Prelude to calculus and linear algebra

Citation preview

JOHN R. HUBBARD AT DEPARTMENT OF MATHEM ALLEGHENY

COLLEGE

ARD JOHN R. HUBB ICS AT DEPT. of MATHEM EGE

- COLUMBUS

COLL

ft

Digitized by the Internet Archive in 2023 with funding from Kahle/Austin Foundation

https://archive.org/details/preludetocalculuO000john

PRELUDE

AND

TO

CALCULUS

LINEAR

ALGEBRA

The

Appleton-Century

Mathematics

Series

Raymond W. Brink and John M. H. Olmsted, Editors

A First Year of College Mathematics, 2nd ed., by Raymond W. Brink Algebra—College Course, 2nd ed., by Raymond W. Brink

Analytic Geometry, rev. ed., by Raymond W. Brink College Algebra, 2nd ed., by Raymond W. Brink Plane Trigonometry, 3rd ed., by Raymond W. Brink

Spherical Trigonometry by Raymond W. Brink An Introduction to Matrices, Vectors, and Linear Programming by Hugh G. Campbell Modern Basic Mathematics by Hobart C. Carter Elementary Concepts of Modern Mathematics by Flora Dinkines Parts also available individually under the following titles: Part Part Part Introduction

I, Elementary Theory of Sets II, Introduction to Mathematical Logic III, Abstract Mathematical Systems to the Laplace Transform by Dio L. Holl, Clair G. Maple, and Bernard

Vinograde Introductory Analysis by V.O. McBrien

College Geometry by Leslie H. Miller Advanced Calculus by John M. H. Olmsted Calculus with Analytic Geometry (2 volumes) by John M. H. Olmsted Intermediate Analysis by John M. H. Olmsted Real Variables by John M. H. Olmsted Solid Analytic Geometry by John M. H. Olmsted The Real Number System by John M. H. Olmsted Analytic Geometry by Edwin J. Purcell Calculus with Analytic Geometry by Edwin J. Purcell

Analytic Geometry and Calculus by Lloyd L. Smail Calculus by Lloyd L. Smail The Mathematics of Finance by Franklin C. Smith

PRELUDE

AND

TO

CALCULUS

LINEAR

ALGEBRA

John

M.

H. Olmsted

Southern

sje

\New York

Illinois University

{AAC} APPLETON-CENTURY-CROFTS C

DIVISION

OF

MEREDITH

CORPORATION

Copyright © 1968 by MEREDITH CORPORATION All rights reserved

This book, or parts thereof, must not be used or reproduced in any manner without written permission. For information address the publisher, Appleton-Century-Crofts, Division of Meredith Corporation, 440 Park Avenue South, New York, N. Y. 10016. 628-1

Library of Congress Card Number: 68-14040

Copyright © 1966 by Meredith Publishing Company under the title of Calculus With Analytic Geometry PRINTED

E 67865

IN THE

UNITED

STATES

OF AMERICA

TO

CYNTHIA

AIMNTMVS

OF:

PREFACE

This volume is the first of a set of three containing the material for a first course in calculus, and the subject matter immediately prerequisite thereto. The other two volumes are entitled Basic Concepts of Calculus and A Second Course in Calculus. The three-volume set is a condensation of the author’s two-volume Calculus with

Analytic Geometry, in this same Appleton-Century Mathematics Series, obtained by deleting from the earlier book all honors sections, most of the especially difficult and sophisticated proofs, and many of the more advanced exercises. For a general discussion of purposes and objectives of the total three-volume set the reader is referred to the Preface to Calculus with Analytic Geometry that is reprinted on the pages immediately following this preface. Compression has not been the only purpose in revising the author’s original text in calculus. A second objective was inspired by the close accord existing between the Calculus with Analytic Geometry and the general spirit of the recommendations of the Committee on the Undergraduate Program in Mathematics (CUPM), of the Mathematical Association of America. This three-volume revision has been fashioned to approximate the first few courses in college mathematics outlined in the 1965 booklet entitled A General Curriculum in Mathematics for Colleges (GCMC). The main deviation from the GCMC recommendations is a more consistent emphasis on

rigor. The GCMC recommends a “spiraling” upward from intuitive discussion to more rigorous content. The present set of books is dedicated to a slightly different form of spiraling, which maintains a substantial flavor of rigor throughout, while progressing steadily from elementary aspects of rigor in the simplest contexts to relatively advanced phases in some of the more difficult settings of analysis. An underlying principle in the present volumes is that it is pedagogically preferable to pursue a joint discussion and development of intuition and rigor, rather than to treat them separately, the intuitive first and the rigorous second. This goal, as explained in the preface of the Calculus with Analytic Geometry, is a realistic one and is accom-

plished by focusing attention first on relatively lightly quantified global concepts, including the Riemann integral, and then undertaking a study of more heavily quantified local concepts, in particular the various forms of the limit idea. This first volume is designed for a 4-hour, 1-semester course of approximately 48 class hours (aside from tests and reviews), although with deletions or reduced em-

phases, or with postponements of certain sections for incorporation into subsequent Vii

Viil

PREFACE

courses, the book can be used for a shorter course, such as a 5-hour, 1-quarter course of around 40 class hours. The first five chapters are sufficient prerequisite for the following volume, Basic Concepts of Calculus, with the exception of a few exercises on the reflecting properties of the conics. The special feature of this volume that should be noted is its unusual suitability as a preparation for modern courses in beth calculus and linear algebra. This is achieved through immediate use of sets and logic and the central role played by the concept of a vector space, exemplified by vector spaces of functions and vector spaces of ordered pairs, triples, and n-tuples, and the concept of an algebra, exemplified by algebras of functions and algebras of matrices. A course based on this volume would replace a traditional course in analytic geometry, while including some topics from college algebra. Such a course would correspond to the Math 0 course of the GCMC,

but at a more strictly college level.

Because of frequent cross references among the three volumes of which this is the first, and the parent two-volume book, we shall use the following abbreviations: CWAG: PCLA: BEG; SCG:

Calculus with Analytic Geometry, Prelude to Calculus and Linear Algebra, Basic Concepts of Calculus, A Second Course in Calculus.

The second volume of this set, BCC, is devoted to functions of a single real variable. The final volume, SCC, contains topics from multivariate calculus, intro-

duced as early as possible, as well as such remaining topics concerning a single variable as special techniques of integration, improper integrals, approximation methods, infinite series, and differential equations. These last two volumes, BCC and

SCC, correspond approximately to the GCMC courses Math 1 and 2, and Math 3 and 4, respectively. Each of the two volumes is designed for a 5-hour, 1-semester course of some 60 class hours, although other arrangements are possible. One alternative distribution of the material of the three volumes, for a sequence of four 5-hour quarters, is as follows: ‘

First quarter:

PCLA, BCC, Second quarter: BCC, Third quarter: PCLA, SCC, Fourth quarter: SCC,

Chapters 1-S, Chapters 1-4. Chapters 5-13. Chapters 6-10, Chapters 1 through §508. §509 through Chapter 12.

For those wishing to use the three-volume set for regular sections of a course, in conjunction with CWAG for honors sections, the following is a list of correspondences: Volume

PCLA

Chapter

1-5 6

7 8 9 10

CWAG

Chapter or Section

Chapters 1-5 Chapter 13 §§2701-2704 §§2801-2806 §§2811-2828 §§2901-2922

PREFACE

BCC

SCC

Chapters Chapters Chapter Chapter Chapters Chapter

§§2701, Chapters Chapter Chapters Chapter

6-12 14-18 21 20, §§2807-2808 30-32 22 2705-2708, 3125-3126 23-24 33 25-26 34

J.M.H.O. Carbondale, Illinois

ne

eqns

| AEE wy nes yale? ~

7. ie

~

enon. i

%.

i

~

=

‘ (a

im

te

oi

: ths

yp.

¢

ViLene Cw

PREFACE

TO

CALCULUS WITH ANALYTIC

GEOMETRY

A perennial problem facing teachers of mathematics is the determination of the proper time for introducing the student to the “delta-and-epsilon” type of mathematical rigor, sometimes referred to informally as “epsilonics.” Experience in recent years has convinced many people that much of the subject matter of “epsilonics” can be effectively taught at a much earlier stage in a student’s academic career than had at one time seemed possible. However, the fact remains that most students do have difficulty when first confronted with the limit concept. What is the best solution for the problem posed above? Should a considerable period of time be invested in a purely intuitive approach before the student is asked to face the harsh realities of mathematical analysis? Or is it possible to present “epsilonics” in such a fashion and with such a timing that the student can assimilate the essential ideas in a natural way and with a reasonably correct intuitive understanding? For purposes of discussion let it be agreed that the limit concept is a complicated and difficult one for most students new to calculus. Explicitly or implicitly the formulation of a limit statement uses not only a universal quantifier and an existential quantifier but also an implication involving two inequalities (or the equivalent in terms of a neighborhood and a deleted neighborhood). One of the principal assumptions determining the character of the present book is that the most appropriate groundwork for the study of limits is a substantial amount of detailed and careful work with inequalities, implications, and quantifiers before these are all combined into the type of compound statement needed for limits. For this reason, and also because of the many variants that involve “neighborhoods of infinity,” the treatment of limits has been placed as late as possible, in Chapter 10. This postponement of limits is made possible, in part, by the earlier discussion, in Chapter 9, of continuity (free from

the complications

of infinities and deleted neighborhoods),

but more

especially by the still earlier focusing of attention on such global concepts as the Riemann integral, in Chapters 6 and 7, and uniform continuity, in Chapter 8. Placing the introduction of the definite integral before that of the derivative has a well-recognized historical background, but it also has a sound logical justification. When the Riemann integral is approached by means of step-functions the definition of integrability involves only two quantifiers and a single inequality, and the definiXI

Xli

PREFACE

tion of the Riemann integral itself is simultaneously the supremum of one set and the infimum of another. The avoidance, at this stage, of implications between pairs of inequalities provides a natural arena for students to get a mathematical workout with quantifiers. Then later, when it is time to struggle with limit statements, the quantified portions have become familiar objects and are no longer a barrier. It is possible that some readers of this preface are not disposed at first to agree with the proposition that the global concept of uniform continuity is a simpler one than the local concept of ordinary continuity. Undoubtedly, if one wishes only to talk intuitively in terms of “approach” or “nearness” with accompanying “handwaving” gestures, uniform continuity is more complicated. However, it is when one

expresses himself in exact terms that this concept becomes simpler. More precisely, it is nearly always simpler to find an explicit 8 in terms of ¢ for a function on a set than for a function at a point. An example may help clarify this statement. For the “squaring” function x — x2 on the compact interval [3, 5] a suitable § can be found quite simply to be ¢/10, whereas for this same function at the point 3, either 8 turns out to be something like min(1, «/7) or else « must be artificially limited in size. In

fine, the position adopted in this book is that to concentrate on relatively simple global concepts before turning attention to point-wise continuity and limits makes good sense, and is preferable in the long run to the more traditional intuitive approach which necessitates an often painful relearning process later. The dominant philosophy of this book emphasizes concepts and structure, with the dual objectives of developing an appreciation for a truly beautiful and well-conceived subject and exploiting the great potentials of these concepts and structures. The concept of vector space should serve to illustrate what is meant. Vector spaces are introduced first for function spaces, with examples such as bounded functions, step-functions, and polynomials. It then becomes natural to develop the Riemann integral as a positive linear functional on the vector space of integrable functions. Then later, many of the standard theorems of continuous functions.and limits find simple expression when cast in the language of vector spaces and algebras of functions. Furthermore, these same ideas become tools when it is shown that the Riemann integral can be obtained as the limit of a sum. Later in the book, generous

attention is again devoted to vector spaces and algebras, this time in the context of matrices and vectors in two and three dimensions. Eigenvalue techniques are used effectively to simplify transformations of equations of the second degree. In order to achieve the goals enumerated above, the first two chapters contain an introduction to sets and logic. Truth sets serve to explain exactly what an implication is. The quantifier symbols V and 3 are used widely, and their role in the formulation of negations is discussed carefully and used frequently. Both the nature of a proof and the meaning of a counterexample are given considerate attention. The groundwork laid in the first two chapters forms the basis for the extensive use of both sets and logic throughout the book. It is felt that the time spent initially in forming a sound foundation for the ideas that are so essential to calculus is well worth the investment, and is amply justified by the ultimate dividends in the form of understanding and intellectual satisfaction. Alternative formulations are sometimes included. For example, several limit

PREFACE

xii

statements are expressed in terms of deltas and epsilons, neighborhoods and deleted neighborhoods, and mappings. These alternatives help provide a broad and varied texture for the underlying flow of ideas, as well as emphasizing a basic unity that pervades many apparently dissimilar settings. Much attention is given to applications. For example, Chapter 7 is devoted entirely to such applications of the Riemann integral as area, volume, work, and distance. Applications of the derivative are scattered through most of the chapters. Differential equations find their first use in Chapter 18, where they are applied to problems involving gravitation, orthogonal trajectories, radioactive decay, bacterial growth, cooling, and mixing. The final Chapter 34 returns to the subject of differential equations, concentrating on linear equations with constant coefficients and their

applications. Extremal problems receive unusually thorough consideration, with full attention given to sufficiency conditions and endpoint extrema. Area and volume are introduced axiomatically, each being a positive additive function on a ring of sets and satisfying a certain completeness axiom. For a function of a single real variable these axioms suffice for establishing the theorem that the ordinate set of a nonnegative Riemann-integrable function f on a compact interval has area equal to the integral of f. Elliptic notation (for example, “the function e* sin 3x”) that the student learns to understand and use both the strict tion. This is important since both notations abound in the of the world. Special attention should be called to the character and sections, suitable for enrichment

content in honors courses.

is discussed clearly so f and the elliptic notamathematical literature

arrangement of honors These honors sections

are planned so as to permit an honors course running concurrently with a standard course to cover the standard as well as the extra material, and also to permit exploratory reading and extra assignments for interested and capable students who are not registered in a special honors class. In order that a student may be able to transfer either out of or into an honors sequence between terms it is desirable that the honors course keep pace with the standard course to which it corresponds. This is made especially feasible for courses using this book by the uniform arrangement of the honors sections, which constitute exactly twenty percent of the sections, as follows: Every chapter contains a multiple of ten sections, including sections of exercises; every section whose number ends with the digit 9 is an honors section, and every section whose number ends with the digit 0 is the accompanying honors section of exercises. All honors material is identified by the letter H. The division of the book into two volumes is done simply to reduce the bulk of the alternative of a single tome. The place of division between Chapters 21 and 22 achieves a local minimum for cross-references between the two volumes. The use of a table of integrals is limited to the second volume. The book is designed for courses of three semesters — or four or five quarters — meeting four or five days a week. Prerequisite is a standard high school mathematical preparation, including trigonometry and college algebra. Review material in trigonometry and such topics as mathematical induction is included. Over 6000 exercises of all levels of difficulty are available, for practice, for chal-

X1V

PREFACE

lenge, and for individual exploration. These exercises form an essential part of the book, and are designed to fortify and deepen, as well as guarantee, learning. Answers to nearly all problems are given in the back portions of the two volumes. Illustrative examples are liberally provided. The author wishes to express his appreciation for the extensive aid and numerous Suggestions given by Professor R. W. Brink. He is also indebted to many others for their helpful comments, given both informally in conversation and more specifically

on paper. J.M.H. O. Carbondale,

Illinois

CONTENTS

ELC CAC Cte

Rae

Sede

RUE

Dh

ear,

Preface to Calculus with Analytic Geometry

1

101 102 103 104 105 106 107 108 109 110 | a 112 ELS 114 115 116

RELATIONS,

AND

es

ANA

....... 0.00 cece ees

Vil Xl

FUNCTIONS

MEVECvETUG (SON ates Fe NP Pre ape Sey be gh osaig sees Gresfaint gens Sra De CSMICTITOCTS TAI SUDSCUS) 5 clog Gieh Sieste cease cued pele Sa a AUS we yslewerisls AG IVE Ese saa aCOMP EINECS bent eke eS con loss ore be Ste Ge wanes as (CUB COTIISE NYS TUT EET EY OTe ve les rae ra ae oan ere NINet sec IONS COMUIMAIONS | 420.0 2. sateen cie eos oon en's Lier eke & Sw Tere SG a Men CIABEAINS TOE StAtCIMeMt(S co. c5e hina ian eae «o's 00>.mS nsete) DG steve 8 SYSLOGTSS? SOUR UTSTaRS © Be aes eek ape Rony em sr ren TEPC) RCPS ES «a ek SihsySin nr RN ea ele cee am eA inG@oduction torvaniables-and TUNCtIONS forall there exists Vr

3

301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 one

os

eee

Eos

THE

REAL

NUMBER

SYSTEM

Introductions binary operavionSi.. 46 4 es we aes ee eee AXIOMS. O0 a:lGld 5 time. eae cesnnn eects hocks howtos) Oca se ee PIXON CISES) «SAORI Pe cabole 8 Gare Feo tO RINE SAE See Ee eee ee ee Ordered fields: sierra ee Ne eR Se Ie cow SRaen ae eens Cae eee |S). Co)he) (on Gentian Ra ne a PI NE GSE meant VyCeara sn ar OCS, ) 6)ts Natural’ numibers.2 3 with domain: x € ®, then p(5) is the true statement 5 > 3 and p(1) is the false statement 1 > 3. If p is the formula fis an even function, where the domain of p= is the set of all real-valued functions of a real variable with domain ®, then P(f) is a true statement if fis the function defined by f(x) = x? for x € ®, and p(f) is a false statement if fis the function defined by f(x) = x3 for x € &.

Example 5. The solution sets of the ten formulas of Example 3 are: Hoes Way for (12): @ (there is no solution that is a member of 9v, the set of natural numbers); for (13): the collection of all nonempty subsets of @; for (14): the collection of all finite subsets of @, including Q; for (15): the set of all points (x, y) of the Cartesian plane ® X ® such that x + y = 4; this set corresponds to a /ine in the Euclidean plane £2 (cf. Chapter 4). for (16):

the subset of the solution set for (15) such that x and y are integers;

tony) 2 ds 2)n@r0)i: fOLLs)eml-4)-(2,.3)(254),.(3, 2), (3,.3),.G, 4)}5 for (19): the collection of all even functions with domain = ® and range C ®&; for (20): @ (there is no odd function with domain = ® whose values are ail positive).

Any two formulas with the same domain and the same truth set are said to be equivalent. Example 6. formulas

(24) are equivalent.

If the range of the variable 7 is the set SU of all natural numbers, then the two

A 2a

nS

ELEMENTARY

32 202

NEGATIONS,

DISJUNCTIONS,

LOGIC AND

IN MATHEMATICS

[$202

CONJUNCTIONS

For simplicity of notation, and not out of any basic necessity, we shall limut our considerations in this section to formulas involving a single variable x. We shall also assume (again for simplicity) that all formulas under consideration have a common domain D. In the main it will be convenient to denote the truth set of a formula by the upper case or capital form of the letter used for the formula. Thus the truth set of the formula p will be denoted P and the truth set of the formula q will be denoted Q: (1)

P=

oa

Ce) Isstries.

OF

Geos

tinue.

If Sis any statement, its negation, denoted ~S, is the statement “‘it is false that S,” although in informal practice any statement logically equivalent to “‘it is false that S”’ is called the negation of S and denoted ~S. For example, if x is a particular real number and S is the statement “‘x is negative,’ then ~S can be formulated in the following fashions (among many): “‘it is false that x is negative,” “‘x is not negative,” “x is nonnegative,” ‘‘x is positive or zero,” and ‘‘x is greater than or equal to 0.” In any case, whenever S is a true statement ~S is a false statement, and whenever S is a false statement ~S is a true statement. Finally, S is the negation of its negation (S = ~-~S), and we have:

ae is true if and only if S is false, ~S is false if and only if S is true.

(2) Example 1.

The following are examples of statements and their negations:

GneeNesey o Hil BG

Wo, Be a

OD)

5341,.2) E12,

al:

(5)

S: Every even function is a polynomial, ~S: There exists an even function that is not a polynomial.

In (3), S is true and ~S is false.

In (4) and (5), S is false and ~S is true.

We now extend the concept of negation to formulas, as follows: Definition I. Jf p is any formula, with domain D, its negation, written ~p, is the formula with domain D whose value at any point x of D is the negation of p(x):

(6)

(~p\(x) = ~(p(x))

(domain: x € D).

Equivalently, ~p is the formula with domain D whose truth set is the complement P' of P in the universe or space D:

(7)

{x |(~p)(x) is true} = {x |p(x) is false} = P’ = D\ P.

(Cf. Fig. 202-1.)

§202]

NEGATIONS,

DISJUNCTIONS

AND

CONJUNCTIONS

33

Figure 202-1

Example 2. Ifp is the formula defined

(8)

2x —j)-(domain:

x €

2, 354),5});

the related truth set is P = {1, 2, 3}. The negation ~p of p may be written (9)

2x)

a(domainaex

© | ly 23e4.5)0))

Wwithetruthyset Pp’ — {455}! Example 3.

The formula

(10)

x 23

(domain:

x € @),

with truth set {x € ®| x = 3} has the negation (11)

x 6 (domain:

x € ®).

22.

36

204

ELEMENTARY

LOGIC

IN MATHEMATICS

[§204

IMPLICATIONS, CONVERSES, AND CONTRAPOSITIVES

As in §202 we shall assume for convenience that all formulas involve a single variable x and have a common domain D. Definition I. Let p and q be formulas with common domain D and with truth sets P and Q, respectively. Then whenever P is a subset of Q, P C Q, p is said to imply q, andq is said to be implied by p, with the notation p = q, read “‘p implies q,” or gq=p, read “‘q is implied by p.”’ In short:

(1)

p=9¢=

F.C.

The statement p does not imply q, or q is not implied by p, denotedp + q, or q # p, is the negation of the statement (1):

(2)

(p#q) = ~(P$9)

and is the same as the statement P is not a subset of Q:

(3)

pHq=P¢

Q.

The statement p = q is called an implication, and the statement p # q is called the negation of an implication. Discussion. The implication p = q states that g must be true whenever p is true. If we write the inclusion P C Q inthe form P\ Q = P(\ Q’ = @, we see that the implication p= q can be expressed by the statement that for no point of D is it possible to have p true and q false. This is sometimes expressed as follows: “‘It is impossible forp to be true and q to be false simultaneously.”” The negation p 4 q, on the contrary, states that the set P is not contained entirely within the set Q and therefore intersects the complement of Q: P(\ Q’ # @. In other words, p 4 q means that there must be a point of D for which p is true and gq is false. These observations point up the fact that an implication expresses nonexistence (of members of the set P (\ Q’) whereas its negation expresses existence (of at least one member of P\ Q’). Above all it should be stressed that the negation of one implication is never another implication. (In particular, the negation of the implication p = q is NOT the implication p = ~q!) (Cf. Fig. 204-1.) In proving that an implication p = q isfalse it is necessary to establish the existence of a point x belonging to P and not to Q@. Sucha

ena!

PAA Figure 204-1

LG

§204]

IMPLICATIONS,

CONVERSES,

AND

CONTRAPOSITIVES

37

point x for which p(x) is true and q(x) is false is called a counterexample. Finally, if explicit expressions for p(x) and q(x) are available, it is often convenient to use the notation p(x) = g(x) to mean p= q. This is illustrated at the end of Example 2, below. Example 1.

Let p and q be the formulas with domain ®, given in terms of the variable x:

(4)

JOR

SEO.

GR

Be 5. Therefore the negation can be expressed in the form (x S$ 1) V (« > 5). Since (4) is a disjunction, its negation is the conjunction of the two negations x = O0and x S 10, which can be written in the concise form 0 S x S 10.

Example 7. Prove the equivalence of the two formulas A C B and B’ C A’, where A and B are subsets of a space S and the common domain of the two formulas is the set of all ordered pairs (A, B) of subsets A and B of S:

(39)

AGB

=

B GAS

Solution. The formula on the left of (39) is equivalent to the formula (that is, the ordered pairs (A, B) for which A C B are precisely those for which (0). Similarly, the formula on the right of (39) is equivalent to the formula B’ ()\ Finally, since B’ ()\ (A’)’ = B’()\ A = AT) B’, it follows that the two formulas equivalent.

4/)\B’ = @ 4 () B’ = (A’)’ = @.

of (39) are

If S and T are statements, the converse of the implication S = T is the implication T= S. If p and qg are formulas having a common domain, the converse of the implication p = q is the implication q => p. The converse of a true implication may or may not be true. In fact, any of the four combinations of truth and falsity for an implication and its converse are possible. An implication and its converse are both true if and only if the two formulas involved are equivalent. On the other hand, an implication and its converse may both be false without having equivalent formulas. (Cf. Example 8.) Example 8. The formulas in the following implications are all assumed to have the common domain &: (40)

x ~x

< 4is true; the converse x < 4= x < 2 is false;

$204]

IMPLICATIONS,

CONVERSES,

AND

CONTRAPOSITIVES

(41)

x x? < 2x is true;

(43)

x 2 is false; the converse

x

x >

4]

x < 4is true;

2=> x < 5 is false.

The contrapositive of the implication p => q, where p and q have a common domain D, is the implication ~g => ~p. If P and Q are the truth sets of p and q, respectively, then their complements P’ and Q’ are the truth sets of the negations ~p and ~g, respectively. Since the implication p = q is equivalent to the set inclusion PC Q, and since the implication ~q = ~p is equivalent to the set inclusion Q’ C P’, it follows from Example 7 that the implication p => q and its contrapositive ~q = ~p are equivalent statements whenever p and q are formulas with a common domain:

(44)

Law of Contraposition: (p =q)

= (~q

> ~p).

The equivalence (44) is of great importance in mathematical proof, which is not surprising in view of the extent to which mathematics involves the establishment of implications. The most common method of establishing the implicationp > q is to assume that x is an arbitrary member of the truth set P of p— that is, that p(x) is true — and then to show that x must consequently be a member ofthe truth set Q of q. On frequent occasions, however, it is more practical — or even essential — to make the assumption that g(x) is false and to pursue logical reasoning from that point of departure in order to infer that p(x) must also be false. This method of proof will be referred to as the contrapositive method. A veriant on the contrapositive method of proof is the indirect method, also called proof by contradiction or reductio ad absurdum.

By this method

one assumes

that the implication p = q is false by assuming the existence of a point x such that p(x) is true and q(x) is false, and then draws an inference incompatible with at least one of the two fundamental assumptions: p(x) is true and q(x) is false. The idea of this form of proof lies in the following way of saying that the truth set P is not a subset of the truth set Q:

(45)

Pi\ OFF

@.

The contradiction obtained from the assumption that there exists a point x belonging to P()\ Q’ is usually either of the form p(x) isfalse or of the form q(x) is true. We conclude this section with the following definition of the compound statement p=>q=", where p, q, and r are formulas with a common domain:

(46)

P=qene@rH=QA gG=”).

This is in keeping with the notation A C B C C for the conjunction of A C B and BCC, and 1 < x < 4 for the conjunction of | < x and x < 4. The notation of (46) extends in like fashion to conjunctions involving four or more formulas, It should be observed that there is no p>q>r>s, p=qa>r—>s=>u,:-:. p= q p> (=> 7"), Since both occasion to confuse (46) with either (p => q) => r or has = and g=>r are statements rather than formulas, and the implication symbol formula. been given no meaning when it is placed between a statement and a

[§205

IN MATHEMATICS

LOGIC

ELEMENTARY

42 EXERCISES

205

In each of the following exercises the formulas are assumed to have

Ik; Bae

The common

SR

Bb JOR PPK 58 eR

domain in every case is ®.

=> sh

Wee

2s JOR SES BR GRse

By,

IR ER

domain.

p= q. If p= q explain why. If p * q find a

In Exercises 1-4, determine whether p > q or counterexample.

a common

So

ADD

> 7.

OS Gna

In Exercises 5-8, determine whether S = T or S # T. Gy OIA SS 3 = Ce USO Ss ©.

OR NEO Say

By S245 3

Ss ©: Ch SASS 3 = S38 USO

Ib SRA se Sy = S28 IO SS Oe

In Exercises

9-11, prove the specified law.

9. (34), §204.

10.

Illustrate with Venn diagrams.

(36), §204.

11. (37), §204.

In Exercises 12-16, prove the stated law of logic. 12. Law of the excluded middle:

13. Commutative laws:

14. Law of addition:

p

p V ~p

= ft.

VqeqVp,pANqeq

p= (p V q).

16. Hypothetical syllogism:

Illustrate with Venn diagrams.

p.

15. Law of contradiction:

p A ~p = f.

(p= q) A G@=>r)>(:p=}/r).

In Exercises 17-20, formulate the contrapositive of the given implication. The common domain throughout is ®. Do not establish truth or falsity of any implication. Wh

0

ae

SS 2

19 (c< QV

GS

UY Se SS Geese SS al

HS 3 S07 7

2055

~p(x) is true)

@ (4x € A 9 p(x) is false);

(14)

~ (Ax € A 35 p(x) is true)= (Vx € A, ~p(x) is true) 0, 8x? — 24x +7

Negation of (10):

Vn € MN, 2" S 33.

= 0;

Implications and their negations (§203) are readily expressed quantifiers V and 3:

(15)

P=qo—(Wx € P,x € Q)

(16)

(Pp¥QgeGxeEPdx ¢€ Q) = (4x 5 p(x) is true and q(x) is false).

in terms of the

= (Vx 5 p(x) is true, g(x) is true);

Once more we see why the negation of an implication cannot be another implication: an implication is a universal statement, while its negation is an existential statement. Example 4. The following two implications (one true and one false) and their negations (one false and one true) are expressed with quantifier symbols (the domain in each case is ®): (17)

XD > BX N. In other words, we are seeking an implication of the type x > N= f(x) > M. The next order of business is therefore to find the number N, and we write

(2)

V OV Tans saat 3G 1) er

The entire statement sought is therefore

(3)

VICES

ING

Roe

N=] fo) > MM,

We now seek a formulation of the negation of (3). This we shall do bit by bit. In the first place. the negation of any statement of the form (1) is of the form

(4)

3) WWE G sees,

where the dots - - - represent the negation of the corresponding part of (1). From (2), we see that (4) then becomes

(5)

IMERIVNER,:::

and the negation of (3) is

(6)

IMECRIVNER,

> M. x >NHSf(xX)

Finally, since the negation of an implication is an existential statement, we formulate (6) in positive terms:

(7)

IMECRIVNER,

SM. Jx>NIfx%)

In words, (7) says: ‘There exists a real number M such that for every real number WN there exists a number x greater than N such that f(x) is less than or equal to M.”

IN MATHEMATICS

LOGIC

ELEMENTARY

46

[$208

Example 2. Give precise meaning to the informal statement that there are arbitrarily large values of x for which the values of the real-valued function f with domain & vanish. Then by means of quantifiers formulate the negation of this statement in positive terms, and express the result in words. We first restate the proposition, still informally but with an attempt to give adSolution. ditional meaning to it: ‘‘No matter how large a positive number N we may be given, there must exist a larger (positive) number x such that f(x) = 0.’ As in Example 1, we start:

(8)

IVECO

and continue:

(9)

WIN EW,

slse S IN Soe,

and finally:

(10)

WINE

OC

Six

WY SiGe) = ©.

The negation of (10) is constructed piecemeal:

(11)

BIIN (EG See.

(12)

INE

PIVx>N,:::,

and finally,

(13)

dN SG Ooi Ne

oO.

In words, (13) states: ““There exists a positive number MN such that for all x greater than N, F(x) is nonzero.”

By reasoning similar to that employed in the two preceding examples, we see that the negation of any statement framed in terms of the quantifiers V and J has a ‘“‘dual”’ formulation obtained by replacing each of the two symbols VY and i by the other and each of the symbols 3 and , (that is, the comma) by the other, and replacing the final portion of the statement by its negation.

208

EXERCISES

In Exercises 1-4, express the given statement in terms of either the universal quantifier or (The statements are not necessarily true.)

the existential quantifier.

1. For every positive number x, x2 — x + 4 is positive. 2. For every real number x, 36x2 + 12x is an integer. 3. There exists a natural number 7 such that 3”2 + 5x is odd. 4. There exists a linear function fsuch that f(1) = 2 and f(2) =

1.

In Exercises 5-8, express the negation of the statement of the indicated exercise in positive terms using quantifiers. 5. Exercise 1.

6. Exercise 2.

7. Exercise

3.

8. Exercise 4.

In Exercises 9-16, express the given statement in terms of quantifiers. Then formulate the negation in positive terms. Finally, express the negation in words. (The given statement is not necessarily true.)

9. For every natural number », there exists a natural number m greater than 72.

§208]

EXERCISES

47

10. For every natural number n, there exists a natural number m such that 2m = n. 11. For every positive number M, there exists a positive number N such that for every number x less than N, x + 1 is less than M.

12. For every real number M, there exists a real number

N such that x > Nimplies x2 > M.

13. There exists a real number x such that every natural number #7 is less than x. 14. There exists a real number x such that for every real number y, xy = x.

15. There exists a positive number M such that for every positive number 6, there exists a number x such that 0 < x < dandxM S11.

16. There exists a positive number M such that for every positive number N there exists a number x such that Mxy ~0. =O

— 0.

(xiv) Zero has no reciprocal; that is, there is no number y such that 0-y = 1. (xo) (x = 0) A & # 0) = Gy)

(xvi) (b # 0) A (dS (xviii) (b # 0)

Ad # >

= x1y-!, oF papsenicd xy eay

= g

(xii) (6

0) A (dx 0)

2.5 =

E45= Ce

(xix) (—1)(-1) = 1.

(xxi) (—x)(—y) = xy. (xxi) y= 0=> a = en = =

(xx) (—1)x = —x.

(xxii) —(xy) = (—x)y = x(—y). (xxiv) x(y — z) = xy — xz.

§304]

ORDERED

FIELDS

53

(xxv) (a+ b\(e + d) = ac + ad+ be + bd. (xxvi) If x2 = x- x, x2 — y2 = (x — y\(x+ y). (xxvii) The general linear equation x = —D/a. 303

ax + b = 0, a #0,

has a unique solution

EXERCISES

In Exercises 1— 4, prove the given statement or establish the given equation for an arbitrary field.

1. The additive inverse of a number is unique.

Ay)

= — xy,

. There is only one number having the property of the number 1 of Axiom II(iii), §203. WNee WwW mp

OR

een,

5. Show that the set of all nonzero members of a field with the operation of multiplication (instead of addition) is a group. 6. Show that the set consisting of the distinct numbers 0 and 1 only, with the following addition and multiplication tables, is a field:

Addition 304

ORDERED

Multiplication

FIELDS

Statements involving such phrases as “‘less than,” “greater than or equal to,” and “cannot exceed” depend on what is called the ordering of the real number system. There is more than one way to describe the properties of ® that are based on ordering, but probably the simplest is in terms of a particular subset © of & consisting of those numbers that we shall call positive. In terms of the properties specified for this set @, we shall be able to define what are called relations defined by inequalities. 99

66

Before formulating Axiom IV which, together with Axioms I-III, states that the real number system is an ordered field, we introduce a definition of closure:

Definition I. Jf T is a binary operation on a nonempty set B and if A is a nonempty subset of B, @# ACB,

then A is closed with respect to T (or closed under T) if and only if

(1)

CA xX A = OG)

¢ 4. 1%)

- Example 1. As will be seen (§309), the set of all integers is closed with respect to addition. However, the set of all odd integers is not closed with respect to addition (for example, 3 and

THE

54

REAL

NUMBER

SYSTEM

(§304

5 are odd integers but their sum is not). The statement of Property (x7), §302, can be expressed: The set of nonzero numbers is closed with respect to multiplication. However, the set of nonzero numbers is not closed with respect to addition. (Why not?)

We are now ready for the next axiom: IV. Order There exists a subset & of & such that (i) & is closed with respect to addition:

(x y)E PX CS>X+YVES. (ii) @ is closed with respect to multiplication:

(x, y)E OX

P= xVE@.

(iii) There is no number x such that both x and —x are members of ®. (iv) If x is any nonzero number, then either x or —x is a member of &. Definition II.

An ordered field is a field § with a subset & satisfying Axiom IV.

Definition III.

A number x is positive if and only ifx © &. A number x is nega-

tive if and only if —x

€ @.

Axioms IV(ii/) and (iv) together state that for an arbitrary number x exactly one of the following three statements is true: x € ®, x = 0, —x € &. In other words, exactly one of the following three statements is correct: x is positive, x is zero, and

x is negative. In particular, 0 is neither positive nor negative, and 0 is the only number that is neither positive nor negative. Definition IV. The relations , S, and = are the following subsets of the Cartesian plane ® X B:

VA IA IV

Wai) aera icaen (Cowie aoe Ch), (a YIG-x€ e)V (x= yy}, (yO Sy Sie) eG

In other words,

< y if and only if y — x is positive,

> y if and only ifx — y is positive, be eesS y if and only if eitherx < yorx = y, x 2 y ifand only if eitherx > yorx = y. Each of the four formulas defined by the four sentences x y, x Vv and x 2 y is called an inequality. These are read “‘x is less than y,” “x is greater

than y,”’ “x is less than or equal to y,” and “‘x is greater than or equal to y,” respectively. In each case the domain is the Cartesian plane ® & ®. The word inequality

§304]

ORDERED

FIELDS

55

is also used for a formula or statement resulting from substituting values for variables in any inequality as defined above. Inequalities of the form x < yorx>y (in distinction to those of the form x < y or x = y)are called strict inequalities. It follows from Definition IVthat the two inequalities x < yand y > xare equivalent and that the two inequalities x < y and y = x are equivalent. The sense of an inequality of the form x < y or x S yis said to be the reverse or opposite of that of an inequality of the form x > y or x = y. The simultaneous inequalities x < y and y < z are usually written together, thus: x < y < z, and the simultaneous inequalities x > y and y > z are usually written Xe Vo aes Lhat,as, (2)

RZ

and =. Since, in formulas (2)-(6), three variables are involved, the domain in each case is implicitly assumed to be the Cartesian product R X R X R. If either (2) or (3) holds for three numbers x, y, and z, the number y is said to be between x and z. Two inequalities of opposite sense should not be combined in the manner of (2 )(6). For example, x < y > zand x > y < zare both combinations of symbols to which we shall not attempt to assign meaning. Example 2. Solution.

Prove thatx

>0@ x € @.

Since, from Definition IV, ce SVeraO

Ee ©,

we have only to show that x — 0 = x, for every x € ®. ties of the type illustrated in §302:

But this follows from field proper-

x—-O=x+(-0)=x+0=x.

Notice that for any x the statement x > 0 is another way of saying “‘x is positive.” Similarly, the statement x < 0 is another way of saying ‘‘x is negative.” Example 3.

Establish the transitive law for each of < and >:

(7)

«Kx=-ZzEP.

THE

56

REAL

[§304

SYSTEM

NUMBER

Each of these is a consequence of the closure of ® with respect to addition. (9), since

(11)

For example, in

GV—-xy+@-y=yt(-—)+2+(-y) = 74+ (—-x) =2-x,

it follows that whenever

y — x and z ~ y are members of @, z — x must also be.

Example 4. Establish the law of trichotomy: following statements holds. (12)

SERS WA

For any numbers x and y exactly one of the

Be NE

Oe IY

Solution. Define z= y — x. Then, from the remarks following Definition III, we can infer that exactly one of the following three statements holds for z:

(13)

“ss,

s=0,

2
y, and x = y, each with domain ® X R:

AX LY)

xX 2 Vs

(x Sy)ax>y, “x>yoxsy, mx

2 y)ex 0.

(vi) The reciprocal of a positive number is positive and the reciprocal of a negative number is negative.

§305]

EXERCISES

57

(vii) The quotient of two numbers is positive if and only if the two numbers are either both positive or both negative. (viii) The quotient of two numbers is negative if and only if one of the two numbers is positive and the other is negative.

(ix) x < Oif and only if x is negative. (x) The sum of any two negative numbers is negative. (xi) Multiplication or division of both members of an inequality by a positive number preserves the order relation: If z > 0, then x < y implies xz < yz and x/z < y/z, and x S y implies xz S yz and x/z S y/z; similarly for Sands.

(xii) Multiplication or division of both members of an inequality by a negative number reverses the order relation: If z < 0, then x < y implies xz > yz and x/z > y/z, and x S y implies xz = yz and x/z = y/z; similarly for >and

305

=.

EXERCISES

In the following exercises, prove the given statement, where all letters designate real numbers. 1. The equation x2 + 1 = 0 has no real root; that is, no member x of ® can satisfy this equation. 2

1 1 Zs — VS — GS

Vier VisS aXe

4.a 0. (ii) If n is a natural number, then n = 1; that is, 1 is the least natural number. (iii) If m and n are natural numbers, then m + nis a natural number; that is, N is closed with respect to addition. (iv) Ifm and n are natural numbers, then mn is a natural number, that is, Nis closed with respect to multiplication. (v) If m and n are natural numbers and m n+

1.

