = 2 − 3i in the form r eiθ, where −π , θ < π. Im O α Sketch the Argand diagram, showing the position of the complex number. 2 Re r 3 z = 2 – 3i ___________ __ r = |z| = √22 + (−3)2  = √ 13  θ = arg z = −arctan (__   32 ) = −0.983 (3 s.f.) __ Here z is in the fourth quadrant so the required argument is −α. Therefore, z = √13 e−0.983i Find r and θ. z = reiθ. 3 M01_ALEVEL_CP2_83343_U01_001-030.indd 3 26/02/2018 09:17 Chapter 1 Example 3 3πi __ ____     Express z = √ 2 e 4 in the form x + iy, where x, y ∈ ℝ. __ ___ __     3π z=√ 2 e 4 , so r = √2  and θ = ___     4 Compare with reiθ. __ 3π 3π z = √2  cos  ___  + i sin  ___   4 4 Write z in modulus–argument form. 3πi ( ( __ ) ) 1__  + i  ___ 1__   = √2  − ___ √ √ 2  2  Therefore, z = −1 + i Example Simplify. 4 23πi _____     Express z = 2e 5 in the form r (cos θ + i sin θ), where −π , θ < π. 23πi  _____   23π z = 2e 5 , so r = 2 and θ = ____     5 23π 13π 13π 3π ____     − 2π = ____    , ____     − 2π = ___     5 5 5 5 3π ___    is in the range −π , θ < π 5 3π 3π So z = 2 cos  ___  + i sin  ___   5 5 ( Compare with reiθ. Problem-solving cos θ = cos (θ + 2π) and sin θ = sin (θ + 2π). 23π Subtract multiples of 2π from ____    until you find a 5 value in the range −π , θ < π. ) Write z in the form r (cos θ + i sin θ ). Example 5 Use eiθ = cos θ + i sin θ to show that cos θ = _ 2 (eiθ + e−iθ ). 1 eiθ = cos θ + i sin θ −iθ i(−θ) e = e (1) = cos (−θ) + i sin (−θ) So e = cos θ − i sin θ −iθ (2) e + e = 2 cos θ iθ Use cos (−θ ) = cos θ and sin (−θ ) = −sin θ. −iθ eiθ + e−iθ ⇒ ________     = cos θ 2 ), as required. Hence, cos θ =  _21   (eiθ + e−iθ Add (1) and (2). Divide both sides by 2. 4 M01_ALEVEL_CP2_83343_U01_001-030.indd 4 26/02/2018 09:17 Complex numbers 1A Exercise 1 Express the following in the form r eiθ, where −π , θ < π. Use exact values of r and θ where possible, or values to 3 significant figures otherwise. __ a −3 b 6i c −2√ 3  − 2i d −8 + i e 2 − 5i f −2√ 3  + 2i√ 3  i 2 ( cos  __π5  − i sin  __π5   ) __ π π g√ 8 ( cos  __  + i sin  __   ) 4 4 __ π π h 8cos  __  − i sin  __   6 6 ( ) 2 Express the following in the form x + iy where x, y ∈ ℝ. πi __     b 4eπi ae 3 πi __     5πi ____     d 8e 6 e 3e 2 ge−πi h 3√ 2 e 4 __ πi __ c 3√ 2  e 4 πi − __      __ __ fe 6 4πi − ____  3π i −___  i 8e 3 3 Express the following in the form r (cos θ + i sin θ), where −π , θ < π. 16πi ____   P 9πi − ___  8 17πi ____       b 4e 5 ae 13 c 5e 1 4 Use e iθ= cos θ + i sin θ to show that sin θ = __     ( e iθ− e−iθ ). 2i 1.2 Multiplying and dividing complex numbers You can apply the modulus−argument rules for multiplying and dividing complex numbers to numbers written in exponential form. Recall that, for any two complex numbers z  1 and z  2, ●|z  1 z  2| = | z  1||z  2| ●arg (z  1 z  2) = arg (z  1) + arg (z  2) z  1| |___ ●     =   z  2 |z  2|  z  1 __ || (z  2 ) = arg (z  1)− arg (z  2) z  1 __ ●arg   Links These results can be proved by considering the numbers z  1 and z  2 in the form r(cos θ + i sin θ) and using the addition formulae for cos and sin. ← Book 1, Section 2.3 Applying these results to numbers in exponential form gives the following result: ■ If z1 = r1e  iθ  and z2 = r2e  iθ  , then: 1 • z1z2 = r1r2e   i(θ1  + θ2  ) z1 __ r1 i(θ  − θ  ) 1 2   =   •  __ z r   e  2 2 2 Watch out You cannot automatically assume the laws of indices work the same way with complex numbers as with real numbers. This result only shows that they can be applied in these specific cases. 5 M01_ALEVEL_CP2_83343_U01_001-030.indd 5 26/02/2018 09:17 Chapter 1 Example 6 __ __ πi πi __ a Express 2e 6 × √ 3 e 3 in the form x + iy. b z = 2 + 2i, Im(zw) = 0 and |zw| = 3|z| Use geometrical reasoning to find the two possibilities for w, giving your answers in exponential form. __ πi __ __ __ πi a 2e 6 × √ 3  e 3 = (2 × √ 3 ) ei ( 6   +  3 ) __ __ π __ π z 1z 2 = r 1r  2e  i(θ1  +θ2  ) __ πi = 2√ 3  e 2 π __ __ Simplify. π __ = 2√ 3 (cos    + i sin   ) 2 2 Convert the complex number to modulus− argument form. __ = 2√ 3 (0 + i) __ = 2i√ 3  b |zw| = 3|z| ⇒ |w| = 3 2 π    ) = __     arg z = arctan (__ 2 4 |zw| = |z||w| = 3|z|. Im(zw) = 0 so arg (zw) = 0 or π 3π π So arg w = ___     or −  __  4 4 Im 3π wz lies on the real axis, so z is rotated ___     4 π clockwise or __    anticlockwise when multiplied by w. 4 z = 2 + 2i 3π 4 zw2 π zw1 π 4 4 O ___ πi Re ___ 3πi w1 = 3e− 4  and w2 = 3e    4 Example 7 π π 2(cos  ___  + i sin  ___ ) 12 12 __________________ Express          in the form reiθ. __ 5π 5π ___ ___ √ 2  (cos     + i sin    ) 6 6 π π __ πi 2(cos  ___  + i sin  ___ ) 12 12 2e 12 ___________________ ______           =   __ ___   5πi __ 5π 5π ___ ___ √ 2 e 6 √ 2 (cos     + i sin    ) 6 6 5π π − __ 2__ i(__   e  12   6 ) =  ___ √ 2  __ Convert the numerator and denominator to exponential form. r  1  r    e  i(θ1  −θ2  )  z    = __ 2 2 z  1 __ 3πi − ___ = √ 2  e   4 Simplify. 6 M01_ALEVEL_CP2_83343_U01_001-030.indd 6 26/02/2018 09:17 Complex numbers 1B Exercise 1 Express the following in the form x + iy, where x, y ∈ ℝ. πi __ __ πi __ __ ae   3 × e   4 πi ___ ___ __ c√ 2 e   3 × e −   3 × 3e   6 √ 5 e  iθ × 3e  3iθ b 2πi 7πi 2 Express the following in the form x + iy where x, y ∈ ℝ. __ ___ 3πi   2 ___ a  ___   9πi 8e   2 __ √ 3 e   7 _____ 7πi 2e ____ b  c    4e −  7 2πi ___ 3 Express the following in the form reiθ. a (cos 2θ + i sin 2θ)(cos 3θ + i sin 3θ) π π c 3   + i sin  ___   ) ( cos  __π4  + i sin  __π4   ) × 2( cos  ___ 12 12 ( ) 15πi ___ √ 2 e −  6 ______ __ ___ 19πi   × √ 2  e   3 2e   3 πi __ )( ( ) 8π 8π 3π 3π bcos  ___  + i sin  ___   cos  ___  + i sin  ___   11 11 11 11 __ __ π π π π d√ 6  cos ( − ___   ) + i sin ( −  ___   )  × √ 3  cos  __  + i sin  __   3 3 12 12 ( ) 4 Express the following in the form reiθ.__ π π √ 2 (cos  __  + i sin  __ ) 2 2 cos 5θ + i sin 5θ a  _____________       b ________________       π π   _1 __ __ cos 2θ + i sin 2θ   2 (cos     + i sin    ) 4 4 π π 3(cos  __  + i sin  __ ) 3 3        c _________________ 5π 5π 4(cos  ___  + i sin  ___ ) 6 6 __ __ 7π     5 z and w are two complex numbers where z = −9 + 3i √ 3 , |w | = √ 3  and arg w = ___ 12 Express the following in the form reiθ, where −π , θ < π. z b w c zw d __ a z w P 6 Use the exponential form for a complex number to show that (cos 9θ + i sin 9θ)(cos 4θ + i sin 4θ) ____________________________         ≡ cos 6θ + i sin 6θ cos 7θ + i sin 7θ E/P __ z  2 z  2 7 z = 1 + i√ 3 , Re(__   w ) = 0 and __   w  = |z| | | Use geometrical reasoning to find the two possibilities for w, giving your answers in exponential form. (4 marks) E/P 8 a Evaluate (1 + i)2, giving your answer in exponential form. nπi _n ___ b Use mathematical induction to prove that (1 + i)  n = 2 2 e 4 for n ∈ ℤ+. c Hence find (1 + i)16. P (2 marks) (4 marks) (1 mark) 9 Use Euler’s relation for eiθ and e−iθ to verify that cos2 θ + sin2 θ ≡ 1. 7 M01_ALEVEL_CP2_83343_U01_001-030.indd 7 26/02/2018 09:17 Chapter 1 Challenge Watch out You cannot assume that the laws of indices will apply to complex numbers. Prove these results using only the properties a Given that n is a positive integer, prove by induction that (re  iθ)  n= r  ne  inθ 1 b Given further that z  −n = __   n  for all z ∈ 픺, show that z   ( re  iθ)  −n= r  −ne  −inθ z 1z  2 = r 1r  2e  i(θ1  +θ2  ) r  1 __ i(θ1  −θ2  ) z    2  = r    2  e  z  1 __ 1.3 De Moivre’s theorem You can use Euler’s relation to find powers of complex numbers given in modulus−argument form. ( r(cos θ + i sin θ))  2 = ( re  iθ)  2 = re  iθ × re  iθ = r  2 e  i2θ = r  2(cos 2θ + i sin 2θ) Similarly, ( r(cos θ + i sin θ))  3 = r  3( cos 3θ + i sin 3θ), and so on. The generalisation of this result is known as de Moivre’s theorem: ■ For any integer n, n (r(cos θ + i sin θ))  n= r   (cos nθ + i sin nθ) You can prove de Moivre’s theorem quickly using Euler’s relation. (r(cos θ + i sin θ))  n= ( re  iθ)  n = r  n e  inθ = r  n( cos nθ + i sin nθ) You can also prove de Moivre’s theorem for positive integer exponents directly from the modulus− argument form of a complex number using the addition formulae for sin and cos. This step is valid for any integer exponent n. ← Exercise 1B, Challenge Links This proof uses the method of proof by induction. ← Book 1, Chapter 8 1. Basis step n = 1; LHS = (r(cos θ + i sin θ))1 = r(cos θ + i sin θ) RHS = r1(cos 1θ + i sin 1θ) = r(cos θ + i sin θ) As LHS = RHS, de Moivre’s theorem is true for n = 1. 2. Assumption step Assume that de Moivre’s theorem is true for n = k, k ∈ ℤ + : (r(cos θ + i sin θ))k = rk (cos kθ + i sin kθ) 8 M01_ALEVEL_CP2_83343_U01_001-030.indd 8 26/02/2018 09:17 Complex numbers 3. Inductive step When n = k + 1, (r(cos θ + i sin θ))k + 1 = (r(cos θ + i sin θ))k × r(cos θ + i sin θ) = r k(cos kθ + i sin kθ) × r(cos θ + i sin θ) By assumption step = r k + 1(cos kθ + i sin kθ)(cos θ + i sin θ) = r k + 1((cos kθ cos θ − sin kθ sin θ) + i (sin kθ cos θ + cos kθ sin θ)) = r k + 1(cos(kθ + θ) + i sin(kθ + θ)) By addition formulae = r k + 1(cos((k + 1)θ) + i sin((k + 1)θ)) Therefore, de Moivre’s theorem is true when n = k + 1. 4. Conclusion step Links The corresponding proof for negative integer exponents is left as an exercise. If de Moivre’s theorem is true for n = k, then it has been shown to be true for n = k + 1. As de Moivre’s theorem is true for n = 1, it is now proven to be true for all n ∈ ℤ+ by mathematical induction. Example → Exercise 1C, Challenge 8 9π 9π     + i sin  ___   ) ( cos ___ 17 17 _________________ Simplify     2π 2π ( cos ___     − i sin ___      ) 17 17 5 3 ( ( ) ) Problem-solving 9π 9π 5 cos  ___  + i sin  ___   17 17 __________________      3  2π 2π cos  ___  − i sin  ___   17 17 ( You could also show this result by writing both numbers in exponential form: ) 9π 9π 5 cos  ___  + i sin  ___   17 17         = _______________________     2π 2π 3 ___ cos −      + i sin −___        17 17 ( e ) ____ e i(___ − − __ ______     =     = e   17 (   17 )) = e  3πi = e  πi = −1 45π 45π cos  ____  + i sin  ____  17 17 ____________________     =        6π 6π ___ ___ cos (−θ) = cos θ and sin (−θ) = −sin θ ( ( ) ( 9πi 5 ___  17 cos −  ( ) ( ) 45π 45π 6π 6π     − (− ___ )  + i sin ____     − ( − ___ )  = cos ____ 17 17 17 17 51π 51π = cos  ____  + i sin  ____  17 17 = cos 3π + i sin 3π   17 45π 6π ( e ) e   17 2πi 3 − ___   17 )) ( 17   ) + i sin( −  17   ) 45πi ___ 6πi − ___ Apply de Moivre’s theorem to both the numerator and the denominator. z1 __     = cos (θ − θ ) + i sin (θ − θ ) z2 1 2 1 2 Simplify. = cos π + i sin π Subtract 2π from the argument. = −1 + i(0) 9π 9π   + i sin  ___   ) ( cos  ___ 17 17 __________________ So        = −1 2π 2π ( cos  ___  − i sin  ___   ) 17 17 5 3 9 M01_ALEVEL_CP2_83343_U01_001-030.indd 9 26/02/2018 09:17 Chapter 1 Example 9 __ 7 Express ( 1 + i√ 3 ) in the form x + iy where x, y ∈ ℝ. First, you__need to find the modulus and argument of 1 + i√ 3  . You may want to draw an Argand diagram to help you. Im 3 θ O 1 _________ __ Re __ r = √ 12 + (√ 3  )2   = √ 4  = 2 Find r and θ. ( ) __ √ 3  π θ = arctan ___      = __     1 3 __ π π So 1 + i√3  = 2( cos  __  + i sin  __   ) 3 3 __ π π 7 ( 1 + i√ 3  )7 = ( 2(cos  __   + i sin  __ ))  3 3 ( __ Write 1 + i√ 3 in modulus–argument form. ) 7π 7π = 27 cos  ___  + i sin  ___   3 3 Apply de Moivre’s theorem. π π = 128( cos  __  + i sin  __   ) 3 3 Subtract 2π from the argument. ( ( )) __ √ 3  = 128 __   1   + i ___        2 2 __ __ Therefore, (1 + i√3  ) = 64 + 64i√3  Exercise 7 1C 1 Use de Moivre’s theorem to express each of the following in the form x + iy, where x, y ∈ ℝ. a (cos θ + i sin θ)6 b (cos 3θ + i sin 3θ)4 π π 8 d cos  __  + i sin  __   3 3 2π 2π 5 e cos  ___  + i sin  ___   5 5 ( ) ( ) π π 5 c cos  __  + i sin  __   6 6 ( ) π π 15 f cos  ___  − i sin  ___   10 10 ( ) 2 Express each of the following in the form einθ. cos 5θ + i sin 5θ        a __________________ (cos 2θ + i sin 2θ)2 (cos 2θ + i sin 2θ)7 __________________ b      (cos 4θ + i sin 4θ)3 1 c __________________   3  (cos 2θ + i sin 2θ) (cos 2θ + i sin 2θ)4   3  d __________________ (cos 3θ + i sin 3θ) cos 5θ + i sin 5θ e __________________        (cos 3θ − i sin 3θ)2 cos θ − i sin θ f __________________      3  (cos 2θ − i sin 2θ) 10 M01_ALEVEL_CP2_83343_U01_001-030.indd 10 26/02/2018 09:17 Complex numbers 3 Evaluate the following, giving your answers in the form x + iy, where x, y ∈ ℝ. 7π 7π 4 (cos  ___  − i sin  ___ ) 13 13 _________________ a      4π 4π 6 ___ ___ (cos     + i sin    ) 13 13 3π 11π 3   (cos  ___  – i sin  ____ 7 7 )       b _________________   15π π 2 ____ __ (cos     + i sin    ) 7 7 4 Express the following in the form x + iy where x, y ∈ ℝ. a (1 + i)5 b (−2 + 2i)8 __ e( _  2  − _  2  i√3   ) 3 d (1 − i√ ) 3 6 E E E __ 9 __ 1 4π 2π 7 (cos  ___  – i sin  ___ ) 3 3 __________________    c      10π 4π 4 ____ ___ (cos     + i sin    ) 3 3 c (1 − i)6 __ f( −2√3  − 2i ) 5 __ 5 5 Express ( 3 + i√  3 ) in the form a + bi√3  where a and b are integers. (2 marks) π π 6 w = 2( cos  __  + i sin  __ ) 6 6 Find the exact value of w4, giving your answer in the form a + ib where a, b ∈ ℝ. (2 marks) __ 3π 3π 7 z = √ 3  (cos  ___  – i sin  ___ ) 4 4 Find the exact value of z6, giving your answer in the form a + ib where a, b ∈ ℝ. (3 marks) __ E/P E/P 1 + i √  3 __ in the form reiθ, where r > 0 and −π , θ < π. 8 a Express _______   1 − i √  3 __ n √ 3  1 + i __    is real b Hence find the smallest positive integer value of n for which ( _______   √3 1 − i   ) and positive. 9 Use de Moivre’s theorem to show that (a + bi)n + (a − bi)n is real for all integers n. Challenge (3 marks) (2 marks) (5 marks) Problem-solving Without using Euler’s relation, prove that if n is a positive integer, (r(cos θ + i sin θ))  −n= r  −n(cos (−nθ) + i sin (−nθ)) You may assume de Moivre’s theorem for positive integer exponents, but do not write any complex numbers in exponential form. 1.4 Trigonometric identities You can use de Moivre’s theorem to derive trigonometric identities. Applying the binomial expansion to (cos θ + i sin θ)n allows you to express cos nθ in terms of powers of cos θ, and sin nθ in terms of powers of sin θ. Links (a + b)  n= a  n+ n   C  1 a  n−1b + n   C  2 a  n−2 b  2 + … + n   C  r a  n−r b  r+ … + b  n, n ∈ ℕ n n! where n   C  r = ( r )  =  ________   ← Pure Year 1, Chapter 8 r!(n − r)! 11 M01_ALEVEL_CP2_83343_U01_001-030.indd 11 26/02/2018 09:17 Chapter 1 Example 10 Use de Moivre’s theorem to show that cos 6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 (cos θ + i sin θ)6 = cos 6θ + i sin 6θ = cos6 θ + 6C1 cos5 θ(i sin θ) + 6C2 cos4 θ(i sin θ)2 + 6C3 cos3 θ(i sin θ)3 + 6C 4 cos2 θ(i sin θ)4 + 6C5 cos θ(i sin θ)5 + (i sin θ)6 = cos6 θ + 6i cos5 θ sin θ + 15i2 cos4 θ sin2 θ + 20i3 cos3 θ sin3 θ + 15i4 cos2 θ sin4 θ + 6i5 cos θ sin5 θ + i6 sin6 θ = cos6 θ + 6i cos5 θ sin θ − 15 cos4 θ sin2 θ − 20i cos3 θ sin3 θ + 15 cos2 θ sin4 θ + 6i cos θ sin5 θ − sin6 θ Equating the real parts gives cos 6θ = cos6 θ − 15 cos4 θ sin2 θ + 15 cos2 θ sin4 θ − sin6 θ = cos6 θ − 15 cos4 θ(1 − cos2 θ) + 15 cos2 θ(1 − cos2 θ)2 − (1 − cos2 θ)3 = cos6 θ − 15 cos4 θ(1 − cos2 θ) + 15 cos2 θ(1 − 2 cos2 θ + cos4 θ) − (1 − 3 cos2 θ + 3 cos4 θ − cos6 θ) = cos6 θ − 15 cos4 θ + 15 cos6 θ + 15 cos2 θ − 30 cos4 θ + 15 cos6 θ − 1 + 3 cos2 θ − 3 cos4 θ + cos6 θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 Apply de Moivre’s theorem. Apply the binomial expansion to (cos θ + i sin θ)6. Simplify. Simplify the powers of i. The real part of cos 6θ + i sin 6θ is cos 6θ. Apply sin2 θ ; 1 − cos2 θ, sin4 θ ; (sin2 θ)2 and sin6 θ ; (sin2 θ)3. Multiply out the brackets. Apply a cubic binomial expansion. Expand the brackets. Simplify. Therefore, cos 6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 You can also find trigonometric identities for sinn θ and cosn θ where n is a positive integer. If z = cos θ + i sin θ, then __  1 = z−1 = (cos θ + i sin θ)−1 z = (cos(−θ) + i sin(−θ)) = cos θ − i sin θ Apply de Moivre’s theorem. Use cos θ = cos (−θ) and −sin θ = sin (−θ). It follows that z + __  1z = cos θ + i sin θ + cos θ − i sin θ = 2 cos θ z − __  1z = cos θ + i sin θ − (cos θ − i sin θ) = 2i sin θ 12 M01_ALEVEL_CP2_83343_U01_001-030.indd 12 26/02/2018 09:17 Complex numbers Also, z n = (cos θ + i sin θ)n = cos nθ + i sin nθ By de Moivre’s theorem. __ = (cos θ + i sin θ)–n   1   = z–n z n = (cos(−nθ) + i sin(−nθ)) Apply de Moivre’s theorem. = cos nθ − i sin nθ Use cos θ = cos (−θ) and sin (−θ) = −sin θ. It follows that   1n  = cos nθ + i sin nθ + cos nθ − i sin nθ = 2 cos nθ z n + __ z z n − __   1n  = cos nθ + i sin nθ − (cos nθ − i sin nθ) = 2i sin nθ z It is important that you remember and are able to apply these results: 1 ■ z + __   z = 2 cos θ 1 ■ zn + __   n = 2 cos nθ Notation In exponential form, these results are equivalent to: z 1 ■ zn − __   n = 2i sin nθ z 1 ■ z − __   z = 2i sin θ 1 1 cos nθ = __    (e  inθ + e  −inθ) sin nθ = __    (e  inθ − e  −inθ). 2 2i Example 11 Express cos5 θ in the form a cos 5θ + b cos 3θ + c cos θ, where a, b and c are constants. Let z = cos θ + i sin θ 5 ( z + _ z1   ) = (2 cos θ)5 = 32 cos5 θ 2 3 = z5 + 5C1 z4( _ z1   ) + 5C2 z3( _ z1   ) + 5C3 z2( _ z1   ) 5 4 + 5C 4 z( _ z1   ) + ( _ z1   ) = z5 + 5z4 _  1   + 10z3 __   12   + 10z2 __   13   z z z 1 1 __ __ + 5z   4   +   5   z z ( ) () ( ) ( ) ( ) 10 5 __ = z5 + 5z3 + 10z + __   + __   3  +   15  z z z ( ) ( ) ( ) = z5 + __   15   + 5 z3 + __   13   + 10 z + _  1   z z z = 2 cos 5θ + 5(2 cos 3θ) + 10(2 cos θ) So, 32 cos5 θ = 2 cos 5θ + 10 cos 3θ + 20 cos θ ⇒ 5 cos5 θ = __  161   cos 5θ + __  16   cos 3θ + __  58  cos θ 1 Use z + __   z  = 2 cos θ. Apply the binomial expansion to 1 5 (z + __   z ) . Simplify. 1 Group zn and __   n  terms. z 1 Use zn + __   n  = 2 cos nθ. z 1 This is in the required form with a = __   16 , 5 5 __ _ b =   16 and c =   8 13 M01_ALEVEL_CP2_83343_U01_001-030.indd 13 26/02/2018 09:17 Chapter 1 Example 12 a Express sin4 θ in the form d cos 4θ + e cos 2θ + f, where d, e and f are constants. ∫ π __     2 b Hence find the exact value of   sin4 θ dθ. 0 a Let z = cos θ + i sin θ 4 ( z − _ z1   ) = (2i sin θ)4 = 16i4 sin4 θ = 16 sin4 θ 2 = z4 + 4C1 z3( − _z1   ) + 4C2 z2 ( − _z1   ) 1 Use z − __   z  = 2i sin θ, noting that i4 = 1 Apply the binomial expansion to 1 4 (z − __   z ) 4 3 + 4C3 z1( − _z1   ) + ( − _z1   ) ( ) ( ) 1    + __ + 4z( − __   1    z ) (z ) = z4 + 4z3 − _1   + 6z2 __   12   z z Simplify. 4 3 4 = z4 − 4z2 + 6 − __   2  + __   14  z z ( ) ( ) = z4 + __   14   − 4 z2 + __   12   + 6 z z 1 Group zn and __   n  terms. z = 2 cos 4θ − 4(2 cos 2θ) + 6 1 Use zn + __   n  = 2 cos nθ. z So, 16 sin4 θ = 2 cos 4θ − 8 cos 2θ + 6 sin4 θ = __  81   cos 4θ − _  21   cos 2θ + __  38  ⇒ ∫ b   0 ∫   (__  cos 4θ − __  cos 2θ + __  ) dθ π __     2 sin4 θ dθ = π __     2 1 0 8 3 8 1 2 π 1 = [___   32 sin 4θ − __   41 sin 2θ + __   38 θ] 02 This is in the required form with d = _  18 , e = −  _12 and f = _  38  Use the answer from part a. __ π 1 = ( ___   32   sin 2π − __   41   sin π + __   38 (__     − 0 2 )) 3π = 0 − 0 + ___     16 3π = ___     16 Exercise 1 cos kθ integrates to __     sin kθ. k 1D Use de Moivre’s theorem to prove the following trigonometric identities: P 1 a sin 3θ ≡ 3 sin θ − 4 sin3 θ c cos 7θ ≡ 64 cos7 θ − 112 cos5 θ + 56 cos3 θ − 7 cos θ e sin5 θ ≡ __  16  (sin 5θ − 5 sin 3θ + 10 sin θ) 1 b sin 5θ ≡ 16 sin5 θ − 20 sin3 θ + 5 sin θ d cos4 θ ≡ _ 8  (cos 4θ + 4 cos 2θ + 3) 1 14 M01_ALEVEL_CP2_83343_U01_001-030.indd 14 26/02/2018 09:17 Complex numbers E/P 2 a Use de Moivre’s theorem to show that cos 5θ ≡ 16 cos5 θ − 20 cos3 θ + 5 cos θ (5 marks) 3 a Show that 32 cos6 θ ≡ cos 6θ + 6 cos 4θ + 15 cos 2θ + 10. (6 marks) b Hence, given also that cos 3θ = 4 cos3 θ − 3 cos θ, find all the solutions of cos 5θ + 5 cos 3θ = 0 (6 marks) in the interval 0 < θ , π. Give your answers to 3 decimal places. E/P ∫ π __     6 __ b Hence find   cos6 θ dθ in the form aπ + b√ 3  where a and b are rational constants to be 0 (3 marks) found. E/P 4 a Show that 32 cos  2 θ sin  4 θ ≡ cos 6θ − 2 cos 4θ − cos 2θ + 2. ∫ (6 marks) π _   3 b Hence find the exact value of   cos  2 θ sin  4 θ dθ. (3 marks) 0 P 5 By using de Moivre’s theorem, or otherwise, compute the following integrals. ∫ a E/P ∫ π _   2   sin6 θ dθ 0 b 0 ∫ π _   4   sin  2 θ cos  4 θ dθ π _   6 c   sin  3 θ cos  5 θ dθ 0 6 a Use de Moivre’s theorem to show that cos 6θ ≡ 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 b Hence find the six distinct solutions of the equation 32x6 − 48x4 + 18x2 − _  2 = 0 3 giving your answers to 3 decimal places where necessary. (5 marks) E/P (5 marks) Problem-solving Use the substitution x = cos θto reduce the equation to the form cos 6θ = k. Find as many values of θas you need to find six distinct values of x. 7 a Use de Moivre’s theorem to show that sin 4θ ≡ 4 cos3 θ sin θ − 4 cos θ sin3 θ. 4 tan θ − 4 tan3 θ    b Hence, or otherwise, show that tan 4θ ≡ ________________      1 − 6 tan2 θ + tan4 θ (4 marks) (4 marks) c Use your answer to part b to find, to 2 decimal places, the four solutions of the equation (5 marks) x4 + 4x3 − 6x2 − 4x + 1 = 0. 15 M01_ALEVEL_CP2_83343_U01_001-030.indd 15 26/02/2018 09:17 Chapter 1 1.5 Sums of series You can use results about the sums of geometric series with complex numbers. ■ For w, z ∈ ℂ, n−1 w(z  n− 1)   •  ∑ wz  r= w + wz + wz  2 + ... + wz  n−1 = ________   z−1 r=0 ∞ w •  ∑ wz  r= w + wz + wz  2 + ... = _____   , |z| < 1 1−z r=0 Links These results match the corresponding results for real numbers. ∞ The infinite series  ∑ wz  rconverges only r=0 when |z| , 1. ← Pure Year 2, Chapter 3 Example 13 π π __ Given that z = cos  __ n  + i sin   n , where n is a positive integer, show that π    ) 1 + z + z2 + … + zn − 1 = 1 + i cot (___ 2n n−1 z  n − 1 1 + z + z2 + … + zn − 1 = ______     z−1 Use the result for  ∑ wz  r with w = 1. r=0 ( e  n ) − 1   ________ =   __ πi e  n − 1 πi n __ πi __ Write z in exponential form as z = e   n , and substitute. eπi − 1   =  ______ πi __ e   n − 1 −2 = ______   __   πi   n e − 1 πi n __ (e  n ) = e  πi = −1 Problem-solving 1 You know that s in nθ = __     (e  inθ − e  −inθ). 2i You can use this result to simplify an expression like e  iθ− 1by writing it in the iθ iθ iθ iθ __ __ __ __ θ form e  2 (e  2 − e−   2 ) = e  2 (2i sin  __ ). In this case 2 this is equivalent to multiplying the top πi − ___ −2e 2n _________ =   ___ πi πi     − ___ e2n − e 2n πi − ___ −2e 2n =  ________ π   ___ 2i sin   2n πi − ___ π   − i ___ ie 2n  ______ = π   sin  ___ 2n and bottom of the fraction by e 2n. π π i(cos(−  ___ ) + i sin(−  ___ )) 2n 2n ______________________    =      π   sin  ___  2n π π     − i sin ___     i cos ___ ( 2n ) ( 2n )) (        =  ___________________ π   ___ sin     2n −2 ___     = i 2i e−iθ = cos (−θ) + i sin (−θ) = cos θ − i sin θ. π π    ) + i cos(___    ) sin(___ 2n 2n _________________    =      π   ___ sin     2n 16 M01_ALEVEL_CP2_83343_U01_001-030.indd 16 26/02/2018 09:17 Complex numbers π π    )    ) i cos(___ sin(___ 2n 2n _________ = _______   π   +   π   sin (___    ) sin (___    ) 2n 2n π    ) as required = 1 + i cot(___ 2n Simplify. The series eiθ + e2iθ + e3iθ + … + eniθ is geometric with first term eiθ, common ratio eiθ and n terms. e  iθ(e  niθ− 1)   iθ   The sum of this series is given by Sn  = __________ e  − 1 Converting the exponential form into modulus−argument form lets you consider the real and imaginary parts of the series separately. eiθ + e2iθ + e3iθ + … + eniθ = (cos θ + i sin θ) + (cos 2θ + i sin 2θ) + (cos 3θ + i sin 3θ) + … + (cos nθ + i sin nθ) = (cos θ + cos 2θ + cos 3θ + … + cos nθ) + i(sin θ + sin 2θ + sin 3θ + … + sin nθ) Therefore, e  iθ( e  niθ− 1) cos θ + cos 2θ + cos 3θ + … + cos nθ = Re(__________   iθ   e  − 1 ) e  iθ(e  niθ− 1)   iθ   sin θ + sin 2θ + sin 3θ + … + sin nθ = Im(__________ e  − 1 ) Example 14 S = eiθ + e2iθ + e3iθ + … + e8iθ, for θ ≠ 2nπ, where n is an integer. 9iθ __ e sin 4θ a Show that S = ________     θ __   2 sin     2 Let P = cos θ + cos 2θ + cos 3θ + … + cos 8θ and Q = sin θ + sin 2θ + sin 3θ + … + sin 8θ θ 9θ b Use your answer to part a to show that P = cos  ___  sin 4θ cosec  __ and find similar expressions for 2 2 Q Q and __     P e  iθ( ( e  iθ)  8 − 1) a eiθ + e2iθ + e3iθ + … + e8iθ = __________       e  iθ − 1 e  iθ( e  8iθ − 1) = _________     e  iθ − 1 __ iθ ( ) __________ =   e    2  e  8iθ − 1   __ __ iθ iθ e    2  − e  −   2  S is the sum of a geometric series. n−1 w(z  n− 1) Use   ∑ wz  r= _______   with w = z = e  iθ and n = 8.   z−1 r=0 (e  iθ)  8 = e  8iθ iθ __ Multiply the numerator and denominator by e −  2 17 M01_ALEVEL_CP2_83343_U01_001-030.indd 17 26/02/2018 09:17 Chapter 1 __ iθ e    2 ( e  8iθ − 1) __________ =     θ 2i sin  __  2 iθ iθ __ __ θ e   2 − e −  2 = 2i sin  __  2 __ iθ e    2  e  4iθ( e  4iθ − e  -4iθ)       =  _______________   θ 2i sin  __  2 Use the relationship e  8iθ − 1 = e  4iθ(e  4iθ − e  −4iθ)to rewrite the numerator. ___ 9iθ e    2 (2i sin 4θ)      = ___________   θ 2i sin  __  2 e   4iθ − e  −4iθ = 2i sin 4θ ___ 9iθ e    2  sin 4θ =  ________   θ __ sin     2 ___ 9iθ e    2  sin 4θ ________ b P = Re(S) = Re     θ __ ( sin   2  ) Simplify. Problem-solving By writing each term of S in modulus– argument form you can see that P is the real part of S and Q is the imaginary part of S. 9θ cos  ___  sin 4θ 2 = __________       θ sin  __  2 9θ θ = cos  ___  sin 4θ cosec  __  2 2 ___ 9iθ e    2  sin 4θ   Q = Im(S) = Im ________   θ __ ( sin   2  ) 9θ sin  ___  sin 4θ 2 __________ =       θ __ sin     2 9θ θ = sin  ___  sin 4θ cosec  __  2 2 9θ 9θ θ sin  ___  sin 4θ cosec  __  sin  ___  Q ___________________ 9θ 2 2 2 ______ __     =       =       = tan  ___  2 P 9θ 9θ θ ___ __ ___ cos     sin 4θ cosec     2 2 Exercise P cos     2 1E πi __ 1 Given z = e  n , where n is a positive integer, show that: a 1 + z + z2 + … + z2n − 1 = 0 πi __ π b 1 + z + z2 + … + zn = i cot(___    ) 2n 12 P 2 Show that if z = e    2 , then   ∑ z  r = 1. P 3 Show that   ∑  (1 + i)  r= −15i. r=0 7 r=0 18 M01_ALEVEL_CP2_83343_U01_001-030.indd 18 26/02/2018 09:17 Complex numbers E/P 4 The convergent infinite series C and S are defined as Hint   27 cos 3θ + … C = 1 + _  3 cos θ + _  9 cos 2θ + __ 1 1 1   27 sin 3θ + … S = _  3 sin θ + _  9 sin 2θ + __ 1 E/P 1 1 The sum of an infinite geometric series with a first term a and common ratio r is S  ∞ = _____     1−r ← Pure Year 2, Chapter 3 3 a Show that C + iS = ______     3 − e  iθ (4 marks) 9 − 3 cos θ , and find a similar expression for S. b Hence show that C = __________   10 − 6 cos θ (4 marks) 5 The series P and Q are defined for 0 , θ , π as P = 1 + cos θ + cos 2θ + cos 3θ + … + cos 12θ Q = sin θ + sin 2θ + sin 3θ + … + sin 12θ 13iθ ___ 13iθ ___ 6iθ(e    2   − e  −   2 ) e____________ a Show that P + iQ =      iθ iθ  __ __ e  2  − e−   2  (4 marks) 13θ θ   cosec  __ and write down the corresponding expression for P. b Deduce that Q = sin 6θ sin  ____ 2 2 e  iθ − e  −iθ e  iθ + e  −iθ You can assume the results sin θ =  ________   and cos θ = ________     2 2i c Hence find the values of θ, in the range 0 , θ , π, for which P + iQ is real. E/P (4 marks) (2 marks) 6 Series C and S are defined as n n n n C = 1 + ( 1)  cos θ + ( 2)  cos 2θ + ( 3)  cos 3θ + … + ( n)  cos nθ n n n n S = ( 1)  sin θ + ( 2)  sin 2θ + ( 3)  sin 3θ + … + ( n)  sin nθ nθ a Show that C = ( 2 cos    )  cos  ___  2 2 (4 marks) S nθ b Show that __     = tan  ___  C 2 (3 marks) θ __ E/P n 7 a Show that (2 + eiθ)(2 + e−iθ) = 5 + 4 cos θ. (2 marks) The convergent infinite series C and S are defined by C = 1 − _  2 cos θ + _  4 cos 2θ − _  8 cos 3θ + … 1 1 1 S = _  2 sin θ − _  4 sin 2θ + _  8 sin 3θ + … 1 1 1 4 + 2 cos θ b By considering C − iS, show that C = _________   and write down the corresponding expression 5 + 4 cos θ for S. (4 marks) 19 M01_ALEVEL_CP2_83343_U01_001-030.indd 19 26/02/2018 09:17 Chapter 1 1.6 nth roots of a complex number You can use de Moivre’s theorem to solve an equation of the form zn = w, where z, w ∈ ℂ. This is equivalent to finding the nth roots of w. __ __ Just as a real number, x, has two square roots, √ x  and −√ x , any complex number has n distinct nth roots. ■ If z and w are non-zero complex numbers and n is a positive integer, then the equation zn = w has n distinct solutions. You can find the solutions to zn = w using de Moivre’s theorem, and by considering the fact that the argument of a complex number is not unique. Note cos (θ + 2kπ) = cos θ and sin (θ + 2kπ) = sin θfor integer values of k. ■ For any complex number z = r(cos θ + i sin θ), you can write z = r(cos (θ + 2kπ) + i sin (θ + 2kπ)), where k is any integer. Example 15 a Solve the equation z3 = 1. b Represent your solutions to part a on an Argand diagram. c Show that the three cube roots of 1 can be written as 1, ω and ω2 where 1 + ω + ω2 = 0. az  3 = 1 z3 = cos 0 + i sin 0 (r(cos θ + i sin θ))3 = cos (0 + 2kπ) + i sin (0 + 2kπ), k ∈ ℤ r3(cos 3θ + i sin 3θ) = cos (0 + 2kπ) + i sin (0 + 2kπ), k ∈ ℤ So r = 1 3θ = 2kπ k = 0 ⇒ θ = 0, so z1 = cos 0 + i sin 0 = 1 2π k = 1 ⇒ θ = ___     3 __ √ 3  2π 2π ___ ___ __1 ___ so z  2 = cos (   ) + i sin (   ) = −   2 + i     2 3 3 2π k = −1 ⇒ θ = −  ___  3 __ √ 3  2π 2π ___ ___ __1 ___ so z 3 = cos (−    ) + i sin (−    ) = −  2 − i     2 3 3 Therefore, __ __ √ √ 3  3  __1 ___ __1 ___ z = 1, z = −   2 + i     or z = −  2 − i     2 2 Start by writing 1 in modulus−argument form. Write z in modulus−argument form, and write the general form of the argument on the right-hand side by adding integer multiples of 2π. Apply de Moivre’s theorem to the left-hand side of the equation. Compare the modulus on both sides to get r = 1. Compare the arguments on both sides. Problem-solving Choose values of k to find the three distinct roots. By choosing values on either side of k = 0 you can find three different arguments in the interval [ −π, π]. These are the cube roots of unity. 20 M01_ALEVEL_CP2_83343_U01_001-030.indd 20 26/02/2018 09:17 Complex numbers b z2 = – Im 3 1 +i 2 2 __ 3 √   Plot the points z = 1, z = −  __1  + i  ___  and 2π 2π 3 3 2π 3 z3 = – z1 = 1 Re 1 2 __ 2 2 3 √   z3 = −  __12  − i  ___ on an Argand diagram: 2 The points z1, z2 and z3 lie on a circle of radius 1 unit. The angles between each of the vectors z1, z2 and 2π z3 are ___     , as shown on the Argand diagram. 3 3 1 +i 2 2 1 __ __ √ 3  ___ ___ 2πi c Let ω = z 2 = −     + i     = e   3 2 2 2 Then, ω  2 = (e   3 ) = e   3 __ 2πi √ 3  1 − ___ __ ___   3 = e = −     − i     = z 3 2 2 ___ 2πi 1 + ω + ω2 = __ 4πi ___ __ 1 + ( −     + i    ) + ( −     − i    ) = 0 2 2 2 2 1 __ √ 3  ___ 1 __ √ 3  ___ Notice that ω* = ω2. Note It can be proved that the sum of the nth roots of unity is zero, for any positive integer n > 2. 2πk 2πk ____ ____   ■ In general, the solutions to z   n = 1 are z = cos (____   n  ) + i sin(  n  ) = e  n for k = 1, 2, … , n and 2πik are known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω = e    n  such that: 2πi __ • the nth roots of unity are 1, ω, ω  2, ⋯ , ω  n−1 • 1, ω, ω  2, ⋯ , ω n−1form the vertices of a regular n-gon • 1 + ω + ω  2 + … + ω  n−1 = 0 21 M01_ALEVEL_CP2_83343_U01_001-030.indd 21 26/02/2018 09:17 Chapter 1 Example 16 __ Solve the equation z4 = 2 + 2i√ 3  . Im 2 + 2i 3 2 3 O θ 2 Re To solve an equation of the form zn = w, start by writing w in modulus−argument form. ___________ __ _______ 2 + (2 √ 3 )2  = √ modulus = 2    4 + 12  = 4 √ argument = arctan (  __ π       = __ 2 ) 3 π π So z4 = 4(cos  __  + i sin  __ ) 3 3 2√ 3  ____ Now let z = r(cos θ + i sin θ), and write the general form of the argument on the RHS by adding integer multiples of 2π. (r(cos θ + i sin θ))4 π π = 4(cos (__     + 2kπ) + i sin (__     + 2kπ)), k ∈ ℤ 3 3 Apply de Moivre’s theorem to the LHS. r4(cos 4θ + i sin 4θ) π π = 4(cos (__     + 2kπ) + i sin (__     + 2kπ)), k ∈ ℤ 3 3 __ __ So r4 = 4 ⇒ r = √ 4  = √ 2  π 4θ =  __  + 2kπ 3 __ π π π k = 0 ⇒ θ =  ___ , so z1 = √ 2 (cos  ___  + i sin  ___ ) 12 12 12 4 7π k = 1 ⇒ θ =  ___ , 12 7π 7π so z2 = √ 2 (cos  ___  + i sin  ___ ) 12 12 __ __ 5π 5π 5π k = −1 ⇒ θ = −  ___ , so z3 = √ 2 (cos (−  ___ ) + i sin (−  ___ )) 12 12 12 11π 11π 11π k = −2 ⇒ θ = −  ___ , so z4 = √ 2 (cos (−  ___ ) + i sin (−  ___ )) 12 12 12 __ __ __ πi __ ___ 7πi __ 5πi ___ __ ___ 11πi or z = √ 2  e 12, z = √ 2  e 12 , z = √ 2  e −   12 or z = √ 2  e −  12 Compare the modulus on both __ sides to get r = √ 2 . Compare the arguments on both sides. π When k = 1, 4θ =  __  + 2π 3 π ___ 2π ___ 7π ___ ⇒ θ =     +     =     12 4 12 Watch out Make sure you choose n consecutive values of k to get n distinct roots. If an argument is not in the interval [ −π, π]you can add or subtract a multiple of 2π. These are the solutions in the form reiθ. 22 M01_ALEVEL_CP2_83343_U01_001-030.indd 22 26/02/2018 09:17 Complex numbers You can also use the exponential form of a complete number when solving equations. Example 17 __ __ Solve the equation z3 + 4√ 2  + 4i√ 2  = 0. __ __ z3 + 4√2  + 4i√2  = 0 __ __ z3 = −4√2  − 4i√2  Im 4 2 O Re θ 4 2 –4 2 – 4i 2 ___________________ __ __ ________ ___ −4√2   ) + ( −4√2   )   = √ 32 + 32  = √64  = 8 modulus = √(   2 2 ( 4 2     ) = −π +  4  = − 4  __ 2  4√__ _____ argument = −π + arctan   i reiθ 3 = 8e (−   4 + 2kπ) ( ) __ 3π √ π __ 3π ___ i r3e3iθ = 8e (−   4 + 2kπ) __ 3π __ So r3 = 8 ⇒ r = √ 8  = 2 3 3π 3θ = −  ___  + 2kπ 4 ___ −πi π k = 0 ⇒ θ = −  __ , so z1 = 2e    4  4 5πi ___ 5π k = 1 ⇒ θ =  ___ , so z2 = 2e    12  12 ____ −11πi 11π k = −1 ⇒ θ = −  ____ , so z3 = 2e    12  12 5π 5π π π or z = 2(cos(−  __ ) + i sin(−  __ )), z = 2(cos  ___  + i sin  ___ ) 4 4 12 12 11π 11π or z = 2(cos(−  ____ ) + i sin(−  ____ )). 12 12 Find the modulus__ and __ argument of –4√ 2  – 4i√ 2 . Write z = reiθ and use ( reiθ)n = rneinθ. Remember to write the general form of the argument on the righthand side by adding integer multiples of 2π. Compare the modulus on both sides to get r = 2. Compare the arguments on both sides. Choose values of k to find three distinct roots. Either choose values that produce arguments in the interval −π , θ < π, or add or subtract multiples of 2π as necessary. 23 M01_ALEVEL_CP2_83343_U01_001-030.indd 23 26/02/2018 09:17 Chapter 1 Exercise 1F 1 Solve the following equations, expressing your answers for z in the form x + iy, where x , y ∈ ℝ. a z4 − 1 = 0 b z3 − i = 0 c z3 = 27 d z4 + 64 = 0 e z4 + 4 = 0 f z3 + 8i = 0 2 Solve the following equations, expressing the roots in the form r (cos θ + i sin θ), where −π , θ < π. a z7 = 1 b z4 + 16i = 0 c z5 + 32 = 0 d z3 = 2 + 2i e z4 + 2i√ 3  = 2 f z3 + 32√ 3  + 32i = 0 __ __ 3 Solve the following equations, expressing the roots in the form r eiθ, where r . 0 and −π , θ < π. Give θ to 2 decimal places. ___ P __ b z3 = √ 11  − 4i a z4 = 3 + 4i c z4 = −√ 7  + 3i 4 a Find the three roots of the equation (z + 1)3 = −1. Give your answers in the form x + iy, where x, y ∈ ℝ. b Plot the points representing these three roots on an Argand diagram. c Given that these three points lie on a circle, find its centre and radius. P 5 a Find the five roots of the equation z5 − 1 = 0. Give your answers in the form r (cos θ + i sin θ), where −π , θ < π. b Hence or otherwise, show that 2π 4π 1 cos  ___   + cos  ___   = − _2 5 5 ( ) E ( ) __ 6 a Find the modulus and argument of −2 − 2i√ 3 . b __ Hence find all the solutions of the equation z4 + 2 + 2i√ 3  = 0. Give your answers in the form r eiθ , where r . 0 and −π , θ < π and illustrate the roots on an Argand diagram. Problem-solving Use the fact that the sum of the five roots of unity is zero. (2 marks) (4 marks) __ E 7 Find the four distinct roots of the equation z4 = 2(1 – i√ 3 ) in exponential form, and show these roots on an Argand diagram. (7 marks) E/P 8 z = √ 6  + i√ 2  __ __ a Find the modulus and argument of z. (2 marks) b Find the values of w such that w3 = z4, giving your answers in the form reiθ, where r . 0 (4 marks) and −π , θ < π. 24 M01_ALEVEL_CP2_83343_U01_001-030.indd 24 26/02/2018 09:17 Complex numbers P 9 a Solve the equation Problem-solving 1 + z + z2 + z3 + z4 + z5 + z6 + z7 = 0 b Hence deduce that (z2 + 1) and (z4 + 1) are factors of 1 + z + z2 + z3 + … + z7 is the sum of a geometric series. 1 + z + z2 + z3 + z4 + z5 + z6 + z7. Challenge a Find the six roots of the equation z6 = 1 in the form eiθ, where −π , θ < π. b Hence show that the solutions to (z + 1)6 = z6 are kπ z = −  _12 + _  12 i cot (___    ), k = 1, 2, 3, 4, 5. 6 1.7 Solving geometric problems You can use properties of complex nth roots to solve geometric problems. ■ The nth roots of any complex number a lie Notation at the vertices of a regular n-gon with its centre at the origin. The orientation and size of the regular polygon will depend on a. The centre of a regular polygon is considered to be the centre of the circle that passes through all of its vertices. Im Re For example, the sixth roots of 7 + 24i form this regular hexagon. Each vertex of the hexagon is equidistant from the origin, which lies at the centre of the circle passing through all six vertices. Online Explore nth roots of complex numbers in an Argand diagram using GeoGebra. You can find the vertices of this regular polygon by finding a single vertex, and rotating that point around the origin. This is equivalent to multiplying by the nth roots of unity. ■ If z1 is one root of the equation zn = s, and 1, ω, ω2, …, ωn−1 are the nth roots of unity, then , z  1ω   2, …, z  1ω   n−1. the roots of zn = s are given by z 1, z  1ω 25 M01_ALEVEL_CP2_83343_U01_001-030.indd 25 26/02/2018 09:17 Chapter 1 Example 18 __ The point P(√ 3 , 1) lies at one vertex of an equilateral triangle. The centre of the triangle is at the origin. a Find the coordinates of the other vertices of the triangle. Problem-solving Consider the Cartesian coordinate plane as an Argand diagram. The vertices of the triangle will __ correspond to the cube roots of (__√ 3  + i)3. You can find these roots by multiplying √ 3  + i by the cube roots of unity. b Find the area of the triangle. a The cube roots of unity are 1, ω, ω2 where ___ 2πi ω = e 3 . __ __ πi √ 3  + i = 2e   6 Use ω = e  n  to write down the cube root of unity. 2πi __ So the vertices are at: __ __ πi 2e   6 = √ 3  + i 2e   6 (e   3 ) = 2e   6 = −√ 3  + i __ πi ___ 2πi __ __ π |√ 3  + i| = 2 and arg (√ 3  + i) = __     6 __ 5πi ___ 2e   6 (e   3 ) = 2e   6 (e   3 ) = 2e   6 = 2e −  2 = −2i ___ 2πi 2 __ πi __ πi ___ 4πi __ πi ___ 9πi __ So the coordinates of the vertices of the triangle are __ __ (√ 3 , 1), (−√ 3 , 1) and (0, −2). Write your answers as Cartesian coordinates. y b __ You know that one cube root of (√ 3  + i)3 is √ 3  + i. Multiply this by the cube roots of unity to find the other roots. ( 3, 1) (– 3, 1) x O (0, –2)   21   × base × height Area = __ __ 1 √ =  __ 2  × 2 3  × 3 __ = 3√ 3  Exercise P 1G 1 Find the coordinates of the vertices of the following regular polygons with centres at the origin. a Equilateral triangle with one vertex at (0, 4) b Square with one vertex at (5, 0) __ c Regular pentagon with one vertex at (−1, √ 3 ) d Regular hexagon with one vertex at (2, 2) P 2 Find the coordinates of the vertices of an equilateral triangle with centre (2, 3) and one vertex at (3, −2). 26 M01_ALEVEL_CP2_83343_U01_001-030.indd 26 26/02/2018 09:17 Complex numbers E/P 3 The triangle OAB in an Argand diagram is equilateral. O is the origin and A corresponds to the __ √ complex number 3 (1 − i). B is represented by the complex number b. Find the two possibilities for b in the form reiθ. Illustrate the two possibilities for OAB in a (5 marks) sketch. E/P 4 a Find the 4th roots of −12i in the form reiθ where r . 0 and −π , θ < π. Illustrate these roots (6 marks) on an Argand diagram. Let the points representing these roots on an Argand diagram, taken in order of increasing θ, be A, B, C, D. The midpoints of the sides of ABCD represent the 4th roots of a complex number w. b Find w. E/P (4 marks) 5 P is one vertex of a regular hexagon in an Argand diagram. The centre of the hexagon is at the origin. P corresponds to the complex number 8 + 8i. a Find, in the form a + bi, the complex numbers corresponding to the other vertices of the (5 marks) hexagon, and illustrate these on an Argand diagram. b The six complex numbers corresponding to the vertices of the hexagon are squared to form the vertices of a new figure. Find, in the form a + bi, the complex numbers corresponding to (4 marks) the other vertices of the new figure. Find the area of the new figure. E/P 2π 6 An ant walks forward one unit and then turns to the right by ___    . It repeats this a further 9 4π    ) sin (___ 9   three times. Show that the distance of the ant from its initial position is _______   (6 marks) π sin (__    ) 9 Mixed exercise P E/P 1 1 a Use eiθ = cos θ + i sin θ to show that cos θ = _  2 ( e  iθ + e  −iθ). cos (A + B) + cos (A − B)        b Hence prove that cos A cos B ≡ ______________________     2 1 2 Given that z = r(cos θ + i sin θ), r ∈ ℝ, prove by induction that zn = rn(cos nθ + i sin nθ), n ∈ ℤ  +. (5 marks) 2 (cos 3x + i sin 3x) ________________    3 Express       in the form cos nx + i sin nx where n is an integer to be determined. cos x − i sin x 4 Use de Moivre’s theorem to evaluate: 1 a (−1 + i)8 b _________   1 1 16  _ (  2  − _ 2 i ) E/P 1   n  = 2 cos nθ. 5 a Given z = cos θ + i sin θ, use de Moivre’s theorem to show that zn + __ z 1 3   2   in terms of cos 6θ and cos 2θ. b Express z2 + __ z ( ) (4 marks) (3 marks) c Hence, or otherwise, find constants a and b such that cos3 2θ = a cos 6θ + b cos 2θ. (3 marks) ∫ π __     6 __ d Hence, or otherwise, show that   cos3 2θ dθ = k√ 3 , where k is a rational constant. (4 marks) 0 27 M01_ALEVEL_CP2_83343_U01_001-030.indd 27 26/02/2018 09:17 Chapter 1 E/P 6 a Show that cos  5 θ ≡ __   16 (cos 5θ + 5 cos 3θ + 10 cos θ) 1 (5 marks) π π The diagram shows the curve with equation y = cos5 x, –  __  < x < __    . The finite region R is 2 2 bounded by the curve and the x-axis. y 1 R y = cos5 x O x b Calculate the exact area of R. E/P (6 marks) 7 a Show that ( ) sin6 θ ≡ −  __ (5 marks) 32 cos 6θ − 6 cos 4θ + 15 cos 2θ − 10 π     − θ), or otherwise, find a similar identity for cos6 θ. (3 marks) b Using the substitution α = ( __ 2 1 a 5π     , find the exact value of a. c Given that ∫   cos  6θ + sin  6 θ dθ = ___ 32 0 E/P (5 marks) 8 Use de Moivre’s theorem to show that sin 6θ ≡ sin 2θ (16 cos4 θ − 16 cos2 θ + 3) E/P (5 marks) 9 a Use de Moivre’s theorem to show that cos 5θ ≡ 16 cos5 θ − 20 cos3 θ + 5 cos θ (5 marks) b Hence find all solutions to the equation 16x5 − 20x3 + 5x + 1 = 0 (5 marks) giving your answers to 3 decimal places where necessary. E/P 10 a Show that sin5 θ ≡ __   16 (sin 5θ − 5 sin 3θ + 10 sin θ) 1 (5 marks) b Hence solve the equation sin 5θ − 5 sin 3θ + 9 sin θ = 0 for 0 < θ , π E/P (4 marks) 11 a Use de Moivre’s theorem to show that cos 5θ ≡ cos θ (16 cos4 θ − 20 cos2 θ + 5) __ 5 + √ 5  π      =  ______   b By solving the equation cos 5θ = 0, deduce that cos2 ___ 10 8 ( ) ( ) (4 marks) ( ) 3π 7π c Hence, or otherwise, write down the exact values of cos2 ___      , cos2 ___      and 10 10 9π ___ 2 cos      . 10 ( ) (5 marks) (3 marks) 28 M01_ALEVEL_CP2_83343_U01_001-030.indd 28 26/02/2018 09:17 Complex numbers E/P 12 a Use de Moivre’s theorem to find an expression for tan 3θ in terms of tan θ. cot  3 θ − 3 cot θ      b Deduce that cot 3θ = _____________      3 cot  2 θ − 1 E/P (4 marks) (2 marks) 13 The infinite series C and S are defined as C = 1 + k cos θ + k2 cos 2θ + k3 cos 3θ + … S = k sin θ + k2 sin 2θ + k3 sin 3θ + … where k is a real number and |k| , 1. 1 − k cos θ By considering C + iS, show that C = ______________      and write down the corresponding    1 + k  2 − 2k cos θ (8 marks) expression for S. E 14 a Express 4 − 4i in the form r (cos θ + i sin θ), where r . 0, −π , θ < π, where r and θ are exact values. (2 marks) b Hence, or otherwise, solve the equation z5 = 4 − 4i, leaving your answers in the form z = Reikπ , (4 marks) where R is the modulus of z and k is a rational number such that −1 < k < 1. c Show on an Argand diagram the points representing the roots. E/P 15 a Find the cube roots of 2 − 2i in the form reiθ where r . 0 and −π , θ < π. (2 marks) (5 marks) These cube roots are represented by points A, B and C in the Argand diagram, with A in the fourth quadrant and ABC going anticlockwise. The midpoint of AB is M, and M represents the complex number w. E/P b Draw an Argand diagram, showing the points A, B, C and M. (2 marks) c Find the modulus and argument of w. (2 marks) d (3 marks) Find w6 in the form a + bi. 16 An equilateral triangle has its centre at the origin and one vertex at the point (2, 1). a Find the coordinates of the other two vertices. ___ b Show that the length of one side of the triangle is √ 15 . (4 marks) (2 marks) Challenge Show that the points on an Argand diagram that represent the roots z+1 6 of (_____   z  )  = 1lie on a straight line. 29 M01_ALEVEL_CP2_83343_U01_001-030.indd 29 26/02/2018 09:17 Chapter 1 Summary of key points 1 You can use Euler’s relation, eiθ = cos θ + i sin θ, to write a complex number z in exponential form: z = reiθ where r = |z| and θ = arg z. 2, 2 For any two complex numbers z1 = r1eiθ1  and z2 = r2eiθ • z1z2 = r1r2ei(θ1 + θ2) z1 ___ r1 i(θ − θ ) 1 2 •  __ z2  =   r2 e 3 De Moivre’s theorem: For any integer n, (r(cos θ + i sin θ))n = rn(cos nθ + i sin nθ) 1 4 • z + __   z  = 2 cos θ 1 • z − __   z  = 2i sin θ 1   n  = 2 cos nθ • zn + __ z 1   n  = 2i sin nθ • zn − __ z 5 For w, z ∈ ℂ, n−1 w(z  n − 1)     • ∑ wz  r= w + wz + wz  2 + … + wz  n−1 = ________ z−1 r=0 ∞ w • ∑   wz  r= w + wz + wz  2 + … = _____     , |z| , 1 1 −z r=0 6 If z and w are non-zero complex numbers and n is a positive integer, then the equation zn = w has n distinct solutions. 7 For any complex number z = r(cosθ + i sinθ), you can write z = r(cos (θ + 2kπ) + i sin (θ + 2kπ)) where k is any integer. 2πk 2πk ___ ____   2πik    n  8 In general, the solutions to z  n = 1 are z = cos (____ ) + i sin (  n  ) = e  n for k = 1, 2, … , n and are known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω = e    n  such that: 2πi __ • The nth roots of unity are 1, ω, ω2, … , ω n−1 • 1, ω, ω2, … , ωn−1 form the vertices of a regular n-gon • 1 + ω + ω2 + … + ω n−1 = 0 9 The nth roots of any complex number s lie on the vertices of a regular n-gon with its centre at the origin. 10 If z 1is one root of the equation z n = s, and 1, ω, ω2, … , ω n−1 are the nth roots of unity, then the roots of z n = s are given by z 1, z 1ω, z 1ω2, … , z 1ω n−1. 30 M01_ALEVEL_CP2_83343_U01_001-030.indd 30 26/02/2018 09:17 2 Series Objectives After completing this chapter you should be able to: ● Understand and use the method of differences to sum finite series → pages 32–37 ● Find and use higher derivatives of functions → pages 38–39 ● Know how to express functions as an infinite series in ascending → pages 40–44 powers using Maclaurin series expansion ● Be able to find the series expansions of compound functions → pages 44–48 Prior knowledge check 1 Find the sums of the following series. 18 a∑  (99 − 4n) n=1 2 16 b∑   _12 (3)n−1 n=6 ← Pure Year 2, Chapter 3 a Show that n Physicists use Maclaurin series in special relativity to approximate the Lorentz factor. The Lorentz factor relates time, length and relativistic mass change for a moving object. Experiments with atomic clocks have shown that time passes more quickly for a stationary observer than for one travelling at high speeds. → Exercise 2D, Challenge ∑  (r2 + 2r + 3) = _  16  n(2n2 + 9n + 25) r=1 30 b Hence find ∑   (r2 + 2r + 3) 10 3 ← Book 1, Chapter 3 Given y = sin 3x, find: dy a ___  dx d 2y b ___2  dx ← Pure Year 2, Chapter 9 31 M02_ALEVEL_CP2_83343_U02_031-051.indd 31 24/09/18 8:22 AM Chapter 2 2.1 The method of differences You can use the method of differences to find the sum of a finite series. ■ If the general term, ur, of a series can be expressed in the form f(r) − f(r + 1) n n You can also start with ur written in the form f(r + 1) − f(r). After adding and cancelling, r=1 r=1 you get ∑  u   r = f(n + 1) − f(1) then ∑  ur = ∑  (f( r) − f(r + 1)) u1 = f(1) − f(2) u2 = f(2) − f(3) u3 = f(3) − f(4) ⋮ un = f(n) − f(n + 1) so n r = f(1) − f(n + 1) Then adding ∑  u r=1 Example n r=1 u 1 + u 2 = f(1) − f(2) + f(2) − f(3) = f(1) − f(3) The f(2) terms cancel. By summing u  1 + u 2 + … + un  all terms cancel except the very first term, f(1), and the very last term, f(n + 1). 1 a Show that 4r 3 ≡ r 2(r + 1)2 − (r − 1)2r 2 b Hence prove, by the method of differences, that n _ ∑ r 3 =   4 n2(n + 1)2 1 r=1 a r 2(r + 1)2 − (r − 1)2r 2 ≡ r 2(r 2 + 2r + 1) − (r 2 − 2r + 1)r 2 ≡ r 4 + 2r 3 + r 2 − r 4 + 2r 3 − r 2 ≡ 4r 3 Start with the RHS. Expand and simplify the brackets. n b Consider ∑  (r 2(r + 1)2 − (r − 1)2r 2) r=1 r = 1: r = 2: r = 3: ⋮ r = n: Let Sum of terms 12(2)2 − (0)212 22(3)2 − (1)222 33(4)2 − (2)223 n2(n + 1)2 − (n − 1)2n2 = n2(n + 1)2 n Then 4 ∑  r 3 = n2(n + 1)2 r=1 n So ∑  r 3 = __   41  n2(n + 1)2 r=1 All the terms cancel except the first and last. Watch out When using the method of differences, be sure to write out enough terms to make it clear which terms cancel. When you cancel terms, make sure that they can still be clearly read. You could cross them out in pencil. The same result could be proved by mathematical induction. ← Book 1, Chapter 8 32 M02_ALEVEL_CP2_83343_U02_031-051.indd 32 24/09/18 8:22 AM Series Example 2 n 1 1 1 1 Verify that _______       _____       __  and hence find ∑  _______  using the method of differences. r(r + 1) ≡ r − r + 1 r(r + 1) r=1 r   − _____   1   ≡ ________   r + 1 −   r(r + 1) r r+1 1 _ ≡ _______   1   r(r + 1) n Write as a single fraction. Simplify. n   1   ≡ ∑  (_ 1 − _____   1  ) ∑   _______ r+1 r = 1 r(r + 1) r=1 r Let r = 1: _1  − __   1   1 2 1   − __ r = 2: __   1   2 3 1   − __ r = 3: __   1   3 4 ⋮ All terms cancel except the first and last.   1   r = n: __1 − _____ n n+1 n 1   = 1 − _____   1   So ∑   _______ n+1 r = 1 r(r + 1) n+1−1 =  ________   n+1 Put over a common denominator. n =  _____  n+1 Example 3 n 1 Find ∑   ______ 2 − 1 using the method of differences. 4r r=1 1 _______   1     ≡ ______________      4r 2 − 1 (2r + 1)(2r − 1) 1 ______________      ≡ ______   A   + ______   B   (2r + 1)(2r − 1) 2r + 1 2r − 1 A(2r − 1) + B(2r + 1) ___________________ ≡           (2r + 1)(2r − 1) so 1 ≡ A(2r − 1) + B(2r + 1) Use the difference of two squares to factorise the denominator. Split the fraction into partial fractions. ← Pure Year 2, Chapter 1 Add the fractions. Set numerators of both sides equal to each other. 33 M02_ALEVEL_CP2_83343_U02_031-051.indd 33 24/09/18 8:22 AM Chapter 2 Let r = _  21  : 1=0+B×2 Let r = −  _21  : 1 = A × −2 B=_  21   Put values of r in to find A and B. A = −  _21   _1  2  −  _21   1   ≡ ______ ______ So _______   +       2 2r + 1 2r −1 4r − 1   1   − ______ ≡_  21  (______   1   2r − 1 2r + 1) n n 1 1 1 _1 ______ ______ ⇒ ∑   _______ 2 − 1  =   2  ∑  (  2r − 1  −   2r + 1 ) 4r r=1 r=1 Let r = 1: _11  − _  31   r = 2: _31   − _  51   _1 _1 r = 3: 5  −  7  ⋮ 1   − ______ r = n: ______   1   2n − 1 2n + 1 All terms cancel except the first and last. 1 1 Substitute the values of r into ______     − ______     2r − 1 2r + 1 1 only. The  _2 is only required later. n 1   = _   1  ) So ∑   _______  21  (1 − ______ 2 2n + 1 r = 1 4r − 1 2n + 1 − 1 =  _21  ( _________   2n + 1 ) n = ______     2n + 1 If the general term of the series is given in the form f(r) − f(r + 2), you need to adapt the method of differences to consider the terms f(1), f(2), f(n + 1) and f(n + 2). Example 4 2 a Express ____________     in partial fractions. (r + 1)(r + 3) b Hence prove by the method of differences that n n(an + b) 2 ∑  ____________     = ______________      6(n + 2)(n + 3) r = 1 (r + 1)(r + 3) where a and b are constants to be found. 30 2 c Find the value of ∑   ____________    to 5 decimal places. (r + 1)(r + 3) r = 21 34 M02_ALEVEL_CP2_83343_U02_031-051.indd 34 24/09/18 8:22 AM Series 2 A   + _____ a ____________     ≡  _____   B   (r + 1)(r + 3) r + 1 r + 3 Split into partial fractions. A(r + 3) + B(r + 1)    ≡ _________________        (r + 1)(r + 3) ⇒ Add the fractions. 2 ≡ A(r + 3) + B(r + 1) Compare numerators. Let r = −3: 2 = −2B ⇒ B = −1 2 = 2A ⇒ A = 1 2 1   − _____ Therefore ____________      ≡  _____   1   (r + 1)(r + 3) r + 1 r + 3 Let r = −1: b Using the method of differences,  41   when r = 1: _21   − _ r = 2: _31   − _  51   Cancel terms. r = 3: _41   − __  61   ⋮ Problem-solving 1 _____ 1 r = n − 1: __ n −  n + 2  r = n:  1 _____ n ∑ (f(r) − f(r + 2))= r=1 f(1) + f(2) − f(n + 1) − f(n + 2) 1 _____     −   n+1 n+3 n 5 2     =  __  − _____ So ∑   ____________   1   − _____   1   6 n+2 n+3 r = 1 (r + 1)(r + 3) Put these four terms over a common denominator. 5(n + 2)(n + 3) − 6(n + 3) − 6(n + 2)          = _________________________________________________     6(n + 2)(n + 3) 5n2 + 25n + 30 − 6n − 18 − 6n − 12          =  _________________________________________________   6(n + 2)(n + 3) 5n2 + 13n       = ______________     6(n + 2)(n + 3) n(5n + 13)           = ______________ 6(n + 2)(n + 3) Factorise. So a = 5 and b = 13. 30 30 20 2 2 2     = ∑    ____________     − ∑    ____________     c ∑    ____________ r = 1 (r + 1)(r + 3) r = 1 (r + 1)(r + 3) r = 21 (r + 1)(r + 3) 20 30 r=1 r=1 Subtract ∑  from ∑  . 30(5 × 30 + 13) _________________ 20(5 × 20 + 13)          = _________________     −        6(30 + 2)(30 + 3) 6(20 + 2)(20 + 3) 815 565   = _____     − ____   1056 759 Evaluate. 665 = _______     = 0.02738 to 5 d.p. 24 288 Give answer to 5 d.p. 35 M02_ALEVEL_CP2_83343_U02_031-051.indd 35 24/09/18 8:22 AM Chapter 2 Exercise 2A 1 a Show that r ≡ _  2 (r(r + 1) − r(r − 1)). 1 n n    (n + 1) using the method of differences. b Hence show that ∑ r = __ 2 r=1 E 1 1 1 2 Given  _____________     ≡ ________          − _____________ r(r + 1)(r + 2) 2r(r + 1) 2(r + 1)(r + 2) n 1     using the method of differences. find ∑ _____________ r(r + 1)(r + 2) r=1 E/P 1 3 a Express  _______ in partial fractions. r(r + 2) n 1    using the method of differences. b Hence find the sum of the series ∑ _______ r(r + 2) r=1 E 1 4 a Express ____________     in partial fractions. (r + 2)(r + 3) (5 marks) (1 mark) (5 marks) (1 mark) n 1 b Hence find the sum of the series ∑ ____________     using the method of differences. (5 marks) (r + 2)(r + 3) r=1 E/P E P 1 1 r     − _______   5 a Show that _______     ≡ __   (r + 1)! r! (r + 1)! n r     b Hence find ∑ _______ (r + 1)! r=1 (2 marks) (5 marks) n 1 _______ 1 2r + 1 2r + 1 __ ________   ≡     −     , find   2   6 Given that  _________ ∑ 2 2 2 2 r (r + 1) r (r + 1) r (r + 1)2 r=1 (6 marks) n n 11 7 a Use the method of differences to prove that ∑   ___________   = ______     , where a and b are an + b r = 1 (2r + 3)(2r + 5) constants to be found. b Prove your result from part a using mathematical induction. E/P n n(an + b) 8 8 8 Prove that ∑   ___________   = ______________      , where a and b are constants to be found. (3n + 1)(3n + 4) r = 1 (3r – 2)(3r + 4) (6 marks) Hint This question can be answered using either the method of differences or proof by induction. In the exam, either method would be acceptable. If you use proof by induction, you will need to substitute values of n to find the values of a and b. E/P n 9 Prove that ∑ ( r + 1)  2 − ( r − 1)  2 = an(n + 1), where a is a constant to be found. (4 marks) r=1 36 M02_ALEVEL_CP2_83343_U02_031-051.indd 36 24/09/18 8:22 AM Series E/P n 3 an 10 a Prove that ∑ ____________       = ______     , where a, b and c are constants to be found. (5 marks) (3r + 1)(3r + 4) bn + c r=1 2n 33 3(n + 1) b Hence, or otherwise, show that ∑   ___________   = ______________       (4 marks) ( 23n + 1)( 3n + 2) r = n (3r + 1)(3r + 4) n E/P 2r + 1 1       = 1 − _____   11 Robin claims that ∑  _______ n + 1 r(r + 1) r=1 His workings are shown below. Explain the error that he has made. Using partial fractions: 2r + 1 _______   r(r + 1) B A   ≡ __     + _____ r r+1 Therefore 2r + 1 ≡ A(r + 1) + Br So A = 1 and B = 1. Using the method of differences, 1 f(1) = 1 +  __ 2 f(2) = __   21 + __   31 f(3) = __   31 + __   41 ⋮ 1 1 f(n − 1) =  _____ + __     n−1 n 1 1 f(n) = __     + _____   n n+1 n 2r + 1 1 _______ _____ Summing the differences: ∑  r(r + 1) = 1 −  n + 1 r=1 E/P E/P (2 marks) an + b 3 1 1 1 1 12 Show that _____        , where a and b are   + _____     + _____     + … + _______     = __     − _____________ 1×3 2×4 3×5 n(n + 2) 4 2(n + 1 ) (n + 2) constants to be found. (6 marks) 4 13 a Express ____________      in partial fractions. (2r + 1)( 2r + 5) 25 4      to 4 decimal places. b Find the value of ∑  ____________ ( )( ) r = 16 2r + 1 2r + 5 (3 marks) (5 marks) Challenge 30 1 a Given that ∑   ln(1 + _____     = ln k, where k is an integer, find k. r + 2) r=1 n n(an2 + bn + c) 18 b Given that ∑      _______  =   _________________       , find a, b and c. (n + 1)(n + 2)(n + 3) r=1 r(r + 3) 37 M02_ALEVEL_CP2_83343_U02_031-051.indd 37 24/09/18 8:22 AM Chapter 2 2.2 Higher derivatives You need to be able to find third, and higher, derivatives of given functions. You already know how to find first and second derivatives. dy     = f9(x), and the second derivative of f(x) is given If y = f(x), the first derivative of f(x) is given by ___ dx dy d  2y ___ d ___ ____ by   2  =    (   ) = f 0(x). dx  dx dx d  3y d ____ d  2y Similarly, the third derivative is given by ____   3  = ___    (  2 ) = f -(x), and so on. dx  dx dx  You can find the nth derivative of f(x) by differentiating n times with respect to x. Example Notation The nth derivative of y = f(x) is d   ny ____ written as   n  = f  (n)(x). dx  5 d  3y 1     Given that y = ln(1 − x), find the value of ____   3  when x = __ 2 dx  Use the chain rule. dy ___   d 1     1   ___   1   × (−1) = −  _____   = _____     (1 − x) = _____ dx 1 − x dx 1−x 1−x ← Pure Year 2, Section 9.3 ( ) 2 d dy d 1   × (−1) = −  _______ 1       ___     = ___     (−(1 − x)−1) =  _______   2  = ___ dx dx dx dx (1 − x)2 (1 − x)2 dy ___ ( ) dy ___ 3 d d2y d 2 2   3  = ___     ___      = ___     (−(1 − x)−2) =  _______   × (−1) = −  _______   dx dx2 dx dx (1 − x)3 (1 − x)3 d y _______ −2 1  , ____ So when x =  __   =     3  = −16 2 dx3 ( 1 − _   1    ) 3 Substitute x = _  12  2 Example 6 f(x) = ex 2 a Show that f 9(x) = 2xf(x). b By differentiating the result in part a twice more with respect to x, show that: i f 0(x) = 2f(x) + 2xf 9(x) ii f -(x) = 2xf 0(x) + 4f 9(x) c Deduce the values of f 9(0), f 0(0), and f -(0). 2 2 d a f9(x) = ex ___    (x2) = 2xex dx = 2xf(x) b i f0(x) = 2f(x) + 2xf9(x) ii f-(x) = 2f9(x) + (2xf0(x) + 2f9(x)) = 2xf0(x) + 4f9(x) If f(x) = eu, then f9(x) = eu ___   du  dx f(x) = ex 2 Use the product rule. ← Pure Year 2, Section 9.4 Differentiate again. 38 M02_ALEVEL_CP2_83343_U02_031-051.indd 38 24/09/18 8:22 AM Series c f(0) = e0 = 1 f9(0) = 2 × 0 × e0 = 0 f0(0) = 2f(0) + 2 × 0 × f9(0) = 2f(0) = 2 f-(0) = 2 × 0 × f0(0) + 4f9(0) = 4f9(0) = 0 Exercise Substitute x = 0 into f0(x). Substitute x = 0 into f-(x). 2B 1 For each of the following functions, f(x), find f9(x), f 0(x), f -(x) and f (n)(x). a e2x P b (1 + x)n c xex d ln(1 + x) d ny 2 a Given that y = e2 + 3x, find an expression, in terms of y, for ____   n  dx 6 dy 1 b Hence evaluate ___   6  when x = ln (_  9 ). dx 3 Given that y = sin2 3x, dy a show that  ___  = 3 sin 6x dx d2y d3y d 4y b find expressions for  ____2  ,  ____3  and ____   4  dx dx dx d4y π c Hence evaluate  ____4  when x = __     6 dx 4 f(x) = x2e−x. a Show that f -(x) = (6x − 6 − x2)e−x. b Show that f -9(2) = 0. 5 Given that y = sec x, d2y a show that  ____2  = 2 sec3 x − sec x dx __ d3y π b show that the value of ____   3  when x = __     is 11√2 . 4 dx P 6 Given that y is a function of x, ( ) d2y dy 2 d2 a show that  ____2  ( y2) = 2y  ____2  + 2 ___      dx dx dx dy d2y d3y d3 b Find an expression, in terms of y, ___    , ____   2  and ___   3  , for ____   3  ( y2). dx dx dx dx ______ 7 Given that f(x) = ln (x + √ 1 + x2  ), show that: ______ a√1 + x2  f 9(x) = 1 b (1 + x2) f 0(x) + xf 9(x) = 0 c (1 + x2) f -(x) + 3xf 0(x) + f 9(x) = 0 d Deduce the values of f 9(0), f 0(0) and f -(0). 39 M02_ALEVEL_CP2_83343_U02_031-051.indd 39 24/09/18 8:22 AM Chapter 2 2.3 Maclaurin series Many functions can be written as an infinite sum of terms of the form axn. You may have already encountered series expansions like these: 1 _____   = 1 + x + x  2 + x  3 + …, |x| , 1   1−x x   2 x 2 8 16 x  2 x  3 x   4 e  x = 1 + x + __     +  __  +  ___  + …, x ∈ ℝ 2 6 24 ____ x  3 √ 1 + x  = 1 + __     − __     +  ___  − …, |x| , 1 Example Links The first two series expansions shown here are examples of the binomial expansion. ← Pure Year 2, Chapter 4 7 Given that f(x)can be differentiated infinitely many times and that it has a valid series expansion of the form f(x) = a 0 + a 1x + a 2x  2 + a 3x  3 + … + ar  x  r + …, where the ai are all real constants, show that the series expansion must be f 0(0)x  2 f  (r) (0)x  r     + … +   _______   + … f(x) = f(0) + f 9(0)x + ______ 2! r! Write f(x) = a 0 + a1x + a2x2 + a3x3 + … + arxr + … f(0) = a 0 Differentiating f(x) gives: f9(x) = a 1 + 2a 2x + 3a 3x  2 + … + rar  x  r−1 + … f0(x) = 2 × 1a 2 + 3 × 2a 3x + … + r (r − 1) ar  x  r−2 + … f-(x) = 3 × 2 × 1a 3 + … + r (r − 1)(r − 2) ar  x  r−3 + … The coefficient of a 0 can be found by setting x = 0. Successively differentiate with respect to x to obtain f9(x), f 0(x) and f -(x). Continuing in this way by differentiating r times: f  (r)( x) = r!ar  + terms in powers of x Evaluate each term at x = 0: f9(0) = a 1 ⇒ a 1 = f9(0) f0(0) f 0(0) = 2!a 2 ⇒ a 2 = ____     2! f-(0)     f-(0) = 3!a 3 ⇒ a 3 = _____ 3! f  (r)(0)     f   (r)(0) = r!ar  ⇒ ar  = ____ r! f0(0) f-(0) 3 f   (r)(0)     x  2 + _____       xr + … Therefore f(x) = f(0) + f9(0)x + _____   x  + … + _____ r! 2! 3! Find the coefficients a1, a2, a3,…, ar,… by substituting x = 0 into each result and rearranging. Substitute a  1 = f9(0),  f0(0) f-(0) a 2 = ____      , a 3 = _____     , … , 2! 3! (r)( ) f  0     , … ar  = ____ r! 40 M02_ALEVEL_CP2_83343_U02_031-051.indd 40 24/09/18 8:22 AM Series In this process, outlined in the worked example above, a polynomial in powers of x is being formed step by step. The process focuses on x = 0; substituting x = 0 into successive derivatives increases the power of the polynomial. For example, if you stop the process after finding f9(0) the polynomial is f0(0) linear, f(0) + f9(0)x, after f0(0) it is quadratic, f(0) + f9(0)x + ____     x  2, after f-(0) it is cubic, 2! f-(0) 3 f0(0) 2 ____ ____ f(0) + f9(0)x +     x  +     x  and so on. Watch out Not all functions satisfy the 2! 3! condition that f(0), f9(0), f0(0), … , f  (r)(0) all The above argument assumes that the function can have finite values. be written in the given form. This is only true if the 1  For example, when f(x) = ln x, f9(x) = __  x given series converges. The above reasoning also only so f9(0)is undefined and therefore does not holds if the function can be differentiated an infinite have a finite value. number of times, and if f(r)(0) is always finite. ■ The Maclaurin series expansion of a function f(x) is given by f0(0) f(r)(0) r f(x) = f(0) + f9(0)x +   _____  x2 + … +   _____   x + … 2! r! The series is valid provided that f(0), f9(0), f 0(0), … , f(r)(0), … all have finite values. The polynomial f(0) + f9(0)xis a Maclaurin polynomial of degree 1. f0(0) The polynomial f(0) + f9(0)x + ____     x  2is a Maclaurin polynomial of degree 2. 2! f0(0) 2 f  (r)( 0) ____     x  ris a Maclaurin polynomial of degree r. The polynomial f(0) + f9(0)x +     x  + … + ____ 2! r! r ( ) Even when f (0) exists and is finite for all r, a Maclaurin series expansion is only valid for 1 values of x that give rise to a convergent series. For example, the Maclaurin series of _____     1−x is 1 + x + x  2 + x   3 + … . 1 But when x = 2, the series gives 1 + 2 + 4 + 8 + … which does not converge to _____     = −1. 1−2 Note The range of validity for some individual Maclaurin series is Example given in the formulae booklet. If no range of validity is given in this chapter, you may assume that the expansion is valid for all x ∈ ℝ. 8 a Express ln(1 + x) as an infinite series in ascending powers of x. b Using only the first three terms of the series in part a, find estimates for: i ln 1.05 ii ln 1.25 iii ln 1.8 Comment on the accuracy of the estimates. a f(x) = ln (1 + x) ⇒ f(0) = ln 1 = 0 f9(x) = _____   1   = (1 + x)−1 ⇒ 1+x f0(x) = −(1 + x)−2 ⇒ f0(0) = −1 f-(x) = (−1)(−2)(1 + x)−3 ⇒ f-(0) = 2! f9(0) = 1 f(r)(x) = (−1)(−2)(−3)…(−(r − 1))(1 + x)−r ⇒ f(r)(0) = (−1)r − 1(r − 1)! Problem-solving The term ( −1)  rcan be used in the general term of alternating sequences, in which the terms are alternately positive and negative. 41 M02_ALEVEL_CP2_83343_U02_031-051.indd 41 24/09/18 8:22 AM Chapter 2 (2!) So ln (1 + x) = 0 + 1x + ____  −1  x2 + ____     x3 + … 2! 3! (−1)r − 1(r − 1)! r + ____________        x + … r! x2 ___ x3 ___ x   + … + (−1)r − 1  __  + … r ln(1 + x) = x −     +   2 3 b i r 0.052 ______ 0.053 ln 1.05 = 0.05 − ______     +     − … 2 3 ≈ 0.0487916… This is correct to 5 d.p. ii 0.252 ______ 0.253 ln 1.25 = 0.25 − ______     +     − … 2 3 Substitute the values for f(0), f9(0), f 0(0) etc. into the Maclaurin series for f(x). Online This expansion is valid for −1 , x < 1. If you use a computer to generate the graphs of the successive Maclaurin polynomials you will see that they converge to the graph of ln (1 + x) between x = −1 and x = 1, but outside that interval they diverge rapidly. Explore this using GeoGebra. ≈ 0.223958… This is correct to 2 d.p. 0.82 _____ iii ln 1.8 = 0.8 −   0.83 _____   +   2   − … 3 ≈ 0.6506666… This is not correct to 1 d.p. Example The further away a value is from x = 0, the less accurate the approximation will be and the more terms of the series you need to take to maintain a required degree of accuracy. 9 a Find the first four terms in the Maclaurin series of sin x. b Using the first two terms of the series find an approximation for sin 10°. a f(x) = sin x f9(x) = cos x f0(x) = −sin x f-(x) = −cos x f-9(x) = sin x ⇒ ⇒ ⇒ ⇒ ⇒ f(0) = sin 0 = 0 f9(0) = cos 0 = 1 f0(0) = −sin 0 = 0 f-(0) = −cos 0 = −1 f-9(0) = sin 0 = 0 (−1)r 2r + 1 So sin x = x + ____  −1 x3 + __   1  x5 + ____  −1 x7 + … + _______   +…   x 5! 7! (2r + 1)! 3! = x − __   1   x3 + __   1   x5 − __   1   x7 + … 5! 7! 3! π π π 3 b sin 10° = sin  ___  ≈ ___     − __   1   ___     18 18 6 (18) ≈ 0.174532925 − 0.000886096 ≈ 0.173646829 f(n) = 0, if n is even, and the cycle of values 0, 1, 0, −1 repeats itself. This expansion is valid for all values of x. Watch out x must be in radians in expansions of trigonometric functions. This estimate is correct to 5 decimal places; even using sin x ≈ x, the approximation is correct to 2 d.p. 42 M02_ALEVEL_CP2_83343_U02_031-051.indd 42 24/09/18 8:22 AM Series Exercise 2C 1 Use the formula for the Maclaurin series and differentiation to show that: Hint The binomial expansions of (1 + x)n , where n is fractional or negative and |x| , 1, are the Maclaurin series of the function. a (1 − x)−1 = 1 + x + x2 + … + xr + … _____ x x2 ___ x3     − __     +     − … b√1 + x  = 1 + __ 2 8 16 2 Use Maclaurin series and differentiation to show that the first three terms in the series x2     expansion of esin x are 1 + x + __ 2 P x2 x4 x2r     + __     + … + (−1)r ____     + … 3 a Show that the Maclaurin series of cos x is 1 − __ 2! 