H2 Mathematics Textbook CHOO YAN MIN
& Answers. Covers both 9740 & 9758 syllabuses. Includes TYS
This version: 24th July 2016. The latest version will always be at this link.
This book is optimised for viewing in PDF format (click the above link). Other existing formats are crude conversions and may be sub-optimal.
Recent changes: First complete draft done. Spent a few days checking for errors. Upcoming changes: None planned.
Page 2, Table of Contents
www.EconsPhDTutor.com
, Errors? Feedback? Email me! , With your help, I plan to keep improving this textbook.
Page 3, Table of Contents
www.EconsPhDTutor.com
Please do not be intimidated by the length of this book (~1,400 pages). The actual main content takes up only about 700+ pages. The other 700 pages are for things like front matter, TYS questions, appendices, reproductions of formula lists and syllabuses, and answers to exercises.
Page 4, Table of Contents
www.EconsPhDTutor.com
This book is licensed under the Creative Commons license CC-BY-NC-SA 4.0.
You are free to: • Share — copy and redistribute the material in any medium or format • Adapt — remix, transform, and build upon the material Under the following terms: • Attribution — You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. • NonCommercial — You may not use the material for commercial purposes. • ShareAlike — If you remix, transform, or build upon the material, you must distribute your contributions under the same license as the original.
Author: Choo, Yan Min. Title: H2 Mathematics Textbook. ISBN: 978-981-11-0383-4 (e-book).
Page 5, Table of Contents
www.EconsPhDTutor.com
The first thing to understand is that mathematics is an art. Paul Lockhart (2009, A Mathematician’s Lament, p. 22).
A mathematician, like a painter or a poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas. ... Beauty is the first test: there is no permanent place in the world for ugly mathematics. - G.H. Hardy (1940 [1967], A Mathematician’s Apology, pp. 84-85).
The scientist does not study nature because it is useful to do so. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful. - Henri PoincarĂŠ (1908 [1914], Science and Method, English trans., p. 22).
Page 6, Table of Contents
www.EconsPhDTutor.com
About This Book This textbook is for Singaporean H2 Maths students (hence the occasional Singlish and TLAs1 ). Of course, I hope that anyone else in the world will also find this useful! I needed a definitive reference for my own teaching needs, but could find nothing satisfying. So I decided to just write my own textbook. This textbook is based exactly on the old (9740) and revised (9758) syllabuses (also reproduced in Part VIII). Do check to make sure which exam you’re taking. The revised syllabus (9758) is the same as the old syllabus (9740), but with noticeable chunks excised and is thus easier.2 9740 (old) examined? 9758 (revised) examined? 2016 Yes. No. 2017 Yes, for the last time. Yes, for the first time. 2018 No. Yes.
SYLLABUS ALERT Where there are any differences between the old and revised syllabuses, I’ll let you know with a yellow box like this.
• FREE! This book is free. But if you paid any money for it, I certainly hope your money is going to me! This book is free because: 1. It is a shameless advertising vehicle for my awesome tutoring services. 2. The marginal cost of reproducing this book is zero. • DONATE! This book may be free, but donations are more than welcome! Donation methods in footnote.3 It’s irrational for Homo economicus to donate. But please consider donating because: 1. You’re a nice human being , [*emotional_manipulation*]. 2. Your donations will encourage me and others to continue producing awesome free content for the world. Three Letter Abbreviations. Indeed, some chunks of the old syllabus (9740) have simply been moved into the syllabus of Further Maths (9649), which the authorities have kindly resurrected for the 2017 exam season. 3 Singapore. POSB Savings Account 174052271 or OCBC Savings Account 5523016383 (Name: Choo Yan Min). International. Bitcoin wallet: 1GDGNAdGZhEq9pz2SaoAdLb1uu34LFwViz. Paypal ychoo@umich.edu (Name: Yan Min Choo, USD preferred because this account was set up in the US). USA. Venmo link (Name: Yanmin Choo).
1
2
Page 7, Table of Contents
www.EconsPhDTutor.com
• HELP ME IMPROVE THIS BOOK! Feel free to email me if: 1. There are any errors in this book. Please let me know even if it’s something as trivial as a spelling mistake or a grammatical error. 2. You have absolutely any suggestions for improvement. 3. Any part of this book is less than crystal clear. Here’s an anecdote about Richard Feynman, the great teacher and physicist: Feynman was once asked by a Caltech faculty member to explain why spin 1/2 particles obey Fermi-Dirac statistics. He gauged his audience perfectly and said, “I’ll prepare a freshman lecture on it.” But a few days later he returned and said, “You know, I couldn’t do it. I couldn’t reduce it to the freshman level. That means we really don’t understand it.” I agree: If you can’t explain something simply, you don’t understand it well enough.4 And as a corollary, the best way to gauge whether you understand something is to see if you can explain it simply to someone else. If at any point in this textbook, you have read the same passage a few times, tried to reason it through, and still find things confusing, then it is a failure on MY part. Please let me know and I will try to rewrite it so that it’s clearer. (There is also the possibility that I simply messed up! So please let me know if there’s anything confusing!) I deeply value any feedback, because I’d like to keep improving this textbook for the benefit of everyone! I am very grateful to all the kind folks who’ve already written in, allowing me to rid this book of more than a few embarrassing errors. • LyX rocks! This book was written using LYX.5 • Is the font size big enough? You’re probably reading this on some device. So I’ve tried to set the font sizes and stuff so that one can comfortably read this on a device as small as a seven-inch tablet. It should also be possible to read this on a phone, though somewhat less comfortably. (Please let me know if you have any feedback about this!) (I’ll probably be contacting some publishers to see if they want to do a print version of this, for anyone who prefers it in print.) This quote or some similar variant is often (mis)attributed to Einstein. But as Einstein himself once said, “73% of Einstein quotes are misattributed.” 5 A L TEX is the typesetting programme used by most economists and scientists. But LATEX can be difficult to use. LYX is a user-friendly GUI version of LATEX. LYX has boosted my productivity by countless hours over the years and you should use LYX too!
4
Page 8, Table of Contents
www.EconsPhDTutor.com
Tips for the Student • Read maths slowly. Reading maths is not like reading Harry Potter. Most of Harry Potter is fluff. There is little fluff in maths. So go slowly. Dwell upon and carefully consider every sentence in this textbook. Make sure you completely understand what each statement says and why it is true. Reading maths is very different from reading any other subject matter. If you don’t quite understand some material, you might be tempted to move forward anyway. Don’t. In maths, later material usually builds on earlier material. So if you simply move forward, this will usually cost you more time and frustration in the long run. Better then to stop right there. Keep working on it until you “get” it. Ask a friend or a teacher for help. Feel free to even email me! (I’m always interested to know what the common points of confusion are and how I can better clear them up.) • Examples and exercises are your best friends. So work through them. A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one.
- Paul Halmos (1983, Google Books). Work through all the examples and exercises. Merely moving your eyeballs is not the same as working. Working means having pencil and paper by your side and going through each example/exercise word-by-word, line-by-line. For example, I might say something like “x2 − y 2 = 0. Thus, (x − y)(x + y) = 0.” If it’s not obvious to you why the first sentence implies the second, stop right there and work on it until you understand why. Don’t just let your eyeballs fly over these sentences and pretend that your brain is “getting” it. I will often not bother to explain some steps, especially if they simply involve some simple algebra. • You get a List of Formulae during the A-level exam. So there’s no need to memorise all the formulae that are already on the list you’re getting. Note that you get a different list depending on which exam you’re taking — List of Formulae (MF15) for the old 9740 exam and List of Formulae (MF26) for the revised 9758 exam. (Both lists are reproduced in Part VIII of this book.) I cannot guarantee though that your JC will give you the List during your JC common tests and exams. Page 9, Table of Contents
www.EconsPhDTutor.com
• Remember your O-Level Additional Mathematics? You’ve probably forgotten some (or most?) of it, but unfortunately, you are still assumed to know EVERYTHING from O-Level ‘A’ Maths. See the lists near the end of either the 9740 (old) or the 9758 (revised) syllabus. Skim through and see if anything looks totally alien to you! Some chapters (e.g. Chapters 5 and 26) in this textbook will give a quick review of some of the O-Level Maths material that you may have forgotten but which we’ll use quite often. • Online Calculators Google is probably the quickest for simple calculations. Type in anything into your browser’s Google search bar and the answer will instantly show up:
Wolfram Alpha is somewhat more advanced (but also slower). Enter “sin x” for example and you’ll get graphs, the derivative, the indefinite integral, the Maclaurin series, and a bunch of other stuff you neither know nor care about. The Derivative Calculator and the Integral Calculator are probably unbeatable for the specific purposes of differentiation and integration. Both give step-by-step solutions for anything you want to differentiate or integrate. Here is a collection of spreadsheets I made. These spreadsheets are for doing tedious and repetitive calculations you’ll often encounter in H2 maths (e.g. with vectors, complex numbers, etc.). As with anything I do, I welcome any feedback you may have about these spreadsheets. Perhaps in the future I will make a more attractive version of it. (Instructions: Click “Make a copy” to open up your own independent copy of this spreadsheet. Enter your input in the yellow cells. Output is produced in the blue cells. If you mess up anything, simply click the same link and “Make a copy” again.)
Page 10, Table of Contents
www.EconsPhDTutor.com
• Other Online Resources There are way too many websites out there catering to primary, secondary, and lower-level undergraduate maths. Unfortunately, some of them can be awful and can get things wrong. Three resources I like (though are probably a bit advanced for JC students) are: 1. Math StackExchange A great resource where you can ask maths questions and often get them answered fairly promptly. Note though that this site is mostly frequented by fairly advanced students of maths (not to mention also mathematicians), so they can be pretty impatient and quick to downvote questions they perceive to be “stupid”. Nonetheless, if you make an effort to write down a carefully-crafted question and show also that you’ve made some effort to look for an answer (either on your own or online), they can be very helpful.6 2. ProofWiki gives succinct and rigorous definitions and proofs. Unfortunately it is very incomplete. 3. Mathworld.Wolfram is also great, but at times excessively encyclopaedic, at the cost of clarity and brevity.
And of course, you can find countless free maths textbooks online (some less legal than others). Two totally illegal7 resources are: LibGen for books and SciHub for articles.8 An old reliable is Bittorrent.
There is an entire StackExchange family of websites. The flagship site is StackOverflow where you can ask any programming question and get it answered amazingly quickly. 7 Well, depending on which jurisdiction you live in. Of course, in Singapore, unless told otherwise, you should assume that everything is illegal. 8 Note though that these sites are constantly playing whac-a-mole with the fascist authorities so the URLs often change — if so, simply google to look up the current URLs.
6
Page 11, Table of Contents
www.EconsPhDTutor.com
Preface Divide students into two extremes: 1. Type #1 is happy to get an A, even if this means learning absolutely nothing. 2. Type #2 would rather learn a lot, even if this means getting a C. The good Singaporean is trained to view pragmatism is the highest virtue (and obedience second). She is thus also trained to be a Type #1 student (and indeed a Type #1 human being). If you’re a Type #1 student, then this textbook may not be the best use of your time (though you may still find the TYS and answers useful). Please use instead these three resources, which are provided with the efficient Type #1 student in mind: • The H2 Mathematics CheatSheet, which contains all the formulae you’ll ever need on two sides of a single A4 sheet of paper.9 • The H2 Maths Exercise Book (coming soon), which teaches you how to mindlessly apply formulae and give the “correct” answer to every exam question. • My totally awesome tuition classes! Of course, it is fully intended that this textbook (complemented by a capable teacher) will help any student get her A. But that for me is quite beside the point. My broader goal in writing this textbook is to impart genuine understanding — or at least as much as is possible, within the stultifying confines of the A-level syllabus. Maths education in Singapore is at least every bit as “stupid and boring” and “formulaic” and “mindless” as in the US.10 But at least the average US student has the consolation that only a very small portion of her life will have been squandered on “mindless” “pseudomaths”. The same cannot be said for the average Singaporean student. By the time she turns 18, she will have — just for the subject of maths alone — clocked many thousands of hours attending classes; doing homework; doing practice exam questions; doing assessment books, Ten Year Series; going to tuition classes; taking common tests, promos, prelims, one big exam after another; etc. This textbook is for the Singaporean A-level student. So a good deal of “mindless formulae” is unavoidable. But at the same time, I try in this textbook to give the student a tiny glimpse of what maths really is — “the art of explanation”.11 So for example, this textbook explains Two things: (1) This CheatSheet does not include many of the formulae already printed in List MF26. (2) It is written for 9758 (revised) students (so 9740 students may find a few things missing). 10 At least as described by Paul Lockhart, in A Mathematician’s Lament. 11 Lockhart, p. 29.
9
Page 12, Table of Contents
www.EconsPhDTutor.com
• A bit of intuition behind differentiation, integration, and the Fundamental Theorems of Calculus. (To get an A, no understanding of these is necessary. Instead, one need merely know how to “do” differentiation and integration problems.) • Why the Central Limit Theorem is so amazing. (To get an A, one need merely treat the CLT as yet another mysterious mathematical trick that helps solve exam questions. No appreciation of why it is wonderful is necessary.) • A bit of intuition behind the Maclaurin series. (To get an A, it suffices to know how to mindlessly apply this strange formula that falls out from the sky.) • Why it is terribly wrong to believe that “a high correlation coefficient means a good model”. (Yet this is exactly what the writers of the A-level exams seem to believe. See Section 73.9.) Once upon a time, I had the misfortune of being a Singaporean JC student myself. I remember being deeply mystified by why the scalar (or dot) product, despite having a simple algebraic definition, could at the same time also tell us about the cosine of the angle between the two vectors. I never figured it out,12 but it didn’t matter, because this was simply “yet another formula” that we were required to know, for the sole purpose of answering exam questions. I remember being confused about the difference between the sample mean, the mean of the sample mean, the variance of the sample mean, and the sample variance. But this confusion didn’t matter, because once again, all we needed to do to get an A was to mindlessly apply formulae and algorithms. Monkey see, monkey do. This textbook is thus partly in response to my unhappy and unsatisfactory experience as a maths student in Singapore. Almost all results are proven. I often try to supply the intuition for each result in the simplest possible terms. Many proofs are relegated to the appendices, but where a proof is especially simple and beautiful, I encourage the student to savour it by leaving it in the main text. In the rare instances where proofs are entirely omitted from this book — usually because they are too advanced — I make sure to clearly state so, lest the student wonder whether the result is supposed to be obvious. Finally, I also hope that this textbook will serve as an authoritative resource to which teachers and students alike can refer.
12
The internet was, at that time, not so well-developed, so one could not easily find answers online.
Page 13, Table of Contents
www.EconsPhDTutor.com
Tuition Ad I give tuition for the following at any level: • Economics. • Mathematics. • Writing, English, General Paper. I have a PhD in economics (University of Michigan) and have been teaching and tutoring since 2010.
For more information, please visit:
www.EconsPhDTutor.com Or email:
DrChooYanMin@gmail.com
Contents About This Book
7
Tips for the Student
9
Preface
12
I
35
Functions and Graphs
1 Sets
36
1.1
In ∈ and Not In ∉ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
1.2
Greater than >, Less Than <, Positive > 0, and Negative < 0 . . . . . . . . . .
38
1.3
Types of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
1.4
The Order of the Elements Doesn’t Matter . . . . . . . . . . . . . . . . . . . .
40
1.5
Repeated Elements Don’t Count . . . . . . . . . . . . . . . . . . . . . . . . . .
41
1.6
Ellipsis . . . Means Continue in the Obvious Fashion . . . . . . . . . . . . . . .
42
1.7
Sets can be Finite or Infinite . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
1.8
Special Names of Sets: Z, Q, R, and C . . . . . . . . . . . . . . . . . . . . . . .
44
1.9
Special Names of Sets: Intervals
. . . . . . . . . . . . . . . . . . . . . . . . . .
45
1.10 Special Names of Sets: The Empty Set ∅ . . . . . . . . . . . . . . . . . . . . .
47
1.11 Subset Of ⊆ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
1.12 Proper Subset Of ⊂ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
1.13 Union ∪ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
1.14 Intersection ∩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
1.15 Set Minus / . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
1.16 Set Complement A′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
1.17 Set-Builder Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
2 Dividing By Zero
Page 15, Table of Contents
55
www.EconsPhDTutor.com
3 Functions
56
3.1
Formal Mathematical Notation for Functions
. . . . . . . . . . . . . . . . . .
58
3.2
EVERY x ∈ D Must be Mapped to EXACTLY ONE y ∈ C . . . . . . . . . . .
62
3.3
Real-Valued Functions of a Real Variable . . . . . . . . . . . . . . . . . . . . .
64
3.4
The Range of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
3.5
Creating New Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66
3.6
One-to-One Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
3.7
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
3.8
Domain Restriction to Create an Invertible Function . . . . . . . . . . . . . .
73
3.9
Composite Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
4 Graphs 4.1
77
Graphing with Your TI84 Graphing Calculator . . . . . . . . . . . . . . . . . .
5 Quick Revision: Exponents, Surds, Absolute Value
83 86
5.1
Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
5.2
Rationalising the Denominator of a Surd . . . . . . . . . . . . . . . . . . . . .
87
5.3
Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
6 Intercepts
90
7 Symmetry
93
7.1
Reflection of a Point in a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
7.2
Reflection of a Graph in a Line . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
7.3
Lines of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
8 Limits, Continuity, and Asymptotes
99
8.1
Limits: Introduction and Examples . . . . . . . . . . . . . . . . . . . . . . . . .
8.2
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
8.3
Limits: More Examples
8.4
Infinite Limits and Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . 109
8.5
Limits at Infinity, Horizontal and Oblique Asymptotes . . . . . . . . . . . . . 111
Page 16, Table of Contents
99
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
www.EconsPhDTutor.com
9 Differentiation
114
9.1
Motivation: The Derivative as Slope of the Tangent . . . . . . . . . . . . . . . 114
9.2
Lagrange’s, Leibniz’s, and Newton’s Notation . . . . . . . . . . . . . . . . . . . 117
9.3
The Derivative is a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
9.4
Second and Higher-Order Derivatives
. . . . . . . . . . . . . . . . . . . . . . . 121
9.5
More About Leibniz’s Notation: The
d Operator . . . . . . . . . . . . . . . . 123 dx
9.6
Standard Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 124
9.7
Differentiable and Twice-Differentiable Functions . . . . . . . . . . . . . . . . 127
9.8
Differentiability Implies (i.e. is Stronger Than) Continuity . . . . . . . . . . . 130
9.9
Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
10 Increasing, Decreasing, and f ′
133
10.1 When a Function is Increasing or Decreasing . . . . . . . . . . . . . . . . . . . 133 10.2 The First Derivative Increasing/Decreasing Test . . . . . . . . . . . . . . . . . 134 11 Extreme, Stationary, and Turning Points
135
11.1 Maximum and Minimum Points . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.2 Global Maximum and Minimum Points . . . . . . . . . . . . . . . . . . . . . . 139 11.3 Stationary and Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 11.4 The Interior Extremum Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 145 11.5 How to Find Maximum and Minimum Points . . . . . . . . . . . . . . . . . . . 147 12 Concavity, Inflexion Points, and the 2DT
151
12.1 The Second Derivative Test (2DT) . . . . . . . . . . . . . . . . . . . . . . . . . 155 12.2 Summary of Points and Venn Diagram
. . . . . . . . . . . . . . . . . . . . . . 157
13 Relating the Graph of f ′ to that of f
159
14 Quick Revision: Quadratic Equations y = ax2 + bx + c
162
Page 17, Table of Contents
www.EconsPhDTutor.com
15 Transformations
166
15.1 y = f (x) + a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 15.2 y = f (x + a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 15.3 y = af (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 15.4 y = f (ax) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 15.5 Combinations of the Above . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 15.6 y = ∣f (x)∣ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 15.7 y = f (∣x∣) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 15.8 y =
1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 f (x)
15.9 y 2 = f (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 16 Conic Sections
176
16.1 The Ellipse x2 + y 2 = 1 (The Unit Circle) . . . . . . . . . . . . . . . . . . . . . . 179 x2 y 2 16.2 The Ellipse 2 + 2 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 a b 16.3 The Hyperbola: y = 1/x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 16.4 The Hyperbola x2 − y 2 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 x2 y 2 16.5 The Hyperbola 2 − 2 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 a b 16.6 The Hyperbola
y 2 x2 − = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 b2 a2
16.7 Long Division of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 16.8 The Hyperbola y =
bx + c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 dx + e
ax2 + bx + c . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 16.9 The Hyperbola y = dx + e 17 Simple Parametric Equations
203
17.1 Eliminating the Parameter t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Page 18, Table of Contents
www.EconsPhDTutor.com
18 Equations and Inequalities
211
18.1
ax + b > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 cx + d
18.2
ax2 + bx + c > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 dx2 + ex + f
18.3 Solving Inequalities by Graphical Methods . . . . . . . . . . . . . . . . . . . . 218 18.4 Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
II
Sequences and Series
19 Finite Sequences
224 225
19.1 A Corresponding Function for a Sequence . . . . . . . . . . . . . . . . . . . . . 226 19.2 Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 19.3 Creating New Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 20 Infinite Sequences
231
20.1 Creating New Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 21 Series
233
21.1 Convergent and Divergent Sequences and Series . . . . . . . . . . . . . . . . . 234 22 Summation Notation Σ
236
23 Arithmetic Sequences and Series
240
23.1 Finite Arithmetic Sequences and Series . . . . . . . . . . . . . . . . . . . . . . 241 24 Geometric Sequences and Series
242
24.1 Finite Geometric Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . 243 24.2 Infinite Geometric Sequences and Series . . . . . . . . . . . . . . . . . . . . . . 244 25 Proof by the Method of Mathematical Induction
245
III
251
Vectors
Page 19, Table of Contents
www.EconsPhDTutor.com
26 Quick Revision of Some O-Level Maths
252
26.1 Lines vs. Line Segments vs. Rays . . . . . . . . . . . . . . . . . . . . . . . . . . 252 26.2 Angles - Acute, Right, Obtuse, Straight, Reflex . . . . . . . . . . . . . . . . . . 253 26.3 Triangles - Acute, Right, Obtuse . . . . . . . . . . . . . . . . . . . . . . . . . . 254 26.4 Sine, Cosine, Tangent - Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.5 Sine, Cosine, Tangent - Values and Graphs . . . . . . . . . . . . . . . . . . . . 256 26.6 Formulae for Sine, Cosine, and Tangent . . . . . . . . . . . . . . . . . . . . . . 257 26.7 Arcsine, Arccosine, Arctangent . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 26.8 The Law of Sines and the Law of Cosines . . . . . . . . . . . . . . . . . . . . . 262 27 Vectors in Two Dimensions (2D)
264
27.1 Sum and Difference of Points and Vectors . . . . . . . . . . . . . . . . . . . . . 269 27.2 Sum, Additive Inverse, and Difference of Vectors . . . . . . . . . . . . . . . . . 272 27.3 Displacement Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 27.4 Length (or Magnitude) of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . 276 27.5 Scalar Multiplication of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . 278 27.6 Unit Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 27.7 The Ratio Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 28 Scalar Product
284
28.1 The Angle between Two Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 28.2 Projection of One Vector on Another . . . . . . . . . . . . . . . . . . . . . . . . 289 28.3 Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 29 Vectors in 3D
293
30 Vector Product
297
30.1 Vector Product in 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 30.2 Areas of Triangles and Parallelograms . . . . . . . . . . . . . . . . . . . . . . . 299 30.3 Vector Product in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 Page 20, Table of Contents
www.EconsPhDTutor.com
31 Lines
305
31.1 Lines on a 2D Plane: Cartesian to Vector Equations . . . . . . . . . . . . . . 305 31.2 Lines on a 2D Plane: Vector to Cartesian Equations . . . . . . . . . . . . . . 310 31.3 Lines in 3D Space: Vector Equations . . . . . . . . . . . . . . . . . . . . . . . . 312 31.4 Lines in 3D Space: Vector to and from Cartesian Equations . . . . . . . . . . 314 31.5 Collinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 32 Planes
323
32.1 Planes: Vector to Cartesian Equations . . . . . . . . . . . . . . . . . . . . . . . 330 32.2 Planes: Hessian Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 33 Distances
333
33.1 Distance of a Point from a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 33.2 Distance of a Point from a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 341 34 Angles
345
34.1 Angle between Two Lines (2D) . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 34.2 Angle between Two Lines (3D) . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 34.3 Angle between A Line and a Plane . . . . . . . . . . . . . . . . . . . . . . . . . 353 34.4 Angle between Two Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 35 Relationships between Lines and Planes
357
35.1 Relationship between Two Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 35.2 Relationship between a Line and a Plane . . . . . . . . . . . . . . . . . . . . . 361 35.3 Relationship between Two Planes . . . . . . . . . . . . . . . . . . . . . . . . . . 363 35.4 Relationship between Three Planes . . . . . . . . . . . . . . . . . . . . . . . . . 368
IV
Complex Numbers
36 Complex Numbers: Introduction
374 375
36.1 The Real and Imaginary Parts of Complex Numbers . . . . . . . . . . . . . . 378 Page 21, Table of Contents
www.EconsPhDTutor.com
37 Basic Arithmetic of Complex Numbers
379
37.1 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 37.2 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 37.3 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 38 Solving Polynomial Equations
384
38.1 Complex Roots to Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . 384 38.2 The Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . . . . 386 38.3 The Complex Conjugate Roots Theorem . . . . . . . . . . . . . . . . . . . . . . 389 39 The Argand Diagram
390
39.1 Complex Numbers in Polar Form . . . . . . . . . . . . . . . . . . . . . . . . . . 392 39.2 Complex Numbers in Exponential Form . . . . . . . . . . . . . . . . . . . . . . 397 40 More Arithmetic of Complex Numbers
398
40.1 The Product of Two Complex Numbers . . . . . . . . . . . . . . . . . . . . . . 398 40.2 The Ratio of Two Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . 401 40.3 Sine and Cosine as Weighted Sums of the Exponential . . . . . . . . . . . . . 403 41 Geometry of Complex Numbers
406
41.1 The Sum and Difference of Two Complex Numbers . . . . . . . . . . . . . . . 406 41.2 The Product and Ratio of Two Complex Numbers . . . . . . . . . . . . . . . . 408 41.3 Conjugating a Complex Number . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 42 Loci Involving Cartesian Equations
411
42.1 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 42.2 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 42.3 Intersection of Lines and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . 420
Page 22, Table of Contents
www.EconsPhDTutor.com
43 Loci Involving Complex Equations
423
43.1 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 43.2 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 43.3 Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 43.4 Quick O-Level Revision: Properties of The Circle . . . . . . . . . . . . . . . . 429 44 De Moivreâ&#x20AC;&#x2122;s Theorem
432
44.1 Powers of a Complex Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 44.2 Roots of a Complex Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
V
Calculus
45 Solving Problems Involving Differentiation
441 442
45.1 Inverse Function Theorem (IFT) . . . . . . . . . . . . . . . . . . . . . . . . . . 442 45.2 Differentiation of Simple Parametric Functions
. . . . . . . . . . . . . . . . . 443
45.3 Equations of Tangents and Normals . . . . . . . . . . . . . . . . . . . . . . . . 444 45.4 Connected Rates of Change Problems . . . . . . . . . . . . . . . . . . . . . . . 445 45.5 Finding Max/Min Points on the TI84 . . . . . . . . . . . . . . . . . . . . . . . 447 45.6 Finding the Derivative at a Point on the TI84 . . . . . . . . . . . . . . . . . . 449 46 The Maclaurin Series
451
46.1 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451 46.2 Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 46.3 The Amazing Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 46.4 Finite-Order Maclaurin Series as Approximations . . . . . . . . . . . . . . . . 456 46.5 Product of Two Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460 46.6 Composition of Two Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462 46.7 How the Maclaurin Series Works (Optional) . . . . . . . . . . . . . . . . . . . . 465
Page 23, Table of Contents
www.EconsPhDTutor.com
47 The Indefinite Integral
466
47.1 The Constant of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 47.2 The Indefinite Integral is Unique Up to the C.O.I. . . . . . . . . . . . . . . . . 469 48 Integration Techniques
470
48.1 Basic Rules of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 48.2 More Basic Rules of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 48.3 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 48.4 Integration by Substitution (IBS) . . . . . . . . . . . . . . . . . . . . . . . . . . 475 48.5 Integration by Parts (IBP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480 49 The Fundamental Theorems of Calculus (FTCs)
481
49.1 The Area Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 49.2 The First Fundamental Theorem of Calculus (FTC1) . . . . . . . . . . . . . . 486 49.3 The Definite (or Riemann) Integral
. . . . . . . . . . . . . . . . . . . . . . . . 490
49.4 The Second Fundamental Theorem of Calculus (FTC2) . . . . . . . . . . . . . 491 50 Definite Integrals
492
50.1 Area between a Curve and Lines Parallel to Axes . . . . . . . . . . . . . . . . 493 50.2 Area between a Curve and a Line . . . . . . . . . . . . . . . . . . . . . . . . . . 494 50.3 Area between Two Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 50.4 Area below the x-Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 50.5 Area under a Parametrically-Defined Curve
. . . . . . . . . . . . . . . . . . . 497
50.6 Volume of Rotation about the y- or x-Axis . . . . . . . . . . . . . . . . . . . . 498 50.7 Finding Definite Integrals on your TI84 . . . . . . . . . . . . . . . . . . . . . . 501 51 Differential Equations dy = f (x) . . . . . . . dx dy 51.2 = f (y) . . . . . . . dx d2 y 51.3 = f (x) . . . . . . . dx2 51.4 Word Problems . . . 51.1
502 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
51.5 Family of Solution Curves to Represent the General Solution . . . . . . . . . 510 Page 24, Table of Contents
www.EconsPhDTutor.com
VI
Probability and Statistics
52 How to Count: Four Principles
511 512
52.1 How to Count: The Addition Principle . . . . . . . . . . . . . . . . . . . . . . . 513 52.2 How to Count: The Multiplication Principle . . . . . . . . . . . . . . . . . . . 516 52.3 How to Count: The Inclusion-Exclusion Principle . . . . . . . . . . . . . . . . 520 52.4 How to Count: The Complements Principle . . . . . . . . . . . . . . . . . . . . 522 53 How to Count: Permutations
523
53.1 Permutations with Repeated Elements . . . . . . . . . . . . . . . . . . . . . . . 527 53.2 Circular Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 53.3 Partial Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536 53.4 Permutations with Restrictions . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 54 How to Count: Combinations
539
54.1 Pascalâ&#x20AC;&#x2122;s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542 54.2 The Combination as Binomial Coefficient . . . . . . . . . . . . . . . . . . . . . 543 54.3 The Number of Subsets of a Set is 2n . . . . . . . . . . . . . . . . . . . . . . . . 545 55 Probability: Introduction
547
55.1 Mathematical Modelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547 55.2 The Experiment as a Model of Scenarios Involving Chance . . . . . . . . . . . 549 55.3 The Kolmogorov Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 55.4 Implications of the Kolmogorov Axioms . . . . . . . . . . . . . . . . . . . . . . 555 56 Probability: Conditional Probability
557
56.1 The Conditional Probability Fallacy (CPF) . . . . . . . . . . . . . . . . . . . . 559 56.2 Two-Boys Problem (Fun, Optional) . . . . . . . . . . . . . . . . . . . . . . . . . 564 57 Probability: Independence
566
57.1 Warning: Not Everything is Independent . . . . . . . . . . . . . . . . . . . . . 571 57.2 Probability: Independence of Multiple Events . . . . . . . . . . . . . . . . . . . 573 Page 25, Table of Contents
www.EconsPhDTutor.com
58 Fun Probability Puzzles
574
58.1 The Monty Hall Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574 58.2 The Birthday Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577 59 Random Variables: Introduction
578
59.1 A Random Variable vs. Its Observed Values . . . . . . . . . . . . . . . . . . . 579 59.2 X = k Denotes the Event {s â&#x2C6;&#x2C6; S â&#x2C6;ś X(s) = k} . . . . . . . . . . . . . . . . . . . . 580 59.3 The Probability Distribution of a Random Variable . . . . . . . . . . . . . . . 581 59.4 Random Variables Are Simply Functions . . . . . . . . . . . . . . . . . . . . . 584 60 Random Variables: Independence
586
61 Random Variables: Expectation
589
61.1 The Expected Value of a Constant R.V. is Constant . . . . . . . . . . . . . . . 592 61.2 The Expectation Operator is Linear . . . . . . . . . . . . . . . . . . . . . . . . 594 62 Random Variables: Variance
596
62.1 The Variance of a Constant R.V. is 0 . . . . . . . . . . . . . . . . . . . . . . . . 603 62.2 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 62.3 The Variance Operator is Not Linear . . . . . . . . . . . . . . . . . . . . . . . . 605 62.4 The Definition of the Variance (Optional) . . . . . . . . . . . . . . . . . . . . . 607 63 The Coin-Flips Problem (Fun, Optional)
608
64 The Bernoulli Trial and the Bernoulli Distribution
609
64.1 Mean and Variance of the Bernoulli Random Variable . . . . . . . . . . . . . 611 65 The Binomial Distribution
612
65.1 Probability Distribution of the Binomial R.V. . . . . . . . . . . . . . . . . . . 614 65.2 The Mean and Variance of the Binomial Random Variable
Page 26, Table of Contents
. . . . . . . . . . 615
www.EconsPhDTutor.com
66 The Poisson Distribution
617
66.1 Formal Definition of the Poisson Random Variable . . . . . . . . . . . . . . . 619 66.2 When is the Poisson Random Variable an Appropriate Model? . . . . . . . . 620 66.3 The Mean and Variance of the Poisson Random Variable
. . . . . . . . . . . 623
66.4 The Poisson Distribution as an Approximation of the Binomial Distribution 624 66.5 The Sum of Two Independent Poisson R.V.’s is a Poisson R.V. . . . . . . . . 627 67 The Continuous Uniform Distribution
630
67.1 The Continuous Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . 631 67.2 The Cumulative Distribution Function (CDF) . . . . . . . . . . . . . . . . . . 633 67.3 Important Digression: P (X ≤ k) = P (X < k) . . . . . . . . . . . . . . . . . . . 634 67.4 The Probability Density Function (PDF) . . . . . . . . . . . . . . . . . . . . . 635 68 The Normal Distribution
636
68.1 The Normal Distribution, in General . . . . . . . . . . . . . . . . . . . . . . . . 642 68.2 Sum of Independent Normal Random Variables . . . . . . . . . . . . . . . . . 651 68.3 The Central Limit Theorem and The Normal Approximation . . . . . . . . . 655 68.3.1 Normal Approximation to the Binomial Distribution 68.3.2 Normal Approximation to the Poisson Distribution 69 The CLT is Amazing (Optional)
. . . . . . . . . 658 . . . . . . . . . . 659 660
69.1 The Normal Distribution in Nature . . . . . . . . . . . . . . . . . . . . . . . . . 660 69.2 Illustrating the Central Limit Theorem (CLT) . . . . . . . . . . . . . . . . . . 664 69.3 Why Are So Many Things Normally Distributed? . . . . . . . . . . . . . . . . 668 69.4 Don’t Assume That Everything is Normal . . . . . . . . . . . . . . . . . . . . . 669 70 Statistics: Introduction (Optional)
675
70.1 Probability vs. Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675 70.2 Objectivists vs Subjectivists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 676
Page 27, Table of Contents
www.EconsPhDTutor.com
71 Sampling
678
71.1 Population . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678 71.2 Population Mean and Population Variance . . . . . . . . . . . . . . . . . . . . 679 71.3 Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 680 71.4 Distribution of a Population . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 681 71.5 A Random Sample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682 71.6 Sample Mean and Sample Variance . . . . . . . . . . . . . . . . . . . . . . . . . 684 71.7 Sample Mean and Sample Variance are Unbiased Estimators . . . . . . . . . 690 71.8 The Sample Mean is a Random Variable
. . . . . . . . . . . . . . . . . . . . . 693
71.9 The Distribution of the Sample Mean . . . . . . . . . . . . . . . . . . . . . . . 694 71.10Non-Random Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695 71.11Stratified, Quota, and Systematic Sampling . . . . . . . . . . . . . . . . . . . . 696 72 Null Hypothesis Significance Testing (NHST)
701
72.1 One-Tailed vs Two-Tailed Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 72.2 The Abuse of NHST (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . 708 72.3 Common Misinterpretations of the Margin of Error (Optional) . . . . . . . . 709 72.4 Critical Region and Critical Value . . . . . . . . . . . . . . . . . . . . . . . . . . 712 72.5 Testing of a Population Mean 2 (Small Sample, Normal Distribution, σ Known) . . . . . . . . . . . . . . . . . 714 72.6 Testing of a Population Mean (Large Sample, Any Distribution, σ 2 Known) . . . . . . . . . . . . . . . . . . . 716 72.7 Testing of a Population Mean 2 (Large Sample, Any Distribution, σ Unknown) . . . . . . . . . . . . . . . . . 718 72.8 Testing of a Population Mean 2 (Small Sample, Normal Distribution, σ Unknown) . . . . . . . . . . . . . . . 720 72.9 Formulation of Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724
Page 28, Table of Contents
www.EconsPhDTutor.com
73 Correlation and Linear Regression
725
73.1 Bivariate Data and Scatter Diagrams . . . . . . . . . . . . . . . . . . . . . . . . 725 73.2 Product Moment Correlation Coefficient (PMCC) . . . . . . . . . . . . . . . . 727 73.3 Correlation Does Not Imply Causation (Optional) . . . . . . . . . . . . . . . . 733 73.4 Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734 73.5 Ordinary Least Squares (OLS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736 73.6 TI84 to Calculate the PMCC and the OLS Estimates . . . . . . . . . . . . . . 741 73.7 Interpolation and Extrapolation . . . . . . . . . . . . . . . . . . . . . . . . . . . 743 73.8 Transformations to Achieve Linearity . . . . . . . . . . . . . . . . . . . . . . . . 751 73.9 The Higher the PMCC, the Better the Model? . . . . . . . . . . . . . . . . . . 755
VII
Ten-Year Series
757
74 Past-Year Questions for Part I: Functions and Graphs
758
75 Past-Year Questions for Part II: Sequences and Series
769
76 Past-Year Questions for Part III: Vectors
779
77 Past-Year Questions for Part IV: Complex Numbers
788
78 Past-Year Questions for Part V: Calculus
794
79 Past-Year Questions for Part VI: Prob. and Stats.
823
VIII
854
Syllabuses and Lists of Formulae
80 Revised (9758) Syllabus
855
81 New List of Formulae (MF26)
876
82 Old (9740) Syllabus
889
Page 29, Table of Contents
www.EconsPhDTutor.com
83 Old List of Formulae (MF15)
908
IX
921
Appendices (Optional)
84 Appendices for Part I: Functions and Graphs
922
84.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922 84.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923 84.3 Reflection in a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924 84.4 The Hyperbola y =
bx + c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926 dx + e
84.5 The Hyperbola y =
ax2 + bx + c . . . . . . . . . . . . . . . . . . . . . . . . . . . 927 dx + e
85 Appendices for Part II: Sequences and Series
929
86 Appendices for Part III: Vectors
930
86.1 Vectors in 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 930 86.2 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 931 86.3 The Ratio Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 932 86.4 Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933 86.5 2D Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936 86.6 3D Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937 87 Appendices for Part IV: Complex Numbers
943
88 Appendices for Part V: Calculus
946
88.1 Limits Formally Defined . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946 88.2 Left- and Right-Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 948 88.3 Infinite Limits and Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . 949 88.4 Limits at Infinity, Horizontal, and Oblique Asymptotes
. . . . . . . . . . . . 950
88.5 Limit Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 951 88.6 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954 Page 30, Table of Contents
www.EconsPhDTutor.com
88.7 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955 88.8 Differentiability Implies Continuity . . . . . . . . . . . . . . . . . . . . . . . . . 961 88.9 Maximum, Minimum, and Turning Points . . . . . . . . . . . . . . . . . . . . . 962 88.10Concavity and Inflexion Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 964 88.11Concavity and Inflexion Points with Differentiability . . . . . . . . . . . . . . 966 88.12Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 969 88.13Parametric Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 970 88.14Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 971 88.15Product of Two Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975 88.16Composition of Two Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977 88.17The Fundamental Theorems of Calculus . . . . . . . . . . . . . . . . . . . . . . 978 88.18The Natural Logarithm and Eulerâ&#x20AC;&#x2122;s Number e . . . . . . . . . . . . . . . . . . 982 88.19Eulerâ&#x20AC;&#x2122;s Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 984 89 Appendices for Part VI: Probability and Statistics
986
89.1 How to Count . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 986 89.2 Circular Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 988 89.3 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 989 89.4 Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 990 89.5 The Poisson Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994 89.6 The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 996 89.7 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999 89.8 Null Hypothesis Significance Testing . . . . . . . . . . . . . . . . . . . . . . . . 1001 89.9 Calculating the Margin of Error . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002 89.10Correlation and Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . 1004 89.10.1 Deriving a Linear Model from the Barometric Formula . . . . . . . . . 1006
X
Answers to Exercises
Page 31, Table of Contents
1007
www.EconsPhDTutor.com
90 Answers to Exercises in Part I: Functions and Graphs
1008
90.1 Answers to Exercises in Ch. 1: Sets . . . . . . . . . . . . . . . . . . . . . . . . . 1008 90.2 Answers to Exercises in Ch. 2: Dividing by Zero . . . . . . . . . . . . . . . . . 1010 90.3 Answers to Exercises in Ch. 3: Functions . . . . . . . . . . . . . . . . . . . . . 1011 90.4 Answers to Exercises in Ch. 4. Graphs . . . . . . . . . . . . . . . . . . . . . . . 1017 90.5 Answers to Exercises in Ch. 5. Quick Revision . . . . . . . . . . . . . . . . . . 1020 90.6 Answers to Exercises in Ch. 6. Intercepts . . . . . . . . . . . . . . . . . . . . . 1022 90.7 Answers to Exercises in Ch. 7. Symmetry . . . . . . . . . . . . . . . . . . . . . 1023 90.8 Answers to Exercises in Ch. 8. Limits, Continuity, and Asymptotes . . . . . 1024 90.9 Answers to Exercises in Ch. 9. Differentiation . . . . . . . . . . . . . . . . . . 1025 90.10Answers to Exercises in Ch. 11. Stationary, Maximum, Minimum, and Inflexion Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1027 90.11Answers to Exercises in Ch. 14. Quadratic Equations . . . . . . . . . . . . . . 1035 90.12Answers to Exercises in Ch. 15. Transformations . . . . . . . . . . . . . . . . 1036 90.13Answers to Exercises in Ch. 16: Conic Sections . . . . . . . . . . . . . . . . . 1038 90.14Answers to Exercises in Ch. 17. Simple Parametric Equations . . . . . . . . . 1050 90.15Answers to Exercises in Ch. 18: Equations and Inequalities . . . . . . . . . . 1055 91 Answers to Exercises in Part II: Sequences and Series
1068
91.1 Answers for Ch. 19: Finite Sequences . . . . . . . . . . . . . . . . . . . . . . . 1068 91.2 Answers for Ch. 20: Infinite Sequences . . . . . . . . . . . . . . . . . . . . . . . 1070 91.3 Answers for Ch. 22: Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1071 91.4 Answers for Ch. 23: Arithmetic Sequences and Series . . . . . . . . . . . . . . 1072 91.5 Answers for Ch. 24: Geometric Sequences and Series . . . . . . . . . . . . . . 1073 91.6 Answers for Ch. 25: Proof by Induction . . . . . . . . . . . . . . . . . . . . . . 1074
Page 32, Table of Contents
www.EconsPhDTutor.com
92 Answers to Exercises in Part III: Vectors
1077
92.1 Answers for Ch. 27: Vectors in 2D . . . . . . . . . . . . . . . . . . . . . . . . . 1077 92.2 Answers for Ch. 29: Vectors in 3D . . . . . . . . . . . . . . . . . . . . . . . . . 1082 92.3 Answers for Ch. 30: Vector Product . . . . . . . . . . . . . . . . . . . . . . . . 1085 92.4 Answers for Ch. 31: Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086 92.5 Answers for Ch. 32: Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1089 92.6 Answers for Ch. 33: Distances . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1091 92.7 Answers for Ch. 34: Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1098 92.8 Answers for Ch. 35: Relationships between Lines and Planes . . . . . . . . . 1101 93 Answers to Exercises in Part IV: Complex Numbers
1105
93.1 Answers for Ch. 36: Introduction to Complex Numbers . . . . . . . . . . . . . 1105 93.2 Answers for Ch. 37: Basic Arithmetic of Complex Numbers . . . . . . . . . . 1106 93.3 Answers for Ch. 38: Solving Polynomial Equations . . . . . . . . . . . . . . . 1108 93.4 Answers for Ch. 39: The Argand Diagram . . . . . . . . . . . . . . . . . . . . 1111 93.5 Answers for Ch. 40: More Arithmetic of Complex Numbers . . . . . . . . . . 1114 93.6 Answers for Ch. 41: Geometry of Complex Numbers . . . . . . . . . . . . . . 1117 93.7 Answers for Ch. 42: Loci Involving Cartesian Equations . . . . . . . . . . . . 1118 93.8 Answers for Ch. 43: Loci Involving Complex Equations . . . . . . . . . . . . . 1122 93.9 Answers for Ch. 44: De Moivreâ&#x20AC;&#x2122;s Theorem . . . . . . . . . . . . . . . . . . . . 1125 94 Answers to Exercises in Part V: Calculus
1128
94.1 Answers for Ch. 45: Solving Problems Involving Differentiation . . . . . . . . 1128 94.2 Answers for Ch. 46: Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . 1130 94.3 Answers for Ch. 47: The Indefinite Integral . . . . . . . . . . . . . . . . . . . . 1133 94.4 Answers for Ch. 48: Integration Techniques . . . . . . . . . . . . . . . . . . . . 1134 94.5 Answers for Ch. 49: The Fundamental Theorems of Calculus . . . . . . . . . 1143 94.6 Answers for Ch. 50: Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . 1144 94.7 Answers for Ch. 51: Differential Equations . . . . . . . . . . . . . . . . . . . . 1147 Page 33, Table of Contents
www.EconsPhDTutor.com
95 Answers to Exercises in Part VI: Probability and Statistics
1152
95.1 Answers for Ch. 52: How to Count: Four Principles . . . . . . . . . . . . . . . 1152 95.2 Answers for Ch. 53: How to Count: Permutations . . . . . . . . . . . . . . . . 1155 95.3 Answers for Ch. 54: How to Count: Combinations . . . . . . . . . . . . . . . . 1157 95.4 Answers for Ch. 55: Probability: Introduction . . . . . . . . . . . . . . . . . . 1160 95.5 Answers for Ch. 56: Conditional Probability . . . . . . . . . . . . . . . . . . . 1164 95.6 Answers for Ch. 57: Probability: Independence . . . . . . . . . . . . . . . . . 1165 95.7 Answers for Ch. 59: Random Variables: Introduction . . . . . . . . . . . . . . 1166 95.8 Answers for Ch. 60: Random Variables: Independence . . . . . . . . . . . . . 1170 95.9 Answers for Ch. 61: Random Variables: Expectation . . . . . . . . . . . . . . 1171 95.10Answers for Ch. 62: Random Variables: Variance . . . . . . . . . . . . . . . . 1173 95.11Answers for Ch. 64: Bernoulli Trial and Distribution . . . . . . . . . . . . . . 1174 95.12Answers for Ch. 65: Binomial Distribution . . . . . . . . . . . . . . . . . . . . 1175 95.13Answers for Ch. 66: Poisson Distribution . . . . . . . . . . . . . . . . . . . . . 1176 95.14Answers for Ch. 67: Continuous Uniform Distribution . . . . . . . . . . . . . 1178 95.15Answers for Ch. 68: Normal Distribution . . . . . . . . . . . . . . . . . . . . . 1179 95.16Answers for Ch. 71: Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186 95.17Answers for Ch. 72: Null Hypothesis Significance Testing . . . . . . . . . . . 1189 95.18Answers for Ch. 73: Correlation and Linear Regression . . . . . . . . . . . . . 1194 96 Answers to Exercises in Part VII (2006-2015 A-Level Exams) 1198 96.1 Answers for Ch. 74: Functions and Graphs . . . . . . . . . . . . . . . . . . . . 1198 96.2 Answers for Ch. 75: Sequences and Series . . . . . . . . . . . . . . . . . . . . . 1225 96.3 Answers for Ch. 76: Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1255 96.4 Answers for Ch. 77: Complex Numbers . . . . . . . . . . . . . . . . . . . . . . 1269 96.5 Answers for Ch. 78: Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1292 96.6 Answers for Ch. 79: Probability and Statistics . . . . . . . . . . . . . . . . . . 1345
Page 34, Table of Contents
www.EconsPhDTutor.com
Part I
Functions and Graphs
Page 35, Table of Contents
www.EconsPhDTutor.com
1
Sets
The glory of [maths] is its complete irrelevance to our lives. That’s why it’s so fun! Paul Lockhart (2009, A Mathematician’s Lament, p. 38).
I have never done anything ‘useful’. No discovery of mine has made, or is likely to make, directly or indirectly, for good or ill, the least difference to the amenity of the world. - G.H. Hardy (1940 [1967], A Mathematician’s Apology, p. 150).
The set is the basic building block of mathematics. Informally, a set is a “container” that usually has some objects in it, but can sometimes also be empty. Each object in a set is called an element (of that set). Example 1. Let A = {3, π 2 , Clementi Mall, Love, the colour green}. Observations: • The name of a set is often an upper-case letter; in this case, it is A. • Mathematical punctuation marks called braces {} are used to denote a set. Listed within these braces are the elements of the set. • Elements of the set are separated by commas (,). This mathematical punctuation mark means “and”. • Thus, {3, π 2 , Clementi Mall, Love, the colour green} is the set consisting of five elements, namely 3 and π 2 and Clementi Mall and Love and the colour green. • Elements in a set can be almost anything whatsoever! In this example, the elements included a building (Clementi Mall), an abstract notion (Love), and even a colour (green). The elements of a set can even be another set! But don’t worry, in the context of A-level maths, the elements of a set will almost always be numbers. • When we talk about a set, we refer to both the “container” itself and all the objects in it.
Exercise 1. B is the set of the first 7 positive integers. Write down B in set notation. (Answer on p. 1008.) Exercise 2. C is the set of even prime numbers. Write down C in set notation. (Answer on p. 1008.)
Page 36, Table of Contents
www.EconsPhDTutor.com
1.1
In ∈ and Not In ∉
The mathematical punctuation mark ∈ means “is in”, while ∉ means “is not in”. Example 2. Let B = {1, 2, 3, 4, 5, 6, 7}. Then 1 ∈ B, 2 ∈ B, 3 ∈ B, etc. You can read these statements aloud as “1 is in B”, “2 is in B”, “3 is in B”, etc. We can also write 1, 2, 3 ∈ B (“1, 2, and 3 are in B”). Also, 8 ∉ B, 9 ∉ B, 10 ∉ B, etc. (“8 is not in B”, “9 is not in B”, “10 is not in B”, etc.). We can also write 8, 9, 10 ∉ B (“8, 9, and 10 are not in B”). Example 3. Cow ∈ {Cow, Chicken} reads aloud as “Cow is in the set consisting of Cow and Chicken”. Cow, Chicken ∈ {Cow, Chicken} reads aloud as “Cow and Chicken are in the set consisting of Cow and Chicken”.
Page 37, Table of Contents
www.EconsPhDTutor.com
1.2
Greater than >, Less Than <, Positive > 0, and Negative < 0
In this textbook: • Greater than means “strictly greater than” (>). So I won’t bother saying “strictly”, unless it’s something I want to emphasise. • Less than means “strictly less than” (<). • If I want to say greater than or equal to (≥) or smaller than or equal to (≤), I’ll say exactly that. • Positive means “greater than zero” (> 0). • Negative means “less than zero” (< 0). • Non-negative means “greater than or equal to zero” (≥ 0). • Non-positive means “less than or equal to zero” (≤ 0). • 0 is neither positive nor negative. Instead, 0 is both non-negative and non-positive.
Page 38, Table of Contents
www.EconsPhDTutor.com
1.3
Types of Numbers
The following taxonomy lists the several types of numbers you’ll encounter in this textbook.
Complex Numbers
Real Numbers
Rational Numbers
Imaginary Numbers
Irrational Numbers
Integers NonIntegers
We’ll study imaginary numbers only later on in Part IV of this textbook. For now, all numbers we’ll consider are real numbers (or reals). We won’t define what real numbers are. Instead, we’ll simply assume (like in secondary school) that “everyone knows” what real numbers are. Infinity (∞) and negative infinity (−∞) are NOT numbers. Informally, ∞ is the “thing” that is greater than any real number. Similarly, −∞ is the “thing” that is smaller than any real number. I repeat: INFINITY IS NOT A NUMBER.13 So what exactly are real numbers, infinity, and negative infinity? This is actually a fascinating question that mathematicians were able to answer satisfactorily only from the 19th century, but is beyond the scope of the A-levels. Definition 1. An integer is any one of these real numbers: . . . , −3, −2, −1, 0, 1, 2, 3, . . .
Definition 2. A rational number (or simply rationals) is any real number that can be expressed as the ratio of two integers. An irrational number (or simply irrationals) is any other real number.
Example 4. The number 16 is an integer, a rational, and a real.
Example 5. The number 1.87 is a rational and a real, but it is not an integer.
Example 6. The number π ≈ 3.14159 is an irrational and a real, but it is neither an integer nor a rational. 13
Actually, the truth is somewhat more complicated. Under certain special contexts in more advanced mathematics, infinity is treated as a number. But in this textbook, I’ll simply keep it simple and insist that infinity is not a number.
Page 39, Table of Contents
www.EconsPhDTutor.com
1.4
The Order of the Elements Doesnâ&#x20AC;&#x2122;t Matter
The order in which we write out the elements of the set does not matter: Definition 3. Two sets are equal (or identical) if both sets contain exactly the same elements.
Example 7. There are at least six equivalent ways to write the set of the 3 smallest positive even numbers: {2, 4, 6} = {2, 6, 4} = {4, 2, 6} = {4, 6, 2} = {6, 2, 4} = {6, 4, 2}. Example 8. {Cow, Chicken} = {Chicken, Cow}.
Page 40, Table of Contents
www.EconsPhDTutor.com
1.5
Repeated Elements Don’t Count
Repeated elements are simply ignored. Example 9. The set of the 3 smallest positive even numbers can be written as {2, 4, 6}. It can also be written as: {2, 2, 4, 6} or {2, 6, 6, 6, 4, 4}. Repeated elements are simply ignored. The notation n({2, 4, 6}) denotes the number of elements in the set of the first 3 even numbers. Hence, n({2, 4, 6}) = 3. And we also have n({2, 2, 4, 6}) = 3 and n({2, 6, 6, 6, 4, 4}) = 3. Example 10. {Cow, Chicken} = {Cow, Cow, Chicken} = {Chicken, Cow, Chicken}. And n({Cow, Chicken}) = n({Cow, Cow, Chicken}) = n({Chicken, Cow, Chicken}) = 2.
Note that more commonly, the number of elements in the set A is written as ∣A∣. But for some reason, the A-level syllabus instead uses the notation n(A), so that’s what we’ll use.
Exercise 3. W = {Apple, Apple, Apple, Banana, Banana, Apple}. What is n(W )? (Answer on p. 1008.) Exercise 4. C is the set of even prime numbers. What is n(C)? (Answer on p. 1008.)
Page 41, Table of Contents
www.EconsPhDTutor.com
1.6
Ellipsis . . . Means Continue in the Obvious Fashion
The mathematical punctuation mark “. . . ” is called the ellipsis and means “continue in the obvious fashion”. Example 11. D is the set of all odd positive integers smaller than 100. So in set notation, we can write D = {1, 3, 5, 9, 11, . . . , 99}.
Example 12. T is the set of all negative integers greater than 100. So in set notation, we can write T = {−99, −98, −97, . . . , −2, −1}.
What is obvious to one person might not be obvious to another. So only use the ellipsis when you’re confident it will be obvious to your reader! And never be shy to write a few more of the set’s elements (as I did with the sets above)!
Exercise 5. Let D and T be as in the above two examples. What are n(D) and n(T )? (Answer on p. 1008.)
Page 42, Table of Contents
www.EconsPhDTutor.com
1.7
Sets can be Finite or Infinite
Example 13. Z+ is the set of all positive integers. So, Z+ = {1, 2, 3, . . . }. And since Z+ is infinite, we write n(Z+ ) = ∞. Example 14. Z is the set of all integers. So, Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }. And since Z is infinite, we write n(Z) = ∞.
Obviously, for an infinite set, we cannot explicitly list out all of its elements. So we’ll often use ellipses to help out, as we did in the above examples. Alternatively, we can use interval notation or set-builder notation, which we’ll learn about shortly.
Exercise 6. H is the set of all prime numbers. Write down H in set notation. (Answer on p. 1008.)
Page 43, Table of Contents
www.EconsPhDTutor.com
1.8
Special Names of Sets: Z, Q, R, and C
The following sets are so common that they have special symbols: 1. Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . } is the set of all integers. (Z is for Zahl, German for number.) 2. Q is the set of all rational numbers. (Q is for quoziente, Italian for quotient.) 3. R is the set of all real numbers. 4. C is the set of all complex numbers. (To be studied only in Part IV of this textbook.) To create a new set that contains only the positive (or negative) elements of the old set, append a superscript plus (+ ) or minus (− ) to the name of a set: 1. Z+ = {1, 2, 3, . . . } is the set of all positive integers. Z− = {. . . , −3, −2, −1} is the set of all negative integers. 2. Q+ is the set of all positive rational numbers. Q− is the set of all negative rational numbers. 3. R+ is the set of all positive real numbers. R− is the set of all negative real numbers. As we’ll learn later, there is no such thing as a positive or negative complex number. Hence, there is no such set named C+ or C− . To add the number 0 to a set, append a subscript zero (0 ) to its name: Example 15. The set A = {3, π 2 , Clementi Mall, Love, the colour green}. And so the set A0 = {3, π 2 , Clementi Mall, Love, the colour green, 0}. Example 16. The set B = {1, 2, 3, 4, 5, 6, 7}. And so the set B0 = {0, 1, 2, 3, 4, 5, 6, 7}.
Adding both a superscript + and a subscript 0 to the name of a set creates a new set that contains all positive elements of the old set and in addition the number 0. Similarly, adding both a superscript − and a subscript 0 to the name of a set creates a new set that contains all negative elements of the old set and in addition the number 0. Example 17. If V = {−2, −1, 3, 4}, then V + = {3, 4}, V − = {−2, −1}, V0+ = {0, 3, 4}, and V0− = {−2, −1, 0}.
Exercise 7. If U = {−1, 0, 2}, then what are U + , U − , U0 , U0+ , and U0− ? (Answer on p. 1008.)
Page 44, Table of Contents
www.EconsPhDTutor.com
1.9
Special Names of Sets: Intervals
Here are four new mathematical punctuation marks: The left-parenthesis: ( The right-parenthesis: )
The left-bracket: [ The right-bracket: ]
Together, () are called parentheses and [] are called brackets. An interval is a (usually infinite) set of real numbers. It is written using parentheses and/or brackets. Let a and b be real numbers where b ≥ a. Then: 1. (a, b) is the set of all real numbers that are greater than a and smaller than b. Such a set is also called an open interval. Example 18. The set I = (0, 3) denotes √ the set of all real numbers that are greater than 0 and smaller than 3. So for example, 2 ≈ 1.41 ∈ I, but 0, 3 ∉ I.
2. [a, b] is the set of all real numbers that are greater than or equal to a and smaller than or equal to b. Such a set is also called an closed interval. Example 19. The set J = [0, 3] denotes the set of all real numbers that are √ greater than or equal to 0 and smaller than or equal to 3. So for example, the numbers 0, 2, 3 ∈ J.
3. (a, b] is the set of all real numbers that are greater than a and smaller than or equal to b. Such a set is also called a half-open interval or a half-closed interval. Example 20. The set K = (0, 3] denotes the set of all real numbers that are greater than √ 0 and smaller than or equal to 3. So for example, the numbers 2, 3 ∈ K, but 0 ∉ K.
4. [a, b) is the set of all real numbers that are greater than or equal to a and smaller than b. Such a set is also called a half-open interval or a half-closed interval. Example 21. The set L = [0, 3) denotes the set of all real numbers that are greater than √ or equal to 0 and smaller than or equal to 3. So for example, the numbers 0, 2 ∈ L, but 3 ∉ L.
Page 45, Table of Contents
www.EconsPhDTutor.com
Exercise 8. How many elements does the set Z = [1, 1] contain? (Answer on p. 1008.) Exercise 9. How many elements does the set Y = (1, 1) contain? (Answer on p. 1008.) Exercise 10. How many elements does the set X = (1, 1.01) contain? (Answer on p. 1008.) Exercise 11. Write down R, R+ , R+0 , Râ&#x2C6;&#x2019; , and Râ&#x2C6;&#x2019;0 in interval notation. (Answer on p. 1008.)
Page 46, Table of Contents
www.EconsPhDTutor.com
1.10
Special Names of Sets: The Empty Set ∅
The empty set is literally the set that contains no elements. Hence the name! Definition 4. The empty set is the set {}. It can also be denoted ∅. Example 22. In 2016, the set of all Singapore Ministers who are younger than 30 is {} or ∅. This means that there is no Singapore Minister who is younger than 30. Example 23. The set of all even prime numbers greater than 2 is {} or ∅. This means that there is no even prime number that is greater than 2.
Example 24. The set of numbers that are greater than 4 and smaller than 4 is {} or ∅. This means that there is no number that is simultaneously greater than 4 and smaller than 4. As already mentioned, in this textbook (and also for the A-levels), the elements in a set will almost always be numbers. But in general, the elements of a set can be (nearly) anything whatsoever. In other words, a set really and simply is a “container” that can “contain” (nearly) anything whatsoever. Indeed, the elements of a set can be other sets, including even the empty set! Here are two examples to illustrate: Example 25. The set {∅} is not the same as the set ∅. The former is a set containing a single element, namely the empty set. The latter is the empty set. It is perhaps clearer if we rewrite them as {∅} = {{}} and ∅ = {} . Now we clearly see that {{}} ≠ {}. Note that the set {∅} = {{}} is certainly not empty, because it contains a single element (namely the empty set). Example 26. The set {∅, 1, {∅}} is the set containing three elements, namely the empty set, the number 1, and a set containing the empty set.
Page 47, Table of Contents
www.EconsPhDTutor.com
1.11
Subset Of ⊆
Definition 5. A is a subset of B (written as A ⊆ B) if every element in A is also an element in B. Not surprisingly, A ⊈/ B denotes that A is not a subset of B. Example 27. Let M = {1, 2}, N = {1, 2, 3}, and O = {1, 2, 4, 5}. Then M ⊆ N , but N ⊈ M . Also, M ⊆ O, but O ⊈ M . Further, N ⊈ O and O ⊈ N .
Exercise 12. State whether Z, Q, and R are subsets of each other. (Answer on p. 1008.) Exercise 13. True or false: “The set of currently-serving Singapore Prime Ministers is a subset of the set of currently-serving Singapore Ministers.” (Answer on p. 1008.)
The next fact is useful for showing that two sets are equal. Fact 1. Two sets are subsets of each other ⇐⇒ They are identical.
Proof. Optional, see p. 922 in the Appendices. The symbol ⇐⇒ stands for is equivalent to or if and only if. The above claim may be decomposed into two separate claims: 1. Two sets are subsets of each other Ô⇒ they are identical. (The symbol Ô⇒ stands for implies or only if.) 2. Two sets are subsets of each other ⇐Ô they are identical. (The symbol ⇐Ô stands for is implied by or if.) Note importantly that A Ô⇒ B is different from its converse B Ô⇒ A. For example, x > 10 Ô⇒ x > 3, but it is certainly not the case that x > 3 Ô⇒ x > 10.
Page 48, Table of Contents
www.EconsPhDTutor.com
1.12
Proper Subset Of ⊂
Definition 6. A is a proper subset of B (written as A ⊂ B) if A ⊆ B but A ≠ B. Not surprisingly, A ⊂/ B denotes that A is not a proper subset of B. Example 28. Let M = {1, 2}, N = {1, 2, 3}, O = {1, 2, 4, 5}, and P = {1, 2, 3}. Then M ⊆ N, O, P and M ⊂ N, O, P . In contrast, N ⊆ P , but N ⊂/ P ; this is because N = P .
Exercise 14. Is the set of all squares (call it S) a proper subset of the set of all rectangles (call it R)? (Answer on p. 1008.) Exercise 15. Does A ⊆ B imply that A ⊂ B? (Answer on p. 1009.) Exercise 16. Does A ⊂ B imply that A ⊆ B? (Answer on p. 1009.) Exercise 17. True or false statement: “If A is a subset of B, then A is either a proper subset of or is equal to B.” (Answer on p. 1009.)
Remark 1. The official A-level syllabus uses the symbol ⊆ to mean “subset of” and ⊂ to mean “proper subset of”. So this is what we’ll use in this textbook. However, confusingly enough, many writers use the symbol ⊂ to mean “subset of” and ⊊ to mean “proper subset of”. We will not follow such practice in this textbook. Just to let you know, in case you get confused while reading other mathematical texts!
Page 49, Table of Contents
www.EconsPhDTutor.com
1.13
Union ∪
Definition 7. The union of the sets A and B (denoted A ∪ B) is the set of elements that are either in A OR B. Tip: “U” for Union. Example 29. Let T = {1, 2}, U = {3, 4}, and V = {1, 2, 3}. Then T ∪ U = {1, 2, 3, 4}, T ∪ V = {1, 2, 3}, and U ∪ V = {1, 2, 3, 4}. And T ∪ U ∪ V = {1, 2, 3, 4}.
Exercise 18. Rewrite each of the following sets more simply: (a) [1, 2] ∪ [2, 3]. (b) (−∞, −3) ∪ [−16, 7). (c) {0} ∪ Z+ ? (Answer on p. 1009.) Exercise 19. What is the union of the set of squares (S) and the set of rectangles (R)? (Answer on p. 1009.) Exercise 20. What is the union of the set of rationals (Q) and the set of irrationals? (Answer on p. 1009.)
Page 50, Table of Contents
www.EconsPhDTutor.com
1.14
Intersection ∩
Definition 8. The intersection of the sets A and B (denoted A ∩ B) is the set of elements that are in A AND B.
Definition 9. Two sets intersect if their intersection contains at least one element (i.e. A ∩ B ≠ ∅).
Definition 10. Two sets are mutually exclusive or disjoint if their intersection is empty (i.e. A ∩ B = ∅). Example 30. Let T = {1, 2}, U = {3, 4}, and V = {1, 2, 3}. Then T ∩ U = ∅, T ∩ V = {1, 2}, and U ∩ V = {3}. And T ∩ U ∩ V = ∅.
Exercise 21. Rewrite each of the following sets more simply: (a) (4, 7] ∩ (6, 9). (b) [1, 2] ∩ [5, 6]. (c) (−∞, −3) ∩ (−16, 7). (Answer on p. 1009.) Exercise 22. What is the intersection of the set of squares (S) and the set of rectangles (R)? (Answer on p. 1009.) Exercise 23. What is the intersection of the set of rationals (Q) and the set of irrationals? (Answer on p. 1009.)
Page 51, Table of Contents
www.EconsPhDTutor.com
1.15
Set Minus /
The set minus (sometimes also called set difference) operator is very convenient. Sadly, it is not in the A-level syllabus and so I’ll avoid using it in this textbook. Nonetheless, it’s worth a quick mention. Definition 11. A set minus B (denoted A/B or A − B) is the set that contains every element in A that is not also in B. Example 31. Let T = {1, 2}, U = {3, 4}, and V = {1, 2, 3}. Then T /U = T , T /V = ∅, and U /V = {4}.
Page 52, Table of Contents
www.EconsPhDTutor.com
1.16
Set Complement A′
Definition 12. The set complement of A (denoted A′ or Ac ) is the set of all elements that are not in A. Example 32. Consider the set of positive integers. Let A = {2, 4, 6, 8, 10, . . . }. Then in this context, A′ = {1, 3, 5, 7, 9, 11, . . . }. Example 33. Consider the set of all reals. Let A = R+ . Then in this context, A′ = R−0 . Example 34. Consider the roll of a die. The set of desired outcomes was A = {1, 6}. Unfortunately, no desired outcome occurred. In other words, the actual outcome was an element in the set A′ = {2, 3, 4, 5}.
Page 53, Table of Contents
www.EconsPhDTutor.com
1.17
Set-Builder Notation
Set-builder notation is an alternative method of writing down sets. In the current context, the mathematical punctuation mark colon “∶” will mean “such that”. Example 35. The set {x ∈ R ∶ x > 0} contains all x ∈ R such that x > 0. In words, this set contains all real numbers that are positive.
What comes after the colon are the conditions or criteria that x must satisfy, in order to qualify as a member of the set. Our sets will usually contain only numbers, but here’s an example to show you how we can write down one particular set of musical artists. Example 36. The set {x ∶ x is an artist that has had a US Billboard Hot 100 #1 Single} contains all the artists who have ever had a US Billboard Hot 100 #1 Single.
It will however be more typical for our sets to be sets such as these: Example 37. {x ∈ R ∶ x > 0} = R+ , Q+ = {x ∈ Q ∶ x > 0}, Z+ = {x ∈ Z ∶ x > 0}, R+0 = {x ∈ R ∶ x ≥ 0}, Q+0 = {x ∈ Q ∶ x ≥ 0}, and Z+0 = {x ∈ Z ∶ x ≥ 0}.
Remark 2. We use the colon ∶ but some writers use instead the pipe ∣. Exercise 24. Write down R− , Q− , Z− , R−0 , Q−0 , and Z−0 in set-builder notation. (Answer on p. 1009.) Exercise 25. Write down (a, b), [a, b], (a, b], and [a, b) in set-builder notation. (Answer on p. 1009.) Exercise 26. Let X = {x ∶ x is a living current or former Prime Minister of Singapore}. Write down the set X so that all its elements are explicitly stated. (Answer on p. 1009.) Exercise 27. Rewrite each of the following sets in set-builder notation: (a) (−∞, −3) ∪ √ (5, ∞). (b) (−∞, 2] ∪ (e, π) ∪ (π, ∞). (c) (−∞, 3) ∩ (0, 7). (Answer on p. 1009.)
Page 54, Table of Contents
www.EconsPhDTutor.com
2
Dividing By Zero
This very brief chapter is to warn you against making a common mistake — dividing by 0. Students have little trouble avoiding this mistake if the divisor is obviously a big fat 0. Instead, students usually make this mistake when the divisor is an unknown constant or variable that might be 0. Example 38. Find the values of x for which x(x − 1) = (2x − 2)(x − 1). Here’s the wrong solution: “Divide both sides by x − 1 to get x = 2x − 2. So x = 2.” Here’s the correct solution: “Case #1. Suppose x − 1 = 0. Then the given equation is satisfied. So x = 1 is one possible value for which x(x − 1) = (2x − 1)(x − 1). Case #2. Now suppose x − 1 ≠ 0. So we can divide both sides by x − 1 to get x = 2x − 2. So x = 2. Conclusion. The two possible values of x for which x(x − 1) = (2x − 1)(x − 1) are x = 1 and x = 2.” Moral of the story. Whenever you divide by a certain quantity, make sure it’s non-zero. If you’re not sure whether it equals 0, then break up your analysis into two cases, as was done in the above example: Case #1 — the quantity equals 0 (and see what happens in this case); Case #2 — the quantity is non-zero (in which case you can go ahead and divide).
By the way, let’s take this opportunity to clear up another popular misconception — You 1 1 may have heard that = ∞. This is wrong. ≠ ∞. Instead, any non-zero number divided 0 0 by 0 is undefined.14 “Undefined” is the mathematician’s way of saying, “You haven’t told me what you are talking about. So what you are saying is meaningless.”15 Exercise 28. What’s wrong with this “proof” that 1 = 0? (Answer on p. 1010.) 1. Let x, y be positive numbers such that x = y. 2. Square both sides: x2 = y 2 . 3. Rearrange: x2 − y 2 = 0 4. Factorise: (x − y)(x + y) = 0.
5. Divide both sides by x − y to get x + y = 0. 6. Since x = y, sub y = x into the above equation to get 2x = 0. 7. Divide both sides by 2x to get 1 = 0. Once again, the truth is actually somewhat more complicated. Under certain special contexts in more advanced mathe1 1 matics, is well-defined. But in this textbook, I’ll simply keep it simple and insist that is undefined. 0 0 0 15 On the other hand, is indeterminate, which means that it’s typically undefined, but can sometimes be defined under 0 certain circumstances.
14
Page 55, Table of Contents
www.EconsPhDTutor.com
3
Functions
Undoubtedly the most important concept in all of mathematics is that of a function — in almost every branch of modern mathematics functions turn out to be the central objects of investigation. - Michael Spivak (1994 [2006], Calculus, p. 39). You are probably familiar from secondary school with such statements as: “Let f (x) = x + 8 be a function.” Strictly speaking, this is not the correct way of describing a function. Here is a more precise definition of a function.16 A function consists of three pieces: 1. A set called the domain; 2. A set called the codomain; and 3. A mapping rule (or simply mapping or simply rule) which specifies how each and every element in the domain is mapped (or assigned) to one (and exactly one) element in the codomain. Remark 3. The codomain is not the same thing as the range. We’ll learn about the range only in the next section. Altogether then, a function simply maps (or assigns) each element in the domain to one (and exactly one) element in the codomain. Example 39. Let f be the function whose ... • Domain is the set {Cow, Chicken}; • Codomain is the set {Produces eggs, Produces milk, Guards the home}; and • Mapping rule is, informally, “match the animal to its role”. According to the mapping rule, “Cow” (in the domain) is mapped to “Produces milk” (in the codomain) and “Chicken” (in the domain) is mapped to “Produces eggs” (in the codomain). Every element in the domain is mapped to exactly one element in the codomain. This is indeed a function, because it has a domain, codomain, and a correctly-specified mapping rule.
16
This definition is still informal. See Definition 135 in the Appendices for the exact, formal definition (optional).
Page 56, Table of Contents
www.EconsPhDTutor.com
Example 40. Let f be the function whose ... • Domain is the set {1, 2}; • Codomain is the set {1, 2, 3, 4, 5}; and • Mapping rule is, informally, “multiply by 2”. According to the mapping rule, “1” (in the domain) is mapped to “2” (in the codomain) and “2” (in the domain) is mapped to “4” (in the codomain). Every element in the domain is mapped to exactly one element in the codomain. This is indeed a function, because it has a domain, codomain, and a correctly-specified mapping rule.
Example 41. Let f be the function whose ... • Domain is the set R; • Codomain is the set R; and • Mapping rule is, informally, “round off to the nearest integer, where half-integers are rounded up” According to the mapping rule, “3” (in the domain) is mapped to “3” (in the codomain), “3.14159” (in the domain) is mapped to “3” (in the codomain), “3.5” (in the domain) is mapped to “4” (in the codomain), and “3.88” (in the domain) is mapped to “4” (in the codomain). Every element in the domain is mapped to exactly one element in the codomain. This is indeed a function, because it has a domain, codomain, and a correctly-specified mapping rule.
Page 57, Table of Contents
www.EconsPhDTutor.com
3.1
Formal Mathematical Notation for Functions
In general, the correct way to describe a function is: The function f ∶ D → C is defined by f ∶ x ↦ f (x) for all x ∈ D. Or alternatively: The function f ∶ D → C is defined by f (x) = ... for all x ∈ D. This says that the function’s name is f , its domain is D, and its codomain is C. The last bit “f ∶ x ↦ f (x)” is the mapping rule and this mapping rule applies to “all x ∈ D” (all elements in the domain). To save ourselves a bit of writing, if it’s clear from the context that we’re talking about the function f , then we’ll omit “f ∶” from the front of the mapping rule. Also, if the mapping rule applies universally to all elements of the domain, then we also omit the “for all x ∈ D” at the end. Altogether then, we will often simply write: The function f ∶ D → C is defined by x ↦ f (x). We will sometimes also denote the domain and codomain of f by Dom(f ) and Cod(f ).
Page 58, Table of Contents
www.EconsPhDTutor.com
Example 40 (revisited). In formal mathematical notation, we write: “the function f ∶ {1, 2} → {1, 2, 3, 4, 5} is defined by x ↦ 2x.” Or alternatively, “the function f ∶ {1, 2} → {1, 2, 3, 4, 5} is defined by f (x) = 2x.” This says that the function has: • Name f ; • Domain {1, 2}; • Codomain {1, 2, 3, 4, 5}; and • Mapping rule: Map every element x in the domain to the element 2x in the codomain.
Let’s examine this formal notation a little more closely. In the context of set-builder notation (Section 1.17), the mathematical punctuation mark colon “∶” stood for “such that”. However, in the context of functions, the colon “∶” stands instead for “from”. Unfortunately there are only so many symbols and punctuation marks, so invariably some symbols will have to play more than one role! The mathematical punctuation mark “→” (right arrow) simply stands for “to”. Altogether then, “f ∶ D → C” reads as “f is the function from domain D to domain C”. The mathematical punctuation mark “↦” stands for “maps to”. Hence, “x ↦ f (x)” reads as “x is mapped to f (x)”.
Page 59, Table of Contents
www.EconsPhDTutor.com
Example 39 (revisited). In formal mathematical notation, we write: “the function f ∶ {Cow, Chicken} → {Produces eggs, Produces milk, Guards the home} is defined by Cow ↦ Produces milk and Chicken ↦ Produces eggs.” Or alternatively, “the function f ∶ {Cow, Chicken} → {Produces eggs, Produces milk, Guards the home} is defined by f (Cow) = Produces milk and f (Chicken) = Produces eggs.” This says that the function has • Name f ; • Domain {Cow, Chicken}; • Codomain {Produces eggs, Produces milk, Guards the home}; and • Mapping rule: Map the element “Cow” in the domain to the element “Produces milk” in the codomain and the element “Chicken” in the domain to the element “Produces eggs” in the codomain.
