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Evaluating Inverse Trig Expressions Puzzle
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PRODUCT DESCRIPTION
This jigsaw puzzle was designed to help students with evaluating expressions dealing with inverse trig functions (including composition).
The activity is designed as a puzzle sort and match. Students are given 32 puzzle pieces to match in sets of 4. Each set will have 3 expressions (some have 1 composition while others have 3) and 1 solution | 677.169 | 1 |
Abstract Algebra: A Geometric Approach
Author:Theodore Shifrin
ISBN 13:9780133198317
ISBN 10:133198316
Edition:1
Publisher:Pearson
Publication Date:1995-08-24
Format:Paperback
Pages:416
List Price:$116.60
 
 
This book explores the essential theories and techniques of modern algebra, including its problem-solving skills, basic proof techniques, many applications, and the interplay between algebra and geometry.
It takes a concrete, example-oriented approach to the subject matter.
Booknews
A textbook that presents the basic ideas of modern algebra, its problem-solving skills and basic proof techniques, and some of its elegant applications. The link with geometry comes from understanding a geometry by means of its group of motions. There are over 225 substantial examples and some 750 exercises. These include computations, proofs varying from the routine to the challenging, and open-ended problems. Assumes some experience with linear algebra (with exposure to matrix algebra and naive vector space theory) and the language of proofs. Annotation c. Book News, Inc., Portland, OR (booknews.com) | 677.169 | 1 |
Resource allocation and power optimization is a new challenge in multimedia services in cellular communication systems. To provide a better end-user experience, the fourth generation (4G) standard Long Term Evolution/Long Term Evolution-Advanced (LTE/LTE-Advanced) has been developed for high-bandwidth mobile access to accommodate today's... more...
This book offers an introduction to harmonic analysis on the simplest symmetric spaces. It places an emphasis on motivation, concrete examples, history, and, above all, applications in mathematics, statistics, physics, and engineering. more...
Teach lessons that suit the individual needs of your classroom with this SQA endorsed and flexibly structured resource that provides a suggested approach through all three units. National 4 Maths Teacher's Book, Answers and Assessment provides detailed answers to all questions contained in National 4 Maths, a book specifically written to meet the... more...
With this book, children can unlock the mysteries of maths and discover the wonder of numbers. Readers will discover incredible information, such as why zero is so useful; what a googol really is; why music, maths and space are connected; why bees prefer hexagons; how to tell the time on other planets; and much much more. From marvellous measurements... more... | 677.169 | 1 |
Based
"As a non-specialist, I found this book very helpful. It gave me a better understanding of the nature of fractals, and of the technical issues involved in the theory. I think it will be valuable as a textbook for undergraduate students in mathematics, and also for researchers wanting to learn fractal geometry from scratch. The material is well-organized and the proofs are clear; the abundance of examples is an asset for a book on measure theory and topology." (Fabio Mainardi, MathDL, February, 2008)
"This is the second edition of a well-known textbook in the field … . The book may serve as a textbook for a one-semester (advanced) undergraduate course in mathematics. … the book is also suitable for readers interested in theoretical fractal geometry coming from other disciplines (e.g. physics, computer sciences) with a basic knowledge of mathematics. The presentation of the material is appealing … and the style is clear and motivating. … the book will remain as a standard reference in the field." (José-Manuel Rey, Zentralblatt MATH, Vol. 1152, 2009)
Dalla quarta di copertina:
For the Second Edition of this highly regarded textbook,.
From reviews of the First Edition:
"...there has been a deluge of books, articles and television programmes about the beautiful mathematical objects, drawn by computers using recursive or iterative algorithms, which Mandelbrot christened fractals. Gerald Edgar's book is a significant addition to this deluge. Based on a course given to talented high-school students at Ohio University in 1988, it is, in fact, an advanced undergraduate textbook about the mathematics of fractal geometry, treating such topics as metric spaces, measure theory, dimension theory, and even some algebraic topology...the book also contains many good illustrations of fractals..."
- Mathematics Teaching
"The book can be recommended to students who seriously want to know about the mathematical foundation of fractals, and to lecturers who want to illustrate a standard course in metric topology by interesting examples."
- Christoph Bandt, Mathematical Reviews
"...not only intended to fit mathematics students who wish to learn fractal geometry from its beginning but also students in computer science who are interested in the subject. [For such students] the author gives the required topics from metric topology and measure theory on an elementary level. The book is written in a very clear style and contains a lot of exercises which should be worked out.", 2007. Condizione libro: New. Brand New, Unread Copy in Perfect Condition. A+ Customer Service! Summary: "For the Second Edition of this textbook, author."--BOOK JACKET. Codice libro della libreria ABE_book_new_0387747486
Descrizione libro Springer-Verlag New York Inc.,47484
Descrizione libro Springer, 2007. Hardback. Condizione libro: NEW. 9780387747484 Hardback, This listing is a new book, a title currently in-print which we order directly and immediately from the publisher. Codice libro della libreria HTANDREE0274283
Descrizione libro Springer-Verlag Gmbh Dez 2007, 2007. Buch. Condizione libro: Neu. 234x167x19 mm. Neuware - 268 pp. Englisch. Codice libro della libreria 9780387747484 | 677.169 | 1 |
arry Schoenborn, a longtime math, science, and technical writer, is the coauthor of Technical Math For Dummies, Medical Dosage Calculations For Dummies, and Physician Assistant Exam For Dummies.
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The easy way to brush up on the math skills you need in real life
Not life. Math For Real Life For Dummies provides you with the simple formulas and theorems that you're likely to encounter in the workplace, the kitchen, and even when playing games.
You can turn to Math For Real Life For Dummies to brush up on your math skills or to handle everyday encounters, like calculating restaurant tips, understanding interest rates, and figuring out percentages and odds. Packed with real-world examples that make sense, Math For Real Life For Dummies takes the stress out of your daily calculation encounters.
Provides tips for understanding and using basic mathematical concepts
Shows you how math helps the mind to reason and organize complicated situations or problems into clear, simple, and logical steps
Covers all of the math skills you're likely to need in everyday situations
If you're looking for a practical, plain-English guide to mastering everyday math skills, Math For Real Life For Dummies has you covered. | 677.169 | 1 |
Here's an innovative shortcut to gaining a more intuitive understanding of both differential and integral calculus. In Calculus Demystified an experienced teacher and author of more than 30 books puts all the math background you need inside and uses practical examples, real data, and a totally different approach to mastering calculus.
With Calculus Demystified you ease into the subject one simple step at a time — at your own speed. A user-friendly, accessible style incorporating frequent reviews, assessments, and the actual application of ideas helps you to understand and retain all the important concepts.
THIS ONE-OF-A-KIND SELF-TEACHING TEXT OFFERS:
Questions at the end of each chapter and section to reinforce learning and pinpoint weaknesses
A 100-question final exam for self-assessment
Detailed examples and solutions
Numerous "Math Notes" and "You Try It" items to gauge progress and make learning more enjoyable
An easy-to-absorb style — perfect for those without a mathematics background | 677.169 | 1 |
Algebraic Proofs
Pupils complete simple algebraic proofs to prepare them for more challenging geometric proofs. Class begins with a demonstration by the teacher, then students work out their own problems and share results with | 677.169 | 1 |
MATLAB: An Introduction with Applications, 4th Edition
4.11 - 1251 ratings - Source
MATLAB: An Introduction with Applications is used by more college students than any other MATLAB text or reference. This concise book is known for its just-in-time learning approach, giving students the information when they need it. The new edition presents the latest MATLAB functionality gradually and in detail. Equally effective as a freshmen-level text, self-study tool, or course reference, the book is generously illustrated through computer screen shots and step-by-step tutorials, with abundant and motivating applications to problems in mathematics, science, and engineering.This concise book is known for its just-in-time learning approach, giving students the information when they need it. The new edition presents the latest MATLAB functionality gradually and in detail.
Title
:
MATLAB: An Introduction with Applications, 4th Edition
Author
:
Amos Gilat
Publisher
:
Wiley Global Education - 2010-12-07
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Synopsis
IGCSE Mathematics Practice for Edexcel, 2nd edition has been updated to ensure that this second edition fully supports Edexcel's International GCSE Specification A and the Edexcel Certificate in Mathematics. Containing a wealth of exam-style questions and written by experienced examiners, teachers and authors, this is the perfect resource for Higher Tier students both during their course and when they are revising for their exams. IGCSE Mathematics Practice for Edexcel, 2nd edition can be used in conjunction with IGCSE Mathematics for Edexcel, 2nd edition and IGCSE Mathematics Teacher's Resource for Edexcel, or as a standalone resource
Reviews
Great addition to the main student textbook. Lots of practice with exam style questions - answers in the back of the book. Good for reinforcing topics. -- Mrs Bridget Jones Abigail Lambert
About the Author
Trevor Johnson is currently a chief examiner for a leading awarding body. He is an experienced author and was formerly Head of Maths in a Staffordshire comprehensive school. Tony Clough is currently a senior examiner for a leading awarding body. He was Head of Maths for 22 years and has written extensively for Key Stage 3, GCSE and A Level. | 677.169 | 1 |
Help students get to grips with the new style examinations in Numeracy and Mathematics with books that build all the necessary skills to progress their learning and develop their problem-solving skills. - Provides one book for each tier ensuring full coverage, with mathematics only questions and chapters clearly highlighted so the specifications... more...
The fusion between graph theory and combinatorial optimization has led to theoretically profound and practically useful algorithms, yet there is no book that currently covers both areas together. Handbook of Graph Theory, Combinatorial Optimization, and Algorithms is the first to present a unified, comprehensive treatment of both graph theory and... more...... more...
This... more...
This text is a systematic, state-of-the-art introduction to the use of innovative computing paradigms as an investigative tool for applications in time series analysis. It includes frontier case studies based on recent research. more... | 677.169 | 1 |
0412406 undergraduate textbook suitable for linear algebra courses. This text develops the linear algebra hand in hand with the geometry of linear (or affine) spaces in such a way that the understanding of each reinforces the other. The text is divided into two parts: Part I is on linear algebra and affine geometry, finishing with a chapter on transformation groups; Part II is on quadratic forms and their geometry (Euclidean geometry), including a chapter on finite subgroups of O(2). Each of the chapters concludes with exercises and a selection of these have solutions at the end of the book. The chapters also contain many examples, both numerical worked examples (mostly two and three dimensions), as well as examples which take some of the ideas further. Many of the chapters contain "complements" which develop more special topics, and which can be omitted on a first reading. The structure of the book is designed to allow as much flexibility as possible in designing a course, both overall by omitting chapters, and within chapters by omitting the complements or specific examples | 677.169 | 1 |
Fifty Lectures for American Mathematics Competitions
4.11 - 1251 ratings - Source
While the books in this series are primarily designed for AMC competitors, they contain the most essential and indispensable concepts used throughout middle and high school mathematics. Some featured topics include key concepts such as equations, polynomials, exponential and logarithmic functions in Algebra, various synthetic and analytic methods used in Geometry, and important facts in Number Theory. The topics are grouped in lessons focusing on fundamental concepts. Each lesson starts with a few solved examples followed by a problem set meant to illustrate the content presented. At the end, the solutions to the problems are discussed with many containing multiple methods of approach. I recommend these books to not only contest participants, but also to young, aspiring mathletes in middle school who wish to consolidate their mathematical knowledge. I have personally used a few of the books in this collection to prepare some of my students for the AMC contests or to form a foundation for others. By Dr. Titu Andreescu US IMO Team Leader (1995 - 2002) Director, MAA American Mathematics Competitions (1998 - 2003) Director, Mathematical Olympiad Summer Program (1995 - 2002) Coach of the US IMO Team (1993 - 2006) Member of the IMO Advisory Board (2002 - 2006) Chair of the USAMO Committee (1996 - 2004) I love this book! I love the style, the selection of topics and the choice of problems to illustrate the ideas discussed. The topics are typical contest problem topics: divisors, absolute value, radical expressions, Veita's Theorem, squares, divisibility, lots of geometry, and some trigonometry. And the problems are delicious. Although the book is intended for high school students aiming to do well in national and state math contests like the American Mathematics Competitions, the problems are accessible to very strong middle school students. The book is well-suited for the teacher-coach interested in sets of problems on a given topic. Each section begins with several substantial solved examples followed by a varied list of problems ranging from easily accessible to very challenging. Solutions are provided for all the problems. In many cases, several solutions are provided. By Professor Harold Reiter Chair of MATHCOUNTS Question Writing Committee. Chair of SAT II Mathematics committee of the Educational Testing Service Chair of the AMC 12 Committee (and AMC 10) 1993 to 2000.While the books in this series are primarily designed for AMC competitors, they contain the most essential and indispensable concepts used throughout middle and high school mathematics.
Title
:
Fifty Lectures for American Mathematics Competitions
Author
:
Jane Chen, Yongcheng Chen, Sam Chen, Guiling Chen
Publisher
:
Createspace Independent Pub - 2013-01-18
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Enrollment Info
Precalculus
Prerequisites:Successful completion of two years of algebra and one year of geometry. Length:Two Semesters
Precalculus is a course that combines reviews of algebra, geometry, and functions into a preparatory course for calculus. The course focuses on the mastery of critical skills and exposure to new skills necessary for success in subsequent math courses. The first semester includes linear, quadratic, exponential, logarithmic, radical, polynomial, and rational functions; systems of equations; and conic sections. The second semester covers trigonometric ratios and functions; inverse trigonometric functions; applications of trigonometry, including vectors and laws of cosine and sine; polar functions and notation; and arithmetic of complex numbers.
Within each Precalculus lesson, students are supplied with a post-study Checkup activity that provides them the opportunity to hone their computational skills by working through a low-stakes problem set before moving on to formal assessment. Unit-level Precalculus assessments include a computer-scored test and a scaffolded, teacher-scored test.
The content is based on the National Council of Teachers of Mathematics (NCTM) standards and is aligned with state | 677.169 | 1 |
Solving Systems of Equations
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1.28 MB | N/A pages
PRODUCT DESCRIPTION
This lesson comes after students have been taught graphing, substitution, and elimination methods to solve systems of equations. In this lesson students choose the method they want to use to solve each system.
This lesson has a PowerPoint with a warm up, four example problems, practice problems and a closure problem. Guided Notes are provided for students to use as they follow along with the PowerPoint. There are also two practice sheets with answers. Each practice sheet has 15 problems | 677.169 | 1 |
Graphing Technology Guide for Calculus
4.11 - 1251 ratings - Source
To graph y = [[x]] , go into the Y= menu, move beside y1 and press CLEAR 2nd
MATH -agt; 6[floor(] X ) ENTER a™c GRAPH. ... Move the cursor down to the second
line and press ENTER or press 2; to have the selected graph plotted in Dot style.
Title
:
Graphing Technology Guide for Calculus
Author
:
Benjamin N. Levy
Publisher
:
Brooks/Cole Publishing Company - 1998-01-06
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
This best-selling "Spreadsheet" book provides excellent coverage of all versions of Excel including the latest version, Excel 2002. It discusses how to use Excel to solve a variety of problems in introductory engineering analysis, such as graphing data, unit conversions, simple statistical analysis, sorting, searching and analyzing data, curve fitting, interpolation, solving algebraic equations, logical decisions, evaluating integrals, comparing economic alternatives, and finding optimum solutions. Numerous examples are included illustrating both traditional and spreadsheet solutions to a variety of problems. The underlying mathematical solution procedures are also discussed, so that the reader is provided with an understanding of what the spreadsheet does and how it does itDescripción McGraw-Hill Science/Engineering/Math, 2005. Estado de conservación: New. Brand New, Unread Copy in Perfect Condition. A+ Customer Service! Summary: Through three editions, this practical text has found a permanent spot in many introductory engineering courses by successfully combining an introduction to Excel fundamentals with a clear presentation on how Excel can be used to solve common engineering problems. The third edition provides beginning engineering students with a strong foundation in problem solving using Excel as the modern day equivalent of the sliderule.As part of McGraw-Hill's BEST series for freshman engineering curricula, this text is particularly geared toward introductory students. The author provides plenty of background information on technical terms, and numerous examples illustrating both traditional and spreadsheet solutions for a variety of engineering problems. The first three chapters introduce the basics of problem solving and Excel fundamentals. Beyond that, the chapters are largely independent of one another. Topics covered include graphing data, converting units, analyzing data, interpolation and curve fitting, solving equations, evaluating integrals, writing macros, and comparing economic alternatives. Nº de ref. de la librería ABE_book_new_0072971843 | 677.169 | 1 |
Learning and Teaching Mathematics in The Global Village
Authors: Danesi, Marcel Provides a fundamental reassessment for math education in the digital era Introduces a new approach for introducing technology into the math classroom Provides state-of-the art literature review of digital math pedagogy Examines the nature of mathematical learning as part of other learning modes This book provides a fundamental reassessment of mathematics education in the digital era. It constitutes a new mindset of how information and knowledge are processed by introducing new interconnective and interactive pedagogical approaches. Math education is catching up on technology, as courses and materials use digital sources and resources more and more. The time has come to evaluate this new dynamic, which transcends all previous use of ancillary devices to supplement classroom math instruction. Interactivity and interconnectivity with the online world of math and math texts (such as television programs and internet sites) can be integrated with our traditional modes for delivery of math instruction. This book looks at how this integration can unfold practically by applying these relevant pedagogical principles to elementary topics such as numeration, arithmetic, algebra, story problems, combinatorics, and basic probability theory. The book further exemplifies how mathematics can be connected to topics in popular culture, information technologies, and other such domains. | 677.169 | 1 |
Hi, sorry this is late but I can't find it anywhere. MathsWatch seem to have removed their predicted topics list for the Edexcel exam board from their website, and I was wondering if anyone had a copy, because I wanted to go through it one more time before the exam and it's GONE!!!?! EDIT: It hasn't been removed, and the situation has been recovered. Thankyou xx
Thank-you so much!!! I can't find it on the site and You have SAVED ME! I'm so glad it's not been removed, but alas I couldn't track it down. Thankyou, thankyou THANKYOU! You have no idea how amazing this is. | 677.169 | 1 |
If you're the parent of a grade schooler or a near graduate or you're preparing to teach a class of your own, you should know that HP has a full line of financial, scientific and graphing calculators that can help start the school year with a stroke of genius.
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College: Figure these into your future
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The HP 50g Graphing Calculator is a powerful, flexible calculator designed for math, science and engineering applications.
Students can confidently tackle problems with 2.5MB total memory(2) and use the available SD card slot(3) to transfer data and expand memory. They can view calculations on a large, high-contrast display, choose between data-entry modes (RPN, algebraic and textbook) and work efficiently with a large equation library and 2300+ built-in functions.
High school and middle school: Solve more simply
A great graphing tool for math and science, the HP 39gs Graphing Calculator is simple to use and displays answers on its large screen in symbolic, numeric and graphing views. Students and teachers can save their work as documents and share them wirelessly(4) to get all students on the same page at once.
The HP 40gs Graphing Calculator is compatible with the HP 39gs, plus it has a full Computer Algebra System (CAS) to perform symbolic operations. The interactive CAS editor lets students write expressions as they appear in text books. The editor also keeps track of all expressions entered and provides one-button access to them.
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The simple-to-use HP 10s Scientific Calculator can go with students as their challenges grow. Its large easy-to-read, two-
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Multiply the reasons
Whether you parent pupils or teach them, you can count on the power, speed and reliability of HP calculators and educational tools to help your star students quickly plow through the toughest problems. Add some to your back-to-school list today! | 677.169 | 1 |
Popular in Mathematics (M)
Reviews for INTRO GROUP 120A An Introduction to Group Theory Neil Donaldson Spring 2009 Text 0 An Introduction to Abstract Algebra John Fraleigh 7th Ed 2003 Adison Wesley optional 0 Also check the library for entries under quotGroup Theoryquot and Abstract Algebra There have been a plethora of textbooks written for Undergraduate group theory courses and the first few chapters of most of them will cover similar ground to the core text 1 Introduction Alongside discussions of Continuity and abstract Calculus Group Theory is usually the first serious taste an Undergraduate gets of Pure mathematics It introduces the idea of axiomatic mathematics a collection of rules and definitions are given and we prove theorems entirely derived from those rules As a result group theory can seem daunting and difficult to the uninitiated Perseverance is the key Usually proofs are not as difficult as they first seem there are relatively few rules governing groups so there is often little choice in how you approach a proof Why study Groups Here are two possible answers Firstly groups are incredibly useful Groups are largely about sym metries and patterns and much of mathematics involves looking for these For example a square has symmetries rotations and re ections these symmetries form a group Here are some further examples of where groups play a role Geometry A large part of modern Geometry involves the study of groups surfaces polyhedra the set of lines in a vector space these sets have many different groups associated to them which help the geometer describe and classify them Combinatorics When studying collections of objects groups of permutations reorderings of sets are widely used Galois Theory Groups describe the roots of polynomial equations Chemistry Groups are used to describe the symmetries of molecules and of crystalline substances Physics Materials science sees group theory in a similar way to Chemistry while modern theories of the nature of the Universe eg string theory rely heavily on groups The second reason to study groups is that they are relatively easy yes really A group is a set with just one operation like the real numbers lR with you are already familiar with objects far more complicated than this eg lR with two operations is a field an is a vector space so it makes sense to begin the study of abstract mathematics at the level of groups Motivational example What do an equilateral triangle and an arbitrary collection 1 2 3 of 3 objects have in common Not much at first glance but both objects have symmetries rotations and re ections of the triangle and permutations of 123 In fact these symmetries are the some The permutations of 123 can be written in cycle notation eg 12 swaps 1 and 2 and leaves 3 alone 123 sends 1 to 2 2 to 3 and 3 to 1 There are 6 permutations Identity Swap 2 Move all3 e 1 23 p1 123 2 13 P2 132 43 12 The permutations interact by composition For instance p1 opz 23132 12 p3 The full list of compositions of the symmetries is shown below left column first then top 8 8 43 3 1 2 p1 pz 8 M1 1 2 2 Now label the vertices of an equilateral triangle 123 The above permutations correspond to symmetries of the triangle phpz are rotations counter clock and clockwise while p is re ection in the altitude through the point i The two sets of symmetries are the same Group theory is about ideas like the above the symmetries and patterns associated to an ob ject are often more important and interesting than the object itself and can lead to unexpected con nections In group notation we have shown that the symmetric group on 3 letters 53 permutations of 123 and the Dihedrol group of order 6 D3 the symmetries of the triangle are isomorphic1 53 E D3 The following sections roughly resemble their counterparts in Fraleigh s book We will sometimes use different notation however eg Fraleigh calls a set S in a definition while we use X Don t let this fox you problems in the exam and in later life are unlikely to use precisely the same notation so make sure you learn the corlcepts rather than photographically memorizing the definitions 1We will explain both examples more fully later and terms such as isomorphic At the moment it is enough to say that isomorphic means the same in a specific way 2 Binary relations De nition 21 A binary relation on a set X is a map X x X a X which is de nedfar all pairs of elements in X We say that X is Closed under Examples 1 Usual multiplication and addition are binary relations on any of the sets R C Q Z lRJ 12 2 and many more The sets R Q C Z with subtraction Any vector space eg Rquot or Cquot has the usual as a binary relation pmm R3 with cross vector product gtlt functions V gt V where V is any space and gtk o composition of functions C with the Hermitian product 2 w 2 8 lN 2 with the least common multiple eg 4 5 20 9 N with the highest common prime factor eg 65 20 5 O The set A 0 123 with a b ab mod 4 ie the remainder upon dividing ab by 4 Remarks There is no requirement for a binary relation on X to have an image that is all of X ie X X need not be all of X see example 8 where a b is always prime The only thing that matters is that the relation is defined everywhere The following is therefore NOT an example Q with pmb ab a 0 is undefined for all a Binary relations on finite sets can be written in multiplication tables Example 9 above can be written where we find a b by taking a in the left column and I along the top WNH096 ooooo WNHOH NONON HNWOU De nition 22 If Y C X is a proper subset and X has a binary relation then we can restrict to Y If Y gtk Y C Y we say that Y is Closed under Otherwise said restricted to Y is a binary relation on Y Examples 1 The negative integers Z are closed under but not under 2 Let Y C Z be the set of integers whose remainder is 1 when divided by 3 ie Y 1 3n n E Z 1471013 72 75 78 ls Y closed under Or Under 1 3n 1 3m 2 3m n has remainder 2 so Y is not closed Under 1 371 1 3m 1 371 3m 9mn 1 3m n 31ml has remainder 1 so Y is closed De nition 23 A binary relation on X is associative if a b c a b c Va b c E X If is associative then the writing a1 a2 an is unambiguous Examples 135789 in the first list in this section are associative De nition 24 A binary relation is commutative on X if a b b a Va b E X Original examples 13789 are commutative Remarks Some algebraists will use for any commutative binary relation While this reminds us of the common addition of numbers which is commutative it can be confusing do not assume that every time you see a in a book that it means addition Observe that the multiplication table of a commutative binary relation has a diagonal symmetry For example our earlier table is symmetric about the principal diagonal Conversely non commutative relations have non symmetric tables 0 l 2 3 a b c d 0 0 0 0 0 a d c d c l 0 l 2 3 b a b a d 2 0 2 0 2 c a c a b 21 3 0 3 2 l d c c a c A commutative relation ex 9 A non commutative relation Observe that the second table is not associative for a b c c c a while a b c a a 01 There are two important examples that it is worth proving the associativity and non commutativity of functions with composition and matrix multiplication Note the similarity of the two proofs Proposition 25 Composition of functions f g V a V is associative where V is any set If V has at least 2 elements V 2 2 then composition is nonicommutative Proof For the first claim we must see that fog ohx f0 go hx VX E V Now fog 01006 foghx fghx and f0 gohx fg0hx fghx Thus 0 is associative To see that o is non commutative we have to exhibit a counter example for any V with at least two elements Leta 7 b be two elements in V and define fg V gt be a gx b VX E V Then f0g2Xgt gtfb g0f2Xl gtg b o is therefore non commutative I Remarks You may think it is enough to choose a specific space say V lR in which to exhibit a counter example but the proof really requires the abstractness It may amuse you to construct pairs of non commuting functions f g lR gt lR There are many large sets of functions that do commute for example in a vector space V the set of scalings fA V gt V X gt gt AX where A E lR form a commuting set Composition of functions between arbitrary sets is actually associative the proof is almost identical but composition only gives a binary relation when maps are from a set to itself Corollary 26 Matrix multiplication of square matrices is associative Proof We will provide three proofs of this fact choose your favourite Using Proposition 25 You may have proved in a linear algebra course that matrix multiplication corresponds to linear transformations of a vector space ie square matrices can be viewed as functions V gt V and multiplication is composition of functions Armed with this knowledge Proposition 25 gives the result straightaway Directly Suppose that A B C are rl gtlt rl matrices Recall that if A has entries aij ith row jth column etc then AB has ikth entry rt ABik Z lyl jk j1 Therefore ABCa lt2m agt Ck Zigwa39bjk kt J r Similarly ABC39 2a El jka Z ijbjkckl f k M These are equal by the associativity of lR For Physicists If you are lucky enough to understand the Einstein Summation Convention observe that the direct proof can be written ABC ijbjk kl ijbjk kl ABC where we have used associativity of multiplication of real complex numbers for the middle ste p I Matrix multiplication is similarly non commutative for example A 3 1 and B 1 3 don t commute We will often ignore associativity and just assume it is true As you can see above it often involves a messy calculation We will certainly never prove associativity for functional composition again any time an example appears we will quote Proposition 25 Make sure you remember what associativity means and that it holds for functions Many binary operations that are not obviously functional composition can be viewed as such when thought about in the correct way so Proposition 25 is extremely useful 3 Morphisms A morphism is simply a function from one set to another often a morphism has a prefix such as homo or iso Recall the following definitions from previous classes De nition 31 Let X Y be sets A function or morphism f X gt Y is said to be 171 iffx fy i x yforallxy E X onto iffX Y Otherwise said f is 1 1 iff each element in the image of f comes from exactly one element of X f is onto iff everything in Y is in the image of f f is a bijectian if it is both 1 1 and onto De nition 32 Let XY be sets with binary relations X y A map 4 X a Y is a homomorphism2 of binary structures ifzpa X b 4101 y 111k Va E X A homomorphism pulls back the structure on Y the binary relation to that on X De nition 33 Ahomomorphism 4 X X a Y y is an isomorphism3 if 4 X a Y is 1 1 and onto We write X E Y or occasionally 4 X E Y 4 X gtY or X g Y Some books eg Fraleigh use X 2 Y Remarks Recall the motivational example where we discussed the symmetries of the equilateral triangle D3 and the permutations of 123 With both sets of transformations labelled the way they are we have an isomorphism 53 E D3 with binary operation of composition on each side Synonyms A 1 1 map is often referred to by the noun injection and adjective injective An onto map is surjective or a surjectian A map that is both 1 1 and onto is a bijectian or a bijective map An isomorphism is therefore a bijective homomorphism We will use bijection but none of the others All are in common usage however so you should be aware of them Remark Given a bijection 4 X gt Y and a binary relation on Y you can always define a relation 9 on X by pulling back to X x1322 zp 1zpx 4X V393 E X Example The map 4 lRJr gt lRJr X gt gt X2 is a bijection Suppose we want 4 to be an isomorphism 4 lRJr gt lRJr Then we must define by x y 4f1lt4gtltxgt w 41408 y2 ixz y2 ZThe prefix homo means similar 3 lso means same Showing Isomorphicity When asked to demonstrate that two structures are isomorphic eg that X E Y you should follow the steps given below 0 Define a function 4 X gt Y 0 Check that 4 is a homomorphism 41X y 1X 1 My 0 Check 4 is 1 14X y x y 0 Check 4 is onto Vz E Y 3X E X such thatz You can perform the final three steps in any order you like but it is often easiest to check the ho momorphism property first as part of the definition This is since you will usually have the homo morphism property at the back of your mind when trying to define 4 in general there are a huge number of functions X gt Y so you need some inspiration in order to pick the right one Sometimes of course the homomorphism property is the hardest one to check Remark Another method to show isomorphisms of finite sets is to write out the multiplication ta bles for each binary relation and compare this is not easy as you will probably have to relabel the elements and swap rows and columns in order to get the two tables to look alike Two tables that look different are not automatically non isomorphic Examples 1 22 E 32 with the binary relation of addition For n E 2 712 nz z E 2 is the set of integers divisible by De ne 4 Let 4 22 a 32 Zm gt gt 3m ie 41x gx Homomorphism 41X y X y gx y 4100 y and so 4 is ahomomorphism 11 4100 470 s x g1 xx ysozpis1 1 Onto An arbitrary element of 32 is y 3m where m E 2 But 3m 412m so 4 is onto 4 is therefore an isomorphism 4 22 E 32 2 22 E 32 with respect to the binary relations x ZZ y xy on 22 and x 32 y xy on 32 and 1X gx 4 is 1 1 and onto by example 1 and is a homomorphism because 412x ZZ 2y 412xy 3xy 3x 32 3y 412x 32 412y 011 E SH Onthe LHS 01 x E R 0 g x lt 1 and x1y Lxyj fractional part of X y ie 03 1 09 L12 02 On the RHS S1 is the unit circle in the complex plane S1 z E C z 1 and is standard multiplication of complex numbers Recall that 2 w 2 MI for any complex numbers 2 w and so is a binary relation on 51 9 4 Z I 51 I I a 42 1 I I i VV I lt15 I O Geometrically we are stretching and then wrapping the interval 01 around the unit circle with thejoin at 1 E C De ne 4 Let 4 01 gt S1 X gt gt 82mquot recall that 82 quot COS27 L39X isin27IX which has modulus cos2 sin2 1 1 1 41X 41y i 82 quot 82 i 82quotix y1 i X E y E Z But Xy E 01so we must have X y 4 is therefore 1 1 Onto An arbitrary element of S1 is z 819 where 0 g 6 lt 27139 is the argument of 2 But 819 ez li39 Now E 01 and so 4 is onto Homomorph X1y LXyD 82nixyj 82nixy 82nix82m39y X 4 y and so 4 is a homomorphism Remember that 82 quot 1 for rl E Z so EMMHW EZMHW Showing nonisomorphicity This is tricky in general you can t try every map X gt Y except for very small sets X Y in order to check that there are no isomorphisms so you have to be a little more cunning and consider various properties that are preserved by isomorphism De nition 34 A structural property of X is a property which is preserved under isomorphisms ie if 4 X gtk gt Y is an isomorphism then X and Y 9K have the same structural proper ties The following are examples of structural properties suppose 4 X gt Y is an isomor phism throughout 0 The elements of X Y are in a pairing X and so the size cardinality of the sets must be the same this is true even for infinite sets 2 Q N0 g 2 lR there are therefore no isomorphisms between Q and lR no matter what binary relations you have 0 Any special elements in X must correspond to special elements in Y eg idempotents if X X X in X then under an isomorphism we must have 41X 9quot 41X 41X in Y Thus if idempotents exist in X but don t in Y there can be no isomorphism o Commutativity amp Associativity If X or Y has one of these properties then so must the other A set with a commutative or associative relation is never isomorphic to a set with a non commutative or non associative relation 0 Identity if X has an identity element 8 e X X e X VX E X then under an isomor phism 4 Y has an identity 418 0 Solutions of equations if a X b has a solution X E X then in the presence of an isomor phism a similar equation must have a solution in Y This example is important and tricky to know when to use There are a couple of examples of this in action below There are many other things that must be preserved by an isomorphism far too many to list The more structure a set and its binary operators get the more properties there are to consider Disprov ing isomorphisms is somewhat of an Art as you will sometimes have to think very hard or have a ash of inspiration in order to think of the correct structural property Note that there are many concepts that are not necessarily preserved by an isomorphism the type of element a number a matrix etc the type of binary operation in the 01 1 E 51 we saw that addition can become multiplication under an isomorphism Again there are others en gage your brain when asking yourself if something is a structural property is the property eXtrinsic depends on how you write down the set and the binary relation or intrinsic independent of how you write things down A loose definition of two binary structures being isomorphic is exactly that one is a relabelling of the other Examples 1 The two binary relations defined by the multiplication tables in 21 are non isomorphic because the first is commutative while the second is not N lR3 SE lR3 gtlt cross product because is commutative and associative while X is nei ther U N0 SE N since lNo contains the additive identity 0 while lN does not Q Recall the 01 S1 example from earlier 01 1 SE lR One reason is because there are two solutions to X 1 X 0 in 01 namely X 0 while there is only one solution to the corresponding equation y y c in lR for any c if 4 01 gt lR was an isomorphism then X 1 X 0 w 41X 41X 410 which is an equation of the form y y c Actually you could assume that c 0 because 0 is the additive identity on both sides U I Q SE Q since yz 2 has no solution in Q but the corresponding equation X X c has a solution in Q for any c 4 Groups De nition 41 A Group is a set G together with an operation G X G gt G which together satisfy the following axioms Closure G is closed under ie is a binary relation on G Associativity is associative Identity 3 identity e E G such that e g g e g Vg E G Inverse Every element has an inverse Vg E G Eg l E G such thatg g 1 g l g e De nition 42 A group G is Abelian if the group operation is commutative4 Remarks You should remember the axioms in the order they are written down so that when you are asked to prove that something is a group you can deal with one property after another Examples of this are given below Although a group is often written as a pair G it is conventional to drop the operation symbol and refer to the group G if there is no ambiguity about the operation Be careful though for sometimes the is needed dropping when multiplying real numbers creates nonsense 3 4 12 becomes the 4After Niels Abel one of the Godfathers of pure mathematics meaningless 34 12 For Abelian groups the operation is often denoted even when it does not denote an addition In such cases we use minus 7 for the inverse Examples 1 Sets such as lRZ Q rl E Z rl even are Abelian groups under addition Check All are closed by assumption Addition is associative 0 is the additive identity 0 rl rl 0 rl The inverse 0er is in E Z N The non zero elements in lR C Q form Abelian groups under multiplication Check Closure ub 7E 0 i uh 7E 0 Multiplication of numbers is associative 1 E Q C lR C C is the multiplicative identity 1 rl rl 1 rl The inverse of rl is rl 1 1 rl which is in lRC Q if rl is These groups are often denoted lRgtltCXQgtlt or lRCQ etc The has a separate meaning recall linear algebra where lR is the alqu space to lR the set of linear maps lR gt lR so in this class we will always use the X notation U Vector spaces any vector space is an Abelian group under addition Note that this includes the set of m X rl matrices under addition as menlR lRm gtlt lRquot g Let P be a regular rl gon Then P has symmetries rotations and re ections The set of symme tries forms a non Abelian group under composition These are the Dihedrul Groups Dn which we will think more in section 9 U1 Permutation groups the symmetric group 5 is the group of rearrangements of the set 1 2 rl for example the element 12 which swaps 1 and 2 is in Sn These groups will also be studied in section 9 O Matrix groups We will later discuss several examples of groups of matrices under multiplicui tiarl Some you will have thought about before like the Orthogonal Group Orl of transfor mations of lRquot which preserve lengths and angles between vectors We have seen the messy calculation that matrix multiplication is associative so the issue with matrix groups is to see that inverses exist and that the set is Closed The matrix with 1 s down its diagonal and 0 s every where else is the multiplicative iderltity matrix and often written In or just I There are many more groups than just these but we shall mostly use the above and subsets of the above as our examples Things that are not groups There are a number of things that can go wrong when trying to make a set into a group even if you overcome the biggest hurdle and start with an associative operation Mostly people forget to check closure and inverses o The set of all real numbers lR as opposed to lRgtlt lR does not form a group under multiplication lR is closed associative and has the identity element 1 but the single element Ohasno inverse Er E leuch that0 r r 0 1 10 The same is true for the other multiplicative sets Q and C are not groups because of the non invertibility of 0 The set of functions f V gt V does not form a group under composition when V has at least two elements We have closure associativity Proposition 25 and an identity map f X gt gt X but in general we have no inverses For example the map f X constant VX E V has no inverse However if the space V is itself a group then the set of maps f X gt V from any set into V forms a group under the operation of V on values for example f X gt lRquot forms a group under addition on lRquot gX gX The set of functions f lR gt lR is not a group under multiplication Any function f which has a 1 zero at any point X has no inverse for m is undefined Remarks A multiplicative group is a group where the operation is written multiplicatively g li gh This is the most common form of notation because we can be maximally lazy in our writing The adjective multiplicative is dangerous however as it doesn t refer to any intrinsic property of the group only in how we write it down Notational laziness goes further for we will often use shorthands uquot Q u and ifquot u 1quot uquot 1 Exponents follow the usual additive law unum WMquot We w mimes also write uo 8 Basic theorems for groups Theorem 43 In it group G identities and inverses are unique Proof Assume 8 7E r are identities Then viewing 8 as an identity we have 8 6 while viewing r as an identity yields 8 8 Therefore 8 r and we have a contradiction Similarly let ghave inverses hl 7E hz Then ghl gh2 8 Thus Mghl h1gh2 li1gh1 h1gl12 by associativity gt8l11 ehz since h is an inverse of g gtl11 hz Again we have a contradiction I Observe that we could have used either hl or hz as the inverse of choice in the second part of the proof Corollary 44 Cancellation laws amp lnverses u ggl ggz i g1 g2 l1 gig gzg i 31 g2 C ghf1 h lg l Proof The first two statements are immediate just multiply on the left resp right by g l For part c note that h lg 1gli h 1g 1gh h leh h lh 8 Thus h lg l is an inverse of gh But inverses are unique by Theorem 43 11 De nition 45 Two groups G H are isomorphic if the group operations on the sets G H are isomor phic binary structures We write G E H Written multiplicatively G E H ltgt 34 G gt H such that 4 is 1 1 and onto and a homomor phism 41glg2 41g14gtg2 Group tables for nite groups Just as for binary relations on finite sets it can be useful to write a group in a multiplication table There are a few rules for group tables however which do not apply to general binary relations In the row and column of the identity e everything remains unchanged You can always start by writing down a group table with the first row and column filled out abc abc w m9 mw mm Lemma 46 Magic square lemma In every row and column each element appears exactly once Proof Suppose element X appears twice in row a Then 31 7E c such that X ab ac But Corollary 44 implies l1 c a contradiction Considering column a is similar Since no element can appear twice we must have all elements appearing exactly once I Any table satisfying the identity and magic square rules defines a binary relation with identity and unique inverses ie the Closure Identity amp Inverse axioms are satisfied The only thing left to check is associativity Unfortunately this is a messy business to see directly from the table and diminishes the use of tables as a good method of defining groups Moreover it is common for people to fail to spot when two tables are isomorphic when the two are identical up to a relabelling of the group elements For larger groups this can be very difficult to spot and you may therefore think that two tables which describe the same group actually describe different groups Group tables for low order groups De nition 47 The order of a group G is the number of elements in G Order 1 in Only 1 group 21 W gtllt 8 J Order 2 g Only 1 group 22 a a m Order 3 Group 23 NWW w mm mw What are 21 2223 Theorem 48 The set 2 012 n 7 1 E Z is un Abeliun group under addition modulo n u 7 I z u b mod We call Zn 11 the cyclic group of order n Proof Let s check the axioms Closure u h mod n E 2 is clear as 2 is the set of all remainders modulo n Associativity u 7 I n c u b n c u b c mod This is symmetric in ubc and therefore n is associative Identity 0 E 2 is the additive identity Inverse 7u E n 7 u mod Thus n 7 u is the inverse of u Zn is therefore a group is moreover commutative since is and so 2 is Abelian Corollary 49 There exists it group of every finite order Le for every positive integer n E 2 there exists at least on group with n elements namely 2 For most orders there are u lot more than just one group 5 Subgroups De nition 51 Let G be a group and H C G a subset H is a subgroup of G if H is a group in its own right A proper subgroup is a subgroup H 7 G e is the trivial subgroup of G for any G All other subgroups are non7triviul We write H g G or H lt G if we want to stress that H is a proper subgroup The very compact definition of subgroup contains several points In particular if H is a subgroup of G then the following are true 0 H C G as sets 0 The operation on G restricts to a binary operation on H H is closed under o The identity e E G is in H o H is closed under inverses h 1 E H Vh E H A priori we only have that h 1 E G for h E H so this really is a condition Examples 1 Z lt Q lt R lt C 2 QX lt lRX lt CX 3 e g Gand G g Gforuny G 4 Rquot lt Cquot 5 Rquot g lRm ifm g rt 6 Zm g 2 if mlrl proof to come later section 7 Sm g 5 if m g rl permutation groups section 9 8 Dm g Dn if mlrl Dihedral groups section 9 9 Sl lt CX recall S1 z E C lzl 1 is the unit circle 10 ClR lt C1lR all differentiable functions are continuous You don t have to check all the axioms of a group to see that a subset is a subgroup indeed you have to check very little as the following two results make clear Don t make the mistake of thinking that Proposition 53 is as useful as it first appears calculating xy l is often more difficult than proving things directly Theorem 52 A subset H of G is u subgroup if H is Closed under the group operutiorl urld inverses Proof If H is closed under the group operation on G and inverses then H satisfies the closure and inverse axioms of a group Moreover we now have that hh l E H for all Ii E H since inverses and products are in H and so the identity 8G is in H Finally H is associative because the group operation is already associative on G and H is just a subset of G H is therefore a subgroup I You should remember the Proposition when showing that a subset is a subgroup it is not required to check the associativity and identity axioms A good exercise is to show directly that examples 1 to 5 on the previous page are subgroups using the Theorem it is very quick Proposition 53 H C G is u subgroup i xy 1 E H Vxy E H Proof i is straightforward since H is a group For we first let X E H But then xx l e E H so that H contains the identity We therefore have ey l y 1 E H Vy E H so that H is closed under inverses Finally xy xy 1 1 E H and so H is closed under products Since the operation restricted to H is automatically associative because it is associative on G we have that H is a group and therefore a subgroup I Subgroups of low order groups Recall the groups 21 2223 under modulo addition It is straightforward to check that the only sub groups these groups have are the identity group 8 E 21 and themselves They are not therefore very interesting groups Only when we reach order 4 do interesting things start to happen It is an exercise to see that the only two groups of order 4 have tables as follows e u b C e u b C e e u b C e e u b C u u b C e u u e C b b b C e u b b C e u C C e u b C C b u e The cycl c group Z4 The llein 4 group V The Klein 4 group can be viewed as the symmetries of a regular rhombus choose one of a b C to be rotation through 180 degrees and the remaining two to be the two re ections across diagonals of the rhombus a D b Observe that V S Z4 since the equation X2 8 has 4 solutions in V while having only 2 solutions in Z4 recall that the existence of solutions to such an equation are a structural property of a group if V and Z4 were isomorphic then both must have the same number of solutions There are many other reasons why V S Z4 but this is one of simplest to exhibit Observe that Z4 has three subgroups Z4 024 E Z2 0 E Z1 02 is a subgroup because 2 4 2 0 E 02 and 2 is its own inverse Note that if we try to make a subgroup out of say 03 then we see that 3 4 3 2 8 03 and so we don t have closure V by contrast has five subgroups V 8 a 8 b 8 C 8 E Z1 the middle three are isomor phic to Z2 Each two element set is a subgroup because all elements of V are their own inverse Conversely if we tried to make a group out of 8 b then we get ab C e 8 b so 8 ab is not closed and not a group Subgroup diagrams We can draw subgroup relations This can be useful in analysing the structure of a group because it makes it extremely easy to see the differences between groups A line in a diagram means that the lower group is a subgroup of the upper 21 22 23 Z4 02 EZZ 8 8b 8C 0 0 0 e 6 Matrix groups We have seen that the groups GLnlR GLnC of matrices with non zero determinant are groups under multiplication There are lots of subgroups of these all of which have some geometric signifi cance 5Most of these groups are for your information only you won t be expected to remember anything about any matrix groups other than GLnlR SLnlR and OnlR for any exam Special linear group 5L 1R A E M7101 detA l the set of determinant 1 matrices forms the special linear group Recall the relation detAB detA detB This says that the product of two determinant 1 matrices also has determinant l and so SLnOR is closed under multiplication Since detA 1 1 detA we also have that inverses have determinant 1 SLnOR is therefore a group as claimed The geometry of SI 1R comes from the fact that an element of determinant l preserves volumes Recall that the signed area 2 dimensional volume of a parallelogram in R2 spanned by vectors uv is Allv u M sin 0W where 0W is the angle between the vectors6 It is an easy calculation to see that if A is a matrix then the signed area of the parallelogram spanned by Au Av is AreaAuAv detA u M sin 0W detAAreauv The area therefore scales by the determinant We can play the same game in R3 Here the signed volume of the parallelepiped spanned by u vw is the triple scalar product uvw Again check that AuAvAw detAuvw VA E M3lR SL3R is therefore the set of signed volume preserving transformations Volumes of parallelepipeds That determinants are related to volume should not be a total surprise if you remember integra tion in several variable calculus you should recall that changing the variables of integration requires you to insert a factor of the determinant of the Iacobian of the transformation into the integral The Orthogonal group The orthogonal group is the set of matrices 01011 A e MnOR ATA In where In is the n X 71 identity matrix diagl 1 By Theorem 7 we need only check that AB 1 ABT E OnlR for A B orthogonal But ABTTABT BTTATABT BET 1 since B 1 BT 0101 is therefore a subgroup of GLn To see the geometry of OnlR recall the standard inner product of vectors on Rquot uv uTv Observe that AuAv AuTAv uTATAv uTv uv 6This is almost the norm of the cross product if the plane R2 is embedded in R3 16 Indeed an alternative definition of the orthogonal matrices is that they are exactly those which pre serve the inner product OnlR A E MnlR AuAv uvVuv E an 61 Theorem 61 OnR is the set of length and angle preserving linear transformations of R Proof Recall that the cosine of the angle between two vectors uv is related to the inner product by uv lul lvlcose Supposing that A E MnlR preserves the lengths of vectors and the angles between them we imme diately have that AuAv uv and so A E OnlR Conversely if A is orthogonal then lAul2 lulz and AuAv 7 uv PM PM T W cos QAUIAV cos Quiv Thus A preserves lengths and angles The Special orthogonal group SOnOR OnlR SLnlR is the subset of determinant 1 orthogonal matrices Taking determinants of ATA I gives detA detAT detA2 1 i detA i1 Thus SOnOR consists of exactly half the orthogonal group Example When n 2 the group SOZOR consists of the rotations around the origin det 1 while 020R is the set of re ections across lines through the origin det 71 Pseudoorthogonal groups The inner product Definition 61 of OnlR can be extended Suppose you put a different non positive definite inner product on lRquot for example 1 0 0 uvuT 0 71 0 v 0 0 71 on lR3 The group which preserves this inner product is the pseudo orthogonal group 01 2 This is not just abstract mathematical nonsense the study of relativity in Physics requires the Lorentz group O31 where the inner product on spacetime lR4 has 3 positive space directions and one negative time direction The Unitary group The unitary group is constructed similarly to the orthogonal group This time we take the Hermitian inner product on Cquot uv uv Tv Then Un A E GLnC ltAuAv uv The special unitary group SUn is the set of determinant 1 unitary matrices that is SUn Un SLnC Remark U1 51 is the unit circle in Cgtlt under multiplication U1 is the set of 1 X 1 complex matrices ie numbers 2 E C which satisfy 22 1 Thus lzl 1 i z E 51 Conversely all numbers on the circle have the form 819 for some 9 E 0271 from which we see that 819 819 8719819 1 A discrete matrix group Not all matrix groups must be continuous like the above The set SLMZ A E SLnlR all entries are integers is a group It is clearly a subset of SLnGR and it is easy to see that the product of two matrices with integer entries has integer entries so it remains to check inverses Recall from linear algebra the formula 1 71 7 T A MAM where adjA is the adjoint of A formed by taking positive and negative multiples of determinant minors of A Observe that if A E SLMZ then adj has integer entries and so therefore has its trans pose Since detA 1 it follows that the inverse not only has determinant 1 but has integer entries 51412 is therefore a group A subgroup diagram for matrix groups For reference the subgroup relations between the various groups are summarised in the following table The complex orthogonal groups are defined exactly as the real orthogonal groups just allow complex matrices with ATA I There are many many more matrix groups that we haven t men tioned Gum GL1R 0C 5Lc U 5L 112 0012 5002 5U sum 50010 7 Cyclic groups Cyclic groups are a very basic class of groups we have already seen some such as Zn 71 Theorem 48 Cyclic groups are relatively easy to work with since their structure is easy to describe The approach in this section is typical of pure mathematics in thatwe prove theorems based on an abstract definition before classifying what it is that we have just defined in this case all cyclic groups This approach may seem bizarre to the uninitiated but it has its logic Thusfar we have been deliberately relaxed about what a group is For example the definition in Theorem 48 of the cyclic group 2 as the set 01 n 7 1 under addition modulo n is not 18 strictly correct If this really was Zn then what would we call the group 0 2 4 2rl 7 1 under addition modulo 2n This new group has the same structure as Zn it is isomorphic so what should we do Which is the cyclic group of order r1 The answer is that both are the group itself is an abstract concept the above two descriptions are simply that they are ways of writing down the group The cyclic group of order rl knows nothing about numbers or sets it just is Any structure it has that is truly intrinsic that comes from the group and not from how we look at it should be discernable without making any choices on how to write it down7 De nition 71 Let G be a group written multiplicatively and X E G The cyclic subgroup of G generated by X is the subgroup X X11 rl E Z The order of an element X E G is the order X of the subgroup generated by X G is a cyclic group iff 3X E G such that G Remarks The groups Zm n and Z are cyclic Both are generated by the element 1 We shall see shortly that up to isomorphism these are the only cyclic groups However we do not need to know this to understand many things about cyclic groups Writing our cyclic groups multiplicatively rather than additively is good practice in thinking abstractly We now have two concepts of order The order of a group is the number of elements in that group while the order of an element is the number of elements in the cyclic group generated by that element It follows that cyclic groups are the only finite groups containing elements having the same order as that of the group Lemma 72 Suppose that G X has order r1 Therl Xk e ltgt k qufor some 11 E Z Proof Divide kby rl with remainder Thus k qr r where qr E Z and 0 g r lt rl Then 8 xk ertr X71qu X7 If r gt 0 this says that the order of X is at most r which is strictly less than rl a contradiction Hence r 0 and k qu l Lemma 73 Every cyclic group is Abeliurl Proof Let G Then any two elements of G can be written Xk X for some kl E Z But then xkxl Xkl Xlk Xlxkl and so G is Abelian I Theorem 74 All subgroups of u cyclic group are cyclic Proof Let G X and let H g G Let rl E ZJr be the smallest positive integer such that Xquot E H Claim H X71 For this let Xm E H be a general element of H But then we may divide m by rl so that there exist integers qr E Z 0 g r lt rl such that m qr r 7This is no different to the use of numbers do you think that 21 cares if it written 10101 or 25 or 15 binary octal or hex notation or in any of the many forms u 39 invented y 39 quot39 39 21 is a concept that transcends notation So it is with the cyclic group of order n Therefore xm 367quot x 7x7 E H But Xquot E H i Xquot 7 E H and so therefore x7 x 7xm E H i r 0 since 0 g r lt n and n E ZJr is smallest such that Xquot E H In summary xm Xquot 7 E Xquot and we are done I Theorem 75 All cyclic groups are isomorphic to one of Zn n or Z Proof Let G There are two possibilities either G n lt 00 or G has infinite order Treat the latter case first Case 1 G 2 N0 Define a map 4 G gt Z by 41Xquot n This is well defined because Xquot xm i JCan e i m n otherwise G would have finite order g m 7 We check that 4 is an isomorphism 1 1 41xm 41xquot i m n i xm xquot onto m 41xm Homomorphism 41xmxquot 410W m 71 41xm 41xquot 4 is therefore an isomorphism 4 G E Z Case 2 G n lt 00 Define a map 4 G gt Zn by 41Xm m mod Wellde ned xk x7 i xk e i k El qn for some 11 E Z by Lemma 72 But then wk W 1 41x 41x 7 i p E 11 mod n i x 7 x i onto 11 41x i Homomorphism 41067367 4106 p 11 mod n p 71 11 zpx 7nzpx 7 4 G E Zn is an isomorphism I The only real difference in the arguments for finite and infinite orders is the extra care that must be taken over well definition in the finite order case Wellde nition as a concept The concept of well definition appears in the above proof for the first time Essentially something is well defined if it makes sense Most times you have written down a function it is completely obvious that you really have defined a function for example f Z gt Z given by f z z2 7 3 is clearly a function However in abstract mathematics you are often asked to define a function from a set for which you haven t got a complete intuitive description For example we know that all the elements in the above cyclic group G X have the form xm for some integer in but it may be that the same element can be written in multiple ways for example X2 38quot 38 quot etc The map f G gt Z given by f Xm in when Xquot e is not well defined since the same element X2 38quot X244quot is mapped to multiple targets 22 712 271 You should get into the habit of checking well definition as the first step of any proof which requires you to construct a function The above example is easy to check but later examples will be less so It is often extremely easy to write down something that looks like a function but fails to be so Subgroups of cyclic groups Since subgroups of cyclic groups are cyclic we know that any H g Z has a generator rl Hence H rlZ Note that H E Z if rl 7E 0 and that by isomorphism it follows that