A First Course in Calculus, 5e [Fifth ed.] 9781441985323

Citation preview

T)

n)

I)

P

1)))

in

M)

ml)

Undergraduate

Texts

in Mathematics) Editors)

S. Axler

F. W. Gehring

K. A.

Springer

New

York

Berlin

Heidelberg

Hong Kong London

Milan

Pa ris Tokyo)))

Ribet)

OF RELATED

BOOKS

INTEREST BY SERGE LANG)

Calculus

Short

2002, ISBN

0-387-95327-2)

Calculusof Several Variables, 1987, ISBN0-387-96405-3)

Second Edition

Analysis,

Undergraduate

Third Edition

1996, ISBN0-387-94841-4) to Linear

Introduction

Algebra

1997, ISBN0-387-96205-0) for Undergraduates

Talks

Math

1999, ISBN0-387-98749-5)

BY LANG

BOOKS

OTHER

VERLAG)

SPRINGER-

Math! Encounters \302\267

in Calculus

Variables

\302\267 Linear

\302\267 Basic

to

Linear

\302\267

Undergraduate

\302\267 Math

Talks

Functional Analysis \302\267 Introduction Differential Geometry \302\267 Algebraic

\302\267 The

Beauty of \302\267 Mathematics Short

Students

School

\302\267 Introduction

Algebra

Analysis

Complex

High

School Course

A High

Geometry:

Course

with

PUBLISHED BY

Algebra

for Undergraduates

Algebra

Manifolds

Number

\302\267

Theory

Cyclotomic

\302\267 SL

\302\267

\302\267

Algebra

Undergraduate \302\267

Several

of

\302\267

Analysis

\302\267 A First

Calculus

\302\267 Calculus

to Differentiable

to Diophantine

Doing Mathematics

\302\267 Real

\302\267 Fundamentals

Fields

I and

Inversion

and of

II

\302\267

on

2 (R) Spherical Approximations \302\267 \302\267 Curves: Functions Elliptic Diophantine SLn(R) (with Elliptic Jay Jorgenson) \302\267 \302\267 Riemann-Roch Arakelov to Introduction Algebra (with Theory Analysis \302\267 \302\267 and Abelian to Algebraic Varieties Introduction William Fulton) Abelian \302\267 \302\267 Modular Introduction to Modular Forms \302\267 Functions Complex Multiplication \302\267 \302\267 to Introduction Daniel Units Complex Hyperbolic Spaces Kubert) (with Introduction

Number

Theory

III

\302\267

Survey

on Diophantine

Geometry)

Introduction to Transcendental the following: Hale in GL2-Extensions Distributions Frobenius I, (with in Basic volume of in Trotter in volume IV, II, Topics Groups Cohomology V (with Jay Jorgenson) in volume Analysis of Regularized Series and Products and Explicit Formulas for Regularized Products and Series (with Jay Jorgenson) in

Collected Papers I-V, in

Numbers

volume

THE

including

volume

V)

FILE

\302\267 CHALLENGES)))

Serge

A

in

Fifth

With

Lang)

First

Calculus)

Edition)

367 Illustrations)

Springer)))

Course

SergeLang of Mathematics

Department

Yale University New

CT 06520

Haven,

USA) Board

Editorial

S. Axler

F. W.

Mathematics

Mathematics

Department

State

Francisco

San

U ni

versi

ty

San Francisco,

East

CA 94132

Arbor,

Department

University

of

Mathematics

Subjects

of Congress

Library

Department of California,

Berkeley

Michigan

MI 48109

Berkeley, CA 94720-3840 USA)

USA)

USA)

Ribet

Mathematics

Hall

University Ann

K.A.

Gehring

Classifications (2000): 26-01,26A06)

Data

Cataloging-in-Publication

Lang, Serge, 1927A first course in calculus.

(Undergraduate texts in index.

mathematics)

Includes

1. Calculus. II. Series. I. Title. 85-17181) QA303.L26 1986 515 on acid-free

Printed

editions

Previous

Publishing

paper.)

of this

Company,

book were

published

in 1978,

1973, 1968,1964by

Addison-Wesley,

Inc.)

1986 Springer-Verlag New York Inc. or copied in whole or in part without the All rights reserved. This work may not be translated written of the publisher (Springer-Verlag,175 Fifth Avenue, New York, New York permission with reviews or scholarly Use 10010,USA), except for brief excerpts in connection analysis. electronic adaptation, comin connection with any form of information storage and retrieval, or by similar or dissimilar methodologynow known or hereafter developed is software, puter

o.)))

an l

integer

xm/n)

> 1,

then)

/\".)

elementary rule. Let We define

a function)

defined

x.

= x3

Show that the

For example, let Thus we obtain

n times.

itself

with

n =

If

the

called

equal

=

all

for

neither?)

integer > 1 and

n be an

product

If

f(x)

if f(x)

function)

= f(x)

we just summarize

this section Let

define

-x)

\037])

POWERS)

\0374.

a2 =

the

= -f(

odd or even. (c)

numbers.

all

for

about

What

h(x)

is

functions are

following

function

even

an

if f(x)

function

odd

an

[I,

2

be any function

f(x) + f(

be

be

is said to

numbers)

It is said to

x.

all

for

FUNCTIONS)

AND

NUMBERS

18)

ami\"

to

be

> 1 integers is which also (al/\m")

m,

n

be

fractional powers,and

gives

us

[I,

now come

We

x

define

D

a is

when

fundamental

the

to powers with negative a negative rational number

This means

true.

be

or O. We

numbers

x >

0 and

or

o.

to

want

want

We

rule) D+ b

X

to

19)

POWERS)

\037])

b)

XDX

define

must

we

that

=

X

O

For instance,

1.

be

to

SInce)

3 =

2

if

2

this example 1. Similarly, in

from

see

we

is

0 =

general,

xD is

true,

then

X

O

m us

Supposefinally number >

=

x

D+ O

=

xDX

this

holds

equation

O)

eq ual to 1. a positive rational

t be

a is

that

let

and

number,

x be

a

define)

We

O.

only way in which if the relation)

the

that

2 3 2 0 ,)

23+0 =

X

1

-D

=-.

x

D)

Thus)

2

We

that

observe

_ 3

=

1

2

1 =

3

in this

=

\037

4 2/ 3

.)

special case,)

(4 In

2/3

4-

and) 8')

-

2/3

2/3)(4

) =

40 =

1.)

general,)

XDX-

D

=

X

O

=

1.)

to define x D

a is not a rational number. even when This is more subtle. For instance, it is absolutely meaningless to say that 2 J2 is the product of 2 square root of 2 times itself. The problem of We

are

2

defining

chapter. assume above

tempted

D

(or

Until that

for

D

a is not

that

when

x ) when

there

rational

chapter, is a function,

numbers,

X

and D+ b

=

rational we deal written

b ,)

be

to a postponed a power, we

such

with

D

x , described the

satisfying

XDX

will

X

O

=

fundamental

1.)))

as we

later shall

have done

relation)

AND FUNCTIONS)

NUMBERS

20)

for all x > O. It is f(x) = xfi defined its values for special numbers, like 2fi. It was a very long time whether 2fi is a rational number or not. is was found in 1927 by the mathematician (it not) only

Example.

to

for

unknown solution

The

Gelfond, who

a function

have

We

hard

actually

[I, 94])

describe

a problem

for solving

famous

became

to

that

was

known

a

function

like

be very hard.) Warning. Do not confuse Given a number e > 0, we can will be discussed in detail in

exponential function.

X

and

e= the

and

than

I,

\0374.

Find

aX

1. a 3.

all x. (It is called an

for

defined

This

function

functions.

exponential

x 2 .

We

eX

as

properties The meaning in Chapter VIII.) special

having

function.

exponential be

will

2.718.. .)

explained

!

the

for

following

values of x

and

x = 4

4. a

9. a

-1

3

and

x =

and

x =

and x

If n is an

odd

4

which

make

of our use of

= -4) integer

numbers?)))

1, 3,

5

x == -

and

x =: 2)

a =

3

and

x == 2

8. a =

-

2

10. a like

a.)

! and

6.

-1

and

a =

2.

-

all

a

x = 3)

7. a =

for

and

function

VIII.) lOX are

and

-!

11.

any

x

and

5. a =

=

a

2

EXERCISES)

= 2

a =

other

\"better\"

word

the

function

exponential

it better

eX as

view Chapter

x

like

function

ber)

anum

select

shall

2

Thus

a

=

=

5, 7,..., can

and

x =

-t

and x =

you

definc\037

an

1)

- 2 9)

n-th

root function

CHAPTER

II)

ments backwards

and

chapter allow us

It is extremely to

intuition

basic for help

what

us solve

and conversely, we can functions to yield results about tions,

II,

\0371.

Once

line.

the

between

forwards

to

certain

translate

of numbers

language

state-

and

the

of geometry.

language

ric

in this

contained

ideas

The

Curves)

and

Graphs

follows,

we can

because

problems concerning

use

theorems

use our geometand

numbers

func-

and

numbers

concerning

geometry.)

COORDINATES)

is selected, we can represent numbers as points a unit length to the plane, and to We shall now extend this procedure

on

pairs

a of

numbers.

We

visualize

a horizontal line

and a

vertical

line

intersecting

origin o.)

o)

These

lines

will be

called coordinate axes or

simply

axes.)))

at

an

GRAPHS

22)

We

lengths

a unit

select

CURVES)

AND

[II,

length and cut the horizontal and the left and to the right,

indicated in visualize the On the vertical line we points

vertical line,

but

spondingto of

the

the same below

the

horizontal

up

line as

corresponding to as

going

figure. below 0

points on We

integers.

negative

as corre-

a thermometer\" where the

in grading

used

are regarded

next

the

See

negative.

of

segments

same to the

the

do

as we visualized

just

integers,

negative

idea as that

zero

down, as

and

into

line

1, 2, 3,... to

\0371])

left

the

follow

numbers

figure.)

3 2)

-2

-3)

We can now cut the

plane

into

have

sides

whose

squares

(3,

length

1.)

4))

4

3

(1, 2)

2

1 4

-. -2 -1 T

I

(-3,

01

I

3

2

4

1

I

-2)

2 3 4)

We

can

describe

each point where

integers. Supposethat we are to the right of the origin 1 unit has been indicated (1, 2) which (3,

4). The diagram is just like Furthermore,

we

vertically There

could

also

(- 3, - 2) we point 2 units. downwards no reason is actually

describethe

lines

two

intersect

given

a pair

of integ{\037rs

and

vertically

up 2

above.

use negative go

why

we

units

the

to

get

the

point

numbers. For instance, to

left of the origin

should

a pair of We go the point

(1, 2).

We have also indicated

a map. to

by

like

limit

3

ourselves

units

to

and

points)))

[II,

are described

which (t,

23)

COORDINATES)

\0371])

-1) and

the

on

3) as

(-)2,

point

instance

For

integers.

by

the

also have

we can

the

point

below.

figure

(- V2, 3)

.------

3)

1/2)

not drawn

have

We the

only

In general,

dicular lines numbers

two

lines

relevant if

take

we

x,

as

y

any

horizontal

the

to

all the squares on the to find our two points.

in

We

plane.

have

drawn

point P in the plane and draw the perpenaxis and to the vertical axis, we obtain

the figure

below.)

p

y)

x)

The

line

perpendicular

number x

is

which

negative

The number

origin.

is

axis

vertical

numbers x, P = (x,y).

yare

y

to

because the

by it

coordinates

because

the lies

axis

horizontal

the

figure

determined

positive called

from P in the

it lies

perpendicular above

the

of the point

a

determines

to the

left

of

the

from P to the origin. The two

P, and

we

can

write

We find of the plane. (x, y) determines a point the 0 in the horizontal a distance x from origin direction and then a distance y in the vertical direction. If x is positive If x is negative, we go to the left of o. If y is we go to the right of o. we go vertically we go vertically upwards, and if y is negative positive call the downwards. The coordinates of the origin are (0, 0). We usually axis the y-axis. If a point P is horizontal axis the x-axis and the vertical to call the first describedby two numbers, say (5, -10), it is customary its x-coordinate and the second number its y-coordinate. Thus number Of 5 and -10 the y-coordinate of our is the x-coordinate, point. and x instance t other besides and for we could use letters s, course, y, Every

the

or

point

u

and

pair

by

v.)))

of numbers

going

AND CURVES)

GRAPHS

24)

is a point in is a y) point in the

If (x, y) (x,

II,

the plane into four quadrants

two axes separate as indicated in the

Our

bered

\0371.

II

I)

III

IV)

first

the

fourth

quadrant,

2. Plot

the

(!,3),

points:

following

3. Let (x, y)

Is

of a

the coordinates

be

Is

negative?

(-:1,

!).

second quadrant.

the

following

points:

(1.2,

6. Plot

the

following

points:

(- 2.5,1),(-

points: (1.5, -1),

3.5,

positive

or

positive

\037:).

(-1.5, -1).)

GRAPHS)

f be

function

Is x

third quadrant.

a function. We define the graph of f to be the coordinate is first of numbers (x,f(x)) whose pairs coordinate is second which is defined and whose f Let

Is x

- 2.3),(1.7,3).

the

\0372.

2), (1,0).

(\037, -2),

the

in

point

5. Plot

the following

in

-

or negative?)

positive

y

point

5,

or negative?

positive

y

(-1, -!),

of a

the coordinates

be

negative?

4. Let (x, y)

II,

If

o.)

EXERCISES)

(-1, 1), (0, 5), (-

Plot

yare > O.

then both x and then x > 0 but Y
0,

looked

a

\0372])

this:)

preceding two functions are

case of

that

this

looks

f(x) = Ixl. When

know

we

[II,

the

ones

instance, we x. In other

x. It is a very simple horizontal line, intersect-

=2)

[II,

If we

took the

line

zontal In

let

f(x) = c is the (0, c). The function

it looks

like

c be

line = c

f(x)

-1,

the

then

The

is called

a

a

few

the

of the

points

be a

would

hori-

1).

of

any

axis

at

function the

point

function.)

constant

function

= l/x

f(x)

graph, you

will

see

(dethat

this:)

(1,

1) (2,

For

graph

the vertical

intersecting

of our examplesis

By plotting

graph

axis at the point (0, -

a fixed number.

horizontal

x =I 0).

=

f(x)

the vertical

The last

5.

Example for

function

intersecting

general,

fined

27)

GRAPHS)

\0372])

you

instance,

can plot the x

1

2 3

1

\"2

1 \"3

becomes

1/2))

points:)

following

l/x

x

l/x

1

-1

-1

21

-2

1

3

-3

2

-2

3

1 1 3)

-211 -3

-2 -3

positive, l/x becomesvery

small. As x apbecomes very large. A similar phenomenon occurs when x approaches 0 from the left; then x is negative and l/x is negative. Hencein that case, l/x is very large negative. In trying to determine how the graph of a function can looks, you alread y watch for the following: As

x

proaches 0

The points What

very large

from

at

happens

the

which

when

right,

l/x

graph intersects x becomes very

the

the two coordinate axes. large positive and very large

negative.

On the nique

the

is

graph

whole, however, in working to plot a lot of points

just

looks

like.)))

out until

the exercises, your main becomes clear to you

it

techwhat

AND CURVES)

GRAPHS

28)

II,

EXERCISES)

\0372.

functions the graphs of the following In each case we give the value

Sketch

each graph.

4. 4x

x

7.

of

and plot at the function

5. 2x

+ 3

8.

three

least

at

points

on

x.)

3. 3x

2. 2x

+ 1

1. x

[II, 92])

1

+

- 3x + 2

6. 5x

+

9. 2.x 2

-

! 1

\"2

10. -

2

3x

11. x

1

+

12. x 4

3

13. fi

14.x- 1/2

15.

16. x + 3

17. Ixl

18. Ix I

19.

20. -Ixl

-Ixl 1

22.

2

26.

x-2

13, 14,

Exercises

x-3

27.

2 -

30.

--

2

x+2

x

x

x+ 1

21

and

1

Ix I)

30, the

through

not

are

functions

defined

for

all

of x.)

values

31. Sketch f(x)

f(x)

the

= 0

32. Sketch 33.

24. --

3 29.

x+5

(In

1

x+2

x+3

-2

28.

+ 2x

21.--

+x

1

23.

x-2

25.

+ x

2x + 1

graph

the

= x

Sketch

of the function = 1 f(x)

f(x)

such

if

x >

of the function = 2. f(O)

f(x)

such

x < O.

if

graph

the graph of the x < o. if

2 f(x) = x

34. Sketch f

(x) =

f(x)= 35. Sketch f(x)

the

the

= x3

f(x) = x2 36. Sketch f(x)

the

[We

leave

such

a way

if

x >

graph 2

such that: x > o.

function f(x) such - 1 < x < 1. > 1. [f(x) is not defined

that:

values of x.]

for other

such

that:

< 2.

0 < x

2.

of the function

if

0
0,

number

any

They constitute

is 2.

origin

then

the

the

of

graph

the equation) x2 is the circle

We have

of radius c, with

x

or

write

2

+

- 1=

y2

our

0 is

equation

in

at the origin. the equation)

that

x2

+

the

of

not

y2 =

any

y = If

x

x

1 and

=1=

=1=

J

two

-

+ 1,

J 1-

-

y =

or)

the

and

function,

graph

is another

defined for

defined of

Neither

We now whose from

radius (1,

2)

can

2

for

and

y

get)

.)

each

of x.

value

indicated on

of this function function)

for

-1 < x

these

functions

=

upper half of

is the

-J

t

graph

is defined for

-

-

our circle. Similarly,

x 2,)

lower half of the

is the

other

ask for the equation of the circle has length 3. It consists of the is 3. These are the points satisfying

- 1)2+ (y

that)

- x2 ,)

J l

< 1, whose

(x

2 ))

- 1 < x < 1,such

g(x) = also

y

x

the points

to

correspond

f(x)

there

of

we get two values values

solve for

we can

- V I-x is a

However, we

diagram.)

following

There

= O.

f(x)

- x 2 .)