Example 41 (revisited). In formal mathematical notation, we write: “the function f ∶ R → R is defined by x ↦ Integer closest to x”. This says that the function has • Name f ; • Domain R; • Codomain R; • Mapping rule: Map every element x in the domain to the closest integer in the codomain.
Students frequently believe that f (x) denotes a function. This is wrong. f and f (x) refer to two different things. f denotes a function. f (x) denotes the value of f at x. This may seem like an excessively pedantic distinction. But maths is precise and pedantic. In maths, what we mean is precisely what we say and what we say is precisely what we mean. There is never any room for ambiguity or alternative interpretations. More examples:
Page 60, Table of Contents
www.EconsPhDTutor.com
Example 42. “The function f ∶ [0, 1] → R is defined by x ↦ 3x + 4.” Or alternatively, “The function f ∶ [0, 1] → R is defined by f (x) = 3x + 4.” This says that the function’s name is f , its domain is [0, 1] (the set of all reals between 0 and 1, including 0 and 1), its codomain is R (the set of all reals), and its mapping rule is that we map each element x in the domain to the element 3x + 4 in the codomain. The value of f at 0.5 is f (0.5) = 3(0.5) + 4 = 5.5. What is f (3)? It is not 3(3) + 4 = 13. This is because 3 is not in the domain of f . Hence, f (3) is simply undefined.
Example 43. “The function f ∶ R+ → R is defined by x ↦ ln x.” Or alternatively, “The function f ∶ R+ → R is defined by f (x) = ln x.” This says that the function’s name is f , its domain is R+ (the set of all positive reals), its codomain is R (the set of all reals), and its mapping rule is that we map each element x in the domain to the element ln x in the codomain. The value of f at 2 is f (2) = ln 2 ≈ 0.693. f (0) is simply undefined, because 0 is not in the domain of f . Likewise, f (a) is undefined, for any a < 0.
Exercise 29. For each of the following functions, write down the value of the function at 1. (a) The function f ∶ R → R is defined by x ↦ x + 1. (b) The function g ∶ [−1, 1] → R is defined by x ↦ 17x. (c) The function h ∶ Z+ → R is defined by x ↦ 3x . (d) The function i ∶ Z− → R is defined by x ↦ 3x . (Answer on p. 1011.)
Page 61, Table of Contents
www.EconsPhDTutor.com
3.2
EVERY x ∈ D Must be Mapped to EXACTLY ONE y ∈ C
This section simply repeats and emphasises what was already said above. Example 44. Say we have ... • Domain {Cow, Chicken}; • Codomain {Produces eggs, Produces milk, Guards the home}; and • Mapping rule: “Chicken” is mapped to “Produces eggs”. Can we define a function using the above domain, codomain, and mapping rule? No. The reason is that the mapping rule fails to specify what “Cow” (an element of the domain) should be mapped to. It thus fails the requirement that every element in the domain be mapped to an element in the codomain. Example 45. Say we have ... • Domain {Cow, Chicken}; • Codomain {Produces eggs, Produces milk, Guards the home}; and • Mapping rule: “Cow” is mapped to both “Produces milk” and “Guards the home”; and “Chicken” is mapped to “Produces eggs”. Can we define a function using the above domain, codomain, and mapping rule? No. The reason is that the mapping rule maps “Cow” (an element of the domain) to more than one element in the codomain. It thus fails the requirement that every element in the domain be mapped to exactly one element in the codomain. Example 46. Say we have ... • Domain R; • Codomain [0, 1]; and • Mapping rule: x ↦ x + 1. Can we define a function using the above domain, codomain, and mapping rule? No. The reason is that the mapping rule fails to map some elements in the domain (e.g. 14) to any element in the codomain. It thus fails the requirement that every element in the domain be mapped to an element in the codomain.
Page 62, Table of Contents
www.EconsPhDTutor.com
Example 47. Say we have ... • Domain R; • Codomain R; and • Mapping rule: x ↦ ±x. Can we define a function using the above domain, codomain, and mapping rule? No. The reason is that the mapping rule maps each element in the codomain (e.g. 14) to more than one element in the codomain (+14 and -14). It thus fails the requirement that every element in the domain be mapped to exactly one element in the codomain.
For Exercises 30-37: (i) State (yes/no) whether we can define a function using the given domain, codomain, and rule. (ii) Explain why or why not. (iii) If we can, then write down the function in formal notation. Exercise 30. Let the domain be {5, 6, 7}, the codomain be Z+ , and the mapping rule be x ↦ 2x (Answer on p. 1011.) Exercise 31. Let the domain be {0, 3}, the codomain be {3, 4}, and the mapping rule be (informally) “any larger number will work”. (Answer on p. 1011.) Exercise 32. Let the domain be {2, 4}, the codomain be {3, 4}, and the mapping rule be (informally) “any smaller number will work”. (Answer on p. 1011.) Exercise 33. Let the domain be {1}, the codomain be {1}, and the mapping rule be (informally) “stay exactly the same”. (Answer on p. 1011.) Exercise 34. Let the domain be {1}, the codomain be {1, 2}, and the mapping rule be (informally) “stay exactly the same”. (Answer on p. 1011.) Exercise 35. Let the domain be {1, 2}, the codomain be {1}, and the mapping rule be (informally) “stay exactly the same”. (Answer on p. 1011.) √ Exercise 36. Let the domain be R, the codomain be R, and the mapping rule be x ↦ x. (Answer on p. 1011.) 1 Exercise 37. Let the domain be R, the codomain be R, and the mapping rule be x ↦ . x (Answer on p. 1011.) Exercise 38. How might you change the domain in Exercise 36 so that a function can be defined? (Answer on p. 1012.) Exercise 39. How might you change the domain in Exercise 37 so that a function can be defined? (Answer on p. 1012.)
Page 63, Table of Contents
www.EconsPhDTutor.com
3.3
Real-Valued Functions of a Real Variable
Definition 13. A function of a real variable is any function whose domain is a subset of R.
Definition 14. A real-valued function is any function whose codomain is a subset of R.
Altogether then, a real-valued function of a real variable is any function both of whose domain and codomain are subsets of R. Example 48. Consider the functions f ∶ R → R, g ∶ R → R, and h ∶ R → R defined by x ↦ x2 . All three are real-valued functions, functions of a real variable, and thus also real-valued functions of a real variable. Consider the function i ∶ {Cow, Chicken} → Z defined by Cow ↦ 5 and Chicken ↦ 32. This is a real-valued function, but not a function of a real variable. Thus, it is not a real-valued function of a real variable. Consider the function j ∶ Z → {Cow, Chicken} defined by x ↦ Cow if x is odd and x ↦ Chicken if x is even. This is a function of a real variable, but not a real-valued function.
Almost all functions considered in H2 Maths are real-valued functions of a real variable. So we’ll see plenty of functions like f , g, and h from the above example, but rarely (if ever) will we see functions like i or j. In this textbook, unless otherwise clearly-stated, it may be assumed that all functions are real-valued functions of a real variable.
Page 64, Table of Contents
www.EconsPhDTutor.com
3.4
The Range of a Function
Informally, the range of a function f ∶ D → C — denoted f (D) — is the set of elements in the codomain C that are “hit” by the function. Formally: Definition 15. The range of a function f ∶ D → C is f (D) = {y ∈ C ∶ There is x ∈ D such that f (x) = y}. The range of f may be denoted Range(f ) or f (D) (if D is the domain of f ). The range is not the same thing as the codomain. Because this is such a common misconception, let me repeat:
♡ The range is not the same thing as the codomain. ♡ Indeed, the range is usually a proper subset of the codomain, as was the case in each of the following examples. Example 49. Define f ∶ [0, 1] → R by x ↦ x + 1. Then Range(f ) = f ([0, 1]) = [1, 2]. Example 50. Define f ∶ {2, 3} → R by x ↦ x + 1. Then Range(f ) = f ({2, 3}) = {3, 4}. Example 51. Define f ∶ R → R by x ↦ ex . Then Range(f ) = f (R) = R+ . The range is often is a proper subset of the codomain, but sometimes they can be equal: Example 52. Define f ∶ R+ → R by x ↦ ln x. Then Range(f ) = f (R+ ) = R = Cod(f ).
Exercise 40. Let the function f ∶ R+0 → R be defined by x ↦ (Answer on p. 1012.)
√
x. What is the range of f ?
Exercise 41. Let the function f ∶ Z → Z be defined by x ↦ x2 . What is the range of f ? (Answer on p. 1012.) Exercise 42. Which of the following statements is/are true? (a) “The range of any function is a subset of its domain.” (b) “The range of any function is a subset of its codomain.” (c) “The range of any function is a proper subset of its codomain.” (Answer on p. 1012.)
Page 65, Table of Contents
www.EconsPhDTutor.com
3.5
Creating New Functions
Let f ∶ A → R and g ∶ B → R be functions with mapping rules f ∶ x ↦ f (x) and g ∶ x ↦ g(x). Let k ∈ R be a constant. Then we can create the function f + g in the “obvious” fashion. f We can also create the functions f − g, f ⋅ g, , and kf in the “obvious” fashions.17 g The symbol ⋅ is an alternative symbol for multiplication. We will often prefer using ⋅ rather than × because there is the slight risk of confusing × with the letter x. As we shall see, f g shall refer to a function that is entirely different from f ⋅ g, so we must really be careful to write f ⋅ g when that is what we mean. Example 53. Let f ∶ R → R be defined by x ↦ 7x + 5 and g ∶ R → R be defined by x ↦ x3 . Let k = 2. Then • f + g is the function with domain R, codomain R, and mapping rule x ↦ 7x + 5 + x3 ; • f − g is the function with domain R, codomain R, and mapping rule x ↦ 7x + 5 − x3 ;
• f ⋅ g is the function with domain R, codomain R, and mapping rule x ↦ (7x + 5) x3 ; and f • is the function with domain R− ∪ R+ , codomain R, and mapping rule x ↦ (7x + 5) /x3 . g • kf is the function with domain R, codomain R, and mapping rule x ↦ 2 (7x + 5). We can of course give these four new functions new names (perhaps a single-letter name for each), but this is not necessary. We can simplywrite: (f + g) (1) = 7(1) + 5 + 13 = 13, (f − g) (1) = 7(1) + 5 − 13 = 11,
(f ⋅ g) (1) = [7 (1) + 5] (1)3 = 12, f ( ) (1) = [7 (1) + 5] /13 = 12, g
(kf )(1) = 2 [7 (1) + 5] = 24, where the pairs of parentheses around each of the five new functions are just to be clear that we are talking about a single, fully-fledged function.
17
Formally, f + g is the function with domain A ∩ B, codomain R, and mapping rule x ↦ f (x) + g(x). Similarly, f − g is the function with domain A ∩ B, codomain R, and mapping rule x ↦ f (x) − g(x). f ⋅ g is the function with domain A ∩ B, codomain R, and mapping rule x ↦ f (x)g(x). f is the function with domain {x ∶ x ∈ A ∩ B, g(x) ≠ 0}, codomain R, and mapping rule x ↦ f (x)/g(x). The set A ∩ g B/ {x ∶ g(x) = 0} is the set of all elements x that are in both A and B, excluding those for which g(x) = 0. This exclusion is necessary, otherwise f (x)/g(x) may sometimes not be well-defined. Finally, kf is simply the function with domain A, codomain R, and mapping rule x ↦ kf (x).
Page 66, Table of Contents
www.EconsPhDTutor.com
3.6
One-to-One Functions
Informally, a function is one-to-one (or invertible) if every element in its range is “hit” exactly once (by exactly one element in the domain). Put another way: every element y in the range corresponds to exactly one element in the domain. Formally: Definition 16. A function f ∶ D → C is one-to-one (or invertible) if for every y ∈ f (D), there is only one x ∈ D such that f (x) = y. Example 54. Consider the function f whose domain is the set {Cow, Chicken}, codomain is the set {Produces eggs, Produces milk, Guards the home}, and mapping rule is Cow ↦ Produces milk and Chicken ↦ Produces eggs. The range is {Produces eggs, Produces milk}. This function is one-to-one because each element in the range is “hit” exactly once, as we can easily verify: Produces eggs is “hit” once by Chicken and Produces milk is “hit” once by Cow.
Example 55. Let f ∶ [0, 1] → R be defined by x ↦ x + 1. The range of f is [1, 2]. To check whether this function is one-to-one, we need to show that every element y in the range corresponds to exactly one element x in the codomain. To this end, let’s pick any element y in the range and write: y = x + 1 ⇐⇒ y − 1 = x. Thus, indeed, this function is one-to-one — every element y in the range corresponds to exactly one element y − 1 in the domain. To show that a function is not one-to-one, simply give a counter-example: Example 56. Let f ∶ R → R be defined by x ↦ x2 . The range of f is R+ . This function is not one-to-one — for example, the element 9 in the range is “hit” twice, once by −3 and again by 3. Remark 4. One-to-one or invertible functions are also known as injective functions (or simply injections), but we won’t use this term in this textbook. Exercise 43. State and explain of the following functions is one-to-one. (a) √ whether each + + f ∶ R0 → R is defined by x ↦ x. (b) g ∶ R0 → R is defined by x ↦ x2 . (c) h ∶ R → R is defined by x ↦ ∣x∣. (d) i ∶ R+0 → R is defined by x ↦ ∣x∣. (e) j ∶ R → R is defined by x ↦ sin x. (Answer on p. 1013.)
Page 67, Table of Contents
www.EconsPhDTutor.com
3.7
Inverse Functions
Definition 17. If f ∶ D → C is invertible, then its inverse function f −1 ∶ f (D) → D is defined by the mapping rule “f −1 (y) = x ⇐⇒ y = f (x)”. Only invertible functions have inverse functions. If a function is not invertible, then its inverse function simply does not exist.
Given a one-to-one (or invertible) function f , to find its inverse function f −1 , follow these steps: 1. Dom (f −1 ) = Range (f ). 2. Cod (f −1 ) = Dom (f ). 3. Write down an expression f −1 (y) that involves only y and show that “f −1 (y) = x ⇐⇒ y = f (x) ”.
Page 68, Table of Contents
www.EconsPhDTutor.com
Example 57. As we showed above, the function f ∶ [0, 1] → R defined by x ↦ x + 1 is one-to-one. So its inverse function f −1 exists. Let’s find it. 1. f has range [1, 2]. So f −1 has domain [1, 2]. 2. f has domain [0, 1]. So f −1 has codomain [0, 1]. 3. Pick any element y in the range of f and write: y =f (x) ⇐⇒ y = x + 1 ⇐⇒ y − 1 = x. ± −1 f
(y)
So f −1 has mapping rule y ↦ y − 1. We’ll actually only formally talk about graphs in the next few chapters. But for now, as a visual aid, I’ll provide the graphs of f −1 (blue) and f (red) anyway. Observe that f −1 is simply the reflection of f in the line y = x (dotted). Section 7.2 (in particular Fact 7) will explain why exactly this is so.
5
f(x), f -1(x)
4 3 2 1 x 0 -5
-4
-3
-2
-1
0
1
2
3
4
5
-1 -2 -3 -4 -5
Page 69, Table of Contents
www.EconsPhDTutor.com
Example 58. You can verify for yourself that the function f ∶ R → R defined by x ↦ 2x is one-to-one. Let’s find its inverse function f −1 . 1. f has range R. So f −1 has domain R. 2. f has domain R. So f −1 has codomain R. 3. As usual, let’s pick any element y in the range of f and write: y =f (x) ⇐⇒ y = 2x ⇐⇒ 0.5y = x. ± −1 f
(y)
So f −1 has mapping rule y ↦ 0.5y.
5
f(x), f -1(x)
4 3 2 1 x 0 -5
-4
-3
-2
-1
0
1
2
3
4
5
-1 -2 -3 -4 -5
Page 70, Table of Contents
www.EconsPhDTutor.com
Example 59. You can verify for yourself that the function f ∶ R+ ∪ R− → R defined by 1 x ↦ is one-to-one. Let’s find its inverse function f −1 . x 1. f has range R+ ∪ R− . So f −1 has domain R+ ∪ R− .
2. f has domain R+ ∪ R− . So f −1 has codomain R+ ∪ R− . 3. As usual, let’s pick any element y in the range of f and write: 1 1 ⇐⇒ y =f (x) ⇐⇒ y = = x (∵ y ≠ 0). x y ® −1 f
(y)
1 So f −1 has mapping rule y ↦ . y (Note that ∵ is the shorthand symbol for because. Similarly, ∴ is the shorthand symbol for therefore.) The condition here that y ≠ 0 is important and goes back to our warning that was Chapter 2 (Dividing by Zero). We know for sure that the range of f does not contain 0. This is why in the last line above, we can safely divide both sides of the equation by y. 5
f(x), f -1(x)
4 3 2 1 x 0 -5
-4
-3
-2
-1
0
1
2
3
4
5
-1 -2 -3 -4 -5
Page 71, Table of Contents
www.EconsPhDTutor.com
Example 60. You can verify for yourself that the function f ∶ R+0 → R defined by x ↦ x2 is one-to-one. Let’s find its inverse function f −1 . 1. f has range R+0 . So f −1 has domain R+0 .
2. f has domain R+0 . So f −1 has codomain R+0 . 3. As usual, let’s pick any element y in the range of f and write: √ y =f (x) ⇐⇒ y = x2 ⇐⇒ ± y = x. ± −1 f
(y)
√ √ Here there are two possibilities for the mapping rule of f −1 , namely y → y and y → − y. We must pick one. We know that the domain of f —and hence the codomain of f −1 — is √ R+0 . So we should pick as the mapping rule of f −1 ∶ y → y.
f(x), f -1(x) 4 3 2 1 x 0 0.0
0.5
1.0
1.5
2.0
+ Exercise 44. Find √ the inverse function for each of the following functions. (a) f ∶ R0 → R defined by x ↦ x. (b) g ∶ [−0.5π, 0.5π] → R defined by x ↦ sin x. (c) h ∶ R → R defined by x ↦ x3 .(Answers on p. 1014.)
Page 72, Table of Contents
www.EconsPhDTutor.com
3.8
Domain Restriction to Create an Invertible Function
We saw that some functions were not one-to-one (or non-invertible). And so for these functions, an inverse function simply does not exist. Nonetheless, we can often transform a non-invertible function into an invertible function. One way to do this is by restricting the domain. The new invertible function will then have an inverse function. Example 61. We saw in Exercise 43 that the function j ∶ R → R defined by x ↦ sin x was not one-to-one. However, we can restrict the domain to [−0.5π, 0.5π] to get a brand new function g ∶ [−0.5π, 0.5π] → R defined by x ↦ sin x. This brand new function g is identical to the original function j except for its domain. g is one-to-one, as you should verify for yourself. We can thus go ahead and construct the inverse function g −1 . Actually, we already did this in Exercise 44.
Example 62. We saw in Example 56 that the function f ∶ R → R defined by x ↦ x2 was not one-to-one. However, we can restrict the domain to R+0 to get a brand new function g ∶ R+0 → R defined by x ↦ x2 . This brand new function g is identical to the original function f except for its domain. g is one-to-one, as we verified in Exercise 43. We can thus go ahead and construct the inverse function g −1 . I leave this as an exercise for you.
There is almost always more than one way to restrict the domain of a non-invertible function to obtain an invertible function. Indeed, a trivial case would be where we restrict its domain to be the empty set! In which case the function thus formed would certainly be invertible, though not very interesting (it would have an empty domain and an empty range — so too would its inverse function).
1 (x − 1)2 is not one-to-one. (b) Show that by restricting its domain to (1, ∞), we can create a new invertible function g (you must prove that this new function is invertible). (c) Then find the inverse function g −1 . (Answer on p. 1015.) Exercise 45. (a) Show that the function f ∶ (−∞, 1) ∪ (1, ∞) → R defined by x ↦
Exercise 46. For the function f in Example 62, let’s instead restrict the domain to [20, 30]. Show that the new function thus obtained is one-to-one and find its inverse. (Answer on p. 1015.)
Page 73, Table of Contents
www.EconsPhDTutor.com
3.9
Composite Functions
Definition 18. Let f and g be functions such that the range of g is a subset of the domain of f . Then the composite function f g is the function with the same domain as g, the same codomain as f , and mapping rule x ↦ f (g(x)). The composite function f g can be read aloud as “f circle g” and is sometimes denoted f ○ g, especially when we want to make clear that we are not talking about f ⋅ g. But we’ll rarely use the f ○ g notation, unless there is some risk of confusion with f ⋅ g. The underlined condition is important: The range of g must be a subset of the domain of f in order for the composite function f g to exist. This condition ensures that given any x from the domain of g, the value g(x) is itself also in the domain of f , so that f (g(x)) is well-defined. If this condition fails, then the composite function f g simply does not exist. Example 63. The functions g, f ∶ R → R are defined by g ∶ x ↦ x + 1 and f ∶ x ↦ 2x. The range of g is R — this is indeed a subset of the domain of f (which is R). So the composite function f g ∶ R → R exists and is defined by x ↦ f (g(x)) = 2 (g(x)) = 2(x + 1). Let’s try computing f g(2). We can use the definition of a composite function: f g(2) = f (g(2)) = f (2 + 1) = f (3) = 6. Alternatively, we can directly use f g(x) = 2(x + 1) to compute f g(2) = 2(2 + 1) = 6.
Notice that for the composite function f g, we apply the function g first before applying the function f . So for example, to compute, say f g(7), we compute g(7) first, then compute f (g(7)). (A common mistake by students is to instinctiely read from left to right, and so apply f first before g.)
Example 64. The functions g, f ∶ R → R are defined by g ∶ x ↦ x2 and f ∶ x ↦ x + 1. The range of g is R+0 — this is indeed a subset of the domain of f (which is R). So the composite function f g ∶ R → R exists and is defined by x ↦ f (g(x)) = g(x) + 1 = x2 + 1. Let’s try computing f g(3). We can use either the definition of a composite function: f g(3) = f (g(3)) = f (32 ) = f (9) = 10. Alternatively, we can directly compute, using f g(x) = x2 + 1: f g(3) = 32 + 1 = 10.
Page 74, Table of Contents
www.EconsPhDTutor.com
Example 65. The function g ∶ R → R is defined by x ↦ x + 1. The function f ∶ R+ → R is defined by x ↦ ln x. The range of g is R, which is not a subset of the domain of f (which is R+ ). Hence, the composite function f g simply does not exist. We saw that if f is non-invertible, then its inverse function f −1 simply does not exist. Nonetheless, we could restrict its domain to create a new invertible function g, whose inverse function g −1 we could then write down. By analogy, suppose we have functions f and g where g’s range is not a subset of f ’s domain. Thus, the composite function f g simply does not exist. But we can play a similar trick: We can restrict the domain of g to create a new function gˆ, so that the range of gˆ is a subset of f ’s domain. We can then write down the composite function f gˆ. Fortunately, this is not in the syllabus, so you don’t need to know how to do this. Yay!
Exercise 47. For each of the following pairs of functions f and g, verify that the composite function f g exists and write it out in full. Also, compute f g(1) and f g(2). (a) The functions g, f ∶ R → R defined by g ∶ x ↦ x2 + 1 and f ∶ x ↦ ex . (b) The functions g, f ∶ R → R defined by g ∶ x ↦ ex and f ∶ x ↦ x2 + 1. (c) The functions g, f ∶ R− ∪ R+ → R defined by g ∶ x ↦ 1/2x and f ∶ x ↦ 1/x. (d) The functions g, f ∶ R− ∪R+ → R defined by g ∶ x ↦ 1/x and f ∶ x ↦ 1/2x. (Answer on p. 1016.) We can of course also build a composite function out of a single function. Example 66. The function f ∶ R → R is defined by x ↦ 2x. The range of f is R and this is indeed a subset of the domain of f (which is R). So the composite function f f ∶ R → R exists and is defined by x ↦ f (f (x)) = 2f (x) = 2(2x) = 4x. And so for example f f (3) = 2(2 × 3) = 12. The composite function f f can instead be written as f 2 . So in the above example, we’d write f 2 (3) = 12. We can, analogously, define the composite function f f 2 and denote it f 3 . Using the above example, f 3 (x) = 8x and f 3 (3) = 24. Of course, there are also f 4 , f 5 , etc.
Page 75, Table of Contents
www.EconsPhDTutor.com
Remark 5. The official A-level syllabus uses f 2 to mean the composite function f f and nothing else. So this is what we’ll do in this textbook. But confusingly enough, some writers use the symbol f 2 to mean “the second derivative of f ”, f 3 to mean “the third derivative of f ”, etc.. We won’t follow such practice. Just to let you know, in case you read other mathematical texts and get confused. However, we will use f (3) to mean “the third derivative of”, f (4) to mean “the fourth derivative of”, etc. This will show up occasionally in Part V (Calculus). Exercise 48. For each of the following functions f , verify that the composite function f 2 exists and write it out in full. Also, compute f 2 (1) and f 2 (2). (a) The function f ∶ R → R defined by x ↦ ex . (b) The function f ∶ R → R defined by x ↦ 3x + 2. (c) The function f ∶ R → R defined by x ↦ 2x2 + 1. (Answer on p. 1016.)
Page 76, Table of Contents
www.EconsPhDTutor.com
4
Graphs
An ordered pair is a mathematical object. Like set of two objects, an ordered pair is, informally, a “container” with two objects, where the objects are listed out with a comma separating them. The only difference between a set of two objects and an ordered pair is that order matters for the latter. To distinguish an ordered pair from a set of two objects, we use parentheses (instead of braces). Example 67. (Cow, Chicken) is an ordered pair. (−5, 4) is an ordered pair. We also refer to (a, b) as ordered set notation. So (Cow, Chicken) and (−5, 4) are both examples of ordered pairs, written out in ordered set notation.
Example 68. Let (Cow, Chicken) and (Chicken, Cow) be ordered pairs. Let {Cow, Chicken} and {Chicken, Cow} be sets. Recall that for sets, order did not matter. Hence, {Cow, Chicken} = {Chicken, Cow}. In contrast, for ordered pairs, order does matter. And so (Cow, Chicken) ≠ (Chicken, Cow). Definition 19. An ordered pair of real numbers is any (x, y) where both x, y are real. Example 69. (−5, 4), (1, 1), and (2, −3) are all ordered pairs of real numbers. Confusingly, above in Section 1.9 (Intervals), we said that (−5, 4) was a set, namely {x ∈ R ∶ −5 < x < 4}. Here we say instead that (−5, 4) is an ordered pair, consisting of two objects (−5 and 4), the order of which matters. Unfortunately this is yet another bit of confusing notation you’ll have to live with. You’ll have to learn to tell, from the context, whether (−5, 4) is a set of infinitely-many real numbers or an ordered pair. But don’t worry, this is usually pretty obvious.
Page 77, Table of Contents
www.EconsPhDTutor.com
Definition 20. In any ordered pair of real numbers, the first real is called the x-coordinate and the second is the y-coordinate.
Definition 21. The cartesian plane is the set of all ordered pairs of real numbers.
In set-builder notation, the cartesian plane can be written as {(x, y) ∶ x ∈ R, y ∈ R}. This reads aloud as “the cartesian plane is the set of ordered pairs of real number (x, y)”. In this textbook, we’ll usually only ever look at ordered pairs of real numbers. Hence, rather than say “ordered pair of real numbers”, we’ll simply say “ordered pair”. And so whenever you see the notation (x, y), it should be understood that this is an ordered pair of real numbers (and not cows or chickens). And so instead of writing the cartesian plane as {(x, y) ∶ x ∈ R, y ∈ R}, we’ll simply write it as {(x, y)}, with the understanding that x, y are reals. In the present context, we’ll also simply call any ordered pair of real numbers a point. (Later on, in the context of three-dimensional geometry, points will also refer to ordered triples of real numbers.) Definition 22. In the context of the cartesian plane, the origin is the point (0, 0). Points are usually given lower-case letters as names.
Page 78, Table of Contents
www.EconsPhDTutor.com
We can illustrate the cartesian plane graphically. The horizontal axis corresponds to the x-coordinate of the points and is thus also called the x-axis. The vertical axis corresponds to the y-coordinate of the points and is thus also called the y-axis. Example 70. The points (or ordered pairs of real numbers) a = (−5, 4), b = (1, 1), and c = (2, −3) are illustrated graphically on the cartesian plane:
5
y
a 3
1
b x
-5
-3
-1
1
3
5
-1
-3
c
-5
Definition 23. A graph (or curve) is any set of points.
Example 71. The set of three points {a, b, c} = {(−5, 4), (1, 1), (2, −3)} is a graph.
Page 79, Table of Contents
www.EconsPhDTutor.com
The graph of a function f is simply the set of points (x, y) that satisfy x ∈ D, y ∈ C, and f (x) = y. Formally: Definition 24. The graph of a function f ∶ D → C is the set {(x, y) ∶ x ∈ D, y ∈ C, y = f (x)}. Given a function that is named with a lower-case letter, we will often use the upper-case version of that same letter to denote that function’s graph. So for example, given the function f , we often give its graph the name F . Example 72. Consider the function f ∶ R → R defined by x ↦ x2 . Its graph may be written as F = {(x, y) ∶ y = x2 }.
4
f(x), y
2
x 0 -2
-1
0
0
1
2
We’ve defined graph as a noun. But at the slight risk of confusion, we’ll also use it as a verb that means “draw in the cartesian plane a given set of points”. So we can say either “we draw the graph of f ” (graph as a noun), or “we graph f ” (graph as a verb).
Page 80, Table of Contents
www.EconsPhDTutor.com
Definition 25. Given a function f ∶ D → C, a point of the function is any element of its graph. That is, it is any ordered pair (x, f (x)), where x ∈ D. To use the above example, we say that (2, 4) and (5, 25) are both points of f . But since x determines f (x), it is nice but not necessary to specify the complete ordered pair (x, f (x)). Instead, we can refer to the point simply as x. So in the above example, we can simply say that “2 and 5 are both points of f ”, with the understanding that what we really mean is “(2, 4) and (5, 25) are both points of f ”. This is a bit sloppy and at the risk of some confusion, but will save us a lot of messy notation. So in the context of functions, x does double duty. It can either refer to an element in the function’s domain OR it can refer to a point of the function. On exams though, it is probably safer to simply list out the full co-ordinates, whenever you’re referring to a point. Just in case your marker is damn niao.
Page 81, Table of Contents
www.EconsPhDTutor.com
We just learnt about the graph of a function. A graph of an equation is very similarly defined: Definition 26. Given an equation involving x and y as the only two variables, the graph of the equation is the set of points (x, y) that satisfy the equation. Example 73. The graph below is of the equation x2 + y 2 = 1. It is simply the set {(x, y) â&#x2C6;ś x2 + y 2 = 1}.
1 f(x), g(x), y p = (x, y)
x
x 2 + y2 = 1
1
y
y x
-1.0
-0.5
0.0
0.5
1.0
(-0, 0) Centre
-1
Exercise 49. (a) Can the equation x2 + y 2 = 1 be rewritten into the form of a single function? (b) Can it be rewritten into the form of two functions? (Answer on p. 1017.) Exercise 50. Draw the graphs of each of the following equations. (a) y = ex . (b) y = 3x + 2. (c) y = 2x2 + 1. (Answers on pp. 1017, 1018, and 1019.)
Page 82, Table of Contents
www.EconsPhDTutor.com
4.1
Graphing with Your TI84 Graphing Calculator
You are required to know how to use a graphing calculator to graph a given function. Pretty bizarre that in this age of the smartphone, they want you to learn how to use these clunky and now-useless devices from the ’80s and ’90s. It is the equivalent of having to learn how to program a VCR. This textbook will give only a very few examples involving graphing calculators. There is no better way of learning to use it than to play around with it yourself. By the time you sit down for your A-level exams, you will have had plenty of practice with it. You can also use any of the seven calculators in the list below (last updated by SEAB on March 1st, 2016, PDF). But this textbook will stick with the TI-84 PLUS Silver Edition (which I’ll simply call the TI84).18
I’ll always start each example with the calculator freshly reset. Example 74. Graph the function f ∶ R → R defined by x ↦ x2 . 1. Press ON to turn on your calculator. 2. Press Y= to bring up the Y= editor. 3. Press X,T,θ,n to enter “X”; then x2 to enter the squared “2 ” symbol. 4. Now press GRAPH and the calculator will graph y = x2 . After Step 1.
18
After Step 2.
After Step 3.
After Step 4.
My understanding is that most students use a TI calculator and that the five approved TI calculators are pretty similar.
Page 83, Table of Contents
www.EconsPhDTutor.com
Example 75. Graph the equation x2 + y 2 = 1. The TI84 requires that we enter equations in a form where y is directly expressed in terms of y. But there is no way to do this here. As explained in Exercise 49, there is no way to rewrite this equation into a single function. However, we can rewrite it into two functions: √ Namely, f ∶ [−1, 1] → R defined by x ↦ √ 2 1 − x2 and g ∶ [−1, 1] → R defined √ by x ↦ − 1 − x√. We can thus tell our calculator to graph two separate equations: y = 1 − x2 and y = − 1 − x2 . 1. Press ON to turn on your calculator. 2. Press Y= to bring up the Y= editor. Most buttons on the TI84 have three different roles. If you simply press a button, then the TI84 executes the role that is printed on the button itself. If you press the blue 2ND and then a button, then the TI84 executes the role printed in blue above the button. And if you press the green ALPHA and then a button, then the TI84 exexcutes the role printed in green above the button. √ (which corresponds to the x2 button) to enter 3. Press the blue 2ND button and then √ “ (”. Next press 1 − X,T,θ,n x2 ) to enter “1 − X 2 )”. Altogether you will have √ entered 1 − x2 . 4. Now press ENTER and the blinking cursor will move down, to the right of “Y2 =”. After Step 1.
After Step 2.
After Step 3.
After Step 4.
(... Example continued on the next page ...)
Page 84, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) After Step 5.
After Step 6.
After Step 8.
After Step 9.
After Step 7.
We’ll now enter the second equation. 4. Press the (-) button. (Warning: This is different from the - button. If you use the button, you will get an error message when you try to generate your graphs later.) Now √ repeat what we did in step 3 above: Press the blue 2ND button and then (which √ corresponds to the x2 button) to enter “ (”. Next press 1 − X,T,θ,n x2 ) to enter √ 2 “1 − X )”. Altogether you will have entered − 1 − x2 . √ √ 5. Now press GRAPH and the calculator will graph both y = 1 − x2 and y = − 1 − x2 . Notice the graphs are very small. To zoom in: 6. Press the ZOOM button to bring up a menu of ZOOM options. 7. Press 2 to select the Zoom In option. Nothing seems to happen. But now press ENTER and the TI84 will zoom in a little for you. We expected to see a perfect circle. Instead we get an elongated oval – what’s going on? The reason is that by default, the x- and y- axes are scaled differently. To set them to the same scale: 8. Press the ZOOM button again to bring up the ZOOM menu of options. Press 5 to select the ZSquare option. Nothing seems to happen. But now press ENTER and the TI84 will adjust the x- and y- axes to have the same scale. And now we have a perfect circle.
Page 85, Table of Contents
www.EconsPhDTutor.com
5
Quick Revision: Exponents, Surds, Absolute Value
Here’s a super quick revision of some O-Level Maths we’ll be using. If you have severe difficulty with these exercises, you should go back and review your O-Level Maths material!
5.1
Laws of Exponents
For all real numbers x, we have x1 = x and x0 = 1.19 For all real numbers x, y, a, and b (provided any denominators are non-zero): x ⋅x a
b
=x
a+b
x a xa ( ) = a, y y
,
xa = xa−b , b x (xa )
b
= xab ,
(xy)a = xa y a ,
x−a =
1 , xa
a1/b =
√ b a,
ac/b =
√ √ c b ac = ( b a) .
Exercise 51. Simplify the two expressions below. (Answer on p. 1020.) (53x ⋅ 251−x ) 52x+1 + 3(25x ) + 17(52x )
(8x+2 − 34(23x )) . √ 2x+1 ( 8)
Exercise 52. Is it true that x(a ) = xab ? (Answer on p. 1021.) b
19
By convention, 00 is usually defined to be equal to 1 – this textbook will follow this practice.
Page 86, Table of Contents
www.EconsPhDTutor.com
5.2
Rationalising the Denominator of a Surd
Example 76. Here’s a case where there’s just a surd in the denominator: √ √ 1 2 2 √ =√ √ = . 2 2 2× 2 For more complicated cases, the trick is to use the fact that (a + b)(a − b) = a2 − b2 . Given a + b, we call a − b the conjugate of a + b. We refer to a + b and a − b as a conjugate pair. Example 77. √ √ √ √ 1− 2 1− 2 1− 2 √ 1 1− 2 √ = √ √ = = = = 2 − 1. √ 1−2 −1 1 + 2 (1 + 2) (1 − 2) 12 − ( 2)2
Exercise 53. Prove the following equality: (Answer on p. 1021.)
x y
Page 87, Table of Contents
+
1 √
x2 y2
√ +1
=
x x2 +1− . 2 y y
www.EconsPhDTutor.com
5.3
Absolute Value
The notation ∣z∣ returns the absolute value of z. ⎧ ⎪ ⎪ ⎪z, ∣z∣ = ⎨ ⎪ ⎪ ⎪ ⎩−z,
if z ≥ 0, if z < 0.
(See footnote for a more formal definition.20 ) √ √ √ √ Example 78. ∣4∣ = 4 and ∣−4∣ = 4. ∣ 2∣ = 2 and ∣− 2∣ = 2.
Fact 2. (a) ∣x∣ < b ⇐⇒ −b < x < b. (b) ∣x∣ ≤ b ⇐⇒ −b ≤ x ≤ b. Proof. (a) ∣x∣ < b ⇐⇒ “0 ≤ x < b OR −b < x < 0” ⇐⇒ −b < x < b. (b) Very similar.
Fact 3. (a) ∣x − a∣ < b ⇐⇒ a − b < x < a + b. (b) ∣x − a∣ ≤ b ⇐⇒ a − b ≤ x ≤ a − b. Proof. (a) By Fact 2, ∣x − a∣ < b if and only if −b < x − a < b. Rearranging the latter set of inequalities yields a − b < x < a + b. (b) Very similar.
20
The absolute value operator ∣⋅∣ is the function with domain R, codomain R+0 , and mapping rule x ↦ x if x ≥ 0 and x ↦ −x if x < 0.
Page 88, Table of Contents
www.EconsPhDTutor.com
Fact 4.
∣a∣ a = ∣ ∣ (provided b ≠ 0). ∣b∣ b
Proof. If a = 0, then clearly this is true. a a a ≥ 0 so that ∣ ∣ = . Moreover, b b b ∣a∣ a ∣a∣ −a a ∣a∣ a either = or = = . Altogether then, indeed = ∣ ∣. ∣b∣ b ∣b∣ −b b ∣b∣ b If a and b have the same signs (and are non-zero), then
a a a < 0 so that ∣ ∣ = − . Moreover, b b b ∣a∣ −a ∣a∣ a ∣a∣ a either = or = . Altogether then, indeed = ∣ ∣. ∣b∣ b ∣b∣ −b ∣b∣ b If a and b have opposite signs (and are non-zero), then
Page 89, Table of Contents
www.EconsPhDTutor.com
6
Intercepts
Example 79. The graph below is of the equation y = x + 3. It has horizontal intercept −3 and vertical intercept 3.
5 y y=x+3 3
1 x -5
-3
-1
1
3
5
-1
-3
-5
Horizontal intercepts are the x-coordinates of the points at which the graph intersects the horizontal or x-axis. Similarly, vertical intercepts are the y-coordinates of the points at which the graph intersects the vertical or y-axis. Definition 27. a is a horizontal intercept (or x-intercept) of a graph G if (a, 0) ∈ G. Definition 28. b is a vertical intercept (or y-intercept) of a graph G if (0, b) ∈ G.
Page 90, Table of Contents
www.EconsPhDTutor.com
Where the graph G is of an equation (or function), we sometimes also call the horizontal intercepts zeros or roots of the equation (or function). (We’ll use the terms zeros and roots interchangeably in this textbook.) Example 80. Consider the equation y = x2 − 1. Its zeros or roots are −1 and 1, because these are the values of x for which y = 0. Of course, −1 and 1 are also the horizontal intercepts (or x-intercepts) of the graph of the equation. Example 81. Consider the the function f with domain R, codomain R, and mapping rule x ↦ x2 − 1. Its zeros or roots are −1 and 1, because these are the values of x for which f (x) = 0. Of course, −1 and 1 are also the horizontal intercepts (or x-intercepts) of the graph of f .
Page 91, Table of Contents
www.EconsPhDTutor.com
Example 82. The graph below is of the function f ∶ R → R defined by x ↦ x2 − 1. It has vertical intercept −1 and two horizontal intercepts, −1 and 1. −1 and 1 are also the zeros or roots of f , because f (−1) = 0 and f (1) = 0.
3
f(x)
f(x) = x2 - 1
2
1
x 0 -2
-1
0
1
2
-1
The A-level exams will often ask you to write down the full co-ordinates of the points at which a graph (or curve) crosses the axes — this means writing down both the x- and y-coordinates, and not just the horizontal intercept or the vertical intercept. Here’s an exercise to help you make this a habit. Exercise 54. Write down in full the point(s) at which the graphs of each the following equations crosses the axes: (a) x2 +y 2 = 1. (b) y = x2 −4. (c) y = x2 +2x+1. (d) y = x2 +2x+2. (Answer on p. 1022.)
Page 92, Table of Contents
www.EconsPhDTutor.com
7 7.1
Symmetry
Reflection of a Point in a Line
A reflection of a point in a line is its mirror image point on that line. Formally: Definition 29. Let a be a point and l1 be a line. Let l2 be the line that is perpendicular to l1 and runs through a. Let x be the point where l1 and l2 intersect. Then the reflection of a in l1 is the point a′ on l2 such that the distances ax and a′ x are equal.
l1 l2
a
x
a'
Fact 5. Let (a, b) be a point. Its reflection in the line y = x is the point (b, a).
Proof. Optional, see p. 924 in the Appendices.
Fact 6. Let (a, b) be a point. Its reflection in the line y = −x is the point (−b, −a).
Proof. Optional, see p. 924 in the Appendices.
Example 83. (a) Given the point (3, 17), its reflection in the line y = x is (17, 3) and its reflection in the line y = −x is (−17, −3). (b) Given the point (−1, 5), its reflection in the line y = x is (5, −1) and its reflection in the line y = −x is (−5, 1). (c) Given the point (0, 0), its reflection in the line y = x is (0, 0) and its reflection in the line y = −x is (0, 0). Exercise 55. For each of the following points, write down their reflections in the lines (i) y = x; and (ii) y = −x. (a) (3, 17). (b) (−1, 5). (c) (0, 0). (Answer on p. 1023.)
Page 93, Table of Contents
www.EconsPhDTutor.com
7.2
Reflection of a Graph in a Line
Definition 30. The reflection of a graph G in a line is the graph G′ where each point in G′ is a reflection of a point in G.
Example 84. The reflection of the graph G = {(x, y) ∶ y = x2 + 4} in the line y = 2 is the graph G′ = {(x, y) ∶ y = −x2 }.
y
G : y = x2 + 4
y=2 line of reflection
x
G ' : y = -x2
Page 94, Table of Contents
www.EconsPhDTutor.com
Example 85. The reflection of the graph G = {(x, y) ∶ y = ln x} in the line x = 0 is the graph G′ = {(x, y) ∶ y = ln(−x)}.
y x=0 line of reflection
G ' : y = ln (-x)
G : y = ln x x
Fact 7 formalises our earlier observation in section 3.7 (Inverse Functions) that the graphs of f and its inverse f −1 are reflections in the line y = x. Fact 7. Let f be an invertible function. Then the reflection of the graph of f in the line y = x is the graph of its inverse function f −1 .
Proof. Optional, see p. 925 in Appendices.
Page 95, Table of Contents
www.EconsPhDTutor.com
The next Fact simply makes the obvious observation that the reflection in the line y = x of any point along the line y = x is itself. Fact 8. Let (a, a) be a point. Its reflection in the line y = x is (a, a).
The above two Facts together imply that Fact 9. Let f be invertible. Suppose f passes through (a, a). Then so too does its inverse f −1 . And hence, f and f −1 intersect at those points where x = f (x).
The above Fact is useful for finding where a function and its inverse intersect. Example 86. Let f ∶ R → R be the invertible function defined by x ↦ 2x. The graph of f intersects the graph of f −1 at the point(s) where x = f (x) ⇐⇒ x = 2x ⇐⇒ x = 0. Notice the intersection point (0, 0) is also on the line y = x. See figure on p. 70. Example 87. Let f ∶ R+0 → R be the invertible function defined by x ↦ x2 . The graph of f intersects the graph of f −1 at the point(s) where x = f (x) ⇐⇒ x = x2 ⇐⇒ x(x − 1) = 0 ⇐⇒ x = 0, 1. Notice the intersection points (0, 0) and (1, 1) are also on the line y = x. See figure on p. 72.
Be careful not to make the mistake of believing that f and f −1 can only intersect at points where x = f (x). A function and its inverse can certainly intersect at points that are not on the y = x line. Example 88. Let f ∶ R+ ∪ R− → R be the invertible function defined by x ↦ 1/x. The graph of f intersects the graph of f −1 at the point(s) where x = f (x) Ô⇒ x = 1/x Ô⇒ x = −1, 1. We have merely found two points at which f and f −1 intersect. There may very well be other intersection points. Indeed, in this example, f and f −1 also intersect at every other x ≠ 0! See figure on p. 71.
Page 96, Table of Contents
www.EconsPhDTutor.com
7.3
Lines of Symmetry
Definition 31. A graph is symmetric in a line if it is unchanged after being reflected in that line.
Example 89. The graph of y = x2 is symmetric in the line x = 0 (which also happens to be the vertical axis).
4 y x=0 Reflection line 3 y = x2
2
1
x 0 -2
Page 97, Table of Contents
-1
0
1
2
www.EconsPhDTutor.com
Example 90. The graph of y =
1 is symmetric in the lines y = x and y = â&#x2C6;&#x2019;x. x
5 y 4 y = -x line
y=x line
3 2 1
y=1/x 0 -5
-4
-3
-2
-1
0 -1
1
2
3
4
5 x
-2 -3 -4 -5
Page 98, Table of Contents
www.EconsPhDTutor.com
8
Limits, Continuity, and Asymptotes
The syllabus makes nearly no mention of limits and none of continuity. Yet differentiation and integration are built entirely on the concept of limits. Continuity is also almost always assumed. It is thus well-worth spending an hour or two on these concepts, especially since they’re not difficult and everything will become that much clearer.
8.1
Limits: Introduction and Examples
Here is a very simple example to illustrate the idea of limits. Example 91. Graphed below is the function f ∶ R → R defined by x ↦ 5x + 2. Observe that “as x approaches 3, f (x) approaches 17”. We write this as: Statement #1. As x → 3, f (x) → 17. (The right arrow symbol “→” means to in the context of functions, but now means approaches in the context of limits.) Equivalently, we may say “The limit of f (x) as x approaches 3 is equal to 17.” We write: Statement #2. lim f (x) = 17. x→3
Statements #1 and #2 are entirely equivalent. Either may be (informally) interpreted thus:
For all values of x that are close to but not equal to 3, f (x) is close to (or possibly even equal to) 17. y
x -5
Page 99, Table of Contents
-3
-1
1
3
5
www.EconsPhDTutor.com
In general, lim f (x) = L is informally interpreted as: x→a
For all values of x that are close to but not equal to a, f (x) is close to (or possibly even equal to) L. This interpretation is informal because the words “close to” are vague. For the formal definitions of limits (optional), see Section 88.1 in the Appendices. The subtle condition “but not equal to” requires emphasis. When considering the limit of f at 3, we do NOT care about the value f (3). Indeed, we do NOT care even if f (3) is undefined! Here’s an example where lim g(x) is well-defined, even though g(3) is not. x→3
Example 92. Graphed below is the function g ∶ (−∞, 3)∪(3, ∞) → R defined by x ↦ 5x+2. It looks almost exactly like that of f (from the previous example), except there is now a “hole” (or more formally, a discontinuity) at x = 3. Nonetheless, it is still true that
For all values of x that are close to but not equal to 3, g(x) is close to (or possibly even equal to) 17. In formal notation, we write “as x → 3, g(x) → 17” or “lim g(x) = 17”. x→3
y
x -5
Page 100, Table of Contents
-3
-1
1
3
5
www.EconsPhDTutor.com
In the next example, both h(3) and lim h(x) are well-defined, but lim h(x) ≠ h(3). x→3
x→3
Example 93. Graphed below is the function h ∶ R → R defined by x ↦ 5x + 2 for x ≠ 3 and h(3) = 0. The graph of h looks almost exactly like those of f and g (from the previous examples), except that now the value of h at x = 3 is, strangely enough, 0. Nonetheless, it is still true that
For all values of x that are close to but not equal to 3, h(x) is close to (or possibly even equal to) 17. In formal notation, we write “as x → 3, h(x) → 17” or “lim h(x) = 17”. x→3
y
x -5
-3
-1
1
3
5
The next example is similar.
Page 101, Table of Contents
www.EconsPhDTutor.com
Example 94. Graphed below is the function i ∶ R → R defined by x ↦ 0 for x ≠ 3 and i(3) = 17. This graph looks very different from those of f , g, and h (from the previous examples). But like f , we again have i(3) = 17. We are tempted to conclude that therefore lim i(x) = 17. This though is wrong, because x→3
we cannot make i(x) as close to 17 as we like by restricting x to values that are close to but not equal to 3. Hence, As x → 3, i(x) → / 17,
or equivalently, lim i(x) ≠ 17. x→3
Instead, lim i(x) = 0, because: x→3
For all values of x that are close to but not equal to 3, i(x) is close to (or possibly even equal to) 0.
y
x -5
-3
-1
1
3
5
Section 8.3 gives more examples of limits. But first, let’s learn about continuity.
Page 102, Table of Contents
www.EconsPhDTutor.com
8.2
Continuity
Informally, a function is continuous at a point a if there is no “hole” or “jump” at a. And so a function is continuous on an interval (of points) if you can smoothly draw its graph for that entire interval without once lifting your pencil. Formally: Definition 32. f ∶ D → R is continuous at a ∈ D if lim f (x) = f (a). x→a
Section 88.6 in the Appendices contains additional definitions and results concerning continuity (optional). Example 91 (revisited). Graphed below is the function f ∶ R → R defined by x ↦ 5x + 2. It is continuous at 3, because lim f (x) = 17 and f (3) = 17. x→3
It is also continuous at 1, because lim f (x) = 7 and f (1) = 7. x→1
Indeed, it is continuous on R, because for any a ∈ R, we have lim f (x) = f (a). x→a
20
y
10 f is continuous everywhere -5
-3
0 -1
x 1
3
5
-10
-20
Page 103, Table of Contents
www.EconsPhDTutor.com
Example 92 (revisited). Graphed below is the function g ∶ (−∞, 3) ∪ (3, ∞) → R defined by x ↦ 5x + 2. It is continuous at 1, because lim g(1) = 7 and g(1) = 7. x→5
However, it is not continuous at 3, because lim g(x) = 17, but g(3) is undefined, and so x→3
lim g(x) ≠ g(3). x→3
Altogether, g is continuous at any a ∈ (−∞, 3)∪(3, ∞), because for any a ∈ (−∞, 3)∪(3, ∞), we have lim g(x) = g(a). But g fails to be continuous at 3. x→a
y
g is continuous everywhere except at x = 3. x -5
Page 104, Table of Contents
-3
-1
1
3
5
www.EconsPhDTutor.com
Example 93 (revisited). Graphed below is the function h ∶ R → R defined by x ↦ 5x + 2 for x ≠ 3 and h(3) = 0. It is continuous at 1, because lim h(x) = 7 and h(1) = 7. x→1
However, it is not continuous at 3, because lim h(x) = 17, but h(3) = 0 and so lim h(x) ≠ x→3
x→3
h(3). Altogether, h is continuous at any a ∈ (−∞, 3)∪(3, ∞), because for any a ∈ (−∞, 3)∪(3, ∞), we have lim h(x) = h(a). But h fails to be continuous at 3. x→a
y h is continuous everywhere except at x = 3. x -5
Page 105, Table of Contents
-3
-1
1
3
5
www.EconsPhDTutor.com
8.3
Limits: More Examples
We now turn to examples where limits do not exist. We start with a trivial example. Example 95. Graphed below is the function f ∶ R+0 → R defined by x ↦ 5x + 2. There is no number L such that for all values of x that are “close to” but not equal to −3, f (x) is also “close to” L. And so we simply say that lim f (x) does not exist. x→−3
This is a trivial example because −3 is “far from” the domain of f . So obviously, for all values of x that are “close to” but not equal to −3, f (x) is undefined and so of course there is no number L that f (x) is always “close to”!
y
x -5
-3
-1
1
3
5
is undefined.
The next example is less trivial.
Page 106, Table of Contents
www.EconsPhDTutor.com
1 x for all x ≠ 0. This is a very strange function. As x gets ever “closer to” 0, g(x) fluctuates ever more rapidly between −1 and 1.
Example 96. Graphed below is the function g ∶ R → R defined by g(0) = 0 and g(x) = sin
y
x
It’s difficult or even impossible to draw an accurate graph of g near the origin. In this example, lim g(x) does not exist. The reason is that for all values of x that are x→0
“close to” but not equal to 0, there is no number L that g(x) is “close to”. When x is “close to” 0, g(x) takes on every value in [−1, 1] infinitely often! And so g(x) can never be said to be “close to” any one single number L. Altogether then, g is not continuous at 0. (With a little work, we can actually prove that g is continuous on R− and also on R+ , but this is beyond the scope of A-levels.)
In the next example, h is nowhere-continuous!
Page 107, Table of Contents
www.EconsPhDTutor.com
Example 97. Graphed below is the function h ∶ R → R defined by x ↦ 1 if the decimal form representation of x contains the digit 7 and x ↦ 2 otherwise. This function is arguably even stranger than the previous one. We have for example, h(7) = h(70) = h(1.27) = h(0.0007) = 1 and h(15) = h(16) = h(16.335) = 2. There are infinitely many points along the line y = 1. And there are also infinitely many points along the line y = 2! It is quite impossible to sketch its graph accurately. Nonetheless, h is a perfectly well-defined function. Indeed, h(3) is well-defined and h(x) is well-defined for any x ∈ R. However, lim h(x) does not exist. However we try to restrict x to values that are close to x→3
(but not equal to) 3, h(x) is never close to any one single value; instead, h(x) switches infinitely often between 1 and 2. Indeed, lim h(x) does not exist for any a ∈ R! However we try to restrict x to values that x→a are close to (but not equal to) a, h(x) is never close to any one single value; instead, h(x) switches infinitely often between 1 and 2. h is nowhere-continuous: For every a ∈ R, h(a) is perfectly well-defined, lim h(x) is not. x→a And so for every a ∈ R, lim h(x) ≠ h(a). x→a
y 2
1 is nowherecontinuous. x
⎧ ⎪ ⎪ ⎪1, Exercise 56. Consider the function f ∶ R → R defined by f (x) = ⎨ ⎪ ⎪2, ⎪ ⎩ lim f (x), lim f (x), and lim f (x)? (Answer on p. 1024.) x→−5
x→0
Page 108, Table of Contents
if x ≤ 0, if x > 0.
What are
x→5
www.EconsPhDTutor.com
8.4
Infinite Limits and Vertical Asymptotes
This section considers infinite limits, i.e. where as x approaches some number, f (x) increases (or decreases) grows without bound . Example 98. Graphed below are the functions f and g, both with domain (−∞, 3)∪(3, ∞) 1 1 and codomain R, defined by f ∶ x ↦ 2 + and g ∶ x ↦ −2 − . 2 (x − 3) (x − 3)2
y
vertical asymptote -2
-1
0
1
2
x 3
4
5
6
7
8
Observe that for all values of x that are “close to” but not equal to 3, there is no number L that f (x) is “close to”. Hence, we say that lim f (x) simply does not exist. Similarly, x→3
lim g(x) does not exist either. x→3
Nonetheless, we observe that as x → 3, f (x) increases without bound, while g(x) decreases without bound. By a very special convention, we are allowed to write these observations as: lim f (x) = ∞ and x→3
lim g(x) = −∞. x→3
We say that x = 3 is a vertical asymptote of both the graph of f and that of g. “lim f (x) = ∞” must NOT be interpreted to mean that there exists something called x→3
“lim f (x)” (no such thing exists); or that this thing is equal to some other thing called “∞” x→3
(recall that ∞ is not a number!). Instead, “lim f (x) = ∞” is interpreted informally as: x→3
f (x) can be made as large as we like, for all values of x that are sufficiently “close to” but not equal to 3. Again, see Section 88.3 in the Appendices (optional) for the formal definitions.
Page 109, Table of Contents
www.EconsPhDTutor.com
Here is another example of vertical asymptotes. Example 99. Graphed below is the equation y = tan x. It has two vertical asymptotes x = ±π/2, because lim = −∞ and lim = ∞. x→−π/2
x→π/2
15
Vertical asymptote x = - π /2
y
10
5
y = tan x x
0 π/2
π/2 -5
Vertical asymptote x = π /2 -10
-15
Page 110, Table of Contents
www.EconsPhDTutor.com
8.5
Limits at Infinity, Horizontal and Oblique Asymptotes
This section considers limits at infinity (not to be confused with the infinite limits discussed in the previous section). That is, the behaviour of f (x) as x increases (or decreases) grows without bound. Example 98 (revisited). Reproduced below are the graphs of the functions f and g, 1 and both with domain (−∞, 3) ∪ (3, ∞) and codomain R, defined by f ∶ x ↦ 2 + (x − 3)2 1 g ∶ x ↦ −2 − . (x − 3)2 We already saw that f and g both have vertical asymptote x = 3, because as x → ∞, f (x) increases without bound and g(x) decreases without bound. We now consider instead what happens as x increases or decreases without bound.
y
horizontal asymptotes
x
As x increases without bound, f (x) → 2 and g(x) → −2. And as x decreases without bound, f (x) → 2 and g(x) → −2. We can write these observations as lim f (x) = 2, x→∞ lim g(x) = −2, lim f (x) = 2, and lim g(x) = −2.
x→∞
x→−∞
x→−∞
We also say that y = 2 is a horizontal asymptote of the graph of f . Similarly, y = −2 is a horizontal asymptote of the graph of g. [See Section 88.4 in the Appendices (optional) for the formal definition of a horizontal asymptote.] Pedantic point: Infinite limits do not exist. In contrast, limits at infinity DO exist. Here in this example, lim f (x) does not exist. In contrast, lim f (x) and lim f (x) both exist x→∞ x→−∞ x→3 (and are both equal to 2).
Page 111, Table of Contents
www.EconsPhDTutor.com
Example 100. Graphed below is the equation y = ex . As x → −∞, y → 0. We can also write this as lim y = 0. And we can also say that this x→−∞ graph has horizontal asymptote y = 0.
20
y
y = ex
15
10
5
Horizontal asymptote x y=0 0 -4
Page 112, Table of Contents
-2
0
2
4
www.EconsPhDTutor.com
The previous two examples were of horizontal asymptotes. Next is an example of an oblique (or slant) asymptote. 1 Example 101. Consider the function f ∶ R− ∪ R+ → R defined by x ↦ x + . x As x increases without bound or decreases without bound, f (x) approaches the line y = x. We can also write these observations as lim f (x) = x and lim f (x) = x. x→∞
x→−∞
We can moreover say that y = x is an oblique asymptote of the graph of g. Again, see Section 88.4 in the Appendices (optional) for the formal definition of an oblique asymptote.
5
y
3
1 x -5
-3
-1
1
3
5
-1 Oblique asymptote y=x -3
-5
Page 113, Table of Contents
www.EconsPhDTutor.com
9 9.1
Differentiation
Motivation: The Derivative as Slope of the Tangent
The problem of finding the derivative is the problem of finding the slope of the tangent to a graph at a given point. Graphed below is some function f ∶ R → R. Pick some point A = (a, f (a)). Draw the line l which is tangent to the graph at the point A. How do we find the slope of l? Unsure of how to proceed, we try a crude approximation. Pick some point X1 = (x1 , f (x1 )) that is also on the graph. Consider the line AX1 . What’s f (x1 ) − f (a) its slope? Slope = Rise ÷ Run and so AX1 has slope . x1 − a This number serves as our first crude approximation of the slope of l. How can we improve on this approximation? Simple — just pick some point X2 = (x2 , f (x2 )) f (x2 ) − f (a) that is closer to A. The line AX2 has slope . x2 − a This number serves as our second, improved approximation of the slope of l.
y f (x1)
X1
l f (x2) X2 A f (a)
y = f (x)
x a
x2
x1
At least in theory, we can keep repeating this procedure, by picking points that are ever closer to A. Our estimates of the slope of l will get ever better. Altogether then, we are motivated to make the following formal definition of the derivative: Page 114, Table of Contents
www.EconsPhDTutor.com
Definition 33. Let f ∶ D → R be a function. Consider f (x) − f (a) . x→a x−a lim
If this limit exists, then we say that f is differentiable at the point a ∈ D and we call this limit the value of f ’s derivative at the point a ∈ D. But if this limit does not exist, then we say that f is not differentiable at the point a ∈ D and the value of f ’s derivative at the point a ∈ D is undefined or does not exist.
Here’s a simple example to illustrate.
Page 115, Table of Contents
www.EconsPhDTutor.com
Example 102. Graphed below is the function f ∶ R → R defined by x ↦ ∣x∣.
y Derivative = -1 for a < 0
Derivative = 1 for a > 0 Derivative does not exist at a = 0.
x
The value of f ’s derivative at the point x = −5 is f (x) − f (−5) ∣x∣ − ∣−5∣ −x − 5 = lim = lim = lim −1 = −1. x→−5 x→−5 x + 5 x→−5 x + 5 x→−5 x − (−5) lim
Similarly, the value of f ’s derivative at the point x = −3 is also ∣x∣ − ∣−3∣ −x − 3 f (x) − f (−3) = lim = lim = lim −1 = −1. x→−5 x + 3 x→−3 x + 3 x→−3 x→−3 x − (−3) lim
Indeed, the value of f ’s derivative at any point a < 0 is −1, because for any a < 0, f (x) − f (a) ∣x∣ − ∣a∣ −x + a = lim = lim = lim −1 = −1. x→a x→a x − a x→a x − a x→a x−a lim
In contrast, for any a > 0, we have f (x) − f (a) ∣x∣ − ∣a∣ x−a = lim = lim = lim 1 = 1. x→a x − a x→a x − a x→a x→a x−a lim
At x = 0, f is not differentiable, as we now prove: ⎧ x ⎪ ⎪ lim = lim 1 = 1, ⎪ ⎪ x→0 x x→0 ⎪ ⎪ f (x) − f (0) ∣x∣ − ∣0∣ ∣x∣ ⎪ lim = lim = lim =⎨ x→0 x − 0 x→0 x x→0 ⎪ x−0 ⎪ ⎪ −x ⎪ ⎪ ⎪ lim = lim (−1) = −1, ⎪ ⎩x→0 x x→0 So as x → 0, there is no one single value towards which the expression proaches. So the limit does not exist. Page 116, Table of Contents
for x > 0, for x < 0. f (x) − f (0) apx−0
www.EconsPhDTutor.com
9.2
Lagrange’s, Leibniz’s, and Newton’s Notation The Value of the Derivative of f at a
Lagrange’s notation:
Leibniz’s notation:
Newton’s notation:
f (x) − f (a) . x→a x−a
f ′ (a) = lim
R f (x) − f (a) df (x) RRRR RRR = lim x→a dx RR x−a Rx=a ⋅
or
R f (x) − f (a) df RRRR RRR = lim . x→a dx RR x−a Rx=a
f (x) − f (a) . x→a x−a
f (a) = lim
Some remarks. • Lagrange’s and Leibniz’s notation are widely-used. Newton’s notation is not. But Newton’s notation is sometimes used in physics (especially when the independent variable is time). You certainly need to know about Newton’s notation because it is on the A-level syllabus. Nonetheless, this textbook will avoid using Newton’s notation.
d • Leibniz’s notation is convenient in that it allows us to interpret as the “differentiate dx with respect to” operator. Section 9.5 will give some examples of how this operator works.
• Here is the motivation behind Leibniz’s notation. Define ∆x to be equal to x − a, and ∆f (x) to be equal to f (x) − f (a), so that we can write: ∆f (x) f (x) − f (a) = . ∆x x−a The limit of this expression as x → a is precisely the value of the derivative of f at a: R f (x) − f (a) ∆f (x) df RRRR = RRR = lim . lim x→a ∆x x→a dx RR x−a Rx=a
Page 117, Table of Contents
www.EconsPhDTutor.com
Example 102 (revisited). Consider again the function f ∶ R → R defined by x ↦ ∣x∣. a = −5
a=2
a=0
Lagrange’s notation: f ′ (−5) = −1,
f ′ (2) = 1,
f ′ (0) is undefined,
Leibniz’s notation:
R df RRRR = −1, R dx RRRR Rx=−5
R df RRRR R = 1, dx RRRR Rx=2
R df RRRR is undefined, R dx RRRR Rx=0
Newton’s notation:
f (−5) = −1,
⋅
⋅
f (2) = 1,
⋅
f (0) is undefined.
Historical Note (optional). Here is a very oversimplified history of Leibniz’s notation, to give you a better sense of why we use it. Leibniz (1646-1716) thought of dx as an “infinitesimal change in x”. And dy was the corresponding “infinitesimal change in y”. Leibniz then defined the derivative to be literally the quotient dy/dx. Unfortunately, the idea of “infinitesimals” was rather vague, imprecise, and non-rigorous. So in the 19th century, mathematicians embarked on a project to put calculus on a firmer footing. In particular, they wished to rid mathematics of all references to “infinitesimals”. Eventually, they settled on the modern notion of limits, in which no reference to “infinitesimals” was necessary. This modern notion of limits is also what you’ve just learnt. So simply put, Leibniz was wrong to think of the derivative as a fraction. And you should be very careful not to think of the derivative as a fraction, even though it looks very much like one. You are now being taught things in the correct order. First you are taught about limits. Next we define the derivative in terms of limits. We are careful to note that the derivative is not a fraction. But if Leibniz was wrong to think of the derivative as a fraction, then why are we still using his notation? The main reason is that it is highly intuitive. In particular, it reminds us of what calculus is really about — how a small change in one variable affects another variable. It also allows us to quickly grasp the intuition behind such results as the Chain Rule, which may informally be stated as: dz dz dy = . dx dy dx It is tempting to naïvely interpret the expressions in the above equation as fractions, naïvely apply simple algebra, naïvely cancel out the dy’s, so that the equation is indeed true. But Page 118, Table of Contents
www.EconsPhDTutor.com
the correct informal interpretation (easily seen when written in Leibniz’s notation) is this: “The change in z caused by a small unit change in x” is equal to “The change in z caused by a small unit change in y” × “The change in y caused by a small unit change in x”. Another result is the Inverse Function Theorem, which may informally be stated as: dy 1 = . dx dx dy dy dx and as fractions, so that indeed by naïve dx dy algebra, the above equation is true. But again, the correct informal interpretation (easily seen when written in Leibniz’s notation) is this: “The change in y caused by a small unit change in x” is equal to “The reciprocal of the change in x caused by a small unit change in y”. Again, the naïve interpretation would be of
For a more detailed discussion, see the leading answer to this question on Math StackExchange.
Page 119, Table of Contents
www.EconsPhDTutor.com
9.3
The Derivative is a Function
Above we defined the value of the derivative at a given point to be a number. In contrast, we now define the derivative to be a function: Definition 34. Let f ∶ D → R be a function and A be the set of points at which f is differentiable. Then the derivative of f is the function with domain A, the same codomain as f (namely R), and mapping rule x ↦ f ′ (x).
The Derivative of f Lagrange’s notation:
f ′.
Leibniz’s notation:
df (x) dx
Newton’s notation:
f.
or
df . dx
⋅
For the next example, I assume you already know that
d 2 d cx = 2cx and cx = c. dx dx
Example 103. Let f ∶ R → R be defined by f (x) = 7x2 . Its derivative is the function f ′ ∶ R → R defined by f ′ (x) = 14x. This derivative may be denoted f ′ or
⋅ df (x) df or or f . dx dx
⋅ df (x) df ∣ = ∣ = f (0.5) = 1.75. dx x=0.5 dx x=0.5 ⋅ df (x) df The value of the derivative of f at 1 is f ′ (1) = ∣ = ∣ = f (1) = 7. dx x=1 dx x=1 ⋅ df (x) df The value of the derivative of f at 2 is f ′ (2) = ∣ = ∣ = f (2) = 28. dx x=2 dx x=2
The value of the derivative of f at 0.5 is f ′ (0.5) =
Page 120, Table of Contents
www.EconsPhDTutor.com
9.4
Second and Higher-Order Derivatives
The derivative is also known as the first derivative. The second derivative is, similarly, also a function: Definition 35. Let f ∶ D → R be a function. The second derivative of f is simply the derivative of the derivative of f .
The Derivative of f Lagrange’s notation:
f ′′ .
Leibniz’s notation:
d2 f (x) dx2
Newton’s notation:
f.
Under Leibniz’s notation, since derivative of f by
d2 d2 f f or . dx2 dx2
or
d2 f . dx2
⋅⋅
d is the operator, it makes sense to denote the second dx
Example 103 (revisited). Let f ∶ R → R be defined by x ↦ 7x2 . Its second derivative is the function with domain and codomain both R, and mapping rule x ↦ 14. This second derivative may be denoted f ′′ or
⋅⋅ d2 f (x) d2 f or or f . dx2 dx2
⋅ df (x) df ∣ = ∣ = f (0.5) = 14. dx x=0.5 dx x=0.5 ⋅ df (x) df The value of the second derivative of f at 1 is f ′ (1) = ∣ = ∣ = f (1) = 14. dx x=1 dx x=1 ⋅ df (x) df The value of the second derivative of f at 2 is f ′ (2) = ∣ = ∣ = f (2) = 14. dx x=2 dx x=2
The value of the second derivative of f at 0.5 is f ′ (0.5) =
Page 121, Table of Contents
www.EconsPhDTutor.com
We similarly define the third, fourth, fifth, etc. derivatives in the “obvious” fashion. Definition 36. Let f ∶ D → R be a function. For n ≥ 3, the nth derivative of f is simply the derivative of the (n − 1)th derivative of f .
The 3rd The 4th Derivative of f Derivative of f Lagrange’s notation:
f (3) .
f (4) .
Leibniz’s notation:
d3 f dx3 .
d4 f dx4
3 ⋅
4 ⋅
Newton’s notation:
f.
Etc.
f.
Example 103 (revisited). Let f ∶ R → R be defined by x ↦ 7x2 . Its first derivative is the function f ′ ∶ R → R defined by x ↦ 14x. Its second derivative is the function f ′′ ∶ R → R defined by x ↦ 14. We have f ′ (2) = 28 and f ′′ (2) = 14. Its third derivative is the function f (3) ∶ R → R defined by x ↦ 0. Its fourth derivative is the function f (4) ∶ R → R defined by x ↦ 0. Observe that f (3) = f (4) . Indeed, the third and all higher-order derivatives are identical functions: f (3) = f (4) = f (5) = . . . We have f (3) (2) = f (4) (2) = f (5) (2) = ⋅ ⋅ ⋅ = 0. Indeed, for any x ∈ R, we have f (3) (x) = f (4) (x) = f (5) (x) = ⋅ ⋅ ⋅ = 0. Exercise 57. Given f ∶ D → R and f ′ ∶ A → R, what is f ′′ ? (Answer on p. 1025.) Exercise 58. (Tedious but easy.) Let g ∶ R → R be defined by x ↦ x4 − x3 + x2 − x + 1. Write down all of its derivatives. Evaluate all of these derivatives at 1. Write your answers in Lagrange’s, Leibniz’s, and Newton’s notation. (Answer on p. 1025.)
Page 122, Table of Contents
www.EconsPhDTutor.com
9.5
More About Leibniz’s Notation: The
Example 104. “
d Operator dx
d 2 x = 2x” is simply shorthand for this statement: dx
The derivative of the function with mapping rule x ↦ x2 is the function with mapping rule x ↦ 2x.
Example 105. “
d f = g” is simply shorthand for this statement: dx The derivative of the function f is the function g.
Example 106. “
d f ⋅ g = g ⋅ f ′ + f ⋅ g ′ ” is simply shorthand for this statement: dx
The derivative of the function f ⋅ g is the function with mapping rule x ↦ g(x) ⋅ f ′ (x) + f (x) ⋅ g ′ (x).
Page 123, Table of Contents
www.EconsPhDTutor.com
9.6
Standard Rules of Differentiation
Proposition 1. Let f ∶ A → R and g ∶ B → R be differentiable functions with derivatives f ′ and g ′ . Suppose also that the composite function f g ∶ A → R is well-defined. Let k ∈ R be a constant. Then: d dx
d sin x = dx
cos x,
d f ± g = f ′ ± g′, dx
d cos x = dx
− sin x,
d dx
kf
=
d dx
f ⋅g
=
g ⋅ f ′ + f ⋅ g′,
d dx
xk
= kxk−1 ,
d dx
f g
=
g ⋅ f ′ − f ⋅ g′ , g⋅g
d dx
ex
=
ex ,
d d (f ○ g) dg f ○g = ⋅ . dx dg dx
d dx
ln x
=
1 , x
k
=
0,
kf ′ ,
(My mnemonic for the Quotient Rule is: “Lo-D-Hi minus Hi-D-Lo; cross over and square the low.”) Proof. Optional, see p. 957 in the Appendices.
Of the above rules, the Chain Rule is the most powerful. We can also write it more elegantly (if a little imprecisely) as dz dz dy = ⋅ . dx dy dx As discussed above in the historical note (p. 118), thus written, the Chain Rule has a beautiful informal interpretation: “The change in z caused by a small unit change in x” is equal to “The change in z caused by a small unit change in y” × “The change in y caused by a small unit change in x”. This makes perfect sense:
Page 124, Table of Contents
www.EconsPhDTutor.com
Example 107. When I add 1 g of Milo (the x-variable) to a cup of water, the volume of dy the water increases by 2 cm3 (the y-variable). That is, = 2 cm3 g-1 dx When the volume of the water increases by 1 cm3 (the y-variable), the water level (in the dz cup) rises by 0.3 cm (the z-variable). That is = 0.3 cm cm-3 = 0.3 cm-2 . dy Altogether then, when I add 1 g of Milo (the x-variable) to a cup of water, I should expect dz the water level to rise by 0.6 cm. That is, = 0.6 cm g-1 . This is indeed consistent with dx dz dz dy = = 2 × 0.3 = 0.6 cm g-1 . dx dy dx In case you’ve forgotten how it works, here are a few examples to illustrate: Example 108. Let h ∶ R → R be defined by x ↦ esin x . desin x desin x d sin x h (x) = = = esin x cos x. dx d sin x dx ′
Another simple example: Example 109. Let g ∶ R → R be defined by x ↦
√
4x − 1.
√ √ d 4x − 1 d 4x − 1 d (4x − 1) −0.5 −0.5 g (x) = = = 0.5 (4x − 1) ⋅ 4 = 2 (4x − 1) . dx d (4x − 1) dx ′
Here’s a more complicated example, where the Chain Rule is applied twice. 3
Example 110. Let f ∶ R → R be defined by x ↦ [sin(2x − 3) + cos(5 − 2x)] . Then 3
d [sin(2x − 3) + cos(5 − 2x)] f (x) = dx 3 d [sin(2x − 3) + cos(5 − 2x)] d [sin(2x − 3) + cos(5 − 2x)] = d [sin(2x − 3) + cos(5 − 2x)] dx d cos(5 − 2x) d(5 − 2x) 2 d sin(2x − 3) d(2x − 3) = 3 [sin(2x − 3) + cos(5 − 2x)] [ + ] d(2x − 3) dx d(5 − 2x) dx ′
2
= 3 [sin(2x − 3) + cos(5 − 2x)] [cos(2x − 3) ⋅ 2 − sin(5 − 2x) ⋅ (−2)] 2
= 6 [sin(2x − 3) + cos(5 − 2x)] [cos(2x − 3) + sin(5 − 2x)] .
Page 125, Table of Contents
www.EconsPhDTutor.com
Exercise 59. For each of the following functions (assume they have a suitably defined domain and codomain), evaluate the first derivative at 0. (a) f (x) = x2 . (b) g(x) = x 2 1 + [x − ln (x + 1)] . (c) h(x) = sin 2 . (Answer on p. 1026.) 1 + [x − ln (x + 1)]
Corollary 1.
d d d tan x = sec2 x, cot x = − csc2 x, and csc x = − csc x cot x. dx dx dx
Proof. Using the quotient rule, d d sin x cos x cos x − sin x(− sin x) cos2 x + sin2 x 1 tan x = = = = = sec2 x. 2 2 2 dx dx cos x cos x cos x cos x For the derivatives of cot x and csc x, see Exercise 60.
SYLLABUS ALERT d csc x = − csc x cot x is in the List of Formulae for 9758 (revised), but not for 9740 (old). dx
Exercise 60. Prove the following: (Answer on p. 1026.) d cot x = − csc2 x and dx
d csc x = − csc x cot x. dx
Exercise 61. (Answer on p. 1026.) (a) Newton’s Second Law of Motion is that force is equal to the rate of change of momentum, where momentum is the product of mass and velocity. Write down this law in mathematical notation, with F , m, v, and t denoting force, mass, velocity, and time. (b) Assume that mass is constant. Explain why Newton’s Second Law then simplifies into the more-familiar F = ma, where a is acceleration (i.e. the rate of change of velocity).