4! (2r)! b Using the first three terms of the series, show that it gives a value for cos 30° correct to 3 decimal places. Hint This expansion is valid for all values of x. 4 Using the series expansions for ex and ln(1 + x) respectively, find, correct to 3 decimal places, the values of: b ln ( _  5   ) 6 a e 5 Use Maclaurin series and differentiation to expand, in ascending powers of x up to and including the term in x 4, b ln (1 + 2x) a e3x P c sin2 x 6 Using the addition formula for cos (A − B) and the series expansions of sin x and cos x, show that x2 x3 ___ x4 π 1    ) = ___   __  (1 + x − __     − __     +     + …) cos (x − __ 4 2 6 24 √ 2 E E/P 7 Given that f(x) = (1 − x)2 ln (1 − x), a show that f 0(x) = 3 + 2ln (1 − x) (2 marks) b find the values of f(0), f 9(0), f 0(0), and f -(0) (1 mark) c express (1 − x)2 ln (1 − x) in ascending powers of x up to and including the term in x3. (3 marks) 8 a Using the series expansions of sin x and cos x, show that   120 x5 + … 3 sin x − 4x cos x + x = _  2 x3 − ___ 3 17 3 sin x − 4x cos x + x b Hence, find the limit, as x → 0, of ____________________        x3 (5 marks) (1 mark) 43 M02_ALEVEL_CP2_83343_U02_031-051.indd 43 24/09/18 8:22 AM Chapter 2 E 9 Given that f(x) = ln cos x, a show that f 9(x) = −tan x (2 marks) b find the values of f 9(0), f 0(0), f -(0) and f -9(0) (1 mark) c express ln cos x as a series in ascending powers of x up to and including the term (3 marks) in x 4 π d show that using the first two terms of the Maclaurin series for ln cos x, with x = __    , gives a 4 π2 π2 ___ ___ (2 marks) value for ln 2 of     (1 +    ). 16 96 E/P  15 x5. 10 Show that the Maclaurin series for tan x, as far as the term in x5, is x + _ 3 x3 + __ 1 Challenge 2 (5 marks) Problem-solving The ratio test is a sufficient condition for the convergence of an infinite ∞ a r+1 series. It says that a series ∑    ar converges if r→∞  lim ____   a    , 1, and diverges r r=1 a r+1 if r→∞  lim ____   a    . 1. r | | | | a r+1 If r→∞  lim ____   a    = 1or does not | | r exist then the ratio test is inconclusive. Use the ratio test to show that a the Maclaurin series expansion of ex converges for all x ∈ ℝ b the Maclaurin series expansion of ln (1 + x) converges for −1 , x , 1, and diverges for x . 1. 2.4 Series expansions of compound functions You can find the series expansions of compound functions using known Maclaurin series. In the last exercise you found the Maclaurin series of simple compound functions, such as e  3x and ln (1 + 2x). However, the resulting series could also be found by replacing x by 3xor x by 2xin the known expansions of e  x and ln (1 + x)respectively. When successive derivatives of a compound function are more difficult, or when there are products of functions involved, it is often possible to use one of the standard results. ■ The following Maclaurin series expansions are given in the formulae booklet:  x   + … + __  x   + … for all x • ex = 1 + x + __ 2! r! 2 r • ln (1 + x) = x − __  x   + __  x  − … + (−1)r + 1 __  xr + … −1 , x < 1 2 3 2 3 r 3 5 2r + 1 • sin x = x − __  x   + __  x   − … + (−1)r ________ x   + … for all x 3! 5! (2r + 1)! 2 4 2r • cos x = 1 − __  x   + __  x   − … + (−1)r ____ x   + … for all x 2! 4! (2r)! 3 5 2r + 1 • arctan x = x − ___  x   + ___  x   − … + (−1)r ______ x   + … −1 < x < 1 5 3 2r + 1 44 M02_ALEVEL_CP2_83343_U02_031-051.indd 44 24/09/18 8:22 AM Series Example 10 Write down the first four non-zero terms in the series expansion, in ascending powers of x, of cos (2x2). (2x2)2 ______ (2x2)4 ______ (2x2)6 cos (2x2) = 1 − _____     +     −     + … 2! 4! 6! Substitute 2x2 for x in the above series for cos x. 4 2 = 1 − 2x4 + __    x8 − ___    x12 + … 45 3 Watch out Make sure you simplify the coefficients as much as possible. Example 11 ______ √1 + 2x  Find the first three non-zero terms in the series expansion of ln (  _______ ), and state the values of 1 − 3x x for which the expansion is valid. ln(   _______ ( ) a   = ln a − ln b Using ln  __ b   = ln √1 + 2x  − ln (1 − 3x) 1 − 3x ) √ (1 + 2x)  _________ _______ = __   21 ln (1 + 2x) − ln (1 − 3x) ( _1 Using ln a 2  = __  12  ln a ) (2x)2 _____ (2x)3 __1 _____ 1  __ 2 ln (1 + 2x) =   2 2x −   2  +   3  − … , −1 , 2x < 1 = x − x2 + __   43 x3 − … (−3x)2 − __   21 , x < __   21 (−3x)3 ln (1 − 3x) = (−3x) − _______       + _______   − …, −1 , −3x < 1 2 3 = −3x − __   92 x2 − 9x3 − … __1 1 −  __ 3 < x ,   3 ______ Substitute 2x for x in the expansion of ln(1 + x). Problem-solving You are substituting 2x into the series expansion of ln (1 + x), so the series is now only valid for −1 , 2x < 1, or − _  12  , x < _  12  √ 1 + 2x  ________ So ln     43 x3 − …)   = (x − x2 + __ 1 − 3x __ 9 2 − (−3x −   2 x − 9x3 − …), −  __1 < x , __   1 3 = 4x + __   72 x2 + __   31 3 x + … , 3 3 __1 1 −  __ 3 < x ,   3 Substitute −3x for x in the expansion of ln(1 + x). You need both intervals to be satisfied. This is the case for −  _13 < x , _  13 45 M02_ALEVEL_CP2_83343_U02_031-051.indd 45 24/09/18 8:22 AM Chapter 2 Example 12 Given that terms in xn with n . 4 may be neglected, use the series expansions for ex and sin x to show that x2 x4     − __     esin x ≈ 1 + x + __ 2 8 Only two terms are used as the next term is kx  5. x   3 sin x = x − __     + … 3! So e  sin x = e  (x −   3!  + …) __ x3 __ x3 = e  x × e  −   6  × … x  2 x  3 ___ x  3 x  4 = ( 1 + x + __     + __     +     + …)(1 + ( −  __ ) + …) … 2 6 24 6 x  2 x  3 x  4 __ x  3 x  4 = 1 + x + __     +  __  + ___     −     − __     + … 2 6 24 6 6 x  2 x  4 ≈ 1 + x +  __  −  __  2 8 Simplify as much as possible. 1 Use the series expansions of ex, ln (1 + x) and sin x to expand the following functions as far as the fourth non-zero term. In each case state the values of x for which the expansion is valid. 1 a __ ex  e2x × e3x b ________ ex  c e1 + x d ln (1 − x) x e sin __      2 f ln (2 + 3x) ( ) E/P x  3 Substitute −  __  for xin the 6 expansion of e  x. 2D Exercise P Use e  a−b = e  a × e  −b Hint For part f, write 2 + 3x as 2(1 + ___   3x ) . 2 2 a Using the Maclaurin series of ln (1 + x), show that ( ) 1+x x3 x5     + …), −1 , x , 1       + __    = 2(x + __ ln _____ 5 3 1−x (4 marks) _____ √ 1+x b Deduce the series expansion for ln _____     , −1 , x , 1 1−x (2 marks) c By choosing a suitable value of x, and using only the first three terms of the series from 2 part a, find an approximation for ln ( _ 3   ), giving your answer to 4 decimal places. (2 marks) d Show that the first three terms of your series from part b, with x = _ 5 , give an approximation for ln 2, which is correct to 2 decimal places. (2 marks) 3 E/P 3 Show that, for small values of x, e2x − e−x ≈ 3x + _ 2 x2. (4 marks) E/P   2 x2 − __   8 x 4 − … 4 a Show that 3x sin 2x − cos 3x = −1 + __ (5 marks) 3 21 ( 59 ) 3x sin 2x − cos 3x + 1 _____________________    b Hence find the lim       x2 x→0 (1 mark) 46 M02_ALEVEL_CP2_83343_U02_031-051.indd 46 24/09/18 8:22 AM Series P 5 Find the series expansions, up to and including the term in x 4, of: a ln (1 + x − 2x2) Note Factorise the quadratic first. b ln (9 + 6x + x2) and in each case give the range of values of x for which the expansion is valid. E/P 6 a Write down the series expansion of cos 2x in ascending powers of x, up to and including the (3 marks) term in x8. b Hence, or otherwise, find the first four non-zero terms in the series expansion for sin2 x. (3 marks) E/P E/P 7 Show that the first two non-zero terms of the series expansion, in ascending powers of x, of ln (1 + x) + (x − 1)(ex − 1) are px3 and qx 4, where p and q are constants to be found. (6 marks) sin x 8 a By considering the product of the series expansions of sin x and (1 − x)−2, expand _______     (1 − x)2 in ascending powers of x as far as the term in x 4. (6 marks) sin x     b Deduce the gradient of the tangent, at the origin, to the curve with equation y = _______ (1 − x)2 (3 marks) P 9 Use the Maclaurin series, together with a suitable substitution, to show that:   3 x3 − 12x 4 + … a (1 − 3x) ln (1 + 2x) = 2x − 8x2 + __ 26   6 x3 + x 4 + … b e2x sin x = x + 2x2 + __ 11 ______ c √1 + x2  e−x = 1 − x + x2 − _ 3 x3 + _  6 x 4 + … E/P 2 1 x2 − __  10 a Write down the first five non-zero terms in the series expansions of e 2 ∫ 1 (3 marks) x2 − __  b Using your result from part a, find an approximate value for   e 2 dx, giving your answer to 3 decimal places. E/P −1 3(p2 − 3) 3 11 a Show that e px sin 3x = 3x + 3px2 + _________   x + … where p is a constant.   2 b Given that the first non-zero term in the expansion, in ascending powers of x, of e px sin 3x + ln (1 + qx) − x is kx3, where k is a constant, find the values of p, q and k. E/P (3 marks) (5 marks) (4 marks) 12 f(x) = ex − ln x sin x, x . 0 a Show that, if x is sufficiently small, x 4 and higher powers of x may be neglected, x2     f(x) ≈ 1 + x + __ 3 (5 marks) b Show that using x = 0.1 in the result from part a gives an approximation for f(0.1) which is correct to 6 significant figures. (2 marks) 47 M02_ALEVEL_CP2_83343_U02_031-051.indd 47 24/09/18 8:22 AM Chapter 2 E/P 13 y = sin 2x − cos 2x d 4y a Show that  ___4  = 16y dx (4 marks) b Find the first five terms of the Maclaurin series for y, giving each coefficient in its simplest form. (4 marks) Challenge Note A light year is the The Lorentz factor of a moving object, λ, is given by the formula 1   γ = _______   ______ √1 − β 2  distance light travels in one year. where β = _ cv  is the ratio of v, the speed of the object, to c, the speed of light (3 × 108 m s−1). 1   in ascending ______ a Find the Maclaurin series expansion of γ =  _______ 1 − β 2  √ powers of β up to the term in β4 The theory of special relativity predicts that a period of time observed as T within a stationary frame of reference will be observed as a period T in a moving frame of reference. of time  __ γ A spaceship travels from Earth to a planet 4.2 light years away. To an observer on Earth, the journey appears to take 20 years. b Use your answer to part a to estimate the observed journey time for a person on the spaceship. c Calculate the percentage error in your estimate. d Comment on whether your approximation would be more or less accurate if the spaceship was travelling at three times the speed. Mixed exercise E/P 2 2 1 a Express  __________ in partial fractions. (r + 2)( r + 4) n 7n  2 + 25n 2       b Hence show that  ∑   __________  = _____________ r = 1 (r + 2)( r + 4) 12(n + 3)( n + 4) E/P 4 2 a Express ____________      in partial fractions. (4r − 1)( 4r + 3) (1 mark) (5 marks) (2 marks) b Using your answer to part a and the method of differences, show that n 4 4n   (   ∑   ____________       = ________ ( ) ( ) 3 4n + 3) 4r − 1 4r + 3 r=1 200 4 c Evaluate   ∑   ____________    giving your answer to 3 significant figures. r = 100 (4r − 1)( 4r + 3) (3 marks) (2 marks) 48 M02_ALEVEL_CP2_83343_U02_031-051.indd 48 24/09/18 8:22 AM Series E 3 a Show that (r + 1)  3 − ( r − 1)  3 = 6r  2 + 2. (2 marks) b Using the result from part a and the method of differences, show that n   ∑ r  2 =  _6 n(n + 1)( 2n + 1) 1 (5 marks) r=1 E/P E/P n n(an + b) 4     = ____________      , where a and b are constants to be found. (5 marks) 4 Prove that   ∑ __________ r = 1 (r + 1)(r + 3) 3(n + 2)(n + 3) 2n 5 Prove that   ∑ ((r + 1)3 − (r − 1)3)= an  3 + bn  2 + cn + d, where a, b, c and d are constants to r=n be found. (5 marks) dny   n  6 a Given that y = e  1−2x, find an expression, in terms of y, for ___ dx d8y e     b Hence show that ___   8  at x = ln 32 is __ dx 4 7 a For the function f(x) = ln (1 + ex), find the values of f9(0) and f 0(0). b Show that f -(0) = 0. c Find the series expansion of ln (1 + ex), in ascending powers of x up to and including the term in x2. E/P 8 a Write down the Maclaurin series of cos 4x in ascending powers of x, up to and including the (3 marks) term in x6. b Hence, or otherwise, show that the first three non-zero terms in the series expansion of sin2 2x are 4x2 − __   3  x4 + ___   45  x6. 16 128 (3 marks) E/P 9 Given that terms in x5 and higher powers may be neglected, use the Maclaurin series for ex and x2 x4 cos x, to show that ecos x ≈ e(1 − __     + __    ). (5 marks) 2 6 E/P 10 Given that |2x| , 1, find the first two non-zero terms in the series expansion of ln ((1 + x)2(1 − 2x)) in ascending powers of x. (5 marks) E/P 11 Use differentiation and Maclaurin series, to express ln (sec x + tan x) as a series in ascending (5 marks) powers of x up to and including the term in x3. P 12 Show that the results of differentiating the standard series expansions of e x,  sin x and cos x agree with the following: d a ___   (ex) = ex dx d b ___  (sin x) = cos x dx d c ___   (cos x) = −sin x dx 49 M02_ALEVEL_CP2_83343_U02_031-051.indd 49 24/09/18 8:22 AM Chapter 2 E x2 x4 x2 5 4 13 a Given that cos x = 1 − __     + __     − …, show that sec x = 1 + __     + ___     x + … . 2 24 2! 4! (4 marks) x3 x5 b Using the result found in part a, and given that sin x = x − __     + __     − …, find the first three 3! 5! non-zero terms in the series expansion, in ascending powers of x, for tan x. (4 marks) E/P 14 By using the series expansions of ex and cos x, or otherwise, find the expansion of ex cos 3x in ascending powers of x up to and including the term in x3. (5 marks) E/P 15 Find the first three derivatives of (1 + x)2 ln (1 + x). Hence, or otherwise, find the expansion of (5 marks) (1 + x)2 ln (1 + x) in ascending powers of x up to and including the term in x3. E/P 16 a Expand ln (1 + sin x) in ascending powers of x up to and including the term in x4. (4 marks) ∫ π _   6  b Hence find an approximation for   ln (1 + sin x) dx giving your answer to 0 3 decimal places. E/P x3 17 a Using the first two terms, x + __     , in the expansion of tan x, show that 3 x2 __ x3 __ x tan = 1 + x +     +     + … e 2 2 (3 marks) (3 marks) b Deduce the first four terms in the series expansion of e−tan x, in ascending powers of x. (3 marks) P x2 x4 18 a Using Maclaurin series, and differentiation, show that ln cos x = − __     − ___     + … 2 12 x      − 1, and the result in part a, show that b Using cos x = 2 cos2 __ 2 ( ) x2 x4 ln (1 + cos x) = ln 2 − __     − ___     + … 4 96 E/P 19 y = e  3x − e  −3x d 4y a Show that  ____4  = 81y. dx (4 marks) b Find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form. (3 marks) c Find an expression for the nth non-zero term of the Maclaurin series for y. (2 marks) Challenge Given that the Maclaurin series of ex is valid for all x ∈ ℂ, show, using series expansions, that eix = cos x + i sin x. 50 M02_ALEVEL_CP2_83343_U02_031-051.indd 50 24/09/18 8:22 AM Series Summary of key points 1 If the general term, ur , of a series can be expressed in the form f(r) − f(r + 1) n n r=1 r=1 then   ∑ u  r =   ∑ (  f(r) − f(r + 1)). so u1 = f(1) − f(2) u2 = f(2) − f(3) u3 = f(3) − f(4) ⋮ un = f(n) − f(n + 1) n Then adding ∑  ur = f(1) − f(n + 1) r=1 2 The Maclaurin series of a function f(x) is given by f(r)(0) f 0(0)     xr + … f(x) = f(0) + f9(0)x + ____     x2 + … + _____ 2! r! The series is valid provided that f(0), f9(0), f 0(0), … , f(r)(0), … all have finite values. 3 The following Maclaurin series are given in the formulae booklet:  x   + … + __  x   + … for all x ex = 1 + x + __ r! 2! r 2 ln (1 + x) = x − __  x   + __  x  − … + (−1)r + 1 __  xr + … −1 , x < 1 2 3 2 3 r 3 5 2r + 1 sin x = x − __  x   + __  x   − … + (−1)r ________ x   + … for all x 3! 5! (2r + 1)! 2 4 2r cos x = 1 − __  x   + __  x   − … + (−1)r ____   x   + … for all x 2! 4! (2r)! 3 5 2r + 1 arctan x = x − ___  x   + ___  x   − … + (−1)r ______ x   + … −1 < x < 1 3 5 2r + 1 51 M02_ALEVEL_CP2_83343_U02_031-051.indd 51 24/09/18 8:22 AM 3 Methods in calculus Objectives After completing this chapter you should be able to: ● Evaluate improper integrals → pages 53–58 ● Understand and evaluate the mean value of a function → pages 58–62 ● Integrate rational functions using trigonometric substitutions → pages 62–69 ● Integrate using partial fractions → pages 69–73 Prior knowledge check 1 2 Find: 5x ____  dx a ∫  _____ √ 3 + x  2  b ∫x  2 e  x dx ← Pure Year 2, Chapter 11 dy Find  ___ in terms of x and y for the following: dx a x  2 + y  2 = 1 b5x  2 + xy + 2y  2 = 11 c x = tan y 3 c ∫ ________   sin x cos x   dx 1 + 3 sin  2 x ← Pure Year 2, Chapter 9 Express in partial fractions: 1 a  _______  x(x + 1) 2x − 1   b _____ x  2 − 4 3x − 2 c __________   2x  2 + 3x + 1 ← Pure Year 2, Chapter 1 The lowest speed necessary for an object to escape from a gravitational field is called the escape velocity. You can use improper integrals to calculate escape velocities. 52 M03_ALEVEL_CP2_83343_U03_052-076.indd 52 24/09/18 9:11 AM Methods in calculus 3.1 Improper integrals If a function f(x) exists and is continuous for all values in the interval [a, b] then the definite integral ∫   f(x) dx represents the area enclosed by the curve y = f(x), the x-axis and the lines x = a and x = b. b a y y = f(x) Notation The interval [a, b] is all the real numbers x satisfying the inequality a < x < b. b A = f(x)dx a O a x b In this section you will consider integrals where one or both of the limits are infinite, or where the function is not defined at some point within the given interval. These are called improper integrals. In these cases it is still possible for the function to enclose a finite area. y y y= y = 12 x O O x 1 1 The area bounded by the curve y = __   2 , x  the x-axis and the line x = 1 is finite. This area is represented by the improper ∞ 1 integral ∫     __2  dx. 1 x  3 1 x x 1 The function f(x) = __   __ is not defined at x = 0. However, the √ x  1 area bounded by the curve y = __   __ , the coordinate axes and √ x  the line x = 3 is finite. This area is represented by the 3 1 improper integral ∫   __   __  dx. x  0 √ ■ The integral ∫ a  f (x) dx is improper if either: b • one or both of the limits is infinite • f(x) is undefined at x = a, x = b or another point in the interval [a, b]. Notation If an improper integral exists then it is said to be convergent. If it does not exist it is said to be divergent. You can determine whether improper integrals are convergent, and evaluate them if so, by considering limits. To find I = ∫   e  −xdx, you need to consider the integral ∫   e  −x dx, for some finite t ∞ 0 0 value t. If this integral tends to a limit as t → ∞ then I is convergent and equal to that limit. If it fails to tend to a limit, then I is divergent. 53 M03_ALEVEL_CP2_83343_U03_052-076.indd 53 24/09/18 9:11 AM Chapter 3 y y 1 1 y = e–x y = e–x O x t As t → ∞, O x t ∫   e  −x dx → ∫   e  −x dx t 0 ∫ ∞ 0 t   e  −x dx = [ − e  −x]  t0   = − e  −t + 1 0 Since e  −t →0 as t →∞, the integral ∫   e  −x dx → 1as t → ∞. So ∫   0 ∞ t 0 e  −x dxis convergent and equal to 1. Notation You can use limit notation to write: ∫  e  −x dx = t→∞ lim  ∫  e  −x dx= t→∞ lim (− e  −t + 1) = 1 ∞ t 0 Example 0 1 Evaluate each improper integral, or show that it is not convergent. ∞ ∞ 1 1   2  dx    dx a∫  __ b∫  __ 1 x  1 x ∞ t 1 1 __ __ a∫    2  dx = t→∞   lim ∫    2  dx x   x  1 1 1 = t→∞   lim [−  __ x ]     t Replace the infinite limit by t, and take the limit as t → ∞. 1 = t→∞   lim (−   + 1) t =1 1 __ ∞ ∞ 1 1 __ __ So ∫     2  dxconverges and ∫     2  dx = 1 x  x  1 1 ∞ t1 1 __ __   lim ∫    dx b∫    dx = t→∞ 1 x 1 x = t→∞   lim [ln x]  t1  1 __     →0 as t → ∞ t 1 The area under the curve y = __   2 from 1 to infinity x  is finite and is exactly equal to 1. y y = 12 x = t→∞   lim (ln t − ln 1) =  t→∞ lim ( ln t) ∞ 1 __ ln t →∞ as t →∞, so ∫     dxdoes not 1 x converge. ` 1 Explore the integral ∫    __      dx 1 x2 using GeoGebra. Online O x Watch out Make sure you show the limiting process clearly in your working. You can’t just ∞ 1 write ∫  __   dx = ln ∞ = ∞. 1 x 54 M03_ALEVEL_CP2_83343_U03_052-076.indd 54 24/09/18 9:11 AM Methods in calculus You need to use a similar limiting process if the function you are integrating is not defined at one or more points in the interval. Example 2 Evaluate each integral, or show that it does not converge. 1 2 x 1 __ _____ b     ____2   dx a     2  dx 4 − x    0 x  0 √ ∫ ∫ 1 1 1 1 __ __ a∫     2  dx = t→0   lim ∫    2  dx x   x  t 0 1 1 = t→0   lim [−  __ x ]  t  1 = t→0   lim (− 1 + __  ) t 1 1 __ 1__ − 1 +   → ∞ as t →0, so     2  dx does t x 0   not converge. 1 y = __   2 is undefined for x = 0, so replace the lower x  limit with t and take the limit as t → 0. ∫ ∫ ∫ 1 __   → ∞as t → 0 t 2 t x x ____   dx =   lim   ____   dx b     ____   ____ 2 t→2 √ √ x    x  2  0 4 − 0 4 − x y = _______   ______ is not defined for x = 2, so replace √ 4 − x2  the upper limit with t. ____ t = t→2   lim [− √ 4 − x  2  ]  0   ____ __ =  t→2 lim (− √ 4 − t  2  − (− √ 4 )) ____ =  t→2 lim(2 − √ 4 − t  2 ) ∫ ____ lt→2 im (√ 4 − t  2  ) = 0 =2 2 x ____   dxconverges and So     ____ √ x  2  0 4 − x   dx = 2 ∫   ____   √ 4 − x    2 0 ____ 2 If both limits of an integral are infinite, then you need to split the integral into the sum of two improper integrals. In other words, you write ∞ Watch out ∞ ∞ c Do not write  ∫  f(x) dx as   f(x) dxfor some value c.   ∫  f(x) dx =   ∫ f  (x) dx + ∫ a −∞ −∞ −∞   lim  ∫  f(x) dx. You must split it into two separate c a→∞ −a If both of these integrals converge, then the original integral converges, but if either diverges, then the original integral is also divergent. Example integrals to determine whether it converges. → Review Exercise 1, Challenge Q3 3 a Find ∫x e  −x  2 dx. b Hence show that  ∫  x e  −x  dx converges and find its value. ∞ 2 −∞ a Let I = ∫xe  −x  2 dx Consider y = e  −x  2 dy ___   = − 2x e  −x  2 dx __1 So I = −   2 e  + c −x  2 Try differentiating e  −x  . 2 The −2x comes from differentiating −x2. Use the reverse chain rule. 55 M03_ALEVEL_CP2_83343_U03_052-076.indd 55 24/09/18 9:11 AM Chapter 3     x e  −x  dx + ∫     x e  −x  dx b Split ∫     x e  −x  dxup as ∫ ∞ 0 2 −∞ ∞ 2 −∞ Consider ∫     x e  0 2 0 dx: −x  2 −t ∫ −t    x e  −x  dx = [ −  __21   e  −x  ]  −t   = −  __21   + __  21 e  −t  0 2 2 0 2 __1 −t  2 __1 1 So t→∞   lim ∫     x e  −x  dx = t→∞   lim (−  __ 2  +   2  e  ) = −   2  0 2 −t 1 So ∫     x e  −x  dxconverges and ∫     x e  −x  dx = −  __ 2  0 0 2 −∞ 2 −∞ Similarly, consider ∫     x e  −x  dx: t 2 t 2 0 So ∫     x e  −x  dxconverges and ∫     x e  −x  dx = __   21   ∞ ∞ 2 0 You need to check that both the integrals converge before you can determine that the original integral converges. 2 0 Since both integrals converge, we know that ∫     x e  −x  dxconverges and ∞ 2 Watch out 2 __1 1 −t  2 __ 1   lim ∫     x e  −x  dx = t→∞   lim (−  __ So t→∞ 2  e  +   2 ) =   2  t To find the integral between –∞ and 0, you should find the integral between –t and 0 and then let t → ∞. 2 ∫ 0    x e  −x  dx = [ −  __21   e  −t  ]  0   = −  __21   e  −t  + __  21   2 You can choose any point at which to split the integral up, but choosing a special value like 0 will often make evaluating the integral easier. Use ∫x e  −x  dx= −  _12  e  −x  + c 2 0 t Problem-solving 2 To find the integral between 0 and ∞, you should find the integral between 0 and t and then let t → ∞. −∞ ∫ −∞    x e  −x  dx = ∫−∞    x e  −x  dx + ∫0    x e  −x  dx ∞ 0 2 2 ∞ 2 __1 1 = −  __ 2  +   2  = 0 Exercise Use ∫x e  −x  dx = −  _12  e  −x  + c 2 2 ∞ c −∞ −∞ ∞ Apply ∫     f(x) dx = ∫   f(x) dx + ∫     f(x) dx c 3A 1 Find the values of the following improper integrals. ∞ 1 a∫     __3  dx x 1 b∫    x − 2 dx ∞ _3 2 c∫    e−3x dx ∞ 0 2 For each of the following, show that the improper integral diverges. ∞ ∞ ∞ 1__ 8x ______  dx   dx c b∫     __ ∫     _______ a∫    ex dx x  0 0 √ 1 √ 1 + x2  3 For each of the following, show that the improper integral converges and find its value. 1 1   __  dx a∫   __ 0 √ x  P _2   3 1 b∫   _______   ______  dx 0 √ 2 − 3x  c∫   ln 3 0 e _______   dx   _____ x √ e − 1  x 4 For each of the following, determine whether the integral converges, and if so, find its value. 2 1   ___  dx a∫   ____ −1 √ |x|  3 x−1 __________  dx b∫   ___________      √ −1 3 + 2x − x2  π c∫   tan x dx 0 56 M03_ALEVEL_CP2_83343_U03_052-076.indd 56 24/09/18 9:11 AM Methods in calculus E 1 5 a Find ∫   _______  2   dx.   ( 7 − 3x)  (2 marks) 2 1     dx converges and b Hence show that ∫   ________ −∞ (7 − 3x)2 find its value. (3 marks) E Watch out Make sure you show the limiting process clearly in your working. 6 a Find ∫   x  2 e  x  3  dx. (2 marks) b Hence show that ∫   x2e  x3 dx converges and find its value. 1 (3 marks) −∞ E ln x 7 a Find ∫   ___   x   dx. (3 marks) ∞ ln x   x   dxis divergent. b Hence show that ∫     ___ E (3 marks) 1 8 a Find ∫   (ln x)  2 dx. (2 marks) Hence show that: b∫  (ln x)  dxis convergent. 1 0 2 (2 marks) c∫  (ln x)  dxis divergent. 0 E ∞ 2 (2 marks) ∫ 2 6x _______ 9 Evaluate    3 ______2   dx. 0 √ 4 − x   ∫ _____ _____ − 3√ 2 + x  2 − x ______ _______________     E/P 10 Evaluate        dx. √ 4 − x2  −2 E/P 2√ (4 marks) 11 The diagram shows the curve with equation y = ln x. Find the shaded area enclosed by the curve and the coordinate axes. You may assume that x ln x →0 as x → 0. (5 marks) (5 marks) y y = ln x O 1 x π __ E 12 a Explain why ∫     tan x  dxis an improper integral.     2 0 (1 mark) π __ b Show that ∫     tan x  dxis divergent.     2 0 (3 marks) 57 M03_ALEVEL_CP2_83343_U03_052-076.indd 57 24/09/18 9:11 AM Chapter 3 E/P π 13 A student writes the following working to evaluate ∫     sec  2 x  dx. 0 π ∫   sec  2 x dx = [ tan x]  π0  0 = tan π − tan 0 = 0 − 0 = 0 a Explain the mistake that the student has made. (1 mark) b Show that ∫     sec  2 x  dxis divergent. (4 marks) π 0 E/P E/P ∞ 1 14 Find all a ∈ ℝfor which ∫     __a  dx converges and find its value in the case when it 1 x converges. k 2k + 1 1   dx = ln(______   ___________        15 a Show that ∫   , 2 k+1) 0 2x  + 3x + 1 where k > 0. (4 marks) b Hence find the exact value ∞ 1 ___________   dx. of       2 0 2x  + 3x + 1 ∫ (7 marks) Problem-solving When an integral is undefined at one or more points within the interval of integration, you need to split the integral and consider each part separately. (3 marks) Challenge ∞ Show that ∫     e  −x sin  2 x  dx = _  25 0 3.2 The mean value of a function You can find the mean of a finite set of values by adding them up, and dividing by the number of values: 1  y¯ = __ n   ( y 1 + y 2 + y 3 + … + yn  ) You can extend this definition to evaluate the mean value (or average value) of a function on a given interval [a, b]. In this case, the function takes an infinite number of values, so you represent their sum by integrating the function between a and b, and you represent the ‘number of values’ by the width of the interval, b − a. ■ The mean value of the function f(x) over the ∫ b 1       f(x) dx. interval [a, b], is given by ____ b − a a Notation The mean value _ _ of f(x) is sometimes written as f, y    or ym   . 58 M03_ALEVEL_CP2_83343_U03_052-076.indd 58 24/09/18 9:11 AM Methods in calculus You can think of the mean value geometrically by considering the area, A, bounded by the curve y = f(x), the x-axis and the _ lines x = a and x = b. If you were to draw a rectangle with its base on the interval [a, b] and height f, then the area of the rectangle would be equal to A. _ y = f(x) y The area of the rectangle is (b − a)f. Setting this equal to the area under the curve gives: _ b   f(x) dx (b − a)f = ∫ a b _ ____ 1     f(x) dx ⇒ f =   f b−a a ∫ Example O 4 a b–a b x 4 Find the mean value of f(x) = _____   _____ over the interval [2, 6]. √ 2 + 3x  ∫ ∫ 6 6 __1 4 ____     ____  dx = 4   (2 + 3x) − 2 dx √ 2 2 + 3x  2 1 = __   83  [(2 + 3x)  2]  2  __ 6 ___ __ __ __ = __   83 (√ 20  − √ 8  ) √ ) (√ = __   16 3  5  − 2  _1 Using the reverse chain rule, if y = ( 2 + 3x) 2 dy _1 then  ___ = _  32  (2 + 3x)− 2 dx Simplify. So the mean value of f(x) on [2, 6] is ∫ 2 __ __ 4 ________ 1  ______      _______  d   16  (√ 5  – √ 2  )) x = __   41  (__ 3 6 – 2 6 √ 2 + 3x  Example __ __ 4 √ √ =  __ 3 ( 5  – 2  ) ∫ b 1 Apply ____     f(x) dx. b−a a 5 4 f( x) = _____     1 + e  x 9 4 ln  _7 _____ a Show that the mean value of f(x) over the interval [ln 2, ln 6] is     ln 3 b Use the answer to part a to find the mean value over the interval [ ln 2, ln 6] of f(x) + 4. c Use geometric considerations to write down the mean value of −f(x) over the interval [ ln 2, ln 6]. a Integrate ∫     ln6 4 _____ x   dxusing substitution. ln2 1 + e  Let u = e  x du du ___   = e  x ⇒ dx = __   u dx x = ln 2  ⇒   u = e  ln 2 = 2 x = ln 6 ⇒ u = e  ln 6 = 6 Calculate the limits in the form u = … . 59 M03_ALEVEL_CP2_83343_U03_052-076.indd 59 24/09/18 9:11 AM Chapter 3 ( ) (u )     x   dx = x=ln2 1 + e  u=2 1 + u 6 ∫ x=ln6 4 _____   ∫ u=6 Transform the integral by substituting u = e  x du and ___   = e  xand substituting the limits. dx 4 du ____ __ = 4 ∫     1 ______     du (1 + u)u Using partial fractions: 1 B A ______       = __   + ____ ( u1 + u) u 1 + u 2 6 1     duit is necessary to To calculate 4∫   ______ 2 (1 + u)u use partial fractions. So A(1 + u) + Bu = 1 When u = 0, A = 1. Find the values of A and B. Equating u terms, A + B = 0, so B = −1. ∫ ∫ 6 6 1 1 1 ______   − ____     du = 4   (__ 4    ( ) 1 + u )  du 1 + u u u 2 2 = 4[ln u − ln (u + 1)]  62  Integrate with respect to u. = 4((ln 6 − ln 7) − (ln 2 − ln 3)) Evaluate the integral using the limits. 18 = 4 ln  __ 14 9 = 4 ln  __ 7 Simplify using the laws of logarithms. So the mean value of f on [ln 2, ln 6] is 9 4 ln  __ 7 1 _________ _____ 9   =     (4 ln  __   7) ln 6 − ln 2 ln 3 b  ∫ 4 4   + 4  ) dx =   ∫ _____ ∫ (______  1 + e     dx +   ∫ 4  dx 1+e    ln6   ln2 ln6 ∫    ln6 x   ln2    ln6   ln2 x     4  dx = [ 4x]  ln6 ln2 ln2 = 4(ln 6 − ln 2) = 4 ln 3 Calculating the mean value, ∫ The original integral can be separated. Calculate the new integral. ln6 4 ______   + 4  ) dx       ln 3   ln2 ( 1 + ex    ln6    ln6 4 _____ 1 1 =  ___     1 + e  x   dx + ___       4  dx ln 3   ln2 ln 3   ln2 1 ___   b 1 Find ____     f(x) dxand use the laws of b−a a logarithms to simplify. =   ∫ 9 4 ln  __ 7 _____ 9 4 ln  __ 7 _____ 1 ___   +   ln 3 ∫   (4 ln 3) ln 3 =     + 4 ln 3 c −f(x) is a reflection in the x-axis of f(x). The mean value of f(x) over the interval 9 4 ln  __ 7 [ln 2, ln 6] was _____     ln 3 Therefore, the mean value of −f(x) over the 9 4 ln  __ 7 _____ interval [ln 2, ln 6] is −     ln 3 Problem-solving Every value of f(x) in the interval has increased by 4, so the mean value has increased by 4. In the example above, you saw that geometric considerations can be used to find mean values of transformed functions, if you already know the mean value of the original function on the same interval. 60 M03_ALEVEL_CP2_83343_U03_052-076.indd 60 24/09/18 9:11 AM Methods in calculus If the function f(x) has mean value ¯f   over the interval [a, b], and k is a real constant, then: ■ f(x) + k has mean value ¯f  + k over the interval [a, b] ■ kf(x) has mean value k ¯fover the interval [a, b] Watch out You cannot deduce the mean value of f(–x) or f(kx) in this way. ■ –f(x) has mean value – ¯fover the interval [a, b]. Exercise 3B 1 For each of the following functions f(x), find the mean value of f(x) on [0, 1]. 1 a f(x) = 1 b f(x) = ____     c f(x) = ex + 1 x+1 2 Find the exact mean value of f(x) over the given interval. e  3x π   ; [0, 2] bf(x) = cos  3 x sin  2 x ; [ 0, __   3x    ] af(x) = _____ 2 e  + 1 5 ___________ −x df(x) =       ; [0, 3] cf(x) = xe  ; [1, 3] (x + 2)( 2x + 1) π __ 2 ef(x) = ( sec x − cos x)  ; [ 0,    ]. 4 3f(x) = x  3 − 3x  2 − 24x + 100 a Find the coordinates of the turning points of f(x). b Sketch the graph of y = f(x). c Without calculation, state an upper and lower bound on the mean value of the function on the interval [−2, 4], giving a reason for your answer. d Calculate the exact mean value of f(x) over the interval [−2, 4]. E sin x cos x π 4 Find the exact mean value of f(x) = _________      ].  over the interval [ 0, __ 2 cos 2x + 2 (4 marks) E 5 Find the exact mean value of f(x) = x √ x + 4 over the interval [0, 5]. (4 marks) E E/P ____ π    ]. 6 Find the exact mean value of f(x) = x sin 2xover the interval [ 0, __ 3 5x ____________ 7f(x) =       (2x − 1) (x + 2) 49 1 a Show that the mean value of f(x) over the interval [1, 5] is _ 4 ln  __ 3 (4 marks) (4 marks) b Hence, or otherwise, find the mean value over the interval [1, 5] of f(x) + ln k where k is a positive constant, giving your answer in the form p ln q, where p and q are constants and (2 marks) q is in terms of k. E/P E/P 8f(x) = x (x  2 − 4)  4 a Show that the mean value of f(x) over the interval [0, 2] is ___   5 (3 marks) b Use the answer to part a to find the mean value over the interval [0, 2] of −2f(x). (2 marks) 256 9 f(x) = ln(kx), where k is a positive constant. Given that the mean value of f(x) on the interval [0, 2] is −2, find the value of k. (4 marks) 61 M03_ALEVEL_CP2_83343_U03_052-076.indd 61 24/09/18 9:11 AM Chapter 3 P 10 Prove that if f(x) has mean value m on the interval [a, b], then f(x) + c has mean value m + c. E/P 1 ____  11 f(x) =  _____ √ 2 − x  Watch out This is an improper integral. (6 marks) Find the exact mean value of f(x) on the interval [0, 2]. P 12 Use geometric reasoning to explain why the mean value of f(x) = sin  5 x on the interval [0, 2π] is 0. E/P cos x     13f(x) = ________ ( 2 + sin x)  2 a Find ∫   f(x) dx. (4 marks) 5π 3 b Hence show that the mean value of f(x) over the interval [ 0,  __ ] is −  ____  (3 + 4 √ 3  ). (2 marks) 3 130π 5π c Hence, or otherwise, find the mean value, over the interval [ 0,  __ ], of f(x) + 3x. (3 marks) 3 P __ 14 a Sketch a graph of f(x) = 1 − 3x − 2x2, finding the coordinates of any turning points. b Calculate ∫   a+1   f(x) dxfor a ∈ ℝ. a c Find the maximum possible mean value of f(x) on any real interval of length 1. 3.3 Differentiating inverse trigonometric functions You can differentiate the inverse trigonometric functions implicitly. Example 6 d 1 Show that  ___  (arcsin x) = _______   ______2   dx √ 1 − x   Let y = arcsin x then sin y = x dy cos y  ___  = 1 dx dy  ___  = _____   1   dx cos y 1 __________ ________ =     1 − sin2 y  √ but sin y = x dy 1   so ___  = _______   ______ dx √1 − x2  Use y = sin (arcsin x) = x. Differentiate implicitly. You could also differentiate dy 1 x = sin y with respect to y then use ___     = ___     dx ___   dx dy ← Pure Year 2, Section 9.6 Divide by cos y. Use cos2 y = 1 − sin2 y and that cos y > 0 when π π y is in the range of arcsin, i.e. −  __  < y < __     2 2 Problem-solving dy 1 ____ , you can Alternatively, since ___     = ±  _____ dx √1 − x  2  conclude the sign is positive since the graph of y = arcsin x shows that the gradient is positive at all points x. 62 M03_ALEVEL_CP2_83343_U03_052-076.indd 62 24/09/18 9:11 AM Methods in calculus You can use similar methods to obtain the following standard results. d 1 ____   ■  ___ (arcsin x) =   _____ 2 √1 − x    d 1 ____   ■  ___ (arccos x) = −   _____ dx √ 1 − x  2  d 1 ■  ___ (arctan x) =   _____2  dx 1 + x  dx You should learn these results, but also be able to derive them as in the example above. d d Notice that  ___ (arcsin x) = −  ___ (arccos x). dx dx Links Recall the domain and range of each of these inverse trigonometric functions: Function Domain arcsin x [−1, 1] arccos x [−1, 1] arctan x (−∞, ∞) Range π π [−  _  ,  _ ] 2 2 [0, π] π _ π _ ( −   2  ,   2 ) ← Pure Year 2, Chapter 6 Example 7 dy Given y = arcsin x2, find ___     dx a using implicit differentiation d b using the chain rule and the formula for  ___  arcsin x. dx a sin y = x2 Use sin (arcsin x) = x. dy cos y  ___  = 2x dx Differentiate. dy 2x  ___  = _____     dx cos y Divide by cos y. dy 2x   ________   ___  = __________ dx √1 − sin2 y  but sin y = x2 dy 2x   so ___  = _______   ______ dx √ 1 − x4  b Let t =x2, then y = arcsin t dt dy 1   Then ___  = 2x and  ___ = _______   _____ dx dt √1 − t2  dt dy dy ___  ___  = ___   ×     dx dt dx 2x   = _______   ______ √ 1 − x4  Use cos2 y = 1 − sin2 y and that cos y > 0 when π π y is in the range of arcsin, i.e. −  __  < y < __     2 2 Substitute t = x2 to get arcsin x2 in the form arcsin x. Differentiate t and y. Use the chain rule. 63 M03_ALEVEL_CP2_83343_U03_052-076.indd 63 24/09/18 9:11 AM Chapter 3 Example 8 dy 1−x Given y = arctan (____        ), find ___ 1+x dx 1−x tan y = ____   1+x Use tan (arctan x) = x. dy −(1 + x) − 1(1 − x)          sec2 y  ___  = ________________ dx (1 + x)2 1−x Differentiate using the quotient rule on _____     1+x 2 = −  ______2  (1 + x)  Simplify. dy 2 1   So ___   = ____ × −  ______  dx sec  2 y ( (1 + x)  2) Divide by sec2 y. 2 1 = _______   × −  ______  1 + tan  2 y ( (1 + x)  2) Use 1 + tan2 y ≡ sec2 y. 2 1 = ___________     2  × ( −  ______2 ) (1 + x)  1_____ −x (1 +   1 + x) Substitute tan y = _____  1 − x  1+x (1 + x)2 2 = ________________           × −  _______  (1 + x)2 + (1 − x)2 ( (1 + x)2) Cancel (1 + x)2. 2 = −  ________2  2 + 2x Problem-solving You could also use the chain rule and the formula d for ___   (arctan x). dx 1   = −  ______ 1 + x2 Exercise 3C 1 Use implicit differentiation to differentiate the following functions. a arctan x b arccos x c arccos x2 1   x ) earcsin (__ d arctan (x3 + 3x) Hint You can check your answers by using the chain rule as well as the results stated earlier in the section. 2 Differentiate y = (arccos x)(arcsin x) E 1 + arctan x   3 Differentiate y = _________   1 − arctan x E/P 4f(x) = arccos x + arcsin x (4 marks) π By considering ∫f ′(x) dx, prove that f(x) = __    for all values of x. 2 (4 marks) 5 Differentiate with respect to x: e arcsin (1 − x2) x b arctan  __  2 f arccos x2 i j a arccos 2x x2 arccos x c arcsin 3x d arcot (x + 1) g ex arccos x h arcsin x cos x earctan x 64 M03_ALEVEL_CP2_83343_U03_052-076.indd 64 24/09/18 9:11 AM Methods in calculus E E dy 6 Given that tan y = x arctan x, find  ___  dx (4 marks) 7 Given that y = arcsin x, prove that d2y dy (1 − x2)   ____2  − x  ___  = 0 dx dx (6 marks) 8 Find an equation of the tangent to the curve with equation y = arcsin 2x at the point where 1 x = _ 4  9 Find the derivatives of the following functions. 1 b ______  a (arctan x)2 arcsin x P 10 Sketch the graphs of the following: a arcsin (arcsin x) P c arctan (arctan x) b arccos (arccos x) 11 Prove each of the following: a ______ sin (arccos x) = √ 1 − x  2  c arctan (arctan x) 1 bcos (arctan x) = _____   ____  √ 1 + x  2  √ 1 __ Hint ______ 1 d sin (arcsec x) = 1 − __   2   x csec (arccos x) =   x  When taking a square root, you should consider the definitions or graphs of the inverse trigonometric functions to determine the sign. 3.4 Integrating with inverse trigonometric functions 1 You can use the results from the previous section to integrate functions of the forms ______   2   a  + x  2 1 and ______   _____  √ a  2 − x  2  Example 9 ∫ x 1 __ By using an appropriate substitution, show that    ______   _____  dx = arcsin (    )+ c, where a is a positive a √ a  2 − x  2  constant and |x| , a. ∫√ ∫√ 1 1 ____________   dx    ____   _____   dx =    ___________      √ a  2 − x  2  x 2 a     2(1 − ( __   a )   )  ∫ 1 _______ 1 ________    __ = dx a      x 2 __ 1 − (   a )     ∫ 1 =    ____   ____   du √ 1 − u  2  x dx Use the substitution u = __   a and du = ___   a = arcsin u + c x   a ) + c = arcsin(__ d 1 Recall that ___   ( arcsin x) = _____   ____  dx √ 1 − x  2  65 M03_ALEVEL_CP2_83343_U03_052-076.indd 65 24/09/18 9:11 AM Chapter 3 The following two results are given in the formula booklet. You can quote them when calculating other integrals, but you should also be able to derive them using substitution as in the example above. x 1 1 ■ ∫    ______   dx =  __  arctan( __ a ) + c, a > 0, |x| < a 2 2 a a  + x  x 1 ■    ________   a ) + c   ______  dx = arcsin(__ 2 2 √ a − x   ∫ Example 10 ∫ Find ________   4   dx. 5 + x2   4   dx = 4∫ ______   1   dx ∫ ______ 5+x 5+x 2 2 x 1 = 4 ___   __  arctan ___   __  + c ( √ 5 )) ( √ 5  __ x 1 1 Use ∫   ______   ) + cwith a = √ 5 .   2   dx = __   arctan(__ 2 a a a  + x  ( ) 4 x__ = ___   __  arctan   ___    + c √ 5  √5  Example 11 1   dx.   ∫ ________ 25 + 9x a Find a __ b Evaluate 2 1 1   dx     dx = ∫ _________   ∫ ________ 25 + 9x 9( __  + x ) 25 9 2 1    , leaving your answer in terms of π. ∫ __ ________ √ 3 − 4x   √ __ 3 4 __ √ 3 −4 ( (  3 ))) x 1 ___   arctan   ___   + c =_  91     __ 5 5 __ ( ) 3x =   15 arctan   ___   + c 5 __ 1 b ∫ ∫ __ √ 3 __ ∫ x 1 1   dx = __ Use  ______   arctan ( __   a ) + c with a = __   5  3 a 2 2 a +x __ √ __   4  _________  43  __________ 1 1 _______ _________   dx =      dx _ 3 √3 − 4x2  √3 √3 __ __ 4(  4  − x2 )  √ −   −   4 4 __ 2 You need to write 25 + 9x2 in the form k(a2 + x2). 2 (( (  3 )) _______ __ ∫ __ √( _    − x )  Write 3 − 4x2 in the form k(a2 − x2). __ √3 __     ________ 1 =     4 _________   dx _1 2 __ − √ 3 4   3 4 2 __ √ __   3 2x __     4 __ =_  21   arcsin   ___ √ 3  √3  −__ [ ( ) ]     4 1 =_  21  arcsin(__   21 ) − _  21  arcsin(−  __ 2 ) π ___ π ___   − −     12 ( 12 ) π = __     6 =   ∫ __ √ 3   x 1 Use    _______   ) + cwith a = __       ______  dx = arcsin(__ 2 2 2 a √ a − x   π − __     < arcsin x < __   π  2 2 π π  1     and arcsin (−  _12 ) = −  __ so arcsin (_  2 ) = __ 6 6 66 M03_ALEVEL_CP2_83343_U03_052-076.indd 66 24/09/18 9:11 AM Methods in calculus Example 12 x+4 Find ∫   ______   _____   dx. √ 1 − 4x  2  x+4 x 1 _____   dx + 4 ∫   _____ ∫   _____   _____   dx = ∫     _____   _____   dx √ 1 − 4x  2  √ 1 − 4x  2  √ 1 − 4x  2  x _____   dxuse the substitution To calculate ∫     _____ 2 √ 1 − 4x    2 u = 1 − 4x  Problem-solving The integral can be split into a fraction which can be integrated using the ‘reverse chain rule’ and one that looks like those covered above. du = − 8x  dx x __ 1        du    dx = −   ∫   __ ∫   _____ u  √ 1 − 4 x    _____ 2 1 8 √ ∫ Substitute for x and dx and adjust for the constant. _ __1 1 − 2 = −  __ 8 u du __ 1   21 = −  __ 4 u + c Rewrite in terms of x by resubstituting u = 1 − 4x  2. _____ 1 2 √ = −  __ 4 1 − 4x    + c 1 1 _____   dx = 4∫   __________ _________   dx 4∫    _____      √ 1 − 4x  2  √ 4(__   41 − x  2)  Write 1 − 4x  2in the form k(a  2 − x  2). 1 = 2 ∫   ________   ______   dx __1 √   4 − x  2  x 1 1 Use ∫   _______   ) + cwith a = __       ______  dx = arcsin(__ 2 2 2 a √ a − x   = 2 arcsin 2x + c ∫ _______ x+4 __1 So    ________   _______  dx = 2 arcsin 2x −   4 √ 1 − 4x2  + c √ 1 − 4x2  Exercise 3D ( ) ∫ x 1 1   dx = __ 1 Use the substitution x = a tan θ to show that  _______   arctan __     + c. a a a2 + x2 ∫ 1   dx = −arccos x + c. 2 Use the substitution x = cos θ to show that  _______ ______ √ 1 − x2  3 Find: a E 3   dx   ______ ∫ _______ √ 4 − x   2 b   4   dx ∫ ______ 5+x 2 c 1   dx   _______ ∫ ________ √ 25 − x   2 d 1   dx   _____ ∫ _______ √ x − 2  2 ∫ 1 4 Find    ______    dx, giving your answer in the form A arctan (Bx) + c where c is an arbitrary   4 + 3x  2 (3 marks) constant and A and B are constants to be found. 67 M03_ALEVEL_CP2_83343_U03_052-076.indd 67 24/09/18 9:11 AM Chapter 3 E ∫ 1 5 Show that    ______   _____    dx = P arcsin (Qx) + c where c is an arbitrary constant and P and Q are √ 3 − 4x  2  constants to be found. (3 marks) 6 Evaluate: ∫ ∫ 3 __ 0.5 3   ________ _______2  dx √ 0.25 1 − 4x   2     dx a   ______ 1 − x2 1 b E 1 π ____   dx = __ 7 Show that ∫ __    _____     2 12 √  2 √ 4 − x    E/P 2 + 3x ______ (x) =   8 f   1 + 3x  2 √  3 (4 marks) __ 3 ) + B ln(1 + 3x  2) + c Show that ∫   f(x)  dx = A arctan (√  x where c is an arbitrary constant and A and B are constants to be found. (4 marks) E/P ∫ 2 1 c   _________   ________2   dx. √ −1 21 − 3x   Hint Start by splitting the fraction into two separate integrals. 2x − 1 _____ (x) =   ____  9 f √ 2 − x  2  ____ x   __  + B √ 2 − x  2  + c where c is an Find ∫   f(x)  dx, giving your answer in the form A arcsin __ ( √ 2  ) arbitrary constant and A and (4 marks) B are constants to be found. E/P 8x − 3 _____ (x) =     10 f 4 + x  2 x Find ∫   f(x)  dx, giving your answer in the form A ln (x  2 + 4) + B arctan (__    ) + c where c is 2 an arbitrary constant and A and B are constants to be found. (4 marks) E/P 4x − 1 ______ (x) =   _____   11 f √ 6 − 5x  2  __ √_6   x)+ c where c is an arbitrary constant Show that ∫   f(x)  dx = P √ 6 − 5x  2  + Q arcsin (   and P and Q are constants to be found. (4 marks) _____ E/P 5 x+5 ______ (x) =   2   12 f x  + 16 x a Find ∫   f(x)  dx, giving your answer in the form A ln(x  2 + 16) + B arctan (__    ) + c where c is 4 an arbitrary constant and A and B are constants to be found. (4 marks) 5π 1 1 b Hence show that the mean value of f(x)over the interval [0, 4] is _ 4 (__     ln 2 + __     ) (3 marks) 2 16 c Hence write down the mean value of −4f(x) over the interval [0, 4]. (1 mark) 68 M03_ALEVEL_CP2_83343_U03_052-076.indd 68 24/09/18 9:11 AM Methods in calculus ∫ P 2 2 13 Use the substitution x = _ 3  tan θ to find _______   x   dx. 9x2 + 4 P __   4  x2 1 1   _______2   dx= ___  192  (2π − 3√3 ). 14 By using the substitution x = _  2  sin θ, show that   ________ 1 − 4x   0 √ ∫ 1 _ Challenge Given that x > 1, use the substitution x = sec θ to find: ∫ ∫ ____ √ x  2 − 1  b    _______     dx x 1 a    ______   ____   dx x √ x  2 − 1  3.5 Integrating using partial fractions In your A level course you used partial fractions to integrate some rational functions. Example 13 Prove that | | 1 a+x 1 ∫   ______   dx =  __  ln ____   a − x  + c   2 2 2a a  − x  Watch out Be careful not to confuse this result x 1 1 with ∫    ______   ) + c   dx = __   arctan ( __ a a a  2 + x  2 where a is a real constant. ∫ ∫ 1 1    ______   2        dx    dx =    ____________ 2 a  − x  (a + x)( a − x) 1 A B ____________       = _____   + _____   a−x (a + x)( a − x) a + x Factorise the denominator in order to integrate using partial fractions. ← Pure Year 2, Section 11.7 A(a − x) + B(a + x) = 1 1 __ x = −a ⇒ A =   2a Split into partial fractions and calculate the values of A and B in terms of a. 1 x = a ⇒ B =  __ 2a ∫ ∫ ∫ 1 1 1 1 1 __ __    ______   2   a + x    dx +      ____   a − x    dx    dx =      ____ 2 2a 2a a  − x  1 1   ln|a − x| + c = __   ln|a + x| − __ 2a 2a Integrate using the reverse chain rule. a+x 1    __ ln ____ = + c 2a a − x Simplify using the law of logarithms. | | 69 M03_ALEVEL_CP2_83343_U03_052-076.indd 69 24/09/18 9:11 AM Chapter 3 If the denominator of a partial fraction includes a quadratic factor of the form (x2 + c), c > 0, you cannot write it as a product of linear factors with real coefficients. However, you can still write it in partial fractions, where the partial fraction corresponding to the quadratic factor has a linear numerator and quadratic denominator. Bx + C 20 A So  ____________      = ____       + ______   2 2 (x + 3)(x  + 1) x + 3 x  + 1 20 = A(x2 + 1) + (Bx + C )(x + 3) Find the values of A, B and C by multiplying both sides by (x + 3)(x2 + 1). Set x = −3: 20 = 10A ⇒ A = 2 Set x = −3 so that (Bx + C )(x + 3) = 0 A + B = 0 ⇒ B = −2 Equate coefficients of x2 on each side. 3B + C = 0 ⇒ C = 6 Equate coefficients of x on each side. 6 − 2x 20 2   = ____     So ____________        + _____   2 2 x + 3 (x + 3)(x  + 1) x  + 1 You can use the techniques from the previous section to integrate the second fraction on the righthand side. Example 14 x   2 x 1+x _____  ) + B arctan(__ Show that    ______    ) + c, where A and B are constants to be found.   3    dx = A ln(  2 3 x  + 9 x  + 9x ∫ ∫ 1+x 1+x A Bx + C    ______   + ______   3   2      dx     dx =    _______    dx =    __ ( x x  2 + 9 ) x  + 9x x(x  + 9) ∫ ∫ A(x  2 + 9) + Bx  2 + Cx = x + 1 Equate x  2 terms: A + B = 0 Equate x terms: C = 1 1 Equate constant terms: 9A = 1 ⇒ A = __   91 ,  B = −  __ 9 Separate into partial fractions to facilitate integration. Calculate the values of the coefficients by equating like terms. ∫ 1 −  __ 1 1+x 9 x + 1 __ ________    ______   +     dx dx =   3     ( 9x x  2 + 9 ) x  + 9x ∫ ∫ x−9 1 =    __   − _______   ( 9x 9(x  2 + 9) )  dx 1 1 1 x 1 dx  __  ∫   __ =     ∫   _____      dx − __   2   2    dx + ∫   _____    9 x 9 x  + 9 x  + 9 __1 ( 2   9 ln x  + 9) __ x 1 1  __  ln x − __________ =       ) + c     +     arctan (__ 2 9 3 3 x 1 1 1 =  __  ln x − ___     arctan (__    ) + c     ln(x  2 + 9) + __ 9 18 3 3 x 1 1 = ___    (2 ln x − ln(x  2 + 9)) + __     arctan (__    ) + c 18 3 3 x  2 x 1 1 = ___     ln(_____       arctan (__    ) + c   + __ 18 x  2 + 9 ) 3 3 Separate into 3 fractions and integrate each separately. Calculate the second integral using the reverse chain rule. Calculate the third integral using x 1 1   dx = __  ______     arctan (__    ) + c with a = 3. a2 + x2 a a ∫ Simplify using the laws of logarithms. 70 M03_ALEVEL_CP2_83343_U03_052-076.indd 70 24/09/18 9:11 AM Methods in calculus Example 15 x  4 + x  as partial fractions. a Express __________   4 x  + 5x  2 + 6 ∫ x   4 + x b Hence find    __________   4   dx. x  + 5x  2 + 6 a x4 + 5x2 + 6 = (x2 + 2)(x2 + 3) Bx + C ______ x  4 + x Dx + E So  ____________      ≡ A + ______   2   +   2   2 (x  + 2)(x  2 + 3) x  + 2 x  + 3 x 4 + x ≡ A(x2 + 2)(x2 + 3) + (Bx + C)(x2 + 3) + (Dx + E)(x2 + 2) Start by factorising the denominator. Since x4 + 5x2 + 6 does not contain an x3 term or an x term, you can write it as u2 + 5u + 6 where u = x2. A=1 The numerator and denominator both have degree 4, so this is an improper fraction. You will need a constant term, and terms with denominators x2 + 2 and x2 + 3. B+D=0 Equate coefficients of x4. 3B + 2D = 1 Equate coefficients of x3 and x. So B = 1 and D = −1 Equate coefficients of x2 and constant terms, and use A = 1. 5A + C + E = 0 Problem-solving 6A + 3C + 2E = 0 There are other ways of determining the coefficients. You could find A = 1 by writing So C = 4 and E = −9 x  4 + x __________ x+4 _____ x+9 _____ So        ≡ 1 +   2   −     x  4 + 5x  2 + 6 x  + 2 x  2 + 3 (x  + 5x  + 6) − (5x  − x + 6) x  + x  __________  = _______________________           4 x  4 + 5x  2 + 6 4 2 2 x  4 + 5x  2 + 6 5x  2 − x + 6 = 1 −  __________   x  4 + 5x  2 + 6 Methods such as this are often quicker than using polynomial__long division. You could also __ substitute x = i √ 2  and x = i √ 3  to eliminate terms. 71 M03_ALEVEL_CP2_83343_U03_052-076.indd 71 24/09/18 9:11 AM Chapter 3 b Using the partial fraction decomposition: ∫ ∫ ∫ x  4 + x x    ____________        2   dx =    1 dx +    _____   dx (x  2 + 2)( x  2 + 3) x  + 2 ∫ ∫ ∫ 4 9 x +    _____   2   2   2   dx −    _____   dx −    _____   dx x  + 2 x  + 3 x  + 3 4 x 1   __  arctan __   __  = x + __     ln|x  2 + 2| + __ ( √ 2 ) 2 √ 2  ( √ 3 ) 9 x 1 −  __  ln|x  2 + 3| − __   __  arctan __   __  + c 2 √ 3  | | x  2 + 2 x 1 = x +  __  ln _____     __    + 2 √ 2  arctan __ ( √ 2 ) 2 x  2 + 3 __ __ x   __  + c − 3 √ 3  arctan __ ( √ 3 ) Exercise Split the integral up into different parts that can be integrated using inverse trigonometric functions or the reverse chain rule. ∫ f9(x) Use    ____     dx = ln|f(x)|+ c and f(x) x 1 1   a ) + c.   a arctan(__    ______   2   dx = __ x  + a  2 ∫ Simplify. 3E 1 Express the following as partial fractions. 1 1   b ___________         a ___________ (x  2 + 1)( x + 3) (x  2 + 2)( x − 1) E/P E/P x−4 c _______   ( xx  2 + 7) ∫ x x   2 − 3x dx, giving your answer in the form A ln |x + 2| + B arctan __   __  + c, 2 Find    ___________         ( √ 6  ) (x  2 + 6)( x + 2) where A and B are constants to be found, and c is an arbitrary constant. (4 marks) (x) = x  4 − x  3 − 4x  2 − 2x − 12 3 f a Given that (x + 2) is a factor of f(x), fully factorise f(x). (2 marks) ∫ x3 − 20x2 + 4x − 24 b Hence find        ____________________   dx, giving your answer in the form    4 x − x3 − 4x2 − 2x − 12 ∙x + 2∙  A x ln _______     + D arctan __   __  + c, where A, B and D are constants to be found, B ( √  2 ) ∙x − 3∙  and c is an arbitary constant. E/P x  2 x 2−x _____  ) + B arctan (__    ) + c, 4 Find    ______    dx, giving your answer in the form A ln (  2   3 2 x   + 4 x  + 4x ∫ where A and B are constants to be found, and c is an arbitary constant. E/P (5 marks) (5 marks) 2x A x   2 + 1 __ __ 5 Find     ________    dx, giving your answer in the form   + B arctan (   ) + c, where A 4 2 3 x 4x  + 9x  ∫ and B are constants to be found, and c is an arbitary constant. (5 marks) 72 M03_ALEVEL_CP2_83343_U03_052-076.indd 72 24/09/18 9:11 AM Methods in calculus E/P E/P ∫ |x − 1|A x  3 + 9x  2 + x + 1 _______ 6 Show that    _____________   + D arctan x + c, where A, B and D are dx = ln          4 | | x + 1 B x  − 1 constants to be found, and c is an arbitary constant. (5 marks) (x) = x  3 − 4x  2 + 6x − 24 7 f a Given that f(4) = 0, fully factorise f(x). (2 marks) 2x  2 − 3x + 24 b Express  ______________  in partial fractions.       x  3 − 4x  2 + 6x − 24 (2 marks) c Use your answer to part b and an appropriate substitution to calculate 2x  − 3x + 24          dx. ∫   ______________ x  − 4x  + 6x − 24 2 3 E/P 2 (4 marks) 1 ___________ (x) =       8 f (x − 2)( 2x − 1) a Calculate ∫   f(x)  dx. (3 marks) 1        dx diverges. b Hence show that ∫ __1    ___________ ( ( )     x − 2 2x − 1) 2 2 E/P (3 marks) x   4 + 5x  2 + 2x ___________  as partial fractions. 9 a Express         x  4 + 10x  2 + 24 ∫ (4 marks) x   4 + 5x  2 + 2x ___________ b Hence find               dx. x  4 + 10x  2 + 24 (5 marks) ∫ E/P x2 + 4x + 10   dx, x . 0. 10 Use the method of partial fractions to find  ___________ x3 + 5x (4 marks) E/P 2 2 1 11 Show that ∫   _____________   dx = _ 4 (π + 2 ln 2).     2 0 (x + 1)(x + 1) (4 marks) E/P x  4 + 1 12 a Express ________    as partial fractions. x (x  2 + 2)  2 (4 marks) ∫ x  4 + 1 b Hence find    ________    dx.   x (x  2 + 2)  2 Challenge Problem-solving Find: ∫ 1 a    _________   2    dx x  − 8x + 8 (5 marks) ∫ 1 b    ___________         dx 2 2x  + 4x + 11 First complete the square in the denominator and then use an appropriate substitution. 73 M03_ALEVEL_CP2_83343_U03_052-076.indd 73 24/09/18 9:11 AM Chapter 3 Mixed exercise E 3 ∫ 1 1 a Using the substitution u = e  x, find    ______   x   dx. e  + e  −x ∞ 1 π __ b Hence show that ∫     ______ x + e  −x  dx =     e   −∞ 2 (3 marks) (3 marks) E 1 − cos x π π 2 Find the exact mean value of f(x) = _______  over the interval [__       ,  __ ]. 2 6 3 sin  x (4 marks) E π 1    ] is _  2 3 Show that the exact mean value of f(x) = x sin 2x over the interval [ 0, __ 2 (4 marks) E 4 a Find the derivative of arccos x  2. (3 marks) 3x b Hence, or otherwise, calculate ∫   ______   _____   dx. √ 16 − x  4  E/P P (1 mark) 2x + 3 _____ (x) = arctan(  5 f   x−1) 1 a Show that f9(x) = −  _________   2 x  + 2x + 2 (4 marks) b Given that −2 < x < 2, show that |f9(x)| < 1 (2 marks) 6 a Explain what it means for an integral to be improper. ∞ 1 __  dxwhich make it an improper integral.     _______ b Identify two features of ∫ ( 0 x + 1) √ x  ∞ __ 1 __  dxis convergent and find     _______ c By differentiating arctan √ x , or otherwise, show that ∫ 0 (x + 1) √ x  its exact value. E/P 1 + 5x ______ (x) =   _____  7 f √ 1 − 5x  2  _____ __ Find ∫   f(x)  dx, giving your answer in the form A √ 1 − 5x  2  + B arcsin (√ 5  x) + c where c (4 marks) is an arbitrary constant and A and B are constants to be found. E t 1    dx = arctan t. 8 a Show that ∫     _____ 0 x  2 + 1 b Hence evaluate: ∞ 1    dx i∫     _____ 0 x  2 + 1 ∞ 1    dx ii∫     _____ −∞ x  2 + 1 (2 marks) (2 marks) (2 marks) 74 M03_ALEVEL_CP2_83343_U03_052-076.indd 74 24/09/18 9:11 AM Methods in calculus E/P 1 + 2x ______ (x) =     9 f 1 + 4x  2 a Find ∫   f(x)  dx, giving your answer in the form A ln (1 + 4x  2) + B arctan (2x) + c where c is an arbitrary constant and A and B are constants to be found. b Hence find the exact value of ∫     f(x)  dx 0.5 0 E (3 marks) 1 10 a Show that ∫   ______   _____   dx = P arcsin Qx + c where c is an arbitrary constant and P and Q are √ 4 − 9x  2  constants to be found. (4 marks) _2 1 π   _____2    dx = __     b Hence show that   ______ 6 0 √ 4 − 9x    ∫   3  (3 marks) _1 __ P (4π − 7 √ 3  ) x  ____   dx = ________   11 Use the substitution x = sin θ to show that    _____   2 64 0 √ 1 − x    E/P x     12 f(x) = _____ 1 + x  4 ∫   2 4 a Use the substitution u = x  2to calculate ∫     f(x)  dx. 1 0 b Hence show that ∫     f(x)  dx converges and state its value. ∞ 0 E/P E/P (4 marks) (4 marks) (3 marks) x B 2x  3 − 2x  2 + 18x + 9 13 Show that ∫    _______________    ) + c, where A, B and D are dx = A ln |x| − __     + D arctan (__          4 2 3 x x  + 9x  constants to be found. (5 marks) x  2 − 3x + 14 _____________ (x) =           14 f x  3 − 4x  2 + 2x − 8 Q P a Express f(x) in the form ____    , where P and Q are constants to be found. (3 marks)   + _____   2 x − 4 x  + 2 x b Find ∫   f(x)  dx, giving your answer in the form A ln |x − 4| + B arctan __   __  + c where ( √ 2  ) A and B are constants to be found. (4 marks) c Hence show that ∫     f(x)  dx diverges. ∞ 4 E/P (2 marks) 2 _____ (x) =   3   15 f x  + x a Find ∫   f(x)  dx. (4 marks) 8 b Hence show that the mean value of f(x) over the interval [1, 2] is ln  __  5 (2 marks) 6 c Hence, or otherwise, find the mean value, over the interval [1, 2], of 2f (x) −  __  x (3 marks) 75 M03_ALEVEL_CP2_83343_U03_052-076.indd 75 24/09/18 9:11 AM Chapter 3 Challenge A function is said to attain its mean value on the interval [a, b] if there b 1     f(x) dx. exists a value c ∈ [a, b] such that f(c) = ____   b−a a ∫ a Show that the function f(x) = x 3 − 2x + 4 attains its mean value on the interval [0, 2], and find the exact values of c ∈ [0, 2] for which f(c) = _  12 ∫    f(x) dx. 2 0 b Give an example of a function which does not attain its mean value on the interval [0, 2], fully justifying your answer. Summary of key points 1 The integral ∫     f(x) dx is improper if either: b a • one or both of the limits is infinite • f(x) is undefined at x = a, x = b or another point in the interval [a, b]. 2 The mean value of the function f(x) over the interval [a, b], is given by ∫ b−a b 1 _____   f(x) dx   a 3 If the function f(x) has mean value ¯f   over the interval [a, b], and k is a real constant, then: • f(x) + k has mean value ¯ f + k over the interval [a, b] • kf(x) has mean value k¯ f over the interval [a, b] • −f(x) has mean value −¯f  over the interval [a, b]. d 1 4 •  ___ (arcsin x) = _____   ____  dx √ 1 − x  2  d 1 (arccos x) = − _____ • ___ ____ dx √ 1 − x  2 d 1 • ___ (arctan x) = _____ dx 1 + x  2 ∫ x 1 • ∫   _______    )+ c     dx = arcsin ( __ a √ a − x x 1 1 5 •    ______   ) + c, a . 0, |x| , a   2   dx =  __ arctan ( __ 2 a a a  + x  ______ 2 2 76 M03_ALEVEL_CP2_83343_U03_052-076.indd 76 24/09/18 9:11 AM Volumes of revolution Objectives After completing this chapter you should be able to: 4 ● Find volumes of revolution around the x-axis → pages 78–80 ● Find volumes of revolution around the y-axis → pages 81–83 ● Find volumes of revolution for curves defined → pages 83–87 parametrically ● Model real-life applications of volumes of revolution → pages 87–89 Prior knowledge check 1 Evaluate: a∫     x2 (3x3 − 6)2 dx 3 0 ∫__π  2   cos  2 x dx b __   π    6  ∫     x  2 e  −2x  dx c 2 0 Volumes of revolution can be used to model objects with circular cross-sections. By defining curves parametrically, you can find volumes of a wider range of objects. → Exercise 4D Q4 ← Pure Year 2, Chapter 11 2 Find the area of the region bounded by the curve y =  _12 sec  4x, the x-axis, the y-axis and the line x = 1. ← Pure Year 2, Chapter 11 3 The region R is bounded by the curve y = 4x2 + 5, the x-axis and the lines x = 2 and x = 4. The region is rotated through 2π radians about the x-axis. Find the volume of the object generated. ← Book 1, Chapter 5 77 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 77 26/02/2018 09:23 Chapter 4 4.1 Volumes of revolution around the x-axis You need to be able to find volumes of revolution of more complicated curves. In this chapter you might need to use any of the functions and integration techniques you encountered in your A level course. Links You have already encountered volumes of revolution with simpler functions. ← Book 1, Section 5.1 ■ The volume of revolution formed when y = f(x) is rotated through 2π radians about the x-axis between x = a and x = b is given by Online Explore volumes of revolution b     y2 dx Volume = π∫ a Example around the x-axis using GeoGebra. 1 The region R is bounded by the curve with equation y = sin 2x, the x-axis and the lines x = 0 and π x = __    . Find the volume of the solid formed when region R is rotated through 2π radians about 2 the x-axis. b π Use V = π ∫     y  2  dx with a = 0, b ≡ __     and y 2 = sin2 2x. a 2 V = π∫ 2    sin22x dx __ π     0 1 = π∫   2  __ 2 (1 − cos 4x) dx __ π     0 Use cos 2A ≡ 1 − 2 sin2A __ π = π[__   21  x − __   81   sin 4x]       2  π2 = ( ___     − 0) − 0 4 π2 ___ =     4 Exercise Rearrange to give sin2A = ... 0 Note that 2 × 2x gives 4x in the cos term. Multiply out and integrate. 4A 1 Find the exact volume of the solid generated when each curve is rotated through 2π radians about the x-axis between the given limits. ________ 4 sin x π 2 _____   a y =     between x = 0 and x = 2 b y = ________    between x = 0 and x = __     2 x+1 1 + cos x ________ __ 5x π c y = √ x  sec x between x = 0 and x = __    between x = 0 and x = 2       d y = ________ 4 10 x   2 + 1 ____ √ ln x  π π     between x = 1 and x = 2 f y = cosec x + cot x between x =  __  and x = __   x  e y = _____ 2 3 √ √ E/P _____ 2 The curve with equation y = cos x √ sin 2x , π    is shown in the diagram. 0 < x < __ 2 The finite region enclosed by the curve and the x-axis is shaded. The region is rotated about the x-axis to form a solid of revolution. Find the volume of the solid generated. (6 marks) y y = cos x sin 2x O π x 2 78 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 78 26/02/2018 09:23 Volumes of revolution E/P 3 The diagram shows the finite region R, which is bounded by the curve y = ln x, the line x = 3 and the x-axis. y The region R is rotated through 2π radians about the x-axis. Use integration to find the exact volume of the (7 marks) solid generated. y = ln x R Problem-solving E a O You will need to find the value of a, where the curve crosses the x-axis 3 x 4 a Using the substitution x = 3sin θ, or otherwise, find the exact value of 3 2  ___   2 1 _________ ______    dx (7 marks)     2 _3 x  √ 9 − x  2    2  ∫ √ __ The diagram shows a sketch of part of the curve with 9 equation y =  __________1  ( x 9 − x  2)   4  The shaded region R, shown in the diagram, is bounded by the curve, the x-axis and the lines with equations __ 3 √  2 3 __ ____ x =     and x =    . The shaded region R is rotated 2 2 through 2π radians about the x-axis to form a solid of revolution. y y= 9 1 x(9 – x2) 4 R O 3 2 x 3 2 2 b Using your answer to part a, find the exact volume of the solid of revolution formed. (2 marks) E/P 4x + 3      is shown in 5 The curve with equation y  2 = _____________ (x + 2 ) (2x − 1) the diagram. y y2 = The shaded region R, bounded by the lines x = 1, x = 4, the x-axis and the curve, is rotated 360° about the x-axis. Use calculus to find the exact volume of the solid generated. E/P (6 marks) R O 1 x 4 y 6 The curve shown in the diagram has equation 2y2 = x sin x + x. 2y2 = x sin x + x a Show that the coordinates of point A are 3π ___ (1 mark) (   2  , 0). The shaded region R is rotated about the x-axis to generate a solid of revolution. b Find the volume of the solid generated. 4x + 3 (x + 2)(2x – 1) R O A x (5 marks) 79 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 79 26/02/2018 09:23 Chapter 4 E/P 10 7 The curve with equation y = ________    is shown in the diagram. ( 35 + 2x) The region R bounded by the curve, the x-axis and the lines x = −1, x = 2 is shown in the diagram. The region is rotated through 360° about the x-axis. a Find the exact volume of the solid generated. (6 marks) y y= R O –1 10   and The region S, bounded by the curves y = ________   3(5 + 2x) 20 y =  ________ , and the lines x = −1 and x = 2, 3(5 + 2x) is shown in the diagram. The region is rotated through 360° about the x-axis. y= y= y = xe–x R in the diagram. Find the volume of the solid of revolution formed. Give your answer correct to 3 significant figures. (8 marks) O 10 3(5 + 2x) 2 x y 1 y = xe−x and the line with equation y = _  4 x, as shown The region is rotated through 2π radians about the x-axis. 20 3(5 + 2x) S –1 8 The region R is bounded by the curve with equation x 2 y b Find the exact volume of the solid generated. (3 marks) E/P 10 3(5 + 2x) O y = 14 x x Challenge The diagram shows the region R, which is bounded by the curve with 1 equation y = sin x, 0 < x < π and the line with equation y = ___   __  √ 2   y y= 1 2 R y = sin x O x 1 Region R is rotated through 2π radians about the line y = ___   __  √ 2   π Show that the solid of revolution formed has area  __ (π − 3) 2 80 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 80 26/02/2018 09:23 Volumes of revolution 4.2 Volumes of revolution around the y-axis You can apply A level integration techniques to volumes of revolution formed when a curve is rotated about the y-axis. Links When you use this formula you are ■ The volume of revolution formed when integrating with respect to y. You might need to x = f( y) is rotated through 2π radians about rearrange functions to get an expression for x2 in the y-axis between y = a and y = b is given by terms of y.   2  dy Volume = π∫     x b a ← Book 1, Section 5.2 Online Explore volumes of revolution around the y-axis using GeoGebra. Example 2 y 4 The diagram shows the curve with equation y = 4 ln x − 1. The finite region R, shown in the diagram, is bounded by the curve, the x-axis, the y-axis and the line y = 4. Region R is rotated by 2π radians about the y-axis. Use integration to show that__the exact value of the volume of the solid generated is 2π√ e (e2 − 1). y = 4 ln x − 1 y+1 ln x = _____     4 y+1 y ___ __ __1   x = e  4 = e   4 e   4 V = π∫ ( 0 y __ ) y 4 __ = 2πe   2 [ e   2]  0  _1 x First rewrite x as a function of y. ∫ y 4 __     e   2  dy 0 = 2πe   2(e  2 − e  0) __1 __ = 2π√ e (e  2 − 1) Exercise y = 4 ln x – 1 O __1 2 _1     e   4  e   4   dy = πe   2  4 R Use V = π∫     x  2  dy with a = 0, b = 4 and x = e   4 e   4. b a _y _1 Integrate with respect to y. Simplify and leave in the correct form. 4B 1 Find the exact volume of the solid generated when each curve is rotated through 2π radians about the y-axis between the given limits. a x = e2y − e−y between y = 0 and y = 1 _______ √ 5 − ln y  between y = 1 and y = 5 c x = ________   y  P __ b x = √ y e  y  2between y = 0 and y = 1 1 _____  between y = e4 and y = e9 d x =  ______ √ y ln y  2 Find the exact volume of the solid generated when each curve is rotated through 2π radians about the y-axis between the given limits. 5 − 2x  2 1   between y = −1 and y = 1 a y = __   x  − 1 between y = 0 and y = 1 b y = _______   2 x  − 1 __ π d y = arccos √ x  between y = 0 and y = __     c y = 2e  x  2 between y = 2 and y = 4 2 81 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 81 26/02/2018 09:23 Chapter 4 E/P E/P 1 3 The diagram shows the curve with equation x = ______     2y + 1 The finite region bounded by the curve, the y-axis and the lines y = 1 and y = b is shown in the diagram. The region is rotated through 2π radians about the y-axis to generate a solid of π    , revolution. Given that the volume of the solid generated is ___ 10 find the value of b. (5 marks) y b x= 1 2y + 1 1 x O __ 4 The curve with equation x = √ y  sin y is shown in the diagram. y b O x= y sin y x The finite region enclosed by the curve and the y-axis is shaded. The region is rotated through 2π radians about the y-axis. E/P a Find the value of b. (1 mark) b Find the volume of the solid generated. (6 marks) 5 The diagram shows the curve with equation y = 3 ln (x − 1). The finite region R, shown shaded in the diagram, is bounded by the curve, the x-axis, the y-axis and the line y = 5. The region R is rotated by 2π radians about the y-axis. Use integration to find the exact value of the volume of the solid generated. (5 marks) Watch out First rearrange the equation to make x the subject. E/P y 5 R O y = 3ln (x – 1) x __ 6 a Express cos y + √  sin 3 y in the form R cos (y − α), where R . 