FHSST Authors The Free High School Science Texts: Textbooks for High School Students Studying the Sciences Mathematics Grades 10 - 12 Version 0 September 17, 2008 ii iii Copyright 2007 “Free High School Science Texts” Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no FrontCover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”. STOP!!!! Did you notice the FREEDOMS we’ve granted you? Our copyright license is different! 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Thousands of hours went into making them and they are a gift to everyone in the education community. iv FHSST Core Team Mark Horner ; Samuel Halliday ; Sarah Blyth ; Rory Adams ; Spencer Wheaton FHSST Editors Jaynie Padayachee ; Joanne Boulle ; Diana Mulcahy ; Annette Nell ; René Toerien ; Donovan Whitfield FHSST Contributors Rory Adams ; Prashant Arora ; Richard Baxter ; Dr. Sarah Blyth ; Sebastian Bodenstein ; Graeme Broster ; Richard Case ; Brett Cocks ; Tim Crombie ; Dr. Anne Dabrowski ; Laura Daniels ; Sean Dobbs ; Fernando Durrell ; Dr. Dan Dwyer ; Frans van Eeden ; Giovanni Franzoni ; Ingrid von Glehn ; Tamara von Glehn ; Lindsay Glesener ; Dr. Vanessa Godfrey ; Dr. Johan Gonzalez ; Hemant Gopal ; Umeshree Govender ; Heather Gray ; Lynn Greeff ; Dr. Tom Gutierrez ; Brooke Haag ; Kate Hadley ; Dr. Sam Halliday ; Asheena Hanuman ; Neil Hart ; Nicholas Hatcher ; Dr. Mark Horner ; Mfandaidza Hove ; Robert Hovden ; Jennifer Hsieh ; Clare Johnson ; Luke Jordan ; Tana Joseph ; Dr. Jennifer Klay ; Lara Kruger ; Sihle Kubheka ; Andrew Kubik ; Dr. Marco van Leeuwen ; Dr. Anton Machacek ; Dr. Komal Maheshwari ; Kosma von Maltitz ; Nicole Masureik ; John Mathew ; JoEllen McBride ; Nikolai Meures ; Riana Meyer ; Jenny Miller ; Abdul Mirza ; Asogan Moodaly ; Jothi Moodley ; Nolene Naidu ; Tyrone Negus ; Thomas O’Donnell ; Dr. Markus Oldenburg ; Dr. Jaynie Padayachee ; Nicolette Pekeur ; Sirika Pillay ; Jacques Plaut ; Andrea Prinsloo ; Joseph Raimondo ; Sanya Rajani ; Prof. Sergey Rakityansky ; Alastair Ramlakan ; Razvan Remsing ; Max Richter ; Sean Riddle ; Evan Robinson ; Dr. Andrew Rose ; Bianca Ruddy ; Katie Russell ; Duncan Scott ; Helen Seals ; Ian Sherratt ; Roger Sieloff ; Bradley Smith ; Greg Solomon ; Mike Stringer ; Shen Tian ; Robert Torregrosa ; Jimmy Tseng ; Helen Waugh ; Dr. Dawn Webber ; Michelle Wen ; Dr. Alexander Wetzler ; Dr. Spencer Wheaton ; Vivian White ; Dr. Gerald Wigger ; Harry Wiggins ; Wendy Williams ; Julie Wilson ; Andrew Wood ; Emma Wormauld ; Sahal Yacoob ; Jean Youssef Contributors and editors have made a sincere effort to produce an accurate and useful resource. Should you have suggestions, find mistakes or be prepared to donate material for inclusion, please don’t hesitate to contact us. We intend to work with all who are willing to help make this a continuously evolving resource! www.fhsst.org v vi Contents I Basics 1 1 Introduction to Book 1.1 II 3 The Language of Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . Grade 10 3 5 2 Review of Past Work 7 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 What is a number? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.4 Letters and Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.5 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.6 Multiplication and Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.7 Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.8 Negative Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.9 2.8.1 What is a negative number? . . . . . . . . . . . . . . . . . . . . . . . . 10 2.8.2 Working with Negative Numbers . . . . . . . . . . . . . . . . . . . . . . 11 2.8.3 Living Without the Number Line . . . . . . . . . . . . . . . . . . . . . . 12 Rearranging Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.10 Fractions and Decimal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.11 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.12 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.12.1 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.2 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.3 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.4 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.13 Mathematical Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.14 Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.15 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3 Rational Numbers - Grade 10 23 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.2 The Big Picture of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.3 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 vii CONTENTS CONTENTS 3.4 Forms of Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.5 Converting Terminating Decimals into Rational Numbers . . . . . . . . . . . . . 25 3.6 Converting Repeating Decimals into Rational Numbers . . . . . . . . . . . . . . 25 3.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.8 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4 Exponentials - Grade 10 29 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.3 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.1 Exponential Law 1: a0 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.2 Exponential Law 2: am × an = am+n . . . . . . . . . . . . . . . . . . . 30 4.3.3 Exponential Law 3: a−n = 4.3.4 4.4 m 1 an , a n 6= 0 . . . . . . . . . . . . . . . . . . . . 31 Exponential Law 4: a ÷ a = am−n . . . . . . . . . . . . . . . . . . . 32 4.3.5 Exponential Law 5: (ab)n = an bn . . . . . . . . . . . . . . . . . . . . . 32 4.3.6 Exponential Law 6: (am )n = amn . . . . . . . . . . . . . . . . . . . . . 33 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5 Estimating Surds - Grade 10 37 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 5.2 Drawing Surds on the Number Line (Optional) . . . . . . . . . . . . . . . . . . 38 5.3 End of Chapter Excercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 6 Irrational Numbers and Rounding Off - Grade 10 41 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6.2 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6.3 Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 6.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 7 Number Patterns - Grade 10 7.1 45 Common Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 7.1.1 Special Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 7.2 Make your own Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . 46 7.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 7.3.1 7.4 Patterns and Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Finance - Grade 10 53 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 8.2 Foreign Exchange Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 8.3 8.2.1 How much is R1 really worth? . . . . . . . . . . . . . . . . . . . . . . . 53 8.2.2 Cross Currency Exchange Rates 8.2.3 Enrichment: Fluctuating exchange rates . . . . . . . . . . . . . . . . . . 57 . . . . . . . . . . . . . . . . . . . . . . 56 Being Interested in Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 viii CONTENTS 8.4 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 8.4.1 8.5 8.6 8.7 CONTENTS Other Applications of the Simple Interest Formula . . . . . . . . . . . . . 61 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 8.5.1 Fractions add up to the Whole . . . . . . . . . . . . . . . . . . . . . . . 65 8.5.2 The Power of Compound Interest . . . . . . . . . . . . . . . . . . . . . . 65 8.5.3 Other Applications of Compound Growth . . . . . . . . . . . . . . . . . 67 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 8.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 8.6.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 9 Products and Factors - Grade 10 71 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.1 Parts of an Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.2 Product of Two Binomials . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.3 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 9.3 More Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 9.4 Factorising a Quadratic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 9.5 Factorisation by Grouping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 9.6 Simplification of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 9.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 10 Equations and Inequalities - Grade 10 83 10.1 Strategy for Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 10.2 Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.3 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 10.4 Exponential Equations of the form ka(x+p) = m . . . . . . . . . . . . . . . . . . 93 10.4.1 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 10.5 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 10.6 Linear Simultaneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.1 Finding solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.2 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.3 Solution by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . 101 10.7 Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 10.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 10.7.2 Problem Solving Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . 104 10.7.3 Application of Mathematical Modelling . . . . . . . . . . . . . . . . . . 104 10.7.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 106 10.8 Introduction to Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . 107 10.9 Functions and Graphs in the Real-World . . . . . . . . . . . . . . . . . . . . . . 107 10.10Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 ix CONTENTS CONTENTS 10.10.1 Variables and Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 10.10.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 10.10.3 The Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 10.10.4 Drawing Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 10.10.5 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . 110 10.11Characteristics of Functions - All Grades . . . . . . . . . . . . . . . . . . . . . . 112 10.11.1 Dependent and Independent Variables . . . . . . . . . . . . . . . . . . . 112 10.11.2 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.11.3 Intercepts with the Axes . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.11.4 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.5 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.6 Lines of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.7 Intervals on which the Function Increases/Decreases . . . . . . . . . . . 114 10.11.8 Discrete or Continuous Nature of the Graph . . . . . . . . . . . . . . . . 114 10.12Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 10.12.1 Functions of the form y = ax + q . . . . . . . . . . . . . . . . . . . . . 116 10.12.2 Functions of the Form y = ax2 + q . . . . . . . . . . . . . . . . . . . . . 120 10.12.3 Functions of the Form y = a x + q . . . . . . . . . . . . . . . . . . . . . . 125 10.12.4 Functions of the Form y = ab(x) + q . . . . . . . . . . . . . . . . . . . . 129 10.13End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 11 Average Gradient - Grade 10 Extension 135 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.2 Straight-Line Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.3 Parabolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 11.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 12 Geometry Basics 139 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 12.2 Points and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 12.3 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 12.3.1 Measuring angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 12.3.2 Special Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 12.3.3 Special Angle Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 12.3.4 Parallel Lines intersected by Transversal Lines . . . . . . . . . . . . . . . 143 12.4 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.4.1 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.4.2 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 12.4.3 Other polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 12.4.4 Extra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 12.5.1 Challenge Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 x CONTENTS 13 Geometry - Grade 10 CONTENTS 161 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 13.2 Right Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 13.2.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 13.2.2 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 13.3 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.3.1 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.4 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4.2 Distance between Two Points . . . . . . . . . . . . . . . . . . . . . . . . 172 13.4.3 Calculation of the Gradient of a Line . . . . . . . . . . . . . . . . . . . . 173 13.4.4 Midpoint of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 13.5 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.5.1 Translation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.5.2 Reflection of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 13.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 14 Trigonometry - Grade 10 189 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 14.2 Where Trigonometry is Used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 14.3 Similarity of Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 14.4 Definition of the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 191 14.5 Simple Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . 195 14.5.1 Height and Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 14.5.2 Maps and Plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 14.6 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 199 14.6.1 Graph of sin θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 14.6.2 Functions of the form y = a sin(x) + q . . . . . . . . . . . . . . . . . . . 200 14.6.3 Graph of cos θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 14.6.4 Functions of the form y = a cos(x) + q . . . . . . . . . . . . . . . . . . 202 14.6.5 Comparison of Graphs of sin θ and cos θ . . . . . . . . . . . . . . . . . . 204 14.6.6 Graph of tan θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 14.6.7 Functions of the form y = a tan(x) + q . . . . . . . . . . . . . . . . . . 205 14.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 15 Statistics - Grade 10 211 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2.1 Data and Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2.2 Methods of Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . 212 15.2.3 Samples and Populations . . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3 Example Data Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 xi CONTENTS CONTENTS 15.3.1 Data Set 1: Tossing a Coin . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3.2 Data Set 2: Casting a die . . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3.3 Data Set 3: Mass of a Loaf of Bread . . . . . . . . . . . . . . . . . . . . 214 15.3.4 Data Set 4: Global Temperature . . . . . . . . . . . . . . . . . . . . . . 214 15.3.5 Data Set 5: Price of Petrol . . . . . . . . . . . . . . . . . . . . . . . . . 215 15.4 Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 15.4.1 Exercises - Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . 216 15.5 Graphical Representation of Data . . . . . . . . . . . . . . . . . . . . . . . . . . 217 15.5.1 Bar and Compound Bar Graphs . . . . . . . . . . . . . . . . . . . . . . . 217 15.5.2 Histograms and Frequency Polygons . . . . . . . . . . . . . . . . . . . . 217 15.5.3 Pie Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 15.5.4 Line and Broken Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . 220 15.5.5 Exercises - Graphical Representation of Data . . . . . . . . . . . . . . . 221 15.6 Summarising Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 15.6.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . 222 15.6.2 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 15.6.3 Exercises - Summarising Data . . . . . . . . . . . . . . . . . . . . . . . 228 15.7 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 15.7.1 Exercises - Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . 230 15.8 Summary of Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 15.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 16 Probability - Grade 10 235 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 16.2 Random Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 16.2.1 Sample Space of a Random Experiment . . . . . . . . . . . . . . . . . . 235 16.3 Probability Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 16.3.1 Classical Theory of Probability . . . . . . . . . . . . . . . . . . . . . . . 239 16.4 Relative Frequency vs. Probability . . . . . . . . . . . . . . . . . . . . . . . . . 240 16.5 Project Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 16.6 Probability Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 16.7 Mutually Exclusive Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 16.8 Complementary Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 16.9 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 III Grade 11 17 Exponents - Grade 11 249 251 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 17.2 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 √ m 17.2.1 Exponential Law 7: a n = n am . . . . . . . . . . . . . . . . . . . . . . 251 17.3 Exponentials in the Real-World . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 17.4 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 xii CONTENTS CONTENTS 18 Surds - Grade 11 18.1 Surd Calculations . . . . . . . . . . √ √ √ 18.1.1 Surd Law 1: n a n b = n ab √ p n a 18.1.2 Surd Law 2: n ab = √ . . n b √ m 18.1.3 Surd Law 3: n am = a n . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 256 18.1.4 Like and Unlike Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 18.1.5 Simplest Surd form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 18.1.6 Rationalising Denominators . . . . . . . . . . . . . . . . . . . . . . . . . 258 18.2 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 19 Error Margins - Grade 11 261 20 Quadratic Sequences - Grade 11 265 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 20.2 What is a quadratic sequence? . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 20.3 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 21 Finance - Grade 11 271 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 21.2 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 21.3 Simple Depreciation (it really is simple!) . . . . . . . . . . . . . . . . . . . . . . 271 21.4 Compound Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 21.5 Present Values or Future Values of an Investment or Loan . . . . . . . . . . . . 276 21.5.1 Now or Later . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 21.6 Finding i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 21.7 Finding n - Trial and Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 21.8 Nominal and Effective Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . 280 21.8.1 The General Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 21.8.2 De-coding the Terminology . . . . . . . . . . . . . . . . . . . . . . . . . 282 21.9 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 21.9.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 21.9.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 21.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 22 Solving Quadratic Equations - Grade 11 287 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 22.2 Solution by Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 22.3 Solution by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . 290 22.4 Solution by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . 293 22.5 Finding an equation when you know its roots . . . . . . . . . . . . . . . . . . . 296 22.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 xiii CONTENTS CONTENTS 23 Solving Quadratic Inequalities - Grade 11 301 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 23.2 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 23.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 24 Solving Simultaneous Equations - Grade 11 307 24.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 24.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 25 Mathematical Models - Grade 11 313 25.1 Real-World Applications: Mathematical Models . . . . . . . . . . . . . . . . . . 313 25.2 End of Chatpter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 26 Quadratic Functions and Graphs - Grade 11 321 26.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 26.2 Functions of the Form y = a(x + p)2 + q . . . . . . . . . . . . . . . . . . . . . 321 26.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 26.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 26.2.3 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 26.2.4 Axes of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 26.2.5 Sketching Graphs of the Form f (x) = a(x + p)2 + q . . . . . . . . . . . 325 26.2.6 Writing an equation of a shifted parabola . . . . . . . . . . . . . . . . . 327 26.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 27 Hyperbolic Functions and Graphs - Grade 11 329 27.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 27.2 Functions of the Form y = a x+p +q . . . . . . . . . . . . . . . . . . . . . . . . 329 27.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 27.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 27.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 27.2.4 Sketching Graphs of the Form f (x) = a x+p + q . . . . . . . . . . . . . . 333 27.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 28 Exponential Functions and Graphs - Grade 11 335 28.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 28.2 Functions of the Form y = ab(x+p) + q . . . . . . . . . . . . . . . . . . . . . . . 335 28.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 28.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 28.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 28.2.4 Sketching Graphs of the Form f (x) = ab(x+p) + q . . . . . . . . . . . . . 338 28.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 29 Gradient at a Point - Grade 11 341 29.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 29.2 Average Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 29.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 xiv CONTENTS 30 Linear Programming - Grade 11 CONTENTS 345 30.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.1 Decision Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.2 Objective Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.3 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.2.4 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.2.5 The Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.3 Example of a Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.4 Method of Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5 Skills you will need . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5.1 Writing Constraint Equations . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5.2 Writing the Objective Function . . . . . . . . . . . . . . . . . . . . . . . 348 30.5.3 Solving the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 30.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 31 Geometry - Grade 11 357 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.2 Right Pyramids, Right Cones and Spheres . . . . . . . . . . . . . . . . . . . . . 357 31.3 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 31.4 Triangle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 31.4.1 Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 31.5 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 31.5.1 Equation of a Line between Two Points . . . . . . . . . . . . . . . . . . 368 31.5.2 Equation of a Line through One Point and Parallel or Perpendicular to Another Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 31.5.3 Inclination of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 31.6 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 31.6.1 Rotation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 31.6.2 Enlargement of a Polygon 1 . . . . . . . . . . . . . . . . . . . . . . . . . 376 32 Trigonometry - Grade 11 381 32.1 History of Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 32.2 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 381 32.2.1 Functions of the form y = sin(kθ) . . . . . . . . . . . . . . . . . . . . . 381 32.2.2 Functions of the form y = cos(kθ) . . . . . . . . . . . . . . . . . . . . . 383 32.2.3 Functions of the form y = tan(kθ) . . . . . . . . . . . . . . . . . . . . . 384 32.2.4 Functions of the form y = sin(θ + p) . . . . . . . . . . . . . . . . . . . . 385 32.2.5 Functions of the form y = cos(θ + p) . . . . . . . . . . . . . . . . . . . 386 32.2.6 Functions of the form y = tan(θ + p) . . . . . . . . . . . . . . . . . . . 387 32.3 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 32.3.1 Deriving Values of Trigonometric Functions for 30◦ , 45◦ and 60◦ . . . . . 389 32.3.2 Alternate Definition for tan θ . . . . . . . . . . . . . . . . . . . . . . . . 391 xv CONTENTS CONTENTS 32.3.3 A Trigonometric Identity . . . . . . . . . . . . . . . . . . . . . . . . . . 392 32.3.4 Reduction Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 32.4 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 399 32.4.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 32.4.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 32.4.3 Solution using CAST diagrams . . . . . . . . . . . . . . . . . . . . . . . 403 32.4.4 General Solution Using Periodicity . . . . . . . . . . . . . . . . . . . . . 405 32.4.5 Linear Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . 406 32.4.6 Quadratic and Higher Order Trigonometric Equations . . . . . . . . . . . 406 32.4.7 More Complex Trigonometric Equations . . . . . . . . . . . . . . . . . . 407 32.5 Sine and Cosine Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 32.5.1 The Sine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 32.5.2 The Cosine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 32.5.3 The Area Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 32.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 33 Statistics - Grade 11 419 33.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2 Standard Deviation and Variance . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2.1 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2.2 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 33.2.3 Interpretation and Application . . . . . . . . . . . . . . . . . . . . . . . 423 33.2.4 Relationship between Standard Deviation and the Mean . . . . . . . . . . 424 33.3 Graphical Representation of Measures of Central Tendency and Dispersion . . . . 424 33.3.1 Five Number Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 33.3.2 Box and Whisker Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 425 33.3.3 Cumulative Histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 33.4 Distribution of Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 33.4.1 Symmetric and Skewed Data . . . . . . . . . . . . . . . . . . . . . . . . 428 33.4.2 Relationship of the Mean, Median, and Mode . . . . . . . . . . . . . . . 428 33.5 Scatter Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 33.6 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 33.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 34 Independent and Dependent Events - Grade 11 437 34.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 34.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 34.2.1 Identification of Independent and Dependent Events . . . . . . . . . . . 438 34.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 IV Grade 12 35 Logarithms - Grade 12 443 445 35.1 Definition of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 xvi CONTENTS CONTENTS 35.2 Logarithm Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446 35.3 Laws of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 35.4 Logarithm Law 1: loga 1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 35.5 Logarithm Law 2: loga (a) = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 35.6 Logarithm Law 3: loga (x · y) = loga (x) + loga (y) . . . . . . . . . . . . . . . . . 448 35.7 Logarithm Law 4: loga xy = loga (x) − loga (y) . . . . . . . . . . . . . . . . . 449 35.8 Logarithm Law 5: loga (xb ) = b loga (x) . . . . . . . . . . . . . . . . . . . . . . . 450 √ 35.9 Logarithm Law 6: loga ( b x) = logab(x) . . . . . . . . . . . . . . . . . . . . . . . 450 35.10Solving simple log equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 35.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 35.11Logarithmic applications in the Real World . . . . . . . . . . . . . . . . . . . . . 454 35.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 35.12End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 36 Sequences and Series - Grade 12 457 36.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 36.2 Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 36.2.1 General Equation for the nth -term of an Arithmetic Sequence . . . . . . 458 36.3 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 36.3.1 Example - A Flu Epidemic . . . . . . . . . . . . . . . . . . . . . . . . . 459 36.3.2 General Equation for the nth -term of a Geometric Sequence . . . . . . . 461 36.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 36.4 Recursive Formulae for Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 462 36.5 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.5.1 Some Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.5.2 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.6 Finite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 36.6.1 General Formula for a Finite Arithmetic Series . . . . . . . . . . . . . . . 466 36.6.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 36.7 Finite Squared Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 36.8 Finite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 36.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 36.9 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 36.9.1 Infinite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . 471 36.9.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 36.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 37 Finance - Grade 12 477 37.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 37.2 Finding the Length of the Investment or Loan . . . . . . . . . . . . . . . . . . . 477 37.3 A Series of Payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 37.3.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 xvii CONTENTS CONTENTS 37.3.2 Present Values of a series of Payments . . . . . . . . . . . . . . . . . . . 479 37.3.3 Future Value of a series of Payments . . . . . . . . . . . . . . . . . . . . 484 37.3.4 Exercises - Present and Future Values . . . . . . . . . . . . . . . . . . . 485 37.4 Investments and Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 37.4.1 Loan Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 37.4.2 Exercises - Investments and Loans . . . . . . . . . . . . . . . . . . . . . 489 37.4.3 Calculating Capital Outstanding . . . . . . . . . . . . . . . . . . . . . . 489 37.5 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 37.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 37.5.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 37.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 38 Factorising Cubic Polynomials - Grade 12 493 38.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 38.2 The Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 38.3 Factorisation of Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 494 38.4 Exercises - Using Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 496 38.5 Solving Cubic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 38.5.1 Exercises - Solving of Cubic Equations . . . . . . . . . . . . . . . . . . . 498 38.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 39 Functions and Graphs - Grade 12 501 39.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.2 Definition of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.3 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 39.4 Graphs of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 39.4.1 Inverse Function of y = ax + q . . . . . . . . . . . . . . . . . . . . . . . 503 39.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.3 Inverse Function of y = ax2 . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.5 Inverse Function of y = ax . . . . . . . . . . . . . . . . . . . . . . . . . 506 39.4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 39.5 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 40 Differential Calculus - Grade 12 509 40.1 Why do I have to learn this stuff? . . . . . . . . . . . . . . . . . . . . . . . . . 509 40.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 40.2.1 A Tale of Achilles and the Tortoise . . . . . . . . . . . . . . . . . . . . . 510 40.2.2 Sequences, Series and Functions . . . . . . . . . . . . . . . . . . . . . . 511 40.2.3 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 40.2.4 Average Gradient and Gradient at a Point . . . . . . . . . . . . . . . . . 516 40.3 Differentiation from First Principles . . . . . . . . . . . . . . . . . . . . . . . . . 519 xviii CONTENTS CONTENTS 40.4 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 40.4.1 Summary of Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . 522 40.5 Applying Differentiation to Draw Graphs . . . . . . . . . . . . . . . . . . . . . . 523 40.5.1 Finding Equations of Tangents to Curves . . . . . . . . . . . . . . . . . 523 40.5.2 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 40.5.3 Local minimum, Local maximum and Point of Inflextion . . . . . . . . . 529 40.6 Using Differential Calculus to Solve Problems . . . . . . . . . . . . . . . . . . . 530 40.6.1 Rate of Change problems . . . . . . . . . . . . . . . . . . . . . . . . . . 534 40.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535 41 Linear Programming - Grade 12 539 41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.2.1 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.3 Linear Programming and the Feasible Region . . . . . . . . . . . . . . . . . . . 540 41.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 42 Geometry - Grade 12 549 42.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2 Circle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2.2 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550 42.2.3 Theorems of the Geometry of Circles . . . . . . . . . . . . . . . . . . . . 550 42.3 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 42.3.1 Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 42.3.2 Equation of a Tangent to a Circle at a Point on the Circle . . . . . . . . 