- 1 and

x 2)

-

then

-1,

Geometrically, these the

t

y

form)

the

of x between

value

1)

type

y2 = 1 For

= c2)

y2

center

remarked

already

+

2)2 =

values

whose points

of

center is (x, y) whose

the equation) 9.)))

circle.

x.

(L 2)

and

distance

AND CURVES)

GRAPHS

38)

I

y-axIs)

[II,

\0376])

Y'-axis

I I I I I I

\037)--

\037\037

x' -axis)

I I)

-aXlS) ;\037

The

equation has been drawn

of this

graph

X' =

In the new coordinate

and)

x-I)

(x',

system

x'

2

We have drawn the (x', y')-axesas dotted To pick another example, we wish

distance2 from the

the

(-1,

point

of the circle is

then)

= 9.)

+ y'2

- 3).

put)

may

y'=y-2)

the equation

y')

also

We

above.

the figure. those determine in

lines

to

the

are

They

points at a

(x, y)

points

satisfying

equation)

(x in other

or,

-

(_1\302\273)2

+

- (-

(y

=

4)

3\302\273)2

words,)

(x + 1)2+

(y

+

3)2 =

4.)

signs!} (Observe carefully the cancellation of minus is the circle of radius 2 and center (this equation

In

general,

circle of

let a, b

radius

r

and

be

(x We

may

in

(a, b) is a)2

+

(y

the

number

o.

Then

of the equation)

graph

- b)2 =

>

3).)

r 2 .)

put)

x' = x Then

-

r a

and

numbers

two

center

the graph

Thus

1,-

the new

-

a)

and)

coordinates x', y' X,2

the

+ y'2

y'

equation

= r 2 .)))

=

y

-

h.)

of the

circle

is)

the

of

Completing the

square)

we are given an equation)

Suppose

Example.

x2 +

where x 2 and the

is

this

of a

then

(a, b) and the

of

0,) to see that wish of completing the

We

method

the

use

we

of the

to be

equation

-

(x

because

5 =

coefficient 1.

the same

circle, and

-

review.

now

we the

want

- 3y

2x

+

y2

with

occur

y2

equation

square, which We

39)

THE CIRCLE)

96])

[II,

a)2 +

form)

-

(y

b)2 =

r 2.)

we know immediately that it represents r. Thus we need x 2 + 2x to be

a circle

radius

centered at terms of

first two

the

expansion)

we

Similarly,

need

-

y2

x

a =

that

means

2 + 2x +

Y

-1

- b)2 =

2-

b =

and

3y

-

=

5

-

2ax +

the first two

be

to

3y

(y This

= x2

- a)2

(x

1)

2-

1+

Y

+

-

2x

y2 +

3y

+

(x

-

5 =

1)2 +

Y

0 is equivalent -

3 2 24.) )

2

(

with

II,

EXERCISES)

\0376.

Sketch

the graph of

1. (a) (x (c) (x 2. (a) x2 + (c)

at

center

J33/4,

+ (y +

2)2

+ (y

2)2

+

(y x 2 + (y

the

(-1,

equations:)

1)2 = 25 1)2 = 1

(b)

- 1)2= 9 - 1)2= 25)

4)

equation

of

- 2)2 + (y

+

a

circle

3/2).)

following

(x

(d) (x (b) x

2

(d) x2

2)2

+ (y + (y

5

33 -9 = -.

4

given equation is the

our

Consequently

= 5 + 1+

)

-9 -

with)

2

3 -

expansion)

-

-

(

Thus x 2

the

Then)

3/2.

(x +

terms of

2 2by + b .)

-

y2

a 2.)

+ (y

1)2 =

4

+ 1)2 =

9

- 1)2= 4 - 1)2= 1)))

of radius

3.

AND CURVES)

GRAPHS

40)

+ 1)2+ y2 (x + 1)2+ y2

[II,

(a) (x

=

1

(b)

(x +

1)2 +

(c)

= 9

(d)

(x +

1)2+ y2

2 4. x +

y2

- 2x

+

-

5. x

+ y2

+ 2x

6. x2

+ y2

+ x

2

2 7. x +

II,

-

3y

3y

2y

- X + 2y

y2

-

10 =

0

-

15 =

0

=

16

=

25)

==

4

==

25)

ELLIPSE)

THE

AND

DILATIONS

\0377.

y2

\0377])

Dilations)

Before

the we have to make some remarks ellipse, to use a more standard word, dilations.

Let

(x,

y)

stretching

by

on

studying

ing,\" or

be a point in the plane. both its coordinates

Fig. 1, where

we

have

Then

drawn (3x,3y)

also

(2x,2y)

is the point obtained

of 2,

factor

a

by

\"stretch-

as illustrated on

and (t, ty).) (3x, 3y))

2x

l.r x

3:t)

1)

Figure

Definition.

the

dilation

Example.

of

c >

In

general,

(x,

y) by a factor

if

the

c.)

Let)

u be

a positive nunlber,

0 is

equation

of the circle x =

2

+

v

2

=

1)

of radius 1.

Put)

cu)

and)

y =

xlc)

and)

v = ylc.)))

CV.)

Then)

u =

we

call

(cx, cy)

[II,

AND

DILATIONS

\0377])

Hence x and

y

the

satisfy

equation)

x2

y2

-+-= 2

1')

c2

c

or

41)

THE ELLIPSE)

equivalently,)

x2 set of

The

we

Thus

points (x, y)

may

+

=

y2

this

satisfying

c2.)

is the circle

equation

of radius c.

say:)

The dilation

of

the

circle

1 by a

radius

of

factor of c >

0

is

circle

the

of radius c.)

This is illustrated on

Fig. 2, with

c =

3.)

x

2

+

= 32)

y2

3)

2)

Figure

The

Ellipse)

is no reason why the same factor. We by

There

may

x = are dilating the first the second coordinate

on

the

circle

by

a

(x, y)

the first factors. y =

and)

2u)

by a factor of 3. In that

factor

of radius

u

Then

different

use

coordinate

we

is a point

dilate

should

we

1, 2

+

other

in

v

2

=

1.)

satisfies the equation) x2 -+-=

4

y2

9

1 .)))

and

second

coordinates

For instance, if

we

put)

3v)

of 2, and case,

words

we

suppose

suppose

are

dilating

that

(u, v) we have)

AND CURVES)

GRAPHS

42)

We

as the

this

interpret

equation of a

[II, 97])

out

\"stretched

circle,\"

as shown

on Fig. 3.)

3)

2)

3)

Figure

More

let a, b

generally,

x =

o.

be numbers>

y =

and)

au)

us

Let

put)

bv.)

(u, v) satisfies

If

(*) then

(x,

y)

u2 +

v

x2

y2

a2 we

Conversely,

mation,

=

1,)

satisfies)

(**))

isfying

2

may

put

u

=

b2

and

x/a

equation (*) correspond and vice versa.)

+

the

to

=

1.)

=

v

y/b

to see

the

that

of (**) under

points

sat-

points this

transfor-

Definition. An ellipse is the set of points satisfying an equation (**) in We have just seen that an some coordinate system of the plane. ellipse of a dilation by factors is a dilated circle, by means a, b > 0 in the first and second coordinates respectively.)

Example. Sketch

the

of the

graph

x2

_

4 This

ellipse

is a

dilated circle by

+

ellipse)

y2 --1

.)

25

the

factors

1')

so

2 and

5, respectively. Note

that)

when

x =

0

we have)

y2 -=

25

y2 =

25)

and)

y

=

+

5.)))

[II,

43)

THE ELLIPSE)

AND

DILATIONS

\0377])

Also)

y =

when

the

Hence

2

x

we

0,

-=

have)

of the ellipse

graph

1')

4

x

so

2

=

x =

and)

4)

+ 2.)

looks like Fig. 4.) y-axIs)

x2

Graph d

5

4

+

y2

25

= 1)

x-axIs) 2)

4)

Figure

the graph

Sketch

Example.

(x

of the

- 1)2

ellipse)

(y +

+

this

In

case,

let us

know

that

in

(u,

and)

x-I)

the

equation

of a

1 .)

y' =

y

circle with u=-

v

2

center

=

2.)

(1,

1)

- 2)

x' and)

V

original equation is

of

the

=

and radius 1. y' 2.)

5)

The

+

v) coordinates)

u2 + is

=

put)

X' =

We

2)2

4

25

form)

X'2

52

+

y'2

=

\302\245

1,)))

Next

we

put)

AND CURVES)

GRAPHS

44)

terms

in

which

of

be

v can

and

u

2

u

our

Thus

ellipse

is obtained u

=

2

=

1.)

u

circle

the

x'j5

\0377])

written)

+ v

from

[II,

2

=

and

v

and

y'

+

v

2

=

1 by

the dilation)

y'/2,

or equivalently,

x' =

5u

= 2v.)

The easiest way to sketch its graph is to draw with coordinates x', y'. To find the intercepts = 0, then) new axes, we see that when y' X,2

-=

1')

52 when

Similarly,

x' =

0, y,2

22

Thus the graph

coordinate

new

the

of the

so

that)

x' =

so

that)

y'

,

.

+

system

ellipse

these

with

5.)

then)

= 1,)

=

+ 2.)

looks like:) y-axIs)

(x

- 1)1

of

Graph

I y-axIs I I I I I)

2S

+

(y

+ 2)2

4

=

1)

x-axIs) I

+ I

-

1(1,

-2)

-... --t-

-....- -f

- -....-

I I I I)

II,

\0377.

the graphs of

Sketch

x2 1.

9

EXERCISES)

+

y2

16

=

the

following

curves.) x2

1)

2.

4

+

y2

9

= 1)))

- -

- - --

x

,

.

-axIs)

x2

3.

5.

THE

98])

[II,

5

+

(x -

y2

7.

1)2

(x +

+

1)2

+

3

9. (x II,

x2

=1

+

1)2

THE

\0378.

A parabola

4\"

(y + 2)2

16 (y + 2)2

4 (y

W

:

6.

=1

8. 25x2

=

is a

is

the

system,

16 y

+

2 = 400)

have

of a

graph = ax

function)

2)

a =1= o.)

with

already seen

the

what

the

graph

of

graph

looks

function

Consider

now)

y

you can

symmetry,

by

2 = 100)

1)

curve which

We Example. 2 looks like. = x y

Then

+ 25 y

1)

PARABOLA)

coordinate

some

2

4x

=

25

= 1

y

in

y2

+

4.

16

9

45)

PARABOLA)

-x 2.)

=

easily see that

the

as on the

figure.)

y-axIs)

x-axIs)

Suppose

that we graph the equation

exactly the same,

it looks

but

as

(1, 0). 2 =

looks

equations

gram.)))

of

these

4)2

as

again

if the

have been

like

origin drawn

y

at

=

that

find

shall

We

-

(4,2).

graphs

- 1)2.

origin were placed

been moved

The

(x

if the

Similarly, the curve y that the whole curve has

(x

=

y

the

x

point

2

except

were the point on

the

next

dia-

AND CURVES)

GRAPHS

46)

these remarks

can formalize

coordinate system we coordinates be x' = x

x' = 0

when

and

pick a

- a

as follows. (a,

point

and

y'

= y

have y' =

= b we

y

y' = the

in

gives

-

old coordinate

of the

terms

as a new

b)

- b.

b)

=

have

we

If

O.

origin.

when

Thus

in our given

that

Suppose

let

We

x =

new

a we have

a curve)

X,2)

whose origin

new coordinate system rise to the equation)

(y in

\0378])

0))

(1,

We

[II,

the point

(a, b),

it

then

- a)2)

(x

system.

at

is

This

type

of curve

is known as

a

parabola. We

can

the

apply

same technique

of

completing

did for the circle.)

Example. What

the

is

- x2

2y the

Completing

our

can be

equation

-

we can

square, x

Thus

of the equation)

graph

2

+ 4x

4x +

6=

write)

= (x +

2)2

-

rewritten)

2y

= (x

+ 2)2 -

2(y

+ 5)

= (x +

10)

or)

We

choose

a new x'

coordinate

= x

+

2)

system)

and)

O?)

y'=y+5)))

2)2.)

4.)

the

square

that

we

[II, 98])

so that

our equation

becomes)

=

2y'

This is a lea ve

whose

function

47)

PARABOLA)

THE

y' =

or)

X,2)

know, and

already

you

graph

tx,2.)

whose sketch we

to you.)

We remark

if

that

an equation

have

we

-

x

y2

=

0)

or)

x = then

we

We can

get a

apply

the

graph

of a

then

see what

the

parabola

can

write

this equation

hence

its graph

+

y2

looks like

5=

2y +

o.)

6) = (y -

1)2)

this:)

y)

Graph

I I I I I I I I)

of (x +

- 6, 1)

x)

Suppose

we

are given the y

like.)

form)

the

in

y'

(

is

of

(x + and

the coordinate system

more general equation

xWe

horizontally.)

of changing

technique

Sketch the graph

Example.

tilted

is

which

y2,)

equation of a

= f(x)

= ax 2

+

parabola)

bx +

c,)))

6) = (y

_

1)2)

to

AND CURVES)

GRAPHS

48)

We

a # O.

with

values for

It is shown formula:) quadratic of f

in

orized, just like the further

without

Example.

to

roots

the

so

times

enough

that

mem-

it is

be used automatically, a quadratic equation.)

It should of

the

1=

equation)

O.)

are)

-

5

:t

x=)

2(

Thus the two roots

-

)25 -

-5 :t J17

8

and

x-axis,

two

:t

J17 4)

are)

5-Ji7

and)

4)

4)

These are the

5

-4)

2))

5+J17

the

by

given

.)

roots of

+ 5x -

of f are

4ac

roots

the

find

- 2X2 The

-

loud

find the

to

want

We

2

table.

multiplication

thinking,

)b

the roots

2a

this formula out

read

should

-b :t

= 0

f(x)

that

school

high

x =

You

which

\0378])

this parabola intersects the and are called the roots

where

determine

to

wish

are the

These

x-axis.

[II,

is shown on the

its graph 5 -

the parabola

where

points

y

=

- 2X2 +

5x - 1 crosses

figure.)

V 17 4)

5+Vi7 4)

Proof quadratic

the quadratic formula. to convince you formula,

of

ax

(*)) I

2

We that

+ bx

now

shall it

+ c=

is true.

o. I)))

give the proof to So we want

of

the

solve)

[II,

a

assumed

we

Since

=1=

obtained

to

means

find

= x2

+ t)2

(x

want

+-x+-=o a a)

Recall the

by a.

dividing

by

t such

2

x

that

formula)

+

+ t

2tx

2.)

has

+

x

form

the

2

+ 2tx.

This)

(:)x we

that

let) b

- = 2t ,

b

t =

is

that

2a

a)

sides of

to both

add

now

We

equation)

c

b

x2

49)

to solving the

amounts

this

0

(**))

We

PARABOLA)

THE

\0378])

.)

and

(**),

equation

obtain)

( \037 y x

be rewritten

can

This

2

b

+

- x

a

2

+-

( 2a)

2

-b

C

=

a

2a)

(

. )

form)

the

in

-b

+

2 X

+

(

or

\037

\037=

+

2a )

\037

a

(

2,

2a))

equivalently,)

+ (x Taking

roots

square

:a Y

=

b

--=C a

( :a Y)

2

- 4ac

4a2)

yields)

b - J

2

b

x+-=+

2a

-

4ac

2a)

whence)

x=

-b

:t Jb

2-

4ac

2a)

thus

proving

the quadratic

Remark. It ratic equation

may

happen

does not

have

,

formula.) that

b

2

-

a solution

4ac


1. We

on the graph then on the graph. So let us when x > 0 and y > o. y)

(x,

quadrant that

follows

it

if

is a point

a point

also

is

53)

x2 -

1 > 0 so x2

claim that

looks

it

>

1.

like this

the

Hence in

first

the

quadrant.)

,,

/

,,

/

/

,,

,,

// /

,, ,, '/

/

//

/

/

// ,

,'

1

''

' \"

'

',

'

,)

see this, we could of coursemake a table of a few values the is like. Do this experimentally what graph yourself. scribe it theoretically. 2 As x increases, the expression x - 1 increases, so J x 2 To

Thus

Also, since first quadrant. y < x

=

y2

x

2

- 1

We have

lies below this

Let us

the

divide

it

the

drawn

in

line

2 y2 < x = line y x.

that

follows

the first

x becomes

large,

x 2)

ratio

Hence this

1 increases.

y/x

is the slope

slope

so

Y


O)

0, we

h >

When

have)

+ h)

I{O and

=

= I{h)

= O. Thus)

I{O)

I{O +

-

h)

I{O) =

0 and

h -+

as

limit

The

0 is

>

h

is the

derivative

left

\037=

h.)

h

The

h,)

1

1.

therefore

limit)

. 11m

I{O

- I{O) .

+ h) h

h\037O

hO)

exists.

Since

1 + h

> 1 we have)

1(1 + Also

1(1)

= 1.

=

1 +

Thus the Newton

quotient

f(l

+ h)

h)

- f(l) h =

trarily function

large. does

Thus not

- 1=

h.)

is)

-

1

= 1_

0 the quotient l/h has no limit the Newton quotient has no have

a

right

derivative

! h.)

h

h

As h approaches

h

when

since

x =

it for

limit 1.)))

arbi0 and the

becomes h >

[III,

93])

III,

\0372.

