y11specialist maths units 1&2 vce

TI-Nspire CAS OS3 and Casio ClassPad version

ESSENTIAL Advanced General Mathematics Third edition

ENHANCED

MICHAEL EVANS SUE AVERY DOUG WALLACE KAY LIPSON CAS calculator material prepared in collaboration with Jan Honnens David Hibbard Russell Brown

ISBN 978-1-107-65235-4 © Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2012 Photocopying is restricted under law and this material must not be transferred to another party.

Cambridge University Press

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CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi, Tokyo, Mexico City Cambridge University Press 477 Williamstown Road, Port Melbourne, VIC 3207, Australia www.cambridge.edu.au Information on this title: www.cambridge.org/9781107652354 C

Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2011

This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2009 Reprinted 2009, 2010 Enhanced version 2011 Typeset by Aptara Corp. Cover design by Marta White Printed in China by Printplus Co. Ltd. National Library of Australia Cataloguing in Publication data Essential advanced general TI-N / CP version / Michael Evans . . . [et al.] 3rd ed. enhanced 9781107652354 (pbk.) Essential mathematics. For secondary school age. Mathematics – Textbooks. Mathematics – Study and teaching (Secondary). Evans, Michael (Michael Wyndham) 510 ISBN 978-1-107-65235-4 Paperback ISBN 978-1-139-09985-1 Electronic version Additional resources for this publication at www.cambridge.edu.au/GO Reproduction and Communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this publication, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Telephone: (02) 9394 7600 Facsimile: (02) 9394 7601 E-mail: [email protected] Reproduction and Communication for other purposes Except as permitted under the Act (for example a fair dealing for the purposes of study, research, criticism or review) no part of this publication may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above. Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables and other factual information given in this work are correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Contents Introduction

xii

CHAPTER 1 — Matrices

1.1 1.2 1.3 1.4 1.5

Introduction to matrices 1 Addition, subtraction and multiplication by a scalar 6 Multiplication of matrices 11 Identities, inverses and determinants for 2 × 2 matrices 14 Solution of simultaneous equations using matrices 19 Chapter summary 23 Multiple-choice questions 24 Short-answer questions (technology-free) 25 Extended-response questions 26

CHAPTER 2 — Algebra I

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

1

28

Indices 28 Standard form 30 Solving linear equations and linear simultaneous equations 33 Solving problems with linear equations 37 Solving problems using simultaneous linear equations 40 Substitution and transposition of formulas 42 Algebraic fractions 44 Literal equations 47 Using the TI-Nspire with algebra 50 Using the Casio ClassPad with algebra 51 Chapter summary 57 Multiple-choice questions 57

ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Contents

Short-answer questions (technology-free) Extended-response questions 60 CHAPTER 3 — Number systems and sets

3.1 3.2 3.3 3.4 3.5

4.1 4.2 4.3 4.4 4.5

Direct variation 89 Inverse variation 93 Fitting data 96 Joint variation 103 Part variation 106 Chapter summary 109 Multiple-choice questions 109 Short-answer questions (technology-free) Extended-response questions 112

6.1 6.2 6.3 6.4

111

115

Introduction to sequences 115 Arithmetic sequences 122 Arithmetic series 125 Geometric sequences 130 Geometric series 135 Infinite geometric series 138 Fixed point iteration 141 Chapter summary 146 Multiple-choice questions 147 Short-answer questions (technology-free) Extended-response questions 148

CHAPTER 6 — Algebra II

84

89

CHAPTER 5 — Sequences and series

5.1 5.2 5.3 5.4 5.5 5.6 5.7

63

Set notation 64 Sets of numbers 67 Surds 69 Natural numbers 74 Problems involving sets 77 Chapter summary 82 Multiple-choice questions 83 Short-answer questions (technology-free) Extended-response questions 85

CHAPTER 4 — Variation

58

147

151

Polynomial identities 151 Quadratics and rates 154 Partial fractions 162 Simultaneous equations 169 Chapter summary 173



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Contents

Multiple-choice questions 174 Short-answer questions (technology-free) Extended-response questions 176 CHAPTER 7 — Revision of chapters 1–6

7.1 7.2

8.7 8.8 8.9 8.10

9.5

190

227

Ratios 227 An introduction to similarity 229 Areas, volumes and similarity 235 Geometric representation of arithmetic operations 242 Golden ratio 244 Chapter summary 248 Multiple-choice questions 248 Short-answer questions (technology-free) Extended-response questions 251

CHAPTER 10 — Circular functions I

10.1 10.2 10.3 10.4 10.5

177

Translations 191 Reflections 194 Dilations from the axes 196 Rules for transformations 197 Composition of transformations 199 Applying transformations to graphs of functions 201 Determining transformations 206 Absolute value function and integer value function 208 Function notation with transformations 213 Summary of transformations 216 Chapter summary 220 Multiple-choice questions 222 Short-answer questions (technology-free) 224 Extended-response questions 224

CHAPTER 9 — Ratios and similarity

9.1 9.2 9.3 9.4

175

Multiple-choice questions 177 Extended-response questions 182

CHAPTER 8 — Transformations

8.1 8.2 8.3 8.4 8.5 8.6

v

249

253

Measuring angles in degrees and radians 253 Defining circular functions: sine and cosine 255 Another circular function: tangent 257 Reviewing trigonometric ratios 258 Symmetry properties of circular functions 259



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Contents

10.6 10.7 10.8 10.9 10.10 10.11 10.12

Exact values of circular functions 261 Graphs of sine and cosine 263 Further transformations of sine and cosine graphs 270 Solution of circular function equations 273 Further sketch graphs 277 Rotation of a point about the origin 280 Applications 283 Chapter summary 287 Multiple-choice questions 289 Short-answer questions (technology-free) 291 Extended-response questions 292

CHAPTER 11 — Circular functions II

11.1 11.2 11.3 11.4 11.5 11.6 11.7

297

Further symmetry properties 297 Addition of ordinates 298 Sketch graphs of the tangent function 300 General solution of circular function equations 301 Trigonometric identities 304 Addition formulas and double angle formulas 309 a cos x + b sin x 316 Chapter summary 320 Multiple-choice questions 321 Short-answer questions (technology-free) 322 Extended-response questions 324

CHAPTER 12 — Trigonometric ratios and

applications 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8

326

Defining sine, cosine and tangent 326 The sine rule 331 The cosine rule 334 Area of a triangle 337 Circle mensuration 340 Angles of elevation and depression and bearings 344 Problems in three dimensions 348 Angles between planes and more difficult 3-D problems 353 Chapter summary 358 Multiple-choice questions 359 Short-answer questions (technology-free) Extended-response questions 362


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Contents CHAPTER 13 — Revision of chapters 8–12

13.1 13.2

375

Angle properties of the circle 375 Tangents 379 Chords in circles 382 Chapter summary 385 Multiple-choice questions 386 Short-answer questions (technology-free) Extended-response questions 388

CHAPTER 15 — Vectors

15.1 15.2 15.3 15.4

363


CHAPTER 14 — Circle theorems

14.1 14.2 14.3

vii

387

390

Introduction to vectors 390 Components of vectors 397 Vectors in three dimensions 401 Applications 403 Chapter summary 406 Multiple-choice questions 407 Short-answer questions (technology-free) Extended-response questions 409

407

CHAPTER 16 — Polar coordinates and complex

numbers 16.1 16.2 16.3 16.4 16.5 16.6

Polar coordinates 411 The set of complex numbers 417 Multiplication and division of complex numbers 420 Argand diagrams 425 Solving equations over the complex field Polar form of a complex number 430 Chapter summary 435 Multiple-choice questions 436 Short-answer questions (technology-free) Extended-response questions 438

CHAPTER 17 — Loci

17.1 17.2 17.3

410

428

437

440

Introduction and parabolas Ellipses 444 Hyperbolas 447 Chapter summary 453 Multiple-choice questions


440

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Contents

Short-answer questions (technology-free) Extended-response questions 456 CHAPTER 18 — Revision of chapters 14–17

18.1 18.2

19.3 19.4

463

Position, velocity and acceleration 463 Using antiderivatives for kinematics problems 468 Constant acceleration 471 Velocity–time graphs 474 Chapter summary 478 Multiple-choice questions 478 Short-answer questions (technology-free) Extended-response questions 482

CHAPTER 20 — Statics of a particle

20.1 20.2

479

485

Triangle of forces 486 Resolution of forces 489 Chapter summary 493 Multiple-choice questions 493 Short-answer questions (technology-free)

CHAPTER 21 — Revision of chapters 19 and 20

21.1 21.2

457


CHAPTER 19 — Kinematics

19.1 19.2

456

494 496


CHAPTER 22 — Describing the distribution of a single

variable 22.1 22.2 22.3 22.4 22.5 22.6 22.7 22.8

500

Types of variables 500 Displaying categorical data—the bar chart 502 Displaying numerical data—the histogram 503 Characteristics of distributions of numerical variables 513 Stem-and-leaf plots 515 Summarising data 520 The boxplot 533 Using boxplots to compare distributions 540 Chapter summary 543 Multiple-choice questions 544 Short-answer questions (technology-free) 546 Extended-response questions 548



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Contents

ix

CHAPTER 23 — Investigating the relationship between two

numerical variables 23.1 23.2 23.3 23.4 23.5

551

Displaying bivariate data 552 The q-correlation coefficient 557 The correlation coefficient 564 Lines on scatterplots 570 The least squares regression line 575 Chapter summary 582 Multiple-choice questions 583 Short-answer questions 585 Extended-response questions 587

CHAPTER 24 — Revision of chapters 22 and 23

24.1 24.2


CHAPTER 25 — Proof and number

25.1 25.2 25.3 25.4

Glossary

589

602

An introduction to proof 602 The principle of mathematical induction Linear Diophantine equations 610 The Euclidean algorithm 613 Chapter summary 617 Multiple-choice questions 617 Short-answer questions (technology-free) Extended-response questions 619

607

618

621

Appendix A: Computer Algebra System (TI-Nspire)

634

Appendix B: Computer Algebra System (ClassPad 330) Answers

650

661

Michael Evans now works at the Australian Mathematical Sciences Institute based at the University of Melbourne. He was Head of Mathematics at Scotch College for many years and has been heavily involved in curriculum development at both Victorian state and national levels. Kay Lipson is Dean of the Faculty of Higher Education, Lilydale at Swinburne University of Technology. She has many years’ experience teaching statistics at undergraduate and graduate level, as well as conducting courses in statistics for industry and government. Doug Wallace has been a teacher of senior mathematics at Wesley College in Melbourne for over 20 years. He is a former Head of Mathematics and is currently the Campus Curriculum and International Baccalaureate Diploma Coordinator. The late Sue Avery was an experienced VCE Mathematics teacher and a key contributor to the Essential VCE Mathematics series.



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Introduction This book provides a complete course for General Mathematics. It has been written as a teaching text, with understanding as its chief aim and ample practice offered through the worked examples and exercises. All the work has been trialled in the classroom, and the approaches offered are based on classroom experience. The book contains five revision chapters which provide short-answer and extended-response questions on each chapter. The use of CAS calculators is referred to throughout the text. Extended-response questions which require the use of a graphics calculator are indicated by the graphics calculator icon shown here. In the following table the contents of this book are assigned to Areas of Study as stipulated in the study design for General Mathematics Units 1 and 2. Area of study 1: Arithmetic Matrices Integer and rational number systems Real and complex number systems Sequences and series Area of study 2: Data analysis and simulation Univariate data Bivariate data Area of study 3: Algebra Linear relations and equations Non-linear relations and equations Algebra and logic Area of study 4: Functions and Graphs Linear graphs and modelling Sketching and interpreting graphs Variation Kinematics Area of study 6: Geometry and trigonometry Shape and measurement Geometry in two and three dimensions Vectors Trigonometric ratios and their applications

Chapter 1, Chapter 10 and Chapter 11 Chapter 3 Chapter 9 and Chapter 16 Chapter 5 Chapter 22 Chapter 23 Chapter 2 Chapter 6 Chapter 25 Chapter 2 and Chapter 23 Chapter 8, Chapter 10, Chapter 16 and Chapter 17 Chapter 4 Chapter 19 Chapter 9 and Chapter 12 Chapter 14 Chapter 1, Chapter 15, Chapter 19 and Chapter 20 Chapter 10, Chapter 11 and Chapter 12

The TI-Nspire calculator instructions have been completed by Jan Honnens and the Casio ClassPad instructions have been completed by David Hibbard. The TI-Nspire instructions are written for the CX CAS model, operating system 3.0.2, but can be used with other versions. Check the companion website for updates. The Casio ClassPad instructions are written for operating system 3 or above. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


Chapter 9 — Linear Relations

In each chapter you will find …

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Revision Enhanced TI-NSpire and Casio ClassPad versions

Enhanced TI-Nspire and Casio ClassPad versions


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C H A P T E R

1 Matrices Objectives To be able to identify when two matrices are equal To be able to add and subtract matrices of the same dimensions To be able to perform multiplication of a matrix and a scalar To be able to identify when the multiplication of two given matrices is possible To be able to perform multiplication on two suitable matrices To be able to find the inverse of a 2 × 2 matrix To be able to find the determinant of a matrix To be able to solve linear simultaneous equations in two unknowns using an inverse matrix

1.1

Introduction to matrices A matrix is a rectangular array of numbers. The numbers in the array are called the entries in the matrix. The following are examples of matrices: ⎤ ⎡ ⎤ ⎡√ 2 ␲ 3 −1 2 ⎥ ⎢ ⎥ ⎢ [5] [2 1 5 6] ⎣−3 4⎦ ⎣ 0 0 1⎦ √ 2 0 ␲ 5 6 Matrices vary in size. The size, or dimension, of the matrix is described by specifying the number of rows (horizontal lines) and columns (vertical lines) that occur in the matrix. The dimensions of the above matrices are, in order: 3 × 2,

1 × 4,

3 × 3,

1 × 1.

The first number represents the number of rows and the second, the number of columns.

1 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Essential Advanced General Mathematics

Example 1 Write down the dimensions of the⎡following matrices. ⎤ 1 ⎢2 ⎥ ⎢ ⎥

1 1 2 ⎥ c 2 2 3 a b ⎢ ⎣3 ⎦ 2 1 0 4 Solution a 2×3

b 4×1

c 1×3

The use of matrices to store information is demonstrated by the following two examples. Four exporters A, B, C and D sell televisions (t), CD players (c), refrigerators (r) and washing machines (w). The sales in a particular month can be represented by a 4 × 4 array of numbers. This array of numbers is called a matrix. r c w A 120 95 370 ⎢ B ⎢430 380 1000 ⎢ C ⎣ 60 50 150 D 200 100 470 column 1 column 2 column 3 ⎡

t ⎤ 250 row 1 900⎥ ⎥ row 2 ⎥ 100⎦ row 3 50 row 4 column 4

From the matrix it can be seen: Exporter A sold 120 refrigerators, 95 CD players, 370 washing machines and 250 televisions. Exporter B sold 430 refrigerators, 380 CD players, 1000 washing machines and 900 televisions. The entries for the sales of refrigerators are made in column 1. The entries for the sales of exporter A are made in row 1. B The diagram on the right represents a section of a road map. The number of direct connecting roads between towns can be represented in matrix form. A B C D

A B C ⎡ 0 2 1 ⎢2 0 1 ⎢ ⎢ ⎣1 1 0 1 0 0

D ⎤ 1 0⎥ ⎥ ⎥ 0⎦ 0

A

C

D

If A is a matrix, aij will be used to denote the entry that occurs in row i and column j of A. Thus a 3 × 4 matrix may be written ⎤ ⎡ a11 a12 a13 a14 ⎥ ⎢ A = ⎣a21 a22 a23 a24 ⎦ a31 a32 a33 a34



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Chapter 1 — Matrices

3

For B, an m × n matrix ⎤ ⎡ b11 b12 . . . . . b1n ⎥ ⎢ ⎢ b21 b22 . . . . . b2n ⎥ ⎥ ⎢ ⎢ . . ⎥ ⎥ ⎢ ⎢ . . ⎥ ⎥ ⎢ B=⎢ ⎥ ⎢ . . ⎥ ⎥ ⎢ ⎢ . . ⎥ ⎥ ⎢ ⎥ ⎢ . ⎦ ⎣ . bm1 bm2 . . . . . bmn Matrices provide a format for the storage of data. In this form the data is easily operated on. Some graphics calculators have a built-in facility to operate on matrices and there are computer packages which allow the manipulation of data in matrix form. A car dealer sells three models of a certain make and his business operates through two showrooms. Each month he summarises the number of each model sold by a sales matrix S: s11 s12 s13 S= , where si j is the number of cars of model j sold by showroom i. s21 s22 s23 So, for example, s12 is the number of sales made by showroom 1, of model 2. If in January, showroom 1 sold three, six and two cars of models 1, 2 and 3 respectively, and showroom 2 sold four, two and one car(s) of models 1, 2 and 3 (in that order), the sales matrix for January would be:

3 S= 4

6 2

2 1

A matrix is, then, a way of recording a set of numbers, arranged in a particular way. As in Cartesian coordinates, the order of the numbers is significant, so that although the matrices 1 2 3 4 , 3 4 1 2 have the same numbers and the same number of elements, they are different matrices (just as (2, 1), (1, 2) are coordinates of different points). Two matrices A, B, are equal, and can be written as A = B when each has the same number of rows and the same number of columns they have the same number or element at corresponding positions. ⎤ ⎡ ⎤ ⎡ 2 1 −1 1 + 1 1 −1 ⎦=⎣ e.g. ⎣ 6⎦ 1−1 1 0 1 3 2



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Example 2 If matrices A and B are equal, find the values of x and y. 2 1 2 1 B= A= −3 y x 4 Solution x = −3 and y = 4 Although a matrix is made from a set of numbers, it is important to think of a matrix as a single entity, somewhat like a ‘super number’. Example 3 There are four rows of seats of three seats each in a minibus. If 0 is used to indicate a seat is vacant and 1 is used to indicate a seat is occupied, write down a matrix that represents a the 1st and 3rd rows are occupied but the 2nd and 4th rows are vacant b only the seat on the front left corner of the bus is occupied. Solution ⎡ 1 1 ⎢0 0 a ⎢ ⎢ ⎣1 1 0 0

⎤ 1 0⎥ ⎥ ⎥ 1⎦ 0

⎡

1 ⎢0 b ⎢ ⎢ ⎣0 0

0 0 0 0

⎤ 0 0⎥ ⎥ ⎥ 0⎦ 0

Example 4 There are four clubs in a local football league. Team A has 2 senior teams and 3 junior teams Team B has 2 senior teams and 4 junior teams Team C has 1 senior team and 2 junior teams Team D has 3 senior teams and 3 junior teams Represent this information in a matrix. Solution ⎡ ⎤ 2 3 ⎢2 4⎥ Note: rows represent teams A, B, C, D and columns represent the number ⎢ ⎥ ⎢ ⎥ ⎣1 2⎦ of senior and junior teams respectively. 3 3



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5


Exercise 1A ⎡ ⎤ p ⎢q ⎥ ⎥ d ⎢ ⎢ ⎥ ⎣r ⎦ s

Example

1

1 Write down the dimensions of the following matrices. 2 1 −1 1 2 c [a b c d ] b a 0 1 3 3 4

Example

3

2 There are 25 seats arranged in five rows and five columns. If 0, 1 respectively are used to indicate whether a seat is vacant or occupied, write down a matrix which represents the situation when a only seats on the two diagonals are occupied b all seats are occupied. 3 If seating arrangements (as in 2) are represented by matrices, consider the matrix in which the i, j element is 1 if i = j, but 0 if i = j. What seating arrangement does this matrix represent?

Example

4

4 At a certain school there are 200 girls and 110 boys in Year 7, 180 girls and 117 boys in Year 8, 135 and 98 respectively in Year 9, 110 and 89 in Year 10, 56 and 53 in Year 11 and 28 and 33 in Year 12. Summarise this information in matrix form.

Example

2

5 From the following, select those pairs of matrices which could be equal, and write down the values of x, y which would make them equal. a 3 , 0 , [0 x ], [0 4 ] 2 x 4 7 1 −2 x 7 b , , , [4 x 1 −2] 1 −2 4 x 1 −2 2 x 4 y 0 4 2 0 4 c , , −1 10 3 −1 10 3 −1 10 3 6 In each of the following find the values of the pronumerals so that matrices A and B are equal. x 3 2 1 −1 x 1 −1 b A= B= a A= B= 2 y 0 1 3 0 1 y

c A = [−3 x] B = [y

4]

1 y 1 −2 B= d A= 4 3 4 x

7 A section of a road map connecting towns A, B, C and D is shown. Construct the 4 × 4 matrix which shows the number of connecting roads between each pair of towns.


D B

A

C


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8 The statistics for the five members of a basketball team are recorded as follows. Player A: points 21, rebounds 5, assists 5 Player B: points 8, rebounds 2, assists 3 Player C: points 4, rebounds 1, assists 1 Player D: points 14, rebounds 8, assists 60 Player E: points 0, rebounds 1, assists 2 Express this data in a 5 × 3 matrix.

1.2 Addition, subtraction and multiplication

by a scalar Addition will be defined for two matrices only when they have the same number of rows and the same number of columns. In this case the sum of two matrices is found by adding corresponding elements. For example, 1 0 0 −3 1 −3 + = 0 2 4 1 4 3 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ b11 b12 a11 + b11 a12 + b12 a11 a12 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ and ⎣a21 a22 ⎦ + ⎣b21 b22 ⎦ = ⎣a21 + b21 a22 + b22 ⎦ a31 a32 b31 b32 a31 + b31 a32 + b32 Subtraction is defined in a similar way. When the two matrices have the same number of rows and the same number of columns the difference is found by subtracting corresponding elements. Example 5 Find a

1 0 2 −1 − 2 0 −4 1 Solution 1 0 2 −1 −1 1 a − = 2 0 −4 1 6 −1

b

2 3 2 3 − −1 4 −1 4

b

2 3 2 3 0 0 − = −1 4 −1 4 0 0

It is useful to define multiplication of a matrix by a real number. If A is an m × n matrix, and k is a real number, then kA is an m × n matrix whose elements are k times the corresponding elements of A. Thus 2 −2 6 −6 3 = 0 1 0 3 These definitions have the helpful consequence that if a matrix is added to itself, the result is twice the matrix, i.e. A + A = 2A. Similarly the sum of n matrices each equal to A is n A (where n is a natural number). The m × n matrix with all elements equal to zero is called the zero matrix. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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7

Example 6 2 3 2 0 5 0 X= ,Y = ,A = ,B = 4 6 −1 2 −2 4

Let

Find X + Y, 2X, 4Y + X, X − Y, −3A, −3A + B. Solution

2 3 5 X+Y= + = 4 6 10 2 4 2X = 2 = 4 8 3 2 12 2 14 4Y + X = 4 + = + = 6 4 24 4 28 2 3 −1 X−Y= − = 4 6 −2 2 0 −6 0 −3A = −3 = −1 2 3 −6 5 0 −1 0 −6 0 = −3A + B = + 1 −2 −2 4 3 −6

Example 7

3 If A = −1

2 0 and B = 1 −2

−4 , find matrices X such that 2A + X = B. 8

Solution If 2A + X = B, then X = B − 2A 0 −4 3 2 ∴X= −2× −2 8 −1 1 0−2×3 −4 − 2 × 2 = −2 − 2 × −1 8−2×1 −6 −8 = 0 6



8


Using the TI-Nspire 2-by-2 matrices are easiest entered using the 2-by-2 matrix template, t (/+r on the clickpad), as shown. Notice that there is also a template for entering m by n matrices. The matrix template can also be obtained using /+b>Math Templates

3 6 To enter the matrix A = , use the 6 7 Nav Pad to move between the entries of the 2 by 2 matrix template and store (/ h) the matrix as a. 3 6 Define the matrix B = in a 5 −6.5 similar way.

Entering matrices directly To enter matrix A without using the template, enter the matrix row by row as [[3,6][6,7]] and store (/ h) the matrix as a.




9

Addition, subtraction and multiplication by a scalar Once A and B are defined as above, A + B, AB and kA can easily be determined.

Using the Casio ClassPad Matrices are entered using the 2D CALC menu on the k . Tap , enter the numbers required then store this as a variable (using VAR). Calculations can be performed as shown in the screen at the far right.

Exercise 1B

Example

6

1 3 1 −1 4 0 ,Y = ,A = ,B = 1 Let X = −2 0 2 3 −1 2 Find X + Y, 2X, 4Y + X, X − Y, −3A and −3A + B. 2 Each showroom of a car dealer sells exactly twice as many cars of each model in February as in January. (See example in section 1.1.) 3 6 2 a Given that the sales matrix for January is , write down the sales matrix for 4 2 1 February.



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b If the sales matrices for January and March (with twiceas many cars of each model 1 0 0 2 1 0 sold in February as January) had been and respectively, find the 4 2 3 6 1 4 sales matrix for the first quarter of the year. c Find a matrix to represent the average monthly sales for the first three months. 1 −1 3 Let A = 0 2 Find 2A, −3A and −6A. 4 A, B, C are m × n matrices. Is it true that a A+B=B+A b (A + B) + C = A + (B + C)? 3 2 0 −3 5 A= and B = −2 −2 4 1 Calculate a 2A b 3B c 2A + 3B 0 4 −1 1 1 0 6 P= ,Q= ,R= 1 1 0 3 2 0

d 3B – 2A

Calculate a P+Q

Example

7

b P + 3Q c 2P − Q + R 0 −10 3 1 and B = , find matrices X and Y such that 7 If A = −2 17 −1 4 2A − 3X = B and 3A + 2Y = 2B. 8 Matrices X and Y show the production of four models a, b, c, d at two automobile factories P, Q in successive weeks. b c d b c d a a P 150 90 100 50 P 160 90 120 40 X= Y= Q 100 0 75 0 Q 100 0 50 0 week 1

week 2

Find X + Y and write what this sum represents.



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1.3

11

Multiplication of matrices Multiplication of a matrix by a real number has been discussed in the previous section. The definition for multiplication of matrices is less natural. The procedure for multiplying two 2 × 2 matrices is shown first. 1 3 5 1 Let A= and B = 4 2 6 3 1 3 5 1 Then AB = 4 2 6 3 1×5+3×6 1×1+3×3 = 4×5+2×6 4×1+2×3 23 10 = 32 10 5 1 1 3 and BA = 6 3 4 2 5×1+1×4 5×3+1×2 = 6×1+3×4 6×3+3×2 9 17 = 18 24 Note that AB = BA. If A is an m × n matrix and B is an n × r matrix, then the product AB is the m × r matrix whose entries are determined as follows. To find the entry in row i and column j of AB single out row i in matrix A and column j in matrix B. Multiply the corresponding entries from the row and column and then add up the resulting products. Note: The product AB is defined only if the number of columns of A is the same as the number of rows of B. Example 8 For A =

2 3

4 6

5 and B = find AB. 3

Solution A is a 2 × 2 matrix and B is a 2 × 1 matrix. Therefore AB is defined. The matrix AB is a 2 × 1 matrix. 2 4 5 2×5+4×3 22 AB = = = 3 6 3 3×5+6×3 33



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Example 9 Matrix X shows the number of cars of models a and b bought by four dealers, A, B, C and D. Matrix Y shows the cost in dollars of model a and model b. Find XY and explain what it represents.

A B X= C D

a 3 ⎢2 ⎢ ⎢ ⎣1 1 ⎡

b ⎤ 1 2⎥ ⎥ ⎥ 4⎦ 1

26 000 a Y= 32 000 b

Solution A B XY = C D

a 3 ⎢2 ⎢ ⎢ ⎣1 1 ⎡

b ⎤ 1 2⎥ ⎥ 26 000 a ⎥ 4⎦ 32 000 b 1

4×2

2×1

The matrix XY is a 4 × 1 matrix ⎡ ⎤ ⎡ ⎤ 3 × 26 000 + 1 × 32 000 110 000 ⎢2 × 26 000 + 2 × 32 000⎥ ⎢116 000 ⎥ ⎢ ⎥ ⎢ ⎥ XY = ⎢ ⎥=⎢ ⎥ ⎣1 × 26 000 + 4 × 32 000⎦ ⎣154 000 ⎦ 1 × 26 000 + 1 × 32 000 58 000 The matrix XY shows dealer A spent $110 000, dealer B spent $116 000, dealer C spent $154 000 and dealer D spent $58 000. Example 10 For A =

2 5

3 6

⎡

4 4 ⎢ and B = ⎣ 1 7 0

⎤ 0 ⎥ 2 ⎦ find AB. 3

Solution A is a 2 × 3 matrix and B is a 3 × 2 matrix. Therefore AB is a 2 × 2 matrix. ⎡ ⎤ 4 0 2 3 4 ⎢ ⎥ AB = ⎣1 2⎦ 5 6 7 0 3 2×4+3×1+4×0 2×0+3×2+4×3 = 5×4+6×1+7×0 5×0+6×2+7×3 11 18 = 26 33 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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13


Exercise 1C

Examples

8,10

2 1 1 1 If X = ,Y = ,A = −1 3 −1

−2 3 ,B = 3 1

2 2 ,C = 1 1

1 1 ,I = 1 0

0 , 1

find the products AX, BX, AY, IX, AC, CA, (AC)X, C(BX), AI, IB, AB, BA, A2 , B2 , A(CA) and A2 C. 2 a Are the following products, of matrices given in 1, defined? AY, YA, XY, X2 , CI, XI 2 0 0 b If A = and B = 0 0 −3

0 , find AB. 2

3 The matrices A and B are 2 × 2 matrices, and O is the zero 2 × 2 matrix. Is the following argument correct? ‘If AB = O, and A = O, then B = O’. 2 4 If L = [2 −1], X = , find LX and XL. −3 5 A and B are both m × n matrices. Are AB and BA defined and, if so, how many rows and columns do they have? 1 0 d −b a b . = 6 Suppose 0 1 c d −c a Show that ad − bc = 1. What is the product matrix if the order of multiplication on the left-hand side is reversed? 7 Using of 6, write down a pair of matrices A, B such that AB = BA = I where the result 1 0 I= . 0 1 8 Select any three 2 × 2 matrices A, B and C. Calculate A(B + C), AB + AC and (B + C) A. Example

9

9 It takes John five minutes to drink a milk shake which costs $2.50, and twelve minutes to eat a banana split which costs $3.00. 5 12 1 Calculate the product and interpret the result in milk bar economics. 2.50 3.00 2 Suppose two friends join John. 5 12 1 2 0 Calculate and interpret the result. 2.50 3.00 2 1 1



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10 The reading habits of five students A, B, C, D and E are shown in the first matrix below where the columns p, q, r, and s represent four weekly magazines. The second matrix shows the cost in dollars of each magazine. Find the product of the two matrices and interpret the result. A B C D E

⎡ p q 0 0 ⎢ ⎢1 0 ⎢ ⎢1 0 ⎢ ⎢ ⎣1 1 0 1

r 1 1 0 1 0

s⎤ 1 ⎥ 1⎥ ⎥ 0⎥ ⎥ ⎥ 1⎦ 1

p q r s

⎡

⎤ 2.00 ⎢3.00⎥ ⎢ ⎥ ⎢ ⎥ ⎣2.50⎦ 3.50

s11 s12 s13 be the sales matrix for two showrooms selling three models of 11 Let S = s21 s22 s23 cars. Here sij is the number of cars of model j sold from showroom i. Let the prices of the three models of cars be $c1 , $c2 , $c3 . ⎡ ⎤ c1 ⎢ ⎥ Call the 3 × 1 matrix, C = ⎣c2 ⎦ the price matrix. c3

a Find SC. b What is the practical meaning of SC? c Suppose the car dealer sells both new and used cars and the price of two-year-old used cars for the three models is $u 1 , $u 2 and $u 3 , respectively. Form a new cost matrix ⎤ ⎡ c1 u 1 ⎥ ⎢ C = ⎣c2 u 2 ⎦ c3 u 3 Find SC and state its meaning. d Suppose the car dealer makes 30% profit on his selling of new cars and 25% on used cars. 0.3 0 If V = , what is the meaning of CV? 0 0.25

1.4 Identities, inverses and determinants

for 2 × 2 matrices Identities

A matrix with the same number of rows and columns is called a square matrix. For square matrices of a given dimension, e.g. 2 × 2, a multiplicative identity I exists. 1 0 For example, for 2 × 2 matrices I = 0 1 ⎡ ⎤ 1 0 0 ⎢ ⎥ and for 3 × 3 matrices I = ⎣0 1 0⎦ 0 0 1 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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2 If A = 1

15

3 , AI = IA = A, and this result holds for any square matrix multiplied by the 4

appropriate multiplicative identity.

Inverses Given a 2 × 2 matrix A, is there a matrix B such that AB = BA = I? x y 2 3 Let B = and A = u v 1 4 2 3 x y 1 0 Then AB = I implies = 1 4 u v 0 1 2x + 3u 2y + 3v 1 0 i.e. = x + 4u y + 4v 0 1

∴ 2x + 3u = 1 and 2y + 3v = 0 x + 4u = 0 y + 4v = 1 These simultaneous equations can be solved to find x, u, y, and v and hence B. 0.8 −0.6 B= −0.2 0.4 B is said to be the inverse of A as AB = BA = I. a b x y Let A be a 2 × 2 matrix with A = and let B = where B is the inverse of A. c d u v ax + bu ay + bv 1 0 Then AB = I. In full this is written = cx + du cy + dv 0 1 Hence ax + bu = 1 ay + bv = 0 cx + du = 0 cy + dv = 1 which form two pairs of simultaneous equations, for x, u and y, v respectively. Taking the x, u pair and eliminating u, (ad − bc)x = d Similarly, eliminating x, (bc − ad)u = c These two equations can be solved for x and u respectively provided ad − bc = 0 i.e.

x=

c −c d and u = = ad − bc cb − ad ad − bc

In a similar way it can be found that −a a −b and v = = ad − bc cb − ad ad − bc ⎤ ⎡ −b d ⎢ ad − bc ad − bc ⎥ ⎥ ⎢ Therefore the inverse = ⎢ −c ⎥ a ⎦ ⎣ ad − bc ad − bc y=



16


The inverse of a square matrix A, is denoted by A−1 . The inverse is unique. ad − bc has a name, the determinant of A. This is denoted det(A). a b i.e. for A = , det(A) = ad − bc c d A 2 × 2 matrix has an inverse only if det(A) = 0 A square matrix is said to be regular if its inverse exists. Those square matrices which do not have an inverse are called singular matrices; i.e. for a singular matrix det(A) = 0.

Using the TI-Nspire The operation of matrix inverse is obtained by raising the matrix to the power of −1. The determinant command (b>Matrices & Vectors>Determinant) is used as shown. 3 6 (a is the matrix A = defined 6 7 on page 8.) Hint: you can also type in det(a)

Using the Casio ClassPad The operation of matrix inverse is obtained by entering A∧−1 in the entry line. The determinant is obtained by entering and highlighting A and tapping Interactive, Matrix-Calc, det.



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17

Example 11

5 For the matrix A = 3

2 find 1 b A−1

a det(A) Solution

b A−1

a det(A) = 5 × 1 − 2 × 3 = −1

1 1 −2 = −1 −3 5 −1 2 = 3 −5

Example 12

3 For the matrix A = 1 a det(A)

2 find 6

5 c X if AX = 7

b A−1

6 2

Solution a det(A) = 3 × 6 − 2 = 16

5 c AX = 7

b A−1

6 2

5 d Y if YA = 7 1 6 = 16 −1

−2 3

6 2

Multiply both sides (from the left) by A−1 . −1 −1 5 6 A AX = A 7 2 1 6 −2 5 6 ∴ IX = X = 16 −1 3 7 2 1 16 30 = 16 16 0 1 2 = 1 0 5 6 d YA = 7 2



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Multiply both sides (from the right) by A−1 1 5 6 6 −2 YAA−1 = 16 7 2 −1 3 1 24 8 ∴ YI = Y = 16 40 −8 ⎤ ⎡ 3 1 ⎢2 2⎥ ⎥ ∴ Y=⎢ ⎣ 5 −1 ⎦ 2 2

Exercise 1D Example

11

1 For the matrices A = a det(A)

Example

12

2 3

1 2

b A−1

−2 and B = 3

−2 find 2

c det(B)

d B−1

2 Find the inverse of the following regular matrices (␪ is any real number, k is any non-zero real number). 3 −1 3 1 1 0 cos ␪ −sin ␪ a b c d 4 −1 −2 4 0 k sin ␪ cos ␪ 2 1 1 0 3 If A, B are the regular matrices A = ,B = , find A−1 , B−1 . 0 −1 3 1 Also find AB and hence find, if possible, (AB)−1 . Also find from A−1 , B−1 , the products A−1 B−1 and B−1 A−1 . What do you notice? 4 3 4 For the matrix A = 2 1 3 4 3 4 a find A−1 b if AX = , find X c if YA = , find Y. 1 6 1 6 3 2 4 −1 3 4 5 If A = ,B = and C = , find 1 6 2 2 2 6 a X such that AX + B = C

b Y such that YA + B = C

6 If A is a 2 × 2 matrix, a12 = a21 = 0, a11 = 0, a22 = 0, then show that A is regular and find A−1 . 7 Let A be a regular 2 × 2 matrix, B a 2 × 2 matrix and AB = 0. Show that B = 0. 8 Find all 2 × 2 matrices such that A−1 = A.



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1.5

19

Solution of simultaneous equations using matrices Inverse matrices can be used to solve certain sets of simultaneous linear equations. Consider the equations 3x − 2y = 5 5x − 3y = 9 This can be written as 3 −2 x 5 = 5 −3 y 9 3 −2 If A = the determinant of A is 3(−3) − 5(−2) = 1 5 −3 which is not zero and so A−1 exists. −3 2 −1 A = −5 3 3 −2 x 5 Multiplying the matrix equation = on the left hand side by A−1 and using 5 −3 y 9 the fact that A−1 A = I yields the following: x −1 −1 5 A =A A y 9 x 5 ∴ I = A−1 y 9 x 3 3 −1 5 ∴ = since A = y 2 9 2 This is the solution to the simultaneous equations. Check by substituting x = 3, y = 2 in the equations. When dealing with simultaneous linear equations in two variables which represent parallel straight lines, a singular matrix results. For example the system x + 2y = 3 −2x − 4y = 6 has associated matrix equation 1 2 x 3 = y 6 −2 −4



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1 2 Note that the determinant of = 1 × −4 − (−2 × 2) = 0. −2 −4 There is no unique solution to the system of equations. Example 13

2 −1 −1 x If A = and K = , solve the system AX = K where X = . 1 2 2 y Solution If AX = K, then X = A−1 K 1 2 1 −1 0 A−1 K = × = 5 −1 2 2 1 0 ∴X= 1 Example 14 Solve the following simultaneous equations. 3x − 2y = 6 7x + 4y = 7 Solution

3 −2 The matrix equation is 7 4

3 −2 A= 7 4

Let

2 3

1 4 x = 26 −7 y

2 3

Then A

and

1 4 = 26 −7

−1

x 6 = y 7

1 6 38 = 26 −21 7

Since any linear system of n equations in n unknowns can be written as ⎡ ⎤ ⎡ ⎤ k1 x1 ⎢ ⎥ ⎢ ⎥ ⎢k 2 ⎥ ⎢ x2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ . ⎥, and K = AX = K, where A is an n × n matrix, X = ⎢ ⎢ ⎥ ⎢. ⎥ ⎢ ⎥ ⎢ ⎥ . ⎣. ⎦ ⎣ ⎦ xn kn ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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21

this method can be applied more generally when A is regular. In fact, as shown, an expression for the solution can be written at once. Multiply AX, and K, on the left by A−1 , and A−1 (AX) = A−1 K and A−1 (AX) = (A−1 A)X = IX = X. Hence X = A−1 K, which is a formula for the solution of the system. Of course it depends on the inverse A−1 existing, but once A−1 is found then equations of the form AX = K can be solved for all possible n × 1 matrices K. Again this process can be completed using a calculator as long as matrices A and K have been entered onto the calculator. Example 15 Consider the system of five equations in five unknowns. 2a + 3b − c + d + 2e = 9 a+b−c =4 a + 2d − 3e = 4 −b + 2c − d + e = −6 a − b + d − 2e = 0 Use a graphics calculator to solve for a, b, c, d and e. Solution Enter 5 × 5 matrix A and 5 × 1 matrix B into the graphics calculator. ⎡ ⎤ ⎡ ⎤ 2 3 −1 1 2 9 ⎢ ⎥ ⎢ ⎥ ⎢1 ⎢ 4⎥ 1 −1 0 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ A = ⎢1 0 0 2 −3 ⎥ B = ⎢ ⎢ 4⎥ ⎢ ⎥ ⎢ ⎥ 2 −1 1⎦ ⎣0 −1 ⎣−6 ⎦ 1 −1 0 1 −2 0 ⎡ ⎤ 4 ⎢ 9 ⎥ ⎢ ⎥ ⎢ 23 ⎥ ⎢ ⎥ ⎢ 9 ⎥ 23 7 2 4 ⎢ ⎥ ∴ a = , b = , c = −1, d = and e = − Then A−1 B = ⎢ −1 ⎥ ⎢ 7 ⎥ 9 9 9 3 ⎢ ⎥ ⎢ ⎥ ⎢ 9 ⎥ ⎣ 2 ⎦ − 3 It should be noted that just as for two equations in two unknowns, there is a geometric interpretation for three equations in three unknowns. There is only a unique solution if the equations represent three planes intersecting at a point.



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Exercise 1E

Example

13

Example

14

3 −1 x 1 If A = , solve the system AX = K where X = , and 4 −1 y −1 −2 a K= b K= 2 3 3 1 2 If A = , solve the system AX = K where −2 4 0 2 a K= b K= 1 0 3 Use matrices to solve the following pairs of simultaneous equations. a −2x + 4y = 6 3x + y = 1

b −x + 2y = −1 −x + 4y = 2

c 2x + 5y = −10 y = x +4

d 1.3x + 2.7y = −1.2 4.6y − 3.5x = 11.4

4 Use matrices to find the point of intersection of the lines given by the equations 2x − 3y = 7 and 3x + y = 5. 5 Two children spend their pocket money buying some books and some CDs. One child spends $120 and buys four books and four CDs. The other child buys three CDs and five books and spends $114. Set up a system of simultaneous equations and use matrices to find the cost of a single book and a single CD. 6 Consider the system 2x − 3y = 3 4x − 6y = 6 a Write this system in matrix form, as AX = K. b Is A a regular matrix? c Can any solutions be found for this system? d How many pairs does the solution set contain? Example

15

7 Consider the system of four equations in four unknowns. p+q −r −s = 5 r +s =1 2 p − q + 2r = −2 p−q +s = 0 Use a graphics calculator to solve for p, q, r and s.



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A matrix is a rectangular array of numbers. Two matrices A and B are equal when: r each has the same number of rows and the same number of columns, and r they have the same number or element at corresponding positions. The size or dimension of a matrix is described by specifying the number of rows (m) and the number of columns (n). The dimension is written m × n. Addition will be defined for two matrices only when they have the same dimension. The sum is found by adding corresponding elements. a b e f a+e b+ f + = c d g h c+g d +h

Review

Chapter summary

23

Subtraction is defined in a similar way. If A is an m × n matrix and k is a real number, kA is defined to be an m × n matrix whose elements are k times the corresponding element of A. a b ka kb k = c d kc kd If A is an m × n matrix and B is an n × r matrix, then the product AB is the m × r matrix whose entries are determined as follows. To find the entry in row i and column j of AB, single out row i in matrix A and column j in matrix B. Multiply the corresponding entries from the row and column and then add up the resulting products. The product AB is defined only if the number of columns of A is the same as the number of rows of B. If A and B are square matrices of the same dimension and AB = BA = I then A is said to the inverse of B and B is said to be the inverse of A. ⎡ ⎤ d −b ⎢ ad − bc ad − bc ⎥ a b ⎥ If A = then A−1 = ⎢ ⎣ −c ⎦ a c d ad − bc ad − bc det(A) = ad − bc is the determinant of matrix A. A square matrix is said to be regular if its inverse exists. Those square matrices which do not have an inverse are called singular matrices. Simultaneous equations can be solved using inverse matrices, for example ax + by = c d x + ey = f

a b can be written as d e

−1 c x c x a b = and = f y f y d e



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Multiple-choice questions ⎡

1

2

3

4

5

⎤ 1 0 ⎢ 2 −1⎥ ⎢ ⎥ The matrix A = ⎢ ⎥ has dimension ⎣−2 3⎦ 3 0 A 8 B 4×2 C 2×4 D 1×4 E 3×4 2 0 1 −3 4 If A = and B = then A + B = −1 3 −1 −3 −1 3 −3 3 4 −1 2 2 1 E Cannot be determined A B C D −2 0 −2 2 2 3 1 −3 1 1 −3 1 2 −3 and D = then D − C = If C = 1 0 −2 2 3 −1 −1 0 0 1 0 0 2 −6 4 C B A −2 0 −4 1 3 1 −1 −3 −1 1 −6 0 E Cannot be determined D 1 3 1 −4 0 If M = then −M = −2 −6 −4 0 0 −4 4 0 0 4 4 0 A B C D E −2 −6 −6 −2 −2 −6 6 2 2 6 0 2 0 4 and N = then 2M − 2N = If M = 3 0 −3 1 0 0 0 −2 0 −4 0 4 0 2 A B C D E −9 2 −6 1 −12 2 12 −2 6 −1

6 If A and B are both m × n matrices, where m = n, then A + B is an A m × n matrix B m × m matrix C n × n matrix D 2m × 2n matrix E Cannot be determined 7 If P is an m × n matrix, and Q is a n × p matrix, the dimension of matrix QP is A n×n B m × p C n × p D m × n E Cannot be determined 2 2 8 The determinant of matrix A = is −1 1 A 4

B 0

C 1 9 The inverse of matrix A = 1 2 1 C A −1 B −1 −1

−4 D 1 E 2 −1 is −2 1 1 1 1 2 −1 D E −1 −2 −1 2 1 −1



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0 −2 0 2 10 If M = and N = then NM = −3 1 3 1 0 −4 −4 −2 0 4 A B C −9 1 2 −8 9 1

D

−6 2 −3 −5

E

6 −2 −3 −5

Review

Short-answer questions (technology-free)

1 0 −1 0 1 If A = and B = , find 2 3 0 1 a (A + B)(A − B)

b A 2 − B2

3 4 8 2 Find all possible matrices A which satisfy the equation A= . 6 8 16 ⎡ ⎤ 5

1 2 6 ⎢ ⎥ , B = [3 −1 2], C = , D = 2 4 and E = ⎣0 ⎦. 3 Let A = 3 −1 1 2

4

5

6

7

8

a State whether or not each of the following products exist: AB, AC, CD, BE b Evaluate DA and A−1 . ⎡ ⎤ 1 −4 1 −2 1 1 2 ⎢ ⎥ If A = , B = ⎣1 −6⎦ and C = , evaluate AB and C−1 . −5 1 2 3 4 3 −8 5 6 1 2 = Find the 2 × 2 matrix A such that A 3 4 12 14 ⎡ ⎤ 2 0 0 ⎢ ⎥ If A = ⎣0 0 2⎦, find A2 and hence A−1 . 0 2 0 1 2 If is a singular matrix, find the value of x. 4 x 2 −1 a If M = , find the value of 1 3 i MM = M2

ii MMM = M3 x 3 b Find x and y given that M = y 5


iii M−1


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Extended-response questions

3 1 A= 1

1 2 −1 ,B = −4 5 2

a Find i A+B ii A − B iii 2A + 3B iv C such that 3A + 2C = B b Find iii X such that AX = B iv Y such that YA = B i AB ii A−1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −2 2 −2 0 1 2 0 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 If A = ⎣ 2 0 −1 ⎦, B = ⎣ 4 2 −2 ⎦ and C ⎣ 3 0 −1 ⎦, find 1 3 4 1 3 3 1 3 1 a AB b AC d X such that AX = C f X such that AXC = CB

c BC e Y such that YA = B g Y such that CYA = BA

3 a Consider the system of equations 2x − 3y = 3 4x + y = 5 i Write this system in matrix form, as AX = K. ii Find detA and A−1 . iii Solve the system of equations. iv Interpret your solution geometrically. b Consider the system of equations 2x + y = 3 4x + 2y = 8 i Write this system in matrix form, as AX = K. ii Find detA and explain why A−1 does not exist. c Interpret your findings in part b geometrically. 4 The final grades for Physics and Chemistry are made up of three components, tests, practical work and exams. Marks out of 100 are awarded in each component each semester. Wendy scored the following marks in each of the three components for Physics. Semester 1: tests 79, practical work 78, exam 80 Semester 2: tests 80, practical work 78, exam 82 a Represent this information in a 2 × 3 matrix. To calculate the final grade for each semester the three components are weighted so that tests are worth 20%. Practical work is worth 30% and the exam is worth 50%. (cont’d)



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b Represent this information in a 3 × 1 matrix. c Calculate Wendy’s final grade for Physics in each semester. Wendy also scored the following marks in each of the three components for Chemistry. Semester 1: tests 86, practical work 82, exam 84 Semester 2: tests 81, practical work 80, exam 70 d Calculate Wendy’s final grade for Chemistry in each semester. Students who gain an aggregate score for Physics and Chemistry of 320 or more over the two semesters are awarded a Certificate of Merit in Science. e Will Wendy be awarded a Certificate of Merit in Science? She asks her teacher to remark her Semester 2 Chemistry Exam hoping that she will gain the necessary marks to be awarded a Certificate of Merit. f How many extra marks does she need?

27

5 A company runs Computing classes and employs full-time and part-time teaching staff as well as technical support staff, cleaners and catering staff. The number of staff employed depends on demand from term to term. In one year they employed the following teaching staff: Term 1: full-time 10, part-time 2 Term 2: full-time 8, part-time 4 Term 3: full-time 8, part-time 8 Term 4: full-time 6, part-time 10 a Represent this information in a 4 × 2 matrix. Full-time teachers are paid $70 per hour and part-time teachers are paid $60 per hour. b Represent this information in a 2 × 1 matrix. c Calculate the cost per hour to the company for teaching staff for each term. In the same year they also employed the following support staff Term 1: technical staff 2, catering staff 2, cleaning staff 1. Term 2: technical staff 2, catering staff 2, cleaning staff 1. Term 3: technical staff 3, catering staff 4, cleaning staff 2. Term 4: technical staff 3, catering staff 4, cleaning staff 2. d Represent this information in a 4 × 3 matrix. Technical staff are paid $60 per hour, catering staff $55 per hour and cleaners are paid $40 per hour. e Represent this information in a 3 × 1 matrix. f Calculate the cost per hour to the company for support staff for each term. g Calculate the total cost per hour to the company for teaching and support staff for each term.



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C H A P T E R

2 Algebra I Objectives To express a number in standard form To solve linear equations To solve problems with linear equations and simultaneous linear equations To use substitution and transposition with formulas To add and multiply algebraic fractions To solve literal equations To solve simultaneous literal equations

2.1 Indices In this section a review of indices is undertaken.

Review of index laws a m × a n = a m+n

a m ÷ a n = a m−n 1 a −n = n a (ab)n = a n bn

(a m )n = a mn 1 √ n a = an a0 = 1 Example 1

Simplify each of the following. 1 4 x4 b c x2 ÷x5 a x2 × x3 2 x

1

d (x 3 ) 2

Solution a x 2 × x 3 = x 2+3 = x 5 1

4

1

4

x4 = x 4−2 = x 2 x2 1 3 d (x 3 ) 2 = x 2 b

3

c x 2 ÷ x 5 = x 2 − 5 = x − 10



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Chapter 2 — Algebra I

29

Example 2 Evaluate a

2 125 3

b

1000 27

2 3

Solution 2 1 2 a 125 3 = (125) 3 = 52 = 25 ⎛ ⎞ 2 1 2 2 10 100 1000 3 ⎝ 1000 3 ⎠ = = = b 27 27 3 9

Example 3

4 Simplify

x 2 y3

1 x2

2 y3

.

Solution

4

x 2 y3

1

2

1

=

x2 y3

(x 2 y 3 ) 4 1

2

x2 y3 2

=

3

x4 y4 1

2

x2 y3 2

1

3

2

= x 4−2 y 4−3 1

= x 0 y 12 1

= y 12

Exercise 2A 1 Simplify each of the following using the appropriate index laws. Example

1a

Example

1b

Example

1c

a x3 × x4 e

x8 x −4 1

m (n 10 ) 5 −

p −5 p2 j (x5 )3 f

i (y−2 )−7

b a 5 × a −3

2 5 5

1

n 2x 2 × 4x 3

c x 2 × x −1 × x 2 1

2

g a2 ÷ a3 3

5

o (a 2 ) 2 × a −4 3

1 a 2 b−3

1 −4

x−2 1 p x −4 l

r x 3 × 2x 2 × −4x − 2

3 4

y3 y7

h (a−2 )4

k (a −20 ) 5

1

q 2n ÷ (4 n ) s (ab3 )2 × a −2 b−4 ×

d

t (22 p −3 × 43 p 5 ÷ (6 p −3 ))0



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Essential Advanced General Mathematics 2

2 Evaluate each of the following. a

1 25 2

e

49 36

b − 1

1 64 3

c

1

2

3

1

1

2

d 16− 2

1

f 27 3 1 81 4 j 16

i 92

16 9

2

g 144 2 0 23 k 5

h 64 3 3

l 128 7

3 Use your calculator to evaluate each of the following, correct to two decimal places. √ a 4.352 b 2.45 c 34.6921 d (0.02)−3 1 √ √ 2 e 3 0.729 f 4 2.3045 g (345.64)− 3 h (4.568) 5 1 i 1 (0.064)− 3 4 Simplify each of the following, giving your answer with positive index. 2a 2 (2b)3 a 2 b3 a −2 b−3 b a c −2 −4 −2 −4 (2a) b a b a −2 b−4 d

a 2 b3 ab × −1 −1 −2 −4 a b a b

5 Write

e

(2a)2 × 8b3 16a −2 b−4

f

2a 2 b3 16ab ÷ −2 −4 8a b (2a)−1 b−1

2n × 8n in the form 2an+b . 22n × 16

6 Write 2−x × 3−x × 62x × 32x × 22x as a power of 6. 7 Simplify each of the following. 1

1

2

a 2 3 × 2 6 × 2− 3 1 2 1 5 d 23 × 22 Example

3

1

2

1

b a 4 × a 5 × a − 10 1 2 1 2 e 2 3 × 2 3 × 2− 5

8 Simplify each of the following. √ √ √ √ 3 3 2 3 a a b ÷ a 2 b−1 b a 3 b2 × a 2 b−1 √ √ √ √ d a −4 b2 × a 3 b−1 e a 3 b2 c−3 × a 2 b−1 c−5 √ √ a 3 b2 a −4 b2 √ 3 −1 g × × a b a 2 b−1 c−5 a 3 b−1

2

5

2

c 2 3 × 2 6 × 2− 3

√ √ 5 3 2 5 a b × a 2 b−1 √ √ 5 3 2 5 f a b ÷ a 2 b−1

c

2.2 Standard form Often when dealing with real world problems, the numbers involved may be very small or very large. For example, the distance from the Earth to the Sun is approximately 150 000 000 kilometres, and the mass of an oxygen atom is approximately 0.000 000 000 000 000 000 000 026 grams. In order to deal with such numbers, a more convenient way to express them can be used. This involves expressing the number as a product of a number between 1 and 10 and a power of ten and is called standard form or scientific notation.



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These examples written in standard form would read 1.5 × 108 kilometres and 2.6 × 10−23 grams respectively. Performing multiplication and division with very small or very large numbers can often be simplified by first converting the numbers into standard form. When simplifying algebraic expressions or manipulating numbers in standard form, a sound knowledge of the index laws is essential. Example 4 Write each of the following in standard form. a 3 453 000 b 0.00675 Solution a 3 453 000 = 3.453 × 106

b 0.00675 = 6.75 × 10−3

Example 5 Find the valueof

32 000 000 × 0.000 004 . 16 000

Solution 32 000 000 × 0.000 004 3.2 × 107 × 4 × 10−6 = 16 000 1.6 × 104 12.8 × 101 = 1.6 × 104 = 8 × 10−3 = 0.008 Example 6 √ 5 a Evaluate 2 if a = 1.34 × 10−10 and b = 2.7 × 10−8 . b Solution √ √ 5 5 a 1.34 × 10−10 = b2 (2.7 × 10−8 )2

1

(1.34 × 10−10 ) 5 = 2.72 × (10−8 )2 = 1.454 43 · · · × 1013 = 1.45 × 1013 to three significant figures. Many calculators have scientific notation. The actual display will vary from calculator to calculator. For example, in standard form 3 245 000 = 3.245 × 106 may appear as 3.245E6 or 3.24506 .



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Using the TI-Nspire The TI-Nspire can be set to express answers in standard form by selecting c>Settings & Status > Settings > General. The number 3 245 000 will then appear as 3.245E6.

Using the Casio ClassPad The Classpad calculator can be set to express decimal answers in various forms. To select a fixed number of decimal places, including specifying scientific notation with and in Basic format tap the arrow to select from the fixed decimal accuracy, tap various Number formats available.

Exercise 2B Example

4

1 Express each of the following numbers in standard form. a 47.8 e 0.0023 i 23 000 000 000

b 6728 f 0.000 000 56 j 0.000 000 0013

c 79.23 g 12.000 34 k 165 thousand

d 43 580 h 50 million l 0.000 014 567

2 Express each of the following in scientific notation. a b c e f

X-rays have a wavelength of 0.000 000 01 cm. The mass of a hydrogen atom is 0.000 000 000 000 000 000 000 0166 g. Visible light has wavelength 0.000 05 cm. d One nautical mile is 1853.18 m. A light year is 9 463 000 000 000 km. The speed of light is 29 980 000 000 cm/s.

3 Express each of the following as an ordinary number. a The star Sirius is approximately 7.5684 × 1013 km from Earth. b A single blood cell contains 2.7 × 108 molecules of haemoglobin. c The radius of an electron is 1.9 × 10−13 cm. Example

Example

5

6

4 Find the value of 324 000 × 0.000 0007 a 4000

b

5 240 000 × 0.8 42 000 000

5 Evaluate the following correct to three significant figures. √ √ 4 3 a a 9 if a = 2 × 1012 and b = 0.05 b a if a = 2 × 10 and b = 3.215 4 4 4b b



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2.3

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Solving linear equations and linear simultaneous equations The solution to many problems may be found by translating them into a mathematical equation which may then be solved using algebraic techniques. An equation is solved by finding the value or values of the unknowns that would make the statement true. Linear equations are simple equations that can be written in the form ax + b = 0. There are a number of standard techniques that can be used for solving linear equations. Example 7 Solve

x x −2= 5 3 Solution x x −2= 5 3 x x × 15 − 2 × 15 = × 15 5 3 Multiply both sides of the equation by the lowest common multiple of 3 and5. 3x − 30 = 5x 3x − 5x = 30 −2x = 30 30 x= −2 x = −15

Example 8

Solve

x − 3 2x − 4 − =5 2 3 Solution 2x − 4 x −3 ×6− ×6=5×6 2 3 3(x − 3) − 2(2x − 4) = 5 × 6 3x − 9 − 4x + 8 = 30 3x − 4x = 30 + 9 − 8 −x = 31 31 x= −1 = −31



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Simultaneous linear equations y

y=

4

4 3

2x –

Finding the intersection of two straight lines can be done graphically, however the accuracy of the solution will depend on the accuracy of the graphs. Alternatively this point of intersection may be found algebraically by solving the pair of simultaneous equations. Three techniques for solving simultaneous equations will be considered.

2 –3 –2 –1

1

1

2

3

x

0 –1 –2

x+

–3 –4

2y

=–

3

(1, –2)

Example 9 Solve the equations 2x − y = 4 and x + 2y = −3. Solution 1: By substitution 2x − y = 4 x + 2y = −3

(1) (2)

First express one unknown from either equation in terms of the other unknown. From equation (2) we get x = −3 − 2y. Then substitute this expression into the other equation. Equation (1) then becomes 2(−3 − 2y) − y = 4 reducing it to one equation in one unknown. −6 − 4y − y = 4 −5y = 10 y = −2 Substituting the value of y into (2) x + 2(−2) = −3 x =1 Check in (1) LHS = 2(1) − (−2) = 4 RHS = 4 Solving (1)

N.B. This means that the point (1, –2) is the point of intersection of the graphs of the two linear relations.




35

2: By elimination 2x − y = 4

(1)

x + 2y = −3

(2)

If the coefficient of one of the unknowns is the same in both equations, we can eliminate that unknown by subtracting one equation from the other. It may be necessary to multiply one of the equations by a constant to make the coefficients of x or y the same for the two equations. To eliminate x multiply equation (2) by 2 and subtract the result from equation (1). Equation (2) becomes Then Subtracting (1) − (2 )

2x + 4y 2x − y 2x + 4y −5y y

=6 =4 = −6 = 10 = −2

(2 ) (1) (2 )

Now substitute for y in (1) to find x, and check as in substitution method.

Using the TI-Nspire The simultaneous equations can be solved in a Calculator application. Use b>Algebra>Solve System of Equations>Solve System of Equations, complete the pop-up screen, and enter the equations as shown to give the solution to the simultaneous equations 2x − y = 4 and x + 2y = −3. The entry solve(2x – 3y = 4 and x + 2y = –3, x,y) also gives the result. The and can either be typed in or found in the Catalog (k 1 A)

The simultaneous equations can also be solved graphically in a Graphs application. The equations are rearranged to make y the subject. The equations in this form are −3 − x . Enter these as y = 2x − 4 and y = 2 shown.



36


If the Entry Line is not visible, press e. Pressing · will hide the Entry Line. /+G will also Hide/Show the Entry Line.

Note:

The intersection point is found by using b>Points and Lines>Intersection Point/s. Use the Touchpad to move the hand to select each of the two graphs.

The intersection point’s coordinates will appear on the screen. to exit the Intersection Point(s) Press menu.

Using the Casio ClassPad The simultaneous equations can also be solved graphically. First, the equations need to be rearranged to make y the subject. In this form the equations are y = 2x − 4 and 3 1 area as y = − x − . Enter these in 2 2 shown. Select both equations by ticking the box to produce the graph. at the left then press To find the solution, click into the graph screen to select it and then click Analysis, G-Solve, Intersect.



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Exercise 2C

Example

7

Example

8

Example

9

1 Solve the following linear equations. x a 3x + 7 = 15 b 8 − = −16 2 2x − 15 = 27 e 5(2x + 4) = 13 d 3 g 3x + 5 = 8 − 7x h 2 + 3(x − 4) = 4(2x + 5) x j 6x + 4 = − 3 3

f −3(4 − 5x) = 24 3 2x − = 5x i 5 4

2 Solve the following linear equations. x 2x x 3x 3x − 2 x a + = 16 − =8 b c + = −18 2 5 4 3 2 4 4 2x 5x x − 4 2x + 5 3 − 3x 2(x + 5) 1 − = d e + =6 f − = 4 3 5 2 4 10 6 20 2(x + 1) 3−x −2(5 − x) 6 4(x − 2) − = −24 g h + = 4 5 8 7 3 3 Solve each of the following pairs of simultaneous equations. a 3x + 2y = 2 2x − 3y = 6 d x + 2y = 12 x − 3y = 2

2.4

c 42 + 3x = 22

b 5x + 2y 3x − y e 7x − 3y x + 5y

=4 =6 = −6 = 10

2x − y = 7 3x − 2y = 2 f 15x + 2y = 27 3x + 7y = 45

c

Solving problems with linear equations Many problems can be solved by translating them into mathematical language and using an appropriate mathematical technique to find the solution. By representing the unknown quantity in a problem with a symbol (called a pronumeral) and constructing an equation from the information, the value of the unknown can be found by solving the equation. Before constructing the equation, state what the pronumeral is and what it stands for (including the unit). It is essential to remember that all elements of the same type in the equation must be in the same units. Example 10 For each of the following, form the relevant linear equation and solve it for x. a The length of the side of a square is (x − 6) cm. Its perimeter is 52 cm. b The perimeter of a square is (2x + 8) cm. Its area is 100 cm2 .



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Solution a The perimeter = 4 × length of a side 4(x − 6) = 52 Therefore x − 6 = 13 and x = 19 b The perimeter of the square is 2x + 8 x +4 2x + 8 = The length of one side = 4 2 x +4 2 Therefore = 100 2 x +4 In this case = 10 as side length must be a positivenumber. 2 Therefore x = 16 Example 11 An athlete trains for an event by gradually increasing the distance she runs each week over a five-week period. If she runs an extra 5 km each successive week and over the five weeks runs a total of 175 km, how far did she run in the first week? Solution Let the distance run in the first week = x km. Then the distance run in the second week = x + 5 km. The distance run in the third week = x + 10 km. So the total distance run = x + x + 5 + x + 10 + x + 15 + x + 20

∴

5x + 50 = 175 5x = 125 x = 25

The distance she ran in the first week was 25 km.

Example 12 A man bought 14 CDs at a sale. Some cost him $15 each and the remainder cost $12.50 each. In total he spent $190. How many $15 CDs and how many $12.50 CDs did he buy?



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Solution Let n equal the number of CDs costing $15. Then 14 – n = the number of CDs costing $12.50.

∴

15n + 12.5(14 − n) = 190 15n + 175 − 12.5n = 190 2.5n + 175 = 190 2.5n = 15 n=6

He bought 6 CDs costing $15 and 8 CDs costing $12.50.

Exercise 2D Example

10

1 For each of the cases below, write down a relevant equation involving the variables defined and solve the equation for parts a, b and c. a The length of the side of a square is (x − 2) cm. Its perimeter is 60 cm. b The perimeter of a square is (2x + 7) cm. Its area is 49 cm2 . c The length of a rectangle is (x − 5) cm. Its width is (12 − x) cm. The rectangle is twice as long as it is wide. d The length of a rectangle is (2x + 1) cm. Its width is (x − 3) cm. The perimeter of the rectangle is y cm. e n persons each has a meal costing $p. The total cost of the meal is $Q. f S persons each has a meal costing $p. 10% service charge is added to the cost. The total cost of the meal is $R. g A machine working at a constant rate produces n bolts in 5 minutes. It produces 2400 bolts in 1 hour. h The radius of a circle is (x + 3) cm. A sector subtending an angle of 60◦ at the centre is cut off. The arc length of the minor sector is a cm.

Example

11

2 Bronwyn and Noel have a women’s clothing shop in Summerland. Bronwyn manages the shop and her sales are going up steadily over a particular period of time. They are going up by $500 a week. If over a five-week period her sales total $17 500, how much did she earn in the first week?

Example

12

3 Bronwyn and Noel have a women’s clothing shop in Summerland and Bronwyn manages the shop. Sally, Adam and baby Lana came into the shop and Sally bought dresses and handbags. The dresses cost $65 each and the handbags cost $26 each. The total number of items was 11 and in total she spent $598. How many dresses and how many handbags did she buy? 4 A rectangular courtyard is three times as long as it is wide. If the perimeter of the courtyard is 67 m, find the dimensions of the courtyard.



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5 A wine merchant buys 50 cases of wine. He pays full price for half of them but gets a 40% discount on the remainder. If he paid a total of $2260, how much was the full price of a single case? 6 A real estate agent sells 22 houses in six months. He makes a commission of $11 500 per house on some and $13 000 per house on the remainder. If his total commission over the six months was $272 500, on how many houses did he make a commission of $11 500? 7 Three boys compare their marble collections. The first boy has 14 less than the second boy, who has twice as many as the third. If between them they have 71 marbles, how many does each boy have? 8 Three girls are playing Scrabble. At the end of the game, the total of their scores adds up to 504. Annie scored 10% more than Belinda, while Cassie scored 60% of the combined scores of the other two. What did each player score? 9 A biathlon event involves running and cycling. Kim can cycle 30 km/h faster than she can run. If Kim spends 48 minutes running and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run? 10 The mass of a molecule of a certain chemical compound is 2.45 × 10−22 g. If each molecule is made up of two carbon atoms and six oxygen atoms and the mass of one 1 oxygen atom is that of a carbon atom, find the mass of an oxygen atom. 3

2.5 Solving problems using simultaneous

linear equations When the relationships between two quantities is linear then the constants which determine the linear relationship can be determined if two sets of information satisfying the relationship are given. Simultaneous linear equations enable this to be done. Another situation in which simultaneous linear equations may be used is where it is required to find the point of the cartesian plane which satisfies two linear relations.

Example 13 There are two possible methods for paying gas bills: Method A: A fixed charge of $25 per quarter + 50c per unit of gas used Method B: A fixed charge of $50 per quarter + 25c per unit of gas used.

Determine the number of units that must be used before method B becomes cheaper than method A.



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Solution

Now

C1 C2 x C1 C2

= charge in $ using method A = charge in $ using method B = number of units of gas used = 25 + 0.5x = 50 + 0.25x

Dollars

Let

C 100

C1 = 0.5x + 25

C2 = 0.25x + 50 50 25

It can be seen from the graph that if the number 0 25 50 75 100 125 150 Units of units exceeds 100 then method B is cheaper. The solution could also be obtained by solving simultaneous linear equations: C1 25 + 0.5x 0.25x x

x

= C2 = 50 + 0.25x = 25 = 100

Example 14 If 3 kg of jam and 2 kg of butter cost $29, and 6 kg of jam and 3 kg of butter cost $54, find the cost of 1 kg of jam and 1 kg of butter. Solution Let the cost of 1 kg of jam = x dollars and the cost of 1 kg of butter = y dollars. Then 1 3x + 2y = 29 6x + 3y = 54

2

Multiply 1 by 2 : 6x + 4y = 58 Subtract 1 from 2 :

1

and

−y = −4 y=4 Substituting in 2 gives: 6x + 3(4) = 54 6x = 42 x =7

∴ The jam costs $7 per kilogram and the butter, $4 per kilogram.

Exercise 2E Example

13

1 A car hire firm offers the option of paying $108 per day with unlimited kilometres, or $63 per day plus 32 cents per kilometre travelled. How many kilometres would you have to travel in a given day to make the unlimited kilometre option more attractive?



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2 Company A will cater for your party at a cost of $450 plus $40 per guest. Company B offers the same service for $300 plus $43.00 per guest. How many guests are needed before Company A’s charge is less than Company B’s? Example

14

3 A basketball final is held in a stadium which can seat 15 000 people. All the tickets have been sold, some to adults at $45 and the rest for children at $15. If the revenue from the tickets was $525 000, find the number of adults who bought tickets. 4 A contractor employed eight men and three boys for one day and paid them a total of $2240. Another day he employed six men and eighteen boys for $4200. What was the daily rate he paid each man and each boy? 5 The sum of two numbers is 212 and their difference is 42. Find the two numbers. 6 A chemical manufacturer wishes to obtain 700 litres of a 24% acid solution by mixing a 40% solution with a 15% solution. How many litres of each solution should be used? 7 Two children had 220 marbles between them. After one child had lost half her marbles and the other had lost 40 marbles they had an equal number of marbles. How many did each child start with and how many did each child finish with? 8 An investor received $31 000 interest per annum from a sum of money, with part of it invested at 10% and the remainder at 7% simple interest. She found that if she interchanged the amounts she had invested she could increase her return by $1000 per annum. Calculate the total amount she had invested. 9 Each adult paid $30 and each student paid $20 to attend a concert. A total of 1600 people attended. The total paid was $37 000. How many adults and how many students attended the concert?

2.6 Substitution and transposition of formulas An equation that states a relationship between two or more quantities is called a formula, e.g. the area of a circle A = ␲r2 . The value of A, the subject of the formula, may be found by substituting a given value of r and the value of ␲. Example 15 Using the formula A = ␲r2 , find the value of A correct to two decimal places, if r = 2.3, ␲ = 3.142 (correct to two decimal places). Solution A = ␲r 2 = 3.142(2.3)2 = 16.621 18 A = 16.62, correct to two decimal places. The formula can also be transposed to make r the subject. When transposing formulas a similar procedure to solving linear equations is followed. Whatever has been done to the pronumeral required is ‘undone’.



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Example 16 Transpose the formula A = ␲r2 to make r the subject and find the value of r, correct to two decimal places, if A = 24.58, ␲ = 3.142 (correct to three decimal places). Solution

∴

∴

A = ␲r 2 A = r2 ␲ A =r ␲ A r= ␲ 24.58 = 3.142 = 2.79697 r = 2.80, correct to two decimal places

Exercise 2F 1 Substitute the specified values to evaluate each of the following, giving the answers correct to two decimal places.

Example

15

a v if v = u + at and u = 15, a = 2, t = 5 PrT and P = 600, r = 5.5, T = 10 b I if I = 100 c V if V = ␲r 2 h and r = 4.25, h = 6 d S if S = 2␲r (r + h) and r = 10.2, h = 15.6 4 e V if V = ␲r 2 h and r = 3.58, h = 11.4 3 1 f s if s = ut + at 2 and u = 25.6, t = 3.3, a = −1.2 2 l and l = 1.45, g = 9.8 g T if T = 2␲ g 1 1 1 h f if = + and v = 3, u = 7 f v u i c if c2 = a 2 + b2 and a = 8.8, b = 3.4 j v if v 2 = u 2 + 2as and u = 4.8, a = 2.5, s = 13.6

Example

16

2 Transpose each of the following to make the symbol in brackets the subject. n a v = u + at (a) b S = (a + l) (l) 2 1 c A = bh (b) d P = I2R (I ) 2 1 1 2 e s = ut + at (a) f E = mv 2 (v) 2 √ 2 (h) h −x y − z = x y + z (x) g Q = 2gh ax + by mx + b = x − b (x) i j =c (x) c x −b



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9C + 32 is used to convert temperatures given in degrees Celsius 3 The formula F = 5 (C ) to degrees Fahrenheit (F). a Convert 28 degrees Celsius to degrees Fahrenheit. b Transpose the formula to make C the subject and find C if F = 135◦ . 4 The sum (S ) of the interior angles of a polygon with n sides is given by the formula S = 180(n – 2). a Find the sum of the interior angles of an octagon. b Transpose the formula to make n the subject and hence determine the number of sides on a polygon whose interior angles add up to 1260◦ . 1 5 The volume (V ) of a right cone is given by the formula V = ␲r 2 h where r is the radius 3 of the base and h is the height of the cone. a Find the volume of a cone with radius 3.5 cm and height 9 cm. b Transpose the formula to make h the subject and hence find the height of a cone with base radius 4 cm and volume 210 cm3 . c Transpose the formula to make r the subject and hence find the radius of a cone with height 10 cm and volume 262 cm3 . n 6 The sum (S ) of a particular sequence of numbers is given by the formula S = (a + l), 2 where n is the number of terms in the sequence, a is the first term and l is the last term. a Find the sum of the sequence of seven numbers whose first term is –3 and whose last term is 22. b What is the first term of a sequence containing thirteen terms, whose last term is 156 and whose sum is 1040? c How many terms are there in the sequence 25 + 22 + 19 + · · · + −5 = 110?

2.7 Algebraic fractions The principles involved in addition, subtraction, multiplication and division of algebraic fractions are the same as for simple numerical fractions. To add or subtract, all fractions must be written with a common denominator. When multiplying, first try to simplify the fractions by cancelling down. This process will involve factorisation of either the numerators or denominators or both.

Addition and subtraction Example 17 Simplify x x + a 3 4 5 4 c − x +2 x −1

2 3a + x 4 7 4 − d x + 2 (x + 2)2 b



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Solution a x + x = 4x + 3x 3 4 12 7x = 12 c

45

b 2 + 3a = 8 + 3ax x 4 4x

5 4 5(x − 1) − 4(x + 2) − = x +2 x −1 (x + 2)(x − 1) 5x − 5 − 4x − 8 = (x + 2)(x − 1) x − 13 = (x + 2)(x − 1)

d

4 7 4(x + 2) − 7 − = x + 2 (x + 2)2 (x + 2)2 4x + 1 = (x + 2)2

Multiplication and division Before multiplying and dividing algebraic fractions, it is best to factorise numerators and denominators where possible so that common factors can be readily identified. Example 18 Simplify 5y 3x 2 × a 10y 2 12x 4x x2 − 1 × 2 c 2x − 2 x + 4x + 3

2x − 4 x 2 − 1 × x −1 x −2 2 x + 3x − 10 x 2 + 6x + 5 ÷ d x2 − x − 2 3x + 3 b

Solution x 3x 2 5y = a × 2 10y 12x 8y b

c

d

2(x − 2) (x − 1)(x + 1) 2x − 4 x 2 − 1 × = × x −1 x −2 x −1 x −2 = 2(x + 1) x2 − 1 4x (x − 1)(x + 1) 4x × = × 2x − 2 x 2 + 4x + 3 2(x − 1) (x + 1)(x + 3) 2x = x +3 (x + 5)(x − 2) 3(x + 1) x 2 + 3x − 10 x 2 + 6x + 5 ÷ = × 2 x −x −2 3x + 3 (x − 2)(x + 1) (x + 1)(x + 5) 3 = x +1

Example 19 √ 3x 3 + 3x 2 4 − x as a single fraction. Express √ 4−x ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Solution

√ √ √ 3x 3 + 3x 2 4 − x 4 − x 3x 3 2 + 3x 4 − x = √ √ 4−x 4−x 3x 3 + 3x 2 (4 − x) = √ 4−x 2 12x = √ 4−x

Example 20 4

1

Express (x − 4) 5 − (x − 4)− 5 as a single fraction. Solution 1

4

1

1

(x − 4) 5 − (x − 4)− 5 = (x − 4) 5 −

4

(x − 4) 5

1

= =

4

(x − 4) 5 (x − 4) 5 − 1 4

(x − 4) 5 x −5 4

(x − 4) 5

Exercise 2G Example

17

1 Simplify each of the following. 3x 2x + a 3 2 3h 5h 3h c + − 4 8 2 2 3 + e x y 2 3 + g x −2 x +1 4 3 i + x + 1 (x + 1)2 6x 2 − 4 k 2x − 5x 2 3 + m (x − 1) (x − 1)(x + 4) 5 3 2 o − 2 + x − 2 x + 5x + 6 x + 3 3 4x q − x −1 1−x

b d f h j

3a a − 2 4 y x 3x − − 4 6 3 2 5 + x −1 x 2x 4x 3 − − x +3 x −3 2 3a a−2 a + + a 4 8

3 2 − 2 x + 4 x + 8x + 16 3 2 4 n − + 2 x −2 x +2 x −4 1 p x−y− x−y 3 2x r + x −2 2−x l



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Chapter 2 — Algebra I Example

18

2 Simplify each of the following. 4y 3 x2 × a b 2y x 3x y x2 ÷ d e 2y 6 (x − 1)2 g h x 2 + 3x − 4 10a 5a 2 ÷ j k 12b2 6b 4x 2 2x ÷ 2 m n (x − 1) x − 1 6x 2 2 3x ÷ × o 9x − 6 x − 2 x + 5

y2 3x 2 × 4y 6x a2 4−x × 3a 4−x x2 − x − 6 x −3 x − 2 x2 − 4 ÷ x 2x 2 2 x − 9 3x + 6 9 × ÷ x +2 x −3 x

3 Express each of the following as a single fraction. 2 1 2 2 + a + b x −3 x −3 x −4 x −3 2x 2 2 1 d + + e 2 x −3 x +4 (x − 5) x −5 1 2 5 2 g − − h x −3 x −3 x −3 x +4 1 2 2 2x j − − k 2 3 (x − 5) x −5 (x − 6) (x − 6)2

12 4x 3 × 4 3 8x 2x + 5 f 4x 2 + 10x x 2 − 5x + 4 i x 2 − 4x 4x + 8 x +2 ÷ 2 l x(x − 3) x − 4x + 3

c

2 3 + x +4 x −3 2 3x + f (x − 4)2 x −4 3x 2x + i x −3 x +3 2x + 3 2x − 4 − l x −4 x −3

c

Example

19

4 Express each of the following as a single fraction. √ 2 2 2 a 1−x + √ + b √ 1−x x −4 3 2 3 √ 3 +√ c √ + x +4 d √ x +4 x +4 x +4 3 √ 3x √ 3x 3 − 3x 2 x + 4 e √ + 3x 2 x + 3 f √ x +4 2 x +3

Example

20

5 Simplify each of the following. 1

2

a (6x − 3) 3 − (6x − 3)− 3 1

1

47

2

b (2x + 3) 3 − 2x(2x + 3)− 3

2

c (3 − x) 3 − 2x (3 − x)− 3

2.8

Literal equations A literal equation in x is an equation whose solution will be expressed in terms of pronumerals rather than numbers. 2x + 5 = 7 is an equation whose solution is the number 1. In the literal c−b . equation ax + b = c, the solution is x = a Literal equations are solved in the same way as solving numerical equations or transposing formulas. Essentially, the literal equation is transposed to make x the subject.



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Example 21 Solve the following for x. a px − q = r

b ax + b = cx + d

c

a b = +c x 2x

Solution a

px − q = r px = r + q r +q x= p

b

ax + b = cx + d ax − cx = d − b x(a − c) = d − b d −b x= a−c

c

b a = +c x 2x Multiply both sides by lowest common denominator (2x) 2a = b + 2xc 2a − b = 2xc 2a − b =x 2c

Simultaneous literal equations Simultaneous literal equations are solved by the same methods that are used for solving simultaneous equations, i.e. substitution and elimination. Example 22 Solve each of the following pairs of simultaneous equations for x and y. a y = ax + c y = bx + d

b ax − y = c x + by = d

Solution a

ax + c = bx + d ∴ ax − bx = d − c x(a − b) = d − c d −c x= a−b and therefore d −c +c y=a a−b ad − bc = a−b

ax − y = c . . . 1 x + by = d . . . 2 Multiply 1 by b abx − by = cb . . . 1 Add 1 and 2

b

abx + x = cb + d x(ab + 1) = cb + d cb + d x= ab + 1 Substitute in 1 cb + d −y=c a ab + 1 cb + d −c ∴ y=a ab + 1 ad − c = ab + 1



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Exercise 2H Example

21

1 Solve each of the following for x. a ax + n = m

b ax + b = bx

d px = q x + 5

e mx + n = nx − m

g

2b b = x −a x +a

i −b(ax + b) = a(bx − a) x x −1= +2 k a b qx − t p − qx +p= m t p

ax +c =0 b b 1 = f x +a x

c

x x +n = +m m n j p 2 (1 − x) − 2 pq x = q 2 (1 + x) x 2x 1 l + = 2 a−b a+b a − b2 1 1 2 n + = x +a x + 2a x + 3a

h

2 For the simultaneous equations ax + by = p and bx – ay = q, show that x = y=

bp − aq . a 2 + b2

3 For the simultaneous equations Example

22

ap + bq and a 2 + b2

x y x y ab + = 1 and + = 1, show that x = y = . a b b a a+b

4 Solve each of the following pairs of simultaneous equations for x and y. a ax + y = c x + by = d c ax + by = t ax − by = s e (a + b)x + cy = bc (b + c)y + ax = −ab

b ax − by = a 2 bx − ay = b2 d ax + by = a 2 + 2ab − b2 bx + ay = a 2 + b2 f 3(x − a) − 2( y + a) = 5 − 4a 2(x + a) + 3( y − a) = 4a − 1

5 Write s in terms of a only in the following pairs of equations. a s = ah h = 2a + 1 c as = a + h h + ah = 1 e s = h 2 + ah h = 3a 2 g s = 2 + ah + h 2 1 h=a− a

b s = ah h = a(2 + h) d as = s + h ah = a + h f as = a + 2h h =a−s h 3s − ah = a 2 as + 2h = 3a



50


Using the TI-Nspire with algebra In this section a demonstration of the basic algebra properties of the TI-Nspire is provided. To access these, open a Calculator application ( >New Document>Add Calculator) and select b >Algebra. The three main commands are solve, factor and expand.

1: Solve( ) This command is used to solve equations, simultaneous equations and some inequalities. An approximate (decimal) answer can be obtained by pressing /+· or by including a decimal number in the expression. The following screens illustrate its use.

2: Factor( ) This command is used for factorisation. Factorisation over the rational numbers is obtained by not specifying the variable, whereas factorisation over the real numbers is obtained by specifying the variable. The following screens illustrate its use.




51

3: Expand( ) This command is used for expansion of expressions. By specifying the variable, the expanded expression will be ordered in decreasing powers of that variable. Symbolic expressions can only be expanded for an appropriate domain. The following screens illustrate its use.

Using the Casio ClassPad with algebra Using the Main menu In this section the Main menu is explored. Work through it to become acquainted with the menu. There are two ways to operate in this menu, but the Interactive menu is the simplest when used in conjunction with the stylus. In this section we shall demonstrate some examples of how this is used.

Solve This is used to solve equation and inequalities. The variables x, y and z are found on the hard keyboard. Other variables may be entered using the k and selecting VAR. Variables are shown in bold keyboard allows you to type italics. The sentences, etc; however, the letters are not always recognised as variables. keyboard, you must type a × x, for example, because If you choose to use the ax will be treated as text.



52


For example: Enter ax + b = 0, highlight it with the stylus, tap Interactive, Equation/inequality, solve and ensure the variable selected is x. Solution b returned is x = − a Enter x 2 + x − 1 = 0 and follow the same instructions as above. The answer is as shown. Note: x 2 + x − 1 = 0 is entered, but the calculator converts this to standard algebraic notation when EXE is pressed. Also note in this example that “= 0” has been omitted. If the right-hand side of an equation is zero, it is unnecessary to enter it. Enter abt − w + t = wt for w: follow the instructions above and select w as the variable. Solve x 3 − x 2 − x + 1 = 0 for x. √ Solve 2x + 2 < 3 for x. See screen at right for solutions returned. is found in when k is . If the activated and is in answer is not in the form required, it is often possible to cut and paste it in the next entry line and use Interactive, Transformation, Simplify as shown here. Solve simultaneous equations uses a dedicated entry style. Note carefully the screen at right. in the window and To enter, tap enter the equations and variables as shown. until the For more than two equations, tap number of equations required is given. Note:

Factor To factorise is to transform the expression to a different form. This command is found in Interactive, Transformation, factor. Examples: To factorise x 3 − 2x over the rational numbers, use factor. To factorise over the real numbers, select rfactor.




53

Some further examples are provided here. See the screens for the results. Factor a 2 − b2 Factor a3 − b3 1 2 + Factor + 1 x − 1 (x − 1)2 Factor (2x 4 − x 2 ) over the rationals Factor (2x 4 − x 2 ) over the reals. This command can also be used to give the prime decomposition (factors) of integers.

Expand To expand expressions, use Interactive, Transformation, expand. For example: Expand (a + b)3 Expand (a + b)2 The expand command can also be used to form partial fractions. In this case, enter and highlight the expression, select Interactive, transformation, expand, select the Partial fraction option and set the variable as x.

For example: 1 x2 − 1 x 3 + 2x + 1 Expand x2 − 1 Note: The top screen shows all the examples, the bottom screen shows how to enter for partial fractions. Expand



54


Zeros To find the zeros of an expression in the menu, use Interactive, Equation/inequality, Solve and ensure you set the variable. The calculator assumes you are solving an equation for which one side is zero. For example: Zeros of x 2 − 1 for x Zeros of x 2 − y 2 for y Zeros of x 2 − y 2 for x Zeros of x 2 − y for y Zeros of x 2 − 4x + 8 for x. No solutions. Zeros of x 2 − 4x + 1 for x. Two solutions. Zeros of x 2 − 4x + 4 for x. One solution.

Approximate Switch mode in the status bar to Decimal. If an answer is given in Standard (Exact) mode, it can be converted by highlighting the in the toolbar. answer and tapping

Combining fractions This command gives an expression with a common denominator and ratio form. The denominator is returned in factored form. For example: Enter and highlight 1/(x − 1) + 1/(x + 1) then select Interactive, Transformation, combine. Enter and highlight y/(x − y) + y/(x + y) then select Interactive, Transformation, combine.




55

Solve numerically There are several ways to find numerical solutions to equations. In each of these ways, only one solution is given. You can vary the guess (Value) or the bounds of the search (Lower and Upper) to find particular solutions. If an expression such as x 2 − x − 2.1 is entered without an equals sign, the calculator will assume the expression is equal to zero and solve the equation. For example: Enter and highlight x 2 − x − 2.1 = 0, select Interactive, Equation/inequality, Solve and tap the solve numerically option. Note that the lower and upper bounds are set to ∞ and a guess of –1 has been entered to return the first solution –1.03297. For the second solution shown, the first line has been copied and pasted (or dragged) to the next entry line and the guess x = 2 has been entered to return the solution x = 2.03297. Alternatively, use from the Main menu and enter Lower and Upper bounds. A guess may also be entered, but is not necessary. (Note that the bounds selected arise from a quick sketch which indicated that the quadratic equation had a solution in the negative and one in the positive domain.) A third method involves using the Graph program and finding the solution using the G-Solve application.



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Exercise 2I This exercise is to provide practice in some of the skills associated with a CAS calculator. Other exercises in this chapter can also be attempted with CAS but it is recommended that you also use this chapter to develop your ‘by hand’ skills. 1 Solve each of the following equations for x. b(b + x) a(a − x) − =x a b a b 2(x − 3) + (x − 2)(x − 4) = x(x + 1) − 33 x +a x +c x +a x + =2 d c =1− x −c x −a x +b x −b 2 Factorise each of the following. a x 2 y2 − x 2 − y2 + 1 c a 4 − 8a 2 b − 48b2

b x 3 − 2 − x + 2x 2 d a 2 + 2bc − (c2 + 2ab)

3 Solve each of the following pairs of simultaneous equations for x and y. a ax y + b = (a + c)y and bx y + a = (b + c)y b x(b − c) + by − c = 0 and y(c − a) − ax + c = 0



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A number is expressed in standard form or scientific notation when written as a product of a number between 1 and 10 and a power of ten; e.g. 1.5 × 108 Index laws r a m × a n = a m+n r a m ÷ a n = a m−n r (a m )n = a mn 1 n r √ r (ab)n = a n bn r a◦ = 1 r a −n = 1 a = an an Linear equations First identify the steps done to construct an equation; the equation is then solved by ‘undoing’ these steps. This is achieved by doing ‘the opposite’ in ‘reverse order’. e.g. Solve for x, 3x + 4 = 16 x has been multiplied by 3 and then 4 has been added Subtract 4 from both sides 3x = 12 Divide both sides by 3 x =4 A formula is an equation that states a relationship between two or more quantities, e.g. the area of a circle A = ␲r 2 . The value of A, the subject of the formula, may be found by substituting a given value of r and the value of ␲. A formula can be transposed to make a different pronumeral the subject using a similar procedure to solving linear equations, i.e. whatever has been done to the pronumeral required is ‘undone’. Literal equations Literal equations are solved using the same techniques as for solving numerical equations or transposing formulas, i.e. by transposing the literal equation to make the required pronumeral the subject.

Review

Chapter summary

57

Multiple-choice questions y 1 For non-zero values of x and y, if 5x + 2y = 0 then the ratio is equal to x 2 2 5 D 1 E C B − A − 5 5 2 2 The solution of the simultaneous equations 3x + 2y = 36 and 3x − y = 12 is 20 A x= y=8 B x =2 y=0 C x = 1 y = −3 3 3 3 20 y=− E x= y=6 D x= 2 2 3 3 The solution of the equation t − 9 = 3t − 17 is 11 A t = −4 B t= C t=4 D t=2 E 2 n−p 4 For m = ,p= n+p n(m − 1) n(1 − m) n(1 + m) n(1 + m) B A E D C 1+m 1+m m−1 1−m


5 4

t = −2

m(n − 1) m+1


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5

2 3 − = x −3 x +3

1 x +3 15 x + 15 E − D C 2 2 9 x −9 x −9 x −9 3 2 3 9x y ÷ (15(x y) ) is equal to 3 3x 3y 18x y 9x E D C B A 5x 5 5x 5 15 1 For the formula V = h(l + w) 3 3V hw 3V − 2w −w B l= A l= C l= h 3V h 1 3V h − w E l = h(V + w) D l= 3 2 (3x 2 y 3 )2 = 2x 2 y 9 6 7 9 6 6 9 6 7 9 2 5 9 2 7 x y E x y D x y C x y B A x y 2 2 2 2 2 If X is 50% greater than Y and Y is 20% less than Z, then A X is 30% greater than Z B X is 20% greater than Z C X is 20% less than Z D X is 10% less than Z E X is 10% greater than Z

A 1 6

7

8

9

B

10 The average of two numbers is 5x + 4. One of the numbers is x. The other number is A 4x + 4 B 9x + 8 C 9x + 4 D 10x + 8 E 3x + 1

Short-answer questions (technology-free) 1 Simplify the following. 3 a (x 3 )4 b ( y −12 ) 4

3

4

d (x 3 ) 3 × x −5

c 3x 2 × −5x 4

2 Express the product 32 × 1011 × 12 × 10−5 in standard form. 3 Simplify the following. 3x y 2x a + − 5 10 5 4 3 + d x +2 x +4 4 Simplify the following. x 2 + 5x x +5 ÷ a 2x − 6 4x − 12

7 4 − x y 4x 5 5x + − e x +4 x −2 2

b

2 5 + x +2 x −1 6 3 − f x − 2 (x − 2)2

c

12x 2 3x ÷ 2 x + 4 x − 16 2 6x 2 2 4x + 20 9 x − 4 3x − 9 × ÷ d × ÷ c 9x − 6 x + 5 3x − 2 x −3 x +2 x +2 5 The human body can produce 2.5 million blood cells per second. If a person donates 500 mL of blood, how long will it take to replace the red blood cells if a litre of blood contains 5 × 1012 red blood cells? b



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7 Swifts Creek Soccer Team has played 54 matches over the past three seasons. They have drawn one third of their games and won twice as many games as they have lost. How many games have they lost? 8 A music store specialises in three types of CDs: classical, blues and heavy metal. In one week they sold a total of 420 CDs. They sold 10% more classical than blues while sales of heavy metal constituted 50% more than the combined sales of classical and blues CDs. How many of each type of CD did they sell?

Review

6 The Sun is approximately 1.5 × 108 km from Earth and a comet is approximately 3 × 106 km from Earth. How many times further from Earth than the comet is the Sun?

59

9 The volume (V) of a cylinder is given by the formula V = ␲r2 h, where r is the radius of the base and h is the height of the cylinder. a Find the volume of a cylinder with base radius 5 cm and height 12 cm. b Transpose the formula to make h the subject and hence find the height of a cylinder with a base radius 5 cm and a volume of 585 cm3 . c Transpose the formula to make r the subject and hence find the radius of a cylinder with a height of 6 cm and a volume of 768 cm3 . 10 Solve for x. a x y + ax = b

b a + =c x x a − dx ax + d d +b = d b b

x x = +2 a b 11 Simplify p q a + p+q p−q x 2 + x − 6 2x 2 + x − 1 × c x +1 x +3 c

2y 1 − x x y − y2 2ab + b2 2a × d 2a + b ba 2 12 A is three times as old as B. In three years’ time, B will be three times as old as C. In fifteen years’ time, A will be three times as old as C. What are their present ages? b

13 a Solve the following simultaneous equations for a and b. 1 1 a − 5 = (b + 3) b − 12 = (4a − 2) 7 5 b Solve the following simultaneous equations for x and y. ( p − q)x + ( p + q)y = ( p + q)2 q x − py = q 2 − pq 14 A man has to travel 50 km in 4 hours. He does it by walking the first 7 km at x km/h, cycling the next 7 km at 4x km/h and motoring the remainder at (6x + 3) km/h. Find x. 15 Simplify each of the following. a 2n × 6nk ÷ (3n) 2

2

16 Solve the equation

1 xy 8c2 x 3 y 2 ÷ b 6a 2 b3 c3 15abc2

x +5 x −5 2x − =1+ 15 10 15



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Extended-response questions 1 Jack cycles home from work, a distance of 10x km. Benny leaves at the same time and drives the 40x km to his home. a Write an expression in terms of x for the time taken for Jack to reach home if he cycles at 8 km/h on average. b Write an expression in terms of x for the time taken for Benny to reach home if he drives at an average speed of 70 km/h. c In terms of x, find the difference in times of the two journeys. d If Jack and Benny arrive at their homes 30 minutes apart i find x, correct to three decimal places ii find the distance from work of each home, correct to the nearest kilometre. 2 Sam’s plastic dinghy has sprung a leak and water is pouring in the hole at a rate of 27 000 cm3 per minute. He grabs a cup and frantically starts bailing the water out at a rate of 9000 cm3 per minute. The dinghy is shaped like a circular prism (cylinder) with a base radius of 40 cm and a height of 30 cm. a How fast is the dinghy filling with water? b Write an equation showing the volume of water, V cm3 , in the dinghy after t minutes. c Find an expression for the depth of water, h cm, in the dinghy after t minutes. d If Sam is rescued after nine minutes, is this before or after the dinghy has completely filled with water? 3 Henry and Thomas Wong collect basketball cards. Henry has five sixths the number of cards that Thomas has. The Wright family also has a collection of cards. George Wright has half as many cards again as Thomas, Sally Wright has 18 less than Thomas and Zeb Wright has one third the number Thomas has. a Write an expression for each child’s number of cards in terms of the number Thomas has. b The Wright family owns six more cards than the Wong family. Write an equation representing this information. c Solve the above equation and use the result to find the number of cards each child has collected. 4 The gravitational force between two objects, F(N), is given by the formula 6.67 × 10−11 m 1 m 2 r2 where m1 and m2 are the mass (in kilograms) of each object and r (in metres) is the distance between them. a What would be the gravitational force between two objects each weighing 200 kg if they are 12 m apart? Express the answer in standard form (to two significant figures). b Transpose the above formula to make m1 the subject. c The gravitational force between a planet and an object 6.4 × 106 m away from the centre of the planet is found to be 2.4 × 104 N. If the object has a mass of 1500 kg, F=



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5 A water storage reservoir is 3 km wide, 6 km long and 30 m deep. (The water storage reservoir is assumed to be a cuboid.) a Write an equation to show the volume of water, V m3 , in the reservoir when it is d metres full. b Calculate the volume of water in the reservoir when it is completely filled. The water flows from the reservoir down a long pipe to a hydro electric power station in a valley below. The amount of energy, E(J), that can be obtained from a full reservoir is given by the formula

Review

calculate the approximate mass of the planet, giving the answer in standard form (two significant figures).

61

E = kVh where k is a constant and h (m) is the length of the pipe. c Find k given E = 1.06 × 1015 when h = 200, expressing the answer in standard form correct to three significant figures. d How much energy could be obtained from a full reservoir if the pipe was 250 m long? e If the rate of water falling through the pipe is 5.2 m3 /s, how many days without rain could the station operate before emptying an initially full reservoir? 6 A new advertising symbol is to consist of three concentric circles as shown, with the outer circle having a radius of 10 cm. It is desired that the three coloured regions cover the same area. Find the radius of the innermost circle in the figure shown.

Yellow

7 Temperatures of Fahrenheit (F ) can be converted to Celsius (C ) by the formula

Blue Red

5 (F − 32) 9 Find the temperature which has the same numerical value in both scales. C=

8 A cyclist goes up a long slope at a constant speed of 15 km/h. He turns around and comes down the slope at constant speed of 40 km/h. Find his average speed over a full circuit. 9 A container has a cylindrical base and a hemispherical top as shown in the figure at right. The height of the container is 20 cm and its capacity is to be exactly 2 litres. Let r cm be the radius length of the base. a Express the height of the cylinder in terms of r. b i Express the volume of the container in terms of r. ii Find r and h if the volume is two litres.

20 cm

r cm



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10 a Two bottles contain mixtures of wine and water. In bottle A there is two times as much wine as water. In bottle B there is five times as much water as wine. Bottle A and bottle B are used to fill a third bottle which has a capacity of one litre. How much liquid must be taken from each of bottle A and bottle B if the third bottle is to contain equal amounts of wine and water? b Repeat for the situation where the ratio of wine to water in bottle A is 1 : 2 and for bottle B the ratio is to 3 : 1. c Generalise the result for the ratio m : n for bottle A and p : q for bottle B. 11 A cylinder is placed so as to fit into a cone as shown. The height of the cone is 20 cm and the radius 10 cm. The radius of the cylinder is r cm and the height h cm.

20 cm h cm

r cm

a Use similar triangles to find h in terms of r.

r cm

20 cm h cm

10 cm

b The volume of the cylinder is given by the formula V = ␲r h. Find the volume of the cylinder in terms of r. c Use a graphics calculator to find the values of r and h for which the volume of the cylinder is 500 cm3 . 2



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C H A P T E R

3 Number systems and sets Objectives To understand and use the notation of sets including the symbols ∈, ⊆, ∩, ∪, Ø, ␰ To be able to identify sets of numbers including natural numbers, integers, rational numbers, irrational numbers, real numbers To know and be able to apply the rules for:

r simplification of surds r operations on surds r rationalisation of surds To know the definition of factor, prime, highest common factor To be able to solve problems with sets

Introduction This chapter introduces set notation and discusses sets of numbers and their properties. Set notation is used widely in mathematics and in this book it is employed where appropriate. A set is a general name for any collection of things or numbers. There must be a way of deciding whether any particular object is a member of the set or not. This may be done by referring to a list of the members of the set or a statement describing them. For example, A = {−3, 3} = {x : x 2 = 9} Note: {x: . . .} is read as ‘the set of objects x such that . . .’.

Number systems Recall that the elements of {1, 2, 3, 4, . . .} are called natural numbers and the elements {. . . , −2, −1, 0, 1, 2, . . .} are called integers. p The numbers of the form with p and q integers, q = 0 , are called rational numbers. q



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The real numbers which are not rational are called irrational, e.g. ␲ and The set of real numbers will be denoted by R. The set of rational numbers will be denoted by Q. The set of integers will be denoted by Z. The set of natural numbers will be denoted by N.

√

2.

R 2 = {(x, y) : x, y ∈ R} These sets of numbers will be discussed further in Sections 3.2 and 3.3.

3.1 Set notation The symbol ∈ means ‘is a member of’ or ‘is an element of’. For example, 3 ∈ {prime numbers} is read ‘3 is a member of the set of prime numbers’. The symbol ∈ / means ‘is not a member of’ or ‘is not an element of’. For example, 4∈ / {prime numbers} is read ‘4 is not a member of the set of prime numbers’. Two sets are equal if they contain exactly the same elements, not necessarily in the same order. For example, if A = {prime numbers less than 10} and B = {2, 3, 5, 7} then A = B. Two sets A and B are equivalent if they have the same number of elements. For example, {1, 2, 3} ⇔ {a, b, c}. A set which has no elements is called the empty or null set and is denoted by Ø. The universal set will be denoted by ␰. The universal set is the set of all elements which are being considered. If all the elements of a set B are also members of a set A, then the set B is called a subset of A. This is written B ⊆ A. For example, {1, 2, 3} ⊆ Z where Z is the set of integers, and {3, 9, 27) ⊆ {multiples of 3}. We note also A ⊆ A and Ø ⊆ A. Venn diagrams are used to illustrate sets. For example, let ␰ denote the set of all real numbers less than 100, A denote the set of real numbers less than 50 and B the set of real numbers between 90 and 100 (non-inclusive). This may be illustrated by a Venn diagram. ξ A B A and B have no elements in common. Two such sets are called disjoint sets.

The union of two sets The set of elements that are in either set A or set B (or both) is the union of sets A and B. The union of A and B is written A ∪ B.



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Chapter 3 — Number systems and sets

Example 1 Let ␰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3} and B = {1, 2, 9, 10}. Find A ∪ B and illustrate on a Venn diagram. Solution A ∪ B = {1, 2, 3, 9, 10}

ξ

5

4

The shaded area illustrates A ∪ B.

A

6

7

1 2

3

8 9 10

B

The intersection of two sets The set of all the elements that are members both of set A and of set B is called the intersection of A and B. The intersection of A and B is written A ∩ B. Example 2 Let ␰ = {prime numbers less than 40}. If A = {3, 5, 7, 11} and B = {3, 7, 29, 37}, find A ∩ B. Solution A ∩ B = {3, 7}

ξ

2

17

A

5 11

19

23 3 7

31 29 37

13 B

Complement If ␰ = {students at Highland Secondary College} and A = {students with blue eyes}, then the complement of A is the set of all members of ␰ that are not members of A. In this case the complement is the set of all students of Highland Secondary College that do not have blue eyes. The complement of A is denoted by A . Similarly, if ␰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9} then A = {2, 4, 6, 8, 10}.

Finite and infinite sets When all the elements of a set may be counted the set is called a finite set, e.g. A = {months of the year}. The number of elements of a set A will be denoted n(A). In this example n(A) = 12. For the set C = {letters of the alphabet}, n(C) = 26. Sets which are not finite are infinite sets. For example, R, the set of real numbers, and Z, the set of integers, are infinite sets.



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Example 3 Given ␰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {odd numbers} = {1, 3, 5, 7, 9} B = {multiples of 3} = {3, 6, 9} show these sets on a Venn diagram. Use the diagram to list the following sets. b B a A d the complement of A ∪ B i.e. (A ∪ B)

c A∪B e A ∩ B ξ

Solution

2

(Each of the numbers in the given sets is placed in the correct position on this Venn diagram.)

A

From the diagram: a A = {2, 4, 6, 8, 10} b B = {1, 2, 4, 5, 7, 8, 10} d (A ∪ B) = {2, 4, 8, 10}

5

4

8

10

1

3 9

6

7

B

c A ∪ B = {1, 3, 5, 6, 7, 9} e A ∩ B = {2, 4, 8, 10}

Exercise 3A Example

1

1 ␰ = {1, 2, 3, 4, 5}, A = {1, 2, 3, 5}, B = {2, 4}. Show these sets on a Venn diagram and use the diagram to find a A

Example

2

3

c A∪B

d (A ∪ B)

e A ∩ B

2 ␰ = {natural numbers less than 17}, P = {multiples of 3}, Q = {evennumbers} Show these sets on a Venn diagram and use it to find a P

Example

b B

b Q

c P∪Q

d (P ∪ Q)

e P ∩ Q

3 ␰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, A = {multiples of 4}, B = {even numbers} Show these sets on a Venn diagram and use this diagram to list the sets a A

b B

c A∪B

d (A ∪ B)

e A ∩ B

4 ␰ = {natural numbers from 10 to 25}, P = {multiples of 4}, Q = {multiples of 5} Show these sets on a Venn diagram and use this diagram to list the sets a P 5

b Q

c P∪Q

d (P ∪ Q)

e P ∩ Q

␰ = {different letters in the word GENERAL}, A = {different letters in the word ANGEL}, B = {different letters in the word LEAN} Show these sets on a Venn diagram and use this diagram to list the sets a A

b B

c A∩B

d A∪B


e (A ∪ B)

f A ∪ B


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6 ␰ = { p, q, r, s, t, u, v, w}, X = {r, s, t, w}, Y = {q, s, t, u, v} Show ␰, X and Y on a Venn diagram, entering all members. Hence list the sets a X b Y c X ∩ Y Which two sets are equal? 7

d X ∪ Y

f (X ∪ Y )

␰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, X = {factors of 12}, Y = {even numbers} Show ␰, X and Y on a Venn diagram entering all members. Hence list the sets b Y c X ∪ Y a X Which two sets are equal?

d X ∩ Y

a A d A ∪ B

b B e A∪B

e X ∪Y

f (X ∪ Y )

ξ

8 Draw this diagram six times. Use shading to illustrate each of the following sets.

9

e X ∪Y

c A ∩ B f (A ∪ B)

A

B

␰ = {different letters in the word MATHEMATICS} A = {different letters in the word ATTIC} B = {different letters in the word TASTE} Show ␰, A and B on a Venn diagram entering all the elements. Hence list the sets a A

3.2

b B

c A∩B

d (A ∪ B)

e A ∪ B

f A ∩ B

Sets of numbers The following notation was introduced earlier in this chapter. R denotes the set of real numbers. Q denotes the set of rational numbers. Z denotes the set of integers. N denotes the set of natural numbers. A geometric construction of a line segment of m where m and n are non-zero integers is length R n Q Z shown in Chapter 9. Constructions of products N and quotients are also shown in that chapter. It is clear that N ⊆ Z ⊆ Q ⊆ R and this may be represented by the diagram shown. The set of all x such that (. . .) is denoted by {x : (. . .)}, where (. . .) stands for some condition. Thus {x : 0 < x < 1} is the set of all real numbers between 0 and 1. {x : x > 0, x ∈ Q} is the set of all positive rational numbers. {2n : n = 1, 2, 3, . . .} is the set of all even numbers.



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√ √ √ √ Examples of irrational numbers are 3, 2, ␲, ␲ + 2, 7 + 6. These numbers cannot be p written in the form where p and q are integers. The decimal representation of these numbers q does not terminate or repeat.

Rational numbers Every rational number can be expressed as a terminating or recurring decimal. For example 1 1 1 1 ˙ 1 = 0.14285 ˙ = 0.5, = 0.2, = 0.1, = 0.3, 7˙ 2 5 10 3 7 m Numbers of the form , where m and n ∈ N have a terminating decimal representation if and n only if n = 2␣ × 5␤ , where ␣ and ␤ are members of the set N ∪ {0}. m In order to find the decimal representation of a rational number , the division m ÷ n is n undertaken. For example 1 0.05 a 1 = 0.05 Therefore 20 20 1.00 20 b 3 7

0.428571428 . . . 7 3.000000000 . . .

Therefore

3 ˙ = 0.42857 1˙ 7

m The method to find a rational number from its decimal representation is demonstrated in the n following example. Example 4 m Write each of the following in the form . n ˙ a 0.05 b 0.42857 1˙ Solution 1 5 = 100 20 ˙ b 0.42857 1˙ = 0.428571428571 . . . Multiply both sides by 106 ˙ 0.42857 1˙ × 106 = 428571.428571428571 . . . a 0.05 =

... 1 ... 2

Subtract 1 from 2 ˙ 0.42857 1˙ × (106 − 1) = 428571 428571 ˙ ∴ 0.42857 1˙ = 6 10 − 1 3 = 7

Irrational numbers The set of irrational numbers has two important subsets, algebraic numbers and transcendental numbers. Algebraic numbers are those which are the solutions of an equation of the form a0 x n + a1 x n−1 + · · · + an = 0, where a0 , a1 , . . . . , an are integers. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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For example,

69

√ 2 is an algebraic number, as it is a solution of the equation

x2 − 2 = 0 ␲ is not an algebraic number, it is a transcendental number. √ The proof that 2 is irrational is presented here. The proof is by contradiction. √ a Assume 2 = , where a, b ∈ N b a and is a fraction in simplest form. b a2 Then 2 = 2 b

∴ ∴ ∴ ∴ ∴

a 2 = 2b2 which implies a 2 is even which implies ais even a = 2k for some k ∈ N a 2 = 4k 2 4k 2 = 2b2 b2 = 2k 2 which implies b2 is even which implies b iseven

a But this contradicts the assumption that is a fraction in simplest form, as a and b are both b divisible by 2.

Exercise 3B 1 Is the a sum b product c quotient (if defined) of two rational numbers rational? 2 Is the a sum b product c quotient of two irrational numbers irrational? Example

4

3 Write each of the following in the form a 0.2˙ 7˙

b 0.12

˙ c 0.28571 4˙

m where m and n are integers. n d 0.3˙ 6˙ e 0.2˙ f 0.45

4 Give the decimal representation of each of the following rational numbers. 2 5 7 4 1 a b c d e 7 11 20 13 17 √ 5 Prove that 3 is not a rational number.

3.3

Surds

√ A number of the form a where a is a rational number which is not a square of another rational number is called a quadratic surd. √ Note: a is taken to mean the positive square root.



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√ If a is a rational which is not a perfect nth power, n a is called a surd of the nth order. √ √ √ 9 1 , are quadratic surds 2, 7, 24, 7 2 √ √ 9 are not surds 9, 16, 4 √ √ 3 3 7, 15 are surds of the third order √ √ 4 4 100, 26 are surds of the fourth order Quadratic surds hold a prominent position in school mathematics. For example, the solutions of quadratic equations often involve surds, e.g. 1 1√ 5 is a solution of thequadratic equation x 2 − x − 1 = 0 x= + 2 2 Values of trigonometric functions sometimes involve surds, e.g. √ √ √ 3 6− 2 ◦ ◦ , sin 15 = sin 60 = 2 4 In Mathematical Methods Units 3 and 4 and Specialist Mathematics Units 3 and 4, exact solutions are often required. √ √ √ Lengths such as 2, 3 or 6 can be constructed geometrically, using a straight edge and a compass. For example, from the right-angled isosceles triangle ABC:

B

B √2

1

C

√3

1

A

1

The length AB =

√ 2

A

√2

C

The length AB =

√ 3

Any quadratic surd (of a natural number) may be constructed in this way. This makes it possible to construct a line segment of the length determined by the solutions of many quadratic equations. 1 1√ 5. For example, one solution of x 2 − x − 1 = 0 is x = + 2 2 The construction of a line segment of this length involves the right angled triangle XYZ. X

1

Y

√5

2

Z

Y

1

X

√5

Z

The length is now achieved by bisecting the line segment. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Properties of square roots The following properties of square roots are often used. √ √ √ √ √ √ √ ab = a b, e.g. 50 = 25 × 2 = 5 2 √ √ √ a a 7 7 7 = √ , e.g. =√ = b 9 3 b 9

Properties of surds Simplest form If possible, a factor is taken ‘out of a square root’. When the number under the square root has no factors which are squares of a rational number, then the surd is said to be in simplest form. Example 5 Write each of the following in simplest form. √ √ 99 700 d c a 72 b 28 64 117 Solution √ √ √ √ √ √ √ a 72 = 36 × 2 = 6 2 b 28 = 4 × 7 = 2 7 √ √ 10 7 700 700 7 × 100 = =√ = √ c 117 3 13 9 × 13 117 √ √ √ 99 99 9 × 11 3 11 d =√ = = 64 8 8 64 Surds which have the same ‘irrational factor’ are called like surds. √ √ √ For example, 3 7, 2 7 and 7 are like surds. The sum or difference of two like surds can be found. √ √ √ √ √ √ i.e. m p + n p = (m + n) p and m p − n p = (m − n) p Example 6 Express each of the following as a single surd in simplest form. √ √ √ √ √ √ √ √ √ 147 + 108 − 363 b 3 + 5 + 20 + 27 − 45 − 48 a √ √ √ √ 1 1 1 d 50 + 2 − 2 18 + 8 − −5 c 8 18 72 Solution √ √ √ a 147 + 108 − 363 = 72 × 3 + 62 × 3 − 112 × 3 √ √ √ = 7 3 + 6 3 − 11 3 √ =2 3



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√ √ √ √ √ 3 + 5 + 20 + 27 − 45 − 48 √ √ √ √ √ √ = 3+ 5+2 5+3 3−3 5−4 3 √ √ =0 3+0 5 =0 √ √ √ √ 1 1 1 d 50 + 2 − 2 18 + 8 c − −5 √ √ √ √ 8 18 72 =5 2+ 2−2×3 2+2 2 √ √ 1 1 1 =8 2−6 2 = − −5 √ 4×2 9×2 36 × 2 =2 2 1 1 1 1 5 1 − − = 2 2 3 2 6 2 3 1 2 1 5 1 − − = 6 2 6 2 6 2 −4 1 = 6 2 −2 1 = 3 2

b

√

Rationalising the denominator In the past, a labour saving procedure with surds was to rationalise any surds in the denominator of an expression. It is still considered to be a neat way of expressing final answers. √ √ √ √ For 5 a rationalising factor is 5 as 5 × 5 = 5 √ √ √ √ For 2 + 1 a rationalising factor is 1 − 2 as (1 + 2)(1 − 2) = 1 − 2 = −1 √ √ √ √ √ √ √ √ For 3 + 6 a rationalising factor is 3 − 6 as ( 3 + 6) ( 3 − 6) = 3 − 6 = −3

Example 7 Rationalise the denominator of each of the following. 1 1 1 c √ √ b √ a √ 3− 6 2− 3 2 7

d

3+ 3−

√ √

8 8

Solution √ √ 7 7 1 a √ ×√ = 14 2 7 7 √ √ √ 1 2+ 3 2+ 3 b =2+ 3 √ × √ = 4−3 2− 3 2+ 3 √ √ √ √ √ 1 3+ 6 3+ 6 −1 √ c √ = ( 3 + 6) √ ×√ √ = 3−6 3 3− 6 3+ 6 √ √ √ √ √ 3+ 8 3+2 2 3+2 2 9 + 12 2 + 8 d = 17 + 12 2 √ = √ × √ = 9−8 3− 8 3−2 2 3+2 2




73

Example 8 Expand the brackets in each of the following and collect like terms, expressing surds in simplest form. √ √ √ b (3 − 2)(1 + 2) a (3 − 2)2 Solution √ a (3 − 2)2 √ √ = (3 − 2)(3 − 2) √ √ √ = 3(3 − 2) − 2(3 − 2) √ √ =9−3 2−3 2+2 √ = 11 − 6 2

√ √ b (3 − 2)(1 + 2) √ √ √ = 3(1 + 2) − 2(1 + 2) √ √ =3+3 2− 2−2 √ =1+2 2

Using the TI-Nspire A CAS calculator can be used to work with irrational numbers. Expressions can be reached and selected using the up arrow (£). This returns the expression to the entry line and modifications can be made. 23 · 22 8 · 2 5 as To illustrate this, evaluate 5 shown.

To find the square root of this expression, type / q and move upwards by pressing the up arrow (£) so that the expression is highlighted.

Press enter (·) to paste this expression into the square root sign and press enter once more to evaluate the square root of this expression.



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Exercise 3C Example

Example

5

6

Example

7

Example

8

1 Express each of the following in terms of the simplest possible surds. √ √ √ √ √ a 8 b 12 c 27 d 50 e 45 √ √ √ √ √ 98 h 108 i 25 j 75 k 512 g

f

√

1210

2 Simplify each of the following. √ √ √ a 8 + 18 − 2 2 √ √ √ 28 + 175 − 63 c √ √ √ 512 + 128 + 32 e

√ √ √ b 75 + 2 12 − 27 √ √ √ d 1000 − 40 − 90 √ √ √ √ f 24 − 3 6 − 216 + 294

3 Simplify each of the following. √ √ √ a 75 + 108 + 14 √ √ √ 720 − 245 − 125 c √ √ √ √ √ 12 + 8 + 18 + 27 + 300 e

√ √ √ b 847 − 567 + 63 √ √ √ √ d 338 − 288 + 363 − 300 √ √ √ √ √ f 2 18 + 3 5 − 50 + 20 − 80

4 Express each of the following with rational denominators. 1 1 1 2 3 a √ b √ c −√ d √ e √ 5 7 2 3 6 1 1 1 1 2 f √ g √ h √ i √ j √ 2 2 2+1 2− 3 4 − 10 6+2 1 3 1 k √ √ l √ √ m √ 5− 3 6− 5 3−2 2 √ 5 Express each of the following in the form a + b c. √ √ √ √ 2 b ( 5 + 2)2 c (1 + 2)(3 − 2 2) d ( 3 − 1)2 a √ 3−2 2 √ √ √ 5+1 8+3 1 1 3+2 g √ h √ −√ e f √ 5−1 3 18 + 2 27 2 3−1 6 Expand and simplify each of the following. √ √ √ b ( x + 1 + x + 2)2 a (2 a − 1)2 7 For real numbers a and b, a > b if and only if a − b > 0. Use this to state the larger of √ √ √ √ a 5 − 3 2 and 6 2 − 8 b 2 6 − 3 or 7 − 2 6

3.4 Natural numbers

Factors and composites The factors of 8 are 1, 2, 4 and 8. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. The factors of 5 are 1 and 5.



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A natural number, a, is a factor of a natural number, b, if there exists a natural number, k, such that b = ak. If a number greater than 1 has only factors 1 and itself, it is said to be prime. Among the first 100 numbers, the following are prime: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 A number, m, is called a composite if it can be written as a product, m = a × b where a and b are numbers greater than 1 and less than m.

Prime decomposition 3000 = 3 × 53 × 23 2294 = 2 × 31 × 37 This method of uniquely expressing a composite in terms of a product of powers of prime numbers is called decomposition. It is useful for finding factors of numbers. For example, the prime decomposition of 12 is given by 12 = 22 × 3 The factors of 12 are 1, 2, 22 , 3, 2 × 3, 22 × 3, i.e. 1, 2, 4, 3, 6, 12. Example 9 Give the prime decomposition of 17 248 and hence list the factors of this number. Solution The prime decomposition can be determined by repeated division. 2 17248 2 8624 2 3112 2 2156 2 1078 7 539 7 77 11 11 1 The prime decomposition of 17248 is 17248 = 25 × 72 × 11 The factors can systematically be determined in the following way. 25 , 24 , 23 , 22 , 2, 1 25 × 7, 24 × 7, 23 × 7, 22 × 7, 2 × 7, 7 25 × 72 , 24 × 72 , 23 × 72 , 22 × 72 , 2 × 72 , 72 25 × 11, 24 × 11, 23 × 11, 22 × 11, 2 × 11, 11 25 × 7 × 11, 24 × 7 × 11, 23 × 7 × 11, 22 × 7 × 11, 2 × 7 × 11, 7 × 11 25 × 72 × 11, 24 × 72 × 11, 23 × 72 × 11, 22 × 72 × 11, 2 × 72 × 11, 72 × 11



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Highest common factor (greatest common divisor) The highest common factor of two natural numbers is the largest natural number which is a factor of both the numbers. For example, the highest common factor of 15 and 24 is 3. The prime decomposition can be used to find the highest common factor of two numbers. Consider the numbers 140 and 110. The prime factorisations of these numbers are 140 = 22 × 5 × 7 and 110 = 2 × 5 × 11. The number which is a factor of 140 and 110 must have prime factors which occur in both factorisations. The exponent (power) of each of these prime factors will be the smaller of the two exponents occurring in the factorisation of 140 and 110. Thus the highest common factor = 2 × 5 = 10. Example 10 Find the highest common factor of 528 and 3168. Solution

∴

528 = 24 × 3 × 11 and 3168 = 25 × 32 × 11 highest common factor = 24 × 3 × 11 = 528

Example 11 Find the highest common factor of 3696 and 3744. Solution

∴

3696 = 24 × 3 × 7 × 11 and 3744 = 25 × 32 × 13 highest common factor = 24 × 3 = 48

Using the TI-Nspire The calculator can be used to factor Natural numbers by using b >Algebra>Factor as shown.

The highest common factor (also called greatest common divisor) of two numbers can be found using the command gcd( ) from b >Number>Greatest Common Divisor, or just typing it in, as shown. Note how ‘nested’ gcd( ) may be used to find the greatest common divisor of several numbers.




77

Using the Casio ClassPad To find the highest common factor, tap Interactive, Calculation, gcd and enter the required numbers in the two lines provided.

Exercise 3D Example

9

1 Give the prime decomposition of each of the following numbers. a 68 640

Examples

10, 11

b 96 096

c 32 032

d 544 544

2 Find the highest common factor of each of the following pairs of numbers. a 4361, 9281 b 999, 2160 c 5255, 716 845 d 1271, 3875 e 804, 2358 Note: Extended-answer questions 5, 6 and 7 are concerned with natural numbers.

3.5

Problems involving sets Sets can be used to help sort information, as each of the following examples demonstrates. Example 12 Two hundred and eighty students either travel by train or tram or both to get to school. One hundred and fifty travel by train and 60 travel by both train and tram. a Show this information on a Venn diagram. b Hence find the number of students who travel by i tram ii train but not tram iii just one of these modes of transport.



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Solution a

b

i n(TRAM) = 130 + 60 = 190 ii n(TRAIN ∩ (TRAM) ) = 90 iii n(TRAIN ∩ (TRAM) ) + n((TRAIN) ∩ TRAM) = 90 + 130 = 220

Example 13 An athletics team consists of 18 members. Each member performs in at least one of three events, sprints (S), jumps (J) and hurdles (H). Every hurdler either jumps or sprints. Also the following information is available. n(S) = 11, n(J ) = 10, n(J ∩ H ∩ S ) = 5, n(J ∩ H ∩ S) = 5 and n(J ∩ H ) = 7 a Draw a Venn diagram. b Find i n(H ) ii n(S ∩ H ∩ J )

iii n(S ∪ J )

iv n(S ∩ J ∩ H )

Solution a

The information given above can be summarised as equations in terms of p, q, r, w, x, y, z. x + y + z + w = 11 as n(S) = 11 p + q + z + w = 10 as n(J ) = 10 x + y + z + w + p + q + r = 18 as all members compete p = 5 as n(J ∩ H ∩ S ) = 5 s = 5 as n(J ∩ H ∩ S) = 5 r = 0 as every hurdler either jumps or sprints w + p = 7 as n(J ∩ H ) = 7 From 4 and 7 ,

... ... ... ... ... ... ...

1 2 3 4 5 6 7

w =2

Equation 3 becomes

i.e.

5 + y + z + 2 + 5 + q = 18 y+z+q =6


... 8


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Equation 1 becomes y+z =4 q=2

Therefore from 8, Equation 2 becomes 5 + 2 + z + 2 = 10 ∴ z=1 ∴ y=3

and

The Venn diagram can be completed.

ξ

S

5 3

2 1 0

b

i n(H ) = 6 iii n(S ∪ J ) = 18

J

5 2 H

ii n(S ∩ H ∩ J ) = 1 iv n(S ∩ J ∩ H ) = 2

Exercise 3E Example

12

1 There are 28 students in a class all of whom take History or Economics or both. Fourteen take History, five of whom also take Economics. a Show this information on a Venn diagram. b Hence find the number of students who take i Economics ii History but not Economics iii just one of these subjects. 2 a Draw a Venn diagram to show three sets, A, B and C in a universal set ␰. Enter numbers in the correct parts of the diagram using the following information. n(A ∩ B ∩ C) = 2, n(A ∩ B) = 7, n(B ∩ C) = 6, n(A ∩ C) = 8, n(A) = 16, n(B) = 20, n(C) = 19, n(␰) = 50 b Use the diagram to find ii n(A ∪ B ) i n(A ∩ C )

iii n(A ∩ B ∩ C )

3 In a border town in the Balkans, 60% of people speak Bulgarian, 40% speak Greek and 20% speak neither. What percentage of the town speak both Bulgarian and Greek? 4 A survey of a class of 40 students showed that 16 own at least one dog and 25 at least one cat. Six students had neither. How many students own both?



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5 At an international conference there were 105 delegates. Seventy spoke English, 50 spoke French and 50 spoke Japanese. Twenty five spoke English and French, 15 spoke French and Japanese and 30 spoke Japanese and English. a How many delegates spoke all three languages? b How many spoke Japanese only? 6 A restaurant serves 350 people lunches. It offers three desserts, profiteroles, gelati and fruit. It is found that 40 people have all three desserts, 70 have gelati only, 50 have profiteroles only and 60 have fruit only. Forty five people have fruit and gelati only, 30 people have gelati and profiteroles only and 10 people have fruit and profiteroles only. How many people do not have a dessert? Example

13

7 Forty travellers were questioned about the various methods of transport they had used the previous day. Every traveller questioned travelled by at least one of the following: car (C), bus (B), train (T). Of those questioned, eight had used all three methods of transport. Four had travelled by bus and car only. Two had travelled by car and train only. The number (x) who had travelled by train only was equal to the number who had travelled by bus and train only. If 20 travellers had used a train and 33 had used a bus, find a the value of x b the number who travelled by bus only c the number who travelled by car only. 8 ␰ is the set of integers and X = {x : 21 < x < 37} Y = {3y : 0 < y ≤ 13} Z = {z 2 : 0 < z < 8} a Draw a Venn diagram representing the information. b Find i X ∩Y ∩ Z ii n(X ∩ Y ) 9 A number of students bought red, green and black pens. Three bought one of each colour. Of the students who bought two colours, three did not buy red, five not green and two not black. The same number of students bought red only as bought red with other colours. The same number bought black only as bought green only. More students bought red and black but not green than bought black only. More bought only green than bought green and black but not red. How many students were there and how many pens of each colour were sold?



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10 For three subsets, B, M and F of a universal set ␰ n(B ∩ M) = 12, n(M ∩ F ∩ B) = n(F ), n(F ∩ B) > n(M ∩ F), n(B ∩ F ∩ M ) = 5, n(M ∩ B ∩ F ) = 5, n(F ∩ M ∩ B ) = 5, n(␰) = 28 Find n(M ∩ F). 11 A group of 80 students were interviewed about which sport they played. It was found that 23 did athletics, 22 swim and 18 play football. If 10 people do athletics and swim only and 11 people do athletics and play football only, six people both swim and play football only and 46 people do none of these activities on a regular basis, how many people do all three? 12 At a certain secondary college students have to be proficient in at least one of the languages Italian, French or German. In a particular group of 33 students, two are proficient in all three languages, three in Italian and French only, four in French and German only and five in German and Italian only. The number proficient in Italian only is x, in French only is x and in German only is x + 1. Find x and the total number proficient in Italian. 13 Two hundred and one students at a certain school studied one or more of Mathematics, Physics and Chemistry. 35 took Chemistry only, 50% more students studied Mathematics only than studied Physics only, four studied all three subjects, 25 studied both Mathematics and Physics but not Chemistry, seven studied both Mathematics and Chemistry but not Physics and 20 studied both Physics and Chemistry but not Mathematics. Find the number of students studying Mathematics.



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Chapter summary Set notation ∈ is a member of ∈ / is not a member of Ø the empty set ␰ the universal set ⊆ subset The union of two sets The set of elements that are in either set A or set B (or both) is the union of set A and B. The union of A and B is written A ∪ B. The intersection of two sets The set of all the elements that are members both of set A and of set B is called the intersection of A and B. The intersection of A and B is written A ∩ B. The complement of A, written A , is the set of all members of ␰ that are not members of A. Sets of numbers R denotes the set of real numbers Q denotes the set of rational numbers Z denotes the set of integers N denotes the set of natural numbers Note: N ⊆ Z ⊆ Q ⊆ R m Numbers of the form , where m, n ∈ N , have a terminating decimal representation if, n and only if, n = 2␣ 5␤ where ␣, ␤ are members of the set N ∪ {0}. Algebraic numbers are those which are the solution(s) of an equation of the form a0 x n + a1 x n−1 + · · · + an = 0, where a0 , a1 , . . . , an are integers. √ A number of the form a where a is a rational number which is not a square of another rational number is called a quadratic surd. √ If a is a rational which is not a perfect nth power n a is called a surd of the nth order. When the number under the square root has no factors which are squares of a rational number, then the surd is said to be in simplest form. Surds which have the same ‘irrational factor’ are called like surds. The sum or difference of two like surds can be found √ √ √ √ √ √ m p + n p = (m + n) p and m p − n p = (m − n) p A natural number, a, is a factor of a natural number, b, if there exists a natural number, k, such that b = ak. If a natural number greater than 1 has only factors 1 and itself, it is said to be prime. A natural number, m, is called a composite if it can be written as a product m = a × b where a and b are natural numbers greater than 1 and less than m. The highest common factor of two natural numbers is the largest natural number which is a factor of both numbers.



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√ √ expressed in the form a + b 2 is 3+2 2 √ √ √ 8√ 3 3 8√ + 2 E 12 + 8 2 D − 2 A 12 − 8 2 B 3+2 2 C 17 17 17 17 2 The prime decomposition of 86400 is B 26 × 33 × 52 C 27 × 33 × 5 A 25 × 32 × 5 E 26 × 33 × 53 D 27 × 33 × 52 √ √ 3 ( 6 + 3)( 6 − 3) is equal to √ √ √ B −3 − 6 6 C −3 + 6 6 D −3 E 3 A 3 − 12 6 1

4

Review

Multiple-choice questions

83

4 For the Venn diagram shown ␰ is the set of positive integers less than 20. A is the set of positive integers less than 10 and B is the set of positive integers less than 20 divisible by 3. The set B ∩ A is ξ 11 10 A {6, 3, 9} 13 14 16 17 19 B {12, 15, 18} B 7 C {10,11,13,14,16,17,19} 12 3 1 2 D {1, 2, 4, 5, 7, 8} 4 5 8 6 15 18 9 E {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18} A 5 (3,∞) ∩ (−∞, 5] = A (−∞, 3) B (−∞, 5] C (3, 5] D R E [3, 5] 6 A bell is rung every 6 minutes and a gong is sounded every 14 minutes. If these occur together at a particular time then the smallest number of minutes until the bell and the gong are again heard simultaneously is A 10 B 20 C 72 D 42 E 84 7 If X is the set of multiples of 2, Y the set of multiples of 7 and Z the set of multiples of 5 then describe X ∩ Y ∩ Z = A the set of multiples of 2 B the set of multiples of 70 C the set of multiples of 35 D the set of multiples of 14 E the set of multiples of 10 8 In a class of students, 50% play football, 40% play tennis and 30% play neither. The percentage that plays both is A B 20 C 30 D 50 E 40 √ 10 √ 7− 6 √ = 9 √ 7+ 6 √ √ √ √ √ A 5+2 7 B 13 + 2 6 C 13 − 2 42 D 1 + 2 42 E 13 − 2 13 10 There are 40 students in a class, all of whom take Literature or Economics or both. Twenty take Literature and five of these also take Economics. The number of students who only take Economics is A 20 B 5 C 10 D 15 E 25



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Short-answer questions (technology-free) 1 Express the following as fractions in their simplest form. a 0.07˙ b 0.4˙ 5˙ c 0.005 d 0.405 e 0.26˙

˙ f 0.171428 5˙

2 Express 504 as a product of powers of prime numbers. 3 Express √ each of the following √ with a rational denominator. √ √ 2 3−1 5+2 3+ 2 a √ b √ c √ √ 2 5−2 3− 2 √ √ 3 + 2 75 4 Express in the form a + b 3 where a, b ∈ R\{0}. √ 3 − 12 5 Express each with a rational denominator. √ of the following √ √ 6 2 a+b− a−b a √ √ b √ √ 3 2−2 3 a+b+ a−b 6 In a class of 100 students, 55 are girls, 45 have blue eyes, 40 are blond, 25 are blond girls, 15 are blue-eyed blonds, 20 are blue-eyed girls, and 5 are blue-eyed blond girls. Find a the number of blond boys b the number of boys who are not blond or blue-eyed. 7 A group of 30 students received prizes in at least one of the subjects of English, Mathematics, and French. Two students received prizes in all three subjects. Fourteen received prizes in English and Mathematics but not French. Two received prizes in English alone, two in French alone and five in Mathematics alone. Four received prizes in English and French but not Mathematics. a How many received prizes in Mathematics and French but not English? b How many received prizes in Mathematics? c How many received prizes in English? 8 Fifty people are interviewed. Twenty-three say they like Brand X, 25 say they like Brand Y, 19 say they like Brand Z. Eleven say they like X and Z. Eight say they like Y and Z. Five say they like X and Y. Two like all three. How many like none of them? 9 Three rectangles A, B and C overlap (intersect). Their areas are 20 cm2 , 10 cm2 and 16 cm2 respectively. The area common to A and B is 3 cm2 , that common to A and C is 6 cm2 , that common to B and C is 4 cm2 . How much of the area is common to all three if the total area covered is 35 cm2 ? √ √ 224 10 Express 112 − 63 − √ in simplest form. 28 √ √ x 7− 3 =√ 11 If √ , find the values of x. x 7+ 3 √ √ √ √ 1+ 2 1− 2 12 Express √ √ +√ √ in the form a 5 + b 6. 5+ 3 5− 3 √ √ √ 48 . 13 Simplify 27 − 12 + 2 75 − 25



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a the number of elements in A ∪ B b the number of elements in C c the number of elements in B ∩ A.

A

B

C

32

0 7 15

3 0

Review

14 A, B and C are three sets and ␰ = A ∪ B ∪ C. The number of elements in the regions of the Venn diagram are as shown. Find

85

√ √ √ √ 15 Using the result that ( a + b)2 = a + b + 2 ab, determine the square root of 17 + 6 8.

Extended-response questions √ √ √ 1 a Show that ( x + y)2 = x + y + 2 x y. b Substitute x = 3 and y = 5 in the identity of a to show √ √ √ ( 3 + 5) = 8 + 2 15 c Use this technique to find the square root of √ i 14 + 2 33 (Hint: use x = 11 and y = 3) √ √ iii 51 − 36 2 ii 15 − 2 56 √ 2 In this question, {a + b 3 : a, b ∈ Q} is considered. Later in this book the set, C, of √ complex numbers is introduced, where C = {a + b −1 : a, b ∈ R}. √ √ √ a If (2 + 3 3) + (4 + 2 3) = a + b 3, find a and b. √ √ √ b If (2 + 3 3)(4 + 2 3) = p + q 3, find p and q. √ 1 c If √ = a + b 3, find a and b. 3+2 3 d Solve each of the following equations for x. √ √ ii (x − 3)2 − 3 = 0 iii (2x − 1)2 − 3 = 0 i (2 + 5 3)x = 2 − 3 √ e Explain why every rational number is a member of {a + b 3 : a, b ∈ Q}. √ 1 3 a Show √ =2− 3 2+ 3 √ x 2 + 3 and the result of a to show that the equation b Use the substitution t = √ x √ x 1 2 + 3 + 2 − 3 = 4 can be written as t + = 4. √ t √ c Show that the solutions of the equation are t = 2 − 3 and t = 2 + 3. √ x √ x d Use this result to solve the equation 2 + 3 + 2 − 3 = 4 4 Use Venn diagrams to illustrate a n(A ∪ B) = n(A) + n(B) − n(A ∩ B) b n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C) √ 5 The number 2 − 3 is a root of the quadratic equation with integer coefficients x 2 + bx + c = 0.



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√ a If x = 2 − 3 is a solution to the equation, find the values of b and c. √ (Hint: Use the result that if m + n 3 = 0 then m = 0 and n = 0, m, n rational.) b Find the other solution to the quadratic. √ c If x 2 + bx + c = 0 and m − n q is a solution, show that i b = −2m ii c = m 2 − n 2 q and hence that √ √ iii x 2 + bx + c = (x − (m − n q))(x − (m + n q)) 6 A triple (x, y, z) is said to be Pythagorean if x 2 + y 2 = z 2 , e.g. (3, 4, 5) is a Pythagorean triple, (5, 12, 13) is a Pythagorean triple. All Pythagorean triples may be generated by the following: x = 2mn, y = m 2 − n 2 , z = m 2 + n 2 where m, n ∈ N e.g. if m = 2, n = 1 then x = 4, y = 3, z = 5 a Find the Pythagorean triple for m = 5, n = 2. b Verify that for x = 2mn, y = m 2 − n 2 , z = m 2 + n 2 where m, n ∈ N , x 2 + y 2 = z 2 . 7 The factors of 12 are 1, 2, 3, 4, 6, 12. a How many factors does each of the following have? ii 37 i 23 b How many factors does 2n have? c How many factors does each of the following have? ii 2n · 3m i 23 · 37 d Every natural number may be expressed as a product of powers and primes. This is called prime factorisation, e.g. 1080 = 23 × 3 × 5. For x, a natural number, let p1␣1 p2␣2 p3␣3 . . . pn␣n be the prime factorisation where ␣i ∈ N and each pi is a prime number. How many factors does x have? (Answer to be given in terms of ␣i .) e Find the smallest number which has eight factors. 8 The least common multiple of natural numbers m and n is the smallest natural number which is a multiple of both m and n, e.g. the least common multiple of 4 and 6 is 12. a Give the prime decomposition of 1080 and 25 200. b In order to find the least common multiple of two numbers a and b, take the prime decomposition of each of the numbers, i.e. a = p1␣1 p2␣2 p3␣3 . . . pn␣n and ␤ ␤ ␤ ␤ b = p1 1 p2 2 p3 3 . . . pmm and then the least common multiple of a and max(␣ ,␤ ) max(␣ ,␤ )

max(␣ ,␤ )

max(␣ ,␤ )

1 1 2 2 n n p2 . . . pn . . . pm m m , where all primes in the prime b = p1 decomposition of either a or b are included in this product, e.g. the least common multiple of 24 = 23 × 3 and 18 = 32 × 2 is 23 × 32 = 72. Find the least common multiple of 1080 and 25 200. c Carefully explain why if m and n are integers mn = least common multiple of m and n × highest common factor of m and n. (cont’d)



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i Find four consecutive even numbers such that the smallest is a multiple of 5, the second a multiple of 7, the third a multiple of 9 and the largest a multiple of 11. ii Find four consecutive natural numbers such that the smallest is a multiple of 5, the second a multiple of 7, the third a multiple of 9 and the largest a multiple of 11.

9 a In the Venn diagram ␰ is the set of all students enrolled at Argos Secondary College. Set R is the set of all students with red hair. Set B is the set of all students with blue eyes. Set F is the set of all female students.

Review

d

87

The numbers on the diagram indicate the eight different regions. i Identify the region in the Venn diagram which represents male students who have neither red hair nor blue eyes. ii Describe the gender, hair colour and eye colour of students represented in region 1 of the diagram. iii Describe the gender, hair colour and eye colour of students represented in region 2 of the diagram. b It is known that at Argos Secondary College, 250 of the students study French (F), Greek (G) or Japanese (J). Forty-one students do not study French. Twelve students study French and Japanese, but not Greek. Thirteen students study Japanese and Greek, but not French. Thirteen students study only Greek. Twice as many students study French and Greek but not Japanese as study all three. The number studying only Japanese is the same as the total of those studying both French and Greek. i How many students study all three languages? ii How many students study only French? 10 Consider the universal set as the set of all students enrolled at Sounion Secondary College. Let B denote the set of students taller than 180 cm at Sounion Secondary College and A denote the set of female students. a Give a brief description of each of the following sets. ii A ∪ B iii A ∩ B i B b Use a Venn diagram to show (A ∪ B) = (A ∩ B ). c Hence show that A ∪ B ∪ C = (A ∩ B ∩ C ), where C is the set of students who play sport.



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11 In a certain city three Sunday newspapers (A, B and C) are available. In a sample of 500 people from this city, it was found that r nobody regularly reads both A and C r a total of 100 people regularly read A r 205 people regularly read only B r of those who regularly read C, exactly half of them also regularly read B r 35 people regularly read A and B, but not C r 35 people don’t read any of the papers at all. a Draw a set diagram showing the number of regular readers for each possible combination of A, B and C. b How many people in the sample were regular readers of C? c How many people in the sample regularly read A only? d How many people are regular readers of A, B and C?



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C H A P T E R

4 Variation Objectives To recognise relationships involving direct variation To evaluate the constant of variation in cases involving direct variation To solve problems involving direct variation To recognise relationships involving inverse variation To evaluate the constant of variation in cases involving inverse variation To solve problems involving inverse variation To establish the relationship that exists between variables from given data To recognise relationships involving joint variation To solve problems involving joint variation To solve problems involving part variation

4.1

Direct variation Emily sets out to drive from her home in Appleton to visit her friend Kim who lives 600 km away in Brownsville. She drives at a constant speed and notes how far she has travelled every hour. The distance and times are represented in the table below. Time (t hours) Distance (d km)

1 100

2 200

3 300

4 400

5 500

6 600

It can be seen that as t increases, d also increases. The rule relating time to distance is d = 100t. This is an example of direct variation and 100 is the constant of variation. In this case d varies directly as t or the distance travelled is proportional to the time spent travelling. The graph of d against t is a straight line passing through the origin. A metal ball is dropped from the top of a tall building and the distance it has fallen is recorded each second. Time (t s) Distance (d km)

0 0

1 4.91

2 19.64

3 44.19

4 78.56

5 122.75



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It can be seen that as t increases, d also increases. This time the rule relating time and distance is d = 4.91t 2 . This is another example of direct variation. In this case, d varies directly as the square of t or the distance travelled is proportional to t 2 . The graph of d against t 2 is a straight line passing through the origin. The symbol used for ‘varies as’ or ‘is proportional to’ is ∝. For example, d varies as t can be written as d ∝ t, and d varies as t 2 can be written as d ∝ t 2 . In the following, a proportional to a positive power of b is considered, i.e. a varies directly as bn , n ∈ R + If a ∝ bn then a = kbn where k is a constant of variation. For all examples of direct variation (where k is positive), as one variable increases the other will also increase. The graph of a against b will show an upwards trend. It should be noted that not all increasing trends will be examples of direct variation. If a ∝ bn then the graph of a against bn is a straight line passing through the origin. Example 1 Use the tables of values below to determine the constant of variation, k, in each case and hence complete each of the tables. 1 √ a y ∝ x2 b y ∝ x i.e. y ∝ x 2 x y

2 12

4

6 108

x y

192

2

Solution a If then When ∴ Check: When

∴

y ∝ x2 y = kx 2 x = 2, y = 12 12 = k(22 ) k =3 x = 6, y = 3(6 ) = 108 y = 3x 2 2

In order to complete the table, consider the following. When x = 4, y = 3(42 ) y = 48 When y = 192, 192 = 3x 2 64 = x 2 x=8

b

If then When ∴ Check: When

4 1

6 1.225

1.414

√ y∝ x √ y =k x x = 4, y = 1 √ 1 = k( 4) k = 0.5 √ x = 6, y = 0.5( 6) ≈ 1.225 √ y = 0.5 x

∴ In order to complete the table, consider the following. When When


√ x = 2, y = 0.5( 2) y ≈ 0.7071 √ y = 1.414, 1.414 ≈ 0.5( x) √ 2.828 ≈ x x ≈8


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Chapter 4 — Variation

x y

2 12

4 48

6 108

8 192

x y

2 0.707

4 1

6 1.225

91

8 1.414

Example 2 In an electrical wire, the resistance (R ohms) varies directly with the length (L m) of the wire. a If a wire 6 m long has a resistance of 5 ohms, what would be the resistance in a wire of length 4.5 m? b How long is a wire for which the resistance is 3.8 ohms? Solution The constant of variation is determined first. R∝L ∴ R = kL When L = 6, R = 5 ∴ 5 = k(6) 5 k= 6 5 i.e. the constant of variation is 6 5L Hence R= 6 5 × 4.5 a When L = 4.5, R = 6 R = 3.75 The resistance of a wire of length 4.5 m is 3.75 ohms.