Page 126, Table of Contents
www.EconsPhDTutor.com
9.7
Differentiable and Twice-Differentiable Functions
Definition 37. A function is differentiable on a set of points if it is differentiable at every point in that set.
Definition 38. A function is differentiable if it is differentiable on its domain.
Definition 39. A function is twice-differentiable on a set of points if it is twice-differentiable at every point in that set.
Definition 40. A function is twice-differentiable if it is twice-differentiable on its domain.
In other words, f is differentiable if and only if f ′ has the same domain as f . Similarly, f is twice-differentiable if and only if f ′′ has the same domain as f . And of course, if a function is twice-differentiable, then it is also differentiable. (The definitions for a function to be thrice-differentiable, four-times-differentiable, etc. are very much analogous, but this textbook will have no reason to use these terms.) The condition that the first derivative (or second derivative) exists at every point in the domain is important. Failing which, we do not consider the function to be differentiable (or twice-differentiable). The three functions in the next example illustrate:
Page 127, Table of Contents
www.EconsPhDTutor.com
Example 111. Consider f ∶ R → R defined by f (x) = x2 . We have f ′ (x) = 2x and f ′′ (x) = 2 for all x ∈ R. And so f is both differentiable and twice-differentiable. Now consider g ∶ R → R defined by g(x) = x ∣x∣ (graphed below). We have g ′ (x) = 2 ∣x∣ for all x ∈ R and ⎧ ⎪ ⎪ ⎪−2, ′′ g (x) = ⎨ ⎪ ⎪ ⎪ ⎩2,
for x < 0, for x > 0.
But g ′′ (0) does not exist. And so g is differentiable but NOT twice-differentiable.
y
, for all . x - 2, for x < 0, 2, for x > 0. is undefined.
Consider h ∶ R → R defined by x ↦ ∣x∣. We have ⎧ ⎪ ⎪ ⎪−2, ′ h (x) = ⎨ ⎪ ⎪2, ⎪ ⎩
for x < 0, for x > 0.
But h′ (0) does not exist. So h is not even once-differentiable. (And thus it is certainly not twice-differentiable either.)
Page 128, Table of Contents
www.EconsPhDTutor.com
We can of course also consider thrice-differentiable, four-times-differentiable, etc. functions. We can even consider infinitely-differentiable functions. Indeed, in the A-levels, most functions are usually infinitely differentiable. For example, all polynomials are infinitelydifferentiable, as illustrated in the next example. Example 112. Consider i ∶ R → R defined by x ↦ x5 − x4 + x3 − x2 + x − 1. We have, for all x ∈ R, i′ (x) = 5x4 − 4x3 + 3x2 − 2x + 1,
i(4) (x) = 120x − 24,
i′′ (x) = 20x3 − 12x2 + 6x − 2,
i(5) (x) = 120,
i(3) (x) = 60x2 − 24x + 6,
i(6) (x) = i(7) (x) = i(8) (x) ⋅ ⋅ ⋅ = 0.
The function i is infinitely-differentiable, with the 6th and higher-order derivatives all having the mapping rule x ↦ 0.
Simple exponential functions are also infinitely differentiable: Example 113. Consider j ∶ R → R defined by x ↦ ex . We have, for all x ∈ R, j ′ (x) = j ′′ (x) = j (3) (x) = j (4) (x) = ⋅ ⋅ ⋅ = ex . The function j is infinitely-differentiable, with every derivative simply being the same function as j.
Page 129, Table of Contents
www.EconsPhDTutor.com
9.8
Differentiability Implies (i.e. is Stronger Than) Continuity
Informally, continuity is a “smoothness” condition — if a function is continuous, then its graph has no “holes” or “jumps” anywhere and can be drawn smoothly without lifting your pencil. Differentiability is a stronger “smoothness” condition. If a function is differentiable, then its graph is continuous (i.e. has no “holes” or “jumps”) and moreover has no “kinks” or other “abrupt turns”. Example 114. Graphed below are the functions f , g, and h. f is both continuous and differentiable. g is continuous — you can draw its entire graph without lifting your pencil. However, it is not differentiable because of the “kink”. h is neither continuous nor differentiable, because of the “hole”.
y h is neither continuous nor differentiable. f is both continuous and differentiable.
x
g is continuous, but not differentiable.
Theorem 1. If f ∶ D → R is differentiable at a ∈ D, then f is continuous at a ∈ D. Proof. Optional, see p. 961 in the Appendices.
Page 130, Table of Contents
www.EconsPhDTutor.com
9.9
Implicit Differentiation
Example 115. Consider the equation x2 + y 2 = 1. What is
dy ? dx
√ Method #1. First write y in terms of x: y = ± 1 − x2 . Then differentiate: dy −2x −x ∓x =± √ = ±√ =√ . dx 2 1 − x2 1 − x2 1 − x2 Method #2 (implicit differentiation). Directly apply
d to the given equation: dx
d d dy dy x (x2 + y 2 ) = (1) ⇐⇒ 2x + 2y = 0 Ô⇒ =− . dx dx dx dx y √ If desired, we can plug in y = ± 1 − x2 to get the same answer as before: ∓x dy x =√ . =− √ dx ± 1 − x2 1 − x2 In the above example, the second method (implicit differentiation) is not obviously superior to the first. However, it is sometimes difficult (or impossible) to express y in terms of x. Nonetheless we might still want to compute dy/dx. In such cases, the method of implicit differentiation is wonderful. The next example illustrates: √ Example 116. Consider the equation x2 y + when evaluated at x = 0)?
y dy dy = 1. What is ∣ (i.e. what is cos x dx x=0 dx
In this example, it’s difficult to express y in terms of x. But this doesn’t matter, because we can use implicit differentiation: √ √ d y d 1 dy y(− sin x) − cos x dx (x2 y + ) = (1) ⇐⇒ 2x y + x2 √ + = 0. dx cos x dx 2 y dx cos2 x dy
Now plug in x = 0: √ 1 dy y(− sin 0) − cos 0 dx dy 2 ⋅ 0 y + 02 √ + = 0 ⇐⇒ = 0. 2 y dx cos2 0 dx dy
Page 131, Table of Contents
www.EconsPhDTutor.com
The four rules of differentiation in the next corollary are in the List of Formulae you get during A-level exams (both 9740 and 9758), so you need not know these by heart. d 1 d −1 d sec x = sec x tan x, sin−1 x = √ , cos−1 x = √ , and dx dx 1 − x2 dx 1 − x2 d 1 tan−1 x = . dx 1 + x2
Corollary 2.
d 1 , first rewrite y = sin−1 x as x = sin y. Next sin−1 x = √ 2 dx 1−x d dy then apply (implicit differentiation) to get 1 = cos y . But sin2 y + cos2 y = 1, so dx √ dx 2 cos y = 1 − x . And so, Proof. To prove that
dy d 1 1 . = sin−1 x = =√ dx dx cos y 1 − x2 Exercise 62 asks you the prove the derivatives of sec x, cos−1 x and tan−1 x are as claimed.
d d −1 d sec x = sec x tan x, cos−1 x = √ , and tan−1 x = 2 dx dx dx 1−x 1 . (Answer on p. 1026.) 1 + x2
Exercise 62. Prove that
Page 132, Table of Contents
www.EconsPhDTutor.com
10 10.1
Increasing, Decreasing, and f ′
When a Function is Increasing or Decreasing
Example 117. Consider the function f ∶ R → R defined by x ↦ x2 . It is decreasing on R−0 , increasing on R+0 , strictly decreasing on R− , and strictly increasing on R+ .
y Decreasing on Strictly decreasing on
Increasing on Strictly decreasing on
x
Both increasing and decreasing at x = 0.
Note: At x = 0, f is both decreasing and increasing, but neither strictly decreasing nor strictly increasing. This follows from the formal definitions (below).
Definition 41. Given a function f and a set of points S, we say that f is ... 1. ... increasing on S if for any x1 , x2 ∈ S with x2 > x1 , we have f (x2 ) ≥ f (x1 ); 2. ... strictly increasing on S if for any x1 , x2 ∈ S with x2 > x1 , we have f (x2 ) > f (x1 ); 3. ... decreasing on S if for any x1 , x2 ∈ S with x2 > x1 , we have f (x2 ) ≤ f (x1 ); 4. ... strictly decreasing on S if for any x1 , x2 ∈ S with x2 > x1 , we have f (x2 ) < f (x1 ); Of course, if a function is strictly increasing on a set of points, then it is also increasing on that set. And if it is strictly decreasing, then it is also decreasing.
Exercise 63. Let g ∶ R → R defined by x ↦ sin x. Identify the sets on which which g is increasing, decreasing, strictly increasing and/or strictly decreasing. (Answer on p. 1027.)
Page 133, Table of Contents
www.EconsPhDTutor.com
10.2
The First Derivative Increasing/Decreasing Test
The derivative is the slope of the tangent. And so not surprisingly, the derivative is intimiately related to whether a function is increasing or decreasing. Formally: Fact 10. Let f ∶ R → R be a differentiable function. Let a, b ∈ R with b > a. Then 1. f is decreasing on (a, b) ⇐⇒ f ′ (x) ≥ 0, for all x ∈ (a, b). 2. f is increasing on (a, b) ⇐⇒ f ′ (x) ≤ 0, for all x ∈ (a, b).
3. f is strictly decreasing on (a, b) ⇐⇒ f ′ (x) < 0, for all x ∈ (a, b). 4. f is strictly increasing on (a, b) ⇐⇒ f ′ (x) > 0, for all x ∈ (a, b). 5. f is both increasing and decreasing at a ⇐⇒ f ′ (a) = 0.
Proof. Optional, see p. 962 in the Appendices.
Example 133 (revisited). Consider again f ∶ R → R defined by x ↦ x2 . 1. f is decreasing on R−0 , and so f ′ (x) ≤ 0 for x ≤ 0.
2. f is increasing on R+0 , and so f ′ (x) ≥ 0 for x ≥ 0. 3. f is strictly decreasing on R−0 , and so f ′ (x) < 0 for x ≤ 0. 4. f is strictly increasing on R+0 , and so f ′ (x) > 0 for x ≥ 0.
5. f is both increasing and decreasing at x = 0, and so f ′ (x) = 0.
y
, for
, for
.
Both increasing and decreasing at x = 0:
Page 134, Table of Contents
.
, for
.
, for
.
x
www.EconsPhDTutor.com
11
Extreme, Stationary, and Turning Points 11.1
Maximum and Minimum Points
Let f ∶ D → R and x ∈ D. Informally:21 1. If f (x) ≥ f (a) for all a ∈ D that are “close to” x, then we call x a maximum point of f and f (x) a maximum value. 2. If f (x) ≤ f (a) for all a ∈ D that are “close to” x, then we call x a minimum point of f and f (x) a minimum value. 3. If f (x) > f (a) for all a ∈ D that are “close to” x, then we call x a strict maximum point of f and f (x) a strict maximum value. 4. If f (x) < f (a) for all a ∈ D that are “close to” x, then we call x a strict minimum point of f and f (x) a strict minimum value. Of course, a strict maximum point is also a maximum point. And a strict minimum point is also a minimum point. Any maximum or minimum point is also known as an extremum (plural: extrema) or an extreme point.
21
See p. 962 in the Appendices for the formal definitions.
Page 135, Table of Contents
www.EconsPhDTutor.com
Example 118. Below is graphed f ∶ R → R defined by f (x) = −(x−1)2 . x = 1 is a maximum point and a strict maximum point of f . The corresponding maximum value (and also strict maximum value) is f (1) = 0. Also graphed is g ∶ R → R defined by g(x) = (x + 1)2 . x = 1 is a minimum point and a strict minimum point of g. The corresponding minimum value (and also strict minimum value) is g(1) = 0.
x = -1 minimum point for g
y
x
x=1 maximum point for f
Page 136, Table of Contents
www.EconsPhDTutor.com
A function can have multiple maximum and multiple minimum points: Example 119. Graphed below is h ∶ R → R defined by x ↦ 6x5 − 15x4 − 10x3 + 30x2 . • x = −1 is a maximum point and a strict maximum point of h. The corresponding maximum value (and also strict maximum value) is h(−1) = 19. • x = 1 is a maximum point and a strict maximum point of h. The corresponding maximum value (and also strict maximum value) is h(1) = 11. • x = 0 is a minimum point and a strict minimum point of h. The corresponding minimum value (and also strict minimum value) is h(0) = 0. • x = 2 is a minimum point and a strict minimum point of h. The corresponding minimum value (and also strict minimum value) is h(2) = −8.
y x = ±1 maximum points
x -2
-1
0
1
2
3
x = 0, 2 minimum points
Page 137, Table of Contents
www.EconsPhDTutor.com
The next example highlights the fact that a maximum point is sometimes not a strict maximum point. Likewise with minimum points. Example 120. Below is graphed i ∶ R → R defined by x ↦ 3. (This is a constant function.) • Every point x ∈ R is a maximum point of i. The corresponding maximum value is always i(x) = 3. • But no point is a strict maximum point. • Every point x ∈ R is a minimum point of i. The corresponding minimum value is always i(x) = 3. • But no point is a strict minimum point.
y
Every point is a maximum point.
Every point is a minimum point. x -2
Page 138, Table of Contents
-1
0
1
2
3
www.EconsPhDTutor.com
11.2
Global Maximum and Minimum Points
Definition 42. Let f ∶ D → R and a ∈ D. 1. If f (a) ≥ f (x) for all x ∈ D, we call a the global maximum point of f and f (a) the global maximum value. 2. If f (a) ≤ f (x) for all x ∈ D, we call a the global minimum point of f and f (a) the global minimum value. 3. If f (a) > f (x) for all x ∈ D/{a}, we call a the strict global maximum of f and f (a) the strict global maximum value. 4. If f (a) < f (x) for all x ∈ D/{a}, we call a the strict global minimum of f and f (a) the strict global minimum value.
The next fact is perhaps obvious: Fact 11. There cannot be more than one strict global maximum point of a function. (Similarly, there cannot be more than one strict global minimum point of a function.)
Proof. Suppose for contradiction that two distinct points x1 and x2 are strict global maximum points of f . Then since x1 is a strict global maximum point, we have f (x1 ) > f (x2 ). Similarly, since x2 is a strict global maximum point, we have f (x2 ) > f (x1 ). The two inequalities are contradictory. So it is impossible that two distinct points x1 and x2 are strict global maximum points of f .
Page 139, Table of Contents
www.EconsPhDTutor.com
Example 119 (revisited). Consider again the function h ∶ R → R defined by x ↦ 6x5 − 15x4 − 10x3 + 30x2 . (Graph reproduced below for convenience.) x = ±1 are maximum points. However, they are not global maximum points. Indeed, h has no global maximum point because lim h(x) = ∞ (“as x increases without bound, h(x) x→∞ also increases without bound”). In other words, there is no x such that h(x) ≥ h(a) for all a ∈ R. Similarly, x = 0, 2 are minimum points. However, they are not global minimum points. Indeed, h has no global minimum point because lim h(x) = −∞ (“as x decreases without x→−∞ bound, h(x) also decreases without bound”). In other words, there is no x such that h(x) ≤ h(a) for all a ∈ R. y x = ±1 maximum points
x -2
-1
0
1
2
3
x = 0, 2 minimum points
We next restrict the domain of h in two ways to create two new functions i and j:
Page 140, Table of Contents
www.EconsPhDTutor.com
Example 119 (revisited). Graphed below (left) is the function i ∶ [−1.5, 2.5] → R defined by x ↦ 6x5 − 15x4 − 10x3 + 30x2 . i has three maximum points in total, namely ±1, 2.5. However, only 2.5 is a global maximum point of i because only i(2.5) ≥ i(x) for all x ∈ [−1.5, 2.5]. Of course, it is also a strict global maximum point because i(2.5) > i(x) for all x ∈ [−1.5, 2.5]. i has three minimum points in total, namely −1.5, 0, 2. However, only −1.5 is a global maximum point of i because only i(−1.5) ≤ i(x) for all x ∈ [−1.5, 2.5]. Of course, it is also a strict global minimum point because i(−1.5) < i(x) for all x ∈ [−1.5, 2.5].
y
x = ±1 max
y
x = 2.5 max and global max
x = -1 max and global max x = 1, 1.2 max
x -2
-1
0 1 2 x = -1.5 min and global min x = 0, 2 min
3
x -2
-1
x = -1.2, 0 min
0
1
2
3
x = 2 min and global min
Also graphed above (right) is the function j ∶ [−1.2, 2.2] → R defined by x ↦ 6x5 − 15x4 − 10x3 + 30x2 . Again, there are three maximum points in total, namely ±1, 2.2. However, only −1 is a global maximum point of j because only j(−1) ≥ j(x) for all x ∈ [−1.2, 2.2]. Of course, it is also a strict global maximum point because j(−1) > i(x) for all x ∈ [−1.2, 2.2]. And again, there are three minimum points in total, namely −1.2, 0, 2. However, only 2 is a global minimum point of j because only j(2) ≤ j(x) for all x ∈ [−1.2, 2.2]. Of course, it is also a strict global minimum point because j(2) < j(x) for all x ∈ [−1.2, 2.2].
Page 141, Table of Contents
www.EconsPhDTutor.com
Note that the A-level syllabuses and exams only ever talk about maximum and minimum points. They do not ever talk about 1. Strict maximum points; 2. Strict minimum points; 3. Global maximum points; 4. Global minimum points; 5. Strict global minimum points; and 6. Strict global maximum points. Nonetheless, these concepts are not difficult to grasp. It is thus well worth learning them, just so you have a better understanding of how to find maximum and minimum points. Note also that what we simply call maximum and minimum points are sometimes instead called local maximum and minimum points, so that they are better contrasted with global maximum or minimum points.
Exercise 64. (Answer on p. 1027.) For each of the following functions, write down, if any of these exist, the (i) maximum points, (ii) minimum points, (iii) strict maximum points, (iv) strict minimum points, (v) global maximum points, (vi) global minimum points, (vii) strict global maximum points, (viii) strict global minimum points; and also all the corresponding values of the function at these points. (a) f ∶ R → R defined by x ↦ 100. (b) g ∶ R → R defined by x ↦ x2 . (c) h ∶ [1, 2] → R defined by x ↦ x2 .
Page 142, Table of Contents
www.EconsPhDTutor.com
11.3
Stationary and Turning Points
Definition 43. A point x is a stationary point of f if f â&#x20AC;˛ (x) = 0. Graphically, a stationary point is where the slope of the tangent is 0 (flat). Definition 44. A turning point is any point that is both a stationary point and a maximum or minimum point.
So every turning point is both a stationary point and an extreme point. But the converse is not true: A stationary point need not always be a turning point. And an extreme point need not always be a turning point.
Page 143, Table of Contents
www.EconsPhDTutor.com
Example 121. Graphed below is the function f ∶ [−1.5, 0.5] → R defined by x ↦ x5 +2x4 +x3 . Five points are labelled. The table below classifies each point. D is a stationary point but not a turning point. (As we shall learn in Section 151, D is an example of an inflexion point.) A is a minimum point and E is a maximum point. But neither is a turning point. Type Max Min Strict Max Strict Min Global Max Global Min Strict Global Max Strict Global Min Stationary Turning
A B C D E ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓
y E
B
D
x
C f (x) = x5 + 2x4 + x3 A
Exercise 65. Is each of the following statements true or false? To show that a statement is false, simply give a counterexample from the above example. If it is true, explain why. (Answer on p. 1028.) (a) Every maximum point or minimum point is a stationary point. (b) Every maximum point or minimum point is a turning point. (c) Every stationary point is a maximum point or minimum point. (d) Every turning point is a maximum point or minimum point. (e) Every turning point is a stationary point. (f) Every stationary point is a turning point.
Page 144, Table of Contents
www.EconsPhDTutor.com
11.4
The Interior Extremum Theorem
Informally, a point x ∈ S is in the interior of a set S if x is not at the “edge” of S. Formally: Definition 45. x ∈ S is in the interior of S if there exists δ such that (x − δ, x + δ) ∈ S. x ∈ S is a non-interior point of S if it is not in the interior of S.
Example 122. Consider the set S = [0, 1]. The points 0.2, 1/3, and 0.775 are all in the interior of S. Indeed, every point x ∈ (0, 1) is in the interior of S. In contrast, the points 0 and 1 are non-interior points of S. Example 123. Consider the set S = [0, 0.5) ∪ (0.5, 1]. The points 0.2, 1/3, and 0.775 are all in the interior of S. Indeed, every point x ∈ (0, 0.5) ∪ (0.5, 1) is in the interior of S. In contrast, the points 0 and 1 are non-interior points of S. The point 0.5 is not in the interior of S. It is not even a non-interior point of S, because it is not in the set S to begin with.
Page 145, Table of Contents
www.EconsPhDTutor.com
The Interior Extremum Theorem (IET) is the fundamental reason why we lurrrve taking derivatives and setting them equal to zero — this is a great way to find maxima and minima! Theorem 2. (Interior Extremum Theorem [IET].) Let f ∶ D → R be a differentiable function. If a is a maximum or minimum point AND in the interior of D, then f ′ (a) = 0 (i.e. c is a stationary point).
Proof. Optional, see p. 963 in the Appendices.
Here’s a non-rigorous explanation of the intuition behind the IET: Example 136 (revisited). Graphed below is f ∶ R → R defined by x ↦ −(x − 1)2 . Here’s the intuition for why f ′ (0) = 0: In order for 1 to be a maximum point of f , it must be that to its left, f is increasing; while to its right, f is decreasing. In other words, to the left of 1, f ′ (x) ≥ 0. While to the right of 1, f ′ (x) ≤ 0. Altogether then, we must have f ′ (1) = 0 — at the maximum point, the slope of the function must be 0.
x = -1 minimum point for g
y
x
x=1 maximum point for f
Exercise 66. Refer to the above Example. Explain the intuition for why g ′ (−1) = 0. (Answer on p. 1028.)
Exercise 67. True or false: “Let f ∶ D → R be a differentiable function. If c is a maximum or minimum point AND in the interior of D, then x is a turning point.” (Answer on p. 1028.)
Page 146, Table of Contents
www.EconsPhDTutor.com
11.5
How to Find Maximum and Minimum Points
In secondary school, you may have been taught that to find the maximum and minimum points of f , simply follow this procedure:
The Incorrect Recipe for Finding Maximum and Minimum Points. Given a differentiable function f ∶ D → R, 1. Compute f ′ (x). Find the points x at which f ′ (x) = 0. 2. These points are also the maximum and minimum points. (If we also want to know which are maximum and which are minimum points, then simply employ some method like sketch-the-graph or the Second Derivative Test.)
Unfortunately, the above procedure (let’s call it the Incorrect Recipe) may sometimes fail. It rests on the false belief that “f ′ (x) = 0 ⇐⇒ x is an extremum”. This is false because 1. The IET does NOT say, “f ′ (x) = 0 Ô⇒ x is an extremum.” It is perfectly possible that f ′ (x) = 0 without x being an extremum.
2. The IET does NOT say, “x is an extremum Ô⇒ f ′ (x) = 0 .” Instead, it says, “x is an extremum AND an interior point Ô⇒ f ′ (x) = 0.” Thus, it is perfectly possible that x is an extremum without f ′ (x) = 0. Here is an example to illustrate these two failings of the Incorrect Recipe.
Page 147, Table of Contents
www.EconsPhDTutor.com
Example 144 (revisited). Graphed below is the function f ∶ [−1.5, 0.5] → R defined by x ↦ x5 + 2x4 + x3 . Five points are labelled. According to the Incorrect Recipe, 1. Compute f ′ (x) = 5x4 + 8x3 + 3x2 = x2 (5x2 + 8x + 3) = x2 (5x + 3)(x + 1). We see that 3 f ′ (x) = 0 ⇐⇒ x = − , −1, 0. 5 3 2. So − , −1, 0 are the maximum and minimum points of f . 5 The Incorrect Recipe does correctly identify the points B = (−1, f (−1)) and C = 3 3 (− , f (− )) as maximum and minimum points, respectively. But it makes two mistakes. 5 5 Mistake #1: D = (0, 0) is neither a maximum nor a minimum point, contrary to the Incorrect Recipe. Mistake #2: A and E are respectively a minimum and a maximum point, but neither is detected by the Incorrect Recipe.
y E
B
D
x
C f (x) = x5 + 2x4 + x3 A
We now give the Correct Recipe for finding maximum and minimum points:
Page 148, Table of Contents
www.EconsPhDTutor.com
The Correct Recipe for Finding Maximum and Minimum Points. Given a differentiable function f ∶ D → R, 1. Identify all the stationary points (i.e. x where f ′ (x) = 0). 2. Identify all the non-interior points. 3. Check to see if each stationary point and each non-interior point is a maximum point, a minimum point, or neither. (To do so, employ some method like sketch-the-graph or the Second Derivative Test.)
The Correct Recipe rectifies the Incorrect Recipe in two ways: 1. The Correct Recipe demands that you also check the non-interior points, which may possibly be maximum or minimum points, but may not be detected by the Incorrect Recipe. 2. The Correct Recipe does not assume that every single one of our shortlist of points (the stationary points and the non-interior points) is either a maximum point or a minimum point. It allows for the possibility that some of these points could be neither. By the way, the condition that f is differentiable is very important. If f is not differentiable, then the above Correct Recipe might not work. But not to worry, since most functions on the A-levels are usually differentiable. Example 124. Consider f ∶ [−1, 1] → R defined by x ↦ x3 . Let’s apply the Correct Recipe. 1. Identify all the stationary points (i.e. x where f ′ (x) = 0). f ′ (x) = 3x2 . So f ′ (x) = 0 ⇐⇒ x = 0. The only stationary point is x = 0. 2. Identify all the non-interior points. Every point x ∈ (−1, 1) is in the interior of [−1, 1]. The only non-interior points are −1 and 1. 3. Check if each of these points is a maximum point, a minimum point, or neither. From a sketch of the graph, we see that the stationary point x = 0 is neither a maximum nor a minimum point. The non-interior point −1 is a minimum point. The non-interior point 1 is a maximum point. Altogether, we conclude that −1 is the only minimum point and 1 is the only maximum point.
Page 149, Table of Contents
www.EconsPhDTutor.com
Exercise 68. For each of the following functions, find all the maximum and minimum points using the Correct Recipe. (Answer on p. 1029.) (a) f ∶ R → R defined by x ↦ x. (b) g ∶ [0, 1] → R defined by x ↦ x. (c) h ∶ R → R defined by x ↦ x4 − 2x2 . Identify also the global minimum point(s) of h (if any exist).
Page 150, Table of Contents
www.EconsPhDTutor.com
12
Concavity, Inflexion Points, and the 2DT
• A function is concave downwards (or simply concave) on an interval if the line segment connecting any two points of the graph in this interval is below the graph. • A function is concave upwards (or simply convex) on an interval if the line segment connecting any two points of the graph in this interval is above the graph. • An inflexion point is any point where the concavity of the function changes, either from downwards to upwards, or upwards to downwards.22
Example 125. Graphed below is f ∶ R → R defined by x ↦ x3 . f is concave downwards on R−0 because there, the line segment connecting any two points on f is below the graph of f .
y Tangent line at x = 0 is concave upwards on x -2
-1
0
1
2
is concave downwards on
In contrast, f is concave upwards on R+0 because there, the line segment connecting any two points on f is above the graph of f . 0 is an inflexion point because this is where the function f changes from being concave downwards to being concave upwards. A test for whether a point is an inflexion point is this: Draw the tangent line to the graph at that point. The point is an inflexion point ⇐⇒ The line is above the graph on one side of the point and below the graph on the other side (see Fact 95 in the Appendices). The tangent line to the graph at the point 0 is drawn in green (it coincides with the horizontal axis). We indeed see that the line is above the graph on the left side of the point and below the graph on the right side of the point. Therefore, 0 is an inflexion point. 22
These are informal definitions. For the formal definitions, see p. 964 in the Appendices (optional).
Page 151, Table of Contents
www.EconsPhDTutor.com
For a graph to be concave downwards, its slope must be decreasing. Conversely, to be concave upwards, its slope must be increasing. Altogether then, the following proposition is intuitively plausible. Proposition 2. Let f ∶ D → R be a twice-differentiable function. (a) f is concave downwards on an interval ⇐⇒ f ′′ (x) ≤ 0 for every x in this interval. (b) f is convex upwards on an interval ⇐⇒ f ′′ (x) ≥ 0 for every x in this interval. (c) x is an inflexion point Ô⇒ f ′′ (x) = 0.
Proof. Optional, see p. 967 in the Appendices.
Example 151 (revisited). Consider f ∶ R → R defined by x ↦ x3 . f is concave downwards on R−0 , concave upwards on R+0 , and has an inflexion point at x = 0. We can verify that, as per the above proposition: ⎧ ⎪ ⎪ < 0, ⎪ ⎪ ⎪ ⎪ f ′ (x) = 3x2 ⎨= 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩> 0,
Page 152, Table of Contents
for x ∈ R−0 , for x = 0, for x ∈ R+0 .
www.EconsPhDTutor.com
It is very tempting to believe that the converse of part (c) of the above proposition is true. That is, it is very tempting to believe that “f ′′ (x) = 0 Ô⇒ x is an inflexion point.” But this is wrong! It is perfectly possible that f ′′ (x) = 0 without x being an inflexion point! Here’s an example: Example 126. Consider g ∶ R → R defined by x ↦ x4 . We have g ′ (x) = 4x3 and g ′′ (x) = 12x2 , so that g ′′ (x) = 0 ⇐⇒ x = 0. We might thus be tempted to conclude that 0 is an inflexion point. However, this is not the case. Although g ′′ (0) = 0, we have g ′′ (x) > 0 for x > 0 and we also have g ′′ (x) > 0 for x < 0, and so the concavity of g does not change at the point 0. To qualify as an inflexion point, the concavity of the function must change. At 0, the concacivty of g does not change. Therefore, 0 is NOT an inflexion point.
y g is concave upwards everywhere
. However, is not an inflexion point of . x
Page 153, Table of Contents
www.EconsPhDTutor.com
Inflexion points may be further sub-divided into stationary points of inflexion and non-stationary points of inflexion. Definition 46. A stationary point of inflexion is simply any point that is both an inflexion point and a stationary point. A non-stationary point of inflexion is simply any point that is an inflexion point, but not a stationary point.
The A-level syllabuses explicitly exclude non-stationary points of inflexion. Nonetheless, there is the temptation to believe that “every inflexion point must also be a stationary point”. Here’s a quick counter-example that dispels this belief: Example 127. The graph below is for the function f ∶ R → R defined by x ↦ x3 + x. We have f ′ (x) = 3x2 + 1 and f ′′ (x) = 6x. The point 0 is not a stationary point because f ′ (0) = 1 ≠ 0. However, 0 is an inflexion point, because to the left of 0, f is concave downwards; and to the right, f is concave upwards. So 0 is a point of inflexion. Indeed, it is a non-stationary point of inflexion. Also illustrated is the tangent line at y = x (whose slope is indeed non-zero). Observe that indeed, to the left of 0, the tangent line is above the graph; while to the right of 0, the tangent line is below the graph. This serves as a second way to verify that 0 is a point of inflexion.
y
Concave upwards on
x Tangent line at 0 Concave downwards on
Page 154, Table of Contents
www.EconsPhDTutor.com
12.1
The Second Derivative Test (2DT)
y
y
x
f must be concave upwards around the minimum turning point 0. So
f must be concave downwards around the maximum turning point 0.
for all x near 0.
So
for all x near 0.
x
From graphs, it looks like around a maximum turning point a, f must be concave downwards, i.e. f ′′ (a) < 0. Similarly, around a minimum turning point b, f must be concave upwards, i.e. f ′′ (b) > 0. The next proposition is thus intuitively plausible.
Proposition 3. (Second Derivative Test [2DT].) Let f be a twice-differentiable function. Let a be a stationary point (i.e. f ′ (a) = 0). 1. If f ′′ (a) < 0, then a is a maximum point. 2. If f ′′ (a) > 0, then a is a minimum point.
3. If f ′′ (a) = 0, then the 2DT is uninformative. That is, a could be a maximum point, a minimum point, an inflexion point, or something else altogether!
Proof. Optional, see p. 968 in the Appendices.
The third part of the above Proposition must be heavily emphasised: If f ′ (a) = 0 and f ′′ (a) = 0, then the 2DT tells us absolutely nothing about a! a could be a maximum point, a minimum point, an inflexion point, or something else altogether! We previously gave the Correct Recipe for finding maximum and minimum points. Let’s now add the 2DT to this recipe:
Page 155, Table of Contents
www.EconsPhDTutor.com
The Enriched Recipe for Finding Maximum and Minimum Points. Given a twice-differentiable function f ∶ D → R, 1. Identify all the stationary points (i.e. a where f ′ (a) = 0). (a) Evaluate f ′′ at each of these points. (b) f ′′ (a) < 0 Ô⇒ a is a maximum point. Conversely, f ′′ (a) > 0 Ô⇒ x is a minimum point. If f ′′ (a) = 0, then we need to determine the nature of a using some other method (e.g. sketch-the-graph). 2. Identify all the non-interior points. (a) Check if each of these points is a maximum point, a minimum point, or neither.
If f is not twice-differentiable, then the Enriched Recipe may not work. Fortunately, most functions in A-levels are twice-differentiable. Example 121 (revisited). Consider f ∶ [−1.5, 0.5] → R defined by x ↦ x5 + 2x4 + x3 . 1. Identify all the stationary points. f ′ (x) = 5x4 + 8x3 + 3x2 = x2 (5x2 + 8x + 3) = 0 ⇐⇒ x = 0 or x = −1, −0.6 (quadratic formula). (a) f ′′ (x) = 20x3 + 24x2 + 6x = 2x(10x2 + 12x + 3). (b) f ′′ (−0.6) > 0 Ô⇒ −0.6 is a minimum point. f ′′ (−1) < 0 Ô⇒ −1 is a maximum point. But f ′′ (0) = 0, so the 2DT tells us nothing. By sketching the graph (nonrigorous method that will suffice for the A-levels), we see that 0 is an inflexion point. 2. The only two non-interior points are −1.5 and 0.5. Again by sketching the graph, we see that −1.5 is a minimum point and 0.5 is a maximum point. Altogether, we conclude that there are two maximum points — −1 and 0.5 — and two minimum points — −0.6 and −1.5. Exercise 69. Use the Enriched Recipe to find the maximum and minimum points of each of the following functions. (Answer on p. 1031.) (a) g ∶ R → R defined by x ↦ x8 + x7 − x6 . π π (b) h ∶ (− , ) → R defined by x ↦ tan x. 2 2 (c) i ∶ [0, 2π] → R defined by x ↦ sin x + cos x.
Page 156, Table of Contents
www.EconsPhDTutor.com
12.2
Summary of Points and Venn Diagram
The Venn diagram below depicts the five types of points you need to know for the A-levels: Inflexion, maximum, minimum, stationary, and turning points. To its right is a graph of a rather-arbitrary function t ∶ D → R designed to illustrate these various points. The x- and y-coordinates of a are denoted ax and ay ; similarly for other points.
a b
Inflexion
All points
c
y e
d
Stationary
j
i f h
e
Turning
g c
g
f h
Max
i
j
Min
b a x
• For most functions you’ll ever encounter, most points are like a. For lack of a better name, we can call such points boring points — a boring point is simply any point that is not an inflexion, maximum, minimum, stationary, or turning point. • b is a non-stationary point of inflexion (explicitly excluded from the A-levels). • c is a stationary point of inflexion. • A point like d (not illustrated) — a stationary point that is not a maximum, minimum, or inflexion point — is extremely unusual. You can find an exotic example on p. 968. • f is both a maximum and minimum point because for all x ∈ D that are “close to” fx ∈ D, we have t(x) ≤ t (fx ) ≤ t(x). • The set of turning points is simply the intersection of the set of stationary points and the set of maximum and minimum points. • h is a maximum point because t(x) ≤ t (hx ) for all x ∈ D that are “close to” hx . • j is a minimum point because t(x) ≥ t (jx ) for all x ∈ D that are “close to” jx . • i is both a maximum and minimum point because there are simply no x ∈ D that are “close to” ix ∈ D, and thus it is trivially or vacuously true that t(x) ≤ t (ix ) ≤ t(x) for x that are “close to” x.23 i is not a stationary point because t′ (ix ) ≠ 0 — indeed, t′ (ix ) is undefined.24 23 24
A point like ix ∈ D that is not “close to” any other x ∈ D is, aptly enough, called an isolated point. ix is an example of a critical point. A critical point is any point that is either stationary or where the derivative is undefined. Don’t worry, not something you need to know for the A-levels.
Page 157, Table of Contents
www.EconsPhDTutor.com
Exercise 70. For each of the following equations, (i) sketch its graph. (ii) Write down the points at which it intersects the axes. (iii) Identify any turning points. (iv) Write down the equations of any lines of symmetry and also (v) asymptotes. (a) y = 2ex + x. (b) x = 3x + 2. (c) y = 2x2 + 1. (Answers on pp. 1032, 1033, and 1034.)
Page 158, Table of Contents
www.EconsPhDTutor.com
13
Relating the Graph of f ′ to that of f
Given the graph of f ′ , you are required to know how to figure out what f looks like. Let’s start with a very simple example. Example 128. Let f ∶ R → R be some differentiable function. Graphed below in blue is its derivative f ′ . You are told also that f (0) = 2. What does the graph of f look like? (Pretend for a moment that you can’t see the red graph.)
y
1
x
The derivative simply gives the slope of f . Since f ′ (x) = 1 for all x, this means that f has constant slope of 1. We are given moreover that f (0) = 2 (i.e. the vertical intercept is 2). Altogether then, f (x) = x + 2 and is graphed in red above.
Page 159, Table of Contents
www.EconsPhDTutor.com
Example 129. Let g ∶ R− ∪ R+ → R be some differentiable function. Graphed below in blue is its derivative g ′ . You are told also that lim g(x) = −2. What does the graph of g look x→0 like? (Pretend for a moment that you can’t see the red graph.)
y
1
x
-1
The derivative simply gives the slope of g. Since g ′ (x) = −1 for all x < 0 and g ′ (x) = 1 for all x > 0, this means that g has constant slope of −1 for x < 0 and constant slope of 1 for all x > 0. We are given moreover that lim g(x) = −2, so the two branches of g nearly meet x→0
at (0, −2), with a hole there. Altogether then, ⎧ ⎪ ⎪ ⎪−x − 2, g(x) = ⎨ ⎪ ⎪ ⎪ ⎩x − 2,
for x < 0, for x > 0.
Or more concisely, g(x) = ∣x∣ − 2. Graphed above in red is g.
Page 160, Table of Contents
www.EconsPhDTutor.com
Example 130. Let h ∶ R → R be some differentiable function. Graphed below in blue is its derivative h′ defined by h′ (x) = x. You are told also that h(x) = 0. What does the graph of h look like? (Pretend for a moment that you can’t see the red graph.)
y
x
The derivative simply gives the slope of h. Since h′ (x) < 0 for all x < 0, h′ (0) = 0, and h′ (x) > 0 for all x > 0, this means that h is strictly decreasing on R− , a turning point at 0, and strictly increasing on R+ . Moreover, the derivative (slope) is increasing (indeed it is increasing at a constant rate) — so the graph of h is concave upwards throughout. Altogether then, even if we don’t know how to figure out what h(x) is, we can at least roughly sketch the graph of h (in red above below). (Of course, you probably already know x2 from secondary school that h(x) = , but we’re not supposed to know this until we learn 2 about integration later in this textbook.)
Page 161, Table of Contents
www.EconsPhDTutor.com
14
Quick Revision: Quadratic Equations y = ax2 + bx + c
Quadratic equations show up very often in various contexts. So here is a fairly complete if brisk review of quadratic equations, which you were supposed to have completely mastered in secondary school. Example 131. Below are the graphs of the equations y = x2 + 3x + 1 (red), y = x2 + 2x + 1 (blue), y = x2 +x+1 (green), y = −x2 +x+1 (red dotted), y = −x2 −2x−1 (blue dotted), and y = −x2 − x − 1 (green dotted).
6 y=
x2
y
+x+1
y = x2 + 2x + 1
4
y = x2 + 3x + 1 2 y = - x2 + x + 1
x
0 -4
-3
-2
-1
0
-2 y=-
x2
1 y = - x2 - x - 1
- 2x - 1
-4
Page 162, Table of Contents
www.EconsPhDTutor.com
Here’s a whirlwind study of the quadratic equation y = ax2 + bx + c. Assume that a ≠ 0, otherwise we are in the trivial case of a linear equation. First, write: ax2 + bx + c =
1 2 b c (x + x + ) . a a a
To complete the square, observe that (x + k)2 = x2 + 2kx + k 2 and so b b 2 b2 x + x = (x + ) − . a 2a 4a 2
b 2 b2 c 1 b 2 b2 − 4ac 1 ]. Hence, ax + bx + c = [(x + ) − 2 + ]= [(x + ) − a 2a 4a a a 2a 4a2 2
What we just did above is called completing the square. We can use this to compute the zeros or roots of the equation ax2 + bx + c = 0. ax2 + bx + c = 0
b 2 b2 − 4ac 1 b 2 b2 − 4ac ] = (x + ) − = [(x + ) − a 2a 4a2 2a 4a2 ⇐⇒
⇐⇒
b 2 b2 − 4ac (x + ) = 2a 4a2 x=
−b ±
√ b2 − 4ac . 2a
This last expression give the roots of the equation ax2 + bx + c = 0. This expression will NOT be printed in the A-Level List of Formulae! So be sure you remember it!
√ −b ± b2 − 4ac x= . 2a
Page 163, Table of Contents
www.EconsPhDTutor.com
We can distinguish between six categories of quadratic equations, based on the signs of a (the coefficient of x2 ) and b2 − 4ac (the discriminant). Each of these six categories was illustrated in the figure above. Category 1. a > 0, b2 − 4ac > 0 2. a > 0, b2 − 4ac = 0 3. a > 0, b2 − 4ac < 0 4. a < 0, b2 − 4ac > 0 5. a < 0, b2 − 4ac = 0 6. a < 0, b2 − 4ac < 0
Features ∪-shaped. Intersects the horizontal axis at two points. ∪-shaped. Just touches the horizontal axis at the minimum point. ∪-shaped. Doesn’t intersect the horizontal axis. ∩-shaped. Intersects the horizontal axis at two points. ∩-shaped. Just touches the horizontal axis at the maximum point. ∩-shaped. Doesn’t intersect the horizontal axis.
The vertical intercept (the value of f at 0) is simply c. The Sign of a. If a > 0, then the graph is ∪-shaped and has a minimum turning point b at x = − . Conversely, if a < 0, then the graph is ∩-shaped and has a maximum turning 2a b point at x = − . 2a The Discriminant. The term b2 − 4ac is called the discriminant. This name makes sense, because it helps us discriminate between several possible cases of the equation ax2 +bx+c = 0: • If b2 − 4ac > 0, then: – There are two real roots (or zeros or horizontal intercepts), namely −b ±
√
b2 − 4ac . 2a
– Moreover, we can write ax2 + bx + c = (x −
−b +
√ √ b2 − 4ac −b + b2 − 4ac ) (x + ). 2a 2a
What we have just done is to factorise the expression ax2 + bx + c. Factorisation is often a useful trick to play. Notice that if you plug in either of the roots into the right hand side (RHS) of the above equation, we do indeed get zero, as expected. Page 164, Table of Contents
www.EconsPhDTutor.com
• If b2 − 4ac = 0, then: – There is only one real root (or zero or horizontal intercept), namely −
b . 2a
– Moreover, we can write −b 2 b 2 ax + bx + c = (x − ) = (x + ) . 2a 2a 2
b – Notice that if you plug x = − into the RHS of the above equation, we do indeed get 2a zero, as expected. • If b2 − 4ac < 0, then: – There are no real roots (or zeros or horizontal intercepts). – There is no way to factorise the expression ax2 +bx+c (unless we use complex numbers, which we’ll learn about only in Part IV).
Exercise 71. For each of the following equations, sketch its graph and identify its intercepts and turning points (if these exist). (a) y = 2x2 + x + 1. (b) y = −2x2 + x + 1. (c) y = x2 + 6x + 9. (Answer on p. 1035.)
Page 165, Table of Contents
www.EconsPhDTutor.com
15
Transformations 15.1
y = f (x) + a
The graph of y = f (x) + a is simply the graph of y = f (x) translated (moved) upwards by a units. Example 132. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red) and y = f (x) + 2 (blue) are shown below. Notice the blue curve is simply the red curve translated upwards by 2 units.
f(x), f(x) + 2
10
8
6
4
2
x 0 -3.0
-1.5
0.0
1.5
3.0
-2
-4
-6
-8
-10
Page 166, Table of Contents
www.EconsPhDTutor.com
15.2
y = f (x + a)
The graph of y = f (x + a) is simply the graph of y = f (x) translated leftwards by a units. Why leftwards (and not rightwards)? The reason is that in order for f (x1 ) and f (x2 + a) to hit the same value, we must have x2 = x1 − a. That is, every x value is moved to the left by a units. Example 133. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red) and y = f (x + 2) (blue) are shown below. (The latter equation is simply y = (x + 2)3 − 1.) Notice the blue curve is simply the red curve translated leftwards by 2 units.
f(x), f(x+2)
10
8
6
4
2
x 0 -4
-2
0
2
4
-2
-4
-6
-8
-10
Page 167, Table of Contents
www.EconsPhDTutor.com
15.3
y = af (x)
The graph of y = af (x) is simply the graph of f (x) vertically-stretched (outwards from the horizontal axis) by a stretching factor of a. Example 134. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red) and y = 2f (x) (blue) are shown below. Notice the blue curve is simply the red curve stretched vertically (outwards from the horizontal axis) by a factor of 2. 10
f(x), 2f(x)
8
6
4
2
x 0 -3.0
-1.5
0.0
1.5
3.0
-2
-4
-6
-8
-10
Page 168, Table of Contents
www.EconsPhDTutor.com
15.4
y = f (ax)
The graph of y = f (ax) is simply the graph of f (x) horizontally-stretched (outwards from the vertical axis) by a stretching factor of 1/a. Or equivalently, the graph of y = f (ax) is simply the graph of f (x) horizontally-compressed (inwards towards the vertical axis) by a compression factor of a. Why a stretching factor of 1/a (and not a)? The reason is that in order for f (x1 ) and f (ax2 ) to hit the same value, we must have x2 = x1 /a. That is, every x value is scaled by a factor of 1/a. Example 135. Define the function f ∶ R → R by x ↦ x3 − 1.The graphs of f (red) and y = f (2x) (blue) are shown below. (The latter equation is simply y = (5x)3 − 1 = 125x3 − 1.) Notice the blue curve is simply the red curve stretched horizontally (outwards from the 1 vertical axis) by a factor of . (Again, the A-level exams might instead word this as a 2 stretch with scale factor 0.5 parallel to the y-axis.) Equivalently, the blue curve is simply the red curve compressed horizontally (inwards towards from the vertical axis) by a factor of 2. 8
f(x), f(2x)
6
4
2
x 0 -2
-1
0
1
2
-2
-4
-6
-8
Page 169, Table of Contents
www.EconsPhDTutor.com
15.5
Combinations of the Above
Example 136. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red), y = 1.1f (x − 1) (blue), and y = f (1.1x) − 1 (green) are shown below. Notice the blue curve is simply the red curve translated rightwards by 1 unit and then stretching it vertically (outwards from the vertical axis) by a factor of 1.1. Notice the green curve is simply the red curve stretched horizontally (outwards from the vertical axis) by a factor of 1/1.1 and then translated downwards by 1 unit. f(x), 1.1f(x-1), f(1.1x)-1 8
6
4
2
x 0 -2
-1
0
2
3
-2
-4
-6
-8
Page 170, Table of Contents
www.EconsPhDTutor.com
15.6
y = ∣f (x)∣
The graph of y = ∣f (x)∣ is simply the graph of f (x), but with all points for which f (x) < 0 reflected in the horizontal axis. Example 137. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red) and y = ∣f (x)∣ (blue) are shown below.
8
f(x), |f(x)|
6
4
2
x 0 -2
-1
0
1
2
-2
-4
-6
-8
Page 171, Table of Contents
www.EconsPhDTutor.com
15.7
y = f (∣x∣)
The graph of y = f (∣x∣) is simply the graph of f (x), but with all points for which x < 0 reflected in the vertical axis. Example 138. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red) and y = f (∣x∣) (blue) are shown below.
8
f(x), f(|x|)
6
4
2
x 0 -2
-1
0
1
2
-2
-4
-6
-8
Page 172, Table of Contents
www.EconsPhDTutor.com
15.8
y=
1 f (x)
Example 139. Define the function f ∶ R → R by x ↦ x3 − 1. The graphs of f (red) and 1 (blue) are shown below. y= f (x) 1 . So f (x) in this case, x = 1 Ô⇒ f (x) = 0 and thus x = 1 is a vertical asymptote for the graph of 1 1 y= . As x approaches 1 from the left, → −∞. And as x approaches 1 from the f (x) f (x) 1 right, → ∞. f (x) 1 Also, if as x → ±∞, f (x) → ±∞, then we also have → 0, so that y = 0 is a horizontal f (x) asymptote. So here, as x → ∞, f (x) approaches 0 from above and as x → −∞, f (x) approaches 0 from below. Notice that wherever f (x) = 0, we have a vertical asymptote for the graph of y =
8
f(x), 1/f(x)
6
4
2
x 0 -2
-1
0
1
2
-2
-4
-6
-8
Page 173, Table of Contents
www.EconsPhDTutor.com
15.9
y 2 = f (x)
Three observations about the graph of y 2 = f (x): 1. It is symmetric in the horizontal axis. This is because if y1 satisfies y 2 = f (x), then so too does −y1 .
2. If f (x) < 0, then there is no value of y for which y 2 = f (x). And so the graph of y 2 = f (x) is empty wherever f (x) < 0.
3. The graph of y 2 = f (x) intersects the horizontal axis at the same point as the graph of y = f (x). Moreover, at any such point, the tangent to the graph of y 2 = f (x) is vertical.
Example 140. Define the function f ∶ R → R by x ↦ x3 − 1.The graphs of f (red) and y 2 = f (x) (blue) are shown below.
8
y
7 y = f(x)
6 5 4 3 2 1
x
0 -1
-1 0 -2
1
2
3
y2 = f(x)
-3 -4 -5 -6 -7 -8
Page 174, Table of Contents
www.EconsPhDTutor.com
Exercise 72. The graph of the function f ∶ R → R is drawn below in red. Graph each of the following equations. (a) y = ∣2f (3x)∣. (b) y = f (∣x − 1∣). (c) y 2 = f (x) + 4. (Answer on p. 1036.)
-5
-4
-3
-2
-1
30 f(x), y 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 -1 0 -2 -3 -4 -5 -6 -7 -8 -9 -10
x 2
3
4
5
1 Exercise 73. Describe a series of transformations that would transform the graph of y = x 1 to y = 3 − . (Answer on p. 1037.) 5x − 2
Page 175, Table of Contents
www.EconsPhDTutor.com
16
Conic Sections
Conic sections are formed from the intersection of a double cone and a 2D cartesian plane. Take an infinitely large double cone (it goes upwards and downwards forever). Use a 2D cartesian plane to slice the double cone from all conceivable positions and at all conceivable angles. The intersection of the plane and the surface of the double cone form curves which, aptly enough, are called conic sections. The figure below25 doesnâ&#x20AC;&#x2122;t show the upper half of the double cone, but you can easily imagine it. Of the four curves depicted, only the hyperbola also cuts the upper half of the double cone.
25
Taken from Wikipedia, which has an excellent page on conic sections.
Page 176, Table of Contents
www.EconsPhDTutor.com
The three types of conic sections are the ellipse (plural: ellipses), the parabola (parabolae), and the hyperbola (hyperbolae). The circle is regarded as a special case of the ellipse.26 Here are the distinguishing characteristics of each: Type Ellipse Parabola Hyperbola
Description Formed from only one half of the double cone. A closed curve.27 Formed from only one half of the double cone. Not a closed curve. Formed from both halves of the double cone and is thus composed of two distinct branches. Not a closed curve.
Arises when B 2 − 4AC < 0 B 2 − 4AC = 0 B 2 − 4AC > 0
We can prove (but do not do so in this textbook) that in general, a conic section is the graph of the equation 1
Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0, where A, B, C, D, E, F are real constants and x and y are the two variables (on the cartesian plane). We refer to the expression B 2 − 4AC as the discriminant of the above equation. It is so named because it discriminates between the three possible types of conic sections. We can prove (but do not do so in this textbook) that if B 2 − 4AC > 0, then we have an ellipse; if B 2 − 4AC = 0, then we have a parabola; and if B 2 − 4AC < 0, then we have a hyperbola. In secondary school, we already learnt in some detail a special case of conic sections — the 1 quadratic y = ax2 + bx + c. This is the special case of the equation = where A = a, B= 0, C = 0, D= b, E = −1, and F = c. The quadratic y = ax2 + bx + c is indeed a parabola, because B 2 − 4AC = 02 − 4(a)(0) = 0. We already reviewed qudratic equations in section 14 and so we won’t talk any more about them in this chapter.
26
Strictly speaking, there are also the so-called degenerate conic sections, but we shall ignore these.
Page 177, Table of Contents
www.EconsPhDTutor.com
For A-levels, we are only required to learn about five more special cases of conic sections, listed below. And so that’s the plan for this chapter. 1.
x2 y 2 + = 1, a2 b2
2.
x2 y 2 − = 1, a2 b2
3.
y 2 x2 − = 1, b2 a2 ax + b , cx + d
4.
y=
5.
ax2 + bx + c y= . dx + e 1
Exercise 74. As per the general form given in =, state for each of the above five equations, what A, B, C, D, E, and F are. Compute the discriminant for each equation. Hence, conclude that first equation is of an ellipse and the remaining four are of hyperbolae. (Answer on p. 1038.)
Page 178, Table of Contents
www.EconsPhDTutor.com
The Ellipse x2 + y 2 = 1 (The Unit Circle)
16.1
x2 y 2 + = 1 describes an ellipse. In this section, we’ll study a special case of a2 b2 this equation, where a = b = 1. The equation then becomes x2 + y 2 = 1, which is the unit circle centred on the origin. The equation
By unit circle, we mean that it has radius of unit length, i.e. length 1.
1 f(x), g(x), y p = (x, y)
x
x2
+
y2
1
y
=1
y x
-1.0
-0.5
0.0
0.5
1.0
(-0, 0) Centre
-1 Why does this equation describe a circle? You can easily see that (1, 0), (0, 1), (−1, 0), and (0, −1) all satisfy the equation and are thus part of its graph. Indeed, these are the horizontal and vertical intercepts. What about elsewhere on the circle? Consider any point p on the unit circle. It forms a triangle — the line connecting it to the origin is the hypothenuse; that connecting it to the horizontal axis is the side; and that Page 179, Table of Contents
www.EconsPhDTutor.com
connecting it to the vertical axis is the base. By the Pythagorean Theorem, x2 + y 2 = 12 = 1. We have just proven that every point (x, y) on the unit circle satisfies the equation x2 +y 2 = 1. We now examine some of its characteristics. 1. Intercepts. The graph intersects the vertical axis at the points (0, −1) and (0, 1) and the horizontal axis at the points (−1, 0) and (1, 0). 2. Turning points. In this case, it is easy to see that there is a maximum turning point at (0, 1) and a minimum turning point at (0, 1). But just as an exercise, let’s also try to find these turning points more rigorously, i.e. through calculus. Exercise 49 showed that although it is impossible to rewrite the equation x2 + y 2 = 1 into the form of a single function, it is nonetheless possible√to and rewrite it into the form of two functions. Namely, f ∶ [−1, 1] → R defined by x ↦ 1 − x2 and g ∶ [−1, 1] → R defined √ by x ↦ − 1 − x2 . Above, the graph of the function f is the upper semicircle (red) and the graph of the function g is the lower semicircle (blue). Let’s compute the first derivative of f and set it equal to 0: f ′ (x) = 0.5(1 − x2 )−0.5 (−2x) = − x(1 − x2 )−0.5 −x(1 − x2 )−0.5 = 0 Ô⇒ x = 0. So the only stationary point of the function f is 0. We must now determine whether it is a maximum, minimum, or inflexion point. Compute the second derivative and evaluate it at the stationary point: f ′′ (x) = −(1 − x2 )−0.5 − x [−0.5(1 − x2 )−1.5 (−2x)] . This second derivative is messy and can be further simplified, but in this case there is no need to simplify it, since all we want is to evaluate it at 0. We have f ′′ (0) = −(1 − 02 )−0.5 − 0 × [−0.5(1 − 02 )−1.5 (−2 × 0)] = −1 < 0. Hence, the point x = 0 is a maximum turning point of f . We should make it a habit to write out the point in full, as (0, f (0)) = (0, 1). Since g = −f , it follows that g ′ (0) = 0 and g ′′ (0) = 1 > 0. That is, the only stationary point of the function g is (0, g(0)) = (0, −1). and it is a minimum point. 3. Asymptotes. By observation, there are no asymptotes. 4. Symmetry. The graph is a perfect circle centred on the origin. So by observation, every line that passes through the origin is a line of symmetry!
Page 180, Table of Contents
www.EconsPhDTutor.com
16.2
x2 y 2 The Ellipse 2 + 2 = 1 a b
f(x), g(x), y
b Vertical Intercept
a Horizontal Intercept
x2 / a2 + y2 / b2 = 1
(-0, 0) Centre y=0 Line of Symmetry
x
x=0 Line of Symmetry -b Vertical Intercept
-a Horizontal Intercept
Squares are a proper subset of rectangles. Similarly, circles are a proper subset of ellipses. The ellipse can be regarded as the generalisation of the circle. Why does the equation x2 /a2 + y 2 /b2 = 1 describe an ellipse? Rewrite the equation as y 2 x 2 ( ) + ( ) = 1. a b Hence, going from x2 + y 2 = 1 to x2 /a2 + y 2 /b2 = 1 involves two transformations: 1. First, stretch the graph horizontally, outwards from the vertical axis, by a factor of a. 2. Then stretch the graph vertically, outwards from the horizontal axis, by a factor of b. This gives us an “elongated circle” that we call an ellipse. 1. Intercepts. The graph intersects the vertical axis at the points (0, −b) and (0, b), and the horizontal axis at the points (−a, 0) and (a, 0). 2. Turning points. Clearly, there are maximum and minimum turning points at (0, b) and (0, −b). Let’s find these rigorously using calculus. Let’s again break the equation√up and rewrite it into the form of two functions. √ Namely, f ∶ [−a, a] → R defined by x ↦ ∣b∣ 1 − x2 /a2 and g ∶ [−a, a] → R defined by x ↦ −∣b∣ 1 − x2 /a2 . These are graphed above. Let’s compute the first derivative of f and set it equal to 0: ′
f (x) = 0.5∣b∣
Page 181, Table of Contents
−0.5
−2x x2 ( 2 ) = −∣b∣x (1 − 2 ) a a
−0.5
a−2 = 0 Ô⇒ x = 0.
www.EconsPhDTutor.com
So the only stationary point of the function f is 0. We can show that it is a maximum point, by computing the second derivative and evaluating it at 0: d x2 f (x) = [−∣b∣x (1 − 2 ) dx a ′′
−0.5
d x2 a ] = −a ∣b∣ [x (1 − 2 ) dx a −2
−2
−0.5
−0.5
]
−1.5
x2 x2 −2x = −a ∣b∣ [(1 − 2 ) − 0.5 (1 − 2 ) ( 2 )] a a a 2 2 0 0 −0x f ′′ (0) = −a−2 ∣b∣ [(1 − 2 ) −0.5 − 0.5 (1 − 2 ) −1.5 ( 2 )] = −a−2 ∣b∣ < 0. a a a −2
So (0, f (0)) = (0, b) is a maximum turning point of f . And since g = −f , g ′ (0) = 0 and g ′′ (0) = a−2 ∣b∣ > 0. That is, the only stationary point of g is (0, −b) and it is a minimum point. 3. Asymptotes. By observation, there are no asymptotes. 4. Symmetry. By observation, there are only two lines of symmetry, namely y = 0 and x = 0 (the horizontal and vertical axes).
Exercise 75. (Answer on p. 1039.) Let a, b, c, d be constants with a, b non-zero. Consider the equation 2
2
(x + c) (y + d) + = 1. a2 b2 (i) Sketch its graph. (ii) Write down the points at which it intersects the axes. (iii) Identify any turning points. (iv) Write down the equations of any lines of symmetry and also (v) asymptotes.
Page 182, Table of Contents
www.EconsPhDTutor.com
16.3
The Hyperbola: y = 1/x
y = 1/x (graphed) is the first hyperbola weâ&#x20AC;&#x2122;ll study. It is also the simplest possible hyperbola.
5
y = -x line of symmetry
y The graph of y = 1 / x has two branches.
4 3
y=x line of symmetry
2 1 x 0 -5
-4
-3
-2
-1
0
1
-1 -2
(0, 0) Centre
2
3
4 y=0 horizontal asymptote
5
-3 -4
x=0 vertical asymptote
-5 It turns out that all hyperbolae weâ&#x20AC;&#x2122;ll study have some common features. They have two branches. In the case of y = 1/x, one branch is top-right and the other is bottom-right.
Page 183, Table of Contents
www.EconsPhDTutor.com
Moreover, for all hyperbolae we’ll study, 1. Intercepts. Hyperbolae may or may not cross the axes. It depends. y = 1/x is an example of a hyperbola that crosses neither the vertical nor the horizontal axis. (But this is not true of all hyperbolae.) 2. Turning points. Hyperbolae may or may have turning points. It depends. y = 1/x is an example of a hyperbola that has no turning points. (But this is not true of all hyperbolae.) 3. Asymptotes. Hyperbolae always have two asymptotes. In the case of y = 1/x, they are y = 0 and x = 0. An interesting feature here is that the two asymptotes are perpendicular. A rectangular hyperbola is any hyperbola whose two asymptotes are perpendicular. And so y = 1/x is an example of a rectangular hyperbola. (But as we’ll see, not all hyperbolae are rectangular.) 4. The centre is the point at which the two asymptotes intersect. In the case of y = 1/x, the centre is (0, 0). 5. Two lines of symmetry. Both pass through the centre. Moreover, each line of symmetry bisects an angle formed by the two asymptotes. In the case of y = 1/x, they are y = x and y = −x.
Page 184, Table of Contents
www.EconsPhDTutor.com
The Hyperbola x2 − y 2 = 1
16.4
x2 − y 2 = 1 is a hyperbola and so it has two distinct branches. Notice also that if x ∈ (−1, 1), then there is no value of y for which x2 − y 2 = 1. Hence, the graph of this equation is empty in the region where x ∈ (−1, 1).
5
f(x), g(x), y
4
x=0 Line of Symmetry
(-0, 0) Centre 3
y = f(x)
2 x2 - y2 = 1
y=0 Line of Symmetry
1
x
0 -5
-4
-3
-2
-1
0
1
2
3
4
5
-1 Horizontal Intercept
-2
Horizontal Intercept
-3 y = g(x) y = x -4 Linear Asymptote -5
y = -x Linear Asymptote
1. Intercepts. The graph crosses the horizontal axis at the points (−1, 0) and (1, 0), but does not intersect the vertical axis. 2. The two turning points — there is a minimum turning point at (0, b) and a maximum turning point at (0, −b). √ √ √ 3. Asymptotes. We have y = ± x2 − 1. So as x → ∞, y = ± x2 − 1 → ± x2 = ±x. (Informally, as x → ∞, the 1 becomes negligible and we can simply ignore it). And so the two asymptotes are y = x and y = −x. The two asymptotes are perpendicular and so this is a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (0, 0). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes. So they must have slope 1 and −1. Moreover, both pass through the centre (0, 0). Altogether, we can work out that the lines of symmetry are y = x and y = −x.
Page 185, Table of Contents
www.EconsPhDTutor.com
16.5
x2 y 2 The Hyperbola 2 − 2 = 1 a b
To get from the equation x2 − y 2 = 1 to the equation x 2 y 2 ( ) −( ) =1 a b involves two simple transformations: 1. First stretch the graph horizontally, outwards from the vertical axis, by a factor of a. 2. Then stretch the graph vertically, outwards from the horizontal axis, by a factor of b.
f(x), g(x), y
y = f(x)
(-0, 0) Centre
x=0 Line of Symmetry
x2 / a2 - y2 / b2 = 1
y=0 Line of Symmetry
x
a Horizontal Intercept
-a Horizontal Intercept y = g(x) y = bx / a Linear Asymptote
Page 186, Table of Contents
y = -bx / a Linear Asymptote
www.EconsPhDTutor.com
y 2 x 2 The graph’s characteristics are similar to before. Again, ( ) − ( ) = 1 is a hyperbola a b and so it has two distinct branches. Notice also that if x ∈ (−a, a), then there is no value x2 y 2 of y for which 2 − 2 = 1. Hence, the graph of this equation is empty in the region where a b x ∈ (−a, a). 1. Intercepts. The graph crosses the horizontal axis at the points (a, 0) and (a, 0), but does not intersect the vertical axis. 2. There are no turning points.
√
2
√
x x 2 3. Asymptotes. We have y = ±∣b∣ ( ) − 1. So as x → ∞, y = ±∣b∣ ( ) − 1 → a a √ x 2 x x x ±∣b∣ ( ) = ±∣b∣ . And so the two asymptotes are y = b and y = −b . The two a a a a asymptotes are perpendicular and so this is a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (0, 0). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes. So they must have slope 1 and −1. Moreover, both pass through the centre (0, 0). Altogether, we can work out that the lines of symmetry are y = x and y = −x.
Exam Tip On the A-level exams, they typically only ask for (i) the intercepts; (ii) the asymptotes; and (iii) turning points. Nonetheless, you might as well know about the centre and the two lines of symmetry, because these concepts are not difficult and will help you to sketch better graphs.
Page 187, Table of Contents
www.EconsPhDTutor.com
16.6
y 2 x2 The Hyperbola 2 − 2 = 1 b a
SYLLABUS ALERT y 2 /b2 − x2 /a2 = 1 is explicitly in the 9758 (revised) but not the 9740 (old) syllabus. But even if you’re taking 9740, you might as well learn to draw y 2 /b2 − x2 /a2 = 1, because it’s really simple (since you now know how to draw x2 /a2 − y 2 /b2 = 1). y 2 x2 The graph of the equation 2 − 2 = 1 is simply the graph we studied in the previous section, b a π but rotated clockwise (or anticlockwise). 2
y = f(x)
x=0 Line of symmetry (-0, 0) Centre
f(x), g(x), y b Vertical Intercept
y2 / b2 - x2 / a2 = 1
y=0 Line of symmetry
x
y = -bx / a Linear asymptote
y = bx / a Linear asymptote y = g(x)
-b Vertical intercept
Let’s summarise the graph’s characteristics. This is a hyperbola and so there are two distinct branches. Notice also that if y ∈ (−b, b), then there is no value of x for which
Page 188, Table of Contents
www.EconsPhDTutor.com
y 2 x2 − = 1. Hence, the graph of this equation is empty in the region where y ∈ (−b, b). The b2 a2 range of y is thus (−∞, b] ∪ [b, ∞). 1. Intercepts. The graph crosses the vertical axis at the points (0, −b) and (0, b), but does not intersect the horizontal axis. 2. The two turning points are (0, b) (minimum) and (0, −b) (maximum). √ √ 2 x x 2 3. Asymptotes. We have y = ±∣b∣ 1 + ( ) . So as x → ∞, y = ±∣b∣ 1 + ( ) → a a √ 2 x x x x ±∣b∣ ( ) = ±∣b∣ . And so the two asymptotes are y = b and y = −b . The two a a a a asymptotes are perpendicular and so this is a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (0, 0). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes. So they must have slope 1 and −1. Moreover, both pass through the centre (0, 0). Altogether, we can work out that the lines of symmetry are y = x and y = −x.
y 2 x2 If we’d like, we can also find the turning points of 2 − 2 = 1 more rigorously, that is, b a through calculus. As with the circle, although it is not possible to rewrite this equation into the form of a single function, it is possible to rewrite √ it into the form of two functions. x 2 Namely, f ∶ (−∞, −a] ∪ [a, ∞) → R defined by x ↦ ∣b∣ ( ) + 1 and g ∶ (−∞, −a) ∪ (a, ∞) → a √ 2 x R defined by x ↦ −∣b∣ ( ) + 1. The graph of the function f is entirely above the horia zontal axis, while that of g is entirely below the horizontal axis. Let’s compute the first derivative of f : x 2 f (x) = 0.5∣b∣ [( ) + 1] a ′
−0.5
(
2x ∣b∣ x √ )= . a2 a x2 + a2
Hence, the only stationary point of f is (0, b). Let’s check what sort of a stationary point this is. ∣b∣ f ′′ (x) = a And so f ′′ (0) =
√ −0.5 x2 + a2 − x(0.5) (x2 + a2 ) (2x) . x2 + a2
∣b∣ > 0. Hence, this is a minimum point. a2
Similarly, by computing the first derivative of g and doing the work, we can find that the only stationary point of g is (0, −b) and that this is a maximum point.
Page 189, Table of Contents
www.EconsPhDTutor.com
16.7
Long Division of Polynomials
Remember long division? Turns out it’ll be useful for dividing polynomials. Here are a couple primary school examples to jog your memory. Example 141. What’s 83 ÷ 7? By long division, the quotient is 11 with a remainder of 6. So, 83 ÷ 7 = 116/7. 11 7 83 77 6 The quotient is the integer portion of the solution and the remainder is the “left-over” integer. Example 142. What’s 470 ÷ 17? By long division, the quotient is 27 with a remainder of 11. So, 470 ÷ 17 = 2711/17. 27 17 470 459 11
Page 190, Table of Contents
www.EconsPhDTutor.com
Long division can be used to divide one polynomial by another. But first of all, in case you don’t remember what a polynomial is... Definition 47. An nth-degree polynomial in one variable is any expression a0 xn + a1 xn−1 + a2 xn−2 + ⋅ ⋅ ⋅ + an−1 x + an where each ai is a constant and x is the variable. In this textbook, we’ll almost always consider only polynomials in one variable. So when I say polynomial, I’ll always mean a polynomial in one variable, unless otherwise stated.28 Example 143. The expressions 7x − 3 and 4x + 2 are 1st-degree polynomials (in one variable). These are also called linear polynomials. (Polynomials of low degree are often also called by such special names.)
Example 144. The expressions 3x2 + 4x − 5 and −x2 + 2x + 1 are 2nd-degree polynomials. These are also called quadratic polynomials.
Example 145. The expressions 2x3 + 2x2 + 3x − 1 and −3x3 + 2x2 + 3x + 1 3rd-degree polynomials. These are also called cubic polynomials.
Example 146. The expressions 5x4 − 2x3 + 2x2 + 3x − 1 and −9x4 + 3x3 + 2x2 + 3x + 1 are 4th-degree polynomials. These are also called quartic polynomials.
28
Actually, we’ve already secretly studied an example of a polynomial in two variables — the expression on the LHS of the equation of the conic section: Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0.
Page 191, Table of Contents
www.EconsPhDTutor.com
Now let’s do some polynomial division. x2 + 3 . We might be perfectly content with Example 147. Say you have an expression x−1 this expression. Or we might try to simplify it through long division: x +1 x − 1 x2 +0x +3 x2 −x +0 x +3 x −1 4 The “quotient” is x + 1 and the “remainder” is 4. Hence, 4 x2 + 3 =x+1+ x−1 x−1 4x3 + 2x2 + 1 Example 148. Let’s simplify through long division: 2x2 − x − 1 2x +2 2x2 − x − 1 4x3 +2x2 4x3 −2x2 4x2 4x2
+0x −2x +2x −2x 4x
+1 +0 +1 +3 +3
The “quotient” is 2x + 2 and the “remainder” is 4x + 3. Hence, 4x3 + 2x2 + 1 4x + 3 = 2x + 2 + . 2x2 − x − 1 2x2 − x − 1
Exercise 76. Simplify each of the following fractions through long division. (a) (b)
4x2 − 3x + 1 x2 + x + 3 . (c) . (Answer on p. 1040.) x+5 −x2 − 2x + 1
Page 192, Table of Contents
16x + 3 . 5x − 2
www.EconsPhDTutor.com
16.8
The Hyperbola y =
bx + c dx + e
In the next section, we’ll study this equation: ax2 + bx + c . y= dx + e In this section, as a warm-up, we’ll study the special case of the above equation, where a = 0: y=
Page 193, Table of Contents
bx + c . dx + e
www.EconsPhDTutor.com
Example 149. Graphed below is the equation y = (2x + 1)/(x + 1). This is the case where b = 2, c = 1, d = 1, and e = 1. Do the long division: 2 x + 1 2x +1 2x +2
Ô⇒
y=
−1 7 y = -x + 1 line of symmetry
5
3
(-1, 2) Centre -6
-4
2x + 1 1 =2− . x+1 x+1
y y=x+3 line of symmetry y=2 horizontal asymptote
1 x -2 0 -1 x = -1 vertical asymptote
2
4
-3
As usual, this is a hyperbola with two distinct branches. Other features: 1. Intercepts. The graph intersects the vertical axis at the point (0, 1) and the horizontal axis at the point (−0.5, 0). 2. There are no turning points. 3. Asymptotes. As x → −1, y → ±∞. And so x = −1 is a vertical asymptote. As x → ±∞, y → 2. And so y = 2 is a horizontal asymptote. The two asymptotes are perpendicular and so this is a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (−1, 2). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes. So they must have slope 1 and −1. Moreover, both pass through the centre (−1, 2). Altogether, we can work out that the lines of symmetry are y = x + 3 and y = −x + 1.
Page 194, Table of Contents
www.EconsPhDTutor.com
Example 150. Graphed below is the equation y = (7x + 3)/(2x + 4). This is the case where b = 7, c = 3, d = 2, and e = 4. Do the long division: 3.5 2x + 4 7x
+3
7x +14 −11
Ô⇒
y=
7x + 3 11 = 3.5 − . 2x + 4 2x + 4
Let’s summarise the graph’s characteristics. This is a hyperbola and so there are two distinct branches. 1. Intercepts. The graph intersects the vertical axis at the point (0, 0.75) and the horizontal axis at the point (−3/7, 0). 2. There are no turning points. 3. Asymptotes. As x → −2, y → ±∞. And so x = −2 is a vertical asymptote. As x → ±∞, y → 3.5. And so y = 3.5 is a horizontal asymptote. The two asymptotes are perpendicular and so this is a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (−2, 3.5). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes. So they must have slope 1 and −1. Moreover, both pass through the centre (−2, 3.5). Altogether, we can work out that the lines of symmetry are y = x + 5.5 and y = −x + 1.5.
Page 195, Table of Contents
www.EconsPhDTutor.com
Let’s now look, more generally, at the equation y = b/d dx + e bx bx
bx + c . By long division, we have: dx + e
+c +be/d c − be/d
The “quotient” is b/d and the “remainder” is c − be/d. Let’s further simplify this so that x has no coefficient. bx + c b c − be/d = + dx + e d dx + e =
b c − be/d 1 + d d x + e/d
=
b cd − be 1 + d d2 x + e/d
We can thus get from y = 1/x to the above equation, through these transformations: 1. Shift the graph leftwards by
1 e units to get the graph of y = . d x + e/d
2. Stretch the graph vertically, outwards from the horizontal axis, by a factor of cd − be get the graph of y = 2 . d (x + e/d) 3. Finally, shift the graph upwards by b/d units to get the final graph.
cd − be to d2
Exam Tip The A-level exams often ask you to list down a series of transformations that will get you from one graph to another, as was just done.
Page 196, Table of Contents
www.EconsPhDTutor.com
bx + c Let’s now summarise the characteristics of the graph of the equation y = . This is dx + e a hyperbola with two distinct branches. 1. Intercepts. If e = 0, then the graph does not cross the vertical axis. If e ≠ 0, then the graph intersects the vertical axis at the point (0, c/e). If b = 0, then the graph does not cross the horizontal axis. If b ≠ 0, then the graph intersects the vertical axis at the point (−c/b, 0). 2. There are no turning points.29 3. Asymptotes. As x → −e/d, y → ±∞. And so x = −e/d is a vertical asymptote. As x → ±∞, y → b/d. And so y = b/d is a horizontal asymptote. The two asymptotes are perpendicular and so this is a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (−e/d, b/d). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes. So they must have slope 1 and −1. Moreover, both pass through the centre (−e/d, b/d). Altogether, we can work out that the lines of symmetry are y = x + e/d + b/d and y = −x − e/d + b/d.
Exercise 77. For each of the following equations, sketch its graph and identify its intercepts, turning points, asymptotes, centre, and lines of symmetry (if there are any of these). x−2 −3x + 1 3x + 2 (a) y = . (b) y = . (c) y = . (Answers on pp. 1041, 1042, and 1043.) x+2 −2x + 1 2x + 3
29
See p. 926 in the Appendices (optional) for a proof that y =
Page 197, Table of Contents
bx + c has no turning points. dx + e
www.EconsPhDTutor.com
16.9
ax2 + bx + c The Hyperbola y = dx + e
We now study the more general equation ax2 + bx + c . y= dx + e We’ll rule out the following cases. • a = 0, because in that case section.
ax2 + bx + c bx + c = and this was already studied in the last dx + e dx + e
ax2 + bx + c ax2 + bx + c • d = 0, because in that case = , which is a quadratic and which dx + e e we already studied in secondary school. ax2 + bx a b • Both c and e are 0, because in that case = x + , which is a linear expression. dx d d We’ll start with the simplest possible case (a = 1, b = 0, c = 1, d = 1, and e = 0). This is the equation y=
Page 198, Table of Contents
x2 + 1 . x
www.EconsPhDTutor.com
Example 151. Graphed below is the equation y = (x2 + 1) /x.