THE

EXERCISES)

of the following (a) the derivatives x-coordinate is 2, and whose point

Find the

69)

DERIVATIVE)

functions, (b) the

find

the

equation

slope

of

the

of the

graph

tangent

line

at

at

that point.)

1. x 2

2. x

1

+

3. 2x 3 5.

x2 -

7.

2X2

8. !x 3

3x

10.

x+

III,

2

6. 2X2 +

5

1 9.

3x

4.

-

3

1)

x

+ 2x

2

x+ 1

LIMITS)

\0373.

the slope of a curve at a point, or the derivative, we used the which we regarded as intuitively It is indeed. You clear. can see in the Appendix at the end of Part how Four one may define limits of numbers, but we do not worry about this using only properties here. we shall make a list of the properties of limits which will However, be used in the sequel, just to be sure of what we assume about them, and also to give you a technique for computing limits. In

defining

of limit,

notion

F(h) defined for all

functions

consider

We

h, except

of

values

small

sufficiently

write)

We

h =1= o.

that

lim F(h)

= L

h-O)

to

that

mean

First,

we

L as h

F(h) approaches note

that

if

constant

is a

F

approaches

lim

F(h)

function,

O.)

F(x)

= c

for all x,

then)

= c

h-O)

is

the

constant

If F(h)

=

h,

itself. then)

F(h) =

lim

o.

h-O)

The

next

relate limits

properties

with

addition,

subtraction,

multiplica-

tion, division, and inequalities.

Supposethat for the

we

have

same numbers.

two Then

functions we

can

F(x) and G(x) which the sum of the

form

are two

defined functions)))

DERIVATIVE)

THE

70)

F + G, whose value and G(x) = 5x3/2 we

value

The

+ G(x) is also the sum of two

F(x)

concerns

limits

1.

Property values

small

= x4

+ G(x)

F(x)

+ G(x). Thus

x is F(x)

a point have)

at

written

(F

[III, 93]) = x4

F(x)

5x 3/2 .)

+

The

+ G)(x).

first

of

property

functions.)

we Suppose that and assume

of h,

limits)

the

lim F(h)

F and G defined for

two functions

have that

lim

and)

G(h)

h-+O)

h-+O)

exist.

when

Then)

+ G(h)]

lim[F(h) h-+O)

exists

and)

+ G)(h) =

lim(F

In

words

other

A similar

the limit of

statement lim

lim

a

the

sum

- G(h)) =

a

x

number

F(x)

= 2X2

2.

Property

Let

two

and

G(x)

= x2

- 2x )(x

functions

and)

F(h)

h-+O)

exist.

G(h).

h-+O)

2

+

+

5x,

Then

the

limit

of

lim(FG)(h)

their

product

product

lim

G(h)

exists and we

have)

= lim[F(h)G(h)] h-+O)

h-+O

=

lim h-+O

the product

values of

h-+O)

the

functions

two

is)

5x).)

small

for

then

that)

lim

limits.)

is)

= (2x 2

F, G be

the

namely)

G,

- lim

F(h)

sum of

= F(x)G(x).)

- 2X

(FG)(x)

sume

lim

h-+O

(FG)(x)

if

to the

equal

we discuss the product. Supposewe have Then we can form the same numbers. for

F and G defined FG whose value at

For instance

is

holds for the differenceF -

(F(h)

G(h).)

h-+O)

sum

h-+O

After

+ lim

F(h)

h-+O)

h-+O)

F(h) .lim h-+O)))

G(h).

h,

and

as-

[III,

can say

In words, we of the product.)

Then we can

the

that

case, suppose

a special

As

71)

LIMITS)

\0373])

constant

is the

F(x)

function

the

form

that

is equal to

of the limits

product

function

F(x)

of the constant

cG, product

the

by

G,

limit

= c. and

we have)

lim cG(h)

= c . Jim

h-+Q

F(h) =

Let

Example.

Then

5.

+

3h

G(h).

h-+Q)

= 5.

lim F(h) h-+Q)

3 F(h) = 4h

Let

Example.

1. Then

5h +

lim

F(h)

= 1.

We can see)

h-+Q)

this

the

considering

by

lim

4h

limits)

3 =

0,

lim

and

taking

the appropriate

sum.)

lim

3xh =

Example.

We

have

5h =

0,

1 =

lim

1,

h-+Q)

h-+Q)

h-+Q)

0, and

h-+Q)

-

lim(3xh

7y) =

-

7y.

h-+Q)

G(x)

=1=

to quotients. Let x. Then we can form any

we

Thirdly, 0

for

value at x

come

before, but function quotient as

be

the

assume that FIG whose

is)

F

G 3 Example. Let F(x) = 2x

F G

Property

F, G

3.

Assume

-

lim h-+Q)

F(h)

F(x)

G(x)')

4x and

(x) =

that the

(x) =

4 G(x) = x

F(x) = 2x3 G(x)

-

4x

x4 +

XI/3.)

+

X

limits)

and)

lim h-+Q)))

G(h)

I /3

. Then)

DERIVATIVE)

THE

72)

\0373])

that)

and

exist,

[III,

lim G(h)

=1=

o.

h-+O)

the

Then

limit of

the

F(h)

lim

In words, the As we have

sake of

done above,

h-+O

G(h)

the

limits

we

=

lim F(h) lim

G(hf)

is equal to the

we

find

be skipped Property assume

is stated here for completeness. property the derivative of sine and cosine,and

F, G be two functions also G(h) < F(h). Assume

4. Let that

lim F(h)

of

writing

the

quotient.)

h -+

0 for the

It will

not be

used

should

consequently

for small

defined that)

lim

and)

values of

h,

and

G(h)

h-+O)

Then)

exist.




to find the

wish

Theorem

gent line at x =

2

when

find

the

the

line

is)

other

2. Find

3.

hand,

X

4. What

.

1.)

J2xJi

=

Then f'(x)

= x.

Considerthe

1.)

lines to certain

of tangent

equations

curve)

= x 5)

= 5x4.

the point

at

line

tangent

f'(x)

Hencethe

(2, 32).

slope

By

tan-

of the

= 25

f(2)

4 = 80.)

= 5.2 =

= 80(x -

- 32

of

the equation

Hence

32.

expression of (x + h)4

out the

the

of

derivative

the

the

tangent

2).)

2/ 3

is

the

the

using

directly,

(c)

tangent

of x

h.

and

Newton

quotient.

functions?

following

of the

equation

of powers

2 X- 3/

(b) the

terms

in

function x 4

of

derivatives

the

are

What

(a)

1/ 2

EXERCISES)

\0374.

1. Write

!X

9/ 4 .

is)

y

III,

=

f'(x)

f(x)

equation of its

f'(2)

On

=

f'(x)

case

then

then

= -iX-

do before.

= x5,

if f(x)

4.1,

0),

f'(x)

y We

= 10x 9 .

then

= X- 5 / 4 , then = x Ji , then

If f(x)

Note

X

3/2

line

7 6 X / = x9

to

the

curve

y

at

the

point

(8,4)?

the

at

point

(1, I)?)

5. What

is

equation

6. Give the point

7. Give point

the

of

slope

whose the

slope the

tangent

the

line

and

equation is

X

2/3

What

is

the

at that point?

equation of x-coordinate is 16.

x-coordinate

y =

curve

and

slope

whose

of

of 3.)))

the

the

tangent

tangent

line

line

to the curve

y

to the curve

= x

y

=

- 3/4 at the

\037 at

the

[III,

8. Give the derivatives 1 4 (a) f(x) = X / at (c) f(x) = xfi at

III,

functions

of the following x = 5 x = 10

We

begin

Definition.

A

is

function

x =

at

= 7 7)

you functions

of

of continuous is continuous. said to be continuous

function

points:

at x

allow

which

functions

definition

a

with

differentiable

a

1/ 4

QUOTIENTS)

In this section we shall derive several rules derivatives for sums, products, and quotients know the derivative of each factor. why

indicated

the

at

(b) f(x) = X(d) f(x) = XX

AND

PRODUCTS,

SUMS,

\0375.

79)

AND QUOTIENTS)

PRODUCTS,

SUMS,

\0375])

the

when

you

and the reason

a point

at

to find

x

if

and

only)

if)

lim

f(x

+ h)

= f(x).

if it

is continuous

h-+O)

A

of

Let f at

to be continuous

is said

function

domain

of its

at every point

definition.)

be a function

a derivative

having

at x. Then

f'(x)

f is continuous

x.)

The

Proof

quotient)

f(x +

-

h)

f(x)

h)

approaches

the

as

limit f'(x) f(x

h

-

+

h

f(x)

= f(x

\037

Therefore approaches

using

rule

the

0, we

the

for

limit

We have)

o.

approaches

of

+

h)

a

product,

lim

is another

+ h)

f(x

way

of

stating

lim h-+O)

In

other

words,

f(x).)

find)

- f(x) = Of'(x)

h-+O)

This

-

f is

continuous.)))

that)

f(x

+ h) =

f(x).

=

O.

and

noting

that

h

THE

80)

Of

0/0,

to

amounts then

taking

to

impossible

procedure of if h

i=

taking

never substitute h = 0 in our quotient, because then which is meaningless. Geometrically, letting h = 0 the two points on the curve equal to each other. It is have a unique straight line one point. Our through the limit of the Newton quotient is meaningful only

in the Newton quotient, both itself, approach O. The quotient

that

denominator

proach

Let

Example.

= Ixl. though it

f(x)

0, even

at

Then

need

and

the

not

ap-

0 as h

approaches

tiable at 0 and

O. left

approaches

we saw

As

on

at

f(x) Fig. 9.)

= 0

if

function

f(x)

= Ixl is

f(x) = 1 if

0, and

that)

true

still

is not

function

the

x >

differen-

differentiable

right

differentiable at

0, but not x
o.)

r is

Pythagoras,

Note

151)

POLAR COORDINATES)

96])

[IV,

=

=

3

coor-

Also)

2.

.)

f

polar coordinatesare (2, n/3).)

we may have

several polar coordinates

whose the same point. The point polar in our same as the point (r, e). Thus the also be polar coordinates for our would

coordinates

to

example

corresponding

are (r, e + above, (2, n/3

point. In practice, we

2n) is +

2n)

usually

0 and 2n. Its position is completely determined if we know the angle e and the distance of the bug from the r from origin, that is if we know the polar coordinates. If the distance the bug is traveling along a of e, then the origin is given as a function curve.) curve and we can sketch this use

the value for the

angle

Suppose a bug is

traveling

lies

which

in

between

the

plane.

r = sin e for 0 < e < n. 2. Sketch the graph of the function e we don't get a hence for such e < 2n, then sin e < 0 and consider We we make a table of values. on the curve. Next, point sin e is always increasing or always intervals of e such that decreasing is moving further the these intervals. This tells us whether over point r is the since from the origin, or coming closer to the a way origin, Example If n


obtained, multiplying

O.

We

both

can

sides

obtain)

+ y2

=

from Chapter II that We recall square.

y.)

is the

this

here

how

form)

x

We

y2

substitution is valid only the equation we have just

x2

You

2 J x

this

course, 2 J x

yj

obtain)

2 J x +

then

=

yjr

2

-

+ y2

equation to be x 2 + (y

of

-

Y

=

form)

the

b)2

O.)

=

c

2 ,)))

this

equation is done.

of a We

circle, write

[IV,

then we know

because

and radius c.

We

x

-

2

_

+ y2

x

is equivalent

is the

corresponding

4.

in polar

origin

Y

2

y

=

angle ray,

fJ

is

2

+ (y

on

2

+

(y

t)2 _

-

1)

-

=

Y

0)

- t)2 = 1.)

simply)

polar 1

and

(0, t)

= 0 is

and radius the

t.

with

point

The

point

rectangular

o.)

of

2 J x

or)

geometrically, as

x

equation)

coordinates is

5. Consider

Example

Thus

=

Then)

+ y2

+

This expresses the condition that the origin is the constant 3. The

with

t.

The equation

r=3)

point

- 2by + b 2 .)

y2

equation of a circle of center to the polar coordinate r

coordinates x = 0 and Example

(0, b)

with)

x

This

=

b)2

Thus the

1 cancels.

the

because

of center

circle

a

IS

that)

= 1 and b =

let 2b

we

this

that

immediately

know

(y Therefore

153)

COORDINATES)

POLAR

\0376])

of

circle

the

y2 = that angle

x

or)

3)

distance fJ

be

can

3 and

radius

2

+

center

at

the

y2 = 9.)

of the point arbitrary.)

(x, y)

from

1 in A coordinates. polar coordinates (r, fJ) satisfies this equation if and only if its there is no restriction on its r-coordinate,i.e.r > this set of points can be described as a half line, or a the

equation

fJ

=

o.

the figure

(a).)

y

x)

(a))

(b))))

\037= x)

tan 8

of the tangent,

definition

the

By

a

AND COSINE)

SINE

154)

ray, and

this

on

point

=f:.

y

y/x =

if

coordinates of

are the ordinary

y)

(x,

[IV, 96])

then)

0,

1)

tan

and)

x >

and)

x > o.)

0,)

whence) =

y

Of course, the whose

point

(tan

x =

with

point

on

tion

=

()

lies on the ray.

0 also

(x, y)

(tan l)x)

=

Instead of 1 we by the equation () of conditions)

could =

n/6

y

n/6 =

tan

=

(tan

x >

and)

1/J3, we

y=-x

the

by

x >

and)

n/6)x)

may

coordinates pair

by the equaof conditions)

number. For instance, the coordinates is also defined

take any in polar

(tan

o)

x > 0.)

and)

l)x)

Conversely, any

satisfy)

the ray. Hence the ray defined in polar 1 is defined in ordinary coordinates y

Since

=

y

coordinates

ordinary

y = lies

l)x)

the

write

the

pair

0.)

pair of conditions)

equivalent

1

defined

ray by

x > 0.)

and)

J3)

tan

a tan

1.

Only of

way

n/6

is no simpler way when dealing with

there

that

Note

the

writing

=

of

trigonometric

1 than just writing of TC do we have

tan

expressing

fractional

multiples

functions

in

of

terms

like

roots,

1/J3.)

Example

6. Let us

sketch the

curve

given

in polar

coordinates

by

the

equation)

r = The

there is a decrease n/2.

value

absolute

Hence

value

of

the

2()

I.)

sign makes the right-hand r for every value of ().

for sin 2() will it is natural

make a table of over such intervals.)))

I sin

occur to

look

increasing

side Regions

>

always

of

0, and

increase

so and

2() ranges over intervals of length at intervals for () of length We now n/4. and decreasing behavior of I sin 2() I and r

when

r=

0 inc.

0 to

therefore

looks like

1 to

0

inc.

0 to

1

dec.

1 to

0

dec.

and so

graph

forth)

this:)

of r =

Graph

of

Because for

value

the

r which

absolute is > O.

value sign, for r = sin

of

tions of

the

value sign, then for which

the

absolute

the

above graph for which)

graph

Isin

201)

0 we

of

value

convention,

if

we

obtain a wanted

would

to omit

have

is negative,

20

those por-

i.e. those portions

n

2 O.

(c) r =

8.

-

1. Just

coordinates.

the following to

assume

Sketch

(d) (2,

- 3)

(4,

(b)

sin

= 2

Change (a)

6.)

coordinates.)

(1, 1)) the

-n/4))

(c) (1,

in Exercise (b)

coordinates

polar

r

6

coordinates:

polar

(3, n/6)

coordinates:)

4. Sketch

in

points

(b)

as

(a) (1, 1) (These are polar

(a)

11 n

and

EXERCISES)

\0376.

1. Plot

3.

7n

also drawn the rays determInIng angles of

We have

IV,

157)

POLAR COORDINATES)

96])

[IV,

f))

f))

- cos f))

16.

r =

1

19.

r =

sin

4f))))

20. r = cos 23.

r =

26. r In

21. r

28)

the next the

30

I

r =

25.

r = 0)

Icos 281

three

1

30. r =

tan

r = 11

put

28. r =

1 - cos0) the following

34. 0 =

problems

the equation

In

coordinates

rectangular

curve.)

Sketch

0 =

I cos

22.

I/O)

27. r =

36.

[IV, 96])

= cos 38

24. r =

Isin 301

=

sketch

32.

AND COSINE

SINE

158)

0)

+ 2 cos 0 I)

curves

given

2

2in

cos)

polar 31.

29. r =

0 coordinates.) r = 5

+ 2 sin

33. (a)

r =

(b)

r =

n

35. 0 =

n/2

-n/2

37. 0 =

5n/4

38. 0 = 3n/2)))

39.

2+ 2-

0 = 3n/4)

0 sin

20

sin

28

4

1

+ 2 cos) 0

and

V)

CHAPTER

Theorem)

Value

Mean

The

to a curve, y = f(x), we shall use the derivative find we shall the tion about the curve. For instance, the curve minimum of the graph, and regions where decreasing. We shall use the mean value theorem, which theory of derivatives.)

give

Given

V,

Definition. Let f

be a

function.

differentiable

f'(c) = The

derivative

and

thus

of

this

f is a

o.)

line

tangent

have

drawn

is 0

three

phenomenon.)

\\) I c

I c

1

Figure

f'(x)

point of

A critical

of the the zero means that slope We line itself is horizontal. tangent

/\037

The

THEOREM)

being the

that

examples

and

is increasing or in the is basic

that)

such

c

number

MINIMUM

AND

MAXIMUM

THE

\0371.

us informa-

maximum

Figure

third

= 3x

2

example and

hence

is

that

when

of

a

2)

function

x = 0, f'(O) = o.)))