0 and α is acute. (4 marks) 1 __  , the y-axis and the The region R is bounded by the curve with equation x = _____________      cos y + √ 3  sin y π __ lines y = 0 and y =     3 b Using your answer to part a, or otherwise, show that the volume __of the solid formed when π √3  the region R is rotated through 2π radians about the y-axis is  ____   (6 marks) 4 82 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 82 26/02/2018 09:23 Volumes of revolution E/P 7 a Using the substitution u = 2y, or otherwise, find the exact value of 1 2   y _______   dy     ( y )2 ___ 0 2  + 1   √ 2  y  ______   The diagram shows part of the curve with equation x =   y 2  + 1 The shaded region R, shown in the diagram, is bounded by the curve, the y-axis and the lines y = 0 and y = 1. The region is rotated through 2π radians about the y-axis to form a solid of revolution. ∫ (6 marks) y x= 1 2y 2y + 1 R O b Using your answer to part a, find the exact volume of solid of revolution formed. E/P x 0.5 (2 marks) 8 a By writing a suitable expansion for sin 5θ, or otherwise, show that   16 (10 sin θ − 5 sin 3θ + sin 5θ) sin  5 θ ≡ __ 1 (3 marks) _____ The curve shown in the diagram has equation x = sin  2 y √ sin y . The finite region bounded by the curve, the y-axis and the line π y = __    is shown in the diagram. 4 The region is rotated through 2π radians about the y-axis to generate a solid of revolution. 15 ( __ 8 ) (7 marks) 43√ 2  π         8 + _____ b Show that the volume of the solid generated is __ y π x = sin2 y sin y π 4 O x 4.3 Volumes of revolution of parametrically defined curves When the equations of curves are given parametrically, you can adjust the formulae for volumes of revolution by using the chain rule. ▪▪ The volume of revolution formed when the parametric curve with equations x = f(t) and y = g(t) is rotated through 2π radians about the x-axis between x = a and x = b is given by t=p dx x=b ___ Volume = π∫   y  2  dx = π     y  2     dt dt x=a t=q ∫ Links For a parametric curve, x and y are each given as a function of a parameter, t. ← Pure Year 2, Chapter 8 Watch out After you have used the chain rule, you are integrating with respect to the parameter, t. Generally if x = a, then t ≠ a. You can evaluate the definite integral by rewriting the limits of the integral in terms of t. ▪▪ The volume of revolution formed by rotating the same curve through 2π radians about the y-axis between y = a and y = b is given by t=p dy y=b ___ Volume = π∫   x  2  dy = π    x  2     dt dt y=a t=q ∫ 83 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 83 26/02/2018 09:23 Chapter 4 Example y 3 1 The curve C has parametric equations x = t(1 + t), y = _____    , 1+t t > 0. The region R is bounded by C, the x-axis and the lines x = 0 and x = 2. Find the exact volume of the solid formed when R is rotated 2π radians about the x-axis. 1 1 y = _____       ⇒ y  2 = ______   1+t (1 + t)  2 dx     = 1 + 2t x = t + t  2 ⇒ ___ dt x=0⇒t=0 C R O 2 x dx Find y2 and ___     dt x=2⇒t=1 1 1 ______   (1 + 2t)  dt V = π     ( 1 + t)  2 0 1 + 2t A B _______     ≡ _______     + ______     (1 + t)2 (1 + t)2 1 + t 1 + 2t = A + B(1 + t) ⇒ B = 2 and A = −1 1 2 1 _____ _______   dt So V = π     (  1 + t  −   (1 + t)2 ) 0 1 1 = π[2 ln|1 + t| + ______          1 + t]0 ∫ ∫ Find the limits in terms of t. You can ignore the second solution to each quadratic equation as the domain of t is given as t > 0. q dx Use V = π ∫     y  2  ___   dt with p = 0, q = 1, p dt dx 1 ______ 2 y  =     and   ___  = 1 + 2t ( 1 + t)  2 dt Use partial fractions. Substitute values of t or compare coefficients. = π((2 ln 2 + __   21 ) − (0 + 1)) = π(2 ln 2 − __   21 ) Exercise 4C 1 The curve C is given by the parametric equations x = t3, y = t2, t ∈ ℝ. The region R bounded by the curve, the coordinate axes and the line x = 8 is rotated through 360° about the x-axis. Find the volume of the solid of revolution formed. ____ 2 The curve C is defined by parametric equations x = et, y = √ t − 1 , t > 1. The finite region bounded by the curve, the x-axis and the lines x = e2 and x = e3 is rotated through 2π radians about the x-axis. a Write down the values of t corresponding to x = e2 and x = e3. b Find the volume of the solid of revolution formed. c Show that a Cartesian equation of C is y2 = ln x − 1. d Evaluate π∫     (ln x − 1)  dx. e   3 e  2 Hint Your answers to parts b and d should be the same. You can find a volume of revolution for a parametric curve by either integrating with dx respect to the parameter using π y2  ___  dt, or dt by converting to Cartesian form and then using π∫   y  2 dx. If you convert to Cartesian form, you must remember to convert the limits of integration to values of x. ∫ 84 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 84 26/02/2018 09:23 Volumes of revolution _______ 3 The curve C is defined by the parametric equations x = √ 1 − sin θ , y = cos θ, 0 < θ < 2π. y C R P O x a Show that a Cartesian equation of the curve is y2 = 2x2 − x4. b Find the coordinates of the point P, where the curve intersects the x-axis. The finite region bounded by the curve is rotated about the x-axis to form a solid of revolution. c Find the volume of the solid formed. π 4 The curve C is given by the parametric equations x = tan θ, y = sec3 θ, 0 < θ , __     2 The region R bounded by the curve, the y-axis and the lines y = 1 and y = 8 is rotated through 2π radians about the y-axis. a Find the values of θ corresponding to y = 1 and y = 8. b Find the volume of the solid of revolution formed. _2 c Show that a Cartesian equation of the curve is x2 = y   3 − 1. d Use π∫     x   2  dy to verify your answer to part b. P b a y _____ π 5 The curve C has parametric equations x = sin  4 θ √ cos θ , y = cos θ, 0 < θ < __     2 The finite region R bounded by the curve and the y-axis is rotated through 360° about the y-axis. C R x O Find the volume of the solid of revolution formed. P 6 The diagram shows the curve C with parametric equations x = 2t, y = t2, −2 < t < 2. The points P and Q correspond to the points where t = −2 and 2 respectively. y P Q a R O C x The region R is bounded by the curve and the line y = a. Region R is rotated about the y-axis to form a solid of revolution. Use parametric integration to show that the volume of the solid formed is 32π. 85 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 85 26/02/2018 09:23 Chapter 4 E/P 7 a Find ∫   cos  2 θ  dθ (2 marks) The diagram shows part of the curve C with parametric π equations x = cot θ, y = 4 sin 2θ, 0 < θ , __    . The finite region 2 1   __ , R shown in the diagram is bounded by C, the lines x = ___ √ 3  __ x = √ 3 and the x-axis. Region R is rotated through 2π radians about the x-axis to form a solid of revolution. y b Show that the volume of the solid of revolution formed b is given by the integral k∫     c os  2 θ   dθ, where a, b and k are a constants to be found. O C R 1 3 x 3 (5 marks) c Hence find the exact value for this volume, giving your answer in the form pπ2, where p is a constant to be found. E/P E/P 1 1 8 The curve C has parametric equations x =  __ , y = ln 2t, t > __     2 2t The finite region R, shown in the diagram, is bounded by C, the x-axis, the y-axis and the line y = a. Region R is rotated through 2π radians about the y-axis. 24π The solid of revolution formed has volume  ____  49 Find the exact value of a. (8 marks) (3 marks) y a C R x O 9 The curve C has parametric equations x = 2 sin t, y = t2, 0 < t < π. y The finite region R, shown in the diagram, is bounded by C and the y-axis. The shaded region is rotated through 2π radians about the y-axis. Use calculus to find the exact volume of the solid generated. (6 marks) C R O E/P x 10 The diagram shows the curve C with parametric equations x = t2 − 2t, y = 1 − t2, −1 < t < 1. y Q C R P O S x C intersects the coordinate axes at points P, Q and S as shown in the diagram. The region R is bounded by the curve and the line segments PQ and QS. Region R is rotated through 2π radians about the x-axis. Find the exact volume of the solid of revolution. (8 marks) 86 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 86 26/02/2018 09:23 Volumes of revolution E/P 11 The diagram shows part of the curve C with parametric equations x = et, y = e−2t, t ∈ ℝ. y 6 The region R is bounded by the curve, the y-axis and the lines y = 1 and y = 6. Region R is rotated through 2π radians about the y-axis. a Use parametric integration to find the volume of the solid (6 marks) of revolution. C R 1 x O The tangent to the curve at the point (1, 1) is shown on the diagram. The region S is bounded by this tangent, the curve, the y-axis and the line y = 6. y 6 b Find the volume of the solid of revolution formed when the region S is rotated through 2π radians about the y-axis. (3 marks) S 1 x O 4.4 Modelling with volumes of revolution Volumes of revolution can be used to model real-life situations. Example 4 The diagram shows a model of a goldfish bowl. The crosssection of the model is described by the curve with π 11π parametric equations x = 2 sin t,  y = 2 cos t + 2,   __  < t < ____    , 6 6 where the units of x and y are in cm. The goldfish bowl is formed by rotating this curve about the y-axis to form a solid of revolution. a Find the volume of water required to fill the model to a height of 3 cm. y 3 cm O 4 cm x The real goldfish bowl has a maximum diameter of 48 cm. b Find the volume of water required to fill the real goldfish bowl to the corresponding height. 87 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 87 26/02/2018 09:23 Chapter 4 a x = 2 sin t ⇒ x  2 = 4 sin  2 t dy y = 2 cos t + 2 ⇒ ___     = − 2 sin t dt y = 0 ⇒ 2 cos t + 2 = 0 ⇒ t = π π 5π y = 3 ⇒ 2 cos t + 2 = 3 ⇒ t = __     ,  ___  3 3 __   5π  V = π∫   3   4 sin  2 t(− 2 sin t) dt π = − 8π∫   3   sin  3 t  dt 5π __     π = − 8π∫   3   sin t(1 − cos2 t) dt 5π __     π = − 8π∫   3   (sin t − sin t cos  2 t)  dt 5π __     π = − 8π[− cos t +   __   5π 3  3   cos  t     1 __ 3 ]π 5π = − 8π((− cos     +     cos  3  ___ ) 3 3 3 1 __ 3 − (− cos π +     cos  π)) 3 1 __ 1 1 1 __ __ = − 8π((−     +    (   )) − ( 1 − __    )) 2 3 8 3 9 = − 8π(−  __ ) = 9π 8 5π ___ 1 __ dy Find x2 and ___     dt Find the limits in terms of t. Watch out The two possible values of t when y = 3 correspond to the two sides of the bowl. Choose one of these, as you need to rotate half of the bowl about the y-axis. ∫ q dy 5π Use V = π     x  2  ___   dt with p = π, q = ___    , 3 dt p dy     = − 2 sin t x  2 = 4 sin  2 t and ___ dt Using the identity sin2t + cos2t ≡ 1 allows you to integrate sin3t. Integrate sin t cos 2 t using the reverse chain rule. Substitute the limits. Simplify. b Linear scale factor = 12 Volume scale factor = 123 = 1728 Volume in actual tank = 1728 × 9π = 48 900 cm3 (3 s.f.) Exercise E/P E/P 4D y 120 2000 1 The diagram shows the curve with equation x = ______    , y . 0. 20 + y A volcano is modelled as the solid of revolution formed when the region R bounded by the x-axis, the y-axis, the line y = 120 and the curve is rotated about the y-axis. O The units of x and y are metres. x = 2000 20 + y R a x a Write down the diameter of the base of the volcano according to this model. (2 marks) b Use this model to estimate the volume of the volcano. (6 marks) 2 The diagram shows the cross-section of a vase, which has a height of 30 cm. 100y The vase is formed by rotating the curve C with equation x2 = ________  ,   10y  2 + 1 0 <y < 30 through 360° about the y-axis. The vase is filled to a height of 20 cm with water. Find the exact volume of water in the vase in the form pπ ln q cm3, where p and q are integers to be found. (6 marks) 88 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 88 26/02/2018 09:23 Volumes of revolution E/P 3 a Prove that cos  3 θ ≡ _  4 cos θ + _  4 cos 3θ 3 1 (3 marks) The diagram shows the cross-section of a domed tent. The tent can be modelled by a solid of revolution of a curve C about the y-axis. Curve C has parametric π equations x = 50 cos θ, y = 30 sin θ, 0 < θ < __     2 b Find the volume of the tent. (5 marks) E/P 30 m 100 m 4 A scale model of a hot-air balloon is modelled as a solid of revolution of a curve C about the y-axis. Curve C has ______ π    where the units of x equation x = sin y √ sin 2y , 0 < y < __ 2 and y are in metres. a Find the volume of the model hot-air balloon. (5 marks) b The real hot-air balloon has a height of 6π metres. Find the volume of this balloon. (2 marks) y E/P 5 The diagram shows the image of a silver earring, which has a height of 3 mm. The earring is modelled by a solid of revolution of a curve C about the y-axis. Curve C has parametric equations x = 2 sin 2θ, y = 3 sin θ, 0 < θ < π. C x O   81 y  2(9 − y  2) a Show that a Cartesian equation of the curve C is x  2 = __ 16 (4 marks) Silver is melted down and cast into a mould to create each earring. b Using the model, estimate the maximum number of earrings that can be manufactured from 300 mm3 of silver. (6 marks) c Give one reason why this might be: i an underestimate ii an overestimate. Mixed exercise E/P (2 marks) 4 1 a Find ∫x cos 2x dx. b The diagram shows part of the curve C with _1 equation y = 2x   2 sin x. The shaded region in the diagram is bounded by the curve, the x-axis π and the line with equation x = __    . This shaded 2 region is rotated through 2π radians about the x-axis to form a solid of revolution. Using calculus, find the volume of the solid of revolution formed, giving your answer in terms of π. (5 marks) y C O π 2 x (4 marks) 89 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 89 26/02/2018 09:23 Chapter 4 E y 2 a Use integration by parts to show that ∫     x sec2 x dx = _  π4 − _  12 ln 2 π _   4 (5 marks) 0 The finite region R, bounded by the curve π _1 with equation y = x   2 sec x, the line x = __     and 4 the x-axis is shown. Region R is rotated through 2π radians about the x-axis. R b Find the volume of the solid of revolution (2 marks) generated. E/P 1 y = x 2 sec x π 4 O x x+2 3 The diagram shows part of the curve C with equation y = _____   x  , x . 0. y C T (1, 3) y= R O x+2 x x 5 2 1 The tangent T to C at the point (1, 3) meets the x-axis at the point ( _  2 , 0). The shaded region is 5 bounded by C, the line x = _  2 and T, as shown in the diagram. 5 The region is rotated by 2π radians about the x-axis to generate a solid of revolution. (10 marks) Find the exact volume of thjis solid. E/P 4 The shaded region R, shown in the diagram, is bounded by the curves y = sec x − cos x and y = cosec x − sin x, and by the x-axis. The shaded region R is rotated through 2π radians about the x-axis to form a solid of revolution. y y = sec x – cos x y = cosec x – sin x R π 2 O x (9 marks) Find the exact volume of the solid. E/P _1 5 The diagram shows part of the curve x = e   2 y − 2 The region R is bounded by the curve, the y-axis and the lines y = 2 and y = 4, as shown in the diagram. Region R is rotated through 2π radians about the y-axis. Use integration to find the exact value of the volume of the solid formed. Leave your answer in the form (6 marks) π(Ae4 + Be3 + Ce2 + De + F ). y 4 2 O R 1 x = e 2y – 2 x 90 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 90 26/02/2018 09:23 Volumes of revolution P y 6 The diagram shows the region R that is bounded by the curve C and the line l. π Curve C has parametric equations x = 2 sin  2 t,  y = 2 cos t,  0 < t < __     2 3 Line l is the tangent to the curve C at the point P(_  2  , 1). l C P a Find the Cartesian equation of the line l. The region R is rotated through 2π radians about the y-axis. E R O b Find the volume of solid generated. x 7 This graph shows, for 0 < t < 2, the curve C with parametric equations x = (t + 1)2, y = _  2 t3 + 3. 1 y C R O 1 9 x The shaded region R is bounded by curve C and the lines x = 1 and x = 9. a Find the area of the region R. (3 marks) The region R is rotated by 2π radians about the x-axis. (5 marks) b Use integration to find the exact value of the volume of the solid formed. P E/P 8 A point on the unit circle has coordinates (cos t, sin t). Use parametric integration to show that 4π the volume of the unit sphere is ___     3 9 a Prove the identity sin  3 θ ≡ _  4 sin θ − _  4 sin 3θ. 3 1 (3 marks) The diagram shows a rugby ball, which has a length of 30 cm and a height of 20 cm. The curve C has parametric equations x = 15 cos θ, y = 10 sin θ, 0 < θ < π The rugby ball is modelled as the volume of revolution formed when the curve C is rotated by 2π radians about the x-axis. b Find the exact volume of the rugby ball according to the model. E/P 10 Part of the outline of a solid glass pendant is shown in the diagram. The outline is modelled by the curve with parametric equations π x = 2 sin 2t,  y = 4 cos t,  0 < t < __    . The piece of jewellery is 2 formed by rotating the shaded region through 2π radians about the y-axis. (6 marks) y R O Use the model to estimate the volume of glass contained in the pendant. x (7 marks) 91 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 91 26/02/2018 09:23 Chapter 4 Challenge The curve C is defined by the parametric equations x = t2, y = 2t, t ∈ ℝ. The diagram shows the finite region R bounded by the curve C and the line with equation y = x. y y=x C R x O A solid of revolution is formed by rotating the region through 2π radians 32π about the line y = x. Show that the volume of this solid is _____   __  15 √   2 Summary of key points 1 The volume of revolution formed when y = f(x) is rotated through 2π radians about the x-axis between x = a and x = b is given by Volume = π ∫     y   2dx b a 2 The volume of revolution formed when x = f( y) is rotated through 2π radians about the y-axis between y = a and y = b is given by   2dy Volume = π ∫     x b a ● The volume of revolution formed when the parametric curve with equations x = f(t) and y = g(t) is rotated through 2π radians about the x-axis between x = a and x = b is given by t=p dx x=b ___ Volume = π∫     y   2  dx = π     y  2     dt dt x=a t=q ∫ ● The volume of revolution formed by rotating the same curve through 2π radians about the y-axis between y = a and y = b is given by t=p dy y=b 2     dy = π     x  2   ___ dt Volume = π∫     x dt y=a t=q ∫ 92 M04A__ALEVEL_CP2_83343_U04A_077-092.indd 92 26/02/2018 09:23 1 Review exercise E 1 Show that cos 2x + i sin 2x ______________       cos 9x − i sin 9x can be expressed in the form cos nx + i sin nx, where n is an integer to be found. (4) E/P C = 1 + cos θ + cos 2θ + … + cos(n – 1)θ S = sin θ + sin 2θ + … sin(n – 1)θ By considering C + iS, show that 1 − cos θ + cos (n − 1) θ − cos nθ C = ___________________________            2 − 2 cos θ and write down the corresponding expression for S. (4) ← Section 1.2 E/P 2 a Use de Moivre’s theorem to show that cos 5θ = 16 cos5 θ − 20 cos3 θ + 5 cos θ. (4) b Hence find 3 distinct solutions of the equation 16x5 − 20x3 + 5x + 1 = 0, giving your answers to 3 decimal places where appropriate. (5) ← Section 1.5 E 7 a Solve the equation z5 = 4 + 4i ← Section 1.4 E/P 6 The convergent infinite series C and S are defined as giving your answers in the form 3 a Use de Moivre’s theorem to show that sin 5θ = sin θ (16 cos4 θ − 12 cos2 θ + 1) (4) z = reikπ, where r is the modulus of z and k is a rational number such that b Hence, or otherwise, solve, for 0 < θ , π, sin 5θ + cos θ sin 2θ = 0.(5) b Show on an Argand diagram the points representing your solutions. (2) (6) 0 < k < 2. ← Section 1.4 E/P 4 a Use de Moivre’s theorem to show that 1 sin5 θ = __  16(sin 5θ − 5 sin 3θ + 10 sin θ). (4) ← Section 1.6 E π _   2 5 0 E/P giving your answers in the form reiθ, (6) where r . 0, −π , θ < π. (6) ← Section 1.4 b Show that your solutions satisfy the equation 5 a Given that z = cos θ + i sin θ, show that (2) zn + z−n = 2 cos nθ. z9 + 2k = 0 b Express cos6 θ in terms of cosines of multiples of θ. (4) c Hence show that π _   2 5π ∫   cos6 θ dθ = ___     32 0 (6) ← Section 1.4 __ z3 = 32 + 32√3 i b Hence, or otherwise, show that 8 ∫   sin θ dθ = __  15  8 a Solve the equation for an integer k, the value of which should be stated. (3) ← Section 1.6 E 9 Solve the equation z5 = i, giving your answers in the form cos θ + i sin θ. (6) ← Section 1.6, 1.7 93 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 93 26/02/2018 09:24 Review exercise 1 E 10 a Find, in the form re  iθ, the solutions to the equation _ (5) z  5 − 16 − 16i √  3 = 0 E/P The solutions form the vertices of a polygon in the Argand diagram. b State the name of the polygon formed. (1) ← Section 1.7 E/P ← Section 2.1 E/P 11 a Write down the five distinct solutions to z5 = 1, giving your answers in exponential form, and show that their sum is 0. (4) b The point (3, 0) lies at one vertex of a regular pentagon. Given that the pentagon has its centre at the point (2, 1), find the coordinates of the (4) other vertices. E/P 12 Prove that n 2  ∑ ___________      = _____     n+2 r = 1 (r + 1)(r + 2) n E/P E n (5)  ∑  r 2 = _ 6n(n + 1)(2n + 1) 14 a Show that r+1 1 r  _____  − _____      , r  핑+      ___________ r + 2 r + 1 (r + 1)(r + 2) (2) 2 15 f(x) =  __________________     (x + 1)(x + 2)(x + 3) (2) b Hence find   ∑   f(r). (3) r=1 1 r=1 40 c Calculate  ∑(3r − 1)2. r=1 E/P (3) (2) ← Section 2.1 20 Prove that 2n n(an + b) 1 ∑      _____________     = ______________       c(n + 1)(2n + 1) r = 1 r(r + 1)(r + 2) where a, b and c are constants to be found. (6) ← Section 2.1 a Express f(x) in partial fractions. n (2) b Hence show that ← Section 2.1 E/P (2) 19 Given that for all real values of r, (2r + 1)3 − (2r − 1)3 = Ar 2 + B where A and B are constants, n b Hence, or otherwise, find n 1  ∑   ___________    , giving your answer r = 1 (r + 1)(r + 2) as a single fraction in terms of n. (3) (2) (5) a find the value of A and the value of B. ← Section 2.1 E/P (5) ← Section 2.1 ← Section 2.1 n(an + b) 2 ∑     ____________     = _____________       c(n + 2)(n + 3) r = 1 (r + 1)(r + 3) where a, b and c are constants to be found. 18 a Prove that n 2 1  ∑  ______   2   = 1 − ______     2n + 1 r = 1 4r − 1 b Hence find the exact value of 20 2   2   ∑ ______   r = 11 4r − 1 (5) 13 Prove that 17 a Prove that n n(an + b) 4 ∑     _______  = ____________       (n + 1)(n + 2) r=1 r(r + 2) where a and b are constants to be found. 100 4 b Find the value of  ∑ _______    , to r = 50 r(r + 2) 4 decimal places. ← Section 2.1 ← Section 1.8 E/P 16 a Express as a simplified single fraction 1 1  _______2  − __     (2) (r − 1) r 2 b Hence prove, by the method of differences, that n 2r − 1 1 __  ∑  ________   2 (3) 2  = 1 −   2  r (r − 1) n r=2 ← Section 2.1 E 21 a Show that r3 − r + 1 1 1 ________        r − 1 + __   r  − _____   r+1 r(r + 1) for r ≠ 0, −1. (2) 94 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 94 26/02/2018 09:24 Review exercise 1 E/P E/P n r3 − r + 1 b Find  ∑   ________ , expressing your r = 1 r(r + 1) answer as a single fraction in its simplest form. (3) b Hence, or otherwise, obtain the first two non-zero terms in the series expansion for ln(sec x) in ascending (4) powers of x. ← Section 2.1 ← Sections 2.3, 2.4 n 2r + 3 22 Find  ∑  ________   r   3 (r + 1) r=1 (5) E ← Section 2.1 23 Given that x is so small that terms in x3 and higher powers of x may be neglected, show that a f9(x) = −2 tan x (2) b f 00(x) = −(f 09(x) f9(x) − (f 0(x)2) (5) c Find the Maclaurin series expansion of f(x), in ascending powers of x, up (4) to and including the term in x4. 11 sin x − 6 cos x + 5 = A + Bx + Cx2 stating the values of the constants A, B and C. 28 Given that π f(x) = ln(1 + cos 2x), 0 < x , __ 2 Show that: (6) ← Sections 2.2, 2.3, 2.4 ← Sections 2.3, 2.4 E/P E 24 Show that for x . 1, ln(x2 − x + 1) + ln(x + 1) − 3 ln x (−1)n − 1 1 1   6  + … + ______   =  __3  − ___   + … (6) x 2x nx3n 25 Given that x is so small that terms in x4 and higher powers of x may be neglected, find the values of the constants A, B, C and D for which e−2x cos 5x = A + Bx + Cx2 + Dx3 29 Evaluate   e  −x sin x dx 0 E ← Sections 2.3, 2.4 E/P ∫ ∞ E ∫ 1 x  2 _______ 30 Evaluate     ______2   dx 1 − x    0 √ ∫ ∫ (3) ∞ 1 _______   dx b Hence show that     x(x + 3) 3 converges and find its value. (6) 26 a Find the first four terms of the expansion, in ascending powers of x, of (2x + 3)−1, |x| ,   _3 2 27 a By using the Maclaurin series for cos x and ln(1 + x), find the series expansion for ln(cos x) in ascending powers of x up to and including the term in x4. (6) 32 Show that ∫   x  3 e  −x  dxconverges and ∞ E 4 1 find its exact value. (5) ← Section 3.1 E ∫ 1 33 a Find ________     dx (5 − 2x)2 (2) 3 1   b Hence show that  ∫  ________ 2 dx −∞ (5 − 2x)  diverges. ← Sections 2.3, 2.4 E/P (3) ← Section 3.1 (3) b Hence, or otherwise, find the first four non-zero terms of the expansion, in ascending powers of x, of sin 2x 2 ______   (5)  , |x| ,   _3 2x + 3 (5) ← Section 3.1 1 31 a Find _______     dx x(x + 3) ← Sections 2.3, 2.4 E/P (5) ← Section 3.1 (4) ← Section 3.1 E 34 Find the exact mean value of π    ]. (4) f(x) = x cos 2x over the interval [0, __ 2 ← Section 3.2 95 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 95 26/02/2018 09:24 Review exercise 1 E 3x 35 f(x) =  ______________     (x − 1)(2x − 3) a Show that the mean value of f(x) over 343 1 (4) the interval [2, 5] is  _2  ln  _ 16  E/P 41 f(x) = arcsin x 1   _______2   a Show that f9(x) = ________ (3) √ 1 − x   dy b Given that y = arcsin 2x, obtain ___     dx as an algebraic fraction. (3) b Use the answer to part a to find the mean value over the interval [2, 5] of f(x) + ln k where k is a positive constant, giving your answer in the form p ln q, where p and q are constants and q is in terms of k. (2) c Using the substitution x = _ 2sin , show that 1 ∫ _1   4 __ 2x x arcsin ________  481 (6 − π√3 ). (4)   dx = __    _________ 2 √ 1 − 4x   0 ← Sections 3.3, 3.4 ← Section 3.2 E/P E/P 36 f(x) = x  2 (x3 − 1)3 42 Show that   2x + 1 dx = A arctan x + ln x ∫______ x +x a Show that the mean value of f(x) over 57122 the interval [1, 3] is _   3  (3) 3 where A and B are constants to be found. b Use the answer to part a to find the mean value over the interval [1, 3] of –2f(x). (2) 1 _____   37 f(x) =  ______ √ 3 − x  Find the exact mean value of f(x) on the interval [1, 3]. (4) E/P ← Section 3.2 E/P (5) ← Section 3.5 ← Section 3.2 E/P + B ln(x2 + 1) + c 38 f(x) = ln kx, where k is a positive constant. Given that the mean value of f(x) on the 1 interval [1, 5] is  _4 (9 ln 5 − 4), find the value of k. (4) 3x  2 + 5x 43 f(x) =  ________________         3 x  − 3x  2 + 5x − 15 a Show that f(x) can be written in the form B A     2   _____  + ______ x − 3 x  + 5 where A and B are constants to be found. (2) b Hence show that ∫   f(x) dx = P ln(x − 3) + Q arctan Rx + c where P, Q and R are constants to be found. (4) ← Section 3.2 ← Section 3.5 E/P 39 Given that y = (arcsin x) , 2 dy (4) a prove that (1 − x2)  ___   = 4y dx dy d2y b deduce that (1 − x2)  ____2  − x  ___  = 2. (2) dx dx 2 ( ) E 44 y y = xex ← Section 3.3 E/P 40 a Given that y = arctan 3x, and assuming the derivative of tan x, prove that dy 3  ___  = _______   (3)   dx 1 + 9x2 b Show that __ √ 3  __ ∫   6 x arctan 3x = _ 19(4π − 3√3 )   3 __ 0 ← Sections 3.3, 3.4 (4) R O 1 3 x The figure shows the finite region R, which is bounded by the curve y = xex, the line x = 1, the line x = 3 and the x-axis. The region R is rotated through 360° about the x-axis. 96 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 96 26/02/2018 09:24 Review exercise 1 Use integration by parts to find the exact volume of the solid generated. (8) E/P y 47 5 ← Section 4.1 E 45 y a y = 3 sin x 2 x= x O 2π x O __  , The curve with equation y = 3 sin  x 2 0 < x < 2π, is shown in the figure. The finite region enclosed by the curve and the x-axis is shaded. 1 The curve shown has equation x = ______     3y − 1 The finite region shaded is bounded by the y-axis, the line y = 5 and the line y = a. The region is rotated about the y-axis through 360°. Given that the volume of the solid 3π generated is ___    , find the value of a. (5) 70 a Find, by integration, the area of the shaded region. (4) This region is rotated through 2π radians about the x-axis. b Find the volume of the solid generated. 46 y ← Section 4.2 (5) ← Section 4.1 E 1 3y – 1 E/P y 48 y = x sin x O k π x O x ____ The figure shows a graph of y = x√ sin x , 0 , x , π. The finite region enclosed by the curve and the x-axis is shaded as shown in the figure. A solid body S is generated by rotating this region through 2π radians about the x-axis. Find the exact volume of S. (8) ← Section 4.1 x = y cos y The curve shown has equation x = y cos y. The finite region shaded is bounded by the curve and the y-axis. The curve intersects the positive y-axis at (0, k). π (2) a Show that k = __     2 The region is rotated through 2π radians about the y-axis. b Show that the volume of the solid generated is aπ4 + bπ2 where a and b are constants to be found. (6) ← Section 4.2 97 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 97 26/02/2018 09:24 Review exercise 1 E 49 y a show that the exact volume of the vase can be written in the form aπ ln b where a and b are integers to be found. (5) C ln 2 O b Suggest a reason why the volume of the vase may actually be less than your answer to part a. (1) x b ← Section 4.4 E/P y 51 The diagram shows the curve C with parametric equations A x = y(3 – 12 ey) O x x = ln t, y = t2 – 2, t . 0 The finite region shown shaded is bounded by the curve, the x-axis and the lines x = ln 2 and x = b. The region is rotated through 2π radians about the x-axis. Given that the volume of the solid generated is 36π + 4π ln 2, find the value of b. (6) ← Section 4.3 E/P 50 y The diagram shows the curve with equation x = √ y (3 − _  2 e  y) __ 1 a Show that the coordinates of the point marked A where the curve crosses the y-axis are (0, ln 6). (2) The solid of revolution formed when the shaded region is rotated through 360° about the y-axis is used to model a prototype of a new type of orthopaedic cushion. The prototype is 3D printed using plastic filament. Given that each unit on the axes is 1 cm, b find, correct to 3 significant figures, the (6) volume of the prototype. C O 4 x The shaded region above is formed by the x-axis, the lines x = 0 and x = 4 and the curve C where the equation of C is 5 ______   y =  _______ √ 1 + 4x  A pottery vase is modelled by rotating the shaded region through 360° about the x-axis. Given that each unit on the axes represents 2 cm, c Suggest a reason why the amount of filament used to print to model may exceed your answer to part b. (1) ← Section 4.4 98 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 98 26/02/2018 09:24 Review exercise 1 E/P 52 y 2 The region bounded by the x-axis and the graph of the function ______ x+3   3   f(x) = _____ x  on the interval [2, ∞) is rotated through 360° about the x-axis. √ C R x O The diagram shows the curve C with parametric equations π x = 2 sin t, y = 3 sin 2t, 0 < t <   __  2 A jewellery pendant is made in the shape of the solid of revolution formed when the region marked R is rotated through 2π radians about the x-axis. Each unit on the axes represents 0.5 cm. a Show that the volume of the pendant can be found by evaluating the integral π _     ∫   sin2 2t cos t dt 4 0 9π ___   2 (4) b Hence show that the exact volume of 6π the pendant is ___     cm3. (6) 5 ← Section 4.4 Challenge 1 a Show that if ω = e   _3 , then n 1_____________ + ωn + (ω 2)n 1 if n is zero or a multiple of 3      = { 0 otherwise 3 Let f(x) be a finite polynomial whose largest power of x is a multiple of 3, so that f(x) = a 0 + a1x + a2x2 + … + a3kx3k where ai ∈ ℝ, k ∈ ℕ. The sum S is given by k S = a 0 + a3 + a6 + … + a3k = ∑a3 2πi r=0 Show that the volume of the described solid is finite and find its value. ← Sections 3.1, 4.1 3 A continuous random variable, X, has probability density function A   , x ∈ ℝ f(x) = ______ 1 + x2 where A is a constant. a Given that   ∫  f(x) dx= 1, show that −∞ 1 A =   __   π The variance of a continuous probability distribution which is defined over the real numbers and is symmetrical about x = 0 is given by ∞   ∫  x  2 f(x) dx ∞ −∞ b Show that X has infinite variance. The mean of a continuous probability distribution which is defined over the ∞ real numbers is given by   ∫  xf(x) dx. −∞ c Show that a 2a x x   dx lim  ∫  _______ lim  ∫   ______   2 dx ≠ a→∞ a→∞ −a −a 1+x 1 + x2 and explain why the mean of X is ← Section 3.1 undefined. r b By considering a general term of f(x), f(1) + f(ω) + f(ω  2)      show that S = ______________   3 c Hence, by considering the binomial expansion of (1 + x)45, show that 15 45 45 2 ∑(  )= ______   2  −  ← Section 1.6 3 r=0 3r 99 M04B_ALEVEL_CP2_83343_U04B_093-099.indd 99 26/02/2018 09:24 5 Polar coordinates Objectives After completing this chapter you should be able to: ● Understand and use polar coordinates → pages 101–104 ● Convert between polar and Cartesian coordinates→ pages 102–104 ● Sketch curves with r given as a function of θ ● Find the area enclosed by a polar curve → pages 104–109 → pages 109–112 ● Find tangents parallel to, or at right angles to, the initial line → pages 113–116 Prior knowledge check 1 Find the exact value of π ∫     sin2θ  dθ 0 ← Pure Year 2, Chapter 11 2 y = cos x + sin x cos x Find, in the interval 0 , x , π, the values dy of x for which ___     = 0. dx ← Pure Year 2, Chapter 9 3 a On an Argand diagram, show the locus of points given by values of z that satisfy |z – 3i| = 3 b Find the area of the region defined by the set of points, R, where π R = {z : |z − 3i| < 3} ∩ {z : 0 < arg z <  __}    2 ← Book 1, Chapter 2 Polar coordinates describe positions in terms of angles and distances. GPS navigation systems use polar coordinates to triangulate the position of a ship or an aircraft. 100 M05_ALEVEL_CP2_83343_U05_100-118.indd 100 02/03/2018 12:40 Polar coordinates 5.1 Polar coordinates and equations Polar coordinates are an alternative way of describing the position of a point P in two-dimensional space. You need two measurements: firstly, the distance the point is from the pole (usually the origin O ), r, and secondly, the angle measured anticlockwise from the initial line (usually the positive x-axis), θ. Polar coordinates are written as (r, θ). Notation When working in polar coordinates y the axes might also be labelled like this: θ= x P(x, y) or (r, θ) r y θ O π 2 O x Initial line The coordinates of P can be written in either Cartesian form as (x, y) or in polar form as (r, θ). You can convert between Cartesian coordinates and polar coordinates using right-angled triangle trigonometry. From the diagram above you can see that: ■ r cos θ = x r sin θ = y Watch out ■ r 2 = x2 + y2 Always draw a sketch diagram to check in which quadrant the point lies, and always measure the polar angle from the positive x-axis. θ = arctan (__    ) Example y x 1 Find polar coordinates of the points with the following Cartesian coordinates. a (3, 4) b (5, −12) __ c (−√ 3 , −1) a y Draw a sketch. r O 4 θ x 3 ________ r = √ 32 + 42  = 5 4 __ θ = arctan   3 = 0.927… So the polar coordinates are (5, 0.927) Use Pythagoras’ theorem to find r. Use trigonometry to find θ. Give your answer in radians. 101 M05_ALEVEL_CP2_83343_U05_100-118.indd 101 02/03/2018 12:40 Chapter 5 b y O Draw a sketch. 5 α r x 12 Use Pythagoras’ theorem to find r. ___________ 2 + (−12)2  = 13 r = √ 5    12 α = arctan  __ 5 = 1.176… So θ = −1.176 So the polar coordinates are (13, −1.176) y c – 3 α r –1 The sketch shows that the point is in the 3rd quadrant. O x ___________ __ √ 3) 2 + (−1)2  = 2 r = √ (−    π 1__ __ α = arctan  ___   =     6 √ 3  π 7π So θ = π +  __  = ___     6 6 7π    ) So the polar coordinates are ( 2, ___ 6 Example Use trigonometry to find θ, taking care to ensure it is in the correct quadrant. You could also write this point as (13, 5.107) since −1.176 + 2π = 5.107 Use Pythagoras’ theorem to find r. 7π The point is in the third quadrant so use θ = ___     6 7π 5π You could also use θ =  ___  − 2π = −  ___  6 6 2 Convert the following polar coordinates into Cartesian form. The angles are measured in radians. 4π 2π a(10, ___    ) b(8, ___    ) 3 3 4π a x = r cos θ = 10 cos  ___  = −5 3 __ 4π ___ y = r sin θ = 10 sin     = −5√ 3  3 __ So the Cartesian coordinates are (−5, −5√ 3 ) 2π b x = r cos θ = 8 cos  ___  = − 4 3 __ 2π ___ y = r sin θ = 8 sin     = 4√ 3  3 __ So the Cartesian coordinates are (−4, 4√ 3 ) Polar equations of curves are usually given in the form r = f(θ). For example, r = 2 cos θ r = 1 + 2θ r = 3 In this example r is constant. You can convert between polar equations of curves and their Cartesian forms. 102 M05_ALEVEL_CP2_83343_U05_100-118.indd 102 02/03/2018 12:40 Polar coordinates Example 3 Find Cartesian equations of the following curves. b r = 2 + cos 2θ a r=5 π c r 2 = sin 2θ, 0 , θ < __     2 a r=5 You need to replace r with an equation in x and y. Use r 2 = x2 + y2. So the equation r = 5 represents a circle with centre O and radius 5. Square both sides to get r 2 = 25 So a Cartesian equation is x2 + y2 = 25 b r = 2 + cos 2θ You need an equation in x and y, so use x = r cos θ. This means first writing cos 2θ in terms of cos θ. r = 1 + (1 + cos 2θ) r = 1 + 2 cos2 θ Now use x = r cos θ and r 2 = x2 + y2. Multiply by r2: r3 = r2 + 2r2 cos2 θ __ 3 (x2 + y2)   2 = x2 + y2 + 2x2 Or Watch out Polar coordinates often give rise to complicated Cartesian equations, which cannot be written easily in the form y = … __ 3 (x2 + y2)   2 = 3x2 + y2 c r2 = sin 2θ, 0 , θ < __  π 2 r2 = 2 sin θ cos θ Problem-solving Multiply by r2: You need to use the substitutions x = r cos θ and y = r sin θ. Use sin 2θ ≡ 2 sin θ cos θ and then multiply by r2. r4 = 2 × r sin θ × r cos θ (x2 + y2)2 = 2xy Example 4 Find polar equations for the following: a y2 = 4x b x2 − y2 = 5 a y2 = 4x r 2 sin2 θ = 4r cos θ r sin2 θ = 4 cos θ 4 cos θ r = _______     = 4 cot θ cosec θ sin2 θ So a polar equation is r = 4 cot θ cosec θ __ c y√ 3  = x + 4 Substitute x = r cos θ and y = r sin θ. Divide by r and simplify. b x2 − y2 = 5 r2 cos2 θ − r2 sin2 θ = 5 Substitute x = r cos θ and y = r sin θ. r2 cos 2θ = 5 Use cos 2θ ≡ cos2 θ − sin2 θ. r2 (cos2 θ − sin2 θ) = 5 So a polar equation is r2 = 5 sec 2θ 103 M05_ALEVEL_CP2_83343_U05_100-118.indd 103 02/03/2018 12:40 Chapter 5 __ c y√ 3  = x + 4 __ r√ 3  sin θ = r cos θ + 4 __ r(√ 3  sin θ − cos θ) = 4 __ √ 3  r(___     sin θ − __   42   21 cos θ) = __ 2 π r sin (θ − __    ) = 2 6 π So a polar equation is r = 2 cosec (θ − __     ) 6 Exercise Substitute x = r cos θ and y = r sin θ and then try to simplify the trigonometric expression. Use the sin(A − B) formula. 5A 1 Find polar coordinates of the points with the following Cartesian coordinates. a (5,12) b (−5, 12) d (2, −3) e (√ 3 , −1) __ c (−5, −12) 2 Convert the following polar coordinates into Cartesian form. 3π π π    ) b(6, − __ ) c(6, ___    ) a(6, __ 4 6 6 5π    ) e (2, π) d(10, ___ 4 3 Find Cartesian equations for the following curves, where a is a positive constant. a r=2 d r = 4a tan θ sec θ g r = 4(1 − cos 2θ) b r = 3 sec θ c r = 5 cosec θ h r = 2 cos2 θ i r2 = 1 + tan2 θ e r = 2a cos θ f r = 3a sin θ 4 Find polar equations for the following curves. a x2 + y2 = 16 b xy = 4 d x2 + y2 − 2x = 0 e (x + y)2 = 4 f x−y=3 g y = 2x h y = −√ 3 x + a i y = x(x − a) c (x2 + y2)2 = 2xy __ Challenge Show that the distance, d, between the two points (r1, θ1) and (r2, θ2) in polar coordinates is ______________________ d = √ r     2  + r2  2  − 2r1r2 cos (θ1 − θ2)  1 5.2 Sketching curves You can sketch curves given in polar form by learning the shapes of some standard curves. ■ r = a is a circle with centre O and radius a. ■ θ = α is a half-line through O and making an angle α with the initial line. ■ r = aθ is a spiral starting at O. 104 M05_ALEVEL_CP2_83343_U05_100-118.indd 104 02/03/2018 12:40 Polar coordinates Example 5 Sketch the following curves. 3π b θ =  ___  4 where a is a positive constant. a r=5 c r = aθ θ=π a 2 5 r =5 θ =0 5 O 5 This is a standard curve: a circle with centre O and radius 5. Initial line 5 b θ = 3π 4 θ= π 2 O This is another standard graph: a half-line. Notice it is only ‘half’ of the line y = −x. The other half of π 7π the line would have equation θ = −__     or θ = ___     4 4 θ=0 Initial line θ=π c 2 r = aθ θ=0 O Initial line You can also sketch curves by drawing up a table of values of r for particular values of θ. It is common to choose only values of θ that give positive values of r. This is another standard curve: a spiral. It crosses the horizontal axis at −aπ, 0 and 2aπ and aπ 3aπ the vertical axis at ___     and – ____    . The curve here 2 2 drawn for values of θ in the range 0 < θ < 2π. Watch out Some graph-drawing programs and graphical calculators will sketch polar curves for negative values of r so take care when using these tools to help you. 105 M05_ALEVEL_CP2_83343_U05_100-118.indd 105 02/03/2018 12:40 Chapter 5 Example 6 Sketch the following curves. a r = a(1 + cos θ) b r = a sin 3θ a r = a(1 + cos θ) θ 0 r 2a π __     2 a 3π ___ π     2 a 0 c r2 = a2 cos 2θ 2π 2a θ= π 2 r = a (1 + cos θ) a When sketching polar curves it is useful to plot points for key values of θ. Make a table of values π for θ at multiples of __    to determine the points 2 at which the curve meets or intersects the coordinate axes. θ=0 2a O Problem-solving Initial line a b r = a sin 3θ Need to consider 4π 5π π 2π 0 < θ < __    , ___     < θ < π and ___     < θ < ___     3 3 3 3 π π __ __         0 θ 6 3 0 0 r a θ= π 2 r = a sin 3θ θ=0 O Initial line This curve is ‘heart’ shaped and is known as a cardioid. Since we only draw the curve when r > 0 you need to determine the values of θ required. Choose values of θ which give exact values of r. The values shown here define the first loop of the curve. The values of r will be the same in the other two loops. Problem-solving The curve given by r = a sin 3θ is typical of the patterns that arise in polar curves for equations of the form r = a cos nθ or r = a sin nθ. They will have n loops symmetrically arranged around O. a c r2 = a2 cos 2θ You need values of θ in the ranges 5π 3π π π − __  < θ < __     and ___     < θ < ___     4 4 4 4 π π 3π 5π __ __     ___     π ___     θ −  4  0 4 4 4 0 0 0 0 r a a Establish the values of θ for which the curve exists. Draw up a table of values and sketch the curve. θ= π 2 r2 = a2 cos 2θ a O a θ=0 Initial line Online Explore curves given in polar form using GeoGebra. 106 M05_ALEVEL_CP2_83343_U05_100-118.indd 106 02/03/2018 12:40 Polar coordinates Curves with equations of the form r = a(p + q cos θ) are defined for all values of θ if p > q. An example of this, when p = q, was the cardioid seen in Example 6a. These curves fall into two types, those that are ‘egg’ shaped (i.e. a convex curve) and those with a ‘dimple’ (i.e. the curve is concave at θ = π). The conditions for each type are given below: Links ‘egg’ shape when p > 2q Example ‘dimple’ shape when q < p , 2q You can prove these conditions by considering the number of tangents to the curve that are perpendicular to the initial line. → Example 14 7 Sketch the following curves. a r = a(5 + 2 cos θ) b r = a(3 + 2 cos θ) a r = a(5 + 2 cos θ) θ 0 r 7a π __     2 5a 3π ___ π     2 5a 3a Draw up a suitable table of values. θ= π 2 5a r = a (5 + 2cos θ) Since 5 . 2 × 2 there is no ‘dimple’. θ=0 3a 7a O Initial line 5a b r = a(3 + 2 cos θ) θ 0 r 5a π __     2 3a 3π ___ π     2 3a a Draw up a suitable table of values. θ=π 2 3a r = a (3 + 2cos θ) θ=0 a O 5a Since 3 , 2 × 2 there will be a ‘dimple’ for θ close to π. Initial line 3a 107 M05_ALEVEL_CP2_83343_U05_100-118.indd 107 02/03/2018 12:40 Chapter 5 You may also need to find a polar curve to represent a locus of points on an Argand diagram. Example Links If the pole is taken as the origin, and the initial line is taken as the positive real axis, then the point (r, θ) will represent the complex number re  iθ ← Section 1.1 8 a Show on an Argand diagram the locus of points given by the values of z satisfying |z − 3 − 4i| = 5 b Show that this locus of points can be represented by the polar curve r = 6 cos θ + 8 sin θ. a Im |z – 3 – 4i| = 5 8 This locus is a circle with centre 3 + 4i and radius 5. 3 + 4i O 6 Re b In Cartesian form, (x − 3)2 + (y − 4)2 = 25 (r cos θ − 3)2 + (r sin θ − 4)2 = 25 r  2 cos  2 θ − 6r cos θ + 9 + r  2 sin  2 θ − 8r sin θ + 16 = 25 2 2 r  (cos  θ + sin  2 θ) − 6r cos θ − 8r sin θ = 0 r  2 = 6r cos θ + 8r sin θ r = 6 cos θ + 8 sin θ Exercise 5B 1 Sketch the following curves. g r = a sin θ h r = a(1 − cos θ) π c θ = − __  4 π    ) f r = 2 sec (θ − __ 3 i r = a cos 3θ m r = a(2 + sin θ) n r = a(6 + sin θ) o r = a (4 + 3 sin θ) a r=6 d r = 2 sec θ j r = a(2 + cos θ) p r = 2θ E Substitute for x and y in polar form. 5π b θ = ___     4 e r = 3 cosec θ k r = a(6 + cos θ) l r = a (4 + 3 cos θ) q r 2 = a2 sin θ r r 2 = a2 sin 2θ 2 Sketch the graph with polar equation π     – θ) r = k sec (__ 4 where k is a positive constant, giving the coordinates of any points of intersection with the coordinate axes in terms of k. (4 marks) 108 M05_ALEVEL_CP2_83343_U05_100-118.indd 108 02/03/2018 12:40 Polar coordinates E 3 a Show on an Argand diagram the locus of points given by the values of z satisfying |z − 12 − 5i| = 13 b Show that this locus of points can be represented by the polar curve r = 24 cos θ + 10 sin θ E 4 a Show on an Argand diagram the locus of points given by the values of z satisfying |z + 4 + 3i| = 5 b Show that this locus of points can be represented by the polar curve r = − 8 cos θ − 6 sin θ (2 marks) (4 marks) (2 marks) (4 marks) 5.3 Area enclosed by a polar curve You can find areas enclosed by a polar curve using integration. θ= π 2 ■ The area of a sector bounded by a polar curve and the half-lines θ = α and θ = β, where θ is in radians, is given by the formula Area β Area = _  1  ∫     r   2 dθ 2 α Example β 9 α O Initial line Find the area enclosed by the cardioid with equation r = a(1 + cos θ). Problem-solving θ= π 2 r = a (1 + cos θ) a θ=0 2a O Start by sketching the curve. You can simplify your calculation by using the fact that the curve is symmetric about the initial line. Hence you can integrate from 0 to π and then double your answer. Initial line a The curve is symmetric about the initial line and so finding the area above this line and doubling it gives: a2 π Area = 2 × ___     ∫     (1 + cos θ)2 dθ 2 0 π = a2∫     (1 + 2 cos θ + cos2 θ) dθ Use the formula for area. Remember to square the expression for r. You can use trigonometric identities for cos 2θ to integrate terms in cos2 θ or sin2 θ: 1 + cos 2θ cos2 θ ≡ _________   ← Pure Year 2, Chapter 11   2 0 π   32  + 2 cos θ + __   21   cos 2θ)dθ = a2∫     (__ 0 π = a [  2 θ + 2 sin θ +   4  sin 2θ]    3 2 __ __1 = a2 ((__   32 π + 0 + 0) − 0) 3a2π =  _____   2 0 Watch out Unlike Cartesian integration, areas in the third and fourth quadrants do not produce negative integrals. You could obtain the same result by integrating between 0 and 2π: 2π 3a  2π  _12  ∫     a2(1 + cos θ )  2 dθ = _____     2 0 109 M05_ALEVEL_CP2_83343_U05_100-118.indd 109 02/03/2018 12:40 Chapter 5 Example 10 Find the area of one loop of the curve with polar equation r = a sin 4θ. Find the values of θ which will give the beginning and end of a loop by solving r = 0. r = a sin 4θ will have one loop for π 0 < θ < __     4 __   π a2 __  π  1 4    r2 dθ = ___    ∫ 4    sin2 4θ dθ Area =  __ 2 ∫ 2 0 0 a2 __ 4π ___ =    ∫     (1 − cos 8θ) dθ 4 0  __π sin 8θ 4 a2 = ___     [θ − ______    ]     4 8 0 2 π sin 2π a =  ___  (__     − ______ − 0     4 4 8 ) 2 a π = ____     16 Watch out Use the area formula. Use the trigonometric identity for cos 2θ. In this 1 − cos 8θ case, sin2 4θ ≡ _________     2 Remember sin 2π = 0. Online Explore the area enclosed by a loop of the polar curve with the form r = a sin θ using GeoGebra. r = sin nθ has n loops and so a simple way of finding the area of one loop would appear to be 2π aπ to find  _1 ∫     r2 dθ and divide by n. This would give ____     2 2 8 0 The reason why this is not the correct answer is because when you take r 2 in the integral you are also including the n loops given by r , 0. You need to choose your limits carefully so that r > 0 for all values within the range of the integral. Example 11 a On the same diagram, sketch the curves with equations r = 2 + cos θ and r = 5 cos θ . b Find the polar coordinates of the points of intersection of these two curves. c Find the exact area of the region which lies within both curves. θ= π a A table of values would consider π 3π θ = 0, __    , π, ___     2 2 2 r = 5 cos θ r = 2 + cos θ θ=0 O Initial line b The points of intersection are given by 2 + cos θ = 5 cos θ So 4 cos θ = 2 π θ = ± __     3 π   52 , ± __    ) So the polar coordinates are ( __ 3 This is the region required in part c. Form a suitable equation to find the points of intersection. Solve for θ and then substitute in r = 5 cos θ to find the value of r. 110 M05_ALEVEL_CP2_83343_U05_100-118.indd 110 02/03/2018 12:40 Polar coordinates c θ= π 2 Remember that the area formula gives areas of sectors. So you need the sector formed by the purple curve and the sector formed by the blue curve. Again you can use symmetry about the initial line. θ=0 O Initial line Area = 2 × __   21 ∫ 3    (2 + cos θ)2 dθ + 2 × __   21 ∫__π  2  (5 cos θ)2 dθ __ π __ π 0   3     = ∫ 3    (4 + 4 cos θ + cos2 θ) dθ + ∫ __π  2  25 cos2 θ dθ __ π   __ π     3 0 __ __   π 9 cos 2θ   π 25   2 + 4 cos θ + ______   (1 + cos 2θ) dθ   ) dθ + ∫ __π  2  ___ =∫ 3    (__ 2__   3 2 0 __   3π   π sin 2θ sin 2θ 2 25 __ ______ ___ ______ 9 = [  2 θ + 4 sin θ +       +   2 [θ +           4 ]0 2 ]__ 3π __ Square and use the trigonometric identity for cos 2θ. __ = (    + 2√ 3  +     ) − (0) + (     + 0) − (     +     2 4 6 8 8 ) 43π √__ = ____     − 3  12 3π ___ Exercise E E/P __ √ 3  ___ 25π ____ 25π ____ 25√ 3  ______ 2π Use the exact value of sin  ___  3 5C 1 Find the area of the finite region bounded by the curve with the given polar equation and the half-lines θ = α and θ = β. π π π π π a r = a cos θ, α = 0, β = __         b r = a(1 + sin θ), α = − __ , β = __     c r = a sin 3θ, α = __    , β = __ 2 2 2 4 6 π π __ __ 2 2 2 2 e r = a tan θ, α = 0, β =     f r = 2aθ, α = 0, β = π d r = a cos 2θ, α = 0, β =     4 4 π g r = a(3 + 2 cos θ), α = 0, β = __     2 2p2 + q2 2 Show that the area enclosed by the curve with polar equation r = a(p + q cos θ) is _______     πa2. 2 3 Find the area of a single loop of the curve with equation r = a cos 3θ. 187π 4 A curve has equation r = a + 5 sin θ, a . 5. The area enclosed by the curve is _____     2 Find the value of a. (5 marks) 5 The diagram shows the curves with equations r = a sin 4θ π and r = a sin 2θ for 0 < θ < __     2 The finite region R is contained within both curves. θ= π 2 r = a sin 2θ Find the area of R, giving your answer in terms of a. (8 marks) R O r = a sin 4θ Initial line 111 M05_ALEVEL_CP2_83343_U05_100-118.indd 111 02/03/2018 12:40 Chapter 5 E/P π 2 θ= 6 The diagram shows the curves with equations r = 1 + sin θ and r = 3 sin θ. The finite region R is contained within both curves. Find the area of R. (8 marks) r = 3 sin θ R r = 1 + sin θ Initial line O E/P 7 The set of points, A, is defined by π     < arg z < 0} ∩ { z : |z − 4 + 3i | < 5} A = {z : − __ 4 a Sketch on an Argand diagram the set of points, A. (4 marks) Given that the locus of points given by the values of z satisfying |z − 4 + 3i| = 5 can be expressed in polar form using the equation r = 8 cos θ − 6 sin θ, b find, correct to three significant figures, the area of the region defined by A. (8 marks) E/P 8 The set of points, A, is defined by π A = {z :  __  < arg z < π} ∩ { z : |z + 12 − 5i | < 13} 2 a Sketch on an Argand diagram the set of points, A. b Find, correct to three significant figures, the area of the region defined by A. E/P E/P 9 The diagram shows the curve C with polar equation π r = 1 + cos 3θ, 0 < θ <  __  __ 3 2 + √ 2  _____ At points A and B, the value of r is     2 Point A lies on C and point B lies on the initial line. Find, correct to three significant figures, the finite area bounded by the curve, the line segment AB and the initial line, shown shaded in the diagram. (9 marks) 10 The diagram shows the curves r = 1 + sin θ and r = 3 sin θ. Find the shaded area, giving your answer correct to two decimal places. (8 marks) θ= (4 marks) (8 marks) π 2 C O A B θ= Initial line π 2 r = 3 sin θ r = 1 + sin θ O Challenge The cross-section of a shell is modelled using the curve with polar equation r = k θ, 0 < θ < 4π, where k is a positive constant. The horizontal diameter of the shell, as shown in the diagram, is 3 cm. a Find the exact value of k. b Hence find the total shaded area of the crosssection. Initial line 3 cm π θ= 2 Initial line r = kθ 112 M05_ALEVEL_CP2_83343_U05_100-118.indd 112 02/03/2018 12:40 Polar coordinates 5.4 Tangents to polar curves If you are given a curve r = f(θ) in polar form, you can write it as a parametric curve in Cartesian form, using θ as the parameter: x = r cos θ = f(θ) cos θ y = r sin θ = f(θ) sin θ By differentiating parametrically, you can find the gradient of the curve at any point: dy __     dy __ dθ ___     =     dx ___ dx     dθ dy When __     = 0, a tangent to the curve will be horizontal. dθ dx When ___     = 0, a tangent to the curve will be vertical. dθ You need to be able to find tangents to a polar curve that are parallel or perpendicular to the initial line. dy ■ To find a tangent parallel to the initial line set ___   = 0. dθ dx ■ To find a tangent perpendicular to the initial line set ___   = 0. dθ Example 12 Find the coordinates of the points on r = a (1 + cos θ) where the tangents are parallel to the initial line θ = 0. y = r sin θ = a(sin θ + sin θ cos θ) dy ___   = a(cos θ + cos2 θ − sin2 θ) dθ So 0 = 2 cos2 θ + cos θ − 1 0 = (2 cos θ − 1)(cos θ + 1) π cos θ = __   21 ⇒ θ = ± __     3 3a     so r = a (1 + __   21 ) = ___ 2 cos θ = −1 ⇒ θ = π, and so r = 0 So the tangents parallel to the initial line 3a π    , ± __ are at ( ___    ) and (0, π). 2 3 dy Find an expression for y and then solve ___     = 0. dθ Solve the equations to find θ and then substitute back to find r. Problem-solving You can see these tangents on a sketch of y = a(1 + cos θ) ( 3a2 , π3 ) (O, π) ( 3a2 , – π3 ) 113 M05_ALEVEL_CP2_83343_U05_100-118.indd 113 02/03/2018 12:40 Chapter 5 Example 13 π Find the equations and the points of contact of the tangents to the curve r = a sin 2θ, 0 < θ <  __  2 that are: a parallel to the initial line b perpendicular to the initial line. Give answers to three significant figures where appropriate. a y = r sin θ = a sin θ sin 2θ dy ___   = a(cos θ sin 2θ + 2 cos 2θ sin θ) dθ = 2a sin θ(cos2 θ + cos2 θ − sin2 θ) dy  ___ = 0 ⇒ sin θ = 0 ⇒ θ = 0 dθ __ or 2 cos2 θ = sin2 θ ⇒ tan θ = ± √ 2  ⇒ θ = 0.955 So θ = 0 or 0.955 __ √ 2  __ 1 ___ and r = 0 or r = 2a ×     × ___   __  √ √ 3  3 __ 2a√ 2     , 0.955) So the points are (0, 0) and ( _____ 3 The equation of the initial line is θ = 0 and that is the tangent through (0, 0). The equation of the tangent through __ √ 2  2a (_____    , 0.955) 3 __ __ __ √ 2  2a√ 2  2a√ 2  ___ __ 4a _____ _____ is y =     × sin θ =     ×     = ____   __  3 3 √ 3  3√ 3  So the equation of the tangent is 4a__ r =  ____   cosec θ 3√ 3  b x = r cos θ = a cos θ sin 2θ dx  ___ = −a sin θ sin 2θ + 2a cos θ cos 2θ dθ = 2a cos θ (−sin2 θ + cos2 θ − sin2 θ) dx π  ___ = 0 ⇒ cos θ = 0 ⇒ θ =  __  2 dθ So the y-axis is a tangent. 1   __  Or cos2 θ − 2 sin2 θ = 0 ⇒ tan θ = ± ___ √ 2  So θ = 0.615__ __ √ 2  ___ __ 2a√ 2  1__ _____ ___   and r = 2a ×     ×     =   √ 3  √ 3 __ 3 2a√ 2    The tangent is at (_____  , 0.615) 3 __ __ __ √ 2  2a√ 2  2a√ 2  ___ __ 4a _____ _____ x =     × cos α =     ×     = ____   __  3 3 √ 3  3√ 3  So the equation of the tangent is: 4a__ r =  ____   sec θ 3√ 3  Form an expression for y and differentiate using the product rule. Use sin 2θ ≡ 2 sin θ cos θ and then take out the common factor. Then use cos 2θ ≡ cos2 θ – sin2 θ Choose values of θ within the range given in the question. __ If tan a =__√ 2 then drawing a triangle shows that √ 2  1 __  and cos α = ___   __  sin α =  ___ √ √ 3  3  Use sin 2A ≡ 2 sin A cos A to find r. Use y = r sin θ to find the equation of the tangent and write it in polar form using r = y cosec θ. Form an expression for x and differentiate using the product rule. Use sin 2θ = 2 sin θ cos θ and then take out the common factor. Then use a formula for cos 2θ. 1 If α = ___   __ then drawing a triangle shows that √ 2 __ √ 2  1 __  and sin α = ___ cos α =  ___   __ . √ √ 3  3  Use sin 2A ≡ 2 sin A cos A to find r. Use x = r cos θ to find the equation of the tangent in the form r = x sec θ. 114 M05_ALEVEL_CP2_83343_U05_100-118.indd 114 02/03/2018 12:40 Polar coordinates Example 14 The curve C has equation r = (p + q cos θ), where p and q are positive constants and p . q. Prove that the curve is convex for p > 2q, and has a dimple for p , 2q. θ= Problem-solving π 2 If the curve is not convex then there will be more than two tangents to the curve that are perpendicular to the initial line. a θ=0 2a O Initial line –a x = r cos θ = p cos θ + q cos2 θ dx ___   = 0 ⇒ 0 = −p sin θ − 2q cos θ sin θ dθ ⇒ 0 = −sin θ(p + 2q cos θ) This has solutions sin θ = 0 when θ = 0 or π p and cos θ = − ___ 2q If p , 2q then there will be two solutions to this equation in the second and third quadrants (the green tangents). In this case the curve is not convex and has a dimple. Find an expression for x and differentiate. Solve the equation and consider all possible cases. The two tangents at the two points represented by these solutions have the same equation. If p = 2q then the solution is θ = π and so there are only two tangents (the blue ones). In this case the curve is convex. If p . 2q then there is no solution to this equation and only the two blue tangents are possible. In this case the curve is convex. Hence the curve is convex for p > 2q, and has a dimple for p , 2q. Exercise 5D 1 Find the points on the cardioid r = a(1 + cos θ) where the tangents are perpendicular to the initial line. 2 Find the points on the spiral r = e2θ, 0 < θ < π, where the tangents are a perpendicular to the initial line b parallel to the initial line. Give your answers to three significant figures. 115 M05_ALEVEL_CP2_83343_U05_100-118.indd 115 02/03/2018 12:40 Chapter 5 π π 3 a Find the points on the curve r = a cos 2θ, − __     < θ < __    , where the tangents are parallel to 4 4 the initial line, giving your answers to three significant figures where appropriate. b Find the equations of these tangents. E E E E/P 4 Find the points on the curve with equation r = a(7 + 2 cos θ) where the tangents are parallel to the initial line. (6 marks) 5 Find the equations of the tangents to r = 2 + cos θ that are perpendicular to the initial (6 marks) line. π    , where the tangent is 6 Find the point on the curve with equation r = a(1 + tan θ), 0 < θ , __ 2 (6 marks) perpendicular to the initial line. 7 The curve C has polar equation π     r = 1 + 3 cos θ, 0 < θ < __ 2 The tangent to C at a point A on the curve is parallel to the initial line. Point O is the pole. (7 marks) Find the exact length of the line OA. E/P 8 The diagram shows a cardioid with polar equation r = 2(1 + cos θ) θ= π 2 The shaded area is enclosed by the curve and the vertical line segment which is tangent to the curve and perpendicular to the initial line. Find the shaded area, correct to three significant figures. (8 marks) Mixed exercise E r = 2 (1 + cos θ) O Initial line 5 1 Determine the area enclosed by the curve with equation r = a(1 +  _2 sin θ), 1 a . 0, 0 < θ , 2π, giving your answer in terms of a and π. (6 marks) E/P 2 a Sketch the curve with equation r = a(1 + cos θ) for 0 < θ < π, where a . 0. (2 marks) π π __ __ b Sketch also the line with equation r = 2a sec θ for −     , θ ,    , on the same diagram. (2 marks) 2 2 π __ c The half-line with equation θ = α, 0 , α ,    , meets the curve at A and the line with equation 2 r = 2a sec θ at B. If O is the pole, find the value of cos α for which OB = 2OA. (5 marks) E/P 3 Sketch, in the same diagram, the curves with equations r = 3 cos θ and r = 1 + cos θ and find the area of the region lying inside both curves. (9 marks) 116 M05_ALEVEL_CP2_83343_U05_100-118.indd 116 02/03/2018 12:40 Polar coordinates E 4 Find the polar coordinates of the points on r2 = a2 sin 2θ where the tangent is perpendicular to the initial line. (7 marks) E/P 5 a Shade the region R for which the polar coordinates r, θ satisfy π π     r < 4 cos 2θ for −  __  < θ < __ 4 4 b Find the area of R. E E/P E (2 marks) (5 marks) 6 Sketch the curve with polar equation r = a(1 − cos θ), where a . 0, stating the polar coordinates of the point on the curve at which r has its maximum value. 7 a On the same diagram, sketch the curve C1 with polar equation π π r = 2 cos 2θ, −  __  , θ < __     4 4 π     and the curve C2 with polar equation θ = ___ 12 b Find the area of the smaller region bounded by C1 and C2. (3 marks) (6 marks) 8 a Sketch on the same diagram the circle with polar equation r = 4 cos θ and the line with polar equation r = 2 sec θ. (4 marks) b State polar coordinates for their points of intersection. E/P (5 marks) 9 The diagram shows a sketch of the curves with polar equations (4 marks) θ=π 2 r = a(1 + cos θ) and r = 3a cos θ, a . 0 P a Find the polar coordinates of the point of intersection P of the two curves. (4 marks) b Find the area, shaded in the figure, bounded by the two curves and by the initial line θ = 0, giving your answer in terms of a and π. (7 marks) E/P E/P O Initial line 10 Obtain a Cartesian equation for the curve with polar equation a r 2 = sec 2θ (4 marks) b r 2 = cosec 2θ (4 marks) 11 a Show on an Argand diagram the locus of points given by the values of z satisfying __ |z − 1 − i | = √ 2  (2 marks) b Show that this locus of points can be represented by the polar curve r = 2 cos θ + 2 sin θ (4 marks) The set of points, A, is defined by __ π π A={ z :  __  < arg z < __    } ∩ { z : |z − 1 − i | < √ 2  } 2 6 c Show, by sketching on your Argand diagram, the set of points, A. (2 marks) d Find, correct to three significant figures, the area of the region defined by A. (5 marks) 117 M05_ALEVEL_CP2_83343_U05_100-118.indd 117 02/03/2018 12:40 Chapter 5 E 12 The diagram shows the curve C with polar equation π r = 4 cos 2θ, 0 < θ <  __  4 At point A the value of r is 2. Point A lies on C and point B lies on the initial line vertically below A. Find, correct to three significant figures, the area of the finite region bounded by the curve, the line segment AB and the initial line, shown shaded in the (9 marks) diagram. E/P 13 The diagram shows the curve with polar equation π r = 4 sin 2θ, 0 < θ <  __  2 The shaded region is bounded by the curve, the initial line and the tangent to the curve which is perpendicular to the initial line. Find, correct to two decimal places, the area of the shaded region. (8 marks) θ=π 2 A O Initial line B θ=π 2 r = 4 sin 2θ O Initial line Challenge __ The curve C has polar equation r = √ 2  θ. Show that an equation for the tangent to the curve at the point where π θ =  __ is 2(π − 4)y + 2(π + 4)x = π2 4 Summary of key points 1 For a point P with polar coordinates (r, θ) and Cartesian coordinates (x, y), •• r cos θ = x and r sin θ = y y •• r2 = x2 + y2, θ = arctan (__   x ) Care must be taken to ensure that θ is in the correct quadrant. 2 • r = a is a circle with centre O and radius a. • θ = α is a half-line through O and making an angle α with the initial line. • r = aθ is a spiral starting at O. 3 The area of a sector bounded by a polar curve and the half-lines θ = α and θ = β, where θ is in radians, is given by the formula Area = _  12 ∫     r2 dθ β α dy 4 • To find a tangent parallel to the initial line set ___    = 0. dθ dx • To find a tangent perpendicular to the initial line set ___    = 0. dθ 118 M05_ALEVEL_CP2_83343_U05_100-118.indd 118 02/03/2018 12:40 Hyperbolic functions Objectives 6 After completing this chapter you should be able to: ● Understand the definitions of hyperbolic functions → pages 120–123 ● Sketch the graphs of hyperbolic functions → pages 121–123 ● Understand and use the inverse hyperbolic functions → pages 123–125 ● Prove identities and solve equations using hyperbolic → pages 125–129 functions ● Differentiate and integrate hyperbolic functions → pages 130–142 Prior knowledge check Hyperbolic curves feature often in architectural modelling. A hanging chain might look like a parabola but it is actually a curve called a catenary which has equation x   a ). → Mixed exercise Q23 y = a cosh ( __ 1 f(x) = 2ex − e−x Solve the equation f(x) = 2. ← Pure Year 1, Chapter 14 2 Show that 1  _____   − tan  2 x ≡ 1 cos  2 x 3 ← Pure Year 2, Chapter 6 π Show that ∫   ex sin x dx = _  12 (1 + e  π). 0 ← Pure Year 2, Chapter 11 119 M06_ALEVEL_CP2_83343_U06_119-146.indd 119 26/02/2018 09:29 Chapter 6 6.1 Introduction to hyperbolic functions Hyperbolic functions have several properties in common with trigonometric functions, but they are defined in terms of exponential functions. ex − e−x   ■ Hyperbolic sine (or sinh) is defined as sinh x ≡ _______   2 ex + e−x ■ Hyperbolic cosine (or cosh) is defined as cosh x ≡ _______     2 sinh x ■ Hyperbolic tangent (or tanh) is defined as tanh x ≡ ______   cosh x You can use the definitions of sinh x and cosh x to write tanh x in exponential form. sinh x e  x − e  −x _______ e  x − e  −x 2 _______     ×   x ≡       tanh x ≡  ______ ≡ _______ e  + e  −x e  x + e  −x 2 cosh x Multiplying the numerator and denominator of the final expression through by e  x gives: e2x − 1     2x ■ tanh x ≡______ e +1 There are also hyperbolic functions corresponding to the reciprocal trigonometric functions: 1 2   cosech x ≡ ______ ≡ _______   x   e   − e  −x sinh x 1 2 sech x ≡ ______   ≡ _______   x   e  + e  −x cosh x coth x Example Note You won’t need to use these functions in your exam, but they are useful to know, and if you are confident with them you can use them to simplify your working. e  2x + 1 1 ≡ ______     ≡ ______   2x tanh x e  − 1 1 Find, to 2 decimal places, the values of: a sinh 3 b cosh 1 c tanh 0.8 e  3 − e  −3 asinh 3 = _______     = 10.02(2 d.p.) 2 e  1 + e  −1 b cosh 1 = _______     = 1.54(2 d.p.) 2 e  1.6 – 1 c tanh 0.8 = ______   1.6   = 0.66(2 d.p.) e  + 1 Example 2 Find the exact value of tanh (ln 4). 2 eln4 − 1 _______ e2ln4 − 1 _______ eln16 − 1   =   tanh (ln 4) = _______   2ln4   =     e + 1 eln42 + 1 eln16 + 1 15 16 − 1 ___ = ______     =     16 + 1 17 Use elnk = k. 120 M06_ALEVEL_CP2_83343_U06_119-146.indd 120 26/02/2018 09:29 Hyperbolic functions Example 3 Use the definition of sinh x to find, to 2 decimal places, the value of x for which sinh x = 5. ex − e−x _______   = 5 ⇒ ex − e−x = 10 2 e2x − 1 = 10ex e2x − 10ex − 1=0 _ x e  = 5 ± √ 26 _ ⇒ e  x = 5 + √ 26  _ So x = ln(5 + √ 26 ) = 2.31 (2 d.p.)   