569 42.4 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 42.4.1 Rotation of a Point about an angle θ . . . . . . . . . . . . . . . . . . . . 571 42.4.2 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573 42.4.3 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573 42.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574 43 Trigonometry - Grade 12 577 43.1 Compound Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577 43.1.1 Derivation of sin(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 577 43.1.2 Derivation of sin(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578 43.1.3 Derivation of cos(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578 43.1.4 Derivation of cos(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.5 Derivation of sin 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.6 Derivation of cos 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.7 Problem-solving Strategy for Identities . . . . . . . . . . . . . . . . . . . 580 43.2 Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . 582 43.2.1 Problems in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 582 xix CONTENTS CONTENTS 43.2.2 Problems in 3 dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 584 43.3 Other Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.1 Taxicab Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.2 Manhattan distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.3 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 43.3.4 Fractal Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 43.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 44 Statistics - Grade 12 591 44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.2 A Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.3 Extracting a Sample Population . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 44.4 Function Fitting and Regression Analysis . . . . . . . . . . . . . . . . . . . . . . 594 44.4.1 The Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . 596 44.4.2 Using a calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 44.4.3 Correlation coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 44.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600 45 Combinations and Permutations - Grade 12 603 45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.1 Making a List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.2 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3.1 The Factorial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.4 The Fundamental Counting Principle . . . . . . . . . . . . . . . . . . . . . . . . 604 45.5 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.1 Counting Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.2 Combinatorics and Probability . . . . . . . . . . . . . . . . . . . . . . . 606 45.6 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 45.6.1 Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 607 45.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 45.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 V Exercises 613 46 General Exercises 615 47 Exercises - Not covered in Syllabus 617 A GNU Free Documentation License 619 xx Part IV Grade 12 443 Chapter 35 Logarithms - Grade 12 In mathematics many ideas are related. We saw that addition and subtraction are related and that multiplication and division are related. Similarly, exponentials and logarithms are related. Logarithms, commonly referred to as logs, are the inverse of exponentials. The logarithm of a number x in the base a is defined as the number n such that an = x. So, if an = x, then: loga (x) = n (35.1) Extension: Inverse Function When we say “inverse function” we mean that the answer becomes the question and the question becomes the answer. For example, in the equation ab = x the “question” is “what is a raised to the power b.” The answer is “x.” The inverse function would be loga x = b or “by what power must we raise a to obtain x.” The answer is “b.” The mathematical symbol for logarithm is loga (x) and it is read “log to the base a of x”. For example, log10 (100) is “log to the base 10 of 100”. Activity :: Logarithm Symbols : Write the following out in words. The first one is done for you. 1. 2. 3. 4. 5. 35.1 log2 (4) is log to the base 2 of 4 log10 (14) log16 (4) logx (8) logy (x) Definition of Logarithms The logarithm of a number is the index to which the base must be raised to give that number. From the first example of the activity log2 (4) (read log to the base 2 of 4) means the power of 2 that will give 4. Therefore, log2 (4) = 2 (35.2) The index-form is then 22 = 4 and the logarithmic-form is log2 4 = 2. 445 35.2 CHAPTER 35. LOGARITHMS - GRADE 12 Definition: Logarithms If an = x, then: loga (x) = n, where a > 0; a 6= 1 and x > 0. Activity :: Applying the definition : Find the value of: 1. log7 343 Reasoning : 73 = 343 theref ore, log7 343 = 3 2. log2 8 3. log4 1 64 4. log10 1 000 35.2 Logarithm Bases Logarithms, like exponentials, also have a base and log2 (2) is not the same as log10 (2). We generally use the “common” base, 10, or the natural base, e. The number e is an irrational number between 2.71 and 2.72. It comes up surprisingly often in Mathematics, but for now suffice it to say that it is one of the two common bases. Extension: Natural Logarithm The natural logarithm (symbol ln) is widely used in the sciences. The natural logarithm is to the base e which is approximately 2.71828183.... e is like π and is another example of an irrational number. While the notation log10 (x) and loge (x) may be used, log10 (x) is often written log(x) in Science and loge (x) is normally written as ln(x) in both Science and Mathematics. So, if you see the log symbol without a base, it means log10 . It is often necessary or convenient to convert a log from one base to another. An engineer might need an approximate solution to a log in a base for which he does not have a table or calculator function, or it may be algebraically convenient to have two logs in the same base. Logarithms can be changed from one base to another, by using the change of base formula: loga x = logb x logb a (35.3) where b is any base you find convenient. Normally a and b are known, therefore logb a is normally a known, if irrational, number. For example, change log2 12 in base 10 is: log2 12 = log10 12 log10 2 446 CHAPTER 35. LOGARITHMS - GRADE 12 35.3 Activity :: Change of Base : Change the following to the indicated base: 1. log2 (4) to base 8 2. log10 (14) to base 2 3. log16 (4) to base 10 4. logx (8) to base y 5. logy (x) to base x 35.3 Laws of Logarithms Just as for the exponents, logarithms have some laws which make working with them easier. These laws are based on the exponential laws and are summarised first and then explained in detail. loga (1) = loga (a) = loga (x · y) = x = loga y loga (xb ) = √ loga b x = 35.4 0 (35.4) 1 loga (x) + loga (y) (35.5) (35.6) loga (x) − loga (y) (35.7) b loga (x) loga (x) b (35.8) Logarithm Law 1: loga 1 = 0 Since Then, a0 = loga (1) = = 1 loga (a0 ) 0 by definition of logarithm in Equation 35.1 For example, log2 1 = 0 and log2 51 = 0 Activity :: Logarithm Law 1: loga 1 = 0 : Simplify the following: 1. log2 (1) + 5 2. log10 (1) × 100 3. 3 × log16 (1) 4. logx (1) + 2xy 5. logy (1) x 447 (35.9) 35.5 CHAPTER 35. LOGARITHMS - GRADE 12 35.5 Logarithm Law 2: loga (a) = 1 Since a1 Then, = loga (a) = = a loga (a1 ) 1 by definition of logarithm in Equation 35.1 For example, log2 2 = 1 and log25 25 = 1 Activity :: Logarithm Law 2: loga (a) = 1 : Simplify the following: 1. log2 (2) + 5 2. log10 (10) × 100 3. 3 × log16 (16) 4. logx (x) + 2xy 5. logy (y) x Important: Useful to know and remember When the base is 10, we do not need to state it. From the work done up to now, it is also useful to summarise the following facts: 1. log 1 = 0 2. log 10 = 1 3. log 100 = 2 4. log 1000 = 3 35.6 Logarithm Law 3: loga (x · y) = loga (x) + loga (y) The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the base a. So, assume that x = am and y = an . Then from Equation 35.1, we have that: and loga (x) = m (35.10) loga (y) = n (35.11) This means that we can write: loga (x · y) = = = = loga (am · an ) loga (am+n ) Exponential Law Equation 4.4 loga (x)+loga (y) loga (a loga (x) + loga (y) ) From Equation 35.10 and Equation 35.11 From Equation 35.1 448 CHAPTER 35. LOGARITHMS - GRADE 12 35.7 For example, show that log(10 · 100) = log 10 + log 100. Start with calculating the left hand side: log(10 · 100) = log(1000) = log(103 ) = 3 The right hand side: log 10 + log 100 = = 1+2 3 Both sides are equal. Therefore, log(10 · 100) = log 10 + log 100. Activity :: Logarithm Law 3: loga (x · y) = loga (x) + loga (y) : Write as seperate logs: 1. log2 (8 × 4) 2. log8 (10 × 10) 3. log16 (xy) 4. logz (2xy) 5. logx (y 2 ) 35.7 Logarithm Law 4: loga x y = loga (x) − loga (y) The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise. 10 For example, show that log( 100 ) = log 10 − log 100. Start with calculating the left hand side: log( 1 ) 10 log(10−1 ) −1 10 ) = 100 = = log( The right hand side: log 10 − log 100 = = 1−2 −1 10 ) = log 10 − log 100. Both sides are equal. Therefore, log( 100 Activity :: Logarithm Law 4: loga seperate logs: 1. log2 ( 85 ) 2. log8 ( 100 3 ) 3. log16 ( xy ) 449 x y = loga (x) − loga (y) : Write as 35.8 CHAPTER 35. LOGARITHMS - GRADE 12 4. logz ( 2y ) 5. logx ( y2 ) 35.8 Logarithm Law 5: loga (xb) = b loga (x) Once again, we need to relate x to the base a. So, we let x = am . Then, ∴ loga (xb ) = = loga ((am )b ) loga (am·b ) (Exponential Law in Equation 4.8) But, m = loga (xb ) = loga (x) (Assumption that x = am ) loga (ab·loga (x) ) = b · loga (x) (Definition of logarithm in Equation 35.1) For example, we can show that log2 (53 ) = 3 log2 (5). log2 (53 ) = log( 5 · 5 · 5) = log2 5 + log2 5 + log2 5 = 3 log2 5 (∵ loga (x · y) = loga (am · an )) Therefore, log2 (53 ) = 3 log2 (5). Activity :: Logarithm Law 5: loga (xb ) = b loga (x) : Simplify the following: 1. log2 (84 ) 2. log8 (1010 ) 3. log16 (xy ) 4. logz (y x ) 5. logx (y 2x ) 35.9 √ Logarithm Law 6: loga ( b x) = loga (x) b The derivation of this law is identical to the derivation of Logarithm Law 5 and is left as an exercise. √ log 5 For example, we can show that log2 ( 3 5) = 32 . √ 3 log2 ( 5) √ Therefore, log2 ( 3 5) = 1 = log( 5 3 ) 1 log2 5 (∵ loga (xb ) = b loga (x)) = 3 log2 5 = 3 log2 5 3 . 450 CHAPTER 35. LOGARITHMS - GRADE 12 35.9 √ Activity :: Logarithm Law 6: loga ( b x) = √ 1. log2 ( 4 8) √ 2. log8 ( 10 10) √ 3. log16 ( y x) √ 4. logz ( x y) √ 5. logx ( 2x y) loga (x) b : Simplify the following: Worked Example 155: Simplification of Logs Question: Simplify, without use of a calculator: 3 log 2 + log 125 Answer Step 1 : Try to write any quantities as exponents 125 can be written as 53 . Step 2 : Simplify 3 log 2 + log 125 = 3 log 2 + log 53 = 3 log 2 + 3 log 5 ∵ loga (xb ) = b loga (x) Step 3 : Final Answer We cannot simplify any further. The final answer is: 3 log 2 + 3 log 5 Worked Example 156: Simplification of Logs Question: Simplify, without use of a calculator: 2 8 3 + log2 32 Answer Step 1 : Try to write any quantities as exponents 8 can be written as 23 . 32 can be written as 25 . Step 2 : Re-write the question using the exponential forms of the numbers 2 2 8 3 + log2 32 = (23 ) 3 + log2 25 Step 3 : Determine which laws can be used. We can use: loga (xb ) = b loga (x) Step 4 : Apply log laws to simplify 2 2 (23 ) 3 + log2 25 = (2)3 3 + 5 log2 2 Step 5 : Determine which laws can be used. 451 The final answer does not have to be that simple. 35.10 CHAPTER 35. LOGARITHMS - GRADE 12 We can now use loga a = 1 Step 6 : Apply log laws to simplify 2 (2)3 3 + 5 log2 2 = (2)2 + 5(1) = 4 + 5 = 9 Step 7 : Final Answer The final answer is: 2 8 3 + log2 32 = 9 Worked Example 157: Simplify to one log Question: Write 2 log 3 + log 2 − log 5 as the logarithm of a single number. Answer Step 1 : Reverse law 5 2 log 3 + log 2 − log 5 = log 32 + log 2 − log 5 Step 2 : Apply laws 3 and 4 = log 32 × 2 ÷ 5 Step 3 : Write the final answer = log 3,6 35.10 Solving simple log equations In grade 10 you solved some exponential equations by trial and error, because you did not know the great power of logarithms yet. Now it is much easier to solve these equations by using logarithms. For example to solve x in 25x = 50 correct to two decimal places you simply apply the following reasoning. If the LHS = RHS then the logarithm of the LHS must be equal to the logarithm of the RHS. By applying Law 5, you will be able to use your calculator to solve for x. Worked Example 158: Solving Log equations Question: Solve for x: 25x = 50 correct to two decimal places. Answer Step 1 : Taking the log of both sides log 25x = log 50 Step 2 : Use Law 5 x log 25 = log 50 Step 3 : Solve for x x = log 50 ÷ log 25 x = 1,21533.... Step 4 : Round off to required decimal place x = 1,22 452 CHAPTER 35. LOGARITHMS - GRADE 12 35.10 In general, the exponential equation should be simplified as much as possible. Then the aim is to make the unknown quantity (i.e. x) the subject of the equation. For example, the equation 2(x+2) = 1 is solved by moving all terms with the unknown to one side of the equation and taking all constants to the other side of the equation 2x · 22 2 x = 1 1 = 22 Then, take the logarithm of each side. 1 ) 22 − log (22 ) log (2x ) = log ( x log (2) = x log (2) = ∴ x = −2 log (2) Divide both sides by log (2) −2 Substituting into the original equation, yields 2−2+2 = 20 = 1 X Similarly, 9(1−2x) = 34 is solved as follows: 9(1−2x) = 34 2(1−2x) 3 32 3−4x = = 34 34 3−4x 3−4x = = 34 · 3−2 32 take the logarithm of both sides log(3−4x ) = −4x log(3) = log(32 ) 2 log(3) −4x = 2 ∴x = − divide both sides by log(3) 1 2 Substituting into the original equation, yields 9(1−2( −1 2 )) = 9(1+1) = 32(2) = 34 X Worked Example 159: Exponential Equation Question: Solve for x in 7 · 5(3x+3) = 35 Answer Step 1 : Identify the base with x as an exponent There are two possible bases: 5 and 7. x is an exponent of 5. Step 2 : Eliminate the base with no x In order to eliminate 7, divide both sides of the equation by 7 to give: 5(3x+3) = 5 Step 3 : Take the logarithm of both sides log(5(3x+3) ) = log(5) Step 4 : Apply the log laws to make x the subject of the equation. 453 35.11 CHAPTER 35. LOGARITHMS - GRADE 12 (3x + 3) log(5) = log(5) divide both sides of the equation by log(5) 3x + 3 = 3x = 1 −2 2 − 3 x = Step 5 : Substitute into the original equation to check answer. 2 7 · 5(−3 3 +3) = 7 · 5(−2+3) = 7 · 51 = 35 X 35.10.1 Exercises Solve for x: 1. log3 x = 2 2. 10log27 = x 3. 32x−1 = 272x−1 35.11 Logarithmic applications in the Real World Logarithms are part of a number of formulae used in the Physical Sciences. There are formulae that deal with earthquakes, with sound, and pH-levels to mention a few. To work out time periods is growth or decay, logs are used to solve the particular equation. Worked Example 160: Using the growth formula Question: A city grows 5% every 2 years. How long will it take for the city to triple its size? Answer Step 1 : Use the formula A = P (1 + i)n Assume P = x, then A = 3x. For this example n represents a period of 2 years, therefore the n is halved for this purpose. Step 2 : Substitute information given into formula 3 = log 3 = n = n = n (1,05) 2 n × log 1.05 (usinglaw5) 2 2 log 3 ÷ log 1,05 45,034 Step 3 : Final answer So it will take approximately 45 years for the population to triple in size. 454 CHAPTER 35. LOGARITHMS - GRADE 12 35.11.1 35.12 Exercises 1. The population of a certain bacteria is expected to grow exponentially at a rate of 15 % every hour. If the initial population is 5 000, how long will it take for the population to reach 100 000 ? 2. Plus Bank is offering a savings account with an interest rate if 10 % per annum compounded monthly. You can afford to save R 300 per month. How long will it take you to save up R 20 000 ? (Answer to the nearest rand) Worked Example 161: Logs in Compound Interest Question: I have R12 000 to invest. I need the money to grow to at least R30 000. If it is invested at a compound interest rate of 13% per annum, for how long (in full years) does my investment need to grow ? Answer Step 1 : The formula to use A = P (1 + i)n Step 2 : Substitute and solve for n 12 000(1 + 0,13)n 5 2 log 2,5 30 000 < 1,13n > n log 1,13 > n n > > log 2,5 ÷ log 1,13 7,4972.... Step 3 : Determine the final answer In this case we round up, because 7 years will not yet deliver the required R 30 000. The investment need to stay in the bank for at least 8 years. 35.12 End of Chapter Exercises 1. Show that loga x = loga (x) − loga (y) y 2. Show that loga 3. Without using a calculator show that: log loga (x) √ b x = b 75 5 32 − 2 log + log = log 2 16 9 243 4. Given that 5n = x and n = log2 y A Write y in terms of n B Express log8 4y in terms of n C Express 50n+1 in terms of x and y 5. Simplify, without the use of a calculator: 455 35.12 CHAPTER 35. LOGARITHMS - GRADE 12 2 A 8 3 + log2 32 B log3 9 − log5 C 4−1 5 − 9−1 √ 5 1 2 + log3 92,12 6. Simplify to a single number, without use of a calculator: log 32 − log 8 log 8 B log 3 − log 0,3 A log5 125 + 7. Given: log3 6 = a and log6 5 = b A Express log3 2 in terms of a. B Hence, or otherwise, find log3 10 in terms of a and b. 8. Given: Prove: pq k = qp−1 k = 1 − 2 logq p 9. Evaluate without using a calculator: (log7 49)5 + log5 1 125 − 13 log9 1 10. If log 5 = 0,7, determine, without using a calculator: A log2 5 B 10−1,4 11. Given: M = log2 (x + 3) + log2 (x − 3) A Determine the values of x for which M is defined. B Solve for x if M = 4. log x 12. Solve: x3 = 10x2 (Answer(s) may be left in surd form, if necessary.) 13. Find the value of (log27 3)3 without the use of a calculator. √ 14. Simplify By using a calculator: log4 8 + 2 log3 27 15. Write log 4500 in terms of a and b if 2 = 10a and 9 = 10b . 16. Calculate: 52006 − 52004 + 24 52004 + 1 17. √ Solve the following equation for x without the use of a calculator and using the fact that 10 ≈ 3,16 : 6 2 log(x + 1) = −1 log(x + 1) 18. Solve the following equation for x: 66x = 66 456 (Give answer correct to 2 decimal places.) Chapter 36 Sequences and Series - Grade 12 36.1 Introduction In this chapter we extend the arithmetic and quadratic sequences studied in earlier grades, to geometric sequences. We also look at series, which is the summing of the terms in a sequence. 36.2 Arithmetic Sequences The simplest type of numerical sequence is an arithmetic sequence. Definition: Arithmetic Sequence An arithmetic (or linear ) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term For example, 1,2,3,4,5,6, . . . is an arithmetic sequence because you add 1 to the current term to get the next term: first term: second term: third term: .. . nth term: 1 2=1+1 3=2+1 n = (n − 1) + 1 Activity :: Common Difference : Find the constant value that is added to get the following sequences and write out the next 5 terms. 1. 2,6,10,14,18,22, . . . 2. −5, − 3, − 1,1,3, . . . 3. 1,4,7,10,13,16, . . . 4. −1,10,21,32,43,54, . . . 5. 3,0, − 3, − 6, − 9, − 12, . . . 457 36.2 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.2.1 General Equation for the nth -term of an Arithmetic Sequence More formally, the number we start out with is called a1 (the first term), and the difference between each successive term is denoted d, called the common difference. The general arithmetic sequence looks like: a1 = a1 a2 a3 = a1 + d = a2 + d = (a1 + d) + d = a1 + 2d a4 ... = a3 + d = (a1 + 2d) + d = a1 + 3d an = a1 + d · (n − 1) Thus, the equation for the nth -term will be: an = a1 + d · (n − 1) (36.1) Given a1 and the common difference, d, the entire set of numbers belonging to an arithmetic sequence can be generated. Definition: Arithmetic Sequence An arithmetic (or linear ) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term: an = an−1 + d (36.2) where • an represents the new term, the nth -term, that is calculated; • an−1 represents the previous term, the (n − 1)th -term; • d represents some constant. Important: Arithmetic Sequences A simple test for an arithmetic sequence is to check that the difference between consecutive terms is constant: a2 − a1 = a3 − a2 = an − an−1 = d (36.3) This is quite an important equation, and is the definitive test for an arithmetic sequence. If this condition does not hold, the sequence is not an arithmetic sequence. Extension: Plotting a graph of terms in an arithmetic sequence Plotting a graph of the terms of sequence sometimes helps in determining the type of sequence involved. For an arithmetic sequence, plotting an vs. n results in: 458 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 a9 36.3 an = a1 + d(n − 1) b a8 b Term, an a7 b a6 b a5 gradient d b a4 b a3 b a2 a1 b y-intercept, a1 1 36.3 2 3 4 5 6 Index, n 7 8 9 Geometric Sequences Definition: Geometric Sequences A geometric sequence is a sequence in which every number in the sequence is equal to the previous number in the sequence, multiplied by a constant number. This means that the ratio between consecutive numbers in the geometric sequence is a constant. We will explain what we mean by ratio after looking at the following example. 36.3.1 Example - A Flu Epidemic Extension: What is influenza? Influenza (commonly called “the flu”) is caused by the influenza virus, which infects the respiratory tract (nose, throat, lungs). It can cause mild to severe illness that most of us get during winter time. The main way that the influenza virus is spread is from person to person in respiratory droplets of coughs and sneezes. (This is called “droplet spread”.) This can happen when droplets from a cough or sneeze of an infected person are propelled (generally, up to a metre) through the air and deposited on the mouth or nose of people nearby. It is good practise to cover your mouth when you cough or sneeze so as not to infect others around you when you have the flu. Assume that you have the flu virus, and you forgot to cover your mouth when two friends came to visit while you were sick in bed. They leave, and the next day they also have the flu. Let’s assume that they in turn spread the virus to two of their friends by the same droplet spread the following day. Assuming this pattern continues and each sick person infects 2 other friends, we can represent these events in the following manner: Again we can tabulate the events and formulate an equation for the general case: 459 36.3 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 Figure 36.1: Each person infects two more people with the flu virus. Day, n 1 2 3 4 5 .. . n Number of newly-infected people 2 =2 4 = 2 × 2 = 2 × 21 8 = 2 × 4 = 2 × 2 × 2 = 2 × 22 16 = 2 × 8 = 2 × 2 × 2 × 2 = 2 × 23 32 = 2 × 16 = 2 × 2 × 2 × 2 × 2 = 2 × 24 .. . = 2 × 2 × 2 × 2 × . . . × 2 = 2 × 2n−1 The above table represents the number of newly-infected people after n days since you first infected your 2 friends. You sneeze and the virus is carried over to 2 people who start the chain (a1 = 2). The next day, each one then infects 2 of their friends. Now 4 people are newly-infected. Each of them infects 2 people the third day, and 8 people are infected, and so on. These events can be written as a geometric sequence: 2; 4; 8; 16; 32; . . . Note the common factor (2) between the events. Recall from the linear arithmetic sequence how the common difference between terms were established. In the geometric sequence we can determine the common ratio, r, by Or, more general, a3 a2 = =r a1 a2 (36.4) an =r an−1 (36.5) Activity :: Common Factor of Geometric Sequence : Determine the common factor for the following geometric sequences: 1. 5, 10, 20, 40, 80, . . . 2. 1 1 1 2,4,8, . . . 3. 7, 28, 112, 448, . . . 4. 2, 6, 18, 54, . . . 460 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.3 5. −3, 30, −300, 3000, . . . 36.3.2 General Equation for the nth -term of a Geometric Sequence From the above example we know a1 = 2 and r = 2, and we have seen from the table that the nth -term is given by an = 2 × 2n−1 . Thus, in general, an = a1 · rn−1 (36.6) where a1 is the first term and r is called the common ratio. So, if we want to know how many people are newly-infected after 10 days, we need to work out a10 : an = a10 = = = = a1 · rn−1 2 × 210−1 2 × 29 2 × 512 1024 That is, after 10 days, there are 1 024 newly-infected people. Or, how many days would pass before 16 384 people become newly infected with the flu virus? an = a1 · rn−1 16 384 = 2 × 2n−1 16 384 ÷ 2 = 2n−1 8 192 = 2n−1 213 = 2n−1 13 = n − 1 n = 14 That is, 14 days pass before 16 384 people are newly-infected. Activity :: General Equation of Geometric Sequence : Determine the formula for the following geometric sequences: 1. 5, 10, 20, 40, 80, . . . 2. 1 1 1 2,4,8, . . . 3. 7, 28, 112, 448, . . . 4. 2, 6, 18, 54, . . . 5. −3, 30, −300, 3000, . . . 36.3.3 Exercises 1. What is the important characteristic of an arithmetic sequence? 461 36.4 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 2. Write down how you would go about finding the formula for the nth term of an arithmetic sequence? 3. A single square is made from 4 matchsticks. Two squares in a row needs 7 matchsticks and 3 squares in a row needs 10 matchsticks. Determine: A the first term B the common difference C the formula for the general term D how many matchsticks are in a row of 25 squares 4. 5; x; y is an arithmetic sequence and 81; x; y is a geometric sequence. All terms in the sequences are integers. Calculate the values of x and y. 36.4 Recursive Formulae for Sequences When discussing arithmetic and quadratic sequences, we noticed that the difference between two consecutive terms in the sequence could be written in a general way. For an arithmetic sequence, where a new term is calculated by taking the previous term and adding a constant value, d: an = an−1 + d The above equation is an example of a recursive equation since we can calculate the nth -term only by considering the previous term in the sequence. Compare this with equation (36.1), an = a1 + d · (n − 1) (36.7) where one can directly calculate the nth -term of an arithmetic sequence without knowing previous terms. For quadratic sequences, we noticed the difference between consecutive terms is given by (??): an − an−1 = D · (n − 2) + d Therefore, we re-write the equation as an = an−1 + D · (n − 2) + d (36.8) which is then a recursive equation for a quadratic sequence with common second difference, D. Using (36.5), the recursive equation for a geometric sequence is: an = r · an−1 (36.9) Recursive equations are extremely powerful: you can work out every term in the series just by knowing previous terms. As you can see from the examples above, working out an using the previous term an−1 can be a much simpler computation than working out an from scratch using a general formula. This means that using a recursive formula when using a computer to work out a sequence would mean the computer would finish its calculations significantly quicker. 462 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.5 Activity :: Recursive Formula : Write the first 5 terms of the following sequences, given their recursive formulae: 1. an = 2an−1 + 3, a1 = 1 2. an = an−1 , a1 = 11 3. an = 2a2n−1 , a1 = 2 Extension: The Fibonacci Sequence Consider the following sequence: 0; 1; 1; 2; 3; 5; 8; 13; 21; 34; . . . (36.10) The above sequence is called the Fibonacci sequence. Each new term is calculated by adding the previous two terms. Hence, we can write down the recursive equation: an = an−1 + an−2 36.5 (36.11) Series In this section we simply work on the concept of adding up the numbers belonging to arithmetic and geometric sequences. We call the sum of any sequence of numbers a series. 36.5.1 Some Basics If we add up the terms of a sequence, we obtain what is called a series. If we only sum a finite amount of terms, we get a finite series. We use the symbol Sn to mean the sum of the first n terms of a sequence {a1 ; a2 ; a3 ; . . . ; an }: S n = a1 + a2 + a3 + . . . + an (36.12) For example, if we have the following sequence of numbers 1; 4; 9; 25; 36; 49; . . . and we wish to find the sum of the first 4 terms, then we write S4 = 1 + 4 + 9 + 25 = 39 The above is an example of a finite series since we are only summing 4 terms. If we sum infinitely many terms of a sequence, we get an infinite series: S ∞ = a1 + a2 + a3 + . . . (36.13) In the case of an infinite series, the number of terms is unknown and simply increases to ∞. 36.5.2 Sigma Notation In this section we introduce a notation that will make our lives a little easier. 463 36.5 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 P A sum may be written out using the summation symbol . This symbol is sigma, which is the capital letter “S” in the Greek alphabet. It indicates that you must sum the expression to the right of it: n X ai = am + am+1 + . . . + an−1 + an (36.14) i=m where • i is the index of the sum; • m is the lower bound (or start index), shown below the summation symbol; • n is the upper bound (or end index), shown above the summation symbol; • ai are the terms of a sequence. The index i is increased from m to n in steps of 1. If we are summing from nP = 1 (which implies summing from the first term in a sequence), then we can use either Sn - or -notation since they mean the same thing: Sn = n X ai = a1 + a2 + . . . + an (36.15) i=1 For example, in the following sum, 5 X i i=1 we have to add together all the terms in the sequence ai = i from i = 1 up until i = 5: 5 X i = 1 + 2 + 3 + 4 + 5 = 15 i=1 Examples 1. 6 X 2i = 21 + 22 + 23 + 24 + 25 + 26 = 2 + 4 + 8 + 16 + 32 + 64 = 126 i=1 2. 10 X (3xi ) = 3x3 + 3x4 + . . . + 3x9 + 3x10 i=3 for any value x. Some Basic Rules for Sigma Notation 1. Given two sequences, ai and bi , n X (ai + bi ) = n X i=1 i=1 464 ai + n X i=1 bi (36.16) CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.6 2. For any constant c, which is any variable not dependent on the index i, n X i=1 c · ai = c · a1 + c · a2 + c · a3 + . . . + c · an = c (a1 + a2 + a3 + . . . + an ) n X ai = c (36.17) i=1 Exercises 1. What is 4 P 2? k=1 2. Determine 3 P i. i=−1 3. Expand 5 P i. k=0 4. Calculate the value of a if: 3 X k=1 36.6 a · 2k−1 = 28 Finite Arithmetic Series Remember that an arithmetic sequence is a set of numbers, such that the difference between any term and the previous term is a constant number, d, called the constant difference: an = a1 + d (n − 1) (36.18) where • n is the index of the sequence; • an is the nth -term of the sequence; • a1 is the first term; • d is the common difference. When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series. The simplest arithmetic sequence is when a1 = 1 and d = 0 in the general form (36.18); in other words all the terms in the sequence are 1: ai = = = {ai } = a1 + d (i − 1) 1 + 0 · (i − 1) 1 {1; 1; 1; 1; 1; . . .} If we wish to sum this sequence from i = 1 to any positive integer n, we would write n X i=1 ai = n X 1 = 1 + 1 + 1 + ...+ 1 i=1 465 (n times) 36.6 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 Since all the terms are equal to 1, it means that if we sum to n we will be adding n-number of 1’s together, which is simply equal to n: n X 1=n (36.19) i=1 Another simple arithmetic sequence is when a1 = 1 and d = 1, which is the sequence of positive integers: ai = = = {ai } = a1 + d (i − 1) 1 + 1 · (i − 1) i {1; 2; 3; 4; 5; . . .} If we wish to sum this sequence from i = 1 to any positive integer n, we would write n X i = 1 + 2 + 3 + ...+ n (36.20) i=1 This is an equation with a very important solution as it gives the answer to the sum of positive integers. teresting Mathematician, Karl Friedrich Gauss, discovered this proof when he was only Interesting Fact Fact 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Karl realised how to do this almost instantaneously and shocked the teacher with the correct answer, 5050. We first write Sn as a sum of terms in ascending order: Sn = 1 + 2 + . . . + (n − 1) + n (36.21) We then write the same sum but with the terms in descending order: Sn = n + (n − 1) + . . . + 2 + 1 (36.22) We then add corresponding pairs of terms from equations (36.21) and (36.22), and we find that the sum for each pair is the same, (n + 1): 2 Sn = (n + 1) + (n + 1) + . . . + (n + 1) + (n + 1) (36.23) We then have n-number of (n + 1)-terms, and by simplifying we arrive at the final result: 2 Sn = Sn = Sn = n X i=1 36.6.1 n (n + 1) n (n + 1) 2 i= n (n + 1) 2 (36.24) General Formula for a Finite Arithmetic Series If we wish to sum any arithmetic sequence, there is no need to work it out term-for-term. We will now determine the general formula to evaluate a finite arithmetic series. We start with the 466 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.6 general formula for an arithmetic sequence and sum it from i = 1 to any positive integer n: n X ai = i=1 = = = = n X i=1 n X i=1 n X i=1 n X i=1 n X i=1 = = = = [a1 + d (i − 1)] (a1 + di − d) [(a1 − d) + di] (a1 − d) + n X (a1 − d) + d (a1 − d) n + (di) i=1 n X i i=1 dn (n + 1) 2 n (2a1 − 2d + dn + d) 2 n (2a1 + dn − d) 2 n [ 2a1 + d (n − 1) ] 2 So, the general formula for determining an arithmetic series is given by Sn = n X i=1 [ a1 + d (i − 1) ] = n [ 2a1 + d (n − 1) ] 2 (36.25) For example, if we wish to know the series S20 for the arithmetic sequence ai = 3 + 7 (i − 1), we could either calculate each term individually and sum them: S20 = 20 X i=1 = [3 + 7 (i − 1)] 3 + 10 + 17 + 24 + 31 + 38 + 45 + 52 + 59 + 66 + 73 + 80 + 87 + 94 + 101 + 108 + 115 + 122 + 129 + 136 = 1390 or, more sensibly, we could use equation (36.25) noting that a1 = 3, d = 7 and n = 20 so that S20 = = = 20 X [3 + 7 (i − 1)] i=1 20 2 [2 1390 · 3 + 7 (20 − 1)] In this example, it is clear that using equation (36.25) is beneficial. 36.6.2 Exercises 1. The sum to n terms of an arithmetic series is Sn = n (7n + 15). 2 A How many terms of the series must be added to give a sum of 425? B Determine the 6th term of the series. 2. The sum of an arithmetic series is 100 times its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero. 467 36.7 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 3. The common difference of an arithmetic series is 3. Calculate the values of n for which the nth term of the series is 93, and the sum of the first n terms is 975. 4. The sum of n terms of an arithmetic series is 5n2 − 11n for all values of n. Determine the common difference. 5. The sum of an arithmetic series is 100 times the value of its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero. 6. The third term of an arithmetic sequence is -7 and the 7t h term is 9. Determine the sum of the first 51 terms of the sequence. 7. Calculate the sum of the arithmetic series 4 + 7 + 10 + · · · + 901. 8. The common difference of an arithmetic series is 3. Calculate the values of n for which the nth term of the series is 93 and the sum of the first n terms is 975. 