the only non trivial subgroups of an infinite order cyclic group are isomorphic to the group itself8 Having completely described the subgroup of infinite cyclic groups we now consider finite cyclic groups Theorem 76 Let G x E Zn The subgroup of G gerleruted by the element y XS is isomorphic to Znd where d gcd51 l Proof We know that y is a cyclic group of finite order as it is a subgroup of G E Zn It is enough to check therefore that yk e ltgt k E 0 mod rld Clearly ifk 11 then yk 05 3 en i e since 3 E Z Conversely ykeExSkeExSke E sk am for some 0c E Z Lemma 72 s E ah 0c since 01 d1v1des srl E d 7 39 i E E k 0 mod rld s1nce gcd d d E 1 Therefore y E Znd Corollary 77 Zn has exactly one subgroup Z M for each divisor d of rt Proof Let d be a divisor of rl Then by the Theorem 01 E Znd Z d A gc r1 Conversely suppose that gcdurl d Then firstly u Ad for some A E Z so that u E d and so u g Secondly by the Euclidean Algorithm9 3A4 E Z such that Au url d and so Au E 01 mod Thus 01 E u and so at lt ii In conclusion if gcdurl 01 then ltagt on e 2m Remark The above Corollary can be phrased in terms of a general cyclic group of finite order If G X rl then the subgroups ltXjgt xkgt are the same iff gcdjrl gcdkrl Examples 1 Z8 0 123 4 5 67 is generated by 135 7 eg 5 52741630 28 The order10 of 6 is 4 8 gcd86 which is exactly what we see 6 6420 8This is a semmingly paradoxical property of infinite sets you can throw away half or more of the set and still have the same number of elements 939You don t need to know what this is for this class 10Recall Definition 71 for the order of an element 2 Find all the subgroups of 230 Note that 30 2 3 5 We list all the numbers in 230 their greatest common divisor with 30 and the subgroup generates recall that 230 has exactly one subgroup for each divisor of 30 Remember each subgroup in the right column is generated by any of the numbers in the left column X gcd30x 30 gcd30x subgroup generated 030 030 1 21 15 15 2 22 1020 10 3 23 6121824 6 5 Z5 525 6 26 392127 3 10 1210 24 81416222628 2 15 1215 17111317192329 1 30 1230 Here is the subgroup diagram for 230 with the obvious generator chosen for each subgroup 230 1 215 2 210 1 3 26 5 25 6 23 lt10 22 lt15 21 1 lt0 8 Generating sets De nition 81 If X C G is a subset of a group G then the subgroup of G generated by X is the subgroup of all combinations and inverses of elements in X G is nitely generated if there exists a finite subset X C G which generates G Remarks The subgroup of G generated by X really is a subgroup it is a subset so we need only check that it is closed under multiplication and inverses but the definition says that we keep throw ing things into the subgroup so that it satisfies precisely these conditions The notation for cyclic groups will be extended for generating sets le the subgroup of G generated by X will be denoted X E X Examples 1 Z 1 23 since the inverse of2 72 E 23 i 3 7 2 1 E 23 2 In general if mn E Z then the subgroup mn Am nn A44 E Z is d all where d gcdmn 3 Q 1 n n E 2 Q is not finitely generated there exists no finite subset which generates Q To see this suppose X piqi1 it is impossible to generate 11 from X if q is any number relatively prime to all the qi this argument is very similar to that of there being an infinite number of primes 9 Permutation groups De nition 91 A permutation of a set A is a bijection 4 A 7 A Theorem 92 The set of permutations S A of any set A forms it group under composition Proof Closure If p 1p are bijective then so is the composition p 0 1p Associativity Permutations are functions the composition of which we know is associative Propo sition 25 Identity ldA x gt gt X is a permutation Inverse If p is a permutation then it is a bijection and so p l exists and is also a bijection S A is therefore a group under composition De nition 93 The symmetric group on n7letters Sn is the group of permutations of 12 rl Proposition 94 Sn hus rt elements unlike Zn where the subscript is the order of the group Proof This is just counting If U E Sn then there are rl choices for the value of 71 Choosing one leaves rt 7 1 choices for 72 since 7 is a bijection lterating this argument we get 2 choices for 0rl 7 l and only 1 possibility for 7rl We therefore have rlrl 7 1 2 1 rt possibilities in total I Permutation groups are fundamental The following Theorem says that in fact all groups are permutation groups Sadly this result is not as useful as it might sound Theorem 95 Cayley s theorem Every group G is isomorphic to u group of permutations Proof Fix it E G and let p11 G 7 G be the map p11 x gt ux ie left multiplication by u Each p11 is a bijection lo1400 u lx and is therefore a permutation of G that is p11 E SG We claim that the set of all such maps under composition forms a group isomorphic to G M to E Go 2 Gly For this define the map 4 G 7 p11 by 1X px Observe that all the p11 are distinct p11 p1 i ux bx VX i u b and so 4 is 1 1 and therefore abijection Moreover 41W ny px Opy 41000 My since pxyz xyz pxyz pxpyz Therefore 4 is a homomorphism and being bijective is an isomorphism I Two notations Standard notation Suppose o E S4 is the following map 3 72 gt gt 1 2 3 4 4 2 We would then write 0 i 1 2 3 4 T 3 1 4 2 where you read columns to find where 7 maps an element Composition is read in the usual way for functions do the right permutation first Thus if 71234th 771234123471234 9 1432 en9 31421432 3241 Writing 7 as a map on the vector 123 4T suggests another method of writing Sn as matrices 0 07 1 T 0 H000 0H00 1 0 0 0 0 is a permutation matrix it swaps the entries of vectors An rl gtlt rl matrix with exactly one entry of 1 in each row and column and the remaining entries 0 is a permutation matrix the set of such forms a group under multiplication isomorphic to Sn Indeed we have proved Proposition 96 Everyfirlite group of permutations is isomorphic to u group of matrices Cyclic notation The above 7 can be more compactly written as o 1 3 4 2 We read from left to right looping back to 1 at the end the next entry telling us where the previous entry is mapped to by 7 Thus 1 3 4 2 maps 1gt gt3gt gt4gt gt2gt gt1 We have shorter cycles if some of the elements are fixed for example in our two notations 1 2 3 4 5 13lt3 2 1 4 5gtESS qutaposition is used for composition lt1342gtlt24gti i i G 13343234134 Remember that multiplication of cycles is really composition of functions thus although each cycle is read from left to right when determining how it acts on elements of 12 rl we multiply cycles by considering the rightmost cycle first For example 1354234 13254 24 is calculated by considering first that 1 is mapped to 3 by 234 followed by 1 354 Then 3 is mapped to 4 by 234 and then 4 is sent to 1 by 1 354 Hence the composition sends 1 to 3 and 3 to 1 yielding the 2 cycle 1 3 The remaining 3 cycle appears similarly It is not always possible to write an element as a single cycle eg the Klein 4 group V can be written in terms of cyclzes if we label the corners of a rhombus 1 D V e132 41324 4 De nition 97 A kecycle is an element a1 a2 wk E Sn k g 71 Two cycles a1 wk and b1 11 are disjoint if no element appears in both cycles11 1 k b1b Dihedral groups De nition 98 The nth Dihedral group Dn is the group of symmetries of the regular n gon regular polygon with n sides Group Now that we have defined permutation groups it is easy to see that the Dihedral groups are indeed groups doing this directly is much harder Observe that a symmetry can be viewed as a permutation of the corners of the n gon so that neighbourliness is preserved For example if we label the corners of the regular hexagon as 123456 then we see that D6 is the set of U E 56 such that 71 is always next to 76 and 72 etc Dn C 56 is a subset To see that D6 is a subgroup note that the composition of two neighbour preserving transforms also preserves neighbours as does the inverse of such a 5 6 map Elements of Du The regular n gon has 271 distinct symmetries and so Dn 271 These consist of n rotations Let p be rotation counter clockwise by radians where 0 n 7 1 n re ections Let y be re ection across the line making angle with the positive X axis make sure you put one of the corners of the n gon on the X axisl Remarks Some authors write D271 instead of Du precisely because Dn 271 we will always have Dn meaning the symmetries of the n gon Note that every Dn is a discrete finite subgroup of the orthogonal group 02lR of length and angle preserving matrices The correspondence is coslt2jgt isinlt27njgt coslt27 jgt sinlt27njgt 39 It is a good exercise to convince yourself that these matrices really do correspond to the rotations and re ections claimed In particular multiply any two of them together and see what you get 11Later we will say that their orbits are disjoint Examples D3 is the group of symmetries of the equilateral triangle See the introduction for the multiplication table Here we write all the elements in permutation notation and exhibit all the subgroups their generators and the subgroup diagram Element Standard notation Cyclic notation 5 1 2 3 a 90 1 2 3 8 o 1 2 3 14 3 41 2 3 1 123 1 quot396 m 1 2 3 p2 lt3 1 2 132 1 2 3 3 M a 1 lt1 3 2 23 p0 identity g 1 2 3 13 p1 rotate counter clockwise by 2713 1 2 3 2 1 p2 rotate clockw1seby 2713 Q 3 1 2 3 12 2 1 3 D3 23 22 22 22 N4 H D3 53 and D4 is the group of symmetries of the square otherwise known as the attic group D4 consists of 4 rotations and 4 re ections the notation 5f for re ection across a diagonal is used rather than calling all re ections m Element Standard notation Cyclic notation 1 2 3 4 8 9 1 2 3 4 m 1 2 3 4 5 pl lt2 3 4 1 1234 g 1 2 3 4 g pz 3 4 1 2 W24 1 2 3 4 p3 lt4 1 2 3 1432 1 2 3 4 141 2 1 4 3 lt12gtlt34gt p0 identity 2 42 1 2 3 4 1423 9 4 3 2 1 p1 rotate counter clockw1se by 712 395 m l 2 3 4 p2 rotate by 71 lg 51 1 4 3 2 24 p3 rotate clockwise by 712 m 6 1 2 3 4 13 2 3 2 1 4 26 D4 VZk 2222 my V D4 where i Subgroup relations 0 Sm g Sn ltgt m g n For example fix the final n E m elements of l n so that Sm oESnoi i Vigtrn In fact Sm is a subgroup of Sn in precisely different way each copy of Sm comes from fixing n E m elements of 1 n and there are ways of choosing these fixed elements C Dm g Dn ltgt Join every nrnth vertex of the regular n gon to get a regular m gon For example the diagram shows that every element of D8 which preserves the dashed square is a subgroup But this subgroup is D4 10 Aside Equivalence relations De nition 10 An equivalence relation on a set A is a binary condition12 which satisfies Re exivity x N x Symmetry x N y i y N x Transitivity x N y amp y N z i x N z The set x y E A y N x is the equivalence class ofx The set of equivalence classes is written A read A mod twiddles or tilde if you prefer We may therefore write X E X E A N This is not paradoxial X is an element of the equivalence class X which in turn is an element of the set of equivalence classes Theorem 102 An equivalence relation on A partitions A into distinct subsets Conversely if A Uid Ai is any partition Ai Aj if i 73 j where I is some possibly infinite indexing set then de ned on A such that x N y ltgt 3i E Isuch thatxy E A is an equivalence relation 12Either x N y x is related to y or x no y x is not related to y This is not a binary relation x N y is either true or false not an element of the set A Proof Suppose we are given an equivalence relation N Observe first that X N X i X E X so that every element is in some equivalence class Now suppose X N y We want to show that X We do this in two stages 1 Let a E Then a N X but by transitivity a N X amp X N y i a N y Therefore a E y and we have X C 2 Now let a E Thus a N y But now by symmetry we have y N X and so transitivity says ayampyX aXThusa E Xandsoy C Since X C y and y C X we conclude that X y and we re done The converse is straightforward X N y ltgt X y in the same Ai is obviously an equivalence relation with all three conditions being clear I 11 Orbits De nition 111 The orbit of o E 5 containingj E 12 n is the set OrbjW 0139 k E Z C 12n Remark Observe that orb7k0o orbjo for any k E Z Be careful each orbit is a subset of the set 12 n not of the group Sn We are really looking at the orbits of the action of 7 on the set 1 2 Group actions will be considered further in section 18 Proposition 112 The orbits ofo partition 12 n Proof Define X N y ltgt y E orbxo ltgt y UkX for some k E Z We claim that N is an equivalence relation Re exivity X N X since X 00X Symmetry X N y i y UkX for some k E Z But then X o ky i y N X Transitivity X N y amp y N z i y UkX amp z 07y for some kl E Z Thus 2 ok X and so X N z l Recall Definition 97 of a cycle Observe that o E 5 is a cycle if it has at most one orbit with more than one element Cycle notation therefore records orbits the cycles are disjoint for example Orbits of 134 E S5 are 134 2 5 Orbits of 1245 E S5 are 12 3 45 Theorem 113 Every permutation can be written as a product of disjoint cycles Proof Write out each of the orbits of o E 5 in order putting each orbit in parentheses Since the orbits of o partition 12 n the cycles obtained are disjoint I If you mechanically follow the algorithm for multiplying cycles together you will automatically end up with a product of disjoint cycles Examples 1 13234 1342 2 13241234 1423 Disjoint cycles commute Eg l423 23l4 Theorem 114 The order of a pennutiztion o is the least common multiple of the lengths of its disjoint cycles Proof A k cycle clearly has order k the least positive integer l such that ill ilk e Suppose the orbits of U have size 0939 E 2 Writing 7 as a product of disjoint cycles 7 71 Um we have that aquot 71 all since disjoint cycles commute All these factors must be the identity for Uquot to be the identity Thus 7quot e ltgt 7 e Vj Thus n is a multiple of 0939 for all The least such n is clearly lCl 1 10 39 I Examples 1 The order OfU l45362789 E 59 is lcm342 12 2 We can easily calculate 03465 for the above 7 Since 3465 12 288 9 we have 03465 01528809 a9 145936279899 362789 since 145 3627 and 89 have orders 3 4 and 2 respectively Transpositions 2 cycles lung are also known as transpositions since they swap two elements of l2n and leave the rest untouched Proposition 115 Every 0 E 5 is the product of transpositions Proof There are many many ways to write out a single permutation An easy one to remember is to first write 7 as a product of disjoint cycles then write each cycle as follows 11 39 k l k 1tlk71quot39 l 2 Just look carefully to see that this works Example 17645 15l416l7 De nition 116 7 E 5 is even odd if it can be written as the product of an even odd number of transpositions Proposition 117 The concept of evennessoddness is wellide ned Proof Recall that any permutation U E 5 can be written as an n X n permutation matrix A 2 cycle is a permutation matrix which swaps two rows it therefore differs from the n X n identity matrix only in that two of its columns are swapped For example 24gt 000DA H000 0H00 00H0 From linear algebra we have that swapping two columns of a matrix changes the sign of its determi nant Hence det2 cycle 71 It follows that d tar 1 if U is the product of an even number of 2 cycles e 71 if U is the product of an odd number of 2 cycles In particular 7 cannot be both odd and even De nition 118 The alternating group An n 2 2 is the group of even permutations in Sn Proposition 119 An has exactly half the elements of Sn that is An Proof Since n 2 2 we have 12 E Sn Define 4 An gt odd permutations by 410 12 o 7 We claim that this is a bijection 1 1 410 41T i 12o 12T i 711 onto If p is an odd permutation then 12 o p is even and so in An Therefore p 4112 o p Since 4 is a bijection it follows that there are exactly the same number of even and odd permutations in Sn Exactly half of them are therefore even I 12 Cosets 8 Lagrange s theorem Take a look at any of the subgroup relations in the notes and observe that the order of every subgroup is a divisor of the order of its parent group This is a general theorem Theorem 121 Lagrange The order of a subgroup H g G divides the order of G Otherwise said H g G gt HlG We will prove this shortly Remarks A similar statement to Lagrange s Theorem is the following the order of an element di vides the order of the group This is really only a statement about the cyclic subgroups of a group G and is not as general as Lagrange proper Lagrange s Theorem only makes complete sense for finite groups as any finite number is a divisor of infinity For such a simple result Lagrange is extremely powerful You may have observed that there is only one group structure of order 2 3 5 and 7 This is a general result Corollary 122 There is only one group up to isomorphism of each prime order p namely 2 Proof Let G p and choose X E G e Then X g G Lagrange tells us that the subgroup X has order that divides p But p is prime and X has at least 2 elements since X 7E e and so X p Therefore G I De nition 123 Let H g G and fix u E G The subsets qu uhzh E H Hu huzh E H are respectively the left and right cosets of H containing u Lemma 124 IfG is art Ahelum group therl the left and right cosets ofurly subgroup H g G are identical Le uH Hu Vu E G For rlorliAheliurl groups this is generally not true Examples 1 The left cosets of the subgroup 23 E 4 g Zn are as follows we write these cosets additively since the group operation is additive 0 4 048 4 1 4 159 24 2610 34 3711 Observe that the right cosets of 4 are the same as the left cosets Here we have an Abelian group and so left and right cosets are always the same 2 Recall the multiplication table for 53 given in the introduction The left and right cosets of the subgroup H e m are as follows Left cosets l Right cosets eHu1HHeu1 l HeHu1Heu1 91H rsH pirs l Hm H142 pir2 PzH th P2442 l HPZ Hhs P1443 The left and right cosets of H are different this time Theorem 125 The left cosets of H purtitiorl G Proof Define by x N y ltgt x ly E H First observe that x N y ltgt x and y are in the same left coset of H For this suppose that Xy E uH Then u lxhl ly E H and so X lu E H Since H is a group we have x luu ly x ly E H Conversely if x ly E H then y E xH Butx E XH so that Xy are in the same left coset We claim that N is an equivalence relation Re exivity x N x since x lx e E H Symmetry x N y i x ly E H x 1y 1 y lx E H i y N X Transitivity x N y amp y N z i x ly ampy 1z E H Therefore x lyy lz x lz E H i x N z The equivalence classes of the left cosets of H thus partition the group G Theorem 102 Remark The right cosets of a subgroup H also partition G In general Lemma 124 they do this is a di ererlt way to the left cosets Observe the earlier example of the cosets of the subgroup e m g 53 Observe that each of the three conditions on N being an equivalence relation corresponds to one of the three properties that characterises H as being a subgroup of G re exivity says that H contains the identity symmetry says that H is closed under inverses and transitivity says that H is closed under composition It is exactly the property that H is a subgroup that guarantees partition if H is only a subset we do not get a partition For example consider the subset 01 C 23 its left cosets are 0 01 01 1 01 12 2 01 21 which are certainly not disjoint ProofofLugmnge s Theorem The map 4 H a uH given by 41h uh is a bijection13 It follows that every left coset of H has the same cardinality as H Therefore G number of left cosets of H H Thus H divides G I Remarks Again we can argue with right cosets there is no difference In particular we see that H has exactly the same number of left and right cosets The argument of the Theorem also holds for infinite groups cosets of subgroups of infinite groups also have the same cardinality De nition 126 If H g G then G H is the number of left or right cosets of H in G G H is the index of H in G Remark If G is a finite group it is immediate that G H This is not true for infinite groups as infinite cardinals do not divide For example it can be shown that 0012 50011 2 and Q Z N0 where all 4 groups listed have the same cardinality N0 Theorem 127 IfK g H g G is u sequence ofsubgroups such that H K 8 G H are both nite then GzK G By the above remark the argument would be trivial if all threee groups were finite Proof Let G H p and H K q where 1111 lt 00 Then 3 distinct g E G such that the left cosets of H in G are ngg2H ng Similarly 3 distinct h E G such that the left cosets of K in H are thh2K h1K Claim that the cosets ofK in G are given by gith where i 1 p and 1 q 13Argument exactly like that in Proposition 119 1 Every X E giH for some i since the left cosets of H partition G Therefore gle E H However every 11 E H is in some th for some j thus 3ij such that X E gith Every X E G is therefore in one of the claimed cosets of K N Now suppose X E givth gah K for some ijtx Then in particular X E giH gaH i X i since the left cosets of H partition G But then gi lX E th h K i since the left cosets of K in H partition H It follows that givth gah K unless 0c i and 8 By 1 amp 2 the sets givth partition G and each is a coset thus they are the cosets of K in G It is clear that there are exactly pg of them I 13 Direct products De nition 131 Let G1 Gn be any groups The direct product of G1 Gn is the group 1 l l H GiG1gtltgtltGn under elementwise multiplication The elements of the direct product are written like vectors g1 gn E G1 gtlt gtlt Gn where each g E G By counting elements it is straightforward to see that the cardinality of a direct product is given by 7t 71 G G1 Gn HGv i1 i1 There is still something important to check Lemma 132 The above de nition really de nes a group Proof Multiplication in 111 G is defined as follows let m b E G for each i then d1 n 171 bn Q1171 nbn We now check the group axioms Closure Since each ibi E Gi this is immediate Associativity Each G is associative it is a short calculation to see that the associativity condition on the whole direct product is equivalent to each individual Gi being associative Identity lf ei is the identity in G then e1 en is the identity in the direct product Inverse d1dn 1 df1d1 I Remarks Really the proof uses nothing but the fact that each factor G is a group in its own right this should not be surprising as we have no other information to work with In particular you should note that the individual groups have no effect on each other all group operations are carried out at the level of the factors This is very similar to thinking about a direct sum of vector space W U 63 V without further information about where U and V sit inside W we can only view U and V as separate 33 spaces Indeed the direct sum of vector spaces is a perfect example of a direct product of groups when the groups G are written additively we usually use the direct sum notation 7i GiG1 Gn i1 For example 71 RquotRRR i1 Examples 1 22 X 23 000102101112 under addition 23 Ob serve that if we choose a 11 then the group can be written in the above order as 6a4o 2o3o a 5d Thus 22 X 23 is generated by a and is therefore cyclic Having order 6 we necessarily have 22 X 23 E 26 N 22 X 22 00 0 1 10 1 Notice that all elements are their own inverses since this is a group of order 4 we see that 22 X 22 E V is the Klein 4 group Remember this definition many people see V for the first time defined this way and are less happy thinking about it as the symmetries of the rhombus or indeed written in permutation notation as in section 9 The point to take from the above two examples is that the direct product of two cyclic groups is sometimes cyclic example 1 and sometimes not example 2 There is a general Theorem to tell you when Theorem 133 Zm X 2 E Zmn ltgt gcdmn 1 Since Zm X 2 automatically has order mn the Theorem can be rephrased to say thath X 2 is cyclic ltgt gcdmn 1 Proof Suppose first that gcdmn 1 Consider the order of 11 E Zm gtlt Zn k11 k k is the identity 00 iff k E 0 mod m and k E 0 mod But then a Am pn for some integers A44 However gcdmn 1 E m divides p and n divides A In particular a is divisible by the product mn Moreover mn11 mnmn is indeed the identity 00 Thus mo 00 ltgt o is a multiple of mn which is the same as saying that a has order mn i1 therefore generates Zm X 2 which is then cyclic Conversely suppose gcdmn d 2 2 Then is an integer divisible by both m and n Hence 7 Q m 7 d 7 lt d ml d n 7 010 for any pq E Zm X 2 since g are integers No element of Zm X 2 has order mn and so Zm X 2 is not cyclic l Corollary 134 The previous proof can be modified to show that 1 l l H Zm E Zmlmn ltgt gcdmvmj 1 Vi 7 1 Moreover if n p11 p is the prime decomposition of an integer n into powers of distinct primes then anqumxzyk 1 17k 17 34 Example Is 25 X Zn gtlt Z43 cyclic By the Corollary we can say yes since no pair of the numbers 51243 have any common factors Indeed there are 15 different ways up to reordering 22580 g 23 X 2860 g Z4 X 2645 g ZS gtlt Z516 g Z43 X 260 g 212 X Z215 g Z15 gtlt Z172 g 220 X Z129 23 X24 X2215 23 X25 X2172 23 X220 X243 24 X25 X2129 24 X215 X243 25 XleXZ43 23 XZ4XZ5 X243 Theorem 135 Suppose u E G has order r for each i Then u1un E G1 gtlt X G has order lcmr1rn Proof The argument is similar to part of the previous Theorem uh unk 0111 u is the identity iff r divides k for every i Thus k is a multiple of all the r the smallest such is the order of u1 won but this is the definition of lcmr1 rn I Example What is the order of 14 32 E Z4 gtlt Z7 gtlt Z5 X 220 Recalling that the order of X E 2 is n gcdxn we see that the above elements have orders 475 amp 10 respectively Thus the order of l432 is lcm47510 140 14 Finitely generated Abelian groups Theorem 141 Fundamental Theorem of finitely generated Abelian groups Every nitely generated Abelian group is isomorphic to u group of the form Zp31gtltgtltZpLKgtltZgtltgtltZ where the p are not necessarily distinct primes the r are positive integers and there are u finite number of factors of Z The proof is far too difficult for this course or indeed most group theory courses Its purpose here is to allow us to classify finite Abelian groups up to isomorphism and we ll do an example of this below Remember Definition 81 for what finitely generated means it is not the same as finite The above direct product is only a finite group if it has no factors of Z Example Find up to isomorphism all Abelian groups of order 450 First note that 450 2 32 52 Now apply the fundamental theorem to see that the complete list is 1 Z450 g 22 X 1232 X 252 2 22 X 23 X 23 X 252 3 22 X 1232 x25 X 25 4 22x23x23x25x25 Since writing a group as a direct product is common we have a name for groups that can be written in such a way De nition 142 G is decomposable if there exists proper subgroups HK lt G such that G E H X K Otherwise G is said to be indecomposable Example V E 22 X 22 is decomposable but Z4 is indecomposable Corollary 143 If G is finite Abelian and indecomposable then G E an for some prime p any positive integer n Proof By the fundamental theorem such an indecomposable G can have only one factor an l Recall Lagrange s Theorem 121 and how it doesn t have a general converse if miG then we cannot in general claim that G has a subgroup of order m The smallest group for which this is evident is A4 which though it has order 12 has no subgroup of order 6 When G has certain special forms however we can find partial converses to Lagrange for example Corollary 77 shows that if G is cyclic then G has exactly one subgroup for every integer dividing G The next Theorem illustrates what happens for finite Abelian groups Theorem 144 Let G befinite Abelian Then G has it subgroup of order rn for every divisor rn of G Proof G is finite Abelian and is therefore isomorphic to some 2 7 X X Z The order of G is p11 p1 Let rn pii pfquot for some integers s g r14 Recall that the cyclic subgroup of va generated by X has order p gcdp x Thus the sub group generated by pg is has order p gcdp pi Ts p The direct product of all such sub groups is then a group of order m ltp117s1gt X X ltp2ws gt gzpil X X Z Bnut this is a subgroup of 2 7 X X Z and is therefore isomorphic to a subgroup of G of order I Theorem 145 If rn is u square free integer Ek E Zgt1 such that kzirn then there is only one Abelian group of order rn up to isomorphism Proof By the fundamental theorem such a group G must be isomorphic to some val X gtlt me 1 n with rn p11 p1quot But rn being square free implies that every s is equal to 1 and all the primes p are distinct By Corollary 134 we have G E 2m Zm I Example As examples of the above we list all the groups of orders 1 through 15 and the Abelian groups of order 16 The Fundamental Theorem gives us all the Abelian groups In particular observe where Theorem 145 applies There are only two new groups we haven t seen before the Quuternion group Q8 and the Dieyelie group Dng look these up on Wikipedia or Q8 on page 226 of Fraleigh if you re interested These groups are definitely non examinable There are 9 non Abelian groups of order 16 up to isomorphism at least 5 of which we have no notation for hence we omit them You may be suspicious from looking at the table that there are no non Abelian groups of any odd order This is not so but you need to go to order 21 before you find one Note that even though the pi do not have to be distinct we still write m in the above manner and distinguish between exponents sisj even if pi pf 36 Order Abelian Non Abelian 1 21 2 22 g 52 3 23 4 Z4 V g 22 X 22 5 Z5 6 Z gsz23 D3 Sg 7 Z7 8 Zg 22 X Z4 22 X 22 X 22 D4 Q8 g DiCz 9 29 23 X 23 10 210 g 22 X Z5 D5 11 Zn 12 Zn g 23 X Z4 22 X Z g 22 X 22 X 23 D6 A4 Dng 13 1213 14 214 g 22 X Z7 D7 15 215 g 23 X Z5 16 216 Z4 X Z4 Zz X Zg Zz X Zz X Z4 Zz X Zz X Zz X Zz Many 15 Homomorphisms 8 Normal subgroups Recall Definition 32 of a homomorphism of binary relations Even though we have been using this definition when the binary relations are group operations we restate for convenience if G1 and G2 are multiplicative groups then a map 4 G1 gt G2 is a homomorphism if 41ab zpazpb Vab E G1 Homomorphisms of groups are far more interesting than those of general binary structures Before we see why it is necessary to formalise some notation for functions and sets De nition 151 Let 4 G1 gt G2 be a function between for the moment sets G1 G2 The image of a subset A C G1 is the subset 47A 47g E Gztg E A C C32 The image of G1 itself is written Im 41 41G1 Given a subset B C G2 the inverse image of B is the subset 1Bg e 61244506 B c Gl If G2 happens to have a well defined zero or identity eG2 for example if G2 is a group then the inverse image of eG2 is the kernel of 4 kart 4771862 g E G1ltPg 862 C G1 Remark The image and inverse image of a set make sense for any functions and sets the kernel of a function only makes sense if the target space has an identity Theorem 152 Let 4 G1 a G2 be a homomorphism of groups Then 1 8G1 EGz39 2 4101 W 3 H g G1 x 4H G2 4 K g G2 x 4r11lt G11 By choosing H 8G1 g G1 and K 8G2 g G2 we immediately see that the image and kernel of a homomorphism are subgroups of the domain and target groups respectively Corollary 153 If 4 G1 a G2 is u homomorphism of groups then ker 4 g G1 and Im 41 g G2 ProofofTheorem 1 Choose any g E G1 Then zpeG1zpg 418G1g 41g Since 418G1 fixes an element of G2 it is therefore the identity 8G2 1 WWW 47861 HGT Thus 4701 470W Let H g G1 be a subgroup and h1h2 E H Since zph1zph2 1h1h2 E we have that is closed under multiplication By 2 we see that zph1 1 41h1 1 E 41H thus is closed under inverses Since 41H is a subset of a group G2 it follows that 41H is a subgroup Of 8G2 RN 5 LetK g 8G2 be a subgroup and k1k2 E 1K C G1 Firstly 1k1k2 1k1zpk2 E K so 1K is closed under multiplication By 2 we have 1kfl 1k1 1 E K and so 1K is also closed under inverses 1K is therefore a subgroup of G1 I De nition 154 A subgroup H g G is normal if its left and right cosets are identical uH Hu Vu E G We write H lt1 G Lemma 155 All subgroups of un Abelian group are normal Conversely we have already seen section 12 that the left and right cosets of H p0 m lt D3 are different this H is not a normal subgroup However D3 does have the subgroup of rotations p0p1p2 g 23 which is normal Theorem 156 Let 4 G1 a G2 be u homomorphism of groups Then ker 4 lt1 G1 is normal Proof We must show that u ker 4 ker 4 u for all u E G However x E it kerzp ltgt u lx E kerzp ltgt 41u 1x 8G2 4gtlta1gt4gtltxgt 8G2 Marlon ea ltgt 470 470 The left coset of ker 4 containing u is therefore the set of elements X E G1 such that 1X 4101 The mirror argument shows that this is also the characterisation of the right coset containing ii We are therefore done I Examples 1 Recall that the trace function tr MnR gt lR is a homomorphism of additive groups These are Abelian groups and so the kernel of tr is automatically normal without needing the above Theorem The additive group of trace free matrices15 is a normal subgroup of MnlR kertr A E M7101 trA 0 4MnlR 2 4 236 gt 220 defined by 4n mod 24 The kernel of 4 is the subgroup kerzp n 4n E 0 mod 24 0612182430lt1236 is isomorphic to 26 1 if 7 even 71 if 7 odd groups Since the identity in the target space is 1 we have ker sgn An the alternating group of even permutations in Sn Indeed An 4 Sn 0 The map sgn 5 gt 1 71 E 22 given by sgnU is a homomorphism of g det GLnlR gt lRgtlt is a homomorphism and so ker det SLnlR is a normal subgroup SLnlR 4 GLnlR Compare this with example 1 U1 7r G1 gtlt Gn gt G defined as projection onto the ith factor 7rg1 gn g is ahomomor phism Therefore ker7IG1gtltgtltG391gtlt8XG1gtltgtltGn4G1gtltgtltGn If we are willing to abuse notation we can say that ker 7r G1 gtlt gtlt Giel X G X gtlt Gn is a normal subgroup of the whole direct product Indeed any combination of projections is a homomorphism eg 711 714 715 G1 gtlt X G gt G1 gtlt G4 gtlt G5 and so we can say that 71 HGj4HGi for any subsetl C 1n jel i1 All these examples should suggest an idea to you Not only are all kernels of homomorphisms normal subgroups the converse is also true any normal subgroup is in fact the kernel of some homo morphism of groups This loosely is the 1st isomorphism Theorem which we will come to shortly For the present it is helpful to record three criteria whereby we can determine if a subgroup is normal since check individual cosets is often tiresome Proposition 157 The following are equivalent 1 H 4 G is a normal subgroup 2i gH Hg VgE G 3 ghg l E H VgE G h E Hi 4 gHg 1 H Vg E G 15This kernel is often written 7013 for the specialilinear algebra and is very much related to the special linear group SLnlR the expression det 8A 8 shows that matrix exponentiation is a map exp 5nlR gt SLnlR This is the beginning of the theory of Lie groups and their Lie algebras 39 Only the last requires any comment the third is equivalent to gHg 1 C H but the map h gt gt ghg 1 is a bijection gHg 1 gt H and thus onto so we have equality we will prove this shortly The final Theorem in this section appears obvious but is actually very useful in determining what can possibly be a homomorphism Theorem 158 Let 4 G1 a G2 be u homomorphism 1 If G1 is u finite group then lmzp is u finite subgroup of G2 and its order divides that of G1 More succinctly G1 lt 00 i lmzplG1 2 Moreover Gz lt 00 i lmzplG2 Note that we are only assuming one of the groups to be finite in each case Proof 1 Recall from the proof of Theorem 156 that 1g 41h ltgt g kerzp h kerzp ie gh E G1 have the same image under 4 iff their left cosets with ker 4 are the same There are therefore exactly as many distinct elements of lmzp as there are left cosets of kerzp Thus lmzp G1 kerzp the index of kerzp in G1 Recall Definition 126 where we saw that G1 lt 00 i G1 ker 4 G1llter 4 This is clearly finite and divides G1 2 This is immediate from Lagrange s Theorem and Theorem 152 lm 4 is a subgroup of a finite group G2 and so the order of lm 4 divides the order of G2 I Examples 1 How many homomorphisms are there 4 Zm gt 2 when gcdmn 1 Suppose that 4 were a homomorphism then the Theorem says that lm 4 divides both m and n But the only integer dividing m and n is 1 therefore lm 4 0 lt 2 the image of a homomorphism always contains the identity In particular 4 must be the function 4 z gt gt 0 V2 E Zm There is therefore only one homomorphism The Proposition below gives the details when gcdmn 3A 1 How many homomorphisms are there 4 Z4 gt 53 Again the Theorem tells us that lmzp divides 4 24 and 6 532 thus lm 4 is either 1 or 2 If lm 4 2 then lm 4 is a subgroup of order 2 of 53 There are exactly 3 of these e u for each i 123 Since a homomorphism must map the identity to the identity we therefore have the homomorphisms N 4110123eu1eu1 4120123eu2eu2 IP3012138i43181i43 Finally if lm 4 has one element then 4 is the trivial homomorphism 412 e Vz E Z4 There are therefore 4 distinct homomorphisms Proposition 159 There are exactly at gcdmn distinct homomorphisms 4 Zm a Zn de ned by 41x igx mod n wherei 0d7 1 Proof Suppose thatzp1 X Then 1X xx mod n sincezp is ahomomorphism This completely determines 4 and it remains only to show which choices of 0c give well defined functions 4 is well defined iff 41X km 1X for all k E Z This is iff xx ukm E xx mod n ltgt ukm E 0 mod nfor all k E Z 40 ltgt am An for some EZ ltgt amiAn d7 a As in Theorem 76 g g have no common factors so this implies that X i g for some i E Z However 0c i kd i kn E i mod n so that onlyi 0 d 7 1 give distinct homomorphisms Conversely if X i then ma ign E 0 mod n and 4 is well defined The homomorphisms 4 Zm gt 2 are thus defined by pix igx fori 0 d 7 1 Example The only homomorphisms 4 Zn gt 216 are 410x 0 411x 4x mod 16 412x 8x mod 16 413x 12x mod 16 while the only homomorphisms 1p 216 gt Zn are 0x 0 1x 3x mod 12 Ipzx 6x mod 12 3x 9x mod 12 16 Centers and automorphisms In this section we describe some groups associated to any group which will give us further examples of homomorphisms and normal subgroups De nition 16 Let G be any group g1g2 E G We say that g1 is conjugate to g2 if 3h E G such that liglli 1 g2 Note that this is exactly the same definition you have already seen for two matrices being conju gate Proposition 162 Conjugacy is an equivalence relation Le de ned by g1 g2 ltgt g1 conjugate to g2 Proof Re exivitiy egle 1 g1 i g1 g1 Symmetry g1 g2 i Eli E G such that lig1li 1 g2 But this implies that h lgzh g1 and so g2 N g1 Transitivity Suppose g1 g2 and g2 g3 Then 3hk E G such that liglli 1 g2 and kgzk 1 g3 But then g3 khg1kli 1 and so g1 g3 I De nition 163 We call the equivalence classes of the conjugacy classes of G Note in particular that elements in the same conjugacy class have the same order hgh 1i ligjh 1 e ltgt V e The converse is not true however in A4 the elements 123 and 132 have the same order 3 but are not conjugate However they are conjugate in S4 since 23123231 132 Examples 1 If G is Abelian then every conjugacy class contains only one element This is since g1 g2 ltgt liglli 1 g2 for some li E G but by commutativity this reads g1 g2 41 2 The conjugacy classes of D3 E 53 are 8 V11 V2 V3 plP2 Theorem 164 The conjugacy Classes of Sn are the cycle types The concept of cycle type sounds vague for example 12345 has the same cycle type as 15623 but not the same as 1234 Just write an element of Sn as a product of disjoint cycles then its cycle type is obvious Proof Suppose ill uk E Sn is a k cycle and p E Sn is any element It is immediate that 1011 WW4 7011 470110 is also a k cycle Let T E Sn be written as the product of disjoint cycles T T1 17 Then pm pTlp 1psz 1New which has the same cycle type as T Conversely let 7 71 7 and T T1 17 be elements of Sn with the same cycle type that is 7139 and Ti are k cycles for the some k Define the permutation 7T by writing 7 and T one on top of the other 7Tlt71 72 7 T1 T2 T where each 7139 is the elements of the cycle 7139 written out in a row Then 7T0 T7T if Tiff is the jth element of the orbit Ti then 7TU7T 1TM 711707 moiH1 quotTiH1 TTvj Example The permutations 145 627 and 165 234 in S7 are conjugate by the permutation 1456273 1234567 lt1652347gtlt1376524gt2374639 7T1456277T 1 23746145627 26473 165234 There are many other possible choice of 7T for example by writing the orbits in different orders De nition 165 An uutomorphism Aut G of a group G is an isomorphism of G with itself The irmer automorphisms Inn G of G are the conjugations lnnG Cg G a Gwhere Cgh ghg l Proposition 166 Both the set of automorphisms Aut G and the irmer automorphisms Inn G form groups under composition Proof We do Aut G first Closure Composition of isomorphisms is an isomorphism this is since the composition of two bi jection is a bijection and the composition of two homomorphisms is also a homomorphism For the latter let 4 1p G gt G be homomorphisms then 47 0 1Pgh WWW g h g h 47 0 1Pg47 0 1PM WM 6 G 42 Associativity Aut G consists of functions under composition this is associative by Proposition 25 Identity Id g gt gt g is clearly an automorphism and moreover 1p 0 Id 1p Inverse If 4 E Aut G then 41 1 is a well defined function since 4 is a bijection Moreover the inverse of an isomorphism is also an isomorphism 4f1lt4gtltggt4gtltmgt 4f1lt4gtltghgtgt gh 4f1lt4gtltggtgt4f1lt4gtlthgtgt thus 41 1 E AutG For Inn G note that each Cg is a homomorphism MUM Mbgb l l abuzwa1 cug It also has inverse U C il since C 71C g u lugu lu g Thus each Cg is an automorphism of G and moreover Inn G is closed under composition and inverses Inn G is therefore a subgroup of Aut G 1 In fact Inn G is a subgroup of Aut G Instead of checking this explicitly we get it for free from the next Theorem Theorem 167 Inn G lt1 Aut G Proof Let T E Aut G Cg E Inn G By Proposition 157 it is enough to see that T o Cg o T 1 E Inn G For this let 1 E G then T 0 Cg 0 TAM TCgT 1h T8T 1hg 1 TgTT 1hTg 1 TghTg 1 since T is an homomorphism W901 1 Therefore T o Cg o T CTg which is in Inn G since Tg E G Example Since conjugation by an element g E G is an automorphism of G conjugation must map subgroups of G to isomorphic subgroups For example let H p0 ul g D3 Then le pinfl papuma par2 Both H and Cle are isomorphic to 22 We say that H Cle are conjugate subgroups The center of a group When a group is non Abelian there are several notions of what constitutes a maximally sized Abelian subset One of these is the center of a group De nition 168 The center of a group G is the subgroup ZGgEGguugVuEGgG The center is the set of elements of G which commute with everything in G As such the center is certamly Abelian even though we ve not shown it is a group yet this will be remedied in Proposi tion 169 Moreover Z G G ltgt G is Abelian itself Examples 1 ZDg e 2 In general ZDZnH e and ZDZn ep2 3 ZGLnlR Aln A E RX 4 ZOnlR iln It is often very difficult to calculate the centers of groups as there are a lot of calculations to check An example argument for the above example 3 runs as follows Let A E GL1 lR be a matrix with only one eigenvector v such A exist just extend v to a basis vvz vn then in this basis A having all entries on or above the leading diagonal equal to 1 and the rest 0 is such a matrix and let Z E ZGLnlR Then AZv ZAv AZv so that Zv is an eigenvector of A and so Zv is parallel to v In particular Zv must be parallel to v for every possible choice of v The only possible such matrices are multiples of the identity Proposition 169 ZG 4 G Le the center ofa group is a normal subgroup in particular it is a subgroup Proof There are two methods at our disposal to prove this Theorem firstly one can check the indi vidual group axioms to see that Z G is a subgroup of G then it is easy to see that its left and right cosets are identical gZG ZGg for any g E G since ZG commutes with everything in G Instead we will prove the Proposition by exhibiting Z G as the kernel of a homomorphism 4 We have already done most of the work Define 4 G gt Inn G by 41g Cg Thus each element 41g is the function conjugation by g The argument given in Proposition 166 shows that 4 is a homomor phism Moreover 4 has kernel ZG since ker39ygEGCgldgE GZCgIi IiVIiEG but CgIi Ii ltgt ing 1 Ii ltgt in Iig Proposition 157 tells us that kerzp ZG 4 G 17 Factor groups As the name suggests factor groups indicate a type of division in group theory The problem with dividing say g by IL is that we don t know whether to multiply by Ii on the left or the right in an Abelian group we have no problem but in general we re stuck It turns out that a sensible notion of division exists when we have normal subgroups we don t divide individual elements we work with their cosets De nition 17 Let H 4 G be normal The cosets16 of H form the factor group written GHgHgE G and where the group operation is defined by aH bH abH 16Since H is normal it is unambiguous whether we mean left or right We will tend to write left cosets 44 Ignore for the moment the problem of whether the group operation is well defined and let us check the group axioms Closure H bH abH E GH is acoset Associativity H bH CH H bCH obcH Similarly H bH CH obCH By the associatiVity of G these are identical Identity 8H H eoH H therefore 8H H is the identity Inverse o lH H o loH 8H H therefore L1H 1 o lH Why do we need H normal In the definition of a factor group we made a definition of the group operation but the definition is not independent of choices H th for any 111 E H thus the multiplication is well defined iff h1H l1th H bH for all hhhz E H ml 6 G Lemma 172 The group operation on G H is wellede 39ned if H 4 G Proof We must see that H bH th bth According to the definition this is iff obH ohlbth ltgt abrlahlbhz e H ltgt b lo lohlb e H 4 14111 6 H V111 e H b e G But this is the definition of H being a normal subgroup of G I Examples 1 mZ is a subgroup of Z and since Z is Abelian mZ is normal Write the cosets of mZ as ZmZ 0 m1 mm71 Addition in Z mZ is then H 7 0 7 i H ltmgt But this is simply addition modulo m Indeed j gt gt mod m is an isomorphism Z mZ E Zm N We can do a similar thing with 271 4 lR Then lR 271 E 0271 2 Recall example 3 of section 3 that this is isomorphic to the circle S1 lt CX U It can be seen that the Klein 4 group V is a normal subgroup of A4 Actually A4 V E Z3 g Writing the cosets of Z4 E 5 in Z20 we get Z2024 0 lt5gt11 lt5gt12 lt5gt13 lt5gtI4 lt5gt1 which is isomorphic to Z5 01 We saw before that G1 4 G1 gtlt G2 Thus G1 gtlt G2G1 is a factor group The map g1g2G1 X 82 gt gt g2 is an isomorphism G1 gtlt G2G1 E G2 45 6 Recall Propositions 166 and 169 where we saw that ZG 4 G by viewing ZG as the kernel of a homomorphism 4 G gt Inn G Thus GZG is a group WIn fact we will see in the next Theorem that GZG E Inn G ie G ker 4 E lmzp 7 We also saw that Inn G is a normal subgroup of Aut G The outer automorphism group of a group G is defined to be the factor group Aut G Inn G Remark lNarningl It may be tempting to take the idea of division of groups too far For example suppose we are given that H 4 G Since we have the concept of direct product it might be reasonable to guess that H X G H E G That life isn t as simple as this is clear from the following example ZS Zz E Z4 but 22 X ZS Zz E 22 X Z4 S Zg Compare this with example 5 above G1 gtlt G2 N G g T1 E G2 is true but H X H E G is false We have seen that all kernels of group homomorphisms are normal subgroups In fact all normal subgroups are the kernel of some homomorphism We state this as a 2 part Theorem Theorem 173 1st lsomorphism Theorem a Let H 4 G Then 7 G a GH de ned by 7g gH is u homomorphism with ker 39y H b Ifq G a G is u homomorphism with kernel H therl u GH 4 Im 41 de ned by ugH 41g is or isomorphism The Theorem can be summarised by the following formula G ker 4 E Im 11 Alternatively we say that the following diagram commutes ie 4 u o 39y GLMG lmzp g G 7 GH De nition 174 Given a normal subgroup H 4 G the homomorphism 39y G gt G H defined above is the canonical homomorphism Proof of Theorem We need to check that the two functions 39y and u defined in the Theorem have the properties we claim a Let H 4 G and define 39y G gt G H by 7g gH This is certainly a well defined function so we need only check that it is a homomorphism with ker 39y H But rg1rg2 ng ng g1g2H glgz by the definition of multiplication in a factor group Thus 39y is indeed a homomorphism More over 39yg gH H ltgt g E H thus the kernel of 39y is H as claimed b Suppose 4 G gt G is a homomorphism with kernel H then H is a normal subgroup of G and hence the factor group G H is well defined We need to check that u G H gt lm 4 defined by ugH 1g is an isomorphism Wellde ned There are choices in the definition of y gH th for any h E H However yth 1gh zpgzph 41g since 4 is ahomomorphism and h E H kerzp Homomorphism ngi4ng 4102047022 41glg2 glng thus l4 is a homomor phism 11 ng ng 3 41g1 41g2 3 gz 1g1 8c and SO gilgl E kerrp H In particular ng ng thus y is 1 1 Onto Every element of lm 4 has the form 1g for some g E G But 41g ygH so y is onto y is a well defined bijective homomorphism and thus an isomorphism y G H E Im 11 Here are several examples where we can calculate all the pieces in the Theorem Examples 1 Let 4 Z10 gt Z20 be the homomorphism with 411 4 Then by the homomor phism property 4n mod 20 In particular kerzp n E Z10 2471 E 0 mod 20 05 and lmzp 41Z10 4n mod 20 n E Z10 0481216 Observe that ker 4 5 lt Z10 is isomorphic to Z2 under addition modulo 10 and 41Z10 4 lt Z20 is isomorphic to Z5 under addition modulo 20 The cosets of ker 4 are as follows Z10kerzp kerzp16273849 The function 05 0 16 4 y Z10kerzp gt 41Z10 27 gt gt 8 38 12 49 16 is the isomorphism from the Theorem N Classify the factor group Z4 gtlt Z8 01 in terms of the Fundamental Theorem 141 of finitely generated Abelian groups le the factor group is a finite Abelian group so we must be able to write it as a direct product of Zi s The cyclic subgroup 0 1 has order 8 and its cosets are Z4 X 200011 00 lt01gt1110 lt01gt1210 lt01gt1310 01gt There are no further cosets either observe that a non zero entry in the Z8 part of i can be subtracted away by an element in the cyclic group 01 or count elements via the index of 01 there are Z4 gtlt Z8 01 488 4 cosets It is clear that the operation 1300011 j10lt01gt ij mod 40 0011 makes the factor group into a copy of Z4 Thus y ik 01 gt gt i mod 4 is an iso morphism Z4 gtlt Z8 01 E Z4 lfzp Z4 gtlt Z8 gt Z4 is the map zpik i then 4 is a homomorphism and y is the corresponding isomorphism in the 1St isomorphism Theorem | 677.169 | 1 |
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Mathematica by Example
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Mathematica by Example, 4e is designed to introduce the Mathematica programming language to a wide audience. This is the ideal text for all scientific students, researchers, and programmers wishing to learn or deepen their understanding of Mathematica. The program is used to help professionals, researchers, scientists, students and instructors solve complex problems in a variety of fields, including biology, physics, and engineering.
Read also publishers, and is now, after many years, back in print. This is a new and freshly published…
- Clear organization, complete topic coverage, and accessible exposition for novices - Fully compatible with Mathematica 6.0 - New applications, exercises and examples from a variety of fields including biology, physics and engineering - Includes a CD-ROM with all Mathematica input appearing in the book, useful to students so they do not have to type in code and commands
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Potty training a child can make even the most patient parents a bit anxious. Children react to the potty in many different ways, and it's important to know why. Before you go crazy trying to get your child to use the toilet, you need to understand…
This edition of Eric Flint's Grantville Gazette Volume 25 is derived directly from the on-line edition.It is different than the editions provided by Baen and webscriptions, has somewhat different content and different formatting.As always with…
About AbellMartha L.
However, our own repository connected with publications not really discovered details about the writer AbellMartha L.. Nevertheless our team is always spending so much time to get along with create brand-new facts. Knowing the data who am I?, you can include it from the variety to add a review. | 677.169 | 1 |
302190
ISBN-13: 9781615302192
DDC: 512
Grade Level Range: 9th Grade
- 12th Grade
288 Pages | eBook
Original Copyright 2011 | Published/Released April588320291540585293655084958686390819
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Calculating and manipulating the unknown has been the enterprise of the field of algebra since its earliest inception in Babylon and ancient Egypt. Trigonometry draws on principles presented in algebra and uses angle measurements to elaborate on geometric calculations. Essential to further mathematical and scientific study, both algebra and trigonometry provide crucial tools in managing variables and understanding the relationships between them. This volume presents the fundamentals of these fascinating areas of mathematics while chronicling their respective histories. | 677.169 | 1 |
The third of three volumes of a mathematics training course for Navy personnel, this text emphasizes topics needed in understanding digital computers and computer programing. The text begins with sequences and series, induction and the binomial theorem, and continues with two chapters on statistics. Arithmetic operations in number systems other than the decimal system are covered. Set theory and Venn diagrams lead into Boolean algebra; the text concludes with a chapter on matrices and determinants. Related documents are SE 014 115 and SE 014 116. (JM) | 677.169 | 1 |
Algebra 2
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Designed for use with overhead projectors, this kit helps you demonstrate concepts in a using probability tools, including algebra tiles, tangrams, a geoboard with rubber bands, a spinner, and pattern blocks.
Product Name: Overhead Manipulatives Kit
Invoice Title: PH MATH OVERHEAD MANIP KIT MAN N/A 2004
ISBN-10: 0131158465
ISBN-13: 9780131158467Spanish Practice Workbook
9780131658394
$5.47
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Provides extra practice for every lesson.
Product Name: Spanish Practice Workbook
Invoice Title: ADV ALGEB ALG 2 SPANISH WORKBOOK SSG 07C
ISBN-10: 0131658395
ISBN-13: 9780131658394
Spanish Vocabulary and Study Skills Workbook
9780131658400
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Focus on key mathematical vocabulary and specific support on study skills so every student can be successful.
Product Name: Spanish Vocabulary and Study Skills Workbook
Invoice Title: ADV ALGEB ALG 2 SPAN VOCAB WKBK SSG 07C
ISBN-10: 0131658409
ISBN-13: 9780131658400
Spanish Workbook Answer Key†
9780131910096
$11.47
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Product Name: Spanish Workbook Answer Key
Invoice Title: ADV ALGEB ALG2 SPAN WKBKS ANSKEY TMK 07C
ISBN-10: 0131910094
ISBN-13: 9780131910096
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† - Items can only be purchased with PO or school credit card.
Spanish Assessment Resources †
9780131658882
$38.97
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Spanish translations of all the Checkpoint Quizzes, Chapter Tests, and Cumulative Assessments. Also includes Alternative Assessment in Spanish.
Quarter, Mid-course, and Final Tests—regular and adapted for special needs
Test-Taking Strategy Practice
Standardized Practice Tests
Comprehensive Report Forms
Product Name: Progress Monitoring Assessments
Invoice Title: ADV ALGEB ALG 2 PMA (PROGMONAS) BLM 07C
ISBN-10: 0132014661
ISBN-13: 9780132014663
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High School Math Skills Review and Practice†
9780131657496
$42.97
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A comprehensive, easy-to-use resource for providing remediation, one skill at a time, for students needing another opportunity for review and practice.
Product Name: High School Math Skills Review and Practice
Invoice Title: ALGEBRA I HS SKILLS&CONCEPTS REV BLM 07C
ISBN-10: 0131657496
ISBN-13: 9780131657496
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† - Items can only be purchased with PO or school credit card.
Test Prep Workbook
9780132503426
$4.47
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Prepare your students for high-stakes with complete practice tests and test-taking strategies for national, ACT, and SAT tests.
Product Name: Test Prep Workbook
Invoice Title: ADV ALGEB ALGEBRA 2 TEST PREP SSG 2007C
ISBN-10: 0132503425
ISBN-13: 9780132503426
Test-Taking Strategies with Transparencies†
9780131657489
$51.47
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Product Name: Test-Taking Strategies with Transparencies
Invoice Title: ALGEBRA I TEST TAKING STRAT W/TR TRP 07C
ISBN-10: 0131657488
ISBN-13: 9780131657489
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Classroom Aid Transparencies†
9780131658714
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ISBN-10: 0131658719
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Additional Examples on Transparencies†
9780131658707
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Invoice Title: ADV ALGEB A2 ADD'L EXAMPLES ON T TRP 07C
ISBN-10: 0131658700
ISBN-13: 9780131658707
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Daily Skills Check and Lesson Quiz Transparencies†
9780131658653
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Product Name: Daily Skills Check and Lesson Quiz Transparencies
Invoice Title: ADV ALGEB A2 DAILYSKILLS/QUIZ TR TRP 07C
ISBN-10: 0131658654
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Student Edition Answers on Transparencies†
9780131658691
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ISBN-10: 0131658697
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Algebra 2 Teacher Center (2009)†
9780133659788
$411.47
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Your Prentice Hall Teacher Center is the one resource you need for complete support for planning your instruction with TeacherEXPRESS™ Powered by LessonView® planning software, to digital presentations of your lessons found on the PresentationEXPRESS™ CD ROM, to a fun way to assess and review with MindPoint™ Quiz Show. Only Prentice Hall provides all this support in one convenient location.
Product Name: Algebra 2 Teacher Center (2009)
Invoice Title: ALGEBRA 2 TEACHER CENTER CD-PACK
ISBN-10: 013365978X
ISBN-13: 9780133659788
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ExamView® 6.1 (2007/2009) †
9780133659924
$142.47
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The most powerful test generator available with the most comprehensive test banks.
QuickTest Wizard t build assessments in seconds
All test item translated into Spanish
Support for modifying tests quickly and easily
Math Art Gallery to import images into your assessments
New Worksheet Banks allow you to edit and modify existing worksheets
Product Name: ExamView® 6.1 (2007/2009)
Invoice Title: ALGEBRA 2 EXAMVIEW CD
ISBN-10: 0133659925
ISBN-13: 9780133659924
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Success Tracker™ 6 year online access
9780138901523
$22.97
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SuccessTracker™ Personalized intervention for each student.
Assess—Students take Diagnostic Tests or Benchmark Tests online.
Diagnose—Based on assessment results, each student automatically recieves an assignment for any skills that have not been mastered.
Notes
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Instant Check System for TI-Navigator CD-ROM†
9780132504607
$273.47
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This product was developed for use with the TI-Navigator System. The CD conatins built-in assessment from your Prentice Hall textbook are provided in a format that can be used with the Navigator System.
Product Name: Instant Check System for TI-Navigator CD-ROM
Invoice Title: ALG1/GEOM/ALG2 INSTANT CK SYS FOR TI-NAV
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Bundle with AGS Mathematics
9780133662856
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Product Name: Bundle with AGS Mathematics
Invoice Title: PEARSON ALGEBRA 2 SE BUNDLE
ISBN-10: 0133662853
ISBN-13: 9780133662856
With text purchase, add All-in-One Student Workbook
9780133660104
$4.97
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This resource provides all the support your students need—in one convenient workbook! The following resources are included:
Daily Notetaking Guide—Provides the structure your students need to take effective notes as you teach. Student notes make a great study guide for quizzes and tests.
Practice—Provides additional practice for every lesson.
Guided Problem Solving—Helps students master word problems by providing step-by-step guidance for a selected problem from every lesson.
Vocabulary & Study Skills—Focus on key mathematical vocabulary and specific support on study skills so every student can be successful.
Notes | 677.169 | 1 |
Calculators: Printing and Display
4.11 - 1251 ratings - Source
This text provides an easy-to-use approach to learning 10-key operation based on current business procedures. Students learn the touch method and solve both business and personal math problems. Step-by-step instructions make learning easy; pictures and explanations for reaches provide good visual cues; and sufficient drills develop workplace skills. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.1 T9915 STEEL-BELTED RADIAL TIRE 82 1996 RO STATE INSPECTION 14 50
7 50 a. b. c. d. 10 00 ADJUST LEFT ... Amount FORD F150 Make and Model Year
T2A-994 License Mileage Name Address Date Phone AUTO CARE aamp; REPAIRanbsp;...
Title
:
Calculators: Printing and Display
Author
:
William R. Pasewark
Publisher
:
Cengage Learning - 2011-03-16
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Algebra and Trigonometry: An Early Functions Approach (2nd
If at the top of the tower shown in Fig. 35, a telescope were to be rotated from the horizontal until it pointed to an object at 0, the angle so formed is called the angle of depression. When the symbol is the only one on the key you simply press that key. Remember that the exponent tells how many times the base is multiplied by itself. [/frame]Subscripts – Your variable goes outside the bracket and the subscript goes inside.
Pages: 1216
Publisher: Pearson; 2 edition (February 26, 2009)
ISBN: 0321588703
A Treatise On Trigonometry, and On Trigonometrical Tables and Logarithms: Together with a Selection of Problems and Their Solutions
ALGEBRA 2 WITH TRIGONOMETRY (PRACTICE MASTERS)
A Treatise on Plane Trigonometry, Containing an Account of Hyperbolic Functions: With Numerous Examples, Vol. 1 of 2 (Classic Reprint)
Elements of Geometry: Containing the First six Books of Euclid : With a Supplement on the Quadrature of the Circle, and the Geometry of Solids : to ... Elements of Plane and Spherical Trigonometry
That makes sense, until you stop and actually count the school subjects you used recently. As a regular Joe, when was the last time you used World History? The fact that many people say this about mathematics, but exclusively about mathematics (and also physics) even though one can theoretically say that about most, if not all school subjects, is illuminating Algebra and Trigonometry: An Early Functions Approach (2nd Edition) online. The Trigonometrical Ratios 59 triangle with the usual construction (Fig. 51) as follows: In continuation of section 48 we note that: or the tangent of an angle is equal to the cotangent of its complement pdf. Regiomontanus had become a pupil of Peuerbach at the University of Vienna in 1450. Later, he undertook with Peuerbach to correct the errors found in the Alfonsine Tables Plane and Spherical Trigonometry. [With] Solutions of Problems. [Followed By] Appendix: Being the Solutions of Problems.
Essentials Of Plane Trigonometry And Analytic Geometry
Exercises in Algebra
The Egyptians divided the 360 degrees of the ecliptic into 36 sections of 10 degrees each. [see Note 1 below] Baron Von Vega's LOGARITHMIC TABLES OF NUMBERS and TRIGONOMETRICAL FUNCTIONS.. One mathematical cat factoring polynomials of the form worksheet answers, make square root a fraction, free printable worksheet on volume of cuboids, college college algebra solver programs, solving 2nd order differential equation Algebra and Trigonometry (4th Edition). This memo focuses on math course enrollment patterns throughout high school by following the 2013-14 twelfth grade cohort. Trigonometry is an important, fundamental step in math education. From the seemingly simple shape, the right triangle, we gain tools and insight that help us in further practical as well as theoretical endeavors Plane and Spherical Trigonometry, second edition. Standards set high expectations for all students to ensure that Tennessee graduates are prepared for the rigorous demands of mathematical understanding for college and career. The Tennessee State Board of Education adopted new state standards in 2010 for mathematics for grades K-8 and the first three high school courses (Algebra 1, Geometry, Algebra 2, and Integrated Math 1, 2, and 3), and in 2014 adopted new standards for the further high school mathematics courses MyMathLab for Trigsted Trigonometry -- Access Card. I'll shorten the names from "hypotenuse", "adjacent", and "opposite" to "hyp", "adj", and "opp": The ratios in the left-hand table are the "regular" trig ratios; the ones in the right-hand table are their reciprocals (that is, the inverted fractions) Mathematics for Self Study (Trigonometry for the Practical Man). S. pennies, nickels, dimes, and quarters. You can play in four modes: (1) Match the Value: Try to make the same value as the coins shown using a different combination of coins. (2) Make the Total: Use coins to make the total shown. (3) How Many Cents?: Type the value of the coins shown. (4) Show Values: Drag any amount of coins and show or hide the value download Algebra and Trigonometry: An Early Functions Approach (2nd Edition) pdf.