10
y
y = (x2 + 1) / x
8 y=x Oblique Asymptote
6 (0, 0) Centre
4
Minimum Turning Point
2 0
-10
-6 Maximum Turning Point
-2
2
-2 -4
x=0 vertical asymptote
-8 -10 x x x2 +1
Do the long division:
10
y = (1 - √2) x Line of Symmetry
-6 y = (1 + √2) x Line of Symmetry
6
x
x2
Ô⇒
y=
x2 + 1 1 =x+ . x x
1 As usual, this is a hyperbola that has two distinct branches. Other features: 1. Intercepts. The graph intersects neither the vertical axis nor the horizontal axis. 2. There are two turning points — (−1, −2) is a maximum turning point and (1, 2) is a minimum turning point. (To find these, compute the first derivative dy/dx = 1 − 1/x2 . Set these equal to 0 for find two stationary points: x = ±1. Use the 2DT to determine that x = −1 and x = 1 are, respectively maximum and minimum turning points.) By observation, y can take on any value except those between these two turning points. The range of y is thus (−∞, −2] ∪ [2, ∞). 3. Asymptotes. As x → 0, y → ±∞. Hence, there is one vertical asymptote: x = 0. As x → ±∞, y → x. Hence, there is one oblique asymptote: y = x. The two asymptotes are not perpendicular and so this is not a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (0, 0). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes and pass through the centre. You don’t need to learn how to figure out their equations (but see pp. 927ff. in the Appendices if you’re interested).
Page 199, Table of Contents
www.EconsPhDTutor.com
x2 + 3x + 1 . Do the long division: Example 152. Graphed below is the equation y = x+1 x +2 x + 1 x2 +3x +1 x2 +x Ô⇒
2x 2x
+2 −1
y = (1 - √2) x + 2 - √2 Line of Symmetry
-11
y=
-7
y = (1 + √2) x + 2 + √2 Line of Symmetry
10 8 6 4 2 0 -3 -2 -4 -6 -8 -10
y
x2 + 3x + 1 1 =x+2− . x+1 x+1
y=x+2 Oblique Asymptote
x 1
5
9
(-1, 1) Centre
x = -1 vertical asymptote
As usual, this is a hyperbola that has two distinct branches. Other features: 1. Intercepts. The graph intersects the vertical axis at the point (0, 1) and the horizontal √ √ axis at the points (0.5(−3 + 5), 0) and (0.5(−3 − 5), 0). (The horizontal intercepts are simply the zeros of the quadratic x2 + 3x + 1.) 2. There are no turning points. (Compute dy/dx = 1 + 1/(x + 1)2 . Set this equal to 0 — there are no stationary points and thus no turning points either.) By observation, y can take on any value. The range of y is thus R. 3. Asymptotes. As x → −1, y → ±∞. Hence, there is one vertical asymptote: x = −1. As x → ±∞, y → x+2. Hence, there is one oblique asymptote: y = x+2. The two asymptotes are not perpendicular and so this is not a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (−1, 1). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes and pass through the centre. Again, you don’t need to know how to find their equations.
Page 200, Table of Contents
www.EconsPhDTutor.com
2x2 + 2x + 1 . Do the long division: Example 153. Graphed below is the equation y = −x + 1 −2x −4 −x + 1 2x2 +2x +1 2x2 −2x 4x
2x2 + 2x + 1 5 5 = −2x − 4 + = −2x − 4 + . −x + 1 −x + 1 −x + 1
Ô⇒
4x −4 5
14 10 Minimum Turning Point -9
6 2
-5 y = (-2 + √5) x - 4 - √5 Line of Symmetry
-1-2 -6 -10 -14
Maximum Turning Point
-18
y
y = (-2 - √5) x - 4 + √5 Line of Symmetry x=1 vertical asymptote x 3
7
11
(1, -6) Centre y = -2x - 4 Oblique Asymptote
-22 -26
As usual, this is a hyperbola that has two distinct branches. Other features: 1. Intercepts. The graph intersects the vertical axis at the point (0, 1), but not the horizontal axis, because there are no real zeros for the quadratic 2x2 + 2x + 1. √ √ 2. There are two turning points — (1 − 2.5, 0.325) and (1 + 2.5, −12.325) are the minimum and maximum turning points. (Verify this.) By observation, y can take on any value except those between these two turning points. The range of y is thus (−∞, −12.325] ∪ [0.325, ∞). 3. Asymptotes. As x → 1, y → ±∞. Hence, there is one vertical asymptote: x = 1. As x → ±∞, y → −2x − 4. Hence, there is one oblique asymptote: y = −2x − 4. The two asymptotes are not perpendicular and so this is not a rectangular hyperbola. 4. The centre (point at which the two asymptotes intersect) is (1, −6). 5. We know that the two lines of symmetry bisect the angles formed by the asymptotes and pass through the centre. Again, you don’t need to know how to find their equations. Page 201, Table of Contents
www.EconsPhDTutor.com
ax2 + bx + c We will not look more generally at the equation y = , because it gets rather dx + e messy. But if you want, you can read about it on pp. 927ff. of the Appendices (optional).
Exercise 78. For each of the following equations, sketch its graph and identify its intercepts, turning points, asymptotes, centre, and lines of symmetry (if any of these exist). (a) −x2 + x − 1 2x2 − 2x − 1 x2 + 2x + 1 . (b) y = . (c) y = . (Answers on pp. 1044, 1046, and y= x−4 x+1 x+4 1048.)
Page 202, Table of Contents
www.EconsPhDTutor.com
17
Simple Parametric Equations
A graph (or curve) is simply a set of points. Parametric equations give us an alternative method to describing the same graph (or curve). Example 154. Recall that the graph of the equation x2 + y 2 = 1 — i.e. the set S = {(x, y) ∶ x2 + y 2 = 1} — is the unit circle centred on the origin.
y
t = 3π / 4, x = - √2 / 2, y = √2 / 2 vx = - √2 / 2 ms-1, vy = - √2 / 2 ms-1 ax = √2 / 2 ms-2, ay = - √2 / 2 ms-2
Arrows indicate the instantaneous direction of travel. x2 + y2 = 1 x t = 0, x = 1, y = 0 vx = 0 ms-1, vy = 1 ms-1 ax = -1 ms-2, ay = 0 ms-2
t = 3π / 2, x = 0, y = -1 vx = 1 ms-1, vy = 0 ms-1 ax = 0 ms-2, ay = 1 ms-2
Observe that if x = cos t and y = sin t, then by a trigonometric identity, x2 + y 2 = 1. As it turns out, this gives us a second way of writing the set S: S = {(x, y) ∶ x = cos t, y = sin t, t ∈ R}. The variable t is called a parameter, hence the name parametric equations. As t increases from 0 to 2π, we trace out, anti-clockwise, a unit circle centred on the origin. t = 0 Ô⇒ (x, y) = (1, 0), √ √ t = π/4 Ô⇒ (x, y) = ( 2/2, 2/2) ,
t = 4π/4 Ô⇒ (x, y) = (−1, 0), √ √ t = 5π/4 Ô⇒ (x, y) = (− 2/2, − 2/2) ,
t = 2π/4 Ô⇒ (x, y) = (0, 1), √ √ t = 3π/4 Ô⇒ (x, y) = (− 2/2, 2/2) ,
t = 6π/4 Ô⇒ (x, y) = (0, −1), √ √ t = 7π/4 Ô⇒ (x, y) = ( 2/2, − 2/2) .
Page 203, Table of Contents
www.EconsPhDTutor.com
A nice interpretation is of the parameter t as time. Example 203 (continued from above). The set S = {(x, y) ∶ x = cos t, y = sin t, 0 ≤ t < 2π} can also be interpreted as tracing the motion of a particle as it moves anti-clockwise around a circle. x and y give the distances of the particle (in metres) from the origin, in the x- and y-directions. We have x = cos t and y = sin t. This says that at any instant of time t, the particle is cos t metres to the east of the origin and sin t metres to the north of the origin. (Note that if cos t < 0, then the particle is to the west of the origin. And if sin t < 0, then the particle is to the south of the origin.) At time t = 0 s, the particle is at the position (x, y) = (1, 0). At time t = 1 s, the particle has moved to the position (x, y) = (0.54, 0.84). At time t = π/2 ≈ 1.07 s, the particle has moved to position (x, y) = (0, 1).
Having interpreted t as time, we can now also easily talk about the velocity and acceleration of the particle at different instants in time. Example 203 (continued from above). We have x = cos t and y = sin t. From this, we can easily compute the particle’s velocity in each direction: vx = dx/dt = − sin t and vy = dy/dt = cos t. This says that at any instant of time t, the velocity of the particle is − sin t ms-1 in the x-direction and cos t ms-1 in the y-direction. (Note that if − sin t < 0, then the particle is moving westwards. And if cos t < 0, then the particle is moving southwards.) √ So for example, at time t = 7π/4, its velocity is − sin (7π/4) = 2/2 ms-1 rightwards and √ cos (7π/4) = 2/2 ms−1 upwards. Similarly, we can compute the particle’s acceleration in each direction: ax = d2 x/dt2 = − cos t and ay = d2 y/dt2 = − sin t. √ So for example,√at time t = 7π/4, its acceleration is − cos (7π/4) = − 2/2 ms-1 rightwards and − cos (7π/4) = 2/2 ms−1 upwards. That is, the particle is travelling rightwards (because its velocity rightwards at this instant in time is positive); however, its rightwards velocity is slowing down.
Exercise 79. (Answer on p. 1050.) Let P be the particle whose position (in metres) is described by the set {(x, y) ∶ x = cos t, y = sin t, t ∈ R}, where t is time (seconds). Let Q be the particle whose position (in metres) is described by the set {(x, y) ∶ x = sin t, y = cos t, t ∈ R}. (a) How does the starting point (when t = 0) of Q differ from that of P ? (b) What about the direction of travel?
Page 204, Table of Contents
www.EconsPhDTutor.com
Example 155. Recall that the graph of the equation x2 /a2 + y 2 /b2 = 1 — i.e. the set T = {(x, y) ∶ x2 /a2 + y 2 /b2 = 1} — is the ellipse centred on the origin, with horizontal intercepts ±a and vertical intercepts ±b.
y t = 3π / 4
Arrows indicate the instantaneous direction of travel. x t = 0, x = 1, y = 0
t = 3π / 2
Observe that if x = a cos t and y = b sin t, then by the same trigonometric identity as before, x2 /a2 + y 2 /b2 = 1. As it turns out, this gives us a second way of writing the set T : T = {(x, y) ∶ x = a cos t, y = b sin t, t ∈ R} . Similar to before, as t increases from 0 to 2π, we trace out, anti-clockwise, an ellipse centred on the origin. At any instant in time t, the particle’s position, velocity, and acceleration are (x, y) = (a cos t, b sin t), (vx , vy ) = (−a sin t, b cos t), and (ax , ay ) = (−a cos t, −b sin t).
Exercise 80. Let P be the particle whose position (in metres) is described {(x, y) ∶ x = a cos t, y = b sin t, t ∈ R}, where t is time (seconds). At each of the following times, state the particle’s position and also its velocity and acceleration in both the x- and π π y- directions. (a) t = ; (b) t = ; (c) t = 2π. (Answer on p. 1050.) 4 2
Page 205, Table of Contents
www.EconsPhDTutor.com
Example 156. Recall that the graph of the equation x2 − y 2 = 1 — i.e. the set U = {(x, y) ∶ x2 − y 2 = 1} — is the rectangular “east-west” hyperbola centred on the origin, with horizontal intercepts ±1 and no vertical intercepts.
Arrows indicate 5 y the instantaneous 4 direction of travel. 3 x2 - y2 = 1 2 1 t=4 0 t=3 -5 -4 -3 -2 -1 -1 0 -2 t=2 -3 -4 -5
t=1 x
t=0 1
2
3
4
5
t=5
Observe that if x = sec t and y = tan t, then by a trigonometric identity, x2 − y 2 = 1. As it turns out, this gives us a second way of writing the set U : U = {(x, y) ∶ x = sec t, y = tan t, t ∈ R, t ≠ kπ/2} . Note that t cannot be a half-integer multiple of π, because then tan t would be undefined. Again, let’s interpret this as the movement of a particle. Interestingly, the particle always moves upwards, as we can easily prove — vy = dy/dt = sec2 t > 0 for all t. At t = 0, the particle is at (x, y) = (1, 0). During t ∈ [0, π/2), the particle moves northeast along the green segment and flies off towards infinity as t → π/2 ≈ 1.57. An instant after π/2 seconds, the particle magically reappears “near” infinity in the southwest. During t ∈ (π/2, π], the particle moves northeast along the blue segment. At t = π, the particle is at (−1, 0). During t ∈ [π, 3π/2), the particle moves northwest along the red segment and flies off towards infinity as t → 3π/2 ≈ 4.71. An instant after 3π/2 seconds, the particle magically reappears “near” infinity in the southeast. During t ∈ (3π/2, 2π], the particle moves northwest along the pink segment.
Page 206, Table of Contents
www.EconsPhDTutor.com
Exercise 81. (Answer on p. 1051.) Suppose that the position of a particle is described by the set {(x, y) ∶ x = tan t, y = sec t, t ∈ R}, where t is time, measured in seconds. (a) Rewrite the set using a single cartesian equation. (b) Compute dx/dt. And hence make an observation about how the particle travels in the x-direction. The graph below indicates six positions of the particle — A, B, C, D, E, and F . (Also indicated are the directions of travel.) The particle is at these positions at times t = 0, 1, 2, 3, 4, and 5 but not necessarily in that order. (c) Using only the graphs of s = tan t and s = sec t (above) to guide you and without using a calculator, state where the particle is, at each of the the times t = 0, 1, 2, 3, 4, and 5.
5
y
4
A
3 C
2 B 1 {(x, y): x = tan t, y = sec t, t -5
-4
-3
-2
0
-1
x
} 0
1
2
3
4
5
-1 E -2 D
F
-3
Arrows indicate -4 the instantaneous direction of travel. -5
Page 207, Table of Contents
www.EconsPhDTutor.com
17.1
Eliminating the Parameter t
Given a pair of parametric equations that describes a set of points, we can often go in reverse: We can eliminate the parameter t and describe the same set of points using a single equation. Example 157. The set {(x, y) ∶ x = t2 + t, y = t − 1, t ∈ R} describes the position (metres) of a particle at time t (seconds).
y Instantaneous Direction of Travel
Instantaneous Direction of Travel
Instantaneous Direction of Travel
x
t = 1, x = 2, y = 0 vx = (2t + 1) ms-1 = 3 ms-1 vy = 1 ms-1, ax = 2 ms-2, ay = 0 ms-2 t = 0, x = 0, y = - 1 vx = (2t + 1) ms-1 = 1 ms-1 vy = 1 ms-1, ax = 2 ms-2, ay = 0 ms-2 t = - 1, x = 0, y = - 2 vx = (2t + 1) ms-1 = - 1 ms-1 vy = 1 ms-1, ax = 2 ms-2, ay = 0 ms-2
x = y2 + 3y + 2
Write t = y + 1 and x = (y + 1)2 + (y + 1) = y 2 + 3y + 2. And so the same set can be rewritten as: {(x, y) ∶ x = y 2 + 3y + 2}. As an exercise, let’s also compute the velocity and acceleration of the particle. vx = dx/dt = 2t + 1 and vy = dy/dt = 1. This says that at any instant in time t, the particle has velocity 2t + 1 ms−1 rightwards and 1 ms−1 upwards. ax = d2 x/dt2 = 2 and ay = d2 y/dt2 = 0. This says that the particle is always accelerating rightwards at the rate 2 ms−2 . Moreover, it is never accelerating upwards (this is consistent with the above finding that its upwards velocity is a constant 1 ms−1 ).
Page 208, Table of Contents
www.EconsPhDTutor.com
Example 158. The set {(x, y) ∶ x = 2 cos t − 4, y = 3 sin t + 1, t ∈ R} describes the position (metres) of a particle at time t (seconds).
5 t = π / 2, x = - 4, y = 4 vx = - 2 sin (t) ms-1 = - 2 ms-1 vy = 3 cos (t) ms-1 = 0 ms-1 ax = - 2 cos (t) ms-2 = 0 ms-2 ay = - 3 sin (t) ms-2 = -3 ms-2 t = π, x = - 6, y = 1 vx = - 2 sin (t) ms-1 = 0 ms-1 vy = 3 cos (t) ms-1 = - 3 ms-1 ax = - 2 cos (t) ms-2 = 2 ms-2 ay = - 3 sin (t) ms-2 = 0 ms-2
y
4 3 2 1 x 0
-7
-5
-3
-1
1 -1
t = 3π / 2 , x = - 4, y = - 2 vx = - 2 sin (t) ms-1 = 2 ms-1 -2 vy = 3 cos (t) ms-1 = 0 ms-1 ax = - 2 cos (t) ms-2 = 0 ms-2 ay = - 3 sin (t) ms-2 = 3 ms-2 -3 Write (x + 4) /2 = cos t and (y − 1) /3 = sin t. Using the trigonometric identity cos2 t+sin2 t = 2 2 1, we can rewrite the set as {(x, y) ∶ x = [(x + 4) /2] + [(y − 1) /3] = 1}. This is the ellipse centred on (−4, 1). As an exercise, let’s also compute the velocity and acceleration of the particle. vx = dx/dt = −2 sin t and vy = dy/dt = 3 cos t. This says that at any instant in time t, the particle has velocity 2 sin t ms−1 leftwards and 3 cos t ms−1 upwards. ax = d2 x/dt2 = −2 cos t and ay = d2 y/dt2 = −3 sin t. This says that at any instant in time t, the particle is accelerating leftwards at the rate −2 cos t ms−2 and upwards at the rate −3 sin t ms−2 .
Page 209, Table of Contents
www.EconsPhDTutor.com
Exercise 82. Each of the following sets describes the position (metres) of a particle at time t (seconds). Rewrite each set into a form where the parameter t is eliminated. Sketch the graph of each. Indicate the particle’s position and direction of travel at t = 0. (Answers on pp. 1052, 1053, and 1054.) (a) {(x, y) ∶ x = 2 sin t − 1, y = 3 cos2 t, t ∈ R}. (b) {(x, y) ∶ x =
1 , y = t2 + 1, t ∈ R}. t−1
(c) {(x, y) ∶ x = t − 1, y = ln(2t + 1), t > −0.5}.
Page 210, Table of Contents
www.EconsPhDTutor.com
18
Equations and Inequalities
N N Given any fraction (where N and D are real numbers with D non-zero), we have >0 D D if and only if one of the following is true: 1. “N > 0 AND D > 0”; OR 2. “N < 0 AND D < 0”. The expressions that are in the numerator (N ) and denominator (D) can get pretty complicated. So here are some very simple examples just to warm you up.
Example 159.
4 > 0 because both the numerator and denominator are positive. 7
Example 160.
−5 > 0 because both the numerator and denominator are negative. −3
Example 161.
−9 < 0 because the numerator is negative but the denominator is positive. 2
Example 162.
1 > 0 because the numerator is positive but the denominator is negative. −8
Page 211, Table of Contents
www.EconsPhDTutor.com
18.1
Example 163.
ax + b >0 cx + d
x+3 > 0 ⇐⇒ one of the following is true: 3x + 2
1. “x + 3 > 0 AND 3x + 2 > 0”; OR 2. “x + 3 < 0 AND 3x + 2 < 0”. Notice that (1) “x + 3 > 0 AND 3x + 2 > 0” ⇐⇒ “x > −3 AND x > −2/3” , which in turn is equivalent to the single inequality “x > −2/3”. Notice that (1) “x + 3 < 0 AND 3x + 2 < 0” ⇐⇒ “x < −3 AND x < −2/3” , which in turn is equivalent to the single inequality “x < −3”. x+3 > 0 ⇐⇒ “x > −2/3 OR x < −3” (equivalently, “x ∈ (−∞, −3) ∪ Altogether then, 3x + 2 2 (− , ∞)”). 3 Note that I use quotation marks “⋅”, but these are not necessary. Instead, they merely help to make especially clear which groups of conditions corresponds to each other.
Example 164.
4x − 1 > 0 ⇐⇒ one of the following is true: x+2
1. “4x − 1 > 0 AND x + 2 > 0” ⇐⇒ “x > 1/4 AND x > −2” ⇐⇒ “x > 1/4” ; OR 2. “4x − 1 < 0 AND x + 2 < 0” ⇐⇒ “x < 1/4 AND x < −2” ⇐⇒ “x < −2”. Altogether then, (1/4, ∞)”). Example 165.
4x − 1 > 0 ⇐⇒ “x > 1/4 and x < −2” (equivalently, “x ∈ (−∞, −2) ∪ x+2
5x + 4 > 0 ⇐⇒ one of the following is true: −2x + 1
1. “5x + 4 > 0 AND −2x + 1 > 0” ⇐⇒ “x > −4/5 AND x < 1/2” ⇐⇒ “x ∈ (−4/5, 1/2)” ; OR 2. “5x + 4 < 0 AND −2x + 1 < 0” ⇐⇒ “x < −4/5 AND x > 1/2”, but these are mutually contradictory and thus impossible. Altogether then,
5x + 4 > 0 ⇐⇒ “x ∈ (−4/5, 1/2)”. −2x + 1
When given any inequality that is of a slightly different form, be sure to always convert it N into what I’ll call the standard form > 0. Strictly speaking, this is not necessary, but D Page 212, Table of Contents
www.EconsPhDTutor.com
if you always do this, you’ll make a habit of solving inequalities in this form, and thus be less likely to make a careless mistake.
Example 166. Consider the inequality
3x − 2 < 3. This inequality is equivalent to −5x + 1
3x − 2 >0 −5x + 1
⇐⇒
3−
⇐⇒
−15x + 3 − (3x − 2) >0 −5x + 1
−18x + 5 > 0. −5x + 1
This last inequality is in turn true ⇐⇒ one of the following is true: 1. “−18x + 5 > 0 AND −5x + 1 > 0” ⇐⇒ “x < −5/18 AND x < 1/5” ⇐⇒ “x < −5/18” ; OR 2. “−18x + 5 < 0 AND −5x + 1 < 0” ⇐⇒ “x > −5/18 AND x > 1/5” ⇐⇒ “x > 1/5”. Altogether then, (1/5, ∞)”).
3x − 2 < 3 ⇐⇒ “x < −5/18 OR x > 1/5” (equivalently, “x ∈ (−∞, −5/18) ∪ −5x + 1
2x + 1 Exercise 83. For what values of x is each of the following inequalities true? (a) > 0. 3x + 2 −1 1 −3x − 18 2x + 3 x−1 > 0. (c) > 0. (d) > 0. (e) > 0. (f) < 9. (Answers on p. (b) −4 −4 −4 9x − 14 −x + 7 1055.)
Page 213, Table of Contents
www.EconsPhDTutor.com
18.2
ax2 + bx + c >0 dx2 + ex + f
ax2 + bx + c . But dx2 + ex + f ax2 + bx + c > 0. This is you are required to know how to find the values of x for which dx2 + ex + f just the same game as before, albeit slightly more complicated.
Don’t worry, you are not required to know how to graph the equation y =
2x2 + x + 3 > 0 ⇐⇒ one of the following is true: −x2 + 3x + 2
Example 167.
1. “2x2 + x + 3 > 0 AND −x2 + 3x + 2 > 0”; OR 2. “2x2 + x + 3 < 0 AND −x2 + 3x + 2 < 0”.
y = 2x2 + x + 3 is a ∪-shaped quadratic and has no real roots (because the discriminant is negative). Hence, it is always positive. It is thus impossible that “2x2 + x + 3 < 0 AND −x2 + 3x + 2 < 0” (Case 2). We need thus only examine Case 1. As we just said, it is always true that 2x2 + x + 3 > 0. So we need only examine when it is true that −x2 + 3x + 2 > 0. The equation y = −x2 + 3x + 2 has a ∩-shaped graph and has two real zeros given by: √
√ √ √ 32 − 4(−1)(2) −3 ± 17 3 ∓ 17 = = = 0.5 (3 ∓ 17) . 2(−1) −2 2 √ √ Hence, the expression −x2 + 3x + 2 > 0 ⇐⇒ “x ∈ (0.5 (3 − 17) , 0.5 (3 + 17))”. −3 ±
Altogether then,
√ √ 2x2 + x + 3 > 0 ⇐⇒ “x ∈ (0.5 (3 − 17) , 0.5 (3 + 17))”. −x2 + 3x + 2
A dirty trick is to use your TI84 to do a quick check that this answer is correct:
Page 214, Table of Contents
www.EconsPhDTutor.com
−x2 + 4x − 1 > 0 ⇐⇒ one of the following is true: Example 168. 2x2 + x + 2 1. “−x2 + 4x − 1 > 0 AND 2x2 + x + 2 > 0”; OR 2. “−x2 + 4x − 1 < 0 AND 2x2 + x + 2 < 0”. The equation y = 2x2 + x + 2 has a ∪-shaped graph and has no real zeros (because the discriminant is negative). Hence, it is always positive. It is thus impossible that “−x2 + 4x − 1 < 0 AND 2x2 + x + 2 < 0” (Case 2). We need thus only examine Case 1. As we just said, it is always true that y = 2x2 +x+2 > 0. So we need only examine when it is true that −x2 + 4x − 1 > 0. The equation y = −x2 + 4x − 1 has a ∩-shaped graph and has two real zeros given by: √
√ √ √ 42 − 4(−1)(−1) −4 ± 12 4 ∓ 12 = = = 2 ∓ 3. 2(−1) −2 2 √ √ Hence, the expression −x2 + 4x − 1 > 0 ⇐⇒ “x ∈ (2 − 3, 2 + 3)”. −4 ±
√ √ −x2 + 4x − 1 Thus, > 0 ⇐⇒ x ∈ (2 − 3, 2 + 3) ≈ (0.268, 3.732). As usual, let’s check 2x2 + x + 2 using our TI84:
Page 215, Table of Contents
www.EconsPhDTutor.com
x2 + 5x + 4 > 0 ⇐⇒ one of the following is true: Example 169. −x2 − 2x + 1 1. “x2 + 5x + 4 > 0 AND −x2 − 2x + 1 > 0”; OR 2. “x2 + 5x + 4 < 0 AND −x2 − 2x + 1 < 0”. The equation y = x2 + 5x + 4 has a ∪-shaped graph and has two real zeros given by: −5 ±
√
√ (5)2 − 4(1)(4) −5 ± 9 −5 ± 3 = = = −4, −1. 2(1) 2 2
Hence, the expression x2 + 5x + 4 > 0 ⇐⇒ “x < −4 OR x > −1”. The equation y = −x2 − 2x + 1 has a ∩-shaped graph and has two real roots given by: √
√ √ (−2)2 − 4(−1)(1) 2 ± 8 = = −1 ∓ 2. 2(−1) −2 √ √ Hence, the expression −x2 + 4x − 1 > 0 ⇐⇒ “x ∈ (−1 − 2, 2 − 1)”. Thus: 2±
√ √ 1. “x2 +5x+4 > 0 AND −x2 −2x+1 > 0” ⇐⇒ “x < −4 OR x > −1 AND x ∈ (−1 − 2, 2 − 1)”. √ √ Since −1 − 2 < −1, this is equivalent to x ∈ (−1, 2 − 1). √ √ 2. “x2 + 5x + 4 < 0 AND −x2 − 2x + 1 < 0” ⇐⇒ “x ∈ (−4, −1) AND x < −1 − 2 or x > 2 − 1”. √ √ Since −1 − 2 < −1, this is equivalent to x ∈ (−4, −1 − 2). √ √ x2 + 5x + 4 Altogether then, > 0 ⇐⇒ x ∈ (−4, −1 − 2) ∪ (−1, 2 − 1). As usual, let’s −x2 − 2x + 1 check using our TI84:
Page 216, Table of Contents
www.EconsPhDTutor.com
x2 − 4x + 3 > 0 ⇐⇒ one of the following is true: Example 170. x2 − 2x 1. “x2 − 4x + 3 > 0 AND x2 − 2x > 0”; OR 2. “x2 − 4x + 3 < 0 AND x2 − 2x < 0”. The equation y = x2 − 4x + 3 has a ∪-shaped graph and has two real zeros given by: 4±
√
√ (−4)2 − 4(1)(3) 4 ± 4 = = 1, 3. 2(1) 2
Hence, the expression x2 − 4x + 3 > 0 ⇐⇒ “x ∈ (1, 3)”. The equation y = x2 − 2x has a ∪-shaped graph and has two real roots given by: 2±
√
√ (−2)2 − 4(1)(0) 2 ± 4 = = 0, 2. 2(1) 2
Hence, x2 − 2x > 0 ⇐⇒ “x ∈ (0, 2)”. Thus: 1. “x2 − 4x + 3 > 0 AND x2 − 2x > 0” ⇐⇒ “x ∈ (1, 3) AND x ∈ (0, 2)” ⇐⇒ ”x ∈ (1, 2)”. 2. “x2 − 4x + 3 < 0 AND x2 − 2x < 0” ⇐⇒ “x < 1 OR x > 3 AND x < 0 OR x > 2” ⇐⇒ “x < 0 or x > 3”. Altogether then, using our TI84:
x2 − 4x + 3 > 0 ⇐⇒ “x ∈ (−∞, 0) ∪ (1, 2) ∪ (3, ∞)”. As usual, let’s check x2 − 2x
Exercise 84. Without using a calculator, find the values of x for which each of the following x2 + 2x + 1 x2 − 1 x2 − 3x − 18 2x + 5 inequalities is true. (a) 2 > 0. (b) 2 > 0. (c) > 0. (d) > x − 3x + 2 x −4 −x2 + 9x − 14 −x + 4 −3x + 1 . (Answers on pp. 1057, 1058, 1059, and 1060.) 6x − 7
Page 217, Table of Contents
www.EconsPhDTutor.com
18.3
Solving Inequalities by Graphical Methods
Example 171. For what values of x is x > sin (0.5πx)? Rewrite the inequality as x − sin(0.5πx) > 0. Graph y = x − sin(0.5πx) on your graphing calculator. Our goal is to first find the horizontal intercepts of this equation; this will let us solve for x > sin (0.5πx). After Step 1.
After Step 2.
After Step 3.
After Step 4.
After Step 5.
After Step 6.
In the TI84: 1. Press ON to turn on your calculator. 2. Press Y= to bring up the Y= editor. 3. Press X,T,θ,n − SIN 0 . 5 . To enter “π”, press the blue 2ND button and then π (which corresponds to the ∧ button). Now press X,T,θ,n ) and altogether you will have entered “x − sin(0.5πx)”. 4. Now press GRAPH and the calculator will graph y = x − sin(0.5πx). It looks like the horizontal intercepts are close to the origin. Let’s zoom in to see better. 5. Press the (ZOOM) button to bring up a menu of ZOOM options. 6. Press 2 to select the Zoom In option. Nothing seems to happen. But now press ENTER and the TI will zoom in a little for you. It looks like there are 3 horizontal intercepts. To find out what precisely they are, we’ll use the TI84’s “zero” option.
(... Example continued on the next page ...)
Page 218, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) After Step 7.
After Step 8.
After Step 9.
After Step 11.
After Step 12.
After Step 13.
After Step 10.
3. Press the blue 2ND button and then CALC (which corresponds to the TRACE button). This brings up the CALCULATE menu. 4. Press 2 to select the “zero” option. This brings you back to the graph, with a cursor flashing. Also, the TI84 prompts you with the question: “Left Bound?” TI84’s ZERO function works by you first specifying a “Left Bound” and a “Right Bound” for x. TI84 will then check to see if there are any horizontal intercepts (i.e. values of x for which y = 0) within those bounds. 5. Using the < and > arrow keys, move the blinking cursor until it is where you want your first “Left Bound” to be. For me, I have placed it a little to the left of where I believe the leftmost horizontal intercept to be. 6. Press ENTER and you will have just entered your first “Left Bound”. TI84 now prompts you with the question: “Right Bound?”. 7. So now just repeat. Using the < and > arrow keys, move the blinking cursor until it is where you want your first “Right Bound” to be. For me, I have placed it a little to the right of where I believe the leftmost horizontal is. 8. Again press ENTER and you will have just entered your first “Right Bound”. TI84 now asks you: “Guess?” This is just asking if you want to proceed and get TI84 to work out where the horizontal intercept is. So go ahead and: 9. Press ENTER . TI84 now informs you that there is a “Zero” at “x = −1”, “y = 0” and places the blinking cursor at precisely that point. This is the first horizontal intercept we’ve found. To find each of the other 2 horizontal intercepts, just repeat steps 3 through 9. You should be able to find that they are at x = 0 and x = 1. Altogether, the 3 intercepts are x = −1, 0, 1. Based on these and what the graph looks like, we conclude: x > sin (0.5πx) ⇐⇒ x ∈ (−1, 0) ∪ (1, ∞).
Page 219, Table of Contents
www.EconsPhDTutor.com
Example 172. For what values of x is x > e + ln x? For this example, I won’t give the full detailed instructions of what to do on the TI84; I’ll only show a few screenshots. First, rewrite the inequality as x − e − ln x > 0 and so graph y = x − e − ln x on your graphing calculator: After Graphing. Zoom In, Adjust Window.
Look for the values of x for which x − e − ln x = 0. They are x = 0.7083, 4.1387: Leftmost horizontal intercept. Rightmost horizontal intercept.
Based on these horizontal intercepts and what the graph looks like, we conclude: x > e+ln x if and only if x ∈ (0, 0.7083) ∪ (4.1387, ∞).
Exercise 85. Use a graphing calculator to find the values of x for which each of the √ 1 > x3 + sin x. following inequalities is true. (a) x3 − x2 + x − 1 > ex . (b) x > cos x. (c) 2 1−x (Answers on pp. 1061, 1061, and 1062.)
Page 220, Table of Contents
www.EconsPhDTutor.com
18.4
Systems of Equations
Warm-up questions: Exercise 86. (PSLE-style question.) When Apu was 40 years old, Beng was twice as old as Caleb. Today, Caleb is 28 years old and Apu is twice as old as Beng. What are the ages of Apu and Beng today? (If necessary, assume that the age of a person is always an integer and is fixed between January 1st and December 31st of each year.) (Answer on p. 1063.) Exercise 87. (O-Level style question.) Planes A and B leave the same point at 12pm. Plane A travels northeast at a constant speed of 100 km/h. Plane B travels south at a constant speed of 200 km/h. At 3pm, both planes make an instant turn and start flying directly towards each other at the same speed. At what time will the two planes collide? (Answer on p. 1063.)
Definition 48. Given an equation involving a single variable x, a real solution to the equation is any value of x ∈ R such that the equation is true. Example 173. The equation x + 5 = 8 has one real solution: 3. The equation x2 − 1 = 0 has two real solutions: −1 and 1. The equation x2 − 1 = 8 has two real solutions: −3 and 3. The equation x3 − 4x = 0 has three real solutions: −2, 0, and 2. Example 174. The equation x2 + 1 = 0 has no real solution.
Definition 49. Given an equation involving a single variable x, a real solution set is the set of values of x ∈ R such that the equation is true. Example 175. The real solution set of the equation x + 5 = 8 is {3}. The real solution set of the equation x2 − 1 = 0 is {−1, 1}. The real solution set of the equation x2 − 1 = 8 is {−3, 3}. The real solution set of the equation x3 − 4x = 0 is {−2, 0, 2}. Example 176. The real solution set of the equation x2 + 1 = 0 is ∅ = {}.
Page 221, Table of Contents
www.EconsPhDTutor.com
Definition 50. Given a system of equations (or more simply a set of equations) involving two variables x and y, a real solution to the set of equations is any point (or ordered pair) (x, y) with x, y ∈ R for which the system of equations is true; and a real solution set is the set of ordered pairs (x, y) for which the system of equations is true. Example 177. Consider the system of equations y = x + 1, y = −x + 3. To solve this system of equations, plug in the second equation into the first to get: −x + 3 = x + 1. Now solve: x = 1. And so y = x + 1 = 2. Altogether, this system of equations has one real solution (1, 2). Its real solution set is thus {(1, 2)}. Example 178. Consider the system of equations y = 0.5x2 − 1.5 and y = x. To solve this system of equations, plug in the second equation into the first to get: x = 0.5x2 − 1.5. Rearranging: x2 − 2x − 3 = 0. Now solve: x = 3, −1. Correspondingly, y = 3, −1. Altogether, this system of equations has two real solutions: (3, 3) and (−1, −1). Its real solution set is thus {(3, 3), (−1, −1)}. A system of equations can have no real solutions. Example 179. Consider the system of equations y = ln x and y = x. Observe that for all x ∈ (0, 1), ln x < 0 and hence x > ln x. Moreover, for x = 1, ln x = 0 < x. Also, for x > 1, 1 d d ln x = < 1 < x = 1, so the slope of y = x is steeper than that of y = ln x. Altogether dx x dx then, for all x > 0, x > ln x. Hence, this system of equations has no real solutions. Its real solution set is thus ∅ = {}. A system of equations can also have infinitely many real solutions. Example 180. Consider the system of equations y = x and 2y = 2x. Observe that this system of equations has infinitely many real solutions, e.g. (1, 1), (2, 2), (2.74, 2.74). There is thus no way to explicitly list out all its real solutions. However, using set-builder notation, we can write down its real solution set as {(x, y} ∶ y = x}. This says that every ordered pair (x, y) such that y = x is a real solution to the given system of equations.
Solve these problems without using a calculator. Exercise 88. The points (1, 2), (3, 5), and (6, 9) satisfy the equation y = ax2 +bx+c. What are a, b, and c? (Answer on p. 1064.) Exercise 89. The point (−1, 2) satisfies the equation y = ax2 + bx + c. Moreover, the minimum point of the equation y = ax2 + bx + c is (0, 0). What are a, b, and c? (Answer on p. 1064.)
Page 222, Table of Contents
www.EconsPhDTutor.com
You are required to know how to use a graphing calculator to find the numerical solution of equations (including system of linear equations). Example 181. Solve the system of equations y = x4 − x3 − 5, y = ln x. One method is to graph both equations on your graphing calculator and then find their intersection points. Here I’ll use another method: First rewrite the two equations as a third equation y = x4 − x3 − 5 − ln x. Our goal is to find the horizontal intercepts of this equation, which will in turn also be the solutions to the above set of equations. Briefly, in the TI84: 1. Graph the equation y = x4 − x3 − 5 − ln x. It looks like there is only one horizontal intercept. 2. Zoom in. 3. Find the horizontal intercept using the “zero” option. Conclusion: There is one solution to this set of equations and its x-coordinate is 1.8658. To find the y-coordinate, we need merely plug in this value of x into either of the equations in the original set of equations: y = ln x = ln 1.8658 ≈ 0.6237. Altogether, this set of equations has one solution: (1.8658, 0.6237). After Step 1.
After Step 2.
After Step 3.
Exercise 90. Using your graphing calculator, solve the following systems of equations. (a) 1 1 √ , y = x5 − x3 + 2. (c) y = x2 + y 2 = 1, y = sin x. (b) y = , y = x3 + sin x. (Answers 2 1 − x 1+ x on pp. 1065, 1066, and 1067.)
Page 223, Table of Contents
www.EconsPhDTutor.com
Part II
Sequences and Series
Page 224, Table of Contents
www.EconsPhDTutor.com
19
Finite Sequences
Recall that an ordered pair (of real numbers) was simply any pair of real numbers, enclosed by parentheses, and whose order matters (and this was the only difference between an ordered pair and a set of two objects). Example 182. (1, 2) and (2, 1) are both ordered pairs with (1, 2) ≠ (2, 1). We can analogously define ordered triples, quadruples, quintuples, etc. Example 183. (1, 2, 3) and (2, 1, 3) are both ordered triples with (1, 2, 3) ≠ (2, 1, 3). (1, 1, 1, 1) and (2, 4, 1, 3) are both ordered quadruples with (1, 1, 1, 1) ≠ (2, 4, 1, 3). (2, 2, 3, 2, 2) and (2, 4, 1, 5, 3) are both ordered quintuples with (2, 2, 3, 2, 2) ≠ (2, 4, 1, 5, 3). We’ll simply call all of these ordered n-tuples or even simply tuples. Hence, Example 184. (1, 2, 3), (2, 1, 3), (1, 1, 1, 1), (2, 4, 1, 3), (2, 2, 3, 2, 2), and (2, 4, 1, 5, 3) are all ordered n-tuples. (1, 2, 3) and (2, 1, 3) are ordered 3-ples or triples. (1, 1, 1, 1) and (2, 4, 1, 3) are ordered 4-tuples or quadruples. (2, 2, 3, 2, 2) and (2, 4, 1, 5, 3) are ordered 5-tuples or quintuples. In fact, when talking about tuples, it will be understood that they are ordered, so we’ll drop the word “ordered” and simply call them tuples (instead of ordered tuples). Definition 51. A finite sequence of length n is any n-tuple.
Example 185. (1, 2, 3) and (2, 1, 3) are 3-ples or, equivalently, finite sequences of length 3. (1, 2, 3, 4) and (2, 4, 1, 3) are 4-tuples or, equivalently, finite sequences of length 4. (1, 2, 3, 4, 5) and (2, 4, 1, 5, 3) are 5-tuples or, equivalently, finite sequences of length 5. We refer to the objects in a sequence as terms. Example 186. Given the sequence (2, 1, 3), 2 is its first term, 1 is its second term, and 3 is its third term.
Page 225, Table of Contents
www.EconsPhDTutor.com
19.1
A Corresponding Function for a Sequence
Another perspective is to think of a finite sequence of length n as a function whose domain is {1, 2, 3, . . . , n} and whose codomain is R.30 Example 187. (2, 4, 6, 8, 10, 12, 14) is a finite sequence of length 7, consisting of the first seven even positive integers. A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7}; • Codomain R; and • Mapping rule f (n) = 2n, for all n. Indeed, the values of the function f (1) = 2, f (2) = 4, f (3) = 6, ..., f (7) = 14 exactly list out the terms in the finite sequence (2, 4, 6, 8, 10, 12, 14). Example 188. (2, 5, 12, 23, 38, 57, 80, 107, 138, 173) is a finite sequence of length 10. A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; • Codomain R; and • Mapping rule : f (n) = 2n2 − 3n + 3, for all n. Indeed, the values of the function f (1) = 2, f (2) = 5, f (3) = 12, f (4) = 23, ..., f (10) = 173 exactly list out the terms in the finite sequence (2, 5, 12, 23, 38, 57, 80, 107, 138, 173).
Exercise 91. (Answer on p. 1068.) For each of the following finite sequences, write down a corresponding function. (a) (1, 4, 9, 16, 25, 36, 49, 64, 81, 100). (b) (2, 5, 8, 11, 14, 17, 20). (c) (0.5, 4, 13.5, 32, 62.5, 108, 171.5). (d) (2, 6, 6, 12, 10, 18, 14, 24, 18, 30, 22, 36, 26, 42). (e) (18, 14.5).
30
Indeed, this is how a sequence is usually formally defined.
Page 226, Table of Contents
www.EconsPhDTutor.com
19.2
Recurrence Relations
SYLLABUS ALERT Recurrence relations are included in the 9740 (old) syllabus, but not in the 9758 (revised) syllabus. So you can skip this section if you’re taking 9758.
Example 189. (1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024) is a finite sequence of length 10. A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; • Codomain R; and • Mapping rule : f (1) = 1 and f (n) = 2f (n − 1) (the recurrence relation), for all n ≥ 2. The equation f (n) = 2f (n−1) is an example of a recurrence relation. That is, it describes how each term in the sequence is generated, depending on what previous terms were. In this particular example of a sequence, we can easily write down another corresponding function that does not involve a recurrence relation: Example 190. (1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024) is a finite sequence of length 10. A corresponding function g for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; • Codomain R; and • Mapping rule : g(n) = 2n−1 (not a recurrence relation), for all n.
If we can describe a sequence without using a recurrence relation, then we can immediately compute what each term in the sequence is. So in the case of the finite sequence just given, we prefer to use the function g rather than the function f as a corresponding function. In contrast, with a recurrence relation, we need to know what some of the previous terms are, in order to compute each term. So if possible, we prefer to describe sequences without using recurrence relations. But sometimes, it is difficult to describe a sequence without using a recurrence relation.
Page 227, Table of Contents
www.EconsPhDTutor.com
Example 191. (1, 4, 10, 22, 46, 94, 190, 382, 766, 1534) is a finite sequence of length 10. A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; • Codomain R; and • Mapping rule : f (1) = 1 and f (n) = 2f (n − 1) + 2 (the recurrence relation), for all n ≥ 2. It is possible to describe the sequence just given without using a recurrence relation, but it does not come obviously (at least to the untrained eye) and takes a little work, as we’ll see.
A recurrence relation can certainly involve more than just the previous term. In the Fibonnaci sequence, each term (from the third term onwards) is the sum of the previous two terms: f (n) = f (n − 2) + f (n − 1). This equation is again a recurrence relation. But in the past ten years’ exams, I haven’t seen a question where the recurrence relation involves more than just the previous term. So we shall not bother doing much of these. Example 192. (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89) is a finite sequence of length 11, consisting of the first 11 Fibonacci numbers. A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}; • Codomain R; and • Mapping rule : f (n) = 1, for n = 1, 2; and f (n) = f (n − 2) + f (n − 1) (the recurrence relation), for all n ≥ 3.
Exercise 92. Each of the following finite sequences involves a recurrence relation. (Hint: Each involves only the previous term and also a squared term.) Write down a corresponding function for each. (a) (3, 4, 9, 64, 3969). (b) (1, 2, 10, 290, 252010). (Answer on p. 1069.)
Page 228, Table of Contents
www.EconsPhDTutor.com
19.3
Creating New Sequences
Notation. Consider the finite sequence of length k (a1 , a2 , a3 , . . . , ak ). A shorthand piece of notation for this sequence is (an )n≤k . We call n the index variable or dummy variable. We’ll assume the index variable always starts from 1, unless otherwise specified. Example 193. (1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193) is a finite sequence of length 11. We can also write it as (an )n≤11 = (a1 , a2 , a3 , . . . , a11 ), where a1 = 1, a2 = 1, a3 = 1, a4 = 3, a5 = 5, ..., a11 = 193. Example 194. (1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151) is a finite sequence of length 20. We can also write it as (bn )n≤20 = (b1 , b2 , b3 , . . . , b20 ), where b1 = 1, b2 = 1, b3 = 1, b4 = 2, b5 = 2, ..., b20 = 151. Example 195. (2, 4, 6, 8, 10, 12, 14) is a finite sequence of length 7. We can also write it as (cn )n≤7 = (c1 , c2 , c3 , . . . , c7 ), where c1 = 2, c2 = 4, c3 = 6, ..., c7 = 14.
Example 196. (1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683) is a finite sequence of length 11. We can also write it as (dn )n≤11 = (d1 , d2 , d3 , . . . , d11 ), where d1 = 1, d2 = 1, d3 = 3, d4 = 5, d5 = 11, ..., d11 = 683. We can create new sequences out of old ones, in the “obvious” fashion: Example 197. Using the sequence (an )n≤11 = (1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193), here are some new sequences we can create: (zn )n≤11 = (an + 1)n≤11 = (a1 + 1, a2 + 1, a3 + 1, . . . , a11 + 1) = (2, 2, 2, 4, 6, 10, 18, 32, 58, 106, 194) = (z1 , z2 , z3 , . . . , z11 ) , (yn )n≤11 = (2an )n≤11 = (2a1 , 2a2 , 2a3 , . . . , 2a11 ) = (2, 2, 2, 6, 10, 18, 34, 62, 114, 210, 386) = (y1 , y2 , y3 , . . . , y11 ) , (xn )n≤11 = (an − 1)n≤11 = (a1 − 1, a2 − 1, a3 − 1, . . . , a11 − 1) = (0, 0, 0, 2, 4, 8, 16, 30, 56, 104, 192) = (x1 , x2 , x3 , . . . , x11 ) , (wn )n≤11 = (an /2)n≤11 = (a1 /2, a2 /2, a3 /2, . . . , a11 /2) = (1/2, 1/2, 1/2, 3/2, 5/2, 9/2, 17/2, 31/2, 57/2, 105/2, 193/2) = (w1 , w2 , w3 , . . . , w11 ) .
Page 229, Table of Contents
www.EconsPhDTutor.com
Moreover, using two (or more) finite sequences that are of the same length, we can likewise create a new finite sequence (also of the same length), in the “obvious” fashion: Example 198. Using the sequences (an )n≤11 = (1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193) and (dn )n≤11 = (1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683), here are some new sequences we can create: (en )n≤11 = (an + dn )n≤11 = (a1 + d1 , a2 + d2 , a3 + d3 , . . . , a11 + d11 ) = (2, 2, 4, 8, 16, 30, 60, 116, 228, 446, 876) = (e1 , e2 , e3 , . . . , e11 ) , (fn )n≤11 = (an ⋅ dn )n≤11 = (a1 ⋅ d1 , a2 ⋅ d2 , a3 ⋅ d3 , . . . , a11 ⋅ d11 ) = (1, 1, 3, 15, 55, 189, . . . , 131819) = (f1 , f2 , f3 , . . . , f11 ) , (gn )n≤11 = (an − dn )n≤11 = (a1 − d1 , a2 − d2 , a3 − d3 , . . . , a11 − d11 ) = (0, 0, −2, −2, −6, −12, −26, . . . , −490) = (g1 , g2 , g3 , . . . , g11 ) , (hn )n≤11 = (an /dn )n≤11 = (a1 /d1 , a2 /d2 , a3 /d3 , . . . , a11 /d11 ) = (1, 1, 1/3, 3/5, 5/11, 9/21, . . . , 193/683) = (h1 , h2 , h3 , . . . , h11 ) .
There are of course many other new sequences we can create, whether using only one sequence, using two sequences, or even using three or more sequences. Remark 6. You cannot create a new sequence using two finite sequences that are of different lengths. For example, given two finite sequences (an )n≤11 = (1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193) and (cn )n≤7 = (2, 4, 6, 8, 10, 12, 14), there is no such sequence as (an + cn )n≤11 or even (an + cn )n≤7 . Either of these supposed sequences is simply undefined. It turns out that we are rarely interested in finite sequences. Instead, we are much more interested in infinite sequences, which is a simple extension of the concept of finite sequences.
Page 230, Table of Contents
www.EconsPhDTutor.com
20
Infinite Sequences
We can easily extend the concept of finite sequences to infinite sequences, which have domain Z+ = {1, 2, 3, 4, . . . } (the entire set of positive integers). Example 199. (2, 4, 6, 8, 10, 12, 14, 16, 18, . . . ) is the infinite sequence consisting of all the even positive integers. A corresponding function f for this sequence has • Domain Z+ ; • Codomain R; and • Mapping rule f (n) = 2n for all n.
Example 200. (1, 3, 6, 10, 15, 21, 28, 36, 45, 55, . . . ) is the infinite sequence consisting of the triangular numbers. A corresponding function f for this sequence has • Domain Z+ ; • Codomain R; and • Mapping rule f (1) = 1 and f (n) = 1 + 2 + ⋅ ⋅ ⋅ + n for all n ≥ 2.
Example 201. The infinite sequence (1, 2, 6, 24, 120, 720, 5040, ...) has the corresponding function f with • Domain Z+ ; • Codomain R; and • Mapping rule f (n) = 1 × 2 × ⋅ ⋅ ⋅ × n = n! for all n.
Exercise 93. For each of the following infinite sequences, write down a corresponding function. (a) (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, . . . ). (b) (2, 5, 8, 11, 14, 17, 20, . . . ). (c) (0.5, 4, 13.5, 32, 62.5, 108, 171.5, . . . ). (d) (2, 6, 6, 12, 10, 18, 14, 24, 18, 30, 22, 36, 26, 42, . . . ). (Answer on p. 1070.)
Page 231, Table of Contents
www.EconsPhDTutor.com
20.1
Creating New Sequences
(an ) is our shorthand notation for an infinite sequence, where (an ) = (a1 , a2 , a3 , . . . ). As stated, we are rarely interested in finite sequences. And so whenever we talk about a sequence, it should be assumed that we are talking about an infinite sequence, unless otherwise clearly stated. The idea of creating new sequences carries over from the finite case in the “obvious” fashion. Example 202.
Let and Then
(an ) = (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . ) (bn ) = (2, 4, 6, 8, 10, 12, 14, 16, 18, 20, . . . ) . (an + bn ) = (3, 5, 8, 11, 15, 20, 27, 37, 52, 75, . . . ) .
Analogous to Remark 6, you cannot create a new sequence using a finite sequence and an infinite sequence. Instead, you can only create one using two infinite sequences. Example 203.
Let and Then
Page 232, Table of Contents
(an ) = (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . ) (bn )n≤7 = (2, 4, 6, 8, 10, 12, 14) . (an + bn ) is undefined.
www.EconsPhDTutor.com
21
Series
Definition 52. Given a finite sequence (an )n≤k , its series is the expression a1 + a2 + a3 + ⋅ ⋅ ⋅ + ak . We refer to a1 as the first term of the sequence and also as the first term of the series. Similarly, a2 is the second term of both the sequence and the series. Etc. Definition 53. Given a finite sequence (an )n≤k , its sum of series is the number S such that S = a1 + a2 + a3 + ⋅ ⋅ ⋅ + ak .
Example 204.
(an )n≤8 = (1, 1, 1, 3, 5, 9, 17, 31) , 1 + 1 + 1 + 3 + 5 + 9 + 17 + 31 68.
Example 205.
(bn )n≤11 = (2, 4, 6, 8, 10, 12, 14) , 2 + 4 + 6 + 8 + 10 + 12 + 14 56.
Given the sequence its series is the expression and its sum of series is the number
Given the sequence its series is the expression and its sum of series is the number
It may seem strange and unnecessary to distinguish between a series and a sum of series. Aren’t they exactly the same thing? It turns out that expressions like a1 + a2 + a3 + ⋅ ⋅ ⋅ + ak play an important role in maths and so we want to reserve a special name for the expression itself and distinguish it from the sum of series. For example, we might be specifically interested in the series 1 + 2 + 3, rather than just the sum of series 6. Clearly, every finite sequence has a well-defined sum of series – simply add up all the terms in the finite sequence! Definition 54. Given an infinite sequence (an ), its series is the expression a1 + a2 + a3 + . . . . A series that corresponds to a finite sequence is called a finite series, while a series that corresponds to an infinite sequence is called an infinite series.
Page 233, Table of Contents
www.EconsPhDTutor.com
21.1
Convergent and Divergent Sequences and Series
Every finite sequence has a sum of series. In contrast, not all infinite sequences do: Example 206. Consider the sequence (an ) = (1, 1, 1, 1, 1, 1, . . . ). Its series is the expression 1 + 1 + 1 + 1 + 1 + . . . . There is no number equal to 1 + 1 + 1 + 1 + 1 + . . . and so a sum of series does not exist for this sequence. But some infinite sequences do have sums of series: Example 207. Consider the sequence (bn ) = (0, 0, 0, 0, 0, 0, . . . ). Its series is the expression 0 + 0 + 0 + 0 + 0 + . . . . The sum of series for this sequence exists and is 0.
Definition 55. An infinite sequence for which a sum of series exists is said to have a convergent series. An infinite sequence for which no sum of series exists is said to have a divergent series. So in the above examples, we say that the sequence (an ) has a divergent series (because its sum of series does not exist), while the sequence (bn ) has a convergent series (because its sum of series exists).
Page 234, Table of Contents
www.EconsPhDTutor.com
But what exactly is a convergent series? When exactly is a series convergent? These are actually fascinating questions, which means, of course, that they’re not in the syllabus. Here is a simple example that gives you a glimpse of the difficulties involved. Chapter 85 in the Appendices (optional) gives the precise definitions of when a series converges or diverges. Example 208. Consider the sequence (cn ) = (1, −1, 1, −1, 1, −1, . . . ), where the terms simply alternate between 1 and −1. Its series is the expression 1 − 1 + 1 − 1 + 1 − 1 + . . . . Is there any number that is equal to 1 − 1 + 1 − 1 + 1 − 1 + . . . ? It’s actually not obvious. On the one hand, we can pair together every two terms like so: 1 − 1 + 1 − 1 + 1 − 1 + . . . = (1 − 1) + (1 − 1) + (1 − 1) + . . . ´¹¹ ¹ ¹ ¸ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¸ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¸ ¹ ¹ ¹ ¶ 0
0
0
= 0 + 0 + 0 + ...
and happily conclude that the sum of series is 0. But wait a minute ... what if we instead pair together every two terms like so: 1 − 1 + 1 − 1 + 1 − 1 + 1 . . . = 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + . . . ´¹¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¶ 0
= 1 + 0 + 0 + 0 + ...
0
0
Then we’d have to conclude that the sum of series is 1! It turns out that the sequence (cn ) = (1, −1, 1, −1, 1, −1, . . . ) is divergent. Or equivalently, a sum of series simply does not exist for this sequence.
Page 235, Table of Contents
www.EconsPhDTutor.com
22
Summation Notation Σ
Σ is the upper-case Greek letter sigma. An enlarged version of that letter ∑, read aloud as “sum”, is used to express series in compact notation: Example 209. Consider the series 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9. Another way to write it is to use summation notation: 9
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = ∑ n. n=1
Let’s examine the expression on the RHS. The variable n below the ∑ is called the index variable or dummy variable. We could have named it p or z or x or any other letter (instead of n) and it wouldn’t have mattered. Hence the name “dummy”. The “= 1” below the ∑ says that we start counting the index variable from n = 1. We call the number “1” the starting point. The “9” above the ∑ is called the stopping point. It says that we should stop adding once we hit n = 9. 9
Altogether, the notation ∑ says that we are adding up 9 terms, namely a1 , a2 , ..., a9 . n=1
The expression to the right of the ∑ tells us what each an is. In this example, it is n, which simply says that for every n, an = n. 9
Altogether, ∑ n says that we add up a1 through a9 , where each an is simply equal to n. n=1
Example 210. The series 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 can be written as 9
∑ 1.
n=1
This says that the starting point is 1 and the ending point is 9. In other words, we add up a1 , a2 , . . . , a9 , where for each n, an = 1. And so a1 = 1, a2 = 1, etc. Altogether: 9
∑ 1 = a1 + a2 + ⋅ ⋅ ⋅ + a9 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1.
n=1
Page 236, Table of Contents
www.EconsPhDTutor.com
Example 211. The series 2 + 4 + 6 + 8 + 10 + 12 + 14 can be written as 7
∑ 2n.
n=1
This says that the starting point is 1 and the ending point is 7. In other words, we add up a1 , a2 , . . . , a7 , where for each n, an = 2n. And so a1 = 2, a2 = 4, etc. Altogether: 7
∑ 2n = a1 + a2 + ⋅ ⋅ ⋅ + a7 = 2 × 1 + 2 × 2 + ⋅ ⋅ ⋅ + 2 × 7 = 2 + 4 + 6 + 8 + 10 + 12 + 14.
n=1
7
The series 3 + 5 + 7 + 8 + 11 + 13 + 15 can be rewritten as ∑ (2n + 1) — the parentheses help n=1
7
to clarify that we are not talking about 1 + ∑ 2n. n=1
This says that the starting point is 1 and the ending point is 7. In other words, we add up a1 , a2 , . . . , a7 , where for each n, an = 2n + 1. And so a1 = 3, a2 = 5, etc. Altogether: 3
5
15
³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ∑ (2n + 1) = a1 + a2 + ⋅ ⋅ ⋅ + a7 = (2 × 1 + 1) + (2 × 2 + 1) + ⋅ ⋅ ⋅ + (2 × 7 + 1). 7
n=1
Page 237, Table of Contents
www.EconsPhDTutor.com
Example 212. The series 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 can be written as 10
∑ 2n .
n=1
This says that the starting point is 1 and the ending point is 10. In other words, we add up a1 , a2 , . . . , a10 , where for each n, an = 2n . And so a1 = 2, a2 = 4, etc. Altogether: 10
∑ 2n = a1 + a2 + ⋅ ⋅ ⋅ + a10 = 21 + 22 + 23 + ⋅ ⋅ ⋅ + 210 = 2 + 4 + 8 + ⋅ ⋅ ⋅ + 1024.
n=1
It’s nice to have 1 as the starting point, but there’s no reason why this must always be so. Example 213. The series 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 can be written as 10
∑ 2n .
n=0
This says that the starting point is 0 and the ending point is 10. In other words, we add up a0 , a1 , a2 , . . . , a10 , where for each n, an = 2n . And so a0 = 1, a1 = 2, a2 = 4, etc. Altogether: 10
∑ 2n = a0 + a1 + a2 + ⋅ ⋅ ⋅ + a10 = 20 + 21 + 22 + ⋅ ⋅ ⋅ + 210 = 1 + 2 + 4 + ⋅ ⋅ ⋅ + 1024.
n=0
Page 238, Table of Contents
www.EconsPhDTutor.com
Exercise 94. Rewrite each of the following in summation notation. (Answer on p. 1071.) (a) 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100. (b) 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23. (c) 0.5 + 4 + 13.5 + 32 + 62.5 + 108 + 171.5. Exercise 95. Find the sum of each of the following series. (Answer on p. 1071.) 5
n
(a) ∑ (2 − n) . n=−2 17
(b) ∑ (4n + 5). n=16 33
(c) ∑ (x − 3). x=31
Here’s the general definition of the summation notation: Definition 56. Let s, k be integers with s ≤ k. Let f be a function whose domain contains s, s + 1, . . . , k and whose codomain is R. Then k
∑ f (n) = f (s) + f (s + 1) + ⋅ ⋅ ⋅ + f (k).
n=s
Page 239, Table of Contents
www.EconsPhDTutor.com
23
Arithmetic Sequences and Series
Example 214. Consider the finite sequence (4, 7, 10, 13, 16, 19, 22). A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7} (a subset of Z+ ); • Codomain R; and • Mapping rule f (1) = 4 and f (n) − f (n − 1) = 3 for all n ≥ 2. This is an example of a finite arithmetic sequence. Example 215. Consider the infinite sequence (4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, . . . ). A corresponding function f for this sequence has • Domain Z+ ; • Codomain R; and • Mapping rule f (1) = 4 and f (n) − f (n − 1) = 3 for all n ≥ 2. This is an example of an infinite arithmetic sequence.
Definition 57. An arithmetic sequence (or an arithmetic progression) is any finite or infinite sequence (an ) where an+1 − an is a constant for all n = 1, 2, 3, . . . . We call an+1 − an the common difference. We call the series for an arithmetic sequence an arithmetic series. And its sum of series (if it exists at all) is called an arithmetic sum of series.
Example 216. The sequence (an ) = (1, 4, 7, 10, 13, 16, 19, . . . ) is an arithmetic sequence because an+1 − an is constant for n = 1, 2, 3, . . . . But the sequence (bn ) = (1, 1, 4, 7, 10, 13, 16, 19, . . . ) is not an arithmetic sequence because a2 − a1 = 0 ≠ a3 − a2 = 3.
The next fact is intuitively obvious. Clearly, there is no number for which, for example, 4 + 7 + 10 + 13 + 16 + 19 + 22 + . . . is equal to. Fact 12. The infinite arithmetic sequence (an ) has no sum of series, except in the trivial case where (an ) = (0, 0, 0, 0, 0, 0, . . . ).
Page 240, Table of Contents
www.EconsPhDTutor.com
23.1
Finite Arithmetic Sequences and Series
Every finite sequence, including arithmetic ones, has a sum of series. Example 217. You’ve probably heard of the apocryphal story about an eight-year-old Gauss adding up the numbers from 1 to 100 in an instant. The trick is to pair the first number with the last, the second number with the second last, etc. then use multiplication. Like this: 50 terms ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹· ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ 1 + 2 + 3 + 4 + ⋅ ⋅ ⋅ + 100 = (1 + 100) + (2 + 99) + (3 + 98) + ⋅ ⋅ ⋅ + (50 + 51) ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ 101
101
101
101
= 101 × 50 = 5050. In general, there is a simple formula for the sum of a finite arithmetic series: (First Term + Last Term) × (Number of Terms) ÷ 2. k Fact 13. The finite arithmetic series a1 + a2 + ⋅ ⋅ ⋅ + ak has sum of series (a1 + ak ) . 2 (We will only prove Fact 13 on p. 249.) Example 218. Consider the arithmetic sequence (7, 17, 27, 37, . . . , 837). Its common difference is 10. The difference between the first and last terms is 830. And so the last term is 830 ÷ 10 = 83 terms after the first. Hence, there are in total 84 terms. By Fact 13, its 84 sum of series is (7 + 837) × = 35488. 2 Example 219. Consider the arithmetic sequence (1, 5, 9, 13, 17, 21, 25, 29, 33, . . . , 393). Its common difference is 4. The difference between the first and last terms is 392. And so the last term is 392 ÷ 4 = 98 terms after the first. Hence, there are in total 99 terms. By Fact 99 13, its sum of series is (1 + 393) × = 19503. 2
Exercise 96. Rewrite each of the following arithmetic series in summation notation and compute their sums. (a) 2+7+12+17+22+27+32+⋅ ⋅ ⋅+997. (b) 3+20+37+54+71+⋅ ⋅ ⋅+1703. (c) 81 + 89 + 97 + 105 + 113 + ⋅ ⋅ ⋅ + 8081 (Answer on p. 1072.)
Page 241, Table of Contents
www.EconsPhDTutor.com
24
Geometric Sequences and Series
Example 220. Consider the finite sequence (1, 2, 4, 8, 16, 32, 64, 128). A corresponding function f for this sequence has • Domain {1, 2, 3, 4, 5, 6, 7, 8} (a subset of Z+ ); • Codomain R; and • Mapping rule f (1) = 1 and f (n + 1) ÷ f (n) = 2 for all n = 1, 2, 3, . . . . This is an example of a finite geometric sequence. Example 221. Consider the finite sequence (1, 2, 4, 8, 16, 32, 64, 128, 256, 512, . . . ). A corresponding function f for this sequence has • Domain Z+ ; • Codomain R; and • Mapping rule f (1) = 1 and f (n + 1) ÷ f (n) = 2 for all n = 1, 2, 3, . . . . This is an example of a infinite geometric sequence. Definition 58. A geometric sequence (or a geometric progression) is any sequence (an ) where an+1 ÷ an is constant for all n = 1, 2, 3, . . . . We call an+1 ÷ an the common ratio. We call the series for a geometric sequence a geometric series. And its sum of series (if it exists at all) is called a geometric sum of series.
Example 222. The sequence (an ) = (1, 2, 4, 8, 16, 32, . . . ) is a geometric sequence because an+1 ÷ an is constant for all n = 1, 2, 3, . . . . But the sequence (bn ) = (1, 1, 2, 4, 8, 16, 32, . . . ) is not a geometric sequence because a2 ÷a1 = 1 ≠ a3 ÷ a2 = 2.
Page 242, Table of Contents
www.EconsPhDTutor.com
24.1
Finite Geometric Sequences and Series
It turns out that just like with finite arithmetic series, there is a nice formula for the finite geometric series. Let’s start with the simple case first where the first term is simply 1.
2
3
Fact 14. 1 + r + r + r + ⋅ ⋅ ⋅ + r
n−1
1 − rn . = 1−r
Proof. Let S = 1 + r + r2 + ⋅ ⋅ ⋅ + rn−1 . Then rS = r + r2 + r3 + ⋅ ⋅ ⋅ + rn . Now take the difference: S − rS = 1 − rn . 1 − rn Hence, S = . 1−r The trick used in the above proof is called the method of differences and the A-level syllabus requires you to know it. The general case of a geometric series follows immediately from the above:
2
3
Fact 15. a1 + a1 r + a1 r + a1 r + ⋅ ⋅ ⋅ + a1 r
n−1
1 − rn = a1 . 1−r
Example 223. Consider the geometric sequence (1, 2, 4, 8, 16, . . . , 1024). Its common ratio is 2. The ratio of the last term to the first is 1024 ÷ 1 = 1024 = 210 . And so the last term is 10 terms after the first. Hence, there are in total 11 terms. Thus, its sum of series is 1 − 211 −2047 1× = = 2047. 1−2 −1 Example 224. Consider the geometric sequence (4, 12, 36, 108, . . . , 8748). Its common ratio is 3. The ratio of the last term to the first is 8748 ÷ 4 = 2187 = 37 . And so the last term is 7 terms after the first. Hence, there are in total 8 terms. Thus, its sum of series is 1 − 38 −6560 4× =4× = 4 × 3280 = 13120. 1−3 −2
Exercise 97. Rewrite each of the following geometric series into summation notation and compute their sums. (a) 7 + 14 + 28 + 56 + ⋅ ⋅ ⋅ + 448 + 896. (b) 20 + 10 + 5 + ⋅ ⋅ ⋅ + 5/8. (c) 1 + 1/3 + 1/9 + ⋅ ⋅ ⋅ + 1/243. (Answer on p. 1073.)
Page 243, Table of Contents
www.EconsPhDTutor.com
24.2
Infinite Geometric Sequences and Series
Perhaps surprisingly, it turns out that under a certain condition, an infinite geometric sequence can have a sum of series. Again, let’s start with the simple case:
Fact 16. If ∣r∣ < 1, then 1 + r + r2 + r3 + ⋅ ⋅ ⋅ =
1 . 1−r
Proof. Write the series as S = 1 + r + r2 + r3 + . . . . Then rS = r + r2 + r3 + r4 + . . . . (By the ∞
∞
way, we can also use summation notation for infinite series: S = ∑ r and S = ∑ rn+1 .) n=0
n
n=0
Since ∣r∣ < 1, it follows that as n → ∞, rn → 0. Hence, if we take the difference, we have simply S − rS = 1. And so, S =
1 . 1−r
The general case follows immediately: Fact 17. If ∣r∣ < 1, then a1 + a1 r + a1 r2 + a1 r3 + ⋅ ⋅ ⋅ =
a1 . 1−r
The converse is also true: Fact 18. If ∣r∣ ≥ 1, then a1 + a1 r + a1 r2 + a1 r3 + . . . diverges.
Proof. Optional, see p. 929 in the Appendices.
Exercise 98. Rewrite each of the following infinite geometric series in summation notation and compute its sum. (a) 6 + 9/2 + 27/8 + . . . . (b) 20 + 10 + 5 + . . . . (c) 1 + 1/3 + 1/9 + . . . . (Answer on p. 1073.)
Page 244, Table of Contents
www.EconsPhDTutor.com
25
Proof by the Method of Mathematical Induction
SYLLABUS ALERT Proof by the method of mathematical induction is included in the 9740 (old) syllabus, but not in the 9758 (revised) syllabus. So you can skip this Chapter if you’re taking 9758.
We’ll now learn a new technique called proof by the method of mathematical induction. It’s pretty difficult, so go real slow.31 Imagine an infinite chain of dominos. Our goal is to knock all of them down. Suppose we manage to do two things: 1. “Knock down the 1st domino” (the base case). 2. Prove that “if the jth domino is knocked down, then so too is the (j + 1)th domino” (the inductive step). Then we will have succeeded. Because once the 1st domino is knocked down, the inductive step implies that the 2nd domino is also knocked down, and now again by the inductive step the 3rd domino is also knocked down, and now again by the inductive step the 4th domino is also knocked down, ..., ad infinitum (to infinity).
31
Which is perhaps why they decided to drop it from the revised 9758 syllabus! It does appear though as the first topic of Further Maths, which will be revived in 2017 and for which a free textbook will soon be appearing!
Page 245, Table of Contents
www.EconsPhDTutor.com
The metaphor of dominos is an apt description of the method of mathematical induction, which I’ll standardise into a three-step recipe:
The Method of Mathematical Induction Step #1. Let P(k) be (shorthand for) the proposition to be proven. Our goal is to show that P(k) is true for all k = 1, 2, 3, . . . Step #2 (the base case). Verify that P(1) is true. Step #3 (the inductive step). Show that P(j) implies P(j + 1) (for all j = 1, 2, 3, . . . ).
Step #1 rarely involves much work. Step #2 is usually, but not always, very easy. Step #3 is usually the hardest part — on the A-level exams, it usually just involves some (or a lot of) algebra. Why does the method of mathematical induction work? Step #2 (the base case) shows that P(1) is true (“knock down the 1st domino”). Step #3 (the inductive step) then implies that P(2) is also true (“the falling 1st domino knocks down the 2nd domino”). Step #3 (the inductive step) then implies that P(3) is also true (“the falling 2nd domino knocks down the 3rd domino”). Step #3 (the inductive step ) then implies that P(4) is also true (“the falling 3rd domino knocks down the 4th domino”). Ad infinitum (to infinity). Thus, we have proven that P(k) is true for all k = 1, 2, 3, . . . , as desired. Too abstract? Work through all the examples and exercises and you should find that it is not very difficult. For our first example, we’ll reprove an earlier fact, but now using the method of mathematical induction.
Page 246, Table of Contents
www.EconsPhDTutor.com
Fact 14 (reproduced from p. 243). 1 + r + r2 + r3 + ⋅ ⋅ ⋅ + rn−1 =
1 − rn . 1−r
Proof. Step #1. Let P(k) be (shorthand for) the proposition that 1 + r + r2 + r3 + ⋅ ⋅ ⋅ + rk−1 =
1 − rk . 1−r
Our goal is to show that P(k) is true for all k = 1, 2, 3, . . .
Step #2. Verify that P(1) is true. 1=
1 − r1 . 1−r
✓
Step #3. Show that P(j) implies P(j + 1) (for all j = 1, 2, 3, . . . ). Assume that P(j) is true. That is, 1
1 + r + r2 + r3 + ⋅ ⋅ ⋅ + rj−1 =
1 − rj . 1−r
Our goal is to show that P(j + 1) is true. That is, 1 + r + r2 + r3 + ⋅ ⋅ ⋅ + rj = To this end, write:
1 − rj+1 . 1−r
1 + r + r2 + r3 + ⋅ ⋅ ⋅ + rj = (1 + r + r2 + r3 + ⋅ ⋅ ⋅ + rj−1 ) + rj j 1 − rj + (1 − r)rj 1 − rj+1 1 1−r j = +r = = , as desired. 1−r 1−r 1−r
In this particular instance, the method of mathematical induction was terribly cumbersome, compared to our earlier four-sentence proof (p. 243). But it turns out that in many other instances, this method is the best and sometimes the only tool to use. Let’s try more examples.
Page 247, Table of Contents
www.EconsPhDTutor.com
n
Example 225. Prove that ∑ r2 = r=1
n(n + 1)(2n + 1) . 6 k
Step #1. Let P(k) be (shorthand for) the proposition that ∑ r2 = r=1
k(k + 1)(2k + 1) . 6
Our goal is to show that P(k) is true for all k = 1, 2, 3, . . . 1
Step #2. Verify that P(1) is true: ∑ r2 = 1 = n=1
1(1 + 1)(2 × 1 + 1) . ✓ 6
Step #3. Show that P(j) implies P(j + 1) (for all j = 1, 2, 3, . . . ). j
1
Assume that P(j) is true. That is, ∑ r2 = n=1
j(j + 1)(2j + 1) . 6
Our goal is to show that P(j + 1) is true. That is, j+1
∑ r2 =
n=1
(j + 1) [(j + 1) + 1] [2(j + 1) + 1] . 6
To this end, write: j+1
2
j
∑ r2 = ∑ r2 + (j + 1)2
n=1
n=1
3
= 6
= 7
= 5
= 4
=
j(j + 1)(2j + 1) + (j + 1)2 6 j+1 [j(2j + 1) + 6(j + 1)] 6 j+1 (2j 2 + 7j + 1) 6 (j + 1)(j + 2)(2j + 3) 6 (j + 1) [(j + 1) + 1] [2(j + 1) + 1] , 6
1
(Using =)
as desired.
I just used the “backwards-forwards method”. The order in which I wrote down each line is given by the numbers above each = sign. Another trick is to exploit the fact that it has got to work out right. So for example, it might not immediately be obvious that 2j 2 + 7j + 1 = (j + 2)(2j + 3), but you know it has got to work out right and thus this must surely be true (unless of course you made some mistake with the algebra somewhere). And if you expand the RHS, you find that this equation is indeed true.
Page 248, Table of Contents
www.EconsPhDTutor.com
Fact 13 (reproduced from p. 241). The finite arithmetic sequence (an )n≤k has sum of k series (a1 + ak ) . 2 Proof. Step #1. Let P(k) be (shorthand for) the proposition that a1 + a2 + ⋅ ⋅ ⋅ + ak =
k(a1 + ak ) . 2
Our goal is to show that P(k) is true for all k = 1, 2, 3, . . .