36.7 Finite Squared Series When we sum a finite number of terms in a quadratic sequence, we get a finite quadratic series. The general form of a quadratic series is quite complicated, so we will only look at the simple case when D = 2 and d = (a2 − a1 ) = 3 in the general form (???). This is the sequence of squares of the integers: ai = i2 {ai } = = {12 ; 22 ; 32 ; 42 ; 52 ; 62 ; . . .} {1; 4; 9; 16; 25; 36; . . .} If we wish to sum this sequence and create a series, then we write Sn = n X i 2 = 1 + 4 + 9 + . . . + n2 i=1 which can be written, in general, as Sn = n X i2 = i=1 n (2n + 1)(n + 1) 6 The proof for equation (36.26) can be found under the Advanced block that follows: Extension: Derivation of the Finite Squared Series We will now prove the formula for the finite squared series: Sn = n X i 2 = 1 + 4 + 9 + . . . + n2 i=1 3 We start off with the expansion of (k + 1) . (k + 1)3 (k + 1)3 − k 3 = k 3 + 3k 2 + 3k + 1 = 3k 2 + 3k + 1 468 (36.26) CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.8 k=1 : 23 − 13 = 3(1)2 + 3(1) + 1 k=2 : 33 − 23 = 3(2)2 + 3(2) + 1 k=3 : .. . k=n : 43 − 33 = 3(3)2 + 3(3) + 1 (n + 1)3 − n3 = 3n2 + 3n + 1 If we add all the terms on the right and left, we arrive at 3 (n + 1) − 1 = n X (3i2 + 3i + 1) i=1 n3 + 3n2 + 3n + 1 − 1 = 3 n3 + 3n2 + 3n = 3 n X n X i2 n X i+ i=1 i=1 i=1 n X i2 + 3 i2 + n X 1 i=1 3n (n + 1) + n 2 = 1 3 3n [n + 3n2 + 3n − (n + 1) − n] 3 2 = 3 3 1 3 (n + 3n2 + 3n − n2 − n − n) 3 2 2 = 1 3 3 2 1 (n + n + n) 3 2 2 = n (2n2 + 3n + 1) 6 i2 = n (2n + 1)(n + 1) 6 i=1 Therefore, n X i=1 36.8 Finite Geometric Series When we sum a known number of terms in a geometric sequence, we get a finite geometric series. We know from (??) that we can write out each term of a geometric sequence in the general form: an = a1 · rn−1 (36.27) where • n is the index of the sequence; • an is the nth -term of the sequence; • a1 is the first term; • r is the common ratio (the ratio of any term to the previous term). By simply adding together the first n terms, we are actually writing out the series Sn = a1 + a1 r + a1 r2 + . . . + a1 rn−2 + a1 rn−1 (36.28) We may multiply the above equation by r on both sides, giving us rSn = a1 r + a1 r2 + a1 r3 + . . . + a1 rn−1 + a1 rn 469 (36.29) 36.8 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 You may notice that all the terms on the right side of (36.28) and (36.29) are the same, except the first and last terms. If we subtract (36.28) from (36.29), we are left with just rSn − Sn = a1 rn − a1 Sn (r − 1) = a1 (rn − 1) Dividing by (r − 1) on both sides, we arrive at the general form of a geometric series: Sn = n X i=1 36.8.1 a1 · ri−1 = a1 (rn − 1) r−1 (36.30) Exercises 1. Prove that a + ar + ar2 + ... + arn−1 = a (1 − rn ) (1 − r) 2. Find the sum of the first 11 terms of the geometric series 6 + 3 + 3 2 + 3 4 + ... 3. Show that the sum of the first n terms of the geometric series 54 + 18 + 6 + ... + 5 ( 31 )n−1 is given by 81 − 34−n . 4. The eighth term of a geometric sequence is 640. The third term is 20. Find the sum of the first 7 terms. 5. Solve for n: n P t=1 8 ( 21 )t = 15 43 . 6. The ratio between the sum of the first three terms of a geometric series and the sum of the 4th -, 5th − and 6th -terms of the same series is 8 : 27. Determine the common ratio and the first 2 terms if the third term is 8. 7. Given the geometric series: 2 · (5)5 + 2 · (5)4 + 2 · (5)3 + . . . A Show that the series converges B Calculate the sum to infinity of the series C Calculate the sum of the first 8 terms of the series, correct to two decimal places. D Determine ∞ X n=9 2 · 56−n correct to two decimal places using previously calculated results. 8. Given the geometric sequence 1; −3; 9; . . . determine: A The 8th term of the sequence B The sum of the first 8 terms of the sequence. 9. Determine: 4 X n=1 3 · 2n−1 470 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.9 36.9 Infinite Series Thus far we have been working only with finite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the first n terms. In this section, we consider what happens when we add infinitely many terms together. You might think that this is a silly question - surely the answer will be ∞ when one sums infinitely many numbers, no matter how small they are? The surprising answer is that in some cases one will reach ∞ (like when you try to add all the positive integers together), but in some cases one will get a finite answer. If you don’t believe this, try doing the following sum, a geometric series, on your calculator or computer: 1 1 1 1 1 2 + 4 + 8 + 16 + 32 + . . . You might think that if you keep adding more and more terms you will eventually get larger and larger numbers, but in fact you won’t even get past 1 - try it and see for yourself! We denote the sum of an infinite number of terms of a sequence by S∞ = ∞ X ai i=1 When we sum the terms of a series, and the answer we get after each summation gets closer and closer to some number, we say that the series converges. If a series does not converge, then we say that it diverges. 36.9.1 Infinite Geometric Series There is a simple test for knowing instantly which geometric series converges and which diverges. When r, the common ratio, is strictly between -1 and 1, i.e. −1 < r < 1, the infinite series will converge, otherwise it will diverge. There is also a formula for working out what the series converges to. Let’s start off with formula (36.30) for the finite geometric series: Sn = n X i=1 a1 · ri−1 = a1 (rn − 1) r−1 Now we will investigate the value of rn for −1 Take r = 21 : n = 1 : rn = r1 = ( 12 )1 = n = 2 : rn = r2 = ( 12 )2 = n = 3 : rn = r3 = ( 12 )3 = 1 2 1 2 1 2 · · 1 2 1 2 = · 1 2 1 4 < = 1 8 1 2 < Since r is a fractional value in the range −1 Therefore, Sn = a1 (rn − 1) r−1 S∞ = a1 (0 − 1) r−1 = −a1 r−1 = a1 1−r 471 for − 1 < r < 1 1 4 36.10 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 The sum of an infinite geometric series is given by the formula S∞ = ∞ X a1 .ri−1 = i=1 a1 1−r for −1 (36.31) where a1 is the first term of the series and r is the common ratio. 36.9.2 Exercises 1. What does ( 52 )n approach as n tends towards ∞? 2. Find the sum to infinity of the geometric series 3 + 1 + 1 3 + 1 9 + ... 3. Determine for which values of x, the geometric series 2+ 2 3 (x + 1) + 2 9 (x + 1)2 + . . . will converge. 4. The sum to infinity of a geometric series with positive terms is 4 61 and the sum of the first two terms is 2 32 . Find a, the first term, and r, the common ratio between consecutive terms. 36.10 End of Chapter Exercises 1. Is 1 + 2 + 3 + 4 + ... an example of a finite series or an infinite series? 2. Calculate 6 X k+2 3 ( 31 ) k=2 3. If x + 1; x − 1; 2x − 5 are the first 3 terms of a convergent geometric series, calculate the: A Value of x. B Sum to infinity of the series. 4. Write the sum of the first 20 terms of the series 6 + 3 + 3 2 5. Given the geometric series: 2 · 55 + 2 · 54 + 2 · 53 + . . . + 3 4 + ... in P -notation. A Show that the series converges. B Calculate the sum of the first 8 terms of the series, correct to TWO decimal places. C Calculate the sum to infinity of the series. D Use your answer to 5c above to determine ∞ X n=9 2 · 5(6−n) correct to TWO decimal places. 6. For the geometric series, 54 + 18 + 6 + ... + 5 ( 31 )n−1 calculate the smallest value of n for which the sum of the first n terms is greater than 80.99. ∞ P 12( 51 )k−1 . 7. Determine the value of k=1 8. A new soccer competition requires each of 8 teams to play every other team once. A Calculate the total number of matches to be played in the competition. 472 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.10 B If each of n teams played each other once, determine a formula for the total number of matches in terms of n. 9. The midpoints of the sides of square with length equal to 4 units are joined to form a new square. The process is repeated indefinitely. Calculate the sum of the areas of all the squares so formed. 10. Thembi worked part-time to buy a Mathematics book which cost R29,50. On 1 February she saved R1,60, and saves everyday 30 cents more than she saved the previous day. (So, on the second day, she saved R1,90, and so on.) After how many days did she have enough money to buy the book? 11. Consider the geometric series: 5 + 2 12 + 1 14 + . . . A If A is the sum to infinity and B is the sum of the first n terms, write down the value of: i. A ii. B in terms of n. B For which values of n is (A − B) < 1 24 ? 12. A certain plant reaches a height of 118 mm after one year under ideal conditions in a greenhouse. During the next year, the height increases by 12 mm. In each successive year, the height increases by 85 of the previous year’s growth. Show that the plant will never reach a height of more than 150 mm. 13. Calculate the value of n if n P a=1 (20 − 4a) = −20. 14. Michael saved R400 during the first month of his working life. In each subsequent month, he saved 10% more than what he had saved in the previous month. A How much did he save in the 7th working month? B How much did he save all together in his first 12 working months? C In which month of his working life did he save more than R1,500 for the first time? 15. A man was injured in an accident at work. He receives a disability grant of R4,800 in the first year. This grant increases with a fixed amount each year. A What is the annual increase if, over 20 years, he would have received a total of R143,500? B His initial annual expenditure is R2,600 and increases at a rate of R400 per year. After how many years does his expenses exceed his income? 16. The Cape Town High School wants to build a school hall and is busy with fundraising. Mr. Manuel, an ex-learner of the school and a successful politician, offers to donate money to the school. Having enjoyed mathematics at school, he decides to donate an amount of money on the following basis. He sets a mathematical quiz with 20 questions. For the correct answer to the first question (any learner may answer), the school will receive 1 cent, for a correct answer to the second question, the school will receive 2 cents, and so on. The donations 1, 2, 4, ... form a geometric sequence. Calculate (Give your answer to the nearest Rand) A The amount of money that the school will receive for the correct answer to the 20th question. B The total amount of money that the school will receive if all 20 questions are answered correctly. 17. The first term of a geometric sequence is 9, and the ratio of the sum of the first eight terms to the sum of the first four terms is 97 : 81. Find the first three terms of the sequence, if it is given that all the terms are positive. 18. (k − 4); (k + 1); m; 5k is a set of numbers, the first three of which form an arithmetic sequence, and the last three a geometric sequence. Find k and m if both are positive. 473 36.10 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 19. Given: The sequence 6 + p ; 10 + p ; 15 + p is geometric. A Determine p. B Show that the common ratio is 45 . C Determine the 10th term of this sequence correct to one decimal place. 20. The second and fourth terms of a convergent geometric series are 36 and 16, respectively. Find the sum to infinity of this series, if all its terms are positive. 21. Evaluate: 5 k(k + 1) P 2 k=2 22. Sn = 4n2 + 1 represents the sum of the first n terms of a particular series. Find the second term. ∞ P 23. Find p if: 27pk = 12 P t=1 k=1 (24 − 3t) 24. Find the integer that is the closest approximation to: 102001 + 102003 102002 + 102002 25. Find the pattern and hence calculate: 1 − 2 + 3 − 4 + 5 − 6 . . . + 677 − 678 + . . . − 1000 26. Determine ∞ P (x + 2)p , if it exists, when p=1 A x=− 5 2 B x = −5 27. Calculate: ∞ P i=1 5 · 4−i 28. The sum of the first p terms of a sequence is p (p + 1). Find the 10th term. 29. he powers of 2 are removed from the set of positive integers 1; 2; 3; 4; 5; 6; . . . ; 1998; 1999; 2000 Find the sum of remaining integers. 30. Observe the pattern below: 474 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 36.10 E A D E C D E B C D E B C D E B C D E C D E D E E A If the pattern continues, find the number of letters in the column containing M’s. B If the total number of letters in the pattern is 361, which letter will the last column consist of. 31. The following question was asked in a test: Find the value of 22005 + 22005 . Here are some of the students’ answers: A Megansaid the answer is 42005 . B Stefan wrote down 24010 . C Nina thinks it is 22006 . D Annatte gave the answer 22005×2005 . Who is correct? (“None of them” is also a possibility.) 32. Find the pattern and hence calculate: 1 − 2 + 3 − 4 + 5 − 6 . . . + 677 − 678 + . . . − 1000 33. Determine ∞ P (x + 2)p , if it exists, when p=1 5 2 B x = −5 A x=− 34. Calculate: ∞ P i=1 5 · 4−i 475 36.10 CHAPTER 36. SEQUENCES AND SERIES - GRADE 12 35. The sum of the first p terms of a sequence is p (p + 1). Find the 10th term. 36. The powers of 2 are removed from the set of positive integers 1; 2; 3; 4; 5; 6; . . . ; 1998; 1999; 2000 Find the sum of remaining integers. 37. A shrub of height 110 cm is planted. At the end of the first year, the shrub is 120 cm tall. Thereafter, the growth of the shrub each year is half of its growth in the previous year. Show that the height of the shrub will never exceed 130 cm. 476 Chapter 37 Finance - Grade 12 37.1 Introduction In earlier grades simple interest and compound interest were studied, together with the concept of depreciation. Nominal and effective interest rates were also described. Since this chapter expands on earlier work, it would be best if you revised the work done in Chapters 8 and 21. If you master the techniques in this chapter, when you start working and earning you will be able to apply the techniques in this chapter to critically assess how to invest your money. And when you are looking at applying for a bond from a bank to buy a home, you will confidently be able to get out the calculator and work out with amazement how much you could actually save by making additional repayments. Indeed, this chapter will provide you with the fundamental concepts you will need to confidently manage your finances and with some successful investing, sit back on your yacht and enjoy the millionaire lifestyle. 37.2 Finding the Length of the Investment or Loan In Grade 11, we used the formula A = P (1 + i)n to determine the term of the investment or loan, by trial and error. In other words, if we know what the starting sum of money is and what it grows to, and if we know what interest rate applies - then we can work out how long the money needs to be invested for all those other numbers to tie up. Now, that you have learnt about logarithms, you are ready to work out the proper algebraic solution. If you need to remind yourself how logarithms work, go to Chapter 35 (on page 445). The basic finance equation is: A = P · (1 + i)n If you don’t know what A, P , i and n represent, then you should definitely revise the work from Chapters 8 and 21. Solving for n: A = (1 + i)n = P (1 + i)n (A/P ) log((1 + i)n ) = n log(1 + i) = log(A/P ) log(A/P ) n = log(A/P )/ log(1 + i) Remember, you do not have to memorise this formula. It is very easy to derive any time you need it. It is simply a matter of writing down what you have, deciding what you need, and solving for that variable. 477 37.3 CHAPTER 37. FINANCE - GRADE 12 Worked Example 162: Term of Investment - Logarithms Question: If we invested R3 500 into a savings account which pays 7,5% compound interest for an unknown period of time, at the end of which our account is worth R4 044,69. How long did we invest the money? How does this compare with the trial and error answer from Chapters 21. Answer Step 1 : Determine what is given and what is required • P =R3 500 • i=7,5% • A=R4 044,69 We are required to find n. Step 2 : Determine how to approach the problem We know that: A = (1 + i)n = P (1 + i)n (A/P ) log((1 + i)n ) = n log(1 + i) = log(A/P ) log(A/P ) n = log(A/P )/ log(1 + i) Step 3 : Solve the problem n = = = log(A/P )/ log(1 + i) log( R4R3044,69 500 ) log(1 + 7,5%) 2.0 Step 4 : Write final answer The R3 500 was invested for 2 years. 37.3 A Series of Payments By this stage, you know how to do calculations such as ”If I want R1 000 in 3 years’ time, how much do I need to invest now at 10% ?” But what if we extend this as follows: If I want R1 000 next year and R1 000 the year after that and R1 000 after three years ... how much do I need to put into a bank account earning 10% p.a. right now to be able to afford that?” The obvious way of working that out is to work out how much you need now to afford the payments individually and sum them. We’ll work out how much is needed now to afford the payment of R1 000 in a year (= R1 000 × (1,10)−1 = R909,0909), the amount needed now for the following year’s R1 000 (= R1 000 × (1,10)−2 = R826,4463) and the amount needed now for the R1 000 after 3 years (= R1 000 × (1,10)−3 = R751,3148). Add these together gives you the amount needed to afford all three payments and you get R2486,85. So, if you put R2486,85 into a 10% bank account now, you will be able to draw out R1 000 in a year, R1 000 a year after that, and R1 000 a year after that - and your bank account will come down to R0. You would have had exactly the right amount of money to do that (obviously!). You can check this as follows: 478 CHAPTER 37. FINANCE - GRADE 12 Amount Amount Amount Amount Amount Amount Amount at Time 0 (i.e. Now) at Time 1 (i.e. a year later) after the R1 000 at Time 2 (i.e. a year later) after the R1 000 at Time 3 (i.e. a year later) after the R1 000 37.3 = = = = = = 2486,85(1+10%) 2735,54 - 1 000 1735,54(1+10%) R1909,09 - 1 000 909,09(1+10%) 1 000 - 1 000 = = = = = = = R2486,85 R2735,54 R1735,54 R1909,09 R909,09 R1 000 R0 Perfect! Of course, for only three years, that was not too bad. But what if I asked you how much you needed to put into a bank account now, to be able to afford R100 a month for the next 15 years. If you used the above approach you would still get the right answer, but it would take you weeks! There is - I’m sure you guessed - an easier way! This section will focus on describing how to work with: • annuities - a fixed sum payable each year or each month either to provide a pre-determined sum at the end of a number of years or months (referred to as a future value annuity) or a fixed amount paid each year or each month to repay (amortise) a loan (referred to as a present value annuity). • bond repayments - a fixed sum payable at regular intervals to pay off a loan. This is an example of a present value annuity. • sinking funds - an accounting term for cash set aside for a particular purpose and invested so that the correct amount of money will be available when it is needed. This is an example of a future value annuity 37.3.1 Sequences and Series Before we progress, you need to go back and read Chapter 36 (from page 457) to revise sequences and series. In summary, if you have a series of n terms in total which looks like this: a + ar + ar2 + ... + arn−1 = a[1 + r + r2 + ...rn−1 ] this can be simplified as: a(rn − 1) r−1 a(1 − rn ) 1−r 37.3.2 useful when r > 1 useful when 0 ≤ r < 1 Present Values of a series of Payments So having reviewed the mathematics of Sequences and Series, you might be wondering how this is meant to have any practical purpose! Given that we are in the finance section, you would be right to guess that there must be some financial use to all this Here is an example which happens in many people’s lives - so you know you are learning something practical Let us say you would like to buy a property for R300 000, so you go to the bank to apply for a mortgage bond. The bank wants it to be repaid by annually payments for the next 20 years, starting at end of this year. They will charge you 15% per annum. At the end of the 20 years the bank would have received back the total amount you borrowed together with all the interest they have earned from lending you the money. You would obviously want to work out what the annual repayment is going to be! Let X be the annual repayment, i is the interest rate, and M is the amount of the mortgage bond you will be taking out. Time lines are particularly useful tools for visualizing the series of payments for calculations, and we can represent these payments on a time line as: 479 37.3 CHAPTER 37. FINANCE - GRADE 12 0 X X X X X 1 2 18 19 20 Cash Flows Time Figure 37.1: Time Line for an annuity (in arrears) of X for n periods. The present value of all the payments (which includes interest) must equate to the (present) value of the mortgage loan amount. Mathematically, you can write this as: M = X(1 + i)−1 + X(1 + i)−2 + X(1 + i)−3 + ... + X(1 + i)−20 The painful way of solving this problem would be to do the calculation for each of the terms above - which is 20 different calculations. Not only would you probably get bored along the way, but you are also likely to make a mistake. Naturally, there is a simpler way of doing this! You can rewrite the above equation as follows: M = X(v 1 + v 2 + v 3 + ... + v 20 ) where v = (1 + i)−1 = 1/(1 + i) Of course, you do not have to use the method of substitution to solve this. We just find this a useful method because you can get rid of the negative exponents - which can be quite confusing! As an exercise - to show you are a real financial whizz - try to solve this without substitution. It is actually quite easy. Now, the item in square brackets is the sum of a geometric sequence, as discussion in section 36. This can be re-written as follows, using what we know from Chapter 36 of this text book: v 1 + v 2 + v 3 + ... + v n = v(1 + v + v 2 + ... + v n−1 ) 1 − vn ) = v( 1−v 1 − vn = 1/v − 1 1 − (1 + i)−n = i Note that we took out a common factor of v before using the formula for the geometric sequence. So we can write: M = X[ This can be re-written: X= (1 − (1 + i)−n ) ] i M −n ) ] [ (1−(1+i) i So, this formula is useful if you know the amount of the mortgage bond you need and want to work out the repayment, or if you know how big a repayment you can afford and want to see what property you can buy. For example, if I want to buy a house for R300 000 over 20 years, and the bank is going to 480 CHAPTER 37. FINANCE - GRADE 12 37.3 charge me 15% per annum, then the annual repayment is: X = = M −n ) ] [ (1−(1+i) i R300 000 −20 ) ] [ (1−(1,15) 0,15 = R47 928,44 This means, each year for the next 20 years, I need to pay the bank R47 928,44 per year before I have paid off the mortgage bond. On the other hand, if I know I will only have R30 000 a year to repay my bond, then how big a house can I buy? That is easy .... (1 − (1 + i)−n ) ] i (1 − (1,15)−20 ) ] = R30 000[ 0,15 = R187 779,90 M = X[ So, for R30 000 a year for 20 years, I can afford to buy a house of R187 800 (rounded to the nearest hundred). The bad news is that R187 800 does not come close to the R300 000 you wanted to buy! The good news is that you do not have to memorise this formula. In fact , when you answer questions like this in an exam, you will be expected to start from the beginning - writing out the opening equation in full, showing that it is the sum of a geometric sequence, deriving the answer, and then coming up with the correct numerical answer. Worked Example 163: Monthly mortgage repayments Question: Sam is looking to buy his first flat, and has R15 000 in cash savings which he will use as a deposit. He has viewed a flat which is on the market for R250 000, and he would like to work out how much the monthly repayments would be. He will be taking out a 30 year mortgage with monthly repayments. The annual interest rate is 11%. Answer Step 1 : Determine what is given and what is needed The following is given: • Deposit amount = R15 000 • Price of flat = R250 000 • interest rate, i = 11% We are required to find the monthly repayment for a 30-year mortgage. Step 2 : Determine how to approach the problem We know that: M X = (1−(1+i)−n ) ] [ i . In order to use this equation, we need to calculate M , the amount of the mortgage bond, which is the purchase price of property less the deposit which Sam pays upfront. M = = R250 000 − R15 000 R235 000 481 37.3 CHAPTER 37. FINANCE - GRADE 12 Now because we are considering monthly repayments, but we have been given an annual interest rate, we need to convert this to a monthly interest rate, i12. (If you are not clear on this, go back and revise section 21.8.) (1 + i12)12 = 12 (1 + i12) = i12 = (1 + i) 1,11 0,873459% We know that the mortgage bond is for 30 years, which equates to 360 months. Step 3 : Solve the problem Now it is easy, we can just plug the numbers in the formula, but do not forget that you can always deduce the formula from first principles as well! X = = = M −n ) ] [ (1−(1+i) i R235 000 −360 ) ] [ (1−(1.00876459) 0,008734594 R2 146,39 Step 4 : Write the final answer That means that to buy a house for R300 000, after Sam pays a R15 000 deposit, he will make repayments to the bank each month for the next 30 years equal to R2 146,39. Worked Example 164: Monthly mortgage repayments Question: You are considering purchasing a flat for R200 000 and the bank’s mortgage rate is currently 9% per annum payable monthly. You have savings of R10 000 which you intend to use for a deposit. How much would your monthly mortgage payment be if you were considering a mortgage over 20 years. Answer Step 1 : Determine what is given and what is required The following is given: • Deposit amount = R10 000 • Price of flat = R200 000 • interest rate, i = 9% We are required to find the monthly repayment for a 20-year mortgage. Step 2 : Determine how to approach the problem We are consider monthly mortgage repayments, so it makes sense to use months as our time period. The interest rate was quoted as 9% per annum payable monthly, which means that the monthly effective rate = 9%/12 = 0,75% per month. Once we have converted 20 years into 240 months, we are ready to do the calculations! First we need to calculate M , the amount of the mortgage bond, which is the purchase price of property less the deposit which Sam pays up-front. M = R200 000 − R10 000 = R190 000 482 CHAPTER 37. FINANCE - GRADE 12 37.3 The present value of our mortgage payments, X, must equate to the mortgage amount that we borrow today, so X × (1 + 0,75%)−1 X × (1 + 0,75%)−2 X × (1 + 0,75%)−3 X × (1 + 0,75%)−4 + + + + ... X × (1 + 0,75%)−239 + X × (1 + 0,75%)−240 But it is clearly much easier to use our formula that work out 240 factors and add them all up! Step 3 : Solve the problem X× 1 − (1 + 0,75%)−240 = 0,75% X × 111,14495 = X = R190 000 R190 000 R1 709,48 Step 4 : Write the final answer So to repay a R190 000 mortgage over 20 years, at 9% interest payable monthly, will cost you R1 709,48 per month for 240 months. Show me the money Now that you’ve done the calculations for the worked example and know what the monthly repayments are, you can work out some surprising figures. For example, R1 709,48 per month for 240 month makes for a total of R410 275,20 (=R1 709,48 × 240). That is more than double the amount that you borrowed! This seems like a lot. However, now that you’ve studied the effects of time (and interest) on money, you should know that this amount is somewhat meaningless. The value of money is dependant on its timing. Nonetheless, you might not be particularly happy to sit back for 20 years making your R1 709,48 mortgage payment every month knowing that half the money you are paying are going toward interest. But there is a way to avoid those heavy interest charges. It can be done for less than R300 extra every month... So our payment is now R2 000. The interest rate is still 9% per annum payable monthly (0,75% per month), and our principal amount borrowed is R190 000. Making this higher repayment amount every month, how long will it take to pay off the mortgage? The present value of the stream of payments must be equal to R190 000 (the present value of the borrowed amount). So we need to solve for n in: R2 000 × [1 − (1 + 0,75%)−n ]/0,75% = −n 1 − (1 + 0,75%) log(1 + 0,75%)−n = = −n × log(1 + 0,75%) = −n × 0,007472 = n = = R190 000 (R190 000/2 000) × 0,75% log[(1 − (R190 000/R2 000) × 0,75%] log[(1 − (R190 000/R2 000) × 0,75%] −1,2465 166,8 months 13,9 years So the mortgage will be completely repaid in less than 14 years, and you would have made a total payment of 166,8× R2 000 = R333 600. Can you see what is happened? Making regular payments of R2 000 instead of the required R1,709,48, you will have saved R76 675,20 (= R410 275,20 - R333 600) in interest, and yet you have only paid an additional amount of R290,52 for 166,8 months, or R48 458,74. You surely 483 37.3 CHAPTER 37. FINANCE - GRADE 12 know by now that the difference between the additional R48 458,74 that you have paid and the R76 675,20 interest that you have saved is attributable to, yes, you have got it, compound interest! 37.3.3 Future Value of a series of Payments In the same way that when we have a single payment, we can calculate a present value or a future value - we can also do that when we have a series of payments. In the above section, we had a few payments, and we wanted to know what they are worth now - so we calculated present values. But the other possible situation is that we want to look at the future value of a series of payments. Maybe you want to save up for a car, which will cost R45 000 - and you would like to buy it in 2 years time. You have a savings account which pays interest of 12% per annum. You need to work out how much to put into your bank account now, and then again each month for 2 years, until you are ready to buy the car. Can you see the difference between this example and the ones at the start of the chapter where we were only making a single payment into the bank account - whereas now we are making a series of payments into the same account? This is a sinking fund. So, using our usual notation, let us write out the answer. Make sure you agree how we come up with this. Because we are making monthly payments, everything needs to be in months. So let A be the closing balance you need to buy a car, P is how much you need to pay into the bank account each month, and i12 is the monthly interest rate. (Careful - because 12% is the annual interest rate, so we will need to work out later what the month interest rate is!) A = P (1 + i12)24 + P (1 + i12)23 + ... + P (1 + i12)1 Here are some important points to remember when deriving this formula: 1. We are calculating future values, so in this example we use (1 + i12)n and not (1 + i12)−n . Check back to the start of the chapter is this is not obvious to you by now. 2. If you draw a timeline you will see that the time between the first payment and when you buy the car is 24 months, which is why we use 24 in the first exponent. 3. Again, looking at the timeline, you can see that the 24th payment is being made one month before you buy the car - which is why the last exponent is a 1. 4. Always check that you have got the right number of payments in the equation. Check right now that you agree that there are 24 terms in the formula above. So, now that we have the right starting point, let us simplify this equation: A = = P [(1 + i12)24 + (1 + i12)23 + . . . + (1 + i12)1 ] P [X 24 + X 23 + . . . + X 1 ] using X = (1 + i12) Note that this time X has a positive exponent not a negative exponent, because we are doing future values. This is not a rule you have to memorise - you can see from the equation what the obvious choice of X should be. Let us reorder the terms: A = P [X 1 + X 2 + . . . + X 24 ] = P · X[1 + X + X 2 + . . . + X 2 3] This is just another sum of a geometric sequence, which as you know can be simplified as: A = = P · X[X n − 1]/((1 + i12) − 1) P · X[X n − 1]/i12 484 CHAPTER 37. FINANCE - GRADE 12 37.4 So if we want to use our numbers, we know that A = R45 000, n=24 (because we are looking at monthly payments, so there are 24 months involved) and i = 12% per annum. BUT (and it is a big but) we need a monthly interest rate. Do not forget that the trick is to keep the time periods and the interest rates in the same units - so if we have monthly payments, make sure you use a monthly interest rate! Using the formula from Section 21.8, we know that (1 + i) = (1 + i12)12 . So we can show that i12 = 0,0094888 = 0,94888%. Therefore, 45 000 = P = P (1,0094888)[(1,0094888)24 − 1]/0,0094888 1662,67 This means you need to invest R1 662,67 each month into that bank account to be able to pay for your car in 2 years time. There is another way of looking at this too - in terms of present values. We know that we need an amount of R45 000 in 24 months time, and at a monthly interest rate of 0,94888%, the present value of this amount is R35 873,72449. Now the question is what monthly amount at 0,94888% interest over 24 month has a present value of R35 873,72449? We have seen this before - it is just like the mortgage questions! So let us go ahead and see if we get to the same answer P = M/[(1 − (1 + i)−n )/i] = R35 873,72449[(1 − (1,0094888)−24)/0,0094888] = R1 662,67 37.3.4 Exercises - Present and Future Values 1. You have taken out a mortgage bond for R875 000 to buy a flat. The bond is for 30 years and the interest rate is 12% per annum payable monthly. A What is the monthly repayment on the bond? B How much interest will be paid in total over the 30 years? 2. How much money must be invested now to obtain regular annuity payments of R 5 500 per month for five years ? The money is invested at 11,1% p.a., compounded monthly. (Answer to the nearest hundred rand) 37.4 Investments and Loans By now, you should be well equipped to perform calculations with compound interest. This section aims to allow you to use these valuable skills to critically analyse investment and load options that you will come across in your later life. This way, you will be able to make informed decisions on options presented to you. At this stage, you should understand the mathematical theory behind compound interest. However, the numerical implications of compound interest is often subtle and far from obvious. Recall the example in section ??FIXTHIS. For an extra payment of R290,52 a month, we could have paid off our loan in less than 14 years instead of 20 years. This provides a good illustration of the long term effect of compound interest that is often surprising. In the following section, we’ll aim to explain the reason for drastic deduction in times it takes to repay the loan. 37.4.1 Loan Schedules So far, we have been working out loan repayment amounts by taking all the payments and discounting them back to the present time. We are not considering the repayments individually. 