Figure

like f(x)

= x3 .

We

3)

have

MEAN

THE

160)

THEOREM)

VALUE

two examples are those of a maximum respectively, we look at the graph of the function c. We shall now formalize these notions. Let a, b be two numbers with a < b. We shall b. Sometimes the interval of numbers between a and the end points a and b, and sometimes we do not.

and a near only

other

The

if

dard

to

want

with

include

We recall the

stan-

The collection of

[a, b]. is

to include only one end point, We have of course two half-closed.

the one

wish

consisting of

Sometimes, if a (or x < a) an open Let f be a function,

is a

say that the interintervals, namely

context will always a number at which f

The

interval. and

c

one

a


is defined.)

the

of

point

function

f

if)

only

>

f(c)

for all numbers x at for

has a

maximum at c 1.

Example

which

x

numbers

all

holds

f(n/2) = 1

and

Note

-

that

shall

a < x < b, and the other b. number, we call the collection of numbers

We shall say

Definition.

we

half-closed

the closed the by symbols

called

x with x


(3.)

a


the

0,

is

the

maxi-

local

Newton

1 ]) \037

quotient

satisfies)

-

+

f(c

f(c)


-

-

k)

- k) > 0)

f(c) - f(c -

- f(c)

-k) Newton

the

f(c

is)

f(c

Thus

O. Then)

proaches0, we

quotient

see

k)

k)

is

> O.

f(c

+ h)

the

Taking

limit

as

h (or

k) ap-

that) . 1 1m

- f(c) > = 0.

h

h-O h 0

in

We state

which

a function

be

throughout can interpret is

function

is

2 ))

as

in

continuous

this as

and positive, a theorem.)

always this

an inter-

some

interval,

the end points).

interval

(excluding

the

interval

(excluding

the end points),

then

f

IS

the

interval

(excluding

the end points),

then

f

is

the

increasing.

strictly

If f'(x) < 0 in strictly decreasing.

=0

If f'(x)

the

interval

(excluding

statement,

the

hypothesis

in

the end points),

then

f

is con-

stant.) In

last

this

means function

that

is

the

of change is To see how

rate

constant.

that

f'(x)

0, and so it is quite these statements fit

= 0

the interval that the plausible into a more formal in

context, see 93.) of parabolas)

Graphs

Application.

Example. Let us

graph

the

y = which

you

should

curve)

2 f(x) = x

know is a

-

3x +

parabola as

in

5,)

Chapter

II.

We

treat

the)))

INCREASING

92])

[V,

here

graph

by

works

which

method

the

167)

FUNCTIONS)

DECREASING

AND

more

in

First, we

cases.

general

have)

-

= 2x

f'(x)

3,)

and)

= 0)

f'(x)

and

if

x =

if)

only

x =

so

3/2,)

cri tical

f'(x) >




f'(x)

if and

0)

only

and

if

if and


0 f'(x)

positive

< 0

x

x

2

2

- a

>0

- a

\037,)

x2 and

if

x

for if

only

x

point of f for x =1= O. Hence the its numerator x 2 - a is critical

only 0

>



consider

we

because x >

=

- a

2 = x

=0

2)

X

2

2

all

> a

< a

x>\037,

x

decreasing

x
0

- cosx.)

> 0 for

increasing

1(0)

Hence

can

Then)

f'(x) First

and decreasingfunctions

= 0

for

because cos x 0 < x < nI2.

- sin 0 0
x >

that

that f'(x)

We

Proof

for prov-

used

interval that)

< g(a),)

< g'(x) throughout

Then

interval.

the

f(x)
0

that h(x)

principle

g(a) the

throughout

stated

just

drawn for the case

when

f(a)

f(a)

- f'(x) > 0,) Since)

interval.

g(x) > The

for

interval.)

the

so

is

f and g over a certain functions are differentiable. Suppose f, g f(a)

and

holds

n12.)

two

have

b] and we assume

\0372])

3.14), and so

inequality

illustrates a technique which example functions. In general:) between inequalities

Suppose we

Then

desired

the

Thus

[V,

is approximately

n

(because

n12.

THEOREM)

VALUE

preceding

certain

[a,

1

x >

n/2, then whenever

simpler reasons when

ing

MEAN

THE

172)

can be

f(a)

>

interval,

0,)

whence)

f(x).) visualized In

the

picture,

following

= g(a).)

= g(a))

a)

Figure

12)

or to f at x = other words, if g is bigger than equal for all x > f(x) f, then g(x) is bigger than grows faster than

In

a, a.)))

and

if

g

[V,

Example. Show one has the inequali

for

that

=

f(x)

-

x\"

1

- n(x

x > 1 it

follows

for x > 1.

increasing

equivalent to

x > 1

number

any

1).)

- 1). Then) -

= nx\"-l

n.)

so f'(x) > f(x) > 0 for

that

X\"-l

> 1 and

O.

f(l)

= O.

Hence

x >

But

desired

the

-

- 1 > n(x

f'(x)

Since

and

ty)

x\"

Let

n > 1

integer

any

173)

FUNCTIONS)

DECREASING

AND

INCREASING

\0372])

is

Hence

f

1.

This is

inequality.)

theorem tells us

On the other hand, the next functions have the same derivative

an

throughout

what

if two

happens

interval.)

Constants)

Theorem 2.2. Let able

in

f(x)

and

in

Then

interval.

the

x

all

for

in

= g(x)

+

C

that)

such

C)

interval.)

the

Let h(x)

Proof

differenti-

= g'(x))

there is a constant

f(x)

are

which

that)

f'(x)

for all x

two functions

be

g(x)

and assume

interval

some

= f(x)

-

of our two

difference

the

be

g(x)

functions.

Then)

is constant Theorem by C and all x. This proves the

Hence

h(x)

number

Remark.

If

theorem

The

use Theorem on

logarithms,

applications

2.2

its

then

in

a

that

IS h(x)

= C

for

some

of

derivative

fundamental

the

statement:)

is equal to way

O.)

when we come

to

the

integration.

For the applications

on

2.1, theorem.)

is the converse

is constant,

a function shall

We chapter

- g'(x) = o.)

= f'(x)

h'(x)

and here.)))

also

of

the

theorem,

the beginning

see

the beginning

of Chapter X, 91. We

of the chapter give

simpler

THE

174)

f(O)

be a

Let f Example. = 2. Determine from

know

We

MEAN

experience

that

is constant

there

C such

function)

the

5x)

also

are

given f(O)

= 2.

= 5x

C =

2. Thus

f(O) =

origin.

A

particle

t =

time

Determine

0+

C.)

finally)

= 5x

f(x)

Example. 5 cm/sec. At

+ C.)

Hence)

2 =

Therefore

5.)

that)

f(x)

the

= 5. Supposethat

f(x) completely. past

g'(x) =

We

92])

derivative)

the

Hence

[V,

that f'(x)

x such

of

function

g(x) = has

THEOREM)

VALUE

+ 2.)

moves on the x-axis toward the left at a rate of 5 the particle is at the point 8 cm to the right of the x-coordinate x = f(t) completely as a function

of time. are

We

given)

dx

dt Let

g(t)

=

- 5t.

Then

g'(t)

= f'(t)

- 5 also.

=

=-

5.)

Hence there

that)

f(t)

But

we

are

also

given f(5) 8 =

Therefore

C = 8

= 8.

=

+ C.)

-5t

Hence)

- 5.5 +

C=

-

25

+ 25 = 33, so finally) f(t)

=

-5t

+

33.)))

+

c.)

is

a constant

C such

[V,

92])

V,

\0372.

INCREASING

FUNCTIONS)

DECREASING

AND

175)

EXERCISES)

the intervals on

Determine

the

which

are

functions

following

and

increasing

de-

creasing.)

1. I(x)

= x3

+

3. I(x)

= x3

+ x

5. I (x) = 2x

3

1

+ 5

= -4x

7. I(x)

Sketch the each case.)

3

= x2

-

15. I (x) = For

19.

4

- 4x

21. 3x -

(a)

x-I

+

6. I(x)

= 5x

2

8. I(x)

= 5x

3

+ 1

+ 2x 1

+

+ 6x)

Determine the

10. I(x)

= x2

12. I(x)

=-x

+ x

point

In

x in

the

+ 1

-

2

critical

x-I

= x2 - 5x + 1 16. I(x) = 2X2 - 4x - 3) functions,

the

find

for

minimum

maximum,

- x2 ,

x3 ,

the

with x
0)

flex)

(1))

for

inequality)

x > o.)

all

prove:)

(b ) 1 -

[Hint:

2

-x -< 2 Let

x for

cos

fix)

x >

-

= cos x

o.

-

-

(1 f 2 (x) >

(2))

(c) x -

1-

and use \0372

(1), to

0)

all x

for

= sin

x

-

2

x +

3\03732

(x

4

4.3.2)

> 0.])

-

[Hint: x2

prove)

)

Let fix)

< sin x. 3\03732

(d) cos x
o.

all

sin

how to

show

steps

following

sin

J3]

[-2,

start

20. x - x2 , 22. (x - 4)5,

- 1, 4]

[

[-

18. x 2 - 2x + 1,

4]

[0,

8,

Let fl(x)

Now

= -x 3

14. I(x)

+ 1

+ 4 x-I)

2X2

4. I(x)

parabolas.

x-I

+ 3x

- 2x -

We

following

= x2 - x + 5

terval.)

in

23. The

the

of the following

each

17. x 2

2x)

of

= x2 11. I(x) = - x2

gi ven

-

graphs

9. I(x)

13. I(x)

- 2

2. I(x)

(e)

sin x -
x

that tan x

Prove

[V,

\0373])

x < n12.

0
-) 2

t+-

t >

for

O.)

t

=

Let f(t)

[Hint:

f is

and

(b) Let a,

two

be

b

+

t

Show that f is strictly for 1 < t. What is

lit.

increasing

strictly

numbers.

positive

= ax

f(x)

Show

26. A

with

box

C.

surface 27.

two

is to have a fixed if it is to have

top

open

and

its

height

C

is a constant

there

function

f(x)

any function

be

g(x)

of

and

such

that

such

that

g(x) =

that

such

nx 2 and

has radius x

are closed at is 2nx. The

container

the

and

x is

radius

base

whose

Let

the box

when

circle of

there is a

f'(x) = f(x).

its

length

y is

height

2

nx

y.)

f(x);/= 0 for all g'(x) = g(x). Show [Hint: Differentiate

Cf(x).

x,

and that the

gl f]

quotient

30. Supposethat

31. Supposethat 32. Supposethat

find

- 3

=

f'(t)

= 2 and

of

t

and

f(O)

f(O) =

f(t) completely.

- 5. Determine

f(t)

the x-axis toward the right is at a distance the particle of t. its x-coordinate as a function time

is

dripping

at

a rate

30 ft.

Find

MEAN

The theorems in omit

this

the

understanding

tank so that tank is full,

of a vertical

of 2 ft/day.

explicitly

THE

t = 9

out

(a) f'(t)

= 1. Determine

on

moving

at

that

such

=

-

3,

say about f(t)?

you

f'(t)

falling

you might wish, after

can

What

33. A particle is

7 ft/sec. If

function

is a differentiable

f

= 2.

f'(t)

\0373.

f is 2j;;b.

radius

problems

cylinder

that

29. Assume

V,

x > o.)

for

of a cylinder with of its base

the

(The area of a

volume of a

34. Water

-b

volume.)

top.

origin,

1

Let)

+

value of

shape

Find

C.

the above

the

(b)

the

in

area

maximum

28. Do





0, we

for

have,

all

inequality)

1-

Therefore

.)

given a small number b

In particular, large,

sufficiently

1.)

x 3)

2

1.

determined

1

-

x2

2

x

-

+ 2x

1+--x approaches

are

form)

( x becomes

x-axis

\0371])

polynomial)

x3 1 + When

[VI,

3

(1

2

x

3)


.

inequality)

-1\302\273
0 and bending only down for x < O. There is an inflection point at x = o. we find that the graph of f looks like this.) all this together, Putting

maximum

graph of 3 f(x) = x -

2x

+ 1)))

193)

CUBIC POLYNOMIALS)

93])

[VI,

Observe how we useda quadratic polynomial, as an intermediate step in the arguments.)

Then

-J2i3,

that

for all x


- J2i3.

x
0) us the

gives

when)

0,

figure.)

of

= x1

j2

x < 1-

when)

regions of

increaseand

x < 1-

For For

1 +

For

2 is 1 >

-

and

-

2x

1 +

1)

j2.

From the graph

I-j2

< {a

AND

word

techniques

two variables.

perhaps

3.

1

x

x)

\037

MAXIMA

APPLIED

This section deals with

2.

-

O.)

\0375.

1.

that

2 2

b +-

= ax

of f(x)

value

Deduce

x

Let)

numbers.

minimum

assertions.

your

x>

the

that

+ 5

l/x.

f{x)

Show

x+l

x2

18.

f(x) = x +

the graph of

[VI, 95])

J x+l

17.

Sketch

19.

X

14.

CURVES)

say,

where

happen the

that

derivative

not apply. Find

the

on

point

to the point

(2,

the graph

3).)

Figure

4)))

of

the

equation

y2 =

4x

[VI,

To to

the distance between a point the square of the distance, which occurs in its formula. Indeed,

root

square

the square

for

for

value

minimum

The

square

Z2 =

that

is

Z2

solve

we can

Substituting tance only

=

y

+

X)2

we find

(2 -

We now determine the

+

X)2

2

x

-

(3 x

+

4x

= 13+

-

2

+

f'(x)

=

of

We

f.

of

=

2\037.

the

4x

x =

have)

, = 6

2x\037

V9.

< 0

f'(x) is strictly

V9. Hence V9 y

x

when

3 3

> 9.


minimum.

a

is

x

increasing

> 6

2x\037

for

y

we have

f'(x) > 0

x


Xl' in

X 2 < Xl' in

=

f (x) between a

is

and

[a,

to

b].

each

and b such

y

that

set

of values of f)

Given a number

be

then)

which case

!(x 2 )

which casef(x

and

Yl

X2

number

another

2

> f(x

) 0

f'(x)

interval

the

Consider

interval,

and so

x -+

it

y

>

00,

that

y

Now Therefore

1 > g(O)

J 2/3 and x

Then f

J2/3. inverse

function

Since f(x) is

g(y)

this

on

increasing

strictly

g is defined.

function

the

is

< -J2/3.) -+

defined

00

when

for

>

= (2/3)3/2

f(J2/3)

so 1 lies

J2/3,

in

the

- 2(2/3)1/2+

interval

x >

1.)

J2/3,

and f(l) = o.

= 1.

Similarly .f(2)

all

is

that

f(J2/3),

x >

inverse

the

follows

x >

=

5

and

2 lies in

the interval x

> J2/3, so g(5)= 2.)))

[VII,

Note

do

we

that

When dealing with

give an explicit

not

> 3,

function.

inverse

our

for

formula

of degree

polynomials

221)

FUNCTIONS)

OF INVERSE

DEFINITION

\0371])

no single formula

be

can

gIven.)

5. On

Example

other

the

take 1

hand, 3 I(x) = x

as a

viewed

but

_ of 1 is given

derivative

The

by

1 is

is quite different defined, in the interval, and 1(0) = h(1) = O.)

1,)

-A 0:) 2

(a) 1 < eX) [Hint: using

17. Let

(b)

1 +

x
0,

are

log u

b =

and

=

then)

uv =

log

u

+

v.)

log

log v. Then)

eae

to be

uv

=

b

=

eloguelogv =

=

uv.)

uv)

a +

b =

log u

u

- 1 =

-log

2.2.

We

If u >

have o =

-log

+

log

v)

shown.)

0,

then)

log

Adding

it

that)

Theorem

Proo.f

x,

only for

log was

O.)

is defined

ea + b

as

TC.)

the relation)

definition,

means

=

for all numbers

positive

ea + b By

TC.

that)

1 means

log

Proof

=

log en

log 1 = all

-)2,

way:)

the

Furthermore,

=

log e-ft

2,

e 1og2

Since

[VIII,

We have

Examples.

And

LOGARITHMS)

AND

1=

uu

log 1 =

u to both

1.

u.)

Hence

log(uu-1) = log

sides proves

the

u +

theorem.)))

1 log u- .

that

follows.

92])

[VIII,

We have)

Examples.

log(lj2) =

Of

we can

course,

just as we can For instance)

terms.

take the log

take

the

is a

if n

the

where

the

e

Q+ Q+

on the

product

We have

=

instance,

2

log(u

3

)

= log(u 2u)

= 2 log u = 3 log And

so forth, to get log if It now follows that

more

more

than

two

than

two

n times.

for the

u\") = n

2.1, we

log(u. u) =

) =

with

sum of

c.) = eQebe

taken

is

rule

corresponding

log(u

product of a

.'. + Q Q = e Qe Q ... e Q = (e )\",)

right

by Theorem

a

of

then)

log(

For

- log 3.)

2

= eQ+bec

positive integer, e\"Q

2)

exponential

eQ + b + c Similarly,

-log

= log

log(2/3)

terms,

249)

LOGARITHM)

THE

\0372])

log

log,

namely)

u.)

find:)

log u

= log u 2

= 210g u.

log u

+

+ log u

log u

+

u.)

u\"

u. a positive

=

n log

n is

integer, then)

1

1/\" = log U -log

u.

n)

Proof

Let

v =

U

1/ \".