5L 6 L = 4.56 The length of a wire of resistance 3.8 ohms is 4.56 m.

b When

R = 3.8, 3.8 =

Example 3 The volume of a sphere varies directly as the cube of its radius. By what percentage will the volume increase if the radius is b increased by 20%? a doubled Solution i.e.

V ∝ r3 V = kr 3

Initially set the radius equal to 1, then

V = k(13 ) = k

a If r is doubled, then set r = 2 Then V = k(23 ) = 8k



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∴ the volume has increased from k to 8k, an increase of 7k 100 7k × k 1 = 700%

∴ % increase of volume =

b If r is increased by 20%, then set r = 1.2. Then V = k(1.23 ) = 1.728k ∴ % increase of volume = 72.8%

Exercise 4A Example

1

1 Determine the value of k, the constant of variation, in each of the following and hence complete the table of values. a y ∝ x2 x y

c y∝

b y∝x 2 8

√

x y

4 32

6 128

x

1 2

y

1 6

3 2

1

1 2

2 3

1

d y ∝ x5

x 4 6

9 9

49 90

x

1 32

1

y

1 5

2 5

32 8 5

2 If V ∝ r 3 and V = 125 when r = 2.5, find b r when V = 200

a V when r = 3.2 3 If a ∝

2 b3

and a =

a a when b = 2 Example

2

2 when b = 1, find 3

b b when a = 2

4 The area (A) of a triangle of fixed base length varies directly as its perpendicular height (h). If the area of the triangle is 60 cm2 when its height is 10 cm, find a the area when its height is 12 cm

b the height when its area is 120 cm2 .

5 The extension in a spring (E) varies directly with the weight (w) suspended from it. If a weight of 452 g produces an extension of 3.2 cm, find a the extension produced by a weight of 810 g b the weight that would produce an extension of 10 cm. 6 The weight (W) of a square sheet of lead varies directly with the square of its side length (L). If a sheet of side length 20 cm weighs 18 kg, find the weight of a sheet that has an area of 225 cm2 . ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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93

7 The volume (V) of a sphere varies directly with the cube of its radius (r). A sphere whose radius is 10 cm has a volume of 4188.8 cm3 . Find the radius of a sphere whose volume is 1 cubic metre. Example

3

8 The time taken for one complete oscillation of a pendulum is called its period. The period (T) of a pendulum varies directly with the square root of the length (L) of the pendulum. A pendulum of length 60 cm has a period of 1.55 seconds. Find the period of a pendulum that is one and a half times as long. 9 The distance (d) to the visible horizon varies directly with the square root of the height (h) of the observer above sea level. An observer 1.8 m tall can see 4.8 km out to sea when standing on the shoreline. a How far could the person see if they climbed a 4 m tower? b If the top of a 10 m mast on a yacht is just visible to the observer in the tower, how far out to sea is the yacht? 10 In each of the following calculate the percentage change in y when x is a doubled d increased by 40% i y ∝ x2

4.2

b halved ii y ∝

√

c reduced by 20% iii y ∝ x 3

x

Inverse variation A builder employs a number of bricklayers to build a brick wall. Three bricklayers will complete the wall in eight hours but if he employs six bricklayers the wall will be complete in half the time. The more bricklayers he employs, the shorter the time taken to complete the wall. The time taken (t) decreases as the number of bricklayers (b) increases. This is an example of inverse variation. The time taken to complete the wall varies inversely as the number of bricklayers employed. t varies inversely as b or t is inversely proportional to b i.e.

t∝

1 b

In general, inverse variation exists if a ∝

1 where n is some positive number bn

a varies inversely as bn . 1 If a∝ n b k then a = n where k is a positive constant called the constant of variation. b For all examples of inverse variation, as one variable increases the other will decrease and vice versa. The graph of a against b will show a downward trend. It should be noted, however, that any graph showing a decreasing trend will not necessarily be an example of inverse variation. 1 1 If a ∝ n then the graph of a against n will be a straight line. b b 1 However, since if b = 0, n is undefined, the line will not be defined at the origin. b i.e.



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Example 4 Use the tables of values below to determine the value of the constant of variation, k, in each case and hence complete each of the tables. 1 1 a y∝ 2 b y∝√ x x x y

2 0.1

5 0.016

10

x y

0.001

1 10

25 5

100 1

Solution a

1 x2 k ∴ y= 2 x When x = 2, y = 0.1 k ∴ 0.1 = 2 2 k = 0.4 i.e. the constant of variation is 0.4 y∝

Check: When

When

∴ x y

When

∴

When

∴

In order to complete the table, consider the following. When

∴

0.4 102 y = 0.004 0.4 y = 0.001, 0.001 = 2 x 0.001x 2 = 0.4 0.4 x2 = 0.001 x = 20 5 0.016

10 0.004

20 0.001

10 x = 100, y = √ 100 =1 10 y=√ x

In order to complete the table, consider the following.

x = 10, y =

2 0.1

1 y∝√ x k y=√ x x = 1, y = 10 k 10 = √ 1 k = 10

Check:

0.4 52 = 0.16 0.4 y= 2 x

x = 5, y =

∴

b

When

When

x y

10 x = 4, y = √ 4 y =5 10 y = 2, 2 = √ x √ 2 x = 10 x = 25 1 10

4 5

25 2

100 1

Example 5 For a cylinder of fixed volume, the height (h cm) is inversely proportional to the square of the radius (r cm). ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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a What percentage change in the height would result if its radius were reduced by 25%? b If a cylinder 15 cm high has a base radius of 4.2 cm, how high would a cylinder of equivalent volume be if its radius were 3.5 cm? Solution 1 r2 k h= 2 r k h= =k (1)2 h∝

a i.e. If r = 1, then

b

If r is reduced by 25%, then set r = 0.75 k Then h= (0.75)2 k = 0.5625 ≈ 1.778k (correct to three decimal places)

k r2 When h = 15, r = 4.2 k ∴ 15 = (4.2)2 k = 15(4.2)2 = 264.6 264.6 ∴ h= r2 h=

Consider a cylinder of radius 3.5 cm. If r = 3.5,

then h =

264.6 (3.5)2

h = 21.6 The height of the cylinder is 21.6 cm.

∴ h is increased by 77.8%

Exercise 4B Example

4

1 Determine the value of k, the constant of variation, in each of the following and hence complete the tables of values. 1 1 b y∝√ a y∝ x x x

2

4

y

1

1 2

c y∝ x y

6 1 16

1 x2

x

1 4

1

y

1

1 2

d y∝

9 1 4

1 1

x3

1

2

3

3 4

3 1 12

x

1 8

1

y

2 3

1 3

125 1 9

√ 1 and a = 4 when b = 2, find 3 b √ 1 b b when a = a a when b = 2 2 16

2 If a ∝



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1 and a = 5 when b = 2, find b4 a a when b = 4

3 If a ∝

b b when a = 20.

4 The gas in a cylindrical canister occupies a volume of 22.5 cm3 and exerts a pressure of 1.9 kg/cm2 . If the volume (V) varies inversely as the pressure (P), find the pressure if the volume is reduced to 15 cm3 . Example

5

5 The current (I amperes) that flows in an electrical appliance varies inversely as the resistance (R ohms) of the appliance. If the current is 3 amperes when the resistance is 80 ohms, find a the current when the resistance is 100 ohms b the increase in resistance required to reduce the current to 80% of its original value. 6 The intensity of illumination (I) of a light varies inversely as the square of the distance (d) from the light. At a distance of 20 m a light has an intensity of 100 candela. Find the intensity of the light at a distance of 25 m. 7 The radius (r) of a cylinder of fixed volume varies inversely as the square root of its height (h). If the height is 10 cm when the radius is 5.64 cm, find the radius if the height is 12 cm. 8 In each of the following, calculate the percentage change in y when x is c reduced by 20%

b halved

a doubled d increased by 40%. 1 i y∝ 2 x

1 ii y ∝ √ x

iii y ∝

1 x3

4.3 Fitting data Sometimes the relationship that exists between two variables a and b is not known. By inspection of a table of values, it is sometimes possible to ascertain whether the relationship between the variables is direct or inverse proportion. Analysis is required to establish the rule that best fits the given data. This may involve graphing the data. Example 6 Establish the relationship between the two variables for each of the following tables of values. a

b a

0 0

2 12

4 48

6 108

8 192

b

x y

1 30

3 10

6 5

12 2.5

15 2

Solution a By inspection it can be conjectured that some type of direct variation exists. As b increases, a also increases and when a = 0, b = 0.



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Assume ∴ i.e.

97

a ∝ bn for some positive number n a = kbn a k= n b

Select a value for n (it must be a positive number) and test each of the pairs of values given in the table (do not use (0, 0)). If the value of k for each pair of values is the same then the choice of n is correct. a a Let n = 2 ∴ k = 2 Let n = 1 ∴ k = b b a a Consider 2 for the values Consider for the values b b given in the table. given in the table. 12 =6 2 48 = 12 4 108 = 18 6 192 = 24 8 Since the quotients differ, n = 1.

Testing:

12 =3 4 48 =3 16 108 =3 36 192 =3 64 The quotients are all equal to 3.

Testing:

∴ k = 3 and n = 2 i.e. a = 3b2 b By inspection it can be conjectured that some type of inverse variation exists. As x increases, y decreases. Assume

∴ i.e.

1 for some positivenumber n xn k y= n x k = yx n y∝

Let n = 1 ∴ k = yx Consider the product yx for the values given in the table. Testing:

∴ i.e.

30 × 1 = 30 10 × 3 = 30 5 × 6 = 30 2.5 × 12 = 30 2 × 15 = 30 k = 30 and n = 1 30 y= x

The type of variation can also be investigated by graphical analysis. By plotting the graph of a against b, an upward trend may indicate direct variation or a downward trend may indicate inverse variation. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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To find the specific type of variation that exists, the following can be used as a guide. If direct variation exists (a ∝ bn ), then the graph of a against bn will be a straight line through the origin. The gradient of this line will be the constant of variation k. 1 1 If inverse variation exists a ∝ n , then the graph of a against n will be a straight line. b b This line will not be defined at the origin. The gradient of this line will be the constant of variation k. Example 7 For the table of values below, plot the graph of a against b2 and hence establish the rule relating a to b. b a

1 0.5

2 2

3 4.5

4 8

5 12.5

16 8

25 12.5

Solution b2 a

1 0.5

4 2

9 4.5

a 10

Since this is a straight line, it can be conjectured that the relationship is a = kb2 where k corresponds to the gradient of the graph.

5

0

From the graph it can be seen that a =

5

10

15

20

25 b2

1 2 b . 2

If it is known that the relationship between two variables x and y is of the form y = kx n where k ∈ R + and n ∈ Q\{0} then a CAS calculator can be used to find n and k is sufficient information is given.

Example 8 The following data was collected recording N, the number of calls to a company, D days after the commencement of an advertising campaign.

Days (D) Number of calls (N )

5 50

10 400

15 1350

Find a relationship between N and D using the graphics calculator.




99

Solution

Using the TI-Nspire Store the x-values and y-values as shown.

Use b >Statistics>Stat Calculations> Power Regression and complete as shown. Press enter, and the result is given as y = a ∗ x ∧ b, a = 0.4, b = 3. Hence y = 0.4x 3 , so the required relationship is N = 0.4D 3 .

The graph and the data can be graphed in a Graphs application (/+I>Add Graphs) as a Function (select f1(x)) and a Scatter Plot (b >Graph Type>Scatter Plot) respectively. To set an appropriate window use b >Window/Zoom>Zoom-Data

Using the Casio ClassPad In the program area, enter the data into list 1 and list 2 then tap Calc, Power Reg and ensure the settings are as shown. Note that selecting y1 will copy the program formula to graph y1 in the area. Note the formula from the Stat Calculation screen before tapping OK. The required relationship is N = 0.4D 3 .



100


Example 9 Establish a rule connecting y and x given the following data. x y

1 5

8 2.5

64 1.25

Solution The solution is given in the screens below.

Using the TI-Nspire

Note that y = 5x − 3 = 1

5 1

x3

Using the Casio ClassPad The solution is given in the screens. Note that y = 5x − 3 = 1

5 1

x3

Exercise 4C Example

6

1 Each of the tables in parts a to e fits one of the following types of variation: 1 1 direct square y ∝ x 2 direct y ∝ x inverse y ∝ inverse square y ∝ 2 x x √ direct square root y ∝ x



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101


Establish the relationship between x and y in each case. a

x y

0 0

x

20 1 4

15 1 3

10 1 2

5

1 4

1.5 1.78

2 1

2.5 0.64

c

e

y x y

3 2

6 4

9 6

12 8

b

x y

1 4

2 16

d

x y

1 2

2 2.828

1

1

5

3 36

4 64

5 100

3 3.464

4 4

5 4.472

3 0.444

2 Which of the following graphs could represent examples of direct variation? y

a

x

0

0

x y

f

1 x

0

y

0

y

e

x

c

x3

0

y

d

y

b

x

0

3 Which of the following graphs could represent examples of inverse variation? y

a

x

0

0

y

e

y

d

x

1 x

0

y

c

y

b

y

f

1 0


x3

x

0

0

x


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4 Give the rule connecting y and x for each of the following. y

a

y

c

y

b

(3, 10) (1, 3)

(2, 6) x

0

(9, 3)

2 9.6

2.5 15

3 21.6

3.5 29.4

6 Plot the graph of y against x y

1 1.5

4 3

9 4.5

16 6

x3

0

x y

0.2 50

0.3 22.2

0.4 12.5

√

4 38.4

x and hence establish the relationship between x and y.

25 7.5

7 Plot the graph of y against

8

√x

0

5 Plot the graph of y against x2 and hence establish the relationship between x and y. x y

Example

(1, 6) 1

√x

0 7

y

f

(1, 2)

Example

x2

0

y

e

y

d

1 x

0

1 and hence establish the relationship between x and y. x2 0.5 8

1 2

8 Given that for each of the following y ∝ ax b use your graphics calculator’s PwrReg function to establish the values of a and b. a

x y

4.00 0.50

8.00 0.71

12.00 0.87

16.00 1.00

b

x y

1 2.00

c

x y

1 3.50

10 8.79

100 22.08

1000 55.47

d

x y

10 46.42

e

x y

1 2.00

2 0.35

3 0.13

f

x y

1 3.20

4 0.06


5 14.95 20 73.68 3 2.06

10 35.57

15 59.04

30 96.55 5 1.68

40 116.96

7 1.47


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Chapter 4 — Variation Example

9

103

9 The concentration of antibodies (C) in an animal’s bloodstream is directly proportional to time (t hours) after which the animal is injected with an antigen (i.e., C = at b ). The following data is collected. t C

1 100

2 114.87

3 124.57

4 131.95 b Find the concentration after 10 hours.

a Find values for a and b.

10 The level of infestation (I) of a pest in a crop is proportional to the time (t days) after which the crop is sprayed with an insecticide. The relationship can be modelled using the rule I = at b , t ≥ 1. The following data is collected. t I

1 1500

2 1061

3 866

a Find values for a and b.

4.4

4 750 b Find the level of infestation after 10 days.

Joint variation There are many situations where one variable depends on more than one other variable. The variable is said to vary jointly as the other variables. For example, the volume of a cylinder varies jointly as the square of the radius and the height. i.e. or

V ∝ r 2h V = kr 2 h (the value of k is known to be ␲)

Example 10 x2 Given that y ∝ , use this table of values to determine z the value of the constant of variation k and hence complete the table.


x z y

2 10 2

4 8

10 50 2.5

4


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Solution x2 z kx 2 y= z x = 2 and z = 10, y = 2 k(22 ) 2= 10 k =5 5x 2 y= z x = 4, z = 8 5(42 ) y= 8 y = 10 y∝

∴ When

∴ i.e. When

∴

When

∴

When

∴

z = 50, y = 2.5 5(x 2 ) 2.5 = 50 2 25 = x x =5 x = 10, y = 4 5(102 ) 4= z 4z = 500 z = 125

x z y

2 10 2

4 8 10

5 50 2.5

10 125 4

Example 11 The speed (s) of a conveyor belt varies jointly as the diameter (d) of the cog around which it passes and the number of revolutions per second (n) the cog makes. The speed of a belt that passes round a cog of diameter 0.3 m, revolving 20 times per second, is 18.85 m/s. Find the value of a the constant of variation b the speed of a belt passing around a cog half as big revolving 30 times per second.



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105


Solution a i.e. When

∴ ∴ b When

s ∝ dn s = kdn n = 20 and d = 0.3, s = 18.85 18.85 = k(0.3)(20) k = 3.142 (correct to three decimal places) s = 3.142dn d = 0.15 and n = 30 s = 3.142 (0.15) (30) s = 14.14 m/s (correct to two decimal places)

Exercise 4D Example

10

x , use this table of values to z determine the value of the constant of variation k and hence complete the table.

1 Given that y ∝

2 Given that y ∝ xz, use this table of values to determine the value of the constant of variation k and hence complete the table.

z , use this table of values to x2 determine the value of the constant of variation k and hence complete the table.

x z y

2 10 1

4 2 10

x z y

2 10 10

4 8

x z

2 10

3 4

y

15 2

4 3

10 60 0.5

4

10 50 25

15

3 Given that y ∝

10 50 6

4

4 a varies directly as b2 and inversely as c. If a = 0.54 when b = 1.2 and c = 2, find a when b = 2.6 and c = 3.5. 5 z varies as the square root of x and inversely as the cube of y. If z = 1.46 when x = 5 and y = 1.5, find z when x = 4.8 and y = 2.3. Example

11

6 The simple interest (I) earned on an investment varies jointly as the interest rate (r) and the time (t) for which it is invested. If a sum of money invested at 6.5% per annum for two years earns $130, how much interest would the same amount of money earn if it were invested at 5.8% for three years? 7 The kinetic energy (E) of an object varies directly as its mass (m) and the square of its velocity (v). If the kinetic energy of an object with a mass of 2.5 kg moving at 15 m/s is 281.25 joules, find the energy of an object with a mass of 1.8 kg moving at 20 m/s.



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8 The resistance (R) in an electrical wire varies directly as its length (l) and inversely as the square of its diameter (d). Find the percentage change in R if a l is increased by 50% and d is reduced by 50% b l is decreased by 50% and d is increased by 50%. 9 The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l). a What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight? b What percentage change in the weight would be capable of being supported by a beam three times as long with twice the diameter? 10 If p varies as the square of q and inversely as the square root of r, what is the effect on p if a both q and r are doubled

b q is doubled and r is halved?

11 a The tension in a spring (T) varies directly with the extension (x) and inversely with the natural length (l) of the spring. Compare the tension in a spring with a natural length of 3 m that is extended by 1 m with the tension in a second spring with a natural length of 2.7 m that is extended by 0.9 m. b The work done (W) in stretching a spring varies directly with the square of the extension (x) and inversely with the natural length of the spring (l). Compare the work done on the two springs in part a.

4.5 Part variation The total cost ($C) of printing cards is made up of a fixed overhead charge ($b) plus an amount that varies directly as the number printed (n). i.e.

C = b + kn

The total surface area (A) of a closed cylinder of fixed height is made up of two parts. The area of the curved surface (2␲r h), which varies as the radius, and the area of the two ends (2␲r 2 ), which varies as the square of the radius. i.e. A = k1r + k2r 2 where k1 = 2␲h and k2 = 2␲ are the constants of variation. These are examples of part variation. Part variation exists when the value of one variable is the sum of two or more quantities each of which varies independently in some way. In some cases, as in the first example above, one of those quantities may be constant. Example 12 A monthly telephone account (A) is made up of a fixed charge (c) for rental and servicing plus an amount that is proportional to the number of calls made (n). In January, 220 calls were made and the account was for $98.20. In February, 310 calls were made and the account was for $120.70. Find the fixed charge and the cost per call. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Solution A = c + kn, where c equals the fixed charge and k equals cost per call 98.20 = c + 220k ... 1 120.70 = c + 310k ... 2 Solving simultaneously, subtract 1 from 2 22.5 = 90k k = 0.25 Substitute in 1 98.20 = c + 220 (0.25) = c + 55 c = 43.2 The fixed charge is $43.20 and the cost per call is $0.25, i.e. 25 cents. Example 13 The stopping distance of a tram (d) (i.e. the distance travelled by the tram after its brakes are applied) varies partly with the speed of the tram (s) and partly with the square of its speed. A tram travelling at 15 km/h can stop in 57 m and at 20 km/h in 96 m. Find the formula that relates s to d and hence the stopping distance of a tram travelling at 18 km/h. Solution d = k1 s + k2 s 2 57 = 15k1 + 225k2 96 = 20k1 + 400k2

... 1 ... 2

Multiply 1 by 4 and 2 by 3 228 = 60k1 + 900k2 288 = 60k1 + 1200k2

... 3 ... 4

Subtract 3 from 4 60 = 300k2 1 k2 = 5 Substitute in 1

1 57 = 15k1 + 225 5 57 = 15k1 + 45 12 k1 = 15



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4 5 4 1 = s + s2 5 5 = 18 1 4 = (18) + (18)2 5 5 = 79.2

k1 =

∴

d

When

s d d

The stopping distance of the tram will be 79.2 m.

Exercise 4E Example

12

1 The cost of a taxi ride (C) is partly constant (b) and partly varies with the distance travelled (d). A ride of 22 km costs $42.40 and a ride of 25 km costs $47.80. Find the cost of a journey of 17 km. 2 The cost of a wedding reception at Hillview Reception Centre includes a fixed overhead charge and an amount per guest. a If a reception for 50 people costs $2625 and a reception for 70 people costs $3575, find the fixed overhead charge and the cost per guest. b Hence find the total cost of a reception for 100 guests.

Example

13

3 p is the sum of two numbers, one of which varies as x and the other, as the square of y. If p = 14 when x = 3 and y = 4, and p = 14.5 when x = 5 and y = 3, find p when x = 4 and y = 5. 4 The cost of running a ferris wheel in an amusement park varies partly as the number of people who ride it and partly as the inverse of the number of people who ride it. If the running cost is $32 when 200 people ride it and $61 if 400 people ride it, find the running cost on a day when 360 people ride it. 5 The distance travelled (s) by a particle varies partly with time and partly with the square of time. If it travels 142.5 m in 3 s and 262.5 m in 5 s, find a how far it would travel in 6 s

b how far it would travel in the sixth second.

6 The time taken (t) to load boxes onto a truck varies partly with the number of boxes (b) and partly with the inverse of the number of men (m) loading the boxes. If it takes one man 45 minutes to load ten boxes and two men 30 minutes to load eight boxes, how long would it take four men to load sixteen boxes?



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Direct variation a ∝ bn , i.e. a varies directly as bn (n ∈ R + ) This implies a = kbn where k is the constant of variation (k ∈ R + ). As b increases, a will also increase. If a ∝ bn , the graph of a against bn is a straight line through the origin. Inverse variation 1 a ∝ n , i.e. a varies inversely as bn (n ∈ R + ) b k This implies a = n where k is the constant of variation (k ∈ R + ). b As b increases, a will decrease. 1 1 If a ∝ n , the graph of a against n is a straight line but is undefined at the origin. b b Joint variation One quantity varies with more than one other variable. This may be a combination of direct and/or inverse variation.

Review

Chapter summary

e.g. V ∝ r 2 h implies V = kr 2 h c kc a ∝ √ implies a = √ b b Part variation The value of one variable is the sum of two or more quantities each of which is determined by a variation. In some cases, one of those quantities may be constant. e.g. A = k1r + k2r 2 where k1 , k2 are constants of variation.

Multiple-choice questions 1 For the values in the table shown, it is known that y ∝ x 2 . The value of k, the constant of variation, is equal to

x y

1 D 2 3 2 For the values in the table shown, it is known that x 1 y ∝ . The value of k, the constant of variation, x y is equal to A 3

A

1 2

B 9

B 1

2 4 3

D 2


6

3

12 4 3

E

C

C 4

3

2 1 4

4 1 8 E

8 1 16 1 4


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3 a ∝ b3 and a = 32 when b = 2. Find a when b = 4. D 16 C 4 B 256 A 64 1 1 4 p ∝ 2 and p = when q = 3. Find q when p = 1 q 3 √ D 1 C 3 B −3 A 3 5 The rule connecting y and x as shown in the graph is √ C y=3 x B y = 3x 2 A y = 3x 1 D y= x E y = x2 + 2 3

E 128

1 3

E y

(2, 6)

x2

6 The rule connecting y and x as shown in the graph is √ 1 C y= x B y = 4x A y= x 4√ D y=4 x E y=x

y (1, 4)

√x 7 For the values in the table shown, it is known that x y ∝ 2 . The value of k, the constant of variation, z is equal to

A 2

B

4 3

C

1 3

D 3

x z

2 2

4 2

8 2

y

1 3

2 3

4 3

E

2 3

8 a varies directly as the square of p and inversely as q and a = 8 when p = 2 and q = 5. If p = 3 and q = 6 then a = 1 E 5 D 15 C 120 B 12 A 2 9 If p ∝ q 2 and q is increased by 10%, p would be A Increased by 10% D Increased by 21%

B Increased by 20% E Remain the same

1 and q is decreased by 20%, p would be q B Increased by 25% A Decreased by 25% E Unchanged D Increased by 20%

C Increased by 100%

10 If p ∝


C Decreased by 20%


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3 when b = 2, find a when b = 4 and b when a = 8. 2 1 1 b If y ∝ x 3 and y = 10 when x = 2, find y when x = 27 and x when y = . 8 1 1 4 1 c If y ∝ 2 and y = when x = 2, find y when x = and x when y = . x 3 2 27 √ 1 d a varies directly as b and inversely as c. If a = when b = 1 and c = 4, find a when 4 4 16 b = and c = . 9 9 2 The distance, d metres, which a body falls varies directly as the square of the time, t seconds, for which it has been falling. If a body falls 78.56 m in 4 s, find a the formula connecting d and t b the distance fallen in 10 s c the time taken to fall 19.64 m. 1 a If a ∝ b2 and a =

Review


3 The velocity of a falling body (v metres per second) varies directly as the square root of the distance (s metres) through which it has fallen. A body has a velocity of 7 metres per second after falling 2.5 m. a Find its velocity after falling 10 m. b Find the distance through which it falls to attain a velocity of 28 metres per second. c What variables would be plotted on the axis to obtain a straight line graph? 4 The time taken for a journey is inversely proportional to the average speed of travel. If it takes 4 hours travelling at 30 km/h, how long will it take travelling at 50 km/h? 5 If y varies inversely as x, what is the effect on a y if x is doubled b x if y is doubled c y if x is halved

d x if y is halved?

6 The cost of running an electric appliance varies jointly as the time it is run, the electrical resistance and the square of the current. It costs 9 cents to use an appliance of resistance 60 ohms, which draws 4 amps of current for 2.5 hours. How much will it cost to use an appliance of resistance 80 ohms, which draws 3 amps of current for 1.5 hours? 7 The cost of printing is made up of two parts: a fixed charge and a charge proportional to the number of copies. If the cost of printing 100 copies is $20 and the cost of printing 500 copies is $30, what would be the cost of printing 700 copies? 8 For a constant resistance, the voltage (v volts) of an electric circuit varies directly as the current (I amps). If the voltage is 24 volts when the current is 6 amps, find the current when the voltage is 72 volts. 9 The intensity of sound varies inversely as the square of the distance of the observer from the source. If the observer moves to twice the distance from the source, compare the second intensity I2 with the first intensity I1 . 10 If y varies directly as x2 and inversely as z, find the percentage change in y when x is increased by 10% and z is decreased by 10%.



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Extended-response questions 1 A certain type of hollow sphere is designed in such a way that the mass varies as the square of the diameter. Three spheres of this type are made. One has mass 0.10 kg and diameter 9 cm, the second has diameter 14 cm and the third has mass 0.15 kg. Find b the diameter of the third sphere. a the mass of the second sphere 2 The height (h m) to which a centrifugal pump raises water is proportional to the square of its speed of rotation (n revs/min). If the pump raises water to a height of 13.5 m when it is rotating at 200 revs/min, find a the formula connecting h and n b the height that the water can be raised to when it is rotating at 225 revs/min c the speed required to raise the water to a height of 16 m. 3 The maximum speed of yachts of normal dimensions varies as the square root of their length. If a yacht 20 m long can maintain a maximum speed of 15 knots, find the maximum speed of a yacht 15 m long. 4 a The air in a tube occupies 43.5 cm3 and the pressure is 2.8 kg/cm2 . If the volume (V cm3 ) varies inversely as the pressure (P), find the formula connecting V and P. b Calculate the pressure when the volume is decreased to 12.7 cm3 . 5 The weight (w kg) which a beam supported at each end will carry without breaking, varies inversely as the distance (d m) between supports. A beam which measures 6 m between supports will just carry a load of 500 kg. a Find the formula connecting w and d. b What weight would a similar beam carry if the distance between the supports were 5 m? c What weight would a similar beam carry if the distance between the supports were 9 m? 6 The relationship between pressure and volume of a fixed mass of gas when the temperature is constant is shown by the following table. Pressure ( p) Volume (v)

12 12

16 9

18 8

a What is a possible equation relating p and v? b Using this equation, find i the volume when the pressure is 72 units ii the pressure when the volume is 3 units. 1 c Sketch the graph relating v and . p 7 The time taken to manufacture particular items of scientific equipment varies partly as the diameter of the item and partly as the number of parts required in the item. If it takes



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8 The cost of decorative wrought iron is the sum of two parts which vary as the length and the square of the length respectively. When the length is 2 m, the cost is $18.40 and when the length is 3 m, the cost is $33.60. Find the cost when the length is 5 m.

Review

30 minutes to make a 3 cm diameter item with eight parts and 38 minutes to make a 5 cm diameter item with ten parts, how long does it take to make a 4 cm diameter item with twelve parts?

9 The sum of the first n natural numbers is equal to the sum of two quantities, the first of which is proportional to n and the second to n2 . Work out the sums of the first three and four natural numbers and hence find the formula for the sum of the first n natural numbers. 10 Data about the number of pies sold at football matches and the size of the crowds attending has been recorded as follows Attendance (N × 1000) Number of pies sold (P)

20 15 650

30 19 170

60 27 110

a Use a graphics calculator to find an approximate relationship between N and P of the form P = a N b . b The crowd predicted for a forthcoming match is 55 000. Assuming the model found in part a applies, how many pies would the caterers anticipate selling on that day? c The caterers have only 25 000 pies available for sale. Again assuming the model found in part a applies, what is the maximum crowd the caterers might be hoping for if they are able to satisfy all customers wanting pies? 11 The effectiveness of an anaesthetic drug is being tested by varying the dose (d mL) given to patients and recording both the time (t min) for the patient to lose consciousness and the time (T min) for the patient to regain consciousness. The following data was recorded: Dosage (d mL) Time to lose consciousness (t min) Time to regain consciousness (T min)

10 36 14

30 4 126

60 1 504

a Establish the relationship between d and t (assume t is proportional to a power of d). b Establish the relationship between d and T (assume T is proportional to a power of d). c If it is desirable to have a particular patient unconscious for no longer than 80 minutes, what is the maximum dosage of the drug that should be given? d How long would it take that patient to lose consciousness? e Another patient is given a dose of 20 mL. How long will it take for the patient to lose consciousness and how long will they remain unconscious? 12 The German astronomer Johannes Kepler collected data on the mean distance from the Sun to the planets (R × 106 km) and the period of the orbit (T years). He was able to establish a relationship between R and T.



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a Using the data below (approximations only) i establish the relationship between R and T (assume T is proportional to a power of R) ii complete the table of values showing the period of orbit of the remaining planets

Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

Approximate radius of orbit (R × 106 km) 58 108 150 228 779 1427 2870 4497 5900

Period of orbit (T years) 0.24 0.61 1

b A comet orbits the sun every 70 years. What is its radius of orbit? 13 To test the effectiveness of an advertising campaign for cheap flights to Hawaii, a travel agent keeps a record of the number of enquiries she receives. It is estimated that the number of enquiries, E, is proportional to the number of times, n, that the advertisement is shown on television. The following data is collected. Number of advertisements (n) Number of enquiries (E)

10 30

20 40

30 47

a Assuming that the relationship between the number of enquiries and the number of advertisements is modelled by the rule E = an b , use your graphics calculator to find values for a and b. b Predict the number of enquiries received if the advertisement is shown 100 times. After two weeks the advertisement has been shown 50 times and the advertising campaign is stopped. The travel agent continues to get enquiries and continues to record them. It is now estimated that the number of enquiries, E, is proportional to the number of days, d, since the advertising campaign stopped. The following data is recorded. Number of days (d) Number of enquiries (E)

3 45

5 25

7 17

10 11

c Assuming that the relationship between the number of enquiries and the number of days is modelled by the rule E = kd p , use your graphics calculator to find values for k and p. d Predict the number of enquiries received on the 14th day after the advertising campaign has finished.



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C H A P T E R

5 Sequences and series Objectives To explore sequences of numbers and their difference equations To use a CAS calculator to generate sequences and display graphs To recognise arithmetic sequences To find the terms, difference equation and number of terms for an arithmetic sequence To calculate the sum of the terms of an arithmetic series To recognise geometric sequences To find the terms, difference equation and number of terms for a geometric sequence To calculate the sum of the terms of a geometric series To recognise and calculate the sum of the terms in an infinite geometric series To use fixed point iteration to generate convergent sequences and hence solve equations To apply sequences and series to solving problems

5.1

Introduction to sequences The following are examples of sequences of numbers: A 1, 3, 5, 7, 9, . . . . . . B 0.1, 0.11, 0.111, 0.1111, . . . . . . 1 1 1 1 C , , , ,...... D 10, 7, 4, 1, −2, . . . . . . E 0.6, 1.7, 2.8, 3.9, . . . . . . 3 9 27 81 Note each sequence is a set of numbers with order being important. For some sequences of numbers a rule can be found for getting from one number to the next. For example: for sequence A, a rule is: add 2 for sequence C, a rule is: multiply by

1 3

for sequence D, a rule is: subtract 3 for sequence E, a rule is: add 1.1



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The numbers of a sequence are called terms. The nth term of a sequence is denoted by the symbol tn . So the first term is t1 , the 12th term is t12 and so on. A sequence may be defined by specifying a rule which enables each subsequent term to be found using the previous term. In this case, the rule specified is called an iterative rule or a difference equation. For example: sequence A may be defined by t1 = 1, tn = tn−1 + 2 1 1 sequence C may be defined by t1 = , tn = tn−1 3 3 Example 1 Use the difference equation to find the first four terms of the sequence t1 = 3, tn = tn−1 + 5 Solution t1 t2 t3 t4

=3 = t1 + 5 = 8 = t2 + 5 = 13 = t3 + 5 = 18

The first four terms are 3, 8, 13, 18. Example 2 Find the first four terms of the sequence defined by the rule tn = 2n + 3. Solution t1 t2 t3 t4

= 2(1) + 3 = 5 = 2(2) + 3 = 7 = 2(3) + 3 = 9 = 2(4) + 3 = 11

The first four terms are 5, 7, 9, 11. Example 3 Find the difference equation for the following sequence. 1 9, −3, 1, − , . . . 3 Solution

∴

1 1 −3 = − × 9 i.e. t2 = − t1 3 3 1 1 i.e. t3 = − t2 1 = − × −3 3 3 1 tn = − tn−1 , t1 = 9 3



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Chapter 5 — Sequences and series

117

Alternatively a sequence may be defined by a rule that is stated in terms of n. For example: tn = 2n tn = 2n−1

defines the sequence t1 = 2, t2 = 4, t3 = 6, t4 = 8, . . . defines the sequence t1 = 1, t2 = 2, t3 = 4, t4 = 8, . . .

Example 4 Find the rule for the nth term for the sequence 1, 4, 9, 16 in terms of n. Solution t1 t2 t3 t4 ∴ tn

=1 = 4 = 22 = 9 = 32 = 16 = 42 = n2

Example 5 At a particular school, the number of students studying General Mathematics increases each year. If in 2006 there are 40 students studying General Mathematics a set up the difference equation if the number is increasing by five students each year b write down an expression for tn in terms of n for the difference equation found in a c find the number of students expected to be doing General Mathematics at the school in 2011. Solution a tn = tn−1 + 5 b t1 = 40, t2 t3 Therefore tn tn c

= t1 + 5 = 45 = 40 + 1 × 5 = t2 + 5 = 50 = 40 + 2 × 5 = 40 + (n − 1) × 5 = 35 + 5n

n=6 t6 = 40 + 5 × 5 = 65

Sixty-five students will study General Mathematics in 2011. Example 6 The height of a sand dune is increasing by 10% each year. It is currently 4 m high. a Set up the difference equation that describes the height of the sand dune. b Write down an expression for tn in terms of n for the difference equation found in a. c Find the height of the sand dune seven years from now. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


118


Solution a tn = tn−1 × 1.1 b t1 = 4 t2 = 4 × 1.1 = 4.4 t3 = 4 × (1.1)2 = 4.84 Therefore

tn = 4 × (1.1)n−1

c Seven years from now implies n = 8 t8 = 4 × (1.1)7 ≈ 7.795 The sand dune will be 7.795 m high in 7 years

Using the TI-Nspire Sequences defined in terms of n This type of sequence is best handled in a Calculator application (c>New Document>Add Calculator). For example, complete as shown to generate the first 10 terms of the sequence of numbers defined by the rule tn = 3 + 4n Storing the resulting list will enable us to graph the sequence. To graph the sequence, open a Graphs application (/+I>Add Graphs) and graph the sequence as a Scatter Plot (b >Graph Type>Scatter Plot), using an appropriate Window (b >Window/Zoom>Zoom-Data). Note that it is possible to see the coordinates of the points using Trace (b >Trace>Graph Trace). This sequence can also be plotted directly in the sequence plotter (b >Graph Type>Sequence>Sequence) with the initial value left blank (or type in 7 in this particular case if desired) as shown.

Note:

Iteratively defined sequences This type of sequence is best handled in a Lists & Spreadsheet application (c >New Doc>Add Lists & Spreadsheet). The sequence plotter shown at the end of the following example is quicker and easier.




119

Example 7 Use a graphics calculator to generate the sequence defined by the difference equation tn = tn−1 + 3, t1 = 1 and sketch the graph of the sequence against n. Solution Use the arrows (¡ ¢ £¤) to name the first two columns n and tn respectively. Enter 1 in cell A1 and enter 1 in cell B1.

Enter = a1 + 1 in cell A2 and = b1 + 3 in cell B2.

Highlight the cells A2 and B2 using (g on the Clickpad) and the Touchpad and fill down using b >Data>Fill to generate the sequence of numbers.

To graph the sequence, open a Graphs application (/+I>Add Graphs) and graph the sequence as a Scatter Plot (b >Graph Type>Scatter Plot), using an appropriate Window (b >Window/Zoom>Zoom-Data). Note that it is possible to see the coordinates of the points using b >Trace>Graph Trace. This sequence can also be plotted directly in the sequence plotter (b >Graph Type>Sequence >Sequence) with rule u1(n)=u1(n-1)+3, with initial value of 1, as shown. Use (/+T) to show table of values. Note:



120


Using the Casio ClassPad The Casio Classpad spreadsheet is an efficient way to produce and graph the sequence. It is similar to operating in a computer spreadsheet such as Microsoft Excel. Enter the values for n in column A, then in B1, enter the formula = 3 + 4∗ A1. Highlight cell B2 and the cells below it and tap Edit, Fill Range to complete the sequence.

Click the arrow beside , select graph type and click on this icon to produce the graph.

For an iteratively defined sequence, the procedure is similar except the cells in column A are each defined in terms of the cell(s) above. The entry for the Fibonacci sequence is A1 = 1, A2 = 1, A3 = A1 + A2. Highlight the formula in A3, together with as many cells as required below, and then tap Edit, Fill Range to complete the operation.

In the screen shown, the formula in cell A3 is displayed in the formula bar at the bottom of the screen.

Note:



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Exercise 5A Example

1

1 In each of the following an iterative definition for a sequence is given. List the first five terms. a t1 = 3, tn = tn−1 + 4 d t1 = −1, tn = tn−1 + 2

Example

Examples

2

3, 4

b t1 = 5, tn = 3tn−1 + 4 c t1 = 1, tn = 5tn−1 e tn+1 = 2tn + tn−1 , t1 = 1, t2 = 3

2 Each of the following is a rule for a sequence. In each case find t1 , t2 , t3 , t4 . 1 a tn = c tn = 2n d tn = 2n b tn = n 2 + 1 n e tn = 3n + 2 f tn = (−1)n n 3 g tn = 2n + 1 h tn = 2 × 3n−1 3 For each of the following sequences i find a possible rule for tn in terms of n ii find the difference equation. a 3, 6, 9, 12

b 1, 2, 4, 8

d 3, −6, 12, −24

e 4, 7, 10, 13

1 1 1 c 1, , , 4 9 16 f 4, 9, 14, 19

4 Consider a sequence for which tn = 3n + 1. Find tn+1 , t2n . Example

5

5 Hamish collects Pokemon cards. He currently has 15 and he adds three to his collection every week. a Set up the difference equation that will generate the number of cards Hamish has in any given week. b Write down an expression for tn in terms of n for the difference equation found in a. c Find the number of cards Hamish should have after another 12 weeks.

Example

6

6 Isobel can swim 100 m in 94.3 s. She aims to reduce her time by 4% each week. a Set up the difference equation that generates Isobel’s time for the 100 m in any given week. b Write down an expression for tn in terms of n for the difference equation found in a. c Find the time in which Isobel expects to be able to complete the 100 m after another 8 weeks. 7 Stephen is a sheep farmer with a flock of 100 sheep. He wishes to increase the size of his flock by both breeding and buying new stock. He estimates that 80% of his sheep will produce one lamb each year and he intends to buy 20 sheep to add to the flock each year. Assuming no sheep die a write the difference equation for the expected number of sheep at the end of each year (let t0 = 100) b calculate the number of sheep at the end of each of the first five years.



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8 Alison invests $2000 at the beginning of the year. At the beginning of each of the following years, she puts a further $400 into the account. Compound interest of 6% p.a. is paid on the investment at the end of each year. a Write down the amount of money in the account at the end of each of the first three years. b Set up a difference equation to generate the sequence for the investment. (Let t1 be the amount in the investment at the end of the first year.) c With a calculator or spreadsheet, use the difference equation to find the amount in the account after ten years. Example

7

9 For each of the following difference equations, use a graphics calculator to find the first six terms of the sequence defined and sketch the graph of these terms against n. a tn = tn−1 + 3, t1 = 1 1 c tn = 2tn−1 , t1 = 2 e tn = (tn−1 )2 , t1 = 1.1 g tn = 2tn−1 + 5, t1 = −1

b tn = tn−1 − 2, t1 = 3 1 d tn = tn−1 , t1 = 32 2 2 f tn = tn−1 , t1 = 27 3 h tn = 4 − tn−1 , t1 = −3

10 a For a sequence for which tn = 2n−1 , find t1 , t2 , t3 . 1 b For a sequence for which u n = (n 2 − n) + 1, find u 1 , u 2 , u 3 . 2 c What do you notice? d Find t4 and u 4 . 11 If Sn = an 2 + bn, a ∈ R, b ∈ R, find S1 , S2 , S3 and Sn+1 − Sn . 1 2 tn + , find t2 , t3 , t4 . The terms of the 12 For the sequence defined by t1 = 1, tn+1 = 2 tn sequence are successive rational approximations of a real number. Can you recognise the number? 13 The Fibonacci sequence is defined by t1 = 1, t2 = 1, tn+2 = tn+1 + tn (n ∈ N ). Use the rule to find t3 , t4 , t5 . Show that tn+2 = 2tn + tn−1 (n ∈ N \{1}).

5.2 Arithmetic sequences A sequence in which each successive term is found by adding a constant amount to the previous term is called an arithmetic sequence. For example, 2, 5, 8, 11. . . . is an arithmetic sequence. An arithmetic sequence can be defined by a difference equation of the form tn = tn−1 + d, where d is a constant. If the first term of an arithmetic sequence t1 = a then the nth term of the sequence can also be described by the rule tn = a + (n − 1)d where a = t1 where d is the common difference.


and

d = tn − tn−1


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Example 8 Find the 10th term of the arithmetic sequence −4, −1, 2, 5 . . . Solution a = −4, d = 3 tn = a + (n − 1)d t10 = −4 + (10 − 1)3 t10 = 23

∴

Example 9 A national park has a series of huts along one of its mountain trails. The first hut is 5 km from the start of the trail, the second is 8 km from the start, the third 11 km and so on. a How far from the start of the trail is the sixth hut? b How far is it from the sixth hut to the twelfth hut? Solution a Distances of the huts from the start of the trail form an arithmetic sequence with a = 5 and d = 3. For the sixth hut t6 = a + 5d t6 = 5 + 5 × 3 = 20 The sixth hut is 20 km from the start of the trail b For the twelfth hut t12 = a + 11d t12 = 5 + 11 × 3 = 38

∴ distance from sixth to the twelfth hut = t12 − t6 = 38 − 20 = 18 km The twelfth hut is 18 km from the sixth hut. Example 10 The 12th term of an arithmetic sequence is 9 and the 25th term is 100. Find a and d. Solution Since

tn = a + (n − 1)d 9 = a + 11d ... 100 = a + 24d ...

1 2

Subtract 1 from 2

∴

91 = 13d d=7



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From 1 we have

∴

9 = a + 11(7) a = −68

Arithmetic mean

a+b . 2 If the numbers a, c and b are consecutive terms of an arithmetic sequence, then

The arithmetic mean of two numbers a and b is defined as c−a = b−c ∴ 2c = a + b a+b ∴ c= 2 i.e. c is the arithmetic mean of a and b.

Exercise 5B Example

8

1 For the arithmetic sequence where tn = a + (n − 1)d, find the first four terms given that √ a a = 0, d = 2 b a = −3, d = 5 c a=d=− 5 d a = 11, d = −2 2 Find a and d and hence find the rule of the arithmetic sequence whose first few terms are √ √ 1 3 7 11 a 3, 7, 11 b 3, −1, −5 c − , , , d 5 − 5, 5, 5 + 5 2 2 2 2 3 a An arithmetic sequence has a first term of 5 and a common difference of −3. Find the thirteenth term. b An arithmetic sequence has a first term of −12 and a common difference of 4. Find the tenth term. c In an arithmetic sequence a = 25 and d = −2.5. Find the ninth term. √ √ d In an arithmetic sequence a = 2 3 and d = 3. Find the fifth term.

Example

9

4 David goes fishing every day for 10 days. On the first day he catches four fish and each day after that he catches two more than the previous day. a How many fish did David catch on the sixth day? b How many fish did he catch on the 10th day? c On which day did he catch 10 fish? 5 An amphitheatre has 25 seats in row A, 28 seats on row B, 31 seats in row C and so on. a How many seats in row P? b How many seats are there in row X? c Which row has 40 seats in it? 6 In each of the following, tn is the nth term of an arithmetic sequence. a Find t5 if t1 = 6, t2 = 10. b Find t12 if t1 = 5, t2 = 2. c Find n if t1 = 16, t2 = 13 and tn = −41. d Find n if t1 = 7, t2 = 11 and tn = 227.



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Example

10

125

1 7 For an arithmetic sequence the first term is 7 and the thirtieth term is 108 . Find the 2 common difference. 8 The number of people who go to see a movie over a period of a week follows an arithmetic sequence. If on the first day only three people go to the movie but on the sixth day 98 go, find the rule for the sequence and hence determine how many attend on the seventh day. 9 For an arithmetic sequence, t3 = 18 and t6 = 486, find the rule for the sequence, i.e. find tn . 10 The number of laps a swimmer swims each week follows an arithmetic sequence. If in the fifth week she swims 24 laps and in the tenth week she swims 39 laps, how many laps did she swim in the fifteenth week? 11 In an arithmetic sequence, t7 = 0.6 and t12 = −0.4. Find t20 . 12 An arithmetic sequence contains 10 terms. If the first is 4 and the tenth is 30, what are the other eight terms? 13 The number of goals kicked by a team in the first six games of a season follows an arithmetic sequence. If the team kicked 5 goals in the first game and 15 in the sixth, how many did they score in each of the other four games? 14 The first term of an arithmetic sequence is a. The mth term is zero. Find the rule for tn for the sequence. √ √ 15 For an arithmetic sequence, find t6 if t15 = 3 + 9 3 and t20 = 38 − 3. 16 Find the arithmetic mean of a 8 and 15 1 1 b √ and √ 2 2−1 2 2+1 17 Find x if 3x − 2 is the arithmetic mean of 5x + 1 and 11. 18 If a, 4a − 4 and 8a − 13 are successive terms of an arithmetic sequence, find a. 19 If tx = y and t y = x, prove that tx+y , = 0 (tx and t y are the xth and yth terms of an arithmetic sequence). 20 If a, 2a and a 2 are consecutive terms of an arithmetic sequence, find a(a = 0).

5.3

Arithmetic series The sum of the terms in a sequence is called a series. If the sequence in question is arithmetic, the series is called an arithmetic series. The symbol Sn is used to denote the sum of n terms of a sequence. i.e. Sn = a + a + d + a + 2d + · · · + a + (n − 1)d



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If this sum is written in reverse order, then Sn = a + (n − l)d + a + (n − 2)d + · · · + a + d + a Adding these two expression together gives 2Sn = n[2a + (n − 1)d] n [2a + (n − 1)d] 2 and since the last term l = tn = a + (n − 1)d

∴

Sn =

Sn =

n (a + l) 2

Example 11 A hardware store sells nails in a range of packets containing different numbers of nails. Packet A contains 50 nails, packet B has 75 nails, packet C has 100 and so on. a Find the number of nails in packet J. b Lachlan buys one of each of packets A to J. How many nails in total does Lachlan have? c Assuming he buys one of each packet starting at A, how many packets does he need to buy to have a total of 1100 nails? Solution a a = 50, d = 25, tn = a + (n − 1)d For packet J, t10 = 50 + 9 × 25 = 275 Packet J contains 275 nails

b a = 50, d = 25 n Sn = (2a + (n − 1)d) 2 10 (2(50) + 11 × 25) S10 = 2 S10 = 1625 Packets A to J contain a total of 1625 nails

c a = 50, d = 25, Sn = 1100 n Sn = (2a + (n − 1)d) 2 n Sn = (2(50) + (n − 1)(25)) = 1100 2 n(100 + 25n − 25) = 2200 25n 2 + 75n − 2200 = 0 n 2 + 3n − 88 = 0 (n + 11)(n − 8) = 0 n = −11 or n = 8 since n > 0, n = 8 If Lachlan buys one of each of the first eight packets (A to H) he will have exactly 1100 nails.



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Example 12 For the arithmetic sequence 3, 6, 9, 12, . . . , calculate a the sum of the first 25 terms b the number of terms in the series if Sn = 1395. Solution a a = 3, d = 3, n = 25 n Sn = [2a + (n − 1)d] 2 25 [2(3) + (24)(3)] = 2 = 975 b a = 3, d = 3, Sn = 1395 n [2a + (n − 1)d] 2 n 1395 = [2(3) + (n − 1)(3)] 2 2790 = n[6 + 3n − 3] 2790 = 3n + 3n 2 3n 2 + 3n − 2790 = 0 n 2 + n − 930 = 0 (n − 30)(n + 31) = 0 ∴ n = 30 since n > 0 ∴ there are 30 terms in the series. Sn =

Example 13 For the arithmetic sequence 27, 23, 19, 15, . . . , −33, find a the number of terms b the sum of the terms. Solution a a = 27, d = −4, l = tn = −33 tn = a + (n − 1)d −33 = 27 + (n − 1)(−4) −60 = (n − 1)(−4) 15 = n − 1 n = 16 There are 16 terms in the series.



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b a = 27, l = tn = −33, n = 16 n Sn = (a + l) 2 16 (27 − 33) S16 = 2 S16 = −48 The sum of the terms is −48. Example 14 3 3 The sum of the first 10 terms of an arithmetic sequence is 48 . If the fourth term is 3 , find 4 4 the first term and the common difference. Solution 3 4 15 a + 3d = 4 3 10 (2a + 9d) = 48 = 2 4 195 10a + 45d = 4 40a + 120d = 150

t4 = a + 3d = 3

∴ S10

∴ 1 × 40

... 1

... 2

2 ×4 ∴

40a + 180d = 195 60d = 45 3 ∴ d= 4 15 3 = Substitute in 1 a + 3 4 4 6 4 3 1 The first term is 1 and the common difference is . 2 4 a=

Exercise 5C 1 For the arithmetic sequences a 8, 13, 18, . . . , find S12 1 √ 3 c √ , 2, √ , . . . , find S15 2 2 Example

11

b −3.5, −1.5, 0.5, . . . , find S10 d −4, 1, 6, . . . ,

find S8

2 Greg goes fishing every day for a week. On the first day he catches seven fish and each day he catches three more than the previous day. How many fish did he catch in total? 3 There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next and so on. How many layers will there be?



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4 Find the sum of the first 16 multiples of 5. 5 Find the sum of all the even numbers between 1 and 99. 6 Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km and the last walk is distance of 27 km and the distances of each of the walks form an arithmetic sequence, a How far is the eighth walk? b How far does the club plan to walk in the first five walks? Dora goes away on holiday and misses the 9th, 10th and 11th walks but completes all other walks. c How far does Dora walk in total? 7 Liz has to proofread 500 pages of a new novel. She plans to do 30 pages on the first day and increase the number by five each day. a How many days will it take her to complete the proofreading? She has only five days to complete the task. She therefore decides to read 50 pages on the first day and increase the number she reads by a constant amount each day. b By how many should she increase the number she reads each day if she is to meet her deadline? 8 For the sequence 4, 8, 12, . . . , find {n : Sn = 180}. Example

12

9 The sum of m terms of an arithmetic sequence with first term −5 and common difference 4 is 660. Find m. 10 An assembly hall has 50 seats in row A, 54 seats in row B, 58 seats in row C, i.e. there are four more seats in each row. a How many seats in row J? b How many seats are there altogether if the back row is row Z? If on a particular day the front four rows are reserved for parents (and there is no other seating for parents) c how many parents can be seated d how many students can be seated? The hall is extended by adding more rows following the same pattern. e If the final capacity of the hall is 3410, how many rows were added? 11 A new golf club is formed with 40 members in its first year. Each following year the number of new members exceeds the number of retirements by 15. Each member pays $120 p.a. in membership fees. Calculate the amount received from fees in the first 12 years of the club’s existence.

Example

14

12 In an arithmetic sequence, t2 = −12 and S12 = 18. Find a, d, t6 and S6 . 13 The sum of the first ten terms of an arithmetic sequence is 120 and the sum of the first twenty terms is 840. Find the sum of the first thirty terms.

Example

13

14 Evaluate 54 + 48 + 42 + · · · + −54.



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15 If t6 = 16 and t12 = 28, find S14 . 16 For an arithmetic sequence, find tn if: √ 6 a t3 = 6.5, S8 = 67 b t4 = √ , S5 = 16 5 5 17 For the sequence with tn = bn (b ∈ R), find a tn+1 − tn

b t1 + t2 + · · · + tn

18 For a sequence where tn = 15 − 5n, find t5 and the sum of the first 25 terms. 19 An arithmetic sequence has a common difference of d. If the sum of 20 terms is 25 times the first term, find, in terms of d, the sum of 30 terms. 20 The sum of the first n terms of a particular sequence is given by Sn = 17n − 3n 2 . a Find an expression for the sum to (n − 1) terms. b Find an expression for the nth term of the sequence. c Show that the corresponding sequence is arithmetic and find a and d. 21 Three consecutive terms of an arithmetic sequence have a sum of 36 and a product of 1428. Find the three terms. 22 Show that the sum of the first 2n terms of an arithmetic sequence is n times the sum of the two middle terms.

5.4 Geometric sequences A sequence in which each successive term is found by multiplying the previous term by a fixed amount is called a geometric sequence. For example, 2, 6, 18, 54, . . . is a geometric sequence. A geometric sequence can be defined by an iterative equation of the form tn = r tn−1 , where r is constant If the first term of a geometric sequence t1 = a then the nth term of the sequence can also be described by the rule tn tn−1 and r is the common ratio. tn = ar n−1 , where r =

Example 15 Calculate the tenth term of the sequence 2, 6, 18, . . . Solution a = 2, r = 3 tn = ar n−1 t10 = 2 × 3(10−1) = 39 366 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 16 Georgina draws a pattern consisting of a number of similar equilateral triangles. The first triangle has sides of length 4 cm and the side length of each successive triangle is one and a half times the side length of the previous one. a How long is the side length of the fifth triangle? 9 b Which triangle has a side length of 45 cm? 16 Solution a

a = 4, r = tn = ar n−1

3 2

4 3 1 t5 = ar = 4 × = 20 2 4 4

1 The fifth triangle has a side length of 20 cm 4 9 3 b a = 4, r = tn = 45 2 16 9 n−1 tn = ar = 45 16 n−1 9 3 = 45 which implies 4 × 2 16 729 = 16 n−1 729 3 = Hence 2 64 Recognising that 729 = 36 and 64 = 26 n−1 6 3 3 yields = 2 2 Therefore n − 1 = 6 and n = 7 9 The seventh triangle will have a side length of 45 cm 16 An application of geometric sequences is compound interest. Compound interest is interest calculated at regular intervals on the total of the amount originally invested and the amount accumulated in the previous years. So $1000 invested at 10% per annum would grow to 1000 + 10%(1000) = $1100 at the end of the first year. At the end of the second year, it will have grown to (1000 + 10%(1000)) + 10%(1000 + 10%(1000)) = $1210 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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The value of the investment at the end of each year forms a geometric sequence. In the above example a = 1000, r = 1.1; i.e. r = 100% + 10% Example 17 Hamish invests $2500 at 7% p.a. compounded annually. Find a the value of his investment after 5 years b how long it takes until his investment is worth $10 000. Solution a = 2500, r = 1.07 a t6 = ar 5 tn is the end of the (n − 1)th year. = 2500(1.07)5 = 3506.38 The value of his investment after 5 years is $3506.38. b tn = ar n−1 = 10 000 2500(1.07)n−1 = 10 000 (1.07)n−1 = 4 log10 (1.07)n−1 = log10 4 (n − 1) log10 (1.07) = log10 4 log10 4 log10 (1.07) n = 21.489 By the end of the 21st year, his investment will be worth in excess of $10 000. Note: The number of years can also be found by trial and error or through using a CAS calculator. n−1=

Example 18 The third term of a geometric sequence is 10 and the sixth term is 80. Find r and the first term. Solution t3 = ar 2 = 10

...

1

t6 = ar 5 = 80

...

2

Divide 2 by 1

∴ ∴

80 ar 5 = ar 2 10 r3 = 8 r =2



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Substitute in 1 to find a. a × 4 = 10 5 ∴ a= 2 5 The first term is 2

Geometric mean

√ The geometric mean of two numbers a and b is ab. Note that if three numbers a, c, b are consecutive members of a geometric sequence √ b c = and c = ab a c

Exercise 5D 1 For a geometric sequence tn = ar n−1 , find the first four terms given that a a = 3, r = 2 Example

15

b a = 3, r = −2

c a = 10 000, r = 0.1

d a=r =3

2 Find the specified term in each of the following geometric sequences. 15 5 5 1 1 , , ,... find t6 a find t5 b 1, − , , . . . 7 7 21 4 16 √ √ c 2, 2, 2 2, . . . find t10 d a x , a x+1 , a x+2 , . . . find t6 3 Find the rule for the geometric sequence whose first few terms are √ 4 b 2, −4, 8, −16 c 2, 2 5, 10 a 3, 2, 3 16 4 For a geometric sequence the first term is 25 and the fifth term is . Find the common 25 ratio. 1 5 A geometric sequence has first term and common ratio 2. Which term of the sequence 4 is 64? 6 If tn is the nth term of the following geometric sequences, find n in each case.

Example

16

a 2, 6, 18, . . .

tn = 486

c 768, 384, 192, . . .

tn = 3

4 2 1 e − , ,− ,... 3 3 3

tn =

1 96

b 5, 10, 20, . . . 8 4 d , , 2, . . . 9 3

tn = 1280 27 tn = 4

7 An art collector has a painting that is increasing in value by 8% each year. If the painting is currently valued at $2500 a how much will it be worth in 10 years b how many years before its value exceeds $100 000?



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8 An algal bloom is growing in a lake. The area it covers triples each day. a If it initially covers an area of 10 m2 , how many square metres will it cover after 1 week? b If the lake has a total area of 200 000 m2 , how long before the entire lake is covered? 3 9 A ball is dropped from a height of 2 m and continues to bounce so that it rebounds to of 4 the height from which it previously falls. Find the height it rises to on the fifth bounce. 10 The Tour de Moravia is a cycling event which lasts for 15 days. On the first day the cyclists must ride 120 km and each successive day they ride 90% of the distance of the previous day. a How far do they ride on the eighth day?

b On which day do they ride 30.5 km?

11 A child negotiates a new pocket money deal with her unsuspecting father in which she receives 1 cent on the first day of the month, 2 cents on the second, 4 cents on the third, 8 cents on the fourth and so on . . . until the end of the month. How much would the child receive on the 30th day of the month? (Give your answer to the nearest thousand dollars.) 12 The number of fish in the breeding tanks of a fish farm follow a geometric sequence. The third tank contains 96 fish and the sixth tank contains 768. a How many fish are in the first tank? b How many fish are in the 10th tank? Example

18

13 The 12th term of a geometric sequence is 2 and the fifteenth term is 54. Find the seventh term. √ 1 14 A geometric sequence has t2 = √ and t4 = 2. Find t8 . 2 2 15 The first three terms of a geometric sequence are 4, 8, 16. Find the first term which exceeds 2000. 16 The first three terms of a geometric sequence are 3, 9, 27. Find the first term in the sequence which exceeds 500.

Example

17

17 $5000 is invested at 6% p.a. compounded annually. a Find the value of the investment after 6 years. b Find how long it will take for the original investment to double in value. 18 How much would need to be invested at 8.5% p.a. compounded annually to yield a return of $8000 after 12 years? 19 What annual compound interest rate would be required to triple the value of an investment of $200 in 10 years? 20 The number of ‘type A’ apple bugs present in an orchard is estimated to be 40 960 and is reducing in number by 50% each week. At the same time it is estimated that there are



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40 ‘type B’ apple bugs whose number is doubling each week. After how many weeks will there be the same number of each type of bug? 21 Find the geometric means of

√ 1 c √ and 3 d x 2 y 3 and x 6 y 11 3 22 The fourth, seventh and sixteenth terms of an arithmetic sequence also form consecutive terms of a geometric sequence. Find the common ratio of the geometric sequence. a 5 and 720

5.5

b 1 and 6.25

Geometric series The sum of the terms in a geometric sequence is called a geometric series. An expression for Sn , the sum of n terms of a geometric sequence, can be found using a similar method to that used in the development of a formula for an arithmetic series. Let

Sn = a + ar + ar 2 + · · · + ar n−1 . . .

1

Then

r Sn = ar + ar + ar + · · · + ar . . .

2

2

3

n

Subtract 1 from 2

∴

r Sn , −Sn = ar n − a Sn (r − 1) = a(r n − 1)

and

Sn =

a(r n − 1) r −1

For values of r such that −1 < r < 1, it is often more convenient to use the alternative formula Sn =

a(1 − r n ) 1−r

which is obtained by subtracting 2 from 1 above. Example 19 1 1 1 1 Find the sum of the first nine terms of the sequence , , , , . . . 3 9 27 81 Solution a=

∴

1 1 ,r = ,n = 9 3 3 1 1 9 − 1 3 3 S9 = 1 −1 3 9 1 −1 −1 = 2 3 1 (0.999949) 2 ≈ 0.499975

≈



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Example 20 For the geometric sequence 1, 3, 9, . . . , find how many terms must be added together to obtain a sum of 1093. Solution

∴ ∴

a = 1, r = 3, Sn = 1093 a(r n − 1) Sn = r −1 1(3n − 1) Sn = = 1093 3−1 3n − 1 = 1093 × 2 3n = 2187

Taking logarithms of both sides gives log10 3n = log10 2187 n log10 3 = log10 2187 log10 2187 n= log10 3 n=7 Seven terms are required to give a sum of 1093. A CAS calculator can also be used. Example 21 In the 15-day Tour de Moravia the cyclists must ride 120 km and each successive day they ride 90% of the distance of the previous day. a How far do they ride in total to the nearest km? b After how many days will they have ridden half that distance? Solution a

a = 120, r = 0.9 a(r n − 1) Sn = r −1 120(1 − (0.9)15 ) S15 = 1 − 0.9 = 952.93 ≈ 953 km

b

a = 120, r = 0.9, Sn = 476.5 km a(r n − 1) Sn = r −1 120(1 − (0.9)n ) = 476.5 Sn = 1 − 0.9



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∴ ∴ ∴

137

476.5 × 0.1 = 0.3971 120 (0.9)n = 1 − 0.3971 (0.9)n = 0.6029

1 − (0.9)n =

Taking logarithms of both sides gives log10 (0.9)n = log10 (0.6029) n log10 (0.9) = log10 (0.6029) log10 (0.6029) n= log10 (0.9) n = 4.8023 ∴ on the fifth day they pass the halfway mark.

Exercise 5E Example

19

1 Find the sum specified for each of the following geometric series. a 5 + 10 + 20 + · · · , find S10 4 2 1 c − + − + · · · , find S9 3 3 3

b 1 − 3 + 9 − ···,

find S6

2 Find a 2 − 6 + 18 − · · · + 1458 c 6250 + 1250 + 250 + · · · + 2

b −4 + 8 − 16 + · · · − 1024

3 Gerry owns a milking cow. On the first day he milks the cow, it produces 600 mL of milk. On each successive day, the amount of milk increases by 10%. a How much milk does the cow produce on the seventh day? b How much milk does it produce in the first week? Example

21

4 An insurance salesman makes $15 000 commission on sales in his first year. Each year, he increases his sales by 5%. a How much commission would he make in his fifth year? b How much commission would he make in total over 5 years? 5 On Monday, William spends 20 minutes playing the piano. On Tuesday, he spends 25 minutes playing and on each successive day he increases the time he spends playing in the same ratio. a For how many hours does he play on Friday? b How many hours in total does he play from Monday to Friday? c On which day of the following week will his total time played pass 15 hours? 6 A ball dropped from a height of 15 m rebounds from the ground to a height of 10 m. With each successive rebound, it rises two-thirds of the height of the previous rebound. What total distance will it have travelled when it strikes the ground for the 10th time?



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7 Andrew invests $1000 at 20% simple interest for 10 years. Bianca invests her $1000 at 12.5% compound interest for 10 years. At the end of 10 years, whose investment is worth more? 8 For the geometric sequence with nth term tn √ a t3 = 20, t6 = 160, find S5 b t3 = 2, t8 = 8, find S8 Example

20

9 a How many terms of the geometric sequence where t1 = 1, t2 = 2, t3 = 4, . . . must be taken for Sn = 255? b Let Sn = 1 + 2 + 4 + · · · + 2n−1 . Find {n: Sn > 1000 000}. 10 Find 1 − x 2 + x 4 − x 6 + · · · + x 2m (m iseven).

5.6 Infinite geometric series If the common ratio of a geometric sequence has a magnitude less than 1, i.e. − 1 < r < 1, then each successive term of the sequence is closer to zero. 1 1 e.g. 4, 2, 1, , , . . . 2 4 When the terms of the sequence are added, the corresponding series a + ar + ar 2 + · · · + ar n−1 will approach a limiting value, i.e. as n → ∞, Sn → a limiting value. Such a series is called convergent. In Example 19 from the previous section, it was found that for the sequence 1 1 1 1 , , , , . . . , the sum of the first nineterms, S9 , was 0.499975 3 9 27 81 For the same sequence, S20 = 0.4999999999 ≈ 0.5 a(1 − r n ) 1−r ar n a − ⇒ Sn = 1−r 1−r

Given that Sn =

as n → ∞, r n → 0 and hence

ar n →0 1−r

It follows then that the limit as n → ∞ of Sn is So

S∞ =

a 1−r

a 1−r

This is also referred to as ‘the sum to infinity’ of the series.



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Example 22 Find the sum to infinity of the series 1 +

1 1 1 + + + ··· 2 4 8

Solution 1 ,a = 1 2 1 =2 = 1 1− 2

r=

∴

S∞

Example 23 A square has a side length of 40 cm. A copy of the square is made so that the area of the copy is 80% of the original. The process is repeated each time with the area of the new square being 80% of the previous one. If this process continues indefinitely, find the total area of all the squares. Solution Area of first square is 402 = 1600 cm2 a = 1600, r = 0.8 a S∞ = 1−r 1600 = 8000 cm2 ∴ S∞ = 1 − 0.8 Example 24 Express the recurring decimal 0.3˙ 2˙ as a ratio of two integers. Solution

∴ and i.e.

0.3˙ 2˙ = 0.32 + 0.0032 + 0.000032 + · · · a = 0.32, r = 0.01 32 0.32 = S∞ = 0.99 99 32 0.3˙ 2˙ = 99

Exercise 5F Example

22

1 Find a 1+

1 1 1 + + + ··· 5 25 125

b 1−

2 4 8 + − + ··· 3 9 27



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2 An equilateral triangle has perimeter p cm. The midpoints of the sides are joined to form another triangle, and this process is repeated. Find the perimeter and area of the nth triangle, and find the limits as n → ∞ of the sums of perimeters and areas of the first n triangles. 3 A rocket is launched into the air so that it reaches a height of 200 m in the first second. Each subsequent second it gains 6% less height. Find how high the rocket will climb. 4 A patient has an infection that, if it exceeds a certain level, will kill him. He is given a drug that will inhibit the spread of the infection. The drug acts in such a way that the level of infection only increases by 65% of the previous day’s level. On the first day, the level of infection is measured at 450. The critical level of infection is 1280. Will the infection kill him? 5 A man can walk 3 km in the first hour of a journey, but in each succeeding hour walks half the distance covered in the preceding hour. Can he complete a journey of 6 km? Where does this problem cease to be realistic? 6 A frog standing 10 m from the edge of a pond sets out to jump towards it. Its first jump is 1 1 2 m, its second jump is 1 m, its third jump is 1 m and so on. Show that the frog will 2 8 never reach the edge of the pond. 7 A computer-generated virus acts in such a way that initially it blocks out a third of the area of the screen of an infected computer. On each successive day, it blocks out a further 1 of the area it blocked the previous day. If the virus continues to act unchecked 3 indefinitely, what percentage of the user’s screen will eventually be blocked out? 8 A stone is thrown so that it skips across the surface of a lake. If each skip is 30% less that the previous skip, how long should the first skip be if the total distance travelled by the stone is 40 m? 9 A ball dropped from a height of 15 m rebounds from the ground to a height of 10 m. With each successive rebound, it rises two-thirds of the height of the previous rebound. If it continues to bounce indefinitely, what is the total distance it will travel?

Example

24

10 Express each of the following periodic decimals as the ratio of a pair of integers. a 0.4˙ b 0.03˙ c 10.3˙ d 0.03˙ 5˙ e 0.9˙ f 4.1˙ 11 The sum of the first four terms of a geometric series is 30 and the sum to infinity is 32. Find the first two terms. 1 12 Find the third term of a geometric sequence that has a common ratio of − and a sum to 4 infinity of 8. 13 Find the common ratio of a geometric sequence that has a first term of 5 and a sum to infinity of 15.



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5.7

141

Fixed point iteration The solution(s) to equations may be found using a numerical method that involves generating a sequence of numbers. This method is particularly useful when solving equations for which using an analytic method may be problematic or impossible. The solution to an equation of the form f (x) = 0 may be found by first finding an approximation to the solution and then using this first approximation to generate a better approximation, which is in turn used to produce an even better approximation and so on. This process of using a previous value to generate the next value is called iteration. If the sequence of numbers produced using the iterative process converges to a limit, this limit will be the solution of the equation in question. Consider the equation f (x) = 0. Begin by rewriting the equation in the form x = g(x). If x1 is the initial approximation for the solution to the equation, the second approximation is found by evaluating x2 = g(x1 ). This value is then used to generate a third approximation x3 = g(x2 ), and so on. If the sequence is convergent, each successive term will be closer to the actual solution of the original equation i.e., the sequence of numbers generated by the equation xn = g(xn−1 ) converges to a value which is the solution to the equation x = g(x) (which is of course the solution to the equation f (x) = 0). The iterative process is continued until the value of xn is equal to xn−1 to a pre-determined level of accuracy such as five decimal places. This type of iteration is called fixed point iteration. Example 25 Write down the first five terms generated by the iterative equation xn = hence state if the sequence produced appears to be convergent.

xn−1 + 1, x1 = 2 and 5

Solution x1 = 2 2 x2 = + 1 = 1.4 5 1.4 + 1 = 1.28 x3 = 5 1.28 x4 = + 1 = 1.256 5 1.256 + 1 = 1.2512 x5 = 5 1.2512 + 1 = 1.25024 x6 = 5 The sequence generated appears to be convergent.



142


Example 26 √ Given that the solution to the equation f (x) = 0 where f (x) = x − x + 4 is approximately 6.5, use fixed point iteration to find the solution to the equation correct to five decimal places. Solution First rearrange the equation into the form x = g(x). Note: There may be a number of ways to re-arrange the original equation producing different functions g(x). The implications of these different forms will be discussed later. √ x −x +4=0 √ √ i.e., g(x) = x + 4 Therefore x = x +4 Using x1 = 6.5 √ Therefore x2 = x1 + 4 √ = 6.5 + 4 = 6.549509757 √ √ x4 = x3 + 4 x3 = x2 + 4 √ √ = 6.559201 + 4 = 6.549509757 + 4 = 6.559201 = 6.561093712 x5 x6 x7 x8

= 6.561463197 = 6.56153532 (= 6.56154 to five decimal places) = 6.561549398 (= 6.56155 to five decimal places) = 6.561552146 (= 6.56155 to five decimal places)

Since the values of x8 = x7 to the required level of accuracy the iteration process is terminated. √ Hence the solution to the equation x − x + 4 = 0 is 6.56155 correct to five decimal places.

Using the TI-Nspire A solution using this technique can be found efficiently using the TI-Nspire calculator. Using the above example, start by entering the initial approximation x1 = 6.5 followed by enter (·). Then enter √ ans + 4 and repeatedly press enter. The successive terms of the sequence will be generated until the required level of accuracy is achieved.




143

Using the Casio ClassPad A solution using this technique can be found efficiently using the CAS calculator. Enter the initial approximation 6.5 in the first entry line. √ ( as Then, in the next line, enter 4 + shown. Repeatedly tapping EXE will produce successive terms in the sequence until the desired accuracy is achieved. Note: It is useful to change the display mode from Standard to Decimal to return answers in the appropriate form.

√ In the above example a decision was made to rearrange the original equation x − x + 4 = 0 √ √ into an equation of the form x = x + 4 so that g(x) = x + 4. An alternative rearrangement could have been used. √ Again consider the equation x − x + 4 = 0 √ x = x −4 i.e. x = (x − 4)2 i.e., in this case g(x) = (x − 4)2 Again using x1 = 6.5 a solution can be sought using the iterative process If x1 = 6.5 Therefore x2 = (x1 − 4)2 = (6.5 − 4)2 = 6.25 x3 = (x2 − 4)2 = (6.25 − 4)2 = 5.0625

x4 = (x3 − 4)2 = (5.0625 − 4)2 = 1.12890625

x5 = 8.243179321 x6 = 18.00457075 x7 = 196.128002 x8 = 36913.16914 It is clear that the sequence of numbers generated by the iterative equation xn = (xn−1 − 4)2 √ is not convergent and the solution to the equation x − x + 4 = 0 cannot be found using the iterative process with this particular rearrangement. This method of finding solutions to equations is not universally applicable. It is, however, possible to establish whether the sequence to be generated will converge, therefore producing a solution, by considering the graph of y = g(x).



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√ In solving the equation x − x + 4 = 0, two different rearrangements were used to produce different functions. √ y First consider the graph of f (x) = x − x + 4. It appears that a solution to the equation f (x) = 0 occurs between x = 6 and x = 7.

6 4 2 0 0 1 2 3 4 5 6 7 8 9 10

–1

x

–2 –4

Now consider the graphs of y = x and y = g(x), √ where g(x) = x + 4.

y 8 y = g (x) 6 4

y=x

2 0 0 1 2 3 4 5 6 7 8 9 10

x

y Also consider the graphs of y = x and y = g(x), 2 8 where g(x) = (x − 4) . In both of these pairs of graphs it is clear that a 6 solution to the equation x = g(x) occurs between y=x x = 6 and x = 7. What is of interest in both 4 y = g(x) cases is the gradient of the graph of y = g(x) in the 2 vicinity of the actual solution. 0 In the first of the two, the gradient of y = g(x) in x 0 1 2 3 4 5 6 7 8 9 10 the vicinity of x = 6.5 is quite small (less than that of the line y = x); however in the second the gradient of y = g(x) in the vicinity of x = 6.5 is quite large (greater than that of the line y = x). It is the gradient of the function of y = g(x) in the vicinity of the solution which will determine if the iterative process is to be successful. (If students have studied differential calculus a more rigorous examination of the gradient of y = g(x) may be done, however an informal recognition of the significance of the gradient of y = g(x) is sufficient for students to appreciate that the iterative process will not always succeed.)



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145

Exercise 5G Example

25

1 a Write down the first six terms generated by the following iterative equations xn−1 2 + 2, x1 = 3 i xn = ii xn = xn−1 − 3, x1 = 1 4 √ 2 + 1, x1 = 2 iv xn = xn−1 + 2 + 1, x1 = 3 iii xn = 3xn−1 b Which of the sequences produced in a are convergent?

Example

26

2 Use fixed point iteration to find a solution to the equations f (x) = 0. In each question, the initial approximation x1 is given. Note: It may be necessary to try more than one re-arrangement of f (x) = 0 before a solution can be found successfully. a c e g

f (x) = x 3 + 4x − 3, x1 = 1 x2 − x − 1, x1 = −1 f (x) = 3 x f (x) = 2 − 4x, x1 = 0.5 f (x) = 4x × 2x − 3, x1 = 1

b

f (x) = x 3 + x − 1, x1 = 1

d f h

f (x) = x 4 − x − 2, x1 = 1 f (x) = −x + log10 x + 2, x1 = 5 f (x) = x 3 − 3x + 1, x1 = 0.5



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Chapter summary Sequences The nth term of a sequence is denoted using the symbol tn . A difference equation enables each subsequent term to be found using the previous term. A rule specified in this way is said to be defined iteratively. e.g. t1 = 1, tn = tn−1 + 2 A sequence may be defined by a rule that is stated in terms of n. e.g. tn = 2n An arithmetic sequence is a sequence where tn = a + (n − 1)d with d = tn − tn−1 where a is the first term and d is called the common difference. a+b The arithmetic mean of two numbers a and b is defined as 2 The sum of the terms in an arithmetic sequence is called an arithmetic series. The sum to n terms of an arithmetic sequence, n Sn = [2a + (n − 1)d] 2 n = [a + l] where l = the last term (l = tn = a + (n − 1)d) 2 A geometric sequence is a sequence where tn = ar n−1 with r =

tn tn−1

a is the first term and r is called the common ratio. The sum of the terms in a geometric sequence is called a geometric series. The sum of n terms of a geometric sequence is a(r n − 1) r = 1 (r − 1) a(1 − r n ) or Sn = (1 − r ) Sn =

If −1 < r < 1, the sequence is convergent and Sn approaches a limiting value. The sum to an infinite number of terms (sum to infinity) is denoted by S∞ and S∞ =

a . 1−r Fixed point iteration can be used to find the solution(s) to equations of the form f (x) = 0 by finding the sequence of numbers generated by the equation xn = g(xn−1 ), as long as the sequence is convergent. The equation xn = g(xn−1 ) is found by an appropriate rearrangement of the equation f (x) = 0.



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Review

Multiple-choice questions 1 The first three terms of the sequence defined by the rule tn = 3n + 2 are A 1, 2, 3 B 2, 4, 6 C 5, 7, 9 D 5, 8, 11 E 5, 8, 10 2 If t1 = 3, tn+1 = tn + 3, then t4 is A 4 B 12 C 9

D 15

E 14

3 For the arithmetic sequence 10, 8, 6 . . . t10 = A −8 B −10 C −12

D 10

E 8

4 For the arithmetic sequence 10, 8, 6 . . . S10 = A 10 B 0 C −10

D 20

E −20

5 If 58 is the nth term of the arithmetic sequence 8, 13, 18 . . . then n = A 12 B 11 C 10 D 5 E 3 16 6 The sixth term of the geometric sequence 12, 8, , . . . . is . . . 3 256 128 64 128 16 E D C B A 81 81 81 27 3 7 For the sequence 8, 4, 2, . . . S6 = 3 7 1 1 D 15 E 15 C 15 B 15 A 8 4 2 4 8 For the sequence 8, 4, 2, . . . S∞ = 1 A B 0 C 16 D 4 E ∞ 2 9 $2000 is invested at 5.5% p.a. compounded annually. The value of the investment after 6 years is A $13 766.10 B $11 162.18 C $2550 D $2613.92 E $2757.69 1 10 If S∞ = 37.5 and r = , then a equals 3 2 2 A B 12.5 C 16 D 25 E 56.25 3 3

Short-answer questions (technology-free) 1 Find the first six terms of the following sequences b t1 = 5, tn = 2tn−1 + 2 a t1 = 3, tn = tn−1 − 4 2 Find the first six terms of the following sequences b tn = −3n + 2 a tn = 2n 3 Nick invests $5000 at 5% p.a. compound interest at the beginning of the year. At the beginning of each of the following years he puts a further $500 into the account. a Write down the amount of money in the account at the end of each of the first two years. b Set up a difference equation to generate the sequence for the investment. 4 The fourth term of an arithmetic sequence is 19 and the seventh term is 43. Find the 20th term. 5 In an arithmetic sequence, t5 = 0.35 and t9 = 0.15. Find t14 .



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6 An arithmetic sequence has t6 = −24 and t14 = 6. Find S10 . 7 For the arithmetic sequence −5, 2, 9, . . . , find {n: Sn = 402}. 8 The sixth term of a geometric sequence is 9 and the tenth is 729. Find the fourth term. 9 One thousand dollars is invested at 3.5% p.a. compounded annually. Find the value of the investment after n years. 10 The first term of a geometric sequence is 9 and the third term is 4. Find the possible values for the second and fourth terms. 11 The sum of three consecutive terms of a geometric sequence is 24 and the sum of the next three terms is also 24. Find the sum of the first 12 terms. 12 Find the sum of the first eight terms of a geometric sequence with first term 6 and common ratio −3. 1 1 1 + ··· 13 Find the sum to infinity of 1 − + − 3 9 27 14 The numbers x, x + 4, 2x + 2 are three successive terms of a geometric sequence. Find the value of x.

Extended-response questions 1 A firm offering a do-it-yourself picture frame kit makes the kit in various sizes. Size 1 contains 0.8 m of ‘moulding’, size 2 contains 1.5 m, size 3 contains 2.2 m, . . . and so on. a Form the sequence of lengths of mouldings. b Is the sequence of lengths of moulding an arithmetic sequence? c Find the length of moulding contained in the largest kit, size 12. 2 A firm proposes to sell coated seeds in packs containing the following number of seeds: 50, 75, 100, 125, a Is this an arithmetic sequence? b Find a formula for the nth term. c Find the number of seeds in the 25th size packet. 3 A number of telegraph poles are to be placed in a straight line between two towns, A and B, which are 32 km apart. The first is placed 5 km from town A, the last is placed 3 km from town B. The poles are placed so that the intervals starting from town A and finishing at town B are 5, 5 − d, 5 − 2d, 5 − 3d, . . . , 5 − 6d, 3. There are seven poles. How far is the fifth pole from town A and how far is it from town B? 4 A new, electronic desk-top telephone exchange, for use in large organisations, is available in various sizes. Size 1 can handle 20 internal lines Size 4 can handle 68 internal lines, and so on . . . Size 2 can handle 36 internal lines Size n can handle Tn internal lines Size 3 can handle 52 internal lines (cont’d)



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5 A firm making nylon thread made it in the following deniers (thicknesses):

Review

a Continue the sequence up to T8 . b Write down a formula for Tn in terms of n. c A customer said he needed an exchange to handle 196 lines. Is there a version of the desk-top exchange which will just do this? If so, which size is it? If not, which is the next largest size? 2, 9, 16, 23, 30, . . . etc. a Find the denier number, Dn , of the firm’s nth thread in order of increasing thickness. A request came in for some very heavy 191 denier thread, but this turned out to be one stage beyond the thickest thread made by the firm. b How many different thicknesses did the firm make? 6 A new house appears to be slipping down a hillside. The first year it slipped 4 mm, the second year 16 mm, the third year 28 mm. If it goes on like this, how far will it slip during the 40th year? 7 Anna sends 16 Christmas cards the first year, 24 the second year, 32 the next year and so on. How many Christmas cards will she have sent altogether after ten years if she keeps increasing the number sent each year in the same way? 8 Each time Lee rinses her hair after washing it, the result is to remove a quantity of shampoo from the hair. With each rinsing the quantity of shampoo removed is a tenth of the previous rinse. a If Lee rinses out 90 mg of shampoo with the first rinse, how much will she have washed out altogether after six rinses? b How much shampoo do you think was present in her hair at the beginning? 9 A prisoner is trapped in an underground cell which is inundated with a sudden rush of water which comes up to a depth of 1 m, a third of the height of the ceiling (3 m). After an 1 hour a second inundation occurs, but this time the water level rises by only m. After a 3 1 second hour another inundation of water raises the level by m. If this process continues 9 for 6 hours, write down a the amount the water level will rise at the end of the sixth hour, b the total height of the water level then. If this process continues, do you think the prisoner, who cannot swim, will drown? Why? 10 After an undetected leak in a storage tank, the staff at an experimental station were subjected to 500 curie hours of radiation the first day, 400 curie hours the second day, 320 the third day and so on. Find the number of curie hours they were subjected to a on the 14th day b during the first 5 days of the leak.



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11 A rubber ball is dropped from a height of 81 m. Each time it strikes the ground, it rebounds two-thirds of the distance through which it has fallen. a Find the height the ball reaches after the sixth bounce. b Assuming the ball continues to bounce indefinitely, find the total distance travelled by the ball. 12 In payment for loyal service to the king, a wise peasant asked to be given one grain of rice for the first square of a chessboard, two grains for the second square, four for the third square and so on for all 64 squares of the board. The king thought this seemed fair and readily agreed, but was horrified when the court mathematician informed him of how many grains of rice he would have to pay the peasant. How many grains of rice did the king have to pay? (Leave your answer in index form.) 13 a In its first month of operation a cement factory, A, produces 4000 tonnes of cement. In each successive month, production rises by 250 tonnes per month. This growth in production is illustrated for the first five months in the table shown. Month number (n) Amount of cement produced (tonnes)

1 4000

2 4250

3 4500

4 4750

5 5000

i Find an expression in terms of (n) for the amount of cement produced in the nth month. ii Find an expression in terms of n for the total amount of cement produced in the first n months. iii In which month is the amount of cement produced 9250 tonnes? iv In month m the amount of cement produced is T tonnes. Find m in terms of T v The total amount of cement produced in the first p months is 522 750. Find the value of p. b A second factory, B, commences production at exactly the same time as the first. In its first month of production it produces 3000 tonnes of cement. In each successive month, production increases by 8%. i Find an expression for the total amount of cement produced by this factory after n months. ii Let Q A be the total amount of cement produced by factory A in the first n months and Q B be the total amount of cement produced by factory B in the first n months. Find an expression in terms of n for Q B − Q A and find the smallest value of n for which Q B − Q A ≥ 0. √ 14 By using fixed point iteration to solve the equation x 2 − 8 = 0, find the value of 8 correct to five decimal places. Hint: Add x 2 to both sides of the equation and then re-arrange to produce an iterative equation of the form xn = g(xn−1 ).



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C H A P T E R

6 Algebra II Objectives To understand equality of polynomials To use equating coefficients to solve problems To solve quadratic equations by various methods To use rates to solve problems To resolve a rational algebraic expression into partial fractions To find the coordinates of the points of intersection of linear graphs with

r parabolas r rectangular hyperbolae r circles

6.1

Polynomial identities Polynomials are introduced in Chapter 3 of Essential Mathematical Method 1 & 2 CAS. A polynomial function has a rule of the type y = an x n + an−1 x n−1 + . . . a1 x + a0 , n ∈ N where a0 , a1 , . . . an are numbers called coefficients. The degree of a polynomial is given by the value of n, the highest power of x with non-zero coefficient. Two polynomials are equal if they give the same value for all x. If two polynomials are equal then they are of the same degree, and corresponding coefficients are equal. For example, if ax + b = cx 2 + d x + e, then c = 0, d = a and e = b if x 2 − x − 12 = x 2 + (a + b)x + ab then ab = −12 and a + b = −1 This process is called equating coefficients. Example 1 If the expressions (a + 2b)x 2 − (a − b)x + 8 and 3x 2 − 6x + 8 are equal for all x, find the values of a and b.



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Solution If (a + 2b)x 2 − (a − b)x + 8 = 3x 2 − 6x + 8 for all x, then a + 2b = 3 and −(a − b) = −6 Solve as simultaneous equations a + 2b = 3 ... 1 −a + b = −6 . . . 2 Add 1 and 2 3b = −3 b = −1

∴

Substitute into 2

∴

a=5

Example 2 Express x 2 in the form c(x − 3)2 + a(x − 3) + d. Solution x 2 = c(x − 3)2 + a(x − 3) + d, x 2 = c(x 2 − 6x + 9) + a(x − 3) + d = cx 2 + (a − 6c)x + 9c − 3a + d which implies c=1 ... 1 a − 6c = 0 ... 2 9c − 3a + d = 0 ... 3 From 2 a=6 and from 3 9 − 18 + d = 0 i.e. d=9 ∴ x 2 = (x − 3)2 + 6(x − 3) + 9 If then

Example 3 Find the values of a, b, c and d such that x 3 = a(x + 2)3 + b(x + 1)2 + cx + d for all x. Solution Expand the right hand side. a(x 3 + 6x 2 + 12x + 8) + b(x 2 + 2x + 1) + cx + d Collect like terms. If

ax 3 + (6a + b)x 2 + (12a + 2b + c)x + (8a + b + d) x 3 = ax 3 + (6a + b)x 2 + (12a + 2b + c)x + (8a + b + d)



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Chapter 6 — Algebra II

then

a 6a + b 12a + 2b + c 8a + b + d

=1 =0 =0 =0

... ... ... ...

153

1 2 3 4

Substituting a = 1 into 2 gives b = −6a = −6 Substituting a = 1 and b = −6 into 3 gives 12 − 12 + c = 0 ∴ c=0 Substituting a = 1 and b = −6 into 4 gives

∴

8−6+d =0 d = −2 x 3 = (x + 2)3 − 6(x + 1)2 − 2

Example 4 Show that 2x 3 − 5x 2 + 4x + 1 cannot be expressed in the form a(x + b)3 + c. Solution Assume that 2x 3 − 5x 2 + 4x + 1 can be expressed in the form a(x + b)3 + c i.e.,

2x 3 − 5x 2 + 4x + 1 = a(x + b)3 + c

Expanding the right hand side 2x 3 − 5x 2 + 4x + 1 = a(x 3 + 3bx 2 + 3b2 x + b3 ) + c = ax 3 + 3abx 2 + 3ab2 x + ab3 + c a=2 3ab = −5 3ab2 = 4 ab3 + c = 1

... ... ... ...

1 2 3 and 4 √ 5 2 6 = From 2 b = − , but from 3 b = 6 3 3 We have a contradiction and therefore have shown that 2x 3 − 5x 2 + 4x + 1 cannot be expressed in the form a(x + b)3 + c. Equating coefficients

Exercise 6A Example

1

1 If ax 2 + bx + c = 10x 2 − 7, find the values of a, b and c. 2 If (2a − b)x 2 + (a + 2b)x + 8 = 4x 2 − 3x + 8, find the values of a and b.



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3 If (2a − 3b)x 2 + (3a + b)x + c = 7x 2 + 5x + 7, find the values of a, b and c. 4 If 2x 2 + 4x + 5 = a(x + b)2 + c, find the values of a, b and c. Example

2

5 Express x 2 in the form c(x + 2)2 + a(x + 2) + d. 6 Express x 3 in the form (x + 1)3 + a(x + 1)2 + b(x + 1) + c.

Example

3

7 Find the values of a, b and c such that x 2 = a(x + 1)2 + bx + c.

Example

4

8 a Show that 3x 3 − 9x 2 + 8x + 2 cannot be expressed in the form a(x + b)3 + c. b If 3x 3 − 9x 2 + 9x + 2 can be expressed in the form a(x + b)3 + c, then find the values of a, b and c. 9 Show that constants a, b, c and d can be found such that n 3 = a(n + 1)(n + 2)(n + 3) + b(n + 1)(n + 2) + c(n + 1) + d 10 a Show that no constants can be found such that n 2 = a(n + 1)(n + 2) + b(n + 2)(n + 3) b Express n 2 in the form a(n + 1)(n + 2) + b(n + 1) + c 11 a Express a(x + b)2 + c in expanded form. b Express ax 2 + bx + c in completed square form. 12 Prove that, if ax 3 + bx 2 + cx + d = (x − 1)2 ( px + q), then b = d − 2a and c = a − 2d 13 If 3x 2 + 10x + 3 = c(x − a)(x − b) for all values of x, find the values of a, b and c. 14 If n is any number, show that n 2 can be expressed in the form a(n − 1)2 + b(n − 2)2 + c(n − 3)2 , and find the values of a, b and c. 15 If x 3 + 3x 2 − 9x + c is of the form (x − a)2 (x − b), show that c = 5 or c = −27 and find a and b for each of these cases.

6.2 Quadratics and rates

Quadratics The general expression of a quadratic function is y = ax 2 + bx + c, x ∈ R, a = 0 The number of solutions to the quadratic equation ax 2 + bx + c = 0 can be determined by the discriminant, , where = b2 − 4ac i If b2 − 4ac > 0, the quadratic equation ax 2 + bx + c = 0 has two real solutions ii If b2 − 4ac = 0, the quadratic equation ax 2 + bx + c = 0 has one real solution iii If b2 − 4ac < 0, the quadratic equation ax 2 + bx + c = 0 has no real solutions A quadratic equation may be solved by factorising, completing the square or using the general √ −b ± b2 − 4ac . The following example demonstrates each method. quadratic formula x = 2a ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.



155

Example 5 Solve the following quadratic equations for x. b 3x 2 + 4x = 2 a x 2 + 3x = 4

c 9x 2 + 6x + 1 = 0

Solution a Rearranging the quadratic equation Factorising Applying the null factor law Therefore Note:

x 2 + 3x − 4 = 0 (x + 4)(x − 1) = 0 x + 4 = 0 or x − 1 = 1 x = −4 or 1

= 32 − 4 × 1 × (−4) = 25, so there are two real solutions.

Using the TI-Nspire Use b >Algebra>Solve as shown.

Using the Casio ClassPad Enter and highlight the equation 2x 2 + 5x − 12 = 0, tap Interactive, Equation/inequality, solve and ensure the variable is set to x.



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b Rearranging the quadratic equation

2 −2=0 3x + 4x 4 3 x2 + x − 2 = 0 3

2 b Add and subtract to ‘complete the square’ 2 2 2 2 4 2 2 − −2=0 3 x + x+ 3 3 3 2 2 4 −2=0 − 3 x+ 3 9 2 2 4 3 x+ − −2=0 3 3 2 2 10 3 x+ = 3 3 2 2 10 x+ = 3 9√ 2 10 x+ =± 3 3 √ 2 10 x=− ± 3 3 √ √ −2 − 10 −2 + 10 or Therefore x= 3 3 Note:

= 42 − 4 × 3 × (−2) = 40, so there are two real solutions.

c Consider 9x 2 + 6x + 1 = 0 Using the general quadratic formula x = √

−b ±

√

b2 − 4ac 2a

62 − 4 × 9 × 1 2×9 √ −6 ± 0 = 18 1 Therefore x = − 3 More simply 9x 2 + 6x + 1 = (3x + 1)2 Note: = 62 − 4 × 9 × 1 = 0, so there is one real solution. x=

−6 ±

Example 6 Consider the quadratic equation x 2 − 4x = t. Make x the subject and give the values of t for which real solution(s) to the equation can be found.




157

Solution x 2 − 4x = t Completing the square x − 4x + 4 = t + 4 (x − 2)2 = t + 4 √ x −2=± t +4 √ x =2± t +4 2

For real solutions to exist, t + 4 ≥ 0, i.e., t ≥ −4

Using the TI-Nspire Use b >Algebra>Solve as shown.

Using the Casio ClassPad Enter and highlight the equation x 2 − 4x = t, tap Interactive, Equation/inequality, solve and ensure the variable is set to x.

Note: Variable t is found in the keyboard screen.

menu in the



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Example 7 A rectangle has an area of 288 cm2 . If the width is decreased by 1 cm, and the length increased by 1 cm, the area would be decreased by 3 cm2 . Find the original dimensions of the rectangle. Solution Let w and l be the width and length, in centimetres, of the original rectangle. Therefore

wl = 288

... 1

The dimensions of the new rectangle are w − 1 and l + 1, and the area is 285 cm2 Therefore

(w − 1)(l + 1) = 285

... 2

Rearranging 1 to make l the subject, and substituting in 2 gives 288 (w − 1) + 1 = 285 w 288 + w − 1 = 285 288 − w 288 w− +2=0 w w 2 + 2w − 288 = 0 √ −b ± b2 − 4ac Using the general quadratic formula x = 2a √ −2 ± 22 − 4 × 1 × −288 w= 2×1 = −18 or 16 But w > 0, so w = 16. The original dimensions of the rectangle are 16 cm by 18 cm.

Rates A rate describes how a certain quantity changes with respect to the change in another quantity (often time). An example of a rate is ‘speed’. A speed of 60 km/h gives us a measure of how fast an object is travelling. A further example is ‘flow’, where a rate of 20 L/min is going to fill an empty swimming pool faster than, say, a rate of 6 L/min. Many problems are solved using rates, which can be expressed as fractions. For example, a 60(km) . speed of 60 km/h can be expressed in fraction form as 1(h) It is often first necessary to add two or more fractions with different denominators, as shown in the following examples.



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159

Example 8

6 6 + as a single fraction. x x +8 6 6 b Solve the equation + = 2 for x. x x +8 a Express

Solution 6x a 6 + 6 = 6(x + 8) + x x +8 x(x + 8) x(x + 8) 6x + 48 + 6x = x(x + 8) 12(x + 4) = x(x + 8) 6 6 12(x + 4) + = x x +8 x(x + 8) 12(x + 4) =2 then x(x + 8)

b Since

12(x + 4) = 2x(x + 8) 6(x + 4) = x(x + 8) 6x + 24 = x 2 + 8x 0 = x 2 + 2x − 24 0 = (x + 6)(x − 4) x +6= 0 or x − 4 = 0 x = −6 or x=4 Example 9 A tank is filled by two pipes. The smaller pipe alone will take 24 minutes longer than the larger pipe alone, and 32 minutes longer than when both pipes are used. How long will each pipe take to fill the tank alone? How long will it take for both pipes used together to fill the tank? Solution Let C cubic units be the capacity of the tank, and x minutes the time it takes for the larger pipe alone to fill the tank. Therefore the average rate of flow for the larger pipe C is cubic units per minute. x Since the smaller pipe alone takes (x + 24) minutes to fill the tank, the average rate C of flow for the smaller pipe is cubic units per minute. x + 24



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The average rate of flow when both pipes are used together is the sum of these two C C cubic units per minute. rates, + x x + 24 C C(x + 24) + C x C + = Expressed as a single fraction, x x + 24 x(x + 24) 2C(x + 12) = x(x + 24) The time taken to fill the tank using both pipes is C÷

x(x + 24) 2C(x + 12) =C× x(x + 24) 2C(x + 12) x(x + 24) = 2(x + 12)

The time taken for the smaller pipe to fill the tank can be also be expressed as x(x + 24) + 32 2(x + 12) i.e.

x(x + 24) + 32 = x + 24 2(x + 12) x(x + 24) = x −8 2(x + 12) x(x + 24) = 2(x + 12)(x − 8) x 2 + 24x = 2x 2 + 8x − 192 0 = x 2 − 16x − 192 0 = (x − 24)(x + 8) x − 24 = 0 or x +8=0 x = 24 or x = −8 (but x > 0)

It takes 24 minutes for the larger pipe alone to fill the tank, and 48 minutes for the smaller pipe alone to fill the tank, and 16 minutes to fill the tank using both pipes.

Exercise 6B Example

Example

5

6

1 Solve the following quadratic equations for x. b x 2 − 6x + 9 = 0 a −x 2 + 2x = 1 e 2x 2 + 4x = 7 d −2x 2 + 4x = 1

c 5x 2 − 10x = 1 f 6x 2 + 13x + 1 = 0

2 Make x the subject in each of the following and give the values of t for which real solution(s) to the equation can be found. b 4x 2 + 4x − 4 = t − 2 a 2x 2 − 4t = x c 5x 2 + 4x + 10 = t d t x 2 + 4t x + 10 = t



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3 a Solve the quadratic equation x 2 + 3x − 9 = 0 for x, giving exact solutions. b i Solve the quadratic equation x 2 + px − 16 = 0 in terms of p. ii Find the value(s) of p, where 0 ≤ p ≤ 10 and p is a non-negative integer, for which the quadratic equation in i has a non-negative integer solution. 4 A pole 10 m long leans against a wall. The bottom of the pole is 6 m from the wall. If the bottom of the pole is pulled away x m so that the top slides down by the same amount, find x.

xm

10 m 8m

6m Example

8

Example

9

xm

6 6 as a single fraction. 5 a Express − x x +3 6 6 b Solve the equation − = 1 for x. x x +3 36 . Form a quadratic 6 The sum of the reciprocals of two consecutive odd numbers is 323 equation and hence determine the two numbers. 7 A car travels from town A to town B, a distance of 600 km, in x hours. A plane, travelling 220 km/h faster than the car, takes five and a half hours less to cover the same distance. a Express, in terms of x, the average speed of the car and the average speed of the plane. b Find the actual average speed of each of them. 8 A car covers a distance of 200 km at a speed of x km/h. A train covers the same distance at a speed of (x + 5) km/h. If the time taken by the car is 2 hours more than that taken by the train, find x. 1 9 A man travels 108 km, and finds that he could have made the journey in 4 hours less had 2 he travelled at an average speed 2 km/h faster. What was the man’s average speed when he made the trip? 10 A bus is due to reach its destination 75 km away at a certain time. The bus usually travels with an average speed of x km/h. Its start is delayed by 18 minutes but, by increasing its average speed by 12.5 km/h, the driver arrives on time. a Find x.

b How long did the journey actually take?

11 Ten minutes after the departure of an express train, a slow train starts, travelling at an average speed of 20 km/h less. The slow train reaches a station 250 km away 3.5 hours after the arrival of the express. Find the average speed of each of the trains. 12 When the average speed of a car is increased by 10 km/h the time taken for the car to make a journey of 105 km is reduced by 15 minutes. Find the original average speed. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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1 13 A tank can be filled with water by two pipes running together in 11 minutes. If the larger 9 pipe alone takes 5 minutes less to fill the tank than the smaller pipe, find the time that each pipe will take to fill the tank. 20 minutes. The rate that 3 water runs through each of the pipes is then adjusted. If one pipe, running alone, takes 1 minute less to fill the tank at its new rate, and the other pipe, running alone, takes 2 minutes more to fill the tank at its new rate, then the two running together will fill the tank in 7 minutes. Find in what time the tank will be filled by each pipe running alone at the new rates.

14 At first two different pipes running together will fill a tank in

15 The journey between two towns by one route consists of 233 km by rail followed by 126 km by sea. By a second route the journey consists of 405 km by rail followed by 39 km by sea. If the time taken for the first route is 50 minutes longer than for the second route, and travelling by rail is 25 km/h faster than travelling by sea, find the average speed by rail and the average speed by sea. 16 A freighter sailing due north at 12 km/h sights a cruiser straight ahead at an unknown distance and speeding due east at unknown speed. After 15 minutes the vessels are 10 km apart and then, 15 minutes later, they are 13 km apart. (Assume both travel at constant speeds.) How far apart are the vessels when the cruiser is due east of the freighter? 17 A cask A, of capacity 20 litres, is filled with wine. A certain quantity of wine from A is poured into a cask B which also has a capacity of 20 litres. Cask B is then filled with 20 litres water. After this cask A is filled with some of the mixture from cask B. A further 3 of the mixture now in A is poured back into B, and the two casks now have the same amount of wine. How much wine was first taken out of cask A?

6.3 Partial fractions A rational function is the quotient of two polynomials. If g(x) and h(x) are polynomials, g(x) is a rational function, f (x) = h(x) e.g.,

f (x) =

x2 + 1 x2 − 1

g(x) is a proper fraction. h(x) g(x) is an improper fraction. If the degree of g(x) ≥ the degree of h(x), then h(x) It is convention to consider rational functions for their maximal domain. For example, x2 + 1 is only considered for R\{−1, 1}. x2 − 1

If the degree of g(x) < the degree of h(x), then



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163

A rational algebraic function may be expressed as a sum of separate functions by resolving it into what are called partial fractions. This can help in the sketching of graphs of these functions or performing other mathematical procedures such as integration.

Proper fractions For proper fractions, the technique used for obtaining partial fractions depends on the type of factors in the denominator of the original algebraic fraction. Only examples where the denominators have factors that are either 1st degree (linear) or 2nd degree (quadratic) will be considered. For every linear factor (ax + b) in the denominator, there will be a partial fraction of the A form ax + b For every repeated linear factor (cx + d)2 in the denominator, there will be partial B C fractions of the form and cx + d (cx + d)2 For every irreducible quadratic factor (ax 2 + bx + c) in the denominator, there will be a Dx + E partial fraction of the form 2 ax + bx + c To resolve an algebraic fraction into its partial fractions, firstly write a statement of identity between the original fraction and a sum of the appropriate number of partial fractions. Then express the sum of the partial fractions as a single fraction and note that the numerators of both sides are equivalent. By choosing an appropriate value(s) for x and/or equating coefficients, the values of the introduced constants A, B, C, etc. can be found. Example 10 Resolve

3x + 5 into partial fractions. (x − 1)(x + 3) Solution Since the denominator has two linear factors, there will be two partial fractions of the A B form and x −1 x +3 Let

3x + 5 A B = + (x − 1)(x + 3) x −1 x +3

(x ∈ R\{1, −3})

Express the right hand side as a single fraction.

∴ ∴

A(x + 3) + B(x − 1) 3x + 5 = (x − 1)(x + 3) (x − 1)(x + 3) (A + B)x + 3A − B 3x + 5 = (x − 1)(x + 3) (x − 1)(x + 3) 3x + 5 = (A + B)x + 3A − B



164


Equate the coefficients, A+B =3 3A − B = 5 Solving these equations simultaneously,

i.e.,

i.e., and Therefore

4A = 8 A=2 B=1

2 1 3x + 5 = + (x − 1)(x + 3) x −1 x +3

Using the TI-Nspire

Use b >Algebra>Expand as shown. Notice that you can access the fraction template by using / p.

Using the Casio ClassPad Tap Interactive, Transformation, expand and the Partial Fraction button. Enter the expression and the variable.



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Example 11 Resolve

2x + 10 into partial fractions. (x + 1)(x − 1)2 Solution Since there is a repeated linear factor and a single linear factor, there are three partial fractions such that B C A 2x + 10 + + = (x + 1)(x − 1)2 x + 1 x − 1 (x − 1)2

∴ Therefore

2x + 10 A(x − 1)2 + B(x + 1)(x − 1) + C(x + 1) = (x + 1)(x − 1)2 (x + 1)(x − 1)2 2x + 10 = A(x − 1)2 + B(x + 1)(x − 1) + C(x + 1)

To find A, B and C, a combination of methods will be used. First let ∴ ∴ ∴ Let ∴ ∴ ∴

x=1 2(1) + 10 = C(1 + 1) 12 = 2C C=6 x = −1 2(−1) + 10 = A(−1 − 1)2 8 = 4A A=2

Substitute these values for A and C.

∴

2x + 10 = 2(x − 1)2 + B(x + 1)(x − 1) + 6(x + 1) . . . 1 = 2(x 2 − 2x + 1) + B(x 2 − 1) + 6(x + 1) = (2 + B)x 2 + 2x + 8 − B

Now by equating coefficients,

and ∴ So

2+ B 8− B B 2x + 10 (x + 1)(x − 1)2

=0 = 10 = −2 2 6 2 − + = x + 1 x − 1 (x − 1)2

The value of B may also be found by substituting x = 0 into equation 1 . In the exercises for this section, the following result is established: that it is impossible to find A and C such that A C 2x + 10 = + 2 (x + 1)(x − 1) (x + 1) (x − 1)2



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Example 12 Resolve

x 2 + 6x + 5 into partial fractions. (x − 2)(x 2 + x + 1) Solution The denominator contains a quadratic factor, which cannot be reduced to linear factors (an irreducible quadratic), as well as a single linear factor.

∴ ∴ ∴ Let

Also

A Bx + C x 2 + 6x + 5 = + 2 2 (x − 2)(x + x + 1) x −2 x +x +1 2 A(x 2 + x + 1) + (Bx + C)(x − 2) x + 6x + 5 = (x − 2)(x 2 + x + 1) (x − 2)(x 2 + x + 1) 2 2 x + 6x + 5 = A(x + x + 1) + (Bx + C)(x − 2) x=2 2 2 + 6(2) + 5 = A(4 + 2 + 1) 21 = 7A A=3 2 x + 6x + 5 = A(x 2 + x + 1) + (Bx + C)(x − 2) . . . 1 = A(x 2 + x + 1) + Bx 2 − 2Bx + C x − 2C = (A + B)x 2 + (A − 2B + C)x + A − 2C

Since A = 3, then x 2 + 6x + 5 = (3 + B)x 2 + (3 − 2B + C)x + 3 − 2C Equating coefficients 3+ B ∴ B and 3 − 2C ∴ C [checking : 3 − 2B + C x 2 + 6x + 5 Therefore (x − 2)(x 2 + x + 1) x 2 + 6x + 5 or (x − 2)(x 2 + x + 1)

=1 = −2 =5 = −1 = 3 − 2(−2) + (−1) = 6] 3 −2x − 1 = + 2 x −2 x +x +1 3 2x + 1 = − 2 x −2 x +x +1

The values of B and C can also be found by substituting x = 0 and x = 1 respectively in equation 1 .

Note:

Improper fractions Improper algebraic fractions can be expressed as a sum of partial fractions by first dividing the denominator into the numerator to produce a quotient and a proper fraction. The resulting proper fraction can then be resolved into its partial fractions using the techniques outlined above.




167

Example 13 Express

x5 + 2 in partial fractions. x2 − 1 Solution Dividing through

x3 + x

x2 − 1 x5 + 2 x5 − x3 x3 + 2 x3 − x

∴ Expressing

x +2 x5 + 2 x +2 = x3 + x + 2 2 x −1 x −1

x +2 x +2 = as partial fractions, 2 x −1 (x − 1)(x + 1) 3 1 x5 + 2 = x3 + x − + 2 x −1 2 (x + 1) 2 (x − 1)

Using the TI-Nspire

Use b >Algebra>Expand as shown. Notice that you can access the fraction template by using / p.

Using the Casio ClassPad Tap Interactive, Transformation, expand and the Partial Fraction button. Enter the expression and the variable.



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Partial fractions are summarised through examples, as follows. Linear factors A B 3x − 4 = + (2x − 3)(x + 5) 2x − 3 x + 5 Repeated linear factors A B C 3x − 4 = + + 2 (2x − 3)(x + 5) 2x − 3 x + 5 (x + 5)2 Irreducible quadratic factors A Bx + C 3x − 4 = + 2 2 (2x − 3)(x + 5) 2x − 3 x +5 g(x) and the degree of g(x) is greater than or equal to the degree of h(x) then h(x) division must be performed first.

If f (x) =

Exercise 6C Example

10

1 Resolve the following rational expressions into partial fractions. 5x + 1 −1 3x − 2 a b c (x − 1)(x + 2) (x + 1)(2x + 1) x2 − 4 7−x 4x + 7 e d 2 (x − 4)(x + 1) x +x −6

Example

11

2 Resolve the following rational expressions into partial fractions. 2x + 3 9 2x − 2 a b c 2 2 (x − 3) (1 + 2x)(1 − x) (x + 1)(x − 2)2

Example

12

3 Resolve the following rational expressions into partial fractions. x 2 + 2x − 13 3x 2 + 2x + 5 3x + 1 c b a 2x 3 + 6x 2 + 2x + 6 (x 2 + 2)(x + 1) (x + 1)(x 2 + x + 1)

Example

13

4 Resolve

3x 2 − 4x − 2 into partial fractions. (x − 1)(x − 2)

5 Find values of A and C such that C A 2x + 10 + = 2 (x + 1)(x − 1) x + 1 (x − 1)2



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6 Express each of the following in partial fractions. x 1 b a (x − 2)(x + 3) (x − 1)(x + 1) 1 3x + 5 d e (2x − 1)(x + 2) (3x − 2)(2x + 1) 3x + 1 3x 2 + 8 g h x3 + x x(x 2 + 4) x +3 x3 − x2 − 1 j k x 2 − 4x x2 − x x2 − x x2 + 2 m n (x + 1)(x 2 + 2) x 3 − 3x − 2 3x 2 − 6x + 2 1 − 2x q p (x − 1)2 (x + 2) 2x 2 + 7x + 6 s

6.4

x 3 − 2x 2 − 3x + 9 x2 − 4

t

3x + 1 (x − 2)(x + 5) 2 2 x −x 1 2 x − 4x x3 − x2 − 6 2x − x 2 2x 2 + x + 8 x(x 2 + 4) 4 2 (x − 1) (2x + 1) 2x − 1 (x + 1)(3x + 2)

c f i l o r

x3 + 3 (x + 1)(x − 1)

169

u

Simultaneous equations In this section, methods for finding the coordinates of the points of intersection of a linear graph with different non-linear graphs are discussed. The non-linear graphs are parabolas, circles and rectangular hyperbolae. The associated relations have been discussed in Essential Mathematical Methods 1 and 2 CAS.

Example 14 Find the coordinates of the points of intersection of the parabola with equation y = x 2 − 2x − 2 with the straight line with equation y = x + 4. Solution Consider x + 4 = x 2 − 2x − 2 Then 0 = x 2 − 3x − 6 √ 3 ± 9 − 4 × −6 × 1 ∴ x= 2 √ 3 ± 33 = 2 The points of intersection have √ √ 3 − 33 11 − 33 coordinates A , and 2 2 √ √ 3 + 33 11 + 33 B , 2 2


y y = x 2 – 2x – 2 y=x+4 A –4

B

4 –2

0

x


170


Using the TI-Nspire

Use b >Algebra>Solve System of Equations>Solve System of Equations as shown. The entry solve (y = x 2 − 2x − 2 and y = x + 4, x, y) also gives the same result. The and can either be typed or found in the catalog (k 1 A). Use the Touchpad to move the cursor up to the solution and see all the solutions.

Using the Casio ClassPad The CAS calculator will give exact values for the points of intersection. Turn on the screen keyboard, tap 2D and select . the simultaneous equations symbol Enter the equations in the spaces provided and the variables x, y as shown.

Example 15 Find the coordinates of the points of intersection of the circle with equation (x − 4)2 + y 2 = 16 and the line with equation x − y = 0. Solution Rearrange x − y = 0 to make y the subject. Substitute y = x into the equation of the circle. i.e., ∴ i.e.,

(x − 4)2 + x 2 = 16 x 2 − 8x + 16 + x 2 = 16 2x 2 − 8x = 0 2x(x − 4) = 0 x = 0 or x = 4

y (4, 4)

0

(4, 0)

x

The points of intersection are (0, 0) and (4, 4) Example 16 1 2 Find the point of contact of the line with equation x + y = and the curve with equation 9 3 x y = 1. ISBN 978-1-107-65235-4 © Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2012 Photocopying is restricted under law and this material must not be transferred to another party.



171

Solution 1 2 1 Rewrite the equations as y = − x + and y = 9 3 x 2 1 1 Consider − x+ = 9 3 x 9 ∴ −x + 6 = x and −x 2 + 6x = 9 Therefore

x 2 − 6x + 9 = 0

y y= 1 2 y=– x+ 9 3 0

1 x x

(x − 3)2 = 0, i.e. x = 3 1 The point of intersection is 3, 3 and

Using the TI-Nspire

Use b >Algebra>Solve System of Equations>Solve System of Equations as shown. The multiplication sign between x and y is required as the calculator will consider x y a variable.

Using the Casio ClassPad Turn on the screen keyboard, tap 2D and select the simultaneous equations symbol . Enter the equations in the spaces provided and the variables x, y as shown.



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Exercise 6D Example

14

1 Find the coordinates of the points of intersection for each of the following. a

Example

Example

15

16

y = x2 y=x

b

y − 2x 2 = 0 y−x =0

c

y = x2 − x y = 2x + 1


x 2 + y 2 = 178 x + y = 16

b

x 2 + y 2 = 125 x + y = 15

d

x 2 + y 2 = 97 x + y = 13

e

x 2 + y 2 = 106 x−y=4

c

x 2 + y 2 = 185 x−y=3


x + y = 28 x y = 187

b

x + y = 51 x y = 518

c

x−y=5 x y = 126

4 Find the coordinates of the points of intersection of the straight line with equation y = 2x and the circle with equation (x − 5)2 + y 2 = 25. 5 Find the coordinates of the points of intersection of the curves with equation 1 y= + 3 and y = x. x −2 6 Find the coordinates of the points A and B for which the line with equation x − 3y = 0 meets the circle with equation x 2 + y 2 − 10x − 5y + 25 = 0. 7 Find the coordinates of the points of intersection of the line with equation the circle with equation x 2 + 4x + y 2 = 12.

y x − = 1 and 4 5

8 Find the coordinates of the points of intersection of the curve with equation 1 y= − 3 and the line with equation y = −x. x +2 9 Find the coordinates of the point where the line 4y = 9x + 4 touches the parabola with equation y 2 = 9x. √ 10 Find the coordinates of the point where the line with equation y = 2x + 3 5 touches the circle x 2 + y 2 = 9. 1 11 Find the coordinates of the point where the straight line with equation y = x + 1 4 1 touches the curve with equation y = − . x 12 Find the coordinates of the points of intersection of the curve with equation y = and the line y = x − 1.


2 x −2


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Review

Chapter summary The general expression of a quadratic function is y = ax 2 + bx + c, x ∈ R A quadratic equation may be solved by r Factorising e.g.,

2x 2 + 5x − 12 = 0 (2x − 3)(x + 4) = 0 ∴ x =

3 or −4 2

r Completing the square

x 2 + 2x − 4 = 0 2 b to ‘complete the square’. Add and subtract 2 e.g.,

x 2 + 2x + 1 − 1 − 4 = 0 ∴ (x + 1)2 − 5 = 0 ∴ (x + 1)2 = 5 √ √ ∴ x + 1 = ± 5 ∴ x = −1 + 5 √ −b ± b2 − 4ac r Using the general quadratic formula x = 2a e.g., −3x 2 − 12x − 7= 0 −(−12) ± (−12)2 − 4(−3)(−7) x= 2(−3) √ 6 ± 15 = −3 The number of solutions to the quadratic equation ax 2 + bx + c = 0 can be determined by the discriminant, , where = b2 − 4ac r If b2 − 4ac > 0, the quadratic equation ax 2 + bx + c = 0 has two real solutions r If b2 − 4ac = 0, the quadratic equation ax 2 + bx + c = 0 has one real solution r If b2 − 4ac < 0, the quadratic equation ax 2 + bx + c = 0 has no real solutions

g(x) , where g(x) and h(x) are polynomials in x, is called a h(x) x +1 rational algebraic function, e.g. f (x) = 2 x −1 Some rational algebraic functions may be expressed as a sum of partial fractions. For every A linear factor (ax + b) in the denominator there will be a partial fraction of the form ax + b For every repeated linear factor (cx + d)2 in the denominator there will be two partial B C fractions of the form and (cx + d) (cx + d)2 A function of the form f (x) =



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For every irreducible quadratic factor (ex 2 + f x + g) in the denominator there will be a Dx + E partial fraction of the form 2 (ex + f x + g) 2x + 10 may be expressed as partial fractions in the form e.g. (x + 1) (x − 1)2 A B C + + (x + 1) (x − 1) (x − 1)2 where A = 2, B = −2 and C = 6

Multiple-choice questions 1 If x 2 is written in the form (x + 1)2 + b(x + 1) + c, then the values of b and c are A b = 0, c = 0 B b = −2, c = 0 C b = −2, c = 1 D b = 1, c = 2 E b = 1, c = −2 2 If x 3 = a(x + 2)3 + b(x + 2)2 + c(x + 2) + d, then the values of a, b, c and d are A a = 0, b = −8, c = 10, d = −6 B a = 0, b = −6, c = 10, d = −8 C a = 1, b = −8, c = 10, d = −6 D a = 1, b = −6, c = 12, d = −8 E a = 1, b = −8, c = 12, d = −6 3 The quadratic equation 3x 2 − 6x + 3 = 0 has A two real solutions, x = ±1 B one real solution, x = −1 C no real solutions D one real solution, x = 1 E two real solutions, x = 1 and x = 2 4 The quadratic equation whose solutions are 4 and −6 is C 2x 2 + 4x = 48 A (x + 4)(x − 6) = 0 B x 2 − 2x − 24 = 0 E x 2 + 2x + 24 = 0 D −x 2 + 2x − 24 = 0 5 3 − is equal to 5 x +4 x −2 −2(x − 7) 2(x + 1) −2 C B A (x + 4)(x − 2) (x + 4)(x − 2) (x + 4)(x − 2) −2(x + 13) 2(4x + 13) E D (x + 4)(x − 2) (x + 4)(x − 2) 2x 4 is equal to + 6 (x + 3)2 x +1 8x 2(3x 2 + x + 18) 3x 2 + 13x + 18 A B C (x + 3)2 (x + 1) (x + 3)2 (x + 1) (x + 3)2 (x + 1) 2 3 2 2(3x + 13x + 18) 2(x + 6x + 11x + 2) D E 2 (x + 3) (x + 1) (x + 3)2 (x + 1) 7x 2 + 13 a bx + c 7 If is expressed in the form + , then the values of a, (x − 1)(x 2 + x + 2) x − 1 x2 + x + 2 b and c are A a = 5, b = 0, c = −13 B a = 5, b = 0, c = −10 C a = 5, b = 2, c = −3 D a = 7, b = 2, c = 3 E a = 7, b = 3, c = 13



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Review

4x − 3 is equal to (x − 3)2 3 1 A + x −3 x −3 4 9 D + x − 3 (x − 3)2 8x + 7 is equal to 9 2 2x + 5x + 2 3 2 − A 2x + 1 x + 2 1 −4 + D 2x + 2 x + 1 8

10

B E

B E

−3x 2 + 2x − 1 is equal to (x 2 + 1)(x + 1) 3 2 + A B 2 x +1 x +1 2 3 − D E 2 x +1 x +1

4x 3 − x −3 x −3 15 4 − x − 3 (x − 3)2

2 3 + 2x + 1 x + 2 4 1 − 2x + 2 x + 1

2 3 − +1 x +1 3 2 + 2 x +1 x +1

C

9 4 + x − 3 (x − 3)2

C

−4 1 − 2x + 2 x + 1

C

x2

x2

175

5 2 + +1 x +1

Short-answer questions (technology-free) 1 If (3a + b)x 2 + (a − 2b)x + b + 2c = 11x 2 − x + 4, find the values of a, b and c. 2 Express x 3 in the form (x − 1)3 + a(x − 1)2 + b(x − 1) + c. 3 Prove that, if ax 3 + bx 2 + cx + d = (x + 1)2 ( px + q), then b = 2a + d and c = a + 2d. 1 4 Prove that, if ax 3 + bx 2 + cx + d = (x − 2)2 ( px + q), then b = −4a + d and 4 c = 4a − d. 5 Solve the following quadratic equations for x. b x2 − 2 = x a x 2 + x = 12 e 3x 2 − 2x + 5 = t d 2x 2 − 4x + 1 = 0 6 Solve the equation

c −x 2 + 3x + 11 = 1 f t x2 + 4 = t x

3 1 2 − = for x. x −1 x +2 2

7 Express the following as partial fractions. 7x + 2 7−x −3x + 4 b c a 2−4 2 + 2x − 15 x x (x − 3)(x + 2) e

3x − 4 (x + 3)(x + 2)2

h

−x + 4 (x − 1)(x 2 + x + 1)

6x 2 − 5x − 16 (x − 1)2 (x + 4) −4x + 5 i (x + 4)(x − 3)

f

8 Express each of the following in partial fractions. 1 14(x − 2) b a (x + 1)(x 2 − x + 2) (x − 3)(x 2 + x + 2)


3x − 9 x 2 − 4x − 5 x 2 − 6x − 4 g (x 2 + 2)(x + 1) −2x + 8 j (x + 4)(x − 3)

c

d

3x 3 x 2 − 5x + 4


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y = x2 y = −x

b

x 2 + y 2 = 16 x+y=4

c

x+y=5 xy = 4

10 Find the coordinates of the points of intersection of the line with equation 3y − x = 1 and the circle with equation x 2 + 2x + y 2 = 9.

Extended-response questions 1 A train completes a journey of 240 km at a constant speed. a If it had travelled 4 km/h slower, it would have taken two hours more for the journey. Find the actual speed of the train. b If it had travelled a km/h slower, and still taken two hours more for the journey of 240 km, what would have been the actual speed? (Answer in terms of a.) Discuss the practical possible values of a and also the possible values for the speed of the train. c If the train had travelled a km/h slower, and taken a hours more for the journey of 240 km, and if a is an integer and the speed is an integer, find the possible values for a and the speed of the train. 2 An upholsterer purchased some fabric for $a. If he had bought the fabric from another supplier who charged $b per metre more he would have received b metres less for the same amount of money. a How many metres did he purchase, in terms of a and b? b If a and b, and the number of metres purchased, are natural numbers, find the possible values of a given a < 100. 3 Two trains are travelling at uniform speeds. The slower train takes a hours longer to cover b km. It travels 1 km less than the faster one in c hours. a What is the speed of the faster train, in terms of a, b and c? b If a, b and c, and the speeds of the trains, are rational numbers, find five sets of values for a, b and c. Choose and discuss two sensible sets of values. 4 A tank can be filled using two pipes. The smaller pipe alone will take a minutes longer than the larger pipe alone to fill the tank. Also the smaller pipe will take b minutes longer to fill the tank than when both pipes are used. a Find, in terms of a and b, how long it will take each of the pipes to fill the tank. b If a = 24 and b = 32, find how long it takes for each of the pipes to fill the tank. c If a and b are consecutive positive integers, find five pairs of values of a and b such that b2 − ab is a perfect square. Interpret these results in the context of this tank problem.



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C H A P T E R

7

Revision

Revision of chapters 1–6 7.1

Multiple-choice questions 1 If P2 = 4I, then P−1 equals 1 1 P B P A 2 4 ⎡ 2 If R = [5

C ⎤

1 I 2

0 ⎢ ⎥ 1] and S = ⎣−1⎦, then RS is 2

3

A not defined

−3

2]

⎡

0 ⎢ C ⎣−5 10

B [−1] ⎡

D [0

D 2P

⎤ 0 ⎢ ⎥ E ⎣−3⎦ 2

0 −3 6

E 4P

⎤ 0 ⎥ −1⎦ 2

9 8 3 If A = , then det (A) equals −11 5 1 1 D 17 E 133 C B − A −43 133 43 ⎡ ⎤ 1 ⎢ ⎥ 4 If A = ⎣2⎦ and B = [−2 6 4], then BA has dimensions 5 A 1×1 B 3×1 C 1×3 D 3×3 E 3×2 5 2 2 −1 5 4 5 Given that A = ,B= and C = , then if AX + B = C, 2 1 6 7 8 9 X equals 1 −2 19 −2 19 −1 1 C B A 20 −2 −2 6 6 4 0 1 1 3 3 −10 D E 20 4 5 −4 10



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2 −1 4 6 Let P = ,Q = 3 2 6 elements in X is A 0

2 2 ,R = 5 −3

1 and X = PQR. The number of zero 2

B 1 C 2 3 5 7 If X = , then X−1 is −1 −2 2 5 −2 −5 A B −1 −3 1 3 −3 −1 3 −1 D E 5 2 −5 −2 4 6 8 The determinant of the matrix is 2 4

D 3

A 16

⎡ 1 ⎢ C ⎣ 3 −1

C −16

B 4

E 4

1 4

D

5 7 9 If S = , then S−1 is 2 2 5 7 5 −7 A − B 2 2 −2 5 1 −2 1 −2 −7 7 D E 4 4 −2 −5 2 −5

1⎤ 5⎥ 1⎦ − 2

E −4

1 −2 C − 4 2

7 −5

10 In algebraic form, five is seven less than three times one more than x can be written as A 5 = 7 − 3(x + 1) D 5 = 7 − 3x + 1 11

2 3 − is equal to x −3 x +3 x + 15 A 1 B x2 − 9

B 3x + 1 = 5 − 7 E 5 = 3x − 4

C

15 x −9

C (x + 1) − 7 = 5

D

x −3 x2 − 9

E −

1 6

12 p varies directly as x and inversely as the square of y. If x is decreased by 30% and y is decreased by 20%, the percentage change in p is best approximated by A increase by 10% D decrease by 9.4%

B decrease by 10% E no change

C increase by 9.4%

13 The sum of the odd numbers from 1 to n inclusive is 100. The value of n is A 13

B 15

C 17

D 19

E 21

14 If the sum of the first n terms of a geometric sequence is 2n+1 − 2 , the nth term of the geometric sequence is A 2n−1

B 2n

C 2n − 1


D 2n−1 + 1

E 2n + 1


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Chapter 7 — Revision of chapters 1–6

179

A {1, 2, 3, 4, 5, 6, 7} D {3, 4}

B {1, 2, 3, 4, 5, 6} E {2, 3, 4, 5, 6, 7}

C {2, 3, 4}

17 The price of painting the outside of a cylindrical tank (the bottom and top are not painted) of radius r and height h varies directly as the total surface area. If r = 5 and h = 4, the price is $60. The price when r = 4 and h = 6 is A $45

B $57.60

C $53.50

D $62.80

Revision

15 If m ∝ n and m = 9 when n = 4, then k, the constant of variation, equals 9 4 A B 13 C 36 D E 5 4 9 16 If A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6} and C = {3, 4, 5, 6, 7} then A ∩ (B ∪ C) is equal to

E $72

18 If x ∝ y and x = 8 when y = 2, the value of x when y = 7 is A 20

B 13

C 11

19 The recurring decimal 0.7˙ 2˙ is equal to 72 72 C A B 101 100

72 99

D 28

D

72 90

E 1.75

E

73 90

20 If x varies directly as y 2 and inversely as z, the percentage increase of x when y is increased by 25% and z is decreased by 20% is best approximated by A 5% B 50% C 85% 3 x −4 − + is equal to 21 x −1 1−x x −1 7x 1 C A 1 B −1 D x −1 1−x x +2 5 − is equal to 22 3 6 x −3 2x + 4 2x − 1 A B C 6 6 6 1 23 If a = 1 + , then b equals 1+b 1 1 1 −1 C A 1− B 1+ a−1 a−1 a−1

D 95%

E 100%

E none of these

D

2x − 5 6

E

x −3 3

D

1 +1 a+1

E

1 −1 a+1

24 When the repeating decimal 0.3˙ 6˙ is written in simplest fractional form, then the sum of the numerator and denominator is A 15

B 45

C 114

D 135

3 x 2x − y = , then equals 2x + y 4 y 3 7 2 C B D A 4 2 7 E Not possible unless the values of x and y are known

E 150

25 If

26 The sum to infinity of the series A 2

B 1

4 3

1 1 1 1 − + − + · · · is 2 4 8 16 1 1 C D 2 3


E

2 3 Cambridge University Press

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27 If x varies directly as y and inversely as the square of z and x = 10 when y = 4 and z = 14, then when y = 16 and z = 7, x equals A 180

B 160

3 = 4, then y equals 3+y 9 1 B − A 4 4

C 154

D 140

E 120

D 0

E −

28 If

C

9 4

4 9

29 The coordinates of the point where the lines with equations 3x + y = −7 and 2x + 5y = 4 intersect are A (3, −16)

B (−3, 2)

C (3, −2)

D (−2, 3)

m+2 2−m 1 − = then m is equal to 4 4 2 1 C A 1 B −1 2

E no solution

30 If

E −

D 0

1 2

31 46 200 can be written as A 2 × 3 × 5 × 7 × 11 C 2 × 32 × 5 × 72 × 11 E 22 × 3 × 53 × 7 × 11

B 22 × 32 × 52 × 7 × 11 D 23 × 3 × 52 × 7 × 11

32 Three numbers, y, y − 1 and 2y − 1, are consecutive numbers of an arithmetic sequence. y equals A −1

B 1

C 0

E −2

D 2

33 If the integers n + 1, n − 1, n − 6, n − 5, n + 4 are arranged in increasing order of magnitude then the middle number is A n+1

B n−1

C n−6

D n−5

1 , and y is multiplied by 5, then x will be y A decreased by 5 B increased by 5 D divided by 5 E none of these

E n+4

34 If x ∝

C multiplied by 5

35 An arithmetic sequence has 3 as its first term and 9 as its fourth term. The eleventh term is A 23 36 The expression A

7n − 1 1 − n2

B 11

C 63

D 21

E none of these

3 4 + is equal to n+1 n−1 B

1 − 7n 1 − n2

C

7n − 1 n2 + 1

D

n2

7 −1

E

7 n

37 If the second number is twice the first number and a third number is half the first number and the three numbers sum to 28, then the numbers are A (8, 16, 4) B (2, 3, 12) √ √ 38 ( 7 + 3)( 7 − 3) is equal to A −2

B 10

C (7, 9, 11)

C

√ 14 − 19


D (6, 8, 16)

E (12, 14, 2)

√ D 2 7−9

E 45 Cambridge University Press

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P Q 13x − 10 = + then the values of P and Q are 2 2x − 9x + 4 x − 4 2x − 1

A P = 1 and Q = 1 D P = −6 and Q = 1

B P = −1 and Q = 1 E P = 1 and Q = −6

C P = 6 and Q = 1

40 The first term of a geometric sequence is a and the infinite sum of the geometric sequence is 4a. The common ratio of the geometric sequence is 3 4 3 A 3 B 4 C E − D − 4 3 4 41 If

Revision

39 If

181

P Q 5x = + , then (x + 2)(x − 3) x +2 x −3

A P = 2, and Q = 3 D P = −2, and Q = −3

B P = 2, and Q = −3 E P = 1, and Q = 1

C P = −2, and Q = 3

42 If n is a perfect square then the next largest perfect square greater than n is A n+1

B n2 + 1

C n 2 + 2n + 1

D n2 + n

√ E n+2 n+1

43 The area of triangle varies directly as the base length provided the altitude is constant. If the area equals 14 when the base is 2.4, then the base length (correct to three decimal places) when the area is 18 will equal A 3.086

B 5.000

C 6.400

44 Which of the following is not a rational number? √ 3 C 5 A 0.4 B 8 1 a 1 45 If = and = a − b, then x + y equals x b y ba − b2 + a a 2 − b2 2 C B D A a(a − b) a a

D 9.600

D

a2

E 0.324

√ 16

2a − b2

E 4.125

E

−2b − b2

a2

46 9x 2 − 4mx + 4 is a perfect square when m equals A 5

B ±12

C 2

D ±1

E ±3

47 If x = (n + 1)(n + 2)(n + 3) where n is a positive integer, then x is not always divisible by A 1

B 2

C 3

D 5

E 6

48 The numbers −4, a, b, c, d, e, f, 10 are consecutive terms of an arithmetic sequence. The sum a + b + c + d + e + f is equal to A 6

B 10

C 18

D 24

E 48

49 If n and p are both odd numbers, which one of the following numbers must be an even number? A n+p

B np

C np + 2


D n+ p+1

E 2n + p Cambridge University Press

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7.2 Extended-response questions 1 The diagram represents a glass containing milk. When the height of the milk in the glass is h cm, the diameter, d cm, of the surface of the milk is given by the formula h d = +6 5 h cm a Find d when h = 10 b Find d when h = 8.5

d cm

c What is the diameter of the bottom of the glass? d The diameter of the top of the glass is 9 cm. What is the height of the glass? 360 gives the size of each interior angle, A◦ , of a regular polygon 2 The formula A = 180 − n with n sides. a Find the value of A when n equals i 180 ii 360 iii 720 iv 7200 b As n becomes very large i what value does A approach?

ii what shape does the polygon approach?

c Find the value of n when A = 162.

d Make n the subject of the formula.

e Three regular polygons, two of which are octagons, meet at a point so that they fit together without any gaps. Describe the third polygon. w

3 The figure shows a solid consisting of three parts, a cone, a cylinder and a hemisphere, all of the same base radius. a Find in terms of w, s, t and ␲ the volume of each part. b

s

i If the volume of each of the three parts is the same, find the ratio w : s : t. ii If also w + s + t = 11, find the total volume in terms of ␲.

t

4 The cost, $C, of manufacturing each jacket of a particular type is given by the formula C = an + b

for

0 < n ≤ 300

where a and b are constants and n is the size of the production run of this type of jacket. For making 100 jackets, the cost is $108 each. For making 120 jackets, the cost is $100 each. a Find the values of a and b. b Sketch the graph of C against n for 0 < n ≤ 300. c Find the cost of manufacturing each jacket if 200 jackets are made. d If the cost of manufacturing each jacket is $48.80, find the size of the production run.



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5 a In the diagram, OPQ is a sector of radius R. A circle, centre C1 and radius r1 is inscribed in this sector.

183

P

r1

r2 r3 O

60°

C3 r3

C1 C2 r2

r1 Q R

i Express OC1 in terms of R and r1 . r1 1 ii Show that = and hence express r1 in terms of R. OC1 2 b Another circle, centre C2 , is inscribed in the sector as shown. i Express OC2 in terms of r2 and R. ii Express r2 in terms of R. c Circles C3 , C4 , . . . are constructed in a similar way. Their radii are r3 , r4 , . . . respectively. It is known that r1 , r2 , r3 , . . . is a geometric sequence. i Find the common ratio. ii Find rn . iii Find the sum to infinity of the sequence, and interpret the result geometrically. iv Find in terms of R and ␲, the sum to infinity of the areas of the circles with radii r1 , r2 , r3 , . . . . 6 At the beginning of 1997, Andrew and John bought a small catering business. The profit, $P, in a particular year is given by P = an + b where n is the number of years of operation and a and b are constants. a Given the table, find the values of a and b. Year Number of years of operation (n) Profit b Find the profit when n = 12.

1997 1 −9000

2001 5 15 000

c In which year was the profit $45 000?

7 Two companies produce the same chemical. For Company A the number of tonnes produced increases by 80 tonnes per month. For Company B production increases by 4% per month. Each company produced 1000 tonnes in January 2003. (Let n be the number of months of production. Use n = 1 for January 2003.) a Find, to the nearest tonne where appropriate, i the production of Company A in the nth month ii the production of each company in December 2004 (i.e. for n = 24) (cont’d)



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iii the total production of Company A over n months (starting with n = 1 for January 2003) iv the total production of each company for the period January 2003 to December 2004 inclusive. b Find in which month of which year the total production of Company A passed 100 000 tonnes. 1 8 The square shown has each side of length one. a The perimeter of the square is denoted by P1 . What is the value of P1 ?

1

1

1 b A new figure is formed by joining two squares of 1 side length to this square, as shown. The 2 perimeter is denoted by P2 . What is the value of P2 ?

1 1

1

2

1

1

2

2 1 2

1

1

1

2

1

c What is the perimeter, P3 , of this figure?

1

1

2

1

1

2

2

2

1

1 1 1 1 4 4 4 4 1

1

1

1 1 1

2

2

4 4 4 4

4

d It is known that P1 , P2 , P3 , . . . are the terms of an arithmetic sequence with first term P1 . What is the common difference ? e

i Find P4 . ii Find Pn in terms of Pn−1 . iii Find Pn in terms of n. iv Draw the diagram of the figure corresponding to P4 .

3x cm 9 A piece of wire 28 cm long is cut into two parts, one to make a rectangle three times as long as it is wide and the other to make a square. a What is the perimeter of the rectangle in terms of x?

x cm

b What is the perimeter of the square in terms of x? c What is the length of each side of the square in terms of x? Let A be the sum of the areas of the two figures. d Show that A = 7(x 2 − 4x + 7) e Use a graphics calculator to help sketch the graph of A = 7(x 2 − 4x + 7) for 0 < x < 5 f Find the minimum value that A can take and the corresponding value of x.



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185

b Let $I be the income from selling x plates. Write an expression for I in terms of x. c On the one set of axes, sketch the graphs of I against x and C against x. d How many plates must be sold for the income to equal the cost of production? e How many plates must be sold for a profit of $2000 to be made?

Revision

10 A particular plastic plate manufactured at a factory sells at $1.50. The cost of production consists of an initial cost of $3500 and then $0.50 per plate. Let x be the number of plates produced. a Let $C be the cost of production of x plates. Write an expression for C in terms of x.

f Let P = I − C. Sketch the graph of P against x. What does P represent? 11 n is a natural number less than 50 such that n + 25 is a perfect square. a Show that there exists an integer a such that n = a(a + 10) b Any natural number less than 100 can be written in the form 10 p + q where p and q are digits. For this representation of n show that q = p2 . c Give all possible values of n. √ √ √ 12 a i For the equation 7x − 5 − 2x = 15 − 7x square both sides to show that this equation implies

8x − 10 = 14x 2 − 10x ii Square both sides of this equation and simplify to form the equation x 2 − 3x + 2 = 0

1

iii The solutions to the equation 1 are x = 1 or x = 2. Test these solutions for the equation √ √ √ 7x − 5 − 2x = 15 − 7x and hence show that x = 2 is the only solution for the original equation. b Use the techniques of a to solve the equations √ √ √ √ √ √ i x +2−2 x = x +1 ii 2 x + 1 + x − 1 = 3 x 13 A geometric series is defined by 1 x +1 1 − ··· − + x2 x x +1 a Let r be the common ratio. Find r in terms of x. b

i Find the infinite sum if x = 1.

1 ii Find the infinite sum if x = − . 4 iii Find the infinite sum if x = 2. c Find the possible values of x for which the infinite sum is defined.



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y

14 a The area, A, of the shaded region varies directly as the cube of a. 4 i If A = when a = 2, find an expression 3 for A in terms of a. ii Find A when a = 3. iii If A = 4500, find a.

y = x(x – a)

x

0 y

b The area, A1 , of the shaded region varies directly as the cube of a i If A1 = 1152 when a = 24, find an expression for A1 in terms of a. ii Find A1 when a = 18. iii Find a when A1 = 3888.

y = x(x – a)

x

0 –a2

a , 2 4 y

c The area, A2 , of the shaded region varies partly as the reciprocal of a and partly as the reciprocal of b.

1 y= 2 x

i Find A2 in terms of a and b if, 1 and 2 1 when a = 3 and b = 4, A2 = 12 Find A2 when a = 1 and b = 6. 1 Find A2 when a = and b = 3. 4 1 Find A2 when a = and b = 100. 100 1 and b = 1000. Find A2 when a = 1000 when a = 1 and b = 2, A2 =

ii iii iv v

a

b

x

15 In a vegetable garden carrots are planted in rows parallel to the fence.

nce

Fe

rabbit burrow

row

0.5 m

1.5 m

1

row

1.5 m

2

row

3

row

4

1.5 m

a Calculate the distance between the fence and the 10th row of carrots. b If tn represents the distance between the fence and the nth row, find a formula for tn in terms of n.