485 37.4 CHAPTER 37. FINANCE - GRADE 12 Think about the time you make a repayment to the bank. There are numerous questions that could be raised: how much do you still owe them? Since you are paying off the loan, surely you must owe them less money, but how much less? We know that we’ll be paying interest on the money we still owe the bank. When exactly do we pay interest? How much interest are we paying? The answer to these questions lie in something called the load schedule. We will continue to use the example from section ??FIXTHIS. There is a loan amount of R190 000. We are paying it off over 20 years at an interest of 9% per annum payable monthly. We worked out that the repayments should be R1 709,48. Consider the first payment of R1 709,48 one month into the loan. First, we can work out how much interest we owe the bank at this moment. We borrowed R190 000 a month ago, so we should owe: I = = = M × i12 R190 000 × 0,75% R1 425 We are paying them R1 425 in interest. We calls this the interest component of the repayment. We are only paying off R1 709,48 - R1 425 = R284.48 of what we owe! This is called the capital component. That means we still owe R190 000 - R284,48 = R189 715,52. This is called the capital outstanding. Let’s see what happens at end of the second month. The amount of interest we need to pay is the interest on the capital outstanding. I = = = M × i12 R189 715,52 × 0,75% R1 422,87 Since we don’t owe the bank as much as we did last time, we also owe a little less interest. The capital component of the repayment is now R1 709,48 - R1 422,87 = R286,61. The capital outstanding will be R189 715,52 - R286,61 = R189 428,91. This way, we can break each of our repayments down into an interest part and the part that goes towards paying off the loan. This is a simple and repetitive process. Table 37.1 is a table showing the breakdown of the first 12 payments. This is called a loan schedule. Now, let’s see the same thing again, but with R2 000 being repaid each year. We expect the numbers to change. However, how much will they change by? As before, we owe R1 425 in interest in interest. After one month. However, we are paying R2 000 this time. That leaves R575 that goes towards paying off the capital outstanding, reducing it to R189 425. By the end of the second month, the interest owed is R1 420,69 (That’s R189 425×i12). Our R2 000 pays for that interest, and reduces the capital amount owed by R2 000 - R1 420,69 = R579,31. This reduces the amount outstanding to R188 845,69. Doing the same calculations as before yields a new loan schedule shown in Table 37.2. The important numbers to notice is the “Capital Component” column. Note that when we are paying off R2 000 a month as compared to R1 709,48 a month, this column more than doubles? In the beginning of paying off a loan, very little of our money is used to pay off the captital outstanding. Therefore, even a small incread in repayment amounts can significantly increase the speed at which we are paying off the capital. Whatsmore, look at the amount we are still owing after one year (i.e. at time 12). When we were paying R1 709,48 a month, we still owe R186 441,84. However, if we increase the repayments to R2 000 a month, the amount outstanding decreases by over R3 000 to R182 808,14. This means we would have paid off over R7 000 in our first year instead of less than R4 000. This 486 CHAPTER 37. FINANCE - GRADE 12 37.4 Time Repayment Interest Component Capital Component 0 1 2 3 4 5 6 7 8 9 10 11 12 R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R 1 1 1 1 1 1 1 1 1 1 1 1 709,48 709,48 709,48 709,48 709,48 709,48 709,48 709,48 709,48 709,48 709,48 709,48 1 1 1 1 1 1 1 1 1 1 1 1 425,00 422,87 420,72 418,55 416,37 414,17 411,96 409,72 407,48 405,21 402,93 400,63 284,48 286,61 288,76 290,93 293,11 295,31 297,52 299,76 302,00 304,27 306,55 308,85 Capital Outstanding R 190 000,00 R 189 715,52 R 189 428,91 R 189 140,14 R 188 849,21 R 188 556,10 R 188 260,79 R 187 963,27 R 187 663,51 R 187 361,51 R 187 057,24 R 186 750,69 R 186 441,84 Table 37.1: A loan schedule with repayments of R1 709,48 per month. Time Repayment Interest Component Capital Component 0 1 2 3 4 5 6 7 8 9 10 11 12 R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R 2 2 2 2 2 2 2 2 2 2 2 2 000,00 000,00 000,00 000,00 000,00 000,00 000,00 000,00 000,00 000,00 000,00 000,00 1 1 1 1 1 1 1 1 1 1 1 1 425,00 420,69 416,34 411,97 407,55 403,11 398,63 394,12 389,58 385,00 380,39 375,74 575,00 579,31 583,66 588,03 592,45 596,89 601,37 605,88 610,42 615,00 619,61 624,26 Capital Outstanding R 190 000,00 R 189 425,00 R 188 845,69 R 188 262,03 R 187 674,00 R 187 081,55 R 186 484,66 R 185 883,30 R 185 277,42 R 184 667,00 R 184 052,00 R 183 432,39 R 182 808,14 Table 37.2: A loan schedule with repayments of R2 000 per month. 487 37.4 CHAPTER 37. FINANCE - GRADE 12 increased speed at which we are paying off the capital portion of the loan is what allows us to pay off the whole load in around 14 years instead of the original 20. Note however, the effect of paying R2 000 instead of R1 709,48 is more significant in be beginning of the loan than near the end of the loan. It is noted that in this instance, by paying slightly more than what the bank would ask you to pay, you can pay off a loan a lot quicker. The natural question to ask here is: why are banks asking us to pay the lower amount for much longer then? Are they trying to cheat us out of our money? There is no simple answer to this. Banks provide a service to us in return for a fee, so they are out to make a profit. However, they need to be careful not to cheat their customers for fear that they’ll simply use another bank. The central issue here is one of scale. For us, the changes involved appear big. We are paying off our loan 6 years earlier by paying just a bit more a month. To a bank, however, it doesn’t matter much either way. In all likelihoxod, it doesn’t affect their profit margins one bit! Remember that a bank calculates repayment amount using the same methods as we’ve been learning. Therefore, they are correct amounts for given interest rates and terms. As a result, which amount is repaid does generally make a bank more or less money. It’s a simple matter of less money now or more money later. Banks generally use a 20 year repayment period by default. Learning about financial mathematics enables you to duplicate these calculations for yourself. This way, you can decide what’s best for you. You can decide how much you want to repay each month and you’ll know of its effects. A bank wouldn’t care much either way, so you should pick something that suits you. Worked Example 165: Monthly Payments Question: Stefan and Marna want to buy a house that costs R 1 200 000. Their parents offer to put down a 20% payment towards the cost of the house. They need to get a moratage for the balance. What are their monthly repayments if the term of the home loan is 30 years and the interest is 7,5%, compounded monthly ? Answer Step 1 : Determine how much money they need to borrow R1 200 00 − R240 000 = R960 000 Step 2 : Determine how to approach the problem Use the formula: P = x[1 − (1 + i)−n ] i Where P = 960 000 n = 30 × 12 = 360months i = 0,075 ÷ 12 = 0,00625 Step 3 : Solve the problem R960 000 = = x = x[1 − (1 + 0,00625)−360] 0,00625 x(143,017 627 3) R6 712,46 Step 4 : Write the final answer The monthly repayments = R6 712,46 488 CHAPTER 37. FINANCE - GRADE 12 37.4.2 37.5 Exercises - Investments and Loans 1. A property costs R1 800 000. Calculate the monthly repayments if the interest rate is 14% p.a. compounded monthly and the loan must be paid of in 20 years time. 2. A loan of R 4 200 is to be returned in two equal annual instalments. If the rate of interest os 10% per annum, compounded annually, calculate the amount of each instalment. 37.4.3 Calculating Capital Outstanding As defined in Section 37.4.1, Capital outstanding is the amount we still owe the people we borrowed money from at a given moment in time. We also saw how we can calculate this using loan schedules. However, there is a significant disadvantage to this method: it is very time consuming. For example, in order to calculate how much capital is still outstanding at time 12 using the loan schedule, we’ll have to first calculate how much capital is outstanding at time 1 through to 11 as well. This is already quite a bit more work than we’d like to do. Can you imagine calculating the amount outstanding after 10 years (time 120)? Fortunately, there is an easier method. However, it is not immediately why this works, so let’s take some time to examine the concept. Prospective method for Capital Outstanding Let’s say that after a certain number of years, just after we made a repayment, we still owe amount Y . What do we know about Y ? We know that using the loan schedule, we can calculate what it equals to, but that is a lot of repetitive work. We also know that Y is the amount that we are still going to pay off. In other words, all the repayments we are still going to make in the future will exactly pay off Y . This is true because in the end, after all the repayments, we won’t be owing anything. Therefore, the present value of all outstanding future payments equal the present amount outstanding. This is the prospective method for calculating capital outstanding. Let’s return to a previous example. Recall the case where we were trying to repay a loan of R200 000 over 20 years. At an interested rate of 9% compounded monthly, the monthly repayment is R1 709,48. In table 37.1, we can see that after 12 month, the amount outstanding is R186 441,84. Let’s try to work this out using the the prospective method. After time 12, there is still 19 × 12 = 228 repayments left of R1 709,48 each. The present value is: n = i = 228 0,75% Y = R1 709,48 × = R186 441,92 1 − 1,0075−228 0,0075 Oops! This seems to be almost right, but not quite. We should have got R186 441,84. We are 8 cents out. However, this is in fact not a mistake. Remember that when we worked out the monthly repayments, we rounded to the nearest cents and arrived at R1 709,48. This was because one cannot make a payment for a fraction of a cent. Therefore, the rounding off error was carried through. That’s why the two figures don’t match exactly. In financial mathematics, this is largely unavoidable. 37.5 Formulae Sheet As an easy reference, here are the key formulae that we derived and used during this chapter. While memorising them is nice (there are not many), it is the application that is useful. Financial 489 37.6 CHAPTER 37. FINANCE - GRADE 12 experts are not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve financial problems. 37.5.1 P i n iT Definitions Principal (the amount of money at the starting point of the calculation) interest rate, normally the effective rate per annum period for which the investment is made Rate the interest rate paid T times per annum, i.e. iT = Nominal Interest T 37.5.2 Equations Present Value - simple Future Value - simple Solve for i Solve for n Present Value - compound Future Value - compound Solve for i Solve for n = P (1 + i · n) = P (1 + i)n Important: Always keep the interest and the time period in the same units of time (e.g. both in years, or both in months etc.). 37.6 End of Chapter Exercises 1. Thabo is about to invest his R8 500 bonus in a special banking product which will pay 1% per annum for 1 month, then 2% per annum for the next 2 months, then 3% per annum for the next 3 months, 4% per annum for the next 4 months, and 0% for the rest of the year. The are going to charge him R100 to set up the account. How much can he expect to get back at the end of the period? 2. A special bank account pays simple interest of 8% per annum. Calculate the opening balance required to generate a closing balance of R5 000 after 2 years. 3. A different bank account pays compound interest of 8% per annum. Calculate the opening balance required to generate a closing balance of R5 000 after 2 years. 4. Which of the two answers above is lower, and why? 5. After 7 months after an initial deposit, the value of a bank account which pays compound interest of 7,5% per annum is R3 650,81. What was the value of the initial deposit? 6. Suppose you invest R500 this year compounded at interest rate i for a year in Bank T. In the following year you invest the accumulation that you received for another year at the same interest rate and on the third year, you invested the accumulation you received at the same interest rate too. If P represents the present value (R500), find a pattern for this investment. [Hint: find a formula] 7. Thabani and Lungelo are both using UKZN Bank for their saving. Suppose Lungelo makes a deposit of X today at interest rate of i for six years. Thabani makes a deposit of 3X at an interest rate of 0.05. Thabani made his deposit 3 years after Lungelo made his first deposit. If after 6 years, their investments are equal, calculate the value of i and find X. if the sum of their investment is R20 000, use X you got to find out how much Thabani got in 6 years. 490 CHAPTER 37. FINANCE - GRADE 12 37.6 8. Sipho invests R500 at an interest rate of log(1,12) for 5 years. Themba, Sipho’s sister invested R200 at interest rate i for 10 years on the same date that her brother made his first deposit. If after 5 years, Themba’s accumulation equals Sipho’s, find the interest rate i and find out whether Themba will be able to buy her favorite cell phone after 10 years which costs R2 000. 9. Moira deposits R20 000 in her saving account for 2 years at an interest rate of 0.05. After 2 years, she invested her accumulation for another 2 years, at the same interest rate. After 4 years, she invested her accumulation for which she got for another 2 years at an interest rate of 5 %. After 6 years she choose to buy a car which costs R26 000. Her husband, Robert invested the same amount at interest rate of 5 % for 6 years. A Without using any numbers, find a pattern for Moira’s investment? B How Moira’s investment differ from Robert’s? 10. Calculate the real cost of a loan of R10 000 for 5 years at 5% capitalised monthly and half yearly. 11. Determine how long, in years, it will take for the value of a motor vehicle to decrease to 25% of its original value if the rate of depreciation, based on the reducing-balance method, is 21% per annum. 12. André and Thoko, decided to invest their winnings (amounting to R10 000) from their science project. They decided to divide their winnings according to the following: Because Andr was the head of the project and he spent more time on it, André got 65,2 % of the winnings and Thoko got 34,8%. So, Thoko decided to invest only 0,5 % of the share of her sum and Andrédecided to invest 1,5 % of the share of his sum. When they calculated how much each contributed in the investment, Thoko had 25 % and André had 75 % share. They planned to invest their money for 20 years , but, as a result of Thoko finding a job in Australia 7 years after their initial investment. They both decided to take whatever value was there and split it according to their initial investment(in terms of percentages). Find how much each will get after 7 years, if the interest rate is equal to the percentage that Thoko invested (NOT the money but the percentage). 491 37.6 CHAPTER 37. FINANCE - GRADE 12 492 Chapter 38 Factorising Cubic Polynomials Grade 12 38.1 Introduction In grades 10 and 11, you learnt how to solve different types of equations. Most of the solutions, relied on being able to factorise some expression and the factorisation of quadratics was studied in detail. This chapter focusses on the factorisation of cubic polynomials, that is expressions with the highest power equal to 3. 38.2 The Factor Theorem The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial. Definition: Factor Theorem For any polynomial, f (x), for all values of a which satisfy f (a) = 0, (x − a) is a factor of f (x). Or, more concisely: f (x) = q(x) x−a is a polynomial. In other words: If the remainder when dividing f (x) by (x − a) is zero, then (ax + b) is a factor of f (x). So if f (− ab ) = 0, then (ax + b) is a factor of f (x). Worked Example 166: Factor Theorem Question: Use the Factor Theorem to determine whether y − 1 is a factor of f (y) = 2y 4 + 3y 2 − 5y + 7. Answer Step 1 : Determine how to approach the problem In order for y − 1 to be a factor, f (1) must be 0. Step 2 : Calculate f (1) f (y) = ∴ f (1) = = = 2y 4 + 3y 2 − 5y + 7 2(1)4 + 3(1)2 − 5(1) + 7 2+3−5+7 7 493 38.3 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 Step 3 : Conclusion Since f (1) 6= 0, y − 1 is not a factor of f (y) = 2y 4 + 3y 2 − 5y + 7. Worked Example 167: Factor Theorem Question: Using the Factor Theorem, verify that y + 4 is a factor of g(y) = 5y 4 + 16y 3 − 15y 2 + 8y + 16. Answer Step 1 : Determine how to approach the problem In order for y + 4 to be a factor, g(−4) must be 0. Step 2 : Calculate f (1) g(y) = 5y 4 + 16y 3 − 15y 2 + 8y + 16 ∴ g(−4) = 5(−4)4 + 16(−4)3 − 15(−4)2 + 8(−4) + 16 = 5(256) + 16(−64) − 15(16) + 8(−4) + 16 = 1280 − 1024 − 240 − 32 + 16 = 0 Step 3 : Conclusion Since g(−4) = 0, y + 4 is a factor of g(y) = 5y 4 + 16y 3 − 15y 2 + 8y + 16. 38.3 Factorisation of Cubic Polynomials Cubic expressions have a highest power of 3 on the unknown variable. This means that there should be at least 3 factors. We have seen in Grade 10 that the sum and difference of cubes is factorised as follows.: (x + y)(x2 − xy + y 2 ) = x3 + y 3 and (x − y)(x2 + xy + y 2 ) = x3 − y 3 We also saw that the quadratic terms do not have rational roots. There are many methods of factorising a cubic polynomial. The general method is similar to that used to factorise quadratic equations. If you have a cubic polynomial of the form: f (x) = ax3 + bx2 + cx + d then you should expect factors of the form: (Ax + B)(Cx + D)(Ex + F ). (38.1) We will deal with simplest case first. When a = 1, then A = C = E = 1, and you only have to determine B, D and F . For example, find the factors of: x3 − 2x2 − 5x + 6. In this case we have a b = 1 = −2 c = −5 d = 6 494 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 38.3 The factors will have the general form shown in (38.1), with A = C = E = 1. We can then use values for a, b, c and d to determine values for B, D and F . We can re-write (38.1) with A = C = E = 1 as: (x + B)(x + D)(x + F ). If we multiply this out we get: (x + B)(x + D)(x + F ) = (x + B)(x2 + Dx + F x + DF ) = x3 + Dx2 + F x2 + Bx2 + DF x + BDx + BF x + BDF = x3 + (D + F + B)x2 + (DF + BD + BF )x + BDF We can therefore write: b c d = −2 = D + F + B = −5 = DF + BD + BF = 6 = BDF. (38.2) (38.3) (38.4) This is a set of three equations in three unknowns. However, we know that B, D and F are factors of 6 because BDF = 6. Therefore we can use a trial and error method to find B, D and F . This can become a very tedious method, therefore the Factor Theorem can be used to find the factors of cubic polynomials. Worked Example 168: Factorisation of Cubic Polynomials Question: Factorise f (x) = x3 + x2 − 9x − 9 into three linear factors. Answer Step 1 : By trial and error using the factor theorem to find a factor Try f (1) = (1)3 + (1)2 − 9(1) − 9 = 1 + 1 − 9 − 9 = −16 Therefore (x − 1) is not a factor Try f (−1) = (−1)3 + (−1)2 9(−1)9 = 1 + 1 + 99 = 0 Thus (x + 1) is a factor, because f (−1) = 0. Now divide f (x) by (x + 1) using division by inspection: Write x3 + x2 − 9x − 9 = (x + 1)( ) The first term in the second bracket must be x2 to give x3 if one works backwards. The last term in the second bracket must be −9 because +1 × −9 = −9. So we have x3 + x2 − 9x − 9 = (x + 1)(x2 ?x − 9). Now, we must find the coefficient of the middelterm (x). (+1)(x2 ) gives x2 . So, the coefficient of the x-term must be 0. So f (x) = (x + 1)(x2 − 9). Step 2 : Factorise fully x2 − 9 can be further factorised to (x − 3)(x + 3), and we are now left with f (x) = (x + 1)(x − 3)(x + 3) In general, to factorise a cubic polynomial, you find one factor by trial and error. Use the factor theorem to confirm that the guess is a root. Then divide the cubic polynomial by the factor to obtain a quadratic. Once you have the quadratic, you can apply the standard methods to factorise the quadratic. For example the factors of x3 − 2x2 − 5x + 6 can be found as follows: There are three factors which we can write as (x − a)(x − b)(x − c). 495 38.4 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 Worked Example 169: Factorisation of Cubic Polynomials Question: Use the Factor Theorem to factorise x3 − 2x2 − 5x + 6. Answer Step 1 : Find one factor using the Factor Theorem Try f (1) = (1)3 − 2(1)2 − 5(1) + 6 = 1 − 2 − 5 + 6 = 0 Therefore (x − 1) is a factor. Step 2 : Division by expection x3 − 2x2 − 5x + 6 = (x − 1)( ) The first term in the second bracket must be x2 to give x3 if one works backwards. The last term in the second bracket must be −6 because −1 × −6 = +6. So we have x3 − 2x2 − 5x + 6 = (x − 1)(x2 ?x − 6). Now, we must find the coefficient of the middelterm (x). (−1)(x2 ) gives −x2 . So, the coefficient of the x-term must be −1. So f (x) = (x − 1)(x2 − x − 6). Step 3 : Factorise fully x2 − x − 6 can be further factorised to (x − 3)(x + 2), and we are now left with x3 − 2x2 − 5x + 6 = (x − 1)(x − 3)(x + 2) 38.4 Exercises - Using Factor Theorem 1. Find the remainder when 4x3 − 4x2 + x − 5 is divided by (x + 1). 2. Use the factor theorem to factorise x3 − 3x2 + 4 completely. 3. f (x) = 2x3 + x2 − 5x + 2 A Find f (1). B Factorise f (x) completely 4. Use the Factor Theorem to determine all the factors of the following expression: x3 + x2 − 17x + 15 5. Complete: If f (x) is a polynomial and p is a number such that f (p) = 0, then (x − p) is ..... 38.5 Solving Cubic Equations Once you know how to factorise cubic polynomials, it is also easy to solve cubic equations of the kind ax3 + bx2 + cx + d = 0 496 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 38.5 Worked Example 170: Solution of Cubic Equations Question: Solve 6x3 − 5x2 − 17x + 6 = 0. Answer Step 1 : Find one factor using the Factor Theorem Try f (1) = 6(1)3 − 5(1)2 − 17(1) + 6 = 6 − 5 − 17 + 6 = −10 Therefore (x − 1) is NOT a factor. Try f (2) = 6(2)3 − 5(2)2 − 17(2) + 6 = 48 − 20 − 34 + 6 = 0 Therefore (x − 2) IS a factor. Step 2 : Division by expection 6x3 − 5x2 − 17x + 6 = (x − 2)( ) The first term in the second bracket must be 6x2 to give 6x3 if one works backwards. The last term in the second bracket must be −3 because −2 × −3 = +6. So we have 6x3 − 5x2 − 17x + 6 = (x − 2)(6x2 ?x − 3). Now, we must find the coefficient of the middelterm (x). (−2)(6x2 ) gives −12x2 . So, the coefficient of the x-term must be 7. So, 6x3 − 5x2 − 17x + 6 = (x − 2)(6x2 + 7x − 3). Step 3 : Factorise fully 6x2 + 7x − 3 can be further factorised to (2x + 3)(3x − 1), and we are now left with x3 − 2x2 − 5x + 6 = (x − 2)(2x + 3)(3x − 1) Step 4 : Solve the equation 6x3 − 5x2 − 17x + 6 = 0 (x − 2)(2x + 3)(3x − 1) = 0 x 1 3 = 2; ; − 3 2 Sometimes it is not possible to factorise the trinomial (”second bracket”). This is when the quadratic formula √ −b ± b2 − 4ac x= 2a can be used to solve the cubic equation fully. For example: Worked Example 171: Solution of Cubic Equations Question: Solve for x: x3 − 2x2 − 6x + 4 = 0. Answer Step 1 : Find one factor using the Factor Theorem Try f (1) = (1)3 − 2(1)2 − 6(1) + 4 = 3 − 2 − 6 + 4 = −1 Therefore (x − 1) is NOT a factor. Try f (2) = (2)3 − 2(2)2 − 6(2) + 4 = 8 − 8 − 12 + 4 = −8 497 38.6 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 Therefore (x − 2) is NOT a factor. f (−2) = (−2)3 − 2(−2)2 − 6(−2) + 4 = −8 − 8 + 12 + 4 = 0 Therefore (x + 2) IS a factor. Step 2 : Division by expection x3 − 2x2 − 6x + 4 = (x + 2)( ) The first term in the second bracket must be x2 to give x3 . The last term in the second bracket must be 2 because 2 × 2 = +4. So we have x3 − 2x2 − 6x + 4 = (x + 2)(x2 ?x + 2). Now, we must find the coefficient of the middelterm (x). (2)(x2 ) gives 2x2 . So, the coefficient of the x-term must be −4. (2x2 −4x2 = −2x2 ) So x3 − 2x2 − 6x + 4 = (x + 2)(x2 − 4x + 2). x2 − 4x + 2 cannot be factorised any futher and we are now left with (x + 2)(x2 − 4x + 2) = 0 Step 3 : Solve the equation (x + 2)(x2 − 4x + 2) (x + 2) = 0 = or 0 (x2 − 4x + 2) = 0 Step 4 : Apply the quadratic formula for the second bracket Always write down the formula first and then substitute the values of a, b and c. √ −b ± b2 − 4ac x = 2a p −(−4) ± (−4)2 − 4(1)(2) = 2(1) √ 4± 8 = 2√ = 2± 2 Step 5 : Final solutions √ x = −2 or x = 2 ± 2 38.5.1 Exercises - Solving of Cubic Equations 1. Solve for x: x3 + x2 − 5x + 3 = 0 2. Solve for y: y 3 − 3y 2 − 16y − 12 = 0 3. Solve for m: m3 − m2 − 4m − 4 = 0 4. Solve for x: x3 − x2 = 3(3x + 2) Important: : Remove brackets and write as an equation equal to zero. 5. Solve for x if 2x3 − 3x2 − 8x = 3 38.6 End of Chapter Exercises 1. Solve for x: 16(x + 1) = x2 (x + 1) 498 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 2. 38.6 A Show that x − 2 is a factor of 3x3 − 11x2 + 12x − 4 B Hence, by factorising completely, solve the equation 3x3 − 11x2 + 12x − 4 = 0 3. 2x3 − x2 − 2x + 2 = Q(x).(2x − 1) + R for all values of x. What is the value of R? 4. A Use the factor theorem to solve the following equation for m: 8m3 + 7m2 − 17m + 2 = 0 B Hence, or otherwise, solve for x: 23x+3 + 7 · 22x + 2 = 17 · 2x 5. A challenge: Determine the values of p for which the function f (x) = 3p3 − (3p − 7)x2 + 5x − 3 leaves a remainder of 9 when it is divided by (x − p). 499 38.6 CHAPTER 38. FACTORISING CUBIC POLYNOMIALS - GRADE 12 500 Chapter 39 Functions and Graphs - Grade 12 39.1 Introduction In grades 10 and 11 you have learnt about linear functions and quadratic functions as well as the hyperbolic functions and exponential functions and many more. In grade 12 you are expected to demonstrate the ability to work with various types of functions and relations including the inverses of some functions and generate graphs of the inverse relations of functions, in particular the inverses of: y = ax + q y = ax2 y = ax; a > 0 . 39.2 Definition of a Function A function is a relation for which there is only one value of y corresponding to any value of x. We sometimes write y = f (x), which is notation meaning ’y is a function of x’. This definition makes complete sense when compared to our real world examples — each person has only one height, so height is a function of people; on each day, in a specific town, there is only one average temperature. However, some very common mathematical constructions are not functions. For example, consider the relation x2 + y 2 = 4. This relation describes a circle of radius 2 centred at the origin, as in figure 39.1. If we let x = 0, we see that y 2 = 4 and thus either y = 2 or y = −2. Since there are two y values which are possible for the same x value, the relation x2 + y 2 = 4 is not a function. There is a simple test to check if a relation is a function, by looking at its graph. This test is called the vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the relation more than once, then the relation is not a function. If more than one intersection point exists, then the intersections correspond to multiple values of y for a single value of x. We can see this with our previous example of the circle by looking at its graph again in Figure 39.1. We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the circle more than once. Therefore this is not a function. 39.2.1 Exercises 1. State whether each of the following equations are functions or not: A x+y =4 501 39.3 CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 2 b 1 −2 −1 1 2 −1 b −2 Figure 39.1: Graph of y 2 + x2 = 4 B y = x4 C y = 2x D x2 + y 2 = 4 2. The table gives the average per capita income, d, in a region of the country as a function of the percent unemployed, u. Write down the equation to show that income is a function of the persent unemployed. u d 39.3 1 22500 2 22000 3 21500 4 21000 Notation used for Functions In grade 10 you were introduced to the notation used to ”name” a function. In a function y = f (x), y is called the dependent variable, because the value of y depends on what you choose as x. We say x is the independent variable, since we can choose x to be any number. Similarly if g(t) = 2t + 1, then t is the independent variable and g is the function name. If f (x) = 3x − 5 and you are ask to determine f (3), then you have to work out the value for f (x) when x = 3. For example, f (x) = f (3) = = 39.4 3x − 5 3(3) − 5 4 Graphs of Inverse Functions In earlier grades, you studied various types of functions and understood the effect of various parameters in the general equation. In this section, we will consider inverse functions. An inverse function is a function which ”does the reverse” of a given function. More formally, if f is a function with domain X, then f −1 is its inverse function if and only if for every x ∈ X we have: f −1 (f (x)) = f (f −1 (x)) = x (39.1) For example, if the function x → 3x + 2 is given, then its inverse function is x → is usually written as: f : f −1 : x → 3x + 2 (x − 2) x→ 3 502 (x − 2) . This 3 (39.2) (39.3) CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 39.4 The superscript ”-1” is not an exponent. If a function f has an inverse then f is said to be invertible. If f is a real-valued function, then for f to have a valid inverse, it must pass the horizontal line test, that is a horizontal line y = k placed anywhere on the graph of f must pass through f exactly once for all real k. It is possible to work around this condition, by defining a multi-valued function as an inverse. If one represents the function f graphically in a xy-coordinate system, then the graph of f −1 is the reflection of the graph of f across the line y = x. Algebraically, one computes the inverse function of f by solving the equation y = f (x) for x, and then exchanging y and x to get y = f −1 (x) 39.4.1 Inverse Function of y = ax + q The inverse function of y = ax + q is determined by solving for x as: y = ax = x = = ax + q y−q y−q a 1 q y− a a (39.4) (39.5) (39.6) (39.7) Therefore the inverse of y = ax + q is y = a1 x − aq . The inverse function of a straight line is also a straight line. For example, the straight line equation given by y = 2x − 3 has as inverse the function, y = 1 3 2 x + 2 . The graphs of these functions are shown in Figure 39.2. It can be seen that the two graphs are reflections of each other across the line y = x. 3 2 f −1 (x) = 12 x + −3 −2 3 2 −1 1 1 −1 2 3 f (x) = 2x − 3 −2 −3 Figure 39.2: The function f (x) = 2x − 3 and its inverse f −1 (x) = 12 x + 23 . The line y = x is shown as a dashed line. 503 39.4 CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 Domain and Range We have seen that the domain of a function of the form y = ax + q is {x : x ∈ R} and the range is {y : y ∈ R}. Since the inverse function of a straight line is also a straight line, the inverse function will have the same domain and range as the original function. Intercepts The general form of the inverse function of the form y = ax + q is y = a1 x − aq . By setting x = 0 we have that the y-intercept is yint = − aq . Similarly, by setting y = 0 we have that the x-intercept is xint = q. It is interesting to note that if f (x) = ax + q, then f −1 (x) = a1 x − aq and the y-intercept of f (x) is the x-intercept of f −1 (x) and the x-intercept of f (x) is the y-intercept of f −1 (x). 39.4.2 Exercises 1. Given f (x) = 2x − 3, find f −1 (x) 2. Consider the function f (x) = 3x − 7. A Is the relation a function? B Identify the domain and range. 3. Sketch the graph of the function f (x) = 3x − 1 and its inverse on the same set of axes. 4. The inverse of a function is f −1 (x) = 2x − 4, what is the function f (x)? 39.4.3 Inverse Function of y = ax2 The inverse function of y = ax2 is determined by solving for x as: y 2 x x = ax2 y = a r y = a (39.8) (39.9) (39.10) We see that the inverse function of y = ax2 is not a function because √ it fails the vertical line test. If we draw a vertical line through the graph of f −1 (x) = ± x, the line intersects the graph more than once. There has to be a restriction on the domain of a parabola for the inverse to also be a function. Consider the function f (x) = −x2 + 9. The inverse of f can be found by witing f (y) = x. Then x = y2 = y = −y 2 + 9 9−x √ ± 9−x √ √ If x ≥ 0, then 9 − x is a function. If the restriction on the domain of f is x ≤ 0 then − 9 − x would be a function. 39.4.4 Exercises 1. The graph of f −1 is shown. Find the equation of f , given that the graph of f is a parabola. (Do not simplify your answer) 504 CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 f (x) = x2 39.4 3 f −1 (x) = 2 √ x 1 −3 −2 −1 1 2 3 −1 √ f −1 (x) = − x −2 −3 √ Figure 39.3: The function f (x) = x2 and its inverse f −1 (x) = ± x. The line y = x is shown as a dashed line. f −1 b (3; 1) b (1; 0) 2. f (x) = 2x2 . A Draw the graph of f and state its domain and range. B Find f −1 and state the domain and range. C What must the domain of f be, so that f −1 is a function ? p 3. Sketch the graph of x = − 10 − y 2 . Label a point on the graph other than the intercepts with the axes. 4. A Sketch the graph of y = x2 labelling a point other than the origin on your graph. B Find the equation of the inverse of the above graph in the form y = . . .. √ C Now sketch the y = x. √ D The tangent to the graph of y = x at the point A(9;3) intersects the x-axis at B. Find the equation of this tangent and hence or otherwise prove that the y-axis bisects the straight line AB. 5. Given: g(x) = −1 + √ x, find the inverse of g(x) in the form g −1 (x). 505 39.4 CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 Inverse Function of y = ax 39.4.5 The inverse function of y = ax2 is determined by solving for x as: y = log(y) = = ∴ x = ax (39.11) x log(a ) x log(a) log(y) log(a) (39.12) (39.13) (39.14) The inverse of y = 10x is x = 10y , which we write as y = logx. Therefore, if f (x) = 10x , then f −1 = logx. f (x) = 10x 3 2 1 f −1 (x) = log(x) −3 −2 −1 1 2 3 −1 −2 −3 Figure 39.4: The function f (x) = 10x and its inverse f −1 (x) = log(x). The line y = x is shown as a dashed line. The exponential function and the logarithmic function are inverses of each other; the graph of the one is the graph of the other, reflected in the line y = x. The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse. 39.4.6 Exercises 1. Given that f (x) = [ 51 ]x , sketch the graphs of f and f −1 on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is f and which is f −1 . 2. Given that f (x) = 4−x , A Sketch the graphs of f and f −1 on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is f and which is f −1 . B Write f −1 in the form y = . . .. √ 3. Given g(x) = −1 + x, find the inverse of g(x) in the form g −1 (x) = . . . 4. A B C D Sketch the graph of y = x2 , labeling a point other than the origin on your graph. Find the equation of the inverse of the above graph in the form y = . . . √ Now, sketch y = x. √ The tangent to the graph of y = x at the point A(9; 3) intersects the x-axis at B. Find the equation of this tangent, and hence, or otherwise, prove that the y-axis bisects the straight line AB. 506 CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 39.5 39.5 End of Chapter Exercises 1. Sketch the graph of x = − 2. f (x) = 1 , x−5 p 10 − y 2 . Is this graph a function ? Verify your answer. A determine the y-intercept of f (x) B determine x if f (x) = −1. 