Then

v\"

log

=

v\"

we have

u, and =

n log

v.)

Hence)

log v =

-1 log v\", n)

which

is precisely

the relation

log U 1/

\"

=

11n log

u.)))

already seen that)

AND LOGARITHMS)

EXPONENTS

250)

The same type

are positive

If m, n

for fractional

holds

rule

of

m

rn /\"

-

=

n)

write

We

exponents, that

is:)

integers, then) log u

Proof

[VIII, 92])

u rn /\"

=

(Urn)l/\".

log

rn /\"

u

log u.

Then) =

log(urn)l/n)

-1 log urn

=

n)

-m

=

log u

n)

Just

cases separately.)

the two

using

by

to

give

approximate

you values:)

log 10

4.6. . . ,)

log 1000= see that if x like an arithmetic

can

You

grows

where log

10 is

In Exercises

Make up a table the can

of

the

en with

growth

of

then

see that

log en

a geometric

progression.

we

log,

give

a

few

above

The

Ion = n

= 11.5...,) =

13.8...

.)

progression, then values

illustrate

log

the

x

rule)

log 10,)

2.3.

approximately

17 and

1,000,000

log

like

log

100,000

log

6.9...,)

grows

the

log 10,000= 9.2...,)

= 2.3...,)

100 =

log

behavior of

a feeling for the

19 of values

log

e\"

grows

have proved that 2.5 < e < 3. = n. You can then compare in a similar way. For positive integers you to en. For instance,) very slowly compared

91

you should en and log

log e log e

log

log e

e

3

=

4= 5

=

10 =

e\"

3,)

4,)

5,)

10.)))

251)

THE LOGARITHM)

92])

[VIII,

Using the fact that e lies between 2 and 3, you e5 or e 10 are quite large compared to the values and

10, respectively,

e

the same

have

We

powers

phenomenon

powers like

2 we

5

are

which

have)

1,000.)

the

in

since e >

instance,

2 10>

lO >

that

log,

direction

opposite

for negative

instance:)

For

e.

of

For

cases.

these

in

see

of the

can

-1 = -1,

log

lo g

e)

-1 = 2

- 2'

-1 = 3

-

e)

lo g

3

'

e)

1

= -10.

log 10 e)

Put

h =

l/e

Y

a feeling

h approaches

As

.

slowly. Make a

for numerical

Observethat y = log x is

if

In

becomes n

log(I/10

n =

1, 2, 3, 4, 5, 6 x =

write

we

and

to

get

eY

then

For instance)

x =

l/e

106 =

x=

1/ e

10100=e) -

e-

106)

10100

then)

log x

then)

log

- 106 ,)

=

_10 100.)

x =

short:)

If

x -4

0

then

log

x -4

- 00.

I

The

but rather

posItIve,

large

) with

small positive number

large negative.

. lf

y

examples.)

is a

if x

0,

for

table

similar

comes

reason

Similarly,

property

if y of

the

\037

00

I)

from the behavior then

e

Y \037

00.

This

of

eYe

translates

If

y

\037

- 00

into the corresponding

function:)

inverse

If

x -4

00

then

log

x -4

then eY \037 o.

00. I)))

AND LOGARITHMS)

EXPONENTS

252)

The derivative

of

log)

we consider the differentiation

Next

y

By

for

rule

the

=

eX)

differentiating

x =

inverse

functions,

1

dyjdx)

dy)

we

have

properties of

and)

dx

Hence

[VIII, 92])

log

1 ----

-) 1)

eX

y)

the

log

function.

Let)

y.)

we find:)

the formula:)

Theorem 2.3.) d log

y

dy)

the

From the

graph function

inverse

the general way of its graph looks like

1) y)

eX takes on all values > O. Hence and for all by log is defined positive real numbers, we see that the graph of an inverse function, finding

of

the

in

that

see

we

eX,

that

figure.)

y-axIs)

x-axIs)

In

the figure, the graph eO =

Note

that

the derivative

d log x dx

so

the

log function

is

strictly

crosses the horizontal axis means)

1)

log

1 =

O.)

satisfies)

=

!>

0

x)

increasing.)))

for all

x >

0,)

at

1,

because)

[VIII,

253)

LOGARITHM)

THE

\0372])

Furthermore) d 2 log

x

= _ \037

dx 2

We

the log

that

conclude

log(f(x)) is defined understoodwhenever in

Thus

when

other

words

we

only we

write

-

with

1/3,

slope

log(x-

=

2)

y

We

begin

-

We must find (5, log 3). This

log 3

= -i(x -

the is

f'(x)

only

- 2) at the

log(x

= 2x

+

of the line

equation

namely:)

easy,

5).)

of the functionf(x) the derivative, namely)

by taking

meaningful

and)

2),

the graph

Sketch

Example.

3.

log

through

passing

=

y

- 2 > 0,

= 1/3.)

f'(5)

When x = 5,

= 1/(x-

is

this

curve

line to the

when x

only

loge sin x), sin x < O.)

when

Then f'(x)

2).

defined

is

this

write

we

defined

tangent

point x = 5.

Let f(x) = log(x

- 2),

When

not

the

Find

Example.

defined

not

log(x

2.

x >

O. It is

when sin x >

shown.)

of the type consider composite functions for numbers < 0, the expression > o. This is to be for numbers x such that f(x) such an expression. write

Since the log is

log(f(x)).

as

down

is bending

function

sometimes

shall

We

Remark.

0




o.

1

-. x)

when is 2x = - llx, that f has a critical point precisely This can never be the case. Hence there is no critical point. in this interval, the func> 0, the derivative is positive. Hence

function

The

2X2 =

-1.

When

x

tion is When

strictly

increasing.

x becomes

large positive, both

x

2

and

log x become

large posi-

Hence)

tive.

if

As

0 from the

x approaches

large negative.

x -+ 00

right,

then

f(x)

x

2

-+

00.)

0, but

approaches

Hence)

if

x -+

0

and x > 0

then

f(x)

-+

-

00.)))

log x becomes

AND LOGARITHMS)

EXPONENTS

254)

the

second

regions where

the

determine

to

Finally,

take

and

derivative,

[VIII, 92])

f is bending up or down,

we

find)

- 1

2X2

1

,

f'(x)=2--= x2

2)

x

.

Then:

f\"(x)

> 0

-

0

is bending

up.

2X2

j'

1 >

x >

1/)2

x


for

that

limit

0,

have)

X

1 +

lim

x-

an interesting

has

This

Example. at

the

to the

increases

following

A + rA

=

1 year:

After

2

years:

(1 + r)A

After

3

years:

(1 +

in

this

ere

per cent of 100 per cent. of 100r

interest

After

Continuing

=

)

r)2A

after

the

(1 +

r)A.

+ r(l +

r(l

+ r)A

=

+ r)2 A

= (1

after

r)n

this

O.

Thus

original

invested

r is the amount

of years:)

(1 +

that

+

r >

r)2 A. r)3 A.)

(1 + =

Then

be

dollars

A

where

number

indicated

way, we conclude An

of

an amount

Let

interest.

interest of

rate

x)

(

application.)

Compound

compound

yearly

ratio of

00

\037

n years

the amount

is)

A.)

cent is compounded to is equivalent This where m is a positive integer. l/m years, every that the rate is 100rlm per cent per every compounded 11m years, saying formula to the case where the unit the preceding 11m years. Let us apply are equal to qm. 11m years. of time is 11m year. Then Therefore,))) q years Now

suppose

that

this

same

interest

rate

100r per

[VIII,

every l/m years, after

is compounded

interest

the

if

Aq.m

1 +

one gets after

The amount

should have

of

limit

the

is

tinuously

if the interest is compounded con00. In light of the limit which you that after q years of continuous com-

see

we

the amount

pounding,

is)

m\037oo

qm

-r

1 +

lim

e'q A.

=

A

m)

(

)

return case, suppose 1,000dollars = r Then After 10 years,the compounded continuously. 15/100.

To

a numerical

give

is)

years m -+

q as

Aq,m

determined,

the amount

A.)

m)

(

q years

qm

r

=

261)

FUNCTION)

EXPONENTIAL

GENERAL

THE

\0373])

15

per

amount

cent will

be)

..il.. 10

e

100

.1,000 =

Since

e is

approximately 2.7 you

VIII,

\0373.

EXERCISES)

1. What

is the

derivative

of

lOX?

What

is the

derivative

of

X 3 ?

2. 3.

the curves

Sketch

4. Sketch

y

y =

curves

the

= 3 x and x

2

can

7 n

y

X

numerical

a definite

get

?)

-

x. Plot

2 -x. Plot

at

least

five points.

at

least

five points.

5.

Find

the

equation

of the

tangent

line

to

the

curve

6.

Find

the

equation

of the

tangent

line

to

the

curve of

derivative

of

7. (a) What

8.

Find (a)

Find

is the

What

(b)

the

the

is

function

the

XX

y

=

lOX at y

of

derivative

the equation of

the

the function

tangent

point x = 1 (b) at x lines of the following tangent

at the

9. y

= xJ;

10.y

= x\037

3-X

It. If a is

(a) (a)

a

number

at

line = 2

(defined

curves:

x = 2

(b) at

x= 5

= 2

(b) at

x = 5

at x >

1 and

x > 0, xQ -

x(XX)?

to the curve y (c) at x = 3.

that)

show

I >

a(x

-

1).)))

x = O.

= n X at

= eX 10 1 X])

XX

answer.)

?)

= 3

y =

and

X

e1.5 1,OOO.)

= XX

for

x =

x >

2.

O)? [Hint:

AND LOGARITHMS)

EXPONENTS

262)

12. Let

a be a

number>

13. Let

0 < r.

Using

O.

Limit

points of

the critical

Find

3, prove

the

the

[VIII, 94]) function

f(x)

= x2/a x .

limit)

X

1 lim x-+ 00 (

Let

[Hint:

14. Show

x =

ry

and

let y

\037

+

\037

= er.

X)

)

00.]

that)

lim

n( \037

- 1) =

log

a.)

n-+oo)

of the log can how approximations 1/n.] This exercise shows k In fact, if we take n = 2 and roots. by taking ordinary n-th the of use large integers k, we obtain log by arbitrarily good approximations of square roots. Do it on a pocket calculator to a succession extracting Let

[Hint:

h =

be obtained

just

check it

VIII,

out.)

SOME

\0374.

APPLICATIONS)

(from experimental data) that when a piece of radium is left amount the rate of disintegration is proportional to the disintegrate, one is a constant of radium left. Two quantities are proportional when multiple of the other. and let f(t) be that at time t = 0 we have 10 grams of radium Suppose at time t. Then) the amount of radium left

It is known to

-df = Kf(t) dt)

some

for

Let

constant K. We take K negative since the physical the amount of substance decreases. that) that there is a constant C such show

that

is

tion

us

f(t)

If

we

take

the derivative

of

the

= Ce Kt .)

quotient)

f(t)

e Kt)))

interpreta-

[VIII,

of a

and use the rule for the derivative f(t)

\037

because

-

eKtf'(t)

e

)

or

t = O.

Let

10

In

= C.

Then f(O)

find)

= 0

IS

that)

CeKt .)

= 10,

C

Thus

Kt

f(t)/e

quotient

if

we

we started

that

assumed

grams.

general, as

stance

KeK'f(t) 2Kt)

Since the derivative is 0, the C such equivalently, there is a constant f(t) =

with

we

quotient,

= Kf(t).

f'(t)

constant,

=

( eKt

dt

263)

APPLICATIONS)

SOME

\037])

if

= Ce

f(t)

Kt

the

function

/(0)

= c,)

is

of time,

a function

of sub-

giving the amount

then)

I

interpreted as original amount.

and C is

the rate of the reaction stance present. If f(t)

when

of substance

amount

t

reaction. It is frequently to the quantity

a chemical

consider

Similarly,

the

of

case

that sub-

reacting

time

left after

of substance

amount

is the

0, that

the

is proportional

denotes the

=

t,

then)

-df =

Kf(t)

dt)

some

for

therefore

constant in

a

similar

K (determined experimentally in situation as before, and) f(t)

C is

where

Example

the amount 1.

f(3) = 5. Find

of substanceat f(t) =

10eKt

K

IS

Assume

constant.

5 =

10e K3)

3K =

TO -

therefore) _

1 2')

whence)

3K

= log(1/2))

and)

are

t = O.)

where

5

We

,)

K.

e

case).

Kt

have)

We

and

Suppose

= Ce

each

K=

2

-log

3

.)))

that

AND LOGARITHMS)

EXPONENTS

264)

2.

Example

amount

still

20

will

cent

per

Let

rate

of the sugar be decomposed? at amount of sugar undecomposed,

\037])

the

to

proportional

sugar reduceto 15lb

lb of

If 50

the

be

S(t)

decomposes at a

in water

Sugar

unchanged.

[VIII,

Then

t.

time

when

3 hr,

in

by

hypothesis,)

S(t) =

C =

50.

have

we

Thus)

= 50e

S(t)

We also

Furthermore, SInce S(O)= C,

C and k.

constants

suitable

for

Ce-kt,)

-kt.)

have)

= 50e- 3k

S(3)

= 15)

so) e we

Thus

can

- 3k -_

solve for k, namely we - 3k =

3

1 5 -_

50

TO.)

take

the

log

and

get)

cent

is left. Note

log(3/10),)

whence)

=

- k

per cent has decomposed then cent of 50 is 40. We want to find

per

other

in

80

per

t such

that

that)

kt

= 50e-

40 or

log(3/10).)

20

When 80

!

,)

words,)

e

- kt

-

40 -50

!

5.)

obtain)

We

-

kt

=

log( 4/5),)

whence) =

t

-

= 3 log(4/5) .

k

log(3/10))

is our answer.)

This

Remark. It does

not

make

S(t) = We

log( 4/5)

could

tions,

when

also

Ce-

any kt)

have worked the

substances

decrease,

whether

difference or)

S(t)

problem it is

the

we

originally

let)

= Ce Kt .) other

way.

For

convenient to use a

applica-

convention)))

[VIII,

that k >

such

0 so

that

But mathematically the Example

amount

APPLICATIONS)

SOME

\037])

the

procedures are

constant k. positive amount left? original

some the

of

To do

this,

we

want

At

we

want

to

that we can

-

k.)

to

the

say)

kt)

time

will there

value of

t

such

be exactly 1/4-th

that)

= C/4.)

solve)

CeNote

K =

putting

time,

what

to know the

t increases.

when

decreases

disintegrates proportionally

= Ce-

f(t)

Thus

- kt

equivalent,

substance 3. A radioactive of substance present at a given f(t)

for

e

expression

265)

cancel C

to

kt

=

e-

get

C/4.) kt =

= -log

-kt

logs yields)

1/4. Taking 4,)

whence)

log 4

t=T.)

Observe that the answer is independent of the original amount C. us to determine the constant k. For instance, if Experiments also allow we can analyze a sample, and determine that is left after 1000 1/4-th years, then we find that) log 4 k=

.

1000)

4. Exponential growth also reflects population is the population as a function of time t, then its rate to the total population, in other words,) proportional Example

P(t)

dP

- = KP(t) dt)

for

some

positive constant

K. It

then

P(t) = for

some

constant

C

which

is

the

follows

that)

CeKt)

population

at time

t

=

o.)))

explosion.

If

of increaseis

AND LOGARITHMS)

EXPONENTS

266)

then

or

the

time

what

must

We

double.

will

population

\0374])

that)

t such

find

at

ask

we

Suppose

[VIII,

Ce Kt

=

2C

e Kt

=

2.)

,)

equivalently)

the log

Taking

yields)

= log

Kt

2,)

whence)

t=-.log

2

K)

VIII,

f(t) =

3. One gram 0.1 gram

5.

10eKt

equal

to

the

of

substance

left

is the

What

chemical

certain

is

radium

of left.

popula-

know

you

that

f(I/2)

the

constant

= 2.

left.

a given

at

to

time

what

6. Suppose K of

third

will

=-

4

be exactly

there in

the

substance

the

rate

the

giving

million

years,

there

C. is

of disintegration?

proportionally

disintegrates time, say) f(t)

At

After one

disintegrate.

formula

Determine

reacts in such a way that the rate of reaction is of substance present. After one there are 20 hour, How much substance was there at the beginning?

substance

radioactive

f(2) = 5.

substance

quantity

stance present

7.

K. Suppose

you know

that

Suppose

grams A

the

C.)

constant

some

for

= Ce 2t .

2. - Let f(t)

A

of

K.)

Find

4.

of change

EXERCISES)

\0374.

1. Let

of

rate

the

on

only

on the original value

not

tion,

time depends

this

that

Note

preceding

to

the

amount

of sub-

= Ce Kt .)

half

the

exercise.

original At

what

amount time

left? will there

be one-

left?)

increase in number at a rate proportional to the number present, if it how long will it take before 1,000,000 bacteria increase to 10,000,000 takes 12 minutes to increase to 2,000,000? If bacteria

8. A

substance

the When

end

at a rate proportional to the amount present. At substance has decomposed. minutes, 10 per cent of the original half the original amount have decomposed?)))

of 3 will

decomposes

[VIII,

9. Let f be a

10. In

t and increasing at the rate t 1 is a fixed value of an = f(nt 1) where is a geometric progression.) of a variable

fun\037tion

k is a constant. that a o, ai'

Let

a 2 ,...

of a

the population

1900

of

of increase

the

was 50,000. In is proportional

city

population

In what year is

1984?

in

population

it

decomposes at a rate sugar reduces to be decomposed?

changed.