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d A systematic rabbit has its burrow under nce Fe the fence as shown in the diagram. It runs to the first row, takes a carrot and returns it to its burrow. It then runs to the 1 2 row row rabbit second row, collects a carrot and burrow Trip 1 returns it to its burrow. Trip 2 It continues in this way until it has 15 carrots. Calculate the shortest distance the rabbit has to run to accomplish this.

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c Given that the last row of carrots is less than 80 m from the fence, what is the largest number of rows possible in this vegetable garden?

16 The potential energy, P joules, of a body varies jointly as the mass, m kg, of the body and the height, h m, of the body above the ground. a For a body of mass 5 kg i find P in terms of h if P = 980 when h = 20 ii sketch the graph of P against h iii find P if h = 23.2. b

i Find P in terms of h and m if P = 980 when h = 20 and m = 5. ii Find the percentage change in potential energy if the height (h m) is doubled and the mass remains constant. iii Find the percentage change in potential energy if a body has a quarter of the original height (h m) and double the original mass (m kg).

c If a body is dropped from a height, h m, above ground level its speed, V m/s, when it √ reaches the ground is given by V = 19.6h. i Find V when h = 10. ii Find V when h = 90. d In order to double the speed a given body has when it hits the ground, by what factor must the height from which it is dropped be increased? 17 In its first month of operation a soft drink manufacturer produces 50 000 litres of a type of soft drink. In each successive month the production rises 5000 litres a month. a i The quantity of soft drink, tn , produced in the nth month can be determined from a rule of the form tn = a + (n − 1)d Find the values of a and d. ii In which month will the factory double its original production? iii How many litres in total will be produced in the first 36 months of operation? b Another soft drink manufacturer sets up a factory at the same time as the first. In the first month the production is 12 000 litres. The production of this factory increases by 10% every month. i Find a rule for qn , the quantity of soft drink produced in the nth month. ii Find the total amount of soft drink produced in the first 12 months. c If the two factories start production in the same month, in which month will the production of the second factory exceed the production of the first factory? ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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18 In a certain country grain production and population statistics are produced. In December 1986 the population of the country was 12.5 million. In 1986 the grain production was 10 million tonnes. It was found that since then the population has grown by 5% each year and grain production has increased by 0.9 million tonnes each year. Let P1 denote the population in December 1986. Let p2 denote the population in December 1987. Pn denotes the population n − 1 years after December 1986. Let t1 denote the grain production in 1986. Let t2 denote the grain production in 1987. tn denotes the grain production in the (n − 1)th year after 1986. a Find, in millions of tonnes, the grain production in i 1992 ii 1999 b Find an expression for tn . c Find the total grain production for the 20 years starting 1986. d How many years does it take for the grain production to double? e Find an expression for Pn . f How many years does it take for the population to double? 19 The diagram shows a straight road OD where OD = 6 km. A hiker is at A, 2 km from O. The hiker can walk at 3 km/h when off-road but at 8 km/h along the road. a Calculate the time taken, in hours and minutes, correct to the nearest minute, if he hikes directly to X then along the road to D where OX = 3 km.

6 km O

x km X

D

2 km A

b Calculate OX, correct to one decimal place (in km) if the total time taken was 1 1 hours. 2 20 Seventy-six photographers submitted work for a photographic exhibition in which they were permitted to enter not more than one photograph in each of the three classes, black and white (B), colour prints (C), transparencies (T ). Eighteen entrants had all their work rejected while 30 B, 30 T and 20 C were accepted. From the exhibitors, as many showed T only as showed T and C. There were three times as many exhibitors showing B only as showing C only. Four exhibitors showed B and T but not C. a Write the last three sentences in symbolic form. b Draw a Venn diagram representing the information. c Find i n(B ∩ C ∩ T )

ii n(B ∩ C ∩ T )



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b with b = 0 and c = 0 d

a Find i A2

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a 21 Let A = c

189

ii 3A

b If A2 = 3A − I, show that i a+d =3 ii det(A) = 1 c If A has the properties r a+d =3 r det(A) = 1 show that A2 = 3A − I. 22 The trace of square matrix A is defined to be the sum of the leading diagonal of A, and it is denoted by Tr(A). 6 −3 For example, if A = , Tr(A) = 8 2 2 a Prove each of the following for any 2 × 2 matrices X and Y. i Tr(X + Y) = Tr(X) + Tr(Y) ii Tr(−X) = −Tr(X) iii Tr(XY) = Tr(YX) b Use the results of a to show that there do not exist 2 × 2 matrices X and Y such that XY − YX = I.



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C H A P T E R

8 Transformations Objectives To define translations To define reflections in the axes and in the line y = x To define dilations from the x axis and the y axis To apply these transformations to points and figures To find algebraic rules for these transformations To find the composition of two transformations and give the rule for this transformation To apply transformations to graphs To determine the rule which transforms one graph to another (within a suitable family of graphs) To sketch the graph of the absolute value function, the integer part function and transformations of these graphs To describe transformations with function notation

Introduction In this chapter three different types of transformations of the cartesian plane are discussed. These are Translations Reflections Dilations. These transformations are very useful in the graphing of functions. A transformation is a rule which ‘associates’ each point in the cartesian plane to another point in the plane. These points uniquely define each other through the rule. For example, one transformation can be defined by: ‘Add 5 to each x coordinate.’ This can be expressed algebraically (x, y) → (x + 5, y) e.g., (1, 6) → (6, 6)



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191

This is read as: ‘The point with coordinates (1, 6) is mapped to the point with coordinates (6, 6)’. A formal definition is the following: A transformation T is a mapping from R 2 to R 2 such that if T (a, b) = T (c, d) then a = c and b = d. The transformation defined above can be written T : R 2 → R 2 , T (x, y) = (x + 5, y) This formal notation is avoided in this book and the transformation will be defined by the rule given in the form (x, y) → (x + 5, y)

8.1

Translations The transformation defined above, i.e. (x, y) → (x + 5, y), is an example of a translation. A translation is a transformation for which each point in the plane is moved the same distance in the same direction. In this section, a vector will mean a column of two numbers. The first number indicates a ‘move’ in the positive or negative direction of the x axis and the second indicates a ‘move’ in the positive or negative direction of the y axis. A directed line segment is used to illustrate a vector. 2 For example, the vector is the vector ‘2 to the right and 3 up’. 3 2 3 The image of the point (1, 2) would be (3, 5) under the translation 2 determined by the vector . 3 The top number gives the displacement in the positive or negative direction of the x axis and the lower number gives the displacement in the positive or negative direction of the y axis.

1 2

–4 3

3 –2

–2 –1

If the top number is negative, the displacement is to the left and if the lower number is negative, the displacement is downwards.



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Vectors can be used to describe translations. They will be studied in a more general context in Chapter 15. Example 1 The point Ahas coordinates (1, 2). Find the image of A under the translation defined by the −4 vector . 2 Solution y A'

4

The image of A is (−3, 4).

2 –4

–2

A 0

2

x

Example 2 Find the vector which defines the translation which takes A(3, 4) to A (2, 6). Solution Let (3, 4) → (2, 6) a The vector defines this translation if (3 + a, 4 + b) = (2, 6) b This implies 3+a =2 and 4 + b = 6 i.e., a = −1 and b=2 −1 The vector is 2 Example 3 A translation is defined by the rule (x, y) → (x − 3, y + 2) and the point A with coordinates (a, b) is mapped to A (7, −1). Find the values of a and b. Solution (a, b) → (a − 3, b + 2) = (7, −1) ∴ a − 3 = 7 and b + 2 = −1 a = 10 and b = −3



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Exercise 8A Example

1

1 Find the images of the points in each of the following under the translations described by the given vectors. 4 2 4 a (3, 1), b (4, 5), c (−2, 4), 2 4 3 −2 −3 d (3, 2), e (4, 5), 3 2

Example

2

2 For each of the following, find the vectors describing the translations that map A to A . a A(1, 2), A (5, 3) d A(−3, 0), A (4, 6)

b A(3, 8), A (2, 9) e A(−4, −3), A (0, 0)

c A(1, 2), A (5, 4)

3 In each of the following the given point A is the image of an object point A under the translation described by the given vector. Find the coordinates of A. 2 1 2 b A (3, 6), c A (0, 6), a A (7, 9), 3 4 3 y

4 Give the vectors describing the translations which map

C N

4 a ABC to PQR A b ABC to LMN L M 2 c XYZ to ABC –4 –2 Z 0 2 d ABC to ABC 5 Draw axes for x and y from −4 to 5. Draw P Y X the following triangles. –4 ABC withA(2, 2), B(4, 2), C(2, 5) PQR with P(1, −2), Q(3, −2), R(1, 1) XYZ with X (−3, 1), Y (−1, 1), Z (−3, 4). Give the vectors describing the translations which map

a ABC to PQR c PQR to XYZ

B R 6

x

Q

b PQR to ABC d ABC to ABC

3 6 a Find the image of the point (2, 3) under the translation determined by the vector 4 −1 . followed by the translation determined by the vector 5 b Find the image of the point (−5, 6) under the translation determined by the vector 3 −1 followed by the translation determined by the vector . 4 5 c Describe how one translation may be used to obtain the final image in a and b.



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7 A translation has a rule (x, y) → (x − 5, y + 3). Find a the image of the point (1, 3) under this translation b a and b if (a, b) → (6, 7) under this translation. 8 A translation has a rule (x, y) → (x + 1, y).

y (3, 9)

a The points (0, 0), (1, 1), (2, 4), (3, 9) all lie on the graph of y = x 2 . Find the image of each of these points under the translation. b Sketch the graph of y = x 2 for x ≥ 0 as shown, and complete. c Describe the image of all the points on the graph of y = x 2 under this transformation.

(2, 4)

(1, 1) (0, 0)

( , )

(3, 4)

( , ) x

(1, 0)

8.2 Reflections m

P1 is the image of P under a reflection in the line m. The line m is the perpendicular bisector of line PP1 . P1

A1 is the image of A under the transformation ‘reflection in the y axis’. A2 is the image of A under the transformation ‘reflection in the x axis’. A3 is the image of A under the transformation ‘reflection in the line y = x’.

P

y y=x 3 A3(1, 2)

2 A1(–2, 1)

A(2, 1)

1

x –3

–2

0

–1 –1

1

2

3 4 A2(2, –1)

–2

For reflection in the x axis the rule is (x, y) → (x, −y) For reflection in the y axis the rule is (x, y) → (−x, y) For reflection in the line y = x the rule is (x, y) → (y, x) Example 4 The triangle ABC has coordinates A(1, 0), B(1, 6) and C(4, 6). Find the image of triangle ABC under a reflection in b the line y = x. a the x axis



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Solution a

A(1, 0) → A (1, 0) B(1, 6) → B (1, −6) C(4, 6) → C (4, −6)

y 6

B(1, 6)

5

C"(6, 4)

4

b

A(1, 0) → A (0, 1) B(1, 6) → B (6, 1) C(4, 6) → C (6, 4)

y=x

C(4, 6)

3 2 A"(0, 1) 1 –1

B"(6, 1)

A(1, 0) 01

2

3

4

5

x

6

–2 –3 –4 –5 –6

B'(1, –6)

C'(4, –6)

Exercise 8B Example

4

1 Draw axes for x from −5 to 5 and for y from 0 to 5. Draw triangle ABC by plotting A(1, 2), B(3, 2) and C(3, 5). Draw the image AB C when ABC is reflected in the y axis. 2 Draw axes for x from 0 to 5 and for y from −2 to 2. Draw triangle PQR where P is (1, −1), Q is (5, −1) and R is (4, 0). Draw the image P Q R when PQR is reflected in the x axis. 3 Draw axes for x and y from −5 to 0. Draw rectangle WXYZ where W is (−3, −1), X is (−3, −2), Y is (−5, −2) and Z is (−5, −1). Draw the mirror line y = x. Draw the image W X Y Z when WXYZ is reflected in the mirror line. 4 Draw axes for x and y from −1 to 8. Plot the points A(2, 1), B(5, 1), C(7, 3) and D(4, 3). Draw the parallelogram ABCD and its image by reflection in the line y = x. 5 Draw axes for x and y from −6 to 7. Draw triangle ABC where A is (−6, −2), B is (−3, −4) and C is (−2, −1). Draw the following images of triangle ABC a triangle A1 B1 C1 by reflection in the y axis b triangle A2 B2 C2 by reflection in the line y = −x (this is the straight line through the points (2, −2), (−4, 4)) c triangle A3 B3 C3 by reflection in the x axis. 6 Find the image of (6, −2) under each of the following a reflection in the line y = x c reflection in the line y = 0

b reflection in the line x = 0



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7 Find the image of (0, −1) under each of the following a reflection in the line y = x c reflection in the line y = 0

b reflection in the line x = 0 d reflection in the line y = −x

8.3 Dilations from the axes For reflections and translations, lengths and angles are preserved. In this section dilations from the axes are introduced. These transformations do not preserve distances or angles. The transformation ‘dilation from the y axis of factor k’, is defined by the rule (x, y) → (kx, y); k ∈ R +

y

For example, for k = 2, the unit square

B(0, 1)

C(1, 1)

C'(2, 1)

D(1, 0)

D'(2, 0)

A(0, 0), B(0, 1), C(1, 1), D(1, 0) is transformed to the rectangle A(0, 0), B(0, 1), C (2, 1), D (2, 0) A(0, 0)

x

Example 5 Triangle ABC has vertices A(1, 2), B(3, 4), C(5, 1). Find the image of the triangle under a dilation of factor 2 from the y axis. Solution (3, 4) → (6, 4) (1, 2) → (2, 2) (5, 1) → (10, 1)

y 5 B(3, 4)

4

B'(6, 4)

3 2

A'(2, 2) A(1, 2)

1 0

C'(10, 1)

C(5, 1)

x 1

2

3

4

5

6

7

8

9

10

The transformation ‘dilation from the x axis of factor k’ is defined by the rule: (x, y) → (x, ky); k ∈ R + For example, a triangle with vertices A(1, 2), B(3, 4), C(5, 1) is mapped to the triangle with vertices A (1, 4), B (3, 8), C (5, 2) under a dilation of factor 2 from the x axis.



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Chapter 8 — Transformations y

The unit square with vertices A(0, 0), B(0, 1), C(1, 1), D(1, 0) is mapped to the rectangle with vertices A(0, 0), B (0, 2), C (1, 2), D (1, 0)

B'(0, 2)

C'(1, 2)

B(0, 1)

C(1, 1)

A(0, 0)

D(1, 0)

x

Exercise 8C Example

5

1 Find the image of the point (1, 3) under each of the following a a dilation of factor 3 from the x axis c a dilation of factor 4 from the y axis.

b a dilation of factor 2 from the y axis

2 Give the rule for the dilation from the x axis which maps (1, 2) → (1, 8). 3 Give the rule for the dilation from the y axis which maps (3, −2) → (9, −2). 4 a Find the image of the unit square A(0, 0), B(0, 1), C(1, 1), D(1, 0) under a dilation from the x axis of factor 3. b Find the image of the unit square A(0, 0), B(0, 1), C(1, 1), D(1, 0) under a dilation from the y axis of factor 3. 5 Triangle ABC has vertices A(0, 0), B(3, 0), C(3, 4). a Find the image of triangle ABC under 1 i a dilation of factor from the x axis 2 1 ii a dilation of factor from the y axis. 2 b Illustrate triangle ABC and its images.

8.4

Rules for transformations Transformations can be described through a rule given in terms of ordered pairs. For example, it has been seen that (x, y) → (x + 2, y + 3) describes the transformation of 2 units in the positive direction of the x axis and 3 units in the positive direction of the y axis.



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Example 6 Find the rules for each of the following transformations. 3 a a translation determined by the vector −2 b a reflection in the line y = −x

c a dilation of factor

1 from the x axis. 4

Solution a (x, y) → (x + 3, y − 2)

b (x, y) → (−y, −x)

1 c (x, y) → x, y 4

A point (a, b) is said to be invariant under a transformation if (a, b) → (a, b) under that transformation. The transformation which maps (x, y) → (x, y) for all (x, y) ∈ R 2 is called the identity transformation. Example 7 A transformation has rule (x, y) → (2x + 3, y + 4). a Find the image of (2, 3) under this transformation. b Find the coordinates of the point which maps to (11, 12). Solution a If x = 2 and y = 3 (x, y) → (2 × 2 + 3, 3 + 4) = (7, 7) i.e., (2, 3) → (7, 7)

b If (2x + 3, y + 4) = (11, 12) Then 2x + 3 = 11 and y + 4 = 12 ∴ x = 4 and y = 8

Exercise 8D Example

6

1 For each of the following transformations find i the rule

ii the invariant points (if they exist). 5 a a translation determined by the vector 6 b a dilation from the x axis of factor 4 d a translation determined by the vector

1 c a dilation from the y axis of factor 3

−2 3

e a reflection in the line y = x Example

7

f a reflection in the y axis

2 A transformation has rule (x, y) → (3 − x, 2y + 1) a Find the image of (2, 3) under this transformation.



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b Find the coordinates of the point which maps to (−6, 12). c Find the coordinates of the point which is invariant under this transformation. 3 A transformation has rule (x, y) → (3 − y, 4 − x). Find the coordinates of the invariant point. 4 A transformation has rule (x, y) → (3 − 2x, −2y + 1). a Find the image of (4, −1) under this transformation. b Find the coordinates of the point which maps to (7, 12). c Find the coordinates of the point which is invariant under this transformation. 5 A transformation has rule (x, y) → (−x, −2y). a Find the image of (−1, 3) under this transformation. b Find the coordinates of the point which maps to (0, 0). c Find the coordinates of the point which is invariant under this transformation.

8.5

Composition of transformations Consider a transformation determined by the vector

2 followed by a reflection in the line 3

x = 0. The rules for these two transformations are (x, y) → (x + 2, y + 3) and (x, y) → (−x, y) respectively. The triangle A(1, 1), B(5, 1), C(5, 2) is mapped to A (−3, 4), B (−7, 4), C (−7, 5) y 6 C"

C'

5

B"

A"

4

A'

B'

3 C

2 1 –7 –6 –5 –4 –3 –2 –1 0

A

B x 1

2

3

4

5

6

7

One rule can be used to describe the transformation of triangle ABC to triangle A B C . Consider (x, y) → (x + 2, y + 3) → (−(x + 2), y + 3) translation reflection The rule is (x, y) → (−(x + 2), y + 3) Note:

(1, 1) → (−3, 4)

(5, 1) → (−7, 4)

(5, 2) → (−7, 5)

This new rule is called the composition of the two transformations. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 8 Find the rule for the transformation defined bya dilation of factor 2 from the x axis followed −1 by a translation determined by the vector . 3 Solution The dilation is given by the rule (x, y) → (x, 2y) and the translation by the rule (x, y) → (x − 1, y + 3). The composition is determined by (x, y) → (x, 2y) → (x − 1, 2y + 3) dilation translation i.e., the rule is (x, y) → (x − 1, 2y + 3)

Exercise 8E Example

8

1 For each of the following find the rule for i transformation of column A followed by transformation of column B ii transformation of column B followed by transformation of column A. Column A

Column B

a

a translation determined by the 2 vector 3

a translation determined by the −3 vector −2

b

a dilation from the x axis of factor 2 a translation determined by the 4 vector 5

a dilation from the y axis offactor 2

d

a translation determined by the −1 vector 2

a reflection in the y axis

e f g

a reflection in the line y = x a reflection in the line x = 2 a dilation from the y axis of 1 factor 2

a reflection in the line y = −x a reflection in the x axis a translationdetermined by the −1 vector 2

h

a dilation from the y axis of factor 2

a translation determined by the 2 vector −3

c

a dilation from the x axis of factor 3



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8.6

Applying transformations to graphs of functions As mentioned in the introduction, transformations have an important role in graphing functions. In this section techniques are developed for graphing functions using transformations. Consider {(x, y) : y = x 2 }. This is the set of points on the graph of y = x 2 . Transformations can be applied to this set of points. y For example, the translation determined consider 2 by the vector applied to this set of points. 0 It can be seen that all the images of the points of (2, 4) (4, 4) (0, 4) (–2, 4) y = x 2 lie on the graph of y = (x − 2)2 The rule for this transformation is (x, y) → (x + 2, y). Each coordinate (x , y ) of the image must be of this (1, 1) form, i.e. x = x + 2, y = y. The relationship between (–1, 1) (3, 1) x and y is known to be y = x 2 . Therefore if (x , y ) is x the image of a point on y = x 2 , and x = x − 2, then (0, 0) (2, 0) y = (x − 2)2 . Example 9 Find the of the image of the line y = x + 1 under the translation determined by the equation 2 vector . Sketch the graph of y = x + 1 and its image. 1 Solution The rule for the translation is (x, y) → (x + 2, y + 1). Let (x , y ) be the coordinates of the image of (x, y)

Thus

x = x + 2, y = y + 1 and y = x + 1 x = x − 2 and y = y − 1

and the relation y = x + 1

y y=x+1

becomes y − 1 = x − 2 + 1 ∴ y = x The points (−1, 0), (0, 1), (1, 2) and their images are shown, i.e.,

(−1, 0) → (1, 1) (0, 1) → (2, 2) (1, 2) → (3, 3)


y=x (3, 3)

(1, 2)

(0, 1)

(2, 2) (1, 1)

(–1, 0) 0

x


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Example 10 Find the equation of the image of the parabola y = x 2 under a dilation from the y axis of factor 1 . Sketch the graph of y = x 2 and its image. 2 Solution

1 x, y The rule for the dilation is (x, y) → 2 Therefore, if (x , y ) is a point of the image of y = x 2 x Then x = and y = y 2 Also y = x 2

Therefore the relationship between x and y is y = (2x )2 = 4(x )2 1 ,1 (1, 1) → 2 (2, 4) → (1, 4) (−2, 4) → (−1, 4) Note:

y y = 4x2

y = x2 (–2, 4) (–1, 4)

(1, 4)

1 ,1 2

(2, 4)

(1, 1) x

0

Example 11 For y = x 2 a find the equation for the image of y = x 2 under a dilation of factor

1 from the x axis 2

b sketch the graph of y = x 2 and its image under this transformation. c i Find the equation for the image of y = x 2 under the dilation of a followed by a 1 translation determined by the vector . 2 ii Sketch the graph of this second image on the same set of axes as b. Solution

1 a The rule for dilation is (x, y) → x, y 2 If (x , y ) is a point on the image 1 y 2 x = x and y = 2y y = x2 x = x and y =

i.e., Also



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∴ The image will have equation 2y = (x )2 i.e.,

y =

1 2 (x ) 2 y

b

1 y = x2 2

y = x2 (2, 4)

(2, 2) (1, 1) 1,

1 2

x

0

c

i The rule becomes 1 1 (x, y) → x, y → x + 1, y + 2 2 2 dilation translation i.e., let (x , y ) be a point on the image. 1 y+2 2 x = x − 1 and y = 2(y − 2) y = x 2 is mapped to 2(y − 2) = (x − 1)2 1 y = (x − 1)2 + 2 2

x = x + 1, y =

∴ ∴ ∴ ii

y

y = 1 x2 2 y = 1 (x – 1)2 + 2 y = x2 2 (2, 4) (1, 2) (2, 2)

(1, 1) 1, 1 2 0

x



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In Exercise 8F, reference is made to the following basic graphs. y

y

y 1 y=x

y=x y = x2 0

x

0

x x

0

y

y x2 + y 2 = 1

0

x

y = 2x (0, 1) 0

x

Example 12 Find the image of each of the following curves under of factor 3 from the y axis a dilation −1 followed by a translation determined by the vector . 2 1 a y = x2 c x 2 + y2 = 1 b y= x Solution (x, y) → (3x, y) → (3x − 1, y + 2) dilation translation ∴ x = 3x − 1 and y = y + 2 x + 1 i.e., x= and y = y − 2 3 x +1 2 a y = x 2 is mapped to y − 2 = 3 x +1 2 The image has equation y = +2 3 1 1 b y = is mapped to y − 2 = x +1 x 3 3 +2 ∴ y = x +1 3 The image has equation y = +2 x +1 The rule is given by



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c x + y = 1 is mapped to 2

2

The image has equation

x + 1 3

205

2 + (y − 2)2 = 1

(x + 1) 2 + (y − 2)2 = 1 9

The image is an ellipse.

Exercise 8F Example

Example

Examples

9

10

11, 12

1 Find the of each of the following curves under the translation determined by the image 1 vector . State the equation of the image and sketch the graph of both the original 4 relation and its image on the one set of axes. 1 a y=x b y = x2 c y= x 1 2 2 e x +y =1 d y= 2 x 2 Find the image of each of the following curves under the dilation from the x axis given by the rule (x, y) → (x, 2y), i.e., a dilation of factor 2 from the x axis. State the equation of the image and sketch the graph of both the original relation and its image on the one set of axes. 1 1 c y= e x 2 + y2 = 1 a y=x b y = x2 d y= 2 x x 3 For each of the curves considered in 1, find the equation of its image under the transformation defined bya reflection in the line x = 0 followed by a translation 3 determined by the vector . 2 4 For each of the curves considered in 1, find the equation of its image under the transformation defined bya dilation of factor 2 from the y axis followed by a translation −3 determined by the vector . 1 5 Find the image of the graph of y = 2x + 3 under the transformation definedby first −4 reflecting in the line y = x and then translating as determined by the vector . 6 6 Find the image of the line {(x, y) : y = x + 2} under each of the following transformations. Sketch a graph showing {(x, y) : y = x + 2} and its image in each case. 2 a a translation determined by the vector b a reflection in the x axis 4 c a dilation of factor 4 from the x axis d a reflection in the line y = x e a reflection in the y axis



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7 Repeat 1 for the circle {(x, y) : x 2 + y 2 = 4}. 8 Consider {(x, y) : y = x 2 }. Find the image after a reflection in the x axis then a dilation of factor 2 from the x axis 1 b translation where (0, 0) → (5, 2) then a dilation of factor from the y axis 2 1 c dilation of factor from the x axis followed by translation where (0, 0) → (5, 2) 2 d translation where (0, 0) → (2, 1) followed by a reflection in the y axis e reflection in the line y = x followed by a translation where (0, 0) → (0, 2). 9 Sketch the graph of the image of {(x, y) : y = x 2 } in 8 a−e above. 10 Consider {(x, y) : y = 2x }. Find the image after a translation where (0, 0) → (1, 0) c d e f

b translation where (0, 0) → (0, 1) 1 dilation of factor 2 from the y axis followed by dilation of factor from the x axis 3 1 dilation of factor from the x axis followed by dilation of factor 2 from the y axis 3 translation where (0, 0) → (2, 4) followed by reflection in the y axis 1 translation where (0, 0) → (0, 2) then dilation of factor from the y axis then 2 reflection in the x axis.

11 Sketch the graph of the image of {(x, y) : y = 2x } in 10 a−f above.

8.7 Determining transformations In the previous section a method was presented for determining the equation of the image of a graph under a given transformation. In this section a procedure for determining the transformations which have produced a particular image is discussed.

Example 13 Find a sequence of transformations which maps y = x 2 to y = 2(x + 3)2 − 4 Solution The composition of transformations maps (x, y) to (x , y )

∴

y = 2(x + 3)2 − 4

Rearrange to make the transformation from y = x 2 more obvious y + 4 = (x + 3)2 2



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It can be seen that to obtain this, take y + 4 and x = (x + 3) 2 y = 2y − 4 and x = x − 3 y=

i.e.,

A dilation of factor 2 from the x axis followed by a translation determined by the −3 vector gives this image. −4 Example 14 Find a sequence of transformations which maps y = 2x to y = 3(2x−2 ) − 4. Solution Assume the composition of transformations maps (x, y) to (x , y ) Write y = 3(2x −2 ) − 4 y + 4 = 2x −2 Rearrange to 3 It can be seen to obtain this, take y + 4 and x = x − 2 3 y = 3y − 4 and x = x + 2 y=

∴

A dilation of factor 3 from the x axis followed by a translation determined by the 2 vector gives this image. −4

Exercise 8G Example

13

1 Find the single transformation which maps a b c d e f g h

Example

14

{(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) :

y y y y y y y y

= x 2} = x 2} = x 2} = 2x } = 2x } = 2x } = 2x } = 2x }

to to to to to to to to

{(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) : {(x, y) :

y y x y y y y y

= 2x 2 } = (x + 2)2 } = y2} = 22x } = 3 × 2x } = 2x−3 } = 2−x } = −2x }

2 Find a sequence of transformations which maps a {(x, y) : y = x 2 } b {(x, y) : y = x 2 } c {(x, y) : y = x 2 }

to to to

{(x, y) : y = 2(x − 3)2 } {(x, y) : y = 2x 2 − 3} {(x, y) : y = 2(x − 3)2 + 1}



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1 x 1 e (x, y) : y = x 1 f (x, y) : y = x

d

(x, y) : y =

to to to

g y = 2x

to

y=2

1 x

to

y=

h y=

x−1 3

2 x −3 2 (x, y) : y = − 3 x 1 (x, y) : y = 3−x (x, y) : y =

+4

3 2x − 4

8.8 Absolute value function and integer

value function y

Absolute value function Let f : R → R be defined as x if x ≥ 0 f (x) = −x if x < 0 This function is written as f (x) = |x| The graph of this function is as shown. It is known as the modulus function or absolute value function.

x

0

The images of this graph under several transformations are considered. Example 15 Find the image of {(x, y) : y = |x|} and sketch the graph of the image for each of the following transformations. 2 a a translation defined by the vector b a dilation of factor 4 from the y axis 3 c a reflection in the line y = x d a reflection in the x axis Solution a (x, y) → (x + 2, y + 3) (x, y) → (x , y ) ∴ x = x + 2, y = y + 3 ∴ x = x − 2 and y = y − 3 {(x, y) : y = |x|} is mapped to {(x , y ) : y − 3 = |x − 2|}

y

(0, 5) y = |x – 2| + 3 (2, 3)

0


x


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b (x, y) → (4x, y) (x, y) → (x , y ) ∴ x = 4x and y = y {(x, y) : y = |x|} is mapped to x (x , y ) : y = 4 c

d

y y= (–4, 1)

(4, 1)

x 4 x

0 y

(x, y) → (y, x) (x, y) → (x , y ) ∴ x = y and y = x {(x, y) : y = |x|} is mapped to {(x , y ) : x = |y |}

x = |y|

x

0

(x, y) → (x, −y) (x, y) → (x , y ) ∴ x = x, y = −y {(x, y) : y = |x|} is mapped to {(x , y ) : −y = |x |}

y

0

x

y = – |x|

Integer value function The integer value function I : R → Z is defined by I (x) = [x] where [x] is the greatest integer not exceeding x. For example, [3.9] = 3, [5] = 5, [−4.1] = −5, [␲] = 3 For all real numbers x, we have x − 1 ≤ [x] ≤ x

y 3 y = I(x)

2 1

x –2

–1

1

2

3

4

–1 –2



210


Using the TI-Nspire Absolute Value function The abs( ) function can be found in the catalog (k 1 A) or typed directly. The template | | can also be obtained using t (/+r on the Clickpad). It works as shown.

The graph of the absolute value function can be obtained in the usual way in a Graphs application (/+I>Add Graphs). The graph of f 1(x) = abs(x) is shown with medium line width. The graph of the composite function f 2(x) = abs(x 2 − 4) is also shown.

Integer Value function The integer value function rounds down to the nearest integer, so on the TI-Nspire it is called the floor( ) function and can be found using b >Number>Number Tools>Floor. It works as shown. (Note the difference from the Integer Part Function (b>Number>Number Tools>Integer Part) that returns the integer part of a number.) The graph of the integer value function can be obtained in the usual way in a Graphs application (/+I>Add Graphs). The graph of f 1(x) = floor(x) is shown with medium line width. The graph ofthe x is composite function f 2(x) = floor 2 also shown.




211

Using the Casio ClassPad To enter the absolute value function, switch on tap . the screen keyboard and in menu. This is entered in the Tick the selection box and then tap $ to produce the graph. Note: The Zoom is set to Quick Initialize for the graph shown.

The absolute value function can be composed with other functions. The small graph shows y = |x 2 − 4|.

The CAS calculator has the absolute value and integer part functions as built-in features.

(absolute value function) Turn on the screen keyboard and tap then enter the number.

(integer part function) This function is found in the catalogue. Tap and use the alphabetic shortcut at the bottom of the screen to find a function beginning with ‘i’. The required function is intg(. This function returns the next whole number which is less than the given number.



212


Using the Casio ClassPad The CAS calculator has the absolute value and integer part functions as built-in features.

(absolute value function) Turn on the screen keyboard and tap then enter the number.

(integer part function) This function is found in the catalog. Tap and use the alphabetic shortcut at the bottom of the screen to find a function beginning with ‘i’. The required function is intg(. This function returns the next whole number which is less than the given number.

Exercise 8H Example

15

1 Find the image of {(x, y) : y = |x|} and sketch the graph of the image for the following. −1 a a translation corresponding to the vector 3 b a reflection in the line y = −x c a dilation from the y axis of factor 4 d a reflection in the x axis −1 e a translation corresponding to the vector followed by a reflection in the line y = x 3 −1 f a reflection in the line y = x then a translation corresponding to the vector 3 g a dilation from the x axis of factor 2 followed by a reflection in the x axis. 2 State the transformation(s) which map(s) a b c d

{(x, y) : {(x, y) : {(x, y) : {(x, y) :

y y y y

= |x|} to {(x, y) : = |x|} to {(x, y) : = |x|} to {(x, y) : = |x|} to {(x, y) :

y y y y

= |x| + 3} = |x − 3| + 3} = |2x|} = −2|x|}

3 Find the image of {(x, y) : y = [x]} and sketch the graph of the image of the following. a a dilation from the y axis of factor 2



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b c d e f g

213

2 a translation corresponding to the vector 0 0 a translation corresponding to the vector 2 a reflection in the line y = x a reflection in the line y = −x 4 a translation corresponding to the vector 0 a dilation from the y axis of factor 2 followed by a reflection in the x axis

0 h a reflection in the x axis followed by a translation corresponding to the vector . 2

8.9

Function notation with transformations In this section the image of the graph y = f (x) (where f is an arbitrary function) under a transformation is considered. This is particularly useful when using a graphics or CAS calculator. Example 16 Let y = f (x) be the equation of a curve. Find the image of y = f (x) under each ofthe following transformations. h b a reflection in the x axis a a translation determined by the vector k c a dilation of factor k from the y axis d a dilation of factor k from the x axis Solution a The rule is (x, y) → (x + h, y + k)

∴ Consider x = x + h and y = y + k ∴ x = x − h and y = y − k ∴ y = f (x) is mapped to y − k = f (x − h) i.e., the image is y = f (x − h) + k b The rule is (x, y) → (x, −y)

∴ Consider x = x and y = −y i.e., the image is y = − f (x) c The rule is (x, y) → (kx, y) ∴ Consider x = kx and y = y x and y = y ∴ x= k x ∴ y = f (x) is mapped to y = f k

x i.e., the image of y = f (x) is y = f k ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


214


d The rule is (x, y) → (x, ky)

∴ Consider ∴ ∴

x = x and y = ky y x = x and y = k y = f (x ) y = f (x) is mapped to k

i.e., the image of y = f (x) is y = k f (x) Example 17 For f (x) = x 2 find 4 f (x + 3) and state the transformations which take y = f (x) to y = 4 f (x + 3) Solution

∴

4 f (x + 3) = 4(x + 3)2 y = x 2 is mapped to y = 4(x + 3)2

If (x, y) is mapped to (x , y ) Then

i.e.,

y = (x + 3)2 4 y x = x + 3 and y = 4 x = x − 3 and y = 4y

−3 ∴ a dilation of factor 4 from the x axis followed by a translation takes 0 y = f (x) to y = 4 f (x + 3)

Using the TI-Nspire The notation for transformations can be used with a calculator. In the screen shown, the transformation which takes the curve of y = f (x) to the curve of y = f (x − 5) is applied to the curve with equation f (x) = x 2 by defining f 1(x) = x 2 and f 2(x) = f 1(x − 5).




215

Consider the following sequence of transformations: dilation of factor 2 from the x-axis translation of ‘3 to the left’ translation ‘4 down’ These transformations will take the curve of y = f (x) to the curve of y = 2 f (x + 3) − 4. In the screen shown, this is applied to the curve with equation f (x) = x 2 .

Using the Casio ClassPad The notation for transformations can be used with the CAS calculator. Consider the transformation which takes the curve of y = f (x) to the curve of y = f (x − 5). The function for y1 can be entered and then y2 can be defined as shown. The y must be selected from the tab on the screen keyboard, not the y on the physical keyboard. 1 A dilation factor from the y-axis is 2 applied to the curve with rule f (x) = x 2 . The rule has been entered as y3 = y1 (2x). Consider the following sequence of transformations: dilation of factor 2 from the x-axis translation of ‘3 to the left’ translation ‘4 down’ These transformations will take the curve of y = f (x) to the curve of y = 2 f (x + 3) − 4. In the screen shown, this is applied to the curve with equation y = x 2. Note:



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Exercise 8I Example

16

1 Let y = f (x) be the equation of a curve. Find the image of y = f (x) under each of the following transformations. 2 b a reflection in the line y = x a a translation determined by the vector 3 1 d a dilation of factor 2 from the x axis c a dilation of factor from the y axis 2

Example

17

2 For f (x) = 2x , find a

f (x + 3)

b

f (2x)

x

c

f

x 2

d 4 f (x − 3)

3 For f (x) = x 2 , find 4 f and state the transformations which take

x 2 y = f (x) to y = 4 f 2 1 find f (2x − 3) + 4 and state the transformations which take x y = f (x) to y = f (2x − 3) + 4

4 For f (x) =

5 For f (x) = x 2 find −3 f (2 − x) and state the transformations which take y = f (x) to y = −3 f (2 − x)

8.10 Summary of transformations A summary of some of the transformations and their rules is presented here. Suppose (x , y ) is the image of (x, y) under the mapping in the first column of the table below. Mapping

Rule

Reflection in the x axis

x y x y

Dilation by factor k from the y axis

Rotation of

␲c about O in an anticlockwise direction 2

Expansion of factor k from the origin Reflection in the line y = x a Translation defined by a vector b

=x = −y = kx =y

= x + 0y = 0x + −y = kx + 0y = 0x + y

x = −y

= 0x + −y

y =x

= x + 0y

x y x y

= kx + 0y = 0x + ky = 0x + y = x + 0y

= kx = ky =y =x

x = x + a y = y + b



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The first five mappings are special cases of a general kind of mapping defined by x = ax + by y = cx + dy where a, b, c, d are real numbers. These equations can be rewritten as x = a11 x + a12 y y = a21 x + a22 y which yields the equivalent matrix equation x a11 a12 x = y a21 a22 y A transformation of the form (x, y) → (a11 x + a12 y, a21 x + a22 y) is called a linear transformation. Example 18 Consider a linear transformation such that (1, 0) → (3, −1) and (0, 1) → (2, −4). Find the image of (−3, 5). Solution a11 a21

a12 a22

1 3 a11 = and 0 −1 a21

a12 a22

0 2 = 1 −4

∴ a11 = 3, a21 = −1 and a12 = 2, a22 = −4

3 i.e., the transformation can be defined by the 2 × 2 matrix −1 Let (−3, 5) → (x , y ) x 3 2 −3 ∴ = −1 −4 5 y 3 × −3 + 2 × 5 = −1 × −3 − 4 × 5 1 = −17

∴

2 −4

(−3, 5) → (1, −17)

The image of (−3, 5) is (1, −17) Note:

Non-linear mappings cannot be represented by a matrix in the way indicated above.



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Thus for the translation defined by (0, 0) → (a, b) x = x + a y = y + b While this cannot be represented by a square matrix, the defining equations x a x = + suggest y b y where the ‘sum’ has definition: x a x +a for each x, y, a, b in R, + = y b y+b

Composition of mappings

a11 Consider a linear transformation defined by the matrix A = a21 b11 b12 linear transformation defined by the matrix B = b21 b22

a12 composed with a a22

The composition consists of the transformation of A being applied first and then the transformation of B. The matrix of the resulting composition is the product b11 a11 + b12 a21 b11 a12 + b12 a22 BA = b21 a11 + b22 a21 b21 a12 + b22 a22

Exercise 8J 1 Find a

2 5

−3 4

−1 2

b

−3 −1

a b

−1 3

Example

18

2 2 If a linear transformation is defined by the matrix A = −4 (1, 0), (0, 1) and (3, 2) under this transformation.

−1 , find the image of 3

3 Find the images of (1, 0) and (−1, 2) under the linear transformation whose matrix is 2 3 −2 0 2 −1 c b a 3 −1 0 1 1 1



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219

4 a Find the matrix of the linear transformation which maps (1, −2) → (−4, 5) and (3, 4) → (18, 5). b The images of two points are given for a linear transformation. Investigate whether this is sufficient information to determine the matrix of the transformation. c Find the matrix of the linear transformation such that (1, 0) → (1, 1) and (0, 1) → (2, 2). What is the range of this transformation? 5 By finding the images of (1, 0) and (0, 1), write down the matrix of each of the following transformations. a b c d e f

reflection in the line x = 0 reflection in the line y = x reflection in the line y = −x dilation of factor 2 from the x axis expansion of factor 3 from the origin dilation of factor 3 from the y axis



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Chapter summary A transformation T is a mapping from R 2 to R 2 such that if T(a, b) = T(c, d) then a = c and b = d A translation is a transformation for which each point in the plane is moved the same distance in the same direction. A translation of 5 units in the positive direction of the x axis can be represented by the rule (x, y) → (x + 5, y) A vector will mean a column of two numbers. The first number indicates a ‘move’ in the positive or negative direction of the x axis and the second indicates a ‘move’ in the positive or negative direction of the y axis. Directed line segments are used to illustrate vectors. 2 For example, the vector is the vector ‘2 to the right and 3 up’. 3 The rule for the translation described by this vector is (x, y) → (x + 2, y + 3) For reflection in the x axis the rule is (x, y) → (x, −y) For reflection in the y axis the rule is (x, y) → (−x, y) For reflection in the line y = x the rule is (x, y) → (y, x) Lengths and angles are preserved by reflections and translations. The transformation ‘dilation from the y axis of factor k’ is defined by the rule (x, y) → (kx, y); k ∈ R + The transformation ‘dilation from the x axis of factor k’ is defined by the rule (x, y) → (x, ky); k ∈ R + Lengths and angles are not preserved by dilations. A point (a, b) is said to be invariant under a transformation if (a, b) → (a, b) under that transformation. The composition of two transformations is defined by applying one transformation, and then the second, to the images of the first. e.g., (x, y) → (x + 2, y + 3) → (−(x + 2), y + 3) translation reflection Applying transformations to graphs of functions e.g., let (x, y) → (2x + 1, y + 3) Let (x , y ) be the image of (x, y) Then x = 2x + 1 and y = y + 3 x − 1 and x = and y = y − 3 2 x −1 ∴ y = f (x) is mapped to y − 3 = f 2 x −1 2 2 If f (x) = x , y = f (x) is mapped to y = +3 2



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Review

Determining transformations x +3 −4 For example, if y = f 2 x +3 then y + 4 = f 2 Let (x, y) → (x , y ) x + 3 =x Then y + 4 = y and 2 Hence y = y − 4 and x = 2x − 3 The transformation is given by the rule (x, y) → (2x − 3, y − 4) y x if x ≥ 0 |x| = −x if x < 0

y = |x| x

0 This is called the absolute value function. The integer value function is defined by

y

I (x) = [x]

2

where [x] is the greatest integer not exceeding x.

1 –2

–1

x

0

1

2

3

–1 –2 Function notation The image of the graph of y = f (x) under the translation (x, y) → (x + h, y + k) is the graph of y = f (x − h) + k The image of the graph of y = f (x) under the reflection (x, y) → (x, −y) is the graph of y = − f (x) The image of the graph of y = f (x) under the reflection (x, y) → (−x, y) is the graph of y = f (−x) The image of the graph of y = f (x) under the dilation

x (x, y) → (kx, y) is the graph of y = f k



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Multiple-choice questions 1

y

y 4

4

y = g(x)

y = f(x) 2

2

–6

–4

–2 0

x 2

4

6

–2 –4

–6

–4

–2 0

x 2

4

6

–2 –4

The transformation which maps the graph of y = f (x) to the graph of y = g(x) is B rotation of 270◦ about the origin A rotation of 180◦ about the origin C reflection in the y axis D reflection in the x axis E reflection in the line y = x 2 If the graph of y = g(x) is obtained by reflecting the graph of y = f (x) in the x axis, the equation relating f (x) and g(x) is A f (x) = g(x) B f (x) = −g(x) C f (x) = g(−x) 1 E f (x) = 2g(x) D f (x) = g(x) 3 The translation that maps the graph of y = x 2 to the graph of y = (x − 5)2 − 2 can be described as A 2 units in the negative direction of the x axis and 5 units in the positive direction of the y axis B 2 units in the positive direction of the x axis and 5 units in the positive direction of the y axis C 2 units in the negative direction of the x axis and 5 units in the negative direction of the y axis D 5 units in the positive direction of the x axis and 2 units in the negative direction of the y axis E 5 units in the negative direction of the x axis and 2 units in the negative direction of the y axis 4 The translation that maps the graph of f (x) = (x + 2)2 + 8 to the graph of g(x) = x 2 is A 8 units in the negative direction of the x axis and 2 units in the positive direction of the y axis B 2 units in the positive direction of the x axis and 8 units in the positive direction of the y axis C 2 units in the negative direction of the x axis and 8 units in the positive direction of the y axis



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y = 3f(x)

y = f(x)

y = 3f(x) x

0 y

D

0

x

0

x

y y = f(x)

y = 3f(x)

y = 3f(x)

y = f(x) x

0

E

y = f(x)

y = f(x)

Review

D 2 units in the positive direction of the x axis and 8 units in the negative direction of the y axis E 2 units in the negative direction of the x axis and 8 units in the negative direction of the y axis 1 5 If y = f (x) = 2 , the graph of y = 3 f (x) is as shown in x y y y B C A

y = 3f(x)

0

x

√ 6 The equation of the image of the graph of y = x after a reflection in the y axis followed by a translation of 2 units in the positive direction of the x axis and 3 units in the positive direction of the y axis is √ √ √ B y = −x − 2 − 3 C y = x +2−3 A y = −x − 2 − 3 √ √ E y = −x + 2 + 3 D y = x −2+3 1 7 The equation of the image of the graph of y = 2 after a reflection in the x axis followed x by a dilation of factor 2 from the x axis is 1 2 2 1 2 D y= 2 E y= A y=− 2 B y=− 2 C y= 2 2x (−x)2 x 2x x 1 8 The equation of the image of the graph of y = |x| after a dilation of factor from the 3 y axis followed by a translation of 5 units in the negative direction of the x axis and 2 units in the positive direction of the y axis is x − 5 +2 A y = 3|x − 5| + 2 B y = 3|x + 5| + 2 C y = 3 x + 5 x + 15 +2 D y = +2 E y= 3 3 9 The value of [−4.6] + [7.2] + [8.7] is A 9 B 10 C 8 D 13 E 11 10 The equation of the image of the graph of y = [x] after a reflection in the x axis followed by a translation of 5 units in the positive direction of the x axis and 4 units in the negative direction of the y axis is A y = [x] + 1 B y = −[x − 5] − 4 C y = [x + 5] − 4 D y = [x] + 9 E y = [x + 5] + 4



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Short-answer questions (technology-free) 1 Find the image of the point (3, −1) under each of the following transformations. a a dilation of factor 3 from the y axis b a dilation of factor 2 from the x axis −3 c a translation determined by the vector d a reflection in the line x = 0 2 e a reflection in the line y = 0 f a reflection in the line y = x 2 Find the equation of the image of the graph of y = x 2 under each of the following transformations. −2 a a translation determined by the vector b a reflection in the line y = x 3 c a reflection in the line y = 0 3 Find the rule for each of the following compositions of transformations. a b c

d

−2 a reflection in the line y = x followed by a translation determined by the vector 3 a dilation of factor 5 from the x axis followed by a reflection in the x axis a dilation offactor 4 from the y axis followed by a translation determined by the −2 vector 3 −2 followed by a dilation of factor 4 from the a translation determined by the vector 3 x axis

4 Find the image of y = 2x − 1 under each of the transformations defined in 1. 5 Find the image of y = 2 − x 2 under each of the transformations defined in 3. 6 Sketch the graph of each of the following. Indicate the coordinates of the vertex and at least one other point. a y = −|x| b y = |2x| + 3 c y = 4 − |2x| 7 Give the rule for the transformations which would map 2 1 +4 a y = to y = b y = x 2 to y = 2(x − 4)2 + 3 x x −3 x +1 1 c y = |x| to y = |3x| − 4 d y = to y = x x −1 e y = x 2 to y = −(x − 2)2 + 3 8 Sketch the graph of each of the following using transformations. b (x − 1)2 = y + 2 c (x − 2)2 = 3(y − 2) a x2 = y − 1 d y = [4x] e y = −3|2x − 1| f y = 2 − 3|x − 2|

Extended-response questions 1 a Find the image of the point with coordinates (2, 6) under reflection in the line x = 3. b Find the rule for the following sequence of transformations.



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−3 r a translation determined by the vector 0 r a reflection in the y axis 3 r a translation determined by the vector 0

c Hence give the rule for the transformation, reflection in the line x = 3. d i Give a sequence of three transformations which would determine a reflection in the line x = m. ii Give the rule for the transformation, reflection in the line x = m. e i Give a sequence of three transformations which would determine a reflection in the line y = n. ii Give the rule for the transformation, reflection in the line y = n. f Find the image of each of the relations under the transformation, reflection in the line x = 3. iv y = (x − 3)2 i y = x −3 ii y = x iii y = x 2

Review

225

y 2 a Find the coordinates of the image of the point A(3, 1) ◦ under a rotation of 90 in an anticlockwise A' direction around the origin. b i State the gradient of line OA. A(3, 1) ii State the gradient of line OA . c A point A has coordinates (p, q). x i State the gradient of line OA. 0 ii The point is rotated about the origin by 90◦ in an anticlockwise direction. Find the coordinates of A , the image of A. d Find the rule for the transformation rotation about the origin by 90◦ in an anticlockwise direction. e Find the equation of the image of each of the following curves under this transformation. 1 iii x 2 + y 2 = 1 i y=x ii y = x 2 iv y = x 3 a Find the image of the point A(1, 3) under a rotation of 180◦ about the origin. b Find the image of the point (a, b) under a rotation of 180◦ about the origin. c If the curve with equation y = f (x) is rotated 180◦ about the origin, find the rule for its image in terms of f. d i Find the rule for the following sequence of transformations. −3 r a translation determined by the vector 0 ◦ r a rotation of 180 about the origin 3 r a translation determined by the vector 0 ii Find the image of the line with equation y = 3x + 1 under the transformation with rule determined in d i. (cont’d.) ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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e Find the rule for rotation of 180◦ in an anticlockwise direction about the point (m, n). f Find the rule for rotation of 90◦ in a clockwise direction about the point (m, n). g Find the rule for rotation of 90◦ in an anticlockwise direction about the point (m, n). h i Find the image of the curve with rule y = x 2 after a rotation of 90◦ in a clockwise direction about the point (0, 1). ii Sketch the graph of the curves on one set of axes. i Find the dilation from the x axis which takes y = x 2 to the parabola with vertex at the origin and which passes through the point (25, 15). ii State the rule which reflects this dilated parabola in the x axis. iii State the rule which takes the reflected parabola of ii to a parabola with x intercepts (0, 0) and (50, 0) and vertex (25, 15). iv State the rule which takes the curve of y = x 2 to the parabola defined in iii. b The plans for the entrance of a new building involve twin parabolic arches as shown in the diagram.

4 a

y

15 m

0

Arch 1

Arch 2

50 m

50 m

x

i From the results of a give the equation for the curve of Arch 1. ii Find the translation which maps the curve of Arch 1 to the curve of Arch 2. iii Find the equation of the curve of Arch 2. c The architect wishes to have flexibility in his planning and so wants to develop an algorithm for determining the equations of the curves given arch width m metres and height n metres. i Find the rule for the transformation which takes y = x 2 to Arch 1 with these dimensions. ii Find the equation for the curve of Arch 1. iii Find the equation for the curve of Arch 2. 5 a Let f (x) = [x] i Find the equation of the image of the graph of y = f (x) under the transformation with rule (x, y) → (2x − 1, y + 3) ii Sketch the graph of the image of y = f (x) under this transformation. iii Describe a sequence of transformations which takes the graph of y = f (x) to y = − f (x − 3) − 2 b Let g(x) = |x| i Sketch the graph of y = −g(x − 3) + 3 and the graph of y = g(2x − 1) ii Solve the equation −g(x − 3) + 3 = g(2x − 1)



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C H A P T E R

9 Ratios and similarity Objectives To divide a quantity in a given ratio To determine the ratio in which a quantity has been divided To apply the transformations which are expansions from the origin To define similarity of two figures To determine when two triangles are similar by using the conditions

r equal angles (AAA) r equal ratios (PPP) r corresponding sides having the same ratio and the included angle equal (PAP) To apply similarity to solve problems To determine and apply expansion factors for areas and volumes

9.1

Ratios This section is revision of work of previous years. Several examples are presented. Example 1 Divide 300 in the ratio 3 : 2. Solution

∴ ∴

one part = 300 ÷ 5 = 60 two parts = 60 × 2 = 120 three parts = 60 × 3 = 180

Example 2 Divide 3000 in the ratio 3 : 2 : 1.



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Solution

∴ ∴

one part = 3000 ÷ 6 = 500 two parts = 500 × 2 = 1000 three parts = 500 × 3 = 1500

Example 3 A day is divided into 10 new-hours, each new-hour is divided into 100 new-minutes and each new-minute is divided into 100 new-seconds. What is the ratio of a new-second to an ordinary second? Solution There are 10 × 102 × 102 new-seconds in a day and 24 × 60 × 60 ordinary seconds in a day ∴ the ratio of new-seconds : ordinary seconds 1 1 = 5 : 10 8.64 × 104 = 864 : 1000 = 108 : 125 Example 4 Two positive integers are in the ratio 2 : 5. If the product of the integers is 40 find the larger integer. Solution Let a and b denote the integers

and From 1

∴ ∴ ∴

2 a = ... 1 b 5 ab = 40 . . . 2 2 a = b Substitute in 2 5 2 2 b = 40 5 b2 = 100 b = ±10

and as b is a positive integer, b = 10 and a = 4 The larger integer is 10.

Exercise 9A Example

1

1 Divide 9000 in the ratio 2 : 7.

Example

2

2 Divide 15 000 in the ratio 2 : 2 : 1.



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Chapter 9 — Ratios and similarity

3 x : 6 = 9 : 15. Find x. 4 The ratio of the numbers of orange flowers to pink flowers in a garden is 6 : 11. There are 144 orange flowers. How many pink flowers are there? 5 15 : 2 = x : 3. Find x. 6 The angles of a triangle are in the ratio 6 : 5 : 7. Find the sizes of the three angles. 7 Three men X, Y and Z share an amount of money in the ratio 2 : 3 : 7. If Y receives $2 more than X, how much does Z obtain? 8 An alloy consists of copper, zinc and tin in the ratios 1 : 3 : 4 (by weight). If there is 10 g of copper in the alloy, find the weights of zinc and tin. 9 In a bag the ratio of red beads to white beads to green beads is 7 : 2 : 1. If there are 56 red beads, how many white beads and how many green beads are there? 10 On a map the length of a road is represented by 45 mm. If the scale is 1 : 125 000, find the actual length of the road. 11 Five thousand two hundred dollars was divided between a mother and daughter in the ratio 8 : 5. Find the difference between the sums they received. 12 Points A, B, C and D are placed in that order on a line so that AB = 2BC = CD. Express BD as a fraction of AD. 13 If the radius of a circle is increased by two units, find the ratio of the new circumference to the new diameter. 14 In a class of 30 students the ratio of boys to girls is 2 : 3. If six boys join the class, find the new ratio of boys to girls in the class. 15 If a : b = 3 : 4 and a : (b + c) = 2 : 5, find the ratio a : c. 16 The scale of a map reads 1 : 250 000. Find the distance, in kilometres, between two towns which are 3.5 cm apart on the map.

9.2

An introduction to similarity The two triangles ABC and AB C are similar. Note: OA = 2OA, OB = 2OB, OC = 2OC. Triangle AB C can be considered as the image of triangle ABC under a mapping of the plane in which the coordinates are multiplied by 2. This mapping is called an expansion from the origin of factor 2. This can be written in transformation notation: (x, y) → (2x, 2y).


12

A'(4, 12)

10 8 6

A(2, 6)

B'(10, 6)

4 B(5, 3)

2 0

C(4, 1) 2 4

C'(8, 2) 6

8

10 12


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There is also a mapping from AB C to ABC which is an expansion from the origin of 1 factor . 2 1 1 The rule for this is (x, y) → x, y . 2 2 Two figures are similar if one is congruent to an image of the other under an expansion from the origin of factor k. For example, the rectangle of side lengths 1 and 2 is similar (3, 9) (6, 9) 9 to the rectangle with side lengths 3 and 6. 8 Note here the expansion factor is 3 and the rule is 7 (x, y) → (3x, 3y). 6 5 4 (1, 3) (2, 3) 3 (3, 3) 2 1 (1, 1) (2, 1) 0 1 2 3 4

Note:

any two circles are similar any two squares are similar any two equilateral triangles are similar For a triangle ABC with side lengths a, b, c and a similar triangle AB C with corresponding side lengths a , b , c it can be seen that

(6, 3)

5

6

a b c = = =k a b c where k is the appropriate expansion factor. Similar statements can be made about other pairs of similar polygons. Note also that the measure of an angle does not change under an expansion: i.e., for two similar figures, corresponding angles are equal.

Similar triangles Two triangles are similar if one of the following conditions holds: triangles have equal angles (AAA)

B'

B

100°

100°

A

45°

35°

C

A'

45°

35°

C'

corresponding sides are in the same ratio (PPP) AB B C AC = = = k, where k is the expansion (enlargement) factor AB BC AC two pairs of corresponding sides have the same ratio and the included angles are equal, (PAP) B' B

45° A

C

A'

45°

C'

A B A C = AB AC ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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two pairs of corresponding sides have the same ratio and two corresponding non-included angles are equal, provided these angles are right angles or obtuse. B'

B

120° A

120° C

C'

A'

Triangle ABC is similar to triangle AB C can be written symbolically as ABC ∼ AB C The triangles are named so that angles of equal magnitude hold the same position i.e., A corresponds to A , B corresponds to B , C corresponds to C . i.e.

AC BC BA BC = or = BC AC BC BA

Example 5 a Give the reason for triangle ABC being similar to triangle AB C . b Find the value of x.

B' B

Solution a Triangle ABC is similar to triangle AB C as 3 5 = = 0.8 6.25 3.75

5 cm

6.25 cm

20°

20° 3.75 cm

3 cm

A

x cm

C

A'

3.013 cm

C'

and the magnitude of ∠ ABC = magnitude of ∠ AB C = 20◦ PAP is the condition for similarity. 5 x b = 3.013 6.25 5 × 3.013 ∴ x= 6.25 = 2.4104 Example 6 a Give the reason for triangle ABC being similar to triangle AXY. b Find the value of x.

B x cm A

Y

C 3 cm


X

6 cm

2.5 cm Cambridge University Press

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Solution a Corresponding angles are of equal magnitude (AAA). b AC AB = AX AY x 3 i.e., = x +6 5.5 5.5x = 3(x + 6) 2.5x = 18

∴

x = 7.2

Exercise 9B Example

5

1 Give reasons why the following pairs of triangles are similar and find the value of x in each case. A'

a

A 82°

x cm

9 cm

5 cm

4 cm

56° B

C

B'

b

C'

D B ×

10 cm

14 cm

C 13 cm

×

x cm

E

12 cm A

c

C E

x cm

2 cm ×

×

A

d

B

6 cm C

Q

x cm

A

8 cm

P

×

6 cm

×

10 cm

D

4 cm

B


R


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Chapter 9 — Ratios and similarity Example

6

2 Give reasons why the following pairs of triangles are similar and find the value of x in each case. a

b

A 12 cm

B x cm

16 cm

D P

Q

x cm B

8 cm

2 cm

C A

c

d

A 2 cm P

E

3 cm

C

2 cm

C 1.5 cm D

x cm Q

x cm 6 cm

8 cm A

B

E 2 cm

10 cm

B

C

3 Given that AD = 14, ED = 12, BC = 15 and EB = 4, find AC, AE and AB.

C D 15 14

12

A

E

4

B

4 A tree casts a shadow of 33 m and at the same time a stick 30 cm long casts a shadow 224 cm long. How high is the tree?

tree 0.3 m

5 A 20 m high neon sign is supported by a 40 m steel cable as shown. An ant crawls along the cable starting at A. How high is the ant when it is 15 m from A?

40 m

20 m

A

6 A hill has gradient of 1 in 20, i.e. for every 20 m horizontally there is a 1 m increase in height. If you go 300 m horizontally, how high up will you be?



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7 A man stands at A and looks at point Y across the river. He gets a friend to place a stone at X so that A, X and Y are collinear. He then measures AB, BX and XC to be 15 m, 30 m and 45 m respectively. Find CY, the distance across the river. A C

45 m

15 m B

X 30 m

Y

8 Find the height, h m, of a tree that casts a shadow 32 m long at the same time that a vertical straight stick 2 m long casts a shadow 6.2 m long. 9 A plank is placed straight up stairs that are 20 cm wide and 12 cm deep. Find x, where x cm is the width of the widest rectangular box of height 8 cm that can be placed on a stair under the plank.

nk

pla

x cm

12 cm

8 cm 20 cm

10 The sloping edge of a technical drawing table is 1 m from front to back. Calculate the height above the ground of a point A, which is 30 cm from the front edge.

11 Two similar rods 1.3 m long have to be hinged together to support a table 1.5 m wide. The rods have been fixed to the floor 0.8 m apart. Find the position of the hinge by finding the value of x.

1m

A 30 cm

92 cm

80 cm

1.5 m xm (1.3 – x) m 0.8 m

12 A man whose eye is 1.7 m from the ground, when standing 3.5 m in front of a wall 3 m high, can just see the top of a tower that is 100 m away from the wall. Find the height of the tower. 13 A man is 8 m up a 10 m ladder, the top of which leans against a vertical wall and touches it at a height of 9 m above the ground. Find the height of the man above the ground. 14 A spotlight is at a height of 0.6 m above ground level. A vertical post 1.1 m high stands 3 m away and 5 m further away there is a vertical wall. How spotlight high up the wall does the shadow reach? 0.6 m 3m


vertical post 1.1 m

wall

5m


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15 Measurements in the diagram shown are in cm. a b c d

A

Prove that ABC ∼ EDC. Find x. Use Pythagoras’ theorem to find y and z. Verify y : z = ED : AB.

z

5

4

B

x

D 2

C y

E

C'

16 Find a.

a C 10

12 7

B'

AA'

B

17 A man who is 1.8 m tall casts a shadow of 0.76 m in length. If at the same time a telephone pole casts a 3 m shadow, find the height of the pole. R

18 In the diagram shown, RT = 4 cm, ST = 10 cm. Find the length NT.

S

T

N

19 ABC is a triangular frame with AB = 14 m, BC = 10 m, CA = 7 m. A point P on AB, 1.5 m from A, is linked by a rod to a point Q on AC, 3 m from A. Calculate the length of PQ. 20 Using this diagram, find a, x and y.

4 6

x

y

a

9.3

Areas, volumes and similarity If two shapes are similar and the expansion (enlargement) factor is k, i.e., for any length AB of one shape, the length of the corresponding length AB of the similar shape has length kAB, then the area of similar shape = k 2 × area of the original shape For two triangles ABC and AB C which are similar, i.e., ABC ∼ AB C with AB = kAB, area of triangle AB C = k 2 × area of triangle ABC B'

B c

h

A b

D

c'

a

a' h'

C

A'


b'

D'

C'


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This can be shown by observing that ABC ∼ AB C and 1 1 × b × h = × kb × kh, 2 2 (where AC = b and AC = b ) 1 = k2 bh 2 = k 2 × area of triangle ABC area of triangle AB C =

Some examples of similar shapes and the ratio of their areas are considered in the following. Similar circles 4 Scale factor = 3

3 cm

Area = ␲.32

4 cm

Area = ␲.42 2 ␲.42 42 4 Ratio of areas = = = ␲.32 32 3

2 cm

Similar rectangles Scale factor = 2

4 cm

3 cm 6 cm Area = 24 cm2

Area = 6 cm

2

Ratio of areas =

24 = 4 = 22 6

10 cm 6 cm

5 cm

3 cm

4 cm

Similar triangles Scale factor = 2

Area = 6 cm

2

Ratio of areas =

8 cm Area = 24 cm2

24 = 4 = 22 6

Example 7 The two rectangles shown below are similar. The area of rectangle ABCD is 20 cm2 . Find the area of rectangle AB C D .



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Chapter 9 — Ratios and similarity B

C

B'

D

5 cm

237

C'

3 cm A

A'

D'

Solution

5 AB = AB 3 5 2 25 Area of AB C D = = The ratio of their areas = Area of ABCD 3 9 The ratio of the length of their bases =

25 × 20 cm2 9 5 = 55 cm2 9

Area of AB C D =

∴

Two solids are considered to be similar if they have the same shape and the ratio of their corresponding linear dimensions are equal. C'

B'

B

C

A

D

Scale factor = 2.5

A'

D'

3 cm F E

G 1 cm

2 cm

H 7.5 cm G'

F' 2.5 cm E'

5 cm

H'

The cuboids ABCDEFGH and AB C D E F G H are similar. For similar solids, if the scale factor is k then the volume of the similar solid = k 3 × volume of the original solid For example, Volume of ABCDEFGH = (3 × 2 × 1) cm3 = 6 cm3 Volume of AB C D E F G H = (5 × 2.5 × 7.5) cm3 = 93.75 cm3 The ratio of volumes =

93.75 = 15.625 = 2.53 6



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Here is another example. V'

V

Scale factor =

5 3

5 cm

3 cm 3

A

B'

B

cm

5 cm

C

3 cm

A'

Ratio of volumes =

5 cm

53 = 33

C'

3 5 3

Example 8 The two square pyramids are similar. VO = 9 cm.

V'

V 9 cm B A

C 4 cm

O

B'

D

C' 5 cm

O'

A'

D'

a Find the ratio of the length of their bases, and hence the height, V O , of the pyramid V AB C D . b The volume of VABCD is 48 cm3 . Find the ratio of their volumes, and hence find the volume of V AB C D . Solution a The ratio of the length of their bases =

∴

5 C D = CD 4

5 ×9 4 45 = 4

VO =

The length of V O is 11.25 cm. b The volume of VABCD is 48 cm3 3 5 Volume of V AB C D 125 = The ratio of their volumes = = Volume of VABCD 4 64

125 × 48 cm3 64 = 93.75 cm3

∴ Volume of V AB C D =



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Exercise 9C 1 These four rectangles are similar.

a Write down the ratio of the lengths of their bases. b By counting rectangles, write down the ratio of their areas. c Is there a relationship between these two ratios? 2 These four parallelograms are similar.

a Write down the ratio of the lengths of their bases. b By counting parallelograms, write down the ratio of their areas. c Is there a relationship between these two ratios? Example

7

3 The two rectangles shown are similar. The area of rectangle ABCD is 7 cm2 . B

B'

C

3 cm A

C'

5 cm D D'

A'

Find the area of rectangle A B C D . YZ ZX XY = = = 2.1 AB BC CA The area of triangle XYZ is 20 cm2 . Find the area of triangle ABC.

4 Triangle ABC is similar to triangle XYZ.

5 Triangles ABC and AB C are equilateral triangles. b Find a. a Find the length of BF. Area of triangle AB C c Find the ratio Area of triangle ABC


B'

B 2 cm

2 cm 2 cm

A

F 2 cm

C

A'

F' a cm

C'


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6 The areas of two similar triangles are 16 and 25. What is the ratio of a pair of corresponding sides? 7 The areas of two similar triangles are 144 and 81. If the base of the large triangle is 30, what is the corresponding base of the smaller triangle? 8 These three solids are similar. a Write down the ratio of the lengths of the bases. b Write down the ratio of the lengths of the heights. A B c By counting cuboids equal in shape and size to the cuboid given in A, write down the ratio of the volumes. d Is there a relationship between the answers to a, b and c?

C

9 These are two similar rectangular blocks.

3 cm 8 cm

4 12 cm

4 cm 12 cm

6 cm

a Write down the ratio of their iii heights. ii depths i longest edges b By counting cubes of side 1 cm, write down the ratio of their volumes. c Is there any relationship between the ratios in a and b? Example

8

10 These three solids are spheres. a Write down the ratio of the radii of the three spheres. b The volume of a sphere of radius r is given by the formula 4 V = ␲r 3 . Express the volume of each sphere as a multiple 3 of ␲. Hence write down the ratio of their volumes. c Is there any relationship between the ratios found in a and b?

3 cm

2 cm

5 cm

In 11 to 20, objects referred to in the same question are mathematically similar. 11 The sides of two cubes are in the ratio 2 : 1. What is the ratio of their volumes? 12 The radii of two spheres are in the ratio 3 : 4. What is the ratio of their volumes? 13 Two regular tetrahedrons have volumes in the ratio 8 : 27. What is the ratio of their sides?



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14 Two right cones have volumes in the ratio 64 : 27. What is the ratio of a their heights

b their base radii?

15 Two similar bottles are such that one is twice as high as the other. What is the ratio of a their surface areas

b their capacities?

1 16 Each linear dimension of a model car is of the corresponding car dimension. 10 Find the ratio of a the areas of their windscreens c the widths of the cars

b the capacities of their boots d the number of wheels they have.

1 17 Three similar jugs have heights 8 cm, 12 cm and 16 cm. If the smallest jug holds litre, 2 find the capacities of the other two. 18 Three similar drinking glasses have heights 7.5 cm, 9 cm and 10.5 cm. If the tallest glass holds 343 millilitres, find the capacities of the other two. 19 A toy manufacturer produces model cars which are similar in every way to the actual cars. If the ratio of the door area of the model to the door area of the car is 1 : 2500, find b the ratio of the capacities of their petrol tanks a the ratio of their lengths c the width of the model, if the actual car is 150 cm wide d the area of the rear window of the actual car if the area of the rear window of the model is 3 cm2 . 20 The ratio of the areas of two similar labels on two similar jars of coffee is 144 : 169. Find the ratio of a the heights of the two jars

b their capacities.

21 a In the figure, if M is the midpoint of AF and K is the midpoint of AB, the area of ABF is how many times as great as the area of AKM? M b If the area of ABF is 15, find the area of AKM.

F

A

22 In the diagram, ABC is equilateral. ∠BDE = ∠CAF and D is the midpoint of AC. Find the ratio of area of BDE : area ofACF.

B

K

B

A D

E

C

F



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23 The areas of two similar triangles are 144 cm2 and 81 cm2 . If the length of one side of the first triangle is 6 cm, what is the length of the corresponding side of the second?

9.4 Geometric representation of

arithmetic operations Simple arithmetic operations correspond to elementary geometrical constructions. In many cases the validity of these constructions can be established through similar triangles. If two segments are given with lengths a and b (as measured by a given unit segment) then √ a a + b, a − b, ra (where r is any rational number), ab, and a can be constructed. b a

b

Construction of a + b Draw a straight line and mark off with a compass, as shown in the diagram, the distance OA and AB where OA = a and AB = b. Then OB = a + b.

O

A

B

a+b

Construction of a -- b Draw a straight line and mark off with a compass the distance OA and AB where OA = a and AB = b, but this time AB is constructed in the other direction. Then OB = a − b.

O

B a

A b

a–b

Construction of ra To construct 3a = a + a + a, three copies of the line segment of length a are constructed. For na = a + a + · · · + a, where n is a natural number, n copies of the line segment of length a are constructed.

Construction of ab Mark off line segments OA and OB of length a units and b units respectively. Construct OC of length 1 unit. Join O points C and B and draw a line parallel to the line CB b through A. The line segment OD has length ab. Note that triangle OAD is similar to triangle OCB and OA = aOC. Therefore OD = aOB = ab.


a C

1

A

B ab D


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Construction of

1 b

This will be done for b = 5. Line segment AB is of unit length. Draw any line AX. Choose a line segment AC and then replicate this line segment four times to form line segments CC , C C , C C and C D. Draw line segment DB and then parallel line segments CY, C Y , C Y and C Y to divide line segment AB into five equal 1 segments. Each of these segments has length of a 5 unit.

X

C A

Y

C'

C''

Y'

C'''

D

Y"

Y"'

B

Note that triangle ACY is similar to triangle ADB. And 5AC = AD. Hence AB = 5AY

Construction of

a b

a is to mark off line b a segments OA and OB of length a units and b units respectively. Construct OD of length 1 unit. Join a C points A and B and draw a line parallel to the line b a D O AB through D. The line segment OC has length 1 b Note that triangle OAB is similar to triangle OCD and a OB = bOD. Therefore OA = bOC and this implies OC = . b One way of constructing

Construction of

A

B

b

√ a

Construct line segments of length a and 1, and a circle of diameter a + 1. In the diagram OA = a and AB = 1. Angle ODB is a right angle (right angle subtended at the circle by a diameter), and OAD is a right angle by construction. Therefore triangle ODB is similar to triangle OAD and to triangle DAB. AD AB AD 1 a √ and hence AD = a

OA = AD a = AD Therefore AD2 = Hence

D

O

A

B

Exercise 9D √ 3 units. √ 2 Construct a line segment of length 5 units. 1 Construct a line segment of length



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3 Draw a line segment of length 10 cm and use a construction described above to divide it into three equal intervals. 4 Draw a line segment of length 20 cm and use a construction described above to divide it into nine equal intervals. 5 Draw two line segments OA and OB of lengths 4 cm and 14 cm respectively. Use a 2 construction described above to construct a line segment of length units. 7 6 Draw two line segments OA and OB of lengths 9 cm and 13 cm respectively. Use a 9 construction described above to construct a line segment of length units. 13 7 Describe the method for constructing a line of length

10 units. 3

8 Illustrate the construction of a line segment of length 3 × 4 units, given line segments of length 3 units, 4 units and 1 unit.

9.5 Golden ratio

c a = then a is said to be the geometric mean of c and b (or sometimes the mean b a proportional of b and c). AB AC Let AB be a line segment length a units and C a point on AB such that = . AC CB Let AC = x. Therefore CB = a − x If

A

C

B

a–x x x a holds. AC is the geometric mean of AB and CB. The relation = x a−x x a = then a(a − x) = x 2 If x a−x Which implies that x 2 + ax − a 2 = 0 √ −a ± a 2 − 4 × 1 × −a 2 Therefore using the general quadratic formula x = √ 2×1 −a ± 5a 2 = 2×1 √ −1 ± 5 x =a× 2 √ −1 + 5 Only one of these is possible as AC is a length. Thus AC = a × (which is positive) 2 √ AB (−1 + 5)a Therefore =a÷ AC 2 2 = √ −1 + 5 √ 2 1+ 5 = √ × √ −1 + 5 1 + 5 √ 1+ 5 = 2 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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AB is independent of the length of AB and is always the same number. This AC √ 1+ 5 number is known as the golden ratio or section and is denoted by ␾, i.e., ␾ = 2 ␾ is the only number which when diminished by one becomes its own reciprocal, 1 i.e., ␾ − 1 = ␾ √ 1+ 5 −1 This is shown as ␾ − 1 = 2 √ −1 + 5 = 2 √ √ −1 + 5 1 + 5 = × √ 2 1+ 5 2 = √ 1+ 5 Hence the ratio

A construction of the golden ratio is as follows. Let AB be a segment of unit length. Draw BD of AB length , perpendicular to AB. Draw line segment AD. 2 With centre D draw an arc of radius DB cutting AD at E. Draw an arc of radius AE with centre A cutting AB at C. 1 AB = =␾ AC x

D E x A

x

C 1–x B

The golden rectangle The rectangle ABHF shown is known as the golden rectangle. The ratio of the side lengths AB : AF = 1 + ␾ : ␾ √ √ 1+ 5 1+ 5 1+␾ = 1+ ÷ and ␾ 2 2 √ 3+ 5 2 = × √ 2 1+ 5 √ 3+ 5 = √ 5+1 √ 2+2 5 = 4 √ 1+ 5 =␾ = 2

A φ F

φ

E L

1

B φ–1 K 1

G

H

That is, the ratio of the side lengths is ␾. This rectangle has some very pleasant properties, as observed in the following explorations.



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Forming a sequence of similar golden rectangles In the golden rectangle ABHF, construct square AEGF with side length ␾. The remaining rectangle EBHG has side lengths 1 and ␾. Construct the square LKHG with side length 1. The sides of the remaining rectangle EBKL are 1 and ␾ − 1. 1 It was established earlier in the section that ␾ − 1 = . ␾ Thus the rectangles ABHF, EBHG, EBKL are all similar as they all have sides in the ratio ␾ : 1. This pattern continues. Consider the golden rectangle EBKL. 1 1 Now rectangle YBKX has sides and 1 − . ␾ ␾ 1 1 : 1− = 1 : ␾−1 ␾ ␾ √ 1+ 5 −1 and as shown previously ␾−1 = 2 √ −1 + 5 = 2 1 = ␾

The ratio

A

φ

E

1

L

φ

B φ–1 K 1

F

H

G

E

Y

B

L

X

K

Therefore 1 : ␾ − 1 = ␾ : 1 It can be shown that all the rectangles formed in this way are similar to each other.

Forming a sequence of squares and rectangles, the areas of 1 which are in geometric sequence with common ratio ␾ The ratio of the areas of the squares and rectangles is also worth considering. The areas in sequence are E Y A B Area rectangle ABHF = (␾ + 1)␾ = ␾3 Area of square AEGF = ␾2 K L X Area of rectangle EBHG = ␾ Area of square LKHG = 1 1 Area of rectangle EBKL = ␾ 1 F H G Area of square EYXL = 2 ␾ 1 Area of rectangle YBKX = 3 ␾



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Exercise 9E 1 For the golden ratio ␾ show that 1 a ␾−1 = b ␾3 = 2␾ + 1 ␾ 1 c 2 − ␾ = (␾ − 1)2 = 2 ␾ 2 ABC is a right-angled triangle with the right angle at C. CX is the altitude of the triangle from C. CX AX = ; i.e., the length CX is the a Prove that A CX XB geometric mean of lengths AX and XB. b Find CX if i AX = 2 and XB = 8 ii AX = 1 and XB = 10. 3 A square is inscribed in a semicircle as shown. BD AD = = ␾ − 1. Prove that BD CD

C

B

X

B

A

D

E

C

4 A regular decagon is inscribed in a circle with unit radius as shown. a Find the magnitude of angle ii OAB i AOB b The line AX bisects angle OAB. Prove that i triangle AXB is isosceles ii triangle AXO is isosceles iii triangle AOB is similar to triangle BXA c Find the length of AB, to two decimal places. d Describe a construction for ii a regular pentagon. i a regular decagon

A

B X O

5 Calculate ␾0 , ␾1 , ␾2 , ␾3 , ␾4 and ␾−1 , ␾−2 , ␾−3 , ␾−4 . Show that each power of ␾ is equal to the sum of the two powers before it, i.e., ␾n+1 = ␾n + ␾n−1 6 The Fibonacci sequence is defined by t1 = t2 = 1 and tn+1 = tn−1 + tn . The sequence is t2 t3 t4 t5 1, 1, 2, 3, 5, . . . Consider the sequence , , , . . . and show that as n gets very large t1 t2 t3 t4 tn+1 (n approaches infinity), approaches ␾. tn



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Chapter summary Two figures are similar to each other if one is congruent to the other under an expansion from the origin of factor k. An expansion of factor k from the origin has rule (x, y) → (kx, ky) Similar triangles Two triangles are similar if one of the following conditions holds. r Triangles have equal angles (AAA) r Corresponding sides are in the same ratio (PPP) r Two pairs of corresponding sides have the same ratio and the included angles are equal (PAP) B B' A

45° AB AC = A'B' A'C'

C A'

45°

C'

If triangle ABC is similar to triangle XYZ, this can be written symbolically as ABC ∼ XYZ. The triangles are named so that angles of equal magnitude hold the same position, i.e., A corresponds to X, B corresponds to Y, C corresponds to Z. If two shapes are similar and the scale factor is k, i.e. for any length AB of one shape, the corresponding length AB of the similar shape has length kAB, then the area of the similar shape = k 2 × area of the original shape. For similar solids, if the scale factor is k, then the volume of the similar solid is k 3 × volume of the original solid.

Multiple-choice questions 1 If 5 : 3 = 7 : x then x is equal to 5 21 35 E D C 5 B A 12 21 5 3 2 Brass is composed of a mixture of copper and zinc. If the ratio copper : zinc is 85 : 15, then the amount of copper in 400 kg of brass is E 150 kg D 380 kg C 360 kg B 340 kg A 60 kg 3 If the total cost of P articles is Q dollars, then the cost of R articles of the same type is P R QR PQ A PQR B E D C QR PQ P R 4 A car is 3.2 m long. The length in cm of a model of the car if the scale is 1 : 100 is E 32 D 320 C 3.2 B 0.32 A 0.032 5 An athlete runs 75 m in 9 seconds. If she were to maintain the same average speed for 100 m her time for 100 m in seconds would be E 12.4 D 12.2 C 11.8 B 12.0 A 11.6



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Review

6 If 50 is divided into three parts in the ratio 1 : 3 : 6 then the largest part is 50 E 3 D 30 C B 15 A 5 3 7 Two similar cylinders are shown. The ratio of the volume of the smaller cylinder to the larger cylinder is 15 cm C 1 : 27 B 1:9 A 1:3 D 1:5

45 cm

E 2:9

4 10 cm 8 The radius of sphere A is times the radius of 5 sphere B. Hence, the ratio of the volume of sphere A to the volume of sphere B is E 64 : 125 D 25 : 16 C 5:4 B 4:5 A 16 : 25 9 Triangles ABC and XYZ are similar isosceles triangles. The length of XY is C 4.2 cm B 5 cm A 4 cm D 2.5 cm

30 cm

Z

C

E 3.6 cm 10 cm

A

10 cm

12 cm

12 cm

B

3 cm

X

1 YX. The area of triangle 3 2 XYZ is 60 cm . The area of triangle X Y Z is 20 C cm2 B 30 cm2 A 20 cm2 9 80 20 cm2 E cm2 D 3 3

10 YZ is parallel to Y Z and Y Y =

Y

X Y'

Z' Z

Y

Short-answer questions (technology-free) 1 In triangle XYZ, P is a point on XY and Q is a point on XZ such that PQ is parallel to YZ. a Show that the two triangles XYZ and XPQ are similar. b If XY = 36 cm, XZ = 30 cm and XP = 24 cm, find ii QZ i XQ c Write down the values of XP : PY and PQ : YZ. 2 Triangles ABC and DEF are similar. If the area of triangle ABC is 12.5 cm2 , the area of triangle DEF is 4.5 cm2 and AB = 5 cm, find c the value of EF : BC. b the value of AC : DF a the length of DE 3 If a 1 m stake casts a shadow 2.3 m long, find the height of a tree (in metres) which casts a shadow 21 m long.



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4 ABC is a right-angled triangle with AB = 4 and AC = 3. If the triangle is folded along the line XY, vertex C coincides with vertex B. Find the length of XY.

A X

C

5 Points A, B and C lie on a straight line. The squares are adjacent and have side lengths 4, 7 and x. Find the value of x.

B

Y

C B A 4

6 Find the value of y in the diagram on the right.

x

7

7.2 2.2 y

6.6

26.4

7 An alloy is produced by mixing metal X with metal Y in the ratio of 5 : 3 by volume. The 8 4 mass of 1 cm3 of metal X is g and of 1 cm3 of metal Y is g. Calculate 5 3 a the mass of a solid cube of alloy of edge 4 cm b the ratio, in the form n : 1, by mass, of metal X to metal Y in the alloy c the volume, to the nearest cm3 , of a cubic block of alloy whose mass is 1.5 kg d the length, in mm, of the edge of this cubic block. 8 ABCD is a rectangle in which AB = 40 cm and AD = 60 cm. M is the midpoint of BC, and DP is perpendicular to AM. a Prove that the triangles BMA and PAD are similar. b Calculate the ratio of the areas of the triangles BMA and PAD. c Calculate the length of PD.

M

B

C

P 40 cm

A

D

60 cm

9 A sculptor is commissioned to create a bronze statue 2 m high. He begins by making a clay model 30 cm high. a Express, in simplest form, the ratio of the height of the completed bronze statue to the height of the clay model. b If the total surface area of the model is 360 cm2 , find the total surface area of the statue. c If the total volume of the model is 1000 cm3 , find the volume of the statue. 10 The radius of a spherical soap bubble increases by 1%. Find, correct to the nearest whole number, the percentage increase in b its volume. a its surface area



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YZ d AD

area triangle AXY e area triangle ABC

B

C

Y

X

area triangle CYZ f area triangle ACD

Z

A

D

Review

11 AC is the diagonal of a rhombus ABCD. The line XYZ is parallel to AD, AX = 3 cm and AB = 9 cm. Find CY AY XY c b a AC AC BC

12 AB and DC are parallel sides of a trapezium and DC = 3AB. The diagonals AC and DB 1 intersect at O. Prove that AO = AC. 4 13 Triangles ABC and PQR are similar. The medians AX and PY are drawn. (X is the midpoint of BC and Y is the midpoint of QR.) Prove BC AX a that triangles ABX and PQY are similar = b PY QR

Extended-response questions 1 a In this diagram which other triangle is similar D to DAC? F y h . b Explain why = p x+y E p c Use another pair of similar triangles to write q h h down an expression for in terms of x and y. q x y A B C 1 1 + = 1. d Explain why h e Calculate h when p = 4 and q = 5. p q 2 ABCDE is a regular pentagon whose sides are each B 1 unit long. Each diagonal is of length d units. 1 1 In a regular pentagon, each diagonal is parallel to one of the sides of the pentagon. d C A a What kind of shape is ABCF and what is the length of CF? F 1 1 b Explain why the length of EF is d − 1. c Which triangle is similar to EFD? E D d Use the pair of similar triangles to write an equation 1 for d and show that the equation can be rewritten as d 2 − d − 1 = 0. e Find d. 3 Place conditions upon x such that DE is parallel to AB given that CD = x − 3, DA = 3x − 19, CE = 4 and EB = x − 4.

C x–3 D 3x – 19 A


4 E x–4 B


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4 a If BR, CS and DT are perpendicular to BD, name the pairs of similar triangles. b Which is correct: p z p z = or = ? y q y p+q c Which is correct: q z q z = or = ? x p x p+q

R T S

x

y z p

B

1 1 1 + = x y z 5 In the diagram, PQ is parallel to BC and PR is parallel to AC. AQ = 2 cm, QC = 6 cm, AP = 3 cm and PQ = 4 cm. a Calculate ii BR i PB area BPR area APQ iv iii area ABC area ABC B b If the area of triangle APQ is a cm2 , express in terms of a: ii area CPQ i area ABC

q

C

D

d Show that

A 3 cm P

2 cm Q 4 cm 6 cm

C

R

6 Construct a triangle ABC such that BC = 10 cm, AC = 9 cm and AB = 6 cm. Find a point D on AB and a point E on AC, such that DE is parallel to BC and the area of ADE is one-ninth the area of ABC. C

7 A triangular lot has boundaries of lengths AB = 130 m, √ BC = 40 10 m and CA = 150 m. The length of CD is 120 m. A fence is to be erected which runs at right angles from AB. If the lot is to be divided into two equal areas, find x. fence A

8 The Greek historian Herodotus wrote that the proportions of the great pyramid at Giza in Egypt were chosen so that the area of a square, for which the side lengths are equal to the height of the great pyramid, is equal to the area of one of the triangular faces. Let h m be the height of the pyramid, k m the altitude of one of the face triangles, and b m be the length of a side of the square base. b Show that Herodotus’ definition gives k : = ␾. 2


B

D

xm V

km D

C

hm

A

bm

B


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C H A P T E R

10 Circular functions I Objectives To use radians and degrees for the measurement of angles To convert radians to degrees and vice versa To define the circular functions sine, cosine and tangent To explore the symmetry properties of circular functions To find standard exact values of circular functions To understand and sketch the graphs of circular functions

10.1

Measuring angles in degrees and radians The diagram shows a unit circle, i.e., a circle of radius one unit.

y

The circumference of the unit circle = 2␲ × 1 = 2␲ units

∴ the distance in an anticlockwise direction around the circle from ␲ A to B = units 2 A to C = ␲ units 3␲ units A to D = 2

1 B

C –1

O

A 1

x

–1 D

Definition of a radian In moving around the circle a distance of one unit from A to P, the angle POA is defined. The measure of this angle is one radian. One radian (written 1c ) is the angle subtended at the centre of the unit circle by an arc of length one unit.



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Angles formed by moving anticlockwise around the circumference of the unit circle are defined as positive. Those formed by moving in a clockwise direction are said to be negative.

Note:

y 1

P 1 unit 1c

–1

O

A 1

x

–1

Degrees and radians The angle, in radians, swept out in one revolution of a circle is 2␲ c . 2␲ c = 360◦ ␲ c = 180◦ 180◦ ␲c or 1◦ = 1c = ␲ 180

∴ ∴ ∴

Example 1 Convert 30◦ to radians. Solution Since

∴

␲c 180 30 × ␲ ␲c 30◦ = = 180 6 1◦ =

Example 2 Convert

␲c to degrees. 4 Solution Since

∴

180◦ ␲ ␲ c ␲ × 180 = = 45◦ 4 4×␲ 1c =

␲ Often the symbol for radian, c , is omitted. For example, angle 45◦ is written as rather 4 ␲c than . 4 Note:



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Chapter 10 — Circular functions I

Exercise 10A Example

1

1 Express the following angles in radian measure in terms of ␲. a 60◦

Example

2

b 144◦

c 240◦

d 330◦

e 420◦

f 480◦

2 Express in degrees the angles with the following radian measures. 5␲ 7␲ 5␲ 2␲ 11␲ 9␲ d 0.9␲ e c b a g f 6 9 6 3 9 5 3 Use a calculator to convert the following angles from radians to degrees. a 0.6

b 1.89

c 2.9

d 4.31

e 3.72

f 5.18

h 1.8␲

g 4.73

h 6.00

4 Use a calculator to express the following in radian measure. a 38◦ e 84◦ 10

b 73◦ f 228◦

c 107◦ g 136◦ 40

d 161◦ h 329◦

5 Express in degrees the angles with the following radian measures. ␲ a − b −4␲ c −3␲ d −␲ 3 5␲ −11␲ −23␲ 23␲ e f h g 3 6 6 6 6 Express each of the following in radian measure in terms of ␲. a −360◦

10.2

b −540◦

c −240◦

d −720◦

e −330◦

f −210◦

Defining circular functions: sine and cosine y

Consider the unit circle. The position of point P on the circle can be described by relating the cartesian coordinates x and y and the angle ␪. The point P on the circumference corresponding to an angle ␪ is written P(␪). Many different angles will give the same point, P, on the circle, so the relation linking an angle to the –1 coordinates is a many-to-one function. There are, in fact, two functions involved and they are called sine and cosine and are defined as follows:

1

P(θ) 1

O

y

θc x

x

1

–1 y

The x coordinate of P, x = cosine ␪, ␪ ∈ R The y coordinate of P, y = sine ␪, ␪ ∈ R

1

These functions are usually written in an abbreviated form as follows:

Note:

x = cos ␪ y = sin ␪ cos (2␲ + ␪) = cos ␪ and sin (2␲ + ␪) = sin ␪, as adding 2␲ results in a return to the same point on the unit circle.

P(θ) = (cos θ, sinθ) 1

c

θ –1

sinθ x

O cosθ

1

Note:


–1


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Example 3 ␲ 3␲ and cos − b sin − 2 2 9␲ and cos 27␲. d sin 2

Evaluate a sin ␲and cos ␲ c sin

5␲ 7␲ and sin 2 2 Solution

a In moving anticlockwise through an angle of ␲, the position is P(␲) which is (−1, 0)

∴

cos ␲ = −1 sin ␲ = 0

3␲ 3␲ , the position is P − which is b In moving clockwise through an angle of 2 2 (0, 1) 3␲ =1 ∴ sin − 2 ␲ =0 cos − 2 1 ␲ ␲ 5␲ c sin = sin 2 ␲ = sin 2␲ + = sin = 1 2 2 2 2 1 3␲ 3␲ 7␲ = sin 3 ␲ = sin 2␲ + = sin = −1 sin 2 2 2 2 ␲ ␲ 9␲ d sin = sin 4␲ + = sin = 1 2 2 2 cos(27␲) = cos (26␲ + ␲) = cos ␲ = −1

Exercise 10B Example

3

1 For each of the following angles, t, determine the values of sin t and cos t. 5␲ 3␲ 3␲ d t= c t =− a t =0 b t= 2 2 2 7␲ 9␲ h t = 4␲ g t= f t= e t = −3␲ 2 2 2 Evaluate, using a calculator (with calculator in Radian mode). a sin 1.9 e cos 2.1

b sin 2.3 f cos (−1.6)

c sin 4.1 g sin (−2.1)

d cos 0.3 h sin (−3.8)

3 For each of the following angles, ␪, determine the values of sin ␪ and cos ␪. 5␲ 9␲ 27␲ a ␪ = 27␲ b ␪=− d ␪=− c ␪= 2 2 2 11␲ e ␪= f ␪ = 57␲ g ␪ = 211␲ h ␪ = −53␲ 2 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Another circular function: tangent

10.3

Again consider the unit circle. If a tangent to the unit circle at A is drawn, then the y coordinate of C, the point of intersection of the extension of OP and the tangent is called tangent ␪ (abbreviated to tan ␪). By considering the similar triangles OPD and OCA:

∴

tan ␪ =

B

y

C(1, y) 1

P(θ)

θ –1

sin ␪ tan ␪ = 1 cos ␪

257

sin ␪ cos ␪

O

cos θ

sin θ

tan θ

A D 1

x

–1

Now when cos ␪ = 0, tan ␪ is undefined. ␲ 3␲ 5␲ Hence tan ␪ is undefined when ␪ = ± , ± , ± , · · · 2 2 2

∴

Domain of tan = R\{␪ : cos ␪ = 0}. Example 4

Use a calculator to evaluate, correct to two decimal places a tan 1.3 b tan 1.9 c tan (−2.8) d tan 59◦

e tan 138◦

Solution a b c d e

tan 1.3 = 3.60 tan 1.9 = −2.93 tan (−2.8) = 0.36 tan 59◦ = 1.66 tan 138◦ = 0.90

(Don’t forget calculator must be in Radianmode.)

(Put calculatorin Degree mode.)

Exercise 10C Example

4

1 Use a calculator to find, correct to two decimal places a tan 1.6 e tan 3.9

b tan (−1.2) f tan (−2.5)

c tan 136◦ g tan 239◦

d tan (−54◦ )

2 Evaluate a tan ␲ d tan (−2␲)

b tan (−␲) 5␲ e tan 2

7␲ 2 ␲ f tan − 2

c tan

3 For each of the following values of ␪ find tan ␪. a ␪ = 180◦ d ␪ = −180◦

b ␪ = 360◦ e ␪ = −540◦

c ␪=0 f ␪ = 720◦



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10.4 Reviewing trigonometric ratios For right-angled triangles opp sin ␪ = hypotenuse hyp opposite side adj cos ␪ = θ hyp opp tan ␪ = adjacent side adj y Applying these trigonometric ratios to the right-angled triangle, OAB, in the unit circle 1 B y opp = =y sin ␪ = 1 hyp 1 y θ adj x x cos ␪ = = =x O x A 1 –1 hyp 1 opp y sin ␪ tan ␪ = = = adj x cos ␪ –1 ␲ For 0 < ␪ < , functions sin, cos and tan are defined by the trigonometric ratios and are the 2 same as the respective circular functions introduced earlier.

Exercise 10D 1 Find the value of the pronumeral for each of the following.

a

b

5

8

c

6

25°

x

25°

x θ 3 d

e

f

θ

10 20°

x

5

10

6

50° x g

h

i

x

5 65°

7

5

40°

x

70°

x



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2 a Use a calculator to find a and b, correct to four decimal places. b Hence find the values of c and d. c i Use a calculator to find cos 140◦ and sin 140◦ . ii Write cos 140◦ in terms of cos 40◦ .

10.5

259

y

(a, b)

(c, d) 140°

1 40°

x

0

Symmetry properties of circular functions y

The coordinate axes divide the unit circle into four quadrants. The quadrants can be numbered anticlockwise from the positive direction of the x axis, as shown.

Quadrant 2

Quadrant 1 x

0 Quadrant 3

Quadrant 4

Relationships, based on symmetry between circular functions, for angles in different quadrants can be determined.

Quadrant 2 By symmetry sin(π – θ) = b = sinθ cos(π – θ) = –a = –cos θ tan(π – θ) = b = –tanθ –a

Quadrant 1

(0, b)

P(π – θ)

P(θ) = (cos θ, sin θ) = (a, b) θ

(–a, 0)

0

P(π + θ) (0, –b) Quadrant 3 sin(π + θ) = –b = –sin θ cos(π + θ) = –a = –cos θ tan(π + θ) = –b –a = tan θ Note:

(a, 0) P(2π – θ)

Quadrant 4 sin(2π – θ) = –b = –sin θ cos(2π – θ) = a = cos θ tan(2π – θ) = –b a = –tanθ

These relationships are true for all values of ␪.



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y

Signs of circular functions These symmetry properties can be summarised for the signs of sin, cos and tan for the four quadrants as follows: 1st quadrant 2nd quadrant 3rd quadrant 4th quadrant

S

All are positive (A) Sin is positive (S) Tan is positive (T) Cos is positive (C)

A x

T

Negative of angles

C

y

By symmetry

1

cos (−␪) = cos ␪ sin (−␪) = − sin ␪ − sin ␪ tan (−␪) = cos ␪ = − tan ␪

P(θ)

–1

0

θ –θ

1

x

P(–θ) –1 Example 5 If sin x = 0.6, find the value of a sin(␲ − x) b sin (␲ + x)

c sin (2␲ − x)

d sin (−x)

Solution a sin (␲ − x) = sin x = 0.6

b sin (␲ + x) = − sin x = −0.6

c sin (2␲ − x) = − sin x = −0.6

d sin (−x) = − sin x = −0.6

Example 6 If cos x ◦ = 0.8, find the value of b cos (180 + x)◦ a cos (180 − x)◦

c cos (360 − x)◦

d cos (−x)◦

Solution a cos (180 − x)◦ = − cos x ◦ = −0.8

b cos (180 + x)◦ = − cos x ◦ = −0.8


c cos (360 − x)◦ = cos x ◦ = 0.8

d cos (−x)◦ = cos x ◦ = 0.8


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261

Exercise 10E Example

5

1 If sin ␪ = 0.42, cos x = 0.7 and tan ␣ = 0.38, write down the values of a sin (␲ + ␪) e sin (␲ − ␪)

Example

6

b cos (␲ − x) f tan (2␲ − ␣)

c sin (2␲ − ␪) g cos (␲ + x)

d tan (␲ − ␣) h cos (2␲ − x)

2 If sin x ◦ = 0.7, cos ␪ ◦ = 0.6 and tan ␣◦ = 0.4, write down the values of a sin (180 + x)◦ e sin (360 − x)◦

b cos (180 + ␪)◦ f sin (−x)◦

c tan (360 − ␣)◦ g tan (360 + ␣)◦

d cos (180 − ␪)◦ h cos (−␪)◦

y

3 Write down the values of a b c d e f

a = cos (␲ − ␪) b = sin (␲ − ␪) c = cos (−␪) d = sin (−␪) tan (␲ − ␪) tan (−␪)

(a, b)

10.6

π–θ θ –θ 1

0

x (c, d )

y

4 Write down the values of a b c d e

1 √3 , 2 2

1 √3 – , 2 2

d = sin (␲ + ␪) c = cos (␲ + ␪) tan (␲ + ␪) sin (2␲ − ␪) cos (2␲ − ␪)

θ x

0 π+θ 1 (c, d )

Exact values of circular functions A calculator can be used to find the values of the circular functions for different values of ␪. For many values of ␪ the calculator gives an approximation. Consider some values of ␪ such that sin, cos and tan can be calculated exactly.

Exact values for 0 (0◦) and y

From the unit circle When ␪ = 0, sin 0 = 0 cos 0 = 1 tan 0 = 0

–1

␲ (90◦) 2

␲ , 2

1

When ␪ =

θ 0 cos θ

␲ =1 2 ␲ cos = 0 2 ␲ tan is undefined 2 sin

sin θ 1

x

–1



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␲ ␲ (30◦) and (60◦) 6 3

Exact values for

Consider an equilateral triangle ABC of side length two units. In √ ACD, by Pythagoras’ theorem DC = AC2 − AD2 = 3 √ 3 1 AD CD ◦ ◦ = sin 60 = = sin 30 = AC 2√ AC 2 2 3 CD AD 1 ◦ ◦ cos 30 = = cos 60 = = AC 2 AC 2√ 1 AD CD 3 √ tan 30◦ = =√ = = 3 tan 60◦ = 60° CD AD 1 3 A

sin 45◦ cos 45◦ tan 45◦

30° 30° 2

60° D

1

1

␲ (45◦) 4

Exact values for AC

C

√ = 12 + 1 2 = 2 BC 1 = =√ AC 2 AB 1 = =√ AC 2 BC =1 = AB

B

C

1

A

45° 1

B

As an aid to memory, the exact values for circular functions can be tabulated.

Summary ␪(␪ ◦ ) 0

sin ␪ 0

␲ ◦ (30 ) 6 ␲ ◦ (45 ) 4

1 2 1 √ 2 √ 3 2

␲ ◦ (60 ) 3 ␲ ◦ (90 ) 2

1

cos ␪ 1 √ 3 2 1 √ 2

tan ␪ 0

1 2

√

0

undefined

1 √ 3 1 3

Example 7 Evaluate 5␲ a cos 4

b sin

11␲ 6



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263

Solution 5␲ 4 ␲ = − cos (bysymmetry) 4 1 = −√ 2

11␲ 6 ␲ = − sin (by symmetry) 6 1 =− 2

a cos

b sin

Exercise 10F Example

7

1 Write down the exact values of 3␲ 5␲ 7␲ 5␲ 2␲ b cos e cos d sin c tan a sin 4 4 6 6 3 11␲ 5␲ 7␲ 4␲ i tan g sin h cos f tan 6 3 4 3 2 Without using a calculator, evaluate the sin, cos and tan of each of the following. a 120◦ f 390◦

b 135◦ g 420◦

c 210◦ h −135◦

3 Write down the exact values of 11␲ 2␲ b cos a sin − 4 3 3␲ 14␲ f cos e cos 4 4

10.7

d 240◦ i −300◦

e 315◦ j −60◦

13␲ c tan 6 11␲ g sin 4

15␲ d tan 6 21␲ h cos − 3

Graphs of sine and cosine Graphs of sine functions The graph of f (x) = sin x is given below. It has been plotted for –␲ ≤ x ≤ 3␲. y 1 1 √2 –3 –π

–2

1

–1

–3π –π –π 4 2 4

π 4

0

2 π 2

3 3π π 4

4

5

5π 3π 7π 4 2 4

–1

6

7

8

9

x

2π 9π 5π 11π 3π 4 2 4 f (x) = sin x

–1 √2

Observations from the graph The graph repeats itself after an interval of 2␲ units, i.e. f (x + 2k␲) = f (x) for all x ∈ R, k ∈ Z . A function which repeats itself regularly is called a periodic function and the interval between the repetitions is called the period of the function. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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The maximum and minimum values of sin x are 1 and −1 respectively. The distance between the mean position and the maximum position is called the amplitude. The graph of f (x) = sin x has an amplitude of 1.

Graphs of cosine functions The graph of g(x) = cos x for –␲ ≤ x ≤ 3␲ is as shown. y 1

–3

–2

–π –3π 4

–1

0

–π –π 2 4

1 π π 4 2

2

3

4

3π π 5π 4 4

5

6

7

8

9

x

3π 7π 2π 9π 5π 11π 3π 4 2 4 4 2 g(x) = cosx

–1

Dilations of sine and cosine functions

A dilation of factor 2 from ␲ they axis has the rule (x, y) → (2x, y). , 1 → (␲, 1) and (␲, 0) → (2␲, 0). Hence (0, 0) → (0, 0), 2 When this transformation is applied to y = sin x, it will be ‘stretched’ away from the y axis. Let (x, y) → (x , y ) under this transformation. x Since (x, y) → (2x, y), then x = 2x and y = y, and thus x = and y = y . 2 x Hence y = sin x is mapped to y = sin . 2 1 A dilation of factor from the y axis will map y = sin x to y = sin 2x. 2 y

y

1

y

1

0

2π

–1

4π

x

0

–1

1

π

2π

x

0

π 2

π

x

–1

x ←−−−−−−−−−−−−−−−− y = sin x −−−−−−−−−−−−−−→ y = sin 2x 2 Period = 4␲ Period = 2␲ Period = ␲ Range = [−1, 1] Range = [−1, 1] Range = [−1, 1] y = sin



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265

In general: f : R → R, f (x) = sin (nx) 2␲ Period = n Amplitude = 1 Range = [−1, 1]

f : R → R, f (x) = cos (nx) 2␲ Period = n Amplitude = 1 Range = [−1, 1]

A dilation of factor 3 from ␲has the rule (x, y) → (x, 3y). ␲ the x axis ,0 → , 0 and (␲, −1) → (␲, −3). Hence (0, 1) → (0, 3), 2 2 When this transformation is applied to y = cos x, it will be ‘stretched’ away from the x axis. y = cos x is mapped to y = 3 cos x

y

y

3

3

2

2

1

1

0 –1

π

π 2

x 3π 2

0

2π

–1

–2

–2

–3

–3 y = cos x Period = 2␲ Range = [−1, 1]

−−−−→

π 2

π

x 3π 2

2π

y = 3 cos x Period = 2␲ Range = [−3, 3]

In general: f : R → R, f (x) = a sin (nx), n, a > 0 2␲ Period = n Amplitude = a Range = [−a, a]

f : R → R, f (x) = a cos (nx), n, a > 0 2␲ Period = n Amplitude = a Range = [−a, a]

Example 8 Sketch the graphs of a y = 2 cos 2 ␪ Show one complete cycle.

b y=

1 x sin 2 2



266


Solution 2␲ =␲ 2 amplitude = 2

1 b period = 2␲ ÷ = 4␲ 2 1 amplitude = 2

a period =

y

y 1 2

2

0

π 4

π 2

3π 2

π

–2

x

0

π

2π

3π

4π

x

–1 2

Using the TI-Nspire Check that the calculator is in Radian mode. Open a Graphs application (c>New Document>Add Graphs), enter the function and select an appropriate window (b >Window/Zoom>Window Settings) as shown. a f 1 (x) = 2 cos (2x) xMin = 0 xMax = ␲ xScale = ␲ ÷ 4 yMin = −2 yMax = 2

b

x 1 sin 2 2 xMin = 0 xMax = 4␲ xScale = ␲ yMin = −0.5 yMax = 0.5 f 1 (x) =




267

Using the Casio ClassPad Ensure that the calculator is set to Radian mode (bottom left of the screen; tap to change). a Enter the function y = 2 cos (2x) in the window. Tap $ to produce the graph.

To show one complete cycle as required, tap 6 and make the window settings ␲ x min = 0, max = ␲ and scale = . 4 The appropriate values for y should also be set. Do not be concerned with the setting for ’dot’, it will take care of itself. ␲ 1 in the b Enter the function y = sin 2 2 window then tap $ to produce the graph. The 6 settings for the graph are shown below.



268


Example 9 Sketch the graph of y = 5 sin 3␪ for −

4␲ ≤ ␪ ≤ 2␲. 3

Solution The amplitude = 5, the period =

2␲ 3

y 5

–4π 3

–π

–2π 3

–π 3

0

π 3

2π 3

π

4π 3

5π 3

2π

x

–5

The x axis intercepts can also be found by observing that one half period is

␲ . 3

Using the TI-Nspire Check that the calculator is in Radian mode. Open a Graphs application (c>New Document>Add Graphs), enter the function and select an appropriate window (b >Window/Zoom>Window Settings) as shown. f 1(x) = 5 sin(3x) 4␲ xMin = − 3 xMax = 2␲ xScale = ␲ ÷ 6 yMin = −5 yMax = 5




269

Using the Casio ClassPad Check that the calculator is in Radian mode. Enter the function as shown with the window settings shown to produce the graph.

Exercise 10G 1 For each of the following, write down i the period and

ii the amplitude.

a 2 sin ␪

b 3 sin 2 ␪

e 4 cos 3 ␪

1 f − sin 4 ␪ 2

1 cos 3 ␪ 2 1 g −2 cos ␪ 2 c

d 3 sin

1 ␪ 2

Example

8

2 Sketch the graph of each of the following, showing one complete cycle. State the amplitude and period. ␪ c y = 4 sin a y = 3 sin 2x b y = 2 cos 3 ␪ 2 1 d y = cos 3x e y = 4 sin 3x f y = 5 cos 2x 2 ␪ ␪ g y = −3 cos i y = −2 sin h y = 2 cos (4␪) 2 3

Example

9

3 Sketch the graph of x for x ∈ [−6␲, 6␲] 3 c f (x) = 2 cos 3x for x ∈ [0, 2␲] d f (x) = −2 sin 3x for x ∈ [0, 2␲] 2x 5 4 Sketch the graph of f : [0, 2␲] → R, f (x) = cos 2 3 Hint: For endpoints find f (0) and f (2␲). a

f (x) = sin 2x for x ∈ [−2␲, 2␲]

b

f (x) = 2 sin

5 a On the one set of axes, sketch the graphs of f : [0, 2␲] → R, f (x) = sin x and g: [0, 2␲] → R, g(x) = cos x b By inspection from the graph state the values of x for which sin x = cos x. ISBN 978-1-107-65235-4 © Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2012 Photocopying is restricted under law and this material must not be transferred to another party.


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10.8 Further transformations of sine and

cosine graphs Reflection in the axes The function with rule f (x) = sin x is described as an odd function, i.e., f (−x) = − f (x). A reflection in the y axis gives the same result as a reflection in the x axis when applied to the graph of y = sin x. The function with rule f (x) = cos x is described as an even function, i.e., f (−x) = f (x). The graph of f (x) = cos x is mapped onto itself when reflected in the y axis.

Example 10 Sketch the graphs of a f (␪) = −3 cos 2␪

for 0 ≤ ␪ ≤ 2␲

b g(␪) = 5 sin(−3␪)

for 0 ≤ ␪ ≤ 2␲

Solution y

a 3

f (θ) = –3 cos 2θ

0

π

2π

θ

Period = π Amplitude = 3 –3 y

b 5

g(θ) = 5 sin(–3θ)

0

π 3

2π 3

π

4π 3

5π 2π 3

θ 2π 3 Amplitude = 5 Period =

–5



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271


Translations Translations in the direction of the y axis The graph of y = sin x + 1 is obtained from the graph of y = sin x by a translation of one unit in the positive direction of the y axis.

y 2 Period = 2π Range = [0, 2] Amplitude = 1

0

The graph of y = cos 2x − 2 is obtained from the graph of y = cos 2x by a translation of two units in the negative direction of the y axis.

x

π

2π

3π

4π

y π

0

2π

3π

4π

x

–1 –2 –3

Translations inthe direction of the x axis

␲ is obtained from 3 ␲ the graph of y = sin x by a translation of 3 in the positive direction of the x axis. The graph of y = sin x −

y

0

␲ is obtained The graph of y = cos 2 x + 3 from the graph of y = cos 2x by a translation ␲ of in the negative direction of the x axis. 3

5π 6

4π 3

11π 6

7π 3

17π 6

y

0 –π –π 3 12


x

π 3

–π 6

π 6

x 5π 2π 11π 12 3 12

7π 6

17π 5π 12 3


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Example 11 On separate axes sketch the graphs of ␲ for −␲ ≤ t ≤ 2␲ a y = 3 sin 2 t − 4

␲ for −␲ ≤ t ≤ ␲ b y = 2 cos 3 t + 3

Solution a The transformations applied to y = sin t are r a dilation of factor 3 from the x axis 1 r a dilation of factor from 2 the y axis ␲ r a translation of in the positive 4 direction of the x axis.

y 3 –3π –π π 7π 3π 5π 4 4 4 4 4 4 –π –π 0 π π 3π 2π 2 2 2 –3

t

Period = π Amplitude = 3 Range = [–3, 3]

This is the graph of y = −3 cos 2t. b Note: The transformations applied to y = cos t are r a dilation of factor 2 from the x axis 1 r a dilation of factor from the y axis 3 ␲ r a translation of in the negative direction of the x axis. 3 Note:

y 2

–π

–5π –π 6 2 –2π 3

–π 3

–π 6 0

π 6

π 2 π 3

5π 6

2π 3

π

t

–2

Exercise 10H Example

10

1 Sketch the graph of each of the following showing one complete cycle. State the period, amplitude and the greatest and least values. √ a y = 4 cos (−2x) b f (␪) = − 2 sin 2␪ c f (x) = 2 sin (−3x)

Example

11

2 Sketch the graph of each of the following showing one complete cycle. State the period, amplitude and the greatest and least values. ␲ ␲ a y = 3 sin ␪ − b y = sin 2(␪ + ␲) c y = 2 sin 3 ␪ + 2 4



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√ ␲ 3 sin 2 ␪ − 2 √ ␲ g y = 2 sin 2 ␪ − 3

d y=

␲ f y = 2 cos 3 ␪ + 4 ␲ i y = −3 cos 2 ␪ + 2

e y = 3 sin 2x h y = −3 sin 2x

␲ 3 For the function f : [0, 2␲] → R, f (x) = cos x − 3 a find f (0), f (2␲) b sketch the graph of f. ␲ 4 For the function f : [0, 2␲] → R, f (x) = sin 2 x − 3 a find f (0), f (2␲) b sketch the graph of f. ␲ 5 For the function f : [−␲, ␲] → R, f (x) = sin 3 x + 4 a find f (−␲), f (␲) b sketch the graph of f.

10.9

Solution of circular function equations Example 12 Find all solutions to the equation sin ␪ =

1 for ␪ ∈ [0, 4␲]. 2

Solution It is clear from the graph that there are four solutions in the interval [0, 4␲]. ␲ ␲ is ␪ = . The solution for ␪ ∈ 0, 2 6 This solution can be obtained from a knowledge of exact values or using sin−1 on a calculator.

y 1

y = sinθ y=

0 π 6

π

2π

3π

1 2 θ

4π

–1 y The second solution is obtained by symmetry. The function is positive in 1 the second quadrant and sin (␲ − ␪) = sin ␪. 5␲ is the second Therefore ␪ = 6 solution. 0 π It can be seen that further solutions 6 can be achieved by adding 2␲, as sin ␪ = sin (␪ + 2␲). –1 17␲ 13␲ and are also solutions. Thus ␪ = 6 6


y = sinθ y=

5π 6

13π 17π 6 6

1 2 θ


274


Using the TI-Nspire Check that the calculator is in Radian mode. Use solve( ) from b >Algebra>Solve as shown. The symbol ≤ can be found in the symbols palette (/+k), or /+= and select ≤ (<= or /+< on the Clickpad) Hint: when solving circular functions it is a good idea to quickly check using a graphical approach to ensure you have all the solutions.

Using the Casio ClassPad Set the calculator to Radian mode. Enter the functions y = sin (x) and y = 1/2. The window settings should be as shown. Tap $ to produce the graph. Ensure the graph window is selected (bold border) and tap Analysis, G-solve, intersect to find decimal approximations for the solutions. The scroll key moves the cursor between solutions. window To find exact solutions, the is used. Enter and highlight the equation 1 sin (x) = . Tap Interactive, 2 Equation/inequality, solve and ensure the variable is set to x. The answer returned is 5␲ ␲ . x = 2␲ ∗ constn (9) + , x = 2␲ ∗ constn (10) + 6 6 This may be read as x = 2m␲ +

5␲ ␲ , 2n␲ + . 6 6

Note: The restricted domain can be entered. See the Appendix for directions.



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It should be clear that there are 4 solutions to the problem. Hence, the values for each of m and n will be required which produce a solution in the domain. In this case the ␲ 11␲ 13␲ 23␲ , , . values are m = 0, 1 and n = 0, 1. The solutions are x = , 6 6 6 6 Example 13 For the following, find two values of x in the range 0 ≤ x ≤ 360. b cos x ◦ = −0.7 a sin x ◦ = −0.3 Solution a First solve the equation sin x ◦ = 0.3. Use a calculator to find the solution for x ∈ [0, 90]; x = 17.46 Now the value of sin is negative for P(x) in the 3rd and 4th quadrants. From the symmetry relationships (or from the graph of y = sin x ◦ ) x = 180 + 17.46 = 197.46 4th quadrant x = 360 − 17.46 = 342.54 ◦ ∴ if sin x = −0.3, x = 197.46 or x = 342.54 3rd quadrant

b First solve the equation cos x ◦ = 0.7. Use a calculator to find the solution for x ∈ [0, 90]; x = 45.57 Now the value of cos is negative for P(x) in the 2nd and 3rd quadrants x = 180 − 45.57 = 134.43 3rd quadrant x = 180 + 45.57 = 225.57 ◦ ∴ if cos x = −0.7, x = 134.43 or x = 225.57 2nd quadrant

Example 14 √ 3 for ␪ ∈ [−␲, ␲]. Solve the equation sin 2␪ = − 2 Solution y

It is clear that there are four solutions. To solve the equation let x = 2␪.

1

if ␪ ∈ [−␲, ␲] then 2␪ = x ∈ [−2␲, 2␲]

y = sin2θ

Note:

√ 3 Consider the equation sin x = − 2 for x ∈ [−2␲, 2␲]

–π

y=– ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

–π 2 √3 2

0

π 2

π

θ

–1 Cambridge University Press

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y The 1st quadrant solution of the √ ␲ 3 equation sin x = is x = 1 2 3 Symmetry gives the solutions to √ 3 sin x = − for x ∈ [0, 2␲] as 2 –2π 0 ␲ ␲ x = ␲ + and x = 2␲ − 3 3 √3 5␲ 4␲ –1 y=– or x = i.e., x = 2 3 3 The other two solutions are obtained by subtracting 2␲, 4␲ 5␲ i.e., − 2␲ and − 2␲ 3 3 −␲ 4␲ 5␲ 2␲ or or or ∴ the required solutions for x are − 3 3 3 3 −␲ 2␲ 5␲ ␲ or or ∴ the required solutions for ␪ are − or 3 6 3 6

y = sinx

2π

x

Exercise 10I Example

12

1 Find, without using a calculator, all the values of ␪ between 0 and 2␲ for each of the following. √ √ a 2 sin x + 1 = 0 b 2 cos x − 1 = 0 2 Find all values of x between −␲ and ␲ for which √ 3 1 b sin x = a cos x = − √ 2 2

c cos x = −

1 2

3 Find, without using a calculator, all the values of ␪ between 0◦ and 360◦ for each of the following. √ 3 1 1 ◦ a cos ␪ = − c cos ␪ ◦ = − b sin ␪ ◦ = 2 2 2 √ √ d 2 cos ␪ ◦ + 1 = 0 e 2 sin ␪ ◦ = 3 f 2 sin ␪ ◦ − 1 = 0 Example

13

4 Find all the values of x between 0 and 2␲ for which a sin x = 0.8 d sin x = 0.4

b cos x = −0.4 e cos x = −0.7

5 Find all the values of x between 0 and 4␲ for which 1 a sin x = 0.6 b sin x = − √ 2


c sin x = −0.35 f cos x = −0.2 √ c sin x =

3 2


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Chapter 10 — Circular functions I Example

14

6 Solve the following equations for ␪ ∈ [0, 2␲]. √ 3 1 b cos 2␪ = a sin 2␪ = − 2√ 2 3 1 e cos 2␪ = − d sin 3␪ = − √ 2 2

c sin 2␪ =

1 2

1 f sin 2␪ = − √ 2

7 Solve the following equations for ␪ ∈ [0, 2␲]. a sin 2␪ = −0.8

b sin 2␪ = −0.6

c cos 2␪ = 0.4

d cos 3␪ = 0.6

8 a Sketch the graph of f : [−2␲, 2␲] → R, f (x) = cos x.

1 b On the graph, mark the points which have y coordinate and give the associated 2 x values. 1 c On the graph, mark the points which have y coordinate − and give the associated 2 x values.

10.10

Further sketch graphs Sketch graphs of y = sin n(t + ε) + b and y = a cos n(t + ε) + b Consider the graph of each of the following functions. ␲ 5␲ ␲ + 2, ≤ t ≤ a y = 3 sin 2 t − 4 4 4 ␲ ␲ ␲ − 1, − ≤ t ≤ b y = 2 cos 3 t + 3 3 3

y

a

y = 3 sin 2 t – 5 4

–1

y = 2 cos 3 t +

1

3 2 1 0

y

b

π +2 4

–π 3 π 4

π 2

3π 4

π

t 5π 4

–π 0 6 –1

π 6

π –1 3 t

π 3

–2 –3

Observations ␲ + 2 is the same shape as the graph of 1 The graph of y = 3 sin 2 t − 4 ␲ y = 3 sin 2 t − but it is translated two units in the positive direction of the y axis. 4 ␲ 2 Similarly, the graph of y = 2 cos 3 t + − 1 is the same shape as the graph of 3 ␲ y = 2 cos 3 t + but it is translated one unit in the negative direction of the y axis. 3



278


In general, the effect of b is to translate the graph b units in the positive direction of the y axis when b > 0, and in the negative direction of the y axis when b < 0.

Finding axis intercepts Example 15 Sketch the graphs of each of the following for x ∈ [0, 2␲]. Clearly indicate axis intercepts. √ ␲ √ b y = 2 cos 2x − 1 c y = 2 sin 2 x − − 3 a y = 2 sin x + 1 3 Solution a To determine the axis intercepts, the equation √ 2 sin x + 1 = 0 1 sin x = − √ 2 ␲ ␲ ∴ x = ␲ + , 2␲ − 4 4 5␲ 7␲ , ∴ x= 4 4 5␲ 7␲ ∴ intercepts ,0 , ,0 4 4

∴

√ 2 sin x + 1 = 0 must be solved.

y √2 + 1 y = √2 sin x + 1 3π 2 x

1 0 – √2 + 1

π 2

π 5π 4

7π 2π 4

Using the TI-Nspire Check that the calculator is in Radian mode. Open a Graphs application (c>New Document>Add Graphs), enter f 1(x) = √ 2 sin(x) + 1 and select the window (b > Window/Zoom>Window Settings) as shown. xMin = 0 xMax = 2␲ yMin = 1 − (2) yMax = 1 + (2) Open a Calculator application (/+*> Add Calculator) to find the exact values of the x-intercepts using solve( ) from the Algebra menu as shown. The symbol ≤ can be found in the symbols palette (/+k), or /+= and select ≤ (<= or /+< on the Clickpad)




279

Using the Casio ClassPad Ensure that the calculator is in Radian mode. Enter and highlight the equation √ 2 sin(x) + 1 = 0. Tap Interactive, Equation/inequality, solve. 7␲ 5␲ The required solutions x = or are found by selecting the appropriate values for 4 4 the constant, ‘constn(2) and constn(2)’. b

2 cos 2x − 1 = 0 1 ∴ cos 2x = 2 ␲ 5␲ 7␲ 11␲ , , ∴ 2x = , 3 3 3 3 ␲ 5␲ 7␲ 11␲ , , ∴ x= , 6 6 6 6 ␲ 5␲ 7␲ 11␲ ,0 , ,0 , ,0 ∴ intercepts , 0 , 6 6 6 6 y

y = 2 cos 2x – 1

1 0 –1

π 6

5π 6

π

x 7π 11π 6 6

2π

–2 –3 ␲ √ 2 sin 2 x − − 3=0 3 √ ␲ 3 = ∴ sin 2 x − 3 2 ␲ ␲ 2␲ 7␲ 8␲ = , , , ∴ 2 x− 3 3 3 3 3 ␲ ␲ 7␲ 4␲ ␲ , ∴ x− = , , 3 6 3 6 3 ␲ 2␲ 3␲ 5␲ ∴ x= , , , 2 3 2 3 ␲ 2␲ 3␲ 5␲ ,0 , ,0 , ,0 , ,0 ∴ axis intercepts 2 3 2 3

c

y 0

π 2

2π 3

3π 5π 2 3 π

y = 2 – √3 x 2π

– √3 y = 2 sin 2 x – (0, –2√3)

π 3

– √3

y = –2 – √3



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Exercise 10J Example

15

1 Sketch the graphs of each of the following for x ∈ [0, 2␲]. List the x axis intercepts of each graph for this interval. √ √ a y = 2 sin x + 1 b y = 2 sin 2x − 3 c y = 2 cos x + 1 √ ␲ +1 d y = 2 sin 2x − 2 e y = 2 sin x − 4 2 Sketch the graphs of each of the following for x ∈ [−2␲, 2␲]. ␲ a y = 2 sin 3x − 2 b y = 2 cos 3 x − 4 c y = 2 sin 2x − 3 d y = 2 cos 2x + 1 ␲ ␲ −1 e y = 2 cos 2 x − +1 f y = 2 sin 2 x + 3 6 3 Sketch the graphs of each of the following for x ∈ [−␲, ␲]. ␲ ␲ a y = 2 sin 2 x + +1 +1 b y = −2 sin 2 x + 3 6 √ ␲ + 3 c y = 2 cos 2 x + 4

10.11 Rotation of a point about the origin Consider the points with coordinates (0, 1) and (1, 0) under a rotation of ␪ c in an anticlockwise direction around the origin. y (0, 1)

π π cos + θ , sin +θ 2 2 θc

From the diagram

0

(1, 0) → (cos ␪, sin ␪) ␲ ␲ + ␪ , sin +␪ and (0, 1) → cos 2 2 ␲ ␲ cos + ␪ , sin + ␪ = (− sin ␪, cos ␪) 2 2

∴ The matrix defining the transformation =

cos ␪ − sin ␪ sin ␪ cos ␪


(cosθ, sin θ)

θc (1, 0)

x


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Hence if

281

(x, y) → (x , y )

x cos ␪ − sin ␪ x = sin ␪ cos ␪ y y

x cos ␪ − y sin ␪ = x sin ␪ + y cos ␪

i.e., x = x cos ␪ − y sin ␪ and y = x sin ␪ + y cos ␪ Example 16 ␲ about O in an anticlockwise direction, and hence find 3 the coordinates of the image of the point (1, 0) under this rotation. ␲ b Find the matrix, B, for rotation of about O in a clockwise direction, and hence find the 3 coordinates of the image of the point (1, 0) under this rotation. a Find the matrix, A, for rotation of

Solution a

⎡1 √ ⎤ ␲⎤ ␲ 3 − sin cos 2√ − 2 ⎥ 3 3⎥ ⎢ ⎢ ⎥ A=⎣ ␲ ␲ ⎦=⎢ ⎣ 3 1 ⎦ sin cos 3 3 2 2 √ ⎡ ⎤ ⎡1 ⎤ ⎡ ⎤ 1 3 ⎡ ⎤ 1 x 2√ − 2 ⎥ ⎢ ⎥ ⎢ 2√ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ =⎢ ⎣ ⎦ = ⎢ ⎣ 3 1 ⎦⎣0⎦ ⎣ 3 ⎦ y ⎡

2 b

2

2

⎡ √ ⎤ 1 −␲ −␲ ⎤ 3 cos − sin ⎥ 2√ ⎥ ⎢ ⎢ 3 3 2 ⎢ ⎥ = B=⎣ ⎦ −␲ −␲ ⎦ ⎣ 1 3 sin cos − 3 3 2 2 ⎡ ⎤ ⎡ √ ⎤ ⎡ ⎤ ⎡ ⎤ 1 1 3 x ⎥ 1 ⎢ ⎥ ⎢ 2 ⎥⎣ ⎦ = ⎢ 2 √ ⎥ ⎣ ⎦=⎢2√ ⎣− 3⎦ ⎣ 3 1 ⎦ 0 y − 2 2 2 ⎡

Example 17 ⎤ 1 1 ⎢ − √2 − √2 ⎥ ⎥ Describe the rotation represented by the matrix ⎢ ⎣ 1 1 ⎦ −√ √ 2 2 ⎡



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Solution The matrix defining a rotation of ␪ c in an anticlockwise direction around the origin is given by

cos ␪ − sin ␪ sin ␪ cos ␪ Let

cos ␪ − sin ␪ sin ␪ cos ␪

⎤ 1 1 − − √ √ ⎢ 2 2⎥ ⎥ =⎢ ⎣ 1 1 ⎦ −√ √ 2 2 ⎡

1 1 Therefore cos ␪ = − √ and sin ␪ = √ , and the smallest positive solution for ␪ is 2 2 3␲ 4 3␲ The matrix represents a rotation of radians about the origin in an anticlockwise 4 direction. 5␲ radians about the origin in a Note: This can also be described as a rotation of 4 clockwise direction.

Exercise 10K Example

16

1 For each of the following, find the matrix associated with the rotation and hence the image of the given point under that transformation. ␲ a rotation of about O in an anticlockwise direction, point (1, 0) 6 ␲ b rotation of about O in a clockwise direction, point (0, 1) 4 ␲ c rotation of about O in an anticlockwise direction, point (−1, 0) 2 2␲ d rotation of about O in a clockwise direction, point (0, −1) 3 1 5␲ 1 about O in an anticlockwise direction, point √ , √ e rotation of 4 2 2 √ 3 1 5␲ about O in a clockwise direction, point , f rotation of 6 2 2

Example

17

2 Describe the rotation represented by each of the following matrices. √ ⎤ ⎡ ⎤ ⎤ ⎡ 1 ⎡√ 3 4 3 3 −1 − − ⎢ ⎥ ⎢ 2 ⎢ 5⎥ 2 ⎥ c ⎣5 b ⎣ √ a ⎣ 2 √2 ⎦ ⎦ 3 4⎦ 3 1 3 1 − − 5 5 2 2 2 2 ␲ 3 a Find the matrix R ␲ of rotation of about O in an anticlockwise direction. 4 4 b Find the image of (2, 1) under this rotation.



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c Let R␪ denote the matrix defining a rotation of ␪ c in an anticlockwise direction about O. Find: i R 3␲ ii R ␲ iii R 5␲ iv R −␲ 4

2

6

4

4 ␲ d If 0 < ␪ < and cos ␪ = , find sin ␪ and write down the matrix R␪ . 2 5

10.12

Applications Example 18 It is suggested that the height h(t) metres of the tide above mean sea level on 1 January at ␲ t where t is the number of hours Warnung is given approximately by the rule h(t) = 4 sin 6 after midnight. a Draw the graph of y = h(t) for 0 ≤ t ≤ 24. b When was high tide ? c What was the height of the high tide ? d What was the height of the tide at 8 am ? e A boat can only cross the harbour bar when the tide is at least 1 m above mean sea level. When could the boat cross the harbour bar on 1 January? Solution y

a 4

0

period = 2␲ ÷

y = h(t)

6

12

18

24

␲ = 12 6

t

–4

b High tide occurs when h(t) = 4 ␲ t =4 4 sin 6 ␲ t =1 implies sin 6 ␲ 5␲ ␲ t= , ∴ 6 2 2 ∴ t = 3, 15 i.e., high tide occurs at 3.00 (3 am) and 15.00 (3 pm) c The high tide has height 4 m above the mean√ height. √ 8␲ 4␲ − 3 d h(8) = 4 sin = 4 sin =4× = −2 3 3 2 √6 The water is 2 3 m below the mean height at 8 am.



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␲ e First consider 4 sin t =1 ␲6 1 t = Thus sin 6 4 ␲ t = 0.2527, 2.889, 6.5359, 9.172 6 ∴ t = 0.48, 5.52, 12.48, 17.52

∴

i.e., the water is at height 1 m at 00:29, 05:31, 12:29, 17:31. Thus the boat can pass across the harbour bar between 00:29 and 05:31 and between 12:29 and 17:31.

Exercise 10L Example

18

1 The number of hours of daylightat a point on the Antarctic Circle is given approximately 1 1 where t is the number of months which have elapsed by d = 12 + 12 cos ␲ t + 6 3 since 1 January. a Find d i on 21 June (t ≈ 5.7) ii on 21 March (t ≈ 2.7). b When will there be five hours of daylight ? 2 The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a ␲t , 0 ≤ t ≤ 24. particular day is given by D(t) = 10 + 3 sin 6 a Sketch the graph of D(t) for 0 ≤ t ≤ 24. b Find the values of t for which D(t) ≥ 8.5. c Boats which need a depth of w metres are permitted to enter the harbour only if the depth of the water at the entrance is at least w metres for a continuous period of one hour. Find, correct to one decimal place, the largest value of w which satisfies this condition. 3 The depth of water at the entrance to a harbour t hours after high tide is D metres, where D = p + q cos (r t)◦ for suitable constants p, q, r. At high tide the depth is 7 m; at low tide, six hours later, the depth is 3 m. a Show that r = 30 and find the values of p and q. b Sketch the graph of D against t for 0 ≤ t ≤ 12. c Find how soon after low tide a ship which requires a depth of at least 4 m of water will be able to enter the harbour. 4 A particle moves on a straight line, Ox, and its distance x metres from O at time t (seconds) is given by x = 3 + 2 sin 3t. a c d e

Find its greatest distance from O. b Find its least distance from O. Find the times at which it is 5 m from O for 0 ≤ t ≤ 5. Find the times at which it is 3 m from O for 0 ≤ t ≤ 3. Describe the motion of the particle.



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5 The temperatureA◦ Cinside a house at t hours after 4 am is given by ␲t A = 21 − 3 cos for 0 ≤ t ≤ 24, and the temperature B ◦ C outside the house at the 12 ␲t same time is given by B = 22 − 5 cos for 0 ≤ t ≤ 24. 12 a Find the temperature inside the house at 8 am. b Write down an expression for D = A − B, the difference between the inside and outside temperatures. c Sketch the graph of D versus t for 0 ≤ t ≤ 24. d Determine when the inside temperature is less than the outside temperature. 6 The high-water mark on a beach wall is a sinusoidal function, i.e., ␲ it has␲arule of the form t− where t is the a sin (nt + ε) + b. In this case the function is d(t) = 6 + 4 cos 6 3 number of hours after midnight and d is the depth of the water in metres. a What is the earliest time of day at which the water is at its highest ? b When is the water 2 m up the wall ? 7 The graph shows the distance d(t) of the top of the hour hand of a large clock from the ceiling at time t hours.

d (m) 5

2

0

12

24

t (h)

a d(t) is the rule of a sinusoidal function. Find i the amplitude ii the period iii the rule for d(t) iv the length of the hour hand. b At what times is the distance less than 3.5 m from the ceiling ? 8 In a tidal river the time between high tide and low tide is 8 hours. The average depth of water in a point on the river is 4 m; at high tide the depth is 5 m. a Sketch a graph of the depth of the water at the point over time if the relationship between time and depth is sinusoidal and there is a high tide at noon. b If a boat requires a depth of 4 m of water in order to sail, how many hours before noon can it enter the point and by what time must it leave to avoid being stranded? c If a boat requires a depth of 3.5 m of water in order to sail, at what time before noon can it enter the point and by what time must it leave to avoid being stranded ?



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9 The population, N, of a particular species of ant varies with time. The population at time t weeks after 1 January 2006 is given by N = 3000 sin

(␲ (t − 10)) + 4000 26

(␲ (t − 10)) + 4000 state 26 i the period ii the amplitude iii the range. i State the values of N(0) and N(100). ii Sketch the graph of y = N (t) for t ∈ [0, 100]. Find the values of t(t ∈ [0, 100]) for which the population is i 7000 ii 1000 Find {t : N (t) > 5500} for t ∈ [0, 100], i.e., find the intervals of time during the first hundred days for which the population of ants is greater than 5500. A second population M(t) of ants also varies with time. The population has the following properties. r minimum population is 10 000 at t = 20

a For N (t) = 3000 sin b c d e

r no maximum or minimum value between t = 10 and t = 20 r maximum population is 40 000 at t = 10

(␲ (t − c)) + d where a, b, c and d are positive constants b Find a set of possible values of a, b, c and d. r M(t) = a sin



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Review

Chapter summary Definition of a radian One radian (written 1c ) is the angle formed at the centre of the unit circle by an arc of length 1 unit. 1c =

180◦ ␲

1◦ =

y 1

␲c 180

1 unit

1 1c –1

x

1

0

–1

Sine and cosine x coordinate of P(Q) in unit circle,

y

x = cosine ␪, ␪ ∈ R

1

y coordinate of P(␪) in unit circle, –1

y = sine ␪, ␪ ∈ R Abbreviated to

P(θ)

1 θ 0 x

y x

1

–1

x = cos ␪, y = sin ␪ y

Tangent If the tangent to the unit circle at A is drawn then the y coordinate of B is called tangent ␪ (abbreviated to tan ␪). Also by similar triangles,

B 1 tan θ 1

sin ␪ tan ␪ = cos ␪

θ –1

0

cosθ

sin θ x A

–1

Circular functions and trigonometric ratios y

1 θ 0 x

opposite side

y x

opp y = =y hyp 1 x adj = =x cos ␪ = hyp 1 opp y sin ␪ tan ␪ = = = adj x cos ␪

sin ␪ =

hypotenuse θ adjacent side



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Symmetry properties of circular functions

y

Quadrant 2 (sin is positive) sin (π – θ) = b = sin θ

Quadrant 1 (all function are positive) e.g. sin θ = b

1 b θ

–1

1

–b sin (π + θ) = –b = –sin θ –1 Quadrant 3 (tan is positive)

x

sin (2π – θ) = –b = – sin θ Quadrant 4 (cos is positive) y

Further symmetry properties Negative angles

1

cos (−␪) = cos ␪ sin (−␪) = − sin ␪ sin ␪ = − tan ␪ tan (−␪) = − cos ␪

θ –θ

–1

1

x

–1

Exact values of circular functions ␪ 0

sin ␪ 0

␲ 6

1 2

␲ 4

1 √ 2 √ 3 2

1 √ 2 1 2

√

1

0

undefined

␲ 3 ␲ 2

cos ␪ 1 √ 3 2

tan ␪ 0 1 √ 3 1

3



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Graphs of sine and cosine functions

y

y 1

π

y = cos θ

1

y = sin θ

2π

–1

θ

π 2

π

θ

3π 2π 2

–1 amplitude = 1 period = 2π

amplitude = 1 period = 2π

Solutions of circular function equations of the type sin x ◦ = a and cos x ◦ = a e.g., If cos x ◦ = −0.7, find the two values of x in the range 0 ≤ x ≤ 360. If cos x ◦ = 0.7, then x = 45.6 Since cos is negative in the 2nd and 3rd quadrants

∴ x = 180 − 45.6 = 134.4 and x = 180 + 45.6 = 225.6 Sketch graphs of circular functions of the type y = a sin n(t + ε) + b and y = a cos n(t + ε) + b ␲ −1 e.g., y = 2 cos 3 t + 3 amplitude, a = 2 2␲ 2␲ period = = n 3

y 1 π 6

–π –π 0 3 6 –1

The graph is the same shape as y = 2 cos 3t but is translated ␲ i units in the negative direction of the x axis and 3 ii 1 unit in the negative direction of the y axis.

π 3

t

–2 –3

Multiple-choice questions 1 In the triangle ABC, cos x is equal to b a B √ A √ 2 2 2 a + b2 a +b √ b a 2 + b2 D E a a

C

a b

A

x C

b

a B



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2 The period of the graph of y = 2 sin (3x − ␲) + 4 is 2␲ A B 2 C 3 D ␲ 3 3 The amplitude of the graph of y = −5 cos 5x + 3 is A −5 B −2 C 2 D 5

E 2␲ E 8

4 The number of solutions of 5 sin (2x − ␲) + 2 = 0 in the interval [0, 2␲] is A 1 B 2 C 3 D 4 E 8 3␲ radians expressed in degrees (correct to two decimal places) is 5 An angle of 11 A 49 B 154.22 C 49.09 D 0.01 E 0.00 √ 5␲ 23␲ 6 The solutions of 2 sin 3x + 2 = 0 in the interval , are 12 12 5␲ 7␲ 13␲ 5␲ 7␲ 23␲ , , , , , B 1.83, 3.40, 3.93, 5.50 A 12 12 12 4 4 4 4␲ 5␲ 10␲ 11␲ 16␲ 17␲ 7␲ 13␲ 5␲ 7␲ , , , , , C D , , , E none of the above 9 9 9 9 9 9 12 12 4 4 −13␲ 7 cos is equal to 6 √ − 3 13␲ −7␲ 2␲ −1 B A cos C cos E sin D 2 6 6 3 2 ◦ 8 tan (180 − ␪) is equal to sin (90 − ␪)◦ cos (180 − ␪)◦ sin (90 + ␪)◦ C B A cos (90 + ␪)◦ sin (180 − ␪)◦ cos (90 − ␪)◦ ◦ cos (90 − ␪) cos (90 + ␪)◦ D E . ◦ sin (90 + ␪) sin (90 − ␪)◦ 9 The period of the graph of f (x) = 4 sin (3␲x) − 3 cos (2␲x) is 2 A 1 B 2 C 3 D 4 E 3 √ ⎤ ⎡ 1 − 3 ⎢2 2 ⎥ ⎥ ⎢ 10 The transformation given by the matrix ⎢ √ ⎥is 3 1 ⎦ ⎣ 2 2 A a rotation of 30◦ about the origin in an anticlockwise direction B a rotation of 60◦ about the origin in an anticlockwise direction 1 C a dilation by factor from the x axis 2 D a reflection in the line y = x, E a rotation of 60◦ about the origin in a clockwise direction



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1 Change each of the following to radian measure in terms of ␲. b 810◦ c 1080◦ d 1035◦ a 330◦ ◦ ◦ ◦ f 405 g 390 h 420 i 80◦

e 135◦

2 Change each of the following to degree measure. 15␲ c 3␲ c 11␲ c 7␲ c 5␲ c e d c b a 2 12 4 4 6 ␲c 23␲ c −3␲ c 11␲ c g − i − f h − 4 4 4 4 3 Give exact values of each of the following. −7␲ 11␲ 11␲ −7␲ d cos c sin a sin b cos 6 6 4 4 −23 23␲ 13␲ −17 ␲ g cos f sin e cos ␲ h sin 3 6 6 4 4 State the amplitude and period of each of the following. ␪ 1 a 2 sin b −3 sin 4␪ c sin 3␪ 2 2 x 2 2x d −3 cos 2x e −4 sin f sin 3 3 3 5 Sketch the graphs of each of the following (showing one cycle). x c y = −2 sin 3x a y = 2 sin 2(2x) b y = −3 cos 3 x ␲ 2␲ d y = 2 sin e y = sin x − f y = sin x + 3 4 3 5␲ ␲ g y = 2 cos x − h y = −3 cos x + 6 6 6 Solve each of the following equations for R. √ √ 3 3 , ␪ ∈ [−␲, ␲] b sin 2␪ = − , ␪ ∈ [−␲, ␲] a sin ␪ = − 2 2 1 ␲ ␲ = − , ␪ ∈ [0, 2␲] c sin ␪ − d sin ␪ + = −1, ␪ ∈ [0, 2␲] 3 2 3 ␲ 1 e sin − ␪ = − , ␪ ∈ [0, 2␲] 3 2 7 Sketch the graphs of each of the following for x ∈ [−␲, 2␲]. ␲ a f (x) = 2 sin 2x + 1 b f (x) = 1 − 2 cos x c f (x) = 3 cos x + 3 ␲ d f (x) = 2 − cos x + e f (x) = 1 − 2 sin 3x 3


Review


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Extended-response questions 1 The depth, D metres, of sea water in a bay, t hours after midnight on a particular day, may be represented by the function with rule 2␲t , where a, b and k are real numbers. D(t) = a + b cos k The water is at a maximum depth of 15.4 m at midnight and noon, and is at a minimum depth of 11.4 m at 6.00 and 18.00 hours. a Find the value of i a ii b iii k b Find the times when the depth of the water is 13.4 m. c Find the values of t for which the depth of the bay is less than 14.4 m. 2 The temperature (◦ ) in a small town in the mountains over a day is modelled by the function with rule ␲t + 6 , where t is the time in hours aftermidnight, 0 ≤ t ≤ 24. T = 15 − 8 cos 12 a b c d

What is the temperature at midnight, correct to two significant figures? What are the maximum and minimum temperatures reached? At what times of the day, to the nearest minute, are temperatures warmer than 20◦ ? Sketch the graph for the temperatures over a day.

3 A particle oscillates back and forth, in a straight line, between points A and B about a point O. Its position, x(t) metres, relative to O at time t seconds is given by the rule x(t) = 3 sin (2␲t − a). The position of the particle when t = 1 is x = −1.5.

A

O

B

␲ a If a ∈ [0, ], find the value of a. 2 b Sketch the graph of x(t) against t for t ∈ [0, 2]. Label maximum and minimum points, axes intercepts and endpoints with their coordinates. c How far from O is point A? d At what time does the particle first pass through A? e How long is it before the particle returns to A? f How long does it take for the particle to go from A to O? g How far does the particle travel in i the first 2 s of its motion ii the first 2.5 s of its motion?