By the induction assumption, and the fact that m => 2, we have m'tl=m-m>nemeand+1)=n+n2n+1,

and therefore (20) is true whenever (19) is. By the fundamental induction (Theorem II, §306) the proof is complete. Example 3.

theorem of mathematical

Prove that ® (or, in general, any ordered field S$) is infinite.

Solution. Assume that & is a finite ordered field, and let x denote its largest member, which exists by Theorem VI. Since 1 > 0, we have by addition of x to both members: x + 1 > x + 0 = x, and consequently $ contains a member x + 1 greater than its greatest member x. This is the desired contradiction to the assumption that $ is finite, and therefore 5 must be infinite. The argument just given establishes the fact that there is no largest real number. In a similar fashion (cf. Ex. 13, §308) it can be shown that there is no largest natural number.

308

EXERCISES

In Exercises 1-6, use mathematical induction to establish the formula for n € 9.

eS wah28 gy waited elegy, Hl

263

Pree

Pea

miriatel) pieced

fail Sieseal is Qn FL" One Cneaa

3.5

3. The sum of the first n odd natural numbers 1, 3,---, (2n — 1) is n?.

4, 124+ 42 +---+ (3n — 2)? = 3n(6n? — 3n — 1). §.1-24+2-34---+

nant

1) = inst

I(r

2).

6. 1-34+2-4+---+

n(n 4+ 2) = n(n t+ 1)(2n+ 7).

7. Prove the following form of the fundamental theorem of mathematical induction (Theorem II, §306): Let k be a fixed natural number and let p be a formula with domain Mt \ {1, 2,--+, k} such that (i) p(k + 1) is true,

q

(ii)

Then

p(n) is true

(OU eeepc

pt.

a

i

p(n+ 1) is true.

That is, p(n) is true for every n © SONAL, Qyeres ck}.

THE

64

REAL

[§309

SYSTEM

NUMBER

8-11, establish the given inequality for the specified values of n € MN (cf.

In Exercises Ex.'7):

Saale 2 nea

OS ag ss Sei

10. 27 > n2,n = S.

o)

11. 3" > 3, n = 4.

12. Illustrate the law of Pascal’s triangle (Theorem XI, §307) forO

Sr 0 for all x of its domain is called positive-valued or strictly positive, and a function fsuch that f(x) < 0 for all x of its domain is negativevalued or strictly negative. Example 3. Let f be the greatest integer function (f(x) = [x]), let g be the absolute value function (g(x) =

|x|), and let / be the zero function (A(x) = 0), all restricted to the domain D = [—2, 2] (cf. Example 2). Then the following inequalities are true:

(21)

if

SG See

The following inequalities are false:

(22)

fah

hss.

Finally, the absolute value of a (real-valued) function fwith domain D is denoted |f| and defined to be the nonnegative function with domain D:

(23)

If] () = [f@)I, x € D.

§505]

EXERCISES 133 Equivalently, ifg is the absolute-value function g(x) = |x| with domain @, then lf] is defined to be the composite of f by g:

(24)

lf| = g°f, domain

Example 4, Since the function f(x) = x2 — x —2 df+ ug, where \ and uw € &.

More generally, a linear combination of the n functions fi, f2,: + , of the form

(4)

fn is any function

Mfi + A2f2 + °°: +Anfns where 1, 2, °°° 5 An E RB.

The first two of the following three theorems are almost immediate consequences of the preceding definitions. The third requires mathematical induction. Theorem I. Every vector space V is closed with respect to (i) the formation of negatives and (ii) subtraction, and contains the zero function:

(aS

Ui

nen,

@) (f,8) © UX U>f —2 = 0, (iii) O€ V. Proof. Part (i) is obtained by letting \ = —1in(2). Iffand g belong to UV, then from part (7), both fand —g belong to VU, and since V is closed with respect to addition, f + (—g) € UV. Since f— g = f + (—g), part (ii) is proved. Part (iii) can be obtained by letting fand g be the same arbitrary member of U(‘U is nonempty) and using (ii).

Theorem II. A nonempty function space © of functions having a common domain D is a vector space ifand only if it is closed with respect to the formation of linear combinations of pairs of its members. (5)

Q)#) © GX

Qand

GF, sre Ul

U0

fe

enn.

§507]

BOUNDED

FUNCTIONS

137

Proof. Assume first that U is a vector space, that \ and uw € ®, and that fand g € U. By (2), both \f and yg € V, and hence by (1) their sum Af + ug € VU. Now assume (5). If \ and yu are both set equal to 1, then (5) states that (1) holds, while if

\ is permitted to be arbitrary and yp is set equal to 0, (5) states that (2) holds. fore is a vector space.

There-

Theorem III. A nonempty function space V0 of functions having a common domain D is a vector space if and only if it is closed with respect to the formation of arbitrary finite linear combinations of its members:

(6)

Nido,An GO and fi,f2,2- >; fs © O NM tan? ae

a An fr 1 0s

Proof. Since (6) implies (5) and (5) implies that U is a vector space, we have only to show that every vector space VU 1s closed with respect to finite linear combinations. Let A be the set of all natural numbers n such that the implication (6) is true. We already know that 1 € A and 2 € A, and wish to establish (7)

NnEASnN+LIEA.

Accordingly, we assume that the implication (6) holds for some particular natural number # and seek to prove as a consequence that this same implication holds for n + 1, or: (8)

M,

WIG? ep

Sift

Nae ntl

ABf2

€

R

oo

AON

[35°

ae Ant

Ss}

ns et

Anti fn

€ x0)

© OU

Since the right-hand sum in (8) can be written

@)

Mie

et

feet = (ATi a>

or Ansa) 4 Anel faqs

by the induction assumption this function is the sum of two members of U and is therefore also a member of U. Therefore the implication (7) is established, and by Theorem I, $306, 4 = % and the proof is complete. 507

BOUNDED

FUNCTIONS

A real-valued function is bounded above, bounded below, or Definition. bounded if and only if its range is a set of real numbers that is bounded above, bounded below, or bounded, respectively.

A similar definition applies to a functionf bounded above, bounded below, or bounded on a set A contained in the domain of f; this is obtained by applying the preceding definition to the restriction offto A. In terms of quantifiers, the three conditions for a real-valued functionf with domain

D to be bounded

above, bounded

below, or bounded, take the following

forms, respectively:

()

D> fG)- 1/x, x > 0, is unbounded above. (Cf. Fig. 507-3.) of

2b

i 2b

o|e

Figure 507-3

Solution.

Negating the existential statement (1) gives the universal statement

(7)

VDC

Gx

D

f(x) Sb,

VRC

Ie) > f(a)

or, in positive terms:

(8)

sd.

COMBINATIONS

140

OF

FUNCTIONS

[§508

For the functionf: x > 1/x, x > 0, (8) becomes

(9)

V bie Reale S 0

1) x8.

Since for positive x the final inequality of (9) is equivalent to xb < 1, we can establish the 6 > 0, choose x = 1/26. In either case, truth of (9) as follows: If 6S 0, choose x = 1, andif x 1s positive and xb < 1.

Of interest within the spirit of the present chapter is the following theorem: Theorem II. The set 8 of all bounded real-valued functions with domain equal to a prescribed nonempty set D is a vector space.

Proof. We wish to show that (/) iffand g € ® thenf+ g¢ Band (i)ifAE R and f© ®thend\f € ®B. Ineach case we use Theorem I for simplicity: (/) If |f(x)| S ¢ and |e(x)| < d for all x € D, then by the triangle inequality (Theorem I(vi/), $311), f(x) + g(x)| S |f0d| + |g@)| S ¢ + d for every x € D, and hence by Theorem I, f+g is bounded. (ii) If [fQ)| Sc, then |Af(x)| = [A|-[ fd] S [Al-e, and Af is bounded. Some of the techniques of vector spaces are exhibited in the proof of the following theorem:

Theorem III. Proof.

Every polynomial is bounded on every bounded interval.

We first show that the function x", where n € 9, is bounded if x is. This

follows from the fact that |x| = |x|" S cif |x| S c(cf. Theorem X, §307, Ex. 31, $313). If Gis the space of all bounded real-valued functions with the given bounded interval J as their common domain, then the functions (10)

DHE

Ne

cs

Gal,

are all members of ®. Since @ is a vector space and every polynomial on Jis a linear combination of the functions of (10), every polynomial on J must be a member of ®, and therefore bounded. Note. IfJisa bounded interval, and if II and @ are the vector spaces of polynomials and bounded functions, respectively, on J, then the preceding Theorem III can be expressed by the inclusion statement

(11)

TT ee:

Example 2. Find an upper bound for the values of |f| where fis the polynomial defined as follows, for the indicated interval:

(12)

fC)

ox ee

Solution. Since |x| S$ 3, we have for |f(x):

(13)

[fOd| S [xP + Slax] +2 52741542 = 44.

Therefore 44 is an upper bound for the values of |f]. 508

EXERCISES

In Exercises 1-6, determine whether the given collection of functions on & is a vector space. Justify your conclusion.

§509]

ABSTRACT

VECTOR

SPACES

141

. All polynomials of the form ax* + bx7, where a and b € &. . All linear functions ax + b, where a and b € J.

. All polynomials vanishing at at least one point. . All polynomials vanishing at 0. . All even functions f(x) such that f(0) = 1. WO ff mn a N =.

All even functions f(x) such that f(1) = 0.

In Exercises 7-12, use (8), §507, to show that the given function is unbounded above on the specified set. 7. x?,x € @.

9.

8. 1/x2,x € @.

1+ x)/x,0 XS) = Avs

(iii)

X € § and(f, 2g) € UX

(jv)

QA wWEFXFandfEe

USAC

+H g) =

fH dg,

VLSAt+ywf=AWt

uf

We shall now formulate the concept of /inear functional for a general vector space, although most applications and illustrative examples will be to vector spaces of realvalued functions.

Definition I]. A linear functional* on a vector space 0 over a field § is a function ¢ on VU into & that has the following two properties:

(i) ¢ is additive: (ii)

fandg€0>¢(f+ 8) = o(f) + 4(g),

¢ is homogeneous:

d € § and fe V=> df) = AP(/).

Nore. It will be assumed henceforth that for every vector space under consideration the field § of Definition II is the real number system @&. With § = ®, a linear functional on a vector space ‘U is simply a real-valued function on ‘U having properties (/) and (ii).

A frequently used property of linear functionals is their subtractivity: Theorem. Jf ¢ is alinear functional on a vector space © over a field §, then & is subtractive there:

(1) Proof.

(2)

fandg €

U=¢(f— g) = ¢(f) — o(g).

Since (f— g) +g =f, and ¢ is additive:

Hf) = $(f — g) + 8) = of — 8) + (8),

and the result follows by subtraction of 4(g). *Also called a linear form.

$509]

ABSTRACT

VECTOR

SPACES

143

Example 1. Let U bea vector space of real-valued functions on a nonempty common main D, and let a be an arbitrary but fixed member of D. For anyf € U, define ¢:

(3)

do-

o(f) = f(a).

Then ¢ is a linear functional since f(a) is a real number, and (i) @ is additive: o(f+ g) =

G+ ga) = f(a) + g@) = $f) + (8);

(ii) > is homogeneous:

(Af) = (f(a) = f(a)

= \d(f). For a specific illustration of the functional of this example, consider the vector space II of all polynomials on the interval J = [1, 4], and let a = 2. If fis the polynomial f(x) = x7 + x — 3 then ¢(f/) = 44 2 —3 = 3, and if g(x) = 2x — 10 then ¢(g) = 4 — 10 = —6. (Cf. Fig. 509-1.) ay

Figure 509-1

Example 2. Let U be a vector space of real-valued functions on a common domain D consisting of at least two distinct members, and let a and 4 be any two distinct members of D, to be held fixed. For anyf € U, define ¢:

(4)

of) =f@ +f).

Then ¢ is a linear functional since f(a) + f (4) is a real number, and (/) ¢ is additive:

¢(/+ g)

= (f+ gia) + F+8)6) = f@ + s@ +f) + 8) = F@ + FO) + [g@ + | = o(f) + o(g); Gi) @ is homogeneous: ¢AS) = AN@ + AAO) = A\F@ + AF) = A f(@ + f(d)] = A¢(f). For a specific illustration of the functional of this example, consider the vector space of all linear functions on the interval [1, 3], and let a = 1 and 5 = 2. If f is the function 2x — 3 then $(f) = (2 — 3) + (4 — 3) = 0, and if g(x) = 3x +5

then $(g) = (3 + 5) + (+ 5) = 19. Definition III. Let U be a vector space of real-valued functions. A linear functional ¢ on V is positive, written ¢ > 0, if and only if

(5)

20 f 20>¢(f)

and ¢ # 0, that is, there exists an f such that $(f) # 9. Example 3. The functional ¢ of Example 1 is positive since the inequality f 2 0 implies the inequality f(x) = 0 for all x of D, and hence the inequality ¢(f) = f(@) 2 0; and furthermore, if f is the constant function 1, ¢(f) = 1 > 0. f2 0 implies Example 4. The functional ¢ of Example 2 is positive since the inequality both f(a) = 0 and f(6) = 0, and hence ¢(f) = f@ + f() 2 0. Furthermore, if f is the constant function 1, ¢(f) = 2 > 0.

COMBINATIONS

144 510

THE SIGMA

OF

[§510

FUNCTIONS

FUNCTIONAL

In §509 two examples are presented to show that if, for any vector space of realvalued functions on a set D that consists of at least two points, ¢(f) is defined to be either the value of fon a fixed point of D or the sum of the values of fon two distinct fixed points of D, then ¢ is a linear functional. This idea can be extended to a domain D consisting of at least n distinct points, for any givenn € 9, with the values of f being added at a prescribed set of n distinct points of D. In the present section we shall be concerned with this general idea, but only for the specific case where D is the set of nonnegative integers: D = U {0}. For present purposes let

(1)

D=NU

{0},

and let U be the vector space of al/ real-valued functions with domain D. Let A = {m,m-+ 1,---,n} be any nonempty set of consecutive integers in D and define, for any fe U:

(2)

o(f) = fm) + flm + 1) +°°+ fn).

Then the real-valued function ¢ with domain VU is a linear functional.

(The proof

follows the lines of the two examples of $509, with the aid of the general associative, commutative, and distributive laws of $307.) This linear functional is called the sigma functional, with the notation:

(3)

b=

= De

= ss

AOL:

The notation (3) is called the sigma summation notation, or simply the sigma notation. 6 Example 1. If ¢ = YS and f(x) = x2, then 3

, DO) oa

(4)

SP

5

Gs 2

10

Example 2. If ¢ = >) and f(n) = 2n — 1, then 1

(5)

10

of) =>

Only

21 499455

E19

S100)

n=1

0 Example 3. If ¢ = DS and g(k) = k!, then 0 0

(6)

$(g) ae kt! =O!=1. =O)

There are two special comments .

be made:

.

regarding the linear functional >) that should .

i

o

(i) The letter k that appears in the last part of (3) may be replaced by any other

letter or symbol that is not already being used; it is what is called a dummy variable:

§510]

THE

()

SIGMA

FUNCTIONAL

145

I= S1o= Ssfo= (ii) If fis the constant function 1, then ¢(f) = ¢(1) can be explicitly evaluated:

tet

Dati

(8)

ian—m41.

In obtaining the results of the following examples it will be helpful to have the theorem: Theorem.

For any natural number n,

(0)

Sw ~flk = n=) — FO.

Proof.

The left-hand member of (9) is

Ol

Ol Drs

).— fn = 11,

which can be rewritten

Ce

(i ete |fre Bl) Ke

Dict

=a Lf(1) — fO)I,

or

et

=)

Ee Nf

2) by = 2h

[= fO)],

which is equal to the right-hand side of (9). Several useful summation formulas will now be derived in the following examples,

using the linearity of the functional >>, property (8), and the preceding theorem. Example 4. Derive the formula

(10) Solution.

Dk H=1t24--:+2 = Gn 4 1).

&=1

If f(n) = n?, formula (9) becomes

(11)

> [k2 — (k — 1)7] = ye (2k —4) = nr’. k=1

By the linearity of pes and property (8), (11) can be written: T

(12)

Dy

=

ka n= 1

Solving for >> k gives (10). k=1

Example 5. Derive the formula

(13) :

Yi

k=1

= PE BH

+

= alt

Qn + I.

Solution.

FUNCTIONS

OF

COMBINATIONS

146

(§511

Let f(n) = n3. Then formula (9) gives n

(14)

ss [k3 — (k — 1)3] = > Bk? — 3k +1) = 23, k=1

k=1

and hence, by properties of >) and formula (10): I

(15)

305-3 k=

> k+n=3) k=1

_

kK —Snnt+ lI t+n=e,

k=1

3D)virkt = nif? + 8 + 1) — 1) = tn? k

+ 3n + 1);

_

and‘(13) follows. Example 6. Derive the formula of Example 1, §307:

(16)

12 + 32 +.--+ (2n — 1)? = gn(4n? — 1).

Solution.

The left-hand member of (16) can be written ye, (2k — 1)?, which by the linearity k=1

nr

of Ds can be expressed I

(17)

3 (4k2 — 4k + 1) = 4 kz — 4k +> ie

By formulas (8), (10), and (13), this becomes

(18)

n(n + 1)(2n + 1) — 2n(n + 1) + 2 = Gn(4n2 + 6n + 2 — 6n — 6 + 3),

which simplifies to the right-hand member of (16).

511

EXERCISES

1, Let U be the vector space consisting of all real-valued functions on an interval /, and let a, b, and c be any three distinct points of J, to be held fixed.. For any f€ V, define ¢:¢(f) =f@+f) +f. Prove that ¢ is a positive linear functional.

In Exercises 2-5, evaluate the given sum.

k=2

n=0

5

if

4 t=22

58) Giese) j=l

6. Show that the formula (1), §307, of the general distributive law, Theorem II, §307, can be written n

n

XLV k=1

k=1

eV:

7. Show that the formula (14), §307, of the binomial theorem, Theorem XII, §307, can be written

@+y=)>) k=0

(;)aen—k yk,

§512]

STEP-FUNCTIONS

147

8. Show that the formula (2) of Exercise 14, §308, for the sum S of an arithmetic progression can be written

S= 2, (4 + (k — 1)d) = na + An(n — 1d. 9. Show that the formula (4) of Exercise 15, §308, for the sum S of a geometric progression with r ~ 1 can be written n

S=

>> af!

=

a — ar"

k=1

:

l—r

In Exercises 10-12, derive the formula. n

n

2

10. 5) 3 = 4n%(n+ 12 = (x) ; k=1

11. >> k4 = Ayn(n + 1)(2n + 1)Gnr? + 3n — 1). 12. >) AS = Ayn2(n + 1)2(2n? + 2n — 1). In Exercises 13-15, use the methods of §510 to derive the formula of the specified exercise of §308. 13. Exercise 4.

14. Exercise 5.

15. Exercise 6.

In Exercises 16-19, derive a formula for the given sum.

16. 2? + 524+ 82+.---+

(3n — 1).

17. 1-24+ 2-34+3-4+---+n(n+

1).

18.12 9)4222.3 432-4 et n2(n +1). 19, 1-22 4°2-32'43:42 +--+ n(n +1). 512

STEP-FUNCTIONS

In this section we shall discuss a vector space that is of especial importance in mathematics because of its fundamental role in the theory of integration. This vector space is the function space whose members are step-functions, which are defined on compact — that is, bounded closed — intervals in terms of nets: Definition I.

Jf J is a compact interval [a, b], a net N on I is any finite subset of I

that contains a and b, with the notation: (1)

Nios (do, Gita

2

IV

I,

where ao = a, a, = b, and the points of the net are arranged in increasing order: (2)

C=

dg

Cay

0, then A fis also increasing (or decreasing) on J. Show that if fis increasing (or decreasing) on J and if \ < 0, then Afis decreasing (or increasing) on J. function there.

Solution. If f and g are increasing functions on J, then f+ g is also, since x1 < x2 => (f+ g)(x1) = fxr) + g(x1) S f(x2) + g(x2) = (f+ g)(x2). If, in addition, at least one of these functions is strictly increasing on / a strict inequality < results, and f+ g is strictly increasing. Similar remarks apply to the sum of two decreasing functions: the sum is also decreasing, and if in addition one of the terms is strictly decreasing so is the sum. For the statements concerning scalar multiples a typical case is sufficient. Assume that fis increasing on J and

§514]

MONOTONIC FUNCTIONS

153

that \ < 0, and let x; < x2. Upon multiplying both members of the inequality f(x1) S f(x2) by the negative number

)\ we obtain Af (x1) 2 Af (x2) (cf. property (xii), §304), and hence

OPS)1) = OAf)(x2) and df is decreasing on I.

Although monotonic functions are important individually in mathematics, they lack a completely satisfactory structure as a class. For example, the space of monotonic functions is not a vector space since it is not closed under addition, as the following example shows: Example 3. Show that although both x? + x and —2x are monotonic on the interval (0, 1], their sum is not.

Solution. By Examples 1 and 2, x? + x is a strictly increasing function on [0, 1], and by elementary considerations —2x is strictly decreasing there. However, the sum of these two functions is f(x) = x? — x, which is neither increasing nor decreasing on (0, 1] (cf. Fig. 514-2): x? — x is not increasing, since 0 < 4 and f(0) = 0 > f(4) = —1; x2 — x is not decreasing

since 3 < 1 and f(4) = — i < f(1) = 0.

af

O

‘

i os ASees 4

x

1

2

ON seh 7

Figure 514-2

Example 4. Show that the product of two increasing functions on an interval may not be monotonic there. Show that the product of two nonnegative increasing functions on an interval is increasing there. Comment on the product of two nonnegative decreasing functions. Solution. The function x and the function x are both (strictly) increasing on the interval [—1, 1], but their product x? is neither increasing nor decreasing there. Assume now that both fand g are increasing on J and are nonnegative there. If x1 < x2 (x1 and x2 € J), then multiplication of both members of the inequality f(x1) S f(x2) by the nonnegative number 2(x1) gives f(xi)g(x1) S f(x2)g(x1). Similarly, multiplication of both members of the inequality g(x1) < g(x2) by the nonnegative number f (x2) gives f(x2)g(«1) S f(x2)g(x2). Combining these inequalities gives f(x1)g(x1) S f(x2)g(x2), as desired. A similar proof shows that the product of two nonnegative decreasing functions is decreasing.

Norte 2. If ¢ is a positive linear functional on a vector space VU of real-valued functions, then ¢ has the following property:

(6)

(f,8)€

UX Vandfs

g>P/(/) S G(Q).

(The reason is that if ¢ is positive andf S g, then ¢(g — f) = ¢(g) — ¢(f) 2 0.) It can also be shown (cf. Ex. 21, §516) that a nonzero linear functional ¢ satisfying (6 ) is positive. For this reason, by analogy with (1) of this section, positive linear functionals are also called in-

_creasing linear functionals.

COMBINATIONS

154

OF

[$514

FUNCTIONS

Example 5. Find a lower bound and an upper bound ) x? — 4x + 11 on the interval {—1, 2].

for the values of the function

Solution. We analyze the given function term by term. In the first place, the function x? is monotonic on (0, 2], and 0 S x? S 4 there, and x? is monotonic on [—1, 0], and 0 S Xe there. Therefore, on the interval [—1, 2], 0 < x2 S 4. The function —4x is monotonic on (—1, 2],and —8 < —4x < 4there. We therefore have the following sets of inequalities, valid on [—1, 2]: O) See! ss 4 —8

s —4x

$4,

job es iiss i, Addition of corresponding members of these inequalities gives

Seek

a

Lies 9:

Therefore 3 and 19 are a lower bound and an upper bound, respectively, for the values of the

given function on the interval {—1, 2].

Example 6. Find an upper bound for the values ofthe function |x3 — x — 8| on the interval

Pi

Solution. Since the functions x3, —x, and —8 are all monotonic on the interval [—1, 2], we have the inequalities

and therefore —ll*

|f@)issie-and.|e(x)|' Sd.

Therefore x € D= |(fg)(x)| = |f@)g(x)| = [fC)|-le@)| S cd, and fg € B. (Cf. Theorem I, $507.)

Theorem II. algebra.

Jf I is a compact interval, the space § of all step-functions on I is an

156

OF

COMBINATIONS

[§515

FUNCTIONS

Proof. If f and g are any two step-functions on J, let $ be a net for which bothf and g are step-functions. Then for any open interval of $, since both f and g are constant there, so is their product fg. Example.

If fand g are the step-functions of Example 2, §512, given by the tables:

Occasionally it is convenient to have available the following theorem as an aid in establishing the fact that a certain vector space of functions is also an algebra:

Theorem III. 4 vector space V of functions is an algebra if and only if it is closed under squaring; that is, if and only if (3)

i SO fA

©.

Proof. Since (3) is a special case of (1) with g = f, any algebra of functions is closed under squaring. Now assume (3) and write one-fourth the difference of the

expansions (f-+ g)? = f2 + 2fg + g? and (f — g)? =f? — 2fg + g2 in the form

(4)

Ser

8) rea

hea

2

Since U is assumed to be a vector space that is closed under squaring we have the two implications

(5)

f and.g © UR

Gd 2g)2-6.0,

(6)

fandge€v>(f-

gyre.

Consequently, whenever f and g belong to U, so does the right-hand member of (4), and hence U is closed under multiplication and the proof is complete. Norte. Much as the concept of vector space can be abstracted (§509), so can that of algebra. An algebra is a vector space UO over a field $ of scalars on which there is defined a binary operation of multiplication, where the product of members f and g of © is denoted fe, such that the following homogeneous and distributive laws hold:

(i)

(Au) € FX F and (f,g)E

UVXV>Af)ug)

= Qw(fe),

§516]

EXERCISES

i)

fadhena

157

vet Daf th (f+ gh = fh + gh.

An associative algebra is an algebra © for which the following associative law holds:

(ii)

f, g, andh € U0 =

f(gh) = (fg)h.

A commutative algebra is an algebra ‘© for which the following commutative

(iv)

law holds:

fand g€ U= fe = gf.

516

EXERCISES

In Exercises 1-4, show that the given function is monotonic on the given interval.

1. sgn x (cf. Example 2, §115), [a, 5].

2. [x] (cf. Example 3, §115), [a, 5}.

3. x2 + 5x — 6, [0, 6].

4. —2x3 — 7x + 12, (—3, 2).

In Exercises 5-8, show that the given function is not monotonic

on the given interval.

5. |x|, [—8, 8].

6. x2 — 12x — 6, (5, 8).

7. |x2 — 8x|, [6, 10].

8. x3 + 100x2 — x, (0, 100).

In Exercises 9-12, find a lower bound and an upper bound for the values of the given function on the specified interval.

9. x3 — 4x + 20, [1, 4].

10. x3 — 5x — 30, (1, 3).

11. x? + 3x — 10, (—2, 1).

12. 25 — 2x — x2, [—2, 3].

In Exercises 13-20, find an upper bound for the values of the given function on the specified interval.

13. |x? — 8x + 10], [1, 4]. 15. |x3 + 5x2 — 5], (—2, 1).

14, |x2 — 3x + 1], (1, 3). 16. [14x — 3x2 — x3|, [—2, 3].

17.

at

[1, 4] (cf. Exs. 9, 13).

18.

et

(1, 3) (cf. Exs. 10, 14).

19. a.

(—2, 1) (cf. Exs. 11, 15).

20. 2)

[—2, 3] (cf. Exs. 12, 16).

21. Prove that a nonzero linear functional ¢ satisfying (6 ), §514, is positive. 22. Use mathematical

induction

to prove

that if U

is an algebra

of functions,

then

In Exercises 23-26, describe the step-function fg, where f and g are the step-functions defined in the indicated exercise of §513, by means of a table. Draw graphs of f, g, and fg. 23. Exercise

1.

24. Exercise 2.

25. Exercise 3.

26. Exercise 4.

158

COMBINATIONS

OF

FUNCTIONS

[§516

In Exercises 27-30, prove the given statement about the composite fo g of two monotonic functions, where the range of g is a subset of the domain off. Show that in each case the adverb strictly can be included. 27. Iff and g are increasing, then fo g is increasing. 28. If fand g are decreasing, then fo g is increasing. 29. If fis increasing and g is decreasing, then fo g is decreasing. 30. If fis decreasing and g is increasing, then fo g is decreasing.

Conic

601

Sections

INTRODUCTION

The mathematical curves known as conic sections, or simply conics, were originally defined and studied by the ancient Greeks in terms of the literal interpretation of the words conic and sections, that is, as (plane) sections of (right circular) cones. The principal types of curves obtained in this fashion are known as parabolas, ellipses, and hyperbolas, although circles as special cases of ellipses, and pairs of straight lines as limiting cases of the other curves, are usually considered to be members of the general family of conics. Since their original introduction, conic sections have proved to be of great interest in themselves, and of the utmost importance scientifically. We mention in passing the central role of conics in planetary, satellite, and stellar motion as studied in astronomy and space physics, and the reflecting properties of these curves and their uses in optics. In order to obtain results more quickly and efficiently than is possible when conic sections are defined and treated as sections of cones, we shall depart from the classical approach and instead define and study the various conic sections as particular types of Joci. For a proof of the equivalence of these two approaches to conic sections, see C. B. Morrey, Jr., University Calculus with Analytic Geometry (Reading, Massachusetts, Addison-Wesley Publishing Co., 1962), pp. 268-270. In this chapter all loci will be assumed to be in the Euclidean plane. 602

THE PARABOLA

Definition. 4 parabola is the locus of a point that is equidistant from a fixed line D called the directrix of the parabola and a fixed pointf not on D called the focus of the parabola.

159

160

CONIC

SECTIONS

[$602

+c = 0 and if the If the directrix D of a parabola C has the equation ax + by C consists of all parabola focus is the point (x1, y1), then by (1) and (3), §413, the points (x, y) such that lax + by + c¢| = V(x — x1)? +O — yi)? Va2 + b2

(1)

Since equation (1) is cumbersome to handle, we choose, for a given parabola C, a coordinate system that gives a much simpler equation. A particularly appropriate set of axes is obtained by letting the x-axis be perpendicular to the directrix D and pass through the focusf, directed positively from D towardf, and by letting the y-axis be parallel to D and halfway between D andf. (Cf. Fig. 602-1.)

Figure 602-1

Letting the coordinates of f be (c, 0), where c > 0, the equation of D becomes

x = —c,orx+cz=0.

Therefore the distance of a point (x, y) from D is |x + c|

and its distance fromf is \(x — c)? + y2, so that equation (1) in this coordinate system is

(2) By Exercise

Poaebetale

Oe au

7, §305, equation (2) is equivalent to that obtained by squaring:

OnE) Pe

0) Taye,

which simplifies to (3)

y? = 4cx.

The parabola of Figure 602-1 is said to “open to the right.” Similar derivations hold in the following cases, where c is assumed to be positive: (i)f= (—e, 0) and D is the line x — c = 0 (cf. Fig. 602-2a); (ii)f= (0, c) and D is the line y + c = 0 (cf. Fig. 602-25); (iii) f= (0, —c) and D is the line Vi C20 (cf. Fig. 602-2c). (Give derivations of the equations appearing in Figure 602-2, in Ex. 23, §604.)

$603]

VERTICES

DIFFERENT

y

FROM

THE

ORIGIN

161

ey

Vy

D

f (0, c) ,0 (= ¢,{ ) Vo

x

v

3

f x=c

y=-c

D

y?= —4cx

x= dey

x? =—4cy

Parabola opening to the left

Parabola opening up

Parabola opening down

(a)

(b)

(c)

Figure 602-2

For any parabola the line through the focus and perpendicular to the directrix is called the axis. From (3) we infer that the axis of any parabola C is a line of symmetry of C. The point on the axis of a parabola halfway between the focus and the directrix is called the vertex of C. In the parabola (3) and in the parabolas of Figure 602-2 the vertex is the origin in each case. Example 1.

Find the focus and the directrix of the parabola x2 = 20y.

Solution. The equation x2 = 20y is of the type x? = 4cy of Figure 602-245, with c = 5S. Therefore the focus is (0, 5) and the directrix is y = —5. The parabola ‘‘opens upward.” Example 2.

Write the equation of the parabola having focus (— 3, 0) and directrix x = 3.

Solution. This parabola is of the type illustrated in Figure 602-2a, therefore has equation y2 = —12x.

603.

VERTICES

DIFFERENT

FROM

with c = 3, and

THE ORIGIN

Consider the parabola C with vertex at the point (A, k), with focus c units to the right at the point (h + c, k), and with directrix vertical and c units to the left with equation

x =

h—c,orx—h+c=0.

(Cf. Fig. 603-1.)

Then from (1), $602,

the equation of C is

(1)

Pie) — NG Se ee (irk).

Squaring gives

Ce

2c)

= (x hh)? — 2x — hb) ce? + (y — ky,

or, equivalently,

(2) for the equation of C.

(y — k)? = 4c(x — h)

162

CONIC

[§603

SECTIONS

oy

ie (hk) (ht c,h) x=h—c

(y — k)’we l| 4¢(x —h) Figure 603-1

Similar derivations hold for parabolas opening to the left, up, or down, as indicated in Figure 603-2, where c is assumed to be positive in each case. (Give derivations of the equations appearing in Figure 603-2, in Ex. 24, § 604.) Example 1.

Write the equation of the parabola with focus (5, —2) and directrix y = 4.

The vertex v must be on the vertical line x = 5, with y-coordinate the average of Solution. —2and 4. Thereforev = (h, k) = (5, 1). Since c must be equal to 3, the equation of Figure 603-2c gives (x — 5)? = —12(y — 1), or x? — 10x + 12y +13 = 0. (Cf. Fig. 603-3.)

If equation (2) or any of the equations of the parabolas of Figure 603-2 is expanded the resulting equation in the two variables x and y has the following two properties: (i) (ii)

There is no term involving the product xy. The equation is quadratic in one of the variables and linear in the other. yy

PS

y

ipa e

(h, k + c)

; y=kr+e

/

t

v|(h, k) U\(h, k)

~G

Oe

x

iS

O

words

(hk —c)

(y—k)?=—4c(x—h)

(x—h)?=4e(y—k)

; (x—h)°=—4c(y—k)

Parabola opening to the left

Parabola opening up

Parabola opening down

(a)

(b)

(c)

Figure 603-2

x

$603]

VERTICES

DIFFERENT

FROM

THE

y

ORIGIN

163

y

(a)

(b) Figure 603-3

In Chapter 9 (cf. §910) we shall study conic sections by means equation of the second degree, having the form: (3)

of the general

ax? + 2cxy + by2 + 2px + 2qyv+d=0.

We can now state a converse to the preceding statement about (7) and (ii). If an equation of the form (3) satisfies (/) and (ii) its graph is a parabola. The technique of showing this combines completing a square (cf. §413) and factoring a constant, and is illustrated in the two examples that follow. Example 2. Solution.

Describe in detail the graph of the equation x? — 6x — 8y — 7 = 0. We complete a square with the variable x:

(4)

Cer

Ot)

=e)

and factor the constant 8 on the right:

(5)

Ca 32 = 80

2):

By comparison with the equation for Figure 603-24, this represents the parabola that has vertex (3, —2) and opens upward (8 is positive) with the distance between the vertex and either the focus or the directrix equal to 2(8 = 4c). Therefore (cf. Fig. 603-35), the focus is (3, 0) and the directrix is y = —4. Example Solution.

3.

Describe

in detail the graph

of the equation

3y? + 15y — 9x —1=0.

Division by 3, completing a square with y, and factoring out the constant 3 give

(6)

(y? + Sy + 48) = 3x +34 38,

(7)

(ye)? = 3c +38).

Therefore the graph is the parabola that has vertex (—4%, — 3) and opens to the right with

c = tr: 8. Its focus is thus so) 1

aleo

(—$2 + 3, — $) = (—1, — 8) and its directrix is x = —3% — 2

[$604

SECTIONS

CONIC

164

The following facts are almost immediate consequences of the basic principles set Norte. forth above: The graph of the quadratic function y = f(x) = ax? + bx + c, wherea ¥ 0, isa parabola. This parabola opens up if a > 0 and opens down if a < 0. The vertex of this : iy b b 4ac — b? ee parabola is the point sare — = - Similar statements hold when

2a

2a ssa

the variable x is given as a quadratic function of the variable y.

604

EXERCISES

A graph should accompany each of the following exercises.

In Exercises 1—4, find the focus and the directrix of the given parabola. Ire

0y:

In Exercises

We WE = Sy. 5-16, write the equation

3. y? =

=x.

of the parabola

4. x2 =

—12y.

satisfying the given conditions.

5. Focus (0, 6), vertex (0, 0).

6. Focus (—5, 0), vertex (0, 0).

7. Directrix x =

8. Focus (0, —4), directrix y = 4.

—1, vertex (0, 0).

9. Vertex (0, 0), focus on the x-axis, passing through (2, 5). 10. Vertex (0, 0), horizontal directrix, passing through (2, —5).

11. Vertex (—1, 4), focus (2, 4).

12. Vertex (3, 1), directrix

13. Focus (2, —1), directrix x = 6.

14. Focus (—5, 3), directrix

y = 0. y =

—1.

15. Vertex (—3, 1), vertical directrix, passing through (0, 0). 16. Vertex (2, 4), focus on the line x = 2, passing through (0, 0). In Exercises 17-22, find the vertex, focus, and directrix of the given parabola.

17.

y2 — 20x + 14y — 11 = 0.

18. x2 + 16x — 36y — 260 = 0.

19. x2 + 18x + 14y + 67 = 0.

20. y2 + 22x

— 8y + 60 = 0.

21. 3y2 + 16x + 36) + 92 = 0.

22. 5x2 — 100x + 28y + 556 = 0.

23. Derive the equations of Figure 602-2. 24. Derive the equations of Figure 603-2. 25.

Prove

the statements

of the Note,

§603,

concerning

the function

ax? + bx +c.

In Exercises 26 and 27, use the definition of a parabola and equation (1), §602, to find the equation of the parabola with the given focus and directrix. Simplify the equation to a form without radicals. 26. Focus (4, 4), directrix x + y = 0. 21. Focus (—%, —+%), directrix x-+ y = 4.

28. A cable of variable density hangs in the form of a parabola with vertical axis of symmetry. If the ends of the cable lie at the endpoints of a horizontal segment 300 feet long and the lowest point of the cable is 100 feet below the midpoint of this segment, what is the vertical distance to the cable from the points of trisection of this segment?

§605]

THE

ELLIPSE

165

29. A parabolic arch with vertical axis of symmetry has a horizontal base of length 2a and altitude 4. Find the equation of the arch in an appropriately chosen coordinate system. 30. The latus rectum of a parabola C is the chord of C through the focus of C perpendicular to the axis of C. Prove that the length of the latus rectum of a parabola is twice the distance between its focus and directrix. 31. Prove that a parabola with the origin as focus and the x-axis as axis of symmetry must have an equation of the form y2 = 4c(x + c),c ¥ 0.

605

THE ELLIPSE

Definition. An ellipse is the locus of a point the sum of whose distances from two fixed points is a constant. The two fixed points are called the foci of the ellipse.

If the two foci of an ellipse are identical, the ellipse is a circle.

We shall assume

henceforth in this chapter that the ellipses under discussion are noncircular, that is,

that the two foci are distinct. For simplicity we shall choose a coordinate system so that the two foci lie on the X-axis equidistant from the origin, with designations:

fi:(c,0), f2:(—c, 0), where c > 0. (1) Since for any point p in the plane other than a point on the x-axis between f; and f2, the sum of the distances of p from fi and from /2 is greater than the distance 2c between these two foci, we shall denote the constant sum of the two distances in

the definition of an ellipse by 2a, where

(2)

Gc.

(By means of (2) we make explicit our assumption that an ellipse is not simply the line segment joining the two foci.) Accordingly, the equation of an ellipse in the coordinate system chosen assumes the form

(3)

Vx — 0)? + y2 + VO + €)? + y? = 2a.

(Cf. Fig. 605-1.)

Figure 605-1

[$605

SECTIONS

CONIC

166

It is our purpose now to simplify equation (3) to a form without radicals. The simplest procedure manipulatively is to transpose a radical, square, and after simplifying the resulting equation, to arrange terms with the single remaining radical on one side, and square again. We shall not follow this procedure since, in general, squaring does not yield a result equivalent to the equation (or inequality) upon which the squaring is applied, and the justification fer squaring for this particular method is difficult. Although the algebraic manipulations are more awkward, we shall proceed by squaring the two members of (3) as they stand. This first squaring step is justified since the two members of (3) are nonnegative (cf. Ex. 7, $305). The result can be written

(x2 — 2ex + €? + y2) + (x2 + 20x +e? + y?)

aPC? Mee

a eo

od

Cp

a eye

= len,

or after simplification, (4)

4c2x2 =

VO? + y2 + c?)2 —

2a? —

(x? + y? 4+ c?).

