Introductory Calculus. With Analytic Geometry and Linear Algebra

I N T R O D U C T O R Y C A L C U L U S Second Edition, with Analytic Geometry and Linear Algebra A WAYNE ROBERTS Macalester College ACADEMIC PRESS New York and London C O P Y R I G H T © 1972, BY A C A D E M I C P R E S S , I N C ALL RIGHTS RESERVED NO PART O F THIS BOOK MAY B E REPRODUCED IN ANY F O R M , BY PHOTOSTAT, M I C R O F I L M , RETRIEVAL SYSTEM, OR ANY OTHER MEANS, W I T H O U T W R I T T E N PERMISSION F R O M T H E PUBLISHERS A C A D E M I C PRESS, I N C I l l Fifth Avenue, New Yo[.]

CALCULUS

Second Edition, with Analytic Geometry and Linear Algebra

A WAYNE ROBERTS Macalester College

ACADEMIC PRESS New York and London

NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM,

BY PHOTOSTAT, MICROFILM, RETRIEVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS

ACADEMIC PRESS, INC

I l l Fifth Avenue, New York, New York 10003

United Kingdom Edition published by

ACADEMIC PRESS, INC (LONDON) L T D

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LIBRARY O F CONGRESS CATALOG CARD N U M B E R : 72-182601 AMS(MOS) 1970 Subject classifications: 26-01, 26A06

PRINTED IN THE UNITED STATES OF AMERICA

PREFACE TO THE SECOND EDITION

doubt if very many chemists chose their field because they found the chart so interesting In the same way, certain essentials, such as limits and the proper­ties of real numbers, must be mastered if one is to understand calculus We doubt, however, that a study of these rather sophisticated notions will be initially responsible for attracting very many people to the study of calculus

In our view, calculus should be presented in a way that appeals to the student's intuitive feeling of what ought to be This book represents such an effort

We have wished to sacrifice neither accuracy nor honesty in our attempt

to be interesting In fact, the definitions we give for function, graph, linear, derivative, and other terms used in advanced mathematics differ from those ordinarily given in elementary texts in that they do not need revision later on Where a proof is not given, its omission is clearly pointed out Often an example is given to show why careful attention to a proof is warranted in a future course

We said in the preface to the first edition that a conscious effort had been made to reverse the trend of calculus texts to become encyclopedic One concession has been made by including in this edition an introductory chapter on elementary analytic geometry (and a concession of another type

was made to the editor when we agreed not to title this chapter The Plane

Facts) We have also included a separate section to discuss limits more fully

Nevertheless, we have adhered to the principle that conciseness should take precedence over the temptation to say everything one can think of about the subject at hand

The first edition stressed the role of the derivative as a tool of approxi­

mation by defining the derivative to be a linear transformation Though the

spirit of this idea is retained in the second edition, formal terminology (and notation toward this end) has been left out of the treatment of one variable except for one optional section Thus, for a traditional course in one variable calculus, this notion can be ignored by those who choose to do so

Keeping in mind the needs of increasing numbers of social science students appearing in the mathematics classroom, we have continued the emphasis on functions of several variables Here there are clear advantages to thinking of

the derivative as a linear transformation, and the presentation is set in this

context The background in linear algebra necessary to this approach, somewhat scattered through the first edition, has in this edition been collected into a separate chapter It is to be stressed that we have included the linear algebra here for the same reason that it was included in the first edition; not

because it is the modern thing to do so, but because it plays an essential role

in what is to follow

The opening chapter on analytic geometry begins with a discussion of vectors in the plane By using vectors to develop some of the results of

—xi—

might otherwise have on those for whom it is essentially a review We also prepare the way for using the same definitions and methods in the later chapter on linear algebra

We distinguish between problems and exercises in this text Problems are scattered throughout the text and are to be attempted by every reader Their purpose is to clinch an idea of a previous paragraph or to prepare the way for what is to come They should be attempted immediately when they are met in reading Because they form an important part of the text, complete solutions are given in the rear of the book The reader should not be dis­couraged if he finds the problems very challenging; some are given more in the spirit of raising questions than in the expectation of getting correct answers

Exercises occur at the end of most sections In the nature of drill problems, they are arranged in what seems to the author to be ascending order of difficulty Some sections include starred exercises at the end which are more difficult and serve to extend the abilities of the more able students Exercises are arranged so that the same skills are required by the even-numbered problems (which have no answers given at the rear) and by the odd-numbered ones (for which the answers are given) Some sections and their corresponding exercise lists from the first edition have been broken into two parts so as to correspond more directly with the amount of material that might normally be covered in one lecture hour