Step #2. Verify that P(1) is true: a1 =
1(a1 + a1 ) . ✓ 2
Step #3. Show that P(j) implies P(j + 1) (for all j = 1, 2, 3, . . . ). Assume that P(j) is true. That is, 1
a1 + a2 + ⋅ ⋅ ⋅ + aj =
j(a1 + aj ) . 2
Our goal is to show that P(j + 1) is true. That is, a1 + a2 + ⋅ ⋅ ⋅ + aj+1 =
(j + 1)(a1 + aj+1 ) . 2
Let’s first observe that aj − a1 = (j − 1) (aj+1 − aj ). In words, this equation says: Consider the difference between the j th term and the first term; it is equal to j −1 times the difference 2 (j − 1)aj+1 + a1 between two consecutive terms. Rearranging, we have aj = . j Now write: a1 + a2 + ⋅ ⋅ ⋅ + aj+1
j(a1 + aj ) + aj+1 2 j {a1 + [(j − 1)aj+1 + a1 ] /j} ja1 + (j − 1)aj+1 + a1 = + aj+1 = + aj+1 2 2 (j + 1)a1 + (j − 1)aj+1 (j + 1)a1 + (j − 1)aj+1 + 2aj+1 = + aj+1 = 2 2 (j + 1)a1 + (j + 1)aj+1 (j + 1)(a1 + aj+1 ) = = , as desired. 2 2 = (a1 + a2 + ⋅ ⋅ ⋅ + aj ) + aj+1
Page 249, Table of Contents
=
www.EconsPhDTutor.com
2
n(n + 1) Exercise 99. Prove that ∑ r = [ ] . (Answer on p. 1074.) By the way, this shows 2 r=1 n
n
n
r=1
r=1
3
2
that ∑ r3 = (∑ r) . Exercise 100. (Answer on p. 1075.) Let a ∈ R. Prove that 1 − (n + 1)an + nan+1 , ∑ ra = a (1 − a)2 r=1 n
r
n
Exercise 101. Prove that ∑ r4 = r=1
Page 250, Table of Contents
n(n + 1)(2n + 1)(3n2 + 3n − 1) . (Answer on p. 1076.) 30
www.EconsPhDTutor.com
Part III
Vectors
Page 251, Table of Contents
www.EconsPhDTutor.com
26
Quick Revision of Some O-Level Maths 26.1
Lines vs. Line Segments vs. Rays
A line is infinite, while a line segment is finite. Example 226. The line running through points a and b goes forever, in both directions (red dotted line). In contrast, the line segment ab is finite. The line ab is a different mathematical object from the line segment ab.
a
b
The length of the line segment ab is thus a well-defined concept. In contrast, it makes no sense to talk about the length of the line ab. A ray is a portion of a line, beginning at some point along the line, then going towards infinity. You can think of a ray as a half-infinite-line. The figure above illustrates in grey the ray that starts from the point a and goes in the direction b.
This textbook will strictly reserve the word ray to mean a half-infinite-line. But you should know that some other writers use ray to mean a (finite) line segment.
Page 252, Table of Contents
www.EconsPhDTutor.com
26.2
Angles - Acute, Right, Obtuse, Straight, Reflex
We will not use the degree ○ as a unit of measurement for angles. In this textbook, the unit of measurement for angles is the radian. As we’ll see in a moment, the radian is actually a “unitless” unit. So we’ll always write, for example, “π/3” instead of “π/3 rad”. π π rad= 45○ , rad= 90○ , π rad= 180○ , and 4 2 2π rad= 360○ . (This last sentence is the one and only time in this textbook that we’ll use degrees as a unit of measurement for angles.)
But just to refresh your memory, 0 rad= 0○ ,
Angles are given different names, depending on their size. θ is the zero angle if θ = 0,
π θ is an obtuse angle if θ ∈ ( , π), 2
π θ is an acute angle if θ ∈ (0, ), 2
θ is a straight angle if θ = π,
θ is a right angle if θ =
π , 2
θ is a reflex angle if θ ∈ (π, 2π).
In the figure below, the angle A is acute, R is right, O is obtuse, S is straight, and X is reflex. The zero angle is not depicted. By convention, every angle is depicted as a sector of a circle, unless it is a right angle, in which case it is depicted by a square.
A
O
R
Page 253, Table of Contents
www.EconsPhDTutor.com
26.3
Triangles - Acute, Right, Obtuse
Triangles are also given different names, depending on the size of their largest angle. A triangle is: • Acute if its largest angle is acute; • Right if its largest angle is right; and • Obtuse if its largest angle is obtuse. In the figure below, the largest angle of each triangle is highlighted.
Obtuse triangle
Acute triangle Right triangle
Page 254, Table of Contents
www.EconsPhDTutor.com
26.4
Sine, Cosine, Tangent - Definitions
Both the sine and cosine functions have domain R and codomain [−1, 1]. The tangent function has domain ⋅ ⋅ ⋅∪(−1.5π, −0.5π)∪(−0.5π, 0.5π)∪(0.5π, 1.5π)∪. . . —i.e. all reals except half-integer multiples of π. And the tangent function’s codomain is R. Draw a unit circle. Then given any point p = (px , py ) on the unit circle and the angle A that the line segment op makes with the positive x-axis, we define sin A = py , cos A = px , py and tan A = . Note that the line segment op has length 1. px
y p
py
x
A px
In the case where A is acute (the point p is in the top-right quadrant of the cartesian plane), one mnemonic is “SOH, CAH, TOA” — Sine is Opposite over Hypothenuse, Cosine is Adjacent over Hypothenuse, and Tangent is Opposite over Adjacent.
Page 255, Table of Contents
www.EconsPhDTutor.com
26.5
Sine, Cosine, Tangent - Values and Graphs
Sine and cosine fluctuate between −1 and 1. We describe their fluctuations as being sinusoidal. In contrast, tangent fluctuates between −∞ and ∞. At half-integer multiples of π, the tangent function is undefined.
2
y
y = tan x
y = cos x
y = sin x
0 3π
2π
x 0
π
π
2π
3π
-2
You don’t need to memorise the following (because you have a calculator). But you will solve problems a little more quickly if you have these memorised. x
0
sin x 0 cos x 1 tan x 0
Page 256, Table of Contents
π 6 1/2 √
√
3/2 3/3
π 4 √
√
2/2 2/2
1
π 3 √
3/2
1/2
√
3
π 2
2π 3
1
√ 3/2
0
−1/2
√ Undefined − 3
3π 4 √
2/2
5π 6
π
1/2
0
√
√
−1
− 3/3
− 2/2 − 3/2 −1 √
0
www.EconsPhDTutor.com
26.6
Formulae for Sine, Cosine, and Tangent
For all x for which all expressions are well defined, we have: tan x =
sin x , cos x
sin(−x) = − sin x, sin(x + 2π) = sin x,
cos(−x) = cos x, cos(x + 2π) = cos x,
tan(−x) = − tan x, tan(x + 2π) = tan x.
The following formulae will appear in the List of Formulae you’ll get during exams, so you don’t need to memorise them. Exam Tip: Whenever you see a question with trigonometric functions, make sure you have this list right next to you! For all A, B, P, Q for which all expressions are well-defined, we have: sin(A ± B) = sin A cos B ± cos A sin B, cos(A ± B) = sin A cos B ∓ cos A sin B, tan(A ± B) =
tan A ± tan B , 1 ∓ tan A tan B
sin 2A = 2 sin A cos A, cos 2A = cos2 A − sin2 A = 2 cos2 A − 1 = 1 − 2 sin2 A, tan 2A =
2 tan A , 1 − tan2 A
sin P + sin Q = 2 sin (
P −Q P +Q ) cos ( ), 2 2
sin P − sin Q = 2 cos (
P +Q P −Q ) sin ( ), 2 2
cos P + cos Q = 2 cos (
P +Q P −Q ) cos ( ), 2 2
cos P − cos Q = −2 sin (
Page 257, Table of Contents
P +Q P −Q ) sin ( ). 2 2
www.EconsPhDTutor.com
26.7
Arcsine, Arccosine, Arctangent
We define sin2 A to be the square of sin A. One might thus suppose that analogously, sin−1 x = 1/ sin x, but this is not so! Instead: Definition 59. The arcsine function, denoted sin−1 , has domain [−1, 1], codomain (and range) [−0.5π, 0.5π], and rule x ↦ y where sin y = x. Below is the graph of the arcsine function. The endpoints (−1, −0.5π) and (1, 0.5π) are marked with red dots.
y 0.5π y = sin-1 x
x -1.0
-0.6
-0.2
0.2
0.6
1.0
-0.5π
Page 258, Table of Contents
www.EconsPhDTutor.com
A remark about principal values. We refer to [−0.5π, 0.5π] as the principal values of the arcsine function. What does this mean? π Angles come full circle every 2π radians. And so for example, sin = 0.5. But also 6 π π π sin ( + 2π) = 0.5. And also sin ( + 4π) = 0.5. And also sin ( − 2π) = 0.5. Indeed, 6 6 6 π sin ( + 2kπ) = 0.5 for any k ∈ Z. We say that the sine function is periodic. 6 π Yet we do not say that sin−1 (0.5) = + 2kπ for any k ∈ Z — because this would mean that 6 sin−1 maps each element in the domain to more than one (indeed infinitely many) elements in the codomain. And so sin−1 wouldn’t be a function. Instead, we define the arcsine function so that its principal values are [−0.5π, 0.5π]. That π is, the codomain of the arcsine function is [−0.5π, 0.5π]. And thus, sin−1 (0.5) = . 6 Note that the choice of [−0.5π, 0.5π] as the principal values of the arcsine function is a somewhat arbitrary convention. We could equally well have chosen, say, [0.5π, 1.5π] as our principal values. It’s nicer though that our principal values are centred on 0.
Page 259, Table of Contents
www.EconsPhDTutor.com
Definition 60. The arccosine function, denoted cos−1 , has domain [−1, 1], codomain (and range) [0, π], and rule x ↦ y where cos y = x. Below is the graph of the arccosine function. The endpoints (−1, π) and (1, 0) are marked with blue dots. Note that [0, π] are the principal values of the arccosine function. Why can’t we select [−0.5π, 0.5π] as the principal values for the arccosine function, like we did for the arcsine function?32
y π
y = cos-1 x
-1.0
32
-0.6
-0.2
0.2
0.6
1.0 x
Because then cos−1 (−1), for example, would be undefined.
Page 260, Table of Contents
www.EconsPhDTutor.com
Definition 61. The arctangent function, denoted tan−1 , has domain R, codomain (and range) (−0.5π, 0.5π), and rule x ↦ y where tan y = x. Below is the graph of the arctangent function. There are two horizontal asymptotes, namely y = 0.5π and y = −0.5π. That is, as x → ±∞, y → ±0.5π. Note that (−0.5π, 0.5π) are the principal values of the arctangent function.
y y = 0.5π horizontal asymptote x -10
-6
y=
tan-1
-2
x
2
6
10
y = -0.5π horizontal asymptote
Remark 7. This notation can be tremendously confusing, which is why many writers prefer to write arcsin x, arccos x, and arctan x instead of sin−1 x, cos−1 x, tan−1 x. But the Singapore Cambridge A-level syllabus does not use the arcsin x, arccos x, or arctan x notation and so neither shall this textbook. Page 261, Table of Contents
www.EconsPhDTutor.com
26.8
The Law of Sines and the Law of Cosines
Consider a triangle with sides of lengths a, b, and c and angles A, B, and C.
B
c
a a sin C
C
A b - a cos C
a cos C b
Proposition 4. A triangle with sides of lengths a, b, and c and angles A, B, and C has area is 0.5ab sin C.
Proof. The triangle has base b and height a sin C. Hence, its area is 0.5ab sin C.
Page 262, Table of Contents
www.EconsPhDTutor.com
Proposition 5. (The Law of Sines.) For a triangle with sides of lengths a, b, and c and angles A, B, and C, a b c = = . sin A sin B sin C
Proof. The area of the above triangle is 0.5ab sin C. By symmetry, it is also 0.5bc sin A and 0.5ac sin B. Equate these and divide by 0.5abc: 0.5ab sin C = 0.5bc sin A = 0.5ac sin B ⇐⇒
b c a = = . sin A sin B sin C
Proposition 6. (The Law of Cosines.) For a triangle with sides of lengths a, b, and c and angles A, B, and C, c2 = a2 + b2 − 2ab cos C.
Proof. (Optional.) By the Pythagorean Theorem, 2
c2 = (a sin C)2 + (b − a cos C) = a2 sin2 C + b2 − 2ab cos C + a2 cos2 C = a2 (sin2 C + cos2 C) + b2 − 2ab cos C = a2 + b2 − 2ab cos C, where the last line uses the identity sin2 C + cos2 C = 1.
One perhaps-obvious implication of the Law of Cosines is that the length of any one side of a triangle is always less than the sum of the lengths of the other two sides. Corollary 3. For a triangle with sides of lengths a, b, and c, a < b + c. Proof. c2 = a2 +b2 −2ab cos C = a2 +b2 −2ab+2ab−2ab cos C = (a−b)2 +2ab(1−cos C) > (a−b)2 . Hence, c > a − b or a < b + c.
Page 263, Table of Contents
www.EconsPhDTutor.com
27
Vectors in Two Dimensions (2D)
Recall that a point is simply an ordered pair of real numbers. Example 227. The points a = (−1, 2), b = (3, −1), c = (−1, 1), and d = (3, −2) can be illustrated graphically on the cartesian plane. The origin (0, 0) is usually named o.
3
y
a 2 c 1
0 -3
-2
-1
x
o 0
1
2
3
4
5
b -1
-2
d
-3
We now introduce an entirely new mathematical object, called a vector. We will not formally define vectors, because to do so would require more maths than is covered at A-level. But informally, a vector is an “arrow” with two properties: direction and length.
Page 264, Table of Contents
www.EconsPhDTutor.com
Ð → Ð → Example 228. In the figure, ab, cd, and u are all vectors. (As we’ll see, there are multiple ways to denote vectors.)
3
y 4
2
a
The vector ab = v = v 3 Length = 5 c 1
0 -3
-2
-1
0
1
2
-1 The vector cd = v = v -2
3
4
5 x
b The vector u d
-3
Ð → Given two points a and b, ab denotes the vector from point a to point b. The word vector means carrier (in Latin). You may have learnt in biology that mosquitoes are vectors, because they carry diseases (to humans). In mathematics likewise, a vector carries us from one point to another. Ð → Ð → Example 229. The vector ab carries us from point a to point b. The vector cd carries us from point c to point d.
Page 265, Table of Contents
www.EconsPhDTutor.com
Like a point, a vector can be described as being an ordered pair of real numbers Ð → Example 230. The vector ab = (4, −3) carries us 4 units to the right and 3 units down. The Ð → vector cd = (4, −3) carries us 4 units to the right and 3 units down. The vector u = (2, −1.5) carries us 2 units to the right and 1.5 units down. Note that we’re now using the (x, y) ordered set notation for the third time!33 Do not confuse a point with a vector! Example 231. The point (4, −3) is a zero-dimensional object. In contrast, the vector (4, −3) is a two-dimensional object.
The vector (x, y) can also be written as
Example 232. We can write (4, −3) =
⎛x⎞ . ⎝y ⎠
⎛ 4 ⎞ ⎛ 2 ⎞ and u = (2, −1.5) = . ⎝ −3 ⎠ ⎝ −1.5 ⎠
⎛a⎞ notation for vectors is very useful, because as we’ll see shortly, we’ll be doing a ⎝ b ⎠ lot of addition and multiplication with vectors, and this notation can help us see better (in a literal sense). But in print, I’ll often prefer using the (a, b) notation, simply because this takes up less space. The
The point a is called the vector’s tail and the point b is called the vector’s head. This is potentially confusing, so always remember: a vector carries us from tail to head and not the other way round! A vector is defined by two characteristics: direction and length. It must be stressed that the tail and head of a vector do not matter. Only the direction and length do. So long as two vectors have the same direction and length, they are considered to be exact same vector. Examples to illustrate
33
So far, we have used (x, y) to denote (i) an open interval — specifically, the set of real numbers greater than x but smaller than y; (ii) the ordered pair of real numbers x and y; and now also (iii) the vector that carries us x units to the right and y units up.
Page 266, Table of Contents
www.EconsPhDTutor.com
Ð → Ð → Ð → Ð → Example 233. Informally, ab, cd, and u all point in the same direction. ab √ and cd have the same length, which we can compute using the Pythagorean Theorem as 32 + 42 = 5. Ð → Ð → Ð → Ð → Hence, ab and cd are considered to be exactly the same: ab = cd. Even though they have Ð → Ð → different heads and tails, both ab and cd carry us 4 units right and 3 units down. The Ð → Ð → vector (4, −3) can carry us from a to b or from c to d. Thus, cd = (4, −3) = ab = (4, −3). They are one and the same vector. Ð → Ð → In contrast, the vector u has only half the length of ab and so u ≠ ab. (Indeed, as we shall Ð → learn later, we can write u = 0.5ab.) Example 234. The vector (0, −1) can carry us from a to c or from b to d. Thus, Ð → → = (0, −1). Thus, bd = (0, −1) = Ð ac But, and
Ð → = (0, 1) ≠ Ð → = (0, −1), ca ac → = (0, −1). (0, −0.5) ≠ Ð ac
Yet another way of denoting vectors is by a single letter, either with a right arrow overhead Ð → Ð → or in bold font. For example, in the figure above, the vector ab or cd is also named using → the letter v, either as Ð v or as boldfont v. Ð → Example 235. So altogether, I can write the vector ab in five different ways: ⎛ 4 ⎞ Ð → → ab = Ð v = v = (4, −3) = . ⎝ −3 ⎠ → Given a choice between writing Ð v or v, the bold font v is preferred in print publications. → But in handwriting, most people prefer Ð v (because writing in bold font is hard).
Exercise 102. Using a, b, c, or d from the above figure as the tail and a distinct point as the head, there are 12 possible vectors. We’ve already written out 4 of these in the last two examples. Write out the other 8 in ordered set notation. (Answer on p. 1077.)
Page 267, Table of Contents
www.EconsPhDTutor.com
The position vector of a point a is simply the vector from the origin o = (0, 0) to the point a. Formally: Definition 62. Given a point a = (a1 , a2 ), its position vector is the vector a = (a1 , a2 ). The position vector of the point a carries us from the origin o to the point a and so it → Take care not to confuse the point a = (a , a ) with the vector can also be denoted Ð oa. 1 2 a = (a1 , a2 ) — they are different objects!
Informally, the zero vector is the vector that carries us nowhere. Formally: Ð → Definition 63. The zero vector is the vector (0, 0) and can be denoted 0 or 0 .
Page 268, Table of Contents
www.EconsPhDTutor.com
27.1
Sum and Difference of Points and Vectors
Here is a quick summary of what you’ll learn in this section. (1) (2) (3) (4)
Point Point Point Point
+ Point = Undefined, − Point = Vector, + Vector = Point, − Vector = Point.
1. Point + Point = Undefined If a and b are points, then there is no such thing as a + b.34 The analogy is to points in the real world — it makes no sense to talk about the sum of two locations: Example 236. Consider the points Paris and Tokyo. The sum Paris + Tokyo = ?? is undefined. It makes no sense to talk about the sum of two locations.
p+v
q
v u b–a p
34
a
b q–u
At least in this textbook (and in the A-levels).
Page 269, Table of Contents
www.EconsPhDTutor.com
2. Point − Point = Vector
Definition 64. Given two points a = (a1 , a2 ) and b = (b1 , b2 ), their difference b−a is defined to be the vector from a to b, i.e., b − a = (b1 − a1 , b2 − a2 ). Example 237. Paris − Tokyo = The journey that carries us from Tokyo to Paris. We might write Paris − Tokyo =(−9000 km, 1000 km), meaning that to get from Tokyo to Paris, we must travel 9, 000 km west and 1, 000 km north. It makes sense to talk about the distance of the journey from Tokyo to Paris. Shortly, we’ll see that it similarly makes sense to talk about the length of the vector from a to b. Example 238. (See figure on p. 265.) Given the points a = (−1, 2) and b = (3, −1), their difference b − a is the vector from a to b, i.e., b − a = (3 − (−1), −1 − 2) = (4, −3).
3. Point + Vector = Point
Definition 65. Given the point p = (p1 , p2 ) and the vector v = (v1 , v2 ), their sum p + v is defined to be the point p + v = (p1 + v1 , p2 + v2 ). Geometrically, if the vector v has tail p, then it also has head p + v. Example 239. Tokyo + (−9000 km, 1000 km) = Paris. This says that starting from Tokyo, if we embark on a journey that carries us 9, 000 km west and 1, 000 km north, then we’ll end up in Paris.
Example 240. (See figure on p. 265.) Consider the vector (4, −3). If its tail is a = (−1, 2), then its head is (−1, 2) + (4, −3) = (3, −1) = b. And if its tail is c = (−1, 1), then its head is (−1, 1) + (4, −3) = (3, −2) = d.
Page 270, Table of Contents
www.EconsPhDTutor.com
4. Point − Vector = Point
Definition 66. Given the point q = (q1 , q2 ) and the vector u = (u1 , u2 ), their difference q − u is defined to be the point q − u = (q1 − u1 , q2 − u2 ). Geometrically, if the vector u has head q, then it also has tail q − u. Example 241. Paris − (−9000 km, 1000 km) = Tokyo. This says that starting from Paris, if we embark on a journey that is the exact opposite of going 9, 000 km west and 1, 000 km north (equivalently, we embark on a journey that goes 9, 000 km east and 1, 000 km south), then we’ll end up in Tokyo.
Example 242. (See figure on p. 265.) Consider again the vector (4, −3). If its head is b = (3, −1), then its tail is (3, −1) − (4, −3) = (−1, 2) = a. And if its head is d = (3, −2), then its tail is (3, −2) − (4, −3) = (−1, 1) = c.
Exercise 103. Consider the vector (4, −3). (a) If it has tail (0, 0), then what is its head? (b) If it has head (0, 0), then what is its tail? (c) If it has tail (5, 2), then what is its head? (d) If it has head (5, 2), then what is its tail? (Answer on p. 1077.)
Page 271, Table of Contents
www.EconsPhDTutor.com
27.2
Sum, Additive Inverse, and Difference of Vectors
Here is a quick summary of what you’ll learn in this section. (1) Vector + Vector = Vector, (2) − Vector = Vector, (additive inverse) (3) Vector − Vector = Vector.
1. Vector + Vector= Vector
Definition 67. If u = (u1 , u2 ) and v = (v1 , v2 ) are vectors, then their sum, denoted u + v, is the vector defined by u + v = (u1 + v1 , u2 + v2 ). Geometrically, if the tail of v is the head of u, then u + v is the vector from the tail of u to the head of v.
u+v
v
u
Ð → Ð → → Example 243. (See figure on p. 265.) ab + bc = (4, −3) + (−4, 2) = (0, −1) = Ð ac. Ð → Ð → Example 244. (See figure on p. 265.) ad + cb = (4, −4) + (4, −2) = (8, −6).
Page 272, Table of Contents
www.EconsPhDTutor.com
2. − Vector= Vector (Additive inverse)
Definition 68. If v = (v1 , v2 ), then its additive inverse, denoted −v, is defined by −v = (−v1 , −v2 ). Geometrically, if the vector v is from point a to point b, then −v is the vector from point b to point a. And so informally, the additive inverse is simply the same vector but flipped in the opposite direction. Ð → Ð → Ð → Ð → Example 245. The additive inverse of ab is ba. That is, −ab = ba. Ð → Ð → Ð → Ð → Example 246. The additive inverse of bc is cb. That is, − bc = cb.
3. Vector − Vector= Vector
Definition 69. Given two vectors u and v, their difference, denoted u − v, is defined to be the sum of the vectors u and −v. Or equivalently, if u = (u1 , u2 ) and v = (v1 , v2 ), then u − v is the vector defined by u − v = (u1 − v1 , u2 − v2 ). Geometrically, if we place the heads of u and v at the same point, then u − v is the vector from the tail of u to the tail of v.
u
u-v
Page 273, Table of Contents
v
www.EconsPhDTutor.com
→ could be written as the In the previous section, we learnt that by definition, the vector Ð pq → = q − p. Now, we’ll prove that Ð → can also be written as the difference of two points: Ð pq pq difference of two vectors: → = q − p. Fact 19. Let p and q be two points with position vectors p and q. Then Ð pq → + (−Ð → =Ð →+Ð →=Ð →+Ð → This is thus the vector that carries Proof. q − p = q + (−p) = Ð oq op) oq po po oq. us first from p to o, then from o to q; in short, it carries us from p to q. So it is simply the → vector Ð pq.
Ð → Example 247. (See figure on p. 265.) b − a = (3, −1) − (−1, 2) = (4, −3) = ab. Ð → Example 248. (See figure on p. 265.) d − c = (3, −2) − (−1, 1) = (4, −3) = cd. Interpreting u − v as the sum of the vectors u and −v is often convenient: Ð → Ð → Example 249. (See figure on p. 265.) Without any numbers, we can compute: ab − cb = Ð → Ð → Ð → Ð → → Ð → → We can verify with numbers that this is correct: Ð ab + (− cb) = ab + bc = Ð ac. ab − cb = → ✓. (4, −3) − (4, −2) = (0, −1) = Ð ac Ð → Ð → Example 250. (See figure on p. 265.) ad − cb = (4, −4) − (4, −2) = (0, −2).
→ Ð → → Ð → Ð → Ð → Ð → Ð → Ð → Ð → Ð → →+Ð Exercise 104. Write down what Ð ac cb, dc + Ð ca, bd + da, ad − cd, −dc − bd, and bd + db are, without writing out any numbers.(Answer on p. 1077.) → Ð → → →−Ð Exercise 105. Using the figure on p. 265, compute each of the following: Ð ac cb, dc − Ð ca, Ð → Ð → Ð → Ð → Ð → Ð → Ð → Ð → bd − da, ad + cd, dc + bd, and bd − db? (Answer on p. 1077.)
Page 274, Table of Contents
www.EconsPhDTutor.com
27.3
Displacement Vectors
Ă? â&#x2020;&#x2019; Definition 70. If a moving particle starts at point a and ends at point b, we call ab its displacement vector.
Example 251. A particle is travelling along the red arc, along the path shown. Its starting point is in blue and its ending point is in purple. Its displacement vector is thus (2, 2).
1
y x
0 -1
0 -1 -2
2 Ending point
1
3
4
Displacement vector (2, 2) Starting point
-3 -4
Page 275, Table of Contents
www.EconsPhDTutor.com
27.4
Length (or Magnitude) of a Vector
The Pythagorean Theorem says that if c is the length of hypothenuse of a right-angled triangle and a, b are the lengths of the other two sides, then a2 + b2 = c2 .
c a
b
As you learnt in secondary school we can calculate the distance between two points using the Pythagorean Theorem: Example be two points. Then the distance between p √252. Let p = (1, 1) and q =√(−1, −1)√ 2 2 and q is [1 − (−1)] + [1 − (−1)] = 4 + 4 = 8.
2
y
1 - (-1)
1
p
1 - (-1)
x
0 -2
-1
q
0
1
2
-1
-2
Page 276, Table of Contents
www.EconsPhDTutor.com
The vector v = (v1 , v2 ) goes v1 units right and v2 units up. We are thus motivated to define its length (or magnitude) as: Definition √ 71. The length (or magnitude) of a vector v = (v1 , v2 ) is denoted ∣v∣ and defined by ∣v∣ = v12 + v22 .
Example 252 (continued). Another way to find the distance between p and q is to first → = (−2, −2). The distance between p find the vector that carries us from p to q. This is Ð pq √ √ 2 2 Ð → and q is thus simply the length (or magnitude) of this vector: ∣pq∣ = (−2) + (−2) = 8. Of course, the distance from p to q is the same as the distance from q to p. So we could → = (2, 2) and gotten the same answer – ∣Ð → = just as well have calculated the length of Ð qp qp∣ √ √ 22 + 22 = 8. → Ð → → →−Ð Exercise 106. Using the figure on p. 265, compute each of the following: ∣Ð ac cb∣, ∣dc − Ð ca∣, Ð → Ð → Ð → Ð → Ð → Ð → Ð → Ð → ∣bd − da∣, ∣ad + cd∣, ∣dc + bd∣, and ∣bd − db∣. Also, find the distance between (18, 4) and (−1, −2). (Answer on p. 1077.) Exercise 107. In general, given any two vectors u and v, is it true that ∣u + v∣ = ∣u∣ + ∣v∣? (Answer on p. 1077.)
Page 277, Table of Contents
www.EconsPhDTutor.com
27.5
Scalar Multiplication of a Vector
Definition 72. A scalar is simply any real number. A scalar is often contrasted with a vector. A vector has both magnitude (or length) and direction. In contrast, a scalar has magnitude but no direction. Definition 73. If v = (v1 , v2 ) is a vector and c ∈ R is a scalar, then cv denotes the vector defined by cv = (cv1 , cv2 ). We call this operation scalar multiplication of a vector. Graphically, cv is simply the vector that has the same direction as v, but with c times the length. This is formally shown in the next fact.
v cv
Fact 20. If v = (v1 , v2 ) is a vector and c ∈ R, then ∣cv∣ = ∣c∣ ∣v∣.
Proof.
√ 2 2 ∣cv∣ = ∣(cv1 , cv2 )∣ = (cv1 ) + (cv2 ) √ √ 2 2 2 2 = c v1 + c v2 = ∣c∣ v12 + v22 = ∣c∣ ∣(v1 , v2 )∣ = ∣c∣ ∣v∣ .
Ð → → Ð → Exercise 108. Using the figure on p. 265, write down 2ab, 3Ð ac, and 4ad in ordered set Ð → Ð → → Ð → → = 3 ∣Ð → and ∣4Ð notation. Verify that ∣2ab∣ = 2 ∣ab∣, ∣3Ð ac∣ ac∣, ad∣ = 4 ∣ad∣. (Answer on p. 1078.)
Page 278, Table of Contents
www.EconsPhDTutor.com
27.6
Unit Vectors
Definition 74. A unit vector is any vector of length 1. √
√ 2 2 Example 253. Let’s verify that the vectors (1, 0), (0, 1), and ( , ) are all unit vectors: 2 2 √ ✓ ∣(1, 0)∣ = 12 + 02 = 1, √ ∣(0, 1)∣ = 02 + 12 = 1, ✓ ¿ √ √ √ 2 Á √ 2 √ Á 2 2 2 2 À( ∣( , )∣ = Á ) +( ) = 2/4 + 2/4 = 1. ✓ 2 2 2 2 Example 254. Let’s verify that the vectors (1, 1) and (−1, −1) are not unit vectors: √ √ 12 + 12 = 2 ≠ 1, ✓ √ √ 2 2 ∣(−1, −1)∣ = (−1) + (−1) = 2 ≠ 1. ✓ ∣(1, 1)∣ =
Ð → We specially reserve the name i (or i ) for the unit vector (1, 0), which is the unit vector that is purely in the direction of the x-axis. Similarly, we specially reserve the name j (or Ð → j ) for the unit vector (0, 1), which is the unit vector that is purely in the direction of the y-axis. And so, using also what we learnt about the sum of and scalar multiplication of vectors, we can rewrite any vector into the sum of i’s and j’s:
Page 279, Table of Contents
www.EconsPhDTutor.com
Example 255. The position vectors for the points a, b, and c (illustrated below) are a = (1, 2) = i + 2j, b = (4, â&#x2C6;&#x2019;3) = 4i â&#x2C6;&#x2019; 3j, and c = (0, 6) = 6j.
7
y c
6 j 5 j 4 j 3 j
a
2 ji 1 j
x
0 -3
-2
-1 -1 -2 -3
0-j
2
3
4
5
-j -j i
Page 280, Table of Contents
1
b i
i
i
www.EconsPhDTutor.com
ˆ — points in the same direction Informally, the unit vector in the direction v — denoted v v, but has length 1. Formally:
Definition 75. The unit vector in the direction v is defined by v ˆ=
1 v. ∣v∣
Ð → → Exercise 109. In the figure on p. 265, what are the unit vectors in the directions ab, Ð ac, Ð → Ð → Ð Ð → → and ad? What are the unit vectors in the directions 2ab, 3ac, and 4ad? (Answer on p. 1078.)
The following fact is an obvious corollary to Fact 20. Fact 21. If c is a scalar and v ˆ is a unit vector, then the vector cˆ v has length c.
Informally, two vectors have the same unit vector ⇐⇒ they both point in the same direction. Formally: ˆ ⇐⇒ a can be written as a scalar ˆ=b Fact 22. Let a and b be any two vectors. Then a multiple of b.
Proof. Optional, see p. 930 in the Appendices. Informally, any vector in the plane can be written as the linear combination of any other two vectors. Formally: Fact 23. Let a and b be any two vectors in the same plane with distinct directions (i.e. ˆ Then every vector in the same plane can be written as αa + βb for some α, β ∈ R. ˆ ≠ b). a Proof. Optional, see p. 930 in the Appendices. See TYS Exercise 338 (i) for an application of the above fact. Exercise 110. Given the vectors a = (1, 3) and b = (7, 5), show that each of the following vectors can be written in the form αa + βb for some α, β ∈ R. (i) (0, 1). (ii) (1, 0). (iii) (1, 1). (Answer on p. 1078.)
Page 281, Table of Contents
www.EconsPhDTutor.com
27.7
The Ratio Theorem a
p
b o
Theorem 3. Ratio Theorem. Let a, b, and p be points, where p is on the line segment ab. Let a, b, and p be the corresponding position vectors. Then Ð → ∣bp∣
→ ∣Ð ap∣ p= → a+ Ð → b. → + ∣Ð → + ∣Ð ∣Ð ∣ap∣ ap∣ bp∣ bp∣ Proof. Optional, see p. 932 (Appendices).
→ → and µ = ∣Ð Or if we let λ = ∣Ð ap∣ bp∣, then the above can be rewritten in a form that is perhaps easier to remember: p=
µa λb µa + λb + = . λ+µ λ+µ λ+µ
By the way, the List of Formulae (p. 4) contains this statement: “The point dividing AB in the ratio λ ∶ µ has position vector
Page 282, Table of Contents
µa + λb .” λ+µ
www.EconsPhDTutor.com
Example 256. Consider the points a = (3, 4) and b = (−1, 2). Find the point p that divides the line segment ab into the ratio 3 ∶ 2. 2 3 2 3 3 14 3 14 We have p = a + b = (3, 4) + (−1, 2) = ( , ). Hence, the point is p = ( , ). 5 5 5 5 5 5 5 5
Example 257. Consider the points a = (8, 3) and b = (2, −6). Find the point p that divides the line segment ab into the ratio 3 ∶ 7. We have p = 0.7a +0.3b = 0.7(8, 3)+0.3(2, −6) = (6.2, 0.3). Hence, the point is p = (6.2, 0.3).
Exercise 111. (a) Consider the points a = (1, 2) and b = (3, 4). Find the point p that divides the line segment ab into the ratio 5 ∶ 6. (b) Consider the points a = (1, 4) and b = (2, 3). Find the point p that divides the line segment ab into the ratio 5 ∶ 1. (c) Consider the points a = (−1, 2) and b = (3, −4). Find the point p that divides the line segment ab into the ratio 2 ∶ 3. (Answer on p. 1078.)
Page 283, Table of Contents
www.EconsPhDTutor.com
28
Scalar Product
Definition 76. Given two 2D vectors u = (u1 , u2 ) and v = (v1 , v2 ), their scalar product (or dot product), denoted u ⋅ v, is defined by u ⋅ v = u1 v1 + u2 v2 .
And so to get the scalar product, simply multiply each term of each vector with the corresponding term of the other, then add these up. It’s that simple! The scalar product is itself simply a scalar (i.e. a real number). Hence the name. Example 258. (5, −3) ⋅ (2, 1) = 5 × 2 + (−3) × 1 = 7. Example 259. (0, 17) ⋅ (−1, 3) = 0 × (−1) + 17 × 3 = 51. Ordinary multiplication is distributive: Example 260. 3 × (5 + 11) = 3 × 5 + 3 × 11 and 18 × (7 − 31) = 18 × 7 − 18 − 31. It turns out that the scalar product is likewise distributive: Fact 24. Let a, b, and c be vectors. Then a ⋅ (b + c) = a ⋅ b + a ⋅ c and (a + b) ⋅ c = a ⋅ c + b ⋅ c.
Proof. Optional, see p. 931 in the Appendices. Here is one use of the scalar product: the length of a vector is simply the square root of its scalar product with itself. Formally: Fact 25. Given a vector v, ∣v∣ =
√
v ⋅ v.
√ Proof. By Definition 71 (length of vector), ∣v∣ = v12 + v22 . By Definition 76 (scalar product), √ v ⋅ v = v1 v1 + v2 v2 = v12 + v22 . Hence, ∣v∣ = v ⋅ v. Next up is a more important use of the scalar product:
Page 284, Table of Contents
www.EconsPhDTutor.com
28.1
The Angle between Two Vectors
Fact 26. Let θ ∈ [0, π] be the angle between two non-zero vectors u and v. Then u ⋅ v = ∣u∣ ∣v∣ cos θ.
v
Ĭ u
Proof. Optional, see p. 931 (Appendices). The above fact35 gives us a very convenient way to calculate the angle between two vectors, because rearranging, we have: θ = cos−1 (
35
u⋅v ). ∣u∣ ∣v∣
We have two possible interpretations of the scalar product that are entirely equivalent. We can use either of these interpretations as our definition and then prove that the other interpretation is true. (1) In this textbook, we first define the scalar product by u ⋅ v = u1 v1 + u2 v2 , then prove that u ⋅ v = ∣u∣ ∣v∣ cos θ. That is, we start with the algebraic definition, then prove a geometric property. (2) In contrast, others may prefer to first define the scalar product by u⋅v = ∣u∣ ∣v∣ cos θ, then prove that u⋅v = u1 v1 +u2 v2 . That is, we start with the geometric definition, then prove an algebraic property. Either way, we first define the scalar product one way or the other. We then prove that the alternative statement is equivalent. (It is possible that your JC teachers take the second approach, rather than the first, as is done in this textbook. Or worse, your teachers simply leave you confused as to why the hell u ⋅ v = ∣u∣ ∣v∣ cos θ and at the same time, magically enough, u ⋅ v = u1 v1 + u2 v2 . This was my experience as a JC student a number of years ago. If this is also your current experience, hopefully this textbook has helped to clear things up!)
Page 285, Table of Contents
www.EconsPhDTutor.com
Example 261. The vector i = (1, 0) points east. The vector (1, 1) points northeast. We π know the angle between these two vectors is . Let’s check and verify that the formula 4 works: θ = cos−1 (
i ⋅ (1, 1) (1, 0) ⋅ (1, 1) ) = cos−1 ( ) ∣i∣ ∣(1, 1)∣ ∣(1, 0)∣ ∣(1, 1)∣
⎛ ⎞ 1×1+0×1 ⎟ = cos ⎜ √ √ ⎝ ( 12 + 02 ) × ( 12 + 12 ) ⎠ −1
= cos−1 (
1+0 1 π √ ) = cos−1 ( √ ) = . 4 1× 2 2
✓
Example 262. The vector i = (1, 0) points east. The vector j = (0, 1) points north. We π know the angle between these two vectors is right (i.e. ). Let’s check and verify that the 2 formula works: θ = cos−1 (
i⋅j (1, 0) ⋅ (0, 1) ) = cos−1 ( ) ∣i∣ ∣j∣ ∣(1, 0)∣ ∣(0, 1)∣
⎛ ⎞ 1×0+0×1 ⎟ = cos ⎜ √ √ ⎝ ( 12 + 02 ) × ( 02 + 12 ) ⎠ −1
= cos−1 (
Page 286, Table of Contents
0+0 π ) = cos−1 0 = 1×1 2
✓
www.EconsPhDTutor.com
Example 263. The angle between the vectors (3, 2) and (−1, −4) is θ = cos−1 (
(3, 2) ⋅ (−1, −4) ) ∣(3, 2)∣ ∣(−1, −4)∣
⎛ ⎞ 3 × (−1) + 2 × (−4) ⎜ ⎟ ⎟ = cos−1 ⎜ √ √ ⎜ 2 2 ⎟ 2 2 ⎝ ( 3 + 2 ) × ( (−1) + (−4) ) ⎠ −11 −3 − 8 √ ) = cos−1 ( √ ) ≈ 2.404 = cos−1 ( √ 13 × 17 221
This is an example where the angle is obtuse, i.e. between π/2 and π.
y (3, 2)
x 2.404 rad
(-1, -4)
Page 287, Table of Contents
www.EconsPhDTutor.com
Recall that the arccosine function is defined to have range [0, π]. That is, cos−1 x ∈ [0, π]. Moreover, π • x > 0 Ô⇒ cos−1 x ∈ [0, ), i.e. cos−1 x is an acute (or zero) angle. 2 π • x = 0 Ô⇒ cos−1 x = , i.e. cos−1 x is a right angle. 2 π • x < 0 Ô⇒ cos−1 x ∈ ( , π], i.e. cos−1 x is an obtuse (or straight) angle. 2
These three observations, together with Fact 26, imply the following Fact, which by the way was already illustrated by the previous three examples: Fact 27. Let u and v be vectors. The angle between u and v is (i) acute (or zero) if u ⋅ v < 0; (ii) right if u ⋅ v = 0; and (iii) obtuse (or straight) if u ⋅ v > 0. We’ll use the words perpendicular, orthogonal, and normal interchangeably: Definition 77. Two vectors are orthogonal (or perpendicular or normal) if the angle beπ tween them is right (i.e. equal to ). 2 I will sometimes write u ⊥ v to mean u is orthogonal (or perpendicular or normal) to v.
Exercise 112. First write down the angle between each of the following pairs of vectors without using the above formula. Then verify that the formula does indeed √ give you these correct angles: (a) (2, 0) and (0, 17); (b) (5, 0) and (−3, 0); (c) i and (1, 3/3); (d) i √ and (1, 3). (Answers on pp. 1079 and 1080.)
Exercise 113. Verify that i and j are orthogonal, by computing their scalar product. (Answer on p. 1081.)
Page 288, Table of Contents
www.EconsPhDTutor.com
28.2
Projection of One Vector on Another
The scalar product also gives a convenient way of computing the length of the projection of one vector on another. Say we have a right triangle (left diagram) where the angle θ and the length a are known. What is the length b? It is simply ∣a∣ cos θ.
a
a
Ĭ
Ĭ
b
a
b
b
Now suppose a (blue) and b (green) are vectors (right diagram). The projection of the vector a on the vector b is denoted a⊥b (red). Note that a⊥b is itself a vector. What is the length of the projection? Well, if ∣a∣ is the length of the vector a and θ is the angle between the two vectors, then the length of the projection is ∣a⊥b ∣ simply a cos θ. Nicely enough, we actually have a quick alternative method of computing this length. Let ˆ be the unit vector for b. Then b ˆ = ∣a∣∣b∣ ˆ cos θ = ∣a∣ × 1 × cos θ = ∣a∣ cos θ = ∣a⊥b ∣. a⋅b ˆ or more correctly ∣a ⋅ b∣, ˆ since a ⋅ b ˆ may sometimes So we have a nice interpretation for a ⋅ b be negative:
ˆ is simply the length of the projection of a on b! ∣a ⋅ b∣ Page 289, Table of Contents
www.EconsPhDTutor.com
Example 264. The length of the projection of (3, 2) on (1, 1) is 1 1 (1, 1)] = √ (3, 2) ⋅ (1, 1) ∣(1, 1)∣ 2 √ 5 5 2 1 = √ (3 × 1 + 2 × 1) = √ = . 2 2 2
̂ (3, 2) ⋅ (1, 1) = (3, 2) ⋅ [
You should verify for yourself that the length of the projection of (3, 2) on (1000, 1000) is √ 5 2 . The length of the vector to be projected — (3, 2) — matters, but the length of also 2 the vector onto which it is projected — be it (1, 1) or (1000, 1000) — doesn’t matter.
Example 265. The length of the projection of (−6, 1) on (2, 0) is 1 1 (2, 0)] = (−6, 1) ⋅ (2, 0) ∣(2, 0)∣ 2 −12 1 = −6. = (−6 × 2 + 1 × 0) = 2 2
̂ (−6, 1) ⋅ (2, 0) = (−6, 1) ⋅ [
Again, you can verify for yourself that the length of the projection of (−6, 1) on (50000, 0) is also −6. Again, the length of the vector to be projected — (−6, 1) — matters, but the length of the vector onto which it is projected — be it (2, 0) or (50000, 0) — doesn’t matter.
Exercise 114. What are the lengths of the projections of (a) (1, 0) on (33, 33) and (b) (33, 33) on (1, 0)? (Answer on p. 1081.)
Page 290, Table of Contents
www.EconsPhDTutor.com
28.3
Direction Cosines
The angle between a vector v and the x-axis is simply the angle between v and i = (1, 0). Similarly, the angle between v and the y-axis is simply the angle between v and j = (0, 1). Example 266. Consider the angle a between the vector (3, 2) and the x-axis. We have: α = cos a =
3×1+2×0 3 (3, 2) ⋅ (1, 0) = √ =√ . √ ∣(3, 2)∣ ∣(1, 0)∣ ( 32 + 22 ) × ( 12 + 02 ) 13
√ We refer to α = 3/ 13 as the x-direction cosine of the vector (3, 2). By computing √ cos−1 α = cos−1 (3/ 13) ≈ 0.588, we find that the angle a between the vector (3, 2) and the x-axis is 0.588.
y (3, 2) 0.983 rad x 2.404 rad
(-1, -4)
Page 291, Table of Contents
www.EconsPhDTutor.com
Example 267. Consider the angle b between the vector (3, 2) and the y-axis. We have: β = cos b =
(3, 2) ⋅ (0, 1) 2 2 3×0+2×1 √ =√ . =√ = √ √ ∣(3, 2)∣ ∣(0, 1)∣ ( 32 + 22 ) × ( 02 + 12 ) 13 × 1 13
√ We refer to β = 2/ 13 as the y-direction cosine of the vector (3, 2). By computing √ cos−1 β = cos−1 (2/ 13) ≈ 0.983, we find that the angle b between the vector (3, 2) and the y-axis is 0.983. Definition 78. Given a vector v, its x-direction cosine α is simply the length of the ˆ on the x-axis. projection of v ˆ on the y-axis. Similarly, its y-direction cosine β is simply the length of the projection of v The next Fact is immediate from the above definition: ˆ = (α, β). Fact 28. Let v be a vector and α and β be its x- and y-direction cosines. Then v Example 268. The x- and y-direction cosines of the vector (3, 2) are 3 α= √ 13
2 and β = √ . 13
3 2 Hence, the unit vector in the direction (3, 2) is ( √ , √ ). 13 13
Exercise 115. For each of the following vectors, find their x- and y-direction cosines. Hence write down their unit vectors. (a) (1, 3). (b) (4, 2). (c) (−1, 2). (Answer on p. 1081.)
Page 292, Table of Contents
www.EconsPhDTutor.com
29
Vectors in 3D
In two dimensions, we had the cartesian (or two-dimensional) plane with x- and y-axes. Informally, the x-axis goes to the right and the y-axis goes up. A point was any ordered pair of real numbers. The origin o = (0, 0) was the intersection point of the two axes. And relative to the origin, the generic point a = (a1 , a2 ) was the point a1 units to the right and a2 units up. In three dimensions, we now instead have the three-dimensional space (3D space). The x- and y-axes are as before. There is an additional z-axis that, informally, comes “out of the paper, perpendicular to the plane of the paper, straight towards your face”. We call this the right hand coordinate system, because if you take your right hand, stick out your thumb, forefinger, and middle finger so that they are perpendicular, your thumb represents the x-axis, your forefinger the y-axis, and your middle finger the z-axis. (Try it!) (If instead the z-axis goes “into the paper”, then we’d have a left hand coordinate system. Can you explain why?)
y
a2
a
x a1
a3 z
Page 293, Table of Contents
www.EconsPhDTutor.com
In the context of 3D space, a point is any ordered triple of real numbers. The origin o = (0, 0, 0) is the point where the x-, y-, and z-axes intersect. And relative to the origin, the generic point a = (a1 , a2 , a3 ) is the point a1 units to the right, a2 units up, and a3 units “out of the paper”. Everything we learnt about 2D vectors finds its analogy in three-dimensional (3D) vectors. Most of the time, the analogy is obvious. Try these exercises.
Exercise 116. (Answer on p. 1082.) (a) Fill in the blanks. A 3D vector is an “arrow” that has two characteristics: __________ and __________. Just like a point, it can be described by an __________ of __________. The vector a = (a1 , a2 , a3 ) carries us from the origin to _______________. (b) What other ways are there to denote the vector a = (a1 , a2 , a3 )? (Hint. The unit vector in the z-axis is now called k.) Ð → → → Ð (c) Let a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) be points. What are (i) a+b; (ii) a+ ob; (iii) Ð oa+ ob; → →−Ð ba? and (iv) Ð oa
Page 294, Table of Contents
www.EconsPhDTutor.com
√ The length (or magnitude) of a 2D vector v = (v1 , v2 ) was defined by v12 + v22 . What then is the length (or magnitude) of a 3D vector? This is the one instance where the analogy from the 2D case to the 3D case is perhaps less than obvious. So let’s explore this issue. Consider the blue point a in the figure below. What is its distance from the origin (0, 0, 0)? In other words, what is the length of the green dotted line?
y
a2
a
x a1
a3 z First let’s calculate the distance of the red point from the origin, in other words the length √ of the red dotted line. By the Pythagorean Theorem, it is a22 + a23 . √ Now, notice the the green dotted line, the red dotted line (length a22 + a23 ), and the blue dotted line (length a1 ) form a right-angled triangle, with the hypothenuse being the green dotted line. Thus, the length of the green dotted line is (again by the Pythagorean Theorem): √ a21
Page 295, Table of Contents
√ √ 2 2 2 + ( a2 + a3 ) = a21 + a22 + a23 .
www.EconsPhDTutor.com
We are thus motivated to define the length (or magnitude) of a 3D vector as follows: Definition 79.√ The length (or magnitude) of a vector a = (a1 , a2 , a3 ) is denoted ∣a∣ and defined by ∣a∣ = a21 + a22 + a23 . This is very much analogous to the definition of the length (or magnitude) of a 2D vector.
Let’s continue with our exercises for 3D vectors:
Exercise 117. (Answer on p. 1083.) (a) Compute the lengths of the vectors a = (1, 2, 3), b = (4, 5, 6), and a − b. (b) Compute the lengths of the vectors 2a = (2, 4, 6), 3b = (12, 15, 18), and 4(a − b). (c) Compute the unit vectors in the directions a = (1, 2, 3), b = (4, 5, 6), and a − b. (d) Compute (1, 2, 3) ⋅ (4, 5, 6) and (−2, 4, −6) ⋅ (1, −2, 3). (e) Compute the angles (i) between the vectors a = (1, 2, 3) and b = (4, 5, 6); and (ii) between the vectors u = (−2, 4, −6) and v = (1, −2, 3). (iii) Are the vectors (−2, 4, −6) and (1, −2, 3) orthogonal? (f) Compute the length of the projection of a = (1, 2, 3) on b = (4, 5, 6). (g) Find the point that divides the line segment ab in the ratio 2 ∶ 3. (h) For each of the following vectors, find their x-, y-, and z-direction cosines. And then write down their unit vectors. (i) (1, 3, −2). (ii) (4, 2, −3). (iii) (−1, 2, −4).
Page 296, Table of Contents
www.EconsPhDTutor.com
30 30.1
Vector Product Vector Product in 2D
Recall that given two 2D vectors u = (ux , uy ) and v = (vx , vy ), their scalar product was the scalar defined by u ⋅ v = ux vx + vx vy . We now define a very similar concept. Definition 80. Given two 2D vectors u = (ux , uy ) and v = (vx , vy ), their vector product (or cross product), denoted u × v, is the scalar defined by u × v = ux vy − uy vx .
Example 269. If u = (1, 2) and v = (3, 4), then u × v = 1 × 4 − 2 × 3 = −2. Example 270. If p = (−3, 5) and q = (6, 1), then p × q = −3 × 1 − 5 × 6 = −33. Example 271. If u = (−1, 4) and v = (2, −3), then u × v = (−1) × (−3) − 4 × 2 = −5. Ordinary multiplication is commutative. This simply means that given any real numbers a, b, we have a × b = b × a. For example, Example 272. 4 × 7 = 7 × 4 and 3 × 5 = 5 × 3. In contrast, the vector product is not commutative because u × v ≠ v × u. This might be the first time in your life that you’re encountering a product that isn’t commutative. In fact, the vector product is anticommutative because u × v = −v × u! For example, Example 273. If u = (1, 2) and v = (3, 4), then u × v = 1 × 4 − 2 × 3 = −2, but v × u = 2 × 3 − 1 × 4 = 2. Example 274. If u = (−1, 4) and v = (2, −3), then u × v = (−1) × (−3) − 4 × 2 = −5, but v × u = 4 × 2 − (−1) × (−3) = 5.
Page 297, Table of Contents
www.EconsPhDTutor.com
Recall that if θ ∈ [0, π] is the angle between two vectors, then based on our definition that u ⋅ v = ux vx + uy vy , we could prove that u ⋅ v = ∣u∣ ∣v∣ cos θ. It turns out based on our definition that u × v = ux vy − uy vx , we can prove a very similar result:36 Fact 29. Let u and v be two non-zero 2D vectors and θ ∈ [0, π] be the angle between them. Then the scalar u × v is equal to either ∣u∣ ∣v∣ sin θ or − ∣u∣ ∣v∣ sin θ.
Proof. Optional, see p. 933 in Appendices.
Earlier we already had one formula for calculating the angle between two vectors. Let θ ∈ [0, π] be the angle between u and v. Then θ = cos−1 (
u⋅v ). ∣u∣ ∣v∣
The above Fact now gives us a second formula. Let θ ∈ [0, π] be the acute or right angle between u and v. Then θ = sin−1 ∣
u×v ∣. ∣u∣ ∣v∣
However, we’ll stick with using only the first cosine formula. We won’t use the second sine formula, mainly because, as we’ll see, computing the vector product is very tedious, especially in the 3D case, where it is a different creature altogether.
36
Footnote 36 explained that the scalar product could be defined in one of two equivalent ways. Similarly, the vector product can be defined in one of two equivalent ways. We can use either definition and then prove that the other is true. (1) In this textbook, we first define the vector product by u × v = ux vy − uy vx ; we then prove that u × v = ± ∣u∣ ∣v∣ sin θ, where θ is the angle between the two vectors. That is, we start with the algebraic definition, then prove a geometric property. The alternative approach is this: π π (2) Define the vector product by u × v = ∣u∣ ∣v∣ sin θ if θ ∈ [0, ] or u × v = − ∣u∣ ∣v∣ sin θ if θ ∈ ( , π] ; then prove that 2 2 u × v = ux vy − uy vx . That is, we start with the geometric definition, then prove an algebraic property.
Page 298, Table of Contents
www.EconsPhDTutor.com
30.2
Areas of Triangles and Parallelograms
SYLLABUS ALERT Calculation of the area of a triangle or parallelogram is included in the 9740 (old) syllabus, but not in the 9758 (revised) syllabus. So you can skip this section if you’re taking 9758.
The vector product is also helpful for computing the area of triangles and parallelograms. Fact 30. The triangle with sides of lengths ∣u∣, ∣v∣, and ∣v − u∣ has area 0.5∣u × v∣.
Case #1.
v
|v| sin (π – Ʌ) v |v| sin Ʌ
v–u
v–u Ʌ u
Case #2.
Ʌ u
Proof. Case #1. If the vectors u and v form an acute or right angle θ, then the area of the triangle is simply 0.5 × Base × Height or 0.5 ∣u∣ ∣v∣ sin θ. And by Fact 29, 0.5 ∣u∣ ∣v∣ sin θ = 0.5∣u × v∣. Case #2. And if the vectors u and v form an obtuse angle θ, then the area of the triangle is again simply 0.5 × Base × Height or 0.5 ∣u∣ ∣v∣ sin(π − θ). Recall that sin(π − θ) = sin π cos θ − sin θ cos π = sin θ. So again the area of the triangle is 0.5 ∣u∣ ∣v∣ sin θ or 0.5∣u × v∣.
Example 275. Consider the triangle formed by the points (0, 0), (3, 4), and (5, 6). Its area is simply 0.5 ∣(3, 4) × (5, 6)∣ = 0.5∣3 × 6 − 4 × 5∣ = 1.
Page 299, Table of Contents
www.EconsPhDTutor.com
Fact 31. The parallelogram with sides of lengths ∣u∣ and ∣v∣, and diagonal of length ∣v − u∣ has area ∣u × v∣.
v–u u
ȣ
v
Proof. Such a parallelogram is simply composed of two of the triangles from Fact 30. And so its area is simply twice the area of the triangle, or 2 × 0.5∣u × v∣ = ∣u × v∣.
Page 300, Table of Contents
www.EconsPhDTutor.com
30.3
Vector Product in 3D
The 3D vector product is very different from the 2D vector product. The latter was simply a scalar (real number); in contrast, the 3D vector product is instead a VECTOR! Also previously, we first started with the algebraic definitions. For example, the 3D scalar product was defined as u⋅v = u1 v1 +u2 v2 +u3 v3 and the 2D vector product as u×v = u1 v2 −u2 v1 . We then showed that these algebraic definitions were equivalent to some geometric interpretations. For the vector product in 3D, I will go the other way round. That is, I will start with the (very long) geometric definition, then show that it is equivalent to some algebraic interpretation.
Page 301, Table of Contents
www.EconsPhDTutor.com
Definition 81. Given two distinct 3D vectors u = (ux , uy , uz ) and v = (vx , vy , vz ), their vector product (or cross product), denoted u × v, is the (unique) vector that satisfies 3 properties: 1. u × v is orthogonal (perpendicular) to both u and v. Let’s see what this first property means. Recall that it doesn’t matter where we put the heads and tails of vectors. So let’s put u and v on the same plane, with their heads at the same point.
u×v
u
Plane
ȣ
v
We see that there are exactly two vectors that are orthogonal to both u and v — the vector pointing up (green) and the vector pointing down (purple). There is thus an ambiguity. Which of these two vectors is u × v? To resolve this ambiguity, we also require that u × v satisfy a second property: 2. u × v satisfies the right-hand rule: Take your right hand, stick out your thumb, forefinger, and middle finger so that they are perpendicular, your thumb represents the vector u × v, your forefinger the vector u, and your middle finger the vector v. Hence, in the figure, u × v points up (green). (Try it yourself!) Note that the right-hand rule is a mere convention, but one that everyone has agreed upon. There is no especially compelling reason for using it, other than the fact that left-handed people are an oppressed minority! (If instead we used the left-hand rule, then u × v would point down (purple) (try it!), but for better or worse, we don’t use the left-hand rule.) The third and last property specifies the length (or magnitude) of u × v. 3. ∣u × v∣ = ∣u∣ ∣v∣ sin θ, where θ ∈ [0, π] is the angle between them.
Page 302, Table of Contents
www.EconsPhDTutor.com
Note that θ ∈ [0, π] Ô⇒ sin θ ≥ 0, so that ∣u∣ ∣v∣ sin θ is never negative. (Otherwise, we’d have the distressing possibility that the length of u × v is sometimes negative!) One implication of this third and last property is that if both u and v point in the same direction (so that θ = 0), then u × v is the zero vector (i.e. u × v = 0). Fact 32. (a) i × j = k; (b) j × k = i; (c) k × i = j; (d) j × i = −k; (e) k × j = −i; and (f) i × k = −j. Proof. In each case, use the right-hand rule to show that properties #1 and #2 of Definition 81 are satisfied. π In each case, the length of the cross product is ∣u∣ ∣v∣ sin θ = 1 × 1 × sin = 1. So indeed, 2 property #3 is also satisfied.
Ordinary multiplication is distributive: Example 276. 3 × (5 + 11) = 3 × 5 + 3 × 11 and 18 × (7 − 31) = 18 × 7 − 18 − 31. It turns out that the vector product is likewise distributive: Fact 33. Let a, b, and c be vectors. Then a × (b + c) = a × b + a × c. Moreover, (a + b) × c = a × c + b × c. Proof. Optional, see p. 934 in the Appendices.
The next proposition gives the promised algebraic interpretation of the vector product. Proposition 7. Given two 3D vectors u = (ux , uy , uz ) and v = (vx , vy , vz ), their vector product is given by: ⎛ uy vz − uz vy u×v=⎜ ⎜ uz vx − ux vz ⎝ ux vy − uy vx
⎞ ⎟. ⎟ ⎠
Proof. Optional, see p. 935 in Appendices. (The proof is actually quite simple. It just involves some tedious algebra.)
Page 303, Table of Contents
www.EconsPhDTutor.com
Example 277. If u = (1, 2, 3) and v = (4, 5, 6), then u × v = (2 × 6 − 3 × 5, 3 × 4 − 1 × 6, 1 × 5 − 2 × 4) = (−3, 6, −3) . Let’s verify that u × v is orthogonal to u, by computing (u × v) ⋅ u = (−3, 6, −3) ⋅ (1, 2, 3) = −3 + 12 − 9 = 0 ✓. Similarly, let’s verify that u × v is orthogonal to v, by computing (u × v) ⋅ v = (−3, 6, −3) ⋅ (4, 5, 6) = −12 + 30 − 18 = 0 ✓. Example 278. If u = (−1, 3, −5) and v = (2, −4, 6), then u × v = (3 × 6 − (−5) × (−4), (−5) × 2 − (−1) × 6, (−1) × (−4) − 3 × 2) = (−2, −4, −2) . Let’s verify that u×v is orthogonal to u, by computing (u × v)⋅u = (−2, −4, −2)⋅(−1, 3, −5) = 2 − 12 + 10 = 0 ✓. Similarly, let’s verify that u × v is orthogonal to v, by computing (u × v) ⋅ v = (−2, −4, −2) ⋅ (2, −4, 6) = −4 + 16 − 12 = 0 ✓. As in the 2D case of the vector product, here again in the 3D case, the vector product is anticommutative, i.e. u × v = −v × u (see Exercise 120). Exercise 118. For each of the following pairs of vectors, compute the vector product and verify that it is orthogonal to each of the two vectors. (a) u = (0, 1, 2) and v = (3, 4, 5). (b) u = (−1, −2, −3) and v = (1, 0, 5). (Answer on p. 1085.) Exercise 119. Verify that in general, u × v is orthogonal to u and v by showing that (u × v) ⋅ u = 0 and that (u × v) ⋅ v = 0. (Answer on p. 1085.) Exercise 120. (a) Given u = (1, 2, 3) and v = (4, 5, 6), show that v × u = −u × v. (b) Prove that in general, i.e. the 3D vector product is anti-commutative, i.e. u × v = −v × u. (Answer on p. 1085.)
Ordinary multiplication is associative. This simply means that (a × b) × c = a × (b × c). Example 279. (2 × 3) × 7 = 2 × (3 × 7) and (8 × 13) × 2 = 8 × (13 × 2). In contrast, the vector product is not associative. Example 280. If u = (1, 2, 3), v = (4, 5, 6), and w = (1, 0, 1), then (u × v)×w = (−3, 6, −3)× (1, 0, 1) = (6, 0, −6), but u × (v × w) = (1, 2, 3) × (5, 2, −5) = (−16, 20, −8).
Page 304, Table of Contents
www.EconsPhDTutor.com
31 31.1
Lines
Lines on a 2D Plane: Cartesian to Vector Equations
In general, a line on a 2D plane can be described by the cartesian equation ax + by + c = 0. This says that the line consists of exactly those points (x, y) that satisfy the equation ax + by + c = 0. (You may be more familiar with describing lines in the form y = mx+d. This simply involves a rearrangement of the above equation. But the above equation is preferred because it is more general — it allows for the possibility that the coefficient on y is 0.) Example 281. Consider the line described by the cartesian equation 3x − y + 2 = 0. Rearranging, we get a more familiar-looking equation: y = 3x + 2. For convenience (but at the cost of some sloppiness), we may even simply identify the line with the cartesian equation. Example 282. Consider the line 3x − y + 2 = 0.
Describing lines using cartesian equations is secondary school stuff. We’ll now learn a second method of describing lines — through vector equations. In general, any line can be described in the form r = p + λv, λ ∈ R, where r is a generic point on the line, p is some known point on the line, v is a direction vector of the line, and λ is a parameter that can take any real value. Here are some examples to make sense of this.
Page 305, Table of Contents
www.EconsPhDTutor.com
Example 283. Consider the line (on a 2D plane) described by the cartesian equation 3x − y + 2 = 0. It runs through the point (0, 2). A vector that points in the same direction as this line is (1, 3). Hence, we can also describe it using the vector equation r = (0, 2) + λ(1, 3), λ ∈ R. This says that the line consists of every point r that can be written as (0, 2) + λ(1, 3) for some real number λ. We call λ a parameter. As λ varies, we get different points of the line. So for example, corresponding to λ = 0, 1, and −1, the line contains the points (0, 2) + 0(1, 3) = (0, 2), (0, 2) + 1(1, 3) = (1, 5), and (0, 2)−1(1, 3) = (−1, −1). Of course, it also contains infinitely many other points, one for each value of λ ∈ R. We call (1, 3) a direction vector of the line. Note that this direction vector is not unique, any scalar multiple thereof, i.e. c(1, 3) with c ∈ R, is also a direction vector of the line! Again, we can either say “the line is described by the vector equation r = (0, 2) + λ(1, 3)”. OR, for convenience (but at the cost of some sloppiness), we can also say “the line is the very equation r = (0, 2) + λ(1, 3)”.
y 4
Line
2
The point (0, 2)
Cartesian equation 3x - y + 2 = 0
The vector (1, 3) x 0
-4
-2
Vector equation r = (0, 2) + ɉ(1, 3)
0
2
4
-2
-4 Page 306, Table of Contents
www.EconsPhDTutor.com
Example 284. Consider the line (on a 2D plane) described by the cartesian equation x + y − 1 = 0. It runs through the point (0, 1). A vector that points in the same direction as this line is (1, −1). Hence, we can also describe it using the vector equation r = (0, 1) + λ(1, −1), λ ∈ R. This says that the line consists of every point r that can be written as (0, 1) + λ(1, −1) for some real number λ. Corresponding to λ = 0, 1, and −1, the line contains the points (0, 1) + 0(1, −1) = (0, 1), (0, 1) + 1(1, −1) = (1, 0), and (0, 1)−1(1, −1) = (−1, 2).
Line
Cartesian equation x+y-1=0
y 4
2 The point (0, 1) x 0
-4
-2
0
-2
2
4
Vector equation r = (0, 1) + ɉ(1, -1)
The vector (1, -1)
-4
Page 307, Table of Contents
www.EconsPhDTutor.com
Example 285. Consider the line (on a 2D plane) described by the cartesian equation y − 3 = 0. It runs through the point (0, 3). A vector that points in the same direction as this line is (1, 0). Hence, we can also describe it using the vector equation r = (0, 3) + λ(1, 0), λ ∈ R. This says that the line consists of every point r that can be written as (0, 3) + λ(1, 0) for some real number λ.Corresponding to λ = 0, 1, and −1, the line contains the points (0, 3) + 0(1, 0) = (0, 3), (0, 3) + 1(1, 0) = (1, 3), and (0, 3)−1(1, 0) = (−1, 3).
y 4
Vector equation r = (0, 3) + ɉ(1, 0)
Line Cartesian equation y-3=0
The point (0, 3) 2
x 0 -4
-2
0
2
4
The vector (1, 0) -2
-4
Exercise 121. Rewrite each of the following lines into vector equation form. (a) −5x+y+1 = 0. (b) x − 2y − 1 = 0. (c) y − 4 = 0. (d) x − 4 = 0. (Answer on p. 1086.)
Page 308, Table of Contents
www.EconsPhDTutor.com
We just learnt how to describe a line using the vector equation 1
r = p + λv, λ ∈ R. Notice that the LHS of this equation is the generic Point r. And the RHS of the equation is the Point p minus the Vector λv, which equals Point (see p. 270). So LHS and RHS do indeed match up. There is another way to describe a line using a vector equation. We can instead write: 2
r = p + λv, λ ∈ R, where now r is the position vector of a generic point r on the line and p is the position vector of some known point p on the line. So now LHS is a Vector and so too is RHS. 1
Equation = said that the line consists of those points r that could be written as p + λv. In 2 contrast, equation = says that the line consists of those points whose position vector r can be written as p + λv. But both equations can equally well describe the very same line. The difference is a fine and pedantic one and really doesn’t matter much.
What matters is that you take care not to write either 3
r = p + λv, λ ∈ R; WRONG! 4
or r = p + λv, λ ∈ R. WRONG! 3
3
3
The LHS of = is a Point while the RHS of = is a Vector. Therefore = cannot possibly be true. 4
4
4
The LHS of = is a Vector while the RHS of = is a Point. Therefore = cannot possibly be true. As usual, this is all very pedantic, but can serve as a useful test of your understanding.
Page 309, Table of Contents
www.EconsPhDTutor.com
31.2
Lines on a 2D Plane: Vector to Cartesian Equations
In the previous section, given the cartesian equation of a line, we worked out its vector equation. Now given its vector equation, we’ll work out its cartesian equation. Suppose a line (on a 2D plane) can be described by the vector equation r = p + λv = (p1 , p2 ) + λ(v1 , v2 ). where λ ∈ R and v is a non-zero vector.37 And so any point (x, y) on this line must satisfy x = p1 + λv1
and y = p2 + λv2 .
The above are the cartesian equations for a line (on a 2D plane)! But wait a minute ... isn’t there supposed to be just one equation? Well, if we’d like, we can quite easily combine them into a single equation by eliminating the parameter λ. In general:
Fact 34. The line with vector equation r = (p1 , p2 ) + λ(v1 , v2 ) (for λ ∈ R) is the line with cartesian equations as given by the 3 cases below. (1)
x − p1 y − p2 = , v1 v2
if v1 , v2 ≠ 0;
(2) x = p1 , y is free,
if v1 = 0, v2 ≠ 0;
(3) x is free, y = p2 ,
if v1 ≠ 0, v2 = 0;
Note that Case (1) is the most common situation. Proof. Optional, see p. 936 in the Appendices.
Some examples:
37
Otherwise we’d simply be describing the single point p!
Page 310, Table of Contents
www.EconsPhDTutor.com
Example 286. The line described by the vector equation r = (1, 2) + λ(1, 1), where λ ∈ R has cartesian equations x = 1 + λ and y = 2 + λ. As λ varies between −∞ and ∞, this pair of equations gives us the points that are on the line. For example, when λ = 1, 17, 33, we have the points (2, 3), (18, 20), and (34, 36). We can eliminate λ and reduce the above pair of equations into the single cartesian equation y = x + 1 or y−1 x = . 1 1 Example 287. Consider the line described by the vector equation r = (0, 0)+λ(4, 5), where λ ∈ R has cartesian equations x = 4λ and y = 5λ. As λ varies between −∞ and ∞, this pair of equations gives us the points that are on the line. For example, when λ = 1, 17, 33, we have the points (4, 5), (68, 85), and (132, 165). Eliminating λ, we can reduce the above pair of equations to y = 1.25x or x y = . 1.25 1 Example 288. Consider the line described by the vector equation r = (3, 1)+λ(0, 2), where λ ∈ R has cartesian equations x = 3 and y = 1 + 2λ. As λ varies between −∞ and ∞, this pair of equations gives us the points that are on the line. So in fact, the above equations say that x must always be 3 and y is free to vary along with λ. For example, when λ = 1, 17, 33, we have the points (3, 3), (3, 25), and (3, 67). Hence, the above pair of equations can be reduced to x = 3. Exercise 122. Rewrite each of the following lines into cartesian equation form. (a) r = (−1, 1) + λ(3, −2), where λ ∈ R. (b) r = (5, 6) + λ(7, 8), where λ ∈ R. (c) r = (0, −3) + λ(3, 0), where λ ∈ R. (Answer on p. 1086.)
Page 311, Table of Contents
www.EconsPhDTutor.com
31.3
Lines in 3D Space: Vector Equations
Lines in 2D space are described by the cartesian equation ax + by + c = 0. A reasonable guess might be that lines in 3D space are analogously described by the cartesian equation ax + by + cz + d = 0. Turns out this is wrong! The equation ax + by + cz + d = 0 actually describes a plane, as we’ll see later (Chapter 32). In the 3D case, it’s easier to start by looking at the vector equation of a line. It turns out to be exactly analogous to the 2D case. It can be written as r = a + λv, where λ ∈ R and v is a non-zero 3D vector. This vector equation says that the line contains every point r can be expressed as (a1 , a2 , a3 )+ λ(v1 , v2 , v3 ), where λ ∈ R is a parameter. Example 289. Consider the line described by the vector equation r = (1, 2, 3) + λ(0, 1, 1), where λ ∈ R. Corresponding to λ = 0, 1, and −1, the line contains the points (1, 2, 3), (1, 3, 4), and (1, 1, 2).
y 3 (1, 2, 3)
2 1 x 1
(1, 3, 4)
1 2 3 (1, 1, 2)
z
Line r = (1, 2, 3) + ɉ(0, 1, 1)
4
Page 312, Table of Contents
www.EconsPhDTutor.com
Example 290. Consider the line described by the vector equation r = (0, 0, 0) + λ(1, 0, 0), where λ ∈ R. Corresponding to λ = 0, 1, and −1, the line contains the points (0, 0, 0), (1, 0, 0), and (−1, 0, 0).
y 2 (-1, 0, 0) 1
Line r = (0, 0, 0) + ɉ(1, 0, 0) x 1 (1, 0, 0)
z
1 2 (0, 0, 0)
Page 313, Table of Contents
www.EconsPhDTutor.com
31.4
Lines in 3D Space: Vector to and from Cartesian Equations
We now try to work out the cartesian equation of a line in 3D space. Suppose a line can be described by the vector equation r = p + λv = (p1 , p2 , p3 ) + λ(v1 , v2 , v3 ). where λ ∈ R and v is a non-zero vector.38 And so any point (x, y, z) on this line must satisfy x = p1 + λv1 , y = p2 + λv2 , and z = p3 + λv3 . The above are the cartesian equations for a line (in 3D space)! These are exactly analogous to the cartesian equations (p. 31.2) in the 2D case. Unlike in the 2D case, it is generally impossible to reduce these equations into a single cartesian equation. However, we can reduce them into two equations. Fact 35. The line with vector equation r = (p1 , p2 , p3 ) + λ(v1 , v2 , v3 ) where λ ∈ R is the line with cartesian equations as given by the 7 cases below.
(1)
x − p 1 y − p2 z − p3 = = v1 v2 v3
if v1 , v2 , v3 ≠ 0;
(2) x = p1 ,
y − p2 z − p3 = , v2 v3
if v1 = 0, v2 , v3 ≠ 0;
(3) y = p2 ,
x − p1 z − p 3 = , v1 v3
if v2 = 0, v1 , v3 ≠ 0;
(4) z = p3 ,
x − p1 y − p2 = , v1 v2
if v3 = 0, v1 , v2 ≠ 0;
(5) x = p1 , y = p2 , z is free,
if v1 , v2 = 0, v3 ≠ 0;
(6) x = p1 , z = p3 , y is free,
if v1 , v3 = 0, v2 ≠ 0;
(most common case)
(7) y = p2 , z = p3 , x is free, if v2 , v3 = 0, v1 ≠ 0. Proof. Optional, see p. 937 in the Appendices.
38
If v is a zero vector, then we are simply describing the single point p!
Page 314, Table of Contents
www.EconsPhDTutor.com
The first two examples are where v1 , v2 , and v3 are non-zero (Case 1 of Fact 35). Example 291. Consider the line described by the vector equation r = (1, 2, 3) + λ(4, 5, 6), where λ ∈ R. It can be described by the cartesian equations x = 1 + 4λ, y = 2 + 5λ, and z = 3 + 6λ. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (5, 7, 9), (13, 17, 21), and (69, 87, 105). By rearranging each equation so that λ is on one side, we can reduce these three equations to just two: x−1 y−2 z−3 = = . 4 5 6 That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations.
Example 292. Consider the line described by the vector equation r = (0, 0, 0) + λ(2, 3, 5), where λ ∈ R. It can be described by the cartesian equations x = 2λ, y = 3λ, and z = 5λ. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (2, 3, 5), (6, 9, 15), and (34, 51, 85). By rearranging each equation so that λ is on one side, we can reduce these three equations to just two: x y z = = . 2 3 5 That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations.
Page 315, Table of Contents
www.EconsPhDTutor.com
We now look at examples where exactly one of v1 , v2 , or v3 is non-zero (Cases 2, 3, and 4 of Fact 35).
In the case where v1 = 0 (but v2 ≠ 0 and v3 ≠ 0), then this is a line that is on the 2D yz plane where x = p1 . Example 293. Consider the line described by the vector equation r = (1, 2, 3) + λ(0, 5, 6), where λ ∈ R. It can be described by the cartesian equations x = 1, y = 2 + 5λ, and z = 3 + 6λ. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (1, 7, 9), (1, 17, 21), and (1, 87, 102). We see that x must always be equal to 1. By rearranging the second and third equations so that λ is on one side, we can reduce these three equations to just two: y−2 z−3 = . x = 1, 5 6 That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations.
Similarly, in the case where v2 = 0 (but v1 ≠ 0 and v3 ≠ 0), then this is a line that is on the 2D xz plane where y = p2 . Example 294. Consider the line described by the vector equation r = (1, 2, 3) + λ(4, 0, 6), where λ ∈ R. It can be described by the cartesian equations x = 1 + 4λ, y = 2, and z = 3 + 6λ. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (5, 2, 9), (13, 2, 21), and (69, 2, 105). We see that y must always be equal to 2. By rearranging the first and third equations so that λ is on one side, we can reduce these three equations to just two: x−1 z−3 = . y = 2, 4 6 That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations.
Page 316, Table of Contents
www.EconsPhDTutor.com
Finally, in the case where v3 = 0 (but v2 ≠ 0 and v3 ≠ 0), then this is a line that is on the 2D xy plane where z = p3 . Example 295. Consider the line described by the vector equation r = (1, 2, 3) + λ(4, 5, 0), where λ ∈ R. It can be described by the cartesian equations x = 1 + 4λ, y = 2 + 5λ, and z = 3. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (5, 7, 3), (13, 17, 3), and (69, 87, 3). We see that z must always be equal to 3. By rearranging the first and second equations so that λ is on one side, we can reduce these three equations to just two: x−1 y−2 = . z = 3, 4 5 That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations.