Squaring the two sides of (4) gives, after simplification:

a(x2 + y2 + ¢2) = c2x2 + af,

(>)

We pause now to show that equations (4) and (5) are equivalent. In the first place (4) implies (5) since u = v always implies u2 = v2. To show that (5) implies (4) we show that (5) implies the inequality (x? + y? + c2) < 2a? in order that we may infer that the squaring of (4) involves only squaring of positive quantities. Accordingly, we assume (5) in the form (a? — c2)x2 — (a? — c?)a2? = —a?’y? to infer (a2 — c?)x? S (a? — c*)a2, and consequently, since a2? — c2? > 0, x? S a2. Therefore, from (5), a2(x2 + y2 + c?) = c?x2 + a4 S c?a2 + a4, and hence

tee) eel ee

Coe

eed tn,

as desired. We continue by writing (5) in the form

(6)

(a? — c2)x? + ary? = aX(a? — 2),

and therefore, with

(7)

b=

GPa?

bb? = a2 = e2, 2b2 4 et =a

the equation of the ellipse simplifies to

(8) or

9 (9)

b?x? + ary? = a2b?2,

tone ptt

|

It is clear from the form of equation (9) (cf. §115) that the ellipse (9) is symmetric with respect to each coordinate axis and the origin. The points (—a, 0) and (a, 0) are called the vertices of the ellipse (9). The segment joining the vertices is called

$605]

THEDELEIPSE

167

the major axis, the segment joining the points (0, —b) and (0, b) is called the minor axis, and (0, 0), the common midpoint of the major and minor axes is the center of (9). The terms major axis and minor axis are also used to designate the lengths of the corresponding segments just defined, 2a and 2, respectively. The numbers a and 6 are called the major semiaxis and the minor semiaxis, respectively. The Pythagorean relationship among the numbers a, 5, and c (a being the hypotenuse) is shown in Figure 605-2a. If the foci are taken on the y-axis, at the points (0, c) and (0, —c), a derivation similar to the one given above leads to the equation (cf. Fig. 605-2): tv

Hy) to

(0, b)

vertex / focus @

2 2 ae

ty

(0, y —b)

vertex

(a)

|(0, —a)

(b) Figure 605-2

The numbers a, b, and c are again

In this case the vertices are (0, a) and (0, —a).

related by (7), and the major and minor axes and semiaxes are defined by analogy with the definitions given above. Example 1. axis 6. Solution.

Find the equation of the ellipse having foci (2, 0) and (—2, 0) and major

With the notation used above, c = 2,a = 3, and b? = at —c?

2

2

=9 —4=5.

é

Therefore the equation is = + = = 1. (Cf. Fig. 605-32.)

Example 2.

Solution. (11)

Find the vertices, major axis, minor axis, and foci of the ellipse 6x2 + 5y? = 90.

Division by 90 puts the equation in the form x2 2 1s + ee

1

CONIC

168

[$606

SECTIONS

y

oy,

V5

x

—3

x

3

=Vi5

—V5 oe

ce

[Ope

5

(a) Figure 605-3

Either from (11) or from the fact that the x-intercepts are +15 and the y-intercepts are se vi8 (as can be seen directly from the original equation), we know that the foci are on the

y-axis, and that a = 18 and 6 = 15. Therefore the vertices are (0, + «/18), the major axis is the segment (0, —18)(0, 18), and the minor axis is the segment (—V15, 015, 0). To find the foci we use (7) and get c2? = a? — b? = 18 — 15 = 3,andc = V3. Therefore the

foci are the points (0, + 3). (Cf. Fig. 605-35.) Norte. The definition of an ellipse given in this section is based on the following property: If two pins are placed in a sheet of paper at a distance 2c, if a closed loop of string of length 2c + 2a is placed around the pins, and if a pencil point is placed within the loop, then the curve traced out by the pencil moving so as to keep the loop taut is an ellipse with major axis 2a and foci at the two pins.

606

ECCENTRICITY

The eccentricity e of an ellipse is defined to be the ratio of (/) the distance between its foci to (i) the length of its major axis. In the notation of §605,

(1)

pags 2a

1s

a

a

a

Since 0 < c < a, e satisfies the inequalities

(2)

O

oral:

A circle can be considered as a special case of an ellipse with eccentricity zero. In this chapter we are eliminating circles from consideration, and therefore we shall adhere to the strict inequalities (2).

$607]

CENTERS

Example 1.

DIFFERENT

FROM

x2

2

5

7

Solution. In the notation of (10), §605, e? = c*/q? = 2/7. Therefore e = 2 = 114.

Solution.

ORIGIN

eal a2 =7

and

42 = 5, so

that

c2 =2

and

Write the equation of the ellipse with eccentricity } and foci (+3, 0).

In the notation

of §605,

c = 3, and therefore

b? = a? — c? = 36 — 9 = 27, the equation can be written

607

169

Find the eccentricity of the ellipse

(3)

Example 2.

THE

CENTERS

DIFFERENT

FROM

a = c/e = 3/3 = 6. D + = = Il

Since

THE ORIGIN

If an ellipse has center at the point (A, k) and foci on a line through (A, k) parallel to one of the coordinate axes, then derivations similar to those of $605, with x and y replaced by x — A and y — k, respectively, lead to the equations given in Figure

607-1. If either of the equations given for the ellipses of Figure 607-1 and put in the form (1)

ax? + 2cxy + by? + 2px + 2qvy

is expanded

+d=0

of the general equation of the second degree (the coefficients a, b, and c in (1) are not to be confused with the constants a, b, and c of Figure 607-1) the following three properties of the resulting equation appear:

vertex | (h, k + a)

focus

| SN

(Ah, k +)

vertex

(h + a, k)

(h +¢,k)

(h—a,k) O

x

b

ee

(h, k —c)

k — a) vertex! (h,

O

(x —h)? ie (y—k)? Mb a2

focus

(x

oo

ib) res

y=

(a)

ee ZA

a

(b) Figure 607-1

ee

x

CONIC

170

(‘) (ii) (iii)

[$607

SECTIONS

There is no term involving the product xy. The equation is quadratic in each of the variables x and y. The coefficients of the terms in x? and in y? are unequal and of the same sign.

As in §603, for parabolas, there is a converse to the preceding statement: If an equation of the form (1) satisfies (/)-(i/i), and if its graph consists of more than one (real) point, then its graph is an ellipse. The procedure is again that of completing squares, and is illustrated in Example 2, below. Example 1. Write the equation of the ellipse with center (—2, 5), one focus at (—2, 7), and one vertex (—2, 2).

Solution.

With the notation of §605,

adapted to the present case, a = |5 — 2| = 3 and

c = |7 — 5| = 2. Therefore b? = a? — c? = 9 — 4 = 5, and the equation of the ellipse is :

;

given by the equation of Figure 607-16:

as

2)2 — 5) . ) aa se 9 ) = 1. (Cf. Fig. 607-2a.)

(—2, 2) ie.

x

(a)

O

f

(b) Figure 607-2

Example 2.

Describe the curve

(2) Solution.

(3)

x? + 5y? + 6x — 20y — 151 = 0. We

prepare

CZ

to complete

Ow

squares,

factoring

Ary

= Ay

5 from

both terms

involving y:

) = 151.

Supplying the missing terms 9 and 4 within the parentheses is equivalent to adding 9 + 5:4 = 29, and therefore the quantity 29 must be added to the right-hand member of (3) as well, giving

(4)

(53)?

5 =)

ed)

Division by 180 gives an equation of the form of that of Figure 607-la, with a= 180 = 13.42 (to two decimal places), b = 6, h = —3, and k = 2. Therefore (2) is an

$608]

EXERCISES

171

ellipse with center (—3, 2), vertices (—3 + 180, 2), major semiaxis 180 = 65, and minor semiaxis 6. Since c? = a? — b? = 180 — 36 = 144, c = 12 and the foci are (33

"0:

Again, we make explicit our assumption that a hyperbola does not consist simply of points on the line through the foci by denoting the absolute value of the difference of the two distances in the definition of a hyperbola by 2a, where (2)

ORGGEAG:

Accordingly, the equation of a hyperbola in the coordinate system chosen assumes the form

(3) (Cf. Fig. 609-1.)

WG =o? + 2 — Va + 0)? + | = 2a.

$609]

THE

HYPERBOLA

173

y

f,: (c, 0)

Figure 609-1

The algebraic simplification of (3) is similar to that of (3), §605. squaring the two nonnegative members of (3) and simplifying:

(4)

We start by

Vx? + y2 + €2)2 = 4c2x2 = (x2 + y2 + c2) — 222.

Squaring the two members of (4) gives, after simplification: (5)

a*(x2 + y2 + c2) = c2x2 + a4.

In order to show that (4) and (5) are equivalent we need to show that (5) implies the inequality (x? + y2+ c?) > 2a2. Accordingly, we assume (5) in the form (c2 — a)x? — (c? — a*)a2 = a?y? to infer (c?2 — a?)x? = (c2 — a?)a2, and consequently, since c? > a2, x2 2 a2. Therefore, from (5), a%(x2 + y2 + c2)2 c2a2 + a‘, and hence x2 + y? + c? 2 c2 + a? > a2 + a? = 2a?, as desired.

We continue by writing (5) in the form

(6)

(c2 — a2)x? — ay? = a%(c? — a2),

and therefore, with

(7)

b=V2—a@,

Bb =2-—a@,

the equation of the hyperbola simplifies to

(8)

b2x2 — a2y2 = a2b?,

or

0)

a Siig=

a+b?

= e?,

174

CONIC

SECTIONS

[$609

It is clear from the form of equation (9) (cf. §115) that the hyperbola (9) is symmetric with respect to each coordinate axis and the origin. The points (—a, 0) and (a, 0) are called the vertices of the hyperbola (9). The segment joining the vertices is called the transverse axis, the segment joining the points (0, —b) and (0, b) is called the conjugate axis, and (0, 0), the common midpoint of the transverse and conjugate axes is the center of (9). The terms transverse axis and conjugate axis are also used to designate the /engths of the corresponding segments just defined, 2a and 2b, respectively.

The numbers a and 6 are called the transverse semiaxis

and the conjugate

semiaxis, respectively. The Pythagorean relationship among the numbers a, 6, and c (c being the hypotenuse) is shown in Figure 609 -2a.

focus ¢ (0, c) vertex

(0, a)

vertex | (0, —a)

focus ¢ (0, —c)

(0) Figure 609-2

§610]

ASYMPTOTES

175

If the foci are taken on the y-axis, at the points (0, c) and (O, —c), a derivation

similar to the one given above leads to the equation (cf. Fig. 609-26):

10

Ape pated

10)

a2

b2

L.

In this case the vertices are (0, a) and (0, —a). The numbers a, b, and c are again related by (7), and the transverse and conjugate axes and semiaxes are defined by analogy with the definitions given above. Example 1. Find the equation of the hyperbola having foci (0, +4) and conjugate axis 6. (For a graph, cf. Example 1, § 610.)

Solution.

The equation is of the form (10) with

6 = 3. Since c = 4,a =

and the equation becomes

16

9 = v7,

Example 2. Find the vertices, transverse axis, conjugate axis, and foci of the hyperbola 3x2 — 5y? = 7. (For a graph, cf. Example 1, § 610.)

Solution.

Division by 7 puts the equation in the form

(11)

pasa md 3

5

We therefore have, in the notation of (9), a? = and 6? = 7, and hence c? = $+ 7 = 38. The vertices are thus (+24; 0) (+2721, 0), the transverse axis is the segment

(—321, 0)-y21, 0), the conjugate axis is the segment (0, — 21V35)(0, 235), and the foci are (+ V$$, 0) = (+7210, 0).

610

ASYMPTOTES

The straight lines that contain the diagonals of the rectangles in Figure 609 -2 have the property that if « is an arbitrary positive number, then any point of the hyperbola concerned that is sufficiently remote from the origin is within a distance less than « from some point on one of these lines. (We shall prove this statement presently.) For this reason, these lines are called asymptotes of the corresponding hyperbolas. To be specific, the equations of the asymptotes of the hyperbolas in Figure 609-2 are as follows: 5 al oa OM

or

Fae

bar

oe

a 0

and

niga= 9 rie pe a = meee 3 — Tee Asymptotes of Wes

OD

he

ak aeem O

and

2

Asymptotes fam

Aa

=e

5 Oe

ee

toe

aah: =)

Us, slegeie = 0 moa

It should be noted that for either type of hyperbola, the equation of the pair of asymptotes is given by replacing the constant term | by the constant term 0, and the - individual equations of the asymptotes are given by factoring.

176

CONIC

In order to prove the statement about of Figure 609-2a, we use symmetry in quadrant, and choose on the hyperbola same x-coordinate. The problem, then,

SECTIONS

[§610

the asymptotes of the hyperbola (9), §609, order to focus our attention on the first and the line y/b = x/a points having the is to prove that the difference

byh = CNb —=— D ee = at Nx PE a

(1)

can be made less than any preassigned positive number e (however small this number may be) by requiring x to be sufficiently large. The quantity (1) can be rewritten as follows:

Fi ope a

Ne? l

Oe EN xtye—@

ee ee @xtye-a@

ab x4+yP-a

If x> a, then yx? — a2 > O and x + yx? — a? > x, and therefore

(2)

D=

CT ce

x+yx2—a2 ~ x

We now insist that x be so large that the quantity ab/x is less than e. Since the inequality ab/x < « is equivalent to x > ab/e, we have only to guarantee that the number x be greater than both a and ab/e to be certain that the difference D of (1) is less thane. This completes the proof for the hyperbola (9), $609. The corresponding proof for the hyperbola (10), $609, is similar and will be omitted. Note. Asymptotes provide a useful aid in drawing hyperbolas. After the numbers a and 5 in equations (9) and (10), § 609 , are found, the most convenient steps in drawing an approximate graph of either of these equations are usually: (i) draw the rectangles indicated in Figure 609 -2, (ii) draw the asymptotes, (i//) plot the vertices, and (iv) sketch the hyperbola. 2 he Example 1. Draw graphs of the hyperbolas ss— 9 1 and 3x2 — S5y2 = 7 of Examples 1 and 2, §609.

Figure 610-1

§611]

ECCENTRICITY

177,

Solution. For the hyperbola of Example 1, §609, we havea = y7 = 2.65 (to two decimal places) and b = 3, and thus the rectangle and vertices shown in Figure 610-la. The hyperbola itself is shown in Figure 610-15. For the hyperbola of Example 2, §609, we have a= 4421 = 1.53 (to two decimal places) and 6 = 435 = 1.18 (to two decimal places), and thus the rectangle and vertices shown in Figure 610-2a. The hyperbola itself is shown in Figure 610-24. Example 2. and y = —2x.

Find the equation of the hyperbola with foci (+3, 0) and asymptotes y = 2x

Solution. From Figure 609-2a we know that the absolute value of the slopes of the asymptotes is b/a, and therefore we have in this case b/a = 2, or

(3)

b = 2a.

Since c = 3, we also have, from the equation c? = a? + 62,

(4) From

(PPD (3) we have b* = 4a?, and hence from (4), Sa? = 9, or a? = 2, and consequently

b2 = 38

Therefore

the equation

of the hyperbola

can

2 2 be written ~ — a = 1, or a. a

202 —5y* = 36.

Figure 610-2

611

ECCENTRICITY

The eccentricity e of a hyperbola is defined to be the ratio of (7) the distance between its two foci to (ii) the length of its transverse axis. In the notation of §609, (1)

c Va + b2 es DC Aenea a =O i amy, a

Since c > a, e satisfies the inequality (2)

Cal.

pe bbee +2

CONIC

178 Example.

Find

the equation

[§612

SECTIONS

of the hyperbola

of eccentricity

2 and vertices (0, +5).

Since a = 5 and e = 2, equation (1) gives c = ae = 10, and Solution. b2 = c2 — @ = 100 — 25 = 75. By (10), §609, the equation can be written

consequently

Oe: Sees 2

Ap)

O55

is the slope of either of the asymptotes of the hyperbola of Figure 609-2a, Note 1. If then the eccentricity e of this hyperbola is equal to e = yi + m2. For the hyperbola of

Figure 609-26 the relationship is e = V1 + (1/m)? = V1 4 m?/m.

Note 2. An equilateral hyperbola is one whose transverse length, or equivalently, in the notation of §609,a = b. Each is necessary and sufficient for a hyperbola C to be equilateral (i) the asymptotes of C are perpendicular, (ii) the eccentricity

612

CONJUGATE

and conjugate axes are of equal of the following two conditions (give a proof in Ex. 37, §614): of C is y2.

HYPERBOLAS

The hyperbolas

Sedeer cue ae 2s alan l a 2 2 pe

(1)

are said to be conjugate. The conjugate axis of each is the transverse axis of the other.

They have identical asymptotes, and their foci are at the same distance from their common

center (cf. Fig. 612-1).

Example.

613.

The hyperbolas 3x2 — 5y? = 2 and 3x2 — Sy? =

CENTERS

DIFFERENT

FROM

THE

—2 are conjugate.

ORIGIN

If a hyperbola has center at the point (A, k) and foci on a line through (A, k) parallel to one of the coordinate axes, then derivations similar to those of §609, with x and y replaced by x — hand y — k, respectively, lead to the equations given in Figure 613-1.

If either of the equations for the hyperbolas of Figure 613-1 put into the form

(1)

ax

26x) 4 Dy?

is expanded and

2px + 2qy -- a = 0

of the general equation of the second degree (the coefficients a, b, and c in (1) are not to be confused with the constants a, b, and c of Figure 613-1)

the following three

properties of the resulting equations appear: (i) There is no term involving the product xy. (ii) The equation is quadratic in each of the variables x and y. (iii) The coefficients of the terms in x? and in y? are of opposite sign.

$613]

CENTERS

DIFFERENT

FROM

Figure 612-1

Figure 613-1

THE

ORIGIN

W729

[§613

SECTIONS

CONIC

180

As in §§ 603 and 607, there is a converse to the preceding statement: If an equation of the form (1) satisfies (/)-(iii), then its graph is either a hyperbola or a pair of intersecting straight lines. The procedure once more is that of completing squares, and is

illustrated in Examples 2 and 3, below. Write the equation of the hyperbola with foci (3, 10) and (3, —2), and eccen-

Example 1. tricity 2.

c= 6. By (1), §6l11, Since the distance between the foci is 2c = 12, Solution. a = c/e = 6/2 = 3. Therefore b? = c? — a? = 36 — 9 = 27. Since the center is midway between the two foci its coordinates are (3, 4), and the equation of the hyperbola —

ee

9

2

i

eee

2

BACH eels)

27

y.

ny.

focuse (3,10) See ee

(3,7)

vertex

ee (5,—4) e focus

center Peay eet

(O94)

vertex (3, 1)

oe

OT focus ¢(3, —2)

(a)

(b) Figure 613-2

Example 2.

Describe the curve

(2) Solution.

(3)

3x2 — Sy? — 12x — 40y — 95 = 0. We prepare to complete squares:

3(x2 — 4x

) — 502 + 8y

=

The next step adds 12 — 80 =

—68 to each side:

(4)

3(x — 2) — Sy + 4)? = 27.

&,

Division by 27 gives an equation of the form of that of Figure 613-la, with a = 3, be V2 = 3/15 = 2.32 (to two decimal places), A = 2, and k = ae Therefore (2) is a hyperbola with center (2, —4), vertices (—1, —4) and (5, —4), transverse semiaxis 3, and

conjugate semiaxis 215.

Since c? = a? + 6? = 9 + 22 = 22, ¢ = 6\2 = $y10 = 3.79 (to

two decimal places), and the foci are (2 + 3.79, —4), or (—1.79, —4) and (5.79, —4). The eccentricity is equal to e = c/a = 24/10 = 1.26 (to two decimal places). The asymptotes are

given by 3(x — 2)? — S(y + 4)? = 0, or

(5)

V5(y + 4) = ++3(x — 2),

and are the two lines through (2, —4) with decimal places). (Cf. Fig. 613-26.)

slopes

«4/2 =

a5 ATS =

+ 0.77 (to two

§614]

EXERCISES

Example 3.

18]

Describe the curve

(6)

3x? — Sy — 12x — 40y — 68 = 0.

Solution.

Completion

of squares as in the solution of Example

(7)

2 gives the equation

3(x — 2? — Sy + 4 = 0.

Therefore the ‘“‘curve” (6) consists of the two asymptotes (5) of the hyperbola (2).

614

EXERCISES

A graph should accompany each of the following exercises. In Exercises 1-12, show that the hyperbola satisfying the given conditions has its center at the origin, and find its equation. fem Vertexa(l10)stoci(=-2.0)) . Vertices (0, +1), eccentricity 3.

. Foci (0, +7), conjugate axis 10. . Vertices (+4, 0), through (5, 3). . Vertex (6, 0), nearer focus (8, 0), eccentricity 4. . Vertex (0, 12), farther focus (0, —13), conjugate semiaxis 5. . Vertex (0, 3), focus (0, 4), transverse semiaxis 3. . Vertex (6, 0), farther focus (—9, 0), eccentricity 3. MN NH eN © & N W . Asymptotes — i—).

y2 = 4x2, conjugate axis 8.

Asymptotes x2 = 7y2, focus (0, 4).

. Asymptotes y* = x2, through (3, 4).

—_a"

12. Difference of distances from (+5, 0) is equal to +4.

In Exercises 13-18, find the vertices, lengths of the transverse and conjugate axes, foci, eccentricity, and asymptotes of the given hyperbola.

x2 Los 15. 9y2 —

yp Got 16x? = 1.

17. x* — 3y? = 15. In Exercises

we 14.

ee 5

16. 25x? — 144)? = 81.

18. 2y2 — x? = 10.

19-30, find the equation of the hyperbola satisfying the given conditions.

19. Foci (—1, 4) and (9, 4), transverse axis 4. 20. Foci (—2, 2) and (—2, 12), eccentricity 3.

21. Vertices (5, —6) and (5, 6), conjugate axis 20. 22. Vertices (—1, —1) and (7, —1), slopes of asymptotes 23. Center (0, 3), vertex (—5, 3), focus (—8, 3).

24. Center (4, 2), vertex (4, 5), focus (4, —3). 25. Center (3, —2), vertex (3, —1), conjugate semiaxis 5.

+3.

CONIC

182

[$615

SECTIONS

26. Center (—2, —1), focus (4, —1), eccentricity 3.

27. Difference of distances from (—2, 7) and (10, 7) is equal to +6.

28. Asymptotes 3x? = 4y?, through (J, 2). 29. Vertex (10, 3), nearer focus (10, 7), conjugate axis 6. 30. Vertex (—7, 1), farther focus (11, 1), conjugate axis 12.

In Exercises 31-36, find the center, vertices, lengths of the transverse and conjugate axes, foci, eccentricity, and asymptotes of the given hyperbola. 31. 9x2 —

16y? — 54x — 64y —

33. x2 — 4)? —

10x —

35. x2 — 2y2 + 12x — 37.

loy

127 = 0.

+ 13 |=

0.

12y + 16 = 0.

32. 16x2 — 9y2 + 32x + 72y + 16 = 0.

34. 9x? — y2 + 54x + 2y + 71 = 0. 36. 3x2 — y2 — 24x + 16y —

13 = 0.

Prove the statement of Note 2, §611.

In Exercises 38-41, write the equation of the hyperbola conjugate to the given hyperbola.

a

See = hye = 3

39, 3y2 — 2x2 = 13.

40. 2x2 — 2 — 8x + 6y — 29 = 0.

41. x2 — 3)? — 8x — 6y + 16 = 0.

42. Prove that the distance between a focus and an asymptote of a hyperbola is equal to the conjugate semiaxis.

43. Prove that the product of the distances from any focus of a hyperbola to the two vertices is equal to the square of the conjugate semiaxis. 44. Prove that the distance between a focus of ahyperbola and the nearer vertex is less than the conjugate semiaxis. 45. A latus rectum of a hyperbola C is a chord of C through a focus of C perpendicular to the transverse axis of C. Prove that the length of a latus rectum ofa hyperbola with transverse and conjugate semiaxes a and J, respectively, is equal to 2b2/a.

615

LOCUS

PROBLEMS

The idea of /ocus problem was defined in §415. We are now able to consider and resolve many new locus problems, with the aid of the conic sections, parabolas, ellipses, and hyperbolas. Once again, there are two principal categories of locus problems, those framed within a given coordinate system, and those for which a coordinate system may be introduced. Example 1. Find the locus of a point p moving so that the sum of the squares of the distances ofp from the origin and from the x-axis is equal to 4. Solution.

If the point p has coordinates

(x, y), then

the required condition can

be ex-

pressed by means of the equation

(1)

(x2 + y*) + 7) = 4.

Therefore the locus is an ellipse with center (0, 0), vertices (+2, 0), and minor axis 2,2. Example 2.

Find the nature of the locus of a point moving in the Euclidean plane so that

its distance from a fixed point is equal to its distance from a fixed line.

§616]

EXERCISES

183

Solution. There are two cases. We first consider the case where the fixed point lies on the fixed line. For simplicity, we choose a coordinate system such that the fixed point is the origin and the fixed line is the y-axis. The given condition is then expressed by the equation New + y2 = |x|, or equivalently, y2 = 0, and the locus consists of the x-axis.

We now consider the case where the fixed point does not lie on the fixed line. We choose a coordinate system such that the fixed point is the origin and the fixed line has the equation x + y = 0, where y > 0. The given condition now is expressed by the equation yx? + y2

= |x + y|, or equivalently, y2 = 2yx + y? = 2y (x + 4y), and the locus is a parabola.

616

EXERCISES

A graph should accompany each of the following exercises. In Exercises 1-8, write the equation of the locus of a point p : (x, y) in the Euclidean plane subject to the given restriction. Then describe the graph. 1. The distance of p from the line x = 3 is twice the distance of p from the point (0, 0). 2. The distance ofp from the line y + 2 = 0 is three times the distance ofp from the point (0, 2). 3. The distance of p from the point (—9, 0) is three times the distance of p from the line x+1=0.

4. The distance of p from the point (0, —3) is twice the distance of p from the x-axis.

5. The sum of the squares of the distances of p from the two x — 3y = 0 is equal to 45.

lines x + 3y = 0 and

6. The square of the distance ofp from the line 2x + y = 0 minus the square of the distance of p from the line x + 2y = 0 is equal to 15. 7. The sum of the squares of the distances of p from the point (2, 0) and the line x + 2 = 0 is equal to 10. 8. The square of the distance ofp from the point (2, 0) minus the square of the distance ofp from the line x + 2 = 0 is equal to 16.

In Exercises 9 and 10, find the nature of the locus of the point p in the Euclidean plane subject to the given restrictions. 9, The sum of the squares of the distances of p from two fixed intersecting lines is a positive constant.

10. The sum of the squares of the distances of p from a fixed point and a fixed line is a constant greater than half the square of the distance between the point and the line. (The point may or may not lie on the line.) 11. A moving circle passing through a fixed point remains tangent to a fixed line not containing the point. Find the locus of the midpoint of the circle.

12, In Example 2, §615, in the case where the fixed point does not lie on the fixed line, prove that the origin is a focus of the parabola.

Polar

701

INFORMAL

REVIEW

Coordinates

OF TRIGONOMETRY

We begin this chapter with a brief review of some of the elementary properties of the standard trigonometric functions. A short description of an analytic development of these properties is given in Chapter 11 of the sequel volume, Basic Concepts of Calculus. A more complete treatment is to be found in Chapter 19, CWAG. We shall now merely list metric functions, show their measure, state fundamental basic approach at present is

the basic relations among the standard six trigonographical representations in terms of angles in radian identities, and give a few illustrative examples. The intuitive and geometrical.

We shall consider the sine and cosine functions as fundamental, connected by the identity (1)

sin2 @ + cos? @ =

l,

and therefore limited in values by the inequalities (2)

=WeStsin ots! if

—l sScoséS1.

The relationship of sin 6 and cos @ to the unit circle x2 + y? =

1, with the angle @in

“standard position” is shown in Figure 701-1, with indications of the four quadrants and the signs of these two functions in each of the quadrants. Because of the relationship exhibited in Figure 701-1, the trigonometric functions are also called the circular functions. 184

§701]

INFORMAL

REVIEW

OF

TRIGONOMETRY

(cos 57,sin 37) =

et y=1

185

(0, 1)

(cos 0, sin @)

Quadrant| 2

Quadrant il

Quadrant}

Quadrant

3

0

4

ant)

(sey'ar)

(=a)

(+,-)

>

>

(cos 0, sin 0)

;

(cos7r, sin7r) = (= 1,0)

=

(1, 0)

ee3 | \= (0, —1) COS oe 5 7, sin

6, sin 0) (cos

Figure 701-1

The graphs of y = sin x and y = cos x in rectangular coordinates are shown in Figure 701-2. These functions are odd and even, respectively: sin (—x) = —sin x, cos (—x) = cos x.

Sea

ae

i

Figure 701-2

The four other trigonometric functions can be defined in terms of sin x and cos x: (3)

tan x=

sin x , COS x

cot x =

cos x —— sin x

(4)

SEC

1 ) COs x

csc x =

1 —— sin x

=

for values of x where the denominator does not vanish, and are related by the identities:

(5) (6)

cot x

=

tan x

)

tan2 x + 1 = sec? x,

tan x

=

cot x

)

cot2 x + 1 = csc? x,

- for values of x where the denominators are nonzero.

of

[§701

COORDINATES

POLAR

186

4

(ese 6, sec @)

(cot 6, tan @)

(b)

(a)

Figure 701-3

The relationships of these functions to the unit circle are shown in Figure 701-3, and their graphs are shown in Figures 701-4 and 701-5. y

y

3 ou

+

|

-

al —

Omi

als.

{

DSN

—

3

3)

Oe

gt

y = tan x

y = cot x Figure 701-4

§701]

INFORMAL

REVIEW

OF

TRIGONOMETRY

Y

187

y

|

| |

|

||

||

| 4 3

O

757

ae

ee

$1

ae Z

2

i

x

i

1

Hi

'

t

sm

7

br

3

ai

Ie

Db:

T

0

5m

TT

|

|

|

|

Vi

SCC XL

y = csc x

Figure 701-5

The addition and subtraction formulas for sine and cosine are

(7)

sin(x + y) = sin x cos y + cos x sin y,

(8)

COs(x =- yp) =(cOS x COS y — sin x sin y,

where either the upper sign is to be used on both sides of the equation, or the lower sign is to be used on both sides. These are valid for all real numbers x and y. From these can be derived the addition and subtraction formulas for the other functions. For example, Ohman

Ge s> y)=

sini _ y) . sil X COS y-== cos x siny cos(x +y) cosxcosy + sin x sin y sin x cos y

cos x sin y

cos xX COS y

cosxcosy

tanx =a tany

_sinxsiny

1 -+tanxtany

~ cCosxcosy COS.X—

y = 2 cos

~F» cos

y = —2 sin ~

ee =a

sin 2.

EXERCISES

In Exercises 1-8, derive the indicated formula or verify the indicated figure of §701.

1. Figure 701-1.

2. Figure 701-3a.

3. Figure 701-30.

4. (9).

5. (10)+(12).

6. (13).

7. (14)-(17).

8. (18)-(21).

9. Derive the identity cot (x + y) =

COURACOLAeeal cot

y + cot x

10. Derive the double-angle identity cot 2x =

cot? x —

|

ZICOUnX

In Exercises 11-14, derive the triple-angle identity. 11° sin 3x = sin x@ cos? x — 1) = 3 sin x — 4 sin?’ x.

IP2, (COS She = ANeose — Si Gos se 133 tans:

is:

3 tan x — tan3 x 1 — 3 tan2 x

14. cot 3x =

3 cot x — cot? x 1 — 3 cot? x

In Exercises 15 and 16, prove the identity.

15. 8 sin* x = 3 — 4 cos 2x + cos 4x.

16. 8 cost x = 3 + 4cos 2x + cos 4x.

§703]

CURVES

IN POLAR

COORDINATES

189

In Exercises 17-27, prove the identity. 17. sin(@r += x) = cos x.

18. cos(47 + x) =

= sin x.

19. sin(ia + x) = 3\/2(cos x + sin x).

20. cos(}r + x) =

34/2(cos x * sin x).

21. tan(x + 7) = tan x.

22. cot(x + 7) = cot x.

233

tan

Xl Ain D2

1 + tan x

25. Bata

1 — tan x

COST% 5 sin x

—

sin x a 1+ cos x

x oy a

1+cosx = sin x

—

; ae a SI

OG ir (x + in).

ies

sinx =

i

°

1 — cosx z

LoSes Peal ar)

sinx — siny

tan 3(x — y)

27. sin? x — sin? y = sin(x + y) sin (x — y). 28. Use mathematical induction to prove that for every natural number 1, it is possible to express COS nx as a polynomial in cos x, and sin nx as sin x times a polynomial in cos x.

703

CURVES

IN POLAR

COORDINATES

If (p, 6) is an arbitrary ordered pair of real numbers, then (p cos 6, p sin @) is an ordered pair of real numbers and therefore determines a point p in the plane with coordinates

(1)

x =p cos 6,

=) pisini’:

The usual geometric interpretation of (1) is given in Figure 703-1, where p is shown as either a positive or a negative number and @ is the measure of an angle in standard position (having its initial side coincide with the oriented positive half of the x-axis. The numbers p and @ are called polar coordinates of the point p; p is called a radial coordinate and @ an angular coordinate of p. This relationship will be written p:[p, 6], with square brackets. The origin, given by p = 0, is called the pole of the polar coordinate system. Let us consider now the reverse problem of finding pairs of polar coordinates for a given point p : (x, y). If we square both members of the equations (1) and add the results we obtain

(2)

pe ES

eee

If p is the origin, or pole, there is no problem; p = 0 and 6 is completely arbitrary. Otherwise, if x2 + y2 > 0, there are precisely two determinations for p, one positive and one negative. For either one of these determinations we have y

ay

(pcos@, psin 8)

(|p| cos6, |p| sin6)

o


0, and 6 ranges from 0 to 27, including 0 but not including 27, then the point (a cos 6, a sin 6) moves around the

circle x? + y? = a?, occupying every point of the circle exactly once. If a < 0, the point (a cos 6, a sin @) is diametrically opposite the point (—a cos 6, —a sin 6), and the graph is therefore again the circle x? + y? = a? (cf. Fig. 703-4a). y

(a cos 6, asin 6)

(—acos@, —asin@)

(a)

(b) Figure

Example 4.

703-4

Describe the graph of the equation @ = a, where a is a constant.

Solution. The graph is a straight line through the origin, the points on one side being given by positive p, those on the other side by negative p, and the origin by p = 0. Ifa = 0, +a, +27,--- the graph of 6 = a is the entire x-axis, called the polar axis. Example 5.

Transform

to polar coordinates:

(i) x = a; (ii) y = 6, where a and 6 are

constants.

(i): Direct substitution from (1) gives the equation p cos 6 Se If a = 0, Solution. this can be written more simply @ = 37. If a ¥ 0, then pcos 6 ¥ 0 for points on the line x = a, and therefore cos 6 # 0 and the equation can be written

p=asec0,ax#0.

(4)

(ii): The equation y = b becomes p sin@ = 4 in polar coordinates. written more simply 0 = 0. If 6 ¥ 0, this equation is equivalent to (5)

p = bcesc 0, b# 0.

Most equations in polar coordinates take the form

(6)

p = f(),

or, occasionally,

(7)

pu

(0):

If 5 = 0 this can

be

[§703

COORDINATES

POLAR

a2

In either case the graph can be found by computing values of the function f(#) and setting up a table of values. In practice, however, it is more expedient to plot only a few “key points,” or to graph a portion of the curve and deduce the full curve from that portion. One feature of a curve that is frequently helpful is the set of values of 6that correspond to p = 0, since these usually disclose the behavior of the curve near the origin.

Example 6. Solution.

Graph p = asin 6, where a > 0, and show that the graph is a circle.

The table of values

gives the points shown in Figure 703-5a. The fact that p = 0 for integral multiples of 7, where sin 6 does not attain extreme values, tells us that the x-axis is a tangent to the graph. The maximum value of p is clearly a, assumed when 6 = 37. Multiplication by sin @ gives p sin @ = a sin? 6 = 0, so that the graph lies entirely above the x-axis, except for the origin. The complete graph is shown in Figure 703-55. To prove that the graph of p = asin @ isa circle, we consider the related graph obtained by multiplying by p: p? = apsin 6, which in rectangular coordinates can be written

(8)

Xe ey?

ay

Or

2X? =) = 3d)

y

Ga). ¥y

ae

(0, a)

p=asin@ x

x

>

(a)

(b) Figure 703-5

The graph of (8) is a circle of radius 3a and center (0, 3a). By factorization of p? — ap sin@ = 0 we see that this circle is the union of the two graphs (/) p = 0 and (ii) p — asin@ = 0. Since the graph of (/) consists of the origin only, and since the origin is a point of the graph of (i), we conclude that the graph of (i) alone is the circle of Figure 703-5d.

Nore. The graph of p = asin@ is a circle whether a > 0 or a < 0 (cf. Ex. 38, §704). The graph of p = acos @ is also a circle, whether a > 0 or a < 0 (cf. Ex. 39, §704).

§703]

CURVES

Example 7. Solution.

IN POLAR

COORDINATES

193

Graph the cardioid p = a(1 — cos 8), where a > 0.

The table of values

6 || 0

an

$1

el] 0 | 3(22— \3a | 3(2- 2a gives the points shown in Figure 703-6a. symmetric with respect to the x-axis.

a4

y

y

ea

@

os 2a

The complete graph, as shown in Figure 703-64, is

t

x

x 2a

a

p = a(1-—cos6) (a)

Example 8. Solution.

Figure 703-6

b

i

Graph the four-leaved rose p = a sin 20, where a > 0.

The table of values

An extension of this table for 37 VAN ) IIA ) go < o n ¥y

at

ail

p= asin 20 Figure 703-7

(b)

POLAR

194

[§704

COORDINATES

negative values for p, and therefore a portion of the curve located in the fourth quadrant. Similarly, for 7 < 6 < 32, 2x S 20 S 32, asin 20 = 0, and this portion of the curve lies in the third quadrant, and for $4 S 0 S 2, 3m S 26 S 4n, a sin 20 S 0, and this portion of

the curve lies in the second quadrant. The fact that the function a sin 26 vanishes at integral multiples of 42 is indicated in Figure 703-7a by short vertical and horizontal line segments

through the origin. The extreme values of a sin 20 at @ = jm, 3a, jm, and 77 give points shown in Figure 703-7a.

Example 9. Solution.

The entire curve is shown in Figure 703-74.

Graph the lemniscate p* = a* cos 26, where a > 0.

The graph is real if and only if cos 20 = 0, and therefore for —ja S 0 S ir

and for 24 < 6 < 2, and is imaginary for +7 < @ < ?n and for ?a < 6 < fa. The vanishing of cos 20 at 6 = {7, {7, 5, and 77 is indicated in Figure 703-8a by short line segments through the origin. Extreme values of p = +a are given by @ = 0 and zw. These points and those corresponding to 6 = +47, 27, and Z7, where p = +42a, are also shown in Figure 703-8a. The entire graph is given in Figure 703-8).

p?= a’cos 20

(a)

(6) Figure 703-8

704

EXERCISES

In Exercises 1-6, one set of polar coordinates of a point is given. Find another set [p, 9] with 0 S @ < 2m, graph the point, and find the rectangular coordinates of the point. 1. [4, $7].

23" al:

5. [V13, Arccos (—3/113)].

3. [—6, 3a].

4. [—2, gn].

6. [V17, 2m — Arctan 4).

In Exercises 7-12, the rectangular coordinates of a point are given. find two sets of polar coordinates [p, 6] with 0 < @ < 2r.

70g) 10. (12, 0).

8. (—5, 53). Tine):

Graph the point and

9, (—4y3, —4). 1255; 1).

§705]

POLAR

In Exercises

EQUATIONS

OF

CONICS

195

13-18, transform the given equation to rectangular coordinates, and graph.

13. 6 = in.

14. p = 2sec 0.

15°"

16.

17.

18. p = 8cos0.

p = —3.

p = —6sin@.

=" —4 esc 0;

In Exercises 19-24, transform the given equation to polar coordinates, and graph. 19k a

=O)

22. x2 ++ y2 = 16.

20. x = —3.

Z1.y = "1.

23.

24

x2 + y2 = 10x.

ny

1)

Ie

In Exercises 25-37, draw a graph of the given equation. 25.

Cardioid,

p = a(l — sin @), a > 0.

26. Cardioid,

p = al + cos @), a > 0.

27. Three-leaved rose, p = asin 36,a > 0.

28... Three-leaved rose, p = acos 36,a > 0.

29.

30. Five-leaved rose,

p = acos 50,a > 0.

31. Lemniscate, p? = a? sin 26, a > 0.

32. Lemniscate, p? =

—a? cos 26,a > 0.

33.

Limacon, p = 3 — 2 cos 0.