This text is intended for a two- or three-semester course in introductory calculus The student is expected to have proficiency at high school algebra and trigonometry, though specific topics are reviewed in the text where experience suggests this to be helpful Chapter 1 includes the analytic geom­etry so many missed in the first edition It may be used with an emphasis suited to the preparation of the class The heart of a traditional one variable calculus course is contained in Chapters 2-8 and 14 with Chapters 9 and 15 being optional A third-semester course focusing on functions of several variables would include Chapters 10-13 with Chapters 9 and 15 again optional In that Chapter 15 has been rewritten to take advantage of the linear algebra terminology, it would be understood more readily by a student having the background of Chapter 10

Once again I wish to take the opportunity afforded by a Preface to thank the students and faculty at Macalester College who have now made substantial contributions to two editions of this text Special thanks are due to Carol Svoboda who helped with many of the mechanical details of the second edition, checked all the answers in the back of the book, and read the galley proof of the entire book Finally, it is a pleasure to acknowledge again the helpfulness and good spirit of the staff of Academic Press

—xii—

ACKNOWLEDGMENTS

The quotations appearing in this book are printed with the permission of the following publishers

Chapter 2, page 44: Phillip E B Jourdain, "The Nature of

Mathematics," The World of Mathematics (James R New­

man, ed.) New York: Simon and Schuster, 1956, p 39

Chapter 5, page 136: Giorgio de Santillana, The Crime of

Galileo New York: Time, Inc., Book Division, 1962, p 18

Reprinted courtesy University of Chicago Press

Chapter 12, page 408: J A Dieudonne, Foundations of Modern

Analysis New York: Academic Press, 1960

Chapter 15, page 508: Albert Einstein, Ideas and Opinions

New York: Crown Publishers, Inc., p 233

— x i i i —

1

SOME ANALYTIC GEOMETRY

mathematician is always happy to have a good problem A good problem is one that stimulates the imagination, relates to other problems that are topics

of lively investigation, and seems to be possible to solve by mental as opposed

to physical effort

We distinguish between problems and exercises in this text Problems are scattered throughout the text and are to be attempted by every reader Their purpose is to clinch an idea of a previous paragraph or to prepare the way for what is to come They should be attempted immediately when they are met in reading Because they form an important part of the text, complete solutions are given in the rear of the book The reader should not be discouraged if he finds the problems very challenging; some are given more in the spirit of raising questions than in the expectation of getting correct answers

1 ■ 1 What Is Analytic Geometry?

Points may be located in a plane with respect to two mutually perpendicular lines in the following way: Draw a vertical and a horizontal line, calling their

intersection the origin Mark off equal divisions on both of these lines It is common to call the vertical axis the y axis, the horizontal axis the x axis

Number the divisions on each axis, using positive numbers to the right and above the origin, negative numbers to the left and below the origin In this way it is possible to specify a certain point in the plane by means of an ordered

pair (x, y) of numbers It is usual to call the first number given (always the distance from the vertical axis) the abscissa The second number, representing the distance from the horizontal axis, is called the ordinate

In Fig 1.1, ( — 2, 3) locates the indicated point The abscissa is —2; the ordinate is 3 Note that (3.1416,7) and (π, 7) are different points even though

the distinction is one that we will have to make in our minds, since the scale chosen on the axes is usually too small to permit us to see any difference in the plotted points

This correspondence between ordered pairs of numbers and points in a plane affords us two ways to describe a certain set We may give a geometric description, or we may give an analytic description by specifying a relation

between the values of x and y For instance, we know from the theorem of

Pythagoras that the set described geometrically as a circle of radius 2

centered at the origin may also be described by the relationship x 2 + y 2 = 4

(Fig 1.2) Geometric pictures of several other sets, together with analytic descriptions of the sets, are given in Fig 1.3 and 1.4

FIG 1.2

The study that relates analytic descriptions of a set to the corresponding

geometric figure in the plane is called analytic geometry It addresses itself to

two principal questions:

(1) Given an equation or an inequality involving two variables x and y, can we give a description of the corresponding set in the

plane ?

(2) Given a geometric description of a set in the plane, can we find a corresponding equation or inequality?