Multiply both sides by ex. The substitution u = ex turns this into the quadratic equation u2 – 10u – 1 = 0. ex cannot be negative. y You can sketch the graphs of the hyperbolic functions by considering the graphs of y = ex and y = e–x. ex + (–e–x) ex – e–x _________   sinh x = _______   =     2 2 so the graph of y = sinh x is the ‘average’ of the graphs of y = ex and y = –e–x. y = ex y = sinh x 1 x O y = – e –x y For the graph of y = sinh x, y = sinh x • when x is large and positive, e−x is small, so sinh x ≈  _12 ex • when x is large and negative, ex is small, so sinh x ≈ −  _12 e−x ■ For any value a, sinh (−a) = −sinh a x O Notation f(x) = sinh x is an odd function since f(−x) = − f(x). Consider the graphs of y = ex and y = e–x. y ex + e–x cosh x = _______     2 so the graph of y = cosh x is the ‘average’ of the graphs of y = ex and y = e–x. y = cosh x 1 O For the graph of y = cosh x, • • y = ex y when x is large and positive, e−x is small, so cosh x ≈  _12 ex when x is large and negative, ex is small, so cosh x ≈  _12 e−x y = e –x x y = cosh x ■ For any value a, cosh (−a) = cosh a Notation f(x) = cosh x is an even function because f(−x) = f(x). 1 O x 121 M06_ALEVEL_CP2_83343_U06_119-146.indd 121 26/02/2018 09:29 Chapter 6 Example 4 Sketch the graph of y = tanh x. sinh x tanh x =   ______   cosh x  01  = 0 When x = 0, tanh x = _ Online Explore graphs of hyberbolic functions using GeoGebra. When x is large and positive, sinh x ≈ _  21  ex and _1 x cosh x ≈   2 e , so tanh x ≈ 1. When x is large and negative, sinh x ≈ −  _21  e−x and cosh x ≈  _21  e−x , so tanh x ≈ −1. As x → ∞, tanh x → 1 and as x → −∞, tanh x → −1 Consider the graphs of y = sinh x and y = cosh x to work out the behaviour of y = tanh x as x → ∞ and x → –∞. For f(x) = tanh x, x ∈ ℝ, the range of f is −1 , f(x) , 1 y = −1 and y = 1 are asymptotes to the curve. y 1 y = tanh x x O You should always include any asymptotes on a sketch graph. –1 Exercise 6A 1 Find, correct to 2 decimal places: a sinh 4 b cosh  _2 c tanh (−2) b cosh 4 c tanh 0.5 b cosh (ln 3) c tanh (ln 2) 1 2 Write, in terms of e: a sinh 1 3 Find the exact values of: a sinh (ln 2) In questions 4 to 6, use the definitions of the hyperbolic functions (in terms of exponentials) to find each answer, then check your answers using an inverse hyperbolic function on your calculator. 4 Find, to 2 decimal places, the values of x for which cosh x = 2. 5 Find, to 2 decimal places, the values of x for which sinh x = 1. 6 Find, to 2 decimal places, the values of x for which tanh x = −  _2 1 122 M06_ALEVEL_CP2_83343_U06_119-146.indd 122 26/02/2018 09:29 Hyperbolic functions 7 On the same diagram, sketch the graphs of y = cosh 2x and y = 2 cosh x. 8 Find the range of each hyperbolic function. a f(x) = sinh x, x ∈ ℝ b f(x) = cosh x, x ∈ ℝ c f(x) = tanh x, x ∈ ℝ E 9 a Sketch the graph of y = 3 tanh x + 2. (3 marks) b Write down the equations of the asymptotes to this curve. Challenge (2 marks) Sketch the graphs of: a y = sech x b y = cosech x c y = coth x 6.2 Inverse hyperbolic functions You can define and use the inverses of the hyperbolic functions. y y = sinh x If f(x) = sinh x, the inverse function f−1 is called arsinh x. The graph of y = arsinh x is the reflection of the graph of y = sinh x in the line y = x. y = arsinh x O x y =x The inverse of a function is defined only if the function is one-toone, so for cosh x the domain must be restricted in order to define an inverse. y y = cosh x, x > 0 y =x For f(x) = cosh x, x > 0, f−1(x) = arcosh x, x > 1 y = arcosh x 1 O 1 x ■ The following table shows the inverse hyperbolic functions, with domains restricted where necessary. Hyperbolic function y = sinh x y = cosh x, x > 0 y = tanh x Inverse hyperbolic function y = arsinh x y = arcosh x, x > 1 y = artanh x, |x| , 1 Notation arsinh, arcosh and artanh are sometimes written as sinh–1, cosh–1 and tanh–1. 123 M06_ALEVEL_CP2_83343_U06_119-146.indd 123 26/02/2018 09:29 Chapter 6 You can express the inverse hyperbolic functions in terms of natural logarithms. Example 5 ______ Show that arsinh x = ln (x + √ x2 + 1 ). Let y = arsinh x x = sinh y ey − e−y x =  _______   2 ey − e−y = 2x e2y − 1 = 2xey e2y − 2xey − 1 = 0 (ey − x)2 − ______ x2 − 1 = 0 ey = x ± √ x2 + 1  Use the definition of sinh. Multiply by ey. Problem-solving ______ ______ ey = x − √ x2 + 1 can be ignored since √x2 + 1  . x, and would give a negative value of ey, which is not possible. ______ So ey =______ x+√ x2 + 1  y = ln(x + √ x2 + 1 ) ______ ⇒ arsinh x = ln(x + √ x2 + 1 ) Example e   2y − 2x e  y − 1 = 0 is a quadratic in ey. You can write it as ( e  y )  2 − 2x e  y − 1 = 0 and then complete the square. 6 ______ Show that arcosh x = ln (x + √ x2 − 1  ), x > 1. Use the definition of cosh. Let y = arcosh x x = cosh y ey + e−y x =  _______   2 ey + e−y = 2x e2y + 1 = 2xey e2y − 2xey + 1 = 0 (ey − x)2 − x2 + 1 = 0 Multiply by ey. Form and solve a quadratic in ey. _____ ______ So ey = x ± √ x2 − 1  ______ y = ln(x ± √ x2 − 1 ) _____ Note that both x + √ x2 – 1  and x – √ x2 – 1  are positive. ______ ⇒ arcosh x = ln(x + √x2 − 1 ) arcosh x is always non-negative. For all values of ______ 2 √ x . 1, x______ − x − 1  , 1so the value of ln(x − √ x2 − 1  )is negative. You can use a similar method to express artanh x in terms of natural logarithms. The following formulae are provided in the formula booklet and can be used directly unless you are asked to prove them . ______ ■ arsinh x = ln( x + √ x2 + 1  ) 1+x ■ artanh x = _  12 ln(  _____), |x| , 1 1−x ______ ■ arcosh x = ln( x + √ x2 − 1  ), x > 1 124 M06_ALEVEL_CP2_83343_U06_119-146.indd 124 26/02/2018 09:29 Hyperbolic functions Example 7 Express as natural logarithms: a arsinh 1 c artanh  _3 1 b arcosh 2 ______ __ a arsinh 1 = ln(1 + √ 12 + 1 ) = ln(1 + √ 2 ) ______ __ b arcosh 2 = ln(2 + √ 22 − 1 ) = ln(2 + √ 3 ) __ 1 + __  31 __1 ______ 1   = __ c artanh  __ =   l n     21 ln 2 = ln √2  2 3 __ 1 ( 1 −   3 ) Exercise Use a ln x = ln xa. 6B 1 Sketch the graph of y = artanh x, |x| , 1. P 2 Sketch the graph of y = (arsinh x)2. E/P 1+x 1     , |x| , 1. 3 Prove that artanh x = _  2 ln (_____ 1 − x) (5 marks) 4 Express as natural logarithms: a arsinh 2 5 Express as natural logarithms: __ a arsinh √2  1 __ b arcosh √5  6 Express as natural logarithms: c artanh 0.1 1__   c artanh  ___ √ 3  b arcosh  _2 a arsinh (−3) E/P c artanh  _2 b arcosh 3 3 __ 2x − 1 7 Given that artanh x + artanh y = ln √3 , prove that y = ______     . x−2 (6 marks) 6.3 Identities and equations You can find and use identities for the hyperbolic functions that are similar to the trigonometric identities. Example 8 Prove that cosh2 A − sinh2 A ≡ 1 ( ) ( ) ) eA + e−A 2 eA − e−A 2 LHS ≡ cosh2 A − sinh2 A ≡ ________        − ________    2 2 e2A + 2 + e−2A e2A − 2 + e−2A _____________ _____________ ≡        −        4 4 4 ≡ __     ≡ 1 ≡ RHS 4 ( ) ( Use the definitions of cosh and sinh. eA eA × e−A = ___   A  = 1. e ■ cosh2 A − sinh2 A ≡ 1 125 M06_ALEVEL_CP2_83343_U06_119-146.indd 125 26/02/2018 09:29 Chapter 6 Example 9 Prove that sinh (A + B) ≡ sinh A cosh B + cosh A sinh B RHS ≡ sinh A cosh B + cosh A sinh B ( )( ) ( )( ) eA − e−A ________ eB + e−B eA + e−A ________ eB − e−B ≡ ________             + ________         2 2 2 2 ( ) ( ) eA+B + eA−B − e−A+B − e−A−B eA+B − eA−B + e−A+B − e−A−B ≡ _________________________                    +  ________________________    4 4 2eA+B − 2e−A−B ____________ eA+B − e−(A+B) ≡ _____________        ≡       ≡ sinh (A + B) ≡ LHS 4 2 You can prove other sinh and cosh addition formulae similarly, giving: ■ sinh (A ± B ) ≡ sinh A cosh B ± cosh A sinh B ■ cosh (A ± B ) ≡ cosh A cosh B ± sinh A sinh B Example 10 Prove that cosh 2A ≡ 1 + 2sinh2 A RHS ≡ 1 + 2 sinh2 A ( )( ) eA − e−A ________ eA − e−A ≡ 1 + 2 ________           2 2 ( ) Use the definition of sinh. ( ) e2A − 2 + e−2A e2A + e−2A _____________ ≡ 1 + 2        ≡ 1 − 1 +  _________    4 2 ≡ cosh 2A ≡ LHS Given a trigonometric identity, it is generally possible to write down the corresponding hyperbolic identity using what is known as Osborn’s rule: • Replace cos by cosh: cos A → cosh A • Replace sin by sinh: sin A → sinh A However … • replace any product (or implied product) of two sin terms by minus the product of two sinh terms: e.g. sin A sin B → −sinh A sinh B tan2 A → −tanh2 A This is the implied product of two sin terms because sin2 A tan2 A ≡ ______   2   cos A 126 M06_ALEVEL_CP2_83343_U06_119-146.indd 126 26/02/2018 09:29 Hyperbolic functions Example 11 Write down the hyperbolic identity corresponding to: tan A − tan B _____________ b tan (A − B) ≡          1 + tan A tan B a cos 2A ≡ 2 cos2 A − 1 a cosh 2A ≡ 2 cosh2 A − 1 Implied product of two sin terms because sin A sin B tan A tan B ≡ __________     cos A cos B tanh A − tanh B _______________ b tanh (A − B) ≡          1 − tanh A tanh B Example 12 Given that sinh x = _  4 , find the exact value of: 3 a cosh x b tanh x c sinh 2x a Using cosh2 x − sinh2 x ≡ 1, 9 cosh2 x − __  16   = 1 ⇒ cosh2 x = __  25 16  ⇒ cosh x = _   54  cosh x > 1, so cosh x = −  _54 is not possible. sinh x b Using tanh x ≡  ______  cosh x   34  ÷ _   54  = _  35  tanh x = _ c Using sinh 2x ≡ 2 sinh x cosh x, sinh 2x = 2 × _  34  × _  45  = __  15 8  You can solve equations involving hyperbolic functions. Example 13 Solve 6 sinh x − 2 cosh x = 7 for real values of x. ( ) ( ) ex − e−x ex + e−x 6 _______        − 2 _______    = 7 2 2 3ex − 3e−x − ex − e−x = 7 There is no hyperbolic identity that will easily transform the equation into an equation in just one hyperbolic function, so use the basic definitions. 2ex − 7 − 4e−x = 0 2e2x − 7ex − 4 = 0 (2ex + 1)(ex − 4) = 0 ex = −  _21  , ex = 4 ex = 4 There are no real values of x for which ex = −  _12 x = ln 4 127 M06_ALEVEL_CP2_83343_U06_119-146.indd 127 26/02/2018 09:29 Chapter 6 Example 14 Solve 2 cosh2 x − 5 sinh x = 5, giving your answers as natural logarithms. Use this identity to transform the equation into an equation in just one hyperbolic function. Using cosh2 x − sinh2 x ≡ 1, 2(1 + sinh2 x) − 5 sinh x = 5 2 sinh2 x − 5 sinh x − 3 = 0 (2 sinh x + 1)(sinh x − 3) = 0 1 So sinh x = −  __ or sinh x = 3 2 Then, 1 x = arsinh −  __   or x = arsinh 3 2_____ _____ 1 1 ⇒ x = ln −  __  + __     + 1   or x = ln(3 + √ 9 + 1  ) 2 4 ( ) ( √ ) √ 5  1   + ___ ⇒ x = ln ( −  __      ) or x = ln(3 + √10  ) 2 2 __ ______ Use arsinh x = ln(x + √ x2 + 1  ). __ Example 15 Solve cosh 2x − 5 cosh x + 4 = 0, giving your answers as natural logarithms where appropriate. Use this identity to transform the equation into an equation in just one hyperbolic function. Using cosh 2x ≡ 2 cosh2 x − 1, 2 cosh2 x − 1 − 5 cosh x + 4 = 0 2 cosh2 x − 5 cosh x + 3 = 0 (2 cosh x − 3)(cosh x − 1) = 0 3 So cosh x = __    or cosh x = 1 2 _____ 3 9 __ ⇒ x = ln       ± __     − 1   or x = 0 2 4 ( √ ( __ ) ______ √ You can use arcosh x = ln(x +______ x2 − 1  ), but remember that both ln(x + √ x2 − 1  ) and ______ ln(x − √ x2 − 1  ) are possible. ) For any value of k greater than 1, cosh x = k will give two values of x, one positive and one negative. 5  3 √ ⇒ x = ln __     ± ___      or x = 0 2 2 Exercise 6C 1 Prove the following identities, using the definitions of sinh x and cosh x. a sinh 2A ≡ 2 sinh A cosh A b cosh (A − B) ≡ cosh A cosh B − sinh A sinh B A−B A+B d sinh A − sinh B ≡ 2 sinh  ______      cosh ______    2 2 ( c cosh 3A ≡ 4 cosh3 A − 3 cosh A ) ( ) 2 Use Osborn’s rule to write down the hyperbolic identities corresponding to the following trigonometric identities. a sin (A − B) ≡ sin A cos B − cos A sin B ( ) ( ) A+B A−B   c cos A + cos B ≡ 2 cos ______    cos  ______    2 2 e cos 2A ≡ cos4 A − sin4 A b sin 3A ≡ 3 sin A − 4 sin3 A 1 − tan2 A d cos 2A ≡ __________   1 + tan2 A 128 M06_ALEVEL_CP2_83343_U06_119-146.indd 128 26/02/2018 09:29 Hyperbolic functions 3 Given that cosh x = 2, find the exact values of: a sinh x b tanh x c cosh 2x 4 Given that sinh x = −1, find the exact values of: a cosh x b sinh 2x c tanh 2x 5 Solve the following equations, giving your answers as natural logarithms. E a 3 sinh x + 4 cosh x = 4 b 7 sinh x − 5 cosh x = 1 c 30 cosh x = 15 + 26 sinh x d 13 sinh x − 7 cosh x + 1 = 0 e cosh 2x − 5 sinh x = 13 f 3 sinh2 x − 13 cosh x + 7 = 0 g sinh 2x − 7 sinh x = 0 h 4 cosh x + 13e−x = 11 i 2 tanh x = cosh x 6 a Starting from the definitions of sinh x and cosh x in terms of exponentials, prove that cosh 2x ≡ 2 cosh  2 x − 1 (3 marks) b Solve the equation cosh 2x − 3 cosh x = 8 (5 marks) giving your answers as exact logarithms. E 7 Solve the equation 2 sinh  2 x − 5 cosh x = 5 (6 marks) giving your answer in terms of natural logarithms in simplest form. P 8 Joshua is asked to prove the following identity: 1 + tanh  2 x __________   ≡ 2 cosh  2 x − 1   1 − tanh  2 x His answer is below. 1 + tanh  2 x __________ sech   2 x           ≡ __________ 2 1 − tanh  x 2 − sech   2 x sech   2 x ≡ ______     − 1 2 2   ≡ ______ − 1 sech   2 x (using sech   2 x ≡ 1 + tanh  2 x: same identity as the trig one) (splitting the fraction up and cancelling) (taking the reciprocal of both terms) ≡ 2 cosh  2 x − 1 Joshua has made three errors. Explain the errors and provide a correct proof. E/P 9 aExpress 10 cosh x + 6 sinh x in the form R cosh (x + a) where R . 0. Give the value of a correct to 3 decimal places. (4 marks) (6 marks) Hint Use the identity for cosh (A + B). b Write down the minimum value of 10 cosh x + 6 sinh x. (1 mark) c Use your answer to part a to solve the equation 10 cosh x + 6 sinh x = 11. Give your answers to 3 decimal places. (4 marks) 129 M06_ALEVEL_CP2_83343_U06_119-146.indd 129 26/02/2018 09:29 Chapter 6 6.4 Differentiating hyperbolic functions You can differentiate hyperbolic functions. d ■ ___ (sinh x) = cosh x dx Watch out The rules for sinh x and tanh x are the same as the corresponding rules for sin x and tan x. However, the derivative of cosh x is positive sinh x. d ■ ___ (cosh x) = sinh x dx d ■ ___ (tanh x) = sech2 x dx Example 16 d Show that ___     (cosh x) = sinh x. dx ex + e−x cosh x = _______     2 ( Use the definition of cosh x. ) d ex + e−x d     _______   So ___  (cosh x) = ___    dx dx 2 ex − e−x _______ =     2 ex − e−x  _______   = sinh x 2 d So  ___  (cosh x) = sinh x dx Differentiate with respect to x: d ___     (e−x) = −e−x dx By definition. Example 17 Differentiate cosh 3x with respect to x. d ___   (cosh 3x) = 3 sinh 3x   dx Use the chain rule. Example 18 Differentiate x2 cosh 4x with respect to x. d ___   d d     (x2) cosh 4x + x2  ___  (cosh 4x)   (x2 cosh 4x) = ___ dx dx = 2x cosh 4x + x2 × 4 sinh 4x dx Use the product rule. = 2x cosh 4x + 4x2 sinh 4x 130 M06_ALEVEL_CP2_83343_U06_119-146.indd 130 26/02/2018 09:29 Hyperbolic functions Example 19 d2y Given that y = A cosh 3x + B sinh 3x, where A and B are constants, prove that ____   2  = 9y. dx dy  ___  = 3A sinh 3x + 3B cosh 3x dx Differentiate y. dy ___ 2     = 9A cosh 3x + 9B sinh 3x dx2 Differentiate again. = 9(A cosh 3x + B sinh 3x) Factorise. = 9y y = A cosh 3x + B sinh 3x You can also differentiate the inverse hyperbolic functions. d 1   ______  ■ ___ (arsinh x) = _______ 2 dx √x + 1  d 1   ______ , x . 1 ■ ___ (arcosh x) = _______ 2 dx √ x − 1  d 1 ■ ___ (artanh x) = ______    , |x| , 1 dx 1 − x2 Example 20 dy Given y = x arcosh x, find  ___  dx dy ___   dx 1     ______   = arcosh x + x × _______ √ x2 − 1  x   = arcosh x + _______   ______ √ x2 − 1  Use the product rule. Example 21 (  dx )   = 4y. Given y = (arcosh x)2, prove that (x2 − 1) dy ___ 1     ______   = 2arcosh x × _______ √ x2 − 1  ______ dy ⇒√ x2 − 1   ___  = 2arcosh x dx dy 2 ⇒ (x2 − 1) ___      = 4(arcosh x)2 dx   dx ( ) dy ___ 2 Use the chain rule. ______ Multiply by √x2 − 1  . Square both sides. But y = (arcosh x)2 ( ) dy 2 so (x2 − 1) ___      = 4y dx 131 M06_ALEVEL_CP2_83343_U06_119-146.indd 131 26/02/2018 09:29 Chapter 6 Example 22 d 1 ______  a Show that ___     (arsinh x) =  _______ dx √ 1 + x2  b Find the first two non-zero terms of the series expansion of arsinh x. The general form for the series expansion of arsinh x is given by ∞ (−1)  n (2n)! x   2n+1 ______     arsinh x = ∑ _________     2n 2 r=0 ( 2  (n!)  ) 2n + 1 c Find, in simplest terms, the coefficient of x5. d Use your approximation up to and including the term in x5 to find an approximate value for arsinh 0.5. e Calculate the percentage error in using this approximation. a Let y = arsinh x then sinh y = x dx ⇒  ___  = cosh y dy _________ =√ sinh2 y + 1  But sinh y = x, ______ dx so  ___  = √x2 + 1  dy dy 1 ______   therefore ___  =  _______ 2 dx √ x + 1  bf(x) = arsinh x ⇒ f(0) = 0 1 f9(x) = _____   _  ⇒ f9(0) = 1 √ 1 + x  2  x f0(x) = −  __________3  ⇒ f0(0) = 0 (1 + x  2) 2 2x  2 − 1   f-(x) = ________   ⇒ f-(0) = −1 5 __ ( 1 + x  2) 2 x  3 1     x3 arsinh x ≈ x − ___     = x − __ 6 3! c The x5 term will be when n = 2: ( −1)  24! ___ 3 x  5 1 × 24 x  5 ___ ________   4     =     x  5       =  ______  × ___ ( 2  (2!)  2) 5 5 16 × 4 40 3 The coefficient of x5 is ___     40 3 1     (0.5)  5 d arsinh 0.5 ≈ 0.5 −  __  (0.5)  3 + ___ 6 40 = 0.48151… Differentiate. Use cosh2 x − sinh2 x ≡ 1. Take the reciprocal of both sides. Find the second derivative using the chain rule or quotient rule. Watch out You need to find the first two nonzero terms. f(0) = 0 and f′′(0) = 0 so there is no constant term and no x2 term. Differentiate again to find the x 3 term. Find the third derivative using the quotient rule. Use the standard Maclaurin series expansion. ← Chapter 2 Problem-solving The general term contains (–1)n and a power of x2n + 1. The non-zero terms will all have odd powers of x and will alternate signs. 0.48151… − arsinh 0.5 e % error =   ______________________          × 100 arsinh 0.5 = 0.062% (3 d.p.) 132 M06_ALEVEL_CP2_83343_U06_119-146.indd 132 26/02/2018 09:29 Hyperbolic functions Exercise 6D 1 Differentiate with respect to x: E E/P E a sinh 2x b cosh 5x c tanh 2x d sinh 3x e coth 4x f sech 2x g e−x sinh x Hint h x cosh 3x sinh x   i ______ 3x j x2 cosh 3x k sinh 2x cosh 3x l ln (cosh x) m sinh x3 n cosh2 2x o ecosh x p cosech x 1 For part e, use coth x = ______     tanh x 1 For part f, use sech 2x = _______     cosh 2x 1 For part p, use cosech x = ______     sinh x d2y 2 If y = a cosh nx + b sinh nx, where a and b are constants, prove that ____   2  = n2y. dx 3 Find the exact coordinates of the stationary point on the curve with equation y = 12 cosh x − sinh x. d2y 4 Given that y = cosh 3x sinh x, find ____   2  dx (4 marks) (7 marks) (6 marks) 5 Differentiate: a arcosh 2x b arsinh (x + 1) c artanh 3x d arsech x e f arcosh 3x g x2 arcosh x j arsinh x arcosh x P arcosh x2 x h arsinh  __  2 k arcosh x sech x E ex 7 Given that y = artanh (__    ) , prove that 2 dy (4 − e2x)  ___  = 2ex dx d 1  , |x| , 1   b ___  (artanh x) = ______ dx 1 − x2 (6 marks) 8 Given that y = arsinh x, show that d 3y d2y dy     = 0 (1 + x2)  ____3  + 3x  ____2  + ___ dx dx dx E l x arcosh 3x 6 Prove that: d 1 a ___  (arcosh x) = _______   ______ , x . 1 2 dx √ x − 1  E i ex3 arsinh x d2y   2  9 If y = (arcosh x)2, find ____ dx (7 marks) (6 marks) 133 M06_ALEVEL_CP2_83343_U06_119-146.indd 133 26/02/2018 09:29 Chapter 6 E 10 Find the equation of the tangent at the point where x = __   13 on the curve with equation y = artanh x. (3 marks) E/P 11 Show that the equation of the normal at the point where x = 2 on the curve with equation y = arcosh 2xcan be written as y = ax + b + ln c, where a, b and c are exact real numbers to (5 marks) be found. E/P 12 a Find the first three non-zero terms of the Maclaurin series for cosh x. 12 (5 marks) b Hence find the percentage error when this approximation is used to evaluate cosh 0.2. (3 marks) E/P 13 a Find the first three non-zero terms of the Maclaurin series for sinh x. b Hence find an expression for the nth non-zero term of the Maclaurin series for sinh x. E/P 14 y = tanh x a Use a Maclaurin series expansion to show that y ≈ x − _  3  x3. 1 (5 marks) (2 marks) (5 marks) b Find the percentage error when the approximation in a is used to evaluate tanh 0.8. (2 marks) E/P 15 y = artanh x a Show that the first three non-zero terms of the Maclaurin series of y are x +  _3 x3 + _  5 x5 1 1 (6 marks) b Hence write down the general term for the nth non-zero term in the series expansion. (1 mark) c Find the first two non-zero terms of the series expansion of cosh x artanh x. (5 marks) E/P 16 y = sinh x cosh 2x Find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form. (7 marks) E/P 17 y = cos x cosh x d4y (4 marks) a Show that ____   4  = − 4y. dx b Hence find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form. (4 marks) ∞ c Hence write the series expansion for y in the form ∑f(r)x4r, where f is function to be r=0 determined. (2 marks) Challenge Find the first three non-zero terms of the series expansion of s ech x. 134 M06_ALEVEL_CP2_83343_U06_119-146.indd 134 26/02/2018 09:29 Hyperbolic functions 6.5 Integrating hyperbolic functions You can integrate hyperbolic functions. ■ You should be familiar with the following integrals: ∫sinh x dx = cosh x + c • ∫cosh x dx = sinh x + c d ( cosh x) = sinh x since  ___ dx • • • d ( sinh x) = cosh x since  ___ dx   1   dx = arsinh x + c ∫_______ √ 1 + x   1 d ( arsinh x) = _______ since  ___   ______  dx √ 1 + x2  ______ 2     dx = arcosh x + c, x > 1 ∫_______ √ x − 1  1 ______ 2 1 d ( arcosh x) = _______ since  ___   ______ , x > 1 2 dx √ x − 1  Example 23 Find ∫cosh (4x − 1) dx. ∫cosh (4x − 1)dx =  __ sinh (4x − 1) + c 1 4 Example 24 Find 2 + 5x   dx. ∫  ________ √ x + 1  _______ 2   2 + 5x   dx = ∫________   2   dx + ∫________   5x   dx ∫________ √x + 1  √x + 1  √x + 1  _______ _______ 2 _______ 2 2 ∫ ∫ 1   dx + 5 x(1 + x2)−   _21  dx = 2 ________   _______ 2 √ x + 1  _______ = 2 arsinh x + 5√1 + x2  + c Splitting the numerator gives two recognisable integrals. This is the standard result for arsinh x. This integral is of the form k∫f9(x)(f(x))n dx. ← Pure Year 2, Chapter 9 Example 25 Find: a ∫cosh 2x sinh 2x dx b ∫tanh x dx 5 a ∫ ∫ cosh5 2x sinh 2x dx =  _21   (cosh 2x)5(2 sinh 2x) dx = __  121  cosh6 2x + c ∫ (f(x))n11 Use (f(x))nf9(x) dx 5 ________     1 c, n11 with f(x) 5 cosh 2x and n 5 5. 135 M06_ALEVEL_CP2_83343_U06_119-146.indd 135 26/02/2018 09:29 Chapter 6 ∫ sinh x   dx ∫tanh x dx = ∫  ______ cosh x b Use ____   f9(x)  dx 5 ln|f(x)| 1 c, f(x) with f(x) 5 cosh x. Modulus signs are not necessary because cosh x . 0, for all x. = ln cosh x + c ■ ∫tanh x dx = ln cosh x + c Example 26 Find the following integrals. a ∫cosh 3x dx b ∫sinh x dx 2 3 6x dx 1 + cosh      ∫cosh 3x dx = ∫ ____________ 2 a 2 ( Use cosh 2A ≡ 2 cosh2 A − 1 with A ≡ 3x. ) sinh 6x =_  21   x +  _______    + c 6 =_  21  x + __  121  sinh 6x + c ∫sinh x dx = ∫sinh x sinh x dx = ∫(cosh x − 1)sinh x dx = ∫cosh x sinh x dx − ∫sinh x dx b 3 2 2 2 =_  31  cosh3 x − cosh x + c Problem-solving For small odd values of n, you can use ∫sinhn x dx = ∫sinhn−1 x sinh x dx. ∫coshn x dx, for odd values of n, can be found similarly. Sometimes the exponential definition of a hyperbolic function can be used to find a given integral. Example 27 Find ∫e sinh x dx. 2x e −e dx   ∫e sinh x dx = ∫e ( _______ 2 ) =_    ∫(e − e ) dx 2x 2x 1 2 −x x 3x ( Use the definition of sinh x. x ) e3x =_  21     ___  − ex + c 3 = __  61  (e3x − 3ex) + c You need to be able to use hyperbolic substitutions to find the integrals of expressions of the forms 1   dx   dx and ∫________   1   ∫________ √ x  + a    √x  − a    _______ 2 2 _______ 2 2 136 M06_ALEVEL_CP2_83343_U06_119-146.indd 136 26/02/2018 09:29 Hyperbolic functions Example 28 By using an appropriate substitution, find   1   dx, x . a. ∫_________ √x − a   ________ 2 2 As cosh2 u − 1 ≡ sinh2 u, it follows that a2 cosh2 u − a2 = a2 sinh2 u This is similar to using the substitution x = a sin u 1 to integrate _______   ______  ← Section 3.4 2 √ a – x2  Use the substitution x = a cosh u, dx  ___ = a sinh u du so dx can be replaced by a sinh u du. 1 1   dx = ∫________________  a sinh u du        ∫_________ √x − a   √a   cosh u − a   ________ 2 ______________ 2 2 = 2 2   1 a sinh u du ∫_______ a sinh u =u+c x = arcosh( __   a   ) + c ■ ■ x x x = a cosh u ⇒ cosh u = __   a  ⇒ u = arcosh (__   a ) x   1   dx = arsinh (__   a ) + c ∫________ √ a + x   ______ 2 2 x   1   dx = arcosh (__   a ) + c, x . a ∫________ √ x − a   ______ 2 2 Example 29 ∫ ( 8 __ ) 2 + √ 3 1 Show that   ________   _______      dx = ln  _______ 2 2 5 √x − 16  1 x   dx = [ arcosh(   __   )  ]   ∫ ________ 4 √ x − 16  8 _______ 2 5 ∫ 1 Use the result for _______   ______  dx with a = 4. 2 √ x − a2  8 5 = arcosh 2 − arcosh(__   54 ) __ __ 9 = ln(2 + √ 3 ) − ln(__   54 + √ __   16  ) ______ Use arcosh x = ln (x + √ x2 − 1 ). __ = ln(2 + √ 3 ) − ln 2 ( __ 2 + √3  _______ = ln   )    2 a Use ln a − ln b = ln (__    ) b 137 M06_ALEVEL_CP2_83343_U06_119-146.indd 137 26/02/2018 09:29 Chapter 6 Example 30 Show that ∫√1 + x   dx = _  arsinh x + _  x√ 1 + x   + c. ______ 2 1 2 ______ 2 1 2 dx x = sinh u ⇒   ___ = cosh u du so dx can be replaced by cosh u du ∫ ______ ∫ = ∫cosh u du =_    ∫(1 + cosh 2u) du _________ √1 + sinh2 u   cosh u du √1 + x2  dx = 2 Use cosh 2u ≡ 2 cosh2 u − 1. 1 2 ( ) sinh 2u =_  21   u + ______      + c 2 You need to be able to use x = sinh u, so use sinh 2u ≡ 2 sinh u cosh u. =_  21  (u + sinh u cosh u) + c _________ ______ u = arsinh x and cosh u ≡ √ 1 + sinh2 u . =_  21  arsinh x + _   21  x√ 1 + x2  + c Example 31 ∫ 6 x3 By using a hyperbolic substitution, evaluate   _______   ______   dx. x2 + 9  0 √ Use the substitution x = 3 sinh u dx   ___ = 3 cosh u du So dx can be replaced by 3 cosh u du You need to reduce x2 + 9 to a single term. Using x = 3 sinh u gives 9 sinh2 u + 9 = 9 (sinh2 u + 1) = 9 cosh2 u x   dx = ∫ ∫     _______   √x + 9  When x = 6, sinh u = 2 ⇒ u = arsinh 2. When x = 0, sinh u = 0 ⇒ u = 0. 6 3 arsinh2 27sinh3 u ________ ______ 0 2   0 ∫ 3 cosh u 3 cosh u du arsinh2 = 27   sinh3 u du 0 Use the method from Example 26. = 27[ _  31  cosh3 u − cosh u ]0 arsinh2 = 27(  __   31   − 1)   − √ 5 ) − 27(_ 5√ 5  ____ __ 3 __ ________ __ As sinh u = 2, cosh u = √ 1 + 22  = √ 5 . = 18√5  + 18 138 M06_ALEVEL_CP2_83343_U06_119-146.indd 138 26/02/2018 09:29 Hyperbolic functions Example 32 Find 1   dx.   ∫__________ √12x + 2x   _________ 2 12x + 2x2 = 2(x2 + 6x) Complete the square. = 2((x + 3)2 − 9) So       dx = ∫  _______________     dx ∫ ___________ √ √ 2((x + 3) − 9)  12x + 2x   1 1 __________ 2 ______________ 2    Let u = x + 3, then du = dx Choose the substitution.   du       dx = ___   1  ∫ ________   ∫ ___________ 2  √ u − 9  √ 12x + 2x   1 __________ 2 1 __ ______ 2 √ ( ) u = ___   1__  arcosh __      + c 3 √ 2  Select the standard form. x+3 = ___   1__  arcosh _____      + c 3 √ 2  Rewrite the answer in terms of x. ( ) Example 33 Use the substitution x = _ 2 (3 + 4 cosh u) to find 1 1   dx.     ∫______________ √ 4   x − 12x − 7  _____________ 2 dx 1 x=_  21  (3 + 4 cosh u) ⇒  ___ = _    (4 sinh u) = 2 sinh u du 2 so dx can be replaced by 2 sinh u du 4x2 − 12x − 7 = 4(__   21  (3 + 4 cosh u)) − 6(3 + 4 cosh u) − 7 2 = 9 + 24 cosh u + 16 cosh2 u − 18 − 24 cosh u − 7 = 16 cosh2 u − 16 = 16 sinh2 u So 1 1       dx = ∫ _______   × 2 sinh u du ∫ _______________ 4 sinh u √ 4x − 12x − 7  Use cosh2 u − sinh2 u ≡ 1 _____________ 2    ∫ =_   21   1du =_   21  u + c ( ) 2x − 3 =_   21  arcosh ______      + c 4 As x = _  12 (3 + 4 cosh u), 2x − 3 cosh u = ( ______     4 ) 139 M06_ALEVEL_CP2_83343_U06_119-146.indd 139 26/02/2018 09:29 Chapter 6 6E Exercise 1 Integrate the following with respect to x. 1   b cosh x − _______   cosh2 x a sinh x + 3 cosh x 2 Find: a ∫sinh 2x dx b ∫cosh( __ x3   )dx b   x − 3   dx ∫_______ √ 1 + x   b tanh 4x dx ∫ c ∫ b x dx  1 + tanh   ∫__________ cosh x c 5 cosh x + 2 sinh x dx       ∫_________________ cosh x c ∫cosh x cosh 3x dx c ∫cosh x dx 3 Find: a   1 + x   dx ∫_______ √ x − 1  ______ 2 4 Find: a ∫ sinh3 x cosh x dx 5 Find: a   sinh x   dx ∫___________ 2 + 3 cosh x 6 Use integration by parts to find ∫e cosh x dx b x ∫ ______ 2 2 _______ √cosh 2x  sinh 2x dx ∫x sinh 3x dx. 7 Find: a sinh x   c _______ cosh2 x ∫e sinh 3x dx −2x 1 1 8 Evaluate   ______________       dx, giving your answer in terms of e. 0 sinh x + cosh x 9 Use appropriate identities to find: a E ∫sinh x dx b 2 ∫ ln2 ∫sinh x cosh x dx 2 2 ( ) x 1 10 Show that   cosh2 __      dx = _  8 (3 + ln 16). 2 0 11 Use suitable substitutions to find: a   1   dx ∫_______ √x − 9  ______ 2 b 5 (7 marks) 1   dx   ∫_________ √4x + 25  ________ 2 12 Write down the results for the following: a   3   dx ∫_______ √ x + 9  ______ 2 b   1   dx ∫_______ √x − 2  ______ 2 140 M06_ALEVEL_CP2_83343_U06_119-146.indd 140 26/02/2018 09:29 Hyperbolic functions 13 Find: a 1   dx   ∫_________ √ 4x − 12  b ________ 2 ∫ 1   dx   ∫_________ √ 9x + 16  ________ 2 2 3 14 Evaluate   ________   _______2   dx. √ 1 1 + 4x   15 Evaluate, giving your answers as a single natural logarithms. ∫ ∫ 4 1   _______   dx a   ________ 2 x + 16  0 √ 15 1 b   _________   ________   dx 2 x − 144  13 √ x      dx, x . 0. ∫√_____ x+1 (6 marks) 2x     dx. ∫   _______ √ x − 1  (6 marks) ______ E 16 Use the substitution x E 17 By using the substitution u = x E 2, evaluate 3 ______ 4 2 18 Use the substitution x = 2 cosh u to show that ∫√x − 4  dx = _   x√ x − 4  − 2 arcosh( __ x2   ) + c ______ E/P = sinh2 u to find 19 a ______ 2 1 2 Show that 1   dx can be written as ∫_______       2e   dx. ∫_______________ 2 cosh x − sinh x e +3 (6 marks) 2 x 2x b Hence, by using the substitution u = ex, find ∫ 1   dx.     ∫_______________ 2 cosh x − sinh x 1 E/P cosh x 2 ____________      20 Using the substitution u = _  3  sinh x, evaluate   _____________   dx. 2 x + 9  √ 4 sinh 0 P 21 Find the following: a c P 1   dx     ∫_____________ √    x + 6x + 10  ____________ b 1     dx ∫____________ 2x + 4x + 7 d 1   dx      ∫_____________ √ 9x − 8x + 1  b 1      dx ∫ ______________ √ 4   x − 12x + 4  2 2 22 Find: a P 1      dx ∫_____________ √   x − 4x − 12  1   dx     ∫_______________ √ 4   x − 12x + 10  ______________ 2 ∫ ____________ 2 ____________ 2 (3 marks) (5 marks) (8 marks) Problem-solving Complete the square in the denominator to identify a suitable substitution. 1 1 ____________ __________  = ____________ ___________  so the           √ x2 – 4x – 12  √ (x – 2)2 – 16  substitution u = x – 2 will reduce the integral in part a to a standard form. ______________ 2 1 1 ___________ 23 Evaluate   ____________       dx 2 √ 0 x + 2x + 5  141 M06_ALEVEL_CP2_83343_U06_119-146.indd 141 26/02/2018 09:29 Chapter 6 ∫ 3 E/P 1 ___________ 24 Evaluate   ____________       dx, giving your answer as a single natural logarithm. 2 √ 1 x − 2x + 2  E/P __ 1 1 ____________        __  ln(2 + √3 ) 25 Show that   _____________   dx = ___ 2 √ 3 3x − 6x + 7  1 √ E/P 26 The curves shown have equations ∫ 3 (8 marks) y y = 5 cosh xand y = 7 − sinh x. a Using the definitions of sinh x and cosh x in terms of ex, find exact values for the x-coordinates of the two points where the curves intersect. (6 marks) y = 5 cosh x R b Use calculus to find the area of the finite region, R, between the two curves, giving your answer in the form ln a + b where a and b are integers. (6 marks) E/P (8 marks) O y = 7 – sinh x x 27 The region bounded by the curve y = sinh x, the line x = 1 and the positive x-axis is rotated through 360° about the x-axis. Show that the volume of the solid of revolution formed is π ___2  (e4 − 4e2 − 1) 8e (5 marks) Challenge 1 Using the substitution x = 1 + sinh θ, show that   + c      dx = ___________   ∫ ____________ √ x − 2x + 2  (x − 2x + 2) x−1 1 __________ 2 _3 2    2 2 By means of a suitable substitution, or otherwise, find: a ∫x cosh (x ) dx 2 Mixed exercise 2 b x     dx ∫ ________ cosh (x ) 2 2 6 1 Find the exact value of: a sinh (ln 3) b cosh (ln 5) c tanh (ln  _4 ) 1 2 Given that artanh x − artanh y = ln 5, find y in terms of x. E/P 3 Using the definitions of sinh x and cosh x, prove that sinh (A − B) ≡ sinh A cosh B − cosh A sinh B (5 marks) 142 M06_ALEVEL_CP2_83343_U06_119-146.indd 142 26/02/2018 09:29 Hyperbolic functions E/P 4 Using definitions in terms of exponentials, prove that 2 tanh  _2 x sinh x ≡ ____________   1   1 − tanh2  _2 x 1 E (5 marks) 5 Solve, giving your answers as natural logarithms 9 cosh x − 5 sinh x = 15 E (6 marks) 6 Solve, giving your answers as natural logarithms 23 sinh x − 17 cosh x + 7 = 0 E (6 marks) 7 Solve, giving your answers as natural logarithms 3 cosh2 x + 11 sinh x = 17 E (6 marks) 8 a On the same diagram, sketch the graphs of y = 6 + sinh x and y = sinh 3x. (2 marks) b Using the identity sinh 3x ≡ 3 sinh x + 4 sinh3 x, show that the graphs intersect where (5 marks) sinh x = 1 and hence find the exact coordinates of the point of intersection. E/P 9 a Given that 13 cosh x + 5 sinh x = R cosh (x + α), R . 0, use the identity cosh (A + B) ≡ cosh A cosh B + sinh A sinh B to find the values of R and α, giving the value of α to 3 decimal places. (4 marks) b Write down the minimum value of 13 cosh x + 5 sinh x. E/P (1 mark) 10 a Express 3 cosh x + 5 sinh x in the form R sinh (x + α), where R . 0. Give α to 3 decimal places. (4 marks) b Use the answer to part a to solve the equation 3 cosh x + 5 sinh x = 8, giving your answer to 2 decimal places. (3 marks) c Solve 3 cosh x + 5 sinh x = 8 by using the definitions of cosh x and sinh x. (4 marks) dy 11 Given y = cosh 2x, find ___     dx 12 Differentiate with respect to x: a arsinh 3x E b arsinh (x2) E d x2 arcosh 2x 13 Given that y = (arsinh x)2, prove that d2y dy (1 + x2)  ____2  + x  ___  − 2 = 0 dx dx E/P x c arcosh  __  2 (5 marks) 14 Given that f(x) = 5 cosh x − 3 sinh x, find: a f9(x) (2 marks) b the turning point on the curve y = f(x) (4 marks) 15 Differentiate arcosh (sinh 2x). (4 marks) 143 M06_ALEVEL_CP2_83343_U06_119-146.indd 143 26/02/2018 09:29 Chapter 6 E/P 16 y = sin x cosh x d  4 x (4 marks) a Show that ____   4  = − 4y. dy  b Hence find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form. (4 marks) E/P 17 y = sinh 2x cosh x Find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form. (7 marks) E/P 18 4x2 + 4x + 17 ≡ (ax + b)2 + c, a . 0. a Find the values of a, b and c. ∫ (3 marks) 3 _   2 1 ____________ b Find the exact value of      _____________      dx 2 1 √   x + 4x + 17  − _2 4 19 Find the following: a E/P ∫sinh 4x cosh 6x dx b (6 marks) ∫e sinh x dx x 20 The diagram shows the cross-section R of an artificial ski slope. y R O 5 x The slope is modelled by the curve with equation 10 y = __________   _________  , 0 < x < 5 √ 4x2 + 9  Given that 1 unit on each axis represents 10 metres, use integration to calculate the area of R. Show your method clearly and give your answer to 2 significant figures. 1     dx ∫  ____________ √ x − 2x + 10  E/P 21 Find E/P 22 a Find ___________ 2 1     dx ∫  ______________ sinh x + 2 cosh x ∫ 4 __ 3x − 1 ____________ √2 b Show that   _____________      x = 9(   − 1) + 2 arsinh 1   d x2 − 2x + 10  1 √ (6 marks) (7 marks) (6 marks) (6 marks) 144 M06_ALEVEL_CP2_83343_U06_119-146.indd 144 26/02/2018 09:29 Hyperbolic functions E/P 23 The diagram shows a hanging chain fixed to two vertical bollards fixed on horizontal ground 7 metres apart. x The chain is modelled by the curve y = cosh  __  3 a Find the acute angle that the chain makes with the right-hand (5 marks) bollard. Give your answer to 3 significant figures. y R O –3 4 x b Use calculus to find, correct to 2 decimal places, the area of the finite region, R, enclosed by the bollards, the chain and the ground. (5 marks) E y 24 The curves shown have equations y = 3 cosh 2x and y = 8 + sinh 2x. y = 8 + sinh 2x a Using the definitions of sinh x and cosh x in terms of ex, find exact values for the x-coordinates of the two (6 marks) points where the curves intersect. R b Use calculus to find the area of the finite region, R, between the two curves, giving your answer correct to (6 marks) 3 significant figures. E/P x O 25 The diagram shows the cross-section of a loaf of bread. Each unit on the axes represents 5 cm. The curved top of the loaf is modelled using the equation 5 y = _______   ______  2 √ x  + 4  y –2 Given that the loaf can be modelled as a prism, with length 30 cm, find, correct to 3 significant figures, the volume of the loaf. E y = 3 cosh 2x O 2 x (8 marks) y 26 The diagram shows the curve y = 3 − sinh x. a Find the exact coordinates of the point where the curve crosses the x-axis. (3 marks) The shaded area is bounded by the curve, the y-axis and the x-axis. y = 3 – sinh x b Find the volume of the solid of revolution formed when the shaded area is rotated through 360° about the x-axis. Give your answer correct to 3 significant figures. (8 marks) Challenge The diagram shows the graph of x y 1 y = sech x. Show that the area bounded by the curve and the x-axis is π. O O y = sech x x 145 M06_ALEVEL_CP2_83343_U06_119-146.indd 145 26/02/2018 09:29 Chapter 6 Summary of key points e  x − e  −x 1 • Hyperbolic sine (or sinh) is defined as sinh x ≡ _______     , x ∈ ℝ 2 e  x + e  −x • Hyperbolic cosine (or cosh) is defined as cosh x ≡ _______     , x ∈ ℝ 2 sinh x ______ e  2x − 1 • Hyperbolic tangent (or tanh) is defined as tanh x ≡ ______     , x ∈ ℝ ≡   2x cosh x e  + 1 2 • The graph of y = sinh x: y • The graph of y = cosh x: y y = sinh x x O y = cosh x 1 x O For any value a, sinh (−a) = −sinh a. For any value a, cosh (−a) = cosh a. 3 The table shows the inverse hyperbolic functions, with domains restricted where necessary. Hyperbolic function y = sinh x y = cosh x, x > 0 y = tanh x Inverse hyperbolic function y = arsinh x y = arcosh x, x > 1 y = artanh x, |x| , 1 ______ ( ______ ) 4 • arsinh x = ln(x + √ x2 + 1  ) • arcosh x = ln(x + √ x2 − 1  ), x > 1 • artanh x = _ 12 ln _____  1 + x   , |x| , 1 1−x 5 cosh2 A − sinh2 A ≡ 1 6 • sinh (A ± B) ≡ sinh A cosh B ± cosh A sinh B • cosh(A ± B) ≡ cosh A cosh B ± sinh A sinh B 7 8 9 d   (sinh x) = cosh x •  ___ dx 1 d   (arsinh x) = _______ •  ___   ______  2 dx √ x + 1  • ∫sinh x dx = cosh x + c ∫tanh x dx = ln cosh x + c 10 • 1   dx = arsinh ( __   x ∫  _______ a   ) + c √ a + x   ______ 2 2 d   (cosh x) = sinh x •  ___ dx 1 d   (arcosh x) = _______ •  ___   ______  2 dx √ x − 1  d   (tanh x) = sech2 x •  ___ dx 1 d   (artanh x) = ______ •  ___     dx 1 − x2 • ∫cosh x dx = sinh x + c • x 1   dx = arcosh (__   a ) + c, x . a ∫  _______ √ x − a   ______ 2 2 146 M06_ALEVEL_CP2_83343_U06_119-146.indd 146 26/02/2018 09:29 Methods in differential equations Objectives 7 After completing this chapter you should be able to: ● Solve first-order differential equations using an integrating factor → pages 148–153 ● Solve second-order homogeneous differential equations using the auxiliary equation → pages 153–157 ● Solve second-order non-homogeneous differential equations using the complimentary function and the particular integral → pages 157–162 ● Find particular solutions to differential equations using given → pages 162–165 boundary conditions Prior knowledge check 1 Find the general solution to the differential dy equation ___     = xex. ← Pure Year 2, Chapter 11 dx Population growth can be modelled by a differential equation. For example, the rate of change of the population of bacteria in a petri dish is proportional to the number of bacteria present, subject to the limiting factor of the amount of space on the dish. 2 Find the particular solution to the dy x differential equation ___     = −  __ y , given that dx y = 0 when x = 2. ← Pure Year 2, Chapter 11 3 Find: ∫ 3 a    ______     dt 50 − 2t b ∫   tan 4x dx ← Pure Year 2, Chapter 11 147 M07_ALEVEL_CP2_83343_U07_147-169.indd 147 02/03/2018 12:43 Chapter 7 7.1 First-order differential equations If a first-order differential equation can be dy written in the form ___     = f(x)g(y), then you can dx 1 solve it by writing    ____     dy = ∫   f(x)dx. g(y) ∫ Example Links This process is called separating the variables. ← Pure Year 2, Section 11.10 1 Find the general solution to the differential equation y dy  ___ = −  __ x dx and sketch members of the family of solution curves represented by the general solution. ∫ ∫ 1 1 __      __ y dy = −      x dx Separate the variables and integrate. ln |y| = −ln |x| + c Collect the ln terms together and combine using the laws of logarithms. ln |y| + ln |x| = c ln |xy| = c |xy| = ec A c y = ±__ x , where A = e Some solution curves corresponding to different values of A are shown below. y A y=± x Problem-solving If c is a constant of integration, then A = ec can also be used as a constant. Writing the equation in this form helps you determine the family of solution curves. A The graphs of y = ±  __ x  form the family of solution curves for this differential equation. O x Online Explore families of solution curves using GeoGebra. In this chapter, you will consider differential equations which cannot be solved by separating variables. The following example uses the product rule ‘in reverse’ to obtain a solution. 148 M07_ALEVEL_CP2_83343_U07_147-169.indd 148 02/03/2018 12:43 Methods in differential equations Example 2 Find the general solution to the differential equation dy x3 ___ + 3x2y = sin x You can use the product rule dx dy x3___   + 3x2y = sin x dx d So ___   (x3y) = sin x dx ⇒ x3y = ∫ sin x dx So = −cos x + c c 1 y = −  ___3 cos x + ___   3  x x dv du d u___    + v___    = ___   ( uv), with u = x3 and v = y, to dx dx dx dy d recognise that x3___   + 3x2y = ___   ( x3y). dx dx Use integration as the inverse process of differentiation. Integrate each side of the equation, including an arbitrary constant on the right-hand side. Make y the subject by dividing each of the terms One side of the differential equation in the on the right-hand side by x3. example above is an exact derivative of a product in the form dy d    (f(x)y) f(x) ___  + f ′(x)y = ___ dx dx You can solve some first-order differential equations by turning them into equations of this form. Consider the differential equation dy 3y ____ sin x ___     + ___    =   3  x  dx x The left-hand side is not a derivative of a product but if you multiply both sides by x3 then the equation becomes the same as that in Example 2: dy Notation x3 is called an integrating factor. x  3  ___  + 3x  2 y = sin x dx In general, if you can write a linear first-order differential equation in the form dy ___     + P(x)y = Q(x) dx where P(x) and Q(x) are functions of x, and the function P(x) can be integrated, then it is possible to find an integrating factor which will convert the left-hand side of the equation into an exact derivative of a product. Suppose that the integrating factor has the form f(x), then dy f(x)  ___  + f(x)P(x)y = f(x)Q(x) Multiply the equation by the integrating factor f(x). dx In order for the left-hand side to be the derivative of the product f(x)y it must have the form dy f(x)  ___  + f ′(x)y. You can see that f(x)P(x) must be equal to f ′(x), so: dx f ′(x) ____   Divide both sides by f(x)and integrate.   dx = ∫P(x)dx f(x) ∫ ⇒ ln|f(x)| = ∫P(x)dx ⇒ f(x) = e  ∫   P(x)dx 149 M07_ALEVEL_CP2_83343_U07_147-169.indd 149 02/03/2018 12:43 Chapter 7 Multiplying every term in the original differential equation by this integrating factor, dy e  ∫   P(x)dx  ___  + e  ∫   P(x)dx P(x)y = e  ∫   P(x)dx Q(x) dx d ___ So    ( e  ∫   P(x)dx y) = e  ∫   P(x)dx Q(x) You can use the ‘product rule in reverse’ on the LHS. dx The equation is now directly solvable by integrating both sides. e   ∫   P(x)dx y = ∫   e  ∫   P(x)dx Q(x) dx Notice that the left-hand side is the integrating factor multiplied by y and the right-hand side is the integral of the integrating factor multiplied by Q(x). This will always be the case. ■■ You can solve a first-order differential dy equation of the form ___     + P(x)y = Q(x) by dx multiplying every term by the integrating factor e  ∫   P(x)dx. Example Watch out You may write down this integrating factor without proof in your exam, but you must show all the subsequent stages of your working clearly. 3 Find the general solution of the differential equation dy  ___ − 4y = ex Find the integrating factor. dx The integrating factor is e∫P(x) = e∫(−4) dx = e−4x dy e−4x ___   − 4e−4x y = exe−4x dx d ⇒ ___ (e−4x y) = e−3x dx ⇒ e−4x y = ∫e−3x dx 1 −3x = −__ +c 3 e 1 x 4x y = −  __ 3 e + ce So Multiply each term by the integrating factor. Express the LHS as the derivative of a product. Integrate to get the general solution. Divide every term, including the constant, by the integrating factor to make y the subject. When you have integrated and found the general solution, you can let the arbitrary constant take different numerical values, thus generating particular solutions. In some questions you will be given a boundary condition, such as y = 1 when x = 0. You can use this to find the value of the constant. Different boundary conditions will give rise to different particular solutions. Example 4 dy a Find the general solution of the differential equation cos x  ___  + 2y sin x = cos4 x. dx b Find the particular solution which satisfies the condition that y = 2 when x = 0. 150 M07_ALEVEL_CP2_83343_U07_147-169.indd 150 02/03/2018 12:43 Methods in differential equations a Divide through by cos x: dy ___   + 2y tan x = cos3 x (1) dx The integrating factor is 2 e∫P (x)dx = e∫2 tan x dx = e2 ln sec x = eln sec x = sec2 x dy sec2 x ___   + 2y sec2 x tan x = sec2 x cos3 x dx d So ___   (y sec2 x) = cos x dx ⇒ y sec2 x = ∫cos x dx ⇒ y sec2 x = sin x + c ⇒ y = cos2 x(sin x + c) Divide by cos x so that equation is in the form dy ___     + P(x) = Q(x) dx Use properties of ln to simplify the integrating factor. Multiply equation (1) by the integrating factor and simplify the right-hand side. Integrate to get the general solution and multiply through by cos2 x. b 2 = cos20(sin 0 + c) ⇒c=2 y = cos2 x(sin x + 2) is the required particular solution. Exercise Substitute y = 2 and x = 0 into the general solution to find c. 7A 1 For each of the following differential equations, find the general solution and sketch the family of solution curves. dy a ___ = 2x dx dy 2y c ___ =  ___ dx x dy  ___  = cos x e dx dy b ___ = y dx dy x d ___ =  __ dx y dy f ___ = y cot x, dx Hint You can solve all of these equations by separating variables. 0,x,π 2 Given that k is an arbitrary positive constant, a show that y2 + kx2 = 9k is the general solution to the differential equation dy −xy     , ∣x∣ , 3  ___ = ______ dx 9 − x2 b find the particular solution, which passes through the point (2, 5) c sketch the family of solution curves for k = _  9 , _  9 , 1 and include your particular solution in the diagram. 1 4 E 3 Find the general solutions to these differential equations. dy dy dy __ __ __ a x     + y = cos x b e−x     − e−x y =xe x c sin x     + y cos x = 3 dx dx dx d y d y d 1 __ __ 1 __y x y=x 2e y __   −     y = e e x   + 2 x e f 4 xy     + 2y2 = x2     d __   2 x dx x dx dx dy 4 a Find the general solution to the differential equation ___     + 2xy = e  −x  2 (4 marks) dx b Describe the behaviour of y as x → ∞. (1 mark) 151 M07_ALEVEL_CP2_83343_U07_147-169.indd 151 02/03/2018 12:43 Chapter 7 5 a Find the general solution to the differential equation dy x2 ___   + 2xy = 2x +1 dx b Find the three particular solutions which pass through the points with coordinates 1 1 1 (−_2 , 0), (−_2 , 3) and (−_2 , 19) respectively and sketch their solution curves for x , 0. 6 a Find the general solution to the differential equation dy y ____________ 1   =       , x . 1 ln x   ___ + __ dx x (x + 1)(x + 2) b Find the particular solution which passes through the point (2, 2). 7 Find the general solutions to these differential equations by using an integrating factor. dy ___   + 2y = ex a dx dy ___   + y sin x = ecos x c dx dy ___ e   + y tan x = x cos x dx dy x3 g x2  ___ − xy =   _____  , x . −2 x+2 dx dy i (x + 2) ___   − y = x + 2 dx E E E E E/P dy ___ b   + y cot x = 1 dx dy ___ d   − y = e2x dx dy y __ 1 f  ___ + __   x =   2  dx x dy h 3x  ___ + y = x dx dy ex j x  ___ + 4y =   __2  dx x dy 8 Find y in terms of x given that x  ___ + 2y = ex and that y = 1 when x = 1. dx dy 9 Solve the differential equation, giving y in terms of x, where x3  ___ − x2y = 1 dx and y = 1 at x = 1. (8 marks) 10 a Find the general solution to the differential equation 1 dy (x + __   x ) ___   + 2y = 2(x2 + 1)2 dx giving y in terms of x. (6 marks) (8 marks) b Find the particular solution which satisfies the condition that y = 1 at x = 1. (2 marks) 11 a Find the general solution to the differential equation dy π π   + y = 1, −__  , x ,  __  cos x ___ 2 2 dx b Find the particular solution which satisfies the condition that y = 2 at x = 0. (6 marks) 12 Find the general solution to the differential equation _____ dy (5 marks)  ___  = x√ y  2 − 4  dx (2 marks) Problem-solving You can use the substitution y = 2coshu 1 to integrate ______   _____  ← Section 6.5 2 √ y – 4  152 M07_ALEVEL_CP2_83343_U07_147-169.indd 152 02/03/2018 12:43 Methods in differential equations E/P E/P 13 a Find the general solution to the differential equation dy ___     = y cosh x dx b Find the particular solution which satisfies the condition that y = e when x = 0. (4 marks) 14 a Find the general solution to the differential equation _____ dy ___     = √ 1 + y  2  dx b Sketch the family of solution curves. (4 marks) 15 a Find the general solution to the differential equation dy cos x  ___  + y sin x = 1 dx b Find the particular solution such that y = 3 when x = π. 3π π    , 1) and ( ___    , −1)lie on all possible solution curves. c Show that the points ( __ 2 2 dy E/P 16 Find a general solution to the equation a  ___  + by = 0in terms of a and b. dx (2 marks) (3 marks) E (6 marks) (2 marks) (3 marks) (6 marks) 7.2 Second-order homogeneous differential equations A second-order differential equation contains second derivatives. Example 5 Find the general solution to the differential equation d  2y ____   2  = 12x dx  Watch out dy ___     = 6x2 + A dx y = 2x3 + Ax + B The general solution to this secondorder differential equation needs two arbitrary constants. If you wanted to find a particular solution you would need to know two boundary conditions. In this section you will look at techniques for solving linear differential equations that are of the form d   2y dy a  ____2  + b  ___  + cy = 0 dx dx  Notation You sometimes see differential where a, b and c are real constants. Equations equations of this type written as of this form (with 0 on the right-hand side) are d  2y dy ay″ + by′ + cy = 0where y″ = ____   2  and y′ = ___     called second-order homogeneous differential dx dx  equations with constant coefficients. dy Using the techniques from the previous section, the general solution of a  ___  + by = 0is of the form dx b   . Notice that k is the solution to the equation ak + b = 0. y = Ae  kx, where k = −  __ a 153 M07_ALEVEL_CP2_83343_U07_147-169.indd 153 02/03/2018 12:43 Chapter 7 This suggests that an equation of the form y = Aekx might also be a solution of the second-order dy d  2y differential equation a  ____2  + b  ___  + cy = 0. But it cannot be the general solution, as it only contains dx dx  one arbitrary constant. Since two constants are necessary for a second-order differential equation, you can try a solution of the form y = Ae  λx + Be  μx, where A and B are arbitrary constants and λ and μ are constants to be determined. dy  ___  = Aλe  λx + Bμe  μx dx ____ d   2y   2  = Aλ  2e  λx + Bμ2e  μx dx  Substituting these into the differential equation gives a(Aλ2eλx + Bμ2eμx) + b(Aλeλx + Bμeμx) + c(Aeλx + Beμx) = 0 aAλ  2 e  λx + aBμ  2e  μx + bAλ e  λx + bBμe  μx + cAe  λx + cBe  μx = 0 Ae  λx(aλ  2 + bλ + c) + Be  μx(aμ2 + bμ + c) = 0 This shows that the equation y = Ae  λx + Be  μxwill satisfy the original differential equation if both λ and μ are solutions to the quadratic equation am  2 + bm + c = 0. The equation am  2 + bm + c = 0is called the auxiliary equation. ■■ The natures of the roots α and β of the auxiliary equation, am2 + bm + c = 0 determine the dy d  2y general solution to the differential equation a  ____2  + b  ___  + cy = 0. dx dx  You need to consider three different cases: •• Case 1: b  2> 4ac The auxiliary equation has two distinct real roots α and β. The general solution will be of the form y = Ae  αx + B e  βx where A and B are arbitrary constants. •• Case 2: b  2 = 4ac The auxiliary equation has one repeated root α. The general solution will be of the form y = (A + Bx) e  αx where A and B are arbitrary constants. •• Case 3: b  2< 4ac Links Case 3 is equivalent to y = Aeαx + Beβx The auxiliary equation has two complex with complex α and β. → Exercise 7B, Challenge conjugate roots α and β equal to p ± q i. The general solution will be of the form Note If the roots are purely imaginary ( p = 0), the y = e  px (A cos qx + B sin qx) where A and B general solution reduces to y = A cos qx + B sin qx are arbitrary constants. Example 6 dy d  2y a Find the general solution to the equation 2  ____2  + 5  ___  + 3y = 0. dx dx  b Verify that your answer to part a satisfies the equation. a 2m  2 + 5m + 3 = 0 3 (2m + 3)(m + 1) = 0 ⇒ m = −  _ 2  or m = − 1 So the general solution is 3 _ y = Ae  − 2 x + Be  −x where A and B are arbitrary constants Write down the auxiliary equation. Solve the auxiliary equation. Write down the general solution. 154 M07_ALEVEL_CP2_83343_U07_147-169.indd 154 02/03/2018 12:43 Methods in differential equations 3 _ b y = Ae  − 2 x + Be  −x dy 3 3 − _  x −x  ___  = −  __ 2 Ae  2 − Be  dx d2y  9 − _3 x  ____2  = _     Ae  2 + Be −x dx  4 3 3 __ 3 9 − _  x −x − _  x −x 2(  __ 4 Ae  2 + Be  ) + 5(−  2 Ae  2 − Be  ) 3 _ + 3(Ae  − 2 x + Be  −x) 3 3 3 _ _ _ 15 __ __ 9 −    x −  =   2 Ae  2 −   2 Ae  2 x + 3Ae  − 2 x + 2Be  −x − 5Be  −x + 3Be  −x = 0 as required. Example Write down expressions for the first and second derivatives. d  2y dy Substitute your expressions for ____   2 , ___     and y into dx  dx the differential equation, expand and simplify. 7 d2y dy Show that y = (A + Bx)e3x satisfies the differential equation ____   2  − 6   ___ + 9y = 0. dx dx Let y = Ae3x + Bxe3x, then dy ___   = 3Ae3x + 3Bxe3x + Be3x dx d2y ____   2  = 9Ae3x + 9Bxe3x + 3Be3x + 3Be3x dx = 9Ae3x + 9Bxe3x + 6Be3x d2y dy   − 6  ___ + 9y = 9Ae3x + 9Bxe3x + 6Be3x dx dx2 − 6(3Ae3x + 3Bxe3x + Be3x) + 9(Ae3x + Bxe3x) = 0 ____   So y = (A + Bx)e3x is a solution to the equation. Example Differentiate the expression for y twice. Substitute into the left-hand side of the differential equation and simplify to show that the result is zero. Note The auxiliary equation m  2 − 6m + 9 = 0has a repeated root at m = 3 so the general solution is in the form you expect. 8 d 2y dy Find the general solution to the differential equation ____   2  + 8  ___ + 16y = 0. dx dx m 2   + 8m + 16 = 0 (m + 4)  2 = 0 ⇒ m = − 4 So the general solution is y = (A + Bx)e  −4x. Example Write down the auxiliary equation. Solve the auxiliary equation. In this case there is a repeated root. 9 d 2y dy Find the general solution to the differential equation ____   2  − 6  ___ + 34y = 0. dx dx   2 − 6m + 34 = 0 ⇒ m = 3 ± 5i m So the general solution is y = e  3x(A cos 5x + B sin 5x) Write down the auxiliary equation and solve it using the quadratic formula or by completing the square. In this case there are two complex conjugate roots. 155 M07_ALEVEL_CP2_83343_U07_147-169.indd 155 02/03/2018 12:43 Chapter 7 Example 10 d2y Find the general solution to the differential equation ____   2  + 16y = 0. dx Write down the auxiliary equation and solve it. In this case there are two purely imaginary roots. m  2 + 16 = 0 ⇒ m = ± 4i So the general solution is y = A cos 4x + B sin 4x Exercise This is in the form y = e  px(A cos qx + B sin qx) with p = 0. 7B 1 Find the general solution to each to the following differential equations. dy d2y ____ a   2  + 5 ___ + 6y = 0 dx dx 2 dy dy c ____2  + 2 ___ − 15y = 0 dx dx 2 dy dy e ____2  + 5 ___ = 0 dx dx 2 dy dy g 4 ____2  − 7 ___ − 2y = 0 dx dx d2y dy b  ____2  − 8 ___ + 12y = 0 dx dx 2 dy dy d ____2  − 3 ___ − 28y = 0 dx dx 2 dy dy f 3 ____2  + 7 ___ + 2y = 0 dx dx 2 dy dy h 15 ____2  − 7 ___ − 2y = 0 dx dx 2 Find the general solution to each of the following differential equations. d2y dy a ____2  + 10  ___ + 25y = 0 dx dx 2 dy dy c ____2  + 2 ___ + y = 0 dx dx 2 dy dy e 16 ____2  + 8 ___ + y = 0 dx dx 2 dy dy g 4 ____2  + 20  ___ + 25y = 0 dx dx d2y dy b ____2  − 18 ___ + 81y = 0 dx dx 2 dy dy d ____2  − 8 ___ + 16y = 0 dx dx 2 dy dy f 4 ____2  − 4 ___ + y = 0 dx dx 2 __ dy dy h ____2  + 2√ 3  ___ + 3y = 0 dx dx 3 Find the general solution to each of the following differential equations. dy ____ 2   + 25y = 0 dx2 d2y ____ c  2  + y = 0 dx d2y dy ____ e  2  + 8 ___ + 17y = 0 dx dx 2 dy dy g ____2  + 20 ___ + 109y = 0 dx dx a  dy ____ 2   + 81y = 0 dx2 d2y ____ d 9   2  + 16y = 0 dx 2 dy dy f ____2  − 4 ___ + 5y = 0 dx dx 2 __ dy dy h ____2  + √ 3   ___ + 3y = 0 dx dx b  156 M07_ALEVEL_CP2_83343_U07_147-169.indd 156 02/03/2018 12:43 Methods in differential equations 4 Find the general solution to each of the following differential equations. dy d2y a ____2  + 14 ___  + 49y = 0 dx dx d2y dy c ____2  + 4 ___  + 13y = 0 dx dx d2y dy e 9 ____2  − 6 ___  + 5y = 0 dx dx E/P d2y dy b ____2  + ___     − 12y = 0 dx dx d2y dy d 16 ____2  − 24 ___  + 9y = 0 dx dx d2y dy f 6 ____2  − ___     − 2y = 0 dx dx d   2x dx 5 Given the differential equation  ____2  + 2k  ___  + 9x = 0, where k is a real constant, dt dt  a find the general solution to the differential equation in each of the following cases: i |k| . 3 ii |k| , 3 iii |k| = 3 (8 marks) b In the case where k = 2, i find the general solution ii describe what happens to x as t → ∞. (4 marks) P 6 Given that am2 + bm + c = 0 has equal roots m = a, prove that y = (A + Bx)eax is a solution d 2y dy to the differential equation a ___2  + b ___  + c = 0. dx dx P Note This result is known as the 7 Given that y = f(x)and y = g(x)are both solutions to the 2 dy   y d principle of superposition. second-order differential equation a  ____2  + b  ___  + cy = 0, dx dx  prove that y = Af(x) + Bg(x), where A and B are real constants, is also a solution. Challenge Let α and β be the roots of a real-valued quadratic equation, so that a = p + iq and β = p − iq, p, q [ R. Show that it is possible to choose A, B [ C such that Aeax + Beax can be written in the form epx(Ccosqx + Dsinqx) where C and D are arbitrary real constants. 7.3 Second-order non-homogeneous differential equations Second-order differential equations of the form d  2y dy a  ____2  + b  ___  + cy = f(x) dx dx  are non-homogeneous. Notation You sometimes see differential equations of this type written as d  2y dy ay″ + by′ + cy = f(x)where y ″ =  ____2  and y ′ =  ___  dx dx  To solve an equation of this type you first find the general solution of the corresponding homogeneous d  2y dy differential equation, a  ____2  + b  ___  + cy = 0. This is called the complementary function (C.F.). dx dx  157 M07_ALEVEL_CP2_83343_U07_147-169.indd 157 02/03/2018 12:43 Chapter 7 You then need to find a particular integral (P.I.), which is a function that satisfies the differential equation. The form of the particular integral depends on the form of f (x). This table provides some particular integrals to try. Form of f(x) p p + qx p + qx + rx2 pekx p cos ωx + q sin ωx Form of particular integral λ λ + μx λ + μx + vx2 λekx λ cos ωx + μ sin ωx Use this form of the P.I. for functions such as 4x2 or 1 – x2. Use this form of the P.I. for functions such as sin 2x or 5 cos x. ■■ A particular integral is a function which satisfies the original differential equation. Example 11 d2y dy Find a particular integral of the differential equation ____   2  − 5 ___  + 6y = f(x) when f(x) is: dx dx x 2 c 3x d e e 13 sin 3x a 3 b 2x dy d2y a Let y = λ, then ___     = 0 and ____   2  = 0 dx dx 2y d dy Substitute into ____   2  − 5 ___  + 6y = 3: dx dx 0 − 5 × 0 + 6λ = 3 ⇒ λ = __   21 So a particular integral is __  21 dy d2y b Let y = λx + μ, then ___     = λ and ____   2  = 0 dx dx 2y d dy Substitute into ____   2  − 5 ___  + 6y = 2x: dx dx 0 − 5 × λ + 6(λx + μ) = 2x ⇒ (6μ − 5λ) + 6λx = 2x ⇒ 6μ − 5λ = 0 and 6λ = 2 5   18 ⇒ λ = __   31 and μ = __ 5 So a particular integral is __  31 x + __   18 c Let y = λx2 + μx + ν dy d2y Then ___     = 2λx + μ and ____   2  = 2λ dx dx d2y dy ____ ___ Substitute into   2  − 5    + 6y = 3x2: dx dx 2 2λ − 5(2λx + μ) + 6(λx + μx + ν) = 3x2 ⇒ (2λ − 5μ + 6ν) + (6μ − 10λ)x + 6λx2 = 3x2 ⇒ 2λ − 5μ + 6ν = 0, 6μ − 10λ = 0 and 6λ = 3 19 ⇒ λ = __   21 , μ = __   56 and ν = ___   36 19 So a particular integral is __  21 x2 + __   56 x + ___   36 When f(x) = 3, which is constant, choose P.I. = λ, which is also constant. Differentiate twice and substitute the derivatives into the differential equation. Solve equation to give the value of λ. When f(x) = 2x, which is a linear function of x, choose P.I. = λx + μ. Differentiate twice and substitute the derivatives into the differential equation. Equate the constant terms and the coefficients of x to give simultaneous equations, which you can solve to find λ and μ. As f(x) = 3x2, which is a quadratic function of x let P.I. = λx2 + μx + ν. Equate the constant terms, the coefficients of x and the coefficients of x2 to give simultaneous equations, which you can solve to find λ, μ and ν. 158 M07_ALEVEL_CP2_83343_U07_147-169.indd 158 02/03/2018 12:43 Methods in differential equations dy d2y d Let y = λex, then ___     = λex and ____   2  = λex dx dx d2y dy ____ ___ Substitute into   2  − 5    + 6y = ex: dx dx λex − 5λex + 6λex = ex ⇒ 2λex = ex ⇒ λ = __   21 So a particular integral is __  21 ex e Let y = λ sin 3x + μ cos 3x dy Then ___     = 3λ cos 3x − 3μ sin 3x dx d2y and ____   2  = −9λ sin 3x − 9μ cos 3x dx d2y dy   2  − 5 ___  + 6y = 13 sin 3x: Substitute into ____ dx dx −9λ sin 3x − 9μ cos 3x − 5(3λ cos 3x − 3μ sin 3x) + 6(λ sin 3x + μ cos 3x) = 13 sin 3x ⇒ (−9λ + 15μ + 6λ) sin 3x + (−9μ − 15λ + 6μ) cos 3x = 13 sin 3x ⇒ −9λ + 15μ + 6λ = 13 and −9μ − 15λ + 6μ = 0 5 __ 1 ⇒ λ = −__ 6 and μ =   6 5 __ 1 So a particular integral is −__ 6 sin 3x +   6 cos 3x As f(x) = ex, which is an exponential function of x let P.I. = λex. Equate coefficients of ex to find the value of λ. As f(x) = 13 sin 3x, which is a trigonometric function of x let P.I. = λ sin 3x + μ cos 3x, also a similar trigonometric function. Problem-solving Equate coefficients of sin 3x and of cos 3x and solve simultaneous equations. dy d  2y ■■ To find the general solution to the differential equation a  ____2  + b  ___  + cy = f(x), dx dx  dy ____ d  2y •• Solve the corresponding homogeneous equation a   2  + b  ___  + cy = 0 to find the dx dx  complementary function (C.F.) •• Choose an appropriate form for the particular integral (P.I.) and substitute into the original equation to find the values of any coefficients. •• The general solution is y = C.F. + P.I. Example 12 d2y dy Find the general solution to the differential equation ____   2  − 5 ___  + 6y = f(x) when f(x) is: dx dx x 2 c 3x d e e 13 sin 3x a 3 b 2x m2 − 5m + 6 = 0 (m − 3)(m − 2) = 0 ⇒ m = 3 or m = 2 Hence the complementary function is y = Ae  3x + Be  2x where A and B are arbitrary constants. Solve the auxiliary equation to find the values of m. 159 M07_ALEVEL_CP2_83343_U07_147-169.indd 159 02/03/2018 12:43 Chapter 7 The particular integrals were already found in example 11 so the general solutions are: a y = Ae3x + Be2x + __   21 5 b y = Ae3x + Be2x + __   31 x + __   18 19 c y = Ae3x + Be2x + __   21 x2 + __   56 x + ___   36 The general solution is y = C.F. + P.I. y = Ae3x + Be2x + __   1 ex d 2 e y = Ae3x + Be2x − __   61 sin 3x + __   56 cos 3x You need to be careful if the standard form of the particular integral contains terms which form part of the complementary function. If this is the case, you need to modify your particular integral so that no two terms in the general solution have the same form. For example, this situation occurs when f(x) is of the form pekx, and k is one of the roots of the auxiliary equation. In this case you can try a particular integral of the form,λxekx. Example 13 d2y dy Find the general solution to the differential equation ____   2  − 5 ___  + 6y = e2x dx dx As in Example 12, the complementary function is y = Ae3x + Be2x. The particular integral cannot be λe2x, as this is part of the complementary function. So let y = λxe2x dy Then ___     = 2λxe2x + λe2x dx and d y ____ The function λe2x is part of the C.F. and satisfies the differential equation d2y dy ____   2  − 5 ___  + 6y = 0, so it cannot also satisfy dx dx dy d2y ____   2  − 5 ___  + 6y = e2x . dx dx 2   = 4λxe2x + 2λe2x + 2λe2x = 4λxe2x + 4λe2x dx2 d2y dy Substitute into ____   2  − 5 ___  + 6y = e2x: dx dx 4λxe2x + 4λe2x − 5(2λxe2x + λe2x) + 6λxe2x = e2x ⇒ −λe2x = e2x ⇒ λ = −1 So a particular integral is −xe2x. The general solution is y = Ae3x + Be2x − xe2x.   Watch out Let the P.I. be λxe2x and differentiate, substitute and solve to find λ. The general solution is y = C.F. + P.I. When one of the roots of the auxiliary equation is 0, the complementary function will contain a constant term. If f(x) is a polynomial, you will need to multiply its particular integral by x to make sure the P.I. does not also contain a constant term. 160 M07_ALEVEL_CP2_83343_U07_147-169.indd 160 02/03/2018 12:43 Methods in differential equations Example 14 Find the general solution to the differential equation dy d2y ____   2  − 2 ___  = 3 dx dx d2y dy First consider the equation  ____2  − 2 ___  = 0 dx dx m2 − 2m = 0 m(m − 2) = 0 ⇒ m = 0 or m = 2 Write down and solve the auxiliary equation. So the complementary function is y = A + Be2x. The particular integral cannot be a constant, as this is part of the complementary function, so let y = λx. dy d2y Then  ___  = λ and ____   2  = 0 dx dx 2y d dy Substitute into ____   2  − 2 ___  = 3: dx dx 0 − 2λ = 3 3 ⇒ λ = − __ 2  3 So a particular integral is −  __ 2 x 3 The general solution is y = A + Be2x −  __ 2 x. Exercise Find the complementary function by putting the right-hand side of the differential equation equal to zero, and solving this new equation. Try to find a particular integral. The righthand side of the original equation was 3, which was a constant and usually this would imply a constant P.I. As the C.F. includes a constant term ‘A’, the P.I. cannot also be constant. A value of λ d2y dy would satisfy  ____2  − 2 ___  = 0 rather than dx dx d2y dy ____ ___   2  − 2    = 3. dx dx Multiply the ‘expected’ P.I. by x and try λx instead. The general solution is y = C.F. + P.I. 7C 1 Solve each of the following differential equations, giving the general solution. d2y dy d2y dy b ____2  − 8 ___  + 12y = 36x a ____2  + 6 ___  + 5y = 10 dx dx dx dx E d2y dy c ____2  + ___     − 12y = 12e2x dx dx d2y dy d ____2  + 2 ___  − 15y = 5 dx dx d2y dy e ____2  − 8 ___  + 16y = 8x + 12 dx dx d2y dy f ____2  + 2 ___  + y = 25 cos 2x dx dx d2y g ____2  + 81y = 15e3x dx d2y h ____2  + 4y = sin x dx d2y dy i ____   2  − 4 ___  + 5y = 25x2 − 7 dx dx d2y dy j ____2  − 2 ___  + 26y = ex dx dx 2 a Find a particular integral for the differential equation dy d  2y ____   2  − 5 ___  + 4y = x  2 − 3x + 2 dx  dx b Hence find the general solution. (6 marks) (3 marks) 161 M07_ALEVEL_CP2_83343_U07_147-169.indd 161 02/03/2018 12:43 Chapter 7 E/P 3 y satisfies the differential equation d  2y dy  ____2  − 6 ___  = 2x  2 − x + 1 dx  dx a Find the complementary function for this differential equation. b Hence find a suitable particular integral and write down the general solution to the differential equation. (7 marks) E/P E/P (3 marks) Hint Try a particular integral of the form λx + μx2 + νx3. 