Trigonometry
Elements of Geometry: Containing the First Six Books of Euclid : With a Supplement On the Quadrature of the Circle, and the Geometry of Solids : To ... Elements of Plane and Spherical Geometry
From Theory of accounts (International Library of Technology). The Cliff's Quick Review Guides are wonderful to have in my "back pocket" when I need to quickly look something up that is covered in dust in the "archives of my brain." I'm a (returning :P) university Freshman preparing for the College Board CLEP tests. I was already familiar with the material covered in this book, but needed to refresh my memory Trigonometry (07) by Young, Cynthia Y [Hardcover (2006)]. Then bring several precalculus skills together by using a decaying exponential term in a sine function to model damped harmonic motion. x For a given trigonometric function, only a small part of its graph qualifies as an inverse function as defined in Lecture 5 Algebra & Trigonometry Second Custom Edition for Tidewater Community College. During this activity, the students will review trigonmetry and various mathematics that are considered prerequisites for the Physics 1500 and Physics 1600 series of courses Student solutions manual to accompany Algebra with trigonometry for college students, third edition [by] Charles P. McKeague. Another brief introduction at the sophomore level with some emphasis on logic and Boolean algebra. Does Algebra;: Structure and method, book 1. The student is expected to: (A) describe and perform transformations of figures in a plane using coordinate notation; (B) determine the image or pre-image of a given two-dimensional figure under a composition of rigid transformations, a composition of non-rigid transformations, and a composition of both, including dilations where the center can be any point in the plane; (C) identify the sequence of transformations that will carry a given pre-image onto an image on and off the coordinate plane; and (D) identify and distinguish between reflectional and rotational symmetry in a plane figure. (4) Logical argument and constructions Student Solutions Manual Algebra and Trigonometry (3rd Edition). Plane trigonometry with practical applications, by Leonard E. Dickson.. Natural Trigonometric Functions,. Among these were the works of Euclid, Archimedes Apollonius and of course, Ptolemy. The Arabs now had two competing versions of astronomy, and soon the Almagest prevailed A Treatise On Surveying: Containing the Theory and Practice : To Which Is Prefixed a Perspicuous System of Plane Trigonometry : The Whole Clearly ... Particularlry Adapted to the Use of S. It should be noted in Fig. 61 that From which it is evident that if we draw a straight line at right angles to PO, the projection of AB upon such a straight line is AB sin e. General questions on the trigonometrical ratios. In a right-angled triangle the two sides containing the right angle are 2.34 m and 1.64 m. Find the angles and the hypotenuse. 2 In a triangle ABC, C being a right angle, AC is 122 ern, AB is 175 cm Trigonometry, plane and spherical. | 677.169 | 1 |
For
The
Special Features:
Children's Strategies and samples of student work are provided in each chapter for teacher analysis.
Activities for practicing the thinking strategies explained in each chapter, designed to be solved without rules or algorithms, using reasoning alone.
Online Resources for Instructors including a sample syllabus, chapter overviews and relevant Big Content ideas for each chapter, and graphics that can be copied and resized for use in instruction and in constructing exams.
An Routledge 2011-118859
Descripción Paperback. Estado de conservación: New. Not Signed; begin's book. Nº de ref. de la librería ria9780415886123_rkm886123
Descripción Routledge. Estado de conservación: New. Pushes readers beyond the limits of their current understanding of rational numbers, challenging them to refine and explain their thinking - without falling back on rules and procedures they have relied on throughout their lives. This book is suitable for researchers and curriculum developers in mathematics education. Num Pages: 160 pages, 300 black & white line drawings. BIC Classification: JN. Category: (UP) Postgraduate, Research & Scholarly. Dimension: 180 x 254 x 14. Weight in Grams: 510. . 2011. 3rd Edition. Paperback. . . . . Books ship from the US and Ireland. Nº de ref. de la librería V9780415886123 | 677.169 | 1 |
It was a basic math course during the summer that you can take as a freshman. It helps refresh your memory of all basic math skills and help you use them for the harder material that comes later.
Course highlights:
The highlight of this course were the real life skills taught. For example, there was a chapter on fiances, loans, insurance and many other real life skills you need that tied into math. It is always helpful to be exposed to material that you will carry with you through life.
Hours per week:
9-11 hours
Advice for students:
I would advise students to take this course as soon as possible so they can avoid losing the information they are bringing fresh from high school. Also, once in the course dont feel like its too hard or you cant remember the material so give up. Utilize your professor, their office hours, peers, and tutoring labs.
I would recommend this course because it helps you set the foundation on important credits you'll need in order to graduate with your class.
Course highlights:
The highlights of this course were the instructors enthusiastic teaching style. The instructor kept me amused which kept me interested in learning more about this course. I learned that math isn't as hard if you have the correct teaching and break down of it.
Hours per week:
0-2 hours
Advice for students:
The course is difficult but Mike is an interesting person with an interesting way of teaching. So my advice to students would be to get to know him personally and understand the type of person he is. That may help in how you receive his lessons. | 677.169 | 1 |
Basic Engineering Mathematics
4.11 - 1251 ratings - Source
Introductory mathematics written specifically for students new to engineering Now in its sixth edition, Basic Engineering Mathematics is an established textbook that has helped thousands of students to succeed in their exams. John that readers can relate theory to practice. The extensive and thorough topic coverage makes this an ideal text for introductory level engineering courses. This title is supported by a companion website with resources for both students and lecturers, including lists of essential formulae, multiple choice tests, full solutions for all 1, 600 further questions contained within the practice exercises, and biographical information on the 25 famous mathematicians and engineers referenced throughout the book. The companion website for this title can be accessed from it is important to understand: Graphical solution of equations It has been
established in previous chapters that the solution of linear, quadratic,
simultaneous and cubic equations occur often in engineering and science and
may be solvedanbsp;...
Title
:
Basic Engineering Mathematics
Author
:
John Bird
Publisher
:
Routledge - 2014-03-26
ISBN-13
:
Continue
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About
Overview
Authors Mark Clark and Cynthia Anfinson have developed several key ideas to make concepts real and vivid for students.
First, the authors place an emphasis on developing strong algebra skills that support the applications, enhancing student comprehension and developing their problem solving abilities.
Second
Third, important concepts are developed as students progress through the course and overlapping elementary and intermediate content in kept to a minimum. Chapter 8 sets the stage for the intermediate material where students explore the "eyeball best-fit" approach to modeling and understand the importance of graphs and graphing including graphing by hand.
Fourth, Mark and Cynthia's approach prepares students for a range of courses including college algebra and statistics.
In short, BEGINNING AND INTERMEDIATE ALGEBRA: CONNECTING CONCEPTS THROUGH APPLICATIONS develops strong mathematical skills using an engaging, application-driven and problem solving-focused approach to algebra.
Additional Product Information
Features and Benefits
Prealgebra Review. The text begins in Chapter R by reviewing some prealgebra concepts, providing students with a review of those topics most necessary for beginning algebra such as operations with integers, operations with fractions, operations with decimals and percents and the real number system.
An Innovative Critical-Thinking Feature: Concept Investigations. These directed-discovery activities called Concept Investigations are ideal as group work during class, incorporated as part of a lecture or as individual assignments to investigate concepts further. Inserted at key points within the chapter, each Concept Investigation helps students explore patterns and relationships such as the graphical and algebraic representations of the concepts being studied.
Worked Examples That Reinforce Problem Solving. This text provides a broad range of examples that give students more practical experience with mathematics. All examples are immediately followed by a practice problem that is similar to that example and all examples are labeled to help students connect the example and concept being studied. Examples of hand-drawn graphs in the intermediate section also help students visualize what their own work should look like.
Integrated "Student" Work. Clearly identifiable examples of "student" work appear throughout the Exercises in the text. These boxes ask students to find and correct common errors in "student" work.
An Eyeball Best-Fit Approach in the intermediate section of the text. Modeling is introduced in the intermediate part of the book as students now have the beginning algebra skills to handle "messy data" at this point in the course. Linear, quadratic and exponential functions are analyzed through modeling data using an eyeball best-fit approach. These models investigate questions in the context of real-life situations. Creating models by hand leads students to analyze more carefully the parts of each function reinforcing solving techniques and making a better connection to the real-life data and how they affect the attributes of the function's graph. Graphing calculators are used to plot data and check the fit of each model.
Margin Notes. The margin contains three kinds of notes written to help the student with specific types of information:
Reinforcement of Visual Learning through Graphs and Tables. Graphs and tables are used throughout the book to organize data, examine trends, and have students gain knowledge of graphing linear and quadratic equations. The graphical and numeric approach helps support visual learners, incorporating realistic situations into the text and reinforcing the graphs and data that students see in their daily lives.
Exercise Sets. The exercise sets include a balance of both applications and skill-based problems developed with a clear level of progression in terms of difficulty level. Some exercise sets begin with a few warm-up problems before focusing on applications. Exercise sets typically end with additional skill practice to help students master the concepts when needed. A balance of graphical, numerical, and algebraic skill problems is included throughout the book to help students see mathematics from several different views.
Flexible Use of the Calculator: Calculator Details. The calculator emphasis across the beginning and intermediate portions of the text differ in keeping with the goals of each course. Calculator Details margin boxes will appear as necessary to instruct students on the correct use of a scientific calculator in the beginning portion of the text though most exercises do not require its use. These boxes also appear in the intermediate portion of the text and include tips on graphing calculators. Modeling is introduced in chapter 8 and though students are required to solve problems, graph and do other algebraic skills by hand, calculators are used to create scatterplots of real data and to check the reasonableness of algebraic models for the data. They are used to check solutions both graphically and numerically and to do some numerical calculations.
Extensive End-of-Chapter Material includes Chapter Summaries, Review Exercises, Chapter Tests, Chapter Projects, Cumulative Reviews and Equation Solving Toolboxes. Chapter Summaries revisit the big ideas of the chapter and reinforce them with new worked out examples. Students can also review and practice what they have learned with the Chapter Review exercises before taking the Chapter Test.
Chapter Projects. To enhance critical thinking, end-of-chapter projects can be assigned either individually or as group work. Instructors can choose which projects best suit the focus of their class and give their students the chance to show how well they can tie together the concepts they have learned in that chapter. Some of these projects include on-line research or activities that students must perform to analyze data and make conclusions.
Cumulative Reviews. Cumulative Reviews group together the major topics across chapters. Answers to all the exercises are available to students in the answer appendix.
Equation Solving Toolbox. The equation-solving toolbox covers the processes for finding models as the models are introduced.
Practical Help for Instructors. Practical tips are provided in the Annotated Instructor's Edition on how to approach and pace chapters as well as integrate features such as Concept Investigations. In addition, for every student example in the student text, there is a different instructor classroom example in the AIE, with accompanying answers that can be used for additional in-class practice and/or homework.
Appendixes. Several useful appendixes are included—on Matrices and Using the Graphing Calculator, and Answers to Practice Problems and Selected Exercises.
Instructor Supplements
This CD-ROM provides the instructor with dynamic media tools for teaching. Create, deliver, and customize tests (both print and online) in minutes with ExamView Computerized Testing featuring algorithmic equations. Microsoft Powerpoint lecture slides, figures, calculator practice, and additional new examples from the annotated instructor's edition are also included on this CD-ROM.
Prepare for exams and succeed in your mathematics course with this comprehensive solutions manual! Featuring worked outPrepare for exams and succeed in your mathematics course with this comprehensive solutions manual! Featuring worked outMark ClarkCynthia Anfinson | 677.169 | 1 |
Algebraic Thinking Word Problems
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.29 MB
PRODUCT DESCRIPTION
Based on the old FCAT 2.0, these word problems require the student to read the questions and then choose the correct expression that will solve the problem. Contains 180 problems. If you wish, you can also have the students solve the problems. Answer key and student answer sheet are included | 677.169 | 1 |
Chapter 1 - Solving One-Variable Equations (Utah) - 8th Grade
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files.
6.76 MB
PRODUCT DESCRIPTION
This unit corresponds with the 8th grade OER (Open Educational Resource) created by Utah Middle School Math Project ( This unit includes animated PowerPoints that demonstrate how Algebra tiles work when representing algebraic expressions and solving algebraic equations. These PowerPoints create clear connections between the concrete model (algebra tiles) and the algebraic process. This unit also includes two interactive foldable notes (Solving Equations and Types of Solutions), card sorts to make the practice more interesting and group oriented, a Jeopardy review game, learning target posters that can be displayed in the classroom, and word wall strips for relevant terms.
Please see the preview document for a specific day-by-day planning guide and description of what is included in this unit. This guide also gives you insight into the homework that can be assigned from the Utah pages | 677.169 | 1 |
How to Prepare for the OGT
4.11 - 1251 ratings - Source
An accredited Ohio high school math teacher who is also a math instructor at OhioA''s Youngstown State University offers Ohio high school students comprehensive preparation for the math exit exam. The test, which is taken in stages in the tenth-through-twelfth grades, is a prerequisite for high school graduation in Ohio. This brand-new manualA''s subject review opens with a chapter on number, number sense, and operations; then goes on to review patterns, functions, and algebra; geometry and spatial sense; measurement; and data analysis and probability. The manual concludes with three full-length sample Ohio Graduation Tests (OGTs) with solutions for all problems. Valuable appendices offer an OGT calculator primer, an OGT mathematics reference sheet, and a glossary of mathematical terminology. The book is an excellent learning tool for individual students or for use in classrooms, featuring more than 800 OGT problems, examples and exercises with an explanation accompanying the solution of every problem.The book is an excellent learning tool for individual students or for use in classrooms, featuring more than 800 OGT problems, examples and exercises with an explanation accompanying the solution of every problem. [back cover] BARRONa#39;S HOW ...
Title
:
How to Prepare for the OGT
Author
:
Tom Reardon
Publisher
:
Barron's Educational Series - 2005
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
After Calculus Analysis
This book is written for a course that serves as a gentle, transition from "cookbook" mathematics to the more rigorous material in post calculus courses. It acquaints students with the language and general methods of contemporary mathematics, within the context of "analysis". Using notation most commonly found in standard algebraic literature, it marks optional material that might be needed for reference or more advanced courses with an asterisk. Over 1200 problems are included and these range from drill exercises to problems that probe for deep understanding. | 677.169 | 1 |
"This is a rigorous and well-written treatment of differential forms with a careful and detailed progression from very basic notions."--MAA.org, 24-Sep-14
From the Publisher:
Differential forms are utilized as a mathematical technique to help students, researchers, and engineers analyze and interpret problems where abstract spaces and structures are concerned, and when questions of shape, size, and relative positions are involved. Differential Forms has gained high recognition in the mathematical and scientific community as a powerful computational tool in solving research problems and simplifying very abstract problems through mathematical analysis on a computer. Differential Forms, 2nd Edition, is a solid resource for students and professionals needing a solid general understanding of the mathematical theory and be able to apply that theory into practice. Useful applications are offered to investigate a wide range of problems such as engineers doing risk analysis, measuring computer output flow or testing complex systems. They can also be used to determine the physics in mechanical and/or structural design to ensure stability and structural integrity. The book offers many recent examples of computations and research applications across the fields of applied mathematics, engineering, and physics. * The only reference that provides a solid theoretical basis of how to develop and apply differential forms to real research problems* Includes computational methods for graphical results essential for math modeling * Presents common industry techniques in detail for a deeper understanding of mathematical applications* Introduces theoretical concepts in an accessible manner1402923944031-RMX Academic Press, 2014. Estado de conservación: New. Brand New, Unread Copy in Perfect Condition. A+ Customer Service! Summary: Put the theory into practice with this informative and accessible reference that teaches you how differential forms can be used as a powerful computational tool in analyzing and solving research problems. Nº de ref. de la librería ABE_book_new_0123944031 Elsevier Science Publishing Co Inc. Hardback. Estado de conservación: new. BRAND NEW, Differential Forms: Theory and Practice (2nd Revised edition), Steven Weintraub, B9780123944030 | 677.169 | 1 |
MATH 2414.S01 Advice
MATH 2414.S01 Documents
Showing 1 to 30 of 44
Student Quick Start Guide
This Quick Start Guide provides basic instructions for some of your most common tasks. For more information, see the Student Online Help at
Enrolling in Class
Your instructor can enroll yo
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
We now have several ways
of testing a series for convergence
or divergence.
The problem is to decide which test to use
on which series.
INFINITE SEQUENCES AND SERIES
In this respect, testing
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.6
Absolute Convergence
and the Ratio and Root tests
In this section, we will learn about:
Absolute convergence of a series
and tests to determine it.
ABSOLUTE CONVERGENCE
Given any series a
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
The convergence tests that we have
looked at so far apply only to series
with positive terms.
INFINITE SEQUENCES AND SERIES
11.5
Alternating Series
In this section, we will learn:
How to deal
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.4
The Comparison Tests
In this section, we will learn:
How to find the value of a series
by comparing it with a known series.
COMPARISON TESTS
In the comparison tests, the idea
is to compar
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
In general, it is difficult to find the exact
sum of a series.
We were able to accomplish this for geometric series
and the series 1/[n(n+1)].
This is because, in each of these cases, we can
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.2
Series
In this section, we will learn about:
Various types of series.
Series 1
SERIES
If we try to add the terms of an infinite
n n =1
sequence cfw_a
we get an expression
of the form
a1
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
Infinite sequences and series were
introduced briefly in A Preview of Calculus
in connection with Zenos paradoxes and
the decimal representation of numbers.
INFINITE SEQUENCES AND SERIES
Their
10
PARAMETRIC EQUATIONS
AND POLAR COORDINATES
AND
PARAMETRIC EQUATIONS & POLAR COORDINATES
10.4
Areas and Lengths
in Polar Coordinates
In this section, we will:
Develop the formula for the area of a region
whose boundary is given by a polar equation.
AREA
10
PARAMETRIC EQUATIONS
AND POLAR COORDINATES
AND
PARAMETRIC EQUATIONS & POLAR COORDINATES
A coordinate system represents
a point in the plane by an ordered
pair of numbers called coordinates.
PARAMETRIC EQUATIONS & POLAR COORDINATES
Usually, we use Carte
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.8
Power Series
In this section, we will learn about:
Power series and testing it
for convergence or divergence.
POWER SERIES
Equation 1
A power series is a series of the form
c x
n =0
n
n
=
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.9
Representations of
Functions as Power Series
In this section, we will learn:
How to represent certain functions as
sums of power series.
FUNCTIONS AS POWER SERIES
We can represent certain
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.11
Applications of
Taylor Polynomials
In this section, we will learn about:
Two types of applications of Taylor polynomials.
APPLICATIONS IN APPROXIMATING FUNCTIONS
First, we look at how th
11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
In section 11.9, we were able to
find power series representations for
a certain restricted class of functions.
INFINITE SEQUENCES AND SERIES
Here, we investigate more general
problems.
Which7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
As we have seen, integration is more
challenging than differentiation.
In finding the derivative of a function, it is obvious
which differentiation formula we should apply.
However, it may not be obv
7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
7.4
Integration of Rational Functions
by Partial Fractions
In this section, we will learn:
How to integrate rational functions
by reducing them to a sum of simpler fractions.
PARTIAL FRACTIONS
We show
7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
7.3
Trigonometric Substitution
In this section, we will learn about:
The various types of trigonometric substitutions.
TRIGONOMETRIC SUBSTITUTION
In finding the area of a circle or an ellipse,
2
2
an i
7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
7.2
Trigonometric Integrals
In this section, we will learn:
How to use trigonometric identities to integrate
certain combinations of trigonometric functions.
TRIGONOMETRIC INTEGRALS
We start with
power
7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
Due to the Fundamental Theorem of Calculus
(FTC), we can integrate a function if we know
an antiderivative, that is, an indefinite integral.
We summarize the most important integrals
we have learned s
6
APPLICATIONS OF INTEGRATION
APPLICATIONS
APPLICATIONS OF INTEGRATION
6.5
Average Value
of a Function
In this section, we will learn about:
Applying integration to find out
the average value of a function.
AVERAGE VALUE OF A FUNCTION
It is easy to calcul
6
APPLICATIONS OF INTEGRATION
APPLICATIONS
APPLICATIONS OF INTEGRATION
6.4
Work
In this section, we will learn about:
Applying integration to calculate the amount of
work done in performing a certain physical task.
WORK
The term work is used in everyday
l | 677.169 | 1 |
Sample records for advanced placement math
Math Sense consists of five books that develop from basic to more advancedmathThe equation is simple: No matter their background, students who take challenging math courses in high school get better jobs and earn more money throughout their entire lives. This paper stresses that: (1) Higher-level math opens doors for any and all postsecondary programs and keeps it open for advancement beyond entry-level jobs; and (2)…
This fact sheet explains that to thrive in today's world, all students will need to graduate with very strong math skills. That can only mean one thing: advancedmath courses are now essential math courses. Highlights of this paper include: (1) Advancedmath equals college success; (2) Advancedmath equals career opportunity; and (3) Advanced math…
This study was designed to examine benefits and challenges of teaching through videoconferencing in the context of students' field placement experiences, particularly as it relates to an inquiry-based approach to teaching and learning math and science. In the context of mathematics and science methods courses, preservice teachers, with the…
The purpose of this action research case study was to explore attitudes and beliefs regarding strategies to improve mathplacement scores for low-socioeconomic community college students at an urban campus. Using a pragmatic worldview, qualitative techniques were used to gather data to explore this problem at a single community college located in…
This study evaluates the effectiveness of mathplacement policies for entering community college students on these students' academic success in math. We estimate the impact of placement decisions by using a discrete-time survival model within a regression discontinuity framework. The primary conclusion that emerges is that initial placement in a scoring method and looks at…
Drawn from surveys completed by 122 students enrolled in developmental math at four community colleges and from seven student focus groups with a total of 34 developmental math students at those same colleges, this research brief illuminates student experiences with and perspectives on the math assessment and placement process. Findings suggest…
Community college students are often placed in developmental math courses based on the results of a single placement test. However, concerns about accurate placement have recently led states and colleges across the country to consider using other measures to inform placement decisions. While the relationships between college outcomes and such…
Given the Common Core State Standards for Mathematics, California's history of math acceleration in the middle grades, and the concern for correct math course placement for all students, this brief examines patterns from the past to shed light on considerations for the future. The brief, written by WestEd's Tony Fong and Neal Finkelstein, presentsChanging placement policy may help to improve developmental education student outcomes in community colleges, but there is little understanding of the impacts of these reforms. We take advantage of heterogeneous placement policy in a large urban community college district in California to compare the effects of math remediation under different…
Using data from the Educational Longitudinal Study of 2002-2006, the authors investigated the effects of advancedmath course taking on math achievement and college enrollment and how such effects varied by socioeconomic status and race/ethnicity. Results from propensity score matching and sensitivity analyses showed that advancedmath courseMiddle school mathplacement and progression are topics that are part of an active policy and practice discussion in California and elsewhere. Beginning in the 2008/09 school year, California's State Board of Education recommended that students complete algebra I by the end of grade 8. Between 2003 and 2009 the proportion of grade 8 students…
Minority and low-income students are less likely to have access to, enroll in and succeed in higher-level math courses in high school than their more advantaged peers. Under these circumstances, higher-level math courses function not as the intellectual and practical boost they should be, but as a filter that screens students out of the pathway to…
Using data from the Educational Longitudinal Study of 2002–2006 (ELS:02/06), this study investigated the effects of advancedmath course taking on math achievement and college enrollment and how such effects varied by socioeconomic status (SES) and race/ethnicity. Results from propensity score matching and sensitivity analyses showed that advancedmath course taking had positive effects on math achievement and college enrollment. Results also demonstrated that the effect of advancedmath forAs the federal government encourages all students to attempt advancedmathNo student who hopes to compete in today's rapidly evolving global economy and job market can afford to graduate from high school with weak mathematical skills, which include the ability to use logic, reason, and solve problems. The benefits associated with improving the math performance of American students also extend to the larger U.S. economy.Answers and justifications for an interesting problem on the AdvancedPlacement Calculus AB Examination are discussed. The problem provides diverse ways in which students can gain appreciation and understanding for the subject. (MNS) Fee call to action is based on a simple but important premise: The nation cannot allow college placement policies, processes, and instruments to undermine promising efforts to increase student success in mathematics and increase attainment of STEM credentials. Efforts to redesign math pathways hold great promise for improving the teaching and…
The Tennessee Department of Education explored course enrollment patterns in an effort to better understand in which courses students are enrolling and whether course enrollment policies and procedures are promoting students' interests. This report focuses on math course enrollment patterns throughout high school by following the 2013-14 twelfth…
This Strategic Action Plan for AdvancingMathEveryone knows that there is racial inequality in achievement returns from advancedmath; however, they do not know why black students and white students taking the same level of math courses are not leaving with the same or comparable skill levels. To find out, the author examines variation in course coverage by the racial composition of the…
Black and Latino students are still underepresented in upper-level math classes in the United States, a fact which has serious implications for their academic achievement and futures. Walker provides six suggestions for how educators can encourage more black and Latino students to successfully take higher level math courses: (1) Expand our…
This article describes the Jaime Escalante Math Program, a system that in 1989 helped an East Los Angeles high school set a record by administering over 450 AdvancedPlacement exams, having administered only 10 tests in 1978. The article is presented in three sections. The first section describes the program, discussing origins and backgrounds:A common argument against raising math course-taking requirements for all students is that it will cause more students to drop out of high school. But most students who drop out for academic reasons do so not because they are being "too challenged," but rather because they are not being challenged enough. It is important to raise the rigor and honors and AdvancedPlacement coursework for high-ability African American students with previously limited access to rigorous courses. The qualitative investigation explores practical solutions fromThis article reports on reactions to the U.S. Department of Education's first time decision to sit out an international study designed to show how advanced high school students around the world measure up in math and science. Mark S. Schneider, the commissioner of the department's National Center for Education Statistics, which normally takes the…
There is growing concern that the remedial math courses taken by most community college students unnecessarily divert some students from earning a degree. Anecdotes of students who thought they had completed their math requirements in high school only to have remedial courses delay their progress through college are common. In addition, researchMath anxiety is a reoccurring problem for many students, and the effects of this anxiety on college students are increasing. The purpose of this study was to examine the association between pre-enrollment math anxiety, standardized test scores, mathplacement scores, and academic success during freshman math coursework (i.e., pre-algebra, college…
Questions about how best to place students into appropriate middle grade math courses have been central to ongoing education policy and practice discussions in California and across the United States. Recent studies have shown that enrolling in algebra I in grade 8 works well for some students but backfires for others. This REL West report writing toAn increasing number of recent high school graduates' college mathplacement test results indicate that they are not ready for college level math. This study examined the perceptions of recent high school graduates enrolled in developmental math classes to determine their level of satisfaction with their mathplacement, the relationship of…
This publication contains materials used in the three phases of the reading and mathematics components of work-specific classes. Each section begins with an overview of developments in that phase. Section 1 focuses on Phase 1 during which math and reading were taught as separate components. It contains a mathplacement appraisal, worksheets and……"Informed math self-placement," a program implemented at American River College in Sacramento, California, to determine students' readiness for college-level math, has been in place for three years. This case study describes the development and implementation of math self-placement at American River. Math self-placement consists of a Web-basedwork and to develop a plan of action to close the achievement gap between White and non-White students. Prior to this study there was no clear, concise data to move this discussion…
While curriculum specialists and committees often decide how mathematics is taught, it is ultimately principals who influence the extent to which these initiatives are carried out. The overall goal of this article is to provide school leaders with classroom-based research that describes one way of improving the math skills of middle school…
How do we determine if incoming students are ready for college-level work? California's community college system is currently working to address this complex question in a more nuanced, comprehensive, and equitable way. This research brief offers insights that can inform these efforts resulting from the Student Transcript-Enhanced Placement Study…
The linear relations between math anxiety and math cognition have been frequently studied. However, the relations between anxiety and performance on complex cognitive tasks have been repeatedly demonstrated to follow a curvilinear fashion. In the current studies, we aimed to address the lack of attention given to the possibility of such complex interplay between emotion and cognition in the math-learning literature by exploring the relations among math anxiety, math motivation, and math cognition. In two samples-young adolescent twins and adult college students-results showed inverted-U relations between math anxiety and math performance in participants with high intrinsic math motivation and modest negative associations between math anxiety and math performance in participants with low intrinsic math motivation. However, this pattern was not observed in tasks assessing participants' nonsymbolic and symbolic number-estimation ability. These findings may help advance the understanding of mathematics-learning processes and provide important insights for treatment programs that target improving mathematics-learning experiences and mathematical skills. PMID:26518438
Like their sighted peers, many blind students in elementary, middle, and high school are naturally interested in space. This interest can motivate them to learn fundamental scientific, quantitative, and critical thinking skills, and sometimes even lead to careers in Science, Technology, Engineering, and Math (STEM) disciplines. However, these students are often at a disadvantage in science because of the ubiquity of important graphical information that is generally not available in accessible formats, the unfamiliarity of teachers with non-visual teaching methods, lack of access to blind role models, and the low expectations of their teachers and parents. We discuss joint efforts by the National Aeronautics and Space Administration (NASA) and the National Federation of the Blind's (NFB) National Center for Blind Youth in Science (NCBYS) to develop and implement strategies to promote opportunities for blind youth in science. These include the development of tactile space science books and curriculum materials, science academies for blind middle school and high school students, and college-level internship and mentoring programs. The partnership with the NFB exemplifies the effectiveness of collaborations between NASA and consumer-directed organizations to improve opportunities for underserved and underrepresented individuals.Advancedplacementmathplacement English, foreign language, and social science) and advanced quantitative (honors and advancedplacementmathThis report presents the description for the NASTRAN finite element for the AMSU-A1 module. The purpose of this report is to document the NASTRAN bulk data deck, transmitted under separate cover. The structural Math Model is to be used by the spacecraft contractor for dynamic loads analysis.
This booklet presents an overview of the Math Science Partnership program (MSP) at the National Science Foundation (NSF). This program responds to a growing national concern--the educational performance of U.S. children in mathematics and science. Through the MSP, NSF awards competitive, merit-based grants to teams composed of institutions of…
This activity for grades 6-12 is designed to promote an increased interest in mathematics and its study. Directions for the game "Math Trivia" are given, with questions ready for cards and additional questions listed. (MNS) study's main objective was to provide a detailed description of math Assessment and Placement (A&P) policies in the Los Angeles Community College District (LACCD). The study was focused on math because a larger proportion of students place into remedial math than remedial reading or English (Bailey, Jeong, & Cho, 2010; Parsad,…
Educators recognize that group work and physical involvement with learning materials can greatly enhance the understanding and retention of difficult concepts. As a result, math manipulatives--such as pattern blocks and number lines--have increasingly been making their way into classrooms and children's museums. Yet without the constant guidance…
Emperor penguins, the largest of all the penguin species, attain heights of nearly four feet and weigh up to 99 pounds. Many students are not motivated to learn mathematics when textbook examples contain largely nonexistent contexts or when the math is not used to solve significant problems found in real life. This article's project explores how…
Discussion in maths lessons has always been something encouraged by ATM but can be difficult to initiate for non-specialist and inexperienced teachers who may feel they need material in books to get them going. In this article, the author describes resources aimed at encouraging discussion among primary mathematicians. These resources include: (1)…
In this paper, we exploit a high school pilot scheme to identify the causal effect of advanced high school math on labor market outcomes. The pilot scheme reduced the costs of choosing advancedmath because it allowed for a more flexible combination of math with other courses. We find clear evidence of a causal relationship between math and…
Although traditional college students are more prepared for college-level math based on college admissions tests, little data have been collected on nontraditional adult learners. The purpose of this study was to investigate relationships between mathplacement tests and community college students' success in math courses and persistence to degree or certificate completion. Guided by Tinto's theory of departure and student retention, the research questions addressed relationships and predictability of math Computer-adaptive Placement Assessment and Support System (COMPASS) test scores and students' performance in math courses, persistence in college, and degree completion. After conducting correlation and regression analyses, no significant relationships were identified between COMPASS Math test scores and students' performance (n = 234) in math courses, persistence in college, or degree completion. However, independent t test and chi-squared analyses of the achievements of college students who tested into Basic Math (n = 138) vs. Introduction to Algebra (n = 96) yielded statistically significant differences in persistence (p = .039), degree completion (p < .001), performance (p = .008), and progress ( p = .001), indicating students who tested into Introduction to Algebra were more successful and persisted more often to degree completion. In order to improve instructional methods for Basic Math courses, a 3-day professional development workshop was developed for math faculty focusing on current, best practices in remedial math instruction. Implications for social change include providing math faculty with the knowledge and skills to develop new instructional methods for remedial math courses. A change in instructional methods may improve community college students' math competencies and degree achievement withTo take math anxiety out of math instruction, teachers need to first know how to easily diagnose it in their students and second, how to analyze causes. Results of a recent study revealed that while students believed that their math anxiety was largely related to a lack of mathematical understanding, they often blamed their teachers for causing…
Why should your school have a Family Math Night?: (1) Help students learn essential math concepts; (2) Give parents a chance to serve as models of motivation, persistence and competence; and (3) Promote math success in a supportive setting. With its step-by-step directions and suggestions for both teachers and parents, this book takes the worry…
More than 40 states have now signed onto the Common Core standards in English language arts and math, which have been both celebrated as a tremendous advance and criticized as misguided and for bearing the heavy thumbprint of the federal government. This article presents an interview with Ze'ev Wurman and W. Stephen Wilson. Wurman, who was a U.S.…
Maintaining America's productivity as a nation depends importantly on developing a highly qualified cadre of scientists, engineers, entrepreneurs, and other professionals. To realize that objective requires a system of schooling that produces students with advancedmath and science skills. To see how well schools in the United States do at…
"When am I ever going to use this?" This question is heard or thought in every middle-level math class across the land. Teachers struggle to apply math lessons to everyday life and make math meaningful and useful for their students. This author, too, struggled with this problem, until she read the book "Math Curse" by Jon Scieszka (Viking Books,…
Presents innovative, multiple intelligence math practices that can boost student learning and promote fun in the classroom. Suggestions include doing picture book math, working on open-ended problems, doing math all day and everywhere, and working on cooperative group math. (SM) studyConcern about students' math achievement is nothing new, and debates about the mathematical training of the nation's youth date back a century or more. In the early 20th century, American high-school students were starkly divided, with rigorous math courses restricted to a college-bound elite. At midcentury, the "new math" movement sought,…
A math review exam, written and administered in conjunction with the Quantitative Assessment Program at the Univ. of Wisconsin-Madison, is used at the beginning of the 1st food engineering course to evaluate math skills needed for successful completion of the course. Students who do not score well on the math exam are targeted for individual…
The main objective of the authors' proposed study is to evaluate the effectiveness of mathplacement policies for entering community college students on these students' academic success in math, and their transfer and graduation rates. The main research question that guides the proposed study is: What are the effects of various basic skills…
Entering college freshmen, in increasing numbers, in practically every public institution of higher learning are in need of one or two remedial math courses. This is particularly a big problem at the Historically Black Colleges and Universities where a large number of remedial math course sections are offered to meet the growing demand for such courses. For most of these students, graduation is delayed by at least a year. In addition, these students continue to be taught by teaching methodologies that did not work for them even in high schools resulting in disgust and hatred for math. This situation makes entry for these students into STEM areas difficult and is the perfect recipe for failure in STEM disciplines if they enroll in college level courses. The University of the District of Columbia (UDC) is no exception. A first attempt was made in summer 2006 to remedy this situation. The problem for this exploratory research study was to ascertain if a short, intensive six-week project in basic math and introductory algebra would produce a recognizable improvement in the math performance of entering UDC freshmen students as measured by the UDC mathplacement test. The results are eye opening. On the pre-test for basic math (005), the mean score for the group (N=10) was 35.6, with the passing score being 70. On the post-test, the mean increased to 63.4 showing an improvement of 78 percent. The authors will present the results of this research study at the conference | 677.169 | 1 |
Learn the concepts and methods of linear algebra, and how to use them to think about computational problems arising in computer science. Coursework includes building on the concepts to write small programs and run them on real data.
When you take a digital photo with your phone or transform the image in Photoshop, when you play a video game or watch a movie with digital effects, when you do a web search or make a phone call, you are using technologies that build upon linear algebra. Linear algebra provides concepts that are crucial to many areas of computer science, including graphics, image processing, cryptography, machine learning, computer vision, optimization, graph algorithms, quantum computation, computational biology, information retrieval and web search. Linear algebra in turn is built on two basic elements, the matrix and the vector.
In this class, you will learn the concepts and methods of linear algebra, and how to use them to think about problems arising in computer science. You will write small programs in the programming language Python to implement basic matrix and vector functionality and algorithms, and use these to process real-world data to achieve such tasks as: two-dimensional graphics transformations, face morphing, face detection, image transformations such as blurring and edge detection, image perspective removal, audio and image compression, searching within an image or an audio clip, classification of tumors as malignant or benign, integer factorization, error-correcting codes, secret-sharing, network layout, document classification, and computing Pagerank (Google's ranking method).
You are not expected to have any background in linear algebra. You need not know Python, but you should have at least some exposure to programming. You should also be prepared to read and write a few simple mathematical proofs. | 677.169 | 1 |
KEY MESSAGE: Tom Carson's Prealgebra, Third Edition responds to individual learning styles with a complete study system. The system begins with a Learning Styles Inventory and then presents targeted learning strategies to guide students to success. Tom speaks to readers in everyday language and walks them through the concepts, explaining not only how to do the math, but also where the concepts come from and why they work. | 677.169 | 1 |
Differential Equations Demystified
Here's the perfect self-teaching guide to help anyone master differential equations, a common stumbling block for students looking to progress to advanced topics in both science and math
Covers First Order Equations, Second Order Equations and Higher, Properties, Solutions, Series Solutions, Fourier Series and Orthogonal Systems, Partial Differential Equations and Boundary Value Problems, Numerical Techniques, and more.
Perfect for a student going on to advanced analytical work in mathematics, engineering, and other fields of mathematical science.
Differential equations is an important subject that lies at the heart of the calculus. Here one sees how the calculus applies to real-world problems. Differential Equations DeMystified, (to use the spelling on the cover) is ...a serious, straightforward work. In style and substance this book is like standard differential equations books...The emphasis is consistently...on the computations needed to find...solutions to specific equations... (Choice 2005-02-01)
Krantz asserts that if calculus is the heart of modern science, differential equations are the guts. Writing for those who already have a basic grasp of calculus, Krantz provides explanations, models, and examples that lead from differential equations to higher math concepts in a self-paced format. He includes chapters on first-order and second-order equations, power series solutions and spatial functions, Fourier series, Laplace transforms, numerical methods, partial equations and boundary value problems. His models come from engineering, physics and other fields in math. He includes solutions to the exercises and a final exam. (Sci-Tech Book News 2004-12-01)
About the Author:
Steven Krantz, Ph.D., is Chairman of the Mathematics Department at Washington University in St. Louis. An award-winning teacher and author, Dr. Krantz has written more than 45 books on mathematics, including Calculus Demystified, another popular title in this series. He lives in St. Louis, Missouri.
Descrizione libro McGraw-Hill Education140808
Descrizione libro McGraw-Hill Education - Europe. Paperback. Condizione libro: new. BRAND NEW, Differential Equations Demystified, S. Krantz, Here's B9780071440257 | 677.169 | 1 |
0078296358: Applications and Concepts, Course 3, Student Edition
Mathematics: Applications and Concepts is a three-text Middle School series intended to bridge the gap from Elementary Mathematics to High School Mathematics. The program is designed to motivate middle school students, enable them to see the usefulness of mathematics in the world around them, enhance their fluency in the language of mathematics, and prepare them for success in Algebra and Geometry | 677.169 | 1 |
Math Triumphs is intended for use as a pull out intervention class (RtI level 3) for students who are struggling in mathematics concepts two or more grade levels below their current grade. It is designed for use in after school, before school, summer school, intercession, tutoring, or pull-out/resource rooms. Math Triumphs is the first intervention program designed around the NCTM Focal Points. The Student Editions are 4-color, consumable Student Study Guide worktexts focusing on the foundational skills and concepts leading up to one of the three grade level Focal Points. Each chapter begins with a diagnostic assessment to identify concepts and skills that students may need to review or reinforce before introducing the new chapter. Author : McGraw-Hill Education ISBN : 0078882028 Language : English No of Pages : 307 Edition : 1 Publication Date : 1/2/2008 Format/Binding : Paperback Book dimensions : 10.6x8.5x0.4 Book weight : 0.01 | 677.169 | 1 |
Elementary Linear Algebra, Textbook and Student Solutions Manual
4.11 - 1251 ratings - Source
Elementary Linear Algebra 10th edition gives an elementary treatment of linear algebra that is suitable for a first course for undergraduate students. The aim is to present the fundamentals of linear algebra in the clearest possible way; pedagogy is the main consideration. Calculus is not a prerequisite, but there are clearly labeled exercises and examples (which can be omitted without loss of continuity) for students who have studied calculus. Technology also is not required, but for those who would like to use MATLAB, Maple, or Mathematica, or calculators with linear algebra capabilities, exercises are included at the ends of chapters that allow for further exploration using those tools.Elementary Linear Algebra 10th edition gives an elementary treatment of linear algebra that is suitable for a first course for undergraduate students.
Title
:
Elementary Linear Algebra, Textbook and Student Solutions Manual
Author
:
Howard Anton, Chris Rorres
Publisher
:
- 2010-06-08
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Popular in Mathematics (M)
Reviews for Advanced topics Vectors vector addition scalar multiplication Linear combinations Matrix algebra addition multiplication properties Interpretations of matrix multiplication eg Each column of AB is a combination of cols of A generated by corresponding column of B Each row of AB is a combination of rows of B generated by corresponding row of A An element of AB is a row ofA times a column of B Subspaces of Rn 7 what is a subspace how do we recognize a subspace Space spanned by a set of vectors 7 matrix form of such a space all vectors Ax same as column space of A same as space spanned by columns of A Space de ned by homogeneous conditions All solutions of Ax0 same as null space of The dimension ofa subspace basis ofa subspace Row echelon and reduced row echelon form of a matrix 7 row operations Four subspaces Column space row space null space left null space null space of AT Determining a basis for any of the subspaces using elimination The rank of a matrix and its quot to the J39 39 of the 39 The general solution of Axb homogeneous plus particular Determining linear independence of a set of vectors nding linear dependencies among vectors Inverses 7 calculating inverses using elimination using inverses in linear equations inverse of a product of square matrices Singularnonsingular matrices Determinants Basic properties and derived properties The big formula Effect of rowcolumn operations on the determinant Laplace expansion formula using minors Evaluating a determinant using row operations and expansion Determinant of a triangular matrix Determinant of A What a determinant determines Formula for the inverse using a determinant Cramer s rule 1 Singular 7 not invertible definition 7 detA0 7 columns dependent 7 rows dependent 7 rankAltn 7 Ax0 has a nonzero solution 7 A not row equivalent to I DetABdetAdetB for square matrices only Review topics for exams Solving rst order DE s Linear separable Plotting simple integral curves Determining where an initial value problem has a unique solution either with an explicit solution for a nonlinear de or in general for a linear de Second order linear de s General principles lnitial value problems where do solutions exist Homogeneous de s superposition of of solutions independent solutions and the Wronskian general solution of a homogeneous 2nd order DE Second order nonhomogeneous de s Homogeneous particular General formula for a particular solution variation of parameters formula yp y1lyWy2lyW Constant coef cient DE s Exponential solutions characteristic equation determination of general solution based on roots ofthe characteristic equation Undetermined coefficients form of the solution for constant coeff de s with special nonhomogeneous terms polynomialexponentialsinc | 677.169 | 1 |
What is the New Calculus?
History: The New Calculus (NC) was conceived in the early part of the second half of the twentieth century. It is the first and only rigorous formulation of calculus in human history. I say in "human" history, because it's quite possible advanced alien civilisations might have discovered it a long time ago. Based only on well-formed concepts, the new calculus does not use any ill-formed concepts such as limit, infinity or infinitesimals.
Due to persecution from mainstream academia, this is the sixth time this site has been rebuilt. In fact, Weebly once took down the site due to false reports of spamming from Google Inc. Weebly subsequently apologised, but I had already created a new site on Wix. As of 9/17/2016, the owners of Wix succumbed to pressure from mainstream academics to take the site offline. It is unbelievable that in this day and age, such Nazi tactics are used by the supposedly enlightened, educated and open-minded academic mainstream community. These efforts were driven by Prof. Gilbert Strang (aka Port563 on sci.math) of MIT and Prof. Jack Huizenga, a Harvard alumnus. Both these academics did not like the comments I made about them. You can read these comments here and here. Neither of the previous two links are on Weebly.
The single variable New Calculus can be learned and mastered in just 2-4 weeks with the only prerequisite being high school mathematics. The multivariable New Calculus can be mastered in 8 weeks. However, the New Calculus contains a new kind of mathematics, one that includes tangent objects. While the single NC is easily extended to the multivariable NC in the same way as mainstream calculus, tangent objects provide a far more efficient way of working with multivariable calculus, including differential equations and partial differential equations.
The purpose of this site is to expose the flaws of mainstream calculus and elaborate on the beauty and light of the New Calculus which contains many features that are not possible using the flawed mainstream formulation:
1. The Auxiliary equation is the first and most powerful feature that a student learns. Not possible in the flawed formulation, this feature has been used by thousands of STEM professionals worldwide in fields as varied as technology, computer aided design, statistics and last but not least, education. Some important uses of the auxiliary equation include: a. Non-linear regression b. Finite difference estimation c. Solution without iteration
and much more. There is no similar or equivalent possible in the flawed mainstream formulation.
2. Systematic integration is possible for any function in the New Calculus.
3. The Gabriel polynomial (single and multivariable forms) far surpasses the Taylor analog in mainstream calculus. It contains a varied number of fixed terms and is a closed form always.
4. Solution of differential equations and use of superior numeric integration techniques.
The entire single variable New Calculus is summarised in the leading graphic on this page.
Here is a short excerpt:
We can prove that if f(x) is a function with tangent line equation t(x)=kx+b and a parallel secant line equation s(x)=[{f(x+n)-f(x-m)}/(m+n)] x + p, then f'(x)={f(x+n)-f(x-m)}/(m+n).
Proof:
Let t(x)=kx+b be the equation of the tangent line to the function f(x).
Then a parallel secant line is given by s(x)=[{f(x+n)-f(x-m)}/(m+n)] x + p.
So, k={f(x+n)-f(x-m)}/(m+n) because the secant lines are all parallel to the tangent line.
But the required derivative f'(x) of f(x), is given by the slope of the tangent line t(x).
Therefore f'(x)={f(x+n)-f(x-m)}/(m+n).
Q.E.D.
Also, m+n is a factor of the expression f(x+n)-f(x-m).
Proof:
From k(m+n)=f(x+n)-f(x-m), it follows that m+n divides the LHS exactly. But since m+n divides the left hand side exactly, it follows that m+n must also divide the RHS exactly. Hence, m+n is a factor of the expression f(x+n)-f(x-m). This means that if we divide f(x+n)-f(x-m) by m+n, the expression so obtained must be equal to k. This is only possible if the sum of all the terms in m and n are 0.
Q.E.D.
Understand why the standard integral is a product of two arithmetic means using one of the following applets: Applet 1. The New Calculus uses no ill-formed concepts such as infinite sums, limits or infinitesimals. Applet 2. This applet explains the new calculus integral in great detail.
The Future. Rather than rebuild this site from scratch again, I have chosen to use the Google New Calculus Site exclusively for the New Calculus and this site to expose the fallacies in mainstream mathematics. You are invited to join my New Calculus group in order to ask questions and to comment.
The ideas that added zero rigour to calculus. The applet explains the ideas that added no rigour to calculus, but in fact made it more obscure and complicated.
In the applet, you can click on the check box called "Limit Magic" to see why it is an ill-formed definition. It fails for many reasons, but the two most important are:
i. There is no valid construction of real numbers. ii. Circular definitions are invalid.
The 13 fallacies of mainstream mathematics. What's kind of sad and also kind of funny is how mainstream academics always ask for sources, usually printed or peer reviewed. The greatest mathematicians never had their work reviewed because they had no peers on their level. It is the same with me. In order for my work to be reviewed, my "peers" (I have none) would need to possess my intelligence and not to be infinitely ignorant like Prof. Gilbert Strang from MIT.
So, courtesy of the internet, I am able to share some of my ideas and educate those aspiring young mathematicians with knowledge that is well formed.
Here are my four essential requirements for any concept to be well defined:
In order to be well defined, a concept
1. Must be reifiable either intangibly or tangibly.
2. Must be defined in terms of attributes which it possesses, not those it lacks.
3. Must not lead to any logical contradictions.
4. Must exist in a perfect Platonic form. What this means is that it exists independently of the human mind or any other mind.
If you can't reify a concept, then it may not exist outside your mind. If a group of mainstream academics get together and claim an infinite sum is possible, even among themselves, they do not think of it the same way. The fallacy that 0.999... and 1 are the same, is a fine example. Some academics think that it is actually possible to sum the series.
Others (such as Rudin) realise that only a limit is possible. Still others believe that it's a good idea to give a sequence a value in terms of its limit even when the limit is not known.
To reify, means to produce an instance of the concept so that someone who knows nothing about it, can understand it exactly the same way as another. Even though the 0.999... fallacy has been around for so many decades, ask yourself how it is that so many students and even educators have different views on it, with most forums split almost evenly among those who acknowledge the fallacy and those who don't.
You can reify a concept without someone else being able to understand it for many other reasons; some include intelligence, ignorance, etc. However, I am talking about all things being equal, in which case the concept can be acknowledged as having been reified.
If you can't get past reification, then your concept is no doubt nonsense. Some examples are:
Irrational numbers. Infinitesimals. Limits when there is in fact no valid construction of "real" numbers. Infinity. Einstein's theories. Hawking's theories. Definitions that are self-referential.
If a concept is not defined in terms of attributes it possesses, then you may as well be talking about innumerably many other concepts. You have endless ambiguity. It is the most important second step after reification. It describes the boundaries or limits, the extent of the instantiated object from the concept. I used to think:
"Mathematicians are like artists, the objects arising from concepts in a mathematician's mind are only as appealing as they are well defined"
Clearly, they are not even usable if they cannot be well defined. A good example is 0.999... - it has no use and nothing worthwhile can be done with such an idiotic definition, that is, S = Lim S. This is what I call the Eulerian Blunder because it was Euler who stated S = Lim S.
Once a concept has been reified and well defined (there are limitations to being well defined and this is why one needs to have checks for contradictions until the concept becomes axiomatic over a long period of vetting), it has to be vetted. This is done by always verifying that any results stemming from its use do not produce logical contradictions.
Finally, the last point which is sufficient for a well-formed concept, is that it must exist outside of the human mind or any other mind. For example, if aliens think of pi, they will think of it in the only logical way: ratio of the circumference magnitude to the diameter magnitude.
Perfect concepts exist whether life exists or not. That is what the Greeks discovered when they studied geometry. The concepts of geometry are all without any exception perfect concepts (Platonic).
The 13 fallacies that form the foundation of mythmatics (mainstream mathematics):
2. There is an infinite set. Proof by Prof. W. Mueckenheim using mainstream theory that a contradiction exists in set theory.Mueckenheim does a great job of showing how infinite set theory invalidates itself. The video is an eye opener.
5. An infinite sum is possible. It is impossible to sum an infinite series according to mainstream academics. The mean value theorem is the reason calculus works and the fundamental theorem (which does not have two parts) is derived directly from it. The single variable New Calculus can be explained in under 25 minutes.
8. The integral is an infinite sum. The definite integral is a product of arithmetic means and has nothing to do with non-existent infinite sums. The mean value theorem is the reason we can evaluate definite integrals. Riemann's ideas in this regard are completely misguided.
9. Numbers can be derived using sets. The von Neumann ordinal approach is a joke. The first major stumbling block is that in order to define rational numbers using set theory, we already need to know how to "count". That's right, you need to be able to compute the cardinality of a given set. Unless you are one of Cantor's delusional followers, cardinal value means number, not bijective cardinality myths involving sets whose members are not distinct, that is, the illusion of infinitely many points. Now, do you have any clue what effort went into deriving the machinery of counting numbers which came long after ratios of magnitudes? Of course you don't. Unless you have read my article, you don't have a clue what it means to be a number. Thousands of worthless books on numbers and history of numbers have been printed - all of them worthless rot.
After reading that article, ask yourself, does set theory require the natural numbers to be in place? Hint: YES
Since there is no valid construction of irrational number, can there be any valid mathematical concept for real number? Hint: NO
11. Natural numbers came first. The real story of numbers is very different to anything ever published on thousands of worthless books on the same. The understanding of mainstream academics is based entirely on that mythical object called a real number line.It is absurd that any rational mind can think that real numbers correspond to points of a line. In fact, points are not part of lines. The original definition of line is a distance between two given points.
12. dy/dx is an instantaneous rate of change. There is no such thing as an instantaneous rate of change. Of all the ideas in calculus, this is the most ill-formed.
13. The "real" numbers can be thought of as points on the number line. Euclid attempted the perfect derivation of number. I am not alone in these opinions. Prof. Wildberger has similar ideas. | 677.169 | 1 |
About The Calculia
Mathematics is not just about numbers and equations, rather its about the real life scenarios that we come across everyday. And so we are here for you to make you acquainted with the relation between mathematics and practical experiences via smart class. We offer non-conventional methodology of classroom teaching through activities and demonstrations with a bit of practice to ensure the understanding of each student. | 677.169 | 1 |
Algebra is a branch of mathematics that deals with the properties of operations and
the structures by which these operations are defined. These pages are an introduction to the basic ideas
of algebra, including the concepts of variables, polynomials, factorization and graphing.
Algebra essentially evolved from the rules of arithmetic which begins with the four operations: addition,
subtraction, multiplication and division. Operations in algebra follow the same rules as those in arithmetic
except Algebra uses variables, which are usually letters, to represent unknown numbers, and expressions which are mathematical
statements that use numbers and/or variables.
Algebra is used for problem solving in engineering, architecture, science, economics, finance, and many other
tasks. We hope the information here will help you better understand this powerful tool.
Originally created November 7, 2007 by Grandpa Freddy, founder of the Weird Grandpas'
Association.™ This site is currently maintained by
Grandpa Freddy's evil twin. | 677.169 | 1 |
Logarithms III
In this logarithm worksheet, learners solve complex logarithm problems containing one or more operations. This one-page worksheet contains four problems. Answers are provided at the bottom of the page. | 677.169 | 1 |
Browse related Subjects ...
Read More' hallmark approach of encouraging mastery of mathematics through careful practice. The text provides detailed, straightforward explanations and accessible pedagogy to help students grow their math skills from the ground up. The authors use a three-pronged approach of communication, pattern recognition, and problem solving to present concepts understandably, stimulate critical-thinking skills, and stress reading and communication skills in order to help students become effective problem-solvers. Features such as Tips for Student Success, Check Yourself exercises, and Activities underscore this approach and the underlying philosophy of mastering math through practice. Exercise sets have been significantly expanded and are now better-organized, and applications are now more thoroughly integrated throughout the text. The text is fully-integrated with McGraw-Hill's online learning system, Connect Math Hosted by ALEKS Corp, and is available with ALEKS 360.
Read Less
Good. Looseleaf. SKU: 9780077574055-4-0-3 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions. ISBN: 9780077574055.
Very good. Paperback. SKU: 9780073384443-3Fair. Paperback. SKU: 9780073384443 Paperback. SKU: 9780073384443-4 Paperback. Instructor Edition: Same as student edition with additional notes or answers. SKU: 9780077573973-2-0-1 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions. ISBN: 9780073384443 | 677.169 | 1 |
Understanding Engineering Mathematics
4.11 - 1251 ratings - Source
Studying engineering, whether it is mechanical, electrical or civil relies heavily on an understanding of mathematics. This new textbook clearly demonstrates the relevance of mathematical principles and shows how to apply them to solve real-life engineering problems. It deliberately starts at an elementary level so that students who are starting from a low knowledge base will be able to quickly get up to the level required. Students who have not studied mathematics for some time will find this an excellent refresher. Each chapter starts with the basics before gently increasing in complexity. A full outline of essential definitions, formulae, laws and procedures are introduced before real world situations, practicals and problem solving demonstrate how the theory is applied. Focusing on learning through practice, it contains examples, supported by 1, 600 worked problems and 3, 000 further problems contained within exercises throughout the text. In addition, 34 revision tests are included at regular intervals. An interactive companion website is also provided containing 2, 750 further problems with worked solutions and instructor materialsWhy it is important to understand: Graphical solution of equations It has been
established in previous chapters that the solution of linear, quadratic,
simultaneous and cubic equations occur often in engineering and science and
may be solvedanbsp;...
Title
:
Understanding Engineering Mathematics
Author
:
John Bird,
Publisher
:
Routledge - 2013-11-20
ISBN-13
:
Continue
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You'll gain access to interventions, extensions, task implementation guides, and more for this lesson.
Big Idea:
Changes to the values of A, B, h, and k in the general form of a trigonometric function, f(x)=A*sin[B(x-h)]+k, alter the graph of the function through both rigid and non-rigid transformations by changing the amplitude, midline, period, etc.
This lesson extends the student's understanding of function transformations in parent functions to the trigonometric functions, and more specifically, the sinusoidal functions of f(x)=sin(x) and f(x)=cos(x). Students will graph several functions in the form f(x)=A*sin(x) and observe how the factor A changes the amplitude of (or reflects) the graph of sin(x). This lesson will work in conjunction with the lessons on understanding the period and shifts to complete the student's study on trigonometric function transformations.
Vocabulary: amplitude, midline, sinusoidal function, non-rigid transformations
Special Materials:
Students may benefit from having trigonometric graph paper.
This lesson can take on more of a "discovery learning" format if students have access to graphing technology with capability of creating sliders (i.e. Desmos or GeoGebra). | 677.169 | 1 |
Numerical Methods, 4th
4.11 - 1251 ratings - Source
NUMERICAL METHODS, Fourth Edition emphasizes the intelligent application of approximation techniques to the type of problems that commonly occur in engineering and the physical sciences. Students learn why the numerical methods work, what kinds of errors to expect, and when an application might lead to difficulties. The authors also provide information about the availability of high-quality software for numerical approximation routines. The techniques are the same as those covered in the authorsa€™ top-selling Numerical Analysis text, but this text provides an overview for students who need to know the methods without having to perform the analysis. This concise approach still includes mathematical justifications, but only when they are necessary to understand the methods. The emphasis is placed on describing each technique from an implementation standpoint, and on convincing the student that the method is reasonable both mathematically and computationally. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.The techniques are the same as those covered in the authorsa€™ top-selling Numerical Analysis text, but this text provides an overview for students who need to know the methods without having to perform the analysis.
Title
:
Numerical Methods, 4th
Author
:
J. Faires, Richard Burden
Publisher
:
Cengage Learning - 2012-04-23
ISBN-13
:
Continue
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A Concrete Approach To Abstract Algebra
4.11 - 1251 ratings - Source
A Concrete Approach to Abstract Algebra begins with a concrete and thorough examination of familiar objects like integers, rational numbers, real numbers, complex numbers, complex conjugation and polynomials, in this unique approach, the author builds upon these familar objects and then uses them to introduce and motivate advanced concepts in algebra in a manner that is easier to understand for most students. The text will be of particular interest to teachers and future teachers as it links abstract algebra to many topics wich arise in courses in algebra, geometry, trigonometry, precalculus and calculus. The final four chapters present the more theoretical material needed for graduate study.Student Solutions Manual Jeffrey Bergen. 73. The leading coefficient is 2 and it is
not divisible by 3. All the other coefficients are divisible by 3 and the constant
term, a€"57, is not divisible by 3Ad. Therefore, exercise 69 implies that the
polynomialanbsp;...
Title
:
A Concrete Approach To Abstract Algebra
Author
:
Jeffrey Bergen
Publisher
:
Academic Press - 2010-05-15
ISBN-13
:
Continue
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Basic Engineering Mathematics
4.11 - 1251 ratings - Source
Introductory mathematics written specifically for students new to engineering Now in its sixth edition, Basic Engineering Mathematics is an established textbook that has helped thousands of students to succeed in their exams. John that readers can relate theory to practice. The extensive and thorough topic coverage makes this an ideal text for introductory level engineering courses. This title is supported by a companion website with resources for both students and lecturers, including lists of essential formulae, multiple choice tests, full solutions for all 1, 600 further questions contained within the practice exercises, and biographical information on the 25 famous mathematicians and engineers referenced throughout the book. The companion website for this title can be accessed from the operation manual for your particular calculator to determine how to
use these two funcNow try the following Practice Exercise ... For example, with
the Casio fx-991ES PLUS calculator, or similar, to change the Cartesian number (
3anbsp;...