In the spirit of our comments above, we illustrate these general questions with two problems

P R O B L E M S

A · Sketch the set of all points (;t, y) satisfying x 2 — y 2 — 0

B · Let S be the set of all points (x, y) that are equidistant from the y axis and the point (2, 0) Find an equation relating x to y for all (x9 y) in S

1 ■ 2 Vectors in the Plane

We wish to discuss a problem involving two tractors being used to pull a tree stump from the ground One tractor is larger, exerting twice as much force

as the other Both tractors are secured to the stump by cables, and the angle between the cables is 40° Our problem is to determine the actual force being exerted on the stump, and the direction of this force (Alternatively, we are to determine how to replace the two tractors with one tractor With what force must the new tractor pull, and in what direction?)

The situation is summarized in Fig 1.5 Note that the arrow representing

B

FIG 1.5

A

tractor B is twice as long as that of A, corresponding to the information that

one tractor exerts twice the force of the other The picture also suggests a geometric solution often taught to students in an elementary physics course The procedure is to draw a parallelogram using the given arrows as two ad­jacent sides Then draw, starting at the common initial point of the given arrows, the diagonal of the parallelogram (Fig 1.6) This diagonal, called the

FIG 1.6

A

resultant, indicates the actual force being exerted upon the stump, and the

direction in which this force acts

Apparently it was Newton who first noticed that the results obtained in constructing the resultant correspond to what actually happens Thus, the use of arrows in this and a variety of other problems is a valuable tool in applied mathematics These arrows, useful because they represent both di­

rection and magnitude (by their length), are called vectors Before proceeding

with a formal presentation of vectors, we pause to state concisely what we have learned about the use of vectors in applications

(1) Addition Given two vectors having the same initial point, their sum

vector (also called the resultant vector) is obtained by drawing from this common initial point the diagonal of the parallelogram having the two given vectors as adjacent sides

(2) Multiplication by a Real Number Given a vector v and a real

number r,

for r > 1, r\ is a vector longer than v;

for re(0, 1), rv is a vector shorter than v;

for r = 0, Ov = 0, the zero vector;

for r < 0, rv is a vector with the same length as | r | v but pointing in

the opposite direction (We say in this case that the vectors have opposite sense.)

(3) Equality of Vectors In our problem concerning the removal of a

tree stump, the effect on the stump should be the same whether the tractors pull or push In fact it should not matter if one pushes while the other pulls Such considerations make it useful in many applications to regard as equal any two vectors that:

(a) are parallel;

(b) have the same sense (point in the same direction);

(c) have the same length

Geometrically we think of two vectors v and w as equal if we can slide v, always keeping it parallel to its original position, so that it will coincide with

w The idea of being able to " slide vectors through parallel displacements " without changing them gives us alternative methods to represent v + w and

v — w Draw v anywhere Beginning at the terminal point of v, draw w Then the vector drawn from the inital point of v to the terminal point of w represents

v + w (Fig 1.7) If v and w are drawn from the same initial point, then the

FIG 1.7 FIG 1.8

vector drawn from the terminal point of w to the terminal point of v must be

v — w (Fig 1.8) (We see this clearly when we note that according to the addi­tion pictured in Fig 1.7, the vector we labeled v — w when added to w must give v.)

Guided by these considerations, we now set about representing vectors in

a plane with respect to a set of coordinate axes We begin by giving names to two special vectors

i is the horizontal vector of length 1 pointing in the direction of the

It is now clear that any vector v may be represented in the form v = a\ + b\

In particular, suppose the vector v has Ρ1 (χ ί , γ γ ) as its initial point, P 2 (x 2»yi)

as its terminal point Then the vector v, also designated by P1P2 (or Pl P 2

when boldface is unavailable), is (Fig 1.11)

and it is clear from what we know about congruent triangles that v = w (that

is, v and w are parallel, having the same sense and length) if and only if

Let v and w be two vectors in the plane We have seen that they may be written in the form (see Figure 1.12)

v = a\ + bl

w = c\ + dy

D E F I N I T I O N A

(Equality) v = w if and only if a — c, b = d

(Scalar Multiplication) rv = rax + rb\ (r real)

(Addition) v + w = (a + c)\ + (b + d)j

With i and j representing the vectors of unit length along the coordinate axes, we find the following definition helpful

D E F I N I T I O N B (Length)