4 Find the general solution to the differential equation dy d  2y ____   2  + 4 ___  = 24x2 dx  dx (10 marks) 5 a Explain why λxe  xis not a suitable form for the particular integral for the differential equation d2y dy ____ (2 marks)   2  − 2 ___  + y = ex dx dx b Find the value of λ for which λx2ex is a particular integral for the differential equation. (5 marks) c Hence find the general solution. E/P (3 marks) d  2y dy ____   2  + 4 ___  + 3y = kt + 5, where k is a constant and t . 0. 6 dt dt  a Find the general solution to the differential equation in terms of k. (7 marks) For large values of t, this general solution may be approximated by a linear function. b Given that k = 6, find the equation of this linear function. (2 marks) Challenge Find the general solution of the differential equation d  2y ____   2  + y = 5xe  2x dx  7.4 Using boundary conditions You can use given boundary conditions to find a particular solution to a second-order differential equation. Since there are two arbitrary constants, you will need two boundary conditions to determine the complete particular solution. Example 15 d2y dy Find y in terms of x, given that ____   2  − y = 2ex, and that ___     = 0 and y = 0 at x = 0. dx dx d2y First consider the equation ____   2  − y = 0. dx m2 − 1 = 0 ⇒ m = ±1 So the complementary function is y = Aex + Be−x. Solve the auxiliary equation to find the values of m. 162 M07_ALEVEL_CP2_83343_U07_147-169.indd 162 02/03/2018 12:43 Methods in differential equations The particular integral cannot be λex, as this is part of the complementary function, so let y = λxex. dy d2y Then  ___  = λxex + λex and ____   2  = λxex + λex + λex dx dx 2y d Substitute into ____   2  − y = 2ex: dx λxex + λex + λex − λxex = 2ex ⇒ λ=1 So a particular integral is xex. The general solution is y = Aex + Be−x + xex Since y = 0 at x = 0, 0 = A + B ⇒ A+B=0 Differentiating y = Aex + Be−x + xex with respect to x gives dy   ___   = Aex − Be−x + ex + xex dx dy Since   ___   = 0 at x = 0, 0 = A − B + 1 dx ⇒ A − B = −1 Solving the simultaneous equations gives __1 1 A = −__ 2 , and B =   2 __1 x __1 −x So y = −2 e +   2 e d2y As λex satisfies ____   2  − y = 0, it cannot also dx d2y satisfy ____   2  − y = 2ex. dx Substitute the boundary condition, y = 0 at x = 0, into the general solution to obtain an equation relating A and B. Substitute the second boundary condition, y d ___     = 0 at x = 0, into the derivative of the dx general solution, to obtain a second equation relating A and B. Solve the two equations to find values for A and B. + xex is the required solution. Example 16 Given that a particular integral is of the form λ sin 2t, find the solution to the differential d2x x  = 1 when t = 0. equation ____   2  + x = 3 sin 2t, for which x = 0 and d  ___ dt dt d2x First consider the equation ____   2  + x = 0. dt m2 + 1 = 0 ⇒ m = ±i So, the complementary function is x = A cos t + B sin t. The particular integral is λ sin 2t, so let x = λ sin 2t. dx d2x Then ___     = 2λ cos 2t and  ____   = −4λ sin 2t dt dt 2 d2x Substitute into ____   2  + x = 3 sin 2t: dt −4λ sin 2t + λ sin 2t = 3 sin 2t ⇒ λ = −1 Solve the auxiliary equation to find the values of m. Watch out Normally you would need to try a particular integral of the form λ sin 2t + μ cos 2t for this equation. However, in this case you are told that there is a particular integral in the form λ sin 2t. 163 M07_ALEVEL_CP2_83343_U07_147-169.indd 163 02/03/2018 12:43 Chapter 7 So a particular integral is −sin 2t. The general solution is x = A cos t + B sin t − sin 2t Since x = 0 at t = 0, A = 0. Differentiating x = B sin t − sin 2t with respect to t gives dx ___     = B cos t − 2 cos 2t dt dx Since  ___  = 1 at t = 0, 1 = B − 2 dt ⇒ B=3 Use general solution = complementary function + particular integral. Substitute the initial condition, x = 0 at t = 0, into the general solution to obtain A = 0. dx Substitute the second initial condition, ___     = 1 at dt t = 0, into the derivative of the general solution, to obtain a second equation leading to B = 3. And so x = 3 sin t − sin 2t is the required solution. Exercise E E E 7D 1 a Find the general solution to the differential equation dy d  2y ____ (5 marks)   2  + 5 ___  + 6y = 12e  x dx  dx dy b Hence find the particular solution that satisfies y = 1 and ___     = 0when x = 0. (4 marks) dx 2 a Find the general solution to the differential equation dy d  2y ____ (5 marks)   2  + 2 ___  = 12e  2x dx  dx dy b Hence find the particular solution that satisfies y = 2 and ___     = 6when x = 0. (5 marks) dx dy 1 3 Given that y = 0 and ___     = _  6 when x = 0, find the particular solution to the differential equation dx dy d  2y ___ ____ (10 marks)   2  −     − 42y = 14 dx  dx 4 a Find the general solution to the differential equation d  2y ____   2  + 9y = 16 sin x dx  dy b Hence find the particular solution that satisfies y = 1 and ___     = 8when x = 0. dx E 5 a Find the general solution to the differential equation dy d  2y 4____   2  + 4 ___  + 5y = sin x + 4 cos x dx  dx dy b Hence find the particular solution that satisfies y = 0 and ___     = 0when x = 0. dx E/P 6 a Find the general solution to the differential equation dx d   2x ____   2  − 3 ___  + 2x = 2t − 3 dt dt  b Given that x = 1 when t = 0, and x = 2 when t = 1, find a particular solution of this differential equation. E (6 marks) (6 marks) (6 marks) (6 marks) (6 marks) (6 marks) 164 M07_ALEVEL_CP2_83343_U07_147-169.indd 164 02/03/2018 12:43 Methods in differential equations E 7 Find the particular solution to the differential equation d  x ____ 2   2  − 9x = 10 sin t dt  dx that satisfies x = 2 and ___     = −1 when t = 0. dt E/P (10 marks) 8 a i Find the value of λ for which y = λt3e2t is a particular solution to the differential equation   x d ____ 2 dx   2  − 4 ___  + 4x = 3te  2t dt dt  ii Hence find the general solution to the differential equation. dx b Find the particular solution that satisfies x = 0 and ___     = 1when t = 0. dt E (6 marks) (6 marks) 9 Find the particular solution to the differential equation d  2x 25 ____2  + 36x = 18 dt  dx that satisfies x = 1 and ___     = 0.6when t = 0. dt E 10 a Find the general solution to the differential equation d  x ____ 2 dx   2  − 2 ___  + 2x = 2t  2 dt dt  dx b Hence find the particular solution that satisfies x = 1 and ___     = 3when t = 0. dt E/P 11 a Find the general solution to the differential equation dy d2y  ____2  – 3 ___  + 2y = 3e2x dx dx dy b Hence find the particular solution that satisfies y = 0, ___     = 0 when x = 0. dx E/P (6 marks) (6 marks) (7 marks) (6 marks) 12 Solve the differential equation d2y ____   2  + 9y = sin 3x dx dy subject to the boundary conditions y = 0, ___     = 0 when x = 0. dx 2 d  x dx ____ E/P 13   2  + 5 ___  + 6x = 2e  −t dt dt  dx Given that x = 0 and  ___  = 2 at t = 0, dt __ a find x in terms of t. 2 √ 3  ____ b Show that the maximum value of x is    and justify that this is a maximum. 9 Mixed exercise E (12 marks) (14 marks) (8 marks) (7 marks) 7 1 Find the general solution to the differential equation dy ___     + y tan x = 2 secx dx giving your answer in the form y = f(x). (7 marks) 165 M07_ALEVEL_CP2_83343_U07_147-169.indd 165 02/03/2018 12:43 Chapter 7 E 2 Find the general solution to the differential equation dy (1 − x2)  ___  + xy = 5x, dx −1 , x , 1 giving your answer in the form y = f(x). E (7 marks) 3 Find the general solution to the differential equation dy x ___     + x + y = 0 dx giving your answer in the form y = f(x). E 4 y satisfies the differential equation __ dy y ___     = √ x      + __ dx x Find y as a function of x. E (7 marks) 6 Find the general solution to the differential equation dy x(1 − x2)  ___  + (2x2 − 1)y = 2x3, 0 , x , 1 dx giving your answer in the form y = f(x). E/P (7 marks) 5 y satisfies the differential equation dy ___     + 2xy = x dx Find y in terms of x. E/P (7 marks) (7 marks) dy 7 Find the general solution to the equation ___     − ay = Q(x), where a is a constant, giving your dx answer in terms of a, when a Q(x) = ke  λx (k and λ are constants). (6 marks) Given that Q(x) = kx  ne  ax, where k and n are constants, b find the general solution to the differential equation. E/P E 8 Find, in the form y = f(x), the general solution to the differential equation dy π tan x  ___  + y = 2 cos x tan x, 0 , x , __     2 dx (6 marks) 9 a Find the general solution to the differential equation dy π π ___         + y tan x = e  x cos x, −  __  , x , __ 2 2 dx giving your answer in the form y = f(x). b Find the particular solution for which y = 1 at x = π. E (7 marks) (6 marks) (3 marks) dy ___ 10     − 3y = sin x dx Given that y = 0 when x = 0, find y in terms of x. (7 marks) 166 M07_ALEVEL_CP2_83343_U07_147-169.indd 166 02/03/2018 12:43 Methods in differential equations E/P 11 a Find the general solution to the differential equation dy  ___  = y sinh x dx b Find the particular solution which satisfies the condition that y = 1 when x = 0. E (4 marks) (2 marks) dy ___     = x (4 − y  2) 12 dx Given that y = 1 when x = 0, find y in terms of x. (7 marks) E d2y dy     + y = 0 13 Find the general solution to the differential equation ____   2  + ___ dx dx (6 marks) E dy d2y 14 Find the general solution to the differential equation ____   2  − 12 ___  + 36y = 0 dx dx (6 marks) E dy d2y   2  − 4  ___  = 0 15 Find the general solution to the differential equation ____ d x dx (6 marks) E/P E E d2y   2  + k2y = 0 where k is a constant, and y = 1 16 Find y in terms of k and x, given that ____ dx dy ___ and     = 1 at x = 0. dx d2y dy   2  − 2 ___  + 10y = 0 for which y = 0 17 Find the solution to the differential equation ____ dx dx dy ___ and     = 3 at x = 0. dx (8 marks) (8 marks) 18 a Find the value of k for which y = ke  2xis a particular integral of the differential equation dy d  2y (4 marks)  ____2  − 4 ___  + 13y = e  2x dx  dx b Using your answer to part a, find the general solution to the differential equation. (5 marks) 19 Find the general solution of the differential equation d  2y  ____2  − y = 4e  x dx  d 2y dy E/P 20 The differential equation ____   2  − 4  ___  + 4y = 4e2x is to be solved. dx dx a Find the complementary function. E/P b Explain why neither λe2x nor λ x e2x can be a particular integral for this equation. (7 marks) (3 marks) (2 marks) A particular integral has the form k x2e2x. c Determine the value of the constant k and find the general solution of the equation. E/P (6 marks) 21 Find the particular solution of the differential equation d  2y ____   2  + 4y = 5 cos 3t dt  dy which satisfies the initial conditions that when t = 0, y = 1 and ___     = 2. dt (12 marks) 167 M07_ALEVEL_CP2_83343_U07_147-169.indd 167 02/03/2018 12:43 Chapter 7 E E/P E/P E/P 22 a Find the values of λ, μ and k such that y = λ + μx + kxe  2xis a particular integral of the differential equation dy d  2y ____ (5 marks)   2  − 3 ___  + 2y = 4x + e  2x dx  dx b Using your answer to part a, find the general solution of the differential equation. (5 marks) dy d2y 23 a Find the solution of the differential equation 16  ____2  + 8  ___  + 5y = 5x + 23 for which y = 3 dx dx dy ___ (8 marks) and     = 3 at x = 0. dx b Show that y ≈ x + 3 for large values of x. (2 marks) d2y dy     − 6y = 3 sin 3x − 2 cos 3x for which 24 Find the solution of the differential equation ____   2  − ___ dx dx y = 1 at x = 0 and for which y remains finite as x → ∞. (8 marks) 25 x satisfies the differential equation dx d   2x ____   2  + 8 ___  + 16x = cos 4t, t > 0 dt dt  a Find the general solution of the differential equation. b 1 Find the particular solution of this differential equation for which, at t = 0, x =  _2  dx and ___     = 0. dt c Describe the behaviour of the function for large values of t. (8 marks) (5 marks) (2 marks) Challenge 1 Use the substitution z = y2 to transform the differential equation dy 1 2(1 + x2) ___   + 2xy =  __ y dx into a differential equation in z and x. By first solving the transformed equation, a find the general solution of the original equation, giving y in terms of x. b Find the particular solution for which y = 2 when x = 0. 2 a Find the general solution of the differential equation d2y dy x2  ____2  + 4x  ___ + 2y = ln x, x . 0, dx dx using the substitution x = eu , where u is a function of x. b Find the equation of the solution curve passing through the point (1, 1) with gradient 1. 168 M07_ALEVEL_CP2_83343_U07_147-169.indd 168 02/03/2018 12:43 Methods in differential equations Summary of key points 1 2 dy You can solve a first-order differential equation of the form ___    + P(x)y = Q(x)by multiplying dx every term by the integrating factor e  ∫   P(x)dx. The natures of the roots α and β of the auxiliary equation determine the general solution dy d2y     + c = 0. to the second-order differential equation a  ____2  + b ___ dx dx You need to consider three different cases: • Case 1: b  2 > 4ac The auxiliary equation has two real roots α and β (α ≠ β). The general solution will be of the form y = Ae  αx + Be  βx where A and B are arbitrary constants. • Case 2: b  2 = 4ac The auxiliary equation has one repeated root α. The general solution will be of the form y = (A + Bx)e  αx where A and B are arbitrary constants. • Case 3: b  2 < 4ac The auxiliary equation has two complex conjugate roots α and β equal to p ± qi. The general solution will be of the form y = e  px(A cos qx + B sin qx) where A and B are arbitrary constants. 3 4 A particular integral is a function which satisfies the original differential equation. dy d  2y     + cy = f(x), To find the general solution to the differential equation a ____   2  + b ___ dx  dx ____ d  2y dy • Solve the corresponding homogeneous equation a  2  + b ___  + cy = 0to find the dx  dx complementary function, C.F. • Choose an appropriate form for the particular integral, P.I., and substitute into the original equation to find the values of any coefficients. • The general solution is y = C.F. + P.I. 169 M07_ALEVEL_CP2_83343_U07_147-169.indd 169 02/03/2018 12:43 8 Modelling with differential equations Objectives After completing this chapter you should be able to: ● Model real-life situations with first-order differential equations → pages 171–175 ● Use differential equations to model simple harmonic motion → pages 175–180 ● Model damped and forced oscillations using differential equations → pages 180–186 ● Model real-life situations using coupled first-order differential → pages 186–190 equations Prior knowledge check 1 Find the particular solution to the dy y differential equation ___     = 40 − ______     50 + x dx given that y = 0 when x = 0. ← Section 7.1 2 Find the general solution to the differential d2y dy equation ____   2  + 5  ___ − 6x = 0 ← Section 7.2 dx dx 3 Find the particular solution to the differential equation dy d2y ____   2  + 2  ___ + 2y = 10 cos x dx dx dy given that ___   = 0 and y = 0 when x = 0. dx ← Sections 7.3, 7.4 Population levels of predators and their prey can be modelled using a pair of coupled first-order differential equations. → Mixed exercise Q16 170 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 170 26/02/2018 09:31 Modelling with differential equations 8.1 Modelling with first-order differential equations First-order differential equations can be used to model problems in kinematics. You have covered some of these contexts in A level mathematics and you have learned the following relationships between displacement, velocity and acceleration: ds dv v = ___   a = ___   dt dt where s is the displacement of a particle, v is the velocity and a is the acceleration at time t. You can use these relationships to construct first-order differential equations which you can solve using the standard methods of direct integration and separation of variables from A level mathematics, or the method of integrating factors covered in Chapter 7 of this book. Example 1 A particle P starts from rest at a point O and moves along a straight line. At time t seconds the acceleration, a m s−2, of P is given by 6 a = ______    , t > 0 (t + 2)2 a Find the velocity of P at time t seconds. b Show that the displacement of P from O when t = 6 is (18 − 12 ln 2) m. 6 dv a ___ = _______     dt (t + 2)2 6 _______   dt =    6(t + 2)−2 dt v =      (t + 2)2 6(t + 2)−1 v = _________     + c −1 6   = c − _____   t+2 When t = 0, v = 0: ∫ ∫ 0 = c − __   6 2  ⇒ c = 3 The velocity of P at time t seconds is 6 (3 − _____     m s−1. t + 2) 6 ds b ___ = 3 − _____     dt t+2 6 _____ s =    (3 −   t + 2 ) dt = 3t − 6 ln (t + 2) + d When t = 0, s = 0: 0 = − 6 ln 2 + d ⇒ d = 6 ln 2 s = 3t − 6 ln (t + 2) + 6 ln 2 ∫ Write the given relationship as a first-order differential equation. Integrate both sides with respect to t. Notice that this equation can be solved by direct integration. Remember to include the constant of integration. Use the initial conditions to find the value of c. Set up another first-order differential equation ds using ___     = v. dt Solve this differential equation, again using standard methods, and include a second constant of integration. You are told that the particle starts from rest at the origin so your initial conditions are s = 0 when t = 0. 171 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 171 26/02/2018 09:31 Chapter 8 When t = 6, s = 18 − 6 ln 8 + 6 ln 2 8 = 18 − 6 ln ( __ 2 ) = 18 − 6 ln 4 = 18 − 12 ln 2 The displacement of P from O when t = 6 is (18 − 12 ln 2) m, as required. Example Use the laws of logarithms to simplify your answer into the form asked for in the question. This can be done in more than one way. The working shown here uses ln 8 − ln 2 = ln (_  82 ) = ln 4 and ln 4 = ln 22 = 2 ln 2. 2 A particle P is moving along a straight line. At time t seconds, the acceleration of the particle is given by 3 a = t + __   t v, t > 0 Given that v = 0 when t = 2, show that the velocity of the particle at time t is given by the equation v = c t  3 − t  2 where c is a constant to be found. 3 − __   v = t t dt 3 __ Integrating factor = e  ∫   Pdt = e  ∫  −  t  dt = e  −3lnt = e  lnt  −3 = t  −3 So the original equation becomes dv t−3  ___ − 3t −4v = t −2 dt d ⇒ ___(t −3v) = t −2 dt ⇒ t −3v = −t −1 + c 2 ⇒ v = − t  + c t  3 Since v = 0 when t = 2, 0 = − 4 + 8c ⇒ c = _   21   So the velocity of the particle at time t is 1 3 2 given by v =  _ 2  t  − t  dv ___   Write the differential equation in the form dv   __   + Pv = Q where P and Q are functions of t. dt To solve this differential equation, you need to use the integrating factor method. ← Section 7.1 Find the integrating factor and use the standard method to find the general solution. Multiply through by t3 to find v. 172 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 172 26/02/2018 09:31 Modelling with differential equations First-order differential equations can also be used to model other real-life situations involving rates of change. You might need to construct or analyse models based on situations other than kinematics in your exam. Example 3 A storage tank initially contains 1000 litres of pure water. Liquid is removed from the tank at a constant rate of 30 litres per hour and a chemical solution is added to the tank at a constant rate of 40 litres per hour. The chemical solution contains 4 grams of copper sulphate per litre of water. Given that there are x grams of copper sulphate in the tank after t hours and that the copper sulphate immediately disperses throughout the tank on entry, a show that the situation can be modelled by the differential equation dx 3x ___     = 160 −  _______ , t > 0 100 + t dt b Hence find the number of grams of copper sulphate in the tank after 6 hours. c Explain how the model could be refined. a Litres of liquid in the tank after t hours is given by 1000 + 40t − 30t = 1000 + 10t Let x grams be the amount of copper sulphate in the tank after t hours. Concentration of copper sulphate after t hours x = ___________     grams per litre. 1000 + 10t Rate copper sulphate in = 40 × 4 = 160 grams per hour x Rate copper sulphate out = 30 × ___________     1000 + 10t 3x =  _______grams per hour 100 + t 3x dx hence ___   = 160 − _______   dt 100 + t 3 dx b ___ + _______   x = 160 dt 100 + t 3 _____ Integrating factor = e  ∫   Pdt = e  ∫    100+t  dt = e  3ln(100+t) = e  ln(100+t)  3 = (100 + t)  3 So the original equation becomes dx (100 + t)3___   + 3(100 + t)2x = 160(100 + t)3 dt d ⇒ ___( (100 + t)  3 x) = 160 (100 + t)  3 dt ⇒(100 + t)  3 x = 40 (100 + t)  4 + c When t = 0, x = 0, 0 = 4 000 000 000 + c ⇒ c = −4 × 109 After 6 hours, (106)  3 x = 40 × (106)  4 − 4 × 10  9 ⇒ x = 882 g (3 s.f.) Write down the amount of liquid in the tank after t hours. Problem-solving Work out the rate that copper sulphate enters the tank and leaves it. The rate at which the copper sulphate leaves the tank will be dependent on the concentration of copper sulphate in the tank. Write down the final differential dx equation. ___   is the rate of change of dt copper sulphate in the tank, so it is equal to the rate of copper sulphate in minus the rate of copper sulphate out. Write the differential equation in the dx form ___   + Px = Q. dt Work out the value of c from the initial conditions. 173 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 173 26/02/2018 09:31 Chapter 8 c The model could be refined to take into account the fact that the copper sulphate does not disperse immediately on entering the tank. Exercise 8A You could give any sensible suggestion here, but make sure it is in the context of the model. For example, the rate at which liquid is removed could be made to vary with the volume of liquid in the tank. 1 A particle P is moving along the x-axis in the direction of x increasing. At time t seconds, the π    , velocity of P is (t sin t) m s−1. When t = 0, P is at the origin. Show that when t = __ 2 P is 1 metre from O. 2 A particle P is moving along a straight line. Initially P is at rest. At time t seconds P has velocity v m s−1 and acceleration a m s−2 where 6t a =  _______  , t > 0 (2 + t 2)2 Find v in terms of t. P 3 A particle P is moving along the x-axis. At time t seconds P has velocity v m s−1 in the direction x increasing and an acceleration of magnitude 4e0.2t m s−2 in the direction x decreasing. When t = 0, P is moving through the origin with velocity 20 m −1 in the direction x increasing. Find: a v in terms of t b the maximum value of x attained by P during its motion. P 4 A particle, P, is moving along the x-axis. At time t seconds, the displacement of the particle from O is x m, and its velocity is v m s−1, where x __ v = e  −  2  Given that the particle is initially at the origin, find x as a function of t. E/P 5 A sports car moves along a horizontal straight road. At time t seconds the acceleration, in m s−2, is modelled using the differential equation dv  ___  − 2vt = t dt where v is the velocity of the car in m s−1. When t = 0, the car is travelling at 1 m s−1. 1 a Show that the velocity of the car at time t seconds can be written as v = __   2  (3 e  t  2 − 1). (5 marks) E/P b Find the velocity of the car after 2 seconds. (2 marks) c State, with a reason, whether the model is suitable when t = 4 seconds. (2 marks) 6 A raindrop falls vertically from rest through mist. Water condenses on the raindrop as it falls. You are given that the motion of the raindrop may be modelled by the equation dv (t + 4)  __ dt  + 4v = 9.8(t + 4) where t is the time in seconds and v is the velocity of the raindrop in m s−1. 49 (t + 4)  5 − c ____________   where c is a constant to be found. a Show that v =      25 (t + 4)  4 b Find the velocity of the raindrop after 5 seconds. (6 marks) (2 marks) c By considering the velocity for large values of t, suggest one criticism of the model. (1 mark) 174 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 174 26/02/2018 09:31 Modelling with differential equations E/P 7 A gas storage tank initially contains 500 cm3 of helium. The helium leaks out at a constant rate of 20 cm3 per hour and a gas mixture is added to the tank at a constant rate of 50 cm3 per hour. The gas mixture contains 5% oxygen and 95% helium. Given that there is x cm3 of oxygen in the tank after t hours and that the oxygen immediately mixes throughout the tank on entry, a show that the situation can be modelled by the differential equation dx 2x  ___  = 2.5 −  _______  50 + 3t dt b Hence find the volume of oxygen in the tank after 4 hours. (4 marks) c Explain how the model could be refined. (1 mark) (5 marks) 8.2 Simple harmonic motion You can use second – order differential equations to model particles moving with simple harmonic motion. ■■ Simple harmonic motion (S.H.M.) is motion in which the acceleration of a particle P is always towards a fixed point O on the line of motion of P. The acceleration is proportional to the displacement of P from O. The point O is called the centre of oscillation. We write ẍ = −ω2x The minus sign means that the acceleration is always directed towards O. This can be shown on a diagram ω2x O P x dv Given that ẍ  =  ___you can use the chain rule to dt derive a relationship between x, v and a which will allow you to solve the second-order differential equation for simple harmonic motion: dv dv dx   ẍ  =  ___=  ___× ___ dt dx dt dx Since ___   = v, you can write acceleration as follows: dt dv ■■ ẍ  = v  ___ dx Example Links The constant of proportionality in this equation is ω 2, where ω is the angular velocity of the particle. → FM2, Chapter 1 Online Explore simple harmonic motion using GeoGebra. Notation You can use ‘dot’ notation to indicate differentiation with respect to time. dx d 2x ẋ  = ____     and ẍ  =  ____2 . So if x is used to denote dt dt displacement, then ẋ represents velocity and ẍ  represents acceleration. 4 A particle P moves with simple harmonic motion about a point O. Given that the maximum displacement of the particle from O is a, a show that v  2 = ω  2 ( a  2 − x  2 ), where v is the velocity of the particle and ω2 is a constant. b show that x = a sin (ωt + α), where α is an arbitrary constant. 175 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 175 26/02/2018 09:31 Chapter 8 Write down the equation for simple harmonic dv motion and use ẍ  = v   __  to write a differential dx equation involving v and x only. dv a ẍ = − ω  2 x and ẍ = v  ___ dx dv So v  ___ = − ω  2 x dx ∫   v dv = − ∫   ω2 x dx Separate the variables. __1 2 2 1 2  __ 2  v  = −   2  ω  x  + c v = 0 when x = a, __1 2 2 1 2 2 therefore 0 = −  __ 2  ω  a  + c ⇒ c =   2  ω  a  __1 2 __1 2 2 __1 2 2 Hence   2  v  = −   2  ω  x  +   2  ω  a  ⇒ v  2 = ω  2 (a  2 − x  2 ) as required. b ẍ + ω2x = 0 d 2x  ____   + ω2x = 0 dt2 m2 + ω2 = 0 ⇒ m = ±ωi x = Acos ωt + β sin ωt = R sin (ωt + α) This has maximum amplitude R. Since the maximum amplitude of the particle is a: x = a sin (ωt + α) as required. If the particle is at its equilibrium position at time t = 0, then α = 0. This is a second-order linear differential equation. d2x dx Write it in the form a  ____2  + b  ___  + cx = 0 and dt dt then solve the corresponding auxiliary equation. ← Secton 7.2 Problem-solving Notation The period of the motion is the time the particle takes to complete one oscillation. The amplitude of the motion is the maximum displacement from the origin. If the particle is at its stationary point, i.e. its π maximum displacement, when t = 0, then α = __    . 2 t a t a –a The velocity of the particle will be zero when the particle is at maximum displacement. You are given information about the maximum displacement, so write the general solution in a form where you can determine the amplitude. The arbitrary constants become R and a, where A R 2 = A2 + B 2 and tan a =  __  B You can see from the above example that the path of a particle P under simple harmonic 2π motion is a sine wave with period ___   ω  and amplitude a. O Integrate both sides and add a constant of integration. 2π x ω O 2π ω x –a 176 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 176 26/02/2018 09:31 Modelling with differential equations Example 5 A particle is moving along a straight line. At time t seconds its displacement, x m from a fixed d   2x point O is such that  ____2  = −4x. dt  Given that at t = 0, x = 1 and the particle is moving with velocity 4 m s−1, a find an expression for the displacement of the particle after t seconds b hence determine the maximum displacement of the particle from O. d  2 x a ____   + 4x = 0 d t  2 m2 + 4 = 0 ⇒ m = ±2i x = A cos 2t + B sin 2t x = 1 when t = 0 ⇒ A = 1 dx v = ___   = − 2 sin 2t + 2B cos 2t dt v = 4 when t = 0 ⇒ B = 2 Hence x = cos 2t + 2 sin 2t __ b x = √ 5  sin (2t + 0.4636...) Hence maximum displacement is __ √ 5  metres. Example 6 This is a second–order linear differential equation. d2y dy Write it in the form a  ____2  + b  ___  + c = 0. dx dx Write down and solve the auxiliary equation. Write down the general solution. There are two purely imaginary roots so the general solution is in the form y = A cos qx + B sin qx ← Section 7.2 Substitute the initial condition for x: 1 = A cos(0) + B sin(0) Differentiate x to find an expression for v and use the initial condition for v: 4 = −2 sin(0) + 2B cos(0) Write your solution to a in the form Rsin(θ + α). A particle, P, is attached to the ends of two identical elastic springs. The free ends of the springs are attached to two points A and B. The point C lies between A and B such that ABC is a straight line and AC ≠ BC. The particle is held at C and then released from rest. At time t seconds, the displacement of the particle from C is x m and its velocity is v m s−1. The subsequent motion of the particle can be described by the differential equation x ̈  = − 25x. a Describe the motion of the particle. Given that x = 0.4 and v = 0 when t = 0, b solve the differential equation to find x as a function of t c state the period of the motion and calculate the maximum speed of P. a The particle moves with simple harmonic motion. b ẍ + 25x = 0 m2 + 25 = 0 ⇒ m = ±5i x = A cos 5t + B sin 5t x = 0.4 when t = 0 ⇒ A = 0.4 ẋ = − 5A sin 5t + 5B cos 5t v = 0 when t = 0 ⇒ B = 0 Hence x = 0.4 cos 5t Write down and solve the auxiliary equation. Then write down the general solution. Find the values of A and B. 177 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 177 26/02/2018 09:31 Chapter 8 2π c Period of motion is  ___  seconds 5 ẋ = − 2 sin 5t, so maximum speed = 2 m s−1. Exercise E 8B 1 A particle is moving along a straight line. At time t seconds its displacement, x m, from a fixed d  2x   2  = −9x. point O is such that ____ dt  a Describe the motion of the particle. (1 mark) Given that at t = 0, x = 2 and the particle is moving with velocity 3 m s−1, E b find an expression for the displacement of the particle after t seconds. (7 marks) c Hence determine the maximum distance of the particle from O. (2 marks) 2 A particle is moving with simple harmonic motion. After t seconds its displacement, x m from a fixed point O is such that ẍ = − 16x. Given that at t = 0, x = 5 and ẋ = 2, E/P a find an expression for the displacement of the particle after t seconds. (7 marks) b Hence determine the period of motion and the maximum distance of the particle from O. (3 marks) 3 A particle moves along a straight line. The particle moves such that its acceleration, in m s−2, acts towards a fixed point O and is proportional to its distance, x m, from O. a Describe the motion of the particle. (1 mark) Given that the acceleration of the particle is −5 m s−2 when x = 1, b write down a differential equation to describe the motion of the particle. (2 marks) If the velocity and displacement of the particle at time t = 0 are 6 m s−1 and 5 m respectively, E/P c find an expression for the displacement of the particle after t seconds. (7 marks) d Hence find the maximum distance of the particle from O. (2 marks) 4 A particle moves along a straight line such that its acceleration, in m s−2, acts towards a fixed point O and is proportional to its distance, x m, from O. d  2x The equation of motion is given as  ____2  = − kx, t > 0 dt  where k is a constant. Given that the acceleration is −7 m s−2 when x = 2, a write down the value of k. (1 mark) b Use your answer to part a to find x as a function of t given that x = 6 and ẋ = 1 at time t = 0. (7 marks) c Hence find, correct to 2 decimal places, the period of the motion. (2 marks) 178 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 178 26/02/2018 09:31 Modelling with differential equations E/P 5 A small rowing boat is floating on the surface of the sea, tied to a pier. The boat moves up and down in a vertical line, such that the vertical displacement, x m, from its equilibrium point satisfies the equation d  2x ____   2  = − 2.25x, t > 0 dt  Given that the maximum displacement is 1.3 m and it occurs when t = 2 seconds, a find an expression for x in terms of t in the form x = A sin ωt + B cos ωt where A, B and ω are constants to be found. Give A and B correct to three decimal places. (7 marks) E/P b State the time elapsed between the boat being at its highest and lowest points. (2 marks) c Criticise the model in terms of the motion of the boat for large values of t. (1 mark) 6 A particle P is attached to one end of a light elastic spring. The other end of the spring is fixed to a point A on the smooth horizontal surface on which P rests. The particle is held at rest with AP = 0.9 m and then released. At time t seconds the displacement of the particle from A is x m. The motion of the particle can be modelled using the equation ẍ  = − 200x. a State the type of motion exhibited by the particle P. (1 mark) Given that x = 0.3 and the particle is at rest when t = 0, E b solve the differential equation to find x as a function of t (7 marks) c find the period and amplitude of the motion (3 marks) d calculate the maximum speed of P. (2 marks) 7 A particle P is attached to one end of a light elastic spring. The other end of the spring is fixed to a point O on the smooth horizontal surface on which P rests with OP = 2.6 m and then 100 released. The motion of P can be described using the equation ẍ  = −  ___ 0.64 x, where x m is the displacement of P from O at time t seconds. Given that x = 1 and the particle is at rest at time t = 0, E a solve the differential equation to find x as a function of t (7 marks) b find the period of motion. (2 marks) 8 A smooth cylinder is fixed with its axis horizontal. A piston of mass 2.5 kg is inside the cylinder, attached to one end of the cylinder by a spring of natural length 50 cm. The piston is held at rest in the cylinder with the spring compressed to a length of 42 cm. The piston is then released. The spring can be modelled as a light elastic spring and the piston can be modelled as a particle. The motion of P can be described using the equation ẍ  = − 320x, where x cm is the compression of the spring from its natural length at time t seconds. Given that x = 8 and v = 0 when t = 0, E/P a solve the differential equation to find x as a function of t (7 marks) b find the period of the resulting oscillations. (2 marks) 9 A pendulum P is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point A on a ceiling. The pendulum hangs in equilibrium at a point B vertically below A. The pendulum is then moved through a horizontal distance of 15 cm and released from rest. 179 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 179 26/02/2018 09:31 Chapter 8 250 The subsequent motion can be modelled using the equation ẍ  = −  ___ 3  x where x cm is the horizontal displacement from the vertical at time t seconds and t > 0. a Solve the differential equation to find x as a function of t. (7 marks) b Find the period and amplitude of the motion. (3 marks) Anton says that since the pendulum describes simple harmonic motion, the model suggests that the pendulum will stay swinging forever. c Suggest a refinement of the model in light of this statement. (1 mark) 8.3 Damped and forced harmonic motion You can refine the model for simple harmonic motion by adding an additional force which is proportional to the velocity of the particle. When this force acts so as to slow the particle down it is known as a damping force, and the motion of the particle is known as damped harmonic motion. ■■ For a particle moving with damped harmonic motion   2x d ___ __  + ω  2x = 0   2  + k   dx dt dt  where x is the displacement from a fixed point at time t, and k and ω  2 are positive Links constants. There are three separate cases corresponding to the auxiliary equation having distinct real, equal or complex roots. Notation You could also write this as x ̈  + kẋ + ω  2x = 0. This is an example of a second-order homogeneous differential equation. You can solve equations like this by considering the auxiliary equation. ← Section 7.2 When k 2 . 4ω 2 there are two distinct real roots for the auxiliary equation. This is known as heavy damping. In this case there will be no oscillations performed as the resistive force is large compared with the restoring force. When k 2 = 4ω 2 the auxiliary equation has equal roots. This is known as critical damping. Again there will be no oscillations performed. When k 2 , 4ω 2 the auxiliary equation has complex roots. This is known as light damping and is the only case where oscillations are seen. The amplitude of the oscillations will decrease exponentially over time. For heavy and critical damping the exact nature of the motion will depend on the initial conditions given. For light damping the period of the observed oscillations can be calculated. Example 7 A particle P of mass 0.5 kg moves in a horizontal straight line. At time t seconds, the displacement of P from a fixed point, O, on the line is x m and the velocity of P is v m s−1. A force of magnitude 8x N acts on P in the direction PO. The particle is also subject to a resistance of magnitude 4v N. When t = 0, x = 1.5 and P is moving in the direction of increasing x with speed 4 m s−1. d2x dx a Show that  ____2  + 8  ___  + 16x = 0 b Find the value of x when t = 1. dt dt 180 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 180 26/02/2018 09:31 Modelling with differential equations a x 8x + 4v O Draw a diagram to show the situation. x F = ma .. −(8x + 4v) = 0.5x . 0.5ẍ + 4x + 8x = 0 d2x dx ____   2  + 8  ___  + 16x = 0 dt dt b Auxiliary equation: m2 + 8m + 16 = 0 (m + 4)2 = 0 m = −4 General solution is x = (A + Bt)e−4t t = 0, x = 1.5 ⇒ 1.5 = A dx v = ___     = Be−4t − 4(A + Bt)e−4t dt t = 0, v = 4 ⇒ 4 = B − 4A B = 4 + 4A = 4 + 4 × 1.5 = 10 So x = (1.5 + 10t)e−4t t = 1 ⇒ x = 11.5e−4 = 0.2106… When t = 1, x = 0.211 (3 s.f.) The auxiliary equation has equal roots. This means that the general solution is in the form x = (A + Bt)eαt, where α is the root. ← Section 7.2 Use the initial conditions given in the question to obtain values for A and B. Problem-solving This is an example of critical damping: y O x Online Explore damped harmonic motion using GeoGebra. Example 8 A particle P hangs freely in equilibrium attached to one end of a light elastic string. The other end of the string is attached to a fixed point A. The particle is now pulled down and held at rest in a container of liquid which exerts a resistance to motion on P. P is then released from rest. While the string remains taut and the particle in the liquid, the motion can be modelled using the equation dx d  2x ____   2  + 6k  ___  + 5k  2x = 0, where k is a positive real constant dt dt  Find the general solution to the differential equation and state the type of damping that the particle is subject to. Auxiliary equation: m2 + 6km + 5k2 = 0 (m + 5k)(m + k) = 0 m = −5k or −k General solution is x = Ae−5kt + Be−kt The auxiliary equation has two distinct real roots so the particle is subject to heavy damping. If the auxiliary equation has distinct real roots α and β, then the general solution is in the form x = Aeαt + Beβt 181 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 181 26/02/2018 09:31 Chapter 8 Example 9 One end of a light elastic spring is attached to a fixed point A. A particle P is attached to the other end and hangs in equilibrium vertically below A. The particle is pulled vertically down from its equilibrium position and released from rest. A resistance proportional to the speed of P acts on P. The equation of motion of P is given as dx d   2x ____   2  + 2k  ___  + 2k  2x = 0 dt dt  where k is a positive real constant and x is the displacement of P from its equilibrium position. a Find the general solution to the differential equation. b Write down the period of oscillation in terms of k. a Auxiliary equation: m2 + 2km + 2k 2 = 0 ______________ 2 − 4 × 2k 2     −2k ± √ 4k m = _____________________           = −k ± ik 2 So x = e−kt (A cos kt + B sin kt) 2π b Period =  ___  k If the auxiliary equation has complex roots p ± qi, then the general solution is in the form x = e  pt (A cos qt + B sin qt) (A cos kt + B sin kt) can be written as R cos (kt + e) 2π to give a period of ___     k Problem-solving This is an example of light damping: y O x The displacement of the particle will oscillate with a reducing amplitude. You can investigate the motion of a particle which is subject to the same two forces as above but is also forced to oscillate with a frequency other than its natural one. This type of motion is called forced harmonic motion. ■■ For a particle moving with forced harmonic motion d   2x dx ____     + k   ___  + ω  2x = f(t) dt dt  2 where x is the displacement from a fixed point at time t, and k and ω  2are positive constants. Links This is an example of a second-order non-homogeneous differential equation. You will need to find the solution to the corresponding homogeneous equation, then add a particular integral, the form of which will depend on f(t). ← Section 7.3 182 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 182 26/02/2018 09:31 Modelling with differential equations Example 10 A particle P of mass 1.5 kg is moving on the x-axis. At time t the displacement of P from the origin O is x metres and the speed of P is v m s−1. Three forces act on P, namely a restoring force of magnitude 7.5x N, a resistance to the motion of P of magnitude 6v N and a force dx of magnitude 12 sin t N acting in the direction OP. When t = 0, x = 5 and ___     = 2. dt dx d2x ____ ___ b Find x as a function of t. a Show that   2  + 4     + 5x = 8 sin t. dt dt c Describe the motion when t is large. a F = ma 2 d x − 7.5x − 6     + 12 sin t = 1.5  ____   dt dt2 2 d x dx  ____   + 5x + 4  ___  = 8 sin t dt dt2 dx d2x ____ ___   2  + 4     + 5x = 8 sin t dt dt dx ___ 0 b Auxiliary equation: m2 + 4m + 5 = ___________ __ 2    − 4 × 5  ________ −4 ± √ 4 −4 ± i√ 4  ________________ m =        =     = −2 ± i 2 2 Complementary function: xc = e−2t (A cos t + B sin t) Particular integral: try x = p sin t + q cos t dx ___     = p cos t − q sin t dt d2x ____   2  = −p sin t − q cos t dt So (−p sin t − q cos t) + 4(p cos t − q sin t) + 5(p sin t + q cos t) = 8 sin t (−p − 4q + 5p) sin t + (−q + 4p + 5q) cos t = 8 sin t Equating coefficients of cos t: 4p + 4q = 0 ⇒ p + q = 0 Equating coefficients of sin t: −4q + 4p = 8 ⇒ p − q = 2 So p = 1 and q = −1 So the particular integral is x = sin t − cos t and the general solution is x = e−2t (A cos t + B sin t) + sin t − cos t t = 0, x = 5 ⇒ 5 = A − 1 so A = 6 x = e−2t (A cos t + B sin t) + sin t − cos t dx ___     = −2e−2t (A cos t + B sin t) + e−2t (−A sin t + B cos t) + cos t + sin t dt dx t = 0, ___     = 2 ⇒ 2 = −2A + B + 1 so B = 2 + 12 − 1 = 13 dt So x = e−2t (6 cos t + 13 sin t) + sin t − cos t c As t → ∞, e−2t → 0 So x → sin t − cos t = R sin (t − a) __ _____ R = √ 1 + 1 = √ 2 , π tan a = 1 ⇒ a = __     2 __ π    ) x = √ 2  sin (t − __ 4 __ For large values of t, the motion is S.H.M. with amplitude √ 2  and period 2π. Online Explore forced harmonic motion using GeoGebra. This is the general solution to the corresponding homogeneous differential equation. General solution = complementary function + particular integral Use the initial conditions given in the question to obtain values for A and B. Write sin t – cos t in the form R sin(θ – α) 183 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 183 26/02/2018 09:31 Chapter 8 Example 11 A particle P is attached to end A of a light elastic string AB. Initially the particle and the string lie at rest on a smooth horizontal plane. At time t = 0, the end B of the string is set in motion and moves with constant speed U in the direction AB, and the displacement of P from A is x. Air resistance acting on P is proportional to its speed. The subsequent motion can be modelled by the differential equation dx d  2x ____   2  + 2k  ___  + k  2x = 2kU dt dt  Find an expression for x in terms of U, k and t. Auxiliary equation: m2 + 2km + k2 = 0 (m + k)2 = 0 m = −k Complementary function: x = (A + Bt)e−kt Particular integral: try x = a . .. x= x = 0 2U So k 2a = 2kU ⇒ a = ____     k 2U     General solution is x = (A + Bt)e−kt + ____ k 2U 2U ____ t = 0, x = 0 ⇒ 0 = A +     ⇒ A = −  ____  k k . x = −k(A + Bt)e−kt + Be−kt . t = 0 ⇒ x= U U = −kA + B B = U + kA = −U 2U 2U So x = ( −  ____ − Ut)e−kt + ____     k k Exercise The right–hand side of the differential equation is constant, so try a constant for the particular integral. Use the initial conditions given in the question to obtain values for A and B. 8C 1 A particle P is moving in a straight line. At time t, the displacement of P from a fixed point on the line is x. The motion of the particle is modelled by the differential equation dx d2x ____   2  + 4  ___  + 8x = 0 dt dt When t = 0, P is at rest at the point where x = 2. a Find x as a function of t. π b Calculate the value of x when t = __     3 c State whether the motion is heavily, critically or lightly damped. 2 A particle P is moving in a straight line. At time t, the displacement of P from a fixed point on the line is x. The motion of the particle is modelled by the differential equation d2x dx ____   2  + 8  ___  + 12x = 0 dt dt When t = 0, P is at rest at the point where x = 4. Find x as a function of t. 184 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 184 26/02/2018 09:31 Modelling with differential equations E/P 3 A particle P is moving in a straight line. At time t, the displacement of P from a fixed point on the line is x. The motion of the particle is modelled by the differential equation d2x dx ____   2  + 2  ___  + 6x = 0 dt dt When t = 0, P is at rest at the point where x = 1. a Find x as a function of t. (6 marks) The smallest value of t, t . 0, for which P is instantaneously at rest is T. b Find the value of T. E/P E/P (2 marks) 4 A particle P is attached to one end of a light elastic spring. The other end of the spring is attached to a fixed point A and P hangs freely in equilibrium vertically below A. At time t = 0, P is projected vertically downwards with speed u. A resistance proportional to the speed of P acts on P. The motion of P can be modelled using the differential equation d  2x dx ____   2  + 4k  ___  + 4k  2x = 0 dt dt  where x is the displacement of P from its equilibrium position at time t and k is a positive constant. a Find an expression for x in terms of u, t and k. (6 marks) b Find the time at which P comes to instantaneous rest. (2 marks) 5 A particle of mass 2 kg moves in a horizontal straight line. At time t seconds, the displacement of P from a fixed point O is x metres and the speed of P is v m s−1. A force of magnitude 6x N acts on P in the direction PO. The particle is also subject to a resistance of magnitude 2v N. When t = 0, x = 1 and P is moving in the direction of increasing x with speed 2 m s−1. a Show that ẍ  + ẋ  + 3x = 0. (2 marks) b Solve the differential equation in part a to find x as a function of t. (8 marks) c Find the value of x when t = 2. (2 marks) d Describe the motion of P for large values of t. (1 mark) E/P 6 A particle P is attached to end A of a light elastic spring AB. The end B of the spring is oscillating. At time t the displacement of P from a fixed point is x. When t = 0, x = 0 and dx k ___     =  __  where k is a constant. Given that x satisfies the differential equation dt 5 d2x  ____2  + 9x = k cos t dt find x as a function of t. (8 marks) E/P 7 A particle P is attached to end A of a light elastic spring AB. Initially the spring and the particle lie at rest on a horizontal surface. The end B of the spring is then moved in a straight line in the direction AB with constant speed U. As P moves it is subject to a resistance proportional to its speed. The extension, x, in the spring can be modelled using the differential equation dx d  2 x ____   2  + 5k  ___  + 6 k  2 x = 5kU dt d t  Find an expression in x in terms of t. (8 marks) 185 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 185 26/02/2018 09:31 Chapter 8 E/P 8 An engineering student is designing an oscillating piston that is attached to a vertical rod. The piston is to be released from rest from a point half way up the rod so that it oscillates in a vertical line. The vertical displacement, x metres, of the top of the piston below its initial position at time t seconds is modelled by the differential equation, dx d  2x 2  ___2  + 3  ___  + x = 100 cos t, t > 0 dt dt  a Show that a particular solution to the differential equation is x = 30 sin t − 10 cos t (3 marks) b Hence find the general solution to the differential equation. (5 marks) c Use the model to find, to the nearest centimetre, the vertical distance of the top of the piston from its initial position 5 seconds after it is released. (4 marks) 8.4 Coupled first-order simultaneous differential equations In some real-life situations rates of change of two variables are connected. For example, in a predator– prey model, the rate of change of the population of bears might be dependent on both the number of bears and the number of fish in a river. Simultaneously, the rate of change of the number of fish might be dependent on both the number of bears and the number of fish. In this case, you have two dependent variables, the number of bears and fish, and one independent variable, time. You can set up two first-order differential equations to model the rates of change of the numbers of bears and fish. Letting the number of bears at time t be x and the number of fish at time t be y, you can write: dx   ___  = ax + by + f(t) Notation If f(t) and g(t) are both zero, the dt system is said to be homogenous. dy   ___  = cx + dy + g(t) dt These equations are called coupled first-order linear differential equations and you can solve them simultaneously to find x and y as functions of t. ■■ You can solve coupled first-order linear differential equations by eliminating one of the dependent variables to form a second-order differential equation. Example 12 At the start of the year 2010, a survey began on the numbers of bears and fish on a remote island in Northern Canada. After t years the number of bears, x, and the number of fish, y, on the island are modelled by the differential equations dx ___ (1)     = 0.3x + 0.1y dt dy ___     = − 0.1x + 0.5y (2) dt 2 dx d  x a Show that ____   2  − 0.8  ___  + 0.16x = 0. dt dt  b Find the general solution for the number of bears on the island at time t. c Find the general solution for the number of fish on the island at time t. d At the start of 2010 there were 5 bears and 20 fish on the island. Use this information to find the number of bears predicted to be on the island in 2020. e Comment in the suitability of the model. 186 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 186 26/02/2018 09:31 Modelling with differential equations a dx y = 10  ___ − 3x dt dy dx d2x ___ ____   = 10   2  − 3  ___ dt dt dt dx   2x d dx ____ ___ 10   2  − 3   = −0.1x + 0.5(10  ___ − 3x) dt dt dt  2 d   x dx 10  ____   − 8  ___ + 1.6x = 0 dt dt  2 d  2x dx ____ ___   2  − 0.8   + 0.16x = 0 dt dt  b m  2 − 0.8m + 0.16 = 0 ⇒ m = 0.4 Hence x = Ae  0.4t + Bte  0.4t c  Rearrange equation (1) to make y the subject. Differentiate both sides with respect to t. dy Substitute your expressions for y and ___     into dt equation (2) to form a second–order differential equation in x and t. Rearrange into the correct form. Write down the auxiliary equation and solve it. dx ___ = 0.4Ae  0.4t + 0.4Bte  0.4t + Be  0.4t dt dx y = 10  ___ − 3x from equation (1) dt y = 10(0.4Ae  0.4t + 0.4Bte  0.4t + Be  0.4t) − 3(Ae  0.4t + Bte  0.4t) y = Ae  0.4t + 10Be  0.4t + Bte  0.4t d At t = 0, x = 5 ⇒ A = 5 At t = 0, y = 20 ⇒ B = 1.5 x = 5e  0.4t + 1.5te  0.4t At time t = 10: x = 5e  4 + 15e  4 = 1092 The model predicts there will be 1092 bears on the island by 2020. e The model predicts the number of bears (and the number of fish) will grow without limit so it is unlikely to be realistic. Use the general form of the solution of a differential equation with a repeated root. Problem-solving You do not need to go through the whole process to find an expression for y in terms of t. If you differentiate your answer to part b with respect dx to t you will have expressions for x and ___     in dt terms of t only. You can substitute these into equation (1) and simplify to obtain the answer. Use the initial conditions to find A and B. Substitute t = 10 into the equation for x. Round to the nearest whole number. Example 13 Two barrels contain contaminated water. At time t seconds, the amount of contaminant in barrel A is x ml and the amount of contaminant in barrel B is y ml. Additional contaminated water flows into barrel A at a rate of 5 ml per second. Contaminated water flows from barrel A to barrel B and from barrel B to barrel A through two connecting hoses, and drains out of barrel A to leave the system completely. The system is modelled using the differential equations dx 4 1 ___ (1)     = 5 +  _9 y − _  7 x dt dy 3 4  ___  =  __ x − _  9 y (2) dt 70 d   2y dy Show that 630  ____2  + 370  ___  + 28y = 135. dt dt  187 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 187 26/02/2018 09:31 Chapter 8 dy __   dt 4 3     x from equation (2) + __     y = ___ 9 70 70 dy 280  ___   __ + _____     y = x (3) 27 3 dt d   2y 280 __ dy dx 70 ___  __ = ___       2  + _____     (4)   dt 3 dt  27 dt dy dy 280 70 d  2y _____ 4 280 ___ 1 70 ___ ___         + _____    (___      ____       = 5 + __    y − __     y   +   2 7 3 dt 27 ) 9 3 dt  27 dt d2y _____ dy 40 4 280 dy 70  ____   +      ___  = 15 +  __ y − 10  ___  −  ___  y dt 9 dt 9 3 dt2 2y dy d dy   + 280  ___  = 135 + 12y − 90  ___  − 40y 630  ____ dt dt dt2 ___ d   2y dy __ ⇒ 630   2  + 370     + 28y = 135 as required. dt dt  Exercise Rearrange (2) to make x the subject. dx Differentiate (3) to find  ___ . dt Substitute (3) and (4) into (1), simplify coefficients and remove fractions. Write your answer in the form required in the question. 8D P 1 Find the particular solutions to the differential equations dx ___     = x + y dt dy ___     = x − y dt given that x = 1 and y = 2 at t = 0. P 2 a Find the general solutions to the differential equations dx ___     = x + 5y dt dy ___     = − 3y − x dt b Given that at time t = 0, x = 1 and y = 2, find the particular solutions. P 3 A system of differential equations is given as dx ___     = 2x − 3y − 2 dt dy ___     = x + y − 1 dt Given that x = 0 and y = 1 when t = 0, find the particular solutions to the system of differential equations. E/P 4 At the start of 2012, a survey began on the number of sand foxes and the number of meerkats on a remote desert island. After t years, the number of sand foxes, x, and the number of meerkats, y, on the island are modelled by the differential equations dx ___     = 0.2x + 0.2y dt dy ___     = − 0.5x + 0.4y dt 188 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 188 26/02/2018 09:31 Modelling with differential equations dx d   2x a Show that ____   2  − 0.6  ___  + 0.18x = 0. dt dt  b Show that the general solution for the number of sand foxes at time t is x = e  αt(A cos βt + B sin βt) where α and β are constants to be found. (3 marks) (4 marks) c Find the general solution for the number of meerkats at time t. Give your answer in the form y = Pe  αt( Q cos βt + R sin βt) where P is a constant to be found (3 marks) and Q and R are functions of A and B. d Given that there were initially 3 sand foxes and 111 meerkats on the island, during which year does the model predict that the meerkats die out? (5 marks) E/P e How many sand foxes will there be when the meerkats die out? (1 mark) f Use your answers to parts d and e to comment on the model. (1 mark) 5 A tank of water contains two different types of chemical that react with each other. The rates of change of each chemical can be modelled using the differential equations dx ___     = − 3x + 2y dt dy ___     = − 2x + y dt where x is the number of litres of chemical X and y is the number of litres of chemical Y at time t hours. Initially there is one litre of chemical X and two litres of chemical Y in the tank. a Show that the solutions to the differential equations can be written as x = Pe  −t y = Qe  −t where P and Q are functions of t to be found. (8 marks) b Find, correct to three significant figures, the amount of each chemical at time t = 2 hours. (2 marks) c Use the model to describe what happens to the amount of each chemical as t gets large. (2 marks) E/P 6 A freely hanging pendulum oscillates in both the x and y directions. At time t, the rates of change of the x and y displacements are given by the differential equations dx ___     = − 4y dt dy ___     = 4x dt a Show that the pendulum describes simple harmonic motion in the y direction. (3 marks) Given that the initial displacement of the pendulum is x = 4 and y = 5, b find the particular solutions to the system of differential equations. (6 marks) 189 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 189 26/02/2018 09:31 Chapter 8 E/P 7 Joanna is investigating how a harmful substance from pollution is absorbed by the human body. She finds that the substance enters the body in the bloodstream and is transferred between the blood and the bodily organs. The harmful substance is then expelled from the body through, for example, sweat. The amount of the substance in the blood, x mg, and the organs, y mg, at time t days, can be modelled using the differential equations dx ___     = − 0.03x + 0.01y + 50 dt dy ___     = 0.01x − 0.03y dt dx d  2x (6 marks) a Show that ____   2  + 0.06  ___  + 0.0008x = 1.5 dt dt  b Hence find the general solutions to the system of differential equations in the form x = f(t) y = g(t) (8 marks) c Describe what happens to the amount of the substance in the blood and the organs as t gets large. (2 marks) E/P 8 A biologist is examining the rates of change of nutrients in both tree roots and the surrounding soil. Nutrients pass from the tree roots into the soil and from the soil into the tree roots. Nutrients also enter the system through both the roots and the soil and escape from the system in the same way. The biologist believes that the amount of nutrients in the roots, x, and the amount of nutrients in the soil, y, at time t hours, can be modelled using the differential equations dx ___     = −2x + y + 1 dt dy ___     = 4x + y + 2 dt d  2x dx (4 marks) a Show that ____   2  +  ___  − 6x = 1. dt  dt b Find the general solutions to the system of differential equations. (6 marks) c Using your answers to part b, comment on the suitability of the biologist’s model. Challenge A closed environment supports populations of owls and field mice. At time t months, the sizes of each population are x and y respectively. The situation is modelled by the pair of differential equations dy y  2   ___  = 2y −   _____  − 60x 6000 dt dx ___     = 0.02y − x dt Find the number of owls and the number of field mice such that the population of both is stable. (2 marks) Hint A population is stable when its rate of growth is zero. 190 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 190 26/02/2018 09:31 Modelling with differential equations Mixed exercise E 8 1 A particle P moves along the x-axis in the direction of x increasing. At time t seconds, the velocity of P is v m s−1 and its acceleration is 20 t e−t2 m s−2. When t = 0 the velocity of P is 8 m s−1. Find: a v in terms of t (4 marks) b the limiting velocity of P. (3 marks) E 2 A particle P moves along a straight line. Initially P is at rest at a point O on the line. At time 18 t seconds, where t > 0, the acceleration of P is ________     m s−2 directed away from O. (2t + 3)3 (5 marks) Find the value of t for which the speed of P is 0.48 m s−1. E 3 A car moves along a horizontal straight road. At time t seconds the acceleration of the car is 100 ________     m s−2 in the direction of motion of the car. When t = 0, the car is at rest. Find: (2t + 5)2 a an expression for v in terms of t (4 marks) b the distance moved by the car in the first 10 seconds of its motion. E 4 A particle P is moving in a straight line with acceleration cos2 t m s−2 at time t seconds. The particle is initially at rest at a point O. a Find the speed of P when t = π. π __ 1 __ b Show that the distance of P from O when t =     is   64 (π 2 + 8) m. 4 E E/P (4 marks) (4 marks) (4 marks) 5 A particle P is moving along the x-axis. At time t seconds, P has velocity v m s−1 in the direction 2t + 3 x increasing and an acceleration of magnitude  ______  m s−2 in the direction x increasing. t+1 When t = 0, P is at rest at the origin O. Find: a v in terms of t (4 marks) b the distance of P from O when t = 2. (4 marks) 6 A colony of bacteria reproduces in a laboratory jar. At time t = 0, the volume of bacteria is 1 cm3. Scientist Steve suggests that the rate of growth of the bacteria can be modelled using the differential equation dV  ___ = 2t + 3V + 5 dt where t is the time in hours and V is the volume in cm3. E/P a Show that V = At + B + C e  3t where A, B and C are exact constants to be found. (6 marks) b Find the volume of bacteria after 2 hours. (2 marks) c Give one criticism of Steve’s model and suggest one refinement he could make. (2 marks) 7 A fluid reservoir initially contains 10 000 litres of unpolluted fluid. The reservoir is leaking at a constant rate of 200 litres per day. It is suspected that contaminated fluid flows into the reservoir at a constant rate of 300 litres per day and that the contaminated fluid contains 4 grams of contaminant in every litre of fluid. 191 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 191 26/02/2018 09:31 Chapter 8 It is assumed that the contaminant instantly disperses throughout the reservoir upon entry. Given that there are x grams of contaminant in the reservoir after t days, a Show that the situation can be modelled by the differential equation dx 2x  ___  = 1200 −  _______  (4 marks) 100 + t dt b Hence find, correct to three significant figures, the number of grams of contaminant in the reservoir after 7 days. (5 marks) c Explain how the model could be refined. (1 mark) E 8 A particle P is attached to one end of a light elastic string of natural length 1.2 m. The other end of the string is attached to a fixed point A. The particle is hanging in equilibrium at the point O, which is vertically below A. The particle is now displaced to a point B, vertically below A, and released from rest. The subsequent motion while the string remains taut can be modelled using the equation ẍ  = − 49x, where x m is the displacement of the particle from O at time t seconds. a Describe the motion of the particle while the string remains taut. (1 mark) b Solve the differential equation and hence find the period of the motion. (6 marks) E 9 A particle P of mass 0.6 kg is attached to one end of a light elastic spring of natural length 2.5 m. The other end of the spring is attached to a fixed point A on the smooth horizontal table on which P lies. The particle is held at the point B where AB = 4 m and released from rest. 50 The motion of the particle can be described using the equation x ̈ = −  __ 3  x, where x m is the displacement of the particle from A at time t seconds. a Describe the motion of the particle. (1 mark) b Solve the differential equation and hence find the period and amplitude of the motion. (7 marks) E/P 10 A fisherman’s float bobs up and down on the surface of the water. The float moves up and down in a vertical line, such that the vertical displacement, x cm, from its equilibrium point satisfies the equation d  2x  ____2  = − 0.25x, t > 0 dt  Given that the maximum displacement is 4 cm and it occurs when t = 2 seconds, a find an expression for x in terms of t in the form x = A sin ωt + B cos ωt where A, B and ω are constants to be found. Give A and B correct to three decimal places. (7 marks) b State the time elapsed between the float being at its highest and lowest points. (2 marks) c Criticise the model in terms of the motion of the float. (1 mark) E/P 11 A particle P of mass m is moving in a straight line. At time t the displacement of P from a fixed point O on the line is x. Given that x satisfies the differential equation dx d2x ____   2  + 2k  ___  + n2x = 0 dt dt where k and n are positive constants with k , n, a find an expression for x in terms of k, n and t. (6 marks) b Write down the period of the motion. (1 mark) 192 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 192 26/02/2018 09:31 Modelling with differential equations E/P 12 A particle P of mass m is attached to one end of light elastic spring. The other end of the spring is attached to a fixed point A and P is hanging in equilibrium with AP vertical. The particle is now projected vertically downwards from its equilibrium position with speed U. A resistance of magnitude 2mkv, where v is the speed of P, acts on P. At time t, t . 0, the displacement of P from its equilibrium position is x. The motion of the particle is modelled by the differential equation d  2x dx ____   2  + 2k  ___  + 2 k  2 x = 0 dt dt  1 a Show that P is instantaneously at rest when kt = (n + _ 4 )π, where n ∈ ℕ. b Sketch the graph of x against t. E/P (9 marks) (3 marks) 13 A particle P of mass m is attached to one end of a light elastic spring. The other end of the spring is attached to the roof of a stationary lift. The particle is hanging in equilibrium with the spring vertical. At time t = 0 the lift starts to move vertically upwards with constant speed U. At time t, t . 