We now look at examples where exactly two of v1 , v2 , or v3 are zero (Cases 5, 6, and 7 of Fact 35). In the case where v1 = 0 and v2 = 0, but v3 ≠ 0, then this is a line that runs through the points (p1 , p2 , λ) for λ ∈ R. Example 296. Consider the line described by the vector equation r = (1, 2, 3) + λ(0, 0, 6), where λ ∈ R. It can be described by the cartesian equations x = 1, y = 2, and z = 3 + 6λ. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (1, 2, 9), (1, 2, 21), and (1, 2, 105). We see that x and y must always be equal to 1 and 2. Hence, the above equations simply reduce to: x = 1,
y = 2.
That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations. These are the points (1, 2, λ), where λ can be any real.
Page 317, Table of Contents
www.EconsPhDTutor.com
Similarly, in the case where v1 = 0 and v3 = 0, but v2 ≠ 0, then this is a line that runs through the points (p1 , λ, p3 ) for λ ∈ R. Example 297. Consider the line described by the vector equation r = (1, 2, 3) + λ(0, 5, 0), where λ ∈ R. It can be described by the cartesian equations x = 1, y = 2 + 5λ, and z = 3. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (1, 7, 3), (1, 17, 3), and (1, 87, 3). We see that x and z must always be equal to 1 and 3. Hence, the above equations simply reduce to: x = 1,
z = 3.
That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations. These are the points (1, λ, 3), where λ can be any real. In the case where v2 = 0 and v3 = 0, but v1 ≠ 0, then this is a line that runs through the points (λ, p2 , p3 ) for λ ∈ R. Example 298. Consider the line described by the vector equation r = (1, 2, 3) + λ(4, 0, 0), where λ ∈ R. It can be described by the cartesian equations x = 1 + 4λ, y = 2, and z = 3. As λ varies between −∞ and ∞, these 3 equations give us the points that are on the line. For example, when λ = 1, 3, 17, we have the points (5, 2, 3), (13, 2, 3), and (69, 2, 3). We see that y and z must always be equal to 2 and 3. Hence, the above equations simply reduce to: y = 2,
z = 3.
That is, this is the line that contains the points (x, y, z) which satisfy the above cartesian equations. These are the points (λ, 2, 3), where λ can be any real.
Exercise 123. Rewrite each of the following vector equation descriptions of lines into cartesian equations describing the same line. (a) r = (−1, 1, 1) + λ(3, −2, 1), where λ ∈ R. (b) r = (5, 6, 1) + λ(7, 8, 1), where λ ∈ R. (c) r = (0, −3, 1) + λ(3, 0, 1), where λ ∈ R. (d) r = (9, 9, 9) + λ(1, 0, 0), where λ ∈ R. (Answer on p. 1087.)
Page 318, Table of Contents
www.EconsPhDTutor.com
SYLLABUS ALERT The 9740 (old) List of Formulae contains the following statement, but not the 9758 (revised) List of Formulae! “If A is the point with position vector a = a1 i + a2 j + a3 k and the direction vector b is given by b = b1 i + b2 j + b3 k, then the straight line through A with direction vector b has cartesian equation x − a1 y − a2 z − a3 = = (= λ).” b1 b2 b3
By the way, the above statement printed in the 9740 (old) List of Formulae is false (*gasp*), because it fails to specify that b1 , b2 , b3 must be non-zero. (The correct statement was just given as Fact 35.) Consider for example, the point (0, 0, 0) and the direction vector b given by b = j + k. Then contrary to the above statement, the straight line through A with direction vector b does not have cartesian equation x y z = = , 0 1 1 x is undefined. This is the common mistake to which I devoted an entire chapter 0 (Chapter 2) earlier in this book. This seems like a very pedantic point to make, but dividing by zero has been the cause of the downfall of many a student (and in this case some folks at MOE). because
Page 319, Table of Contents
www.EconsPhDTutor.com
In the examples and exercise we just went through, we started with a vector equation of a line and then wrote down the line’s cartesian equations. Now we’ll go the other way round, starting with the line’s cartesian equations, then write down the vector equations. Example 299. Consider a line described by the cartesian equations 3x − 4 2y − 18 z − 1 = = . 6 5 3 In order to directly apply Fact 35, you must make sure that the coefficients on x, y, and z are all 1! So first rewrite the above into x − 4/3 y − 9 z − 1 = = . 2 2.5 3 And now by Fact 35, we can immediately describe this line by the vector equation r = (4/3, 9, 1) + λ(2, 2.5, 3), for λ ∈ R. Example 300. Consider a line described by the cartesian equations 5x y − 13 3z − 14 = = . 2 6 8 Rewrite the above equations into x y − 13 z − 14/3 = = 8 . 2/5 6 /3 And so by Fact 35, we can immediately describe this line by the vector equation r = (0, 13, 14/3) + λ(2/5, 6, 8/3), for λ ∈ R. Example 301. Consider a line described by the cartesian equations 2x = 17 and 3z = 4. Rewrite them into x = 8.5 and z = 4/3. And so by Fact 35, we can immediately describe this line by the vector equation r = (8.5, 0, 4/3) + λ(0, 1, 0), for λ ∈ R.
Exercise 124. Rewrite each of the following cartesian equation descriptions of lines into 7x − 2 0.3y − 5 8z a vector equation describing the same line. (a) = = . (b) 2x = 3y = 5z. (c) 5 7 7 3y − 1 x − 3 5z − 2 17x − 4 = = 3z. (d) = , 3y = 11. (Answer on p. 1088.) 2 2 7
Page 320, Table of Contents
www.EconsPhDTutor.com
31.5
Collinearity
Definition 82. A set of points are said to be collinear if there is some line that contains all of these points. Any two points are always collinear â&#x20AC;&#x201D; simply take the line that passes through both of them. In contrast, three points may not be collinear. To check whether three points are collinear, 1. First take the line that passes through two of the points. 2. Then check whether the third point is on this line.
c
b
c
a a
a, b, and c are not collinear.
a, b, and c are collinear.
b
Page 321, Table of Contents
www.EconsPhDTutor.com
Example 302. Are the points a = (1, 2, 3), b = (4, 5, 6), and c = (7, 8, 9) collinear? First take the line through a and b. The vector from a to b is (3, 3, 3) and the line passes through a. Hence, the line can be written as r = (1, 2, 3) + λ(3, 3, 3) (λ ∈ R). Then check whether c is on the line: Is there λ such that c = (7, 8, 9) = (1, 2, 3) + λ(3, 3, 3)? Rearranging, we have (6, 6, 6) = λ(3, 3, 3), which we can write out as: 6 = 3λ,
6 = 3λ.
6= 3λ,
Clearly, all three of the above equations are true if λ = 2. And so c is also on the line. Hence, the three points are collinear.
a, b, and c are collinear.
d, e, and f are not collinear. f = (0, 0, 1)
c = (7, 8, 9)
= (-1, 1, 0)
= (3, 3, 3)
e = (0, 1, 0)
b = (4, 5, 6) a = (1, 2, 3)
d = (1, 0, 0)
Example 303. Are the points d = (1, 0, 0), e = (0, 1, 0), and f = (0, 0, 1) collinear? First take the line through d and e . The vector from d to e is (−1, 1, 0) and the line passes through d. Hence, the line can be written as r = (1, 0, 0) + λ(−1, 1, 0) (λ ∈ R). Then check whether f is on the line: Is there λ such that f = (0, 0, 1) = (1, 0, 0)+λ(−1, 1, 0)? Rearranging, we have (−1, 0, 1) = λ(−1, 1, 0), which we can write out as: −1 = −λ,
0= λ,
1 = 0.
Clearly, there is no λ such that the above three equations can be true. And so the point f is not on the line through d and e. Hence, the three points are not collinear.
Exercise 125. Determine whether each of the following set of three points are collinear. (a) a = (3, 1, 2), b = (1, 6, 5), and c = (0, −1, 0). (b) a = (1, 2, 4), b = (0, 0, 1), and c = (3, 6, 10). (Answer on p. 1088.)
Page 322, Table of Contents
www.EconsPhDTutor.com
32
Planes
Informally, a plane is a flat 2D surface in 3D space. How do we describe it using equations? Here are two very useful clues: 1. u ⊥ v ⇐⇒ u ⋅ v = 0. In words: Two vectors are orthogonal if and only if their scalar product is 0. 2. Since the plane is a flat surface, there must be some vector n that is orthogonal (perpendicular) to this plane. That is, n is orthogonal to every vector on the plane. We call n the plane’s normal vector (hence the use of the letter n). Is the normal vector unique? No, because any other vector cn (where c is any scalar) serves equally well as a normal vector. In the figure below, n is a normal vector to the illustrated plane. So too is 0.5n. And so too is −n. But otherwise, besides cn, there are no other vectors that are also orthogonal to the plane. That is, any vector that cannot be written in the form cn is not orthogonal to the plane.
Black vectors Plane are on the plane n (a normal 0.5n (Also a vector) normal vector) -n (Also a normal vector)
Page 323, Table of Contents
www.EconsPhDTutor.com
Suppose a plane contains some point p = (p1 , p2 , p3 ) and has a normal vector n = (a, b, c). Now consider any point r on the plane. We can construct the vector r − p. This vector r − p lies on the plane and must therefore be orthogonal to n, the plane’s normal vector. So, for any point r that lies on the plane, we have (r − p) ⋅ n = 0.
q (point not on the plane)
n is normal to r – p , but not to q – p.
q – p (vector not on the plane) n (a normal vector)
p (point on the plane) r – p (vector on the plane)
Plane r1 (point on the plane)
Now consider any point q that is not on the plane. We can construct the vector q − p. This vector q − p does not lie on the plane and must therefore not be orthogonal to n, the plane’s normal vector. So, for any point q that does not lie on the plane, we have (q − p) ⋅ n ≠ 0. Altogether then, we conclude: A point r is on the plane if and only if Fact 36. Suppose a plane contains point p and has normal vector n. Then the plane contains exactly those points r such that (r − p) ⋅ n = 0.
Page 324, Table of Contents
www.EconsPhDTutor.com
Recall that a line can be described by the vector equation r = p + λv where r is a generic point on the line and p is a known point on the line. Alternatively, it can also be described by the vector equation r = p + λv, where r is the position vector of r and p is the position vector of p. On the previous page, we proved that if a plane that contains point p and has normal vector n, then it may be described by the vector equation (r − p) ⋅ n = 0, where r is a generic point on the plane. Similar to a line, the same plane can also be described by the vector equation (r − p) ⋅ n = 0. By the distributivity of the scalar product, we have: (r − p) ⋅ n = 0 ⇐⇒ r ⋅ n − p ⋅ n = 0 ⇐⇒ r ⋅ n = p ⋅ n. Now, p is known (it is the position vector of a point p known to be on the plane). So too is n (it is the plane’s normal vector). Thus, p ⋅ n is simply some known number. So we can describe the plane even more simply by the vector equation
r ⋅ n = d, where d = p ⋅ n.
Page 325, Table of Contents
www.EconsPhDTutor.com
Example 304. Consider a plane that contains the point p = (1, 2, 3) and has normal vector (1, 1, 0). We compute ⎛1⎞ ⎛1 ⎟ ⎜ d=p⋅n=⎜ ⎜ 2 ⎟⋅⎜ 1 ⎝3⎠ ⎝0
⎞ ⎟ = 1 × 1 + 2 × 1 + 3 × 0 = 3. ⎟ ⎠
We thus conclude that the plane may be described by the vector equation r ⋅ (1, 1, 0) = 3. This says that the plane contains exactly every point r, whose position vector r satisfies the above equation. For example, the points r1 = (3, 0, 0), r2 = (0, 3, 5), and r3 = (1, 2, −1) are on the plane, because their position vectors r1 = (3, 0, 0), r2 = (0, 3, 5), and r3 = (1, 2, −1) satisfy the above equation, as we can easily verify: ⎛3⎞ ⎛1 ⎟ ⎜ r1 ⋅ n = ⎜ ⎜ 0 ⎟⋅⎜ 1 ⎝0⎠ ⎝0
⎞ ⎟ = 3 × 1 + 0 × 1 + 0 × 0 = 3. ⎟ ⎠
⎛0⎞ ⎛1 ⎟ ⎜ r2 ⋅ n = ⎜ ⎜ 3 ⎟⋅⎜ 1 ⎝5⎠ ⎝0
⎞ ⎟ = 0 × 1 + 3 × 1 + 5 × 0 = 3. ⎟ ⎠
⎛ 1 ⎞ ⎛1 ⎟ ⎜ r3 ⋅ n = ⎜ ⎜ 2 ⎟⋅⎜ 1 ⎝ −1 ⎠ ⎝ 0
⎞ ⎟ = 1 × 1 + 2 × 1 + (−1) × 0 = 3. ⎟ ⎠
Lest you be sceptical that a plane could be described so simply, let’s verify that two vectors on the plane are indeed orthogonal to the normal vector n. First consider r2 −r1 = (0, 3, 5)− (3, 0, 0) = (−3, 3, 5) — this is a vector on the plane. We can verify that indeed ⎛ −3 ⎞ ⎛ 1 ⎟ ⎜ (r2 − r1 ) ⋅ n = ⎜ ⎜ 3 ⎟⋅⎜ 1 ⎝ 5 ⎠ ⎝0
⎞ ⎟ = −3 × 1 + 3 × 1 + 5 × 0 = 0. ⎟ ⎠
Next consider p − r3 = (1, 2, 3) − (1, 2, −1) = (0, 0, 4) — this is also a vector on the plane. We can verify that indeed ⎛0⎞ ⎛1 ⎟ ⎜ (p − r3 ) ⋅ n = ⎜ ⎜ 0 ⎟⋅⎜ 1 ⎝4⎠ ⎝0 Page 326, Table of Contents
⎞ ⎟ = 0 × 1 + 0 × 1 + 4 × 0 = 0. ⎟ ⎠ www.EconsPhDTutor.com
Example 305. Consider a plane that contains the point p = (0, 0, 1) and has normal vector (2, −1, 1). We compute ⎛0⎞ ⎛ 2 ⎟ ⎜ d=p⋅n=⎜ ⎜ 0 ⎟ ⋅ ⎜ −1 ⎝1⎠ ⎝ 1
⎞ ⎟ = 0 × 2 + 0 × (−1) + 1 × 1 = 1. ⎟ ⎠
We thus conclude that the plane may be described by the vector equation r ⋅ (2, −1, 1) = 1. This says that the plane contains exactly every point r, whose position vector r satisfies the above equation. For example, the points r1 = (1, 1, 0), r2 = (0, 1, 2), and r3 = (1, 2, 1) are on the plane, because their position vectors r1 = (1, 1, 0), r2 = (0, 1, 2), and r3 = (1, 2, 1) satisfy the above equation, as we can easily verify: ⎛1⎞ ⎛ 2 ⎟ ⎜ r1 ⋅ n = ⎜ ⎜ 1 ⎟ ⋅ ⎜ −1 ⎝0⎠ ⎝ 1
⎞ ⎟ = 1 × 2 + 1 × (−1) + 0 × 1 = 1. ⎟ ⎠
⎛0⎞ ⎛ 2 ⎟ ⎜ r2 ⋅ n = ⎜ ⎜ 1 ⎟ ⋅ ⎜ −1 ⎝2⎠ ⎝ 1
⎞ ⎟ = 0 × 2 + 1 × (−1) + 2 × 1 = 1. ⎟ ⎠
⎛1⎞ ⎛ 2 ⎟ ⎜ r3 ⋅ n = ⎜ ⎜ 2 ⎟ ⋅ ⎜ −1 ⎝1⎠ ⎝ 1
⎞ ⎟ = 1 × 2 + 2 × (−1) + 1 × 1 = 1. ⎟ ⎠
Lest you be sceptical that a plane could be described so simply, let’s verify that two vectors on the plane are indeed orthogonal to the normal vector n. First consider r2 −r1 = (0, 1, 2)− (1, 1, 0) = (−1, 0, 2) — this is a vector on the plane. We can verify that indeed ⎛ −1 ⎞ ⎛ 2 ⎟ ⎜ (r2 − r1 ) ⋅ n = ⎜ ⎜ 0 ⎟ ⋅ ⎜ −1 ⎝ 2 ⎠ ⎝ 1
⎞ ⎟ = (−1) × 2 + 0 × (−1) + 2 × 1 = 0. ⎟ ⎠
Next consider p − r3 = (0, 0, 1) − (1, 2, 1) = (−1, −2, 0) — this is also a vector on the plane. We can verify that indeed ⎛ −1 ⎞ ⎛ 2 ⎟ ⎜ (p − r3 ) ⋅ n = ⎜ ⎜ −2 ⎟ ⋅ ⎜ −1 ⎝ 0 ⎠ ⎝ 1 Page 327, Table of Contents
⎞ ⎟ = (−1) × 2 + (−2) × (−1) + 0 × 1 = 0. ⎟ ⎠ www.EconsPhDTutor.com
We just learnt how to write down the vector equation of a plane, given a point on the plane and its normal vector. We now do the same, but given instead three points on a plane. Example 306. A plane contains the points a = (1, 2, 3), b = (4, 5, 8), and c = (2, 3, 5). Ð → → = (1, 1, 2) are on the plane. Hence, a normal vector to the Both vectors ab = (3, 3, 5) and Ð ac Ð → → plane is ab × Ð ac = n = (1, −1, 0). Since a ⋅ n = −1, the plane can be described by the vector equation r ⋅ (1, −1, 0) = −1. Example 307. A plane contains the points a = (1, 0, 0), b = (0, 1, 0), and c = (0, 0, 1). Ð → → = (−1, 0, 1) are on the plane. Hence, a normal vector to Both vectors ab = (−1, 1, 0) and Ð ac Ð → → the plane is ab × Ð ac = n = (1, 1, 1). Since a ⋅ n = 1, the plane can be described by the vector equation r ⋅ (1, 1, 1) = 1.
Page 328, Table of Contents
www.EconsPhDTutor.com
We jnow write down the vector equation of a plane, given two points and a vector on the plane. Example 308. A plane contains the points a = (0, 0, 3) and b = (1, 4, 5), and the vector v = (3, 2, 1). Ð → Both vectors ab = (1, 4, 2) and v = (3, 2, 1) are on the plane. Hence, a normal vector to the Ð → plane is ab × v = n = (0, 5, −10). Since a ⋅ n = −30, the plane can be described by the vector equation r ⋅ (0, 5, −10) = −30. Example 309. A plane contains the points a = (8, −2, 0) and b = (3, 6, 9), and the vector v = (0, 1, 1). Ð → Both vectors ab = (−5, 8, 9) and v = (0, 1, 1) are on the plane. Hence, a normal vector to Ð → the plane is ab × v = n = (−1, 5, −5). Since a ⋅ n = −18, the plane can be described by the vector equation r ⋅ (−1, 5, −5) = −18.
Page 329, Table of Contents
www.EconsPhDTutor.com
32.1
Planes: Vector to Cartesian Equations
Let n = (a, b, c) be the normal vector of a plane. Let p = (p1 , p2 , p3 ) be a point on the plane. Then the plane can be described by the vector equation r ⋅ n = p ⋅ n, where r = (x, y, z) is the position vector of a generic point on the plane. Writing out the vectors in the above equation explicitly, we have: ⎛x⎞ ⎛a ⎜ y ⎟⋅⎜ b ⎜ ⎟ ⎜ ⎝z ⎠ ⎝ c or
⎞ ⎛ p1 ⎞ ⎛ a ⎟=⎜ p ⎟⋅⎜ b ⎟ ⎜ 2⎟ ⎜ ⎠ ⎝ p3 ⎠ ⎝ c
⎞ ⎟, ⎟ ⎠
ax + by + cz = ap1 + bp2 + cp3 .
This last equation is the cartesian equation description of the same plane. Note, once again, that d = ap1 + bp2 + cp3 is simply some known number. So this cartesian equation simply says that the plane contains exactly those points (x, y, z) that satisfy the equation ax + by + cz = d. Example 310. The plane with vector equation r ⋅ (1, 1, 0) = 3 has cartesian equation x + y = 3. Example 311. The plane with vector equation r ⋅ (2, −1, 1) = 1 has cartesian equation 2x − y + z = 1. Example 312. The plane with vector equation r ⋅ (1, −1, 0) = −1 has cartesian equation x − y = −1. Example 313. The plane with vector equation r ⋅ (1, 1, 1) = 1 has cartesian equation x + y + z = 1. Example 314. The plane with vector equation r ⋅ (0, 5, −10) = −30 has cartesian equation 5y − 10z = −30. Example 315. The plane with vector equation r ⋅ (−1, 5, −5) = −18 has cartesian equation −x + 5y − 5z = −18.
Page 330, Table of Contents
www.EconsPhDTutor.com
It’s thus surprisingly easy to go back and forth between a plane’s vector and cartesian equations: r ⋅ (a, b, c) = d
⇐⇒
ax + by + cz = d.
Example 316. Given a plane with cartesian equation 2x + 3y + 5z = −7, we immediately know that it has vector equation r ⋅ (2, 3, 5) = −7. Example 317. Given a plane with cartesian equation 2x + 3z = −5, we immediately know that it has vector equation r ⋅ (2, 0, 3) = −5.
Here’s a nice observation: Every plane that contains the origin (0, 0, 0) can be written in the form ax + by + cz = 0. Conversely, every plane that does not contain the origin can be written in the form ax + by + cz = 1. Formally: Fact 37. A plane r ⋅ n = d contains the origin ⇐⇒ d = 0. Proof. Given a plane r ⋅ n = d, the origin is on the plane (and thus satisfies this equation) ⇐⇒ 0 ⋅ n = d = 0.
SYLLABUS ALERT The 9740 (old) List of Formulae contains the following statement, but not the 9758 (revised) List of Formulae! “The plane through A with normal vector n = n1 i + n2 j + n3 k has cartesian equation n1 x + n2 y + n3 z + d = 0
where
d= −a ⋅ n.”
Exercise 126. Find the vector and cartesian equations that describe the planes containing each of the following set of three points: (a) a = (7, 3, 4), b = (8, 3, 4), and c = (9, 3, 7). (b) a = (8, 0, 2), b = (4, 4, 3), and c = (2, 7, 2). (c) a = (8, 5, 9), b = (8, 4, 5), and c = (5, 6, 0). (Answer on p. 1089.)
Exercise 127. Write down the vector equations of the planes whose cartesian equations are as given: (a) 3x + 2y + 5z = −3. (b) 2y + 5z = −3. (c) 5z = −3. (Answer on p. 1090.)
Page 331, Table of Contents
www.EconsPhDTutor.com
32.2
Planes: Hessian Normal Form
Example 318. The plane with vector equation r ⋅ (1, 0, 1) = 11 or cartesian equation x + z = 11 can be described in an infinite number of ways. For example, the same plane can also be described by any of the following four equations: r ⋅ (2, 0, 2) = 22, r ⋅ (1/11, 0, 1/11) = 1, 2x + 2z = 22, and x/11 + z/11 = 1. If you talk about the plane r ⋅ (2, 0, 2) = 22 and I talk about the plane x/11x + z/11z = 1, it make take us a moment to realise that we are talking about the exact same plane. To save ourselves such trouble, it may be desirable to describe planes in a standardised form, called the Hessian normal form. ˆ as our normal vector. However, This involves simply picking the unit normal vector n there are two possible unit normal vectors, one pointing “up” and the other pointing “down”. ˆ ≥ 0, so that the RHS of our We will choose the unit normal vector that ensures that p ⋅ n vector or cartesian equation in Hessian normal form is always non-negative. ̂ Example 319. Consider the plane r ⋅ (1, 0, 1) = 11 or x + z = 11. We have (1, 0, 1) = √ √ ( 2/2, 0, 2/2). And so the plane can be rewritten in Hessian normal form as r ⋅ √ √ √ √ √ √ ( 2/2, 0, 2/2) = 11 2/2 or ( 2/2) x + ( 2/2) z = 11 2/2. ˆ is uniquely defined. Indeed, Notice that in the Hessian normal form, the number dˆ = p ⋅ n it is the distance of the plane from the origin! (We’ll prove this in section 33.2.) ̂ Example 320. Consider the plane r ⋅ (8, 1, 3) = −3 or 8x + y + 3z = −3. We have (8, 1, 3) = √ √ √ (8/ 74, 1// 74, 3// 74). Note though that right now, the RHS is negative. So in order to ensure that dˆ ≥ 0 (as required by the Hessian normal form), we need simply reverse the √ √ √ sign of our unit normal vector — that is, we should pick (−8/ 74, −1/ 74, −3/ 74) as our unit normal then, can be rewritten √ vector. √ Altogether √ √ the plane √ √ in Hessian √ normal form √ as r ⋅ (−8/ 74, −1/ 74, −3/ 74) = 3/ 74 or (−8/ 74) x − (1/ 74) y − (3/ 74) z = (3/ 74).
Exercise 128. Rewrite each of the following planes’ vector equation into Hessian normal form. (a) r ⋅ (3, 6, 2) = 4. (b) r ⋅ (1, 2, 2) = −1. (c) r ⋅ (8, 1, 4) = 0. (Answer on p. 1090.)
Page 332, Table of Contents
www.EconsPhDTutor.com
33
Distances
Before we proceed, here are some useful things to remember. A line can be fully determined by 1. Any two distinct points. 2. Any vector and a point. Similarly, a plane can be fully determined by 1. Any three distinct points. 2. Any two distinct points and a distinct vector.39 3. Two distinct vectors and a point.
39
Ð → If the two points are a and b, then the vector must be distinct from cab, for any c ∈ R.
Page 333, Table of Contents
www.EconsPhDTutor.com
33.1
Distance of a Point from a Line
Definition 83. The foot of the perpendicular from a point a to a line l is the point b on the line l that is closest to the point a. The distance between the point a and the line l is the length of the line segment ab.
Distance between a and b
a
b p
Note that the line ab must be perpendicular to the line l. Hence the name foot of the perpendicular.
It is easier to remember how the proof of the following proposition works, than to try to memorise the proposition itself: Proposition 8. Given a point a and a line r = p +√ λv (for λ ∈ R), 2 → 2 − (Ð →⋅v ˆ ) ; and (a) The distance between the point and the line is ∣Ð pa∣ pa
→⋅v ˆ) v ˆ. (b) The foot of the perpendicular from the point to the line is the point p + (Ð pa
Proof. Let b be the foot of the perpendicular from the point to the line. (a) Pick any known point on the line — here the obvious choice is p. Consider the right→ and base of length ∣Ð →⋅v ˆ ∣ (refer angled triangle △bpa — it has hypothenuse of length ∣Ð pa∣ pa Page 334, Table of Contents
www.EconsPhDTutor.com
to the diagram above). Hence, by the Pythagorean Theorem, the length of line segment ab (or the distance between the point a and the line l) is: √ 2 → 2 − (Ð →⋅v ˆ ) , as desired. ∣Ð pa∣ pa Ð → →⋅v ˆ ∣ away from the point p, heading in the direction pb. pa (b) The point b is a distance ∣Ð ̂ Ð → ∗ →⋅v ˆ ∣ pb. There are two possible cases to examine. pa Hence b = p + ∣Ð Ð → ˆ is pointing in the same direction as pb. Case #1 : v ̂ Ð → ∗ →⋅v →⋅v →⋅v ˆ and Ð ˆ > 0, so that ∣Ð ˆ∣ = Ð ˆ . Altogether then, = becomes b = Then pb = v pa pa pa →⋅v →⋅v ˆ∣ v ˆ = p + (Ð ˆ) v ˆ , as desired. ✓ p + ∣Ð pa pa Ð → ˆ and pb are pointing in opposite directions. Case #2 : v ̂ Ð → ∗ →⋅v →⋅v →⋅v ˆ ∣ = −Ð ˆ . Altogether then, = becomes b = ˆ < 0, so that ∣Ð pa pa Then pb = −ˆ v and Ð pa →⋅v →⋅v ˆ ) (−ˆ ˆ ) (ˆ p + (−Ð pa v ) = p + (Ð pa v), as desired. ✓
On p. 938 in the Appendices (optional), I give another proof of the above Proposition using calculus. The idea of this second proof will be illustrated in the last two examples of this section.
Page 335, Table of Contents
www.EconsPhDTutor.com
Example 321. Consider the point a = (1, 2, 3) and the line r = (0, 1, 2) + λ(9, 1, 3) (λ ∈ R). → = (1, 1, 1) and so ∣Ð → 2 = 12 +12 +12 = 3. Pick a point on the line – say p = (0, 1, 2). We have Ð pa pa∣ Also, (1, 1, 1) ⋅ (9, 1, 3) 13 Ð →⋅v ˆ= √ pa =√ . 91 92 + 12 + 32 2 →⋅v ˆ ) = 169/61. Hence, the length of the side is pa And so (Ð
√ 3−
169 = 91
√
104 = 91
√
8 ≈ 1.069. 7
This is the distance between point a and the line l. Moreover, 13 (9, 1, 3) 1 √ b = (0, 1, 2) + √ = (9, 8, 17) . 7 91 91
Not to scale.
a = (1, 2, 3) Distance between a and b is 1.069
l b=
(9, 8, 17)
p = (0, 1, 2) Ð → →⋅v ˆ > 0. Note that in this example, v and pb do point in the same direction and we have Ð pa Ð → In contrast, in the next example, v and pb will point in opposite directions and we will →⋅v ˆ < 0. have Ð pa Page 336, Table of Contents
www.EconsPhDTutor.com
Example 322. Consider the point a = (−1, 0, 1) and the line r = (3, 2, 1)+λ(5, 1, 2) (λ ∈ R). → = (−4, −2, 0) and so ∣Ð → 2 = 42 +22 +02 = Pick a point on the line – say p = (3, 2, 1). We have Ð pa pa∣ 20. Also, (−4, −2, 0) ⋅ (5, 1, 2) 22 Ð →⋅v ˆ= √ pa = −√ . 30 52 + 12 + 22 Ð → →⋅v ˆ < 0 and sure enough, v and pb point in the opposite directions.) So (As noted, Ð pa 2 →⋅v ˆ ) = 484/30. Hence, the length of the side is (Ð pa √ 20 −
484 = 30
√
116 = 30
√
58 ≈ 2.823. 15
This is the distance between point a and the line l. Moreover, 22 (5, 1, 2) 1 √ (−10, 19, −7) . = b = (3, 2, 1) − √ 15 30 30
Not to scale.
a = (-1, 0, 1) Distance between a and b is 2.823
l b=
(-10, 9, -7)
p = (3, 2, 1)
Page 337, Table of Contents
www.EconsPhDTutor.com
Exercise 129. For each of the following, find (i) the distance between the given point a and the given line l; and also (ii) the point b on the line that is closest to a. (a) The point a = (7, 3, 4) and the line l described by r = (8, 3, 4) + λ(9, 3, 7). (b) The point a = (8, 0, 2) and the line l described by r = (4, 4, 3) + λ(2, 7, 2). (c) The point a = (8, 5, 9) and the line l described by r = (8, 4, 5) + λ(5, 6, 0). (Answers on pp. 1091, 1092, and 1093.)
Page 338, Table of Contents
www.EconsPhDTutor.com
We now learn a second method for finding the foot of a perpendicular and hence the distance of a point to a line. This second method involves calculus and finding the minimum point. It also occasionally features on the A-level exams. Example 323. Consider the point a = (1, 2, 3) and the line r = (0, 1, 2) + λ(9, 1, 3) (λ ∈ R). The distance between a and a generic point r on the line is RRR RRR⎛ 1 ⎞ ⎛ 9λ ∣a − r∣ = RRRR⎜ 2 ⎟−⎜ 1+λ RRR⎜ ⎟ ⎜ RRR⎝ 3 ⎠ ⎝ 2 + 3λ =
√
R R ⎞RRRR RRRR⎛ 1 − 9λ ⎟RRRR = RRRR⎜ 1 − λ ⎟RR RR⎜ ⎠RRRR RRRR⎝ 1 − 3λ R R
R ⎞RRRR ⎟RRRR ⎟RR ⎠RRRR R
(1 − 9λ)2 + (1 − λ)2 + (1 − 3λ)2 =
√ 91λ2 − 26 + 3.
Recall what λ means. It is a parameter — as λ varies, the vector equation r = (0, 1, 2) + λ(9, 1, 3) gives us another point of the line. √ So now the expression 91λ2 − 26 + 3 tells us: √ As λ varies, the distance between the point a and the corresponding point r on the line is 91λ2 − 26λ + 3. Our goal is to find the point √ r on the line that is closest to the point a. In other 2 words, √ our goal is to minimise 91λ − 26λ + 3. So we can look for the minimum point of 91λ2 − 26λ + 3. √ To simplify matters, note that minimising 91λ2 − 26λ + 3 is the same as minimising 91λ2 − 26λ + 3. So we might as well look for the minimum point of 91λ2 − 26λ + 3. To this end: d set (91λ2 − 26 + 3) = 182λ − 26 = 0 dλ
⇐⇒
λ=
26 1 = . 182 7
Altogether then, the point b on the line l that is closest to the point a has parameter λ = 1/7. So b = (0, 1, 2) + 1/7(9, 1, 3) = 1/7(9, 8, 17). And the distance between a and l (or equivalently, the length of the line segment ab) is √ 91λ2 − 26λ + 3 =
√
1 2 1 91 ( ) − 26 ( ) + 3 = 7 7
√
8 . 7
Of course, these are the same as what we found in Example 321 a few pages ago.
Page 339, Table of Contents
www.EconsPhDTutor.com
Example 324. Consider the point a = (−1, 0, 1) and the line described by the vector equation r = (3, 2, 1) + λ(5, 1, 2) (λ ∈ R). The distance between a and a generic point r on the line is RRR RRR⎛ −1 ⎞ ⎛ 3 + 5λ ⎜ ∣a − r∣ = RRRR⎜ 0 ⎟ ⎟−⎜ 2+λ RRR⎜ RRR⎝ 1 ⎠ ⎝ 1 + 2λ =
R R ⎞RRRR RRRR⎛ −4 − 5λ ⎟RRRR = RRRR⎜ −2 − λ ⎟RR RR⎜ ⎠RRRR RRRR⎝ −2λ R R
R ⎞RRRR ⎟RRRR ⎟RR ⎠RRRR R
√ √ (−4 − 5λ)2 + (−2 − λ)2 + (−2λ)2 = 30λ2 + 44λ + 20.
Now look for the minimum point of 30λ2 + 44λ + 20: d set (30λ2 + 44λ + 20) = 60λ + 44 = 0 dλ
⇐⇒
λ=
−44 −11 = . 60 15
So b = (3, 2, 1) − 11/15(5, 1, 2) = 1/15(−10, 19, −7). And the distance between a and l (or equivalently, the length of the line segment ab) is √ 30λ2 + 44λ + 20 =
√
−11 2 −11 30 ( ) + 44 ( ) + 20 = 15 15
√
58 . 15
Of course, these are the same as what we found in Example 322 a few pages ago.
Exercise 130. For each of the following, use the second method (calculus) to find (i) the distance between the given point a and the given line l; and also (ii) the point b on the line that is closest to a. (a) The point a = (7, 3, 4) and the line l described by r = (8, 3, 4) + λ(9, 3, 7). (b) The point a = (8, 0, 2) and the line l described by r = (4, 4, 3) + λ(2, 7, 2). (c) The point a = (8, 5, 9) and the line l described by r = (8, 4, 5) + λ(5, 6, 0). (Answers on p. 1094.)
Page 340, Table of Contents
www.EconsPhDTutor.com
33.2
Distance of a Point from a Plane
This is very much analogous to the distance of a point from a line. Definition 84. The foot of the perpendicular from a point a to a plane P is the point b on the plane P that is closest to the point a. The distance between the point a and the plane P is the length of the line segment ab.
a
Distance between a and b
Plane p b
Proposition 9. Given a point a (with position vector a) and a plane given in Hessian ˆ ˆ = d, normal form r ⋅ n ˆ ∣; and (a) The distance between the point and the plane is ∣dˆ − a ⋅ n ˆ) n ˆ. (b) The foot of the perpendicular from the point to the plane is the point a + (dˆ − a ⋅ n
Proof. Let b be the foot of the perpendicular from the point to the line. (a) Pick any point p on the plane. The length of the line segment ab — and hence also the → on distance between the point and the plane — is simply the length of the projection of Ð ap →⋅n ˆ ∣ = ∣(p − a) ⋅ n ˆ ∣ = ∣d − a ⋅ n ˆ ∣, as desired. the plane’s normal vector, which is simply ∣Ð ap Ð → ˆ ∣ away from a, heading in the direction ab. Hence, (b) The point b is a distance ∣d − a ⋅ n ̂ Ð → ∗ ˆ ∣ ab. There are two possible cases to examine. b = a + ∣d − a ⋅ n ̂ Ð → Ð → →⋅n ˆ is pointing in the same direction as pb, then n ˆ = ab. Moreover Ð ˆ = Case #1 : If n ap ∗ ˆ > 0, so that ∣d − a ⋅ n ˆ∣ = d − a ⋅ n ˆ . Altogether then, = becomes b = a + (d − a ⋅ n ˆ) n ˆ , as d−a⋅n desired. ✓ ̂ Ð → Ð → →n ˆ and pb are pointing in opposite directions, then n ˆ = −ab. Moreover Ð ˆ = d− Case #2 : If n ap⋅ ∗ ˆ < 0, so that ∣d − a ⋅ n ˆ ∣ = − (d − a ⋅ n ˆ ). Altogether then, = becomes b = a−(d − a ⋅ n ˆ ) (−ˆ a⋅ n n) = ˆ) n ˆ , as desired. ✓ a + (d − a ⋅ n Page 341, Table of Contents
www.EconsPhDTutor.com
Example 325. Consider the point a = (1, 2, 3) and the plane described by the vector equation r ⋅ (1, 1, 1) = 3. Convert the vector equation of the plane to Hessian normal form: √ 3 1 1 1 r ⋅ ( √ , √ , √ ) = √ = 3. 3 3 3 3 √ ˆ= So n
√ √ 3 ˆ = 2 3. (1, 1, 1), dˆ = 3, and a ⋅ n 3
√ √ ˆ ∣ = ∣ 3 − 2 3∣ = Altogether then, the distance between the point and the plane is ∣dˆ − a ⋅ n √ 3 and the foot of the perpendicular is √
√ ˆ) n ˆ = (1, 2, 3) + ( 3 − 2 3) a + (dˆ − a ⋅ n
√
3 (1, 1, 1) = (0, 1, 2). 3
Ð → By the way, notice that in this example, n points in the opposite direction from ab. And ̂ Ð → ˆ < 0. n. And moreover, dˆ − a ⋅ n so ab = −ˆ
a = (1, 2, 3) Not to scale.
Plane p = (0, 1, 2)
Page 342, Table of Contents
Distance between a and b b = (0, 1, 2)
www.EconsPhDTutor.com
Example 326. Consider the point a = (0, 0, 0) and the plane described by the vector equation r ⋅ (1, 2, 3) = 32. Convert the vector equation of the plane to Hessian normal form: 2 3 32 1 r ⋅ (√ , √ , √ ) = √ . 14 14 14 14 32 1 ˆ = 0. ˆ = √ (1, 2, 3), dˆ = √ , and a ⋅ n So n 14 14 32 ˆ ∣ = ∣ √ − 0∣ = Altogether then, the distance between the point and the plane is ∣dˆ − a ⋅ n 14 32 √ and the foot of the perpendicular is 14 32 1 16 ˆ) n ˆ = (0, 0, 0) + ( √ − 0) √ (1, 2, 3) = (1, 2, 3). a + (dˆ − a ⋅ n 7 14 14 Ð → By the way, notice that in this example, n points in the same direction as ab. And so ̂ Ð → ˆ . And moreover, dˆ − a ⋅ n ˆ > 0. ab = n
a = (0, 0, 0) Not to scale.
Distance between a and b
Plane p = (4, 5, 6)
Page 343, Table of Contents
b=
(1, 2, 3)
www.EconsPhDTutor.com
Exercise 131. For each of the following, find (i) the distance between the given point a and the given plane P ; and also (ii) the point b on the line that is closest to a. (a) a = (7, 3, 4), P ∶ r ⋅ (9, 3, 7) = 109. (b) a = (8, 0, 2), P ∶ r ⋅ (2, 7, 2) = 42. (c) a = (8, 5, 9), P ∶ r ⋅ (5, 6, 0) = 64. (Answers on pp. 1095, 1096, and 1097.)
Page 344, Table of Contents
www.EconsPhDTutor.com
34 34.1
Angles
Angle between Two Lines (2D)
Consider two lines on the 2D cartesian plane that are parallel (and thus either do not intersect or are identical). We define the angle between them to be 0.
We define the angle between two parallel lines to be 0.
Now consider two lines that intersect (see diagram below). Taking their intersection point to be the vertex, A and B are, respectively, the acute and obtuse angles between the two lines. Of course, there is the possibility that the two lines are perpendicular, in which case A and B are both right (i.e. equal to π/2).
So when talking about “the angle between two lines”, there is some potential for confusion. Are we talking about angle A or angle B? By convention, the angle between two lines is the smaller angle. (Also, on the A-level exams, they are usually quite careful to specifying that they want the acute angle, so that there is no confusion.) Page 345, Table of Contents
www.EconsPhDTutor.com
If we have the direction vectors of the lines, then we can simply use what we learnt about the scalar product to compute the angle between them. Example 327. Consider the lines (on the 2D cartesian plane) r = (1, 3) + λ(2, 1) and r = (−1, −1) + λ(1, 3) (λ ∈ R). The angle θ between their direction vectors v1 = (2, 1) and v2 = (1, 3) is given by θ = cos−1 (
v1 ⋅ v2 (2, 1) ⋅ (1, 3) 5 ) = cos−1 ( ) = cos−1 ( √ √ ) ≈ 0.785. ∣v1 ∣ ∣v2 ∣ ∣(2, 1)∣ ∣(1, 3)∣ 5 10
So the acute angle between the two lines is 0.785.
y 4 A = 0.785 Vector equation r = (1, 3) + ɉ(2, 1) 2
x 0 -4
-2
Vector equation r = (-1, -3) + ɉ(1, 3)
0
2 The vector (1, 3)
4
-2 A = 0.785 The vector (2, 1) -4
Page 346, Table of Contents
www.EconsPhDTutor.com
Example 328. Consider the lines (on the 2D cartesian plane) r = (0, 0) + λ(−2, 3) and r = (1, 0) + λ(3, 1) (λ ∈ R). The angle θ between their direction vectors v1 = (−2, 3) and v2 = (3, 1) is given by θ = cos−1 (
(−2, 3) ⋅ (3, 1) −3 v1 ⋅ v2 ) = cos−1 ( ) = cos−1 ( √ √ ) ≈ 1.837. ∣v1 ∣ ∣v2 ∣ ∣(−2, 3)∣ ∣(3, 1)∣ 13 10
This is the obtuse angle between the two lines. So the acute angle between the two lines is A = π − 1.837 = 1.305.
y 4
The vector (-2, 3) B = 1.837
Vector equation r = (0, 0) + ɉ(-2, 3)
The vector (3, 1)
2 A = 1.305 B = 1.837 x 0
-4
-2
0
2
4
A = 1.305 Vector equation r = (1, 0) + ɉ(3, 1)
-2
-4
Page 347, Table of Contents
www.EconsPhDTutor.com
Example 329. Consider the lines (on the 2D cartesian plane) r = (2, −2) + λ(3, 3) and r = (1, 1) + λ(−1, −1) (λ ∈ R). The angle θ between their direction vectors v1 = (3, 3) and v2 = (−1, −1) is given by θ = cos−1 (
(3, 3) ⋅ (−1, −1) −6 −6 v1 ⋅ v2 ) = cos−1 ( ) = cos−1 ( √ √ ) = cos−1 ( ) = π. ∣v1 ∣ ∣v2 ∣ ∣(3, 3)∣ ∣(−1, −1)∣ 6 18 2
So the two vectors are parallel. Which means that the two lines are parallel and so by definition, the angle between the two lines is 0.
y 4
The lines are parallel and thus the angle between them is 0.
The vector (3, 3)
2
The vector (-1, -1)
x 0
-4
-2
0
2
4
Vector equation r = (2, -2) + ɉ(3, 3) -2 Vector equation r = (1, 1) + ɉ(-1, 1) -4
Page 348, Table of Contents
www.EconsPhDTutor.com
Exercise 132. Find the acute angle between each of the following pairs of lines. (a) r = (−1, 2) + λ(−1, 1) and r = (0, 0) + λ(2, −3) (λ ∈ R). (b) r = (−1, 2) + λ(1, 5) and r = (0, 0) + λ(8, 1) (λ ∈ R). (c) r = (−1, 2) + λ(2, 6) and r = (0, 0) + λ(3, 2) (λ ∈ R). (Answer on p. 1098.)
Page 349, Table of Contents
www.EconsPhDTutor.com
34.2
Angle between Two Lines (3D)
Visualising lines in 3D space is difficult. Which is why we tackled the 2D case first. It turns out that we compute angles between two lines in 3D space in exactly the same way as in the 2D case. 1. If two lines are parallel, then again we define the angle between them to be 0. 2. If two lines intersect, then again we take their intersection point to be the vertex and take the smaller angle formed to be the angle between the two lines. On the 2D cartesian plane, the above were the only two possibilities â&#x20AC;&#x201D; two lines either are parallel or intersect. In contrast, in 3D space, there is the third possibility that two lines neither are parallel nor intersect! As weâ&#x20AC;&#x2122;ll learn in section 35.1, any two lines that neither are parallel nor intersect are called skew lines. What is the angle between two skew lines, given that they do not intersect? 3. Given two skew lines, translate one of them so that they intersect. Examine the angle between the two now-intersecting lines. This is defined to be the angle between the two skew lines. The next example illustrates.
Page 350, Table of Contents
www.EconsPhDTutor.com
Example 330. Below, the red line and pink line are skew lines, i.e., they neither intersect nor are parallel. To find the angle between them, translate the red line upwards so that the new red dotted line intersects the pink line at the purple dot. The angle A is then defined to be the angle between the two skew lines.
Skew lines (lines that neither intersect nor are parallel) y Translate one of the lines so that they intersect
3
A
2 1 x 1 1
2 3 z
4
So once again, given any two lines, the angle between them is simply the angle between their direction vectors. So again the scalar product comes in handy.
Page 351, Table of Contents
www.EconsPhDTutor.com
Example 331. Consider the lines r = (0, 1, 2)+λ(9, 1, 3) and r = (4, 5, 6)+λ(3, 2, 1) (λ ∈ R). The angle θ between their direction vectors v1 = (9, 1, 3) and v2 = (3, 2, 1) is given by θ = cos−1 (
(9, 1, 3) ⋅ (3, 2, 1) 32 v1 ⋅ v2 ) = cos−1 ( ) = cos−1 ( √ √ ) ≈ 0.459. ∣v1 ∣ ∣v2 ∣ ∣(9, 1, 3)∣ ∣(3, 2, 1)∣ 91 14
So the acute angle between the two lines is 0.459. Example 332. Consider the lines r = (−1, 2, 3) + λ(0, 1, 0) and r = (0, 0, 0) + λ(8, −3, 5) (λ ∈ R). The angle θ between their direction vectors v1 = (0, 1, 0) and v2 = (8, −3, 5). Thus, θ = cos−1 (
(0, 1, 0) ⋅ (8, −3, 5) −3 v1 ⋅ v2 ) = cos−1 ( ) = cos−1 ( √ √ ) ≈ 1.879. ∣v1 ∣ ∣v2 ∣ ∣(0, 1, 0)∣ ∣(8, −3, 5)∣ 1 98
So the obtuse angle between the two lines is 1.879. And the angle between the two lines is 1.263. Example 333. Consider the lines r = (1, 3, 3) + λ(1, 5, 3) and r = (7, 4, 7) + λ(7, −2, 1) (λ ∈ R). The angle θ between their direction vectors v1 = (1, 5, 3) and v2 = (7, −2, 1). Thus, θ = cos−1 (
v1 ⋅ v2 (1, 5, 3) ⋅ (7, −2, 1) ) = cos−1 ( ) = cos−1 (0) ≈ 0.5π. ∣v1 ∣ ∣v2 ∣ ∣(1, 5, 3)∣ ∣(7, −2, 1)∣
So the two lines are perpendicular and the angle between them is right (i.e. π/2).
Exercise 133. Find the angle between each of the following pairs of lines. (a) r = (−1, 2, 3)+ λ(−1, 1, 0) and r = (0, 0, 0) + λ(2, −3, 4) (λ ∈ R). (b) r = (−1, 2, 3) + λ(1, 5, 6) and r = (0, 0, 0) + λ(8, 1, 1) (λ ∈ R). (c) r = (−1, 2, 3) + λ(2, 6, 7) and r = (0, 0, 0) + λ(3, 2, 1) (λ ∈ R). (Answer on p. 1098.)
Page 352, Table of Contents
www.EconsPhDTutor.com
34.3
Angle between A Line and a Plane
Fact 38. The angle A between the line r = p + λv and the plane r ⋅ n = d is given by A = sin−1 ∣
n
v⋅n ∣. ∣v∣ ∣n∣
n
Line
Line
ȣ is obtuse
ȣ is acute ȣ A = 0.5π – Ʌ
Plane
ȣ
A = Ʌ – 0.5π
Plane
Direction vector of line
Direction vector of line
Proof. Consider the angle θ between the line’s direction vector and the plane’s normal v⋅n . vector. By the scalar product, we have cos θ = ∣v∣ ∣n∣ Case #1: If θ is acute or right, i.e. θ ∈ (0, π/2], then the angle A between the line and the plane is A = π/2 − θ (see figure). And so π π π v⋅n sin A = sin ( − θ) = sin ( ) cos θ − sin θ cos ( ) = cos θ = . 2 2 2 ∣v∣ ∣n∣ Note that if θ ∈ (0, π/2], then v ⋅ n ≥ 0, so that ∣v ⋅ n∣ = v ⋅ n. Altogether, we indeed have sin A = ∣
v⋅n ∣ ∣v∣ ∣n∣
or A = sin−1 ∣
v⋅n ∣. ∣v∣ ∣n∣
Case #2: If θ is obtuse or straight, i.e. θ ∈ (π/2, π], then the angle A between the line and the plane is A = θ − 0.5π (see figure). And so π π π −v ⋅ n sin A = sin (θ − ) = sin θ cos ( ) − sin ( ) cos θ = − cos θ = . 2 2 2 ∣v∣ ∣n∣ Note that if θ ∈ (π/2, π], then v ⋅ n < 0, so that ∣v ⋅ n∣ = −v ⋅ n. Altogether, we indeed have sin A = ∣
Page 353, Table of Contents
v⋅n ∣ ∣v∣ ∣n∣
or A = sin−1 ∣
v⋅n ∣. ∣v∣ ∣n∣
www.EconsPhDTutor.com
Example 334. The angle between the line r = (0, 1, 2) + λ(9, 1, 3) (λ ∈ R) and the plane r ⋅ (1, 1, 1) = 3 is √ 13 v ⋅ n (9, 1, 3) ⋅ (1, 1, 1) 13 sin−1 ∣ ∣ = sin−1 ∣ ∣ = sin−1 ∣ √ √ ∣ = sin−1 ∣ √ √ ∣ ≈ 0.906. ∣v∣ ∣n∣ ∣(9, 1, 3)∣ ∣(1, 1, 1)∣ 91 3 7 3 Example 335. The angle between the line r = (4, 2, 3) + λ(1, 0, 1) (λ ∈ R) and the plane r ⋅ (−1, −1, 1) = 5 is sin−1 ∣
(1, 0, 1) ⋅ (−1, −1, 0) −1 v⋅n ∣ = sin−1 ∣ ∣ = sin−1 ∣ √ √ ∣ = sin−1 (1/2) = π/6. ∣v∣ ∣n∣ ∣(1, 0, 1)∣ ∣(−1, −1, 0)∣ 2 2
Example 336. The angle between the line r = (5, 5, 5) + λ(1, 0, 1) (λ ∈ R) and the plane r ⋅ (0, 1, 0) = 3 is sin−1 ∣
(1, 0, 1) ⋅ (0, 1, 0) v⋅n ∣ = sin−1 ∣ ∣ = sin−1 ∣0∣ = 0. ∣v∣ ∣n∣ ∣(1, 0, 1)∣ ∣(0, 1, 0)∣
Exercise 134. Find the angle between the given line and plane. (a) r = (−1, 2, 3) + λ(−1, 1, 0) (λ ∈ R) and r ⋅ (3, 4, 5) = 0. (b) r = (−1, 2, 3) + λ(0, 2, 6) (λ ∈ R) and r ⋅ (1, 3, 5) = 2. (c) r = (−1, 2, 3) + λ(1, 9, 8) (λ ∈ R) and r ⋅ (2, 8, 2) = 3. (Answer on p. 1099.)
Page 354, Table of Contents
www.EconsPhDTutor.com
34.4
Angle between Two Planes
Given two planes P1 and P2 , the angle between them is simply the angle between any two vectors v1 and v2 on the two planes.
n1 v2
n2
Angle between the two planesâ&#x20AC;&#x2122; normal vectors
Angle between the two planes
P2
v1
P1
But the normal vector n1 of the first plane is orthogonal to v1 ; similarly, the normal vector n2 of the second plane is orthogonal to v2 . And so the angle between v1 and v2 is equal to the angle between n1 and n2 . Altogether then, the angle between two planes is simply the angle between their normal vectors. Again, there are two possible angles â&#x20AC;&#x201D; by convention, we take the smaller one.
Page 355, Table of Contents
www.EconsPhDTutor.com
Example 337. Consider the planes r ⋅ (1, 1, 1) = 12 and r ⋅ (−1, −1, 0) = −1. The angle θ between the two planes is: θ = cos−1 (
= cos−1 (
n1 ⋅ n2 ) ∣n1 ∣ ∣n2 ∣ (1, 1, 1) ⋅ (−1, −1, 0) ) ∣(1, 1, 1)∣ ∣(−1, −1, 0)∣
−2 −2 = cos−1 ( √ √ ) = cos−1 ( √ ) 3 2 6 ≈ 2.526. This is the obtuse angle. So the acute angle between the two planes is π − 2.526 = 0.615 radian. Example 338. Consider the planes r ⋅ (2, 1, 3) = 26 and r ⋅ (−3, 0, 5) = −25. The angle θ between the two planes is θ = cos−1 (
= cos−1 (
n1 ⋅ n2 ) ∣n1 ∣ ∣n2 ∣ (2, 1, 3) ⋅ (−3, 0, 5) ) ∣(2, 1, 3)∣ ∣(−3, 0, 5)∣
9 = cos−1 ( √ √ ) ≈ 1.146. 14 34
Exercise 135. Find the angle between the two given planes. (a) r ⋅ (−1, −2, −3) = 1 and r ⋅ (3, 4, 5) = 2. (b) r ⋅ (1, −2, 3) = 3 and r ⋅ (5, 1, 1) = 4. (c) r ⋅ (1, 1, 8) = 5 and r ⋅ (−3, 0, 10) = 6. (Answer on p. 1100.)
Page 356, Table of Contents
www.EconsPhDTutor.com
35
Relationships between Lines and Planes 35.1
Relationship between Two Lines
Definition 85. Two lines are parallel if their direction vectors can be written as scalar multiples of each other.
Example 339. The lines r = (0, 0, 0) + λ(0, 1, 0) and r = (4, 17, 0) + λ(1, 0, 0) (λ ∈ R) are not parallel, because (0, 1, 0) cannot be written as a scalar multiple of (1, 0, 0). Example 340. The lines r = (8, 1, 1) + λ(3, 6, 9) and r = (4, 5, 6) + λ(1, 2, 3) (λ ∈ R) are parallel, because (3, 6, 9) = 3(1, 2, 3).
Page 357, Table of Contents
www.EconsPhDTutor.com
Definition 86. A set of points are coplanar if there is some plane that contains all of these points.
Any two points are always coplanar â&#x20AC;&#x201D; indeed, they are collinear (p1 and p2 in the figure below). Three points are also always coplanar, although they may not be collinear (p1 , p2 , and p3 in the figure below). But four points may not be coplanar (p1 , p2 , p3 , and p4 in the figure below).
Line 2 Two points are coplanar. They also lie on the same line.
Three points are coplanar, though not necessarily on the same line.
p1 Plane
p3 p2 Line 1
p4
Four points may not be coplanar.
Definition 87. Two lines are coplanar if there is some plane on which both lie. Two lines that are not coplanar are called skew lines.
Example 341. In the figure above, Line 1 and Line 2 are skew lines. Line 1 lies on the plane illustrated. Line 2 cuts through the plane and does not intersect Line 1.
Page 358, Table of Contents
www.EconsPhDTutor.com
How do we determine whether two lines l1 and l2 are coplanar or skew? Well, 1. If they are parallel, then obviously we can construct a plane that contains both lines. And so the two lines are coplanar. 2. If they are not parallel and they lie on the same plane, then they must intersect. This is just the familiar fact you learnt in primary school — two non-parallel lines on the plane must definitely intersect. Altogether we conclude: Fact 39. Two lines are coplanar if and only if they (i) are parallel; OR (ii) intersect. Equivalently, two lines are skew if and only if they (i) are not parallel; AND (ii) do not intersect.
Example 342. Consider the lines r = (8, 1, 1)+λ(3, 6, 9) and r = (4, 5, 6)+λ(1, 2, 3) (λ ∈ R). The direction vector of one can be written as the scalar multiple of the other, so they are parallel. Hence, they are also coplanar; or equivalently, they are not skew.
Example 343. Consider the lines r = (0, 0, 0) + λ(0, 1, 0) and r = (4, 17, 0) + λ(1, 0, 0) (λ ∈ R). The direction vector of one cannot be written as the scalar multiple of the other, so they are not parallel. If they intersect, then there are reals α and β such that (0, 0, 0) + α(0, 1, 0) = (4, 17, 0) + β(1, 0, 0), or 0 = 4 + β, α = 17, and 0 = 0. α = 17, β = −4 solves the above equations. (What does this mean? This means that the first line goes through the point (0, 0, 0) + α(0, 1, 0) = (0, 17, 0) and the second line also goes through the same point (4, 17, 0) + β(1, 0, 0) = (0, 17, 0).) The two lines intersect at (0, 17, 0). And so they are coplanar — or equivalently, they are not skew. If we’d like, we can easily find the plane on which these two lines lie. Remember: All we need are two distinct vectors and a point to determine a plane. We already have two distinct vectors, namely the direction vectors of the two lines. Using these, we can find a normal vector for the plane — namely (0, 1, 0) × (1, 0, 0) = (0, 0, −1). Noting also that the origin is on the first line and therefore on the plane, we conclude that the plane is r ⋅ (0, 0, −1) = 0.
Page 359, Table of Contents
www.EconsPhDTutor.com
Example 344. Consider the lines r = (0, 1, 2)+λ(9, 1, 3) and r = (4, 5, 6)+λ(3, 2, 1) (λ ∈ R). They are not parallel. Let’s see if they have an intersection point. If they intersect, then there are reals α and β such that (0, 1, 2) + α(9, 1, 3) = (4, 5, 6) + β(3, 2, 1), or 1
2
3
9α = 4 + 3β, 1 + α = 5 + 2β, and 2 + 3α = 6 + β. 3
2
Take 2× = minus = to get (4 + 6α) − (1 + α) = (12 + 2β) − (5 + 2β) or 3 + 5α = 7 or α = 0.8. 2 Now from =, this means that β = −1.6. These do not work if we try plugging them into 1 =. Hence, there are no reals α and β that solve the above system of equations. In other words, the two lines do not intersect. And so the two lines are not coplanar — or equivalently, they are skew.
Exercise 136. Determine whether each of the following pairs of lines is coplanar or skew. If they are coplanar, find the plane that contains both of them. (a) r = (8, 1, 5) + λ(3, 2, 1) and r = (1, 2, 3) + λ(5, 6, 7) (λ ∈ R). (b) r = (0, 0, 6) + λ(3, 9, 0) and r = (1, 1, 1) + λ(1, 3, 0) (λ ∈ R). (c) r = (6, 5, 5) + λ(1, 0, 1) and r = (8, 3, 6) + λ(0, 1, 1) (λ ∈ R). (Answer on p. 1101.)
Page 360, Table of Contents
www.EconsPhDTutor.com
35.2
Relationship between a Line and a Plane
Definition 88. A line with direction vector v and a plane with normal vector n are parallel if v ⋅ n = 0 (i.e. v and n are perpendicular). The above definition makes sense, because if the line is perpendicular to the plane’s normal vector, then the line must be parallel to the plane itself. Fact 40. Given a plane and a line, there are three possible cases (illustrated below): 1. The line and plane are parallel and do not intersect at all. 2. The line and plane are parallel and the line lies completely on the plane. 3. The line and plane are not parallel and intersect at exactly one point.
Line 1
Line 3
Plane Line 2
Proof. Optional, see p. 939 in the Appendices. Note that if a line and a plane are parallel, then either (i) they do not intersect at all; or (ii) the line lies completely on the plane. • So if a line and a plane are parallel and you can prove that they share at least one intersection point, then it must be that the line lies completely on the plane. • Conversely, if a line and a plane are parallel and you can prove that there is at least one point on the line that is not on the plane (or that there is at least one point on the plane that is not on the line), then it must be that they do not intersect at all.
Page 361, Table of Contents
www.EconsPhDTutor.com
Example 345. Consider the line r = (3, 5, 5)+λ(9, 1, 3) (λ ∈ R) and the plane r⋅(1, 1, 1) = 3. We have (9, 1, 3) ⋅ (1, 1, 1) = 13 ≠ 0 and so they are not parallel. They must therefore intersect at exactly one point. Let’s find it. Plug in a generic point of the line into the equation for the plane: [(3, 5, 5) + λ(9, 1, 3)] ⋅ (1, 1, 1) = 3 ⇐⇒ 3 + 9λ + 5 + λ + 5 + 3λ = 3 ⇐⇒ 13 + 13λ = 3 ⇐⇒ λ = −10/13. So the intersection point is (3, 5, 5) − 10/13(9, 1, 3). Example 346. Consider the line r = (3, 5, 5) + λ(9, 1, 3) (λ ∈ R) and the plane r ⋅ (1, 0, −3) = −6. We have (9, 1, 3) ⋅ (1, 0, −3) = 0 and so they are parallel. There are two possibilities. Either they do not intersect at all OR the line lie completely on the plane. The point (3, 5, 5) is on the line but is not on the plane, as we can easily verify — (3, 5, 5) ⋅ (1, 0, −3) = −12 ≠ −6. And so the line and plane do not intersect at all. Example 347. Consider the line r = (3, 5, 3) + λ(9, 1, 3) (λ ∈ R) and the plane r ⋅ (1, 0, −3) = −6. We have (9, 1, 3) ⋅ (1, 0, −3) = 0 and so they are parallel. There are two possibilities. Either they do not intersect at all OR the line lie completely on the plane. The point (3, 5, 3) on the line is also on the plane: (3, 5, 3) ⋅ (1, 0, −3) = −6. Since they are parallel and share at least one intersection point, it must be that the line lies completely on the plane.
Exercise 137. Determine whether the given line and plane are (i) parallel but do not intersect; (ii) parallel with the line lying completely on the plane; or (iii) intersect at exactly one point. (a) r = (4, 5, 6) + λ(2, 3, 5) (λ ∈ R) and r ⋅ (−10, 0, 4) = −26. (b) r = (5, 5, 6) + λ(2, 3, 5) (λ ∈ R) and r ⋅ (−10, 0, 4) = −26. (c) r = (4, 5, 6) + λ(2, 3, 5) (λ ∈ R) and r ⋅ (−10, 0, 3) = −26. (Answer on p. 1102.)
Page 362, Table of Contents
www.EconsPhDTutor.com
35.3
Relationship between Two Planes
Definition 89. Two planes are parallel if their normal vectors can be written as scalar multiples of each other. (Note that an alternative definition is this: “Two planes are parallel if they do not intersect.” We will show that these two definitions are equivalent.) Imagine that two planes intersect at some line, which we’ll call the intersection line. Since this intersection line is on both planes, it must also be perpendicular to the normal vectors of both planes. In other words, it must have direction vector n1 × n2 . The next fact is thus not surprising (although actually proving it takes a little work). Fact 41. Two non-parallel planes with normal vectors n1 and n2 intersect at all if and only if they intersect along a line with direction vector n1 × n2 (i.e. the line is perpendicular to both n1 and n2 ).
Proof. Optional, see p. 940 in the Appendices.
Fact 42. Given two planes, there are three possible cases: 1. The two planes are parallel and exactly identical. 2. The two planes are parallel and do not intersect at all. 3. The two planes are not parallel and share an intersection line with direction vector n1 × n2 (where n1 , n2 are the normal vectors of the plane).
Proof. Optional, see p. 942 in the Appendices.
Page 363, Table of Contents
www.EconsPhDTutor.com
Example 348. In the figure below, planes P1 and P2 are parallel and do not intersect at all. Planes P2 and P3 are not parallel and share an intersection line with direction vector n2 × n3 .
n2 P3
n3
P2 Intersection line of P2 and P3
n2 × n3
P1
Note that analogous to our study of two lines, if two planes are parallel, then either (i) they do not intersect at all; or (ii) they are identical. • So if two planes are parallel and you can prove that they share at least one intersection point, then it must be that the two planes are identical. • Conversely, if two planes are parallel and you can prove that there is at least one point on one plane that is not on the other plane, then it must be that they do not intersect at all. And in the case where they are not parallel, to find the intersection line, simply find a point p where the two planes intersect. Then the intersection line is simply r = p + λ(n1 × n2 ) (λ ∈ R).
Page 364, Table of Contents
www.EconsPhDTutor.com
Example 349. Consider the planes r ⋅ (7, 1, 1) = 42 and r ⋅ (1, 1, 2) = 6. Clearly, (7, 1, 1) cannot be written as a scalar multiple of (1, 1, 2). So the two planes are not parallel and share an intersection line whose direction vector is (7, 1, 1) × (1, 1, 2) = (1, −13, 6). Find a point p = (x, y, z) where the two planes intersect: 1
7x + y + z = 42,
2
x + y + 2z = 6.
There are infinitely many points where the two planes intersect. So why not we look for an intersection point where x = 0. I’ll call this the “plug in x = 0” trick. 2
1
In which case = minus = yields z = −36 and y = 78. Hence, the intersection line is r = (0, 78, −36) + λ(1, −13, 6) (λ ∈ R).
Example 350. Consider the planes r ⋅ (1, 1, 1) = 12 and r ⋅ (−1, −1, 0) = −1. Clearly, (1, 1, 1) cannot be written as a scalar multiple of (−1, −1, 0). So the two planes are not parallel and share an intersection line whose direction vector is (1, 1, 1) × (−1, −1, 0) = (1, −1, 0). Find a point p = (x, y, z) where the two planes intersect: 1
x + y + z = 12,
2
−x − y = −1. 2
Again, we can play the “plug in x = 0” trick. In which case = says that y = 1 and now from 1 =, we have z = 11. And so (0, 1, 11) is an intersection point of the two planes. Hence, the intersection line is r = (0, 1, 11) + λ(1, −1, 0) (λ ∈ R).
Page 365, Table of Contents
www.EconsPhDTutor.com
You can use the “plug in x = 0” trick whenever the intersection line has direction vector with a x-coordinate that is not equal to 0. But the “plug in x = 0” trick may not work if the intersection line has direction vector with x-coordinate equal to 0.
Example 351. Consider the planes r ⋅ (0, 1, 3) = 0 and r ⋅ (−1, 1, 3) = 2. Clearly, (0, 1, 3) cannot be written as a scalar multiple of (−1, 0, 5). So the two planes are not parallel and share an intersection line whose direction vector is (0, 1, 3) × (−1, 1, 3) = (0, −3, 1). Find a point p = (x, y, z) where the two planes intersect: 1
y + 3z = 0,
2
−x + y + 3z = 2.
Here the direction vector of the intersection line has x-coordinate 0. So the “plug in x = 0” 1 2 trick might not work. And indeed it doesn’t, because if we plug in x = 0, then = and = are contradictory. So let’s try the “plug in y = 0” trick instead, which I know will work because the ycoordinate of the direction vector of the interesction line is non-zero (it’s −3). Then from 1 2 = we have z = 0 and now from = we have x = −2. And so (−2, 0, 0) is an intersection point of the two planes. Hence, the intersection line is r = (−2, 0, 0) + λ(0, −3, 1) (λ ∈ R). Alternatively, we could also have used the “plug in z = 0” trick instead, which again I know will work because the z-coordinate of the direction vector of the interesction line is 1 2 non-zero (it’s 1). Then from = we have y = 0 and now from = we have x = −2. And so again we find that (−2, 0, 0) is an intersection point of the two planes. And so again we would have concluded that the intersection line is r = (−2, 0, 0) + λ(0, −3, 1) (λ ∈ R).
Page 366, Table of Contents
www.EconsPhDTutor.com
Example 352. Consider the planes r ⋅ (4, 0, 3) = 5 and r ⋅ (−8, 0, −6) = −64. Clearly, (4, 0, 3) can be written as a scalar multiple of (−8, 0, −6) and so the two planes are parallel. Let’s check if they are identical. The point (5/4, 0, 0) is on the first plane. However, it is not on the second plane because (5/4, 0, 0) ⋅ (−8, 0, −6) = −10 ≠ −64. So the two planes are not identical and do not intersect at all.
Example 353. Consider the planes r ⋅ (4, 0, 3) = 32 and r ⋅ (−8, 0, −6) = −64. Clearly, (4, 0, 3) can be written as a scalar multiple of (−8, 0, −6) and so the two planes are parallel. Let’s check if they are identical. The point (8, 0, 0) is on the first plane. And it is also on the second plane because (8, 0, 0) ⋅ (−8, 0, −6) = −64. Since the two planes are parallel and share at least one intersection point, it must be that the two planes are exactly identical.
Exercise 138. Determine whether the given pair of planes are parallel and identical, parallel and do not intersect, or are not parallel. If they are not parallel, determine also their intersection line. (a) r⋅(4, 9, 3) = 61 and r⋅(1, 1, 2) = 19. (b) r⋅(1, 1, 0) = 4 and r⋅(1, 6, 8) = 60. (c) r ⋅ (4, 4, 8) = 56 and r ⋅ (1, 1, 2) = 12. (d) r ⋅ (4, 4, 8) = 48 and r ⋅ (1, 1, 2) = 12. (Answer on p. 1103.)
Page 367, Table of Contents
www.EconsPhDTutor.com
35.4
Relationship between Three Planes
SYLLABUS ALERT The relationship between three planes is included in the 9740 (old) syllabus, but not in the 9758 (revised) syllabus. So you can skip this section if you’re taking 9758.
Given 3 planes P1 , P2 , and P3 , we can form 3 pairs of planes: • P1 and P2 ; • P1 and P3 ; and • P2 and P3 . To find the relationship between the 3 planes is simply to find the relationships between each of these 3 pairs of planes. This can be insanely tedious, but there is nothing new here. Everything follows from what you learnt in the previous sections. Let’s nonetheless give a summary of the possibilities. Given three planes, we have 3 possible cases, each of which can be broken up into several sub-cases, for a total of 8 distinct possibilities.
Page 368, Table of Contents
www.EconsPhDTutor.com
1. All 3 planes are parallel to each other. There are 3 possible sub-cases: (a) All 3 planes are identical. (b) Only 2 planes are identical. (c) No 2 planes are identical.
P1 , P2 , P3
3 parallel, identical planes
P3
P2 P3 P1 P1 , P2
3 parallel, nonintersecting planes
3 parallel planes, where P1 and P2 are identical
Page 369, Table of Contents
www.EconsPhDTutor.com
2. Two planes are parallel to each other, but the 3rd plane is not parallel to either of the first 2 planes. There are 2 possible sub-cases: (a) The first 2 planes are identical. And so here we are really back to the situation of two non-parallel planes, which we already covered in detail in the previous section. They intersect along a line. (b) The first 2 planes are not identical. And so the non-parallel plane intersects each of the other two planes along a separate line of intersection.
P3
P1 and P2 are parallel and identical. P3 intersects both at the same line. P1 , P2
P3 P2
P1 and P2 are parallel but non-identical. P3 intersects both at separate lines.
Page 370, Table of Contents
P1
www.EconsPhDTutor.com
3. No 2 planes are parallel. Each pair of planes intersects along a line. There are thus three intersection lines (though possibly some may be identical). It is possible to prove (but we wonâ&#x20AC;&#x2122;t do so in this book) that there are only 3 possible sub-cases: (a) None of the intersection lines intersect with each other. That is, each pair of planes simply intersects along some distinct intersection line. (b) All 3 intersection lines are identical. So all 3 planes intersect along the same intersection line. (c) The 3 intersection lines and thus all 3 planes intersect at a single point. To determine which of the above sub-cases weâ&#x20AC;&#x2122;re in, we must determine the relation between each pair of intersection lines. This is tedious, but nothing new.
P3
3 non-parallel planes that intersect at different lines P2
P1
3 non-parallel planes that intersect at the same line P3 P2
P1 P3 P1
P2
Page 371, Table of Contents
3 non-parallel planes that intersect at only one point
www.EconsPhDTutor.com
Example 354. Consider the planes P1 , P2 , and P3 , given by r ⋅ (1, 0, 0) = 0, r ⋅ (0, 1, 0) = 0, and r ⋅ (0, 0, 1) = 0. Step #1. Check if any two planes are parallel. By observation, no plane’s normal vector can be written as a scalar multiple of another plane’s normal vector. So no two planes are parallel. (So we are in Case 3.) Step #2. Find the 3 intersection lines along which each pair of planes intersect. By observation, all three planes contain the origin. The planes P1 and P2 share an intersection line with direction vector (1, 0, 0) × (0, 1, 0) = (0, 0, 1) and so their intersection line is r = (0, 0, 0) + λ(0, 0, 1) (λ ∈ R). Call this line l1 . The planes P1 and P3 share an intersection line with direction vector (1, 0, 0) × (0, 0, 1) = (0, −1, 0) and so their intersection line is r = (0, 0, 0) + λ(0, −1, 0) (λ ∈ R). Call this line l2 . The planes P2 and P3 share an intersection line with direction vector (0, 1, 0) × (0, 0, 1) = (1, 0, 0) and so their intersection line is r = (0, 0, 0) + λ(1, 0, 0) (λ ∈ R). Call this line l3 . Step #3. Determine where, if at all, the 3 intersection lines intersect. l1 and l2 are not parallel, but they do intersect at the point (0, 0, 0) and so that is also their only intersection point. l1 and l3 are not parallel, but they do intersect at the point (0, 0, 0) and so that is also their only intersection point. l2 and l3 are not parallel, but they do intersect at the point (0, 0, 0) and so that is also their only intersection point. Conclusion. Altogether, we conclude that the 3 intersection lines intersect at a single point. Hence, the 3 planes also intersect at a single point. (So we are in Case 3c.)
Page 372, Table of Contents
www.EconsPhDTutor.com
Example 355. Consider the planes P1 , P2 , and P3 , given by r⋅(1, 1, 0) = 1, r⋅(−2, −2, 0) = −4, and r ⋅ (0, 1, 1) = 1. Step #1. Check if any two planes are parallel. By observation, P1 ’s normal vector (1, 1, 0) can be written as a scalar multiple of P2 ’s normal vector (−2, −2, 0). And so these two planes are parallel. P3 is not parallel to either of the first two planes, since (0, 1, 1) cannot be written as a scalar multiple of (1, 1, 0) or (−2, −2, 0). Altogether then, we are in Case 2. Step #2. Check if the two parallel planes are identical. They are not, because (1, 0, 0) is on P1 but is not on P2 , as we can easily verify — (1, 0, 0) ⋅ (−2, −2, 0) = −2 ≠ −4. So we are in Case 2b. Step #3. Find the intersection lines. There are two — one shared by P1 and P3 and the other shared by P2 and P3 . (P1 and P2 are distinct, parallel planes and thus do not intersect at all.) The intersection line of P1 and P3 has direction vector (1, 1, 0) × (0, 1, 1) = (1, −1, 1). Let’s 1 find a point (x, y, z) at P1 and P3 intersect: the equations for the planes are x + y = 1 and 2 y + z = 1. Using the “plug in x = 0” trick, we see that they intersect at (0, 1, 0). Hence, their intersection line is r = (0, 1, 0) + λ(1, −1, 1) (λ ∈ R). Call this line l1 . The intersection line of P2 and P3 must also have direction vector (1, −1, 1). Let’s find a 1 point (x, y, z) at P2 and P3 intersect: the equations for the planes are −2x − 2y = −4 and 2 y + z = 1. Using the “plug in x = 0” trick, we see that they intersect at (0, 2, −1). Hence, their intersection line is r = (0, 2, −1) + λ(1, −1, 1) (λ ∈ R). Call this line l2 . The lines l1 and l2 are parallel.
Exercise 139. What is the relationship between the 3 planes P1 , P2 , and P3 , given by r ⋅ (1, 0, 1) = 1, r ⋅ (0, 1, −1) = −1, and r ⋅ (1, 1, 0) = 2? (Answer on p. 1104.)
Page 373, Table of Contents
www.EconsPhDTutor.com
Part IV
Complex Numbers
Page 374, Table of Contents
www.EconsPhDTutor.com
36
Complex Numbers: Introduction
Here’s a brief motivation of complex numbers: 1. Solve x − 1 = 0. Easy; the answer is a natural number: x = 1. 2. To solve x + 1 = 0, we must invent negative numbers. The answer is x = −1. √ 3. To solve x2 = 2, we must invent irrational numbers. The answer is x = ± 2.