The length of vector v = a\ + b\ is | v| = -Ja 2 + b 2

The length of the vector P ^ according to this definition corresponds to the

usual definition for the distance between the points Ρλ and P2 A unit vector

is a vector of length one The vector 0 having the property that 0 4- v = v for

all v is called the zero vector Note that Ov = 0

If an arbitrary nonzero vector v is multiplied by the scalar 1/1 v |, the result­ing vector will be a unit vector parallel to v Since two parallel nonzero vectors

v and w must be scalar multiples of the same unit vector, it is clear that v and w

will be parallel if and only if there is a scalar r such that v = rw Such vectors are said to be linearly dependent

Note that if v = ax + b], and w = c\ + d\ are linearly dependent, then a = re and b = rd If a and c are nonzero, then b/a = d/c This ratio is called the

slope of the vector Thus, vector v = a\ + b\ has slope bja We have seen that

vectors are parallel if and only if they have the same slope Vectors for which the slope is undefined (that is, where the coefficient of i is 0) are said to be vertical

P R O B L E M S

A · Find a unit vector pointing in the direction of v = 3i + 5j

B · Show that (1, - 9 ) , (3, - 4 ) , and (7, 6) are collinear

There is a third definition that will be most useful to us We shall motivate it

by an example, but we need to pause to recall from trigonometry the law of

cosines This is the formula that enables one to determine the angles of a triangle if the sides are given Specifically, with notation chosen as in Fig 1.13,

FIG 1.13

the law of cosines says

a = b + c — 2bc cos A

E X A M P L E A

Given the two vectors v = a\ -f b\ and w = c\ + d\, find the angle

between them (Fig 1.14)

C · A triangle has vertices A(\, f), Β(\ Ί -, f), and C(f, - f ) Find LABC

D · Prove that for any vector v = a\ + &j, v · v = | v|2

We see immediately (2) that two nonzero vectors are perpendicular if and

only if their dot product is zero (so the cosine of their included angle is zero)

Notethati· j = 0

Given a vector to represent graphically, we may place its initial point any­

where Since we often wish to specify that the initial point is to be taken at the

origin, such a vector is given the name radius vector Thus, the radius vector

a\ + frj has (a, b) as its terminal point This fact, together with the other

properties of vectors, can be used to solve certain kinds of problems

EXAMPLE B

Find the point of intersection of the medians of the triangle having

as its vertices the points A(-l, 4), 5(3, - 3 ) , and C(4, 2)

A theorem from plane geometry informs us that the medians intersect in a

common point two-thirds of the way along any median from the corresponding

vertex We begin by finding the coordinates of M From Fig 1.15, since M

is the midpoint between A and C,

OM = OA + iAC

OA is a radius vector, so OA = — i + 4j, AC = 5i — 2j (Caution: A common

mistake is to get the negative result here Remember that you subtract the

coordinates of A from those of C.)

OM = - i + 4j + i(5i - 2j) = fi + 3j

FIG 1.15

Since OM is a radius vector, the coordinates of M are (|, 3) This looks

reasonable from our figure, so we proceed (If we make the common mistake mentioned above, our answer will not even look reasonable This is a good place to check on ourselves.)

We now use the theorem from plane geometry and the same ideas already employed to write

OK = OB + |BM,

OK = 3i - 3j + f ( - f i + 6j) = 2i + j

Therefore K is at (2, 1) |

The reader should verify his understanding of the preceding example by

finding K from the relation

OK = ON + iNA

We may use our knowledge of vectors to prove some of the theorems of elementary geometry

EXAMPLE C

Prove that the diagonals of a rhombus are perpendicular

We take the base of the rhombus to be the radius vector a\ Then another of the sides must be a radius vector, say b\ + c\ (Fig 1.16) Let C designate the

vertex opposite the origin

Exercises 1-14 all refer to the points A(-3, 4), £(3, 2), C(—2, 2)

1 · Write vector AC in the form a\ + b]

2 · Write vector AB in the form a\ + b j

3 · Find a unit vector pointing in the direction of CA

4 · Find a unit vector pointing in the direction of BA

5 · Find a point D so that ABCD are the consecutive vertices of a parallelogram

6 · Find a point D so that AC and BD have the same sense and are opposite

sides of a parallelogram

7 · Find the coordinates of a point two-thirds of the way from A to C

8 · Find the coordinates of a point three-quarters of the way from B to C

9 · The vector AC is placed so that its initial point is at (1, 2) Where is the terminal point?

10· The vector BC is placed so that its initial point is at (—3, 1) Where is the terminal point?