3. Below, you are given 3 graphs and 5 equations. Graph 1 y Graph 2 y Graph 3 y x x x A y = log3 x B y = − log3 x C y = log3 (−x) D y = 3−x E y = 3x Write the equation that best describes each graph. 4. The graph of y = f (x) is shown in the diagram below. y f (x) 2 −2 A Find the value of x such that f (x) = 0. B Evaluate f (3) + f (−1). 507 x 39.5 CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 5. Given g(x) = −1 + √ x, find the inverse of g(x) in the form g −1 (x) = . . . 6. Given the equation h(x) = 3x A Write down the inverse in the form h−1 (x) = ... B Sketch the graphs of h(x) and h−1 (x) on teh same set of axes, labelling the intercepts with the axes. C For which values of x is h−1 (x) undefined ? 7. A Sketch the graph of y = x2 , labelling a point other than the origin on your graph. B Find the equation of the inverse of the above graph in the form y = . . . √ C Now, sketch y = x. √ D The tangent to the graph of y = x at the point A(9; 3) intersects the x-axis at B. Find the equation of this tangent, and hence, or otherwise, prove that the y-axis bisects the straight line AB. 508 Chapter 40 Differential Calculus - Grade 12 40.1 Why do I have to learn this stuff? Calculus is one of the central branches of mathematics and was developed from algebra and geometry. Calculus is built on the concept of limits, which will be discussed in this chapter. Calculus consists of two complementary ideas: differential calculus and integral calculus. Only differential calculus will be studied. Differential calculus is concerned with the instantaneous rate of change of quantities with respect to other quantities, or more precisely, the local behaviour of functions. This can be illustrated by the slope of a function’s graph. Examples of typical differential calculus problems include: finding the acceleration and velocity of a free-falling body at a particular moment and finding the optimal number of units a company should produce to maximize its profit. Calculus is fundamentally different from the mathematics that you have studied previously. Calculus is more dynamic and less static. It is concerned with change and motion. It deals with quantities that approach other quantities. For that reason it may be useful to have an overview of the subject before beginning its intensive study. Calculus is a tool to understand many natural phenomena like how the wind blows, how water flows, how light travels, how sound travels and how the planets move. However, other human activities such as economics are also made easier with calculus. In this section we give a glimpse of some of the main ideas of calculus by showing how limits arise when we attempt to solve a variety of problems. Extension: Integral Calculus Integral calculus is concerned with the accumulation of quantities, such as areas under a curve, linear distance traveled, or volume displaced. Differential and integral calculus act inversely to each other. Examples of typical integral calculus problems include finding areas and volumes, finding the amount of water pumped by a pump with a set power input but varying conditions of pumping losses and pressure and finding the amount of rain that fell in a certain area if the rain fell at a specific rate. teresting Both Isaac Newton (4 January 1643 – 31 March 1727) and Gottfried Liebnitz Interesting Fact Fact (1 July 1646 – 14 November 1716 (Hanover, Germany)) are credited with the ‘invention’ of calculus. Newton was the first to apply calculus to general physics, while Liebnitz developed most of the notation that is still in use today. When Newton and Leibniz first published their results, there was some controversy over whether Leibniz’s work was independent of Newton’s. While Newton derived his results years before Leibniz, it was only some time after Leibniz published in 1684 that Newton published. Later, Newton would claim that Leibniz 509 40.2 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 got the idea from Newton’s notes on the subject; however examination of the papers of Leibniz and Newton show they arrived at their results independently, with Leibniz starting first with integration and Newton with differentiation. This controversy between Leibniz and Newton divided English-speaking mathematicians from those in Europe for many years, which slowed the development of mathematical analysis. Today, both Newton and Leibniz are given credit for independently developing calculus. It is Leibniz, however, who is credited with giving the new discipline the name it is known by today: ”calculus”. Newton’s name for it was ”the science of fluxions”. 40.2 Limits 40.2.1 A Tale of Achilles and the Tortoise teresting Zeno (circa 490 BC - circa 430 BC) was a pre-Socratic Greek philosopher of Interesting Fact Fact southern Italy who is famous for his paradoxes. One of Zeno’s paradoxes can be summarised by: Achilles and a tortoise agree to a race, but the tortoise is unhappy because Achilles is very fast. So, the tortoise asks Achilles for a head-start. Achilles agrees to give the tortoise a 1 000 m head start. Does Achilles overtake the tortoise? We know how to solve this problem. We start by writing: xA = vA t (40.1) xt = 1000 m + vt t (40.2) where xA vA t xt vt distance covered by Achilles Achilles’ speed time taken by Achilles to overtake tortoise distance covered by the tortoise the tortoise’s speed If we assume that Achilles runs at 2 m·s−1 and the tortoise runs at 0,25 m·s−1 then Achilles will overtake the tortoise when both of them have covered the same distance. This means that 510 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.2 Achilles overtakes the tortoise at a time calculated as: xA = xt (40.3) vA t −1 = = (40.4) (40.5) (2 m · s−1 − 0,25 m · s−1 )t = 1000 + vt t 1000 m + (0,25 m · s−1 )t t = (2 m · s )t = = = = 1000 m 1000 m 1 43 m · s−1 1000 m 7 −1 4 m·s (4)(1000) s 7 4000 s 7 3 571 s 7 (40.6) (40.7) (40.8) (40.9) (40.10) (40.11) However, Zeno (the Greek philosopher who thought up this problem) looked at it as follows: Achilles takes 1000 = 500 s t= 2 to travel the 1 000 m head start that the tortoise had. However, in this 500 s, the tortoise has travelled a further x = (500)(0,25) = 125 m. Achilles then takes another 125 = 62,5 s 2 to travel the 125 m. In this 62,5 s, the tortoise travels a further t= x = (62,5)(0,25) = 15,625 m. Zeno saw that Achilles would always get closer but wouldn’t actually overtake the tortoise. 40.2.2 Sequences, Series and Functions So what does Zeno, Achilles and the tortoise have to do with calculus? Well, in Grades 10 and 11 you studied sequences. For the sequence 1 2 3 4 0, , , , , . . . 2 3 4 5 which is defined by the expression 1 n the terms get closer to 1 as n gets larger. Similarly, for the sequence an = 1 − 1 1 1 1 1, , , , , . . . 2 3 4 5 which is defined by the expression 1 n the terms get closer to 0 as n gets larger. We have also seen that the infinite geometric series has a finite total. The infinite geometric series is an = S∞ = ∞ X i=1 a1 .ri−1 = a1 1−r for −1 where a1 is the first term of the series and r is the common ratio. 511 40.2 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 We see that there are some functions where the value of the function gets close to or approaches a certain value. Similarly, for the function: x2 + 4x − 12 x+6 The numerator of the function can be factorised as: y= y= (x + 6)(x − 2) . x+6 Then we can cancel the x − 6 from numerator and denominator and we are left with: y = x − 2. However, we are only able to cancel the x + 6 term if x 6= −6. If x = −6, then the denominator becomes 0 and the function is not defined. This means that the domain of the function does not include x = −6. But we can examine what happens to the values for y as x gets close to -6. These values are listed in Table 40.1 which shows that as x gets closer to -6, y gets close to 8. Table 40.1: Values for the function y = x y= -9 -8 -7 -6.5 -6.4 -6.3 -6.2 -6.1 -6.09 -6.08 -6.01 -5.9 -5.8 -5.7 -5.6 -5.5 -5 -4 -3 (x + 6)(x − 2) as x gets close to -6. x+6 (x+6)(x−2) x+6 -11 -10 -9 -8.5 -8.4 -8.3 -8.2 -8.1 -8.09 -8.08 -8.01 -7.9 -7.8 -7.7 -7.6 -7.5 -7 -6 -5 The graph of this function is shown in Figure 40.1. The graph is a straight line with slope 1 and intercept -2, but with a missing section at x = −6. Extension: Continuity We say that a function is continuous if there are no values of the independent variable for which the function is undefined. 40.2.3 Limits We can now introduce a new notation. For the function y = lim x→−6 (x + 6)(x − 2) , we can write: x+6 (x + 6)(x − 2) = −8. x+6 512 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 4 3 2 1 −9 −8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 bc Figure 40.1: Graph of y = This is read: the limit of (x+6)(x−2) x+6 1 2 3 4 (x+6)(x−2) . x+6 as x tends to -6 is 8. Activity :: Investigation : Limits If f (x) = x + 1, determine: f(-0.1) f(-0.05) f(-0.04) f(-0.03) f(-0.02) f(-0.01) f(0.00) f(0.01) f(0.02) f(0.03) f(0.04) f(0.05) f(0.1) What do you notice about the value of f (x) as x gets close to 0. Worked Example 172: Limits Notation Question: Summarise the following situation by using limit notation: As x gets close to 1, the value of the function y =x+2 gets close to 3. 513 40.2 40.2 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Answer This is written as: lim x + 2 = 3 x→1 in limit notation. We can also have the situation where a function has a different value depending on whether x approaches from the left or the right. An example of this is shown in Figure 40.2. 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 1 2 3 4 5 6 7 −2 −3 −4 Figure 40.2: Graph of y = x1 . As x → 0 from the left, y = x1 approaches −∞. As x → 0 from the right, y = +∞. This is written in limits notation as: lim x→0− 1 x approaches 1 = −∞ x for x approaching zero from the left and lim x→0+ 1 =∞ x for x approaching zero from the right. You can calculate the limit of many different functions using a set method. Method: Limits If you are required to calculate a limit like limx→a then: 1. Simplify the expression completely. 2. If it is possible, cancel all common terms. 3. Let x approach the a. Worked Example 173: Limits 514 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Question: Determine lim 10 x→1 Answer Step 1 : Simplify the expression There is nothing to simplify. Step 2 : Cancel all common terms There are no terms to cancel. Step 3 : Let x → 1 and write final answer lim 10 = 10 x→1 Worked Example 174: Limits Question: Determine lim x x→2 Answer Step 1 : Simplify the expression There is nothing to simplify. Step 2 : Cancel all common terms There are no terms to cancel. Step 3 : Let x → 2 and write final answer lim x = 2 x→2 Worked Example 175: Limits Question: Determine x2 − 100 x→10 x − 10 lim Answer Step 1 : Simplify the expression The numerator can be factorised. (x + 10)(x − 10) x2 − 100 = x − 10 x − 10 Step 2 : Cancel all common terms x − 10 can be cancelled from the numerator and denominator. (x + 10)(x − 10) = x + 10 x − 10 Step 3 : Let x → 1 and write final answer x2 − 100 = 20 x→10 x − 10 lim 515 40.2 40.2 40.2.4 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Average Gradient and Gradient at a Point In Grade 10 you learnt about average gradients on a curve. The average gradient between any two points on a curve is given by the gradient of the straight line that passes through both points. In Grade 11 you were introduced to the idea of a gradient at a single point on a curve. We saw that this was the gradient of the tangent to the curve at the given point, but we did not learn how to determine the gradient of the tangent. Now let us consider the problem of trying to find the gradient of a tangent t to a curve with equation y = f (x) at a given point P . tangent P b f (x) We know how to calculate the average gradient between two points on a curve, but we need two points. The problem now is that we only have one point, namely P . To get around the problem we first consider a secant to the curve that passes through point P and another point on the curve Q. We can now find the average gradient of the curve between points P and Q. secant P b f (a) f (a − h) b Q f (x) a a−h If the x-coordinate of P is a, then the y-coordinate is f (a). Similarly, if the x-coordinate of Q is a − h, then the y-coordinate is f (a − h). If we choose a as x2 and a − h as x1 , then: y1 = f (a − h) y2 = f (a). We can now calculate the average gradient as: y2 − y1 x2 − x1 = = f (a) − f (a − h) a − (a − h) f (a) − f (a − h) h (40.12) (40.13) Now imagine that Q moves along the curve toward P . The secant line approaches the tangent line as its limiting position. This means that the average gradient of the secant approaches the gradient of the tangent to the curve at P . In (40.13) we see that as point Q approaches point P , h gets closer to 0. When h = 0, points P and Q are equal. We can now use our knowledge of limits to write this as: gradient at P = lim h→0 f (a) − f (a − h) . h (40.14) and we say that the gradient at point P is the limit of the average gradient as Q approaches P along the curve. 516 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.2 Activity :: Investigation : Limits The gradient at a point x on a curve defined by f (x) can also be written as: lim h→0 f (x + h) − f (x) h (40.15) Show that this is equivalent to (40.14). Worked Example 176: Limits Question: For the function f (x) = 2x2 − 5x, determine the gradient of the tangent to the curve at the point x = 2. Answer Step 1 : Calculating the gradient at a point We know that the gradient at a point x is given by: lim h→0 f (x + h) − f (x) h In our case x = 2. It is simpler to substitute x = 2 at the end of the calculation. Step 2 : Write f (x + h) and simplify f (x + h) = = = 2(x + h)2 − 5(x + h) 2(x2 + 2xh + h2 ) − 5x − 5h 2x2 + 4xh + 2h2 − 5x − 5h Step 3 : Calculate limit f (x + h) − f (x) h→0 h lim = = = = = = 2x2 + 4xh + 2h2 − 5x − 5h − (2x2 − 5x) h 2x2 + 4xh + 2h2 − 5x − 5h − 2x2 + 5x lim h→0 h 4xh + 2h2 − 5h lim h→0 h h(4x + 2h − 5) lim h→0 h lim 4x + 2h − 5 h→0 4x − 5 Step 4 : Calculate gradient at x = 2 4x − 5 = 4(2) − 5 = 3 Step 5 : Write the final answer The gradient of the tangent to the curve f (x) = 2x2 − 5x at x = 2 is 3. 517 40.2 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Worked Example 177: Limits Question: For the function f (x) = 5x2 − 4x + 1, determine the gradient of the tangent to curve at the point x = a. Answer Step 1 : Calculating the gradient at a point We know that the gradient at a point x is given by: f (x + h) − f (x) h→0 h lim In our case x = a. It is simpler to substitute x = a at the end of the calculation. Step 2 : Write f (x + h) and simplify 5(x + h)2 − 4(x + h) + 1 f (x + h) = 5(x2 + 2xh + h2 ) − 4x − 4h + 1 5x2 + 10xh + 5h2 − 4x − 4h + 1 = = Step 3 : Calculate limit f (x + h) − f (x) h→0 h lim = = = = = = 5x2 + 10xh + 5h2 − 4x − 4h + 1 − (5x2 − 4x + 1) h 5x2 + 10xh + 5h2 − 4x − 4h + 1 − 5x2 + 4x − 1 lim h→0 h 2 10xh + 5h − 4h lim h→0 h h(10x + 5h − 4) lim h→0 h lim 10x + 5h − 4 h→0 10x − 4 Step 4 : Calculate gradient at x = a 10x − 4 = 10a − 5 Step 5 : Write the final answer The gradient of the tangent to the curve f (x) = 5x2 − 4x + 1 at x = 1 is 10a − 5. Exercise: Limits Determine the following 1. x2 − 9 x→3 x + 3 lim 2. lim x→3 3. x+3 x2 + 3x 3x2 − 4x x→2 3 − x lim 518 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 4. 40.3 x2 − x − 12 x→4 x−4 lim 5. lim 3x + x→2 40.3 1 3x Differentiation from First Principles The tangent problem has given rise to the branch of calculus called differential calculus and the equation: f (x + h) − f (x) lim h→0 h defines the derivative of the function f (x). Using (40.15) to calculate the derivative is called finding the derivative from first principles. Definition: Derivative The derivative of a function f (x) is written as f ′ (x) and is defined by: f (x + h) − f (x) h→0 h f ′ (x) = lim (40.16) There are a few different notations used to refer to derivatives. If we use the traditional notation y = f (x) to indicate that the dependent variable is y and the independent variable is x, then some common alternative notations for the derivative are as follows: f ′ (x) = y ′ = dy df d = = f (x) = Df (x) = Dx f (x) dx dx dx d are called differential operators because they indicate the operation of The symbols D and dx differentiation, which is the process of calculating a derivative. It is very important that you learn to identify these different ways of denoting the derivative, and that you are consistent in your usage of them when answering questions. dy is a limit and Important: Though we choose to use a fractional form of representation, dx dy dy is not a fraction, i.e. dx does not mean dy ÷ dx. dx means y differentiated with respect to dp d means p differentiated with respect to x. The ‘ dx ’ is the “operator”, operating x. Thus, dx on some function of x. Worked Example 178: Derivatives - First Principles Question: Calculate the derivative of g(x) = x − 1 from first principles. Answer Step 1 : Calculating the gradient at a point We know that the gradient at a point x is given by: g(x + h) − g(x) h→0 h g ′ (x) = lim Step 2 : Write g(x + h) and simplify 519 40.3 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 g(x + h) = x + h − 1 Step 3 : Calculate limit g ′ (x) = = = = = g(x + h) − g(x) h x + h − 1 − (x − 1) lim h→0 h x+h−1−x+1 lim h→0 h h lim h→0 h lim 1 lim h→0 h→0 = 1 Step 4 : Write the final answer The derivative g ′ (x) of g(x) = x − 1 is 1. Worked Example 179: Derivatives - First Principles Question: Calculate the derivative of h(x) = x2 − 1 from first principles. Answer Step 1 : Calculating the gradient at a point We know that the gradient at a point x is given by: g(x + h) − g(x) h→0 h g ′ (x) = lim Step 2 : Write g(x + h) and simplify g(x + h) = x + h − 1 Step 3 : Calculate limit g ′ (x) = = = = = g(x + h) − g(x) h x + h − 1 − (x − 1) lim h→0 h x+h−1−x+1 lim h→0 h h lim h→0 h lim 1 lim h→0 h→0 = 1 Step 4 : Write the final answer The derivative g ′ (x) of g(x) = x − 1 is 1. 520 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.4 Exercise: Derivatives 1. Given g(x) = −x2 g(x + h) − g(x) A determine h B hence, determine lim h→0 g(x + h) − g(x) h C explain the meaning of your answer in (b). 2. Find the derivative of f (x) = −2x2 + 3x using first principles. 1 3. Determine the derivative of f (x) = using first principles. x−2 4. Determine f ′ (3) from first principles if f (x) = −5x2 . 5. If h(x) = 4x2 − 4x, determine h′ (x) using first principles. 40.4 Rules of Differentiation Calculating the derivative of a function from first principles is very long, and it is easy to make mistakes. Fortunately, there are rules which make calculating the derivative simple. Activity :: Investigation : Rules of Differentiation From first principles, determine the derivatives of the following: 1. f (x) = b 2. f (x) = x 3. f (x) = x2 4. f (x) = x3 5. f (x) = 1/x You should have found the following: f (x) b x x2 x3 1/x = x−1 f ′ (x) 0 1 2x 3x2 −x−2 If we examine these results we see that there is a pattern, which can be summarised by: d (xn ) = nxn−1 dx (40.17) There are two other rules which make differentiation simpler. For any two functions f (x) and g(x): d [f (x) ± g(x)] = f ′ (x) ± g ′ (x) (40.18) dx 521 40.4 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 This means that we differentiate each term separately. The final rule applies to a function f (x) that is multiplied by a constant k. d [k.f (x)] = kf ′ (x) dx (40.19) Worked Example 180: Rules of Differentiation Question: Determine the derivative of x − 1 using the rules of differentiation. Answer Step 1 : Identify the rules that will be needed We will apply two rules of differentiation: d (xn ) = nxn−1 dx and d d d [f (x) − g(x)] = [f (x)] − [g(x)] dx dx dx Step 2 : Determine the derivative In our case f (x) = x and g(x) = 1. f ′ (x) = 1 and g ′ (x) = 0 Step 3 : Write the final answer The derivative of x − 1 is 1 which is the same result as was obtained earlier, from first principles. 40.4.1 Summary of Differentiation Rules d dx b =0 d n dx (x ) = nxn−1 d dx (kf ) df = k dx d dx (f + g) = Exercise: Rules of Differentiation x2 − 5x + 6 . x−2 √ 2. Find f ′ (y) if f (y) = y. 1. Find f ′ (x) if f (x) = 3. Find f ′ (z) if f (z) = (z − 1)(z + 1). √ x3 + 2 x − 3 dy 4. Determine dx if y = . x 522 df dx + dg dx CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 5. Determine the derivative of y = 40.5 √ x3 + 40.5 1 . 3x3 Applying Differentiation to Draw Graphs Thus far we have learnt about how to differentiate various functions, but I am sure that you are beginning to ask, What is the point of learning about derivatives? Well, we know one important fact about a derivative: it is a gradient. So, any problems involving the calculations of gradients or rates of change can use derivatives. One simple application is to draw graphs of functions by firstly determine the gradients of straight lines and secondly to determine the turning points of the graph. 40.5.1 Finding Equations of Tangents to Curves In section 40.2.4 we saw that finding the gradient of a tangent to a curve is the same as finding the slope of the same curve at the point of the tangent. We also saw that the gradient of a function at a point is just its derivative. Since we have the gradient of the tangent and the point on the curve through which the tangent passes, we can find the equation of the tangent. Worked Example 181: Finding the Equation of a Tangent to a Curve Question: Find the equation of the tangent to the curve y = x2 at the point (1,1) and draw both functions. Answer Step 1 : Determine what is required We are required to determine the equation of the tangent to the curve defined by y = x2 at the point (1,1). The tangent is a straight line and we can find the equation by using derivatives to find the gradient of the straight line. Then we will have the gradient and one point on the line, so we can find the equation using: y − y1 = m(x − x1 ) from grade 11 Coordinate Geometry. Step 2 : Differentiate the function Using our rules of differentiation we get: y ′ = 2x Step 3 : Find the gradient at the point (1,1) In order to determine the gradient at the point (1,1), we substitute the x-value into the equation for the derivative. So, y ′ at x = 1 is: 2(1) = 2 Step 4 : Find equation of tangent y − y1 y−1 y y = = m(x − x1 ) (2)(x − 1) = 2x − 2 + 1 = 2x − 1 523 40.5 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Step 5 : Write the final answer The equation of the tangent to the curve defined by y = x2 at the point (1,1) is y = 2x − 1. Step 6 : Sketch both functions y = x2 4 3 2 1 −4 −3 −2 −1 b (1,1) 1 2 3 4 −1 −2 y = 2x − 1 −3 −4 40.5.2 Curve Sketching Differentiation can be used to sketch the graphs of functions, by helping determine the turning points. We know that if a graph is increasing on an interval and reaches a turning point, then the graph will start decreasing after the turning point. The turning point is also known as a stationary point because the gradient at a turning point is 0. We can then use this information to calculate turning points, by calculating the points at which the derivative of a function is 0. Important: If x = a is a turning point of f (x), then: f ′ (a) = 0 This means that the derivative is 0 at a turning point. Take the graph of y = x2 as an example. We know that the graph of this function has a turning point at (0,0), but we can use the derivative of the function: y ′ = 2x and set it equal to 0 to find the x-value for which the graph has a turning point. 2x = 0 x = 0 We then substitute this into the equation of the graph (i.e. y = x2 ) to determine the y-coordinate of the turning point: f (0) = (0)2 = 0 This corresponds to the point that we have previously calculated. 524 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.5 Worked Example 182: Calculation of Turning Points Question: Calculate the turning points of the graph of the function f (x) = 2x3 − 9x2 + 12x − 15 . Answer Step 1 : Determine the derivative of f (x) Using the rules of differentiation we get: f ′ (x) = 6x2 − 18x + 12 Step 2 : Set f ′ (x) = 0 and calculate x-coordinate of turning point 6x2 − 18x + 12 = 2 x − 3x + 2 = (x − 2)(x − 1) = 0 0 0 Therefore, the turning points are at x = 2 and x = 1. Step 3 : Substitute x-coordinate of turning point into f (x) to determine y-coordinates f (2) = 2(2)3 − 9(2)2 + 12(2) − 15 = 16 − 36 + 24 − 15 = −11 f (1) = 2(1)3 − 9(1)2 + 12(1) − 15 = 2 − 9 + 12 − 15 = −10 Step 4 : Write final answer The turning points of the graph of f (x) = 2x3 − 9x2 + 12x − 15 are (2,-11) and (1,-10). We are now ready to sketch graphs of functions. Method: Sketching GraphsSuppose we are given that f (x) = ax3 + bx2 + cx + d, then there are five steps to be followed to sketch the graph of the function: 1. If a > 0, then the graph is increasing from left to right, and has a maximum and then a minimum. As x increases, so does f (x). If a < 0, then the graph decreasing is from left to right, and has first a minimum and then a maximum. f (x) decreases as x increases. 2. Determine the value of the y-intercept by substituting x = 0 into f (x) 3. Determine the x-intercepts by factorising ax3 + bx2 + cx + d = 0 and solving for x. First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary. 525 40.5 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 4. Find the turning points of the function by working out the derivative zero, and solving for x. df dx and setting it to 5. Determine the y-coordinates of the turning points by substituting the x values obtained in the previous step, into the expression for f (x). 6. Draw a neat sketch. Worked Example 183: Sketching Graphs Question: Draw the graph of g(x) = x2 − x + 2 Answer Step 1 : Determine the y-intercept y-intercept is obtained by setting x = 0. g(0) = (0)2 − 0 + 2 = 2 Step 2 : Determine the x-intercepts The x-intercepts are found by setting g(x) = 0. g(x) = 0 = x2 − x + 2 x2 − x + 2 which does not have real roots. Therefore, the graph of g(x) does not have any x-intercepts. Step 3 : Find the turning points of the function dg Work out the derivative dx and set it to zero to for the x coordinate of the turning point. dg = 2x − 1 dx dg = dx 2x − 1 = 2x = x = 0 0 1 1 2 Step 4 : Determine the y-coordinates of the turning points by substituting the x values obtained in the previous step, into the expression for f (x). y coordinate of turning point is given by calculating g( 21 ). 1 g( ) = 2 = = 1 1 ( )2 − ( ) + 2 2 2 1 1 − +2 4 2 7 4 The turning point is at ( 12 , 47 ) Step 5 : Draw a neat sketch 526 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.5 y 9 8 7 6 5 4 3 2 1 b b (0.5,1.75) x −3 −2 −1 1 2 3 4 Worked Example 184: Sketching Graphs Question: Sketch the graph of g(x) = −x3 + 6x2 − 9x + 4. Answer Step 1 : Calculate the turning points Find the turning points by setting g ′ (x) = 0. If we use the rules of differentiation we get g ′ (x) = −3x2 + 12x − 9 g ′ (x) −3x2 + 12x − 9 = = 0 0 x2 − 4x + 3 = (x − 3)(x − 1) = 0 0 The x-coordinates of the turning points are: x = 1 and x = 3. The y-coordinates of the turning points are calculated as: g(x) = g(1) = = = g(x) = g(3) = = = −x3 + 6x2 − 9x + 4 −(1)3 + 6(1)2 − 9(1) + 4 −1 + 6 − 9 + 4 0 −x3 + 6x2 − 9x + 4 −(3)3 + 6(3)2 − 9(3) + 4 −27 + 54 − 27 + 4 4 Therefore the turning points are: (1,0) and (3,4). 527 40.5 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Step 2 : Determine the y-intercepts We find the y-intercepts by finding the value for g(0). = −x3 + 6x2 − 9x + 4 g(x) yint = g(0) = −(0)3 + 6(0)2 − 9(0) + 4 = 4 Step 3 : Determine the x-intercepts We find the x-intercepts by finding the points for which the function g(x) = 0. g(x) = −x3 + 6x2 − 9x + 4 Use the factor theorem to confirm that (x − 1) is a factor. If g(1) = 0, then (x − 1) is a factor. g(x) −x3 + 6x2 − 9x + 4 = −(1)3 + 6(1)2 − 9(1) + 4 −1 + 6 − 9 + 4 g(1) = = = 0 Therefore, (x − 1) is a factor. If we divide g(x) by (x − 1) we are left with: −x2 + 5x − 4 This has factors −(x − 4)(x − 1) Therefore: g(x) = −(x − 1)(x − 1)(x − 4) The x-intercepts are: xint = 1, 4 Step 4 : Draw a neat sketch y 9 8 7 6 5 4 (3,4) b b 3 2 1 (1,0) (4,0) x b −1 −1 1 b 2 528 3 4 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.5 Exercise: Sketching Graphs 1. Given f (x) = x3 + x2 − 5x + 3: A Show that (x − 1) is a factor of f (x) and hence fatorise f (x) fully. B Find the coordinates of the intercepts with the axes and the turning points and sketch the graph 2. Sketch the graph of f (x) = x3 − 4x2 − 11x+ 30 showing all the relative turning points and intercepts with the axes. 3. A Sketch the graph of f (x) = x3 − 9x2 + 24x − 20, showing all intercepts with the axes and turning points. B Find the equation of the tangent to f (x) at x = 4. 40.5.3 Local minimum, Local maximum and Point of Inflextion dy ) is zero at a point, the gradient of the tangent at that point is zero. It If the derivative ( dx means that a turning point occurs as seen in the previous example. y 9 8 7 6 5 4 (3;4) b b 3 2 1 (1;0) (4;0) x b −1 −1 1 b 2 3 4 From the drawing the point (1;0) represents a local minimum and the point (3;4) the local maximum. A graph has a horizontal point of inflexion where the derivative is zero but the sign of the sign of the gradient does not change. That means the graph always increases or always decreases. 529 40.6 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 y b (3;1) x From this drawing, the point (3;1) is a horizontal point of inflexion, because the sign of the derivative stays positive. 40.6 Using Differential Calculus to Solve Problems We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. However, determining stationary points also lends itself to the solution of problems that require some variable to be optimised. For example, if fuel used by a car is defined by: f (v) = 3 2 v − 6v + 245 80 (40.20) where v is the travelling speed, what is the most economical speed (that means the speed that uses the least fuel)? If we draw the graph of this function we find that the graph has a minimum. The speed at the minimum would then give the most economical speed. fuel consumption (l) 60 50 40 30 20 10 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 speed (km·hr−1 ) We have seen that the coordinates of the turning point can be calculated by differentiating the function and finding the x-coordinate (speed in the case of the example) for which the derivative is 0. Differentiating (40.20), we get: 3 v−6 40 If we set f ′ (v) = 0 we can calculate the speed that corresponds to the turning point. 530 f ′ (v) = CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.6 3 v−6 40 3 v−6 = 40 6 × 40 = 3 = 80 f ′ (v) = 0 v This means that the most economical speed is 80 km·hr−1 . Worked Example 185: Optimisation Problems Question: The sum of two positive numbers is 10. One of the numbers is multiplied by the square of the other. If each number is greater than 0, find the numbers that make this product a maximum. Answer Step 1 : Examine the problem and formulate the equations that are required Let the two numbers be a and b. Then we have: a + b = 10 (40.21) We are required to minimise the product of a and b. Call the product P . Then: P =a·b (40.22) We can solve for b from (40.21) to get: b = 10 − a (40.23) Substitute this into (40.22) to write P in terms of a only. P = a(10 − a) = 10a − a2 Step 2 : Differentiate The derivative of (40.24) is: P ′ (a) = 10 − 2a Step 3 : Find the stationary point Set P ′ (a) = 0 to find the value of a which makes P a maximum. P ′ (a) = 0 = 2a = a = a = 10 − 2a 10 − 2a 10 10 2 5 Substitute into (40.27) to solve for the width. b = 10 − a = 10 − 5 = 5 Step 4 : Write the final answer The product is maximised if a and b are both equal to 5. 531 (40.24) CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Worked Example 186: Optimisation Problems Question: Michael wants to start a vegetable garden, which he decides to fence off in the shape of a rectangle from the rest of the garden. Michael only has 160 m of fencing, so he decides to use a wall as one border of the vegetable garden. Calculate the width and length of the garden that corresponds to largest possible area that Michael can fence off. wall garden length, l 40.6 width, w Answer Step 1 : Examine the problem and formulate the equations that are required The important pieces of information given are related to the area and modified perimeter of the garden. We know that the area of the garden is: A= w·l (40.25) We are also told that the fence covers only 3 sides and the three sides should add up to 160 m. This can be written as: 160 = w + l + l (40.26) However, we can use (40.26) to write w in terms of l: w = 160 − 2l (40.27) Substitute (40.27) into (40.25) to get: A = (160 − 2l)l = 160l − 2l2 (40.28) Step 2 : Differentiate Since we are interested in maximising the area, we differentiate (40.28) to get: A′ (l) = 160 − 4l Step 3 : Find the stationary point To find the stationary point, we set A′ (l) = 0 and solve for the value of l that maximises the area. A′ (l) = 0 = ∴ 4l = l = l = 160 − 4l 160 − 4l 160 160 4 40 m Substitute into (40.27) to solve for the width. w = = = = 160 − 2l 160 − 2(40) 160 − 80 80 m 532 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 Step 4 : Write the final answer A width of 80 m and a length of 40 m will yield the maximal area fenced off. Exercise: Solving Optimisation Problems using Differential Calculus 1. The sum of two positive numbers is 20. One of the numbers is multiplied by the square of the other. Find the numbers that make this products a maximum. 2. A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides 3x, 4x and 5x. The length of the block is y. The total surface area of the block is 3 600 cm2 . 