34. Limagon, p = 3 — cos 6.

35.

Limacon, p =

36.

Four-leaved rose,

p = acos 26,a > 0.

1 — 2cos @.

37. Spiral of Archimedes,

Limagon, p =

p = a0,a > 0. (Take a =

1 — 2sin 6.

1.)

38. Draw the graph of p = a sin 6, a < 0, and show that the graph is a circle. equation of this circle in rectangular coordinates.

Write the

39. Draw the graph of p = acos 6, a ¥ 0, and show that the graph is a circle. Consider the two cases a > 0 and a < 0. Write the equation of this circle in rectangular coordinates.

705

POLAR

EQUATIONS

OF CONICS*

The conic sections — parabolas, ellipses, and hyperbolas — are graphs of equations that assume a particularly simple form when expressed in terms of polar coordinates, with a focus at the origin or pole. As we shall see, the equation can be put in a form where p is a simple rational function of either cos @ or sin 9, and all three types of conics can be treated by a single unifying form of expression. This polar form is especially suitable for study of space trajectories and celestial mechanics. We consider the three classes of conics in turn. I.

Parabola.

As shown in §602, the equation of the parabola with vertex at the origin, focus at (c, 0), and directrix x + c = 0, where cc > 0, is y? = 4cx. By §603, the equation of the “shifted” parabola with vertex (—c, 0), focus (0, 0), and directrix x + 2c = 0

is

(1)

y? = 4c(x + ¢)

*This section, together with the Exercises of §706, can be omitted without disturbing the continuity of the remaining material. It is not prerequisite to any portions of this book or of the sequel volumes, Basic Concepts of Calculus and A Second Course in Calculus.

196

POLAR

COORDINATES

(cf. Fig. 705-1). Equation (1) can (x + 2c)?, or with p? = x2 + y?,

(2)

be

written

[§705 x? + y2 = x? + 4cx + 4c? =

p? = (x + 2c).

Extracting square roots and using polar coordinates to express x = pcos 6, we observe that the parabola (1) is the union of the graphs of the two equations p = pcos@+ (3)

2c and p = Pre

=

2c Cane—

—p cos 6 — 2c or, equivalently, of the two equations iS GOS = lls fp

|

—2c Meena

SS)

cos 6 #

2

il.

We now make a simplifying observation: whenever [p, 6] is a pair of polar coordinates satisfying either one of the equations (3), then [—p, 6 + 7] is a pair of polar coordinates satisfying the other equation, and since the two points [p, 6] and [—p, 9 + 7] are the same point, the two equations (3) have the same graph. For simplicity we choose the first equation of (3), for which p > 0, and with the notation p = 2c for the. (positive) distance between the focus and directrix (cf. Fig. 705-lc), we have for the equation of the parabola (1) in polar coordinates: ye P ray

(4)

6~2nnr,

y

n€

J,

y

y

D

x+2c=0

v

=e

f

/

1(0,0)

(—c,0)\) t—

(a)

De

x

vy?=4e(x +c) —}

(b)

(c)

Figure 705-1

Example 1. Write the equation in polar coordinates of the parabola with focus at the origin and vertex on the x-axis 3 units to the left. Solution. tion

Since the vertex is midway between the focus and the directrix, p = 6, and equa6

(4) iS Di=

II.

i

ee

Ellipse.

The equation of the ellipse with foci (—c, 0) and (c, 0) and major and minor semiaxes a and 6, respectively, is (x2/a?) + (y2/b2) = 1, with a2 = b2 + ¢

§705]

POLAR

EQUATIONS

OF

CONICS

197

(cf. $605). Therefore the equation of the “shifted” ellipse with foci (0, 0) and (2c, 0) and center (c, 0) is (§607): _

(5)

a9

2

2

+5-1

@=R +e

(cf. Fig. 705-2a). Equation (5) can be written a2x2 + a2y2 = c2x2 + 2b2cx + b4 or, with p2 = x? 4+ y2,

(6)

a2p? = (cx + b2)2.

1(0,0)

(0)

(2c, 0)

x is

p

_._ €P ~ 1-ecosé

(0) Figure 705-2

The graph of (6) is the union of the graphs of the two equations ap = cx + b? and ap = —cx — b?, but since these two graphs are identical, for the same reasons as those used in connection with equations (3), we choose one, ap = cx + b? = cp cos @ + b2, or

(7)

p=

b2 a—ccosé@

It is not hard to show that (7) gives positive values to p for all values of #6. Recalling from §606 the definition of the eccentricity e of an ellipse: dé;

C=

e=>-

(8)

we can rewrite equation (7) by dividing by a and using (8):

,

@)

eee

P=

1 —(c/aycos@

1—ecosé

and therefore, after multiplication by 1 — ecos@: (10)

p — ep cosé = p — ex = b?2/a,

from which we have p = ex + (b2/a) = e(x + (6?/ae)), or

(11)

oe

POLAR

198

[§705

COORDINATES

Interpreting equation (11) as a description of a locus, we have an interesting new formulation of an ellipse as the locus of a point moving so that the ratio of its distance from a fixed point to its distance from a fixed line not passing through thefixed point is a constant.

The fixed point is a focus, the fixed line is a directrix, and the constant

ratio is the eccentricity of the ellipse. From (11), the equation of the directrix is x + (b2/ae) = 0, so that the directrix is a vertical line located at a (positive) distance b2/ae to the left of the origin or focus. If we label by p this distance between the

directrix D and the focusf, we have

(12)

Pema

wes ae cle

so that the equation (9) of the ellipse can be put in the form

pe ORS oat Ea cond

(9)

Example 2. Show that for the ellipse (13) the distance between the center and a directrix is a/e and the distance between a vertex and the corresponding directrix is (a — c)/e.

Solution.

The distance between the center and a directrix is c + p (cf. Fig. 705-26), b2 be a ap : and by (8) and (12),c + p=c+ ae -.ne = = - Therefore the distance between a :

Cs

peta

2!

vertex and the corresponding directrix is — — a = -

=

e

Example 3.

Find the rectangular coordinates of the foci and vertices of the ellipse

20

14

Cy

p

Beet eg Tet 3 — 2cos 0

Also find the major and minor axes and the eccentricity. Solution.

Substitution

p = 4: (20, 0), (—4, 0).

of 6 = 0 and

@ = m gives the two

vertices,

with

p = 20 and

(Cf. Fig. 705-3.) Therefore the major axis is 2a = 24, anda = 12. 20

Division in (14) by 3 gives p =

1 — 3 cos 0

» and e = 3. Therefore c = ae = 8. Since the

center is (8,0)the foci are (0, 0) and (16, 0). Therefore 6? = a? — c? = 144 — 64 = 80,b = 45, and the minor axis is 8Y5. y

(—4, 0)

Figure 705-3

§705] III.

POLAR

EQUATIONS

OF

CONICS

199

Hyperbola.

The equation of the hyperbola with foci (—c, 0) and (c, 0) and transverse and conjugate semiaxes a and J, respectively, is (x2/a2) — (y2/b2) = 1, with c2 = a2 + b2 (cf. $609). Therefore the equation of the “shifted” hyperbola with foci (—2e, 0) and (0, 0) and center (—c, 0) is (§ 613):

(15)

iar cea ei aera

Mies

(cf. Fig. 705-4a). |Equation (15) can be simplified in much the same way that (5) was, to the same form a’p? = (cx + b2)? as (6). (Give details in Ex. 15, §706.) By the same reasoning as that used for (6) and (7) we can extract a single square root to write the equation of the hyperbola (15) in the form (7) or, with the notation e = c/aand p = b?/ae, in the form (13). The only difference in the present case is that p, given in (7) or (13), is no longer restricted to positive values. In fact, positive values of p (for which cos 6 < a/c) correspond to points of the right-hand branch of the hyperbola and negative values of p (for which cos @ > a/c) correspond to points on the left-hand branch. The asymptotes correspond to cos @ = a/c. The italicized statement above defining an ellipse as a locus applies in equal measure to a hyperbola, it being understood that absolute values should be taken in order that the distances involved are positive.

Za

p

=

ep

~ 1—ecos 0

(b) Figure 705-4

Example 4. Write the equation in polar coordinates of the hyperbola with a focus at the origin and with vertices at the points whose rectangular coordinates are (— 12, 0) and (—2, 0). The major axis is 2a = 10, and a = 5. The center is (—7, 0) and therefore Solution. 3: tA, CAS , From the equation c = 7 and the eccentricity See 1

(16)

p : wel

ep 3P te Tp "e.cogdemels—=* 7 cos:-0

we find p by substitution of 6 = 7:

p=2=

Tp

ae

:

Tp = 24, p = 24/7, and the equation

sought is p = 24/(5 — 7 cos 6). An easy check on our work is given by substitution of 6 = 0 in this last equation: p = 24/(5 — 7) = —12, in agreement with the vertex (—12, 0).

[§705

COORDINATES

POLAR

200 IV. Synthesis.

If the eccentricity e of a parabola is defined to be 1, then the locus definition given for an ellipse, as well as equation (13), applies to all three classes of conic sections. The three types of curve, restricted to points near the focus in question, are indicated in Figure 705-5.

alo)

Ey

Oe

6=0

py ae

6=0

O

al

P~1—ecos6’ ©

UWE. :

d=0

O

a

2p

UGE

iep ere

>

P=

cp

=

T=ecos6’©

Ellipse

Parabola

Hyperbola

(a)

(b)

(c)

Figure 705-5

For conic sections with different orientation relative to the focus — that is, opening to the left, upward, or downward instead of to the right in the neighborhood of the focus located at the origin — equations similar to (13) apply. These can be obtained from (13) by replacing @ in turn by 6 + 7, 6 — 4m, and 6 + 47m. (Cf. Fig. 705-6.)

To the left

ne

ep 1+ ecos 6 (a)

Upward

Be

EBs 1—esin# (b)

Downward

Ae

ep_ 1+esin#é (c)

Figure 705-6

Example 5. Write the equation in polar coordinates of the ellipse with focus at the origin, eccentricity 3, and directrix y + 2 = 0.

§706]

EXERCISES

Solution.

There are two cases.

at the origin, then 5

Poin 0

(i) If y+ 2 = 0 is the directrix corresponding to the focus

from Figure

bole

P=

2 —sin

201

705-64,

with p = 2, the equation

is p = ee as = 1 —esin 0

r (ii) If y+ 2 = Ois the directrix corresponding to the focus that is

not at the origin, then the equation of Figure 705-6c is appropriate. In this case, the distance from one focus to the directrix of the opposite focus is equal to the sum of the distance between the foci and the distance p between a focus and the corresponding directrix.

this equation takes the form

U7)

In this case

2=2c+p=2ae+tp=actp.

From the equation p = ep/(1 + e sin @) we find the two vertices by substituting @ = 32 and § = $m, the corresponding values of p being

(18)

4

=

ep l+e

;

=

Z

ep l—e

Since the major axis is the sum of these distances, 2a =

2P

2p

coy

Ee

l+e

+

Sa

1—e

aid

ee

1 — e

From (17) we have 2 = 2p + p = 3p, p = $. The equation sought is therefore

p

706

:

ip eS

ep

et

ee

35

ee

1+esn@

1+4sin0

$5

=

=

2+sin 0

6

.

10 + 5 sin6

EXERCISES

In Exercises 1—6, draw the graph of the given polar equation. Give the rectangular coordinates of each focus and vertex. Locate the center of each central conic, and draw the asymptotes of each hyperbola.

‘a

Cetel i Tree ea 9 Li ag prey: 2

baipartemuanet > ee ee 6 ee ee eee Peps Tan Sat

Rr e e e Ser = Riemer aeatG 10 686 eee DN ere 6

In Exercises 7-12, find the polar equation of the conic section having one focus at the origin (pole), and satisfying the additional given conditions. Draw a graph. All points are specified in rectangular coordinates. 7. Parabola, vertex (0, 6).

9. Vertices (0, 7), (0, —3). 11. Focus (—8, 0), vertex (—S, 0).

8. Eccentricity 1, directrix

x + 8 = 0.

10. Focus (—3, 0), vertex (2, 0). 12. Directrices

y = 4, y = 6.

In Exercises 13 and 14, find the polar equation of a conic having one focus at the origin (pole), and satisfying the additional given conditions. Draw a graph. All points are specified in rectangular coordinates. (In each case there are two solutions.)

13. Eccentricity 4, vertex (0, 3).

14. Eccentricity 2, vertex (—4, 0).

15. Give the details in the reduction of equation (15), §705, to the form a*p? = (cx + 6*). 16. Show that for the hyperbola p = ep/(1 — e cos 6) the distance between the center anda directrix is a/e and the distance between a vertex and the corresponding directrix is (c — a)/e.

Vectors

801

VECTORS

IN TWO

in the

Plane

DIMENSIONS

The concept of a two-dimensional vector is presented in the following Definition. All vectors considered in this chapter will be assumed to be two-dimensional.

In

the following Chapter 9, n-dimensional vectors are introduced so that they may be used in connection with systems of linear equations. Applications are then made to the two-dimensional Euclidean plane in the latter sections of Chapter 9, and to three-dimensional Euclidean space in Chapter 10. Definition.

4 two-dimensional vector is an ordered pair (h, k) of real numbers

h and k. The numbers h and k are called the components or coordinates of the vector (h, k); h is the x-component or x-coordinate and k is the y-component or y-coordinate. Two vectors (h, k) and (c,d) are equal, written (h, k) = (c, d), if and only if their corresponding components are equal: h = c andk = d. Otherwise they are unequal, written (h, k) ¥ (c,d). The vector (h, k) is represented by the directed line segment joining any point (a, b) to the point (a +h, b + k) (cf. Fig. 801-1). A vector may also be denoted by a single letter with an arrow, thus: 6. The magnitude or absolute value of a vector 6 = (h, k) is denoted and defined:

(1)

|| = |(A, k)] = VA? + 2.

The zero vector is the vector 0 = (0, 0) having zero magnitude.

A unit vector is

a vector with magnitude equal to 1. A unit vector is also called a direction. If p: (x1, yi) and q : (x2, y2) are any two points of the plane, the term the vector pq 202

$802]

LINEAR

COMBINATIONS

OF

VECTORS

203

means

the vector (x2 — x1, y2 — yi) represented by the directed line segment pq. The radius vector ee the point p : (x1, y1) is the vector Op = (%1, 1), where O is the origin. The vector pq, as represented by the directed line segment pq, is said to have initial point p and terminal point q (cf. Fig. 801-1).

(h, k)

q

|

| (0,ieee

The vector (h, k)

Dp

The vector pq

(a+th,b+k)

y D

;

fee 7

O (a, b) The vector (h, k)

The radius vector Op Figure

Example.

801-1

If p = (3, —1),¢ = (1, 5), r = (6, 4), and s = (4, 10), then

pq = rs = (—2, 6). Nore: The preceding example illustrates the fact that two directed line segments represent the same vector if and only if they are parallel, similarly directed, and of equal length.

802

LINEAR

COMBINATIONS

OF VECTORS

Addition of vectors and multiplication of a vector by a real number are denoted and defined:

(1)

(u1, U2) + (v1, v2) = (ur + U1, U2 + v2),

(2)

Mui, U2) = (Aut, Au2).

That is, if # = (wi, u2) and 6 = (v1, v2) are any two vectors, then the sum @ + 6 of ii and @ is the vector W = (w1, W2), where wi = uw + 1 and w2 = u2 + v2, and the product \i is the vector whose coordinates are Au; and du2. Any real number used in the context (2) is called a scalar, and the vector Au is called a scalar multiple of a.

204

IN THE

VECTORS

[§802

PLANE

Vector addition can be represented either by means of the diagonal of a parallelogram (Fig. 802-la) or the completion of a triangle (Fig. 802-15). Multiplication of a vector ii = pg by a scalar gives a vector represented by a line segment parallel to the original segment pq (cf. Fig. 802-1c).

(b) Figure

(c)

802-1

Since vectors represented by parallel directed line segments can also be represented by directed line segments on the same line, such vectors are called collinear. Thus, for all scalars \, the vectors # and dw are collinear. If \ > 0, % and du are similarly directed; if \ < 0, # and Xu are oppositely directed. If \ = —1, the vector (—1)u, also written —i, is called the negative of uw. Any vector and its negative are therefore oppositely directed collinear vectors of the same magnitude. Vectors that are not collinear are called noncollinear. Subtraction of vectors is denoted and defined:

(3)

a—b=a+4+

(3D).

If u and 6 are represented by directed line segments with a common initial point, the vector “ — & is represented by the directed line segment from the terminal point of & to the terminal point of u (cf. Fig. 802-2).

Figure

Division of a vector by a nonzero scalar is defined: 4 (4)

ae

|

£ = -0Dv. ear

The result of dividing a nonzero vector by its magnitude:

(5)

oe

|O|

802-2

§803]

LINEAR

DEPENDENCE

AND

INDEPENDENCE

205

is a unit vector collinear with it, and similarly directed. The vector (5) is the result of normalizing 3. Vector magnitude has some of the properties of absolute value in the real number system (Theorem I, §311). Proofs of the following properties, for two-dimensional vectors, are requested in Exercises 32-35, §804, where suggestions are provided: Properties of magnitude

(Z)

|é| = 0; |é| = Oif and only if 6 = 6.

(ii) [dO] = |A|6], A € AB. (iii) al=_ [al Rp d€ B\ 10) (iv)

The triangle inequality* holds:

| + 6| < |a| + |al.

(oe) |e] = lol; la — o| = |6 —al.t

(vi)

|a — 6] S |a| + fd}.

(vii)

||a| — |é|| S la — dj.

The vector \u% + ub, where d and yw are scalars, is called a linear combination of the vectors «7 and 6. More generally, a linear combination of a set of vectors 01, 02,°**, 0, is a vector of the form

(6)

NET

A202

NO

where Aq, A2,°°*, An are scalars. Note. the set of addition, MG + ©)

It is not difficult to see that the conditions of Definition I, $509, are satisfied by U2 of all two-dimensional vectors (U2 is a commutative group with respect to 16 = 6, and the homogeneous law (ui) = (Aw)éi and the distributive laws = AX + AD and (A + pw)#* = AX + pH hold), and therefore U2 is a vector space

over §&.

Example.

(3, 5) + (—1, 4) = (, 9),

Gi.)

— 4, 4) = a1),

62; =3) = (12, =18): =—(=3,

i= G01),

|(2, —6)| = V4 + 36 = 2y10, |3(1, —2)| = |G, —6)| = 45 = 3V5 = 3|(1, —2)J, iat) + (1, 4)

803

LINEAR

DEPENDENCE

14, 3)| = 10'S V5 + yt7 = (2, —1)| + (1, 4).

AND

INDEPENDENCE

The following fundamental definition applies to ah arbitrary space U (of vectors) over a field of scalars 5: *In geometrical terms, no side of a triangle can exceed the sum of the other two (cf. Fig. 802-14). +The absolute value of the difference between two vectors having a common initial point is the dis‘tance between their terminal points (cf. Fig. 802-2).

[§803

VEGTORSAINSTHERRLANE

206

Definition I.

The vectors 61, 62,:: +, 6, are linearly dependent if and only if there

exist scalars \1, \2,°**, \n not all equal to 0 such that

(1)

181 + dade ++ +++ Andn = 8.

If no such scalars exist, that is, if (1) holds only for scalars \1, X2,°**, Xn all of which

are zero, then the vectors 61, 62,:**, 6, are linearly independent.*

We begin our discussion of linear dependence and independence by considering sets {01, d2,--°, 8,} Of two-dimensional vectors for the special cases n = 1, n = 2, and n = 3,inturn.

I.

We start with n =

1.

A single vector 6 is linearly dependent if and only if it is the zero vector: b = 0.

Proof. If 6 = 0, then there exists a nonzero scalar 1 such that ld = 6 = 0. Conversely, if there exists a nonzero scalar \ such that \d = 0, then division by d gives ee = xOr2)

~1%

| : (;\)e

; Ila Daas es

.

and 6 must be the zero vector.

II. Two vectors ui and 6 (in two dimensions) are linearly dependent if and only if they are collinear or, equivalently, if and only if one is a scalar multiple of the other. If one of the two vectors i and b is linearly dependent — that is, equal to the zero vector — then the two vectors tu and 0 are linearly dependent. If tu is a linearly independent (nonzero) vector, and if ti and t are linearly dependent (6 is a vector collinear with u), then 0 is uniquely representable as a linear combination (scalar multiple) of i. Proof. Assume first that uw and 6 are linearly dependent, that is, that there exist scalars \ and uw not both of which are zero such that

(2)

hu + pd = O.

If « + 0 we can solve (2) for 3b:

(3)

b= —*G, bb

and w and 0 are collinear (even in the extreme case when d = 0). Similarly, if \ ~ 0, we can solve for w in terms of 6. Conversely, if i and 6 are collinear it is possible to write one as a scalar multiple of the other, say: (4)

D =

pi

where vy is an appropriate scalar multiplier depending on the lengths and relative sense of u and 8, and consequently w and 6 are linearly dependent:

(5)

vi + (—1)6 = 0,

*It should be appreciated that the concepts of linear dependence and linear independence apply to sets of vectors rather than to the vectors as individuals. Our terminology is well established in general mathematical usage. For example, we speak of “two disjoint sets” (although neither set is “disjoint” by itself), “‘similar triangles” (although no triangle is “similar” by itself), and “perpendicular lines” (although no line is “perpendicular” by itself). In contrast to these examples displaying properties of sets are the following examples displaying properties of individuals: “bounded sets,” “equilateral

triangles,” and ‘‘vertical lines.”

$803]

LINEAR

DEPENDENCE

AND

INDEPENDENCE

207

where at least the coefficient of 6 is nonzero. Next, assume that one of the vectors u and 0 is the zero vector — for definiteness let z = 0. Then

(6)

la +0-5=

1:04+06—6,

and uw and © are linearly independent since at least one (exactly one in this case) of the two scalars | and 0 in (6) is nonzero. Finally, if # and é are collinear and a # 0, then u in equation (2) must be nonzero, so that it is possible to write é as a scalar multiple of u: 6 = vi. To show that this representation is unique, we write 6 = ru = y2u, and

by subtraction

and

a distributive

law (cf. the Note, §802)

obtain

(v1 — v2) = 0 and consequently, since # # 0, ») — v2 = 0, and ») = 12. III. Three vectors ty, 6, and w (in two dimensions) are always linearly dependent. For any three vectors it is always possible to express one as a linear combination of the other two. If u and 6 are any two linearly independent (noncollinear) vectors, then any vector W is uniquely representable as a linear combination of i and 6b.

Proof. We shall prove the third statement first, using a geometrical argument (for a purely algebraic proof, cf. Example 3, $907). Accordingly, let a and 6 be any two noncollinear vectors, and let w be an arbitrary vector. Since u@ and 0 are noncollinear, both are nonzero and can be represented by nonzero noncollinear directed line segments having their initial points at the origin O (cf. Fig. 803-1). If # is also represented by a directed line segment from the origin, let lines L and M be drawn through the terminal point of w parallel to the two vectors 6 and 4d, respectively. Let U and V be the vectors collinear with # and b, respectively, having as initial point the origin and as terminal points the appropriate intersections of the =

3

f

lines L and M with the lines containing # and 6. Then U and V can be written as scalar multiples of # and 3d, respectively:

)

=

have w represented as a linear combination of a and 0:

(7)

E

U = \u, V = wo, and since

Ww = Ad + po.

Figure

803-1

Y

w =

>

>

U + V, we

IN THE

VECTORS

208

[§803

PLANE

To prove the uniqueness of the representation (7), we assume written:

(8)

that w can be

W = iW + w1H = 2d + pad.

Then, by subtraction and factoring by a distributive law, we have (A; — d2)u + (u1 — y2)d = O, and since w and 6 are linearly independent (by II, above), A1 — A2 = uw) — w2 = 0, A1 = A2, and yw = pz, and the representation (7) is unique. Furthermore, as a result of (7), we can say that a, 6, and W are linearly dependent since (7) can be put in the form

(9)

hi + pb + (—1)¥ = 0,

and at least the scalar coefficient —1 is nonzero. Now let i, 6, and w be any three vectors. If # and @ are linearly independent we have shown in (9) and (7) that

i, b, and Ww are linearly dependent and that w can be expressed as a linear combination of u and o. Assume finally that # and 6 are linearly dependent, and let \ and yu be scalars, not both zero, such that \w + yo =

(10)

0. Then,

hi + pb + Ow = O,

and since X, uw, and O are not all equal to 0, uw, 6, and w are linearly dependent. Furthermore, it is possible to solve for a or 6 or both (depending on the location of

a nonzero coefficient) in terms of the other two vectors. Definition I].

Let S =

This completes the proof.

{01, b2,---,6,} be a set of vectors in a vector space V over

a field of scalars §. Then the vectors ofS are said to span the space V if and only if every vector b ofU is expressible as a linear combination of 61, 62,° ++, bn: UO =

{drb1

+ Azd2

+--+ ++ Anda | Ar € F, A2 € Fy +, An E SF}.

The set S is a basis or a base set of vectors for O if and only if the vectors 61, 62,°*-, 0, span ‘0 and are linearly independent. Owing to the property expressed in the third sentence of III, above, any set of two linearly independent (noncollinear) vectors % and @ in the plane is a basis for the space U2 of two-dimensional vectors. The two linearly independent vectors # and é span the space U2, since {Az + wo|r and wp € R} = V2. Similarly, any single nonzero

vector % spans the space Uj; consisting

of all scalar

multiples \w of i,

and {iz} is a basis for the space U1. In the two-dimensional vector space U2, the standard basic pair of unit vectors are denoted and defined:

(11)

i= (1,0), HO).

The vectors i and j are also called the standard coordinate vectors of U>.

For this

basis the representation of an arbitrary vector 6 = (uv, v2) is particularly simple: (12)

v=

vi AF va].

We conclude this section by establishing three fundamental theorems that underlie the entire theory of linear dependence and independence. These incorporate

$803]

LINEAR

DEPENDENCE

AND

INDEPENDENCE

209

some of the results derived above, but are so expressed that they are available for vector spaces in general. Theorem I. /f the vectors of a finite set S are linearly independent, then the vectors of any subset R of S are also linearly independent. Equivalently, if the vectors of a finite set R are linearly dependent, then the vectors of any finite superset S of R are also linearly dependent. Proof. Let R consist of the m linearly dependent vectors 61, d2,---, im, and let S consist-of the--vectors 01, 02;-* >, Gm,°>:, On, wWhetre-n 2 m. If \1, 2,-**; Am are

scalars not all of which are 0 such that \18; + A2d2 +---+ Anim = 6, then 0’s can be adjoined to produce the set of numbers \j, \2,°°-, Am, 0,°--, 0 not all of which are O such that didi + Azb2 +°°>+ AmOm + Odme1 +:--+ 05, = 0, and therefore the vectors of S are linearly dependent. Theorem II.

The vectors 61, 62,:: +, tn, where n > 1, are linearly dependent if and

only if one of the vectors is a linear combination of the others. Proof. “If: Assume that the vector combination of the others:

(13)

Ue = MOT

0, where

Nea iteet

k € {1, 2,---,n},

Ang bet

is a linear

PA, 0.

Then since

(14)

Nor

Bees + Ag tbe

+ (Lb

+ Nee tBee1 + °°

+ Ande = 8,

and not every coefficient in (2) is equal to 0 (—1 # 0), the vectors are linearly dependent. “Only if’: Assume that the vectors 01, 52, :-: , 0, are linearly dependent, and in the equation

(15) assume

A181 + Azd2 + +++ + Anon = O, that d, is a nonzero

Nor fees

Ht Agr beer

coefficient.

Then

+ Angi Oe41 + °°* +Andn,

from (15) we can

and by dividing

write

—A,d, =

by —)x

solve

for ox:

(16)

; (BY

Theorem III.

eee (2s tice dett7 et

The vectors

Ha

Sse A

Vk+1

ee

01, 62,°::, 6, are linearly independent

i Dp.

if and only if

whenever a vector b is represented as a linear combination of 01, 62,°**, Un, this representation is unique.

Proof. “If”: The vector 0 can be represented as a linear combination (1) with all coefficients equal to 0. If this representation is unique, then (1) holds only for scalars \1, \2,°°*, An all of which are 0, and the vectors &1, d2,---, 5, are linearly independent by definition. “Only if”: Assume that 01, 02,°°*, Un are linearly independent, and that

+--+ + Ande = wib2 + w2b2 + °°

B= dyO1 + Azd2

(17)

[§803

PLANE

IN THE

VECTORS

210

+ Madr.

By subtraction we can write

(18)

(Ar — mi)dir + (2 — pa)i2 +--+

+ On = Ma)on = 0,

and by the linear independence of 61, 52, ---, d, every coefficient on the left of (18) must be 0.

Therefore

\1 = ui, \2 = #2,

°°°, An = Mn, and the representation (17)

as a linear combination of 1, 2, -+-, 0, is unique.

of

Example 1. The vector (0, 3) is linearly independent since it is nonzero. The vectors (3, 9) and (2, 6) are linearly dependent since 2(3, 9) + (—3)(2, 6) = (0, 0). The vectors (2, 5) and (1, 3) are linearly independent since neither is a scalar multiple of the other; for example, there is no real number X such that (2, 5) = A(1, 3), or 2 = \ and 5 = 3X. If (x, y) is an arbitrary two-dimensional vector, then (x, y) is a unique linear combination of (2, 5) and (1, 3):

(19)

(xy y= ACPD)

ny 3);

since the corresponding system in ) and yn,

2r SUG iptcee = teed 56,

(20)

SA + 3u = y, has the unique solution \ = 3x — y, uw=

Example 2.

—5x + 2y.

Prove that the diagonals of a parallelogram bisect each other.

First solution.

Let ABCD

(Fig. 803-2a)

midpoint of the diagonal from A to C.

be the given parallelogram, and let M

By assumption, 4M

= MC.

be the

We wish to prove that

M is also the midpoint of the diagonal from D to B or, in vector notation, that DVM = MB. —>

Using the equality CB = DA we have —

DM

—

MB

aa

—

2

—

= DA + AM, =

—

—

MC + CB = AM

+ DA.

By the commutative law (Note, §802), the desired equality is obtained.

A

B

Ee

(a)

(b) Figure 803-2

Second solution.

Let P be the point of intersection of the diagonals (Fig. 803-24). _—

—

—

—

—>

—-

—

—>

—

Then

—

AB = AP + PB = DC = DP + PC,

and consequently by subtraction,

(21)

MP = IKE = IDP = JB.

Since the vectors on the two sides of (21) are collinear with AC and DB, respectively, and since —_ 4 Z AC and DB are linearly independent, each of the two differences in (21) must be the zero > ~ —> — _ 4 vector 0, and hence 4P = PC and DP = PB. In other words, P is the midpoint of each diagonal.

—_

§804]

804

EXERCISES

211

EXERCISES

In Exercises 1-6, find the point s from the given data.

1. pg = 73,p= (3, 8),q = (4, 5), r= (6, 7). 2. Pg = =7rs,p= (1, 2),q= (6, —5), 7 = G, 4). SoA 3%

Pq

— =

Os, P =

Es 4.

Op

(—7,

THA) Ng

(—7,

11), q
|3; —4)|:

a

6i+4j|

a,

—————

|

14. |—4(5i + f)|= |—4]-|5i +J].

|6i+ 4y|

| (3, —4)

ry

16.

iG, —4)|

—

1

17. |(4i — 7/) — 2 + 3/)| = \(2i + 37) — Gi - FI. Teo a 7) (Sal) 1G — 7), 19. |(5i — 6j) + (i+ 2j)| S |5i — Oy] + [37+ 2i].

20. ||(6, 2)! — |G, 3)l] S |, 2) — @, 3). In Exercises 21-26, determine whether the vectors are linearly dependent or linearly independent. Give reasons for your conclusion.

21. 3i. pan lB) (3). — 9):

22. (0, 0). 2A Gi 5y) 0:

7500.15), (255):

26. i+), 21+ 4), 31+ 9.

In Exercises 27-29, express one of the vectors as a linear combination of the remaining vectors.

Be; = 30), — 181-1527):

28. (0, 0), (6, 17).

py

OFA. Vie ae

In Exercises 30 and 31, use vector methods to prove the given geometrical theorem.

30. The segment joining the midpoints of two sides of any triangle is parallel to the third side and half as long. 31. The medians of a triangle (that is, the line segments joining the vertices to the midpoints of opposite sides) intersect in a point of trisection of each.

In Exercises 32-35, prove the specified properties of magnitude of vectors, given in § 802.

32. (i)-(iii).

Suggestion for (iii):

1D)

Use (ii) and |é| = Ari

33. (iv).

lent

to

Suggestions:

By Exercise

[$805

PLANE

IN THE

VECTORS

212

7, §305, if a = (ui, u2) and 6 = (v1, v2), this is equiva-

(ua + v1)? + (v2 + v2)? S ui? + up? + 2Vur2 ae uz2\o12 + v22 + vi? + 22,

uv) + u2v2 S Vui2 ae uy

Vo12 + 22.

|u101 4 4202| and (ujv2 —

u2v)? = 0.

34. (v)(vi).

35. (vii).

Suggestion for (vi):

Suggestion:

From

Square

again,

making

use

of

mv

+ u2v2

OF S

|a — 6| = |a+ (—d)| S [a] + |—-F.

(iv),

|a| = |(@ — 6) + 8] S |a— 8] + jd]

and

[a] =

(6 — @) + a| S |a — b| + [a]. Use Theorem I(v), §311. 36. Prove that if p is the midpoint of a line segment pip2, and if O is the origin (all in the Euclidean plane), then —

=>

OPiri Op? Op = PS.

a

In particular, deduce that the radius vector of the midpoint p of the segment joining the two points (x1, v1) and (x2, y2) is

On 2: (3 ar x2 V1 i ). 2 2 37. Prove that the sum of any number of vectors (in the plane) is the zero vector if and only if they can be represented by the sides of a closed polygon (possibly self-crossing) directed in cyclic sequence.

805

DIRECTION

ANGLES,

COSINES,

AND

NUMBERS

For certain purposes it is more convenient to study a general line or directed line in the plane in terms of two angles, rather than the single angle known as the inclination whose tangent is the s/ope of the given line or directed line. For our present objective we must redefine the measure of an angle. If Land M are any two directed lines in the plane, we shall define the measure of the angle 6 between L and M as follows: If L and M are parallel and similarly directed, then 6 = 0; if L and M are parallel and oppositely directed, then 6 = 7. Otherwise, @ is the least positive measure of the angle between the rays issuing from the point of intersection of L and M, in the directions ofL and M (cf. Figure 805-1). In any case, then, the angle 0 between two directed lines is symmetric — the angle between L and M is equal to the angle between M and L — and satisfies the inequalities:

(1)

Ors" 7:

Figure

805-1

§805]

DIRECTION

ANGLES,

COSINES,

AND

NUMBERS

213

Any directed line L in the plane has two direction angles: (7) the angle a between L and the (positively) directed x-axis, and (ii) the angle 6 between L and the (positively) directed y-axis. (Cf. Fig. 805-2, where the line L is taken through the origin for convenience.) The cosines of these directions angles, (2)

A=cosa,

wu = cos 8,

are called the direction cosines of L.

E

1

+e)

&

Ye

First quadrant

Second quadrant

lid

ie R

YR

Third quadrant

Fourth quadrant Figure

805-2

Direction angles and direction cosines of any nonzero vector or directed line segment are defined as with a directed line. If d is a nonzero vector represented by the directed line segment from the point p: : (x1, y1) to the point p2 : (x2, y2), and if d is the (positive) distance between p; and pz, then the direction cosines of 6 are equal to

h=cosa==—

(3)

"|, p= sos 6 =

—

=

=— (01, 02) = (2 = X1, y2 — Vi):

OreiromessO2,

ie

(4)

= cOS a = A oo

VI =) a

v2 a (= = COS [83 =5—,

D Fi (A,Ns 2) =, ae,

(Cf. Fig. 805-3.) Since d2 = (x2 — x1)? + (v2 — y1)?, we have from (3):

(5)

2 + p2 = 1.

That is, for any directed line L in the plane, the sum of the squares of the direction cosines is equal to 1. In other words, if \ and yp are the direction cosines of a directed line L, then (d, ») is aunit vector. This unit vector (A, ~), sometimes called the direction of L, is collinear with L and similarly directed.

VECTORS

214

Py: (%4, 4)

IN THE

[§805

PLANE

Do: (Xo, Yo)

Po:(Xo, Y2)

e

e

Uy= X_— Xy

Dy: (x1, yy)

Do: (X9, Yo)

Pg: (Xo, Yq) Figure

805-3

Any ordered pair (/, m) of numbers / and m not both of which are zero is a set of direction numbers of a directed line or line segment L if and only if the vector (/, m) is collinear with L, or equivalently, if and only if the vector (/, m) is collinear with the vector (\, ). This condition of collinearity holds if and only if there exists a nonzero constant k such that (/, m) = kQ), uw) = (kA, ku), or 1]= kX and m = ky. From (3), with k = d, we see that x2 — x; and y2 — y; are direction numbers of the directed line segment from (x1, yi) to (2, y2). Since a vector (/,m) collinear with a direction (A, u) is also collinear with the opposite direction (—A, — yz), / and m are called direction numbers of the undirected line L as well as of the directed line L. Example. The vector (—2, 3) represented by the directed line segment joining the point (—1, 1) to the point (—3, 4) has direction cosines

(6) (Cf. Fig. 805-4.)

—p) \ = cosa= ae

fi

y13

cos B i :

y13

With the notation used above, d = v13, identity (5) becomes ;“5 + 7°5 = 1, y

Figure

805-4

$806]

THE

SCALAR

OR

INNER

OR

DOT

PRODUCT

215

and the following are ordered pairs of direction numbers for the line L through (—1, 1) and (—3,4):* (—2, 3), 2, —3), (46); (6,-—9), (—20, 30). The two directions for this line are 3

( ee =.) a( 2 Ve) icy aa Eien eye

806

THE

SCALAR

OR

INNER

OR

DOT

PRODUCT

The scalar product or inner product or dot product of two vectors D = (Yj, v2) is denoted and defined:

(1)

.

ab

T S =

+

=

# = (uw, w2) and

S

In this section we shall discuss the geometric significance of this concept.* If either @ or o is the zero vector, w-6 = 0. Assuming for present purposes that neither @ nor 0 is the zero vector, let 6 be the angle between w and 6, and let (Ai, 1) and (A2, u2) be the unit vectors whose components are the direction cosines of a and 0, respectively. Applying the law of cosines to the triangle whose vertices are the origin O and the two points pi : (Ai, u1) and pz: (A2, uw2), and denoting by c the distance between p; and pz (cf. Fig. 806-la), we have:

eee Ce

Ee

eGo

2A

© 2A? ==

Fait)

ss

te A) rats (A

or after simplification (since \1;2 + wi2 = A22 + p22 =

(2)

D

Oe? 5

22)

22) = 201A2- + wha)

1):

COS 6 = d1A2 + pie.

y

(a) Figure

From

806-1

(4), §805, we have wm = dla, u2 = mila, v1 = d2[6|, and vz = u2|6|, and

hence wv;

+ w2v2

=

(AiA2 + uim2)|i|

|.

By (2), then, wv;

+ u2v2 =

|| - [5| “COS 0.

If, for the case where either @ or & is the zero vector we permit the indeterminate angle @ to be arbitrarily assigned, we have from (1) the general theorem: *This product is called a scalar product because the result is always a real number.

VECTORS

216

Theorem I.

IN THE

[$806

PLANE

The scalar product of two vectors i and & is equal to the product of

their magnitudes and the cosine of the angle between them:

ub = |ii|-|b|-cos 6.

(3)

If we agree to say that the zero vector is orthogonal (perpendicular) to every vector, we have the following corollary to (3):

Theorem II.

Two vectors are orthogonal if and only if their scalar product is zero.

Example 1. The vectors (10, —15) and (12, 8) are orthogonal since 10-12 — 15:8 = 0. The vectors (11, —5) and (6, 13) are not orthogonal since 11:6 — 5-13 = 1 # 0.

We now assemble a number of facts concerning the scalar product, whose proofs are left to the exercises (Exs. 38-41, $808):

(i) The dot product of a vector and itself is often denoted by means exponent 2. In this special case (% = 6), formula (3) gives

of an

w= ii = |al?.

(4) (ii)

If the component of w in the direction of 6, denoted comp, wu, is defined to be

the number || cos 6 (cf. Fig. 806-16),

then z:-6 = (comp,a)|6|.

In particu-

(iii)

lar, if 6 is a unit vector, w7:-6 = comp,w, and if u@ = (uw, u2), comp; and compjt = uz. (Cf. (11), $803.) The following commutative law holds for all vectors # and ob:

(iv)

The following two distributive laws hold for all vectors w, 6, and w:

= m4

uv = Bu.

u(6+ WW) =a (i + 3):0 | (v)

b+ a-w, uw + vw.