11 · Find the acute angle of AABC at A

12 · Find the acute angle of AABC at B

13 · Find a unit vector that is perpendicular to AC

14· Find a unit vector that is perpendicular to AB

15 · Show that (2, 2), (4, 1), and (3, 4) are vertices of a right triangle

16· Show that (2, 2), (0, 3), and (—1, —4) are vertices of a right triangle

Use vectors to prove the following theorems of plane geometry

17· The midpoint of the hypotenuse of a right triangle is equidistant from the

vertices

18 · The sum of the lengths of the bases of a trapezoid is twice the distance between

the midpoints of the nonparallel sides

19· The line segments joining the midpoints of the adjacent sides of an arbitrary

quadrilateral form a parallelogram

20 · Any triangle inscribed in a semicircle must be a right triangle

21 · If the diagonals of a parallelogram are perpendicular, then the parallelogram

must be a rhombus

22 · An isosceles trapezoid has equal diagonals

23 · The perpendicular bisectors of the sides of a triangle are concurrent

24 · In any triangle the sum of the squares of the medians is equal to three-fourths

the sum of the squares of the three sides

1 ■ 3 The Straight Line

Two points P^Xi, y^) and P 2 (x 2 > ϊι) determine a straight line The slope

m of such a line is defined to be the slope of the vector P i P2 ;

x 2 X\

Note that the slope is undefined if and only if the line is vertical An analytic

description of the line through P 1 and P 2 is easily determined Choose any

point P(x, y) on the line (Fig 1.17) Vector P: P, being linearly dependent upon

vector Ρ Χ Ρ 2 , has the same slope Hence,

Using this form, we can immediately write the equation of a line if we know

its slope and a point through which the line passes

The point at which a straight line cuts the y axis, called the y intercept,

is usually designated by (0, b) If in (1) we set (x l9 yj = (0, b), then (1) may

be written in the form

y = mx + b

of the line in the form y = mx + b You now have the equation of the same

line written in two forms Verify that they are equivalent

Any equation of the form Ax + By + C = 0 can be put into the form

y = mx + b unless B = 0 (in which case x = — C/A, and the graph is a vertical

line) Since this tells us the slope m and the place b where the line cuts the y axis, it is easy to graph any set of points (x, y) satisfying Ax + By + C = 0

EXAMPLE A

Plot those points (x, y) that satisfy 3x + 5y = 12

It is a simple mental exercise to write (or even to think without writing)

y = -%χ + -ψ

This line intersects the y axis at \2- The slope is — f Count 5 units (of any convenient length) from the right of (0, - ^ ) Then go down 3 units (of the same length, of course) Alternately, go to the left 5 units and up 3 units Sketch in the line (see Fig 1.18) |

By using the method of the preceding example, we immediately find the y

intercept A closely related method for graphing a straight line is to rewrite the equation in the form

a b

so the x intercept is 4 and the y intercept is (still) -^2-

We may summarize what we have learned about the straight line as follows

Standard Form

Ax + By + C = 0 is the equation of a straight line

Slope-Intercept Form

y = mx 4- b is the equation of a straight line in a form where we may

recognize the slope m — sjr and the y intercept b (Fig 1.19a)

Point-Slope Form

y — y 0 = m(x — x0 ) is the equation of the straight line that passes through

(*o, Jo) wit n a slope of m = s/r (Fig 1.19b)

. I

;

Intercept Form

(x/a) + (y/b) = 1 is the equation of the straight line that cuts the x axis at

x = a, the y axis at y = b (Fig 1.19a)

Given a nonvertical line L with the standard equation Ax + By + C = 0,

it is easily seen that its slope is m = —A/B Thus, it is parallel to the vector

v = B\ — A] having the same slope

A second line Αγ χ + B x y + Q = 0 is similarly parallel to vector BJ — Α χ \

Then the two lines are parallel if and only if the two vectors are linearly dependent; that is, the lines are parallel if and only if their respective co­

efficients of x and y are proportional

Returning to our line L with slope m, let us consider a line perpendicular to

L meeting it at point P 1 (x u y^) Note that vector v = i + m\ is parallel to L

(Fig 1.20) Choosing a point P on the perpendicular line, we see that

P U, y)

FIG 1.20

PXP = (x — ;q)i + (y — y^)} and v must satisfy v · P ^ = 0 Hence, (x — xx )