3x 4x y 300 − x2 . x B Find the value of x for which the block will have a maximum volume. (Volume = area of base × height.) A Show that y = 3. The diagram shows the plan for a verandah which is to be built on the corner of a cottage. A railing ABCDE is to be constructed around the four edges of the verandah. y C D x verandah F B A E cottage If AB = DE = x and BC = CD = y, and the length of the railing must be 30 metres, find the values of x and y for which the verandah will have a maximum area. 533 40.6 40.6 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.6.1 Rate of Change problems Two concepts were discussed in this chapter: Average rate of change = f (b)−f (a) b−a and Instan- (x) limh→0 f (x+h)−f . h taneous rate of change = When we mention rate of change, the latter is implied. Instantaneous rate of change is the derivative. When Average rate of change is required, it will be specifically refer to as average rate of change. Velocity is one of the most common forms of rate of change. Again, average velocity = average rate of change and instantaneous velocity = instantaneous rate of change = derivative. Velocity refers to the increase of distance(s) for a corresponding increade in time (t). The notation commonly used for this is: v(t) = ds = s′ (t) dt Acceleration is the change in velocity for a corersponding increase in time. Therefore, acceleration is the derivative of velocity a(t) = v ′ (t) This implies that acceleration is the second derivative of the distance(s). Worked Example 187: Rate of Change Question: The height (in metres) of a golf ball that is hit into the air after t seconds, is given by h(t) = 20t = 5t2 . Determine 1. the average velocity of the ball during the first two seconds 2. the velocity of the ball after 1,5 seconds 3. when the velocity is zero 4. the velocity at which the ball hits the ground 5. the acceleration of the ball Answer Step 1 : Average velocity Ave velocity = = = = h(2) − h(0) 2−0 [20(2) − 5(2)2 ] − [20(0) − 5(0)2 ] 2 40 − 20 2 10 ms−1 Step 2 : Instantaneous Velocity v(t) = = dh dt 20 − 10t Velocity after 1,5 seconds: v(1,5) = 20 − 10(1,5) = 5 ms−1 Step 3 : Zero velocity 534 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 v(t) = 0 20 − 10t = 10t = t 40.7 0 20 = 2 Therefore the velocity is zero after 2 seconds Step 4 : Ground velocity The ball hits the ground when h(t) = 0 20t − 5t2 = 0 = 0 t=0 or 5t(4 − t) t=4 The ball hits the ground after 4 seconds. The velocity after 4 seconds will be: v(4) = h′ (4) = 20 − 10(4) = 20 ms−1 The ball hits the gound at a speed of 20ms−1 Step 5 : Acceleration a = v ′ (t) = −10 ms−1 40.7 End of Chapter Exercises 1. Determine f ′ (x) from first principles if: f (x) = x2 − 6x f (x) = 2x − x2 2. Given: f (x) = −x2 + 3x, find f ′ (x) using first principles. 3. Determine dx dy if: A y = (2x)2 − B 1 3x √ 2 x−5 √ y= x 4. Given: f (x) = x3 − 3x2 + 4 A Calculate f (−1), and hence solve the equationf (x) = 0 B Determine f ′ (x) C Sketch the graph of f neatly and clearly, showing the co-ordinates of the turning points as well as the intercepts on both axes. 535 40.7 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 D Determine the co-ordinates of the points on the graph of f where the gradient is 9. 5. Given: f (x) = 2x3 − 5x2 − 4x + 3. The x-intercepts of f are: (-1;0) ( 21 ;0) and (3;0). A Determine the co-ordinates of the turning points of f . B Draw a neat sketch graph of f . Clearly indicate the co-ordinates of the intercepts with the axes, as well as the co-ordinates of the turning points. C For which values of k will the equation f (x) = k , have exactly two real roots? D Determine the equation of the tangent to the graph of f (x) = 2x3 − 5x2 − 4x + 3 at the point where x = 1. 6. A Sketch the graph of f (x) = x3 − 9x2 + 24x − 20, showing all intercepts with the axes and turning points. B Find the equation of the tangent to f (x) at x = 4. 7. Calculate: 1 − x3 x→1 1 − x lim 8. Given: f (x) = 2x2 − x A Use the definition of the derivative to calculate f ′ (x). B Hence, calculate the co-ordinates of the point at which the gradient of the tangent to the graph of f is 7. √ 9. If xy − 5 = x3 , determine dx dy 10. Given: g(x) = (x−2 + x2 )2 . Calculate g ′ (2). 11. Given: f (x) = 2x − 3 A Find: B Solve: f −1 (x) f −1 (x) = 3f ′ (x) 12. Find f ′ (x) for each of the following: √ 5 x3 + 10 A f (x) = 3 (2x2 − 5)(3x + 2) B f (x) = x2 13. Determine the minimum value of the sum of a positive number and its reciprocal. 14. If the displacement s (in metres) of a particle at time t (in seconds) is governed by the equation s = 21 t3 − 2t, find its acceleration after 2 seconds. (Acceleration is the rate of change of velocity, and velocity is the rate of change of displacement.) 15. A After doing some research, a transport company has determined that the rate at which petrol is consumed by one of its large carriers, travelling at an average speed of x km per hour, is given by: P (x) = 55 x + 2x 200 litres per kilometre i. Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per hour (travelling time). Now deduce that the total cost, C, in Rands, for a 2 000 km trip is given by: 256000 + 40x C(x) = x ii. Hence determine the average speed to be maintained to effect a minimum cost for a 2 000 km trip. 536 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 40.7 B During an experiment the temperature T (in degrees Celsius), varies with time t (in hours), according to the formula: 1 T (t) = 30 + 4t − t2 2 t ∈ [1; 10] i. Determine an expression for the rate of change of temperature with time. ii. During which time interval was the temperature dropping? 16. The depth, d, of water in a kettle t minutes after it starts to boil, is given by d = 86 − 81 t − 41 t3 , where d is measured in millimetres. A How many millimetres of water are there in the kettle just before it starts to boil? B As the water boils, the level in the kettle drops. Find the rate at which the water level is decreasing when t = 2 minutes. C How many minutes after the kettle starts boiling will the water level be dropping at a rate of 12 81 mm/minute? 537 40.7 CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 538 Chapter 41 Linear Programming - Grade 12 41.1 Introduction In Grade 11 you were introduced to linear programming and solved problems by looking at points on the edges of the feasible region. In Grade 12 you will look at how to solve linear programming problems in a more general manner. 41.2 Terminology Here is a recap of some of the important concepts in linear programming. 41.2.1 Feasible Region and Points Constraints mean that we cannot just take any x and y when looking for the x and y that optimise our objective function. If we think of the variables x and y as a point (x,y) in the xyplane then we call the set of all points in the xy-plane that satisfy our constraints the feasible region. Any point in the feasible region is called a feasible point. For example, the constraints x≥0 y≥0 mean that every (x,y) we can consider must lie in the first quadrant of the xy plane. The constraint x≥y means that every (x,y) must lie on or below the line y = x and the constraint x ≤ 20 means that x must lie on or to the left of the line x = 20. We can use these constraints to draw the feasible region as shown by the shaded region in Figure 41.1. Important: The constraints are used to create bounds of the solution. 539 41.3 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 y 20 y 10 = x x = 20 15 5 x 5 10 15 20 Figure 41.1: The feasible region corresponding to the constraints x ≥ 0, y ≥ 0, x ≥ y and x ≤ 20. Important: ax + by = c ax + by ≤ c If b 6= 0, feasible points must lie on the line y = − ab x + bc If b = 0, feasible points must lie on the line x = c/a If b 6= 0, feasible points must lie on or below the line y = − ab x + cb . If b = 0, feasible points must lie on or to the left of the line x = c/a. When a constraint is linear, it means that it requires that any feasible point (x,y) lies on one side of or on a line. Interpreting constraints as graphs in the xy plane is very important since it allows us to construct the feasible region such as in Figure 41.1. 41.3 Linear Programming and the Feasible Region If the objective function and all of the constraints are linear then we call the problem of optimising the objective function subject to these constraints a linear program. All optimisation problems we will look at will be linear programs. The major consequence of the constraints being linear is that the feasible region is always a polygon. This is evident since the constraints that define the feasible region all contribute a line segment to its boundary (see Figure 41.1). It is also always true that the feasible region is a convex polygon. The objective function being linear means that the feasible point(s) that gives the solution of a linear program always lies on one of the vertices of the feasible region. This is very important since, as we will soon see, it gives us a way of solving linear programs. We will now see why the solutions of a linear program always lie on the vertices of the feasible region. Firstly, note that if we think of f (x,y) as lying on the z axis, then the function f (x,y) = ax + by (where a and b are real numbers) is the definition of a plane. If we solve for y in the equation defining the objective function then f (x,y) = ax + by ∴ y= f (x,y) −a x+ b b (41.1) What this means is that if we find all the points where f (x,y) = c for any real number c (i.e. f (x,y) is constant with a value of c), then we have the equation of a line. This line we call a level line of the objective function. 540 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 41.3 Consider again the feasible region described in Figure 41.1. Lets say that we have the objective function f (x,y) = x − 2y with this feasible region. If we consider Equation ?? corresponding to f (x,y) = −20 then we get the level line 1 x + 10 2 which has been drawn in Figure 41.2. Level lines corresponding to y= x +5 2 x f (x,y) = 0 or y = 2 x f (x,y) = 10 or y = − 5 2 x f (x,y) = 20 or y = − 10 2 f (x,y) = −10 or y = have also been drawn in. It is very important to realise that these are not the only level lines; in fact, there are infinitely many of them and they are all parallel to each other. Remember that if we look at any one level line f (x,y) has the same value for every point (x,y) that lies on that line. Also, f (x,y) will always have different values on different level lines. y f (x,y) = −20 20 f (x,y) = −10 15 f (x,y) = 0 10 f (x,y) = 10 5 f (x,y) = 20 x 5 10 15 20 Figure 41.2: The feasible region corresponding to the constraints x ≥ 0, y ≥ 0, x ≥ y and x ≤ 20 with objective function f (x,y) = x − 2y. The dashed lines represent various level lines of f (x,y). If a ruler is placed on the level line corresponding to f (x,y) = −20 in Figure 41.2 and moved down the page parallel to this line then it is clear that the ruler will be moving over level lines which correspond to larger values of f (x,y). So if we wanted to maximise f (x,y) then we simply move the ruler down the page until we reach the “lowest” point in the feasible region. This point will then be the feasible point that maximises f (x,y). Similarly, if we wanted to minimise f (x,y) then the “highest” feasible point will give the minimum value of f (x,y). Since our feasible region is a polygon, these points will always lie on vertices in the feasible region. The fact that the value of our objective function along the line of the ruler increases as we move it down and decreases as we move it up depends on this particular example. Some other examples might have that the function increases as we move the ruler up and decreases as we move it down. It is a general property, though, of linear objective functions that they will consistently increase or decrease as we move the ruler up or down. Knowing which direction to move the ruler in order to maximise/minimise f (x,y) = ax + by is as simple as looking at the sign of b (i.e. “is b negative, positive or zero?”). If b is positive, then f (x,y) increases as we move the ruler up and f (x,y) decreases as we move the ruler down. The opposite happens for the case when b is negative: f (x,y) decreases as we move the ruler up and f (x,y) increases as we move the ruler down. If b = 0 then we need to look at the sign of a. 541 41.3 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 If a is positive then f (x,y) increases as we move the ruler to the right and decreases if we move the ruler to the left. Once again, the opposite happens for a negative. If we look again at the objective function mentioned earlier, f (x,y) = x − 2y with a = 1 and b = −2, then we should find that f (x,y) increases as we move the ruler down the page since b = −2 < 0. This is exactly what we found happening in Figure 41.2. The main points about linear programming we have encountered so far are • The feasible region is always a polygon. • Solutions occur at vertices of the feasible region. • Moving a ruler parallel to the level lines of the objective function up/down to the top/bottom of the feasible region shows us which of the vertices is the solution. • The direction in which to move the ruler is determined by the sign of b and also possibly by the sign of a. These points are sufficient to determine a method for solving any linear program. Method: Linear Programming If we wish to maximise the objective function f (x,y) then: 1. Find the gradient of the level lines of f (x,y) (this is always going to be − ab as we saw in Equation ??) 2. Place your ruler on the xy plane, making a line with gradient − ab (i.e. b units on the x-axis and −a units on the y-axis) 3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to check whether b is negative, positive or zero. A If b > 0, move the ruler up the page, keeping the ruler parallel to the level lines all the time, until it touches the “highest” point in the feasible region. This point is then the solution. B If b < 0, move the ruler in the opposite direction to get the solution at the “lowest” point in the feasible region. C If b = 0, check the sign of a i. If a < 0 move the ruler to the “leftmost” feasible point. This point is then the solution. ii. If a > 0 move the ruler to the “rightmost” feasible point. This point is then the solution. Worked Example 188: Prizes! Question: As part of their opening specials, a furniture store has promised to give away at least 40 prizes with a total value of at least R2 000. The prizes are kettles and toasters. 1. If the company decides that there will be at least 10 of each prize, write down two more inequalities from these constraints. 2. If the cost of manufacturing a kettle is R60 and a toaster is R50, write down an objective function C which can be used to determine the cost to the company of both kettles and toasters. 542 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 41.3 3. Sketch the graph of the feasibility region that can be used to determine all the possible combinations of kettles and toasters that honour the promises of the company. 4. How many of each prize will represent the cheapest option for the company? 5. How much will this combination of kettles and toasters cost? Answer Step 1 : Identify the decision variables Let the number of kettles be xk and the number of toasters be yt and write down two constraints apart from xk ≥ 0 and yt ≥ 0 that must be adhered to. Step 2 : Write constraint equations Since there will be at least 10 of each prize we can write: xk ≥ 10 and yt ≥ 10 Also the store has promised to give away at least 40 prizes in total. Therefore: xk + yt ≥ 40 Step 3 : Write the objective function The cost of manufacturing a kettle is R60 and a toaster is R50. Therefore the cost the total cost C is: C = 60xk + 50yt Step 4 : Sketch the graph of the feasible region yt 100 90 80 70 60 50 40 30 B 20 A 10 xk 10 20 30 40 50 60 70 80 90 100 Step 5 : Determine vertices of feasible region From the graph, the coordinates of vertex A is (3,1) and the coordinates of vertex B are (1,3). Step 6 : Draw in the search line The seach line is the gradient of the objective function. That is, if the equation C = 60x + 50y is now written in the standard form y = ..., then the gradient is: 6 m=− , 5 which is shown with the broken line on the graph. 543 41.3 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 yt 100 90 80 70 60 50 40 30 B 20 A 10 xk 10 20 30 40 50 60 70 80 90 100 Step 7 : Calculate cost at each vertex At vertex A, the cost is: C = = 60xk + 50yt 60(30) + 50(10) = = 1800 + 500 2300 = = 60xk + 50yt 60(10) + 50(30) = = 600 + 1500 2100 At vertex B, the cost is: C Step 8 : Write the final answer The cheapest combination of prizes is 10 kettles and 30 toasters, costing the company R2 100. Worked Example 189: Search Line Method Question: As a production planner at a factory manufacturing lawn cutters your job will be to advise the management on how many of each model should be produced per week in order to maximise the profit on the local production. The factory is producing two types of lawn cutters: Quadrant and Pentagon. Two of the production processes that the lawn cutters must go through are: bodywork and engine work. • The factory cannot operate for less than 360 hours on engine work for the lawn cutters. • The factory has a maximum capacity of 480 hours for bodywork for the lawn cutters. 544 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 • Half an hour of engine work and half an hour of bodywork is required to produce one Quadrant. • One third of an hour of engine work andone fifth of an hour of bodywork is required to produce one Pentagon. • The ratio of Pentagon lawn cutters to Quadrant lawn cutters produced per week must be at least 3:2. • A minimum of 200 Quadrant lawn cutters must be produced per week. Let the number of Quadrant lawn cutters manufactured in a week be x. Let the number of Pentagon lawn cutters manufactured in a week be y. Two of the constraints are: x ≥ 200 3x + 2y ≥ 2 160 1. Write down the remaining constraints in terms of x and y to represent the abovementioned information. 2. Use graph paper to represent the constraints graphically. 3. Clearly indicate the feasible region by shading it. 4. If the profit on one Quadrant lawn cutter is R1 200 and the profit on one Pentagon lawn cutter is R400, write down an equation that will represent the profit on the lawn cutters. 5. Using a search line and your graph, determine the number of Quadrant and Pentagon lawn cutters that will yield a maximum profit. 6. Determine the maximum profit per week. Answer Step 1 : Remaining constraints: 1 1 x + ≤ 480 2 5 3 y ≥ x 2 Step 2 : Graphical representation y 2400 1080 0 200 Step 3 : Profit equation 720 960 P = 1 200x + 400y Step 4 : Maximum profit P = 1 200(600) + 400(900) 545 41.3 41.4 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 P = R1 080 000 41.4 End of Chapter Exercises 1. Polkadots is a small company that makes two types of cards, type X and type Y. With the available labour and material, the company can make not more than 150 cards of type X and not more than 120 cards of type Y per week. Altogether they cannot make more than 200 cards per week. There is an order for at least 40 type X cards and 10 type Y cards per week. Polkadots makes a profit of R5 for each type X card sold and R10 for each type Y card. Let the number of type X cards be x and the number of type Y cards be y, manufactured per week. A One of the constraint inequalities which represents the restrictions above is x ≤ 150. Write the other constraint inequalities. B Represent the constraints graphically and shade the feasible region. C Write the equation that represents the profit P (the objective function), in terms of x and y. D On your graph, draw a straight line which will help you to determine how many of each type must be made weekly to produce the maximum P E Calculate the maximum weekly profit. 2. A brickworks produces “face bricks” and “clinkers”. Both types of bricks are produced and sold in batches of a thousand. Face bricks are sold at R150 per thousand, and clinkers at R100 per thousand, where an income of at least R9,000 per month is required to cover costs. The brickworks is able to produce at most 40,000 face bricks and 90,000 clinkers per month, and has transport facilities to deliver at most 100,000 bricks per month. The number of clinkers produced must be at least the same number of face bricks produced. Let the number of face bricks in thousands be x, and the number of clinkers in thousands be y. A List all the constraints. B Graph the feasible region. C If the sale of face bricks yields a profit of R25 per thousand and clinkers R45 per thousand, use your graph to determine the maximum profit. D If the profit margins on face bricks and clinkers are interchanged, use your graph to determine the maximum profit. 3. A small cell phone company makes two types of cell phones: Easyhear and Longtalk. Production figures are checked weekly. At most, 42 Easyhear and 60 Longtalk phones can be manufactured each week. At least 30 cell phones must be produced each week to cover costs. In order not to flood the market, the number of Easyhear phones cannot be more than twice the number of Longtalk phones. It takes 23 hour to assemble an Easyhear phone and 12 hour to put together a Longtalk phone. The trade unions only allow for a 50-hour week. Let x be the number of Easyhear phones and y be the number of Longtalk phones manufactured each week. A Two of the constraints are: 0 ≤ x ≤ 42 Write down the other three constraints. 546 and 0 ≤ y ≤ 60 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 41.4 B Draw a graph to represent the feasible region C If the profit on an Easyhear phone is R225 and the profit on a Longtalk is R75, determine the maximum profit per week. 4. Hair for Africa is a firm that specialises in making two kinds of up-market shampoo, Glowhair and Longcurls. They must produce at least two cases of Glowhair and one case of Longcurls per day to stay in the market. Due to a limited supply of chemicals, they cannot produce more than 8 cases of Glowhair and 6 cases of Longcurls per day. It takes half-an-hour to produce one case of Glowhair and one hour to produce a case of Longcurls, and due to restrictions by the unions, the plant may operate for at most 7 hours per day. The workforce at Hair for Africa, which is still in training, can only produce a maximum of 10 cases of shampoo per day. Let x be the number of cases of Glowhair and y the number of cases of Longcurls produced per day. A Write down the inequalities that represent all the constraints. B Sketch the feasible region. C If the profit on a case of Glowhair is R400 and the profit on a case of Longcurls is R300, determine the maximum profit that Hair for Africa can make per day. 5. A transport contracter has 6 5-ton trucks and 8 3-ton trucks. He must deliver at least 120 tons of sand per day to a construction site, but he may not deliver more than 180 tons per day. The 5-ton trucks can each make three trips per day at a cost of R30 per trip, and the 3-ton trucks can each make four trips per day at a cost of R120 per trip. How must the contracter utilise his trucks so that he has minimum expense ? 547 41.4 CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 548 Chapter 42 Geometry - Grade 12 42.1 Introduction Activity :: Discussion : Discuss these Research Topics Research one of the following geometrical ideas and describe it to your group: 1. taxicab geometry, 2. sperical geometry, 3. fractals, 4. the Koch snowflake. 42.2 Circle Geometry 42.2.1 Terminology The following is a recap of terms that are regularly used when referring to circles. arc An arc is a part of the circumference of a circle. chord A chord is defined as a straight line joining the ends of an arc. radius The radius, r, is the distance from the centre of the circle to any point on the circumference. diameter The diameter, , is a special chord that passes through the centre of the circle. The diameter is the straight line from a point on the circumference to another point on the circumference, that passes through the centre of the circle. segment A segment is the part of the circle that is cut off by a chord. A chord divides a circle into two segments. tangent A tangent is a line that makes contact with a circle at one point on the circumference. (AB is a tangent to the circle at point P . 549 42.2 CHAPTER 42. GEOMETRY - GRADE 12 segment chord a rc rad i us b O diameter b A B P tangent Figure 42.1: Parts of a Circle 42.2.2 Axioms An axiom is an established or accepted principle. For this section, the following are accepted as axioms. 1. The Theorem of Pythagoras, which states that the square on the hypotenuse of a rightangled triangle is equal to the sum of the squares on the other two sides. In △ABC, this means that AB 2 + BC 2 = AC 2 A B C Figure 42.2: A right-angled triangle 2. A tangent is perpendicular to the radius, drawn at the point of contact with the circle. 42.2.3 Theorems of the Geometry of Circles A theorem is a general proposition that is not self-evident but is proved by reasoning (these proofs need not be learned for examination purposes). Theorem 6. The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord. Proof: bO A P 550 B CHAPTER 42. GEOMETRY - GRADE 12 42.2 Consider a circle, with centre O. Draw a chord AB and draw a perpendicular line from the centre of the circle to intersect the chord at point P . The aim is to prove that AP = BP 1. △OAP and △OBP are right-angled triangles. 2. OA = OB as both of these are radii and OP is common to both triangles. Apply the Theorem of Pythagoras to each triangle, to get: OA2 = OP 2 + AP 2 OB 2 = OP 2 + BP 2 However, OA = OB. So, OP 2 + AP 2 = OP 2 + BP 2 ∴ AP 2 and AP = = BP 2 BP This means that OP bisects AB. Theorem 7. The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the chord. Proof: bO A P B Consider a circle, with centre O. Draw a chord AB and draw a line from the centre of the circle to bisect the chord at point P . The aim is to prove that OP ⊥ AB In △OAP and △OBP , 1. AP = P B (given) 2. OA = OB (radii) 3. OP is common to both triangles. ∴ △OAP ≡ △OBP (SSS). ˆ OAP ˆ + OBP ˆ OAP ˆ ∴ OAP ∴ OP = ˆ OBP = = 180◦ (AP B is a str. line) ˆ = 90◦ OBP ⊥ AB 551 42.2 CHAPTER 42. GEOMETRY - GRADE 12 Theorem 8. The perpendicular bisector of a chord passes through the centre of the circle. Proof: bQ A P B Consider a circle. Draw a chord AB. Draw a line P Q perpendicular to AB such that P Q bisects AB at point P . Draw lines AQ and BQ. The aim is to prove that Q is the centre of the circle, by showing that AQ = BQ. In △OAP and △OBP , 1. AP = P B (given) 2. ∠QP A = ∠QP B (QP ⊥ AB) 3. QP is common to both triangles. ∴ △QAP ≡ △QBP (SAS). From this, QA = QB. Since the centre of a circle is the only point inside a circle that has points on the circumference at an equal distance from it, Q must be the centre of the circle. Exercise: Circles I 1. Find the value of x: 552 CHAPTER 42. GEOMETRY - GRADE 12 42.2 a) b) O O x x 4 R 5 P Q R Q P PR=6 PR=8 c) d) R Q 6 2 10 S 6 P O x O x R Q P PR=8 e) f) x S 24 T 5 P Q Q U 8 T R P R 5 10 x O 25 O S R Theorem 9. The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference of the circle. Proof: P bO A R B Consider a circle, with centre O and with A and B on the circumference. Draw a chord AB. Draw radii OA and OB. Select any point P on the circumference of the circle. Draw lines P A and P B. Draw P O and extend to R. ˆ = 2 · APˆ B. The aim is to prove that AOB ˆ = P AO ˆ + APˆ O (exterior angle = sum of interior opp. angles) AOR ˆ = APˆ O (△AOP is an isosceles △) But, P AO 553 42.2 CHAPTER 42. GEOMETRY - GRADE 12 ˆ = 2APˆ O ∴ AOR ˆ = 2BP ˆ O. Similarly, BOR So, ˆ AOB = = ˆ + BOR ˆ AOR ˆO 2APˆ O + 2BP ˆ O) 2(APˆ O + BP = 2(APˆ B) = Exercise: Circles II 1. Find the angles (a to f ) indicated in each diagram: 1. 3. 2. J a J K 45◦ b H Ob K O Ob K b c J H K 4. 5. 20◦ H 6. K K 30◦ J 100◦ O b Ob Ob d 120◦ f e J H H H J Theorem 10. The angles subtended by a chord at the circumference of a circle on the same side of the chord are equal. Proof: Q P O b A B Consider a circle, with centre O. Draw a chord AB. Select any points P and Q on the circumference of the circle, such that both P and Q are on the same side of the chord. Draw lines P A, P B, QA and QB. 554 CHAPTER 42. GEOMETRY - GRADE 12 42.2 ˆ = APˆ B. The aim is to prove that AQB ˆ AOB ˆ and AOB ˆ ∴ 2AQB ˆ ∴ AQB ˆ ∠ at centre = twice ∠ at circumference = 2AQB ˆ = 2AP B ∠ at centre = twice ∠ at circumference = 2APˆ B = APˆ B Theorem 11. (Converse of Theorem 10) If a line segment subtends equal angles at two other points on the same side of the line, then these four points lie on a circle. Proof: P Q R A B Consider a line segment AB, that subtends equal angles at points P and Q on the same side of AB. The aim is to prove that points A, B, P and Q lie on the circumference of a circle. By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut AP (or AP extended) at point R. ˆ AQB ˆ but AQB ˆ ∴ ARB ˆ but this cannot be true since ARB ˆ ∠s on same side of chord = ARB = APˆ B (given) = APˆ B ˆ (ext. ∠ of △) = APˆ B + RBP ∴ the assumption that the circle does not pass through P , must be false, and A, B, P and Q lie on the circumference of a circle. Exercise: Circles III 1. Find the values of the unknown letters. 555 42.2 CHAPTER 42. GEOMETRY - GRADE 12 A 1. E 2. a F B 21◦ 15◦ I D G b H C 3. 4. N J O c K 17◦ M Q 24◦ d L 5. P S T 6. W R 35◦ 45◦ 35◦ X bO 12◦ f e Y Z U V Cyclic Quadrilaterals Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The vertices of a cyclic quadrilateral are said to be concyclic. Theorem 12. The opposite angles of a cyclic quadrilateral are supplementary. Proof: 556 CHAPTER 42. GEOMETRY - GRADE 12 42.2 Q P Ob 1 2 A B Consider a circle, with centre O. Draw a cyclic quadrilateral ABP Q. Draw AO and P O. ˆ + AQP ˆ = 180◦ and QAB ˆ + QPˆ B = 180◦. The aim is to prove that ABP Ô1 Ô2 But, Ô1 + Ô2 ˆ + 2AQP ˆ ∴ 2ABP ˆ + AQP ˆ ∴ ABP ˆ + QPˆ B Similarly, QAB ˆ ∠s at centre = 2ABP ˆ ∠s at centre = 2AQP = 360◦ = 360◦ = 180◦ = 180◦ Theorem 13. (Converse of Theorem 12) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Proof: Q R P A B ˆ + AQP ˆ = 180◦ and QAB ˆ + QPˆ B = 180◦ . Consider a quadrilateral ABP Q, such that ABP The aim is to prove that points A, B, P and Q lie on the circumference of a circle. By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut AP (or AP extended) at point R. Draw BR. ˆ + QRB ˆ QAB ˆ + QPˆ B but QAB ˆ ∴ QRB ˆ but this cannot be true since QRB = 180◦ opp. ∠s of cyclic quad. = 180◦ (given) = QPˆ B ˆ (ext. ∠ of △) = QPˆ B + RBP ∴ the assumption that the circle does not pass through P , must be false, and A, B, P and Q lie on the circumference of a circle and ABP Q is a cyclic quadrilateral. 557 42.2 CHAPTER 42. GEOMETRY - GRADE 12 Exercise: Circles IV 1. Find the values of the unknown letters. 1. 2. Y X 106◦ 87◦ Ob P 34 a Q b ◦ W a b c S Z L 3. 4. K R U a 86◦ X H a 114◦ 57◦ I J V W Theorem 14. Two tangents drawn to a circle from the same point outside the circle are equal in length. Proof: A O b P B Consider a circle, with centre O. Choose a point P outside the circle. Draw two tangents to the circle from point P , that meet the circle at A and B. Draw lines OA, OB and OP . The aim is to prove that AP = BP . In △OAP and △OBP , 558 CHAPTER 42. GEOMETRY - GRADE 12 42.2 1. OA = OB (radii) 2. ∠OAP = ∠OP B = 90◦ (OA ⊥ AP and OB ⊥ BP ) 3. OP is common to both triangles. △OAP ≡ △OBP (right angle, hypotenuse, side) ∴ AP = BP Exercise: Circles V 1. Find the value of the unknown lengths. A 1. 2. AE=5cm AC=8cm CE=9cm a G 5c m bF B bJ b d E c b D b I 8cm H C 3. 4. K Rb 2c m 6c m O b 3cm S P e f N M L b Q LN=7.5cm Theorem 15. The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment. Proof: T Q A O b P S R B 559 42.2 CHAPTER 42. GEOMETRY - GRADE 12 Consider a circle, with centre O. Draw a chord AB and a tangent SR to the circle at point B. Chord AB subtends angles at points P and Q on the minor and major arcs, respectively. Draw a diameter BT and join A to T . ˆ and AQB ˆ = ABS. ˆ The aim is to prove that APˆ B = ABR ˆ = ABS ˆ as this result is needed to prove that APˆ B = ABR. ˆ First prove that AQB ˆ + ABT ˆ ABS ˆ BAT ˆ + ATˆ B ∴ ABT ˆ ∴ ABS ˆ However, AQB ˆ ∴ AQB ˆ + QBR ˆ SBQ ˆ APˆ B + AQB ˆ + QBR ˆ ∴ SBQ ˆ From (42.1), AQB ∴ APˆ B = 90◦ (T B ⊥ SR) = 90◦ (∠s at centre) = 90◦ (sum of angles in △BAT ) ˆ = ABT = ATˆ B( angles subtended by same chord AB) ˆ = ABS = 180◦ (SBT is a str. line) = 180◦ (ABP Q is a cyclic quad.) ˆ = APˆ B + AQB ˆ = ABS ˆ = ABR 560 (42.1) CHAPTER 42. GEOMETRY - GRADE 12 42.2 Exercise: Circles VI 1. Find the values of the unknown letters. P 1 2 Q R 33◦ O d b c S R a S P e Q O 3 O 4h-70◦ 8◦ g S R f S R 4 P Q 17◦ 3h S h+50◦ S P 2h-20◦ Q O 5 6 R k O l j S R P ◦ 19 i 121◦ R Q Q O 7 8 O R Q m n R Q o 52◦ p T T r b O 34◦ q S O S Theorem 16. (Converse of 15) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle. Proof: Q A O b Y S X B R 561 42.2 CHAPTER 42. GEOMETRY - GRADE 12 Consider a circle, with centre O and chord AB. Let line SR pass through point B. Chord AB ˆ = AQB. ˆ subtends an angle at point Q such that ABS The aim is to prove that SBR is a tangent to the circle. By contradiction. Assume that SBR is not a tangent to the circle and draw XBY such that XBY is a tangent to the circle. ˆ ABX ˆ ABS = = ˆ AQB ˆ AQB ˆ ∴ ABX ˆ But since, ABX ˆ (42.2) can only be true if, XBS = = ˆ ABS ˆ + XBS ˆ ABS = 0 However, (tan-chord theorem) (given) (42.2) ˆ is zero, then both XBY and SBR coincide and SBR is a tangent to the circle. If XBS Exercise: Applying Theorem 9 1. Show that Theorem 9 also applies to the following two cases: A P O bO R R b P B A B 562 CHAPTER 42. GEOMETRY - GRADE 12 42.2 Worked Example 190: Circle Geometry I BD is a tangent to the circle with centre O. BO ⊥ AD. Prove that: 1. CF OE is a cyclic quadrilateral O E A D Question: 2. F B = BC 3. △COE///△CBF F C 4. CD2 = ED.AD 5. OE BC = CD CO B Answer 1. Step 1 : To show a quadrilateral is cyclic, we need a pair of opposite angles to be supplementary, so lets look for that. ˆ F OE ˆ F CE = 90◦ (BO ⊥ OD) = 90◦ (∠ subtended by diameter AE) ∴ CF OE is a cyclic quadrilateral (opposite ∠’s supplementary) 2. Step 1 : Since these two sides are part of a triangle, we are proving that triangle to be isosceles. The easiest way is to show the angles opposite to those sides to be equal. ˆ = x. Let OEC ∴ ∴ ∴ ˆ = x (∠ between tangent BD and chord CE) F CB ˆ C = x (exterior ∠ to cyclic quadrilateral CF OE) BF BF = BC (sides opposite equal ∠’s in isosceles △BF C) 3. Step 1 : To show these two triangles similar, we will need 3 equal angles. We already have 3 of the 6 needed angles from the previous question. We need only find the missing 3 angles. ˆ CBF OC 180◦ − 2x (sum of ∠’s in △BF C) OE (radii of circle O) ˆ = x (isosceles △COE) ECO ˆ = 180◦ − 2x (sum of ∠’s in △COE) COE = = ∴ ∴ ˆ = CBF ˆ • COE ˆ ˆ • ECO = F CB ˆ = CFˆ B • OEC ∴ △COE///△CBF (3 ∠’s equal) 563 42.2 CHAPTER 42. GEOMETRY - GRADE 12 4. Step 1 : This relation reminds us of a proportionality relation between similar triangles. So investigate which triangles contain these sides and prove them similar. In this case 3 equal angles works well. Start with one triangle. In △EDC ˆ CED ˆ ECD 180◦ − x (∠’s on a straight line AD) = 90◦ − x (complementary ∠’s) = Step 2 : Now look at the angles in the other triangle. In △ADC ˆ ACE ˆ CAD ˆ and ECO) ˆ 180◦ − x (sum of ∠’s ACE ◦ 90 − x (sum of ∠’s in △CAE) = = Step 3 : The third equal angle is an angle both triangles have in common. ˆ = EDC ˆ since they are the same ∠. Lastly, ADC Step 4 : Now we know that the triangles are similar and can use the proportionality relation accordingly. ∴ △ADC///△CDE (3 ∠’s equal) ED CD ∴ = CD AD ∴ CD2 = ED.AD 5. Step 1 : This looks like another proportionality relation with a little twist, since not all sides are contained in 2 triangles. There is a quick observation we can make about the odd side out, OE. OE = CD (△OEC is isosceles) Step 2 : With this observation we can limit ourselves to proving triangles BOC and ODC similar. Start in one of the triangles. In △BCO ˆ OCB ˆ CBO = 90◦ (radius OC on tangent BD) = 180◦ − 2x (sum of ∠’s in △BF C) Step 3 : Then we move on to the other one. In △OCD ˆ OCD ˆ COD = = 90◦ (radius OC on tangent BD) 180◦ − 2x (sum of ∠’s in △OCE) Step 4 : Again we have a common element. Lastly, OC is a common side to both △’s. Step 5 : Then, once we’ve shown similarity, we use the proportionality relation , as well as our first observation, appropriately. 