Title
:
Basic Engineering Mathematics
Author
:
John Bird
Publisher
:
Routledge - 2014-03-26
ISBN-13
:
Continue
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MAT-129 Precalculus
Prepares students for courses in calculus and higher mathematics and for courses in science and technology where knowledge of precalculus is required. Topics include exponential and logarithmic functions and equations; trigonometric functions, identities and equations; applications of trigonometry; systems of equalities and inequalities; series and sequences; and analytic geometry.
Advisory: it is advisable to have knowledge in a course equivalent to MAT-121 College Algebra with a grade of C or better to succeed in this course. Students are responsible for ensuring that they have the required knowledge. BSAST and ASAST students should take MAT-121 and MAT-129. | 677.169 | 1 |
"Altogether the book gives a comprehensive introduction to graphs, their theory and their application…The use of the text is optimized when the exercises are solved. The obtained skills improve understanding of graph theory as well… It is very useful that the solutions of these exercises are collected in an appendix." (Simulation News Europe)
From the reviews of the second edition:
"This book is a gentle introduction to graph theory, presenting the main ideas and topics … . It is accessible to everyone … . This introductory book is addressed to a mixed audience – undergraduate mathematics majors, computer scientists, engineers … . this book is ideal as well for self-reading. The style is always concise and the essential techniques are well highlighted … . It is highly recommended to any student, or working scientist, wishing to explore for the first time this fascinating area of mathematics." (Fabio Mainardi, The Mathematical Association of America, August, 2009)
"This book is intended as an introductory course in Graph Theory, one of the fastest growing disciplines of modern Mathematics. … The book is nicely written, the presentation is comprehensible but at the same time mathematically precise. The text is supplemented with many figures, with historical notes to many topics and with many examples. … Summarizing, this is a nice book, useful not only as an introductory reading for 'beginners' in Graph Theory, but also for those who teach introductory courses in Graph Theory." (Zden?k Ryjá?ek, Mathematica Bohemica, Issue 2, 2010)
From the Publisher:
Concisely written, gentle introduction to graph theory suitable as a textbook or for self-study4898064
Descripción Birkhäuser. Estado de conservación: New. 081764481764484095083639456136256071 | 677.169 | 1 |
Projective Geometry An introduction
This lucid, accessible text provides an introductory guide to projective geometry, an area of mathematics concerned with the properties and invariants of geometric figures under projection. Including numerous examples and exercises, this text is ideal for year 3 and 4 mathematics undergraduates | 677.169 | 1 |
Hercules AlgebraMike N.
...Articulation is an essential skill to teach. Many professors don't thoroughly explain how to derive efficient and effective solutions to problems and instead assign projects with little to no background on why the process of solving problem is important. If you have a genuine interest in comput... | 677.169 | 1 |
Solving Logarithm Equations with Properties We Have a Match Partner Activity
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
3.48 MB | 4 pages
PRODUCT DESCRIPTION
Solving equations is a foundational skill needed in algebra. Being able to solve a variety of equations enables students to analyze functional relationships.
Students need practice in solving equations in order to develop fluency in uncovering unknown values.
We Have a Match: Students work in pairs to solve 8 logarithm equations that require an understanding of log properties; partner solutions should match making the activity self-checking. Solutions are included. | 677.169 | 1 |
eCompanion for Aufmann/Lockwood's Beginning Algebra, 1st
4.11 - 1251 ratings - Source
This new text is a companion to the traditional and comprehensive print and eBook versions of the best-selling Beginning Algebra with Applications text by the Aufmann/Lockwood team. The eCompanion provides a telescopic view of the core concepts for introductory algebra as a slim portable inexpensive print option that provides the traditional and online student the summary per learning objective they require. For many students, the format and functionality of the traditional print model has been a hurdle since it doesna€™t match or reflect their busy lifestyles. Students today need something in print but want the delivery to be digital as well. With that in mind, the eCompanion will provide both for them; a portable tool that is not 1000 pages long but under 250 pages, black-a-white, paperback providing students with what they need to learn and which accompanies the digital version of the more comprehensive text. Together, the print and digital combination helps them study and prepare for mastering the introductory algebra course. With the eCompanion, when accompanied by the entire traditional textbook whether in print or digital medium, they now have a tool that summarizes and presents the key learning objectives for their course. In addition, the eCompanion comes to life when students use it with Enhanced WebAssign our online homework system for Beginning Algebra with Applications. The Enhanced WebAssign program contains over 2, 300 exercises from Beginning Algebra with Applications that students can solve either as practice or assessed homework. The EWA content also includes links to video examples, problem-specific tutorials, and more Example Simplify: 532 Solution 532 516 2 516 2 5
42 202 Simplify Variable Radical Expressions [10.1.2] A variable or a product of
variables written in ... Simplify: 24 37 7 2 xy 3 x y Solution 24 3 8 37 7 2 5 4 xy
x y y x 8 5 4 y x 424 4 yy x 222 2 yy x A radical expression is not in simplest form if
a.
Title
:
eCompanion for Aufmann/Lockwood's Beginning Algebra, 1st
Author
:
Richard Aufmann, Joanne Lockwood
Publisher
:
Cengage Learning - 2010-04-08
ISBN-13
:
Continue
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CHAPTER 1 INTRODUCTION
Mathematics pervades all aspects of life, whether at home, in civic life or in the workplace. It has been central to nearly all major scientific and technological advances. Also, many of the developments and decisions made in industry and commerce, the provision of social and community services as well as government policy and planning, rely to an extent on the use of mathematics. It is important for our students to gain experience and build up the foundation skills and knowledge in mathematics that can facilitate their future development in various aspects. It is also important that our students are able to value mathematics and appreciate the beauty of mathematics after mathematics education in school. In the information explosion era, there are drastic changes both in our society and in the background of our students. It is vital that the curriculum should undergo continuous review and renewal in order to meet the needs of our students and the community. In reviewing the mathematics curriculum, there are several principles to be followed. The needs of our students and the community are important considerations in developing the aims of the secondary school mathematics education. The worldwide trends of mathematics curriculum have also been taken into consideration in deciding the principles. This syllabus presented is a revised edition of the version published in 1985. It has been scheduled for implementation in schools with effect from September 2001 at Secondary 1.
Principles of Curriculum Design
The following principles are used to guide the evolution of the Aims and Objectives, the structure of the curriculum, and the identification of objectives in each module and unit in the syllabus. l Target-oriented To ensure that learners will spend their time and effort meaningfully and for maximum benefits, there must be a plan for them to work according to specific Learning Targets and Objectives which are geared towards the Aims and Objectives of the school mathematics curriculum. They are organized progressively across four Key Stages in primary and secondary schooling: Key Stage 1 (Primary 1 to 3), Key Stage 2 (Primary 4 to 6), Key Stage 3 (Secondary 1 to 3) and Key Stage 4 (Secondary 4 to 5). The learning targets and objectives for Key Stages 1 and 2 can be found in Annex I. Continuing the learning in primary schooling, the overall Aims and Objectives for Key Stages 3 and 4 are stated in Chapter 2. The Learning Targets and Objectives for each dimension in each learning stage are further elaborated in Chapter 4 to spell out the specific learning objectives for each learning area. All learning and assessment activities fulfil the learning objectives of that particular unit and are geared towards the maximum learning effectiveness for achieving the Aims and Objectives.
1
a wider range of students gain access to secondary mathematics than have been in the past. exposure to concrete objects and personal experience is planned to support abstract discussion as far as possible. A new module "Further Applications" which includes the application of mathematics in more complex real-life situations requires students to integrate their knowledge and skills from various disciplines to solve problems. teachers can judge for themselves the suitability and relevance of other topics in the Whole Syllabus for their own students. The Foundation Part is the essential part of the Syllabus which ALL students should strive to learn. Daily life applications are emphasized in the curriculum.
2
. For instance. teachers can adopt some enrichment topics at their discretion to extend these students' horizon and exposure in mathematics. Besides the various ways of organizing students' activities in the class to cater for learner differences.l Catering for learner differences Upon the implementation of universal education in Hong Kong. The school mathematics curriculum should cater for the diversity of students' needs and for the wide spectrum of ability among them. the learning units and modules for each dimension are subdivided into key stages (KS). Students who find the study relevant to their experience will be motivated to learn the subject. In order to provide further flexibility for teachers to organize the teaching sequences to meet individual teaching situations. i. KS3 for S1-S3. the Foundation Part of the curriculum is identified. Great care is taken to ensure that the curriculum is organized with a cognitive developmental perspective. Apart from the Foundation Part. the knowledge and skills to be learnt should be determined by the activities deemed suitable for the age-group concerned. KS4 for S4-S5.e. Teachers are free to design their school based mathematics curriculum for each year level with all learning areas suggested for each key stage in mind. l Relevance of study to students In order that mathematics learning efforts are effective. Stories of historical development of mathematics knowledge are included to enable students to understand mathematics knowledge evolved from reallife problems and refined after years. For more able students.
A variety of learning activities should be planned and geared towards the development of these general abilities and skills. to access information and process it effectively. On the other hand. but fostering these general abilities and skills are strongly advocated for all students in the revised curriculum. Acquiring mathematics knowledge has always been emphasized. students are expected to use these learning strategies to construct their mathematics knowledge. the positive role that calculators could play in the mathematics learning is generally aware of. to think logically and creatively. reasoning. communicating.l Impact of information technology The tools for solving mathematical problems change from time to time. and to communicate with others so that they can meet the challenges that confront them now and in the future. It is important that students need to develop their capabilities to learn how to learn. the availability of computers and other information technology aids in the classrooms will have impact on the mathematics curriculum in terms of contents and strategies for teaching and learning of mathematics. conceptualizing and problem-solving are considered important in mathematics education. including data analysis.
3
. students are expected to learn mathematics to enhance the development of these skills. The popularity of graphing calculators. graphical presentation. In the curriculum. There are ranges of ways in which information technology may be used in mathematics classes. There are different roles electronic calculators can play. to develop and use knowledge. On the one hand. symbolic manipulation and observing patterns. simulation device. Today we are confronted with a similar situation. l Fostering general abilities and skills Knowledge is expanding at an ever faster pace and new challenges are continually being posed by the rapid changes in technology and in the way society evolves. With years of experience in the classrooms of various countries. The introduction of electronic calculators in 1980's has influenced the teaching of secondary school mathematics. The appropriate use of information technology in the teaching and learning of mathematics becomes one of the emphases in the mathematics curriculum. to analyze and solve problems. fundamental and intertwining ways of learning and using knowledge such as inquiring. A general worry among teachers and parents is that the unwise use of calculators by students would hamper their development of computational skills. | 677.169 | 1 |
Institutional Access
Description
Designed for those who want to gain a practical knowledge of modern computational techniques for the numerical solution of linear algebra problems, Numerical Linear Algebra with Applications contains all the material necessary for a first year graduate or advanced undergraduate course on numerical linear algebra with numerous applications to engineering and science.
With a unified presentation of computation, basic algorithm analysis, and numerical methods to compute solutions, this book is ideal for solving real-world problems. It provides necessary mathematical background information for those who want to learn to solve linear algebra problems, and offers a thorough explanation of the issues and methods for practical computing, using MATLAB as the vehicle for computation. The proofs of required results are provided without leaving out critical details. The Preface suggests ways in which the book can be used with or without an intensive study of proofs.
Key Features
Six introductory chapters that thoroughly provide the required background for those who have not taken a course in applied or theoretical linear algebra
Detailed explanations and examples
A through discussion of the algorithms necessary for the accurate computation of the solution to the most frequently occurring problems in numerical linear algebra
Examples from engineering and science applications
Readership
Graduate or advanced undergraduate students in engineering, science, and mathematics, professionals in engineering and science, such as practicing engineers who want to see how numerical linear algebra problems can be solved using a programming language such as MATLAB, MAPLE, or Mathematica.
Table of Contents
Dedication
List of Figures
List of Algorithms
Preface
Topics
Intended Audience
Ways to Use the Book
Matlab Library
Supplement
Acknowledgments
Chapter 1: Matrices
Abstract
1.1 Matrix Arithmetic
1.2 Linear Transformations
1.3 Powers of Matrices
1.4 Nonsingular Matrices
1.5 The Matrix Transpose and Symmetric Matrices
1.6 Chapter Summary
1.7 Problems
Chapter 2: Linear Equations
Abstract
2.1 Introduction to Linear Equations
2.2 Solving Square Linear Systems
2.3 Gaussian Elimination
2.4 Systematic Solution of Linear Systems
2.5 Computing the Inverse
2.6 Homogeneous Systems
2.7 Application: A Truss
2.8 Application: Electrical Circuit
2.9 Chapter Summary
2.10 Problems
Chapter 3: Subspaces
Abstract
3.1 Introduction
3.2 Subspaces of n
3.3 Linear Independence
3.4 Basis of a Subspace
3.5 The Rank of a Matrix
3.6 Chapter summary
3.7 Problems
Chapter 4: Determinants
Abstract
4.1 Developing the Determinant of A 2 × 2 and A 3 × 3 matrix
4.2 Expansion by Minors
4.3 Computing a Determinant Using Row Operations
4.4 Application: Encryption
4.5 Chapter Summary
4.6 Problems
Chapter 5: Eigenvalues and Eigenvectors
Abstract
5.1 Definitions and Examples
5.2 Selected Properties of Eigenvalues and Eigenvectors
5.3 Diagonalization
5.4 Applications
5.5 Computing Eigenvalues and Eigenvectors Using Matlab
5.6 Chapter Summary
5.7 Problems<
Details
Reviews
"An important part of the book deals with iterative methods for solving large sparse systems. We can find here Jacobi, Gauss-Seidel and successive overrelaxation (SOR) methods, as well as Krylov subspace methods..." --Zentralblatt Math
"...this is a book that merits careful consideration as a possible text for a course in numerical linear algebra, particularly one stressing applications to engineering and other areas of science." --MAA Reviews, January 2015 | 677.169 | 1 |
Explore More Items Similar to Mathematical Circles
Overview: Mathematical Circles
The book, Mathematical Circles: Russian Experience (Mathematical World) tells readers about the Russian mathematical culture and shows how mathematics can be a fun activity. Mathematical circles were an innovative programme initiated in Soviet Russia to promote interest in studying mathematics among secondary school students by making associations of students and teachers. They worked on the principle that the complex subject of mathematics can be made simple by looking at it as a team sport and dealing with it collectively rather than individually. This book can be seen as a guide of mathematical recreations and, at the same time, a text that contains vast theoretical and problem material in main areas of extracurricular mathematics. This book also reflects the experience gained by several generations of Russian educators and scholars. Mathematical Circles: Russian Experience (Mathematical World) is a two-year course book in mathematics. The book is divided into two parts, and the first part contains simple and basic problems that serve as the foundation. The problems of the second part are more complex and require a better understanding. The second part is designed for the second year. Plus there are three appendices in the end. The first appendix talks about the types of mathematical contests that were popular in the former Soviet Union. The second appendix deals with the answers or hints and solutions to the problems that are discussed in the book. The third appendix is for the references that are used in this book. Dmitri Fomin studied at St. Petersburg State University and is a software engineer by profession. Sergey Genkin joined Microsoft in 1995 and is a part of the development team major products like Word, Internet Explorer and Silverlight. He is regarded as a lead software developer. Ilia Itenberg is a noted academician and is also a famous writer. . For online shopping, this book is available in paperback binding. It has an ISBN 10 number of 8173711151 and ISBN 13 number of 978-8173711152. | 677.169 | 1 |
Description of the book "Introduction to the Analysis of Metric Spaces":
This is an introduction to the analysis of metric and normed linear spaces for undergraduate students in mathematics. Assuming a basic knowledge of real analysis and linear algebra, the student is exposed to the axiomatic method in analysis and is shown its power in exploiting the structure of fundamental analysis, which underlies a variety of applications. An example is the link between normed linear spaces and linear algebra; finite dimensional spaces are discussed early. The treatment progresses from the concrete to the abstract: thus metric spaces are studied in some detail before general topology is begun, though topological properties of metric spaces PDF are explored in the book. Graded exercises are provided at the end of each section; in each set the earlier exercises are designed to assist in the detection of the structural properties in concrete examples while the later ones are more conceptually sophisticated.
Reviews of the Introduction to the Analysis of Metric Spaces
Until now regarding the ebook we have now Introduction to the Analysis of Metric Spaces comments end users have never nevertheless eventually left their own writeup on the overall game, or you cannot make out the print however. But, if you have previously see this book and you're prepared to help make the conclusions convincingly expect you to spend time to exit an evaluation on our site (we can easily distribute both bad and the good reviews). Put simply, "freedom of speech" We all wholeheartedly supported. Your current opinions to lease Introduction to the Analysis of Metric Spaces -- various other followers can determine in regards to a ebook. These assistance is likely to make people more U . s .!
John R. Giles
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MATH pure maths Documents
Showing 1 to 20 of 20
w
w
Syllabus
9709
Marks are of the following three types:
M
Method mark, awarded for a valid method applied to the problem. Method
marks are not lost for numerical errors, algebraic slips or errors in units.
However, it is not usually sufficient for a can | 677.169 | 1 |
Math 2210
Final exam
Tuesday 14 December 2010
2:004:30 pm
No books or electronic devices allowed. You are allowed a two-sided letter size paper of
notes.
Please show all work and justify your answers.
Put your name and student ID on the front cover of you
Math 2210 - Linear Algebra
Second prelim - 25 October 2012 - 7:30 to 9:00pm
Name and NetID:
What time do you attend lecture? Circle one. 9:05-9:55 10:10-11 11:15-12:05 12:20-1:10
INSTRUCTIONS
OFFICIAL USE ONLY
This test has 6 problems on 6 pages, worth a
October 27, 2011
Math 2210
Prelim 2
No books or electronic devices allowed. You are allowed a one-sided letter size paper of
notes.
Please show all work and justify your answers.
Put your name and student ID on the front cover of your exam booklet and sig
1.8 SOLUTIONS
20. Use the basic denition of As to construct A. Write
Ttx}-xv+rv[v v]x]3 Fl); A 3 Fl
It-22I2Jr25_2.5_2
31}. Given any x in R. there are constants cl. cp such that x = cm + (Fr... because v]. vp span
R. Then. from property (5) of a lin | 677.169 | 1 |
Fun Self-Discovery Tools
2.2 - Absolute Value (Part A)
Rating:
Description:
In this section we will review the concept of absolute value and examine inequalities and equations involving absolute value. Recall that in the previous unit, you worked with inequalities and equations from both a graphical and algebraic perspective, and now we will see how the inclusion of absolute value symbols can change the solution set for an equation or inequality. | 677.169 | 1 |
This ebook is available for the following devices:
iPad
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iRiver Story
"General Relativity Without Calculus" offers a compact but mathematically correct introduction to the general theory of relativity, assuming only a basic knowledge of high school mathematics and physics. Targeted at first year undergraduates (and advanced high school students) who wish to learn Einstein's theory beyond popular science accounts, it covers the basics of special relativity, Minkowski space-time, non-Euclidean geometry, Newtonian gravity, the Schwarzschild solution, black holes and cosmology. The quick-paced style is balanced by over 75 exercises (including full solutions), allowing readers to test and consolidate their understanding.
In the press
From the reviews:
"The book is a short, clear and elementary introduction to the main ideas of Special Relativity and General Relativity. The primary target was an audience of well-trained high school students, interested in a course beyond the popular science expositions. The contents reveals the aim of the author, to cover both the classic theory and the modern history of the Relativity Theory … ." (Gabriel Teodor Pripoae, Zentralblatt MATH, Vol. 1244, 2012) | 677.169 | 1 |
books.google.com - Mathemat... Techniques
Mathematical Techniques: An Introduction for the Engineering, Physical, and Mathematical Sciences
Mathemat all the essential topics with which a physical sciences or engineering student should be familiar. By With a huge array of end of chapter problems, and new self-check questions, the fourth edition of Mathematical Techniques provides extensive opportunities for students to build their confidence in the best way possible: by using the maths for themselves. Online Resource Centre The Online Resource Centre features the following materials for all users of the book: · Figures from the book in electronic format, ready to download · A downloadable solutions manual, featuring worked solutions to all end of chapter problems · Mathematica-based programs, relating to the Projects featured at the end of the book
LibraryThing Review
User Review - biffbogan - LibraryThing
A straightforward introduction to beginning University-level mathematics and a good companion volume to Stroud for this level. Working from this book will take some of the sting out of the transition from 'A'-Level to more advanced work. BGNRead full review | 677.169 | 1 |
diffrence between algabra 1 and pre-algabra 1 you go father deep and add more letters. Pre algebra teaches you how to solve problems with one letter in them pretty much. Pre algebra prepares you for algebra 1. | 677.169 | 1 |
An Introduction to Mathematical Analysis for Economic Theory and Econometrics
4.11 - 1251 ratings - Source
qUnlike other mathematics textbooks for economics, An Introduction to Mathematical Analysis for Economic Theory and Econometrics takes a unified approach to understanding basic and advanced spaces through the application of the Metric Completion Theorem. This is the concept by which, for example, the real numbers complete the rational numbers and measure spaces complete fields of measurable sets. Another of the book's unique features is its concentration on the mathematical foundations of econometrics. To illustrate difficult concepts, the authors use simple examples drawn from economic theory and econometrics.q qAccessible and rigorous, the book is self-contained, providing proofs of theorems and assuming only an undergraduate background in calculus and linear algebra.q--BOOK JACKET.This is the concept by which, for example, the real numbers complete the rational numbers and measure spaces complete fields of measurable sets.
Title
:
An Introduction to Mathematical Analysis for Economic Theory and Econometrics
Author
:
Dean Corbae, Maxwell B. Stinchcombe, Juraj Zeman
Publisher
:
Greenwood Publishing Group - 2009-03-09
ISBN-13
:
Continue
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An introduction to one of the major branches of modern mathematics covering
axiomatic development of topological spaces and metric spaces, and the
concepts of convergence, continuity, separation, compactness and connectedness.
OBJECTIVES:
It is the purpose of this course to introduce the student to one of
the major branches of modern mathematics. As such, it is of itself worthy
of his attention. In addition, several important areas of mathematics,
in particular modern analysis, including elementary calculus, depend upon
or are clarified by certain topological concepts. To this end the course
is designed to emphasize these concepts and to advance his mathematical
maturity by exposing the student to other mathematical rigor and abstraction.
CONTENT:
1. Introduction
1.1 Set theory
1.2 Indexing notation
1.3 Function theory
2. Topologies and Topological Spaces
2.1 Open and closed sets
2.2 Metric spaces
2.3 Neighborhoods
2.4 Bases for a topology
2.5 Closure, interior and boundary
2.6 Continuity
2.7 Homeomorphism
2.8 Separation axioms - T sub 0, T sub 1, T sub 2
3. Connectedness
3.1 Connectedness or R
3.2 Applications
3.3 Separated sets
4. Compactness
4.1 Compact spaces
4.2 Compact subsets of R
4.3 Heine-Borel Theorem
TEXTS:
Texts, presenting this material for a first course of general topology,
such as the following may be appropriate: | 677.169 | 1 |
For any high school, community college or vocational/technical college student needing mathematics as a prep. for a career in electronics and computers.Best-selling author Nigel Cook's new text Mathematics for Electronics and Computers provides a complete math course for technology students. Since mathematics is interwoven into the very core of science, an understanding is imperative for anyone pursuing a career in technology. Employing an "integrated math applications" approach, this text reinforces all math topics with extensive electronic and computer applications to show a student the value of math as a tool. Author : Nigel P. Cook ISBN : 0130811629 Language : English No of Pages : 784 Edition : 1st Publication Date : 7/29/2002 Format/Binding : Paperback Book dimensions : 11.02x8.34x1.1 Book weight : 0.03 | 677.169 | 1 |
Amazing Profit
Students explore the concept of linear equations. In this linear equations lesson plan, students use linear equations to determine eBay profit on new technology such as a PS3. Students discuss how eBay works and how it illustrates supply and demand. | 677.169 | 1 |
Showing 1 to 3 of 3
This course really seeks to improve your refine math skills and problem solving. It prepares you for life outside of school as most of the math learned in this class will stay with you your entire life.
Course highlights:
Problem solving, graphing, and working with equations were the main points of this course. Not only did you learn the skills, but you also learned to keep persevering. If one equation didn't work, try another.
Hours per week:
9-11 hours
Advice for students:
Study efficiently and practice often to maintain a strong understanding of the work. Don't be afraid to ask for help if you need it.
I recommend taking it because it gives you the foundation for pre-cal and physics which both use applied mathematics in real world situations, which is also the foundation for many types of engineering, a large and lucrative career field.
Course highlights:
I learned many formulas for new concepts such as conics and parabolas and I learned to manipulate them in order to isolate the variable.
Hours per week:
9-11 hours
Advice for students:
You need to pay attention to the details on how to solve all of the equations you learn and fill in the blanks of all of the notes.
Course Term:Spring 2016
Professor:Liebsch
Course Required?Yes
Course Tags:Math-heavyGo to Office HoursMany Small Assignments
Mar 21, 2016
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
Not only is this class the perfect challenge, the teacher, Tracy Popescu, is great at explaining and describing the solution of each and every problem. Mrs. P. has a gift of teaching and is always helping students understand the reason why to every problem.
Course highlights:
I learned most of the high school basics in this class. Mrs Popescu taught us all the typical fundamentals of math needed for everyday lives.
Hours per week:
3-5 hours
Advice for students:
Stay on your feet it's a great course! As long as you pay attention in class, do your homework, and ask questions in times of confusion, you will be just fine. | 677.169 | 1 |
Issues in school mathematics programs are considered in this book. In the first section, mathematics education at the start of the decade is examined through two articles: a discussion of the status of mathematics programs and of prospects for the 80s, and an interpretation of the results of the second national mathematics assessment of the National Assessment of Educational Progress (NAEP). The next five articles discuss the basics: the role of computation in the curriculum, the basic nature of measurement, the increasing importance of estimation, experiences with finding and using data, and the scope of problem solving in the curriculum. Section 3, on the tools of technology, contains suggestions for using calculators and a discussion of the increasing importance of computers in school mathematics. In section 4, mathematics as a critical filter is considered in two final articles: one on women and mathematics and one on the case for a new high school mathematics curriculum. (MNS) | 677.169 | 1 |
...The business development requires significant advanced mathematics, statistics, and multiple-variable analysis for a large range of business, economic, engineering and design issues. Sincerely,
Dr. Bob I am a sophisticated instructor of advanced science courses with all the complementary mathematics | 677.169 | 1 |
Building a toolbox for science and math literacy
Miki Kelley
A report issued in June by the National Research Council emphasizes the importance of science education and recommends ways to improve K-12 education in science, technology, engineering and math, the so-called STEM subjects.
"A growing number of jobs – not just those in professional science – require knowledge of STEM fields," says Adam Gamoran, chair of the NRC committee that produced the report. "We need to help all students become scientifically literate because citizens are increasingly facing decisions related to science and technology – whether it's understanding a medical diagnosis or weighing competing claims about the environment."
But recent test scores indicate that California's students perform abysmally in science, especially economically disadvantaged and minority students, according to Maria Simani, the executive director of the California Science Project and a physicist at the University of California Riverside. She laments that in many schools she sees a lack of advanced science courses and too few students taking them.
Highly integrated with science – whether we like it or not – is a proficiency in math. And learning math is like buying tools for a project. The American physicist and Nobel laureate Richard Feynman even referred to math methods as "tools."
You would never attempt to build a house without investing in the right tools. In the same way, you wouldn't want to solve a difficult math problem without first understanding some basic concepts and patterns. These are your tools, your tricks of the trade.
Yet students arrive each year in college with math toolboxes so spare that we are learning backward. As a physics major, I didn't have the basic concepts of linear algebra neatly arranged in my toolbox, and neither did many of my peers. Instead, we learned those concepts out of necessity while tackling our quantum mechanics homework. It was like being asked to cook a gourmet meal when we barely knew how to peel potatoes.
At the very least, high school students need to be challenged to be math literate and well prepared for college. Where will the math help you down the line? Calculus is useful to solve standard mechanics problems. A solid foundation in math makes understanding electricity and magnetism much easier.
I would have appreciated statistics much more had I known that it was largely the foundation of artificial intelligence (AI). In college I took an AI introductory course that I soon found fascinating. We wrote computer programs that took data from the real world and produced a meaningful output. But without a solid grasp of basic statistics, I struggled through the class.
What could have helped me? The answer, I believe, is challenge and early exposure, which often come from mentors, the kind that coach open-ended group activities like Math or Science Olympiad, a robotics club, fieldwork outings, and hands-on labs.
What did help me in the end was my modern physics course professor, who challenged me with a goal.
One day he told me: "All of the excitement about physics really comes from modern physics, so if you do well in the class and you like the material, you should major in physics."
But then he went further: "What's your grade in this class right now?"
"I think I have a B."
"That's not good enough," he said. "If you bring your grade up to an A by the end of the course, I'll buy you a copy of "Mr. Tompkins in Paperback." (In the book, Mr. Tompkins is an everyman bank clerk who encounters the concepts of modern physics through a series of fantastic dreams and adventures.)
Before my professor challenged me, I could have cared less whether I got an A or B. Passing is passing. But now I had to prove to him – and myself – that I could get an A. I studied harder with the intent to get an A (and a free book), steadily sharpening my math tools. I practiced the problems again and again, and would not budge until I got them right.
I would not have put in the effort without my professor's timely challenge. What it engendered in me was the curiosity to delve further into physics. And practicing math wasn't so bad. In fact, it paid off.
I got an A in the course, and soon afterward I chose to major in physics. | 677.169 | 1 |
Numerical Methods - Solution of equations & Eigen value problems
Introduction :
Numerical Methods play an indispensible role in solving real life Mathematical, physical & Engg. Problems. Sometimes, the analytical method fails or unable to give desirable solutions to such problems. In such cases, we go for numerical methods.
The aim of numerical analysis is therefore, to provide constructive methods for obtaining approximate answers to such problems, using only simple arithmetic operations.
The advent of digital computers has, however, enhanced the speed and accuracy of numerical computations. | 677.169 | 1 |
Combinatorial Geometry in the Plane
4.11 - 1251 ratings - Source
Advanced undergraduate-level text discusses theorems on topics restricted to the plane, such as convexity, coverings, and graphs. Two-part treatment begins with specific topics followed by an extensive selection of short proofs. 1964 edition.Based on classical principles, this book is intended for a second course in
Euclidean geometry and can be used as a refresher. More than 200 problems
include hints and solutions. 1968 edition. 272pp. 5 3/8 x 8 1/2. 0-486-47720-7
TOPOLOGY OF3-MANIFOLDS AND RELATED TOPICS, Edited ... Additional
enrichment materials make it equally valuable as a reference. 1964 edition.
336pp. 5 3/8 x 8 1/2.
Title
:
Combinatorial Geometry in the Plane
Author
:
Hugo Hadwiger, Hans Debrunner, Victor Klee
Publisher
:
Courier Corporation - 2015-01-15
ISBN-13
:
Continue
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Principles of Linear Algebra with Mathematica
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Many topics in linear algebra are simple, yet computationally intensive, and computer algebra systems such as Mathematica® are essential not only for learning to apply the concepts to computationally challenging problems, but also for visualizing many of the geometric aspects within this field of study. Principles of Linear Algebra with Mathematica uniquely bridges the gap between beginning linear algebra and computational linear algebra that is often encountered in applied settings, and the commands required to solve complex and computationally challenging problems using Mathematica are provided.
The book begins with an introduction to the commands and programming guidelines for working with Mathematica. Next, the authors explore linear systems of equations and matrices, applications of linear systems and matrices, determinants, inverses, and Cramer's rule. Basic linear algebra topics, such as vectors, dot product, cross product, and vector projection are explored, as well as a unique variety of more advanced topics including rotations in space, 'rolling' a circle along a curve, and the TNB Frame. Subsequent chapters feature coverage of linear transformations from Rn to Rm, the geometry of linear and affine transformations, with an exploration of their effect on arclength, area, and volume, least squares fits, and pseudoinverses.
Mathematica is used to enhance concepts and is seamlessly integrated throughout the book through symbolic manipulations, numerical computations, graphics in two and three dimensions, animations, and programming. Each section concludes with standard problems in addition to problems that were specifically designed to be solved with Mathematica, allowing readers to test their comprehension of the presented material. All related Mathematica code is available on a corresponding website, along with solutions to problems and additional topical resources.
Extensively class-tested to ensure an accessible presentation, Principles of Linear Algebra with Mathematica is an excellent book for courses on linear algebra at the undergraduate level. The book is also an ideal reference for students and professionals who would like to gain a further understanding of the use of Mathematica to solve linear algebra problems Miss Billie's Decision by Eleanor HodgmanPrison is an unknown world for most of us. It is a place where time stops and lives are held in suspension, taken out of circulation. Amongst the jail population are the dangerous inmates: killers and rapists, gang 'hit-men' and…
About Kenneth M. Shiskowski, Karl Frinkle
Regrettably, your database involving ebooks not observed specifics of this author Kenneth M. Shiskowski. Although our company is always spending so much time to seek out and add brand-new information. Knowing the info the author, you can include the idea with the type to provide an overview. | 677.169 | 1 |
This traditional treatment of abstract algebra is designed for the particular needs of the mathematics teacher. Readers must have access to a Computer Algebra System (C. A. S.) such as Maple, or at minimum a calculator such as the TI 89 with C. A. S. capabilities. Includes "To the Teacher" sections that Draw connections from the number theory or abstract algebra under consideration to secondary mathematics. Provides historical context with "From the Past" sections in each chapter. Features "Worksheets" that outline the framework of a topic in most chapters. A useful reference for mathematics teachers who need to brush up on their abstract algebra skills.
An Introduction to Abstract Algebra with Notes to the Future Teacher, 1/E | 677.169 | 1 |
Students often get lost in multi-step math problems. This PowerPoint lesson is unique because it uses a flow-through technique, guided animation, that helps to eliminate confusion and guides the student through the problem. The lesson highlights each step of the problem as the teacher is discussing it, and then animates it to the next step within the lesson. Every step of every problem is shown, even the minor or seemingly insignificant steps. A helpful color-coding technique engages the students and guides them through the problem (Green is for the answer, red for wrong or canceled numbers, & blue, purple & sometimes orange for focusing the next step or separating things.) Twice as many examples are provided, compared to a standard textbook. All lessons have a real-world example to aid the students in visualizing a practical application of the concept.
This lesson applies to the Common Core Standard:
High School: Algebra » Creating Equations A.CED.1, A.CED.3
Create equations that describe numbers or relationships.
1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.
Algebra » Reasoning with Equations & Inequalities A.REI.3, A.REI.12
Solve equations and inequalities in one variable.
3. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Represent and solve equations and inequalities graphically.
12.
Please note that these PowerPoints are not editable.
If you need an alternative version because your country uses different measurements, units, slight wording adjustment for language differences, or a slide reordering just ask.
Are you looking for the Algebra 1 Curriculum Bundle? Coming Soon!Click here!
This resource is for one teacher only. You may not upload this resource to the internet in any form. Additional teachers must purchase their own license. If you are a coach, principal or district interested in purchasing several licenses, please contact me for a district-wide quote at prestonpowerpoints@gmail.com. This product may not be uploaded to the internet in any form, including classroom/personal websites or network drives | 677.169 | 1 |
Mechanics: v.3 (Advanced Modular Mathematics) (Vol 3)
Complete coverage of the Applied Mathematics required for the 'Mechanics 3' module of the new A-level syllabuses. A student-friendly handbook designed to support practice, revision, and exam success. Covers all the content required to complete the Mechanics 3 module for A-level Mathematics706 | 677.169 | 1 |
two-fifths the size of Stewart's's website, Despite the more compact size, the book has a modern flavor, covering technology and incorporating material to promote conceptual understanding, though not as prominently as in Stewart's other books. SINGLE VARIABLE ESSENTIAL CALCULUS: EARLY TRANSCENDENTALS features the same attention to detail, eye for innovation, and meticulous accuracy that have made Stewart's textbooks the best-selling calculus texts in the world. | 677.169 | 1 |
Linear Programming
In this linear programming worksheet, 11th graders solve and complete 4 different word problems that include various applications of linear programming. First, they find the maximum earning of a situation given the time worked. Then, students write the inequalities that represent the described information | 677.169 | 1 |
PISA Learning Mathematics for Life A Perspective from PISA
4.11 - 1251 ratings - Source
Learning Mathematics for Life examines the link between the PISA test requirements and student performance. It focuses specifically on the proportions of students who answer questions correctly across a range of difficulty. The questions are classified by content, competencies, context and format.THE OECD PROGRAMME FOR INTERNATIONAL STUDENT ASSESSMENT (PISA) PISA is a collaborative process among the 30 member countries of the OECD and nearly 30 partner countries and economies.
Title
:
PISA Learning Mathematics for Life A Perspective from PISA
Author
:
OECD
Publisher
:
OECD Publishing - 2010-03-04
ISBN-13
:
Continue
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In the first term of my Calculus, I had a hard time to review all my previous Math subjects, like Trigonometry and Algebra. I don't have any idea in Calculus; I just know that it's harder and more complicated than my previous Math subjects. After the midterm of this Semester I somehow know what Calculus for is, I learned that Differential Calculus can be use in finding the change in ratio. I also know that Calculus can be use to find the maximum and minimum rate of objects. It can also apply in other situation that involve changing and math. Using the concept of function derivatives, it studies the behavior and rate on how different quantities change. Using the process of differentiation, the graph of a function can actually be computed, analyzed, and predicted.
Though it is complicated to use well, calculus does have a lot of practical uses - uses that you probably won't comprehend at first. The most common practical use of calculus is when plotting graphs of certain formulae or functions. Using methods such as the first derivative and the second derivative, a graph and its dimensions can be accurately estimated. These 2 derivatives are used to predict how a graph may look like, the direction that it is taking on a specific point, the shape of the graph at a specific point (if concave or convex), just to name a few. When do you use calculus in the real world? In fact, you can use calculus in a lot of ways and applications. Among the disciplines that utilize calculus include physics, engineering, economics, statistics, and medicine. It is used to create mathematical models in order to arrive into an optimal solution. For example, in physics, calculus is used in a lot of its concepts. Among the physical concepts that use concepts of calculus include motion, electricity, heat, light, harmonics, acoustics, astronomy, and dynamics. In fact, even advanced physics concepts including electromagnetism and Einstein's theory of...
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Calculus, historically known as infinitesimal calculus, is a mathematical discipline focused on limits, functions, derivatives, integrals, and infinite series. Ideas leading up to the notions of function, derivative, and integral were developed throughout the 17th century, but the decisive step was made by Isaac Newton and Gottfried Leibniz. Publication of Newton's main treatises took many years, whereas Leibniz...
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Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, Differential Calculus and Integral Calculus, which are related by the fundamental theorem of calculus.
Calculus is the...
...deep realization about marketing begins its existence when I was browsing on the Internet. An author named Seth Godin captures my oblivious being in view of marketing with an aphorism that goes this way, "marketing is a contest for people's attention". This line must be too short to consider but it carries a lot of denotations which could be branched out to many ideas. Thinking, with respect to my instructor's introduction of Marketing Management, I...
...THE HISTORY OF CALCULUS
The discovery of calculus is often attributed to two men, Isaac Newton and Gottfried Leibniz, who independently developed its foundations. Although they both were instrumental in its creation, they thought of the fundamental concepts in very different ways. While Newton considered variables changing with time, Leibniz thought of the variables x and y as ranging over sequences of infinitely close values. He introduced dx and dy as...
...The Simple Ledger: Ledger Accounts
You will now learn the system used to maintain an up to date financial position.
They use an account and ledger.
Account:
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Ledger:
All the accounts together are called the ledger
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Teaching Mathematics in Colleges and Universities
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Progress in mathematics frequently occurs first by studying particular examples and then by generalizing the patterns that have been observed into far-reaching theorems. Similarly, in teaching mathematics one often employs examples to motivate a general principle or to illustrate its use. This volume uses the same idea in the context of learning how to teach: By analyzing particular teaching situations, one can develop broadly applicable teaching skills useful for the professional mathematician. These teaching situations are the Case Studies of the title. Just as a good mathematician seeks both to understand the details of a particular problem and to put it in a broader context, the examples presented here are chosen to offer a serious set of detailed teaching issues and to afford analysis from a broad perspective. Each case raises a variety of pedagogical and communication issues that may be explored either individually or in a group facilitated by a faculty member. Teaching notes for such a facilitator are included for each Case in the Faculty Edition. The methodology of Case Studies is widely used in areas such as business and law. The consideration of the mathematics cases presented here will help readers to develop teaching skills for their own classrooms. See the graduate edition at Teaching Mathematics in Colleges and Universities: Case Studies for Today's Classroom: Graduate Student EditionCase Studies for Todaya#39;s Classroom Solomon Friedberg ... A linear algebra
course? 2. ... (Answers include bringing in connections to other disciplines such
as physics, computer science, and economics, mentioning connections to otheranbsp;...
Title
:
Teaching Mathematics in Colleges and Universities
Author
:
Solomon Friedberg
Publisher
:
American Mathematical Soc. - 2001
ISBN-13
:
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Vector
Calculus
Michael Corral
Vector Calculus
Michael Corral
Schoolcraft College
About the author:
Michael Corral is an Adjunct Faculty member of the Department of Mathematics
at Schoolcraft College. He received a B.A. in Mathematics from the University
of California at Berkeley, and received an M.A. in Mathematics and an M.S. in
Industrial & Operations Engineering from the University of Michigan.
This text was typeset in L
A
T
E
X2
ε
with the KOMA-Script bundle, using the GNU
Emacs text editor on a Fedora Linux system. The graphics were created using
MetaPost, PGF, and Gnuplot.
Copyright c 2008 Michael Corral.
and no Back-Cover Texts. A copy of the license is included in the section entitled
"GNU Free Documentation License".
Preface
This book covers calculus in two and three variables. It is suitable for a one-semester
course, normally known as "Vector Calculus", "Multivariable Calculus", or simply
"Calculus III". The prerequisites are the standard courses in single-variable calculus
(a.k.a. Calculus I and II).
I have tried to be somewhat rigorous about proving results. But while it is impor-
tant for students to see full-blown proofs - since that is how mathematics works - too
much rigor and emphasis on proofs can impede the flow of learning for the vast ma-
jority of the audience at this level. If I were to rate the level of rigor in the book on a
scale of 1 to 10, with 1 being completely informal and 10 being completely rigorous, I
would rate it as a 5.
There are 420 exercises throughout the text, which in my experience are more than
enough for a semester course in this subject. There are exercises at the end of each
section, divided into three categories: A, B and C. The A exercises are mostly of a
routine computational nature, the B exercises are slightly more involved, and the C
exercises usually require some effort or insight to solve. A crude way of describing A,
B and C would be "Easy", "Moderate" and "Challenging", respectively. However, many
of the B exercises are easy and not all the C exercises are difficult.
There are a few exercises that require the student to write his or her own com-
puter program to solve some numerical approximation problems (e.g. the Monte Carlo
method for approximating multiple integrals, in Section 3.4). The code samples in the
text are in the Java programming language, hopefully with enough comments so that
the reader can figure out what is being done even without knowing Java. Those exer-
cises do not mandate the use of Java, so students are free to implement the solutions
using the language of their choice. While it would have been simple to use a script-
ing language like Python, and perhaps even easier with a functional programming
language (such as Haskell or Scheme), Java was chosen due to its ubiquity, relatively
clear syntax, and easy availability for multiple platforms.
Answers and hints to most odd-numbered and some even-numbered exercises are
provided in Appendix A. Appendix B contains a proof of the right-hand rule for the
cross product, which seems to have virtually disappeared from calculus texts over
the last few decades. Appendix C contains a brief tutorial on Gnuplot for graphing
functions of two variables.
This book is released under the GNU Free Documentation License (GFDL), which
allows others to not only copy and distribute the book but also to modify it. For more
details, see the included copy of the GFDL. So that there is no ambiguity on this
iii
iv Preface
matter, anyone can make as many copies of this book as desired and distribute it
as desired, without needing my permission. The PDF version will always be freely
available to the public at no cost (go to Feel free to
contact me at mcorral@schoolcraft.edu for any questions on this or any other
matter involving the book (e.g. comments, suggestions, corrections, etc). I welcome
your input.
Finally, I would like to thank my students in Math 240 for being the guinea pigs
for the initial draft of this book, and for finding the numerous errors and typos it
contained.
January 2008 MICHAEL CORRAL
Contents
Preface iii
1 Vectors in Euclidean Space 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.4 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5 Lines and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.6 Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1.7 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
1.8 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
1.9 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
2 Functions of Several Variables 65
2.1 Functions of Two or Three Variables . . . . . . . . . . . . . . . . . . . . . 65
2.2 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
2.3 Tangent Plane to a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . 75
2.4 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . 78
2.5 Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
2.6 Unconstrained Optimization: Numerical Methods . . . . . . . . . . . . . 89
2.7 Constrained Optimization: Lagrange Multipliers . . . . . . . . . . . . . . 96
3 Multiple Integrals 101
3.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
3.2 Double Integrals Over a General Region . . . . . . . . . . . . . . . . . . . 105
3.3 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
3.4 Numerical Approximation of Multiple Integrals . . . . . . . . . . . . . . 113
3.5 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . 117
3.6 Application: Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . 124
3.7 Application: Probability and Expected Value . . . . . . . . . . . . . . . . 128
4 Line and Surface Integrals 135
4.1 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
4.2 Properties of Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 143
4.3 Green's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
v
vi Contents
4.4 Surface Integrals and the Divergence Theorem . . . . . . . . . . . . . . . 156
4.5 Stokes' Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
4.6 Gradient, Divergence, Curl and Laplacian . . . . . . . . . . . . . . . . . . 177
Bibliography 187
Appendix A: Answers and Hints to Selected Exercises 189
Appendix B: Proof of the Right-Hand Rule for the Cross Product 192
Appendix C: 3D Graphing with Gnuplot 196
GNU Free Documentation License 201
History 209
Index 210
1 Vectors in Euclidean Space
1.1 Introduction
In single-variable calculus, the functions that one encounters are functions of a vari-
able (usually x or t) that varies over some subset of the real number line (which we
denote by ). For such a function, say, y = f (x), the graph of the function f con-
sists of the points (x, y) = (x, f (x)). These points lie in the Euclidean plane, which,
in the Cartesian or rectangular coordinate system, consists of all ordered pairs of
real numbers (a, b). We use the word "Euclidean" to denote a system in which all the
usual rules of Euclidean geometry hold. We denote the Euclidean plane by
2
; the
"2" represents the number of dimensions of the plane. The Euclidean plane has two
perpendicular coordinate axes: the x-axis and the y-axis.
In vector (or multivariable) calculus, we will deal with functions of two or three vari-
ables (usually x, y or x, y, z, respectively). The graph of a function of two variables, say,
z = f (x, y), lies in Euclidean space, which in the Cartesian coordinate systemconsists
of all ordered triples of real numbers (a, b, c). Since Euclidean space is 3-dimensional,
we denote it by
3
. The graph of f consists of the points (x, y, z) = (x, y, f (x, y)). The
3-dimensional coordinate system of Euclidean space can be represented on a flat sur-
face, such as this page or a blackboard, only by giving the illusion of three dimensions,
in the manner shown in Figure 1.1.1. Euclidean space has three mutually perpendic-
ular coordinate axes (x, y and z), and three mutually perpendicular coordinate planes:
the xy-plane, yz-plane and xz-plane (see Figure 1.1.2).
x
y
z
0
P(a, b, c)
a
b
c
Figure 1.1.1
x
y
z
0
yz-plane
xy-plane
xz-plane
Figure 1.1.2
1
2 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
The coordinate system shown in Figure 1.1.1 is known as a right-handed coordi-
nate system, because it is possible, using the right hand, to point the index finger in
the positive direction of the x-axis, the middle finger in the positive direction of the
y-axis, and the thumb in the positive direction of the z-axis, as in Figure 1.1.3.
Figure 1.1.3 Right-handed coordinate system
An equivalent way of defining a right-handed system is if you can point your thumb
upwards in the positive z-axis direction while using the remaining four fingers to
rotate the x-axis towards the y-axis. Doing the same thing with the left hand is what
defines a left-handed coordinate system. Notice that switching the x- and y-axes
in a right-handed system results in a left-handed system, and that rotating either
type of system does not change its "handedness". Throughout the book we will use a
right-handed system.
For functions of three variables, the graphs exist in 4-dimensional space (i.e.
4
),
which we can not see in our 3-dimensional space, let alone simulate in 2-dimensional
space. So we can only think of 4-dimensional space abstractly. For an entertaining
discussion of this subject, see the book by ABBOTT.
1
So far, we have discussed the position of an object in 2-dimensional or 3-dimensional
space. But what about something such as the velocity of the object, or its acceleration?
Or the gravitational force acting on the object? These phenomena all seem to involve
motion and direction in some way. This is where the idea of a vector comes in.
1
One thing you will learn is why a 4-dimensional creature would be able to reach inside an egg and
remove the yolk without cracking the shell!
1.1 Introduction 3
You have already dealt with velocity and acceleration in single-variable calculus.
For example, for motion along a straight line, if y = f (t) gives the displacement of
an object after time t, then dy/dt = f
′
(t) is the velocity of the object at time t. The
derivative f
′
(t) is just a number, which is positive if the object is moving in an agreed-
upon "positive" direction, and negative if it moves in the opposite of that direction. So
you can think of that number, which was called the velocity of the object, as having
two components: a magnitude, indicated by a nonnegative number, preceded by a
direction, indicated by a plus or minus symbol (representing motion in the positive
direction or the negative direction, respectively), i.e. f
′
(t) = ±a for some number a ≥ 0.
Then a is the magnitude of the velocity (normally called the speed of the object), and
the ± represents the direction of the velocity (though the + is usually omitted for the
positive direction).
For motion along a straight line, i.e. in a 1-dimensional space, the velocities are
also contained in that 1-dimensional space, since they are just numbers. For general
motion along a curve in 2- or 3-dimensional space, however, velocity will need to be
represented by a multidimensional object which should have both a magnitude and a
direction. A geometric object which has those features is an arrow, which in elemen-
tary geometry is called a "directed line segment". This is the motivation for how we
will define a vector.
Definition 1.1. A (nonzero) vector is a directed line segment drawn from a point P
(called its initial point) to a point Q (called its terminal point), with P and Q being
distinct points. The vector is denoted by
−−→
PQ. Its magnitude is the length of the line
segment, denoted by
_
_
_
−−→
PQ
_
_
_, and its direction is the same as that of the directed line
segment. The zero vector is just a point, and it is denoted by 0.
To indicate the direction of a vector, we draw an arrow from its initial point to its
terminal point. We will often denote a vector by a single bold-faced letter (e.g. v) and
use the terms "magnitude" and "length" interchangeably. Note that our definition
could apply to systems with any number of dimensions (see Figure 1.1.4 (a)-(c)).
0
x P Q R S
−−→
PQ
−−→
RS
(a) One dimension
x
y
0
P
Q
R
S
−
−
→
P
Q
−−→
RS
v
(b) Two dimensions
x
y
z
0
P
Q
R
S
−
−
→
P
Q
−−→
R
S
v
(c) Three dimensions
Figure 1.1.4 Vectors in different dimensions
4 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
A few things need to be noted about the zero vector. Our motivation for what a
vector is included the notions of magnitude and direction. What is the magnitude of
the zero vector? We define it to be zero, i.e. 0 = 0. This agrees with the definition of
the zero vector as just a point, which has zero length. What about the direction of the
zero vector? A single point really has no well-defined direction. Notice that we were
careful to only define the direction of a nonzero vector, which is well-defined since the
initial and terminal points are distinct. Not everyone agrees on the direction of the
zero vector. Some contend that the zero vector has arbitrary direction (i.e. can take
any direction), some say that it has indeterminate direction (i.e. the direction can not
be determined), while others say that it has no direction. Our definition of the zero
vector, however, does not require it to have a direction, and we will leave it at that.
2
Now that we know what a vector is, we need a way of determining when two vectors
are equal. This leads us to the following definition.
Definition 1.2. Two nonzero vectors are equal if they have the same magnitude and
the same direction. Any vector with zero magnitude is equal to the zero vector.
By this definition, vectors with the same magnitude and direction but with different
initial points would be equal. For example, in Figure 1.1.5 the vectors u, v and w all
have the same magnitude
√
5 (by the Pythagorean Theorem). And we see that u and
w are parallel, since they lie on lines having the same slope
1
2
, and they point in the
same direction. So u = w, even though they have different initial points. We also see
that v is parallel to u but points in the opposite direction. So u v.
1
2
3
4
1 2 3 4
x
y
0
u
v
w
Figure 1.1.5
So we can see that there are an infinite number of vectors for a given magnitude
and direction, those vectors all being equal and differing only by their initial and
terminal points. Is there a single vector which we can choose to represent all those
equal vectors? The answer is yes, and is suggested by the vector w in Figure 1.1.5.
2
In the subject of linear algebra there is a more abstract way of defining a vector where the concept of
"direction" is not really used. See ANTON and RORRES.
1.1 Introduction 5
Unless otherwise indicated, when speaking of "the vector" with a given magnitude
and direction, we will mean the one whose initial point is at the origin of the
coordinate system.
Thinking of vectors as starting from the origin provides a way of dealing with vec-
tors in a standard way, since every coordinate system has an origin. But there will
be times when it is convenient to consider a different initial point for a vector (for
example, when adding vectors, which we will do in the next section).
Another advantage of using the origin as the initial point is that it provides an easy
correspondence between a vector and its terminal point.
Example 1.1. Let v be the vector in
3
whose initial point is at the origin and whose
terminal point is (3, 4, 5). Though the point (3, 4, 5) and the vector v are different ob-
jects, it is convenient to write v = (3, 4, 5). When doing this, it is understood that the
initial point of v is at the origin (0, 0, 0) and the terminal point is (3, 4, 5).
x
y
z
0
P(3, 4, 5)
(a) The point (3,4,5)
x
y
z
0
v = (3, 4, 5)
(b) The vector (3,4,5)
Figure 1.1.6 Correspondence between points and vectors
Unless otherwise stated, when we refer to vectors as v = (a, b) in
2
or v = (a, b, c)
in
3
, we mean vectors in Cartesian coordinates starting at the origin. Also, we will
write the zero vector 0 in
2
and
3
as (0, 0) and (0, 0, 0), respectively.
The point-vector correspondence provides an easy way to check if two vectors are
equal, without having to determine their magnitude and direction. Similar to seeing
if two points are the same, you are now seeing if the terminal points of vectors starting
at the origin are the same. For each vector, find the (unique!) vector it equals whose
initial point is the origin. Then compare the coordinates of the terminal points of
these "new" vectors: if those coordinates are the same, then the original vectors are
equal. To get the "new" vectors starting at the origin, you translate each vector to
start at the origin by subtracting the coordinates of the original initial point from the
original terminal point. The resulting point will be the terminal point of the "new"
vector whose initial point is the origin. Do this for each original vector then compare.
6 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Example 1.2. Consider the vectors
−−→
PQand
−−→
RS in
3
, where P = (2, 1, 5), Q = (3, 5, 7), R =
(1, −3, −2) and S = (2, 1, 0). Does
−−→
PQ =
−−→
RS ?
Solution: The vector
−−→
PQ is equal to the vector v with initial point (0, 0, 0) and terminal
point Q − P = (3, 5, 7) − (2, 1, 5) = (3 − 2, 5 − 1, 7 − 5) = (1, 4, 2).
Similarly,
−−→
RS is equal to the vector w with initial point (0, 0, 0) and terminal point
S − R = (2, 1, 0) − (1, −3, −2) = (2 − 1, 1 − (−3), 0 − (−2)) = (1, 4, 2).
So
−−→
PQ = v = (1, 4, 2) and
−−→
RS = w = (1, 4, 2).
∴
−−→
PQ =
−−→
RS
y
z
x
0
−−→
P
Q
−−→
R
S
Translate
−−→
PQ to v
Translate
−−→
RS to w
P
(2, 1, 5)
Q
(3, 5, 7)
R
(1, −3, −2)
S
(2, 1, 0)
(1, 4, 2)
v = w
Figure 1.1.7
Recall the distance formula for points in the Euclidean plane:
For points P = (x
1
, y
1
), Q = (x
2
, y
2
) in
2
, the distance d between P and Q is:
d =
_
(x
2
− x
1
)
2
+ (y
2
− y
1
)
2
(1.1)
By this formula, we have the following result:
For a vector
−−→
PQ in
2
with initial point P = (x
1
, y
1
) and terminal point
Q = (x
2
, y
2
), the magnitude of
−−→
PQ is:
_
_
_
−−→
PQ
_
_
_ =
_
(x
2
− x
1
)
2
+ (y
2
− y
1
)
2
(1.2)
1.1 Introduction 7
Finding the magnitude of a vector v = (a, b) in
2
is a special case of formula (1.2)
with P = (0, 0) and Q = (a, b) :
For a vector v = (a, b) in
2
, the magnitude of v is:
v =
_
a
2
+ b
2
(1.3)
To calculate the magnitude of vectors in
3
, we need a distance formula for points
in Euclidean space (we will postpone the proof until the next section):
Theorem 1.1. The distance d between points P = (x
1
, y
1
, z
1
) and Q = (x
2
, y
2
, z
2
) in
3
is:
d =
_
(x
2
− x
1
)
2
+ (y
2
− y
1
)
2
+ (z
2
− z
1
)
2
(1.4)
The proof will use the following result:
Theorem 1.2. For a vector v = (a, b, c) in
3
, the magnitude of v is:
v =
_
a
2
+ b
2
+ c
2
(1.5)
Proof: There are four cases to consider:
Case 1: a = b = c = 0. Then v = 0, so v = 0 =
√
0
2
+ 0
2
+ 0
2
=
√
a
2
+ b
2
+ c
2
.
Case 2: exactly two of a, b, c are 0. Without loss of generality, we assume that a =
b = 0 and c 0 (the other two possibilities are handled in a similar manner). Then
v = (0, 0, c), which is a vector of length |c| along the z-axis. So v = |c| =
√
c
2
=
√
0
2
+ 0
2
+ c
2
=
√
a
2
+ b
2
+ c
2
.
Case 3: exactly one of a, b, c is 0. Without loss of generality, we assume that a = 0,
b 0 and c 0 (the other two possibilities are handled in a similar manner). Then
v = (0, b, c), which is a vector in the yz-plane, so by the Pythagorean Theorem we have
v =
√
b
2
+ c
2
=
√
0
2
+ b
2
+ c
2
=
√
a
2
+ b
2
+ c
2
.
x
y
z
0
a
Q(a, b, c)
S
P
R
b
c
v
Figure 1.1.8
Case 4: none of a, b, c are 0. Without loss of generality, we can
assume that a, b, c are all positive (the other seven possibil-
ities are handled in a similar manner). Consider the points
P = (0, 0, 0), Q = (a, b, c), R = (a, b, 0), and S = (a, 0, 0), as shown
in Figure 1.1.8. Applying the Pythagorean Theorem to the
right triangle △PS R gives |PR|
2
= a
2
+b
2
. A second application
of the Pythagorean Theorem, this time to the right triangle
△PQR, gives v = |PQ| =
_
|PR|
2
+ |QR|
2
=
√
a
2
+ b
2
+ c
2
.