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h(t)

Review

4 The depth of water, h(t) m, at a particular jetty in a harbour at time t hours after midnight is ␲t given by the rule h(t) = p + q sin . The graph of h(t) against t for t ∈ [0, 24] is as 6 shown.

10.2 6

(24, 6)

1.8 O

t

The maximum depth is 10.2 m and the minimum depth is 1.8 m. a Find the values of p and q. b State the times at which the depth of water is a maximum for the interval of time [0, 24]. c What is the average depth of the water in the time interval [0, 24]? d At what times in the time interval [0, 24] is the depth of the water 3.9 m? e For how long in the 24-hour period from midnight is the water more than 8.1 m in depth? 5 For the function f : [0, 2␲] → R with rule f (x) = 2 sin(3x) + 1 a Find the values of k such that the equation f (x) = k has i six solutions for x ∈ [0, 2␲] ii three solutions for x ∈ [0, 2␲] iii no solutions for x ∈ [0, 2␲] b Find a sequence of transformations which takes the graph of y = f (x) to the graph of y = sin x. c Find the values of h ∈ [0, 2␲] such that ␲ ,3 i there is a maximum for the graph of y = f (x + h) at the point ␲3 , −1 ii there is a minimum for the graph of y = f (x + h) at the point 3 6 a Find a sequence of transformations which takes the graph of y = cos x to the graph of y = sin x. b Find a sequence of transformations which takes the graph of y = 2 cos x to the graph of 1 y = − sin 2x. 2 2 c i Find the rule for the image of the graph of f (x) = sin x under a dilation of factor ␲ from the y axis, followed by reflection in the line y = 2. ii Find the range and period of the new function.



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7 Two pistons A and B move backwards and forwards in a cylinder as shown. A

O

B

x cm y cm

The distance x centimetres of the right hand end of piston A from the point O at time t seconds is modelled by the rule x = 4 sin (3t) + 4 and the distance y centimetres of the left hand end of piston B from the point O at time t seconds is modelled by the rule ␲ + 10 y = 2 sin 2t − 6 The pistons are set in motion at time t = 0. a State the value of x and the value of y when t = 0 b i State the amplitude of the motion of piston A. ii State the amplitude of the motion of piston B. c i State the maximum and minimum values of x. ii State the maximum and minimum values of y. d i State the period of the motion of piston A. ii State the period of the motion of piston B. e Find the time(s) in the first cycle of A that its distance from O is a maximum. f Find the next four values of t for which x takes its maximum value. g Find the values of t, 0 ≤ t ≤ 4␲, for which y attains its minimum value. h On the one set of axes draw the graphs of x = 4 sin (3t) + 4 and ␲ + 10 over the interval [0, ␲]. y = 2 sin 2t − 6 i State the time when the pistons first touch each other. j How many seconds are there between the first and second times the pistons touch? 8 The pistons A and B (from 7) are adjusted so that the distance x cm of the right hand end of piston A from point O at time t seconds is modelled by the formula x = a sin (nt) + b and the distance y centimetres of the left hand end of piston B from the point O at time t seconds is modelled by the formula y = c sin (mt) + d The pistons meet every second at a point 8 cm from O. The right hand end of piston A cannot go to the left of the point O. a Find one possible set of values of a, b, n and c, m, d and explain your solution. b Using the set of values found in a, sketch the graphs of x against t and y against t on the one set of axes.



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9 The population, N, of a particular species of ant varies with the seasons. The population is ␲(t − 1) modelled by the equation N = 3000 sin + 4000, where t is the number of 6 months after 1 January in a given year. The population, M, of a second species of ant also varies with time. Its population is ␲ (t − 3.5) modelled by the equation M = 3000 sin + 5500, where t is again the 5 number of months after 1 January in a given year. Use a graphics calculator to sketch the graphs of both equations over a period of one year on the same axes and find a the maximum and minimum populations of both species and the months in which those maxima and minima occur b during which month of the year the populations of both species are equal and the population of each species at that time c by formulating a third equation, when the combined population of species N and M is at a maximum and what that maximum is d by formulating a fourth equation, when the difference between the two populations is a maximum. 10 Passengers on a ferris wheel ride access their seats from a platform 5 m above the ground. As each seat is filled the ferris wheel moves around so that the next seat can be filled. Once all seats are filled the ride begins and lasts for 6 minutes. The height h m of Isobel’s seat above the ground t seconds after the ride has begun is given by the equation h = 15 sin (10t − 45)◦ + 16.5. a Use a graphics calculator to sketch the graph of the equation for the first 2 minutes of the ride. b How far above the ground is Isobel’s seat at the commencement of the ride? c After how many seconds does Isobel’s seat pass the access platform? d How many times will her seat pass the access platform in the first 2 minutes? e How many times will her seat pass the access platform during the entire ride? Due to a malfunction the ride stops abruptly 1 minute and 40 seconds into the ride. f How far above the ground is Isobel stranded? g If Isobel’s brother Hamish had a seat 1.5 m above the ground at the commencement of the ride, how far above the ground was Hamish stranded? 11 The transformation of rotation about the origin in an anticlockwise direction by ␪ ◦ is defined by the matrix

cos ␪ − sin ␪ R␪ = sin ␪ cos ␪ a Find the matrices i R␪ ·R␺ and hence deduce identities for cos (␪ + ␺ ) and sin (␪ + ␺ ) ii R␪ ·R␪ = R2␪ and hence deduce identities for cos 2␪ and sin 2␪ iii R2␪ ·R␪ = R3␪ and hence deduce identities for cos 3␪ and sin 3␪



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b Find the matrix R−1 ␪ . What does this matrix represent? c Rotation about another point with coordinates (a, b) can be defined by the following sequence of transformations:

−a r translation defined by the vector −b r rotation defined by the matrix R ␪

a r translation defined by the vector b i Find the image of the point (x, y) following rotation about the point (a, b). ii Find the image of the point (x, y) following rotation about the point (1, 1).

x x d i If R ␪ × = , find x and y in terms of x and y. y y ii Find the image of the curve of y = x 2 under a rotation of 45◦ around the origin in an anticlockwise direction. iii Find the image of the line y = x under a rotation of ␪ ◦ around the origin in an anticlockwise direction. y e i Find the image of the points with coordinates (1, 0) and (0, 1) after reflection in the line with equation y = (tan ␪)x. y = (tan θ)x (0, 1) ii Given that this transformation may be represented by a 2 × 2 matrix a b , find the values of a, b, c and d θ° 0 c d x (1, 0) in terms of ␪. iii Use a matrix technique to show that a rotation of ␪ degrees about the origin can be achieved through the composition of two reflections. f Find the image of the curve with equation y = x 2 under a reflection in the line y = x.



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C H A P T E R

11 Circular functions II Objectives To further explore the symmetry properties of circular functions To further understand and sketch the graphs of circular functions To solve circular function equations To evaluate simple trigonometric expressions using trigonometric identities To prove simple trigonometric identities To apply addition theorems for circular functions To apply double angle formulas for circular functions To simplify expressions of the form a cos x + b sin x To sketch graphs of functions of the form f (x) = a cos x + b sin x To solve equations of the form a cos x + b sin x = c

11.1

Further symmetry properties

y

Complementary relationships ␲

−␪ =a 2 and since a = cos ␪ ␲ sin − ␪ = cos ␪ 2 sin

Similarly cos

␲

b a

π –θ 2

θ

−␪ =b

P(θ) b

θ

2 and since b = sin ␪ ␲ cos − ␪ = sin ␪ 2 ␲ sin + ␪ = a = cos ␪ 2 ␲ + ␪ = −b = −sin ␪ cos 2

π –θ 2

P

x

a

y

π P +θ 2 b

a

π +θ 2

P(θ)

θ θ

b a

x



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Example 1 If sin ␪ = 0.3 and cos ␣ = 0.8, find the values of ␲ ␲ a sin −␣ b cos +␪ 2 2 Solution ␲ a sin − ␣ = cos ␣ 2 = 0.8

b cos

c sin(−␪)

+ ␪ = −sin ␪ 2 = −0.3

␲

c sin (−␪) = − sin ␪ = −0.3


1

1 If sin x = 0.3, cos ␣ = 0.6 and tan ␪ = 0.7, find the values of ␲ +␣ c tan(−␪) a cos(−␣) b sin 2 ␲ ␲ g cos +x −␪ e sin(−x) f tan 2 2 3␲ 3␲ −x j cos i sin +␣ 2 2

−x ␲2 h sin −␣ 2

d cos

␲

11.2 Addition of ordinates Example 2 Using the same scale and axes, sketch the graphs of y1 = 2 sin x and y2 = 3 cos 2x for 0 ≤ x ≤ 2␲. Use addition of ordinates to sketch the graph of y = 2 sin x + 3 cos 2x. Solution The graphs of y1 = 2 sin x and y2 = 3 cos 2x are shown below. To obtain points on the graph of y = 2 sin x + 3 cos 2x the process of addition of ordinates is used. Let y = y1 + y2 when y1 = 2 sin x and y2 = 3 cos 2x e.g., at x = 0, y ␲ x= , y 4 ␲ x= , y 2 x = ␲, y 3␲ ,y x= 2 and so on.

y

=0+3=3 √ 2 2 = √ +0= √ = 2 2 2 = 2 − 3 = −1 =0+3=3 = −2 − 3 = −5

3 2 1 0 –1 –2 –3 –4 –5


y1 = 2sinx

π 2

π

x 3π 2

2π

y2 = 3cos2x y = 2sinx + 3cos 2x


Chapter 11 — Circular functions II

299

Using the TI-Nspire Check that the calculator is in Radian mode. Open a Graphs application (c>New Document>Add Graphs) and enter the functions f 1 (x) = 2 sin (x) f 2 (x) = 3 cos (2x) f 3 (x) = f 1 (x) + f 2 (x) The graph of f 3(x) is shown with medium line width. Add a table of values (b>View>Show Table or press /+T) and split the screen >Page Layout>Select using Layout>Layout 3 (or /+c>Page Layout>Select Layout>Layout 3 on the Clickpad) to see that the values of f 1(x) and f 2(x) add to give f 3(x). Use the down arrow (¤) to see more x-values. Alternatively show a table of values on a full screen by adding a Lists & Spreadsheet(/+I >Add Lists & Spreadsheet) application and pressing /+T.

Note:

Using the Casio ClassPad Check that the calculator is Radian mode. Choose Graphs & Tab. from the Menu. Enter y1 = 2 sin(x) y2 = 3 cos(2x) y3 = 2 sin(x) + 3 cos(2x). Tap the graph icon to obtain the graph. Tap the table icon to obtain a table and then the table input icon to obtain the desired table.

Graph icon Table icon

Table input



300


The table input window has been completed as shown.

Another way to achieve this addition of functions is to Define f (x) = 2 sin(x) and g(x) = 3 cos(2x). Then enter the functions as shown. Note also that the graph of ( f + g)(x) has been drawn with a thick line. The menu shown to choose this as shown. It can be accessed by taping on the line segment to the right of y3.


2

1 Use addition of ordinates to sketch the graphs of a y = 2 sin ␪ + cos ␪ 1 c y = sin 2␪ − cos ␪ 2 e y = 4 sin ␪ − 2 cos ␪

11.3

b y = 3 cos 2␪ + 2 sin 2␪ d y = 3 sin ␪ + cos 2␪

Sketch graphs of the tangent function A table of values for y = tan x is given below. Use a calculator to check these values. x −␲ − y

0

3␲ 4 1

−

␲ 2

−

␲ ␲ 0 4 4

␲ 2

undefined −1 0 1 undefined

3␲ 5␲ ␲ 4 4 −1

0

1

3␲ 2 undefined

7␲ 9␲ 2␲ 4 4 −1


0

1

5␲ 2 undefined

11␲ 3␲ 4 −1

0


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The graph of y = tan x is given below.

y

–π 2 2 1

π 2 π

–π –4

–3

–2

0

–1

3π 2

1

2

3

5π 2 2π

4

5

6

3π 7

8

9

x 10

–1 –2

Note:

␲ ␲ 3␲ 5␲ x =− , , and are asymptotes. 2 2 2 2

Observations from the graph 1 The graph repeats itself every ␲ units, i.e., the period of tan is ␲. 2 The range of tan is R.

Exercise 11C 1 Sketch the graph of each of the following, showing one complete cycle. ␲ a y = tan 2x b y = 2 tan 3x c y = 2 tan x + 4␲ ␲ +1 d y = 3 tan x + 1 e y = 2 tan x + f y = 3 tan 2 x − −2 2 4

11.4

General solution of circular function equations The solution of circular function equations has been discussed in Section 10.9 for functions over a restricted domain. In this section, we consider the general solutions of such equations over the maximal domain for each function. If a circular function equation has one or more solutions in one ‘cycle’, then it will have corresponding solutions in each ‘cycle’ of its domain, i.e., there will be an infinite number of solutions. For example, if cos x = a, then the solution in the interval [0, ␲] is given by: x = cos−1 (a) By the symmetry properties of the cosine function, other solutions are given by: −cos−1 (a), ±2␲ + cos−1 (a), ±2␲ − cos−1 (a), ±4␲ + cos−1 (a), ±4␲ − cos−1 (a), . . . and so on. In general, if cos (x) = a, x = 2n␲ ± cos−1 (a),

where n ∈ Z and a ∈ [−1, 1]



302


Similarly, if tan (x) = a, x = n␲ + tan−1 (a),

where n ∈ Z and a ∈ R

If sin (x) = a, x = 2n␲ + sin−1 (a) or x = (2n + 1)␲ − sin−1 (a), Note:

where n ∈ Z and a ∈ [−1, 1]

An alternative and more concise way to express the general solution of sin (x) = a is: x = n␲ + (−1)n sin−1 (a), where n ∈ Z and a ∈ [−1, 1]

Example 3 Find the general solution to each of the following equations. √ 3 tan (3x) = 1 a cos (x) = 0.5 b

c 2 sin (x) =

√

2

Solution 1 b tan (3x) = √ 3

a x = 2n␲ ± cos−1 (0.5) ␲ = 2n␲ ± 3 (6n ± 1)␲ ,n ∈ Z = 3

c

1 sin (x) = √ 2 −1

3x = n␲ + tan

−1

1 √ 3

␲ 6 (6n + 1)␲ = 6 (6n + 1)␲ ,n ∈ Z x= 18 = n␲ +

1 √ 2

x = 2n␲ + sin ␲ = 2n␲ + 4 (8n + 1)␲ ,n ∈ Z = 4

or

−1

x = (2n + 1)␲ − sin ␲ = (2n + 1)␲ − 4 (8n + 3)␲ = ,n ∈ Z 4

1 √ 2

Using the TI-Nspire Check that the calculator is in Radian mode. a Use solve( ) from the Algebra menu and complete as shown. Note the use of 12 rather than 0.5 to ensure that the answer is exact.




303

b Complete as shown.

c Complete as shown.

Using the Casio ClassPad a Enter and highlight the equation cos (x) = 0.5, tap Interactive, Equation/inequation, solve and ensure the variable is set to x. b Enter and highlight the equation √ 3 tan (3x) = 1, tap Interactive, Equation/inequation, solve and ensure the variable is set to x. √ c Enter and highlight the equation 2 sin (x) = 1, tap Interactive, Equation/inequation, solve and ensure the variable is set to x.

Example 4 Find the first three positive solutions to each of the following equations. √ √ 3 tan (3x) = 1 c 2 sin (x) = 2 a cos (x) = 0.5 b



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Solution

(6n ± 1)␲ ,n ∈ Z 3 5␲ 7␲ ␲ or x = When n = 0, x = ± , and when n = 1, x = 3 3 3 ␲ 5␲ 7␲ , The first three positive solutions of cos (x) = 0.5 are x = , 3 3 3 (6n + 1)␲ b The general solution (from Example 3) is given by x = ,n ∈ Z 18 ␲ 7␲ 13␲ When n = 0, x = , and when n = 1, x = , and when n = 2, x = 18 18 18 √ ␲ 7␲ 13␲ , , The first three positive solutions of 3 tan (3x) = 1 are x = 18 18 18 (8n + 1)␲ or c The general solution (from Example 3) is given by x = 4 (8n + 3)␲ x= ,n ∈ Z 4 ␲ 3␲ 9␲ 11␲ When n = 0, x = or , and when n = 1, x = or x = 4 4 4 4 √ ␲ 3␲ 9␲ , The first three positive solutions of 2 sin (x) = 2 are x = , 4 4 4 a The general solution (from Example 3) is given by x =

Exercise 11D Example

Example

3

4

1 Find the general solution to each of the following equations. √ a sin (x) = 0.5 b 2 cos (3x) = 3

c

√

3 tan (x) = −3

2 Find the first two positive solutions to each of the following equations. √ √ a sin (x) = 0.5 b 2 cos (3x) = 3 c 3 tan (x) = −3 √ ␲ 3 Find the general solution to 2 cos 2x + = 2, and hence find all the solutions for x in 4 the interval (−2␲, 2␲). ␲ √ − 3x − 1 = 0, and hence find all the solutions for 4 Find the general solution to 3 tan 6 x in the interval [−␲, 0]. √ 5 Find the general solution to 2 sin (4␲x) + 3 = 0, and hence find all the solutions for x in the interval [−1, 1].

11.5 Trigonometric identities

Reciprocal functions The functions sin, cos, and tan can be used to form three other functions called the reciprocal circular functions. 1 1 (sin ␪ = 0) cosec ␪ = (cos ␪ = 0) sec ␪ = sin ␪ cos ␪ cos ␪ (sin ␪ = 0) cot ␪ = sin ␪ 1 1 and tan ␪ = Note: For cos ␪ = 0 and sin ␪ = 0, cot ␪ = tan ␪ cot ␪ ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.



305

Example 5 Find the exact value of each of the following. 5␲ 2␲ b cot a sec 4 3 Solution 1 2␲ cos 3 1 = ␲ cos ␲ − 3 1 = ␲ −cos 3 1 = 1 − 2 = −2

2␲ a sec = 3

c cosec

7␲ 4

5␲ cos 7␲ 5␲ 4 c cosec b cot = 5␲ 4 4 sin 1 4 = ␲ ␲ sin 2␲ − cos ␲ + 4 4 = 1 ␲ ␲ = sin ␲ + 4 −sin 4 −1 −1 = √ ÷√ 1 2 2 = 1 =1 −√ 2 √ =− 2

Example 6 Find the value(s) of x between 0 and 2␲ for which a sec x = −2 b cot x = −1 Solution a

∴ ∴ ∴ i.e.,

b cot x = −1 sec x = −2 1 implies = −2 cos x −1 tan x = −1 cos x = ␲ ␲ 2 ␲ ∴ x = ␲ − or x = 2␲ − ␲ 4 4 x = ␲ − or x = ␲ + 3␲ 7␲ 3 3 i.e. x = or x = 4␲ 2␲ 4 4 or x = x = 3 3

Using the TI-Nspire Check that the calculator is in Radian mode. Use solve( ) from the Algebra menu as shown. a b



306


Using the Casio ClassPad 1 = −2, cos(x) tap Interactive, Equation/inequation, solve and ensure the variable is set to x. 1 = −1, b Enter and highlight the equation tan(x) tap Interactive, Equation/inequation, solve and ensure the variable is set to x. Note: The restriction to the domain can be used as shown in the Appendix. a Enter and highlight the equation

The Pythagorean identity Consider a point, P(␪), on the unit circle. By Pythagoras’ theorem:

∴

OP2 = OM 2 + MP2 1 = (cos ␪)2 + (sin ␪)2

Now (cos ␪)2 and (sin ␪)2 may be written as cos2 ␪ and sin2 ␪. –1 Since this is true for all values of ␪ it is called an identity. In particular this is called the Pythagorean identity.

y 1

P(θ) sin θ

O cos θ M 1

x

–1

cos2 ␪ + sin2 ␪ = 1 Other forms of the identity can be derived. Dividing both sides by cos2 ␪ gives: sin2 ␪ 1 cos2 ␪ + = 2 2 cos ␪ cos ␪ cos2 ␪ 1 + tan2 ␪ = sec2 ␪ Dividing both sides by sin2 ␪ gives: 1 cos2 ␪ sin2 ␪ + = 2 2 sin ␪ sin ␪ sin2 ␪ cot2 ␪ + 1 = cosec2 ␪



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Example 7 a If cosec x =

7 , find cos x. 4

3 ␲ b If sec x = − , find sin x where ≤ x ≤ ␲. 2 2

Solution 4 7 , sin x = 4 7 cos2 x + sin2 x = 1 16 =1 cos2 x + 49 33 cos2 x = 49 √ 33 cos x = ± 7

a Since cosec x = Now so

∴ ∴

2 3 b Since sec x = − , cos x = − 2 3

∴ ∴

cos2 x + sin2 x = 1 4 + sin2 x = 1 9 √ 5 sin x = ± 3

For P(x) in the 2nd quadrant, sin x is positive √ 5 ∴ sin x = 3

Example 8 If sin ␪ =

␲ 3 and < ␪ < ␲, find the value of cos ␪ and tan ␪. 5 2

Solution Since then

∴

∴ ∴

cos2 ␪ + sin2 ␪ = 1 32 cos2 ␪ + 2 = 1 5 9 2 cos ␪ = 1 − 25 16 = 25 ␲ 4 cos ␪ = − since < ␪ < ␲ 5 2 sin ␪ 3 tan ␪ = − as tan ␪ = 4 cos ␪



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Example 9 Prove the identity

1 1 + = 2 cosec2 ␪ 1 − cos ␪ 1 + cos ␪

Solution 1 1 + 1 − cos ␪ 1 + cos ␪ 1 + cos ␪ + 1 − cos ␪ = 1 − cos2 ␪ 2 = 1 − cos2 ␪ 2 = sin2 ␪ = 2 cosec2 ␪ = RHS

LHS =


5

1 Find the exact value of each of the following. 3␲ 5␲ 5␲ a cot b cot c sec 4 4 6 4␲ 13␲ 7␲ e sec f cosec g cot 3 6 3

␲ 2 5␲ h sec 3 d cosec

2 Without using a calculator write down the exact value of each of the following. a cot 135◦ f sec 330◦ Example

6

Example

7

Example

8

b sec 150◦ g cot 315◦

c cosec 90◦ h cosec 300◦

d cot 240◦ i cot 420◦

3 Find the values of x between 0 and 2␲ for which √ √ a cosec x = 2 b cot x = 3 c sec x + 2 = 0 4 If sec ␪ = a cos ␪ 5 If tan ␪ =

−17 ␲ and < ␪ < ␲, find 8 2 b sin ␪

e cosec 225◦

d cosec x = sec x

c tan ␪

−7 3␲ and < ␪ < 2␲, find cos ␪ and sin ␪. 24 2

6 Find the value of sec ␪ if tan ␪ = 0.4 and ␪ is not in the 1st quadrant. 7 If tan ␪ =

3␲ sin ␪ − 2 cos ␪ 4 and ␲ < ␪ < , evaluate . 3 2 cot ␪ − sin ␪

2 and ␪ is in the 4th quadrant, find the simplest expression in surd form for 3 tan ␪ − 3 sin ␪ . cos ␪ − 2 cot ␪

8 If cos ␪ =



Chapter 11 — Circular functions II Example

9

9 Prove each of the following identities for suitable values of ␪ and ␾. a (1 − cos2 ␪)(1 + cot2 ␪) = 1 tan ␪ tan ␪ + cot ␾ c = tan ␾ cot ␪ + tan ␾ 1 + cot2 ␪ = sec ␪ e cot ␪ cosec␪

11.6

309

b cos2 ␪ tan2 ␪ + sin2 ␪ cot2 ␪ = 1 d (sin ␪ + cos ␪)2 + (sin ␪ − cos ␪)2 = 2 f sec ␪ + tan ␪ =

cos ␪ 1 − sin ␪

Addition formulas and double angle formulas Addition formulas y

Consider a unit circle. Let arc length AB = v units

u–v

arc length AC = u units ∴ arc length CB = u − v units

1

(cos u, sin u) C

(cos v, sin v) B

Rotate OCB so that B is coincident with A. The point P has coordinates

A 1

O

–1

y

(cos (u − v), sin (u − v)). Since the triangles CBO and PAO are congruent,

x

(cos(u – v), sin(u – v)) P

1

u–v

CB = PA Applying the coordinate distance formula CB = (cos u − cos v) + (sin u − sin v) = 2 − 2(cos u cos v + sin u sin v) 2 PA = (cos (u − v) − 1)2 + (sin (u − v) − 0)2 = 2 − 2(cos (u − v)) 2

2

2

–1

O

A (1, 0)

x

Equating these 2 − 2(cos u cos v + sin u sin v) = 2 − 2(cos (u − v))

∴

cos (u − v) = cos u cos v + sin u sin v

Using the TI-Nspire Access the tExpand( ) command from b>Algebra>Trigonometry>Expand and complete as shown.



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Replacing v with −v cos (u − (−v)) = cos u cos (−v) + sin u sin (−v) From symmetry properties cos (−␪) = cos ␪ sin (−␪) = − sin ␪ cos (u + v) = cos u cos v − sin u sin v Example 10 Evaluate cos 75◦ . Solution cos 75◦ = cos (45◦ + 30◦ ) ◦ cos 30◦ − sin 45◦ sin 30◦ = cos 45√ 1 1 3 1 −√ · =√ · 2 2 2 √2 3−1 = √ 2 √ √ 2 3−1 2 = √ ×√ 2 √2 2 √ 6− 2 = 4 ␲ − u in cos (u − v) 2 ␲ ␲ ␲ ∴ cos − u − v = cos − u cos v + sin − u sin v 2 2 2 Applying symmetry properties ␲ −␪ sin ␪ = cos ␲2 and cos ␪ = sin −␪ 2 ␲ ∴ cos − (u + v) = sin u cos v + cos u sin v 2 ∴ sin (u + v) = sin u cos v + cos u sin v Replacing u with

Replacing v with −v sin (u − v) = sin u cos (−v) + cos u sin (−v)

∴

sin (u − v) = sin u cos v − cos u sin v

Example 11 Evaluate a sin 75◦

b sin 15◦ .



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Solution a sin 75◦ = sin (30◦ + 45◦ ) = sin 30◦ cos 45◦ + cos 30◦ sin 45◦ √ 1 1 3 1 ·√ = ·√ + 2 2 2 2 √ 1+ 3 = √ 2 2 √ √ 1+ 3 2 = √ ×√ 2 2 2 √ √ 2+ 6 = 4 ◦ b sin 15 = sin (45◦ − 30◦ ) = sin 45◦ cos 30◦ − cos 45◦ sin 30◦ √ 1 1 1 3 −√ · = √ · 2 2 2 2 √ 3−1 = √ 2 2 √ √ 3−1 2 = √ ×√ 2 2 2 √ √ 6− 2 = 4

Addition formula for tangent Also

sin (u + v) cos (u + v) sin u cos v + cos u sin v = cos u cos v − sin u sin v

tan (u + v) =

Divide the numerator and denominator by cos u cos v = 0 tan (u + v) =

tan u + tan v 1 − tan u tan v

Similarly it can be shown that tan (u − v) =

tan u − tan v 1 + tan u tan v

Example 12 If tan u = 4 and tan v =

␲ 3 and u and v are acute angles, show that u − v = . 5 4



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Solution tan u − tan v 1 + tan u tan v 3 4− 5 = 3 1+4× 5 17 = 5 =1 17 5 ␲ ∴ u−v = 4 ␲ Note: tan ␪ is a one-to-one function for 0 < ␪ < 2 tan (u − v) =

Derivation of the addition formulas using rotation about the origin y The use of matrices to describe rotations about the origin has been discussed in Section 10.11. (cos(u + v), sin(u + v)) (cos u, sin u) We can use matrices as an alternative method to derive the addition formulas. Consider, for v example, the point with coordinates u x (cos (u + v), sin (u + v)), which can be regarded O as the image of a point with coordinates (cos u, sin u) under a rotation of v c in an anticlockwise direction around the origin. The matrix that defines a rotation of v radians anticlockwise about the origin is given by cos v −sin v sin v cos v x cos v −sin v x i.e., = y sin v cos v y cos (u + v) cos v −sin v cos u becomes = sin (u + v) sin v cos v sin u

i.e., cos (u + v) = cos v cos u − sin v sin u and sin (u + v) = sin v cos u + cos v sin u

or cos (u + v) = cos u cos v − sin u sin v or sin (u + v) = sin u cos v + cos u sin v

Similarly, consider a point (cos (u − v), sin (u − v)), which can be regarded as the image of a point with (cos u, sin u) coordinates (cos u, sin u) under a rotation of v c in a clockwise direction around the origin.


y

v u O

(cos (u – v), sin (u – v)) x


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The matrix that defines a rotation of v radians clockwise about the origin is given by cos (−v) −sin (−v) cos v sin v = sin (−v) cos (−v) −sin v cos v x cos v sin v x i.e., = −sin v cos v y y cos (u − v) cos v sin v cos u becomes = sin (u − v) −sin v cos v sin u i.e., cos (u − v) = cos v cos u + sin v sin u and sin (u − v) = −sin v cos u + cos v sin u

or cos (u − v) = cos u cos v + sin u sin v or sin (u − v) = sin u cos v − cos u sin v

Double angle formulas cos (u + v) = cos u cos v − sin u sin v Replacing v with u cos (u + u) = cos u cos u −sin u sin u cos 2u = cos2 u −sin2 u = 2 cos2 u − 1 = 1 − 2 sin2 u

since since

sin2 u = 1 − cos2 u cos2 u = 1 −sin2 u

Similarly, replacing v with u in sin (u + v) = sin u cos v + cos u sin v

∴

sin 2u = sin u cos u + cos u sin u sin 2u = 2 sin u cos u

Replacing v with u in tan (u + v) =

tan u + tan v 1 − tan u tan v

∴ tan (u + u) =

tan u + tan u 1 − tan u tan u

tan 2u =

2 tan u 1 − tan2 u

Example 13 4 ␲ and 0 < ␪ < , evaluate 3 2 a sin 2␪ b tan 2␪ If tan ␪ =



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Solution a sin ␪ =

2 tan ␪ 1 − tan2 ␪ 4 2× 3 = 16 1− 9 8 = 3 −7 9 24 =− 7

3 4 and cos ␪ = 5 5

∴ sin 2␪ = 2 sin ␪ cos ␪ 4 3 =2× × 5 5 24 = 25

b tan 2␪ =

5

4

θ 3

Example 14 Prove each of the following identities. 2 sin ␪ cos ␪ cos ␪ 2 sin (␪ + ␾) sin ␪ = tan 2␪ a + = b 2 2 sin ␾ cos ␾ sin 2␾ cos ␪ −sin ␪ 1 1 c + = tan 2␪ cosec ␪ cos ␪ + sin ␪ cos ␪ −sin ␪ Solution

2 sin ␪ cos ␪ cos2 ␪ −sin2 ␪ sin 2␪ = cos 2␪ = tan 2␪ = RHS

a LHS =

Note:

sin ␪ cos ␪ + sin ␾ cos ␾ sin ␪ cos ␾ + cos ␪ sin ␾ = sin ␾ cos ␾ sin (␪ + ␾) = 1 sin 2␾ 2 2 sin (␪ + ␾) = sin 2␾

b LHS =

Identity holds when cos 2␪ = 0 Note:

c

Identity holds when sin 2␾ = 0

1 1 + cos ␪ + sin ␪ cos ␪ −sin ␪ cos ␪ −sin ␪ + cos ␪ + sin ␪ = cos2 ␪ −sin2 ␪ 2 cos ␪ = cos 2␪ sin 2␪ 2 sin ␪ cos ␪ = But 2 cos ␪ = sin ␪ sin ␪ sin 2␪ ∴ LHS = cos 2␪ sin ␪ tan 2␪ = sin ␪ = tan 2␪ cosec ␪ LHS =

Note:

Identity holds when cos 2␪ = 0



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Sometimes the easiest way to prove two expressions are equal is to simplify each of them. This is demonstrated in the following example. Example 15 Prove that (sec A − cos A)(cosecA −sin A) =

1 tan A + cot A

Solution LHS

1 tan A + cot A 1 = cos A sin A + cos A sin A 1 = sin2 A + cos2 A cos A sin A cos A sin A = sin2 A + cos2 A = cos A sin A

= (sec A − cos A)(cosecA −sin A) 1 1 − cos A −sin A = cos A sin A =

RHS =

1 − cos2 A 1 −sin2 A × cos A sin A

sin2 A cos2 A cos A sin A = cos A sin A =

LHS = RHS


10

1 By making use of the appropriate addition formula find the exact values for each of the following. a cos 15◦

Example

11

b cos 105◦

2 By making use of the appropriate addition formula find exact values for each of the following. a sin 165◦

b tan 75◦

3 Find exact values of 5␲ 11␲ a cos b sin 12 12 Example

12

4 If sin u =

c tan

−␲ 12

3 12 and sin v = , evaluate sin (u + v). (Note: There is more than one answer.) 13 5

5 Simplify the following. ␲ ␲ a sin ␪ + b cos ␾ − 6 4

␲ c tan ␪ + 3

␲ d sin ␪ − 4

6 Simplify a cos (u − v) sin v + sin (u − v) cos v

b sin (u + v) sin v + cos (u + v) cos v



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13

−5 −3 and ␪ is in the 3rd quadrant and cos ␾ = and ␾ is in the 2nd quadrant, 5 13 evaluate each of the following without using a calculator.

7 If sin ␪ =

a cos 2␾ e sin (␪ + ␾)

b sin 2␪ f cos (␪ − ␾)

c tan 2␪ g cosec (␪ + ␾)

d sec 2␾ h cot 2␪

5 4 and tan v = and both u and v are acute angles evaluate: 3 12 a tan (u + v) b tan 2u c cos (u − v) d sin 2u

8 If tan u =

3 24 ␲ and sin ␤ = and < ␤ < ␣ < ␲ evaluate 5 25 2 a cos 2␣ b sin (␣ − ␤) c tan (␣ + ␤) √ 1 3 10 If sin ␪ = − and cos ␪ = evaluate 2 2 a sin 2␪ b cos 2␪ 9 If sin ␣ =

d sin (2␤)

11 Simplify each of the following expressions. a (sin ␪ − cos ␪)2 Examples

14, 15

11.7

b cos4 ␪ −sin4 ␪

12 Prove the following identities, √ ␲ = sin ␪ − cos ␪ a 2 sin ␪ − 4 ␲ ␲ c tan ␪ + tan ␪ − = −1 4 4 ␲ 1 + tan ␪ e tan ␪ + = 4 1 − tan ␪ tan u + tan v sin (u + v) g = tan u − tan v sin (u − v) i sin 4␪ = 4 sin ␪ cos3 ␪ − 4 cos ␪ sin3 ␪

␲ ␲ b cos ␪ − + cos ␪ + = cos ␪ 3 3 √ ␲ ␲ d cos ␪ + + sin ␪ + = 3 cos ␪ 6 3 sin (u + v) f = tan v + tan u cos u cos v h cos 2␪ + 2 sin2 ␪ = 1 1 −sin 2␪ j = sin ␪ − cos ␪ sin ␪ − cos ␪

a cos x + b sin x In Section 11.2 the method of addition of ordinates was used in the plotting of the sums of circular functions. In this section it will be shown how functions with rule of the form f (x) = a cos x + b sin x may have the rule written in terms of a single circular function. First write √ a b cos x + √ sin x a cos x + b sin x = a 2 + b2 √ a 2 + b2 a 2 + b2 √ = a 2 + b2 (cos ␣ cos x + sin ␣ sin x) a b where cos ␣ = √ and sin ␣ = √ a 2 + b2 a 2 + b2 √ Let r = a 2 + b2 and thus a cos x + b sin x = r cos (x − ␣)




317

Similarly it may be shown that a cos x + b sin x = r sin (x + ␤) where r =

√ a b a 2 + b2 , sin ␤ = √ and cos ␤ = √ 2 2 2 a +b a + b2

Example 16 √ Express cos x − 3 sin x in the form r cos (x − ␣) and hence find the range of the function √ with rule f (x) = cos x − 3 sin x, and the maximum and minimum values of the function. Solution

also

√ √ a = 1, b = − 3 ∴ r = 1 + 3 = 2√ a 1 b − 3 cos ␣ = = and sin ␣ = = r 2 r 2 ␲ ∴ ␣=− 3 √ ␲ ∴ cos x − 3 sin x = 2 cos x + 3

∴ Range of f is [−2, 2] The maximum and minimum values of f are 2 and −2 respectively.

Using the TI-Nspire Access the tCollect( ) command from b>Algebra>Trigonometry>Collect and complete as shown.

Example 17 Solve cos x −

√

3 sin x = 1 for x ∈ [0, 2␲].

Solution From Example 16,

∴

√ ␲ cos x − 3 sin x = 2 cos x + 3 ␲ =1 2 cos x + 3 ␲ 1 cos x + = 3 2 ␲ ␲ 5␲ 7␲ x+ = , , 3 3 3 3 4␲ x = 0, , 2␲ 3



318


Using the TI-Nspire Use solve( ) from the Algebra menu as shown. The symbol ≤ can be found in the symbols palette (/+k), or /+= and select ≤ (<= or /+< on the Clickpad)

Example 18 Express

√

3 sin 2x − cos 2x in the form r sin (2x + ␣).

Solution A slightly different technique is used. √ 3 sin 2x − cos 2x = r sin (2x + ␣) Let √ 3 sin 2x − cos 2x = r [sin 2x cos ␣ + cos 2x sin ␣] Then This is to hold for all x. ␲ √ If x = , 3 = r cos ␣ 4 If

x = 0, −1 = r sin ␣

...

1

...

2

Squaring and adding 1 and 2 gives r 2 cos2 ␣ + r 2 sin2 ␣ = 4 i.e., r2 = 4 ∴ r = ±2 The positive solution is taken. Substituting in 1 and 2 gives √ 1 3 = cos ␣ and − = sin ␣ ∴ 2 2 ␲ ∴ ␣=− 6 √ ␲ ∴ 3 sin 2x − cos 2x = 2 sin 2x − 6 Expand the right hand side of the equation to verify.



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Exercise 11G Example

16

1 Find the maximum and minimum values of the following. √ a 4 cos x + 3 sin x b 3 cos x + sin x √ d cos x + sin x e 3 cos x + 3 sin x √ h 5 + 3 sin x − 2 cos x g cos x − 3 sin x + 2

Example

17

2 Solve each of the following for x ∈ [0, 2␲] or ␪ ◦ ∈ [0, 360]. √ a sin x − cos x = 1 b 3 sin x + cos x = 1 √ √ d 3 cos x − 3 sin x = 3 c sin x − 3 cos x = −1 √ f 2 2 sin ␪ ◦ − 2 cos ␪ ◦ = 3 e 4 sin ␪ ◦ + 3 cos ␪ ◦ = 5 √ 3 Write 3 cos 2x −sin 2x in the form r cos (2x + ␣).

Example

18

4 Write cos 3x −sin 3x in the form r sin (3x − ␣).

c cos x −sin x √ f sin x − 3 cos x

5 Sketch the graphs of the following, showing one cycle. √ a f (x) = sin x − cos x b f (x) = 3 sin x + cos x √ c f (x) = sin x + cos x d f (x) = sin x − 3 cos x



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Chapter summary Further symmetry properties: complementary angles ␲ − ␪ = cos ␪ sin 2 ␲ + ␪ = cos ␪ sin 2␲ − ␪ = sin ␪ cos 2 ␲ + ␪ = −sin ␪ cos 2 Addition of ordinates y

3 2 1 0 –1 –2 –3 –4 –5

y1 = 2 sin x

π

π 2

y2 = 3 cos 2x

θ

2π

3π 2

y = 2 sin x + 3 cos 2x

Graph of tangent function

y y = tan θ period = π

1 –π 2

–1

π π 4 2

π

3π 2

2π

5π 2

θ

General solution of circular function equations If cos (x) = a, If tan (x) = a, If sin (x) = a,

where n ∈ Z and a ∈ [−1, 1] x = 2n␲ ± cos−1 (a), where n ∈ Z and a ∈ R x = n␲ + tan−1 (a), −1 x = 2n␲ + sin (a), or x = (2n + 1)␲ −sin−1 (a), where n ∈ Z and a ∈ [−1, 1]

Reciprocal circular functions secant ␪ = sec ␪ =

1 cos ␪

1 sin ␪ 1 cotangent ␪ = cot ␪ = , sin ␪ = 0 and cos ␪ = 0 tan ␪ cosecant ␪ = cosec ␪ =



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cos2 ␪ + sin2 ␪ = 1 1 + tan2 ␪ = sec2 ␪ cot2 ␪ + 1 = cosec2 ␪ Addition formulas

Review

Pythagorean identity

321

cos (u − v) = cos u cos v + sin u sin v cos (u + v) = cos u cos v −sin u sin v sin (u + v) = sin u cos v + cos u sin v sin (u − v) = sin u cos v − cos u sin v tan u + tan v tan (u + v) = 1 − tan u tan v tan u − tan v tan (u − v) = 1 + tan u tan v Double angle formulas cos 2u = cos2 u −sin2 u = 2 cos2 u − 1 = 1 − 2 sin2 u sin 2u = 2 sin u cos u 2 tan u tan 2u = 1 − tan2 u a cos x + b sin x can be written as r cos (x − ␣) a b where r = a 2 + b2 and cos ␣ = √ and sin ␣ = √ 2 2 2 a +b a + b2 It can also be written as r sin (x + ␤) a b and cos ␤ = √ where r = a 2 + b2 and sin ␤ = √ 2 2 2 a +b a + b2

Multiple-choice questions 1 cosec x −sin x is equal to A cos x cot x B cosec x tan x C 1 −sin2 x 1 −sin x D sin x cosec x E sin x −1 , the possible values of sin x are 2 If cos x = √ 3√ −2 2 2 2 −2 2 −8 8 , A , B , C 3 3 3 3 9 9 √ √ − 2 2 1 −1 D , , E 3 3 2 2



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a 3 If cos ␪ = and 0 < ␪ < b √ a 2 + b2 A B b a E D √ 2 b + a2

␲ , then tan ␪ in terms of a and b is 2 √ a b2 − a 2 C √ a b2 − a 2 a √ b b2 + a 2

4 The magnitude of ∠ABX is ␪, AX = 4 cm, XC = x cm and BC = 2 cm. In terms of x, tan ␪ is equal to 4 8 B A 2 (x + 2) x C 8−x D 8+x 8 E √ 2 x +4

A 4 cm X x cm

θ B

C

2 cm

3␲ ␲ < A < ␲ and ␲ < B < , with cos A = t and sin B = t, sin (B + A) is equal to 2 2 D 1 − 2t 2 E −1 A 0 B 1 C 2t 2 − 1

5 For

6

sin 2A is equal to cos 2A − 1 A cot 2A − 1

B sin 2A + sec 2A

C

D sin 2A − tan 2A E −cot A ␲ 7 sin − x is not equal to 2 3␲ A cos (2␲ − x) B −sin +x C sin x 2 8 (1 + cot x)2 + (1 − cot x)2 is equal to A 2 + cot x + 2 cot 2x B 2 C −4 cot x

sin A cos A − 1

D cos (−x) D 2 + cot 2x

E sin

␲ 2

+x

E 2cosec2 x

9 If sin 2A = m and cos A = n, tan A in terms of m and n is equal to 2n 2n m n 2n 2 C D A B E m2 m 2n 2 m m 10 −cos x + sin x, in the form r sin (x + ␣) where r > 0, is √ √ ␲ 5␲ ␲ A 2 sin x + C 2 sin x + B −sin x + 4 4 4 √ √ 3␲ 7␲ 2 sin x + E D 2 sin x + 4 4

Short-answer questions (technology-free) 1 Prove each of the following identities. tan2 ␪ + cos2 ␪ a sec ␪ + cosec ␪ cot ␪ = sec ␪ cosec2 ␪ b sec ␪ −sin ␪ = sec ␪ + sin ␪ 2 Find the maximum and minimum values of each of the following. 3 a 3 + 2 sin ␪ b 1 − 3 cos ␪ c 4 sin ␪ 2 1 1 d 2 sin2 ␪ e 2 2 + cos ␪ ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Review

3 Find the values of ␪, ␪ ∈ [0, 2␲], for which 1 1 b sin 2␪ = a sin2 ␪ = 2 4 1 d sin2 2␪ = 1 e tan2 ␪ = 3√ g sin 3␪ = −1 h sec 2␪ = 2

√ c cos 3␪ =

3 2

f tan 2␪ = −1

4 Solve the equation tan ␪ = 2 sin ␪ for values of ␪ from 0◦ to 360◦ . 8 5 , sin B = where A and B are acute, find 5 If sin A = 13 17 a cos (A + B) b sin (A − B) c tan (A + B) 6 Find a cos 80◦ cos 20◦ + sin 80◦ sin 20◦ ␲ , find the value of 2 a sin A cos B + cos A sin B

b

323

tan 15◦ + tan 30◦ 1 − tan 15◦ tan 30◦

7 If A + B =

b cos A cos B −sin A sin B

8 Find the maximum and minimum values of the function with rule a 3 + 2 sin ␪ b 4 − 5 cos ␪ 9 Prove each of the following. a sin2 A cos2 B − cos2 A sin2 B = sin2 A −sin2 B

b

sin ␪ 1 + cos ␪ 2 + = 1 + cos ␪ sin ␪ sin ␪

sin ␪ − 2 sin3 ␪ = tan ␪ 2 cos3 ␪ − cos ␪ √ 5 10 Given that sin A = and that A is obtuse, find the value of each of the following: 3 a cos 2A b sin 2A c sin 4A c

11 Prove 1 − tan2 A = cos 2A a 1 + tan2 A

b

sin A 1 + cos A 2 + = 1 + cos A sin A sin A

12 a Find tan 15◦ in simplest surd form. b Using the identities for sin (u ± v), express 2 sin x cos y as the sum of two sines. √ 13 Given f : [0, 2␲] → R, f (x) = 2 3 cos x − 2 sin x, find the coordinates of a the y intercept b the x intercepts c the maximum point d the minimum point. √ Hence sketch the graph of f (x) = 2 3 cos x − 2 sin x 14 Solve for x, 0 ≤ x ≤ 2␲. a sin x + cos x = 1 c 3 tan 2x = 2 tan x

√ 3 e sin 3x cos x − cos 3x sin x = 2

1 1 1 b sin x cos x = − 2 2 4 d sin2 x = cos2 x + 1 √ ␲ =− 3 f 2 cos 2x − 3



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15 Sketch graphs of a y = 2 cos2 x

x c f (x) = tan 2x 2 2 16 It is given that tan A = 2. Find the exact value of tan ␪, given that tan (␪ + A) = 4. ␲ 17 a Express 2 cos ␪ + 9 sin ␪ in the form r cos (␪ − ␣), where r > 0 and 0 < ␣ < 2 b i Give the maximum value of 2 cos ␪ + 9 sin ␪ ii Give the cosine of ␪ for which this maximum occurs. iii Find the smallest positive solution of the equation 2 cos ␪ + 9 sin ␪ = 1 b y = 1 − 2 sin

␲

−

Extended-response questions 1 The diagram shows a rectangle ABCD inside a semicircle, centre O and radius 5 cm. ∠BOA = ∠COD = ␪ ◦ a Show that the perimeter, P cm, of the rectangle C is given by B P = 20 cos ␪ + 10 sin ␪ b Express P in the form r cos (␪ − ␣) and hence find the value of ␪ for which P = 16.

5 cm

5 cm

θ°

θ°

A O c Find the value of k for which the area of the rectangle is k sin 2␪ cm2 . d Find the value of ␪ for which the area is a maximum. 2 The diagram shows a vertical section through a tent in which AB = 1 m, BC = 2 m and ∠BAD = ∠BCD = ␪. CD is horizontal. The diagram is symmetrical about the vertical AD. B a Obtain an expression for AD in terms of ␪. b Express AD in the form 2m r cos (␪ − ␣), where r is positive.

D

A 1m

θ

θ c State the maximum length of AD and the corresponding value of ␪. C D d Given that AD = 2.15 m, find the value of ␪ for which ␪ > ␣. 1 − tan2 ␪ 3 a Prove the identity cos 2␪ = 1 + tan2 ␪ √ √ 1 ◦ 2 2 b i Use the result of a to show 1 + x = 2x − 2 where x = tan 67 2 √ 1 ◦ ii Hence find the values of integers a and b such that tan 67 =a+b 2 2 ◦ 1 c Find the value of tan 7 . 2



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A

θ h2 h1

h3

B cos ␪ b Show that the infinite sum h 1 + h 2 + h 3 + . . . = 1 −sin ␪ √ c If the infinite sum = 2, find ␪.

C

Review

4 In the diagram triangle ABC has a right angle at B. Length of BC = 1 unit. a Find in terms of ␪ ii h 2 iii h 3 iv h n i h1

325

5 ABCD is a regular pentagon with side length one unit. 2π B The exterior angles of a regular pentagon each have 5 2␲ magnitude . 5 ␲ a i Show that the magnitude of ∠BCA is C A 5 P Q R ii Find the length of CA 2␲ b i Show the magnitude of ∠DCP is 5 ii Use the fact that AC = 2CQ = 2CP + PR ␲ 2␲ to show that 2 cos = 2 cos +1 5 5 E D 2 iii Use the identity cos 2␪ = 2 cos ␪ − 1 to form a quadratic equation in terms of ␲ cos 5 ␲ iv Find the exact value of cos 5 6 a Prove each of the identities ␪ ␪ 1 − tan2 2 tan 2 2 i cos ␪ = ii sin ␪ = ␪ ␪ 2 1 + tan 1 + tan2 2 2 ␪ b Use the result of a to find the value of tan , given 8 cos ␪ − sin ␪ = 4 2



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C H A P T E R

12 Trigonometric ratios and applications Objectives To solve practical problems using the trigonometric ratios To use the sine rule and the cosine rule to solve problems To find the area of a triangle given two sides and an included angle To find the area of a sector and a segment of a circle To find the length of an arc To solve problems involving angles of depression and angles of elevation To identify the line of greatest slope of a plane To solve problems in three dimensions including determining the angle between planes

12.1 Defining sine, cosine and tangent The unit circle is a circle of radius 1 with centre at the origin.

y (0, 1)

(–1, 0)

(0, 0) (1, 1)

x

(0, –1)

Sine and cosine may be defined for any angle through the unit circle. For the angle of ␪ ◦ , a point P on the unit circle is defined as illustrated here. The angle is measured in an anticlockwise direction from the positive direction of the x axis.

y P(cos(θ°), sin(θ°)) θ° (0, 0)

x



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Chapter 12 — Trigonometric ratios and applications

327

Cos (␪ ◦ ) is defined as the x coordinate of the point P and sin (␪ ◦ ) is defined as the y coordinate of P. A calculator gives approximate values for these coordinates. y

y

y

(–0.1736, 0.9848) (–0.7071, 0.7071)

(0.8660, 0.5)

135° 30°

100° x

x

x

1 sin 135◦ = √ ≈ 0.7071 2 1 cos 135◦ = − √ ≈ −0.7071 2

sin 30◦ = 0.5 (exact value) √ 3 ◦ cos 30 = ≈ 0.8660 2

cos 100◦ = −0.1736 sin 100◦ = 0.9848

In this chapter, angles greater than 180◦ or less than 0◦ will not be considered. For a right-angled triangle OBC, a similar triangle OBC can be constructed that lies in the unit circle. From the diagram,

B

B'

= sin (␪ ◦ ) OC = cos (␪ ◦ ) and C B

The scale factor is the length OB. Hence BC = OB sin (␪ ◦ ) and OC = OB cos (␪ ◦ ) This implies

1 θ° O

C'

C

OC BC = sin (␪ ◦ ) and = cos (␪ ◦ ) OB OB This gives the ratio definition of sine and cosine for a right-angled triangle. The naming of sides with respect to an angle ␪ ◦ is as shown. B

hypotenuse

opposite

θ° O

C

adjacent

opp sin ␪ = hyp ◦

opposite hypotenuse

adjacent hypotenuse opp opposite tan ␪ ◦ = adj adjacent

adj cos ␪ = hyp ◦



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From the unit circle, note that sin (␪ ◦ ) = sin (180 − ␪)◦ , e.g. sin 45◦ = sin 135◦ and cos (␪ ◦ ) = −cos (180 − ␪)◦ , e.g. cos (45◦ ) = −cos (135◦ )

y

(cos (180 – θ)°, sin(180 – θ)°)

(180 – θ)°

(cos(θ°), sin(θ°))

θ°

x

0

This result will be used later in this chapter. Example 1 Find the value of x correct to two decimal places.

C 80 cm

Solution

∴ ∴

x A = sin 29.6◦ 80 ◦ x = 80 sin 29.6 = 39.5153 . . . x = 39.52 correct to two decimal places

x cm

29.6°

B

Example 2 Find the length of the hypotenuse correct to two decimal places.

B

Solution

∴ ∴

10 = cos 15◦ AB 10 = AB cos 15◦ 10 AB = cos 15◦ = 10.3527 . . .

A

15° 10 cm

C

The length of the hypotenuse = 10.35 cm correct to two decimal places. Example 3 Find the magnitude of ∠ABC.

A

Solution 11 11 cm 3 11 x = tan−1 x° 3 B C 3 cm ∴ x = (74.74 . . .)◦ ∴ x = 74◦ 44 42 (to the nearest second). Remember that this is read as 74 degrees, 44 minutes and 42 seconds. tan x =



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329


1

1 Find the value of x in each of the following. b

a

c

10 cm

x cm

x cm

5°

5 cm

20.16° 8 cm

35° x cm

e

d

f 10 cm

10 cm

x cm x° 30°15' 7 cm

40°

15 cm

x cm

Example

2

2 An equilateral triangle has altitudes of length 20 cm. Find the length of one side.

Example

3

3 The base of an isosceles triangle is 12 cm long and the equal sides are 15 cm long. Find the magnitude of each of the three angles of the triangle. 4 A pole casts a shadow 20 m long when the altitude of the sun is 49◦ . Calculate the height of the pole.

pole

49° 20 m

5 This figure represents a ramp.

A 6m

a Find the magnitude of angle ACB. b Find the distance BC.

1m

C

B P

6 This figure shows a vertical mast PQ, which stands on horizontal ground. A straight wire 20 m long runs from P at the top of the mast to a point R on the ground, which is 10 m from the foot of the mast.

20 m

◦

a Calculate the angle of inclination, ␪ , of the wire to the ground. b Calculate the height of the mast. θ∞ R ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

10 m

Q


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7 A ladder leaning against a vertical wall makes an angle of 26◦ with the wall. If the foot of the ladder is 3 m from the wall, calculate a the length of the ladder

b the height it reaches above the ground.

8 An engineer is designing a straight concrete entry ramp, 60 m long, for a car park 13 m above street level. Calculate the angle of the ramp to the horizontal. 9 A vertical mast is secured from its top by straight cables 200 m long fixed at the ground. The cables make angles of 66◦ with the ground. What is the height of the mast? 10 A mountain railway rises 400 m at a uniform slope of 16◦ with the horizontal. What is the distance travelled by a train for this rise? 11 The diagonals of a rhombus bisect each other at right angles.

B

C

If BD = AC = 10 cm, find a the length of the sides of the rhombus b the magnitude of angle ABC.

A

D

12 A pendulum swings from the vertical through an angle of 15◦ on each side of the vertical. If the pendulum is 90 cm long, what is the distance x cm between its highest and lowest point?

90 cm

90 cm

x cm

13 A picture is hung symmetrically by means of a string passing over a nail with its ends attached to two rings on the upper edge of the picture. The distance between the rings is 30 cm and the angle between the two portions is 105◦ . Find the length of the string.

105° 30 cm

14 The distance AB = 50 m. If the line of sight of a person standing at A to the tree makes an angle of 32◦ with the bank, how wide is the river?

B


32° 50 m

A


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331


15 A ladder 4.7 m long is placed against a wall. The foot of the ladder must not be placed in a flower bed, which extends a distance of 1.7 m from the foot of the wall. How high up the wall can the ladder reach? B

16 A river is known to be 50 m wide. A swimmer sets off from A to cross the river and the path of the swimmer AB is as shown. How far does the person swim?

50 m 60° A

12.2

The sine rule In Section 12.1, methods for finding unknown lengths and angles for right-angled triangles were discussed. In this section and the next, methods for finding unknown quantities in non-right-angled triangles are discussed. The sine rule is used to find unknown quantities in a triangle when one of the following situations arises: one side and two angles are given two sides and a non-included angle are given. In the first of the two cases, a unique triangle is defined, but for the second it is possible for two triangles to exist.

Labelling convention The following convention is followed in the remainder of this module. Interior angles are denoted by upper case letters and the length of the side opposite an angle is denoted by the corresponding lower case letter. For example, the magnitude of angle BAC is denoted by A, and the length of side BC is denoted by a. The sine rule states that for triangle ABC a b c = = sin A sin B sin C

B a

c A

C

b

B

a

c A

C

b

A proof will only be given for the acute-angled triangle case. The proof for obtuse-angled triangles is similar.

Proof In triangle ACD,

∴ In triangle BCD,

∴ ∴ i.e.,

h b = b sin A h = a = a sin B = b sin A b = sin B

sin A = h sin B h a sin B a sin A


C b

A

h

D

a

B


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Similarly, starting with a perpendicular from A to BC would give c b = sin B sin C Example 4 Use the sine rule to find the length of AB. Solution

∴ ∴

10 c = ◦ sin 31 sin 70◦ 10 × sin 31◦ c= sin 70◦ c = 5.4809 . . .

B 70° c 31° A

C

10 cm

The length of AB is 5.48 cm correct to two decimal places. Example 5 Use the sine rule to find the magnitude of angle XZY in the triangle, given that Y = 25◦ , y = 5 cm, and z = 6 cm.

Z

5 cm

25°

X

6 cm

Y

Solution

∴ ∴

6 5 = sin 25◦ sin Z sin 25◦ sin Z = 6 5 6 × sin 25◦ sin Z = 5 = 0.5071 . . . −1

Z1 30° 28' 25" 149° 31' 35" 5 cm 5 cm

Z2 25°

X

6 cm

∴

Z = sin

∴

Z = (30.4736 . . .)◦ or (180 − 30.4736 . . .)◦

∴

Z = 30◦ 28 25 or Z = 149◦ 31 35 (to the nearest second)

(0.5071 . . .)

Y

Remember: sin (180 − ␪) = sin ␪ There are two solutions for the equation sin Z = 0.5071 . . . Note: When using the sine rule in the situation where two sides and a non-included angle are given, the possibility of two such triangles existing must be considered. Existence can be checked through the sum of the given angle and the found angle not exceeding 180◦ .



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333


4

1 Find the value of the pronumeral for each of the following triangles. a

b

Y

Z

70°

65° x cm

X

50°

Z

10 cm

c

5

Y

6 cm

d

12 cm

Y

X

38° x cm

5.6 cm

90°

100°

28°

Y

37°

X

Z x cm

Example

y cm

Z

X

2 Find the value of ␪ for each of the following triangles. a

b

C

A θ°

72° 7 cm

9.4 cm θ°

A

8 cm

42° B C

d

c C

B

8.3 cm

B

10 cm 8 cm

8 cm

A

108°

θ°

B

A

θ°

38° 9 cm

C

3 Solve the following triangles (i.e. find all sides and angles). a a = 12, B = 59◦ , C = 73◦

b A = 75.3◦ , b = 5.6, B = 48.25◦

c A = 123.2◦ , a = 11.5, C = 37◦

d A = 23◦ , a = 15, B = 40◦

e B = 140◦ , b = 20, A = 10◦ 4 Solve the following triangles (i.e. find all sides and angles). a b = 17.6, C = 48.25◦ , c = 15.3

b B = 129◦ , b = 7.89, c = 4.56

c A = 28.35◦ , a = 8.5, b = 14.8 5 A landmark A is observed from two points B and C, which are 400 m apart. The magnitude of angle ABC is found to be 68◦ and the magnitude of angle ACB is 70◦ . Find the distance of A from C. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Essential Advanced General Mathematics P

6 P is a point at the top of a lighthouse. Measurements of the length of AB and angles PBO and PAO are taken and are as shown in the diagram. Find the height of the lighthouse. 27.6° 46.2° B 34 m

A

O

7 A and B are two points on a coastline. They are 1070 m apart. C is a point at sea. The angles CAB and CBA have magnitudes of 74◦ and 69◦ respectively. Find the distance of C from A. Y

8 Find a AX

b AY

X

88°

32°

A

89°

20°

B

50 m

12.3 The cosine rule The cosine rule is used to find unknown quantities in a triangle when one of the following situations arises: two sides and an included angle are given three sides are given.

The cosine rule states that for triangle ABC

c

B

A

a b

C

a 2 = b2 + c2 − 2bc cos A or equivalently cos A =

b2 + c2 − a 2 2bc

The symmetrical results also hold, i.e. b2 = a 2 + c2 − 2ac cos B c2 = a 2 + b2 − 2ab cos C The result will be proved for an acute-angled triangle. The proof for obtuse-angled triangles is similar.

Proof In triangle ACD

C

b = x + h (Pythagoras’ theorem) x cos A = and therefore x = b cos A b 2

2

2

a

b

h x

A


D

c

B


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335


In triangle BCD a 2 = (c − x)2 + h 2 (Pythagoras’ theorem) Expanding gives

∴

a 2 = c2 − 2cx + x 2 + h 2 = c2 − 2cx + b2 (as x 2 + h 2 = b2 ) a 2 = b2 + c2 − 2bc cos A (as x = b cos A)

Example 6 B

For triangle ABC, find the length of AB in centimetres correct to two decimal places.

5 cm

c

Solution

67°

c = 5 + 10 − 2 × 5 × 10 cos 67 = 85.9268 . . . c ≈ 9.2697 2

∴

2

2

◦

10 cm

A

C

The length of AB is 9.27 cm correct to two decimal places. Example 7 Find the magnitude of angle ABC for triangle ABC. Solution a 2 + c2 − b2 2ac 2 12 + 62 − 152 = 2 × 12 × 6 = −0.3125

B

cos B =

∴

B = (108.2099 . . .)◦

∴

B ≈ 108◦ 12 35.845

12 cm

6 cm A

15 cm

C

The magnitude of angle ABC is 108◦ 12 36 (to the nearest second).


6

1 Find the length of BC. A


B 10 cm 15° 15 cm

C


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2 Find the magnitude of angles ABC and ACB.

B 8 cm A

3 For triangle ABC with a A = 60◦ b = 16 ◦

b a = 14

B = 53

c a = 27

b = 35 ◦

find a

c = 12,

find b

c = 46,

find the magnitude of angle ABC find b

B = 120

c = 63,

e a = 31

b = 42

C = 140◦ , find c

f a = 10

b = 12

c = 9,

g a = 11

b=9

C = 43.2◦ , find c

h a=8

b = 10

c = 15,

C

10 cm

c = 30,

d a = 17

5 cm

find the magnitude of angle BCA

find the magnitude of angle CBA

4 A section of an orienteering course is as shown. Find the length of leg AB.

B 4 km A

5 Two ships sail from point O. At a particular time their positions A and B are as shown. Find the distance between the ships at this time.

20°

C

6 km N A

B 6 km 4 km 30° O 6 ABCD is a parallelogram. Find the length of the diagonals: a AC

B

5 cm

C

b BD

4 cm 48° A


D


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7 A weight is hung from two hooks in a ceiling by strings of length 54 cm and 42 cm, which are inclined at 70◦ to each other. Find the distance between the hooks.

42 cm

54 cm 70°

B

8 a Find the length of diagonal BD. 4 cm 5 cm

b Use the sine rule to find the length of CD. A 92°

88° C 6 cm

D

9 Two circles of radius 7.5 cm and 6 cm have a common chord of length 8 cm. a Find the magnitude of angle AO B.

A 7.5 cm 8 cm

O

6 cm O'

b Find the magnitude of angle AOB. B

10 Two straight roads intersect at an angle of 65◦ . A point A on one road is 90 m from the intersection and a point B on the other road is 70 m from the intersection, as shown on the diagram. a Find the distance of A from B.

A 90 m O

65° 70 m

C B

b C is the midpoint of AB. Find the distance of C from the intersection.

12.4

Area of a triangle It is known that the area of a triangle is given by the formula 1 bh 2 1 Area = × base length × height 2

B

Area =

h

A

b

C

By observing that h = c sin A the following formula can be found.



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1 bc sin A 2 i.e., the area is given by half the product of the length of two sides and the sine of the angle included between them. Area of triangle =

Example 8 B

Find the area of triangle ABC shown in the diagram.

7.2 cm

140°

6.5 cm

A

Solution

C

1 × 7.2 × 6.5 sin 140◦ 2 = 15.04 cm2

Area =

The area of triangle ABC is 15.04 cm2 correct to two decimal places. Example 9 Find the area of each of the following triangles, correct to three decimal places. a b c A

D

8.2 cm

E

70°

G

8 cm 85° 10 cm

B

10 cm

7 cm

F I

6.4 cm

12° H

C

Solution a Using the cosine rule, 82 = 6.42 + 102 − 2 × 6.4 × 10 cos C 64 = 140.96 − 128 cos C cos C = −0.60125 C ◦ = (126.95 . . .)◦ (the exact value can be stored on the graphics calculator as C, say) 1 Area of triangle ABC = × 6.4 × 10 × sin C 2 = 25.570 cm2 , correct to three decimal places.



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E ◦ = (180 − (70 + 85))◦ = 25◦

b

Using the sine rule, 8.2 sin (85◦ ) = 3.47 . . . (the exact value can be stored on the graphics calculator as E, say) 1 Area of triangle DEF = × 8.2 × E × sin (70◦ ) 2 = 13.403 cm2 , correct to three decimal places. DF = sin (25◦ ) ×

c Using the sine rule, sin (12◦ ) sin I = 10 × 7 = 0.2970 . . . I ◦ = (180 − 17.27 . . .)◦ since I is an obtuse angle = (162.72 . . .)◦ (the exact value can be stored on the graphics calculator as I, say) ◦ G = (180 − (12 + I ))◦ = (5.27 . . .)◦ (the exact value can be stored on the graphics calulator as G, say) 1 Area of triangle GHI = × 10 × 7 × sin (G ◦ ) 2 = 3.220 cm2 , correct to three decimal places.


8

1 Find the area of each of the following triangles. a

b

C 6 cm

X 72.8°

70°

6.2 cm

4 cm 5.1 cm Z

A

B Y

c

3.5 cm M 130°

N

B

d 25°

8.2 cm

5 cm

C L

5 cm A



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2 Find the area of each of the following triangles, correct to three decimal places. a A

b A 9 cm

5.9 cm C

4.1 cm

C

7 cm 100°

3.2 cm

B

B

c

E

d

E

6.3 cm D

5.7 cm D

65°

5.9 cm 5.1 cm 55° F

F

e G

f G

12 cm 24°

5 cm

H

10° 19°

4 cm

H

I

I

12.5 Circle mensuration

Terminology

A

In this circle with centre O, the interval AB is called a chord of the circle. A chord is an interval with endpoints on the circle. D C If the centre of the circle is on the chord, the interval is called O a diameter, e.g. interval CD in the diagram. Any two points on a circle divide the circle into arcs. The shorter arc is called the B minor arc, the longer is the major arc, e.g. arc ACB is a minor arc and ADB is a major arc in this diagram. Note that arc DBC and arc DAC are semicircular arcs in this diagram. Every chord divides the interior of a circle into two regions called segments. The smaller is called the minor segment, the larger is the major segment. In the above diagram the minor segment has been shaded. Two radii and an arc define a region called a sector. In this diagram with circle centre O, the shaded region is a minor sector and the unshaded region is a major sector. Formulas to find arc lengths, chord lengths and areas of regions inside a circle will now be developed.

A

D

C

O B



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Arc length

A

The arc ACB and the corresponding chord AB are said to subtend the angle ∠AOB at the centre of the circle. If the magnitude of D ∠AOB = ␪ ◦ and radius length is r units, then l units, the length of arc ACB, will be a fraction of the circumference.

r O θ∞

C

B

Since circumference = 2␲r ␪ × 2␲r ∴ l= 360 ␲r ␪ = 180 ␲␪ = ␾ where ␪ ◦ = ␾c Now since 180 l = r ␾ where ␾c = mag ∠AOB

Chord length From the diagram, the cosine rule gives

A r

AB = r + r − 2r cos ␪ 2 (1 − cos ␪) = 2r ∴ AB = 2r 2 (1 − cos ␪) 2

2

2

2

O θ r B

In triangle OAP,

␪ 2 ␪ ∴ AB = 2r sin 2 ␪ 2 ␪ = 2 sin2 Note: 1 − cos ␪ = 1 − 1 − 2 sin 2 2 AP = r sin

O θ 2

B

r A

P

Area of sector

A

If mag ∠AOB = ␪ ◦ the area of the sector is a fraction of the area of the circle. Now area of circle is given by area of circle = ␲r 2

r O θ∞

∴ area of sector = fraction of ␲r 2 ␲r 2 ␪ = 360 ␲␪ Again using ␾ = 180 1 Area of sector = r 2 ␾ where ␾c = mag ∠AOB 2

B

Example 10

A 10 cm

In this circle, centre O, radius length 10 cm, the angle subtended at O by arc ACB has magnitude 120◦ . Find a the exact lengths of i the chord AB

O 120°

C

10 cm

ii the arc ACB


B Cambridge University Press

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b the exact area of the minor sector AOB c the magnitude of angle AOC, in degrees and minutes, if the minor arc AC has length 4 cm. Solution a

i Use chord length = 2r sin

␪ where r = 10 and ␪ = 120◦ 2

chord length = 20 sin√ 60◦ 3 = 20 × √ 2 = 10 3 √ Length of chord is 10 3 cm. 2␲ ii Use l = r ␪ where r = 10, ␪ = (note use of radians) 3 2␲ = 10 × 3 20␲ = 3 20␲ cm. Length of arc is 3 (Verify that length of arc is greater than length of chord as a check.) 1 2␲ b Use area of sector = r 2 ␪ where r = 10, ␪ = (note use of radians) 2 3 2␲ 1 = × 102 × 2 3 100␲ = 3 100␲ So area of minor sector AOB = cm2 . 3 c Use arc length = r ␪

∴

∴ ∴

4 = 10␪ 4 ␪= 10 180 angle AOC = 0.4 × ␲ = (22.918 . . .)◦

A 10 cm O

4 cm

θ

C

= 22◦ 55 (to the nearest minute)

Area of segment Area of segment shaded = area of minor sector OAB – area of AOB ␲r 2 ␪ 1 2 − r sin ␪ 360 2 Where mag ∠AOB = ␪ ◦ ␲␪ but if mag ∠AOB = ␾c , ␾ = 180 1 1 A = r 2 ␾ − r 2 sin ␾ 2 2 1 2 = r (␾ − sin ␾) 2 So

A

A=


r O θ∞

B


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Generally speaking the formulas are simpler if mag ∠AOB is measured in radians. The following formulas assume ␪ is in radians. Arc length = r ␪

␪ Chord length = 2r sin 2 1 Area of sector = r 2 ␪ 2 1 Area of segment = r 2 (␪ − sin ␪) 2 Example 11 A circle, centre O, with radius length 20 cm has a chord AB that is 10 cm from the centre of the circle. Calculate the area of the minor segment formed by this chord. Solution 1 Now area of segment = r 2 (␪ − sin ␪) 2 r = 20 but ␪ needs to be calculated. A 10 ␪◦ = In OCB, cos 20 cm 2 20 ␪ = 60 ∴ 2 and ␪ = 120 2␲ c Hence mag ∠AOB = 3 2␲ 1 2 2␲ − sin cm2 Area of segment = × 20 2 3√ 3 2␲ 3 = 200 − cm2 3 2 √ 4␲ − 3 3 cm2 = 200 6 √ 100 4␲ − 3 3 cm2 = 3

C B θ

10 cm

O


10

1 Find the arc length which subtends an angle of magnitude 105◦ at the centre of a circle of radius length 25 cm. 2 Find the magnitude, in degrees and minutes, of the angle subtended at the centre of a circle of radius length 30 cm, by b a chord of length 50 cm. a an arc of length 50 cm

Example

11

3 A chord of length 6 cm is drawn in a circle of radius 7 cm. Find a the length of the minor arc cut off by the chord b the area of the smaller region inside the circle cut off by the chord.



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4 Sketch, on the same set of axes, the graphs of A = {(x, y) : x 2 + y 2 ≤ 16} and B = {(x, y) : y ≥ 2} Find the area measure of the region A ∩ B. 5 Use results from Chapter 11 to show that ␪ 2r 2 (1 − cos ␪) = 2r sin 2 6 Find the area of the region between an equilateral triangle of side length 10 cm and the circumcircle of the triangle (the circle that passes through the three vertices of the triangle). 7 A person stands on level ground 60 m from the nearest point of a cylindrical tank of radius length 20 m. Calculate a the circumference of the tank b the percentage of the circumference that is visible to the person. 8 The minute hand of a large clock is 4 m long. a How far does the tip of the minute hand move between 12.10 p.m. and 12.35 p.m? b What is the area covered by the minute hand between 12.10 p.m. and 12.35 p.m? 9 Two circles of radii 3 cm and 4 cm have their centres 5 cm apart. Calculate the area of the region common to both circles. 10 A sector of a circle has perimeter of 32 cm and an area of 63 cm2 . Find the radius length and the magnitude of the angle subtended at the centre of the two possible sectors. 11 Two wheels (pulleys) have radii of length 15 cm and 25 cm and have their centres 60 cm apart. What is the length of the belt required to pass tightly around the pulleys without crossing? 12 A frame in the shape of an equilateral triangle encloses three circular discs of radius length 5 cm so that the discs touch each other. Find a the perimeter of the smallest frame which can enclose the discs b the area enclosed between the discs.

12.6 Angles of elevation and depression

and bearings The angle of elevation is the angle between the horizontal and a direction above the horizontal. eye level

The angle of depression is the angle between the horizontal and a direction below the horizontal.

eye level angle of depression cliff


ght of si line angle of elevation

line

of s

igh

t


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Example 12 The pilot of a helicopter flying at 400 m observes a small boat at an angle of depression of 1.2◦ . Calculate the horizontal distance of the boat to the helicopter.

H 1.2° (angle of depression) 400 m

Solution

∴

A

AH = tan 1.2◦ AB 400 = tan 1.2◦ AB 400 AB = tan 1.2◦ AB = 19 095.800 56 . . .

B (diagram not to scale)

The horizontal distance is 19 100 m to the nearest 10 m. Example 13 L

The light on a cliff-top lighthouse, known to be 75 m above sea level, is observed from a boat at an angle 75 m of elevation of 7.1◦ . Calculate the distance of the boat from the lighthouse.

7.1°

A

B

Solution

∴

75 = tan (7.1◦ ) AB 75 AB = tan (7.1◦ ) = 602.135 . . .

The distance of the boat from the lighthouse is 602 m to the nearest metre. Example 14 From the point A, a man observes that the angle of elevation of the summit of a hill is 10◦ . He then walks towards the hill for 500 m along flat ground. The summit of the hill is now 166° at an angle of elevation of 14◦ . 14° 10° A Find the height of the hill above the level of A. B 500 m

H 4°

C

Solution The magnitude of angle HBA = (180 − 14)◦ = 166◦ The magnitude of angle AHB = [180 − (166 + 10)]◦ = 4◦ ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Using the sine rule in triangle ABH:

∴

HB 500 = ◦ sin 4 sin 10◦ 500 × sin 10◦ HB = sin 4◦ = 1244.67 . . .

In triangle BCH:

∴

HC = sin 14◦ HB HC = HB sin 14◦ = 301.11 . . .

The height of the hill is 301 m to the nearest metre.

Bearings

N

The bearing (or compass bearing) is the direction measured from north clockwise. The bearing of A from O is 030◦ The bearing of B from O is 120◦ The bearing of C from O is 210◦ The bearing of D from O is 330◦

A

D 30°

120° O W

E

330° 210° B C S

Example 15 The road from town A runs due west for 14 km to town B. A television mast is located due south of B at a distance of 23 km. Calculate the distance and bearing of the mast from the centre of town A.

N B

14 km θ

A

23 km

Solution 23 14 ∴ ␪ = 58.67◦ (to two decimal places) ∴ bearing = 180◦ + (90 − 58.67)◦ = 211.33◦ By Pythagoras’ theorem tan ␪ =

∴

T

AT 2 = AB2 + BT 2 = 142 + 232 = 725 AT = 26.925 . . .

∴ The mast is 27 km from the centre of town (to the nearest kilometre) and on a bearing of 211.33◦ . ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 16 B

A yacht starts from a point A and sails on a bearing of 038◦ for 3000 m. It then alters its course to one in a direction with a bearing of 318◦ and after sailing for 3300 m it reaches a point B. Find a the distance AB b the bearing of B from A.

3300 m

N

42° C 318° N 38°

Solution

3000 m

a The magnitude of angle ACB needs to be determined A so that the cosine rule can be applied in triangle ABC. ◦ ◦ The magnitude of angle ACB = (180 − (38 + 42)) = 100 In triangle ABC

∴

AB2 = 30002 + 33002 − 2 × 3000 × 3300 cos (100◦ ) = 23 328 233.92 AB = 4829.931 04 . . .

The distance of B from A is 4830 m (to the nearest metre). b To find the bearing of B from A, the magnitude of angle BAC must first be found. The sine rule may be used.

∴ ∴ ∴ ∴

AB 3300 = sin A sin 100◦ 3300 × sin 100◦ sin A = AB sin A = 0.672 8 . . .

B

N 42°

◦

A = (42.288 . . .)

C

The bearing of B from A ◦

◦

N 38°

◦

= 360 − (42.29 − 38 ) = 355.71◦ .

38°

A

◦

The bearing of B from A is 356 to the nearest degree.


12

1 From the top of a vertical cliff 130 m high the angle of depression of a buoy at sea is 18◦ . What is the distance of the buoy from the foot of the cliff?

Example

13

2 The angle of elevation of the top of an old chimney stack at a point 40 m from its base is 41◦ . Find the height of the chimney.



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3 A man standing on top of a mountain observes that the angle of depression to the foot of a building is 41◦ . If the height of the man above the foot of the building is 500 m, find the horizontal distance from the man to the building. 4 A man lying down on top of a cliff 40 m high observes the angle of depression to a buoy in the sea below to be 20◦ . If he is in line with the buoy, calculate the distance between the buoy and the foot of the cliff, which may be assumed to be vertical. Example

14

5 A man standing on top of a cliff 50 m high is in line with two buoys whose angles of depression are 18◦ and 20◦ . Calculate the distance between the buoys.

Example

15

6 A ship sails 10 km north and then 15 km east. What is its bearing from the starting point? 7 A ship leaves port A and steams 15 km due east. It then turns and goes 22 km due north. a What is the bearing of the ship from A? b What is the bearing of port A from the ship?

Example

16

8 A yacht sails from point A on a bearing of 035◦ for 2000 m. It then alters course to a direction with bearing of 320◦ and after sailing for 2500 m it reaches point B. a Find the distance AB.

b Find the bearing of B from A.

9 The bearing of a point A from a point B is 207◦ . What is the bearing of B from A? 10 The bearing of a ship S from a lighthouse A is 055◦ . A second lighthouse B is due east of A. The bearing of S from B is 302◦ . Find the magnitude of angle ASB. 11 A yacht starts from L and sails 12 km due east to M. It then sails 9 km on a bearing of 142◦ to K. Find the magnitude of angle MLK. 12 The bearing of C from A is 035◦ . The bearing of B from A is 346◦ . The distance of C from A is 340 km. The distance of B from A is 160 km. a Find the magnitude of angle BAC. b Use the cosine rule to find the distance from B to C.

C

N B

340 km 160 km 346°

35° A

13 From a ship S two other ships P and Q are on bearings 320◦ and 075◦ respectively. The distance PS = 7.5 km and the distance QS = 5 km. Find the distance PQ.

12.7 Problems in three dimensions Problems in three dimensions are solved by picking out triangles from a main figure and finding lengths and angles through these triangles.



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Example 17 ABCDEFGH is a cuboid. Find a distance DB b distance HB c the magnitude of angle HBD d the magnitude of angle HBA.

H

G

E

F

7 cm

D

A

C 8 cm B

10 cm

Solution a

DB2 = 82 + 102 = 164 √ ∴ DB = 164 = 12.806 . . .

D 8 cm A

The length of DB is 12.81 cm correct to two decimal places. b HB2 = HD2 + DB2 H = 72 + 164 = 49 + 164 7 cm = 213 √ ∴ HB = 213 D = 14.59 . . . The length of HB is 14.59 cm correct to two decimal places. HD H c tan ␪ = BD 7 7 cm =√ 164 = 0.5466 . . . D ◦

B

10 cm

␪ = 28 40 to the nearest minute.

θ° √164 cm

B

θ° √164 cm

B

H

d From triangle HBA

√ 10 10 213 cos B = √ 213 213

B = 46◦ 45 to the nearest minute.

A

B

10 cm

Example 18 The figure shows a pyramid with a square base. The base has sides 6 cm long and the edges VA, VB, VC, VD are each 10 cm long. a Find the length of DB. b Find the length of BE. c Find the length of VE. d Find the magnitude of angle VBE. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

V

10 cm A D

6 cm

B E C


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Essential Advanced General Mathematics A

Solution a

DB2 = 62 + 62 = 72 √ ∴ DB = 6 2 = 8.4852 . . .

B 6 cm

E D

C

6 cm

The length of DB is 8.49 cm to two decimal places. 1 DB 2 1√ BE = 72 2√ =3 2 = 4.2426 . . . BE =

b

∴

The length of BE is 4.24 cm correct to two decimal places. c

VE2 = VB2 − EB2 1 = 100 − × 72 4 = 100 − 18 = 82 √ ∴ VE = 82 = 9.0553 . . .

V 10 cm

B

E

The length of VE is 9.06 cm correct to two decimal places. d

V

VE VB √ 82 = 10 = 0.9055 . . . ␪ = 64◦ 54

sin ␪ =

∴

10 cm

θ∞

E

B

The magnitude of angle VBE is 64◦ 54 to the nearest minute.

Example 19 A communications mast is erected at the corner, A, of a rectangular courtyard ABCD whose sides measure 60 m and 45 m. If the angle of elevation of the top of the mast 12° from C is 12◦ , find a the height of the mast C b the angle of elevation of the top of the mast from B (where AB = 45 m).


H

D

A 45 m

60 m

B


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Solution a

AC 2 = AB2 + CB2 = 452 + 602 = 5625 ∴ AC = 75 HA = tan 12◦ 75 ∴ HA = 75 tan 12◦ = 15.9417

A 45 m C

H

The height of the mast is 15.94 m, correct to two decimal places. b

B

60 m

12° 75 m

C

A

HA 45 = 0.3542 . . .

tan ␪ =

∴

H

␪ ≈ 19◦ 30

θ°

B

A

45 m

The angle of elevation of the top of the mast, H, from B is 19◦ 30 to the nearest minute.


17

1 ABCDEFGH is a cuboid with dimensions as shown. Find a b c d

Example

18

19

B

C

D E

8 cm

F 5 cm

H

G

12 cm

2 VABCD is a right pyramid with a square base. The sides of the base are 8 cm in length. V The height, VF, of the pyramid is 12 cm. Find a b c d e f

Example

the length of FH the length of BH the magnitude of angle BHF the magnitude of angle BHG.

A

the length of EF the magnitude of angle VEF the length of VE the length of a sloping edge the magnitude of angle VAD the surface area of the pyramid.

F

E A

3 A tree stands at the corner of a square playing field. Each side of the square is 100 m long. At the centre of the field the tree subtends an angle of 20◦ . What angle does it subtend at each of the other three corners of the field?

B

8 cm

A 20°

T 100 m

B


C

D

100 m

C


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4 Suppose that A, C, and X are three points in a horizontal plane and B is a point vertically above X. If the length of AC = 85 m and the magnitudes of angles BAC, ACB and BCX are 45◦ , 90◦ and 32◦ respectively, find a the distance CB b the height XB.

B

X 32° 45° 85 m

A

C

5 Standing due south of a tower 50 m high, the angle of elevation of the top is 26◦ . What is the angle of elevation after walking a distance 120 m due east? 6 From the top of a cliff 160 m high two buoys are observed. Their bearings are 337◦ and 308◦ . Their respective angles of depression are 3◦ and 5◦ . Calculate the distance between the buoys. 7 Find the magnitude of each of the following angles for the cuboid shown.

H E

a ACE b HDF c ECH

6 cm A

G F

D

C

12 cm

B

5 cm

8 From a point A due north of a tower, the angle of elevation to the top of the tower is 45◦ . From point B, 100 m on a bearing of 120◦ from A, the angle of elevation is 26◦ . Find the height of the tower. 9 A and B are two positions on level ground. From an advertising balloon at a vertical height of 750 m, A is observed in an easterly direction and B at a bearing of 160◦ . The angles of depression of A and B as viewed from the balloon are 40◦ and 20◦ respectively. Find the distance between A and B. 10 A right pyramid, height 6 cm, stands on a square base of side 5 cm. Find a the length of a sloping edge

b the area of a triangular face.

11 A light aircraft flying at a height of 500 m above the ground is sighted at a point A due east of an observer stationed at a point O on the ground, measured horizontally to be 1 km from the plane. The aircraft is flying south west (along B' AB ) at 300 km/h. 500 m B a How far will it travel in one minute? b Find its bearing from O (O ) at this time. c What will be its angle of elevation from O at this time?


O' O

1000 m

45°

A' A


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12.8

353

Angles between planes and more difficult 3-D problems Angles between planes A

Consider any point P on the common line of two planes 1 and 2 . If PA and PB are drawn at right angles to the common line so that PA is in 1 and PB is in 2 , then angle APB is the angle between 1 and 2 .

Π1 P

θ

Π2

B

A

If one of the planes, 2 say, is horizontal, then PA is called a line of greatest slope in the plane 1 .

Note:

Π1 Π2 P

lines of greatest slope

angle of greatest slope

Example 20 D'

Given the cuboid shown in the diagram, find a the angle between AC and the plane ABBA b the angle between the planes ACD and DCD .

D

C' a

C A'

A

3a

B' B

3a

Solution a To find the angle between AC and the plane ABB A , D' we need the projection of AC in the plane. So we C D drop a perpendicular from C to the plane, i.e. the A' θ line C B , and join the foot of the perpendicular to 3a A B 3a A, i.e. B A. The required angle lies between C A and B A. Drawing separate diagrams showing the base and the section through A, C and we have B'

A'

3a

Thus and

a B'

B

C' a

and A

A

C'

θ

B'

B

3a

√ (3a)2 + (3a)2 = 3a 2 a 1 tan ␪ = √ = √ 3a 2 3 2

AB =

Hence the required angle, ␪, is 13.3◦ .



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b The line common to the planes ACD and DCD is CD . If M is the midpoint of this line, then MD is perpendicular to D C in plane DCD and MA is perpendicular to D C in plane D C A. Thus ␾ is the angle between the planes DCD and D CD. D' D

φ

C'

M C

3a

M

φ

a B'

A' A

D

A

3a B

√ 1 1 DC = (3a 2) 2 √ 2 √ 3a 2 2 Hence tan ␾ = a ÷ = 2 3

But

DM =

i.e. the required angle is 25.2◦ Example 21 Three points A, B and C are on a horizontal line such that AB = 70 m, and BC = 35 m. The angles of elevation of 1 , the top of a tower are ␣, ␤ and ␥ where tan ␣ = 13 1 1 tan ␤ = and tan ␥ = (as shown in the diagram). 15 20 The foot of the tower is at the same level as A, B and C. A Find the height of the tower. 1 tan ␣ = 13 1 tan ␤ = 15 1 tan ␥ = 20

P

α

Q

β

γ B

70 m

35 m

C

Solution Q

Let the height of the tower, PQ, be h m. Then h = QA tan ␣ = QB tan ␤ = QC tan ␥ which implies QA = 13h, QB = 15h, QC = 20h Now consider the base triangle ABCQ. Using the cosine formula in AQB and CQB,

and

13h

20h 15h θ

A

70 m

cos ␪ =

(70)2 + (15h)2 − (13h)2 2(70)(15h)

cos (180 − ␪) = −cos ␪ =

(35)2 + (15h)2 − (20h)2 2(35)(15h)


B

35 m

C


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(20h)2 − (15h)2 − (35)2 (70)2 + (15h)2 − (13h)2 = 2(70)(15h) 2(35)(15h)

Hence

∴ ∴ Hence

4900 + 56h 2 = 2(175h 2 − 1225) 7350 = 294h 2 h=5

The height of the tower is 5 m. Example 22 A sphere rests on the top of a vertical cylinder which is open at the top. The inside diameter of the cylinder is 8 cm. The sphere projects 8 cm above the top of the cylinder. Find the radius length of the sphere. Solution This 3-D problem can be represented by a 2-D diagram without loss of information. From the diagram, in OBC, if radius length of sphere is r cm, OC = (8 − r ) cm, OB = r cm, BC = 4 cm Using Pythagoras’ theorem (8 − r )2 + 42 = r 2 64 − 16r + r 2 + 16 = r 2 −16r + 80 = 0 r =5

O A

C

8 cm B

So radius length of sphere is 5 cm. 8 cm

Example 23 A box contains two standard golf balls that fit snugly inside. The box is 85 mm long. What percentage of the space inside the box is air? Solution 2-D diagrams may be used to represent the 3-D situation.

85 mm

Now

∴ i.e.

side view Use r mm = radius length of a ball length of box = 85 mm = 4r mm 85 r= 4 r = 21.25


end view


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So box has dimensions 85 mm by 42.5 mm by 42.5 mm Now

So

volume of box in mm3 = 42.52 × 85 (using V = Ah) 4 4 3 3 volume of two golf balls = 2 × × ␲ × 21.25 using V = ␲r 3 3 8 = ␲ × 21.253 3 8 100 42.52 × 85 − ␲ × 21.252 3 percentage air = 2 42.5 × 85 = 47.6% (to one decimal place)


20

1 The diagram shows a rectangular prism. AB = 4a units, BC = 3a units, GC = a units.

H

G

E

F D

C

a Calculate the areas of the faces ABFE, BCGF, ABCD. A B b Calculate the magnitude of the angle which plane GFAD makes with the base. c Calculate the magnitude of the angle which plane HGBA makes with the base. d Calculate the magnitude of the angle which AG makes with the base. 2 VABCD is a right pyramid with square base ABCD.

V

AB = 2a and OV = a. a Find the slope of edge VA, i.e., the magnitude of ∠VAO. b Find the slope of the face VBC. 5 3 A hill has gradient . If BF makes an angle of 45◦ with 12 the line of greatest slope, find a the gradient of BF b the magnitude of ∠FBD.

D A

C O B

F

E 5 C

D 12 A

B

4 The cross-section of a right prism is an isosceles triangle ABC. AB = BC = 16 cm and the magnitude of ∠ABC = 58◦ . The equal edges AD, BE and CF are parallel and each of length 12 cm. Calculate a the length of AC b the length of AE c the magnitude of the angle between AE and EC. Example

21

5 A vertical tower, AT, of height 50 m, stands at a point A on a horizontal plane. The points A, B, and C lie on the same horizontal plane, B is due west of A and C is due south of A. The angles of elevation of the top, T, of the tower from B and C are 25◦ and 30◦ respectively. a Calculate, giving answers to the nearest metre, the distances i AB ii AC iii BC b Calculate the angle of elevation of T from the midpoint, M, of AB.



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6 A right square pyramid, vertex O, stands on a square base ABCD. The height is 15 cm and base side length is 10 cm. Find a the length of the slant edge b the inclination of a slant edge to the base c the inclination of a sloping face to the base d the magnitude of the angle between two adjacent sloping faces. 7 A post stands at one corner of a rectangular courtyard. The elevations of the top of the post from the nearest corners are 30◦ and 45◦ . Find the elevation from the diagonally opposite corner. 8 VABC is a regular tetrahedron with base ABC. (All faces are equilateral triangles.) Find the magnitude of the angle between a a sloping edge and the base

b adjacent sloping faces.

9 An observer at a point A at sea level notes an aircraft due east at an elevation of 35◦ . At the same time an observer at B, 2 km due south of A, reports the aircraft on a bearing of 50◦ . Calculate the altitude of the aircraft. Example

23

10 Four congruent spheres, radius length 10 cm, are placed on a horizontal table so that each touches two others and their centres form a square. A fifth congruent sphere rests on them. Find the height of the top of this fifth sphere above the table. A

11 ABFE represents a section of a ski run which has a uniform inclination of 30◦ to the horizontal.

B

D

AE = 100 m, AB = 100 m.

C

E

A skier traverses the slope from A to F. Calculate

F

a the distance that the skier has traversed b the inclination of the skier’s path to the horizontal. Example

22

12 A sphere of radius length 8 cm rests on the top of a hollow inverted cone of height 15 cm whose vertical angle is 60◦ . Find the height of the centre of the sphere above the vertex of the cone. 13 A cube has edge length a cm. What is the radius length, in terms of a, of a the sphere that just contains the cube

b the sphere that just fits inside the cube?

14 In this diagram AB is vertical and BCD is horizontal. ∠CBD is a right angle. AB = 20 m, BD = 40 m, BC = 30 m. Calculate the inclination to the horizontal of a AD b AE where AE is the line of greatest slope c AE where E is the midpoint of CD.

A

B C


D E


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Chapter summary The sine rule is used to find unknown quantities in a triangle when one of the following situations arises: r one side and two angles are given r two sides and the non-included angle are given. In the first of the two cases a unique triangle is defined but for the second it is possible for two triangles to exist. Labelling convention B The following convention is followed. Interior angles are a c denoted by upper case letters and the length of the side opposite an angle is denoted by the corresponding A C b lower case letter. e.g. The magnitude of angle BAC is denoted by A. The length of side BC is denoted by a. B The sine rule states that for a triangle ABC a c b c a = = sin A sin B sin C A C b

The cosine rule is used to find unknown quantities in a triangle when one of the following situations arises: r two sides and an included angle are given r three sides are given. B a The cosine rule states that for a triangle ABC c C

a 2 = b2 + c2 − 2bc cos A or equivalently b2 + c2 − a 2 cos A = 2bc The symmetrical results also hold, i.e.

A

b

b2 = a 2 + c2 − 2ac cos B c2 = a 2 + b2 − 2ab cos C It is known that the area of a triangle is given by the formula 1 bh 2 1 Area = × base length × height 2

B

Area =

c A

h b

a C

By observing that h = c sin A the following formula can be found: 1 bc sin A 2 i.e. the area is given by half the product of the length of two sides and the sine of the angle included between them. Area of triangle =



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A

l = r␪

r O θc

The area of sector AOB (shaded) is given by the formula

l

1 Area = r 2 ␪ 2 Chord length (red line) is given by

B

␪ 2 The area of a segment (shaded) is given by l = 2r sin

A r O θc

1 Area = r 2 (␪ − sin ␪) 2

B

A

Angle between planes Consider any point P on the common line of two planes 1 and 2 . If PA and PB are drawn at right angles to the common line so that PA is in 1 and PB is in 2 then angle APB is the angle between 1 and 2 . Note: If one of the planes, 2 say, is horizontal, then PA is called a line of greatest slope in the plane 1 .

Review

The length of the minor arc AB (red line) is given by the formula

Π1 P

θ

Π2

B A

Π1 Π2 P

lines of greatest slope

angle of greatest slope

Multiple-choice questions 1 In a triangle XYZ, x = 21 cm, y = 18 cm and ∠YXZ = 62◦ . The magnitude of ∠XYZ, correct to one decimal place, is B 0.8◦ C 1.0◦ D 49.2◦ E 53.1◦ A 0.4◦ 51 . The value of c, to the nearest whole 2 In a triangle ABC, a = 30, b = 21 and cos C = 53 number, is A 9 B 10 C 11 D 81 E 129 3 In a triangle ABC, a = 5.2 cm, b = 6.8 cm and c = 7.3 cm. The magnitude of ∠ACB, correct to the nearest degree, is B 63◦ C 74◦ D 82◦ E 98◦ A 43◦ 4 The area of the triangle ABC, where b = 5 cm, c = 3 cm, ∠ A = 30◦ and ∠ B = 70◦ , is B 3.75 cm2 C 6.50 cm2 D 7.50 cm2 E 8 cm2 A 2.75 cm2



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5 The length of the radius of the circle shown, correct to two decimal places, is A 5.52 cm B 8.36 cm C 9.01 cm 130∞ D 12.18 cm 10 cm E 18.13 cm 6 A chord of length 5 cm is drawn in a circle of radius 6 cm. The area of the smaller region inside the circle cut off by the chord, correct to one decimal place, is B 2.3 cm2 C 3.9 cm2 D 13.6 cm2 E 15.5 cm2 A 1.8 cm2 7 From a point on a cliff 500 m above sea level, the angle of depression to a boat is 20◦ . The distance from the foot of the cliff to the boat, to the nearest metre, is A 182 m B 193 m C 210 m D 1374 m E 1834 m 8 A tower 80 m high is 1.3 km away from a point on the ground. The angle of elevation to the top of the tower from this point, correct to the nearest degree, is B 4◦ C 53◦ D 86◦ E 89◦ A 1◦ 9 A man walks 5 km due east followed by 7 km due south. The bearing he must take to return to the start is B 306◦ C 324◦ D 332◦ E 348◦ A 036◦ 10 A boat sails at a bearing of 215◦ from A to B. The bearing it must take from B to return to A is B 055◦ C 090◦ D 215◦ E 250◦ A 035◦

Short-answer questions (technology-free) B

1 a Find x. b Find y.

10 cm

6 cm y∞

30∞

A

x cm

2 Find a AH, where AH is the altitude b CM, where CM is the median.

C A

30∞ 40 cm B

40 cm

C

◦

3 From a port P, a ship Q is 20 km away on a bearing of 112 , and a ship R is 12 km away on a bearing of 052◦ . Find the distance between the two ships. 4 In a quadrilateral ABCD, AB = 5 cm, BC = 5 cm, CD = 7 cm, B = 120◦ and C = 90◦ . Find a the length of the diagonal AC b the area of triangle ABC c the area of triangle ADC d the area of the quadrilateral.



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361

Review

5 If sin x = sin 37◦ and x is obtuse, find x. 6 A point T is 10 km due north of a point S, and a point R, which is east of a straight line joining T and S, is 8 km from T and 7 km from S. Calculate the cosine of the bearing of R from S. 7 In ABC, AB = 5 cm, magnitude of ∠BAC = 60◦ and AC = 6 cm. Calculate the sine of ∠ABC. 8 The area of a sector of a circle with radius 6 cm is 33 cm2 . Calculate the angle of the sector. 9 The diagram shows two survey points, A and B, which are on an east–west line on level ground. From point A, the bearing of the foot of a tower is 060◦ , while from B the bearing of the tower is 045◦ . Find a i the magnitude of ∠TAB ii the magnitude of ∠ √ATB √ 6− 2 b Given that sin 15◦ = , find 4 AT and BT.

N

60° A

T

N

45° 300 m

B

10 A boat sails 11 km from a harbour on a bearing of 220◦ . It then sails 15 km on a bearing of 340◦ . How far is the boat from the harbour? 11 A helicopter leaves a heliport A and flies 2.4 km on a bearing of 150◦ to a check point B. It then flies due east to its base C. a If the bearing of C from A is 120◦ , find the distances AC and BC. b The helicopter flies at a constant speed throughout and takes five minutes to fly from A to C. Find its speed. 12 The diagram shows a circle of radius length 13 cm and a chord AB of length 24 cm. Calculate a the length of arc ACB b the area of the shaded region.

C

A 24 13 O

B

13

13 A sector of a circle has an arc length of 30 cm. If the radius of the circle is 12 cm, find the area of the sector. 14 A chord PQ of a circle, radius 5 cm, subtends an angle of two radians at the centre of the circle. Taking ␲ to be 3.14, calculate, correct to one decimal place, the length of the major arc PQ. 15 From a cliff top 11 m above sea level, two boats are observed. One has an angle of depression of 45◦ and is due east, the other an angle of depression of 30◦ on a bearing of 120◦ . Calculate the distance between the boats.



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Extended-response questions A

1 AB is a tower 60 m high on top of a hill. The magnitude of ACO is 49◦ and the magnitude of BCO is 37◦ . a Find the magnitude of angles ACB, CBO and CBA. b Find the length of BC. c Find the height of the hill, i.e. the length of OB.

B

C O

2 The angle of a sector of a circle, centre O and radius length 12 cm has magnitude 2.5 radians. The sector is folded so that OA and OA are joined to form a cone. Calculate a the base radius length of the cone b the curved surface area of the cone c the shortest distance between two points diametrically opposed on the edge of the base. 3 A tower 110 m high stands on the top of a hill. From a point A at the foot of the hill the angle of elevation of the bottom of the tower is 7◦ , and that of the top is 10◦ . a Find the magnitude of angles TAB, ABT and ATB. b Use the sine rule to find the length of AB. c Find CB, the height of the hill.

O 2.5c A

A'

T 110 m B

A

C

10°

7°

B

4 Point S is a distance of 120 m from the base of a building. On the building is an aerial, AB. The angle of elevation from S to A is 57◦ . The angle of elevation from S to B is 59◦ . Find a the distance OA b the distance OB c the distance AB.

A

59° 57° S

5 From the top of a communications tower, the angles of depression of two points A and B on a horizontal line through the foot of the tower are 30◦ and 40◦ . The distance between the points is 100 m. Find A B a the distance AT 100 m b the distance BT c the height of the tower. 6 Angles VBC, VBA and ABC are right angles. Find a the distance VA b the distance VC c the distance AC d the magnitude of angle VCA.

base of tower

V

8 cm

8 cm B A


O 120 m T top of tower 40° 30°

6 cm C


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C H A P T E R

13

Revision


Multiple-choice questions 1 A ladder 2.6 m long rests with one end on horizontal ground while the other end rests against a vertical wall at a point which is 2.1 m from the ground. The angle between the ladder and the wall, to the nearest degree, is ◦

◦

B 39

A 36

◦

C 51

◦

D 54

B 3 E 2␲

2.1 m

◦

E 63

y

2 The graph shown has amplitude A 2 D 6

2.6 m

C 4

2 0

x 2π

3 Which one of the following equations gives the –4 correct value for c? ◦ ◦ 58 cos 38 58 sin 38 A c= B c= A ◦ cos 130 sin 130◦ ◦ 130° 58 cos 130 C c = 58 sin 38◦ D c= 38° 12° ◦ cos 38 B C 58 cm 58 sin 130◦ E c= sin 38◦ 4 A map is drawn so that a wall 17.1 m long is represented by a line 45 mm long. The scale is A 1 : 3.8

B 1 : 38

C 1 : 380

D 1 : 3800

E 1 : 38000

5 The point (5, −2) is reflected in the line y = x. The coordinates of its image are A (5, −2) 6 If sin A = A

140 221

B (−5, 2)

C (2, −5)

D (−2, 5)

E (−5, −2)

8 5 , sin B = where A and B are acute, then sin (A − B) is given by 13 17 107 −107 34 209 −21 E D C B 140 140 23 560 221



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7 In triangle ABC, c = 5, b = 9 and A = 43◦ . Which of the following statements are correct? i With the information we can find the area of triangle ABC. ii With the information given we can find angle B. iii With the information given we can find side a. A i and ii only D i, ii and iii

B i and iii only E none of these

C ii and iii only

8 In the figure, AB = 15, CD = 5, BF = 6, GD = 6, EG = 9. x is equal to A 3 D 4.75

B 4 E 5

B

6

F 6

15

C 4.5

E

A

9 Gx

D 5

C

9 The point (2, −6) is reflected in the line y = −x. The coordinates of its image are A (2, −6)

B (−2, 6)

C (6, −2)

E (−2, −6)

D (−6, 2) y

10 The graph shown is best described by A y = sin (a) C y = sin (a) + 1 E y = cos (a) + 1

B y = 2 cos (a) D y = cos (2a)

2 1 0

π 2

π

a 3π 2

2π

8 5 , sin B = where A and B are acute, then tan (A + B) is given by 13 17 171 −171 34 209 −21 140 E D C B A 140 140 23 560 221 221 2 12 P is the point (5, −4). After translation by and reflection in the line y = 1, the −3 coordinates of the image of P are

11 If sin A =

A (7, 7)

B (7, 9)

C (−5, −7)

D (7, 11)

E (7, 10)

13 A model car is 8 cm long and the real car is 3.2 m long. The scale factor is A 1:8 B 1 : 32 C 1 : 24 D 1 : 400 E 1 : 40 √ ␲ 14 If 2 sin x − = 3 and 0 ≤ x ≤ 2␲, then x is equal to 6 ␲ 5␲ ␲ 5␲ C ␲ or ␲ ␲ 2␲ ␲ or ␲ E or D or B or A 3 2 6 6 2 6 6 3 3 15 Which one of the following expressions will give the area of triangle ABC? 1 1 7 cm A × 6 × 7 sin 48◦ A × 6 × 7 cos 48◦ B 2 2 48° 1 1 × 6 × 7 sin 52◦ C × 6 × 7 cos 52◦ D 6 cm 2 2 1 C × 6 × 7 tan 48◦ E 2


52°

B


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365

r range is [−4, 2]

r y = −1 when a = 0

This graph would be described by the equation A y = 3 sin (a)◦ + 1 D y = 3 sin (3a)◦ + 1

B y = 3 cos (3a)◦ − 1 E y = cos (3a)◦ − 2

C y = −3 sin (3a)◦ − 1

Revision

16 Given that cos ␪ = c and that ␪ is acute, cot ␪ can be expressed in terms of c as √ √ √ c 1 E 2c 1 − c2 D √ C √ A c 1 − c2 B 1 − c2 2 2 1−c 1−c 17 A trigonometric graph has the following characteristics: r period of 120◦ r amplitude is 3

18 The point (a, b) is reflected in the line with equation x = m. The image point has coordinates A (2m − a, b) D (a, b − m)

B (a, 2m − b) E (2m + a, b)

C (a − m, b)

19 A child on a swing travels through an arc of length 3 m. If the ropes of the swing are 4 m in length, the angle which the arc makes at the top of the swing (where the swing is attached to the support) is best approximated by A 135◦

B 75◦

E 42◦ 58 1 ␪ has 20 Compared with the graph of y = sin ␪, the graph of y = sin 2 A the same amplitude but double the period B the same amplitude but half the period C double the amplitude but the same period D half the amplitude but the same period 1 E the same amplitude but shifted a unit to the left. 2 1 21 The image of the line {(x, y) : x + y = 4} after a dilation of factor from the y axis 2 followed by a reflection in x = 4 is A {(x, y) : y = 2x} D {(x, y) : x + y = 0}

C 12◦

D 75c

B {(x, y) : y + 2 = 0} C {(x, y) : y + 2x − 16 = 0} E {(x, y) : y = 2x − 12}

␲ , the value of cos A cos B − sin A sin B is 2 A −2 B 1 C −1 D 0 E 2 √ 5 23 Given that sin A = and that A is obtuse, the value of sin 2A is √ 3 √ √ 4 5 5 5 8 1 16 5 E − D C − B − A 9 9 27 9 243 24 A ladder rests against a wall, touching the wall at a height of 5.6 m. The bottom of the ladder is 2 m from the wall. The distance (to the nearest centimetre) that a person, of height 1.6 m, must be from the wall to just fit under the ladder is 22 If A + B =

A 1.43 m

B 0.57 m

C 1.75 m


D 0.25 m

E 1.2 m


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Essential Advanced General Mathematics y 25 A possible equation of the graph shown is ␲ ␲ 1 A y = sin 2 x − B y = cos 2 x − 12 12 ␲ ␲ C y = sin 2 x + D y = cos 2 x + x 12 12 0 π π π 3π ␲ 12 4 2 4 E y = − sin 2 x − 12 26 Let ABC and DEF be similar triangles such that AB = 4 cm and DE = 10 cm. If the area of ABC is 24 cm2 , then the area of DEF, in cm2 , is

A 60

B 240

C 150

D 96

E none of these

27 Which of the following statements is true for f (x) = −2 tan (3x)◦ ? i The period is 60. ii The amplitude is 2. iii The period is 120. iv The graph is a reflection of the graph of h(x) = 2 tan (3x)◦ in the x axis. v The graph is a reflection of g(x) = tan (x)◦ in the y axis. A i and iv only D ii and iv only

B i, ii and iv only E iii and iv only

C i, iv and v only

3 followed 28 The image of {(x, y) : y = x } under a translation determined by the vector 2 by a reflection in the x axis is 2

A {(x, y) : y = (x − 3)2 + 2} C {(x, y) : y = (x + 3)2 + 2}

B {(x, y) : −(x − 3)2 = y + 2} D {(x, y) : −y + 2 = (x − 3)2 }

E none of these

29 The area, in cm2 correct to two decimal places, of a sector with included angle of 60◦ in a circle of diameter 10 cm is A 104.72 cm2 B 52.36 cm2

C 13.09 cm2

30 KLMN is a parallelogram and OQ is parallel to KL. If O divides KN in the ratio of 1 : 2, areaKOP is equal to the ratio areaKLMN 1 1 1 C B A 12 9 4

D 26.16 cm2

E 750 cm L

K P

O

Q M

N

D

1 18

E

31 VABCD is a right, square pyramid with base length 80 mm and perpendicular height 100 mm. The angle between a sloping face and the base ABCD, to the nearest degree, is A 22◦ D 61◦

B 29◦ E 68◦

1 20 V

C 51◦

D

E

O A

C

θ B

32 Given that cos ␪ = c and that ␪ is acute, sin 2␪ can be expressed in terms of c as √ √ √ 1 c C √ A c 1 − c2 D √ B 1 − c2 E 2c 1 − c2 1 − c2 1 − c2



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34 The angles between 0◦ and 360◦ which satisfy the equation 4 cos x − 3 sin x = 1, given correct to two decimal places, are A 53.13◦ and 126.87◦ D 131.59◦ and 334.67◦

B 48.41◦ and 205.33◦ E 154.67◦ and 311.59◦

C 41.59◦ and 244.67◦

35 In the figure, the volume of the shaded solid B is 49 cm3 . The volume of A is A 19.5 cm3 D 12.5 cm3

B 17.3 cm3 E 10.5 cm3

Revision

33 The image of {(x, y) : y = 2x } after a dilation of factor 2 from the x axis followed by a 1 dilation of factor from the y axis is 3 x 1 B y = 3 × 22 C y = 2 × 23x A y = × 23x 3 x D y = 2 × 23 E none of these

3 cm

C 13.5 cm3

A 2 cm

B

36 The area of the shaded region in the diagram, in cm2 (to the nearest cm2 ), is A 951 D 2895

B 992 E 110 424

110°

C 1944

45 cm

37 The expression 8 sin ␪ cos3 ␪ − 8 sin3 ␪ cos ␪ is equal to A 8 sin ␪ cos ␪

B sin 8␪

C 2 sin 4␪

D 4 cos 2␪

E 2 sin 2␪ cos 2␪

y 38 A possible equation for the graph shown is 1 ␲ A y = tan x− +3 4 3 2 2π 1 ␲ x B y = tan x+ −3 –π π 3π 2 4 2 –3 2 1 ␲ C y = 3 tan x− 4 2 1 ␲ D y = 3 tan x+ E y = tan 3x 2 4 39 If the ratio volume of the hemisphere : volume of the right circular cone equals 27 : 4 where r is the radius of the base of the cone and R is the radius of the hemisphere, then R : r is equal to R r r

A 1:2

B 2:3

C 3:

√ 2

D 27 : 8


E 3:2 Cambridge University Press

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2 40 Let T be the translation determined by the vector and S the transformation, reflection 3 in the line with equation x = 2. The rule for the composition TS is given by A TS(x, y) = (2 − x, y + 3) C TS(x, y) = (x + 2, y + 3)

B TS(x, y) = (−x, y + 3) D TS(x, y) = (6 − x, y + 3)

E none of these y

41 The square shown is subject to successive transformations. −1 0 The first transformation has matrix and the second 0 1 0 −1 transformation has matrix . −2 1

1

(1, 1)

0

x

1

Which one of the following shows the image of the square after these two transformations? y

A

y

B 3

2

1

3

x

0 –1

2

3 2

1

1 x

–1 0 y

D

2

3 x

2

1

1

–1 –2

1

y

E

0

x

–1 0

2

1

1 –1

y

C

–1 0

x 1

13.2 Extended-response questions 1 a Find the rule of the transformation which maps triangle ABC to triangle A B C . b On graph paper, draw triangle ABC and its image under reflection in the x axis. The C coordinates of A, B and C are (−4, 1), (−2, 1) and (−2, 5) respectively. c On the same set of axes draw the image of ABC under a dilation of factor 2 from A B the y axis. –2 –4 d Find the image of the parabola y = x 2 under a dilation of factor 2 from the x axis −3 followed by a translation defined by the vector . 2

y C'

8

5 4

A'

B'

1 0

x 2

4

(cont’d.) ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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369

2 In ABC, A = 30◦ , a = 60 (i.e., for the diagram BC = BC = 60) and c = 80.

B 80

a Find the magnitudes of angles 60 60 i ABC and BCA ii ABC and BC A 30° b Find the length of line segment A C' C iii CC i AC ii AC c i Show that the magnitude of ∠CBC is 96.38◦ (correct to two decimal places). Using this value, iii the area of the shaded sector ii find the area of triangle BCC iv the area of the shaded segment.

Revision

e Find the rule for the transformation which maps the graph of y = x 2 to y = −2(x − 3)2 + 4 f If f (x) = x 3 − 2x use a graphics calculator to help sketch the graph of y = 3 f (x − 2) + 4

3 a Find the image of the point (1, 1) under a dilation D, of factor 4 from the y axis. b i Describe the image of the square with vertices A(0, 0), B(0, 1), C(1, 1), E(1, 0) under the dilation D. ii Find the area of the square ABCE. iii Find the area of the region defined by the image of ABCE. iv If the dilation had been of factor k, what would the area of this region be? c State the rule for the dilation. d i Find the equation of the image of the curve with equation y = x 2 under the dilation D. ii Find the equation of the image of the curve with equation y = x 2under the 2 dilation D followed by the translation defined by the vector . −1 iii Sketch the graph of y = x 2 and of its image defined in ii on the one set of axes. State the coordinates of the vertex and of the axes intercepts. e State the rule for the transformation which maps the curve with equation y = 5(x + 2)2 − 3 to the curve with equation y = x 2 . ⎤ ⎡ 3 4 ⎢ 5 5⎥ ⎥ 4 A transformation is represented by the matrix M = ⎢ ⎣ 4 3⎦ − 5 5 a Describe the transformation. b Let C be the circle which passes through the origin and which has as its centre the point (0, 1). i Find the equation of C. ii Find the equation of C , the image of C under the transformation determined by M. c Find the coordinates of the points of intersection of C and C .



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4 5 A transformation is defined by the matrix M = 2

1 . 3

a Find the image of (−2, 5) under this transformation. b Find the inverse of M. c Given that the point (11, 13) is the image of the point (a, b), find the values of a and b. d Find in terms of a, the image of the point (a, a). a ␭a e If M = , find the possible values of ␭ and the relationship between b ␭b a and b in each of these cases. ␲ in an anticlockwise direction. 4 4 Give the 2 × 2 matrix associated with this transformation. Find the inverse of this matrix. If the image of (a, b) is (1, 1), find the value of a and b. If the image of (c, d) is (1, 2), find the value of c and d. i If (x, y) → (x , y ) under this transformation, use the result of b to find x and y in terms of x and y . ii Find the image of y = x 2 under this transformation.

6 Let R ␲ be the matrix of the transformation, rotation of a b c d e

7 A particle oscillates along a straight line. Its displacement x (m), at time t(s), from a point ␲ t . O is given by x = 5 + 3 sin 6 a Find its displacement at time i t =0 ii t = 3 b Sketch the graph of x against t for t ∈ [0, 24], labelling clearly all turning points. c State i the maximum distance of the particle from O ii the minimum distance of the particle from O. d At what times (t ∈ [0, 24]) is the particle i 5 m from O ii 6 m from O, correct to two decimal places? 8 A logo for a Victorian team is as shown here. O is the centre of the circle and A, B and C are points on the circle. OC = OA = OB = 10 cm. i Convert 30◦ to radians. ii Find the length of the minor arc AB. b The magnitude of ∠AOC is 167◦ and the magnitude of ∠BOC is 163◦ . Find the length of chord BC, correct to two decimal places.

B

A

30°

a

O 167°

163°

C

c Find, correct to two decimal places, i the area of triangle BOC ii the area of triangle AOC iii the shaded area of the logo.



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Chapter 13 — Revision of chapters 8–12 L

a Prove PRQ ∼ LMN. b State the scale factor. X R P a units c Find the area of triangle PQR in terms of a. Z Y X, Y and Z are the midpoints of PR, PQ and RQ respectively. N M Q a units XYZ is similar to LMN. d State the scale factor. e Find the area of triangle XYZ in terms of a. f Let A1 be the area of triangle LMN. Let A2 be the area of triangle PQR. Let A3 be the area of triangle XYZ. The process of forming triangles by joining midpoints of the previous triangle is continued to form a sequence of triangles, 1 , 2 , 3 , . . . , n , . . . and associated areas A1 , A2 , A3 , . . . , An , . . . i Find An in terms of a and n. ii Find in terms of a, the sum to infinity of the series A1 + A2 + · · · + An + · · ·

Revision

9 Triangle LMN is an isosceles, right-angled triangle. P, Q and R are midpoints of LM, MN and LN respectively.

371

1 10 It is known that y varies partly as x and partly as 2 ; i.e. there exist constants k1 and k2 x k2 such that y = k1 x + 2 . x a When x = −1, y = 1 and when x = 1, y = 5. Find the values of k1 and k2 . b The graph of y against x is as shown. y i Sketch the graph of the image of k2 k2 y = k1x + 2 y = k1 x + 2 under the transformation x x determined by reflection in the x axis followed by a translation determined 3 by the vector . 0 (e, f ) (The answers to parts ii, iii and iv (c, 0) below may help you answer this.) x 0 ii Find the value of c and hence the x axis intercept of the image. y = k1x iii The image of the point with coordinates (e, f ) is (e , f ). Find e and f in terms of e and f. k2 iv Find the equation of the image of the curve with equation y = k1 x + 2 under x this transformation. 11 Let M be the transformation ‘reflection in the line y = x . a

i Find the coordinates of the image of the point A (1, 3) under this transformation.



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ii Find the coordinates of the triangle which is the image of the triangle with vertices A (1, 3), B (1, 5), C (3, 3). iii Illustrate triangle ABC and its image on a set of axes scaled from −5 to 5 on both axes. b i Show that the equation of the image of the curve with equation y = x 2 − 2 under the transformation M is x = y 2 − 2. ii Find the coordinates of the points of intersection of the curve y = x 2 − 2 with the line y = x. iii Show that the x coordinates of the points of intersection of y = x 2 − 2 and its image may be determined by the equation x 4 − 4x 2 − x + 2 = 0. √ 1 iv Two solutions of the equation x 4 − 4x 2 − x + 2 = 0 are x = (−1 + 5) and 2 √ 1 x = (−1 − 5). 2 Use this result and the result of b ii to find the coordinates of the points of intersection of y = x 2 − 2 and its image under M. 12 In the figure, AE = BE = BD = 1 unit. ∠BCD is a right angle.

D E

1 a Show that the magnitude of ∠BDE is 2␪. 1 1 b Use the cosine rule in triangle BDE to show that θ 3θ A C DE = 2 cos 2␪. B c Show that i DC = sin 3␪ sin 3␪ ii AD = sin ␪ d Use the results of b and c to show sin 3␪ = 3 sin ␪ − 4 sin3 ␪ 13 a Adam notices a distinctive tree while orienteering on a flat horizontal plane. He discovers that the tree is 200 m from where he is standing on a bearing of 050◦ . Two other people, Brian and Colin, who are both standing due east of Adam, claim the tree is 150 m away from them. Given that their claim is true and that Brian and Colin are not standing in the same place, how far apart are they? Give your answer to the nearest metre. T b From the top of a vertical tower of height 10 m, standing in the corner of a rectangular courtyard, the angles of depression to the nearest corners A D (B and D) are 32◦ and 19◦ respectively. Find B i AB, correct to two decimal places C ii AD, correct to two decimal places iii the angle of depression of corner C diagonally opposite the tower from T, correct to the nearest degree. c Two circles, each of radius length 10 cm, have their centres 16 cm apart. Calculate the area common to both circles, correct to one decimal place.



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Chapter 13 — Revision of chapters 8–12 S satellite 30°

direction of travel

8000 km

a If the tracking station antenna is aimed O at 30◦ above the horizon, at what time will the satellite pass through the beam of the antenna? b Find the distance between the satellite and the tracking station at 12.06 p.m. c At what angle above the horizon should the antenna be pointed so that its beam will intercept the satellite at 12.06 p.m.? 15 An athlete in a gymnasium is training on an exercise bike. Pedal direction of At time t = 0, the position of the pedal is as shown. 60° movement 25 cm The height of the pedal, h cm, from the floor at time t seconds, is given by h(t) = a + b cos (␲(t + c)) 35 cm where a and b are in centimetres. a Find the values of a, b and c. Floor b i Find the times at which the height of the pedal above the floor is 60 cm, 0 ≤ t ≤ 4. ii Find the times at which the height of the pedal above the floor is 37.5 cm, 0 ≤ t ≤ 4. c Sketch the graph of h against t for 0 ≤ t ≤ 4. A D 16 ABCD is a parallelogram whose diagonals ◦ intersect at angle ␪ at the point E. θ Let AB = CD = x, AD = BC = y, BD = p, E AC = q. B

Revision

14 A satellite travelling in a circular orbit 1600 km above the Earth is due to pass directly over a tracking station at noon. Assume that the T satellite takes two hours to make an orbit and 6400 km the radius of the Earth is 6400 km.

373

C

a Apply the cosine rule to triangle DEC to find x in terms of p, q and ␪. b Apply the cosine rule to triangle DEA to find y in terms of p, q and ␪. c Use the results of a and b to show that 2(x 2 + y 2 ) = p 2 + q 2 d A parallelogram has sides 8 cm and 6 cm and one diagonal of 13 cm. Find the length of the other diagonal. X 17 The figure shows the circular cross section of a uniform log of radius 40 cm floating in water. The 8 cm A B points A and B are on the surface of the water and the highest point X is 8 cm above the surface. a Show that the magnitude of ∠AOB is approximately 1.29 radians. b Calculate i the length of arc AXB


40 cm

O


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ii the area of the cross section below the surface iii the percentage of the volume of the log below the surface. 18 In ABC, AB = 7 cm , BC = 9 cm and ∠ABC = ␪. a Show that AC2 = 130 − 126 cos ␪. D is the point on the opposite side of AC from B such that ABCD is a cyclic quadrilateral in which CD = 6 cm and DA = 5 cm. 23 . b Obtain another expression for AC2 in terms of ␪ and prove that cos ␪ = 62 c Calculate i the length of AC ii the area of quadrilateral ABCD.



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C H A P T E R

14 Circle theorems Objectives To establish the following results and use them to prove further properties and solve problems:

r The angle subtended at the circumference is half the angle at the centre subtended by the same arc

r Angles in the same segment of a circle are equal r A tangent to a circle is perpendicular to the radius drawn from the point of contact r The two tangents drawn from an external point to a circle are the same length r The angle between a tangent and a chord drawn from the point of contact is equal to any angle in the alternate segment

r A quadrilateral is cyclic (that is, the four vertices lie on a circle) if and only if the sum of each pair of opposite angles is two right angles

r If AB and CD are two chords of a circle which cut at a point P (which may be inside or outside a circle) then PA · PB = PC · PD

r If P is a point outside a circle and T, A, B are points on the circle such that PT is 2

a tangent and PAB is a secant then PT = PA · PB

These theorems and related results can be investigated through a geometry package such as Cabri Geometry. It is assumed in this chapter that the student is familiar with basic properties of parallel lines and triangles.

14.1

Angle properties of the circle

P x° O

Theorem 1 The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.

2x° A

B



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Proof Join points P and O and extend the line through O as shown in the diagram. Note that AO = BO = PO = r the radius of the circle. Therefore triangles PAO and PBO are isosceles. Let ∠APO = ∠PAO = a ◦ and ∠BPO = ∠PBO = b◦ Then angle AOX is 2a ◦ (exterior angle of a triangle) and angle BOX is 2b◦ (exterior angle of a triangle)

P a°

b° r

a°

O

A

b°

r

r

B

X

∠AOB = 2a ◦ + 2b◦ = 2(a + b)◦ = 2∠APB Note: In the proof presented above, the centre and point P are considered to be on the same side of chord AB. The proof is not dependent on this and the result always holds. The converse of this result also holds: i.e., if A and B are points on a circle with centre O and angle APB is equal to half angle AOB, then P lies on the circle. E A segment of a circle is the part of the plane bounded by an arc and its chord. Arc AEB and chord AB define a major segment which is shaded. B Arc AFB and chord AB define a minor segment which is not A shaded. F ∴

∠AEB is said to be an angle in segment AEB.

E

O B

A

Theorem 2 Angles in the same segment of a circle are equal. Proof Let ∠AXB = x ◦ and ∠AYB = y ◦ Then by Theorem 1 ∠AOB = 2x ◦ = 2y ◦ Therefore x = y

Y X

y° x°

O

B

A

Theorem 3 The angle subtended by a diameter at the circumference is equal to a right angle (90◦ ). Proof The angle subtended at the centre is 180◦ . Theorem 1 gives the result.


E

A

O

B


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377

Chapter 14 — Circle theorems

A quadrilateral which can be inscribed in a circle is called a cyclic quadrilateral.

Theorem 4 The opposite angles of a quadrilateral inscribed in a circle sum to two right angles (180◦ ). (The opposite angles of a cyclic quadrilateral are supplementary). The converse of this result also holds. Proof B O is the centre of the circle b° A By Theorem 1 y Also x+y Therefore 2b + 2d i.e. b+d

= 2b and x = 2d = 360 = 360 = 180

C

x° y° O d° D

The converse states: if a quadrilateral has opposite angles supplementary then the quadrilateral is inscribable in a circle. Example 1 y°

z°

Find the value of each of the pronumerals in the diagram. O is the centre of the circle and ∠AOB = 100◦ .

O 100°

Solution A Theorem 1 gives that z = y = 50 The value of x can be found by observing either of the following. Reflex angle AOB is 260◦ . Therefore x = 130 (Theorem 1) or y + x = 180 (Theorem 4) Therefore x = 180 − 50 = 130

B

x°

Example 2 A, B, C, D are points on a circle. The diagonals of quadrilateral ABCD meet at X. Prove that triangles ADX and BCX are similar. Solution

∠DAC and ∠DBC are in the same segment. Therefore m = n ∠BDA and ∠BCA are in the same segment. Therefore p = q Also ∠AXD = ∠BXC (vertically opposite). Therefore triangles ADX and BCX are equiangular and thus similar.


B A

n° m° X q° D

p° C


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Example 3

A

An isosceles triangle is inscribed in a circle. Find the angles in the three minor segments of the circle cut off by the sides of this triangle.

32° O 74°

74°

B

Solution

C A

First, to determine the magnitude of ∠AXC cyclic quadrilateral AXCB is formed. Thus ∠AXC and ∠ABC are supplementary. Therefore ∠AXC = 106◦ . All angles in the minor segment formed by AC will have this magnitude.

X O 74° B

C

In a similar fashion it can be shown that the angles in the minor segment formed by AB all have magnitude 106◦ and for the minor segment formed by BC the angles all have magnitude 148◦ .


1

1 Find the values of the pronumerals for each of the following, where O denotes the centre of the given circle. a

c

b y°

50°

z° y°

O x°

z° O

O 35°

108°

y°

x°

e

d

f

3x°

y°

25°

y° O z°

O x°

125° y°

x° O

x°

2 Find the value of the pronumerals for each of the following. a

b

c

59°

x°

130° y°

y°

112°

70° 93°

y° x° ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

x° 68° Cambridge University Press

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A

3 An isosceles triangle ABC is inscribed in a circle. What are the angles in the three minor segments cut off by the sides of the triangle?

40°

B

C ◦

4 ABCDE is a pentagon inscribed in a circle. If AE = DE and ∠BDC = 20 , ∠CAD = 28◦ and ∠ABD = 70◦ , find all of the interior angles of the pentagon. Example

2

5 If two opposite sides of a cyclic quadrilateral are equal, prove that the other two sides are parallel. 6 ABCD is a parallelogram. The circle through A, B and C cuts CD (produced if necessary) at E. Prove that AE = AD.

Example

3

7 ABCD is a cyclic quadrilateral and O is the centre of the circle through A, B, C and D. If ∠AOC = 120◦ , find the magnitude of ∠ADC. 8 Prove that if a parallelogram is inscribed in a circle it must be a rectangle. 9 Prove that the bisectors of the four interior angles of a quadrilateral form a cyclic quadrilateral.

14.2

Tangents Line PC is called a secant and line segment AB a chord. If the secant is rotated with P as the pivot point a sequence of pairs of points on the circle is defined. As PQ moves towards the edge of the circle the points of the pairs become closer until they eventually coincide. When PQ is in this final position (i.e., where the intersection points A and B collide) it is called a tangent to the circle. PQ touches the circle. The point at which the tangent touches the circle is called the point of contact. The length of a tangent from a point P P outside the tangent is the distance between P and the point of contact.

B

A

P

C

Q B1 B2

A1 A2 A3

Q

B3

Q

A4 A5

B5

B4

Q

Q

Theorem 5 A tangent to a circle is perpendicular to the radius drawn to the point of contact.



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Proof Let T be the point of contact of tangent PQ. O P Let S be the point on PQ, not T, such that OSP is a right angle. Triangle OST has a right angle at S. Therefore OT > OS as OT is the hypotenuse of triangle OTS. T S Q ∴ S is inside the circle as OT is a radius. ∴ The line through T and S must cut the circle again. But PQ is a tangent. A contradiction. Therefore T = S and angle OTP is a right angle.

Theorem 6 The two tangents drawn from an external point to a circle are of the same length. Proof P Triangle XPO is congruent to triangle XQO as XO is r a common side. X

∠XPO = ∠XQO = 90◦ OP = OQ (radii) Therefore XP = XQ

O r Q

Alternate segment theorem The shaded segment is called the alternate segment in relation to ∠STQ. The unshaded segment is alternate to ∠PTS

S

P

Q

T

Theorem 7 The angle between a tangent and a chord drawn from the point of contact is equal to any angle in the alternate segment. Proof Let ∠STQ = x ◦ , ∠RTS = y ◦ and ∠TRS = z ◦ where RT is X a diameter. Then ∠RST = 90◦ (Theorem 3, angle subtended by a diameter) Also ∠RTQ = 90◦ (Theorem 5, tangent is perpendicular to radius) P Hence x + y = 90 and y + z = 90 Therefore x = z But ∠TXS is in the same segment as ∠TRS and therefore ∠TXS = x ◦

z° O

S y°

x° Q

T

T

Example 4 Find the magnitude of the angles x and y in the diagram.

R

Q x°

30°

P

y° S ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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381

Solution Triangle PTS is isosceles (Theorem 6, two tangents from the same point) and therefore ∠PTS = ∠PST Hence y = 75. The alternate segment theorem gives that x = y = 75 Example 5 T

Find the values of x and y. PT is tangent to the circle centre O

y° x° P

Solution

O

x = 30 as the angle at the circumference is half the angle subtended at the centre and y = 60 as ∠OTP is a right angle.

60°

Example 6 Y

The tangents to a circle at F and G meet at H. If a chord FK is drawn parallel to HG, prove that triangle FGK is isosceles.

F

K

Solution Let ∠XGK = y ◦ Then ∠GFK = y ◦ (alternate segment theorem) and ∠GKF = y ◦ (alternate angles) Therefore triangle FGK is isosceles with FG = KG

y°

H

X

G


4

1 Find the value of the pronumerals for each of the following. T is the point of contact of the tangent and O the centre of the circle. a

x°

c

b y°

C z°

O

y°

33°

B

x° 73°

81°

q°

T

d

e

w°

Q z°


x°

T BC = BT T

w°

x° 80° 40° y°

74°

T

S and T are points of z°

54°

y° x° S

P contact of tangents

from P. TP is parallel to QS


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2 A triangle ABC is inscribed in a circle, and the tangent at C to the circle is parallel to the bisector of angle ABC. a Find the magnitude of ∠BCX. b Find the magnitude of ∠CBD, where D is the point of intersection of the bisector of angle ABC with AC. c Find the magnitude of ∠ABC.

3 AT is a tangent at A and TBC is a secant to the circle. Given ∠CTA = 30◦ , ∠CAT = 110◦ , find the magnitude of angles ACB, ABC and BAT.

B A

X

40° D C

Y

C

B

T

A

4 If AB and AC are two tangents to a circle and ∠BAC = 116◦ , find the magnitudes of the angles in the two segments into which BC divides the circle. Example

6

5 From a point A outside a circle, a secant ABC is drawn cutting the circle at B and C, and a tangent AD touching it at D. A chord DE is drawn equal in length to chord DB. Prove that triangles ABD and CDE are similar. 6 AB is a chord of a circle and CT, the tangent at C, is parallel to AB. Prove that CA = CB. 7 Through a point T, a tangent TA and a secant TPQ are drawn to a circle AQP. If the chord AB is drawn parallel to PQ, prove that the triangles PAT and BAQ are similar. 8 PQ is a diameter of a circle and AB is a perpendicular chord cutting it at N. Prove that PN is equal in length to the perpendicular from P on to the tangent at A.

14.3 Chords in circles Theorem 8 If AB and CD are two chords which cut at a point P (which may be inside or outside the circle) then PA · PB = PC · PD. Proof CASE 1 (The intersection point is inside the circle.) Consider triangles APC and BPD. ∠APC = ∠BPD (vertically opposite) ∠CDB = ∠CAB (angles in the same segment) ∠ACD = ∠DBA (angles in the same segment) Therefore triangle CAP is similar to triangle BDP. Therefore CP AP = PD PB and AP · PB = CP · PD, which can be written PA · PB = PC · PD


A

D P C B


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CASE 2 (The intersection point is outside the circle.) Show triangle APD is similar to triangle CPB Hence PD AP = CP PB

A B C

P

D

AP · PB = PD.CP

i.e.

which can be written PA · PB = PC · PD

Theorem 9 If P is a point outside a circle and T, A, B are points on the circle such that PT is a tangent and PAB is a secant then PT 2 = PA · PB T Proof

∠PTA = ∠TBA (alternate segment theorem) ∠PTB = ∠TAP (angle sum of a triangle) Therefore triangle PTB is similar to triangle PAT

P A B

PB PT = which implies PT 2 = PA · PB PA PT

∴

Example 7 The arch of a bridge is to be in the form of an arc of a circle. The span of the bridge is to be 25 m and the height in the middle 2 m. Find the radius of the circle. Solution

R

By Theorem 8 RP · PQ = MP · PN

M

12.5 m

2m P 12.5 m

Therefore

N

O

2PQ = 12.52

Also

12.52 2 PQ = 2r − 2 where r is the radius of the circle.

Hence

2r − 2 =

∴

and

PQ =

Q

12.52 2 1 12.52 r= +2 2 2 =

641 m 16



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Example 8 If r is the radius of a circle, with center O, and if A is any point inside the circle, show that the product CA · AD = r 2 − OA2 , where CD is a chord through A. Solution Let PQ be a diameter through A Theorem 8 gives that

Also ∴

CA · AD = QA · AP QA = r − OA and PA = r + OA CA · AD = r 2 − OA2

C

P O A

Q

D


7

1 Two chords AB and CD intersect at a point P within a circle. Given that a AP = 5 cm, PB = 4 cm, CP = 2 cm, find PD b AP = 4 cm, CP = 3 cm, PD = 8 cm, find PB. 2 If AB is a chord and P is a point on AB such that AP = 8 cm, PB = 5 cm and P is 3 cm from the centre of the circle, find the radius. 3 If AB is a chord of a circle with centre O and P is a point on AB such that BP = 4PA, OP = 5 cm and the radius of the circle is 7 cm, find AB.

Example

8

4 Two circles intersect at A and B and, from any point P on AB produced tangents PQ and PR are drawn to the circles. Prove that PQ = PR. 5 PQ is a variable chord of the smaller of two fixed concentric circles. PQ produced meets the circumference of the larger circle at R. Prove that the product RP.RQ is constant for all positions and lengths of PQ. 6 ABC is an isosceles triangle with AB = AC. A line through A meets BC at D and the circumcircle of the triangle at E. Prove that AB2 = AD · AE.



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Review

Chapter summary The angle subtended at the circumference is half the angle at the centre subtended by the same arc.

385

θ° O 2θ° A

B

Angles in the same segment of a circle are equal.

θ°

θ°

θ°

O B

A

A tangent to a circle is perpendicular to the radius drawn from the point of contact.

P

O

T T

The two tangents drawn from an external point are the same length i.e. PT = PT . O

P

T'

The angle between a tangent and a chord drawn from the point of contact is equal to any angle in the alternate segment.

A

θ°

θ° B

A quadrilateral is cyclic if and only if the sum of each pair of opposite angles is two right angles. If AB and CD are two chords of a circle which cut at a point P then PA · PB = PC · PD. A

B

C

B P

D

P D

A ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

C Cambridge University Press

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Multiple-choice questions 1 In the diagram A, B, C and D are points on the circumference of a circle. ∠ABC = 115◦ , ∠BAD = 70◦ and AB = AD. The magnitude of ∠ACD is E 50◦ D 70◦ C 40◦ B 55◦ A 45◦

B C

115° A

70°

D

2 In the diagram, PA and PB are tangents to the circle centre O. Given that Q is a point on the minor arc AB and that ∠AOB = 150◦ the magnitudes of ∠APB and ∠AQB are A ∠APB = 30◦ and ∠AQB = 105◦ A B ∠APB = 40◦ and ∠AQB = 110◦ P Q C ∠APB = 25◦ and ∠AQB = 105◦ O 150° D ∠APB = 30◦ and ∠AQB = 110◦ E ∠APB = 25◦ and ∠AQB = 100◦ B C

3 A circle centre O, passes through A, B and C. AT is the tangent to the circle at A. CBT is a straight line. Given that ∠ABO = 68◦ and ∠OBC = 20◦ the magnitude of ∠ATB is E 66◦ D 70◦ C 65◦ B 64◦ A 60◦

O

20° 68°

B T

A A

4 In the diagram the points A, B and C lie on a circle centre O. ∠BOC = 120◦ and ∠ACO = 42◦ . The magnitude of ∠ABO is E 26◦ D 24◦ C 22◦ B 20◦ A 18◦

O 120°

42° C

B D

5 ABCD is a cyclic quadrilateral with AD parallel to BC. ∠DCB = 65◦ . The magnitude of ∠CBE is ◦ E 122◦ A 100◦ B 110◦ C 115◦ D 120 C

A

B E

◦

6 A chord AB of a circle subtends an angle of 50 at a point on the circumference of the circle. The acute angle between the tangents at A and B has magnitude E 82◦ D 85◦ C 75◦ B 65◦ A 80◦



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Review

7 Chords AB and CD of circle intersect at P. If AP = 12 cm, PB = 6 cm and CP = 2 cm, the length of PD in centimetres is E 56 D 48 C 36 B 24 A 12 P 8 In the diagram AB is the diameter of a circle with centre O. PQ is a chord perpendicular to AB. N is the point of intersection of AB and PQ and ON = 5 cm. If the radius of the circle is 13 cm 13 cm O B A the length of chord PB is, in centimetres, 5 cm N √ √ E 8 D 14 C 2 13 B 4 13 A 12 Q B

9 In the diagram A, B, C and D are points on the circumference of a circle. ∠ABD = 40◦ and angle ∠AXB = 105◦ . The magnitude of ∠XDC is E 55◦ D 50◦ C 45◦ B 40◦ A 35◦

C

40° 105° X A D

10 A, B, C and D are points on a circle, centre O such that AC is a diameter of the circle. If ∠BAD = 75◦ and ∠ACD = 25◦ The magnitude of ∠BDC is E 30◦ D 25◦ C 20◦ B 15◦ A 10◦

B C 75°

25° O

A D

Short-answer questions (technology-free) 1 Find the value of the pronumerals in each of the following. b a y°

50° x°

O

75°

140°

O y°

x°

d

c

x°

x°

53°

z°

47° z°

y° y°

70°

30°

2 O is the centre of a circle and OP is any radius. A chord BA is drawn parallel to OP. OA and BP intersect at C. Prove that b ∠PCA = 3∠PBA a ∠CAB = 2∠CBA



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3 A chord AB of a circle, centre O, is produced to C. The straight line bisecting ∠OAB meets the circle at E. Prove that EB bisects ∠OBC 4 Two circles intersect at A and B. The tangent at B to one circle meets the second again at D, and a straight line through A meets the first circle at P and the second at Q. Prove that BP is parallel to DQ. 5 Find the values of the pronumerals for each of the following: b c a 64° x°

x°

x°

y°

57°

48°

6 Two circles intersect at M and N. The tangent to the first at M meets the second circle at P, while the tangent to the second at N meets the first at Q. Prove that MN2 = NP · QM. 7 If AB = 10 cm, BE = 5 cm and CE = 25 cm, find DE. A

B E D

C

Extended-response questions 1 The diagonals PR and QS of a cyclic quadrilateral PQRS intersect at X. The tangent at P is parallel to QS. Prove that R a PQ = PS b PR bisects ∠QRS.

Q X

P

S

2 Two circles intersect at A and B. The tangents at C and D intersect at T on AB produced. If CBD is a straight line prove that a TCAD is a cyclic quadrilateral b ∠TAC = ∠TAD c TC = TD.

A C

B

D

T ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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389


b the circle through A, P and D touches BA at A c ABCD is a cyclic quadrilateral.

4 PQRS is a square of side length 4 cm inscribed in a circle with centre O. M is the midpoint of the side PS. QM is produced to meet RS produced at X. a Prove that: X i XPR is isosceles ii PX is a tangent to the circle at P.

Review

3 ABCD is a trapezium in which AB is parallel to DC and the diagonals meet at P. The circle through D, P and C touches AD, BC at D and C respectively. A Prove that P a ∠BAC = ∠ADB

B

D

C

Q

P O M S

4 cm

R

b Calculate the area of trapezium PQRX. 5 a An isosceles triangle ABC is inscribed in a circle. AB = AC and chord AD intersects BC at E. Prove that AB2 − AE2 = BE · CE b Diameter AB of circle with centre O is extended to C and from C a line is drawn tangent to the circle at P. PT is drawn perpendicular to AB at T. Prove that CA · CB − TA · TB = CT 2



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C H A P T E R

15 Vectors Objectives To understand the concept of vector To apply basic operations to vectors To understand the zero vector To use the unit vectors i and j to represent vectors in two dimensions To use the fact that, if a and b are parallel, then a = kb for a real value k. The converse of this also holds. To use the unit vectors i, j and k to represent vectors in three dimensions

15.1 Introduction to vectors For experiments in science or engineering some of the things which are measured are completely determined by their magnitude. For example, mass, length or time are determined by a number and an appropriate unit of measurement. e.g.,

length: time:

30 cm is the length of the page of a particular book 10 s is the time for one athlete to run 100 m

More is required to describe velocity, displacement or force. The direction must be recorded as well as the magnitude. e.g.,

displacement: velocity:

30 km in a direction north 60 km/h in a direction south east

Quantities that have direction as well as magnitude are represented by arrows that point in the direction of the action and whose lengths give the magnitude of the quantity in terms of a suitably chosen unit. Arrows with the same length and direction are regarded as equivalent. These arrows are directed line segments and the sets of equivalent segments are called vectors.



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391

Chapter 15 — Vectors

y

The five directed line segments shown all have the same magnitude and direction. A directed line segment from a point A to a point B is − → denoted by AB. − → For simplicity of language this is also called vector AB, i.e., the set of equivalent segments can be named through one member of that set. − → − → − → − → − → AB = CD = OP = EF = GH Note:

B A

D

C

P

O

x H

F E

G

In Chapter 8, vectors were introduced in the context of y translations (in two dimensions). A column of numbers was introduced to represent the translation and it was B called a vector. This is consistent with the approach here as the column 2 units of numbers corresponds to a set of equivalent directed A line segments. 3 units 3 The column corresponds to the directed line segment x 2 0 which goes 3 across and 2 up. This notation will be used to represent a directed line segment in the first section of this chapter. The notation is widely used to represent vectors but not to a large extent in Victorian schools and so the notation, although useful, will be abandoned in the latter sections of the chapter. Vectors are often denoted by a single bold face roman letter. The vector from A to B can be − → − → denoted by AB or by a single v. That is, v = AB. When a vector is handwritten the notation is v . ∼

Addition of vectors Two vectors u and v can be added geometrically by drawing a line segment representing u from A to B and then a line segment from B to C representing v. The sum u + v is the vector from A to C. − → That is, u + v = AC. The same result is achieved if the order is reversed. This is represented in the following diagram − → i.e. u + v = AC and u + v = v + u

C v B

u+v u

A C v u B

u+v D u v A



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The addition can also be achieved with the column vector notation. −1 4 e.g., if u= and v = 1 3 4 −1 3 then u+v= + = 1 3 4

1

3

4

v 3

u+v

1

u 4

Scalar multiplication Multiplication by a real number (scalar) changes the length of the vector. 2u = u + u and 1 1 u+ u=u u 2 2 2u is twice the length of u and 1 u is half the length of u. 2 The vector ku, k ∈ R + , with u = 0, has the same direction as u, but its length is multiplied by a factor of k. When a vector is multiplied by −2 the vector’s direction is reversed and its length doubled. u When a vector is multiplied by −1 the vector’s direction is reversed and the length remains the same. 3 −3 6 −6 If u= , −u = , 2u = and −2u = 2 −2 4 −4 e.g.,

If

2u

1 u 2

–2u

− → − → − → u = AB then −u = − AB = BA.