13.

the

sugar

A

particle

14.

with speed

moves

constant. 2 min,

30 lb of

If

If

value

the

find

initial

the

speed of t when

that the difference air decreases surrounding 100\302\260 when t = 0, and x = Assume

x =

when

(b)

given

any

is

that

takes

it

that

ft

what

long

in

=

ds/dt

10units/min.

rate equal to have lost half

of the

- ks, where k is some in speed is halved

In

this

eX grows

we

section

much

x becomes We

OF

large

consider

analyze

faster

than

of

If x

=

=

70\302\260,

of his 5568

half-life

of

amount

to the

the

he owns at

amount

the

initial

capital?

years, meaning to

amount

decompose.

present, so

formula)

for this amount, where C and K are constants. K explicitly. (a) Find the constant is found in a cave, and an (b) Some decomposedcarbon one-fifth of the original amount has decomposed. carbon been in the cave?)

ORDER

when x

t (a)

that

= CeKr)

f(t)

\0375.

and

body

difference.

original

Also, the rate of decomposition is proportional that we have seen in the text, we have by what

VIII,

cent of

if the

and

has a

carbon

un-

still

per

t = 20.

when

one-half

95

will

the temperature of a

at a gambling time t will he

amount

the

to

4 hr, when

to this proportional t = 40 minutes, find

40\302\260 when

for

1950 it was 100,000. If the rate to the population, what is the

proportional

10lb

units/min

rate

Show

to height of atmospheric prespressure there. If the barometer above sea level, find the barometric

the speed is

between a

t,

the

satisfying

value of x

radioactive

that

known

in

At

time.

x at

the

loses money

15. A moron

16. It

16\302\260, (c)

s(t) is 16

t 1 > O.

respect

water

in

Sugar

kf where

=

df/dt

200,000?

that the rate of change with at any height is proportional sure to reads 30 at sea level and 24 at 6000 sea level. reading 10,000 ft above

11. Assume

12.

267)

OF MAGNITUDE)

ORDER

\0375])

analysis How

shows that has the

long

MAGNITUDE)

more closely what we mean when we say that x than much slower x, and x, when log grows

positive.

the quotient) eX) x)))

AND LOGARITHMS)

EXPONENTS

268)

[VIII, 95])

as x becomeslarge positive. Both the numerator and the denominator become large, and the question is, what is the behavior of the quotient? values First let us make a table for simple 2n/n when n is a positive becomes integer, to see that 2n/n large as n becomes large, experimenthat n always denotes a positive intetally. We agree to the convention otherwise

unless

ger,

specified.)

2

n

n

2n/n

1

2

2

2

4

2

3

8

8/3

4

16

4

5

32

32/5

10

1,024

102.4 >

100

20

1,048,576

52,428.8

>

>

6

104)

5 x

e, we have 2n/n < en/n, and we see experimentally that to prove this fact. We first becomes We now wish prove large. for eX. We use techniques from the exercises of 91. We inequalities We consider x > o. ceed stepwise. 2
0 for x

increasing

for x

> 0,

means)

x) > 0,)

(1 +

which

or

o.)

x >

for

0)

in other

> 0.

words,)

show

that)

I +

12(X)

=

eX

>

eX

shown.

we

x +

x;




for

I

Let

Since11(0)

-

(1 + x

o. I)

+ x 2 /2). I\037(x)

Then = eX

12(0) =

- (1

+

O.

x).)))

Furthermore,)

=

1 +

0,

x,)

we con-

[VIII,

we know

By part (a), creasing, and

it

that

eX _

f2(X) >

+

+ x

the desired

proves

5.1.

Theorem

\037/x

of

both sides 1

becomes

large,

large

inequality

(b)

x

+-

for

> 0)

is

\0372

(1 This

for x > O. Hencef2 0 for x > 0, or in other

> 0

f\037(x)

that

follows

269)

MAGNITUDE)

OF

ORDER

\0375])

Theorem

and

side,

5.1

IS

proved.)

5.2.

Theorem More

generally,

The let

2

X

e /x

function

large as x

becomes

a positive integer.

m be

becomes large.

Then)

eX

-\037oo m)

x

as

\037

00.)

x We use the

Proof

that

by definition,

x

2!

3!)

x

x

= O. Furthermore,

f3(0)

f\037(x)

= eX

-

1 +x+

f3(X)

proves inequality If we

is strictly

. 2

= 6. X

inequality)

+ x+

=fz{x)

\037:

(b)

>

O.)

This

we

time

.)

we

find)

x >

for

0)

O.)

)

increasing, and

therefore

f3(X)

> 0

for x >

(c).

divide both sides

let)

3

2! +3! )

inequality

using

(1 Hence

x>

X2

-

( Then

the

prove

for

3! = 3

2! = 2 and

f3(x)=e

we

3

x2 l+x+-+-

for

eX)

O.)

=

-

eX

P n + 1 (x))

In(x) =

and)

eX

-

Pn(x)o)

= 0 and I\037+ l(X) = In(x) > 0 for x > O. Hence In and therefore In + 1 (X) > 0 for x > 0, as desired. our integer m, we have an inequality) given

1 (0)

increasing, Therefore,

1 +

We

x >

for

let)

In + 1 (x) Then

x

prove)

P n + 1 (x) To

n!o)

n!)

Pn(x) < then

-

that)

words,

We shall

+

n

+x+-+...+- 0)

for

eX)

stepwise

an

using

general

that)

4

x

+

becomes

5.2.

of Theorem

statement

first

so eX/x2

side becomes large,

the left-hand

large,

[VIII, 95])

sum

x +

2

- + ... 2

x

sides of

this

xm+l +




O.)))

the is)

+

1

is

strictly

o.)

left-hand

side

[VIII,

Since the left-hand sidebecomeslarge when 5.2 is proved.) right-hand side, and Theorem We sketch

Example.

the

= xeX

Since eX

>

0 for

+

becomes

x

= xe X .

of f(x)

graph

f'(x)

271)

OF MAGNITUDE)

ORDER

\0375])

eX =

eX(x +

large,

so does the

have)

We

1).)

all x, we get:)

f'(x) = 0

)

x+1=0

)

x

f'(x)

> 0

)

x+1>0

)

x>

f'(x)

< 0

)

x+1 0 0

1
0

(Remember

aX)

see that)

has exactly

function

f'(x)

function

.)

a +

log

aX(x log

= 0

f'(x)

Thus

X

the derivative:)

take

Since

the maximum

Find

number.

fixed

\0375])

function)

f(x) First,

[VIII,

of f at

value

The

maximum.

critical

this

to)

a) =

f( -ljlog

-

-

1 =

a-l/loga

-

a

log

-

1 e-Ioga/loga

log

a)

5x has

at

1

e log

a.)

Example. Show that the equation Let f(x) = 3 x - 5x. Then f(O) = value where f is negative, namely) f(2)

By

the

2 and

intermediate

0 such

that

f(x)

value theorem, = 0, and this

From Theorems 5.1 and analyze

what

happens

= 9

when

5.2,

by

x 3 =

1,

and

by

trial

one

least

and

error

solution.

we

a

find

- 10< o.) there existssome number number

means

comparing

fulfills

our

of a change of log x with powers

x

between

requirement.) we

variable,

of

x.)))

can

273)

ORDER OF MAGNITUDE)

95])

[VIII,

Theorem 5.3.

becomes

x

As

the

large,

becomes

also

x

x/log

quotient

large.)

Proof. Our strategy a change of tient has the form)

make

is

y = log

Let

x

log

that

We

know

by Theorem

log x becomes large when 5.1. This proves the theorem.)

becomeslarge, the

As x

5.1.

Theorem

= eY

our

and

We

quo-

y)

=

y

5.4.

Corollary

x. Then x

e Y)

x

eYj y

to

this statement

reduce

to

variables.

becomes

x

x

function

So does

large.

- log

x also becomes

large.)

write)

We

Proof.

x

- log x = x 1 - log X x (

factor

we

is

that

(log x)jx

x

in

approaches 0

the

- log x.

x

expression

')

)

as x becomeslarge. Hence

Theorem

By

the

factor)

large. Hence the

product

5.3,

log x

1 _

x)

approaches

large.

1. The factor x becomes This proves the corollary.)

Remark. used

have

We

in analyzing

just

used

the behavior

x 3 -2x

2

the same

of polynomials,

+5=x

2

to

see

the x 3

that

becomes

the behavior

determines

that

We

5.5.

As

x

becomes

large,

5 x

3)

)

of the polynomial when

xlix approaches 1

write) Xlix

=

e(logx)/x.)))

was

we wrote)

large.)

Corollary Proof.

term

x

technique

when

as

1--+-

3

(

x

factoring

becomes

as a

limit.)

By

AND LOGARITHMS)

EXPONENTS

274)

5.3 we know

Theorem

that

0

approaches

x)/x

(log

[VIII,

becomes

x

when

\0375])

Hence)

large.

e(logx)/x)

Remark.

the graph of

1. Sketch the

8. 11.

the

x)

the

eX I x)

9.

eX

functions.

Exercises

(In

4 .xe)

7.

-

10.

x)

except

some

6 through

8,

2 - x2

eX /X2)

eX + x)

n be a

17. Sketch

(a)

y

(c)

y

Show

19.Sketch Sketch

the

= x =

x log

has at least

x

when

x =

x)\"lx

one

for

solution

any

!, i,...,

!,

in

when

general

number

x =

--+

X)2

curve

the curve

x

0 as

(d)

y =

f(x)

=

f(x) XX

that

--+

O?

about

What

x2

log

for

= x- X

for

x)\"

--+

0 as

x -+ O.

> O.

xllog XX

x(1og

00.

= x 2 log

y

f(x) =

--+

for x

(b)

function

x

as

limit

become large.]

curves

following

that the

a

integer. Prove

log x

x(log

the

let y

and

positive

that (log

Prove

= ax

positive integer x log x approach

Let x = e Y

15. Let

eX

a

1/2\"

for

n.

Does

(b)

x.)

log

< e.

of

values

-

= x

equation

0 < a

when

14. (a) Give

20.

6.

of f(x)

graph

that the

13. Show

18.

approaches

e-x+x)

12. Sketch

16.

u

eO = 1.)

-x2

3 . xe)

3)

eX Ix

then

is

ap-

xe 2x . In this and other exercises, you may as optional, but it usually comes out easily.

the following

of

graphs

- x)

2e-

(log x)jx,

u

y =

curve

the

properties

convexity

Sketch x =/; 0.))

5. x

u =

if

Thus

If

EXERCISES)

\0375.

treat

e UO.

approaches

e\"

function

is continuous.

function

becomes large, so e\"

0 as x

approaches

2. xe

e\"

the

that

fact

the

used

differentiable

any

then

Uo

proaches

5.5 we

In Corollary because

continuous,

VIII,

desired.)

1, as

approaches

x x)

is strictly

x >

O.

x >

O.)))

increasing

for x >

lie.

x?

[Hint:

[VIII,

\0376J

21. Let

f(x)

22. Find (a)

the

(c) (n/e

VIII,

= 2 x x x . Show

n

The

[(log n)/n]1/n

)1/n

(d)

(n log

AS

is interesting

insight

for its It also

crete introduction to integration which We shall give an interpretation part.

a

AREA

THE 1/x)

CURVE

the logarithm.

into

> 1/2e.

n)1/n)

LOGARITHM

section

present

further

for x

increasing

275)

CURVE l/x

--. 00

n

(b)

UNDER THE

THE

UNDER

AREA

strictly

n)l/n

THE

\0376.

that f is

limits as

following

(log

AS THE

LOGARITHM

THE

is of

own

provides a to be

going the

it gives us and con-

because

sake,

nice

very

covered

in

logarithm

next

the

area under

as the

curve.

We a function L(x) to be the area define under the curve l/x between 1 and x if x > 1, and the negative of the area under the curve l/x between 1 and x if 0 < x < 1. In particular, L(I) = O.

shaded

The

between

of f(x) =

Graph

0
1. that

If

right. the

area.

l/x. log x.

The first assertion this chapter, and Theorem

we

would said

L(x) < 0 if 0 We shall prove:

1. L'(x) =

x >

we

have

Thus

2.

x

for

L(x)

follows left

Graph

l/x)

x)

1)

o
0 for all y. Thus E numbers. is the set of positive of L, which = that the graph of all for and is strictly y shows E\"(y) E(y) increasing, so L(x) Now

E bends

up.

have

We

we

Then

L(l) =

= 1 because

E(O)

can

O.

prove) + v)

E(u

= E(u)E(v). I)

I

L(a) and

u =

E(u) and b =

a =

let

Namely,

v

=

the

By

E(v).

of inverse

meaning

function,

Then:)

L(b).

L(ab) =

L(a) +

E(u)E(v)

= ab

=

L(b)

u +

v.)

Hence)

as

to be

was

= 1.

v),)

shown.)

e=

define

now

We

L(e)

= E(u +

From the

E(I). SinceE E(u

we now get for

any

inverse

the

is

function

rule)

E( 1 + 1 +

E(u)E(v))

n that)

integer

positive

E( n) =

+ v) =

... + 1)

=

E( l)n

=

en.)

Similarly,)

= E(u)n.)

E(nu)

Put

u

=

I/n.

Then)

e =

Hence E(l/n) is the

n-th

E( 1) = root

e

Next

we deal

with

the

U

general

n

E(

of e.

instead

.

= \037)

EG

r)

From now on we of

exponential

E(u).) function.)))

write)

of L we

have

SYSTEMATIC

ApP.])

[VIII,

a positive number,

a be

Let

281)

PROOF)

and

x any

aX =

eX loga.)

We

number.

define)

Thus)

a-vT =

If

we

put

u = x

U log a and use log e =

log

For

log

eft

aX =

a.)

u, we find

x log

the

formula)

a.)

instance,)

log 3-vT =

cases

instance

in of aX, we must be sure that the general definition when we have a preconceived idea of what aX should be, for x = n is a positive integer, then)

made

Having

those

when

en loga

For instance,

so

and

forth.

is the

take x = 2.

of a

product Then)

= e loga + loga = elogaeloga

e310ga

= e loga + loga + loga

For

any

log a = =

positive

elog

a

=

if n

is a

= a

. a,) = a

elogaelogaeloga

. a . a)

n we have)

integer

+ log a + ... + log a)

elogaeloga . . . e loga)

(prod uct

=a.a...a)

Therefore,

n times.)

itself

with

e210ga

en

itself

J2 log 3.)

positive integer,

taken

n

a en log means

times).)

the

product

of a

with

n times.

Similarly,) (e(l/n)loga)n

=

=

e(l/n)logae(l/n)loga e(l/n)loga+(l/n)loga+

= e loga) =

a.)))

...

e(l/n)loga)

... +(l/n)loga)

(product

taken

n times))

AND LOGARITHMS)

EXPONENTS

282)

the

Hence

e(

that

shows

This

e(1/n )loga

power of

n-th

1/n) loga

exloga is

we

so)

x is a

when

expect

ApP.])

of a.)

n-th root

is the

what

to a,

is equal

[VIII,

positive integer

or

a fraction.

other propertiesof

Next we prove

aO

definition,

Proof

By

For all

numbers

=

aO

We

= 1.)

y we have)

x,

with

start

1.)

eOloga = eO

aX + Y

Proof

=

=

aXaY.)

side to

the right-hand = exlogaeyloga

aXaY

=

get:)

yloga)

exloga+

= e(x + y)log =

This

X+

a

a)

y.)

the formula.)

proves

For all

First:)

aX.

function

the

numbers

x,

y,)

(aX)Y =

a XY.)

Proof)

(aX)Y =

eY log aX)

(because

=

eyxloga)

(becauselog aX

= a

th us

At ponential

proving this

XY)

the desired

point

we have

function

which

uY

=

(because at = special value

eylogu =

for u >

x log

etloga, with t

=

0)

a) the

= yx)) xy

property.)

recovered all were

used

in

five

of the

properties

\0371, \0372,

and

\0373.)))

general ex-

SYSTEMATIC

ApP.])

[VIII,

EXERCISE)

APPENDIX.

VIII,

Suppose you

not

did

know

are given a function

You

E

the

about

anything

exponential

and log

functions.

that)

such

= E(x))

E'(x)

283)

PROOF)

all numbers

for

E(O) =

and)

x,)

1.)

Prove:

(a)

=1= 0

E(x) this

product

be a

(b) Let f

C

constant

(c)

For

all

x. [Hint:

for all

is

constant.

function

such

numbers

Using

such

E(O) = 1,

that f'(x)

Fix

the

number

u and let

product

what

= f(x)

that f(x) = CE(x). u, v the function E E(u

[Hint:

the

Differentiate

for

that E(x)E( - x) to show constant?] x. Show that there exists a

is this all

satisfies)

+ v) =

f(x) =

E(u)E(v).) E(u

+ x).

Then apply

(b).J)))

Part

Integration)))

Three)

CHAPTER

IX)

Integration)

In

this

we

chapter,

or less

more

solve,

the

simultaneously,

following

problems:)

a

(1) Given

function

(2) Given a

of

under the curve

in this

Actually,

f(x)

f

IX,

by

\0371.

Let f(x)

THE

be a

Definition.

of

little

by horizontal

f

function

of our

solutions

compute effectively to the next chapter.

idea of functions,

Archimedes. It is to and the area under

INTEGRAL)

indefinite

F'(x) =

defined

over

interval.)

for f is

integral

f(x))

some

for

all

x in

area

to

us

rectangles.)

INDEFINITE

An

integration.

a definition of the to geometric intuition.)

ideas behind the allow

which

be postponed follow an

will

shall

we

function

the

the sum

we give the

are given

In carrying out (2)

approximate

which

f(x)

chapter,

data

specific

is > 0, give does not appeal

which

The techniques

two problems. when

=

y

is called

and

differentiation,

function

that)

= f(x).)