0, the displacement of P from its initial position is x. The motion of the particle is modelled by the differential equation d  2x  ____2  + n  2x = n  2Ut dt  a find an expression for x in terms of t and n. (8 marks) At time t = T, the particle is instantaneously at rest. Find: E/P b the smallest value of T (3 marks) c the displacement of P from its initial position at this time. (1 mark) 14 A theme park ride designer is designing a new ride where the passengers will be in an enclosed car attached to a horizontal bar. The car will be released from rest from a point half way along the bar so that it oscillates in a horizontal line. The horizontal displacement, x metres, of the centre of the car relative to its initial position at time t seconds is modelled by the differential equation, dx d  2x  ____2  + 4  ___  + 3x = 150 cos t, t > 0 dt dt  a Show that a particular solution to the differential equation is x = 30 sin t + 15 cos t b Hence find the general solution to the differential equation. (3 marks) (5 marks) c Use the model to find, to the nearest metre, the horizontal distance of the centre of the car from its initial position 10 seconds after it is released. (4 marks) E/P d2x dx 15 a Find the general solution to the differential equation ____   2  + 2  ___  + 10x = 27 cos t − 6 sin t. dt dt (8 marks) The equation is used to model water flow in a reservoir. At time t days, the level of the water above a fixed level is x m. When t = 0, x = 3 and the water level is rising at 6 metres per day. b Find an expression for x in terms of t. (2 marks) c Show that after about a week, the difference between the lowest and highest water level is approximately 6 m. (3 marks) 193 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 193 26/02/2018 09:31 Chapter 8 E/P 16 At the start of 2008 a survey began on the number of hedgehogs and the number of slugs in a closed ecosystem. After t years, the number of hedgehogs, x, and the number of slugs, y, in the ecosystem are modelled by the differential equations dx  ___  = 2x + y dt dy ___     = − 2x + 4y dt dx d  2 x (3 marks) a Show that ____   2  − 6  ___  + 10x = 0. dt d t  b Show that the general solution for the number of hedgehogs at time t is x = e  αt( A cos βt + B sin βt) where α and β are constants to be found. (4 marks) c Find the general solution for the number of slugs at time t. (3 marks) d Given that there were initially 10 hedgehogs and 20 slugs in the ecosystem, during (5 marks) which year does the model predict that the slugs die out? e How many hedgehogs will there be when the slugs die out? (1 mark) (1 mark) f Use your answers to parts d and e to comment on the model. E/P 17 A sealed tank of bionutrient contains two different types of organism that interact with each other. The rates of change of each organism can be modelled using the differential equations dx  ___  = − 4x + 3y dt dy ___     = − 3x + 2y dt where x is the number of organism M and y is the number of organism N at time t days. Initially x = 10 and y = 20. a Show that the solutions to the differential equations can be written as x = Ate  −t + Be  −t y = Cte  −t + De  −t where A, B, C and D are constants to be found. (8 marks) b Find, correct to the nearest organism, the number of each organism at time t = 2.1 days. (2 marks) c Use the model to describe what happens to the numbers of each type of organism as t gets large. (2 marks) 18 An industrial process consists of two linked tanks, A and B, containing a chemical solution. The solution is free to pass between the tanks but it flows from A to B at a different rate than it flows from B to A. The solution also enters both tanks, and flows directly out of tank B. The situation is modelled using the differential equations dx 1 1 ___    x (1)     = 2 +  __ y − __ 3 2 dt dy 2 1 ___     = 1 +  __ x − __    y (2) 2 3 dt where x litres is the amount of solution in tank A and y litres is the amount of solution in tank B at time t minutes. d2y dy a Show that 6 ___2  + 7 ___  + y = 9 (6 marks) dt dt b Given that both tanks initially contain 8 litres of solution, find x and y as functions of t. (7 marks) c State, with a reason, the approximate amount of solution in each tank after the system has been running for a long time. (2 marks) 194 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 194 26/02/2018 09:31 Modelling with differential equations Challenge a Three water tanks are positioned as shown in the diagram. Water flows from tank X to tank Y, and from tank Y to tank Z, by means of identical taps. Each tap allows water to flow at a rate of r gallons/hour, where r is the amount of water in the corresponding tank in gallons. Initially, tank X contains 300 gallons of water, tank Y contains 200 gallons of water, and tank Z contains 100 gallons of water. The taps are opened. i Show that after t hours have elapsed, the amount of water in tank X is 300e−t gallons. ii Find the number of minutes after which tanks X and Y contain the same amount of water. iii Find an expression for the amount of water in tank Z after t hours. b A second identical tap is attached to tank X, which is filled to the brim. Tanks Y and Z are emptied, and all three taps are opened. i Show that the amount of water in tank Y is at a maximum after 42 minutes have elapsed, to the nearest minute. ii Find the exact time elapsed before the amounts of water in tanks Y and Z are equal. X Y Z X Y Z Summary of key points 1 2 Simple harmonic motion (S.H.M.) is motion in which the acceleration of a particle P is always towards a fixed point O on the line of motion of P. The acceleration is proportional to the displacement of P from O. dv x¨ = v  ___  dx 3 For a particle moving with damped harmonic motion d  2x dx  ____2  + k   ___  + ω  2x = 0 dt dt  where x is the displacement from a fixed point at time t, and k and ω  2are positive constants. 4 For a particle moving with forced harmonic motion d  2x dx  ____2  + k   ___  + ω  2x = f(t) dt dt  where x is the displacement from a fixed point at time t, and k and ω  2are positive constants. 5 You can solve coupled first-order linear differential equations by eliminating one of the dependent variables to form a second-order differential equation. 195 M08A_ALEVEL_CP2_83343_U08A_170-195.indd 195 26/02/2018 09:31 2 E Review exercise 1 Relative to the origin O as pole and initial line θ = 0, find an equation in polar coordinate form for: a a circle, centre O and radius 2 (1) b a line perpendicular to the initial line and passing through the point with polar coordinates (3, 0) (2) c a straight line through the points with π      . (2) polar coordinates (4, 0) and 4, __ 3 ( E/P a sketch, on the same diagram, the graphs of C and D, indicating where each curve cuts the initial line. (4) ) The graphs of C and D intersect at the pole O and at the points P and Q. ← Section 5.2 E b Find the polar coordinates of P and Q. 2 a Sketch the curve with polar equation (2) r = a cos 3θ, 0 < θ , 2π b Find the area enclosed by one loop of this curve. (6) 3 a Sketch the curve with polar equation π π r = 3 cos 2θ, − __  < θ , __     (2) 4 4 b Find the area of the smaller finite region enclosed between the curve and π (6)     the half-line θ = __ 6 c Find the exact distance between the two tangents which are parallel to the initial line. (6) ← Sections 5.2, 5.3, 5.4 E/P 4 a Sketch, on the same diagram, the curves defined by the polar equations r = a and r = a(1 + cos θ), where a is a positive constant and −π , θ < π. (4) b By considering the stationary values of r sin θ, or otherwise, find equations of the tangents to the curve r = a(1 + cos θ) which are parallel to the initial line. (6) c Show that the area of the region for which a , r , a(1 + cos θ) is (π + 8)a2 ________     (6) 4 ← Sections 5.2, 5.3, 5.4 (3) c Use integration to find the exact value of the area enclosed by the curve D π and the lines θ = 0 and θ =  __ . (6) 3 The region R contains all points which lie outside D and inside C. ← Sections 5.2, 5.3 E/P 5 The curve C has polar equation π π r = 3a cos θ, − __  < θ , __    . The curve D has 2 2 polar equation r = a(1 + cos θ), −π < θ , π. Given that a is positive, Given that the value of the smaller area π     enclosed by the curve C and the line θ = __ 3 is __ 3a2  ___  (2π − 3√3 ), 16 d show that the area of R is πa2. (6) ← Sections 5.2, 5.3, 5.4 E 6 a Show on an Argand diagram the locus of points given by the values of z satisfying |z − 3 + 4i | = 5 (2) b Show that this locus of points can be represented by the polar curve r = 6 cos θ − 8 sin θ.(4) The set of points A is defined by π A = {z : −  __  < arg z < 0} ∩ {z:|z − 3 + 4i| < 5} 2 c Find, correct to three significant figures, the area of the region defined by A. (4) ← Sections 5.2, 5.3 196 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 196 26/02/2018 09:36 Review exercise 2 E/P 7 a Sketch the curve with polar equation π π r = cos 2θ, − __  < θ < __     (2) 4 4 At the distinct points A and B on this curve, the tangents to the curve are parallel to the initial line, θ = 0. b Determine the polar coordinates of A and B, giving your answers to 3 significant figures. (6) The figure shows a sketch of the curve C with polar equation π r2 = a2 sin 2θ, 0 < θ < __    , 2 where a is a constant. Find the area of the shaded region enclosed by C. ← Section 5.3 E/P ← Sections 5.2, 5.4 E/P 11θ = π 2 8 a Sketch the curve with polar equation π r = sin 2θ, 0 < θ < __     (2) 2 At the point A, where A is distinct from O, on this curve, the tangent to the curve π is parallel to θ = __     2 b Determine the polar coordinates of the point A, giving your answer to (6) 3 significant figures. 9 The curve C has polar equation π π r = 6 cos θ, − __  < θ , __     2 2 and the line D has polar equation π 5π π r = 3 sec(__     − θ), − __  < θ , ___     6 6 3 a Find a Cartesian equation of C and a Cartesian equation of D. (4) b Sketch on the same diagram the graphs of C and D, indicating where each cuts the initial line. (4) The graphs of C and D intersect at the points P and Q. c Find the polar coordinates of P and Q. E 10 m Initial line The figure shows a curve C with polar π equation r = 4a cos 2θ, 0 < θ < __    , and a 4 π __ line m with polar equation θ =    . The 8 shaded region, shown in the figure, is bounded by C and m. Use calculus to show that the area of the shaded region 1 (6) is  _2a2(π − 2). ← Section 5.3 E/P θ= 12 π 2 a r = a(1 + – 0.5a 1 cosθ) 2 1.5a Initial line O –a (3) ← Sections 5.1, 5.2 π θ= 2 C O ← Sections 5.2, 5.4 E (6) The curve shown in the figure has polar equation r = a (1 + _ 2 cos θ), a . 0, 0 , θ < 2π. 1 Determine the area enclosed by the curve, giving your answer in terms of a and π. (6) C ← Section 5.3 O Initial line 197 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 197 26/02/2018 09:36 Review exercise 2 E/P 13 θ = π E/P 2 15 θ= W π 2 X r = sin 2θ A r=1 2 O Initial line B O Z Initial line π The figure show the half-lines θ = 0, θ = __     , 2 and the curves with polar equations π 1 r =  _2 , 0 < θ < __    , and 2 π     r = sin 2θ, 0 < θ < __ 2 a Find the exact values of θ at the two (4) points where the curves cross. b Find by integration the area of the shaded region, shown in the figure, which is bounded by both curves. (6) ← Sections 5.2, 5.3 E/P 14 θ = π 2 P C O Initial line Q The curve C, shown in the figure, has polar equation E/P __ r = a(3 + √5  cos θ), −π < θ , π a Find the polar coordinates of the points P and Q where the tangents to C are parallel to the initial line. (6) The curve C represents the perimeter of the surface of a swimming pool. The direct distance from P to Q is 20 m. b Calculate the value of a. (2) c Find the area of the surface of the pool. (6) ← Sections 5.2, 5.3, 5.4 Y The figure shows a sketch of the cardioid C with equation r = a(1 + cos θ), −π , θ < π. Also shown are the tangents to C that are parallel and perpendicular to the initial line. These tangents form a rectangle WXYZ. a Find the area of the finite region, shaded in the figure, bounded by the curve C. (6) b Find the polar coordinates of the points A and B where WZ touches the curve C. (6) (2) c Hence find the length of WX. __ 3√3 a _____ Given that the length of WZ is    , 2 d find the area of the rectangle WXYZ.(2) A heart-shape is modelled by the cardioid C, where a = 10 cm. The heart shape is cut from the rectangular card WXYZ, shown the figure. e Find a numerical value for the area of card wasted in making this heart shape. (3) 16 C1 ← Sections 5.3, 5.4 θ=π 2 B C2 O Initial line A The figure is a sketch of two curves C1 and C2 with polar equations C1 : r = 3a(1 − cos θ), −π < θ , π and C2 : r = a(1 + cos θ), −π < θ , π 198 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 198 26/02/2018 09:36 Review exercise 2 The curves meet at the pole O and at the points A and B. a Find, in terms of a, the polar coordinates of the points A and B. (2) b Show__ that the length of the line AB 3√ 3  (3) is ____     a. 2 The region inside C2 and outside C1 is shaded in the figure. c Find, in terms of a, the area of this region. (6) A badge is designed which has the shape of the shaded region. Given that the length of the line AB is 4.5 cm, d calculate the area of this badge, giving your answer to 3 significant figures. (3) E ← Section 6.1 E 17 r = a(5 – 2 cos θ) E 20 The curves with equations y = 5 sinh x and y = 4 cosh x meet at the point A(ln p, q). Find the exact values of p and q. (4) ← Section 6.3 E θ=π 2 21 Find the values of x for which 5 cosh x − 2 sinh x = 11, giving your answers as natural logarithms. (5) ← Section 6.3 E D O 19 Starting from the definition of cosh x in terms of exponentials, find, in terms of natural logarithms, the values of x for (4) which 5 = 3 cosh x. ← Section 6.1 ← Sections 5.2, 5.3 E/P 18 Find the value of x for which 2 tanh x − 1 = 0, giving your answer in terms of a natural logarithm. (4) r = a(3 + 2 cos θ) B A Initial line 22 By expressing sinh 2x and cosh 2x in terms of exponentials, find the exact values of x for which 6 sinh 2x + 9 cosh 2x = 7, 1 giving each answer in the form  _2 ln p, where p is a rational number. (5) ← Section 6.3 A logo is designed which consists of two overlapping closed curves. The polar equations of these curves are r = a(3 + 2 cos θ), 0 < θ , 2π and r = a(5 − 2 cos θ), 0 < θ , 2π The figure is a sketch (not to scale) of these two curves. a Write down the polar coordinates of the points A and B where the curves meet the initial line. (2) b Find the polar coordinates of the points C and D where the two curves meet. (4) c Show that the area of the overlapping region, which is shaded in the figure, is __ 2 (6)   a__  (49π − 48√3 ) 3 E 23 Given that sinh x + 2 cosh x = k, where k is a positive constant, a find the set of values of k for which at least one real solution of this equation (4) exists b solve the equation when k = 2. (3) ← Section 6.3 E 24 Using the definitions of cosh x and sinh x in terms of exponentials, (3) a prove that cosh2 x − sinh2 x ≡ 1 1 2 b solve the equation _____     − ______     = 2, sinh x tanh x giving your answer in the form k ln a, where k and a are integers. (5) ← Section 6.3 ← Sections 5.2, 5.3 199 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 199 26/02/2018 09:36 Review exercise 2 E 25 a From the definition of cosh x in terms of exponentials, show that (3) cosh 2x ≡ 2 cosh2 x − 1. b Solve the equation cosh 2x − 5 cosh x = 2, giving the answers in terms of natural logarithms. (5) E 26 a Using the definition of cosh x in terms of exponentials, prove that (4) 4 cosh3 x − 3 cosh x ≡ cosh 3x. b Hence, or otherwise, solve the equation cosh 3x = 5 cosh x, giving your answer as natural logarithms. (4) ← Section 6.3 E 27 a Starting from the definitions of cosh x and sinh x in terms of exponentials, prove that cosh (A − B) ≡ cosh A cosh B − sinh A sinh B. (4) b Hence, or otherwise, given that cosh(x − 1) = sinh x, show that e2 + 1 tanh x = _________   2   (5) e + 2e − 1 ) c Solve the equation 3 tanh2 x − 4 sech x + 1 = 0, giving exact answers in terms of natural logarithms. (4) ← Sections 6.2, 6.3 E/P 31 a Express cosh 3 and cosh 5 in terms (4) of cosh . b Hence determine the real roots of the equation 2 cosh 5x + 10 cosh 3x + 20 cosh x = 243, giving your answers to 2 decimal places. (6) ← Section 6.3 E/P 32 a Show that, for x = ln k, where k is a positive constant, k4 + 1     cosh 2x = _____ 2k2 (4) b Given that f(x) = px − tanh 2x, where p is a constant, find the value of p for which f(x) has a stationary value at x = ln 2, giving your answer as an exact fraction. (6) _______ arsinh x = ln(x + √ (1 + x2) ). (4) b Hence, or otherwise, prove that, for 0 , θ , π, θ arsinh(cot θ) = ln (cot  __ ). (5) 2 E/P ( ______ 28 a Starting from the definition ey − e−y sinh y = ______    , 2 prove that, for all real values of x, ← Sections 6.2, 6.3 ) _______ 1 + √ 1 − x  2  1   x ) = ln(__________   (3) arcosh(__ ) x  ← Section 6.3 E/P ( _______ 1−√ (1 − x2)  1+√ (1 − x2)  ___________     = −ln         ln ___________ x x    . (2) b Using the definitions of cosh x and sinh x in terms of exponentials, show that, for 0 , x < 1, ← Section 6.3 E 30 a Show that, for 0 , x < 1, ← Sections 6.1, 6.4 E 33 The curve with equation 29 a Starting from the definition of tanh x in terms of ex, show that 1+x 1 artanh x = _ 2 ln _____      , 1x1 , 1 (5) 1−x b Sketch the graph of y = artanh x. (2) ____ c Solve the equation x = tanh(ln√6x  ) for (5) 0 , x , 1. b Show__ that the y-coordinate of A is __ _1 √ √  4 (2 3  − ln(2 + 3 )). (3) ← Sections 6.2, 6.3 ← Section 6.4 ( ) y = −x + tanh 4x, x > 0, has a maximum turning point A. a Find, in exact logarithmic form, the x-coordinate of A. (5) 200 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 200 26/02/2018 09:36 Review exercise 2 E 34 y = sinh 2x cosh 2x E a Find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form.(7) a Find the values of the constants p, q and r. (2) b Hence, or otherwise, find 1     dx. ∫ ___________ 4x + 4x + 5 b Find an expression for the nth non‑zero term of the Maclaurin series for y. (2) c Show that 2      dx ∫ _____________ √4x + 4x + 5  ____________ 35 f(x) = cos 2x cosh x 2 ___________ a Find the first two non-zero terms of the Maclaurin series for f(x), giving each coefficient in its simplest form. (6) E/P (4) 2 ← Section 6.4 E 39 4x2 + 4x + 5 ≡ (px + q)2 + r = ln((2x + 1) + √4x2 + 4x + 5  ) + k, ← Section 6.5 b Hence find, correct to two significant figures, the percentage error when this approximation is used to evaluate f(0.1). (3) E ← Section 6.4 E 36 Use the substitution x =   a _____ , where a is sinh  a constant, to show that, for x . 0, a . 0, a 1 1 _________   ______   arsinh (__   x )+ constant.   dx = −  __ a 2 2 √ x x + a   (6) ∫ (5) where k is an arbitrary constant. ∫ x+2   dx 40 Find ________   _______ √ 4x2 + 9  (7) ← Section 6.5 ∫ 5 1 ___________   dx= arsinh k, 41 Show that     __________ 2 √ 2 x  − 4x + 8  where k is a rational constant to be found. (6) ← Section 6.5 E/P 42 y ← Section 6.5 E/P 37 a Prove that the derivative of artanh x, 1 −1 , x , 1, is ______    . (4) 1 − x2 b Find E/P ∫ artanh x dx. (4) ← Sections 6.4, 6.5 38 a Starting from the definition of sinh x in terms of ex, prove that ______ arsinh x = ln(x + √ x2 + 1 ). (2) R O 1 2 x b Prove that the derivative of arsinh x _1 (4) is (1 + x2)−   2. The figure above shows a sketch of the curve with equation c Show that the equation d2y dy (1 + x2) ___2  + x  ___  − 2 = 0 dx dx is satisfied when y = (arsinh x)2. The region R, shaded in the figure, is bounded by the curve, the x-axis and the line x = 2. (4) d Use integration by parts to find ∫   arsinh x dx, giving your answer in 1 0 terms of a natural logarithm. (5) y = x arcosh x, 1 < x < 2. Show that the area of R is __ __ √ 3  _7 ___ √   4ln(2 + 3 ) −     2 (8) ← Section 6.5 ← Sections 6.2, 6.4, 6.5 201 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 201 26/02/2018 09:36 Review exercise 2 E/P 43 The diagram shows the cross-section of a new greenhouse. Each unit on the axes represents 1 m. E The curved top of the roof of the greenhouse is modelled using the equation 10 ______  y =  _______ √ x  2 + 9  y 48 Find the general solution to the differential equation dy π      ___  + 2y cot 2x = sin x, 0 , x , __ 2 dx giving your answer in the form y = f(x). (5) ← Section 7.1 E 49 Solve the differential equation dy (1 + x)  ___  − xy = xe−x dx given that y = 1 at x = 0. (6) ← Section 7.1 –2 O E x 2 Given that the greenhouse can be modelled as a prism, length 55 m, find, correct to three significant figures, the volume of the greenhouse. (8) ← Section 6.5 E c Sketch the graph, 0 < x < 2π, of the particular solution for which y = 0 at x = 0. (3) 44 Find, in the form y = f(x), the general solution to the differential equation dy 4 (5)     y = 6x − 5, x . 0  ___  + __ dx x ← Section 7.1 ← Section 7.1 E 45 Solve the differential equation dy y  ___  − __     = x2, x . 0 dx x giving your answer for y in terms of x. (5) E ← Section 7.1 E 46 Find the general solution to the differential equation dy 1  x , x . 0 (x + 1) ___  + 2y = __ dx giving your answer in the form y = f(x). 51 a Find the general solution to the differential equation dy  ___  + 2y = x dx Given that y = 1 at x = 0, (5) b find the exact values of the coordinates of the minimum point of the particular solution curve, (3) c draw a sketch of the particular solution curve. (2) ← Section 7.1 (5) 47 Obtain the solution to dy π      ___  + y tan x = e2x cos x, 0 < x , __ 2 dx for which y = 2 at x = 0, giving your answer in the form y = f(x). (6) 52 a Find the general solution to the differential equation dy (4)  ___  = y sinh x dx b Find the particular solution which satisfies the condition that y = e3 when x = 0. (2) ← Section 7.1 ← Section 7.1 ← Section 7.1 E 50 a Find the general solution to the differential equation dy (5) cos x  ___  + (sin x)y = cos3 x dx b Show that, for 0 < x < 2π, there are two points on the x-axis through which all the solution curves for this (4) differential equation pass. E 202 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 202 26/02/2018 09:36 Review exercise 2 E/P 53 Given that θ satisfies the differential equation b Find the particular solution to this differential equation for which y = 1 dy (2) and ___     = 1 at t = 0. dt dθ dθ ____ + 4___ + 5θ = 0 2 dt dt 2 dθ and that, when t = 0, θ = 3 and  ___  = −6, dt express θ in terms of t. (8) ← Sections 7.3, 7.4 E/P ← Sections 7.2, 7.4 E/P dx d 2x (8)  ___2  + 2  ___  + 5x = 0 dt dt dx ___ b Given that x = 1 and     = 1 at t = 0, dt find the particular solution to the differential equation, giving your answer in the form x = f(t). (2) 54 Given that 3x sin 2x is a particular integral of the differential equation d2y ____2 + 4y = k cos 2x dx where k is a constant, a calculate the value of k c Sketch the curve with equation x = f(t), 0 < t < π, showing the coordinates, as multiples of π, of the points where the curve cuts the t-axis. (2) (3) b find the particular solution of the differential equation for which at x = 0, π π y = 2, and for which at x = __    , y = __     (8) 2 4 ← Sections 7.3, 7.4 E/P ← Sections 7.3, 7.4 dy dy ____ ___   + 5y = 65 sin 2x, x . 0 E/P 56  2  + 4   dx dx 2 a Find the general solution to the differential equation. (8) b Show that for large values of x this general solution may be approximated by a sine function and find this sine function. (2) ← Section 7.3 E/P ← Sections 7.2, 7.4 E/P 55 Given that a + bx is a particular integral of the differential equation d2y dy ____ 2 − 4 ___ + 4y = 16 + 4x dx dx a find the values of the constants a and b (3) b find the particular solution to this differential equation for which y = 8 dy     = 9 at x = 0. and ___ (8) dx 57 a Find the general solution to the differential equation dy d2y  ___2  + 2  ___  + 2y = 2e−t dt dt (8) 58 a Find the general solution to the differential equation 59 a Find the general solution to the differential equation dy d2y (8) 2  ___2  + 7  ___  + 3y = 3t2 + 11t dt dt b Find the particular solution to this differential equation for which y = 1 dy and ___     = 1 when t = 0. (2) dt c For this particular solution, calculate the value of y when t = 1. (2) ← Sections 7.3, 7.4 E/P 60 a Find the value of λ for which λx cos 3x is a particular integral of the differential equation d2y (3)  ___2  + 9y = −12 sin 3x dx b Hence find the general solution to this differential equation. (6) The particular solution of the differential dy equation for which y = 1 and ___     = 2 at dx x = 0, is y = g(x). c Find g(x). (2) d Sketch the graph of y = g(x), 0 < x < π. (2) ← Sections 7.3, 7.4 203 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 203 26/02/2018 09:36 Review exercise 2 E d2y dy 61  ___2  − 6   ___  + 9y = 4e3t, t > 0 dt dt a Show that Kt 2 e3t is a particular integral of the differential equation, where K is a constant to be found. (3) b Find the general solution to the differential equation. E/P (6) Given that a particular solution satisfies dy y = 3 and ___     = 1 when t = 0, dt c find this solution. Another particular solution which satisfies dy y = 1 and ___     = 0 when t = 0, has equation dt y = (1 − 3t + 2t 2)e3t(2) ← Sections 7.3, 7.4 E d For this particular solution, draw a sketch graph of y against t, showing where the graph crosses the t-axis. Determine also the coordinates of the minimum point on the sketch graph. (4) ← Sections 7.3, 7.4 E/P 62 a Find the general solution to the differential equation dx d2x 2  ___2  + 5  ___  + 2x = 2t + 9 dt dt (8) b Find the particular solution of this differential equation for which x = 3 ___  = −1 when t = 0. (2) and   dx dt The particular solution in part b is used to model the motion of the particle P on the x-axis. At time t seconds (t > 0), P is x metres from the origin O. c Show that the minimum distance 1 between O and P is _ 2 (5 + ln 2) m and justify that the distance is a minimum. (4) ← Sections 7.3, 7.4, 8.3 63 Given that x = At 2 e−t satisfies the differential equation dx d2x ___   2  + 2  ___  + x = e−t, dt dt a find the value of A. (3) b Hence find the solution to the differential equation for which x = 1 dx (7) and ___     = 0 at t = 0. dt c Use your solution to prove that for (2) t > 0, x < 1. 64 Given that y = kx is a particular solution of the differential equation d2y  ___2  + y = 3x, dx a find the value of the constant k. (3) b Find the most general solution to this differential equation for which y = 0 at x = 0. (6) c Prove that all curves given by this solution pass through the point (π, 3π) and that they all have equal gradients π (3)    . when x = __ 2 d Find the particular solution to the differential equation for which y = 0 at π x = 0 and at x = __    . (2) 2 e Show that a local minimum value of the solution in part d is ______ 2 3  π   )− _ 2√ π 2 − 4  (4) 3 arccos( __ ← Sections 7.3, 7.4 E/P 65 During an industrial process, the mass of salt, S kg, dissolved in a liquid, t minutes after the process begins, is modelled by the differential equation 2S dS + ______ ___ = _1 , 0 < t , 120 dt 120 − t 4 Given that S = 6 when t = 0, a find S in terms of t (6) b calculate the maximum mass of salt that the model predicts will be dissolved in the liquid at any one time during the process. (3) ← Section 8.1 204 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 204 26/02/2018 09:36 Review exercise 2 E/P 66 A fertilized egg initially contains an embryo of mass m0 together with a mass 100m0 of nutrient, all of which is available as food for the embryo. At time t, the embryo has mass m and the mass of nutrient which has been consumed is 5(m − m0). a Show that, when three-quarters of the nutrient has been consumed, m = 16m0. c Find the limiting value of v. b This differential equation is used to model the motion of a particle which has speed v m s−1 at time t seconds. When t = 2, the speed of the particle is 3 m s−1. Find, to 3 significant figures, the speed of the particle when t = 4. (4) ← Section 8.1 E/P 68 A particle P moves in a straight line. At time t seconds, the acceleration of P is e2t m s−2, where t > 0. When t = 0, P is at rest. Show that the speed, v m s−1, of P at time t seconds is given by v =  _2 (e2t − 1) 1 (2) ← Section 8.1 E/P 70 A particle P moves along the x-axis. At time t seconds its acceleration is (−4e−2t) m s−2 in the direction of x increasing. When t = 0, P is at the origin O and is moving with speed 1 m s−1 in the direction of x increasing. a Find an expression for the velocity of P at time t. (4) b Find the distance of P from O when P comes to instantaneous rest. (2) (5) 67 a Find the general solution to the differential equation t ___ dv − v = t, t . 0 dt and hence show that the solution can be written in the form v = t(ln t + c), where c is an arbitrary constant. (5) (4) b Find, to 3 significant figures, the speed of P when t = 3. (2) ← Section 8.1 E 69 A particle P moves along the x-axis in the positive direction. At time t seconds, the velocity of P is v m s−1 and its acceleration 1 _1 is _ 2 e−   6 t m s−2. When t = 0 the speed of P is 10 m s−1. a Express v in terms of t. (3) The rate of increase of the mass of the embryo is a constant μ multiplied by the product of the mass of the embryo and the mass of the remaining nutrient. dm b Show that ___    = 5 μm (21m0 − m). (4) dt The egg hatches at time T, when three-quarters of the nutrient has been consumed. c Show that 105 μm0T = ln 64. E/P ← Section 8.1 E/P 71 A water droplet falls vertically from rest through low cloud. Water condenses on the droplet as it falls. You are given that the motion of the water droplet may be modelled by the equation dv (t + 3)  ___  + 3v = 9.8(t + 3) dt where t is the time in seconds and v is the velocity of the droplet in m s–1. 49(t + 3)  4 + c    a Show that v =  ____________   where c is 20(t + 3)  3 a constant to be found. (6) b Find the velocity of the water droplet after 6 seconds. (2) c By considering the velocity for large values of t, suggest one criticism of the model. (1) ← Section 8.1 (5) ← Section 8.1 205 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 205 26/02/2018 09:36 Review exercise 2 E/P 72 A water bottle initially contains 400 ml of distilled water. The water leaks out at a constant rate of 30 ml per minute and a mixture is added to the bottle at a constant rate of 40 ml per minute. The mixture contains 10% acid and 90% distilled water. E/P Given that there is x ml of acid in the bottle after t minutes and that the acid immediately disperses on entry, a show that the situation can be modelled by the differential equation dx 3x (4)  ___  = 4 −   ______  40 + t dt b Hence find the amount of acid in the bottle after 7 minutes. (5) c Explain how the model could be refined. a solve the differential equation and hence find, correct to three significant figures, the maximum displacement of the boat. (7) b State the time elapsed between the boat being at its highest and lowest points. (2) c Criticise the model in terms of the motion of the boat for large values of t. (1) (1) ← Section 8.1 E/P 73 A particle, P, is attached to the ends of two identical elastic springs. The free ends of the springs are attached to two points A and B. The point C lies between A and B such that ACB is a straight line and AC ≠ BC. The particle is held at C and then released from rest. The subsequent motion of the particle can be described by the differential equation ẍ = − 49x. a Describe the motion of the particle. (1) Given that x = 0.3 and v = 0 when t = 0, b solve the differential equation to find x (7) as a function of t. c State the period of the motion and calculate the maximum speed of P. (2) ← Section 8.2 74 A small boat is floating on the surface of a river, tied to a jetty. The boat moves up and down in a vertical line, such that the vertical displacement, x m, from its equilibrium point satisfies the equation d  2x  ____2  = −1.6x, t > 0 dt  Given that the displacement is zero and the boat is moving at a velocity of 1 m s–1 at time t = 0 minutes, ← Section 8.2 E/P 75 A particle P moves in a straight line. At time t seconds its displacement from a fixed point O on the line is x metres. The motion of P is modelled by the differential equation dx d2x  ___2  + 2  ___  + 2x = 12 cos 2t − 6 sin 2t dt dt When t = 0, P is at rest at O. a Find, in terms of t, the displacement of P from O. (8) b Show that P comes to instantaneous π rest when t = __     (2) 4 c Find, in metres to 3 significant figures, the displacement of P from O when π t =  __  (2) 4 d Find the approximate period of the motion for large values of t. (2) ← Section 8.3 206 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 206 26/02/2018 09:36 Review exercise 2 E/P 76 A particle P of mass m is suspended from a fixed point by a light elastic spring. At time t = 0 the particle is projected vertically downwards with speed U from its equilibrium position. At time t, the displacement of P downwards from its equilibrium position is x. The motion of the particle can be modelled using the differential equation 2 x  + 2ω  dx d ___  + 2ω2x = 0  ____ dt dt 2 Given that the solution to this differential equation is x = e−ωt(A cos ωt + B sin ωt), where A and B are constants, a find A and B. (8) b Find an expression for the time at which P first comes to rest. (2) a find x in terms of t b show that the particle continues to move down through the liquid throughout the motion. 77 A particle P is attached to one end A of a light elastic string AB and is free to move on a horizontal table. At time t = 0, P is at rest and the end B of the string is forced to move horizontally away from P with speed V. After t seconds the displacement of P from its initial position is x metres. The motion of P can be modelled by the differential equation 2 x  + 2k  dx d ___  + 10k2 x = 10k2Vt  ____ dt dt2 Find an expression for x in terms of t, k and V. (8) ← Section 8.3 E/P 78 A particle P is attached to one end of a light elastic string. The other end of the string is fixed to a point vertically above the surface of a liquid. The particle is held on the surface of the liquid and then released from rest. At time t seconds, the distance travelled down by P is x metres. Given that the motion of P can be modelled using the differential equation 2 x  + 6  dx d ___  + 8x = 2.4  ____ dt dt2 (2) ← Section 8.3 79 Colonies of angler fish and angel fish live in an ocean cave sealed off from the rest of the ocean by a thick wall of seaweed. After t years the number of angler fish, x, and the number of angel fish, y, in the cave are modelled by the differential equations dx (1)   ___  = 0.1x + 0.1y dt dy ___     = − 0.025x + 0.2y (2) dt dx d  2x a Show that  ____2  − 0.3   ___  + 0.0225x = 0 dt dt  (3) ← Section 8.3 E/P (8) b Find the general solution for the number of angler fish in the cave at time t. (4) c Find the general solution for the number of angel fish in the cave at time t. (4) At the start of 2010 there were 20 angler fish and 100 angel fish in the cave. d Use this information to find the number of angler fish predicted to be in the cave in 2017. (4) e Comment in the suitability of the model. (1) ← Section 8.4 E/P 80 An industrial chemist is examining the rates of change of gases in two connected tanks, A and B. Gas passes between the two tanks. Gas also enters the system into both tanks and escapes from the system in the same way. The chemist believes that the amount of gas in tank A, x litres, and the amount of gas in tank B, y litres, at time t hours, can be modelled using the differential equations dx (1)   ___  = 2x + y + 1 dt dy ___     = 4x − y + 1 (2) dt 207 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 207 26/02/2018 09:36 Review exercise 2 d  2x dx a Show that ____   2  −   ___  − 6x = 2 (4) dt  dt b Given that tank A initially contains 20 litres of gas and tank B initially contains 60 litres of gas, find expressions for the amount of gas in each tank at time t hours. (8) The tanks have a maximum capacity of 500 litres. c Comment on the suitability of the model after one hour. (2) ← Section 8.4 Challenge 1 Given that n  Z+, x  R and cosh2 x cosh2 x M = (    , −sinh2 x −sinh2 x) prove that Mn = M. ← Section 6.3 2 a A system of differential equations is given as dy ___  = y  ___  = x,   dx dt dt Given that y = 0 and x = 1 when t = 0, show that y = sinh t and x = cosh t. b Another system of differential equations is given as dp ___     = p − q dt dq ___     = p + q dt dr   __  = p + 2q + r dt Given that p = q = r = 1 when t = 0, show that r = et(3 sin t – cos t + 2). ← Section 8.4 3 The diagram shows the curve C with polar equation r = f(θ). The line l is a tangent to the curve at the point P(r, θ), and α is the acute angle between l and the radial line at P. θ= π 2 l C α P (r, θ) O θ r Show that tan α = _____     dr ( ___) dθ Initial line ← Section 5.4 208 M08B_ALEVEL_CP2_83343_U08B_196-208.indd 208 26/02/2018 09:36 Exam-style practice Further Mathematics A Level Paper 1: Core Pure Mathematics Time: 1 hour 30 minutes You must have: Mathematical Formulae and Statistical Tables, Calculator 1 The diagram shows a section of a curve C with parametric equations x = t + t  2, y = t − t  2 + 1, y 0 < x < 3.75 The curve is rotated through 360° about the x-axis and the volume of revolution formed is used to model a clay pottery vase. Given that each unit on the axes represents 10 cm, 2 C x O a find, correct to three significant figures, the volume of the vase (7) b give one criticism of the model used. (1) The plane Π  1has vector equation r.(2i + 3j − k) = 10 _ a Show that the perpendicular distance from the plane Π  1to the point (1, 4, 7) is a √ 14  where a is a rational constant to be found. (3) The plane Π  2has vector equation r = λ(i − 3j + 2k) + μ(ai + 2j − 3k) Given that the vector 5i + j − k is perpendicular to the plane Π  2, b find the value of a (2) c find, in degrees correct to 1 decimal place, the acute angle between Π  1 and Π  2.(3) 3 a Prove by induction that for all positive integers n, 1 2 2 n 1 2n 2n  2   0    1   2   = 0      1     2n (0 0 1) (0 0 1 ) 1 k 4 b M = 2    − 2     0 where k is an integer. (3 − 1 1) i Find the value of k for which M–1 does not exist. ii Given that M is non-singular, find M in terms of k. –1 (6) (2) (4) 209209 M09_ALEVEL_CP2_83343_PP1_209-210.indd 209 26/02/2018 09:37 Exam-style practice 4 A complex number z has argument θ and modulus 1. 1   n  = 2i sin nθ, n ∈ ℤ  + a Show that z  n− __ z  b Hence, show that 8 sin  4 θ = cos 4θ − 4 cos 2θ + 3 5 (2) (5) An oil vat initially contains 500 litres of pure corn oil. The oil drains out at a constant rate of 15 litres per minute and a corn oil and sugar mixture is added to the vat at a constant rate of 30 litres per minute. The mixture contains 25 grams of sugar per litre of oil. Given that there is x grams of sugar in the vat after t minutes and that the sugar immediately disperses throughout the vat on entry, 6 a show that the situation can be modelled by the differential equation dx 3x   ___  = 750 −   ________  100 + 3t dt b Hence find the number of grams of sugar in the vat after 10 minutes. (4) c Explain how the model could be refined. (1) (5) a Show that the locus of points given by the values of z satisfying |z + 12 + 5i| = 13 can be represented by the polar curve with equation r = −2(12 cos θ + 5 sin θ) (4) b Show on an Argand diagram the set of points A defined by A = {z : |z + 12 + 5i| < 13} ∩{z : −π < arg z < −  __ 4 } 3π c Find, correct to three significant figures, the area of the region defined by A. 7 (4) (5) A scientific experiment looks at the concentration of glucose, dissolved in water, on either side of an osmotic membrane. The glucose solution passes through the membrane in both directions and also enters the system through a tube on the left side. A researcher believes that the concentration of glucose on the left side of the membrane, x, and the concentration of glucose on the right side of the membrane, y, at time t hours, can be modelled by the differential equations dx   ___  = 0.3x + 0.2y + 1 dt dy ___     = −0.2x + 0.3y dt dx d  2x (3) a Show that 100  ____2  − 60   ___  + 13x + 30 = 0. dt dt  b Find the general solution for the concentration of glucose on the left side of the membrane at time t. (6) c Hence find the general solution for the concentration of glucose on the right side of the membrane at time t. (3) At time t = 0, the concentration of glucose on the left side is 10 and on the right side is 5. d Find the particular solutions to the system of differential equations. (3) e By considering the concentrations on each side on the membrane predicted by the model after 3 hours, comment on the suitability of the model. (2) 210 M09_ALEVEL_CP2_83343_PP1_209-210.indd 210 26/02/2018 09:37 Exam-style practice Further Mathematics A Level Paper 2: Core Pure Mathematics Time: 1 hour 30 minutes You must have: Mathematical Formulae and Statistical Tables, Calculator 1 a Prove that n n(pn + q) 1   ____________ ∑     = ______________       r=1 (r + 2)(r + 4) 24(n + 3)(n + 4) where p and q are constants to be found. (5) b Prove that, for all positive integers n, f(n) = 2  n+2 + 3  2n+1 is divisible by 7.(6) 2 f(x) = z4 + az3 + 30x2 + bz + 85 where a and b are real constants. z = 1 + 4i is a root of the equation f(x) = 0. 3 a Write down another root of the equation f(x) = 0. (1) b Hence solve the equation f(x) completely. (6) c Show the roots of f(x) = 0 on a single Argand diagram. (2) The diagram shows the curve C with polar equation π r = 6 sin 2θ 0 < θ <   __  2 The line segment AB is tangent to the curve at A and perpendicular to the initial line. The finite region R, shown shaded in the diagram, is bounded by the curve, the initial line and the line segment AB. Find, correct to three significant figures, the area of the shaded region R. 4 θ= π 2 A R O B Initial line (9) f(x) = cos x sinh 2x a Find the first three non-zero terms of the Maclaurin series for f(x), giving each coefficient in its simplest form. (8) b Hence find, correct to four significant figures, the percentage error when the approximation is used to evaluate f(0.1). (2) 211211 M10_ALEVEL_CP2_83343_PP1_211-212.indd 211 26/02/2018 09:37 Exam-style practice 5 6 ∞ π 1   dx = __     Prove that ∫  ______   2 4 0 x  + 4 (5) A triangle T has vertices at (0, 2), (k, 0) and (0, 8). Triangle T is transformed onto triangle T ′ by the matrix 2 2 (      ) −3 5 The area of triangle T ′ is 456 units2, find the value of k. 7 8 (5) 1 f(x) = ___________   _   2 √ x  + 2x + 2  a Find the mean value of f(x) over the interval [–1, 1], giving your answer correct to 3 decimal places. (6) b Hence find the mean value of f(x) + 2 on the same interval. (1) y − 2 _____ x − 3 _____ z−1 The line l1 has equation  _____   =     =     −2 3 −1 The plane Π has equation 2x − y + 3z = 4. Point A on l1 has x-coordinate x = 4. Line l1 is reflected in the plane Π and the image of point A is A′. Find the exact distance AA′. 9 (7) The differential equation d  2x dx ____   2  + 2   ___  + 3x = 21 + 15 cos t dt dt  is used to model the flow of water through a pump. x is the volume of water, in litres, at time t seconds. a Show that a particular solution to the differential equation is x = 7 +  __ 4 (sin t + cos t) 15 (3) Given that the initial flow is two litres and the initial rate of change of flow is three litres per second, b Find the solution to the differential equation. (8) c State what happens to the flow of water as t gets large. (1) 212 M10_ALEVEL_CP2_83343_PP1_211-212.indd 212 26/02/2018 09:37 Answers Answers CHAPTER 1 Prior knowledge check π b  __  3 e 4 1 a 8 π d  __  2 2 (cos 9θ + i sin 9θ)(cos 4θ + i sin 4θ) ______   4iθ e  9iθe 6  _______________________________            =   7iθ  = e  9iθ+4iθ−7iθ cos 7θ + i sin 7θ e  = e  6iθ= cos 6θ + i sin 6θ π 5πi ___ 7 2e 6 i, 2e −  6 πi __   2 8 a 2e c 16 π f __  6 Im b n = 1: LHS = (1 + i) 1= 1 + i _ _ πi 1 __ __ π π 1_ 1 RHS = 2   2 e       4 = √ 2   cos  __  + i sin  __  = √ 2   ___   _  + i  ___ ( ( √ 2 4 4)   √ 2   ) =1+i As LHS = RHS, the equation holds for n =1. Assume the equation holds for n = k, k ∈ ℤ  +. k __ kπi ___ i.e. (1 + i)  k = 2  2e     4  With n = k + 1, the equation becomes k __ kπi ___ (1 + i)  k+1 = (1 + i)  k× (1 + i) = 2  2e     4 × (1 + i) k __ πi 1 __ __   2 ___   kπi           = 2 e   4 × 2  2e   4 (k + 1)πi k kπi πi k+1 1 __ __ ___ __ ____ _____ = 2   2 +  2 e    4   +   4  = 2    2 e     4  Therefore, the equation holds when n = k + 1. If the equation holds for n = k, then it has been shown to be true for n = k + 1. As the equation holds for n = 1, it is now also true for all n ∈ ℤ  +by mathematical induction. c 256 9e  iθ= cos θ + i sin θ, e  −iθ= cos θ − i sin θ So e  iθe   −iθ= (cos θ + i sin)(cos θ − i sin θ) LHS = e  i(θ−θ) = e  0= 1 RHS = cos  2 θ − i  2 sin  2 θ = cos  2 θ + sin  2 θ Hence cos  2 θ + sin  2 θ ≡ 1 2 1 –1 –2 O 2 Re 1 –1 –2 3 4032 Exercise 1A πi __ 1 a 3eπi b 6e    2  ___ 5πi  −  ____ c 4e 6 ___ √ e 29  e−1.19i 3.02i d √ 65  __ e____  3πi  √ f 2 6 e 4 g 2√2 e 4 h 8e 6 __ i πi  −  __ __  πi  πi  −  __ 2e 5 __ √ 3   1  + ___  __ 2 a     i 2 2 c 3 + 3i b −4 Challenge a n = 1; LHS = (re  iθ)  1 = re  iθ RHS = r  1e   iθ = re  iθ As LHS = RHS, the equation holds for n = 1. Assume the equation holds for n = k, k ∈ ℤ  +. i.e. (re  iθ)  k= r  ke   ikθ With n = k + 1, the equation becomes (re  iθ)  k+1 = (re  iθ)  k × re  iθ= r  ke   ikθ × re  iθ= r  k+1e   i(kθ+θ)= r  k+1e   i(k+1)θ Therefore, the equation holds when n = k + 1. If the equation holds for n = k, then it has been shown to be true for n = k + 1. As the equation holds for n = 1, it is now also true for all n ∈ ℤ  +by mathematical induction. 1 1 b Given n ∈ ℤ  +, we have: (re  iθ)  −n = _______    = r  −ne   iθ n  = ______   −inθ (re  )  r  ne   inθ __   + 4i d 4√3 __ √   1 3 ___     i f −     + __ 2 2 h −3 − 3i e −3i g −1 __   i −4 + 4i√3 10π 10π ____ 3 a cos(−     + i sin(−  ____ ) 13 ) 13 3π 3π ___ b 4 cos(−    ) + i sin (−  ___ ) ( 5 5 ) 7π 7π ___ ___ c 5 cos(   ) + i sin(   ) ( 8 8 ) 4 eiθ = cos θ + i sin θ e−iθ = cos (−θ) + i sin (−θ) = cos θ − i sin θ (1) − (2): eiθ − e−iθ = 2i sin θ (1) (2) Exercise 1C 1  __ (eiθ − e−iθ) = sin θ 2i 1 ⇒ sin θ = __    (eiθ − e−iθ) (as required) 2i Exercise 1B _ 7π 7π 1 a cos  ___  + i sin  ___  12 12 _ 2 c 3i√   _ b e  πi 4 a e   3iθ _ 5 a 6√ 3 e   6 5πi ___ _ √     3 √ 3 5π 5π b  ___  cos  ___  + i  ___  sin  ___  7 7 4 4 2 a −  __14 c −i 3 a e  5iθ _ b 3 √ 5  cos 4θ + 3i √ 5  sin 4θ πi __ _ c 6e   3 _ πi __ d 3√ 2 e   4 πi πi __ __ b 2√ 2 e   4 c  __34 e −  2 _ ___ 7πi 7πi πi ___ __ − √ b 3 e   12 c 18e   12 d 6e   4 1 a cos__6θ + i sin 6θ √ 3   1 c −  ___  + __     i 2 2 e 1 2 a e   iθ b e   2iθ d e   −iθ e e  11iθ b cos 12θ__+ i sin 12θ 3 1 √   d −  __  + ___    i 2 2 f i c e   −6iθ fe  5iθ 3 a 4 a c e c 1 8 b (−2 + 2i) = 4096 __ (1 − i√ 3  )6 = 64 d 1 b −1 (1 + i)5 = −4 − 4i (1 − i)6__= 8i __ 3 1 __ __  i)9 = 81i√ 3   (  2  −   2 √ 3 __ __ f (−2√ 3 − 2i)5 = 512√ 3 − 512i __ __ 5 (3 + √ 3  i)5 = −432 + 144i√ 3   __ 6 −8 + 8i √ 3   7 −27i 2π __ 8 a e    3 i b 3 213 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 213 22/09/18 3:09 PM Answers 9 Write a + bi and a − bi as r(cos θ + i sin θ) and r(cos θ − i sin θ) respectively. Then by de Moivre’s theorem, (a + bi)n + (a − bi) n = rn(cos nθ + i sin nθ) + rn(cos nθ − i sin nθ) = 2rncos nθ which is always real. e Let z = cos θ + i sin θ ( z − _ z )  = ( 2i sin θ)  5 = 32i  5 sin  5 θ = 32i sin  5 θ 3 2 = z  5+ 5z  4(−  _1z ) + 10z  3 (−  _1z )   + 10z  2 (−  _1z )   1 5 + 5z (−  _1z )   + ( −  _1z )   4 5 1 _ _ = z  5− 5z  3+ 10z − _  10 z +   z  3  −   z  5  = (z  5 − _   z1  5 )− 5(z  3 − _   z1  3 )+ 10(z − _  1z ) = 2i sin 5θ − 5(2i sin 3θ) + 10(2i sin θ) 32i sin  5 θ = 2i sin 5θ − 10i sin 3θ + 20i sin θ 1 ⇒ sin  5 θ = __  16   (sin5θ − 5 sin 3θ + 10 sin θ) Challenge Given n ∈ ℤ  +, we have: 1 1 (r(cos θ + i sin))  −n = _______________     n  = _________________       (r(cos θ + i sin))  r  n(cos nθ + i sin nθ) by de Moivre’s theorem for positive integer exponents. cos nθ − i sin nθ 1    = _________________        × ______________       r  n(cos nθ + i sin nθ) cos nθ − i sin nθ cos nθ − i sin nθ cos nθ − i sin nθ         = __________________       = ____________________      r  n(cos  2 nθ − i  2 sin  2 nθ) r  n(cos  2 nθ + sin  2 nθ) 2 a(cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5C1 cos  4 θ(i sin θ) + 5C2 cos  3 θ (i sin θ)  2 + 5C3 cos  2 θ (i sin θ)  3 + 5C4 cos θ (i sin θ)  4 + (i sin θ)  5 = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ −10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ + i sin  5 θ Equating the real parts gives cos 5θ = cos  5 θ − 10 cos  3 θ sin  2 θ + 5 cos θ sin  4 θ = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ (1 − cos  2 θ)  2 = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ(1 − 2 cos  2 θ + cos  4 θ) = 16 cos  5 θ − 20 cos  3 θ + 5 cos θ b 0.475, 1.57, 2.67 (3 s.f.) 1 3 a Let z = cos θ + i sin θ, then 2 cos θ = z + __     z 6 1 (z + __    )  = ( 2 cos θ)  6 = 64 cos  6 θ z 1 1 1 3 ___ = z  6 + 6 z  5(__    )+ 15 z  4 ___ (  z  2 )+ 20 z  (  z  3 ) z 1 1 1 ___ ___ + 15 z  2 ___ (  z  4 )+ 6z(  z  5 )+ (   z  6 ) 1 1 1   6 )+ 6 z  4 + ___   4  + 15 z  2 + ___   2  + 20 = (z  6 + ___ ( ( z  z  ) z  ) = r  −n(cos nθ − i sin nθ) = r  −n(cos (−nθ) + i sin (−nθ)) Exercise 1D 1 a ( cos θ + i sin θ)  3= cos 3θ + i sin 3θ = cos  3 θ + 3i cos  2 θ sin θ + 3i  2 cos θ sin  2 θ + i  3 sin  3 θ = cos  3 θ + 3i cos  2 θ sin θ − 3 cos θ sin  2 θ − i sin  3 θ ⇒ cos 3θ + i sin 3θ = cos  3 θ + 3i cos  2 θ sin θ − 3 cos θ sin  2 θ − i sin  3 θ Equating the imaginary parts: sin 3θ = 3 cos  2 θ sin θ − sin  3 θ = 3 sin θ(1 − sin  2 θ) − sin  3 θ = 3 sin θ − 4 sin  3 θ b (cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ + 10 i  2 cos  3 θ sin  2 θ + 10 i  3 cos  2 θ sin  3 θ + 5 i  4 cos θ sin  4 θ + i  5 sin  5 θ ⇒ cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ − 10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ + i sin  5 θ Equating the imaginary parts: sin 5θ = 5 cos  4 θ sin θ − 10 cos  2 sin  3 θ + sin  5 θ = 5(1 − sin  2 θ )  2 sinθ − 10(1 − s in 2 θ)sin  3 θ + sin  5 θ = 16 sin  5 θ − 20 sin  3 θ + 5 sin θ c (cos θ + i sin θ)  7= cos 7θ + i sin 7θ = cos  7 θ + 7i cos  6 θ sin θ + 21 i  2 cos  5 θ sin  2 θ + 35 i  3 cos  4 θ sin  3 θ + 35 i  4 cos  3 θ sin  4 θ + 21 i  5 cos  2 θ sin  5 θ + 7 i  6 cos θ sin  6 θ + i  7 sin  7 θ ⇒ cos 7θ + i sin 7θ = cos  7 θ + 7i cos  6 θ sin θ − 21 cos  5 θ sin  2 θ − 35i cos  4 θ sin  3 θ + 35 cos  3 θ sin  4 θ + 21i cos  2 θ sin  5 θ − 7 cos θ sin  6 θ − i sin  7 θ Equating the real parts: cos 7θ = cos  7 θ − 21 cos  5 θ sin  2 θ + 35 cos  3 θ sin  4 θ − 7 cos θ sin  6 θ = cos  7 θ − 21 cos  5 θ (1 − cos  2 θ) + 35 cos  3 θ (1 − cos  2 θ)  2− 7 cos θ (1 − cos  2 θ)  3 = 64 cos  7 θ − 112 cos  5 θ + 56 cos  3 θ − 7 cos θ d Let z = cos θ + i sin θ 1 4 ( z + _ z )  = ( 2 cos θ)  4 = 16 cos  4 θ = z  4+ 4z  3( _   1z ) + 6z  2(_   z1  2 )+ 4z(_   z1  3 ) + _   z1  4  = ( z  4 + _   z1  4 )+ 4(z  2 + _   z1  2 )+ 6 = 2 cos 4θ + 4(2 cos 2θ) + 6 16 cos  4 θ = 2 cos 4θ + 4(2 cos 2θ) + 6 = 2(cos 4θ + 4 cos 2θ + 3) ⇒ cos  4 θ = __  18  (cos 4θ + 4 cos 2θ + 3) 214 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 214 5 4 = 2 cos 6θ + 6(2 cos 4θ) + 15(2 cos 2θ) + 20 64 cos  6 θ = 2 cos 6θ + 6(2 cos 4θ) + 15(2 cos 2θ) + 20 32 cos  6 θ = cos 6θ + 6 cos 4θ + 15 cos 2θ + 10 __   6π 9 __ 5π b ∫   cos6 θ dθ = ___     + ___       √ 3 0 96 64 1 a If z = cos θ + i sin θ, then 2 cos θ = z +  __  and z 1 __ 2i sin θ = z −     z 2 4 1 1 So, 2  2 cos  2 θ × (2i)  4 sin  4 θ = ( z + __    )  (z − __    )  z z 2 2 2 2 1 1 1 1 1    )(z − __    ))  (z − __    )  = z  2 − ___   2    (z − __    )  = ( (z + __ ( z z z z z  ) 1 1   4  z  2− 2 + ___   2  = z  4− 2 + ___ ( z  )( z  ) 1 2 1   2  − ___     + ___     = z  6− 2z  4− z  2+ 4 − ___ z  z  4 z  6 1 1 1   6  − 2 z  4 + ___   4  − z  2 + ___   2  + 4 = z  6 + ___ ( ( z  ) ( z  ) z  ) = 2 cos 6θ − 2(2 cos 4θ) − 2 cos 2θ + 4 So, 64 cos  2 θ sin  4 θ = 2 cos 6θ − 4 cos 4θ − 2 cos 2θ + 4 ⇒ 32 cos  2 θ sin  4 θ = cos 6θ − 2 cos 4θ − cos 2θ + 2 π b ___     48 π 5π 1 67 ____ b  ___  + ___ c   6144     5 a  ___  32 64 48 6 a (cos θ + i sin θ)  6= cos 6θ + i sin 6θ = cos  6 θ + 6C1 cos  5 θ(i sin θ) + 6C2 cos  4 θ (i sin θ)  2 + 6C3 cos  3 θ (i sin θ)  3 + 6C4 cos  2 θ (i sin θ)  4 + 6C5 cos θ (i sin θ)  5 + (i sin θ)  6 = cos  6 θ + 6i cos  5 θ sin θ − 15 cos  4 θ sin  2 θ − 20i cos  3 θ sin  3 θ + 15 cos  2 θ sin  4 θ + 6i cos θ sin  5 θ − sin  6 θ Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers Equating the real parts gives cos 6θ = cos  6 θ − 15 cos 4 θ sin  2 θ + 15 cos 2 θ sin  4 θ − sin  6 θ = cos  6 θ − 15 cos  4 θ(1 − cos  2 θ) + 15 cos  2 θ (1 − cos  2 θ)  2 − (1 − cos  2 θ)  3 = cos  6 θ − 15 cos  4 θ(1 − cos  2 θ) + 15 cos  2 θ(1 − 2 cos  2 θ + cos  4 θ) − (1 − 3 cos 2 θ + 3 cos  4 θ − cos  6 θ) = 32 cos  6 θ − 48 cos  4 θ + 18 cos  2 θ − 1 π 5π 7π b cos  ___ ≈ 0.985, cos  ___ ≈ 0.643, cos  ___ ≈ 0.342, 18 18 18 13π 17π 11π cos  ____ ≈ −0.342, cos  ____ ≈ −0.643, cos  ____ ≈ −0.985 18 18 18 5 a P + iQ = (1 + cos θ + cos 2θ + … + cos 12θ) + i(sin θ + sin 2θ + … + sin 12θ) = 1 + (cos θ + i sinθ) + (cos 2θ + i sin 2θ) + … + (cos 12θ + i sin 12θ) ( e  iθ)  13− 1 = 1 + e  iθ + e  2iθ + … + e  12iθ = __________     iθ e  − 1 13iθ _____ b 7 a cos 4θ + i sin 4θ = ( cos θ + i sin θ)4 = cos4θ + 4i cos3θ sin θ + 6 i2cos2θ sin2θ + 4 i3cos θ sin3θ + i4sin4θ = cos4θ + 4i cos3θ sin θ − 6 cos2θ sin2θ − 4i cos θ sin3θ + sin4θ Equating the imaginary parts: sin 4θ = 4 cos3θ sin θ − 4 cos θ sin3θ b Equating the real parts: cos 4θ = cos4θ − 6 cos2θ sin2θ + sin4θ sin 4θ 4 cos3 θ sin θ − 4 cos θ sin3 θ         tan 4θ =  ______ =  __________________________ cos 4θ cos4 θ − 6 cos2 θ sin2 θ + sin4 θ 1 ( ______ (4 cos3 θ sin θ − 4 cos θ sin3 θ) cos4 θ ) c 6 a           =   ___________________________________ 1 ( ______ (cos4 θ − 6 cos2 θ sin2 θ + sin4 θ) cos4 θ ) 4 tan θ − 4 tan3 θ = __________________         1 − 6 tan2 θ + tan4 θ c x = 0.20, 1.50, −5.03, −0.67 (2 d.p.) Exercise 1E 1 a 1 + z + … + z  ( e )  − 1 z  − 1 _________ = _______       =   2n−1 __  πi  2n 2n n πi __ z−1 e  n − 1 _______ e   2πi− 1 ______ 1−1   =   __   = 0 =   __ πi πi e  n − 1 e  n − 1 n+1 __  πi n )   − 1 z  n+1− 1 ( e_________       =   b 1 + z + … + z  n = ________ πi __ z−1 e  n − 1 π πi   ___ − 2 cos  ___    πi   − ___ πi πi __ __ − e    n − 1 __________ − e2n− e 2n _________ 2n e  πi+  n − 1 ________ ________   =   __   =   ___   =   =   __ πi πi πi   π     πi   − ___ 2i sin  ___  e    n − 1 e    n − 1 e − e 2n π i cos  ___  2n π ___ = _______   π   = i cot  2n  sin  ___  2n πi 13 __ ( e  2 )  − 1 _____ z  13− 1 _________ i−1   =       =   __   = 1 2 1 + z +… + z  12 = _______  πi  z−1 i−1 2 − 1 e 8πi 8 __     ____     4 = 2  4e   2πi= 16 3 z  8 = 22e z  8− 1 __________ 16 − 1 15 1 + z + … + z  7 = ______    =     = ___    = −15i z − 1 (1 + i) − 1 i 2n 2n  13 cos θ + __  19 cos 2θ + …) 4 a C + iS = (1 + __ + i(__   31  sin θ + __  19  sin 2θ + …) = 1 +  __13 (cos θ + i sin θ) + __  19 (cos 2θ + i sin 2θ) + … 3 1 1 = 1 +  __13 e  iθ + ___     2  e  2iθ+ … = ________     = _______   iθ 1 3  1 − __     e  iθ 3 − e  3 3(3 − e  −iθ) 3(3 − e  −iθ) 3 _______________  iθ = _______________        =             b  _______ 3 − e  (3 − e  iθ)(3 − e  −iθ) 10 − 3(e  iθ + e  −iθ) 3(3 − (cos θ − i sin θ)) __________________ 9 − 3 cos θ + 3i sin θ       = ___________________       =       10 − 3(2 cos θ) 10 − 6 cos θ 9 − 3 cos θ 3 sin θ         So C = ___________      and S = ___________ 10 − 6 cos θ 10 − 6 cos θ ( (e  2 e   2  − e−   2 ) _______________ e6iθ (e 2 − e 2 ) e   13iθ− 1 _______________   =             =         = ________   iθ iθ  iθ iθ iθ __ __ __ __  iθ  − __         −    e  − 1 e 2 − e 2 e 2 ( e 2 − e 2 ) 13θ 13θ 13iθ _____ e  6iθ(2i sin  ____  ) e  6iθ sin  ____    13iθ − _____     e  6iθ( e 2 − e 2 ) _____________ 2 2 _______________ __________     =              =     iθ  __  iθ  − __ θ θ 2i sin  __  sin  __  e 2 − e 2 2 2 13θ   (cos 6θ + i sin 6θ) sin  ____ 2 =  ______________________          θ sin  __  2 13θ θ θ 13θ ____   cosec  __  = cos 6θ sin     cosec  __  + i sin 6θ sin  ____ 2 2 2 2 θ 13θ θ 13θ   cosec  __ , Q = sin 6θ sin  ____   cosec  __  So, P = cos 6θ sin  ____ 2 2 2 2 4π π ___ 6π π ___ 8π 2π ____ 10π 5π ____ 12π 2π π ___    ,    , __    ,    , __    ,    , ___    ,    , ___    ,      ___ , __ 13 6 13 3 13 2 13 3 13 6 13 n n C + iS = 1 + (  )  (cos θ + i sin θ) + (  )  (cos 2θ + i sin 2θ) 1 2 n + … + ( )  (cos nθ + i sin nθ) n n iθ n 2iθ n niθ   + (  )  e   + … + ( )  e   = 1 + (  )  e n 1 2 n iθ n iθ 2 n iθ n   + (  )  ( e  )  + … + ( )  ( e  )  = 1 + (  )  e n 1 2 n n iθ  __ ___  iθ  __  iθ  −  __  niθ  θ iθ n 2 2 2 = (1 + e  )  = ( e (e + e ))  = e 2 (2 cos  __ )  2 n n nθ nθ θ θ = ( 2 cos  __ )  cos  ___  + i (2 cos  __ )  sin  ___  2 2 2 2 n nθ θ __ ___ So, C = ( 2 cos    )  cos     2 2 n nθ θ nθ __ ___ 2 cos    )  sin     sin  ___   2 2 nθ S ( 2 ______          =     = tan  ___   __  = _______________ n nθ 2 C ___ nθ θ (2 cos  __ )  cos  ___  cos     2 2 2 (2 + e  iθ)(2 + e  −iθ) = 4 + 2e  iθ + 2e  −iθ+ 1 = 5 + 2(e iθ + e  −iθ) = 5 + 2(2 cos θ) = 5 + 4 cos θ b 7 a 13iθ _____ 13iθ _____ _____  13iθ   13iθ − _____   b C − i S = 1 −  __12 (cos θ + i sin θ) + __  14 (cos 2θ + i sin 2θ) −  __18 (cos 3θ + i sin 3θ) + … = 1 − __  12 e  iθ + __   14 e  2iθ − __   18 e  3iθ+ … 1 1   2  (e  iθ)  2 − ___   3  (e  iθ)  3+ … = 1 −  __12 e  iθ + ___ 2  2  2(2 + e  −iθ) 2 1   = ________________          = ______      = ________   iθ iθ −iθ 1 iθ __ 1 +   e  2 + e  (2 + e  )(2 + e  ) 2 2(2 + e  −iθ) 2(2 + (cos θ − i sin θ))       =  __________  = ___________________     5 + 4 cos θ 5 + 4 cos θ 4 + 2 cos θ − 2i sin θ         =  __________________ 5 + 4 cos θ 2 sin θ 4 + 2 cos θ     So, C =  __________ , S = __________ 5 + 4 cos θ 5 + 4 cos θ Exercise 1F 1 a z = 1,__i, −1, −i __ √ √     3 3 b z =  ___  + _ 12 i, −  ___  + _ 12 i, −i 2 __2 __     3√3 3√3     i, −  _32 − ____     i c z = 3, −  _32 + ____ 2 2 d z = 2 + 2i, −2 + 2i, 2 − 2i, −2 − 2i 215 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 215 22/09/18 3:09 PM Answers e z = 1__+ i, −1 + i, 1__ − i, −1 − i f z=√   3 − i, 2i, −√3   − i 2π  + i sin  ___ 2π , 2 a z = cos 0 + i sin 0, cos  ___ 7 7 4π , cos  ___ 6π  6π  + i sin  ___ 4π  + i sin  ___ cos  ___ 7 7 7 7 2π   + i sin −  ___ 2π   , cos −  ___ 4π   + i sin −  ___ 4π   , cos −  ___ 7 7 7 7 6π   + i sin −  ___ 6π   cos −  ___ 7 7 π π     , __ b z = 2 cos −      + i sin −  __ 8 8 3π   + i sin   ___ 3π     , 2 cos    ___ 7π   + i sin   ___ 7π     , 2 cos    ___ 8 8 8 8 5π   + i sin −  ___ 5π     2 cos −  ___ 8 8 π  + i sin  __ π   , 2 cos  ___ 3π  + i sin  ___ 3π   , c z = 2 cos  __ 5 5 5 5 π   + i sin −  __ π     , 2(cos π + i sin π), 2 cos −  __ 5 5 3π   + i sin −  ___ 3π     2 cos −  ___ 5 5 __ __ π   + i sin  ___ π    , √ 3π  + i sin  ___ 3π   ,   2 cos  ___ 2   cos  ___ d z=√ 12 12 4 4 __ − 7π     7π   + i sin   ____ √ 2 cos −  ___   12 12 __ π    + i sin −  ___ π      ,   2 cos −  ___ e z=√ 12 12 __ __ 5π   + i sin    ___ 5π     , √ 11π   + i sin    ____ 11π     , 2 cos   ___ 2   cos   ____ √  12 12 12 12 __ 7π   + i sin −  ___ 7π     2 cos −  ___ √  12 12 5π 5π     , ___ f z = 4 cos −      + i sin −  ___ 18 18 7π   + i sin    ___ 7π     , 4 cos −  ____ 17π   + i sin −  ____ 17π     4 cos    ___ 18 18 18 18 ( ( ) ) ( ( ) ( ( ( ( ) ( ( ) ( ( ( ) )) ( ) ( ( )) ( ( ) ) ( ( ( ( )) ( ( ) ( )) ) ( ( ( ( ) ) ) ( ) ( )) 1   __ 4 1.80i 1   __ 4 −1.34i Im ( )) 5πi ( )) ( 2e 6 ) )) 1   __ 4 −2.91i πi 2e 3 ) ( )) ( ( ) ( ( ) ( )) ( ( ) ( ( ) ( )) ( )) ( ( ) ( ( ) 1   __ 4 0.23i ) 2π 2π 4π 4π ⇒ 1 + cos  ___  + i sin  ___  + cos  ___  + i sin  ___  5 5 5 5 2π 2π 4π 4π ___ ___ ___    )= 0 + cos(   )− i sin(   )+ cos(   )− i sin(___ 5 5 5 5 4π 2π ⇒ 1 + 2 cos  ___  + 2 cos  ___ = 0 5 5 4π 2π cos  ___  + cos  ___  = −  __12 5 5 2π 6 a r = 4, θ = −  ___  3__ πi __ 5πi __ 2πi __ −  __ πi   ____   __  −  ____  2 e 6 , √ 2 e 3 , √ 2 e 6 , √ 2 e 3 b z=√ ( ( , 5 e __ , 5 e __ , 5 e 3 a z = 5__e −2.39i √ b z=√ __  e 3 −0.29i, √ 3  e1.80i ,  e 3 __ __ __ c z=√  e 2 0.57i__ ,z=√ 2  e2.14i, z__= √ 2  e−1.00i, z = √ 2  e−2.57i √ √   3   3 4 a z = −   __12 +   ___  i, −2, −  __12 −   ___  i 2 2 Im b 3 1 – + i 2 2 )) )) – 2e – πi 3 Re 2πi 2e 3 __ 7πi  −  ____ __ __ πi   −  ___ ____  5πi  __ _____  11πi  7 √ 2 e 12 , √ 2  e 12, √ 2  e 12 , √ 2  e 12 Im 5πi 2e 12 11πi π 2 π 2 π 2 π 2 2e 12 – Re – 2e πi 12 7πi 2e 12 –2 – O __ π 8 a r=√ 8  , θ = __     6 2πi  8πi  4πi   ____  ____ −  ____ b w = 4e 9 , w = 4e 9 , w = 4e 9 Re __  πi  _ c centre (−1, 0) radius 1 2π 2π 4π 4π 5 a z = 1, cos (___    ) + i sin (___    ), cos (___    ) + i sin (___    ), 5 5 5 5 2π 2π 4π 4π ___ ___ ___ ___ cos (−    ) + i sin (−    ), cos (−    ) + i sin (−    ) 5 5 5 5 b z  1 + z  2 + z  3 + z  4 + z  5 = 0 2π 2π 2π 4π 4π 1 + cos  ___  + i sin  ___  + cos  ___  + i sin  ___ + cos(−  ___ ) 5 5 5 5 5 2π 4π 4π + i sin(−  ___ )+ cos(−  ___ )+ i sin(−  ___ )= 0 5 5 5 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 216 ____  5πi  ____  7πi  b Expressing as a product_of the linear factors: _ _ _ √ √ √   ___   2 √ 2   2   2 ___ (z + 1)(z − i)(z + i)(z −     −     i)(z − ___     + ___     i) 2 2 2 2 3 1 – i 2 2 216 ____  3πi  9 ae 4 , i, e 4 , −1, e 4 , −i, e 4 _ _ _ √ √   √ 2   2 √   2 2       + ___     − ___     i)(z + ___     i) (z + ___ 2 2 2 2 _ _   z + 1)(z  2+ √ 2   z + 1) = (z + 1)(z  2+ 1)(z  2− √ 2 = (z + 1)(z  2+ 1)(z  4+ 1) Therefore (z2 + 1) and (z4 + 1) are factors. Challenge __  πi  πi  −  __ ____  2πi  2πi  −  ____ a 1, e3 , e 3 , e 3 , e 3 , e  πi 1 b Rewrite the equation as ( 1 + __    )  = 1. z kπi  ___   1 Then 1 + __     = e 3 for some k ∈ ℤ, by a. z ___  kπi  1 So,  __  = e 3 − 1 z 6 Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers _ 1 1   = _____________       z = _______   kπi kπi  ___ ___      kπi  ___  kπi  −  ___ e 3 − 1 e 6 (e 6 − e 6 ) kπ kπ − i(cos  ___  − i sin  ___ ) kπi  −  ___ 6 6 ie 6 1 =  _____________     = −  ________  = ___________________           ___  kπi  kπ kπ kπ ___ ___ ___ 6 2 sin     2 sin     e 2i sin     ( 6) 6 6 _  )_i, 5 a −8 − 8i,_4(−1 + √ 3 )_− 4(1 + √ 3 _ 4(1 − √ 3 _ ) + 4(1 + √ 3 _ )i, 4(1 + √ 3 ) + 4(1 − √ 3  )i, −4(1 + √ 3 ) − 4(1 − √ 3  )i Im 4(1 – 3) + 4(1 + 3)i kπ kπ sin  ___  + i cos  ___  8 + 8i kπ 6 6         12 i cot  ___  = −  _______________   = −  __12 − __ 6 kπ 2 sin  ___  –4(1 + 3) – 4(1 – 3)i 6 kπ   12 i cot(−  ___ ) = −  __12 + __ 6 so take k to be −1, −2, −3, −4, −5. Exercise 1G _ ___   πi   _ Re 3) + 4(1 – 3)i _ 1 a (0,4), (−2√ 3 , −2), (2√ 3  , −2) b (5,0), (−5,0), (0,5), (0, −5) _ 4π 16π 4π 16π c (2 cos  ___ , 2 sin  ___ ), (−1,√ 3  ),(2 cos  ____ , 2 sin  ____ ), 15 15 15 15 22π 28π 22π 28π ____ ____ ____ ____ 2 cos   , 2 cos    , 2 sin      , 2 sin     ( 15 15 ) ( 15 15 ) __ _ 7π 7π d (2, 2), (2√ 2  cos  ___ , 2√ 2  sin  ___ ), 12 12 _ _ 11π 11π (2√ 2  cos  ____ , 2√ 2  sin  ____ ), 12 12 _ _ 15π 15π (2√ 2  cos  ____ , 2√ 2  sin  ____ ), 12 12 _ _ 19π 19π (2√ 2  cos  ____ , 2√ 2  sin  ____ ), 12 12 _ _ 23π 23π (2√ 2  cos  ____ , 2√ 2  sin  ____ ) 12__ 12 _ __ _ 2 (3, −2), (__   12 (3 + 5√ 3 ), __   12 (11 + √ 3  )), (  __12 (3 − 5√ 3 ), __   12 (11 − √  ) 3 ) _ π π 3 π 3 3 π π 3 π 3 3 4(1 + 7πi  −  ____ 3 √ 6  e12, √ 6  e 12 Im πi 6e 12 O –8 – 8i 4(–1 + 3) – 4(1 + 3)i _ _ _ b 128i, −64(√ 3 + i), 64(√ 3 − i), Area = 12288√ 3   6 Let the position of the ant be denoted by a + bi, where a in the number of units forward and b is the number of units to the right from its initial position. In this notation, walking forwards one unit at an angle of θ to the right corresponds to adding e iθto its position. 2π 4π 6π Since the angles are 0, ___    , ___     and ___    , the final position of 9 9 9 the ant is: ____ ____ ____ ____ ____ ____ ( e 9 ) − 1  2πi   4πi   6πi   2πi   2πi   2πi    ____   1 + e 9 + e 9 + e 9 = 1 + e 9 + ( e 9 ) + ( e 9 ) = _________  2πi  e 9 − 1 ____  4πi  4π  4π  4πi  8πi  ____ e 9 (2i sin  ___  4πi  ____  4πi  −  ____ sin  ___  ____ 9 ( 9 − e 9 ) 9  πi  9 ) ______ e e 9   e__ e − 1 _______ ______________ _____________ 3 =   ____      =            =     =    π  πi  __ __ __  πi   2πi   πi  __  πi  −  __ π sin   __ 9 9 e (2i sin    ) e − 1 e 9 (e 9 − e 9 ) 9 9 sin 4π  ______   9 ______ So the distance from its initial position is   π   sin  __  9 2 3 4 ____  2πi  Mixed exercise 1 Re 1 a e  iθ= cos θ + i sin θ, e  −iθ= cos θ − i sin θ e  iθ + e  −iθ= 2 cos θ, so cos θ = __  12 (e  iθ + e  −iθ) b cos A cos B =  __12 (e   iA + e  −iA) × __   12 (e   iB + e  −iB) 3(1 3 – i) =  __14 (e   iA + e  −iA)(e  iB + e  −iB) 7πi – 12 =  __14 (e   i(A+B) + e  i(A−B) + e  i(B−A) + e  −i(A+B)) 6e 4 ___ 5πi  4 −  ____ ___ πi  4 ___ −  __ ____  3πi  4 ___ = __   14 ((e  i(A+B) + e  −i(A+B)) + (e  i(A−B) + e  −i(A−B))) ____  7πi  4 a √ 12  e 8 , √ 12  e 8 , √ 12  e 8 , √ 12  e 8 Im 4 4 7πi π 2 π 2 π 2 π 2 12e 8 4 – 5πi 12e 8 12e 3πi 8 Re 4 – πi 12e 8 =  __14 (2 cos(A + B) + 2 cos(A − B)) cos(A + B) + cos(A − B)          =  _______________________ 2 2 n = 1; LHS = r(cos θ + i sin θ) RHS = r  1(cos θ + i sin θ) = r(cos θ + i sin θ) As LHS = RHS, the equation holds for n = 1. Assume the equation holds for n = k, k ∈ ℤ  +. i.e. z  k= r  k (cos kθ + i sin kθ) With n = k + 1, the equation becomes: z  k+1= z  k × z = rk(cos kθ + i sin kθ) × r(cos θ + i sin θ) = rk+1((cos kθ cos θ − sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)) = rk+1(cos(k + 1) θ + i sin (k + 1) θ) by the addition formulae. Therefore, the equation holds when n = k + 1. b 3i 217 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 217 22/09/18 3:09 PM Answers If the equation holds for n = k, then it has been shown to be true for n = k + 1. As the equation holds for n = 1, it is now also true for all n ∈ ℤ  +by mathematical induction. 3 cos 7x + i sin 7x 4 a 16 b 256 5 a Let z = cos θ + i sin θ z  n = (cos θ + i sin θ)  n= cos nθ + i sin nθ 1 ___   n  = z  −n = (cos θ + i sin θ)  −n= cos (−nθ) + i sin (−nθ) z  = cos nθ − i sin nθ 1 ⇒ z  n + ___   n  = cosnθ + i sinnθ + cosnθ − i sinnθ = 2 cosnθ z  3 1   2  = 2 cos 6θ + 6 cos 2θ b z2 + __ ( ) z c a = _ 14 , b = _  34  ∫  __6π ∫ __   π 3 1 d   cos32 θ dθ =   (_  4cos 6θ + _ 4cos 2 θ)dθ 0 6 0 _   π 3 √ 1 sin 6θ +  _38 sin 2θ]    = __  16   3  = [  __ 24 __ 6 0 1 6 a If z = cos θ + i sin θ,then 2 cos θ = z + __     z 5 1 So 2  5 cos  5 θ = ( z + __    )  z 1 1 1 1 1 ___ ___ 5 = z  5 + 5C1z  4( __ )+ 5C2z  3  ___2  + 5C3z  2 ___ ( z  ) (  z  3 ) + C4z(  z  4 ) +   z  5  z 10 5 ___ 1    + ___   3  +   5  = z  5+ 5z  3+ 10z + ___ z  z  z 1 1 1   5  + 5 z  3 + ___   3  + 10(z + __    ) = z  5 + ___ ( ( z z  ) z  ) = 2 cos 5θ + 5(2 cos 3θ) + 10(2 cos θ) So 32 cos  5 θ = 2 cos 5θ + 10 cos 3θ + 20 cos θ 1 cos  5 θ = __  16 (cos 5θ + 5 cos 3θ + 10 cos θ) __ b   16 15 1 7 a If z = cos θ + i sin θ, then 2i sin θ = z − __     z 6 1 So (2i)  6 sin  6 θ = ( z − __    )  z 1 1 1 6 6 5 __ 6 3 ___ = z  − C1z  (   )+ 6C2z  4 ___ (  z  2 ) − C3z  (  z  3 ) z 1 1 1 ___ ___ 6 + 6C4z  2 ___ (  z  4 ) − C5z(  z  5 ) +   z  6  15 6 1   2  −  ___4  + ___     = z  6− 6z  4+ 15z  2− 20 + ___ z  z  z  6 1 1 1   6  − 6 z  4 + ___   4  + 15 z  2 + ___   2  − 20 = z  6 + ___ ( ( ( z  ) z  ) z  ) = 2 cos 6θ − 6(2 cos 4θ) + 15(2 cos 2θ) − 20 So, − 64 sin  6 θ = 2 cos 6θ − 12 cos 4θ + 30 cos 2θ − 20 1 b cos  6 θ ≡ __   32   (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10) π __ c     4 8(cos θ + i sin θ)  6= cos 6θ + i sin 6θ = cos  6 θ + 6C1 cos  5 θ(i sin θ) + 6C2 cos  4 θ (i sin θ)  2 + 6C3 cos  3 θ (i sin θ)  3 + 6C4cos  2 θ (i sin θ)  4 + 6C5cos θ (i sin θ)  5 + (i sin θ)  6 = cos  6 θ + 6i cos  5 θ sin θ − 15 cos  4 θ sin  2 θ − 20i cos  3 θ sin  3 θ + 15 cos  2 θ sin  4 θ + 6i cos θ sin  5 θ − sin  6 θ Equating imaginary parts gives sin 6θ = 6 cos  5 θ sin θ − 20 cos  3 θ sin  3 θ + 6 cos θ sin  5 θ = 2 sin θ cos θ(3 cos  4 θ − 10 cos  2 θ sin  2 θ + 3 sin  4 θ) = sin 2θ(3 cos  4 θ − 10 cos  2 θ(1 − cos  2 θ) + 3 (1 − cos  2 θ)  2) = sin 2θ(3 cos4 θ − 10 cos2 θ(1 − cos2 θ) + 3(1 − 2 cos2 θ + cos4 θ)) = sin 2θ(16 cos  4 θ − 16 cos  2 θ + 3) 218 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 218 9 a(cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5C1cos  4 θ(i sin θ) + 5C2cos  3 θ (i sin θ)  2 + 5C3cos  2 θ (i sin θ)  3 + 5C4cos θ (i sin θ)  4 + (i sin θ)  5 = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ − 10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ + i sin  5 θ Equating real parts gives cos 5θ = cos  5 θ − 10 cos  3 θ sin  2 θ + 5 cos θ sin  4 θ = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ (1 − cos  2 θ)  2 = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ(1 − 2 cos  2 θ + cos  4 θ) = 16 cos  5 θ − 20 cos  3 θ + 5 cos θ _ _ b –1, __   14 (1 + √ 5 ) ≈ 0.809,  __14 (1 − √ 5 ) ≈ − 0.309 10 a Let z = cos θ + i sin θ 1 5 __ 5 5 5 5 ( z −  z ) = (2i sin θ)  = 32 i  sin  θ = 32i sin  θ 2 1 1 = z  5+ 5 z  4(−  __ )+ 10 z  3 (−  __ ) z z 5 3 1 1 4 1 + 10 z  2 (−  __ ) + 5z (−  __ ) + ( −  __ )  z z z 10 5 ___ 1    + ___   3  −   5  = z  5− 5 z  3+ 10z − ___ z z  z  1 1 1   5 )− 5(z  3 − ___   3 )+ 10(z − __    ) = ( z  5 − ___ z z  z  = 2i sin 5θ − 5(2i sin 3θ) + 10(2i sin θ) So 32i sin  5 θ = 2i sin 5θ − 10i sin 3θ + 20i sin θ 1 ⇒ sin  5 θ = __  16 (sin 5θ − 5 sin 3θ + 10 sin θ) π 5π b 0, __    , ___     6 6 11 a ( cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ + 10 i  2 cos  3 θ sin  2 θ + 10 i  3 cos  2 θ sin  3 θ + 5 i  4 cos θ sin  4 θ + i  5 sin  5 θ ⇒ cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ − 10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ − i sin  5 θ Equating the real parts: cos 5θ = cos  5 θ − 10 cos  3 θ sin  2 θ + 5 cos θ sin  4 θ = cos θ(cos  4 θ − 10cos  2 θ(1 − cos  2 θ) + 5(1 − cos  2 θ)  2) = cos θ(16 cos  4 θ − 20 cos  2 θ + 5) b If cos 5θ = 0, then cos θ(16 cos  4 θ − 20 cos  2 θ + 5) = 0 If x = cos θ, then x(16x  4− 20x  2_ + 5) = 0 which has _ √ √ 80  5   20 ± 5 ± solutions x = 0 and x  2 = _________     = _______    , by the 32 8 quadratic formula. π π Since θ = ___    is a solution to cos 5θ = 0, x = cos  ___  10 10 must be a solution to x(16x  4− 20_ x  2+ 5) = 0. √   5 5 ± π Since x ≠ 0, cos  2  ___  = x  2 = _______    , for some choice ( 10 ) 8 of sign. 3π To find which, note that θ = ___    gives another 10 3π π solution and cos  ___  . cos  ___ by looking at the graph. 