+ m(y — y^) — 0, and the slope of vector PXP is given by

y - yi = i

In words, this says the slope of the perpendicular line is the negative reciprocal

of the slope of line L

EXAMPLE B

Find the equation of the perpendicular bisector of the line segment

f r o m / ^ - 3 , - 2 ) t o P2( l , - 4 )

The line passes through the midpoint M(— 1, —3) of the given segment

(Fig 1.21) Since P ^ = 4i — 2j has a slope of —2/4, the slope of the desired line is 2 Using the point-slope form of the straight line,

It is instructive to note that we could also solve this problem directly

with vector methods Choose P(x, y) on the desired line Vector P i P ^

The lines are respectively parallel to the vectors v = i — 2j and w = 3i — j

The angle Θ between the lines is determined from

v w 3 + 2 1

cos Θ = = —=—— = —=

M|w| J5y/l0 yjl Thus, θ = π/4 I

We wish now to find the distance from a point to a line We begin with the special case in which the point is the origin Since distance is measured along

a perpendicular, we draw a perpendicular from the origin meeting the given

line at Ρχ(χΐ9 J>i) a nd making an angle of a with the positive x axis (Fig 1.22)

u = (cos a)i + (sin a)j, are perpendicular, so their dot product is zero:

(x — xjcos a + (y — yx )un a = 0,

(cos oc)x + (sin a)>> = xt cos a + y1 sin a

Using the expressions (2) for xx andy 1 together with the trigonometric identity

cos2 a + sin2 a = 1,

(cos a)x + (sin ct)y = p

This is called the normal form of a straight line

JV*!/**

P (x, y)

To put Ax + By + C = 0 into normal form, we need to find a unit vector

u that is perpendicular to the line Since vector v = Ai + B\ is perpendicular

to the line, it is clear that u may be chosen as either

Ambiguity is removed by agreeing to choose the sign so as to make the coef­

ficient of y (the sine of a) positive This limits a e (0, π) for nonvertical lines, and gives p > 0 for lines above the origin, p < 0 for lines below the origin

It is now easy, given a line, to determine its distance from the origin But we began with the more ambitious goal of finding the distance between any point and any line This is now possible, however, as an easy consequence of our work

EXAMPLE D

Find the distance from the point ( — 3, 1) to the line 3x — 4y

- 10 = 0

In this case, we divide the equation of the given line by — -JA 2 + B 2 =

— yJ9 + 16= —5 to make the coefficient oiy positive This gives — \x + fy =

— 2 The graph is indicated in Fig 1.23

Now let us pass a line through (—3,1) that is parallel to the given line It has

an equation of the form 3x — 4y = Z>, and we easily see that D = —13 by

noting that ( — 3, 1) is to be on the line Normal form for this line is — fx + y j = -^ We now see that the distance from the given point to the given line

i s Y - ( - 2 ) = ^ |

The method of the previous example may be used to obtain a general

formula for the distance d from a point (xu y x ) to a line Ax + By + C = 0 The

line parallel to the given line and through (xu y x ) is Ax + By = Ax x + Byt

Putting both equations into normal form gives

Reflection (or perhaps just faith in the printed word) will convince the reader

that d is positive if and only if the given point lies above the given line

6· 2x+3y = S

7 2x-4y = 5

8 3x- 6y = 10

9 · Find a vector v parallel to and a vector w perpendicular to the line of Exercise 1

10 · Find a vector v parallel to and a vector w perpendicular to the line of Exercise 2

11 · Find the acute angle formed by the intersection of the lines of Exercises 1

and 3

12 · Find the acute angle formed by the intersection of the lines of Exercises 2 and 4

In Exercises 13-30, write the equation of the line

Perpendicular to 3x + y — 4, 3 units above the origin

Parallel to 2x = 5y, tangent to a circle of radius 4 centered at the origin

The altitude through A of the triangle with vertices A(5, 3), £(—1, 1), and

C(3, -2)

The altitude through B of the triangle of Exercise 21

Bisects the acute angle formed by the lines of Exercises 1 and 3

Bisects the acute angle formed by the lines of Exercises 2 and 4

Through (5, 0), tangent to a circle of radius Vl centered at (1, 2)

Through (7, 0), tangent to a circle of radius 5 centered at (0, 1)