This proves the theorem. QED
8 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Example 1.3. Calculate the following:
(a) The magnitude of the vector
−−→
PQ in
2
with P = (−1, 2) and Q = (5, 5).
Solution: By formula (1.2),
_
_
_
−−→
PQ
_
_
_ =
_
(5 − (−1))
2
+ (5 − 2)
2
=
√
36 + 9 =
√
45 = 3
√
5.
(b) The magnitude of the vector v = (8, 3) in
2
.
Solution: By formula (1.3), v =
√
8
2
+ 3
2
=
√
73.
(c) The distance between the points P = (2, −1, 4) and Q = (4, 2, −3) in
2
.
Solution: By formula (1.4), the distance d =
_
(4 − 2)
2
+ (2 − (−1))
2
+ (−3 − 4)
2
=
√
4 + 9 + 49 =
√
62.
(d) The magnitude of the vector v = (5, 8, −2) in
3
.
Solution: By formula (1.5), v =
_
5
2
+ 8
2
+ (−2)
2
=
√
25 + 64 + 4 =
√
93.
3
, i.e. equations of the form
Ax
2
+ By
2
+ Cz
2
+ Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0 (1.33)
for some constants A, B, . . . , J. If the above equation is not that of a sphere, cylinder,
plane, line or point, then the resulting surface is called a quadric surface.
y
z
x
0
a
b
c
Figure 1.6.4 Ellipsoid
One type of quadric surface is the ellipsoid,
given by an equation of the form:
x
2
a
2
+
y
2
b
2
+
z
2
c
2
= 1 (1.34)
In the case where a = b = c, this is just a sphere.
In general, an ellipsoid is egg-shaped (think of
an ellipse rotated around its major axis). Its
traces in the coordinate planes are ellipses.
Two other types of quadric surfaces are the hyperboloid of one sheet, given by
an equation of the form:
x
2
a
2
+
y
2
b
2
−
z
2
c
2
= 1 (1.35)
and the hyperboloid of two sheets, whose equation has the form:
x
2
a
2
−
y
2
b
2
−
z
2
c
2
= 1 (1.36)
y
z
x
0
Figure 1.6.5 Hyperboloid of one sheet
y
z
x
0
Figure 1.6.6 Hyperboloid of two sheets
44 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
For the hyperboloid of one sheet, the trace in any plane parallel to the xy-plane is
an ellipse. The traces in the planes parallel to the xz- or yz-planes are hyperbolas (see
Figure 1.6.5), except for the special cases x = ±a and y = ±b; in those planes the traces
are pairs of intersecting lines (see Exercise 8).
For the hyperboloid of two sheets, the trace in any plane parallel to the xy- or xz-
plane is a hyperbola (see Figure 1.6.6). There is no trace in the yz-plane. In any plane
parallel to the yz-plane for which | x| > |a|, the trace is an ellipse.
y
z
x
0
Figure 1.6.7 Paraboloid
The elliptic paraboloid is another type of quadric sur-
face, whose equation has the form:
x
2
a
2
+
y
2
b
2
=
z
c
(1.37)
The traces in planes parallel to the xy-plane are ellipses,
though in the xy-plane itself the trace is a single point. The
traces in planes parallel to the xz- or yz-planes are parabo-
las. Figure 1.6.7 shows the case where c > 0. When c < 0 the
surface is turned downward. In the case where a = b, the
surface is called a paraboloid of revolution, which is often
used as a reflecting surface, e.g. in vehicle headlights.
10
A more complicated quadric surface is the hyperbolic paraboloid, given by:
x
2
a
2
−
y
2
b
2
=
z
c
(1.38)
-10
-5
0
5
10
-10
-5
0
5
10
-100
-50
0
50
100
z
x
y
z
Figure 1.6.8 Hyperbolic paraboloid
10
For a discussion of this see pp. 157-158 in HECHT.
1.6 Surfaces 45
The hyperbolic paraboloid can be tricky to draw; using graphing software on a com-
puter can make it easier. For example, Figure 1.6.8 was created using the free Gnuplot
package (see Appendix C). It shows the graph of the hyperbolic paraboloid z = y
2
− x
2
,
which is the special case where a = b = 1 and c = −1 in equation (1.38). The mesh lines
on the surface are the traces in planes parallel to the coordinate planes. So we see
that the traces in planes parallel to the xz-plane are parabolas pointing upward, while
the traces in planes parallel to the yz-plane are parabolas pointing downward. Also,
notice that the traces in planes parallel to the xy-plane are hyperbolas, though in the
xy-plane itself the trace is a pair of intersecting lines through the origin. This is true
in general when c < 0 in equation (1.38). When c > 0, the surface would be similar to
that in Figure 1.6.8, only rotated 90
◦
around the z-axis and the nature of the traces in
planes parallel to the xz- or yz-planes would be reversed.
y
z
x
0
Figure 1.6.9 Elliptic cone
The last type of quadric surface that we will consider is
the elliptic cone, which has an equation of the form:
x
2
a
2
+
y
2
b
2
−
z
2
c
2
= 0 (1.39)
The traces in planes parallel to the xy-plane are ellipses,
except in the xy-plane itself where the trace is a single
point. The traces in planes parallel to the xz- or yz-planes
are hyperbolas, except in the xz- and yz-planes themselves
where the traces are pairs of intersecting lines.
Notice that every point on the elliptic cone is on a line
which lies entirely on the surface; in Figure 1.6.9 these
lines all go through the origin. This makes the elliptic
cone an example of a ruled surface. The cylinder is also a ruled surface.
What may not be as obvious is that both the hyperboloid of one sheet and the hy-
perbolic paraboloid are ruled surfaces. In fact, on both surfaces there are two lines
through each point on the surface (see Exercises 11-12). Such surfaces are called
doubly ruled surfaces, and the pairs of lines are called a regulus.
It is clear that for each of the six types of quadric surfaces that we discussed, the
surface can be translated away from the origin (e.g. by replacing x
2
by (x − x
0
)
2
in
its equation). It can be proved
11
that every quadric surface can be translated and/or
rotated so that its equation matches one of the six types that we described. For ex-
ample, z = 2xy is a case of equation (1.33) with "mixed" variables, e.g. with D 0 so
that we get an xy term. This equation does not match any of the types we considered.
However, by rotating the x- and y-axes by 45
◦
in the xy-plane by means of the coor-
dinate transformation x = (x
′
− y
′
)/
√
2, y = (x
′
+ y
′
)/
√
2, z = z
′
, then z = 2xy becomes
the hyperbolic paraboloid z
′
= (x
′
)
2
− (y
′
)
2
in the (x
′
, y
′
, z
′
) coordinate system. That is,
z = 2xy is a hyperbolic paraboloid as in equation (1.38), but rotated 45
◦
in the xy-plane.
11
See Ch. 7 in POGORELOV.
46 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
2
which is tangent to the level curve f (x, y) = c (see Figure 2.4.1).
x
y
0
v ∇f
f (x, y) = c
Figure 2.4.1
5
Sometimes the notation grad( f ) is used instead of ∇f .
2.4 Directional Derivatives and the Gradient 81
The value of f (x, y) is constant along a level curve, so since v is a tangent vector to
this curve, then the rate of change of f in the direction of v is 0, i.e. D
v
f = 0. But we
know that D
v
f = v ··· ∇f = v ∇f cos θ, where θ is the angle between v and ∇f . So
since v = 1 then D
v
f = ∇f cos θ. So since ∇f 0 then D
v
f = 0 ⇒ cos θ = 0 ⇒ θ = 90
◦
.
In other words, ∇f ⊥ v, which means that ∇f is normal to the level curve.
In general, for any unit vector v in
2
, we still have D
v
f = ∇f cos θ, where θ is the
angle between v and ∇f . At a fixed point (x, y) the length ∇f is fixed, and the value
of D
v
f then varies as θ varies. The largest value that D
v
f can take is when cos θ = 1
(θ = 0
◦
), while the smallest value occurs when cos θ = −1 (θ = 180
◦
). In other words, the
value of the function f increases the fastest in the direction of ∇f (since θ = 0
◦
in that
case), and the value of f decreases the fastest in the direction of −∇f (since θ = 180
◦
in that case). We have thus proved the following theorem:
Theorem 2.4. Let f (x, y) be a continuously differentiable real-valued function, with
∇f 0. Then:
(a) The gradient ∇f is normal to any level curve f (x, y) = c.
(b) The value of f (x, y) increases the fastest in the direction of ∇f .
(c) The value of f (x, y) decreases the fastest in the direction of −∇f .
Example 2.16. In which direction does the function f (x, y) = xy
2
+ x
3
y increase the
fastest from the point (1, 2)? In which direction does it decrease the fastest?
Solution: Since ∇f = (y
2
+ 3x
2
y, 2xy + x
3
), then ∇f (1, 2) = (10, 5) 0. A unit vector in
that direction is v =
∇f
∇f
=
_
2
√
5
,
1
√
5
_
. Thus, f increases the fastest in the direction of
_
2
√
5
,
1
√
5
_
and decreases the fastest in the direction of
_
−2
√
5
,
−1
√
5
_
.
Though we proved Theorem 2.4 for functions of two variables, a similar argument
can be used to show that it also applies to functions of three or more variables. Like-
wise, the directional derivative in the three-dimensional case can also be defined by
the formula D
v
f = v··· ∇f .
Example 2.17. The temperature T of a solid is given by the function T(x, y, z) = e
−x
+
e
−2y
+e
4z
, where x, y, z are space coordinates relative to the center of the solid. In which
direction from the point (1, 1, 1) will the temperature decrease the fastest?
Solution: Since ∇f = (−e
−x
, −2e
−2y
, 4e
4z
), then the temperature will decrease the fastest
in the direction of −∇f (1, 1, 1) = (e
−1
, 2e
−2
, −4e
4
).
82 CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES
R
f (x, y) dA is defined as follows. Assume that f (x, y)
is a nonnegative real-valued function and that R is a bounded region in
2
, so it can
be enclosed in some rectangle [a, b] × [c, d]. Then divide that rectangle into a grid of
subrectangles. Only consider the subrectangles that are enclosed completely within
the region R, as shown by the shaded subrectangles in Figure 3.2.5(a). In any such
subrectangle [x
i
, x
i+1
] ×[y
j
, y
j+1
], pick a point (x
i∗
, y
j∗
). Then the volume under the surface
z = f (x, y) over that subrectangle is approximately f (x
i∗
, y
j∗
) ∆x
i
∆y
j
, where ∆x
i
= x
i+1
− x
i
,
108 CHAPTER 3. MULTIPLE INTEGRALS
∆y
j
= y
j+1
−y
j
, and f (x
i∗
, y
j∗
) is the height and ∆x
i
∆y
j
is the base area of a parallelepiped,
as shown in Figure 3.2.5(b). Then the total volume under the surface is approximately
the sum of the volumes of all such parallelepipeds, namely
j
i
f (x
i∗
, y
j∗
) ∆x
i
∆y
j
, (3.6)
where the summation occurs over the indices of the subrectangles inside R. If we
take smaller and smaller subrectangles, so that the length of the largest diagonal of
the subrectangles goes to 0, then the subrectangles begin to fill more and more of
the region R, and so the above sum approaches the actual volume under the surface
z = f (x, y) over the region R. We then define
R
f (x, y) dA as the limit of that double
summation (the limit is taken over all subdivisions of the rectangle [a, b] ×[c, d] as the
largest diagonal of the subrectangles goes to 0).
x
y
0
d
c
y
j
y
j+1
a
b
x
i
x
i+1
(x
i∗
, y
j∗
)
(a) Subrectangles inside the region R
y
z
x
0
R
x
i
x
i+1
y
j
y
j+1
z = f (x, y)
∆y
j
∆x
i
(x
i∗
, y
j∗
)
f (x
i∗
, y
j∗
)
(b) Parallelepiped over a subrectan-
gle, with volume f (x
i∗
, y
j∗
) ∆x
i
∆y
j
Figure 3.2.5 Double integral over a general region R
A similar definition can be made for a function f (x, y) that is not necessarily always
nonnegative: just replace each mention of volume by the negative volume in the de-
scription above when f (x, y) < 0. In the case of a region of the type shown in Figure
3.2.1, using the definition of the Riemann integral from single-variable calculus, our
definition of
R
1 dA gives the area of the region R. For sim-
plicity, you can assume that R is a region of the type shown in Figure 3.2.1(a).
C
b
c
a
Figure 3.2.6
12. Prove that the volume of a tetrahedron with mutually per-
pendicular adjacent sides of lengths a, b, and c, as in Figure
3.2.6, is
abc
6
. (Hint: Mimic Example 3.5, and recall from
Section 1.5 how three noncollinear points determine a plane.)
13. Show how Exercise 12 can be used to solve Exercise 10.
110 CHAPTER 3. MULTIPLE INTEGRALS
3.3 Triple Integrals
Our definition of a double integral of a real-valued function f (x, y) over a region R in
2
can be extended to define a triple integral of a real-valued function f (x, y, z) over
a solid S in
3
. We simply proceed as before: the solid S can be enclosed in some
rectangular parallelepiped, which is then divided into subparallelepipeds. In each
subparallelepiped inside S , with sides of lengths ∆x, ∆y and ∆z, pick a point (x
∗
, y
∗
, z
∗
).
Then define the triple integral of f (x, y, z) over S , denoted by
R
1 dA.)
156 CHAPTER 4. LINE AND SURFACE INTEGRALS
4.4 Surface Integrals and the Divergence Theorem
In Section 4.1 we learned how to integrate along a curve. We will now learn how to
perform integration over a surface in
3
, such as a sphere or a paraboloid. Recall
from Section 1.8 how we identified points (x, y, z) on a curve C in
3
, parametrized by
x = x(t), y = y(t), z = z(t), a ≤ t ≤ b, with the terminal points of the position vector
r(t) = x(t)i + y(t)j + z(t)k for t in [a, b].
The idea behind a parametrization of a curve is that it "transforms" a subset of
1
(normally an interval [a, b]) into a curve in
2
or
3
(see Figure 4.4.1).
1
a
t
b
y
z
x
0
(x(a), y(a), z(a))
(x(t), y(t), z(t))
(x(b), y(b), z(b)) r(t)
C
x = x(t)
y = y(t)
z = z(t)
Figure 4.4.1 Parametrization of a curve C in
3
Similar to how we used a parametrization of a curve to define the line integral along
the curve, we will use a parametrization of a surface to define a surface integral. We
will use two variables, u and v, to parametrize a surface Σ in
3
: x = x(u, v), y = y(u, v),
z = z(u, v), for (u, v) in some region R in
2
(see Figure 4.4.2).
u
v
R
2
(u, v)
y
z
x
0
Σ
r(u, v)
x = x(u, v)
y = y(u, v)
z = z(u, v)
Figure 4.4.2 Parametrization of a surface Σ in
3
In this case, the position vector of a point on the surface Σ is given by the vector-
valued function
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k for (u, v) in R.
4.4 Surface Integrals and the Divergence Theorem 157
Since r(u, v) is a function of two variables, define the partial derivatives
∂r
∂u
and
∂r
∂v
for (u, v) in R by
∂r
∂u
(u, v) =
∂x
∂u
(u, v)i +
∂y
∂u
(u, v)j +
∂z
∂u
(u, v)k , and
∂r
∂v
(u, v) =
∂x
∂v
(u, v)i +
∂y
∂v
(u, v)j +
∂z
∂v
(u, v)k .
The parametrization of Σ can be thought of as "transforming" a region in
2
(in the
uv-plane) into a 2-dimensional surface in
3
. This parametrization of the surface is
sometimes called a patch, based on the idea of "patching" the region R onto Σ in the
grid-like manner shown in Figure 4.4.2.
In fact, those gridlines in R lead us to how we will define a surface integral over Σ.
Along the vertical gridlines in R, the variable u is constant. So those lines get mapped
to curves on Σ, and the variable u is constant along the position vector r(u, v). Thus, the
tangent vector to those curves at a point (u, v) is
∂r
∂v
. Similarly, the horizontal gridlines
in R get mapped to curves on Σ whose tangent vectors are
∂r
∂u
.
Now take a point (u, v) in R as, say, the lower left corner of one of the rectangular grid
sections in R, as shown in Figure 4.4.2. Suppose that this rectangle has a small width
and height of ∆u and ∆v, respectively. The corner points of that rectangle are (u, v),
(u +∆u, v), (u +∆u, v +∆v) and (u, v +∆v). So the area of that rectangle is A = ∆u ∆v. Then
that rectangle gets mapped by the parametrization onto some section of the surface
Σ which, for ∆u and ∆v small enough, will have a surface area (call it dσ) that is very
close to the area of the parallelogram which has adjacent sides r(u + ∆u, v) − r(u, v)
(corresponding to the line segment from (u, v) to (u + ∆u, v) in R) and r(u, v + ∆v) − r(u, v)
(corresponding to the line segment from (u, v) to (u, v + ∆v) in R). But by combining our
usual notion of a partial derivative (see Definition 2.3 in Section 2.2) with that of the
derivative of a vector-valued function (see Definition 1.12 in Section 1.8) applied to a
function of two variables, we have
∂r
∂u
≈
r(u + ∆u, v) − r(u, v)
∆u
, and
∂r
∂v
≈
r(u, v + ∆v) − r(u, v)
∆v
,
and so the surface area element dσ is approximately
_
_
_(r(u + ∆u, v) − r(u, v)) ××× (r(u, v + ∆v) − r(u, v))
_
_
_ ≈
_
_
_
_
_
_
(∆u
∂r
∂u
) ××× (∆v
∂r
∂v
)
_
_
_
_
_
_
=
_
_
_
_
_
_
∂r
∂u
×××
∂r
∂v
_
_
_
_
_
_
∆u ∆v
by Theorem 1.13 in Section 1.4. Thus, the total surface area S of Σ is approximately
the sum of all the quantities
_
_
_
∂r
∂u
×××
∂r
∂v
_
_
_ ∆u ∆v, summed over the rectangles in R. Taking
the limit of that sum as the diagonal of the largest rectangle goes to 0 gives
S26)
158 CHAPTER 4. LINE AND SURFACE INTEGRALS
We will write the double integral on the right using the special notation
_
Σ
dσ27)
This is a special case of a surface integral over the surface Σ, where the surface area
element dσ can be thought of as 1 dσ. Replacing 1 by a general real-valued function
f (x, y, z) defined in
3
, we have the following:
Definition 4.3. Let Σ be a surface in
3
parametrized by x = x(u, v), y = y(u, v),
z = z(u, v), for (u, v) in some region R in
2
. Let r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k be the
position vector for any point on Σ, and let f (x, y, z) be a real-valued function defined on
some subset of
3
that contains Σ. The surface integral of f (x, y, z) over Σ is
_
Σ
f (x, y, z) dσ =
_
R
f (x(u, v), y(u, v), z(u, v))28)
In particular, the surface area S of Σ is
S =
_
Σ
1 dσ . (4.29)
Example 4.9. A torus T is a surface obtained by revolving a circle of radius a in the
yz-plane around the z-axis, where the circle's center is at a distance b from the z-axis
(0 < a < b), as in Figure 4.4.3. Find the surface area of T.
y
z
0
a
(y − b)
2
+ z
2
= a
2
u
b
(a) Circle in the yz-plane
x
y
z
v
a
(x,y,z)
(b) Torus T
Figure 4.4.3
Solution: For any point on the circle, the line segment from the center of the circle
to that point makes an angle u with the y-axis in the positive y direction (see Figure
4.4 Surface Integrals and the Divergence Theorem 159
4.4.3(a)). And as the circle revolves around the z-axis, the line segment from the origin
to the center of that circle sweeps out an angle v with the positive x-axis (see Figure
4.4.3(b)). Thus, the torus can be parametrized as:
x = (b + a cos u) cos v , y = (b + a cos u) sin v , z = a sin u , 0 ≤ u ≤ 2π , 0 ≤ v ≤ 2π
So for the position vector
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
= (b + a cos u) cos v i + (b + a cos u) sin v j + a sin u k
we see that
∂r
∂u
= −a sin u cos v i − a sin u sin v j + a cos u k
∂r
∂v
= −(b + a cos u) sin v i + (b + a cos u) cos v j + 0k ,
and so computing the cross product gives
∂r
∂u
×××
∂r
∂v
= −a(b + a cos u) cos v cos u i − a(b + a cos u) sin v cos u j − a(b + a cos u) sin u k ,
which has magnitude
_
_
_
_
_
_
∂r
∂u
×××
∂r
∂v
_
_
_
_
_
_
= a(b + a cos u) .
Thus, the surface area of T is
S =
_
Σ
1 dσ
=
_
2π
0
_
2π
0
_
_
_
_
_
_
∂r
∂u
×××
∂r
∂v
_
_
_
_
_
_
du dv
=
_
2π
0
_
2π
0
a(b + a cos u) du dv
=
_
2π
0
_
abu + a
2
sin u
¸
¸
¸
¸
u=2π
u=0
_
dv
=
_
2π
0
2πab dv
= 4π
2
ab
Since
∂r
∂u
and
∂r
∂v
are tangent to the surface Σ (i.e. lie in the tangent plane to Σ at each
point on Σ), then their cross product
∂r
∂u
×××
∂r
∂v
is perpendicular to the tangent plane to
160 CHAPTER 4. LINE AND SURFACE INTEGRALS
the surface at each point of Σ. Thus,
_
Σ
f (x, y, z) dσ =
_
R
f (x(u, v), y(u, v), z(u, v)) n dσ ,
where n =
∂r
∂u
×××
∂r
∂v
. We say that n is a normal vector to Σ.
y
z
x
0
Figure 4.4.4
Recall that normal vectors to a plane can point in two
opposite directions. By an outward unit normal vector
to a surface Σ, we will mean the unit vector that is normal
to Σ and points away from the "top" (or "outer" part) of the
surface. This is a hazy definition, but the picture in Figure
4.4.4 gives a better idea of what outward normal vectors
look like, in the case of a sphere. With this idea in mind,
we make the following definition of a surface integral of a
3-dimensional vector field over a surface:
Definition 4.4. Let Σ be a surface in
3
and let f(x, y, z) = f
1
(x, y, z)i + f
2
(x, y, z)j +
f
3
(x, y, z)k be a vector field defined on some subset of
3
that contains Σ. The surface
integral of f over Σ is
_
Σ
f ··· dσ =
_
Σ
f ··· ndσ , (4.30)
where, at any point on Σ, n is the outward unit normal vector to Σ.
Note in the above definition that the dot product inside the integral on the right is
a real-valued function, and hence we can use Definition 4.3 to evaluate the integral.
Example 4.10. Evaluate the surface integral
Σ
f··· dσ is often referred to as the flux
of f through the surface Σ. For example, if f represents the velocity field of a fluid,
then the flux is the net quantity of fluid to flow through the surface Σ per unit time.
A positive flux means there is a net flow out of the surface (i.e. in the direction of the
outward unit normal vector n), while a negative flux indicates a net flow inward (in
the direction of −n).
The term divergence comes from interpreting div f as a measure of how much a
vector field "diverges" from a point. This is best seen by using another definition of
div f which is equivalent
4
to the definition given by formula (4.32). Namely, for a point
(x, y, z) in
3
,
div f(x, y, z) = lim
V→0
1
V
_
Σ
f ··· dσ , (4.33)
3
See TAYLOR and MANN, § 15.6 for the details.
4
See SCHEY, p. 36-39, for an intuitive discussion of this.
4.4 Surface Integrals and the Divergence Theorem 163
where V is the volume enclosed by a closed surface Σ around the point (x, y, z). In the
limit, V → 0 means that we take smaller and smaller closed surfaces around (x, y, z),
which means that the volumes they enclose are going to zero. It can be shown that this
limit is independent of the shapes of those surfaces. Notice that the limit being taken
is of the ratio of the flux through a surface to the volume enclosed by that surface,
which gives a rough measure of the flow "leaving" a point, as we mentioned. Vector
fields which have zero divergence are often called solenoidal fields.
The following theorem is a simple consequence of formula (4.33).
Theorem 4.9. If the flux of a vector field f is zero through every closed surface con-
taining a given point, then div f = 0 at that point.
Proof: By formula (4.33), at the given point (x, y, z) we have
div f(x, y, z) = lim
V→0
1
V
_
Σ
f ··· dσ for closed surfaces Σ containing (x, y, z), so
= lim
V→0
1
V
(0) by our assumption that the flux through each Σ is zero, so
= lim
V→0
0
= 0 . QED
Lastly, we note that sometimes the notation
Σ
f (x, y, z) dσ and
Σ
f ··· dσ
is used to denote surface integrals of scalar and vector fields, respectively, over closed
surfaces. Especially in physics texts, it is common to see simply
_
Σ
instead of
Σ
(curl f ) ··· ndσ, as predicted by Stokes' Theorem.
The line integral in the preceding example was far simpler to calculate than the
surface integral, but this will not always be the case.
Example 4.15. Let Σ be the elliptic paraboloid z =
x
2
4
+
y
2
9
for z ≤ 1, and let C be its
boundary curve. Calculate
_
C
f ··· dr for f(x, y, z) = (9xz + 2y)i + (2x + y
2
)j + (−2y
2
+ 2z)k,
where C is traversed counterclockwise.
Solution: The surface is similar to the one in Example 4.14, except now the boundary
curve C is the ellipse
x
2
4
+
y
2
9
= 1 laying in the plane z = 1. In this case, using Stokes'
Theorem is easier than computing the line integral directly. As in Example 4.14, at
each point (x, y, z(x, y)) on the surface z = z(x, y) =
x
2
4
+
y
2
9
the vector
n =
−
∂z
∂x
i −
∂z
∂y
j + k
_
1 +
_
∂z
∂x
_
2
+
_
∂z
∂y
_
2
=
−
x
2
i −
2y
9
j + k
_
1 +
x
2
4
+
4y
2
9
,
is a positive unit normal vector to Σ. And calculating the curl of f gives
curl f = (−4y − 0)i + (9x − 0)j + (2 − 2)k = −4y i + 9x j + 0 k ,
so
(curl f ) ··· n =
(−4y)(−
x
2
) + (9x)(−
2y
9
) + (0)(1)
_
1 +
x
2
4
+
4y
2
9
=
2xy − 2xy + 0
_
1 +
x
2
4
+
4y
2
9
= 0 ,
and so by Stokes' Theorem
_
C
f ··· dr =
_
Σ
(curl f ) ··· ndσ =
_
Σ
0 dσ = 0 .
174 CHAPTER 4. LINE AND SURFACE INTEGRALS
In physical applications, for a simple closed curve C the line integral
_
C
f··· dr is often
called the circulation of f around C. For example, if E represents the electrostatic
field due to a point charge, then it turns out
8
that curl E = 0, which means that the
circulation
_
C
E ··· dr = 0 by Stokes' Theorem. Vector fields which have zero curl are
often called irrotational fields.
In fact, the term curl was created by the 19
th
century Scottish physicist James Clerk
Maxwell in his study of electromagnetism, where it is used extensively. In physics,
the curl is interpreted as a measure of circulation density. This is best seen by using
another definition of curl f which is equivalent
9
to the definition given by formula
(4.46). Namely, for a point (x, y, z) in
3
,
n··· (curl f )(x, y, z) = lim
S →0
1
S
_
C
f ··· dr , (4.50)
where S is the surface area of a surface Σ containing the point (x, y, z) and with a
simple closed boundary curve C and positive unit normal vector n at (x, y, z). In the
limit, think of the curve C shrinking to the point (x, y, z), which causes Σ, the surface it
bounds, to have smaller and smaller surface area. That ratio of circulation to surface
area in the limit is what makes the curl a rough measure of circulation density (i.e.
circulation per unit area).
x
y
0
f
Figure 4.5.6 Curl and rotation
An idea of how the curl of a vector field is related
to rotation is shown in Figure 4.5.6. Suppose we
have a vector field f(x, y, z) which is always parallel
to the xy-plane at each point (x, y, z) and that the vec-
tors grow larger the further the point (x, y, z) is from
the y-axis. For example, f(x, y, z) = (1 + x
2
) j. Think
of the vector field as representing the flow of wa-
ter, and imagine dropping two wheels with paddles
into that water flow, as in Figure 4.5.6. Since the
flow is stronger (i.e. the magnitude of f is larger) as
you move away from the y-axis, then such a wheel
would rotate counterclockwise if it were dropped to
the right of the y-axis, and it would rotate clockwise if it were dropped to the left of
the y-axis. In both cases the curl would be nonzero (curl f(x, y, z) = 2x k in our example)
and would obey the right-hand rule, that is, curl f(x, y, z) points in the direction of your
thumb as you cup your right hand in the direction of the rotation of the wheel. So
the curl points outward (in the positive z-direction) if x > 0 and points inward (in the
negative z-direction) if x < 0. Notice that if all the vectors had the same direction and
the same magnitude, then the wheels would not rotate and hence there would be no
curl (which is why such fields are called irrotational, meaning no rotation).
8
See Ch. 2 in REITZ, MILFORD and CHRISTY.
9
See SCHEY, p. 78-81, for the derivation.
4.5 Stokes' Theorem 175
Finally, by Stokes' Theorem, we know that if C is a simple closed curve in some solid
region S in
3
and if f(x, y, z) is a smooth vector field such that curl f = 0 in S , then
_
C
f ··· dr =
_
Σ
(curl f ) ··· ndσ =
_
Σ
0 ··· ndσ =
_
Σ
0 dσ = 0 ,
where Σ is any orientable surface inside S whose boundary is C (such a surface is
sometimes called a capping surface for C). So similar to the two-variable case, we
have a three-dimensional version of a result from Section 4.3, for solid regions in
3
which are simply connected (i.e. regions having no holes):
The following statements are equivalent for a simply connected solid region S in
Σ
∂u
∂n
dσ = 0. (Hint: Use Green's second identity.)
Bibliography
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Classic tale about a creature living in a 2-dimensional world who encounters a higher-
dimensional creature, with lots of humor thrown in.
Anton, H. and C. Rorres, Elementary Linear Algebra: Applications Version, 8th edi-
tion. New York: John Wiley & Sons, 2000
Standard treatment of elementary linear algebra.
Bazaraa, M.S., H.D. Sherali and C.M. Shetty, Nonlinear Programming: Theory and
Algorithms, 2nd edition. New York: John Wiley & Sons, 1993
Thorough treatment of nonlinear optimization.
Farin, G., Curves and Surfaces for Computer Aided Geometric Design: A Practical
Guide, 2nd edition. San Diego, CA: Academic Press, 1990
An intermediate-level book on curve and surface design.
Hecht, E., Optics, 2nd edition. Reading, MA: Addison-Wesley Publishing Co., 1987
An intermediate-level book on optics, covering a wide range of topics.
Hoel, P.G., S.C. Port and C.J. Stone, Introduction to Probability Theory, Boston, MA:
Houghton Mifflin Co., 1971
An excellent introduction to elementary, calculus-based probability theory. Lots of good
exercises.
Jackson, J.D., Classical Electrodynamics, 2nd edition. New York: John Wiley & Sons,
1975
An advanced book on electromagnetism, famous for being intimidating. Most of the
mathematics will be understandable after reading the present book.
Marion, J.B., Classical Dynamics of Particles and Systems, 2nd edition. New York:
Academic Press, 1970
Standard intermediate-level treatment of classical mechanics. Very thorough.
O'Neill, B., Elementary Differential Geometry, New York: Academic Press, 1966
Intermediate-level book on differential geometry, with a modern approach based on dif-
ferential forms.
187
188 Bibliography
Pogorelov, A.V., Analytical Geometry, Moscow: Mir Publishers, 1980
An intermediate/advanced book on analytic geometry.
Press, W.H., S.A. Teukolsky, W.T. Vetterling and B.P. Flannery, Numerical Recipes
in FORTRAN: The Art of Scientific Computing, 2nd edition. Cambridge, UK:
Cambridge University Press, 1992
An excellent source of information on numerical methods for solving a wide variety of
problems. Though all the examples are in the FORTRAN programming language, the code
is clear enough to implement in the language of your choice.
Protter, M.H. and C.B. Morrey, Analytic Geometry, 2nd edition. Reading, MA:
Addison-Wesley Publishing Co., 1975
Thorough treatment of elementary analytic geometry, with a rigor not found in most
recent books.
Ralston, A. and P. Rabinowitz, A First Course in Numerical Analysis, 2nd edition.
New York: McGraw-Hill, 1978
Standard treatment of elementary numerical analysis.
Reitz, J.R., F.J. Milford and R.W. Christy, Foundations of Electromagnetic Theory,
3rd edition. Reading, MA: Addison-Wesley Publishing Co., 1979
Intermediate text on electromagnetism.
Schey, H.M., Div, Grad, Curl, and All That: An Informal Text on Vector Calculus, New
York: W.W. Norton & Co., 1973
Very intuitive approach to the subject, from a physicist's viewpoint. Highly recom-
mended.
Taylor, A.E. and W.R. Mann, Advanced Calculus, 2nd edition. New York: John Wiley
& Sons, 1972
Excellent treatment of n-dimensional calculus. A good book to study after the present
book. Many intriguing exercises.
Uspensky, J.V., Theory of Equations, New York: McGraw-Hill, 1948
A classic on the subject, discussing many interesting topics.
Weinberger, H.F., A First Course in Partial Differential Equations, New York: John
Wiley & Sons, 1965
A good introduction to the vast subject of partial differential equations.
Welchons, A.M. and W.R. Krickenberger, Solid Geometry, Boston, MA: Ginn & Co.,
1936
A very thorough treatment of 3-dimensional geometry from an elementary perspective,
includes many topics which (sadly) do not seem to be taught anymore.
Appendix A
Answers and Hints to Selected Exercises
Chapter 1
Section 1.1 (p. 8)
1. (a)
√
5 (b)
√
5 (c)
√
17 (d) 1
(e) 2
√
17 2. Yes 3. No
Section 1.2 (p. 14)
1. (a) (−4, 4, −3) (b) (2, 6, −1)
(c)
_
−1
√
30
,
5
√
30
,
−2
√
30
_
(d)
√
41
2
(e)
√
41
2
(f) (14, −6, 8) (g) (−7, 3, −4)
(h) (−1, −6, 1) (i) (−2, −4, 2) (j) No.
3. No. v + w is larger.
Section 1.3 (p. 18)
1. 10 3. 73.4
◦
5. 90
◦
7. 0
◦
9. Yes, since v··· w = 0.
11. |v··· w| = 0 <
√
21
√
5 = v w
13. v+w =
√
26 <
√
21 +
√
5 = v +w
15. Hint: use Definition 1.6.
24. Hint: See Theorem 1.10(c).
Section 1.4 (p. 29)
1. (−5, −23, −24) 3. (8, 4, −5) 5. 0
7. 16.72 9. 4
√
5 11. 9 13. 0
and (8, −10, 2) 15. 14
Section 1.5 (p. 39)
1. (a) (2, 3, 2) + t(5, 4, −3) (b) x = 2 + 5t,
y = 3 + 4t, z = 2 − 3t (c)
x−2
5
=
y−3
4
=
z−2
−3
3. (a) (2, 1, 3) + t(1, 0, 1) (b) x = 2 + t,
y = 1, z = 3 + t (c) x − 2 = z − 3, y = 1
5. x = 1 + 2t, y = −2 + 7t, z = −3 + 8t
7. 7.65 9. (1, 2, 3)
11. 4x − 4y + 3z − 10 = 0
13. x − 2y − z + 2 = 0
15. 11x − 24y + 21z − 26 = 0 17. 9/
√
35
19. x = 5t, y = 2+3t, z = −7t 21. (10, −2, 1)
Section 1.6 (p. 46)
1. radius: 1, center: (2, 3, 5) 3. radius: 5,
center: (−1, −1, −1) 5. No intersection.
7. circle x
2
+ y
2
= 4 in the planes z = ±
√
5
9. lines
x
a
=
y
b
, z = 0 and
x
a
= −
y
b
, z = 0
13.
_
2a
2−c
,
2b
2−c
, 0
_
Section 1.7 (p. 50)
1. (a) (4,
π
3
, −1) (b) (
√
17,
π
3
, 1.816)
3. (a) (2
√
7,
11π
6
, 0) (b) (2
√
7,
11π
6
,
π
2
)
5. (a) r
2
+ z
2
= 25 (b) ρ = 5
7. (a) r
2
+9z
2
= 36 (b) ρ
2
(1 +8 cos
2
φ) = 36
10. (a, θ, a cot φ) 12. Hint: Use the dis-
tance formula for Cartesian coordinates.
Section 1.8 (p. 57)
1. f
′
(t) = (1, 2t, 3t
2
); x = 1 + t, y = z = 1
3. f
′
(t) = (−2 sin 2t, 2 cos 2t, 1); x = 1,
y = 2t, z = t 5. v(t) = (1, 1 − cos t, sin t),
a(t) = (0, sin t, cos t)
9. (a) Line parallel to c (b) Half-line
189
190 Appendix A: Answers and Hints to Selected Exercises
parallel to c (c) Hint: Think of the
functions as position vectors.
15. Hint: Theorem 1.16
Section 1.9 (p. 63)
1.
3π
√
5
2
3.
2
27
(13
3/2
− 8) 5. Replace
t by
__
27s+16
2
_
2/3
− 4
__
9 6. Hint: Use
Theorem 1.20(e), Example 1.37, and
Theorem 1.16 7. Hint: Use Exercise 6.
9. Hint: Use f
′
(t) = f(t)T, differ-
entiate that to get f
′′
(t), put those ex-
pressions into f
′
(t) ××× f
′′
(t), then write
T
′
(t) in terms of N(t). 11. T(t) =
1
√
2
(−sin t, cos t, 1), N(t) = (−cos t, −sin t, 0),
B(t) =
1
√
2
(sin t, −cos t, 1), κ(t) = 1/2
Chapter 2
Section 2.1 (p. 70)
1. domain:
2
, range: [−1, ∞) 3. domain:
{(x, y) : x
2
+ y
2
≥ 4}, range: [0, ∞)
5. domain:
3
, range: [−1, 1] 7. 1
9. does not exist 11. 2 13. 2 15. 0
17. does not exist
Section 2.2 (p. 74)
1.
∂f
∂x
= 2x,
∂f
∂y
= 2y 3.
∂f
∂x
= x(x
2
+y +4)
−1/2
,
∂f
∂y
=
1
2
(x
2
+ y + 4)
−1/2
5.
∂f
∂x
= ye
xy
+ y,
∂f
∂y
= xe
xy
+ x 7.
∂f
∂x
= 4x
3
,
∂f
∂y
= 0
9.
∂f
∂x
= x(x
2
+ y
2
)
−1/2
,
∂f
∂y
= y(x
2
+ y
2
)
−1/2
11.
∂f
∂x
=
2x
3
(x
2
+ y + 4)
−2/3
,
∂f
∂y
=
1
3
(x
2
+ y + 4)
−2/3
13.
∂f
∂x
= −2xe
−(x
2
+y
2
)
,
∂f
∂y
= −2ye
−(x
2
+y
2
)
15.
∂f
∂x
= y cos(xy),
∂f
∂y
= x cos(xy) 17.
∂
2
f
∂x
2
= 2,
∂
2
f
∂y
2
= 2,
∂
2
f
∂x ∂y
= 0 19.
∂
2
f
∂x
2
= (y + 4)(x
2
+ y + 4)
−3/2
,
∂
2
f
∂y
2
= −
1
4
(x
2
+ y + 4)
−3/2
,
∂
2
f
∂x ∂y
= −
1
2
x(x
2
+ y + 4)
−3/2
21.
∂
2
f
∂x
2
= y
2
e
xy
,
∂
2
f
∂y
2
= x
2
e
xy
,
∂
2
f
∂x ∂y
= (1 + xy)e
xy
+ 1 23.
∂
2
f
∂x
2
= 12x
2
,
∂
2
f
∂y
2
= 0,
∂
2
f
∂x ∂y
= 0 25.
∂
2
f
∂x
2
= −x
−2
,
∂
2
f
∂y
2
= −y
−2
,
∂
2
f
∂x ∂y
= 0
Section 2.3 (p. 77)
1. 2x + 3y − z − 3 = 0 3. −2x + y − z − 2 = 0
5. x + 2y = z 7.
1
2
(x − 1) +
4
9
(y − 2) +
√
11
12
(z −
2
√
11
3
) = 0 9. 3x + 4y − 5z = 0
Section 2.4 (p. 82)
1. (2x, 2y) 3. (
x
√
x
2
+y
2
+4
,
y
√
x
2
+y
2
+4
)
5. (1/x, 1/y) 7. (yz cos(xyz), xz cos(xyz), xy cos(xyz))
9. (2x, 2y, 2z) 11. 2
√
2 13.
1
√
3
15.
√
3 cos(1) 17. increase: (45, 20),
decrease: (−45, −20)
Section 2.5 (p. 88)
1. local min. (1, 0); saddle pt. (−1, 0)
3. local min. (1, 1); local max. (−1, −1);
saddle pts. (1, −1), (−1, 1) 5. local min.
(1, −1); saddle pt. (0, 0) 7. local min. (0, 0)
9. local min. (−1, 1/2) 11. width = height
= depth=10 13. x = y = 4, z = 2
Section 2.6 (p. 95)
2. (x
0
, y
0
) = (0, 0) : → (0.2858, −0.3998);
(x
0
, y
0
) = (1, 1) : → (1.03256, −1.94037)
191
Section 2.7 (p. 100)
1. min.
_
−4
√
5
,
−2
√
5
_
; max.
_
4
√
5
,
2
√
5
_
3. min.
_
20
√
13
,
30
√
13
_
; max.
_
−
20
√
13
, −
30
√
13
_
4. min.
_
−9
√
5
, 0,
2
√
5
_
; max.
_
9
8
,
√
59
4
,
−1
4
_
5.
8abc
3
√
3
Chapter 3
Section 3.1 (p. 104)
1. 1 3.
7
12
5.
7
6
7. 5 9.
1
2
11. 15
Section 3.2 (p. 109)
1. 1 3. 8 ln 2 − 3 5.
π
4
6.
1
4
7. 2 9.
1
6
10.
6
5
Section 3.3 (p. 112)
1.
9
2
3. (2 cos(π
2
) + π
4
− 2)/4 5.
1
6
7. 6
10.
1
3
Section 3.4 (p. 116)
1. The values should converge to ≈ 1.318.
(Hint: In Java the exponential function
e
x
can be obtained with Math.exp(x).
Other languages have similar functions,
otherwise use e = 2.7182818284590455 in
your program.)
2. ≈ 1.146 3. ≈ 0.705 4. ≈ 0.168
Section 3.5 (p. 123)
1. 8π 3.
4π
3
(8 − 3
3/2
) 7. 1 −
sin 2
2
9. 2πab
Section 3.6 (p. 127)
1. (1, 8/3) 3. (0,
4a
3π
) 5. (0, 3π/16)
7. (0, 0, 5a/12) 9. (7/12, 7/12, 7/12)
Section 3.7 (p. 134)
1.
√
π 2. 1 6. Both are
n
(n+1)
2
(n+2)
7.
1
n
Chapter 4
Section 4.1 (p. 142)
1. 1/2 3. 23 5. 24π 7. −2π 9. 2π
11. 4π
Section 4.2 (p. 149)
1. 0 3. No 4. Yes. F(x, y) =
x
2
2
−
y
2
2
5. No 9. (b) No. Hint: Think of how F is
defined. 10. Yes. F(x, y) = axy +bx +cy +d
Section 4.3 (p. 155)
1. 16/15 3. −5π 5. Yes. F(x, y) = xy
2
+ x
3
7. Yes. F(x, y) = 4x
2
y + 2y
2
+ 3x
Section 4.4 (p. 163)
1. 216π 2. 3 3. 12π/5 7. 15/4
Section 4.5 (p. 175)
1. 2
√
2 π
2
2. (17
√
17 − 5
√
5)/3 3. 2/5
4. 1 5. 2π(π − 1) 7. 67/15 9. 6
11. Yes 13. No 19. Hint: Think of
how a vector field f(x, y) = P(x, y) i+Q(x, y) j
in
2
can be extended in a natural way to
be a vector field in
3
.
Section 4.6 (p. 186)
1. 0 3. 12
_
x
2
+ y
2
+ z
2
5. 6(x + y + z)
7. 12ρ 8. (4ρ
2
− 6)e
−ρ
2
9. −
2z
r
3
e
r
+
1
r
2
e
z
11. div f =
2
ρ
−
sin θ
sin φ
+ cot φ;
curl f = cot φ cos θ e
ρ
+ 2e
θ
− 2 cos θ e
φ
25. Hint: Start by showing that e
r
=
cos θ i + sin θ j, e
θ
= −sin θ i + cos θ j, e
z
= k.
Appendix B
We will prove the right-hand rule for the cross product of two vectors in
3
.
For any vectors v and w in
3
, define a new vector, n(v, w), as follows:
1. If v and w are nonzero and not parallel, and θ is the angle between them, then
n(v, w) is the vector in
3
such that:
(a) the magnitude of n(v, w) is v w sin θ,
(b) n(v, w) is perpendicular to the plane containing v and w, and
(c) v, w, n(v, w) form a right-handed system.
2. If v and w are nonzero and parallel, then n(v, w) = 0.
3. If either v or w is 0, then n(v, w) = 0.
The goal is to show that n(v, w) = v ××× w for all v, w in
3
, which would prove the
right-hand rule for the cross product (by part 1(c) of our definition). To do this, we will
perform the following steps:
Step 1: Show that n(v, w) = v××× w if v and w are any two of the basis vectors i, j, k.
This was already shown in Example 1.11 in Section 1.4.
Step 2: Show that n(av, bw) = ab(v ××× w) for any scalars a, b if v and w are any two of
the basis vectors i, j, k.
If either a = 0 or b = 0 then n(av, bw) = 0 = ab(v ××× w), so the result holds. So assume
that a 0 and b 0. Let v and w be any two of the basis vectors i, j, k. For example,
we will show that the result holds for v = i and w = k (the other possibilities follow in
a similar fashion).
For av = ai and bw = bk, the angle θ between av and bwis 90
◦
. Hence the magnitude
of n(av, bw), by definition, is ai bk sin 90
◦
= |ab|. Also, by definition, n(av, bw) is
perpendicular to the plane containing ai and bk, namely, the xz-plane. Thus, n(av, bw)
must be a scalar multiple of j. Since its magnitude is |ab|, then n(av, bw) must be
either |ab|j or −|ab|j.
There are four possibilities for the combinations of signs for a and b. We will con-
sider the case when a > 0 and b > 0 (the other three possibilities are handled simi-
larly).
192
193
In this case, n(av, bw) must be either abj or −abj. Now, since i, j, k form a right-
handed system, then i, k, j form a left-handed system, and so i, k, −j form a right-
handed system. Thus, ai, bk, −abj form a right-handed system (since a > 0, b > 0, and
ab > 0). So since, by definition, ai, bk, n(ai, bk) form a right-handed system, and since
n(ai, bk) has to be either abj or −abj, this means that we must have n(ai, bk) = −abj.
But we know that ai ××× bk = ab(i ××× k) = ab(−j) = −abj. Therefore, n(ai, bk) = ab(i ××× k),
which is what we needed to show.
∴ n(av, bw) = ab(v××× w)
Step 3: Show that n(u, v + w) = n(u, v) + n(u, w) for any vectors u, v, w.
If u = 0 then the result holds trivially since n(u, v +w), n(u, v) and n(u, w) are all the
zero vector. If v = 0, then the result follows easily since n(u, v + w) = n(u, 0 + w) =
n(u, w) = 0+n(u, w) = n(u, 0) = n(u, w) = n(u, v) +n(u, w). A similar argument shows
that the result holds if w = 0.
So now assume that u, v and ware all nonzero vectors. We will describe a geometric
construction of n(u, v), which is shown in the figure below. Let P be a plane perpen-
dicular to u. Multiply the vector v by the positive scalar u, then project the vector
u v straight down onto the plane P. You can think of this projection vector (denoted
by pro j
P
u v) as the shadow of the vector u v on the plane P, with the light source
directly overhead the terminal point of u v. If θ is the angle between u and v, then
we see that pro j
P
u v has magnitude u v sin θ, which is the magnitude of n(u, v).
So rotating pro j
P
u v by 90
◦
in a counter-clockwise direction in the plane P gives a
vector whose magnitude is the same as that of n(u, v) and which is perpendicular to
pro j
P
u v (and hence perpendicular to v). Since this vector is in P then it is also per-
pendicular to u. And we can see that u, v and this vector form a right-handed system.
Hence this vector must be n(u, v). Note that this holds even if u v, since in that case
θ = 0
◦
and so sin θ = 0 which means that n(u, v) has magnitude 0, which is what we
would expect.
u
v
pro j
P
u v
u v
n(u, v)
θ
θ
P
Now apply this same geometric construction to get n(u, w) and n(u, v + w). Since
u (v + w) is the sum of the vectors u v and u w, then the projection vector
194 Appendix B: Proof of the Right-Hand Rule for the Cross Product
pro j
P
u (v + w) is the sum of the projection vectors pro j
P
u v and pro j
P
u w (to
see this, using the shadow analogy again and the parallelogram rule for vector addi-
tion, think of how projecting a parallelogram onto a plane gives you a parallelogram in
that plane). So then rotating all three projection vectors by 90
◦
in a counter-clockwise
direction in the plane P preserves that sum (see the figure below), which means that
n(u, v + w) = n(u, v) + n(u, w).
u
v
w
v + w
u (v + w)
pro j
P
u v
pro j
P
u w
pro j
P
u (v + w)
u v
u w
n(u, v) n(u, w)
n(u, v + w)
θ
θ
P
Step 4: Show that n(w, v) = −n(v, w) for any vectors v, w.
If v and w are nonzero and parallel, or if either is 0, then n(w, v) = 0 = −n(v, w), so
the result holds. So assume that v and w are nonzero and not parallel. Then n(w, v)
has magnitude w v sin θ, which is the same as the magnitude of n(v, w), and hence
is the same as the magnitude of −n(v, w). By definition, n(v, w) is perpendicular to
the plane containing w and v, and hence so is −n(v, w). Also, v, w, n(v, w) form a
right-handed system, and so w, v, n(v, w) form a left-handed system, and hence w,
v, −n(v, w) form a right-handed system. Thus, we have shown that −n(v, w) is a vec-
tor with the same magnitude as n(w, v) and is perpendicular to the plane containing
w and v, and that w, v, −n(v, w) form a right-handed system. So by definition this
means that −n(v, w) must be n(w, v).
Step 5: Show that n(v, w) = v××× w for all vectors v, w.
Write v = v
1
i + v
2
j + v
3
k and w = w
1
i + w
2
j + w
3
k. Then by Steps 3 and 4, we have
195
n(v, w) = n(v
1
i + v
2
j + v
3
k, w
1
i +) + n(v
1
i + v
2
j + v
3
k, w
3
k)
= −n(w
1
i, v
1
i + v
2
j + v
3
k) + −n(w
2
j, v
1
i + v
2
j + v
3
k) + −n(w
3
k, v
1
i + v
2
j + v
3
k).
We can use Steps 1 and 2 to evaluate the three terms on the right side of the last
equation above:
−n(w
1
i, v
1
i + v
2
j + v
3
k) = −n(w
1
i, v
1
i) + −n(w
1
i, v
2
j) + −n(w
1
i, v
3
k)
= −v
1
w
1
n(i, i) + −v
2
w
1
n(i, j) + −v
3
w
1
n(i, k)
= −v
1
w
1
(i ××× i) + −v
2
w
1
(i ××× j) + −v
3
w
1
(i ××× k)
= −v
1
w
1
0 + −v
2
w
1
k + −v
3
w
1
(−j)
−n(w
1
i, v
1
i + v
2
j + v
3
k) = −v
2
w
1
k + v
3
w
1
j
Similarly, we can calculate
−n(w
2
j, v
1
i + v
2
j + v
3
k) = v
1
w
2
k − v
3
w
2
i
and
−n(w
3
j, v
1
i + v
2
j + v
3
k) = −v
1
w
3
j + v
2
w
3
i .
Thus, putting it all together, we have
n(v, w) = −v
2
w
1
k + v
3
w
1
j + v
1
w
2
k − v
3
w
2
i − v
1
w
3
j + v
2
w
3
i
= (v
2
w
3
− v
3
w
2
)i + (v
3
w
1
− v
1
w
3
)j + (v
1
w
2
− v
2
w
1
)k
= v××× w by definition of the cross product.
∴ n(v, w) = v××× w for all vectors v, w.
So since v, w, n(v, w) form a right-handed system, then v, w, v ××× w form a right-
handed system, which completes the proof.
Appendix C
3D Graphing with Gnuplot
Gnuplot is a free, open-source software package for producing a variety of graphs.
Versions are available for many operating systems. Below is a very brief tutorial on
how to use Gnuplot to graph functions of several variables.
INSTALLATION
1. Go to and followthe links to down-
load the latest version for your operating system. For Windows, you should get the
Zip file with a name such as gp420win32.zip, which is version 4.2.0. All the
examples we will discuss require at least version 4.2.0.
2. Install the downloaded file. For example, in Windows you would unzip the Zip file
you downloaded in Step 1 into some folder (use the "Use folder names" option if
extracting with WinZip).
RUNNING GNUPLOT
1. In Windows, run wgnuplot.exe fromthe folder (or bin folder) where you installed
Gnuplot. In Linux, just type gnuplot in a terminal window.
2. You should now get a Gnuplot terminal with a gnuplot> command prompt. In
Windows this will appear in a new window, while in Linux it will appear in the
terminal window where the gnuplot command was run. For Windows, if the font
is unreadable you can change it by right-clicking on the text part of the Gnuplot
window and selecting the "Choose Font.." option. For example, the font "Courier",
style "Regular", size "12" is usually a good choice (that choice can be saved for
future sessions by right-clicking in the Gnuplot window again and selecting the
option to update wgnuplot.ini).
3. At the gnuplot> command prompt you can now run graphing commands, which
we will now describe.
GRAPHING FUNCTIONS
The usual way to create 3D graphs in Gnuplot is with the splot command:
splot <range> <comma-separated list of functions>
196
197
For a function z = f (x, y), <range> is the range of x and y values (and optionally the
range of z values) over which to plot. To specify an x range and a y range, use an
expression of the form [a : b][c : d], for some numbers a < b and c < d. This will cause
the graph to be plotted for a ≤ x ≤ b and c ≤ y ≤ d.
Function definitions use the x and y variables in combination with mathematical op-
erators, listed below:
Symbol Operation Example Result
+ Addition 2 + 3 5
− Subtraction 3 − 2 1
* Multiplication 2*3 6
/ Division 4/2 2
** Power 2**3 2
3
= 8
exp(x) e
x
exp(2) e
2
log(x) ln x log(2) ln 2
sin(x) sin x sin(pi/2) 1
cos(x) cos x cos(pi) −1
tan(x) tan x tan(pi/4) 1
Example C.1. To graph the function z = 2x
2
+ y
2
from x = −1 to x = 1 and from y = −2
to y = 2, type this at the gnuplot> prompt:
splot [−1 : 1][−2 : 2] 2*x**2 + y**2
The result is shown below:
-1
-0.5
0
0.5
1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0
1
2
3
4
5
6
7
2*(x**2) + y**2
198 Appendix C: 3D Graphing with Gnuplot
Note that we had to type 2*x**2 to multiply 2 times x
2
. For clarity, parentheses can
be used to make sure the operations are being performed in the correct order:
splot [−1 : 1][−2 : 2] 2*(x**2) + y**2
In the above example, to also plot the function z = e
x+y
on the same graph, put a
comma after the first function then append the new function:
splot [−1 : 1][−2 : 2] 2*(x**2) + y**2, exp(x+y)
By default, the x-axis and y-axis are not shown in the graph. To display the axes, use
this command before the splot command:
set zeroaxis
Also, by default the x- and y-axes are switched from their usual position. To show the
axes with the orientation which we have used throughout the text, use this command:
set view 60, 120, 1, 1
Also, to label the axes, use these commands:
set xlabel "x"
set ylabel "y"
set zlabel "z"
To show the level curves of the surface z = f (x, y) on both the surface and projected
onto the xy-plane, use this command:
set contour both
The default mesh size for the grid on the surface is 10 units. To get more of a col-
ored/shaded surface, increase the mesh size (to, say, 25) like this:
set isosamples 25
Putting all this together, we get the following graph with these commands:
set zeroaxis
set view 60, 120, 1, 1
set xlabel "x"
set ylabel "y"
set zlabel "z"
set contour both
set isosamples 25
splot [−1 : 1][−2 : 2] 2*(x**2) + y**2, exp(x+y)
199
-1
-0.5
0
0.5
1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0
5
10
15
20
25
z
2*(x**2) + y**2
6
5
4
3
2
1
exp(x+y)
20
15
10
5
x
y
z
The numbers listed below the functions in the key in the upper right corner of the
graph are the "levels" of the level curves of the corresponding surface. That is, they
are the numbers c such that f (x, y) = c. If you do not want the function key displayed,
it can be turned off with this command: unset key
PARAMETRIC FUNCTIONS
Gnuplot has the ability to graph surfaces given in various parametric forms. For
example, for a surface parametrized in cylindrical coordinates
x = r cos θ , y = r sin θ , z = z
you would do the following:
set mapping cylindrical
set parametric
splot [a : b][c : d] v*cos(u),v*sin(u),f(u,v)
where the variable u represents θ, with a ≤ u ≤ b, the variable v represents r, with
c ≤ v ≤ d, and z = f (u, v) is some function of u and v.
Example C.2. The graph of the helicoid z = θ in Example 1.34 from Section 1.7 (p. 49)
was created using the following commands:
200 Appendix C: 3D Graphing with Gnuplot
set mapping cylindrical
set parametric
set view 60, 120, 1, 1
set xyplane 0
set xlabel "x"
set ylabel "y"
set zlabel "z"
unset key
set isosamples 15
splot [0 : 4*pi][0 : 2] v*cos(u),v*sin(u),u
The command set xyplane 0 moves the z-axis so that z = 0 aligns with the xy-plane
(which is not the default in Gnuplot). Looking at the graph, you will see that r varies
from 0 to 2, and θ varies from 0 to 4π.
PRINTING AND SAVING
In Windows, to print a graph from Gnuplot right-click on the titlebar of the graph's
window, select "Options" and then the "Print.." option. To save a graph, say, as a PNG
file, go to the File menu on the main Gnuplot menubar, select "Output Device ...", and
enter png in the Terminal type? textfield, hit OK. Then, in the File menu again, select
the "Output ..." option and enter a filename (say, graph.png) in the Output filename?
textfield, hit OK. Now run your splot command again and you should see a file called
graph.png in the current directory (usually the directory where wgnuplot.exe is lo-
cated, though you can change that setting using the "Change Directory ..." option in
the File menu).
In Linux, to save the graph as a file called graph.png, you would issue the following
commands:
set terminal png
set output 'graph.png'
and then run your splot command. There are many terminal types (which determine
the output format). Run the command set terminal to see all the possible types. In
Linux, the postscript terminal type is popular, since the print quality is high and
there are many PostScript viewers available.
To quit Gnuplot, type quit at the gnuplot> command prompt.
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History
This section contains the revision history of the book. For persons making modifications
to the book, please record the pertinent information here, following the format in the
first item below.
1. VERSION: 1.0
Date: 2008-01-04
Author(s): Michael Corral
Title: Vector Calculus
Modification(s): Initial version
209
Index
Symbols
D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
M
x
, M
y
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
M
xy
, M
xz
, M
yz
. . . . . . . . . . . . . . . . . . . . . . . 126
∆ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
About the author: Michael Corral is an Adjunct Faculty member of the Department of Mathematics at Schoolcraft College. He received a B.A. in Mathematics from the University of California at Berkeley, and received an M.A. in Mathematics and an M.S. in Industrial & Operations Engineering from the University of Michigan.