10 10 π Hence θ = ___    corresponds to the larger of the two 10 _   5 + √ 5 π     = _______     solutions and cos  2 ___ ( 10 ) 8 __ __ 5 5 5 − √   5 − √   3π 7π      , cos2( ___ ) = _______  , c cos2( ___ ) = _______ 10 8 __ 10 8 5 5 + √   9π    ) = _______     cos2(___ 10 8 3 tan θ − tan3 θ      12 a tan3θ ≡ ______________      1 − 3 tan2 θ Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers 1 − 3 tan2 θ 1 − 3 cot−2 θ b cot 3θ = ______________         = _______________          3 tan θ − tan3 θ 3 cot−1 θ − cot−3 θ cot3 θ − 3 cot θ         =  _____________ 3 cot2 θ − 1 Challenge 6 1 Rewrite the equation as ( 1 + __    )  = 1. z ___  kπi  1 Then 1 + __     = e 3 for some k ∈ ℤ, since it is a sixth root of z unity. 13 C + i S = 1 + k(cos θ + i sinθ) + k  2 (cos 2θ + i sin 2θ) + k  3 (cos 3θ + i sin 3θ) + … = 1 + ke  iθ+ k  2e   2iθ+ k  3e   3iθ+ …= 1 + ke  iθ + (ke  iθ)  2 + (ke  iθ)  3+ … 1 =  ________  , since |ke  iθ| = |k| , 1. 1 − ke  iθ 1 − ke  −iθ 1 − ke  −iθ         = ___________________       = ___________________      (1 − ke  iθ)(1 − ke  −iθ) 1 + k  2 − k(e  iθ + e  −iθ) 1 − k(cos θ − i sin θ) _______________ 1 − k cos θ     =          = __________________     1 + k  2− 2k cos θ 1 + k2 − 2k cos θ k sin θ        + i  _______________ 1 + k2 − 2k cos θ k sin θ 1 − k cos θ        and S = _______________          So C =  _______________ 1 + k2 − 2k cos θ 1 + k  2− 2k cos θ __ π π 2 cos −  __  + i sin −  __  14 a 4√   ( 4 )) ( ( 4) __ πi   −  ___ __ ____  7πi  __ ____  3πi  __ 9πi  −  ____ __ b z = √ 2 e 20, √ 2 e 20 , √ 2 e 4 , √ 2 e 20 , √ 2 e 17πi  −  _____ 20 Im c z2 z3 O z1 Re z5 15 a √ 2 e iπ   −  ___ 12 _ ____  7iπ  __ , √ 2 e , √ 2 e 12 CHAPTER 2 Prior knowledge check 1 a 1098 b 10 761 619.5 2 a Use the following: n n(n + 1)(2n + 1)   ∑ r2 = _______________       6 r=1 n n(n + 1)       ∑ r2 = ________ 2 r=1 and simplify to get the answer. b 10 073 dy d2y ____ 3 a  ___  = 3 cos 3x b   2  = −9 sin 3x dx dx Exercise 2A 1 a __   12  (r(r + 1) − r(r − 1)) = __  12  (r  2 + r − r  2 + r) = __  12  (2r) = r n n n r=1 r=1 r=1 b ∑   r =  __12  ∑  r (r + 1) −  __12  ∑  r(r − 1) r = 1:  __12 × 1 × 2 − __  12 × 1 × 0 z4 _ ___  kπi  1 So  __  = e 3 − 1 z kπi  −  ___ ie 6   1 1 1   = _____________ _____________ _______   =     = −           z =  _______ kπi  ___ ___ ___  kπi   kπi  kπ   kπi  ___  kπi  −  ___ kπ  2 sin  ___ e 3 − 1 e 6 (e 6 − e 6 ) e 6 (2i sin  ___ 6 6) kπ kπ kπ kπ ___ ___ ___ ___ i(cos     − i sin    ) sin     + i cos     kπ 6 6 6 6 1  − __             = −  _________________   1  i cot  ___    = −  _______________   = −  __ 2 2 6 kπ kπ 2 sin  ___  2 sin  ___  6 6 1  + it for t ∈ ℝ. So the points lie on the straight line z = −  __ 2 b r = 2:  __12 × 2 × 3 − __  12 × 2 × 1 9iπ  −  ____ 12 r = 3:  __12 × 3 × 4 − __  12 × 3 × 2 ⋮ Im r = n − 1:  __21 × (n − 1)(n) − __  12  (n − 1)(n − 2) B r = n:  __12  n(n + 1) − __  12  n(n − 1) When you add, all terms cancel except __  12  n(n + 1) n M 2π 3 Hence ∑  r =  __12  n(n + 1) 2π 3 r=1 2π 3 Re A C _ √ 2   π c r = ___    , θ = __     4 2 i d −  __  8 _ _ √ √   1 √_   _ 1 3 3    , − __   ), ( −1 − ___    , √ 3  − __    )  − 3 16 a (−1 + ___ 2 2 2 2 the two_vertices above respectively, b If z1 and z2 are _ then z1 − z2 = √ 3 − 2i √ 3   So the length of this side of the triangle is ______________ _ _ _ _ √3 |z1 − z2| = √ (     )2 + (−2√ 3 )  2  = √ 3 + 12 = √ 15  n(n + 3) 2 ______________         4(n + 1)(n + 2) n(3n + 5) 1 1            b ______________ 3 a  ___  − ________ 2r 2(r + 2) 4(n + 1)(n + 2) 1 1 1     = _____   − _____       4 a ____________ (r + 2)(r + 3) r + 2 r + 3 n   b ________   3(n + 3) (r + 1)! − r! ___________ r!(r + 1 − 1) _______ 1 r 1        ≡ ___________     ≡        ≡     5 a __  − _______ r! (r + 1)! r!(r + 1)! r!(r + 1)! (r + 1)! 1   b 1 − _______   (n + 1)! n(n + 2)   6  ________ (n + 1)2 1 1    , which 7 a Method of differences yields  ___  − _________ 10 2(2n + 5) n  , so a = 10 and b = 25. simplifies to _________   10n + 25 219 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 219 22/09/18 3:09 PM Answers 1 1 b For n = 1, _____     = ________    . Assume true for n = k. 5 × 7 10 + 25 Let n = k + 1, then k+1 1 k 1       = _________  ∑______________     + _______________       10k + 25 (2k + 5)(2k + 7) r = 1 (2r + 3)(2r +5) k+1 2k2 + 7k + 5        = _____________       =  _______________ (2k + 1)(2k + 7) 10(k + 1) +25 Therefore true for all values of n. 1 4 1   8 Method of differences yields __    (______   − ______     , 3 3r − 2 3r + 4 ) n(15n + 17)    which simplifies to _______________     , so a = 15 and (3n + 1)(3n + 4) b = 17. 9 Method of differences yields (n + 1)  + n  − 1  − 0  = 2n  2+ 2n = 2n(n + 1) so a = 2. 3n 10 a Method of differences yields _________    , 12n + 16 so a = 3, b = 12 and c = 16. 2 2 2 2 3n − 3 6n   − ________     b Simplify: _________   24n + 16 12n + 4 1 1    is a sum, not a difference, 11 The general term __     + _____ r r+1 so the terms will not cancel out, and the method of differences cannot be used in this case. n 1  and apply method of 12 Recognise this is ∑ _______ r = 1 r(r + 2) 1 1 1 1       − ________   + ________     to differences. Simplify __     + __ 2 4 ( 2(n + 1) 2(n + 2)) 2n + 3 3      , stating a = 2 and b = 3. obtain __     − ______________ 4 2(n + 1)(n + 2) 1 1   − ______     13 a ______   2r + 1 2r + 5 b 0.0218 Challenge a k = 11 b a = 11, b = 48, c = 49 Exercise 2B 1 a f  ′(x) = 2e2x, f  ″( x) = 4e2x, f‴ (x) = 23e2x = 8e2x, f (n)(x) = 2ne2x b f  ′(x) = n(1 + x)n − 1, f  ″(x) = n(n − 1)(1 + x)n − 2, f‴ (x) = n(n − 1)(n − 2)(1 + x)n − 3, f (n)(x) = n! c f  ′(x) = ex + xex, f  ″(x) = 2ex + xex, f‴ (x) = 3ex + xex, f (n)(x) = nex + xex d f  ′(x) = (1 + x)−1, f  ″(x) = −(1 + x)−2, f‴ (x) = 2(1 + x)−3, f (n)(x) = (−1)n − 1(n − 1)!(1 + x)−n dny 2 a ____n = 3ne2+3x = 3ny b e2 dx dy 3 a ___  = 3 × cos 3x × 2 sin 3x dx = 6 sin x cos x = 3 sin 6x d2y d3y d4y ____ b  2  = 18 cos 6x, ____   3  = −108 sin 6x, ____   4  = −648 cos 6x dx dx dx c 648 4 af′ (x) = 2xe  −x− x  2e   −  x f″( x) = (2e  −x− 2xe  −  x) − (2xe  −x− x  2e   −  x) = e  −  x(2 − 4x + x  2) f‴(x) = e  −  x(−4 + 2x) − e  −  x(2 − 4x + x  2) = e  −  x(−6 + 6x − x  2) bf″  ″ (x) = e  −  x(6 − 2x) − e  −  x(−6 + 6x − x  2) = e  −  x(12 − 8x + x  2) so f″  ″ (2) = e  −  2(12 − 16 + 4) = 0 220 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 220 dy 5 a Given that y = sec x,  ___  = sec x tan x dx 2 d  y ____   2 = sec x(sec  2 x) + (sec x tan x)tan x dx  = sec x(sec  2 x + tan  2 x) = 2 sec  3 x − sec x ____ d  3y b  3  = 6 sec  2 x(sec x tan x) − sec x tan x dx  = sec x tan x(6 sec  2 x − 1) _ π d  y (√_)  = 2   (1)(6(2) − 1) = 11√   2   When x =  __ , ____ 4 dx  2 2 2y 2 dy d   dy d d     (2y ___ )= 2y ____2  + 2(___    )  6 a ____2 (y  2) = ___ dx dx dx dx  dx  3 dy ____ d  3y d  2y ____ ___ b 2 (y  dx  3  + 3  dx  ×  dx  2 ) x 1 1 _ × 1 + _______      _  = _______   _  7 af′ (x) =  ___________ 2  ( √ √ 1 + x  2  ) √ 1 + x  2  x + 1 + x   _ So √ 1 + x  2  f′ (x) = 1 b Differentiating this equation w.r.t. x, _ x √ 1 + x  2  f″ (x) + _______   _ f′ (x) = 0 √ 1 + x  2  ⇒ (1 + x  2) f″ (x) + xf′ (x) = 0 c Differentiating this equation w.r.t. x ((1 + x  2) f‴ (x) + 2xf″ (x))+ (f′ (x) + xf″ (x))= 0 ⇒ (1 + x  2) f‴ (x) + 3xf″ (x) + f′ (x) = 0 d f9(0) = 1, f 0(0) = 0, f -(0) = −1 Exercise 2C 1 a f(x) = (1 − x)  −  1 ⇒ f(0) = 1 f′ (x) = −1(1 − x)  −  2(−1) = (1 − x)  −  2 ⇒ f′ (0) = 1 f″ (x) = −2(1 − x)  −  3(−1) = 2(1 − x)  −  3 ⇒ f″ (0) = 2 f‴ (x) = − (3 × 2)(1 − x)  −  4(−1) = (3 × 2)(1 − x)  −  4 ⇒ f‴ (0) = 3! General term: f  (r)(x) = r(r − 1)…2(1 − x)  −(r+1) = r!(1 − x)  −(r+1) ⇒ f  (r)(0) = r! f″ (0) f  (r)(0) Using f(x) = f(0)+ f′ (0)x + _____    x  2+ … + ______    x  r+ … 2! r! r! 2    x  2+ … + __    x  r+ … (1 − x)  −1= 1 + x + ___ 2! r! = 1 + x + x  2+ … + x  r+ … _ b f(x) = √ 1 + x = (1 + x)    2  ⇒ f(0) = 1 ( ) ( )( ⇒ f″ (0) = −  _14  1 _ 1 _ f′ (x) =  _12 (1 + x)  −  2  3 _ f″ (x) = _  12  −  _12  ( 1 + x)  −  2  5 _ 3 1 1 _ _ _ f‴ (x) =  2  −   2  −   2  ( 1 + x)  −  2  ) ⇒ f′ (0) = _   12  ⇒ f‴ (0) =  _38  Using Maclaurin’s expansion, 3 1 _ _ _ (  8 ) (−   4 ) √ 1 + x = 1 + _  1 x + ____    x  2 + ___    x  3− … 2 2! 3! x   x x  ___     +    − … = 1 + __     − ___ 2 3 2 8 16 2 f(x) = e  sinx ⇒ f(0) = 1 f′ (x) = cos xe  sinx ⇒ f′ (0) = 1 f″ (x) = cos  2 xe  sinx− sin xe  sinx ⇒ f″ (0) = 1 Using Maclaurin’s expansion, 1 e  sinx = 1 + x + ___    x  2+ … = 1 + x + _  12 x  2… 2! 3 a f(x) = cos x ⇒ f(0) = 1 f′ (x) = − sin x ⇒ f′ (0) = 0 f″ (x) = − cos x ⇒ f″ (0) = −1 f‴ (x) = sin x ⇒ f‴ (0) = 0 f″  ″ (x) = cos x ⇒ f″  ″(0)   = 1 Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers The process repeats itself every 4th derivative. Using Maclaurin’s expansion, (−1)r 2r −1 1 cos x = 1 + ___     x  2 + ___     x  4+ … + _____     x  + … 2! 4! (2r)! r 2 4 (−1) 2r x  x     + … + _____     x  + … = 1 − ___     + ___ 2! 4! (2r)! x  2 x  4 π     with x = __    , b Using cos x ≈ 1 −  ___  + ___ 2! 4! 6 π  4 π  2    = 0.86605… which is     + _______ cos x ≈ 1 − ___ 72 31104 correct to 3 d.p. 4 a e = 2.718 (3 d.p.) b ln (  __65 )= 0.182 (3 d.p.) 5 a 1 + 3x +  __92 x 2 + __   92 x 3 + __   27 x4 + . 8 b 2x − 2x2 + __   83 x 3 − 4x4 + … 4 x c x2 − __    + … 3 π π π __ 6 cos x −     = cos x cos  __ + sin x sin  __  ( 4) 4 4 1 = ___   _  (cos x + sin x) √   2 x  2 x  4 x  3 x  5 1_ ___     + ___     + ___ =    ((1 − ___    − …)+ ( x − ___    − …)) 2! 4! 3! 5! √   2 2 3 4 x  x  x  1_   1 + x − ___     −  ___  +  ___ − …) =  ___ 2 24 √ 6   2( 7 a f(x) = (1 − x) ln (1 − x) −1 f9(x) = (1 − x)  2 × _____     + 2(1 − x)(−1) ln(1 − x) 1−x = x − 1 − 2(1 − x) ln(1 − x) −1 f 0(x) = 1 − 2((1 − x) × _____     + (−1) ln(1 − x)) 1−x = 3 + 2 ln(1 − x) b f(0) = 0, f9(0) = −1, f 0(0) = 3, f-(0) = −2 2 c −x + __   32 x 2 − __  13 x3   3 x  5 x 1 8 a sin x = x − ___     +  ___ − … = x − __   3 + ___   120 x   5− …  16 x 3! 5! 2 4   x  x 1 4   2 + __   24 x   − … cos x = 1 − ___     +  ___ − … = 1 −  __12 x 2! 4! 3 sin x − 4x cos x + x 1 1 5 = 3(x − __  16 x   3 + ___   120 x   5− …)− 4(x − __  12 x   3 + __   24 x   − …) + x 17 5 = __   32 x   3 − ___   120 x   + … b __   32 1 9 a f9(x) =  _____  × (− sin x) = − tan x cos x b f9(0) = 0, f 0(0) = −1, f-(0) = 0, f -(0) = −2 −x2 x4 c ____     −  ___  2 12 1 __ π __ d ln cos    = ln(2  −  2 ) = −  __12 ln 2 ( 4) And by the Maclaurin series we have also π 2 π 4  __     __    (4) (4) π __ _____ _____   ln cos    ≈ −     −   ( 4) 2 12 π2 π2     ln 2 ≈ ___     1 + ___ 16 ( 96 ) 10 f(x) = tan x ⇒ f(0) = 0 f′ (x) = sec  2 x ⇒ f′ (0) = 1 f″ (x) = 2 sec  2 x tan x ⇒ f″ (0) = 0 f‴ (x) = 4 sec2 x tan2 x + 2 sec4 x ⇒ f‴ (0) = 2 f″  ″ (x) = 16 sec4 x tan x + 8 sec2 x tan3 x ⇒ f″  ″(0)   = 0 f‴  ″ (x) = 16 sec2 x tan4 x + 88 sec4 x tan2 x +16 sec6 x ⇒ f‴  ″(0)   = 16 So the Maclaurin series is 0 0 16 2 0 + 1x + ___     x  2 + ___     x  3 + ___     x  4 + ___     x  5+ … 2! 3! 4! 5! 2 5 1 3 ___ __ = x +    x +    x + … 3 15 Challenge x   2 x  3 x  r ae  x= 1 + x + ___     +  ___ … +  ___ + … 2! 3! r! x  r x  r+1    ; ar+1 = _______     ar = ___ (r + 1)! r! |x| ar+1 x  r+1 r! ____ im  _______     × ___     = lim  _____  , 1 lr→∞ im      = lr→∞ ar (r + 1)! x  r r→∞ r + 1 x  2 x x  r   3 (1 + x) = x − ___     +  ___ − … + (−1)  r+1___    + … b ln r 2 3 x x  r   r+1    ; ar+1 = (−1)  r+2_____     ar = (−1)  r+1___ r r+1 a ( −1)  r+2x   r+1 _________ rx r+1 r ____ __________ im        ×     = lim  _____   lr→∞ im      = lr→∞ ar ( −1)  r+1x r+1   r r→∞ r + 1 x ______ = lr→∞ im      = |x| 1 1 + __     r So ln(1 + x) converges for −1 , x , 1 and diverges for x . 1. | | | | | | | | | | Exercise 2D | | x  + … valid for all values of x 1 a 1 − x + __  x   −  __ 2 6 3 b 1 + 4x + 8x2 + _____  32x + … valid for all values of x 3 2 3 x  + … valid for all values of x c e(1 + x + __  x   +  __ ) 2 6 2 3 x   −  __ x  − … d −x − __  x   −  __ 2 3 4 2 3 4 −1 < x , 1 x x  − ___   x   + _____  + … valid for all values of x   x   −  ________ e __ 2 48 3840 645 120 3 f 5 7 2 9x 3 + … ln 2 + ___  3x  − ___   9x  +  ___ 2 8 8 − _23  , x < _ 23  x  2 x  3 x  4 x  5 2 a ln (1 + x) = x −  ___  +  ___  −  ___  +  ___ − …, −1 , x < 1 2 3 4 5 x  3 ___ x  4 ___ x  5 x  2 ___ ___ ln (1 − x) = −x −     −     −     +    − …, −1 < x , 1 2 3 4 5 1+x x  3 x  5 _____     +  ___ + …)  )= ln (1 + x) − ln (1 − x) = 2(x + ___ ln(  1−x 3 5 As x must be in both the intervals −1 , x < 1 and −1 < x , 1, x must be in the interval −1 , x , 1. ___ ___ b (x +  3  +   5 + …), −1 , x , 1 c x = − _15; −0.0027% (4 d.p.) x3 x5 __   1  d 2 1 + __  35     = __     1 ln (4) = ln (2) ln _____ ( 1 − __ 3 ) 2 5 and the series from b gives 3 3 __ __ (   5 )  (   5 )      + ____    + 0.69 …  __35 + ____ 3 5 Which is ln 2 correct to 2 dp 3 5 (2x)  2 (2x)  3 4x  3     +  _____    + …  + … = 1 + 2x + 2x  2 + ____ 3 e   2x= 1 + 2x + _____ 3 2! 3! 2 3 2 3 (−x) (−x) x   x   _____ _____ ___ ___   +    + … = 1 − x +     −    + … e  −x= 1 − x +   2 2! 3! 6  32 x   2if terms in x  3and above may be So e  2x − e  −x≈ 3x + __ neglected. 221 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 221 22/09/18 3:09 PM Answers 4x3 4 a 3x sin 2x = 3x(2x − ____    + …) 3 2 4 9x 27x     +  _____  − …) cos 3x = ( 1 − ____ 2 8 So we get that 9x2 27x4 3x sin 2x − cos 3x = 6x2 – 4x4 – ( 1 – ____     +  _____   + … 2 8 ) b 21 2 x – __   59 x 4 + … = –1 +  __ 2 8 __   21   43 x   3 − __   23 x   4+ … b y = −1 + 2x + 2x  2 − __ 2 5x2 7x3 17x4   12  5 a x – ____     +  ____  –  _____  + …,  __12  , x < __ 2 3 4 2x x2 ____ 2x3 x4 b 2 ln 3 +  ___  – __     +     –  ____ + …, –3 , x < 3 3 9 81 162 2x4 4x6 2x8     –  ____  +  ____  – … 6 a 1 – 2x2 + ____ 3 45 315 x4 ____ 2x6 ____ x8 __ 2 b x –     +     –     + … 3 45 315 2 1 __ __ 7 p =   3 , q = –   8 17x3 11x4 8 a x + 2x2 + _____     +  _____   + … 3 6 b 1 9 a (1 − 3x) ln (1 + 2x) 8x  3 = (1 − 3x)(2x − 2x  2 + ____    − 4x  4+ …) 3 3− 12x  4+ … = 2x − 8x  2 + _   26   x   3 b e  2x sin x ( 2x)  2 (2x)  3 _____ (2x)  4   3 x =( 1 + 2x + _____     + _____    + …)     +    + …)(x − ___ 2! 3! 4! 3! x  3 4x  3 2x  4     + ____    + …)    + …)(x − ___ = (1 + 2x + 2x  2 + ____ 3 3 6 = x + 2x  2 + _   11   x  3+ x  4+ … 6 _ e   0.1 sin 0.1 b f(0.1) =  __________  = 1.103329… 0.1 Using the approximation in part a, f(0.1) = 1 + 0.1 + 0.00333333 = 1.103333…. This result is correct to 6 s.f. d  4y 13 a ____   4 = 16(sin 2x − cos 2x) = 16y dx  √ 1 + x  2  e  −x= (1 + x  2 )    2  e  −x c 1 _ ( x  2)  2   2 x  3 x  4 x = (1 +   _12  x  2+ ( _   12 )(−  _12 )  _____     −  ___  +  ___ + …)  + …)(1 − x + ___ 2! 2! 3! 4!   2 ___   2 ___ x x   4   3 ___ x x   4 x ___ ___ = ( 1 +     −    + …)(1 − x +     −     +    + …) 2 8 2 24 6 Challenge a γ = 1 + __  12 β   2 + __   38 β   4 b 19.6 years (3 s.f.) c 0.0027% d As β is larger, the error in γ is larger, so the approximation would be less accurate. Mixed exercise 2 2 1 1 1 a ___________       = _____     − _____     (r + 2)( r + 4) r + 2 r + 4 n n 1 1 2     − _____          =  ∑(_____ b  ∑____________ r + 4) r = 1 (r + 2)(r + 4) r=1 r + 2 1 1 1 _____ 1 =  __  + __     −     − _____     3 4 n+3 n+4 2 7n + 25n    = _______________       12(n + 3)(n + 4) 1 1 4   − ______       2 a ______________       = ______ (4r − 1)(4r + 3) 4r − 1 4r + 3 n 4 4n + 3 − 3 1 _______ 1     =  __   −     = __________     b ∑  ______________ 3 4n + 3 r=1 (4r − 1)(4r + 3) 3(4n + 3) 4n   = _________   3(4n + 3) c 0.00126 3 a (r + 1)  3 − (r − 1)  3 = (r  3+ 3r  2+ 3r + 1) − (r  3− 3r  2+ 3r − 1) = 6r  2+ 2 n n n b  ∑(6r2 + 2) = 6∑   r  2+ ∑   2 = 2n  3+ 3n  2+ 3n r=1 10 a 6 8 2 8 48 384 r=1 n r=1 r=1 = n(n + 1)(2n + 1) n So  ∑r = _ 16 n(n + 1)(2n + 1) b 1.711 (3 d.p.) 2 r=1 n (r + 1 r + 3) n(5n + 13) 2 2 2 2 =  __  + __        − _____     − _____     = _____________       2 3 n + 2 n + 3 3(n + 2)(n + 3) n 2 2 4     − _____     4 ∑ ____________     =  ∑ _____ 11 a e   px sin 3x (px)  (px)  (3x)  = (1 + px + _____     + _____    + …)    + …)(3x − _____ 2! 3! 3! 2 ( r=1 n r=1 r=1 x x x x     – … 1 –  __  +  __  –  ___  + ____ 4 n So 6 ∑r 2 = 2n  3+ 3n  2 + n = n(2n  2+ 3n + 1)   23  x  3 + _   16  x  4+ … = 1 − x + x  2 − _ 2 n So 6∑   r  2+ ∑   2 = 6∑   r  2+ 2n = 2n  3+ 3n  2+ 3n 3 3 )( )   p   p  x   x 9x      +  _____    + …  + … 3x − ____ = 1 + px + _____ 2 2 2 3 3 6 3(p2 − 3)x3     + … = 3x + 3px2 + __________ 2 13   b q = –2 p = __  32  k = –  __ 2 3 2 e  x 12 a e   x−lnx = e  x × e  −lnx = e  x × e  lnx  −1 = ___     x e   x sin x     ⇒ e  x−lnx sin x = _______ x x  2 ___ x   3 x  3 ___    + …) (1 + x +     +    + …)(x − ___ 2 6 6           ,x.0 f(x) =  ________________________________ x x  2 ___ x   2 x  3 ___ ___ = (1 + x +     +    + …)(1 −    + …) 2 6 6 x  2 = 1 + x + ___    ignoring terms in x  4and above 3 222 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 222 r=1 n (r + 1)(r + 3) r=1 5  ∑((r + 1)3 − (r − 1))3 = n(2n2 + 3n + 3) r=1 Calculate (2n)(2(2n)2 + 3(2n) + 3) − (n − 1)(2(n − 1)2 + 3(n − 1) + 3) Which gives that a = 14, b = 15, c = 3, d = 2 d  ny 6 a ____   n  = ( −2)  ne   1−2x dx  d  8y   1−2ln32 = 256e  1+ln32  b ____   8  = ( −2)  8e dx  1   ln _____   256  e  1 = __   e   = 256(e  1)( e 1024) = _____ 1024 4 7 a f ′(0) = __  12 , f″(0) = __  14 −2   x − e  x2(1 + e  x)e   x e   x(1 − e  x) (1 + e  x)  2e          = __________     b f ′′′(x) =  __________________________ x 4 (1 + e  )  (1 + e  x)  3 f ′′′(0) = 0 x x2 c ln 2 +  __  + __     + … 2 8 Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers 8 a 1 – 8x2 + __   32 x 4 – ___   256 x 6 + … 3 45 x + __  x  + … 3 b cos 4x = 1 − 2 sin 2 2x so 2 sin  2 2x = 1 − cos 4x = 8 x  2 − __   32 x  4 + ___   256 x  6+ … 3 45 so sin  2 2x = 4x  2 − __   16 x  4 + ___   128 x  6+ … 3 45 x  2 x  3 x  4     +  ___  +  ___ + … and 9Using e  x= 1 + x + ___ 2 6 24 x  4 x  2 ___ ___ cos x = 1 −     +    − … 2 24   cosx = e( e 4 x2  + ___ 1 −  __   x   2 24) x   −  __ 2 x    ___ 4 = e × e 2 × e24 2 x  2 __1 ___ x  2 x  4 ___ = e(1 + ( −    ) +   2 (−    )  + …)(1 + ___    + …) 2 2 24 x   2 x  4 x  4 x  2 x  4     +  ___  +  ___ + …)= e(1 − ___     +  ___ + …) = e(1 − ___ 2 8 24 2 6 10 –3x2 – 2x3 – … x3 11 x + __     + … 6 d 12 a ___    (e  x) dx x  2 x   3 x  4 x  r x  r+1 d     +  ___  +  ___ + … +  ___  + ________    + … = ___     1 + x + ___ ) 2! 3! 4! r! (r + 1)! dx ( r 2 3 (r + 1)x  3x  ____ 4x  2x ____ ___ _________  + … = 0 + 1 +     +     +    + … +   2! 3! 4! (r + 1)! x  3 ___  r x ___ x  2 ___ = 1 + x +     +    + … +     + … = e  x 2! 3! r! x  3 x  5 x  2r+1 d d     +  ___  − … + (−1)  r _________ + …     x − ___ b ___    (sin x) = ___ ) 3! 5! (2r + 1)! dx dx ( 2r 2 4 (2r + 1) x   5 x   3 x    + … = 1 −  ____  +  ____  − … + (−1)  r __________ 3! 5! (2r + 1)! x  2r x  2 x  4    … = cos x = 1 − ___     +  ___  − … + (−1)  r_____ 2! 4! (2r)! x  2 x  4 x  6 x  2r d d     +  ___  −  ___ … + (−1)  r _____      1 − ___ c  ___ (cos x) = ___ 2! 4! 6! (2r)! dx dx ( x  2r+2    + … + (−1)  r+1_________ ) (2r + 2)! 3 5 2r−1 2rx  4x  ____ 6x  2x ____ ___ r_______   = −     +     −    … + (−1)    2! 4! (2r)! 6! 2r+1 (2r + 2)x       + . + (−1)  r+1____________ (2r + 2)! 3 5 x  x  x  2r+1     +  ___  − … + (−1)  r_________     = − sin x = − x − ___ ( 3! 5! (2r + 1)! ) 2 4   x  x     −  ___ + …) 13 a cos x = 1 − (___ 2 24 −1 x  2 x  4 1 _____     −  ___ + …))    = 1 − ( ___ ⇒ sec x =   cos x ( 2 24 x  2 x   4     −  ___ )) sec x = 1 + (−1)(− (___ 2 24 2 (−1)(−2) x  2 x  4 ________     −  ___ ))  + …   (− (___ +   2 24 2! 5 4   2 + __   24 x   + … = 1 +  __12 x 3 x 2 5 __ __ b x +     +   15 x + … 3 13 3 __ 2 14 1 + x – 4x –   3 x + … 15 f ′(x) = (1 + x)(1 + 2 ln (1 + x)) f ′′(x) = 3 + 2 ln (1 + x) 2 f ′′′(x) = _____     1+x   32 x 2 + __   13 x 3 + … (1 + x)2 ln (1 + x) = x + __ x2 x3 x4 16 a x – __     +  __  –  ___  + … 2 6 12 b 0.116 (3 d.p.) __  x   3 3 17 a f(x) = e  tanx = e = e  x… × e 3 2 x  x  3 x  3 = (1 + x + ___     +  ___ + …)(1 + ___    + …) 3 2! 3! x  2 x  3 = 1 + x +  ___  +  ___ + … 2 2 x2 x3 b 1 – x + __     –  __  + … 2 2 18 a f(x) = ln cos x f(0) = 0 − sin x f ′(x) =  ______ = − tan x f ′(0) = 0 cos x 2 f ′′(x) = − sec  x f ′′(0) = − 1 f ′′′(x) = − 2 sec  2 x tan x f ′′′(0) = 0 f ′′′′(x) = − 2 sec  4 x − 4 sec  2 x tan  2 x f ′′′′(0) = −2 Substituting into Maclaurin, x  2   4 x   4 x   2 x ln cos x = (−1)  ___  + (−2)  ___ + … = −  ___  −  ___ + … 2 12 2! 4! x b Using 1 + cos x = 2 cos  2  __ , 2 x x ln (1 + cos x) = ln 2 cos  2  __ = ln 2 + 2 ln cos  __  2 2 x 2 1 __ x 4    )  − _   12  (   )  − …) so ln (1 + cos x) = ln 2 + 2(−  _12  (__ 2 2 x  2 x  4 = ln 2 − ___     −  ___ − … 4 96 d2y dy ___ 3x −3x ____ 3x 19 a     = 3(e + e ),   2 = 9(e + e−3x), dx dx d4y d3y ____ 3x −3x ____   3 = 27(e + e ),   4  = 81(e3x – e–3x) = 81y dx dx b y = 6x + 9x  3 + __   81 x   5+ … 20 2(3)  2n−1x   2n−1 c  ____________      (2n − 1)! Challenge x x x __ x ___ e = 1 + x +   __ +   +   __ +   120 … 2 24 6 (ix) ___ (ix) ___ (ix) ___ (ix) = 1 + (ix)+   ___ e +   +   +   + … 2 24 120 6 x x x x __ __ ___ __ + i   120 … = 1 + ix −   − i   +   2 24 6 x x __ __ − … cos x = 1 −   +   2 24 x x ___ __ − … sin x = x −   +   120 6 x x ___ __ − …) i sin x = i(x −   +   120 2 3 4 5 x 2 3 4 5 ix 2 3 2 4 3 5 3 4 5 5 6 Match up the items to show eix = cos x + i sin x CHAPTER 3 Prior knowledge check _ 1 a 5√ 3 + x  2  + c c b (x  2− 2x + 2)e  x + c __   1 ln(1 + 3 sin  2 x) + c 6 10x + y 1 b −  ________    c ______   x + 4y sec  2y 5 7 3 5 1 1     b ________     − _______       + ________     c _____   3 a __  − _____ x x+1 x + 1 2x + 1 4(x + 2) 4(x − 2) x 2 a −  __  y Exercise 3A 1 a __   12 __ b 2√ 2  ∞ c  __13 2 a ∫   ex dx = lt→∞ im ∫   ex dx = lt→∞ im (e   t− 1) 0 t 0 and e  t → ∞ as t → ∞, so the integral diverges 223 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 223 22/09/18 3:09 PM Answers ∫ ∫ ∞ t _ 1 1 im    ___ im (2√ t  − 2) b   ___   __  dx = lt→∞   __  dx = lt→∞ √ x  1 x  1 √ _ and √ t → ∞ as t → ∞, so the integral diverges ∞ t _ 8x 8x c   _______ im    _______ im (8√ t  2+ 1 − 8)   ______  dx = lt→∞   ______  dx = lt→∞ 2 2 0 √ 0 √ 1_ + x   1 + x   and √ t  2 + 1  → ∞ as t → ∞, so the integral diverges _ 1 1 __ 1 1 1   __  d im    ___ x = lt→0 im  [2 √ x ]  t  = lt→0 __  dx = lt→0 im (2 − 2 √ t ) = 2 3 a    ___ x  x  t √ 0 √ ∫ ∫ ∫ ∫ ∫ _  2 ∫ t 3 _______ t 1 1   _______ dx = lim2    ________   _______ dx = lim2  [−  _2 √ 2 − 3x ]    b   ________ 3 t→ _3 0 t→ _3 √ 2 − 3x  0 2 − 3x  0 √ __ __ 2 √ 2  __ 2 √ 2  2 √______ ____ ____ = lim_2 (    −     2 − 3t ) =     t→ 3 3 3 3 x x ln3 ln3 ______ ln3 e e _______ _______   dx = lim        ______  dx = lim  [2 √ ex − 1 ]  t  c      ______ x x t→0 t→0 √ √ e − 1  e − 1  t 0 ∫ ∫ __ ______ __ = lt→0 im  (2√ 2 − 2√ et − 1  ) = 2√ 2  _ 4 a Integral converges; 2(1 + √ 2  ) b Integral converges; 0 c Integral diverges 1 5 a _________  + c 3(7 − 3x) 2 2 1   1 2  dx = lr→∞ im  _________         = lr→∞ im (__   1  − _______  1    1   = __ b   ________ [ ] 3(7 − 3x) 6 3 21 − 9t) 3 −∞(7 − 3x) ∫ 6 a __13 e  x  + c 3 e 1 t  __ e 1  lim[__    e  x  ]    = t→−∞   lim (__     − __    e  =       2e  x  dx = t→−∞ b   x 3 3 3 ) 3 −∞ t (ln x)  2   + c 7 a ______ 2 ∞ t (ln t)  2 ln x ln x     dx = lt→∞     dx = lt→∞   im    ____ im (______   b   ____ x x 2 ) 1 1 and (ln t)  2 → ∞ as t → ∞, so the integral diverges 8 a x((ln x)  2− 2 ln x + 2) + c ∫ 1 1 3 ∫ 3 3 ∫ 14 The integral converges precisely when a > 1, in which 1 case its value is _____     a−1 1 1 2 1 ____________ ______________   =       = _______   − _____       15 a     2x  2+ 3x + 1 (2x + 1)(x + 1) 2x + 1 x + 1 in partial fractions. k 1   dx = [ln|2x + 1| − ln|x + 1|]  k       So,   ____________ 0 x  2+ 3x + 1 0 2 2k + 1 _______ = ln   ( k + 1 ) b ln 2 ∫ Challenge ∫   e −xsin  2x dx =  __2 ∫   e−x(1 − cos 2x) dx 1 1 −x = −  __12 e  −x + __   10 e  (−2 sin 2x + cos 2x) + c, where the last step is done by using repeated integration by parts. ∞ t So, ∫   e  −xsin  2 x dx = lt→∞ im ∫   e  −xsin  2 x dx 0 0 1 −x = l t→∞ im[__   10   e  (−2 sin 2x + cos 2x − 5)]     0 1 −t = lt→∞ im (__   10   e  (−2 sin 2t + cos 2t − 5) − ( −  __25 )) =  __25 since e  sin 2t → 0, e  −t cos 2t → 0 as t → ∞ −t Exercise 3B 1 a 1 1 + e  6 1 __     ln(______ 2 a     2 ) 6 b ln 2 4 b  ____  15π 5 3 14 e __     − __     d  __59 ln __ 5 π 2 3 a (−2, 128), (4, 20) y b c e e  2− 2   c ______   e  3 (–2, 128) y = f(x) b ∫   (ln x)  2 dx = lt→0 im ∫  (ln x)  2 dx 1 1 100 t 0 = lt→0 im (2 − t((ln t)  2− 2 ln t + 2)) = 2 since t(ln t) → 0 and t(ln t)  2 → 0 as t → 0, so the integral converges ∞ ∞ 1 c Split ∫   (ln x)  2 dx up as ∫   (ln x)  2 dx + ∫   (ln x)  2 dx 0 0 (4, 20) 1 and show the second integral diverges. ∞ t 1 = lt→∞ im  (t((ln t)  2− 2 ln t + 2) − 2) and t(ln t) → ∞, t(ln t)  2 → ∞ as t → ∞, so the integral diverges 3 _ 9 9√  2 10 − 8 11 1 π 12 a tan x is undefined at the upper limit x = __     2 π __   2 t b ∫   tan x dx = lt→  im  ∫   tan x dx = lt→  im__π (−ln cos t) and π __ 0 2 0 2 π π __ cos t → 0 as t →    , so −ln cos t → ∞ as t → __    , 2 2 so the integral diverges. π 13 a s ec  2 x is undefined at the point x = __    in the domain of 2 the integral __   π π π b Split ∫   sec  2 x dx up as ∫   sec  2 x dx + ∫__π  sec  2 x dx and x O ∫   (ln x)  2 dx = lt→∞ im ∫   (ln x)  2 dx 1 t 2 0 0 show the first integral diverges. __   π   2 t 224 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 224 _   1 3√ 3     6  __  + _____ 4 8π 5x 2 1     = _______   + _____    in partial fractions.   7 a ______________ (2x − 1)(x + 2) 2x − 1 x + 2 5 5x 1 49      dx = __  14 ln  __     ______________ So, _____   3 5 − 1 1 (2x − 1)(x + 2) 1 49 k  4 __   b     ln  _____ 4 3 2 1 2 1 __ 2  12 [ __ (x   2 − 4)  5]    = ___   256 8 a   2 ∫   x (x  − 4)  4 dx = __ 10 5 ∫ 0 ∫   sec  2 x dx = lt→  im im__π (tan t)  ∫   sec  2 x dx= lt→  π __ 0 2 0 2 π and tan t → ∞ as t → __    , so the integral diverges 2 2 c A lower bound is 20 and a upper bound is 128, since these are the minimum and maximal values attained on [−2, 4] respectively d 74 ln 3 4 ____     2π 5 ___   506 75 0 512 b −  ___ 5 1 ___ 9     2e Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers b 1 10 If _____     ∫   f(x) dx = m, then b−a a b b b 1 1     ∫   f(x) dx + ∫   cdx)  _____  ∫   (f(x) + c) dx = _____ a b−a a b − a( a 1 = m +  _____  (c(b − a)) = m + c _ b−a 11 √   2 10 a 12 The graph of f(x) between 0 and π is the negative of the graph between π and 2π and so the area under the graph between 0 and 2π must be zero. 1 13 a −  ________ + c 2 + sin x ∫ __   5π   3 cos x 3 ________ 3 3     1 ___    ]     [−   b ___       ___________ 2 dx =   5π 0 (2 + sin x)  5π 2 + sin x 0 _ _ 3 3 __1 __ 2 ___ 3 ) = −  _____  (3 + 4√ 3  ) =    (  2 +   13 (4 + √  ) 130π 5π _ 5π 3  ) c ___     −  _____  (3 + 4√ 3 2 130π –sin 1 y π 2 y = arcsin(arcsin x) O sin 1 x 5π __ – b 17 ) 14 a Turning point is at (−  __34 , __ 8 π 2 y π 2 y = arccos(arccos x) b − 2a  2− 5a − __  76 c __   47 24 Exercise 3C 1 1 a  ______2  1 + x  1 _  b −  _______ √ 1 − x  2  3(x  2+ 1) 1 _         e −  _________ d  _____________ 1 + (x  3+ 3x)  2 x√ x  2− 1  2x _  c −  _______ √ 1 − x  4  arccos_ x − arcsin x         2 __________________   √ 1 − x  2  2         3  _____________________ (x  2+ 1)(1 − arctan x)  2 O cos 1 c y 1 d 1 1 _  + _______   _ = 0 4  ___  (arccos x + arcsin x) = −  _______ dx √ 1 − x  2  √ 1 − x  2  So f(x) = arccos x + arcsin x is constant. π π π f(0) = __    + 0 = __    , so f(x) = __    for all x 2 2 2 2 2 _  b  ______2  5 a −  _________ 4 + x  √ 1 − 4x  2  3 _  c  _________ √ 1 − 9x  2  2x _     e −  ___________ √ x  2(2 − x  2)  x y = arctan(arctan x) x O 1 d −  ___________     1 + (x + 1)  2 2x _  −  _______ √ 1 − x  4  cos x _ − sin x arcsin x h _______ √ 1 − x  2  e  arctanx j ______2  1 + x  –1 f 1   _  ge  x arccos x − _______ ( √ 1 − x  2  ) x   _  i x 2 arccos x − _______ ( √ 1 − x  2  ) x + (1 + x  2) arctan x       6 _________________________      (1 + x  2)(1 + x  2( arctan x)  2) d  2y dy x 1   2  = _________       _  and ____ 7 If y = arcsin x, then  ___  = _______ dx √ 1 − x  2  dx  (1 − x  2)  __ 23  d  2y dy So, (1 − x  2)  ____2  − x  ___  = 0 dx dx  _ 1 _ 4√   3   π √ 3         − ___ 8 y = _____     x + __ 3 3 6 2 arctan x   9 a __________ 1 + x  2 1      c _______________________     (1 + x  2)(1 + (arctan x)  2) 1 _      b −  _________________ √ 1 − x  2 (arcsin x)  2 _ _ 11 a If y = arccos x, then sin y = ± √ 1 − cos  2y = ± √ 1 − x  2 . Since arccos x has range [0,π] and sin y_ is positive on this domain, we must have sin y = √ 1 − x  2  b If y = arctan x, then 1 1 1 _  = ±  ________ _    cos y = _____      = ±  ___________ sec y √ 1 + x  2  √ 1 + tan  2y π π Since arctan x has range −  __ ,  __  and cos y is positive ( 2 2) 1 _  on this range, we must have cos y =  ________ √ 1 + x  2  1 1 c If y = arccos x, then sec y =  _____  = __     cos y x d If y = arcsec x, then __________ _ 1 sin y = ± √ 1 − cos  2y = ± 1 − ______   2    sec  y √ ______ 1   2   = ± 1 − ___ x  √ 225 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 225 22/09/18 3:09 PM Answers [ ] π π Since arcsecx has range 0, __     ∪ __    , π and sin y is ( 2) (2 ) Challenge a arcsec x + c _ positive on this domain, we must have √ b √ x  2− 1 − arcsec x + c ______ 1   2   sin y = 1 − __ x Exercise 3D ∫ Exercise 3E ∫ a sec  2 θ x θ 1 1 1 _______   2          arctan  __  + c   dx = _____________   dθ = __     + c = __ a a a a  + x  2 a  2 + (a tan θ)  2 − sin θ 1   _  dx = __________   _  dθ = − θ + c = − arccosx + c 2 _______ √ √ _ _ 1 − x  2  1 − cos  2 θ      4√ 5 x√ 5 x b _____     arctan  ____ 3 a 3 arcsin  __  + c   + c 2 5 5 _ x d ln |x + √ x  2− 2 | + c c arcsin  __  + c 5 _ _ √ 3 3 x√       + c 4 ___     arctan  ____ 2 6 1 1 2x 1 _______ 1   _______  dx =  __ __  + c       ______  dx =  __  arcsin  ___ 5   ________ √ 2 √  __3 − x2  2 √ 3 − 4x2  3  4 π     6 a 2arctan 3 − __ ( 4) π __ b   2 ∫ ∫ ∫ ∫ c √ 3   2_ 1_ ___   + arcsin  ___       arcsin  ___ 3( √ √ 7   7   ) _ ∫ __ __ __ __ ∫ ∫ ∫ ∫ ∫ 3 _ | ∫ ∫ ∫ √ 2 ∫ x−2 t−2 1 _______ 1 ______ 1 __ im  __ = __ [  3  ln   2x − 1  ]  1  = lt→2 (  3  ln   2t − 1  ) +   3  ln 1 t−2 and ln  ______  → −∞ as t → 2, so the integral diverges 2t − 1 x−2 x + 3 ______   +    + 1 9 a −  ______ x  2+ 6 x  2+ 4 __ Z01_ALEVEL_CP2_83343_ANS_213-255.indd 226 | | | t | | | √ __ x  2+ 4 x x_ 3 1 __   + c        arctan  ___   + x − arctan  __  − __ b     ln ______ 2 2 ( x  2+ 6 ) 2 √ 6   x_ x   4 1 __   + c     _  arctan  ___   + ___ 10     ln ______ 2 ( x  2+ 5 ) √ 5   √ 5   1 − x _____ 1 2   = ______   2   +    in partial fractions. 11 ______________      (x + 1)(x  2+ 1) x  + 1 x + 1 1 2       dx So,   ______________ x  2+ 1) 0 (x + 1)( 1 1 1 x 1 1   dx −   ______   dx +   _____   dx   2   2   =   ______ x   + 1 x   + 1 0 0 0 x + 1 4 ∫ ∫ ∫ ∫ π 1 1     ln (x  2+ 1) + ln |x + 1|]    = __     − __     ln 2 + ln 2 = [ arctan x − __ 4 2 2 0 1 __ =     (π + 2 ln 2) 4 1 __   π 6 1 1 1 6     [x + __     sin 2x]    = ___       (1 + cos 2x) dθ = ___ 2 16 0 16 0 _ _ √   3 π ___ 1 1 __ ___ ____ √  )   (2π − 3 3 =    (    +    ) =   4 192 16 6 226 ∫ 6 ∫ 1 ∫ __   π __   π 2 __   1 2 ∫ x   1   dx =   __     sin  2 θ dθ _ 14    _________ √ 1 − 4x  2  0 0 8 1 the second integral diverges. t 2 1 1   ______________      dx = lt→2      dx im    ______________ 1 (x − 2)(2x − 1) 1 (x − 2)(2x − 1) 5 1 4   __  arcsin( __      x) + c = −  __  √ 6 − 5x2  − ___ 5 6 √ 5  x 5 1     arctan  __  + c 12 a  __  ln (x  2+ 16) + __ 4 2 4 4 x 5 1 1 1 4  x + 5   dx __ 2     ln (x  2+ 16) + __     arctan  __ ]    =    [__ b  __    _______ 40 4 0 x  + 16 4 2 4 5π 1 1 __ __ ___ =    (    ln 2 +    ) 4 2 16 5π 1 __ ___ c −     ln 2 −     2 16 3x x ___ 2 __ 13     −     arctan  ___  + c 9 27 2 _______ | 2 ∫ ∫ __ 3 2 b _____       − ______   x − 4 x  2+ 6 1 1 ∫   ______________      dx + ∫      dx and show   ______________ (x − 2)(2x − 1) (x − 2)(2x − 1) x_  − 2√ 2 − x  2  + c 9 − arcsin  ___ √   2 x  32 arctan  __  + c 10 4 ln (x  2+ 4) − __ 2 4x 1   _______2   dx −   ________   _______2   dx 11   f(x) dx =   ________ √ √ 6 − 5x   6 − 5x   1  − __ 1 1   ______ 2  dx   __    _______ = −  __    u 2 du − ___ 5 √ 5  √  __65 − x2  ∫ √ ∫ ∫ 2 _ ∫ x_ 3   + c      arctan  ___ c 2 ln|x − 4| − __ 2 √ 6   x−2 1 __     + c 8 a     ln _______ 3 2x − 1 2 1      dx up as b Split   ______________ (x − 2)(2x − 1) __   1 __   2√ 3 1     ln (1 + 3x2) + c = _____     arctan (√ 3 x) + __ 3 ∫ 7 a (x − 4)(x  2+ 6) _ √ 3    π __ √ 2 π π         −     = ___ = arcsin  ___  − arcsin  ___  = __ 3 4 12 2 2 2 + 3x 3x 2   dx =   _______   dx +   _______   dx       8   _______ 1 + 3x2 1 + 3x2 1 + 3x2 2 1 1  __ 2   dx +  __ 1      du =  __     ______ 3 __ 2 u   + x2 ∫ __ |x + 2|4 x   __  + c b ln  _______3  − √ 2  arctan ___ ( √ 2 ) |x − 3| x   2 x 1 1 __       arctan  __  + c   − __ 4     ln ______ 2 4 ( x  2+ 4 ) 2 2x 5 1     arctan  ___  + c 5 −  ____  + ___ 3 9x 54 x3 + 9x2 + x + 1 3 2 4 _______________               2   dx   dx   dx −   _____   dx +   ______ =   _____ 6   x−1 x+1 x4 − 1 x +1 = 3 ln | x − 1 | − 2 ln |x + 1 | + 4 arctan x + c |x − 1 |3 = ln  _______2 + 4 arctan x + c |x + 1 | 1 x √3    _  dx = arcsin  __       7     _______ [ √ 4 − x  2  2 ]√2  √ 2   4 √ __   √ 3   __   1 3−x 1 1 a __________     + _________     10(x  2+ 1) 10(x + 3) x+1 1   + ________     b −  _________ 3(x  2+ 2) 3(x − 1) 4x + 7 4   − ___     c _________   7(x  2+ 7)__ 7x x_ 3   + c 2 ln |x + 2| −  __   arctan  ___ 2 √ 6   3 a (x + 2)(x − 3)(x  2+ 2) Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers 3x 5x 1 12 a _________     − __________     + ___     4(x  2+ 2) 2(x   2 + 2)  2 4x 1 10 __   + 3 ln (x  2+ 2) + 2 ln |x| + c b      ______ ) 8 ( x  2+ 2 Challenge _ | _ |   x − 4 − 2√ 2 1_ _  + c   ln  ____________    a _____ 4√   2 x − 4 + 2√ 2   √ 2  (x + 1) 1_   arctan  _________  _____   + c 3 3√ 2   b Mixed exercise 3 1 a arctan e  x + c 1 ∫   ________     dx up as e  + e  ∞ b Split  ∫ −∞ x −x ∫ ∫ ∫ ∫ ∫ __   π ∫  __π 2 2 __ 1 1 2 2 __ __ 3  __ π  0  x sin 2x dx =  π  [  4  (sin 2x − 2x cos 2x)]  0  =   2  x  2 2x 3 _  b −  __  arccos  ___  + c 4 a −  _______ 4 2 4 √ 1 − x    −5 −5 1 _____________ ________ __________________   ×          =    5 af′(x) =     2 (x − 1)  2 (2x + 3)  2 + (x − 1)  2 2x + 3 _______ (      + 1 x−1) 5 1     = −  ___________     = −  _______________ 5x  2+ 10x + 10 x  2+ 2x + 2 b x  2+ 2x + 2 = (x + 1)  2+ 1 > 1, 1 so |f(x)| =  ___________     < 1 x  2+ 2x + 2 | | 6 a We say an integral ∫   f(x) dx is improper if one or b a both of the limits are infinite or if f(x) is undefined at some point in [a, b] b It in undefined at x = 0 and has upper limit ∞ c π _ _ 1 7 − √ 1 − 5x  2  + ___   _  arcsin √ 5  x + c √ 5   8 a Use the substitution x = tan θ and identify sec2 θ ≡ 1 + tan2 θ so that the integral becomes   ∫ arctan t 1dθ = arctan t 0 π b i __  2 π ii   12 arctan 2x + c 9 a __   14 ln (1 + 4x  2) + __ b  __18 (π + 2 ln 2) 1 1 3 1   ______ 1   dx = __ 10 a   ________       _______   _______2   dx = __     arcsin(  __ x) + c √ 3 √ __ 3 2 4 − 9x     4 − x2  ∫ ∫ __   23 9 1 3x 3 __ π 1 ___   dx = __   _ b   _________ [  3  arcsin   2 ]  0  =   6  4 − 9x  2  0 √ ∫ ∫ __   1 ∫ __   π __   2 ∫ __   π x  1   _  dx =   sin4 θ dx =   __     (3 − 4 cos 2x + cos 4x) dx 11   ________ √ 1 − x  2  0 0 0 8 2 4 6 ∫ ∫ ∫ ∫ ∫ 4    ∞ 1 1   ________   x  e  x + e  −x  dx.   dx +    ________   −x −∞ e  + e  0 0 0 1 1    ________     x  lim    ________  lim[ arctan ex]0t   x −x dx = t→−∞ −x dx = t→−∞ −∞ e  + e  t e  + e  π π π = t→−∞   lim __     − arctan e  t = __    − arctan 0 = __     (4 ) 4 4 ∞ ∞ 1 π π 1 __ ________ __   x   Similarly   ________ e  + e  −x dx =  4 , so   −∞    e  x + e  −x  dx =  2  0 6 4_  − 2 2 __   π   ___ (√ )   3 0 Here we know sin  4 x = __  18 (3 − 4 cos 2x + cos 4x) by de Moivre’s theorem or by using the formula for sin  2 x twice. π 12 a __     8 ∞ t π 1 1 im  [__     arctan x2]    = lt→∞ im (__     arctan t2 − 0) = __     b   f(x) dx = lt→∞ 2 2 4 0 0 3 2 1   dx    dx +   __   2  dx −   ______   2 13   f(x) dx =   __ x x x +9 x 1 = 2 ln | x | −  __  − arctan __ (  3 ) + c x x_ 3 3 1   + c   b ln |x − 4| − ___   _  arctan  ___   14 a _____  − ______ x − 4 x  2+ 2 √ √ 2   2   ∞ ∞ 5 c Split ∫   f(x) dx up as ∫   f(x) dx + ∫   f(x) dx and show 6 __   π 6 _ 1 1 1     sin 4x]    = ___     (4π − 7√ 3  ). = __     [3x − 2 sin 2x + __ 8 4 64 0 4 the first integral diverges. 5 ∫   f(x) dx = lt→4 im ∫  f(x) dx t 4 t_ 5_ 3 = lt→∞ im  ln |t − 4| − ___   + arctan  ___     _  arctan  ___ ( √ 2   ( √ 2   √ 2   )) and ln |t − 4| → − ∞ as t → 4, so the integral diverges   2 x 15 a ln  ______   + c x  2+ 1 2 2 2 x  2 8  dx = ln  ______   3       = ln  __  b   ______ 2 [ ] x   + x 5 x   + 1 1 1 c − 2 ln 5 5 5 | | ∫ | | Challenge __   14 x4 − x2 + 4x]    = __   82 = 4 a   12 ∫   f(x) dx =  __12 [__ 2 2 0 0 _ f(c) = c3 − 2c + 4 = 4 ⇒ c = 0 or c = √ 2   b One example is f(x) = 0 for x < 1 and f(x) = 1 for x . 1. This has mean value  __12 on [0,2] but the function only attains the values 0 and 1. CHAPTER 4 Prior knowledge check 1 a 15 633 2 1.41 (3 s.f.) 59 566π 3 ________   15 __ π √ 3  b __  – ___     6 8 1 – 13e–4 c _________   4 Exercise 4A 8π π 1 a ___  b 4π ln 2 c  __ (π – ln 4) 3 4 __ π π π __     e  __  (1 − ln 2) f π 2√ 3 – 2 – __ d     ln 41 ( 4 2 6) π 2 __  2 3 π(3 (ln 3)2 – 6 ln 3 + 4) __ __ 4 a __19 (√ 3  – 1) b 9π (√ 3  – 1) 5 π ln 14 3π 6 a x = 0 or sin x + 1 = 0 ⇒ x = 0 or ___     2 2 π 9π b  __  (____    − 1) 2 8 100π 100π   b _____   7 a _____ 81 27 8 0.237 (3 s.f.) Challenge __ 4 4 1 1 1 1   __    dx = π   (__     − __    )dx     cos 2x − √ 2  sin x + __ π   (sin x − ___ ) π π 2 2 2 √ __ __ 2    4   4 __   3π ∫ 2 __   3π ∫ __ 4 π 1     sin 2x + √ 2  cos x]    = __    (π − 3) = π [x − __ π __ 4 2   4 __   3π 227 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 227 22/09/18 3:09 PM Answers Exercise 4B 8 ln 7 1 a π ( __14 e4 –2e –  __12 e–2 + __  94 ) π __ c     (ln 5 + 16) 5 b  __14 π (e2 – 1) 2 a __12 π b π (2 + 3 ln 3) 3π2 d  ____  16 9 2π3 4π 10 ___  5 11 a π ln 6 d π ln  __94 c π (4 ln 2 – 2) 3 b = __   13 4 Exercise 4D 1 200 000π 3 b ___________      m 7 1 a 200 m π b ___  3 4 a π 10 25π ln 4001 cm3 3 a cos3θ = cos(2θ + θ) = cos2θ cos θ − sin2θ sin θ = (cos2 θ − sin2 θ) cos θ − 2 sin2 θ cos θ = cos3 θ − (1 − cos2 θ) cos θ − 2(1 − cos2 θ)cos θ = 2 cos3 θ − cos θ + cos3 θ − 2 cos θ = 4 cos3 θ − 3 cos θ ⇒ cos3 θ = __   34 cos θ + __   14 cos 3θ 4 __ __ 5 π (__   32 e   3 + 6e  3 – __   25 ) 5 π 6 a 2 cos y – __     ( 3) __  __3π 1 π√ 3  π π π     dy = __   tan 0 – tan –  __  = ____     b π    __  sec2 y – __ ( ( 3 )) 3) 4( 4 0 4 ∫ π b  _____  6 ln 2 1 7 a _____  6 ln 2 8 a By de Moivre’s theorem, sin 3θ = 3cos2θ sin θ − sin3θ = 3sin θ − 4sin3θ sin 5θ = 5cos4θ sin θ − 10cos2θ sin3θ + sin5θ = 5sin θ − 20sin3θ + 16sin5θ = 5sin θ − 5(3sin θ − sin 3θ) + 16sin5θ = −10sin θ + 5sin 3θ + 16sin5θ 1 ⇒ sin5θ = __   16 ( 10sin θ − 5sin 3θ + sin 5θ) ∫ ∫ π π π       (10 sin y − 5 sin 3y + sin 5y)dy b π   s in5 y dy = ___ 16 __ 4π __   4π π 5 1     cos 3y − __     cos 5y]    =  ___  [− 10 cos y + __ __ 3 5 16   4π __ √ 2  43 π     =  ___ (8 + ______ 8 ) 15 Exercise 4C 384π 1 _____   7 2 a 2, 3 b πe3 c ln x = t ⇒ y2 = t – 1 = ln x – 1 d πe3 3 a x2 = 1 – sin θ ⇒ sin θ = 1 – x2, y2__= cos2 θ = 1 – sin2 θ = 1 – (1 – x2)2 = 2x2 – x4 b (√ 2 , 0) 8 √ c π (__   15 2 ) π 58π __ 4 a 0,     b ____   3 5 2 2 2 _   32  c tan θ = x , sec θ = y  2 _ so tan2θ ≡ sec2θ − 1 ⇒ x2 = y   3 − 1 8 8 2 _ 58π 3 _5 d V = π   ( y   3  − 1) dy = π [__     y   3  − y]    = ____     5 5 1 1 1 __ 5  10 π 2 2 2 dy 6 π   x2  ___  dt = π   4t2 × 2t dt = 8π   t3 dt = 32π dt 0 0 0  12 θ + c 7 a  __14 sin 2θ + __ b y = 4 sin 2θ = 8 sin θ cos θ ⇒ y2 = 64 sin2 θ cos2 θ dx ___   = – cosec2 θ dθ __ π π 1__   ⇒ θ = __    , x = √ 3  ⇒ θ = __     x =  ___ 3 6 √ 3  π π __ __ __ ∫ ∫ ∫ ∫ b π (ln 6 – __  23 )   6 ∫ ∫   6 π   y2 dx/dθ dθ = – 64π   cos2 θ dθ __   3π __   3π 16 2 π  c  __ 3 228 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 228 π b 50 000π m3 π 4 a __     b 864π m3 2 5 a x2 = 16 sin2 θ cos2 θ = 16 sin2 θ(1 – sin2 θ) y 2 y 2 16 = 16 ( __ ) (1 – (  __ ) ) = ___     y2(9 – y2) 3 3 81 b 14 c i e.g. Patterned earring may mean that earring requires less material ii e.g. Wasted material upon transfer to mould Mixed exercise 4 π2 b π (___    + 1) 4 1 a __12 x sin 2x + __   14 cos 2x + c dv 2 a Use integration by parts with u = x and ___     = sec2 x dx π __   4  __π to get   x sec2 x dx = [ x tan x + ln cos x]  04   ∫ 0 π 1     – __     ln 2) b π (__ 4 2 π 1     – __     ln 2) =(__ 4 2   35 ) 3 π(4 ln  __52 – __ π 4 __ (10 − 3π) 4 5 π(e4 −9e2 + 8e + 8) 6 a y = −x + __   52 172 7 a  ___ 5 ∫ 1 ∫  __π 2 ∫  __π 113π b  _____  120 6196 b  ____ π 35 2 8 π     x2 dy = π     cos3 t dt = π     cos t(1 − sin2 t) dt −1 − __2π − __2π Using the substitution u = sin t, this becomes 1 4π π         (1 − u2 ) du = ___ 3 −1 9 a sin 3θ = sin (2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = (2 sin θ cos θ) cos θ + (1 − 2sin2 θ) sin θ = 2 sin θ cos2 θ + sin θ − 2 sin3 θ = 2 sin θ (1 − sin2 θ) + sin θ − 2 sin3 θ = 2 sin θ − 2 sin3 θ + sin θ − 2 sin3 θ = 3 sin θ − 4 sin3 θ 3 ⇒ sin θ = __   34 sin θ – __   14 sin 3θ b 2000π cm3 128π 10 _____   15 ∫ Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers Challenge π Rotate C by −   __ anticlockwise to the x-axis using the matrix 4 ⎛ ___ 1⎞ 1 π π ⎛cos −   __   _    _  ___   −sin −   __  ⎞ ( 4) ( 4) √ 2   √ 2          π     π   =   ___ 1_ 1_ ___ sin −   __  cos −   __  ( 4) ( 4 ) ⎠ ⎝−   √     √  ⎠ ⎝ 2   2   ⎛ ___ 1_ ⎞ 1_ ___         √   2 √   2 t  2 ___ 1 t   2+ 2t   =   _    2       1_ (2t) √ 2 1_ ___ ___   (− t  + 2t) −         ⎝ √ 2   √   2⎠ So new parametric equations become 1 1 x = ___   _ (t  2+ 2t) and y = ___   _ (− t  2+ 2t) √ √   2   2 1_ ( 2 ___ ) y = 0 ⇒     − t  + 2t = 0 ⇒ t = 0, 2 √   2 2 2 1 dx ___ V = π ∫       _ (− t  2+ 2t)     ___ dt ) dt 0 (√   2 2 2 1 1_ (   _  (− t  2+ 2t)    ___   2t + 2) dt = π ∫     ___ ( ) (√ ) √ 0   2 2   2 t  6 ____ 3t  5 4t  3 π_ __ π_ 2 (t  5− 3 t  4+ 4 t  2) ___ ___ dt =     [    −     +  ____ ]    =     ∫     3 0 5 √ √   2 0 2   6 96 ___ 32π π_ ___ 32 ___ 16 ___ ______ =    (    −     +    )− 0 =   _  3 5 √   2 3 15√ 2   ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ Review exercise 1 cos 2x + i sin 2x cos 2x + i sin 2x 1 ______________     = __________________         cos 9x − i sin 9x cos (−9x) + i sin (−9x) e  2xi   = e  11xi= cos 11x + i sin 11x =  ____ e  −9xi Hence n = 11 2 a cos 5θ + i sin 5θ = (cos θ + i sin θ)5 = cos5 θ + 5i cos4 θ sin θ – 10 cos3 θ sin2 θ – 10i cos2 θ sin3 θ + 5 cos θ sin4 θ + i sin5 θ Equating real parts: cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ = cos5 θ – 10 cos3 θ(1 – cos2 θ) + 5 cos θ(1 – cos2 θ)2 = 16 cos5 θ – 20 cos3 θ + 5 cos θ b x = 0.809, −0.309, −1 3 a cos 5θ + i sin 5θ = (cos θ + i sin θ)5 = cos5 θ + 5i cos4 θ sin θ – 10 cos3 θ sin2 θ – 10i cos2 θ sin3 θ + 5 cos θ sin 4θ + i sin 5θ Equating imaginary parts: sin 5θ = 5 cos4 θ sin θ – 10 cos2 θ sin3 θ + sin5 θ = sin θ(5 cos4 θ – 10 cos2 θ sin2 θ + sin4 θ) = sin θ(5 cos4 θ – 10 cos2 θ(1 – cos2 θ) + (1 – cos2 θ)2) = sin θ(16 cos4 θ – 12 cos2 θ + 1) π 3π b 0,  __ ,  ___ , 1.209 (3 d.p.) and 1.932 (3 d.p) 4 4 z − z  −1 e  iθ − e  −iθ    , if z = e  iθ, then sin θ = _______   4 asin θ = ________   2i 2i 5 −1 z − z        sin  5 θ = ( _______ 2i ) 1 ( 5   z  − 5z  3+ 10z − 10 z  −1+ 5z  −3− z  −5) = ____   32i 5(z  3– z  −3) __________ 10(z − z  −1) 1 z   5− z  −5 __________     −     +     = ___    (________ ) 2i 2i 2i 16 1 = ___     (sin 5θ − 5 sin 3θ + 10 sin θ) 16 ∫ π   __  2 ∫ π __     2   1   (sin 5θ − 5 sin 3θ + 10 sin θ)dθ b   sin   5 θ dθ = __ 16 0 0 π   __  2 1 [−  __15 cos 5θ + __  53 cos 3θ − 10 cos θ]    = __   16 0 8 1 = __   16 (0 − ( −  __15 + __   53 − 10)) = __   15 5 a z−n = cos (−nθ) + i sin (−nθ) = cos nθ – i sin nθ zn + z−n = cos nθ + i sin nθ + cos nθ – i sin nθ = 2 cos nθ 1 b ___ (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10) 32 ∫ π  __  2 c   cos6 θ dθ 0 ∫ π __     2 1 = __   32   (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10)dθ 0 π   __  2 1 __   1 sin 6θ + __  64 sin 4θ + __  15 sin 2θ + 10θ]    =  __ 32 [ 6 2 0 5π 1 = ___    (5π − 0) =   ___  32 32 6 C + iS = 1 + (cos θ + i sin θ) + (cos 2θ + i sin 2θ) + … + (cos (n − 1) θ + i sin (n − 1) θ) = e  0 + e  iθ + e  2iθ + . . . + e i(n−1)θ a(1 − r  n) Using the sum of a geometric series S n = _________     and 1−r iθ −iθ   + e  e  , cos θ = ________   2 (1 − e  inθ)(1 − e  −iθ) 1 − e  inθ ________________   =          C + iS = _______   iθ 1 − e  (1 − e  iθ)(1 − e  −iθ) 1 − e  inθ − e  −iθ + e  i(n−1)θ 1 − e  inθ − e  −iθ + e  i(n−1)θ _____________________            =       = _____________________      2 − 2 cos θ 1 − e  iθ − e  −iθ+ 1 Equating real parts and using cos (−θ) = cos θ, 1 − cos nθ − cos (− θ)+ cos (n − 1) θ           C = ________________________________ 2 − 2 cos θ 1 − cos θ + cos (n − 1) θ − cos nθ         = _____________________________   2 − 2 cos θ Equating imaginary parts and using sin (−θ) = − sin θ, − sin nθ − sin ( − θ) + sin (n − 1) θ         S = ______________________________   2 − 2 cos θ sin θ + sin (n − 1) θ − sin nθ          = _________________________   2 − 2 cos θ __ ___   πi   __ ____   9πi  __ _____   17πi  __ 7πi  −  ____ __ 3πi  −  ____ 7 a z = √ 2 e 20, √ 2 e 20 , √ 2 e 20 , √ 2 e 20 , √ 2 e 4 Im b 9πi 2e 20 17πi 2e 20 πi 20 Re 2e – π  i __ 3πi 4 – 7πi 2e 20 7π  i ___ 5π  −i ___ 9 8 a z = 4e 9, 4e 9 , 4e 229 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 229 22/09/18 3:09 PM Answers b z  9 = ( 4e 9) , ( 4e 9 ) , (4e π  9 i __ 7π  9 i ___ ) 5π  9 −i ___ 9 = 4  9 e  iπ, 4 9 e  7iπ, 4 9 e  −5iπ The value of all three of these expressions is −49 = −218 Hence the solutions satisfy z9 + 2k = 0, where k = 18. 9 z = cos θ + i sin θ, where π 9π 7π  3π , −  ___ π  ,  5π ___   =   __ θ =  ___   ,   ___  , −  ___ 10 10 ( 2 ) 10 10 10 ___   πi   ____  7πi  πi  − ___ _____  13πi  11πi  −  _____ 10 a 2e15, 2e 15 , 2e 15 , 2e 3 , 2e 15 b Vertices of a regular pentagon, inscribed in a circle radius 2, centred on the origin. 4πi  −  ____ 2πi  −  ____ ____  2πi  ____  4πi  11 a 5 , e 5 , 1, e 5 , e 5 e 4πi  −  ____ 2πi  −  ____ ____  2πi  ____  4πi  e 5 + e 5 + 1 + e 5 + e 5 = e 5 ( 1 + e5 + e( 5 ) + e( 5 ) + e( 5 ) ) 4πi  −  ____ ____  2πi  2 ____  2πi  3 ____  2πi  4 ____  2πi  4πi  2πi 4πi  4πi  5 − 1 −  ____ −  ____ −  ____   = e 5 _______   e2πi− 1  = e 5 × 0 = 0   e ____ = e 5 ________ 2πi     ( e  5 − 1 ) (e____ ) 5 − 1 b (3.26, 1.64), (1.78, 2.40), (0.60, 1.22), (1.36, −0.26) 12 Using partial fractions, 2 2 2  ____________     = _____   − _____       (r + 1)(r + 2) r + 1 r + 2 _____  10πi   Using the method of differences, n n 2 2 2 2 2     − _____     = __       =  ∑(_____     − _____      ∑____________ r + 2) 2 n + 2 r=1 r + 1 r=1 (r + 1)(r + 2) n + 2 − 2 _____ n 2   = _________     =     = 1 − _____   n+2 n+2 n+2 Hence a = 3, b = 5 b 0.0398 18 a Using partial fractions, 2 1 1  _______   = ______     − ______     4r  2− 1 2r − 1 2r + 1 Using the method of differences, n n 1 1 1 2     − ______     = 1 − _______     = ∑  ______ ∑   _______   2n + 1 r=1 4 r  2− 1 r=1 ( 2r − 1 2r + 1 ) 20 b ___ 861 19 a A = 24, B = 2 b Using the identity from part a, n n r=1 r=1  ∑(24r  2+ 2) = ∑  ((2r + 1)  3 − (2r − 1)  3) n n n r=1 r=1 r=1 Using the method of differences, n Using the method of differences, 24  ∑r  2 = 8n  3+ 12n  2+ 6n + 1 − 1 − 2n n n 1 1 2     − _____        = ∑  (_____ ∑ ____________ r + 3) r=1 r + 1 r=1 (r + 1)(r + 3) r=1 = 8n  3+ 12n  2+ 4n = 4n(n + 1)(2n − 1) 5 1 1 1 1 1 1 _____     −       − _____     = __     − _____   − _____     = __     + __ 2 3 n+2 n+3 6 n+2 n+3 5(n + 2)(n + 3) − 6(n + 3) − 6(n + 2)            = __________________________________   6(n + 2)(n + 3) 5n  2+ 25n + 30 − 6n − 18 − 6n − 12            = __________________________________   6(n + 2)(n + 3) n(5n + 13) 5 n  2+ 13n             = ______________       = ______________ 6(n + 2)(n + 3) 6(n + 2)(n + 3) Hence, a = 5, b = 13, c = 6 (r + 1)( r + 1) − r(r + 2) r r+1            14 a _____  −   _____  = ____________________ r+2 r+1 (r + 1)(r + 2) 1 r  2+ 2r + 1 − r  2− 2r ____________         =       = ___________________   (r + 1)(r + 2) (r + 1)(r + 2) n   b  ________ 2(n + 2) 1   −   _____ 2   +   _____ 1   15 a f(x) =   _____ x+1 x+2 x+3 4n(n + 1) (2n − 1) __ 1      =     n(n + 1)(2n + 1) ∑   r  2 =  _________________ 24 6 c 194 380 20 Using partial fractions, 1 1 1 1  _____________     = ___     + ________       − _____   r(r + 1)(r + 2) 2r r + 1 2(r + 2) n r=1 Using the method of differences, 2n 2n 1 1 1 1     − _____     + ________          = ∑ ___ ∑ _____________ r + 1 2(r + 2) ) r=1 ( 2r r=1 r(r + 1)(r + 2) 1 1 1 _________ =  __   −     + ________     4 2(2n + 1) 4(n + 1) (n + 1)(2n + 1) − 2(n + 1) + (2n + 1)            = __________________________________   4(n + 1)(2n + 1) 2n  2+ 3n + 1 − 2n − 2 + 2n + 1           = _____________________________   4(n + 1)(2n + 1) n(2n + 3) 2n  2+ 3n             = _______________       = _______________ 4(n + 1)(2n + 1) 4(n + 1)(2n + 1) 1   +   _____ 1   __   −   _____ b  1 6 n+2 n+3 2r − 1   16 a   ________ r2(r − 1)2 b Using the method of differences, n n 1 1 2r − 1 1 1 1   =  ∑ _______   − ___   2  = ___         − ___    = 1 − ___   2   ∑ _________ r  ) 1  2 n  2 r=2 r  2 (r − 1)  2 r=2 ( (r − 1)  2 n  Z01_ALEVEL_CP2_83343_ANS_213-255.indd 230 Using the method of differences, n n 2 2 4 2 __ 2 2 2     − _____     =  __  = ∑ (__     +     − _____   ∑  _______   − _____     r + 2) 1 2 n + 1 n + 2 r=1 r r=1 r(r + 2) 2 2   − _____     = 3 − _____   n+1 n+2 3(n + 1)(n + 2) − 2(n + 2) − 2(n + 1)             = __________________________________ (n + 1)(n + 2) 2 3n  + 9n + 6 − 2n − 4 − 2n − 2           = _____________________________   (n + 1)(n + 2) n(3n + 5) 3n  2+ 5n             = _____________       = _____________ (n + 1)(n + 2) (n + 1)(n + 2)   2 =  ∑((2r + 1)  3 − (2r − 1)  3) 24  ∑r  2 + ∑ 13 Using partial fractions, 2 1 1 ____________       = _____   − _____       (r + 1)(r + 3) r + 1 r + 3 230 17 a Using partial fractions, 4 2 2  _______  = __         − _____ r(r + 2) r r + 2 Hence a = 2, b = 3, c = 4 (r − 1) r(r + 1) + (r + 1) − r 1 1            = _________________________       − _____   21 aRHS = r − 1 + __ r(r + 1) r r+1 2 r( r  − 1) + 1 _________ r  3 − r + 1      =     = LHS =  ____________ r(r + 1) r(r + 1) n(n2 + 1) b _________  2(n + 1) Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers 1 22 1 −   _________   3n(n + 1) 1 33 a _________  + c 2(5 − 2x) x  2 x  2 x  4 23cos x = 1 − ___     +  ___ − … = 1 −  ___  ,neglecting terms in x3 2! 4! 2! and higher powers x  3 x   5 sin x = x − ___     +  ___ − … = x,neglecting terms in x3 and 3! 5! higher powers x  2 11 sin x − 6 cos x + 5 = 11x − 6(1 − ___    )+ 5 2 = −1 + 11x + 3x  2 b The integral is infinite at the end points and not defined at the point x = __ 52 , so split up the integral: 5 __     3 0 1 1 1 ∫     ________2  dx= ∫     ________2  dx+ ∫   2   ________2  dx −∞ (5 − 2x)  −∞ (5 − 2x)  0 (5 − 2x)  3 1 +∫ __5     ________2  dx     ( 5 − 2x)  2 5 __     1 Consider the integral I = ∫   2  ________2  dx 0 (5 − 2x)  5 __ t     1 1          I = ∫   2   ________2  dx= lim  _________ 5 [ 2(5 − 2x) ] __ 0 (5 − 2x)  t→    0 A = −1, B = 11, C = 3 24 LHS = ln (x  2 − x + 1)+ ln (x + 1) − 3 ln x = ln((x  2 − x + 1)(x + 1))− ln x  3 x  3+ 1 1 = ln ______ = ln 1 + ___   3  (  x  3  ( ) x  ) 1 Substituting  ___3  for x and n for r in the series x  (−1)  r + 1 x  r x  2 x   3     +  ___ + … +  ___________ ln (1 + x) = x − ___   +… r 2 3 n − 1 ( −1)  1 1    + …  + … + _________   LHS = ___   3  − ____ x  2 x  6 n x  3n 21 , D = __   71 25 A = 1, B = −2, C = −  __ 2 3 8 3 4 2 26 a __13 − __   29 x + __   27 x − __   81 x + … 8 4 4 3 b __23 x − __   49 x2 − __   27 x + __   81 x + … x2 x 4 27 a −  __  −  ___  − … 2 12 x2 x 4 b  __  +  ___  + … 2 12 28 a Let u = 1 + cos 2x, then f(x) = ln u du ___     = −2 sin 2x dx du 1 du 1 f′ (x) = f′ (u)  ___  =  __   ___  =  __________  × (− 2 sin 2x) dx u dx 1 + cos 2x − 2 sin x − 4 sin x cos x ________   =       = − 2 tan x = _____________   cos x 2 cos  2 x 2 1 1     − ___     = lim  _________ 5 10 ) t→ __  ( 2(5 − 2t) 2 5 1 1     − ___     → ∞ As t →  __  ⇒ _________ 2 2(5 − 2t) 10 3 1     dxalso diverges. Therefore, I diverges, so ∫    ________ −∞ (5 − 2x)  2 1 34 −  __ π 5 3x 1     dx 35 aMean value =  _____  ∫     ______________ 2 5 − 2 (x − 1)(2x − 3) Using partial fractions, 9 3x 3     = _______   − _____        ______________ (x − 1)(2x − 3) 2x − 3 x − 1 3 1 5   9   − _____     dx Mean value =  __  ∫     (_______ 3 2 2x − 3 x − 1 ) 5 3 1     − _____     dx = ∫     (_______ 2x − 3 x − 1 ) 2 3     ln (2x − 3) − ln (x − 1)]    = [ __ 2 2 3 3 __ __ = (    ln 7 − ln 4)− (     ln 1 − ln 1) 2 2 3 __ = (    ln 7 − ln 4)− (0 − 0) 2 1 1 343 1 __ =     ln 343 −  __  ln 16 =  __  ln  ____  2 2 2 16 5 b f″ (x) = − 2 sec  2 x f‴ (x) = − 4 sec  2 x tan x f″  ″ (x) = − 8 sec x sec x tan x tan x − 4 sec  2 x sec  2 x = − 8 sec  2 x tan  2 x − 4 sec  4 x c 3 1   3 − 1)  3 dx 36 aMean value =  _____  ∫     x  2 (x 3−1 1 ln 2 − x − __   16 x4 + … 2 29 __12 π 30 __  4 1 1 3     (x  − 1)  4]    = __     [___ 2 12 1 3 3 57122 1 1   3 − 1)  4]  1  = ___    ((26)  4− 0) = _______     = ___     [(x 24 24 3 x 1     + c 31 a __  ln _____ 3 (x + 3) t ∞ t 1 1 x 1 _____ b∫     ________  dx = lt→∞ im  ∫     ________  dx = lt→∞ im  __ [  3  ln(  x + 3 )]  3   3 x(x + 3) 3 x(x + 3) t 1 1 __ 1 _____ __ = lt→∞ im  __ (  3  ln(  t + 3 )) −   3  ln   2  3 1 1 __ 1 _____ __ = lt→∞ im  __ (  3  ln(1 −  t + 3 )) −   3  ln   2  1 1 1     ln 2 = 0 −  __  ln  __  = __ 3 2 3 32 Using repeated integration by parts, 1 ∫   x   3 e  −x  4 dx = −  __  e  −x  4 + c 4 t 114 244 b−  ______ 3 __ 37√ 2  38 5 39 a y = (arcsin x)  2 Let u = arcsin x and y = u  2 dy du 1 _   ___  = 2u and  ___  =  ________ du dx √ 1 − x  2  dy dy ___ 2 arcsin x du 1 _  = __________   _      ×     = 2u ×  ________ ⇒  ___  = ___ dx du dx √ 1 − x  2  √ 1 − x  2  1 −x  4   3 e  −x  4 dx = lim  ∫     x  3 e  −x  4 dx = lim  −  __ ∫     x   e  ]     t→∞ 1 t→∞ [ 4 1 1 1 1 1 1     = ___     = lt→∞ im (−  __  e  −t  4)− ( −  __  e  −1)= 0 + ___ 4 4 4e 4e ∞ 1 343 k  2   b __  ln  ______ 2 16 t _ dy √ 1 − x  2   ___  = 2 arcsin x dx 2 dy 2 (1 − x  ) (___    )  = 4 (arcsin x)  2= 4y dx 231 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 231 22/09/18 3:09 PM Answers b Differentiating the results from part a implicitly with respect to x, dy dy dy d  2 y − 2x (___    )  + (1 − x  2)2  ___   ____2  = 4  ___  dx dx d x  dx d  2 y dy ___ ____ 2 − x     + (1 − x  )   2 = 2 dx d x  dy d  2 y (1 − x  2)  ____  − x  ___  = 2 dx d x  2 2 ⇒ 3x2 + 5x = A(x2 + 5) + B(x − 3) x = 3: 42 = 14A ⇒ A = 3 b Using integration by parts and the result from part a, __ __ ∫ √ 3  ___     3 ∫ 3 3     dx     6x arctan3x dx = 3x2 arctan3x −     3x2 × ________ 1 + 9x  2 0 0 √ √ __ __ 3  ___     ∫ 3  ___     ∫ 1   dx   = 3x  2 arctan 3x −     1 dx +     _______ 2 0 0 1 + 9x 3 3 1     arctan 3x = 3x  2 arctan 3x − x + __ 3 _ √   3 ___     _ √ 3   3 1 4     arctan3x]    = __     arctan √ 3  − ___     [3x2 arctan3x − x + __ 3 3 3 0 _ 1  ) = __     (4π − 3 √ 3 9 41 a Let y = f(x) = arcsin x, then sin y = x Differentiating implicitly with respect to x, dy cos y  ___  = 1 dx dy 1 1 1   _    = ___________   _  = ________    ___  = _____ dx cos y √ 1 − sin  2 y √ 1 − x  2  2 _______  b ________ √ 1 − 4x2  dx 1 1 c x =  __  sin θ ⇒  ___  =  __  cos θ 2 dθ 2 π 1 __ 1 1 __ __     At x =     ,     =     sin θ ⇒ θ = __ 4 4 2 6 1 At x = 0, 0 =  __  sin θ ⇒ θ = 0 2 1 π __ 1 __ __         sin θ arcsin (sin θ)   6 4 dx x arcsin 2x 2 __________ __________________ _________   ___ dθ        _______2   dx =       dθ √ √ 1 − 4x   1 − sin2 θ 0 0 1 π __ π __ __         sin θ × θ     6 2 1 1 6 =    __________     cos θ)dθ = __       θ sin θ dθ  (__ cos θ 2 4 0 0 _ ∫ ∫ ∫ ∫ π __     ∫ 1 6 1 1 1       cos θ dθ = −  __  θ cos θ + __     sin θ = −  __  θ cos θ + __ 4 4 0 4 4 ∫ π __     1 __     6 x arcsin 2x 1 1 __   dx = −  __   _ ⇒   __________ [ 4  θ cos θ +  4  sin θ]  0  1 − 4x  2  0 √ _ π π π 1 1     sin  __  − 0 = ___     (6 − π √ 3  ) = −  ___  cos  __  + __ ( 24 48 6 4 6) 4 2x + 1 2 − x __ 2x + 1 _________ 1   =     = ______     +     42 Using partial fractions, _______   3 x  + x x(x  2+ 1) x  2+ 1 x x 2x + 1 2 − x __ 1 1 2 _______ ______ ______ __   dx = ______   3 (  x  2+ 1  +   x )dx = (  x  2+ 1  −   x  2+ 1  +   x )dx x  + x 1 = 2 arctan x − __     ln (x  2+ 1) + ln x + c 2 1 Hence A = 2, B = −  __  2 ∫ ∫ 232 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 232 ∫ x = 0: 0 = 15 − 3B ⇒ B = 5 5 3x2 + 5x b_________________          3   + ______   dx = _____   dx   ( x − 3 x2 + 5 ) x3 − 3x2 + 5x − 15 _ x_   + c = 3 ln (x − 3) + √ 5  arctan  ___ √   5 _ 1   _  Hence P = 3, Q = √ 5 and R = ___ √   5 π 13     e2 44 ___  πe6 − __ 4 4 45 a 12 b 9π2 46 π3 − 4π 47 a = 2 π 48 a0 = y cos y ⇒ 0 = cos k ⇒ k = __     2 π π __ __         ∫ 40 a y = arctan 3x ⇒ tan y = 3x Differentiating implicitly with respect to x, dy sec  2 y  ___  = 3 dx dy 3 3 3     = __________     = ________      ___  = ______ dx sec  2 y 1 + tan  2 y 1 + 9 x  2 √ 3   ___  3x2 + 5x 3x2 + 5x A B 43 a _________________          = _____________          = _____     + ______   x3 − 3x2 + 5x − 15 (x − 3)(x2 + 5) x − 3 x  2+ 5 ∫ ∫ ∫ 2 2 b V = π   (y cos y)  2 dy= π   y2 cos2 y dy ∫ π __     0 0 π 2   2( cos 2y + 1) dy = __       y 2 0 1 y2 cos 2y dy =  __  y  2 sin 2y − y sin 2y dy 2 1 1 1     y cos 2y − __     cos 2y dy =  __  y  2 sin 2y + __ 2 2 2 1 1 1     y cos 2y −  __  sin 2y + c =  __  y  2 sin 2y + __ 2 2 4 π __     π __ 1 2 1 1 1 3 2 __ __ __ __ V =     [    y  sin 2y +     y cos 2y −     sin 2y +     y  ]     2 2 2 4 3 0 π π  3 π 1 1    − 0 + ___    )− 0) = ___     π  4 − __     π  2 = __    ((0 − __ 2 4 8 24 48 1 1 Hence a = ___     , b = −  __  8 48 49 ln 4 ∫ ∫ ∫ 50 a Length scale factor 2 ⇒ volume scale factor 8 2 4 4 5 25   _    dx = 8π   _______   dx V = 8π   (________   ) √ 1 + 4x 0 0 1 + 4x 4 25     ln (1 + 4x)]    = 8π [___ 4 0 25 ___ = 8π(    ln 17 − 0)= 50π ln 17 4 b The thickness of the vase has not been taken into account. ∫ ∫ 51 a0 = √ y (3 − __  12 e  y) ⇒ 0 = 3 − __  12 e  y ⇒ e  y= 6 ⇒ y = ln 6 _ b 9.65 cm3 c e.g. Filament may be wasted, or the shape may not exactly match the model. 52 a Length scale factor 0.5 ⇒ volume scale factor 0.125 ∫ π  __  ∫ π  __  dx π π V = __       (3 sin 2t)  2  ___ dx = __ 8 sin2  2t cos t dt       1 2 8 0 ∫ dt π  __  2 8 0 2 9π = ___       sin  2 2t cos t dt 4 0 π __     2 ∫ b V = 9π   sin2 t cos3 t dt 0 π __     2 ∫ = 9π   sin  2 t(1 − sin2 t) cos t dt 0 π __     2 ∫ = 9π   (cos t sin2 t − cos t sin4 t)dt 0 [3 π __     ]0 (3 ) ( 15 )   1 sin3 t − __  1 sin5 t   2  = 9π __   1 − __   1 = 9π __   2 = __   6 π cm2 = 9π __ 5 5 5 Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers Challenge 1 a n will be of one of the forms 3k, 3k + 1, 3k − 1: 1 1 ∞ x  2 1 ∞ 1 − ______     dx b Var(X ) = __  π  ∫   ______2  dx =  __ π  ∫−∞  ( −∞ 1 + x  1 + x  2) 1 1 1 ∞ 1 0       dx+ __   dx  π  ∫     (1 − ______ = __   π  ∫   (1 − ______ 2) −∞ 0 1 + x  1 + x  2) n = 3k: ( ) ( ) 1 + e  2πki + e  4πki _________ 1+1+1              = ______________     =    = 1 3 3 3 n = 3k + 1: 2πi 3k 2πi 6k ____ ____ 1  3k + e    3    + e    3    ____________________ + ( e  3 )  + ( e  3 )  1 + e  1  3 + e  3 _________________________                = ____________________       2πi  ____  3k+1 3k+1 2πi  ____  6k+2 2πi 2πki+ ____  3 + ( e  3 )  + ( e  3 )  1 + e  1  3 + e  3 _________________________                 = ____________________     3k−1 2πi ____     6k−2 2πi 2πki− ____  3 4πi 4πki− ____  3 2πi − ____  4πi − ____  1 + e  3 + e  3     = 0   = _______________ 3 b Consider jth term of f(x), ajx . f(1) + f(ω) + f(ω2) The corresponding terms in  _______________      are: 3 j j 2 j aj(1) aj(ω) aj(ω ) _____     +  _____   +  ______   3 3 3 From part a, this expression is equal to aj if j is 0 or a multiple of 3, and 0 otherwise. f(1) + f(ω) + f(ω2)  _______________     is the sum of all such expressions 3 for all terms in f(x), so is equal to the sum of all aj where j is 0 or a multiple of 3, as required. 45  ) x  r. c (1 + x)45 = ∑(45 r=0 r So the sum of the coefficients of powers of x that 15 45 are 0 or multiples of 3 is ∑(  ) . r = 0 3r From part b, this is equal to (1 + 1)45 + (1 + ω)45 + (1 + ω2)45  ____________________________          3 245 + (−ω2)45 + (−ω)45 _______ 245 − 2      =     = ___________________   3 3 j √ = lt→∞ im  [x − arctan x]  0−t   4πi ____ 1 + e    3  + e    3      = 0   = _____________ 3 n = 3k − 1: 2πi ____     0 0 1 1       dx= lim  ∫     1 − ______   dx I=∫   (1 − ______ t→∞ −t ( −∞ 1 + x  2) 1 + x  2) 4πi 4πki+ ____  3 2πi ____ 3k−1 0 1     dx Consider I = ∫   (1 − ______ −∞ 1 + x  2) ______ 2 ∞ ∞ ∞ x+3 x+3 3 1 _____   3  dx = π ∫     ___   3  )  dx = π ∫   _____   2  + ___     dx 2V = π ∫     ( 2 2 2 ( x  x  x  x  3) = 0 − arctan 0 − lt→∞ im (t − arctan t) As t → ∞, I diverges, so Var(X ) is infinite. ∫ ∫ ∫ a a 0 x x x   dx =     ______   dx +   ______   dx       c    ______ 2 2 2 −a 1 + x −a 1 + x 0 1 + x a a x 1 1   dx = __     [ln (1 + x2)]  0 = __     ln (1 + a2)    ______ 2 2 2 0 1 + x 2a x 1 and similarly,     ______   dx = __     ln (1 + 4a2)   2 2 0 1 + x 2a a x x ______ ______ So la→∞ im          dx −       dx 2 ( 0 1 + x2 ) 0 1 + x ∫ ∫ ∫ ∫ 1 + 4a2 1 1 __ lim ln  _______ = __      a→∞ ( 1 + a2 ) =   2  ln 4 ≠ 0 2 2a a x x Therefore     ______   dx ≠   ______   dx     2 2 1 + x 0 0 1 + x 2a a x x ⇒      ______   dx ≠      ______   dx 2 2 −a 1 + x −a 1 + x 0 x 0 1   dx = __ l a→∞ im      ______       la→∞ im  [ln (1 + x2)]  −a,  which diverges 2 −a 1 + x 2 a as a→∞, so the mean of X, l a→∞ im     x f(x) dx is not −a defined. ∫ ∫ ∫ ∫ ∫ ∫ CHAPTER 5 Prior knowledge check 1 0.5π π 5π 2 __ , ___     6 6 3 a Circle centre (0,3), radius 3 t 3 1 3 1 ____   2  + ___   3  dx = lim  π −  __   −   2      = lt→∞ im (π ∫     (___ ) ) t→∞ 2 x  x  ( [ x 2x ] 2) Im t 3 1 1 3   2  − ( −  __  − __     = lt→∞ im  π −  __  − ___ ( ( t 2 8) 2t )) 7π 1 1     As t → ∞ ,  __  → 0 and  __2  → 0, so V = ___ t 8 t  3i ∞ ∞ 0 A A A 3 a V = ∫   ______2  dx= ∫     ______2  dx+ ∫     ______2  dx −∞ 1 + −∞ 1 + 0 1 + x  x  x  t 0 A A im  ∫     ______2  dx = lt→∞ im  ∫     ______2  dx + lt→∞ −t 1 + 0 1 + x  x  O Re im  [arctan x]     + lt→∞ im  [arctan x]  )  = A(lt→∞ 0 −t t 0 9π ___ = A(arctan 0 − lt→∞ im  arctan (−t) + lt→∞ im  arctan t − arctan 0) b     2 = A(lt→∞ im  arctan t − lt→∞ im  arctan (−t)) Exercise 5A π π As t → ∞ , arctan t → __    and arctan (−t) → −  __  1 a (13, 1.176) 2 2 π π c (13, –1.966) __ __ V = A    − −     = Aπ π ( 2 ( 2 )) e 2, −  __  ( 6) 1 __ Hence if V = 1, then A =  π  b (13, 1.966) _ d (√ 13 , – 0.983) 233 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 233 22/09/18 3:09 PM Answers 2 a c e 3 a c e f g i 4 a c e g i __ __ √3 , −3) (3√3 __ , 3) __ b (3 __ __ √2 , 3√2 ) √2 , −5√2 ) (−3 d (−5 (−2, 0) x2 + y2 = 4 b x=3 2 y=5 d x2 = 4ay or y = ___  x 4a 2 2 2 2 2 x + y = 2ax or (x − a) + y = a 2 2 x2 + y2 = 3ay or x2 + y − ___  3a   = ____   9a  2 4 _3 _3 (x2 + y2) 2= 8y2 h (x2 + y2) 2= 2x2 2 x =1 r=4 b r2 = 8 cosec 2θ r2 = sin 2θ d r = 2 cos θ 4 r2 = _________    π     f r = ___   3__   sec θ + __ 1 + sin 2θ 4 √ 2  a  cosec θ + __  π   θ = arctan 2 h r =  __ 2 3 r = tan θ sec θ + a sec θ ( e θ=π 2 3 ) ( ( ) ) r = 3 cosec θ θ=0 O Initial line θ=π 2 f Challenge Consider the triangle formed by the two points and the origin and use the cosine rule to find d. 4 3 r = 2 sec (θ – O 4 π ) 3 θ=0 Initial line Exercise 5B 1 a θ=π 2 g θ=π 2 r=6 6 O r = a sin θ a 2 θ=0 Initial line θ=0 O b θ=0 θ=0 Initial line O 2a Initial line a θ=π 2 O d θ=π 2 r = a(1 – cos θ) a 5π 4 O θ = 5π 4 c h θ=π 2 Initial line i π 4 r = a cos 3θ θ=0 θ=π 6 θ = 5π 6 Initial line θ = –π 4 θ=0 O θ=π 2 θ=0 O θ=π 2 2 Initial line 7π 6 Initial line θ =– π 6 3π 2 r = 2 sec θ 234 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 234 Online Full worked solutions are available in SolutionBank. 