The following homogeneous law holds for all scalars i and ob:

\ and yw and vectors

(Ad): (ub) = (Au)(Ud). (vi) (vii)

(5)

The associative law for scalar products is meaningless product of two vectors is not a vector.

since the scalar

If % = (a, b) is any nonzero vector, if 7; = (x1, y1) is the radius vector of a fixed point (x1, y1), and if 7 = (x, y) is the radius vector of a general point (x, y), then the equation of the line L through (x1, y1) orthogonal to a can be written

u-(F — 71) = Ax — x1) + 460 — yi) = 0. Consequently, if ax + by +c =0 is the equation of any line L, the coefficients of x and y are the components of a vector orthogonal to L. (Cf. Fig. 806-2a.)

(viii)

If 6 = (J, m) is any nonzero vector, if 7; = (x1, y1) is the radius vector of a fixed point (x1, yi), and if 7 = (x, y) is the radius vector of a general point (x, y), then the point (x, y) lies on the line M through (x1, y1) parallel to é

$806]

THE

SCALAR

OR

INNER

OR

DOT

y

PRODUCT

217

y

u

3

Se

L

(x1, Y;)

M

(a, 92)

(X4,1)

=

x

x

(a)

(b) Figure

806-2

if and only if 7 — 7; and 6 are linearly dependent; that is, if and only if the components of 7 — r; and those of é are proportional: Of 2,

(6)

Sr

=

Va

say

ere

where a zero denominator in (6) is to be interpreted to mean corresponding numerator is also zero. (Cf. Fig. 806-20.)

that the

Example 2. The equation of the line through the point (3, 4) perpendicular to the vector (5, —2) can be written (5, —2)-(x — 3, y— 4) = 5(x — 3) — 2(y — 4) = 0, or 5x — 2y = 7. Example 3.

The equation of the line through the point (3, 4) parallel to the vector (5, —2) x-—-3 y-—4 can be written » or 2x + Sy = 26. The equation of the line through the point

—

=2

. Ray (3, 4) parallel to the vector (5, 0) can be written a

oa Sane

or

y—4=0,

or y = 4.

Note. A vector, as defined in $801, is an ordered pair of real numbers. This ordered pair is necessarily associated with a particular rectangular coordinate system. A change of coordinate system may produce a change in the ordered pair. (An example is indicated in Figure 806-3.) However, since the magnitude of any vector and the angle between any two

v2, -v3)

Vibe (a)

(b) Figure

806-3

IN THE

VECTORS

218

[$807

PLANE

vectors are unchanged by any rectangular coordinate transformation, Theorem I establishes the fact that the scalar product of any two vectors is independent of the rectangular coordinate system. (In Figure 806-3, the magnitudes of # and @ are 2 and 2, respectively, in either coordinate system, the angle 6 is 47 in either coordinate system, and the scalar products are

2:14+0-:1=2

and

y2:y2— y2-0 = 2,

in coordinate systems (a) and (4), respectively.)

Example 4. Simple Simon defines what he calls the circle product of two vectors u = (u1, u2) and 6 = (v1, v2), thus: + u20Y. HO’ = uv2

(7)

Show that the circle product is not independent of the coordinate system. Solution. The vectors # and 6 of Figure 806-3a have the circle product 2, while the vectors a and t of Figure 806-35 have the circle product —2.

Example 5.

Discuss the cancellation law for scalar products:

(8)

(@@-6 = 0-0) > db = W.

Solution. The example di = i+j, 6 = i,and v=] shows that (8) is false as it stands, since (i+ 7): =i-i+/7-i=1 and @+/)-j =i-j+ 7-7 = 1. However, if the equation u-b = a-w holds for a// vectors 4, then 6 and W must be equal. This follows from the fact that if 7-6 = u-W for all a, then #-(6 — Ww) = 0 for all @ In particular, this last equation holds when u# = 6 — wW, and therefore, by (4), b — W =

807

: 0, and

(i — w)-(6 — w) = |b — we = 0,

d = Ww.

EXERCISES

In Exercises 1-3, find the direction angles of the directed line segment from the first point to the second.

1 ON4) C7):

(Os

DE(HZ, 3).

3. (3, —2, (—1, 2).

In Exercises 4-8, find the direction cosines of the given directed line or line segment

4. Positively directed x-axis.

5. Negatively directed y-axis.

6. From the origin into the first quadrant, making equal acute angles with the coordinate axes. Thy Jertoan (SO) ior, 3):

8. From (1, 5) to (4, 4).

In Exercises 9-11, use (5), §805, to show that the two given angles do not form a pair of direction angles for any directed line.

7, at47. 9. zis

1 10. 0, 4r.

ie

In Exercises 12-14, find three distinct sets of direction numbers for the directed line segment from the first point to the second.

1b4, @, 3), @& 12).

13. (—1, 6), G, 1).

14. (2, 4), (—5S, 5).

In Exercises 15-20, find the scalar product of the two given vectors. 15. i and i.

16. i and j.

ig fy 7 and i — i

$807]

EXERCISES

18. (3, —5) and (2, —3).

219

19. 61 — j and 2i + 7/.

20. (—3, 1) and (2, 4).

In Exercises 21-23, use (3), §806, to find the cosine of the angle between the two vectors.

2A Oe

f) (= 21).

9 Fah) SE CRG 6

BEES

2G Op.

In Exercises 24-26, determine whether or not the two vectors are orthogonal.

2AM(G7410), (6.13):

26. (7, 9), (14, —11).

255121 sy, 121 4.9),

In Exercises 27-29, find the two unit vectors that are orthogonal to the given vector.

Sie

2a).

29:0, —6).

ONE

In Exercises 30 and 31, find the component of the first vector in the direction of the second.

207687).

1.

BI 6

ee:

32. Verify the two distributive laws (/v), §806, for the three vectors # = (2,

—4), 6 = (—3, 1),

im = (UO)

33. Verify the homogeneous law (v), §806, for the given scalars and vectors. (A OL Di

\ =

—3,

In Exercises 34 and 35, use (vii), §806, to obtain the equation of the line through the given point (x1, yi) and orthogonal to the given vector i.

34. (x1, v1) = (2, 0), &@ = 5i+ 7).

35. (x1, y1) = U, —4), @ = (8, —3).

In Exercises 36 and 37, use (viii), §806, to obtain the equation of the line through the given point (x1, v1) and parallel to the given vector 0.

36. (x1, 11) = (—2, 1), 6 = ©, 4).

37. (Hi, yi) = 3,4), b = 21 —j.

In Exercise 38-41, prove the indicated facts listed in §806. 38. (i)-(ii).

39. (éii)-(iv).

40. (v)-(vi).

41. (vii)-(viii).

42. Give an algebraic proof of the inequality

(1)

|a-6) < |a|- la], or luv + wove] S Yur? + u2? Vor? + m2,

and relate the result to the trigonometric inequality |cos 6| < 1. Suggestion: suggestion given in Ex. 33, §804.

Use the last

43. Prove that equality in (1) holds if and only if the vectors # and 0 are linearly dependent.

44. Prove the general triangle inequality:

jo: + bo +---+8,|

S bi] + leo] +--++ lel.

45. Prove that any two nonzero orthogonal vectors are linearly independent.

46. Prove the parallelogram law for two-dimensional vectors: The sum of the squares of the diagonals of any parallelogram is equal to the sum of the squares of the four sides,

@ + ty + a — 0? = 207

0).

47. Simple Simon defines what he calls the square product of two vectors 4 = (u1, u2) and @L] 6 = wuz + viv. Show that the square product is not independent of b = (v4, v2) thus: the coordinate system.

Matrices

901

MATRICES

AND

THEIR

LINEAR

COMBINATIONS

Two important generalizations of the concept (ordered pair of real numbers) are: (i)

of a two-dimensional

vector

ann-dimensional vector, or ordered n-tuple of real numbers, written Vi v2

(1)

Di

(C0On)

On

—

;

)

Un

consisting of the m components or coordinates v1, v2,:--, U, € ®, arranged either ina horizontal row or in a vertical column;

(ii)

(2)

anm

Xn (read “‘m by n’’) rectangular matrix

A

=

a\1

ai2

ae

Aln

a21

a22

sarees

a2

Ami

Am2

ah

fa

:

)

Amn

consisting of the components or coordinates of the m row vectors (a11, @12,°**, Gin), (21, G22, ***, G2n), ***, (Ami, Gm2,°**, Amn) Or equivalently of the components coordinates of the n column vectors 220

or

§901]

MATRICES

AND

THEIR

LINEAR

COMBINATIONS

a1

ai2

Ain

Ami

m2

Gran

221

The double subscript notation a;; of (2) indicates the element of 4 in the ith row and jth column, the matrix A also being denoted 4 = (a;,) if the numbers m and n are otherwise indicated or clear from context. If a matrix A is denoted (a;,), the element ai; 18 called the general element of 4. An n-dimensional row vector can also be thought of as a 1 X n matrix, and an m-dimensional column vector can also be thought of as an m X 1 matrix. The row vectors and column vectors of a matrix are also called rows and columns for simplicity. Two matrices have the same shape if and only if they have the same number of rows and the same number of columns. An m X n matrix is square if and only if m = n. Equality between two matrices A and B, written A = B, holds if and only if A and B are identical element by element: A and B are of the same shape, m X n, and a;; = 6;; fori = 1, ---, m andj = 1, *++, n. Inequality, written A ~ B, means that equality A = B is false. The main or principal diagonal of a square matrix A = (a;;) consists of the elements aj;, i= 1,---,n. The order of a square nm X n matrix is n. Although we shall obtain results for general m X n matrices, most of the concrete

applications of the general theory will be to matrices where m and n are small, usually less than or equal to 3 in this chapter, and less than or equal to 4 in the following Chapter 10. All that is important in the present section is that the matrices being considered have the same shape. We now State the fundamental definition that permits us to form linear combinations of matrices having the same shape, and to regard the set of all such matrices as a vector space: Definition. Jf A = (a;;) and B = (b;;) are any two m X n matrices and if d is any real number, called a scalar, then the sum C = A + B of A and B and the scalar multiple D = \A of\ andA are them X nmatrices C = (ci;) and D = (dij) defined by (3)

Ciy

= ay

+

Oj,

Oi) =

Gry,

if =

l1,225

(pas

Ohi e

or, in more complete form:

Qi1*** Gin Dut

if

ees ine

bi - + Din Dr


u%,, the vector Ayu) + A2w2 +--+ + Anu, Can be the zero vector all of whose components are 0 only incase \; = A2 =:::=A, = 0. The n vectors (5) are the standard basic unit vectors of U,1.

902

LINEAR

TRANSFORMATIONS

AND

PRODUCTS

OF MATRICES

Let T be a function whose domain is a vector space 0 and whose range is a subset of a vector space WW. Such a function is called a linear transformation of 0 into W if and only if it has the following two properties:

(1)

Tis additive:

wandobc

U>

Ti + &) = TH) + To), (2)

T is homogeneous:

\ € ® and #@ €

VU=> T(Au) = AT(U).

For reasons of notational simplicity rather than of conceptual necessity, we shall now particularize both spaces U and ‘W to the space U2 of two-dimensional vectors. Let T represent an arbitrary linear transformation of U2 into U2. We shall show that T corresponds in a specific fashion to a unique 2 X 2 matrix A. It turns out to be more convenient if the space U2 is thought of as the space V2: of two-dimensional column vectors than it is if U2 is regarded as the space U12 of two-

dimensional vector:

row vectors.

ae [Ves Thus, we shall write 7 = (5):j=

0 (‘) and a general

[§902

224

MATRICES

(3)

= wi + Wj = a0) pee 8

%

_

0

(3) a fe 5s(":)

We now construct the matrix 4 as follows: Let the first column of A be the image under T of the vector i, and let the second column of A be the image under T of the vector /: a

.

isa

l

Q2\

0

a

i) a r(4) a 7: r(?)

ail

a12\

a ie i

Using the linearity of T we can write out the image 6 of u = ui + ujin terms of the elements of A:

(5)

6 = T(i) = AG) = T(mi + uj) = wT) + wT) ey

ail a2\

ahs ie a22

Cie uia21

a ee u2Q22

a (ae =. ai2u2\_ a21u, + anu)

In other words, if T(@@) = 6 = on ae v2J, then

(6)

Vr = a1

+ Qy2uW2,

v2 = a2iu1 + a22U2.

Since the basic unit vectors i andjdetermine the elements of the matrix A given by (4), the matrix A that defines the mapping 7 according to the equations (6) is unique. To express this idea more specifically, assume there is a 2 X 2 matrix B = (6;;) such that if 6 = T(%) then (7)

Oy = biiuy + Di2t2,

v2.= bait) + boar.

Then if uw; = 1 and w2 = O equations (6) and (7) give ai1 = bi; and az; = 621, while if uy = O and uz = | equations (6) and (7) give diz = bj2 and a22 = b22, whence

AmB. The converse of the preceding result is immediate: If A is any 2 X 2 matrix, then the transformation T defined by (6) is a linear transformation of U2 into U2. In other words, the linear transformations of U2 into U2 and the 2 & 2 matrices are in

one-to-one correspondence. Let us now look at the composite of two linear transformations T and U, where the matrix corresponding to Tis A = (a;;) and that corresponding to U is B = (b;,). That is, let the function

W = ToU

be defined:

(8)

W(u) = T(U(@)) for every i € U2.

In the first place, W is a linear transformation of U2 into itself:

[ W is additive: (9)

Wi + 6) = T(U(a + 8)

= T(U(4) + Ul) = TUM) + T(UG)) = Wi) + WOE); W is homogeneous:W(\ui) = T(U(U))

= T(AU(a)) = AT(U@)) = WH). Therefore W is given by a2 X 2 matrix C = (c;;) whose elements we shall now find. The first column of C is the image of i:

oe) me w (6) a a) C21 0 bay

=

aiibi1 + ai2b21 d21b11 + ar2b2,

§902]

LINEAR

TRANSFORMATIONS

AND

PRODUCTS

OF

MATRICES

225

and the second column of C is the image ofj:

ee a w(?) 4 ave C22 l b22

_

(411b12 + a12b22 a21b12 + a22b22

so that

fie an wT (ae ai2b21)

C=

(10)

C21

€22

421611 + a22b21

axib1> + eee) a21b12 + ar2b22

In other words, the composite T° U of the two linear transformations T and U has a matrix C = (c;;) whose general element c;; is obtained from the elements of the matrix A = (a;;) of T and those of the matrix B = (b;,) of U by the formula: 2

Gh)

cj=>>

daar

Oe

1

2.

If A and B are two given 2 X 2 matrices, the matrix C defined by the formula (11) is called the product AB of the matrices A and B, and we have the theorem:

Theorem. Jf T and U are any two linear transformations of U2 into V2, with matrices A and B, respectively, then the composite T° U of T and U is a linear transformation of 02 into 02 whose matrix is the product AB of A and B. Nore 1. The rule for forming the product AB of two matrices A and B can be remembered in terms of dot products of row vectors of A and column vectors of B: the element in the ith row and jth column of the product AB is the dot product of the ith row of A and the jth column of B. This is shown schematically:

su toed) Matlaan 3) nc) Mools:) (oo) s) PNOZOYNO@

pee

pe

le oon

2.

QPOU

(521456

NE ©

G) © (19. 22).

Nad JT eR oe Se gh tteee 43 50 SIME TN (ET? 0\ R/ she MG MAN 5% 2\" 1"

Note

TF

92)\ 284) PONS 4 ees

—3

Matrix multiplication is not in general commutative.

10 That is, if A and B are

general 2 X 2 matrices, the two products AB and BA are not the same:

(13)

AB # BA.

This can be shown by a particular example, thus: ee

(14)

Nie52 6\ . (198922

€ Ae

fee

:)ia Ge 4 is a

eee

=) 7 ( 8/\3

|)

4

Suppose now that 7 is a linear transformation from the n-dimensional vector space U, = Uni into the m-dimensional vector space Um = Umi. If i, t2,°**, Un denote the n unit column vectors of (5), §901, constituting the standard basic n-tuple for V,, we can define in the manner of (4), above, n column vectors of Um:

[$902

MATRICES

226 aii

i

a2

0

Qin

0

art

0)

a22

1

Oy

0

l

Qmn

0

QAm2

0

Qm1

(led bagcarl peweabel hee

=v fa

Y

ape?

Seapensi

(15)

As in (5), it follows from the linearity of 7 that if 4 is the matrix (a;;) whose column vectors are given by (15), and if

ii =

:

is an arbitrary vector of Up,

Un

then & = T(i) is the vector of U,, represented:

(16)

U1 v2 U3

2 j=

Opi Wate Cle ates SUS eet et (OCaten ODUM A222 OpsUs alate ae Gants AS1UT G3 22 te 3 3Usy tae a) cia alla .

=

bs

Um

Ami Uy +

Qm2u2

ei QAm3U3

—- ca,

AP AmnUn

n

Or i = > ajjuj,i = 1, 2, --+, m. That is, every linear transformation from 0, into j=1 Um corresponds uniquely toanm X nmatrix A = (ai;) by means of the equations (16).

Let us examine the composite of two linear transformations T and U, where U maps U, into U, and T maps VU, into Un. That is, W = ToU is the linear transformation mapping VU, into U, defined: (17)

W(a) = T(U(@)) for every @ € U>.

Then, as in (9), W is additive and homogeneous, so that all three transformations T, U, and W correspond to rectangular matrices, thus: aii

(ig) 7

ai2

se

Qivd22

Ain

4" a

Ami Gm2

***

gl

bi

biel Oi

Amn

b12

see

baba

bip

ers be

DrinDam*

P|,

Ww:

Dnp

(Ol

C{2

see

C21

C22

°°" C

oF a

Cm1

Cm2°**

Clip

*P

Cmp

By the same reasoning that led to equation (11) we infer from (16) that the matrix C = (c;;) 1s given by (19)

hy,

=

> ODEs k=1

i=

Il, ROP

ONTTE

i

l, Dea

aD

The m X p matrix C = (c¢;;) defined by (19) is called the product of the m x n matrix A = (a;;) and the n X p matrix B = (6;;), written C = AB. Since the ele-

$903]

ALGEBRA

OF

MATRICES

227

ment c;; of C = AB in the ith row and jth column is the scalar product of the ith row vector of A and the jth column vector of B, the product AB of two rectangular matrices can be formed only when the number of columns of the first factor A is equal to the number of rows of the second factor B. Equations (16) defining the linear transformation 7 can be written in the form of the product of two matrices, where the second column vector: U1 (20)

v2

Um

=

/

factor is a one-column

Qit

Giz

°°

24h

ut

a21

a22

=

Q2n

u2

aml

Am2

ay

Amn

Un

matrix, or

;

or more compactly, 6 = Au. In particular, in the case of a linear transformation of U2 into U2, the equations given by (5) become

(21)

UA) v2

iB

Gates a21

012 Ur \ _ ( @iits + ai2u2 azz 7\ U2 a21u, + a22U2 iL QNAS € :) (5) =

Example 2.

\

17 (35)

1 1 3

Nv OA =

903

ALGEBRA

|

|

a) |

WwW — ie) aS

a |

=ee)

Ss

NO Ww MO

OF MATRICES

We first consider in this section the collection 9%2 of all 2 X 2 matrices. We have already seen ($901) that S112 is a vector space. We shall now see that it is in addition an associative algebra (cf. Note, §515), that is, that the following laws hold for all 2 X 2 matrices A, B, and C and real numbers ) and yu: (1)

Homogeneous law:

(2)

Distributive laws:

(3)

Associative law:

(AA)(uB) = (u)(AB).

A(B + C) = AB+ AC, (A + B)C = AC+ BC. A(BC) = (AB)C.

Establishing laws (1) and (2) gives no difficulty, and the proofs are left to Exercise 31, $904. The associative law (3) is more difficult and we give two proofs:

[§903

MATRICES

228

First proof of (3): If A = (aij), B = (b:)), and C = (c;,), then the element in the ith row and jth column of A(BC) is 2 (4)

ye Qik k=1

2 SS DnCns m=1

=

ain(biic1;

SF b12¢2;)

a

aj2(b21¢1 ; ae b22¢2/).

The element in the ith row and jth column of (AB)C is 2 2 OS & aubin) Cmj = (aitbi1 + aizb21)c1; + (aitb12 + ai2b22)c2;. k=1 m=1

(5)

Since the right-hand members of (4) and (5) are equal, (3) is proved.

Second proof of (3): Let T, U, and V be the linear transformations of U2 into U2 that correspond (uniquely) to the matrices A, B, and C, respectively. Then the associative law (3) corresponds to the associative law To(UcoV) = (ToU)oV of transformation composition. Since al/ transformations obey the associative law whenever the composites involved are defined (Example 5, §501), (3) is established. Norte. The laws (1), (2), and (3) always hold for a// rectangular matrices and real numbers \ and uw whenever the sums and products involved exist. Consequently, the set SI, of all n X nmatrices is an algebra, called a matrix algebra, for every n € 9t. Since the commutative law for multiplication of square matrices fails when there is more than one row or column (Note 2, $902), the matrix algebra 91, for n > 1, is noncommutative. This is in contrast to the function-space algebras considered earlier in this book, all of which are commutative. All algebras treated in this book are associative.

Two special matrices play distinctive roles in the algebra Sz of 2 & 2 matrices: (6)

; The zero matrix

7 0 =

al

(9

(7) The identity matrix J = &

:)

These matrices obey the following laws in M2 (cf. Ex. 32, $904):

(8)

The zero matrix:

0--

(9)

The identity matrix:

A= 4-0 {4 ="Ar=

= 4, 0A=

40 = 0,

A)

The matrices 0 and / therefore correspond to the numbers 0 and 1 in the real number

system.

The linear transformation that corresponds to the identity matrix is the

identity transformation that maps every point of U2 onto itself (Example

13, §111).

The determinants of 0 and / are 0 and 1, respectively: |0| = 0, |Z| = 1. (Cf. §905.) Statements similar to the preceding ones apply to the algebra o1,, for any positive integer n, where the zero and identity matrices, denoted 0 and J, respectively, are defined: rere 0 0:--0

(10)

ON

eee 0. OPEeG

(Cf. Ex. 32, $904.)

eee Goro

Het

§904]

904

EXERCISES

229

EXERCISES

In Exercises 1-4, form the indicated linear combination.

1. 6(7) (_}).

2 a} ‘)- (3 =)

Bhs 08) —84(-4910),

4. (4 : ;)= ae es A

In Exercises 5-8, show that the vectors are orthogonal, and find the magnitude of each.

5. Gd; 3, —2), (—7,-5-4).

6. (3, —5, 1), (4, 2, —2).

7.. 6, 25—4, 5G,

8. (1, —2, 7, —6), (—3, 3, 3, 2).

=1, 3,1).

In Exercises 9 and 10, Tis a linear transformation ofU2 into U2 that transforms the vectors (V)and (5)as indicated.

Find the 2 < 2 matrix that represents T.

+0)-() QQ)

)-()70)-0)

In Exercises 11 and 12, T is a linear transformation of U2 into U2 that transforms the vectors (‘)and a) as indicated.

Find the 2 x 2 matrix that represents 7.

at) eG) ee ser) (27) O\nfl

3

\iea

oN

ON

ees

3

ye)

3\_

4.0\;

In Exercises 13-15, find the indicated product. 2

-1\/3

Aye (ies 6

5

1

NG s)

2

ri € pa

i

1\/2

-8

aes it NG ef)

©

te )

In Exercises 16 and 17, Tis a linear transformation of U2 into ‘U3 that transforms the vectors (5)and ( . as indicated.

1 16. (5) =

Find the 3 x 2 matrix that represents 7.

1 2 1 r(_7) =

3 2 |:

1

1

1 17. r(5) =

0) 2 1 7(_1) = 4

3 0 | =

In Exercises 18 and 19, Tis a linear transformation of 03 into U2 that transforms the vectors 1 2 3 0,11), and{ 2 | as indicated. Find the 2 x 3 matrix that represents 7.

0

0

1

[§905

MATRICES

230

In Exercises 20-23, find the indicated product.

1

1

avn

HK:

3

23. |=2

OG

0}:

1 35 5 N/a

6

Ts gs ae ee

dob

a

@):

ly)

20. (5c

=a Oya

YO

ee

eo ee

Onna

In Exercises 24 and 25, verify the homogeneous law for the indicated product.

L(t IG ~o)} |

2

2

[a(t 3)[PG =)

-1

5

©

1

3

In Exercises 26 and 27, verify the distributive law for the indicated product.

(5 OG Ge vl ia IG SIS) 6

—4

3-4

—2

3

—4

7

5

-2

Sek GN.

In Exercises 28 and 29, verify the associative law for the indicated product.

4

0\/2

28. (1 ale

—3\/-1

0

ml 2 at

—3

am" &

2\/1

6

ale gee

2

-—4

5)

30. Prove properties of magnitude (7), (i/), (iii), and (v) $802, for n-dimensional vectors. Assuming the triangle inequality (iv) (cf. Ex. 5, H$520, CWAG), prove properties (vi) and (vii). 31. Prove the homogeneous and distributive laws (1) and (2), $903, for 2 < 2 matrices. 32. Establish the laws (8) and (9), $903, and their analogues for 91,,.

905

DETERMINANTS

In our further work with vectors and matrices we shall find it helpful to use determinants. Although it will be assumed that the student has already had some previous experience with determinants, we shall review the basic definitions and properties sufficiently to permit anyone familiar with the subject to proceed in the text. Most of the determinants to be used in this chapter will be of order 2 or 3, although we shall make occasional contact with determinants of order higher than 3. Let us recall first the expansion of a determinant of order 2, written in the form of a 2 X 2 matrix except that vertical lines are used in place of parentheses:

(1)

4i1

ai2

a21

422

= 411422 —

412421.

The number @11@22 — ai2a21 on the right-hand side of (1) is called the determinant :

of the matrix

E xamplele 1.

ait

aj2

a21

422

‘

Lie ; i =4-6=— 4

2.

0 -—3 (oe x 15. ; 3 (ee 0 — (—15) =

$905]

DETERMINANTS

231

The expansion of a determinant of order 3 has 6 terms:

(2)

Example 2.

Gite

Gio.

ais

421 431

422 @32

423) a33

= @11022033 + a12423431 + a13a21a32 — 411023032 — @12421433 — Q13022431.

1

2

3

0

4

5} = 1-4-6

Oe 3

2—

0

gle

—I

+ 2-5-0

+ 3-0-0

00.68 3-4-0 4

— 1-5-0

0

5 4) = 3-5-1 + (—1)(—4)(=4) 4-0-2-2 = 3-(—4).2

—4

y

1

—

(—1)-2-1

— 0-5-(—4)

= 15

— 164+ 24+

2 = 25,

Note 1. The “arrow” method indicated below can be used as an aid in evaluating 2 x 2 and 3 X 3 determinants, but is not appropriate for any determinants of order higher than 3:

The word determinant in its strictest sense means a certain real-valued function on the set of all square matrices. The determinant of a matrix A is written det(A), det A, or |A|. We shall also find it convenient to use the word determinant to mean the ordered pair (A, det A), to call the number det A its value, and to interpret the notation of the elements of a matrix contained within vertical bars in this sense. Thus,

@11422 — ai2a21

1s the value of the

determinant

ai1

12

a22

of the matrix

ait 412 ies oo) Listed below are nine properties of determinants, usually proved in any good course in college algebra, which we shall assume. Before listing these properties let us recall the two related concepts of minor and cofactor of an arbitrary element of a determinant. The minor of any element of a determinant of order n is the determinant of order n — 1 (or its value) obtained by removing all elements of the original determinant that are located in the same row or column as the element in question.

Example 3.

wa

i

cats

a

The

3

minors

of the

elements

BALE

: are, in left-to-right

in

the

second

row

i iziee=36 |=33, 81 -36

order:

7

of the

determinant

2

If an element of a determinant is located in the ith row and jth column, then its cofactor is equal to the product of its minor and the number (—1)*”.

232

MATRICES

[$905

Example 4. The cofactors of the elements in the second row of the determinant in Example 3 are, in left-to-right order:

(—1)2+!.33 = —33, (—1)?*?-30 = 30, (—1)?*3-(—9) = 9.

Nine Properties of Determinants:

(i) The value of a determinant is not changed if the rows and columns are interchanged. (ii) If the elements of one row (or column) of a determinant are all zero, the value of the determinant is zero. (iii) If the elements of one row (or column) of a determinant are multiplied by the same constant factor, the value of the determinant is multiplied by this factor. (iv) If one determinant is obtained from another by interchanging any two rows (or columns), the value of either is the negative of the value of the other. (v) If two rows (or columns) of a determinant are identical, the value of the determinant is zero. (vi) If two determinants are identical except for one row (or column), the sum of their values is given by a single determinant obtained by adding corresponding elements of the dissimilar rows (or columns) and leaving unchanged the remaining elements. (vii) The value of a determinant is not changed if to the elements of any row (or column) are added a constant multiple of the corresponding elements of any other row (or column). (viii) The value of a determinant is equal to the sum of the products obtained by multiplying each element of any row (or column) by its cofactor. (ix) If each element of any row (or column) of a determinant is multiplied by the cofactor of the corresponding element of a different row (or column), the sum of these products is zero. Nore 2. When (viii) is used for the evaluation of a determinant, the process is called expansion by, or with respect to, the elements of the row (or column) under consideration.

Example 5.

The matrix of cofactors of the elements of the determinant

1 |—-4 8

2 -3 —9 11 if 6

of Example 3 (with elements in corresponding positions) is as follows:

—131 —33 —5

112 30 |

44 9 |-

Evaluation of the determinant by the elements of the first row gives —131 + 224 — 132 = —39. Similarly, evaluations with respect to the elements of the second and third rows are 132 — 270 + 99 = —39 and —40 + 7 — 6 = —39, respectively. The corresponding evaluations by columns are, from left to right, —131 + 132 — 40 = —39, 224 — 270 +7 = —39, and —132 + 99 — 6 = —39. The value of the determinant by any one of the six preceding evaluations is therefore —39.

$905]

DETERMINANTS

233

Nore 3. The pattern of factors (—1)‘+/ that distinguish cofactors from minors is similar to that of a checkerboard. For a 3 X 3 determinant:

+

—

+

+-— + The following example illustrates the manner in which (viii), in the preceding list of properties, can be used to simplify the evaluation of a determinant. This is the method that should be used for evaluating any determinant of order greater than 3. Example 6. By (vii), in the preceding list of properties, the value of the determinant of Examples 3, 4, and 5 is the same as that of the determinant obtained by adding to the second column minus twice the first:

1 =G 8

(3)

2 SMa =

23 & 1 NG 6

1 Do ee =e

8 tae 6

It is also equal to the value of the determinant obtained from (3) by adding to the third column three times the first:

1 Om Ate Sicko

(4)

405513 ela 6+ 24

eee = Se

6 a 95030

Expanding (4) with respect to the elements of the first row gives

Example 7.

=)

Sn

rie9

39

alge aed

ei

8

Rae

9

= =32:

Evaluate the fourth-order determinant

it i i il ee ee i235 23h) Ly 2a band Solution.

Subtraction of the first column from each of the others, followed by expansion

with respect to the elements of the first row, gives 1 heer ty te?

1 Zien 2 en

hea! ie tk epee Breads oe ioatmed este lhe 4

i

oO @ @ 111 led at De lg eh HET IT BE Ps iby 3 del ee 3

Again, subtraction of the first column from each of the others, followed by expansion with respect to the elements of the first row, gives

me 2 ee

ais—sl pty i 2S = 1 23> J 38

Le'O:0 11 =|) = |) tse ls 32

The following theorem is a basic and important one in the algebra of matrices. - We shall give a proof only for the 2 X 2 case. For a general proof see §2815, CWAG, or a good book on linear algebra.

[$906

MATRICES

234

The determinant of the product of two n X n matrices A andB is equal

Theorem.

to the product of their determinants: |AB| = |A|-|B|. Proof forn = 2. Let A = (aij), B = (6;;), and C = (cij) = AB. Then routine expansions of the two members of the following equation establishes the desired equality:

aiibi1 + ai2b21 =ai1bi2 + ai2b22 = (11422 — a12a21)(b11b22 — bi 2b21). azibi1 + az2b21 a21bi2 + a22b22 This verification is left to the reader. Example 8.

Verify the preceding Theorem for

a

4=(}

: Solution. i

ak:3)and

e/a ve (2 2)

B =

5A ily ©) : The matrix AB is Ge a whose determinant

‘ = 102 — 32 = 70.

17 19—17| _ is equal to ieee ia =

This is equal to the product of the determinants of A and B,

which are, respectively, equal to 10 and 7.

A corollary that will be found useful in later sections is based on the definition: Definition. A matrix is nonsingular if and only if it is square and its determinant is nonzero. Any other matrix is singular. Corollary.

906

Any product of nonsingular matrices is nonsingular.

EXERCISES

In Exercises 1-8, evaluate the given determinant. 1

a+b

Gap

“la—b

3.

5

a+b

@—-—B

3 4 —1

=7 1 0

5

2,

3,

1

1 =f —

—4 3 1 ae Ce

= 6

10 CES) 6

5). 6 =I

1

20 iS 16

30 WS 26

Oy

th

tke

al

os

i | |

@ 2 3

2 @ 4

8 a ©

85)

i a il i al laee3t 9 Oe ah IG iS) PS)

ae al GE)

Tellme Dene S se8ipe 23): it 2 3 a Al i @ B ae &

@a—b

l= BO O83 SB)

1 1 SS ilé peli GA DEE 1B G25

$907 |

SYSTEMS

OF

LINEAR

EQUATIONS;

CRAMER'S

RULE

235

. In Exercises 9-16, verify the Theorem, §905, for the product of matrices given in the indicated exercise of §904, or for the product given here. 9. Exercise 13. ee

—

1}, 10)

eed

Ne

7

11. Exercise 15.

O

QYENG

pea)

0

14. | —1

4

ipsa! s

the a

i

Ola

@W° BD 3

Onn

esan 4)

SYSTEMS

OF

Nee

1

—-1

—3}{ -1

—3

4

0)

29

0-08

3

O

EQUATIONS;

©

©

Le

0O

2

O}:

0

4

7

| A2eee Ay0/1098

1G) | 0e6— 7

S84

LINEAR

1 2

1

PONY ets at

15.

12. Exercise 23.

9

3% SiO—l > S6\5

0)

907.

10. Exercise 14.

5, 4 0-6

i

0) ey Dh

Oo ©

©

i

RULE

CRAMER’S

A system of m linear equations in n variables or unknowns

x1, X2,°°°, Xn has

the form

QW api22 tt A inknt = KI, Q2xiX1 4- G22%2 °° + ad5X, =k,

(1)

Lami X1

4

QAm2X2

ar hes =F AmnXn

=

kee

or, in matrix form,

(2)

AX = K,

where A is the coefficient matrix (a;;), i = the n- and m-dimensional column vectors

(3)

EST

1,--:,m,j =

X1

ky

x2

k2

POR ene Xn

1,-:-,”, and X and K are

er Kin

A set of nm numbers x1, X2,:**, Xn that satisfies all m equations (1) — or equivalently a column vector X satisfying (2) — is called a solution of the system of equations (1). In this section we shall be interested first in the case m = n, that is, where A is square and the number of equations in (1) is equal to the number of unknowns. For such systems we now state Cramer’s rule, usually included with work on determinants in courses in college algebra.

Theorem I.

Cramer’s Rule.

Jf the coefficient matrix A of the system (1), with

m = n, is nonsingular, then the system has one and only one solution. :

In this solution

* After the Swiss mathematician Gabriel Cramer (1704-1752). Cramer’s rule is important more for

its theoretical content than as a practical means of obtaining solutions to systems of linear equations.

[$907

MATRICES

236

the value of each variable is equal to the quotient of two determinants; the denominator is the determinant |A\, and the numerator is obtained from it by replacing the column of coefficients of the variable in question by the column K:

xy =

(4)

Example 1.

Ai

Gree Kt vs Gi,

Qo)

Gia

Gy

Gn2)

CU

TEN Ie © 2

a

G21 G22

je

An1

°°

Qn2

Ky

07,

k,

Sana

‘

feel

PO

(8

OSL o> = ili

2p *° Anj

C2; °°"

Gnn

Solve for x in the system Sx -— yt

z= 2,

3x + 4y.— 2z = 7,

x — 2y — 4z = 6. Solution.

By Cramer’s rule, and properties of determinants, §905,

ee

2 healt (oy ie se (2s PS pe ice ee eee ea eer see yl (vine 232), = 120660 A) hd 9 23: dl aview=94 6 I SO a

In case all of the constants ki, k2,---, km are equal to 0 the equations of (1) are called homogeneous. A system of homogeneous linear equations always has the so-called trivial solution, which is the zero vector 0 every element of which is equal to 0: x1 = x2 =:**= X, = 0. The big question here is, ‘‘Does a system of homogeneous linear equations have any nontrivial solutions, that is, solutions other than the trivial solution?” The following theorem and its corollary give the general statements. We present the proofs for the cases nm= 1 and n = 2. For the cases n > 2, see $2816, CWAG. Theorem II.

The system

Gi ier cle Oiahs oe (5)

Q21

ile anne Shey

aie iO) Sean,

IO)

Lanixt + ai2xr +:°++ danXn = 0 of n homogeneous linear equations in the n unknowns x, X2,°**, Xn has a nontrivial solution if and only if the coefficient matrix A = (a;,) is singular:

(6)

PA "or

20:

§907]

SYSTEMS

OF

LINEAR

EQUATIONS;

CRAMER’S

RULE

237

More generally, the system

MINN GHAI

(7)

At2k2) 4 2 22X20 4

Gmi Xie + Gm2X7 +° ofm homogeneous

1 Cink, = 0; a7 X= 0, "+ dmnxX, = 0

linear equations in the n unknowns x1, X2,‘**, Xn, where m = n,

has a nontrivial solution if and only if every n X n determinant whose rows are an arbitrarily selected set of n rows of the coefficient matrix A = (a;;) vanishes. Proof forn = 1: Ifn = 1, the statement of Theorem II simplifies to the following: The system of equations a\x = 0, a2x = 0,°+-, amx = 0 has a nontrivial solution if and only if every coefficient a1, a2,°**, Am is equal to 0. This is obvious. Proof for n = 2: We know from Cramer’s rule that if there exists a nontrivial solution for the system (7), then the subsystem made up of any two of the equations from (7) must have a singular coefficient matrix (if the coefficient matrix were nonsingular there would be only the trivial solution x1 = x2 = 0 found by substituting 0 for the two k’s in (4)). The “‘only if” part follows immediately. For the “‘if” part, ,

Qj

for n = 2, we assume that every determinant of the form (:

aj

2

to the additional assumption that the system (7) has only the trivial solution x1 = x2 = 0. It follows, then, that the related system of m equations in the single unknown x1,

(8)

aiiX1 = ax1

=*** = Amx1 = 0,

has only the trivial solution x; = 0, for otherwise, if x; were a nontrivial solution of (8), (x1, 0) would be a nontrivial solution of (7). By the present theorem for the case n = 1, there must exist an integer j € {1, 2,---,m} such that aj, + 0. We now define X1 = 4j2,

(9)

x2 =

—aj1.

Then (x1, x2) is a nontrivial solution of (7) since for every i =

. oe aiix1 + ai2x2

(10)

ait

nay

= Ai2

ey

1, 2,°°-, m:

= 10)

This is the desired contradiction, and the proof for n = 2 is complete. The following corollary is an easy consequence of Theorem II: Corollary.

(1)

The system aiiX1 az1x1

+ ai2x2 +:°++ atnxn = DO, + a22xX2 +°°:+ arnx, = 0,

AmiX1

ae Am2X2

++ aN: ai AmnXn

=

0

6

es*| where i and/

jl @j2 are distinct integers of the set {1, 2,---, m}, is equal to 0, and seek a contradiction

[$907

MATRICES

238

of m homogeneous linear equations in the n unknowns X1, X2,° °°, Xn has a nontrivial

solution if m Tp = 5: Dee

7S

Za

4y +

4.

3x — Ty

=

il,

In Exercises 5- 8, use Theorem II, § 907, to show that the system of homogeneous linear equations has a nontrivial solution, and obtain the solutions by solving for x in terms of y in Exercises 5 and 6, and for x and y in terms of z in Exercises 7 and 8. 5 ~

8x + 12y = 0, {@ess Gy = 0. 5x + 3y — 4z = 0, 3x +2y+ z=0,

Ue

Ilx

6

20x — 15y = 0, 165 — 12)7 = 0:

6x — Sy — 15z = 0, 8. < 4x — 3y — 1lz = 0,

+ 7y — 2z = 0.

ee = U3)

era

(0)

In Exercises 9-12, use Example 2, § 907, to determine whether the two given vectors are linearly independent or linearly dependent.

‘ ¢

10

GN

AG BP orog 11. (17, 13), (0, 0).

oe

10.5 \36) Nor 32) 12912513) it):

In Exercises 13-18, find the inverse A~! of the given matrix A, and check your answer by forming either the product AA~ or the product A! A.