A This text was typeset in LTEX 2ε with the KOMA-Script bundle, using the GNU Emacs text editor on a Fedora Linux system. The graphics were created using MetaPost, PGF, and Gnuplot.
Copyright c 2008 Michael Corral
Appendix B contains a proof of the right-hand rule for the cross product.too much rigor and emphasis on proofs can impede the flow of learning for the vast majority of the audience at this level. which in my experience are more than enough for a semester course in this subject. B and C. The code samples in the text are in the Java programming language. the B exercises are slightly more involved. B and C would be "Easy". I have tried to be somewhat rigorous about proving results. the Monte Carlo method for approximating multiple integrals. The prerequisites are the standard courses in single-variable calculus (a.k. divided into three categories: A.4). or simply "Calculus III". A crude way of describing A. There are exercises at the end of each section. "Moderate" and "Challenging". Those exercises do not mandate the use of Java.since that is how mathematics works . hopefully with enough comments so that the reader can figure out what is being done even without knowing Java. The A exercises are mostly of a routine computational nature. There are 420 exercises throughout the text. There are a few exercises that require the student to write his or her own computer program to solve some numerical approximation problems (e. I would rate it as a 5. However. This book is released under the GNU Free Documentation License (GFDL). Appendix C contains a brief tutorial on Gnuplot for graphing functions of two variables. so students are free to implement the solutions using the language of their choice. relatively clear syntax. Calculus I and II). It is suitable for a one-semester course. respectively. But while it is important for students to see full-blown proofs . So that there is no ambiguity on this
iii
. see the included copy of the GFDL. Java was chosen due to its ubiquity. For more details.g. If I were to rate the level of rigor in the book on a scale of 1 to 10.a. normally known as "Vector Calculus". with 1 being completely informal and 10 being completely rigorous. which allows others to not only copy and distribute the book but also to modify it. many of the B exercises are easy and not all the C exercises are difficult. which seems to have virtually disappeared from calculus texts over the last few decades. and easy availability for multiple platforms. Answers and hints to most odd-numbered and some even-numbered exercises are provided in Appendix A. "Multivariable Calculus".Preface
This book covers calculus in two and three variables. While it would have been simple to use a scripting language like Python. and perhaps even easier with a functional programming language (such as Haskell or Scheme). and the C exercises usually require some effort or insight to solve. in Section 3.
Feel free to contact me at mcorral@schoolcraft.net). corrections. I welcome your input. suggestions. I would like to thank my students in Math 240 for being the guinea pigs for the initial draft of this book. The PDF version will always be freely available to the public at no cost (go to without needing my permission.mecmath.g. and for finding the numerous errors and typos it contained.edu for any questions on this or any other matter involving the book (e. anyone can make as many copies of this book as desired and distribute it as desired.iv
Preface
matter. January 2008 M ICHAEL C ORRAL
. etc). comments. Finally.
which. z) = (x. b. y = f (x). In vector (or multivariable) calculus. we will deal with functions of two or three variables (usually x. consists of all ordered pairs of real numbers (a.1 Vectors in Euclidean Space
1. say. b. z = f (x. y)).1 Introduction
In single-variable calculus.
z c P(a. y. Euclidean space has three mutually perpendicular coordinate axes (x. in the Cartesian or rectangular coordinate system. y. lies in Euclidean space. we denote it by 3 . y or x. yz-plane and xz-plane (see Figure 1. in the manner shown in Figure 1.2). y). and three mutually perpendicular coordinate planes: the xy-plane. We denote the Euclidean plane by 2 .2
1
.1. These points lie in the Euclidean plane. the "2" represents the number of dimensions of the plane. f (x. only by giving the illusion of three dimensions. y) = (x. Since Euclidean space is 3-dimensional. c) b 0 a x x y
xz-plane yz-plane
z
y 0
xy-plane
Figure 1. The 3-dimensional coordinate system of Euclidean space can be represented on a flat surface.1. y.1.1. respectively). The Euclidean plane has two perpendicular coordinate axes: the x-axis and the y-axis. which in the Cartesian coordinate system consists of all ordered triples of real numbers (a. c). b). We use the word "Euclidean" to denote a system in which all the usual rules of Euclidean geometry hold. For such a function. the functions that one encounters are functions of a variable (usually x or t) that varies over some subset of the real number line (which we denote by ).1. the graph of the function f consists of the points (x. z. f (x)). y and z). The graph of a function of two variables.1
Figure 1. The graph of f consists of the points (x. say. such as this page or a blackboard.
This is where the idea of a vector comes in. VECTORS IN EUCLIDEAN SPACE
The coordinate system shown in Figure 1. But what about something such as the velocity of the object. as in Figure 1.1. let alone simulate in 2-dimensional space. For functions of three variables. or its acceleration? Or the gravitational force acting on the object? These phenomena all seem to involve motion and direction in some way. we have discussed the position of an object in 2-dimensional or 3-dimensional space. Notice that switching the x. see the book by A BBOTT.1. which we can not see in our 3-dimensional space.
Figure 1.
1 One thing you will learn is why a 4-dimensional creature would be able to reach inside an egg and remove the yolk without cracking the shell!
. 4 ).e. So we can only think of 4-dimensional space abstractly. the middle finger in the positive direction of the y-axis. For an entertaining discussion of this subject.3 Right-handed coordinate system
An equivalent way of defining a right-handed system is if you can point your thumb upwards in the positive z-axis direction while using the remaining four fingers to rotate the x-axis towards the y-axis. and the thumb in the positive direction of the z-axis.1 So far. using the right hand. Throughout the book we will use a right-handed system.1 is known as a right-handed coordinate system. the graphs exist in 4-dimensional space (i.3. to point the index finger in the positive direction of the x-axis.and y-axes in a right-handed system results in a left-handed system. and that rotating either type of system does not change its "handedness".1. Doing the same thing with the left hand is what defines a left-handed coordinate system. because it is possible.2
CHAPTER 1.
Note that our definition could apply to systems with any number of dimensions (see Figure 1. v) and use the terms "magnitude" and "length" interchangeably. f ′ (t) = ±a for some number a ≥ 0. For motion along a straight line. however.1. To indicate the direction of a vector. and the ± represents the direction of the velocity (though the + is usually omitted for the positive direction). in a 1-dimensional space. the velocities are also contained in that 1-dimensional space. i.g. and it is denoted by 0. The derivative f ′ (t) is just a number.or 3-dimensional space. Its magnitude is the length of the line − − → segment. We will often denote a vector by a single bold-faced letter (e. denoted by PQ .1.e. For general motion along a curve in 2. Then a is the magnitude of the velocity (normally called the speed of the object). then dy/dt = f ′ (t) is the velocity of the object at time t. preceded by a direction. Definition 1. we draw an arrow from its initial point to its terminal point. velocity will need to be represented by a multidimensional object which should have both a magnitude and a direction. indicated by a nonnegative number. The zero vector is just a point.e. since they are just numbers.
y z
− −→Q P
v − − → RS 0 R − − → PQ P Q x 0 P R − − → RS
Q S x S x
→ − −S R
R
Q
−→ − PQ
y
0 P v
S
(a) One dimension
(b) Two dimensions
(c) Three dimensions
Figure 1. as having two components: a magnitude. for motion along a straight line.4
Vectors in different dimensions
. A (nonzero) vector is a directed line segment drawn from a point P (called its initial point) to a point Q (called its terminal point). The vector is denoted by PQ. which in elementary geometry is called a "directed line segment". if y = f (t) gives the displacement of an object after time t.4 (a)-(c)).1 Introduction
3
You have already dealt with velocity and acceleration in single-variable calculus. i. This is the motivation for how we will define a vector.1. and its direction is the same as that of the directed line segment. with P and Q being − − → distinct points. For example. indicated by a plus or minus symbol (representing motion in the positive direction or the negative direction. respectively). which was called the velocity of the object. So you can think of that number. and negative if it moves in the opposite of that direction.1. which is positive if the object is moving in an agreedupon "positive" direction. A geometric object which has those features is an arrow.
4
CHAPTER 1. What about the direction of the zero vector? A single point really has no well-defined direction. and is suggested by the vector w in Figure 1. By this definition. Some contend that the zero vector has arbitrary direction (i. So u v.1. however. we need a way of determining when two vectors are equal.1.5 the vectors u. For example. which is well-defined since the initial and terminal points are distinct. This agrees with the definition of the zero vector as just a point.
y 4 3 2 1 0 1 v w x 2 3 4 u
Figure 1. v and w all √ have the same magnitude 5 (by the Pythagorean Theorem). So u = w.2 Now that we know what a vector is.
2 In the subject of linear algebra there is a more abstract way of defining a vector where the concept of "direction" is not really used. even though they have different initial points. 0 = 0. vectors with the same magnitude and direction but with different initial points would be equal. This leads us to the following definition. those vectors all being equal and differing only by their initial and terminal points. Our definition of the zero vector. does not require it to have a direction. and they point in the 2 same direction.
. which has zero length. VECTORS IN EUCLIDEAN SPACE
A few things need to be noted about the zero vector. can take any direction). Not everyone agrees on the direction of the zero vector.2. the direction can not be determined). Two nonzero vectors are equal if they have the same magnitude and the same direction. in Figure 1.e. i. since they lie on lines having the same slope 1 .e. Our motivation for what a vector is included the notions of magnitude and direction. some say that it has indeterminate direction (i. Notice that we were careful to only define the direction of a nonzero vector. What is the magnitude of the zero vector? We define it to be zero. And we see that u and w are parallel.5
So we can see that there are an infinite number of vectors for a given magnitude and direction. See A NTON and R ORRES. Is there a single vector which we can choose to represent all those equal vectors? The answer is yes. Any vector with zero magnitude is equal to the zero vector.e.5. We also see that v is parallel to u but points in the opposite direction. and we will leave it at that.1. while others say that it has no direction. Definition 1.
0. when adding vectors. To get the "new" vectors starting at the origin. 0) and the terminal point is (3. 5) z v = (3. 0). which we will do in the next section). 4. it is convenient to write v = (3. find the (unique!) vector it equals whose initial point is the origin. 4. without having to determine their magnitude and direction.6
Correspondence between points and vectors
Unless otherwise stated. The resulting point will be the terminal point of the "new" vector whose initial point is the origin. you translate each vector to start at the origin by subtracting the coordinates of the original initial point from the original terminal point. Also. when we refer to vectors as v = (a. Similar to seeing if two points are the same. 0) and (0. 4.
z P(3. 5). When doing this. then the original vectors are equal. 5). c) in 3 .
. 5) and the vector v are different objects. Another advantage of using the origin as the initial point is that it provides an easy correspondence between a vector and its terminal point.1. it is understood that the initial point of v is at the origin (0.4. we will mean the one whose initial point is at the origin of the coordinate system. Let v be the vector in 3 whose initial point is at the origin and whose terminal point is (3. respectively. b) in 2 or v = (a. 5)
y 0 x
(a) The point (3. 4.
5
Thinking of vectors as starting from the origin provides a way of dealing with vectors in a standard way. we will write the zero vector 0 in 2 and 3 as (0. Though the point (3. But there will be times when it is convenient to consider a different initial point for a vector (for example.1.1. 4. 4. Then compare the coordinates of the terminal points of these "new" vectors: if those coordinates are the same. when speaking of "the vector" with a given magnitude and direction. you are now seeing if the terminal points of vectors starting at the origin are the same. For each vector.5)
Figure 1. 0. b. 5). Example 1.4.5)
y 0 x
(b) The vector (3. since every coordinate system has an origin. Do this for each original vector then compare.1 Introduction Unless otherwise indicated. The point-vector correspondence provides an easy way to check if two vectors are equal. we mean vectors in Cartesian coordinates starting at the origin.
is obtained by translating w so that its initial point is at the terminal point of v.2. The sum of vectors v and w.3 Examples of scalar quantities are mass. We are now ready to define the sum of two vectors. Definition 1. denoted by v + w. For the zero vector 0. Definition 1. and speed (not velocity). subtraction). Definition 1.2.1).3. rotations). denoted by kv.1. where it means "carrier". and is the zero vector 0 if k = 0. a quantity that is not affected is a scalar. and as flipping the vector in the opposite direction if the scalar is a negative number (see Figure 1. points in the opposite direction as v if k < 0. to convey the sense of something that could be represented by a point on a scale or graduated ruler.
3 The term scalar was invented by 19th century Irish mathematician. we will introduce the notion of a scalar.2 Vector Algebra
9
1. scalars will always be real numbers. used in physics. The word vector comes from Latin. addition. You can think of scalar multiplication of a vector as stretching or shrinking the vector.
v
2v
3v
0. the initial point of v + w is the initial point of v.4 We can now define scalar multiplication of a vector.g. electric charge.2 Vector Algebra
Now that we know what vectors are. is the vector whose magnitude is |k| v . physicist and astronomer William Rowan Hamilton.4.g. we can start to perform some of the usual algebraic operations on them (e. and its terminal point is the new terminal point of w. See M ARION for details.5v
−v
−2v
Figure 1. Two vectors v and w are parallel (denoted by v w) if one is a scalar multiple of the other. we define k0 = 0 for any scalar k. For a scalar k and a nonzero vector v.1
Recall that translating a nonzero vector means that the initial point of the vector is changed but the magnitude and direction are preserved. is that under certain types of coordinate transformations (e.5.
. 4 An alternate definition of scalars and vectors. Before doing that. points in the same direction as v if k > 0. For our purposes. A scalar is a quantity that can be represented by a single number. the scalar multiple of v by k. while a quantity that is affected (in a certain way) is a vector.
v w w+v v+w v
(a) Add vectors
w w v−w −w v+w v v−w w v−w v
(c) Combined add/subtract
(b) Subtract vectors
Figure 1.
w w v
(a) Vectors v and w
v+w w v
(c) The sum v + w
v
(b) Translate w to the end of v
Figure 1. (a) shows that v + w = w + v for any vectors v. we can define vector subtraction as follows: v − w = v + (−w). it uses laws from elementary geometry to prove statements about vectors. Also. VECTORS IN EUCLIDEAN SPACE
Intuitively. it is easy to see that v + (−v) = 0.2.2. In general.4
"Geometric" vector algebra
Notice that we have temporarily abandoned the practice of starting vectors at the origin.2. as we would expect. And (c) shows how you can think of v − w as the vector that is tacked on to the end of w to add up to v.
v −w
(b) Translate −w to the end of v
w v
(a) Vectors v and w
v v−w −w
(c) The difference v − w
Figure 1.3. adding w to v means tacking on w to the end of v (see Figure 1. and hence can be translated).2
Adding vectors v and w
Notice that our definition is valid for the zero vector (which is just a point.2. w. that is. and so we see that v + 0 = v = 0 + v for any vector v. In particular.3
Subtracting vectors v and w
Figure 1. since the scalar multiple −v = −1 v is a well-defined vector.2.4 shows the use of "geometric proofs" of various laws of vector algebra. we have not even mentioned coordinates in this section so far.2). the following two theorems are useful for performing vector algebra on vectors in 2 and 3 starting at the origin. See Figure 1. In fact.
.2. Since we will deal mostly with Cartesian coordinates in this book. 0 + 0 = 0. For example.10
CHAPTER 1.
Proof: (a) Without loss of generality, we assume that v1 , v2 > 0 (the other possibilities are handled in a similar manner). If k = 0 then kv = 0v = 0 = (0, 0) = (0v1 , 0v2 ) = (kv1 , kv2 ), which is what we needed to show. If k 0, then (kv1 , kv2 ) lies on a line with slope kv2 = v2 , which is the same as the slope of the line on which v (and hence kv) lies, kv1 v1 and (kv1 , kv2 ) points in the same direction on that line as kv. Also, by formula (1.3) the magnitude of (kv1 , kv2 ) is (kv1 )2 + (kv2 )2 = k2 v2 + k2 v2 = k2 (v2 + v2 ) = |k| v2 + v2 = 1 2 1 2 1 2 |k| v . So kv and (kv1 , kv2 ) have the same magnitude and direction. This proves (a).
Proof: (a) We already presented a geometric proof of this in Figure 1.2.4(a). (b) To illustrate the difference between analytic proofs and geometric proofs in vector algebra, we will present both types here. For the analytic proof, we will use vectors in 3 (the proof for 2 is similar).
A unit vector is a vector with magnitude 1. Notice that for any nonzero vector v, v 1 the vector v is a unit vector which points in the same direction as v, since v > 0
v and v = v = 1. Dividing a nonzero vector v by v is often called normalizing v. v There are specific unit vectors which we will often use, called the basis vectors: i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) in 3 ; i = (1, 0) and j = (0, 1) in 2 . These are useful for several reasons: they are mutually perpendicular, since they lie on distinct coordinate axes; they are all unit vectors: i = j = k = 1; every vector can be written as a unique scalar combination of the basis vectors: v = (a, b) = a i + b j in 2 , v = (a, b, c) = a i + b j + c k in 3 . See Figure 1.2.7.
C
6. We know that every vector in 3 can be written as a scalar combination of the vectors i, j, and k. Can every vector in 3 be written as a scalar combination of just i and j, i.e. for any vector v in 3 , are there scalars m, n such that v = m i + n j? Justify your answer.
6. does not hold for the dot product of vectors.6)
the dot product is: (1. w3 ) be vectors in The dot product of v and w.1. called the dot product. Why? Because for vectors u. w2 . denoted by v · w. so that 0◦ ≤ θ ≤ 180◦ . v3 ) and w = (w1 . The angle between two nonzero vectors with the same initial point is the smallest angle between them. Definition 1.7. Let v = (v1 . Any two nonzero vectors with the same initial point have two angles between them: θ and 360◦ − θ. for vectors v = (v1 . We will always choose the smallest nonnegative angle θ between them. We will now see one type of multiplication of vectors.
θ θ 360 − θ
(a) 0◦ < θ < 180◦
◦
360◦ − θ θ
360◦ − θ
(b) θ = 180◦ (c) θ = 0◦
Figure 1.e. v. We do not define the angle between the zero vector and any other vector. not a vector. by referencing vector coordinates.
. i.3 Dot Product
You may have noticed that while we did define multiplication of a vector by a scalar in the previous section on vector algebra. Also notice that we defined the dot product in an analytic way. 3. So the associative law that holds for multiplication of numbers and for addition of vectors (see Theorem 1. the dot product u · v is a scalar.1 Angle between vectors
We can now take a more geometric view of the dot product by establishing a relationship between the dot product of two vectors and the angle between them. w. w2 ) in v · w = v1 w1 + v2 w2
2.3 Dot Product
15
1. There is a geometric way of defining the dot product.3. For vectors v = v1 i + v2 j + v3 k and w = w1 i + w2 j + w3 k in component form. is given by: v · w = v1 w1 + v2 w2 + v3 w3 Similarly. and so (u · v) · w is not defined since the left side of that dot product (the part in parentheses) is a scalar and not a vector. which we will now develop as a consequence of the analytic definition. v2 . we did not define multiplication of a vector by a vector. Definition 1. the dot product is still v · w = v1 w1 + v2 w2 + v3 w3 .5(b).1. See Figure 1.7)
Notice that the dot product of two vectors is a scalar.3. v2 ) and w = (w1 .(e)).
(1.
then
By the discussion in Example 1. So we see that b= v and h = w sin θ 1 v w sin θ 2 1 v× w = 2 of the parallelogram PQRS is twice the area of the triangle APQR = APQRS = v w sin θ
So since the area APQRS △PQR. respectively. we have proved the following theorem: Theorem 1. Let θ be the angle between v and w.3 Q S w θ v R P h S
Think of the triangle as existing in 3 .
P h θ Q b R Figure 1. VECTORS IN EUCLIDEAN SPACE
If θ is the angle between nonzero vectors v and w in v× w = v w sin θ
3.11)
It may seem strange to bother with the above formula. The area APQR of 1 △PQR is 2 bh.8. where b is the base of the triangle and h is the height. Area of triangles and parallelograms (a) The area A of a triangle with adjacent sides v. Let △PQR and PQRS be a triangle and parallelogram. The formula is more useful for its applications in geometry. like for any other vector.4.4. when the magnitude of the cross product can be calculated directly. as shown in Figure 1.22
CHAPTER 1. as in the following example.13. Example 1. w (as vectors in A = v× w
is:
.
then (1. in 3 . w (as vectors in A= 1 v× w 2
3) 3)
is:
(b) The area A of a parallelogram with adjacent sides v. respectively. and identify the sides QR and QP with vectors v and w.3.8.
then the volume of the parallelepiped is |u · (v × w)|.13(b).1.12) will give you the negative of the volume of the parallelepiped. then repeating the same steps using the base determined by u and v (since w is on the same side of that base's plane as u × v).12 the height h of the parallelepiped is u cos θ. And we can see that since v× w is θ perpendicular to the base parallelogram dew termined by v and w. Hence.4. v.15. Volume of a parallelepiped: Let the vectors u.6 we know that u · (v × w) .)
(1. u v× w vol(P) = A h u u · (v × w) = v× w u v× w = u · (v × w) cos θ =
h
v Parallelepiped P
Figure 1. Repeating this with the base determined by w and u. The proof of the following theorem is left as an exercise for the reader:
. we have the following result: For any vectors u. because the vector u is on the same side of the base parallelogram's plane as the vector × v× w (so that cos θ > 0). where θ is the angle between u and v × w.12. Show that the volume of P is the scalar triple product u · (v × w).7
In Example 1. w in 3 represent any three adjacent sides of a parallelepiped. v. So taking the absolute value of the scalar triple product for any order of the three adjacent sides will always give the volume: Theorem 1.
u · (v × w) = w · (u × v) = v · (w × u) (Note that the equalities hold trivially if any of the vectors are 0. w in
3. then picking the wrong order for the three adjacent sides in the scalar triple product in formula (1. By Theorem u 1. Another type of triple product is the vector triple product u × (v × w). w in 3 represent adjacent sides of a parallelepiped P. the area A of the base parallelogram × × is v× w . By Theorem 1.4 Cross Product
25
Example 1. w in 3 . as in Figure 1. If vectors u. the volume is w · (u × v). then the height h is u cos θ.4. v. Solution: Recall that the volume of a par.12)
Since v × w = −w × v for any vectors v.v × w allelepiped is the area A of the base parallelogram times the height h.7. and not − u cos θ. Since the volume is the same no matter which base and height we use.
z0 ) is the vector pointing to P. z0 + ct). y0 .1 0 v tv r + tv t>0 y L
Let r = (x0 . y0 . y0 . Since v = (a. b.5 Lines and Planes
31
1. Since multiplying the vector v by a scalar t lengthens or shrinks v while preserving its direction if t > 0. and let L be the line through P which is parallel to v (see Figure 1. y0 .16) where r = (x0 . then the terminal point of the vector r + tv is (x0 + at. The reason for doing this is simple: using vectors makes it easier to study objects in 3-dimensional Euclidean space. z0 ) and nonzero vector v = (a. y0 . c). c) be a nonzero vector.5 Lines and Planes
Now that we know how to perform some operations on vectors. z0 ) and nonzero vector v in 3 . like lines and planes. We can summarize the vector representation of L as follows: For a point P = (x0 . y = y0 + bt.
the line L through P
z = z0 + ct.5. for − ∞ < t < ∞ (1. the line L through P parallel to v is given by r + tv. Note that we used the correspondence between a vector and its terminal point. b. z0 ) r r + tv t<0 x Figure 1. b.5.1).1 that every point on the line L can be obtained by adding the vector tv to the vector r for some scalar t.
3. z0 ) be a point in 3 . then we see from Figure 1.17)
Note that in both representations we get the point P on L by letting t = 0.1. as t varies over all real numbers. z0 ) be the vector pointing from the origin to P. the vector r + tv will point to every point on L. We then get the parametric representation of L with the parameter t: For a point P = (x0 . in the language of vectors. y. and reversing its direction if t < 0.5. let v = (a. for − ∞ < t < ∞
(1.
. y0 + bt. We will first consider lines. That is.
z P(x0 . we can start to deal with some familiar geometric objects. y0 . parallel to a vector
Let P = (x0 . c) in parallel to v consists of all points (x. z) given by x = x0 + at.
Line through a point.
though. 3. 17) on L.19. the parametric representation always gives just the points on L and nothing else. the vector representation gives us the vectors whose terminal points make up the line L. say.2). You may have noticed that the vector representation of L in formula (1. 6). called the symmetric representation of L: For a point P = (x0 . y0 . and so x = x0 + 0t = x0 . 5). z) such that x = 2 + 4t. y = 3 − t. 5) + t(4. y.18)
Note that this says that the line L lies in the plane x = x0 .2 can be derived for the cases when b = 0 or c = 0. which x is parallel to the yz-plane (see Figure 1. (c) symmetric. We can also solve for t in terms of y and in terms of z if neither b nor c.5.19)
x0 z L y 0 x = x0
(1. but we do know that x = x0 + at. −1. Example 1. 1. c) in 3 with a. z) such that x−2 y−3 z−5 = = 4 −1 6 (d) Letting t = 1 and t = 2 in part(b) yields the points (6. is zero: t = (y − y0 )/b and t = (z − z0 )/c. 6). the line L through P parallel to v consists of all points (x.16) is more compact than the parametric and symmetric formulas. Solution: (a) Let r = (2. for − ∞ < t < ∞
(c) L consists of the points (x. Then the symmetric representation of L would be: x = x0 . y − y0 z − z0 = b c (1.5.
. VECTORS IN EUCLIDEAN SPACE
In formula (1. −1.17). y. 3. b and c all nonzero. Write the line L through the point P = (2. z0 ) and vector v = (a. in the following forms: (a) vector. y. Technically. Then by formula (1. That is an advantage of using vector notation. z = 5 + 6t. b. z) given by the equations x − x0 y − y0 z − z0 = = a b c What if. Similar equations Figure 1.16). 3. for − ∞ < t < ∞ (b) L consists of the points (x. L is given by: r + tv = (2. so we can write the following system of equalities. respectively. Lastly: (d) find two points on L distinct from P. These three values all equal the same value t. then we can solve for the parameter t: t = (x − x0 )/a. 2. 5) and parallel to the vector v = (4.32
CHAPTER 1. 11) and (10. if a 0. So you have to remember to identify the vectors r + tv with their terminal points. a = 0 in the above scenario? We can not divide by zero. On the other hand. not just L itself. (b) parametric.
1. y. z − z0 ). Let P be a plane in 3 .6 The plane P
Conversely.23 is 2x + 4y + 8z − 22 = 0. or equivalently: a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 The above equation is called the point-normal form of the plane P. z) lies in P. if (x. y0 .
n
r (x.
.25) and combine the constant terms. z) satisfying the vector equation: n· r = 0 (1.25)
For example. 4. z − z0 ) lies in the plane P (see Figure 1.26) (1. z − z0 ) 0 and n · r = 0. and let n = (a. the plane P consists of all points (x. This proves the following theorem: Theorem 1.
35
Plane through a point. y − y0 . b. y.24) where r = (x − x0 . Then the vector r = (x − x0 . let (x0 .25). z0 )
Figure 1. perpendicular to a vector
Let P be a plane in 3 . z0 ).6). z) is any point in 3 such that r = (x − x0 . z) be any point in the plane P. then r ⊥ n and hence n · r = 0. z0 ) be a point in P. y. c) be a nonzero vector which is perpendicular to the plane P.5 Lines and Planes We will now consider planes in 3-dimensional Euclidean space. the normal form of the plane in Example 1. 8). z) such that: 2(x + 3) + 4(y − 1) + 8(z − 3) = 0 If we multiply out the terms in formula (1. y − y0 . z) (x0 .5. y. Such a vector is called a normal vector (or just a normal) to the plane. y0 .18. we get an equation of the plane in normal form: ax + by + cz + d = 0 (1.23. c) be a nonzero vector which is perpendicular to P. y0 . Find the equation of the plane P containing the point (−3. then r ⊥ n and so (x. So if r 0. Now let (x. 3) and perpendicular to the vector n = (2. b. Solution: By formula (1. Example 1. Then P consists of the points (x.1. and suppose it contains a point P0 = (x0 . And if r = 0 then we still have n · r = 0. y.5. y. y − y0 . Let n = (a.
3). the plane P consists of all points (x. y. 1. −2.
. as in Example 1. 2) and S = (3. 1).36
CHAPTER 1.7).e.
− − → −→ − n = QR × QS − − → QR R S
Q −→ − QS
Figure 1. (1. 1. −2. 2) and (3. −1) and QS = (1. 2. VECTORS IN EUCLIDEAN SPACE
Plane containing three noncollinear points
In 2-dimensional and 3-dimensional space. −1. −2). Then for the vectors QR = −→ − (−1. In fact. We will leave examples of this as exercises for the reader. R = (1. −1) × (1.5. R and S ) lie in the − − → −→ − plane through the point Q with normal vector n = QR × QS (see Figure 1. and so their cross product QR × QS − − → −→ − − − → −→ − is perpendicular to both QR and QS . −3. However. − − → Solution: Let Q = (2. then QR and QS are nonzero − − −→ → − vectors which are not parallel (by noncollinearity). an infinite number of planes would contain the line on which those three points lie. So two skew lines do not determine a plane. 3).24.e. 2. then use the technique above. R. 1) So using formula (1. Find the equation of the plane P containing the points (2. parallel planes. So QR and QS (and hence Q. But two (nonidentical) lines which either intersect or are parallel do determine a plane. the plane P has a normal vector − − → −→ − n = QR × QS = (−1. one point from one line and two points from the other). two points determine a line. R and S are noncollinear points in 3 . three collinear points (i. simply pick from the two lines a total of three noncollinear points (i. Two points do not determine a plane in 3 . For if Q. S
Example 1. z) such that: 5(x − 2) − 3(y − 1) + (z − 3) = 0 or in normal form. 1. 5x − 3y + z − 10 = 0 We mentioned earlier that skew lines in 3 lie on separate. to write the equation.7
Noncollinear points Q.5. In both cases. −1. all on the same line) do not determine a plane. 1). 1. to find the equation of the plane that contains those two lines. three noncollinear points do determine a − − → −→ − plane.25) with the point Q (we could also use R or S ). −2) = (5.24.
intersect in a line L. they do so in L a line (see Figure 1. 31) is on L. Likewise. then L is given by: r + t(n1 × n2 ) = (0. Figure 1. n1 × n2 is parallel to L. we can write L in the following vector form: L : r + t(n1 × n2 ) . and so the point (0. 26). then the two planes are not parallel and hence will intersect. 1) and the plane 2x + 4y − z + 3 = 0 has normal vector n2 = (2. for − ∞ < t < ∞ (1. n1 × n2 is parallel to the intersection of P1 and P2 . −3. i. Then z = 31. and the planes are perpendicular if their normal vectors are perpendicular. 7.9 n1 × n2 ⊥ n2 means that n1 × n2 is also parallel to P2 . Since n1 × n2 = (−1.38
CHAPTER 1.26. y = 7 + 7t. for − ∞ < t < ∞ or in parametric form: x = −t. 26). If two planes do intersect. for − ∞ < t < ∞
. 7. z = 31 + 26t. To find a point in both planes. 31) + t(−1. Since n1 and n2 are not scalar multiples. 7. Suppose that two planes P1 and P2 with normal vectors n1 and n2 . Thus. which leaves you to solve two equations in just two unknowns. VECTORS IN EUCLIDEAN SPACE
Line of intersection of two planes
Note that two planes are parallel if they have normal vectors that are parallel. substituting that into the first equation gives y = 7. This can often be made easier by setting one of the coordinate variables to zero. Example 1. then n1 × n2 is parallel to the plane P1 .
Solution: The plane 5x − 3y + z − 10 = 0 has normal vector n1 = (5.28)
where r is any vector pointing to a point belonging to both planes.9). 7. z) on both planes will satisfy the following system of two equations in three unknowns: 5x − 3y + z − 10 = 0
2x + 4y − z + 3 = 0
Set x = 0 (why is that a good choice?). Since n1 × n2 ⊥ n1 .e. respectively. z) to the two normal form equations of the planes. 4. −1).5. Thus. y.5. Then the above equations are reduced to: −3y + z − 10 = 0
4y − z + 3 = 0
The second equation gives z = 4y + 3. Find the line of intersection L of the planes 5x − 3y + z − 10 = 0 and 2x + 4y − z + 3 = 0. A point (x. find a common solution (x. y.
1
Spheres in
3
Note in Figure 1. z0 ) x0
y
x
(a) radius r. a plane given by ax+by+cz+d = 0 is the solution set of F(x. The plane is the simplest surface. which we will define informally8 as the solution set of the equation F(x. y0 . Definition 1. Figure 1. For example. a plane intersects a sphere either at a single point or in a circle.9.
.
8
See O'N EILL for a deeper and more rigorous discussion of surfaces. z) = ax+by+cz+d. given by x2 + y2 = r2 as a subset of 2 ). In general. y. Surfaces are 2-dimensional. this can be written in the equivalent form: S = { x : x − x0 = r } where x = (x. y. 0.6 Surfaces
In the previous section we discussed planes in Euclidean space. A plane is an example of a surface.
z z
(1.40
CHAPTER 1. Similarly for the intersections with the xz-plane and the yz-plane. z) = 0 for the function F(x. z) in 3 which are a fixed distance r (called the radius) from a fixed point P0 = (x0 .6. y0 .e. z0 ) (called the center of the sphere): S = { (x.1 illustrates the vectorial approach to spheres. center (x0 . z) : (x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r2 } (1.1(a) that the intersection of the sphere with the xy-plane is a circle of radius r (i. y0 .29) Using vector notation. for some real-valued function F. z) and x0 = (x0 . since it is "flat". y. y. A sphere S is the set of all points (x. a great circle. In this section we will look at some surfaces that are more complex. center (0. z0 ) are vectors. y. VECTORS IN EUCLIDEAN SPACE
1. z0 )
Figure 1.30)
x =r x
y 0
x − x0 = r x x − x0 (x0 .6. the most important of which are the sphere and the cylinder.6. 0)
0 x
(b) radius r. y. z) = 0 in 3 . y0 .
The cylinders that we will consider are right circular cylinders. These are cylinders obtained by moving a line L along a circle C in 3 in a way so that L is always perpendicular to the plane containing C. We will only consider the cases where the plane containing C is parallel to one of the three coordinate planes (see Figure 1.6.3).
z r z r y 0 x
(a) x2 + y2 = r2 , any z
z
y x
r 0
y
0 x
(b) x2 + z2 = r2 , any y
3
(c) y2 + z2 = r2 , any x
Figure 1.6.3 Cylinders in
For example, the equation of a cylinder whose base circle C lies in the xy-plane and is centered at (a, b, 0) and has radius r is (x − a)2 + (y − b)2 = r2 , (1.32)
where the value of the z coordinate is unrestricted. Similar equations can be written when the base circle lies in one of the other coordinate planes. A plane intersects a right circular cylinder in a circle, ellipse, or one or two lines, depending on whether that plane is parallel, oblique9 , or perpendicular, respectively, to the plane containing C. The intersection of a surface with a plane is called the trace of the surface.
9
for some constants A, B, . . . , J. If the above equation is not that of a sphere, cylinder, plane, line or point, then the resulting surface is called a quadric surface. One type of quadric surface is the ellipsoid, given by an equation of the form: x2 y2 z2 + + =1 a2 b2 c2 (1.34) a In the case where a = b = c, this is just a sphere. In general, an ellipsoid is egg-shaped (think of an ellipse rotated around its major axis). Its traces in the coordinate planes are ellipses.
0
c
z
y
b
x Figure 1.6.4
Ellipsoid
Two other types of quadric surfaces are the hyperboloid of one sheet, given by an equation of the form: x2 y2 z2 + − =1 (1.35) a2 b2 c2 and the hyperboloid of two sheets, whose equation has the form: x2 y2 z2 − − =1 a2 b2 c2
z z
(1.36)
y 0 0
y
x Figure 1.6.5 Hyperboloid of one sheet
x Figure 1.6.6 Hyperboloid of two sheets
44
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
For the hyperboloid of one sheet, the trace in any plane parallel to the xy-plane is an ellipse. The traces in the planes parallel to the xz- or yz-planes are hyperbolas (see Figure 1.6.5), except for the special cases x = ±a and y = ±b; in those planes the traces are pairs of intersecting lines (see Exercise 8). For the hyperboloid of two sheets, the trace in any plane parallel to the xy- or xzplane is a hyperbola (see Figure 1.6.6). There is no trace in the yz-plane. In any plane parallel to the yz-plane for which | x| > |a|, the trace is an ellipse. The elliptic paraboloid is another type of quadric surface, whose equation has the form: x2 y2 z + 2 = 2 c a b (1.37)
z
The traces in planes parallel to the xy-plane are ellipses, though in the xy-plane itself the trace is a single point. The y traces in planes parallel to the xz- or yz-planes are parabo0 las. Figure 1.6.7 shows the case where c > 0. When c < 0 the x surface is turned downward. In the case where a = b, the surface is called a paraboloid of revolution, which is often Figure 1.6.7 Paraboloid used as a reflecting surface, e.g. in vehicle headlights.10 A more complicated quadric surface is the hyperbolic paraboloid, given by: x2 y2 z − 2 = 2 c a b (1.38)
100 50 0 -10 z -50 -5 -100-10 -5 0 y 5 10 10 5 0 x
Figure 1.6.8
10
Hyperbolic paraboloid
For a discussion of this see pp. 157-158 in H ECHT.
1.6 Surfaces
45
The hyperbolic paraboloid can be tricky to draw; using graphing software on a computer can make it easier. For example, Figure 1.6.8 was created using the free Gnuplot package (see Appendix C). It shows the graph of the hyperbolic paraboloid z = y2 − x2 , which is the special case where a = b = 1 and c = −1 in equation (1.38). The mesh lines on the surface are the traces in planes parallel to the coordinate planes. So we see that the traces in planes parallel to the xz-plane are parabolas pointing upward, while the traces in planes parallel to the yz-plane are parabolas pointing downward. Also, notice that the traces in planes parallel to the xy-plane are hyperbolas, though in the xy-plane itself the trace is a pair of intersecting lines through the origin. This is true in general when c < 0 in equation (1.38). When c > 0, the surface would be similar to that in Figure 1.6.8, only rotated 90◦ around the z-axis and the nature of the traces in planes parallel to the xz- or yz-planes would be reversed. The last type of quadric surface that we will consider is the elliptic cone, which has an equation of the form: x2 y2 z2 + − =0 a2 b2 c2 (1.39)
z
y The traces in planes parallel to the xy-plane are ellipses, 0 except in the xy-plane itself where the trace is a single point. The traces in planes parallel to the xz- or yz-planes are hyperbolas, except in the xz- and yz-planes themselves where the traces are pairs of intersecting lines. x Notice that every point on the elliptic cone is on a line which lies entirely on the surface; in Figure 1.6.9 these Figure 1.6.9 Elliptic cone lines all go through the origin. This makes the elliptic cone an example of a ruled surface. The cylinder is also a ruled surface. What may not be as obvious is that both the hyperboloid of one sheet and the hyperbolic paraboloid are ruled surfaces. In fact, on both surfaces there are two lines through each point on the surface (see Exercises 11-12). Such surfaces are called doubly ruled surfaces, and the pairs of lines are called a regulus. It is clear that for each of the six types of quadric surfaces that we discussed, the surface can be translated away from the origin (e.g. by replacing x2 by (x − x0 )2 in its equation). It can be proved11 that every quadric surface can be translated and/or rotated so that its equation matches one of the six types that we described. For example, z = 2xy is a case of equation (1.33) with "mixed" variables, e.g. with D 0 so that we get an xy term. This equation does not match any of the types we considered. However, by rotating the x- and y-axes by 45◦ in the xy-plane by means of the coor√ √ dinate transformation x = (x′ − y′ )/ 2, y = (x′ + y′ )/ 2, z = z′ , then z = 2xy becomes the hyperbolic paraboloid z′ = (x′ )2 − (y′ )2 in the (x′ , y′ , z′ ) coordinate system. That is, z = 2xy is a hyperbolic paraboloid as in equation (1.38), but rotated 45◦ in the xy-plane.
11
can be extended naturally to vector-valued functions. This is the equation of a helix (see Figure 1. as in the following definition. f3 (t).1).1
It may help to think of vector-valued functions of a real variable in 3 as a generalization of the parametric functions in 2 which you learned about in single-variable calculus. 1).and y-coordinates of f(t) are x = cos t and y = sin t. At t = 1 the value of the function is the vector i + j + k. sin t. 1.8.8 Vector-Valued Functions
51
1. we can begin discussing functions whose values are vectors. f2 (t). As the value of t increases. Definition 1. A vector-valued function of a real variable is a rule that associates a vector f(t) with a real number t. and the second form is useful when considering just the terminal points of the vectors. The concept of a limit. called the component functions of f. f(t) = ti + t2 j + t3 k is a vector-valued function in 3 . We write f : D → 3 to denote that f is a mapping of D into 3 . The first form is often used when emphasizing that f(t) is a vector. For each t.
2 2 2 2
f(2π) f(0)
x
y 0
Figure 1. it will sometimes be the case that results from singlevariable calculus can simply be applied to each of the component functions to yield a similar result for the vector-valued function. the curve lies on the surface of the right circular cylinder x2 + y2 = 1. which in Cartesian coordinates has the terminal point (1. We would write f : → 3 . there are times when such generalizations do not hold (see Exercise 13). t).1. the x. However.10.
.
z
Example 1. By identifying vectors with their terminal points.8 Vector-Valued Functions
Now that we are familiar with vectors and their operations. A vector-valued function of a real variable can be written in component form as f(t) = f1 (t)i + f2 (t)j + f3 (t)k or in the form f(t) = ( f1 (t). where t is in some subset D of 1 (called the domain of f). the terminal points of f(t) trace out a curve spiraling upward. f2 (t).35. For example. though. defined for all real numbers t.8. f3 (t)) for some real-valued functions f1 (t). Define f : → 3 by f(t) = (cos t. Since each of the three component functions are real-valued. Thus. a curve in space can be written as a vector-valued function. Much of the theory of real-valued functions of a single real variable can be applied to vector-valued functions of a real variable. so x + y = cos t + sin t = 1.
t). Let f(t) = (cos t. f3 ′ (a)).52
CHAPTER 1. 1) for all t. if lim f(t) − c = 0. then
t→a t→a
lim f(t) = lim f1 (t). cos t. The tangent line L to the curve at f(2π) = (1. or in parametric form: x = 1. Equivalently. Then f ′ (t) = (− sin t.8.36. z = 2π + s for −∞ < s < ∞. 2π) + s(0. Let f(t) = ( f1 (t). 0. df The derivative of f(t) at a.8.
z
f(a)
f(a
f ′ (a)
+ h) −
L f(t)
y
f(a ) f(a + h)
0 x Figure 1. The above definition shows that continuity and the derivative of vector-valued functions can also be defined in terms of its component functions. f ′ (a) = ( f1 ′ (a). Definition 1.
t→a
f(t) is continuous at a if and only if f1 (t). VECTORS IN EUCLIDEAN SPACE
Definition 1. the derivative of a vector-valued function is a tangent vector to the curve in space which the function represents. let a be a real number and let c be a vector. We say that f(t) is differentiable at a if f ′ (a) exists. is the limit dt f ′ (a) = lim f(a + h) − f(a) h→0 h
if that limit exists. representing the slope of the tangent line to the graph of the function at a point. f2 (t). 1). denoted by f ′ (a) or (a). Equivalently. f2 ′ (a). f2 (t).2). 0. Let f(t) be a vector-valued function. if the component derivatives exist. Recall that the derivative of a real-valued function of a single variable is a real number. 2π) is L = f(2π) + s f ′ (2π) = (1. Then we say that the limit of f(t) as t approaches a equals c.12. and let a be a real number in its domain. lim f3 (t)
t→a t→a t→a t→a
provided that all three limits on the right side exist. Then f(t) is continuous at a if lim f(t) = f(a). written as lim f(t) = c. f3 (t)) be a vector-valued function. lim f2 (t). If f(t) = ( f1 (t).
. f3 (t)).2 Tangent vector f ′ (a) and tangent line L = f(a) + sf ′ (a)
Example 1. and it lies on the tangent line to the curve (see Figure 1.11. and f3 (t) are continuous at a. Similarly. 1. sin t. f2 (t). y = s.
Find the length L of the helix f(t) = (cos t. Then the arc length L of the curve from t = a to t = b is
b b
L=
a
f ′ (t) dt =
a
x ′ (t)2 + y ′ (t)2 + z ′ (t)2 dt
(1.41)
A real-valued function whose first derivative is continuous is called continuously differentiable (or a C1 function). Solution: By formula (1.
(1.9 Arc Length
Let r(t) = (x(t). we have
2π 2π 2π
L=
(− sin t)2 + (cos t)2 + 12 dt = 0 √ √ = 2(2π − 0) = 2 2π
sin2 t + cos2 t + 1 dt =
0 0
√ 2 dt
Similar to the case in 2 . Suppose that in the interval (a. and that no section of the curve is repeated. Duhamel's principle is needed. See the proof in T AYLOR and M ANN.41. normally glossed over in calculus texts.2. z(t)) be a curve in 3 whose domain includes the interval [a.2 and § 18. b]. A rigorous proof requires dealing with some subtleties. it seems natural to define the distance s traveled by the object from time t = a to t = b as the definite integral
b b
s=
a
v(t) dt =
a
x ′ (t)2 + y ′ (t)2 + z ′ (t)2 dt . Let f(t) = (x(t). the arc length of a curve in 3 . sin t. in general. All the functions we will consider will be smooth. z(t)) be the position vector of an object moving in 3 . The sum of the arc lengths over the subintervals will be the arc length over [a. if there are values of t in the interval [a.1. This is indeed how we will define the distance traveled and. b] where the derivative of a component function is not continuous then it is often possible to partition [a. Note that we did not prove that the formula in the above definition actually gives the length of a section of a curve. § 14. Definition 1. and a function whose derivatives of all orders are continuous is called smooth (or a C∞ function).41). y(t). b]. Since v(t) is the speed of the object at time t.
. which can be ignored). A smooth curve f(t) is one whose derivative f ′ (t) is never the zero vector and whose component functions are all smooth.15 Example 1. which are beyond the scope of this book.13. y(t). y(t) and z(t) exists and is continuous.40)
which is analogous to the case from single-variable calculus for parametric functions in 2 . b] into subintervals where all the component functions are continuously differentiable (except at the endpoints. t) from t = 0 to t = 2π. b) the first derivative of each component function x(t).9 Arc Length
59
1.
15
In particular.
b]. Then α is smooth.42. the speeds of f(t) and g(t) are f ′ (t) = 2 and g ′ (t) = 2 2. respectively. and is strictly increasing (since α ′ (s) = 2 > 0 for all s).14. VECTORS IN EUCLIDEAN SPACE
Notice that the curve traced out by the function f(t) = (cos t. and t = α(s) is a differentiable scalar function of s. Chain Rule: If f(t) is a differentiable vector-valued function of t. d] → 3 defined by g(s) = f(α(s)) is a parametrization of C with parameter s. sin 2s. t) for t in [0. d] α t [a. Likewise. b]. Definition 1. defining α : [0. one-to-one. 2π] by α(s) = 2πs shows that h(s) is equivalent to f(t). sin t.
. b] f(t)
3
f
g(s) = f(α(s)) = f(t) Note that the differentiability of g(s) follows from a version of the Chain Rule for vector-valued functions (the proof is left as an exercise): Theorem 1. sin 2πs. d] → [a. s [c. d] onto [a. b] be a smooth one-to-one mapping of an interval [c. over the interval [0. 2πs) for s in [0.42) ds dt ds for any s where the composite function f(α(s)) is defined. define α : [0. 1] To see that g(s) is equivalent to f(t). π] h(s) = (cos 2πs. Example 1. π] → [0. 2π] by α(s) = 2s. and let α : [c. 2π]. 2π] g(s) = (cos 2s. and df df dt = (1. 2π]. 1] → [0. t) from Example 1. We say that g(t) and f(t) are different parametrizations of the same curve.41 is also traced out by the function g(t) = (cos 2t. d] then we say that g(s) is equivalent to f(t). this says that g(t) traces the curve twice as fast as f(t). sin t. g(t) traces out the same section of the curve as f(t) does over the interval [0. viewing the functions as position vectors and their derivatives as velocity √ √ vectors. sin 2t. This makes sense since. Then the function g : [c. π] onto [0.21. maps [0. The following are all equivalent parametrizations of the same curve: f(t) = (cos t. 2s) for s in [0. π]. Let C be a smooth curve in 3 represented by a function f(t) defined on an interval [a. 2t). For example. If α is strictly increasing on [c. then f(s) = f(α(s)) is a differentiable vector-valued function of s. Intuitively.60
CHAPTER 1.
we know that this means that there exists an inverse function α : [0. then f ′ (t) > 0 for all t in [a. And we know that the derivative of α is 1 1 α ′ (s) = ′ = ′ s (α(s)) f (α(s)) So define the arc length parametrization f : [0. b] onto the interval [s(a). L]. That is. There is a natural correspondence between s and t: from a starting point on the curve. L] → [a. so 1 = f ′ (α(s)) ′ . differentiable mapping onto the interval [0.
(1. L]
t [a.
Since f(t) is smooth. by the Chain Rule. with different speeds. b] there is a unique s in [0. So the new parameter will be distance instead of time.43)
In terms of motion along a curve. then it is a function of t. b] → [0.1. b]. s is the distance traveled along the curve after time t has elapsed. b]
s(t)
Figure 1. b].
. L]. t] for each t in [a. the distance traveled along the curve (in one direction) is uniquely determined by the amount of time elapsed. But we see that
a b
s(a) =
a
f ′ (u) du = 0 and
s(b) =
a
f ′ (u) du = L = arc length from t = a to t = b α(s)
So the function s : [a.1 t = α(s)
by
f(s) = f(α(s)) for all s in [0. Recall that this means that s is a one-to-one mapping of the interval [a. From single-variable calculus. f(s) has unit speed: f ′ (s) = f ′ (α(s)) α ′ (s) by the Chain Rule. so which one is the best to use? In some situations the arc length parametrization can be useful. for any given smooth parametrization f(t) defined on [a. b]. L] →
3
s [0. L]. In fact. b]. by the parameter s given by
t
s = s(t) =
a
f ′ (u) du. L]. s(b)]. By the Fundamental Theorem of Calculus. so f (α(s)) f ′ (s) = 1 for all s in [0. Since s is the arc length of the curve over the interval [a. Thus s ′ (t) > 0 and hence s(t) is strictly increasing on the interval [a.9. Then f(s) is smooth. b] → [0. So the arc length parametrization traverses the curve at a "normal" rate. L] such that s = s(t) and t = α(s). b] that is differentiable and the inverse of s : [a. its derivative is s ′ (t) = ds d = dt dt
t
f ′ (u) du = f ′ (t)
a
for all t in [a.9 Arc Length
61
A curve can have many parametrizations. for each t in [a. L] is a one-to-one. and vice versa. b]. The idea behind this is to replace the parameter t.
sin √ . y(t) = r(t) sin θ(t). and these definitions can be shown to be equivalent to those using arc length. 2 2π]. We will leave this to the exercises.8.43. Example 1.
s So we can solve for t in terms of s: t = α(s) = √ . the usual parametrizations of Bézier curves. b]. is desirable. arc length parametrizations are more useful for theoretical purposes than for practical computations. which makes their computation much easier. they are in fact usually impossible to calculate at all.44)
Proof: The Cartesian coordinates (x(t). for t in [a. in CAD. in the field of mathematics known as differential geometry. which we discussed in Section 1. you would then substitute the expression for t in terms of s (which we called α(s)) into the formula for f(t) to get f(s). Suppose that r = r(t). for t in [0. The arc length for curves given in other coordinate systems can also be calculated: Theorem 1.
16 17
y ′ (t) = r ′ (t) sin θ(t) + r(t)θ ′ (t) cos θ(t)
See O'N EILL for an introduction to elementary differential geometry. z(t)) of a point on the curve are given by x(t) = r(t) cos θ(t). For example. we have
t t
s=
0
f ′ (u) du =
0
√ √ 2 du = 2 t for all t in [0. 2π]. are polynomial functions in 3 . sin t. θ = θ(t) and z = z(t) are the cylindrical coordinates of a curve f(t). b] is
b
L=
a
r ′ (t)2 + r(t)2 θ ′ (t)2 + z ′ (t)2 dt
(1.62
CHAPTER 1. 2 √ s s s ∴ f(s) = cos √ . y(t). VECTORS IN EUCLIDEAN SPACE
In practice. Then the arc length L of the curve over [a. But their arc length parametrizations are not only not polynomials. z(t) = z(t)
so differentiating the above expressions for x(t) and y(t) with respect to t gives x ′ (t) = r ′ (t) cos θ(t) − r(t)θ ′ (t) sin θ(t).
. This makes their computation relatively simple. Parametrize the helix f(t) = (cos t. not the norm. √ for all s in [0. t).16 The methods involve using an arc length parametrization. 2π].41 and formula (1. Note that f ′ (s) = 1. 2 2 2 Arc length plays an important role when discussing curvature and moving frame fields. Solution: By Example 1. by arc length. If that can be done. The simple integral in Example 1. In general.17 Curvature and moving frame fields can be defined without using arc length.22. parametrizing a curve f(t) by arc length requires you to evaluate the t integral s = a f ′ (u) du in some closed form (as a function of t) so that you could then solve for t in terms of s. which often leads to an integral that is either difficult or impossible to evaluate in a simple closed form.43). which.43 is the exception.
Example 2. the domain is the set D = {(x. The domain of the function f (x. The domain of the function f (x.
and the range of f is all of
. and there will also be times when it will be convenient to think of the points as vectors (or terminal points of vectors). but there will be times when we will use points in 3 . y) = 1 − x2 − y2
is the set D = {(x. and the range of f is the set of all real numbers f (x. y. y). The largest possible set D in 2 on which f is defined is called the domain of f .1 Functions of Two or Three Variables
In Section 1. since the quantity inside the square root is nonnegative if and only if 1 − (x2 + y2 ) ≥ 0.1. We will now examine real-valued functions of a point (or vector) in 2 or 3 . y. 1] in . Example 2. The domain of the function f (x. The range of f is all real numbers except 0. y) = xy is all of
2.8 we discussed vector-valued functions of a single real variable.
Example 2. A similar definition holds for functions f (x. y) as (x. y) = 1 x−y
is all of 2 except the points (x. We see that D consists of all points on and inside the unit circle in 2 (D is sometimes called the closed unit disk). A real-valued function f defined on a subset D of 2 is a rule that assigns to each point (x. That is. The range of f is the interval [0.
65
.3. y) for which x = y. y) in D a real number f (x. y) varies over the domain D. For the most part these functions will be defined on sets of points in 2 . z) in 3 . y) : x y}.2 Functions of Several Variables
2.2. z) defined on points (x. y) : x2 + y2 ≤ 1}.
6 0.4 z 0. where c varies over . We will now state explicitly what is meant by the limit of a function of two variables.5. y).5 at the point (x.
.2 -0. y. y) = sin x2 + y2 x2 + y2
is shown below. since it satisfies an equation of the form F(x. y) defined in 2 is often written as z = f (x. Level curves are often projected onto the xy-plane to give an idea of the various "elevation" levels of the surface (as is done in topography). The function is not defined at (0. y) is the set {(x.1.4. the level curves are the solution sets of the equations f (x. z) = f (x.66
CHAPTER 2.
and the range of f is all positive real numbers. y.
1 0. The traces of this surface in the planes z = c. so that the graph of f (x.2 0 -0. FUNCTIONS OF SEVERAL VARIABLES
Example 2. y) − z). y) = √ 2 2
x +y
You may be wondering what happens to the function in Example 2. The domain of the function f (x. y)} in 3 . F(x. as was mentioned in Section 1.8 0.
A function f (x. y) approaches (0.1.4 -5 -10 -5 y 0 0 5 5 10 10 x -10
√ sin x2 +y2 Figure 2. So we see that this graph is a surface in 3 . y. y) = (0. z) = 0 (namely.1 The function f (x. z) : z = f (x. 0). z) = e x+y−z is all of
3. but the limit of the function exists (and equals 1) as (x. for c in . 0). y) = c. since both the numerator and denominator are 0 at that point. are called the level curves of the function. y. The graph of the function f (x. Example 2. 0). Note that the level curves (shown both on the surface and projected onto the xy-plane) are groups of concentric circles. Equivalently.
y) sufficiently close to (a. In the single-variable case. The idea behind the above definition is that the values of f (x. you may remember how awkward they can be.1)
if given any ǫ > 0.
(2.b)
lim
f (x.y)→(1. 2). b) (e. b)
x
(a) x → a in
Figure 2. y) = (a. b) in
2
(a. Example 2. inside a circle centered at (a. is not some indeterminate form like 0/0) then you can just substitute (x. y) − L| < ǫ whenever 0 < (x − a)2 + (y − b)2 < δ.2)
lim
x2
xy (1)(2) 2 = 2 = 2 2 5 +y 1 +2
since f (x. the statement "x → a" means that x gets closer to the value a from two possible directions along the real number line (see Figure 2. b).1. b) be a point in 2 . Let (a. If you recall the "epsilon-delta" proofs of limits of real-valued functions of a single variable. within ǫ of L) if we pick (x.2(a)). b) itself). In two dimensions. y) equals L as (x. b) with some sufficiently small radius δ). Then we say that the limit of f (x.
(x. b) (i. y) be a real-valued function defined on some set containing (a. y) is given by a single formula and is defined at the point (a.2. y) can get arbitrarily close to L (i.2(b)). we will simply state that when the function f (x.2 "Approaching" a point in different dimensions
.
y x 0 a x x 0
(b) (x. there exists a δ > 0 such that | f (x. Instead. b) (but not necessarily defined at (a.1. written as
(x.1. y) = L . In general. y) to find the limit.g. y) can approach a point (a. y) =
xy x2 +y2
is properly defined at the point (1. so we will not bother with such proofs. however. and let f (x. y) approaches (a.
The major difference between limits in one variable and limits in two or more variables has to do with how a point is approached.y)→(a. the multivariable cases are at least equally awkward to go through.1. b) along an infinite number of paths (see Figure 2.6. and how they can usually only be done easily for simple functions.
A similar definition can be made for functions of three variables. y) → (a. (x.e.1 Functions of Two or Three Variables
67
Definition 2. b) into the formula for f (x.e.
Clairaut and L. b) with respect to x. y) in the (positive) x or y direction. we can start to develop an idea of a derivative of a function of two or more variables. b) = lim h→0 ∂x h and the partial derivative of f at (a. y) be a real-valued function with domain D in 2 .