564 CHAPTER 42. GEOMETRY - GRADE 12 ∴ ∴ ∴ 42.2 △BOC///△ODC (common side and 2 equal ∠’s) CD CO = BC CO CD OE = (OE = CD isosceles △OEC) BC CO Worked Example 191: Circle Geometry II F D is drawn parallel to the tangent CB Prove that: 1. F ADE is cyclic F Question: C 2. △AF E///△CBD A 3. G E F C.AG GH = DC.F E BD H D B Answer 1. Step 1 : In this case, the best way to show F ADE is a cyclic quadrilateral is to look for equal angles, subtended by the same chord. Let ∠BCD = x ∴ ∠CAH = x (∠ between tangent BC and chord CE) ∴ ∠F DC = x (alternate ∠, F D k CB) ∴ F ADE is a cyclic quadrilateral (chord F E subtends equal ∠’s) 2. Step 1 : To show these 2 triangles similar we will need 3 equal angles. We can use the result from the previous question. Let ∠F EA = y ∴ ∠F DA = y (∠’s subtended by same chord AF in cyclic quadrilateral F ADE) ∴ ∠CBD = y (corresponding ∠’s, F D k CB) ∴ ∠F EA = ∠CBD Step 2 : We have already proved 1 pair of angles equal in the previous question. ∠BCD = 565 ∠F AE (above) 42.3 CHAPTER 42. GEOMETRY - GRADE 12 Step 3 : Proving the last set of angles equal is simply a matter of adding up the angles in the triangles. Then we have proved similarity. ∠AF E ∠CBD = = ∴ 180◦ − x − y (∠’s in △AF E) 180◦ − x − y (∠’s in △CBD) △AF E///△CBD (3 ∠’s equal) 3. Step 1 : This equation looks like it has to do with proportionality relation of similar triangles. We already showed triangles AF E and CBD similar in the previous question. So lets start there. DC BD FA FE DC.F E = FA BD = ∴ Step 2 : Now we need to look for a hint about side F A. Looking at triangle CAH we see that there is a line F G intersecting it parallel to base CH. This gives us another proportionality relation. AG GH = ∴ FA (F G k CH splits up lines AH and AC proportionally) FC F C.AG FA = GH Step 3 : We have 2 expressions for the side F A. ∴ F C.AG DC.F E = GH BD 42.3 Co-ordinate Geometry 42.3.1 Equation of a Circle We know that every point on the circumference of a circle is the same distance away from the centre of the circle. Consider a point (x1 ,y1 ) on the circumference of a circle of radius r with centre at (x0 ,y0 ). b P (x ,y ) 1 1 (x0 ,y0 ) b O Q Figure 42.3: Circle h with centre (x0 ,y0 ) has a tangent, g passing through point P at (x1 ,y1 ). Line f passes through the centre and point P . 566 CHAPTER 42. GEOMETRY - GRADE 12 42.3 In Figure 42.3, △OP Q is a right-angled triangle. Therefore, from the Theorem of Pythagoras, we know that: OP 2 = P Q2 + OQ2 But, PQ = OQ = OP = 2 = r ∴ y1 − y0 x1 − x0 r (y1 − y0 )2 + (x1 − x0 )2 But, this same relation holds for any point P on the circumference. In fact, the relation holds for all points P on the circumference. Therefore, we can write: (x − x0 )2 + (y − y0 )2 = r2 (42.3) for a circle with centre at (x0 ,y0 ) and radius r. For example, the equation of a circle with centre (0,0) and radius 4 is: (y − y0 )2 + (x − x0 )2 (y − 0)2 + (x − 0)2 y 2 + x2 = r2 = = 42 16 Worked Example 192: Equation of a Circle I Question: Find the equation of a circle (centre O) with a diameter between two points, P at (−5,5) and Q at (5, − 5). Answer Step 1 : Draw a picture Draw a picture of the situation to help you figure out what needs to be done. P b 5 O −5 5 −5 b Q Step 2 : Find the centre of the circle We know that the centre of a circle lies on the midpoint of a diameter. Therefore the co-ordinates of the centre of the circle is found by finding the midpoint of the line between P and Q. Let the co-ordinates of the centre of the circle be (x0 ,y0 ), let the co-ordinates of P be (x1 ,y1 ) and let the co-ordinates of Q be (x2 ,y2 ). Then, 567 42.3 CHAPTER 42. GEOMETRY - GRADE 12 the co-ordinates of the midpoint are: x0 = = = y0 = = = x1 + x2 2 −5 + 5 2 0 y1 + y2 2 5 + (−5) 2 0 The centre point of line P Q and therefore the centre of the circle is at (0,0). Step 3 : Find the radius of the circle If P and Q are two points on a diameter, then the radius is half the distance between them. The distance between the two points is: r= 1 PQ 2 = = = = = = 1p (x2 − x1 )2 + (y2 − y1 )2 2 1p (5 − (−5))2 + (−5 − 5)2 2 1p (10)2 + (−10)2 2 1√ 100 + 100 2 r 200 4 √ 50 Step 4 : Write the equation of the circle x2 + y 2 = 50 Worked Example 193: Equation of a Circle II Question: Find the center and radius of the circle x2 − 14x + y 2 + 4y = −28. Answer Step 1 : Change to standard form We need to rewrite the equation in the form (x − x0 ) + (y − y0 ) = r2 To do this we need to complete the square i.e. add and subtract ( 21 cooefficient of x)2 and ( 21 cooefficient of y)2 Step 2 : Adding cooefficients x2 − 14x + y 2 + 4y = −28 ∴ x2 − 14x + (7)2 − (7)2 + y 2 + 4y + (2)2 − (2)2 = −28 Step 3 : Complete the squares ∴ (x − 7)2 − (7)2 + (y + 2)2 − (2)2 = −28 Step 4 : Take the constants to the other side ∴ (x − 7)2 − 49 + (y + 2)2 − 4 = −28 ∴ (x − 7)2 + (y + 2)2 = −28 + 49 + 4 ∴ (x − 7)2 + (y + 2)2 = 25 Step 5 : Read the values from the equation ∴ center is (7; −2) and the radius is 5 units 568 CHAPTER 42. GEOMETRY - GRADE 12 42.3.2 42.3 Equation of a Tangent to a Circle at a Point on the Circle We are given that a tangent to a circle is drawn through a point P with co-ordinates (x1 ,y1 ). In this section, we find out how to determine the equation of that tangent. g b P (x ,y ) 1 1 h f (x0 ,y0 ) b Figure 42.4: Circle h with centre (x0 ,y0 ) has a tangent, g passing through point P at (x1 ,y1 ). Line f passes through the centre and point P . We start by making a list of what we know: 1. We know that the equation of the circle with centre (x0 ,y0 ) is (x − x0 )2 + (y − y0 )2 = r2 . 2. We know that a tangent is perpendicular to the radius, drawn at the point of contact with the circle. As we have seen in earlier grades, there are two steps to determining the equation of a straight line: Step 1: Calculate the gradient of the line, m. Step 2: Calculate the y-intercept of the line, c. The same method is used to determine the equation of the tangent. First we need to find the gradient of the tangent. We do this by finding the gradient of the line that passes through the centre of the circle and point P (line f in Figure 42.4), because this line is a radius line and the tangent is perpendicular to it. mf = y1 − y0 x1 − x0 The tangent (line g) is perpendicular to this line. Therefore, mf × mg = −1 So, mg = − 569 1 mf (42.4) 42.3 CHAPTER 42. GEOMETRY - GRADE 12 Now, we know that the tangent passes through (x1 ,y1 ) so the equation is given by: y − y1 y − y1 y − y1 y − y1 = m(x − x1 ) 1 = − (x − x1 ) mf 1 = − y1 −y0 (x − x1 ) x1 −x0 x1 − x0 (x − x1 ) = − y1 − y0 For example, find the equation of the tangent to the circle at point (1,1). The centre of the circle is at (0,0). The equation of the circle is x2 + y 2 = 2. Use y − y1 = − with (x0 ,y0 ) = (0,0) and (x1 ,y1 ) = (1,1). y − y1 x1 − x0 (x − x1 ) y1 − y0 x1 − x0 (x − x1 ) y1 − y0 1−0 − (x − 1) 1−0 1 − (x − 1) 1 −(x − 1) + 1 −x + 1 + 1 = − y−1 = y−1 = y y = = y = −x + 2 Exercise: Co-ordinate Geometry 1. Find the equation of the cicle: A B C D E with with with with with center center center center center (0; 5) and radius 5 (2; 0) and radius 4 (5; 7) and radius 18 (−2; 0) and radius 6 √ (−5; −3) and radius 3 2. A Find the equation of the circle with center (2; 1) which passes through (4; 1). B Where does it cut the line y = x + 1? C Draw a sketch to illustrate your answers. 3. A Find the equation of the circle with center (−3; −2) which passes through (1; −4). B Find the equation of the circle with center (3; 1) which passes through (2; 5). C Find the point where these two circles cut each other. 4. Find the center and radius of the following circles: A B C D E (x − 9)2 + (y − 6)2 = 36 (x − 2)2 + (y − 9)2 = 1 (x + 5)2 + (y + 7)2 = 12 (x + 4)2 + (y + 4)2 = 23 3(x − 2)2 + 3(y + 3)2 = 12 570 CHAPTER 42. GEOMETRY - GRADE 12 42.4 F x2 − 3x + 9 = y 2 + 5y + 25 = 17 5. Find the x− and y− intercepts of the following graphs and draw a scetch to illustrate your answer: A B C D (x + 7)2 + (y − 2)2 = 8 x2 + (y − 6)2 = 100 (x + 4)2 + y 2 = 16 (x − 5)2 + (y + 1)2 = 25 6. Find the center and radius of the following circles: A B C D E F x2 + 6x + y 2 − 12y = −20 x2 + 4x + y 2 − 8y = 0 x2 + y 2 + 8y = 7 x2 − 6x + y 2 = 16 x2 − 5x + y 2 + 3y = − 34 x2 − 6nx + y 2 + 10ny = 9n2 7. Find the equations to the tangent to the circle: A B C D x2 + y 2 = 17 at the point (1; 4) x2 + y 2 = 25 at the point (3; 4) (x + 1)2 + (y − 2)2 = 25 at the point (3; 5) (x − 2)2 + (y − 1)2 = 13 at the point (5; 3) 42.4 Transformations 42.4.1 Rotation of a Point about an angle θ First we will find a formula for the co-ordinates of P after a rotation of θ. We need to know something about polar co-ordinates and compound angles before we start. Polar co-ordinates b r P y α Notice that : sin α = yr ∴ y = r sin α and cos α = xr ∴ x = r cos α so P can be expressed in two ways: x P (x; y) rectangular co-ordinates or P (r cos α; r sin α) polar co-ordinates. Compound angles (See derivation of formulae in Ch. 12) sin (α + β) = sin α cos β + sin β cos α cos (α + β) = cos α cos β − sin α sin β 571 42.4 CHAPTER 42. GEOMETRY - GRADE 12 Now consider P ′ after a rotation of θ P (x; y) = P (r cos α; r sin α) P ′ (r cos (α + θ); r sin (α + θ)) P′ Expand the co-ordinates of P ′ b P = (r cos α; r sin α) b x − co-ordinate of P ′ = r cos (α + θ) = = r [cos α cos θ − sin α sin θ] r cos α cos θ − r sin α sin θ = y − co-ordinate of P ′ θ α x cos θ − y sin θ = = r sin (α + θ) r [sin α cos θ + sin θ cos α] = = r sin α cos θ + r cos α sin θ y cos θ + x sin θ which gives the formula P ′ = [(x cos θ − y sin θ; y cos θ + x sin θ)]. √ So to find the co-ordinates of P (1; 3) after a rotation of 45◦ , we arrive at: P′ = = = = [(x cos θ − y sin θ); (y cos θ + x sin θ] i h √ √ (1 cos 45◦ − 3 sin 45◦ ); ( 3 cos 45◦ + 1 sin 45◦ !# " √ √ ! 1 1 3 3 √ −√ ; √ +√ 2 2 2 2 ! √ √ 1− 3 3+1 √ ; √ 2 2 572 CHAPTER 42. GEOMETRY - GRADE 12 42.4 Exercise: Rotations Any line OP is drawn (not necessarily in the first quadrant), making an angle of θ degrees with the x-axis. Using the co-ordinates of P and the angle α, calculate the co-ordinates of P ′ , if the line OP is rotated about the origin through α degrees. 1. 2. 3. 4. 5. 6. 42.4.2 P (2, 6) (4, 2) (5, -1) (-3, 2) (-4, -1) (2, 5) b P θ α 60◦ 30◦ 45◦ 120◦ 225◦ -150◦ O Characteristics of Transformations Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and size, and that enlargement preserves shape but not size. 42.4.3 Characteristics of Transformations Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and size, and that enlargement preserves shape but not size. Activity :: : Geometric Transformations 15 10 Draw a large 15×15 grid and plot △ABC with A(2; 6), B(5; 6) and C(5; 1). Fill in the lines y = x and y = −x. A(2; 6) B(5; 6) b b 5 b C(3; 4) −15 Complete the table below , by drawing the images of △ABC under the given transformations. The first one has been done for you. −10 −5 5b −5 b A′ 10 C′ b B′ −10 y = −x y=x −15 573 15 42.5 CHAPTER 42. GEOMETRY - GRADE 12 Transformation (x; y) → (x; −y) Description (translation, reflection, rotation, enlargement) reflection about the x-axis Co-ordinates Lengths Angles A′ (2; −6) B ′ (5; −6) C ′ (5; −2) A′ B ′ = 3 B′C ′ = 4 A′ C ′ = 5 B̂ ′ = 90◦ tan Â = 4/3 ∴ Â = 53◦ , Ĉ = 37◦ (x; y) → (x + 1; y − 2) (x; y) → (−x; y) (x; y) → (−y; x) (x; y) → (−x; −y) (x; y) → (2x; 2y) (x; y) → (y; x) (x; y) → (y; x + 1) A transformation that leaves lengths and angles unchanged is called a rigid transformation. Which of the above transformations are rigid? 42.5 Exercises 1. ∆ABC undergoes several transformations forming ∆A′ B ′ C ′ . Describe the relationship between the angles and sides of ∆KLM and ∆A′ B ′ C ′ (e.g., they are twice as large, the same, etc.) Transformation Reflect Reduce by a scale factor of 3 Rotate by 90◦ Translate 4 units right Enlarge by a scale factor of 2 Sides Angles Area 2. ∆DEF has Ê = 30◦ , DE = 4 cm, EF = 5 cm. ∆DEF is enlarged by a scale factor of 6 to form ∆D′ E ′ F ′ . A Solve ∆DEF B Hence, solve ∆D′ E ′ F ′ 3. ∆XY Z has an area of 6 cm2 . Find the area of ∆X ′ Y ′ Z ′ if the points have been transformed as follows: A (x, y) → (x + 2; y + 3) B (x, y) → (y; x) 574 CHAPTER 42. GEOMETRY - GRADE 12 42.5 C (x, y) → (4x; y) D (x, y) → (3x; y + 2) E (x, y) → (−x; −y) F (x, y) → (x; −y + 3) G (x, y) → (4x; 4y) H (x, y) → (−3x; 4y) 575 42.5 CHAPTER 42. GEOMETRY - GRADE 12 576 Chapter 43 Trigonometry - Grade 12 43.1 Compound Angle Identities 43.1.1 Derivation of sin(α + β) We have, for any angles α and β, that sin(α + β) = sin α cos β + sin β cos α How do we derive this identity? It is tricky, so follow closely. Suppose we have the unit circle shown below. The two points L(a,b) and K(x,y) are on the circle. y K(x; y) b L(a; b) b 1 1 (α − β) α O b β b a M (x; y) x We can get the coordinates of L and K in terms of the angles α and β. For the triangle LOK, we have that b 1 a cos β = 1 sin β = =⇒ b = sin β =⇒ a = cos β 577 43.1 CHAPTER 43. TRIGONOMETRY - GRADE 12 Thus the coordinates of L are (cos β; sin β). In the same way as above, we can see that the coordinates of K are (cos α; sin α). p The identity for cos(α − β) is now determined as follows: Using the distance formula (i.e. d = (x2 − x1 )2 + (y2 − y1 )2 or d2 = (x2 − x1 )2 + (y2 − y1 )2 ), we can find KL2 . T R2 = = = = = (cos α − cos β)2 + (sin α − sin β)2 cos2 α − 2 cos α cos β + cos2 β + sin2 α − 2 sin α sin β + sin2 β (cos2 α + sin2 α) + (cos2 β + sin2 β) − 2 cos α cos β − 2 sin α sin β 1 + 1 − 2(cos α cos β + sin α sin β) 2 − 2(cos α cos β + sin α sin β) Now using the cosine rule for △KOL, we get KL2 = = = KO2 + LO2 − 2 · KO · LO · cos(α − β) 12 + 12 − 2(1)(1) cos(α − β) 2 − 2 · cos(α − β) Equating our two values for T R2 , we have 2 − 2 · cos(α − β) =⇒ cos(α − β) = = 2 − 2(cos α cos β + sin α · sin β) cos α · cos β + sin α · sin β Now let α → 90◦ − α. Then cos(90◦ − α − β) = cos(90◦ − α) cos β + sin(90◦ − α) sin β = sin α · cos β + cos α · sin β But cos(90◦ − (α + β)) = sin(α + β). Thus sin(α + β) = sin α · cos β + cos α · sin β 43.1.2 Derivation of sin(α − β) We can use sin(α + β) = sin α cos β + sin β cos α to show that sin(α − β) = sin α cos β − sin β cos α We know that sin(−θ) = − sin(θ) and cos(−θ) = cos θ Therefore, sin(α − β) 43.1.3 = sin(α + (−β)) = sin α cos(−β) + sin(−β) cos α = sin α cos β − sin β cos α Derivation of cos(α + β) We can use sin(α − β) = sin α cos β − sin β cos α to show that cos(α + β) = cos α cos β − sin α sin β We know that sin(θ) = cos(90 − θ). 578 CHAPTER 43. TRIGONOMETRY - GRADE 12 43.1 Therefore, cos(α + β) = sin(90 − (α + β)) = sin((90 − α) − β)) = sin(90 − α) cos β − sin β cos(90 − α) = cos α cos β − sin β sin α 43.1.4 Derivation of cos(α − β) We found this identity in our derivation of the sin(α + β) identity. We can also use the fact that sin(α + β) = sin α cos β + sin β cos α to derive that cos(α − β) = cos α cos β + sin α sin β As cos(θ) = sin(90 − θ), we have that cos(α − β) = sin(90 − (α − β)) = sin((90 − α) + β)) = sin(90 − α) cos β + sin β cos(90 − α) = cos α cos β + sin β sin α 43.1.5 Derivation of sin 2α We know that sin(α + β) = sin α cos β + sin β cos α When α = β, we have that sin(α + α) 43.1.6 = sin α cos α + sin α cos α = = 2 sin α cos α sin(2α) Derivation of cos 2α We know that cos(α + β) = cos α cos β − sin α sin β When α = β, we have that cos(α + α) = cos α cos α − sin α sin α = cos2 α − sin2 α = cos(2α) However, we can also write cos 2α = 2cos2 α − 1 and cos 2α = 1 − 2sin2 α by using sin2 α + cos2 α = 1. 579 43.1 CHAPTER 43. TRIGONOMETRY - GRADE 12 Activity :: cos 2α Identity : Use sin2 α + cos2 α = 1 to show that: cos 2α = 43.1.7 2 cos2 α − 1 1 − 2 sin2 α Problem-solving Strategy for Identities The most important thing to remember when asked to prove identities is: Important: Trigonometric Identities Never assume that the left hand side is equal to the right hand side. You need to show that both sides are equal. A suggestion for proving identities: It is usually much easier simplifying the more complex side of an identity to get the simpler side than the other way round. Worked Example 194: Trigonometric Identities 1 √ √ Question: Prove that sin 75◦ = 2( 43+1) without using a calculator. Answer Step 1 : Identify a strategy We only know the exact values of the trig functions for a few special angles (30◦ , 45◦ , 60◦ , etc.). We can see that 75◦ = 30◦ + 45◦. Thus we can use our double-angle identity for sin(α + β) to express sin 75◦ in terms of known trig function values. Step 2 : Execute strategy sin 75◦ = sin(45◦ + 30◦ ) = sin(45◦ ) cos(30◦ ) + sin(30◦ ) cos(45◦ ) √ 1 1 3 1 √ · +√ · 2 2 2 2 √ 3+1 √ 2 2 √ √ 3+1 2 √ ×√ 2 2 2 √ √ 2( 3 + 1) 4 = = = = Worked Example 195: Trigonometric Identities 2 580 CHAPTER 43. TRIGONOMETRY - GRADE 12 43.1 Question: Deduce a formula for tan(α + β) in terms of tan α and tan β. Hint: Use the formulae for sin(α + β) and cos(α + β) Answer Step 1 : Identify a strategy We can reexpress tan(α + β) in terms of cosines and sines, and then use the doubleangle formulas for these. We then manipulate the resulting expression in order to get it in terms of tan α and tan β. Step 2 : Execute strategy tan(α + β) = = = = sin(α + β) cos(α + β) sin α · cos β + sin β · cos α cos α · cos β − sin α · sin β sin α·cos β cos α·cos β cos α·cos β cos α·cos β + − sin β·cos α cos α·cos β sin α·sin β cos α·cos β tan α + tan β 1 − tan α · tan β Worked Example 196: Trigonometric Identities 3 Question: Prove that sin θ + sin 2θ = tan θ 1 + cos θ + cos 2θ For which values is the identity not valid? Answer Step 1 : Identify a strategy The right-hand side (RHS) of the identity cannot be simplified. Thus we should try simplify the left-hand side (LHS). We can also notice that the trig function on the RHS does not have a 2θ dependance. Thus we will need to use the doubleangle formulas to simplify the sin 2θ and cos 2θ on the LHS. We know that tan θ is undefined for some angles θ. Thus the identity is also undefined for these θ, and hence is not valid for these angles. Also, for some θ, we might have division by zero in the LHS, which is not allowed. Thus the identity won’t hold for these angles also. Step 2 : Execute the strategy LHS = = = = = sin θ + 2 sin θ cos θ 1 + cos θ + (2 cos2 θ − 1) sin θ(1 + 2 cos θ) cos θ(1 + 2 cos θ) sin θ cos θ tan θ RHS We know that tan θ is undefined when θ = 90◦ + 180◦n, where n is an integer. The LHS is undefined when 1 + cos θ + cos 2θ = 0. Thus we need to solve this equation. =⇒ 1 + cos θ + cos 2θ = 0 cos θ(1 + 2 cos θ) = 0 581 43.2 CHAPTER 43. TRIGONOMETRY - GRADE 12 The above has solutions when cos θ = 0, which occurs when θ = 90◦ + 180◦n, where n is an integer. These are the same values when tan θ is undefined. It also has solutions when 1 + 2 cos θ = 0. This is true when cos θ = − 21 , and thus θ = . . . − 240◦ , −120◦, 120◦, 240◦ , . . .. To summarise, the identity is not valid when θ = . . . − 270◦, −240◦, −120◦ , −90◦,90◦ , 120◦ , 240◦ , 270◦, . . . Worked Example 197: Trigonometric Equations Question: Solve the following equation for y without using a calculator. 1 − sin y − cos 2y = −1 sin 2y − cos y Answer Step 1 : Identify a strategy Before we are able to solve the equation, we first need to simplify the left-hand side. We do this using the double-angle formulas. Step 2 : Execute the strategy =⇒ =⇒ =⇒ =⇒ 1 − sin y − (1 − 2 sin2 y) 2 sin y cos y − cos y 2 sin2 y − sin y cos y(2 sin y − 1) sin y(2 sin y − 1) cos y(2 sin y − 1) tan y ◦ ◦ y = 135 + 180 n; n ∈ Z = −1 = −1 = −1 = −1 43.2 Applications of Trigonometric Functions 43.2.1 Problems in Two Dimensions Worked Example 198: Question: For the figure below, we are given that BC = BD = x. Show that BC 2 = 2x2 (1 + sin θ). 582 CHAPTER 43. TRIGONOMETRY - GRADE 12 b O b D 43.2 A θ b x x b C B b Answer Step 1 : Identify a strategy We want CB, and we have CD and BD. If we could get the angle B D̂C, then we could use the cosine rule to determine DC. This is possible, as △ABD is a rightangled triangle. We know this from circle geometry, that any triangle circumscribed by a circle with one side going through the origin, is right-angled. As we have two angles of △ABD, we know AD̂B and hence B D̂C. Using the cosine rule, we can getBC 2 . Step 2 : Execute the strategy AD̂B = 180◦ − θ − 90◦ = 90◦ − θ Thus B D̂C = = = 180◦ − AD̂B 180◦ − (90◦ − θ) 90◦ + θ Now the cosine rule gives BC 2 = CD2 + BD2 − 2 · CD · BD · cos(B D̂C) = x2 + x2 − 2 · x2 · cos(90◦ + θ) = 2x2 + 2x2 [ sin(90◦ ) cos(θ) + sin(θ) cos(90◦ )] = 2x2 + 2x2 [ 1 · cos(θ) + sin(θ) · 0] = 2x2 (1 − sin θ) Exercise: 1. For the diagram on the right, A Find AÔC in terms of θ. C b b i. cos θ ii. sin θ iii. sin 2θ D Now do the same for cos 2θ and tan θ. 583 b C Using the above, show that sin 2θ = 2 sin θ cos θ. A O b B E θ b B Find an expression for: 43.2 CHAPTER 43. TRIGONOMETRY - GRADE 12 2. DA is a diameter of circle O with radius r. CA = r, AB = DE and DÔE = θ. Show that cos θ = 14 . E b D b θ b b B O b C b A 3. The figure on the right shows a cyclic quadrilateral with BC CD = AD AB . A Show that the area of the cyclic quadrilateral is DC · DA · sin D̂. B Find expressions for cos D̂ and cos B̂ in terms of the quadrilateral sides. C Show that 2CA2 = CD2 + DA2 + AB 2 + BC 2 . D Suppose that BC = 10, CD = 15, AD = 4 and AB = 6. Find CA2 . E Find the angle D̂ using your expression for cos D̂. Hence find the area of ABCD. D b C b b A b B 43.2.2 Problems in 3 dimensions Worked Example 199: Height of tower Question: D is the top of a tower of height h. Its base is at C. The triangle ABC lies on the ground (a horizontal plane). If we have that BC = b, DB̂A = α, DB̂C = x and DĈB = θ, show that b sin α sin x h= sin(x + θ) 584 CHAPTER 43. TRIGONOMETRY - GRADE 12 43.2 D b h C b b θ b A α x b B Answer Step 1 : Identify a strategy We have that the triangle ABD is right-angled. Thus we can relate the height h with the angle α and either the length BA or BD (using sines or cosines). But we have two angles and a length for △BCD, and thus can work out all the remaining lengths and angles of this triangle. We can thus work out BD. Step 2 : Execute the strategy We have that h BD =⇒ h = sin α = BD sin α Now we need BD in terms of the given angles and length b. Considering the triangle BCD, we see that we can use the sine rule. sin θ DB = DB = =⇒ sin(DB̂C) b b sin θ sin(DB̂C) But DB̂C = 180◦ − α − θ, and sin(180◦ − α − θ) = − sin(−α − θ) = sin(α + θ) So DB = = b sin θ sin(DB̂C) b sin θ sin(α + θ) Exercise: 1. The line BC represents a tall tower, with C at its foot. Its angle of elevation from D is θ. We are also given that BA = AD = x. 585 43.3 CHAPTER 43. TRIGONOMETRY - GRADE 12 C b B θ b α x b D x b A A Find the height of the tower BC in terms of x, tan θ and cos 2α. B Find BC if we are given that k = 140m, α = 21◦ and θ = 9◦ . 43.3 Other Geometries 43.3.1 Taxicab Geometry Taxicab geometry, considered by Hermann Minkowski in the 19th century, is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates. 43.3.2 Manhattan distance The metric in taxi-cab geometry, is known as the Manhattan distance, between two points in an Euclidean space with fixed Cartesian coordinate system as the sum of the lengths of the projections of the line segment between the points onto the coordinate axes. For example, in the plane, the Manhattan distance between the point P1 with coordinates (x1 , y1 ) and the point P2 at (x2 , y2 ) is |x1 − x2 | + |y1 − y2 | (43.1) Figure 43.1: Manhattan Distance (dotted and solid) compared to Euclidean Distance √ (dashed). In each case the Manhattan distance is 12 units, while the Euclidean distance is 36 586 CHAPTER 43. TRIGONOMETRY - GRADE 12 43.3 The Manhattan distance depends on the choice on the rotation of the coordinate system, but does not depend on the translation of the coordinate system or its reflection with respect to a coordinate axis. Manhattan distance is also known as city block distance or taxi-cab distance. It is given these names because it is the shortest distance a car would drive in a city laid out in square blocks. Taxicab geometry satisfies all of Euclid’s axioms except for the side-angle-side axiom, as one can generate two triangles with two sides and the angle between them the same and have them not be congruent. In particular, the parallel postulate holds. A circle in taxicab geometry consists of those points that are a fixed Manhattan distance from the center. These circles are squares whose sides make a 45◦ angle with the coordinate axes. 43.3.3 Spherical Geometry Spherical geometry is the geometry of the two-dimensional surface of a sphere. It is an example of a non-Euclidean geometry. In plane geometry the basic concepts are points and line. On the sphere, points are defined in the usual sense. The equivalents of lines are not defined in the usual sense of ”straight line” but in the sense of ”the shortest paths between points” which is called a geodesic. On the sphere the geodesics are the great circles, so the other geometric concepts are defined like in plane geometry but with lines replaced by great circles. Thus, in spherical geometry angles are defined between great circles, resulting in a spherical trigonometry that differs from ordinary trigonometry in many respects (for example, the sum of the interior angles of a triangle exceeds 180◦). Spherical geometry is the simplest model of elliptic geometry, in which a line has no parallels through a given point. Contrast this with hyperbolic geometry, in which a line has two parallels, and an infinite number of ultra-parallels, through a given point. Spherical geometry has important practical uses in celestial navigation and astronomy. Extension: Distance on a Sphere The great-circle distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere’s interior). Because spherical geometry is rather different from ordinary Euclidean geometry, the equations for distance take on a different form. The distance between two points in Euclidean space is the length of a straight line from one point to the other. On the sphere, however, there are no straight lines. In non-Euclidean geometry, straight lines are replaced with geodesics. Geodesics on the sphere are the great circles (circles on the sphere whose centers are coincident with the center of the sphere). Between any two points on a sphere which are not directly opposite each other, there is a unique great circle. The two points separate the great circle into two arcs. The length of the shorter arc is the great-circle distance between the points. Between two points which are directly opposite each other (called antipodal points) there infinitely many great circles, but all have the same length, equal to half the circumference of the circle, or πr, where r is the radius of the sphere. Because the Earth is approximately spherical (see spherical Earth), the equations for great-circle distance are important for finding the shortest distance between points on the surface of the Earth, and so have important applications in navigation. Let φ1 ,λ1 ; φ2 ,λ2 , be the latitude and longitude of two points, respectively. Let ∆λ be the longitude difference. Then, if r is the great-circle radius of the sphere, the great-circle distance is r∆σ, where ∆σ is the angular difference/distance and can be determined from the spherical law of cosines as: ∆σ = arccos {sin φ1 sin φ2 + cos φ1 cos φ2 cos ∆λ} 587 43.3 CHAPTER 43. TRIGONOMETRY - GRADE 12 Extension: Spherical Distance on the Earth The shape of the Earth more closely resembles a flattened spheroid with extreme values for the radius of curvature, or arcradius, of 6335.437 km at the equator (vertically) and 6399.592 km at the poles, and having an average great-circle radius of 6372.795 km. Using a sphere with a radius of 6372.795 km thus results in an error of up to about 0.5%. 43.3.4 Fractal Geometry The word ”fractal” has two related meanings. In colloquial usage, it denotes a shape that is recursively constructed or self-similar, that is, a shape that appears similar at all scales of magnification and is therefore often referred to as ”infinitely complex.” In mathematics a fractal is a geometric object that satisfies a specific technical condition, namely having a Hausdorff dimension greater than its topological dimension. The term fractal was coined in 1975 by Benot Mandelbrot, from the Latin fractus, meaning ”broken” or ”fractured.” Three common techniques for generating fractals are: • Iterated function systems - These have a fixed geometric replacement rule. Cantor set, Sierpinski carpet, Sierpinski gasket, Peano curve, Koch snowflake, Harter-Heighway dragon curve, T-Square, Menger sponge, are some examples of such fractals. • Escape-time fractals - Fractals defined by a recurrence relation at each point in a space (such as the complex plane). Examples of this type are the Mandelbrot set, the Burning Ship fractal and the Lyapunov fractal. • Random fractals, generated by stochastic rather than deterministic processes, for example, fractal landscapes, Lvy flight and the Brownian tree. The latter yields so-called mass- or dendritic fractals, for example, Diffusion Limited Aggregation or Reaction Limited Aggregation clusters. Fractals in nature Approximate fractals are easily found in nature. These objects display self-similar structure over an extended, but finite, scale range. Examples include clouds, snow flakes, mountains, river networks, and systems of blood vessels. Trees and ferns are fractal in nature and can be modeled on a computer using a recursive algorithm. This recursive nature is clear in these examples - a branch from a tree or a frond from a fern is a miniature replica of the whole: not identical, but similar in nature. The surface of a mountain can be modeled on a computer using a fractal: Start with a triangle in 3D space and connect the central points of each side by line segments, resulting in 4 triangles. The central points are then randomly moved up or down, within a defined range. The procedure is repeated, decreasing at each iteration the range by half. The recursive nature of the algorithm guarantees that the whole is statistically similar to each detail. 588 CHAPTER 43. TRIGONOMETRY - GRADE 12 43.4 Summary of the Trigonomertic Rules and Identities Pythagorean Identity Cofuntion Identities Ratio Identities cos2 θ + sin2 θ = 1 sin(90◦ − θ) = cos θ cos(90◦ − θ) = sin θ tan θ = Odd/Even Identities Periodicity Identities Double angle Identities sin(−θ) = − sin θ cos(−θ) = cos θ tan(−θ) = − tan θ sin(θ ± 360◦ ) = sin θ cos(θ ± 360◦) = cos θ tan(θ ± 180◦ ) = tan θ sin(2θ) = 2 sin θ cos θ cos (2θ) = cos2 θ − sin2 θ cos (2θ) = 2 cos2 θ − 1 2 tan θ tan (2θ) = 1−tan 2θ Addition/Subtraction Identities Area Rule Cosine rule sin (θ + φ) = sin θ cos φ + cos θ sin φ sin (θ − φ) = sin θ cos φ − cos θ sin φ cos (θ + φ) = cos θ cos φ − sin θ sin φ cos (θ − φ) = cos θ cos φ + sin θ sin φ tan φ+tan θ tan (θ + φ) = 1−tan θ tan φ tan φ−tan θ tan (θ − φ) = 1+tan θ tan φ Area = 12 bc sin A Area = 21 ab sin C Area = 21 ac sin B a2 = b2 + c2 − 2bc cos A b2 = a2 + c2 − 2ac cos B c2 = a2 + b2 − 2ab cos C Sine Rule sin A a = 43.4 sin B b = sin C c End of Chapter Exercises Do the following without using a calculator. 1. Suppose cos θ = 0.7. Find cos 2θ and cos 4θ. 2. If sin θ = 74 , again find cos 2θ and cos 4θ. 3. Work out the following: A cos 15◦ B cos 75◦ C tan 105◦ D cos 15◦ E cos 3◦ cos 42◦ − sin 3◦ sin 42◦ F 1 − 2 sin2 (22.5◦ ) 4. Solve the following equations: A cos 3θ · cos θ − sin 3θ · sin θ = − 21 B 3 sin θ = 2 cos2 θ C 5. Prove the following identities A sin3 θ = 3 sin θ−sin 3θ 4 589 sin θ cos θ 43.4 CHAPTER 43. TRIGONOMETRY - GRADE 12 B cos2 α(1 − tan2 α) = cos 2α C 4 sin θ · cos θ · cos 2θ = sin 4θ D 4 cos3 x − 3 cos x = cos 3x E tan y = sin 2y cos 2y+1 6. (Challenge question!) If a + b + c = 180◦, prove that sin3 a + sin3 b + sin3 c = 3 cos(a/2) cos(b/2) cos(c/2) + cos(3a/2) cos(3b/2) cos(3c/2) 590 Chapter 44 Statistics - Grade 12 44.1 Introduction In this chapter, you will use the mean, median, mode and standard deviation of a set of data to identify whether the data is normally distributed or whether it is skewed. You will learn more about populations and selecting different kinds of samples in order to avoid bias. You will work with lines of best fit, and learn how to find a regression equation and a correlation coefficient. You will analyse these measures in order to draw conclusions and make predictions. 44.2 A Normal Distribution Activity :: Investigation : You are given a table of data below. 75 80 91 67 75 81 70 77 82 71 78 82 71 78 83 73 78 86 74 78 86 75 79 87 1. Calculate the mean, median, mode and standard deviation of the data. 2. What percentage of the data is within one standard deviation of the mean? 3. Draw a histogram of the data using intervals 60 ≤ x < 64, 64 ≤ x < 68, etc. 4. Join the midpoints of the bars to form a frequency polygon. If large numbers of data are collected from a population, the graph will often have a bell shape. If the data was, say, examination results, a few learners usually get very high marks, a few very low marks and most get a mark in the middle range. We say a distribution is normal if • the mean, median and mode are equal. • it is symmetric around the mean. • ±68% of the sample lies within one standard deviation of the mean, 95% within two standard deviations and 99% within three standard deviations of the mean. 591 44.2 CHAPTER 44. STATISTICS - GRADE 12 68% 95% 99% x̄ − 3σ x̄ − 2σ x̄ − σ x̄ x̄ + σ x̄ + 2σ x̄ + 3σ What happens if the test was very easy or very difficult? Then the distribution may not be symmetrical. If extremely high or extremely low scores are added to a distribution, then the mean tends to shift towards these scores and the curve becomes skewed. If the test was very difficult, the mean score is shifted to the left. In this case, we say the distribution is positively skewed, or skewed right. Skewed right If it was very easy, then many learners would get high scores, and the mean of the distribution would be shifted to the right. We say the distribution is negatively skewed, or skewed left. Skewed left Exercise: Normal Distribution 1. Given the pairs of normal curves below, sketch the graphs on the same set of axes and show any relation between them. An important point to remember is that the area beneath the curve corresponds to 100%. A Mean = 8, standard deviation = 4 and Mean = 4, standard deviation = 8 B Mean = 8, standard deviation = 4 and Mean = 16, standard deviation = 4 C Mean = 8, standard deviation = 4 and Mean = 8, standard deviation = 8 2. After a class test, the following scores were recorded: Test Score 3 4 5 6 7 8 9 Total Mean Standard Deviation A B C D E Frequency 1 7 14 21 14 6 1 64 6 1,2 Draw the histogram of the results. Join the midpoints of each bar and draw a frequency polygon. What mark must one obtain in order to be in the top 2% of the class? Approximately 84% of the pupils passed the test. What was the pass mark? Is the distribution normal or skewed? 