Biba

15. | —3 1

eed)

6

—-ll

pe 0

{) 4}-

5) ©

a)

Ane

16.

2 0 =>)

0 4 |

—3 6 |1

$910] 17.

THE

GENERAL

Lapel 50" One ise? De Ose.) O00

EQUATION

oO 0) 33 2

OF

THE 18.

SECOND 0 0 Oma —1 1 2 1

DEGREE -1 1 3 0

243

1 4 0 0

In Exercises 19-22, find 4~!, B-', AB, BA, (AB)~', and (BA)~', and check your answer by verifying that (4B)-! = B-!4~! and (BA)! = 4-!B"!,

9.

{34 4 3-7 4= (5 pB=(_3 3)

7h,

Ales

Ll) 25 1 OQ iWeb 73) = ||O Omer 0

-3 1 0

j= 20.4 =(_3 -—4 —6 1

22.

2 A =| —3 1

3 B= =4 1 0

Smez (4 3)

1 OP B= 0

Oo @ il (0 1 4 1225

In Exercises 23-26, solve the system of equations of the indicated exercise by use of matrices and equation (5), §908. 23.

Exercise 1.

24. Exercise

2.

25. Exercise 3.

26. Exercise 4.

27. In the matrix algebra 312, prove that the additive cancellation law A ++ B= A+C3 B=C holds. Prove that the multiplicative cancellation law in the form (A #0) A

(AB = AC) => B = C fails, but that this law in the form (|A| # 0) A (AB =

AC) SB=C

holds.

0 28. Prove that the set of all diagonal 2 * 2 matrices of the form (5 y)where x and y € ®, is an algebra, but not a field. 29. Prove that the set of all nonsingular 2 X 2 matrices with the operation of matrix multiplication is a noncommutative group (cf. Definition II, §302).

910

THE GENERAL

EQUATION

OF THE SECOND

DEGREE

The general equation of the second degree in the two variables x and y can be written in the form (1)

S (x y) = ax? + 2cxy + by? + 2px + 2qy +d = 0,

where a, b, and c are not all equal to zero. In the remaining sections of this chapter we shall study this equation and learn how its graph is related to the graphs of conic sections studied in Chapter 6. In this study we shall make repeated use of techniques of matrix algebra. The function on the left of (1) is called a quadratic polynomial in x and y. To see how matrices are related in a natural way to equation (1), we write the left-hand member of this equation in the form x(ax + cy + p)

(2)

(ny)

eyex SD

q) by Gyaate@).

Guided by the suggestive form of (2) we define two matrices:

[$910

MATRICES

244

beri a= owpiap

(3)

Se be ae: Be Ota:

The matrix E is called the matrix of the equation (1) and e is called the matrix of the second degree terms of (1). Example 1.

For the equation 5x? — 7y? — 12 = 0, the matrices of (3) are

5 (fo

0

5

‘i a1),

0

0

Q

=122

Ee,

0

0}:

For the equation x2 — 12xy + 8y? — 6x + S5y + 11 = 0, the matrices of (3) are 1 e=

—6

6 4;

8

1 )

Jos

—6

-6

—3

8

ee

3

eres6

As an aid in AT of anm X n and jth column n,j = 1,:-+,m.

our study of equation (1) we give two definitions: The transpose matrix A = (a;;) is then X m matrix whose element in the ith row is the element a;; in the jth row and ith column of A, i = l,:--, That is, to put it informally, the transpose of a matrix is obtained by changing its rows into columns and its columns into rows. A matrix A is symmetric if and only ifit is (square and) equal to its transpose: A = A’. The matrices e and E of (3) are thus both symmetric matrices. Example

2.

1

2N\2

lies 1)

(; :) -(

1 8

i —-1 5)

8

a:

— ey Wey

Ip

(*) =e

5

The second degree part, ax? + 2cxy + by?, of the expression f(x, y) of (1) and (2) can be given a more specifically matrix formulation in terms of the two-dimensional column vector

o

Pe ()

With the notation of(3) and (4), the second degree part of f(x, y) is equal to (5)

ax + 2cxy + by? = XT eX,

since (x (2 ‘).) = (x (2 Pe = x(ax + cy) + ylex ++ by) = ax? + 2cxy + by?.* “Technically, X7e X isa 1 X 1 matrix rather than the single element ax? + 2cxy + by? contained within it. However, for simplicity of notation we shall not attempt to distinguish between a singleelement matrix (A) and its lone member \. It is for this reason that we avoid placing parentheses around the left-hand member of equation (5).

§910]

THE

GENERAL

EQUATION

OF

THE

SECOND

DEGREE

245

In general, if A = (a;;) is a symmetric n X n matrix and X is an n-dimensional column vector with elements x1, x2,---, x, in its n rows, then the expression

(6)

XTAX

is called a quadratic form in the n variables x1, x2,-**, Xp. In later sections we shall make use of the uniqueness of the representation of any quadratic form: Theorem

I.

Jf A and B are symmetric n X n matrices, and if the corresponding

quadratic forms have identical values: AT AX = XT BX

(7)

for all n-dimensional column vectorsX with elements x1, X2,°**, Xn, then the matrices A and B are equal: A = B.

Proof forn = 2:

Let A and B be written: Gere

2

aac

a (2 p Bare (¢ oy)

Then, ifX =

Gh we have the equation

(9)

ax? + 2cexy + by? = a’x? + 2c’xy + b'y?,

true for all numbers x and y. With x = 1, y = 0, we have a = a’, with x = 0, y = 1, we have b = b’, and with x = 1, y = 1, we havea + 2c + 6 =a’ + 2c’+ bi, and hence e-= c’. We close this section with a fundamental property of transposes: Theorem II. Jf the product AB of the two matrices A andB can be formed, then the transpose of the product is equal to the product of the transposes in reverse order:

(10)

(AB)? = BTA,

This rule extends to products of any number of factors: (ABC)! = CTBTAT, (ABCD)? ERDLGUBTATA wri.

Proof. In the first place, since the number of columns of A is equal to the number of rows of B, the number of columns of B7 is equal to the number of rows of AT and the product B7A7 exists. Let A be the m X n matrix (a;;), let B be the n X p matrix (b;;), let G = (gi;) be the p X n matrix BT’, and let H = (h;;) be the n Xm matrix AT. Since the element in the ith row and jth column of (AB)? is the element in the jth row and ith column of AB:

a

k=1

oy eee

a P= 1 een.

and the element in the ith row and jth column of B’A7 = GH is de Bik k=1

hy

=

> k=1

bxi a jk

=») k=1

jk

be, J =

1G Wa? Oh

Maal

fleas lab Ake Soa

sy

[$911

MATRICES

246

the desired equation (10) is established. The rule for more than two factors follows from that for two by the associative law. For example, (ABC)! = [(AB)C]? = CYABYE = CBA) = CBA. Example 3.

a=

it

2

(j :)and

B=

ie

Sy

(5 4

g) then

5 i ueAVAlacE CMe mesic: 2IAE caay =|(5 Al 4 ®)| “(i 31 “i = Dist arool ee jal Brat =| 5 4}; > 133! 9 6 215i vf

911

TRANSLATIONS

AND

=

—_

ROTATIONS

If two distinct rectangular coordinate systems in the plane are given, as indicated in Figure 91l-la, then any point p in the plane has two sets of coordinates, one for each coordinate system. The problem of expressing each set of coordinates in terms of the other presents itself. We shall resolve this problem by reducing ay

vy

(a)

ay

(b) Figure 911-1

it to two simpler parts by means of introducing a third coordinate system with axes parallel to those of one of the original coordinate systems and with origin coincident with that of the other system, as shown in Figure 911-15 and c. The transformations corresponding to sets of axes related in these two ways will be called translations and rotations, respectively. Thus, in Figure 911-15 we think of first translating the xy-axes into coincidence with the x’’y’’-axes, and then rotating the x’’y’’-axes into coincidence with the x’y’-axes. In Figure 911-lc we reverse the order, first rotating the xy-system into the x’’y’’-system and then translating the latter to the x’y’-system. I.

Translations

Let the x’-axis be parallel to the x-axis and similarly directed, and let the y’-axis be parallel to the y-axis and similarly directed (cf. Fig. 911-2), and let the origin O’

$911]

TRANSLATIONS

AND

ROTATIONS

247

Figure 911-2

of the x’y’ coordinate system p have coordinates (x, y) in system. Then its two radius systems are similarly denoted regarded as the radius vector

(1)

have coordinates (A, k) in the xy-system. Let a point the xy-system and coordinates (x’, y’) in the x’y’vectors (or radii vectores) in these two coordinate (x, y) and (x’, y’), respectively. Since (h, k) can be of O’ in the xy-system, we have by vector addition

(x,y) = (x,y’) + (, ),

or in terms of numbers, Xe

@

th,

e = Vat K-

The equations (1) or (2) are called the equations of the translation represented in Figure 911-2. Solving (1) and (2) for x’ and y’ in terms ofx and y, we have the equations of the inverse translation: x =x

(3)

—h,

x’,y’) = (% y) — A, &) ‘ ee)

ee

/

In §§603, 607, and 613 we obtained equations of parabolas whose vertices are not at the origin, and of ellipses and hyperbolas whose centers are not at the origin. In every case the transformation of the equations can be considered as the result of a translation of coordinates. In cases where the nature and graph of a conic section are determined by the process of completing a square, we can think of the result as a simplification by translation. We illustrate this idea by two further examples. Example 1.

Solution.

Sketch a graph of the equation y3 — 3y? + 3y = x? + 4x + 5.

Completion of a cube and of a square give

y3 — 3y2 + By —1 =x? 4+4e4+5-1L=?+4e4+4,

vy— 18 =

4+ 2),

or yi

Therefore

the graph

=x",

is a semicubical

X=

eX

parabola

2,

Yo=y

—

1,

with cusp at (—2,

1).

(Cf. Fig. 911-3.)

[$911

MATRICES

248

G1)? (x42) ye=x?

(—2, 1)

1

+

Figure 911-3

Example 2. Sketch a graph of the equation xy — 5x + 6y + 2 = 0 by means of a simplifying translation.

Solution. Since (x + 6)(y — 5) = xy — 5x + 6y — 30, the given equation can be written (x + 6)(y— 5) + 32 = 0, or SE

BY

The graph is shown in Figure 911-4.

ee se St,

SW ey = 5,

(Cf. Example 6.)

(x + 6)(y—5) = —382 xy = —32

(—6, 5)

Figure 911-4

x!

$911]

II.

TRANSLATIONS

AND

ROTATIONS

249

Rotations

Let the x’y’ coordinate system be obtained from the xy coordinate system by a rotation about the origin O through an angle 6, and let the standard basic unit vectors in the x’y’- and xy-systems be /’, ib i, and i as shown in Figure 911-5. Ifp is an arbitrary point, with coordinates (x, y) in the xy-system and coordinates (x’, x’) in the x’y’-system, then the radius vector Op can be simultaneously represented:

(4)

Op = xit y= x’ + yf.

Figure 911-5

The vector i’ has x- and y-components cos @ and sin 6, respectively, and the vector j', similarly, has x- and y-components cos(@ + 37) = —sin @ and sin(@ + 37) = cos 6, respectively, so that (5)

i’ = cos 6i+ sin 6),

iy =

—sin 6i + cos 6 j.

Substitution of (5) into the right-hand member of (4) gives

(6)

xi + yj = x(cos 0 i+ sin 6/) + y(—sin 6 i + cos 0 /) = (x cos 6 — y sin 6)i + (x’ sin 6 + y’ cos 6)j.

Since i and j are linearly independent vectors, the representation in (6) is unique (cf. Theorem III, $803), and consequently

7 (7)

=x

cos 6 = yy sin 6,

y = x’ sin 6 + y’ cos 8.

Solving (7) for x’ and y’ in terms of x and y, or by independent derivation, we have

(8)

x’ = xcosé+ysin8,

‘ y'

=

—xsiné@+ y cos 6.

The linear coordinate-transformation (7) is called a rotation about the origin, and its matrix

[$911

MATRICES

250

—sin 6

cos@

©)

a

is called a rotation matrix.

& On

ghacos )

With the notation X¥ =

ee the trans-

G and x =

formation (7) and its inverse (8) can be written in matrix form: (10)

ms

Kw

Me == TEX,

Formulas (9) and (10) point up the following fact about rotation matrices: Every rotation matrix r is nonsingular, and its inverse is equal to its transpose: Chie)

ase

(The matrix r is nonsingular

since its determinant is equal to cos? @ + sin? 6 =

I,

and equation (11) follows from

wt = (6088

~

\sing

—sin 0

cos?

cosé/\—siné@

«sin?\ —

fl #0

cosé/

\O 1) |

ou,

(cf. Theorem I, §908).) It should also be observed that the two rows of r are orthogonal unit vectors, and the two columns of r are orthogonal unit vectors. Example 3. gm are

The rotation equations for a rotation of coordinate axes through an angle of

— y), x= Uy3x' y3y¥), \y=40'+

2s

ee ly = i—x + v3»).

Example 4. Simplify the equation 13x? — 8xy + 7y2 = 30 by means of a rotation through the angle 6 = Arctan 2. Then identify the graph. Solution. 1

—

NG

Since sin @ = 2/5 and cos @ = 1/15, the rotation 1

equations (7) are x =

(x’ — 2y’), y = —=(2x’ + y’), and the given equation becomes

V5

7

AB(x!2 — 4x'y! + 4y"2) — $x"? — 3x'y! — 2y) + FAX? + Ax'y! + y'2) = 30, 5113 — 16 + 28)x’2 + (—52 + 24 + 28)x’y’ + (52 + 16 4 7)y’2] = 30, 5/27) 15/2230) x’2 + 3y'2 = 6,

The graph is an ellipse with major semiaxis 6 and minor semiaxis \/2. Example 5.

By means of the equations for a rotation through an angle of 17,

;x = 4y2("' — y’), ;x’ = yx + y), y= W20'+y),

ly’ =W2C% 4»).

the equation (12)

SVS,

WES)

is transformed to 3(x’2 — y’?) = X, or

(13)

x'2 — yl2 = 2).

§912]

THE

GENERAL

EQUATION

AND

CONIC

SECTIONS

Since (13) is an equilateral hyperbola (cf. Note 2, §611) with asymptotes y’ = equilateral hyperbola with asymptotes x = 0 and y = 0. Example 6.

251 +x’, (12) is an

Describe the graph of

(14)

(Ce = Ip = 1) = Xs,

NO.

Solution. A translation to the point (4, k) followed by a rotation through 17, as in Example 5, shows that (14) is an equilateral hyperbola with asymptotes x = A and y = k. Nore. The remarks made concerning equation (11) and the rotation matrix r extend to n dimensions, where an orthogonal matrix is defined as any nonsingular matrix whose inverse is

the same as its transpose. If A is an orthogonal matrix, then since the determinant |A7| of the transpose is equal to that of A, it follows from the equation AA” = I that |AA?”| = |.4|-|A7| = |A|? = |/|.= 1, and the determinant |A| of A is equal to +1. The linear transformations whose matrices are orthogonal with determinant | are called rotations, and those whose matrices are orthogonal with determinant —1 are called skew-rotations.

912

THE GENERAL

EQUATION

AND

CONIC

SECTIONS

Before stating our goals and outlining our program for the remainder of this chapter, we should first understand exactly what are meant by a degenerate graph and an imaginary graph.” \f f(x, y)is a nonconstant polynomial in the two variables x and y, then the graph of the equation f(x, y) = 0 is degenerate if and only if f(x,y) can be expressed as the product of two nonconstant polynomials, St (x, y) = 2(x, y)h(x, y), where the coefficients in the terms of g(x, y) and A(x, y) may be either real or imaginary. In this case the degenerate graph of f(x, y) = 0 consists of the graphs of the two equations g(x, y) = 0 and A(x, y) = 0. If there are no ordered pairs (x, y) of real numbers x and y such that f(x, y) = 0, the graph of f(x, y) = 0 is imaginary. Example 1. andx+y=

The graph of x2 — y? = O is degenerate, consisting of the two lines x — y = 0 0.

Example 2. The graph of x? + y? = 0 is degenerate, consisting of the two “imaginary lines” x — iy = Oand x + iy = 0. Its real graph consists of one point, (0, 0). It is sometimes called a ‘‘point-circle with radius 0.” Example 3. The graph of x? + y? + 1 = Ois imaginary. One of its “imaginary points” is (0, i). Its real graph is empty. It is sometimes called an “imaginary circle.”

We shall show in this section that the graph of any equation of the second degree is a conic section, or is degenerate or imaginary. In the following section we shall find means of determining the general nature of the graph of a given second degree equation, that is, of determining whether the graph is an ellipse, hyperbola, parab*In the remaining sections of this chapter we shall find occasional use for both real and imaginary numbers. The properties of complex numbers needed for present purposes are no more than those normally studied in high school algebra. For a more formal and extended treatment of complex numbers, see Chapter 12 of A Second Course in Calculus.

ola, degenerate, or techniques, with the of the graph of any Our first task will

(1)

[$912

MATRICES

252

we shall develop refined imaginary. In §§$915 and 916 exact location and shape the finding aid of matrix algebra, for second degree equation in x and y. be to show that if

fy) = ax? = exy + by7 a 2px

gy

ed

0

is an arbitrary equation of the second degree (a, b, and c are not all equal to zero), there exists a rotation r such that the equation in the new variables x’ and y’, (2)

g(x’, y’) =

q’x'2

a

2c'x'y’ ae b’y'2 ans pix

at 2q'y’ = des

0,

has no term in x’y’, that is, c’ = 0. If c in equation (1) is zero, no change need be made, so we assume now that c # 0, and substitute into (1):

(3)

x = x’ cos 6

—.ysin.0,

y=x'sinéd+ y’ cos 6.

The result of this substitution is

(4)

a(x’2 + + +

cos? 6 — 2c(x’2 sin b(x’2 sin? 2p(x’ cos

2x’y’ sin 6 cos 6 + y” sin? @) 6 cos 6 + x’y’(cos? 6 — sin2 0) — y’”2 sin 6 cos 6) 6+ 2x’y’ sin 6 cos 6 + y’2 cos? @) 6 — y’ sin 6) + 2q(x’ sin @ + y’ cos 0) + d= 0.

The coefficient a’ of the terms of (4) in x’? is equal to a cos? 6+ 2c sin @ cos 6 + b sin? 6, and similarly for the coefficients of the terms in x’y’, y’2, x’, and y’, and for the constant term d’. The full set of coefficients is: (a’ a cos? @ + 2c sin 6 cos 6 + 6 sin? 6, b’ = asin? 6 — 2c sin 6 cos 6 + b cos? 4,

(9)

c’ = —(a — b)sin @ cos 6 + c(cos2 6 — sin? 6), p’ / = pcos é+qsin 6, q’ = —psinéd+qcos0,d' =d.

The third equation of (5) involving c’ can be written c’ = —3(a — b)sin 26 + c cos 26, so that c’ is zero for any angle @ such that 2c cos 26 = (a — b)sin 26, or for any angle @ such that

(6)

Colne —b DE

Before continuing, we pause to draw a useful inference from the first three of equations (5). It is clear from the form of these equations that if a = 6 = c = O then a’ = b’ =’ = 0, and therefore that whenever equation (1) fails to be of the second degree, then equation (2) also fails to be of the second degree. In other words, by the

contrapositive, if equation (2) is of the second degree, then so is equation (1). The converse is also true since equation (1) can be obtained from (2) by means of the

rotation 7~! that is the inverse of the rotation r, and the argument just presented applies in equal measure to this inverse rotation: Equation (2) is of the second degree whenever (1) is. (Cf. Ex. 37, $917.) Since equation (1) was assumed to be of the second degree we can now proceed with the knowledge that (2) is of the second degree too.

§912]

THE

GENERAL

EQUATION

AND

CONIC

SECTIONS

253

We now drop primes from equation (2), for simplicity of notation, and assume that a coordinate system has been chosen such that c = 0:

(7)

ax? + by? + 2px + 2qy + d= 0.

There are two main cases to consider:

Case (i): ab ¥ 0. If a and b are both nonzero, then the method

of completing

squares discussed in §§607 and 613 permits the application of a suitable translation that simplifies the equation (7) to a form where the coefficients of the variables x and y vanish:

(8)

ax? + by2+d=0.

If a and b are of one sign and d is of the opposite sign, then the graph of (8) is an ellipse (possibly a circle). If a, b, and d are all of the same sign, then the graph of (8) is imaginary (sometimes called an imaginary ellipse). If a and 6 are of one sign and d = Q, then the real graph of (8) consists of one point only (the origin), and the graph is called degenerate (if imaginary quantities are permitted, the graph of ax* + by? = 0 reduces to two imaginary lines meeting in the real point (0, 0); for

instance, if a and b are positive, ax? + by? factors into (Jax + ivby)(yax — ivby), and the two “imaginary lines” are given by the two equations Vax + If a and b have opposite signs and d # 0, then the graph of (8) is a Finally, if a and b have opposite signs and d = 0, then the graph of (8) two intersecting straight lines (for example, if a > 0 and 6 < 0, then

ivby = 0). hyperbola. consists of ax? + by?

factors into (Vax + —by)(Vax — \—by), and the two intersecting lines are given by the two equations Vax + \—by = 0). Case (ii): ab = 0. Since not both a and b can be equal to zero (the equation (7) is quadratic rather than linear), we shall assume for definiteness that a # O and b = 0, and by means of a translation (found by completing a square) we can assume further that p = O so that (7) reduces to

(9)

ax? + 2qgy

+d=0.

If g # 0, an additional translation simplifies the equation to one without a constant term (cf. § 603): (10) whose graph is a parabola.

(11)

ax2 + 2qy = 0, Ifq = 0, (9) becomes

ax2+d=

If a and d have opposite signs, then the graph sisting of the two parallel lines x = +\/—d/a. graph of (11) is imaginary (also degenerate and nary lines x = +\/—d/a). Finally, if d = 0, consists of the two identical lines x = 0 and x -

0. of (11) is If a and consisting the graph

the degenerate graph cond have the same sign, the of the two parallel imagiof (11) is degenerate and

= 0.

Example 4. In Example 4, § 911, for the equation 13x? — 8xy + We = 30 SQ, @ Sal b =7,andc = —4. With @ = Arctan 2, from (12), §701,

[$913

MATRICES

254 76

1

1 — tan2 6

CCl et arte)

il = al

AG

Sed idee

4

Ait

De

in agreement with (6).

913.

EXERCISES

In Exercises 1-4, write down the matrix E for the given quadratic polynomial. 1. 3x2 —

2. 5x2 + y?.

5)? — 2.

4. 6xy + 3y2 — 2x 4+ 12y 4+ 9.

3. x2 — 4xy + 10x — 7.

In Exercises 5 and 6, write down as a quadratic polynomial the quadratic form given in matrix form.

so o(3 aC)

ac a(t 4)())

In Exercises 7-10, verify that (AB)” = BTAT for the given product AB.

8.5 -9(} ~9)

~G ~)G) 2

—5\/6

3

acer ee te aA 9, la Tr s) = eo aes F bu Lie 3

—4

Db = in ees 10204 6 52 Tiel ii

pass) peak NA |

In Exercises 11 and 12, write down the equations for the translation to parallel axes passing through the given point (A, k) in the original coordinate system. Also give the equations for the inverse transformation.

1G

1k) = 4, =7):

12. (h, k) = (—3, 10).

In Exercises 13 and 14, write down the equations for the rotation about the origin through the given angle. Also give the equations for the inverse transformation.

13. Arcsin (— #).

14. Arccos (— 2).

In Exercises 15-18, with the aid of the formulas cot 20 = (2 tan 6)/(1 — tan? 4) or cot 20 = (cot? @ — 1)/2 cot 6, show through the given angle @ would transform the given equation the term in x’y’ is 0. Do not actually perform the rotation

(a — b)/2c and either tan 20 = that a rotation about the origin to one in which the coefficient of substitutions.

15. @ = Arctan 3; 3x2 — 6xy — S5y2 — 20 = 0. 16. 6 = Arccot 3; 7x? + 6xy — y2 — 15 = 0.

17. 6 = Arccot3; 8x2 + 12xy + 3y2 — 12 = 0. 18. 6 = Arctan 3; 7x? — 12xy + 2y? — 18 = 0.

914

INVARIANTS

Whenever a translation degree, of the form

or a rotation is applied, any equation of the second

§914]

INVARIANTS

(1)

255

f(x, y) = ax? + 2cxy + by? + 2px + 2qy +d =0,

with matrix

GG E=\|{c

(2)

Fe 6b qh

pqed is transformed into a new equation of the form (3)

2(x’, y’) =

a’x!2

aL NEC Ue +

b’y’2 ae 2p’x!

ae 2q'y' ae d' =

0,

with matrix /BY

(4)

=

a’

c’

p’

c’

b’

q’

p’

q’

al

B

The new coefficients a’, b’, c’,--- may in some cases be entirely distinct from their

original counterparts a, b, c,---, and in other cases be only partially altered. Example 1.

In Example 2, § 603, the equation

(5)

x? — 6x — By —7=0

was altered in form by a completion of a square and factoring, to (x — 3)? = 8(y + 2). In other words, by means ofthe translation x’ = x — 3, y’ = y + 2, or equivalently, x = x’ + 3, y = y’ — 2, equation (5) is simplified to (6)

Xe

The matrices E of (5) and E’ of (6) are

(7)

E=

a

1

053

0

0

rae

—4)

1 E’=]0

Oey

0

0

QO

—4}-

On =a)

20

In this example the coefficients a, b, c, and g are unchanged; that is,a = a’ = 1,b = b’ = 0, c=c' = 0, and g =q' = —4. However, p = —3 is changed to p’ = 0, and d = —7 is changed to d’ = 0.

Example 2.

In Example 2, §607, the equation

(8)

x? + 5y2? + 6x — 20y — 151 = 0

was altered by two completions of squares to (x + 3)? + S(y — 2)? = 180. In other words, by means of the translation x’ = x + 3,y’ = y — 2, or equivalently, x = x’ — 3,y = y’ + 2, equation (8) is simplified to

x'2 4+. 5y’2 = 180.

(9)

The matrices E of (8) and E’ of (9) are

1 (10)

E=|0 3

0 5 —10

3 —10} —151

i

E’=[{0

@ 5 02-0

0 0 |}: 2150

In this example the coefficients a, 6, and c are unchanged: a = a’ = 1, Db = bY = 5, aval c = c' = 0, andp = 3 is changed to p’ = 0, g = —10 is changed to q’ = 0, and d = —151 is changed to d’ = —180.

[$914

MATRICES

256 Example 3.

In Example 4, §911, the equation

(11)

13x2 — 8xy + 7y? = 30

is transformed by a rotation to

5x!2 + 15y’2 = 30

(12)

(and then by division by 5 to x’? + 3y’2 = 6). The matrices E of (11) and E’ of (12) are

13 (13)

—4

E=\-4

0

7 0

0

Ol; —30

J

5

0

0

=O 0

is 0

0 }—30

In this example, a = 13 is changed to a’ = 5, b = 7 is changed is changed to c’ = 0. The other coefficients are unchanged:

to b’ = 15, and c = —4 p = p’ = 0, g=q' = 0,

Gl = Gh es 30):

The three preceding examples illustrate the idea of invariance when an equation is transformed by a change of coordinates. In Examples | and 2 the coefficients a, b, and c of the terms of second degree are all individually invariant. In Example 3, every coefficient except for a, b, and c is invariant. In all three of these examples there are further invariants that involve the coefficients in combination rather than individually. We mention one of these now: the determinant of the matrix E is

invariant.

That is, |E| = |E’| in every case:

3 0 3 NB —4 0

0

OF QO

3 —4

dT

0 Sa LOM

3 ior

ees 0 5 7 O}=)0 O30 0

=

I (0

0 0)

Oy

fo Ol ON 0

0 = 4) =

ee 00

ae0 ee 0

0 Oo = —180

0 0 -15 Oe 030

16,

=900,

71530

2250;

Our primary goal in this section will be to establish certain invariants in general, for both translations and rotations. We start by observing from the last equation of (5), $912, d = d’, that d is an invariant for rotations. More precisely: Theorem I. For any rotation about the origin, the constant term d in the general equation of the second degree is an invariant: d= d’. (in Examples3,.d = "ds — -430)) We

now

turn

our

attention

to translations.

If the

translation

x = x’ + 4h,

y = y’ + kis applied, substitution into the general equation of the second degree, (1), followed by expansion and collection of terms, produces a set of relationships between the primed and the unprimed coefficients, somewhat similar to the equations (5), $912, for a rotation. (Give a proof of the following formulas in Ex. 35

§917.)

:

§914]

INVARIANTS

257

(a’ = a, Di=b, Cr & p’ I =ah+ck+p, q’ =ch+bk +4,

14 Ge,

|d’ = ah? + 2chk + bk2 + 2ph + 2gk +. We conclude:

Theorem If. For any translation applied to the general equation of the second degree, the following are invariant: a, b, and c. In other words, the matrix e of the second degree terms is invariant:

a o

¢ Pisa Le s)=e 26

2a

ccs ih

The simplest invariant involving more than one coefficient is the sum a + 6 of the terms on the main diagonal of the matrix e, called the trace of e: Theorem III. of e is invariant:

For any translation and for any rotation about the origin, the trace

(16)

at+b=a

+5

Proof. The fact that a + 6 is invariant with respect to translations follows from the fact that a and 6 are individual invariants, a = a’ and 6 = b’. To prove that a + 6 is invariant with respect to rotations is only slightly more difficult. If we add the first two lines of (5), $912, we get (17)

a’ +b’ = (acos26 + 2c sin 6 cos 6 + b sin20) + (a sin20 — 2c sin 6 cos 6 + 6b cos?@)

= a(cos26 + sin2@) + b(sin2@ + cos?6) = a+ b, and the invariance of a + b is established. This particular invariant will be exploited especially in §916. (In Example 3,@ +b = 134+7=a' +0’ =5+ 15.) Finally, two invariants that will be of more immediate concern are the determinants of the two matrices e and E. We shall establish the invariance of the determinant D of e. Proof of the invariance of the determinant A of the matrix E is given in §2824, CWAG. Theorem IV. For any translation and for any rotation about the origin, each of the following determinants is invariant:

(18)

D=lel\=[ 5

Proof for D for translations:

(Oh

XG

Since e is invariant so is D.

Proof for D for rotations: With the notation X7 e X of (5), $910, and with that for the rotation r, given by (10), §911, X = rX’, we have by substitution, and Theorem II, $910,

[§914

MATRICES XT eX = (PX e (rX’) = XT ev)X’,

258 (19)

and therefore, by the uniqueness ofthe representation of a quadratic form (Theorem I, § 910), the transformed matrix of e is

ol bi)

(29)

|

sin@\/a

cos@

en

ot eee

CN

0

—sin 6)

c\/cosé

\sinie | aces2

\eib)

©" | \ sin 6 cos)

and consequently, when we take determinants (§ 905): (21)

D

pret a c!

S|) Senet b! ci |7 | le|

Ir|

Mr Ley

pee

C

el = b

D,

and the invariance of D is established.

Before drawing up our conclusions that enable us to identify graphs of second degree equations we need to show that two additional properties are invariant, this time with respect to multiplying all members of an equation by a nonzero constant (give a proof in Ex. 36, $917): Theorem V. Jf the equation (1) is altered by multiplying both members by the nonzero constant k, so that

(22)

a! = ka, 6! = kb, c! = ke, p! = kp,’ = ka, d’ = kd,

then the following are invariant:

(1) (ii)

the sign of D, whether A vanishes.

We are now and V:

ready to put to work

the invariants identified

in Theorems

IV

Theorem VI. Depending on the sign of D = ab — c? and the vanishing of A, as defined in (18), above, the nature of the graph of the equation (1) of the second degree is given in the table: ID = Gh) =

(23)

A #0 A=

€ > OID

=

oy

=

© 0 and D < 0. It should be appreciated that both

§914]

INVARIANTS

299

real and imaginary ellipses are included under D > Oand A # 0. The more delicate analysis of -§916 is needed to distinguish between real and imaginary ellipses. We now consider the case D = 0, which corresponds to case (ii), §912. Since the matrices of (10) and (11), §912, are

0810 quo"0 On 0F gle and s1.0-70 0); Oman 00° od

(24)

respectively, the case A # 0 corresponds to a parabola, and the case A = 0 corresponds to a pair of parallel (possibly identical) lines. Corollary. A necessary and sufficient condition for the graph of the equation (1) to be degenerate is that the determinant A = |E| vanish. Note 1. Because of Theorem VI, a point (or pair of intersecting imaginary lines) is also called a degenerate ellipse, a pair of (real) intersecting lines is also called a degenerate hyperbola, and a pair of parallel lines is also called a degenerate parabola. Furthermore, in this same spirit, the graph of any equation of the second degree in x and y is often called a conic section. Example 4. the form:

The invariance of the determinant

(25)

13

=

Solution.

7),P 10) 15),

Identity the conic section 2x? — 4xy + 3y? — 6x + 10y —1=0. Since

puuliea A 2

the curve is an ellipse.

—2

Since

SSO

2

—2

the ellipse is nondegenerate.

-3

rey re

—3

Example 6.

3 takes

S)e4)5.40

—4> Example 5.

D of the matrix e for Example

5s

ee Cay

1

For further information see Example 3, §916.

Identify the conic section 2x2 — xy — 6y? + Sx + lly — 3 = 0.

We first double all coefficients in order to avoid fractions in the matrices e and E: Solution. 4x2 — 2xy — 12y2 + 10x + 22y — 6 = 0. Since

4

-—-1

lie Is S eet the curve is a hyperbola.

Since

4 A=j=1 5

—-1 —=12 Li

5 11)=0, 6

[$915

MATRICES

260

the hyperbola is degenerate, and consists of two straight lines. lines, see Example 3, §915.

Identify the conic section x? — 6xy + 9y? + 8x — 4y +5

Example 7. Solution.

For determination of these

= 0.

Since

1

be e the curve is a parabola.

—3

A =i

Since

= the parabola is nondegenerate.

1 |-3 4

—3 9 -2

4 —2) = —100 ¥ 0, 5

For more information, see Example 4, §916.

Note 2. As illustrated in Example 7, a second degree curve is a parabola (possibly degenerate) if and only if the second degree terms form either a perfect square or a constant multiple of a perfect square. The reason for this is that the simplified equations represented by the matrices (24) have only the single term ax’? of the second degree. The second degree part of the original equation, then, must be equal to the product of the constant a and the square of the linear expression x’ in terms of x and y.

915

SIMPLIFICATIONS

BY TRANSLATIONS

Ellipses and hyperbolas are called the central conics because each has a center of symmetry (§§605, 609). In order to study central conics in the context of the general equation of the second degree, (1)

ax? + 2cxy + by? + 2px + 2qgy +d =

0,

we see first of all what it means for the origin to be a center of (1). In §115 we defined the origin to be a center, or point of symmetry, of a graph I if and only if whenever the point (xo, yo) € T, then the point (—xo, —yo) € I also. We retain this definition, but must now permit the numbers xo and yo to be complex, in order to include the degenerate and imaginary graphs that may arise. With this under-

standing we have the following theorem, for which we shall present here simply the “if? part of the proof.

Theorem I.

(For the “only if” part, cf. $2826, CWAG.)

The origin is a center of the graph of (1) if and only if p= q = 0.

Proof of “if” part:

Ifp = q = 0, then the origin is a center of (1) since

(2) a(—xo0)? + 2c(—x0)(— yo) + b(—yo)? + d = axo? + 2cxoyo + byo? + d. We now have the means of setting up a procedure for finding the coordinates of the center of any central conic:

Theorem II. The point (h, k) is a center of the graph of (1) ifand only if the following two equations obtained from the first two rows of the matrix E, of (3), §910, are satisfied:

§915]

SIMPLIFICATIONS

(3)

BY TRANSLATIONS

Ne

261

micah

ch + bk +q = 0. Furthermore, if (h, k) is a center of (1), and if a translation to this point as the origin of an x'y'-coordinate system is made, the new constant term d’ of the transformed equation is obtained from the third row of E as follows:

(4)

d' = ph+qk +d.

Proof. If a translation to the point (h, k) (whether (A, k) is a center or not) is made, the coefficients of x’ and y’ in the transformed equation relative to the coordinate system having (A, k) as origin, are 2p’ and 2q’, respectively, where p’ and q’, as given by (14), §914, are the left-hand members of the equations (3) above. Therefore, by Theorem I, this point (A, k) is a center if and only if p’ = q’ = 0; that is, if and only if the system (3) is satisfied. Assume now that (A, k) is a center of (1), and write the last equation of (14), §914, as follows: (cf. (2), §910):

(5)

d= h(ah + ck p)'-+ k(ch + bk + q)-+ (ph + qk + d).

By (3), the quantities in the first two pairs of parentheses are zero, and the righthand member of (5) reduces to (4). This completes the proof. Example 1. Simplify the equation x2 + Sy? + 6x — 20y — 151 = O of Example 2, § 607, by means of a translation to its center.

1 Solution.

0

Since E = | 0

3

5

3

-10

—10

}, equations (3) for the center become

—151 h+

(6)

3=0,

Ne eee

with the solution h = —3,k = 2. The new constant term d’ is equal to d’ = 3(—3) — 10(2) — 151 = —180, and the simplified equation is x’* + Sy’? = 180 (cf. Example 2, § 914). Example 2. Simplify the equation 2x? — 4xy + 3y? — 6x + 10y — 1 = 0 of Example 5, §914, by means of a translation to its center.

2 Solution.

Since

E =|

—2

—2

3

—3

5

-3 5 }) the equations (3) for the center become —-1

2h — 2k —3

7)

= 0,

Ey eeket 0,

with the solution

A=

—4,k

=

—2.

The new constant term is d’ = (—3)(—5) + 5(—2),-

1

= — 4, and the simplified equation is therefore (8)

Qx'2 — 4x'y! + 3y’2 = 412.

Example 3. Simplify the equation 4x2 — 2xy — 12y? + 10x + 22y —6=0 6, §914, by means of a translation to its center. Solution.

Since E =

4

—-1

5

{—1 5

—12 11

11 —6

of Example

J, the equations (3) for the center become

[$916

MATRICES

262

4h— —h—

(9) with the solution

A=

k+ 5=0, 12k + 11 = 0,

—1, k = 1. The new constant term is d’ = 5(—1) + 1101) — 6 = 0,

and the simplified equation is 4x’2 — 2x’y’ — 12y’ = 0, or with x’ = x + land y’ = y — (10)

2x’2

Therefore 2x + 3y

916

—x’y’ — 6y'2 = (2x44 3y)Q! — 2y’) = Qe + 3y —

the degenerate hyperbola of Example x —2y +3 =0. —1 =Oand

SIMPLIFICATIONS

6, §914,

Ix — 2y $3)

consists

of the

1,

= 0.

two

lines

BY ROTATIONS

In order to simplify an equation (1)

ax? + 2cxy + by? + 2px + 2qy

+d = 0,

where c 0, by means of a suitable rotation, we may use equation (6), $912, in conjunction with the formula (cf. (12), $701) for the tangent of 20 in terms of ¢t= tan@,

(2)

cot.20 =

Cs

Xe

> Dele9

aye

and solve the resulting quadratic equation in /: (3)

ct?

— (6 — a)jt—c = 0.

Since the product of the roots of (3) is —1, one root must be positive and one negative. For definiteness we shall denote by the letter ¢ the positive root of (3). In other words, we shall consider a rotation through a /first-quadrant angle 6 such that (4)

tan 6 = ¢ = the positive root of (3).

From this point we could sin 6 = t/y1 + 12), and x = x’ cos@ — y’ siné, y ous and cumbersome and,

proceed by finding cos @ and sin @ (cos 6 = 1/y1 + #2, substitute into (1) the rotation formulas ((7), §911), = x’ sin @ + y’ cos 6. However, this procedure is laborifortunately, unnecessary.

Associated

with the matrix e = (: 5)is the following quadratic equation, b known as the characteristic equation of e:

(5)

PanelGime

ats C= =O.

where D = |e| = ab — c?. The important property of the characteristic equation (5) that will be of value to us is that it is invariant with respect to both translations and rotations, that is, the two coefficients —(a + 6) and D are both invariant (Theorems III and IV, §914). To put it in more specific terms, if equation (1) is transformed into a new equation as the result of any combination of translations and rotations, and if the notation for the transformed equation is the same as for (1) except for the use of primes, then for every real number £,

(6)

le-H=P-@+b}%E+D = P-@tbE+D=le— Hl,

where D’ = |e’| = a’b’ — c’2.