Through (2, 3), having an x intercept twice as large as the y intercept

Through (3, 2), forming with the positive axes a triangle of area 16

Through (6, —5), the segment intercepted by the axes having length V34

Through (—3, 10), the product of the intercepts on the axes being 15

1 ■ 4 The Conic Sections

We said that one of the principal questions of analytic geometry is, " Given

an equation, can we give a geometric description of the corresponding set in

the plane?" In the last section we answered this question for any equation of

the form Ax + By + C = 0 In this section we shall answer it for the equation

For reasons we give at the end of this section, the corresponding curves in the plane are called conic sections

We have already studied the special case of (1) in which A = B = C = 0;

the desired graph is a straight line Another special case of (1) is obtained by

deriving the equation of a circle of radius r centered at the origin Directly from the theorem of Pythagoras (Fig 1.24) we have x 2 + y 2 = r 2 This is in

P U, y)

FIG 1.24

the form of (1), where A = C = 1, F = - r2, and B = D = E = 0 Moreover,

it is clear that whenever A = C > 0 and B = D = E = 0, then (1) may be

written

Ax 2 + Ay 2 = -F

If F > 0, there is no graph; and if F < 0, the graph is a circle of radius r where

r 2 = —F/A When F= 0, the graph reduces to a point which we view in this

context as a degenerate circle

One final special case is instructive Consider a circle of radius r centered

at (h, k) (Fig 1.25) Impose a second coordinate system on the plane, denoting

Ix.y) (u,v)

FIG 1.25

the line y — k as the u axis and the line x = h as the v axis Then any point P

in the plane has two "addresses," (Λ:, y) and («, v) related by

x = u + h, y = v + k

Now we know that the equation of the circle with respect to the (w, v) coordi­

nates is u 2 + v 2 = r 2 Hence, by substitution, the equation with respect to the

(x, y) coordinates is

And, of course, any equation that can be written in this form describes a circle

of radius r centered at (/?, k) This means we are able to describe completely

the locus of any equation (1) in which A = C > 0 and B = 0

EXAMPLE A

Graph 4x 2 - \2x + Ay 2 + Ay + 6 = 0

Since the coefficients of x 2 and y 2 are equal, we know that the graph must be a

circle (perhaps degenerate or nonexistent in the case of a negative radius) To

write this in the form (2), we use the technique of completing the square (To do

this, manipulate as in the first step below to get a positive 1 as the coefficient of

the squared term; then add the square of half the coefficient of the first degree

9 · Passes through (4, 0) and (8, 0) and is tangent to the y axis

10 · Passes through (4, 7) and (—3, 0) and has its center on the line x = 1

11 · Has its center on the line y = — \x + 1 and is tangent to both axes

12 · Has its center on the line y = — x + 4 and is tangent to both axes

13 Passes through (6, 10), (0, 2), and (2, 8)

14· Passes through (0, 6), (2, 2), and (3, 5)

15 · Touching the line x — 2y = 3 at (— 1, —2), and having radius V5

16 · Touching the y axis and passing through the points (4, — 1) and (—3, —2)

A Locus Problem

A locus is a collection of all those points and only those points which satisfy

a given condition Thus, a circle is the locus of all points in a plane that are some prescribed distance from a fixed point called the center

Consider now a fixed vertical line L and a fixed point Fnot on the line We

shall be interested in the following locus problem (see Fig 1.27):

/ FIG 1.27 /

Find all points P in the plane of the line L and the point F for which the ratio of the distance from Fto P and from P to L remains

constant In symbols,

— = e (e some constant)

The fixed line L is called the directrix The fixed point F is called the focus, and the constant e is called the eccentricity The line perpendicular to the directrix passing through the focus is called the major axis, though this term is also

applied to just a segment of this line at times

The solution to the problem is most conveniently obtained if we consider

three separate cases: e < 1, e > 1, and e = 1 In each case it is easy to find

several points that must be on the locus The general idea is illustrated by the

three choices of e indicated in Fig 1.28 In each case, the points V-t are on the

Case I ; e < 1

As we did in Fig 1.28a, we may always locate two points V1 and V2 on the major axis of the locus Denote the midpoint between Vx and V2 by 0, and label the distances from 0 to Vl9 V 2, F, and the line L as indicated in Fig 1.29 Since Vx and V2 are members of the locus,

Thus, if we impose a coordinate system on Fig 1.29, we see that the coor­

dinates of Fare ( — ae, 0), and L is described by x = — a/e (Fig 1.30)