F'(x)

This is the inverse

F(x) such

a function

find

I(x),

a

function

the

interval.)))

F

such

that)

INTEGRATION)

288)

another

If G(x) is

F-

- G)'(x)= F'(x)-

(F

=

G'(x)

3.3 of

Corollary

by

Consequently,

of f, then is 0:) G

integral

difference

of the

derivative

the

indefinite

[IX, 91])

f(x)

Chapter

G'(x) =

f(x) also. Hence

- f(x)

=

constant

is a

there

V,

O.)

C such

that)

for all x

Example sin x + 5 is

1.

indefinite

An

also

for cos

integral

indefinite

cos x would

for

integral

indefinite

an

log x is an

2.

Example

logx -

interval.)

the

in

+ C)

= G(x)

F(x)

for

integral

be

SIn

x.

But

x.)

So is log x

l/x.

+ 10 or

n.)

indefinite chapter, we shall develop techniques for finding we observe that time we a formula Here, integrals. merely every prove for a derivative, it has an analogue for the integral. I t is customary of a function f by) to denote an indefinite integral In

next

the

In

this

second

expression

J

substitution practicality We

shall

information

Let

n

be

f(x) dx the

in

next

now

make we

of

-1.

we

xn

dx

=

If

n =

. 1)

- 1, then) dx

I

= log

is true

only

in

the

interval

x.

\037

I

(This

have)

xn+l n +

I

further

I)

x >

0.))))

study

the

confirmation

integrals,

derivatives.

about

Then

we

indefinite

some

obtained

n -#

get

It is

itself.

by

When

we shall

a table

have

integer,

meaningless

meaningful.

chapter,

notation.

an

is

which

of our which

I

the dx is

notation,

dx.)

f(x)

or)

If)

only the method for

full

of the

using the

[IX,

x >

In the interval

0

have)

also

we

XC

XC

=

dx

indefinite

following

are valid for all

integrals

= sin x,

cos x dx

sin x dx

f =

dx

1 eX,)

f 1+

f for

=

x.)

- cosx,

f eX

Finally,

1)

-1.

c #

number

any The

+ 1

c +

f

for

289)

INTEGRAL)

INDEFINITE

THE

\0371])

-1

< x
0, and if x understood that The may be positive or negative. differentiable and left a entiable at for the moment that Assume numbers a, x, x + h we conclude

h

reduces

This

our

between x and x

Let s be a in

this

small

f + (I

a

(I

_

f, I

a

\037

O.

then

(If

h
1, then)

-n+ 1

n =

u =

x

- a)-n+1 -

1

= U

-

1. Then the integral

du

f

n

if

n = 1.)

n

1,)

#=

we

have)

duo

f

- a,

(n

- 1)

\037

has = log

du

1

-n+l(x-a)n-1)

=

1 -.n-1 U)

Suppose

if

u-

because) U

Factoring

afterwards

describe

then

=

dx

a)-n

-n+)

this

be written)

can

to do it. In fact,

substitution

(x-a)-ndx=

case.

1

a))

by

this

-n+l(x-a)n-l

dx =

1. Then

in g(x)

integer

1

a)n)

quotient

of

denominator)

the

= log(xWe

a

g. We assume

these.)

to

and

dx =

This is an old story.

Suppose

in

that

cases,

special

(x-a)n

f (x

we consider

when

terms.)

lower

+

be reduced

number,

1 f

of the

that

assume

also

we

necessary,

We shall begin by discussing how the general case can part.

on the numerator is less

the quotient

and

x,

degree

the degree of f is less than we shall describe works only

d g(x) = x

First

1)

integrate

the

that

+

denominator.

we assume

on,

the

because

of the

degree now

2

the

u,)))

form)

=

dx, 1

we get)

[XI,

94])

and

hence)

1

- a).

log(x

integrals of

we consider

Next

2.

dx =

x-a)

f

Case

expressionslike) x+l

1 f

d x)

(x - 2)(x-

some

for

a l ,...

numbers, to

- al)\".(x -

d x,)

- 1)2(x_ 2) of

of the

terms

form)

an))

need not be distinct. under the integral as

which

,an

the expression

writing

(x

f

(x

amounts

or)

3)

consists of a product

the denominator

where

359)

PARTIAL FRACTIONS)

The procedure a

of

sum

terms,

as in Case 1.)

Example.

the

find

to

wish

We

integral)

1 f

To

do

this,

we

1

numbers

some

on the

expression

C

1

x-2 Thus

(x

- 2)(x

- 3) is

numerator We

want

tor

must

equal

C2

C2

+

the

,

x-3)

+

that

is

x-3)

we have denominator. +

3)

(x -

2)(x

C2(X -

c 2 (x

we

(C1 + c2 )x

We

Put

the

find)

- 2) 3))

- 2) = (ci + c2 )x

1/(x-

to solve.

and)

denominator,

to be equal to

to 1,

C2

+

which

for

- C1(x

common

3)

1

x-2

3))

a common

over

= c1(x -

the fraction be

and

1

right

C

C

- 2)(x-

(x with

3))

write)

to

want

dx.

(x - 2)(x-

2)(x

must

have)

- 3c 1

2c2

=

-

1.)))

3).

- 3c 1

2c2 .)

Thus the numera-

360)

OF INTEGRATION)

TECHNIQUES

Therefore it

to

suffices

the simultaneous equations)

solve

1 +

C

- 3c 1 for

Solving

1 and

C

C2

=

C2

gives 1

f (x

- 2)(x-

that

into (x

-

2))

original

dx

f (x

3))

+ log(x-

- 2)

3).)

1)2

in

their

other

hand, so quotient,

(x it

dx.)

1

2) that)

such

+

C2

+ (x - 1)2

necessary to

- 2)

c2

(x -

2

original quotient. with

terms

two

include

above as)

.

the

in

to one

only

. 2)

1))

only

appears

rise

gives

C3 -

appearing

denominators,

+

x

of the

denominator

the

it is

account,

C1

the

+

2)

-

1)2(x

C

x-I On

(x

x-I

- 1)2 appears in

(x

Hence)

1

x+l

x+l

To take this (x - 1) and

-1. dx

-

C1, C2 , C3

(x - 1)2(xNote

1.)

-log(x

(x -

numbers

find

to

=

the integral)

Find

want

0,)

-1 f

f We

=

C1 =

dx = 3)

C2

- 2c2

1 and

=

Example.

[XI, 94])

first power

In

the

term)

C3

x-2)

in the

partial fraction

end of

the

now

We

rela ti

decomposition.(The general

rule

is stated

at the

section.)

describe

how to

find

constants

the

C

1 , C2,

C3, satisfying

on)

x+l

(x - 1)2(x-

C 2))

1

+

x-I C1(x

-

C2

+ (x - 1)2 1)(x

- 2) +

x

C3 -

C2(X

2)

-

(x - 1)2(x-

2) + 2))))

c 3 (x

-

1)2

the

[XI,

Here we

on the

fraction

the

put

2).)

have)

We

c1(x - 1)(x - 2) + C2(X = (c1 + C 3 )X 2 + (-3c 1 +

= numerator =

x + 1

Thus

the constants

find

to

have to

C

C 2'

l'

simultaneous

the

solve

C3

C2

2c1 - 2c 2

=

linear

c 1, c2 ,

-

dx =

-

2)

\037)\037(\037

a theorem

illustrated

the

in

coefficientsc 1, but

numerically

dx

f x-=-\\

2c 3)x the

=

+ 2c1 -

desired

2c

+

c 3 .)

relation,

we

2

0,

2c 3 =

-

1,

c 3 = 1.)

+

in three unknowns,

One

finds

C

C2,

C3'....

=21)2

which

- 3,

=

1

if you

dx +

you

C2

=

- 2,

be

2).

the method to solve for the to

according

you will always The proof cannot

dx 2)

procedure to

follow the above

fractions

\037

Lx

+ 310g(x

- 1)+ 2 x-I)

of simpler

examples,

+

f (x

that

algebra

have

we

then

on the right-hand

Example.

can

We

able

be given at the level of this you or I make a mistake, we just solve have higher powers of some factor in the to use higher powers also in the simpler side.)

decompose)

x + 1

(x - I)3(xPutting

C2

1)2

in practice, unless in each case. If we

denominator, fractions

in

in terms

fraction

a

course,

C3

equations and c 3.

= -310g(x It is

-

-

c 3 (x

Hence)

3.

f (x

write

2) +

satisfying

+

- 3C + 1

This is a system of three solve to determine can

-

equations) C1

c3

denominator)

common

the

over

right

- 1)2(x-

(x

361)

FRACTIONS)

PARTIAL

\037])

the

the numerator

c 1 2))

side

right-hand with

x-I

x +

+

over a

I, we

C2

(x-I)

common

can solve for c 1,

C2,c3 ,

C 4 .)))

C4

C3

2 +

(x-I)

3+

denominator,

the

coefficients)

x-2) and

equating

362)

OF INTEGRATION)

TECHNIQUES

Second

3. We

Case

factors

Quadratic

part.

the integral)

find

to

want

n

is a

x.

is arctan twist

slight

natural

the

us do the

1 dx f (x + 1)\

--

In

when

2

the integral new, positive integer. If n = 1, there is nothing n > 1 we shall use integration by There will be a parts. on the usual procedure, because if we integrate 1,. by parts in that the exponent n increases by one unit. Let way, we find If

case n

=

so we

an example,

1 as

1

11 =

x2

f

with)

start

dx.

+

1)

Let) 1

u=

x

2

=(x

2

1)

+

+1)-1,

2

ex

1)2

+)

=

dv

-2x

du =

v =

dx,

dx,)

x.)

Then)

I1 =

=

(*))

In the

last integral on X2 2

f

(x

+

dX = 1)2

x x

2

x

2

x2

1

+

x

the

1

+

1

we

now

substitute

1

dX

11 =

in

expression x

X

2

+)))

1

1)2)

X2

x

write

-

dx

+

2

f (x

f (x2 + 1)2

this

2

f (x

+ 2

right,

+

_2X2

-

dx.

+

1)2)

2= 2 x

+ 1-

1

=

f =

If

\0374])

denominator)

the

in

[XI,

x2 + 1

dx -

- 12 .)

arctan

x

(*) we

obtain)

+ 2 arctan x

- 21 . 2

1.

Then)

1 dx)

f (x2 + 1)2

[XI,

Therefore

we can

solve for

x

2

2

+

arctan x

- 11

+ 2

arctan x

-

x.

arctan

+

2

1)

value for 12 :)

2 yields the

1

1

dx = 2 x2 (x + 1)2

The

In -

works

method

same

in

1

x +

2

f

x

arctan

1)

x + by

find)

we

x

=

Dividing

and

+ 2 1

+)

X

x

of 11,

terms

1 2 in

x

21 2 =

363)

FRACTIONS)

PARTIAL

\0374])

2

1

+

We

general.

arctan

want

x.)

to reduce

to

In

finding

1 , where)

=

1n-l

f (x

1

2

+\\r-

dx.)

Let) 1

u=

2

(x

+ l)n

-

dv =

and)

1)

dx.)

Then)

2x

du=-(n-l)

(x

and)

dx

2

+)

v

=

x.)

l)n

Thus)

We

write

1n-l =

or

x 2 = x2

x

=

1n-l

+

2 (X + 1

-

1. We

obtain)

- 1)

2

x

(x

in other

2

+

l)n-l

l)n-l

+ 2(n

+ 2(n

-

1 f

(x

+

X2

1)

f (x2 +

dx

-

dx.)

It

-

2(n

l)n-l

1 1)

f (x

words:)

1n-l =

x 2

(x

+

+ 2(n l)n-l

- 1)1n-l -

2(n

-

l)In.)))

2

+

dx) l)n

364)

OF INTEGRATION)

TECHNIQUES

[XI,

94J)

Therefore)

2(n

- 1)ln

x

=

2

(x

+ (2n -

+ 1)n-l

3)1n-l')

whence)

1

(x

f

2

+

1

dx =

- 1)(x

2( n

1)\"

x 2

-

(2n

+

or

the abbreviation

using

denominator

2 2n _ 2(x +

we reach

until

If

you

to

want

find

plete formula for

say to

Eliminating Sometimes

not

method

the

remember

finding

it

13. To

course you should cases,

1 3 , use

to reduce

again

2

dx =

by

2n

-

2n

_

lowers

3

2

In-I')

the exponent

case, we

arctan

n

in

the

that)

know

x.)

to reduce it is arctan x.

which

11' a

get

dx,)

1

the formula

memorize

+

In that

1.

\037

to

f (x + l)n-l

which

n =

f x

formula

1)n-l

a recursion formula

This gives us

1)

x

1

=

In

1 2

find:)

we

1\",

1)

1)\"

3)

-

2(n

+

formula

complete the

which

above obtained

is

it

for

1 2 , then

to This

gives

use

th\037

a com-

n steps. Of you should only to apply it to special 1\"

takes

formula;

1 3 , 14.)

extra constants

by

substitution)

variation of the one just is a slight For instance, if b is a number, find)

we meet an integral which constant. an extra

considered,with

f (x

2

:

b

2

dx.)))

)n

[XI,

the

Using

x =

substitution

= b dz reduces the

bz, dx

=

b dz

2 2 2 f (b z \037 b )n

to)

+ It 1

1

dz.)

- 1

b 2n

integral

dz)

2n f b (}

=

f (Z2 + It

have)

We

1 2

(z

f two

the

because

a

to use

how b

365)

FRACTIONS)

PARTIAL

\037])

to

+

integrals

the integral

b =

when

Case 4. Find

the

dx

+

l)n)

a change

by

only

the computation

reduce

to

2

f (x

l)n

differ

substitution

1

dz =

of letters. This shows of the integral with

above.)

1 treated

integral)

f (x

This is an old story.

We

make

u = x2

+

2

:

b

dx.) t

substitution)

the b

2

2)

du =

and)

2x

dx.)

Then)

f

which

how to

know

we

(x

b2t

r)

Example.

\302\267

2( -

un

f

we

thus

2

+

n

du,)

find)

2

b )

if

n = 1,)

if

n i= 1.)

I

I

1) {x

+

2

-

2 + b )n

1)

Find)

5x f (x2

We

2

! log(x

x

f

=

dx

evaluate, and

2 dx = (x2 + b r

l

1

x

2+

-

3

dx.) + 5)2

write)

5x f (x

2

+

3

5)2

x

dx = 5 f (x

2

+ 5)2

dx

-

1 3

2

f

(x

dx. +

5)2)))

366)

OF INTEGRATION)

TECHNIQUES

[XI,

\0374])

Then:)

x

5

(x

f

2+

2x

5

=

2

f (x + 5)2

2

5)2

second

the

For

dx

on the

integral x =

dx =

J5t)

2f

we

right,

5 u

1

5

du = 2

u

may

-1

2

1

5

=-

2

x

.)

2

+ 5

put) =

dx

and)

-1

dt.)

J5

Then:)

1 2 f (x

have

we

and

+ 5)2

2

-

(x

f

Third

+

part.

3

\037

The general

t =

using

2

numbers

a,

In factor

x

x

2.)

Case has

1, we degree

In Case 2, we variables, we can

1)2)

2

2

-

t})

J5, we

find:)

xlJ5

1

1

x

+ arctan

2x + 5

have factored the

2

+ bx + c, then be written

thus

can

or)

(x

cases

arise.

6 =

-

x

type

fJ))

X

-

of

Two

{3.

xl

+ arctan

.)

J5

)

f(x)/g(x))

quotient

- a)(x -

Case 1.) Case

f (t +

25 2 ( (X/J5)2 +

5

+

are given a polynomial complete the square. The polynomial

suitable

dt

2

25

: 1

- 3 J5

If you

with

2

C

1

2x

(x

1

-J5

=

dt

5)2

\037

1)2

5

-

5)2

5

J5

dt =

2

together,

dx =

2+

computed)

previously

everything 5x

(5t

f

f (t Putting

1

dx =

(x +

= (x -

- a)2

fJ2)

+

3).)

2

2

.)

and each

two factors,

into

polynomial

or

form)

the

For example:)

2)(x1)2

+

factor

you in

1. not

have it

turn

into

the factored an expression

=

2t)

- 1)2 +

22

x-I

polynomial. By (2 + 1. Namely,

x =

so)

2t +

1.)

Then)

(x

=

22t

2

+

2

2

=

2

22(t

+

1).)))

a

change let)

of

[XI,

We made the We

change of

The

be a

Let g(x) can

(x n, m

being integers

This

can

as a

-

and)

x

2

+

2x +

+

f3)2

1'2],\",)

factor.)

explicitly,

but

the

in

the

exercises,

is easy.)

square, we

write)

(x + 1)2+ 2 = (x +

+

1)2

(J2)2.)

integral:)

1

x2

I

Let

x + 1

=

J2t

and dx = J2

1 I

x

2

+ 2x

+ 3

+ 2x +

1

=

dx

d x) 3

Then)

dt.

I (x +

dx =

1)2+ (J2)2

1

-I

2I

t

= J2

2

+

1

J2) 2 dt

arctan t

2)

= Let

Example.

us

x I We

x+l

J2

.)

arctan

2

J2

find)

dx.