$916]

SIMPLIFICATIONS

BY ROTATIONS

263

Now suppose that equation (1) is transformed by the rotation about the origin through the angle @ prescribed in (4), so that c’ = 0. Then, in terms of the new primed coefficients, the characteristic equation of e’ becomes

(7)

ei

2 — (a

be + ab’ = (6

aye — bY) = 0,

and a’ and b’ are the roots of the characteristic equation (5). These roots are called the characteristic roots or eigenvalues” of the matrix e (and also of the transformed matrix e’). Thus, it appears that in order to obtain the simplified matrix e’ all that is necessary is to solve the characteristic equation (5), and place the two roots Ie a 0 oh , . a’ and b’ on the main diagonal: e’ = ( ) However, it is not quite that simple

Ob’ because we do not yet know which eigenvalue is a’ and which is 5’! The answer to this question is given in the following theorem, whose proof is givenin §2827 CWAG:

Theorem. /f equation (1) is transformed by the rotation about the origin through the angle 6 prescribed in (4), then the coefficient a’ of x'? is the larger of the two characteristic roots of the matrix e in case c > 0, and is the smaller of the two roots in case c < 0. Furthermore, the value of t = tan 0 of (4) is equal to (8)

i

a—a' Cc

pat ~

change in either diagonal element C

C

We summarize procedures for simplifying equations of the second degree. There are two principal cases.

I. Central conics. matrix E to the form (9)

First translate axes to the center in order to simplify the

Fa

@

@¢@

C 0

De} © @

W

Then solve the characteristic equation &2 — (a + b)§ + D = 0, and denote by a’ the larger or smaller root according as c is positive or negative. The simpli-

fied equation is then

(10)

a’x’2 + b’'y’2 +d =),

obtained by rotation through the first-quadrant angle Arctan 7, where / is given by

(8). II. Parabolas and parallel lines. First rotate axes through Arctan /, where f is given by (8). The algebraic details can be simpiified by using the fact that the second degree part ax? + 2cxy + by? of (1) — at least after multiplication of (1) by a suitable constant — is a perfect square. After the rotation simplification is complete, a suitable translation can be found by the method of completing squares in order to place the vertex at the new origin in case the curve is nondegenerate. (meaning *The word eigenvalue is an interlingual hybrid, made up of the German word eigen characteristic, proper, peculiar, or own) and the English word value.

’

[$916

MATRICES

264

Example 1. Simplify the equation 13x? — 8xy + 7»? = 30 of Example 4, §911, by means of the eigenvalue techniques of the present section.

13 The matrix e is ee

Solution.

—4 ne } : ‘)and the characteristic equation of e is & — 20 +

75 = 0. Since the characteristic roots are 5 and 15 and since c = —4 < 0, we place 5 and 15 on the main diagonal of the new matrix e’ with the smaller root 5 appearing first:

SX agit : : Lge is) Since the constant term d is invariant under rotations about the origin, the 0 simplified equation is 5x’? + 15y’2 — 30 = 0, or x’2 + 3y’2 = 6. The acute angle of the 13 —5 7—15 = 2. The graph is shown in rotation, by (13), has tangent equal to 4 4 Figure 916-la. Y=

(

Example 2.

Simplify the equation 2x2 + 4xy — y2 — 12 = 0 by means of the eigenvalue

techniques of the present section.

Solution.

SAR i 2 met ; : The matrix e is (; ee:)and the characteristic equation ofe is

—

§ -6 = 0.

Since the characteristic roots are 3 and —2 and since c = 2 > 0, we place 3 and —-2 on the a . ; 3 0

main diagonal of the new matrix e’ with the /arger root 3 appearing first: e’ =

0

oh.

Since the constant term d is invariant under rotations about the origin, the simplified equation

is 3x’ — 2y’2 = 12. The acute angle of the rotation, by (13), has tangent equal to —1l+2 2

1 =D

7|-

The graph, as shown in Figure 916-14, is a hyperbola with transverse axis 4

and conjugate axis 2/6.

(a) Figure 916-1

Example 3. Complete the analysis of the ellipse 2x2 — 4xy + 3y2 — 6x + 10y —1=0 of Example 5, §914, and Example 2, §915, by means of a rotation about the origin in the x’y'-system. Draw a graph of the curve, showing all three coordinate systems.

$916] Solution.

SIMPLIFICATIONS

BY

ROTATIONS

265

We first simplify the equation by a translation to the point (— 3, — 2), as in

Example 2, § 915: 2x’2 — 4x’y’ + 3y’2 = 4%. The characteristic equation is

— 5+ 2 = 0,

and the eigenvalues are therefore 3(5 + 17) = 0.44 and 4.56, to two decimal places. Since c = —2 < 0 we place 0.44 and 4.56 on the main diagonal of the new matrix e’’ with the smaller root 0.44 appearing first: e’’ = ( es

new third coordinate system.

hee where double primes are used for the

Since the constant term is invariant under rotations about the

origin the simplified equation is $(5 — V17)x’”? + $65 + V17)y'? = 42, or approximately 0.44x'’2 + 4.56y’2 = 4%. The acute angle 6 of the rotation has tangent equal to wee! = 0.78 (approximately), and @ = 38° (to the nearest 10’). The major and minor semiaxes are approximately 4.65 and 1.44, respectively, to two decimal places. The graph is shown in Figure 916-2a. Example 4. Use the eigenvalue techniques of this section to find the vertex and axis and simplified equation of the parabola x2 — 6xy + 9y? + 8x — 4y + 5 = 0 of Example 7, §914. Draw its graph. : 5 ae a, 1 -3 Solution. We start with a rotation of axes about the origin. The matrix e is (ne a:

the characteristic equation is £ — 10 = 0, and the eigenvalues are 10 and 0. Since c = —3 < 0 we place 0 and 10 on the main diagonal of the new matrix e’ with the smaller 0 @ ima root 0 appearing first: e’ = ( io) The acute angle of the rotation is 6 = Arctan 3, so that the x’-axis is the line x — 3y = 0 and the y’-axis 916-26). The rotation equations (7) and (8), § 911, are

(

if (

| )

3x’ — y’

10

a

c

e tenry

—

*i x!

+

, )

xXx

3y’

pe

3x+y =

ee.

a

(a) Figure 916-2

is the line 3x + y = 0 (Cf. Fig.

10 —

=e

a

meio

3y

[$917

MATRICES

266

The original equation can now be written

(x — 3y)2 + 8x — 4y +5 = (10 y’P +

20x’ — 20y’

Vio

+5=0,

or

Cae)

rl a)

or approximately,

(y’ — 0.32)? = —0.63(x’ + 0.63). With x” = x’ + +10, or x’ + 0.63, approximately, and with yp” = y’ — zovi0, or y’ — 0.32, approximately, the simplified equation can be written

yee approximately.

—sNtOx

or

3/2 =) — 0.63 x”,

The vertex of the parabola in the x’y’-coordinate system is at the point

x’! = — $\10, y’ = Zoy10. In the original xy-coordinate system, by (16), the vertex is at the point x = — 75,» = yy. The axis of the parabola is the line through the vertex with slope 3, and therefore (in the original coordinate system) is the line with equation y — 75 = 4(x + 75), or x — 3y + 1 = 0. The graph is shown in Figure 916-26.

917

EXERCISES

In Exercises 1 and 2, transform the given equation by means of the given translation, and verify the invariants described in Theorems II and IV, §914. 1. 3x2? — 4xy

+ y?+2x

—4V4+5=0;

2. 2x? = Dey a 3y? — 4x 2 2y

x =2%'4+3,y=y' -—4.,

6 = 0; 4 =x

— 2

= yy 3:

In Exercises 3 and 4, transform the given equation by means of a rotation about the origin through the given angle, and verify the invariants described in Theorems I, III, and IV, §914.

3. 7x2 + 6xy — 4y? — 10 = 0; Arctan 2.

4, 3x2 — 2xy + 2y? — 7 = 0; Arctan $. In Exercises 5-12, use Theorem VI, §914, to identify the given conic section. 5, 2x2 + 6xy

+ y? —4x —2y+12=0.

7. 9x2 — 12xy + 4y? 4+ 2x 4+ 2y = 0. 9, 3x? — 4xy + 4y? + 2x — 6y — 30 =0. 11. x2 — 4xy + 4y? + 2x — 4y = 0.

6. 2x? + 6xy + 5y? — 2x — 4y — 20 = 0.

8. 3x2 + 2xy — y2 + 16x 10. 3x2 +

—4y +5 =0.

2xy — 2y2 — 6x + 2y + 10 =0.

12. 3x2 + 12xy + 12y? — 2x —6y

+5 =0.

In Exercises 13-16, find the center, and simplify the given equation by means of a translation to this center as the new origin.

13. 2x2 — 4xy + 5y? — 12x + 24y + 24 =0.

§917]

EXERCISES

14, x? + 4xy

+ y? + 10x

15, x2 — 2xy

—y?— 2x 4+10y-7=0.

267

+ 8y + 16 = 0.

16. 3x2? + 2xy + y2?-—4y+4=0.

In Exercises 17-26, use eigenvalue techniques to simplify the given equation by means of a rotation about the origin through an acute angle 6. Identify the conic section. Find tan 6, and draw a figure, showing both sets of axes and a graph of the conic.

17. 2x2 — 4xy + 5y? 19. x2 — 2xy

—6 = 0.

— y? = 0.

18.

x2 ++ 4xy

+ y?+3=0.

20. 3x2 + 2xy + y2?—2=0.

21. 10x? — 12xy + 5y? — 14 = 0.

22. 8x2 + 8xy — Ty? = 0.

23. x2 — 4xy + 3y?+1=0.

24.

25. 11x? + 10xy — 13)? — 84 = 0.

26. 30x? — 20xy + 9)? = 0.

x2 + 2xy + 5y? —4=0.

In Exercises 27—30, draw a graph of the conic section in the specified exercise with the aid of the simplification achieved in the exercise indicated within parentheses. Show three sets of axes. DPE XCleisenlon(E

xen li):

28. Exercise 14 (Ex. 18).

29. Exercise 15 (Ex. 19).

30. Exercise 16 (Ex. 20).

In Exercises 31-34, use both a rotation about the origin and a translation to simplify the given equation. Identify the conic section. Give the tangent of the acute angle of rotation 0. In case of a parabola, write the equation of the axis of symmetry and give the coordinates of the vertex in the original coordinates. In case of degeneracy, write the equations of the two lines in the original coordinates. Draw a figure, showing three sets of axes and a graph of the conic.

31. x2 — 2xy + yp? — 8x — 8) + 16 = 0.

32. 4x2 + 12xy + 9y? — 20x — 30y + 25 = 0. 33. 9x?

— Oxy

34. 4x2 + 4xy

+ py? + 6x — 2yv —3 = 0. + y? — 14x —

12y + 22 = 0.

35. Prove formulas (14), §914.

36. Prove Theorem V, §914. 37. Show that if r is the rotation matrix (9), §911, then its inverse r~! is given by a matrix of the same form, where @ is replaced by —@.

38. Prove that the set of all rotation matrices of the form of (9), §911, with the operation of matrix multiplication, form a commutative group (cf. Definition I, §302). cos 0 sin 0 )is orthogonal with determinant —1 for every 39. Prove that the matrix ( sin@ —cos@ value of 6. Show that the set of all matrices having this form do not forma group. (Cf. Ex. 38.)

10

Vectors

1001

RECTANGULAR

COORDINATES

in Space

IN SPACE

For the analytic study of figures in Euclidean space £3, it is helpful to introduce a rectangular coordinate system, somewhat similar to that of the plane, but consisting of three mutually perpendicular directed lines meeting in a point O called the origin. Let a uniform number scale be assigned to each axis, with the number 0 corresponding to the origin in each case. Although it is not essential for all purposes to have

equal scales on the three axes, we shall assume unless a statement to the contrary is specifically made that the three number scales have equal units. In close analogy with the plane, where there exists a one-to-one correspondence between points and ordered pairs of numbers, in space there exists a one-to-one correspondence between points and ordered triplets or triads of numbers, called the coordinates of the points. If the coordinate axes are labeled x, y, and z, the corresponding coordinates of any

point p are called the x-coordinate, the y-coordinate, and the z-coordinate, respectively, ofp and defined as follows: The x-coordinate (or y-coordinate or z-coordinate) ofp is the number of the number scale on the x-axis (or y-axis or z-axis, respectively) that corresponds to the point of intersection of that axis and the plane through p parallel to the other two axes. This correspondence is expressed symbolically thus: p: (x, y, z) means thatp is the point with coordinates x, y, and z. This point is also labeled (x, y, z) for simplicity. (Cf. Fig. 1001-1.) The coordinate planes are the three planes each of which contains two of the coordinate axes, the xy-plane containing the x- and y-axes, the xz-plane containing the x- and z- axes, and the yz-plane containing the y- and z-axes. Orthogonal projections of the point (x, y, z) are defined as follows: the orthogonal projection of (x, y, z) on the xy-plane (or xz-plane or yz-plane) is the point (x, y, 0) (or (x, 0, z) or (0, y, z), respectively). The orthogonal projection of (x, y, z) on the 268

$1001]

RECTANGULAR

COORDINATES

IN SPACE

269

X-axis (Or y-axis or z-axis) is the point (x, 0, 0) (or (0, y, 0) or (0, 0, z), respectively).

(Cf. Fig. 1001-1.) Although the following definition has no absolute mathematical meaning* it is a convenient one to have because ofits close relation to physical pictures or models of coordinate systems and its help in visualization: Definition. A rectangular coordinate system is right-handed if and only if the axes are arranged so that if the xy-plane is viewed from a point on the positive half of the z-axis, a counterclockwise ninety-degree rotation in this plane about the origin carries points on the positive half of the x-axis into points on the positive half of the y-axis. The system is left-handed if and only if it is not right-handed.

(0, y, 2)

(0, y, 0)

y

Figure 1001-1

Note 1. Using the concept of a right-handed screw (that is, one way that when it is turned about its axis in a clockwise sense it moves we can describe a right-handed rectangular coordinate system as one screw whose axis coincides with the z-axis advances in the positive through the right angle xOy.

that is threaded in such a away from the observer), for which a right-handed z direction when turned

Nore 2. The relationship among the axes expressed in this definition is a cyclic one. The x-, y-, and z-axes could be replaced (in the definition) by the y-, z-, and x-axes, respectively or by the z-, x-, and y-axes, respectively, without changing the effect of the definition.

Examples of right-handed coordinate systems are shown in Figure 1001-2. *Without reference to the physical universe it would be impossible to explain to an extraterrestrial being what we mean by clockwise motion, a right-handed screw, or a southpaw pitcher.

fra bie VECTORS

[$1002

IN SPACE

x

Figure 1001 -2

The open first octant is x > 0, y > 0,andz of whose coordinates are words first octant usually tive:

1002

VECTORS

the set of points (x, y, z) all of whose coordinates are posi> 0. The closed first octant is the set of points (x, y, z) all nonnegative: x = 0, y = 0, and z = 0. The unmodified refer to the open first octant.

IN SPACE

We incorporate the ideas of the Definition, $801, and some of those of $901, into a new formulation designed especially for the space £3: Definition. 4 three-dimensional vector 6 is an ordered triad or triplet (v1, v2, 03) of real numbers. The numbers v4, v2, and v3 are called the components or coordinates of the vector 6 = (v4, 02, v3); v1 is the x-component or x-coordinate, v2 is the y-component or y-coordinate, and v3 is the z-component or z-coordinate. Two vectors t = (U1, U2, U3) and 6 = (U4, v2, v3) are equal, written t = 6, if and only

if their corresponding components are equal: uy; = U1, Uz = U2, and u3 = v3. Otherwise they are unequal, written t # 6. The vector (v1, v2, 03) is represented by the directed line segment joining any point (a1, a2, a3) to the point (a1 + v1, a2 + 02, a3 + v3). The magnitude or absolute value of a vector 6 = (U1, v2, v3) is denoted and defined

|5] = |i, v2, v3)| = Yor? + v22 + 032.

(1)

The zero vector is the vector 0 = (0, 0, 0) having zero magnitude.

A unit vector

is a vector with magnitude 1. A unit vector is also called a direction. If p: (x1, ¥1, 21) andg : (X2, y2, Z2) are any two points of space, the term the vector pq means the vector (x2 — X1, y2 — yi, Z2 — 21) represented by the directed line segment pq. The radius vector of the point p: (x1, ¥1, 21) is the vector Op = (x1, y1, 21) where O is the origin. The vector PQs as represented by the directed line segment pq, is said to have initial point p and terminal point q (cf. Fig. 1002-1).

Unless a statement to the contrary is made, all vectors considered in this chapter will be assumed to be three-dimensional. Notre. As in E2, two directed line segments represent the same vector if and only if they are parallel, similarly directed, and of equal length. Example

1.

If p aa qe

0, 6), C=

—

(6, 35 4), Iie =e

>

(5, =1;, 4), and

jag, = mR = (GS, 3 SZ

5

=

(10,

2. 2»). then

)

Related to the definition (1) of the magnitude of a vector is the following extension to three dimensions of the distance formula (x2 — x1)? + (v2 — yi)? of

§1002]

VECTORS

IN SPACE

271

2o—

2]

42 — Xy

x

Radius vector Op

Vector pq

D:(%1,Y4) 2)

DX

Vee)

(a)

C= Xo; Vos 25)

(b) Figure 1002-1

the plane. This formula states in particular that the magnitude of any vector is equal to the length of any line segment that represents that vector. Distance formula.

(2)

The distance between

the points (x1, y1, 21) and (x2, y2, Z2) is

d = Ve — 11)? + G2 — ni? + @ = 21).

Proof. Formula (2) is a consequence of the Pythagorean theorem applied twice. Let a, b, and c be the lengths of the edges of the parallelepiped R illustrated in Figure 1002-15 having (x1, 1, 21) and (x2, y2, z2) as Opposite vertices:

(3)

ee

eX

oe yy ec = |72'— v2|,

The problem is to show that

(4)

ad = a2 +b? + c?.

If a diagonal of length e is drawn in the base rectangle of R, as shown in Figure 1002-2, then (4) is a consequence of the two Pythagorean equations

(5)

a2+$62=e2,

¢2 = d, e+

the first related to a right triangle in the base of R, and the second related to the right triangle marked with heavy lines in Figure 1002-2.

b Figure 1002-2

IN

VECTORS

272

[$1003

SPACE

The distance between the points p : (6, 3, —1) andgq: (Gains) as

Example 2.

+6 — (—-))? = V4 + 16 + 36 = 2y14,

V4— 62 + (7 — 32

and is equal to the magnitude of the vector pq = (—2, 4, 6).

1003.

LINEAR

COMBINATIONS

OF VECTORS and

Nearly all the definitions and statements of §§802

803

have their counter-

parts in three-dimensional space. We give a few of these, but shall feel free to use three-dimensional analogues of the plane cases when the meaning is clear from context or close parallel. For the sake of easy reference and completeness, we permit present definitions to particularize (and hence duplicate) some of those of §§901 and 902, with n = 3. Addition and multiplication by a scalar are defined as follows:

(1) (2)

(ui, 43, U3) aa (Ui, 02,03) = (Ue 01, te 7 02, U3

03),

A(u1, U2, U3) = (Aut, AU2, AU3).

The vector (—1)i, also written — i, is called the negative of 7. Subtraction of vectors

is denoted and defined as follows:

(3)

Gee)

Sums, differences, and scalar multiples of vectors in space are represented in terms of segments, triangles, and parallelograms in the same fashion as sums, differences, and scalar multiples of vectors in the plane, as indicated in Figures 802-1 and 802-2. The vectors of a set are coplanar if and only if they can be represented by directed line segments lying in a plane. Vectors that are not coplanar are called noncoplanar. Vector magnitude for vectors in space satisfies the seven properties (i)-(vii) listed in § 802. A linear combination of a set of vectors 01, 62 ,:--, 6, is a vector of the form

(4)

NiBina= AaDs SPP

sano

where di, A2,°°°, An are scalars.

The set U3 of all three-dimensional vectors is a vector space over &. The vectors 01, &2,°--+, 0, are linearly dependent if and only if there exist scalars

M1, A2,°°*, An not all equal to O such that (5)

A101

+

A202

+-:---+

},v,

—

0.

If no such scalars exist, that is, if (5) holds only for scalars \1, \2,:-*, Xn all of which are zero, then the vectors 01, d2,:--, 6, are linearly independent. The three general theorems of §803 apply without change to the space 3 of three-dimensional vectors. For purposes of concrete clarification, as we did in $803, we shall consider sets {&1, 62,---, ,} of three-dimensional vectors for various special values of n. In this case, we take n = 1, 2, 3, and 4, in turn. We start with Ves

$1003]

I.

b= 0:

LINEAR

COMBINATIONS

OF

VECTORS

273

A single vector 6 is linearly dependent if and only if it is the zero vector:

Proof.

The proof is the same as with I, §803.

II. Two vectors # and 6 (in three dimensions) are linearly dependent if and only if they are collinear or, equivalently, if and only if one is a scalar multiple of the other. If one of the two vectors u and 6 is linearly dependent — that is, equal to the zero vector — then the two vectors w and 6 are linearly dependent. If % is a linearly independent (nonzero) vector, and if # and 6 are linearly dependent (é is a vector collinear with u), then 6 is uniquely represented as a linear combination (scalar multiple) of i. Proof.

The proof is the same as with II, $803.

III. Three vectors u#, 6, and w (in three dimensions) are linearly dependent if and only if they are coplanar. If @ and 6 are linearly independent and if is any vector coplanar with w and 8, then w is uniquely represented as a linear combination of @ and 0. Proof.

Let u, 6, and

w be any three coplanar

vectors,

and consider

the two

vectors # and v. If w and o are linearly dependent, then so are i, 6, and W by Theorem I, $803. If and 6 are linearly independent vectors, then the geometrical argument given in the proof of III, $803, and illustrated in Figure 803-1, can be repeated to give a representation of w as a linear combination of % and 6, and hence a proof that uw, 0, and Ww are linearly dependent. Conversely, if %, 6, and w are any three linearly dependent vectors, then one (say wW to be specific) must be a linear combination of the other two (# and 6) and hence is representable by means of a directed line segment lying in the same plane as segments representing w and 0. Finally, if a and & are linearly independent and d, 0, and w coplanar, then u, 6, and w are linearly dependent so that there exist constants a, 8, and y not all zero such that ai + Bb + yw = 0. In this equation y cannot be 0 since a and 6 are linearly independent, and therefore # = (—a/y)iu + (—6/y)é. This representation of as

a linear combination of u@ and 6 is unique by Theorem III, $803. IV. Four vectors i, 6, W, and £(in three dimensions) are always linearly dependent. If a, 0, and w are any three linearly independent (noncoplanar) vectors, then any vector £ is uniquely representable as a linear combination of i, 0, and W.

We shall give an algebraic proof for the following statement, leaving the Proof. remaining few details as an exercise: Jf i, 0, and W are linearly independent vectors, then any vector & is representable as a linear combination of u, 6, and w. (Give a geometric proof of this same statement, similar to that of III, §803, iO EX. 39, §1004.)

Let a = (m1, u2, U3), D =

(U1, V2, U3), W = (w1, W2, w3), and — = (&, £2, £3),

and consider the set of three homogeneous equations in the four unknowns x1, x2, 3, XA: UyxX1 + Vix2 + wix3 + £1x4 I

(6)

u2X1 + v2x2 + wox3 + b2Xx4 | U3x1 + 3x2 + W3x3 + §3%4

0, 0, 0.

IN

VECTORS

274

§1003}

SPACE

By the Corollary to Theorem II, §907, this system, which is equivalent to

+ x4é = 0,

+ x26 + x3

xii

(7)

has a nontrivial solution. Since a, 6, and W are linearly independent, x4 ~ 0, and (7) can be solved for — as a linear combination of i, 6, and w.

Owing to the property expressed in the second sentence of IV, above, any three linearly independent (noncoplanar) vectors in space form a basis for the space The three vectors #, 6, and w of IV span the space

03 of three-dimensional vectors.

U3 since {Ai + wd + vw |, w, and »y € R} =

U3.

In the vector space U3 the standard basic triad of unit vectors are denoted and defined:

(8)

P=

ONO) (0,150)

ke—=1

(O50 1).

The vectors i,j, and k are called the standard coordinate vectors of U3. For this basis the representation of an arbitrary vector 6 = (v1, v2, v3) is particularly simple:

(9)

b= vi + voj + o3k.

Example 1. Prove that the two vectors #@ = (uw, u2, 43) and 6 = (v1, v2, v3) are linearly dependent if and only if

(10)

U2

U3

U3

«COU

U2

U3

U3

Uj

UL

U2

Solution. Linear dependence of u# and ¢ is equivalent to the existence of scalars \ and p not both 0 such that Aw + yo = 0 or, equivalently, to the existence of a nontrivial solution (A, ) for the system uid =

(11)

Uh

=

0,

u2d + van = 0,

U3 + v3 = 0, of three homogeneous linear equations.

By Theorem II, §907, where m = 3 and n = 2, the

system (11) has a nontrivial solution if and only if the three 2 * 2 determinants of (10) vanish. For example,

3

[ome

Die 7

Il

as

il

the vectors (1, 2, 3) and (4, 8, 12) are linearly dependent

2 esi 0, whereas the vectors (1, 2, 3) and (1, 7, 3) are linearly independent and

3 d ie? 3) and], | are nonzero Example

and ; Ul =

2.

Prove

that

although

the

three

3 il za4bs vectors

0 )-

w=

(uj, u2 , 43),

© = (v1, v2,

v3), and

W = (1, W2, w3) are linearly independent if and only if

(12)

uj

U2

U3

U1

v2

v3)

Wi

W2

W3

~ 0.

Solution. The vectors 4, ¢, and w are linearly independent if and only if the only scalars r, mu, and y such that Ad + wo + vw = OareX = uw = v = 0, or, equivalently, if and only if the only solution of the system

§1004]

EXERCISES mA + vip + wy

(13)

275 = 0,

u2r + vy + wo = 0, u3A + v3 + w3y = 0,

of three homogeneous linear equations in \, u, and » is the trivial solution \ = je = jp By Theorem II, §907, the system (13) has only the trivial solution if and only if the coefficient determinant — whose value is equal to that in (12) — is nonzero. For example, the three vectors (1, 2, 3), (0, 1, 4), and (0, 0, 1) are linearly independent since

1? 2as3 01 4=1+0 = net

(14)

whereas the three vectors (3, —1, 2), (5, 4, —1), and (11, 19, —10) are linearly dependent since

cs | 2| See he) i ett9 —/10

(15)

Example 3. Show that the three vectors (3, —1, 2), (5, 4, —1), and (11, 19, —10) discussed in connection with (15) are linearly dependent by representing (11, 19, —10) as a linear combination of (3, —1, 2) and (5, 4, —1). Solution. The problem is to solve the system

3X + Su = 11, —\ + 4u = 19,

(16)

2.—

pw =

—10

for \ and yp. From the first two equations of(16) we find \ = —3 and yp = 4. Substitution in the third equation of (16) verifies that \ = —3 and » = 4 isa solution of that equation as well, and hence of (16). Therefore

(11, 19, —10) =

1004

—3(3, —1, 2) + 4(5, 4, —1) = (—9, 3, —6) + (20, 16, —4).

EXERCISES

A figure should accompany each of Exercises

1-6.

1. Find the orthogonal projection of the point (5, —1, 6) on each coordinate axis.

2. Find the orthogonal projection of the point (5, —1, 6) on each coordinate plane.

3. A rectangular parallelepiped with faces parallel to the coordinate planes has opposite vertices at the origin and the point (5, 4, —3). Find the other six vertices. 4. A rectangular parallelepiped with faces parallel to the coordinate planes has opposite vertices at (2, 3, 5) and (8, 5, 7). Find the other six vertices. 5. Find the coordinates of the point where the line through (— 3, 4, 1) parallel to the X-axis meets the plane through (7, 3, —1) parallel to the yz-plane.

6. A plane parallel to one of the coordinate planes passes through three of the following points:

Pi:

G; 4, =),

P2:

©; 3% 6),

P3+

(7,

3, =),

P+:

(=,

Find the three points and name the parallel coordinate plane. In Exercises 7-10, find the point s from the given data.

7. pg = 7s, p =

= ©, 3, 2), r = (0,8, 4): (2,7, 1),¢

4, 6),

and

JOS. (=o: 3; 2).

VECTORS

276

[$1004

IN SPACE

8. = sp = 3; —2, —5),9 =(G 7p slr = 9. Op = gs,p = (1, 4, —1),¢ = (6, —2, —3). 10. ps = 5g, p = (8, 0, 5), q = (2, 6, —3).

4 1,2):

In Exercises 11-14, find the distance between p and q in the indicated exercise. 14. Exercise 4.

13. Exercise 3.

12. Exercise 2.

11. Exercise 1.

In Exercises 15-18, perform the indicated steps.

1556558) 2G) 1-2): 17.12) 5j — 7) — G8

16.74; = j = 3k). 18) 342 — 20) S42

40).

a:

In Exercises 19-24, verify the given equality or inequality.

19. |-5G —f + &] = |—5]-|i -—F+ &l.20. |(4, 7, —2) — 3, —1, ©| = |G, —1, © — , 7, —2)]. 21.

‘s48) =

=

2G pe See 22. |(27 — 27 4- kl

1G; 4, 8)! |—2|

23. |3i+ 4) + (57 — 2k)| S [31+ &| + [57 — 24].

In Exercises 25-31, determine whether the given vectors are linearly dependent or linearly independent. Give reasons to validate your conclusion.

25. 0.

26. (0, 0, 1).

PT OMA GYR (1 e259):

28) 12/5 432)

29, (15.159),

3.0) 1) Onl):

BI) i 4 2y.3)

4k 5k

ee

30. (1,,—1,,.0) 20) 0, — 1) (Oe):

6 Ti = 8)

Ok.

In Exercises 32-34, express the first vector as a linear combination of the remaining vectors.

32. (1, —1, 0), (1, 0, —1), (0, 1, —1).

B30 ak 349 27) 2374 Sie ty

eee Ane ee ka

eee

In Exercises 35-38, determine whether the vectors are (/) collinear, (i) coplanar but not collinear, or (iii) noncoplanar. 35.

(ile il. =).

CG

=I,

4), G; ile sal

36; 10; 2 Sj 15, 377614310). 4h i =), 4 38.

(5, eh

7), (3, Ss)

4), C1

ah

6) 3)

Oe

eee naa i 72)).

39. Give a geometric proof of the statement set forth at the beginning of the proof of

IV, §1003.

40. If p is the midpoint of the segment joining the points p; : (1, y1, z1) and p2 : (x2, y2, Z2), and if O is the origin, prove that

$1005]

DIRECTION

Op =

()

COSINES;

THE

al om yi

x1

Opi + Op2 =

SCALAR

a

,

+

PRODUCT

y2 ZI

5)

277

on Z2

D

(Cf. Ex. 36, §804.) In Exercises 41 and 42, use (1) to find the midpoint of the segment joining the two points. (GiExe405)

ALS G7i=12)'8), (4.6), 1005

DIRECTION

COSINES;

42. (6,4, —11), 0, —9, —3). THE SCALAR

PRODUCT

If L and M are any two directed lines in space that pass through the origin, then a unique 6 between L and M and satisfying the inequalities (1)

OS

657

is determined as in §805. If L and M are arbitrary directed lines in space, not necessarily passing through the origin, then the angle between L and M is defined to be the angle between the lines L’ and M’ that are parallel to L and M, respectively, and are similarly directed, and that pass through the origin. In space any directed line L has three direction angles: (i) the angle a between L and the (positively) directed x-axis, (ii) the angle 8 between L and the (positively) directed y-axis, and (iii) the angle y between L and the (positively) directed z-axis (cf. Fig. 1005-1a). The cosines of these direction angles, (2)

X=cosa,

»=cosf,

v =cos 7,

\z

z

y

ey,

Vector pq

Directed line

Ge

YA

Di (X45.Nr 24)s @: (Kop Voie)

(a)

(b) Figure 1005-1

are called the direction cosines of L. Direction angles and direction cosines of any nonzero vector or directed line segment are defined as with a directed line. The direction cosines of the directed line segment from p : (x1, y1, 21) tog : (X2, 2, 22), of length d (cf. Fig. 1005-15), are equal to (3)

X=

sa

==

mos

ba = cos

B =

Dae

yl

=

Cs

2)

7 =

ee!

7

VECTORS

278

[$1005

IN SPACE

or, if 6= (v1, v2, v3) = pg = (x2 — x1, y2 — Vi, 22 — 21): (4)

A = COSo

Dv

= Joy # = COS B=

y)

D

aly = (G08 y=

|S Lert}

Since d2 = (x2 — x1)?

+ (2 — yi)? + (22 — 21)%, we have from (3):

(5)

Mf yet?=1.

That is, for any directed line L, the sum of the squares of the direction cosines is equal to 1. In other words, if \, u, and v are the direction cosines of a directed line L, then (A, uw, v) is a unit vector. This unit vector (A, u, v), sometimes called the direction of L, is collinear with LZ and similarly directed. If (J, m, n) is a nonzero vector collinear with (A, u, v), where A, u, and v are the direction cosines of a directed line L, then /, m, and n are a set of direction numbers of L. In this case there exists a nonzero number k such that / = k\, m = ku, and n = ky. From (3) with k = d, we see that x2 — x1, y2 — y1, and zz — z are direction numbers of the directed line from (x1, y1, 21) to (x2, y2, 22). Since a vector (/, m, n) collinear with a direction (A, », v) is also collinear with the opposite direction (—\, —, —yv), /, m, and n are called direction numbers of the undirected line L as well as of the directed line L. Example 1. The vector (8, —1, —3), represented (—5, 4, 1) to (3, 3, —2), has direction cosines

\ = cOS

@ =

8 ——= p = cOS 8B =

74

—

by the directed

line segment

from

1 3} ——] vy = COSY = — ——:

v74

74

With the notation used above, d = 74, (5) becomes $$ + 7; + 7% = 1, and the following are sets of direction numbers for the line L through (—S, 4, 1) and (3, 3, —2): (8, —1, —3), (—8, 1, 3), (24, —3, —9), (—40, 5, 15). The two directions for L are + Arv74 (8, —1, —3). The scalar product or inner product or dot product of two vectors i = (u1, u2, U3) and 6 = (v1, v2, v3) is denoted and defined: (6)

u: B= mv, + u2v2 + U303.

Most of the statements concerning the scalar product given in §806 transfer without change. The proofs involve such minor alteration that they are left as exercises .

Theorem I. The scalar product of two vectors i and & is equal to the product of their magnitudes and the cosine of the angle between them:

(7)

ib =

ii|-|5|- cos 0.

In particular, if (1, “1, v1) and (X2, 2, v2) are unit vectors,

(8) Theorem II. Example 2. 2740)

cos 8 = Airh2 + mime + 172. Two vectors are orthogonal if and only if their scalar product is zero. The vectors (3, 7, —2) and (5, —1, 4) are orthogonal since 3-5 — 7-1 —

§1006]

EXERCISES

279

Note 1. Theorem II shows that orthogonality as defined after (3), §901, for n-dimensional vectors is consistent with Euclidean orthogonality when n = 3.

Further properties:

OR (ii) (iii)

= 4-4-= ul: If comp, &@ = |u| cos 6, then w-6 = (comp, #)|dl. Commutative law: i-6 = 6-i.

(iv)

Distributive laws:

(v)

Homogeneous law:

w-(6 + W) = a-6 + -w,(4 + 6)-w = aw

(Ai)-(uo) = (Au)(-6),

+ ow.

X and p € @.

(vi)

The associative law for scalar products is meaningless.

Note

2.

The scalar product of any two vectors is independent

of the rectangular co-

ordinate system.

Example 3.

Show that the only vector orthogonal to each of three linearly independent

vectors i, 6, and W is the zero vector 0. Solution.

Assume the contrary, and let ¢ be a nonzero vector orthogonal to each of 4, ¢,

and w: (-4 = ¢-6 = €-w = 0. Then € is orthogonal to every linear combination of 4, 6, and w:

&-(Qa + wi + vw) = AE-4 4+ we-b + ve-wW = 0, A, wu, andy € R. By IV, §1003, ¢ itself can be represented as a linear combination of 4, 6, and w, and therefore

¢€ is orthogonal to itself: ¢-¢ = ||? = 0. Consequently, |¢| = 0 and ¢ = 0.

1006

EXERCISES

In Exercises 1-3, determine which ordered triads are sets of direction angles.

1. 67, 47, 37. 1

i:

i

2. ET, 27, BT. 1

at

3. 27, 0, a7.

2

if

3

In Exercises 4— 9, find the direction cosines of the given directed line or line segment. 4. Positively directed x-axis.

5. Negatively directed x-axis.

6. Positively directed z-axis.

7. From the origin into the first octant (cf. $1001), making equal angles with the coordinate axes.

8. From (3, 8, 1) to (/, 2, 10).

9. From (0, —5, 13) to (6, 4, —5).

In Exercises 10-12, find three distinct sets of direction numbers for the line segment joining the two points. 10.

ce il, 0), as 0, =i):

11. ry

il —1), (1,

1

12. (9, 1, —7), (4, —8, 2).

In Exercises 13-20, find the scalar product of the two given vectors.

13. jand i.

14. Kand k.

1).

15. jand k.

16. i and k. 18. 2) — 7 — 7k and —6i + 5; — k.

17, 4,4, —7) and 6, —4, —2)

20. (6, 0, —3) and(4, —7, 1).

ya

sian 3)

1990)

[$1007

IN SPACE

VECTORS

280

In Exercises 21-23, find the cosine of the angle between the two vectors.

DING 23.

Q,

04 0)) (dno a): 35 )5 G=3;

DoS]

ed

era

D5 2).

In Exercises 24-26, determine whether or not the two vectors are orthogonal.

24. (2, 8, 3), (5, 1, —6).

25.31 =) 40k, Sho

26. (—7, 5, 11), (—3, —13, 4). In Exercises 27 and 28, find the dot product, cos 6 (where @ is the angle between the two vectors), and the component of each vector in the direction of the other.

27. 4 = (6, 3,,—3), b = (—6, 2, 9).

28.-4= (=2,.3, —1), 0 = Q; 27 —1):

29. Let a, 6, and y be three angles such that O SaS7,05 687, and0S yr. Prove that a, 8, and y are direction angles for some directed line in three dimensions if and only if cos? a + cos? B + cos? y = l.

30. Give an algebraic proof of the inequality

[uz-t] < |a|-[é ,

or

juror + 202 + U303| Ss Vur2 + U2? + U3? Vor? + v2 + 032,

and relate the result to the trigonometric inequality |cos 6] S 1. (Cf. Ex. 42, §808.) 31. Prove that the vectors of any set of mutually orthogonal nonzero independent.

1007

EQUATIONS

OF PLANES

AND

vectors are linearly

LINES

Let II be an arbitrary plane, let p : (xo, yo, zo) be any fixed point of II, and let f = (a, b, c) be any nonzero vector normal (perpendicular) to Il. Then a point q: (x, y, z) lies in the plane I if and only if the vector Pq = (x — x0, y — yo, Z — Zo) iS perpendicular to the vector fi (cf. Fig. 1007-1). Since two vectors are perpendicular if and only if their scalar product is zero (Theorem II, §1005) and since hi pq = a(x — xo) + b(y — yo) + c(z — zo), we have established the theorem:

SY

Q1

Figure 1007-1

$1007]

EQUATIONS

OF

PLANES

AND

LINES

281

Theorem I. The plane through the point (xo, yo, zo) and perpendicular to the nonzero vector (a, b, c) is the graph of the equation (1)

ax — xo)

+ by — yo) + c(z — zo) = 0.

If equation (1) is expanded and if the substitution d = —axo — byo — czo is made, we also have the first part of the theorem: Theorem II. (2)

Every plane 1 is the graph of a linear equation of the form ax + by+cz+d=0,

in which the coefficients a, b, and c form a set of direction numbers for the lines normal to II. Conversely, the graph of every equation of the form (2), where a, b, and c are not all zero, is a plane perpendicular to the vector (a, b, c).

Proof. The only part remaining to be proved every coefficient is zero, and therefore we can find for (2) (for example, if a # 0, we can take xo = plane II through (xo, yo, zo) and perpendicular to hence of (2) as well.

is the last. By assumption, not a solution x = x0, y = yo, Z = Zo —d/a, yo = 0, and zo = 0). The (a, b, c) is the graph of (1), and

Nore 1. Although the equation (2) is not the only linear equation whose graph is II (the equation could be multiplied by an arbitrary nonzero constant, for example), we shall find it convenient to refer to (2) as “‘the equation of II.’”’ We shall also speak elliptically of “the planeax + by + cz+d= 0” to mean the plane that is the graph of ax + by + cz+d=0.

Example 1. Find the equation of the plane parallel to 3x — 2y + 4z + 3 = 0 passing through the point (1, 1, —2). First solution.

By (1) the equation is 3(x — 1) — 2(y — 1) + 4(z + 2) = 0, or 3x — 2y +

4z+7=0.