− → The directed line segment − AB goes from B to A.

Zero vector The zero vector is denoted by 0 and represents a line segment of zero length. The zero vector has no direction.

Subtraction of vectors In order to subtract v from u, add −v to u. For example –v v

u v

u–v


u


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Example 1 Draw the directed line segment defined by

3 . −2

Solution 3 is the vector ‘3 across to the right and 2 down’ −2 Note: Here the vector starts at (1, 1) and goes to (4, −1). It can start at any point.

y

1

A x

0

1

2

3

–1

4 B

Example 2 The vector u is defined by the directed line segment from (2, 6) to (3, 1). y a If u = , find a and b. b Solution

3− 2 1 The vector u = = 1− 6 −5 Hence a = 1 and b = −5

A (2, 6)

B (3, 1) 0

x

Example 3 If the vector u = Solution

3 −1

and the vector v =

−2 , find 2u + 3v. 2

3 −2 2u + 3v = 2 × +3× −1 2 6 −6 = + −2 6 0 = 4

B

Polygons of vectors

C

− → − → − → − → − → For two vectors AB and BC, AB + BC = AC A



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− → − → − → − → − → − → For a polygon ABCDEF, AB + BC + CD + DE + EF + FA = 0

B

C

A

D

F

E

Example 4 − → − → − → Illustrate the vector sum AB + BC + CD, where A, B, C and D are points in the plane. C

Solution − → − → − → − → AB + BC + CD = AD

B

A

D

Parallel vectors The non-zero u and vectors v are said to be parallel if there exists k ∈ R\{0} such that u = kv. −2 −6 If u = and v = then vector u is parallel to v as v = 3u. 3 9

Position vectors A point O, the origin, can be used as a starting point for a vector to indicate the position of a point in space relative to that point. In this chapter, O is the origin for a cartesian plane (three dimensional work is considered briefly). − → For a point A the position vector is OA.

Linear combinations of non-parallel vectors If a and b are non-zero, non-parallel vectors, then ma + nb = pa + qb implies m = p and n = q An argument is as follows:

∴ ∴

ma − pa = qb − nb (m − p)a = (q − n)b q −n b or a= m−p

b=

m−p a q −n

but a and b are not parallel and not zero

∴

q = n and m = p



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Example 5 Let A, B and C be the vertices of a triangle. Let D be the midpoint of BC. − → − → Let AB = a and BC = b. Find the following in terms of a and b. − → − → − → − → a BD b DC c AC d AD Solution → 1 − → 1− a BD = BC = b 2 2 − → − → 1 b DC = BD = b 2 − → − → − → c AC = AB + BC = a + b − → − → − → 1 d AD = AB + BD = a + b 2 − → − → e CA = − AC = −(a + b)

D

B

C

A

− → e CA

(same direction and half the length) (equivalent vectors)

− → − → ( CA + AC = 0)

Example 6

p

− → In the figure, DC = kp where k ∈ R\{0}. a Express p in terms of k, q and r. − → b Express FE in terms of k and p to show FE is parallel to DC. − → − → c If FE = 4 AB, find the value of k.

A q

B r

D

C r

q

E

F

Solution

− → − → − → − → p = AB = AD + DC + CB = q + kp − r ∴ (1 − k)p = q − r 1 and hence p = (q − r) 1−k − → b FE = −2q + p + 2r = 2(r − q) + p But r − q = kp − p = (k − 1)p − → FE = 2kp − 2p + p ∴ = (2k − 1)p a

. . . from a

− → − → c If FE = 4 AB (2k − 1)p = 4p 2k − 1 = 4 5 k= 2


1

1 On the same graph, draw the arrows which represent the following vectors. 1 0 −1 −4 a b c d 5 −2 −2 3



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2

a 2 The vector u is defined by the directed line segment from (1, 5) to (6, 6). If u = , b find a and b. 3 The vector v is defined by the directed line segment from (−1, 5) to (2, −10). If a v= , find a and b. b

Example

4

4

A = (1, −2), B = (3, 0), C = (2, −3) and O is the origin. Express the following vectors a in the form . b − → − → − → − → − → a OA b AB c BC d CO e CB

5

A = (2, −3), B = (4, 0), C = (1, −4) and O is the origin. Sketch the following vectors. − → − → − → − → − → a OA b AB c BC d CO e CB

6 On graph paper, sketch the vectors joining the following pairs of points in the direction indicated. a (0, 0) → (2, 1) d (2, 4) → (4, 3)

b (3, 4) → (0, 0) e (−2, 2) → (5, −1)

c (1, 3) → (3, 4) f (−1, −3) → (3, 0)

7 Identify vectors from 6 which are parallel to each other.

Example

3

8 a Plot the points A (−1, 0) , B (1, 4) , C (4, 3) , D (2, −1) on a set of coordinate axes. − → − → − → − → b Sketch the vectors AB, BC, AD, and DC. c Show that − → − → − → − → i AB = DC ii BC = AD d Describe the shape of the quadrilateral ABCD. 1 1 −2 9 Let a = ,b = and c = . 2 −3 1 a Find i a+b ii 2c − a b Show that a + b is parallel to c.

iii a + b − c

3 2 −19 10 Find the values of m and n such that m +n = −3 4 61 11 In the figure A, B, C, D are the vertices of a parallelogram. − → − → AB = a and AD = b

M

A

B

M, N are the midpoints of AB and DC respectively. a Express the following in terms of a and b. − → − → i MD ii MN − → − → b Find the relationship between MN and AD. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

D

N

C


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Example

Example

5

6

− → 12 The figure represents the triangle ABC with AB = a − → and AC = b. − → − → M, N are midpoints of AB and AC respectively. − → − → a Express CB and MN in terms of a and b. b Hence describe the relation between the two vectors (or directed line segments). 13 The figure represents the regular hexagon ABCDEF − → − → with vectors AF = a and AB = b. Express the following vectors in terms of a and b. − → − → − → − → a CD b ED c BE d FC − → − → − → g FE e FA f FB

C N

A

B

M C

D

B

E

A

F

− → − → 14 In parallelogram ABCD, AB = a and BC = b. Express each of the following vectors in terms of a and b. − → − → − → − → − → a DC b DA c AC d CA e BD − → − → − → − → 15 In triangle OAB, OA = a and OB = b. P is a point on AB such that AP = 2 PB and Q is a − → − → point such that OP = 3 PQ. Express each of the following in terms of a and b. − → − → − → − → − → a BA b PB c OP d PQ e BQ − → − → − → 16 PQRS is a quadrilateral in which PQ = u, QR = v, RS = w. Express the following vectors in terms of u, v and w. − → − → − → a PR b QS c PS − → − → 17 OABC is a parallelogram. OA = u, OC = v. M is the midpoint of AB. − → −→ a Express OB and OM in terms of u and v. − → b Express CM in terms of u and v. → − → − → − → 2 − c If P is a point on CM and CP = CM, express CP in terms of u and v. 3 − → − → d Find OP and hence show that P lies on OB. e Find the ratio OP : PB.

15.2

Components of vectors − →

The vector AB illustrated opposite can be described 3 by the column vector . From the diagram it is 4 − → possible to see that AB can be expressed as the sum − → − → of two vectors AX and XB, − → − → − → i.e., AB = AX + XB.


B(5, 7)

y

A(2, 3)

O

X x


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In column vector notation 3 3 0 = + 4 0 4 This suggests the introduction of two important vectors. Let i be the vector of unit length with direction the positive direction of the x axis. Let j be the vector of unit length with direction the positive direction of the y axis. 1 0 Note that using the column notation, i = and j = . 0 1 − → − → For the example above, AX = 3i and XB = 4j − → Therefore AB = 3i + 4j. It is possible to describe any two-dimensional vectors in this way. x For a vector u = , u = xi + yj. It is said that u is y the sum of two components xi and yj. The magnitude of vector u = xi + yj is denoted by |u| and |u| = x 2 + y 2 Operations with vectors now look more like basic algebra. (xi + yj) + (mi + nj) = (x + m) i + (y + n) j k (xi + yj) = kxi + kyj For

y

j x

i

O

y u

yj xi

O

x

a = xi + yj and b = mi + nj, a = b if x = m and y = n

Example 7 − → − → − → a Find AB if OA = 3i and OB = 2i − j b Find |2i − 3j|. Solution − → − → − → a AB = − OA + OB = −3i + (2i − j) = −i − j √ √ b |2i − 3j| = 4 + 9 = 13 Example 8 − → − → A, B are points on the cartesian plane such that OA = 2i + j and OB = i − 3j. Find − → − → AB and | AB|.



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399

Solution − → − → − → AB = AO + OB − → − → = − OA + OB − → ∴ AB = −(2i + j) + i − 3j = −i − 4j √ √ − → ∴ | AB| = 1 + 16 = 17

Unit vectors A unit vector is a vector of length one unit (i and j are unit vectors). The unit vector in the direction of a is aˆ (pronounced ‘a hat’). |â| = 1 so |a| aˆ = a 1 a or a aˆ = |a| |a| Example 9 Let a = 3i + 4j. Find |a|, the magnitude of a, and hence the unit vector in the direction of a. Solution a = 3i + 4j so |a| =

x 2 + y2 = 32 + 42

So

|a| = 5 1 aˆ = a |a| 1 aˆ = (3i + 4j) 5


7

− → − → − → 1 A, B are points on the cartesian plane such that OA = i + 2j and OB = 3i − 5j. Find AB. − → − → 2 OAPB is a rectangle in which the vector OA = 5i and the vector OB = 6j. Express the following vectors in terms of i and j. − → − → − → a OP b AB c BA

Example

8

3 Determine the magnitude of the following vectors. a 5i

b −2j

c 3i + 4j

d −5i + 12j

4 The vectors u and v are given by u = 7i + 8j and v = 2i − 4j. a Find |u − v|. b Find constants x and y such that xu + yv = 44j.



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− → − → −→ 5 In the triangle OAB, OA = 10i and OB = 4i + 5j. If M is the midpoint of AB, find OM in terms of i and j. − → − → 1 6 In the rectangle OPAQ, OP = 2i and OQ = j. M is the point on OP such that OM = OP. 5 1 N is the point on MQ such that MN = MQ. 6 a Find the following vectors in terms of i and j. −→ − → − → − → − → i OM ii MQ iii MN iv ON v OA b i Hence show that N is on the diagonal OA. ii State the ratio of the lengths ON : NA. − → − → 1 5 7 The position vectors of A and B are given by OA = and OB = . 3 −1 Find the distance between A and B. 8 Find the pronumerals in the following equations. a 2i + 3j = 2 (li + kj) c (x + y) i + (x − y) j = 6i

b (x − 1) i + yj = 5i + (x − 4) j d k (i + j) = 3i − 2j + l (2i − j)

9 Let A = (2, 3) and B = (5, 1). Find − → − → a AB b | AB| 10 Let OA = 3i, OB = i + 4j and OC = −3i + j. Find − → − → − → a AB b AC c |BC| 11 Let A = (5, 1) , B = (0, 4) and C = (−1, 0). Find − → − → − → − → a D such that AB = CD b F such that AF = BC

− → − → c G such that AB = 2GC − → − → 12 Let a = i + 4j and b = −2i + 2j. A, B and C are points such that AO = a, OB = b, − → BC = 2a and O is the origin. Find the coordinates of A, B and C. 13 A, B, C and D are the vertices of a parallelogram and O is the origin. A = (2, −1), B = (−5, 4) and C = (1, 7). a Find − → − → − → − → i OA ii OB iii OC iv BC b Hence find the coordinates of D.

− → v AD

14 The diagram shows a parallelogram OPQR. The points P and Q have coordinates (12, 5) and (18, 13) respectively. Find −→ − → − → − → a OP and PQ b |RQ | and |OR| 15

A(1, 6), B(3, 1), C(13, 5) are the vertices of a triangle ABC

y Q R

O

P

x

a Find − → − → − → i | AB| ii |BC| iii | CA| b Hence show that ABC is a right-angled triangle. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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16

A(4, 4), B(3, 1) and C(7, 3) are the vertices of the triangle ABC. a Find the vector − → − → − → i AB ii BC iii CA b Find − → − → − → i | AB| ii |BC| iii | CA| c Hence show that triangle ABC is an isosceles right-angled triangle.

17

A(−3, 2), B(0, 7) are points on the cartesian plane. O is the origin. M is the midpoint of AB. a Find − → − → − → − → i OA ii OB iii BA iv BM −→ − → − → b Hence find the coordinates of M. (Hint: OM = OB + BM)

Example

9

18 Find the unit vector in the direction of each of the following vectors. a a = 3i + 4j d d=i−j

15.3

b b = 3i − j 1 1 e e= i+ j 2 3

c c = −i + j f f = 6i − 4j

Vectors in three dimensions Points in three dimensions (3-D) are represented in axes as shown. Vectors in 3-D are of the form ⎡ ⎤ x ⎢ ⎥ a = ⎣ y ⎦ = xi + yj + zk where z ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ i = ⎣ 0 ⎦ , j = ⎣ 1 ⎦ and k = ⎣ 0 ⎦ 0 0 1

z (x, y, z)

y x z k 1

i, j and k are represented in the figure. − → OA = xi + yj + zk

j i

By Pythagoras’ theorem

and

∴

y 1

1

x

z

OB2 = x 2 + y 2 OA2 = OB2 + BA2 = x 2 + y2 + z2 − → | OA| = x 2 + y 2 + z 2

A xi x

yj

O zk

y

B

Example 10 Let a = i + j − k and b = i + 7k. Find a a+b b b − 3a

c |a|



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Solution a a + b = i + j − k + i + 7k = 2i + j + 6k b b − 3a = i + 7k − 3(i + j − k) = −2i − 3j + 10k √ c |a| = 3 Example 11 − → − → − → OABCDEFG is a cuboid. OA = 3j, OC = k, OD = i. a Express each of the following in terms of i, j, k. − → − → − → − → i OE ii OF iii GF iv GB − → − → b M, N are the midpoints of OD and GF respectively. Find MN.

C

B F

G O D

A E

Solution − → − → − → − → − → a i OE = OA + AE = 3j + i ( AE = OD) − → − → − → − → − → ii OF = OE + EF = 3j + k + i ( EF = OC) − → − → iii GF = OA = 3j − → − → − → − → iv GB = DA = DO + OA = −i + 3j − → − → − → − → − → − → b MN = MD + DN = MD + DG + GN 3 1 = i+k+ j 2 2 √ − → 9 1 14 +1+ = ∴ |MN| = 4 4 2 Example 12 If a = 3i + 2j + 2k, find aˆ . Solution

√ 9 + 4 + 4 = 17 1 aˆ = √ (3i + 2j + 2k) 17 2 2 3 = √ i +√ j +√ k 17 17 17

|a| =

∴

√


10

1 Let a = i + j + 2k, b = 2i − j + 3k, c = −i + k. Find a a−b

Example

12

b 3b − 2a + c

c |b|

d |b + c|

e 3(a − b) + 2c

2 If a = 3i + j − k a find i aˆ ii −2â b find the vector b in the direction of a such that |b| = 5.



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3 If a = i − j + 5k and b = 2i − j − 3k find the vector c in the direction of a such that |c| = |b|. 4 P and Q are points defined by the position vectors i + 2j − k and 2i − j − k respectively. − → M is the midpoint of PQ. Find − → − → −→ a PQ b |PQ| c OM Example

11

5 OABCDEFG is a cuboid. − → − → − → OA = 2j, OC = 2k, OD = i

C

B

G

F

Express the following vectors in terms of i, j and k. − → − → − → − → a OB b OE c OG d OF − → − → − → − → e ED f EG g CE h BD

O

A

D

6 OABCDEFG is a cuboid. − → − → − → OA = 3j, OC = 2k, OD = i → −→ 1 − M is such that OM = OE. 3 N is the midpoint of BF. Find − → − → a MN b |MN|

15.4

E C F

G O D

B

A E

Applications Example 13 Three points P, Q, and R have position vectors p, q, and k(2p + q) respectively, relative to a fixed origin O. O, P and Q are not collinear. Find the value of k if − → − → a QR is parallel to p b PR is parallel to q c P, Q and R are collinear. Solution − → − → − → a QR = QO + OR = −q + k(2p + q) = −q + 2kp + kq − → If QR is parallel to p, there exists ␭ ∈ R\{0} such that (k − 1)q + 2kp = ␭p ∴ k − 1 = 0 and 2k = ␭ ∴ k=1 − → − → − → b PR = PO + OR = −p + k(2p + q) = (2k − 1)p + kq



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− → If PR is parallel to q there exists m ∈ R\{0} such that (2k − 1)p + kq = mq 1 ∴ k= 2 c If P, Q and R are collinear there exists n ∈ R\{0} such that − → − → n PQ = QR ∴ n(−p + q) = (k − 1)q + 2kp ∴ k − 1 = n and 2k = −n which implies 3k − 1 = 0 i.e., k =

1 3

Exercise 15D − → − → 4 1 In the diagram OR = OP, OP = p, OQ = q 5 and PS : SQ = 1 : 4.

O p

q

R

a Express each of the following in terms of p and q. − → − → − → P S i OR ii RP iii PQ − → − → iv PS v RS b What can be said about line segments RS and OQ? c What type of quadrilateral is ORSQ? d The area of triangle PRS is 5 cm2 . What is the area of ORSQ?

Q

2 The position vectors of three points A, B and C relative to an origin O are a, b and ka respectively. The point P lies on AB and is such that AP = 2PB. The point Q lies on BC and is such that CQ = 6QB. a Find in terms of a and b i the position vector of P ii the position vector of Q b Given that OPQ is a straight line, find i the value of k OP ii the ratio PQ 7 c The position vector of a point R is a. Show that PR is parallel to BC. 3 Example

13

3 The position vectors of two points A and B relative to an origin O are 3i + 3.5j and 6i − 1.5j respectively. − → 1− → − → 1− → a i Given that OD = OB and AE = AB, write down the position vectors of 3 4 D and E. − → ii Hence find |ED|.



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405

− → − → − → − → b Given that OE and AD intersect at X and that OX = p OE and XD = q AD, find the position vector of X in terms of i p ii q c Hence determine the values of p and q. 4 The position vectors of P, Q with reference to an origin O are p and q and M is the point on PQ such that − → − → ␤ PM = ␣ MQ a Prove that the position vector of M is m where m=

␤ p + ␣q ␣+␤

b The vector p = ka and the vector q = lb where k and l are positive real numbers and a and b are unit vectors. i Prove that the position vector of any point on the internal bisector of ∠POQ has the form ␭ (a + b) ii If M is the point where the internal bisector of ∠POQ meets PQ, show that ␣ k = ␤ l 5 ORST is a parallelogram. U is the midpoint of RS and V is the midpoint of ST. Relative to the origin O, r, s, t, u and v are the position vectors of R, S, T, U and V respectively. a Express s in terms of r and t. b Express v in terms of s and t. c Hence or otherwise show that 4 (u + v) = 3 (r + s + t)



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Chapter summary A vector is a set of equivalent directed line segments. − → A directed line segment from a point A to a point B is denoted by AB.

2 In two dimensions, a vector can be represented by a column of numbers, e.g. is the 3 vector ‘2 across’ and ‘3 up’. C The sum u + v can be shown diagrammatically v u+v

B u

u + v = v + u a c a+c If u = and v = then u + v = b d b+d

A

The vector ku, k ∈ R + and u = 0, has the same direction as u but its length is multiplied by a factor k. The vector −v is in the opposite direction to v but it has the same length. u − v = u + (−v) Two non-zero vectors u and v are said to be parallel if there exists k ∈ R\{0} such that u = kv. − → For a point A, the position vector of A is OA where O is the origin. y Every vector u can be expressed as the sum of two vectors xi and yj, where i is the unit vector in the positive direction of the x axis and j is the unit vector in the positive direction u yj of the y axis.

xi x

O

The magnitude of vector u = xi + yj is denoted by |u| and |u| = x 2 + y 2 a . This vector is denoted by aˆ . The unit vector in the direction of a is |a| In three dimensions a vector u can be written as u = xi + yj + zk, where i, j and k are unit vectors as shown. z

z (x, y, z) k j

y

i

x

If u = xi + yj + zk, |u| =

y

x

x 2 + y2 + z2



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407

1 The vector v is defined by the directed line segment from (1, 1) to (3, 5). If v = ai + bj then A a = 3 and b = 5 B a = −2 and b = −4 C a = 2 and b = 4 D a = 2 and b = 3 E a = 4 and b = 2 − → − → − → 2 If vector AB = u and vector AC = v then vector CB is equal to A u+v B v−u C u−v D u×v E v+u 1 2 3 If vector a = and vector b = then a + b = −2 3 1 1 1 1 3 A B C D E −5 5 −1 1 1 3 −1 and vector b = then 2a − 3b = 4 If vector a = −2 3 9 3 9 3 9 A E B D C −13 5 7 −13 −7 − → − → − → 5 PQRS is a parallelogram. If PQ = p and QR = q, then in terms of p and q, SQ equals A p+q B p−q C q−p D 2q E 2p

Review


6 |3i − 5j| = √ 34 C 34 D 8 E −16 A 2 B − → − → − → 7 If OA = 2i + 3j and OB = i − 2j then AB equals A −i − 5j B −i + 5j C −i − j D −i + j E i+j − → − → − → 8 If OA = 2i + 3j and OB = i − 2j then AB equals √ √ A 6 B 26 C 26 D 24 E 36 9 If a = 2i + 3j then the unit vector parallel to a is 1 A 2i + 3j B (2i + 3j) 13 √ 1 E 13(2i + 3j) D √ (2i + 3j) 13 10 If a = −3i + j + 3k then aˆ is 1 1 B √ (−3i + j + 3k) A (−3i + j + 3k) 7 7 1 D (−3i + j + 3k) E −3i + j + 3k 19

C

1 √ (2i + 3j) 5

C

1 √ (−3i + j + 3k) 19

Short-answer questions (technology-free) 1 Given that a = 7i + 6j and b = 2i + xj, find the values of x for which a a is parallel to b b a and b have the same magnitude. − → − → − → 2 ABCD is a parallelogram where OA = 2i − j, AB = 3i + 4j and AD = −2i + 5j. Find the coordinates of the four vertices of the parallelogram.



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3 If a = 2i − 3j + k, b = 2i − 4j + 5k and c = −i − 4j + 2k, find the values of p and q if a + pb + qc is parallel to the x axis. 4 The position vectors of P and Q are 2i − 2j + 4k and 3i − 7j + 12k respectively. Find − → − → a |PQ| b a unit vector parallel to PQ. 5 The position vectors of A, B and C are 2j + 2k, 4i + 10j + 18k and xi + 14j + 26k respectively. Find x if A, B and C are collinear. − − → → 16 . 6 OA = 4i + 3j and C is a point on OA such that OC = 5 − → a Find the unit vector in the direction of OA. − → b Hence find OC. − → − → − → 7 In the diagram, ST = 2TQ, PQ = a, SR = 2a and SP = b. R a Find each of the following in terms of a and b. − → − → − → i SQ ii TQ iii RQ − → − → 2a Q iv PT v TR T

b Show that P, T and R are collinear.

a P

b

S

8 If a = 5i − sj + 2k and b = ti + 2j + uk are equal vectors, find a i s ii t iii u b |a| 9 The vector p has magnitude 7 units and bearing 050◦ and the vector q has magnitude 12 units and bearing 170◦ . (These are compass bearings on the horizontal plane.) Draw a diagram (not to scale) showing p, q and p + q. Calculate the magnitude of p + q. 10 If a = 5i + 2j + k and b = 3i − 2j + k, find a a + 2b b |a| c aˆ d a−b 11 O, A and B are the points (0, 0), (3, 4) and (4, –6) respectively. − → − → − → a C is the point such that OA = OC + OB. Find the coordinates of C. − → − → − → b D is the point (1, 24) and OD = h OA + k OB. Find the values of h and k. 12 Given that p = 3i + 7j and q = 2i − 5j, find the values of m and n such that mp + nq = 8i + 9j. 13 The points A, B and C have position vectors a, b and c relative to an origin O. Write down an equation connecting a, b and c for each of the following cases. a OABC is a parallelogram. b B divides AC in the ratio 3 : 2, i.e., AB : BC = 3 : 2.



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y 1 0 1 Let represent a displacement 1 km due east and 30 0 1 20 represent a displacement 1 km due north. The diagram shows 10 a circle of radius 25 km with a centre at O(0, 0). A lighthouse 0 –30 –20 –10 O 0 10 20 30 entirely surrounded by sea is located at O. –10 The lighthouse is not visible from points outside the circle. P –20 The ship is initially at P, 31 km west and 32 km south of the lighthouse. –30 − → a Write down the vector OP. 4 The ship is travelling parallel to vector u = with speed 20 km/h. An hour after 3 leaving P the ship is at R. − → − → 16 b Show that the vector PR = , and hence find the vector OR. 12 c Show that when the ship reaches R, the lighthouse first becomes visible.

x

Review


2 Given that p = 3i + j and q = −2i + 4j find a | p − q| b | p| − |q| c r, such that p + 2q + r = 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −2 11 7 26 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 3 Let a = ⎣ 1 ⎦ , b = ⎣ 7 ⎦ , c = ⎣ 9 ⎦ and d = ⎣ 12 ⎦ 2 3 7 2 a Find the value of the scalar k such that a + 2b − c = kd b Find the scalars x and y such that xa + yb = d c Use your answers to a and b to find scalars p, q and r (not all zero) such that pa + qb − r c = 0 4 The quadrilateral PQRS is a parallelogram. The point P has coordinates (5, 8), the point R − → − → 20 has coordinates (32, 17) and the vector PQ is given by PQ = . −15 − → a Find the coordinates of Q, and write down the vector QR. − → b Write down the vector RS , and show that the coordinates of S are (12, 32). 5 The diagram shows the path of a light beam from its source 3 P at O in the direction of the vector r = . At P the beam 1 θ is reflected by an adjustable mirror and meets the x axis at M. The position of M varies, depending on the adjustment r of the mirror at P. O M − → a Given that OP = 4r, find the coordinates of P. − → b The point M has coordinates (k, 0). Find in terms of k an expression for the vector PM. − → −→ − → c Find the magnitudes of vectors OP, OM and PM, and hence find the value of k for which ␪ is equal to 90◦ . d Find the value ␪ for which M has coordinates (9, 0).



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C H A P T E R

16 Polar coordinates and complex numbers Objectives To describe points on the plane using polar coordinates To describe graphs with polar coordinates To transform polar coordinates to cartesian coordinates To transform cartesian coordinates to polar coordinates To understand the imaginary number i To understand the set of complex numbers C To understand the real-valued functions of the complex numbers, Re(z) and Im(z) To represent complex numbers graphically on an Argand diagram To understand the rules which define equality, addition, subtraction and multiplication of complex numbers To understand the concept of the complex conjugate To understand the operation of division by complex numbers To understand the modulus-argument form of a complex number and the basic operations on complex numbers in that form To understand the geometrical significance of multiplication and division of complex numbers in the modulus-argument form To be able to factorise quadratic polynomials over C To be able to solve quadratic polynomials over C

A new set of numbers called Complex numbers is introduced in this chapter. The need for this new set of numbers can be equated to the need for a solution of the equation x 2 + 1 = 0. A geometric interpretation is also shown to be useful. Complex numbers can be expressed in two ways, cartesian form and polar form. As a preliminary to this, polar coordinates are introduced.



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Chapter 16 — Polar coordinates and complex numbers

16.1

411

Polar coordinates In previous work the cartesian coordinate system has been used to represent points in two-dimensional space. The point (x, y) is described in terms of its horizontal displacement (x) and its vertical displacement (y) from a fixed point called the origin (O). An alternative way of locating the point P is to describe it in terms of its polar coordinates [r, ␪] where r specifies the distance from the origin or pole and ␪ specifies the angle of the line OP relative to the line OZ which extends to the right from O and is called the polar axis. P[r, θ] Note: An angle in an anticlockwise direction from OZ is considered to be positive. r O

θ

Z P[4, 60°]

For example, the point P[4, 60◦ ] is located a distance of 4 units along a line forming an angle of 60◦ with the polar axis.

4 O

60°

Z

Using this system it is clear that any point can be specified in a number of different ways. For example, the point [4, 60◦ ] may also be specified by [−4, −120◦ ]. The angle ␪ = 120◦ is measured in a clockwise direction from O. The diagram below and to the left illustrates the point P [4, −120◦ ] and the diagram to the right, [−4, −120◦ ]. P O 4

Z 4

120° O

60°

Z

120°

P'

P[4, 60◦ ] may also be specified by P[4, −300◦ ] or P[−4, 240◦ ]. Example 1 Plot the point P with coordinates [3, −30◦ ]. Solution

O

Z 30° 3 P



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The relationship between cartesian and polar coordinates y

If a set of cartesian axes is superimposed over a polar axis, the relationship between cartesian and polar coordinates can be established. From the diagram

and

x = r cos ␪

1

y = r sin ␪

2

P(x, y) r

r sin (θ)

θ O

x(Z)

r cos (θ)

The angle ␪ can be found by finding a solution which satisfies both equations 1 and 2 Squaring both sides of equations 1 and 2 and adding yields x 2 + y 2 = r 2 cos2 ␪ + r 2 sin2 ␪ = r 2 (cos2 ␪ + sin2 ␪) i.e. x 2 + y 2 = r 2 Using these relationships, coordinates can be converted from cartesian to polar and vice versa.

Example 2 √ a Express ( 3, 1) in polar form.

√ b Express [ 2, 45◦ ] in cartesian form.

Solution 1 r 2 = x 2 + y2 sin ␪ = a 2 √ = ( 3)2 + (1)2 =4

√ 3 and cos ␪ = 2 ∴ ␪ = 30◦

∴r =2 √ ∴ ( 3, 1) specifies the same point as [2, 30◦ ] √ ◦ b r = √2 and ␪ = 45√ 1 x = 2 cos 45◦ = 2 × √ = 1 2 √ √ 1 y = 2 sin 45◦ = 2 × √ = 1 2 √ ∴ [ 2, 45◦ ] specifies the same point as (1, 1)




413

Curve sketching using polar coordinates In the same way that graphs of relationships in cartesian form can be sketched, relationships expressed in polar form can also be sketched. Some very interesting curves result from simple polar equations. It is recommended that to sketch these graphs a graphics calculator or a computer graphing package be used. Graphs can of course be plotted using a table of values. A sheet of polar graph paper is also useful although a sheet of blank paper will suffice as long as a ruler and a protractor are used.

Using the TI-Nspire Example 3 Plot the graph of r = 3 (1 − cos ␪). Solution Open a Graphs application (b>New Doc>Add Graphs) and set to polar using b>Graph Type>Polar Enter r 1 (␪) = 3 (1 − cos (␪)) as shown. The symbol ␪ can be obtained from the and select Symbols palette (/+k) or ␪ (f key on the Clickpad) Note that the domain of ␪ as well as the step size can be adjusted in this window. The graph is shown using b>Window/Zoom>Zoom-Fit to set the scale.

Note that every point on the graph satisfies r = 3 (1 − cos (␪)). For example, for ␪ = 60◦ 3 1 ◦ = r = 3 (1 − cos (60 )) = 3 1 − 2 2 For ␪ = 180◦ , r = 3 (1 − cos (180◦ )) = 3 (1 − (−1)) = 6 For ␪ = −90◦ , r = 3 (1 − cos (−90◦ )) = 3



414


Note that Trace (b>Trace>Graph Trace) can be used to show the coordinates of the points on the graph in the form [r, ␪]. To go to the point where ␪ = ␲, simply type ␲ followed by enter. The cursor will then move to the point [r, ␪] = [6, ␲] as shown.

Example 4 Sketch the graph of r = ␪. Solution Open a Graphs application (b>New Doc>Add Graphs) and set to polar using b>Graph Type>Polar Enter r 1 (␪) = ␪. The graph is shown.

If the domain of ␪ is extended, the graph continues to spiral out. This can be observed by extending the domain to 0 < ␪ < 6␲. The resulting graph is shown using b >Window/Zoom>Zoom-Out




415

Using the Casio ClassPad Solution for Example 3 Plot the graph of r = 3 (1 − cos ␪). Ensure that the mode is set to radians. tap and from the menu select In . Enter the equation r 1 = 3 (1 − cos (␪)) and tap $ to produce the graph.

In the screen shown, the window was selected by tapping Zoom, quick initialize.

Example 5 Find the polar equation of the circle whose cartesian equation is x 2 + y 2 = 4x Solution Let x = r cos ␪ and y = r sin ␪ Then ∴ ∴ ∴ ∴

r 2 cos2 ␪ + r 2 sin2 ␪ = 4r cos ␪ r 2 (cos2 ␪ + sin2 ␪) = 4r cos ␪ r 2 − 4r cos ␪ = 0 r (r − 4 cos ␪) = 0 r = 0 or r = 4 cos ␪

∴ r = 4 cos ␪ is the polar equation of the circle Example 6 Find the cartesian equation corresponding to each of the following polar equations. 1 a r =3 b r= c r = 3(1 − cos ␪) 1 + sin ␪



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Solution a

r =3 2 x + y2 = 3 x 2 + y2 = 9 The circle with centre (0, 0) and radius 3

b

1 1 + sin ␪ implies r (1 + sin ␪) = 1 i.e. r+ r sin ␪ = 1 ∴ x 2 + y2 = 1 − y x 2 + y 2 = 1 − 2y + y 2 1 2 ∴ x = −2 y − 2 r=

∴ y=− c

x2 1 + 2 2

r = 3(1 − cos ␪) ∴ r 2 = 3r− 3r cos ␪ (Multiplying both sides of equation by r ) ∴ x 2 + y 2 = 3x 2 + y 2 − 3x 2 x + y 2 + 3x = 3 x 2 + y 2


1

1 Plot each of the following points using a polar axis. a A[2, 30◦ ] e E[5, 50◦ ]

b B[3, 45◦ ] f F[−5, −50◦ ]

c C[−2, 60◦ ] g G[−5, 130◦ ]

2 Plot each of the following points using a polar axis. ␲ a A[1, ␲] b B[−1, −␲] c C 2, 2

d D[4, −30◦ ] h H [5, −130◦ ] 3␲ d D 3, 4

Example

2a

3 Convert the following cartesian coordinates to polar coordinates. (Remember to note which quadrant each point is in.) √ √ a (4, 4) b (1, − 3) c (2 3, −2) d (−5, 12) √ g (−5, −12) h (4, 3) e (6, −5) f ( 3, 1)

Example

2b

4 Convert the following polar coordinates to cartesian coordinates. ␲ 5␲ ◦ a [−2, 30 ] b −4, d [4, −2␲] c −1, 2 4 7␲ g [2, 180◦ ] h [1, −120◦ ] f [5, 240◦ ] e 2, − 6

Examples

3, 4

5 Plot each of the following polar graphs. 4 3 b r= a r= sin ␪ cos ␪ ␲ ␲ d r = 2␪, 0 ≤ ␪ ≤ 6␲ e r = , ≤ ␪ ≤ 4␲ ␪ 6 g r = 5(1 + cos ␪) h r = 2(1 − sin ␪) √ ␲ , 0 ≤ ␪ ≤ 6␲ k r= j r = ± cos 2␪ ␪


c r = 2 cos ␪ f r = cos 2␪ i r = 3 cos ␪ + 2 l r = 2 sin 2␪


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Chapter 16 — Polar coordinates and complex numbers Example

5

6 Obtain the polar equations of each of the following. a x 2 + y 2 = 16

Example

6

b x+y=6

c x2 = y

d

x2 + y2 = 1 4

7 Obtain the cartesian equations of each of the following. a r =2 d r = 2a(1 + sin 2␪)

16.2

417

b r = a(1 + cos ␪) a e r= 1 + cos ␪

c r = a cos ␪ a f r= 1 + sin ␪

The set of complex numbers In earlier work in mathematics it was assumed that an equation of the form x 2 = −1 had no solutions. Mathematicians of the eighteenth century introduced the imaginary number i with √ the property i 2 = −1. i is defined as i = −1 and the equation x 2 = −1 has two solutions, i and −i. By considering i such that i 2 = −1 then the square roots of all negative numbers may be found. √ √ For example −4 = 4 × −1 √ √ = 4 × −1 = 2i Imaginary numbers led to the introduction of complex numbers, which further broadened the scope of mathematical thinking. Today complex numbers are widely used in engineering, the study of aerodynamics and many other branches of physics. Consider the equation x 2 + 2x + 3 = 0. Using the quadratic formula to solve yields: √ −2 ± 4 − 12 x= 2 √ −2 ± −8 = 2√ = −1 ± −2 This equation has no real solutions since the discriminant = b2 − 4ac is less than zero. However, for complex numbers √ x = −1 ± 2i A complex number is an expression of the form a + bi, where a and b are real numbers. C is the set of complex numbers, i.e. C = {a + bi : a, b ∈ R}. The letter often used to denote a complex number is z. Therefore z ∈ C implies z = a + bi where a, b ∈ R If a = 0, z is said to be imaginary. If b = 0, z is real. Real numbers and imaginary numbers are subsets of C.



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Functions of complex numbers Let z = a + bi Re (z) is a function which defines the real component of z. Im (z) is a function which defines the value of the imaginary component of z. Re (z) = a and Im (z) = b Re (z) and Im (z) are both real-valued functions of z, i.e. Re : C → R and Im : C → R. So for the complex number z = 2 + 5i, Re (z) = 2 and Im (z) = 5.

Note:

Equality of complex numbers Two complex numbers are equal if and only if both their real and imaginary parts are equal. i.e.

x1 + y1 i = x2 + y2 i

if and only if

x1 = x2 and y1 = y2

Example 7 If 4 − 3i = 2a + bi find the values of a and b. Solution 2a = 4 a=2

and

b = −3

Example 8 Find the values of a and b such that (2a + 3b) + (a − 2b)i = −1 + 3i Solution 2a + 3b a − 2b 2 × 2 gives 2a − 4b 1 − 3 gives 7b

∴

= −1 1 =3 2 =6

3

= −7

b = −1 and a = 1

Operations with complex numbers Addition and subtraction If z 1 = a + bi and z 2 = c + di Then z 1 ± z 2 = (a ± c) + i(b ± d)

(a, b, c, d ∈ R)

i.e. Re (z 1 ± z 2 ) = Re (z 1 ) ± Re (z 2 ) and Im (z 1 ± z 2 ) = Im (z 1 ) ± Im (z 2 ) ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 9 If z 1 = 2 − 3i and z 2 = −4 + 5i find a z1 + z2

b z1 − z2

Solution a z 1 + z 2 = 2(2 + −4) + (−3 + 5)i = −2 + 2i

b z 1 − z 2 = (2 − −4) + (−3 − 5)i = 6 − 8i

Multiplication by a real constant z = a + bi and kz = k(a + bi) = ka + kbi

If then

k∈R

For example, if z = 3 − 6i then 3z = 9 − 18i

Multiplication by powers of i Successive multiplication by powers of i gives the following: i1 = i i 2 = −1 i 3 = −i i 4 = (−1)2 = 1 i5 = i and so on In general, for n = 0, 1, 2, 3, . . . i 4n = 1 =i i 4n+2 = −1 i 4n+3 = −i i 4n+1

When multiplying by powers of i, the usual index laws apply. Example 10 Simplify a i 13

b 3i 4 × (−2i)3

Solution a i 13 = i 4×3+1 =i

b 3i 4 × (−2i)3 = 3 × (−2)3 × i 4 × i 3 = −24i 7 = 24i



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Exercise 16B 1 State the values of Re (z) and Im (z) for each of the following. 1 3 a 2 + 3i b 4 + 5i − i c 2 2 √ √ 2 − 2 2i d −4 e 3i f Examples

7, 8

2 Find the values of a and b in each of the following if a 2a − 3bi = 4 + 6i c 2a + bi = 10

Example

9

b a + b − 2abi = 5 − 12i d 3a + (a − b)i = 2 + i

3 Simplify the following. a c e g

(2 − 3i) + (4 − 5i) (−3 − i) − (3 + i) (1 − i) − (2i + 3) 4(2 − 3i) − (2 − 8i)

(4 + i) + (2 − 2i) √ √ (2 − 2i) + (5 − 8i) (2 + i) − (−2 − i) −(5 − 4i) + (1 + 2i) 3 1 (4 − 3i) − (2 − i) j 2 2

b d f h

i 5(i + 4) + 3(2i − 7) Example

10

4 Simplify √ a −16 20 f i

√ b 2 −9 g −2i × i 3

√ c −2 h 4i 4 × 3i 2

5 Simplify a i(2 − i)

b i 2 (3 − 4i)

c

√

2i(i −

d i3 √ 5 √ i 8i × −2

2)

e i 14

√ √ √ d − 3( −3 + 2)

16.3 Multiplication and division of

complex numbers Multiplication of complex numbers If z 1 = a + bi and z 2 = c + di(a, b, c, d ∈ R) Then

z 1 × z 2 = (a + bi) (c + di) = ac + bci + adi + dbi 2 = (ac − bd) + (bc + ad)i

(bdi 2 = −bd)

Example 11 If z 1 = 3 − 2i and z 2 = 1 + i, find z 1 z 2 . Solution z 1 z 2 = (3 − 2i)(1 + i) = 3 − 2i + 3i − 2i 2 = 5+i



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Conjugate of a complex number If z = a + bi then the conjugate of z denoted by the symbol z is z = a − bi For example, the conjugate of −4 + 3i is −4 − 3i and vice versa. Note that zz = (a + bi) (a − bi) = a 2 + abi − abi − b2 i 2 = a 2 + b2 which is a real number Using this result, a 2 + b2 can now be factorised over the set of complex numbers. Example 12 If z 1 = 2 − 3i and z 2 = −1 + 2i find a (z 1 + z 2 ) and z 1 + z 2

b z 1 z 2 and z 1 z 2

Solution

a

z 1 = 2 + 3i and z 2 = −1 − 2i z 1 + z 2 = (2 − 3i) + (−1 + 2i) b z 1 z 2 = (2 − 3i)(−1 + 2i) = 4 + 7i = 1−i z 1 z 2 = 4 − 7i (z 1 + z 2 ) = 1 + i z 1 z 2 = (2 + 3i)(−1 − 2i) = 4 − 7i z 1 + z 2 = 2 + 3i + −1 − 2i = 1+i

In general it can be stated that the conjugate of the sum of two complex numbers is equal to the sum of the conjugates the conjugate of the product of two complex numbers is equal to the product of the conjugates. i.e. (z 1 + z 2 ) = z 1 + z 2 and (z 1 z 2 ) = z 1 z 2

Division of complex numbers Division of one complex number by another relies on the fact that the product of a complex number and its conjugate is a real number. If Then

z 1 = a + bi and z 2 = c + di (a + bi) z1 = z2 (c + di)

(a, b, c, d ∈ R)

If the numerator and denominator are multiplied by the conjugate of z 2 then z1 (a + bi) (c − di) × = z2 (c + di) (c − di) ac + bci − adi − bdi 2 = c2 + d 2 (ac + bd) (bc − ad)i = 2 + 2 c + d2 c ©+ d 2 Evans et al. 2011 Michael

ISBN 978-1-107-65235-4 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 13 If z 1 = 2 − i and z 2 = 3 + 2i, find

z1 . z2

Solution 3 − 2i 2−i z1 × = z2 3 + 2i 3 − 2i 6 − 3i − 4i + 2i 2 32 + 22 4 − 7i = 13 1 = (4 − 7i) 13

=

Example 14 Solve for z the equation (2 + 3i) z = −1 − 2i Solution (2 + 3i)z = −1 − 2i −1 − 2i z= 2 + 3i 2 − 3i −1 − 2i × = 2 + 3i 2 − 3i −8 − i z= 13 There is an obvious similarity in the process of expressing a complex number with a real denominator and the process of rationalising the denominator of a surd expression. Example 15 If z = 2 − 5i find z −1 and express with a real denominator. Solution z −1 =

1 z

1 2 − 5i 2 + 5i 1 × = 2 − 5i 2 + 5i 2 + 5i = 29 1 = (2 + 5i) 29 =




423

Using the TI-Nspire The TI-Nspire can be used to deal with complex numbers. Press c>Settings >Settings>General and select Rectangular from the Real or Complex field to perform calculations on complex numbers in the form a + bi. Note: Real mode (default setting) will allow complex numbers in the form a + bi to be computed and displayed if i is entered as a part of the input or if commands from b >Algebra>Complex are used. The square root of a negative number can now be performed as shown.

The results of the operations +, −, × and ÷ are illustrated using the two complex numbers 2 + 3i and 3 + 4i. Note: Do not use the text i for the imaginary and select i (or j on the constant. Use Clickpad). Also available in the Symbols palette (/+k)

It is possible to perform arithmetic operations with complex numbers as shown. The real(..) command (b >Number> Complex Number Tools>Real Part) can be used as shown to find the real part of a complex number.

The magnitude command (b >Number> Complex Number Tools>Magnitude) can be used as shown to find the modulus of a complex number. Alternatively use abs(..). The conj(..) command (b >Number> Complex Number Tools>Complex Conjugate) can be used as shown to find the complex conjugate of a complex number. ISBN 978-1-107-65235-4 © Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2012 Photocopying is restricted under law and this material must not be transferred to another party.


424


There are also commands for factorising polynomials over the complex numbers and for solving polynomial equations over the complex numbers. These are available from the Complex submenu of the Algebra menu.

Using the Casio ClassPad In tap Real in the status bar at the bottom of the screen to enter Cplx mode. In this √ √ to obtain the answer i. Similarly, −16 will return the mode enter −1 and tap answer 4i.

Operations i is found in in the on-screen keyboard. With the calculator set to Complex mode, a number of arithmetic operations can be carried out, as shown in the screen at right using options from Interactive, Complex. Polynomials can be factorised and solved over the complex number field using Interactive, transformation and Equation/inequality, solve.


11

1 Expand and simplify a (4 + i)2 d (−1 − i)2

b (2 − 2i)2 √ √ √ √ e ( 2 − 3i)( 2 + 3i)

c (3 + 2i)(2 + 4i) f (5 − 2i)(−2 + 3i)



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2 Write down the conjugate of each of the following complex numbers. √ a 2 − 5i b −1 + 3i c 5 − 2i d −5i Example

12

3 If z 1 = 2 − i and z 2 = −3 + 2i find a z1 e z1 z2

Example

15

b z2 f z1 + z2

c z1 z2 g z1 + z2

d z1 z2 h z1 + z2

4 If z = 2 − 4i express each of the following in the form x + yi. a z

b zz

c z+z

e z−z

f i(z − z)

g z −1

d z(z + z) z h i

5 Find the values of a and b such that (a + bi)(2 + 5i) = 3 − i Example

13

6 Express in the form x + yi 3 + 2i 2−i b a 2 − 3i 4+i 2 − 2i 1 d e 4i 2 − 3i

4 + 3i 1+i i f 2 + 6i

c

7 Find the values of a and b if (3 − i)(a + bi) = 6 − 7i Example

14

8 Solve each of the following for z. a (2 − i)z = 4 + 2i d 2(4 − 7i)z = 5 + 2i

16.4

b (1 + 3i)z = −2 − i e z(1 + i) = 4

c (3i + 5)z = 1 + i

Argand diagrams An Argand diagram is a geometrical representation of the set of complex numbers. In a vector sense, a complex number has two dimensions; the real part and the imaginary part. Therefore a plane is required to represent C. Im(z) An Argand diagram is drawn with two 3 perpendicular axes. The horizontal axis represents Re(z), z ∈ C, and the vertical 2 axis represents Im(z), z ∈ C. (3 + i) (–2 + i) 1 Each point on an Argand diagram represents a complex number. The complex number a + bi 0 1 –3 –2 –1 2 3 Re(z) is situated at the point (a, b) on the equivalent –1 cartesian axes as shown by the examples in this –2 figure. The number written as a + bi is called the cartesian form of the complex number. (2 – 3i) –3



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Example 16

Im(z) 5 4 B

Write down the complex number represented by each of the points A to F on this Argand diagram.

A

D 5

C 0

–5 E

Re(z) F

–5 Solution A : 2 + 3i C : −5 E : −5 − 2i

B : 4i D : −1 + i F : 1 − 3i

Geometrical representation of the basic operations on complex numbers The addition of two complex numbers is similar to a vector sum and follows the triangle of vectors rule. The multiplication by a scalar follows vector properties of parallel position vectors.

z1 + z2

Im (z)

Im(z)

az

z2

z 0

z1 0

Re (z)

bz Re(z)

cz

a>1 0

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Solution a

Im (z)

b

4 z2 3 2 1

Im(z) 3 –1 + 2i

2 1

2

–3 –2 –1 0 1 –1

z1 + z2

z1

–4 –3 –2 –1 0 1 2 3 4 Re (z) –1 z1 – z2 –2 –z2 –3 –4

2 3 Re (z) 2–i

–2 –(2 + 3i)

427

–3 –3i

z 1 + z 2 = (2 + i) + (−1 + 3i) = 1 + 4i z 1 − z 2 = (2 + i) − (−1 + 3i) = 3 − 2i

Rotation about the origin When the complex number 2 + 3i is multiplied by –1 the result is −2 − 3i. This can be considered to be achieved through a rotation of 180◦ about the origin. When the complex number 2 + 3i is multiplied by i, i(2 + 3i) = 2i + 3i 2 = 2i − 3 = −3 + 2i

i.e.

Im(z) 2 + 3i –3 + 2i

the result can be seen to be achieved through a rotation of 90◦ in an anticlockwise direction about the origin. If −3 + 2i is multiplied by i the result is −2 − 3i. This is again achieved through a rotation of 90◦ in an anticlockwise direction about the origin.

0

Re(z)

–2 – 3i

Example 18 If z 1 = 1 − 4i and z 2 = −2 + 2i, find z 1 + z 2 algebraically and illustrate z 1 + z 2 on an Argand diagram. Solution Im(z)

z 1 + z 2 = (1 − 4i) + (−2 + 2i) = −1 − 2i

3 2

z2

1 0 –4

–3

–2

–1

z1 + z2

–1 1

2

3

4 Re(z)

–2 –3 –4

z1



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16

Im(z)

1 Write down the complex numbers represented on the following Argand diagram.

3 B 2 1

E

–5 –4 –3 –2 –1 0 –1 1 F –2 –3 –4 C Example

17

18

2

3

4

5 Re (z)

D

2 Represent each of the following complex numbers as points on an Argand plane. a 3 − 4i

Example

A

b −4 + i

c 4+i

d −3 + 0i

e 0 − 2i

f −5 − 2i

3 If z 1 = 6 − 5i and z 2 = −3 + 4i, find algebraically and represent on an Argand diagram. a z1 + z2

b z1 − z2

4 If z = 1 + 3i, represent on an Argand diagram a z

b z

c z2

d −z

e

1 z

5 If z = 2 − 5i, represent on an Argand diagram a z

b zi

c zi 2

d zi 3

e zi 4

16.5 Solving equations over the complex field Quadratic equations for which the discriminant is less than zero have no solutions for the real numbers. The introduction of the complex number enables such quadratic equations to be solved. Further solutions to higher degree polynomials may also be found using complex numbers. Solution of higher degree polynomials appears in the Specialist Mathematics course. In this chapter only quadratics will be considered.

Sum of two squares Earlier it was seen that the product of a complex number a + bi and its conjugate a − bi yielded the result (a + bi)(a − bi) = a 2 + b2 Hence sums of two squares can be factorised enabling equations of the form z 2 + a 2 = 0 to be solved.




429

Example 19 Solve the equations a z 2 + 16 = 0

b 2z 2 + 6 = 0

Solution a

z 2 + 16 = 0 ∴ z − 16i 2 = 0 (z + 4i)(z − 4i) = 0 ∴ z = ±4i

b

2

2z 2 + 6 = 0 ∴ 2(z 2 + 3) = 0 ∴ 2(z 2 − 3i 2 ) = 0 √ √ ∴ 2(z + 3i)(z − 3i) = 0 √ ∴ z = ± 3i

Solution of quadratic equations To solve quadratic equations where the discriminant is less than zero, still use the quadratic formula in the usual way. Example 20 Solve the equation 3z 2 + 5z + 3 = 0 Solution Using the quadratic formula √ −5 ± 25 − 36 z= √6 −5 ± −11 = 6 √ 1 = (−5 ± 11i) 6

Using the TI-Nspire Each of the expressions in the above examples can be factorised using cFactor from the Complex submenu of Algebra. For example cfactor (z 2 + 16, z). Each of the equations in the above examples can be solved using cSolve from the Complex submenu of Algebra, for example, cSolve (3z 2 + 5z + 3 = 0, z).



430


Using the Casio ClassPad To factorise in the above examples, ensure the mode is set to Cplx. Enter and highlight the expression z 2 + 16 then tap Interactive, Transformation, rFactor.

To solve in the above examples, the usual method for solving equations is used. For example, enter and highlight 3z 2 + 5z + 3 = 0 then tap Interactive, Equation/inequality, solve and ensure that the variable selected is z.

Exercise 16E Examples

19, 20

1 Solve each of the following equations over C. a d g j

16.6

z2 + 4 = 0 (z − 2)2 + 16 = 0 z 2 + 3z + 3 = 0 2z = z 2 + 5

b e h k

2z 2 + 18 = 0 (z + 1)2 = −49 2z 2 + 5z + 4 = 0 2z 2 − 6z = −10

c f i l

3z 2 = −15 z 2 − 2z + 3 = 0 3z 2 = z − 2 z 2 − 6z = −14

Polar form of a complex number Earlier in this chapter it was shown that points on a cartesian plane (x, y) may be represented in terms of polar coordinates [r, ␪]. Similarly, complex numbers may be represented in polar form. Im (z) Recalling that x = r cos ␪ and y = r sin ␪ where r 2 = x 2 + y 2 then the point P in the complex plane P(x + iy) corresponding to the complex number in cartesian form, r y z = x + yi, may be represented as shown in the diagram. θ z = r cos ␪ + r sin ␪ i x 0 Re (z) = r (cos ␪ + sin i) The polar form is abbreviated to z = r cis ␪. r = x 2 + y 2 is called the absolute value or modulus of z. It is denoted by mod z or |z|. Remember that ␪ is measured in an anticlockwise direction from the horizontal axis. The same point may be represented a number of ways in polar form, since cos ␪ = cos(␪ ± 2n␲) and sin ␪ = sin (␪ ± 2n␲), where n ∈ Z , the polar form of a complex number is not unique.

Note:

i.e.

z = r cis ␪ = r cis (␪ + 2n␲), n ∈ Z



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Usually the interval −␲ < ␪ ≤ ␲ is used. The corresponding value of ␪ is called the principal value of the argument of z and is denoted by Arg z. i.e.

− ␲ < Arg z ≤ ␲

Example 21 Express in polar form the following complex numbers √ b z = 2 − 2i a z = 1 + 3i Solution

√ 3 i is a point in the 1st quadrant. ␲ ∴ 0< ␪ < 2 √ Now x = 1 and y = 3 √ Therefore r = 1 + 3 =2 √ 1 3 ␲ ) also ␪= (since cos ␪ = and sin ␪ = 3 √ 2 2 ∴ z = 1 + 3i ␲ = 2 cis 3

a First note that z = 1 +

b z = 2 − 2i is a point in the 4th quadrant. −␲ <␪<0 2 Now x = 2 and y = 2 √ Therefore r = 4+4 √ = 8 √ =2 2 1 −1 −␲ (since cos ␪ = √ and sin ␪ = √ ) Also ␪= 4 2 2 ∴ z = 2 − 2i √ −␲ = 2 2 cis 4

∴



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Example 22

−2␲ Express in cartesian form z = 2 cis 3 Solution

−2␲ x = r cos ␪ = 2 cos 3 1 =2 − 2 = −1

−2␲ y = r sin ␪ = 2 sin 3

√ − 3 =2 2 √ =− 3 −2␲ ∴ z = 2 cis 3 √ = −1 − 3i

Multiplication and division in polar form z 1 = r1 cis ␪ and z 2 = r2 cis ␪2

If Then and

z 1 z 2 = r1r2 cis (␪1 + ␪2 ) r1 z1 = cis (␪1 − ␪2 ) z2 r2

These results may be proved using the addition formulas for sine and cosine established in Chapter 11. This is left as an exercise for the reader. Example 23 If z 1 = 2 cis 30◦ and z 2 = 4 cis 20◦ find the product z 1 z 2 and represent it on an Argand diagram. Solution

Im (z) z 1 z 2 = r1r2 cis (␪1 + ␪2 ) = 2 × 4 cis (20◦ + 30◦ ) = 8 cis 50◦

z1z2 50°

z1

30° z2 Re (z)

0 20° ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 24 ␲ 5␲ , find the product z 1 z 2 . If z 1 = 3 cis and z 2 = 2 cis 2 6 Solution z 1 z 2 = r1r2 cis (␪1 + ␪2 ) ␲ 5␲ = 6 cis + 2 6 4␲ = 6 cis 3 −2␲ since −␲ < Arg z ≤ ␲ ∴ z 1 z 2 = 6 cis 3 Example 25 √ √ z1 If z 1 = − 3 + i and z 2 = 2 3 + 2i, find the quotient and express it in cartesian form. z2 Solution First express z 1 and z 1 in polar form. |z 1 | =

√

3+1

=2 √ |z 2 | = 12 + 4 =4

√ 5␲ 1 − 3 , since sin ␪1 = and cos ␪1 = Arg z 1 = 6 2 2 where Arg z 1 = ␪1 √ ␲ 1 3 Arg z 2 = , since sin ␪2 = and cos ␪2 = 6 2 2 where Arg z 2 = ␪2

5␲ ␲ and z 2 = 4 cis z 1 = 2 cis 6 6 r1 z1 = cis (␪1 − ␪2 ) z2 r2 2 5␲ ␲ = cis − 4 6 6 1 2␲ = cis 2 3

∴

1 2␲ 1 2␲ z1 = cos + sin i In cartesian form z2 2 3 2 3

√ 1 3 1 −1 + i = 2 2 2 2 √ 1 = − (1 − 3)i 4



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21

1 Express each of the following in the simplest polar form. √ √ a 1 + 3i b 1−i c −2 3 + 2i √ 1 1 d −4 − 4i e 12 − 12 3i f − + i 2 2

Example

22

2 Express each of the following in the form x + yi. √ ␲ ␲ ␲ 3␲ c 2 cis 2 cis b a 3 cis d 5 cis 6 3 2 4 5␲ √ 4␲ −2␲ −␲ e 12 cis g 5 cis h 5 cis f 3 2 cis 6 3 3 4 3 Simplify the following and express the answers in cartesian form. ␲ ␲ ␲ ␲ . 3 cis b 4 cis . 3 cis a 2 cis 12 4 6 12

␲ 5␲ 2␲ −␲ c cis . 5 cis . 3 cis d 12 cis 4 12 3 3 √ √ 5␲ ␲ −3␲ e 12 cis . 3 cis f ( 2 cis ␲). 3 cis 6 2 4

−␲ ␲ 12 cis 10 cis 3 4 g h ␲ 2␲ 5 cis 3 cis 12 3 √ 3␲ −␲ 12 8 cis 20 cis 4 6

i √ j ␲ 5␲ 3 2 cis 8 cis 12 6

Examples

23, 24, 25



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The polar coordinates [r, ␪] may be represented as follows. ␪ is measured in an anticlockwise direction from the polar axis.

P[r, θ]

θ O pole

Review

Chapter summary

polar axis

For conversion of coordinates from cartesian to polar and vice versa x = r cos ␪, y = r sin ␪ and hence x 2 + y 2 = r 2 . ␲ √ ␲ ␲ describes the same point as 2 cos , 2 sin = (1, 3) Therefore 2, 3 3 3 √ For (1, −1), r= 2 √ √ 1 = 2 cos ␪ and − 1 = 2 sin ␪ −1 1 Therefore cos ␪ = √ and sin ␪ = √ 2 2 −␲ and ␪ = 4 √ −␲ (1, −1) = 2, 4 For conversion of an equation from cartesian to polar use x = r cos ␪, y = r sin ␪ and x 2 + y 2 = r 2 Therefore x + y = 1 becomes r cos ␪ + r sin ␪ = 1 i.e. r (cos ␪ + sin ␪) = 1 Consider y = x 2 This becomes

∴ ∴

r sin ␪ = r 2 cos2 ␪ r sin ␪ − r 2 cos2 ␪ = 0 r (sin ␪ − r cos2 ␪) = 0 r = 0 or sin ␪ = r cos2 ␪ r = 0 is the pole. The second equation becomes r =

tan ␪ cos ␪

i is an imaginary number with the property i 2 = −1. C, the set of complex numbers, is defined by C = {a + bi : a, b ∈ R}. Real numbers and Imaginary numbers are subsets of C. Re (z) is the real component of z. Im (z) is the value of the imaginary component of z.



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z 1 = z 2 ⇔ Re (z 1 ) = Re (z 2 ) and Im (z 1 ) = Im (z 2 ) The Argand diagram is a geometrical representation of C. Let z 1 = a + bi and z 2 = c + di, then z 1 z 2 = (ac − bd) + (ad + bc)i The modulus of z, |z|, is the distance from the origin of the point represented by z. The argument of z, arg z, is an angle measured anticlockwise about the origin from the positive direction of the x axis to the line joining the origin to z. The Argument of z, Arg z, is arg z expressed as an angle in the interval (−␲, ␲]. The modulus-argument form of the complex number z is given as: z = r (cos ␪ + i sin ␪) where r = |z| Re (z) Im (z) cos ␪ = and sin ␪ = |z| |z| r (cos ␪ + i sin ␪) is usually written as r cis ␪ The complex conjugate of z is denoted by z, where z = Re (z) − Im (z)i; zz, (z + z) ∈ R z1 z1 z2 The division of complex numbers: = z2 |z 2 |2 Multiplication and division of the modulus-argument form Let z 1 = r1 cis ␪1 , z 2 = r2 cis ␪2 Then z 1 z 2 = r1r2 cis (␪1 + ␪2 ),

z1 r1 = cis (␪1 − ␪2 ) z2 r2

Multiple-choice questions 1 The polar coordinates [−3, 30◦ ] define a point that can also be described by B [3, −30◦ ] C [3, 150◦ ] D [3, −150◦ ] E [−3, 150◦ ] A [3, 30◦ ] 2 The polar coordinates of point A are B [2, −40◦ ] C [2, 140◦ ] A [2, 40◦ ] ◦ ◦ E [2, −140 ] D [4, 40 ]

A 2 40°

O √ 3 The polar coordinates of the point with cartesian coordinates (−1, − 3) are ␲ 2␲ ␲ −4␲ ␲ C 2, D 2, E −2, − A 2, B −2, 3 3 3 3 3 ␲ 4 The cartesian coordinates of the point with polar coordinates 3, are

√ 6

√ √ √ 3 3 3 3 3 3 1 3 3 , C , D B 3, √ A (3, 3) , E 2 2 2 2 2 2 3 ␲ 5 The polar equation of the circle with centre 3, and radius 3 is 2 √ A r = 3 sin ␪ B r = 3 C r =3 D r = 3 cos ␪ E r = 6 sin ␪


Z


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6 The graph of r cos ␪ = 2 is A B

437

C O O

O

D

E

O

O

7 The polar equation of the circle with cartesian equation x 2 + y 2 = 16 is D r = 4 cos ␪ E r = 4 A r = 16 B r = 4 sin ␪ C r 2 + cos2 ␪ = 4 1 8 If u = 1 + i, then = 2−u 1 1 1 2 1 1 1 1 A − − i + i B + i C E 1 + 5i D − + i 2 2 5 5 2 2 2 5 9 The point C on the Argand diagram represents the complex number z. Which point represents the complex number i × z? Im (z) A A B B C C D D E E

B C

A Re (z) D

1 10 If |z| = 5 then = z 1 1 A √ B −√ 5 5

C

1 5

D −

1 5

E

E

√ 5

Short-answer questions (technology-free) 1 Graph each of the following. ␲ a [3, ␲] b 2, 3

◦

c [−2, 210 ]

d

11␲ −3, 6

2 Find the cartesian coordinates of the points in 1.



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3 Graph each of the following. ␲ a {[r, ␪] : r = 3} b [r, ␪] : ␪ = 3 −5␲ c {[r, ␪] : r = −4} d [r, ␪] : ␪ = 4 4 Express each of the following in polar form.

√ √ √ √ −5 −5 3 3 −1 a (3, 3) c , d (4 2, −4 2) , b 2 2 2 2 5 Transform each of the following equations from cartesian to polar form. b x 2 + y2 = 9 c y 2 = 8x a x 2 + y 2 = 16 e x 2 + 4y 2 = 64 f 2x − y + 2 = 0 d x 2 = 4y 6 Transform each of the following equations from polar to cartesian form. a r =5 b r = 3 sin ␪ c r 2 cos 2␪ = 9 d r (1 − 2 cos ␪) = 8 e r (2 − cos ␪) = 7 f r (1 − sin ␪) = 1 7 For z 1 = m + in and z 2 = p + iq, express each of the following in the form a + ib. b z2 c z1 z2 a 2z 1 + 3z 2 z1 e z1 + z1 f (z 1 + z 2 )(z 1 − z 2 ) d z2 z2 1 3z 1 h g i z1 z1 z2 √ 8 In the following, z = 1 − 3i. Express each in the form a + ib and mark each of the following on an Argand diagram. 1 1 e z c z3 d f a z b z2 z z 9 Write each of the following in polar form. √ √ c 2 3+i a 1+i b 1 − 3i √ √ √ √ √ e −3 2 − 3 2i f 3−i d 3 2 + 3 2i 10 Write each of the following in cartesian form. ␲ ␲ a −2 cis b 3 cis 3 4 −3␲ −5␲ d −3 cis e 3 cis 4 6

3␲ 4 √ −␲ f 2 cis 4

c 3 cis

Extended-response questions 1 a Plot the graphs of r = 2 sin ␪ and r = 2 cos ␪. b Write the corresponding cartesian equation for each of these relations. c Describe the curves you obtain from the polar equations r = 2a sin ␪ or r = 2a cos ␪ where a is a non-zero constant.



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Review

2 Investigate each of the families of graphs defined by: a r = a + b sin ␪ or r = a + b cos ␪ where a and b are non-zero constants b r 2 = a 2 sin 2␪, r 2 = −a 2 sin 2␪, r 2 = a 2 cos 2␪, r 2 = −a 2 cos 2␪ where a is a positive constant c r = a␪ d r = a sin n␪ and r = a cos n␪ where a is a non-zero constant √ 3 a Find the exact solutions in C for the equation z 2 − 2 3z + 4 = 0. b i Plot the two solutions from a on an Argand diagram. ii Find the equation of the circle, with centre the origin, which passes through these two points. iii Find the value of a ∈ Z such that the circle passes through (0, ±a) 4 Let z be a complex number with |z| = 6. Let A be the point representing z. Let B be the point representing (1 + i)z. Find a |(1 + i)z| b |(1 + i)z − z| c Prove that OAB is an isosceles right-angled triangle. 1 1 5 Let z = √ + √ i 2 2 a On an Argand diagram O, A, Z , P, Q represent the complex numbers 0, 1, z, 1 + z and 1 − z respectively. Show these points on a diagram. |OP| ␲ . b Prove that the magnitude of ∠POQ = . Find the ratio |OQ| 2



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C H A P T E R

17 Loci Objectives To find the cartesian equation of a locus, where each point P of the locus satisfies the following properties:

r that P is equidistant from two given points A and B, i.e., PA = PB r that P is a fixed distance from a given point A, i.e., there is a positive number k such that PA = k

r that PA = kPB, i.e., the distance from a fixed point A is k times its distance from a fixed point B

r that the sum of the distances from two points A and B is always a constant, i.e. PA + PB = k

r that the difference of the distances from two points A and B is always a constant, i.e. PA − PB = k To sketch the graphs of circles, ellipses and hyperbolas To consider the asymptotes of hyperbolas

17.1 Introduction and parabolas

Introduction In order to find the equation of a curve, some condition must be given which establishes which points are on a curve. Up to now in this book, curves have been described through a relationship between the x and y coordinates (and in Chapter 16, between the polar coordinates). For example, y = 2x is the straight line through the origin with gradient 2 (and for polar coordinates, r = 5 is the circle with center the origin and radius 5). Many curves can also be described through a geometric description. For example, the set of points equidistant from the points A(4, 0) and B(2, 0) lie on the line with equation x = 3. A locus is a set of points which satisfy a condition. For the above example, the locus of the points which are equidistant from A and B is the straight line with equation x = 3. Note that every point which lies on the line x = 3 satisfies this condition. This is an important observation which should be thought about with every locus problem.



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Chapter 17 — Loci

Example 1 Find the equation of the locus of points P satisfying PA = PB, where A is the point with coordinates (3, 0) and B is the point with coordinates (6, 4). Solution Let (x, y) be the coordinates of point P. If PA = PB Then (x − 3)2 + (y − 0)2 = (x − 6)2 + (y − 4)2 Squaring both sides and expanding

y 43 8

(6, 4)

x 2 − 6x + 9 + y 2 = x 2 − 12x + 36 + y 2 − 8y + 16 −6x + 9 = −12x − 8y + 52 8y + 6x = 43

43 6

O (3, 0) The locus is a straight line as shown. Every point P on the line also satisfies the property that PA = PB.

x

Example 2 Find the equation of the locus of points P satisfying PA = 3, where A is the point with coordinates (2, 1). Solution Let (x, y) be the coordinates of point P. If PA = 3 Then (x − 2)2 + (y − 1)2 = 3 Squaring both sides (x − 2)2 + (y − 1)2 = 9 The locus is a circle with centre (2, 1) and radius 3. Every point P on the circle satisfies the property that PA = 3. Example 3 Find the equation of the locus of points P satisfying PO = 2PA, where A is the point with coordinates (4, 0) and O is the origin. Solution Let (x, y) be the coordinates of point P. If PO = 2PA Then x 2 + y 2 = 2 (x − 4)2 + y 2



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Squaring both sides x 2 + y 2 = 4[(x − 4)2 + y 2 ] Expanding x 2 + y 2 = 4[x 2 − 8x + 16 + y 2 ] x 2 + y 2 = 4x 2 − 32x + 64 + 4y 2 0 = 3x 2 − 32x + 64 + 3y 2 64 32 2 + 3y 2 0=3 x − x+ 3 3 Completing the square 256 64 256 32 2 + − + 3y 2 0=3 x − x+ 3 9 3 9 16 2 64 0=3 x− + 3y 2 − 3 3 2 64 16 = x− + y2 9 3 8 16 , 0 and radius . The locus is a circle with centre 3 3 Every point P on this circle satisfies the property that PO = 2 PA.

Parabolas Example 4 Find the equation of the locus of points P satisfying PM = PF, where F is the point with coordinates (3, 0) and PM is the perpendicular distance from P to the line with equation x = −3. y

Solution Let (x, y) be the coordinates of point P. If PF = PM Then (x − 3)2 + y 2 = (x + 3)2 Squaring both sides

x = –3 M(–3, y)

(x − 3)2 + y 2 = (x + 3)2

P(x, y)

O

Hence

x

F(3, 0)

y x 2 − 6x + 9 + y 2 = x 2 + 6x + 9

M(–3, y)

Therefore y 2 = 12x

P(x, y) O F(3, 0)

x

This is a parabola.

x = –3 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Exercise 17A 1 Sketch the locus of points P(x, y) for each of the following and hence write down its cartesian equation. Example

1

Example

2

a b c d

P is equidistant from the points A(3, 0) and B(6, 0) P is equidistant from the points A(0, 8) and B(0, 12) P is always three units from the origin A triangle OAP has vertices O(0, 0), A(4, 0) and P(x, y). The triangle has area 12 square units. Find the locus of P as it moves.

2 Find the locus of a point P(x, y) which moves so that it is equidistant from the origin and the point (−2, 5). 3 Find the locus of a point P(x, y) which moves so that it is equidistant from the points (0, 6) and (−2, 4). 4 Find the locus of a point P(x, y) which moves so that the sum of the squares of its distances from the points (−2, 0) and (2, 0) is 26 units. 5 A point P(x, y) moves so that its distance from the point K (2, 5) is twice its distance from the line x = 1. Find its locus. Example

3

6 A point P moves so that its distance from the point (0, 20) is twice its distance from B(−4, 5). What is the locus of P? 7 Find the locus of a point P(x, y) which moves so that it is equidistant from the points (1, 2) and (−2, −1). 8 A point P(x, y) moves so that its distance from the point K (4, −2) is twice its distance from the origin. Find its locus. 9 Determine the locus of a point P which moves so that the difference of the squares of its distances from the points A(4, 0) and B(−4, 0) is 16. 10 Determine the locus of a point P which moves so that the square of its distance from the origin is equal to the sum of its coordinates. 11

A(0, 0) and B(4, 0) are two of the vertices of a triangle ABP. The third vertex P is such that PA : PB = 2. Find the locus of P.

12 Find the locus of the point P which moves so that it is always equidistant from two fixed points A(1, 2) and B(−1, 0). 13 Given two fixed points A(0, 1) and B(2, 5) find the locus of P if the slope of AB equals that of BP. 14 P moves so that its distance from the line y = 3 is always 2 units. Find the locus of P. Example

4

15 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(2, 0) is equal to the distance PM, the perpendicular distance to the line with equation x = −4. That is, PF = PM.



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16 Find the equation of the locus of points P(x, y) which satisfy the property that the distance from P to the point F(0, −4) is equal to the distance PM, the perpendicular distance to the line with equation y = 2. That is, PF = PM. 17 Describe the locus, in terms of equal distance from a line and a point, of a parabola with equation y 2 = 3x.

17.2 Ellipses The equation for an ellipse can be found in a similar way to those loci considered in Section 17.1.

Example 5 Find the equation of the locus of points P satisfying PA + PB = 8, where A is the point with coordinates (−2, 0) and B is the point with coordinates (2, 0). Solution Let (x, y) be the coordinates of point P. If PA + PB = 8 Then (x + 2)2 + y 2 + (x − 2)2 + y 2 = 8 (x + 2)2 + y 2 = 8 − (x − 2)2 + y 2 Then Squaring both sides (x + 2)2 + y 2 = 64 − 16 (x − 2)2 + y 2 + (x − 2)2 + y 2 Expanding and simplifying

and

x 2 + 4x + 4 + y 2 = 64 − 16 (x − 2)2 + y 2 + x 2 − 4x + 4 + y 2 x − 8 = −2 (x − 2)2 + y 2

Squaring both sides and expanding x 2 − 16x + 64 = 4(x 2 − 4x + 4 + y 2 ) Simplifying yields y2 x2 + =1 16 12 This is an ellipse with centre the origin, x axis intercepts 4 and −4 and y axis √ √ –4 intercepts 2 3 and −2 3. Every point on the ellipse satisfies the property that PA + PB = 8. 48 = 3x 2 + 4y 2

or


y –

2√3

O

4

x

–

–2√3


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Chapter 17 — Loci

In general, an ellipse can be defined as the locus of the point P so that, as it moves, PA + PB = k for some k greater than the distance between A and B. This is shown in the diagram.

445

y P1

P2

O F1

x

F2

P1F1 + P1F2 = P2F1 + P2F2 = P3F1 + P3F2

P3

This can be pictured as a string of length P1 F1 + P1 F2 being attached by nails to a board at F1 and F2 and, considering the path mapped out by a pencil, extending the string so that it is taut, and moving ‘around’ the two points. Example 6 Find the image of the circle x 2 + y 2 = 1 under each of the following transformations. a a dilation of factor 4 from the x axis followed by a dilation of factor 5 from the y axis b a dilation of factor 4 from the x axis followed by a dilation of factor 5 from the y axis and then a translation of 4 units in the positive direction of the x axis and 3 units in the negative direction of the y axis Solution a The transformation is defined by the rule (x, y) → (5x, 4y) Therefore let x = 5x and y = 4y where (x , y ) is the image of (x, y) under the transformation. y x and x = . The image is Hence x = 5 4 –5 (x )2 (y )2 + = 1. This is an ellipse with centre the 25 16 origin, x axis intercepts 5 and −5 and y axis intercepts 4 and −4. b The transformation is defined by the rule (x, y) → (5x + 4, 4y − 3) Therefore let x = 5x + 4 and y = 4y − 3 where (x , y ) is the image of (x, y) under the transformation. x − 4 y + 3 Hence x = and y = . The image is 5 4 2 2 (y + 3) (x − 4) + = 1. This an ellipse with 25 16 centre (4, −3).

y 4 O

5

x

–4

y 4 x

O (4, –3) –4

Example 7 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(1, 0) is half the distance PM, the perpendicular distance to the line with 1 equation x = −2. That is, PF = PM. 2 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Essential Advanced General Mathematics y

Solution Let (x, y) be the coordinates of point P. 1 If PF = PM 2 1 (x − 1)2 + y 2 = (x + 2)2 2 Squaring both sides (x − 1)2 + y 2 4(x 2 − 2x + 1) + 4y 2 4x 2 − 8x + 4 + 4y 2 3x 2 − 12x + 4y 2

x=–2

M(–2, y)

P(x, y)

O

1 = (x + 2)2 4 = x 2 + 4x + 4 = x 2 + 4x + 4 =0

F(1, 0)

x

Completing the square 3[x 2 − 4x + 4] + 4y 2 = 12 3(x − 2)2 + 4y 2 = 12

y2 (x − 2)2 + =1 4 3

or equivalently

This is an ellipse with centre (2, 0). It can be shown that the locus of points P(x, y) satisfying PF = ePM, where 0 < e < 1, F is a fixed point and PM is the perpendicular distance from P to a fixed line l, is an ellipse. From the symmetry of the ellipse it is clear that there is a second point F and a second line l such that PF = ePM , where PM is the perpendicular distance from P to l , that defines the same locus.

l M

l' P

M' F

F'

Exercise 17B 1 Sketch the graph of each of the following ellipses. Label axes intercepts. x2 y2 x2 y2 b + =1 a + =1 25 100 9 64 y2 x2 c + =1 d 25x 2 + 9y 2 = 225 9 64



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2 Sketch the graph of each of the following ellipses. State the centre and label the axes intercepts. (x + 3)2 (y + 4)2 (y − 4)2 (x − 3)2 b + =1 + =1 a 9 25 9 64 (y − 3)2 (x − 2)2 c + =1 d 25(x − 5)2 + 9y 2 = 225 16 4 3 Sketch the graph of the image of the circle with equation x 2 + y 2 = 1 transformed by a dilation of factor 3 from the x axis and a dilation of factor 5 from the y axis. Give the equation of this image. Example

5

4 Find the locus of the point P as it moves such that the sum of its distances from two fixed points F(4, 0) and F (−4, 0) is 10 units. 5 Sketch the graph of the image of the circle with equation x 2 + y 2 = 1 under the transformation, dilation of factor 4 from the x axis and a dilation of factor 8 from the y axis. Give the equation of this image.

Example

7

6 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(2, 0) is half the distance PM, the perpendicular distance to the line with 1 equation x = −4. That is, PF = PM. 2 7 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(0, 8) is half the distance PM, the perpendicular distance to the line with 1 equation y = 4. That is, PF = PM. 2

17.3

Hyperbolas x2 y2 − = 1 is a hyperbola with centre at the origin. The axis a2 b2 b b intercepts are (a, 0) and (−a, 0). The hyperbola has asymptotes y = x and y = − x. An a a argument for this is as follows. The equation y The curve with equation

y2 x2 − =1 a2 b2 may be rearranged

∴

y2 x2 − =1 a2 b2

x2 y2 = −1 b2 a2 a2 b2 x 2 2 y = 2 1− 2 a x

a2 →0 x2 b2 x 2 y2 → 2 a bx y→± a

b y= ax

–b y= a x

(–a, 0)

0

(a, 0)

x

But as x → ±∞,

∴ i.e.



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The general equation for a hyperbola is formed by suitable translations. The curve with equation (y − k)2 (x − h)2 − =1 a2 b2 is a hyperbola with centre (h, k). The asymptotes are b y − k = ± (x − h) a

x2 y2 This hyperbola is obtained from the hyperbola with equation 2 − 2 = 1 by the a b translation defined by (x, y) → (x + h, y + k). Example 8 For each of the following equations, sketch the graph of the corresponding hyperbola, give the coordinates of the centre, the axes intercepts and the equations of the asymptotes. a

y2 x2 − =1 9 4

y2 x2 − =1 9 4 (y − 1)2 (x + 2)2 d − =1 4 9 b

c (x − 1)2 − (y + 2)2 = 1 Solution a

∴

y2 x2 − =1 9 4 9 4x 2 2 1− 2 y = 9 x

y 2 y=– x 3

y=

2 x 3

Equations of asymptotes 2 y=± x 3

(–3, 0)

(3, 0)

0

x

When y = 0, x 2 = 9 and therefore x = ±3 Axes intercepts (3, 0) and (−3, 0), centre (0, 0) b

x2 x2 y2 y2 − = 1 is the reflection of − =1 9 4 9 4 in the line y = x

∴

i.e.

asymptotes are 2 x =± y 3 3 y=± x 2

The y axis intercepts are (0, 3) and (0, −3)


y 3 y=– x 2

3 y= x 2 (0, 3)

0

x

(0, –3)


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c (x − 1)2 − (y + 2)2 = 1. The graph of x 2 − y 2 = 1 is sketched first. The asymptotes are y = x and y = −x. This hyperbola is called a rectangular hyperbola as its asymptotes are perpendicular. The centre is (0, 0) and the axes intercepts are at (1, 0) and (−1, 0). A translation of (x, y) → (x + 1, y − 2) is applied. The new centre is (1, −2) and the asymptotes have equations y + 2 = ±(x − 1), i.e., y = x − 3 and y = −x − 1. When x = 0, y = −2 and when y = 0, (x − 1) = 5 √ x =1± 5 2

y y = –x

y=x

(–1, 0) 0

(1, 0)

x

y y = –x – 1

y=x–3

–1 0 3 x –1 (1 + √5, 0) (1 – √5, 0) (1, –2) (0, –2) (2, –2) –3

d

(x + 2)2 (y − 1)2 − =1 4 9

x2 y2 − = 1 through the 4 9 translation defined by (x, y) → (x − 2, y + 1)

This is obtained by translating the hyperbola

y

√52 +1 3

2 y= x 3

y=

7 2 x+ 3 3

(–2, 3)

(0, 2) y2 x2 – =1 4 9

y

x

(–2, 1)

0

( y – 1)2 (x + 2)2 – =1 4 9 x

0 (–2, –1) (0, –2)

Note:

the asymptotes for

2 y=– x 3

1–

√52 3

1 2 y=– x– 3 3

y2 x2 − = 1 are the same as for those of the hyperbola 4 9

y2 x2 − = 1. The two hyperbolae are called conjugate hyperbolae. 9 4 Example 9

Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(1, 0) is twice the distance PM, the perpendicular distance to the line with equation x = −2. That is, PF = 2PM.



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Solution Let (x, y) be the coordinates of point P. If PF = 2PM (x − 1)2 + y 2 = 2 (x + 2)2

y x = –2

Squaring both sides

P(x, y) M(–2, y)

(x − 1)2 + y 2 = 4(x + 2)2 x 2 − 2x + 1 + y 2 = 4(x 2 + 4x + 4) x 2 − 2x + 1 + y 2 = 4x 2 + 16x + 16 0 = 3x 2 + 18x − y 2 + 15

x

O

F(1, 0)

Completing the square 0 = 3[x 2 + 6x + 9] − y 2 + 15 − 27 3(x + 3)2 − y 2 = 12

y2 (x + 3)2 − =1 4 12

or equivalently

y

This is a hyperbola with centre (−3, 0) y = –√3(x + 3)

–5

–3

–1 O

x

y = √3(x + 3)

It can be shown that the locus of points P(x, y) satisfying PF = ePM, where e > 1, F is a fixed point and PM is the perpendicular distance from P to a fixed line l, is a hyperbola. From the symmetry of the hyperbola it is clear that there is a second point F and a second line l such that PF = ePM , where PM is the perpendicular distance from P to l , that defines the same locus. l M

l' P

M'

F

F'

Hyperbolas may be defined in a manner similar to the methods discussed earlier in this section for circles and ellipses. Consider the set of all points, P, such that PF1 − PF2 = k where k is a suitable constant and F1 and F2 are points with coordinates (m, 0) and (−m, 0) respectively. Then the equation of the curve defined in this way is y2 x2 − 2 = 1, 2 a m − a2

k = 2a



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Example 10 Find the equation of the locus of points P satisfying PA − PB = 3 where A is the point with coordinates (−2, 0) and B is the point with coordinates (2, 0). Solution Let (x, y) be the coordinates of point P. If PA − PB = 3 Then

(x + 2)2 + y 2 −

Then

(x − 2)2 + y 2 = 3 (x + 2)2 + y 2 = 3 +

(x − 2)2 + y 2

Squaring both sides

(x + 2)2 + y 2 = 9 + 6 (x − 2)2 + y 2 + (x − 2)2 + y 2

Expanding and simplifying

x 2 + 4x + 4 + y 2 = 9 + 6 (x − 2)2 + y 2 + x 2 − 4x + 4 + y 2 9 and 8x − 9 = 6 (x − 2)2 + y 2 . Note that this only holds if x > 8 Squaring both sides 64x 2 − 144x + 81 = 36[x 2 − 4x + 4 + y 2 ] Simplifying yields 28x 2 − 36y 2 = 63 4y 2 4x 2 − =1 9 7

x≥

3 2

3 This is the right branch of a hyperbola with centre the origin, x axis intercept . 2 √ 7x The equations of the asymptotes are y = ± . 3


8

1 Sketch the graph of each of the following hyperbolas. Label axes intercepts and give the equation of the asymptotes. x2 y2 x2 y2 b − =1 a − =1 25 100 9 64 x2 y2 − =1 c d 25x 2 − 9y 2 = 225 9 64



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2 Sketch the graph of each of the following hyperbolas. State the centre and label axes intercepts and asymptotes. (x + 3)2 (y + 4)2 (y − 4)2 (x − 3)2 b − =1 − =1 a 9 25 9 64 (y − 3)2 (x − 2)2 c − =1 d 25(x − 5)2 − 9y 2 = 225 16 4 f 2x 2 − y 2 = 4 e x 2 − y2 = 4 2 2 h 9x 2 − 25y 2 − 90x + 150y = 225 g x − 4y − 4x − 8y − 16 = 0 Example

10

Example

9

3 Find the locus of the point P as it moves such that the difference of its distances from two fixed points F(4, 0) and F (−4, 0) is 6 units. 4 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(2, 0) is twice the distance PM, the perpendicular distance to the line with equation x = −4. That is, PF = 2PM. 5 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(0, 8) is four times the distance PM, the perpendicular distance to the line with equation y = 4. That is, PF = 4PM. 6 Find the equation of the locus of points P(x, y) satisfying PA − PB = 4 where A is the point with coordinates (−3, 0) and B is the point with coordinates (3, 0).



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Lines The general equation of a straight line may be written as ax + by = c. For fixed points A and B, the locus of P(x, y), as P moves such that PA = PB, is a straight line. Circles The circle with centre the origin and radius a is the graph of the equation x 2 + y 2 = a 2 . The circle with centre (h, k) and radius a is the graph of the equation (x − h)2 + (y − k)2 = a 2 . For a fixed point A, the locus of P(x, y) as P moves such that PA = k, where k > 0, is a circle. Ellipses y2 x2 The curve with equation 2 + 2 = 1 is an ellipse with centre the origin, x axis intercepts a b (−a, 0) and (a, 0), and y axis intercepts (0, −b) and (0, b). For a > b, the ellipse will appear as shown in the diagram to the left. If b > a, the ellipse is as shown in the diagram to the right. y

y

b B

B b A' –a

O

Review

Chapter summary

A a

x

A' –a

O

A a

x

B' –b –b B'

(x − h)2 (y − k)2 + = 1 is an ellipse with centre (h, k). a2 b2 For fixed points A and B, the locus of P(x, y) as P moves such that PA + PB = k, where k is greater than the distance between A and B, is an ellipse. Hyperbolas y y2 x2 The curve with equation 2 − 2 = 1 is a hyperbola a b b –b y= x with centre the origin. The axis intercepts are (a, 0) a y= x a and (−a, 0). The hyperbola has asymptotes b b y = x and y = − x. a a (y − k)2 (x − h)2 x − =1 The curve with equation (–a, 0) (a, 0) a2 b2 0 is a hyperbola with centre (h, k). The hyperbola b b has asymptotes y − k = (x − h) and y − k = − (x − h). a a For fixed points A and B, the locus of P(x, y) as P moves such that |PA − PB| = k, where k is a suitable constant, is a hyperbola. The curve with equation



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Multiple-choice questions y

1 The equation of the ellipse shown is A 5x 2 + y 2 = 5 B 5x 2 + y 2 = 25 C x 2 + 5y 2 = 25 D x 2 + 5y 2 = 5 x 2 + y2 = 1 E 5

3 2 1 0 1 –6 –5 –4 –3 –2 –1 –1 –2

x 2

3

4

5

6

–3 –4

2 The coordinates of the x axis intercepts of the graph of the ellipse with equation y2 x2 + = 1 are 25 9 A (−5, 0) and (−3, 0) B (−3, 0) and (3, 0) C (0, −5) and (0, 5) D (−5, 0) and (5, 0) E (5, 0) and (3, 0) x 2 y 2 3 The graph of the ellipse with equation + = 1 is 9 4 y y B A 9

5 4 3 2 1

6 3

–7 –6 –5 –4 –3 –2 –1

01 2 3 4 5 6 7

–9 –8 –7 –6 –5–4 –3 –2 –1 –1 –2 –3 –4 –5

x

–3

01 2 3 4 5 6 7 8 9

x

–6 –9

y

C

x 2

3

–3 –2 –1–1 0 1 –2 –3 –4

y

E

3 2 1

2 1 –4 –3 –2 –1–1 0 1 –2

y

D

x 2

3

3 2 1 –4 –3 –2 –1–1 0 1 2 3 4 –2 –3 –4

x

4 The of the y axis intercepts of the graph of the ellipse with equation 2 coordinates (y + 2)2 x + = 1 are 9 4 A (−2, 0) and (2, 0) D (0, 0) and (0, −4)

B (−4, 0) and (4, 0) E (3, 0) and (0, 2)


C (0, −4) and (0, 4)


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6 The circle with equation (x − a)2 + (y − b)2 = 16 has its centre on the y axis and passes through the point with coordinates (4, 4). Then A a = 0 and b = 4 B a = 0 and b = 0 C a = 2 and b = 0 D a = −4 and b = 0 E a = 4 and b = 0

Review

2 2 5 The graph of the equation ax + by = 8 has y axis intercept 2 and passes through the √ 1 point with coordinates 1, 10 . Then 2 √ A a = 2 and b = 3 B a = 4 and b = 3 C a = 3 and b = 2 D a = 3 and b = 2 E a = 2 and b = 2

7 The circle with equation x 2 + y 2 = 1 is transformed to an ellipse through the following sequence of transformations: r dilation of factor 4 from the x axis r dilation of factor 3 from the y axis r translation of 4 units in the positive direction of the x axis r translation of 3 units in the positive direction of the y axis The equation of the resulting ellipse is (y − 3)2 (x − 4)2 (y − 3)2 (x − 4)2 + =1 B + =1 A 9 16 16 9 (y − 3)2 (x − 4)2 (y + 3)2 (x + 4)2 + =1 D + =1 C 36 16 3 4 (y + 3)2 (x + 4)2 + =1 E 20 48 8 The equation of the graph shown is y

(x + 2)2 y2 A − =1 27 108 (x − 2)2 y2 B − =1 9 34 4 (x + 2)2 y2 –7 11 C − =1 x O 81 324 2 –4 (x − 2)2 y2 D − =1 81 324 (x + 2)2 y2 E − =1 9 36 9 The locus of points P(x, y) which satisfy the property that PA = PB where A is the point with coordinates (2, −5) and B is the point with coordinates (−4, 1) is described by the equation A y = x −1 B y = x −6 C y = −x − 3 D y = x +1 E y =3−x 10 The locus of points P(x, y) which satisfy the property that PA = 2PB where A is the point with coordinates (2, −5) and B is the point with coordinates (−4, 1) is A a straight line B an ellipse C a circle D a parabola E a hyperbola ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Short-answer questions (technology-free) 1 A circle has equation x 2 + 4x + y 2 + 8y = 0. Find the coordinates of the centre and radius of the circle. 2 An ellipse has equation x 2 + 4x + 2y 2 = 0. Find the coordinates of the centre and the axes intercepts of the ellipse. 3 Find the locus of the point P(x, y) such that PA = PB, where A is the point with coordinates (0, 2) and B is the point with coordinates (6, 0). 4 Find the locus of the point P(x, y) such that PA = 6, where A is the point with coordinates (3, 2). y2 x2 − = 1. 5 State the equations of the asymptotes of the hyperbola with equation 9 4 6 Find the locus of the point P(x, y) such that PA = 2PB, where A is the point with coordinates (0, 2) and B is the point with coordinates (6, 0). (x − 2)2 y2 7 Sketch the graph of the ellipse with equation + = 1 and state its centre. 9 4 8 Determine the locus of a point P which moves so that the difference of the squares of its distances from two fixed points P1 (4, 0) and P2 (−4, 0) is constant.

Extended-response questions 1 Let A, B and C be points with coordinates (6, 0), (−6, 0) and (0, 6) respectively. Find the locus of the points P which satisfy each of the following. a PA = PC b PA = 6 c PA = 2PC 1 f PA + PB = 20 d PA = 2PB e PA = PB 2 g PA + PB = 12 h PA − PB = 5 i PB − PA = 5 2 Find the equation of the locus of points P(x, y) which satisfy the property that the distance of P to the point F(0, 4) is a equal to PM, the perpendicular distance to the line with equation y = 2 b half the distance PM, the perpendicular distance to the line with equation y = 2 c twice the distance PM, the perpendicular distance to the line with equation y = 2. 3 a The base of a triangle is fixed and the distance from one end of the base to the midpoint of the opposite side is a constant. Find the locus of the vertex joining the other two sides. b The base of a triangle is fixed and the ratio of the lengths of the other two sides is a constant. Find the locus of the vertex joining the other two sides. c Three vertices of a convex quadrilateral are fixed. Find the locus of the fourth vertex if the area of the quadrilateral is a constant.



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18

Revision


Multiple-choice questions 1 ST is a tangent at T to the circle with center O. If angle QOT = 150◦ then the magnitude of ∠QTS is A 70◦ D 105◦

B 75◦ E 150◦

C 95◦

O 150°

Q

T

S

2 ML and MN are tangents to the circle at L and N. The magnitude of angle LMN is A 80◦ D 110◦

B 90◦ E 140◦

M

L

C 100◦ O 40°

N

P

3 TS is a tangent at X and ZX bisects angle TXY. Given these facts it can be proved that A YZ = XT C ∠YZX = ∠ZXT E TX 2 = XY .YZ

Z T

B YZ = XZ D ∠SXY = ∠ZXY

X Y S

4 POQ is a diameter of the circle centre O. The size of angle QRS is A 90◦ D 125◦

B 100◦ E 160◦

P

C 110◦ 70°

O

S Q R



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5 In the figure O is the centre of the circle and D is the midpoint of AB. If AB = 8 cm, and CD = 2 cm, the radius of the circle is A 3 cm D 6 cm

B 4 cm E 7 cm

A 30 D 55◦

◦

B 40 E 65◦

A

O

B T C

◦

C 45

A

P

7 R, S and T are three points on the circumference of a circle, with ∠RST equal to 30◦ . The tangent to the circle at T meets the line segment SR produced at P and ∠RPT is equal to 40◦ . ∠RTS is equal to ◦

A 70 D 100◦

◦

B 80 E 110◦

R

B 90◦ E 120◦

40° T

◦

C 90

30° S

8 If AB = AC, ∠ADB = 60◦ and ∠CAD = 50◦ then ∠ABD is equal to A 80◦ D 110◦

B

D

C 5 cm

6 In the figure, TA and TB are tangents to the circle. If TA is perpendicular to TB and TA is perpendicular to AC then the magnitude of ∠BCA is ◦

C

A

C 100◦

B

50° 60°

D C

9 The unit vector in the direction of vector a = 3i − 4 j is 1 1 (3i − 4j) C i+j B A i−j (3i − 4 j) D 5 25 − → − → − → 10 If OA = 2i − 4 j + k and OB = 3i + 4 j + k , then AB equals A 5i + 2k

B −i − 8 j

C i + 8 j + 2k

E

4 3 i+ j 5 5

D i + 8j

E i

D 5i + 2 j

E i − 6j

11 If a = 2i + 4 j and b = 3i − 2 j, then a − b equals A 5i − 6 j

B −i + 6 j

C 5i − 2 j

12 The magnitude of vector a = 2i − j + 4k is √ A 21 B 21 C 19

D

√ 19

E 7

13 AB is parallel to OC, DC is parallel to OB, − → − → OB = b, OC = c, and AB = OB = OC = DC. − → AD is equal to A b+c D 2b + 2c

B 2(c − b) E |b + c|

C 2(b − c)


B

C b c

A

O

D


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459


A 3i − j + 5k D 5i − 3 j − k

B 3i − 3 j − k E 6i − 4 j − 4k

C 5i − j + 5k R

15 PQR is a straight line and PQ = 2QR. − → − → − → OQ = 2i − 3 j and OR = i + 2 j. OP is equal to A 4i − 13 j D 3i + j

B 3i − j E i − 5j

Q

C 2i − 10 j

P

Revision

14 If r = 2i − j + k and s = −i + j + 3k, then 2r − s equals

16 Let u = i + a j − 5k and v = bi − 3j + 6k. Vectors u and v are parallel when 5 −6 A a = −3 and b = −1 B a = and b = C a = 3 and b = −1 2 5 −5 6 5 2 D a= and b = E a = and b = 6 5 5 6 17 Given that a = 3i + 4j, b = 2i − j, x = i + 5j and x = sa + tb, then the scalars s and t are given by A s = −1 and t = −1 D s = 1 and t = 1

B s = −1 and t = 1 E s = 5 and t = 5

C s = 1 and t = −1 C

18 In this diagram, OABC is a trapezium. − → − → − → − → − → OA = a, OC = c and OA = 3 CB. AB equals 2 A 3c B c − a C 3c − 2a 3 2 4 a−c E D a+c 3 3 − → − → − → 19 In this diagram, OC = c, OA = a, OB = b and AC : AB = 2 : 1. c is equal to A a + 2b D 2a − b

B

c O

A

a C

c

B 3a − 2b C 2a + b E 3a + b

A

a

O

B

b

20 If v, w and z are complex numbers such that v = 4 cis (−0.3␲), w = 5 cis (0.6␲) and z = vw, then Arg z is equal to B −0.9␲ C 0.3␲ 2␲ 21 In cartesian form, 2 cis is equal to 3 √ √ √ A 3−i B − 3 + i C 1 − 3i A 0.9␲

D −0.3␲

D −1 +

√ 3 1 − i then Arg z is equal to 22 If z = − 2 2 2␲ −␲ 7␲ 4␲ D − C B A 3 6 6 3 23 The imaginary part of the complex number −2 − 3i is A −3

B −3i

C 3

D −2

√

E 1.8␲

3i

√ 3 1 − i 3 2

E

E −

5␲ 6

E 3i

2␲ ␲ and v = 5 cis then uv is equal to 2 3 ␲2 ␲ ␲2 −5␲ B 15 cis A 15 cis D 8 cis C 15 cis 3 3 3 6

24 If u = 3 cis


E 8 cis

7␲ 6


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25 The modulus of 12 − 5i is A 169

B 7

C 13

√ 119

D

√ 7

E

26 Let z = x + i y, where x and y are real numbers which are not both zero. Which one of the following expressions does not necessarily represent a real number? A z2

C z −1 z

B zz

E z+z

D Im(z)

27 If z = −14 − 7i then the complex conjugate of z is equal to A 7 − 14i

B 14 + 7i

C −14 + 7i

D 14 − 7i

E −7 + 14i

28 The expression 3z 2 + 9 is factorised over C. Which one of the following is a factor? √ A 3z B z+3 C z + 3i D z − 3i E z + 3i 29 (1 + 2i)2 is equal to A −3

B −3 + 2i

C −3 + 4i D −1 + 4i E 5 + 4i 4␲ 30 A point has polar coordinates 2, . What are its cartesian coordinates, referring to the 3 same origin and with the polar axis in the direction of the positive x axis? √ √ √ √ √ A (1, − 3) B (−1, 3) C (− 3, 1) D (−1, − 3) E (− 3, −1) √ √ 31 The point with cartesian coordinates (4 2, −4 2) has polar coordinates √ ␲ √ 5␲ 3␲ 7␲ −␲ B 8, C −4 2, D −8, E 8, A 8 2, 4 4 4 4 4 ␲ and radius 32 The polar equation of the circle with centre given by polar coordinates 2, 2 2 units is A r = 2 sin ␪

B r = 2 cos ␪

C r = 4 sin ␪

D r = 4 cos ␪

E r =2

33 Which one of the following graphs is the graph of r cos ␪ = 4? A

B O

[4, π]

[4, 0] Z

C

D O

[4, 0]

E

[ 4, π2 ] O

Z

Z

O

Z

[4, 0] Z

O

34 The circle with cartesian equation x 2 + y 2 = 9 has polar equation A r cos ␪ + r sin ␪ = 3

B r =3

C r =9

D r = 3 cos ␪

E r = 3 sin ␪

35 The curve with polar equation r 2 = 1 − r 2 sin2 ␪ has cartesian equation A x 2 + y 2 + 2y = 1 D x 2 + 2y 2 = 1

B x + 2y = 1 E x + 2y 2 = 1


C 2x 2 + y 2 = 1


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O

Which one of the following graphs best represents the relation r = A O

Z

D

O

[ 12 , 0]

Z

[ 12 , 0]

Z

O

[ 12 , 0]

Z

E O

18.2

1 ? 1 + cos ␪

C

B [2, π]

[2, 0] Z

Revision

36 The graph of the relation r = 1 + cos ␪ is the cardioid shown.

[12 , 0]

Z O

Extended-response questions 1 D is the midpoint of AC and E is a point on BC B such that BE : EC = 1 : t, where t > 0. DE is produced to a point F such that E DE : EF = 1 : 7. − → − → Let AD = a and AB = b. − → A C D a Express AE in terms of t, a and b. − → 9 − 7t − → − → 8t c Show that AF = a+ b b Express AE in terms of a and AF. 1+t 1+t d If A, B and F are collinear, find the value of t. B 2 ABC is a triangle whose vertices have position vectors a, b and c respectively, relative to an origin in the plane ABC. a Show that an arbitrary point P on the segment A AB has position vector ma + nb where m ≥ 0, n ≥ 0 and m + n = 1. (Hint: Assume P divides AB in the ratio x : y.) − → b Find PC in terms of a, b and c. c Let Q be an arbitrary point on line segment PC. Show that Q has position vector

C

␭a + ␮b + ␥ c, where ␭ ≥ 0, ␮ ≥ 0, ␭ ≥ 0 and ␭ + ␮ + ␥ = 1. → − → − → − → 4− 3 OA = a, OB = b, OP = OA and Q is the midpoint of AB. 5 − → − → a Express AB and PQ in terms of a and b. − → − → − → b PQ is produced to meet OB produced at R so that QR = n PQ and BR = kb. Express − → QR in terms of i n, a and b ii k, a and b c Find the value of n and k. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Essential Advanced General Mathematics D

4 In the figure, O is the centre of a circle. TD and AC are parallel. TA and TB are tangents to the circle. Let ∠BPT = x ◦ .

B

P x°

C O

a Prove that TBOA is a cyclic quadrilateral. b Find ∠BCA, ∠BOA, ∠TAB and ∠TBA in terms of x. A T

5 a A man walks north at a rate of 4 km/h and notices that the wind appears to blow from the west. He doubles his speed and now the wind appears to blow from the north-west. What is the velocity of the wind? (Note: Direction and magnitude must be given.) b A river 400 m wide flows from east to west at a steady speed of 1 km/h. A swimmer, whose speed in still water is 2 km/h, starts from the south bank and heads north across the river. Find the swimmer’s speed over the river bed and how far downstream he is when he reaches the north bank. c To a motorcyclist travelling due north at 50 km/h, the wind appears to come from the north-west at 60 km/h. What is the true velocity of the wind? d A dinghy in distress is 6 km on a bearing of 230◦ from a lifeboat and drifting in a direction of 150◦ at 5 km/h. In what direction should the lifeboat travel to reach the dinghy as quickly as possible if the maximum speed of the lifeboat is 35 km/h? − → − → − → 6 a Given that O, A, B and C are coplanar, OA = a, OB = b, OC = c and A, B and C are collinear with c = ␣a + ␤b where ␣, ␤ ∈ R, show that ␣ + ␤ = 1. b In the figure, G is the centroid of a triangle (i.e. the point where the lines joining each vertex to the midpoint of the opposite side meet). A line passing through G meets ZX and ZY at points H and K respectively, such that ZH = hZX and ZK = kZY . Z → − → 2− i Prove that ZG = ZM 3 − → − → − → ii Express ZG in terms of h, k, ZHand ZK 1 1 H iii Find the value of + (use the result from a). h k G K iv If h = k, find the value of h and describe X Y geometrically what this implies. M 2 v If the area of triangle XYZ is 1 cm , what is the area of triangle HZK when h = k? vi If k = 2h, find the value of h and describe geometrically what this implies. vii Describe the restrictions on h and k and sketch the graph of h against k for suitable values of k. viii Investigate the area, A cm2 , of triangle ZHK as a ratio with respect to the area of triangle XYZ, as k varies. Sketch the graph of A against k. Be careful of the domain.



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C H A P T E R

19 Kinematics Objectives To model motion in a straight line and to use calculus to solve problems involving motion in a line with constant and variable acceleration To use graphical methods to solve problems involving motion in a straight line

Introduction Kinematics is the study of motion without reference to the cause of the motion. In this chapter we will consider the motion of a particle in a straight line only. Such motion is called rectilinear motion. When referring to the motion of a particle we may in fact be referring to a body of any size. However for the purposes of studying its motion we can consider that all forces that act upon the body, causing it to move, act through a single point. Hence we can consider the motion of a car or train in the same way as we would consider the motion of a dimensionless particle. It is important to make a distinction between vector and scalar quantities when studying motion. Quantities such as displacement, velocity and acceleration must be specified by both magnitude and direction. They are vector quantities. Distance, speed and time on the other hand are specified by their magnitude only. They are scalar quantities. Since we are considering movement in a straight line only, the direction of all vector quantities is simply specified by the sign of the numerical value.

19.1

Position, velocity and acceleration Position The position coordinate of a particle moving in a straight line is determined by its distance from a fixed point O on the line, called the origin, and whether it is to the right or left of O. Conventionally the direction to the right of the origin is considered to be positive.

x O

P

X



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Consider a particle which starts at O and begins to move. The position of a particle is determined by a number, x, called the position coordinate. If the units are metres and if x = −3, the position is 3 m to the left of O, while if x = 3, the displacement is 3 m to the right of O. The displacement is defined as the change in position of the particle relative to O. Sometimes there is a rule which enables the position coordinate, at any instant, to be calculated. In this case x is redefined as a function of t. Hence x(t) is the displacement at time t. Specification of a displacement function together with the physical idealisation of a real situation constitute a mathematical model of the situation. An example of a mathematical model is the following. A stone is dropped from the top of a vertical cliff 45 m high. Assume that the stone is a particle travelling in a straight line. Let x(t) be the downwards position of the particle from O, the top of the cliff, t seconds after the particle is dropped. If air resistance is neglected, an approximate model for the displacement is x(t) = 5t 2 for 0 ≤ t ≤ 3 It is important to distinguish between the scalar quantity distance and the vector quantity displacement. Consider a particle that starts at O and moves firstly five units to the right to point P, and then seven units to the left to point Q.

Q –4

–3

–2

O –1

0

P 1

2

3

4

5

6

Its final position is x = −2. However the distance the particle has moved is 12 units. Example 1 A particle moves in a straight line so that its position x cm relative to O at time t seconds is given by x = t 2 − 7t + 6, t ≥ 0. Find a its initial position b its position at t = 4. Solution a At t = 0, b At t = 4,

x = +6 i.e. the particle is 6 cm to the right of O. x = (4)2 − 7(4) + 6 = −6 i.e. the particle is 6 cm to the left of O.

Velocity You should already be familiar with the concept of a rate of change through your studies in Mathematical Methods. The velocity of a particle is defined as the rate of change of its position with respect to time. We can consider the average rate of change, the change in position over a period of time, or



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we can consider the instantaneous rate of change, which specifies the rate of change at a given instant in time. If a particle moves from x1 at time t1 to x2 at time t2 , then its average velocity =

x2 − x1 t2 − t1

Velocity may be positive, negative or zero. If the velocity is positive the particle is moving to the right and if it is negative the direction of motion is to the left. A velocity of zero means the particle is instantaneously at rest. The instantaneous rate of change of position with respect to time is the instantaneous velocity. If the position, x, of the particle at time t is given as a function of t, then the velocity of the particle at time t is determined by differentiating the rule for position with respect to time. Common units of velocity (and speed) are: 1 metre per second = 1 m/s 1 centimetre per second = 1 cm/s 1 kilometre per hour = 1 km/h The first and third are connected in the following way: 1 km/h = 1000 m/h 1000 m/s = 60 × 60 5 m/s 18 18 ∴ 1 m/s = km/h 5 Note the distinction between velocity and speed. Speed is the magnitude of the velocity. =

Average speed for a time interval [t1 , t2 ] is equal to Instantaneous velocity v =

distances travelled t2 − t1

dx where x is a function of time. dt

Example 2 A particle moves in a straight line so that its position x cm relative to O at time t seconds is given by x = t 2 − 7t + 6, t ≥ 0. Find a its initial velocity b when and where its velocity equals zero c its average velocity for the first 4 s d its average speed for the first 4 s. Solution a

x = t 2 − 7t + 6 dx v= = 2t − 7 dt at t = 0, v = −7 i.e. the particle is moving to the left at 7 cm/s.



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2t − 7 = 0 implies t = 3.5 When t = 3.5, x = (3.5)2 − 7(3.5) + 6 = −6.25 The particle is 6.25 cm to the left of O. change in position c average velocity = change in time at t = 4, x = −6 −6 − +6 ∴ average velocity = 4 = −3 cm/s

b

d average speed =

distance travelled change in time

t=4

t = 3.5

t=0

O 1

–6 4

–6

–5

–4

–3

–2

–1

0

1

2

3

4

5

6

Since the particle has stopped at t = 3.5 and begun to move in the opposite direction, we must consider the distance travelled in the first 3.5 s (from x = 6 to x = −6.25) and then the distance travelled in the final 0.5 s (from x = −6.25 to x = −6). total distance travelled = 12.25 + 0.25 = 12.5 average speed =

12.5 = 3.125 cm/s 4

Acceleration The acceleration of a particle is defined as the rate of change of its velocity with respect to time. v2 − v1 Average acceleration for the time interval [t1 , t2 ] is defined by where v 2 is the t2 − t1 velocity at time t2 and v 1 is the velocity at time t1 . dv d dx d2x Instantaneous acceleration a = = = 2 dt dt dt dt d2x is denoted by x (t) or x¨ (t). dt 2 Acceleration may be positive, negative or zero. Zero acceleration means the particle is moving at a constant velocity. Note that the direction of motion and the acceleration need not coincide. For example, a particle may have a positive velocity indicating it is moving to the right, but a negative acceleration indicating it is slowing down. Also, although a particle may be instantaneously at rest its acceleration at that instant need not be zero. If acceleration has the same sign as velocity then the particle is ‘speeding up’. If the sign is opposite the particle is ‘slowing down’. The most commonly used units for acceleration include cm/s2 and m/s2 . For kinematics, the second derivative



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Example 3 A particle moves in a straight line so that its position x cm relative to O at time t seconds is given by x = t 3 − 6t 2 + 5, t ≥ 0. Find a its initial position, velocity and acceleration and hence describe its motion b the times when it is instantaneously at rest and its position and acceleration at those times. Solution a

for x = t 3 − 6t 2 + 5, v = 3t 2 − 12t and a = 6t − 12 t =0 x = 5, v=0 and a = −12

Particle is instantaneously at rest 5 cm to right of O with an acceleration of −12 cm/s2 . b v = 3t 2 − 12t = 0 3t(t − 4) = 0 t = 0 or t = 4 Particle is initially at rest and stops again after 4 s. At t = 0, x = 5 At t = 4, x = (4)3 − 6(4)2 + 5 = −27

and a = −12 and a = 6(4) − 12 = 12

After 4 s the position of the particle is 27 cm to the left of O and its acceleration is 12 cm/s.

Exercise 19A Examples

1, 2

1 A particle moves in a straight line so that its position x cm relative to O at time t seconds (t ≥ 0) is given by x = t 2 − 7t + 12. Find a its initial position c its initial velocity e its average velocity in the first 5 s

Example

3

b its position at t = 5 d when and where its velocity equals zero f its average speed in the first 5 s.

2 The position x metres at time t seconds (t ≥ 0) of a particle moving in a straight line is given by x = t 2 − 7t + 10. Find a when its velocity equals zero c the distance travelled in the first 5 s

b its acceleration at this time d when and where its velocity is −2 m/s.

3 A particle moving in a straight line is x cm from the point O at time t seconds (t ≥ 0) where x = t 3 − 11t 2 + 24t − 3. Find a c e g

its initial position and velocity b its velocity at any time at what times the particle is stationary d where the particle is stationary for how long the particle’s velocity is negative f its acceleration at any time when the particle’s acceleration is zero and its velocity and position at that time.



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4 A particle moves in a straight line so that its position x cm relative to O at time t seconds (t ≥ 0) is given by x = 2t 3 − 5t 2 + 4t − 5. Find a when its velocity is zero and its acceleration at that time b when its acceleration is zero and its velocity at that time. 5 A particle moving in a straight line is x cm from the point O at time t seconds (t ≥ 0) where x = t 3 − 13t 2 + 46t − 48. Find when it passes through O and its velocity and acceleration at those times. 6 Two particles are moving along a straight path so that their displacements, x cm from a fixed point P at any time, are given by x = t + 2 and x = t 2 − 2t − 2. Find a the time when the particles are at the same position b the time when they are moving with the same velocity.

19.2 Using antiderivatives for kinematics problems So far we have considered examples where the equation of motion has defined the position of the particle in terms of time and from it we have derived equations for the velocity and the acceleration by differentiation. We may be given a rule for acceleration at time t, and by the use of antidifferentiation with respect to t and some additional information we can deduce rules for both velocity and position. Example 4 A body starts from O and moves in a straight line. After t seconds (t ≥ 0) its velocity (v cm/s) is given by v = 2t − 4. Find a its position x in terms of t b its position after 3 s c its average velocity in the first 3 s d the distance travelled in the first 3 s e its average speed in the first 3 s. Solution a Antidifferentiate with respect to t to find the expression for position x m at time t seconds x = t 2 − 4t + c When t = 0, x = 0 and therefore c = 0 x = t 2 − 4t b When t = 3, x = −3. The body is 3 units to the left of O −3 − 0 c Average velocity = = −1 m/s 3



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d v = 0 when 2t − 4 = 0, i.e., when t = 2 When t = 2, x = −4 Therefore the body goes from x = 0 to x = −4 in the first 2 s, then back to −3 in the next second. It has travelled 5 m in the first 3 s. 5 e Its average speed is m/s. 3 Example 5 A particle starts from rest 3 m from a fixed point and moves in a straight line with an acceleration of a = 6t + 8. Find its position and velocity at any time t seconds. Solution dv = 6t + 8 dt 2 by antidifferentiating v = 3t + 8t + c at t = 0, v = 0 and so c = 0 ∴ v = 3t 2 + 8t a=

by antidifferentiating again x = t 3 + 4t 2 + d at t = 0, ∴

x = 3 and so d = 3 x = t 3 + 4t 2 + 3

Example 6 A stone is projected vertically upward from the top of a building 20 m high with an initial velocity of 15 m/s. Find a the time taken for the stone to reach its maximum height b the maximum height reached by the stone c the time taken for the stone to reach the ground d the velocity of the stone as it hits the ground. In this case we only consider the stone’s motion in a vertical direction so we can consider it as rectilinear motion. Also we will assume that the acceleration due to gravity is approximately −10 m/s2 (note that downward is considered the negative direction). Solution Given that ∴ at t = 0, ∴ ∴ at t = 0, ∴

a v v v x x x

= −10 = −10t + c = 15 = −10t + 15 = −5t 2 + 15t + d = 20 = −5t 2 + 15t + 20



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a The stone will reach its maximum height when v = 0

∴ −10t + 15 = 0 which implies t = 1.5 b

At t = 1.5,

x = −5(1.5)2 + 15(1.5) + 20 = 31.25

The maximum height reached by the stone is 31.25 m. c The stone reaches the ground when x = 0

∴

−5t 2 + 15t + 20 = 0 −5(t 2 − 3t − 4) = 0 −5(t − 4)(t + 1) = 0 t = 4 (solution of t = −1 is rejected since t ≥ 0)

i.e. the stone takes 4 s to reach the ground. d

At t = 4,

v = −10(4) + 15 = −25 i.e. velocity on impact is −25 m/s.


4

1 A body starts from O and moves in a straight line. After t seconds (t ≥ 0) its velocity (v cm/s) is given by v = 4t − 6. Find a its position x in terms of t c the distance travelled in the first 3 s e its average speed in the first 3 s.

b its position after 3 s d its average velocity in the first 3 s

2 The velocity (v m/s) at time t seconds (t ≥ 0) of a particle is given by v = 3t 2 − 8t + 5. It is initially 4 m to the right of a point O. Find a its displacement and acceleration at any time b its displacement when the velocity is zero c its acceleration when the velocity is zero. Example

6

3 A body moves in a straight line with an acceleration of 10 m/s2 . If after 2 s it passes through O and after 3 s it is 25 m from O, find its initial displacement relative to O. 4 A body moves in a straight line so that its acceleration a m/s2 after time t seconds (t ≥ 0) is given by a = 2t − 3. If the initial position of the body is 2 m to the right of a point O and its velocity is 3 m/s, find the particle’s position and velocity after 10 s.

Example

5

5 A body is projected vertically upwards with a velocity of 25 m/s. (Its acceleration due to gravity is −10 m/s2 .) Find a the particle’s velocity at any time b its height above the point of projection at any time



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471

c the time it takes to reach its maximum height d the maximum height reached e the time taken to return to the point of projection. 6 In a tall building the lift passes the 50th floor with a velocity of −8 m/s and an acceleration 1 of (t − 5) m/s2 . If each floor spans a distance of 6 m, find at which floor the lift 9 will stop.

19.3

Constant acceleration When considering motion of a particle due to a constant force, e.g. gravity, the acceleration is constant. There are a number of rules that we may establish by considering the case where acceleration remains constant or uniform. dv =a Given that dt by antidifferentiating we have v = at + c where c is the initial velocity. Using the symbol u for initial velocity we have v = u + at

1

dx =v dt by antidifferentiating a second time we have

Now given that

1 x = ut + at 2 + d, where d is the initial position. 2 If we consider s = x − d as the change in position of the particle from its starting point, i.e. the particle’s displacement from its initial position, we have 1 s = ut + at 2 2 2 If we transform the formula v = u + at so that t is the subject we have v −u a 1 By substitution in s = ut + at 2 2 u(v − u) a(v − u)2 + s= a 2a 2 2as = 2u(v − u) + (v − u)2 = 2uv − 2u 2 + v 2 − 2uv + u 2 = v 2 − u2 t=

i.e.

v 2 = u 2 + 2as

3



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Also we know that displacement travelled = average velocity × time. 1 (u + v)t 4 2 These four formulas are very useful but it must be remembered that they only apply when dealing with constant acceleration. When approaching problems involving constant acceleration it is a good idea to list the quantities you are given, establish which quantity or quantities you require and then use the appropriate formula. Ensure that all quantities are converted to compatible units. s=

i.e.

Constant acceleration summary If acceleration is constant, the following formulas may be applied, where u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement. v = u + at

1 s = ut + at 2 2

v 2 = u 2 + 2as

s=

1 (u + v)t 2

Example 7 A body is moving in a straight line with uniform acceleration at an initial velocity of 12 m/s. After 5 s its velocity is 20 m/s. Find a the acceleration b the distance travelled in this time c the time taken to travel a distance of 200 m. Solution Given u = 12 v = 20 t =5 a

c

Find a using v = u + at 20 = 12 + 5a a = 1.6 The acceleration is 1.6 m/s2 .

1 b Find s using s = ut + at 2 2 1 = 12(5) + (1.6)52 2 = 80 The distance travelled is 80 m.

1 s = ut + at 2 gives 2 1 200 = 12t + × (1.6) × t 2 2 4 200 = 12t + t 2 5 1000 = 60t + 4t 2 250 = 15t + t 2 i.e. t 2 + 15t − 250 = 0 (t − 10)(t + 25) = 0 ∴ t = 10 or t = −25 As t ≥ 0, t = 10 is the acceptable solution.

Using the formula

The body takes 10 s to travel a distance of 200 m. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Exercise 19C 1 How long does it take for a body at rest to travel a distance of 30 m if it is accelerated at 1.5 m/s2 ? 2 A car is travelling at 25 m/s when the brakes are applied. It is brought to rest with uniform deceleration in 3 s. How far would it travel after the brakes were applied? Example

7

3 A motor cycle accelerates uniformly from 3 m/s to 30 m/s in 9 s. Find a b c d

the acceleration the time it will take to increase in speed from 30 m/s to 50 m/s the distance travelled in the first 15 s (assuming it starts from rest) the time taken to reach a speed of 200 km/h (assuming it starts from rest).

4 A car accelerating uniformly from rest reaches a speed of 45 km/h in 5 s. Find a its acceleration b the distance travelled in the 5 s. 5 A train starts from rest at a station and accelerates uniformly at 0.5 m/s2 until it reaches a speed of 90 km/h. a How long does the train take to reach this speed? b How far does the train travel in reaching this speed? 6 A train travelling at 54 km/h begins to climb an incline of constant gradient that produces a deceleration of 0.25 m/s2 . a How long will the train take to travel a distance of 250 m? b What will the train’s speed be then? For 7 to 11 assume that the acceleration due to gravity is − 9.8 m/s2 and ignore air resistance. Upward motion is considered to be in the positive direction. 7 A stone is projected vertically upwards from O with a speed of 20 m/s. Find a the velocity of the stone after 4 s b the distance of the stone from O after 4 s. 8 Repeat 7 for the stone being projected downwards from O with the same speed. 9 A body is projected vertically upwards with a velocity of 49 m/s. a After what time will the body return to the point of projection? b When will the body be at a height of 102.9 m above the point of projection? 10 A man dives from a springboard where his centre of gravity is initially 3 m above the water and his initial velocity is 4.9 m/s upwards. Regarding the diver as a particle at his centre of gravity, and assuming that the diver’s motion is vertical, find a the diver’s velocity after t seconds ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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b the diver’s height above the water after t seconds c the maximum height of the diver above the water d the time taken for the diver to reach the water. 11 A stone is thrown vertically upwards from the top edge of a cliff 24.5 m high with a speed of 19.6 m/s. Find a b c d

the time taken for the stone to reach its maximum height the maximum height reached from the foot of the cliff the time taken for the stone to return to the point of projection the time taken for the stone to reach the foot of the cliff.

12 A body is travelling at 20 m/s when it passes point P and 40 m/s when it passes point Q. Find its speed when it is halfway from P to Q, assuming uniform acceleration.

19.4 Velocity–time graphs Many kinematics problems can be solved using velocity–time graphs. These are particularly useful if acceleration is constant but with a broader knowledge of integral calculus they can also be used when acceleration is variable. First, we understand that if the acceleration is constant then v = u + at. This constitutes a linear relationship between v and t where a is the gradient of the corresponding velocity–time graph. t2 dx v(t)dt = x2 − x1 it follows that Since v = dt t1 where x1 is the position at time t1 and x2 is the position at time t2 . Then the total area of the region(s) between v the velocity–time graph and the t axis corresponds to the distance travelled by the particle between times t1 and t2 . area = displacement Consideration of the velocity–time graph is particularly useful in situations where t t2 t1 there are several stages to the particle’s motion. Example 8 A car starts from rest and accelerates uniformly for 25 s until it is travelling at 25 m/s. It then maintains this velocity for 3 minutes before decelerating uniformly until it stops in another 15 s. Construct a velocity–time graph and use it to determine the total distance travelled in kilometres.



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Solution From the graph we can calculate the area of the trapezium. (a + b) h Area = 2 1 = (220 + 180) 25 2 = 5000 m

v (m/s) 25

O

25

205 220

t (s)

= 5 km The total distance travelled is 5 km.

Example 9 A motorist is travelling at a constant speed of 120 km/h when he passes a stationary police car. He continues at that speed for another 15 s before uniformly decelerating to 100 km/h in 5 s. The police car takes off after the motorist the instant it passes. It accelerates uniformly for 25 s by which time it has reached 130 km/h. It continues at that speed until it catches up to the motorist. After how long does the police car catch up to the motorist and how far has he travelled in that time? Solution v (km/h) We start by representing the information on a velocity–time graph. 130 120 The distance travelled by the motorist and the police car will be the same so 100 the areas under each of the velocity–time graphs will be equal. This fact can be O 15 20 25 used to find T, the time taken for the police car to catch up to the motorist. For the motorist, the distance travelled after T seconds 5 1 = 120 × 15 + (120 + 100) × 5 + 100(T − 20) 2 18 5 = (1800 + 550 + 100T − 2000) 18 5 = (100T + 350) 18 5 changes velocities from km/h to m/s. Note: The factor 18


police car motorist

T

t (s)


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5 1 × 25 × 130 + 130(T − 25) 2 18 5 = (1625 + 130T − 3250) 18 5 = (130T − 1625) 18 When the police car catches the motorist

Police car:

100T + 350 = 130T − 1625 30T = 1975 395 T = 6 The police car catches the motorist after 65.83 s.

∴

distance = (100T + 350)

395 5 where T = 18 6

52 000 m 27 distance = 1.926 km =

The police car has travelled 1.926 km when it catches the motorist.


8

It is suggested that you draw a velocity–time graph for each of these questions. 1 A particle starts from rest and accelerates uniformly for 5 s until it reaches a speed of 10 m/s. It immediately decelerates uniformly until it comes to rest after a further 8 s. How far did it travel? 2 A car accelerates uniformly from rest for 10 s to a speed of 15 m/s. It maintains this speed for a further 25 s before decelerating uniformly to rest after a further 15 s. Find a the total distance travelled by the car b the distance it had travelled when it started to decelerate c the time taken for it to reach the halfway point of its journey. 3 A particle starts from rest and travels 1 km before coming to rest again. For the first 5 s it accelerates uniformly. It next maintains a constant speed for 500 m, and then decelerates uniformly for the last 10 s. Find the maximum speed of the particle. 4 A car passes point P with a speed of 36 km/h and continues at this speed for 12 s before accelerating to a speed of 72 km/h in 6 s. How far from P is the car when it reaches a speed of 72 km/h?



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5 A tram decelerates uniformly from a speed of 60 km/h to rest in 60 s. Find a the distance travelled by the tram b how far it had travelled by the time it had reduced its speed by half c the time taken for it to travel half the total distance. Example

9

6 A car passes a point A with a speed of 15 m/s and continues travelling at that speed. A second car starts from rest and accelerates uniformly until it reaches a speed of 25 m/s in 10 s. Both cars continue with a constant speed on to point B which they reach at the same time. a How long does it take for both cars to reach point B? b How far is it from A to B? 7 Two stations A and B are 14 km apart. A train passes through station A, heading towards B, maintaining a constant speed of 60 km/h. At the instant it passes through A, a second train on the same track leaves station B, heading towards A, and accelerates uniformly. After 5 minutes the alarm is raised at both stations simultaneously that a collision is imminent. Both trains are radioed and told to brake. The first train decelerates uniformly so that it will stop in 2.5 minutes. The second train, which has reached a speed of 80 km/h, will take 4 minutes to stop. Will they collide? 8 Two tram stops are 800 m apart. A tram starts at rest from the first stop and accelerates at a constant acceleration of a m/s2 for a certain time and then decelerates at a constant rate of 2a m/s2 , before coming to rest at the second stop. The time taken to travel between the stops is 1 min 40 seconds. Find a the maximum speed reached by the tram in km/h b the time at which the brakes are applied c the value of a.



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Chapter summary The position coordinate of a particle moving in a straight line is determined by its distance from a fixed point O, called the origin, and whether the particle is to the right or left of O. Conventionally, the direction to the right of the origin is considered to be positive. r displacement (x) is the position of the particle relative to O r velocity (v) is the rate of change of its position with respect to time, i.e. v = d x dt r speed is a scalar quantity and refers to the distance travelled per unit time r average velocity = change in position change in time r average speed = distance travelled change in time r acceleration (a) is the rate of change of its velocity with respect to time, i.e. a=

d2x dv = 2 dt dt

Constant acceleration If acceleration is constant, the following formulae may be applied where u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement r v = u + at r s = ut + 1 at 2 r v 2 = u 2 + 2as r s = 1 (u + v) t 2 2 Velocity–time graphs The area of the region(s) between the velocity–time (v against t) graph and the t axis between t = t1 and t = t2 corresponds to the distance travelled by the particle between times t1 and t2 .

Multiple-choice questions 1 A particle moves in a straight line so that its position x cm from a fixed point O at time t seconds (t ≥ 0) is given by x = −t 3 + 7t 2 − 12t. The initial position of the particle relative to O is A 0 cm B −6 cm C 12 cm D −20 cm E 5 cm 2 A particle moves in a straight line so that its position x cm from a fixed point O at time t seconds (t ≥ 0) is given by x = −t 3 + 7t 2 − 12t. The average velocity of the particle in the first 2 s correct to two decimal places is A 4 cm/s B −4 cm/s C 2 cm/s D 4.06 cm/s E −2 cm/s 3 A particle moves in a straight line with acceleration of 4 − 6t m/s2 at time t seconds. The particle has an initial velocity of −1 m/s and an initial position of 4 m from a fixed point O. The velocity of the particle when t = 1 is A −1 m/s B 6 m/s C 0 m/s D 4 m/s E −2 m/s



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4 A body starts from rest with a uniform acceleration of 1.8 m/s2 . The time it will take for the body to travel 90 m is √ √ √ 10 s C 10 s D 10 E 10 2 s A 5s B 5 A car accelerating uniformly from rest reaches a speed of 60 km/h in 4 s. The car’s acceleration is 25 25 m/s2 E km/h2 D B 15 m/s2 C 54 m/s2 A 15 km/h2 6 6 6 A car accelerating uniformly from rest reaches a speed of 60 km/h in 4 s. The distance travelled by the car in the 4 s is 100 m D 100 m E 360 m A 200 m B 100 km C 3 v (m/s) 7 A car’s motion is represented by the velocity–time graph shown. 25 The total distance travelled by the car 20 over the 15 s is 15 A 75 m B 315 m C 182.5 m 10 D 167.5 m E 375 m 5 0

t (s) 4 56

10

15

8 A rock falls from the top of a cliff 40 m high. The rock’s speed just before it hits the ground in m/s g = 9.8 m/s2 is A 20 B 22 C 24 D 26 E 28 9 A body initially travelling at 20 m/s is subject to a constant deceleration of 4 m/s2 . The time it takes to come to rest (t seconds) and the distance travelled before it comes to rest (s metres) is given by A t = 5, s = 50 B t = 5, s = 45 C t = 4, s = 20 D t = 5, s = 40 E t = 4, s = 35 10 A particle moves in a straight line with acceleration of 12t − 5 m/s2 at time t seconds. The particle has an initial velocity of 1 m/s and an initial position of 0 m from a fixed point O. Find the velocity of the particle at t = 1. A 1 m/s B −5 m/s C 7 m/s D 2 m/s E 3 m/s

Short-answer questions (technology-free) 1 A particle moves in a straight line so that its position x cm relative to O at time t seconds (t ≥ 0) is given by x = t 2 − 4t − 5. Find a its initial position b its position at t = 3 c its initial velocity d when and where its velocity equals zero e its average velocity in the first 3 s f its average speed in the first 3 s. 2 A particle moves in a straight line so that its position x cm relative to O at time t seconds (t ≥ 0) is given by x = t 3 − 2t 2 + 8. Find



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a its initial position, velocity and acceleration and hence describe its motion b the times when it is stationary and its positions and acceleration at those times. 3 A particle moving in a straight line is x cm from the point O at time t seconds (t ≥ 0), where x = −2t 3 + 3t 2 + 12t + 7. Find a when the particle passes through O and its velocity and its acceleration at those times b when the particle is at rest c the distance travelled in the first 3 s. 4 Two particles A and B are moving in a straight line such that their displacements x cm from the point O at time t seconds are given by x 1 and x2 respectively, where x1 (t) =t 3 − t 2 t ≥0 t ≥0 x2 (t) =t 2 a Find 1 1 ii the acceleration of A after s i the displacement of A after s 2 2 1 iii the velocity of B after s. 2 b Find i the times when A and B collide (i.e., have the same displacement) ii the maximum distance between A and B during the first 2 s of motion. 5 A particle moving in a straight line has acceleration of 6t m/s2 at time t seconds (t ≥ 0). If the particle starts from rest at the origin O, find a the velocity after 2 s b the displacement at any time t. 6 A particle moving in a straight line has acceleration of (3 − 2t) m/s2 at time t seconds (t ≥ 0). If the particle starts at the origin O with a velocity of 4 m/s, find a the time when the particle comes to rest b the position of the particle at the instant it comes to rest c the acceleration at this instant d the time when the acceleration is zero e the velocity at this time. 7 A particle moves in a straight line and, t seconds after it starts from O, its velocity is 2 2t − 3t 3 m/s. Find a the displacement after 1 s b the velocity after 1 s c the acceleration after 1 s. 8 For a particle moving in a straight line, the velocity function is v : R + → R, where 1 v (t) = 2 and t is the time in seconds. Find 2t a the acceleration after t seconds b the displacement at time t seconds, given that the particle is at O when t = 1. 9 The velocity, v m/s, of a body t seconds after it starts moving from O along a straight line is given by v = t 3 − 11t 2 + 24t, t ≥ 0.



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481

10 A car is travelling at 20 m/s when the brakes are applied. It is brought to rest with uniform deceleration in 4 s. How far did it travel after the brakes were applied?

Review

a Find the acceleration at time t. b Find the acceleration at the instant when the body first changes direction. c Find the displacement of the body from O after 5 s, and the total distance travelled in the first 5 s.

11 A car accelerates uniformly from 0 to 30 m/s in 12 s. Find a the acceleration b the time it will take to increase in speed from 30 m/s to 50 m/s c the distance travelled in the first 20 s d the time taken to reach a speed of 100 km/h. 12 A train starts from rest at a station and accelerates uniformly at 0.4 m/s2 until it reaches a speed of 60 km/h. a How long does the train take to reach this speed? b How far does the train travel in reaching this speed? For questions 13 and 14 assume that the acceleration due to gravity is −9.8 m/s2 and ignore air resistance. Upward motion is considered to be in the positive direction. 13 A body is projected vertically upward with a velocity of 35 m/s. a After what time will the body return to the point of projection? b When will the body be at a height of 60 m above the point of projection? 14 A stone is projected vertically upward from the top of a cliff 20 m high with a speed of 19.6 m/s. Find a the time taken for the stone to reach its maximum height b the maximum height reached, with respect to ground level c the time taken for the stone to return to the point of projection d the time taken for the stone to reach the foot of the cliff. It is suggested that you draw a velocity–time graph for each of the questions 15 to 18. 15 A particle starts from rest and accelerates uniformly for 15 s until it reaches a speed of 25 m/s. It immediately decelerates uniformly until it comes to rest after a further 20 s. How far did it travel? 16 A car accelerates uniformly from rest for 8 s to a speed of 12 m/s. It maintains this speed for a further 15 s before decelerating uniformly to rest after a further 10 s. Find a the total distance travelled by the car b the time taken for it to reach the halfway point of its journey. 17 A vehicle starts from rest and travels 1 km before coming to rest again. For the first 15 s it accelerates uniformly, before maintaining a constant speed for 800 m then finally decelerating uniformly to rest in 10 s. Find the maximum speed of the vehicle. 18 A car travels at a constant speed of 12 m/s along a straight road. It passes a second stationary car which sets off in pursuit 3 s later. Find the constant acceleration required for the second car so that it catches the first car after a further 27 s has passed. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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19 A particle moves in a straight line so that t seconds after passing a fixed point O in the line, t2 its velocity, v metres per second, is given by v = − 3t + 5. 4 Calculate: a the velocity after 10 s b the acceleration when t = 0 c the minimum velocity d the distance travelled in the first 2 s e the distance travelled in the 3rd second. 20 A spot of light moves along a straight line so that its acceleration t seconds after passing a fixed point O on the line is (2 − 2t) cm/s2 . Three seconds after passing O the spot has a velocity of 5 cm/s. Find, in terms of t, an expression for a the velocity of the spot of light after t seconds b the distance of the spot from O after t seconds. 21 A point P is moving along a straight line. It passes through a point O with a velocity 6 m/s and, t seconds after passing through O, its acceleration is (4 − 4t) m/s2 . a Show that, t seconds after passing through O, the velocity of P is 6 + 4t − 2t 2 m/s. b Calculate i the maximum velocity of P ii the value of t when the velocity of P is again 6 m/s iii the distance OP when the velocity of P is zero. 22 A particle travelling in a straight line passes a fixed point O with a velocity 5 m/s. Its acceleration, a m/s2 , is given by a = 27 − 4t 2 , where t seconds is the time after passing O. Calculate a the acceleration of the particle as it passes O b its velocity when t = 3 c the value of t when its velocity is again 5 m/s. 23 A particle passes a fixed point O with a velocity of 2 m/s and moves in a straight line with acceleration of 3 (1 − t) m/s2 , where t is the time in seconds after passing O. Calculate a the velocity when t = 4 b the position of the particle at this instant. 24 A particle P travels in a straight line from a fixed point O so that its velocity, v m/s, is given by v = t 2 − 10t + 24, where t is the time in seconds after leaving O. Find a the values of t for which P is instantaneously at rest b the distance OP when t = 3 c the range of values of t for which the acceleration is negative.

Extended-response questions 1 A particle moves in a straight line so that its displacement x cm relative to O at time 1 1 t seconds is given by x = t 3 − 2t 2 + 4t − 2 . Find 3 3 a its initial displacement b its initial velocity c its acceleration after 3 s d the time when its velocity equals zero



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483

2 A particle moves in a straight line so that its displacement, x cm, relative to O at time t seconds (t ≥ 0) is given by x = t 4 + 2t 2 − 8t. Show that a the particle moves first to the left b the greatest distance of the particle to the left of O occurs after 1 s c after this time, the particle always moves to the right.

Review

e the displacement when its velocity equals zero f the time when its displacement is zero.

3 A defective rocket rises vertically upwards into the air and then crashes back to the ground. The rocket’s height above the ground, at time t seconds after take-off, is h metres where h = 6t 2 − t 3 (an approximate model). a When does the rocket crash and what is its velocity at this time? b At what time is the speed of the rocket zero, and what is its maximum height? c When does the acceleration of the rocket become negative? 4 A body is projected vertically upwards at 20 m/s from the top of a tower 10 m high on the edge of a vertical cliff. The upward displacement, x(t) metres, of the body from ground level O at time t seconds after projection (t ≥ 0) is given by x (t) = −4.9 t 2 + 20t + 10. Use a graphics calculator or otherwise to evaluate the values x (1) − x (0) , x (2) − x (1) , x (3) − x (2) , . . . , x (10) − x (9) . Analyse your results and draw some inference about the motion of the body. 5 A particle is projected vertically upwards with a speed of u m/s. Prove that u a the time taken by the particle to reach its highest point is seconds g b the total time taken for the particle to return to the point of projection is

2u seconds g

c the speed of returning to the point of projection is u m/s. 6 A stone is projected vertically upwards with a speed of 14 m/s, from a point O at the top of a mine shaft. Five seconds earlier, a lift began to descend the mine shaft from O with a constant speed of 3.5 m/s. Find the depth of the lift (to the nearest metre) at the instant when the stone falls on it. (Neglect air resistance and take the acceleration due to gravity as 9.8 m/s2 .) 7 A car is travelling along a straight road at 90 km/h when the brakes are applied. The car comes to rest in 5 s and during this time the velocity decreases linearly with time. Find: a the rule for the velocity–time function after the brakes are applied b the distance travelled in the five seconds. 8 A particle moves in a straight line so that its displacement x cm relative to O at time t seconds (t ≥ 0) is given by x = 3t 4 − 4t 3 + 24t 2 − 48t. Show that the particle moves at first to the left, comes to rest at a point A and then moves always to the right. Find the position of A.



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9 A particle is projected vertically upwards with a velocity of u m/s from a point O on the ground. T seconds later a second particle is projected vertically upwards from O with the same velocity. T u + seconds after the first Prove that the time taken for the particles to collide is g 2 particle was launched, and that the height of the particles at the instant they collide is 4u 2 − g 2 T 2 metres above O. 8g 2u 2u Interpret the case where T = . What happens if T > ? g g



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C H A P T E R

20 Statics of a particle Objectives To be able to use a triangle of forces to solve problems To be able to resolve forces, acting in a plane, in two directions which are at right angles

Students should have completed some of the chapter on vectors, Chapter 15, before attempting this chapter.

Introduction A force is a measure of the strength of a push or pull. When considering the forces that act upon a body, it is convenient to consider the forces acting on a single particle. The single particle may be considered as a point at which the entire mass of the body is concentrated. The unit of force used in this chapter is the kilogram weight (kg wt). If a body has a mass of one kilogram then the force due to gravity acting on the body is one kilogram weight. When a number of forces act simultaneously on a body, their combined effect is called the resultant force. If the resultant force acting on a body is zero, the body will remain at rest or continue moving with constant velocity. The body is said to be in equilibrium. Every object near the surface of the Earth is subject to the force of gravity. We refer to this force as the ‘weight’ of the body. Weight is a force that acts vertically downwards on a body (actually towards the centre of the Earth). When an object such as a book is placed on a table it remains at rest. The book is obviously being subjected to a force due to gravity but the fact that it does not fall to the ground indicates that there must be a second force, equal in magnitude and opposite in direction, being applied to the book. The table is exerting a force equal in magnitude to gravity on the book but in the opposite direction. Hence it remains at rest. The book is said to be in equilibrium. Any mass placed on a surface, either horizontal or inclined, experiences a force perpendicular to the surface. This force is referred to as a normal force.



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N

W kg wt

N

W kg wt

Since a force is a quantity that is defined by both magnitude and direction, it may be represented by a vector. Hence, if a particle under the action of a number of forces is in equilibrium, it is clear that the vector sum of all forces acting must be zero.

20.1 Triangle of forces If three forces are acting on a point in equilibrium, they can be represented by three vectors forming a triangle. F2 F1 If three forces F1 , F2 and F3 act on a particle in equilibrium as shown in the figure they can be rearranged into a triangle.

F3

Since the particle is in equilibrium F1 + F2 + F3 = 0

F1 F3 F2

The magnitudes of the forces acting and the angles between the forces can be found using trigonometric ratios (if the triangle contains a right angle) or using the sine or cosine rule. In the following examples and exercises, strings and ropes are considered to have negligible mass. A smooth light pulley is considered to have negligible mass and the friction between a rope and pulley is considered to be negligible. Example 1 A particle of mass 8 kg is suspended by two strings attached to two points in the same horizontal plane. If the two strings make angles of 30◦ and 40◦ to the horizontal, find the tension in each string.

30°

40° T2 kg wt

T1 kg wt

8 kg wt



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Chapter 20 — Statics of a particle

Solution Representing the forces in a triangle and using the sine rule gives 60°

T1 T2 8 = = ◦ ◦ sin 50 sin 60 sin 70◦ 8 T1 = ◦ sin 50 sin 70◦ 8 T1 = × sin 50◦ ≈ 6.52 kg wt sin 70◦ 8 × sin 60◦ ≈ 7.37 kg wt T2 = sin 70◦

T1 kg wt 70°

8 kg wt 50°

T2 kg wt

Example 2 P

A particle of mass 15 kg is suspended vertically from a point P by a string. The particle is pulled horizontally by a force of F kg wt so that the string makes an angle of 30◦ with the vertical. Find the value of F and the tension in the string.