It is obvious from the equation that the desired locus of points is symmetric

with respect to the y axis (since replacing x by —x does not alter the equation), and is similarly symmetric with respect to the x axis The locus clearly passes

through (±a, 0) and (0, ±b) Plotting a few points reveals the shape of the

locus(Fig 1.31)

y

This curve is called an ellipse It is obvious from the symmetry of the curve

that we would have obtained the same locus if we had started with a focus

at (ae, 0) and a directrix x = a/e For this reason, it is customary to speak of

the foci and directrices of an ellipse

Related to the two foci of an ellipse is another important geometrical fact

Let P be a point on the ellipse (Fig 1.32) Using Fx and L1? we have, by

Adding, F X P + F 2 P = 2a Thus, the ellipse may be characterized as the locus

of all points P for which the sum of the distances to two fixed points (the foci)

is a constant

Note that the major axis of the ellipse (the line segment from ( — a, 0) to

(a, 0)) is longer than the line (called the minor axis) from (0, b) to (0, —b) This follows from the definition of b, since b 2 = a 2 (l — e 2 )< a 2 If we had started

out with a horizontal directrix, then of course the major axis would have been

along the y axis In a particular example we determine whether the directrix

is vertical or horizontal by noting the direction of the major (longer) axis

EXAMPLE B

x 2 y 2

Graph — + — = 1 Locate the foci and directrices

4 9

The graph is most easily sketched (Fig 1.33) Since 4 = 9(1 — e2 ), and since

e > 0, e = >/5/3 The distance of the foci from the center is ae = 3^/5/3 = ^ 5

and the distance from the center to the directrices is

y = 9/ {5

FIG 1.34

y = - 9 / V5

The principal facts about ellipses may now be summarized:

(1) The major axis, length 2a, is the longest axis of the ellipse; (2) The length of the minor axis, 2b, is determined from

b 2 =a 2 (\ -e 2 )',

(3) The distance from the center to the directrices is a/e;

(4) The distance from the center to the foci is ae;

(5) Two geometrical descriptions of an ellipse may be given:

i The locus of all points P for which EP/LP = e < 1

ii The locus of all points P so that F t P + F 2 P is constant

(and in fact equal to the length of the major axis) Something more has now been learned about the curves described by (1)

We know what the curve looks like if B = D = E = 0, while A and C are both

positive Moreover, if we profit from what we learned about the circle, it is

clear that we may drop the restriction on D and E since if they are nonzero

we can complete the square in x and y, obtaining either of the standard forms

indicated in Fig 1.35

Graph x 2 + 6x + 4y 2 - iy + 9 = 0 Locate the foci and directrices

We first perform the algebra necessary to put the given equation in standard form:

ae = 2^/3/2 = ^/3,

- - — =± /3 and the graph is as indicated in Fig 1.36 |

13 · Directrices y = — 2 and y = 6; vertices (—2, — 1) and (—2, 5)

14· Foci (1, 3) and (5, 3); vertices (0, 3) and (6, 3)

Case II; e > 1

This case parallels in many ways the one just considered In particular, notation may be chosen to minimize the number of new formulas to be memorized

As we did in Fig 1.28b, we may always locate two points V1 and V2 of the locus on the major axis Denote the midpoint between Vx and V2 by 0, and label the distances from 0 to Vl9 V 2 , F, and the line L as indicated in Fig 1.37

Thus, if we impose a coordinate system on Fig 1.37, we see that the coor­

dinates of F are (ae, 0) and L is described by x = a/e For an arbitrary P (Fig

Guided by the results obtained for an ellipse, we are tempted to set the

denominator of the second term equal to b 2 But 1 — e 2 < 0 here, so we must

set — b 2 = a 2 (\ — e 2 ) The resulting equation is

r = 1

It is obvious that the desired locus is again symmetric with respect to both

the x andy axes The locus clearly passes through (±a, 0), the vertices It does not pass through (0, ±b) or any other point for which the x coordinate

is between — a and a Plotting a few points reveals the shape of the locus

(Fig 1.39)

FIG 1.39

σ| F(ae,0)

This curve is called a hyperbola It is obvious from the symmetry of the

curve that we would have obtained the same curve if we had started with a

focus at ( — ae, 0) and a directrix x = — a/e We again, therefore, speak of the

foci and directrices

A remark about the dotted lines in Fig 1.39 is in order They are best

understood in light of the following question Among all lines y = mx through

the origin, which ones intersect the hyperbola? The answer is obtained by solving simultaneously the equations

a 2 b 2 and y = mx

Substitution gives

= 1,