. denoted by defined as ∂f f (a. y) = x2 y + y3 .1 Recall that the derivative of a function f (x) can be interpreted as the rate of change of that function in the (positive) x direction. y) = 2xy ∂x and treating x as a constant and differentiating f (x. h→0 ∂y h (2. y) = x2 + 3y2 . b) with respect to y. using the usual rules from single-variable calculus. we can see that the partial derivative of a function f (x. y) for the function f (x. The symbol was first used by the mathematicians A. is ∂y (2.2) ∂f (a. b) be a point in D. and then simply differentiating f (x.3)
Note: The symbol ∂ is pronounced "del". b) (a. Then the partial derivative of f at (a. to distinguish it from the letter d used for the "usual" derivative. ∂f denoted by (a. y) and (x.2 Partial Derivatives
71
2. y) with respect to y gives ∂f (x. b) = lim .10. the partial derivative of f (x. ∂y
1 It is not a Greek letter. y) as if it were a function of y alone. is defined as ∂x f (a + h. We will start with the notion of a partial derivative. y) with respect to x can be calculated by treating the y variable as a constant. y) as if it were a function of x alone. y) with respect to x or y is the rate of change of f (x. b) ∂f (a.2. Euler around 1740. respectively. y) with respect to x gives ∂f (x. b) − f (a.2 Partial Derivatives
Now that we have an idea of what functions of several variables are. and let (a. b). Definition 2. Likewise. y) with respect to y is obtained by treating the x variable as a constant and then differentiating f (x. Find ∂f ∂f (x. Example 2.3. and what a limit of such a function is. ∂x ∂y Solution: Treating y as a constant and differentiating f (x. b). From the definitions above. b + h) − f (a. Let f (x. What this means is that the partial derivative of a function f (x.
b)). b.3.4. b.1 are contained in
.3. b) in the domain D of f (x.2.3. If the (acute) angle between the vector − − → PQ and the plane T approaches zero as the point Q approaches P along the surface S . c) be a point on S . b. Note that since two lines in 3 determine a plane. f (a. b). y) (a. respectively. f (a. y) in the plane y = b is a curve in 3 through the point (a. ∂ f (a. y) be the equation of a surface S in 3 . This indeed turns out to be the case. y). b)) Ly
Lx b 0 (a. you might expect that partial derivatives can be used to define a tangent plane to the graph of a surface z = f (x. The formal definition mimics the intuitive notion of a tangent line to a curve. y. y) ∂y in the plane x = a (see Figure 2. ∂x Similarly.
z z = f (x. b) is the slope of the tangent line Ly to the trace of the surface z = f (x. b)) slope =
∂f ∂x (a. f (x)) in 2 . y). y) in the positive x and y direcdy tions. f (a. Let T be a plane which contains the point P. b)
z = f (x. y) (a.
Definition 2. and the slope of the tangent line L x to that curve at that point is ∂ f (a. we need a definition of a tangent plane. then the two tangent lines to the surface z = f (x. and let Q = (x. b) D
y
x
(b) Tangent line Ly in the plane x = a
Figure 2. There is a similar geometric meaning to the partial derivatives ∂ f and ∂ f of ∂x ∂y a function z = f (x. b. z) represent a generic point on the surface S . b)
z
slope =
∂f ∂y (a. First. then we call T the tangent plane to S at P. The intuitive idea is that a tangent plane "just touches" a surface at a point. the trace of the surface described by z = f (x. namely as the slope of the tangent line to the graph of f at the point (x. y) in the x and y directions described in Figure 2.1
Partial derivatives as slopes
dy Since the derivative dx of a function y = f (x) is used to find the tangent line to the graph of f (which is a curve in 2 ). Let z = f (x. y): given a point (a. Recall that the derivative dx of a function y = f (x) has a geometric meaning.1).3 Tangent Plane to a Surface
In the previous section we mentioned that the partial derivatives ∂ f and ∂ f can be ∂x ∂y thought of as the rate of change of a function z = f (x. b) D
(a) Tangent line L x in the plane y = b
y 0 a x (a.3 Tangent Plane to a Surface
75
2. and let P = (a.
b)) ∂x
(a.6) ∂x (a.
z Suppose that we want an equation of the tangent plane T to the surface z = f (x. ∂ f (a.3. Then the equation for T is T
y
A(x − a) + B(y − b) + C(z − f (a. FUNCTIONS OF SEVERAL VARIABLES
the tangent plane at that point. b)) is ∂x ∂x parallel to Lx (since vx lies in the xz-plane and lies in a line with slope ∂x 1 = ∂ f (a. Hence. Similarly. b). b)) is parallel to Ly . B. the vector ∂y i j n = vx × vy = 1 0 0 1 k
∂f ∂x (a. Since T contains the lines Lx and Ly . b)) = 0
(2. 0.3. b. then the vector vx = (1.3. f (a. b) i ∂x
−
∂f ∂y (a. if the tangent plane exists at that point. b)). § 6. y) will exist at the point (a. it turns out4 that if ∂ f and ∂ f exist in a region around a point (a. Luckily. C) is a normal vector to the plane T .4. b) (x
− a) −
∂f ∂y (a. or it may not have a tangent line at all at that point. b) (y
− b) + z − f (a.3
=
− ∂ f (a. b) = 0
4
See T AYLOR and M ANN.b)
∂f ∂x (a.4)
0 x Figure 2.
z vx = (1. the resulting curve in that plane may have a tangent line which is not in the plane determined by the other two tangent lines. It is possible that if we take the trace of the surface in the plane x − y = 0 (which makes a 45◦ angle with the positive x-axis). y) and Ly be the tangent lines to the traces of the surface (a. b)
x 0 1 Figure 2. Thus the equation of T is −
∂f ∂x (a. b.2).76
CHAPTER 2. then all we need are vectors vx and vy that are parallel to Lx and Ly .3. b) = 0 .
. ∂ f (a. b. b) (y − b) − z + f (a. b) and are continuous at (a. and then let n = vx × vy . b)) L in the planes y = b and x = a. 0. the vector ∂x vy = (0. we have the following result: The equation of the tangent plane to the surface z = f (x. In this text. b. y) at a point (a. The existence of those two tangent lines does not by itself guarantee the existence of the tangent plane. Let Lx z = f (x. f (a. See Figure 2.5)
Multiplying both sides by −1. b) then the ∂x ∂y tangent plane to the surface z = f (x. b). 1. ∂ f (a. respectively. b) j
+k
is normal to the plane T . b)). b) (x − a) + ∂y (a. f (a. y) at the point (a.3). and suppose that the conditions for T to exist do Lx hold. b)
∂f
where n = (A. respectively (as in Figure y 2.2 Tangent plane
Since the slope of Lx is ∂ f (a. f (a.
(2. b) ∂f ∂y (a. those conditions will always hold. b)) is ∂f ∂f (2.
b) = ∂ f (a. b) be a point in D.78
CHAPTER 2. denoted by Dv f (a. b) as a vector. b) . b) = v1 ∂f ∂f (a. 0) then the above formula reduces to Dv f (a.4 Directional Derivatives and the Gradient
For a function z = f (x. ∂x which we know is true since Di f = ∂ f . as that represents a "standard" vector for a given direction. which the reader should be used to by now. Similarly. b) = lim f ((a. then we
. b). If we were to write the vector v as v = (v1 . b + hv2 ) − f (a. b) = lim f (a + hv1 . Then Dv f (a. Let v be a unit vector in 2 . FUNCTIONS OF SEVERAL VARIABLES
2.9)
h→0
From this we can immediately recognize that the partial derivatives ∂ f and ∂ f are ∂x ∂y special cases of the directional derivative with v = i = (1. we use a unit vector in the definition. Let f (x.5. Since there are many vectors with the same ∂x ∂y direction. b).10)
Proof: Note that if v = i = (1. since we are adding the vector hv to it. y) has continuous partial derivatives ∂ f and ∂ f (which will always be the case ∂x ∂y in this text). b) + hv) − f (a. If f (x. b) be a point in D. v2 ) be a unit vector in 2 . then Dv f (a. we learned that the partial derivatives ∂ f and ∂ f represent ∂x ∂y the (instantaneous) rate of change of f in the positive x and y directions. for v = j = (0. 1). Let f (x. That is. But this is just the usual idea of identifying vectors with their terminal points. 1) are the only unit vectors in 2 with a zero component.2. ∂x ∂y and let v = (v1 . b) . v2 ). b). b) = ∂ f (a. y) be a real-valued function with domain D in 2 . h (2. as we noted earlier. and let (a. What about other directions? It turns out that we can find the rate of change in any direction using a more general type of derivative called a directional derivative. 0) and v = j = (0. So since ∂y ∂y i = (1. y). ∂ f = Di f and ∂ f = Dj f . 0) and j = (0. b) + v2 (a. y) be a real-valued function with domain D in 2 such that the partial derivatives ∂ f and ∂ f exist and are continuous in D. then there is a simple formula for the directional derivative: Theorem 2.8)
h→0
Notice in the definition that we seem to be treating the point (a. 1) ∂x the formula reduces to Dv f (a. Let (a. ∂x ∂y (2. respectively. b) in the direction of v. b) h (2. Then the directional derivative of f at (a. Definition 2. which is true since Dj f = ∂ f . is defined as Dv f (a. respectively.
The temperature T of a solid is given by the function T (x. we still have Dv f = ∇ f cos θ. But we know that Dv f = v · ∇ f = v ∇ f cos θ. √ 5 5
. a similar argument can be used to show that it also applies to functions of three or more variables. 1. ∇ f ⊥ v. Dv f = 0. y) decreases the fastest in the direction of −∇ f . with ∇ f 0.4.16. We have thus proved the following theorem: Theorem 2. Then: (a) The gradient ∇ f is normal to any level curve f (x. 1) will the temperature decrease the fastest? Solution: Since ∇ f = (−e−x . (c) The value of f (x.17. y.2.
. (b) The value of f (x. 4e4z ).e. while the smallest value occurs when cos θ = −1 (θ = 180◦ ). 2e−2 . At a fixed point (x. so since v is a tangent vector to this curve. 1. A unit vector in
=
1 2 √ . 5)
∇f ∇f
0. the value of the function f increases the fastest in the direction of ∇ f (since θ = 0◦ in that case).
Though we proved Theorem 2. for any unit vector v in 2 . y) is constant along a level curve. Example 2. y) be a continuously differentiable real-valued function. where θ is the angle between v and ∇ f . −2e−2y . √ 5 5
Solution: Since ∇ f = (y2 + 3x2 y. In which direction from the point (1. 2)? In which direction does it decrease the fastest? that direction is v =
2 1 √ . In general. Thus. Example 2. y) the length ∇ f is fixed.4 Directional Derivatives and the Gradient
81
The value of f (x. 2xy + x3 ). where x. In other words. then ∇ f (1. Let f (x. and the value of f decreases the fastest in the direction of −∇ f (since θ = 180◦ in that case). √ 5 5
and decreases the fastest in the direction of
. z) = e−x + e−2y +e4z . which means that ∇ f is normal to the level curve. y) increases the fastest in the direction of ∇ f . y) = xy2 + x3 y increase the fastest from the point (1. In which direction does the function f (x. −4e4 ). In other words. then the rate of change of f in the direction of v is 0. f increases the fastest in the direction of
−2 −1 √ . So since ∇ f 0 then Dv f = 0 ⇒ cos θ = 0 ⇒ θ = 90◦ . i. where θ is the angle between v and ∇ f . z are space coordinates relative to the center of the solid. 1) = (e−1 . y) = c. So since v = 1 then Dv f = ∇ f cos θ. 2) = (10. and the value of Dv f then varies as θ varies. Likewise.4 for functions of two variables. The largest value that Dv f can take is when cos θ = 1 (θ = 0◦ ). y. the directional derivative in the three-dimensional case can also be defined by the formula Dv f = v · ∇ f . then the temperature will decrease the fastest in the direction of −∇ f (1.
y) goes in all directions from the point (a. Since g ′ (x) = ∂ f (x. y) in the domain of f . b). We know that f (a. The function f (x. y) ≤ f (a. b) for all (x. b) is that ∇ f (a. that is. b). 0): ∂ f = y = 0 ⇒ y = 0. 0) is the only critical point. We say that f has a local maximum at (a. y) = 0 and ∂y (x. functions of three or more variables require methods using linear algebra. b)). b) and ∂ f (a. b) ∂x ∂y exist. b).7. the necessary condition that ∇ f (a. ∂x ∂x Similarly. b). so (0. So given a function f (x. We will consider only functions of two variables. y) = xy has a critical point at (0. Example 2. y). b) in the y direction and so ∂ f (a. Similar to the single-variable case. then ∂ f (a. Then a necessary condition for f (x. y) ≤ f (a. then f has a global maximum at (a. Let f (x. 0) since any disk around (0. 0) contains points (x. f (a. b). If f (x. we say that f has a local minimum at (a.5 Maxima and Minima
The gradient can be used to find extreme points of real-valued functions of several variables.5 can be extended to apply to functions of three or more variables.2. y) where the values of x and y have the same sign (so that f (x. b) is the largest value of f (x. Note: Theorem 2. y).e. b) be a point in the domain of f . b) if f (x. Suppose that (a. 0)) and different signs (so that f (x. But clearly f does not have a ∂y local maximum or minimum at (0. then f has a global minimum at (a. b) is the largest value of f near (a. Likewise. that is. In particular. y) = 0 simultaneously for (x. y) for which (x − a)2 + (y − b)2 < r2 . ∂y We thus have the following theorem: Theorem 2. and that the first-order partial derivatives of f exist at (a. y) ≤ f (a. and let (a. In fact.18. So we know that g ′ (a) = 0. y) be a real-valued function. f (a. points where the function has a local maximum or local minimum. b). b) if f (x. y) = xy > 0 = f (0. y). b) is a local maximum point for f (x. b) is the largest value of f in the x direction (around the point (a. If f (x. y) in the domain of f . y) ≥ f (a. b) = 0. y) = xy < 0 = f (0. in some sufficiently small disk centered at (a. b) where ∇ f (a. y) inside some disk of positive radius centered at (a. i. b) = 0. ∂x and ∂ f = x = 0 ⇒ x = 0. y). b) for all (x. 0)). b) for all (x. y) ≥ f (a. b). b). the single-variable function g(x) = f (x. y) inside some disk of positive radius centered at (a. there is some sufficiently small r > 0 such that f (x. b) = 0. b) for all (x. Definition 2. b) has a local maximum at x = a. y) as (x. to find the critical points of f you have to solve the equations ∂f ∂f ∂x (x.5. b) for all (x. b) = 0 is called a critical point for the function f (x. A point (a. y) be a real-valued function such that both ∂ f (a. y) to have a local maximum or minimum at (a. b) = 0 is not always sufficient to guarantee that a critical point is a local maximum or minimum.5 Maxima and Minima
83
2. Let f (x. along the path y = x
.
which has a local minimum at (0. ∇ f (a. b) = 0). then f has a local maximum at (a. So (0. i. b) (i. FUNCTIONS OF SEVERAL VARIABLES
in 2 . saddle point at (0.1
f (x. y) = −x2 . y) = xy.6. b) 2 (a.
100
50
0 z -50 -10 -100-10 -5 -5 y 0 0 5 5 10 10 x
Figure 2.
. b)
(c) if D < 0. § 7. b) ∂x2
∂2 f ∂2 f ∂2 f (a. b) < 0. y) = x2 . then f has neither a local minimum nor a local maximum at (a.1.5. f (x. 0)
The following theorem gives sufficient conditions for a critical point to be a local maximum or minimum of a smooth function (i.
6
See T AYLOR and M ANN. Define D= Then (a) if D > 0 and (b) if D > 0 and
∂2 f (a. The graph of f (x. while along the path y = −x we have f (x. Let f (x.e. which is a hyperbolic paraboloid. 0).6. then the test fails. y) is shown in Figure 2.6 Theorem 2.e. a function whose partial derivatives of all orders exist and are continuous). b) − ∂y ∂x ∂x2 ∂y
2
> 0. y) be a smooth real-valued function. it is a local maximum in one direction and a local minimum in another direction. b) (a. which we will not prove here. which has a local maximum at (0. b) ∂x2 ∂2 f (a. then f has a local minimum at (a. 0). 0) is an example of a saddle point. with a critical point at (a.e.5. b) (d) if D = 0.84
CHAPTER 2.
That is. or complicated expressions involving trigonometric. let alone two. There are formulas for solving polynomial equations of degree 4. Trial and error would not help 3 √ 3 √ 28 − 1. y) 2 (x. especially since the only real solution8 turns out to be 28 + 1− situation such as this. The method we used required us to find the critical points of f . yn )∞ converges to a critical point. 2. which in general is a system of two equations in two unknowns (x and y). yn ) ∂2 f ∂x ∂y (xn . y ) ∂x2 n n ∂f ∂x (xn . yn ) ∂f ∂y (xn . 1. y) could be any points in the domain of f . . you may have a hard time getting the exact solutions. y ) ∂y2 n n ∂f ∂y (xn . where the points (x. yn ) ∂2 f (x . the only choice may be to find a solution using some numerical method which gives a sequence of numbers which converge to the actual solution. which you probably learned in single-variable calculus. and define D(x. y) (x. which are not quite as simple as the familiar quadratic formula. which meant having to solve the equation ∇ f = 0. yn )
xn+1 = xn −
D(xn . yn ) ∂2 f ∂x ∂y (xn .
yn+1 = yn −
D(xn . if one of the equations that had to be solved was x3 + 9x − 2 = 0 .
Newton's algorithm: Pick an initial point (x0 . y) be a smooth real-valued function. could be impossible by elementary means. yn )
(2. exponential. yn ) ∂f ∂x (xn . While this was relatively simple for the examples we did.6 Unconstrained Optimization: Numerical Methods
89
2.
This is also a problem for the equivalent method (the Second Derivative Test) in single-variable calculus.6 Unconstrained Optimization: Numerical Methods
The types of problems that we solved in the previous section were examples of unconstrained optimization problems. Cubic polynomial equations in one variable can be solved using Cardan's formulas. then you will have to try different initial points to find them. For example. 3. In this section we will describe another method of Newton for finding critical points of real-valued functions of two variables. For n = 0. If there are several n=1 critical points. define:
∂2 f (x . then solving even one such equation. y). in general this will not be the case. y) − ∂y ∂x ∂x2 ∂y
2
. Let f (x. .2.7 For example. 8 There are also two nonreal. In a much. though one that is not usually emphasized. . If the equations involve polynomials in x and y of degree three or higher. but it can be proved that there is no general formula for solving equations for polynomials of degree five or higher. complex number solutions.
7
.14)
Then the sequence of points (xn . y0 ). yn )
. or logarithmic functions. y) = ∂2 f ∂2 f ∂2 f (x. we tried to find local (and perhaps even global) maximum and minimum points of real-valued functions f (x. Newton's method for solving equations f (x) = 0. See U SPENSKY for more details. .
6. y0 ) for our algorithm. Since it may take a large number of iterations of Newton's algorithm to be sure that we are close enough to the actual critical point.
z 50000 0 -50000 -100000 -150000 -200000 -250000 -300000 -350000-20
-20 -15 -10 -5 -15 -10 0 -5 5 0 5 10 10 15 15 20 20 x
y
Figure 2. In each iteration the new point will be printed. which in this case can not be done easily. we will let a computer do the computing. using the Java programming language. FUNCTIONS OF SEVERAL VARIABLES
Solution: First calculate the necessary partial derivatives:
Example 2. which will take a given initial point as a parameter and then perform 100 iterations of Newton's algorithm.23. And we can see that D(0. so that we can see if there is convergence.6.1 below). ∂f ∂f = 3x2 − y − 1 + y3 . so we should pick an initial point where D is not zero. Looking at the graph of z = f (x. Find all local maxima and minima of f (x. y) = x3 − xy − x + xy3 − y4 . = 6xy − 12y2 . We need to pick an initial point (x0 . y) = x3 − xy − x + xy3 − y4 for −20 ≤ x ≤ 20 and −20 ≤ y ≤ 20
Notice in the formulas (2. so take (0.14) that we divide by D.1. y) over a large region may help (see Figure 2.90
CHAPTER 2. 0) = (0)(0) − (−1)2 = −1 0. = −x + 3xy2 − 4y3 ∂x ∂y ∂2 f ∂2 f ∂2 f = −1 + 3y2 = 6x . and since the computations are quite tedious. 0) as our initial point. ∂y ∂x ∂x2 ∂y2
Notice that solving ∇ f = 0 would involve solving two third-degree polynomial equations in x and y. though it may be hard to tell where the critical points are. we will write a simple program. For this.1
f (x.
. The full code is shown in Listing 2.
92
CHAPTER 2.-0. −0.4711356343449874.484506572966545.0.4711356343449874.sun. −0.0. we appear to have converged fairly quickly (after only 8 iterations) to what appears to be an actual critical point (up to Java's level of precision).5) n = 3: (0.39636433796318005) = −8.39636433796318005) n = 100: (0.4711356343449874.0) n = 2: (1.4711356343449874.
9
Available for free at × 10−17 ∂x ∂f (0. 0) with this command: java newton 0 0 Below is the output of the program using (0.java is saved..4711356343449705. FUNCTIONS OF SEVERAL VARIABLES
To use this program. −0.47113558510349535.-0.-0.44194107452339687) n = 4: (0.39636450001936047) n = 7: (0.-0.-0.0.0.4711356343449874.47123972682634485. You will need the Java Development Kit9 to compile the code. n = 96: (0.-0. It turns out that both partial derivatives are indeed close enough to zero to be considered zero: ∂f (0. In the directory where newton..-0.39636433796318005) n = 10: (0.39636433796318005).com/javase/downloads
.3963643379632247) n = 8: (0. either by evaluating ∂ f and ∂ f at the point ourselves or by modifying our ∂x ∂y program to also print the values of the partial derivatives at the point.39636433796318005) n = 97: (0.-0.4711356343449874.39636433796318005) = −8.326672684688674 × 10−17 ∂y We also have D(0.39636433796318005) n = 9: (0.39636433796318005) As you can see.-0.4711356343449874.4711356343449874.39636433796318005) n = 98: (0.-0.4711356343449874. you should first save the code in Listing 2.776075636032301 < 0.6065857885615251. −0. It is easy to confirm that ∇ f = 0 at this point.39636433796318005) . so by Theorem 2.6 we know that (0.39636433796318005) n = 99: (0.4711356343449874.java Then run the program with the initial point (0.39636433796318005) = 4. run this command at a command prompt to compile the code: javac newton. −0. 0) as the initial point. truncated to show the first 10 lines and the last 5 lines: java newton 0 0 Initial point: (0.39636433796318005) is a saddle point.0) n = 1: (0.-0.java.-0.405341511995805) n = 5: (0.-1.-0. namely the point (0.1 in a plain text file called newton.3966334583092305) n = 6: (0.
6 -0.4 0.4 -0. FUNCTIONS OF SEVERAL VARIABLES
. and the proof that it converges (given a "reasonable" choice for the initial point) requires techniques beyond the scope of this text.4711356343449874. See R ALSTON and R ABINOWITZ for more detail and for discussion of other numerical methods.2 -0. global maxima and minima tend to be more interesting than local versions.6703832459238667.2 0 -0.42.8
(-0.6 0. y) = x3 − xy − x + xy3 − y4 for −1 ≤ x ≤ 0 and 0 ≤ y ≤ 1
We can summarize our findings for the function f (x.94
CHAPTER 2.2 1 0 x
Figure 2. Many of these methods are based on the steepest descent technique. The crux of the steepest descent idea. which is based on an idea that we discussed in Section 2.4. y) = x3 − xy − x + xy3 − y4 : (−0.42501465652420045) : local maximum (−7.6 y 0. −0. In general. −5.8 -0.39636433796318005) : saddle point
.
0.595509445899435) : saddle point The derivation of Newton's algorithm.540962756992551. is that starting from some initial point.6. A maximization problem can always be turned into a minimization problem (why?).2 0. 0. then. so a large number of methods have been developed to find the global minimum of functions of any number of variables.4 z -0.8 -1 0 0. In the case of functions which have a global maximum or minimum.2
f (x.0. Our description of Newton's algorithm is the special two-variable case of a more general algorithm that can be applied to functions of n ≥ 2 variables. This field of study is called nonlinear programming. you move a certain amount in the (0.4 0. at least in practical applications. Newton's algorithm can be used to find those points. Recall that the negative gradient −∇ f gives the direction of the fastest rate of decrease of a function f .67.6 -0.57)
-1 -0.0.
for solving this problem. Since we must have 2x + 2y = 20. respectively.96
CHAPTER 2.
. Solution: The area A of a rectangle with width x and height y is A = xy. y) with the condition that they satisfy the constraint equation g(x. Notice in the above example that the ease of the solution depended on being able to solve for one variable in terms of the other in the equation 2x + 2y = 20. This is now a function of x alone. z)) given : g(x. find the dimensions that will maximize the area. and we say that x and y are constrained by g(x. Since f ′ (x) = 10 − 2x = 0 ⇒ x = 5 and f ′′ (5) = −2 < 0. Points (x. which we then substitute into f to get f (x.5 and 2. y) = xy given : 2x + 2y = 20 The reader is probably familiar with a simple method. 10] (since f = 0 at the endpoints of the interval). say. 10]. y) (or f (x.7 Constrained Optimization: Lagrange Multipliers
In Sections 2. so we now just have to maximize the function f (x) = 10x − x2 on the interval [0. for solving constrained optimization problems: Maximize (or minimize) : f (x. y) which are maxima or minima of f (x. Example 2.6 we were concerned with finding maxima and minima of functions without any constraints on the variables (other than being in the domain of the function). using single-variable calculus.24. FUNCTIONS OF SEVERAL VARIABLES
2. y. For a rectangle whose perimeter is 20 m. then the maximum area occurs for a rectangle whose width and height both are 5 m. y) = c. y) = c are called constrained maximum or constrained minimum points. So since y = 10 − x = 5. called the Lagrange multiplier method10 . then we can solve for. Since we are given that the perimeter P = 20. y) = c is called the constraint equation. But what if that were not possible (which is often the case)? In this section we will use a general method. The Lagrange multiplier method for solving such problems can now be stated:
10
Named after the French mathematician Joseph Louis Lagrange (1736-1813). then the Second Derivative Test tells us that x = 5 is a local maximum for f . y) = xy = x(10 − x) = 10x − x2 . What would we do if there were constraints on the variables? The following example illustrates a simple case of this type of problem. y) = c (or g(x. y in terms of x using that equation. and hence x = 5 must be the global maximum on the interval [0. this problem can be stated as: Maximize : f (x. z) = c) for some constant c The equation g(x. y. Similar definitions hold for functions of three variables. The perimeter P of the rectangle is then given by the formula P = 2x + 2y. This gives y = 10 − x.
y) for some λ means solving the equations
11 12
See T AYLOR and M ANN. A rigorous proof of the above theorem requires use of the Implicit Function Theorem.7 Constrained Optimization: Lagrange Multipliers
97
Theorem 2. y) that satisfies ∇ f (x.24 it was clear that there had to be a global maximum.
. y) and g(x. § 6. For instance. y) will occur either at a point (x. y) = c (plus any hidden constraints) describes a bounded set B in 2 . which is beyond the scope of this text. there are "hidden" constraints. see T AYLOR and M ANN. which is bounded. in Example 2. y) = c .24. and suppose that c is a scalar constant such that ∇g(x. namely 0 ≤ x. y) given : g(x. It can be shown12 that if the constraint equation g(x. Then to solve the constrained optimization problem Maximize (or minimize) : f (x. In Example 2. y) for some constant λ (the number λ is called the Lagrange multiplier). this problem can be stated as: Maximize : f (x. Whether a point (x.24 the constraint equation 2x + 2y = 20 describes a line in 2 . y) for some λ actually is a constrained maximum or minimum can sometimes be determined by the nature of the problem itself. y) that solve the equation ∇ f (x. find the points (x. y) satisfying ∇ f (x. Solution: As we saw in Example 2.7. y) = λ∇g(x. which by itself is not bounded. together with any implicit constraints. then the constrained maximum or minimum of f (x. y) = λ∇g(x. y) be smooth functions. y) = λ∇g(x. y) = xy given : g(x. y ≤ 10. y) or at a "boundary" point of the set B. y) = λ∇g(x.11 Note that the theorem only gives a necessary condition for a point to be a constrained maximum or minimum.8 for more detail. of the rectangle. y). y) that satisfy the equation g(x. So how can you tell when a point that satisfies the condition in Theorem 2. which cause that line to be restricted to a line segment in 2 (including the endpoints of that line segment). If there is a constrained maximum or minimum. then it must be such a point. due to the nature of the problem.25.2.7 really is a constrained maximum or minimum? The answer is that it depends on the constraint function g(x. with x and y representing the width and height. Let f (x. y) = c. For a rectangle whose perimeter is 20 m. use the Lagrange multiplier method to find the dimensions that will maximize the area. However. Again. respectively. y) 0 for all (x. y) = 2x + 2y = 20 Then solving the equation ∇ f (x. Example 2.
then set those expressions equal (since they both equal λ) to solve for x and y. 2). Example 2. y) = x2 + y2 = 80 Solving ∇ f (x. y) to the point (1. Similarly. 2(y − 2) = 2λy Note that x 0 since otherwise we would get −2 = 0 in the first equation. So we can solve both equations for λ as follows: x−1 y−2 =λ= x y ⇒ xy − y = xy − 2x ⇒ y = 2x
. y) = 2x + 2y = 2x + 2x = 4x ⇒ x=5 ⇒ y=5
There must be a maximum area.98
CHAPTER 2. 2) is d= (x − 1)2 + (y − 2)2 . since the minimum area is 0 and f (5. 5) = 25 > 0. y 0.
and minimizing the distance is equivalent to minimizing the square of the distance.26. Find the points on the circle x2 + y2 = 80 which are closest to and farthest from the point (1. Thus the problem can be stated as: Maximize (and minimize) : f (x. Doing this we get x y =λ= 2 2 ⇒ x=y. Solution: The distance d from any point (x. y) = (x − 1)2 + (y − 2)2 given : g(x. 5) that we found (called a constrained critical point) must be the constrained maximum. namely: ∂x ∂x ∂y ∂y y = 2λ . ∴ The maximum area occurs for a rectangle whose width and height both are 5 m. x = 2λ The general idea is to solve for λ in both equations.
so now substitute either of the expressions for x or y into the constraint equation to solve for x and y: 20 = g(x. FUNCTIONS OF SEVERAL VARIABLES
∂f ∂g ∂f ∂g =λ and = λ . y) = λ∇g(x. so the point (5. y) means solving the following equations: 2(x − 1) = 2λx .
It turns out that λ gives an approximation of the change in the value of the function f (x. y) = x2 + y2 = 80 yields 5x2 = 80.7. −1 √ 2 2
. then we were guaranteed that the constrained critical points we found were indeed the constrained maximum and minimum. y) = c is changed by 1. y. 2). √ 2 2 1 1 √ .7 Constrained Optimization: Lagrange Multipliers Substituting this into g(x. 0. 0. −8) Figure 2.7. Since f (4.2. So the two constrained critical points are (4.
x2 + y2 = 80 y
99
(4. We needed λ only to find the constrained critical points. and since there must be points on the circle closest to and farthest from (1. Notice that since the constraint equation x2 + y2 = 80 describes a circle. y.1). Since f
> f
. but made no use of its value. −8) = 125. z) = λ∇g(x. 0. −8) is the farthest from (1. Maximize (and minimize) : f (x. 8) is the point on the circle closest to (1. √ 2 2
is the
is the constrained minimum point. y. −1 √ 2 2
then
1 1 √ . 0. √ 2 2 −1 √ . y) that we wish to maximize or minimize. when the constant c in the constraint equation g(x.
So far we have not attached any significance to the value of the Lagrange multiplier λ.27. −1 √ 2 2 1 1 √ .1
The Lagrange multiplier method can be extended to functions of three variables. 2) and (−4.
x2 + y2 + z2 = 1 describes a sphere (which is bounded) in constrained maximum point and
−1 √ . then it must be the case that (4. so x = ±4. z) = x2 + y2 + z2 = 1 Solution: Solve the equation ∇ f (x. z) = x2 + y2 + z2 = 1 yields the constrained critical points and
−1 √ . 0. y. 0. z): 1 = 2λx 0 = 2λy 1 = 2λz The first equation implies λ 0 (otherwise we would have 1 = 0). Substituting these expressions into the constraint equation g(x. so we can divide by λ in the second equation to get y = 0 and we can divide by λ in the first and 1 third equations to get x = 2λ = z. Example 2. 8) (1. which is a bounded set in 2 . −8). z) = x + z given : g(x. and since the constraint equation
3.
. 8) and (−4. 8) = 45 and f (−4. 2) (see Figure 2. 2) 0
x
(−4. y.
d] then depends only on the value of x∗. let A(x) be that area (see Figure 3. the area of the region between the curve and the xy-plane) as y varies over the interval [c. Then the trace of the surface in that plane is the curve f (x∗. that is.
d
101
.
z z = f (x. As we will now see. the double integral of a nonnegative real-valued function f (x. where x∗ is fixed and only y varies. slice the surface z = f (x. y)
c a x b x R 0 A(x)
d
y
Figure 3. For any number x∗ in the interval [a. differentiation and integration are thought of as inverse operations. We will often write this as R = [a. y). b]. c ≤ y ≤ d} in 2 . to integrate a function f (x) it is necessary to find the antiderivative of f .1 Double Integrals
In single-variable calculus. y). Recall also that the definite integral of a nonnegative function f (x) ≥ 0 represented the area "under" the curve y = f (x). y) with the plane x = x∗ parallel to the yz-plane. Let f (x. This makes sense since for a fixed x the function f (x. So using the variable x instead of x∗. Is there a similar way of defining integration of real-valued functions of two or more variables? The answer is yes.1). d]. d].1. and only y varies.1. y) dy since we are treating x as fixed. For instance. y) ≥ 0 for all (x.3 Multiple Integrals
3. y) ≥ 0 represents the volume "under" the surface z = f (x. y) is a continuous function of y over the interval [c.1 The area A(x) varies with x
Then A(x) = c f (x. b] × [c. y) be a continuous function such that f (x. y) : a ≤ x ≤ b. y) on the rectangle R = {(x. another function F(x) whose derivative is f (x).e. as we will see shortly. The area A under that curve (i. so we know that the area under the curve is the definite integral.
which would then depend only on the variable y. treating the variable x as a constant (this is called integrating with respect to y).1)
We will always refer to this volume as "the volume under the surface".
. Find the volume V under the plane z = 8x + 6y over the rectangle R = [0. Also. we will usually discard the brackets and simply write
d b
V=
c a
f (x. we could just as easily have taken the area of cross-sections under the surface which were parallel to the xz-plane. and the last expression in equation (3.1) is called a double integral. That is what occurs in the "inner" integral between the square brackets in equation (3. the result is then an expression involving only x. Example 3. Also.
(3.3)
where it is understood that the fact that dx is written before dy means that the function f (x. so that the volume V would be
d b
V=
c a
f (x. y) is first integrated with respect to x using the "inner" limits of integration a and b. This process of going through two iterations of integrals is called double integration. This order of integration can be changed if it is more convenient. See Ch.1). and then the resulting function is integrated with respect to y using the "outer" limits of integration c and d. 1] × [0. The final result is then a number (the volume). y) dx dy . 18 in T AYLOR and M ANN. The above expression uses what are called iterated integrals. y) is integrated as a function of y. This is the first iterated integral. That is what occurs in the "outer" integral above (the second iterated integral). Once that integration is performed. y) dx dy .
(3. First the function f (x. b] of that crosssectional area A(x):
b b d
V=
a
A(x) dx =
a c
f (x. MULTIPLE INTEGRALS
The area A(x) is a function of x. so by the "slice" or cross-section method from singlevariable calculus we know that the volume V of the solid under the surface z = f (x.
1
due to Fubini's Theorem. y) with respect to y is the inverse operation of taking the partial derivative of f (x.102
CHAPTER 3. 2]. y) dy dx
(3. Notice that integrating f (x. which can then be integrated with respect to x. y) with respect to y.1. y) but above the xy-plane over the rectangle R is the integral over [a.2)
It turns out that in general1 the order of the iterated integrals does not matter.
Similarly. is given by
f (x. with functions of x as the limits of integration. with the A signifying area. The symbol dA is sometimes called an area element or infinitesimal. if we have a region R in the xy-plane that is bounded on the left by a curve x = h1 (y). and bounded above by a curve y = g2 (x). Suppose that we have a region R in the xy-plane that is bounded on the left by the vertical line x = a.1
Double integral over a nonrectangular region R
Then using the slice method from the previous section. y) dA. y) dA =
R a g1 (x)
f (x. b) (they could intersect at the endpoints x = a and x = b. though).3. This makes sense since the result of the first iterated integral will have to be a function of x alone. y) dy dx
R 0
(b) Horizontal slice:
d c h2 (y) h1 (y)
x
(a) Vertical slice:
f (x. y) over the region R.2 Double Integrals Over a General Region
In the previous section we got an idea of what a double integral over a rectangle represents. bounded below by the horizontal
. y) over more general regions in 2 .1(a). y) is first integrated with respect to y.
y y = g2 (x) d y
R c x 0 a
b a g2 (x) g1 (x)
x = h1 (y) x = h2 (y)
y = g1 (x) b
f (x. the double integral of a real-valued function f (x. bounded on the right by the vertical line x = b (where a < b).4)
This means that we take vertical slices in the region R between the curves y = g1 (x) and y = g2 (x). y) dx dy
Figure 3.2.2 Double Integrals Over a General Region
105
3. y) dy dx
(3. which then allows us to take the second iterated integral with respect to x. as in Figure 3. We will assume that g1 (x) and g2 (x) do not intersect on the open interval (a.2. denoted by
R b g2 (x)
f (x. Note that f (x. We can now define the double integral of a real-valued function f (x. bounded below by a curve y = g1 (x). bounded on the right by a curve x = h2 (y).
5
Double integral over a general region R
A similar definition can be made for a function f (x. namely f (xi∗ . as shown in Figure 3. We then define
R
summation (the limit is taken over all subdivisions of the rectangle [a.
y d z ∆xi y j+1 (xi∗ . y) that is not necessarily always nonnegative: just replace each mention of volume by the negative volume in the description above when f (x. If we take smaller and smaller subrectangles.1. y j∗ ) yj c 0 a
(a)
∆y j
z = f (x. y j∗ ) ∆xi ∆y j .2. d] as the largest diagonal of the subrectangles goes to 0). our f (x. b] × [c.2. y) < 0. the region R does not have to be bounded. MULTIPLE INTEGRALS
∆y j = y j+1 − y j .
j i
(3.
. y j∗ ) ∆xi ∆y j
Figure 3. using the definition of the Riemann integral from single-variable calculus. with volume f (xi∗ . We can evaluate improper double integrals (i. and so the above sum approaches the actual volume under the surface f (x. y) f (xi∗ .6)
where the summation occurs over the indices of the subrectangles inside R.5(b). y) is not defined) as a sequence of iterated improper single-variable integrals. definition of
R
Finally.108
CHAPTER 3. y) dA as the limit of that double z = f (x. and f (xi∗ . so that the length of the largest diagonal of the subrectangles goes to 0.2. over an unbounded region.e. y) over the region R. In the case of a region of the type shown in Figure 3. then the subrectangles begin to fill more and more of the region R. Then the total volume under the surface is approximately the sum of the volumes of all such parallelepipeds. y j∗ ) is the height and ∆xi ∆y j is the base area of a parallelepiped. y j∗ ) R
y
x xi xi+1 b
Subrectangles inside the region R
xi+1 x
(b) Parallelepiped over a subrectangle. y) dA reduces to a sequence of two iterated integrals. y j∗ )
y j y j+1 0 xi (xi∗ . or over a region which contains points where the function f (x.
y. Then define the triple integral of f (x. z) dV. S = {(x. y) over a region R in 2 can be extended to define a triple integral of a real-valued function f (x. by
S
f (x. y. y). z) over S . MULTIPLE INTEGRALS
3. In general. the triple integral is a sequence of three iterated integrals. y. the word "volume" is often used as a general term to signify the same concept for any n-dimensional object (e. namely
z2 y2 y1 x2
f (x. denoted by f (x.8)
where the order of integration does not matter.
(3. z1 ≤ z ≤ z2 }. which is then divided into subparallelepipeds. Then
b h2 (x) h1 (x) g2 (x. Physically. y. We simply proceed as before: the solid S can be enclosed in some rectangular parallelepiped. and x varies between a and b. ∆y and ∆z. pick a point (x∗ . what does the triple integral represent? We saw that a double integral could be thought of as the volume under a two-dimensional surface. y. bounded above by a surface z = g2 (x. that is. It may be hard to get a grasp on the concept of the "volume" of a four-dimensional object. with sides of lengths ∆x. y1 ≤ y ≤ y2 . y. which then leaves you with a
. It can be shown that this limit does not depend on the choice of the rectangular parallelepiped enclosing S .y)
f (x. It turns out that the triple integral simply generalizes this idea: it can be thought of as representing the hypervolume under a three-dimensional hypersurface w = f (x.7)
where the limit is over all divisions of the rectangular parallelepiped enclosing S into subparallelepipeds whose largest diagonal is going to 0. A more complicated case is where S is a solid which is bounded below by a surface z = g1 (x. z) whose graph lies in 4 . The symbol dV is often called the volume element.3 Triple Integrals
Our definition of a double integral of a real-valued function f (x. z) : x1 ≤ x ≤ x2 . length in 1 . y.g. y). z) dV =
S z1 x1
f (x. z∗ ) ∆x ∆y ∆z . y is bounded between two curves h1 (x) and h2 (x). In each subparallelepiped inside S . z) over a solid S in 3 . z∗ ).y)
f (x.110
CHAPTER 3.
(3. y. y∗ . x2 ] × [y1 . y2 ] × [z1 . area in 2 ).
(3. y∗ . z) dV =
S a g1 (x. z2 ]. z) dx dy dz . This is the simplest case. but at least we now know how to calculate that volume! In the case where S is a rectangular parallelepiped [x1 . y. y. and the triple summation is over all the subparallelepipeds inside S .9)
Notice in this case that the first iterated integral will result in a function of x and y (since its limits of integration are functions of x and y). z) dV = lim
S
f (x∗ . z) dz dy dx .
Since the function being integrated is the constant 1, then the above triple integral reduces to a double integral of the types that we considered in the previous section if the solid is bounded above by some surface z = f (x, y) and bounded below by the xy-plane z = 0. There are many other possibilities. For example, the solid could be bounded below and above by surfaces z = g1 (x, y) and z = g2 (x, y), respectively, with y bounded between two curves h1 (x) and h2 (x), and x varies between a and b. Then
b h2 (x) h1 (x) g2 (x,y) b h2 (x) h1 (x)
the order of integration in the triple integral changes the limits of integration.)
3.4 Numerical Approximation of Multiple Integrals
113
3.4 Numerical Approximation of Multiple Integrals
As you have seen, calculating multiple integrals is tricky even for simple functions and regions. For complicated functions, it may not be possible to evaluate one of the iterated integrals in a simple closed form. Luckily there are numerical methods for approximating the value of a multiple integral. The method we will discuss is called the Monte Carlo method. The idea behind it is based on the concept of the average value of a function, which you learned in single-variable calculus. Recall that for a continuous function f (x), the average value f¯ of f over an interval [a, b] is defined as f¯ = 1 b−a
b
f (x) dx .
a
(3.11)
The quantity b − a is the length of the interval [a, b], which can be thought of as the "volume" of the interval. Applying the same reasoning to functions of two or three variables, we define the average value of f (x, y) over a region R to be f¯ = 1 A(R)
R
f (x, y) dA ,
(3.12)
where A(R) is the area of the region R, and we define the average value of f (x, y, z) over a solid S to be 1 f (x, y, z) dV , (3.13) f¯ = V(S )
S
where V(S ) is the volume of the solid S . Thus, for example, we have f (x, y) dA = A(R) f¯ .
R
(3.14)
The average value of f (x, y) over R can be thought of as representing the sum of all the values of f divided by the number of points in R. Unfortunately there are an infinite number (in fact, uncountably many) points in any region, i.e. they can not be listed in a discrete sequence. But what if we took a very large number N of random points in the region R (which can be generated by a computer) and then took the average of the values of f for those points, and used that average as the value of f¯? This is exactly what the Monte Carlo method does. So in formula (3.14) the approximation we get is f (x, y) dA ≈ A(R) f¯ ± A(R)
R
namely √ x = g(u) = u + 1 . 3]) and hence has an inverse function (defined on the interval [0. which will give some motivation for how substitution works in multiple integrals. 3] we can define x as a function of u. 2] onto [0. That is.
then you would make the substitution u = x2 − 1 ⇒ x2 = u + 1
du = 2x dx which changes the limits of integration
x=2⇒u=3 so that we get
2 1 2
x=1⇒u=0
x3 x2 − 1 dx = =
1 3 0
1 2 2x 1 2 (u 3
· 2x x2 − 1 dx √ + 1) u du
= =
1 2
u3/2 + u1/2 du . The answer is yes. On the interval of integration [1. √ Then substituting that expression for x into the function f (x) = x3 x2 − 1 gives √ f (x) = f (g(u)) = (u + 1)3/2 u . on [0.5 Change of Variables in Multiple Integrals
Given the difficulty of evaluating multiple integrals. for example. the reader may be wondering if it is possible to simplify those integrals using a suitable substitution for the variables. though it is a bit more complicated than the substitution method which you learned in single-variable calculus. 2]. which can be easily integrated to give .
.
0 √ 14 3 5
Let us take a different look at what happened when we did that substitution.5 Change of Variables in Multiple Integrals
117
3. First. the function x → x2 − 1 is strictly increasing (and maps [1. Recall that if you are given.3. we let u = x2 − 1. 3]). the definite integral
2 1
x3 x2 − 1 dx .
MULTIPLE INTEGRALS
dx = g ′ (u) ⇒ dx = g ′ (u) du du dx = 1 (u + 1)−1/2 du . respectively. differentiable function from an interval [c. b] (on the x-axis). The proof of the following theorem is beyond the scope of the text.32 and § 15. which means f (g(u)) g ′ (u) du . then c = g−1 (a) and d = g−1 (b).2
2
See T AYLOR and M ANN. d).
g−1 (2)
f (x) dx =
1
g−1 (1)
In general. d] (which you can think of as being on the "u-axis") onto an interval [a. We will assume that all the functions involved are continuously differentiable and that the regions and solids involved all have "reasonable" boundaries. which can be written as
=
0 2
√ 1 (u + 1)3/2 u · 2 (u + 1)−1/2 du . which means that g ′ (u) 0 on the interval (c. 2 so since g(0) = 1 ⇒ 0 = g−1 (1) g(3) = 2 ⇒ 3 = g−1 (2) then performing the substitution as we did earlier gives
2 2
f (x) dx =
1 1 3
x3 x2 − 1 dx
1 2 (u 3
=
0
√ + 1) u du .62 for all the details. and it is what you were implicitly using when doing integration by substitution.118 and we see that
CHAPTER 3. so that a = g(c) and b = g(d). if x = g(u) is a one-to-one.
(3.17)
g−1 (a)
This is called the change of variable formula for integrals of single-variable functions. § 15.
. This formula turns out to be a special case of a more general formula which can be used to evaluate multiple integrals. We will state the formulas for double and triple integrals involving real-valued functions of two and three variables. and
b g−1 (b)
f (x) dx =
a
f (g(u)) g ′ (u) du .
. To see this. MULTIPLE INTEGRALS
3.1). y) : a ≤ x ≤ b. 0 ≤ y ≤ f (x)} in 2 that represents a thin. where f (x) is a continuous function on [a.
M=
R
δ(x. Example 3. 2
b
b
My =
a
x f (x) dx .e the mass of R is uniformly distributed over the region.29). y) of points inside R (where R can be any region in 2 ) the coordinates ( x. y) of the coordinates (x.13. 0 ≤ y ≤ 2x2 }. M
(3.
(3.
Mx =
R
yδ(x.27)
assuming that R has uniform density.6. y) = 1 throughout R in the formulas in (3. the center of mass of R has coordinates ( x. for some point (x∗ . In the general case where the density of a region (or lamina) R is a continuous function δ = δ(x. y) is δ(x. i.1
Mx =
a
( f (x))2 dx .
R
Note that the formulas in (3. y∗ ) in that rectangle. respectively. M
Figure 3. y) dA.28)
xδ(x. which is the double integral δ(x. y) given by ¯ ¯ x= ¯ where
b
y y = f (x) R ( x . Find the center of mass of the region R = {(x.27) represent a special case when δ(x. The mass of that rectangle is approximately δ(x∗ . In this case the area M of the region is considered the mass of R (the density is constant. The quantity M is the mass of the region R. y) dA . y) ¯ ¯ 0 a b Center of mass of R x
My M
and
y= ¯
Mx .29)
The quantities M x and My are called the moments (or first moments) of the region R about the x-axis and y-axis. y∗ )∆x ∆y. y) dA . and taken as 1 for simplicity). y) = x + y.6.6 Application: Center of Mass
Recall from single-variable calculus that for a region R = {(x.
M=
a
f (x) dx .124
CHAPTER 3. y) dA . b]. think of taking a small rectangle inside R with dimensions ∆x and ∆y close to 0. Then the mass of R is the limit of the sums of the masses of all such rectangles inside R as the diagonals of the rectangles approach 0. y) of the center of mass of R are given by ¯ ¯ x= ¯ where My =
R
My M
and
y= ¯
Mx . flat plate (see Figure 3.
(3. if the density function at (x. y) : 0 ≤ x ≤ 1.
For a continuous random variable. MULTIPLE INTEGRALS
3. 2. 9-10 in K AMKE . P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) in the die example). Then the probability of rolling a 3.
. 1). 3. 2. and for x in (0. the event X ≤ 3 is the set {1. since of the six numbers 6 2 on the die. 1) the interval (0. Note that the set of all real numbers between 0 and 1 is not a discrete (or countable) set of values. Theory of Sets. 4. 1).. 1). 1).128
CHAPTER 3. 5. it makes no sense to consider P(X = x) since it must be 0 (why?). the probability of an event will instead be the integral of a function. Now let X be a variable representing a random real number in the interval (0. we saw how the probability of an event was the sum of the probabilities of the individual outcomes comprising that event (e.g. length of (0. For example. there are three equally likely numbers (1. E. Note that P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3). which we will now describe. i. 6}. which is given by P(X ≤ x) = x. written as P(X ≤ 3). 1) has length 1.
Probability
Suppose that you have a standard six-sided (fair) die. since there are six sides on the die and each one is equally likely to be rolled. So since X represents a random number in (0.7 Application: Probability and Expected Value
In this section we will briefly discuss some applications of multiple integrals in the field of probability theory. in the case of the die. and you let a variable X represent the value rolled. 1) 1
P(X ≤ x) =
In the case of a discrete random variable. then length of (0. x) x = =x. and 3) that are less than or equal to 3. Let X be a continuous real-valued random variable on a sample space Ω in
3
We call X a continuous random variable on the sample space Ω = (0. 1 is 6 .
. in our case the event X ≤ x is the set (0. The reasoning is this: the interval (0. Instead. For
For a proof see p. and hence is uniformly distributed over (0. 1). x). For example. An event A is a subset of the sample space. Likewise the probability of rolling at most a 3. In particular we will see ways in which multiple integrals can be used to calculate probabilities and expected values. x) has length x. An event A is a subset of the sample space.3 In this case. written as P(X = 3). In our case. we consider the probability P(X ≤ x). 2. We call X a discrete random variable on the sample space (or probability space) Ω consisting of all possible outcomes. it can not be put into a one-to-one correspondence with the set of positive integers. for any real number x in (0. Ω = {1. is 3 = 1 . New York: Dover. 1950. 3}. and hence in particular the 3 has a one out of six chance of being rolled.e.
In addition to individual random variables, we can consider jointly distributed random variables. For this, we will let X, Y and Z be three real-valued continuous random variables defined on the same sample space Ω in (the discussion for two random variables is similar). Then the joint distribution function F of X, Y and Z is given by F(x, y, z) = P(X ≤ x, Y ≤ y, Z ≤ z) ,
z y −∞ ∞ −∞ x
with the ≤ and < symbols interchangeable in any combination. A triple integral, then, can be thought of as representing a probability (for a function f which is a p.d.f.). Example 3.17. Let a, b, and c be real numbers selected randomly from the interval (0, 1). What is the probability that the equation ax2 + bx + c = 0 has at least one real solution x? Solution: We know by the quadratic formula that there is at least one real solution if b2 − 4ac ≥ 0. So we need to calculate P(b2 − 4ac ≥ 0). We will use three jointly distributed random variables to do this. First, since 0 < a, b, c < 1, we have √ √ b2 − 4ac ≥ 0 ⇔ 0 < 4ac ≤ b2 < 1 ⇔ 0 < 2 a c ≤ b < 1 , where the last relation holds for all 0 < a, c < 1 such that 0 < 4ac < 1 ⇔ 0 < c < 1 . 4a
c 1 c= R1 0
1 4 1 4a
R2 1
a
Figure 3.7.1 Region R = R1 ∪ R2
Considering a, b and c as real variables, the region R in the ac-plane where the above 1 relation holds is given by R = {(a, c) : 0 < a < 1, 0 < c < 1, 0 < c < 4a }, which we can see is a union of two regions R1 and R2 , as in Figure 3.7.1 above. Now let X, Y and Z be continuous random variables, each representing a randomly selected real number from the interval (0, 1) (think of X, Y and Z representing a, b and c, respectively). Then, similar to how we showed that f (x) = 1 is the p.d.f. of the
So, for example, if you were to repeatedly take samples of n = 3 random real numbers from (0, 1), and each time store the minimum and maximum values in the sample, 1 then the average of the minimums would approach 4 and the average of the max3 imums would approach 4 as the number of samples grows. It would be relatively simple (see Exercise 4) to write a computer program to test this.
Find Var(X) and Var(Y) for X and Y as in Example 3.18. 7. Continuing Exercise 6, the correlation ρ between X and Y is defined as ρ=
∞ ∞
E(XY) − (EX)(EY) Var(X) Var(Y)
,
where E(XY) = −∞ −∞ xy f (x, y) dx dy. Find ρ for X and Y as in Example 3.18. (Note: The quantity E(XY) − (EX)(EY) is called the covariance of X and Y.) 8. In Example 3.17 would the answer change if the interval (0, 100) is used instead of (0, 1)? Explain.
You may also recall that if f (x) represented the force applied along the x-axis to an object at position x in [a.1 Line Integrals
In single-variable calculus you learned how to integrate a real-valued function f (x) over an interval [a. We will begin with real-valued functions of two variables.1 below). This definition will be motivated by the physical notion of work. b]
We will assume for now that the function f (x. then the work W done in moving that object from position x = a to x = b was defined as the integral:
b
W=
a
f (x) dx
In this section. for some integer n ≥ 2
135
. In physics. y = y(t) for t in [a.e. with a force f (x. so we only consider the magnitude of the force.4 Line and Surface Integrals
4. since an interval (or collection of intervals) is really the only kind of "path" in 1 . y) is continuous and real-valued.1 Curve C : x = x(t).1. a curve) in 2 . Suppose that we want to find the total amount W of work done in moving an object along a curve C in 2 with a smooth parametrization x = x(t). b] as follows: a = t0 < t1 < t2 < · · · < tn−1 < tn = b . b]. b] in 1 . a ≤ t ≤ b. the intuitive idea of work is that Work = Force × Distance .1. y) which varies with the position (x.
y
C t=a
t = ti ∆yi ∆si ≈ ∆xi
∆xi 2 + ∆yi 2
t = ti+1 t=b
x 0 Figure 4. we will see how to define the integral of a function (either realvalued or vector-valued) of two variables over a general path (i. y = y(t). This integral (usually called a Riemann integral) can be thought of as an integral over a path in 1 . Partition the interval [a. y) of the object and is applied in the direction of motion along C (see Figure 4.
y) along the curve C. yi∗ ) = (x(ti ∗). by the Pythagorean Theorem.1. if the subinterval is small enough then the work done in moving the object along that piece of the curve is approximately Force × Distance ≈ f (xi∗ . y(t)).136
CHAPTER 4.1. the line integral of f (x. y) and a curve C in 2 . yi∗ ) ∆xi 2 + ∆yi 2 . LINE AND SURFACE INTEGRALS
As we can see from Figure 4. yi∗ ) becomes f (x(t). for any real-valued function f (x. y) ds =
C a
f (x(t).4)
The integral on the right side of the above equation gives us our idea of how to define. (4. ti+1 ]. Thus.1. y(t))
x ′ (t)2 + y ′ (t)2 dt . yi∗ )
i=0
∆xi 2 + ∆yi 2
(4.
W≈
f (xi∗ . respectively. so that
b
W=
a
f (x(t). But since ∆xi 2 + ∆yi 2 = where ∆ti = ti+1 − ti . the sum over all subintervals becomes the integral from t = a to t = b. over a typical subinterval [ti .
(4. y) along C with respect to arc length s is
b
f (x.6)
. y(t))
x ′ (t)2 + y ′ (t)2 dt . called a line integral: Definition 4.3)
Taking the limit of that sum as the length of the largest subinterval goes to 0.5)
The symbol ds is the differential of the arc length function
t
s = s(t) =
a
x ′ (u)2 + y ′ (u)2 du . ∆xii and ∆yii become x ′ (t) ∆t ∆t and y ′ (t).
(4. a ≤ t ≤ b. the integral of f (x. then
n−1
∆xi ∆ti
2
+
∆yi ∆ti
2
∆ti . y = y(t). parametrized by x = x(t). For a real-valued function f (x. and f (xi∗ . ti+1 ] the distance ∆si traveled along the curve is approximately ∆xi 2 + ∆yi 2 .1)
where (xi∗ .
(4.2)
is approximately the total amount of work done over the entire curve.
(4. y). y(ti ∗)) for some ti ∗ in [ti . and so
n−1
W≈
f (xi∗ . yi∗ )
i=0
∆xi ∆ti
2
+
∆yi ∆ti
2
∆ti .
(4. y) along C with respect to y. y) on 2 . a ≤ t ≤ b. y) defined on 2 by f(x. For a curve C with a smooth parametrization x = x(t). Also.138
CHAPTER 4. (4.e. y) and Q(x. y) ds unchanged. y = y(t) = r sin(2π − t) . a circle of radius r). let r(t) = x(t) i + y(t) j
.
a≤t≤b. LINE AND SURFACE INTEGRALS
Note in Example 4. y) dx =
C a
f (x(t).