Second solution. The equation can be written in the form 3x — 2y + 4z + d = 0. Substitution gives 3 — 2 —8+d=0,ord=7. Example 2. Find the equation of the plane p2:(2, 7, —1), and p3 : (—4, 5, 3).

through

the three

points p: : (1, 12, 1),

Let the equation of the plane be ax + by + cz +d = 0. Substitution of the Solution. coordinates of the three points gives the following system of three equations in the four unknowns 4a, b, c, d:

(3)

a+12b6+ c+d=0, 2a+ 7b—- c+d=0, —4a+ 564+ 3c+d=0.

Elimination of d by subtraction (the first and second equations and the first and third equations) gives the system 1 a — 56 — 2c = 0,

(4)

5a

Th

2c = 0.

VECTORS

282

IN SPACE

[$1007

Subtraction once more gives 4a + 125 = 0. Therefore a solution of (3) is given by a = 3, b = —1,c = 4, and d = 5, and the equationis 3x — y+ 4z + 5=0. Another solution is given in Example 4, $1010.

Example 3. Show that the plane ax + by + cz + d = Ois (i) parallel to the x-axis if and only if a= 0, (ii) parallel to the y-axis if and only if b= 0, and (iii) parallel to the z-axis if and Onilyather— 0: Solution. Saying that a plane II and a line L are parallel is equivalent to saying that lines perpendicular to II are perpendicular to L. Therefore the plane ax + by + cz + d = 0 is parallel to the x-axis if and only if the vectors (a, 6, c) and (1, 0, 0) are orthogonal, and these vectors are orthogonal if and only if their dot product 1-a¢ + 0 + 0 = ais equal to 0. This establishes part (i). Parts (ii) and (iii) are proved similarly with the vector (1, 0, 0) replaced by the vector (0, 1, 0) or the vector (0, 0, 1).

Norte 2. The result proved in Example 3 can be stated in condensed form: A plane is parallel to a coordinate axis if and only if the variable of that axis is missing from its equation.

Now let L be an arbitrary line in space, let p : (xo, yo, Zo) be an arbitrary point on L, and let (/, m, n) be a set of direction numbers of L, that is, a nonzero vector collinear with L. Then a point qg : (x, y, z) lies on L if and only if the vector Pq is collinear with L, or equivalently, if and only if Pq = (x — Xo, y — yo, Z — Zo) and (/, m, n) are collinear. One way to express this is as follows: Theorem III.

A set of equations of the line L through the point (xo, yo, zo) and

having direction numbers 1, m, and n is

(5)

xo)

= (V — yom

= @ — zo) 4

or 2S mee)

)

A

We

RD)

eC

Os

where both (5) and (6) are to be interpreted as statements of proportionality. In other words, any vanishing denominator in (6) means the vanishing of the corresponding numerator, and not “division by zero.” Note 3. Whereas a plane is specified by means of a single linear equation in x, y, and z, a line requires more than one such equation. Equations (5) or (6) can be regarded as equivalent to the system of three equations

(7)

m(x — xo) = ly — yo), n(x — xo) = Kz — zo), nly — yo) = m(z — zo),

or to an appropriate pair of (7). Each equation (7) is an equation of a plane containing L. Equations (6) are called equations of L in symmetric form.

Example 4. can be written

Equations of the line L through (6, 7, —3) parallel to the vector (4, —5, —1)

§1007]

(8)

EQUATIONS

OF

PLANES

AND

LINES

283

—5x — 6) = 4 — 7), —@ — 6) = 42 + 3), -V — 7) = —S(z + 3),

or any pair of these last three, such as (simplified)

(9)

5x + 4y — 58 =0,

x+4z+6=0.

Equations (8) are equations of three planes containing L, each of which is parallel to one of the coordinate axes. Equations (9) are equations of the two that are parallel to the z-axis and the y-axis, respectively. Example 5. Write equations of the line L through (1, 5, —2) perpendicular to the plane 11 = 0.

7x — 8z+

Solution. Since (7, 0, —8) is a vector perpendicular to the given plane it is parallel to L. Therefore the equations of L can be written Meee

i,

es

ee

> ee

ee:

)

or x — 1 7

z+2 =

koe

=o

=O}

or (10)

8& +7z+6=0,

y—S5=0.

The first plane of (10) is the plane through L parallel to the y-axis. The second plane of (10) is the plane through L perpendicular to the y-axis or, equivalently, parallel to both the x-axis and the z-axis. The line L is parallel to the xz-plane since the plane y — 5 = 0 is parallel to the xz-plane.

Example 6. 7) O), Bs 10 Solution.

Write equations of the line L through the two

points p:(9,

5, —1) and

Since Pa = (0, 0, 2), the equations of L can be written Kee?

Vie

he Z er,

(Poe Ka ee or (11)

x—-9=0,

y=i=0.

The planes (11) are the planes through L perpendicular to the x-axis and the y-axis, respectively. Example 7.

The equation of the plane through the point (11, 6, —7) and perpendicular to

the line Mir 20. sie

Ones

7

ey

—4

can be written 0(x — 11) + 5(v — 6) — 42 + 7) = 0, or Sy — 4z —

58 = 0.

A second way to express the statement that the point q : (x, y, z) lies on the |line L through p:(xo, yo, Zo) with direction numbers /, m, and n if and only if Pq and (1, m, n) are collinear is that there exists a real number ¢ such that Pq ===), Tit), cOL:

VECTORS

284

IN

[$1008

SPACE

(x — x0, y — Yo, Z — 20) = (It, mt, nt).

(12)

Solving (12) for x, y, and z in terms of #, and regarding ¢ as a parameter for the line L (cf. §411), we have the following parametric equations for L:

Theorem IV. A set of parametric equations of the line L through (xo, yo, Zo) and having direction numbers 1, m, and n is

the point

x = xo + I, mt, y=yot+ Z = Zo + Nt.

(13)

Example 8.

The equations of the lines of Examples 4, 5, and 6 can be written, respectively,

(14)

SOQ eh,

Wey — si,

BS SS = iE

yo

eS he =

=

fe

ek ish OE yr . Se

SP

Sp

The parametric form (13) for the equations of a line is often convenient for finding points of intersection. Example 9.

Find the point where the line

(15)

es

es

meets the plane

(16)

3x —y+7z+5=0.

Solution.

We write equations (15) in parametric form and substitute in (16):

3(3 + 21) — (-2 + 5) + 7-1

—- 490 +5 =0,

or 9 — 271 = 0. Therefore r = 4, and the point is

1008

DISTANCE

BETWEEN

A PLANE

AND A POINT

The analogue of Theorem II, §413, for three dimensions is the following: Theorem. Jf ax + by +cz+d=0 is the equation of a plane 1 and if pi: (x1, 1, Z1) is any point, then the distance d between I and p, is given by the formula

(1) Proof.

(2)

d — ax t by + cz1 +d) Va? + B+ 0? The parametric equations of the line L through p: perpendicular to II are

X=X+at

y=yt+bt

z=21 + ct.

§1009]

EXERCISES

285

To find the point p : (x, y, z) where the line L meets II we substitute the expressions (2) for x, y, and z in ax + by + cz +d = 0, and solve for f:

(3)

a(x1 + at) + bG1 + bf + c(z1 + ct) + d = 0,

whence ¢ = —(ax1 + by: + cz; + d)/(a2 + b2 + c2). following expression for d?:

(4)

A(X

From

(2),

we

have

the

— x1)? te) — yi)? + @ — 21)

Se? (Ce

2,2 ye

2

— (ax1_teee by + cz + d)

and (1) follows immediately. Example.

The distance between the plane 5x — y — 6z —

11 = Oand the point (1, —8, 2)

is

[5-1 + (—1)-(—8) + (—6):2 — 11] oe

425 -F 1e-36

1009

62

EXERCISES

In Exercises 1 and 2, find the equation of the plane through the given point parallel to the given plane. 1. (3, —1, 8); 2x

—7y —z+11 =0.

2. (4, —7, —2);

x + 3y — 4z

—17

= 0.

In Exercises 3-6, find the equation of the plane through the three given points. 3.

(1, 0, 0), (0, V3 0), (0,

Ss

(1, 2, 0), (3;

=,

=i \ 6).

0), (0, De

4.

(3, 0, 0), (0, —

6.

a25

—4).

0,

=s)}

05 0), (0, oF 5).

(0, =e,

2), (0, 4,

=I).

In Exercises 7 and 8, show that the plane is parallel to a coordinate axis, and identify that axis.

7. Ix

— 13y +3 = 0.

8. 2y + 9z — 10 = 0.

In Exercises 9 and 10, show that the two given planes are perpendicular. 9, 3x + 2y + 7z — 10 = 0, 5x + 3y — 3z = 0. 10. 6x — 3y — 4z4+ 15 = 0, 5x — 2y + 9z 4+ 16 = 0. In Exercises 11-13, find the equation of the plane through the given point perpendicular to the given line or vector. Sha al le

(Qt U, —1);

13. (3, —1, —5);

3

Ze —|

4

12.

(3, 0, 4);

x 5 a

oe

=

radius vector of (3, —1, —5).

In Exercises 14 and 15, find the equation of the plane that is the perpendicular bisector of the segment joining the two given points.

14, (6, —1, —4), (2, 7, 8).

15. (9, 2, —5), (—3, 4, 1).

[§1009

IN SPACE

VECTORS

286

In Exercises 16 and 17, find the equations of the line through the given point perpendicular to the given plane. Write these equations in symmetric form, and also in the form of a pair of linear equations corresponding to two planes.

16. (4,6, —3);

2x —-y —5z+7=0.

17. (1, —2, 4); 4x

—2y

+ 3z

-—8=0.

In Exercises 18-20, find the equations of the line through the two given points. Write these equations in symmetric form, and also in the form of a pair of linear equations corresponding to two planes. 18. @G; i

—5), (2; ip 3) f

19.

(8, 2, —4), ©; a

=i):

20. (7, 6, 8), (2, 6, 8). In Exercises 21 and 22, write the equations in symmetric form of the line through the given point that is parallel to the line whose equations are given in parametric form.

21. (—1,0, 3); x = —8+ 44,y = —1 22520450) ec — Oa

a

— 24,2

= 44 7t.

Oe

In Exercises 23-29, write the equations in parametric form for the line whose equations are requested in the specified exercises. 23. Exercise 16.

24. Exercise 17.

25. Exercise 18:

DT WEXeLcise 20:

28. Exercise 21.

29. Exercise 22)

In Exercises 30 and 31, show that the two lines are parallel. same or distinct. 305)

26. Exercise 19.

Determine whether they are the

ae

de Pe ag

1

—3

ee:

—2’

Vin Ome Z

-1

3

2

In Exercises 32 and 33, show that the two given lines are perpendicular. Mi

Bc

i

3

eee

—2

33, x — 6-46

ta ee

—4°

y=

—74 2 =

y+3

Zz

—4

5

4 —7 + 3ts

x =

—of, yy = 10 — 247 =

1a

2e

In Exercises 34 and 35, find the point where the line meets the plane.

34.

XZ

a

Ag

5

Opeizeta)

Sey

35, oe GaGa

baits 7 .

aad

j

wi

}

vail’

ire

P

eit >

b

, ue

he

ein.

ae i) sae 4

9@.

tal : aber)

Ainadia

Uh

aidtins wham :

’

of 5

bale

AMIE

Ale

‘

ii

FPR

lt

ups,

rue rT, uF

.

(

@')

“yt

(a

3B

re Ha

nein 4+H4

ont. (a >

ly

On

4+ 42 + 16n >

+13

=

3n2 + 3n + 1.

. Hints: Extend the domain of f to 9t by means of the formula (1), §308, and show that (2)

holds for all € 9U. Part of the problem is to show that (na + $n(n — 1)d) + (a +- nd) = (n + loa + 3(n + Ind. ; a — ar"

15. Hints: Cf. Ex. 14, §308.

If r #1, part of the problem

is to show

A + ar" =

al A

a — ar'tl

ab =geu

16. 1.

Hint: x"*! — yt) = x(x — y)

+ yx" — y"). Hint: Let p be the formula with domain 9 such that p() is the statement of the exercise for eachn € MN.

306

ANSWERS

§313. 1.

union of sets intersection of sets slash, set difference the set of all---such that: -ordered pair Cartesian product 14, Cartesian plane relation domain of p, range of p function domain of f, range of f value of f at x function one-to-one correspondence n factorial image of A underf identity function on A characteristic function of A value of fat (x, y) elliptic notation for a function point p with coordinates x and y 319

SPECIAL

SYMBOLS

Meaning

signum function greatest integer or bracket function formula truth set statement

negation disjunction conjunction implies (does not imply) is (is not) implied by is (is not) equivalent to true formula false formula universal quantifier existential quantifier such that sum of two numbers additive identity additive inverse of a number difference of two numbers product of two numbers

multiplicative identity multiplicative inverse of a number quotient of two numbers field square of inequality minimum maximum nth power

a number symbols for numbers of a finite set of a finite set of a number

binomial coefficient

Page

7, oi 31 31 31 ZO) 33 33 36, 38 36, 38 38, 39 37 37 43 43 43 49 49 49 50 50 50 50 50

51 53, 61 54 61 61 61, 65 62

(a, b), (a, 5], [a, 5), a, 5]

ordered field of rational numbers bounded intervals

+o

plus infinity, minus infinity

65 66-67 67

unbounded intervals absolute value of a number neighborhood of a point max of x (—x) and 0 supremum of A least upper bound of 4 infimum of A greatest lower bound of A square root function

67-68 68 70 74 1 75 76 76 78

Q 5 =

2

(=, 2), (—@, a); (a, + &), [a, +2);

[>|

(= © 5 +o)

Ne

Kx) sup A = l.u.b.(A) inf A = g.l.b.(A)

Vx

sup (A) = L.u.b.4 inf(A) = g.l.b.4

SPECIAL Symbol

SYMBOLS

Meaning

2

square root of 2 line, line segment, or directed line segment directed line segment slope differences x2 — x1, y2 — y1 inclination distance between two points composite of two functions inverse function sum of two functions product of two functions difference of two functions quotient of two functions scalar multiple of a function additive inverse of a function constant function zero function unity function square of a function inequality symbols for functions absolute value of a function

Pip2

Pip2 m

Ax, Ay (84

Api, p2) Pe

pre IV GS IIA

vector space

V i=) oagagnac see

= iS)—

OR x© i tsi cr | ~ wy

SL

£ ces

Ler}

—

vector space of constant functions vector space of polynomials vector space of real-valued functions vector space of bounded functions linear functional positivity for a linear functional

sigma functional nets on an interval vector space of step-functions point with polar coordinates two-dimensional vector magnitude of a vector zero vector

vector radius vector sum of two vectors scalar multiple of a vector negative of a vector difference of two vectors

321 Page

78 86 86 86 87 91 113 123 126 129 129 129 130 130 130-131 130 130 130 130 32 132 135, 142 135 135 Iso 140 142 143 144

147-148 149 189 202 202 202 202, 270 203, 270 203, 272 203029. 204, 272 204, 272

quotient of a vector and a scalar

204

standard basic unit vectors

208 243

direction angles

SPECIAL

322

Meaning

Symbol A = COS a, wp = Cos B

l,m °

Olewt)

Beh PS RS SS

det(A) = det A = |A| Av E €

AT XT AX

D = |e| A=

|E|

&b = (v1, v2, v3)

| = [Ce, e2, 09)

Or= (020, 0)

MOL a

SYMBOLS

Teh

Nl,

direction cosines direction numbers scalar or inner or dot product circle product of two vectors square product of two vectors n-dimensional vector matrix sum of two matrices scalar multiple of a matrix difference of two matrices set of all m X n matrices n-dimensional vector space magnitude of an n-dimensional vector zero n-dimensional vector composite of two transformations product of two matrices algebra of m X n matrices zero matrix identity matrix determinant of a matrix inverse of a matrix matrix of a second degree equation matrix of second degree terms transpose of a matrix quadratic form determinant of e determinant of E vector in space magnitude of a vector zero vector linear combination of vectors standard coordinate vectors direction angles direction cosines direction numbers

scalar or inner or dot product

Page

213 214 21S 218 219 220 220 221 221 ny | 222 222 227 222 224 225-226 221-228 228 228 Zo 240 244 244 244 245 2d Zon 270 270 270 PUG(8: 274 PAG) Pais) 278 278

INDEX (The numbers refer to pages)

INL IN IL. SP

Associative algebra, 157

Abelian group, 52

Associative law, 5, 8, 49, 50, 60, 125,

Abscissas, axis of, 22 Absolute value, of a function, 132 of a number, 68 of a vector, 202, 222, 270

ISOM owe227, Asymptote of a hyperbola, 175 Asymptotic cone, 298 Axis, 25

properties of, 69 Acute angle between lines, 94 Addition, law of, 42 (Ex. 14)

conjugate, 174, 296, 297 coordinate, 22, 268

Addition, of functions, 129 of matrices 221 of numbers, 49 Of Vectors, 2039221, 272

minor, 167 of abscissas, 22

major, 167

of ordinates, 22 of a quadric surface, 296-297 of revolution, 292

Addition and subtraction formulas of trigonometry, 187 Additive identity, for functions, 131 for numbers, 50 Additive inverse, for functions, 131 for numbers, 50 Additivity for a linear functional, 142 Algebra, abstract, 156 associative, 157 commutative, 157 of bounded functions, 155 of constant functions, 155 of functions, 155-157 of matrices, 227-228 of polynomials, 155 of step-functions, 155 Altitude of a triangle, 119 (Ex. 52) Analytic geometry, 24 Angles, direction, 213, 277 Angular coordinate, 189 Answers and hints, 303 Archimedean property, 77 Arithmetic mean, 57 (Ex. 14)

of symmetry, 25 polar, 191 reali transverse, 174, 296, 297

Base set of vectors, 208, 223, 274

Basis, 208, 223, 274 Between, 55 Binary operation, 49 Binomial coefficient, 62 Binomial theorem, 62 Boldface type, 2

Bound, greatest lower, 76 least upper, 75 lower, 66 upper, 66 Bounded above, 66 Bounded below, 66 Bounded function, 137-140 Bounded interval, 67 323

324 Bounded set, 66

Bracket function, 27

Cancellation, law of, 51, 52, 218, 243 (Bos, 27) Cardioid, 193, 195 (Exs. 25, 26) Cartesian plane, 14 (COnicn, 22. 25) of acircle, 115

of a conic section, 260-262 of an ellipse, 167 of a hyperbola, 174

INDEX Composite of transformations, 224

Composite propositions, 45 Composition of functions, 123 Condition, necessary, 37 necessary and sufficient, 39 sufficient, 37 Cone, asymptotic, 298 quadric, 297 Conic sections, 159-183, 195-201,

251-266 Conicoid, 295-299 Conjugate axis, 174, 296, 297

of symmetry, 22, 25 Central conic, 260 Centroid of a triangle, 112 (Ex. 21) Characteristic equation, 262 Characteristic function, 18 Characteristic roots, 263

Conjugate hyperbolas, 178 Conjugate semiaxis, 174, 296, 297 Conjunction, 33, 34 Consecutive natural numbers, 59 Constant, 12 Constant function, 16, 130

Circles ial

Constant term, 19

in polar coordinates, 192 Circle product, 218 Circular disk, 115 Circular functions, 184-188 Classification of conic sections, 258

Contains, 3 Contradiction, law of, 42 (Ex. 15) Contradiction, proof by, 41 Contraposition, law of, 41 Contrapositive, 41 Converse of an implication, 40

Closed disk, 115

Coordinate, angular, 189

Class, 2

Closed half-plane, 84, 105

radial, 189

Closed interval, 67, 68

Coordinate axes, 22, 268

Closed under an operation, 53

Coordinate planes, 268

Closure, 53 Coefficient, 19 binomial, 62 leading, 19

Coordinates, of an ordered pair, 13 of a vector, 202, 220, 270 polar, 189-201 rectangular, 21, 23, 268

Coefficient matrix, 235

Coplanar vectors, 272

Collection, 2

Correspondence,

Collinear points, 91 Collinear vectors, 204

one-to-one, 16 Cosine of the angle between two

Column, 221

13

vectors, 215, 278

Column vector, 220

Cosines, direction, 213,277

Combination, linear, 136, 205, 222, 272

Counterexample, 37

Commutative algebra, 157 Commutative group, 52

Cramer’s rule, 235 Cylinder, 291

Commutative law, 5, 8, 42 (Ex. 13), 49,

50,60, 125, 130, 157, 216.225, 279 Compact interval, 68 Complement, 4 Complete ordered field, 75 Completeness of the real number system, 74-77 Components of a vector, 202, 220, 270 Composite function, 123

Decreasing function, 151 strictly, 152 Definition by induction, 60 Degenerate conic section, 259 Degenerate ellipse, 259 Degenerate graph, 251 Degenerate hyperbola, 259 Degenerate interval, 67

INDEX Degenerate parabola, 259 Degree of a polynomial, 19

325 Eccentricity, of a conic, 168, 177, 198, 200

De Morgan, A., 8

of an ellipse, 168

De Morgan laws, 8, 39 Dense set, 78

of a hyperbola, 177 of a parabola, 200

of the set of irrational numbers, 79

Density,

Eigenvalue, 263

of the set of rational numbers, 78 Dependence, linear, 206, 272 Dependent variable, 16 Descartes, René, 14, 24

Ellipse, 165-171 degenerate, 259 imaginary, 253 in polar coordinates, 195-201

Determinant, 230-234

Ellipsis, 20

of the product of two matrices, 234 Diagonal, main, 221 principal, 221 Diagonal matrix, 243 (Ex. 28) Difference, of functions, 129 of matrices, 221 of numbers, 50 of sets, 9 of vectors, 204, 272 Directed distance, 22 Directed line, 91 Directed line segment, 86 Direction, 202, 213, 270,278 Direction angles, 213, 277 Direction cosines, 213, 277 Direction numbers, 214, 278 Directrix, of a conic, 198, 200 of a cylinder, 291 of a parabola, 159 Disjoint sets, 9 Disjunction, 33, 34 DiskeelelS

Ellipsoid, 295 Elliptic paraboloid, 298 Empty set, 3 Endpoint, of an interval, 67 of a line segment, 86 leg belniny, IL, 2,3, 3, I, AOR, 271 27 Equation, characteristic, 262 Equation of a line, general, 99-102 intercept form, 102 point-slope form, 97 slope-intercept form, 98 two-point form, 97 Equation of a plane, 281 Equations of a line, 282, 284 parametric, 110, 284 Equivalent formulas, 31, 39 Equivalent inequalities, 70 Equivalent statements, 38 Euclid, fundamental theorem of, 62, 65 Euclidean plane, 22 Even function, 26 Even integer, 65

Distance between a line and a point, 114 Distance between a plane and a point, 284

Even natural number, 62 Excluded middle, law of the, 42 (Ex. 12)

Element of a set, 2

Distance between points, 22, 113, 271

Existential quantifier, 43

directed, 22 Distinct, 2

Exponent, 61, 65 Exponents, laws of, 61, 65

Distributive law, 8, 12, 51, 60, 130, (Sais

7- 21

6,227,279

Division, of functions, 130 of numbers, 50 of a vector by a scalar, 204, 222 Domain, of a function, 14-16

of a relation, 14

Extended real number system, 67 Extension ofa function, 17-18

Factorial function, 18 False formula, 37

Family, 2

of a variable, 12 (footnote) Dot product, 215, 223, 278

Field, 49-51 complete ordered, 75

Double-angle identities, 187-188 Double negation, law of, 39 Dummy variable, 5, 144

finite, 53 (Ex. 6) ordered, 53-57 Finite induction, 60-64

326 Finite set, 60 First octant, 270 Focus, of an ellipse, 165, 198

of a hyperbola, 172, 199 of a parabola, 159, 195 Form, linear, 142 (footnote) quadratic, 245 Formula, 30

INDEX positive linear, 143 sigma, 144 Functions, difference of, 129 product of, 129

quotient of, 130 sum of, 129 Fundamental theorem, of Euclid, 62, 65 of mathematical induction, 59

false oi true, 37

Formulas, equivalent, 31, 39 Function, 12, 15 bounded, 137-140 bracket, 27

characteristic, 18 circular, 184-188 composite, 123 constant, 16, 130

decreasing, 151 even, 26

factorial, 18 greatest integer, 27 identity, 18

increasing, 151 inverse, 126-128 linear, 19 monotonic, 151 multiple-valued, 15 (footnote)

negative-valued, 132 nonnegative, 132 odd, 26 of areal variable, 16 of two variables, 19

on-into, on—onto, 15, 16 polynomial, 19 positive-valued, 132 power, 61, 65 rational, 19 real-valued, 16 signum, 27 single-valued, 15 (footnote) square-root, 78 statement, 30 strictly decreasing, 152 strictly increasing, 152

strictly negative, 132 strictly positive, 132 unity, 130 Function space, 135

Functional, increasing linear, 153 linear, 142

General element of matrix, 221 General equation of the first degree, 99-103 General equation of the second degree, 163, 243-246 Generator of a cylinder, 291 Generatrix of a surface of revolution, 292 Geometry, analytic, 24

Graph, degenerate, 251 imaginary, 251 in polar coordinates, 191

of an equation, 24 of a formula, 81 of a function, 24

of an inequality, 81 of a relation, 81

of a system, 107 Greatest integer function, 27 Greatest lower bound, 76

Group, 52 Abelian (commutative), 52 Half-plane, closed, 84, 105 open, 84, 105

Hints and answers, 303 Homogeneous law, 131, 142, 216,

DT, DYE Homogeneous linear equation, 236 Horizontal line, 83 Hyperbola, 172-181 degenerate, 259 in polar coordinates, 199 Hyperbolic paraboloid, 298 Hyperboloid, of one sheet, 296 of two sheets, 297 Hypothetical syllogism, 42 (Ex. 16) Identities, trigonometric, 185-188 Identity, 3

additive, 50, 131 multiplicative, 50, 131

INDEX Identity function, 18

Identity matrix, 228 If and only if, 39

lieeeathenssi7 Image of a set, 18

Imaginary circle, 251 Imaginary ellipse, 253 Imaginary graph, 251 Implication, 36-38 Inclination, of a directed line, 91 of an undirected line, 92 Include, 3

Incomplete ordered field, 78 Increasing function, 151

strictly, 152 Increasing linear functional, 153 Independence, linear, 206, 272 Independent variable, 16 Indirect method of proof, 41 Induction, mathematical, 59-64

Inductive definition, 60 Inductive set, 58 Inequalities, equivalent, 70 simultaneous, 107 solving, 70-72 Inequality, 54 linear, 105-107 strict, 55 triangle, 69, 74 (Ex. 33), 205, 219 (Ex. 44) Infimum, 76 Infinite set, 60 Infinity, minus, 67 plus, 67 Inner product, 215, 223, 278

Input-output machine, 13 Integer, 65 even, 65 negative, 65 odd, 65 positive, 65 Intercept, 98 of a plane, 237 (Ex. 39)

of a surface, 290 Intercept equation, 102, 237 (Ex. 39) Intermediate-membership property, 77 Intersection of sets, 7-9 of vector spaces, 141 (Ex. 18) Interval, 66-68 bounded, 67 closed, 67-68

327 compact, 68

degenerate, 67 finite, 68 half-open, 67 infinite, 68 open, 66-68 unbounded, 67-68 Intervals of a net, 148 Invariants, 254-260 Inverse, additive, 50, 131 multiplicative, 50 Inverse function, 126-128 Inverse of a matrix, 240 Irrational number, 65 Irrationality of V2, 78

Latus rectum, of an ellipse, 172 (Ex. 39)

of a hyperbola, 182 (Ex. 45) of a parabola, 165 (Ex. 30) Law, associative, 5, 8, 49, 50, 60, 125,

NBO. US 227 cancellation, 51, 52, 2] Ce, 4S} (ae, 227/)) commutative, 5, 8, 42 (Ex. 13), 49,

SO), GO), AS, WSO IS7/, AKG, ALS, 2S distributive, 8, 12, 51, 60, 130, 131,

US UO, 227, LYS homogeneous, 131, 142, 216, 227, 279 of trichotomy, 56

parallelogram, 219 (Ex. 46) transitive, 55, 60 Laws, De Morgan, 8, 39

of exponents, 61, 65 of logic, 39, 41, 42 (Exs. 12-16) Leading coefficient, 19 Least upper bound, 75 Left-handed coordinate system, 269

Lemniscate, 194, 195 (Exs. 31-32) Length of an interval, 67

Limagon, 195 (Exs. 33-36) Line, 81-114

directed, 91

equation of general, 99-103 equations of, 282-284 horizontal, 83 in polar coordinates, 191 intercept form, 102 of symmetry, 25 parametric equations, 110, 284 point-slope form, 97 slope-intercept form, 98

INDEX

328

Member of a set, 2

two-point form, 97 undirected, 91 vertical, 83

Line segment, 86 directed, 86 Linear combination, 136, 205, 222, 272

Midpoint, of an interval, 67 of asegment, 109, 212 (Ex. 36), 276 (Ex. 40) Minor axis, 167 Minor semiaxis, 167

Linear dependence and independence,

Minus infinity, 67

AVG; Linear Linear Linear

Modus tollendo ponens, 39

LUZ equation, homogeneous, 236 equations, systems of, 235-239 expression, 82

Linear form, 142 (footnote) Linear function, 19 Linear functional, 142

function, 151 (footnote)

Monotonically increasing (decreasing ) 151 (footnote)

Multiplication, of functions, 129 of a matrix by a scalar, 221 of numbers, 50

increasing, 153

of a vector by a scalar, 203, 272

positive, 143

Linear inequality, 105-107 Linear space of functions, 135 Linear transformation, 223 Locus problems,

Monotone

Monotonic function, 151

119-122,

182-183

Multiplicative identity, 50, 131 Multiplicative inverse, 50 Multiple, scalar, 130, 142, 203, 221, 272 Multiple-valued function, 15 (footnote)

Logic, laws of, 39, 41, 42 (Exs. 12-16) Lower bound, 66 greatest, 76

Natural number, 58

even, 62

Machine, input-output, 13 Magnitude of a vector, 202, 222, 270 properties of, 205 Main diagonal, 221 Major axis, 167 Major semiaxis, 167 Mapping, 13, 15 Mathematical induction, 59-64

Matrix, coefficient, 235 diagonal, 243 (Ex. 28)

identity, 228

odd, 62 Necessary and sufficient condition, 39 Necessary condition, 37 Negation, 32 of a formula, 32 of an implication, 36 of a quantified statement, 43-46 of a statement, 32 Negative, of a function, 131 of a matrix, 221 of a number, 50 of a vector, 204, 272

of second degree terms, 244

Negative number, 54 Negative reciprocal, 92 Negative slope, 88

of the general equation of the second degree, 244

Negative-valued function, 132 Net, 147

orthogonal, 251 rectangular, 220 rotation, 250, 301 singular, 234

Non-Abelian group, 52 Noncollinear vectors, 204 Noncommutative group, 52

inverse, 240

nonsingular, 234

square, 221

symmetric, 244 ZETOMe eeeS

Matrix algebra, 228 Mean, arithmetic, 57 (Ex. 14) Median of a triangle, 112 (Ex. 21)

Noncoplanar vectors, 272 Nondecreasing function, 151 (footnote) Nonempty set, 3 Nonincreasing function, 151 (footnote) Nonnegative function, 132 Nonnegative number, 56 Nonpositive number, 56

INDEX Nonsingular matrix, 234

329 hyperbolic, 298

Nontrivial solution, 236

Parallel lines, 89-91, 101, 259

Nonzero number, 50

Parallelogram law, 219 (Ex. 46)

n-tuple, ordered, 15, 220

Parameter,

Number, irrational, 65

Parametric equations of a line, 110, 284

110, 284

natural, 58

Partition, 147

negative, 54 nonnegative, 56

Pascal, Blaise, 62 (footnote) Pascal’s triangle, 62

nonpositive, 56

Perpendicular

positive, 54

Plane, Cartesian, 14

rational, 65

equation of, 281 Euclidean, 22

real, 48-80 Number scale, 21

Numbers,

lines, 93, 101

Plus infinity, 67 Point, 2, 21-24, 268

direction, 214, 278

of symmetry, 22, 25 of trisection, 110 Point-circle, 116, 251

Oblate spheroid, 296 Octant, first, 270 Odd function, 26

Point-slope equation, 97 Polaraxis, 191 Polar coordinates, 189-201 and conic sections, 195-201 Pole, 189 Polynomial, 19

Odd integer, 65 Odd natural number, 62 On-—into function, 15 On-into relation, 15 On-onto function, 16 On-—onto relation, 15 One, 50, 130 One-to-one correspondence, :

16

Only if, 37

quadratic, 243 Polynomial function, 19 A Polynomials, vector space of, 135 Ns : Positive linear functional, 143 Positive number, 54

Open disk, 115

Open half-plane, 84, 105 ;

ere: e,

Positiv

88

Positive terms, 44

Open interval, 66-68 ; ; Operation, binary, 49 i Opposite sense, 55 : Order of a square matrix, 221

ae ; Positive-valued function, 132 Power, 61, 65 Power function, 61, 65 Prcipaldiiconal 221

Oe complete: ue ae Ordered n-tuple, 15, 220 : Ordered pair, 13 ‘ ; 5 Ordered triple (triad, triplet), 15, 268 Order for f P 132 We ea,

Product, circle, 218 dot, inner, or scalar, 215, 223, 278 of functions, 129 ; of numbers, 50 : of a scalar and a matrix, 221 of ascalar and a vector, 203, 272 of two matrices, 225, 226

= mere,

is

me ace ee

s

square, 219 (Ex. 47)

Product formulas, trigonometric, 188

eee

Product of matrices, determinant of, 234

ee

oes

DN

a

2 A 268

Orthogonal vectors, 216, 223, 278

Pair, ordered, 13

Parabola, 159-164, 195, 259, 263 Paraboloid, elliptic, 298

Projection, orthogonal, 23, 268

Prolate spheroid, 296

pioorbyconeedicion. 41

Proper extension, 18 Proper restriction, 18 Proper subset, 3 Properties of absolute value, 69

INDEX

330 Quadratic form, 245 Quadratic polynomial, 243 Quadric cone, 297 Quadric surface, 295-299 Quantifier, existential, 43 universal, 43 Quotient, of functions, 130 of numbers, 50

Radial coordinate,

189

Radius, of a circle, 115 of an interval, 67 Radius vector, 203, 270

Range, of a function, 14-16 of arelation, 14 of a variable, 12, 29 Rational function, 19 Rational number, 65 Real axis, 21 Real number system, 48-80 extended, 67 Real-valued function, 16 Real variable, function of a, 16 Reciprocal, 50 Reciprocal, negative, 92

Rectangular coordinates, 21, 23, 268 Rectangular matrix, 220 Reductio ad absurdum, 41 Relation, 14 on-into, on-onto, 15

Representation of a vector, 15 Restriction of a function, 17 Reverse sense, 55 Right-handed coordinate system, 269 Root, characteristic, 263 square, 78 Rose, four-leaved, 193, 195 (Ex. 29) five-leaved, 195 (Ex. 30) three-leaved, 195 (Exs. 27, 28) Rotation, 249-251, 301 simplification by, 262-266 Rotation matrix, 250, 301 Row, 221 Row vector, 220 Ruling of a cylinder, 291

Scalar, 203 Scalar multiple, 130, 142, 203, 221, 272

Scalar product, 215, 223, 278 Second degree, general equation, 163,

243-246 Section of a surface, 289-290 Segment, 86

Semiaxis, conjugate, 174, 296, 297 major, 167 minor, 167 of a quadric surface, 296-297

transverse, 174, 296, 297 Sense of an inequality, 55 Setar bounded, 66 dense, 78

empty, 3 finite, 60 inductive, 58

infinite, 60 nonempty, 3 solution, 31, 70 jnautdel, Sil, 740)

Set-builder notation, 10 Set difference, 9 Set membership, 2 Set subtraction, 9 Sigma functional, 144 Sigma summation notation, 144 Signum function, 27 Simplification, law of, 39 Simplifications by rotations, 262-266

Simplifications by translations, 260-262 Simultaneous inequalities, 107 Simultaneous system, graph of, 107 Sine of the angle between two vectors, 288 Single-valued function, 15 (footnote)

Singular matrix, 234 Skew rotation, 251

Slope, in tangent form, 91-92 negative, 88 of a directed line segment, 86 of a line, 89 of a line segment, 87 positive, 88 Slope-intercept equation, 98 Solution, nontrivial, 236 of a formula, 31

of an inequality, 70 of a system, 107 trivial, 236 Solution set, 31, 70

Solving inequalities, 70-73

INDEX Space, 4, 268 function, 135 Span, 208, 274

Special symbols, list, 319 Spheroid, 296 Spiral of Archimedes, 195 (Ex. 37) Square, of a function, 130 of a number, 53, 61 Square matrix, 221 Square product, 219 (Ex. 47) Square root function, 78 Square root of 2, 78 Statement, 29 Statement function, 30 Step-function, 147-150 Straight line in polar coordinates, 191 Strict inequality, 55 Strictly decreasing function, 152 Strictly increasing function, 152 Strictly monotonic function, 152 Strictly negative function, 132 Strictly positive function, 132 Subnet, 148 . Subset, 3 proper, 3 Subtraction, of functions, 129 of matrices, 221 of numbers, 50 of sets, 9 of vectors, 204, 272 Subtractivity of linear functionals, 142 Sufficient condition, 37 Sum, of functions, 129 of matrices, 221 of numbers, 49 of vectors, 203, 221, 272 Sums and differences of sines and cosines, 187 Supernet, 148 Supremum, 75 Surface, quadric, 295-299 Surface of revolution, 292

Syllogism, hypothetical, 42 (Ex. 16) Symbols, special list of, 319 Symmetric form for equations of a line, 282

Symmetric matrix, 244 Symmetric set, 22, 25 Symmetry, axis of, 25 center of, 22, 25

331 line of, 25 point of, 22, 25 with respect to a line, 25

with respect to a point, 22, 25 Systems of linear equations, 235-239 homogeneous linear, 236-238

Tangent formula for slope, 91-92 Tautology, 39 (footnote) Term, constant, 19 Terminal point, 203, 270 Trace, of a matrix, 257 of a surface, 290 Transformation, 13, 15 additive, 223 homogeneous, 223 linear, 223

Transitive law, 55, 60 Translation, 246-247, 300

simplification by, 262-266 Transpose of a matrix, 244 Transverse axis, 174, 296, 297 Transverse semiaxis, 174, 296, 297 Triangle, Pascal’s, 62

Triangle inequality, 69, 74 (Ex. 33), 205, 219 (Ex. 44) Trichotomy, law of, 56 Trigonometric functions, 184-188 Trigonometry, informal review, 184-188 Triple (triad, triplet), ordered, 15, 268 Trisection, points of, 110 Trivial solution, 236 True formula, 37 Truth set, 31, 70 Two-point equation, 97

Unbounded, 66 Unbounded above, 66 Unbounded below, 66 Unbounded interval, 67-68 Undirected line, 91 Unequal, 2, 3, 13, 202, 221, 270 Union of sets, 5-7 Uniqueness of inverse, 52, 53 (Ex. 1) Uniqueness of unity, 53 (Ex. 3) Uniqueness of zero, 51 Unit vector, 202, 222, 270 Unit vectors, standard basic, 208, 223 74:

332 Unity, 50 Unity function, 130 Universal quantifier, 43 Universal set of a variable, 12

INDEX linearly independent, 206, 272 noncollinear, 204 noncoplanar, 272 orthogonal, 216, 223, 278

unequal, 202, 270

Universe, of discourse, 4 of a variable, 12

Venn, John, 5

Upper bound, 66

Venn diagram, 5-10 Vertex, of an ellipse, 166

Values, of a function, 16 of a variable, 12, 29 Variable, 12, 29

dependent, 16

of a hyperbola, 174 of a parabola, 161 of a quadric surface, 297-298 Vertical line, 83

dummy, 5, 144

independent, 16 Vector, column, 220

n-dimensional, 220 radius, 203, 270 row, 220 three-dimensional, 270 two-dimensional, 202 unit, 202, 222, 270 zero, 202, 222, 270 Vector space, abstract, 141-142 of bounded functions, 140 of constant functions, 135 of functions, 135

Weakly increasing (decreasing), 151 (footnote ) Weakly monotonic, 151 (footnote)

Well-ordering principle, 59

x-axis, 22, 268 x-coordinate, 23, 268

y-axis, 22, 268

y-coordinate, 23, 268

of polynomials, 135 of step-functions, 149 Vectors, collinear, 204

coplanar, 272 equal, 202, 270 linearly dependent, 206, 272

Zero, 50

Zero function, 130 Zero matrix, 228 Zero vector, 202, 222, 270

JOHN R. HUBBAPD DEPT. of MATHEMATICS

COLUMBUS COLLEGs

QA

39 mo 5Y.

Olmsted, J.M.H. Prelude to calculus linear algebra

Columbus College Libragy Columbus, George

and

ieesite