4. To solve x2 = −1, we must invent complex numbers.
We’ll start by defining the imaginary unit, then work our way to complex numbers. Definition 90. The imaginary unit, denoted i, is a number that satisfies i2 = −1. Using the imaginary unit, we can construct other purely imaginary numbers: Definition 91. A purely imaginary number is any real, non-zero multiple of the imaginary unit. That is, a purely imaginary number is any bi, where b ∈ R with b ≠ 0. (We specify that b ≠ 0 because 0i = 0 is not a purely imaginary number, but a real number.) √ √ √ Example 356. i + i = 2i = 2 −1 is purely imaginary. So too are −i = − −1 and πi = π −1. i is both the imaginary unit and a purely imaginary number. We can add real numbers to purely imaginary numbers to form imaginary numbers: Definition 92. An imaginary number is any a + bi, where a, b ∈ R with b ≠ 0. Again, we specify that b ≠ 0 because otherwise a + 0i would not be an imaginary number, but a real number. Example 357. 3 + 2i is imaginary, but not purely imaginary. In contrast, 2i is both imaginary and purely imaginary.
Page 375, Table of Contents
www.EconsPhDTutor.com
A complex number is simply any real number or imaginary number. Definition 93. A complex number is any a + bi, where a, b ∈ R. Notice that here in contrast, we do not specify that b ≠ 0. The reason is that complex numbers include all real numbers. Example 358. 10 and 17 are complex and real. 2+9i and 3−2i are complex and imaginary. 2i is complex, imaginary, and purely imaginary. i is complex , imaginary, purely imaginary, and also the imaginary unit. We denoted the set of real numbers by the symbol R. We now denote the set of complex numbers by the symbol C. Definition 94. The set of all complex numbers, denoted C, is defined as {a + bi∣a, b ∈ R}.
The set of reals is a proper subset of the set of complex numbers — formally, Fact 43. R ⊂ C. Proof. Every element a ∈ R can be written as a + 0i and is thus an element of C. So R is a subset of C. Moreover R is not equal to C, because for example 3 + 7i ∈ C but 3 + 7i ∉ R. Altogether then, R is a proper subset of C.
Complex numbers are thus the extension of the concept of real numbers. On the next page is a modified version of our taxonomy of numbers from p. 39, with the complex numbers fleshed out:
Page 376, Table of Contents
www.EconsPhDTutor.com
Complex Numbers
Real Numbers
“Impure” Imaginary Numbers
Imaginary Numbers
Purely Imaginary Numbers The Imaginary Unit
Note: there is no such thing as a positive or negative complex number. To fully appreciate why is beyond the scope of the A-levels. But for now here is a very simple example just to illustrate the point. Example 359. 1, −1, i, and −i are all complex numbers. We do say that 1 is a positive real number and −1 is a negative real number. But we do not say that i is a positive complex number or that −i is a negative complex number. In fact, we do not even say that 1 is a positive complex number or that −1 is a negative complex number.
Exercise 140. Fill in the following table. The first column has been done for you. (Answer on p. 1105.) Is this ... 13 − 2i A complex number? Yes A real number? No An imaginary number? Yes A purely imaginary number? No The imaginary unit? No
Page 377, Table of Contents
√ √ 3i 0 4 4 + 2i i 3
www.EconsPhDTutor.com
36.1
The Real and Imaginary Parts of Complex Numbers
Definition 95. Given a complex number z = a + bi, its real part is a and is denoted Re(z). Similarly, its imaginary part is b and is denoted Im(z).
Example 360. Re(3 + 2i) = 3 and Im(3 + 2i) = 2. Example 361. Re(7) = 7 and Im(7) = 0. Example 362. Re(19i) = 0 and Im(19i) = 19. It is also often convenient to write complex numbers in ordered pair notation, with the first term being the real part and the second term being the imaginary. Example 363. Given z = 3 + 2i, we can also write z = (3, 2). Example 364. Given z = 7, we can also write z = (7, 0). Example 365. Given z = 19i, we can also write z = (0, 19). Of course, two complex numbers z and w are equal if and only if (i) their real parts are equal; AND (ii) their imaginary parts are equal. Example 366. Suppose z = 3 + bi and w = a − 17i are equal. Then it must be that a = 3 and b = −17.
Exercise 141. Exactly two of the following complex numbers are identical. Find out which two. (Answer on p. 1105.) √ 1 2 a= √ − i, 2 2
Page 378, Table of Contents
3 1 b = √ − √ i, 2 2
π π c = sin − sin i, 3 3
√ d=
3 π − cos (− ) i. 2 4
www.EconsPhDTutor.com
37
Basic Arithmetic of Complex Numbers
The familiar arithmetic operations work the same way on imaginary numbers as they do on real numbers. Addition and subtraction are especially simple.
37.1
Addition and Subtraction
Example 367. Let z = −2 + i and w = 3i. Then z + w = −2 + 4i and z − w = −2 − 2i. We can also write z = (−2, 1) and w = (0, 3), so that z + w = (−2 + 0, 1 + 3) = (−2, 4) and z − w = (−2 − 0, 1 − 3) = (−2, −2). Example 368. Let z = 7 − i and w = 2 + 5i. Then z + w = 9 + 4i and z − w = 5 − 6i. We can also write z = (7, −1) and w = (2, 5), so that z + w = (7 + 2, −1 + 5) = (7 + 2, −1 + 5) and z − w = (7 − 2, −1 − 5) = (5, −6).
In general, Fact 44. If z = (a, b) and w = (c, d), then z + w = (a + c, b + d), z − w = (a − c, b − d).
Exercise 142. For each of the following, compute √ z +w and z −w. (a) z = −5+2i, w = 7+3i. (b) z = 3 − i, w = 11 + 2i. (c) z = 1 + 2i, w = 3 − 2i. (Answer on p. 1106.)
Page 379, Table of Contents
www.EconsPhDTutor.com
37.2
Multiplication
Below are listed the powers of i. Note that the cycle repeats after every fourth power, because i4 = 1. i = i,
i2 = i × i = −1,
i3 = i × i2 = −i,
i4 = i × i3 = 1,
i5 = i × i4 = i,
i6 = i × i = −1,
i7 = i × i2 = −i,
i8 = i × i3 = 1,
i9 = i × i8 = i,
i10 = i × i = −1,
i11 = i × i2 = −i,
i12 = i × i3 = 1,
etc. Example 369. Let z = i and w = 1 + i. Then zw = i × (1 + i) = i × 1 + i × i = i − 1. Example 370. Let z = −2 + i and w = 3i. Then zw = (−2 + i) × (3i) = (−2) × (3i) + i × (3i) = −6i + 3i2 = −6i + 3(−1) = −3 − 6i. Example 371. Let z = 2 − i and w = −1 + i. Then zw = (2 − i)(−1 + i) = −2 + 2i + i − i2 = −2 + 3i − i2 = −2 + 3i − (−1) = −1 + 3i. Example 372. Let z = 3 + 2i and w = −7 + 4i. Then zw = (3 + 2i)(−7 + 4i) = −21 + 12i − 14i + (2i)(4i) = −21 − 2i + 8i2 = −21 − 2i + 8 (−1) = −29 − 2i.
In general, Fact 45. If z = (a, b) and w = (c, d), then zw = (ac − bd, ad + bc) .
Page 380, Table of Contents
www.EconsPhDTutor.com
Exercise 143. Prove Fact 45. (Answer on p. 1106.)
Exercise 144. For each of the following, √ compute zw. (a) z = −5 + 2i, w = 7 + 3i. (b) z = 3 − i, w = 11 + 2i. (c) z = 1 + 2i, w = 3 − 2i. (Answer on p. 1106.) Exercise 145. Given that z = 2 + i and az 3 + bz 2 + 3z − 1 = 0, find a and b. (Answer on p. 1106.)
Page 381, Table of Contents
www.EconsPhDTutor.com
37.3
Division
Recall that to rationalise a surd in the denominator (section 5.2), we used a trick involving conjugate pairs. Example 373.
√ √ √ ( (1 3 3 − 5) 5 − 1) 3 3 1− 5 √ = √ × √ = = . 1−5 4 1+ 5 1+ 5 1− 5
We called a + b and a − b a conjugate pair because (a + b)(a − b) = a2 − b2 . If b is the square root of some number, then this is a rationalisation (“make rational”) that helps get rid of an ugly surd. Now, given z = a + bi, we call z ∗ = a − bi its conjugate. And we call a + bi and a − bi a conjugate pair, because (a + bi)(a − bi) = a2 − (bi) 2 = a2 − b2 i2 = a2 + b2 . This is a realisation (“make real”) that helps get rid of any complex numbers. Example: Example 374. (1 + i)∗ = 1 − i, i∗ = −i, and (1 − i)∗ = 1 + i. Thus: (a)
1−i 1 1 1−i 1−i = × = 2 2= = 0.5 − 0.5i. 1+i 1+i 1−i 1 −i 1+1 1 1 −i −i −i = × = = = −i. i i −i −i2 1
(b)
(c)
1 1 1+i 1+i 1+i = × = 2 2= = 0.5 + 0.5i. 1−i 1−i 1+i 1 −i 1+1
In general, Fact 46. If z = (a, b), then ∗
z = (a, −b),
∗
2
2
zz = a + b ,
1 1 z∗ z∗ a −b = × ∗ = 2 2 = ( 2 2, 2 2). z z z a +b a +b a +b
Exercise 146. For each of the following z, write down its conjugate z ∗ and hence compute its reciprocal (i.e. 1/z). (a) z = −5 + 2i. (b) z = 3 − i. (c) z = 1 + 2i. (Answer on p. 1106.)
Page 382, Table of Contents
www.EconsPhDTutor.com
We now divide one complex number by another. Example 375. (a)
(b)
−2 + i −2 + i −3i 6i − 3i2 6i + 3 = × = = . 3i 3i −3i −9i2 9 3 + i 3 + i 1 + i (3 + i)(1 + i) 3 + 3i + i + i2 2 + 4i = × = = = = 1 + 2i. 1−i 1−i 1+i 12 − i2 1+1 2
(c)
1+i 1 + i 3 + 2i 3 + 2i + 3i + 2i2 1 + 5i = × = = . 3 − 2i 3 − 2i 3 + 2i 9+4 13
(d)
2 − i −1 − i −2 − 2i + i + i2 −3 − i 2−i = × = = = −1.5 − 0.5i. −1 + i −1 + i −1 − i 1+1 2
(e)
3 + 2i 3 + 2i −7 − 4i −21 − 12i − 14i − 8i2 −13 − 26i = × = = = −0.2 − 0.4i. −7 + 4i −7 + 4i −7 − 4i 49 + 16 65
(f)
−3 + 6i −3 + 6i 2 − πi −6 + 3πi + 12i − 6πi2 6π − 6 + (3π + 12)i = × = = . 2 + πi 2 + πi 2 − πi 22 − π 2 i 2 4 + π2
In general, Fact 47. If z = (a, b) and w = (c, d) with w ≠ 0, then z z w∗ zw∗ ac + bd bc − ad = × ∗= 2 = ( , ). w w w c + d2 c2 + d2 c2 + d2
Exercise 147. Rewrite each of the following fractions into the form a + bi. (a) √ 2 − 3i 11 + 2i 2 − πi −3 7 − 2i √ . (d) . (c) . (e) . (f) . (Answer on p. 1107.) 1+i i 2+i 5+i 3 − 2i
Page 383, Table of Contents
1 + 3i . (b) −i
www.EconsPhDTutor.com
38 38.1
Solving Polynomial Equations Complex Roots to Quadratic Equations
In section 14 (quadratic equations review), we saw that if ax2 + bx + c = 0 has non-negative discriminant (i.e. b2 − 4ac ≥ 0), then its real roots are given by x=
−b ±
√
b2 − 4ac . 2a
Example 376. Consider the equation x2 − 3x + 2 = 0. Its discriminant is positive: b2 − 4ac = (−3)2 − 4(1)(2) = 1 > 0. Hence, it has two real roots, given by x=
−b ±
√ √ b2 − 4ac 3 ± 1 = = 1, 2. 2a 2
Now, armed with our new concept of imaginary numbers, we can completely dispense with the requirement that b2 − 4ac ≥ 0. We can simply say that ax2 + bx + c = 0 ALWAYS has complex roots, given by x=
−b ±
√
b2 − 4ac . 2a
Example 377. Consider the equation x2 −2x+2 = 0. Its discriminant is negative: b2 −4ac = (−2)2 − 4(1)(2) = −4 < 0. It has two imaginary (and thus also complex) roots, given by x=
−b ±
√
√ √ √ b2 − 4ac 2 ± −4 4 × −1 2i = =1± = 1 ± = 1 ± i. 2a 2 2 2
Notice that 1 + i was a root to the given quadratic equation. And interestingly enough, so too was 1 − i. It turns out that in general, a quadratic equation with real coefficients has roots that come in conjugate pairs. That is, if x + yi is a root, then so too is its conjugate x − yi. 40 More examples:
40
√ This is not terribly surprising if you examine the general solution for the quadratic equation — the ± b2 − 4ac bit corresponds precisely to the imaginary part.
Page 384, Table of Contents
www.EconsPhDTutor.com
Example 378. Consider the equation 3x2 +x+1 = 0. Its discriminant is negative: b2 −4ac = (1)2 − 4(3)(1) = −11 < 0. It has two imaginary (and thus also complex) roots, given by x=
−b ±
√ √ √ √ √ b2 − 4ac −1 ± −11 11 × −1 11 1 1 = =− ± =− ± i. 2a 6 6 6 6 6
Example 379. If 3 + 2i is a root to the quadratic equation x2 + bx + c = 0 (where b and c are both real), then what are b and c? Well, we know that 3 − 2i is also a root to the equation. And so x2 + bx + c = [x − (3 + 2i)] [x − (3 − 2i)] = (x − 3)2 − (2i)2 = x2 − 6x + 13. Hence, b = 6 and c = 13.
Exercise 148. Find the roots for each of the following quadratic equations. (a) x2 +x+1 = 0. (b) x2 + 2x + 2 = 0. (c) 3x2 + 3x + 1 = 0. (Answer on p. 1108.) Exercise 149. If 1 − i is a root to the quadratic equation x2 + bx + c = 0 (where b and c are both real), then what are b and c? (Answer on p. 1108.)
Page 385, Table of Contents
www.EconsPhDTutor.com
38.2
The Fundamental Theorem of Algebra
Recall from p. 47 that a polynomial of degree n in one variable is any expression a0 xn + a1 xn−1 + a2 xn−2 + ⋅ ⋅ ⋅ + an−1 x + an where each ai is a constant and x is the variable. Theorem 4. The Fundamental Theorem of Algebra. A polynomial of degree n in one variable has exactly n zeros (though some may be repeated). That is, there are exactly n (possibly repeated) solutions to the equation a0 xn + a1 xn−1 + a2 xn−2 + ⋅ ⋅ ⋅ + an−1 x + an = 0. Proof. The proof of this theorem is way too advanced and so omitted from this book.41
Example 380. x2 − 1 is a polynomial of degree 2. And indeed, x2 − 1 = 0 has two solutions, namely 1 and −1. Example 381. x2 + 1 is a polynomial of degree 2. And indeed, x2 + 1 = 0 has two solutions, namely i and −i. There are sometimes “repeated solutions” or what are more formally called multiple roots, as the next example illustrates. Example 382. x2 − 2x + 1 = 0 has two (repeated) solutions, namely 1 and 1. We call 1 a multiple root (indeed a double root).
Example 383. x3 − 6x2 + 12x − 8 = 0 has three (repeated) solutions, namely 2, 2, and 2. We call 2 a multiple root (indeed a triple root).
41
But see this MathOverflow Q&A if you’re interested.
Page 386, Table of Contents
www.EconsPhDTutor.com
The Fundamental Theorem of Algebra can be useful even if we have no idea how to find the solutions to an equation. Example 384. x17 + 3x4 − 2x + 1 is a polynomial of degree 17. I may not know what the solutions to x17 + 3x4 − 2x + 1 = 0 are, but I know from the Fundamental Theorem of Algebra that there MUST be 17 solutions (though some may possibly be repeated).
Example 385. x4 + x3 − 5x2 + x − 6 is a polynomial of degree 4 and so it must have four zeros. Suppose we are given as a hint that two of them are i and −i. Then how would we go about finding the other two zeros? The problem of finding the zeros of a polynomial is really the same as the problem of factorising a polynomial. This is because a is a zero of a polynomial if and only if (x − a) is a factor of the polynomial. So (x − i) and (x + i) are factors for the polynomial. Now, (x − i)(x + i) = x2 − i2 = x2 + 1. So find (x4 + x3 − 5x2 + x − 6) ÷ (x2 + 1) through long division: x2 +x −6 x2 + 1 x4 +x3 −5x2 x4 +0 +x2 x3 −6x2 x3 +0 −6x2 −6x2
+x −6
+x +0 −6 +0 −6 0.
Hence, (x4 + x3 − 5x2 + x − 6) = (x2 + 1) (x2 + x − 6). By observation, x2 + x − 6 = (x − 2)(x + 3). Hence, (x4 + x3 − 5x2 + x − 6) = (x2 + 1) (x2 + x − 6) = (x − i)(x + i)(x − 2)(x + 3). Altogether, the four zeros of the given polynomial are ±i, 2, and −3.
Page 387, Table of Contents
www.EconsPhDTutor.com
Example 386. x3 − 3x2 − 5x − 25 is a polynomial of degree 3 and so it must have three zeros. As a hint, we are told that one of them is 5. What are the other two? x2 +2x +5 x − 5 x3 −3x2 −5x −25 x3 −5x2 2x2 2x2 −10x 5x −25 5x −25 0. So x3 − 3x2 − 5x − 25i = (x − 3) (x2 + 2x + 5). I’m unable to easily see how x2 + 2x + 5 can be factorised. So let me just use the quadratic formula:
x=
−2 ±
√
√ √ 22 − 4(1)(5) = −1 ± 1 − 5 = −1 ± −4 = −1 ± 2i. 2
Altogether then, x3 − 3x2 − 5x − 25i = (x − 5) (x2 + 2x + 5) = (x − 5) [x − (−1 + 2i)] [x − (−1 − 2i)] . So the three zeros of the polynomial are 5 and −1 ± 2i.
Exercise 150. Each of the following polynomials has 1 as a zero. Find the other zeros. (a) x3 + x2 − 2. (b) x4 − x2 − 2x + 2. (Answer on p. 1109.)
Page 388, Table of Contents
www.EconsPhDTutor.com
38.3
The Complex Conjugate Roots Theorem
We saw above that if c + di is a root to a quadratic equation ax2 + bx + c = 0 (where a, b, and c are real), then so too is its conjugate c − di. What is perhaps surprising is that this generalises to the case of any polynomial, provided that all coefficients of the polynomial are real. Example 387. If told that 2 − i solves x3 − x2 − 7x + 15 = 0, we know immediately that its conjugate 2 + i also solves the same equation. Example 388. If told that i solves 4x4 +5x2 +1 = 0, we know immediately that its conjugate −i also solves the same equation. Similarly, if told also that 0.5i solves the same equation, we know immediately that its conjugate −0.5i also solves the same equation.
Theorem 5. (Complex Conjugate Roots Theorem.) Let a0 , a1 , . . . , ak be real. If a + bi solves an xn + an−1 xn−1 + an−2 xn−2 + ⋅ ⋅ ⋅ + a1 x + a0 = 0, then so does a − bi.
Proof. Optional, see p. 944 in Appendices. The condition that all coefficients ak are real is important. The above theorem does not apply if any of the coefficients are imaginary. Example 389. i solves x2 + ix + 2 = 0. However, its conjugate −i does not solve the same equation (verify this yourself!). √ √ √ √ 2 2 2 2 Example 390. + i solves x2 = i. However, its conjugate − i does not solve 2 2 2 2 the same equation (verify this yourself!).
Example 391. 2 + i solves x3 − (i + 2)x2 + 2x − 2(2 + i) = 0. However, its conjugate 2 − i does not solve the same equation (verify this yourself!).
Exercise 151. Each of the following polynomials has 2 − 3i as a zero. Find the other zeros. (a) x4 − 6x3 + 18x2 − 14x − 39. (b) −2x4 + 21x3 − 93x2 + 229x − 195. (Answer on p. 1110.)
Page 389, Table of Contents
www.EconsPhDTutor.com
39
The Argand Diagram
The complex plane (or Argand diagram) gives us a nice geometric interpretation: The complex numbers are simply points on the plane. The real axis is the horizontal or x-axis. The imaginary axis is the vertical or y-axis. Example 392. In the figure below, marked in red are the real numbers −3, 0, π, and 2, which may be written in ordered pair notation as (−3, 0), (0, 0), (π, 0), (2, 0). Points on the horizontal axis are real numbers. In blue are the purely imaginary numbers −4i and 3i, which may be written in ordered pair notation as (0, −4) and (0, 3). Points on the vertical axis are purely imaginary numbers. In green are the “impure” imaginary numbers 1 + i, −3 + 2i, 1 − 3i, and −4 − i, which may be written in ordered pair notation as (1, 1), (−3, 2), (1, −3), and (−4, −1). Points not on either axis are “impure” imaginary numbers.
5
y
4 3 2 1 x 0 -5
-4
-3
-2
-1
0
1
2
3
4
5
-1 -2 -3 -4 -5
Page 390, Table of Contents
www.EconsPhDTutor.com
For our purposes, we’ll regard the complex plane C as being exactly identical to the cartesian plane {(x, y) ∶ x ∈ R, y ∈ R}. Both are represented graphically as a two-dimensional plane. The only difference is that we interpret points on each plane differently: Points on the complex plane are complex numbers, while points on the cartesian plane are ordered pairs of real numbers.42
Exercise 152. Illustrate the complex numbers 1, −3, 2i, 1 + 2i, and −1 − 3i on a single Argand diagram. (Answer on p. 1111.)
42
The differences between C and R2 in fact run deeper. See e.g. this discussion..
Page 391, Table of Contents
www.EconsPhDTutor.com
39.1
Complex Numbers in Polar Form
To write a complex number in standard form — i.e. z = x + iy, we need only two pieces of information: its real part (x) and its imaginary part (y). We now write a complex number in polar form. Again, we need only two pieces of information: the modulus, denoted ∣z∣, and the argument, denoted arg z. Informally, the modulus is the length of the position vector of z; the argument is the angle the position vector of z makes with the positive x-axis. Example 393. The complex number −3 = (−3, 0) has modulus ∣ − 3∣ = 3 and argument arg 3 = π. The complex number −4i = (0, −4) has modulus ∣−4i∣ = 4 and argument arg√ (−4i) = √ 2 2 −π/2. The complex number 3 + 3i = (3, 3), has modulus ∣3 + 3i∣ = 3 + 3 = 3 2 and argument arg(3 + 3i) = π/4.
5
y
4 3 + 3i = (3, 3) 3 2 1 -3 = (-3, 0)
x 0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1 -2 -3 -4
-4i = (0, -4)
-5
Page 392, Table of Contents
www.EconsPhDTutor.com
The formal definition of the modulus function is simple. Definition 96. The modulus function has domain C, codomain R, and mapping rule z ↦ √ x2 + y 2 . The modulus of z is denoted ∣z∣.
In contrast, it is tricky to write down a formal definition of the argument function. One problem is this: Angles are periodic. Example 394. Consider again the complex number 3 + 3i = (3, 3). The angle it makes with the positive x-axis is π/4. But angles are periodic. Equivalently, angles come full circle 2π radians. So it would make just as much sense to say that the angle is 9π/4. Or 17π/4. Or −7π/4. Or indeed any π/4 + 2kπ, where k is any inteer. To overcome this problem, we shall somewhat arbitrarily choose (−π, π] as our principal values. Thus, arg(3 + 3i) shall be uniquely defined to be the value π/4 and nothing else. Another problem is this: We are tempted to simply define arg(x + yi) = tan−1 (y/x). Unfortunately, the tan−1 function has codomain ge (−π/2, π/2). Whereas, as we just decided, arg should have codomain (−π, π]. To overcome this, altogether, the argument function is defined as follows: Definition 97. The argument function has domain C, codomain (−π, π], and mapping rule as given below: ⎧ ⎪ ⎪ tan−1 (y/x) , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Undefined, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪π/2, arg z = ⎨ ⎪ ⎪ −π/2, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ tan−1 (y/x) + π, ⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎪ ⎪ ⎩tan (y/x) − π,
if x > 0 (top-right and bottom-right quadrants), if x = 0 = y (the origin), if x = 0, y > 0 (the positive y − axis), if x = 0, y < 0, (the negative y − axis) if x < 0, y ≥ 0 (top-left quadrant, including the negative x-axis), if x < 0, y < 0 (bottom-left quadrant).
The argument of z is denoted arg z. y (My mnemonic for the above: “arg z = tan−1 . Top left +π. Bottom left−π.”) x We now illustrate and explain the above definition:
Page 393, Table of Contents
www.EconsPhDTutor.com
y arg z =
arg z = x arg z =
• If x > 0 (top-right and bottom-right quadrants), then define arg(x + yi) = tan−1 (y/x). The green point in the figure above illustrates. The angle that the position vector of the green (x, y) makes with the positive x-axis is indeed simply tan−1 (y/x). • If x = 0, y = 0 (the origin), then arg(x + yi) is undefined. In other words, we leave arg 0 undefined.43 • If x = 0, y > 0 (positive vertical axis), then define arg(x + yi) = arg(yi) = π/2. • If x = 0, y < 0 (negative vertical axis), then define arg(x + yi) = arg(yi) = −π/2. • If x < 0, y ≥ 0 (top-left quadrant plus the negative horizontal axis), then define arg(x + yi) = tan−1 (y/x) + π. The red point illustrates. The angle its position vector makes with the negative x-axis is tan−1 (y/∣x∣). And so arg(x + yi) = π − tan−1 (y/∣x∣). Observe that tan−1 (y/∣x∣) = tan−1 (y/ − x) = − tan−1 (y/x). Thus, arg(x + yi) = π − tan−1 (y/∣x∣) = tan−1 (y/x) + π. • If x < 0, y < 0 (bottom-left quadrant), then define arg(x + yi) = tan−1 (y/x) − π. The blue point illustrates. The angle its position vector makes with the negative x-axis is tan−1 (∣y∣/∣x∣). And so arg(x + yi) = tan−1 (∣y∣/∣x∣) − π. Observe that (∣y∣/∣x∣) = (−y/ − x) = (y/x). Thus, arg(x + yi) = π − tan−1 (∣y∣/∣x∣) = tan−1 (y/x) − π. 43
Some writers define arg 0 = 0, but we shall not do this.
Page 394, Table of Contents
www.EconsPhDTutor.com
The following fact is sometimes useful. Fact 48. (a) z is purely imaginary (z is on the vertical axis) ⇐⇒ arg z = ±π/2. (a) z is real (z is on the horizontal axis) ⇐⇒ arg z = 0, π.
Proof. Immediate from the definition of the arg function.
Exercise 153. Compute the modulus and argument of 4, −3, 2i, 1+2i, and −1−3i. Illustrate these numbers and their arguments on a single Argand diagram. (Answer on p. 1112.)
Exercise 154. Where on the complex plane must a complex number be, if its argument is π π π π ... (a) Positive? (b) Negative? (c) 0? (d) ? (e) − ? (f) > ? (g) < − ? (Answer on p. 2 2 2 2 1113.)
Page 395, Table of Contents
www.EconsPhDTutor.com
Armed with the modulus and the argument, we have a nice geometric interpretation: Fact 49. Let z be a complex number with ∣z∣ = r and arg z = θ. Then z = r (cos θ + i sin θ). We call r (cos θ + i sin θ) the polar form representation of z.
y z = r (cos Ʌ + i sin Ʌ) = (r cos Ʌ, r sin Ʌ)
r r sin Ʌ x r cos Ʌ
√ √ −2 2 ≈ Example 395. For z = 5 − 2i = (5, −2), ∣z∣ = 52 + (−2) = 29 and arg z = tan−1 5 √ −0.381. So in polar form, z = 29 [cos (−0.381) + i sin (−0.381)]. √ √ 3 Example 396. For z = 1 + 3i = (1, 3), ∣z∣ = 12 + 32 = 10 and arg z tan−1 ≈ 1.249. So in 1 √ polar form, z = 10 [cos (1.249) + i sin (1.249)]. √ √ 7 Example 397. For z = −4 + 7i = (−4, 7), ∣z∣ = (−4)2 + 72 = 65 and arg z = tan−1 +π ≈ −4 √ 2.090. So in polar form, z = 65 [cos (2.090) + i sin (2.090)].
Exercise 155. Rewrite each of the following complex numbers in polar form: 1, −3, 2i, 1 + 2i, and −1 − 3i. (Answer on p. 1113.)
Page 396, Table of Contents
www.EconsPhDTutor.com
39.2
Complex Numbers in Exponential Form
Theorem 6. Euler Formula. For any θ ∈ R, eiθ = cos θ + i sin θ.
Proof. Optional, see p. 944 in the Appendices.
Fact 50. Let z be a complex number with ∣z∣ = r and arg z = θ. Then z = reiθ .
1
2
1
Proof. ∣z∣ = r and arg z = θ Ô⇒ z = r(cos θ + i sin θ) ⇐⇒ z = reiθ , where ⇐⇒ uses Fact 2 49 and ⇐⇒ uses the Euler Formula. We call reiθ the exponential form representation of z. Euler’s identity is one of the most extraordinary and beautiful equations in all of mathematics. It links together five fundamental mathematical constants: e, i, π, 1, and 0. Corollary 4. (Euler’s identity.) eiπ + 1 = 0.
Proof. By Theorem 6, eiπ = cos π + i sin π = −1 + 0 = −1. Hence, eiπ + 1 = 0. √ √ 2 Example 398. The number z = 5 − 2i = (5, −2) has modulus 52 + (−2) = 29 and √ argument tan−1 (−2/5) ≈ −0.381. Hence, we can also write z = 29ei(−0.381) . √ √ 2 + 32 = Example 399. The number z = 1 + 3i = (1, 3) has modulus 1 10 and argument √ i(1.249) −1 tan (3/1) ≈ 1.249. Hence, we can also write z = 10e . √ √ Example 400. The number z = −4 + 7i = (−4, 7) has modulus√ (−4)2 + 72 = 65 and argument tan−1 (7/ − 4) + π ≈ 2.090. Hence, we can also write z = 65ei(2.090) .
Exercise 156. Rewrite each of the following complex numbers in exponential form: 1, −3, 2i, 1 + 2i, and −1 − 3i. (Answer on p. 1113.)
Page 397, Table of Contents
www.EconsPhDTutor.com
40
More Arithmetic of Complex Numbers
Now that we know how to write complex numbers in polar and exponential forms, the arithmetic of complex numbers becomes even easier.
40.1
The Product of Two Complex Numbers
Fact 51. Product of two complex numbers. Let z and w be complex numbers. Then ∣zw∣ = ∣z∣ ∣w∣ ,
and
arg (zw) = arg z + arg w + 2kπ,
(where k = −1, 0, 1 ensures that arg z + arg w + 2kπ ∈ (−π, π]). Proof. Let z = r(cos θ + i sin θ) and w = s(cos φ + i sin φ). Then zw = rs(cos θ + i sin θ)(cos φ + i sin φ) = rs(cos θ cos φ + i sin θ cos φ + i cos θ sin φ − sin θ sin φ) = rs [cos (θ + φ) + i sin (θ + φ)] . This is the complex number with modulus rs and which makes an angle θ + φ with the positive x-axis. Note though that θ + φ may not be in (−π, π]. Thus, rather than say that arg(zw) = arg z + arg w, we instead say that arg (zw) = arg z + arg w + 2kπ (where k = −1, 0, 1 ensures that arg z + arg w + 2kπ ∈ (−π, π]).
Here is an alternative quicker proof of the above fact, using the exponential form. Proof. Let z = reiθ and w = seiφ . Then zw = rsei(θ+φ) . This is the complex number with modulus rs and which makes an angle θ + φ with the positive x-axis.
Page 398, Table of Contents
www.EconsPhDTutor.com
Example 401. Let z = 5 − 2i and w = 1 + 3i. Then √ √ 2 ∣z∣ = 52 + (−2) = 29, and arg z = tan−1 (−2/5), √ √ ∣w∣ = 12 + 32 = 10, and arg w = tan−1 (3/1), √ √ √ Ô⇒ ∣zw∣ = 29 × 10 = 290, and arg (zw) = tan−1 (−2/5) + tan−1 (3/1) + 2kπ ≈ 0.869 + 2kπ = 0.869 (k = 0). Notice that here arg z + arg w ≈ 0.869 ∈ (−π, π]. So arg z + arg w is already a principal value and we can simply set k = 0 or arg(zw) = arg z + arg w. √ √ So zw ≈ 290 (cos 0.869 + i sin 0.869) = 290ei(0.869) . To get zw in standard form, use a calculator: You’ll get √ −2 3 290 cos [tan−1 + tan−1 ] = 11, 5 1
and
√ −2 3 290 sin [tan−1 + tan−1 ] = 13. 5 1
And indeed zw = (5 − 2i)(1 + 3i) = 11 + 13i.
Example 402. Let z = −4 + 7i and w = 1 − 6i. Then ∣z∣ =
√
(−4)2 + 72 =
√
65, and
arg z = tan−1 [7/ (−4)] + π,
√ √ ∣w∣ = 12 + (−6)2 = 37, and arg w = tan−1 (−6/1), √ √ √ Ô⇒ ∣zw∣ = 65 × 37 = 2405, and arg (zw) = tan−1 [7/ (−4)] + π + tan−1 (−6/1) + 2kπ ≈ 0.684 + 2kπ = 0.684 (k = 0). Notice that here arg z + arg w ≈ 0.684 ∈ (−π, π]. So arg z + arg w is already a principal value and we can simply set k = 0 or arg(zw) = arg z + arg w. √ √ So zw ≈ 2405 (cos 0.684 + i sin 0.684) = 2405ei(0.684) . To get zw in standard form, use a calculator: You’ll get √ −7 −6 2405 cos [tan−1 + π + tan−1 ] = 38, 4 1
and
√ −7 −6 2405 sin [tan−1 + π + tan−1 ] = 31. 4 1
And indeed zw = (−4 + 7i)(1 − 6i) = 38 + 31i. Page 399, Table of Contents
www.EconsPhDTutor.com
Example 403. Let z = −3 + 4i and w = −5 + 2i. Then √
(−3)2 + 42 = 5, and arg z = tan−1 [4/ (−3)] + π, √ √ ∣w∣ = (−5)2 + 22 = 29, and arg w = tan−1 [2/ (−5)] + π. ∣z∣ =
Ô⇒
√ ∣zw∣ = 5 29, arg (zw) = (tan−1
and 2 4 + π) + (tan−1 + π) + 2kπ ≈ 4.975 + 2kπ = −1.308 (k = −1). −3 −5
Notice that here arg z + arg w ≈ 4.975 ∉ (−π, π]. So we need to set k = −1 to get arg(zw) = arg z + arg w − 2kπ ≈ −1.308 ∈ (−π, π], so that arg(zw) is indeed a principal value. √ √ So zw ≈ 5 29 [cos (−1.308) + i sin (−1.308)] = 5 29ei(−1.308) . To get zw in standard form, use a calculator: You use a calculator, you’ll get √ √ −2 −2 −4 −4 5 29 × cos [tan−1 + π + tan−1 ] = 7 and 5 29 × sin [tan−1 + π + tan−1 ] ≈ −26. 3 5 3 5 And indeed zw = (−3 + 4i)(−5 + 2i) = 7 − 26i.
Exercise 157. Write down zw in polar and exponential forms, for each of the following pair of z and w. (a) z = 1, w = −3. (b) z = 2i, w = 1 + 2i. (c) z = −1 − 3i, w = 3 + 4i. (Answer on p. 1114.)
Page 400, Table of Contents
www.EconsPhDTutor.com
40.2
The Ratio of Two Complex Numbers
Fact 52. Ratio of two complex numbers. Let z and w be complex numbers. Then z ∣z∣ ∣ ∣= , w ∣w∣
and
arg
z = arg z − arg w + 2kπ, w
(where k = −1, 0, 1 ensures that arg z + arg w + 2kπ ∈ (−π, π]). Proof. Let z = r(cos θ + i sin θ) and w = s(cos φ + i sin φ). Then z r(cos θ + i sin θ) r cos θ + i sin θ cos φ − i sin φ = = × w s(cos φ + i sin φ) s cos φ + i sin φ cos φ − i sin φ =
r cos θ cos φ + sin θ sin φ + i sin θ cos φ − i cos θ sin φ s cos2 φ + sin2 φ
=
r cos θ cos φ + sin θ sin φ + i sin θ cos φ − i cos θ sin φ s 1
=
r [cos (θ − φ) + i sin (θ − φ)] . s
This is the complex number with modulus r/s and argument θ − φ + 2kπ (where k is the unique integer such that θ − φ + 2kπ ∈ (−π, π]).
Here is an alternative quicker proof of the above fact, using the exponential form. Proof. Let z = reiθ and w = seiφ . Then z/w = ei(θ−φ) (r/s). This is the complex number with modulus r/s and argument θ − φ + 2kπ (where k is the unique integer such that θ − φ + 2kπ ∈ (−π, π]).
I’ll recycle the examples used in the previous section.
Page 401, Table of Contents
www.EconsPhDTutor.com
Example 404. Let z = 5 − 2i and w = 1 + 3i. Then ∣z∣ =
√ 29,
arg z = tan−1
√ z ∣ ∣ = 2.9, w
Ô⇒
Ô⇒
zw ≈
−2 , 5
∣w∣ =
√
3 arg w = tan−1 . 1
10,
−2 3 z − tan−1 + 2kπ ≈ −1.630 (k = 0). arg ( ) = tan−1 w 5 1
√ √ 2.9 [cos (−1.630) + i sin (−1.630)] = 2.9ei(−1.630) .
Example 405. Let z = −4 + 7i and w = 1 − 6i. Then ∣z∣ = Ô⇒
z ∣ ∣= w
√
65,
arg z = tan−1
7 + π, −4
∣w∣ =
√ 37,
arg w = tan−1
−6 . 1
√
z 65 7 −6 , arg ( ) = tan−1 +π−tan−1 +2kπ ≈ 3.496+2kπ ≈ −2.788 (k = −1). 37 w −4 1 √ √ z 65 65 i(−2.788) ≈ [cos (−2.788) + i sin (−2.788)] = e . Ô⇒ w 37 37
Example 406. Let z = −3 + 4i and w = −5 + 2i. Then ∣z∣ = 5, Ô⇒
z 5 ∣ ∣= √ , w 29 Ô⇒
arg z = tan−1
4 + π, −3
∣w∣ =
√ 29,
arg w = tan−1
2 + π. −5
z 4 2 arg ( ) = tan−1 − tan−1 + 2kπ ≈ −0.547 + 2kπ = −0.547 (k = 0). w −3 −5 z 5 5 ≈ √ [cos (−0.547) + i sin (−0.547)] = √ ei(−0.547) . w 29 29
z in polar and exponential forms, for each of the following w pairs of z and w. (a) z = 1, w = −3. (b) z = 2i, w = 1 + 2i. (c) z = −1 − 3i, w = 3 + 4i. (Answer on p. 1114.) Exercise 158. Write down
Page 402, Table of Contents
www.EconsPhDTutor.com
40.3
Sine and Cosine as Weighted Sums of the Exponential
Fact 53 expresses the sine and cosine functions as weighted sums of the exponential functions. It is not in the syllabus, but made a sudden first-time appearance on the 2015 A-level exams (Exercise 356), just to screw students over.
Fact 53. cos θ =
eiθ + e−iθ eiθ − e−iθ and sin θ = . 2 2i
Proof. By the Euler Formula, eiθ = cos θ+i sin θ. Moreover, e−iθ = cos (−θ)+i sin (−θ) = cos θ− i sin θ, where the second equality uses the properties cos x = cos(−x) and sin(−x) = − sin x. Hence,
eiθ + e−iθ cos θ + i sin θ + cos θ − i sin θ = = cos θ, as desired. 2 2
eiθ − e−iθ cos θ + i sin θ − cos θ + i sin θ = = sin θ, also as desired. Similarly, 2i 2
The 2015 question was about the sum (or difference) of two complex numbers that have the same modulus. Here’s a similar example:
Page 403, Table of Contents
www.EconsPhDTutor.com
Example 407. Let z = 5eiπ and w = 5e0.4iπ . What, exactly, are the modulus and arguments of z + w and z − w? Without the above fact, it’s not obvious. With the above fact, it’s easy. First, observe that 0.7 is the average of 1 and 0.4. Then factorise 5eiπ + 5e0.4iπ into a form where we can exploit the above fact. Like so: z + w = 5eiπ + 5e0.4iπ = 5e0.7iπ (e0.3iπ + e−0.3iπ ) = 5e0.7iπ × 2 cos(0.3π), where the last = uses Fact 53. And thus: arg (z + w) = arg [5e0.7iπ × 2 cos(0.3π)] = arg 5 + arg (e0.7iπ ) + arg 2 + arg [cos(0.3π)] + 2kπ = 0 + 0.7π + 0 + 0 + 2kπ = 0.7π (k = 0). ∣z + w∣ = ∣5e0.7iπ × 2 cos(0.3π)∣ = ∣5∣ ∣e0.7iπ ∣ ∣2∣ ∣cos(0.3π)∣ = 5 × 1 × 2 × cos(0.3π) = 10 cos(0.3π). Altogether then, z + w = 10 cos(0.3π)ei(0.7π) . We can play a similar trick to figure out the modulus and argument of z − w: z − w = 5eiπ − 5e0.4iπ = 5e0.7iπ (e0.3iπ − e−0.3iπ ) = 5e0.7iπ × 2i sin(0.3π). where again the last = uses Fact 53. And thus: arg (z − w) = arg [5e0.7iπ × 2i sin(0.3π)] = arg 5 + arg (e0.7iπ ) + arg 2 + arg i + arg [sin(0.3π)] + 2kπ = 0 + 0.7π + 0 + 0.5π + 0 + 2kπ = −0.8π (k = −1), ∣z − w∣ = ∣5e0.7iπ × 2i sin(0.3π)∣ = ∣5∣ ∣e0.7iπ ∣ ∣2∣ ∣i∣ ∣sin(0.3π)∣ = 5 × 1 × 2 × 1 × sin(0.3π) = 10 sin(0.3π). Altogether then, z − w = 10 sin(0.3π)ei(−0.8π) .
Page 404, Table of Contents
www.EconsPhDTutor.com
In general, Fact 54. eiθ + eiφ = 2 cos
θ − φ i( θ+φ +2kπ) e 2 2
and
eiθ − eiφ = 2 sin
θ − φ i( θ+φ+π +2mπ) e 2 , 2
where k, m = −1, 0, 1 ensure that 0.5(θ + φ) + 2kπ ∈ (−π, π] and 0.5(θ + φ + π) + 2mπ ∈ (−π, π]. Proof. See Exercise 160.
Exercise 159. Let z = 3ei(0.2π) and w = 3ei(−0.9π) . By mimicking the steps in Example 407, find z + w and z − w in exact polar and exponential forms. (Answer on p. 1115.)
Exercise 160. Prove Fact 54. (Answer on p. 1116.)
SYLLABUS ALERT If you’re taking the 9758 (revised) exam, you are done with Part IV: Complex Numbers. The remaining chapters in Part IV covers the following, which are on the 9740 (old) syllabus but not on the 9758 (revised) syllabus: • geometrical effects of conjugating a complex number and of adding, subtracting, multiplying, dividing two complex numbers • loci such as ∣z − c∣ ≤ r, ∣z − a∣ ≤ ∣z − b∣ and arg (z − a) = α • use of de Moivre’s theorem to find the powers and nth roots of a complex number.
Page 405, Table of Contents
www.EconsPhDTutor.com
41
Geometry of Complex Numbers
In secondary school, we learnt to do some geometry using cartesian equations. And in Part III (Vectors), we learnt to do some geometry using vector equations. Now, weâ&#x20AC;&#x2122;ll learn to do some geometry using complex equations!
41.1
The Sum and Difference of Two Complex Numbers
Given two complex numbers z = x + iy and w = a + ib, their sum is simply the complex number z + w = (x + a) + (y + b)i. We already know how to interpret z = (x, y) and w = (a, b) as points on the plane. This gives us a nice geometric interpretation: z +w = (x+a, y +b) is likewise a point on the plane. We can also interpret z = (x, y) and w = (a, b) as position vectors. And thus as usual, the sum of two vectors is itself a vector: z + w = (x + a, y + b).
y
z + w = (x + a, y + b) z = (x, y)
w
z+w
w = (a, b)
x
Page 406, Table of Contents
www.EconsPhDTutor.com
Similarly, their difference z − w is simply the point (x − a, y − b). This corresponds to the vector z − w = (x − a)i + (y − b)j.
y
z = (x, y) z - w = (x - a, y - b)
z-w
w = (a, b) x
Note that in general, ∣z + w∣ ≠ ∣z∣ + ∣w∣ or that ∣z − w∣ ≠ ∣z∣ − ∣w∣. This is perhaps obvious from the above figures and also bearing in mind Corollary 3 (the sum of the lengths of any two sides of a triangle is always greater than the length of the third side).
Page 407, Table of Contents
www.EconsPhDTutor.com
41.2
The Product and Ratio of Two Complex Numbers
With sums and differences, there was an exact analogy to vectors. In contrast, with products and ratios of complex numbers, there is no analogy to vectors. In particular, the product of two complex numbers has nothing to do with the scalar product or vector product of their position vectors. Nonetheless, we do have nice geometric interpretations. We already know from Fact 51 that the product of two complex numbers z and w is simply the complex number zw with 1. ∣zw∣ = ∣z∣ ∣w∣; and 2. arg (zw) = arg z + arg w + 2kπ, where k = −1, 0, 1 ensures that arg z + arg w + 2kπ ∈ (−π, π]. So geometrically, to get zw, we take z and 1. First multiply its length by a factor equal to the length of w; 2. Then rotate it anti-clockwise by the angle arg w.
y zw = (ac - bd, ad + bc)
z = (a, b)
w = (c, d) x
Page 408, Table of Contents
www.EconsPhDTutor.com
Similarly, we already know from Fact 51 that the ratio of complex numbers z to w is simply the complex number z/w with 1. ∣z/w∣ = ∣z∣ / ∣w∣; and 2. arg (z/w) = arg z −arg w +2kπ, where k = −1, 0, 1 ensures that arg z −arg w +2kπ ∈ (−π, π]. So geometrically, to get z/w, we take z and 1. First compress its length by a factor equal to the length of w; 2. Then rotate it anti-clockwise by the angle − arg w. (Or equivalently, clockwise by the angle arg w.)
y
z = (a, b)
w = (c, d)
x
Page 409, Table of Contents
www.EconsPhDTutor.com
41.3
Conjugating a Complex Number
If z = x + yi = (x, y), then z ∗ = x − yi = (x, −y). So the geometric effect of conjugating a complex number is simply to reflect it in the horizontal axis. Fact 55. Complex conjugate. Let z be a complex number. Then ∣z ∗ ∣ = ∣z∣ ,
and
arg (z ∗ ) = − arg z + 2kπ,
where k = −1, 0, 1 ensures that − arg z + 2kπ ∈ (−π, π]. Proof. Let z = r(cos θ + i sin θ). Then z∗ = r(cos θ − i sin θ) = r [cos (−θ) − i sin (−θ)] . This is the complex number with modulus r and angle −θ with the positive x-axis. So given z = r(cos θ + i sin θ) = reiθ , its conjugate is simply z ∗ = r [cos (−θ) + i sin (−θ)] = rei(−θ) . y z* = (x, -y)
x
z = (x, y)
Page 410, Table of Contents
www.EconsPhDTutor.com
42
Loci Involving Cartesian Equations
A locus (plural: loci) is a set of points that satisfy some condition (or conditions). Weâ&#x20AC;&#x2122;ve actually already encountered plenty of loci in Part I (Functions and Graphs), so this is nothing new. This chapter reviews loci involving cartesian equations (and inequalities). The goal is to prepare you for the next chapter, where we look at loci involving complex equations (and inequalities).
42.1
Circles
Example 408. {(x, y) â&#x2C6;ś x2 + y 2 = 1} is the set of all points (x, y) in the cartesian plane that satisfy the condition x2 + y 2 = 1. Graphically, this locus describes describing the unit circle centred on the origin. (To be clear, it includes only the circumference of the circle.)
y
{(x, y): x2 + y2 = 1}
x
Page 411, Table of Contents
www.EconsPhDTutor.com
Example 409. The locus {(x, y) ∶ x2 + y 2 ≤ 1} describes the entire interior of the unit circle centred on the origin, including the circumference of the circle.
{(x, y):
x2
+
y2
≤ 1}
y
x
Page 412, Table of Contents
www.EconsPhDTutor.com
Example 410. The locus {(x, y) â&#x2C6;ś x2 + y 2 < 1} describes the entire interior of the unit circle centred on the origin, excluding the circumference of the circle.
{(x, y):
x2
+
y2
< 1}
y
x
Page 413, Table of Contents
www.EconsPhDTutor.com
Example 411. The locus {(x, y) ∶ x2 + y 2 ≥ 1} describes everything outside the unit circle centred on the origin, including the circumference of the circle.
y
{(x, y): x2 + y2 ≥ 1} x
Page 414, Table of Contents
www.EconsPhDTutor.com
Example 412. The locus {(x, y) ∶ x2 + y 2 > 1} describes everything outside the unit circle centred on the origin, excluding the circumference of the circle.
y
{(x, y): x2 + y2 > 1} x
Exercise 161. Sketch the following loci: (a) {(x, y) ∶ (x − a)2 + (y − b)2 = r2 }. (b) {(x, y) ∶ (x − a)2 + (y − b)2 ≤ r2 }. (c) {(x, y) ∶ (x − a)2 + (y − b)2 < r2 }. (d) 2 2 2 2 2 2 {(x, y) ∶ (x − a) + (y − b) ≥ r }. (e) {(x, y) ∶ (x − a) + (y − b) > r }. (Answer on p. 1118.)
Page 415, Table of Contents
www.EconsPhDTutor.com
42.2
Lines
Example 413. The locus {(x, y) â&#x2C6;¶ y = x} describes the line y = x.
y
{(x, y): y = x}
x
Page 416, Table of Contents
www.EconsPhDTutor.com
Example 414. The locus {(x, y) ∶ y ≤ x} describes the set of all points under the line y = x, including the line itself. It contains literally half the plane, so we call this a halfplane. We can also specify that this is a closed half-plane — the word closed means that it includes also the line y = x. Graphically, the locus {(x, y) ∶ y < x} describes the set of all points under the line y = x, excluding the line itself. This is an open half-plane. The word open means that it excludes the line y = x. y
y {(x, y): y ≤ x }
{(x, y): y < x }
x
x
Example 415. Graphically, the locus {(x, y) ∶ y ≥ x} describes the set of all points above the line y = x, including the line itself. Again, this is a closed half-plane. Graphically, the locus {(x, y) ∶ y > x} describes the set of all points above the line y = x, but excluding the line itself. Again, this is an open half-plane. y
y
x
{(x, y): y ≥ x}
Page 417, Table of Contents
x
{(x, y): y > x}
www.EconsPhDTutor.com
The locus of points that are equidistant to two points is simply a line. Example 416. Let (a, b) and (c, d) be points. The locus of points that are equidistant to (a, b) and (c, d) is the line illustrated below. This is because if you pick any point (e.g. P ) on the line, it is indeed equidistant to (a, b) and (c, d). And if you pick any point (e.g. Q) not on the line, it must be either closer to (a, b) or closer to (c, d) â&#x20AC;&#x201D; in this case, Q is closer to (a, b) than to (c, d).
Any point not on the line must be closer to one of the two points. Q
y P Any point on the line is equidistant to the two points.
(a, b) x
(c, d)
Page 418, Table of Contents
www.EconsPhDTutor.com
Let (a, b) and (c, d) be points. We now prove that the locus {(x, y) ∶ ∣(x − a, y − b)∣ = ∣(x − c, y − d)∣} simply describes a line: ∣(x − a, y − b)∣ = ∣(x − c, y − d)∣ (x − a)2 + (y − b)2 = (x − c)2 + (y − d)2 x2 − 2ax + a2 + y 2 − 2by + b2 = x2 − 2cx + c2 + y 2 − 2dy + d2 −2ax + a2 − 2by + b2 = −2cx + c2 − 2dy + d2
⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒
2(d − b)y + 2(c − a)x + a2 + b2 − (c2 + d2 ) = 0.
This last equation is of the form αx + βy + γ = 0 — this is simply a line.44
Exercise 162. (a) Find the cartesian equation of the line that is equidistant to the points (1, 4) and (−5, 0). (b) Describe in words the set {(x, y) ∶ ∣(x − 17, y − 3)∣ = ∣(x + 2, y + 11)∣}. Then rewrite the cartesian equation ∣(x − 17, y − 3)∣ = ∣(x + 2, y + 11)∣ into the form ay + bx + c = 0. (Answer on p. 1120.)
44
If we’d like, we can further simplify this equation. If d − b ≠ 0, then it can be rewritten as
y=
c2 + d2 − (a2 + b2 ) a−c x+ . d−b 2(d − b)
And if d − b = 0, then this is a vertical line whose equation may be rewritten as
x=
Page 419, Table of Contents
c2 + d2 − (a2 + b2 ) 2(c − a)
.
www.EconsPhDTutor.com
42.3
Intersection of Lines and Circles
Example 417. {(x, y) ∶ x2 + y 2 = 1, y = x} is the set of points satisfying two conditions, namely the equation x2 +y 2 = 1 and the equation y = x. This locus describes the intersection points of the circle x2 + y 2 = 1 and the line y = x. By plugging the equation of the line into the equation of the circle, we can show that this locus consists of only two points: √ √ √ √ 2 2 2 2 {(x, y) ∶ x2 + y 2 = 1, y = x} = {(− ,− ),( , )}. 2 2 2 2
y
{(x, y): y = x}
{(x, y): x2 + y2 = 1}
x
{(x, y): y = x, x2 + y2 = 1}
Page 420, Table of Contents
www.EconsPhDTutor.com
Example 418. {(x, y) ∶ x2 + y 2 ≤ 1, y = x} is the portion of the line y = x that is within the interior of the circle, including the endpoints. It is illustrated in green in the figure below.
y
{(x, y): y = x}
{(x, y): x2 + y2 ≤ 1}
x
{(x, y): y = x, x2 + y2 ≤ 1}
Page 421, Table of Contents
www.EconsPhDTutor.com
Example 419. The locus {(x, y) â&#x2C6;ś x2 + y 2 > 1, y > x} describes the region above both the circle x2 + y 2 = 1 and y = x, excluding the circumference of the circle and the line. It is illustrated in green in the figure below.
y
{(x, y): y = x} x
{(x, y): x2 + y2 = 1}
{(x, y): y > x, x2 + y2 > 1}
Exercise 163. Sketch on a cartesian plane the locus {(x, y) â&#x2C6;ś x2 + y 2 = 1, x > 0}. (Answer on p. 1121.)
We now turn to loci involving complex equations (and inequalities).
Page 422, Table of Contents
www.EconsPhDTutor.com
43
Loci Involving Complex Equations 43.1
Circles
On an Argand diagram (or complex plane), the locus {z ∈ C ∶ ∣z∣ = 1} simply describes the unit circle centred on the origin, as we now prove: √ √ Let z = (x, y). Then ∣z∣ = x2 + y 2 and so the equation ∣z∣ = 1 is equivalent to x2 + y 2 = 1 or x2 + y 2 = 1. Hence, {z ∈ C ∶ ∣z∣ = 1} = {(x, y) ∶ x2 + y 2 = 1} . But we already saw in the previous chapter that the locus {(x, y) ∶ x2 + y 2 = 1} describes the unit circle centred on the origin. Loci involving complex equations (or inequalities) can usually be easily transformed into a familiar cartesian equation (or inequality).
y {z : |z | = 1} = {(x, y): x2 + y2 = 1}
x
Page 423, Table of Contents
www.EconsPhDTutor.com
Exercise 164. (a) Prove that the locus {z ∈ C ∶ ∣z∣ = r} describes the circle of radius r centred on the origin. (b) Let c be some fixed complex number. Prove that the locus {z ∈ C ∶ ∣z − c∣ = r} is the circle of radius r centred on the point c. (c) What does the locus {z ∈ C ∶ ∣z − c∣ ≤ r} describe? (d) What does the locus {z ∈ C ∶ ∣z − c∣ < r} describe? (Answer on p. 1122.)
Page 424, Table of Contents
www.EconsPhDTutor.com
43.2
Lines
Let b and c be fixed complex numbers. The equation ∣z − c∣ = ∣z − b∣ is simply the condition that z is equidistant to b and c. Hence, the locus {z ∈ C ∶ ∣z − c∣ = ∣z − b∣} simply describes the points that are equidistant to b and c. And as we showed earlier, such a locus is simply a line.
Any point not on the line must be closer to one of the two points. e
y d Any point on the line is equidistant to the two points.
b x
c {z : |z – b | = |z – c |}
Exercise 165. Let b and c be fixed complex numbers. What is the locus of complex numbers z that satisfy each of the following inequalities? (a) ∣z − c∣ ≤ ∣z − b∣. (b) ∣z − c∣ < ∣z − b∣. (c) ∣z − c∣ ≥ ∣z − b∣. (d)∣z − c∣ > ∣z − b∣. (Answer on p. 1122.)
Page 425, Table of Contents
www.EconsPhDTutor.com
43.3
Rays
The locus {z ∈ C ∶ arg z = α} describes the set of points z whose argument is α. It is thus the ray (or half-line) which starts from but excludes the origin and which makes an angle α with the positive x-axis. The figure below illustrates. The point A is in the locus, because indeed arg A = α. In contrast, the point B is not in the locus, because its argument is not arg B ≠ α. Note importantly that points along the dotted red ray, such as C, are not in the locus, because arg C = α − π ≠ α. Moreover, the origin is not in the locus, because arg 0 is undefined.
y
{z : arg z = Ƚ}
B
A Ƚ x C
If we really wanted to, we could rewrite the complex equation arg z = α into cartesian form. But it turns out that in this case, the cartesian form is more complicated. And so we’ll just stick with the equation arg z = α.
Page 426, Table of Contents
www.EconsPhDTutor.com
Let a be a complex number. Then the graph of arg (z − a) = α is simply the translation of the graph of arg z = α. And so arg (z − a) = α is the ray (or half-line) which starts from but excludes the point a and which makes an angle α with the positive x-axis. The point b is in the locus, because indeed arg(b − a) = α. In contrast, the point c is not in the locus, because its argument is not arg(c − a) ≠ α. Note importantly that points along the dotted red ray, such as d, are not in the locus, because arg(d − a) = α − π ≠ α. Moreover, the point a is not in the locus, because arg(a − a) = arg 0 is undefined.
y
c
{z : arg (z – a) = Ƚ} b
Ƚ a d
Page 427, Table of Contents
x
www.EconsPhDTutor.com
The 9740 syllabus doesn’t mention loci of the form α ≤ arg z ≤ β. Unfortunately, such loci have occasionally appeared on the A-level exams,45 which means you have to learn it. The locus α ≤ arg z ≤ β is simply the region bounded by (and including) the rays arg z = α and arg z = β.
y
{z : arg z = Ⱦ} Ƚ {z : arg z = Ƚ}
Ⱦ x
Exercise 166. What is {z ∈ C ∶ ∣z∣ = 1, −π < arg z < 0}? (Answer on p. 1122.)
45
See Exercises 360 (2013), 364 (2011), and 370 (2008).
Page 428, Table of Contents
www.EconsPhDTutor.com
43.4
Quick O-Level Revision: Properties of The Circle
Definition 98. A chord is a line segment connecting any two points on a circle’s circumference. Here are a few properties of the circle (which you are supposed to still remember from O-levels) and which would definitely have been useful in some complex loci questions in the past ten years’ A-levels. Fact 56. Let A be a point exterior to a circle. Let B and C be the points at which the tangents from A touch the circle. Let O be the centre of the circle. (a) The line through A and O (i) bisects the angle ∠BAC; (ii) is the perpendicular bisector of the chord BC; and (iii) passes through the points D and E, which are the points on the circle that are respectively that closest to and furthest from A. (b) The lengths AB and AC are equal. (c) The angles ∠OBA and ∠OCA are right.
Perpendicular bisector of chord B
Chord E O
A
D
Tangents C
Proof. See p. ??? of my O-Level Maths Textbook (coming soon)! Here’s an example that illustrates the uses of the above properties of the circle. Page 429, Table of Contents
www.EconsPhDTutor.com
Example 420. The complex number z satisfies the equation ∣z + 4 + 2i∣ = 1. (a) What are the maximum and minimum possible values of ∣z∣? (b) For what values of z is ∣z∣ maximised and minimised? ∣z + 4 + 2i∣ = 1 describes a unit circle centred on the point C = (−4, −2). Even if not asked for, you should make a quick sketch to help yourself see better. By the above fact, ∣z∣ is maximised at F and minimised at N , where F and N lie on the line through the origin and the circle’s centre. √ √ (a) The maximum value of ∣z∣ is the length OF = OC + CF = (−4)2 + (−2)2 + 1 = 20 + 1. √ √ The minimum value of ∣z∣ is the length ON = OC − CN = (−4)2 + (−2)2 − 1 = 20 − 1. (b) Consider △CAN . The line through F , C, N , and the origin is y = 0.5x. So AN = 0.5CA. Moreover, CA2 + AN 2 = CN 2 = 12 = 1. Altogether then, CA2 + 0.25CA2 = 1 or CA2 =
2 1 4 or CA = √ . And AN = √ . Hence, 5 5 5
2 1 N = (−4 + √ , −2 + √ ) 5 5 1 2 Symmetrically, F = (−4 − √ , −2 − √ ). 5 5
y O
|z + 4 + 2i | = 1
x
U N
C
A
F D
y = 0.5 x (Line through the origin and the centre of the circle.)
(... Example continued on the next page ...)
Page 430, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) z satisfies ∣z + 4 + 2i∣ = 1. (c) What are the maximum and minimum possible values of arg z? (d) For what values of z is arg z maximised and minimised? (c) The points U and D at which arg z is maximised and minimised are also where the tangents OU and OD from the origin touch the circle. By the above fact, OU is perpendicular to CU . Similarly, OD is perpendicular to CD. The angle the lower half of the√line y = 0.5x makes with the positive x-axis is θ = tan−1 0.5−π. √ The angle ∠COU is sin−1 (1/ 20). Hence, arg U = θ+∠COU = tan−1 0.5−π−sin−1 (1/ 20). √ Symmetrically, arg D = θ − ∠COD = θ − ∠COU = sin−1 0.5 − π + tan−1 (1/ 20). √ √ (d) △ODC is right. So OD2 +CD2 = OC 2 . OC = 20 and CD = 1. Hence, OD = 20 − 1 = √ √ √ 19. Altogether then ∣D∣ = 19 and arg D = tan−1 0.5 − π + sin−1 (1/ 20). √ √ Symmetrically, we also have ∣U ∣ = 19 and arg U = sin−1 0.5 − π + tan−1 (1/ 20). (Figure reproduced for convenience.)
y O
|z + 4 + 2i | = 1
x
U N
C
A
F D
y = 0.5 x (Line through the origin and the centre of the circle.)
Exercise 167. The complex number z satisfies the equation ∣z − 2 − 2i∣ = 1. (a) What are the maximum and minimum possible values of ∣z∣? (b) For what values of z is ∣z∣ maximised and minimised? (c) What are the maximum and minimum possible values of arg z? (d) For what values of z is arg z maximised and minimised? (Answer on p. 1123f.)
Page 431, Table of Contents
www.EconsPhDTutor.com
44
De Moivre’s Theorem n
Theorem 7. De Moivre’s Theorem. (cos θ + i sin θ) = cos (nθ) + i sin (nθ). Proof. cos θ + i sin θ is the complex number with modulus 1 and argument θ. So by Fact 51, n (cos θ + i sin θ) is the complex number with modulus 1n = 1 and argument nθ + 2kπ (where k is the unique integer such that nθ + 2kπ is a principal value) — this complex number can be written as cos (nθ) + i sin (nθ). Here is an alternative proof that uses the Euler Formula: n 2
n 1
3
1
3
Proof. (cos θ + i sin θ) = (eiθ ) = ei(nθ) = cos (nθ) + i sin (nθ), where = and = use the Euler 2
b
Formula (Theorem 6) and = uses the law of exponents (xa ) = xab , which applies even when a is imaginary.
This is a totally free, bonus exercise on mathematical induction! Exercise 168. Prove de Moivre’s Theorem using the method of mathematical induction. (Answer on p. 1125.)
As stated, de Moivre’s Theorem applies only to complex numbers with modulus 1. It is easy to rewrite it so that it applies more generally to any complex number with modulus r: n
Corollary 5. [r (cos θ + i sin θ)] = rn [cos (nθ) + i sin (nθ)]. n
Or equivalently, (reiθ ) = rn ei(nθ) . Or equivalently, if ∣z∣ = r and arg z = θ, then ∣z n ∣ = rn and arg z n = nθ + 2kπ (where k is the unique integer such that nθ + 2kπ is a principal value).
Page 432, Table of Contents
www.EconsPhDTutor.com
44.1
Powers of a Complex Number
On the Argand diagram, the powers of a complex number form a “spiral”. Example 421. Let z = 1 + i. Then: ∣z∣ =
√ 2,
arg z =
π π + 2kπ = (k = 0), 4 4
√ 2 ∣z 2 ∣ = ( 2) = 2,
π π arg z 2 = 2 ( ) + 2kπ = (k = 0), 4 2
√ 3 √ ∣z 3 ∣ = ( 2) = 2 2,
π 3π arg z 3 = 3 ( ) + 2kπ = (k = 0), 4 4
√ 4 ∣z 4 ∣ = ( 2) = 4,
π arg z 4 = 4 ( ) + 2kπ = π (k = 0), 4
√ 5 √ ∣z 5 ∣ = ( 2) = 4 2,
3π π (k = −1), arg z 5 = 5 ( ) + 2kπ = − 4 4
etc.
The powers of z = 1 + i, up to the 13th, are illustrated in the figure below.
z11
32 y z10 16
z12 -64
z9
z3 z2 z8 z x 0 -16 z5 0 16 7 z 6 z -16 z4
-48
-32
-32 -48 z13
Page 433, Table of Contents
-64
www.EconsPhDTutor.com
Example 422. Let z = 1 + 0.4i. Then: ∣z∣ =
√ √ 12 + 0.42 = 1.16,
arg z = tan−1 (0.4) + 2kπ,
∣z 2 ∣ = 1.16,
arg z 2 = 2 tan−1 (0.4) + 2kπ,
√ ∣z 3 ∣ = 1.16 1.16,
arg z 3 = 3 tan−1 (0.4) + 2kπ,
∣z 4 ∣ = 1.162 ,
arg z 4 = 4 tan−1 (0.4) + 2kπ,
√ ∣z 5 ∣ = 1.162 1.16,
arg z 5 = 5 tan−1 (0.4) + 2kπ,
etc.
The powers of z = 1 + 0.4i, up to the 14th, are illustrated in the figure below.
z5
z6
z4
z3
y z2 z1 z
z7
x
z8 z14 z9 z13 z10 z12 z10
Page 434, Table of Contents
z11
www.EconsPhDTutor.com
Exercise 169. (a) Given z = 3 − 4i, find ∣z∣ and arg z. Hence find ∣z 7 ∣ and arg z 7 . Write down (3 − 4i)7 in exponential form. (b) Given z = −5 + 12i, find ∣z∣ and arg z. Hence find ∣z 8 ∣ and arg z 8 . Write down (−5 + 12i)8 in exponential form. (Answer on p. 1125.) Exercise 170. For each of the given values of z, compute z 10 , expressing your answer in all three forms (polar, exponential, and standard). (a) z = −1 − i. (b) z = 2 + i. (c) z = 1 − 3i. (Answer on p. 1126.)
Page 435, Table of Contents
www.EconsPhDTutor.com
44.2
Roots of a Complex Number
a
Example 423. What are the roots to the equation z 3 = 1 + i? That is, for what values of z is the given equation true? A naïve application of de Moivre’s Theorem might suggest that ∣z 3 ∣ = 21/2 and arg z 3 = π/4
Ô⇒
1/3
∣z∣ = (21/2 )
= 21/6 and arg z = (π/4)/3 = π/12.
This is not incorrect, but it gives us only one root to the equation z 3 = 1 + i, namely z = 21/6 ei(π/12) . In contrast, the Fundamental Theorem of Algebra tells us that since the equation z 3 = 1 + i involves a degree-3 polynomial, it should have 3 roots. We’ve just found one root. How do we find the other two? The trick is to recognise that z 3 = 21/2 eiπ/4 can also be written as z 3 = 21/2 ei(π/4+2kπ) , for any integer k. This is because if you plug in any integer k, you will always get 21/2 ei(π/4+2kπ) = 21/2 ei(π/4) . The reason is that ei(2π) = 1. 1/3
1/3
We then have z = (z 3 ) = [21/2 ei(π/4+2kπ) ] = 21/6 ei(π/4+2kπ)/3 , for any integer k. Now in contrast to before, different integers k will yield us distinct values for z = 21/6 ei(π/4+2kπ)/3 . In particular, if we pick values of k so that the values of (π/4 + 2kπ) /3 are principal values, that is, if we pick k = 0, ±1, we have z = 21/6 ei(π/12) ,
21/6 ei(11π/12) ,
21/6 ei(−7π/12) .
Observe that beautifully enough, the roots of the equation z 3 = 1 + i lie on a circle — in particular, the circle of radius 21/6 centred on the origin. Moreover, each root can be 2π obtained by rotating another root radians about the origin. 3
y
x
Page 436, Table of Contents
www.EconsPhDTutor.com
∣z n ∣ arg z n + 2kπ In general, given z , we have ∣z∣ = and arg z = , where k are those integers n n n arg z such that arg z = + 2kπ ∈ (−π, π]. n n
The annoying part is to figure out the appropriate values of k. So here’s how to do it: 1. If n is odd, then simply pick k = 0, ±1, ±2, . . . , ± 0, ±1, ±2, . . . , ±7.)
n−1 . (E.g., if n = 15, then pick k = 2
n 2. If n is even AND arg z n > 0, then simply pick k = 0, ±1, ±2, . . . , − . (E.g., if n = 16 and 2 arg z n > 0, then pick k = 0, ±1, ±2, . . . , ±7, −8.) n 3. If n is even AND arg z n ≤ 0, then simply pick k = 0, ±1, ±2, . . . , . (E.g., if n = 16 and 2 arg z n ≤ 0, then pick k = 0, ±1, ±2, . . . , ±7, 8.) You can easily verify that in each case, we do indeed have n roots (just count them). See Fact 90 in the Appendices for a proof (or explanation) of why the above values of k ensure that we have k distinct principal values for arg z.
More examples ...
Page 437, Table of Contents
www.EconsPhDTutor.com
Example 424. Consider z 4 = −5 + 12i. −1
We have z 4 = 13ei[π−tan (12/5)+2kπ] , for k ∈ Z. Hence, z = 131/4 ei[π−tan Since 4 is even and arg z 4 > 0, we should pick k = 0, ±1, −2 to get −1 ⎧ ⎪ 131/4 ei[π−tan (12/5)+2kπ]/4 , ⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎪ ⎪ ⎪131/4 ei[3π−tan (12/5)]/4 , z=⎨ −1 ⎪ ⎪ 131/4 ei[−π−tan (12/5)]/4 , ⎪ ⎪ ⎪ ⎪ ⎪ 1/4 i[−3π−tan−1 (12/5)]/4 ⎪ ⎪ , ⎩13 e
−1
(12/5)+2kπ]/4
, for k ∈ Z.
(k = 0), (k = 1), (k = −1), (k = −2).
y
x
Page 438, Table of Contents
www.EconsPhDTutor.com
−1
Example 425. Consider z 7 = 3 − 4i. We have z 7 = 5ei[tan (−4/3)+2kπ] , for k ∈ Z. Hence, −1 z = 51/7 ei[tan (−4/3)+2kπ]/7 , for k ∈ Z. Since 7 is odd, we should pick k = 0, ±1, ±2, ±3. −1
Now consider w8 = 3 − 4i. We have w8 = 5ei[tan (−4/3)+2mπ] , for m ∈ Z. Hence, w = −1 51/8 ei[tan (−4/3)+2mπ]/8 , for m ∈ Z. Since 8 is even and arg w8 ≤ 0, we should pick m = 0, ±1, ±2, ±3, −4. Altogether then, the possible values of z and w are given by: ⎧ 1/8 i[tan−1 (−4/3)]/8 ⎪ ⎪ 5 e , ⎧ −1 ⎪ ⎪ ⎪ 1/7 i[tan (−4/3)]/7 ⎪ ⎪ 5 e , (k = 0), ⎪ ⎪ −1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 51/8 ei[tan (−4/3)+2π]/8 , −1 ⎪ ⎪ 1/7 i[tan (−4/3)+2π]/7 ⎪ ⎪ ⎪ ⎪ 5 e , (k = 1), ⎪ ⎪ ⎪ ⎪ 1/8 i[tan−1 (−4/3)+4π]/8 ⎪ ⎪ 5 e , ⎪ ⎪ −1 ⎪ ⎪ 1/7 i[tan (−4/3)+4π]/7 ⎪ ⎪ ⎪ ⎪ 5 e , (k = 2), ⎪ ⎪ −1 ⎪ ⎪ ⎪ ⎪ ⎪51/8 ei[tan (−4/3)+6π]/8 , ⎪ 1/7 i[tan−1 (−4/3)+6π]/7 and w= ⎨ z= ⎨5 e , (k = 3), ⎪ ⎪ 1/8 i[tan−1 (−4/3)−2π]/8 ⎪ ⎪ 5 e , ⎪ ⎪ −1 ⎪ ⎪ 1/7 i[tan (−4/3)−2π]/7 ⎪ ⎪ ⎪ ⎪ 5 e , (k = −1), −1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 51/8 ei[tan (−4/3)−4π]/8 , ⎪ ⎪ −1 ⎪ ⎪ 1/7 i[tan (−4/3)−4π]/7 ⎪ ⎪ 5 e , (k = −2), ⎪ ⎪ ⎪ ⎪ 1/8 i[tan−1 (−4/3)−6π]/8 ⎪ ⎪ ⎪ ⎪ 5 e , ⎪ ⎪ −1 ⎪ ⎪ 1/7 i[tan (−4/3)−6π]/7 ⎪ ⎪ 5 e , (k = −3), ⎪ ⎪ −1 ⎪ ⎩ 1/8 i[tan (−4/3)−8π]/8 ⎪ ⎪ , ⎩5 e
(m = 0), (m = 1), (m = 2), (m = 3), (m = −1), (m = −2), (m = −3), (m = −4).
y
x /
Notice that the eight possible values of w are on a circle whose radius is just slightly shorter than the red circle. (Only the red circle is illustrated.)
Page 439, Table of Contents
www.EconsPhDTutor.com
Exercise 171. Find the roots of each of the following equations. (a) z 10 = −1 − i. (b) z 11 = 2 + i. (c) z 12 = 1 − 3i. (Answer on p. 1127.)
Page 440, Table of Contents
www.EconsPhDTutor.com
Part V
Calculus
Page 441, Table of Contents
www.EconsPhDTutor.com
45
Solving Problems Involving Differentiation
Part I already covered differentiation. This chapter merely ties up some loose ends.
45.1
Inverse Function Theorem (IFT)
The Inverse Function Theorem (IFT) simply says that “The change in y caused by a small unit change in x (dy/dx)” is the inverse of “the change in x caused by a small unit change in y (dx/dy)”.46 That is, dy 1 = . dx dx dy Example 426. Suppose that adding 1 g of Milo (the x-variable) to a cup of water increases the volume of water by 2 cm3 (the y-variable). That is, dy/dx = 2 cm3 g-1 . Then dx/dy = 0.5 g cm-3 . That is, if instead we had wanted to increase the volume of water by 1 cm3 , we should have added 0.5 g of Milo to the water.
Here’s a more typical use of the IFT: Example 427. Let x ∈ [−π/2, π/2]. Let y = sin x. Suppose we wish to find dx/dy in terms of x. Method #1 (longer method using Corollary 2 ). y = sin x Ô⇒ x = sin−1 y. So 1 dx d 1 1 = sin−1 y = √ =√ = dy dy 1 − y2 1 − sin2 x cos x Method #2 (quicker method using the IFT).
dy dx 1 = cos x Ô⇒ = . dx dy cos x
dy dx . Hence write down . (You may leave dx dy your answers expressed in terms of x and y.) (Answer on p. 1128.)
Exercise 172. Suppose x2 y + sin x = 0. Find
46
This is informal. For the formal statement of the IFT (optional), see p. 969 in the Appendices.
Page 442, Table of Contents
www.EconsPhDTutor.com
45.2
Differentiation of Simple Parametric Functions
“Informal Fact”.
dx dy dy dx = ÷ (provided ≠ 0). dx dt dt dt
Here is an informal “proof” of the above informal fact. By the Chain Rule, dy 1 dy dt = . dx dt dx By the IFT, dt 2 1 = . dx dx dt 2
1
Plugging = into = yields the desired result: dy dy dx = ÷ . dx dt dt See p. 970 in the Appendices for a formal version of the above Fact.
Example 428. Let x = t5 + t and y = t6 − t. Find
dy ∣ . dx t=0
dy dy dx 6t5 − 1 = ÷ = . dx dt dt 5t4 − 1 dy ∣ = 1. It would be much more difficult (perhaps even impossible) if instead we first dx t=0 dy tried to express y in terms of x, then compute . dx
So
Exercise 173. Let x = cos t + t2 and y = et − t3 . Find
Page 443, Table of Contents
dy . (Answer on p. 1128.) dx
www.EconsPhDTutor.com
45.3
Equations of Tangents and Normals
Recall from secondary school the following two facts: Fact 57. The line with slope m through the point (a, b) has equation y − b = m(x − a).