(4. y). i. It is defined at points in 2 . we know that force is actually a vector.10)
Notice that our definition of the line integral was with respect to the arc length parameter s. we used the idea of work as force multiplied by distance. i. 0 ≤ t ≤ 2π . and we have f (x. using the parametrization x = x(t) = r cos(2π − t) . y) = P(x. In the derivation of the formula for a line integral. Such a function f is called a vector field on 2 . y) dy =
C a
f (x(t). In general. So it would be helpful to develop a vector form for a line integral.1 that if we had traversed the circle C twice. However.9)
f (x.12)
as the line integral of f (x. then we would have gotten an area of 4πrh. suppose that we have a function f(x.8)
then it is easy to verify (see Exercise 12) that the value of the line integral is unchanged. for any f (x. y) along C with respect to x. y) j for some continuous real-valued functions P(x. If we had gone in the clockwise direction.11)
as the line integral of f (x. If a curve C has a parametrization x = x(t). For this. let t vary from 0 to 4π. Then −C is parametrized by x = x(a + b − t) . y) i + Q(x. it can be shown (see Exercise 15) that reversing the direction in which a curve C is traversed leaves C f (x. y(t)) x ′ (t) dt
(4. a ≤ t ≤ b. even though the curve itself is still the same (namely. twice the desired area. notice that we traversed the circle in the counter-clockwise direction. y) ds . y = y(t). y) ds =
C −C
y = y(a + b − t) . We can also define
b
f (x. then denote by −C the same curve as C but traversed in the opposite direction. and
b
f (x. and its values are vectors in 2 .e. y(t)) y ′ (t) dt
(4. y = y(t).
y(t)) x ′ (t) + Q(x(t). y) dx + Q(x.14)
f(x(t). the line integral of f along C is
C
f · dr = =
P(x. y) j and a curve C with a smooth parametrization x = x(t).1 Line Integrals be the position vector for a point (x(t). y). Then r ′ (t) = x ′ (t) i + y ′ (t) j and so
b b
139
P(x. y) dy =
C
P(x. This leads us to the following definition: Definition 4. For a realvalued function F(x. y) dy is known as a differential form. y) dy is called exact if it equals dF for some function F(x. y) dy =
a b
P(x(t). The quantity P(x. y) i + Q(x. Putting Definitions 4. y) dx + Q(x. so the last integral on the right looks somewhat similar to our earlier definition of a line integral. y(t)) · r ′ (t) dt
=
a
by definition of f(x. Recall that if the points on a curve C have position vector r(t) = x(t) i+y(t) j. b]. y(t)) on C. a ≤ t ≤ b. We use the notation dr = r ′ (t) dt = dx i + dy j to denote the differential of the vectorvalued function r. then r ′ (t) is a tangent vector to C at the point (x(t). y(t)) · r ′ (t) is a real-valued function on [a. For convenience we will often write P(x. y) dx +
C b a C
Q(x. The line integral in Definition 4. y = y(t). y) dy .2. the differential of F is dF = ∂F dx + ∂F dy. Since C is a smooth curve. y). y) dx +
C C
Q(x. For a vector field f(x.2 together we get the following theorem:
. y(t)) y ′ (t) dt
=
a b
(P(x(t).4.1 and 4. b] and hence T(t) = r ′ (t) r ′ (t)
is the unit tangent vector to C at (x(t).1 which is called a line integral of a scalar field.
where it is understood that the line integral along C is being applied to both P and Q. y) dy
(4. y). y(t)) in the direction of increasing t (which we call the direction of C). y(t)) x ′ (t) dt +
a
Q(x(t). then r ′ (t) 0 on [a. y) = P(x. y(t)). y) dx + Q(x. A differential form ∂x ∂y P(x. y) dx +
C C
Q(x.
where r(t) = x(t) i + y(t) j is the position vector for points on C. y(t)) · r ′ (t) dt . Notice that the function f(x(t).2 is often called a line integral of a vector field to distinguish it from the line integral in Definition 4. y(t)) y ′ (t)) dt f(x(t).13) (4.
y) dy has the same value for both parametrizations.144
CHAPTER 4. y = y(t). and du = α ′1 . it turns out that the value of a line integral of a vector field is unchanged as long as the direction of the curve C is preserved by whatever parametrization is chosen: Theorem 4. such that a = α(c). dt = α ′ (u) du. b] such that c = α−1 (a). y) j be a vector field. and let C be a smooth curve parametrized by x = x(t). The preceding discussion shows the importance of always taking the direction of the curve into account when using line integrals of vector fields. y) i + Q(x. Let f(x. Then C f· dr has the same value for the parametrizations x = x(t). a ≤ t ≤ b. y = y(t). direction is accounted for. b]. QED and hence C f · dr has the same value. a ≤ t ≤ b and x = x(u) = x(α(u)). c ≤ u ≤ d. the curves in line integrals are sometimes referred to as directed curves or oriented curves. d = α−1 (b). y(α(u)))
α−1 (a) d
x ′ (u) ′ ˜ (α (u) du) α ′ (u)
=
c
P( x(u). and by the Chain Rule dt (u) x ′ (u) = ˜ d dx dt dx ˜ = (x(α(u))) = = x ′ (t) α ′ (u) du du dt du ⇒ x ′ (t) = x ′ (u) ˜ α ′ (u)
so making the susbstitution t = α(u) gives
b
P(x(t). Suppose that t = α(u) for c ≤ u ≤ d. d] onto [a. Recall that our definition of a line integral required that we have a parametrization x = x(t). this would mean that our definition is not well-defined. say. y = y(t). and then back to the initial point moving backwards along the same path. Also. y) (treated as a vector) moving an object along a curve C: the total work performed moving the object along C from its initial point to its terminal point.2. y = y(u) = y(α(u)). ˜ ˜ ˜
which shows that C P(x. and α ′ (u) > 0 on the open interval (c. α(u) is strictly · increasing on [c. y) = P(x. b = α(d). c ≤ u ≤ d ? If so. y = y(u). x = x(u). any curve has infinitely many parametrizations.2 means that the two parametrizations move along C in the same direction. y(u)) x ′ (u) du . ˜ ˜ Proof: Since α(u) is strictly increasing and maps [c. For this reason. then we know that t = α(u) has an inverse function u = α−1 (t) defined on [a. This is because when force is considered as a vector. A similar argument shows that C Q(x. Notice that the condition α ′ (u) > 0 in Theorem 4. is zero. y(t)) x (t) dt =
a
′
α−1 (b)
P(x(α(u)). d) (i. LINE AND SURFACE INTEGRALS
The above formula can be interpreted in terms of the work done by a force f(x. But as we know. So could we get a different value for a line integral using ˜ ˜ some other parametrization of C. Luckily. a ≤ t ≤ b for the curve C.e. d]).
. y) dx has the same value for both parametrizations. That was not the case with the "reverse" parametrization for −C: for u = a + b − t we have t = α(u) = a + b − u ⇒ α ′ (u) = −1 < 0.
This proves path independence. In a region R. and so
f · dr =
C2
f · dr. LINE AND SURFACE INTEGRALS
So far. the line integral has been independent of the path joining the two points.g. Example 4.
Conversely. Let C be a closed curve contained in R. Then C = C1 ∪ −C2 is a closed curve in R (from P1 to P1 ). and let C2 be the remaining part of C that goes from P1 to P2 . the examples we have seen of line integrals (e.2
=
C1 C1
f · dr .2.
Clearly. Let C1 be a curve in R going from P1 to P2 . 0=
C
C1
◮
P1 P2
f · dr f · dr + f · dr −
−C2 C2
=
C1
f · dr
◮
C2 Figure 4. and let C2 be another curve in R going from P1 to P2 . again as in Figure 4. this is not always the case.2. Let P1 and P2 be two distinct points in R. As we mentioned before. Then by path independence we have f · dr = f · dr = 0 f · dr = 0 . The following theorem gives a necessary and sufficient condition for this path independence: Theorem 4.3. Proof: Suppose that C f · dr = 0 for every closed curve C which is contained in R. the above theorem does not give a practical way to determine path inde-
. That is. as in Figure 4.2.146
CHAPTER 4. Let P1 and P2 be two distinct points on C.2) have had the same value for different curves joining the initial point to the terminal point. the line integral C f · dr is independent of the path between any two points in R if and only if C f · dr = 0 for every closed curve C which is contained in R.2. suppose that the line integral C f · dr is independent of the path between any two points in R.2. Thus. so f · dr = 0
QED
C1
C2
f · dr
C1
f · dr −
C2
C1
f · dr +
−C2 C
since C = C1 ∪ −C2 . Let C1 be a part of the curve C that goes from P1 to P2 . and so C f · dr = 0.
4)
=
a
= F(x(t). Let f(x. For a more practical method for determining path independence.20)
where A = (x(a). since it is impossible to check the line integrals around all possible closed curves in a region. a ≤ t ≤ b. y = y(t). Suppose that there is a real-valued function F(x. since it depends only on the values of F at those endpoints. we first need a version of the Chain Rule for multivariable functions: Theorem 4. y) i + Q(x. Let C be a smooth curve in R parametrized by x = x(t). with P and Q continuously differentiable functions on R.
. (Chain Rule) If z = f (x.4 (which uses the Mean Value Theorem). y) such that ∇F = f on R.1 We will now use this Chain Rule to prove the following sufficient condition for path independence of line integrals: Theorem 4. The proof is virtually identical to the proof of Theorem 2. y) j be a vector field in some region R. y) = P(x. y(t)) y ′ (t) dt
a b a b
∂F ∂F ∂F dx ∂F dy dt (since ∇F = f ⇒ + = P and = Q) ∂x dt ∂y dt ∂x ∂y F ′ (x(t). y(t)) dt (by the Chain Rule in Theorem 4. and dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt at all points where the derivatives on the right are defined.
1
See T AYLOR and M ANN. y(a)) and B = (x(b). y(t)) x ′ (t) + Q(x(t). Proof: By definition of
b C C
f · dr. y) is a continuously differentiable function of x and y.5. What it mostly does is give an idea of the way in which line integrals behave. and how seemingly unrelated line integrals can be related (in this case. y(t))
b a
= F(B) − F(A)
QED
by the Fundamental Theorem of Calculus. Then
C
(4.2 Properties of Line Integrals
147
pendence.5.2 from Section 2.4.
(4.4. a specific line integral between two points and all line integrals around closed curves). y(b)) are the endpoints of C. so we omit it. we have
f · dr = =
P(x(t).19)
f · dr = F(B) − F(A) . the line integral is independent of the path between its endpoints. Thus. and both x = x(t) and y = y(t) are differentiable functions of t. § 6. then z is a differentiable function of t.
namely F(x. Note that we can also verify that the value of the line integral of f along any curve C going from (0. Example 4. then
C
f · dr = 0 for
any closed curve C in R (i. Solution: We need to find a real-valued function F(x. y) = x2 + g(y) .
C
∇F · dr = 0 for any real-valued function F(x. y) = f(x.6. y): 1 ∂F = x ⇒ F(x.e. so that the endpoints A and B are the same point. y = 3 sin t. y) such that ∇F(x. f · dr = F(1. Use Theorem 4. 2).e. 2) − F(0. ∂y
Suppose that ∂F = x2 + y2 .5 3 1 13 1 . we pick K = 0. y) such that ∂F = x2 + y2 ∂x and ∂F = 2xy . Since any choice for K will do (why?).5 in the special case where C is a closed curve. 0) = (1)3 + (1)(2)2 − (0 + 0) = + 4 = 3 3 3 C A consequence of Theorem 4.
Solution: The vector field f(x. 3
Hence the line integral C (x2 + y2 ) dx + 2xy dy is path independent. 0 ≤ t ≤ 2π. a potential F(x.5 to show that this line integral is indeed path independent. 2) will always be 13 .148
CHAPTER 4. Then we must have F(x. Thus. since by Theorem 4. Evaluate
C
x dx + y dy for C : x = 2 cos t.
Example 4. A conservative vector field is one which has a potential. y)).2 and 4. y) = 1 3 x + xy2 . 0) to the point (1.3 in Section 4. y) = (x2 + y2 ) i + 2xy j exists. 0) to (1.6. LINE AND SURFACE INTEGRALS
Theorem 4. y) is called a potential for f. where ∂y ∂y K is a constant. i. y) = x i + y j has a potential F(x. Recall from Examples 4.5. y) for f(x.1 that the line integral (x2 + y2 ) dx + 2xy dy was found to have the value 13 for three different curves C going 3 C from the point (0. y) = 1 x3 + xy2 + g(y) for some function ∂x 3 g(y). is the following important corollary: Corollary 4. so ∂x 2 ∂F 1 = y ⇒ g ′ (y) = y ⇒ g(y) = y2 + K ∂y 2
. A real-valued function F(x. So ∂F = 2xy + g ′ (y) satisfies the condition ∂F = 2xy if g ′ (y) = 0. If a vector field f has a potential in a region R.5 can be thought of as the line integral version of the Fundamental Theorem of Calculus. g(y) = K.
and (4.3 Green's Theorem
We will now see a way of evaluating the line integral of a smooth vector field around a simple closed curve.23) and
. y) and Q(x. y) = P(x.150
CHAPTER 4. See Figure 4.24) (4. where the boundary curve C can be written as C = C1 ∪ C2 in two distinct ways: C1 = the curve y = y1 (x) from the point X1 to the point X2 C2 = the curve y = y2 (x) from the point X2 to the point X1 . Let f(x.23)
where X1 and X2 are the points on C farthest to the left and right. respectively. y) are smooth. ∂x ∂y
(4. that is. y) j is smooth if its component functions P(x. Then f · dr =
R
C
∂Q ∂P − dA .25)
where Y1 and Y2 are the lowest and highest points. y) i + Q(x. C1 = the curve x = x1 (y) from the point Y2 to the point Y1 C2 = the curve x = x2 (y) from the point Y1 to the point Y2 .7.1 b
Integrate P(x. (Green's Theorem) Let R be a region in 2 whose boundary is a simple closed curve C which is piecewise smooth. on C. y) = P(x.1.
y y = y2 (x)
d
◭
x = x1 (y) X1 Y1
Y2 X2 x = x2 (y) R
◮ C
x
c
y = y1 (x) a Figure 4. y) i+ Q(x. A vector field f(x. y) around C using the representation C = C1 ∪ C2 given by (4. y) j be a smooth vector field defined on both R and C.3. respectively. We will use Green's Theorem (sometimes called Green's Theorem in the plane) to relate the line integral around a closed curve with a double integral over the region inside the curve: Theorem 4. LINE AND SURFACE INTEGRALS
4.3. (4.22) (4. Proof: We will prove the theorem in the case for a simple region R.21)
where C is traversed so that R is always on the left side of C.
which are divided by the slits indicated by the dashed lines.
C1 R1 C2 R1 C3 C2
◮
◭C
◮ ◮ ◭
R2
1
◭
◮ ◭
R2
◮
(a) Region R with one hole
(b) Region R with two holes
Figure 4. then it can be shown (see Exercise 8) that
C
153
y 1 C1
◭
R C2 0 1/2 1 x
1/2
f · dr = 0 . and take the "boundary" C of R to be C = C1 ∪ C2 . that is.3.3. in Figure 4. − ∂x ∂y
which shows that Green's Theorem holds for the annular region R.
∂Q ∂x
We would still have
R
−
∂P ∂y
dA = 0. as opposed to discrete points being cut out.4 above. the "outer" boundary and the "inner" boundaries are traversed so that R is always on the left side. The idea is to cut "slits" between the boundaries of a multiply connected region R so that R is divided into subregions which do not have any "holes".3.3 Green's Theorem If we modify the region R to be the annulus R = { (x.4
Multiply connected regions
The intuitive idea for why Green's Theorem holds for multiply connected regions is shown in Figure 4.8. regions like the annulus in Example 4.3. It turns out that Green's Theorem can be extended to multiply connected regions. where C1 is the unit circle x2 + y2 = 1 traversed counterclockwise and C2 is the circle x2 + y2 = 1/4 traversed clockwise.4.4(a) the region R is the union of the regions R1 and R2 .3 The annulus R
R we would have f · dr =
R
C
∂Q ∂P dA . and we traverse then in the manner indicated by
◭
◮
. so for this
Figure 4. For such regions. which have one or more regions cut out from the interior.3). Those slits are part of the boundary of both R1 and R2 . y) : 1/4 ≤ x2 + y2 ≤ 1 } (see Figure 4.3. For example.
then Green's Theorem holds in each subregion. And if the potential F(x. a region with no holes).154
CHAPTER 4. we have
C1 ∪C2
f · dr =
bdy of R1
f · dr +
bdy of R2
f · dr . − ∂x ∂y
R
and so
C1 ∪C2
f · dr =
R1
∂Q ∂P dA + − ∂x ∂y
R2
∂Q ∂P dA = − ∂x ∂y
which shows that Green's Theorem holds in the region R. and so we know that ∂y ∂2 F ∂2 F ∂P ∂Q = ⇒ = in R. y) = P(x.e. ∂y ∂x ∂x ∂y ∂y ∂x Conversely. y) is smooth in R. Since R1 and R2 do not have holes in them. y) i+Q(x.3. the differential form P dx + Q dy is exact)
(c)
C
(d)
∂P ∂Q = in R ∂y ∂x
. then ∂F = P ∂x and ∂F = Q. the following can be shown: The following statements are equivalent for a simply connected region R in (a) f(x. simple closed curve C) has a potential in R. We know from Corollary 4. − ∂x ∂y
But since the line integrals along the slits cancel out. Notice that along each slit the boundary of R1 is traversed in the opposite direction as that of R2 . so that
bdy of R1
f · dr =
R1
∂Q ∂P dA − ∂x ∂y
and
bdy of R2
f · dr =
R2
∂Q ∂P dA . y) j on a region R (whose boundary is a piecewise smooth. which means that the line integrals of f along those slits cancel each other out. y) in R (b)
C 2:
f · dr is independent of the path for any curve C in R f · dr = 0 for every simple closed curve C in R (in this case.6 that when a smooth vector field f(x. y) j has a smooth potential F(x. then C f · dr = 0. y) i + Q(x. ∂Q ∂P dA . LINE AND SURFACE INTEGRALS
the arrows. if
∂P ∂y
=
∂Q ∂x
in R then f · dr =
R
C
∂Q ∂P dA = − ∂x ∂y
R
0 dA = 0 . A similar argument shows that the theorem holds in the region with two holes shown in Figure 4. y) = P(x.
For a simply connected region R (i.4(b).
4. v) y
Parametrization of a surface Σ in
In this case.1
Parametrization of a curve C in
3
Similar to how we used a parametrization of a curve to define the line integral along the curve.1). v) 0 u Figure 4. The idea behind a parametrization of a curve is that it "transforms" a subset of (normally an interval [a. v) in some region R in 2 (see Figure 4.
. v)j + z(u. z = z(u.4 Surface Integrals and the Divergence Theorem
In Section 4.4. z(b)) y
a
t
b
Figure 4. b]) into a curve in 2 or 3 (see Figure 4. such as a sphere or a paraboloid. to parametrize a surface Σ in 3 : x = x(u. z(a)) x = x(t) y = y(t) z = z(t)
1
1
C r(t) 0 x (x(b). b]. y(a). v).
z (x(t). Recall from Section 1.8 how we identified points (x. y = y(u. We will now learn how to perform integration over a surface in 3 . y(b).2 x
3
r(u. u and v. with the terminal points of the position vector r(t) = x(t)i + y(t)j + z(t)k for t in [a. z = z(t). LINE AND SURFACE INTEGRALS
4.156
CHAPTER 4.
v
2
z Σ R (u. We will use two variables. parametrized by x = x(t). y(t).4. a ≤ t ≤ b.4. for (u. v). v) = x(u.1 we learned how to integrate along a curve.2). v)k for (u. y = y(t). y. v) x = x(u. v). v) y = y(u. v)i + y(u. z(t)) (x(a). we will use a parametrization of a surface to define a surface integral. the position vector of a point on the surface Σ is given by the vectorvalued function r(u. v) in R. v) z = z(u. z) on a curve C in 3 .
v) ≈ . Along the vertical gridlines in R. So the area of that rectangle is A = ∆u ∆v. the tangent vector to those curves at a point (u. v + ∆v) − r(u. respectively. Suppose that this rectangle has a small width and height of ∆u and ∆v. v) is ∂r . v) in R as. the variable u is constant. v + ∆v). v)i + ∂v ∂y ∂z (u. will have a surface area (call it dσ) that is very close to the area of the parallelogram which has adjacent sides r(u + ∆u. v) − r(u. based on the idea of "patching" the region R onto Σ in the grid-like manner shown in Figure 4.8) applied to a function of two variables.2. those gridlines in R lead us to how we will define a surface integral over Σ. Thus. ∂v ∂v
∂r ∂u
157 and
∂r ∂v
The parametrization of Σ can be thought of as "transforming" a region in 2 (in the uv-plane) into a 2-dimensional surface in 3 . define the partial derivatives for (u. ∂u ∂v
(4. v)) ≈ (∆u ∂r ∂r ∂r ∂r ) × (∆v ) = × ∆u ∆v ∂u ∂v ∂u ∂v
by Theorem 1.3 in Section 2. we have ∂r r(u + ∆u. So those lines get mapped to curves on Σ. In fact.4.2. v). v)k .4. v) = ∂v ∂x (u. (u + ∆u.12 in Section 1. Taking ∂u ∂v the limit of that sum as the diagonal of the largest rectangle goes to 0 gives S =
R
∂r ∂r × du dv . v + ∆v) − r(u.13 in Section 1. But by combining our usual notion of a partial derivative (see Definition 2. v) to (u + ∆u. v)k . the horizontal gridlines ∂v in R get mapped to curves on Σ whose tangent vectors are ∂r .4. v)j + (u. v + ∆v) − r(u.4 Surface Integrals and the Divergence Theorem Since r(u. and ∂u ∂u ∂y ∂z (u. ∂v ∆v and so the surface area element dσ is approximately (r(u + ∆u. ∂u Now take a point (u. the total surface area S of Σ is approximately the sum of all the quantities ∂r × ∂r ∆u ∆v. v) (corresponding to the line segment from (u. for ∆u and ∆v small enough. v + ∆v) in R). v). v) is a function of two variables. say. v) in R by ∂r (u. Thus. summed over the rectangles in R. v) in R) and r(u. Similarly. Then that rectangle gets mapped by the parametrization onto some section of the surface Σ which. v) (corresponding to the line segment from (u. v) = ∂u ∂r (u.2) with that of the derivative of a vector-valued function (see Definition 1. v). v)i + ∂u ∂x (u. v) − r(u. v + ∆v) and (u. and ∂u ∆u ∂r r(u.4. as shown in Figure 4. v)j + (u. v) ≈ . This parametrization of the surface is sometimes called a patch. (u + ∆u.26)
. v) to (u. v) − r(u. and the variable u is constant along the position vector r(u. v)) × (r(u. The corner points of that rectangle are (u. the lower left corner of one of the rectangular grid sections in R.
z) over Σ is f (x.
(4. y.29)
Example 4.27)
This is a special case of a surface integral over the surface Σ. y. Find the surface area of T .28)
In particular. and let f (x. Replacing 1 by a general real-valued function f (x. z) be a real-valued function defined on some subset of 3 that contains Σ. ∂u ∂v
(4. v)k be the position vector for any point on Σ. The surface integral of f (x.4.z) a
b
(a) Circle in the yz-plane
x
(b) Torus T
Figure 4. y. v). v)i + y(u.4.3
Solution: For any point on the circle. v) = x(u. y = y(u. z = z(u. Let r(u. as in Figure 4. where the circle's center is at a distance b from the z-axis (0 < a < b). LINE AND SURFACE INTEGRALS
We will write the double integral on the right using the special notation dσ =
Σ R
∂r ∂r × du dv . v) in some region R in 2 . y(u. where the surface area element dσ can be thought of as 1 dσ. ∂u ∂v
(4. A torus T is a surface obtained by revolving a circle of radius a in the yz-plane around the z-axis. z(u. v). v).y. the line segment from the center of the circle to that point makes an angle u with the y-axis in the positive y direction (see Figure
. v).158
CHAPTER 4. z) dσ =
Σ R
f (x(u. z) defined in 3 .3. v)j + z(u. Let Σ be a surface in 3 parametrized by x = x(u. y. for (u.9. the surface area S of Σ is S =
Σ
1 dσ . we have the following: Definition 4. v). v))
∂r ∂r × du dv .
z
z (y − b)2 + z2 = a2 a 0 u y
y v (x.3.
(4. Namely. the surface integral
Σ
f · dσ is often referred to as the flux
of f through the surface Σ. Then f · dσ =
Σ S
div f dV . The term divergence comes from interpreting div f as a measure of how much a vector field "diverges" from a point. bounded below by another surface. See S CHEY.162
CHAPTER 4. z)j + f3 (x. Evaluate x2 + y2 + z2 = 1. y. § 15.e. z)i + f2 (x. for a point (x. and let f(x. For example. and bounded laterally by one or more surfaces. where f(x.
. p.32)
The proof of the Divergence Theorem is very similar to the proof of Green's Theorem. y. y. LINE AND SURFACE INTEGRALS
Theorem 4. This is best seen by using another definition of div f which is equivalent4 to the definition given by formula (4.8. for an intuitive discussion of this. z) = xi + yj + zk and Σ is the unit sphere
Solution: We see that div f = 1 + 1 + 1 = 3. then the flux is the net quantity of fluid to flow through the surface Σ per unit time. while a negative flux indicates a net flow inward (in the direction of −n). 36-39.
Σ
f · dσ.e. it is first proved for the simple case when the solid S is bounded above by one surface.32). y.6 for the details. z) in 3 . y.3 Example 4. z) = lim f · dσ . 1 div f(x. if f represents the velocity field of a fluid.
∂ f1 ∂ f2 ∂ f 3 + + ∂x ∂y ∂z
(4. A positive flux means there is a net flow out of the surface (i. y.33) V→0 V
Σ
3 4
See T AYLOR and M ANN.31)
where div f = is called the divergence of f. z)k be a vector field defined on some subset of 3 that contains Σ. The proof can then be extended to more general solids.
(4. in the direction of the outward unit normal vector n). so f · dσ =
Σ S
div f dV =
S
3 dV 4π(1)3 = 4π . 3
=3
S
1 dV = 3 vol(S ) = 3 ·
In physical applications. i.11. y. z) = f1 (x. (Divergence Theorem) Let Σ be a closed surface in 3 which bounds a solid S .
The ellipsoid
x2 a2
+
y2 b2
+
z2 c2
= 1 can be parametrized using ellipsoidal coordinates
x = a sin φ cos θ . where f(x.164
CHAPTER 4.
. Note that there will be a different outward unit normal vector to each of the six faces of the cube. b and c it can be evaluated using numerical methods. and z ≥ 0. (Hint: Use the parametrization x = r cos θ. § III. (Hint: Use spherical coordinates to parametrize the sphere. for 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π.3 to prove that the surface area S over a region R in surface z = f (x. Use a surface integral to show that the surface area of a right circular cone of √ radius R and height h is πR h2 + R2 .
(Hint: Think of the parametrization of the surface. h y = r sin θ. Evaluate the surface integral
Σ
f · dσ.10. y.) 9. For specific values of a.. New York: Dover. with Applications. 7. z) = x2 i + xyj + zk and Σ is the
part of the plane 6x + 3y + 2z = 6 with x ≥ 0.) 10. 8.5 )
C
11. F.
(Note: The above double integral can not be evaluated by elementary means.)
5
B OWMAN. z = c cos φ .e. LINE AND SURFACE INTEGRALS
B
5. z = R r. 6. Show that the flux of any constant vector field through any closed surface is zero. An alternative is to express the surface area in terms of elliptic integrals. Introduction to Elliptic Functions. y = b sin φ sin θ . with the outward unit normal n pointing in the positive z direction. Use a surface integral to show that the surface area of a sphere of radius r is 4πr2 . Evaluate the surface integral from Exercise 2 without using the Divergence Theorem.3. using only Definition 4. for 0 ≤ r ≤ R and 0 ≤ θ ≤ 2π. 1961. Use Definition 4. as in Example 4. y ≥ 0. y) is given by the formula S =
R 2
of a
1+
∂f 2 ∂x
+
∂f 2 ∂y
dA . i. Show that the surface area S of the ellipsoid is
π 2π
S =
0 0
sin φ a2 b2 cos2 φ + c2 (a2 sin2 θ + b2 cos2 θ) sin2 φ dθ dφ .7.
(4. y. y. y. z) dz
(4. z) dz =
C a
f (x(t).5. y(t). y = y(t). y.36)
The line integral of f (x. y. y.
(4. But the definitions and properties which were covered in Sections 4.39)
C
f(x(t).5 Stokes' Theorem
165
4. y(t). z) dy =
C a
f (x(t). y. y = y(t). z) dx +
C b C
Q(x. z) ≥ 0 then the line integral C f (x. z) dy +
C
R(x.37)
Similar to the two-variable case. z) i + Q(x. z) along C with respect to arc length s is
b
f (x.5 Stokes' Theorem
So far the only types of line integrals which we have discussed are those along curves in 2 .6. y. z(t)) x ′ (t) dt . z) j + R(x. y. if f (x. y. z) k and a curve C in 3 with a smooth parametrization x = x(t).38) (4. For a real-valued function f (x. y(t).
(4.
. parametrized by x = x(t). z(t))
x ′ (t)2 + y ′ (t)2 + z ′ (t)2 dt .
(4. y. y. z = z(t). z(t)) z ′ (t) dt . z) along C with respect to x is
b
f (x. Definition 4. z) ds can be thought of as the total area of the "picket fence" of height f (x. Vector fields in 3 are defined in a similar fashion to those in 2 . the line integral of f (x.1 and 4. y. z(t)) · r ′ (t) dt . y. y.
where r(t) = x(t) i + y(t) j + z(t) k is the position vector for points on C. z) and a curve C in 3 .4.2 can easily be extended to include functions of three variables. y. z) ds =
C a
f (x(t). y(t). a ≤ t ≤ b.34)
The line integral of f (x. y(t). the line integral of f along C is f · dr = =
a
P(x. z = z(t). z(t)) y ′ (t) dt . z) along C with respect to z is
b
f (x. Definition 4. z) = P(x. which allows us to define the line integral of a vector field along a curve in 3 . y.35)
The line integral of f (x. z) along C with respect to y is
b
f (x. y. a ≤ t ≤ b. For a vector field f(x. z) dx =
C a
f (x(t). y. z) at each point along the curve C in 3 . so that we can now discuss line integrals along curves in 3 .
y(8π). paraboloids. We say that such an N is a normal vector field. and 0 −N(x. These "outward" and "inward" normal vector fields on the sphere correspond to an "outer" and "inner" side. resulting in a "twisted" strip (see Figure 4. y. x Figure 4. LINE AND SURFACE INTEGRALS
∇F = f).12 we know that
C
f · dr = F(B) − F(A) .2).5.5.5. That is. In fact. your orientation changed even though your motion was continuous
A
. 0 cos 0.5. since N the continuous vector field N(x. z) is another normal vector field (see Figure 4. y. 0) =0+ = F(8π sin 8π. Other examples of two-sided. ellipsoids. z(0)) and B = (x(8π). 8π) − F(0 sin 0. An example is the Möbius strip. z(8π)). the unit sphere x2 +y2 +z2 = 1 is orientable. as in Figure 4. perpendicular to the tangent plane) at each point of Σ. You may be wondering what kind of surface would not have two sides. Roughly. y. z) is what we have called an outward normal vector. y.2 we say that the sphere is a two-sided surface. where A = (x(0).3(b).3).e. 2
We will now discuss a generalization of Green's Theorem in 2 to orientable surfaces in 3 . so = F(0. z) is an inward normal vector. then you arrive back at the same place from which you started but upside down! That is. A surface Σ in 3 is orientable if there is a continuous vector field N in 3 such that N is nonzero and normal to Σ (i. y(0). −N(x. respectively. So by Theorem 4. called Stokes' Theorem. and hence orientable. surfaces are cylinders. We see in this case that y N(x. z) = x i+y j+z k is nonzero and −N normal to the sphere at each point. 0) (8π)2 + (8π)2 − (0 + 0 + 0) = 96π2 . 0.5. "twosided" means "orientable".168
CHAPTER 4. 8π. A A B
→
→ −→
B
A
(a) Connect A to A and B to B along the ends (b) Not orientable
Figure 4. which is constructed by taking a thin rectangle and connecting its ends at the opposite corners.
z For example.3
Möbius strip
If you imagine walking along a line down the center of the Möbius strip. and planes. 8π) − F(0. 8π cos 8π. of the sphere.
y = y(t) . y) varying over a region D in 2 . with (x. see O'N EILL. then the surface would be on your left. y). a ≤ t ≤ b . z = z(x(t). y). we see that the closed curve C (the boundary curve of Σ) projects onto a closed curve C D which is the boundary curve of D (see Figure 4. and let f(x.2). We can now state Stokes' Theorem: Theorem 4. Then f · dr =
Σ
C
(curl f ) · n dσ .5. Now. z)k be a smooth vector field defined on some subset of 3 that contains Σ.4 in Section 4.45)
where curl f =
∂P ∂R ∂Q ∂P ∂R ∂Q − i+ − j+ − k. Projecting Σ onto the xy-plane. and so C can be parametrized (in
3)
z
Σ : z = z(x.4.
(4. and hence is nonorientable.7 For an orientable surface Σ which has a boundary curve C. ∂x ∂y
For further discussion of orientability. § IV. we know that z ′ (t) =
7
∂z ′ ∂z ′ x (t) + y (t) . y.14.
.4). y(t)) as a function of t.4
as
C : x = x(t) . z)j + R(x. y) for some smooth real-valued function z(x. y.7.5. by the Chain Rule (Theorem 4. there is a discontinuity at your starting point (and. We say in this situation that n is a positive unit normal vector and that C is traversed n-positively. we will prove the theorem only for the special case where Σ is the graph of z = z(x. Assuming that C has a smooth parametrization. its projection C D in the xy-plane also has a smooth parametrization. for z = z(x(t).46)
n is a positive unit normal vector over Σ. y. z) = P(x. thinking of your vertical direction as a normal vector field along the strip. z)i + Q(x. y(t)) .
since the curve C is part of the surface z = z(x. y)
n
C y 0 x D (x. at every point) since your vertical direction takes two different values there. (Stokes' Theorem) Let Σ be an orientable surface in 3 whose boundary is a simple closed curve C. y. say C D : x = x(t) . in fact. ∂y ∂z ∂z ∂x ∂x ∂y
(4. Informally. y) CD Figure 4. a ≤ t ≤ b . pick a unit normal vector n such that if you walked along C with your head pointing in the direction of n.5 Stokes' Theorem
169
along that center line. Proof: As the general case is beyond the scope of this text. The Möbius strip has only one side. y = y(t) . and C is traversed n-positively.
6. for a simple closed curve C the line integral C f· dr is often called the circulation of f around C. Namely. circulation per unit area). z). 78-81. z) and that the vectors grow larger the further the point (x. the surface it bounds. In the limit. that is. z) in 3 . y. See S CHEY.e. LINE AND SURFACE INTEGRALS
· In physical applications.46). for a point (x. Think of the vector field as representing the flow of water. which means that the circulation C E · dr = 0 by Stokes' Theorem. z) which is always parallel f to the xy-plane at each point (x. y. to have smaller and smaller surface area.
8 9
See Ch.e. y. Suppose we have a vector field f(x.
y An idea of how the curl of a vector field is related to rotation is shown in Figure 4. and it would rotate clockwise if it were dropped to the left of the y-axis.6. For example. z) points in the direction of your thumb as you cup your right hand in the direction of the rotation of the wheel.50)
C
where S is the surface area of a surface Σ containing the point (x. In fact. which causes Σ.6 Curl and rotation would rotate counterclockwise if it were dropped to the right of the y-axis. Since the 0 flow is stronger (i. for the derivation.5. M ILFORD and C HRISTY. y.174
CHAPTER 4. z). Notice that if all the vectors had the same direction and the same magnitude. z) and with a simple closed boundary curve C and positive unit normal vector n at (x. y. the magnitude of f is larger) as you move away from the y-axis. the term curl was created by the 19th century Scottish physicist James Clerk Maxwell in his study of electromagnetism. y. curl f(x. This is best seen by using another definition of curl f which is equivalent9 to the definition given by formula (4. then the wheels would not rotate and hence there would be no curl (which is why such fields are called irrotational. if E represents the electrostatic field due to a point charge. as in Figure 4. y. Vector fields which have zero curl are often called irrotational fields. In physics. and imagine dropping two wheels with paddles x into that water flow. then such a wheel Figure 4. y. In both cases the curl would be nonzero (curl f(x. think of the curve C shrinking to the point (x. then it turns out8 that curl E = 0. (4.5. z) = (1 + x2 ) j. n · (curl f )(x. z) = lim 1 S →0 S f · dr . y. z) is from the y-axis. z) = 2x k in our example) and would obey the right-hand rule. y. y. f(x.
. p. where it is used extensively. That ratio of circulation to surface area in the limit is what makes the curl a rough measure of circulation density (i. For example. So the curl points outward (in the positive z-direction) if x > 0 and points inward (in the negative z-direction) if x < 0.5. meaning no rotation). 2 in R EITZ. the curl is interpreted as a measure of circulation density.
where n is any unit vector.10
10
In Gaussian (or CGS) units.18. For instance. × · Since the choice of Σ was arbitrary. Since ∇ f is a vector field.18. Namely. Σ is orientable and its boundary is C). y.15 which can be useful.180
CHAPTER 4. to prove Theorem 4. Example 4. z) and produces an electrostatic field E(x. The proof is not trivial. A system of electric charges has a charge density ρ(x.e. with S being the solid region enclosed by Σ. y.13. y.
(by Theorem 4. assume that f (x. Let C be a simple closed curve in 3 and let Σ be any capping surface for C (i. then we must have (∇× (∇ f ))· n = 0 throughout 3 . y. which completes the proof. z) through any closed surface is zero. Show that ∇ · E = 4πρ. z) is a smooth real-valued function on 3 . This is one of Maxwell's Equations. and is often used in physics. Proof: Let Σ be a closed surface which bounds a solid S . we see that we must have ∇ × (∇ f ) = 0 in 3 . and physicists do not usually bother to prove it. j and k in place of n. z) in space. z) dσ = 0 for all surfaces Σ in some solid region (usually all of 3 ). y. z) at points (x. z) = 0 throughout that region. then we must have f (x. so
= 0 by Corollary 4.
. LINE AND SURFACE INTEGRALS
Corollary 4.17)
QED
There is another method for proving Theorem 4. Gauss' Law states that E · dσ = 4π
Σ S
ρ dV
for any closed surface Σ which encloses the charges. Using i. y. if the surface integral f (x. The flux of the curl of a smooth vector field f(x. y. But the result is true. and can also be applied to double and triple integrals. The flux of ∇ × f through Σ is (∇ × f ) · dσ =
Σ S
∇ · (∇ × f ) dV 0 dV
S
(by the Divergence Theorem)
= =0. then (∇ × (∇ f )) · n dσ =
Σ C Σ
∇ f · dr by Stokes' Theorem.15.
φ) in spherical coordinates is: 1 ∂F 1 ∂F ∂F eρ + eθ + eφ ∇F = ∂ρ ρ sin φ ∂θ ρ ∂φ Idea: In the Cartesian gradient formula ∇F(x. we will solve for k. but we will do it by combining the formulas for eρ and eφ to eliminate k. which will give us an equation involving just i and j. we get:
so using x = ρ sin φ cos θ. j. eθ . and we see that a vector perpendicular to that is − sin θ i + cos θ j + 0 k. y. then in particular eθ ⊥ eρ when eρ is in the xy-plane. z) = ∂F i + ∂F j + ∂F k. eθ .6 Gradient. we get: eφ = cos φ cos θ i + cos φ sin θ j − sin φ k Step 2: Use the three formulas from Step 1 to solve for i.2 that the unit vector eρ in the ρ direction at a general r point (ρ. j. eθ . This. then the unit vector eθ in the θ direction must be parallel to the xy-plane. ∂F . First. with the formula for eθ . will then leave us with a system of two equations in two unknowns (i and j). ∂F in terms of ∂F . Putting φ = π/2 into the formula for eρ gives eρ = cos θ i + sin θ j + 0 k. That is. This comes down to solving a system of three equations in three unknowns. φ) is eρ = r . k. note that sin φ eρ + cos φ eφ = cos θ i + sin θ j
. y = ρ sin φ sin θ. ∂x ∂y ∂z ∂ρ ∂F ∂F . We can see from Figure 4. θ. ∂φ and functions of ρ. θ and φ. eφ in terms of i. and ρ =
eρ = sin φ cos θ i + sin φ sin θ j + cos φ k Now. Lastly.6.4. eφ and functions of ρ. k in terms of the spherical coordinate basis vectors eρ . j. eρ = r xi + yj + zk = . eθ is of the form a i + b j + 0 k. Then put the partial derivatives ∂F . To figure out what a and b are. Divergence. Since this vector is also a unit vector and points in the (positive) θ direction. There are many ways of doing this. eφ . k in terms of eρ . ∂θ Step 1: Get formulas for eρ . r x2 + y2 + z2 x2 + y2 + z2 . θ and φ. where r = x i + y j + z k is the position vector of the point in Cartesian coordinates. it must be eθ : eθ = − sin θ i + cos θ j + 0 k Lastly. since eφ = eθ × eρ . z = ρ cos φ. note that since eθ ⊥ eρ . which we will use to solve first for j then for i. That occurs when the angle φ is π/2. since the angle θ is measured in the xy-plane. Thus. θ. Curl and Laplacian
183
Goal: Show that the gradient of a real-valued function F(ρ. put the Carte∂x ∂y ∂z sian basis vectors i.
We will consider the case when a > 0 and b > 0 (the other three possibilities are handled similarly). If either a = 0 or b = 0 then n(av. Since its magnitude is |ab|.
192
. j. j. bw) = ab(v × w) for any scalars a. w) = 0. by definition. and θ is the angle between them. which would prove the right-hand rule for the cross product (by part 1(c) of our definition). w. 3. bw) = 0 = ab(v × w). Hence the magnitude of n(av. n(v. w) form a right-handed system. we will show that the result holds for v = i and w = k (the other possibilities follow in a similar fashion). bw) is perpendicular to the plane containing ai and bk. If v and w are nonzero and not parallel. we will perform the following steps: Step 1: Show that n(v. bw) must be a scalar multiple of j. Let v and w be any two of the basis vectors i. namely.Appendix B
We will prove the right-hand rule for the cross product of two vectors in For any vectors v and w in
3. w). as follows:
1. 2. w) is perpendicular to the plane containing v and w. by definition. then n(v. n(av. so the result holds. w) is v w sin θ. There are four possibilities for the combinations of signs for a and b. w) = 0. and (c) v. w) is the vector in 3 such that: (a) the magnitude of n(v.
(b) n(v. j.4. So assume that a 0 and b 0. Also. then n(v. The goal is to show that n(v. For av = ai and bw = bk. then n(v.11 in Section 1. n(v. b if v and w are any two of the basis vectors i. w) = v × w if v and w are any two of the basis vectors i. k. w in 3 . This was already shown in Example 1. bw) must be either |ab|j or −|ab|j. 3. To do this. Thus. bw). For example.
define a new vector. then n(av. k. Step 2: Show that n(av. w) = v × w for all v. n(av. If either v or w is 0. k. If v and w are nonzero and parallel. is ai bk sin 90◦ = |ab|. the angle θ between av and bw is 90◦ . the xz-plane.
You can think of this projection vector (denoted by pro jP u v) as the shadow of the vector u v on the plane P. bk. Note that this holds even if u v. bk) = −abj. w) = n(u. n(ai. and since n(ai. w) and n(u. Thus. v) + n(u. then the result follows easily since n(u. −abj form a right-handed system (since a > 0. If v = 0. We will describe a geometric construction of n(u. which is the magnitude of n(u. w) = n(u. bk. ∴ n(av. w). v and w are all nonzero vectors. v + w) = n(u. since in that case θ = 0◦ and so sin θ = 0 which means that n(u. v + w). Since u (v + w) is the sum of the vectors u v and u w. j. v and this vector form a right-handed system. So since. k form a righthanded system. k. 0 + w) = n(u. Multiply the vector v by the positive scalar u . then project the vector u v straight down onto the plane P. then i. n(ai. j form a left-handed system. bk) = ab(i × k). w. v) has magnitude 0. which is shown in the figure below.
u v θ v θ P n(u. n(u. this means that we must have n(ai. If u = 0 then the result holds trivially since n(u.193 In this case. bk) form a right-handed system. v). But we know that ai × bk = ab(i × k) = ab(−j) = −abj. bw) must be either abj or −abj. w) are all the zero vector. then the projection vector
. Since this vector is in P then it is also perpendicular to u. k. v). −j form a righthanded system. v + w). If θ is the angle between u and v. w) = 0 + n(u. and so i. 0) = n(u. bk) has to be either abj or −abj. v) + n(u. by definition. v) and n(u. with the light source directly overhead the terminal point of u v. bw) = ab(v × w) Step 3: Show that n(u. Let P be a plane perpendicular to u. v + w) = n(u. ai. which is what we would expect. Now. v). And we can see that u. So now assume that u. v) u
pro jP u v
Now apply this same geometric construction to get n(u. Hence this vector must be n(u. Therefore. A similar argument shows that the result holds if w = 0. v. and ab > 0). b > 0. v) and which is perpendicular to pro jP u v (and hence perpendicular to v). So rotating pro jP u v by 90◦ in a counter-clockwise direction in the plane P gives a vector whose magnitude is the same as that of n(u. since i. w) for any vectors u. which is what we needed to show. n(av. ai. then we see that pro jP u v has magnitude u v sin θ.
v. and so w. v) has magnitude w v sin θ.
u v u (v + w) v+w pro jP u v pro jP u (v + w) u w w θ θ v u
pro jP u w n(u. −n(v. so the result holds. Also. By definition. Then by Steps 3 and 4. w. then n(w. which means that n(u. w) for any vectors v. v) and is perpendicular to the plane containing w and v. v) = 0 = −n(v. v. w) is a vector with the same magnitude as n(w. So by definition this means that −n(v. v + w) = n(u. v) + n(u. we have shown that −n(v. w). w) form a right-handed system. and hence w. w. n(v. and hence so is −n(v. −n(v. using the shadow analogy again and the parallelogram rule for vector addition.
If v and w are nonzero and parallel. v). and hence is the same as the magnitude of −n(v. w) = v × w for all vectors v. w) form a left-handed system. w) form a right-handed system. we have
. Then n(w. n(v. So then rotating all three projection vectors by 90◦ in a counter-clockwise direction in the plane P preserves that sum (see the figure below). Step 5: Show that n(v. w) form a right-handed system.194
Appendix B:
Proof of the Right-Hand Rule for the Cross Product
pro jP u (v + w) is the sum of the projection vectors pro jP u v and pro jP u w (to see this. v) n(u. w) must be n(w. think of how projecting a parallelogram onto a plane gives you a parallelogram in that plane). So assume that v and w are nonzero and not parallel. w). v + w)
Step 4: Show that n(w. w). or if either is 0. v) = −n(v. w) P
n(u. v. n(v. and that w. w. w). v. which is the same as the magnitude of n(v.
Write v = v1 i + v2 j + v3 k and w = w1 i + w2 j + w3 k. Thus. w) is perpendicular to the plane containing w and v. w).
exe from the folder (or bin folder) where you installed Gnuplot. just type gnuplot in a terminal window." option. For Windows. open-source software package for producing a variety of graphs.0.ini). run wgnuplot. Below is a very brief tutorial on how to use Gnuplot to graph functions of several variables. INSTALLATION 1. In Linux. 2. which is version 4. 3. You should now get a Gnuplot terminal with a gnuplot> command prompt. GRAPHING FUNCTIONS The usual way to create 3D graphs in Gnuplot is with the splot command: splot <range> <comma-separated list of functions>
196
. Go to style "Regular".. you should get the Zip file with a name such as gp420win32.info/download. the font "Courier".2. Versions are available for many operating systems. RUNNING GNUPLOT 1.2. For example. For example. size "12" is usually a good choice (that choice can be saved for future sessions by right-clicking in the Gnuplot window again and selecting the option to update wgnuplot. which we will now describe. All the examples we will discuss require at least version 4. while in Linux it will appear in the terminal window where the gnuplot command was run.Appendix C
3D Graphing with Gnuplot
Gnuplot is a free. At the gnuplot> command prompt you can now run graphing commands. In Windows this will appear in a new window.0. Install the downloaded file. In Windows.zip. For Windows.gnuplot. in Windows you would unzip the Zip file you downloaded in Step 1 into some folder (use the "Use folder names" option if extracting with WinZip).html and follow the links to download the latest version for your operating system. if the font is unreadable you can change it by right-clicking on the text part of the Gnuplot window and selecting the "Choose Font. 2.
use this command before the splot command: set zeroaxis Also. increase the mesh size (to. To display the axes. to label the axes. use this command: set view 60. the x-axis and y-axis are not shown in the graph. to also plot the function z = e x+y on the same graph. by default the x. use these commands: set xlabel "x" set ylabel "y" set zlabel "z" To show the level curves of the surface z = f (x. exp(x+y)
. To get more of a colored/shaded surface. 1 set xlabel "x" set ylabel "y" set zlabel "z" set contour both set isosamples 25 splot [−1 : 1][−2 : 2] 2*(x**2) + y**2.and y-axes are switched from their usual position. use this command: set contour both The default mesh size for the grid on the surface is 10 units. put a comma after the first function then append the new function: splot [−1 : 1][−2 : 2] 2*(x**2) + y**2. we get the following graph with these commands: set zeroaxis set view 60. 25) like this: set isosamples 25 Putting all this together. 120. For clarity. exp(x+y) By default. 1 Also.198
Appendix C:
3D Graphing with Gnuplot
Note that we had to type 2*x**2 to multiply 2 times x2 . y) on both the surface and projected onto the xy-plane. parentheses can be used to make sure the operations are being performed in the correct order: splot [−1 : 1][−2 : 2] 2*(x**2) + y**2 In the above example. 1. say. To show the axes with the orientation which we have used throughout the text. 120. 1.
v) is some function of u and v.5 1 0.5 y 0 0. with c ≤ v ≤ d.f(u. with a ≤ u ≤ b.v) where the variable u represents θ. That is. it can be turned off with this command: unset key PARAMETRIC FUNCTIONS Gnuplot has the ability to graph surfaces given in various parametric forms. The graph of the helicoid z = θ in Example 1.5 1.34 from Section 1.5 -2 -1. the variable v represents r.5 2 1 x
The numbers listed below the functions in the key in the upper right corner of the graph are the "levels" of the level curves of the corresponding surface. If you do not want the function key displayed. Example C.5 -1 0 -0.
z=z
you would do the following: set mapping cylindrical set parametric splot [a : b][c : d] v*cos(u). y) = c. they are the numbers c such that f (x. for a surface parametrized in cylindrical coordinates
x = r cos θ .199
25 20 15 z 10 5 0
2*(x**2) + y**2 6 5 4 3 2 1 exp(x+y) 20 15 10 5
-1 -0.v*sin(u). and z = f (u. 49) was created using the following commands:
.2.7 (p. For example.
y = r sin θ .
v*sin(u). In Linux. you would issue the following commands: set terminal png set output 'graph. Now run your splot command again and you should see a file called graph." option and enter a filename (say.. in the File menu again.." option. as a PNG file. and enter png in the Terminal type? textfield. go to the File menu on the main Gnuplot menubar. to print a graph from Gnuplot right-click on the titlebar of the graph's window. you will see that r varies from 0 to 2.".png) in the Output filename? textfield.u The command set xyplane 0 moves the z-axis so that z = 0 aligns with the xy-plane (which is not the default in Gnuplot). though you can change that setting using the "Change Directory . to save the graph as a file called graph. Looking at the graph.. PRINTING AND SAVING In Windows.. hit OK.200
Appendix C:
3D Graphing with Gnuplot
set mapping cylindrical set parametric set view 60. say. select "Options" and then the "Print." option in the File menu). In Linux..
.. To quit Gnuplot. since the print quality is high and there are many PostScript viewers available.png' and then run your splot command.exe is located. hit OK. the postscript terminal type is popular. 1. Then. There are many terminal types (which determine the output format). 1 set xyplane 0 set xlabel "x" set ylabel "y" set zlabel "z" unset key set isosamples 15 splot [0 : 4*pi][0 : 2] v*cos(u). type quit at the gnuplot> command prompt. Run the command set terminal to see all the possible types. select the "Output .png. To save a graph. graph.. 120.png in the current directory (usually the directory where wgnuplot. select "Output Device . and θ varies from 0 to 4π.
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History
This section contains the revision history of the book.0 Date: 2008-01-04 Author(s): Michael Corral Title: Vector Calculus Modification(s): Initial version
209
. please record the pertinent information here. 1. following the format in the first item below. For persons making modifications to the book. VERSION: 1. | 677.169 | 1 |
Glyphs!: Data Communication for Primary Mathematicades 1-3 ~ Glyphs are a way to represent information in picture or graphic form and an exciting way for students to collect, display, and interpret data. With a step-by-step approach, elementary students are introduced to data analysis and enhance...
The book, Glyphs!: Data Communication for Primary Mathematicians [Bulk, Wholesale, Quantity] ISBN# 9781564176639 in Paperback by O'Connell, Susan may be ordered in bulk quantities. Minimum starts at 25 copies. Availability based on publisher status and quantity being ordered. | 677.169 | 1 |
This text is for courses that are typically called (Introductory) Differential Equations, (Introductory) Partial Differential Equations, Applied Mathematics, and Fourier Series. Differential Equations is a text that follows a traditional approach and is appropriate for a first course in ordinary differential equations (including Laplace transforms) and a second course in Fourier series and boundary value problems. Some. Because many students need a lot of pencil-and-paper practice to master the essential concepts, the exercise sets are particularly comprehensive with a wide range of exercises ranging from straightforward to challenging. Many different majors will require differential equations and applied mathematics, so there should be a lot of interest in an intro-level text like this. The accessible writing style will be good for non-math students, as well as for undergrad classes. * Provides the foundations to assist students in learning how to read and understand the subject, but also helps students in learning how to read technical material in more advanced texts as they progress through their studies.* Exercise sets are particularly comprehensive with a wide range of exercises ranging from straightforward to challenging.* Includes new applications and extended projects made relevant to "everyday life" through the use of examples in a broad range of contexts.* Accessible approach with applied examples and will be good for non-math students, as well as for undergrad classes.
"synopsis" may belong to another edition of this title.
Product Description:
This text is for courses that are typically called (Introductory) Differential Equations, (Introductory) Partial Differential Equations, Applied Mathematics, and Fourier Series. The text is appropriate for two semester courses: the first typically emphasizes ordinary differential equations and their applications while the second emphasizes special techniques (like Laplace transforms) and partial differential equations. The texts follows a "traditional" curriculum and takes the "traditional" (rather than "dynamical systems") approach. Differential Equations is a text that follows a traditional approach and is appropriate for a first course in ordinary differential equations (including Laplace transforms) and a second course in Fourier series and boundary value problems. Note that some depending on the school, course, or instructor. * Provides the foundations to assist students in learning how to read and understand the subject, but also helps students in learning how to read technical material in more advanced texts as they progress through their studies.* The exercise sets offer a wide range of easy and difficult problems.* Includes new applications and extended projects made relevant to "everyday life" through the use of examples in a broad range of contexts.192 27572
Book Description Book Condition: Brand New. New. SoftCover International edition. Different ISBN and Cover image but contents are same as US edition. Customer Satisfaction guaranteed!!. Bookseller Inventory # SHAK143362 | 677.169 | 1 |
Friday, July 1, 2016
Workshop: New Approach for learning Maths for engineers and scientists, Ahmedabad
Workshop: New Approach for learning Maths for engineers and scientists
TWO DAY WORKSHOP ON NEW APPROACH FOR LEARNING MATHEMATICS FOR ENGINEERS & SCIENTISTS
Introduction:
Gujarat State Centre of The Institution of Engineers (India) in association with DASS Scientific Research Labs Private Limited (DASS SRL) is going to organize a two-day mathematics workshop specially to focus on a new system for mathematics Teaching and Demonstration replacing the conventional practices.
About the WORKSHOP:
Math is the key subject in every sphere of life. Be it solving complex equations, bargaining, natural phenomenon or simple house hold work, everywhere math plays its essential role. The conventional methods of solving problems involve high end calculations and complex concepts. Many times without understanding the actual problem, we follow and apply the mathematical concepts mechanically.
This workshop is meant to help participants and professionals to visualize few important mathematical concepts, so that they can correlate these into any other form and develop a quick understanding of a problem and instantly find a method for its solution logically. The workshop provides few tricks that will help analyze a problem without any complexity. | 677.169 | 1 |
Quadrature manufacturer descriptionQuadrature categories
What is new in 2.0 changeinfo log
Given stickers for Quadrature & download buttons
Smarter Choice award means that the price for the value and functions of Quadrature product is rational. Don't pay more than is necessary for the other software. This price is good.
DownloadAtlas.com guarantees that Quadrature was tested by antivirus program and is absolutely clean, which means it does not contain any form of malware, including computer viruses, adware, trojans, spyware, rootkits, badware and other malicious and unwanted software. | 677.169 | 1 |
Selected Materials from a Calculator Enhanced Instruction Project by an Expanded Consortium of New Jersey and Pennsylvania Educational Institutions.
Lane, Jean
This booklet contains a representative sample of the efforts of colleagues at 11 institutions to use graphing calculators to enhance the teaching of calculus and precalculus. The first section contains examples of graphs for teachers to choose from for presentations, including: simple examples to illustrate some standard ideas in precalculus, examples of graphs for which the window choice is critical and a knowledge of mathematics is essential for predicting hidden behavior, and examples that produce interesting shapes. The next sections contain generic and machine specific worksheets for calculus and precalculus. Next is a section that contains original programs for both the TI-81 and HP-48S calculators. A section to acquaint students with the use of the TI-81, HP-48G, and HP-48S concludes the booklet. (MKR) | 677.169 | 1 |
Teachers! Prepare Your Students for the Mathematics for SAT*I: Methods and Problem-Solving Strategies.
Posamentier, Alfred S.; Krulik, Stephen
This book focuses on curricular issues involved in preparing students for taking the SAT I--Reasoning Test using a problem-solving focus. There is a particular philosophy with which this book is presented. First, the illustrations have been selected to demonstrate in dramatic form the power of the procedure presented. Second, the problems both in this book and the companion student book are slightly more difficult than the items on the SAT I. Components of the book include: an overview of the SAT I, selective review of the mathematics taught through elementary algebra and geometry with particular attention to problem solving, less well known mathematics facts and problem solving short cuts, discussion on advising students on strategies for taking the SAT I, and detailed presentations of 10 specific problem-solving strategies. (MKR) | 677.169 | 1 |
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Mathematics is the language of science, data, and logic. Businesses and organizations can find themselves working with a wide variety of mathematical techniques and methods no matter what industry they are in. Mathematics, then, provides the equations that define and represent the physics and mechanics of physical products. Proper use of mathematics generates analytics that can be used to take the enormous amount of data available to modern businesses and provide commercial predictive power, which can drive and generate sales. Or mathematics can be used to create a host of other solutions, from security needs to general timeline extrapolation for projects and businesses.
Mathematics specialists on Upwork are highly-skilled professionals from a variety of backgrounds. Mathematics specialists can be highly educated, holding any number or type of advanced degrees, or they can be self-taught through experience. Your particular needs determine the type of specialist you will look for on Upwork. If you are seeking to develop highly theoretical predictive models, or just develop simple algorithms that can automate sales and inventory processes, a mathematics specialist can provide you with efficient and cost-effective solutions. As the nature of mathematics is generally abstract, a specialist can work remotely, either independently or with a team, and provide you and your organization with highly-flexible and dependable results. No matter what need you have in your organization, a math specialist can work with you to provide precise solutions that will maximize and streamline any aspect of production | 677.169 | 1 |
This text-workbook, along with the support of software, provides an effective tool for teaching the fundamentals of using a computer calculator. Approximately 15 hours of instructional material and application activities are combined to achieve basic mastery of the ten-key touch method and standard math calculations. Author : Barbara Henry ISBN : 0538695447 Language : English No of Pages : 96 Edition : 2 Publication Date : 9/1/2000 Format/Binding : Spiral-bound Book dimensions : 10.6x8.6x0.4 | 677.169 | 1 |
In this best selling Precalculus text, the authors explain concepts simply and clearly, without glossing over difficult points. This comprehensive, evenly-paced book provides complete coverage of the function concept and integrates substantial graphing calculator materials that help students develop insight into mathematical ideas. This author team invests the same attention to detail and clarity as Jim Stewart does in his market-leading Calculus text. Author : James Stewart ISBN : 0534385419 Language : English No of Pages : 877 Edition : 4th Publication Date : 12/14/2001 Format/Binding : Hardcover Book dimensions : 9.5x8.6x1.5 Book weight : 0.04 | 677.169 | 1 |
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