30°

T kg wt F kg wt

Solution

15 kg wt

Representing the forces in a triangle gives F = tan 30◦ 15 ∴ F = 15 × tan 30◦ √ =5 3 15 Also = cos 30◦ T 15 ∴ T = cos 30◦ √ = 10 3 √ The tension in the string is 10 3 kg wt.

30° T kg wt

15 kg wt

F kg wt

Example 3 A body weighing 20 kg is placed on a smooth inclined plane which is inclined at 30◦ to the horizontal. A string is attached to a point further up the plane which prevents the body from moving. Find the tension in the string and the magnitude of the force exerted on the body by the plane.


N F

30°

20 kg wt


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Solution The triangle of forces appears as shown

N kg wt

F kg wt

90° 150° 120°

20 kg wt

Rearrange into a triangle of forces as the body is in equilibrium Therefore F = 20 sin 30◦ = 10 kg wt and √ N = 20 cos 30◦ = 10 3 kg wt

F kg wt 60° 90° N kg wt

30° 20 kg wt


1

1 A particle of mass 5 kg is suspended by two strings attached to two points in the same horizontal plane. If the two strings make angles of 45◦ with the horizontal, find the tension in each string.

45°

45°

T1 kg wt

T2 kg wt

5 kg wt

2 Using strings and pulleys, three weights of mass 6 kg, 8 kg and 10 kg are suspended in equilibrium as shown in the diagram. Calculate the magnitude of the angle ACB.

B

A

C 6 kg

10 kg 8 kg

3 A weight of 20 kg is suspended from two strings of length 10 cm and 12 cm, the ends of the strings being attached to two points in a horizontal line, 15 cm apart. Find the tension in each string.

15 cm 10 cm

T1

T2

12 cm

20 kg wt Example

2

4 If a boat is being pulled by two forces, of 40 kg wt towards the east and 30 kg wt to the north-west, what third force must be acting on the boat if it remains stationary? Give the magnitude and direction.



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Chapter 20 — Statics of a particle Example

3

5 A body weighing 104 kg is placed on a smooth, inclined plane which rises 5 cm vertically for every 12 cm horizontally. A string is attached to a point further up the plane which prevents the body from moving. Find the tension in the string and the magnitude of the force exerted on the body by the plane.

489

N T

5

12 104 kg wt

6 A body weighing 12 kg is kept at rest on a smooth, inclined plane of 30◦ by a force acting at an angle of 20◦ to the plane. Find the magnitude of the force.

N

F 20

°

30° 12 kg wt

7 In each of the following cases, determine whether the particle is in equilibrium. a

10

6

b

4 2√3

80°

150° 120°

160°

8

2

8 Three forces of magnitude 4 kg wt, 7 kg wt and 10 kg wt are in equilibrium. Determine the magnitude of the angles between the forces. 9 A mass of 15 kg is maintained at rest on a smooth, inclined plane by a string that is parallel to the plane. Determine the tension in the string if: a the plane is at 30◦ to the horizontal b the plane is at 40◦ to the horizontal c the plane is at 30◦ to the horizontal, but the string is held at an angle of 10◦ to the plane. 10 A string is connected to two points A and D in a horizontal line and weights of 12 kg and W kg are attached at points B and C. If AB, BC and CD make angles of 40◦ , 20◦ and 50◦ respectively with the horizontal, calculate the tensions in the string and the weight W.

20.2

Resolution of forces Obviously there are many situations where more than three forces (or in fact only two forces) will be acting on a body. An alternative method is required to solve such problems.



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If all forces under consideration are acting in the same plane, then each of the forces and the resultant force can be expressed as a sum of its i and j components. If a force F acts at an angle of ␪ to the x axis, then F can be written as the sum of two components, one ‘vertical’ and the j y other ‘horizontal’. So

F = |F| cos ␪i + |F| sin ␪ j

F

The force F is resolved into two components, one parallel to the x axis (i component) and the other parallel to the y axis (j component).

i

|F| sin θj θ

x

|F| cosθi

For a particle to be in equilibrium, the sum of the i components and the sum of the j components are each zero. Example 4 A particle of mass 8 kg is suspended by two strings attached to two points in the same horizontal plane. If the two strings make angles of 30◦ and 60◦ to the horizontal, find the tension in each string.

30°

60°

T1 kg wt

T2 kg wt

Solution

8 kg wt T2 kg wt T1 kg wt j

30° 30°

60°

i

8

Resolution in the j direction T1 sin 30◦ + T2 sin 60◦ − 8 = 0 √ 1 3 T1 + T2 −8=0 2 2

1

Resolution in the i direction

From 2

−T1 cos 30◦ + T2 cos 60◦ = 0 √ 1 3 −T1 + T2 =0 2 2 √ 3 T1 = T2


2


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491

Substituting in 1 gives √ √ 3 1 T1 + 3 T1 −8=0 2 2 16 = 4T1 ∴ T1 = 4 √ ∴ T2 = 4 3 √ The tensions in the strings are 4 kg wt and 4 3 kg wt. Example 5 A body of mass 10 kg is held at rest on a smooth, inclined plane by a string with tension 5 kg wt as shown here. Find the angle between the string and the inclined plane.

5 kg wt

N

j

θ

i

20° 10 kg wt

Solution

By resolving the forces parallel and perpendicular to the plane it can be seen that N has no parallel component, since N is perpendicular to the plane. Resolving in the i direction 5 cos ␪ ◦ − 10 sin 20◦ = 0 10 sin 20◦ ∴ cos ␪ ◦ = 5 −1 ␪ = cos (0.684) = 46.84◦

Exercise 20B For the following questions give answers correct to two decimal places. Example

3

1 A force F kg wt makes an angle of 40◦ with the horizontal. If its horizontal component is a force of 10 kg wt, find the magnitude of F.

F

40° Example

4

2 Find the magnitude of a force, acting on a smooth, inclined plane of angle 35◦ , required to support a mass of 10 kg resting on the plane.

F

N

35° 10 kg wt ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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3 A body weighing 8 kg rests on a smooth, inclined plane of angle 25◦ under the action of a horizontal force. Find the magnitude of the force and the reaction of the plane on the body.

N

F 25° 8 kg wt

4 A body weighing 10 kg rests on a smooth, inclined plane of angle 20◦ . Find the force that will keep it in equilibrium when it acts at an angle of 54◦ with the horizontal.

F

N

20°

54° 10 kg wt

5 If a body suspended by a string weighs 12 kg, find the horizontal force required to hold it at an angle of 30◦ from the vertical. 6 A force of 20 kg wt acting directly up a smooth plane inclined at an angle of 40◦ maintains a body in equilibrium on the plane. Calculate the weight of the body and the pressure it exerts on the plane. 7 Two men are supporting a block by ropes. One exerts a force of 20 kg wt, his rope making an angle of 35◦ with the vertical, and the other exerts a force of 30 kg wt. Determine the weight of the block and the angle of direction of the second rope. 8 A body A of weight 10 kg is supported against a smooth plane of angle 50◦ . N Find the pressure of the body on the A plane and the tension in the string which is parallel to the slope. A second body B on a second plane of angle 40◦ 50° is connected to A by a string passing 10 kg wt over a smooth pulley on the ridge. If the system is in equilibrium, what is the weight of B? 9 A sphere of radius 9 cm is attached to a point A on a vertical wall by a string of length 15 cm. If the weight of the sphere is 3 kg, find the tension in the string.

B

40°

A 15 cm T 3 kg 9 cm



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493

When a number of forces act simultaneously on a body, their combined effect is called the resultant force. If the resultant force acting on a body is zero, the body will remain at rest or continue moving with constant velocity. If three forces act upon a particle and the resultant force is zero, then vectors representing the forces may be arranged to form a triangle. The magnitudes of the forces acting and the angles between the forces can often be found using trigonometric ratios (if the triangle contains a right angle) or using the sine or cosine rule. Any mass placed on a surface, either horizontal or inclined, experiences a force perpendicular to the surface called a normal force. Resolution of forces y If all forces under consideration are acting in two dimensions, it is possible to express the forces in terms of their components in the i direction and j direction. So

F = |F| cos ␪i + |F| sin ␪ j

Review

Chapter summary

F

|F| sinθj θ

j x

|F| cosθi

i

For a particle to be in equilibrium, the sum of all i components must be zero and the sum of all j components must be zero.

Multiple-choice questions 1 The magnitude of the component of force F in the i direction is E 25 N D 20 N C 40 N B 50 N A 300 N

50 kg wt

60°

Questions 2 and 3 refer to the following information. A 20 kg weight is resting on a smooth plane inclined at 30◦ to the horizontal and is prevented from slipping down the plane by a string as shown in the diagram.

N

30°

√ 3 C kg wt D 60 kg wt 2 3 The magnitude of the tension in the string √ is approximately 3 B 20 kg wt A 10 kg wt C kg wt D 60 kg wt 2

2 The magnitude of N is B 20 kg wt A 10 kg wt


i F

20 kg wt

√ E 10 3 kg wt

E 10 kg wt


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4 Two perpendicular forces have magnitudes 5 kg wt and 4 kg wt. The magnitude of the resultant force is √ √ E 9 kg wt D 1 kg wt C 41 kg wt 11 kg wt B A 3 kg wt 5 The diagram represents a particle in equilibrium acted on by three forces in a plane of magnitudes A, B and C. Which one of the following statements is not true? C cos 60◦ B A= A A = B cos 60◦ cos 30◦ A D B = A cos 60◦ + C cos 30◦ C B = A cos 60◦ E C = B cos 30◦

C

90° 150° B

6 A particle is acted on by a force of magnitude 7 kg wt acting on a bearing of 45◦ , and another force of magnitude a kg wt acting on a bearing of 135◦ . If the magnitude of the resultant force is 9 kg wt, the value of√a must be √ E 32 D 16 C 130 B 4 2 A 2 20N

7 Two forces of magnitude 20 kg wt act on a particle at O as shown. The magnitude of the resultant force in kg wt is √ C 0 B 20 3 A 40 E 10 D 20

60° 20N

O

8 The resultant force when two forces of magnitude 300 kg wt and 200 kg wt act at an angle of 60◦ to each other is √ C 100 kg wt B 436 kg wt A 100 19 kg wt E 500 kg wt D 350 kg wt 9 Two perpendicular forces have magnitudes 16 kg wt and 30 kg wt. The magnitude of the resultant force is C 34 kg wt B 10 kg wt A 50√kg wt E 2 kg wt D 6 35 kg wt 10 A particle is acted on by a force of magnitude 8 kg wt acting on a bearing of 30◦ , and another force of magnitude a kg wt acting on a bearing of 120◦ . If the magnitude of the resultant force is 12 kg wt, the value of √ √ a must be √ E 4 13 D 20 C 130 B 4 5 A 2

Short-answer questions (technology-free) 1 A mass of 15 kg is suspended from two strings of length 6 cm and 8 cm, the ends of the strings being attached to two points in a horizontal line, 10 cm apart. Find the tension in each string.

10 cm 8 cm

6 cm

15 kg wt



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10 kg wt 60°

F

10 kg wt

3 A body of mass 70 kg is placed on a smooth, inclined plane which rises 6 cm vertically for every 12 cm horizontally. A string is attached to a point further up the plane which prevents the body from moving. Find the tension in the string and the magnitude of the force exerted on the body by the plane.

70 kg wt 6 cm

12 cm F

4 A body of mass 15 kg is kept at rest on a smooth, inclined plane of 30◦ by a force acting at an angle of 30◦ to the plane. Find the magnitude of the force.

30° 15 kg wt

30° 5 Three forces of magnitude 5 kg wt, 8 kg wt and 12 kg wt are in equilibrium. Determine the cosine of the angle between the 5 kg wt and 12 kg wt forces.

Review

2 An object is being pulled by two forces of 10 kg wt as shown in the diagram. What is the magnitude and direction of the third force acting on the object if it remains stationary?

495

5 kg wt

12 kg wt

8 kg wt 6 A force of F kg wt makes an angle of 30◦ with the vertical. If its vertical component is a force of 20 kg wt, find the magnitude of F.

F 30°

20 kg wt

7 Find the magnitude of the force, acting up a smooth, inclined plane of angle 45◦ , required to support a mass of 15 kg resting on the plane.

F

45°

15 kg wt

8 A force of 14 kg wt acting directly up a smooth plane inclined at an angle of 30◦ maintains a body in equilibrium on the plane. Calculate the weight of the body and the force it exerts on the plane. 9 A body of mass 12 kg is kept at rest on a smooth incline of 30◦ by a horizontal force. Find the magnitude of the force.



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C H A P T E R

21

Revision

Revision of chapters 19 and 20 21.1 Multiple-choice questions Questions 1 to 9 refer to a particle that moves in a straight line so that its position x cm relative to O at time t seconds (t ≥ 0) is given by x = 2t2 − 5t − 12. 1 The initial displacement of the particle is A 12 cm

B 0 cm

C −15 cm

D −12 cm

E −3 cm

D 3 cm/s

E −5 cm/s

C −4 cm/s2

D 3 cm/s2

E −1 cm/s2

C after 5 s

D after 0.8 s

E never

D −4 s

E never

D 13 cm

E 21 cm

2 The initial velocity of the particle is A 0 cm/s

B 1 cm/s

C −1 cm/s

3 The initial acceleration of the particle is A 0 cm/s2

B 4 cm/s2

4 The particle is stationary A after 1 s

B after 1.25 s

5 The particle passes through O at time t equals A 1s

B −1 s

C 4s

6 The particle’s displacement after 3 s is A 9 cm

B 0 cm

C −9 cm

7 The average velocity of the particle over the first 3 s is A 1 cm/s

B −1 cm/s

C 7 cm/s

D −7 cm/s

E 0 cm/s

D 9.25 cm

E 9 cm

8 The distance travelled by the particle in the first 3 s is A 3 cm

B −3 cm

C −9 cm



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Chapter 21 — Revision of chapters 19 and 20

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A 1 cm/s

B −1 cm/s

1 C 3 12 cm/s

E −3 cm/s

D 3 cm/s

Questions 10 to 13 refer to a body projected upward from the ground with a velocity of 15 m/s. Its acceleration due to gravity is − 10 m/s2 . 10 The body’s velocity at t = 3 s is A 15 cm/s

B −15 cm/s

C 0 cm/s

D 30 cm/s

E −30 cm/s

D 1.5 s

E never

D 11.25 m

E 20 m

D 1.5 s

E never

Revision

9 The average speed of the particle over the first 3 s is

11 The body’s velocity is 0 m/s at time t equals A 1s

B 2s

C 0s

12 The maximum height reached by the body is A 11 m

B 15 m

C 10 m

13 The body returns to the ground after A 2s

B 4s

C 3s

Questions 14 and 15 refer to the following velocity–time graph for a moving vehicle.

v (m/s) 20 0

t (s) 5

11 14

14 The distance travelled by the vehicle over the 14 s is A 100 m

B 150 m

C 160 m

D 180 m

E 200 m

D 4 m/s2

E −4 m/s2

15 The acceleration of the vehicle over the first 5 s is A 20 m/s2

B 10 m/s2

C 2.5 m/s2

Questions 16 and 17 refer to this system of forces which is in equilibrium.

8 kg wt F1 100° 120° F2

16 The magnitude of force F1 is approximately A 10.78 kg wt B 5.94 kg wt

C 9.10 kg wt

D 12.26 kg wt E 7.04 kg wt

17 The magnitude of force F2 is approximately A 10.78 kg wt B 5.94 kg wt

C 9.10 kg wt


D 12.26 kg wt E 7.04 kg wt


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Questions 18 and 19 refer to the following information. A 10 kg weight is resting on a smooth plane inclined at 25◦ to the horizontal and is prevented from slipping down the plane by a string, as shown in the diagram.

N

F

25° 10 kg wt

18 The magnitude of N is approximately A 4.23 kg wt

B 9.06 kg wt

C 8.19 kg wt

D 2.59 kg wt

E 10 kg wt

19 The magnitude of the tension in the string is approximately A 4.23 kg wt

B 9.06 kg wt

C 8.19 kg wt

20 If this system of forces is in equilibrium, ␣ is approximately A 120◦ D 110◦

B 136◦ E 100◦

D 2.59 kg wt

E 10 kg wt 14 kg wt

10 kg wt

C 102◦

α

12 kg wt

21 If this system of forces is in equilibrium, ␪ is approximately A 138◦ D 100◦

B 130◦ E 90◦

12 kg wt

15 kg wt

C 123◦

138° θ

17.5 kg wt

22 The component of force F in the i direction is A 34.64 kg wt D 20 kg wt

B 30 kg wt E 51.96 kg wt

C 40 kg wt

|F| = 60 kg wt

30°

i

23 The component of force F in the j direction is A 34.64 kg wt D 20 kg wt

B 30 kg wt E 51.96 kg wt

j

F

C 40 kg wt

|F| = 60 kg wt F

j

30°

i

21.2 Extended-response questions 1 The velocity, v m/s, of a particle moving in a straight line is given by v = 6 + pt + qt 3 where t is the time in seconds after the particle passes through a fixed point O. Given that when t = 2 s the distance of the particle from O is 16 m and its acceleration is 32 m/s2 , calculate



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b the velocity of the particle at the instant when the acceleration is zero. 2 A stone is projected vertically upwards from the top of a cliff 20 m high. After a time of 3 s it passes the edge of the cliff on its way down. Calculate a the speed of projection b the speed when it hits the ground c the times when it is 10 m above the top of the cliff

Revision

a the value of p and of q

d the time when it is 5 m above the ground. 3 Trials are being undertaken on a horizontal road to test the performance of an electrically powered car. The car has a top speed V. In a test run, the car moves from rest with uniform acceleration a and is brought to rest with uniform retardation r. a If the car is to achieve top speed during a test run, show that the length of the test run must be at least V 2 (a + r ) 2ar b Find the least time taken for a test run of length 2V 2 (a + r ) 2V 2 (a + r ) i ii 9ar 3ar c Find, in terms of V, the average speed of the car for the test run described in b ii. 4 A particle X is projected vertically upwards from the ground with a velocity of 80 m/s. Calculate the maximum height reached by X. A particle Y is held at a height of 300 m above the ground. At the moment when X has dropped 80 m from its maximum height, Y is projected downwards with a velocity of v m/s. The particles reach the ground at the same time. Calculate the value of v. 5 a A particle X moves along a horizontal straight line so that its displacement, s m, from a fixed point O, t seconds after the motion has begun, is given by s = 28 + 4t − 5t 2 − t 3 . Find expressions in terms of t for i velocity ii acceleration. b State i the initial velocity of X ii the initial acceleration of X. c A second particle Y moves along the same horizontal straight line as X and starts from O at the same instant that X begins to move. The initial velocity of Y is 2 m/s, and its acceleration a m/s2 , t seconds after motion has begun is given by a = 2 − 6t. Find the value of t at the instant when X and Y collide. d Find the velocity of X and the velocity of Y at this instant and comment on the direction and motion of each of the particles.



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C H A P T E R

22 Describing the distribution of a single variable Objectives To introduce the two main types of data—categorical and numerical To use bar charts to display frequency distributions of categorical data To use histograms and frequency polygons to display frequency distributions of numerical data To use cumulative frequency polygons and cumulative relative frequency polygons to display cumulative frequency distributions To use the stem-and-leaf plot to display numerical data To use the histogram to display numerical data To use these plots to describe the distribution of a numerical variable in terms of symmetry, centre, spread and outliers To define and calculate the summary statistics mean, median, range, interquartile range, variance and standard deviation To understand the properties of these summary statistics and when each is appropriate To construct and interpret boxplots, and use them to compare data sets

22.1 Types of variables A characteristic about which information is recorded is called a variable, because its value is not always the same. Several types of variable can be identified. Consider the following situations.



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Chapter 22 — Describing the distribution of a single variable

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Students answer a question by selecting ‘yes’, ‘no’ or ‘don’t know’. Students say how they feel about a particular statement by ticking one of ‘strongly agree’, ‘agree’, ‘no opinion’, ‘disagree’ or ‘strongly disagree’. Students write down the size shoe that they take. Students write down their height. These situations give rise to two different types of data. The data arising from the first two situations are called categorical data, because the data can only be classified by the name of the category from which they come; there is no quantity associated with each category. The data arising from the third and fourth examples is called numerical data. These examples differ slightly from each other in the type of numerical data they each generate. Shoe sizes are of the form . . . , 6, 6.5, 7, 7.5, . . . . These are called discrete data, because the data can only take particular values. Discrete data often arise in situations where counting is involved. The other type of numerical data is continuous data where the variable may take any value (sometimes within a specified interval). Such data arise when students measure height. In fact, continuous data often arise when measuring is involved.

Exercise 22A 1 Classify the data which arise from the following situations into categorical, or numerical. a Kindergarten pupils bring along their favourite toy, and they are grouped together under the headings: ‘dolls’, ‘soft toys’, ‘games’, ‘cars’, and ‘other’. b The number of students on each of twenty school buses are counted. c A group of people each write down their favourite colour. d Each student in a class is weighed in kilograms. e Each student in a class is weighed and then classified as ‘light’, ‘average’ or ‘heavy’. f People rate their enthusiasm for a certain rock group as ‘low’, ‘medium’, or ‘high’. 2 Classify the data which arise from the following situations as categorical or numerical. a The intelligence quotient (IQ) of a group of students is measured using a test. b A group of people are asked to indicate their attitude to capital punishment by selecting a number from 1 to 5 where 1 = strongly disagree, 2 = disagree, 3 = undecided, 4 = agree, and 5 = strongly agree. 3 Classify the following numerical data as either discrete or continuous. a b c d e

The number of pages in a book. The price paid to fill the tank of a car with petrol. The volume of petrol used to fill the tank of a car. The time between the arrival of successive customers at an autobank teller. The number of tosses of a die required before a six is thrown.



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22.2 Displaying categorical data—the bar chart Suppose a group of 130 students were asked to nominate their favourite kind of music under the categories ‘hard rock’, ‘oldies’, ‘classical’, ‘rap’, ‘country’ or ‘other’. The table shows the data for the first few students. Student’s name Daniel Karina John Jodie

Favourite music hard rock classical country hard rock

The table gives data for individual students. To consider the group as a whole the data should be collected into a table called a frequency distribution by counting how many of each of the different values of the variable have been observed. Counting the number of students who responded to the question on favourite kinds of music gave the following results in each category. Hard rock 62

Other 27

Oldies 20

Classical 15

Rap 3

Country 3

Number of students

While a clear indication of the group’s preferences can be seen from the table, a visual display may be constructed to illustrate this. When the data are categorical, the appropriate display is a bar chart. The categories are indicated on the horizontal axis and the corresponding numbers in each category shown on the vertical axis. 70 60 50 40 30 20 10 0

Hard rock

Other

Oldies Classical Type of music

Rap

Country

The order in which the categories are listed on the horizontal axis is not important, as no order is inherent in the category labels. In this particular bar chart, the categories are listed in decreasing order by number. From the bar chart the music preferences for the group of students may be easily compared. The value which occurs most frequently is called the mode of the variable. Here it can be seen that the mode is hard rock.



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Exercise 22B 1 A group of students were asked to select their favourite type of fast food, with the following results. a Draw a bar chart for these data. b Which is the most popular food type?

Food type hamburgers chicken fish and chips Chinese pizza other

2 The following responses were received to a question regarding the return of capital punishment. a Draw a bar chart for these data. b How many respondents either agree or strongly agree? 3 A video shop proprietor took note of the type of films borrowed during a particular day with the following results. a Construct a bar chart to illustrate these data. b Which is the least popular film type? 4 A survey of secondary school students’ preferred ways of spending their leisure time at home gave the following results. a Construct a bar chart to illustrate these data. b What is the most common leisure activity?

22.3

Number of students 23 7 6 7 18 8

strongly agree agree don’t know disagree strongly disagree

21 11 42 53 129

comedy drama horror music other

53 89 42 15 33

watch TV read listen to music watch a video phone friends other

42% 13% 23% 12% 4% 6%

Displaying numerical data—the histogram In previous studies you have been introduced to various ways of summarising and displaying numerical data, including dotplots, stem-and-leaf plots, histograms and boxplots. Constructing a histogram for discrete numerical data is demonstrated in Example 1.



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Example 1 The numbers of siblings reported by each student in Year 11 at a local school is as follows: 2 0 2

3 2 3

4 1 4

0 1 1

3 4 1

2 5 0

3 3 9

0 2 0

4 5 1

1 6 1

0 1 1

0 1 1

1 1 1

2 0 0

3 2 1

Construct a frequency distribution of the number of siblings. Solution To construct the frequency distribution count the numbers of students corresponding to each of the numbers of siblings, as shown. Number Frequency

0 9

1 15

2 7

3 6

4 4

5 2

6 1

7 0

8 0

9 1

A histogram looks similar to a bar chart, but because the data are numeric there is a natural order to the plot which may not occur with a bar chart. Usually for discrete data the actual data values are located at the middle of the appropriate column, as shown.

Frequency

15 10 5 0

0

1

2

3 4 5 6 Number of siblings

7

8

9

An alternative display for a frequency distribution is a frequency polygon. It is formed by plotting the values in the frequency histogram with points, which are then joined by straight lines. A frequency polygon for the data in Example 1 is shown by the red line in this diagram.

Frequency

15

10

5

0

0

1

2

3 4 5 6 Number of siblings

7

8

9

When the range of responses is large it is usual to gather the data together into sub-groups or class intervals. The number of data values corresponding to each class interval is called the class frequency. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Class intervals should be chosen according to the following principles: Every data value should be in an interval The intervals should not overlap There should be no gaps between the intervals. The choice of intervals can vary, but generally a division which results in about 5 to 15 groups is preferred. It is also usual to choose an interval width which is easy for the reader to interpret, such as 10 units, 100 units, 1000 units etc (depending on the data). By convention, the beginning of the interval is given the appropriate exact value, rather than the end. For example, intervals of 0–49, 50–99, 100–149 would be preferred over the intervals 1–50, 51–100, 101–150 etc. Example 2 A researcher asked a group of people to record how many cups of coffee they drank in a particular week. Here are her results. 0 5 8

0 9 10 23 25 0 17 14 3 6 19 25 25 0 0

0 0 34 32 0 0 30 0 4 0 33 23 0 32 13 21 22 6 0 2 28 25 14 20 12 17 16

Construct a frequency distribution and hence a histogram of these data. Solution

Frequency

Because there are so many different results and they are spread over a wide range, the data are summarised into class intervals. As the minimum value is 0 and the Number of Frequency maximum is 34, intervals of width 5 cups of coffee would be appropriate, giving the 0–4 16 frequency distribution shown in the table. 5–9 5 10–14 5 The corresponding histogram 15–19 4 may then be drawn. 20–24 5 25–29 5 20 30–34 5 15 10 5 0

5

15 20 25 30 10 Number of cups of coffee

35

Example 2 was concerned with a discrete numerical variable. When constructing a frequency distribution of continuous data, the data are again grouped, as shown in Example 3.



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Example 3 The following are the heights of the players in a basketball club, measured to the nearest millimetre. 178.1 183.3 192.4 196.3

185.6 180.3 203.7 189.6

173.3 182.0 191.1 183.9

193.4 183.6 189.7 177.7

183.1 184.5 191.1 184.1

193.0 185.8 180.4 183.8

188.3 189.1 180.0 174.7

189.5 184.6 202.4 170.9 178.6 194.7 185.3 188.7 180.1 170.5 179.3 193.8 178.9

Construct a frequency distribution and hence a histogram of these data. Solution From the data it seems that intervals of width 5 will be suitable. All values of the variable which are 170 or more, but less than 175, have been included in the first interval. The second interval includes values from 175 to less than 180, and so on for the rest of the table.

The histogram of these data is shown here.

Player heights 170 – 175 – 180 – 185 – 190 – 195 – 200 –

Frequency 4 5 13 9 7 1 2

Frequency

15 10 5 0

170

175

180

185 190 195 Player heights

200

205

The interval in a frequency distribution which has the highest class frequency is called the modal class. Here the modal class is 180.0–184.9.

Using the TI-Nspire The calculator can be used to construct a histogram for numerical data. This will be illustrated using the basketball player height data from Example 3.




507

The data is entered in a Lists & Spreadsheet application (c>New Document>Add Lists & Spreadsheet). Firstly, use the up/down arrows (£ ¤) to name the first column height. Then enter each of the 41 numbers as shown.

Open a Data & Statistics application (/+I>Add Data & Statistics) to plot the data. At first the data displays as shown. This is not a statistical plot.

Press e to show the variable list for the x-axis and select the variable, height. Alternatively, press b>Plot Properties>Add X Variable or use the touchpad to move down below the x-axis and click to add the x variable.

The data now displays as shown. Note: if you wish to change the colour (CX only), move the cursor over a data point and press (/+b) to access the colour palette. Change to a histogram using b>Plot Type>Histogram. The data now displays as shown. Change the bin width and alignment using b>Plot Properties>Histogram Properties>Bin Settings Let Width = 5 and Alignment = 170. Finally, select Zoom – Data from the Window/Zoom menu to have the data displayed as shown. To show values, move the cursor over the columns, or use b>Analyze>Graph Trace.



508


Using the Casio ClassPad The calculator can be used to construct a histogram for numerical data. This will be illustrated using the basketball player height data from Example 3. enter the data into list1, tapping EXE to enter and move down the column. In Tap SetGraph, Setting . . . and the tab for Graph 1, enter the settings shown and tap SET. Tap SetGraph, StatGraph1 and then tap the box to tick and select the graph. to produce the graph selecting HStart Tap = 4 (the left bound of the histogram) and HStep = 4 (the desired interval width) when prompted. The histogram is produced as shown. With the graph window selected (bold border) tap 6 to adjust the viewing window for the graph.

Tap Analysis, Trace and use the navigator key to move from column to column and display the count for that column.



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Relative and percentage frequencies When frequencies are expressed as a proportion of the total number they are called relative frequencies. By expressing the frequencies as relative frequencies more information is obtained about the data set. Multiplying the relative frequencies by 100 readily converts them to percentage frequencies, which are easier to interpret. An example of the calculation of relative and percentage frequencies is shown in Example 4. Example 4 Construct a relative frequency distribution and a percentage frequency distribution for the player height data. Solution From this table it can be seen, for example, that nine out of forty-one, or 22% of players, have heights from 185 cm to less than 190 cm.

Player heights (cm)

Frequency

170 –

4

175 –

5

180 –

13

185 –

9

190 –

7

195 –

1

200 –

2

Relative frequency 4 41 5 41 13 41 9 41 7 41 1 41 2 41

Percentage frequency

= 0.10

10%

= 0.12

12%

= 0.32

32%

= 0.22

22%

= 0.17

17%

= 0.02

2%

= 0.05

5%

Both the relative frequency histogram and the percentage frequency histogram are identical to the frequency histogram—only the vertical scale is changed. To construct either of these histograms from a list of data use a graphics calculator to construct the frequency histogram, and then convert the individual frequencies to either relative frequencies or percentage frequencies one by one as required.

Cumulative frequency distribution To answer questions concerning the number or proportion of the data values which are less than a given value a cumulative frequency distribution, or a cumulative relative frequency distribution can be constructed. In both a cumulative frequency distribution and a cumulative relative frequency distribution, the number of observations in each class are accumulated from low to high values of the variable.



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Example 5 Construct a cumulative frequency distribution and a cumulative relative frequency distribution for the data in Example 4. Solution Player heights (cm) <170 <175 <180 <185 <190 <195 <200 <205

Frequency

Cumulative frequency

Cumulative relative frequency

0 4 5 13 9 7 1 2

0 4 9 22 31 38 39 41

0 0.10 0.22 0.54 0.76 0.93 0.95 1.00

Each cumulative frequency was obtained by adding preceding values of the frequency. In the same way the cumulative relative frequencies were obtained by adding preceding relative frequencies. Thus it can be said that a proportion of 0.54, or 54%, of players are less than 185 cm tall.


A graphical representation of a cumulative frequency distribution is called a cumulative frequency 40 polygon and has a distinctive appearance, as it 30 always starts at zero and is non-decreasing. This graph shows, on the vertical axis, the 20 number of players shorter than any height 10 given on the horizontal axis. The cumulative relative frequency distribution could also be 0 170 175 180 185 190 195 200 plotted as a cumulative relative frequency Player heights polygon, which would differ from the cumulative frequency polygon only in the scale on the vertical axis, which would run from 0 to 1.

205


1

1 The number of pets reported by each student in a class is given in the following table: 2 0

3 2

4 1

0 1

3 4

2 5

3 3

0 2

4 5

1 6

0 1

Construct a frequency distribution of the numbers of pets reported by each student.



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2 The number of children in the family for each student in a class is shown in this histogram.

Number of students

10

5

0

a b c d

1

2

3

4 5 6 7 Size of family

8

9

10

How many students are the only child in a family? What is the most common number of children in the family? How many students come from families with six or more children? How many students are there in the class?

Number of students

3 The following histogram gives the scores on a general knowledge quiz for a class of Year 11 students.

10

5

0

10

20

30

40

50 60 Marks

70

80

90 100

a How many students scored from 10–19 marks? b How many students attempted the quiz? c What is the modal class? d If a mark of 50 or more is designated as a pass, how many students passed the quiz? 4 The maximum temperatures for several capital cities around the world on a particular day, in degrees Celsius, were: 17 16 17 31 Example

2

Example

4

a b c d

26 15 23 19

36 18 28 25

32 25 36 22

17 30 45 24

12 23 17 29

32 33 19 32

2 33 37 38

Use a class interval of 5 to construct a frequency distribution for these data. Construct the corresponding relative frequency distribution. Draw a histogram from the frequency distribution. What percentage of cities had a maximum temperature of less than 25◦ C?



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5 A student purchases 21 new text books from a school book supplier with the following prices (in dollars). 21.65 7.80 8.90 Example

3

Example

5

14.95 3.50 17.15

12.80 7.99 4.55

7.95 42.98 21.95

32.50 18.50 7.60

23.99 19.95 5.99

23.99 3.20 14.50

a Draw a histogram of these data using appropriate class intervals. b What is the modal class? c Construct a cumulative frequency distribution for these data and draw the cumulative frequency polygon. 6 A group of students were asked to draw a line which they estimated to be the same length as a 30 cm ruler. The lines were then measured (in cm) with the following results. 30.3 32.2 32.1

30.9 30.1 31.2

31.2 31.6 30.7

32.3 32.1 32.1

31.3 31.4 30.8

30.7 31.8 29.7

32.8 32.9 30.1

31.0 31.9 28.9

33.3 29.4

30.7 31.6

a Construct a histogram of the frequency distribution. b Construct a cumulative frequency distribution for these data and draw the cumulative frequency polygon. c Write a sentence to describe the students’ performance on this task. 7 The following are the marks obtained by a group of Year 11 Chemistry students on the end of year exam. 21 33 47

49 52 52

58 59 63

68 68 71

72 82 92

31 47 48

49 52 53

59 59 65

68 70 71

72 91 99

a Using a graphics calculator, or otherwise, construct a histogram of the frequency distribution. b Construct a cumulative frequency distribution for these data and draw the cumulative frequency polygon. c Write a sentence to describe the students’ performance on this exam. 8 The following 50 values are the lengths (in metres) of some par 4 golf holes from Melbourne golf courses. 302 371 376 366 398

272 334 332 361 407

311 369 338 299 337

351 334 320 321 371

338 320 321 361 266

325 374 364 312 354

314 364 317 305 331

307 353 362 408 409

336 366 310 245 385

310 260 280 279 260

a Construct a histogram of the frequency distribution. b Construct a cumulative frequency distribution for these data and draw the cumulative frequency polygon.



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c Use the cumulative frequency polygon to estimate: i the proportion of par 4 holes below 300 m in length ii the proportion of par 4 holes 360 m or more in length iii the length which is exceeded by 90% of the par 4 holes.

22.4

Characteristics of distributions of numerical variables Distributions of numerical variables are characterised by their shapes and special features such as centre and spread. Two distributions are said to differ in centre if the values of the variable in one distribution are generally larger than the values of the variable in the other distribution. Consider, for example, the following histograms shown on the same scale. a

b

0

5

10

15

0

5

10

15

It can be seen that plot b is identical to plot a but moved horizontally several units to the right, indicating that these distributions differ in the location of their centres. The next pair of histograms also differ, but not in the same way. While both histograms are centred at about the same place, histogram d is more spread out. Two distributions are said to differ in spread if the values of the variable in one distribution tend to be more spread out than the values of the variable in the other distribution. c

d

0

5

10

15

0

5

10

15

A distribution is said to be symmetric if it forms a mirror image of itself when folded in the ‘middle’ along a vertical axis; otherwise it is said to be skewed. Histogram e is perfectly symmetrical, while f shows a distribution which is approximately symmetric. f e

0

5

10

15

0


5

10

15


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If a histogram has a short tail to the left and a long tail pointing to the right it is said to be positively skewed (because of the many values towards the positive end of the distribution) as shown in the histogram g. If a histogram has a short tail to the right and a long tail pointing to the left it is said to be negatively skewed (because of the many values towards the negative end of the distribution), as shown in histogram h. g h negatively skewed

positively skewed

0

5

10

15

0

5

10

15

Knowing whether a distribution is skewed or symmetric is important as this gives considerable information concerning the choice of appropriate summary statistics, as will be seen in the next section.

Exercise 22D 1 Do the following pairs of distributions differ in centre, spread, both or neither? a

b

0

0

0

0

c



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2 Describe the shape of each of the following histograms. a

b

0

0

c

0

3 What is the shape of the histogram drawn in 6, Exercise 22C? 4 What is the shape of the histogram drawn in 7, Exercise 22C? 5 What is the shape of the histogram drawn in 8, Exercise 22C?

22.5

Stem-and-leaf plots An informative data display for a small (less than 50 values) numerical data set is the stem-and-leaf plot. The construction of the stem-and-leaf plot is illustrated in Example 6.

Example 6 By the end of 2004 the number of test matches played, as captain, by each of the Australian cricket captains was: 3 10 1

16 11 39

2 2 2

1 5 25

8 25 1

3 5 30

6 24 48

4 1 7

8 24 28

21 2 93

2 17 50

15 1 57

10 5 9

6 28 6

Construct a stem-and-leaf plot of these data.



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Solution To make a stem-and-leaf plot find the smallest and the largest data values. From the table above, the smallest value is 1, which is given a 0 in the ten’s column, and the largest is 93, which has a 9 in the ten’s column. This means that the stems are chosen to be from 0–9. These are written in a column with a vertical line to their right, as shown.

0 1 2 3 4 5 6 7 8 9

The units for each data point are then entered to the right of the dividing line. They are entered initially in the order in which they appear in the data. When all data points are entered in the table, the stem-and-leaf plot looks like this. 0 1 2 3 4 5 6 7 8 9

3 6 1 9 8 0

2 1 8 3 6 4 8 2 6 2 5 5 1 2 1 5 1 2 1 7 9 6 5 0 0 1 7 5 4 4 8 5 8 0 7

3

To complete the plot the leaves are ordered, and a key added to specify the place value of the stem and the leaves. 0 1 2 3 4 5 6 7 8 9

1 0 1 0 8 0

1 1 1 1 2 2 2 2 2 3 3 4 5 5 5 6 6 6 7 8 8 9 0 1 5 6 7 4 4 5 5 8 8 9 7 3 | 9 indicates 39 matches

3

It can be seen from this plot that one captain has led Australia in many more test matches than any other (Allan Border, who captained Australia in 93 test matches). When a value sits away from the main body of the data it is called an outlier.



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Stem-and-leaf plots have the advantage of retaining all the information in the data set while achieving a display not unlike that of a histogram (turned on its side). In addition, a stem-and-leaf plot clearly shows: the range of values where the values are concentrated the shape of the data set whether there are any gaps in which no values are observed any unusual values (outliers). Grouping the leaves in tens is simplest—other convenient groupings are in fives or twos, as shown in Example 7. Example 7 The birth weights, in kilograms, of the first 30 babies born at a hospital in a selected month are as follows. 2.9 3.7 2.8

2.7 3.6 3.5

3.5 3.2 3.3

3.6 2.9 3.1

2.8 3.2 3.0

3.6 2.5 4.2

3.7 2.6 3.2

3.6 3.8 2.4

3.6 3.0 4.3

2.9 4.2 3.2

Construct a stem-and-leaf plot of these data. Solution A stem-and-leaf plot of the birth weights, with the stem representing units and the leaves representing one-tenth of a unit, may be constructed. 2 4 5 6 7 8 8 9 9 9 3 0 0 1 2 2 2 2 3 5 5 6 6 6 6 6 7 7 8 3 | 0 indicates 3.0 kilograms 4 2 2 3 The plot, which allows one row for each different stem, appears to be too compact. These data may be better displayed by constructing a stem-and-leaf plot with two rows for each stem. These rows correspond to the digits {0, 1, 2, 3, 4} in the first row and {5, 6, 7, 8, 9} in the second row. 2 2 3 3 4

4 5 6 7 8 8 9 9 9 0 0 1 2 2 2 2 3 5 5 6 6 6 6 6 7 7 2 2 3

8 3 | 0 indicates 3.0 kilograms

The only other possibility for a stem-and-leaf plot is one which has five rows per stem. These rows correspond to the digits {0, 1}, {2, 3}, {4, 5}, {6, 7} and {8, 9}.



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2 2 2 3 3 3 3 3 4 4

4 6 8 0 2 5 6 8

5 7 8 0 2 5 6

9 1 2

9

9

2

3

6

6

6

2

2

3

7

7 3 | 0 indicates 3.0 kilograms

None of the stem-and-leaf displays shown are correct or incorrect. A stem-and-leaf plot is used to explore data and more than one may need to be constructed before the most informative one is obtained. Again, from 5 to 15 rows is generally the most helpful, but this may vary in individual cases. When the data have too many digits for a convenient stem-and-leaf plot they should be rounded or truncated. Truncating a number means simply dropping off the unwanted digits. So, for example, a value of 149.99 would become 149 if truncated to three digits, but 150 if rounded to three digits. Since the object of a stem-and-leaf display is to give a feeling for the shape and patterns in the data set, the decision on whether to round or truncate is not very important; however, generally when constructing a stem-and-leaf display the data is truncated, as this is what commonly used data analysis computer packages will do. Some of the most interesting investigations in statistics involve comparing two or more data sets. Stem-and-leaf plots are useful displays for the comparison of two data sets, as shown in the following example. Example 8 The following table gives the number disposals by members of the Port Adelaide and Brisbane football teams, in the 2004 AFL Grand Final. Port Adelaide 25 12

20 11

19 11

18 11

18 11

17 10

16 10

15 9

14 9

13 7

12 7

Brisbane 25 19 19 18 17 16 15 15 13 13 13 10 10 9 9 8 8 7 6 5 4 0 Construct back to back stem-and-leaf plots of these data.



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Solution To compare the two groups, the stem-and-leaf plots are drawn back to back, using two rows per stem. Port Adelaide

Brisbane

9 9 7 7 4 3 2 2 1 1 1 1 0 0 9 8 8 7 6 5 0 5 0 | 2 represents 20 disposals

0 0 1 1 2 2

0 5 0 5

4 6 7 8 8 9 9 0 3 3 3 5 6 7 8 9 9

5 2 | 0 represents 20 disposals

The leaves on the left of the stem are centred slightly higher than the leaves on the right, which suggests that, overall, Port Adelaide recorded more disposals. The spread of disposals for Port Adelaide appears narrower than that of the Brisbane players.


6

1 The monthly rainfall for Melbourne, in a particular year, is given in the following table (in millimetres). Month

J

Rainfall (mm)

F

M A M

J

J

A

S

O

N

D

48 57 52 57 58 49 49 50 59 67 60 59

a Construct a stem-and-leaf plot of the rainfall, using the following stems. 4 5 6 b In how many months is the rainfall 60 mm or more? Example

7

2 An investigator recorded the amount of time 24 similar batteries lasted in a toy. Her results in hours were: 25.5 4.2

39.7 25.6

29.9 16.9

23.6 18.9

26.9 46.0

31.3 33.8

21.4 36.8

27.4 27.5

19.5 25.1

29.8 31.3

33.4 41.2

21.8 32.9

a Make a stem-and-leaf plot of these times with two rows per stem. b How many of the batteries lasted for more than 30 hours? 3 The amount of time (in minutes) that a class of students spent on homework on one particular night was: 10 39

27 70

46 19

63 37

20 67

33 20

15 28


21 23

16 0

14 29

15 10 Cambridge University Press

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a Make a stem-and-leaf plot of these times. b How many students spent more than 60 minutes on homework? c What is the shape of the distribution? 4 The cost of various brands of track shoes at a retail outlet are as follows. $49.99 $75.49 $68.99 $164.99 $75.99 $210.00 $84.99 $36.98 $95.49 $28.99 $46.99 $76.99 $82.99 $79.99 $149.99 a Construct a stem-and-leaf plot of these data. b What is the shape of the distribution? Example

8

$39.99 $25.49

$35.99 $78.99

52.99 $45.99

5 The students in a class were asked to write down the ages of their mothers and fathers. Mother’s age 49 50 43 44 Father’s age 50 51 43 46

43 40

50 39

47 40

50 41

40 43

46 45

49 48

49 38

42 43

44 37

38 43

41 44

55 44

51 48

48 43

47 48

47 43

52 46

54 48

41 49

44 45

40 46

a Construct a back to back stem-and-leaf plot of these data sets. b How do the ages of the students’ mothers and fathers compare in terms of shape, centre and spread? 6 The results of a mathematics test for two different classes of students are given in the table. Class A 22 19 85 79

48 45

39 82

68 81

47 80

58 91

77 99

76 55

89 65

85 79

82 71

Class B 12 13 74 76

80 80

81 81

83 82

98 84

70 84

70 88

71 69

72 73

72 88

73 91

a Construct a back to back stem-and-leaf plot to compare the data sets. b How many students in each class scored less than 50%? c Which class do you think performed better overall on the test? Give reasons for your answer.

22.6 Summarising data A statistic is a number that can be computed from data. Certain special statistics are called summary statistics, because they numerically summarise special features of the data set under consideration. Of course, whenever any set of numbers is summarised into just one or two figures much information is lost, but if the summary statistics are well chosen they will also help to reveal the message which may be hidden in the data set. Summary statistics are generally either measures of centre or measures of spread. There are many different examples for each of these measures and there are situations when one of the measures is more appropriate than another. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Measures of centre Mean The most commonly used measure of centre of a distribution of a numerical variable is the mean. This is calculated by summing all the data values and dividing by the number of values in the data set. Example 9 The following data set shows the number of premierships won by each of the current AFL teams, up until the end of 2004. Find the mean of the number of premiership wins. Team Carlton Essendon Collingwood Melbourne Fitzroy/Lions Richmond Hawthorn Geelong Kangaroos Sydney West Coast Adelaide Port Adelaide W Bulldogs St Kilda Fremantle

Premierships 16 16 14 12 11 10 9 6 4 3 2 2 1 1 1 0

Solution mean =

16 + 16 + 14 + 12 + 11 + 10 + 9 + 6 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 0 = 6.8 16

¯ which is called ‘x bar’. The mean of a sample is always denoted by the symbol x, In general, if n observations are denoted by x1 , x2 , . . . ., xn the mean is x1 + x2 + · · · · · · + xn n or, in a more compact version x¯ =

n 1 xi n i=1 where the symbol is the upper case Greek sigma, which in mathematics means ‘the sum of the terms’.

x¯ =



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Essential Advanced General Mathematics Note: The subscripts on the x’s are used to identify all of the n different values of x. They do not

mean that the x’s have to be written in any special order. The values of x in the example are in order only because they were listed in that way in the table.

Median Another useful measure of the centre of a distribution of a numerical variable is the middle value, or median. To find the value of the median, all the observations are listed in order and the middle one is the median. The median of 2

3

4

5

median 6 7 5

7

8

8

11

is 6, as there are five observations on either side of this value when the data are listed in order. Example 10 Find the median number of premierships in the AFL ladder using the data in Example 9. Solution As the data are already given in order, it only remains to decide which is the middle observation. 0

1

1

1

2

2

3

4

6

9

10

11

12

14

16

16

Since there are 16 entries in the table there is no actual middle observation, so the median is chosen as the value half way between the two middle observations, in this 1 case the eighth and ninth (6 and 4). Thus the median is equal to (6 + 4) = 5. The 2 interpretation here is that of the teams currently playing in the AFL, half (or 50%) have won the premiership 5 or more times and half (or 50%) have have won the premiership 5 or less times.

In general, to compute the median of a distribution: Arrange all the observations in ascending order according to size. n + 1 th If n, the number of observations, is odd, then the median is the 2 observation from the end of the list. If n, the number of observations, is even, then the median is found by averaging the nth and the two middle observations in the list. That is, to find the median the 2 n th + 1 observations are added together, and divided by 2. 2 The median value is easily determined from a stem-and-leaf plot by counting to the required observation or observations from either end.



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From Examples 10 and 11, the mean number of times premierships won (6.8) and the median number of premierships won (5) have already been determined. These values are different and the interesting question is: why are they different, and which is the better measure of centre for this example? To help answer this question consider a stem-and-leaf plot of these data. 0 0 1 1

0 6 0 6

1 9 1 6

1

1

2

4

2

2

3

4

From the stem-and-leaf plot it can be seen that the distribution is positively skewed. This example illustrates a property of the mean. When the distribution is skewed or if there are one or two very extreme values, then the value of the mean may be quite significantly affected. The median is not so affected by unusual observations, however, and is thus often a preferable measure of centre. When this is the case, the median is generally preferred as a measure of centre as it will give a better ‘typical’ value of the variable under consideration.

Mode The mode is the observation which occurs most often. It is a useful summary statistic, particularly for categorical data which do not lend themselves to some of the other numerical summary methods. Many texts state that the mode is a third option for a measure of centre but this is generally not true. Sometimes data sets do not have a mode, or they have several modes, or they have a mode which is at one or other end of the range of values.

Measures of spread Range A measure of spread is calculated in order to judge the variability of a data set. That is, are most of the values clustered together, or are they rather spread out? The simplest measure of spread can be determined by considering the difference between the smallest and the largest observations. This is called the range. Example 11 Consider the marks, for two different tasks, awarded to a group of students. Task A 2 35

6 38

9 38

10 39

11 42

12 46

13 47

22 47

23 52

24 52

26 56

26 56

27 59

33 91

34 94

16 59

19 63

21 65

23 68

28 71

31 72

31 73

33 75

38 78

41 78

49 78

52 86

53 88

54 91

Task B 11 56

Find the range of each of these data sets.



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Solution For Task A, the minimum mark is 2 and the maximum mark is 94. Range for Task A = 94 − 2 = 92 For Task B, the minimum mark is 11 and the maximum mark is 91. Range for Task B = 91 − 11 = 80 The range for Task A is greater than the range for Task B. Is the range a useful summary statistic for comparing the spread of the two distributions? To help make this decision, consider the stem-and-leaf plots of the data sets:

7 9

6 8 9

3 6 8 7 6

9 2 4 5 7 6

Task A 6 2 1 0 3 2 4 3 6 2 2 2

4

1

Task B 0 1 2 3 4 5 6 7 8 9

1 1 1 1 2 3 1 6 1

6 3 1 9 3 5 2 8

9 8 3 4 8 3

8 6

9

5

8

8

8

From the stem-and-leaf plots of the data it appears that the spread of marks for the two tasks is not well described by the range. The marks for Task A are more concentrated than the marks for Task B, except for the two unusual values for Task A. Another measure of spread is needed, one which is not so influenced by these extreme values. For this the interquartile range is used.

Interquartile range To find the interquartile range of a distribution: Arrange all observations in order according to size. Divide the observations into two equal-sized groups. If n, the number of observations, is odd, then the median is omitted from both groups. Locate Q 1 , the first quartile, which is the median of the lower half of the observations, and Q 3 , the third quartile, which is the median of the upper half of the observations. The interquartile range IQR is defined as the difference between the quartiles. That is IQR = Q 3 − Q 1



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Definitions of the quartiles of a distribution sometimes differ slightly from the one given here. Using different definitions may result in slight differences in the values obtained, but these will be minimal and should not be considered a difficulty. Example 12 Find the interquartile ranges for Task A and Task B data given in Example 11. Solution For Task A the marks listed in order are: 2 35

6 38

9 38

10 39

11 42

12 46

13 47

22 47

23 52

24 52

26 56

26 56

27 59

33 91

34 94

Since there is an even number of observations, then the lower ‘half’ is: 2

6

9

10

11

12

13

22

23

24

26

26

27

33

34

The median of this lower group is the eighth observation, 22, so Q 1 = 22. The upper half is: 35

38

38

39

42

46

47

47

52

52

56

56

59

91

94

The median of this upper group is 47, so Q 3 = 47 Thus, the interquartile range, IQR = 47 − 22 = 25 Similarly, for Task B data, the lower quartile = 31 and the upper quartile = 73, giving an interquartile range for this data set of 42. Comparing the two values of interquartile range shows the spread of Task A marks to be much smaller than the spread of Task B marks, which seems consistent with the display. The interquartile range is a measure of spread of a distribution which describes the range of the middle 50% of the observations. Since the upper 25% and the lower 25% of the observations are discarded, the interquartile range is generally not affected by the presence of outliers in the data set, which makes it a reliable measure of spread. The median and quartiles of a distribution may also be determined from a cumulative relative frequency polygon. Since the median is the observation which divides the data set in half, this is the data value which corresponds to a cumulative relative frequency of 0.5 or 50%. Similarly, the first quartile corresponds to a cumulative relative frequency of 0.25 or 25%, and the third quartile corresponds to a cumulative relative frequency of 0.75 or 75%.



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Example 13 Use the cumulative relative frequency polygon to find the median and the interquartile range for the data set shown in the graph. % 100 75 50 25 0 2

4

6

8

10

12

14

16

18

Solution From the plot of the data it can be seen that the median is 10, the first quartile is 8, the third quartile is 12 and hence the interquartile range is 12 − 8 = 4.

Standard deviation Another extremely useful measure of spread is the standard deviation. It is derived by considering the distance of each observation from the sample mean. If the average of these distances is used as a measure of spread it will be found that, as some of these distances are positive and some are negative, adding them together results in a total of zero. A more useful measure will result if the distances are squared (which makes them all positive) and are then added together. The variance is defined as a kind of average of these squared distances. When the variance is calculated from a sample, rather than the whole population, the average is calculated by dividing by n − 1, rather than n. For the remainder of this discussion it will be assumed that the data under consideration are from a sample. Since the variance has been calculated by squaring the data values it is sensible to find the square root of the variance, so that the measure reverts to a scale comparable to the original data. This results in measure of spread which is called the standard deviation. Standard deviation calculated from a sample is denoted s. Formally the standard deviation may be defined as follows. If a data set consists of n observations denoted x1 , x2 , . . . , xn , the standard deviation is

1 ¯ 2 + (x2 − x) ¯ 2 + · · · + (xn − x) ¯ 2 (x1 − x) s= n−1 or, in more compact notation, n 1 ¯ 2 (xi − x) s= n − 1 i=1



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Example 14 Calculate the standard deviation of the following data set. 13

12

14

6

15

12

7

6

7

8

Solution Construct a table as shown. ¯ 2 (xi − x) 9 4 16 16 25 4 9 16 9 4 ¯ ¯ 2 = 112 (xi − x) 112 √ = 12.44 = 3.53 From the table, the standard deviation s is: s = 9 xi 13 12 14 6 15 12 7 6 7 8 ¯ xi = 100

xi − x¯ 3 2 4 −4 5 2 −3 −4 −3 −2

Interpreting the standard deviation The standard deviation can be made more meaningful by interpreting it in relation to the data set. The interquartile range gives the spread of the middle 50% of the data. Can similar statements be made about the standard deviation? It can be shown that, for most data sets, about 95% of the observations lie within two standard deviations of the mean. Example 15 The cost of a lettuce at a number of different shops on a particular day is given in the table: $3.85 $3.81

$2.65 $1.69

$1.90 $3.66

$2.95 $2.60

$2.40 $2.70

$2.42 $3.10

$2.63 $2.80

$3.20 $1.80

$4.20 $2.88

$2.33 $1.40

$0.85

Calculate the mean cost, the standard deviation and the interval equivalent to two standard deviations above and below the mean.



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Solution The mean cost is $2.66 and the standard deviation is $0.84. The interval equivalent to two standard deviations above and below the mean is: [2.66 − 2 × 0.84, 2.66 + 2 × 0.84] = [0.98, 4.34]. In this case, 20 of the 21 observations, or 95% of observations, have values within the interval calculated. Example 16 The prices of forty secondhand motorbikes listed in a newspaper are as follows: $5442 $2220 $3457 $6469 $5294

$5439 $1356 $4689 $7148 $3847

$2523 $738 $8218 $10 884 $4219

$2358 $656 $11 091 $14 450 $4786

$2363 $715 $11 778 $15 731 $2280

$2244 $1000 $11 637 $13 153 $3019

$1963 $1214 $8770 $10 067 $7645

$2142 $1788 $8450 $9878 $8079

Determine the interval equivalent to two standard deviations above and below the mean. Solution The mean price is $5729 and the standard deviation is $4233 (to the nearest whole dollar). The interval equivalent to two standard deviations above and below the mean is: [5729 − 2 × 4233, 5729 + 2 × 4233] = [−2737, 14 195]. The negative value does not give a sensible solution and should be replaced by 0. 38 of the 40 observations, or 95% of observations, have values within the interval. The exact percentage of observations which lie within two standard deviations of the mean varies from data set to data set, but in general it will be around 95%, particularly for symmetric data sets. It was noted earlier that even a single outlier can have a very marked effect on the value of the mean of a data set, while leaving the median unchanged. The same is true when the effect of an outlier on the standard deviation is considered, in comparison to the interquartile range. The median and interquartile range are called resistant measures, while the mean and standard deviation are not resistant measures. When considering a data set it is necessary to do more than just compute the mean and standard variation. First it is necessary to examine the data, using a histogram or stem-and-leaf plot to determine which set of summary statistics is more suitable.




529

Using the TI-Nspire The calculator can be used to calculate the values of all of the summary statistics in this section. Consider the data from Example 16. Enter the data in a Lists & Spreadsheet application (c>New Document>Add Lists & Spreadsheet). Firstly, use the up/down arrows (£ ¤) to name the first column bike. Then enter each of the 40 numbers as shown.

Open a Calculator application (/+I>Add Calculator) and use b>Statistics>Stat Calculations >One-Variable Statistics to calculate the summary statistics. Hint: use e to move between fields. Specify in the dialog box that there is only one list, and then complete the final dialog box as shown. Hint: when in the X1 List field press ¢ to access the list of variables available. Press enter to calculate the values of the summary statistics.

Use the up arrow (£) to view the rest of the summary statistics.



530


The calculator can also be used to determine the summary statistics when the data is given in a frequency table such as: x Frequency

1 5

2 8

3 7

4 2

Enter the data in a Lists & Spreadsheet application. Firstly, use the up/down arrows (£ ¤) to name the first column x and the second column freq. Then enter the data as shown. Open a Calculator application (/+I>Add Calculator) and use b>Statistics>Stat Calculations >One-Variable Statistics to calculate the summary statistics. Specify in the dialog box that there is only one list, and then complete the final dialog box as shown. Hint: when in the X1 List field press ¢ to access the list of variables available. Press enter to calculate the values of the summary statistics.

Using the Casio ClassPad Consider the following heights in cm of a group of eight women. 176, 160, 163, 157, 168, 172, 173, 169 Enter the data into list1 in the module. Tap Calc, One-Variable and when prompted ensure that the XList is set to list1 and the Freq = 1 (since each score is entered individually). The calculator returns the results as shown and all univariate statistics can be viewed by using the scroll bar. Note that the standard deviation is given by x␴n−1 . Where data is grouped, the scores are entered in list1 and the frequencies in list2. In this case, in Set Calculation use the drop-down arrow to select list2 as the location for the frequencies.



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Exercise 22F 1 Find the mean and the median of the following data sets. Examples

9, 10

a 29 14 11 24 14 14 28 14 18 22 14 b 5

9

11

3

12

13

12

6

13

7

3

15

12

15

5

6

c 8.3 5.6 8.2 6.5 8.2 7.0 7.9 7.1 7.8 7.5 d 1.5 1.0

0.2 3.4

0.7 1.3

0.7 0.9

0.2 1.1

0.2 5.8

0.1 2.7

1.7 3.2

0.5 0.6

1.2 4.6

2.0 0.5

1.7 3.1

2 Find the mean and the median of the following data sets. x 1 2 3 4 5 a Frequency 6 3 10 7 8 −2 5

x Frequency

b

−1 8

0 11

1 3

2 2

3 The price, in dollars, of houses sold in a particular suburb during a one-week period are given in the following list. $187 500 $133 500

$129 500 $135 500

$93 400 $140 000

$400 000 $186 000

$118 000 $140 000

$168 000 $204 000

$550 000 $122 000

Find the mean and the median of the prices. Which do you think is a better measure of centre of the data set? Explain your answer. 4 Concerned with the level of absence from his classes a teacher decided to investigate the number of days each student had been absent from the classes for the year to date. These are his results. No. of days missed No. of students

0 1 2 3 4 5 6 9 21 4 2 14 10 16 18 10 2 1

Find the mean and the median number of days each student had been absent so far that year. Which is the better measure of centre in this case? Examples

11, 12

5 Find the range and the interquartile range for each of the following data sets. a 718

630

1002

b 0.7

−1.6

c 8.56

8.51

d 20

19

16

715 −1.2

0.2 8.96

18

560

8.39 16

1085 −1.0

8.62 18

750

8.51 21


20

510

3.4

3.7

8.58

8.82

17

15

1112

1093

0.8 8.54 22

19


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6 The serum cholesterol levels for a sample of twenty people are: 231 190

159 192

203 209

304 161

248 206

238 224

209 276

193 196

225 189

244 199

a Find the range of the serum cholesterol levels. b Find the interquartile range of the serum cholesterol levels. 7 Twenty babies were born at a local hospital on one weekend. Their birth weights, in kg, are given in the stem-and-leaf plot below. 2 2 3 3 4 4

1 5 1 5 1 5

7 3 6 2

9 3 7 2

9 4 7 3

4 9 3|6 represent 3.6 kg

a Find the range of the birth weights. b Find the interquartile range of the birth weights. Example

14

8 Find the standard deviation for the following data sets. a 30

16

b $2.52 $4.32 c 200

22 $4.38 $5.65

300

23

18

$3.60 $6.89 950

200

18 $2.30 $1.98 200

14

56

$3.45 $4.60 300

13 $5.40 $5.12

840

26

9

$4.43 $3.79 350

31 $2.27 $4.99

200

$4.50 $3.02

200

d 86 74 75 77 79 82 81 75 78 79 80 75 78 78 81 80 76 77 82 9 For each of the following data sets Example

15

a calculate the mean and the standard deviation b determine the percentage of observations falling within two standard deviations of the mean. i 41 16 6 21 1 21 5 31 20 27 17 10 3 32 2 48 8 12 21 44 1 56 5 12 3 1 13 11 15 14 10 12 18 64 3 10 ii 141 152

Example

13

260 141

164 239

235 145

167 134

266 150

150 237

255 254

168 150

245 265

258 140

239 132

10 A group of university students was asked to write down their ages with the following results. 17 17 17 17 17 17 17 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 19 19 19 20 20 20 21 24 25 31 41 44 45 a Construct a cumulative relative frequency polygon and use it to find the median and the interquartile range of this data set. b Find the mean and standard deviation of the ages. c Find the percentage of students whose ages fall within two standard deviations of the mean.



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11 The results of a student’s chemistry experiment are as follows. 7.3

8.3

5.9

7.4

6.2

7.4

5.8

6.0

i Find the mean and the median of the results. ii Find the interquartile range and the standard deviation of the results. b Unfortunately when the student was transcribing his results into his chemistry book he made a small error, and wrote: a

7.3

8.3

5.9

7.4

6.2

7.4

5.8

60

i Find the mean and the median of these results. ii Find the interquartile range and the standard deviation of these results. c Describe the effect the error had on the summary statistics calculated in parts a and b. Example

17

12 A selection of shares traded on the stock exchange had a mean price of $50 with a standard deviation of $3. Determine an interval which would include approximately 95% of the share prices. 13 A store manager determined the store’s mean daily receipts as $550, with a standard deviation of $200. On what proportion of days were the daily receipts between $150 and $950?

22.7

The boxplot Knowing the median and quartiles of a distribution means that quite a lot is known about the central region of the data set. If something is known about the tails of the distribution then a good picture of the whole data set can be obtained. This can be achieved by knowing the maximum and minimum values of the data. These five important statistics can be derived from a data set: the median, the two quartiles and the two extremes. These values are called the five-figure summary and can be used to provide a succinct pictorial representation of a data set called the box and whisker plot, or boxplot. For this visual display, a box is drawn with the ends at the first and third quartiles. Lines are drawn which join the ends of the box to the minimum and maximum observations. The median is indicated by a vertical line in the box. Example 17 Draw a boxplot to show the number of hours spent on a project by individual students in a particular school. 24 59 9

4 102 3

166 13 48

147 108 27

97 2 264

90 71 86


36 102 9

92 147 40

226 56 146

37 181 19

111 35 76


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Solution First arrange the data in order. 2 3 4 9 37 40 48 56 102 102 108 111

9 59 146

13 71 147

19 76 147

24 86 166

27 90 181

35 92 226

36 97 264

From this ordered list prepare the five-figure summary. median, m = 71 24 + 27 = 25.5 first quartile, Q 1 = 2 108 + 111 third quartile, Q 3 = = 109.5 2 minimum = 2 maximum = 264 The boxplot can then be drawn.

0 min = 2 Q1 = 25.5

100

200

300

m = 71 Q3 = 109.5

max = 264

In general, to draw a boxplot: Arrange all the observations in order, according to size. Determine the minimum value, the first quartile, the median, the third quartile, and the maximum value for the data set. Draw a horizontal box with the ends at the first and third quartiles. The height of the box is not important. Join the minimum value to the lower end of the box with a horizontal line. Join the maximum value to the upper end of the box with a horizontal line. Indicate the location of the median with a vertical line.



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The symmetry of a data set can be determined from a boxplot. If a data set is symmetric, then the median will be located approximately in the centre of the box, and the tails will be of similar length. This is illustrated in the following diagram, which shows the same data set displayed as a histogram and a boxplot.

A median placed towards the left of the box, and/or a long tail to the right indicates a positively skewed distribution, as shown in this plot.

A median placed towards the right of the box, and/or a long tail to the left indicates a negatively skewed distribution, as illustrated here.



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A more sophisticated version of a boxplot can be drawn with the outliers in the data set identified. This is very informative, as one cannot tell from the previous boxplot if an extremely long tail is caused by many observations in that region or just one. Before drawing this boxplot the outliers in the data set must be identified. The term outlier is used to indicate an observation which is rather different from other observations. Sometimes it is difficult to decide whether or not an observation should be designated as an outlier. The interquartile range can be used to give a very useful definition of an outlier. An outlier is any number which is more than 1.5 interquartile ranges above the upper quartile, or more than 1.5 interquartile ranges below the lower quartile. When drawing a boxplot, any observation identified as an outlier is indicated by an asterisk, and the whiskers are joined to the smallest and largest values which are not outliers.

Example 18 Use the data from Example 17 to draw a boxplot with outliers. Solution median = 71 interquartile range = Q 3 − Q 1 = 109.5 − 25.5 = 84 An outlier will be any observation which is less than 25.5 − 1.5 × 84 = −100.5, which is impossible, or greater than 109.5 + 1.5 × 84 = 235.5. From the data it can be seen that there is only one observation greater than this, 264, which would be denoted with an asterisk. The upper whisker is now drawn from the edge of the box to the largest observation less than 235.5, which is 226. * 0

100


200

300



537

Using the TI-Nspire The calculator can be used to construct a boxplot. Consider the data from Example 17. The data is entered in a Lists & Spreadsheet application (c>New Document>Add Lists & Spreadsheet). Firstly, use the up/down arrows (£ ¤) to name the first column hours. Then enter each of the 33 numbers as shown.


Press e to show the variable list for the x-axis and select the variable, hours. Alternatively, press b>Plot Properties>Add X Variable or use the touchpad to move down below the x-axis and click to add the x variable. Note: if you wish to change the colour (CX only), move the cursor over a data point and press (/+b) to access the colour palette. Change to a boxplot using b>Plot Type>Box Plot. The data now displays as shown. Notice how the calculator, by default, shows outlier(s), if present.



538


To not show the outlier(s), use b>Plot Properties>Extend Box Plot Whiskers. The data now displays as shown. Note: It is possible to show the values of the five-point summary by moving the cursor over the boxplot or using b>Analyze>Graph Trace.

Using the Casio ClassPad In the following consider the set of marks: 28 21 21 3 22 31 35 26 27 33 36 35 23 24 43 31 30 34 48 enter the data into list1. Tap SetGraph, Setting . . . and the tab for Graph 2, enter In and tap SET. (Note the settings shown including the tick box that on the Classpad you can store settings for a number of different graphs and return to them quickly.) Tap SetGraph, StatGraph2 and tap the box to tick and select the graph (de-select any other to produce the graph. The graphs). Tap boxplot is produced as shown. With the graph window selected (bold border), tap 6 to adjust the viewing window for the graph. Tap Analysis, Trace and use the navigator key to move between the outlier(s), Minimum, Q1, Median, Q3 and Maximum scores. Starting from the left of the plot, we see that the: Minimum value is 3: min X = 3. It is also an outlier Lower adjacent value is 21: X = 21 First quartile is 23: Q1 = 23 Median is 30: Med = 30 Second quartile is 35: Q3 = 35 Maximum value is 48: max X = 48.



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17

1 The heights (in centimetres) of a class of girls are 160 154

165 159

123 149

143 167

154 176

180 163

133 154

123 167

157 168

157 132

135 145

140 143

140 157

150 156

a Determine the five-figure summary for this data set. b Draw a boxplot of the data. c Describe the pattern of heights in the class in terms of shape, centre and spread. Example

18

2 A researcher is interested in the number of books people borrow from a library. She decided to select a sample of 38 cards and record the number of books each person has borrowed in the previous year. Here are her results. 7 28 0 2 38 18 0 0 4 0 0 2 1 1 14 1 8 27 0 52 4 0 12 28 10 1 0 2 0 1 11 5 11 0 13 0 a Determine the five-figure summary for this data set. b Determine if there are any outliers. c Draw a boxplot of the data, showing any outliers. d Describe the number of books borrowed in terms of shape, centre and spread.

13 15

3 The winnings of the top 25 male tennis players in 2004 are given in the following table. Player Roger Federer Lleyton Hewitt Andy Roddick Marat Safin Guillermo Coria Gaston Gaudio Tim Henman Carlos Moya Andre Agassi David Nalbandian Jonas Bjorkman Tommy Robredo Nicolas Massu

Winnings 6 357 547 2 766 051 2 604 590 2 273 283 1 697 155 1 639 171 1 508 177 1 448 209 1 177 254 1 045 985 927 344 861 357 854 533

Player Joachim Johansson Jiri Novak Dominik Hrbaty Guillermo Canas Fernando Gonzalez Sebastian Grosjean Feliciano Lopez Max Mirnyi Juan Ignacio Chela Mikhail Youzhny Radek Stepanek Vincent Spadea

Winnings 828 744 813 792 808 944 780 701 766 416 755 795 748 662 742 196 727 736 725 948 706 387 704 105

a Draw a boxplot of the data, indicating any outliers. b Describe the data in terms of shape, centre, spread and outliers. 4 The hourly rate of pay for a group of students engaged in part-time work was found to be: $4.75 $8.50 $17.23 $9.00 $12.00 $11.69 $6.25 $7.50 $8.89 $6.75 $7.90 $12.46 $10.80 $8.40 $12.34 $10.90 $11.65 $10.00 $10.00 $13.00 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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a Draw a boxplot of the data, indicating any outliers. b Describe the hourly pay rate for the students in terms of shape, centre, spread and outliers. 5 The daily circulation of several newspapers in Australia is: 570 000 217 284 98 158

327 654 214 000 77 500

299 797 212 770 56 000

273 248 171 568 43 330

258 700 170 000 17 398

230 487 125 778

a Draw a boxplot of the data, indicating any outliers. b Describe the daily newspaper circulation in terms of shape, centre, spread and outliers.

22.8 Using boxplots to compare distributions Boxplots are extremely useful for comparing two or more sets of data collected on the same variable, such as marks on the same assignment for two different groups of students. By drawing boxplots on the same axis, both the centre and spread for the distributions are readily identified and can be compared visually. Example 19 The number of hours spent by individual students on the project referred to in Example 17 at another school were: 53 57 136 24

152 106 226 80

82 14 17 54

30 18 9 16

16 173 156 106

136 102 19 6

21 86 107 38

11 227 24 3

1 48 42

55 12 21

128 45 176

Use boxplots to compare the time spent on the project by students at this school with those in Example 17. Solution The five-figure summary for this data set is: median, m = 48; first quartile, Q 1 = 17.5; third quartile, Q 3 = 106.5; minimum = 1; maximum = 227 In order to compare the time spent on the project by the students at each school, boxplots for both data sets are drawn on the same axis.

School 1

*

School 2

0

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200

300


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From the boxplots the distributions of time for the two schools can be compared in terms of shape; centre, spread and outliers. Clearly the two distributions for both schools are positively skewed, indicating a larger range of values in the upper half of the distributions. The centre for School 1 is higher than the centre for School 2 (71 hours compared to 48 hours). As can be seen by comparing the box widths, which indicate the IQR, the spread of the data is comparable for both distributions. There is one outlier, a student who attended School 1 and spent 264 hours on the project. The boxplot is useful for summarising large data sets and for comparing several sets of data. It focuses attention on important features of the data and gives a picture of the data which is easy to interpret. When a single data set is being investigated a stem-and-leaf plot is sometimes better, as a boxplot may hide the local detail of the data set.


19

1 To test the effect of a physical fitness course the number of sit-ups that a person could do in 1 minute, both before and after the course, were recorded. Twenty randomly selected participants scored as follows. Before

29 23

22 22

25 26

29 26

26 30

24 12

31 17

46 21

34 20

28 30

After

28 25

26 24

25 30

35 34

33 30

36 15

32 29

54 21

50 19

43 34

a Construct boxplots of these two sets of data on the same axis. b Describe the effect of the physical fitness course on the number of sit-ups achieved in terms of shape, centre, spread and outliers. 2 The number of hours spent on homework per week by a group of students in Year 8 and a group of students in Year 12 are shown in the tables. Year 8

1 1

2 3

4 4

2 3

4 3

4 1

5 7

3 2

7 1

7 3

2 1

4 4

3 1

3 0

Year 12

1 2

2 3

3 1

5 1

6 4

7 7

7 8

6 9

7 6

8 7

7 8

5 7

4 2

1 3

Draw boxplots of these two sets of data on the same axis and use them to answer the following questions. a Which group does the most homework? b Which group varies more in the number of hours homework they do?



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3 The ages of mothers at the birth of their first child were noted, for the first forty such births, at a particular hospital in 1970 and again in 1990. 1970

21 37 24 16

29 22 21 21

25 26 22 25

32 31 36 26

37 26 22 34

30 27 25 27

24 19 31 18

36 21 20 39

23 33 18 24

19 17 20 21

1990

24 19 26 25

22 33 18 35

35 44 28 31

32 24 32 23

17 18 43 19

28 27 28 46

38 24 26 29

20 33 28 23

30 29 41 34

39 23 28 29

a Construct boxplots of these two sets of data on the same axis. b Compare the ages of the mothers in 1970 and 1990 in terms of shape, centre, spread and outliers.



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Variables may be classified as categorical or numerical. Numerical data may be discrete or continuous. Examination of a data set should always begin with a visual display. A bar chart is the appropriate visual display for categorical data. When a data set is small, a stem-and-leaf plot is the most appropriate visual display for numerical data. When a data set is larger, a histogram, frequency polygon or boxplot is a more appropriate visual display for numerical data. Cumulative frequency distributions and cumulative relative frequency distributions are useful for answering questions about the number or proportion of data values greater than or less than a particular value. These are graphically represented in cumulative frequency polygons or cumulative relative frequency polygons. From a stem-and-leaf plot, histogram or boxplot, insight can be gained into the shape, centre and spread of the distribution, and whether or not there are any outliers. An outlier is a value which sits away from the main body of the data in a plot. It is formally defined as a value more than 1.5IQR below Q1 , or more than 1.5IQR above Q3 . For numerical data it is also very useful to calculate some summary statistics. n 1 The mean is defined as x¯ = xi . n i=1 n + 1 th If n, the number of observations, is odd, then the median is the observation 2 from the end of the ordered list. If n is even, then the median is found by averaging the two n th th n middle observations in the list, i.e., the + 1 observations are added and the 2 2 together and divided by 2. The mode is the most common observation in a group of data. The most useful measures of centre are the median and the mean. To find the interquartile range of a distribution: r Arrange all observations in order according to size. r Divide the observations into two equal sized groups. If n, the number of observations, is odd, then the median is omitted from both groups. r Locate Q , the first quartile, which is the median of the lower half of the 1 observations, and Q3 , the third quartile, which is the median of the upper half of the observations. r The interquartile range IQR is defined as the difference between the quartiles. That is IQR = Q 3 − Q 1 n 1 ¯ 2. The standard deviation is defined as s = (xi − x) n − 1 i=1 The most useful measures of spread are the interquartile range and the standard deviation.


Review

Chapter summary


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The five-figure summary of a set of data consists of the minimum, Q 1 , median, Q 3 , and the maximum. A boxplot is a diagrammatic representation of this, e.g.

min

median Q1

max

Q3

When the data set is symmetric any of the summary statistics are appropriate. When the data set is not symmetric or when there are outliers the median and the interquartile range are the preferred summary statistics. In general, 95% of the values of the data set will fall within two standard deviations of the mean. When comparing the distribution of two or more data sets the comparison should be made in terms of the shape, centre, spread and outliers for each distribution.

Multiple-choice questions 1 In a survey a number of subjects were asked to indicate how much they exercise by selecting one of the following options. 1 Never 2 Seldom 3 Occasionally 4 Regularly The resulting variable was named Level of Exercise, and the level of measurement of this variable is A variable B numerical C constant D categorical E metric Questions 2, 3 and 4 relate to the following information. The numbers of hours worked per week by employees in a large company are shown in this percentage frequency histogram. 40

Percentage Frequency

30

20

10

0

20 40 60 Hours worked weekly

80



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A 1%

B 2%

C 6%

D 10%

E 33%

3 The median number of hours worked is in the interval A 10 to less than 20 B 20 to less than 30 C 30 to less than 40 D 40 to less than 50 E 50 to less than 60

Review

2 The percentage of employees who work from 20 to less than 30 hours per week is closest to

Questions 4 and 5 relate to the following information. A group of 19 employees of a company was asked to record the number of meetings that they attended in the last month. Their responses are summarised in the following stem-and-leaf plot. 0 1 2 3 4

1 0 2

1 2 3

2 4

3 4

4 6

5

5

6

6

7

9

4

4 The median number of meetings is A 6 B 6.5 C 7 D 7.5

E 9

6 The cumulative frequency polygon shown gives the examination scores in Mathematics for a group of 200 students. The number of students who scored less than 70 on the examination is closest to A 30 B 100 C 150 D 175 E 200

Number of students

5 The interquartile range (IQR) of number of meetings is A 0 B 4 C 9.5 D 10 E 14 200

100

0 40

50

60 70 80 Exam Score

90

Questions 7 and 8 relate to the following information. The number of years that a sample of people has lived that their current address is summarised in this boxplot.

Years lived this address

0

10

20

30

40

50

7 The shape of the distribution of years lived at this address is: A positively skewed B negatively skewed C bimodal D symmetric E symmetric with outliers 8 The interquartile range years lived at this address is approximately equal to: A 5 B 8 C 17 D 12 E 50 ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Questions 9 and 10 relate to the following data. The amount paid per week to the employees of each of five large companies are shown in the boxplots:

Company 1

Company 2

Company 3

0

20000

40000 60 000 80 000 Yearly income

100 000 120 000

9 The company with the lowest typical wage is A Company 1 B Company 2 C Company 3 D Company 1 and Company 2 E Company 2 and Company 3 10 The company with the largest variation in wage is A Company 1 B Company 2 C Company 3 D Company 1 and Company 2 E Company 2 and Company 3

Short-answer questions (technology-free) 1 Classify the data which arise from the following situations as categorical or numerical. a The number of phones calls a hotel receptionist receives each day. b Interest in politics on a scale from 1 to 5 where 1 = very interested, 2 = quite interested, 3 = somewhat interested, 4 = not very interested, and 5 = uninterested.


50 40 Percent

2 This bar chart shows the percentage of people working who are employed in private companies, work for the Government or are self-employed in a certain town. a What kind of measurement is the ‘Type of company worked for’? b Approximately what percentage of the people are self-employed?

30 20 10 0

Private Government Self-employed Type of company worked for


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0 5

0 0

9 17

10 14

23 3

25 6

0 0

0 33

34 23

32 0

0 32

0 13

30 21

0 22

4 6

Using an appropriate class interval, construct a histogram of these data.

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3 A researcher asked a group of people to record how many cigarettes they had smoked on a particular day. Here are her results:

4 A teacher recorded the time taken (in minutes) by each of a class of students to complete a test. 56 54

57 52

47 69

68 72

52 65

51 45

43 44

22 55

59 56

51 49

39 50

a Make a stem-and-leaf plot of these times, using one row per stem. b Use this stem-and-leaf-plot to find the median and quartiles for the time taken. 5 The weekly rentals, in dollars, for apartments in a particular suburb are given in the following table. 285 265

185 300

210 210

215 270

320 190

680 245

280 315

Find the mean and the median of the weekly rental. 6 Geoff decided to record the time it takes him to complete his mail delivery round each working day for four weeks. His data are recorded in the following table. 170 164 182

189 176 167

201 161 188

183 187 211

168 180 174

182 201 193

161 147 185

166 188 183

167 186

173 176

¯ is 179 and the standard deviation, s, is 14. The mean of the time taken, x, a Determine the percentage of observations falling within two standard deviations of the mean. b Is this what you would expect to find? 7 A group of students were asked to record the number of SMS messages that they sent in one 24-hour period, and the following five-figure summary was obtained from the data set. Use it to construct a simple boxplot of these data. Min = 0,

Q 1 = 3,

Median = 5,

Q 3 = 12,

Max = 24

8 The following table gives the number of students absent each day from a large secondary college on each of 36 randomly chosen school days. 7 7 15

22 3 16

12 21 13

15 30 21

21 13 10

16 2 16

23 7 11

23 12 4

17 18 3

23 14 0

8 14 31

16 0 44

Construct a boxplot of these data, with outliers.



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Extended-response questions 1 The divorce rates (in percentages) of 19 countries are 27 26

18 8

14 14

25 5

28 15

6 32

32 6

44 19

53 9

0

What is the level of measurement of the variable, ‘divorce rate’? Construct an ordered stem-and-leaf plot of divorce rates, with one row per stem. What shape is the divorce rates? What percentage of countries have divorce rates greater than 30? Calculate the mean and median of the divorce rates for the 19 countries. Construct a histogram of the data with class intervals of width 10. i What is the shape of the histogram? ii How many countries had divorce rates from 10% to less than 20%? g Construct a cumulative percentage frequency polygon of divorce rates. i What percentage of countries has divorce rates less than 20%? ii Use the cumulative frequency distribution to estimate the median percentage divorce rate.

a b c d e f

2 Hillside Trains have decided to improve their service on the Lilydale line. Trains were timed on the run from Lilydale to Flinders Street, and their times recorded over a period of six weeks at the same time each day. The time taken for each journey is shown below. 60 90 63 58

61 59 67 64

70 86 74 69

72 70 78 59

68 77 65 62

80 64 68 63

76 57 82 89

65 65 89 74

69 60 75 60

79 68 62

82 60 64

a Construct a histogram of the times taken for the journey from Lilydale to Flinders Street, using class intervals 55–59, 60–64, 65–69 etc. i On how many days did the trip take from 65–69 minutes? ii What shape is the histogram? iii What percentage of trains took less than 65 minutes to reach Flinders Street? b Calculate the following summary statistics for the time taken (correct to two decimal places). x

s

Min

Q1

M

Q3

Max

c Use the summary statistics to complete the following report. i The mean time taken from Lilydale to Flinders Street (in minutes) was . . . ii 50% of the trains took more than . . . minutes to travel from Lilydale to Flinders Street. iii The range of travelling times was . . . minutes while the interquartile range was . . ... minutes. (cont’d) ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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549

Min = 55

Q 1 = 65

Median = 70

Q 3 = 89

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iv 25% of trains took more than . . . minutes to travel to Flinders Street. v The standard deviation of travelling times was . . . vi Approximately 95% of trains took between . . . and . . . minutes to travel to Flinders St. d Summary statistics for the year before Hillside Trains took over the Lilydale line from the Met are indicated below: Max = 99

Draw simple boxplots for the last year the Met ran the line and the data from Hillside trains on the same axis. e Use the information from the boxplots to compare travelling times for the two transport corporations in terms of shape, centre and spread. 3 In a small company, upper management wants to know if there is a difference in the three methods used to train its machine operators. One method uses a hands-on approach. A second method uses a combination of classroom instruction and on-the-job training. The third method is based completely on classroom training. Fifteen trainees are assigned to each training technique. The following data are the results of a test undertaken by the machine operators after completion of one of the different training methods. Method 1 98 100 89 90 81 85 97 95 87 70 69 75 91 92 93

Method 2 79 62 61 89 69 99 87 62 65 88 98 79 73 96 83

Method 3 70 74 60 72 65 49 71 75 55 65 70 59 77 67 80

a Draw boxplots of the data sets, on the same axis. b Write a paragraph comparing the three training methods in terms of shape, centre, spread and outliers. c Which training method would you recommend?



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4 It has been argued that there is a relationship between a child’s level of independence and the order in which they were born in the family. Suppose that the children in thirteen three-children families are rated on a 50-point scale of independence. This is done when all children are adults, thus eliminating age effects. The results are as follows. Family First-born Second-born Third-born

1 38 9 12

2 45 40 12

3 30 24 12

4 29 16 25

5 34 16 9

6 19 21 11

7 35 34 20

8 40 29 12

9 25 22 10

10 50 29 20

11 44 20 16

12 36 19 13

13 26 18 10

a Draw boxplots of the data sets on the same axis. b Write a paragraph comparing the independence scores of first-, secondand third-born children.



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C H A P T E R

23 Investigating the relationship between two numerical variables Objectives To use scatterplots to display bivariate (numerical) data To identify patterns and features of sets of data from scatterplots To identify positive, negative or no association between variables from a scatterplot To introduce the q-correlation coefficient to measure the strength of the relationship between two variables To introduce Pearson’s product-moment correlation coefficient r to measure the strength of the linear relationship between two variables To fit a straight line to data by eye, and using the method of least squares To interpret the slope of a regression line and its intercept, if appropriate To predict the value of the dependent (response) variable from an independent (explanatory) variable, using a linear equation

In Chapter 22 statistics of one variable were discussed. Sometimes values of a variable for more than one group have been examined, such as age of mothers and age of fathers, but only one variable was considered for each individual at a time. When two variables are observed for each subject, bivariate data are obtained. For example, it might be interesting to record the number of hours spent studying for an exam by each student in a class and the mark they achieved in the exam. If each of these variables were considered separately the methods discussed earlier would be used. It may be of more interest to examine the relationship between the two variables, in which case new bivariate techniques are required. When exploring bivariate data, questions arise such as, ‘Is there a relationship between two variables?’ or ‘Does knowing the value of one of the variables tell us anything about the value of the other variable?’



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Consider the relationship between the number of cigarettes smoked per day and blood pressure. Since one opinion might be that varying the number of cigarettes smoked may affect blood pressure, it is necessary to distinguish between blood pressure, which is called the dependent or response variable, and the number of cigarettes, which is called the independent or explanatory variable. In this chapter some techniques are introduced which enable questions concerning the nature of the relationship between such variables to be answered.

23.1 Displaying bivariate data As with data concerning one variable, the most important first step in analysing bivariate data is the construction of a visual display. When both of the variables of interest are numerical then a scatterplot (or bivariate plot) may be constructed. This is the single most important tool in the analysis of such bivariate data, and should always be examined before further analysis is undertaken. The pairs of data points are plotted on the cartesian plane, with each pair contributing one point to the plot. Using the normal convention, the variable plotted horizontally is denoted as x, and the variable plotted vertically as y. The following example examines the features of the scatterplot in more detail. Example 1 The number of hours spent studying for an examination by each member of a class, and the marks they were awarded, are given in the table. Student Hours Mark

1 4 27

2 36 87

3 23 67

4 28 84

5 25 66

6 11 52

7 18 61

8 13 43

9 4 38

10 8 52

Student Hours Mark

11 4 41

12 19 54

13 6 57

14 19 62

15 1 23

16 29 65

17 33 75

18 36 83

19 28 65

20 15 55

Construct a scatterplot of these data. Solution The first decision to be made when preparing this scatterplot is whether to show Mark or Hours on the horizontal (x) axis. Since a student’s mark is likely to depend on the hours that they spend studying, in this case Hours is the independent variable and Mark is the dependent variable. By convention, the independent variable is plotted on the horizontal (x) axis, and the dependent variable on the vertical (y) axis, giving the scatterplot shown.



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Chapter 23 — Investigating the relationship between two numerical variables

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Mark y 80 60 40 20

0

10

30

20

40 Hours x

From this scatterplot, a general trend can be seen of increasing marks with increasing hours of study. There is said to be a positive association between the variables. Two variables are positively associated when larger values of y are associated with larger values of x, as shown in the previous scatterplot. Examples of variables which exhibit positive association are height and weight, foot size and hand size, and number of people in the family and household expenditure on food. Example 2 The age, in years, of several cars and their advertised price in a newspaper are given in the following table. Age (years) 4 6 5 Price ($) 13 000 9 800 11 000 Age (years) 7 Price ($) 9 700

7 4 2 3 3 8 300 10 500 15 800 14 300 13 800

6 4 6 4 8 9 500 13 200 10 000 11 800 8 000

6 12 200

Construct a scatterplot to display these data. Solution In this case the independent variable is the age of the car, which is plotted on the horizontal axis. The dependent variable, price, is plotted on the vertical axis. Price ($) y 16 000 14 000 12000 10000 8 000 1

2

3

4

5

6

7 8 Age (years) x



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From the scatterplot a general trend of decreasing price with increasing age of car can be seen. There is said to be a negative association between the variables. Two variables are negatively associated when larger values of y are associated with smaller values of x, as shown in the scatterplot above. Examples of other variables which exhibit negative association are weight and number of weeks spent on a diet program, hearing ability and age, and number of cold rainy days per week and sales of ice creams. The third alternative is that a scatterplot shows no particular pattern, indicating no association between the variables.

y 8 6 4 2

0

1

2

3

4

5

6

7

8

x

There is no association between two variables when the values of y are not related to the values of x, as shown in the preceding scatterplot. Examples of variables which show no association are height and IQ for adults, price of cars and fuel consumption, and size of family and number of pets. When one point, or a few points, do not seem to fit with the rest of the data they are called outliers. Sometimes a point is an outlier, not because its x value or its y value is in itself unusual, but rather because this particular combination of values is atypical. Consequently such an outlier cannot always be detected from single variable displays, such as stem-and-leaf plots. For example, consider this scatterplot. While the y variable plotted on the horizontal axis takes values 8 from 1 to 8 and the variable plotted on the vertical 6 axis takes values from 2 to 8, the combination (2, 8) 4 is clearly an outlier. 2

0


1

2

3

4

5

6

7

8

x



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Using the TI-Nspire The calculator can be used to construct a scatterplot of statistical data. The procedure is illustrated using the age and price of car data from Example 2. The data is entered in a Lists & Spreadsheet application (c>New Document>Add Lists & Spreadsheet). Firstly, use the up/down arrows (£ ¤) to name the first column age and the second column price. Then enter the data as shown.


Press e to show the variable list for the x-axis and select the variable, age. Press e again to show the variable list for the y-axis and select the variable, price. Alternatively, press b>Plot Properties>Add X Variable to add the x variable, age, then b>Plot Properties>Add Y Variable to add the y variable, price. The data now displays as shown. Note: if you wish to change the colour (CX only), move the cursor over a data point and press (/+b) to access the colour palette.



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Using the Casio ClassPad The table represents the results of 12 students in two tests. Test 1 score Test 2 score

10 12

18 20

13 11

6 9

8 6

5 6

12 12

15 13

15 17

Enter the data into list1 (x) and list2 (y) in the module. Tap SetGraph, Setting . . . and select the tab for Graph3. (Note: Following on from the types of graphs in univariate statistics, this allows the Scatterplot settings to be remembered and called upon when required.) Ensure that all other graphs are de-selected and to produce the graph shown in the full tap screen.

Select the graph window (bold border) and tap Analysis, Trace to scroll from point to point and display the coordinates at the bottom of the graph.

Exercise 23A Save your data for 1–4 in named lists as they will be needed for later exercises. 1 The amount of a particular pain relief drug given to each patient and the time taken for the patient to experience pain relief are shown.

Note:

Patient Drug dose (mg) Response time (min)

1 0.5 65

2 1.2 35

3 4.0 15

4 5.3 10

5 2.6 22

6 3.7 16

7 5.1 10

8 1.7 18

9 0.3 70

10 4.0 20

a Plot the response time against drug dose. b From the scatterplot, describe any association between the two variables. c Identify outliers, if any, and interpret. 2 The proprietor of a hairdressing salon recorded the amount spent advertising in the local paper and the business income for each month of a year, with the following results.



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Month Advertising ($) Business ($) 1 350 9 450 2 450 10 070 3 400 9 380 4 500 9 110 5 250 5 220 6 150 3 100

557


a Plot the business income against the advertising expenditure. b From the scatterplot, describe any association between the two variables. c Identify outliers, if any, and interpret. 3 The number of passenger seats on the most commonly used commercial aircraft, and the airspeeds of these aircraft, in km/h, are shown in the following table. Number of seats Airspeed (km/h)

405 830

296 797

288 774

258 736

240 757

230 765

193 760

188 718

Number of seats Airspeed (km/h)

148 683

142 666

131 661

122 378

115 605

112 620

103 576

102 603

a Plot the airspeed against the number of seats. b From the scatterplot, describe any association between the two variables. c Identify outliers, if any, and interpret. 4 The price and age of several secondhand caravans are listed in the table. Age (years) 7 7 8 9 4 8 1

Price ($) 4 800 3 900 4 275 3 900 6 900 6 500 11 400

Age (years) 10 9 9 11 3 4 7

Price ($) 8 700 1 950 3 300 1 650 9 600 8 400 6 600

a Plot the price of the caravans against their age. b From the scatterplot, describe any association between the two variables. c Identify outliers, if any, and interpret.

23.2

The q-correlation coefficient If the plot of a bivariate data set shows a basic trend, apart from some randomness, then it is useful to provide a numerical measure of the strength of the relationship between the two variables. Correlation is a measure of strength of a relationship which applies only to



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numerical variables. Thus it is sensible, for example, to calculate the correlation between the heights and weights for a group of students, but not between height and gender, as gender is not a numerical variable. There are many different numerical measures of correlation, and each has different properties. In this section the q-correlation coefficient will be introduced. Consider the scatterplot of the number of hours spent by each member of a class when studying for an examination, and the mark they were awarded, from Example 1. This shows a positive association. To calculate the q-correlation coefficient, first find the median value for each of the variables separately. This can be done from the data, but it is usually simpler to calculate directly from the plot. There are 20 data points, and the median values are halfway between the 10th and 11th points, both vertically and horizontally. A vertical line is then drawn through the median x value, and a horizontal line through the median y value. The effect of this is to divide the plot into four regions, as shown. Marks y 80 60 40 20

0

10

30

20

40 Hours x

Each of the four regions which have been created in this way is called a quadrant, and it can be noticed immediately that most of the points in this plot are in the upper right and lower left quadrants. In fact, wherever there is a positive association between variables this will be the case. Consider the scatterplot of the age of cars and the advertised price from Example 2, which shows negative association. Again the median value for each of the variables is found separately. There are 15 data points, giving the median values at the 8th points, both vertically and horizontally. A vertical line is then drawn through the median x value, and a horizontal line through the median y value. In this particular case they are coordinates of the same point, but this need not be so. Price ($) y 16 000 14 000 12 000 10 000 8 000 1

2

3

4

5

6


7 8 Age (years) x Cambridge University Press

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It can be seen in this example that most points are in the upper left and the lower right quadrants, and this is true whenever there is a negative association between variables. These observations lead to a definition of the q-correlation coefficient. The q-correlation coefficient can be determined from the scatterplot as follows. Find the median of all the x values in the data set, and draw a vertical line through this value.

y

Find the median of all the y values in the data set, and draw a horizontal line through this value.

B

A

C

D

0

x

The plane is now divided into four quadrants. Label the quadrants A, B, C and D as shown in the diagram. Count the number of points in each of the quadrants A, B, C and D. Any points which lie on the median lines are omitted. Let a, b, c, d represent the number of points in each of the quadrants A, B, C and D respectively. Then the q-correlation coefficient is given by q=

(a + c) − (b + d) a+b+c+d

Example 3 Use the scatterplot from Example 1 to determine the q-correlation coefficient for the number of hours each member of a class spent studying for an examination and the mark they were awarded. Solution There are nine points in quadrant A, one in quadrant B, nine in quadrant C and one in quadrant D. Thus

(a + c) − (b + d) a+b+c+d (9 + 9) − (1 + 1) = 9+1+9+1 18 − 2 = 20 16 = 20 = 0.8

q=



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Example 4 Use the scatterplot from Example 2 to determine the q-correlation coefficient for the age of cars and their advertised price. Solution There is one point in quadrant A, six in quadrant B, one in quadrant C and six in quadrant D. (a + c) − (b + d) a+b+c+d (1 + 1) − (6 + 6) = 1+6+1+6 2 − 12 = 14 −10 = 14 = −0.71

q=

From Examples 3 and 4 it can be seen that q-correlation coefficients may take both positive and negative values. Consider the situation when all the points are in the quadrants A and C. (a + c) − (b + d) a+b+c+d a+c (since b and d are both equal to zero) = a+c =1

q=

Thus the maximum value the q-correlation coefficient may take is 1, and this indicates a measure of strong positive association. Suppose all the points are in the quadrants B and D. (a + c) − (b + d) a+b+c+d −(b + d) (since a and c are both equal to zero) = b+d = −1

q=

Thus the minimum value the q-correlation coefficient may take is −1, and this indicates a measure of strong negative association.



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When the same number of points are in each of the quadrants A, B, C and D then: (a + c) − (b + d) a+b+c+d 0 (since a = b = c = d) = a+b+c+d =0

q=

This value of the q-correlation coefficient clearly indicates that no association exists. q-correlation coefficients can be classified as follows: −1 ≤ q −0.75 < q −0.50 < q −0.25 < q 0.25 ≤ q 0.50 ≤ q 0.75 ≤ q

≤ −0.75 ≤ −0.50 ≤ −0.25 < 0.25 < 0.50 < 0.75 ≤1

strong negative relationship moderate negative relationship weak negative relationship no relationship weak positive relationship moderate positive relationship strong positive relationship

Exercise 23B 1 Use the table of q-correlation coefficients to classify the following. a q = 0.20 e q = 0.95 i q = −1

b q = −0.30 f q = −0.75 j q = 0.25

c q = −0.85 g q = 0.75 k q=1

d q = 0.33 h q = −0.24 l q = −0.50

2 Calculate the q-correlation coefficient for each pair of variables shown in the following scatterplots. a

y 36

24

12

15.0

20.0

25.0

30.0


35.0

x


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b

y

280

210

140

120

c

160

200

240

280

x

y 8 6 4 2

0

d

1

2

3

4

5

6

7

8

x

y 80 60 40 20

0

Example

4

10

20

30

40

x

3 The amount of a particular pain relief drug given to each patient and the time taken for the patient to experience pain relief are shown. Patient Drug dose (mg) Response time (min)

1 0.5 65

2 1.2 35

3 4.0 15

4 5.3 10

5 2.6 22

6 3.7 16

7 5.1 10

8 1.7 18

9 0.3 70

10 4.0 20

a Use your scatterplot from 1, Exercise 23A to find the q-correlation coefficient for response time and drug dosage.



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b Classify the strength and direction of the relationship between response time and drug dosage according to the table given. Example

3

4 The proprietor of a hairdressing salon recorded the amount spent advertising in the local paper and the business income for each month of a year, with the following results. Month Advertising ($) Business ($) 1 350 9 450 2 450 10 070 3 400 9 380 4 500 9 110 5 250 5 220 6 150 3 100


a Use your scatterplot from 2, Exercise 23A to find the q-correlation coefficient for advertising expenditure and total business conducted. b Classify the strength and direction of the relationship between advertising expenditure and business income according to the table given. 5 The number of passenger seats on the most commonly used commercial aircraft, and the airspeeds of these aircraft, in km/h, are shown in the following table. Number of seats Airspeed (km/h)

405 830

296 797

288 774

258 736

240 757

230 765

193 760

188 718


148 683

142 666

131 661

122 378

115 605

112 620

103 576

102 603

a Use your scatterplot from 3, Exercise 23A to find the q-correlation coefficient for the number of seats on an airline and the air speed. b Classify the strength and direction of the relationship between the number of seats on an airline and the air speed according to the table given. 6 The price and age of several secondhand caravans are listed in the table. Age (years) 7 7 8 9 4 8 1

Price ($) 4 800 3 900 4 275 3 900 6 900 6 500 11 400

Age (years) 10 9 9 11 3 4 7


Price ($) 8 700 1 950 3 300 1 650 9 600 8 400 6 600


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a Use your scatterplot from 4, Exercise 23A to find the q-correlation coefficient for price and age of secondhand caravans. b Classify the strength and direction of the relationship between price and age of secondhand caravans according to the table given.

23.3 The correlation coefficient When a relationship is linear the most commonly used measure of strength of the relationship is Pearson’s product-moment correlation coefficient, r. It gives a numerical measure of the degree to which the points in the scatterplot tend to cluster around a straight line. Pearson’s product-moment correlation is defined to be r=

degree which the variables vary together degree which the two variables vary separately

Formally, if we call the two variables x and y and we have n observations then Pearson’s product-moment correlation for this set of observations is n 1 xi − x¯ yi − y¯ r= n − 1 i=1 sx sy where x¯ and sx are the mean and standard deviation of the x scores and y¯ and s y are the mean and standard deviation of the y scores. There are two key assumptions made in calculating Pearson’s correlation coefficient r. They are the data is numerical the relationship being described is linear. Pearson’s correlation coefficient r has the following properties: If there is no linear relationship, r = 0. y

0

x r=0

For a perfect positive linear relationship, r = +1.

For a perfect negative linear relationship, r = −1. y

y

0

r = +1

x

0


r = –1

x Cambridge University Press

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Otherwise, −1 ≤ r ≤ +1 Pearson’s correlation coefficient r can be classified as follows: −1 ≤ r −0.75 ≤ r −0.50 ≤ r −0.25 < r 0.25 ≤ r 0.50 ≤ r 0.75 ≤ r

≤ −0.75 strong negative linear relationship ≤ −0.50 moderate negative linear relationship ≤ −0.25 weak negative linear relationship < 0.25 no linear relationship < 0.50 weak positive linear relationship < 0.75 moderate positive linear relationship ≤1 strong positive linear relationship

The following scatterplots show linear relationships of various strengths together with the corresponding value of Pearson’s product-moment correlation coefficient. 30 25 12

Age

CO level

14

20

10 15 100 150 200 250 300 350 400 Traffic volume

Carbon monoxide level in the atmosphere and traffic volume: r = 0.985

800 1000 1200 Testosterone level

600

1400

Age first convicted and testosterone (a male hormone) level of a group of prisoners: r = −0.814 110

160

120

Score

Mortality

140 100

100 80

90

60 80

100 Smoking ratio

120

700

20

600

15

500

10 12 14 16 18 20 Age 1st word

Score on aptitude test (taken later in life) and age (in months) when first word spoken: r = −0.445

Calf

Verbal

Mortality rate due to lung cancer and smoking ratio (100 average): r = 0.716

8

10 5

400 500

700 600 Mathematics

30

800

Scores on standardised tests of verbal and mathematical ability: r = 0.275

40 Age

50

Calf measurement and age of adult males: r = −0.005



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Using the TI-Nspire The value of Pearson’s product-moment correlation coefficient, r, can be calculated using the calculator. This will be illustrated using the age and price of car data from Example 2. With the data entered as the two lists age and price respectively, open a Calculator application (/+I>Add Calculator) to calculate Pearson’s product-moment correlation coefficient. Use b>Statistics >Stat Calculations>Linear Regression (mx + b) and complete the dialog box as shown. Press enter to obtain the regression information including the value for r as shown. Note: Linear Regression (a + bx) can also be used.

Using the Casio ClassPad Consider the following set of data x y

1 2

3 5

5 7

4 2

7 9

Enter the data into list1 (x) and list2 (y). Tap Calc, Linear Reg and select the settings shown. Tap OK to produce the results shown in the second screen. The value of r is shown in the answer box. Tap OK to produce a scatterplot.



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After the mean and standard deviation, Pearson’s product-moment correlation is one of the most frequently computed descriptive statistics. It is a powerful tool but it is also easily misused. The presence of a linear relationship should always be confirmed with a scatterplot before Pearson’s product-moment correlation is calculated. And, like the mean and the standard deviation, Pearson’s correlation coefficient r is very sensitive to the presence of outliers in the sample.

Exercise 23C 1 Use the table of Pearson’s correlation coefficients r to classify the following. a r = 0.20 e r = 0.95 i r = −0.50

b r = −0.30 f r = −0.75 j r = 0.25

c r = −0.85 g r = 0.75 k r =1

d r = 0.33 h r = −0.24 l r = −1

2 By comparing the plots given to those on page 538 estimate the value of Pearson’s correlation coefficient r. a

b

y 36

280

24

210

12

140 15.0

c

20.0

25.0

30.0

35.0

120

x

d

y 8

160

200

240

280

x

y 80

6

60

4

40

2

20

0

e

y

1

2

3

4

5

7

6

8

x

0

y

f y

14 000

8

12 000

6

10 000

4

8 000

2 1

2

3

4

5

6

7

8 x


0

10

1

2

30

20

3

4

5

6

7

40

x

8

x


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3 The amount of a particular pain relief drug given to each patient and the time taken for the patient to experience relief are shown. Patient Drug dose (mg) Response time (min)

1 0.5 65

2 1.2 35

3 4.0 15

4 5.3 10

5 2.6 22

6 3.7 16

7 5.1 10

8 1.7 18

9 0.3 70

10 4.0 20

a Determine the value of Pearson’s correlation coefficient. b Classify the relationship between drug dose and response time according to the table given. 4 The proprietor of a hairdressing salon recorded the amount spent on advertising in the local paper and the business income for each month for a year, with the following results. Month Advertising ($) Business ($) 1 350 9 450 2 450 10 070 3 400 9 380 4 500 9 110 5 250 5 220 6 150 3 100


a Determine the value of Pearson’s correlation coefficient. b Classify the relationship between the amount spent on advertising and business income according to the table given. 5 The number of passenger seats on the most commonly used commercial aircraft, and the airspeeds of these aircraft, in km/h, are shown in the following table. Number of seats Airspeed (km/h)

405 830

296 797

288 774

258 736

240 757

230 765

193 760

188 718


148 683

142 666

131 661

122 378

115 605

112 620

103 576

102 603

a Determine the value of Pearson’s correlation coefficient. b Classify the relationship between the number of passenger seats and airspeed according to the table given.



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6 The price and age of several secondhand caravans are listed in the table. Age (years) 7 7 8 9 4 8 1

Price ($) 4 800 3 900 4 275 3 900 6 900 6 500 11 400

Age (years) 10 9 9 11 3 4 7

Price ($) 8 700 1 950 3 300 1 650 9 600 8 400 6 600

a Determine the value of Pearson’s correlation coefficient. b Classify the relationship between price and age according to the table given. 7 The following are the scores for a group of ten students who each had two attempts at a test (out of 70). Attempt 1 Attempt 2

53 63

56 66

57 67

49 58

44 54

69 70

66 70

40 55

53 63

43 53

68 70

64 70

a Construct a scatterplot of these data, and describe the relationship between scores on attempt 1 and attempt 2. b Is it appropriate to calculate the value of Pearson’s correlation coefficient for these data? Give reasons for your answer. 8 This table represents the results of two Student Test 1 Test 2 different tests for a group of students. 1 214 216 a Construct a scatterplot of these data, and 2 281 270 describe the relationship between scores on 3 212 281 Test 1 and Test 2. 4 324 326 b Is it appropriate to calculate the value of 5 240 243 Pearson’s correlation coefficient for these 6 208 213 data? Give reasons for your answer. 7 303 311 c Determine the values of the q-correlation 8 278 290 coefficient and Pearson’s correlation 9 311 320 coefficient r. d Classify the relationship between Test 1 and Test 2 using both the q-correlation coefficient and Pearson’s correlation coefficient r, and compare. e It turns out that when the data was entered into the student records, the result for Test 2, Student 9 was entered as 32 instead of 320. i Recalculate the values of the q-correlation coefficient and Pearson’s correlation coefficient r with this new data value. ii Compare these values with the ones calculated in c.



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23.4 Lines on scatterplots If a linear relationship exists between two variables it is possible to predict the value of the dependent variable from the value of the independent variable. The stronger the relationship between the two variables the better the prediction that is made. To make the prediction it is necessary to determine an equation which relates the variables and this is achieved by fitting a line to the data. Fitting a line to data is often referred to as regression, which comes from a Latin word regressum which means ‘moved back’. The simplest equation relating two variables x and y is a linear equation of the form y = a + bx where a and b are constants. This is similar to the general equation of a straight line, where a represents the coordinate of the point where the line crosses the y axis (the y axis intercept), and b represents the slope of the line. In order to summarise any particular (x, y) data set, numerical values for a and b are needed that will make the line pass close to the data. There are several ways in which the values of a and b can be found, of which the simplest is to find the straight line by placing a ruler on the scatter diagram, and drawing a line by eye, which appears to follow the general trend of the data. Example 6 The following table gives the gold medal winning distance, in metres, for the men’s long jump for the Olympic games for the years 1896 to 1996. (Some years were missing owing to the two world wars.) Find a straight line which fits the general trend of the data, and use it to predict the winning distance in the year 2008. Year Distance (m) Year Distance (m)

1896 6.35 1960 8.13

1900 7.19 1964 8.08

1904 7.34 1968 8.92

1908 7.49 1972 8.26

1912 7.59 1976 8.36

1920 7.16 1980 8.53

1924 7.44 1984 8.53

1928 7.75 1988 8.72

1932 7.65 1992 8.67

1936 8.05 1996 8.50

1948 7.82 2000 8.55

1952 1956 7.57 7.82 2004 8.59

Solution

9.0

Distance

8.5 8.0 7.5 7.0 6.5 6.0 1900 1920 1940 1960 1980 2000 Year ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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571

Note that this scatterplot does not start at the origin. Since the values of the coordinates that are of interest on both axes are a long way from zero, it is sensible to plot the graph for that range of values only. In fact, any values less than 1896 on the horizontal axis are meaningless in this context. The line shown on the scatterplot is only one of many which could be drawn. To enable the line to be used for prediction it is necessary to find its equation. To do this, first determine the coordinates of any two points through which it passes on the scatterplot. Appropriate points are (1932, 7.65) and (1976, 8.36). The equation of the straight line is then found by substituting in the formula which gives the equation for a straight line between two points. y2 − y1 y − y1 = x − x1 x2 − x1 8.36 − 7.65 y − 7.65 = x − 1932 1976 − 1932 0.71 = 0.016 = 44 y − 7.65 = 0.016(x − 1932) y = 0.016x − 23.26 or distance = −23.26 + 0.016 × year The intercept for this equation is −23.26 m. In theory, this is the winning distance for the year 0! In practice, there is no meaningful interpretation for the y axis intercept in this situation. But the same cannot be said about the slope. A slope of 0.016 means that on average the gold medal winning distance increases by about 1.6 centimetres at each successive games. Using this equation the gold medal winning distance for the long jump in 2008 would be predicted as y = −23.26 + 0.016 × 2008 = 8.87 m Obviously, attempting to project too far into the future may give us answers which are not sensible. When using an equation for prediction, derived from data, it is sensible to use values of the explanatory variable which are within a reasonable range of the data. The relationship between the variables may not be linear if we move too far from the known values.



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Example 7 The following table gives the alcohol consumption per head (in litres) and the hospital admission rate to each of the regions of Victoria in 1994–95.

Region Loddon–Mallee Grampians Barwon Gippsland Hume Western Metropolitan Northern Metropolitan Eastern Metropolitan Southern Metropolitan

Per capita consumption (litres of alcohol) 9.0 8.4 8.7 9.1 10.0 9.0 6.7 6.2 8.1

Hospital admissions per 1000 residents 42.0 44.7 38.6 44.7 41.0 40.4 36.2 32.3 43.0

Find a straight line which fits the general trend of the data, and interpret the intercept and slope. Solution

Admissions

45.0 40.0 35.0 30.0 0

6.0

7.0 9.0 8.0 Alcohol consumption

10.0

One possible line passes through the points (7, 36) and (9, 42). y2 − y1 y − y1 = x − x1 x2 − x1 42 − 36 y − 36 = x −7 9−7 6 = 2 =3 y − 36 = 3(x − 7) y = 3x + 15 or admission rate = 15 + 3 × alcohol consumption Thus

The intercept for this equation is 15, implying that we predict a hospital admission rate would be 15 per 1000 residents for a region with 0 alcohol consumption. While this is interpretable, it would be a brave prediction as it is well out of the range of the



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data. A slope of 3 means that on average the admission rate rises by 3 per 1000 residents for each additional litre of alcohol consumed per capita.


6

1 Plot the following set of data points on graph paper. x y

0 1

1 3

2 6

3 7

4 7

5 11

6 13

7 18

8 17

Draw a straight line which fits the data by eye, and find an equation for this line. 2 Plot the following set of points on graph paper. x y

−3 5

−2 2

−1 0

0 −6

1 −7

2 −11

3 −13

4 −20

Draw a straight line which fits the data by eye, and find an equation for this line. Example

7

3 The numbers of burglaries during two successive years for various districts in one state are given in the following table. a Make a scatterplot of the data. District Year 1 (x) Year 2 (y) b Find the equation of a straight line which A 3233 2709 relates the two variables. B 2363 2208 c Describe the trend in burglaries in this state. C 4591 3685 D E F G H I J K

4317 2474 3679 5016 6234 6350 4072 2137

4038 2792 3292 4402 5147 5555 4004 1980

4 The following data give a girl’s height (in cm) between the ages of 36 months and 60 months. Age (x) Height (y)

36 84

40 87

44 90

52 92

56 94

60 96

Make a scatterplot of the data. Find the equation of a straight line which relates the two variables. Interpret the intercept and slope, if appropriate. Use your equation to estimate the girl’s height at age i 42 months ii 18 years e How reliable are your answers to d?

a b c d



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5 The following table gives the adult heights (in cm) of ten pairs of mothers and daughters. Mother (x) Daughter (y)

170 178

163 175

157 165

165 173

175 168

160 152

164 163

168 168

152 160

173 178

a Make a scatterplot of the data. b Find the equation of a straight line which relates the two variables. c Estimate the adult height of a girl whose mother is 170 cm tall. 6 The manager of a company which manufactures MP3 players keeps a weekly record of the cost of running the business and the number of units produced. The figures for a period of eight weeks are shown in the table. Number of MP3 players produced (x) Cost in 000’s $ (y) a b c d

100 160 80 100 220 150 170 200 2.5 3.3 2.4 2.6 4.1 3.1 3.5 3.8

Make a scatterplot of the data. Find the equation of a straight line which relates the two variables. What is the manufacturer’s fixed cost for operating the business each week? What is the cost of production of each unit, over and above this fixed operating cost?

7 The amount of a particular pain relief drug given to each patient and the time taken for the patient to experience pain relief are shown. Patient Drug dose (mg) Response time (min)

1 0.5 65

2 1.2 35

3 4.0 15

4 5.3 10

5 2.6 22

6 3.7 16

7 5.1 10

8 1.7 18

9 0.3 70

10 4.0 20

a Find the equation of a straight line which relates the two variables. b Interpret the intercept and slope if appropriate. c Use your equation to predict the time taken for the patient to experience pain relief if 6 mg of the drug is given. Is this answer realistic? 8 The proprietor of a hairdressing salon recorded the amount spent on advertising in the local paper and the business income for each month for a year, with the results shown. a Find the equation of a straight line which relates the two variables. b Interpret the intercept and slope if appropriate. c Use your equation to predict the business income which would be attracted if the proprietor of the salon spent the following amounts on advertising: i $1000 ii $0


Month Advertising ($) Business ($) 1 350 9 450 2 450 10 070 3 400 9 380 4 500 9 110 5 250 5 220 6 150 3 100 7 350 8 060 8 300 7 030 9 550 11 500 10 600 12 870 11 550 10 560 12 450 9 850


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23.5

575

The least squares regression line Fitting a line to a scatterplot by eye is not generally the best way of modelling a relationship. What is required is a method which uses a more objective criterion. A simple method is two divide the data set into two halves on the basis of the median x value, and to fit a line which passes through the mean x and y values of the lower half, and the mean x and y values of the upper half. This is called the two-mean line, and while easy to determine, it is not very widely used. The most common procedure is the method of least squares. The least squares regression line is the line for which the sum of squares of the vertical deviations from the data to the line (as indicated in the diagram) is a minimum. These deviations are called the residuals. y 35 30

y = a + bx

25 20 (xi, yi)

15 10 5 0

1

2

3

4

5

6

7

8

9

10

x

Consider the line y = a + bx We would like to find a and b such that the sum of the residuals is zero. That is, n (yi − a − bxi ) = 0

1

i=1

and the sum of residuals square is as small as possible. That is, n (yi − a − bxi )2 is a minimum

2

i=1

We will use the symbol S to denote

n (yi − a − bxi )2 i=1

n

From 1 ,

∴

(yi − a − bxi ) = 0

n i=1

∴ ∴

i=1

yi − na − b

n

xi = 0

i=1

y¯ − a − b x¯ = 0 a = y¯ − b x¯

3



576


Substituting this relationship in 2 S=

n &

¯ − bxi ]2 [yi − ( y¯ − b x)

i=1

=

n &

¯ 2 [(yi − y¯ ) − b(xi − x)]

i=1

=

n &

¯ i − y¯ ) + b2 (xi − x) ¯ 2] [(yi − y¯ )2 − 2b(xi − x)(y

i=1

This can be thought of as a quadratic expression in b. In order to find the value of b which minimises S, we will differentiate with respect to b and set the derivative equal to zero. n n & & dS ¯ i − y¯ ) + 2b ¯ 2 = −2 (xi − x)(y (xi − x) db i=1 i=1 =0 n & ¯ i − y¯ ) (xi − x)(y

Simplifying gives

b=

i=1

n & ¯ 2 (xi − x)

4

i=1

Equations 3 and 4 can then be used to calculate the least squares estimates of the y axis intercept and the slope.

Using the TI-Nspire The calculator can be used to construct the least squares regression line. The procedure is illustrated using the age and price of car data from Example 2. It has previously been illustrated how to enter the data as two lists, age and price respectively, and from these construct the scatterplot in a Data & Statistics application resulting in the data displayed as shown.




577

Use b>Analyze>Regression>Show Linear (mx + b) to place the regression line on the scatterplot as shown.

Using the Casio ClassPad The following data gives the heights (in cm) and weights (in kg) of 11 people. Height (x) Weight (y)

177 74

182 75

167 62

178 63

173 64

184 74

162 57

169 55

164 56

170 68

180 72

Enter the data into list1 (x) and list2 (y).

Tap Calc, Linear Reg and select the settings shown. Tap OK to produce the results shown in the second screen. The format of the formula, y = ax + b is shown at the top and the values of a, b are shown in the answer box. Tap OK to produce a scatterplot showing the regression line. The formula can be automatically copied into a selected entry line if desired by selecting a graph number, e.g. y1 in the Copy Formula line.

Note:

After the equation of the least squares line has been determined, we can interpret the intercept and slope in terms of the problem at hand, and use the equation to make predictions. The method of least squares is also sensitive to any outliers in the data.



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Example 8 Consider again the gold medal winning distance, in metres, for the men’s long jump for the Olympic games for the years 1896 to 2004. Find the equation of the least squares regression line for these data, and use it to predict the winning distance for the year 2008. Year Distance (m) Year Distance (m)

1896 6.35 1960 8.13

1900 7.19 1964 8.08

1904 7.34 1968 8.92

1908 7.49 1972 8.26

1912 7.59 1976 8.36

1920 7.16 1980 8.53

1924 7.44 1984 8.53

1928 7.75 1988 8.72

1932 7.65 1992 8.67

1936 8.05 1996 8.50

1948 7.82 2000 8.55

1952 1956 7.57 7.82 2004 8.59

Solution Using a calculator or computer the equation is found to be distance = −23.87 + 0.0163 × year which is quite similar to the equation to the line fitted by eye. The predicted distance for the year 2008 is distance = −23.87 + 0.0163 × 2008 = 8.86 m Example 9 Consider again the data from Example 7 which related alcohol consumption per head (in litres) and the hospital admission rate to each of the regions of Victoria in 1994–95. Region Loddon–Mallee Grampians Barwon Gippsland Hume Western Metropolitan Northern Metropolitan Eastern Metropolitan Southern Metropolitan

Per capita consumption (litres of alcohol) 9.0 8.4 8.7 9.1 10.0 9.0 6.7 6.2 8.1

Hospital admissions per 1000 residents 42.0 44.7 38.6 44.7 41.0 40.4 36.2 32.3 43.0

Find the equation of the least squares regression line which fits these data. Solution Using a calculator or computer the equation is found to be admissions = 19.9 + 2.45 × alcohol which is slightly different from the line fitted by eye.



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Correlation and causation The existence of even a strong linear relationship between two variables is not, in itself, sufficient to imply that altering one variable causes a change in the other. It only implies that this might be the explanation. It may be that both the measured variables are affected by a third and different variable. For example, if data about crime rates and unemployment in a range of cities were gathered, a high correlation would be found. But could it be inferred that high unemployment causes a high crime rate? The explanation could be that both of these variables are dependent on other factors, such as home circumstances, peer group pressure, level of education or economic conditions, all of which may be related to both unemployment and crime rates. These two variables may vary together, without one being the direct cause of the other.


8

1 The following data give a girl’s height (in cm) between the ages of 36 months and 60 months. Age (x) Height (y)

36 84

40 87

44 90

52 92

56 94

60 96

a Using the method of least squares find the equation of a straight line which relates the two variables. b Interpret the intercept and slope, if appropriate. c Use your equation to estimate the girl’s height at age i 42 months ii 18 years d How reliable are your answers to part c? 2 The number of burglaries during two successive years for various districts in one state are given in the following table. Using the method of least squares find the equation of a straight line which relates the two variables.

District A B C D E F G H I J K


Year 1 (x) 3 233 2 363 4 591 4 317 2 474 3 679 5 016 6 234 6 350 4 072 2 137

Year 2 (y) 2 709 2 208 3 685 4 038 2 792 3 292 4 402 5 147 5 555 4 004 1 980


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3 The following table gives the adult heights (in cm) of ten pairs of mothers and daughters. Mother (x) Daughter (y)

170 163 157 165 175 160 164 168 152 173 178 175 165 173 168 152 163 168 160 178

a Using the method of least squares find the equation of a straight line which relates the two variables. b Interpret the slope in this context. c Estimate the adult height of a girl whose mother is 170 cm tall. 4 The manager of a company which manufactures MP3 players keeps a weekly record of the cost of running the business and the number of units produced. The figures for a period of eight weeks are: Number of MP3 players produced (x) Cost in 000’s $ (y)

100 160 80 100 220 150 170 200 2.5 3.3 2.4 2.6 4.1 3.1 3.5 3.8

a Using the method of least squares find the equation of a straight line which relates the two variables. b What is the manufacture’s fixed cost for operating the business each week? c What is the cost of production of each unit, over and above this fixed operating cost? 5 The amount of a particular pain relief drug given to each patient and the time taken for the patient to experience pain relief are shown. Patient Drug dose (mg) Response time (min)

1 2 3 4 5 6 7 8 9 10 0.5 1.2 4.0 5.3 2.6 3.7 5.1 1.7 0.3 4.0 65 35 15 10 22 16 10 18 70 20

a Using the method of least squares find the equation of a straight line which relates the two variables. b Interpret the intercept and slope if appropriate. c Use your equation to predict the time taken for the patient to experience pain relief if 6 mg of the drug is given. Is this answer realistic?



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6 The proprietor of a hairdressing salon recorded the amount spent on advertising in the local paper and the business income for each month for a year, with the results shown. a Using the method of least squares find the equation of a straight line which relates the two variables. b Interpret the intercept and slope if appropriate.

Month 1 2 3 4 5 6 7 8 9 10 11 12

Advertising ($) 350 450 400 500 250 150 350 300 550 600 550 450

581

Business ($) 9 450 10 070 9 380 9 110 5 220 3 100 8 060 7 030 11 500 12 870 10 560 9 850

c Use your equation to predict the volume which would be attracted if the proprietor of the salon spent the following amounts on advertising. i $1000 ii $0



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Chapter summary Bivariate data arises when measurements on two variables are collected for each subject. A scatterplot is an appropriate visual display of bivariate data if both of the variables are numerical. A scatterplot of the data should always be constructed to assist in the identification of outliers and illustrate the association (positive, negative or none). Two variables are positively associated when larger values of y are associated with larger values of x. Two variables are negatively associated when larger values of y are associated with smaller values of x. There is no association between two variables when the values of y are not related to the values of x. When constructing the scatterplot, the independent or explanatory variable is plotted on the horizontal (x) axis, and the dependent or response variable is plotted on the vertical (y) axis. If a linear relationship is indicated by the scatterplot a measure of its strength can be found by calculating the q-correlation coefficient, or Pearson’s product-moment correlation coefficient, r. If the values on a scatterplot are divided by lines representing the median of x and the median of y into four quadrants A, B, C and D, with a, b, c, d representing the number of points in each quadrant respectively, then the q-correlation coefficient is given by (a + c) − (b + d) q= a+b+c+d Pearson’s product-moment correlation, r, is a measure of strength of linear relationship between two variables, x and y. If we have n observations then for this set of observations n xi − x¯ yi − y¯ 1 r= n − 1 i=1 sx sy where x¯ and sx are the mean and standard deviation of the x scores and y¯ and s y are the mean and standard deviation of the y scores. For these correlation coefficients −1 ≤ q ≤ 1 −1 ≤ r ≤ 1 with values close to ±1 indicating strong correlation, and those close to 0 indicating little correlation. If a linear relationship is indicated from the scatterplot a straight line may be fitted to the data, either ‘by eye’ or using the least squares regression method. The least squares regression line is the line for which the sum of squares of the vertical deviations from the data to the line is a minimum. The value of the slope (b) gives the extent of the change in the dependent variable associated with a unit change in the independent variable. Once found, the equation to the straight line may be used to predict values of the response variable (y) from the explanatory variable (x). The accuracy of the prediction depends on how closely the straight line fits the data, and an indication of this can be obtained from the correlation coefficient.



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583

1 For which one of the following pairs of variables would it be appropriate to construct a scatterplot? A eye colour (blue, green, brown, other) and hair colour (black, brown, blonde, red, other) B score out of 100 on a test for a group of Year 9 students and a group of Year 11 students C political party preference (Labor, Liberal, Other) and age in years D age in years and blood pressure in mm Hg E height in cm and gender (male, female)

Review


2 For which one of the following plots would it be appropriate to calculate the value of the q-correlation coefficient? A

B

D

E

C

3 A q-correlation coefficient of 0.32 would describe a relationship classified as C strong positive B moderate positive A weak positive E moderately strong D close to zero 4 The scatterplot shows the relationship between age and the number of alcoholic drinks consumed on the weekend by a group of people. The value of the q-correlation coefficient is closest to 7 5 7 E 1 D C − A −1 B − 9 6 9

20

No. of drinks

15

10

5

0 10

20

30

40 Age

50

60


70


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5 The following scatterplot shows the relationship between height and weight for a group of people. The value of the Pearson’s product-moment correlation coefficient r is closest to E 0 D 0.3 C 0.5 B 0.8 A 1

220 200

Weight

180 160 140 120 100 80 140

150

160

170 180 Height (cm)

190

200

Questions 6 and 7 relate to the following information. The weekly income and weekly expenditure on food for a group of 10 university students is given in the following table. Weekly income ($) 150 250 300 600 300 380 950 450 850 1000 Weekly food expenditure ($) 40 60 70 120 130 150 200 260 460 600 6 The value of the Pearson product-moment correlation coefficient r for these data is closest to E 0.8 D 0.7 C 0.6 B 0.4 A 0.2 7 The least squares regression line which would enable expenditure on food to be predicted from weekly income is closest to B 0.482 − 42.864 × weekly income A 0.482 + 42.864 × weekly income D 239.868 + 1.355 × weekly income C −42.864 + 0.482 × weekly income E 1.355 + 239.868 × weekly income



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Weekly expenditure on entertainment = 40 + 0.10 × Weekly income 8 Using this rule the expenditure on entertainment by an individual with an income of $600 per week is predicted to be E $240 D $46 C $100 B $24 060 A $40

Review

Questions 8 and 9 relate to the following information. Suppose that the least squares regression line which would enable expenditure on entertainment (in dollars) to be predicted from weekly income is given by

9 From this rule which of the following statements is correct? A On average for each extra dollar of income an extra 10 cents is spent on entertainment B On average for each extra 10 cents in income an extra $1 is spent on entertainment C On average for each extra dollar of income an extra 40 cents is spent on entertainment D On average people spend $40 per week on entertainment E On average people spend $50 per week on entertainment 10 For the scatterplot shown the line of best fit would have a slope closest to: C 10 B −0.1 A 0.1 E 200 D −10

250 200 150 100 50 10

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Short-answer questions Technology is required to answer some of the following questions. 1 The following table gives the number of times Inside 50 the ball was inside the 50 metre line in an AFL 64 football game, and the team’s score in that game 57 a Plot the score against the number of Inside 50s. 34 b From the scatterplot, describe any association 61 between the two variables. 51 2 Use the scatterplot constructed in 1 to determine q-correlation between the score and the number of Inside 50s.


52 53 51 64 55 58 71

Score (points) 90 134 76 92 93 45 120 66 105 108 88 133


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3 The distance traveled to work and the time taken for a group of company employees are given in the following table. Determine the value of the Pearson product-moment correlation r for these data. Distance (kms) Time (mins)

12 15

50 75

40 50

25 50

45 80

20 50

10 10

3 5

10 10

30 35

4 The following scatterplot shows the relationship between height and weight for a group of people. Draw a straight line which fits the data by eye, and find an equation for this line.

220 200

weight

180 160 140 120 100 80 140

150

160

170 180 height (cm)

190

5 The time taken to complete a task, and the number of errors on the task, were recorded for a sample of 10 primary school children. Determine the equation of the least squares regression line which fits these data. 6 For the data in 5: a Interpret the intercept and slope of the least squares regression line. b Use the least squares regression line to predict the number of errors which would be observed for a child who took 10 seconds to complete the task.


200

Time (seconds) 22.6 21.7 21.7 21.3 19.3 17.6 17.0 14.6 14.0 8.8

Errors 2 3 3 4 5 5 7 7 9 9


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1 A marketing company wishes to predict the likely number of new clients each of its graduates will attract to the business in their first year of employment, by using their scores on a marketing exam in the final year of their course. a Which is the independent variable Number of new and which is the dependent variable? Exam score clients b Construct a scatterplot of these data. 65 7 c Describe the association between the 72 9 Number of new clients and Exam 68 8 score. 85 10 d Determine the value of the q-correlation 74 10 coefficient for these data, and classify 61 8 the strength of the relationship. 60 6 e Determine the value of the Pearson 78 10 product-moment correlation coefficient 70 5 for these data and classify the strength 82 11 of the relationship. f Determine the equation for the least squares regression line and write it down in terms of the variables Number of new clients and Exam score. g Interpret the intercept and slope of the least squares regression line in terms of the variables in the study. h Use your regression equation to predict to the nearest whole number the Number of new clients for a person who scored 100 on the exam. i How reliable is the prediction made in h?

Review


2 To investigate the relationship between marks on an assignment and the final examination mark a sample of 10 students was taken. The table indicates the marks for the assignment and the final exam mark for each individual student. a Which is the independent variable Assignment mark Final exam mark and which is the dependent variable? (max = 80) (max = 90) b Construct a scatterplot of these data. 80 83 c Describe the association between the 77 83 assignment mark and exam mark. 71 79 d Determine the value of the q-correlation 78 75 coefficient for these data, and classify the 65 68 strength of the relationship. 80 84 e Determine the value of the Pearson 68 71 product-moment correlation coefficient 64 69 for these data and classify the strength of 50 66 the relationship. 66 58



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f Use your answer to d to comment on the statement: ‘Good final exam marks are the result of good assignment marks.’ g Determine the equation for the least squares regression line and write it down in terms of the variables Final exam mark and Assignment mark. h Interpret the intercept and slope of the least squares regression line in terms of the variables in the study. i Use your regression equation to predict the Final exam mark for a student who scored 50 on the assignment. j How reliable is the prediction made in i? 3 A marketing firm wanted to investigate the relationship between airplay and CD sales (in the following week) of newly released songs. Data was collected on a random sample of 10 songs. a Which is the independent variable and which No. of times the Weekly sales is the dependent variable? song was played of the CD b Construct a scatterplot of these data. 47 3950 c Describe the association between the number 34 2500 of times the song was played and weekly sales. 40 3700 d Determine the value of the q-correlation 34 2800 coefficient for these data, and classify the strength 33 2900 of the relationship. 50 3750 e Determine the value of the Pearson 28 2300 product-moment correlation coefficient for these 53 4400 data and classify the strength of the relationship. 25 2200 f Determine the equation for the least squares 46 3400 regression line and write it down in terms of the variables Number of times the song was played and Weekly sales. g Interpret the intercept and slope of the least squares regression line in terms of the variables in the study. h Use your regression equation to predict the weekly sales for a song which was played 60 times. i How reliable is the prediction made in h?



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24

Revision

Revision of chapters 22 and 23 24.1

Multiple-choice questions 1 For which of the following variables is a bar chart an appropriate display? A weight (kg) B maximum daily temperature (C) C exam results (A, B, C, D, E) D distance travelled to school each day (km) E time taken to travel to school (mins) 2 For which of the following variables is a histogram an appropriate display? A weight (kg) B religious affiliation D weight (light, medium, heavy) 3 The shape of the distribution shown in this histogram is best described as A C D E

negatively skewed B symmetric positively skewed negatively skewed with an outlier positively skewed with an outlier

C exam results (A, B, C, D, E) E gender (male, female) 10 9 8 7 6 5 4 3 2 1 0

4 The following histogram shows the distance travelled to school by a group of students (in kilometres). From the histogram we can say:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Count 10 9 8 7 6 5 4 3 2 1

A Most students travel from 10 to less than 15 kilometres to school. 0 5 10 B Most students travel from 15 to less than 20 kilometres to school. C Most students travel less than 15 kilometres to school. D Most students travel less than 20 kilometres to school. E Most students travel more than 20 kilometres to school.

15 20 25 30 35 40 45 50 Distance (km)



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Questions 5–8 refer to the following information. A class of students scored marks on a mathematics test as shown in the stem-and-leaf plot.

2 3 4 5 6 7 8 9

9 3 0 0 0 2 3 0

8 1 3 3 3 4 3

4 5 3 4 5

9 9 5 4 6

7 4

7 7

9

5 The number of students in the class is A 8

B 26

C 30

D 38

E 40

6 If the pass mark for the test is a score of 50, the number of students who failed is A 6

B 7

C 8

D 10

E 11

7 The teacher decides to award the top 10% of students on this test an A. The lowest mark to score an A is thus A 74

B 77

C 85

D 86

E 90

8 From the stem-and-leaf plot we can say that the shape of the distribution of test scores is A positively skewed D symmetrically skewed

B symmetric C negatively skewed E unable to determine from the information available.

9 The following data are the driving-test scores for a group of people before and after they completed an advanced driving course.

before 7 2 9 9 8 6 6 6 5 4 3 2 2 1 0 4 1 0 0 6

1 2 3 4 5

after 5 9 1 4 5 5 6 8 9 0 0 2 3 4 4 5 6 3 0 4

From the stem-and-leaf plots we could say: A B C D E

The course has no effect on driving scores. Driving scores have in general increased after the course. Driving scores have in general decreased after the course. The course has increased some people’s score but not others. We are unable to determine any effect of the course from the information available.



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Group 1

4 4 4 2 8 8 8 8 8 8 8 4 4 3 3 3 3 2 9 9 9 4 4

2 8 2 8 4 8

1 6 2 8 3 7

8 0 6 2 8 3 7 4

8 0 6 2 7 2 7 4

4 7 0 6 0 7 2 6 2 6

Group 2 8 4 5 0 5 0 6 1 6 2 6 1

4 5 5 6 6 7 7 8 8 9 9 10

4 8 0 6 0 6 0 6 0

8 0 6 0 6 0 8 2

0 6 0 6 0

0 6 0 6 2

1 6 0 6 4

2 8 0 8 4

2 8 2 8

2 8 2 8

2 8 2 8

2 2 4 4 4 4 8 8 8 8 8 8 8 2 4 4 4 4 4 8

Revision

10 The following stem-and-leaf plot shows the resting pulse rates for two groups of people: Group 1, who exercise occasionally, and Group 2, who exercise regularly.

4

From the stem-and-leaf plots we could say: A There does not seem to be any relationship between resting pulse rate and the amount of exercise undertaken. B The pulse rates of those who exercise regularly are in general lower than those who exercise only occasionally. C The pulse rates of those who exercise regularly are in general higher than those who exercise only occasionally. D The pulse rates of those who exercise regularly are in general less variable than those who exercise only occasionally. E The pulse rates of those who exercise regularly are in general more variable than those who exercise only occasionally. Questions 11–14 relate to the following information. An absent-minded professor calculated the following statistics for an examination. mean = 50 range = 50 number of cases = 99 minium = 20 maximum = 70 median = 50 standard deviation = 12.0 He then found an additional exam with a score of 50, and recalculated the statistics. 11 The new value of the median is A 21

B 50

C 51

D 71

E 100

C 51

D 71

E 100

C 51

D 70

E 100

12 The new value of the range is A 21

B 50

13 The new value of the mean is A 20

B 50



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14 The new value of the standard deviation is A 11.0

B 11.9

C 12.0

D 13.1

E 13.0

15 The boxplots shown summarise the test scores for two classes of students, Class X and Class Y. Class X* Class Y

* 25

50

75

100

From the boxplots we can say that, in general, A C D E

Class X scored better than Class Y. B Class Y scored better than Class X. Class X scores are more variable than Class Y scores. Class Y scores are more variable than Class X scores. The scores for the two classes are about the same.

16 The median age of a group of subjects is 33 and the mean is 36. From this we can say that the shape of the age distribution A may be symmetric D has no outliers

B may be positively skewed C may be negatively skewed E may be symmetric with a very low outlier.

17 From the boxplot shown we can conclude: A B C D E

The distribution is symmetric. The distribution is positively skewed. The distribution is negatively skewed. Nothing can be determined about the skewness of a distribution from the boxplot. The distribution is symmetric with an outlier.

18 For the distribution shown in the following histogram we could say:

A B C D E

The mode is larger than the mean. The mode is larger than the median. The mean and the median are about the same. The mean is smaller than the median. The mean is larger than the median.



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A

B

D

E

C

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19 Select which of the histograms A to E best corresponds to the boxplot given below.

20 Suppose a woman is on the 3rd quartile for height. This means that A B C D E

about 75% of women are taller than her about 75% of women are shorter than her about 50% of women are taller than her about 50% of women are shorter than her about 25% of women are shorter than her.

21 For which one of the following pairs of variables would it be appropriate to construct a scatterplot? A B C D E

gender (male or female) and height (in cm) height (in cm) and weight (in kg) for a class of students attitude to abortion (for, against, no opinion) and age (in years) test scores for a group of male students and a group of female students religious affiliation and age (in years).

22 For which one of the following scatterplots would it not be informative to calculate the q-correlation coefficient? B

A 200

100

150

80

100 60 50 50 60 70 80 90 100 –10

C

–5

0

5

10

D 35

0.8

30

0.6

25

0.4

20

0.2

80

100

120

140

0


0.2

0.4

0.6

0.8 Cambridge University Press

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23 The q-correlation for the scatterplot shown here is closest to A −1 D 0.8

B −0.8 E 1

20

C 0

10

0

24 For the scatterplot shown, the line of best fit would have equation closest to A B C D E

y y y y y

=x = 2x = 2x + 10 = x + 10 = −x + 8

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4

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8

9

10 x

y 28 24 20 16 12 8 4 1

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x

25 The Pearson correlation coefficient r relating the number of cigarettes smoked per capita and the rate of heart disease in the population for several countries is 0.7. We may interpret this as meaning: A B C D E

70% of the people who smoke will get heart disease. Smoking causes heart disease. Countries with high smoking rates also tend to have high rates of heart disease. Countries with high smoking rates also tend to have low rates of heart disease. The probability of a smoker getting lung cancer is 70% more than that of a non-smoker.

26 Jane calculates the Pearson correlation coefficient r for the data shown in the scatter diagram as 0.35. We may best interpret this by saying

y 100 50

A There is only a weak relationship between x and y. 4 8 12 B There is obviously a relationship between x and y, so Jane’s calculation must be wrong. C A relationship exists between x and y but it is not linear. D Knowing x does not tell us much about the corresponding value of y. E There is a positive association between x and y.

16

20 x

27 If the correlation between x and y is negative then which of the following statements must be true? A B C D E

y = −x The values of y are in general more than the corresponding values of x. As the values of x decrease the values of y tend to increase. As the values of x increase the values of y tend to increase. The values of y are in general less than the corresponding values of x.



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70 74 78 82 86 90 94 98 x 29 The production costs ($C) of a company which manufactures a particular computer game is related to the number of games produced (N) as follows.

Revision

y 28 For the scatterplot shown, 98 the line of best fit would 94 have equation closest to 90 A y=x 86 82 B y = −x 78 C y = x + 70 74 D y = x − 70 70 E y = −x + 70

C = 12 000 + 35 × N From this equation we can say, on average, A B C D E

cost of production increases at a rate of $12 000 per unit cost of production increases at a rate of $N per unit cost of production increases at a rate of $35 per unit cost of production is fixed at $12 000 cost of production decreases as the number of units produced increases

30 In calculating the least squares regression line, we choose the line which minimises A the sum of the shortest distances from each point on the scatterplot to the line B the sum of the horizontal distances from each point on the scatterplot to the line C the sum of the squared horizontal distances from each point on the scatterplot to the line D the sum of the squared vertical distances from each point on the scatterplot to the line E the sum of the squared perpendicular distances from each point on the scatterplot to the line.

24.2

Extended-response questions 1 A researcher believes that the manner in which an individual is approached by another (i.e. in an aggressive or passive attitude) influences the emotional reaction of the person approached. A researcher approached each of 15 subjects on two separate occasions; once in an aggressive manner and once in a passive manner. He observed the distance he was permitted to approach the subject before he or she stepped back. This distance of approach was recorded in centimetres for the 15 subjects. a Make ordered back to back stem-and-leaf plots for the each of the approaches. b Give the five-figure summaries for each of the approaches.


Subject 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Manner of approach Aggressive Passive 33 13 43 30 25 25 39 15 55 30 30 27 45 18 37 33 30 25 43 15 58 18 45 22 45 22 43 17 60 13 Cambridge University Press

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c Draw boxplots for each of the approaches on the same scale. d Calculate the mean and standard deviation for each of the groups. e Compare the distributions of aggressive and passive approach distances. 2 To test the effect of alcohol on coordination twenty randomly selected participants were timed to complete a task with both 0% blood alcohol and 0.05% blood alcohol. The times taken are shown in the accompanying table. 0% blood alcohol 38 36 35 35 34 40

a b c d e

35 44

43 30

46 25

42 39

64 31

40 29

48 44

0.05% blood alcohol 39 32 35 39 43 42 46 46

36 50

34 32

41 32

56 41

44 40

38 50

Make ordered back to back stem-and-leaf plots for the each of the sets of scores. Give the five-figure summaries for each of the sets of scores. Draw boxplots for each of the sets of scores on the same scale. Calculate the mean and standard deviation for each of the sets of scores. Compare the distributions of the before and after scores.