Fact 58. Given a line with slope m, its perpendicular has slope −
1 . m
Example 429. The curve C has parametric equations x = t5 + t and y = t6 − t, t ∈ R. Consider the normal line at the point where t = 0. Find any point(s) at which the normal line intersects the curve C again. First, note that t = 0 Ô⇒ (x, y) = (0, 0). Next, R R R dy dx RRRR 6t5 − 1 RRRR dy RRRR R = R = 1. ÷ R = 4 dx RRRR dt dt RRRR 5t − 1 RRRR Rt=0 Rt=0 Rt=0 So the tangent line at the point t = 0 or (0, 0) has slope 1. Thus, the normal line at this point has slope −1. Its equation is thus y − 0 = 1(x − 0) or more simply y = x. The points where this normal line intersects the curve is thus given by the system of equations y = x, x = t5 + t, and y = t6 − t. Putting these together, we have t5 + t = t6 − t ⇐⇒ t (t5 − t4 − 2) = 0. So t = 0 or t ≈ 1.45 (calculator). (We know by the Fundamental Theorem of Algebra that there must be six roots altogether — in this case, only two are real, while the other four are complex.) So the normal line intersects the curve C again at the point where t ≈ 1.45 or where (x, y) ≈ (7.88, 7.88). Exercise 174. A curve C is described by the pair of parametric equations x = t5 + t and y = t4 − t. Find the tangent lines to the curve at the points where t = 0 and t = 1. Find the intersection point of these two tangent lines.(Answer on p. 1128.)
Page 444, Table of Contents
www.EconsPhDTutor.com
45.4
Connected Rates of Change Problems
Example 430. Sand is being unloaded onto a flat surface at a steady rate of 0.01 m3 s-1 . Assume that the unloaded sand always forms a perfect cone with equal height and base diameter. Find the rate at which the area of the base of the cone increasing at the instant t = 20 s. 1 First, recall that a cone has volume V = πr2 h, where r is the radius of the base and h is 3 the height. Since the base diameter equals the height (or h = 2r), we can rewrite this as 2 d dV dr V = πr3 . Applying , we have = 2πr2 . 3 dt dt dt The base area is A = πr2 . So the rate at which the base area is increasing is dA dr dV = 2πr = ÷ r. dt dt dt dV = 0.01 at all times. dt 3V 1/3 0.3 1/3 V ∣t=20 = 20 × 0.01 = 0.2, so that r∣t=20 = ( ) ∣ = ( ) . Altogether then, t=20 2π π The volume of the sand is increasing at a rate 0.01, i.e.
.
Page 445, Table of Contents
dA 0.3 1/3 ∣ = 0.01 ÷ ( ) = 0.0219 m2 s−1 . dt t=20 π
www.EconsPhDTutor.com
Exercise 175. (Answer on p. 1129.) Illustrated below is a cone with lateral l, base radius r, and height h. You are given that such a cone has total external surface area (excluding 1 the base) πrl and volume πr2 h. 3
A manufacturer wishes to manufacture a cone whose volume is fixed at 1 m3 and whose total external surface area (excluding the base) is minimised. Find out what its height should be. (You can follow the steps below.) (a) Express r in terms of h. (b) Use the Pythagorean Theorem to express l in terms of r and h. Hence express l solely in terms of h. (c) Now express the total external surface area A (excludes the base) solely in terms of h. dA 3 π − h63 6 1/3 (d) Show that = . Hence conclude that the only stationary point is h = ( ) . dh 2 A π (e) Use the quotient rule to show that 12
6
2
d2 A 9 h4 A2 − (π − h3 ) = . dh2 4 A3 d2 A . Replace A2 with the expression for A that you found dh2 in (c). Now fully expand this numerator. Observe that it is a quadratic and prove that it is always positive.
(f) Consider the numerator of
(g) Hence conclude that the stationary point we found is indeed the global minimum.
Page 446, Table of Contents
www.EconsPhDTutor.com
45.5
Finding Max/Min Points on the TI84
Example 431. Define f ∶ [0, 2] → R by x ↦ x−sin (0.5πx). We can easily find the minimum point of f analytically: π π df = 1 − cos ( x) = 0 dx 2 2
⇐⇒
π 2 cos ( x) = 2 π
⇐⇒
x=
2 2 cos−1 ≈ 0.560664181. π π
But as an exercise, let’s find it using our TI84. After Step 1.
After Step 2.
After Step 3.
After Step 4.
After Step 5.
After Step 6.
1. Press ON to turn on your calculator. 2. Press Y= to bring up the Y= editor. 3. Press X,T,θ,n − SIN 0 . 5 . To enter “π”, press the blue 2ND button and then π (which corresponds to the ∧ button). Now press X,T,θ,n ) and altogether you will have entered “x − sin(0.5πx)”. 4. Now press GRAPH and the calculator will graph y = x − sin(0.5πx). Note that in the question given, the domain is actually [0, 2], but we didn’t bother telling the calculator this. So the calculator just went ahead and graphed the equation y = x − sin(0.5πx) for all possible real values of x and y. No big deal, all we need to do is to zoom in to the region where 0 ≤ x ≤ 2. 5. Press the (ZOOM) button to bring up a menu of ZOOM options. 6. Press 2 to select the Zoom In option. Using the < and > arrow keys, move the cursor to where X = 1.0638298, Y = 0. Now press ENTER and the TI will zoom in a little, centred on the point X = 1.0638298, Y = 0. (... Example continued on the next page ...)
Page 447, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) It looks like starting at x = 0, the function is decreasing, then hits a minimum point, then keeps increasing. Our goal now is to find out what that minimum point is. After Step 7.
After Step 8.
After Step 9.
After Step 11.
After Step 12.
After Step 13.
After Step 10.
4. Press the blue 2ND button and then CALC (which corresponds to the TRACE button). This brings up the CALCULATE menu. 5. Press 3 to select the “minimum” option. This brings you back to the graph, with a cursor flashing. Also, the TI84 prompts you with the question: “Left Bound?” TI84’s MINIMUM function works by you first choosing a “Left Bound” and a “Right Bound” for x. TI84 will then look for the minimum point within your chosen bounds. 6. Using the < and > arrow keys, move the blinking cursor until it is where you want your first “Left Bound” to be. For me, I have placed it a little to the left of where I believe the minimum point to be. 7. Press ENTER and you will have just entered your first “Left Bound”. TI84 now prompts you with the question: “Right Bound?”. 8. So now just repeat. Using the < and > arrow keys, move the blinking cursor until it is where you want your first “Right Bound” to be. For me, I have placed it a little to the right of where I believe the minimum point to be. 9. Again press ENTER and you will have just entered your first “Right Bound”. TI84 now asks you: “Guess?” This is just asking if you want to proceed and get TI84 to work out where the minimum point is. So go ahead and: 10. Press ENTER . TI84 now informs you that there is a “Zero” at “X = .56066485”, “Y = −.2105137” and places the cursor at precisely that point. This is our desired minimum point. (Notice there’s a slight error, because the TI84 uses slightly-imprecise numerical methods. Analytically, we found that the minimum point was x ≈ 0.560664181, while the TI84 claims it is “X = .56066485”.) Page 448, Table of Contents
www.EconsPhDTutor.com
45.6
Finding the Derivative at a Point on the TI84
This example will also illustrate how to graph parametric equations on the TI84. Example 432. The curve C has parametric equations x = t5 + t and y = t6 − t, t ∈ R. We’ll find dy/dx∣ using our TI84, even though this is easily found analytically: t=1
dy 6t5 − 1 5 ∣ = 4 ∣ = . dx t=1 5t + 1 t=1 6 After Step 1.
After Step 2.
After Step 3.
After Step 4.
After Step 5.
After Step 6.
After Step 7.
After Step 8.
1. Press ON to turn on your calculator. 2. Press MODE to bring up a menu of settings that you can play with. In this example, all we want is to plot a curve based on parametric equations. So: 3. Using the arrow keys, move the blinking cursor to the word PAR (short for parametric) and press ENTER . 4. Now as usual, we’ll input the equations of our curve. To do so, press Y= to bring up the Y= editor. Notice that this screen looks a little different from usual, because we are now under the parametric setting. 5. Press X,T,θ,n ∧ 5 + X,T,θ,n and altogether you will have entered “T 5 + T ” in the first line. 6. Now press ENTER to go to the second line. 7. Press X,T,θ,n ∧ 6 - X,T,θ,n and altogether you will have entered “T 6 − T ” in the second line. 8. Now press GRAPH and the calculator will graph the given pair of parametric equations. Notice that strangely enough, the graph seems to be empty for the region where x < 0. But clearly there are values for which x < 0 — for example, t = −1.1 Ô⇒ (x, y) ≈ (−2.71, 2.87). So why isn’t the TI84 graphing this? (... Example continued on the next page ...)
Page 449, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) The reason is that by default, the TI84 graphs only the region for where 0 ≤ t ≤ 2π (at least this is so for my particular calculator). We can easily adjust this: 4. Press the WINDOW button to bring up a menu of WINDOW options. 5. Using the arrow keys, the number pad, and the ENTER key as is appropriate, change Tmin and Tmax to your desired values. In my case, I decided somewhat randomly to enter Tmin = −10 and Tmax= 10. 6. Then press GRAPH again and the calculator will graph the given pair of parametric equations, now for the region Tmin ≤ t ≤ Tmax, where Tmin and Tmax are whatever you chose. After Step 9.
After Step 10.
After Step 11.
After Step 13.
After Step 14.
After Step 15.
After Step 12.
dy ∣ , Actually, the last few steps were really not necessary, if all we wanted was to find dx t=1 as we do now: 7. Press the blue 2ND button and then CALC (which corresponds to the TRACE button). This brings up the CALCULATE menu, which once again looks a little different under the current parametric setting. 8. Press 2 to select the “dy/dx” option. This brings you back to the graph. Nothing seems to be happening. But now, simply ... 9. Press 1 and now the bottom left of the screen changes to display “T = 1”. 10. Hit ENTER . What you’ve just done is to ask the calculator to calculate point where t = 1. The calculator tells you that “dy/dx = .83333528”.
dy at the dx
dy 5 Again, there’s a slight error — the exact correct answer is = = 0.8333..., so again the dx 6 TI84 is a tiny bit off.
Page 450, Table of Contents
www.EconsPhDTutor.com
46
The Maclaurin Series 46.1
Power Series
Recall what polynomials are: Example 433. 4 + x + 3x2 is a 2nd-degree polynomial. 18 + 5x − x2 + x4 is a 4th-degree polynomial. You can easily imagine what a “∞-degree polynomial” is. Only we don’t call it a “∞-degree polynomial”. Instead, we call it a power series. Definition 99. A power series is simply any infinite series ∞
∑ ai xi = a0 + a1 x + a2 x2 + . . . , i=0
where each ai is a real constant and x is the variable. Example 434. 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + . . . is a power series, with a0 = 1, a1 = 2, a2 = 3, . . . , ak = k + 1, . . . So too is 1 − x + x2 − x3 + x4 − x5 + . . . , with a0 = 1, a1 = −1, a2 = 1, . . . , ak = (−1)k+1 , . . . As we learnt before, a series can either be convergent or divergent. Example 435. 1 + x + x2 + x3 + x4 + x5 + . . . is a power series, with a0 = a1 = ⋅ ⋅ ⋅ = ak = ⋅ ⋅ ⋅ = 1. It is, moreover, a convergent power series, provided ∣x∣ < 1. Indeed, provided ∣x∣ < 1, we 1 know that this is an infinite geometric series that converges to and we may write 1−x 1 + x + x2 + x3 + x4 + x5 + ⋅ ⋅ ⋅ =
1 . 1−x
In contrast, if ∣x∣ ≥ 1, then 1 + x + x2 + x3 + x4 + x5 + . . . is a divergent power series.
For H2 Maths, the only power series we’ll be interested in is called the Maclaurin series.
Page 451, Table of Contents
www.EconsPhDTutor.com
46.2
Maclaurin Series
Definition 100. Let f be an infinitely-differentiable function. The Maclaurin series of f at x is denoted M (x) and is defined to be the power series M (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ + an xn + . . . , f (3) (0) f (n) (0) f ′′ (0) , a3 = , ..., an = , ... where a0 = f (0), a1 = f (0), a2 = 2! 3! n! ′
Written out explicitly or in summation notation, we have: f ′′ (0) 2 f (3) (0) 3 f (4) (0) 4 f (n) (0) n M (x) = f (0) + f (0)x + x + x + x + ⋅⋅⋅ + x + .... 2! 3! 4! n! ∞ (i) f (0) i =∑ x. i! i=0 ′
We are often interested in finite-order Maclaurin series: Definition 101. Let f be a n-times differentiable function. The nth-order Maclaurin series of f at x is denoted Mn (x) and is defined as the nth-degree polynomial (or finite series) Mn (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ + an xn , f ′′ (0) f (3) (0) f (n) (0) , a3 = , ..., an = . where a0 = f (0), a1 = f (0), a2 = 2! 3! n! ′
Example 436. Let’s compute the Maclaurin series for f ∶ R → R defined by x ↦ ex . We have f (0) = e0 = 1. We also have f ′ (x) = ex , so that f ′ (0) = e0 = 1. Similarly, f ′′ (x) = ex , so that f ′′ (0) = e0 = 1. Indeed, for any k ∈ Z+ , f (k) (x) = ex , so that f (k) (0) = e0 = 1. Hence, the Maclaurin series for f is M (x) = 1 + 1x +
1 2 1 3 x2 x2 x + x + ⋅⋅⋅ = 1 + x + + + ... 2! 3! 2! 3!
Here are the first four finite-order Maclaurin series for f : M0 (x) = 1,
M1 (x) = 1 + x,
Page 452, Table of Contents
M2 (x) = 1 + x +
x2 , 2!
M3 (x) = 1 + x +
x2 x3 + . 2! 3!
www.EconsPhDTutor.com
Exercise 176. Write down the third-order Maclaurin series for each of the following functions: (Answer on p. 1130.) (a) f ∶ R → R defined by x ↦ (1 + x)n , (b) g ∶ R → R defined by x ↦ sin x, (c) h ∶ R → R defined by x ↦ cos x, (d) i ∶ R → R defined by x ↦ ln(1 + x).
Remark 8. The A-level syllabuses make no mention of the Taylor series and so we won’t talk about it. But just so you know, the Maclaurin series is simply a special case of the Taylor series — specifically, it is the Taylor series about 0.
Page 453, Table of Contents
www.EconsPhDTutor.com
46.3
The Amazing Maclaurin Series
The Maclaurin series is simply an (infinite) series. And as we saw in Part II, an infinite series may or may not be convergent. The following is very powerful theorem: “Informal Theorem”. If f satisfies a “nice” property at a, then M (a) converges to f (a). That is, M (a) = f (a). (See section 971 in the Appendices for a more thorough and formal discussion of this theorem.) This table is in the List of Formulae you get, so no need to memorise.
f (x)
=
f (0)
+
xf ′ (0)
+
x2 ′′ f (0) 2!
+
...
xn (n) f (0) n!
+
...
(1 + x)n
=
1
+
nx
+
n(n − 1) 2 x 2!
+
...
n(n − 1) . . . (n − r + 1) r x r!
+
...
(∣x∣ < 1)
ex
=
1
+
x
+
x2 2!
+
...
xr r!
+
...
(all x)
sin x
=
x
−
x3 3!
+
x5 5!
+
...
(−1)r x2r+1 (2r + 1)!
+
...
(all x)
cos x
=
1
−
x2 2!
+
x4 4!
+
...
(−1)r x2r (2r)!
+
...
(all x)
ln(1 + x)
=
x
−
x2 2
+
x3 3
+
...
(−1)r+1 xr r
+
...
(−1 < x ≤ 1)
1. The first row of the above table says that if x is a value at which the function f satisfies the “nice” property, then f (x) is equal to the Maclaurin series of f at x. 2. The second row says that g ∶ R → R by x ↦ (1 + x)n satisfies the “nice” property for all n(n − 1) 2 x ∈ (−1, 1). Thus, for all x ∈ (−1, 1), we have (1 + x)n = 1 + nx + x + . . . We 2! say that (−1, 1) is the range of values for which g has a convergent Maclaurin series.47 3. The third row says that h ∶ R → R by x ↦ ex satisfies the “nice” property for all x ∈ R. x2 Thus, for all x ∈ R, we have ex = 1 + x + + . . . We say that R is the range of values 2! for which h has a convergent Maclaurin series. 47
We should be careful to state that if n < 0, then the domain should be restricted to exclude 0.
Page 454, Table of Contents
www.EconsPhDTutor.com
4. The fourth row says that i ∶ R → R by x ↦ sin x satisfies the “nice” property for all x ∈ R. x3 x5 Thus, for all x ∈ R, we have sin x = x − + . . . We say that R is the range of values 3! 5! for which i has a convergent Maclaurin series. 5. The fifth row says that j ∶ R → R by x ↦ cos x satisfies the “nice” property for all x ∈ R. x2 x4 Thus, for all x ∈ R, we have cos x = 1 − + . . . We say that R is the range of values 2! 4! for which j has a convergent Maclaurin series. 6. The sixth row says that k ∶ (−1, ∞) → R by x ↦ ln(1 + x) satisfies the “nice” property for x2 x3 all x ∈ (−1, 1]. Thus, for all x ∈ (−1, 1], we have ln(1 + x) = x − + − . . . We say that 2 3 (−1, 1] is the range of values for which k has a convergent Maclaurin series. In the syllabus, these five particular Maclaurin series are called standard series. x2 x3 + + . . . for all x ∈ R. 2! 3! Instead, we will merely verify that this equation is “plausible”, for x = 0, 1, 5. (Try these out yourself using the sheet “Maclaurin series” at the usual link.) Example 437. Here we will not rigorously prove that ex = 1 + x +
For x = 0, we have ex = e0 = 1. And M0 (0) = 1, M1 (0) = 1, M2 (0) = 1, and indeed Mn (0) = 1 for all n. So it does appear “plausible” that e0 = M (0). 1 For x = 1, we have ex = e1 ≈ 2.718. And M0 (1) = 1, M1 (1) = 1 + 1 = 2, M2 (1) = 1 + 1 + = 2.5, 2 ⋅ 1 1 M3 (1) = 1 + 1 + + = 2.67, ..., M7 (1) ≈ 2.718. It appears that Mn (1) ≈ 2.718 for all n ≥ 7. 2 6 So it does appear “plausible” that e1 = M (1). 25 For x = 5, we have ex = e5 ≈ 148.413. And M0 (5) = 1, M1 (5) = 1 + 5 = 6, M2 (5) = 1 + 5 + = 2 1 25 125 = 39 , ..., M18 (5) ≈ 148.413. It appears that Mn (5) ≈ 148.413 18.5, M3 (1) = 1 + 5 + + 2 6 3 for all n ≥ 18. So it does appear “plausible” that e5 = M (5).
Exercise 177. (Tedious, use the sheet named “Maclaurin series” at the usual link.) Verify π x3 x5 that for x = 0, , 2π, it is similarly “plausible” that sin x = x − + . . . (Answer on p. 2 3! 5! 1131.)
Page 455, Table of Contents
www.EconsPhDTutor.com
46.4
Finite-Order Maclaurin Series as Approximations
One important practical use of Maclaurin series is that finite-order Maclaurin series can be used as approximations. Example 438. Consider h ∶ R → R defined by x ↦ ex . We have h(1) = e ≈ 2.718. The 0th-order Maclaurin series is a pretty terrible approximation: M0 (1) = 1. The 1st-order Maclaurin series is slightly better: M1 (1) = 1 + x = 1 + 1 = 2. The 2nd-order Maclaurin series is even better: M2 (1) = 1 + x + 0.5x2 = 1 + 1 + 0.5 = 2.5. The 3rd-order Maclaurin series is ⋅ ⋅ very good: M3 (1) = 1 + x + 0.5x2 + x3 /3! = 1 + 1 + 0.5 + 0.16 = 2.66. We see that it tends to be that the higher the order of the Maclaurin series, the better the approximation.
I emphasise the phrase tends to be, because the approximation can sometimes get worse before it gets better, especially if we’re looking at a value that is far from 0. The next example illustrates.
Page 456, Table of Contents
www.EconsPhDTutor.com
Example 439. Consider i ∶ R → R defined by x ↦ sin x. We have h(2π) = sin(2π) = 0. If we do the tedious computations, we find that M0 (2π) = 0,
M1 (2π) = M2 (2π) = 2π ≈ 6.283,
M3 (2π) = M4 (2π) ≈ −35.059,
M5 (2π) = M6 (2π) ≈ 46.547.
The 0th-order Maclaurin series gets it exactly right. But each subsequent finite-order Maclaurin series then drifts ever further from 0! Having computed the 5th- and 6th-order Maclaurin series, it certainly does not look like the approximations will get any better. Yet if we perservere, we find that M7 (2π) = M8 (2π) ≈ −30.159,
M9 (2π) =M10 (2π) ≈ 11.900,
M11 (2π) =M12 (2π) ≈ −3.195,
M13 (2π) =M14 (2π) ≈ 0.625,
M15 (2π) =M16 (2π) ≈ −0.093, M17 (2π) =M18 (2π) ≈ 0.011,
M19 (2π) =M20 (2π) ≈ −0.001,
M21 (2π) =M22 (2π) ≈ 0.000,
...
Indeed, Mn (2π) ≈ 0.000 for all n ≥ 21. So it does indeed look like the Maclaurin series for sin x converges. Graphed below are sin x and M21 (x). We see that M21 (x) almost perfectly approximates sin x for x ∈ [−7, 7]. But for larger values, M21 (x) veers far away from sin x.
y
x -12
-7
-2
2
7
12
(... Example continued on the next page ...)
Page 457, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) Graphed below are y = sin x, M1 (x), . . . , M10 (x). We see that the 1st-order Maclaurin series M1 (x) = x is indeed a good approximation for values of x that are close to 0, but terrible for larger values. Low-order Maclaurin series work well as approximations, provided we are looking at small values of x (i.e. values that are close to 0). But for large x, even if the Maclaurin series eventually converges, low-order Maclaurin series may fare very poorly as approximations. Indeed, as we saw on the previous page, for sufficiently large values of x, even a relatively-high-order Maclaurin series like M21 (x) will fare poorly as an approximation!
y
x
Page 458, Table of Contents
www.EconsPhDTutor.com
If a is not within the range of values for which the Maclaurin series for the function f converges, then M (a) ≠ f (a). That is, the (infinite) Maclaurin series does not converge. Hence, there is no reason to expect that Mi (a) ≈ f (a) (i.e. that any finite-order Maclaurin series will serve as a good approximation). Example to illustrate. Example 440. Consider k ∶ R → R defined by x ↦ ln(1 + x). The range of values for which the Maclaurin series converges is (−1, −1]. Suppose we pick x = 2, which is certainly outside this range. Then we have k(2) = ln 3 ≈ 1.099. Let’s see what the finite-order Maclaurin series look like: M0 (2) = 0,
M1 (2) = 2,
1 M4 (2) = −1 , 3
M5 (2) = 5
M8 (2) = −19.314,
M9 (2) = 37.575,
1 , 15
M2 (2) = 0,
2 M3 (2) = 2 , 3
M6 (2) = −5.6,
M7 (2) ≈ 12.686,
M10 (2) = −64.825,
M11 (2) ≈ 121.356.
Unlike before, further perserverance will not pay off here. Indeed, the Maclaurin series will grow without bound. For example, M50 (2) ≈ −14.9 trillion! The Maclaurin series simply does not converge for x = 2. So there is no reason to expect any finite-order Maclaurin series to be a good approximation.
Page 459, Table of Contents
www.EconsPhDTutor.com
46.5
Product of Two Power Series
Informally, if two power series converge, then so too does their product; and to get this product, simply multiply the two series together as if they were finite polynomials.48 1 1 3 x + . . . and cos x = 1 + 0 − x2 + 0 + . . . . 3! 2! Thus, for all x ∈ R, we have sin x cos x = c0 + c1 x + c2 x2 + c3 x3 + . . . , where Example 441. For all x ∈ R, sin x = 0 + 1x + 0 −
Constant Term ∶
c0 = 0 × 1 = 0,
Coefficient on x ∶
c1 = 0 × 0 + 1 × 1 = 1,
Coefficient on x2 ∶
1 c2 = 0 × (− ) + 1 × 0 + 0 × 1 = 0, 2!
Coefficient on x3 ∶
1 2 1 c3 = 0 × 0 + 1 × (− ) + 0 × 0 + (− ) × 1 = − 2! 3! 3 ⋮
Take a moment to convince yourself that c0 , c1 , c2 , c3 are as stated. So for all x ∈ R, 2 2 sin x cos x = 0 + 1x + 0x2 + (− ) x3 + ⋅ ⋅ ⋅ = x − x3 + . . . 3 3 The expression on the RHS is, of course, simply also the Maclaurin series for sin x cos x. You are asked to show this in Exercise 178. Exercise 178. Let f ∶ R → R be defined by x ↦ sin x cos x. Evaluate f (0), f ′ (0), f ′′ (0), and f (3) (0). Hence, write down the 3rd-order Maclaurin series for f and verify that this is consistent with what we found in Example 441. (Answer on p. 1132.)
The next example illustrates that one must be careful about when the Maclaurin series is convergent:
48
This assertion is formally stated and proven at Fact 97 in the Appendices (optional).
Page 460, Table of Contents
www.EconsPhDTutor.com
1 Example 442. For all x ∈ R, we have sin x = 0 + 1x + 0 − x3 + . . . For all x ∈ (−1, 1], 3! 1 2 1 3 we have ln(1 + x) = 1x − x + x + . . . And so for x ∈ (−1, 1], we have sin x ln(1 + x) = 2 3 c0 + c1 x + c2 x2 + c3 x3 + . . . , where Constant Term ∶
c0 = 0 × 0 = 0,
Coefficient on x ∶
c1 = 0 × 1 + 1 × 0 = 0,
Coefficient on x2 ∶
1 c2 = 0 × (− ) + 1 × 1 + 0 × 0 = 1, 2
Coefficient on x3 ∶
c3 = 0 ×
1 1 1 1 + 1 × (− ) + 0 × 1 + (− ) × 0 = − 3 2 3! 2
⋮ 1 1 And so sin x ln(1 + x) = 0 + 0x + 1x2 + (− ) x3 + ⋅ ⋅ ⋅ = x2 − x3 + . . . , for x ∈ (−1, 1] — this set 2 2 is simply the intersection of R and (−1, 1], which are respectively the ranges of values on which the Maclaurin series for sin x and ln x converge. The expression on the RHS is, of course, simply also the Maclaurin series for sin x ln(1 + x). You are asked to show this in Exercise 178. Exercise 179. Let f ∶ R → R be defined by x ↦ sin x ln(1+x). Evaluate f (0), f ′ (0), f ′′ (0), and f (3) (0). Hence, write down the 3rd-order Maclaurin series for f and verify that this is consistent with what we found in Example 442. (Answer on p. 1132.)
Page 461, Table of Contents
www.EconsPhDTutor.com
46.6
Composition of Two Functions
Informally, if f (x) = a0 + a1 x + a2 x2 + . . . for x ∈ S and g(c) ∈ S, then 2
f (g(c)) = a0 + a1 g(c) + a2 [g(c)] + . . . That is, to get f (g(c)), simply “plug in” g(c) into the power series for f .49 Example 443. Define f ∶ R− ∪ R+ → R by f (x) = (1 + x)−1 and g ∶ R → R by g(x) = 2x. We know that for all x ∈ (−1, 1), we have f (x) = (1 + x)−1 = 1 − x + x2 − x3 + . . . . Thus, f (g(x)) = (1 + 2x)−1 = 1 − (2x) + (2x)2 − (2x)3 + . . . for all g(x) = 2x ∈ (−1, 1). Equivalently, f (g(x)) = (1 + 2x)−1 = 1 − 2x + 4x2 − 8x3 + . . . for all x ∈ (−0.5, 0.5). Example 444. Define f ∶ R → R by f (x) = ex and g ∶ R → R by g(x) = x2 . We know that x2 x3 + + .... for all x ∈ R, we have f (x) = ex = 1 + x + 2! 3! 2
3
(x2 ) (x2 ) Thus, f (g(x)) = e = 1 + x + + + . . . for all g(x) ∈ R. Equivalently, 2! 3! 2
x2
x4 x6 + + . . . for all x ∈ R. f (g(x)) = e = 1 + x + 2! 3! x2
2
In the case where g also has a convergent Maclaurin series, we can likewise also simply “plug in” the Maclaurin series for g.50 Example:
49 50
For a more careful and formal version of this assertion, see Fact 98 in the Appendices (optional). Again, for a more careful and formal version of this assertion, see Fact 99 in the Appendices (optional).
Page 462, Table of Contents
www.EconsPhDTutor.com
Example 445. Define f ∶ (−∞, −1) ∪ (−1, ∞) → R by f (x) = 1/(1 + x) and g ∶ R → R by g(x) = sin x. Write down the Maclaurin series for f ○ g, up to the 4th-order term. Method #1 (composition method). We know that for all x ∈ (−1, 1), we have f (x) = x3 1−x+x2 −x3 +. . . . And for all x ∈ R, we have g(x) = x− +. . . Hence, for all g(x) ∈ (−1, 1), 3! i.e. for all x ≠ kπ/2 (for k ∈ Z), we have 1 2 3 = 1 − g(x) + [g(x)] − [g(x)] + . . . 1 + sin x 2 3 x3 x3 x3 + . . . ) + (x − + . . . ) − (x − + ...) ... = 1 − (x − 3! 3! 3! 1 5 = 1 − x + x2 + x3 ( − 1) + ⋅ ⋅ ⋅ = 1 − x + x2 − x3 + . . . 3! 6
f (g(x)) =
Find the general term for a Maclaurin series is explicitly excluded from the A-level syllabuses. Usually you’ll just have to write down the first few terms. Method #2 (direct method). Let h(x) = 1/(1 + sin x). We have h(0) = 1. We also have R R − cos x RRRR dh RRRR R = = −1, 2 RRR dx RRRR R (1 + sin x) RRx=0 Rx=0 R R R 2 (1 + sin x) sin x + 2 cos2 x RRRR d2 h RRRR (1 + sin x) sin x + 2 cos2 x(1 + sin x) RRRR RRR = RRR R = 4 3 dx2 RRRR R RRR (1 + sin x) (1 + sin x) R Rx=0 Rx=0 x=0 R sin x + 2 − sin2 x RRRR = RR = 2, 3 (1 + sin x) RRRRx=0 3 2 R (1 + sin x) (cos x − 2 sin x cos x) − (sin x + 2 − sin2 x) 3 (1 + sin x) cos x RRRR d3 h RRRR RRR R = 6 RRR dx3 RRRR (1 + sin x) Rx=0 Rx=0 1−2⋅3⋅1 = = −5. 1
Thus,
2 −5 5 1 = 1 + (−1)x + x2 + x3 + ⋅ ⋅ ⋅ = 1 − x + x2 − x3 + . . . 1 + sin x 2! 3! 6
In the above example, I gave two methods. Use whichever seems to be easier or quicker. Here’s another example:
Page 463, Table of Contents
www.EconsPhDTutor.com
Example 446. Write down the Maclaurin series for sec x up to the 4th-order term. Method #1 (composition method). x2 x4 1 1 = [1 − ( sec x = = − + . . . )] cos x 1 − x2!2 + x4!4 − . . . 2! 4!
−1
2
x2 x4 x2 x4 =1+( − + ...) + ( − + ...) + ... 2! 4! 2! 4! 1 1 2 x2 5x4 x2 4 + x [− + ( ) ] + ⋅ ⋅ ⋅ = 1 + + + ... =1+ 2! 4! 2! 2 24 Method #2 (direct method). Let f (x) = sec x. Then f (0) = sec 0 = 1. And f ′ (x) = sec x tan x
Ô⇒ f ′ (0) = 0,
f ′′ (x) = sec x tan2 x + sec3 x f (3) (x) = 6 sec2 xf ′ (x) − f ′ (x) 2
Ô⇒ f ′′ (0) = 1, Ô⇒ f (3) (0) = 0,
f (4) (x) = 12 sec x [f ′ (x)] + 6 sec2 xf ′′ (x) − f ′′ (x) Thus,
Ô⇒ f (4) (0) = 5.
1 2 0 3 5 4 x2 5x4 sec x = 1 + 0x + x + x + x + ⋅ ⋅ ⋅ = 1 + + + ... 2! 3! 4! 2! 24
Exercise 180. Write down the third-order Maclaurin series for sin [ln(1 + x)]. State also the range of values for which the Maclaurin series converges. (Answer on p. 1132.)
Page 464, Table of Contents
www.EconsPhDTutor.com
46.7
How the Maclaurin Series Works (Optional)
Theorem 8. Let f ∶ [−c, c] → R be an infinitely-differentiable function. Suppose that for all x ∈ (−c, c), we have f (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 . . . Then the coefficients in the above power series are as given by the Maclaurin series. That is, for each i = 0, 1, 2, . . . , we have ai =
f (i) (0) . i!
Proof. Observe that f ′ (x) = a1 + 2a2 x + 3a3 x2 + 4a4 x3 . . . ,
f ′′ (x) = 2a2 + (3 × 2)a3 x + (4 × 3)a4 x2 . . . ,
f (3) (x) = (3 × 2)a3 + (4 × 3 × 2)a4 x . . . , f (4) (x) = (4 × 3 × 2)a4 + . . . Thus, f (0) = a0 and f ′ (0) = a1 ,
f ′′ (0) = 2!a2 ,
f (3) (x) = 3!a3 , f (4) (x) = 4!a4 . Rearranging, we have a0 = f (0), a1 = f ′ (0), a2 = f ′′ (0)/2!, a3 = f (3) (0)/3!, a4 = f (4) (0)/4!, ..., ai = f (i) (0)/i!, ..., as desired. The above theorem is merely a tantalising hint of why the Maclaurin series “works”. This is because the theorem merely says this: If we make the very big assumption that the infinitelydifferentiable function f can be written down as a power series, then the coefficients of the power series are as given by the Maclaurin series. But this is not very useful, because — how do we know that the function can be written down as a power series? For a continuation of this discussion, see section 88.14 in the Appendices.
Page 465, Table of Contents
www.EconsPhDTutor.com
47
The Indefinite Integral
Definition 102. Given functions f and F , we call F an indefinite integral (or antiderivative or primitive) of f if for all x in the domain of f , F ′ (x) = f (x), In Leibniz’s notation, we may write F = ∫ f (x) dx or more simply F = ∫ f dx. The statement “F = ∫ f dx” is thus completely equivalent to the statement “F ′ = f ”. Example 447. Consider the functions f, F ∶ R → R defined by f (x) = 2x and F (x) = x2 . We see that F is an indefinite integral of f , because F ′ (x) = 2x = f (x) for all x. We can equivalently say that f is the derivative of F . We can also write F = ∫ f dx or
dF = f. dx
The statement “the value of F at 5 is 25” can be written as F (5) = 25. It can also be written as ∫ f dx∣x=5 = 25 or
∫ f (x) dx∣x=5 = 25.
The symbol ∫ is called the integration sign — it is an elongated S. The symbol dx is called the differential of the variable x — it informs us that the variable of integration is x. The function f to be integrated is called the integrand. Just like with summation, x is a “dummy” variable. We can replace x with any other letter and the function F will still remain exactly the same function.
Page 466, Table of Contents
www.EconsPhDTutor.com
Example 448. The following two expressions are equal because i on the LHS and r on the RHS are simply “dummy” variables. n
n
i=1
r=1
∑ i = ∑ r. Similarly, the statement F = ∫ f (x) dx is equivalent any of the following three statements, because the letters x, a, b, c, etc. are merely “dummy” variables: F = ∫ f (a) da,
or F = ∫ f (b) db,
or F = ∫ f (c) dc.
So the statement “the value of F at 5 is 25” can also be written F (5) = 25 or any of the following four statements: ∫ f (x) dx∣x=5 = 25, ∫ f (a) da∣a=5 = 25, ∫ f (b) db∣b=5 = 25,
Page 467, Table of Contents
∫ f (c) dc∣c=5 = 25.
www.EconsPhDTutor.com
47.1
The Constant of Integration
Example 449. Consider f ∶ R → R defined by f (x) = sin x. F ∶ R → R defined by F (x) = − cos x is an indefinite integral of f , because F ′ (x) = sin x = f (x) for all x ∈ R. Are there any other indefinite integrals of f ? Yes, certainly. For example, G ∶ R → R defined by G(x) = − cos x + 200 is also an indefinite integral of f , because G′ (x) = sin x = f (x) for all x ∈ R. Indeed, any H ∶ R → R defined by H(x) = − cos x + C where C ∈ R is also an indefinite integral of f , because H ′ (x) = sin x = f (x) for all x ∈ R.
In general: Fact 59. If F is an indefinite integral of f , then so too is G defined by G(x) = F (x) + C, for any C ∈ R. We call C the constant of integration. Proof. Since F ′ (x) = f (x) for all x, we also have G′ (x) = F ′ (x) + C ′ = F ′ (x) + 0 = f (x) for all x. And so by definition, G is also an indefinite integral of f .
Page 468, Table of Contents
www.EconsPhDTutor.com
47.2
The Indefinite Integral is Unique Up to the C.O.I.
The indefinite integral is unique up to the constant of integration. That is, if F and G are both indefinite integrals of f , then it must be that F and G differ only by a constant. Example 450. Say f has indefinite integral F defined by F (x) = sin (ex −3x+5 ). Suppose G is another indefinite integral of f . Then it must be that F (x) = G(x) + C, for some C ∈ R. 2
Formally: Fact 60. If F and G are both indefinite integrals of f , then there exists some C ∈ R such that F (x) = G(x) + C for all x. Proof. Since F and G are both indefinite integrals of f , by definition, F ′ (x) = G′ (x) for all x. And thus (F − G)′ (x) = 0 for all x. But the only functions whose derivative is always 0 are constant functions.51 Thus, F (x) − G(x) = C, for all x, for some C ∈ R.
Exercise 181. (Answer on p. 1133.) Let f ∶ R → R, F ∶ R → R, G ∶ R → R be defined by f (x) = 4 sin 4x, F (x) = − cos 4x, and G(x) = 8 sin2 x cos2 x . (a) Show that F and G are both indefinite integrals of f . (b) F and G seem to be very different functions. Yet both are indefinite integrals of f . Why does this not contradict our assertion that “the indefinite integral is unique up to a constant”?
51
The alert reader will note that this assertion has not actually been proven in this textbook. We’ll simply take it for granted that “the only functions whose derivative is 0 are constant functions”.
Page 469, Table of Contents
www.EconsPhDTutor.com
48
Integration Techniques
As before with our notation for differentiation, let’s be clear (pedantic). To take an example, the notation ∫ sin x dx = − cos x + C is simply shorthand for the following long-winded statement: “Consider a function with mapping rule x ↦ sin x. Its indefinite integrals are functions, all of which have the mapping rule x ↦ − cos x + C.”52
52
This shorthand statement fails to to mention the domain and codomains of the function and its indefinite integral. However, the careful writer will of course have specified these nearby.
Page 470, Table of Contents
www.EconsPhDTutor.com
48.1
Basic Rules of Integration
Proposition 10. Let k, n ∈ R be constants with n ≠ −1. Let f and g be functions with indefinite integrals F and G. Then ∫ k dx n
∫ x dx
=
kx + C,
xn+1 = + C, n+1
−1 ∫ x dx = ln ∣x∣ + C, x ∫ e dx
=
(x ≠ 0 if n < 0) (x ≠ 0)
∫ sin x dx
= − cos x + C,
∫ cos x dx
= − sin x + C,
∫ f (x) ± g(x) dx = F + G + C,
ex + C,
∫ kf (x) dx
=
kF + C,
where in each case, C is the constant of integration. Proof. In general, to prove that ∫ f (x) dx = F , it suffices to prove that F ′ (x) = f (x) for all x. d And so to prove that ∫ x−1 dx = ln ∣x∣ + C, it suffices to prove that (ln ∣x∣ + C) = x−1 for dx all x ≠ 0. This we now do. First note that ⎧ ⎪ ⎪ ⎪ln x + C, ln ∣x∣ + C = ⎨ ⎪ ⎪ ⎪ ⎩ln (−x) + C, ⎧ 1 ⎪ ⎪ , ⎪ ⎪ ⎪ x ⎪ d ⎪ Thus, (ln ∣x∣ + C) = ⎨ ⎪ dx ⎪ ⎪ −1 ⎪ ⎪ ⎪ = ⎪ ⎩ −x And so indeed
for x ≥ 0, for x < 0. for x ≥ 0, 1 , x
for x < 0.
d (ln ∣x∣ + C) = x−1 for all x ≠ 0. dx
You are asked to prove the remaining rules of integration in Exercise 182.
Exercise 182. Prove the remaining rules of integration listed in Proposition 10. (Answer on p. 1134.)
Page 471, Table of Contents
www.EconsPhDTutor.com
48.2
More Basic Rules of Integration
No need to memorise the following rules of integration, because the List of Formulae contains a (slightly less general) version. Proposition 11. Let a ≠ 0. Then (a)
(b)
1 ∫ x2 + a2 dx
=
1 dx = ∫ √ 2 a − x2
1 x tan−1 ( ) + C, a a x sin−1 ( ) + C, a
for ∣x∣ < a,
(c)
1 ∫ x2 − a2 dx
=
1 x−a ln ∣ ∣ + C, 2a x+a
for x ≠ a,
(d)
1 ∫ a2 − x2 dx
=
1 a+x ln ∣ ∣ + C, 2a a−x
for x ≠ a,
(e)
∫ tan x dx
=
ln ∣sec x∣ + C,
(f)
∫ cot x dx
=
ln ∣sin x∣ + C,
(g)
∫ csc x dx
= − ln ∣csc x + cot x∣ + C,
(h)
∫ sec x dx
=
ln ∣sec x + tan x∣ + C,
for x not an odd multiple of
π , 2
for x not an multiple of π, for x not an multiple of π, for x not an odd multiple of
π , 2
where in each case, C is the constant of integration. Proof. We prove only (a), (c), and (e). (You are asked to prove the remaining rules of integration in Exercise 183.) d 1 d 1 x 1 1 1 tan−1 x = 2 . Hence, [ tan−1 ( ) + C] = ⋅ = dx x +1 dx a a a ( x )2 + 1 a a a 1 1 x . So indeed ∫ 2 dx = tan−1 ( ) + C. 2 2 2 x +a x +a a a
(a) By Corollary 2,
(... Proof continued on the next page ...)
Page 472, Table of Contents
www.EconsPhDTutor.com
(... Proof continued from the previous page ...)
(c) Let x ≠ a. Case #1:
x−a ≥ 0. x+a
d 1 x−a d 1 x−a 1 d ( ln ∣ ∣ + C) = ( ln + C) = [ln(x − a) − ln(x + a)] dx 2a x+a dx 2a x + a 2a dx 1 1 1 1 x + a − (x − a) 1 2a 1 = ( − )= = = , 2a x − a x + a 2a (x − a) (x + a) 2a x2 − a2 x2 − a2 1 1 x−a so that indeed ∫ 2 dx = ln ∣ ∣ + C. x − a2 2a x+a Case #2:
x−a < 0. x+a
x−a d 1 a−x 1 d d 1 ( ln ∣ ∣ + C) = ( ln + C) = [ln(a − x) − ln(x + a)] dx 2a x+a dx 2a x + a 2a dx 1 −1 1 1 1 1 1 = ( − )= ( − )= 2 , 2a a − x x + a 2a x − a x + a x − a2 1 x−a 1 so that again ∫ 2 dx = ln ∣ ∣ + C. 2 x −a 2a x+a (e) Let x not be an odd integer multiple of π/2, so that sec x ≠ 0. Case #1: sec x ≥ 0. d d sec x tan x (ln ∣sec x∣ + C) = (ln sec x + C) = = tan x, dx dx sec x so that indeed ∫ tan x dx = ln ∣sec x∣ + C. Case #2: sec x < 0. d − sec x tan x d (ln ∣sec x∣ + C) = [ln (− sec x) + C] = = tan x, dx dx − sec x so that again ∫ tan x dx = ln ∣sec x∣ + C.
Exercise 183. Prove the remaining rules of integration listed in Proposition 11. (Answers on pp. 1135, 1136, 1137, and 1138.)
Page 473, Table of Contents
www.EconsPhDTutor.com
48.3
Trigonometric Functions
The following indefinite integrals are NOT on the List of Formulae and you are definitely required to know how to derive them on your own! Fact 61. Let m, n ∈ R. Then (a)
2 ∫ sin x dx
=
1 sin 2x x− + C, 2 4
(b)
2 ∫ cos x dx
=
sin 2x 1 x+ + C, 2 4
(c)
2 ∫ tan x dx
=
tan x + x + C,
(d)
1 cos [(m − n)x] cos [(m + n)x] + } + C, ∫ sin(mx) cos(nx) dx = − 2 { m−n m+n
(e)
∫ sin(mx) sin(nx) dx =
1 sin [(m − n)x] sin [(m + n)x] { − } + C, 2 m−n m+n
(f)
∫ cos(mx) cos(nx) dx =
1 sin [(m − n)x] sin [(m + n)x] { + } + C, 2 m−n m+n
where C is the constant of integration. Proof. (a) The trick is to recall the trigonometric identity cos 2x = 1 − 2 sin2 x (this is in the List of Formulae, as are several other trig identities). And so: 2 ∫ sin x dx = ∫
1 − cos 2x 1 sin 2x dx = x − + C. 2 2 4
You are asked to prove the remaining rules of integration in Exercise 184.
Exercise 184. Prove the remaining rules of integration listed in Fact 61. (Answer on p. 1139.)
Page 474, Table of Contents
www.EconsPhDTutor.com
48.4
Integration by Substitution (IBS)
The method of integration by substitution (IBS) is the Chain Rule in reverse. Before we explain why it works, here are two examples of how it works.53 cos x . Next, observe that Example 451. Let’s find ∫ cot x dx. First, observe that cot x = sin x d du sin x = cos x. Let u = sin x (this is our substitution), so that we also have = cos x. dx dx So: cos x 1 du cot x dx = dx = ∫ ∫ sin x ∫ u dx dx. So far, nothing unusual has happened. Now we’re going to do something strange, which is to take that last expression and merrily cancel out the dx’s: 1 du 1 ∫ u dx dx = ∫ u du + C1 . du is NOT a fraction? So why are we Didn’t we repeatedly insist earlier that the derivative dx allowed to “merrily” cancel out the dx’s!? Shortly we’ll explain why this move is legitimate. For now, let us blindly perservere: 1 ∫ u du + C1 = ln ∣u∣ + C = ln ∣sin x∣ + C.
Another example, before we explain why exactly we can “merrily” cancel out the dx’s: du Example 452. Let’s find ∫ 2x cos x2 dx. Let u = x2 , so that we also have = 2x. Now, dx du 2 ∫ 2x cos x dx = ∫ dx cos u dx. Again, we merrily cancel out the dx’s and write: du 2 ∫ dx cos u dx = ∫ cos u du + C1 = sin u + C = sin x + C.
53
Actually we secretly already used this method a few times above, though not very explicitly.
Page 475, Table of Contents
www.EconsPhDTutor.com
We now explain why it is OK to “merrily cancel out the dx’s”. In fact, saying that we “merrily cancel out the dx’s” is merely a mnemonic (memory device). We are not actually “cancelling out the dx’s”. Instead, we are appealing to the following result: Theorem 9. Let f ∶ D → R be any continuous function. Let u be a real-valued differentiable function. Assume Range(u) ⊆ D, so that the composite function f ○ u exists. Then du ∫ f ⋅ dx dx = ∫ f du + C. dP 1 du dQ 2 du dx and Q = ∫ f du. In other words, =f⋅ and = f. Proof. Let P = ∫ f ⋅ dx dx dx du 2
Using first the Chain Rule and then =, we have
dQ dQ du 3 du = ⋅ =f⋅ . dx du dx dx
du . And so dx by Fact 60 (uniqueness of the indefinite integral up to a constant), P and Q must be equal (or differ by at most a constant). That is, P = Q + C or 1
3
Examining = and =, we see that P and Q are both indefinite integrals for f ⋅
du ∫ f ⋅ dx dx = ∫ f du + C.
The above result says that when doing integration, we are allowed to “merrily” do two things: du du dx with du (“cancel out the dx’s from dx to get du”); dx dx du dx du 2. Replace du with dx (“multiply du by = 1 to get dx”). dx dx dx
1. Replace
Of course, we are not actually doing any such things as “cancelling out the dx’s” or “muldx tiplying by = 1” — these are merely mnemonics. Instead, all we are doing is appealing dx to the above theorem.54 Let’s try more examples, now that we have a better understanding of how this works:
54
dy dx dy = 1/ . The IFT is true NOT because and dx dy dx dx dy dx are fractions. Nonetheless, as a convenient mnemonic, we can pretend that the IFT holds because and are dy dx dy fractions — even though strictly speaking, such thinking is wrong.
This is analogous to the Inverse Function Theorem, which states that
Page 476, Table of Contents
www.EconsPhDTutor.com
du Example 453. Let’s find ∫ esin x cos x dx. Let u = sin x, so that we also have = cos x. dx Now we can write du 1 sin x cos x dx = ∫ eu dx = ∫ eu du + C1 = eu + C = esin x + C, ∫ e dx 1
1
where = uses Theorem 9. Purely as a mnemonic, we may think of this step = as “cancelling out the dx’s”, even though strictly speaking, we are doing no such thing; instead, we are appealing to Theorem 9.
50
Example 454. Let’s find ∫ (x3 + 5x2 − 3x + 2) (3x2 + 10x − 3) dx. One method would be to fully expand the integrand to get a 152nd-degree polynomial, then integrate this polynomial term-by-term. This is doable, but absurdly tedious. A better method is to observe that 3x2 + 10x − 3 = x3 + 5x2 − 3x + 2. Then we can write
d (x3 + 5x2 − 3x + 2). Thus, let u = dx
50 3 2 2 50 du ∫ (x + 5x − 3x + 2) (3x + 10x − 3) dx = ∫ u dx dx 51
u51 (x3 + 5x2 − 3x + 2) = ∫ u du + C1 = +C = 51 51 1
50
+ C,
1
where once again = uses Theorem 9.
In the next three examples, we go in the “opposite direction”. That is, instead of “cancelling dx out the dx’s” as was done in the previous few examples, we instead “multiply by = 1”. dx
Page 477, Table of Contents
www.EconsPhDTutor.com
√ Example 455. Let’s find ∫ 1 − u2 du. We’ll use the substitution u = sin x. Note that √ √ du 2 1 − u = 1 − sin2 x = cos x. Moreover, = cos x. So dx √ dx du 1 1 − u2 du = ∫ cos x du = ∫ cos x du = ∫ cos x dx + C1 ∫ dx dx sin 2x 2 1 = ∫ cos x cos x dx + C1 = ∫ cos2 x dx + C1 = x + + C, 2 4 √ 1 −1 2 sin x cos x sin−1 u + u 1 − u2 = sin u + = , 2 4 2 1
2
where = uses Theorem 9 and = uses Fact 61. x2 Example 456. Let’s find ∫ √ dx. We’ll use the substitution u3 = 1 + 2x. Note that 3 1 + 2x 1 3 dx 3 2 x = (u − 1) and = u . So 2 du 2 2
2 2 [ 12 (u3 − 1)] x2 (u3 − 1) (u3 − 1) du √ dx = ∫ dx = ∫ dx = ∫ dx ∫ √ 3 3 4u 4u du 1 + 2x u3 2
2
(u3 − 1) dx (u3 − 1) 3 2 3u (1 − 2u3 + u6 ) =∫ du + C1 = ∫ ( u ) du + C1 = ∫ du + C1 4u du 4u 2 8 1
=
3 3 u2 2u5 u8 3 2 1 2u3 u6 4 7 u − 2u + u du + C = ( − + ) + C = u ( − + )+C 1 8∫ 8 2 5 8 8 2 5 8
3 2(1 + 2x) (1 + 2x)2 2/3 1 = (1 + 2x) [ − + ]+C 8 2 5 8 3 20 − 16 − 32x + 5 + 20x + 20x2 3 20x2 − 12x + 9 + C = (1 + 2x)2/3 + C, = (1 + 2x)2/3 8 40 8 40 1
where = uses Theorem 9. (The last line is just further simplification, which is nice but not necessary.)
Page 478, Table of Contents
www.EconsPhDTutor.com
1 Example 457. Let’s find ∫ dx. We’ll use the substitution u = tan x or x = 1 + 3 cos2 x tan−1 u. Note that cos2 x = So
∫
1 1+
3 1+u2
dx = ∫
1 1 1 = = 2 sec2 x 1 + tan x 1 + u2 1 1+
3 1+u2
and
dx 1 = . du 1 + u2
du 1 1 dx 1 1 dx = ∫ du + C1 = ∫ du + C1 3 3 du 1 + 1+u2 du 1 + 1+u2 1 + u2
1 1 1 2 1 −1 u −1 tan x du + C = du + C = tan ( ) + C = tan ( ) + C, 1 1 ∫ 1 + u2 + 3 22 + u2 2 2 2 2
=∫ 1
where = uses Theorem 9 (“multiply by
du 2 = 1”) and = uses Proposition 11. du
Usually, the hard part is to figure out the appropriate substitution to make. Fortunately, in the A-level exams, you’ll always be told what substitution to make.
Exercise 185. (Answers on pp. 1140 and 1141.) (a) (i) Use the substitution x = 3 sec u 9 √ to find ∫ dx. x2 x2 − 9 √ 9 9 √ (ii) Now use instead the substitution x = to find ∫ dx. 2 x2 − 9 1−u x √ 1 (iii) Show that sin (sec−1 y) = 1 − 2 . Then explain why your answers in (i) and (ii) are y consistent. x3 3 tan u to find ∫ dx. 3/2 2 (4x2 + 9) x3 2 (ii) Now use instead the substitution u = 4x + 9 to find ∫ dx. 3/2 (4x2 + 9) 1 (iii) Show that cos (tan−1 y) = √ . Then explain why your answers in (i) and (ii) are 1 + y2 consistent. (b) (i) Use the substitution x =
Page 479, Table of Contents
www.EconsPhDTutor.com
48.5
Integration by Parts (IBP)
The method of integration by parts (IBP) is the Product Rule in reverse. Theorem 10. (Integration by Parts.) Let u and v be differentiable functions, which have continuous derivatives u′ and v ′ . Then ∫ uv ′ dx = uv − ∫ u′ v dx.
Proof. Differentiate the RHS to get u′ v + uv ′ − u′ v = uv ′ . This shows that ∫ uv ′ dx = uv − ∫ u′ v dx, as desired. dv , Exponential, Trig, dx Algebraic, Inverse trig, Log. (This is because exponential functions are easiest to integrate, followed by trigonometric functions, etc.)
To choose v ′ , use the rule of thumb DETAIL — D stands for
Example 458. Find ∫ xex dx. By the DETAIL rule of thumb, we should choose v ′ = ex . Now, ′ u v ©©x
′ x x x x x ∫ x e dx = uv − ∫ u v dx = xe − ∫ e dx = xe − e = e (x − 1).
Sometimes we need to apply IBP more than once:
Example 459. Find ∫ x2 ex dx. By the DETAIL rule of thumb, we should choose v ′ = ex . Now, u v′
©© 2 x ′ 2 x x ∫ x e dx = uv − ∫ u v dx = x e − ∫ 2xe dx = x2 ex − 2ex (x − 1) = ex (x2 − 2x + 2). (Use the previous example.)
Exercise 186. Find ∫ x sin x dx and ∫ x2 sin x dx. (Answer on p. 1142.)
Page 480, Table of Contents
www.EconsPhDTutor.com
49
The Fundamental Theorems of Calculus (FTCs)
The problem of finding the definite integral is the problem of finding the area under a curve. The problem of finding the derivative is the problem of finding the slope of the tangent. The two Fundamental Theorems of Calculus (FTCs) show that, surprisingly enough, these two problems are intimately (indeed inversely) related. This chapter is a largely-informal discussion of the intuition behind the FTCs.
49.1
The Area Function
Given a continuous real-valued function f , its area function is denoted A and is, informally, defined by the mapping A(c) = “Area bounded by the graph of f , the horizontal axis, and the vertical lines x = 0 and x = c”. Example 460. Graphed below is the continuous function f ∶ R+0 → R defined by f (x) = √ x + 1. The area A(6) is highlighted in red. It is the area bounded by the graph of f , the horizontal axis, and the vertical lines x = 0 and x = 6. Using a graphing calculator, A(6) = 15.79795897... Is there a way I can figure this out without a graphing calculator? Here’s one possible approach — let’s approximate the area by using three rectangles.
(... Example continued on the next page ... )
Page 481, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) We’ll use three rectangles of equal width — so each rectangle has width 2. The leftmost rectangle will occupy the interval [0, 2], the middle rectangle will occupy [2, 4], and the rightmost rectangle will occupy [4, 6]. For each rectangle, we choose its height to be the lowest value attained by the function in that interval. In the interval [0, 2], the lowest value attained by f is f (0). So the leftmost blue rectangle has height f (0) and thus area Base × Height = 2f (0). Similarly, the middle green rectangle has height f (2), because in the interval [2, 4], the lowest value attained by f is f (2). Hence, it has area Base × Height = 2f (2). The rightmost grey rectangle has height f (4), because in the interval [4, 6], the lowest value attained by f is f (4). Hence, it has area Base × Height = 2f (4).
y
x -1
0
1
2
3
4
5
6
7
8
9
Altogether, the total area of these three rectangles is √ √ √ SL3 = 2f (0) + 2f (2) + 2f (4) = 2 [( 0 + 1) + ( 2 + 1) + ( 4 + 1)] ≈ 12.828, where SL3 stands for “Lower Sum in the case of 3 rectangles with equal width”. This is our very first approximation of the area A(6). We see that this is a fairly poor approximation, because the true area is A(6) = 15.79795897... Nonetheless, it is useful — we know that SL3 is a lower bound for A(6). That is, we know that SL3 ≤ A(6). We’ll next try a different approximation — SU 3 . Can you guess what this involves? (... Example continued on the next page ... )
Page 482, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ... ) We’ll again use three rectangles of equal width (width 2), occupying intervals [0, 2], [2, 4], and [4, 6]. The difference now is that for each rectangle, we choose its height to be the highest value attained by the function in that interval. In the interval [0, 2], the highest value attained by f is f (2). So the leftmost blue rectangle has height f (2) and thus area Base × Height = 2f (2). Similarly, the middle green rectangle has height f (4), because in the interval [2, 4], the highest value attained by f is f (4). Hence, it has area Base × Height = 2f (4). The rightmost grey rectangle has height f (4), because in the interval [4, 6], the highest value attained by f is f (6). Hence, it has area Base × Height = 2f (6).
y
x -1
0
1
2
3
4
5
6
7
8
9
Altogether, the total area of these three rectangles is √ √ √ SU 3 = 2f (2) + 2f (4) + 2f (6) = 2 [( 2 + 1) + ( 4 + 1) + ( 6 + 1)] ≈ 17.727, where SU 3 stands for “Upper Sum in the case of 3 rectangles with equal width”. This is our second approximation of the area A(6). We see that again, this is a fairly poor approximation, because the true area is A(6) = 15.79795897.... Nonetheless, it is again useful — we know that SU 3 is an upper bound for A(6). That is, we know that A(6) ≤ SU 3 . Altogether, we know that 12.828 ≈ SL3 ≤ A(6) ≤ SU 3 ≈ 17.727. Can we do better than this? Yes, certainly. An obvious follow-up would be to increase the number of rectangles we use. Let’s next use 6 rectangles instead. (... Example continued on the next page ... ) Page 483, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ... ) We’ll now use six rectangles of equal width (width 1), occupying intervals [0, 1], [1, 2], [2, 3], [3, 4], [4, 5], and [5, 6]. To calculuate the Lower Sum SL6 , we give the first rectangle height of f (0), the second f (1), ..., the sixth f (5). So each rectangle has, respectively, area 1×f (0), 1×f (1), ..., and 1×f (5). Hence, SL6 = f (0)+f (1)+f (2)+f (3)+f (4)+f (5) ≈ 14.382.
y
y
x
x
-1 0 1 2 3 4 5 6 7 8 9 -1 0 1 2 3 4 5 6 7 8 9 Analogously, to calculuate the Upper Sum SU 6 , we give the first rectangle height of f (1), the second f (2), ..., the sixth f (6). So each rectangle has, respectively, area 1 × f (1), 1 × f (2), ..., and 1 × f (6). Hence, SU 6 = f (1) + f (2) + f (3) + f (4) + f (5) + f (6) ≈ 16.832. Once again, A(6) has lower and upper bounds SL6 and SU 6 . That is, 14.382 ≈ SL6 ≤ A(6) ≤ SU 6 ≈ 16.832. You can see where this is going. We can get ever better lower and upper bounds, by increasing the number of rectangles we use. (... Example continued on the next page ... ) Page 484, Table of Contents
www.EconsPhDTutor.com
Exercise 187. Continuing with the above example, find SL12 and SU 12 . Hence give lower and upper bounds for A(6). (Answer on p. 1143.)
(... Example continued from the previous page ... ) Let n be the number of rectangles we use. We will always have SLn ≤ A(6) ≤ SU n . As n increases, we have increasingly-many, increasingly-slim rectangles. As n → ∞, we have infinitely-many, infinitely-slim rectangles, whose total area should approach A(6). Indeed, this “slim rectangles approach” is exactly how the area function is formally and rigorously defined — see Section 88.17 in the Appendices for the details (optional).
y
x -1
0
1
2
3
4
5
6
7
8
9
It appears then that we need to do more maths to figure out how to add up all these “infinitely-many, infinitely-slim rectangles”. ... But it turns out though that there is an absolutely-fantastic shortcut we can use.
Page 485, Table of Contents
www.EconsPhDTutor.com
49.2
The First Fundamental Theorem of Calculus (FTC1)
Given a function f , we sketched an idea of how to find its area function A — approximate the area under the curve using “infinitely-many, infinitely-slim rectangles” and add up the total area of these rectangles. This though was merely a sketch of an idea. How do we go about adding up the area of these “infinitely-many, infinitely-slim rectangles”? Easier said than done! It turns out though that we’ll take an entirely different approach. Strangely enough, instead of finding the area function A, we shall try to find the the area function’s derivative A′ . This seems utterly bizarre. If we don’t know what A is in the first place, how could we possibly figure out what A′ is? This is analogous to asking someone, who has no idea where Singapore is, to find the Singapore Flyer! But surprisingly, it turns out to be much easier to find A′ than it is to find A! We’ll recycle the example from the last section:
Page 486, Table of Contents
www.EconsPhDTutor.com
Example 460 (continued from the previous section). Pick some x that’s just a little larger than 6. A(x) is the area bounded by the graph of f , between the vertical lines x = 0 and x = 6. And so A(x) is just slightly larger than A(6).
y
x -1
0
1
2
3
4
5
6
7
8
9
Consider the thin green vertical strip. This green strip is roughly rectangular in shape — its left, right, and bottom edges are all straight. Only its upper edge is not straight. This green strip’s area is exactly A(x) − A(6). Moreover, we know that its base is x − 6, its left side is f (6), and its right side is f (x). Hence, (... Example continued on the next page ...)
Page 487, Table of Contents
www.EconsPhDTutor.com
(... Example continued from the previous page ...) Area of rectangle with Area of thin green Area of rectangle with base x − 6 and height f (6) vertical strip base x − 6 and height f (x) ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ (x − 6) × f (6) < A(x) − A(6) < (x − 6) × f (x) . Rearranging, we have f (6) <
A(x) − A(6) < f (x). x−6
Now consider what happens if we pick another x that is slightly smaller but still larger than 6. Then the above pair of inequalities will still hold. Indeed, for all x > 6, the above pair of inequalities hold. If we let x approach 6, the above pair of inequalities becomes A(x) − A(6) ≤ lim f (x). x→6 x→6 x−6
lim f (6) ≤ lim x→6
(For why the strict inequalities < became weak inequalities ≤, either you simply trust me or see Fact 7 in the Appendices.) Of course, lim f (6) is simply f (6). And by the continuity of f , lim f (x) = f (6). Hence, x→6
x→6
A(x) − A(6) ≤ f (6), x→6 x−6
f (6) ≤ lim
A(x) − A(6) = f (6). But wait a second ... x→6 x−6 A(x) − A(6) lim ? It is simply the value of the derivative of A at 6!! x→6 x−6 which means of course that lim
what is
A(x) − A(6) Def A′ (6) = lim . x→6 x−6 We thus conclude that astonishingly enough, A′ (6) = f (6). And this is more generally true — given a continuous function f , the derivative of its area function is simply the original function itself! This is the First Fundamental Theorem of Calculus.
Page 488, Table of Contents
www.EconsPhDTutor.com
Note that earlier we defined the area function so that we started counting the area from x = 0 (vertical axis). But this was just to keep the above arguments and diagrams simple. It makes no difference if we start counting the area from any other x = a instead. Theorem 11. (First Fundamental Theorem of Calculus [FTC1], informal statement.) Let f be a real-valued continuous function with area function A. Then Aâ&#x20AC;˛ = f .
In words, the FTC1 says that the area function of a continuous function is simply the function itself ! Equivalently, an indefinite integral (or antiderivative) of a continuous function is the area function.55
Exercise 188. Why did I use the indefinite article an, rather than the definite article the, in the last sentence above? (Answer on p. 1143.)
Next is a familiar example from physics to illustrate the FTC1. Example 461. Graphed below is the velocity v (ms-1 ) of a car as a function of time t (s). Recall that the area under the graph is the distance travelled by the car. For example, the shaded red area A(5) is the total distance travelled by the car after 5 s. But the derivative of the distance travelled with respect to time is precisely the velocity! Hence, this example illustrates the FTC1: the derivative of the area under the graph of a function is precisely the function itself!
Velocity (ms-1)
Time (s) 0
55
1
2
3
4
5
6
7
8
For a formal, rigorous statement of FTC1 and its proof, see section 88.17 in the Appendices.
Page 489, Table of Contents
www.EconsPhDTutor.com
49.3
The Definite (or Riemann) Integral
q
∫p f (x) dx is the area under under the graph of f , between p and q. (Compare this to the q
area function: A(k) is the area under the graph of f , between 0 and k.) We call ∫ the definite (or Riemann) integral of f between p and q.
56
p
f (x) dx
q √ x + 1. The definite integral ∫ f dx (simply p the area under f , between 1 and 3) is highlighted in blue. Similarly, the definite integral
Example 462. Define f ∶ R+0 → R by f (x) = q
∫p f dx (simply the area under f , between 5 and 8) is highlighted in red.
y x 0
1
2
3
4
5
6
7
8
9
IMPORTANT REMARK q
The indefinite integral ∫ f dx and the definite integral ∫ f dx have very similar p names and notation. But do not make the mistake of believing that we’ve simply defined them so that they’re similar — we have not. • The indefinite integral ∫ f (x) dx is an antiderivative of f . (It is also a function.) b
• The definite integral ∫ f (x) dx is the area under the graph of f , between a and b. (It a is also a number.) A priori, there is no reason whatsoever to believe that some antiderivative of f and some area under the graph of f have anything in common. It is the two FTCs that establish the connection between the two. This is what makes the FTCs remarkable and surprising. And it is because of this connection that we give these two distinctly-defined mathematical objects such similar names and notation. 56
For the formal definition of the definite integral, see section 88.17 in the Appendices.
Page 490, Table of Contents
www.EconsPhDTutor.com
49.4
The Second Fundamental Theorem of Calculus (FTC2)
Our (informal) definitions tell us that: • A(q) is the area under the curve between 0 and q; • Similarly, A(p) is the area under the curve between 0 and p; q
• And ∫ f dx is the area under the curve between p and q. p q
Thus, ∫ f dx = A(q) − A(p). From this and also with the aid of the FTC1, we can easily p prove the FTC2. Theorem 12. (The Second Fundamental Theorem of Calculus [FTC2].) Let f ∶ [a, b] → R be a continuous function and p, q ∈ [a, b]. Then q
∫p f dx = ∫ f dx∣x=q − ∫ f dx∣x=p . Proof. By Theorem 11, the area function A is an indefinite integral of f . And so by Fact 60, A and ∫ f dx differ by at most a constant. That is, for all r ∈ [a, b], we have A(r) = ∫ f dx∣
x=r
+ C. Hence, q
∫p f dx = A(q) − A(p) = [∫ f dx∣
x=q
= ∫ f dx∣
x=q
Page 491, Table of Contents
+ C] − [∫ f dx∣
− ∫ f dx∣
x=p
+ C]
.
x=p
www.EconsPhDTutor.com
50
Definite Integrals
To repeat: • The indefinite integral ∫ f dx is an antiderivative of f . b
• The definite integral ∫ f dx is an area under the graph of f . a A priori, there is no reason to believe that the two are in any way related. It is the two FTCs that establishes their remarkable relationship: b
∫a f dx = ∫ f dx∣x=b − ∫ f dx∣x=a , b
To compute ∫ f dx, one method would have been to painfully add up the area of the a “infinitely-many infinitely-slim rectangles”. Thanks to the FTCs, we have a wonderful alternative method that is much easier: 1. Find any indefinite integral of f . 2. The difference of the values of this indefinite integral at b and a is our desired area. We can simply apply all the rules of integration we learnt earlier. b
Notation: We’ll let [f (x)]a be shorthand for f (b) − f (a).
Page 492, Table of Contents
www.EconsPhDTutor.com
50.1
Area between a Curve and Lines Parallel to Axes
Example 463. Find the exact area bounded by the curve y = x2 and the horizontal lines y = 1 and y = 2. It’s always helpful to make a quick sketch (given below). Our desired area is labelled A below. To find a desired area, there are usually multiple methods, some quicker than others. √ √ Method #1. The entire rectangle A + B + C + D has area 2 × 2 2 = 4 2. B has area √ √ 3 −1 x 1 2 2 2 2−1 2 . ∫−√2 x dx = [ 3 ] √ = − 3 − (− 3 ) = 3 − 2 −1
By symmetry, D has the same area as B. C has area 1 × 2. Hence, A has area √ √ √ 2 2−1 4 √ 2 2−1 +2+ ) = (2 2 − 1) . A + B + C + D − (B + C + D) = 4 2 − ( 3 3 3 √ Method #2. The right branch of the parabola y = x2 has equation x = y. The right half y=2 √ y=2 2 2 √ 2 4 √ x dy = ∫ of the area A is ∫ y dy = [y 3/2 ]1 = (2 2 − 1). Hence, A = (2 2 − 1). 3 3 3 y=1 y=1
y y=2 A y=1 B
D
C
x -2
-1
0
1
2
Exercise 189. Find the exact area bounded by the curve y = x3 , the horizontal lines y = 1 and y = 2, and the vertical axis. (Answer on p. 1144.)
Page 493, Table of Contents
www.EconsPhDTutor.com
50.2
Area between a Curve and a Line
Example 464. Find the area A bounded by the curve y = x2 and the line y = x + 1. √ 1± 5 . By the quadratic formula, the curve and line intersect at the points x = 2 √ (1+ 5)/2
√ (1+ 5)/2 3 x
x2
2 ∫(1−√5)/2 x + 1 − x dx = [ 2 + x − 3 ] √ (1− 5)/2
√ 3 √ 3 √ 2 √ √ ⎡ (1 + √5)2 (1 + 5) ⎤⎥ ⎡⎢ (1 − 5) (1 − 5) ⎤⎥ ⎢ 1 + 5 1 − 5 ⎥−⎢ ⎥ + − + − = ⎢⎢ 23 2 3 ⋅ 23 ⎥⎥ ⎢⎢ 23 2 3 ⋅ 23 ⎥⎥ ⎢ ⎦ ⎣ ⎦ ⎣ √ √ √ √ √ √ 6 + 2 5 1 + 5 16 + 8 5 6 − 2 5 1 − 5 16 − 8 5 =[ + − ]−[ + − ] 8 2 24 8 2 24 √ √ √ √ √ √ √ √ √ 3− 5 1− 5 2− 5 7+5 5 7−5 5 5 5 3+ 5 1+ 5 2+ 5 + − ]−[ + − ]= − = . =[ 4 2 3 4 2 3 12 12 6
y
A x Exercise 190. Find the exact area bounded by the curve y = sin x and the line y = 0.5, for x ∈ (0, π/2).(Answer on p. 1145.)
Page 494, Table of Contents
www.EconsPhDTutor.com
50.3
Area between Two Curves
Example 465. Find the area A bounded by the curves y = x2 − 2x − 1 and y = 1 − x2 . √ 1± 5 . So By the quadratic formula, the curves intersect at x = 2 A=∫
√ 0.5(1+ 5) √ 0.5(1− 5)
1 − x2 − (x2 − 2x − 1) dx = 2 ∫
√ 0.5(1+ 5) √ 0.5(1− 5)
1 − x2 + x dx
√ 5 5 + ] = , = 2 [x − 3 2 0.5(1−√5) 3 x3
√ 0.5(1+ 5) 2 x
where we’ve simply recycled our tedious calculations from the previous example.
y
x A
Exercise 191. Find exact area bounded by the curves y = 2 − x2 and y = x2 + 1. (Answer on p. 1145.)
Page 495, Table of Contents
www.EconsPhDTutor.com
50.4
Area below the x-Axis
Example 466. Find the area A bounded by x2 − 4 and the x-axis. The definite integral calculates the signed area under the curve and above the x-axis. So if the curve is under the x-axis, the computed area will be negative, as we now see: 2
∫−2
2
8 −8 −32 x3 . x − 4 dx = [ − 4x] = ( − 8) − ( + 8) = 3 3 3 3 −2 2
But of course, an area is simply a magnitude, so we’ll take the absolute value and conclude 32 that the desired area is . 3
y
x A
Exercise 192. Find the exact area bounded by x4 − 16 and the x-axis. (Answer on p. 1146.)
Page 496, Table of Contents
www.EconsPhDTutor.com
50.5
Area under a Parametrically-Defined Curve
Example 467. Consider the curve described by the equations x = t3 −2 and y = 4−t5 . Find the exact area bounded by the curve, the lines x = −2 and x = −1, and the horizontal axis. It helps to graph this curve on your graphing calculator:
Note that x = −1 ⇐⇒ t = 1, x = −2 ⇐⇒ t = 0, and dx/dt = 3t2 . So the area can be computed as: x=−1
x=−1
5
x=−1
5
2
t=1
∫x=−2 y dx = ∫x=−2 4 − t dx = ∫x=−2 (4 − t ) 3t dt = ∫t=0
1
3t8 (4 − t ) 3t dt = [4t − ] = 4. 8 0 5
2
3
Exercise 193. Consider the curve described by the equations x = t2 + 2t and y = t3 − 1. Find the exact area bounded by the curve, the lines y = 1 and y = 2, and the vertical axis. (Answer on p. 1146.)
Page 497, Table of Contents
www.EconsPhDTutor.com
50.6
Volume of Rotation about the y- or x-Axis
Example 468. Consider the line y = 1. Rotate it about the x-axis to form an (infinite) 3D cylinder. Now consider the finite portion of the cylinder between x = 1 and x = 2. By a primary school formula, its volume is Base Area × Height = π12 × (2 − 1) = π.
Height
Radius Volume We can also compute this same volume using integration. The intuition is that we’re adding up infinitely-many infinitely-thin circle-shaped slices, laid on their sides, from x = 1 to x = 2 (left to right). The face of each of these circles has area πy 2 . In this particular example, y is constant (simply 1). Thus, the total volume is ∫1
Page 498, Table of Contents
2
πy 2 dx = ∫
2 1
2
π dx = [πx]1 = π.
www.EconsPhDTutor.com
Example 469. Rotate the line y = 3x about the x-axis to form an infinite double cone. Consider the finite portion of the cone between x = 0 and x = 2. By the formula for the 1 1 volume of a cone, we know its volume is πr2 h = π62 × 2 = 24π. 3 3 We can also compute this same volume using integration. Again, the intuition is that we’re adding up infinitely-many infinitely-thin circle shaped slices, from x = 0 to x = 2. Again, the face of each of these circles has area πy 2 . In this particular example, y = 3x. Thus, the total volume is 2
2
∫0 πy dx = ∫0
2
2
x3 π(3x) dx = 9π [ ] = 24π. 3 0 2
Height Radius
Volume
Now consider instead the finite portion of the cone between x = 3 and x = 5. This looks like a pedestal tilted sideways (not illustrated). We can easily compute its volume using integration: 5
2
∫3 πy dx = ∫3
5
5
x3 π(3x) dx = 9π [ ] = 294π. 3 3 2
Computing its volume using geometric formulae is possible, if slightly more tedious. The 1 1 finite portion of the cone between x = 0 and x = 3 is V1 = πr2 h = π92 × 3 = 81π. The finite 3 3 1 2 1 portion of the cone between x = 0 and x = 5 is V2 = πr h = π152 × 5 = 375π. Hence, the 3 3 desired volume is V = V2 − V1 = 375π − 81π = 294π. Page 499, Table of Contents
www.EconsPhDTutor.com
We can just as easily find the volume of rotation about the y-axis. Example 470. Consider the curve y = x2 . Find its volume of rotation about the y-axis, from y = 0 and y = 5. In this case, there are no familiar geometric formulae we can apply. So we really just have to compute this same volume using integration. Again, the intuition is that we’re adding up infinitely-many infinitely-thin circle-shaped slices, but this time these circle-shaped slices are stacked from bottom to top, from y = 0 to y = 5. The face of each of these circles has area πx2 , where in this particular example, x2 = y. Thus, the total volume is 5
2
∫0 πx dy = ∫0
5
5
y2 πy dy = π [ ] = 12.5π. 2 0
Volume
Exercise 194. Compute the volume of rotation of y = sin x about the x-axis from x = 0 to x = π. (Answer on p. 1146.)
Page 500, Table of Contents
www.EconsPhDTutor.com