592 CHAPTER 44. STATISTICS - GRADE 12 44.3 3. In a road safety study, the speed of 175 cars was monitored along a specific stretch of highway in order to find out whether there existed any link between high speed and the large number of accidents along the route. A frequency table of the results is drawn up below. Speed (km.h−1 ) 50 60 70 80 90 100 110 120 Number of cars (Frequency) 19 28 23 56 20 16 8 5 The mean speed was determined to be around 82 km.h−1 while the median speed was worked out to be around 84,5 km.h−1 . A Draw a frequency polygon to visualise the data in the table above. B Is this distribution symmetrical or skewed left or right? Give a reason fro your answer. 44.3 Extracting a Sample Population Suppose you are trying to find out what percentage of South Africa’s population owns a car. One way of doing this might be to send questionnaires to peoples homes, asking them whether they own a car. However, you quickly run into a problem: you cannot hope to send every person in the country a questionnaire, it would be far to expensive. Also, not everyone would reply. The best you can do is send it to a few people, see what percentage of these own a car, and then use this to estimate what percentage of the entire country own cars. This smaller group of people is called the sample population. The sample population must be carefully chosen, in order to avoid biased results. How do we do this? First, it must be representative. If all of our sample population comes from a very rich area, then almost all will have cars. But we obviously cannot conclude from this that almost everyone in the country has a car! We need to send the questionnaire to rich as well as poor people. Secondly, the size of the sample population must be large enough. It is no good having a sample population consisting of only two people, for example. Both may very well not have cars. But we obviously cannot conclude that no one in the country has a car! The larger the sample population size, the more likely it is that the statistics of our sample population corresponds to the statistics of the entire population. So how does one ensure that ones sample is representative? There are a variety of methods available, which we will look at now. Random Sampling. Every person in the country has an equal chance of being selected. It is unbiased and also independant, which means that the selection of one person has no effect on the selection on another. One way of doing this would be to give each person in the country a number, and then ask a computer to give us a list of random numbers. We could then send the questionnaire to the people corresponding to the random numbers. Systematic Sampling. Again give every person in the country a number, and then, for example, select every hundredth person on the list. So person with number 1 would be selected, person with number 100 would be selected, person with number 200 would be selected, etc. 593 44.4 CHAPTER 44. STATISTICS - GRADE 12 Stratified Sampling. We consider different subgroups of the population, and take random samples from these. For example, we can divide the population into male and female, different ages, or into different income ranges. Cluster Sampling. Here the sample is concentrated in one area. For example, we consider all the people living in one urban area. Exercise: Sampling 1. Discuss the advantages, disadvantages and possible bias when using A systematic sampling B random sampling C cluster sampling 2. Suggest a suitable sampling method that could be used to obtain information on: A passengers views on availability of a local taxi service. B views of learners on school meals. C defects in an item made in a factory. D medical costs of employees in a large company. 3. 5% of a certain magazines’ subscribers is randomly selected. The random number 16 out of 50, is selected. Then subscribers with numbers 16, 66, 116, 166, . . . are chosen as a sample. What kind of sampling is this? 44.4 Function Fitting and Regression Analysis In Grade 11 we recorded two sets of data (bivariate data) on a scatter plot and then we drew a line of best fit as close to as many of the data items as possible. Regression analysis is a method of finding out exactly which function best fits a given set of data. We can find out the equation of the regression line by drawing and estimating, or by using an algebraic method called “the least squared method”, or we can use a calculator. The linear regression equation is written ŷ = a + bx (we say y-hat) or y = A + Bx. Of course these are both variations of a more familiar equation y = mx + c. Suppose you are doing an experiment with washing dishes. You count how many dishes you begin with, and then find out how long it takes to finish washing them. So you plot the data on a graph of time taken versus number of dishes. This is plotted below. t 200 Time taken (seconds) 180 b 160 140 b 120 100 b 80 b 60 b 40 b 20 d 0 0 1 2 3 4 Number of dishes 594 5 6 CHAPTER 44. STATISTICS - GRADE 12 44.4 If t is the time taken, and d the number of dishes, then it looks as though t is proportional to d, ie. t = m · d, where m is the constant of proportionality. There are two questions that interest us now. 1. How do we find m? One way you have already learnt, is to draw a line of best-fit through the data points, and then measure the gradient of the line. But this is not terribly precise. Is there a better way of doing it? 2. How well does our line of best fit really fit our data? If the points on our plot don’t all lie close to the line of best fit, but are scattered everywhere, then the fit is not ’good’, and our assumption that t = m · d might be incorrect. Can we find a quantitative measure of how well our line really fits the data? In this chapter, we answer both of these questions, using the techniques of regression analysis. Worked Example 200: Fitting by hand Question: Use the data given to draw a scatter plot and line of best fit. Now write down the equation of the line that best seems to fit the data. x y 1,0 2,5 2,4 2,8 3,1 3,0 4,9 4,8 5,6 5,1 6,2 5,3 Answer Step 1 : Drawing the graph The first step is to draw the graph. This is shown below. y 7 6 b b b 5 4 3 b b b 2 1 x 0 0 1 2 3 4 5 6 Step 2 : Calculating the equation of the line The equation of the line is y = mx + c From the graph we have drawn, we estimate the y-intercept to be 1,5. We estimate that y = 3,5 when x = 3. So we have that points (3; 3,5) and (0; 1,6) lie on the line. The gradient of the line, m, is given by m = = = y2 − y1 x2 − x1 3,5 − 1,5 3−0 2 3 So we finally have that the equation of the line of best fit is y= 2 x + 1,5 3 595 44.4 CHAPTER 44. STATISTICS - GRADE 12 44.4.1 The Method of Least Squares We now come to a more accurate method of finding the line of best-fit. The method is very simple. Suppose we guess a line of best-fit. Then at at every data point, we find the distance between the data point and the line. If the line fitted the data perfectly, this distance should be zero for all the data points. The worse the fit, the larger the differences. We then square each of these distances, and add them all together. y b b b b b y The best-fit line is then the line that minimises the sum of the squared distances. Suppose we have a data set of n points {(x1 ; y1 ), (x2 ; y2 ), . . . , (xn ,yn )}. We also have a line f (x) = mx + c that we are trying to fit to the data. The distance between the first data point and the line, for example, is distance = y1 − f (x) = y1 − (mx + c) We now square each of these distances and add them together. Lets call this sum S(m,c). Then we have that S(m,c) = = (y1 − f (x1 ))2 + (y2 − f (x2 ))2 + . . . + (yn − f (xn ))2 n X (yi − f (xi ))2 i=1 Thus our problem is to find the value of m and c such that S(m,c) is minimised. Let us call these minimising values m0 and c0 . Then the line of best-fit is f (x) = m0 x + c0 . We can find m0 and c0 using calculus, but it is tricky, and we will just give you the result, which is that P P P n ni=1 xi yi − ni=1 xi ni=1 yi m0 = Pn Pn 2 n i=1 (xi )2 − ( i=1 xi ) n n 1X m0 X c0 = yi − xi = ȳ − m0 x̄ n i=1 n i=0 Worked Example 201: Method of Least Squares Question: In the table below, we have the records of the maintenance costs in Rands, compared with the age of the appliance in months. We have data for 5 appliances. appliance age (x) cost (y) 1 5 90 2 10 140 Answer 596 3 15 250 4 20 300 4 30 380 CHAPTER 44. STATISTICS - GRADE 12 appliance 1 2 3 4 5 Total b a 44.4.2 44.4 x 10 10 15 20 30 80 y 15 140 250 300 380 1160 x2 30 100 225 400 900 1650 xy 20 1400 3750 6000 11400 23000 P P xy − x y 5 × 23000 − 80 × 1160 = = 12 P 2 P 2 = 5 × 1650 − 802 n x − ( x) 1160 12 × 80 − = 40 = ȳ − bx̄ = 5 5 ∴ ŷ = 40 + 12x n P Using a calculator Worked Example 202: Using the Sharp EL-531VH calculator Question: Find a regression equation for the following data: Days (x) Growth in m (y) 1 1,00 2 2,50 3 2,75 4 3,00 5 3,50 Answer Step 1 : Getting your calculator ready Using your calculator, change the mode from normal to “Stat xy”. This mode enables you to type in bivariate data. Step 2 : Entering the data Key in the data as follows: 1 (x,y) 1 DATA n=1 2 (x,y) 2,5 DATA n=2 3 (x,y) 2,75 DATA n=3 4 (x,y) 3,0 DATA n=4 5 (x,y) 3,5 DATA n=5 Step 3 : Getting regression results from the calculator Ask for the values of the regression coefficients a and b. RCL RCL a b gives gives a = 0,9 b = 0,55 ∴ ŷ = 0,9 + 0,55x 597 44.4 CHAPTER 44. STATISTICS - GRADE 12 Worked Example 203: Using the CASIO fx-82ES Natural Display calculator Question: Using a calculator determine the least squares line of best fit for the following data set of marks. Learner 1 2 3 4 5 Chemistry (%) 52 55 86 71 45 Accounting (%) 48 64 95 79 50 For a Chemistry mark of 65%, what mark does the least squares line predict for Accounting? Answer Step 1 : Getting your calculator ready Switch on the calculator. Press [MODE] and then select STAT by pressing [2]. The following screen will appear: 1 3 5 7 1-VAR + CX2 eˆX A . XˆB 2 4 6 8 A + BX ln X A . BˆX 1/X Now press [2] for linear regression. Your screen should look something like this: x y 1 2 3 Step 2 : Entering the data Press [52] and then [=] to enter the first mark under x. Then enter the other values, in the same way, for the x-variable (the Chemistry marks) in the order in which they are given in the data set. Then move the cursor across and up and enter 48 under y opposite 52 in the x-column. Continue to enter the other y-values (the Accounting marks) in order so that they pair off correctly with the corresponding x-values. x 52 55 1 2 3 y Then press [AC]. The screen clears but the data remains stored. 1: 3: 5: 7: Type Edit Var Reg 2: 4: 6: Data Sum MinMax Now press [SHIFT][1] to get the stats computations screen shown below. Choose Regression by pressing [7]. 1: 3: 5: A r ŷ 2: 4: B x̂ Step 3 : Getting regression results from the calculator a) Press [1] and [=] to get the value of the y-intercept, a = −5,065.. = −5,07(to 2 d.p.) Finally, to get the slope, use the following key sequence: [SHIFT][1][7][2][=]. The calculator gives b = 1,169.. = 1,17(to 2 d.p.) The equation of the line of regression is thus: ŷ = −5,07 + 1,17x 598 CHAPTER 44. STATISTICS - GRADE 12 44.4 b) Press [AC][65][SHIFT][1][7][5][=] This gives a (predicted) Accounting mark ofˆ= 70,938.. = 71% Exercise: 1. The table below lists the exam results for 5 students in the subjects of Science and Biology. Learner Science % Biology % 1 55 48 2 66 59 3 74 68 4 92 84 5 47 53 A Use the formulae to find the regression equation coefficients a and b. B Draw a scatter plot of the data on graph paper. C Now use algebra to find a more accurate equation. 2. Footlengths and heights of 7 students are given in the table below. Height (cm) Footlength (cm) 170 27 163 23 131 20 181 28 146 22 134 20 166 24 A Draw a scatter plot of the data on graph paper. B Indentify and describe any trends shown in the scatter plot. C Find the equation of the least squares line by using algebraic methods and draw the line on your graph. D Use your equation to predict the height of a student with footlength 21,6 cm. E Use your equation to predict the footlength of a student 176 cm tall. 3. Repeat the data in question 2 and find the regression line using a calculator 44.4.3 Correlation coefficients Once we have applied regression analysis to a set of data, we would like to have a number that tells us exactly how well the data fits the function. A correlation coefficient, r, is a tool that tells us to what degree there is a relationship between two sets of data. The correlation coefficient r ∈ [−1; 1] when r = −1, there is a perfect negative relationship, when r = 0, there is no relationship and r = 1 is a perfect positive correlation. y y b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b bb b bb b b b b b b b bb y b b b b b b b b b b b b b b b b b bb bb b b b b b b b b b x b b b Positive, strong r ≈ 0,9 y b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 599 b b b b b b b bb b b b b b b b b b b b b b bb b b b b b b b b b b b b bb b b b bb b b b b b bb b b x No association r=0 b b b b b x Negative, fairly strong r ≈ −0,7 We often use the correlation coefficient r2 in order to work with the strength of the correlation only (no whether it is positive or negative). In this case: b b b x Positive, weak r ≈ 0,4 b bb b bb b b b b b b bb b b b b b b b b b bb b b x Positive, fairly strong r ≈ 0,7 b bb b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b bb b b b b b b b b b b bb b b b b b b b b b bb b b b b y b b bb b b b b b bbb b b 44.5 CHAPTER 44. STATISTICS - GRADE 12 r2 = 0 0 < r2 < 0,25 0,25 < r2 < 0,5 0,5 < r2 < 0,75 0,75 < r2 < 0,9 0,9 < r2 < 1 r2 = 1 no correlation very weak weak moderate strong very strong perfect correlation The correlation coefficient r can be calculated using the formula y − ȳ 1 X x − x̄ r= n−1 sx sy • where n is the number of data points, • sx is the standard deviation of the x-values and • sy is the standard deviation of the y-values. This is known as the Pearson’s product moment correlation coefficient. It is a long calculation and much easier to do on the calculator where you simply follow the procedure for the regression equation, and go on to find r. 44.5 Exercises 1. Below is a list of data concerning 12 countries and their respective carbon dioxide (CO2 ) emmission levels per person and the gross domestic product (GDP - a measure of products produced and services delivered within a country in a year) per person. South Africa Thailand Italy Australia China India Canada United Kingdom United States Saudi Arabia Iran Indonesia A B C D E CO2 emmissions per capita (x) 8,1 2,5 7,3 17,0 2,5 0,9 16,0 9,0 19,9 11,0 3,8 1,2 GDP per capita (y) 3 938 2 712 20 943 23 893 816 463 22 537 21 785 31 806 6 853 1 493 986 Draw a scatter plot of the data set and your estimate of a line of best fit. Calculate equation of the line of regression using the method of least squares. Draw the regression line equation onto the graph. Calculate the correlation coefficient r. What conclusion can you reach, regarding the relationship between CO2 emission and GDP per capita for the countries in the data set? 2. A collection of data on the peculiar investigation into a foot size and height of students was recorded in the table below. Answer the questions to follow. Length of right foot (cm) 25,5 26,1 23,7 26,4 27,5 24 22,6 27,1 600 Height (cm) 163,3 164,9 165,5 173,7 174,4 156 155,3 169,3 CHAPTER 44. STATISTICS - GRADE 12 44.5 A Draw a scatter plot of the data set and your estimate of a line of best fit. B Calculate equation of the line of regression using the method of least squares or your calculator. C Draw the regression line equation onto the graph. D Calculate the correlation coefficient r. E What conclusion can you reach, regarding the relationship between the length of the right foot and height of the students in the data set? 3. A class wrote two tests, and the marks for each were recorded in the table below. Full marks in the first test was 50, and the second test was out of 30. A Is there a strong association between the marks for the first and second test? Show why or why not. B One of the learners (in row 18) did not write the second test. Given their mark for the first test, calculate an expected mark for the second test. Learner 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Test 1 (Full marks: 50) 42 32 31 42 35 23 43 23 24 15 19 13 36 29 29 25 29 17 30 28 Test 2 (Full marks: 30) 25 19 20 26 23 14 24 12 14 10 11 10 22 17 17 16 18 19 17 4. A fast food company produces hamburgers. The number of hamburgers made, and the costs are recorded over a week. Hamburgers made Costs 495 R2382 550 R2442 515 R2484 500 R2400 480 R2370 530 R2448 585 R2805 A Find the linear regression function that best fits the data. B If the total cost in a day is R2500, estimate the number of hamburgers produced. C What is the cost of 490 hamburgers? 5. The profits of a new shop are recorded over the first 6 months. The owner wants to predict his future sales. The profits so far have been R90 000 , R93 000, R99 500, R102 000, R101 300, R109 000. A For the profit data, calculate the linear regression function. 601 44.5 CHAPTER 44. STATISTICS - GRADE 12 B Give an estimate of the profits for the next two months. C The owner wants a profit of R130 000. Estimate how many months this will take. 6. A company produces sweets using a machine which runs for a few hours per day. The number of hours running the machine and the number of sweets produced are recorded. Machine hours 3,80 4,23 4,37 4,10 4,17 Sweets produced 275 287 291 281 286 Find the linear regression equation for the data, and estimate the machine hours needed to make 300 sweets. 602 Chapter 45 Combinations and Permutations Grade 12 45.1 Introduction Mathematics education began with counting. At the beginning, fingers, beans, buttons, and pencils were used to help with counting, but these are only practical for small numbers. What happens when a large number of items must be counted? This chapter focuses on how to use mathematical techniques to count combinations of items. 45.2 Counting An important aspect of probability theory is the ability to determine the total number of possible outcomes when multiple events are considered. For example, what is the total number of possible outcomes when a die is rolled and then a coin is tossed? The roll of a die has six possible outcomes (1, 2, 3, 4, 5 or 6) and the toss of a coin, 2 outcomes (head or tails). Counting the possible outcomes can be tedious. 45.2.1 Making a List The simplest method of counting the total number of outcomes is by making a list: 1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T or drawing up a table. die 1 1 2 2 3 3 4 4 5 5 6 6 coin H T H T H T H T H T H T Both these methods result in 12 possible outcomes, but both these methods have a lot of repetition. 603 45.3 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 45.2.2 Tree Diagrams One method of eliminating some of the repetition is to use tree diagrams. Tree diagrams are a graphical method of listing all possible combinations of events from a random experiment. 1 2 3 H T H T H T die coin 4 5 6 T H T H T H Figure 45.1: Example of a tree diagram. Each possible outcome is a branch of the tree. 45.3 Notation 45.3.1 The Factorial Notation For an integer n, the notation n! (read n factorial) represents: n × (n − 1) × (n − 2) × . . . × 3 × 2 × 1 with the special case of 0! = 1. The factorial notation will be used often in this chapter. 45.4 The Fundamental Counting Principle The use of lists, tables and tree diagrams is only feasible for events with a few outcomes. When the number of outcomes grows, it is not practical to list the different possibilities and the fundamental counting principle is used. The fundamental counting principle describes how to determine the total number of outcomes of a series of events. Suppose that two experiments take place. The first experiment has n1 possible outcomes, and the second has n2 possible outcomes. Therefore, the first experiment, followed by the second experiment, will have a total of n1 × n2 possible outcomes. This idea can be generalised to m experiments as the total number of outcomes for m experiments is: n1 × n2 × n3 × . . . × nm = Q is the multiplication equivalent of m Y ni i=1 P . Note: the order in which the experiments are done does not affect the total number of possible outcomes. Worked Example 204: Lunch Special Question: A take-away has a 4-piece lunch special which consists of a sandwich, soup, dessert and drink for R25.00. They offer the following choices for : Sandwich: chicken mayonnaise, cheese and tomato, tuna, and ham and lettuce Soup: tomato, chicken noodle, vegetable Dessert: ice-cream, piece of cake Drink: tea, coffee, coke, Fanta and Sprite. How many possible meals are there? 604 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 45.5 Answer Step 1 : Determine how many parts to the meal there are There are 4 parts: sandwich, soup, dessert and drink. Step 2 : Identify how many choices there are for each part Meal Component Number of choices Sandwich 4 Soup 3 Dessert 2 Drink 5 Step 3 : Use the fundamental counting principle to determine how many different meals are possible 4 × 3 × 2 × 5 = 120 So there are 120 possible meals. 45.5 Combinations The fundamental counting principle describes how to calculate the total number of outcomes when multiple independent events are performed together. A more complex problem is determining how many combinations there are of selecting a group of objects from a set. Mathematically, a combination is defined as an un-ordered collection of unique elements, or more formally, a subset of a set. For example, suppose you have fifty-two playing cards, and select five cards. The five cards would form a combination and would be a subset of the set of 52 cards. In a set, the order of the elements in the set does not matter. These are represented usually with curly braces, for example {2, 4, 6} is a subset of the set {1,2,3,4,5,6}. Since the order of the elements does not matter, only the specific elements are of interest. Therefore, {2, 4, 6} = {6, 4, 2} and {1, 1, 1} is the same as {1} because a set is defined by its elements; they don’t usually appear more than once. Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important (two lists with the same elements in different orders are considered to be the same combination). Also, the elements cannot be repeated in a combination (every element appears uniquely once). 45.5.1 Counting Combinations Calculating the number of ways that certain patterns can be formed is the beginning of combinatorics, the study of combinations. Let S be a set with n objects. Combinations of k objects from this set S are subsets of S having k elements each (where the order of listing the elements does not distinguish two subsets). Combination without Repetition When the order does not matter, but each object can be chosen only once, the number of combinations is: n! n = r!(n − r)! r where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have 10 numbers and wish to choose 5 you would have 10!/(5!(10 - 5)!) = 252 ways to choose. For example how many possible 5 card hands are there in a deck of cards with 52 cards? 52! / (5!(52-5)!) = 2 598 960 combinations 605 45.6 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 Combination with Repetition When the order does not matter and an object can be chosen more than once, then the number of combinations is: (n + r − 1)! n+r−1 n+r−1 = = r!(n − 1)! r n−1 where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have ten types of donuts to choose from and you want three donuts there are (10 + 3 - 1)! / 3!(10 - 1)! = 220 ways to choose. 45.5.2 Combinatorics and Probability Combinatorics is quite useful in the computation of probabilities of events, as it can be used to determine exactly how many outcomes are possible in a given event. Worked Example 205: Probability Question: At a school, learners each play 2 sports. They can choose from netball, basketball, soccer, athletics, swimming, or tennis. What is the probability that a learner plays soccer and either netball, basketball or tennis? Answer Step 1 : Identify what events we are counting We count the events: soccer and netball, soccer and basketball, soccer and tennis. This gives three choices. Step 2 : Calculate the total number of choices There are 6 sports to choose from and we choose 2 sports. There are 6 2 = 6!/(2!(6 − 2)!) = 15 choices. Step 3 : Calculate the probability The probability is the number of events we are counting, divided by the total number of choices. 3 Probability = 15 = 15 = 0,2 45.6 Permutations The concept of a combination did not consider the order of the elements of the subset to be important. A permutation is a combination with the order of a selection from a group being important. For example, for the set {1,2,3,4,5,6}, the combination {1,2,3} would be identical to the combination {3,2,1}, but these two combinations are permutations, because the elements in the set are ordered differently. More formally, a permutation is an ordered list without repetitions, perhaps missing some elements. This means that {1, 2, 2, 3, 4, 5, 6} and {1, 2, 4, 5, 5, 6} are not permutations of the set {1, 2, 3, 4, 5, 6}. Now suppose you have these objects: 1, 2, 3 Here is a list of all permutations of those: 1 2 3; 1 3 2; 2 1 3; 2 3 1; 3 1 2; 3 2 1; 606 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 45.6.1 45.6 Counting Permutations Let S be a set with n objects. Permutations of k objects from this set S refer to sequences of k different elements of S (where two sequences are considered different if they contain the same elements but in a different order, or if they have a different length). Formulas for the number of permutations and combinations are readily available and important throughout combinatorics. It is easy to count the number of permutations of size r when chosen from a set of size n (with r ≤ n). 1. Select the first member of all permutations out of n choices because there are n distinct elements in the set. 2. Next, since one of the n elements has already been used, the second member of the permutation has (n − 1) elements to choose from the remaining set. 3. The third member of the permutation can be filled in (n − 2) ways since 2 have been used already. 4. This pattern continues until there are r members on the permutation. This means that the last member can be filled in (n − (r − 1)) = (n − r + 1) ways. 5. Summarizing, we find that there is a total of n(n − 1)(n − 2)...(n − r + 1) different permutations of r objects, taken from a pool of n objects. This number is denoted by P (n, r) and can be written in factorial notation as: P (n,r) = n! . (n − r)! For example, if we have a total of 5 elements, the integers {1, 2, 3,4,5}, how many ways are there for a permutation of three elements to be selected from this set? In this case, n = 10 and r = 3. Then, P (10,3) = 10!/7! = 720. Worked Example 206: Permutations Question: Show that a collection of n objects has n! permutations. Answer Proof: Constructing an ordered sequence of n objects is equivalent to choosing the position occupied by the first object, then choosing the position of the second object, and so on, until we have chosen the position of each of our n objects. There are n ways to choose a position for the first object. Once its position is fixed, we can choose from (n-1) possible positions for the second object. With the first two placed, there are (n-2) remaining possible positions for the third object; and so on. There are only two positions to choose from for the penultimate object, and the nth object will occupy the last remaining position. Therefore, according to the multiplicative principle, there are n(n − 1)(n − 2)...2 × 1 = n! ways of constructing an ordered sequence of n objects. 607 45.7 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 Permutation with Repetition When order matters and an object can be chosen more than once then the number of permutations is: nr where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have the letters A, B, C, and D and you wish to discover the number of ways of arranging them in three letter patterns (trigrams) you find that there are 43 or 64 ways. This is because for the first slot you can choose any of the four values, for the second slot you can choose any of the four, and for the final slot you can choose any of the four letters. Multiplying them together gives the total. Permutation without Repetition When the order matters and each object can be chosen only once, then the number of permutations is: n! (n − r)! where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have five people and are going to choose three out of these, you will have 5!/(5-3)! = 60 permutations. Note that if n = r (meaning number of chosen elements is equal to number of elements to choose from) then the formula becomes n! n! = = n! (n − n)! 0! For example, if you have three people and you want to find out how many ways you may arrange them it would be 3! or 3 × 2 × 1 = 6 ways. The reason for this is because you can choose from three for the initial slot, then you are left with only two to choose from for the second slot, and that leaves only one for the final slot. Multiplying them together gives the total. 45.7 Applications Extension: The Binomial Theorem In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Its simplest version reads n X n k n−k (x + y) = x y k n k=0 608 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 45.7 Whenever n is any positive integer, the numbers n! n = k k!(n − k)! are the binomial coefficients (the coefficients in front of powers). For example, here are the cases n = 2, n = 3 and n = 4: (x + y)2 = x2 + 2y + y 2 (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 (x + y)4 = x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4 1 1 The coefficients form a triangle, where each number is the sum of the two numbers above it: 1 3 1 1 1 2 1 3 6 4 b b 1 4 1 b This formula, and the triangular arrangement of the binomial coefficients, are often attributed to Blaise Pascal who described them in the 17th century. It was, however, known to the Chinese mathematician Yang Hui in the 13th century, the earlier Persian mathematician Omar Khayym in the 11th century, and the even earlier Indian mathematician Pingala in the 3rd century BC. Worked Example 207: Number Plates Question: The number plate on a car consists of any 3 letters of the alphabet (excluding the vowels and ’Q’), followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that the number plate starts with a ’Y’ and ends with an odd digit? Answer Step 1 : Identify what events are counted The number plate starts with a ’Y’, so there is only 1 choice for the first letter, and ends with an even digit, so there are 5 choices for the last digit (1,3,5,7,9). Step 2 : Find the number of events Use the counting principle. For each of the other letters, there are 20 possible choices (26 in the alphabet, minus 5 vowels and ’Q’) and 10 possible choices for each of the other digits. Number of events = 1 × 20 × 20 × 10 × 10 × 5 = 200 000 Step 3 : Find the number of total possible number plates Use the counting principle. This time, the first letter and last digit can be anything. Total number of choices = 20 × 20 × 20 × 10 × 10 × 10 = 8 000 000 Step 4 : Calculate the probability The probability is the number of events we are counting, divided by the total number of choices. 1 000 Probability = 8200 000 000 = 40 = 0,025 Worked Example 208: Factorial 609 45.8 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 Question: Show that n! =n (n − 1)! Answer Method 1: Expand the factorial notation. n × (n − 1) × (n − 2) × ... × 2 × 1 n! = (n − 1)! (n − 1) × (n − 2) × ... × 2 × 1 Cancelling the common factor of (n − 1) × (n − 2) × ... × 2 × 1 on the top and bottom leaves n. n! So (n−1)! =n n! Method 2: We know that P (n,r) = (n−r)! is the number of permutations of r objects, taken from a pool of n objects. In this case, r = 1. To choose 1 object from n objects, there are n choices. n! =n So (n−1)! 45.8 Exercises 1. Tshepo and Sally go to a restaurant, where the menu is: Starter Main Course Dessert Chicken wings Beef burger Chocolate ice cream Mushroom soup Chicken burger Strawberry ice cream Greek salad Chicken curry Apple crumble Lamb curry Chocolate mousse Vegetable lasagne A How many different combinations (of starter, main meal, and dessert) can Tshepo have? B Sally doesn’t like chicken. How many different combinations can she have? 2. Four coins are thrown, and the outcomes recorded. How many different ways are there of getting three heads? First write out the possibilites, and then use the formula for combinations. 3. The answers in a multiple choice test can be A, B, C, D, or E. In a test of 12 questions, how many different ways are there of answering the test? 4. A girl has 4 dresses, 2 necklaces, and 3 handbags. A How many different choices of outfit (dress, necklace and handbag) does she have? B She now buys two pairs of shoes. How many choices of outfit (dress, necklace, handbag and shoes) does she now have? 5. In a soccer tournament of 9 teams, every team plays every other team. A How many matches are there in the tournament? B If there are 5 boys’ teams and 4 girls’ teams, what is the probability that the first match will be played between 2 girls’ teams? 6. The letters of the word ’BLUE’ are rearranged randomly. How many new words (a word is any combination of letters) can be made? 7. The letters of the word ’CHEMISTRY’ are arranged randomly to form a new word. What is the probability that the word will start and end with a vowel? 610 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 45.8 8. There are 2 History classes, 5 Accounting classes, and 4 Mathematics classes at school. Luke wants to do all three subjects. How many possible combinations of classes are there? 9. A school netball team has 8 members. How many ways are there to choose a captain, vice-captain, and reserve? 10. A class has 15 boys and 10 girls. A debating team of 4 boys and 6 girls must be chosen. How many ways can this be done? 11. A secret pin number is 3 characters long, and can use any digit (0 to 9) or any letter of the alphabet. Repeated characters are allowed. How many possible combinations are there? 611 45.8 CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 612 Appendix A GNU Free Documentation License Version 1.2, November 2002 c 2000,2001,2002 Free Software Foundation, Inc. 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