3 The prices of several cars when new and when two years old are shown in the following tables.

a b c d e

New price 10 817 16 664 17 200 11 490 18 050 17 999

12 477 15 070 10 700

15 450 16 660 14 955

13 239 12 049 11 595

11 585 15 424

16 391 13 410

Secondhand price 9 950 15 850 16 600 11 550 17 250 14 650

11 900 13 650 10 450

14 600 15 700 15 900

12 550 10 900 11 850

10 650 14 950

15 750 11 750

Make ordered back to back stem-and-leaf for the each of the sets of car prices. Give the five-figure summaries for each of the sets of car prices. Draw boxplots for each of the sets of car prices on the same scale. Calculate the mean and standard deviation of each of the sets of car prices. Compare the distributions of new prices and two year old prices.

4 A study was conducted to determine the effect of choice on performance on student assignments. A random sample of students were allowed to choose their assignment topics from a long list of possibilities. Another random sample of students were given the same assignments but without any choice of topic. The marks, out of 50, obtained by the students are given in the table. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.

No choice 44 36 38 35 26 42 21 17 44 24 49 48 28 42 35

Choice 36 36 26 30 40 35 35 25 48 36 50 25 44 30 50

(cont’d) Cambridge University Press

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Make ordered back to back stem-and-leaf for the each of the sets of marks. Give the five-figure summaries for each of the sets of marks. Draw boxplots for each of the sets of marks on the same scale. Calculate the mean and standard deviation for each of the sets of marks. Discuss the effect of students’ choice on the distribution of marks.

5 Consider the scores on a manual dexterity test for two groups of students. One group was randomly selected from the population of normal children. The other group was randomly selected from the population of children with a particular learning disability. The results are shown in the following table. Normal children (x1 ) Learning disabled children (x2 )

32 30 23 25

26 28 21 19

28 29 11 20

28 27 21 30

28 28 29 27

31 30 29 26

24 26 30

Revision

a b c d e

597

24

a Draw boxplots of the scores on the same scale. b Compare the distributions for normal children and learning disabled children. 6 A researcher noted that loss of sleep affected the number of dreams experienced by an individual. He also noted that as soon as people started to dream they exhibited rapid eye movement (REM). To examine this apparent relationship be kept a group of volunteers awake for various lengths of time by reading them spicy chapters from a statistics book. After they fell asleep he recorded the number of times REM occurred. The following data was obtained. Subject Hours of sleep deprivation Number of times REM occurred

1 1.0 10

2 1.5 20

3 2.0 15

4 2.5 30

5 3.0 20

6 3.5 20

7 4.0 25

8 4.5 35

a Construct a scatterplot of these data, and use it to describe the relationship between hours of sleep deprivation and number of times REM occurred. b Determine the value of the q-correlation coefficient. c Calculate the value of Pearson’s r for these data. d Determine the equation for the least squares regression line which relate hours of sleep deprivation and number of times REM occurred. e Interpret the slope, b, of this equation. f Use the equation to predict the number of times REM occurred if the subject experiences five hours of sleep deprivation. 7 To test the effect of driving instruction on driving skill, ten randomly selected learner drivers were given a driving skills test. The number of hours instruction for each learner was also recorded. The results are displayed in the table shown.



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Learner Hours Test score

A 19 32

B 2 12

C 5 17

D 9 19

E 16 23

F 4 16

G 19 28

H 26 36

I 14 30

J 8 23

a Construct a scatterplot of these data, and use it to describe the relationship between the number of hours instruction and the score. b Determine the value of the q-correlation coefficient. c Calculate the value of Pearson’s r for these data. d Determine the equation for the least squares regression line which relates number of hours instruction and score. e Interpret the slope, b, of this equation. f Use the equation to predict the score after 10 hours of instruction. 8 Consider the data in the table, which shows the number of government schools in Victoria over the period 1981–1992. Year Number of schools a Construct a scatterplot of these data, and use it to describe the relationship between the number of government schools and year. b Determine the value of the q-correlation coefficient. c Calculate the value of Pearson’s r for these data. d Determine the equation for the least squares regression line which relates number of government schools and year. e Interpret the slope, b, of this equation. f Use the equation to predict the number of government schools in the year 2000. g Comment on the reliability of the prediction in f.

1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992

2149 2140 2124 2118 2118 2114 2091 2064 2059 2038 2029 2013

9 The table shows the weights (kg) and blood glucose levels (mg/100 mL) of 16 apparently healthy adult males. Weight Glucose

64.0 108

75.2 109

73.1 104

82.1 102

76.3 105

95.7 120

59.4 78

93.4 110

Weight Glucose

82.1 101

78.9 85

76.6 98

82.1 100

83.9 108

73.2 104

64.4 102

77.5 89

a Construct a scatterplot of these data, and use in a describe the relationship between weight and blood glucose level. b Calculate the value of Pearson’s r for these data. c Determine the equation for the least squares regression line which relates weight and blood glucose level. d Interpret the slope, b, of this equation. e Use the equation to predict the blood glucose levels for a male who weighs 70 kg.



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Per capita income ($) 14 198 14 008 16 197 15 343 15 928 13 764 13 244 14 612 14 918 12 194

Expenditure per student ($) 1871 1850 2210 2188 2547 1834 1623 2052 2256 1476

a Construct a scatterplot of these data, and use it to describe the relationship between expenditure and per capita income. b Determine the value of the q-correlation coefficient. c Calculate the value of Pearson’s r for these data. d Determine the equation for the least squares regression line which relates expenditure and per capita income. e Interpret the slope, b, of this equation. f Use the equation to predict the expenditure if the per capita income is $16 000. g Comment on the reliability of the prediction made if the per capita income is $30 000.

Revision

10 The data shown in this table was collected in an investigation to determine if the amount spent by a school district on public education is dependent on the per capita income of the school district.

599

11 The owner of a fruit store records the number of boxes of mangoes she sells each month, together with the average price of the mangoes in that month, giving the following data. Month Jan Feb Mar April May June

Sold 25 30 20 18 11 8

Price ($) 0.95 0.90 1.20 1.40 2.00 2.40

Month July Aug Sep Oct Nov Dec

Sold 6 5 5 7 15 15

Price ($) 2.50 3.00 3.20 2.40 1.90 1.75

a Construct a scatterplot of these data, and use it to describe the relationship between the price of mangoes and the number of boxes sold. b Determine the value of the q-correlation coefficient. c Calculate the value of Pearson’s r for these data. d Determine the equation for the least squares regression line which relates number of boxes sold and price. e Interpret the slope, b, of this equation. f Use the equation to predict the number of boxes of mangoes sold if the price of mangoes is $ 1.40 each. g Comment on the reliability of the predicted number of boxes of mangoes sold if their price is $0.05 (5 cents) each.



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12 To study the effect of water hardness on taste, the data in this table were obtained from specimens of drinking water from eight country towns (a higher taste rating means better tasting water). a Construct a scatterplot of these data, and use Amount of it to describe the relationship between taste magnesium (mg) Taste rating rating and the amount of magnesium in the 8.7 25 water. 9.0 25 b Determine the value of the q-correlation 11.0 26 coefficient. 8.5 48 c Calculate the value of the correlation coefficient 9.2 65 r for these data. 12.0 87 d Determine the equation for the least squares 12.0 90 regression line which relates taste rating and the 18.0 100 amount of magnesium. e Interpret the slope, b, of this equation. f Use the equation to predict the taste rating if the amount of magnesium is 80 mg. 13 The following table gives details of Australian Test Cricket Captains, based on all tests up to and including the Third Test, January 1998. Captain D W Gregory W L Murdoch T P Horan H H Massie J M Blackham H J H Scott P S McDonnell G Griffen G H S Trott J Darling H Trumble M A Noble C Hill S E Gregory W W Armstrong H L Collins W Bardsley

Tenure 1876–79 1880–90 1884–85 1884–85 1884–95 1886 1886–88 1894–95 1896–98 1899–05 1901–02 1903–09 1910–12 1912 1920–21 1921–26 1926

Tests 3 16 2 1 8 3 6 4 8 21 2 15 10 6 10 11 2


W 2 5 0 1 3 0 1 2 5 7 2 8 5 2 8 5 0

L 1 7 2 0 3 3 5 2 3 4 0 5 5 1 0 2 0

D 0 4 0 0 2 0 0 0 0 10 0 2 0 3 2 4 2

T 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (cont’d)


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a

i ii iii b i ii iii c i ii iii iv v vi

Tenure 1928–29 1930–34 1935–36 1936–48 1945–46 1949–53 1951–55 1954–57 1956–57 1957–58 1958–64 1961 1963–78 1965–66 1967–71 1968 1970–75 1975–83 1978–79 1978–85 1984–94 1994–1999 1999–2004 2004–

Tests 5 25 5 24 1 24 2 17 1 5 28 1 39 2 25 1 30 48 7 28 93 50 57 8

W 1 14 4 15 1 14 0 7 0 3 12 1 12 0 9 0 15 21 1 4 32 26 41 6

L 4 7 0 3 0 4 2 5 0 0 4 0 12 1 8 0 5 13 6 13 22 13 9 1

D 0 4 1 6 0 6 0 5 1 2 11 0 15 1 8 1 10 14 0 11 38 11 7 1

T 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0

Revision

Captain J Ryder W M Woodfull V Y Richardson D G Bradman W A Brown A L Hassett A R Morris I W Johnson R R Lindwall I D Craig R Benaud R N Harvey R B Simpson B C Booth W M Lawry B N Jarman I M Chappell G S Chappell G N Yallop K J Hughes A R Border M A Taylor S R Waugh R Ponting

601

Draw a boxplot of the number of tests captained by each captain. Use the boxplot to describe the distribution of the tests captained. Who do you think is Australia’s most successful captain from this plot? Draw a boxplot of the percentage of matches won by each captain. Use the boxplot to describe the distribution of percentage of matches won. Who do you think is Australia’s most successful captain from this plot? Plot the number of matches won against the number of matches played for each captain. Describe the relationship between number of matches played and number of matches won. Calculate the correlation coefficient r for the two variables. Determine the equation of the least squares regression line. Write a sentence interpreting the slope of the regression line. Add the least squares regression line to the plot, and consider the residuals. Who do you think has been Australia’s most successful cricket captain from this plot?



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C H A P T E R

25 Proof and number Objectives To understand implication To understand converse and equivalence To understand the purpose of counter examples To construct proofs To understand the principle of mathematical induction To solve linear Diophantine equations To apply the Euclidean algorithm to find the highest common factor of two numbers To apply the Euclidean algorithm in the solution of linear Diophantine equations

25.1 An introduction to proof

Implication, converse and equivalence For two statements p and q, ‘ p ⇒ q’ is read ‘p implies q’. For example, x −3=4⇒ x =7 x is divisible by 3 ⇒ 2x is divisible by 6 x = 3 ⇒ x 2 = 9. In the first two examples the converse statements also hold. q ⇒ p is the converse of p ⇒ q The converse of the two examples are x =7⇒ x −3=4 2x is divisible by 6 ⇒ x is divisible by 3. Combining p ⇒ q and q ⇒ p we can write p ⇔ qwhich may be read ‘p is equivalent to q’ Thus we can write x −3=4⇔ x =7 x is divisible by 3 ⇔ 2x is divisible by 6.



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Chapter 25 — Proof and number

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An equivalent symbol for ⇔ is iff which is read ‘if and only if’. However, x 2 = 9 does not imply that x = 3, i.e., the converse does not hold. In fact, x 2 = 9 ⇔ x = 3 or x = −3 There are many other examples where the implication only holds one way. For example, x > 3 ⇒ x 2 > 9, but x 2 > 9 does not imply x > 3 x rational ⇒ x 2 rational, but x 2 rational does not imply x rational x and y even ⇒ x + y even, but x + y even does not imply x and y even a = b ⇒ a 2 = b2 , but a 2 = b2 does not imply a = b Quadrilateral ABCD is a square ⇒ the sides of quadrilateral ABCD are of equal length, but the sides of the quadrilateral ABCD are of equal length does not imply quadrilateral ABCD is a square.

Counter example Consider numbers of the form n 2 + n + 11 where n is a natural number. n n 2 + n + 11

1 13

2 17

3 23

4 31

5 41

6 53

7 67

8 83

This table of values might lead to the conjecture: ‘Numbers of the form n 2 + n + 11 are prime numbers’. However, if n = 11 it is evident that n 2 + n + 11 is divisible by 11 and hence n 2 + n + 11 is not prime. The value of n 2 + n + 11 when n = 11 provides a counter example to our conjecture. The conjecture has been shown to be false. Consider the conjecture:

Consider the conjecture:

Consider the conjecture:

(a + b)2 = a 2 + b2 for all a and b ∈ R. The values 2 and 3 for a and b respectively provide a counter example. x 2 > x for all x ∈ R. 1 The value for x provides a counter example. 3 The cube of a natural number is greater than the natural number. The natural number 1 provides a counter example.

This final conjecture can be ‘fixed up’ by making the conjecture ‘the cube of a natural number greater than 1 is greater than the natural number.’ ‘Fix up’ the first two conjectures.

Proof In Year 10 you may have come across proofs in geometry or used them in your problem solving. Proofs are a very important part of Mathematics. If we have a conjecture which we suspect to be true and for which a counter example cannot be found, then we try to construct a ‘chain of reasoning’ which will enable us to deduce the result from assumptions which are as simple as possible.



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Example 1 Consider this pattern. 42 − 32 + 22 − 12 = 10 52 − 42 + 32 − 22 = 14 62 − 52 + 42 − 32 = 18 From this it could be conjectured that ‘If a, b, c, d are consecutive natural numbers with a < b < c < d then d 2 − c2 + b2 − a 2 = d + c + b + a’. Prove this conjecture is true. Solution The result is not immediately obvious. Let the numbers a, b, c, d be n − 1, n, n + 1, n + 2 respectively. Then and

d +c+b+a =n+2+n+1+n+n−1 = 4n + 2 2 2 2 2 d − c + b − a = (n + 2)2 − (n + 1)2 + n 2 − (n − 1)2 = n 2 + 4n + 4 − n 2 − 2n − 1 + n 2 − n 2 + 2n − 1 = 4n + 2

The result has been proved. In fact, it has been proved for any four consecutive integers. It can be seen from the proof that a more general result is true. If a, b, c, d are consecutive integers then d 2 − c2 + b2 − a 2 = a + b + c + d The proof has actually added to the understanding of the problem by leading to a generalisation of the original conjecture. Example 2 Consider the following. 321 941 − 123 − 149 198 792 + 891 + 297 1089 1089

987 − 789 198 + 891 1089

980 − 089 891 + 198 1089

Conjecture Take any three digit number whose digits decrease as you read them from left to right. Make another number by reversing the order of the digits and subtract the smaller from the larger. Reverse the order of the digits of the difference and add the number so formed to the difference. The result will always be 1089. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Solution Let the number be a × 102 + b × 10 + c. Reverse the digits to get the number c × 102 + b × 10 + a. It will be assumed a > c without loss of generality. a × 102 + b × 10 + c − c × 102 − b × 10 − a = (a − c) × 102 + (c − a) Now c − a is negative, but (10 + c − a) is positive. Write (a − c) × 102 + (c − a) = (a − c − 1) × 102 + 90 + (10 + c − a) The digits are now correctly displayed. Reverse the order to obtain (10 + c − a) × 102 + 90 + (a − c − 1) and add. (10 + c − a + a − c − 1) × 102 + (90 + 90) + (10 + c − a + a − c − 1) = 9 × 102 + 180 + 9 = 900 + 180 + 9 = 1089 We can attempt to generalise in this case. The question arises ‘What happens when a four digit number is considered?’ The proof also indicates that the base chosen is important. What is the result for different bases?

Exercise 25A 1 Insert ⇒ or ⇐ to make the following into true statements about integers. a p is even . . . pq is even c x = 0 . . . xy = 0

b p + q is odd . . . pq is even d ab = ac . . . b = c

2 State with reasons whether the following statements are true or false. a n (A) = 5 and n (B) = 3 ⇒ n (A ∪ B) = 8 c A ∩ B = ∅ ⇒ A = ∅ or B = ∅

b A⊂ B ⇒ A∩B = A d A = ∅ ⇐ A = ␰

3 Write down the converses of the statements given in 2 and state whether each new statement is true or false. 4 State the converse of each of the following statements and also state whether the converse is true or false. a If n is odd then n 2 is odd. c x 2 > 4 ⇐ x < −2

b N is divisible by 3 ⇒ N 2 is divisible by 9

5 The sum of two consecutive odd numbers is divisible by 4. Can you make similar statements about: a the sum of three consecutive odd numbers b the sum of four consecutive odd numbers? Prove your assertions.



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6 Prove that the sum of the squares of five consecutive integers is divisible by 5. (Take n − 2, n − 1, n, n + 1, n + 2 as the integers.) 7 For each of the following statements give a counter example which proves that the given statement is false. a The sum of the squares of two numbers is equal to the square of the sum of these two numbers. b If a number is even then it is not divisible by 7. √ √ √ a + b = a + b for all a, b ∈ R c b+c 1 a+b 1 a+ = + c for all a, b, c ∈ R d 2 2 2 2 e The sum of two prime numbers is a prime number. 1 1 1 = + for all s, t ∈ R\ {0} f s+t s t 8 Write the converse of each of the following and state whether it is true or false. a If a − b is positive then a > b b If x = 0 and y = 0 then x = y c If x + y = 0 then x = −y d If x is even and y is odd then xy is even. e The square of an even number is even. 9 For each of the following conjectures for the set of natural numbers either prove or provide a counter example. a b c d

A number N has an odd number of divisors if and only if it is a perfect square. For any odd number, N, there is a number with exactly N divisors. For any number, N, there is a number with exactly N divisors. There are infinitely many numbers with exactly N divisors (N = 1).

10 a Show that a 2 + b2 ≥ 2ab for all real a and b. √ u+v ≥ uv b Hence, or otherwise, prove that if u ≥ 0, v ≥ 0, then 2 1 1 + ≥4 11 Prove that for all positive integers a and b, (a + b) a b 12 Prove each of the following for a, b ∈ Z . a c e f

If a and b are even, a + b is even. b If a and b are odd, a + b is even. If a is even and b is odd, ab is even. d If a and b are odd, ab is odd. a + b is even if and only if a − b is even. a + b even and a − b even implies ab is a difference of perfect squares.

13 The floor of a rectangular room is covered with square tiles. The room is m tiles wide and n tiles long with m ≤ n.If exactly half of the tiles are on the perimeter, find all possible values of m and n. 14 Prove that no positive integer (except 1) all of whose digits are 1s is a perfect square.



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15 Find all right-angled triangles which have integer length sides and for which the area has the same numerical value as the perimeter. 16 If a, b and c are positive integers such that no integer greater than 1 divides them all and 1 1 1 + = , prove that a + b is a perfect square. a b c Note: This is a difficult question.

25.2

The principle of mathematical induction Suppose that P(k) is a statement for each positive integer k. For example, the statement could be ‘k 2 − 2k + 1 ≥ 0 for each positive integer k’ or ‘k 2 − k is an even integer for each positive integer k’. The method of mathematical induction is used to prove P(k) for all k as follows. i Show P(1) is true. ii Show P(k) ⇒ P(k + 1) for every positive integer k. From this it can be seen that if both of the statements hold, then P(1) ⇒ P(2), P(2) ⇒ P(3), P(3) ⇒ P(4) and continuing in this fashion it can be seen that P(k) is true for every positive integer k. The method of proof by induction can be illustrated by a line of dominoes (starting with domino 1) stretching away without an end. To be sure that all dominoes will be knocked over it is enough to know that i the first domino is knocked over and ii if one domino falls it will certainly knock over the next. Another integer, besides 1, may be chosen for the starting value. Example 3 Prove that the sum of the first n integers is

n (n + 1) . 2

Solution Let P(n) be the statement that the sum of the first n integers is

1×2 2 k (k + 1) i.e. that P(k) is true 1 + 2 + ··· + k = 2 k (k + 1) 1 + 2 + ··· + k + k + 1 = + (k + 1) 2 (k + 1)(k + 2) = 2

P(1) is certainly true as Assume then

n (n + 1) 2

1=

∴ P(k + 1) is true and the principle of induction gives that P(n) is true for all n. ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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It is sometimes convenient to begin an induction at a point other than k = 1. In the following example, the starting point is at k = 5. Example 4 Prove that 2n > 1 + n 2 for all n > 4, n ∈ N . Solution Let P(n) be the statement that 2n > 1 + n 2 . If n = 5, 25 = 32 and 1 + 52 = 26 As 25 > 26, P(5) is true. Assume true for k. i.e. 2k > 1 + k 2 For 2k+1 = 2.2k > 2(1 + k 2 ) = 2 + k 2 + k 2 1 2 But k > 2k for k > 2 Therefore 2 + k 2 + k 2 > 2 + 2k + k 2 = 1 + 1 + 2k − k 2 = 1 + (1 + k)2 2

From inequalities 1 and 2 2k+1 > 1 + (1 + k)2

∴ P(k + 1) is true and the principle of induction gives that P(n) is true for all n ≥ 5. Example 5 Prove that 32n − 1 is divisible by 8 for all n ∈ N . Solution Let P(n) be the statement that 32n − 1 is divisible by 8. If n = 1, 32 − 1 = 8 is divisible by 8 i.e. P(l) is true. Assume true for k. i.e. P(k) is true and 32k − 1 is divisible by 8 Consider

32(k+1) − 1 = 32k+2 − 1 = 32 · 32k − 32 + 32 − 1 = 32 (32k − 1) + 8

Since 32k − 1 is divisible by 8 and 8 is divisible by 8, 32(k+1) − 1 is divisible by 8. ∴ P (k + 1) is true, ∴ P(n) is true for all n ∈ N . ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 6 Prove that 2n > n 3 for n ≥ 10. Solution P(10) is true as 210 > 103 (1024 > 1000) Assume P(k) is true i.e. 2k > k 3 for k > 9 then ∴ ∴ ∴ ∴ ∴

2k+1 = 2 × 2k > 2 × k 3 2k+1 > (k + 1)3 + k 3 − 3k 2 − 3k − 1 2k+1 > (k + 1)3 + k 3 − 3k 2 − 3k − 459 2k+1 > (k + 1)3 + (k 2 + 6k + 51)(k − 9) 2k+1 > (k + 1)3 for k > 9 (Note: k 2 + 6k + 51 > 0 for all k) P (k + 1) is true and P(n) is true for all n ≥ 10

Exercise 25B Sums 1 Prove each of the following by induction, for all natural numbers n. 1 a 1 + 2 + · · · + n = n (n + 1) 2 1 b 12 + 22 + · · · + n 2 = n (n + 1) (2n + 1) 6 1 1 1 n c + + ··· + = (2n − 1) (2n + 1) 1×3 3×5 2n + 1 1 d 1 × 2 + 2 × 3 + 3 × 4 + · · · + n(n + 1) = n(n + 1)(n + 2) 3 e 1 × 4 + 2 × 7 + 3 × 10 + · · · + n(3n + 1) = n(n + 1)2 f 1 + 2 × 21 + 3 × 22 + · · · + n × 2n−1 = 1 + (n − 1) 2n 1 1 1 1 1 1 1 + + ··· + − g = (2n + 1) (2n + 3) 3×5 5×7 7×9 2 3 2n + 3 h 13 + 33 + 53 + · · · + (2n − 1)3 = n 2 (2n 2 − 1) 1 i 2 × 1 + 3 × 2 + · · · + n(n − 1) = n(n 2 − 1) 3

Divisibility 2 Prove by induction that, for all natural numbers, a c e g

n(n + 1)(n + 2) is divisible by 3 8n + 2 × 7n − 1 is divisible by 7 23n − 1 is divisible by 22 5n 3 − 3n 2 − 2n is divisible by 6

b 4n 3 − 4n is divisible by 3 d n(n 2 + 2) is divisible by 3 f 8n − 5n is divisible by 3



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3 a Prove that 4n + 5n is divisible by 9 for all positive odd integers, n. b Prove that 3n − 1 is divisible by 8 for all positive even integers, n. c Prove that 6n+1 − 5 (n + 1) − 1 is divisible by 25 for all positive integers, n.

Inequalities 4 a Prove that 3n > n 3 for all positive integers greater than 3. n4 b Prove that 13 + 23 + · · · + n 3 > for all positive odd integers. 4 n 2 c Prove that 2 > n for all positive integers greater than 4. d Prove that 2n > 3n for all positive integers greater than 3. e Prove that 2n ≤ n! for all n ≥ 4. 5 Prove that 1 1 1 1 n a + + + ··· + = ,n ∈ N 1×2 2×3 3×4 n (n + 1) n+1 1 b 13 + 23 + · · · + n 3 = n 2 (n + 1)2 , n ∈ N 4 c 33n+1 + 9 × 2n+3 is divisible by 25, n ∈ N 6 Prove by induction that if S is a set of n elements, then S has 2n subsets.

25.3 Linear Diophantine equations Consider the equation 3x + 4y = 1. This equation defines a straight line. If the values of x and y are integers, a family of solutions may be described. This equation is called a linear Diophantine equation. On the graph, a family of solutions is illustrated.

y (–5, 4)

4 3 2 (–1, 1) 1

–5 – 4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 –1 (3, –2) –2 –3 –4 –5 (7, –5) –6 –7 –8 (11, –8) –9 –10 –11 (15, –11)

x

Notice that as the integer solutions for x increase by 4, the y integer solutions decrease by 3.



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The solutions may be built up in the following way using (−1, 1) as the starting point. x −1 + 4 −1 + 2 × 4 −1 + 3 × 4

y 1−3 1−2×3 1−3×3

i.e.

x 3 7 11

y −2 −5 −8

The family of solutions may be described as x = −1 + 4t, y = 1 − 3t where t ∈ Z Using set notation the solution is {(x, y) : x = −1 + 4t, y = 1 − 3t, t ∈ Z } If a linear Diophantine equation has one solution then it has infinitely many. If ax + by = c is a linear Diophantine equation in two unknowns and (x0 , y0 ) is found to be one solution, the general solution is given by b t d a y = y0 − t, where t ∈ Z d and d is the highest common factor of a and b. x = x0 +

Proof Suppose x1 , y1 are solutions to the equation. Then

ax1 + by1 = c

1

and

ax0 + by0 = c

2

Subtracting 2 from 1 a(x1 − x0 ) = b(y0 − y1 ) Divide both sides by d a b (x1 − x0 ) = (y0 − y1 ) d d a b and have no common factors. d d b Hence x1 − x0 must be divisible by d and Similarly

b t d a y1 = y0 − t d

x1 = x0 +

It can be proved by substitution that x = x0 +

b a t, y = y0 − t, is a solution of the equation for any t ∈ Z . d d



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Example 7 A man has $200 in his wallet. This is made up of $50 and $20 notes. What are the possible numbers of each of these types of notes? Solution Let x, y be the number of $50 and $20 notes respectively. The linear Diophantine equation is 50x + 20y = 200 5x + 2y = 20 By inspection a solution is x = 4, y = 0 The general solution will be x = 4 + 2t, y = 0 − 5t, t ∈ Z We are only interested in the case where x, y ≥ 0 Thus 4 + 2t ≥ 0 and 0 − 5t ≥ 0 Hence −2 ≤ t ≤ 0 For t = −2, x = 0, y = 10 t = −1, x = 2, y = 5 t = 0, x = 4, y = 0 Hence the man can have ten $20 notes or two $50 notes and five $20 notes or four $50 notes.

Exercise 25C 1 Find all solutions of the following Diophantine equations a 11x + 3y = 1 d 22x + 6y = 2

b 2x + 7y = 2 e 2x + 7y = 22

c 24x + 63y = 99 f 10x + 35y = 110

2 For a, c, e in 1 find the solutions for which x, y are both positive. 3 Prove that if ax + by = c and the highest common factor of a and b does not divide c, then there is no solution to the Diophantine equation. 4 A student puts a number of spiders (with eight legs) and a number of beetles (with six legs) in a box. She counted 54 legs in all. a Form a Diophantine equation. b Find the number of spiders and the number of beetles in the box.



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5 Helena has a number of coins in her wallet. They are all either 20c or 50c coins. The total value of the coins is $5.00. What are the possible numbers of each type of coin? 6 One of the solutions of the equation 19x + 83y = 1983 in positive integers x and y is obviously x = 100, y = 1. Show that there is only one other pair of positive integers which satisfy this equation and find it. Consider the equation 19x + 98y = 1998. 7 A man has $500 in his wallet made up of $50 and $10 notes. Find the possible combinations of notes that he could have. 8 There are seven coconuts and 63 heaps of pineapples. Each heap has exactly the same number of pineapples. The fruit is to be divided equally between 23 people. Let x be the number of pineapples in each heap and y the number of pieces of fruit that each person receives. Form a Diophantine equation and find the possible values for x and y. 9 A dealer spent $10 000 buying cattle, some at $410 each and the rest at $530 each. How many of each sort did she buy? 10 Find the smallest positive number which, when divided by 7, leaves a remainder of 6, and when divided by 11 leaves a remainder of 9. Also find the general form of such numbers. 11 Given a 3 litre jug and a 5 litre jug can I measure exactly 7 litres of water? If it is possible, explain how this may be done as efficiently as possible. 12 The Guadeloupe Post Office has only 3c and 5c stamps. What amounts of postage can the post office sell? 13 A man spent $29.60 buying party hats. There were two types of party hat. Type A cost $1.70 while type B cost $1.00. How many of each type did he buy?

25.4

The Euclidean algorithm The Euclidean algorithm provides a method for finding the highest common factor of two numbers and also a method for solving linear Diophantine equations.

Theorem 1 If a, b are integers with a > 0, then there are unique integers q, r such that b = aq + r with 0 ≤ r < a. Proof Suppose there exists another pair of integers, q1 and r1 with b = aq1 + r1 and 0 ≤ r1 < a. Suppose that r > r1 . Then by subtraction

∴ ∴

0 = aq1 + r1 − (aq + r ) 0 = a(q1 − q) + (r1 − r ) r − r1 = a(q1 − q)



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Now since the right hand side is an integer and is a multiple of a, then a divides the left hand side. But the left hand side is an integer which is greater than zero and less than a. Therefore the assumption, that r > r1 , must be false. But if it is assumed r < r1 , then consider 0 = (aq + r ) − (aq1 + r1 ) and a similar contradiction will arise Thus so

r = r1 . In that case r − r1 = 0 = a(q1 − q) q1 = q and r1 = r

and the uniqueness of integers q, r has been proved. Example 8 Express −45 in the form 6q + r where 0 ≤ r < 6. Solution Here −45 = 6(−8) + 3 Note that −45 = 6(−7) − 3 is not a correct answer since the remainder −3 is less than zero. In the following, (a, b) denotes the highest common factor of the integers a and b.

Theorem 2 If a and b are two integers, a = 0 and b = aq + r where q, r are integers, then (a, b) = (a, r ). (This theorem may be used to determine the highest common factor of any two given integers.) Proof If d is a common divisor of a and r, then d divides the right-hand side of equation b = aq + r and so d divides b. This proves that all common divisors of a and r will be common divisors of a and b. But (a, r ) is a common divisor of a and r, and so (a, r ) must divide a and b. It follows that (a, r ) must divide (a, b). That is (a, b) = (a, r )x where x is an integer Now rewrite equation b = aq + r as r = b − aq

1

All common divisors of a and b divide the right hand side of this relation and so divide r. Thus (a, b) must divide r as well as a. Hence (a, b) must divide (a, r ). It has now been proved

(a, r ) = (a, b)y, where y is an integer


2


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From equations 1 and 2 obtain (a, r ) = [(a, r )x]y 1 = xy

Hence

This equation in integers x, y is possible only if both x, y are +1 or −1. Hence (a, b) = +(a, r ), since −1 is inappropriate to this problem. Since the integer r is less than the integer b, the calculation of (a, r ) is easier than the calculation of (a, b). Example 9 Find (1271, 3875). 3875 = 1271 × 3 + 62 Solution and (1271, 3875) = (1271, 62) by theorem 2. Now 1271 = 62 × 20 + 31 ∴ (62, 1271) = (62, 31) again by theorem 2. As 62 = 31 × 2 + 0 we have (1271, 3875) = 31 by using the theorem a final time. This procedure (algorithm) is called the Euclidean algorithm.

Method for finding a solution of a linear Diophantine equation The method presented here uses the Euclidean algorithm. Example 10 Find a, b ∈ Z such that 22a + 6b = 2 Solution Apply the division algorithm to 22 and 6

i.e.

22 = 3 × 6 + 4 6=1×4+2 4=2×2 (22, 6) = 2

1 2 3

Using these results

∴ ∴ ∴ ∴ or

2=6−1×4 from 2 2 = 6 − 1(22 − 3 × 6) from 1 2 = 6 − 22 + 3 × 6 2 = 4 × 6 − 1 × 22 −1 × 22 + 4 × 6 = 2

A solution is (−1, 4) ISBN 978-1-107-65235-4 © Michael Evans et al. 2011 Photocopying is restricted under law and this material must not be transferred to another party.


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Example 11 Find a, b ∈ Z such that 125a + 90b = 5 Solution Divide by 5. 25a + 18b = 1 Apply the division algorithm. 25 = 1 × 18 + 7 18 = 2 × 7 + 4 7=1×4+3 4=1×3+1 3=3×1

1 2 3 4 5

i.e., (25, 18) = 1 1=4−1×3 ∴ 1 = 4 − 1(7 − 1 × 4) ∴ 1=2×4−1×7 ∴ 1 = 2 × (18 − 2 × 7) − 1 × 7 ∴ 1 = 2 × 18 − 5 × 7 ∴ 1 = 2 × 18 − 5(25 − 18) ∴ 1 = 7 × 18 − 5 × 25 ∴ a = −5 and b = 7 is one solution.

from 4 from 3 from 2 from 1

∴ Solution is given by a = −5 + 18t and b = 7 − 25t; t ∈ Z .

Exercise 25D 1 For the following, express b in the form b = aq + r with 0 ≤ r < a and show in each case (a, b) = (a, r ). a a = 5, b = 43

b a = 13, b = 39

c a = 17, b = 37

d a = 16, b = 128

2 If d is a common factor of a and b, prove that d is a common divisor of a + b and a − b. 3 Use the Euclidean algorithm to find a (4361, 9284)

b (999, 2160)

c (−372, 762)

d (5255, 716 485)

4 Solve in the integers the equations a 804x + 2358y = 6 c 3x + 4y = 478 e 804x + 2688y = 12

b 18x + 24y = 6 d 3x − 5y = 38 f 1816x + 2688y = 8



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For two statements p and q, ‘ p ⇒ q’ is read ‘p implies q’. q ⇒ p is the converse of p ⇒ q Combining p ⇒ q and q ⇒ p we can write p ⇔ q which may be read ‘p is equivalent to q’ An equivalent symbol for ⇔ is iff which is read ‘if and only if ’. A counter example is an example which proves a conjecture to be false. The method of mathematical induction is used to prove P(k) for all k as follows i Show P(1) is true. ii Show P(k) ⇒ P(k + 1) for every positive integer k.

Review

Chapter summary

Consider the equation 3x + 4y = 1. This equation defines a straight line. The coefficients of the left hand side of the equation and the right hand side are integers. If the values of x and y are integers, a family of solutions may be described. This equation is called a linear Diophantine equation. If a linear Diophantine equation has one solution then it has infinitely many. If ax + by = c is a linear Diophantine equation in two unknowns and (x0 , y0 ) is found to b a be one solution, the general solution is given by x = x0 + t, y = y0 − t , where t ∈ Z d d If a, b are integers with a > 0, then there are unique integers q, r such that b = aq + r with 0 ≤ r < a. The Euclidean algorithm In the following, (a, b) denotes the highest common factor of the integers a and b. If a and b are two integers, a = 0 and b = aq + r where q, r are integers, then (a, b) = (a, r ). (This result may be used to determine the highest common factor of any two given integers, and to solve any linear Diophantine equation.)

Multiple-choice questions 1 If m is a positive even integer and n is a positive odd integer the statement which is false is A m + 2n is even B m + n is odd C 3m + 2n is even 2 D m × n is even E m + n is even 2 The statement below which is true is A x − 3 > 0 ⇒ (x − 3)(2 − x) > 0 C (x − 3)(2 − x) > 0 ⇔ x − 3 > 0 E x − 3 > 0 ⇔ (x − 3)(x − 2) > 0

B (x − 3)(2 − x) > 0 ⇒ x − 3 > 0 D x − 3 > 0 ⇒ (x − 3)(x − 2) > 0

3 If p and q are positive real numbers, and p > q with p + q = 1, the largest quantity from the following is 1 1 1 C D E pq B A p pq p q



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4 If p > q and pq = 0, it is true that

1 1 < p q

A always B never C only when p and q are positive D for all p and q except when both are negative E whenever pq > 0 5 The number of factors that the integer 2 p 3q 5r has is ( p + q + r )! A B pqr C p+q +r q!q!r ! D ( p + 1)(q + 1)(r + 1) E p+q +r +1 6 If an integer of two digits is k times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by A 9−k B 10 − k C 11 − k D k−5 E k+8 7 The number of pairs of integers (m, n) which satisfy the equation m + n = mn is A 1 B 2 C 3 D 4 E more than 4 8 If a, b, c are any real numbers, and a > b, the statement which must be true is 1 1 1 1 E < > B ac > bc C a 2 > b2 D a+c >b+c A a b a b 9 The number of solutions of the Diophantine equation 3x + 5y = 1008, where x and y are positive integers, is A 1 B 134 C 68 D 67 E infinite 10 If y = (n − 1)(n − 2)(n − 3), where n is a positive integer, then y is not always divisible by A 6 B 5 C 3 D 2 E 1

Short-answer questions (technology-free) 1 Find the highest common factor of 1885 and 365 using the Euclidean algorithm. 2 Consider 9x + 43y = 7. Solve for x and y where a x ∈ Z, y ∈ Z b x ∈ Z +, y ∈ Z + 3 Prove that the product of two consecutive odd integers is odd. (You may assume that the sum and product of any two integers is an integer.) 4 Using the Euclidean algorithm, find the highest common factor of 10 659 and 12 121. 5 a Solve the Diophantine equation 5x + 7y = 1 b Hence solve the Diophantine equation 5x + 7y = 100 c Find {(x, y) : 5x + 7y = 1; y ≥ x, x, y ∈ Z }. 6 The sum of the ages of Tom and Fred is 63. Tom is twice as old as Fred was when Tom was as old as Fred is now. What are the ages of Tom and Fred?



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1 The strips shown below are formed by joining an odd number of triangles together. The triangles are formed on 1 cm isometric paper as shown.

Review


619

A strip of length 3 A strip of length 5 A strip of length 7

In the strip of length 3 there are two parallelograms as shown below.

In the strip of length 5 there are six parallelograms.

a b c d

Find and draw all the possible parallelograms in a strip of length 7. How many parallelograms are there in a strip of length 11? Find the number of parallelograms in a strip of length n where n is an odd number. Prove your result.

2 A common error for adding fractions is the following: c a+c a + = b d b+d Investigate this error for a, b, c, d ∈ N . 3 Twelve cubes are used to build a tower three cubes high. This is shown with the aid of isometric dot paper in the diagram.



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a How many cubes are needed to build a tower of this type which is eighty cubes high? Work out an expression for the number of cubes needed to build a tower of this type which is n cubes high. b The number of cubes needed to build a tower n cubes high, Tn , is related to the number of cubes needed to build a tower (n − 1) cubes high, Tn−1 , by the formula Tn = Tn−1 + kn, where k is a constant. What is the value of k? c How many dots on the isometric dot paper are needed to draw a tower of this type n cubes high? d Investigate other types of towers. 4 You have an inexhaustible supply of 5¢ and 8¢ stamps. a List all possible ways of obtaining a total value of 38¢ with these stamps. b List all possible ways of obtaining a total of $1.20 with these stamps. 5 The digits of a three-digit number are interchanged so that none of the digits has retained its original place. Then the new number is subtracted from the original. If the difference is a two-digit number which is also a perfect square, find all such two-digit numbers. 6 Find the positive integer n such that n and n + 100 have an odd number of divisors. 1 7 Prove that if a > 0, then a + ≥ 2. a 1 2 Hint: Use the result that a − ≥ 0 for all a ∈ R. a 8 Show that the sum of three consecutive positive integers divides the sum of the cubes of these three integers. Hint: Consider the numbers n − 1, n, n + 1. 9 Add 1 to the product of four consecutive positive integers. Prove that the result is a perfect square. 10 Choose any two positive integers which are not divisible by 3. Prove the difference between their squares is divisible by 3. 11 Prove that for every positive integer n, the expression n 2 (n 4 − 1) is divisible by 60. 12 Choose any six consecutive positive integers greater than 3. Prove that there are at most two prime numbers among them. 13 Find all six-digit numbers with the property: If the first and last digits are interchanged, then the new six-digit number is six times the original number. 14 Prove by mathematical induction 1 a 12 + 42 + 72 + · · · + (3n − 2)2 = n(6n 2 − 3n − 1) 2 b 6n + 4 is divisible by 10 for any natural number n.



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Glossary A absolute value function: [p. 208] Let f : R → R be defined as f (x) = and is denoted by |x| Also known as modulus function

x if x ≥ 0 −x if x < 0

acceleration: [p. 466] The acceleration of a particle is defined as the rate of change of its velocity with respect to time. acceleration, average: [p. 466] The average acceleration of a particle for the time interval [t1 , t2 ] v2 − v1 is defined by where v 2 is the velocity at time t2 − t1 t2 and v 1 is the velocity at time t1 . dv acceleration, instantaneous: [p. 466] a = dt addition formulas: [p. 309] cos (u − v) = cos u cos v + sin u sin v cos (u + v) = cos u cos v − sin u sin v sin (u + v) = sin u cos v + cos u sin v sin (u − v) = sin u cos v − cos u sin v addition of complex numbers: [p. 418] If z 1 = a + bi and z 2 = c + di, then z 1 + z 2 = (a + c) + (b + d)i. addition of vectors: [p. 391] Let a = a1 i + a2 j + a3 k, and b = b1 i + b2 j + b3 k Then a + b = (a1 + b1 )i + (a2 + b2 ) j + (a3 + b3 )k algebraic numbers: [p. 68] Algebraic numbers are those which are the solution(s) of an equation of the form a0 x n + a1 x n−1 + · · · + an = 0, where a0 , a1 . . . , an are integers. amplitude of circular functions: [p. 264] The distance between the mean position and the maximum position, e.g., the graph of y = a sin x has an amplitude of |a|. angle between a vector and the i direction: [p. 402] For vector a = a1 i + a2 j + a3 k, where ␣ is the angle between the vector a and the i direction, a1 . cos ␣ = |a|

angle between a vector and the j direction: [p. 402] For vector a = a1 i + a2 j + a3 k, where ␤ is the angle between the vector a and the j direction, a2 cos ␤ = . |a| angle between a vector and the k direction: [p. 402] For vector a = a1 i + a2 j + a3 k, where ␥ is the angle between the vector a and the k direction, a3 cos ␥ = . |a| angle between planes: [p. 353] If P is any point on the common line of two planes 1 and 2 and PA and PB are lines at right angles to the common line in 1 and 2 respectively, then ∠APB is the angle between 1 and 2 . A Π1 P Π2

θ B

angle of depression: [p. 344] The angle between the horizontal and a direction below the horizontal. eye level angle of depression cliff

line

of s

ight

angle of elevation: [p. 344] The angle between the horizontal and a direction above the horizontal.

eye level

ght of si line angle of elevation

arc: [p. 340] Any two points on a circle divide the circle into arcs. The shorter arc is called the minor arc, the longer is the major arc, e.g., arc ACB is a minor arc and ADB is a major arc in this diagram.

A

D

C

O B



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Essential Advanced General Mathematics arc length, l: [p. 341] The length of arc ACB is given by: l = r ␪ where ␪ c = mag ∠AOB

A r D

O

θ

C

B

1 area of a triangle: [p. 337] Area = bh 2 1 Area of triangle = bc sin A, i.e., 2 B the area is given by half the product of the length of two h sides and the sine of the angle included between A them. b

B C

Argand diagram: [p. 425] A geometrical representation of the Im(z) P z = a + bi set of complex numbers. b 0

θ a

asymptote: [p. 447] A line which a graph 1 approaches, e.g. y = has asymptotes at x = 0 and x y = 0. asymptotes of hyperbolas: [p. 447] The hyperbola x2 y2 b with equation 2 − 2 = 1 has asymptotes y = x a b a −b x. and y = a

Re(z)

argument of a complex number, arg (z): [p. 430] Im(z) arg (z) = ␪, where sin ␪ = and |z| Re(z) cos ␪ = |z| arg (z) is not defined uniquely. Argument of a complex number, arg (z): [p. 430] The single value of arg (z) in the interval (−␲, ␲] argument, properties of: [p. 430] The argument of the product of two complex numbers is the sum of their arguments, i.e., arg (z 1 z 2 ) = arg (z 1 ) + arg (z 2 ) Argument, properties of: [p. 431] Arg (z 1 z 2 ) = Arg (z 1 ) + Arg (z 2 ) + 2k␲ where k = 0, 1 or −1 z1 = Arg (z 1 ) − Arg (z 2 ) + 2k␲ Arg z2 where k= 0, 1 or −1 1 = −Arg (z) Arg z arithmetic sequence: [p. 122] A sequence in which each successive term is found by adding a constant value to the previous term, e.g., 2, 5, 8, 11, . . . An arithmetic sequence can be defined by a difference equation of the form: tn = tn−1 + d, where d is the common difference. The nth term of the sequence can be found using: tn = a + (n − 1)d, where a = t1 arithmetic series: [p. 125] The sum of the terms in an arithmetic sequence. The sum of the first n terms, Sn , is given by the rule: n Sn = [2a + (n − 1)d], where a = t1 2 and d = tn − tn−1

bar chart: [p. 502] A visual display of a frequency distribution when the data are categorical. Frequencies (vertical axis) are represented by vertical bars corresponding to each category (horizontal axis). bearing (or compass bearing): [p. 346] The direction measured from north clockwise. bivariate data: [p. 551] Data which arises when two variables are observed for each subject, such as height and weight. box-and-whisker plot: see boxplot boxplot: [p. 533] A visual display of a numerical data set, an alternative to a histogram or stem-and-leaf plot. Once the five-figure summary has been determined the boxplot is constructed by drawing a horizontal box with the ends at the first and third quartiles. The minimum value is joined to the lower end of the box with a horizontal line and the maximum value joined to the upper end of the box with a horizontal line. The median is located with a vertical line. boxplot with outliers: [p. 536] A more complex version of the boxplot, with outliers shown. Outliers are determined according to the formal definition of an outlier, and illustrated on the plot with an asterisk. The boxplot is constructed as before but the lower end of the box is joined by a horizontal line to the smallest value which is not an outlier, and the upper end of the box is joined by a horizontal line to the largest value which is not an outlier.

C C: [p. 417] The set of complex numbers, i.e., C = {a + bi : a, b ∈ R}. cartesian equation: [p. 444] An equation connecting two variables, often called x and y. cartesian form of a complex number: [p. 417] Im(z) P z = a + bi A complex number expressed in the form a + bi, represented by the ordered b pair (x, y), where x is the θ real part of z and y is the a 0 Re(z) imaginary part of z.



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Glossary categorical data: [p. 501] Data collected about a variable which takes values that are categories (not quantities). For example, data from the variable gender which can takes the values male and female. causation: [p. 579] A relationship between two variables where it is known that a change in the independent variable causes a change in the dependent variable. This cannot be determined from the value of the correlation coefficient. centre: [p. 521] A measure of the location of the data values. chord: [p. 341] An interval with end points on a circle. ␪ chord length: [p. 341] AB = 2r sin 2 where ␪ c = mag ∠AOB A r O θ r B

circle, general cartesian equation of: [p. 441] (x − h)2 + (y − k)2 = r 2 . The centre of the circle is the point (h, k) and the radius is r. circular function equations, solution of: [p. 301] If cos (x) = a, x = 2n␲ ± cos−1 (a), where n ∈ Z and a ∈ [−1, 1] If tan (x) = a, x = n␲ + tan−1 (a), where n ∈ Z and a ∈ R If sin (x) = a, x = 2n␲ + sin−1 (a) or x = (2n + 1)␲ − sin−1 (a), where n ∈ Q and a ∈ [−1, 1] circular functions: [pp. 255–57] The sine, cosine and tangent functions. cis ␪: [p. 430] cos ␪ + i sin ␪ class frequency: [p. 506] The number of data values corresponding to each class interval. class intervals: [p. 505] A range of values which forms a sub-group of the values that a variable might take such as 0–9. common difference, d: [p. 122] The difference between two consecutive terms of an arithmetic sequence, i.e., d = tn − tn−1 common ratio, r: [p. 130] The quotient of two consecutive terms of a geometric sequence, tn i.e., r = tn−1 compass bearing (or bearing): [p. 346] The direction measured from north clockwise. complement of a set: [p. 65] The complement of A is the set of all members of ␰ that are not members of A.

623

complementary angles: [p. 298]␲ ␲ − ␪ = cos ␪ cos − ␪ = sin ␪ sin 2 2 ␲ ␲ + ␪ = cos ␪ cos + ␪ = −sin ␪ sin 2 2 complementary relationships: [p. ␲ ␲ 298] sin − ␪ = cos ␪ cos − ␪ = sin ␪ 2 2 ␲ ␲ sin + ␪ = cos ␪ cos + ␪ = −sin ␪ 2 2 complex conjugate, z¯ : [p. 421] If z = a + bi, then z¯ = Re(z) − Im(z)i. If z = r cis ␪, then z¯ = r cis (−␪). complex conjugate, properties of: [p. 421] Let z = a + bi, then z¯ = a − bi. z + z¯ = 2 Re(z) z z¯ = |z|2 z1 + z2 = z1 + z2 z1 z2 = z1 z2 complex number: [p. 417] An expression of the form a + bi, where a and b are real numbers. composite: [p. 74] A natural number, m, is called a composite if it can be written as a product m = a × b where a and b are natural numbers greater than 1 and less than m. compound angle formulas: [pp. 310–11] cos (x − y) = cos x cos y + sin x sin y cos (x + y) = cos x cos y − sin x sin y sin (x − y) = sin x cos y − cos x sin y sin (x + y) = sin x cos y + cos x sin y tan x − tan y tan (x − y) = 1 + tan x tan y tan x + tan y tan (x + y) = 1 − tan x tan y conjugate factor theorem: [p. 428] If the coefficients of P(z) = an z n + an−1 z n−1 + · · · + a1 z + a0 , an = 0, where n is a natural number and an , an−1 , . . . , a1 , a0 are real numbers, then the complex roots occur in conjugate pairs, i.e., if (z − ␣1 ) is a factor, so is (z − ␣1 ) constant acceleration (or kinematics) formulas: v = u + at [p. 471] 1 s = ut + at 2 2 v 2 = u 2 + 2as 1 s = (u + v)t 2 continuous data: [p. 501] Data which can take any value (sometimes within a specified interval), such as height, often arises from measuring. convergent series: [p. 138] A geometric series is an example of a convergent series with a common ratio −1 < r < 1 which will approach a limiting value as successive terms are added to it, i.e., as n → ∞, a tn , where a = t1 and r = Sn → 1−r tn−1 converse: [p. 602] q ⇒ p is the converse of p ⇒ q



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Essential Advanced General Mathematics conversion of polar to cartesian and vice versa: [p. 412] For conversion of a complex number from cartesian to polar and vice versa, x = r cos ␪, y = r sin ␪ and hence x 2 + y 2 = r 2 . correlation coefficient: see Pearson’s product-moment correlation coefficient 1 cosecant function: [p. 304] cosec ␪ = , sin ␪ provided sin ␪ = 0 cosine function: [p. 255] Cosine ␪, or cos ␪, is y defined as the x coordinate of the point P on the unit circle 1 P(θ) = (cos θ, sin θ) where OP forms an angle of ␪ radians with sin θ the positive ray of the θ x x axis. –1 1 0 cos θ

–1

cosine function, graph of: [p. 263] y y = cos θ

1

π 2

π

3π 2π 2

θ


cosine rule: [p. 334] For triangle ABC B a c

A

b

C

a 2 = b2 + c2 − 2bc cos A or, equivalently, b2 + c2 − a 2 cos A = 2bc The cosine rule is used to find unknown quantities in a triangle when either two sides and an included angle are given, or three sides are given. cos ␪ , cotangent function: [p. 304] cot ␪ = sin ␪ provided sin ␪ = 0 cumulative frequency distribution: [p. 510] A table showing the number of values in the data set less than or equal to the upper value of each class interval. cumulative frequency polygon: [p. 510] The graph of a cumulative frequency distribution, with horizontal axis showing the values of the variable and the vertical axis showing cumulative frequency. cumulative relative frequency distribution: [p. 509] A table showing the proportion of values in the data set less than or equal to the upper value of each class interval.

cumulative relative frequency polygon: [p. 510] The graph of a cumulative relative frequency distribution, with horizontal axis showing the values of the variable and the vertical axis showing cumulative relative frequency. Identical in shape to the cumulative frequency polygon but with a vertical axis scaled from 0 to 1.

D data: [p. 501] Information collected about group, such as height, or hair colour or age. degree of a polynomial: [p. 151] Given by the value of n, the highest power of x with non-zero coefficient. dependent variable: [p. 554] The variable in a bivariate situation which cannot be manipulated by a researcher, or which could be considered changed as a result of a change in the independent variable. determinant ofa matrix: [p. 16] For a 2 × 2 matrix a b for A = , det (A) = ad − bc c d diameter: [p. 376] A chord which includes the centre of a circle. difference equation (or iterative rule): [p. 116] A rule which enables each subsequent term of a sequence to be found using the previous term, e.g., t1 = 1, tn = tn−1 + 2 dilation from the x axis: [p. 196] In general a dilation of a units from the x axis is described by the rule (x, y) → (x, ay) In general the curve with equation y = f (x) is mapped to the curve with equation y = a f (x) by the transformation with rule (x, y) → (x, ay). dilation from the y axis: [p. 196] In general a dilation of a units from the y axis is described by the rule (x, y) → (ax, y) In general the curve with equation y = f (x) xis by the mapped to the curve with equation y = f a transformation with rule (x, y) → (ax, y). dimension of a matrix: [p. 1] The size, or dimension, of the matrix is described by specifying the number of rows (horizontal lines) and columns (vertical lines) that occur in the matrix. A matrix with m rows and n columns is said to be an m × n matrix direct variation: [p. 89] a ∝ bn , i.e. a varies directly as bn (n ∈ R + ) This implies a = kbn , where k is the constant of variation (k ∈ R + ). discrete data: [p. 501] Data that can only take particular values (often whole numbers), such as number of children in a family, often arises from counting.



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Glossary discriminant, : [p. 154] = b2 − 4ac i If b2 − 4ac > 0, the quadratic equation ax 2 + bx + c = 0 has two real solutions ii If b2 − 4ac = 0, the quadratic equation ax 2 + bx + c = 0 has one real solution

exact values of circular functions: [p. 261] ␪(␪ ◦ ) 0 ␲ (30◦ ) 6 ␲ (45◦ ) 4

iii If b2 − 4ac < 0, the quadratic equation ax 2 + bx + c = 0 has no real solutions

␲ (60◦ ) 3 ␲ (90◦ ) 2

displacement: [p. 464] The displacement of a particle moving in a straight line is defined as the change in position of the particle. distribution: [p. 513] A term used to indicate the pattern of the data values. division of complex numbers: [p. 421] For z 1 z¯ 2 z¯ z1 = and z −1 = 2 cartesian form: z2 |z 2 |2 |z| For polar form: z1 r1 1 = cis (␪1 − ␪2 ) and z −1 = cis (−␪) z2 r2 r double angle formulas: [p. 313] sin 2x = 2 sin x cos x cos 2x = cos2 x − sin2 x = 1 − 2 sin2 x = 2 cos2 x − 1 tan 2x =

2 tan x 1 − tan2 x

E ellipse, general cartesian equation of: [p. 446] (y − k)2 (x − h)2 + =1 2 a b2 The centre of the ellipse is the point (h, k), the axis of the ellipse parallel to the x axis is of length 2a units, and the axis of the ellipse parallel to the y axis is of length 2b units. equal complex numbers: [p. 418] If z 1 = a + bi and z 2 = c + di, then z 1 = z 2 if and only if a = c and b = d equilibrium: [p. 485] A particle is said to be in equilibrium if the resultant force acting on it is zero, i.e., if F = 0. In this case the particle has zero acceleration. If the particle is at rest it remains at rest and if it is moving it will continue to move with constant velocity. equivalence of vectors: [p. 398] Let a = a1 i + a2 j + a3 k, and b = b1 i + b2 j + b3 k If a = b then a1 = b1 , a2 = b2 and a3 = b3 Euclidean algorithm: [p. 613] A method for finding the highest common factor of two numbers and for solving linear Diophantine equations. even function: [p. 263] A function f for which f (−x) = f (x).

625

sin ␪ 0 1 2 1 √ 2 √ 3 2

cos ␪ 1 √ 3 2 1 √ 2 1 2

tan ␪ 0 1 √ 3

1

0

undefined

1 √ 3

F factor: [p. 74] A natural number, a, is a factor of a natural number, b, if there exists a natural number, k, such that b = ak. factorise: [p. 74] Express as a product of factors five-figure summary: [p. 533] The minimum, first quartile, median, third quartile, and maximum of a data set. fixed point iteration: [p. 141] Fixed point iteration can be used to solve equations of the form f (x) = 0 by finding the sequence of numbers generated by the equation x n = g(xn−1 ), as long as this sequence is convergent. formula: [p. 42] An equation containing symbols that states a relationship between two or more quantities is called a formula. A = lw (Area = length × width) is an example of a formula. The value of A, called the subject of the formula, can be found by substituting in given values of l and w. frequency distribution: [p. 502] A table showing each value that a variable can take, and how many times each of the different values of the variable was observed in the data set. frequency histogram: see histogram frequency polygon: [p. 504] An alternative to a histogram formed by plotting the values in the frequency histogram with points, which are then joined by straight lines.

G g: [p. 473] The acceleration of a particle owing to gravity. Close to the Earth’s surface, the value of g is approximately 9.8 m/s2 . c a geometric mean: [p. 133] If = , then a is said b a to be the geometric mean of c and b (or sometimes the mean proportional of b and c). geometric sequence: [p. 130] A sequence in which each successive term is found by multiplying the previous term by a fixed value, e.g. 2, 6, 18, 54, . . .



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Essential Advanced General Mathematics A geometric sequence can be defined by an iterative equation of the form tn = r tn−1 , where r is the common ratio. The nth term of the sequence can be found using: tn = ar n−1 , where a = t1 geometric series: [p. 135] The sum of the terms in a geometric sequence. The sum of the first n terms, Sn , is given by the rule: a(r n − 1) tn Sn = , where a = t1 and r = r −1 tn−1 golden ratio√(or golden section), ␾: [p. 244] 1+ 5 ␾= 2 golden rectangle: [p. 245] A rectangle with ratio of side lengths 1 + ␾ : ␾ golden section √ (or golden ratio), ␾: [p. 245] 1+ 5 ␾= 2

H highest common factor: [p. 76] The highest common factor of two natural numbers is the largest natural number which is a factor of both numbers. histogram: [p. 503] A visual display of a frequency distribution when the data are numerical, an alternative to a stem-and-leaf plot or boxplot. Frequencies (vertical axis) are represented by vertical bars corresponding to each number or class interval (horizontal axis). Sometimes called a frequency histogram. hyperbola, general cartesian equation of: [p. 447] (y − k 2 ) (x − h)2 − =1 2 a b2 The centre of the hyperbola is the point (h, k), and the equations of the asymptotes are: b y − k = ± (x − h) a

I identities derived from the Pythagorean identity: [p. 307] 1 + tan2 x = sec2 x 1 + cot2 x = cosec2 x cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x = cos2 x − sin2 x imaginary number: [p. 417] The imaginary number i has the property i 2 = −1. imaginary part of a complex number: [p. 417] Im(z) is a function which defines the value of the imaginary component of z = a + bi, i.e., Im(z) = b

independent variable: [p. 552] The variable in a bivariate situation which can be manipulated by a researcher, or which could be considered to be the cause of a change in the dependent variable. index laws: [p. 28] a m × a n = a m+n 1 a −n = n a (a m )n = a mn (ab)n = a n bn

a m ÷ a n = a m−n √ 1 n a = an a0 = 1

infinite geometric series (or sum to infinity), S∞ : a [p. 138] S∞ = , where a = t1 and 1−r tn r= , |r | < | tn−1 integer value function: [p. 208] The integer value function I : R → Z is defined by l(x) = [x] where [x] is the greatest integer not exceeding x. For example, [3, 9] = 3, [−4.1] = −5 integers: [p. 63] The elements of {. . . ,−2, −1, 0, 1, 2, . . .} are called integers. b integrand: [p. 474] In the expression f (x) d x, the a

function to be integrated, f, is called the integrand. interquartile range or IQR: [p. 524] A measure of the spread or variability of the distribution of numerical data, equal to the difference between the quartiles. That is, IQR = Q 3 − Q 1 . intersection of sets: [p. 64] The set of all the elements that are members both of set A and of set B is called the intersection of A and B. The intersection of A and B is written A ∩ B. 1 inverse variation: [p. 93] a ∝ n , i.e., a varies b k where inversely as bn (n ∈ R + ). This implies a = bn + k is the constant of variation (k ∈ R ). irrational numbers: [p. 68] The real numbers which √ are not rationals are called irrational (e.g. ␲ and 2). iterative rule (or difference equation): [p. 116] A rule which enables each subsequent term of a sequence to be found using the previous term, e.g., t1 = 1, tn = tn−1 + 2

J joint variation: [p. 103] One quantity varies with more than one other variable. This may be a combination of direct and/or inverse variation. e.g., V ∝ r 2 h implies V = kr 2 h

K kilogram weight, kg wt: [p. 485] A unit of force. If a body has mass of one kilogram then the



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Glossary gravitational force acting on this body is one kilogram weight. kinematics (or constant acceleration) formulas: [p. 471] v = u + at 1 s = ut + at 2 2 v 2 = u 2 + 2as 1 s = (u + v)t 2

L P Lami’s theorem: [p. 487] Lami’s theorem is a trigonometrically based identity r° q° which simplifies problems p° involving three forces acting R on a particle in equilibrium when the angles between the forces are known. Q R P = = sin p ◦ sin q ◦ sin r ◦

Q

least squares regression line: [p. 575] The rule for a linear relationship determined by minimising the sum of the vertical deviations of the data points from the line, that is, such that: n

(yi − a − bxi )2 is a minimum, i=1

where xi and yi are the data values. like surds: [p. 71] Surds with the same irrational factor. linear Diophantine equation: [p. 610] An equation of the form ax + by = c, where a, b and c are integers, for which a solution (x0 , y0 ) may exist for (x0 , y0 ) integers. If one solution (x0 , y0 ) exists then the general solutions are b a x = x0 + t, y = y0 − t, d d where t ∈ Z and d is the highest common factor of a and b. linear equation: [p. 33] A linear equation is a polynomial equation of degree 1, e.g., 2x + 1 = 0 linear relationship: [p. 567] A relationship which can be described by a rule of the form y = a + bx where y is the dependent variable, x is the independent variable, a is the x intercept of the line and b is the slope of the line. literal equation: [p. 47] An equation for the variable x in which the coefficients of x (including the constants) are pronumerals is known as a literal equation. e.g., ax + b = c. locus (plural loci): [p. 440] A set of points which satisfies a given condition, e.g., the locus of points P

627

which satisfy PO = 3, where O is the origin, is the circle with centre the origin and radius 3. lower limit of integration: [p. 474] In the b f (x) d x, the number a is called the expression a

lower limit of integration.

M magnitude of a vector: [p. 383] The length of a directed line segment corresponding to thevector. x − → If AB is represented by the column vector , then y − → the magnitude, |AB|, is equal to x 2 + y 2 .⎡ ⎤ x − → If AB is represented by the column vector ⎣ y ⎦, then z − → 2 2 the magnitude, |AB|, is equal to x + y + z 2 . mass: [p. 485] The mass of an object is the amount of matter it contains. Mass is not the same as weight. matrices addition and subtraction: [p. 6] Addition will be defined for two matrices only when they have the same number of rows and the same number of columns. In this case the sum of two matrices is found by adding corresponding elements. For example, 1 0 0 −3 1 −3 + = 0 2 4 1 4 3 matrices, equal: [p. 3] Two matrices A, B, are equal, and can be written as A = B when each has the same number of rows and the same number of columns they have the same number or element at corresponding positions. matrix multiplication: [p. 11] If A is an m × n matrix and B is an n × r matrix, then the product AB is the m × r matrix whose entries are determined as follows. To find the entry in row i and column j of AB single out row i in matrix A and column j in matrix B. Multiply the corresponding entries from the row and column and then add up the resulting products. Note: The product AB is defined only if the number of columns of A is the same as the number of rows of B. matrix multiplication by a scalar: [p. 6] If A is an m × n matrix, and k is a real number, then kA is an m × n matrix whose elements are k times the corresponding elements of A. For example, 2 −2 6 −6 3 = 0 1 0 3 matrix, inverse of a square matrix: [p. 15] B is said to be the inverse of A if AB = BA = I. The inverse of a square matrix A, is denoted by A−1 . The



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Essential Advanced General Mathematics inverse is unique. It does not exist for every square matrix. matrix, multiplicative identity: [p. 14] For square matrices of a given dimension, e.g. 2 × 2, a multiplicative identity I exists. 1 0 For example, for 2 × 2 matrices I = 0 1 AI = IA = A, and this result holds for any square matrix multiplied by the appropriate multiplicative identity. matrix, regular: [p. 16] A square matrix is said to be regular if its inverse exists. Those square matrices which do not have an inverse are called singular matrices. matrix, square: [p. 14] A matrix with the same number of rows and columns is called a square matrix. matrix, zero: [p. 6] The m × n matrix with all elements equal to zero is called the zero matrix. mean: [p. 521] The most commonly used measure of centre of numerical data, calculated by summing all the data values and dividing by the number of values in the data set. Always denoted x, called ‘x-bar’. median: [p. 522] A measure of centre of the distribution of numerical data, found by listing all the observations in order and locating the middle value, such that 50% of the data values are less than this value, and 50% are above. Usually denoted m. modal class: [p. 506] The interval which has the highest class frequency. mode: [p. 523] The value which occurs most frequently, usually denoted M. modulus, properties of: [p. 432] The modulus of the product of two complex numbers is the product of their moduli, i.e., |z 1 z 2 | = |z 1 ||z 2 | The modulus of the quotient of two complex numbers is the quotient of their moduli, i.e., z 1 |z 1 | = z |z | 2 2 modulus function: [p. 208] see absolute value function modulus of a complex number, |z|: [p. 431] The distance of the complex number from the origin, also known as the magnitude or√absolute value of z. If z 1 = a + bi, then |z| = a 2 + b2 modulus-argument (or polar) form of a complex number: [p. 431] A complex number expressed in the form r cis ␪, P z = a + bi Im(z) represented by the r ordered pair [r, ␪], b where r is the θ modulus of z and ␪ a 0 Re(z) is an argument of z.

multiplication of a complex number by a real number: [p. 419] If z = a + bi, then kz = ka + kbi, k ∈ R. If z = r cis ␪, then ⎧ k>0 ⎨ kr cis ␪ kz = kr cis (␪ + ␲) k < 0 and −␲ < ␪ < 0 ⎩ kr cis (␪ − ␲) k < 0 and 0 < ␪ ≤ ␲ multiplication of a complex number by i: [pp. 419, 432] Geometrically, a 90◦ rotation of the complex number about the origin in an anticlockwise direction, i.e., if z 1 = a + bi, then i z 1 = i(a + bi) = −b + ai multiplication of a vector by a scalar: [pp. 392, 432] If a = a1 i + a2 j + a3 k, then ma = ma1 i + ma2 j + ma3 k, m ∈ R. multiplication of complex numbers: [pp. 420, 432] If z 1 = a + bi and z 2 = c + di, then z 1 z 2 = (ac − bd) + (ad + bc)i If z 1 = r1 cis ␪1 and z 2 = r2 cis ␪2 , then z 1 z 2 = r1 r2 cis (␪1 + ␪2 ) Geometrically, the effect of multiplying z1 by z2 is to produce an enlargement of Oz 1 , where O is the origin, by a factor r2 and an anticlockwise turn through an angle ␪2 about the origin.

N natural numbers: [p. 63] The elements of {1, 2, 3, 4, . . .} are called the natural numbers. negative association: [pp. 561, 565] When smaller values of y are associated with larger values of x, identified by a downward trend in a scatterplot. negatively skewed distribution: [p. 514] A distribution which has a long tail to the left and a short tail pointing to the right. no association: [p. 564] When the values of y are not related to the values of x, identified by no trend in a scatterplot. normal reaction force: [p. 485] If a particle lies on a smooth surface and exerts a force on the surface, then the surface exerts a force R N on the particle, which acts at right angles to the surface and is called the normal reaction force. numerical data: [p. 501] Data collected about a variable which takes values that are quantities (not categories). For example, data from the variable height (cm).

O odd function: [p. 263] A function f for which f (−x) = − f (x).



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Glossary ordered pair: [p. 64] An ordered pair, denoted (x, y), is a pair of elements x and y in which x is considered to be the first element and y the second. outlier: [p. 536] A value which sits away from the main body of the data in a plot. Formally defined as a value more than 1.5 IQR below Q 1 , or more than 1.5 IQR above Q 3 .

P part variation: [p. 106] The value of one variable is the sum of two or more quantities each of which is determined by a variation. In some cases, one of those quantities may be constant. e.g. A = k1 r + k2 r 2 where k1 , k2 are constants of variation. partial fractions: [p. 162] Some rational algebraic functions may be expressed as a sum of partial fractions, e.g. B C Dx + E A + + + 2 2 ax + b (cx + d) (cx + d) (ex + f x + g) particle model: [p. 486] This means that an object is considered as a point. This can be done when the size (dimension) of the object can be neglected in comparison with other lengths in the problem being considered, or when rotational motion effects can be ignored. Pearson’s product-moment correlation coefficient: [p. 564] Also called the correlation coefficient, denoted r, where 0 ≤ r ≤ 1 and defined as: n xi − x¯ yi − y¯ 1 r= n − 1 i=1 sx sy where x¯ and sx are the mean and standard deviation of the x scores, y¯ and s y are the mean and standard deviation of the y scores, and n is the number of data values. percentage frequency: [p. 509] The frequency expressed as a percentage of the total number of data values, obtained by multiplying the relative frequency by 100. period of a function: [p. 262] The period of a function f with domain R is the smallest positive number a such that f (x + a) = f (x) for a in R. For example, the period of the sine function is 2␲ as sin (x + 2␲) = sin x. For functions of the form y = a cos(nx + ␰) + c or 2␲ . y = a sin (nx + ␰) + c the period is given by n For functions of the form y = a tan(nx + ␰) + c the ␲ period is given by . n periodic function: [p. 262] A function which repeats itself regularly.

polar axis: [p. 411] The polar axis is a ray from a point O from which the angle of a polar coordinate is measured. polar coordinates: [p. 411] The position of a point in the plane is determined through an ordered pair [r, ␪], the polar coordinates. The first entry is the distance from the pole and the second is the angle measured from the polar axis. ␪ is measured in an anticlockwise direction from the polar axis

P[r, θ] θ O pole

polar axis

polar (or modulus-argument) form of a complex P z = a + bi number: [p. 430] Im(z) A complex number r b expressed in the form θ r cis ␪, represented by a 0 Re(z) the ordered pair [r, ␪], where r is the modulus of z and ␪ is an argument of z. pole: [p. 411] A point from which the polar axis emanates. polynomial function: [p. 151] A rule of the type y = an x n + an−1 x n−1 + · · · a1 x + a0 , n ∈ N where a0 , a1 , . . . an are numbers called coefficients. position: [p. 463] The position of a particle moving in a straight line is determined by its distance from a fixed point O on the line, called the origin, and whether it is to the right or left of O. Conventionally the direction to the right of the origin is considered to be positive. −→ position vector: [p. 384] A position vector, O P, indicates the position in space of the point P relative to the origin O. positive association: [pp. 561, 565] When larger values of y are associated with larger values of x, identified by an upward trend in a scatterplot. positively skewed distribution: [p. 514] A distribution which has a short tail to the left and a long tail pointing to the right. prediction: [p. 571] Substituting a value of the independent variable in the rule for the relationship between two variables to determine the value of the dependent variable. prime: [p. 75] If a natural number greater than 1 has only factors 1 and itself, it is said to be prime. prime decomposition: [p. 75] This method of expressing a composite in terms of a product of powers of prime numbers is called prime decomposition. Pythagoras’ theorem: [p. 70] For a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, i.e., (hyp)2 = (opp)2 + (adj)2 Pythagorean identity: [p. 306] cos2 ␪ + sin2 ␪ = 1



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Q Q1 or first quartile: [p. 524] The value such that 25% of the data values are less than this value, and 75% are above, found by locating the median of the lower half of the data set. Also called the lower quartile. Q2 : [p. 524] The median. Q3 or third quartile: [p. 524] The value such that 75% of the data values are less than this value, and 25% are above, found by locating the median of the upper half of the data set. Also called the upper quartile. q-correlation coefficient: [p. 557] A measure of strength of a relationship between two numerical variables, with 0 ≤ q ≤ 1 and defined as (a + c) − (b + d) q= a+b+c+d where a is the number of data points in quadrant A, b is the number of data points in quadrant B, d is the number of data points in quadrant C, d is the number of data points in quadrant D. quadrant: [p. 558] A section of the scatterplot found by finding the median of all the x values in the data set, and drawing a vertical line through this value, and finding the median of all the y values in the data set, and drawing a horizontal line through this value. The resulting four areas are labelled quadrant A, quadrant B, quadrant C, and quadrant D as shown. y

B

A

C

D x

0

quadratic formula: [p. 154] An equation of the form az 2 + bz + c = 0 may be solved quickly by using the quadratic formula: √ −b ± b2 − 4ac z= 2a quadratic function: [p. 154] A polynomial function of degree two with general rule: y = ax 2 + bx + c, x ∈ R quadratic surd: [p. 69] A number of the form √ a, where a is a rational number which is not the square of another rational number.

R R + : [p. 64] R + = {x : x > 0} R − : [p. 64] R − = {x : x < 0}

R \{0}: [p. 64] R\{0} is the set of real numbers excluding 0. R2 : [p. 64] R 2 = {(x, y) : x, y ∈ R}. That is, R 2 is the set of all ordered pairs of real numbers. radian: [p. 253] One radian (written 1c ) is the angle subtended at the centre of the unit circle by an arc of length 1 unit. range of a set of data: [p. 523] The difference between the smallest and the largest observations. rate: [p. 158] Describes how a certain quantity changes with respect to the change in another quantity (often time). rational algebraic functions: [p. 162] Functions which have a rule of the form: P(x) , whereP(x) and Q(x) f (x) = Q(x) are polynomials. rational numbers: [p. 67] The numbers of the form p with p and q integers, q = 0, are called rational q numbers. real part of a complex number: [p. 417] Re(z) is a function which defines the real component of z = a + bi, i.e., Re(z) = a. reciprocal circular functions: [p. 304] The cosecant, secant and cotangent functions. reciprocal functions: [p. 162] Functions which have a rule of the form: 1 , where P(x) is a polynomial. f (x) = P(x) rectangular hyperbola: [p. 204] The basic 1 rectangular hyperbola has equation y = . x reflection in the x axis: [p. 194] In general a reflection in the x axis is described by the rule (x, y) → (x, −y) In general the curve with equation y = f (x) is mapped to the curve with equation y = − f (x) by the transformation with rule (x, y) → (x, −y). reflection in the y axis: [p. 194] In general a reflection in the y axis is described by the rule (x, y) → (−x, y) In general the curve with equation y = f (x) is mapped to the curve with equation y = f (−x) by the transformation with rule (x, y) → (−x, y). reflection in y = x: [p. 194] In general a reflection in line y = x is described by the rule (x, y) → (y, x) In general the curve with equation y = f (x) is mapped to the curve with equation x = f (y) by the transformation with rule (x, y) → (y, x). relative frequency: [p. 509] The frequency expressed as a proportion of the total number of data values, obtained by dividing the class frequency by the sample size.



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Glossary resultant force: [p. 485] The vector sum of the forces acting at a point.

S scatterplot: [p. 552] A visual display of bivariate numerical data, with the independent variable on the horizontal (x) axis and the dependent variable (y) on the vertical axis. Individual data values are represented by a point. scientific notation: see standard form 1 provided secant function: [p. 304] sec ␪ = cos ␪ cos ␪ = 0 sector: [p. 341] Two radii and A an arc define a region called a sector. In this diagram with circle centre O, the D O θ C shaded region is a minor sector and the B unshaded region is a major sector. 1 Area of sector = r 2 ␪ where ␪ c = mag ∠AOB 2 segment: [p. 342] Every chord divides the interior of A a circle into two regions called r segments. The smaller is called the minor segment, the larger O θ is the major segment. In this diagram the minor segment B has been shaded. 1 2 Area of segment = r (␪ − sin ␪) 2 where ␪ c = mag ∠AOB sequence: [p. 115] A set of numbers for which order is important. series: [p. 125] The sum of the terms in a sequence. set notation: [p. 64] ∈ means ‘is an element of’ ∈ / means ‘is not an element of’ Ø is the empty or null set, containing no elements. ⊆ means ‘is a subset of’ ∩ means ‘the intersection of’ ∪ means ‘the union of’ ε means the universal set, or the set of all elements being considered

signs of circular functions: [p. 259] These symmetry properties can be summarised y for the signs of sin, cos and tan for the four quadrants as follows: S

A

T

C

x

1st quadrant 2nd quadrant 3rd quadrant 4th quadrant

all are positive (A) sin is positive (S) tan is positive (T) cos is positive (C)

similar figures: [p. 229] Two figures are similar if one is congruent to an image of the other under an expansion from the origin of factor k. similar triangles, conditions of: [p. 230] Two triangles are similar if one of the following conditions holds: triangles have equal angles (AAA) corresponding sides are in the same ratio (PPP) AB B C AC = = = k, where k is the AB BC AC expansion (enlargement) factor two pairs of corresponding sides have the same ratio and the included angles are equal (PAP) AB AC = AB AC two pairs of corresponding sides have the same ratio and two corresponding non-included angles are equal, provided these angles are right angles or obtuse. simplest form: [p. 71] When the number under the square root has no factors which are squares of a rational number, then a surd is in its simplest form. simultaneous equations: [pp. 34, 169] Equations of two or more lines or curves in a cartesian plane, the solution of which is the point of intersection of the pairs of lines or curves. y sine function: [p. 255] Sine ␪, or sin ␪, defined as the 1 P(θ) = (cos θ, sin θ) y coordinate of the point sinθ P on the unit circle where θ OP forms an angle of x –1 0 1 ␪ radians with the cosθ positive ray of the x axis. –1

sine function, graph of: [p. 263]

sets of numbers: [p. 67] R is the set of real numbers Q is the set of rational numbers Z is the set of integers N is the set of natural numbers C is the set of complex numbers

sigma: [p. 521] Greek letter, when denoted used

to mean ‘add up’, so that the symbol x means ‘add up all the data values’.


y 1

y = sinθ

π

2π

θ



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Essential Advanced General Mathematics of another rational number is called a quadratic surd.

sine rule: [p. 331] For triangle ABC

B a

c A

b

C

a b c = = sin A sin B sin C

surd (surd of the nth order): [p. 70] If a√is a rational which is not a perfect nth power, n a is called a surd of the nth order. symmetric distribution: [p. 514] A distribution which forms a mirror image of itself when folded in the ‘middle’ along a vertical axis.

The sine rule is used to find unknown quantities in a triangle when either one side and two angles are given, or two sides and a non-included angle are given. speed: [p. 465] The magnitude of velocity. speed, average: [p. 465] The average speed of a particle for a time interval [t1 , t2 ] is equal to distance travelled t2 − t 1

T tangent function: [p. 257] If a tangent to the unit circle is drawn at A then the y coordinate of C, the B point of intersection of the extension of OP and the C(1, y) y tangent is called tangent ␪, or tan ␪.

standard form: [p. 30] Standard form involves expressing the number as a product of a number between 1 and 10 and a power of ten and is also called scientific notation. statistic: [p. 520] Any value computed from a set of data. stem-and-leaf plot: [p. 515] A visual display of a numerical data set, an alternative to a histogram or boxplot when the data set is small (less than 50). Leading digits are shown as the stem, and the final digit as the leaf in the plot. Also called a stemplot. stemplot: see stem-and-leaf plot subtraction of complex numbers: [p. 418] If z 1 = a + bi and z 2 = c + di then z 1 − z 2 = (a + bi) − (c + di) = (a − c) + (b − d)i subtraction of vectors: [p. 392] Let a = a1 i + a2 j + a3 k, and b = b1 i + b2 j + b3 k Then a − b = (a1 − b1 )i + (a2 −b2 ) j + (a3 − b3 )k sum to infinity (or infinite geometric series), S∞ : a tn [p. 138] S∞ = , where a = t1 and r = 1−r tn−1 summary statistics: [p. 520] Statistics which numerically summarise special features of a data set, such as centre and spread. surd √ (quadratic surd): [p. 69] A number of the form a where a is a rational number which is not a square

P(θ)

1

spread: [p. 523] A measure of how spread out the data values are; a measure of the variability of the data. standard deviation: [p. 526] A measure of the spread or variability of the distribution of numerical data about the mean, denoted s and defined as n 1 ¯ 2 (xi − x) s= n − 1 i=1

tanθ sinθ θ –1

0

D cosθ

A 1

x

–1

transformation: [p. 191] A transformation T is a mapping from R 2 to R 2 such that if T (a, b) = T (c, d) then a = c and b = d. translation: [p. 191] A translation is a transformation for which each point in the plane is moved the same distance in the same direction. trigonometric ratios: [p. 327] opposite sin ␪ = hypotenuse adjacent hypotenuse cos ␪ = opposite side hypotenuse opposite tan ␪ = θ adjacent

adjacent side

U union of sets: [p. 64] The set of elements that are in either set A or set B (or both) is the union of sets A and B. The union of A and B is written A ∪ B. unit vector: [p. 399] A vector of magnitude 1. For a given vector a the unit vector with the same direction a . as a is denoted by aˆ and aˆ = |a| i, j and k are unit vectors in the positive directions of the x, y and z axes respectively.



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Glossary

V variance: [p. 526] A measure of the spread or variability of the distribution of numerical data about the mean, denoted s 2 and defined as n 1 (xi − x) ¯ 2 s2 = n − 1 i=1 vector: [p. 390] A set of equivalent directed line segments. vector quantity: [p. 390] A quantity determined by its magnitude and direction, e.g., displacement, velocity, acceleration, force. vectors, properties of: [pp. 390–95] a+b =b+a commutative law for vector addition (a + b) + c = a + (b + c) associative law for vector addition a+0=a zero vector a + −a = 0 −a is the opposite or inverse vector m(a + b) = ma + mb distributive law where m ∈ R a is parallel to b if there exists k ∈ R\{0} such that a = kb velocity: [p. 465] The velocity of a particle is defined as the rate of change of its position with respect to time.

633

velocity, average: [p. 465] The average velocity of a particle is the change in its position over a period of x2 − x1 . time, i.e., t2 − t 1 velocity, instantaneous: [p. 465] The instantaneous dx velocity of a particle, v = where x is a function of dt time, specifies the rate of change at a given instant in time. velocity–time graphs: [p. 475] These graphs present information about acceleration (gradient) velocity (ordinates) displacement (signed area or definite integral) distance travelled (area ‘under’ curve)

W weight [p. 485] A mass of m kg, on the Earth’s surface, has a force of m kg wt or mg Newtons acting on it. This force is known as the weight.

Z zero vector, 0 [p. 392] A line segment of zero length with no direction.



A P P E N D I X

A

Computer Algebra System (TI-Nspire OS3.0) A.1

Introduction to the TI-Nspire The appendix is written for the TI-Nspire Clickpad, Touchpad and CX CAS calculators. The CX calculator and the function of some of the keys is shown in the diagram below. Similar diagrams for the Clickpad and Touchpad calculators are available at www.cambridge.edu.au/go/.

634 ISBN 978-1-107-65235-4 © Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2012 Photocopying is restricted under law and this material must not be transferred to another party.


Appendix A — Computer Algebra System (TI-Nspire)

635

The Home screen When pressing the home key (c) you always arrive at the Home screen which displays the Scratchpad and Documents platforms. The Scratchpad platform enables you to: A. Open the Calculate screen in the Scratchpad. B. Open the Graph screen in the Scratchpad. The Documents platform enables you to: 1. Open a New Document 2. Go to the My Documents folder 3. Access Recent Documents 4. Return to the current document. 5. View the System Information and change the Settings Below the Scratchpad and Documents sections of the Home screen are icons that add Calculator, Graphs, Geometry, Lists & Spreadsheet, Data & Statistics and Notes pages to an existing document.

System Settings To change the system settings go to the Home screen (c) and select Settings. Select Settings>General. Use the tab key (e) to move between the different categories and use the selection tool (x) to open a category. Use the up arrow (£) or down arrow (¤) on the touchpad to move between the different options within a category. When finished, move to Make Default and press enter (·) to confirm. The settings shown below are the recommended settings for General Mathematics.

Scratchpad All instructions given in the text, and in the Appendices, are based on the Documents platform. However, all instructions relating to a Calculator application (page) or a Graphs application (page) can be transferred to the Scratchpad:Calculate or Scratchpad:Graph platform respectively.



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The Scratchpad consists of two applications; A: Calculate – this is a fully functional CAS calculation platform that allows for quick and easy access to the home screen and menus.

B: Graph – this is a fully functional graphing platform that allows for quick and easy access to the graphing screen and menus.

when using Scratchpad:Calculate you need to be careful of previously defined variables. If in doubt, press h, to see what variables are defined, and delete them if necessary using b>Actions>Clear a-z or b>Actions>Delete Variable e. g. DelVar f Note that b>Actions>Clear History only clears the screen and does not clear stored variables. It is possible to switch between the Scratchpad and the Documents platforms by pressing c (also » on the Touchpad & CX). This might be useful to do a quick trial, or check, calculation or graph in the Scratchpad whilst working in the Documents platform. Select Current Document on the Home page to return to the current document. It is possible to complete the majority of the VCE General Mathematics course using only the Scratchpad platform if preferred. It should be noted that any defined variables created in the Scratchpad are not available in the Documents platform. To move between the Calculate and Graph screens whilst in the Scratchpad either or / (also » on the Touchpad & press c and select from this screen or press / CX). Note:




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Documents – organisation and navigation The calculator allows you to save documents very much like on a computer. Documents are saved in the My Documents folder (c>My Documents). In this folder you can create your own folders. Each document can contain a number of problems, and each problem can contain a number of pages.

Select New Document (c>New Document) from the home screen to create a new document. You will be prompted to select a Calculator application, a Graphs application, a Geometry application, a Lists & Spreadsheet application, a Data & Statistics application or a Notes application for the first page of this document.

The numbering 1.1 in the top left hand corner indicates that you are in problem 1, on page 1. In general, the numbering m · n indicates problem m, on page n. There are several ways of adding another page (application) to the document. Either go to the home screen (c) and select the appropriate application, / and I and select the (Touchpad only & CAS). application, or / and To insert an additional five pages, each one with one of the remaining five applications, c c c c . The five pages will look as shown below. press c Note: the DataQuest application has not been included here. It is used for datalogging.



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Notice how the new pages have numbers 1.2, 1.3, 1.4, 1.5 and 1.6 as explained above. In General Mathematics we will mainly be using the Calculator application, the Graphs application and the Lists & Spreadsheet application. Navigation within a document can be done in two ways. to move to the page on the One way is to press / to move to the page on the right. left; and press / Alternatively use the cursor (Touchpad & CX only) to point to the page tabs and click to open. Another way is to get an overview of the document by pressing / and navigate using the arrows (¡ ¢ £ ¤) on the Touchpad. Press enter (·) to select a highlighted page. For the six pages above, the overview will look as shown at right. From this view it is easy to insert (/ I) a new page (after a highlighted page) and delete (. or . on the Clickpad) a highlighted page. To get back to the individual page view either press enter (·) on a highlighted page or press / ¨. >Insert>Problem There is also the opportunity to add another problem (/+c>Insert>Problem on the Clickpad) to a document. Within this problem new pages can be inserted as explained above. However, despite all the opportunities for sophisticated organisation of documents, most students of General Mathematics can, to begin with, create one document, with one Calculator application and one Graphs application, and use that for all their calculations and graphing (or use the Scratchpad as mentioned earlier)




A.2

639

Using the Calculator Application The Calculator Application works like a basic calculator and it is here that you will complete most of your arithmetical calculations. The following instructions are also suitable for the Scratchpad:Calculate platform.

Entering and editing expressions The following example shows how to enter, evaluate and edit the expression 100(1.15 − 1) , To enter and evaluate the expression 0.1 press the following keys: /p100r(1 1l5e-1¤0 1·

100(1.15 − 1) . 0.1

Notice the use of the division template and that each opening bracket is automatically followed by a closing bracket. 100(1.25 − 1) , To enter and evaluate the expression 0.2 notice the similarity with the one we entered above and move upwards by pressing the up arrow (£) twice until the previous expression is highlighted as shown.

Press enter (·) to paste this expression in the new line as shown.

Use the arrows (¡ ¢ £ ¤) on the touchpad to move into the expression and use the delete key (. or . on Clickpad) to delete the two 1s before replacing them by 2s. Press enter (·) to evaluate this new expression.

Rather than copy and paste an entire expression as explained above, it is also possible to copy and paste a part of an expression. To do this, move to the expression as explained above, move into it using the arrows (¡ ¢ £ ¤) on the touchpad and select the part you want to copy by pressing and holding down the g (or g on Clickpad) key while using the arrows (¡ ¢ £ ¤) on the touchpad to select the part you want to copy. Press enter (·) to paste this selection in the new line. ISBN 978-1-107-65235-4 © Michael Evans, Sue Avery, Doug Wallace, Kay Lipson 2012 Photocopying is restricted under law and this material must not be transferred to another party.


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The calculator can also perform more complicated mathematical calculations, all of which can be found in the menu (b). The most frequently used in Mathematical Methods are explained below, in order of appearance.

Operations in the Actions (Tools) menu Define This command is used for defining functions of one or more variables. There are two other ways of defining functions; either use the notation := using /t on the Touchpad or : = on the Clickpad or use store (/ h). The following screens illustrate their use.

Clear a-z This command is used to clear the one-character variables a-z. It is highly recommended to use mainly one-character variables and use this command frequently to clear these variables.

Operations in the Algebra menu Solve This command is used to solve equations, simultaneous equations and some inequalities. An approximate (decimal) answer can be obtained by pressing / + · or by including a decimal number in the expression. The following screens illustrate its use.




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Factor This command is used for factorisation. Factorisation over the rational numbers is obtained by not specifying the variable, whereas factorisation over the real numbers is obtained by specifying the variable. The following screens illustrate its use.

Expand This command is used for expansion of expressions. By specifying the variable, the expanded expression will be ordered in decreasing powers of that variable. Symbolic expressions can only be expanded for an appropriate domain. The following screens illustrate its use.

Operations in the Calculus menu Derivative & Derivative at a Point These operations are used to differentiate expressions and/or give numerical derivatives. The following screens illustrate their use.

The differentiation template can also be accessed from the templates menu (t or /+r on Clickpad). In the templates menu there is also a template for the nth derivative.



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Integral This operation is used to integrate expressions. The following screens illustrate its use.

Limit This operation is used to find limits of expressions. The following screens illustrate its use.

Function Minimum and Function Maximum These operations return the value for which the minimum or maximum value (or the greatest lower bound or least upper bound) of a function occurs. The following screens illustrate their use.

Operations in the Probability menu Factorial (!), Permutations and Combinations These functions are used to solve counting method problems. The following screens illustrate their use.




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Distributions This menu contains a list of distributions. For us, especially the Binomial Distribution and the Normal Distribution are of interest. For calculations using the Binomial Distribution we use either Binomial Pdf (binomial probability distribution function) or Binomial Cdf (binomial cumulative distribution function). Use the tab key (e) to move between the number fields. The following screens illustrate their use.

Unfortunately, these functions can only return an approximate (decimal) answer, so to obtain an exact answer consider defining the functions bipdf and bicdf as shown below.

For calculations using the Normal Distribution we use either Normal Cdf (normal cumulative distribution function) or Inverse Normal. When using these functions you can solve for any of the variables using nSolve() (Numerical Solve) and an appropriate domain. The following screens illustrate their use.



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Operations in the Statistics menu Stat Calculations This menu contains a list of operations on data. The following screens illustrate the use of the One-Variable Statistics command.

Note that the commands to find the mean, median, variance and standard deviation of a list of numbers are available from b>Statistics>List Maths The following screens illustrate the use of the Exponential Regression command. The last screen shows that the exponential function going through the points (1,6), (2,12) and (3,24) has equation f (x) = 3 × 2x .

A.3

Using the Graphs Application The following instructions are also suitable for the Scratchpad:Graph platform. As the Points & Lines menu is not available in the Scratchpad, use Analyze Graph to find the intersection point/s.

Plotting functions To graph a function simply type the expression (in terms of x) in the Entry Line for one of the predefined functions f 1(x) − f 99(x) followed by enter as shown in the screens below. If necessary, you can hide/show the entry line by pressing /+ G.




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Plotting hybrid functions In a Graphs application with the cursor in the entry line, select the hybrid (piecewise) function template, t (/ +r on the Clickpad) and enter as shown.

Finding intersection points Use b>Points & Lines>Intersection Point/s to display the approximate (decimal) coordinates of the intersection point(s). Select each of the two graphs (or a graph and an axes) using the arrow keys (¡ ¢ £ ¤) (or the touchpad) followed by enter. Press escape (d) to exit the command. On the screen to the right, Intersection Point(s) has been used to find the intersection points of f1 and f2. Alternatively use b>Analyze Graph>Intersection.

Finding zero(s) and local minimum(s)/maximum(s) Either b>Trace>Graph Trace or b>Analyze Graph>. . . . . can be used to display the approximate (decimal) coordinates of points on the graph. In Graph Trace the tracing point (×) can either be moved using the arrow keys (¡ ¢) or by typing a specific x-value followed by enter. Use £ ¤ to navigate between graphs. When trace reaches a zero it displays zero , when it reaches a local minimum it displays minimum etc. Press escape (d) to exit the command. On the screen to the right, Graph Trace has been used to find the turning point of f2. When using Analyze Graph you select the lower bound (to the left of the key point) by clicking (x) and move to the right (¢) of the key point (to select the upper bound) and click.



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Finding integrals Use b>Analyze Graph>Integral to display the approximate (decimal) integral. Select the graph, if prompted, then type an open parenthesis and type the lower limit, followed by enter, move cursor to the right, type an open parenthesis again and type the upper limit followed by enter. On the screen to the right, Integral has been used to find the integral of f2 between −8 and 0.

Plotting probability distributions To graph a (discrete) Binomial Distribution first create a sequence of the outcomes in a Calculator application and use the binomial probability distribution function command binomPdf(n,p,x) on that list. In a Graphs application select Scatter Plot using b>Graph Type>Scatter Plot and define the x-values to be the sequence of the outcomes and the y-values to be the corresponding binomial probability distribution function values. Finally, use b>Window/Zoom>Zoom – Data to get a window that best fits the data. The following screens illustrate how to graph the binomial distribution with parameters n = 20 and p = 0.1.

To graph a (continuous) Normal Distribution curve we use the normal probability distribution function command normPdf(x,␮,␴) from the catalog (k1N). The following screens illustrate how to graph the standard normal distribution in the window −5 ≤ x ≤ 5 and −0.1 ≤ y ≤ 0.5.




A.4

647

Using the Lists & Spreadsheet Application The Spreadsheet is in many ways similar to a Microsoft ExcelC spreadsheet. Cells can be filled down, sorted etc., and it will handle both absolute and relative references.

Absolute and relative references Simple interest and compound interest is used here to illustrate the use of absolute and relative cell references. Enter 0 in cell A1, 100 in cell B1 and 100 in cell C1. Use the arrows (¡ ¢ £ ¤) on the NavPad to move between cells. Enter =a1 + 1 in cell A2, =b1 + 0.1 × b$1 in cell B2 and = c1 + 0.1 × c1 in cell C2. The $ symbol is found using / k (also on the Touchpad & CX) Highlight the cells A2, B2 and C2 using g (g on the Clickpad) and the arrows (¡ ¢ £ ¤) and use b>Data>Fill and the down arrow key (¤), followed by enter, to generate the numbers in the next three rows. The following screens illustrate the process.

The resulting numbers are the value of 100 dollars invested at 10% p.a. simple interest (column B) and 10% p.a. compounded interest (column C) over the next four years. Note the important difference in the formulas in cells B5 (= B4 + 0.1 × B$1) and C5 (=C4 + 0.1 × C4). In the B-column, the use of the $ symbol in front of 1 in cell B2 fixed this value such that when filled down it did not change, still adding 10% of the value in cell B1. This is an example of an absolute reference. In the C-column, the cell references are slightly different, always adding 10% of the value in the previous cell. This is an example of a relative reference.

A.5

Dynamic Geometry The calculator can dynamically change objects and corresponding calculations and thus provide ample opportunities for teachers and students to dynamically illustrate mathematical concepts and ideas. The Geometry tools are not available in the Scratchpad:Graph platform.

Example 1 – a moving tangent This example shows how the calculator can be used to dynamically illustrate the concept of instantaneous rate of change as the gradient of the tangent.



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1. Open a Graphs application (c ). 2. Type the expression of the function f (x) = x(x − 7)(x + 7)/40 in the Entry Line and press ·. 3. Hide the function label using b>Actions>Hide/Show 4. Place a tangent (b>Points & Lines>Tangent) at any point on the graph using the NavPad. Press · twice. 5. Show the equation of the tangent (b>Actions>Coordinates & Equations). Double click on the tangent line. Press d to exit the show equation tool. 6. Move the open hand ({) to the point. Press and hold down x when the point is flashing in the hand. The hand is now closed ({). 7. Move the point using the touchpad to dynamically illustrate the gradient of the curve. The following screens show the tangent placed at a point on the curve and the point moved along the curve.

Example 2 – a parabola with a point This example shows how the calculator can be used to dynamically illustrate why parabolas play an important part in the design of torches and satellite dishes. ). Open a Graphs application (c Type in the expression of the quadratic f (x) = 0.1x 2 in the Entry Line. Press · Select the first quadrant window (b>Window/Zoom>Zoom-Quadrant 1). Place a ray (b>Points & Lines>Ray) on a point on the graph, pointing upwards, parallel to the y-axis. Hold down g (g on the Clickpad) to move in increments of 15◦ . 5. Place a tangent (b>Points & Lines>Tangent) at the point on the graph using the NavPad. 6. Construct the perpendicular line (b>Construction>Perpendicular) to the tangent at this point. 7. Reflect (b>Transformations>Reflection) the ray in this line. Select the line of reflection first. 8. Place a point of intersection (b>Points & Lines>Intersection Point/s) between the y-axis and this new ray. 9. Show the coordinates (b>Actions>Coordinates & Equations) of this point. (double click) 10. Hide (b>Actions>Hide/Show) the tangent and the perpendicular line. 1. 2. 3. 4.




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11. Change the attribute (b>Actions>Attributes) of the ray and the reflected ray to dashed line style 12. Move the open hand ({) to the point. If necessary, press e until point on is displayed. Press and hold down x when the point is flashing in the hand. The hand is now closed ({). 13. Move the point using the touchpad to dynamically illustrate the unchanging focal point. The following screens show the ray, tangent and reflected ray at a point on the curve and the point moved along the curve.

A.6

Learning More There is much to be learned from reading the manual that comes with the calculator, but there are also some highly recommended free online tutorials on the internet, including: Atomic Learning TI-Nspire tutorials http://www.atomiclearning.com/k12/en/ti nspire/



A P P E N D I X

B

Computer Algebra System (ClassPad 330) B.1

Introduction For reference material on basic operations of the calculator, refer to the free downloadable documentation provided by Casio on the website: http://www.casioed.net.au/downloads/classpad download.php. The Classpad calculator operates in many ways like your computer, with various drop down menus which are accessed using the stylus and tapping items on the Menu bar. Shortcuts in particular applications are found in the pictorial toolbar. You will find that Cut, Copy and Paste in the Edit menu very useful.

Application selector

Icons that access working zones

The screen provides access to various in-built programs of the machine, selected by tapping the icons on the touch-sensitive screen using the stylus. To stop a process that is in . action, tap

Icon panel (Master toolbar)



Appendix B — Computer Algebra System (ClassPad 330)

651

The screen The various areas of the screen referred to in these notes are shown in this diagram. In some programs, such as the screen is split into two halves. If you wish one of these to fill the whole screen, tap to ensure it is selected (bold border) and tap on the Master toolbar. Tapping it again will re-split the screen.

Menu bar

Toolbar

Work area—input displayed on left, output displayed on the right.

Status bar—displays current mode settings

Settings The basic settings for the calculator are managed by tapping from any of the program screens. Quick changes can be made by tapping the appropriate item on the Status Bar. In

they appear as shown at right.

Tapping any of them toggles between the modes. Alg—algebra mode automatically simplifies expressions. (Alternative Assist—assistant mode does not automatically simplify expressions.) Standard—results given in exact form (fraction, square root, etc.) (Alternative Decimal—converts all results to decimal (approximate value).) Real—real mode for real number calculations. (Alternate Cplx—for complex number calculations.) Rad—angles displayed in radians. (Alternative Deg—angles displayed in degrees.)

Executing in commands in Operator entries are shown on the left of the screen. Once entered, press and the answer will be displayed right-justified on the screen. Scroll arrows will appear if your entry, or the answer, is too wide for a single screen. Additional input options, such as square roots, alternative variables and trigonometric or calculus operations, are accessed by pressing the button on the calculator’s keypad. If an answer is given in exact mode and you wish . To convert from decimal to a decimal answer, highlight the answer with the stylus and tap exact, follow the same steps.

Errors, deleting, inserting and clearing The button deletes the last character entered. You can use the central multi-directional arrow key, or the stylus, to return the cursor to the place where you wish to insert extra



652


characters. If you have already executed a command, you can still use the stylus to go back and make a correction and then execute again. To clear the screen (in any program section of the calculator, select Edit—Clear all. To clear values or functions which have been saved in memory locations, select Edit—Clear all variables. To clear individual variables, without —Variable manager. You can double-click on various subfolders to clearing them all, use expand them and then select items to delete using the Edit submenu items. Note also that, like using a Windows computer program, you can highlight, copy, paste and drag-and-drop expressions, or parts of expressions, to create new expressions in the entry line.

B.2

Using the Main menu In this section the Main menu is explored. Work through it to become acquainted with the menu. There are two ways to operate in this menu, but the Interactive menu is the simplest when used in conjunction with the stylus. In this section we shall demonstrate some examples of how this is used.

Solve This is used to solve equation and inequalities. The variables x, y and z are found on the hard keyboard. Other variables may be entered using the and selecting VAR. Variables are shown in bold italics. The abc keyboard allows you to type sentences, etc.; however, the letters are not always recognised as variables. If you choose to use the abc keyboard, you must type a × x, for example, as ax will be treated as text. For example: Enter ax + b = 0, highlight it with the stylus, tap Interactive—Equation/inequality—solve and ensure the variable selected is x. Solution returned b is x = − . a Enter x 2 + x − 1 = 0, and follow the same instructions as above. The answer is as shown. Note: x ∧ 2 + x − 1 = 0 is entered, but the calculator converts this to standard algebraic notation when EXE is pressed. Also note in this example that ‘= 0’ has been omitted. If the right-hand side of an equation is zero, it is unnecessary to enter it.




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Enter abt − w + t = wt for w; follow the instructions above and select w as the variable. Solve x 3 − x 2 − x + 1 = 0 for x. √ Solve 2x + 2 < 3 for x. See the screen above for solutions returned. is found in when is activated and < is in . If the answer is not in the form required, it is often possible to cut and paste it in the next entry line and use Interactive—Transformation—simplify as shown here. Solve simultaneous equations uses a dedicated entry style. Note carefully the screen at right. in the window and enter To enter, tap the equations and variables as shown. For until the more than two equations, tap number of equations required is given. Note:

Factor To factorise is to transform the expression to a different form. This command is found in Interactive—Transformation—factor. Examples: To factorise x 3 − 2x over the rational numbers, use factor. To factorise over the real numbers, select rFactor.

Some further examples are provided here. See the screens for the results. factor a 2 − b2 . factor a3 − b3 . 1 2 + factor + 1 . 2 x 4− 1 2 (x − 1) factor 2x − x over the rationals. factor 2x 4 − x 2 over the reals. This command can also be used to give the prime decomposition (factors) of integers.



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Expand To expand expressions, use Interactive— Transformation—expand. For example: expand (a + b)3 √ 2 expand a + 2b . The expand command can also be used to form partial fractions. In this case, enter and highlight the expression, select Interactive— Transformation—expand, select the Partial Fraction option and set the variable as x. For example: 1 expand 2 x −1 x 3 + 2x + 1 . expand x2 − 1 Note: The top screen shows all the examples; the bottom screen shows how to enter for partial fractions.

Zeros To find the zeros of an expression in the menu, use Interactive—Equation/inequality—solve and ensure you set the variable. The calculator assumes you are solving an equation for which one side is zero. For example: Zeros of x 2 − 1 for x. Zeros of x 2 − y 2 for y. Zeros of x 2 − y 2 for x. Zeros of x 2 − y for y. Zeros of x 2 − 4x + 8 for x. No solutions. Zeros of x 2 − 4x + 1 for x. Two solutions. Zeros of x 2 − 4x + 4 for x. One solution.

Approximate Switch mode in the status bar to Decimal. If an answer is given in Standard (Exact) mode, it in the toolbar. can be converted by highlighting the answer and tapping




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Combining fractions This command gives an expression with a common denominator. The denominator is returned in factored form. For example: Enter and highlight 1/(x − 1) + 1/(x + 1). Then select Interactive—Transformation—combine. Enter and highlight y/(x − y) + y/(x + y). Then select Interactive—Transformation—combine.

Solve numerically There are several ways to find numerical solutions to equations. In each of these ways, only one solution is given. You can vary the guess (Value) or the bounds of the search (Lower and Upper) to find particular solutions. If an expression such as x 2 − x − 2.1 is entered without an equals sign, the calculator will assume the expression is equal to zero and solve the equation. For example: Enter and highlight x 2 − x − 2.1 = 0, select Interactive—Equation/inequality—solve and tap the solve numerically option. Note that the lower and upper bounds are set to ∞ and a guess of –1 has been entered to return the first solution: 1.03297. For the second solution shown, the first line has been copied and pasted (or dragged) to the next entry line and the guess x = 2 has been entered to return the solution x = 2.03297. Alternatively, use from the Main menu and enter Lower and Upper bounds. A guess may also be entered, but is not necessary. Note: The bounds selected arise from a quick sketch which indicated that the quadratic equation has one positive and one negative solution. A third method involves using the Graph program and finding the solution using the GSolve application.



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B.3

Graphing Graphing using the Casio Classpad is usually done in menu. The screen is split with the equation the entry screen and the graph screen both shown. The normal configuration is the y = form; however, in the menu bar, alternative by tapping the arrow forms of equation and inequality may be selected. After entering the equation, tap the select box to produce to the left of the equation and then tap the graph in the lower screen. More than one equation can be entered and you will note tabs allowing more than one set of equations to be entered and stored for other uses. Note: In the example, the upper box is selected—shown by a bold boundary. Tapping the lower half of the screen will select the graph section and the menu bar will change to the operations available on the graph. To alter the viewing window for the graph, various Zoom . menus are available or the window can be manually set using Tap on the icon panel to allow the selected window (graph or equation) to be re-sized to fill the screen. Tapping again returns both windows to the screen.

To find features of the graph, ensure the graph window is selected (bold frame), then tap Analysis—GSolve and select the menu item you wish. If seeking the root, the calculator will return the left-hand root shown in the graph window by default. To find the other root, use Zoom-Box to zoom into an area containing only the desired root. Analysis—GSolve will not return this root. Use Zoom—Previous to return to the previous view. Analysis—Trace allows use of the round arrow cursor on the calculator to move along the curve and display grid points in the lower part of the screen.




B.4

657

Defining functions Expressions can be stored in memory location. is used from and the For this the . variable A is entered using 2 For example x − 3x A. The operations discussed previously can now be applied to A.

B.5

Probability and counting Basic probability functions are found using menu. the n n To find Cr and Pr tap the key and then enter the values of n and r as shown. To find n!, type the value of n and tap the ! key. Generate a random number between 0 and 1 by typing rand(. to produce more numbers. Continue to tap

The function randList(20) will create a set of 20 random numbers between 0 and 1. To generate a random integer between 25 and 50, type rand(2,20) as shown. To create a random list of numbers between 0 and 1, line the Stat editor can also be used. In the for the required list type randList(20) to create a list of 20 random numbers. Note: The mode should be set to Decimal in the status bar at the bottom of the screen. The functions referred to here can be found in the and scroll to and catalogue. Go to select the function required.



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To find the value of a term in a binomial distribution, go to the Stat screen and enter Calc, Distribution and scroll to Binomial PD. Enter the values for n, p and x and tap . To find more values, use and enter the required values. Alternatively, define a function bi(n, p, x) = nCr(n, x) × p ∧ x × (1 − p)∧ (n − x) This can be used for any binomial distribution problem.

B.6

Trigonometric functions The choice of radian or degree measure is made by tapping the appropriate section of the Status bar at the bottom of the screen. Trigonometric equations are solved using the . Initially, we shall Interactive functions in consider the solution of the equation sin x = 0.5. 5␲ or The exact solution is given as x = 2k␲ + 6 ␲ x = 2k␲ + . The parameter k is shown as 6 constn(1) and constn(2) on the screen. This indicates that these are the first and second parameters to be used. The screen shows the solution when the status bar is set to Standard or to Decimal. Tap on to scroll across for the rest of the solution. In cases where the domain is limited, enter the equation with the domain sin(x) = 0.5|0 ≤ x ≤ 2␲, highlight the equation part only and then tap Interactive, Equation/inequality, solve and ensure the calculator the variable is set to x. After tapping should read solve (sin(x) = 0.5, x)|0 ≤ x ≤ 2␲.

B.7

Using the Calculus menu Four of the operations of the calculus menu are considered here. These functions appear on the screen and variables and expressions can be entered directly. (It is also keyboard possible to use the options in the Interactive, Calculation drop-down menu.)




659

Differentiate This operation is use to differentiate expressions. Several examples are shown in the screen here. For example: In tap , enter x in the darkened entry box and the expression x 2 + 2x in the empty box. Change the variable in the darkened box when differentiating with respect to a variable other than x. If a second derivative is required, tap set the order to 2 in the small raised entry box beside d.

Integrate For indefinite integrals, tap and in 2 enter ax + bx + c in the space provided for the expression, then enter x as the variable and tap EXE. For definite integrals, the same process is followed except that the upper and lower bounds are entered.

Limit The limits considered in this section will be those that are considered in the senior years of school mathematics. Right and left limits are shown in the sample screen. , tap and enter To enter limits use the required information in the entry boxes provided. Remember to use the VAR screen to enter variables. x ∧ 3 − 3x ∧ 2.



660


Minimum (fMin) and Maximum (fMax) These operations return the value for which the maximum or minimum value (or the least upper bound or greatest lower bound of a function occurs). fMax returns the value for which a local maximum occurs only if this is the actual maximum for the interval being considered. Similarly for fMin. Use Interactive—Calculate—fMax or fMin A number of examples are show in the screen on the right. The entries for the example seeking the maximum value of x 3 − 3x 2 √ for x < 3 are show below.



Answers Chapter 1

Exercise 1B " " ! 2 4 , , 2X = −4 −2 " ! " ! −2 13 , ,X − Y = 4Y + X = −2 −2 " " ! ! 1 3 −3 3 , −3A + B = −3A = −7 −7 −6 −9 " " ! ! 5 1 0 6 12 4 b 2a 18 7 13 8 4 2 ⎡ ⎤ 5 1 0 ⎢3 3 ⎥ c⎣ 7 13 ⎦ 6 3 3 " " ! ! −3 3 2 −2 , , −3A = 3 2A = 0 −6 0 4 " ! −6 6 −6A = 0 −12 !

Exercise 1A 1a2×2 ⎡ 1 0 ⎢0 1 ⎢ 2a⎢ ⎢0 0 ⎣0 1 1 0 ⎡

1 ⎢0 ⎢ 3⎢ ⎢0 ⎣0 0 ! 4

0 1 0 0 0

1X+Y=

b2×3 ⎤ 0 0 1 0 1 0⎥ ⎥ 1 0 0⎥ ⎥ 0 1 0⎦ 0 0 1 0 0 1 0 0

0 0 0 1 0

c1×4 ⎡ 1 1 ⎢1 1 ⎢ b⎢ ⎢1 1 ⎣1 1 1 1

d4×1 ⎤ 1 1 1 1 1 1⎥ ⎥ 1 1 1⎥ ⎥ 1 1 1⎦ 1 1 1

⎤ 0 0⎥ ⎥ 0⎥ ⎥ Only the seats for top-left 0 ⎦ to bottom-right diagonal 1 are occupied.

200 180 135 110 56 28 110 117 98 89 53 33

"

5 a [0 x] = [0 4] if x = 4 " " ! ! x 7 4 7 if x = 4 = b 1 −2 1 −2 ! " ! " 2 x 4 y 0 4 c = −1 10 3 −1 10 3 " ! 2 0 4 if x = 0, y = 2 = −1 10 3 6 a x = 2, y = 3 c x = 4, y = −3 ⎡

0 ⎢3 ⎢ 7⎣ 1 0

3 0 2 1

1 2 0 1

⎤

0 1⎥ ⎥ 1⎦ 0

4 a Yes b Yes " " ! ! 0 −9 6 4 b 5a 12 3 −4 −4 " " ! ! −6 −13 6 −5 d c 16 7 8 −1 !

b x = 3, y = 2 d x = 3, y = −2 ⎤ ⎡ 21 5 5 ⎢ 8 2 3⎥ ⎥ ⎢ ⎥ 8⎢ ⎢ 4 1 1⎥ ⎣ 14 8 60 ⎦ 0 1 2

6a

0 1 2 3

"

! b

" −2 3 6 3 ⎡ −9

! c

3 3 −1 7

"

−23 ⎤ 2 4 ⎢ 2 ⎥ 7X= ,Y = ⎣ 2 ⎦ 0 −3 −1 11 2 " ! 310 180 220 90 , representing 8X+Y= 200 0 125 0 the total production at two factories in two successive weeks. !

"



Answers

662

Essential Advanced General Mathematics ⎤ 6.00 ⎢ 8.00 ⎥ ⎥ ⎢ ⎥ 10 ⎢ ⎢ 2.00 ⎥ represents how much each student ⎣ 11.00 ⎦ spends in a week on magazines. 6.50 ! " s c + s12 c2 + s13 c3 11 a SC = 11 1 s21 c1 + s22 c2 + s23 c3 ⎡

Exercise 1C " ! " ! " −5 4 4 , , AY = , BX = 1 AX = 8 1 −5 " " ! ! 0 −1 2 , , AC = IX = 1 2 −1 " ! " ! 1 1 −1 , , (AC)X = CA = 0 0 1 " ! ! " 1 −2 9 , , AI = C(BX) = −1 3 5 " " ! ! 1 0 3 2 , , AB = IB = 0 1 1 1 " ! " ! 1 0 3 −8 , A2 = BA = , 0 1 −4 11 " " ! ! 1 −3 11 8 , B2 = , A(CA) = −1 4 4 3 " ! −2 −5 A2 C = 3 7 !

2 a AY, CI are defined, YA, XY, X2 , XI are not defined. " ! 0 0 b AB = 0 0 3 No " ! 4 −2 4 LX = [7], XL = −6 3 5 AB and BA are not defined unless m = n. " ! 1 0 6 0 1 7 One possible ! " answer!is " 1 2 −2 1 A= ,B = 3 4 1.5 −0.5 ! " 1 2 8 One possible answer is A = , 4 3 ! " ! " 0 1 −1 2 B= ,C = , 2 3 −2 1 ! " −1 11 A(B + C) = , −4 24 ! " −1 11 AB + AC = , −4 24 ! " 11 7 (B + C)A = 16 12 ! " 29 9 represents John spending 29 minutes 8.50 consuming food which " cost him $8.50. ! 29 22 12 John’s friends spent 8.50 8.00 3.00 $8.00 and $3.00 and took 22 and 12 minutes respectively to consume their food.

b SC represents the income from car sales for each showroom. c!SC = " s11 c1 +s12 c2 +s13 c3 s11 u 1 +s12 u 2 +s13 u 3 s21 c1 +s22 c2 +s23 c3 s21 u 1 +s22 u 2 +s23 u 3 represents the income for each showroom for new car sales and used car sales. d CV gives the profit on each new car and each used car for the three models.

Exercise 1D

" 2 −2 −3 2 " ! 1 2 2 c2 d 2 −3 −2 ⎡ 2 −1 ⎤ " ! −1 1 ⎢ ⎥ 2a b ⎣ 7 14 ⎦ −4 3 1 3 7 14 " ! 1 0 cos ␪ sin ␪ 1 c d 0 −sin ␪ cos ␪ k⎡ ⎤ " ! 1 1 1 0 , 3 A−1 = ⎣ 2 2 ⎦ , B−1 = −3 1 0 −1 ⎡ 1 1⎤ " ! 5 1 ⎢ 2 ⎥, , (AB)−1 = ⎣ 2 AB = −3 −5 ⎦ −3 −1 2 2 1 −1 A−1 B−1 = 2 , 3 −1 ⎤ ⎡ 1 1 ⎢ 2 2⎥ −1 −1 −1 ⎥ B−1 A−1 = ⎢ ⎣ −3 −5 ⎦ , (AB) = B A 2 2 ⎤ ⎡ 3 −1 0 7 2⎦ b 4a⎣ 2 1 −8 1 −2 ⎡ 5 ⎤ −7 ⎢ 2 2 ⎥ c⎣ 11 −21 ⎦ 2 2 ⎡ −3 11 ⎤ ⎡ −11 17 ⎤ ⎢ ⎥ ⎢ 8 ⎥ 5a⎣ 8 b ⎣ 16 16 ⎦ 1 7 ⎦ −1 3 16 16 4 4 1a1


!

b


663

Answers ⎤ 0

⎥ 1 ⎦ 0 a22 " " ! " ! " ! ! −1 0 1 0 −1 0 1 0 ; , ; , 8 k 1 k −1 0 −1 0 1 " " ! ! −1 k 1 k ,k ∈ R , 0 1 0 −1 ⎡ ⎤ a b ⎣ 1 − a2 ⎦ , b = 0 −a b

Exercise 1E

" ! " 5 3 b 1a 17 10 ⎡ −1 ⎤ ⎡4⎤ ⎢ ⎥ ⎢ ⎥ 2 a ⎣ 14 ⎦ b⎣7⎦ 3 2 14 7 1 10 3ax =− ,y = b x = 4, y = 1.5 7 7 30 2 cx =− ,y =− 7 7 d x = −2.35, y = 0.69 4 (2, −1) 5 books $12, CDs $18 "! " ! " ! 3 x 2 −3 = 6a 6 y 4 −6 " ! 2 −3 is a singular matrix, not a b 4 −6 regular matrix. c There is no unique solution for this system, but a solution can be found. d The solution set contains an infinite number of pairs. 7 p = 2, q = 4, r = −1, s = 2 !

Multiple-choice questions 1B 6A

2E 7E

3C 8A

4E 9E

Short-answer questions (technology-free) " " ! ! 1a ⎡ 2⎣

0 0 12 8 a

b

0 0 8 8

⎤

3 ⎦,a ∈ R 2− a 4 3 a AB does not exist, AC, CD, BE exist. " !

1 1 2 b DA = 14 0 , A−1 = 7 3 −1

5C 10 D

⎡ " −2 2 0 , C−1 = ⎣ 3 4 AB = 2 −2 2 " ! −1 2 5 −3 5 ⎡1 ⎡ ⎤ 4 0 0 ⎢2 ⎢ ⎥ −1 ⎢ ⎢ 2 ⎥ 0 6A =⎢ ⎣0 4 0⎦,A = ⎢ ⎢ ⎣ 0 0 4 0 78 " ! ! 1 3 −5 8a i ii 18 5 8 " ! 1 3 1 iii 7 −1 2 b x = 2, y = 1 !

⎤ 1 1⎦ − 2 ⎤ 0

0

⎥ 1⎥ ⎥ 0 ⎥ 2⎥ ⎦ 1 0 2 " −18 19

Answers

⎡ 1 ⎢ 6 ⎣ a11

Extended-response questions " " ! !

1 2 −4 −6 ⎡ ⎤ " ! −7 −2 12 −1 ⎦ iii iv ⎣ 2 17 −6 1 7 " " ! ! 1 4 1 11 −1 b i ii −18 −9 13 1 −3 " " ! ! 1 1 13 −2 7 5 iii iv 13 −13 −7 13 22 −1 ⎤ ⎤ ⎡ ⎡ −2 6 6 −8 2 11 b ⎣ 3 −3 3 ⎦ 2 a ⎣ −5 −3 −1 ⎦ 15 12 3 14 18 7 ⎤ ⎡ −3 3 −3 4⎦ c ⎣ 12 −6 14 9 2 ⎡ 50 2 −2 ⎤ ⎤ ⎡ ⎢ 33 11 11 ⎥ 0 −33 0 ⎢ ⎥ 1 ⎣ ⎢ −7 5 −5 ⎥ d⎢ −18 70 10⎦ e ⎥ ⎢ 33 11 11 ⎥ 33 −6 5 29 ⎣ 1 ⎦ 4 7 33 11 11 f A−1 CBC−1 g C−1 B "! " ! " ! 3 x 2 −3 3a i = 5 y 4 1 " " ! ! 1 1 1 3 9 ii 14, iii −4 2 7 −1 14 9 1 iv ,− is the point of intersection 7 7 of the two lines "! " ! " ! 3 x 2 1 b i = 8 y 4 2 ii 0; A is a singular matrix c lines represented by the equations are parallel 1a i

5 0 6 −2


ii


Answers

664

Essential Advanced General Mathematics ⎤ 0.2 b ⎣ 0.3 ⎦ 0.5 ⎡

"

!

79 78 80 4a 80 78 82

Exercise 2B 1 a 4.78 × 10 b 6.728 × 103 c 7.923 × 10 d 4.358 × 104 −3 e 2.3 × 10 f 5.6 × 10−7 g 1.200 034 × 10 h 5.0 × 107 i 2.3 × 1010 j 1.3 × 10−9 5 k 1.65 × 10 l 1.4567 × 10−5 −8 2 a 1.0 × 10 b 1.66 × 10−23 −5 c 5 × 10 d 1.85318 × 103 12 e 9.463 × 10 f 2.998 × 1010 3 a 75 684 000 000 000 b 270 000 000 c 0.000 000 000 000 19 262 4 a 0.000 0567 b 2625 5 a 11.8 b 4.76 × 107

c Semester 1: 79.2, Semester 2: 80.4 d Semester 1: 83.8, Semester 2: 75.2 e No aggregate is 318.6 f 3 marks ⎤ ⎡ 10 2 ! " ⎢ 8 4⎥ 70 ⎥ ⎢ 5a⎣ b 60 8 8⎦ 6 10 c Term 1: $820, Term 2: $800, Term 3: $1040, Term 4: $1020 ⎤ ⎡ ⎡ ⎤ 2 2 1 60 ⎢2 2 1⎥ ⎥ e ⎣ 55 ⎦ d⎢ ⎣3 4 2⎦ 40 3 4 2

Exercise 2C

f Term 1: $270, Term 2: $270, Term 3: $480, Term 4: $480 g Term 1: $1090, Term 2: $1070, Term 3: $1520, Term 4: $1500

8 3 d x = 63

1ax =

Chapter 2 Exercise 2A 1 a x7 fp

b a2

−7

g

k a −12 px

1 a −6

l x2 1 q 6 2n b4

4

2a5 f3 k1

c x3

d y −4

e x 12

h a −8

i y 14

j x 15

m n2

n 8x

7 2

g 12 l8

3 a 18.92 d 125 000 g 0.14

b 79.63 e 0.9 h 1.84

c 5.89 f 1.23 i 0.4

4 a a 4 b7

b 64a 4 b7

cb a 2 b5 f 128

6 9

4 7

da b

e 2a b

5 22n−4 1 1 6 7a 2

11

5

b a 20

c 26

19

3

e 25

8a

5

b 1

e a 2 b 2 c−4

5 1 a2b2

c 1

3

f a5b5

1 ab 5

e x = −0.7

−20 3 f x = 2.4 −15 ix= 92

cx=

g x = 0.3 h x = −6 21 jx =− 17 160 68 b x = 19.2 cx =− 2ax = 9 7 80 85 dx= e x = 6.75 fx =− 51 38 487 191 gx= hx= 13 91 14 18 ,y =− 3ax = 13 13 16 18 bx= ,y =− 11 11 c x = 12, y = 17 d x = 8, y = 2 e x = 0, y = 2 f x = 1, y = 6

Exercise 2D

6 63x

d26 1 a3b

oa

−2 5

r −8x s a b t 1 6 1 4 e d c 7 4 3 3 j h 16 i 27 2 2

b x = 48

1 b 2 d a 7

g a −4 b 2 c5

1 a 4(x − 2) = 60; x = 17 2x + 7 2 b = 49; x = 10.5 4 29 c x − 5 = 2(12 − x) ; x = 3 d y = 6x − 4 e Q = np 60n f R = 1.1 pS g = 2400 5 ␲ h a = (x + 3) 3 2 $2500 3 Eight dresses and three handbags 4 8.375 m by 25.125 m 5 $56.50 6 Nine 7 20, 34 and 17 8 Annie, Belinda and Cassie scored 165, 150 and 189 respectively 9 15 km/h 10 2.04 × 10−23 g



665

Answers xy c 8 a 1 f g 3 2x a x −1 j k 4b x 1 x +1 n x(x + 3) 3 2x x −2 3x(3x − 2)(x + 5)

2 a 2x y 2

1 140.625 km 2 50 3 10 000 adults and 5000 children 4 Men $220, boys $160 5 127 and 85 6 252 litres of 40% solution and 448 litres of 15% solution 7 120 and 100, 60 8 370 588 9 500 adults, 1100 students

e i m o 3a

Exercise 2F c 1 a 25 b 330 c 340.47 d 1653.48 e 612.01 f 77.95 g 2.42 h 2.1 i 9.43 j 9.54 2S v −u bl = −a 2aa= n t 2A P cb= dI =± h R 2(s − ut) 2E ea= fv =± t2 m 2 −z Q hx= gh= y 2g −b(c + 1) −b(c + y) jx= ix= m−c a−c 5(F − 32) ◦ ; 57.22◦ C 3 a 82.4 F b C = 9 S 4 a 1080 bn= +2:9 180 3 5 a 115.45 cm b 12.53 cm c 5.00 cm 6 a 66.5 b4 c 11

e g i k

b

3 x −3 5x − 1 2 x + x − 12 2x − 9 x 2 − 10x + 25 1 3−x 5x 2 − 3x x2 − 9 12 (x − 6)3

3−x 4a √ 1−x 5 c√ x +4 12x 2 e −√ x +4 6x − 4

5a

(6x −

2 3) 3

b d f h j l b d f

b

2 x x −1 x +4 2x x +2

d

x y2

hx +2 l

x −1 4x

4x − 14 x 2 − 7x + 12 2x 2 + 10x − 6 x 2 + x − 12 5x − 8 (x − 4)2 23 − 3x x 2 + x − 12 11 − 2x x 2 − 10x + 25 9x − 25 2 x − 7x + 12 √ 2 x −4+6 √ 3 x −4 x +7 √ x +4 9x 2 (x + 2) √ 2 x +3

3 (2x +

2 3) 3

c

Answers

Exercise 2E

3 − 3x 2

(x − 3) 3

Exercise 2H Exercise 2G 13x 6 5x − 2y d 12

1a

g

5a 4 3y + 2x e xy

b

5x − 1 (x − 2)(x + 1)

4x + 7 i (x + 1)2 4(x 2 + 1) k 5x m

3x + 14 (x − 1)(x + 4)

h

−h 8 7x − 2 f x(x − 1)

c

−7x 2 − 36x + 27 2(x + 3)(x − 3)

5a 2 + 8a − 16 j 8a 2x + 5 l (x + 4)2 n

x + 14 (x − 2)(x + 2)

7x 2 + 22x + 28 (x − 2)(x + 2)(x + 3) 4x + 3 (x − y)2 − 1 q p x −1 x−y

cx ex gx ix kx mx 4ax bx

o

r

3 − 2x x −2

m−n b bx= a b−a bc 5 =− dx= a p−q m+n ab = fx= n−m 1−b = 3a h x = −mn p−q a 2 − b2 jx= = p+q 2ab 1 3ab lx = = 3a − b b−a p2 + p2 t + t 2 5a = nx =− q( p + t) 3 d − bc c − ad = ,y = 1 − ab 1 − ab ab a 2 + ab + b2 ,y = = a+b a+b t +s t −s = ,y = 2a 2b = a + b, y = a − b

1ax =

cx dx



Answers

666

Essential Advanced General Mathematics e x = c, y = −a f x = a + 1, y = a − 1 2a 2 1−a a ds = (a − 1)2 3a fs = a+2 5a 2 hs = 2 a +6 bs =

5 a s = a(2a + 1) cs =

a2 + a + 1 a(a + 1)

e s = 3a 3 (3a + 1) g s = 2a 2 − 1 +

1 a2

Exercise 2I 1ax =a−b bx =7 √ a ± a 2 + 4ab − 4b2 cx =− 2 a+c dx= 2 2 a (x − 1)(x + 1)(y − 1)(y + 1) b (x − 1)(x + 1)(x + 2) c (a 2 − 12b)(a 2 + 4b) d (a − c)(a − 2b + c) a+b+c a+b 3ax = y= a+b c −(a − b − c) a−b+c bx= y= a+b−c a+b−c

Multiple-choice questions 1A 6E

2A 7B

3C 8B

4A 9B

5B 10 B

Short-answer questions (technology-free) 11 1 a x 12 b y −9 8 2 3.84 × 10 2x + y 3a 10 7x − 1 c (x + 2)(x − 1) 13x 2 + 2x + 40 e 2(x + 4)(x − 2)

c −15x

2

d x −1

4y − 7x xy 7x + 20 d (x + 2)(x + 4) 3(x − 4) f (x − 2)2 b

x −4 x2 − 4 c d 4x 2 4x 3 31 days 5 106 seconds or 11 54 6 50 7 12 8 88 classical, 80 blues and 252 heavy metal 9 a 300␲ cm3 V 117 bh= ; cm 2 5␲ ␲r √ 8 2 V 128 cr = ; cm = √ cm ␲h ␲ ␲ a+b b bx= 10 a x = c a+y 4a

2 x

b

ab + b2 d − d 2 d(a + b) x+y b x(y − x) 2 d c (x − 2)(2x − 1) a 12 A, B and C are 36, 12 and 2 years old respectively. 13 a a = 8, b = 18 b x = p + q, y = 2q 14 x = 3.5 40cx 2 b 15 a 4n 2 k 2 ab2 16 x = −1

2ab b−a p2 + q 2 11 a 2 p − q2 cx=

dx=

Extended-response questions 5x 4x 19x hours b hours hours c 4 7 28 14 d i x= ≈ 0.737 19 140 km ≈ 7 km ii distance from Jack = 19 560 km ≈ 29 km distance from Benny = 19 3 2 a 18 000 cm per minute b V = 18 000t 45t d after ch= 4␲ 3a cards; Sally, 3 a Thomas, a cards; George, 2 a 5a a − 18 cards; Zeb, cards; Henry, cards 3 6 3a a 5a b + a − 18 + = a + +6 2 3 6 c a = 24; Thomas 24 cards, Henry 20 cards, George 36 cards, Sally 6 cards, Zeb 8 cards Fr 2 1011 b m1 = 4 a 1.9 × 10−8 N 6.67m 2 c 9.8 × 1024 kg 5 a V = 1.8 × 107 d b 5.4 × 108 m3 c k = 9.81 × 103 d 1.325 × 1015 J e 1202 days (to the nearest day) √ 240 10 3 7 −40◦ km/h 8 6 11 3 9 a 20 − r r3 b i V = 20r 2 − ␲ 3 ii r = 5.94 and h = 14.06 2 1 10 a litre from A, litre from B 3 3 b 600 mL from A; 400 mL from B ( p − q)(n + m) c litres from A and 2(np − qm) (n − m)( p + q) n q litres from B and = . 2(np − qm) m p n q Also or ≥ 1 and the other ≤ 1. m p 11 a h = 2(10 − r ) b V = 2␲r 2 (10 − r ) c r = 3.4985 and h = 13 or r = 9.022 and h = 1.955 1a



Answers 7

Exercise 3A 1

a {4} dØ

b {1, 3, 5} eØ

c {1, 2, 3, 4, 5} or ␰

a {5, 7, 8, 9, 10, 11} b {1, 3, 5, 7, 9, 11} c {1, 3, 5, 7, 8, 9, 10, 11} d {5, 7, 9, 11} e {1, 2, 3, 4, 6, 8, 10, 12} f {5, 7, 9, 11} 8a ξ A B A'

Answers

Chapter 3

667

2 b ξ

A

B

B'

a {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16} b {1, 3, 5, 7, 9, 11, 13, 15} c {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16} d {1, 5, 7, 11, 13} e {1, 5, 7, 11, 13} 3ξ

1 B

2 6

4 8 A 10 12

c

ξ

A

B

A' B'

3 5 7 9

d

11

ξ A

B

A' B'

a {1, 2, 3, 5, 6, 7, 9, 10, 11} b {1, 3, 5, 7, 9, 11} c {2, 4, 6, 8, 10, 12} d {1, 3, 5, 7, 9, 11} e {1, 3, 5, 7, 9, 11}

e

4

ξ

A

B

AB

a {10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25} b {11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24} c {10, 12, 15, 16, 20, 24, 25} d {11, 13, 14, 17, 18, 19, 21, 22, 23} e {11, 13, 14, 17, 18, 19, 21, 22, 23} 5

a {R} b {G, R} c {L, E, A, N} d {A, N, G, E, L} e {R} f {G, R}

f

ξ

A

B

(A B)'

9

a {E, H, M, S} c {A, T} e {C, E, H, I, M, S}

b {C, H, I, M} d {H, M} f {H, M}

6

Exercise 3B a { p, q, u, v} c { p} e {q, r, s, t, u, v, w}

b { p, r, w} d { p, q, r, u, v, w} f { p}

1 a Yes b Yes c Yes 2 a No b No c No 9 2 4 2 3 3 f e d c b 3a 20 9 11 7 25 11 ˙ 4 a 0.28571 4˙ b 0.4˙ 5˙ c 0.35 ˙ ˙ ˙ d 0.307692 e 0.058823529411764 7˙



Answers

668


Exercise 3C √ √ √ √ 1a2 2 b2 3 c3 3 d5 2 √ √ √ √ e3 5 f 11 10 g 7 2 h6 3 √ √ i5 j5 3 k 16 2 √ √ √ b6 √ 3 c4 7 2 a 3√2 d5 √ 10 √ e 28 2 √f0 3 a 11 3 + 14 b5 7 √ √ c0 d 2+ 3 √ √ √ √ e 5 2 + 15 3 f 2+ 5 √ √ √ √ 5 7 − 2 2 3 b 4a c d 2 3 √5 √7 √ √ 6 2 g 2−1 h2+ 3 e f 2 √ 4 √ √ √ 5+ 3 4 + 10 j 6−2 k i 2 √ √6 √ m3+2 2 l 3( 6 + 5) √ √ √ b 9√ +4 5 c −1 + 2 5a6+4 2 √ √ 2 3 8+5 3 d4−2 3 e f 9 √ 11 √ 3+ 5 6+5 2 g h 2 √ 14 1 6 a 4a − 4 a + b 3 + 2x + 2 (x + 1)(x + 2) √ √ b7−2 6 7a5−3 2

Exercise 3D 1 a 25 × 3 × 5 × 11 × 13 b 25 × 3 × 7 × 11 × 13 c 25 × 7 × 11 × 13 d 25 × 7 × 11 × 13 × 17 2a1 b 27 c5 d 31

Exercise 3E 1a

e6

7ax =5 8aξ X

b 16 22 35

23 34

26 29

31

c0 21 3

28 32 25

36 1

Z

12

24 27 33 30

4 49

Y 6

15 18 39 9 16

b i X ∩ Y ∩ Z = {36} ii n(X ∩ Y ) = 5 9 31 students; 15 black, 12 green, 20 red 10 n(M ∩ F) = 11 11 1 12 x = 6, 16 proficient in Italian 13 102

Multiple-choice questions 1A 6D

2D 7B

3D 8B

4D 9C

5C 10 A

Short-answer questions (technology-free) 1 5 7 c b 200 11 90 6 4 81 f e d 35 15 200 2 7 2 23 × √3 ×√ √ √ 2 6− 2 b4 5+9 c2 6+5 3a 2 √ 4 −23 − 12 3 √ √ a − a 2 − b2 b 5a2 6+6 b 6 a 15 b 15 7a1 b 22 c 22 √ 85 9 2 cm2 10 −15 7 √ √ √ 51 3 11 ±2 12 5 − 6 13 5 14 a√ 57 b3 c 32 15 2 2 + 3 1a

Extended-response questions √ √ b i 19 2aξ

ii 9

iii 23

A

B

14 3

5 6

9

2 4 7

C

b i 23 3 x = 20 5a5 6 45

ii 37 iii 9 4x =7 b 10

1 c i 11 + 3 √ √ √ √ ii 2 2 − 7 or 7 − 2 2 √ √ √ √ iii 3 3 − 2 6 or 2 6 − 3 3 2 a a = 6 and b = 5 b p = 26, q = 16 2 c a = −1, b = 3 √ √ 12 3 − 19 d i ii 3 ± 3 71 √ 1± 3 iii 2 √ e Q = {a + 0 3; a ∈ Q}



669

Answers ck =3 x y dk =

4 6

9 9

49 21

900 90

2 5 1 32 1 5

x y

1

32

1024

2 5

4 5

8 5

Answers

3 d ±2 √ 5 a b = −4, c = 1 b2+ 3 6 a (20, 21, 29) 7 a i 4 factors ii 8 factors b n + 1 factors c i 32 factors ii (n + 1)(m + 1) d (␣1 + 1)(␣2 + 1) . . . (␣n + 1) e 24 8 a 1080 = 23 × 33 × 5, 25 200 = 24 × 32 × 52 × 7 b 75 600 d i 3476, 3474, 3472, 3470 ii 1738, 1737, 1736, 1735 9 a i region 8 ii male, red hair, blue eyes iii male, not red hair, blue eyes bi 5 ii 182 10 a i students shorter than or equal to 180 cm ii students who are female or taller than 180 cm iii students who are male and shorter than or equal to 180 cm b A B

2 a V = 262.144 b r ≈ 2.924 3 a a ≈ 1.058 b b ≈ 5.196 4 a 72 cm2 b 20 cm 648 5a cm b 1412.5 g 113 6 10.125 kg 7 62.035 cm 8 1.898 s 9 a 8.616 km b 14.221 km 10 a i 300% increase ii 41% increase iii 700% increase b i 75% decrease ii 29% decrease iii 87.5% decrease c i 36% decrease ii 11% decrease iii 48.8% decrease d i 96% increase ii 18% increase iii 174.4% increase

(A ∪ B)' = A' ∩ B' is shaded

11 a

A

Exercise 4B

B 35

65

1ak =2

205 x

x

35

c 65

bk =

d0

Exercise 4A

2 8

4 32

6 72

8 128 dk =

1 bk = 3

y

6 1 3

32 1 16

y

1

x

1 4

1

4

9

y

1

1 2

1 4

1 6

x

1

y

3

2 3 4

3 1 3

6 1 12

1

27

125

1 3

1 9

1 15

ck =3

1ak =2

x

4 1 2

1 2

Chapter 4

x y

2

C

n(A ∩ C ) = 0

b 160

x

1 2 1 6

1 1 3

3 2 1 2

1 3 x

2 2 3

y


1 8 2 3


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Essential Advanced General Mathematics 1 b b ≈ 5.657 2 √ 3 a a = 0.3125 bb= 2 2 4 2.85 kg/cm 5 a 2.4 amperes b 25% 6 64 candela 7 5.15 cm 8 a i 75% decrease ii 29% decrease iii 87.5% decrease b i 300% increase ii 41% increase iii 700% increase c i 56.25% increase ii 12% increase iii 95% increase d i 49% decrease ii 15% decrease iii 64% decrease

2aa=

2k =

x z y

6 4 2 4 8 12 16 y=

2.4x2

0

1 2 3 4 5 y = 1.5√x

0

5 10 15 20 25 y=

2

3

5

z

10

4

50

10 400 3

y

15 2

4 3

6

4

√x

b $5000

2A 7E

3B 8D

4C 9D

5B 10 B


1 x2

1 a When b = 4, a = 6;

2 x2

1√ 8ay= x 4 c y = 3.5x 0.4

b y = 2x

5 4 2

d y = 10x 3

5 2x − 2

ey= f y = 3.2x −0.4 9 a a = 100, b = 0.2 b 158.49 10 a a = 1500, b = −0.5 b 474.34

Exercise 4D 1k =5 x z y

x

1 $33.40 2 a overhead charge = $250, cost per guest = $47.50 3 p = 20.5 4 $55.11 5 a 330 m b 67.5 m 6 45 minutes

1C 6D

50 40 30 20 10

10 3 15


y

7

1 50 25

Exercise 4E

8

x2

4 8 16

4 a ≈ 1.449 5 z ≈ 0.397 6 $174 7 360 joules 8 a 500% increase b 78% decrease 9 a 41% increase b 33% increase 10 a 183% increase b 466% increase 11 a The tensions are the same. b The work done by the second spring is 90% that of the first.