22/09/18 3:09 PM Answers j θ=π 2 o θ=π 2 r = a(2 + cos θ) 2a 7a r = a(4 + 3 sin θ) θ=0 aO 3a Initial line θ=0 2a θ=π 2 6a k O a 4a π θ=0 7a O 5a Initial line Initial line θ=π 2 p r = a(6 + cos θ) 4a r = 2θ θ=0 O 2π 4π Initial line 6a l θ=π 2 3π 4a r = a(4 + 3 cos θ) a O 7a q θ=0 θ=π 2 a r2 = a2 sin θ Initial line θ=0 O θ=π 2 m 3a r Initial line θ=π 2 r = a(2 + sin θ) r2 = a2 sin 2θ θ=0 2a O 2a θ=0 O Initial line Initial line a θ=π 2 n 7a 2 r = a(6 + sin θ) θ=0 6a O 5a θ=π 2 6a Initial line k 2 θ=0 O k 2 Initial line 235 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 235 22/09/18 3:10 PM Answers 3 a 3 ____  πa   12 4 a=9 __ 2 3√3  __  __ 5 a   π  −   ____ ) ( 4 4 16 5π ___ 6     4 7 a Im Im 2 |z – 12 – 5i| = 13 10 12 + 5i |z – 4 + 3i| = 5 O Re 24 O Re 8 A 4 – 3i –6 b Cartesian equation is (x − 12)2 + (y − 5)2 = 169 Convert to polar coordinates: (r cos θ − 12)2 + (r sin θ − 5)2 = 169 Then rearrange this to get r = 24 cos θ + 10 sin θ 4 a arg z = π 4 Im b 35.1 8 a |z + 4 + 3i| = 5 O –8 Im Re –4 – 3i 10 A –12 + 5i –6 O –26 b Cartesian equation is (x + 4)2 + (y + 3)2 = 25 Convert to polar coordinates: (r cos θ + 4)2 + (r sin θ + 3)2 = 25 Then rearrange to get r = −8 cos θ − 6 sin θ 2 1 a ____  πa  8 (π + 2)a2   c ________   48 2 b _____   3πa  4 e __ √2  2 a2 ln  _______ or  ______  a ln   g a   (11π + 24)  __ 2 2 4 ( ∫ π 2a   f ______ 3 ( π 1 2 Area = 2 × __     a2(p + q cos θ)2 dθ 2 0 ∫ π ∫ π a2q2 π     (cos 2θ + 1) dθ = a2[p2θ + 2pq sin θ]  0  + ____ 2 0 π 2 2 aq 1     [ __ sin 2θ + θ]    = a2p2π + ____ 2 2 0 a2q2π ________ 2p2 + q2 2 ______ 2 2   =     πa = a p π +   2 2 236 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 236 ( ) ) 5 r cos θ = 3 r cos θ = −1 r = 3 sec θ r = −sec θ 6 2a, __  π   4 _ 3 + √ 73  ________   7  4 8 0.212 ( = a2   (p2 + 2pq cos θ + q2 cos2 θ) dθ 0 ) __  , ___ __  , ____ 1 (2a, 0),   a     2π   and   a  −2π 2 3 2 3 2 a (9.15, 1.11) b (212, 2.68) __ a√6  2a ___ 3 a   , ±0.421 b r = ±   ____   cosec θ 3 9 15 __ 4 (   2 a, ±1.32 ) 3 2 4 12π b ____     7 Exercise 5D 2 d __   a   4 2 b 385 9 0.0966 10 0.79 Challenge 3 a k = ___     7π Exercise 5C Re ) Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers Mixed exercise 5 7 a θ=π 2 1 _____  9πa  2 8 2 a, b π 4 C2 θ=π 2 θ=0 O Initial line r = 2a sec θ a C1 –π 4 __ θ=0 2a Initial line O π √ 3  b __  −   ___  8 6 8 a θ=π 2 a r = a (1 + cos θ) 2 __ √ 5 − 1 _______ c cos α =   θ=0 O Initial line ( ) √ ( ) 4) 5π  a2 b ___ 8 1   b y =  ___ 2x Im |z – 1 – i| = 2 ___ __ ( ) r = 4 cos 2θ R θ=0 O 4 O c Im 2 O a θ=0 O Re Initial line b 2π θ=π 2 r = a(1 – cos θ) 2 b Cartesian equation is (x − 1)2 − (y + 1)2 = 2 Convert to polar coordinates: (r cos θ − 1)2 + (r sin θ − 1)2 = 2 Then rearrange to get r = 2 cos θ + 2 sin θ θ = –π 4 2a ( 2 √3  π √ 3  7π     , ___  π       and 0, __    , a ___ 4 a   ___  ,  __ 2 6 ( 2 6) 2 5 a π θ= 2 θ=π 4 6 ) 4 __   a, __  π   9 a  3 2 3 10 a y2 = x2 − 1 5π Area = ___     (√ r = 2 sec θ __ __ π  b2√2 , __  π   , 2√2 , − __ 11 a r = 1 + cos θ 4 Initial line 2 r = 3 cos θ ___ __ θ=0 4 2   θ=π 2 3 r = 4 cos θ 2 Initial line a Maximum value at (2a, π) arg z = π 6 A 2 Re d 3.59 12 2.09 13 1.52 Challenge _ _ x = r cos θ = √ 2   θ cos θ, y = r sin θ = √ 2   θ sin θ _ _ _ dy dx √_ ___ ___     = 2   cos θ − √   2 θ sin θ,     = √   2 sin θ + √ 2   θ cos θ dθ dθ dy sin θ + θ cos θ        So ___     = ____________ dx cos θ − θ sin θ 237 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 237 22/09/18 3:10 PM Answers π 1 +  __  π 4 4+π At θ = __    the gradient of the tangent is _____   π  = _____     4 1 −  __  4 − π 4 4+π    )x + c So the tangent is of the form y = (_____ 4−π π π π  2     ,   __  , c = _______     Substituting in the point __ (4 4) 2(π − 4) So the equation for the tangent is 4+π π2 y = ( _______ ) x + ________     4−π 2(π − 4) Rearranging, this is 2(π − 4)y + 2(π + 4)x = π  2 Challenge a y 1 y = sech x CHAPTER 6 Prior knowledge check _ x O 3 1 + √   _______ 1 ln (  2  ) 1   − tan  2 x ≡ sec2x − tan2x ≡ (1 + tan2x) − tan2x ≡ 1 2 ______ cos  2 x ∫ ∫   e sin x dx = [ − e  cos x]    + [e sin x]    − ∫ 1 __ ⇒ ∫   e sin x dx =   2 [− e  cos x + e  sin x]    π π π 3   e  x sin x dx = [− e  x cos x]  0  +   e  x cos x dx 0 x π π x 0 π π x 0 0 x 0 x b π x y 0 0 =  __12 ((− e  π (−1) + e  π (0)) − (−1 + 0)) = __   12 (1 + e  π) y = cosech x Exercise 6A 1 a 27.29 (2 d.p.) b 1.13 (2 d.p.) –1 e – e  e–4 e4 +   2 a _______ b ________ 2 2 3 a  _34  b _53  c –0.96 (2 d.p.) e – 1   c _____ e+1 c _35  4 x = 1.32 (2 d.p.) x = –1.32 (2 d.p.) 5 x = 0.88 (2 d.p.) 6 x = – 0.55 (2 d.p.) 7 y x O y = 2cosh x 2 1 y = cosh 2x c x O 8 a f(x)  R 9 a b f(x) > 1 y y c –1 , f(x) , 1 y = coth x 1 y = 3tanh x + 2 5 O –1 2 O –1 x x b y = 5, y = –1 238 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 238 Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers Exercise 6B 1 y y = artanh x O –1 ( ) ( ) ≡ e3A + 3eA + 3e–A + e–3A 3A 3(eA + e–A) + 3eA +   3e–A + e–3A –  __________ e RHS ≡ ____________________         2 2 3A –3A e + e ≡ _________     ≡ cosh 3A ≡ LHS 2 A –   B cosh  ______ A +   B d RHS ≡ 2 sinh  ______ 2 2 x 1 c RHS ≡ 4 cosh3 A – 3 cosh A A A e–A 3 – 3________ e–A ≡ 4 ________   e +      e +    2 2 (eA + e–A)3 ≡ e3A + 3e2Ae–A + 3eAe–2A + e–3A ( ( ) ____  A–B   )( ______  –A+B   ( ) _____  A+B   ) _____  –A–B   e __________ e   e –      e +    ≡ 2 __________ 2 2 2 ≡  _12 (e 2 2 ____  A–B   + _____  A+B   2 2 2 ______   –A+B  + _____   A+B   – e 2 2 A–B ____  A–B   + – ____   + e 2 2 ) ______  –A+B   + _____  –A–B   – e 2 2 ≡  _1 (eA – eB + e–B – e–A) 2 2 ≡  _12 (eA – e–A) –  _12 (eB – e–B) y 2 a b c y = (arsinh x)2 d e O 3 a x 4 a e  − 1 3 Let y = artanh x. Then x = tanh y = _______   2y  and |x| , 1. e  + 1 2y 2y 2y So (e + 1)x = e − 1 ⇒ e (1 − x) = 1 + x 1+x 1+x ⇒ e2y = _____       ⇒ y = __  12 ln(_____   for |x| , 1. 1 −__x 1 − x ) __ √ 4 a ln (2 + 5 ) b ln (3 + 2√2 ) c  _12 ln 3 2y 5 a 6 a __ __ ln (√2 + √ 3 ) ___ ln (–3 + √ 10 ) b __ ln (2 + √5 ) __ 3+√ 5  ln  _______    ( b 2 _ 1 + y 1+x 1 1 __ _____ __ _____ √     +   ln     = ln 3 7  2 ln(  1 − x) 2 (1 − y) √ (9) __   c  _1 ln   11 ) 2 __ c  _1 ln (2 + √ 3 ) 2 √ ( c x = ln (  _72   ), x = ln 4 ( ___ ) ) f x = ln (4 6 √15 ) h x = ln  _5   , x = ln 3 ( 2) ) d x = ln (  _53   ) ___ –3 + √ 13    17 )    , x = ln (4 + √ e x = ln _________ 2 ___ ( ( __ ) 7 6 3√5  g x = 0, x = ln   ________    2 __ √ i x = ln (1 + 2 ) 6 a RHS ≡ 2(__   14 (e2x + 2 + e–2x)) – 1 ≡ __   12 (e2x + e–2x) ≡ LHS __ b ln(3 ± 2√ 2  ) 7 ln(__   72 ± __   32 √   5 ) _ ______ _ 1+y 1+x          × _____    = ln √ 3 ⇒ ln _____ ( 1−x 1−y) ⇒ (1 + x)(1 + y) = 3(1 − x)(1 − y) ⇒ xy − 2x − 2y + 1 = 0 2x − 1 ⇒ y(x − 2) = 2x − 1 ⇒ y = _______     x−2 ______ 5 a ≡ sinh A – sinh B ≡ LHS sinh (A – B) ≡ sinh A cosh B – cosh A sinh B sinh 3A ≡ 3 sinh A + 4 sinh3 A A +    B cosh ______ B cosh A + cosh B ≡ 2 cosh  ______   A –    2 2 1 + tanh2 A     cosh 2A ≡  ___________ 1 – tanh2 A cosh 2A ≡ cosh4 A – sinh4 A __ __ √ 3  sinh x ≡ 6√3  b tanh x ≡ 6   ___  c cosh 2x ≡ 7 __ 2__ __ 2√2  cosh x ≡ √2  b sinh 2x ≡ –2√2  c tanh 2x ≡ –  ____   3 x = ln (  _17   ), x = 0 b x = ln 3 Exercise 6C 8 He has not applied Osborn’s rule in line 1 – correct identity should be sech  2 x = 1 − tanh 2 x since implied sin2 term; he has split the denominator of the fraction in line 2 which is invalid; he has taken the reciprocal of both terms in line 3 – this is mathematically incorrect. 1 + tanh  2 x ___________ 2 − sech  2 x _______ 2 Correct proof: ___________        ≡        ≡     − 1 1 − tanh  2 x sech  2 x sech  2 x 1 a RHS ≡ 2 sinh A cosh A ≡ 2 cosh  2 x − 1 A –A A –A 9 a 8 cosh (x + 0.693) – e + e e e _______ _______ ≡ 2           2 2 b 8 1 2A _ ≡   2 (e – 1 + 1 – e–2A) c 0.148, –1.534 2A –2A e – e ________ ≡     2 Exercise 6D ≡ sinh 2A ≡ LHS 1 a 2cosh 2x b 5sinh 5x b RHS ≡ cosh A cosh B – sinh A sinh B c 2sech2 2x d 3cosh 3x A B A B e–A _______ e–B – _______ e–A _______ e–B e –4cosech2 4x f –2tanh 2x sech 2x ≡ _______   e +     e +     e –      e –    2 2 2 2 g e–x(cosh x – sinh x) h cosh 3x + 3x sinh 3x A+B –A+B A–B –A–B A+B –A+B A–B –A–B + e + e + e – e – e + e e e ______________________ _____________________ x cosh x –   sinh x         –         i _______________ ≡         j x(2cosh 3x + 3x sinh 3x) 4 4 3x2 –A+B A–B + e ) 2(e      ≡ _____________   k 2cosh 2x cosh 3x + 3sinh 2x sinh 3x 4 l tanh x m 3x2 cosh x3 A–B –(A–B) e + e     ≡  ___________ n 4cosh 2x sinh 2x o sinh x ecosh x 2 p – coth x cosech x ≡ cosh (A – B) ≡ LHS ( ( )( ) )( ) ( )( ) 239 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 239 22/09/18 3:10 PM Answers 2 y = a cosh nx + b sinh nx Differentiate with respect to x: dy ___     = an sinh nx + nb cosh nx dx d2y ____   2  = an2 cosh nx + bn2 sinh nx dx = n2 (a cosh nx + b sinh nx) 2 d y ____   2  = n2y dx 13 √____ 1 ___ 3__ (  2  ln(  11 ), 143 ) d 2y 4  ____2  = 2(5cosh 3x sinh x + 3sinh 3x cosh x) dx 1 2 _______  ___________     b ____________ 5 a ________ √ √ (x    + 1)2 + 1  4x2 – 1  3 1 ______   d −  _________ c _______2  1 – 9x 1 2 __ x   2  −1  x 2x 3 ______  _______  f ________ e _______ √ √ x4 – 1  9x2 – 1  2 x 1 _______ _______ h  ______  g 2x arcosh x +   ______  √ √ x2 – 1  x2 + 4  ex3   ______  i 3x2ex3 arsinh x + _______ √ x2 + 1  1   arsinh x 1   arcosh x + _______   ______ j _______   ______ √ √ x2 + 1  x2 – 1  sech x _______ k   ______  − arcosh x sech x tanh x √ x2 − 1 3x l arcosh 3x + ________   _______  √ 9x2 – 1  6 a y = arcosh x, then cosh y = x ___________ _ dx  ___  = sinh y = √ c   osh  2 y − 1 = √ x  2− 1  dy d 1   _  So  ___ (arcosh x) = ________ dx √ x  2− 1  b y = artanh x, then tanh y = x dx ___     = sech2y = 1 − tanh2y = 1 − x2 dy d 1 So ___     (artanh x) = ______     dx 1 − x  2 dy e  x _________ 2e  x 1   = _______           7 Using the chain rule, ___     = ___ 2 x dx 2 4 − e  2x e  ___ 1 −       ( (2) ) dy So (4 − e2x) ___  = 2ex, as required. dx d  2y dy 3 3 1 __ __ __   2 = 2x(−  __12 )( 1 + x  2)  − 2= − x (1 + x  2)  − 2 8 ___  = (1 + x  2)  − 2, ____ dx dx  d  3y 5 3 __ __ ____   3 = − x(2x(−  __32 ) ( 1 + x  2)  − 2) − (1 + x  2)  − 2 dx  5 3 __ __ = 3x  2 (1 + x  2)  − 2− (1 + x  2)  − 2 3 2 d   y d  y dy So (1 + x2) ____3 + 3x ____2  + ___     dx  dx  dx 3 1 __ __ 2 2 − __32  2 − __12  = 3x  (1 + x  )  − (1 + x  )  − 3x  2 (1 + x  2)  − 2  + (1 + x  2)  − 2 = 0 2x arcosh x ______ 2 9 –  ___________   +   2      3 __ x –1 (x2 – 1)  2 √ 10 25y – 25ln5 = 169x – 156 _ _ _ dy √ 15  2 11 ___  = ________   _______  ⇒ y = −  _____   x + √ 15 + ln(4 + √ 15  ) 2 dx √ 4x2 − 1  1 4   24 x  12 a 1 + __  12 x  2 + __ 13 a x + __  1  x  3 + ___   1   x  5 6 120 240 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 240 b 8.7 × 10−6 % x  2n−1 b _________     (2n − 1)! 14 a y = tanh x, tanh(0) = 0 dy dy  ___  = sech2 x, at 0 ___     = 1 dx dx d2y d2y   2 = 0  ____2 = –2 sech2 x tanh x, at 0 ____ dx dx d3y d3y   3 = –2  ____3 = –2 sech4 x + 4 sech2 x tanh2 x, at 0 ____ dx dx Now use a Maclaurin expansion to obtain the approximation b 5.23% (3 s.f.) 15 a y = artanh x, artanh0 = 0 dy dy  ___  = (1 – x2)–1, at 0 ___     = 1 dx dx d2y d2y   2 = 0  ____2 = 2x(1 – x2)–2, at 0 ____ dx dx d3y d3y   3 = 2  ____3 = 2(1 – x2)–2 + 8x2(1 – x2)–3, at 0 ____ dx dx d4y ____ 2 –3 2 –3   4 = 8x(1 – x ) + 16x(1 – x ) + 48x3(1 – x2)–4, dx d4y at 0  ____4 = 0 dx d5y ____   5 = 8(1 – x2)–3 + 48x2(1 – x2)–4 + 16(1 – x2)–3 + …, dx d5y at 0  ____5 = 24 since remaining terms all involve x dx Now use a Maclaurin expansion to obtain the approximation x  2n−1 b  _______  c x + __  56  x  3 2n − 1 x  3 + ___   121 x  5 16 x + __  13 120 6 dy 17 a  ___  = cos x sinh x – sin x cosh x dx d2y ____   2 = –2sin x sinh x dx d3y  ____3 = –2cos x sinh x – 2sin x cosh x dx d4y  ____4 = –2y + 2sin x sinh x – 2sin x sinh x − 2y = −4y dx x  4n−4 1   2520 x  8 c (−4)  n−1_________     b 1 − __  16 x  4 + ____ (4n − 4)! Challenge 5 4   24 x  1 − __  12 x  2 + __ Exercise 6E 1 a cosh x + 3 sinh x + c c –sech x + c 2 a _12 cosh 2x + c b 3 sinh __   x   + c 3 ______ 3 a arcosh x + √ x2 − 1 + c ( ) 4 a _14 sinh4 x + c _32 c _13 (cosh 2x) + c b sinh x – tanh x + c ______ b√ 1 + x2 − 3arsinh x + c b _14 ln cosh 4x + c 5 a _13 ln (2 + 3 cosh x) + c b tanh x + _ 12 tanh2 x + c or tanh x – _ 12 sech2 x + c c 5x + 2 ln cosh x + c 6 _13 x cosh 3x – _ 19 sinh 3x + c 7 a _14 e2x + _ 12 x + c 1 –5x b _12 ex + __  10 e + c Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers Challenge 1 4x 1 –4x c __ e – __  16 e + _ 18 e2x – _ 18 e–2x + c 16 or _ 18 sinh 4x + _ 14 sinh 2x + c 1  8 1 –  __ e 9 a _14 sinh 2x – _ 12 x + c ∫ c sinh x + _ 23 sinh3 x + _ 15 sinh5 x + c ∫   ln2   ln2 ln2 x  12 sinh x + x    ) dx = __   12     (cosh x + 1) dx = __ 10    cosh2(__ 0 2 0 0 2 − __  12 e   ln2 − e  −ln2     12 ln 2 + _____      + ln 2 − (0 + 0)) = __ = ( __________ 2 4 = __   18 (4ln 2 + 3) = __  18 (3 + ln 16) 11 a arcosh __   x   + c b _12 arsinh ___  2x   + c 3 5 x   + c b arcosh ___   x__    + c 12 a 3 arsinh  __ 3 √ 2  x 1 1 _ ___ _ __ b 3 arsinh ___  3x   + c 13 a  2 arcosh      + c 4 √ 3  14 0.977 (3 s.f.) __ (1 + √ 2 ) b ln (  _43  ) 15 a ln __ ______ __ 16√ x √ 1 + x  – arsinh √ x  + c 17 0.824 (3 s.f.) _ ___________ 18 ∫   √ x  2 − 4  dx= ∫   √ 4    cosh  2 u − 4 × 2 sinh u du 2 = 4∫   sinh  u du= 2∫   (cosh 2u − 1) du= 2(__   12  sinh 2u − u)+ c [ ] ( ) ( ) ( ) ( ) ( ) ( ) ___________ = 2 sinh u cosh u − 2u + c = 2 cosh u √ c   osh  2 u − 1 − 2u + c _ x 1 √ 2 __ __ =   2 x x  − 4 − 2arcosh(   ) + c 2 1 1 _______________       dx = ____________________         dx 19 a 2 cosh x − sinh x (ex + e−x) − __  12 (ex + e−x) x 2e 1     dx = _______   2x   dx = __________ __ e +3   12 ex + __  32 e−x 2ex 2 2   = ________     = _______    = RHS LHS = __________________     2ex + 2e–x + e–x – ex ex + 3e–x e2x + 3 e__x 2__   arctan  ___   + c b ___ ( √ 3 ) √ 3  20 0.360 (3 s.f.) x –    2 +c 21 a arcosh  _____ b arsinh (x + 3) + c 4 __ ___ √ 2  (x__+ 1) √10     + c c ____  arctan  _________ 10 √ 5  –    4 +c 9x __ d _13 arcosh  ______ √ 7  –    3 +c b _12 arcosh ______  2x __ 22 a _12 arsinh (2x – 3) + c √ 5  23 0.400 (3 s.f.) __ 24 ln (2 + √ 5 ) 3 1 1 1 3     ____________  dx= ___ 25  _____________   _    ____________     ___________  dx 2 3    x  − 6x + 7  1 √ √   3 1 √ (x    − 1)2 + __  4   ∫ ∫ ∫ ∫ ( ( ( ) ) ∫ ∫ ∫ 1 b –  _18 x + __  32 sinh 4x + c ∫ dx 1 x2 − 2x + 2 = 1 + sinh2 θ = cosh2 θ, ___   = cosh θ dθ cosh θ 1   3  dx = _______     dθ So  _____________ __ cosh  3 θ ( x  2− 2x + 2)   2  sinh θ   + c =   sech  2 θ dθ = tanh θ + c = ______ cosh θ x−1 ___________  + c = _____________      √ x      2− 2x + 2  2  x   + c 2 a _18 sinh (2x2) + __ b _12 tanh (x2) + c 4 ) ( ∫ Using u = x − 1, this becomes _ 2 √ 1   u 3 1_ 2_______ 1       ______  du= ___          ___   _  [arsinh(_____ 2 )] 0 √ √   3 0 √ u  2 + __ 3     43   ) Mixed exercise 6 13 15 1 a _43 b __ c –  __ 17 5 12 – 13x 2 y =  _________  12x – 13 3 RHS = sinh A cosh B – cosh A sinh B B A B e–A _______ e–B – _______ e–A _______ e–B eA –    =  _______   e +      e +      e –    2 2 2 2 +  eA–B – e–A–B –  ______________________ –  eA–B – e–A–B eA+B + e–A+B    eA+B – e–A+B            =  ______________________ 4 4 A–B –(A–B) 2(eA–B – e–A+B) ___________      =      e – e  = ____________   4 2 = sinh (A – B) = LHS 2 tanh  _12 x   4 RHS = ___________     1 – tanh2  _12 x ( )( ) ( ________ ( 5 x = ln (  _12   ), x = ln 7 6 x = ln (  _53   ) ( ____ ∫ ) –14 + √ 205         7 x = ln ___________ __3 x = ln (1 + √ 2 ) 8 a y y = 6 + sinh x 6 y = sinh 3x x O 2 ∫ π       (cosh 2x − 1) dx 27 Volume = π   sinh  2 x dx= __ 2 0 0 1 π π 1   sinh 2x − x]    = __     (sinh 2 − 2 − sinh 0 + 0) = __     [__ 0 2 2 4   2 − e  −2 π e π   2  (e  4 − 4e  2− 1)  − 2) = ____ = __    ( ________ 4 2 8e  1 ) 3 √ 3u   3 u  2 1_   ln _____    + ____    + 1 )]    =  ___ 4 √   3[ ( 2 0 _ 1 1_ (√_ √_) ___   _  ln(2 + √ 3   ) =    (ln 3 + 3 + 1 − ln (0 + 1)) = ___ √ √   3 3   26 a –ln3, ln2 b ln279 936 – 10 √ ) 2(ex – 1)     x 2 tanh  _12 x = ________ e +1 2 x (ex + 1)2 – (ex – 1)2   e – 1   = _________________          1 – tanh2  _12 x = 1 – ______ ex + 1 (ex + 1)2 x 4e ________   =   x (e + 1)2 x (ex – 1)(ex + 1) (ex + 1)2 _____________ 2(e – 1) ________   =          ×   So RHS = ________   x x e +1 4e 2ex 2x x 1 = _______ e–x = sinh x = LHS  e –   = ______  e –x  2e 2 ∫ _ )( 1 b 6 + sinh x = 3 sinh x + 4 sinh3 x ⇒ 2sinh3 x + sinh x − 3 = 0 __ ⇒ sinh x = 1 ⇒ x = ln(1 + √ 2 ) __ sinh (3ln(1 + √ 2 )) = 7, __ so (ln (1 + √ 2 ), 7)is the point of intersection. 9 a R = 12, a = 0.405 b 12 241 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 241 22/09/18 3:10 PM Answers 10 a 4 sinh (x + 0.693) b x = 0.75 (2 d.p.) c 0.75 (2 d.p.) 11 2sinh 2x 2x 3 _______ ______  12 a ________   b _______ √ √ 9x2 + 1  x4 + 1  x 1 ______  d 2x arcosh 2x + ________   _______  c _______ ( √ x2 – 4  √ 4x2 – 1 ) d  2y dy 2arsinh x ____ 2x arsinh x 2        − ___________   ______     _ ,   2  = ______ 13 ___  = _________ dx √ 1 + x  2  dx  1 + x  2 ( √ 1 + x2 )3 d   2y dy So (1 + x  2)  ____2  + x  ___  − 2 dx dx  x 2x arsinh x 2x arsinh   _ − 2       = 2 − ___________   _  + ___________ √ √ 1 + x  2  1 + x  2  which cancels to 0. dy 14 a ___  = 5sinh x – 3cosh x b (ln 2, 4) dx 2cosh 2x   ___________       15 _____________ 2 √sinh    2x – 1  Challenge   sech x    dx ∫   s echx dx = 2 ∫   s echx dx = 2 lim  ∫   _______ sech x ∞ ∞ −∞ 0 t t→∞ Using the substitution u = tanh x, this becomes 2 lt→∞ im    ∫   tanh t 1 ______  du = 2 lim   [arcsin u]  0   _______ t→∞ 2 √ 1 − u   = 2 lt→∞ im (arcsin (tanh t)) π = 2 arcsin 1 = 2 × __     = π 2 tanh t 0 CHAPTER 7 Prior knowledge check 1 y = (x − 1)ex + c 2 y2 = 4 − x2 3 a−  __32 ln | 50 − 2t | + c b−  __14 ln | cos (4x) | + c Exercise 7A 1 a y = x2 + c where c is constant y y = x2 + 3 dy ___ 16 a     = cos x cosh x + sin x sinh x dx d2y  ____2  = 2cos x sinh x dx d3y ____   3  = 2cos x cosh x – 2sin x sinh x dx d4y  ____4  = 2cos x sinh x – 2sin x cosh x – 2sin x cosh x dx – 2cos x sinh x = –4y 1 5 b x + __  13 x  3 − __   30 x  18 a a = 2, b = 1, c = 16 –1 y = x2 + 2 y = x2 + 1 17 2x +  __73 x  3 + __   61 x  5 60 __ b __1 l n(1 + √ 2 ) ∫ y = x2 – 1 1 y = x2 – 2 O x y y = 3ex y = 2ex y = ex 4 10 − 5arsinh 0 = 9.5944… = 5arsinh  __ 3 So area in real life is 960 m2 (2 s.f.). x –    1 +c 21 arsinh  _____ 3 __ 2__   arctan (√3  e x) + c 22 a ___ √ 3  b Use substitution x − 1 = 3sinh u. Then integral becomes ) 3 2 1 x O arsinh1 9 sinh u + 2 _________ × 3 cosh u du  ___________       √ 9 cosh  2 u  0 = 2 b y = Ae where A is constant 20 Area on graph is 5 5 5 2x 10 1   dx= 5   _______    )       ______  dx = 5 arsinh(___   _________   _ [ 3 ]0 4x  2+ 9  0 √ 0 √x  2 + __  9   ∫ y = x2 x b _14 e2x – _ 12 x + c ( 3 –2 2 1 19 a __ cosh 10x – _ 14 cosh 2x + c 20 ∫ 2 0 ∫ arsinh1 –1 arsinh1 (9 sinh u + 2) du = [9 cosh u + 2u]0 –2 0 = (9cosh(arsinh 1) + 2arsinh 1) − (9cosh 0 + 0) _ Since arsinh 1 = ln (1 + √ 2  ), _ 9cosh(arsinh 1) = 9cosh(ln(1 + √ 2 ) and the integral simplifies to _ _ 9√ 2 + 2arsinh 1 − 9 = 9(√ 2 − 1) + 2arsinh 1 23 a 59.5° b 8.82 m2 _ 24 a __   12 ln (4 ± √ 14 ) b 6.12 25 6610 cm3 ___ 26 a (ln(3 + √ 10 ), 0) 242 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 242 –3 y = –ex y = –3ex y = –2ex b 22.7 Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers c y = Ax2, where A is constant y dy x dx 2 a ___     =  ______   y x2 – 9 y = 2x2 2x     dx ∫  __1y dy =  __∫______ x –9 1 2 y = x2 2   1 ln(x2 – 9) + c lny = __ 2______ 1 y = ec√ x2 – 9  y2 = e2c(x2 – 9) Let k = –e2c, then y2 + kx2 = 9k b y2 + 5x2 = 45 y c y = – 12 x2 7 y = 2 x2 x O 6 5 y = –x2 4 y = –2x2 3 d y − x = c, where c is constant 2 2 y2 – x2 = 16 1 y2 – x2 = 1 4 –2 3 –3 2 1 3 4 x –2 –3 –6 –7 y2 – x2 = –1 c __ 3 a y = __  1 x sin x +  x  c y = 3x cosec x + c cosec x d y = xex_________ + cx e y = ln(__   1  + __ f y = ± __   1  x2 + ___   c     c2 ) 2 x 2x 6 x+c b y→0 4 a y = _____   x   e 5 a y = 1 + __  1 + __   c   x x2 b 1 + __  1 + ____   1   x 4x2 1 + __  1 + __   1   x x2 y = 1 + __  1 + __   5   x x2 y 2 y = sin x + 1 x y = sin x y = sin x – 1 O y = sin x – 2 3 b y = xe2x − e2x + cex √ e y = sin x + c, where c is constant y y 4 3 x –5 y2 – x2 = –4 –4 f k = 19 –4 y2 – x2 = –16 2 k = 49 O –1 –3 y2 – x2 = 0 y 1 k=1 2 y2 – x2 = 4 –4 –3 –2 –1 O –1 y2 + 5x2 = 45 1 y = 1 + 1x + 4x 2 y = 1 + 1x + x52 y = A sin x, where A is constant y = 1 + 1x + x12 y = 2sin x y = sin x y = 3sin x 2 y=1 1 O π –2 –3 –4 π x 2 –1 (–2, 34 ) y = –3sin x y = –sin x O – 12 y = –2sin x 6 a A(x + 1) ln  ________  (x + 2) __________ y =   ln x b x (x + 1) ___   _______   ln   16 3 (x + 2) ____________ y =     ln x 243 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 243 22/09/18 3:10 PM Answers 7 a y = _ 13 ex + ce−2x b y = −cot x + c cosec x c y = xecos x + cecos x 2 e y = __  x   + c cos x 2 g y = x ln (x + 2) + cx ( ) 3 a y = A cos 5x + B sin 5x b y = A cos 9x + B sin 9x c y = A cos x + B sin x d y = A cos  _43 x + B sin  _43 x e y = e−4x(A cos x + B sin x) f y = e2x(A cos x + B sin x) g y = e−10x(A cos 3x + B sin 3x) 3 __ h y = e −2 x(A cos  _32 x + B sin  _32 x) 4 a y = (A + Bx)e−7x b y = Ae−4x + Be3x c y = e−2x(A cos 3x + B sin 3x) d y = (A + Bx)e 3 _ 4 x _1 e y = e 3x(A cos  _23 x + B sin  _23 x) _2 f y = Ae 3x+ Be− 1 _ 2 x 5 a i x = Ae  (–k+√k – 9 )t______ + Be  (–k–√k – 9 )t ______ –t √ ii x = e (A sin(( 9 – k2 )t) + B cos((√ 9 – k2 )t) –kt iii x = (A + Bt)e __ __ b i x = e–2t( A cos (√ 5 t)+ B sin (√ 5 t)) ii x → 0 6 From auxiliary equation: b α = –  ___  (using quadratic formula) 2a b2 = 4ac (setting discriminant = 0) y = (A + Bx)eax dy ___     = αeαx (A + Bx) + Beαx dx d2y ____   2  = Bαeαx + α2eαx (A + Bx) + αBeαx dx Substituting these 5 relationships into d2y dy a ____   2  + b  ___  + cy dx dx yields a result of 0, so (A + Bx)eαx is a solution d y = e2x + cex c __ f y = __  1 x ln x +  x h y =  _14 x + cx− 3 1 _ y = (x + 2) ln (x + 2) + c (x + 2) 1   e x − __ y =  __  c4   14  e x + __ x x x3 1 x __ 1 __ x 8 y = __  1 x e −  x2  e +  x2  1   + ___  4x  9 y = −  ____ 3 3x2   2 c   10 a y = _ 13 (x2 + 1)2 + _______ (x + 1)   22   b y =  _13 (x2 + 1)2 − ________ 3(x + 1) c ____________ 11 a y = 1 +     sec x + tan x 1 or y = 1 + ________   cos x b y = 1 + ____________     sec x + tan x 1 + sin x 1 2 __ 12 y = 2 cosh(    x  + c) 2 13 a y = e  sinh x+c b y = e  sinh x+1 14 a y = sinh (x + c) b y = sinh x i j √ ______ 2 y y = sinh (x – 1) –2 O y = sinh (x + 1) 2 y = sinh (x – 2) 15 a y = A cos x + sin x b y = − 3 cos x + sin x π c x = __     ⇒ y = A × 0 + 1 = 1 2 3π x = ___     ⇒ y = A × 0 − 1 = −1 2 π 3π    , 1 and ( ___    , 1)lie on all possible solution so __ (2 ) 2 curves. bx ___   16 y = e a + c Exercise 7B 1 a c e g 2 a c e g y = Ae−3x + Be−2x y = Ae−5x + Be3x y = A + Be−5x _1 y = Ae− 4x+ Be2x y = (A + Bx)e−5x y = (A + Bx)e−x y = (A + Bx)e− 1 _ 4 x y = (A + Bx)e− 5 _ 2 x 244 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 244 b d f h b d f h y = Ae2x + Be6x y = Ae7x + Be−4x _1 y = Ae− 3x+ Be−2x −_15 x y = Ae + Be  2 _ 3 x y = (A + Bx)e9x y = (A + Bx)e4x y = (A + Bx)e  1__ _ 2 x √ y = (A + Bx)e− 3 x __ x ______ 2 7 Substitute y = Af(x) + Bg(x) into differential equation: a(Af″(x) + Bg″(x)) + b(Af′(x) + Bg′(x)) + c(Af(x) + Bg(x)) = A(af″(x + bf′(x) + cf(x)) + B(ag″(x) + bg′(x) + cg(x)) = A(0) + B(0) =0 Challenge Aeαx + Beβx = Aepx eqi x + Bepx e–qi x = epx((A + B) cosqx + i(A − B) sinqx) Set B = A*, so that A = λ + μi and B = λ − μi, λ, μ ∈ ℝ. Then A + B = 2λ and i(A − B) = −2μ Hence setting λ =  __12 C and μ = − __12 D gives the required result. Exercise 7C 1 a b c d e f g y = Ae−x + Be−5x + 2 y = Ae6x + Be2x + 2 + 3x y = Ae−4x + Be3x − 2e2x y = Ae−5x + Be3x −  _13 y = (A + Bx)e4x + 1 +  _12 x y = (A + Bx)e−x + 4 sin 2x − 3 cos 2x y = A cos 9x + B sin 9x +  _16 e3x h y = A cos 2x + B sin 2x + _ 13 sin x i y = e2x(A cos x + B sin x) + 3 + 8x + 5x2 1 x j y = ex(A cos 5x + B sin 5x) + __  25 e 7 1 2 1 __ __ __ 2 a  4 x  −   8 x +  32 7 b y = Ae  4x + Be  x + __   14 x  2 − __   18 x + __  32 3 a Ae  6x + B 17 1 2 b y = Ae  6x + B − __  19 x  3 + __   36 x  − ___   108 x Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers 4 y = A + Be−4x + 2x3 − __   32 x2 − __   34 x 5 a C.F. contains a term in xex. Results in setting up equation in the form of ex = 0. Not possible. b λ =  _12 c y = (A + Bx +  _12 x2) ex 4k kt 6 a y = Ae–t + Be–3t + __   53 – ___     + ___     9 3 b y = 2t –1 Challenge y = A cosx + B sinx + (x − __  45 )e 2x Exercise 7D 1 a y = A e  −2x + B e  −3x+ e  x 2 a y = A + Be  −2x + __   32 e  2x b y = e−3x − e−2x + ex b y = 2 − _ 32e−2x + _  32 e2x 3 y = _  16 e−6x + _ 16e7x − _ 13 4 a y = A cos 3x + B sin 3x + 2 sin x b y = cos 3x + 2 sin 3x + 2 sin x 1 __ 5 a y = e  − 2 x(A cos x + B sin x) + sin x _1 b y = sin x (1 − e− 2 x) 6 a x = Aet + Be2t + t b x = et + e2t + t 7 x = e3t + e−3t − sin t 8 a x = Ae  2t + Bte  2t + __   12 t  3e   2t b x = (t + _ 12 t3)e2t 6 6 1 _ _ _ 9 x =   2 (cos  5t + sin   5t + 1) 10 a x = e  t(A cos t + B sin t) + t  2+ 2t + 1 b x = et sin t + 1 + 2t + t2 or x = et sin t + (1 + t)2 11 a y = Aex + Be2x + 3xe2x b y = 3ex − 3e2x + 3xe2x x 1 12 y = ___     sin 3x − __     cos 3x 18 6 13 a x = e−t − e−3t dx 1 b Setting ___     = 0 gives t = __ substituting     ln3, then __ 2 dt 2√ 3 this into x = e−t − e−3t gives x = ____   9 d2x Since ____   2 < 0, this is the maximum. dt Mixed exercise 7 1 y = 2 sin x + c cos x 1 _ 2 y = 5 + c (1 − x2) 2 x c 3 y = − __  + __   2 x 3 _ c   2 4 y =  _5 x2 + __  x 5 y =  _12 + ce−x ______ 6 y = 2x + c x√ 1 − x2  keλx kxn+1 ax 7 a y =  _____  + ceax b y = _____     e + ceax λ–a n+1 8 y = sin x + A cosec x 9 a y = e  x cos x + A cos x b y = e  x cos x − (1 + e π) cos x 1 1 3x 10 y = −  __ (3 sin x + cos x) + __  10 e  10 coshx+c 11 a y = e  b y = e  coshx−1 2x  2(3e − 1) 12 y =  ___________      3e2x  + 1 __ __ 1 _ √ √ 3 3     x + B sin  ___  x) 13 y = e −2 x(A cos ___ 2 2 14 y = (A + Bx)e6x 15 y = A + Be4x 16 y = cos kx + __  1 sin kx k x 17 y = e sin 3x 18 a k =  __19 b y = e2x (A cos 3x + B sin 3x) + _ 19 e2x 19 y = Aex + Be−x + 2xex 2 2 2 20 a y = (A + Bx)e2x b They are part of the complementary function. c k = 2 and y = (A + Bx + 2x2)e2x 21 y = sin 2t + 2 cos 2t − cos 3t 22 a k = 1, μ = 2, λ = 3 b y = Aex + Be2x + xe2x + 2x + 3 1 __ 23 a y = 4e  –  4x sin  __12 x + x + 3 1 1 _ _ b As x →∞, e− 4 x→ 0, so 4e− 4 xsin  _12 x → 0 and y ≈ x + 3. 5 −2x 1 _ _ 24 y =   6 e +   6 (cos 3x − sin 3x) 1 25 a x = Ae  −4t + Bt e  −4t + __   32 sin 4t 1 b x = __  12 e  −4t + __   15 t e  −4t + __   32 sin 4t 8 1 c Will oscillate as a sine wave with amplitude  __ and 32 π period __     2 Challenge √ ______ x+c      1 a y = ± ______ 1 + x2 B _1 3 __ _ 2 a y = __  A x +  x2  +  2 ln x −  4 √ ______ x+4 b y = ______      1 + x2 9 4 − ____ 3 1 _ _ b y =  __ x  4x2  +  2 ln x −  4 CHAPTER 8 Prior knowledge check 20 x  2+ 2000x 1 y =  ______________      x + 50 2 y = Ae–6x + B 3 y = 4 sin x + 2 cos x + e  −x (−2 cos x − 6 sin x) Exercise 8A 1 s = −t cos t + sin t + c When t = 0, s = 0, so c = 0. π So t = __     ⇒ s = 0 + 1 = 1 m 2  3  – ______   3   2 v = __ 2 2 + t 2 3 a v = 40 – 20 e0.2t b 200 ln 2 – 100 t 4 x = 2 ln(__    + 1) 2 5 a Integrating factor is e  −t  2, so equation becomes ve  −t  2 = ∫   t e  −t  2 dt ⇒ v = e  t  2( c − __12 e  −t  2) v = 1 when t = 0 ⇒ c = __   32 , so v = __   12 (3 e  t  2 − 1) –1 b 81.4 m s c No; velocity would be over 13 million m s–1. 6 a Integrating factor is (t + 4)4, so equation becomes 49 v(t + 4)4 = 9.8   (t + 4)4 dt = ___    (t + 4)5 + b 25 5 49 (t + 4)   − c      ⇒ v = _____________      25 (t + 4)  4 v = 0 when t = 0 ⇒ c = 50 176, 49 (t + 4)  5 − 50 176         so v =  __________________ 25 (t + 4)  4 –1 b 17.3 m s c Velocity will increase without limit – unlikely to be the case 7 a The 2.5 (cm3 hr–1) comes from 5% of the 50 cm3 gas mixture being added. The volume of the tank at time t is (500 + 30t) cm3, so the amount of oxygen leaking out is x 20 ×  __________  cm3 hr–1. 500 + 30t b 9.34 cm3 c e.g. The model should take into account the fact that the oxygen does not mix throughout immediately on entering the tank ∫ 245 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 245 22/09/18 3:10 PM Answers Exercise 8B 1 a Simple harmonic motion __ √ b x = 2 cos 3t + sin 3t c 5  ____ π √ 101  2 a x = 5 cos 4t + __  12 sin 4t b __ , _____     2 2 3 a Simple harmonic motion b x¨ = –5x _ _ _ c x_ = 5 cos (√ 5   t) + __  65 √ 5  sin ( √ 5  t) √ 805  d  _____   5 4 a  __72 __ __ __ 7 7 2 b x = 6 cos(  __   t)+  __   sin(  __   t) 7 2 2 c 3.36 seconds 3 3 5 a x = 0.183 sin(__     t) − 1.287 cos(__     t) 2 2 2π b  ___  3 c The model does not allow for any changes in amplitude over time. 6 a Simple harmonic motion __ √ 2 t) b x = 0.3 cos(10 __ c ___   π   √ 2 s, 0.3 m 10 __ d 4.24 m s–1 (3 s.f.) or 3√ 2  m s–1 4π 7 a x = cos (12.5t) b ___  __ 25 8 a x = 8 cos(8√ 5 t) b 0.351 s (3 s.f.) _ 9 a x = 15 cos(__   53  √ 10  t) cm √ √ √ b 0.688 s, 15 cm c Incorporate a damping effect, for example air resistance. This could take the form of a factor in the x formula which equals 1 at t = 0, and decreases down to zero as t → ∞ Exercise 8C 1 a 2e–2t (cos 2t + sin 2t) b 0.0901 (3 s.f.) c Lightly damped 2 6e–2t – 2e–6t __ __ 1__  sin √ 3 a e–t cos √ 5 t +  ___ 5 t √ 5  π__  or 1.40 (3 s.f.) b  ___ √ 5  1 4 a x = ut e–2kt b t = ___     2k 5 a F = ma ⇒ −2v − 6x = 2a ⇒ ẍ  + ẋ  + 3x = 0 ( ) ___ _ _ √ √ 11  11  __t 5 √     t)+ __  11 11  sin(____     t)) b x = e  − 2(cos(____ 2 2 c – 0.459 m (3 s.f.) d Maximum displacement will decrease exponentially (light damping). k  cos 3t + ___ k cos t 6 x = –  __   k   sin 3t + __ 8 8 15 5U 5U 5U ___ ___ ___ –2kt –3kt 7 x = –   e +    e +   2k 3k 6k d  2 x dx   2  = −30 sin t + 10 cos t 8 a ___ = 30 cos t + 10 sin t, ____ dt 2 dt  dx d   x ___   + 3    + x = 100 cos t ⇒ 2 ____ dt d t  2 1 __ b x = Ae  −  2t+ Be  −t+ 30 sin t − 10 cos t c –31.53 m Exercise 8D _ _ _ _ 1 x = __  14 (2 + 3√ 2  )e  √ 2 t + __   14 (2 − 3√ 2  )e  −√2 t _ _ _ 2 a x = e  −t (A cos t + B sin t) y = __   15 e–t((B – 2A) cos t – (A + 2B) sin t) OR y = e–t(A cos t + B sin t) x = –e–t((2A + B) cos t + (2B – A) sin t) depending on your method b x = e  −t(cos t + 12 sin t), y = e___   −t(2 cos t − 5 sin t) ___ 3t √ √ __ 11  11  –7 ____ ____   2 ____ 3 x = e   ___  sin (    t)– cos(    t) + 1 ( √ 11  ) 2 2 ___ ___ 3t √ √ __ 11  11  3     t)– ____   ___  sin(____     t) y = e   2 cos(____ ( ) 2 2 √ 11  2 dy dx d  x ___ dx   −         = 5 ____ 4 a y = 5 ___  − x ⇒ ___ dt dt d t  2 dt dx dx d  2 x ___   −    = −0.5x + 0.4(5 ___  − x) ⇒ 5 ____ dt d t  2 dt dx d  2 x ____ ___ ⇒   2  − 0.6    + 0.18x = 0 dt d t  b m2 − 0.6m + 0.18 = 0 ⇒ m = 0.3 ± 0.3i So x = e0.3t(A cos 0.3t + B sin 0.3t) c y = 0.1e0.3t ((5A + 15B)cos 0.3t) + (5B – 15A)sin 0.3t) d t = 6.17…, during 2018 e 441 f Model seems reasonable for the first few years, but becomes unsuitable in the longer term as x and y oscillate, sometimes giving negative values. dy 1 ____ d  2 x __ 3 dx 5 a ___  = __   +     ___    ,       dt 2 d t  2 2 dt 3 dx 1 dx 1 d  2 x __   +      ___ = −2x + __      ( ___  + 3x) so __     ____ 2 d t  2 2 dt 2 dt dx d  2 x   + 2 ___  + x = 0 ⇒ x = (A + Bt)e−t ⇒  ____ dt d t  2 dx 1 ___    = (B − A − Bt)e−t ⇒ y = ( __    B + A + Bt)e −t 2 dt 1 When t = 0, x = A = 1 and y = __    B + 1 = 2 ⇒ B = 2 2 So x = (1 + 2t)e−t and y = (2 + 2t)e−t. b x = 0.677 litres, y = 0.812 litres c The amount of both chemicals tends to zero d  2 y d  2 y 1 d  2 y dx 1 ____ 6 a ___  = __   = −16y   2  = −4y ⇒  ____       2 ⇒  __  ____ dt 4 d t  4 d t  d t  2 ⇒ ÿ  = − 4  2 y. This is SHM in the y direction. b x = 4 cos 4t − 5 sin 4t, y = 4 sin 4t + 5 cos 4t dy dx dx d  2 x 7 a y = 100 ___ + 3x − 5000 ⇒ ___   + 3 ___      = 100 ____ dt dt dt d t  2 dx dx d  2x ____ ___ ___ ⇒100  2  + 3   = 0.01x − 0.03(100     + 3x − 5000) dt dt dt  dx d  2 x ____ ___ ⇒   2  + 0.06   + 0.0008 = 1.5 dt d t  1 1 1 1 __ __ __ __ b x = 1875 + Ae  −  25t+ Be  −  50t, y = 625 − Ae  −  25t+ Be  −  50t c x tends to a limiting value of 1875 and y tends to a limiting value of 625. dy d  2 x dx dx 8 a y = ___    + 2x − 1 ⇒ ___   + 2 ___      = ____   dt dt d t  2 dt dx dx d   2 x ___ ___   + 2      = 4x +   + 2x − 1 So  ____ ( dt ) + 2 dt d t  2 2 dx d  x ___   +    − 6x = 1 ⇒  ____ d t  2 dt   16 , y = Ae  2t− 4Be  −3t + __   13 b x = Ae  2t + Be  −3t − __ 6 c Model not suitable since for large values of t, the amount of nutrients grows exponentially without limit. _ y = __  14 (4 − √ 2  )e  √ 2 t + __   14 (4 + √ 2  )e  −√2 t 246 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 246 Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers Challenge 96 owls and 4800 field mice. 13 a x = Ut – __  U n sin nt Mixed exercise 8 1 a v = 18 – 10e–t b 18 m s–1 2 6 3 a v = 10 – ______   50   b (100 – 25 ln 5) m ≈ 59.8 m 2t + 5 π 4 a  __  m s–1 2 1 1 1 b x = __    t2 − __    cos 2t + c. x = 0 when t = 0 ⇒ c = __     4 8 8 π  2 π 1 1     − 0 + __    (π2 + 8)     = ___ t = __     ⇒ x = ___ 4 8 64 64 5 a v = 2t + ln(t + 1) b (2 + 3 ln 3) m 6 a Integrating factor is e–3t, then solve the differential equation to get V = −  __23 t − __   17   + __   26 e3t 9 9 2 b 1162.2 cm3 c Bacteria will reproduce without limit – build in some limiting/decay factor 7 a The 1200 comes from 4g per incoming litre of water, per day. The fraction is from the leaking, which is the total grams x, times the proportion of 200 the reservoir leaking out, which is ______________      10 000 + 100t b 7860 g c e.g. The model should take into account the fact that the contaminant does not mix throughout immediately on entry. 8 a Simple harmonic motion b General solution is x = A cos 7t + B sin 7t At rest at t = 0 gives B = 0, and A is the displacement 2π of B from A. The period is then ___     (seconds) 7 9 a Simple harmonic motion __ 2 b x = 4 cos(5  __   t) 3 Period is 1.54 s (3 s.f.) Amplitude is (4 – 2.5) m = 1.5 m 10 a x = 2.161 cos(__   12 t)   12 t)+ 3.366 sin(__ b 2π c The model does not allow for any changes in amplitude over time or the effect of oscillations in the fisherman’s_______ line, for example._______ 11 a x = e–kt (A cos(t√ n2 – k2  ) + B sin(t√ n2 – k2  )) 2π _______  b ________ √ n2 – k2  12 a Solving the equation gives x = e−kt(A cos kt + B sin kt) dx and ___     = ke−kt((−A + B)cos kt + (−B − A)sin kt) dt dx U When t = 0, x = 0 and ___     = U, so A = 0 and B = __     dt k So when P is instantaneously at rest, dx  ___  = Ue−kt(cos kt − sin kt) = 0 ⇒ tan kt = 1 dt ⇒ kt = ( n + __  14 )π , n ∊ ℕ. √ d x π k O 3π 2k π 2k 3π k 2π k 4π k t 2Uπ 2π b  ___ c  ____ n n 2 d  x dx   2  = −30 sin t − 15 cos t 14 a ___ = 30 cos t − 15 sin t, ____ dt d t  dx d  2 x ____ ___ So   2  + 4   + 3x = 150 cos t dt d t  b x = Ae–t + Be–3t + 30 sin t + 15 cos t c –29 m 15 a x = e−t(A cos 3t + B sin 3t) + 3 cos t b x = 2e−tsin 3t + 3 cos t c When t = 7, the first term in the expression for x is very small compared to the second, so can be ignored. So x ≈ 3 cos t, which has a distance of 6 between maximum and minimum values. dy d  2x dx dx 16 a y = ___    − 2x ⇒ ___   − 2 ___      = ____   dt dt dt  2 dt dx dx d  2x   2  − 2 ___ = −2x + 4( ___  − 2x) So ____ dt dt dt  dx d  2 x ____ ___ ⇒   2  − 6   + 10x = 0 dt d t  b Auxiliary equation has solutions m = 3 ± i, so the equation has solution x = e3t(A cos t + B sin t). c y = e  3t((A + B) cos t + (B − A) sin t) d 2009 e 1113 f The model predicts a huge amount of hedgehogs when the slugs die out so it might not be sensible. dy d  2x dx dx d   2x 17 a 3 ___  = ____   + 4 ___ , so ____     2  + 2 ___  + x = 0 dt dt  2 dt dt dt  This has general solution x = (A + Bt)e−t. dx 1  ___  = Be−t − (A + Bt)e−t ⇒ y = __    ((B + 3A) + 3Bt)e−t 3 dt Using the initial conditions, A = 10 and B = 30, so x = 30te−t + 10e−t and y = 30te−t + 20e−t. b 9 of organism M and 10 of organism N c The numbers of each organism tend to zero dy 2 dy d2y 4 ___ dx 1 18 a  __ x = ___             + __     = 2 ____2  + __    y − 1 ⇒ ___ 2 3 dt dt 3 dt dt dy 2 dy d2y 4 ___ 1       = 2 + __  ___  + __   y − (   y − 1) 2 ____2 + __ 3 dt 3 dt 3 dt 2 2d y __ 7 dy __ 1 +   ___     +   y = 3 ⇒  _____ 3 dt 3 dt2 dy d2y     + y = 9 6____   2 + 7___ dt dt t t   −  __ −  __ −t − __   85 e 6 + 10, y = __   35 e –t − __   85 6 + 9 b x = − __25 e c As t becomes large e−t → 0, so there will be approximately 10 litres in tank A and 9 litres in tank B. Challenge ∫ ∫ dX dX a i ___ = −X ⇒    ___     = −    dt ⇒ X = ce−t X dt X = 300 when t = 0 ⇒ c = 300, so X = 300e−t ii 20 minutes iii 600 – 100e–t (3t + 5) or equivalent b i x′ = –2x, y′ = 2x – y, x(0) = V, y(0) = 0 ⇒ x = Ve–2t, y = 2V(e–t – e–2t). y′ = 0 ⇒ 2x = y ⇒ e–2t = e–t – e–2t ⇒ t = ln 2 hours = 42 mins (nearest minute) ii ln 3 hours Review exercise 2 b r = 3 sec θ __ p   c r = 2√3  sec θ −  __ 6 1 a r=2 ( ) 247 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 247 22/09/18 3:10 PM Answers 2 a r = a cos 3θ θ= θ=π 2 5 a π 6 P C D O a Initial line θ=– p   a2 b ___ 12 3 a π 6 3 θ=– __ __ θ=π 2 4 a Initial line π 4 3 __ b  32 (2π − 3√3 ) 3a Initial line Q ( ) ( ) __ a   (4π + 9√3 ) c  ___ 16 d Let the smaller area enclosed by C and the half-line π θ =  __  be A2: 3 2 3a    )  − 2A1  − 2A2  R = π (___ 2 _ _ 9 a  2 π ____ 6 a  2( 2 a  2( ______ 3 ) − ____     )   −     4π + 9 √     2π − 3 √ 3 =   4 16 16 _ _ 2 2 2 2 9 √ 3  a  2 9 a  π ____ 3 a  π _______ a  π 9 √ 3  a  ______   −     −  _______   −     +     = πa  2 =  ______ 4 2 8 4 8 6 a Im 2 r = 3 cos 2θ O 2a π   __   a, __ __   a, −  __ b P   3  π   , Q   3 2 3 2 3 θ=π 6 π θ= 4 O 2√6  c  ____   3 r = a (1 + cos θ) O Re 6 r=a O a 3 – 4i 2a Initial line –8 __ 3√3 a b r =   _____   cosec θ 4 __ 3√3 a   cosec θ r = −  _____ 4 θ = π c The circle and the cardioid meet when π a = a(1 + cos θ) ⇒ cos θ = 0 ⇒ θ = ±  __  2 π  __  1 2 A = 2 × __     ∫     r  2 dθ 0 2 π __     ∫     0 π  __  = ∫     2 2 r  dθ 2 0 b I n Cartesian form: (x − 3)  2 + ( y + 4)  2= 25 ⇒ ( r cos θ − 3)  2 + ( r sin θ + 4)  2= 25 ⇒ r  2 cos  2 θ − 6r cos θ + 9 + r  2 sin  2 θ + 8r sin θ + 16 = 25 ⇒ r  2(cos  2 θ + sin  2 θ)− 6r cos θ + 8r sin θ = 0 ⇒ r  2(cos  2 θ + sin  2 θ) = r(6 cos θ − 8 sin θ) ⇒ r = (6 cos θ − 8 sin θ) c 63.3 7 a θ= a  2 (1 + cos θ)  2 dθ π __       2 (1 + 2 cos θ + cos  2 θ) dθ =∫     a 2 π 4 A O 1 0 π __     2 2 1 1   (1 + 2 cos θ + __     cos 2θ + __    ) dθ =∫     a 2 2 0 π __     2 3 1     cos 2θ + __    ) dθ = a  2 ∫     ( 2 cos θ + __ 2 2 0 r = cos 2θ θ=– π 4 Initial line B b (0.667, 0.421) and (0.667, −0.421) π __     3θ 2 3π 1     sin 2θ + ___    ]    = a  2(___    + 2) = a  2 [2 sin θ + __ 4 2 0 4 The required area is A less half the circle 3π π+8 2 1    + 2) − __     πa  2= ( _____   a      2(___ a 2 4 4 ) 248 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 248 Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers 8 a π 1 π cA 1= 2 × __     ∫__  π     r  2 dθ= ∫ __  π     a  2 (3 + 2 cos θ)  2 dθ 2 3 3 θ=π 2 π = a  2 ∫__  π     (9 + 12 cos θ + 4 cos  2 θ) dθ r = sin 2θ 3 π = a  2 ∫__  π     (11 + 12 cos θ + 2 cos 2θ) dθ A 3 _ π   22π 13 √ 3 2     − ______     = a  [11θ + 12 sin θ + sin 2θ]   __π    = a  2(____ 3 3 2 ) π __     1 3 a  2 (5 − 2 cos θ)  2 dθ __ A 2 = 2 ×     ∫     2 0 O Initial line = a  2 ∫     (25 − 20 cos θ + 4 cos  2 θ) dθ 3 b (0.943, 0.615) 0 3 0 [ C P __ 20 p = 3, q =   20 3 Initial line Q 21 ln  _13 , ln 7 22 x = __   12 ln __13 , __  12 l n __35 __ 23 a k > √3   c P 3, __  π   , Q(6, 0) 3 10  _12 a2 ( ) ∫ π __     ∫ 24 a π __     2   8  π __     π __     4 1     sin 4θ]    = 4 a  2 ∫__  π     (1 + cos 4θ) dθ= 4 a  2 [θ + __ π __ 4     8 4 12 _98 πa2 __ ______   m2 c  2875π 32 15 a _32 πa2 2π   2π   , B:   1 __   a,  ___ __   a, − ___ b A:   1 2 3 2 3 c  _94 a __ 27√3  2 d  _____   a 8 e 113 cm2 (3 s.f.) π   , B:(  3 __ 2  a,   π __ 3  ) __   a, −  __ 16 a A:   3 2 3 _   π 3 √ 3 3     a b AB = 2 ×  __  a sin  __  = _____ 2 3 2 __ √3  − 4π)a2 c (9 d 9.07 cm2 (3 s.f.) ( ) ) 17 a A:(5a, 0), B:(3a, 0) 5π   , D: 4a, __ b C: 4a,  ___  π   3 3 ( ) ( ) 25 a b 3  ___ 5π  π   −   √ 13 a θ = ___   π  ,  ___ b ___   12 12 12 16 14 a P(4a, 1.107), Q(4a, −1.107) __ 5√5  b  ____   m 4 ) ( b 8 π 1 1     − __     = __     a  2(π − 2) = 4 a  2(__ 8 4) 2 ( b 0, −ln 3 ( ) −x 2 ( ) −x 2 e  + e  e  – e  cosh  2 x − sinh  2 x = ________     − ________         x 2 x 2 e  2x + 2 + e  −2x − ( e  2x − 2 + e  −2x ) 1 4 1 4 11A = __     __π     r  2 dθ= __     __π     16 a  2 cos  2 2θ dθ 2   8  _ 19 ln (  _13   ), ln 3 6 O ] π __     3 3 27π 19 √   = a  2 27θ − 20 sin θ + sin 2θ     = a  2(____     −  _____   3 _ 2 ) 0 _     13 √ 3 19 √ 3 22π _____ 27π _____ A 1   + A 2 = a  2( ____   −     + a  2( ____   −     3 2 ) 3 2 ) _   2 a = ___    (49π + 48 √ 3   ) 3 1 _ 18  2 ln 3 D A π __     = a  2 ∫     (27 − 20 cos θ + 2 cos 2θ) dθ 2 2 9 a (x − 3) __ + y = 9 √ x + 3 y = 6 b θ=π 2 2 3 π __     26 a 4          =  __ = 1 =  _____________________________ 4 4 k = −1, a = 2 e   x + e  −x 2 e  2x + 2 + e  −2x 2 cosh  2 x − 1 = 2 (________     − 1 = _____________        − 1 2 ) 2 2x −2x e   + e   = cosh 2x = _________   2 __ ±ln (3 + √ 8 ) e  x + e  −x 3 e  x + e  −x 3 4 cosh  x − 3 cosh x = 4 (________     − 3(________       2 ) 2 ) 3 e  x + 3 e  −x _________ e  3x + e  −3x e   3x + 3 e  x + 3 e  −x + e  −3x ___________          −     =     = ______________________   2 2 2 = cosh 3x __ b ln (√2  ± 1) 27 a cosh A cosh B − sinh A sinh B e  A + e  −A ________ e  B + e  −B e  A − e  −A ________ e   B − e  −B ________ = ( ________     )(  2  )(  2  )− (   2  ) 2 1 = __    (e  A+B + e  −A+B + e  A−B + e  −A−B − e  A+B + e  −A+B 4 + e  A−B − e  −A−B) e  A−B + e  −(A−B) 1      = cosh (A − B) = __    (2 e  −A+B + 2 e  A−B) = ____________ 4 2 bcosh x cosh 1 − sinh x sinh 1 = sinh x cosh x cosh 1 = sinh x(1 + sinh 1) e + e  −1 _______     e + e  −1 cosh 1 2 __________ ___________       = __________   =      tanh x =   −1 1 + sinh 1 e − e  2 + e − e −1 1 + _______     2 e   2+ 1 ___________   =      e  2+ 2e − 1 249 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 249 22/09/18 3:10 PM Answers e  y − e  −y 28 a L et arsinh x ⇒ x = sinh y = ________     2 ⇒ 2x = e  y − e  −y ⇒ e  2y− 2x e  y− 1 = 0 _ _ 2x + √ 4 x  2+ 4        = x + √ x  2+ 1  ⇒ e  y = ______________ 2 _ ⇒ y = ln(x + √ x  2+ 1 ) _ barsinh(cot θ)= ln[cot θ + √ 1 + cot  2 θ ] = ln(cot θ + cosec θ) ⎜ ⎛ ⎟ θ ⎞ 2 cos  2  __  cos θ + 1 θ 2 _________ ____________ = ln      = ln(cot  __ )   = ln(  θ θ sin θ ) 2 __ __ 2 sin     cos     ⎝ 2 2⎠ b 0.0029% sinh  2 θ dθ a cosh θ   −  ________ sinh  2 θ 1 1 _  dx =   _______________ _   dθ = −  __  1 dθ  ___________       a x √ x  2+ a  2  a  2 √ 1 + sinh  2 θ _______________        sinh  2 θ ∫ ∫ ∫ a 1 1 = −  __  θ + c = −  __  arsinh(__    )+ c a a x 37 a y = artanhx ⇒ tanh y = x y y = artanh x Differentiate implicitly with respect to x dy dy 1 1 1 sech  2 y  ___  = 1 ⇒ ___   = ___________      = ______         = _______   dx dx sech  2 y 1 − tanh  2 y 1 − x  2 O b x artanh x +  _12 ln (1 − x2) + A x 1 35 a1 −  __32 x  2 sinh θ e  2y − 1   x = tanh y = _______   2y e   + 1 1+x 1+x e   2y = _____       ⇒ 2y = ln(_____   1−x 1 − x) 1+x 1 ⇒ y =  __  ln( _____ ) for | x| < 1 2 1−x –1 64 16     x  5 34 a 2 x +  ___  x  3 + ___ 3 15 2  4n−3 b _________  x  2n−1 (2n − 1)! dx a cosh θ a   ⇒ ___     = −  ________  36 x = ______   29 a L et y = artanh x b __ 3 ) 33 a   _14 ln (2 + √ btanh  2 4x = 1 − sech 2 4x = 1 − __  14 = __   34 _ √   3 As x > 0, tanh 4x = ___     2 _ At x = __  14 ln(2 + √   3 ), _ _ √   3 y = − x + tanh 4x = −  __14 ln(2 + √   3 ) + ___     2 _ _ = __   14 ( 2 √ 3 − ln(2 + √ 3   )) e  y − e  −y 38 aLet y = arsinh x ⇒ x = sinh y = ________     2 ⇒ 2x = e  y − e  −y ⇒ e  2y− 2x e  y− 1 = 0 _ _ 2x + √ 4 x  2+ 4        = x + √ x  2+ 1  ⇒ e  y = ______________ 2 c _12 ,  _13 _ _ _ 1 − √ 1 − x  2  1 + √ 1 − x  2  ____________         30 a l n(____________ )+ ln(    ) x x ( _ )( _ ) 1 − √ 1 − x  2  ____________ 1 + √ 1 − x  2               = ln ____________ x x       = ln  ___2 = ln 1 = 0 = ln(___________ ) x  2 x  1 − (1 − x  2 ) x  2 e  y + e  −y __ 1 1 1    )⇒ cosh y = __     ⇒ ________   b y = arcosh(__   =     x x x 2 ⇒ x e  y + x e  −y− 2 = 0 ⇒ x e  2y − 2 e  y + x = 0 _ _ 2 ± √ 4 − 4 x  2  ____________ 1 ± √ 1 − x  2        =        ⇒ e  y = _____________ x 2x _ 1 + √ 1 − x  2        ⇒ y = ln(____________ ) x __ √ 3 + 5     c ± ln  _______ 2 31 a cosh 3θ = 4 cosh3 θ − 3 cosh θ cosh 5θ = 16 cosh5 θ − 20 cosh3 θ + 5 cosh θ b ±0.96 1 ___ e  lnk  2 + e  ln  k  2  e  2x + e  −2x ____________ e  2lnk + e  −2lnk ___________ 32 a c osh 2x = _________     =        =     2 2 2 k  4+ 1 1 1     2  = ______   =  __  k  2 + ___ 2( k  ) 2 k  2 128 ____ b    289 ( ) 250 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 250 ⇒ y = ln(x + √ x  2+ 1 ) b y = arsinh x ⇒ sinh y = x Differentiating implicitly with respect to x dy dy 1 1 1 _  = ________ cosh y  ___  = 1 ⇒ ___   _      = ______     = ____________     dx dx cosh y √ 1 + sinh  2 y √ 1 + x  2  1 __ d ⇒ ___    (arsinh x) = ( 1 + x  2)− 2 dx dy 1 __ c y = (arsinh x)  2,  ___  = 2arsinh x (1 + x  2) −  2 dx 2 d  y 3 __ ____   = 2 (1 + x  2) −1− 2xarsinh x (1 + x  2)− 2   d x  2 dy d   2 y ⇒ (1 + x  2)  ____2  + x  ___  − 2 dx d x  1 1 __ __ = 2 − 2xarsinh x (1 + x  2)− 2 + 2xarsinh x (1 + x  2)− 2 − 2 =0 __ __ d ln (1 + √ 2 ) − √ 2  + 1 39 a p = 2, q = 1, r = 4 1 +c b  _14 arctan _______   2x +   2 2 2 ___________  dx =  ______________ ___________  dx  _______________       c √ √ (   4    x  2+ 4x + 5  2x + 1)  2+ 4  dx Let 2x + 1 = 2sinh θ ⇒ ___     = cosh θ dθ 2 2 ___________  dx =  ______________ ____________  cosh θ d θ = ∫1d θ  ______________       √ √ (   2x + 1)  2+ 4  4    sinh  2 θ + 4  2x + 1   +c   = θ + c = arsinh(_______ 2 ) ∫ ∫ ( ) ∫ ∫ Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers _ Using arsinh x = ln(x + √ x  2+ 1 ) √ _____________ 2 2x + 1 2x + 1 2 _______ ___________  dx = ln _______  _______________        + 1 ) + c   (  2 + (   2 ) √ 4    x  2+ 4x + 5  ∫ ( ) c y 1 2 2x + 1 __ 1 ___________      x  2+ 4x + 5  + c   +     √ 4 = ln _______ 2 2 ( π 2 O ___________    x  2+ 4x + 5 − ln 2 + c = ln2x + 1 + √ 4 ) ___________ π 3π 2 x 2π = ln(2x + 1 + √ 4    x  2+ 4x + 5 ) + k _ √ 4 x  2+ 9  2x    ) + c  + arsinh(___ 40  __________ 4 3 b (  _12 ln 5,  _14 ln 5 ) x __ 51 a y =   __  −   1   + ce−2x 2 4 c y 5 5 1 1 ___________  dx= ∫     ______________ ____________  dx       41∫     _____________ 2 √ 2 √ (   x      2− 4x + 8  x − 2)  2 + 2  2  x−2 3      )− arsinh0       = arsinh(__ = [arsinh(_____ 2 2 )] 2 5 1 ∫ 1 ∫ x  2 _  dx Use integration by substitution to evaluate  __________ 2 √ x  2− 1  dx Let x = cosh θ ⇒ ___     = sinh θ dθ (cosh θ)  2 x   2 _  dx =  ________________ ___________  sinh θ dθ  __________       2 2 √ x  − 1  2 √ (   cosh θ)  2− 1  ∫ ∫ ∫ ∫ 1 1     (cosh 2θ + 1) dθ =  __  cosh  2 θdθ = __ 2 4 sinh 2θ __ θ sinh θ cosh θ __ θ       +     = ____________   +     =  _______ 4 4 8 4 _ x √ x  2− 1  __ 1   +     arcosh x =  _________ 4 4 ______ x √ x  2− 1  __ x2 1     arcosh x −  _________   −     arcosh x] Area of R = [ __ 2 4 4 1 _ √   3 7     arcosh 2 − ___    )− 0 =( __ 4 2 _ _ √   3 7   ) − ___     =  __  ln(2 + √ 3 4 2 43 688 m3 c 44 y = x2 − x + __4 x x3  + cx 45 y =   __ 2 c x + ln x +     46 y =   ___________ (x + 1)2 47 y =  _12 (e2x + 3) cos x sin x   +   ______ c 48 y =   2_______ 3 sin 2x sin 2x xe−x   −   ________ 5ex   −   ________ e−x   49 y =   ________ 4(1 + x) 2(1 + x) 4(1 + x) 3 50 a y = sin x cos x + c cos x π 3π bcos x = 0, 0 < x < 2π ⇒ x = __     ,  ___  2 2 π 3π     , 0 and ( ___     , 0)lie on all of the The points __ (2 ) 2 solution curves for the differential equation. 1 (2 ln 5, 4 ln 5) 3 3    ), so k = __     = arsinh(__ 2 2 x  2 x  2 _  dx 42 ∫x arcosh x dx =  ___  arcosh x −  __________ 2 2 √ x  2− 1  2 x O 52 a y = e  coshx+c b y = e  coshx+2 53 θ = 3 e−2t cos t 54 a k = 12 b y = 2 cos 2x − __   π  sin 2x + 3x sin 2x 4 55 a a = 5, b = 1 b y = e2x (3 + 2x) + 5 + x 56 a y = e−2x (A cos x + B sin x) + sin 2x − 8 cos 2x b As x → ∞, e  −kx → 0 ⇒ y → sin 2x − 8 cos 2x Let sin 2x − 8 cos 2x = R sin (2x − α) = R sin 2x cos α − R cos 2x sin α Equating the coefficients of cos 2x and sin 2x _ ⇒ R = √ 65  , tan α = 8 x, y can be approximated by the Hence, for large _ sine function √ 65  sin (2x − α), where tan α = 8 (α ≈ 82.9°) 57 a y = e−t (A cos t + B sin t) + 2e−t b y = e−t (2 sin t − cos t) + 2e−t 58 a x = e−t (A cos 2t + B sin 2t) b x = e−t (cos 2t + sin 2t) c x 1 x = f(t) O 3π 8 7π 8 t _1 2 t+ B e−3t + t2 − t + 1 59 a y = A e− b y =  _45 ( e − e−3t )+ t2 − t + 1 −_12 t c 1.45 (3 s.f.) 251 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 251 22/09/18 3:10 PM Answers 60 a λ = 2 b y = A cos 3x + B sin 3x + 2x cos 3x c y = (1 + 2x) cos 3x d y y = g(x) 1 O π 6 π 2 5π 6 x dy d  2y 61 a y = Kt  2 e  3t ,  ___  = 2Kt e  3t+ 3K t  2 e  3t ,  ____2 = 2K e  3t dt dt  + 12Kt e  3t+ 9K t  2 e  3t Substituting into the differential equation 2K e  3t+ 12Kt e  3t+ 9Kt  2 e  3t− 12Kt e  3t− 18Kt  2 e  3t + 9Kt  2 e  3t = 4 e  3t ⇒ 2K = 4 ⇒ K = 2 2t  2 e  3tis a particular integral of the differential equation b y = (A + Bt + 2t2)e3t c y = (3 − 8t + 2t2)e3t d y 5 1 5 ( 6 ,– 9 e 2 ) O 1 2 1 t _1 62 a x = A e− 2 t+ Be−2t + t + 2 −2t b x=e +t+2 dx 1 c ___  = −2e  −2t+ 1 = 0 ⇒ t = __     ln 2 2 dt d  2 x ____   2  = 4e  −2t> 0, for any real t dt  So stationary value is a minimum. 5 1 1 When t =  __  ln 2 ⇒ x = __     + __     ln 2 2 2 2 1 The minimum distance is  __ (5 + ln 2). 2 63 a A =  _12 b x = ( 1 + t +  _12 t2 )e−t dx 1 1 c ___  = (1 + t) e  −t− ( 1 + t + __     t  2) e  −t = −  __  t  2 e  −t < 0, 2 2 dt for all real t When t = 0, x = 1 and x has a negative gradient for all positive t, x is a decreasing function of t. Hence, for t > 0, x < 1. 64 a k = 3 b y = A sin x + 3x c At x = π, y = A sin π + 3π = 3π This is independent of the value of A. Hence, all curves given by the solution in part a pass through (π, 3π). 252 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 252 dy  ___  = A cos x + 3 dx π π dy At x = __     ,  ___  = A cos  __ + 3 = 3 2 dx 2 This is independent of the value of A. Hence, all curves given by the solution in part a have an equal π gradient of 3 at x = __     2 3π  sin x d y = 3x −  ___ 2 dy 3π     cos x = 0 e For a minimum  ___  = 3 − ___ 2 dx 2 2    π  ⇒ x = arccos  __ cos x = __ π d   2 y 3π ____     sin x   2  = ___ d x  2 2 π d  y     ,  ____2  > 0 ⇒ minimum In the interval 0 < x < __ 2 dx  π  2− 4 4       2  = ______ sin  2 x = 1 − cos 2 x = 1 − ___ π  π  2 _ √ π  2− 4  π     In the interval 0 < x < __     , sin x = ________ 2 π  2 _ 3 2 __ √ 2 y = 3 arccos  __ π  −  2  π  − 4  (120 − t)2 − t −   _________ _______     b 9  _38 kg 65 a S =   120 4 600 66 a Three-quarters of the nutrient = 75m0 At time t, the nutrient consumed is 5(m – m0) 75m0 = 5(m – m0) ⇒ m = 16m0 bRate of increase of mass = μ × mass × mass of nutrient remaining dm ____     = μm(100m 0 − 5m + 5m 0 ) = 5μm(21m 0 − m) dt dm c ____  = 5μm(21m 0 − m) dt 1     dm ∫ 5μ dt =  _____________ m(21 m 0 − m) ∫ Using partial fractions 1 1 1 1     = ______     + __________       __  _____________   m(21 m 0 − m) 21 m 0 ( m 21 m 0 − m) 1 1 1 ⇒ 5μt =    ______       + __________   __     dm 21 m 0 ( m 21 m 0 − m) ∫ ⇒ 105μm 0 t = ln m − ln (21 m 0 − m) + c When t = 0, m = m 0 ⇒ 0 = ln m 0 − ln 20m 0 + c 20 m 0 ⇒ c = ln 20 m 0 − ln m 0 = ln  ______  = ln 20 m 0 ⇒ 105μm 0 t = ln m − ln (21 m 0 − m) + ln 20 20m     = ln __________ (21 m 0 − m) From a, when t = T, m = 16m 0 320 m 0    = ln 64 105μm 0 T = ln _______ ( 5 m 0 ) dv v dv     − __     = 1 67 a t  ___  − v = t ⇒ ___ dt dt t 1 1 __ __ 1     Integrating factor is e∫  −   t dt = e  −lnt = e  ln  t = __ t d v 1 1 1 dv v __    (__    )= __       2  =     ⇒ ___  __   ___  − __ t dt t  t t dt t ∫ v 1 ⇒ __     =  __  dt = ln t + c t t ⇒ v = t(ln t + c) b 8.77 m s−1 (3 s.f.) Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers dv 68 a = ___     = e  2t dt   12 e  2t + A v = ∫ e  2t dt = __ When t = 0, v = 0 ⇒ 0 = __  12 + A ⇒ A = −  __12 1 2t __ Hence v =   2 (e  − 1) _1 69 a v = 13 − 3e− 6 t b 11.2 m s−1 (3 s.f.) c 13 70 a v = 2e−2t − 1 b  _12 (1 − ln 2) m 49 dv dv 3v 71 a (t + 3)  ___  + 3v = 9.8(t + 3) ⇒ ___     + _____   = ___       5 dt dt t + 3 3 dt     ∫ ____ Integrating factor e  t+3 = e  3ln(t+3) = e  ln(t+3)  3 = ( t + 3)  3 49 dv     (t + 3)  3 ⇒ ( t + 3)  3  ___  + 3v (t + 3)  2 = ___ 5 dt 49 d ( ___ ___ 3 3 ⇒     ( t + 3)  v) =     (t + 3)  5 dt 49     (t + 3)  4 + c ⇒ ( t + 3)  3 v = ___ 20 3969 49       × 3  4 + c ⇒ c = −  _____ When t = 0, v = 0 ⇒ 0 = ___ 20 20 49 3969     (t + 4)  4 − _____     ⇒ ( t + 3)  3 v = ___ 20 20 49 (t + 4)  4− 3969       ⇒ (t + 3)  3 v = _________________ 20 49 (t + 4)  4− 3969         , c = −3969 ⇒ v =  _________________ 20 (t + 3)  3 b 21  __79 m s−1 c The speed continues to increase as t increases. This is unlikely to happen – terminal velocity, etc. 72 a Volume, ml of distilled water in the bottle after t minutes is given by 400 + 40t – 30t = 400 + 10t x Concentration of acid after t minutes= __________     400 + 10t Rate acid in = 4 ml per minute x 3x Rate acid out = 30 × __________     = ______     400 + 10t 40 + t 3x dx ___ ______   Hence     = 4 −   40 + t dt b 22.3 ml (3 s.f.) c It is unlikely that the acid disperses immediately so this could be factored in. 73 a Simple harmonic motion b x = 0.3 cos (7t) 2π c Period of motion = ___    , Maximum speed = 2.1 m s−1 7 74 a 0.791 m b 2.48 minutes c Boat is unlikely to continue oscillating with such regularity. 75 a 3 sin 2t − 6e−t sin t . b x = 6 e  −t sin t − 6 e  −t cos t + 6 cos 2t π π −__ −__ π . π π π t = __     , x = 6(e  4  sin  __  − e   4  cos  __  + cos  __ ) = 0 4 4 4 2 π ∴ P comes to instantaneous rest when t = __     4 c 1.07 m (3 s.f.) d π π __ 76 a A = 0 B =   U b ___ v 4v 4V   sin 3kt + Vt − ___ 77 e−kt ___   V   cos 3kt −  ____   V   5k 15k 5k 78 a 0.3e−4t − 0.6e−2t + 0.3 ( ) . b x = −1.2 e  −4t + 1.2 e  −2t = 1.2e  −4t (e  2t− 1) e  2t . 1 for all t . 0 . ⇒ x . 0 throughout the motion (expect for t = 0) i.e. the particle continues to move down through the liquid throughout its motion. 79 a Differentiating (1) with respect to t: dy d  2x dx  ____2  = 0.1  ___  + 0.1  ___  dt dt dt  dx d   2x ____ Substituting (2):   2  = 0.1  ___  + 0.1(−0.025x + 0.2y) dt dt  dx d  2x ____ ___ ⇒   2  = 0.1     − 0.0025x + 0.02y dt dt  dx dx d  2x ____ ⇒   2  = 0.1  ___  − 0.0025x + 0.2( ___  − 0.1x) dt dt dt  dx dx d  2x ____ ___ ___ ⇒   2  = 0.1     − 0.0025x + 0.2     − 0.02x dt dt dt  dx d  2x ____ ___ ⇒   2  − 0.3     + 0.0225x = 0 dt dt  b x = Ae  0.15t + Bte  0.15t c y = 0.5Ae  0.15t+ 0.5Bte  0.15t+ 10Be  0.15t d 237 e The number of angler fish and angel fish will both increase without limit so the model is unlikely to be suitable for large t. d  2x dx dy 80 a Differentiating (1) with respect to t:  ____2  = 2  ___  + ___     dt dt dt  dx d  2 x   = 2  ___  + 4x − y + 1 Substituting (2):  ____ dt d t  2 dx dx d  2x ____ ___ ___ ⇒   2  = 2     + 4x − (     − 2x − 1)+ 1 dt dt dt  dx d  2x ___ ____ ⇒   2  =     + 6x + 2 dt  dt d  2x dx     − 6x = 2 ⇒ ____   2  − ___ dt  dt 85 85 1 1    , y = ___         e3t + 32e2t − __ b x = ___     e3t − 8e−2t − __ 3 3 3 3 c As t gets large, the e3t term dominates and suggests the amount of gas increases without limit in both tanks. This is unlikely to be the case, for example size of tank will be a limiting factor. Challenge 1 Let n =1: The result Mn = M becomes M1 = M, which is true. Assume the result is true for n = k. That is cosh  2 x cosh  2 x Mk = M =     (− sinh  2 x    − sinh  2 x) cosh  2 x cosh  2 x cosh  2 x cosh  2 x          k+1= M  k M =     M (− sinh  2 x    − sinh  2 x)( − sinh  2 x − sinh  2 x) cosh  4 x − cosh  2 x sinh  2 x cosh  4 x − cosh  2 x sinh  2 x M  k = M =       ( sinh  4 x − cosh  2 x sinh  2 x  s inh  4 x − cosh  2 x sinh  2 x) cosh  2 x( cosh  2 x − sinh  2 x) cosh  2 x( cosh  2 x − sinh  2 x)       =        (sinh  2 x( − cosh  2 x + sinh  2 x) sinh  2 x( − cosh  2 x + sinh  2 x)) cosh  2 x cosh  2 x =     (− sinh  2 x    − sinh  2 x) and this is the result for n = k + 1. The result is true for n = 1, and if it is true for n = k, then it is true for n = k + 1. By mathematical induction the result is true for all positive integers n. 253 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 253 22/09/18 3:10 PM Answers dy d  2x ____ d   2x 2 a ___  = ____     ⇒     − x = 0 dt dt  2 dt  2 m  2 − 1 = 0 ⇒ m = ± 1 ⇒ x = Ae  t + Be  −t dx y = ___     = Ae  t − Be  −t dt When t = 0, x = A + B = 1 and y = A − B = 0, so A = B =  __12 1 n = k + 1: 0   (0 1 2k + 2 2 k  2+ 4k + 2 = 0           1    2k +  2 (0 ) 0 1 1 2(k + 1) 2 (k + 1)  2 =0          1    2(k +  1) (0 ) 0 1 So x =  __12 ( et + e−t) = cosht and y = __   12 ( et − e−t) = sinht. dp d   2p ___ dq   +     b ___  = −  ____ dt dt  2 dt dp dp dp d   2p   2p ___ d   +     = p −  ___  + p ⇒  ____   − 2 ___  + 2p = 0 −  ____ 2 dt dt dt  dt dt  2 m2 − 2m + 2 = 0 ⇒ m = 1 ± i so p = et (A cost + B sint) dp ___     = e  t(A cos t + B sin t) + e  t(B cos t − A sin t) dt dp q = p − ___    = et (−Bcost + Asint) dt When t = 0, p = A = 1 and q = −B = 1, so A = 1 and B = −1. dr  ___  − r = et(cost − sint) + 2et (cost + sint) dt = et(3cost + sint) Use integrating factor e−t, r = e  t∫    (3 cos t + sin t) dt = e  t(3 sin t − cos t + C) When t = 0, r = C − 1 = 1 ⇒ C = 2 and so r = et(3sint − cost + 2) dr dx x = rcosθ ⇒  ___  = −rsinθ +  ___  cosθ dθ dθ dy dr y = rsinθ ⇒ ___     = rcosθ +  ___  sinθ dθ dθ dr r cos θ +  ___ sin θ dθ         = tan(α + θ) So l has gradient ________________   dr − r sin θ +  ___ cos θ dθ dr r cos θ +  ___ sin θ dθ tan α + tan θ _____________ ________________    Thus          =        1 − tan α tan θ dr − r sin θ +  ___ cos θ dθ Rearrange and cancel to get dr dr     cos2θ tanα = rcos2θ − ___     tanα sin2θ −rsin2θ + ___ dθ dθ dr r ⇒ ___     tanα = r ⇒ tan α =  ___  dθ dr ___     dθ 3 Exam-style practice: Paper 1 1 a 11 200 cm3 b Does not take into account the thickness of the clay. 3 2 a a = __   14 _ |2 × 1 + 3 × 4 + −1 × 7 − 10| ______ 3 √ 14  3 ____________   =             , a =  ___  d = ___________________________   14 14 √ 2      2 + 3  2 + ( − 1)  2  b −1 c 43.9° 1 2 2 1 2 × 1 2 × 1  2     1  2 = 0       3 a Let n = 1: 0 1    2 ×  1 , (0 0 1) (0 0 1 ) which is true. Assume the result is true for n = k. 2 2 1 2k 2 k  2         1    2k  1  2 0 ) 0 1)(0 0 1 and this is the result for n = k + 1. The result is true for n = 1, and if it is true for n = k, then it is true for n = k + 1. By mathematical induction the result is true for all positive integers n. b i 7 − 2 − (k + 4) 8 1 ii ________  −       2    − 11      8 14 − 2k( 4 3k + 1 − 2(k + 1)) 4 a z = cos θ + i sin θ z  n = ( cos θ + i sin θ)  n z  n= cos nθ + i sin nθ 1  ___n = cos nθ − i sin nθ z  1   n = 2i sin nθ z  n − ___ z  4 1 1 1     (2i sin θ)  4= __     (z − __    )  b8 sin  4 θ = __ z 2 2 1 4 ___ 1 4 __ ___ 2 =     z  − 4 z  + 6 −   2  +   4  2( z  z  ) 1 =  __  (2 cos 4θ − 8 cos 2θ + 6) 2 = cos 4θ − 4 cos 2θ + 3 5 a Volume = 500 + 15t x Concentration of sugar = __________     500 + 15t Rate of sugar into mixture = 30 × 25 = 750 x 3x   Lose sugar at rate 15  __________  = _________   500 + 15t 100 + 3t 3x dx     ⇒ ___     = 750 − _________ 100 + 3t dt b 6635 g c Rate of leaking could vary with volume of oil, or model could take into account the fact that the sugar does not disperse throughout the vat on entry. 6 a(x + 12)  2 + ( y + 5)  2= 169 (r cos θ + 12)  2 + ( r sin θ + 5)  2= 169 r  2 cos  2 θ + 24r cos θ + 144 + r  2 sin  2 θ + 10r sin θ + 25 = 169 r  2 = − 24r cos θ − 10r sin θ r = − 2(12 cos θ + 5 sin θ) 1 2k 2 k  2         1    2k  n = k: 0 (0 0 ) 1 254 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 254 Online Full worked solutions are available in SolutionBank. 22/09/18 3:10 PM Answers b |z + 12 + 5i| = 13 –24 O The result is true for n = 1, and if it is true for n = k, then it is true for n = k + 1. By mathematical induction the result is true for all positive integers n. 2 a 1 – 4i b 1 ± 4i, 2 ± i c Im Im Re 5 –12 – 5i 1 + 4i 4 3 –10 2 –4 arg z = – 3π 4 –2 1 2+i O –1 2 2–i –2 c 252 dy dx 7 a 0.3 × ___     − 0.2  ___  = 0.3 × 0.3x + 0.3 × 0.2y + 0.3 dt dt + 0.2 × 0.2x − 0.2 × 0.3y –3 dy dx 0.2  ___  = 0.3  ___  − (0.3  2 + 0.2  2)x − 0.3 dt dt dy dx dx dx d  2 x ____ ___   2  = 0.3     + 0.2  ___  = 0.3  ___  + 0.3  ___  dt dt dt dt d t  − (0.3  2 + 0.2  2)x − 0.3 d  2x dx  ____2  − 0.6  ___  + 0.13x + 0.3 = 0 dt dt  dx d  2x ____ 100   2  − 60  ___  + 13x + 30 = 0 dt dt  30     b x = e  0.3t(A cos 0.2t + B sin 0.2t) − ___ 13 20     c y = e  0.3t(B cos 0.2t − A sin 0.2t) − ___ 13 160 30 85         cos 0.2t + ___     sin 0.2t)− ___ d x = e  0.3t(____ 13 13 13 85 20 160         sin 0.2t)− ___ y = e  0.3t( ___  cos 0.2t − ____ 13 13 13 –5 e Concentration on right side predicted to be negative which isn’t possible. Therefore, model is not suitable. Exam-style practice: Paper 2 1 a p = 7, q = 25 Using partial fractions, 1 1 1  ___________     = _______     − _______     (r + 2)( r + 4) 2(r + 2) 2(r + 4) Using the method of differences, n 4 Re n 1 1 1 ∑  ___________     = ∑  _______  − _______     2(r + 4) r=1 (r + 2)( r + 4) r=1 2(r + 2) –4 1 – 4i 3 3.42 (3 s.f.)   19 x  5 4 a2x + __  13 x  3 − __ 60 5∫     x  2 + 4 ∞ 0   ∫     x  2 + 4 = lim a→∞ 1  _   dx a 0 b 2.754 × 10–6 % (4 s.f.) a   _   1  arctan(_   2x )]    = lim a→∞ [ 2 _   1   dx x 1 __ π π 1     ×     = __ lim  arctan  __  = __     =  __  a→∞ 2 2 2 4 2 0 6 k = 9.5 7 a _ 0.722 b 2.722 √ 6 14  8 ______   7 (−B sin t − C cos t)+ 2(B cos t − C sin t) 9 a + 3(A + B sin t + C cos t) = 21 + 15 cos t 2(B − C ) sin t + 2(B + C ) cos t + 3A = 21 + 15 cos t 3A = 21 ⎫⎪ 15 B − C = 0 ⎪⎬ ⇒ A = 7, B = C = ___     4 15 ⎪⎪ ___ B + C =     2 ⎭ 15 ⇒ x = 7 +  ___ (sin t + cos t) 4 15 ( ___ b x = 7 +     sin t + cos t) 4 _ _ _   19 √ 2 35 2 t) + ___     cos(√ 2   t))   sin(√   − e  −t( ______ 4 4 c The flow will stabilise and oscillate evenly about x = 7. n(7n + 25) 1 1 1 1 ________     −        −  ________  = ______________       = __     + __ 6 8 2(n + 3) 2(n + 4) 24(n + 3)(n + 4) b Let n = 1 : f (1) = 2  3 + 3  3= 35 which is divisible by 7 Assume the result is true for n = k. n = k: f (k) = 2  k+2 + 3  2k+1is divisible by 7 n = k + 1 : f(k + 1) = 2  k+3 + 3  2k+3 = 2(2  k+2) + 3  2(3  2k+1) = 2(2  k+2)+ 9(3  2k+1) = 2(2  k+2)+ 2(3  2k+1)+ 7(3  2k+1) = 2f(k) + 7(3  2k+1)which is divisible by 7 and this is the result for n = k + 1. 255 Z01_ALEVEL_CP2_83343_ANS_213-255.indd 255 22/09/18 3:10 PM Index Index acceleration 171, 175 amplitude 176 angular velocity 175 arccos x, differentiating 63 arcosh x definition 123 differentiating 131 graph 123 integral for 135 in terms of natural logarithms 124 arcsin x, differentiating 62–3 arctan x differentiating 63, 64 Maclaurin series 44 areas enclosed by polar curves 109–11 arsinh x definition 123 differentiating 131 graph 123 integral for 135 in terms of natural logarithms 124 artanh x definition 123 differentiating 131 in terms of natural logarithms 124 auxiliary equation 154, 180 average value see mean value binomial expansion 40 boundary conditions 150, 153 using to find particular solutions 162–4 calculus methods 52–76 differentiating inverse trigonometric functions 62–4 improper integrals 53–6 integrating with inverse trigonometric functions 65–7 integrating using partial fractions 69–72 mean value of function 58–61 cardioid 106, 109 Cartesian coordinates 101, 102 Cartesian equations 103 centre of oscillation 175 circles 104, 105 complementary function (C.F.) 157, 159–61 complex conjugate roots 154 complex numbers 1–30 division 5–6 exponential form 2–4 modulus–argument form 2 multiplication 5–6 nth roots 20–3, 25–6 solving geometric problems 25–6 sums of series 16–18 compound functions, series expansions 44–6 convergent integrals 53–6 convex curves 107, 115 cos nθ 11–14 cos x, Maclaurin series 44 cosnθ 12–14 cosech x, definition 120 cosh x addition formulae 126 definition 120 differentiating 130 as even function 121 graph 121 integrating 135 cosh−1 x see arcosh x coth x, definition 120 critical damping 180, 181 curves sketching 104–8 see also parametric curves; polar curves damped harmonic motion 180–2 damping force 180 de Moivre’s theorem 8–10 derivatives first 38 higher 38–9 second 38 differential equations coupled first-order linear 186–8 first-order 148–51, 171–4 general solution 154, 159 linear 153 methods in 147–69 modelling with see modelling with differential equations second-order homogeneous 153–6 second-order nonhomogeneous 157–61 using boundary conditions 162–4 differentiation hyperbolic functions 130–2 inverse hyperbolic functions 131–2 inverse trigonometric functions 62–4 ‘dimple’ shaped curves 107, 115 displacement 171, 175 divergent integrals 53–6 division, complex numbers 5–6 ‘dot’ notation 175 ex, Maclaurin series 44 ‘egg’ shaped (convex) curves 107, 115 Euler’s identity 2 Euler’s relation 2 even functions 121 exponential form 2–4 forced harmonic motion 182–4 functions, mean value 58–61 geometric problems, solving 25–6 graphs, hyperbolic functions 121– 2 half-lines 104, 105 harmonic motion damped 180–2 forced 182–4 simple (S.H.M.) 175–8 heavy damping 180 homogenous systems 186 hyperbolic cosine see cosh hyperbolic functions 119–46 differentiating 130–2 equations involving 127–8 graphs 121–2 identities 125–7 integrating 135–9 introduction to 120–2 inverse see inverse hyperbolic functions hyperbolic sine see sinh hyperbolic tangent see tanh improper fractions 71 improper integrals 53–6 initial line 101 integrating factors 149–51 integration hyperbolic functions 135–9 improper integrals 53–6 with inverse trigonometric functions 65–7 using partial fractions 69–72 interval 53 inverse hyperbolic functions 123–5 differentiating 131–2 inverse trigonometric functions differentiation 62–4 integrating with 65–7 light damping 180, 182 limit notation 54 ln(1 + x), Maclaurin series 44 loops 106, 110 Lorenz factor 48 Maclaurin polynomials 41 Maclaurin series 40–2, 44 mean value attaining 76 of function 58–61 method of differences 32–5 modelling with differential equations 170–95 coupled first-order simultaneous 186–8 damped harmonic motion 180–2 first-order 171–4 forced harmonic motion 182–4 simple harmonic motion (S.H.M.) 175–8 modelling with volumes of revolution 87–8 modulus–argument form 2 multiplication, complex numbers 5–6 nth roots of complex numbers 20–3, 25–6 of unity 21, 25–6 odd functions 121 Osborn’s Rule 126 parametric curves, volumes of revolution 83–4 partial fractions, integrating using 69–72 particular integral (P.I.) 158, 159–61, 182 period 176 polar coordinates 100–18 polar curves areas enclosed by 109–11 sketching 104–8 tangents to 113–15 polar equations 103–4 pole 101 predator–prey model 186 product rule 149 ratio test 44 regular polygons 25–6 sech x definition 120 graph 145 sech2x, integrating 146 sector, area 109 separating the variables 148 series 31–51 convergence 44 expansions of compound functions 44–6 higher derivatives 38–9 Maclaurin 40–2, 44 sums of see sums of series shell, cross-section 112 simple harmonic motion (S.H.M.) 175–8 sin nθ 11–14 sin x, Maclaurin series 44 sinnθ 12–14 sine wave 176 sinh x addition formulae 126 definition 120 differentiating 130 graph 121 integrating 135 as odd function 121 sinh−1 x see arsinh x spirals 104, 105 sums of series complex numbers 16–18 method of differences 32–5 tangents parallel to initial line 113–14 perpendicular to initial line 113, 114 to polar curves 113–15 tanh x definition 120 differentiating 130 exponential form 120 graph 122 integrating 136 tanh-1 x see artanh x trigonometric functions differentiating inverse 62–4 integrating with inverse 65–7 trigonometric identities 11–14 unity, nth roots of 21, 25–6 velocity 171, 175 volumes of revolution 77–92 around x-axis 78 around y-axis 81 modelling with 87–8 of parametric curves 83–4 256256 Z02_ALEVEL_CP2_83343_IND_256-258.indd 256 26/02/2018 16:18 11 - 19 PROGRESSION Edexcel A level Further Mathematics Core Pure Mathematics Book 2 Series Editor: Harry Smith • Fully updated to match the 2017 specifications, with more of a focus on problem-solving and modelling. • FREE additional online content to support your independent learning, including full worked solutions for every question in the book (SolutionBank) and GeoGebra interactives. • Includes access to an online digital edition (valid for 3 years once activated). • Includes worked examples with guidance, lots of exam-style questions, practice papers, and plenty of mixed and review exercises. Pearson Edexcel AS and A level Mathematics books Year 1/AS Year 2 Options For more information visit: www.pearsonschools.co.uk/edalevelmaths2017 www.pearsonschools.co.uk myorders@pearson.com 2 Edexcel A level Further Mathematics Core Pure Mathematics Book 2 line digital ion Compulsory on it ed Pearson Edexcel AS and A level Further Mathematics books Edexcel A level Further Mathematics Core Pure Maths Book 2 This book can be used alongside Book 1 to cover all the content needed for the compulsory Edexcel A level Core Pure Mathematics exams. ludes a inc n Pearson’s market-leading books are the most trusted resources for Edexcel AS and A level Further Mathematics.

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