1 a direct b direct square c inverse d direct square root e inverse square 2 b, e 3 a, b, e 3 10 2 by= cy= x 4 a y = 3x x 3 √ 1 ey= √ f y = 6x 3 dy=2 x 3 x 5 y 6 y

0

2 10 10

3k =3

Exercise 4C

40 30 20 10

1 2

2 10 1

4 2 10

6 60 0.5

10 12.5 4

8 when a = 8, b = ± √ 3 30 b When x = 27, y = 1 ; 23 1 1 when y = , x = 8 256 000 1 16 4 c When x = , y = ; when y = , 2 3 27 x = ±3 1 d 6 2 a d = 4.91t 2 b 491 m c 2 s, correct to one decimal place. √ 3 a 14 m/s b 40 m c v and s 4 2.4 hours 5 a y is halved b x is halved c y is doubled d x is doubled



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Answers 7 $35

10 34% increase

1 a 0.24 kg b 11 cm 2 a h = 0.0003 375 n2 b 17.1 m c 218 revs/min 3 13 knots 121.8 b 9.6 kg/cm2 4av = P 3000 5aw = b 600 kg c 333 kg d 144 6av = p bi v =2 ii p = 48 c y 12 9 8

7 44.8 min

1 1 1 18 16 12

8 $76

1 p

9 Sn =

ii tn = 2tn−1 , t1 = 1

1 n2 (n − 1)2 ii tn = tn−1 , t1 = 1 n2 n−1 d i tn = 3(−2) ii tn = −2tn−1 , t1 = 3 e i tn = 3n + 1 ii tn = tn−1 + 3, t1 = 4 f i tn = 5n − 1 ii tn = tn−1 + 5, t1 = 4 4 tn+1 = 3n + 4, t2n = 6n + 1 5 a t1 = 15, tn = tn−1 + 3 b tn = 12 + 3n c t13 = 51 6 a t1 = 94.3, tn = 0.96tn−1 b tn = 94.3(0.96)n−1 c t9 = 68.03 7 a tn = 1.8tn−1 + 20, t0 = 100 b t1 = 200, t2 = 380, t3 = 704, t4 = 1287, t5 = 2336 8 a $2120 at end of 1st year, $2671.20 at end of 2nd year, $3255.47 at end of 3rd year b tn = 1.06(tn−1 + 400), t1 = 2120 c $8454.02 9 a 1, 4, 7, 10, 13, 16 b 3, 1, −1, −3, −5, −7 1 c , 1, 2, 4, 8, 16 2 d 32, 16, 8, 4, 2, 1 e 1.1, 1.21, 1.4641, 2.144, 4.595, 21.114 16 32 f 27, 18, 12, 8, , 3 9 g −1, 3, 11, 27, 59, 123 h −3, 7, −3, 7, −3, 7 10 a t1 = 1, t2 = 2, t3 = 4 b u 1 = 1, u 2 = 2, u 3 = 4 c t1 = u 1 , t2 = u 2 , t3 = u 3 d t4 = 8, u 4 = 7 11 S1 = a + b, S2 = 4a + 2b, S3 = 9a + 3b, Sn+1 − Sn = 2an + a + b √ 3 17 577 12 t2 = , t3 = , t4 = . The number is 2. 2 12 408 13 t3 = 2, t4 = 3, t5 = 5 c i tn =


0

b i tn = 2n−1

8 18 amps

1 n(n + 1) 2

10 a P = 3498.544 × N 0.5 b 25 956 c 51 023 3600 b T = 0.14d 2 11 a t = d2 c 23.9 mL d 6.3 min e 9 min; 56 min 12 a i T = 0.000 539 × R 1.501 ii Mars 1.87; Jupiter 11.86; Saturn 29.45; Uranus 84.09 Neptune 165.05; Pluto 248.16 b 2.540 × 109 km 13 a a = 11.7, b = 0.41 b 77 c k = 163, p = −1.167 d7

Chapter 5 Exercise 5A 1 a 3, 7, 11, 15, 19 b 5, 19, 61, 187, 565 c 1, 5, 25, 125, 625 d −1, 1, 3, 5, 7 e 1, 3, 7, 17, 41 1 1 1 2 a t1 = 1, t2 = , t3 = , t4 = 2 3 4 b t1 = 2, t2 = 5, t3 = 10, t4 = 17 c t1 = 2, t2 = 4, t3 = 6, t4 = 8 d t1 = 2, t2 = 4, t3 = 8, t4 = 16 e t1 = 5, t2 = 8, t3 = 11, t4 = 14 f t1 = −1, t2 = 8, t3 = −27, t4 = 64 g t1 = 3, t2 = 5, t3 = 7, t4 = 9 h t1 = 2, t2 = 6, t3 = 18, t4 = 54 3 a i tn = 3n ii tn = tn−1 + 3, t1 = 3

Answers

6 4.05 cents 1 9 I2 = I1 4

671

Exercise 5B 1 a t1 = 0, t2 = 2, t3 = 4, t4 = 6 b t1 = −3, √ t2 = 2, t3 = √7, t4 = 12 √ c t1 = − √ 5, t2 = −2 5, t3 = −3 5, t4 = −4 5 d t1 = 11, t2 = 9, t3 = 7, t4 = 5 2 a a = 3, d = 4, tn = 4n − 1 b a = 3, d = −4, tn = 7 − 4n 1 5 c a = − , d = 2, tn = 2n − 2√ 2 √ d a = 5√− 5, d = √5, tn = 5n + 5 − 2 5 √ 3 a −31 b 24 c5 d6 3



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Essential Advanced General Mathematics 4 a 14 b 22 c 4th day 5 a 70 b 94 c Row F 6 a 22 b −28 c 20 d 56 7 8 117 9 tn = 156n − 450 7 2 10 54 11 −2 62 88 38 140 166 64 218 244 , , , , , , , 12 9 9 3 9 9 3 9 9 a(n − 1) 13 7, 9, 11, 13 14 tn = a − m−1 √ 15 27 3 − 60 √ 2 2 b 16 a 11.5 7 17 16 18 5 20 3

Exercise 5C √ 1 a 426 b 55 c 60 2 d 108 2 112 3 11 4 680 5 2450 6 a 16.5 km b 45 km c 189 km 7 a 10 days b 25 per day 8 {n : n = 9} 9 20 10 a 86 b 2600 c 224 d 2376 e 5 extra rows 11 $176 400 12 a = −15, d = 3, t6 = 0, S6 = −45 13 2160 14 0 15 266 √ √ 46 5 11 5 − 2 5n b tn = 16 a tn = n + 5 4 4 n 17 a b b (b + bn) 2 18 t5 = −10, S25 = −1250 19 1575d 20 a Sn−1 = 23n − 3n 2 − 20 b tn = 20 − 6n c a = 14, d = −6 21 7, 12, 17

Exercise 5D 1 a 3, 6, 12, 24 b 3, −6, 12, −24 c 10 000, 1000, 100, 10 d 3, 9, 27, 81 1 5 c 32 d a x+5 b 2a 256 567 2 n−1 3 a tn = 3 b tn = 2(−2)n−1 3 √ c tn = 2( 5)n−1 2 5 t9 4± 5 6a6 b9 c9 d6 e8 7 a $5397.31 b 48th year 8 a 21 870 m2 b 10th day 9 47.46 cm 10 a 57.4 km b 14th day 11 $5 369 000

12 a 24 b 12 288 √ 2 13 5 14 16 2 3 16 t6 = 729 15 t10 = 2048 17 a $7092.60 b 12 years 18 $3005.61 19 11.6% p.a. 20 5 weeks 21 a 60 b 2.5 c1 d x 4 y7 22 3

Exercise 5E 57 64 2 a 1094 b −684 c 7812 3 a 1062.9 mL b 5692.27 mL 4 a $18 232.59 b $82 884.47 5 a 49 minutes (to nearest minute) b 164 minutes c Friday 481 835 2882 6 = 73 6561 6561 7 Bianca’s is worth $3247.32, Andrew’s is worth √ $3000 15 2 8 a 155 b + 15 2 9a8 b {n : n > 19} x 2m+2 + 1 10 x2 + 1 1 a 5115

b −182

c−

Exercise 5F 1a

5 4

b

3 5

n−1 2 Perimeter of nth triangle = 1 p, 2 √ p2 3 Area of nth triangle = , 4n Sum to infinity of perimeters = √ 2 p, p2 3 Sum to infinity of areas = 3 1 3 3333 4 Yes 3 5 yes, as the number of hours approaches infinity, but problem becomes unrealistic after 4 to 5 hours 1 8 12 m 9 75 m 7 6 S∞ = 8 2 4 1 31 b c 10 a 9 30 3 37 7 f e1 d 9 198 1 11 r = , t1 = 16, t2 = 8 or 2 1 r = − , t1 = 48, t2 = −24 2 5 2 12 13 8 3



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1 a i 3, 2.75, 2.688, 2.672, 2.668, 2.667 ii 1, −2, 1, −2, 1, −2 iii 2, 13, 508, 774193, 1.8 × 1012 , 9.7 × 1024 iv 3, 3.236, 3.288, 3.300, 3.302, 3.303 b i and iv converge 2 a 0.6736 b 0.6823 c −0.7913 d 1.3532 e 0.3099 f 2.3758 g 0.5222 h 0.3473

iii n = 22 T − 4000 +1 iv m = 250 v p = 51 b i Sn = 37 500(1.08n − 1) ii Q B − Q A = 37500(1.08n − 1) − 3875n − 125n 2 ; n = 18 14 2.828 43

Answers

Exercise 5G

673

Chapter 6 Exercise 6A

Multiple-choice questions 1D 6D

2B 7E

3A 8C

4A 9E

5B 10 D

Short-answer questions (technology-free) 1 a 3, −1, −5, −9, −13, −17 b 5, 12, 26, 54, 110, 222 2 a 2, 4, 6, 8, 10, 12 b −1, −4, −7, −10, −13, −16 3 a $5250, $6037.50 b t1 = 5250, tn = 1.05(tn−1 + 500) 4 147 5 −0.1 6 −258.75 7 {n : n = 12} 81 9 1000 × 1.035n 8 8 10 t2 = 6, t4 = or t2 = −6, t4 = − 3 3 3 11 96 12 −9840 13 4 14 x = 8 or x = −2

Extended-response questions 1 a 0.8, 1.5, 2.2, . . . b Yes c 8.5 m 2 a Yes b tn = 25n + 25 c 650 1 3 The fifth pole is 22 km from town A and 7 6 9 km from town B. 7 4 a 20, 36, 52, 68, 84, 100, 116, 132, . . . b Tn = 16n + 4 c Yes, size 12 5 a Dn = 7n − 5 b 27 6 472 mm 7 520 8 a 99.9999 mg b 100 mg 1 m 9a 729 b 1.499 m. No, the maximum height the water will reach is 1.5 m. 10 a 27.49 b 1680.8 1 b 405 m 11 a 7 m 9 64 12 2 − 1 = 1.845 × 1019 13 a i tn = 3750 + 250n ii Sn = 3875n + 125n 2

1 a = 10, b = 0, c = −7 2 a = 1, b = −2 3 a = 2, b = −1, c = 7 4 a = 2, b = 1, c = 3 5 (x + 2)2 − 4(x + 2) + 4 6 (x + 1)3 − 3(x + 1)2 + 3(x + 1) − 1 7 a = 1, b = −2, c = −1 8 a It is impossible to find a, b and c to satisfy a = 3, 3ab = −9, 3ab2 = 8 and ab3 + c = 2 b a = 3, b = −1, c = 5 9 a = 1, b = −6, c = 7, d = −1 5 10 a a = − b and a = −3b are not possible 3 unless a and b are both zero, but a + b = 1 so no constants can be found. b (n + 1)(n + 2) − 3(n + 1) + 1 11 a ax 2 + 2abx + ab2 + c b 2 b2 ba x+ +c− 2a 4a −1 −1 13 a = −3, b = , c = 3 or a = , 3 3 b = −3, c = 3 14 a = 3, b = −3, c = 1 15 If c = 5, a = 1, b = −5; if c = −27, a = −3, b = 3

Exercise 6B 1ax =1 cx ex 2ax


bx cx dx

bx =3 √ √ 30 2 =1± dx =1± 5√ 2√ 3 2 −13 ± 145 = −1 ± fx= 2 12 √ −1 1 ± 32t + 1 ,t ≥ = 4 32 √ −1 ± t + 3 , t ≥ −3 = 2√ 46 −2 ± 5t − 46 ,t ≥ = 5 5 5t(t − 2) , t < 0 or t ≥ 2 = −2 ± t


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Essential Advanced General Mathematics √ −3(1 ± 5) 2 − p ± p 2 + 64 bi x = 2 ii When p = 0, x = 4 and when p = 6, x = 2 4x =2 18 5a b x = −6, 3 x(x + 3) 6 17 and 19 600 km/h, 7 a Average speed of car = x average speed of plane 600 = + 220 km/h x b Average speed of car = 80 km/h, average speed of plane = 300 km/h 8 x = 20 9 6 km/h 10 a x = 50 b 72 minutes 11 Average speed of slow train = 30 km/h, average speed of express train = 50 km/h 12 60 km/h 13 Smaller pipe will take 25 minutes and the larger pipe, 20 minutes 14 Each pipe running alone will fill the tank in 14 minutes 15 Average speed by rail = 43 km/h, average speed by sea = 18 km/h 16 22 km 17 10 litres 3ax =

Exercise 6C 2 3 2 1 + − b x −1 x +2 x + 1 2x + 1 2 1 3 1 c + + d x +2 x −2 x +3 x −2 3 8 e − 5(x − 4) 5(x + 1) 2 9 + 2a x − 3 (x − 3)2 2 3 4 + + b 1 + 2x 1−x (1 − x)2 4 2 −4 + + c 9(x + 1) 9(x − 2) 3(x − 2)2 −2 2x + 3 + 2 3a x +1 x +x +1 x +1 2 1 x −2 b 2 + − c 2 x +2 x +1 x + 1 2(x + 3) 3 2 43+ + x −1 x −2 5 It is impossible to find A and C to satisfy A = 0, C − 2A = 0 and A + C = 10. 1 1 − 6a 2(x − 1) 2(x + 1) 2 3 b + 5(x − 2) 5(x + 3) 1a


2 1 + x −2 x +5 1 2 − d 5(2x − 1) 5(x + 2) 1 3 − e 3x − 2 2x + 1 2 2 3−x 1 − f g + 2 x −1 x x x +1 x 2 1 1 h + 2 − i x x +4 4(x − 4) 4x 3 7 1 1 − j kx+ − 4(x − 4) 4x x x −1 1 3 l −x − 1 − − x 2−x x −4 2 + m 3(x + 1) 3(x 2 + 2) 2 1 1 n + − 3(x − 2) 3(x + 1) (x + 1)2 1 2 5 8 o + 2 − p x x +4 2x + 3 x +2 26 1 1 q + − 9(x + 2) 9(x − 1) 3(x − 1)2 8 4 16 − + r 9(2x + 1) 9(x − 1) 3(x − 1)2 3 1 + sx −2+ 4(x + 2) 4(x − 2) 1 2 tx− + x +1 x −1 7 3 − u x + 1 3x + 2 c

Exercise 6D 1 a (1, 1) (0, 0) c

3+

√

13

2 3−

1 1 , 2 2

√ , 4 + 13

√ 2

b (0, 0)

13

√ , 4 − 13

2 a (13, 3) (3, 13) b (10, 5) (5, 10) c (−8, −11)(11, 8) d (9, 4) (4, 9) e (9, 5)(−5, −9) 3 a (11, 17) (17, 11) b (37, 14) (14, 37) c (14, 9)(−9, −14) 4 (0, 0) (2, 4) √ √ √ √ 5+ 5 5+ 5 5− 5 5− 5 5 , , 2 2 2 2 15 5 6 , (3, 1) 2 2 √ √ −130 + 80 2 60 + 64 2 7 , 41 41 √ √ −130 − 80 2 60 − 64 2 and , 41 41


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Answers 1+

√

2 3 − x − 1 (x − 1)2 6 g x2 + 2 x +3 h x2 + x + 1 3 i x +4 16 j − 7(x + 4) x − 10 8a x2 + x + 2 x −2 b − 4(x 2 − x + 2) 1 64 − c 3x + 15 + x −4 x −1 9 a (0, 0), (−1, 1) b (0, 4), (4, 0) c (1, 4), (4, 1) 10 (−4, −1), (2, 1) f

Multiple-choice questions 1C 6E

2D 7C

3D 8D

4C 9B

5E 10 B

Short-answer questions (technology-free) 1 a = 3, b = 2, c = 1 2 (x − 1)3 + 3(x − 1)2 + 3(x − 1) + 1 5 a x = −4 or 3 b x = −1 or 2 √ 2± 2 c x = −2 or 5 dx= 2 √ √ 1 ± 3t − 14 t ± t 2 − 16t ex= f 3 2t √ −3 ± 73 6x= 2 −1 2 4 3 − + b 7a x −3 x +2 x +2 x −2 1 3 c − 2(x − 3) 2(x + 5) 1 2 d + x −5 x +1 13 13 10 e − − x +2 x + 3 (x + 2)2

b

a 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

b 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

x 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

a 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59

b 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

x a 43 84 44 85 45 86 46 87 47 88 48 89 49 90 50 91 51 92 52 93 53 94 54 95 55 96 56 97 57 98 58 99 59 100

b x 83 84 84 85 85 86 86 87 87 88 88 89 89 90 90 91 91 92 92 93 93 94 94 95 95 96 96 97 97 98 98 99 99 100

4 + x +4 1 − x +1 1 − x −1 1 − 3−x 2 7(x − 3) 1 − x −3 1 4(x + 1)

Answers

√ 21 −1 − 21 , 8 2 2 √ √ 1 − 21 −1 + 21 and , 2 2 √ √ −6 5 3 5 4 10 , 9 ,2 5 5 9 1 11 −2, 12 (0, −1), (3, 2) 2

675

Extended-response questions a(a + 480) , a > 0. When 2 a = 60, speed = 120 km/h, a very fast constant speed for a train. If we choose this as an uppermost value for the speed, 0 < a < 60 and 0 < speed < 120

1 a 24 km/h

b Speed =

a+

a 1 8 14 22 34 43 56 77 118 speed 16 20 24 30 40 48 60 80 120 √ b + b2 + 4a m 2a 2 c

a 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84

b 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40


x 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

a 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99

b 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

20

1

x a 19 45 20 50 21 55 22 60 23 65 24 70 25 75 26 80 27 85 28 90 29 95 30 100 31 32 42 33 48 54 5 60

b 4 5 6 7 8 9 10 11 12 13 14 15

x 9 10 11 12 13 14 15 16 17 18 19 20

1 2 3 4

7 8 9 10


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Essential Advanced General Mathematics a 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Answers

676

CUAU033-EVANS

3a

a+

b 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41

x 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

a 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83

b 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82

x 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83

a

b

x

6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

√ a 2 + 4abc 2ac

√ 4 a Smaller pipe will take (b + b2 − ab) minutes to fill the tank, √ larger pipe will take (b − a + b2 − ab) minutes to fill the tank b Smaller pipe will take 48 minutes to fill the tank, larger pipe will take 24 minutes to fill the tank a b

3 4

8 9

15 16

b 41 42 43 44 45 46 47 48

x 43 44 45 46 47 48 49 50

12 15 18 21 24 27 30 33 36 39 42 45 48 51 54

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

24 25

35 36

7.1 Multiple-choice questions 2B 7A 12 C 17 B 22 C 27 B 32 A 37 A

3E 8B 13 D 18 D 23 C 28 B 33 B 38 A

b 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

x 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

30 35 40

1 2 3

6 7 8

4A 9D 14 B 19 C 24 A 29 B 34 D 39 C

42 E 47 D

a b 66 5 72 6 78 7 84 8 90 9 96 10

x 11 12 13 14 15 16

56 63 70 77 84 91 98

1 2 3 4 5 6 7

8 9 10 11 12 13 14

72 80 88 96

1 2 3 4

9 10 11 12

90 99

1 2

10 11

43 A 48 C

44 C 49 A

45 C

7.2 Extended-response questions

Chapter 7 1A 6C 11 B 16 C 21 A 26 D 31 D 36 B

a 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100

41 A 46 E

4 b a = 3, b = 1, c = etc. 3

c

a 86 88 90 92 94 96 98 100

5B 10 E 15 A 20 D 25 B 30 A 35 A 40 C


1 a 8 cm b 7.7 cm c 6 cm d 15 cm 2 a i 178 ii 179 iii 179.5 iv 179.95 b i 180 ii circle 360 dn= e square c 20 180 − A 2 3 3 a Volume of hemisphere = ␲t , 3 Volume of cylinder = ␲t 2 s 1 Volume of cone = ␲t 2 w 3 b i 6:2:3 ii 54␲ cubic units 4 a a = −0.4, b = 148 b C ($) 148

0

c $68 d 248 5 a i OC1 = R − r1

(300, 28) n

ii r1 =

R 3


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Answers R 9 1 R ii rn = n c i r= 3 3 R ␲ R2 iii S∞ = iv S∞ = 2 8 6 a a = 6000, b = −15 000 b $57 000 c 2006 7 a i 80n + 920 ii A: 2840 tonnes, B: 2465 tonnes iii 40n(n + 24) iv A: 46 080 tonnes, B: 39 083 tonnes b April 2006 8a4 b6 c8 d2 e i 10 ii Pn = Pn−1 + 2 iii Pn = 2n + 2 1 iv 1 2

1

1

ii r2 =

1 2

1 2

1 2

1 2

1 1 4 4

1 4

1 4

1 4

1 4

1 4

b 28 − 8x cm

9 a 8x cm e A

1 4 11111111 88888888 1 1 1 111111 1 8 4 8 888888 8

c 7 − 2x cm (5, 84)

49

(2, 21) x

0

1

2

3

4

5

f A = 21 when x = 2 10 a C = 3500 + 0.5x b I = 1.5x I c I/C ($) C

c i A2 =

0

0

iii

11 3

h (m)

b i P = 9.8 mh ii 100% increase iii 50% decrease c i 14 ii 42 d4 17 a i a = 50 000, d = 5000 ii The 11th month iii 4 950 000 litres b i qn = 12 000 (1.1)n−1 ii 256 611 litres c The 31st month 18 a i 15.4 million tonnes ii 21.7 million tonnes b tn = 0.9n + 9.1 c 371 million tonnes d 12.1 years e Pn = 12.5 (1.05)n−1 f 15 years 19 a 1 hr 35 mins b 2.5 km 20 a n(B ∩ C ∩ T ) = n(C ∩ T ), n(B ∩ C ∩ T ) = 3n(B ∩ C ∩ T ), n(B ∩ C ∩ T ) = 4 b n(ξ) = 76

T 13 B 21 4 5 8 7 C

x

18

e 5500

3500

5 6

(1, 49)

3500

d 3500 f P ($)

ii

iv 99.99 v 999.999 15 a 14 m b tn = 1.5n − 1 c 53 d 330 m 16 a i P = 49h P (joules) ii iii 1136.8

(3500, 5250) 0

1 1 − a b

Answers

R − r2 3

b i OC2 =

x

–3500

P represents profit 11 c 11, 24 and 39 25 1 ii x = 12 b i x = 24 24 x ; x = −1 13 a r = − x +1 4 9 ii S∞ = 18 iii S∞ = b i S∞ = 3 20 −1 cx> and x = 0 2 a3 ii 4.5 iii 30 14 a i A = 6 a3 ii 486 iii 36 b i A1 = 12

ci 5 ii 0 2 a + bc ab + bd 21 a i ac + cd bc + d 2 3a 3b ii 3c 3d

Chapter 8 Exercise 8A 1 a (7, 3) b (6, 9) d (1, 5) e (1, 7) 4 −1 4 c b 2a 2 1 1 3 a (5, 6) 0 4a −5


b (2, 2) 5 −6 c b 6 −1

c (2, 7) d

7 6

e

4 3

c (−2, 3) 0 d 0


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678

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5 Z

X

Y

6 5 4 3 2 1

−1 −4

8

C

6 A

B

5

R x

D' D

3

Q

2

c

−4 3

d

b (−3, 15)

0 0

c Use

–1

–1

A

B x

0 1

2

3

5 a, b, c 2 9

C

A'

1

1 4

6 a (4, 12)

B'

4

2 3 4 5 6 7

–2 P –3 –4 –5 –6

b

C'

7

–6 –5 –4 –3 –2 –1–1 0 1

a

y

4

4

6

7

8

y 7

A2

6 5

B3

7 a (−4, 6) b (11, 4) 8 a (1, 0), (2, 1), (3, 4), (4, 9) y b (3, 9) (4, 9)

5

4 B2

3 2

A3 C3

C2

1 x

–6 –5 –4 –3 –2 –1 0 C –1 –2 A

1

2 C1

3

4

5

6

7

8

A1

–3

(2, 4)

(3, 4)

B

(2, 1) (0, 0)

y C

4 3 A'

A

B

1 x

–5 –4 –3 –2 –1 0

1 2 3 4 5

1 a (1, 9) b (2, 3) c (4, 3) 2 (x, y) → (x, 4y) 3 (x, y) → (3x, y) 4 a A(0, 0), B (0, 3), C (1, 3), D(1, 0) b A(0, 0), B(0, 1), C (3, 1), D (3, 0) 2) 5 a i A (0, 0) , B (3, 0) ,C (3, 3 3 ii A(0, 0), B ,0 ,C ,4 2 2 b y

y

2

4

2 1

C"

C

3 P'

0 –1

c (6, 2)

Exercise 8C

5

2

b (−6, −2) b (0, −1) d (1, 0)

6 a (−2, 6) 7 a (−1, 0) c (0, 1)

Exercise 8B C'

y = –x

x

c y = (x − 1)2

B'

B1

–6

(1, 0)

1

–4 –5

(1, 1)

Q'

1

2

3

RR' 4 5

P

2

C'

1

x

0 A

Q

B 1 B" 2

x

3

–2 y

3 –5 Z Y

–4

–3 –2 W

–1

0 –1

X

–2 X'

W'

Y'

Z'

–3

Exercise 8D

x

(x, y) → (x + 5, y + 6) no invariant points (x, y) → (x, 4y) {(x, 0) : x ∈ R} 1 c i (x, y) → x, y 3 ii {(0, y) : y ∈ R}

1a i ii b i ii

–4 –5



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679

Answers cy=

1 +4 x −1

Answers

d i (x, y) → (x − 2, y + 3) ii no invariant points e i (x, y) → (y, x) ii {(x, x) : x ∈ R} f i (x, y) → (−x, y) ii {(0, y) : y ∈ R} 3 11 , −1 c b 9, 2 a (1, 7) 2 2 3 An invariant point does not exist. −11 1 b −2, 4 a (−5, 3) c 1, 2 3 5 a (1, −6) b (0, 0) c (0, 0)

y 1 +4 x–1

y= 4 3

1 y= x

0

x

1

dy=

Exercise 8E

2

1 x −1

+4 y

(x, y) → (x − 1, y + 1) (x, y) → (x − 1, y + 1) (x, y) → (2x, 2y) (x, y) → (2x, 2y) (x, y) → (x + 4, 3 (y + 5)) (x, y) → (x + 4, 3y + 5) (x, y) → (−(x − 1), y + 2) (x, y) → (−x − 1, y + 2) (x, y) → (−x, −y) (x, y) → (−x, −y) (x, y) → (4 − x, −y) (x, y) → (4 − x, −y) 1 g i (x, y) → x − 1, y + 2 2 1 (x − 1) , y + 2 ii (x, y) → 2 h i (x, y) → (2x + 2, y − 3)

1a i ii b i ii c i ii d i ii e i ii f i ii

2

y=

1 +4 x–1

5 4 y=

1 x2 0

x

1

e (x − 1)2 + (y − 4)2 = 1 y 5 4 3 (x – 1)2 + ( y – 4)2 = 1 1

x2 + y2 = 1

ii (x, y) → (2(x + 2), y − 3) 0

–1

x 1

2

–1

Exercise 8F

b y = 2x 2

2 a y = 2x

1a y = x +3

y

y

y

y = 2x2

y = 2x

y=x+3

y=x

y=x

y = x2

3 x

0

–3

–1 0

x

0

cy= b y = (x − 1)2 + 4 y = (x – 1)2 + 4

y

dy= y

2 y= x 2 –11 0 x –1 1 1 y= x –2

y = x2 4

2 x

5

x 1

2 x2 y

2 2 y= 2 x 1 –10 1 y = 1 x2

x

x 1



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Essential Advanced General Mathematics d {(x, y) : y = x − 2}

y2 =1 4

e x2 +

y y=x+2

y y2 2 x2 + =1 4

y=x–2 2

1 x2 + y2 = 1 x 1

0

–1

x

0

–2

2

–2

–1

e {(x, y) : y = −x + 2}

–2

3a y =5−x 1 cy= +2 3−x

e (x − 3) + (y − 2) = 1 1 1 5 b y = (x + 3)2 + 1 4ay= x+ 4 2 2 2 4 cy= +1 +1 dy= x +3 (x + 3)2 1 e (x + 3)2 + (y − 1)2 = 1 4 1 5 y = (x + 13) 2 6 a {(x, y) : y = x + 4} 2

y

b y = (x − 3) + 2 1 +2 dy= (x − 3)2 2

y=x+2 y = –x + 2 2

2

y=x+2

x 2

–2

7 {(x, y) : (x − 1)2 + (y − 4)2 = 4} y (x – 1)2 + ( y – 4)2 = 4

4 2

y y=x+4

0

–2

0

–2

4

–2

x 1

2

x 2 + y2 = 4

2 –4

0

–2

x

b {(x, y) : y = − (x + 2)} y

8 a {(x, y) : y = −2x 2 } b {(x, y) : y = (2x − 5)2 + 2} 1 c (x, y) : y = (x − 5)2 + 2 2 d {(x, y) : y = (x +√2)2 + 1} e {(x, y) : y = 2 ± x} y 9a b y

y=x+2 0

x

2

–2

y = –2x2

x

0

–2

2

y = –(x + 2) 0

c {(x, y) : y = 4 (x + 2)} y

y

c

d

y = (2x – 5)2 + 2 x 2.5 y

y = (x + 2)2 + 1

1 y = (x – 5)2 + 2 2

y = 4x + 8

5

8 2

1

y=x+2

x

0

–2

0

x

y

e

2

5

4

–2

0

y=2+ – √x

x 2 0


x 4


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y

y = 2x–1

y = 2x + 1

1 2

2 x

0

1 x

0 y

c

y

d

x

y=

1 3

y=

x (2 2 )

1 3

1 3 x

0

e y=

1 2x + 2

y

y x

0 –2

1 4

–3

4 0

x

0

f

+4

4

1 (2 2) 3

b A dilation of factor 2 from the x axis followed by a translation determined by the 0 vector −3 c A dilation of factor 2 from the x axis followed a translation determined by the by 3 vector 1 d A dilation of factor 2 from the x axis followed a translation determined by the by 3 vector 0 e A dilation of factor 2 from the x axis followed by a translation determined by the 0 vector −3 f A reflection in the y axis followed bya 3 translation determined by the vector 0 g A dilation of factor 3 from the y axis followed a translation determined by the by 1 vector 4 3 h A dilation of factor from the x axis 2 followed a translation determined by the by 2 vector 0

Answers

10 a {(x, y); y = 2x−1 } b {(x, y) : y = 2x + 1} 1 x c (x, y) : y = 22 3 1 x d (x, y) : y = 22 3 1 e (x, y) : y = x+2 + 4 2 f {(x, y) : y = −(22x + 2)} y 11 a b

681

Exercise 8H y = – (22x + 2)

x

1 a {(x, y) : y = |x + 1| + 3} y

Exercise 8G 1 1 a A dilation of factor √ from the y axis or a 2 dilation of factor 2 from the x axis b A translation determined by the −2 vector 0

(–2, 4) 4 (–1, 3) x

0

b {(x, y) : x = −|y|}

c Reflection in the line y = x 1 d A dilation of factor from the y axis 2 e A dilation of factor 3 from the x axis f A translation determined by the 3 vector 0 g Reflection in the y axis h Reflection in the x axis 2 a A dilation of factor 2 from the x axis followed a translation determined by the by 3 vector 0


y (–5, 5)

0

x

(–5, –5)

x c (x, y) : y = 4 y

(–4, 1)

(4, 1) 0

x


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Essential Advanced General Mathematics b {(x, y) : y = [x − 2]}

d {(x, y) : y = −|x|} y

y 4

0

x

2 0 (–3, –3)

–4

(3, –3)

2

x

–2

4

6

–2

e {(x, y) : x = |y + 1| + 3}

–4

y

c {(x, y) : y = [x] + 2} y 4 0

x

4

(3, –1)

2

(4, –2)

–4 –2

f {(x, y) : x = |y − 3| − 1} y

2

4

6

4

6

–4

d {(x, y) : x = [y]}

4

y

(–1, 3) 2

4

0

x

2

2 –4

g {(x, y) : y = −2|x|}

–2 2

–4

0 (–1, –2)

x

0 –2

y x

e {(x, y) : x = − [−y]}

(1, –2)

y

2 a A translation determined by the 0 vector 3 b A translation determined by the 3 vector 3 1 c A dilation of factor from the y axis 2 d A reflection in the x axis followed by a dilation of x axis factor 2 from xthe 3 a (x, y) : y = 2 y

4

–4

–2

2 0

x 2

4

6

–2 –4

f {(x, y) : y = [x − 4]} y 4 2

4

–4

2 –4 –2

x

0 –2

0

x 2

4

6

–2

–2

4 0 –2

2

x 6

–4

–4



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Answers

y

4 2 2 –4

4

6

x

–2 0 –2

b Yes, if the points are not collinear with the origin. 1 2 , range is {(x, y) : y = x} c 1 2 0 −1 0 1 −1 0 c b 5a −1 0 1 0 0 1 3 0 3 0 1 0 f e d 0 1 0 3 0 2

–4


h {(x, y) : y = −[x] + 2}

1C 6E

y

2B 7A

3D 8B

4C 9B

Answers

x g (x, y) : y = − 2

5D 10 B

4 2 4 –4

–2

0 –2

6

x

2

–4

Exercise 8I 1 a y = f (x − 2) + 3 c y = f (2x)

b x = f (y) d y = 2 f (x) x

b 4x c 22 d 2x−1 2 a 2x+3 x = x 2 , a dilation of factor 2 from the y 34f 2 axis and a dilation of factor x4from the x axis take y = f (x) to y = 4 f , a sequence of 2 transformations that results in the original function. 1 4 f (2x − 3) + 4 = + 4, a dilation of 2x − 3 1 factor from the y axis followed by a 2 ⎡ ⎤ 3 translation⎣ 2 ⎦ 4 5 −3 f (2 − x) = −3(x − 2)2 , a reflection in the x axis, a dilation of factor 3 from the x axis and the 2 translation 0

Short-answer questions (technology-free) 1 a (9, −1) b (3, −2) c (0, 1) d (−3, −1) e (3, 1) f (−1, 3) 2 a y = (x√ + 2)2 + 3 by=± x c y = −x 2 3 a (x, y) → (y − 2, x + 3) b (x, y) → (x, −5y) c (x, y) → (4x − 2, y + 3) d (x, y) → (x − 2, 4 (y + 3)) 2 4a y = x −1 b y = 4x − 2 3 c y = 2x + 7 d 2x + y + 1 = 0 1 e 2x + y = 1 f y = (x + 1) 2 √ 5 a y = 3 ± −x b y = 5(x 2 − 2) 1 c y = 5 − (x + 2)2 16 d y = 20 − 4(x + 2)2 6 a y = − |x| b y = |2x| + 3 y –1

y

0

1

x

5 3

–1

–1 0

1

x

c y = 4 − |2x| y

Exercise 8J −8 3

1a

b

−3a − b −a + 3b

2 (1, 0) → (2, −4) , (0, 1) → (−1, 3), (3, 2) → (4, −6) 3 a (1, 0) → (2, 1) , (−1, 2) → (−4, 1) b (1, 0) → (−2, 0) , (−1, 2) → (2, 2) c (1, 0) → (2, 3) , (−1, 2) → (4, −5) 2 3 4a 3 −1

4

–2

0

2

x

7 a (x, y) → (x + 3, 2y + 4) b (x, y) → (x x+ 4, 2y+ 3) c (x, y) → ,y−4 3 (x, (x d y) → + 1, 2y + 1) e (x, y) → (x + 2, 3 − y)



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Essential Advanced General Mathematics b (x − 1)2 = y + 2

8 a x2 = y − 1

2 1 0 –1 1

x

c (x − 2)2 = 3(y − 2)

–1 0 1 2 –1 –2 (1, –2)

(–1, 1)

2 1 0 –1 –2

1

(2, 2) 0

x

4 a i Dilation from the x axis of factor x 1 4

1 2

3 4

y 2 − 3|x − 2| fy=

e y = −3|2x − 1| y

2 1 2

1

x

0

x 2

x

0

y

33

–3

x

d y = [4x]

y

0

y

ii

y

y

4

–4

Extended-response questions 1 a (4, 6) b (x, y) → (x − 3, y) → (3 − x, y) → (6 − x, y) c (x, y) → (−x + 6, y) −m d i translation , reflection in the y axis, 0 m translation 0 ii (x, y) → (−x 2m, y) + 0 e i translation , reflection in the x axis, −n 0 translation n ii (x, y) → (x, −y + 2n) f i y = −x + 3 ii y = −x + 6 iii y = (6 − x)2 iv y = (3 − x)2 2 a A (−1, 3) 1 bi ii −3 3 q ci ii A (−q, p) p d (x, y) → (−y, x) e i y = −x ii x = −y 2 −1 iv y = iii x 2 + y 2 = 1 x 3 a (−1, −3) b (−a, −b) c y = − f (−x) d i (x, y) → (6 − x, −y) ii y = 3x − 19 e (x, y) → (2m − x, 2n − y) f (x, y) → (y − n + m, n + m − x) g (x, y) → (n + m − y, x − m + n) h i x = (1 − y)2 − 1

3 125

ii (x, y) → (x, −y) iii (x, y) → (x + 25, y + 15) −3 y + 15 iv (x, y) → x + 25, 125 −3 b i y= (x − 25)2 + 15 125 ii (x, y) → (x + 50, y) −3 (x − 75)2 + 15 iii y = 125 m −4ny c i (x, y) → x + , + n 2 m2 −4n m 2 ii y = +n x− m2 2 −4n 3m 2 iii y = x− +n 2 m 2 x +1 5a i y= +3 2 y ii 4 3 2 1 –7

–5

–3

–1

01

x 3

iii reflection in the x axis, translation 3 determined by the vector −2 y b i

1 0

x 1 2

1

2

ii x = 1 or x =

3

1 3

Chapter 9


Exercise 9A 1 2 parts = 2000, 7 parts = 7000 2 1 part = 3000, 2 parts = 6000 3 3.6 4 264 5 22.5 6 60◦ , 50◦ and 70◦ 7 $14


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10 5.625 km

11 $1200

13 ␲ : 1 16 8.75 km

14 1 : 1

3 5 15 6 : 7

12

Exercise 9B 1 a AAA, 11.25 cm c AAA, 3 cm 2 a AAA, 6 cm

2 cm 3 d AAA, 7.5 cm 1 b AAA, 1 cm 3

16 a 1 : 100 b 1 : 1000 c 1 : 10 d1:1 27 17 and 4 litres. 16 18 125 and 216 mL 19 a 1 : 50 b 1 : 125 000 c 3 cm d 7500 cm2 20 a 12 : 13 b 1728 : 2197 21 a 4 b 3.75 22 3 : 4 23 4.5 cm

Answers

8 30 g zinc, 40 g tin 9 16 white and 8 green beads

685

b AAA, 11

2 d AAA, 7.5 cm cm 3 3 AC = 17.5, AE = 16, AB = 20 4 4.42 m 5 7.5 m 6 15 m 10 2 8 10 m 7 22.5 m 9x =6 31 3 39 1 10 83.6 cm 11 x = 12 40 m 46 7 14 14 1 m 13 7.2 m 15 √ √ 15 b x = 10 c y = 2 5, z = 5 5 36 17 7.11 m 18 1.6 m 16 a = 7 √ √ 1 19 2 m 20 a = 3 5, x = 5, y = 2 5 7 c AAA, 2

Exercise 9C 1a1:2:3:4 b 1 : 4 : 9 : 16 c Yes, the second ratio is the square of the first. 2a1:2:3:4 b 1 : 4 : 9 : 16 c Yes, the second ratio is the square of the first. 4 3 19 cm2 4 4.54 cm2 √ 9 √ 4 4 3 c b cm 5 a 3 cm 3 3 64:5 7 22.5 8a1:2:3 b1:2:3 c 1 : 8 : 27 d Yes, the third ratio is the cube of the first. 9ai 2:3 ii 2 : 3 iii 2 : 3 b 8 : 27 c Yes, the ratios in a are cubed to form the ratios in b. 10 a 3 : 2 : 5 32 500 b Volumes are 36␲, ␲ and ␲ cm3 . 3 3 Ratio of volumes is 27 : 8 : 125 c Yes, the ratios in a are cubed to form the ratios in b. 11 8 : 1 12 27 : 64 13 2 : 3 14 a 4 : 3 b4:3 15 a 4 : 1 b8:1

Exercise 9E √ ii 10 ii 72◦ √ √ 1+ 5 3+ 5 2 0 1 5 ␾ = 1, ␾ = , ␾ = , 2 2 √ √ 5 7 + 3 ␾3 = 2 + 5, ␾4 = , 2 √ √ 3− 5 5−1 ␾− 1 = , ␾−2 = , 2 2√ √ 5 7 − 3 ␾−3 = 5 − 2, ␾−4 = 2

2b i 4 4 a i 36◦ c 0.62

Multiple-choice questions 1D 6D

2B 7C

3D 8E

4C 9E

5B 10 E

Short-answer questions (technology-free) 1 b i 20 cm ii 10 cm c XP : PY = 2 : 1, PQ : YZ = 2 : 3 2 a 3 cm b5:3 c3:5 15 210 5 12.25 6 12 3 m 4 8 23 7 a 96 g b2:1 c 1000 cm3 d 100 mm 8 b 25 : 36 c 48 cm 8 3 m c 9 a 20 : 3 b 1.6 m2 27 10 a 2% b 3% 4 1 2 2 1 1 f e d c b 11 a 9 9 3 3 3 3

Extended-response questions x h 20 = e q x+y 9 2 a Rhombus, CF = 1 c ACF √ 1+ 5 e 2 3 x = 8 or x = 11 4 a BDR and CDS, BDT and BCS, RSB and DST 1 a EBC


c


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Essential Advanced General Mathematics z p = y p+q 5 a i 9 cm 1 iii 16 b i√16a cm2 7 15 26 m b

z q = x p+q ii 12 cm 9 iv 16 ii 3a cm2

2 a a = 0.7660, b = 0.6428 b c = −0.7660, d = 0.6428 c i cos 140◦ = −0.7660, sin 140◦ = 0.6428 ii cos 140◦ = −cos 40◦

c

Exercise 10E

Chapter 10 Exercise 10A ␲ 3 11␲ d 6 2 a 120◦ e 100◦ 3 a 34.38◦ d 246.94◦ g 271.01◦ 4 a 0.66 e 1.47 5 a −60◦ d −180◦ g 690◦ 6 a −2␲ 1a

d −4␲

4␲ 5 7␲ e 3 b 150◦ c 210◦ f 324◦ g 220◦ b 108.29◦ e 213.14◦ h 343.77◦ b 1.27 c 1.87 f 3.98 g 2.39 b −720◦ e 300◦ h −690◦ b −3␲ b

e−

11␲ 6

4␲ 3 8␲ f 3 d 162◦ h 324◦ c 166.16◦ f 296.79◦ c

d 2.81 h 5.74 c −540◦ f −330◦ 4␲ 3 7␲ f− 6

c−

b −1, 0 e 0, −1 h 0, 1 b 0.75 e −0.50 h 0.61 b −1, 0 e −1, 0 h 0, −1

c 1, 0 f 1, 0 c −0.82 f −0.03 c −1, 0 f 0, −1

Exercise 10C 1 a −34.23 b −2.57 c −0.97 e 0.95 f 0.75 g 1.66 2a0 b0 c undefined e undefined f undefined 3a0 b0 c0 d0 e0

d −1.38 d0 f0

Exercise 10D 1 a 67◦ 59 d 6.4279 g 2.3315

b 4.5315 e 50◦ 12 h 6.5778

c −0.42 g −0.7 c −0.4 g 0.4√ 1 3 bb= 3aa=− 2 2√ 1 − 3 cc= dd = 2 2 √ e tan(␲ − ␪) =√− 3 f tan(−␪) = − 3 √ 1 3 b 4a− 2 2 √ 1 3 e− d− 2 2

c 2.5357 f 3.4202 i 6.5270

b −0.7 f −0.38 b −0.6 f −0.7

d −0.38 h 0.7 d −0.6 h 0.6

√ c− 3

Exercise 10F √ 3 2 1 d− 2 √

1a

1 b −√ 2 1 e −√ 2 1 h√ 2

1 c −√ 3 √ f 3

1 3 i −√ 2 √ 3 √ 1 3 , cos = − , tan = − 3 2 a sin = 2 2 1 1 b sin = √ , cos = − √ , tan = −1 2 2 √ 1 3 1 , tan = √ c sin = − , cos = − 2 2 3 √ √ 1 3 d sin = − , cos = − , tan = 3 2 2 1 1 e sin = − √ , cos = √ , tan = −1 2 2 √ 3 1 1 , tan = √ f sin = , cos = 2 2 3 √ √ 3 1 g sin = , cos = , tan = 3 2 2 1 1 h sin = − √ , cos = − √ , tan = 1 2 2 √ √ 3 1 , cos = , tan = 3 i sin = 2 2 √ √ 1 3 , cos = , tan = − 3 j sin = − 2 2 g−

Exercise 10B 1 a 0, 1 d 1, 0 g −1, 0 2 a 0.95 d 0.96 g −0.86 3 a 0, −1 d −1, 0 g 0, −1

1 a −0.42 e 0.42 2 a −0.7 e −0.7



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Answers √ 3 2

d not defined

e0

1 g√ 2

h −1

e

1 c√ 3 1 f −√ 2

1 b −√ 2

y

4

2π

0

x

4π 3

2π 3

Exercise 10G –4

b ␲ and 3

1 a 2␲ and 2 1 2␲ and c 3 2 2␲ and 4 e 3 g 4␲ and 2 2a y

d 4␲ and 3 f

2␲ 3

Amplitude = 4, Period = f

␲ 1 and 2 2

y

Answers

3a−

5

0

π 4

π 2

3π 2

π

2π

5π 4

3π 4

x

7π 4

3 –5 0

π

Amplitude = 5, Period = ␲

x

2π

g

–3

y

3

Amplitude = 3, Period = ␲ b

y 0

π

x 2π

3π

4π

2

0

4π 3

2π 3

–3

θ

2π

Amplitude = 3, Period = 4␲

–2

Amplitude = 2, Period = c

y

2␲ 3

2 π 4

0

4

0

π

2π

3π

4π

θ


π

5π 8

y

x

7π 8

␲ 2

2

y

1 2

0

x 3π 2

2π 3

4π 3

2π

3π

9π 2

6π

θ

–1 2

Amplitude =

3π 8

3π 4

Amplitude = 2, Period = i

0

π 8

π 2

–2

–4

d

y

h

2␲ 1 , Period = 2 3


–2



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3a

Exercise 10H 1

y

1a π 2

–π 2

–3π 2

4

3π 2

π

0

–π

–2π

y = 4 cos(–2x)

x 2π

π 4

0

–1

y

b

π 2

π

3π 4

x

–4 2

Period = ␲, Amplitude = 4, y = ±4 –6π

0

–3π

3π

x

6π

y

b

f(θ) = –√2sin (2θ) √2

–2

π 4

0 y

c

π 2

π

3π 4

–√2

2

0

π 3 π 6

2π 3

4π 3

π

π 5π 2 6

5π 3

7π 6 3π 2

Period = ␲, Amplitude = 2π

y

c

x

11π 6

√ √ 2, y = ± 2

f (x) = 2 sin(–3x)

2

–2

2

0

π 3

π 6

0 y

d

θ

π 2

2π 3

x

–2

π 2π 3 3

π 4π 3

Period =

x

5π 2π 3

y

2a

2␲ , Amplitude = 2, y = ±2 3

–2

3

y

4 5 2

0 x

0

3π 2

2π,

–5 4

π

3π 2

2π

5π 2

θ

–3

3π

Period = 2␲, Amplitude = 3, y = ±3

–5 2

5a

π 2

y

b y y = sin x

1 y = cos x

0 0

π

2π

x

π

2π

θ

–1

b

␲ 5␲ , 4 4


Period = ␲, Amplitude = 1, y = ±1


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Answers y

h

Answers

y

c

3

2 0

5π 12

13π 12

0

θ

π 2

θ

π

–3

–2

Period =


2␲ , Amplitude = 2, y = ±2 3

y

i

y

d

3 √3 0

π 2

π

3π 2

–3

– √3

Period = ␲, Amplitude =

√ √ 3, y = ± 3

y

e

θ

π 2

0

–π 2

θ

3

Period = ␲, Amplitude = 3, y = ±3 1 1 3 a f (0) = , f (2␲) = 2 2 b π 3

1 0

π 2

π

θ

,1

2π,

1 2

0,

0


–1

4π , –1

3 √ √ 3 3 , f (2␲) = − 4 a f (0) = − 2 2 y b

y

f

x

11π 6

5π 6

–3

1 2

2

1

0

–π 4

π 12

θ

5π 12

0

– √3

–2

Period = y

g

2␲ , Amplitude = 2, y = ±2 3

√2

2 –1

π 3

5π 6

x

4π 11π 3 6

√3 2π, – 2

1 1 5 a f (−␲) = − √ , f (␲) = − √ 2 2 y b 1

0

π 3

5π 6

4π 3

1 √2

π

–π

x

0

–√2

Period = ␲, Amplitude =

0,

θ

√ √ 2, y = ± 2


– π, – 1 √2

–1

1 π, _ √2


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Exercise 10I

Exercise 10J

␲ 7␲ 5␲ 7␲ b and and 4 4 4 4 ␲ 2␲ 3␲ 3␲ b , ,− 2a 3 3 3 4 2␲ 2␲ c ,− 3 3 3 a 150 and 210 b 30 and 150 c 120 and 240 d 120 and 240 e 60 and 120 f 45 and 135 4 a 0.93 and 2.21 b 4.30 and 1.98 c 3.50 and 5.93 d 0.41 and 2.73 e 2.35 and 3.94 f 1.77 and 4.51 5 a 0.64, 2.498, 6.93, 8.781 5␲ 7␲ 13␲ 15␲ b , , , 4 4 4 4 ␲ 2␲ 7␲ 8␲ c , , , 3 3 3 3 7␲ 11␲ 19␲ 23␲ , , , 6a 12 12 12 12 ␲ 11␲ 13␲ 23␲ b , , , 12 12 12 12 ␲ 5␲ 13␲ 17␲ c , , , 12 12 12 12 5␲ 7␲ 13␲ 15␲ 21␲ 23␲ d , , , , , 12 12 12 12 12 12 5␲ 7␲ 17␲ 19␲ e , , , 12 12 12 12 5␲ 7␲ 13␲ 15␲ f , , , 8 8 8 8 7 a 2.034, 2.678, 5.176, 5.820 b 1.892, 2.820, 5.034, 5.961 c 0.580, 2.562, 3.721, 5.704 d 0.309, 1.785, 2.403, 3.880, 4.498, 5.974 y 8

1a

1a

3

1 0

– 4π , – 1 2 3

π 1 , 3 2

– π, 1 3 2 0 –1 1 – 2π , – 2 3

x

y

b

π π 2 – √3 6 3 0

7π 4π 6 3 x

– √3 –2 – √3 y

c 1 + √2

0 1 – √2

3π 4

5π 4

2π

x

y

d

0

π 4

5π 4

x

–2 –4 y

e 1 + √2

5π 1 , 3 2 4π , – 1 2 3

11π 6

7π 6

–1

1 – 5π , 1 3 2

y

x (2π, 0) 0 1 – √2

2π , – 1 2 3

3π 2

x

y

2a –11π –2π 6

–7π 6

–π 2

π 6

–π 6

π 2

5π 6

3π 2

2π

0

x

(2π, –2) (–2π, –2)

–2

–4



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Answers

–23π –19π –15π –11π 12 12 12 12

–7π 12

–3π 12

2

9π 12

5π 12

13π 12

17π 12

21π 12

x

π

0

–π

–2π

π 12

Answers

y

b

2π (2π, –1.414)

(–2π, –1.414)

–2

c

y –π

–2π

π

0

2π

691

x

–1

(2π, –3)

–3

(–2π, –3)

–5

y

d

3 (2π, –3)

(–2π, 3)

1 2π –2π

–5π 3

–4π 3

–2π 3

π 3

–π 0 3 –1

e

2π 3

4π 3

x

5π 3

y

–3π 2

–11π 6

–5π 6

1

–π 2

π 6

π 2

7π 6

3π 2

2π x

0

–2π

(2π, –2)

–2 (–2π, –2) –3

y

f –19π 12

–5π 4

–7π 12

5π 12

17π 12

3π 4

7π (2π, 1+ √3) 4

1 + √3

(–2π, 1 + √3)

–2π

–π 4 3

π

–π 0

2π

x

–1



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3a –3π 4

(–π, 1 + √3)

–5π 12

3

π 4

2 a rotation of

7π 12

(π, 1 + √3)

1 + √3

π

–π

x

0 –1

b

y –3π 4

–π 12 3

π 4

11π 12

π

–π (–π, –√3 + 1)

x

0

–√3 + 1

(π, –√3 + 1)

–1

y

c

2 + √3 (–π, √3) √3

–5π –2π 6 3

√3 − 2

direction 4␲ b rotation of about O in an anticlockwise 3 direction 4 −1 c rotation of ␪ = cos , where 5 ␲ ␪ ∈ 0, about O in an anticlockwise 2 direction ⎡ ⎤ 1 −1 √ √ ⎢ 2 3 1 2⎥ ⎥ 3a⎢ , b √ √ ⎣ 1 ⎦ 1 2 2 √ √ 2 2 ⎤ ⎡ −1 −1 ⎢ √2 √2 ⎥ 0 −1 ⎥ ⎢ c i ii ⎣ 1 −1 ⎦ 1 0 √ √ 2 2 ⎡ √ ⎤ ⎤ ⎡ − 3 −1 1 1 √ √ ⎢ 2 2√ ⎥ ⎥ iv ⎢ 2⎥ ⎥ ⎢ 2 iii ⎢ ⎣1 ⎣ −1 1 ⎦ − 3⎦ √ √ 2 2 2 2

π

−3 ⎤ 5 ⎥ 4⎦ 5

⎡4

(π, √3 )

0

–π

␲ about O in an anticlockwise 6

d sin ␪ =

x

π π 6 3

3 ⎢5 ,⎣ 5 3 5

Exercise 10L Exercise 10K ⎡√ ⎤ 3 1 − ⎥ √ ⎢ 3 1 2 2⎥ 1a⎢ √ , ⎢ ⎥, 3⎦ ⎣ 1 2 2 2 2 ⎡ ⎤ 1 1 √ √ ⎢ 2 2⎥ ⎥ , √1 , √1 b⎢ ⎣ 1 1 ⎦ 2 2 −√ √ 2 2 0 −1 , (0, −1) c 1 0 ⎡ √ ⎤ 1 3 √ ⎢ − ⎥ 3 1 2 ⎥ ⎢ √2 d⎢ , ⎥, − 3 1⎦ ⎣ 2 2 − − 2 2 ⎡ ⎤ 1 1 − √ √ ⎢ 2 2⎥ ⎢ ⎥ e⎢ 1 ⎥ , (0, −1) 1 ⎣− √ −√ ⎦ 2 2 ⎤ ⎡ √ 3 1 √ ⎥ ⎢− 3 1 2 ⎥ ⎢ 2 √ ⎥, − ,− f⎢ 1 3 ⎦ ⎣ 2 2 − − 2 2

1 a i 1.83 × 10−3 hours ii 11.79 hours b April 26 (t = 3.856), August 14 (t = 7.477) 2a D 13 10 7

0

3

6

12

24

18

t

b {t : D (t) ≥ 8.5} = {t : 0 ≤ t ≤ 7} ∪ {t : 11 ≤ t ≤ 19} ∪ {t : 23 ≤ t ≤ 24} c 12.9 m 3 a p = 5, q = 2 b D


7 5 3

0

6

12

t

c A ship can enter 2 hours after low tide.


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Answers

1 0 –1

6

18

12

24

t

–2 –3

d {t : 4 < t < 20} 6 a 2 am b 8 am and 8 pm 3 7a i ii 12 2 ␲ 7 3 iii d (t) = − cos t iv 1.5 m 2 2 6 b Between 9 pm and 3 am, and 9 am and 3 pm, each day. 8a D

Short-answer questions (technology-free) 3␲ 23␲ 9␲ e c 6␲ d 4 4 2 4␲ 7␲ 13␲ i h g 9 3 6 ◦ b 315 c 495◦ e 1350◦ f −135◦ h −495◦ i −1035◦ √ 1 − 3 1 c− d b√ 2 2 2 −1 1 1 h√ g f− 2 2 2 1 2␲ ␲ 4 a 2, 4␲ c , b 3, 2 3 2 2 f , 3␲ d 3, ␲ e 4, 6␲ 3 5 a y = 2 sin 2(2x) 11␲ 6 9␲ f 4 2 a 150◦ d 45◦ g −45◦ 1 3a √ 2 √ 3 e 2

y 2

5 4 3 2 1 0

b

1a

Answers

4a5m b1m c t = 0.524 s, 2.618 s, 4.712 s d t = 0 s, 1.047 s, 2.094 s e Particle oscillates about the point x = 3 from x = 1 to x = 5 ␲t b D = −1 + 2 cos 5 a 19.5◦ C 12 c D

693

0

π 8

π 4

3π 8

π 2

x

–2

4

8

12

16

20

24

b y = −3 cos

t

y

b The boat can enter 4 hours before noon and must leave by 4 pm. c The boat can enter at 6.40 am and must leave by 5.20 pm. 9 a i 52 weeks ii 3000 iii [1000, 7000] b i N (0) = 1194.95; N (100) = 1021.87 ii N

x 3

3

x

0 3π 2

3π

9π 2

6π

π 2

2π 3

–3

c y = −2 sin 3x

7000

y 2

(0, 1194.95) 1000

(100, 1021.87) 23

52

75

100

0

t (weeks)

c i t = 23, 75 ii 49 1 2 1 2 d 14 , 31 ∪ 66 , 83 3 3 3 3 e d = 25 000, a = 15 000, b = 10, c = 5

π 6

π 3

x

–2

d y = 2 sin y

x 3

2

Multiple-choice questions 1B 6D

2A 7E

3D 8E

4D 9B

5C 10 B


0

3π 3π 2

6π

x

9π 2

–2


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Essential Advanced General Mathematics ␲ e y = sin x − 4

b f (x) = 1 − 2 cos x y

y

3

1 3π 2

π

0

π π 3π 4 2 4

5π 4

1

x

–π

7π 2π 4

π

π 3

–π 0 3 –1

x 2π

␲ c f (x) = 3 cos x + 3

–1

2␲ f y = sin x + 3

y

y

3

1

0

π 3

–π –5π –π 0 π 6 6 3 –3

x

π 4π 3

π 7π 10π 2π 6 6

x

␲ d f (x) = 2 − cos x + 3

2π

–1

y

g y = 2 cos x −

5␲ 6

3

y

1.5

1

2

–π

0

π 3

5π 6

2π 11π 6

4π 3

0 –π 3 –1

2π π 3

x 5π 2π 3

e f (x) = 1 − 2 sin 3x

x

y 3

–2

h y = −3 cos x + y

17π 18

␲ 1

6 0 –1

–π

3

0 –3√3 2 –3

x 2π

π 5π 18 18

x

π 4π 3

π 3

π

2π


−␲ −␲ 2␲ 5␲ −2␲ −␲ b , , , , 3 6 3 6 3 3 ␲ 7␲ 7␲ ␲ 3␲ e , d c , 2 6 6 6 2 7 a f (x) = 2 sin 2x + 1 6a

1 a i 13.4 ii 2 iii 12 b 3 am, 9 am, 3 pm, 9 pm c 2 < t < 10, 14 < t < 22 2 a 7.3◦ b min = 7◦ , max = 23◦ c Between 9.40 am and 4.30 pm y d (13.08, 23)

y 3 1 –π

2π 0 –1

π

x

(0, 7.32) (24, 7.32) (1.08, 7) 0

3a


x

␲ 6


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Answers

5 s 6 1 f s e1s 4 g i 24 m ii 30 m 4 a p = 6, q = 4.2 b 3 am, 3 pm c 6 m d 7 am, 11 am, 7 pm, 11 pm e 8 hours 5 a i −1 < k < 1, 1 < k < 3 ii k = −1 or k = 3 iii k < −1 or k > 3 b A translation of 1 unit in the negative direction of the y axis, followed by a dilation 1 of factor from the x axis, and a 2 dilation of factor 3 from the y axis. ␲ ␲ ii h = ci h= 6 2 ␲ 6 a A translation of units in the positive 2 direction of the x axis 1 b A dilation of factor from the y axis, 2 ␲ followed by a translation of units in the 4 negative direction of the x axis, and a dilation 1 of factor from the x axis. 4 ␲x c i y = −sin +4 2 ii range = [3, 5], period = 4 7 a x = 4, y = 9 bi 4 ii 2 c i 8 and 0 ii 12 and 8 2␲ di ii ␲ 3 5␲ 3␲ 13␲ 17␲ ␲ f , , , e s 6 2 6 6 6 5␲ 11␲ 17␲ 23␲ , , , g 6 6 6 6 h π , 12

4

x = 4 sin(2πt) + 4 0

31.5 16.5

(0, 5.89) 1.5 13

0

b 5.89 m e 20 times

5π ,8 (π, 4) 6

i

5␲ s 6

t

π 2

c 27.51 s f 4.21 m

t (minutes)

d 6 times g 13.9 m

11 a i R␪ R␺

x = 4sin(3t) + 4 0

t

3 4

9 a For N : max = 7000, min = 1000 minimum occurs in Ocober (when t = 10) maximum occurs in April (when t = 4) For M: max = 8500, min = 2500 minimum occurs at end of January and November (t = 1 and t = 11) maximum occurs in June (when t = 6) b t = 4.31 (April) population is 6961 t = 0.24 (Jan) population is 2836 c 145 556 in May (t = 5.19) d t = 7.49 (July) h (m) 10 a (120, 30.99)

(π, 9) π, 6 8

(1, 4)

4

y = 2sin 2t – π + 10 6

9

(1, 11) 1, 8 4

d

3

3 , 14 4

11

11 , –3 6

5 , –3 6

c3m

y = –3 sin(2πt) + 11

.5 85 . 10 5 3. 5 12 0

t

.5

3 2, – 2

67

19 12

.5

13 12

49

7 12

.5

0 3 1 – 2 12

8 a One possible set of values is a = 4, b = 4, n = 2␲ and c = −3, d = 11, m = 2␲ b

31

4 , 3 3

1 , 3 3

Answers

x

b

cos ␪ cos ␺ − sin ␪ sin ␺ sin ␪ cos ␺ + cos ␪ sin ␺ −cos ␪ sin ␺ − sin ␪ cos ␺ −sin ␪ sin ␺ + cos ␪ cos ␺ cos (␪ + ␺ ) = cos ␪ cos ␺ − sin ␪ sin ␺ , sin (␪ + ␺ ) = sin ␪ cos ␺ + cos ␪ sin ␺ ii R2␪ cos2 ␪ − sin2 ␪ −2 sin ␪ cos ␪ = 2 2 2 sin ␪ cos ␪ cos ␪ − sin ␪ 3 iii R = ␪ 3 4 cos ␪ − 3 cos ␪ − (3 sin ␪ − 4 sin3 ␪) 4 cos3 ␪ − 3 cos ␪ 3 sin ␪ − 4 sin3 ␪ cos ␪ sin ␪ b R−1 ␪ = −sin ␪ cos ␪ cos (−␪) − sin (−␪) , which = sin (−␪) cos (−␪) ◦ represents a rotation of ␪ about O in a clockwise direction =

j 2␲ s



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Essential Advanced General Mathematics c i x = (x − a) cos ␪ − (y − b) sin ␪ + a, y = (x − a) sin ␪ + (y − b) cos ␪ + b ii x = (x − 1) cos ␪ − (y − 1) sin ␪ + 1, y = (x − 1) sin ␪ + (y − 1) cos ␪ + 1 d i x = x cos ␪ − y sin ␪, y = x sin ␪ + y cos ␪ √ √ 2 − 2x ± 2 − 8 2x ii y = 2 cos ␪ + sin ␪ iii y = x cos ␪ − sin ␪ e i (cos 2␪, sin 2␪) and (sin 2␪, − cos 2␪) ii a = cos 2␪, b = sin 2␪, c =√ sin 2␪, d = − cos 2␪ fy=± x

y

c

y= 1 1 2 0 –1 2 –1

1 sin2θ – cosθ 2 2π

θ

π

y

d 3

y = 3sinθ + cos2θ

2 1 0

π

–1

θ

2π

–2 –3

e

y y = 4sin θ – 2cos θ

4

Chapter 11

2

Exercise 11A

0

π

θ

2π

–2

1 a 0.6 e −0.3 h 0.6

c −0.7

b 0.6 10 f 7 i −0.6

d 0.3

–4

g −0.3 j −0.3

Exercise 11C 1a

Exercise 11B

y 4 2

1a

0

y = 2 sinθ + cosθ

2

π 2

–1

π

2π 3π 2

θ

b

y

2 0

π 6

π 3

π 4

3π π 4

π 2

2π 3

x

–2

y

–4

y = 3 cos 2θ + 2 sin 2θ

c

2 1 –1

x

4

–2

0

π

–4

0

3

3π 4

–2

1

b

π 2

π 4

y

y 4

π 2

π

θ

2

–2

0

–3

–2


x 5π 4

7π 2π 4


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Answers

4 3 2 1 0 –1 –2

e

π 2

π

x 3π 2

2π

y 4 2 π 2

0 –2

π

x 3π 2

2π

y

f 2

0

π 4

–2

π 2

3π 4

π

x

–4

Exercise 11F

Exercise 11D 1 ax= bx= 2 ax= bx= cx=

(12n + 5)␲ (12n + 1)␲ or x = 6 6 (3n + 2)␲ (12n ± 1)␲ cx= 3 18 ␲ 5␲ or x = 6 6 ␲ 11␲ or x = 18 18 5␲ 2␲ or x = 3 3

(4n − 1)␲ 3 x = n␲ or x = ; 4 5␲ ␲ 3␲ 7␲ x = − , −␲, − , 0, , ␲ or 4 4 4 4 4x=

2␲ ␲ n␲ ; x = −␲, − , − ,0 3 3 3

3n + 2 6n − 1 or x = ; 12 6 2 7 1 1 1 5 5 11 x =− , − , − ,− , , , , 3 12 6 12 3 12 6 12

5x=

Exercise 11E 1 a −1

b1

e −2

f2

√ −2 −2 3 c√ = 3 √3 3 1 g√ = 3 3

Answers

√ −2 −2 3 b√ = 2 a −1 3 √3 3 1 d√ = c1 3 3 √ √ 2 3 2 g −1 e− 2 f√ = 3 3 √ √ 2 −2 3 3 1 h −√ = i√ = 3 3 3 3 ␲ 5␲ ␲ 7␲ 3 a , b , 6 6 6 6 3␲ 5␲ ␲ 5␲ c , d , 4 4 4 4 8 15 4 a cos ␪ = − b sin ␪ = 17 17 15 c tan ␪ = − 8 24 7 5 cos ␪ = , sin ␪ = − 25 25 √ 8 29 7 6− 31 5 √ 15(6 − 5) 15 8 √ = 124 4(6 + 5)

y

d

697

d1 h2

1 a 2 a 3 a b c

√ √ √ √ √ 2− 6 1− 3 2+ 6 = b √ 4 4 √2 2 √ √ √ 3+1 6− 2 b√ =2+ 3 4 3−1 √ √ √ 3−1 6− 2 = √ 4 2 2 √ √ √ 3−1 6− 2 = √ 4 2 2 √ √ 1− 3 √ = −2 + 3 1+ 3

63 ␲ , 4 0 < u, v < , sin (u + v) = 2 65 ␲ 63 < u, v < ␲, sin (u + v) = − , 2 65 ␲ ␲ 0 < u < , < v < ␲, 2 2 33 ␲ sin (u + v) = − , < u < ␲, 65 2 ␲ 33 0 < v < , sin (u + v) = 2 65 √ 3 1 sin ␪ + cos ␪ 5 a 2 2 1 b √ (cos ␾ + sin ␾) 2 √ 1 tan ␪ + 3 d √ (sin ␪ − cos ␪) c √ 2 1 − 3 tan ␪ 6 a sin u


b cos u


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Essential Advanced General Mathematics 169 24 24 119 d− c b 119 7 25 169 7 65 16 33 h g− f− e− 24 33 65 65 24 56 24 63 d c b− a 25 65 7 16 −336 117 3 7 d c b a 625 44 5 25 √ 1 3 5␲ b− for ␪ = a− 2 2 3 a 1 − sin 2␪ b cos 2␪

7 a−

8 9 10 11

Exercise 11G 1 a max = 5, min = −5 b max = 2, √ min = −2 √ c max = √2, min = −√2 d max = √ 2, min = − 2√ e max = 2 3, min = −2 3 f max = 2, min = −2 g max = 4, min √=0 √ h max = 5 + 13, min = 5 − 13 ␲ 3␲ 2␲ ␲ b 0, , 2␲ c , ,␲ 6 2 3 2 5␲ ◦ d 0, , 2␲ e 53.13 f 95.26◦ , 155.26◦ 3 ␲ 3 2 cos 2x + 6 √ 5␲ 4 2 sin 3x − 4 √ 3␲ 5 a f (x) = sin x − cos x = 2 cos x − 4 √ 7␲ = 2 sin x + 4 √ ␲ = 2 sin x − 4

␲ 2 cos x − 4 √ ␲ = 2 sin x + 4

c f (x) = sin x + cos x =

y √2 1

π 4

–1 – √2

b f (x) =

5π 4

7π 4

3π π 4

√

π 3

4π 3

2 1 0

π

2π

π 3

5π 6

4π 3 π

11π 6 2π

x

– √3 –2

Multiple-choice questions 1A

2A

3B

4A

5C

6E

7C

8E

9A

10 D


x 2π

11π 6

x

3π π 4

y

␲ 3 sin x + cos x = 2 cos x − 3 ␲ = 2 sin x + 6

5π 6

7π 4

√ d f (x) = sin x − 3 cos x 5␲ = 2 cos x − 6 5␲ = 2 sin x + 3 ␲ = 2 sin x − 3

y 2 1 0

π 4

– √2

y

0

5π 4

0

2 a

√2

√

x 2π

–2


b 4, −2 c 4, −4 1 e 1, d 2, 0 3 ␲ 5␲ 7␲ 11␲ , , 3 a , 6 6 6 6 ␲ 5␲ 13␲ 17␲ , , , b 12 12 12 12 ␲ 11␲ 13␲ 23␲ 25␲ 35␲ , , , , , c 18 18 18 18 18 18 ␲ 3␲ 5␲ 7␲ , , d , 4 4 4 4 ␲ 5␲ 7␲ 11␲ , , e , 6 6 6 6 3␲ 7␲ 11␲ 15␲ , , , f 8 8 8 8 ␲ 7␲ 11␲ ␲ 7␲ 9␲ 15␲ , g , h , , , 2 6 6 8 8 8 8

2 a 5, 1

4 60◦ , 300◦ , 0◦ , 180◦ , 360◦ 140 −21 5 a b 221 221 1 6 a b1 2

c

171 140


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√

b0 b 9, −1 √ −4 5 b 9

√

c

8 5 81

b sin (x + y) + sin (x − y) ␲ 4␲ √ b ,0 , ,0 13 a (0, 2 3) 3 3 11␲ 5␲ c ,4 d , −4 6 6 12 a 2 −

3

y 4 2√3 0

x

π 3

4π 11π 2π 3 6

5π 6

–4

√ f (x) = 2 3 cos x − 2 sin x ␲ 14 a x = 0, 2 , 2␲ 7␲ 11␲ c x = 0, ␲, 2␲ bx= , 6 6 ␲ ␲ 7␲ 4␲ ␲ 3␲ ex= , , , dx= , 6 3 6 3 2 2 7␲ 3␲ 19␲ 7␲ fx= , , , 12 4 12 4 15 a y = 2 cos2 x y

2

x

π

π 2

0

b y = 1 − 2 sin

␲

y

2

3π 2

−

2π

x 2

3

√

85 cos (␪− ␣) where 2 ␣ = cos−1 √ 85 √ 2 b i 85 ii √ 85 2 1 −1 −1 iii ␪ = cos + cos √ √ 85 85

17 a

Extended-response questions √

1 b P = 10 5cos (␪− ␣) where 2 ␣ = cos−1 √ ; ␪ = 70.88◦ 5 c k = 25 d ␪ = 45◦ 2 a AD = cos √ ␪ + 2 sin ␪ b AD = 5 cos (␪ − 63)◦ √ 1 = 5 cos (␪ − ␣) where ␣ = cos−1 √ 5 √ c maximum length of AD is 5 m when ␪ is 63◦ d ␪ = 79.38◦ 3 b ii a √ = 1, b√= 1 1+ 2− 3 c √ √ 1 + 3√+ 6√ 2 2− 3−1 = √ 3−1 √ √ √ = 6+ 2− 3−2 4 a i h 1 = cos ␪ ii h 2 = cos ␪ sin ␪ iii h 3 = sin2 ␪ cos ␪ iv h n = sinn−1 ␪ cos ␪, n ∈ N c 19.47◦ ␲ 5 a ii 2 cos 5 ␲ ␲ 2 b iii 4 cos − 2 cos − 1 = 0 5 5 √ 1+ 5 iv 4 2 1 6 b − or 3 2

Answers

7 a1 8 a 5, 1 −1 10 a 9

699

Chapter 12 0 –1

x 2π

2π 3

10π 3

4π

c f (x) = tan 2x y

1 –π

16

2 9

–π 2

0 –1

π 2

π

3π 2

2π

x

Exercise 12A 1 a 4.10 b 0.87 c 2.94 d 4.08 e 33.69◦ f 11.92 √ 40 3 3 66.42◦ , 66.42◦ and 47.16◦ cm 2 3 4 23 m √ 5 a 9.59◦ b 35 m 6 a 60◦ b 17.32 m 7 a 6.84 m b 6.15 m ◦ 8 12.51 9 182.7 m 10 1451 m √ 11 a 5 2 cm b 90◦ 12 3.07 13 37.8 cm 14 31.24 m 15 4.38 m 16 57.74 m



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Exercise 12B 1 a 8.15 b 3.98 c 11.75 d 9.46 2 a 56.32◦ b 36.22◦ c 49.54◦ d 98.16◦ or 5.84◦ 3 a A = 48◦ , b = 13.84 cm, c = 15.44 cm b a = 7.26, C = 56.45◦ , c = 6.26 c B = 19.8◦ , b = 4.66, c = 8.27 d C = 117◦ , b = 24.68, c = 34.21 e C = 30◦ , a = 5.40, c = 15.56 4 a B = 59.12◦ , A = 72.63◦ , a = 19.57 or B = 120.88◦ , A = 10.87◦ , a = 3.87 b C = 26.69◦ , A = 24.31◦ , a = 4.18 c B = 55.77◦ , C = 95.88◦ , c = 17.81 or B = 124.23◦ , C = 27.42◦ , c = 8.24 5 554.26 m 6 35.64 m 7 1659.86 m 8 a 26.60 m b 75.12 m

Exercise 12C 1 5.93 cm 2 ∠ABC = 97.90◦ , ∠ACB = 52.41◦ 3 a 26 b 11.74 c 49.29◦ d 73 e 68.70 f 47.22◦ g 7.59 h 38.05◦ 4 2.626 km 5 3.23 km 6 a 8.23 cm b 3.77 cm 7 55.93 cm 8 a 7.326 cm b 5.53 cm 9 a 83.62◦ b 64.46◦ 10 a 87.61 m b 67.7 m

Exercise 12D 1 a 11.28 cm2 b 15.10 cm2 2 c 10.99 cm d 9.58 cm2 2 2 a 6.267 cm c 19.015 cm2 e 24.105 cm2 or 29.401 cm2

b 15.754 cm2 d 13.274 cm2 f 2.069 cm2

Exercise 12E 1 45.81 cm 2 a 95◦ 30 3 a 6.20 cm 4

b 112◦ 53 b 2.73 cm2 y 4

B y=2 2 –4

4

0

x

A –4

Area of A ∩ B = 9.83 square units

6 61.42 cm2 7 a 125.66 m 8 a 10.47 m 9 6.64 cm2

b 41.96% b 20.94 m2 c 18 10 r = 7 cm, ␪ = or r = 9 cm, 7 c 14 ␪= 9 11 247.33 cm 12 a 81.96 cm b 4.03 cm2

Exercise 12F 1 400.10 m 2 34.77 m 3 575.18 m 4 109.90 m 5 16.51 m 6 056◦ 7 a 034◦ b 214◦ 8 a 3583.04 m b 353◦ ◦ 9 027 10 ∠ASB = 113◦ 11 22.01◦ 12 a ∠BAC = 49◦ b 264.24 km 13 10.63 km

Exercise 12G 1 a 13 cm b 15.26 cm c 31.61◦ d 38.17◦ 2 a 4 cm b 71.57◦ c 12.65 cm d 13.27 cm e 72.45◦ f 266.39 cm2 ◦ ◦ 3 10.31 at B, 14.43 at A and C 4 a 85 m b 45.04 m 5 17.58◦ 6 1702.55 m 7 a 24.78◦ b 65.22◦ c 20.44◦ 8 42.40 m 9 1945.54 m 10 a 6.96 cm b 16.25 cm2 11 a 5 km b 215.65◦ c 6◦ 33

Exercise 12H 1 a 4a 2 , 3a 2 and 12a 2 square units respectively b 14.04◦ c 18.43◦ d 11.31◦ ◦ ◦ 2 a 35.26 b 45 3 a 0.28 b 15.78◦ 4 a 15.51 cm b 20 cm c 45.64◦ 5 a i 107 m ii 87 m iii 138 m ◦ b 43.00 √ 6 a 5 11 cm b 64.76◦ c 71.57◦ d 95.74◦ 7 26.57◦ 8 a 54.74◦ b 70.53◦ 9 1.67 km 10 34.14 cm 11 a 141.42 m b 20.70◦ 12 16 cm √ a a 3 cm b 13 a 2 2 14 a 26.57◦ b 39.81◦ c 38.66◦



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1D

2C

3C

4B

5A

6A

7D

8B

9C

10 A

13.2 Extended-response questions 1 a (x, y) → (x + 6, y + 3) b, c


C''

√ √ 5 5 1 a 5 3 ± 11 b sin−1 or ␲ − sin−1 6 6 √ b 20 cm 2 a 20 3 cm √ 3 4 19 km √ √ 25 3 b cm2 4 a 5 3 cm 4 √ 105 5(21 + 5 3) cm2 c cm2 d 4 √4 11c 3 93 17 8 7 6 5 143◦ 6 31 28 ◦ 9 a i 30◦ ii 15 √ b AT = 300(1√+ 3) √ m, √ BT = 150( 6 + 2) m 10 181 km √ 12 3 km, BC = 2.4 km 11 a AC = 5 b 57.6 km/h 12 cm 12 a 26 tan−1 5 12 b 169 ␲ − tan−1 cm2 5 13 180 cm2 14 21.4 cm 15 11 m

Extended-response questions 1 a ∠ACB = 12◦ , ∠CBO = 53◦ , ∠CBA = 127◦ b 189.33 m c 113.94 m 2 a 4.77 cm b 180 cm2 c 9.55 cm 3 a ∠TAB = 3◦ , ∠ABT = 97◦ , ∠ATB = 80◦ b 2069.87 m c 252.25 m 4 a 184.78 m b 199.71 m c 14.93 m 5 a 370.17 m b 287.94 m c 185.08 m √ 6 a 8 2 cm b 10 cm c 10 cm d 68.90◦

A'' –8

B'' A –4 A'

–6

y

C

5

B

1

–2 B'

0

C'

x –1

–5

d y = 2(x + 3)2 + 2 e (x, y) → (x + 3, −2y + 4) f y 16 (1.18, 7.27) (2.82, 0.73) x 3 4

0.33 0

1

2

2 a i ∠ABC = 11.81◦ , ∠BCA = 138.19◦ ii ∠ABC = 108.19◦, ∠BC A = 41.81◦ b i 24.56 units ii 114.00 units iii 89.44 units c ii 1788.85 sq units iii 3027.9 sq units iv 1239.0 sq units 3 a (4, 1) b i The image is a 4 × 1 rectangle ii 1 sq unit iii 4 sq units iv k sq units c (x, y) → (4x, y) 1 2 d i y= x 16 1 (x − 2)2 − 1 ii y = 16 y iii y = x2 y=

1 (x – 2)2 – 1 16

–0.75 –2 –10

Chapter 13

1A 7D 13 E 19 E 25 A 31 E 37 C

2B 8B 14 D 20 A 26 C 32 E 38 C

3B 9C 15 A 21 E 27 A 33 C 39 E

4C 10 E 16 D 22 D 28 B 34 C 40 D

5D 11 E 17 C 23 E 29 C 35 C 41 E

(2, –1)

4 a Rotation by cos−1 6B 12 B 18 A 24 A 30 D 36 B


x 6

e (x, y) → x + 2,

13.1 Multiple-choice questions

Answers


701

y+3 5 3 clockwise 5

about the origin. b i x 2 + (y − 1)2 = 1 4 2 3 2 ii x − + y− =1 5 5 4 8 c (0, 0) and , 5 5


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1 3 −1 4 10 −2 c a = 2, b = 3 d (5a, 5a) e␭ 5, a = b ⎡ = 2, b = −2a ⎤ or ␭ = ⎡ ⎤ 1 −1 1 1 √ √ √ √ ⎢ ⎢ 2 2⎥ 2⎥ ⎥ 6 a⎢ 2 ⎥ b⎢ ⎣ 1 1 ⎦ ⎣ −1 1 ⎦ √ √ √ √ 2 2 2 2 √ c a = 2, b = 0

5 a (−3, 11)

−k2 , the x axis intercept k1 −k2 of the image is 3 +3 k1 iii e = e + 3, f = − f k2 iv y = −k1 (x − 3) − (x − 3)2 11 a i (3, 1) ii A (3, 1), B (5, 1), C (3, 3) b ii (−1, −1), (2, 2) y iii ii c =

b

3 1 d c = √ ,d = √ 2 2 1 1 e i x = √ (x + y ), y = √ (y − x ) 2 2

(3, 8)

y=x C C'

3 A 1

A'

B' x

0

–5

5

3

1

(15, 8) –5

5 2 0

B

5

1 ii y − x = √ (x + y)2 2 7 ai 5m ii 8 m b x (m) 8

3

(9, 2)

(21, 2)

3 6 9 12 15 18 21 24 t (s)

c i 8m ii 2 m d i t = 0, 6, 12, 18, 24 ii t = 0.65, 5.35, 12.65, 17.35 5␲ ␲c 8 a i ii cm 3 6 b 19.78 cm ii 11.25 cm2 c i 14.62 cm2 2 iii 288.29 cm a2 1 a2 1 e d c 9 b 32 4 8 2 1 2 1 n−1 f i An = a = a 2 (2)1−2n 2 4 2 ii a 2 3 10 a k1 = 2, k2 = 3 y b i

3k1

60 50 47.5 40 30 20 10 0

3

3

(4, 47.5) h(t) = 35 + 25 cos π t + 1 3 2 , 10 3 1

2

8 , 10 3 3

4

t (s)

1 2 p + q 2 − 2pq cos ␪ 2 1 2 p + q 2 + 2 pq cos ␪ by= 2 √ d 31 cm 17 b i 51.48 cm ii 4764.95 cm2 iii 94.8% 18 b AC 2 = 61 + 60 cos ␪ c i 9.12 cm ii 43.18 cm2 16 a x =

x=3

3 0

iv (−1, −1), (2, 2), √ 1 √ 1 −1 − 5 , −1 + 5 2 2 √ 1 √ 1 and −1 + 5 , −1 − 5 2 2 13 a 155 m b i 16.00 m ii 29.04 m iii 17◦ 2 c 32.7 cm 14 a 12.05 pm b 2752 km c 26.05◦ 1 15 a a = 35, b = 25, c = 3 5 11 b i , s 3 3 ii 0.13, 1.20, 2.13, 3.20 s c h (cm) 5 , 60 11, 60

x

y = –k1(x – 3)



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Answers

Chapter 15

Exercise 14A

Exercise 15A

1 ax bx cy dx ex fx

= 100, y = 50 = 126, y = 252, z = 54 = 145, z = 290 = 180, y = 90 = 45, y = 90, z = 270 = 110, y = 100

2 a x = 68, y = 121 c x = 50, y = 110

1 a

b 5

}

Answers

Chapter 14

2

1

b x = 112, y = 87

c

d

1

3

3 110◦ , 110◦ and 140◦ 4 ∠ABC = 98◦ , ∠BCD = 132◦ , ∠CDE = 117◦ , ∠DEA = 110◦ , ∠EAB = 83◦ 7 60◦ or 120◦

Exercise 14B 1 a x = 73, y = 81 b x = 57, q = 57 c x = 53, y = 74, z = 53 d x = 60, y = 60, z = 20, w = 100 e w = 54, x = 54, y = 72, z = 54

2

4

2 a = 5, b = 1 1 4 a −2 −2 d 3

3 a = 3, b = −15 −1 2 c b −3 2 1 e 3

y

5 a

2 a 40◦ b 40◦ c 80◦ ◦ ◦ 3 ∠ACB = 40 , ∠ABC = 70 , ∠BAT = 40◦ 4 32◦ and 148◦

y

b 2

4

2

x

0

0

x

–3 –3

Exercise 14C

y

c 1 a 10 cm 2 7 cm

b 6√ cm 3 5 6 cm

0

d 1

4

y 0

x

1

x

Multiple-choice questions 1B 6A

2A 7C

3E 8B

4A 9A

5C 10 A e

Short-answer questions (technology-free) 1 ax bx cx dx

y 0

= 110, y = 70 = 35, y = 35 = 47, y = 53, z = 100 = 40, y = 40, z = 70

5 a x = 66 b x = 116 c x = 66, y = 114

–4

–4

1

4

x

–4

6 a y

y

b

7 3 cm

Extended-response questions 4 b 24 cm2

4

1 0

x 2 0


x 3


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c

d

3 a5

y

b2

c5

d 13

b x = 2, y = −7 5 5 7i + j 2 −2 2 6 a i i ii i+j 5 5 1 1 1 2 iv i + j iii − i+j 3 6 6 5 v 2i + j → − → 1− b i ON = OA ii 1 : 5 6 √ 7 4 2 units 3 b x = 6, y = 2 8 a k = ,l = 1 2 5 1 c x = 3, y = 3 d k = − ,l = − 3 3 √ 9 a 3i − 2 j b 13 4 a 13

4 3

4 3

0

x 1 y

e

2 –2

5

–1

2

4

y

f

0

x

0

3

–1

x

0

3

x –3

7 a and c 8 a, b y B 4 C 3 2 A 1 0 2 x –1 3 4 1 –1 D

10 a −2i + 4 j

d parallelogram 4 −5 2 9 ai iii ii −2 0 −1 b a + b = −c 10 m = −11, n = 7 1 11 a i b − a ii b 2 − → − → b MN = AD − → − → 1 12 a CB = a − b, MN = (b − a) 2 − → − → b CB = −2MN 13 a a e −a

bb fb−a

c 2a ga+b

d 2b

b −b ca+b eb−a 1 1 c (a + 2b) b (b − a) 15 a a − b 3 3 1 1 d (a + 2b) e (4a − b) 9 9 16 a u + v bv +w cu+v +w − → −→ 1 17 a OB = u + v, OM = u + v 2 2 1 1 c u− v bu− v 3 2 2 − → 2 → 2− d OP = (u + v) = OB 3 3 e2:1 14 a a d −a − b

Exercise 15B 1 2i − 7 j 2 a 5i + 6 j

b −5i + 6 j

c 5i − 6 j

b −6i + j

c5

11 a D = (−6, 3) b F = (4, −3) 3 3 ,− cG= 2 2 12 A = (−1, −4), B = (−2, 2), C = (0, 10) 13 a i 2i − j ii −5i + 4 j iii i + 7 j iv 6i + 3 j v 6i + 3 j b D = (8, 2) − → − → 14 a OP = 12i + 5 j, PQ = 6i + 8 j b 13, 10 √ √ √ ii 116 iii 145 15 a i 29 √ √ √ b ( 29)2 + ( 116)2 = ( 145)2 16 a i −i − 3 j iii√−3i + j √ b i 10 ii 2 5

ii 4i + 2 j

17 a i −3i + 2 j

ii 7 j 1 iv (−3i − 5 j) 2

iii

√

10

iii −3i − 5 j −3 9 bM= , 2 2 1 1 18 a (3i + 4 j) b √ (3i − j) 5 10 1 1 d √ (i − j) c √ (−i + j) 2 2 1 1 6 1 f √ (3i − 2 j) i+ j e√ 3 13 13 2

Exercise 15C 1 a√ −i + 2 j − k√ b 3i − 5 j + 6k c 14 d3 2 e −5i + 6 j − k 1 1 3 2 a i √ i+√ j−√ k 11 11 11 6 2 2 ii − √ i − √ j + √ k 11 11 11



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Exercise 15D 4 1 iii q − p p ii p 5 5 1 1 v q iv (q − p) 5 5 b RS and OQ are parallel c ORSQ is a trapezium d 120 cm2 6 k 2 1 2 a i a+ b ii a + b 7 7 3 3 7 b i 3 ii 2 −→ 15 −→ 9 i+ j 3 a i O D = 2i − 0.5 j, O E = 4 4 √ 170 ii 4 9 15 b i p i+ j 4 4 ii (q + 2)i + (4q − 0.5) j 2 1 c p = ,q = 3 2 1 5 ar +t b (s + t) 2 1 a i

1 (a + b) 3 1 iv (2a − b) 3

7 a i a+b

ii

iii b − a 2 v (2a − b) 3 − → − → b TR = 2 PT ∴ P, T and R are collinear. 8 a√ i −2 ii 5 iii 2 b 33 √ 9 109 units √ 10 a 11i − 2 j + 3k b 30 1 c √ (5i + 2 j + k) d 2i + 4 j 30 11 a (−1, 10) b h = 3, k = −2 12 m = 2, n = 1 13 a b = a + c

bb=

−15 −31 b −20 −32 c |OR| = 25 √ √ √ a 34 b 10 − 20 c r = i − 9j 1 a b x = −2, y = 2 2 c p = 4, q = 2, r = 2 −20 7 b a (25, −7) 15 24 k − 12 a (12, 4) b −4 √ 40 c 160, k, (k − 12)2 + 16, k = 3 d 34.7◦

1 a

2 3

4 5

Chapter 16

1C 6B

Exercise 16A

3E 8C

4A 9D

5B 10 C

3 2 a+ c 5 5


Multiple-choice questions 2C 7A

Answers

15 5 5 b√ i+√ j−√ k 11 11 11 √ 14 3 √ (i − j + 5k) 3 3 √ 1 3 4 a i − 3 j b 10 c i + j −k 2 2 5 a 2 j + 2k b i + 2j c i + 2k d i + 2 j + 2k e −2 j f −2 j + 2k g i + 2 j − 2k h i − 2 j − 2k 1 17 6 a i + 2 j + 2k b 6 6

1a

Short-answer questions (technology-free) 12 1 a b ±9 7 2 A(2, −1), B(5, 3), C(3, 8), D(0, 4) 11 1 3 p= ,q =− 6 12 √ 1 b √ (i − 5 j + 8k) 4 a 3 10 3 10 56 1 16 b (4i + 3 j) 6 a (4i + 3 j) 5 25


b

A

B

2

3

30°

O

Z

45° O

c

dO 2

Z

4 O 2

60°

Z

30° D

Z

120°

C


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Essential Advanced General Mathematics e

c r = 2 cos ␪

fF

E

O 5

5

50° O

1

[2, 0]

Z

130° O

Z

5

Z

50°

d r = 2␪, 0 ≤ ␪ ≤ 6␲ 9π,

9π 2 [4π, 2π]

g 130°

5

h Z

O

5

2a

bO

π A

2

Z

O

[12π, 6π]

130°

1 B

3π 4

3 O

11π 2

␲ ␲ , ≤ ␪ ≤ 4␲ ␪ 6

er =

dD

π 2

11π,

Z

Z

1 O

2, Z

1 , 4π 4

√ ␲ √ = [4 2, 45◦ ] 3 a 4 2, 4 −␲ b 2, = [2, −60◦ ] 3 −␲ c 4, = [4, −30◦ ] 6 −5 d 13, cos−1 ≈ [13, 113◦ ] 13 √ √ −5 e 61, sin−1 √ ≈ [ 16, −40◦ ] 61 ␲ f 2, = [2, 30◦ ] 6 5 g 13, −␲ + cos−1 ≈ [13, −113◦ ] 13 4 h 5, cos−1 ≈ [5, 37◦ ] 5 √ b (0, −4) 4 a (− 3, −1) 1 −1 c √ ,√ d (4, 0) 2 2 √ √ −5 −5 3 e (− 3, 1) , f 2 2 √ 1 − 3 g (−2, 0) h − , 2 2

3

[1, π]

O

Z

1, 2π 2

2 3π , 3 2

f r = cos 2␪ 1,

[1, π]

π 2

O

[1, 0]

Z

1, 3π 2

g r = 5(1 + cos ␪) 5,

π 2

O

[10, 0]

Z

5, 3π 2

h r = 2(1 − sin ␪)

4 br = sin ␪

3 5ar = cos ␪

6, π 6

π 2

Z

O

O

[8π, 4π]

[10π, 5π]

Z

H

G

c C

O

[2, 0] [2, π]

π 4, 2

O

Z

Z O

Z


4, 3π 2


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Answers

2,

π 2 [–1, π]

O

kr =

Z

2, 3π 2

√ j r = ± cos 2␪ [1, π]

[5, 2π]

O

[1, 0]

Z

␲ , 0 ≤ ␪ ≤ 6␲ ␪ √2,

π 2

π 4

2,

Exercise 16C

O

[1, π]

√2

, 2π 2

√6

3

√6

, 3π

6

2

Z

, 6π

l r = 2 sin 2␪ π 2, 4

–2, 7π 4

Z O

2, 5π 4

2 a a = 2, b = −2 b a = 3, b = 2 or a = 2, b = 3 1 2 c a = 5, b = 0 d a = ,b = − 3 3 3 a 6 − 8i b6−i√ c −6 − 2i d 7 − 3 2i e −2 − 3i f 4 + 2i g 6 − 4i h −4 + 6i i −1 + 11i j −1 √ 4 a 4i b 6i c 2i d −i e −1 f1 g −2 h −12 i −4 5a1+ 2i b −3 + 4i √ √ c − 2 − 2i d − 6 − 3i

1 a 15 + 8i b −8i d 2i e5 2a√ 2 + 5i b −1 − 3i c 5 + 2i d 5i 3 a2+i b −3 − 2i d −4 − 7i e −4 − 7i g −1 − i h −1 − i 4 a 2 + 4i b 20 d 8 − 16i e −8i 1 g (1 + 2i) h −4 − 2i 10 17 1 ,b = − 5a= 29 29 7 6 6a − i bi 17 17

–2, 3π 4

6 cos ␪ + sin ␪ sin ␪ 4 cr = d r2 = cos2 ␪ 1 + 3 sin2 ␪ 7 a x 2 + y2 = 4 b x 2 + y 2 − ax = a x 2 + y 2 c x 2 + y 2 = ax

6ar =4

c

br =

e 7a 8a

3

d (x 2 + y 2 ) 2 = 2a(x 2 + y 2 + 2x y) 1 2 e y 2 = a(a − 2x) fy= (a − x 2 ) 2a

Exercise 16B 1 a b c d e f

Re(z) 2 4 1 2 −4 √0 2

Answers

i r = 3 cos ␪ + 2

707

Im(z) 3 5 3 − 2 0 3√ −2 2

c

c −2 + 16i f −4 + 19i c −4 + 7i f −1 + i c4 f8

1 1 7 1 d− − i − i 2 2 2 2 3 1 3 2 f + i + i 20 20 13 13 3 5 = ,b = − 2 2 1 2 (3 + 4i) b − (1 − i) 2 5 1 1 (6 + 43i) (4 + i) d 130 17

e 2 − 2i

Exercise 16D 1 A 3+i D 2 − 2i 2


b d

C −3 − 4i F −1 − i

B 2i E −3 y Im(z) 1

c

x –5 –4 –3 –2 –1 0 1 2 3 4 Re(z) –1 f –2 e –3 –4 a


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Essential Advanced General Mathematics 3 a z1 + z2 = 3 − i

Exercise 16F

Im(z) 4

z2

–3 –2 –1 0 –1

x 6 Re(z)

3 z1 + z 2

–5

z1

b z 1 − z 2 = 9 − 9i z2 Im(z) 4

–3

0

6

z1 – z2

–9

4 c (–8 + 6i)


Im(z) 6

e 0

–8

1D 6D

a (1 + 3i)

3

1

d (–1 – 3i) –3

c (–2 + 5i)

x Re(z)

9

z1

–5

5

␲ √ ␲ b 2 cis − 4 3 √ −3␲ 5␲ d 4 2 cis c 4 cis 4 6 1 3␲ −␲ f √ cis e 24 cis 4√ 3 2 √ √ 2 1 2 a 3i (1 + 3i) b √ (1 + 3i) or 2 2 √ √ −5 5 2 c 3+i d− (1 − i) or √ (1 − i) 2 2 √ f 3(1 − i) e −6( 3 − i) √ √ 5 5 g − (1 + 3i) h − (1 + 3i) 2 √ √2 3 a 3 2(1 + i) b 6(1 + 3i) √ √ 5 c − (1 − 3i) d 18(1 + 3i) 2 √ √ e√ −18(1 + 3i) f 3(1 + i) g 3+i h −4 √ 5 i −4(1 − 3i) j− 2 1 a 2 cis

1 – 3 i 10 10 x Re(z)

2C 7E

3B 8C

Im(z) 5

1 a [3, π]

Z

O b (5 + 2i)

2 –5 d (–5 – 2i)

0 –2

b

2,

x 5 Re(z)

2

π 3

π 3

Z

O –5

5E 10 C

Short-answer questions (technology free)

b (1 – 3i)

–2

4D 9D

a, e (2 – 5i)

c

[–2, 210∞] 210∞

Exercise 16E 1 a ±2i d 2 ± 4i √ 1 g (−3 ± 3i) 2 √ 1 i (1 ± 23i) 6 √ 1 k (3 ± 11i) 2

√ c ± 5i √ f 1 ± 2i

b ±3i e −1 ± 7i √ 1 h (−5 ± 7i) 4 j 1 ± 2i l3±

√

Z

O

d −3,

11␲ 6


11π 6 O

2 a (−3, 0) √ c ( 3, 1)

5i

Z

√ b (1, 3) √ −3 3 3 d , 2 2


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Answers

O

Z

␲ b [r, ␪] : ␪ = 3 π 3

Z

O

c {[r, ␪] : r = −4} O

−5␲ d [r, ␪] : ␪ = 4

Z

4

O

Z

–5π 4

√ ␲ −␲ 4 a 3 2, b 1, 4 6 4␲ −␲ c 5, d 8, 3 4 5ar =4 br =3 8 cos ␪ = 8 cot ␪ cosec ␪ c r = 0 or r = sin2 ␪ 4 sin ␪ d r = 0 or r = = 4 tan ␪ sec ␪ cos2 ␪ 8 e r = ± 1 + 3 sin2 ␪ −2 fr = 2 cos ␪ − sin ␪ 3 2 9 2 6 a x 2 + y 2 = 25 = bx + y− 2 4 c x 2 − y2 = 9 16 2 64 d3 x+ − y2 = 3 3 7 2 196 2 e3 x− + 4y = 3 3 1 2 f y = (x − 1) 2 7 a (2m + 3 p) + i(2n + 3q) b p − iq c (mp + nq) + i(np − mq) (mp + nq) + i(np − mq) d p2 + q 2 e 2m f (m 2 − n 2 − p 2 + q 2 ) + i(2mn − 2pq) m − in g 2 m + n2 (mp + nq) + i(mq − np) h m 2 + n2 3[(mp + nq) + i(np − mq)] i p2 + q 2 8 Im(z) 2 c

d

√ b −2 − 2 3i √ 1 c −8 d (1 + 3i) 4 √ √ 1 e 1 + 3i f (1 − 3i) 4 ␲ √ ␲ 9 a 2 cis b 2 cis − 4 3 √ √ ␲ 3 −1 d 6 cis c 13 cis tan 4 6 −3␲ −␲ e 6 cis f 2 cis 4 6 √ √ √ 3 2 3 2 10 a −1 − 3i + i b 2 2 √ √ √ √ −3 2 3 2 3 2 3 2 c + i + i d 2√ 2 2 2 −3 3 3 f1−i e − i 2 2

Extended-response questions 1a

Z

O

Z

O

b x 2 + (y − 1)2 = 1 and (x − 1)2 + y 2 = 1 c x 2 + (y − a)2 = a 2 , a circle with centre (0, a) and radius a and (x − a)2 + y 2 = a 2 , a circle with√ centre (a, 0) and √ radius a 3 a z = 3 + i or z = 3 − i b i Im(z) z = √3 + i

1 0

√3

Re(z) z = √3 – i

–1

ii√x 2 + y 2 = 4 4a6 2 b6 5a

iii a = 2

Im(z) Z

P

A

O

e

–8 –6 –4 –2 0 f 2 a –2 b –4

√ 3i

a1−

3

Answers

3 a {[r, ␪] : r = 3}

709

Re(z)

Q

z 4 Re(z)

b


√ 2+1


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Chapter 17

Exercise 17B

Exercise 17A 9 1ax = 2

y

1a

8

y

x

3

–3 O –8

O

b y = 10

9 2

x

y

b

10

y 10

–5 O

x

5

–10

x

O

c x 2 + y2 = 9

c

y 3

y –8

3

–3

O

3

x

x

d

y 5

–3

d y = −6 or y = 6

8

O –3

y –3 O

6

3

x

–5

O

x

2 a Centre = (3, 4)

y

–6 12

2 10y = 4x + 29 3x+y=4 4 x 2 + y2 = 9 (3x − 2)2 3 (y − 5)2 − =1 5 4 4 2 16 964 6 x+ + y2 = 7 y = −x 3 9 2 2 80 4 2 + y− = 8 x+ 3 3 9 9 x = −1 or x = 2 1 2 1 1 1 10 x − + y− = 2 2 2 16 2 64 12 x + y = 1 11 x − + y2 = 3 9 13 y = 2x + 1 14 y = 1 or y = 5 1 2 15 y = 12 (x + 1) 16 y = − x 2 − 1 12 17 The parabola with equation y 2 = 3x can be described as the locus of points P(x, y) which satisfy the that the distance of P to the property 3 point F , 0 is equal to the distance PM, the 4 perpendicular distance to the line with equation 3 x =− 4


3√3 2

3– 4

x

O –4

3√3 2

3+

b Centre = (−3, −4)

y –24 5

1 x

O –4 –6 5

c Centre = (2, 3)

–9

y 7

2 – √7 2

3 x

O –1 2+

√7 2


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Answers 2 a Centre = (3, 4)

y

y y = –8 x + 12 3

5

O 2

8

x 3–

3√5 2

2

3+

x y2 4 + =1 25 9

y

–5 y= 3x–9

x2 y2 + =1 64 16

x

8

O –4

3√5 2 x

b Centre = (−3, −4)

4 –8

O –4

2

y x + =1 3 25 9 y 5

8 x–4 3

y=

4

–5 2

Answers

d Centre = (5, 0)

O

(x − 4)2 y2 + =1 6 16 12 (3y − 28)2 3x 2 + =1 7 16 64

y=

711

5 x+1 3

x –4

–3√41 –3 5

3√41 –3 5 –9

c Centre = (2, 3)

Exercise 17C

y

1a

y 8 y=– x 3

y = 2x – 1

8 y= x 3 3 + 4√2

O –3

x

3

b y = –2x y

O

2 4

x

3 – 4√2

y = –2x + 7

y = 2x

d Centre = (5, 0)

O –5

x

5

25 3

20 3

y

c

5 25 y= x– 3 3

y

O 3 O

3 y=– x 8

2 5

3 y= x 8 x

– 20 3

8

x 5 25 y=– x+ 3 3

– 25 3

–3

e Centre = (0, 0) d

5 y=– x 3

y

y 5 y= x 3

y = –x

y=x

O –3

O

3

x


–2

2

x


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Essential Advanced General Mathematics f Centre = (0, 0)

Answers

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7 Centre (2, 0) y

y y = –√2x

y = √2x 2√5 3

O x

√2

– √2

8x =±

g Centre = (2, −1) y y= O 2

4

–2 1 y=– x 2

h Centre = (5, 3) y 3 y= x 5 6 5 – 5√2

5 + 5√2

3 O

5

x

10 3 y=– x+6 5

(x + 6)2 y2 − =1 16 48 x2 y2 (15y − 56)2 15x 2 6 − =1 5 − =1 4 5 256 256

3

y2 x2 − =1 9 7

4

Multiple-choice questions 1C 6A

2D 7B

3B 8D

4D 9A

5

x

k , where k is the given constant. 16


1 x–2 2 x 2 + 2√5

–1

2 – 2√5

O

–1 – 2√5 3

5D 10 C

Short-answer questions (technology free) √

1ay=x b (x − 6)2 + y 2 = 36 2 c (x + 2) + (y − 8)2 = 32 d (x + 10)2 + y 2 = 64 e (x − 10)2 + y 2 = 64 y2 x2 + = 1 g y = 0, −6 ≤ x ≤ 6 f 100 64 2 2 4y 5 4x − = 1, x ≤ − h 25 119 2 4x 2 4y 2 5 i − = 1, x ≥ 25 119 2 (3y − 14)2 3x 2 1 2 a y = x2 + 3 b + =1 4 16 4 c x 2 − 3y 2 + 8y = 0 3 a Let triangle OAB have vertices with coordinates O (0, 0), A (a, 0) and B (x, y). The locus of the point B is a circle with equation (x + a)2 + y 2 = 4k 2 where k is a constant. b Let triangle OAB have vertices with coordinates O (0, 0), A (a, 0) and B (x, y). The locus of the point B is a circle with 2 ak 2 a2k2 equation x − 2 + y2 = 2 k −1 (k − 1)2 where k is a constant. (k = 1, a straight line) c There are three cases to consider: Case 1: Let quadrilateral OBAC have vertices with coordinates O (0, 0), A(a, 0), B(b, c) and C(x, y). The locus of the 2k point C is y = − where k is a a constant.

1 Centre = (−2, −4) , radius = 2 5 2 Centre = (−2, 0), axes intercept at (−4, 0) and (0, 0) 3 y = 3x − 8 4 (x − 3)2 + (y − 2)2 = 36 2x 2x 5 y =− ,y = 3 3 2 2 160 2 6 (x − 8) + y + = 3 9


Case 2: Let quadrilateral OABC have vertices with coordinates O (0, 0), A(a, 0), B(b, c) and C(x, y). The locus of the 2k c where k is a point C is y = x − b b constant. Case 3: Let quadrilateral OACB have vertices with coordinates O (0, 0), A(a, 0), B(b, c) and C(x, y). The locus of the point C is bc − 2ac − 2k c x+ y= b−a b−a where k is a constant.


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Answers

1B 6C 11 B 16 B 21 D 26 A 31 E 36 E

2C 7B 12 A 17 C 22 E 27 C 32 C

3B 8D 13 B 18 B 23 A 28 E 33 C

4D 9B 14 D 19 B 24 C 29 C 34 B

5C 10 D 15 A 20 B 25 C 30 D 35 D

18.2 Extended-response questions − → 1 a AE =

1 (2a + tb) t +1 − → 1 − → 9 b AE = (7a + AF) d t = 8 7 2 b (n − 1) a − nb + c − → − → −3 1 3 a AB = b − a, PQ = a+ b 10 2 1 −3 1 1 b i n a+ b ii k + b− a 10 2 2 2 5 1 c n = ,k = 3 3 4 b ∠BCA = x ◦ , ∠BOA = 2x ◦ ◦ ◦ ∠TAB √ = x , ∠TBA = x 5 a 4 2 blowing from SW b 200 m downstream c true velocity = 43.1 km/h at bearing 80◦ d 222◦ → 1 − → − → 1 − 6 b ii ZG = h ZH + k ZK 3 3 1 1 2 iii + = 3 iv h = (similarity) h k 3 4 2 v cm 9 − → 1 vi h = ; H is the midpoint of Z X , K = Y 2 vii h 1

1 ≤ k ≤ 1; 2 1 ≤h≤1 2

1 2 1 2

viii

1

k

A 2, 4 3 9

1 1 2 1 2

1 k

Chapter 19

b 2 cm to the right of O c moving to the left at 7 cm/s d when t = 3.5 s and the particle is 0.25 cm to the left of O e −2 cm/s f 2.9 cm/s 2 a after 3.5 s b 2 m/s2 c 14.5 m d when t = 2.5 s and the particle is 1.25 m to the left of O 3 a 3 cm to the left of O moving to the right at 24 cm/s 4 b v = 3t 2 − 22t + 24 c after s and 6 s 3 22 d 11 cm to the right of O and 39 cm to the left 27 of O 2 e4 s f a = 6t − 22 3 11 s and the particle is g when t = 3 16 13 cm left of O moving to the left at 27 1 16 cm/s 3 2 4 a when t = s and a = −2 cm/s2 , and when 3 t = 1 and a = 2 cm/s2 5 b when t = s and the particle is moving to the 6 1 left at cm/s 6 5 when t = 2 s, v = 6 cm/s, a = −14 cm/s2 , when t = 3, v = −5 cm/s, a = −8 cm/s2 , when t = 8 s, v = 30 cm/s, a = 22 cm/s2 3 6 a t = 4 s and t = −1 bt = s 2

Answers

Chapter 18 18.1 Multiple-choice questions

713

Exercise 19B 1 a x = 2t 2 − 6t b at the origin O c 9 cm d 0 cm/s e 3 cm/s 2 a x = t 3 − 4t 2 + 5t + 4, a = 6t − 8 5 23 b when t = 1, x = 6, when t = , x = 5 3 27 c when t = 1, a = −2 cm/s2 , when 5 t = , a = 2 cm/s2 3 1 3 20 m to the left of O 4 x = 215 , v = 73 3 5 a v = −10t + 25 b x = −5t 2 + 25t 1 c 2.5 s d 31 m e5s 4 6 the 29th floor

Exercise 19C √ 1 2 10 s 2

3 a 3 m/s

Exercise 19A c 337.5 m 1 a 12 cm to the right of O


2 37.5 m 2 s 3 500 d s 27 b6


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Essential Advanced General Mathematics 4 a 2.5 m/s2 5 a 50 s 6 a 20 s 7 a −19.2 m/s 8 a −59.2 m/s 9 a 10 s 10 a 4.9(1 − 2t) m/s

b 31.25 m b 625 m b 10 m/s b 1.6 m b −158.4 m b after 3 s and 7 s b 4.9t(1 − t) + 3 m 10 d s c 4.225 m 7 11 a 2√s b 44.1 m c4s d5s 12 10 10 m/s

Exercise 19D 1 65 m 2 a 562.5 m b 450 m c 23.75 s 200 m/s 3 4 210 m 3 5 a 500 m b 375 m c 17.57 s 6 a 12.5 s b 187.5 m 7 No, the first train will stop after 6.25 km and the second train will stop after 6 km. 2 b 1 min 6 s 8 a 57.6 km/h c 0.24 3

1 1 −1 bs = − 3 2 2t t b −15 m/s2 9 a a = 3t 2 − 22t + 24 7 1 m to the left, 60 m c2 12 12 10 40 m 100 11 a 2.5 m/s2 b 8 s c 500 m s d 9 2 2 12 a 41 s b 347 m 3 9 2 6 13 a 7.143 s b 2 s, 4 s 7 7 14 a 2 s b 39.6 m c4s d 4.84 s 15 437.5 m 16 a 288 m b 16 s 80 17 16 m/s 18 m/s 81 19 a 0 m/s b −3 m/s2 c −4 m/s 11 2 e m to the left d4 m 12 3 t3 b t 2 − + 8t 20 a 2t − t 2 + 8 3 21 a 4t − 2t 2 + 6 b i 8 m/s ii 2 s iii 18 m 22 a 27 m/s2 b 50 m/s c 4.5 s 23 a −10 m/s b 0 m 24 a 4 s, 6 s b 36 m c0≤t <5 8aa=

Multiple-choice questions 1A 6C

2E 7D

3C 8E

4C 9A

5E 10 D

Short-answer questions (technology-free) 1 a 5 cm to the left of O b 8 cm to the left of O c −4 cm/s d t = 2 s, 9 cm to the left e −1 cm/s 2 f 1 cm/s 3 2 a 8 cm to the right, 0 cm/s, −4 cm/s2 b at 0 s, 8 cm to the right, 22 4 −4 cm/s2 and at s, 6 to the right, 3 27 4 cm/s2 3 a 3.5 s, −40.5 cm/s, −36 cm/s2 b2s c 31 cm 1 4 a i cm to the left 8 ii 1 cm/s2 iii 1 cm/s 32 cm ii b i 0 s, 2 s 27 3 5 a 12 m/s bs =t c 17.57 s 2 6a4s b 18 m to the right 3 1 2 d 1.5 s c −5m/s e 6 m/s 4 1 7a m to the left b −1 m/s c −5 m/s2 12

Extended-response questions 1 cm to the left of O 3 b 4 cm/s c 2 cm/s2 d at 2 s 1 e cm to the right of O 3 f at 1 s 3 a after 6 s at −36 m/s b when t = 0 or 4, when t = 4 the maximum height is 32 m c after 2 s 4 x(1) − x(0) = 15.1, x(2) − x(1) = 5.3, x(3) − x(2) = −4.5, x(4) − x(3) = −14.3, x(5) − x(4) = −24.1, x(6) − x(5) = −33.9, x(7) − x(6) = −43.7, x(8) − x(7) = −53.5, x(9) − x(8) = −63.3, x(10) − x(9) = −73.1 The constant difference between successive numbers is −9.8, the acceleration due to gravity. 6 33 m 7 a v = −5t + 25, 0 ≤ t ≤ 5 b 62.5 m 8 25 m to the left of O 2u 9 When T = , the second particle is projected g upwards at the instant the first particle lands. 2u When T > , the second particle is projected g upwards after the first particle has landed. Hence 2u there is no collision. for T > g 1a2



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Answers

21.2 Extended-response questions

Exercise 20A 5 2 kg wt 2 90◦ 2 3 T1 = 14.99 kg wt, T2 = 12.10 kg wt 4 28.34 kg wt, W48.5◦ S 5 T = 40 kg wt, N = 96 kg wt 6 F = 6.39 kg wt 7 a No b Yes 1 T1 = T2 =

c 7.62 kg wt

10 32.97 kg wt, 26.88 kg wt, 39.29 kg wt, W = 39.29 kg

c 1.043 s, 1.957 s

2 5.74 kg wt

3 3.73 kg wt, 8.83 kg wt

4 4.13 kg wt

6 31.11 kg, 23.84 kg wt

7 44.10 kg, 22◦ 29 to the vertical

1 a categorical

b numerical

c categorical

d numerical

10 B

b categorical

3 a discrete

b discrete

c continuous

Short-answer questions (technology-free) 1 T1 √ = 9 kg wt, T2 = 12 kg wt ◦ 2 10√3 kg wt, 150 √ to the 10 kg wt 3 14 √5 kg wt, 28 5 kg wt −7 15 3 5 4 kg wt 8 3 √ √ 40 3 15 2 kg wt 6 7 kg wt 3 2 √ 8 28 kg, 14 3 kg wt √ 9 4 3 kg wt

e discrete

Exercise 22B

er

a

th

zz Pi

O

s

e

ip

es in

ch & Fi sh

Ch

ke n ic

rg e bu am H


di ron sa gl gr y ee

ee

50

St

5C 10 B 15 D 20 C

100

D kn on’t ow D isa gr ee

4B 9C 14 E 19 A

150

gr

3B 8D 13 C 18 B 23 B

Food type

b Hamburgers 2 a

A

2E 7A 12 D 17 D 22 E

25 20 15 10 5

rs

1 a

Chapter 21 21.1 Multiple-choice questions 1D 6C 11 D 16 A 21 C

d continuous

Ch

9C

f categorical

2 a numerical

ro ag ng re ly e

8A

e categorical

5C

No. of students

7B

4C

ii −10 − 6t

Chapter 22

No. of responses

6B

3A

7V (a + r ) 6ar

b i 4 m/s ii −10 m/s2 1 c2 s 3 2 2 d Y : −9 m/s, X : −35 m/s, X and Y 3 3 travelling in the same direction, X moving faster catches up to Y.

Multiple-choice questions 2E

ii

Exercise 22A

8 6.43 kg wt, 7.66 kg wt, 11.92 kg 9 3.24 kg wt

1E

d 3.81 s

2V (a + r ) 3 b i 3ar 4 c V 7 16 000 4 m; 52.567 m/s 49 5 a i 4 − 10t − 3t 2

Exercise 20B 1 13.05 kg wt

b4

2 a 14.7 m/s

St

8 146◦ 53 , 51◦ 19 , 161◦ 48 9 a 7.5 kg wt b 9.64 kg wt

5 6.93 kg wt

2 m/s 9 b 24.66 m/s

1 a p = −4, q = 3

√

Answers

Chapter 20

715

Attitude to capital punishment

b 32


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c

100

er

ic

th O

H

M

or

ra D

Co

us

ro

a m

ed m

r

50

y

3 a

No. of borrowers


Type of film

b Music 4 a 50

Prices ($) Cumulative frequency less than 5 3 less than 10 9 less than 15 12 less than 20 15 less than 25 19 less than 30 19 less than 35 20 less than 40 20 less than 45 21

O

th


er

s rie

on Ph

W at

ch

ef

av

m

id

nd

ic us

ad Re

to

Li

ste

n

W at

eo

25

ch

Percentage of students

Answers

716

CUAU033-EVANS

TV

P1: FXS/ABE

Leisure activity

b Watching TV

20 15 10 5 0

Exercise 22C 0 4

1 4

2 4

3 4

4 3

5 2

2 a4

b2

c5

d 28

3 a0

b 48

c 60–69

d 33

6 1

4 a, b Temperatures Relative (◦ C) Frequency frequency 0– 1 0.03 5– 0 0 10 – 1 0.03 15 – 9 0.28 20 – 4 0.13 25 – 5 0.16 30 – 7 0.22 35 – 4 0.13 40 – 0 0 45 – 1 0.03 c

10 8

8

b

30 31 32 Measurement

33

34

28

29

30 31 32 Measurement

33

34

6 4 2 0

6 4 2

5 10 15 20 25 30 35 40 45 Price

b


10 15 20 25 30 35 40 45 Temperature

d 47% 5 a No. of books

29

c The students’ estimates ranged from 28.9 cm to 33.3 cm, with most students (86%) over-estimating the 30 cm measure. 7 a 8

2 5

28

28

0

4

0

4

0

6

0

6

2

No. of students

No. of cities

10

6 a Frequency

Number Frequency


1

5 10 15 20 25 30 35 40 45 Price

10 20 30 40 50 60 70 80 90 100 Marks

30 25 20 15 10 5 0

10 20 30 40 50 60 70 80 90 100 Marks

c The students’ marks ranged from 21 to 99, with most students (over 70%) scoring more than 50% on the test.

b $5.00–$5.99



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Answers

No. of holes Cumulative frequency

b

14 12 10 8 6 4 2 0

240 260 280 300 320 340 360 380 400 420 Length of hole

50 40 30 20 10 0

240 260 280 300 320 340 360 380 400 420 Length of hole

4 25 ii no. of holes ≥ 360 m = 17, 17 proportion = 50 iii approximately 305 m

c i below 300 m =

5 a

Exercise 22D 1 a centre

b neither

2 a positively skewed b negatively skewed 3 symmetric

4a 2 5 8 3 5 6 9 4 5 6 9 5 2 6 8 7 5 5 6 8 9 8 2 4 9 5 10 16 | 4 represents $164 11 (truncated) 12 13 14 9 15 16 4 17 18 19 20 21 0 b approximately symmetric

c both c symmetric

4 symmetric

5 symmetric

Exercise 22E 1 a 4 899 5 | 0 represents 50 mm 5 02 7 7899 6 07 b two months 2 a 0 4 1 1 689 2 | 5 represents 25 hours 2 1 1 3 (truncated) 2 55567 799 3 1 1 2 33 3 69 4 1 4 6 b nine batteries 3a 0 0 1 0045569 2 001 37 89 3 37 9 4 6 4 | 6 represents 46 minutes 5 6 37 7 0 b three students c positively skewed

Father's age 3 4 4 5 5

Answers

8 a

717

Mother's age 7 889 0001 2 3333344 567 8999 000

4443331 1 0 988887 7 6665 42 1 1 0 5 0 | 4 represents 40 years 4 | 0 represents 40 years b Both distributions are approximately symmetric. Fathers, with ages centred in the late forties, tend to be older than mothers, with ages centred in the early forties. The spread is similar for both distributions. 6 a Class B Class A 32 1 9 2 2 3 9 4 57 8 5 58 9 6 58 643322100 7 1 67 99 8844321 100 8 01 2 2 559 81 9 1 9 9 | 6 represents 69 marks 7 | 1 represents 71 marks b Six students in class A and two in class B c Class B performed better as more students scored in the higher values of 70’s to 90’s.

Exercise 22F 1 a mean = 18.36, median = 14 b mean = 9.19, median = 10 c mean = 7.41, median = 7.65 d mean = 1.62, median = 1.15



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718

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Essential Advanced General Mathematics 2 a mean = 3.24, median = 3 b mean = −0.38, median = 0 3 mean = $193 386, median = $140 000; the median is a better measure of centre as it is typical of more house prices. 4 mean = 4.06, median = 4; both are reasonable measures of centre in this example. 5 a range = 602, IQR = 455 b range = 5.3, IQR = 3.2 c range = 0.57, IQR = 0.21 d range = 7, IQR = 3.5 6 a 145 b 42 7 a 2.4 kg

c 0

b 1.33

c 281.24

20

3 a

30

40

*

*

b The distribution of winnings is extremely positively skewed with a median value of $854 533. The middle 50% of players won from $752 228.50 to $1 573 674. There were two outliers: Lleyton Hewitt, who is only just an outlier, winning $2 766 051, and Roger Federer, clearly an outlier, winning $6 357 547.

d 3.04

1.0 0.8 0.6 0.4

4 a

0.2 0

0 17

22

27

32 Age

37

42

5

10

15

20

47

median 18, IQR 2 b mean = 20.97, s = 7.37 c 92% 11 a i mean = 6.79, median = 6.75 ii IQR = 1.45, s = 0.93 b i mean = 13.54, median = 7.35 ii IQR = 1.7, s = 18.79 c The error does not affect the median or interquartile range very much. It doubles the mean and increases the standard deviation by a factor of 20. 12 Approximately 95% of share prices lie between $44 and $56. 13 About 95% of days lie in this interval.

b The distribution is symmetric, centred at $10.00. The middle fifty percent of students earn between $8.15 and $11.85 per hour. 5 a

*

0

100

200

300

400

Exercise 22H After

*

Before

Exercise 22G

0

1 a m = 154, Q 1 = 141.5, Q 3 = 161.5, min = 123, max = 180 b 130

140

150

160

170

, 600 000

500

d The distribution is approximately symmetric, centred at about 210 000, with an outlier at 570 000. The middle 50% of papers have circulations from about 88 000 to 270 000.

1 a

120

50

0 1000 000 3 000 000 5 000 000 7 000 000 2 000000 4 000 000 6 000 000 winnings

9 a i mean = 17.61, s = 15.96 ii mean = 195.3, s = 52.9 b i 94% ii 100% Cumulative relative frequency

10

*

d The distribution of number of books borrowed is positively skewed, centred at 3. While 75% of people borrowed 13 books or less, one student borrowed 38 books and another borrowed 52.

b 1.0 kg

8 a 12.39

10 a

*

180

c The distribution of heights is slightly negatively skewed, centred at 154 cm, with the middle 50% of heights ranging from 141.5 cm to 161.5 cm. 2 a m = 3, Q 1 = 0, Q 3 = 13, min = 0, max = 52 b 38, 52


*

* 10

20

30

40

50

b The distribution for the number of sit-ups is negatively skewed before the course, centred at 26. After the course, the distribution is more symmetric, centred at 30, indicating that the course has been effective. The distribution after the course is more variable than before the course showing the course has not had the same effect on all participants. There is one outlier in the before group, who can achieve 46 sit-ups, and two in the after group, recording 50 and 54 sit-ups respectively.


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Answers 8

*

Year 8 0 0

1

2

3

4

5

6

7

8

10

20

30

40

50

9

Extended-response questions a Year 12

b Year 12

1 a numerical

3 a 1990

b 0 056689 1 4 4 5 8 9 3 | 2 represents 32% 2 567 8 3 2 2 4 4 5 3 c positively skewed d 21.0%

1970 15

20

25

30

35

40

45

50

b The distributions of ages in both groups are slightly positively skewed, with the mothers in 1970 (median = 24.5) generally younger than the mothers in 1990 (median = 28). The variability in both groups is the same (IQR = 10 for both groups).

e x = 20.05, m = 18 f

Frequency

2B

3D

4C

5D

6C

7A

8D

9C

10 A

6 5

Multiple-choice questions 1D

Answers

2 Year 12

719

4 3 2 1 0

Short-answer questions (technology-free) 1 a numerical

b categorical

2 a categorical

b 7.5%

12 Frequency

20 30 40 Divorce rate

8 6 4 2

50

60

ii 5

100.0 Cumulative frequency %

g

10

0

10

i positively skewed

80.0 60.0 40.0 20.0 0.0

10 20 30 40 0 No. of cigarettes smoked

0

4 a 2 2 3 9 4 | 7 represents 47 minutes 4 34579 5 01 1 22 456679 6 589 7 2 b m = 52, Q 1 = 47, Q 3 = 57 5 x = $283.57, m = $267.50 6 a x = 178.89, s = 13.990 b 92.9%

2 a

7

10 20 30 40 50 60 Divorce rate

ii ≈ 17%

i 58% 12 10 Frequency

3

0

8 6 4 2 0

0

5

10

15

20

25

60

i 21.4% iii 38.1%


70 80 Travel time

90

ii positively skewed


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Essential Advanced General Mathematics b x = 69.60, s = 9.26, min = 57, Q 1 = 62, m = 68, Q 3 = 76, max = 90

Chapter 23

c i 69.60 iii 33, 14 v 9.26

Exercise 23A

d

ii 68 iv 76 vi 51.08, 88.12

1 a

Met Hillside 50

60

70

80

90

100

e The distributions of travel times are both positively skewed. The travel times for the Met (median = 70) tend to be longer than the travel times for Hillside trains (median = 68). The spread of times is also longer for the Met (IQR = 24) than the travel times for Hillside trains (IQR = 14).

Response time (min)

Answers

720

CUAU033-EVANS

70 60 50 40 30 20 10 0

1.0 2.0 3.0 4.0 5.0 6.0 Drug dose (mg)

b negative association 2 a

Business ($’000)

P1: FXS/ABE

3 a Method 1

15 10 5 0

Method 2

100 200 300 400 500 600 Advertising ($)

3 a 60

70

80

90

100

b The distributions of scores are negatively skewed for methods 1 and 3 and symmetric for method 2. The scores for method 1 are higher than for methods 2 and 3 (90, 79 & 70 respectively), and are also less variable than method 2. They show similar variation to the scores for method 3. c Thus training method 1 would be recommended, as it consistently produces higher scores.

Second born

30

40

50

b The distribution for the first-born is symmetrical, while for the second and third-born the distributions are positively skewed. The centre for the first born is higher than for the second which is higher than the third (35, 21 & 12 respectively), whilst the variability is most for the first-born, followed by the second-born and then the third-born.

400

15 10 5 0

Third born 20

200 300 No. of seats

b positive association c (122, 378) is an outlier 4 a

4 a First born

10

c no outliers

800 700 600 500 400 100

Price ($’000)

50

Airspeed (km/h)

b positive association

Method 3

c no outliers

2

4

6 8 10 12 Age (years)

b negative association c (10, 8700) is an outlier

Exercise 23B 1 a no correlation b weak negative correlation c strong negative correlation d weak positive correlation e strong positive correlation f strong negative correlation g strong positive correlation



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Answers

4 a 0.67

b moderate positive correlation

5 a1

b perfect positive correlation

6 a −0.43

b weak negative correlation

a There is a strong positive linear relationship between the scores on Test 1 and Test 2. b Yes, the data is numeric and the relationship is linear. c q = 0.71, r = 0.87 d q: moderate positive linear relationship r: strong positive linear relationship e i q = 0.43, r = −0.004 ii The error in the data has a much greater effect on Pearson’s correlation coefficient.

Exercise 23D

Exercise 23C

Answers

h no correlation i strong negative correlation j weak positive correlation k strong positive correlation l moderate negative correlation 2 a 0.71 b 0.78 c 0.82 d 0.92 3 a −0.6 b moderate negative correlation

721

Note: Answers will vary for lines drawn by eye.

1 a no linear relationship b weak negative linear relationship c strong negative linear relationship d weak positive linear relationship e strong positive linear relationship f strong negative linear relationship g strong positive linear relationship h no linear relationship i moderate negative linear relationship j weak positive linear relationship k perfect positive linear relationship l perfect negative linear relationship 2 a 0.8 b 0.8 c 0.7 e −0.7

15

5 0

1 2 3 4 5 6 7 8

–5

4

x

4000

J

D

E F C K B A 2000

4000 Year 1

I G H

6000

b y = 424 + 0.794x c The positive slope indicates that districts with high rates in Year 1 also had high rates in Year 2. 4 a

330 310 290 270 250 230 210 0 200 220 240 260 280 300 320 340 Test 1


Height (cm)

Attempt 2 Test 2

3

6000 2000

a a strong positive relationship b Yes, the data is numeric and the relationship is linear. There are no outliers. 8

2

3 a

0

40 45 50 55 60 65 70 Attempt 1

y = – 4.5 – 3.75x

0 1 –3 –2 –1 –5 –10 –15 –20

7 70 65 60 55 50

x

y

2

f −0.2

3 a −0.86 b strong negative linear relationship 4 a 0.95 b strong positive linear relationship 5 a 0.77 b strong positive linear relationship 6 a −0.77 b strong negative linear relationship

y = 1 + 2x

10

Year 2

d 0.8

1 y

95 90 85 36 40 44 48 52 56 60 Age (months)

b y = 72 + 0.4x c The intercept (72 cm) is the predicted height at age 0. The slope predicts an increase of 0.4 cm in height each month. d i 89 cm ii 158 cm


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Essential Advanced General Mathematics e Part i is reasonable as it is a value close to the data. Part ii is not reliable as the relationship may no longer be linear here. 5 a 180

Daughter

Answers

722

CUAU033-EVANS

170 160 150 150

170 160 Mother

b y = 18.3 + 0.91x

180

c 173 cm

Cost ($’000)

6 a 4.0 3.5 3.0 2.5 100 120 140 160 180 200 Number of MP3 players

b y = 1300 + 13x d ≈ $13

3 a y = 50.2 + 0.72x b An increase of 1 cm in the mother’s height is associated with an increase of 0.72 cm in the daughter’s height, on average. c 172 cm (to the nearest cm) 4 a y = 1330 + 12x b $1330 c $12 5 a response time = 57.0 − 10.2 × drug dose b The intercept of 57.0 minutes is the predicted time for pain relief when no drug is given. From the slope, we predict a 10.2 minute decrease in response time for each 1 mg of drug given. c −4.2 minutes, which is not a realistic answer. 6 a business = 1123.8 + 18.9 × advertising b Intercept is the volume of business with no advertising. From the slope we predict an increase in business of $18.90 for every dollar spent on advertising. c i $20 044 (to the nearest dollar) ii $1124 (to the nearest dollar)

c ≈ $1300

7 a y = 70 − 14x b The intercept is the predicted time taken to experience pain relief if no drug is given. From the slope we predict a reduction of 14 minutes in time taken to experience pain relief for each mg of drug administered. c −14 mins, which is not a realistic answer 8 a y = 18.2x b Intercept predicts zero sales if nothing is spent on advertising. The slope means that on average, each $1 spent on advertising is associated with an increase of $18.20 in sales. c i $18 200 ii $0

Multiple-choice questions 1D

2E

3A

4B

5B

6E

7C

8C

9A

10 D

Short-answer questions 1 a

140 120 Score (points)

P1: FXS/ABE

100 80 60 40 30

Exercise 23E 1 a y = 68.2 + 0.46x b The y intercept is the predicted height at birth. From the slope, we predict an increase in height of 0.46 cm each month. c i 88 cm ii 168 cm d The height at 42 months is reliable since this in within the range of data given (interpolation). The height at 18 years is less reliable since this is outside the range of data given (extrapolation). 2 y = 487.6 + 0.77x

40

50 60 Inside 50

70

b Positive 20 3 0.927 4 weight = −200 + 2 × height, approximately 5 errors = 14.9 − 0.533 × time 6 a Intercept: no sensible interpretation. Slope: For each additional second taken to complete the task on average the number of errors is 1 reduced by about an error. 2 b 9.6

Extended-response questions 1 a IV = Exam score, DV = Number of new clients



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Answers b

4500

11.00 Weekly sales

Number of new clients

4000

10.00 9.00 8.00

2500

6.00

2000 25

60.00 65.00 70.00 75.00 80.00 85.00 Exam score

f Number of new clients = −4.00 + 0.173 × Exam score g Intercept: no sensible interpretation. Slope: On average each extra 1 mark in the final exam is associated with an increase of 0.173 clients. h 13 i Not very reliable as it is outside the range of the data. 2 a IV = Assignment mark, DV = Final exam mark 85.00 80.00 75.00 70.00 65.00 60.00 55.00 60.00 50.00 80.00 70.00 55.00 65.00 75.00 Assignment mark

c positive d 1, strong positive e 0.758, strong positive

30 35 40 45 50 55 Number of times played

c positive d 1, strong positive e 0.946, strong positive f Weekly sales = 293.06 + 74.281 × Number of times song played g Intercept: On average songs which get no airplay will have sales of around 293. Slope: On average each extra 1 play of the CD is associated with an increase of about 73 sales per week. h 4750 i Not reliable as it is outside the range of the data.

c positive d 1, strong positive e 0.748, moderate positive

Final exam mark

3000

7.00

5.00

b

3500

Answers

b

723

Chapter 24 Multiple-choice questions 1C

2A

3E

4D

5C

6B

7D

8C

9B

10 D

11 B

12 B

13 B

14 B

15 D

16 C

17 B

18 E

19 B

20 B

21 B

22 A

23 A

24 C

25 C

26 C

27 C

28 A

29 C

30 D

Extended-response questions 1 a

f The statement is incorrect as it is a causal statement. It can only be said that there is an association between the Assignment mark and the Final exam mark. g Final exam mark = 24.74 + 0.699 × Assignment mark h Intercept: On average those who score 0 on the assignment will score about 25 on the final exam. Slope: On average each extra 1 mark in the assignment is associated with an increase of 0.7 marks on the final exam. i 60 j Reliable, as it is within the range of the data. 3 a IV = Number of times the song was played, DV = Weekly sales

Aggressive 1 2 3 4 5 6

5 97300 555333 85 0 5 | 2 represents 25

Passive 3355788 2 2 557 003

1 | 5 represents 15

Passive b Aggressive min = 25 min = 13 Q 1 = 33 Q 1 = 15 median = 43 median = 22 Q 3 = 45 Q 3 = 27 max = 60 max = 33 c Passive Aggressive 10


20

30

40

50

60


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Answers

724

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5:6

Essential Advanced General Mathematics d Aggressive : mean = 42.07 cm, s = 10.25 cm; Passive : mean = 21.53 cm, s = 6.58 cm e Both distributions are approximately symmetrical. The centre for passive is much lower than the centre for aggressive which indicates that individuals allow a passive person to approach closer than an aggressive person. The variability for both groups is similar. 2 a

0.05% blood alcohol 42 2 2 99865 43 2 1 10 66 00 6

2 3 3 4 4 5 5 6

0% blood alcohol 59 01 4 555689 002 344 68

4 0 | 4 represents 40 4 | 0 represents 40 b 0.05% 0% min = 32 min = 25 Q 1 = 35.5 Q 1 = 34.5 median = 40.5 median = 38.5 Q 3 = 45 Q 3 = 43.5 max = 56 max = 64 c 0% *

c

New Secondhand 9 10 11 12 13 14 15 16 17 18 19

d New: mean = $14 275, s = $2508; Secondhand: mean = $13 497, s = $2349 e The distribution of the secondhand price is symmetrical while that of the new price is negatively skewed. The centre for new cars is higher than for secondhand, indicating they are more expensive and new prices are more variable than secondhand prices. 4 a

Choice

655 6665500 840 00 5 | 2 represents 25 b No choice min = 17 Q 1 = 26 median = 36 Q 3 = 44 max = 49 c

0

40

50

60

70

d 0%: mean = 38.9, s = 8.4 cm; 0.05%: mean = 40.8, s = 6.6 e The distributions are both approximately symmetrical. The 0% reading has one outlier. The centre for 0% is slightly lower than the centre for 0.05% and the spread of each is similar. The effect of 0.05% alcohol on performance has been minimal. 3 a Secondhand New 9 7 5 0 10 7 8 9 9 8 6 11 5 6 6 6 12 0 5 7 13 2 4 7 6 14 9 9 8 7 0 15 0 1 4 5 6 16 4 7 7 3 17 2 18 0 1 0 | 15 represents 150 15 | 0 represents 150 b New Secondhand min = 10 700 min = 9950 Q 1 = 11 595 Q 1 = 11 550 median = 14 995 median = 13 650 Q 3 = 16 660 Q 3 = 15 750 max = 18 050 max = 17 250 Prices rounded to nearest hundred


3 | 5 represents 35 Choice min = 25 Q 1 = 30 median = 36 Q 3 = 44 max = 50

10

20

30

40

50

d No choice: mean = 35.3, s = 10.0; Choice: mean= 36.4, s = 8.6 e Giving students a choice of assignment has not affected the centre of the distribution. The typical mark for both groups is around 36. However, marks are less variable when students are allowed to choose. 5 a Normal Disabled 0

10

20

30

40

b The distribution of scores for normal children is symmetric and tightly clustered around the centre at 28. The distribution for learning disabled children is negatively skewed, centred lower at 25 and shows much more variability in scores. 6 a Number REM

30

No choice 7 1 468 5568 2 2 4 4 89

Choice No choice

0.05% 20

1 2 3 4 5

40 30 20 10 0

1 2 3 4 5 Sleep deprivation (hrs)

The scatterplot shows a moderate positive linear relationship.


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Answers

d From the slope, we predict an increase in glucose of 0.55 mg/100 mL for each kg increase in weight. e Glucose = 97.7 mg/100 mL Expenditure ($’00)

10 a

Test score

7 a 35 30 25 20 15 10 4

8 12 16 20 24 28 Hours

12

2150

11 a

30

2050 2000 0 81

86 Year

Mangoes

No. of schools

8 a 2100

13 14 15 16 Income ($’000)

The graph shows a strong positive linear relationship. bq =1 c r = 0.94 d expenditure = −1547 + 0.24 × income e From the slope, we predict that for every extra $1 in income expenditure increases by 24 c. f Expenditure = $2373 g Since $30 000 is well outside our range of data the prediction would not be reliable.

The scatterplot has a strong positive linear relationship. bq =1 c r = 0.94 d test score = 12.3 + 0.93 × hours e From the slope, we predict an increase of about 1 mark in test score for each additional hour of study. f Test score = 22

91

20

10

The scatterplot has a strong negative linear relationship. b q = −1 c r = −0.98 d schools = 2169.4 − 12.5 × year where year is the no. of years since 1980, i.e. 1981 = 1 e The slope tells us that over the time period in question, an average of 12.5 schools closed each year. f schools = 1919 g This is not a reliable estimate because it is a long way outside the range of data we have. The government may decide to change policy with regard to closing schools. 120 115 110 105 100 95 90 85 80 75

0

1

2 Price

3

The graph shows a strong negative linear relationship. bq =1 c r = −0.95 d boxes = 34 − 10.3 × price e Slope tells us that for every dollar increase in price, 10 fewer boxes are sold on average. f 20 boxes g 5 cents each is outside the range of data that we have and any prediction would be unreliable. 12 a

100 80 Taste

Glucose

20

15

0

9 a

25

Answers

b q = 0.6 c r = 0.748 d REM = 8.45 + 4.88 × sleep e From the slope, we predict an increase of 4.88 in REM for each additional hour of sleep deprivation. f REM = 33

725

60 40 20

60

70 80 Weight

90

The scatterplot has a weak positive linear relationship. b r = 0.60 c glucose = 59.2 + 0.55 × weight


0

8

10 12 14 16 18 Magnesium (mg)

The graph shows a moderate positive linear relationship.


P2: FXS

9780521740494ans-20-25.xml

Answers

726

CUAU033-EVANS

July 15, 2011

5:6

Essential Advanced General Mathematics b q = 0.5 c r = 0.73 d taste = −22.4 + 7.3 × magnesium e Slope tells us that each 1 mg of magnesium in the water predicts an increase in the taste rating of 7.3 f taste rating = 562 13 a i

*

vi

50 Steve Waugh 40

Tests won

P1: FXS/ABE

30 20

* 10

0

20

40

60

80

100 0 0

ii The boxplot shows that the distribution of matches won is positively skewed, centred at 8 with an outlier at 93. The middle 50% of captains led their teams about 2–24 times. iii Allan Border

20

40 60 Tests played

80

100

Steve Waugh has the largest positive residual, so if this is used as a measure of success, he is the most successful captain.

Chapter 25

b i 0

10 20 30 40 50 60 70 80 90 100

ii The distribution is approximately symmetric, centred at about 44%. The middle 50% of captains have won from about 18% to 63% of their matches. iii Four captains, H H Massie, H Trumble, W A Brown and R N Harvey, have a success rate of 100%. c i

50

Tests won

40 30 20 10 0 0

20

40 60 Tests played

80

100

ii There is a strong positive linear association between the number of matches played and the number of matches won. iii r = 0.915 iv Matches won = 0.220 + 0.437 × matches played v Approximately, for each extra five test matches played, we predict that two tests are won.


Exercise 25A 1 a⇒ b⇒ c⇒ d⇐ 2 a False: Consider A = {2, 4, 6, 8, 10} and B = {6, 8, 10} , n (A ∪ B) = 5 b True: x ∈ A ⇒ x ∈ B ∴x ∈ A∩B ⇒x ∈ A and x ∈ A ⇒ x ∈ A ∩ B c False: A = {2, 4, 6} and B = {8, 10} , A ∩ B = Ø d True 3 a n (A ∪ B) = 8 ⇒ n (A) = 5 and n (B) = 3: False b A ∩ B = A ⇒ A ∈ B: True c A = Ø or B = Ø ⇒ A ∩ B = Ø: True d A = ␰ ⇐ A = Ø: True 4 a n 2 is odd implies n is odd: True if n is an integer b N 2 is divisible by 9 ⇒ N is divisible by 3: True if n is an integer c x 2 > 4 ⇒ x < −2: False 5 a The sum of three consecutive odd numbers is divisible by 3. (Hint: consider 2n − 1, 2n + 1, 2n + 3) b The sum of four consecutive odd numbers is divisible by 8. 7 a 32 + 42 = 5√2 , (3 + 4)2√= 72 √ b 14 c 1 + 4 = 1 + 4 = 3 1 2+4 d 1+ = 2 and 2 2 1 1+2 11 +4 = 2 2 4 e 7 + 3 = 10 1 1 1 1 11 f = ; + = 100 + 10 110 100 10 100


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Answers d 128 = 16 × 8 + 0 (128, 16) = (16, 0) = 16 3 al 4 ax bx cx dx ex fx

b 27

c6

d5

= 44 + 393t, y = −15 − 134t; t ∈ Z = −1 + 4t, y = 1 − 3t; t ∈ Z = 1434 + 4t, y = −956 − 3t; t ∈ Z = 1 + 5t, y = −7 + 3t; t ∈ Z = 107 + 224t, y = −32 − 67t; t ∈ Z = −37 + 336t, y = 25 − 227t; t ∈ Z

Exercise 25C 1 a x = 2 + 3t, y = −7 − 11t, t ∈ Z b x = 8 + 7t, y = −2 − 2t, t ∈ Z c x = 264 + 21t, y = −99 − 8t, t ∈ Z (Hint: Use your graphics calculator with 33 −8 Y1 = x+ and find the first solution 21 21 through Table.) x = −9 + 21t, y = 5 − 8t, t ∈ Z is also a solution. d x = 2 + 3t, y = −7 − 11t, t ∈ Z e x = 4 + 7t, y = 2 − 2t, t ∈ Z f x = 4 + 7t, y = 2 − 2t, t ∈ Z 4 a 8s + 6b = 54 b s = 6, b = 1 or s = 3, b = 5 5

50c coins 20c coins

0 25

2 20

4 15

6 10

8 5

10 0

6 x = 17, y = 20. The only solution to 19x + 98y = 1998 is x = 2, y = 20 for x, y ∈ Z + besides x = 100, y = 1 7 (10, 0), (9, 5), (8, 10), (7, 15), (6, 20), (5, 25), (4, 30), (3, 35), (2, 40), (1, 45), (0, 50) 8 63x − 23y = −7; x = 5 + 23t, y = 14 + 63t; t ≥ 0 and t ∈ Z 9 5 and 15 10 20; 20 + 77t for t ∈ N ∪ {0} 11 Pour two full 5 litre jugs into a container and remove one 3 litre jugful. 12 All amounts in excess of 3 c except 4 c and 7 c. 13 8 of type A and 16 of type B

Exercise 25D 1 a 43 = 8 × 5 + 3 (43, 5) = (5, 3) = 1 b 39 = 3 × 13 + 0 (39, 13) = (13, 0) = 13 c 37 = 2 × 17 + 3 (37, 17) = (17, 3) = 1

Answers

8 a a > b implies a − b is positive: True b x = y implies x = 0 and y = 0: False c x = −y implies x + y = 0: True d x y is even implies x is even and y is odd: False e A perfect square being even implies its square root is even: True 13 m = 5 and n = 12 or m = 6 and n = 8 15 6, 8, 10 and 5, 12, 13

727

Multiple-choice questions 1E

2D

3C

4E

5D

6C

7B

8D

9D

10 B

Short-answer questions (technology-free) 15 2 a x = 39 + 43t, y = −8 − 9t; t ∈ Z b no solutions 4 17 5 a x = 3 + 7t, y = −2 − 5t; t ∈ Z b x = 300 + 7t, y = −200 − 5t; t ∈ Z c x = 3 + 7t, y = −2 − 5t; t ∈ Z and t ≤ 0 6 Tom is 36 and Fred is 27

Extended-response questions 1 a

n2 − 1 4 d Proof is by induction or by observing an arithmetic sequence is formed. c a < where 2 It is constructive to show if b d a c a+c c a , ∈ Q+; < < b d b b+d d 2 3 a 6480 cubes b k = 2 c n + 6n + 3 b 30

c

4 a six 5c stamps and one 8c stamp b

5c 8 16 0 24

8c 10 5 15 0

5 36, 81 6 576 13 No such numbers exist.



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y11specialist maths units 1&2 vce

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