2x +

+

Xl

3)

write)

x 2 I x

+

2x +

3

dx =

--

! 2

2x

Ix

2

x

2

+

2

- 2

2x +

I

+

2x +

dx

3)

2x + 2

! 2

+

g(x)

type)

of

-

[(x

do

to it

3=

evaluate the

We can then

factor.

as coefficients. Then

terms

> 0, and some constant

By completing the

Example.

the

but

numbers

product of

rx)\

be quite difficult is fixed up so that

situation

proved,

real

with

polynomial

written

be

always

as a

any further. is long, and proof

polynomial

course.)

this

in

the

factor

can be

result

general

following

cannot be given

2 2 would come out

that

so

variables

2, we cannot

in Case

that

note

367)

FRACTIONS)

PARTIAL

\0374])

3

dx -

1

Ix

2

+

2x +

3

dx .)))

368)

OF INTEGRATION)

TECHNIQUES

[XI,

\0374J)

Then:)

2x +

1

2f x

2

this

Putting

f

2x +

x +

Find

Example.

dx =

2

t log(x

3

the

2f

1

=

du

2

\037

the previous

with

x

1

1

dx =

3

2x +

+

together

2

2

3)

(x

f

can

find numbers

2x

(x

2

+

C1 ,

2

+

1)2(x

C 2 ,...

such

+ 5

+C

C1

1)2(x -

2

x

3))

2 X

2X

find:)

J2 - _

X

(

vi

h.2)

)

1

+

1

3)

that the quotient is C4

C3+

+

(x

+ C2

2

+

X

x

+

you can always

that

get

such a

right,

which

to)

-

3)

1

+ 1

C3

+ Cs (x2 + 1)2

algebra

x

1)2

X 2

equal

Cs

+

x

+ C4

theorem of

1

+

arctan

2

d x.)

_

1

+

C

-

It is a

3).)

integral)

2x + 5

We

2 + 2x +

example, we

2x +

+

log (x

x

(x

2

+)

1)

2

1 - 3.)

solve for the constants

C 1,

the fraction into original the partial fraction decomposition. 2 x to the term with Observe that corresponding + 1 you need several x in the numerathose with an terms on the right-hand side, especially tor. If you do not include these, then you would get an incomplete the conwould not work out. You could not which formula, compute stants. side of the We put the right-hand We now compute the constants. over the common denominator) decomposition

C2,

C3

, C4 , C s

the sum

on

the

to

decomposition of is called

(x

2

+

1)2(x

-

3).)))

[XI,

The numerator is

equal

-

+ c4 x(x We

and

be

can

but we

of five It is tedious to

solved.

the

down

C

-3c the

For

integral,

f

(x +

C s = 0)

+

C3

-

3c 3)

we

1)2 (x -

- 3c 4)

C

=

0)

=2)

Cs =

+

5)

x +

l arctan

we

left

the

desired

which

how

to

find

There is a

Example.

x 2

(x

4 +

+ 2x 2)3(X

1

of x

(coefficient

of x

(coefficient

of x),)

(coefficient

of

3

),)

2 ),)

1).)

log(x

+ 1)

tedious to

compute

-

dx

1

1

i C4

x + 2

that of

is just

+ 1)

- 3).

C s log(x

Case 3, so

we

integral.)

fraction

decomposition.)

Cl +C 2 X 2 (x + 2)

+

It would be

2

standing

partial

- 5)2)

(x + 1) 2

f

integral

(coefficient

2

tC2

1

shown

of x

3))

+ C3

The

4 ),)

(coefficient

obtain:)

then

=

dx

+ 2c s

C4

+

2x + 5 2

in five unknowns, which we leave it as an exercise,

=0)

C2)

l)

3))

equations:)

+

- 3c 2

l

-

respective con-

x and the

equations here and

do it

- 3) + c3 (x

1)2.)

- 3c2)

l +

-3c

+ 1)(x

3 2 x , x ,

linear

C2)

Cl

+

of x4 ,

a system

get

write

cs(x2

3) +

2

C 2 X(X

+

3)

the coefficients

equate

stants,

to)

+ 1)(x -

2 + 5 = C1(X

2x

369)

FRACTIONS)

PARTIAL

\037])

C7

xthe

+

2

(x

+

cg

+ 5

C4

C3+

(x

constants,

-

X

2)2

+

C S +C6

(x

2

+

X 2)3)

2. 5)) and

we don't

do

it.)))

have

370)

OF INTEGRATION)

TECHNIQUES

The

with

f(x)/g(x)

possibleinto

degree

m being

< degree of

factor

We

g.

quotient

g as far

as

like)

terms

-

(x n,

of f

Suppose we have a

follows:

as

is

rule

general

[XI, 94])

> O.

integers

-f(x)

and)

rx)n)

=

y2]m,)

Then)

.

of the

of terms

sum

- {3)2+

[(x

type:

followIng

g(x))

C

1

-

x

C2

+ rx

-

(x

Cn

+...+

2

(x

IX)

-

rx)n)

+

suitable constants

with

Once us

allow

the

A rational

\0374.

(x - l)(x + 7)

(a)

f (x

_

dx 3;(X

(b)

y2]m)

,e1, e2 ,....) Cases 1, 2, and 3 the integral involves

then

above,

that

find

(x + l)(x

(x + l)(x

Write

out

2 \037

dx W

dx)

+

x 2)\037

(c)

+ 1)

f

X+2

+ 2)(x +

+ 2)2 in

f (x

f (x

+ 2)

x

9.

2.

dx

x

f

+

integrals.)

2x-3

7.

emx

{3)2

EXERCISES)

1.

f

as then

We

+

function

the following

4.

,d 1, d 2 ,...

written

[(x

y2

-

type:

following

Find

3.

+

{3)2

dm

+...+

t terms.)

Arctangen

f

e1x

terms

Log

XI,

(x -

1 , C 2 ,...

f(x)/g(x) is each term.

the quotient to integrate

functions of

C

d 1 +

full

dx 3)

5. x

f

2

f (x the

f (x

x)

8.

dx

2x - 3 - 1)(x -

integral)

f (x

2

6.

dx

+

dx.)))

: 1)2

x 2 \037 1

dx 2))

:1

)2

dx

dx)

10. Either

formula

(a)

2

the following

Find

2

(x

f

13.

- 3

2x

11.

+

(b)

: 1)3

f (x

2

1)4

integrals.)

1)2 14.

in the

constants

the

2x

2

+ 5

prove

x

f

2

+ b

=

2 dx

f

18. (a)

f

x2

4 x-I x3

\037

- 2x

(b)

dx

(b)

1

3

+) b

2

\0375.

2

c4x

+

+

1)2

c5 X

-

3)

X b')

dx

= -

x+a

arctan

b

. b

1 into

irreducible

factors.)

dx

dx)

+ 1)

+1 x

dx

SUBSTITUTIONS)

EXPONENTIAL

several purposes.

section has we

First,

4\037 1 fx 2

c3 +

(x

text:)

arctan

- 1 and x 4 -

f x(x

in the

- 2

x-I)

This

x

dx

3

f

XI,

factor

2

1

17. (a)

19.

a)

next problems,

the

For

f (x +

1)

example

1

b

1

1 (b))

dx +

formulas:)

two

1 (a))

+

x2 + 1

3))

the

the

+ c2 x

C1

+ 1)2(x-

substitution,

Using

expression from

dx

+ l);x 2

f (x

(x 16.

x+1 2 f (x + 9)2

12.

dx

dx

Find

the

dx.

:

2 f (x : 16)2

15.

into

plugging

integrals:

full

in

dx

f (x

or by repeatedly the following

the integration by parts the text, write out in

doing

by

general

371)

SUBSTITUTIONS)

EXPONENTIAL

95])

[XI,

expand

our

techniques

of integration,

using

by

the

exponen-

tial function.

Second, rithm

for

in

having

a

this

gives

new

context,

used

them.)))

practice

which

in the will

exponential

make

you

learn

function

and

the

these functions

loga-

better

TECHNIQUES OF

372)

two new

introduce

shall

we

Third,

[XI,

\0375])

functions)

X)

e-

eX +

INTEGRATION)

- e-

eX

X)

and) 2)

2)

In

next

the

which

rings.

give the

a

cable,

hanging

to

used

be

also

some physical

or find

Here we just use

curves.

various

of

length

applied to

functions

these

see

will

the equation of Such functions will

describe

two

between

you

chapter to

situations,

a soap the them

film

integrals system-

to find integrals.

atically

We start

us

Let

Example.

a simple substitution.)

to make

how

showing

by

find)

J 1

I =

dx.)

eX

f We

u =

put

= eX dx

du

eX,

so

I =

f Now Then

1

-

u = 1

-

put

u v

= 2

v

2

dv

Then)

duo)

u

\037

to

rid of the

get

1

-- 2

-

v

V2

v

f =

dv

2

2v

1 + 1

2

last

integral

can

-)

f

+

2

v

21-)

dv

dv

1

+

f

This

(-2v)dv=2 f

2[f =

v

2

dv \037

1

]

(v+ 1;(V-1))

be integrated

by

partial

dv.

fractions,

answer.

We

have

learned

sign.

V2

v

1=

f

square root

obtain)

we

and

J 1-

= 2v

-du

and

= duju.

dx

that

how to

integrate expressionsinvolving) J l

- x 2 .)))

to give the

final

[XI,

x = sin

We substitute

sign into

a

the

make

to

()

But

square.

perfect

373)

SUBSTITUTIONS)

EXPONENTIAL

\0375])

under the square root

expression

to deal

have

we

if

what

an

with

integral

like) 2 J l+x

f need

We

the expression under the makes two possible types of There are perfect square. which we can use. First, let us try to substitute x = tan () to the square root. We find)

square root sign functions

of

rid

get

a

to make

which

substitution

a

into

an

over

cos)

where cos () is positive. so nice. Even worse,)

interval

is not

dx =

1

see0 =

2 = J l + tan 0

cosine

dx?)

sec2

0 a negative

Already

power

of

() d(),)

so)

J l +

x2

see 3 0 dO =

=

dx

f be done,

can

which

Here we give sign. We need

not

but

a better

pleasantly,

so we

of

rid

way

Such

are

functions

we

you

the

il (t)2

found

easily

dO,)

0

don't do it. of

the

horrible

11(t) and 12(t)such

square root that)

= 12(t)2.) the exponential

using

by

e'.

function

let)

= il (t)

If

getting

pair of functions

a better

1 +

Namely,

3 f eo:

f

e

t

- e- t

and)

2)

f2(t)

out, you will find immediately multiply relation. These functions have

desired

hyperbolic sine and hyperbolic (sinh is pronounced cinch,

and

cosine, while

cosh

=

e -,

et

.)

+2 that

these functions

a name: they

are denoted

by

called

sinh

and

t =

e

t

- e- t 2)

and)

cosht

=

the cosh.

cosh.) Thus

is pronounced

define)

sinh

satisfy

are

e' + e

-, .)))

2

we

374)

OF INTEGRATION)

TECHNIQUES

the manipulation

out

Carry

-

cosh 2 t

d cosh

t

2

sinh

rules for

standard

the

Furthermore,

t =

1.)

d sinh t h

SIn

and)

t)

dt

dt

except for

similar to those for the ordinary sine and cosine, reversals of sign. They allow us to treat some cases of could not be done before, and in particular get rid of

some

integrals which square root signs as

follows.)

Find)

Example.

2 J 1+ x

I =

f make

We

the

1 +

dx.)

substitution)

x =

Then

= cosh t.)

are very

formulas

These

that)

show

differentiation

.

=

95J)

that)

shows

which

[XI,

sinh

2

t =

sinh

cosh 2 t,

I=

dx =

and)

t)

so

2 J l + x

that

e

et +

-_

t

t cosh

cosh

f

-t

e

t

=

(e

+ 2

dt.)

2 J cosh

=

t =

cosh t.

Hence)

-t dt)

2

21

t

dt)

+ e

2

f

cosh

+e

21)

dt)

\037

f

e

=1 4

The

answer we

is, of need

course, to

+ 2t

( 2

in

given

- 2' - e

terms

inverse

the

x, study arcsinh (hyperbolicarcsine), and then

2t

we

t =

may

arcsinh

2 )

.)

of t. If we function,

write)

x.)))

want

which

it

in terms

we

may

of call

[XI,

At first

cosine,

tion.

we

We

that

those we

here,

x =

If

give a

can

sinh

t =

then

t

x= u =

arcsine and as

log(x +

of

that

a

and

sine

inverse

the

funcis

It

rccosine.

follows.)

Jx

2

+

1).)

We have)

Proof

Let

for

formula

formula

I

to

similar

situation

a

in

give explicitly a inverse functions

not

could

called

just

remarkable

we are

that

seems

it

when

375)

SUBSTITUTIONS)

EXPONENTIAL

\0375])

t

t (e -

e

t).)

et. Then)

x=\037(u-\037}) We

this

multiply

by 2u

equation

u

We

can

then

solve for

of x

+

2x

equation)

- 1 = o.)

- 2ux

terms

in

u

2

and get the

J

u=

by

the

4X2

+

formula,

quadratic

and

get)

4

2)

so) =

u

But

u =

ha ve the

et

>

0 for

Since

all t.

minus sign

e

Now

we

take

the log

to

t =

proves

the desired

2 J x

+

+

1.)

1 >

x, it follows

Hence

u =

x+

log(x

+

Jx

2

that

we

cannot

here

an

explicit

finally)

+

1.)

find)

t =

This

2 J x

relation.

this

in

x +

2 J x

+

1).)

formula.)

Thus unlike the case of sine and formula for the inverse function of the

cosine, we hyperbolic

get sine.)))

376)

OF INTEGRATION)

TECHNIQUES

now

we

If

found

above,

substitute we get the

J 1+

x2

t

=

+

in the

1)

2 J x

(x +

\037

integral

+ Jx

2 Iog(x

1)2 +

+

2

+

1)

D

f

J

-\037(x+

We

indefinite

\0375])

explicit answer:)

=

dx

2 J x

+

log(x

[XI,

to

want

also

may

Let B >

Example.

definite

a

find

x2+1)-21)

integral.)

Find)

O.

J 1+

x2

dx.)

f:

B

substitute

We

indefinite

the

In

we substitute

integral,

B

f

0

2

J l +x

1

e

2f

IOg(B+JB2+1>

e-2f

-+2t--

dx=-

422 [ =

J

+

![!(B

0)

210g(B+ JB2 + 1)

J B2 + 1)2+

because when

0 for

substitute

we

In

cases the

on

For

t

given

>

when you have to assertion.) following

0,

the function

x =

is proved

and 6, them

the

in

expression

1)-2]) brackets

we

for cosh, you

can

in

use

an

cosh t

function

inverse

an

has

inverse

function,

which

IS

by)

t =

This

t

J B2 +

+

1(B

O.

find

rely

sub-

0, and

find:)

to

tract,

which

before

log(x

+

just like the similar are

actually

looking

better for doing

so.)))

up

worked

the answer

2 J x

-

1).)

statement out

in

for

section,

sinh.

answer

the you

will

Do

learn

5 But do

Exercises

section. the

subject

[XI,

Remark.

Integrals

like)

x3

J 1+

4 J 1+ x

and)

dx)

complicated, and

more

much

elementary functions

XI,

by

means

of the

the integrals.)

Jl

1.

f

+

f

eX

X)

+ e-

Sketch

(c)

For

4.

= t(e X - e- X) = sinh x = y. that f is strictly increasing

Let x =

arcsinh

graph of f. y be the inverse

which numbers

(d) Let g(y)

\037eX

1

dx

the

dx)

f 1

J e

x

f

Let f(x) (a) Show

(b)

2.

dx)

eX

1

3.

= arcsinh

Show

y defined?

that)

1 .

J was shown

in

the

x=

text that

=

g(y)

6. Letf(x) = t(e X + e- X) = cosh x = y. (a) Show thatfis strictly increasing the

Then function

Sketch

(c)

For

(d)

Let g(y)

which

= arccoshy.

+

log(y

1 + y)

J

for

arccosh y

this

that x

explicit

This is simply form

ula

=

g(

y) =

expression another

than for

for

way

sine and the arcsine

1).

interval.

Denote

this Inverse

defined?

1 .

J Show

+

that)

Show

g'(y) =

(e)

y2

2

x > O.

for

exists

inverse function = arccosh y. by x

the graph of f. numbers y is

(b)

1)

function.

is arcsinh

y

y.

dx +

x.

for all

g'(y) =

It

found

be

cannot

course.)

this

of

EXERCISES)

\0375.

Find

5.

dx)

f

f are

377)

SUBSTITUTIONS)

EXPONENTIAL

\0375])

-

Y

2

-

1)

an 1). Thus you can actually give function in terms of the logarithm. in which the hyperbolic functions behave more because not give an explicit cosine, w\037 could

log(

y

+

J

y2

this

inverse

and

arccosine.)))

378)

the following

Find

X2 2

+ 4

X2

+

f

J x

f

x - J x2

9.

dx

the

For the

f and the

x-axis

the

between

12.

first quadrant the

of

graph

the area

Find

x =

between

the

13. Let

first quadrant,

a be a

positive

number,

dx 2

y = a

[This

is the

\0373 of

the

14. Verify

differential

l+

=! a

J

of the

equation

B >

1.

1)

= B. cosh(x/a).

2

d y

dx)

hyperbola)

_ x2 =

let

and

1

1)

0 and x

x =

between

-

II, 99.

and the

x-axis

the

between

1

x = B, with

see Chapter

hyperbola,

+

hyperbola)

=

1 and

y2 in

2 J x

10.

1)

2 x _ y2 in

dx

2

Jx

1

the area

Find

d

(

y

dx)

Show

that)

2. )

hanging cable.

Seethe

appendix

next chapter.] for any

that

2 J a

f

\0375])

1

8. f

+

[XI,

integrals.)

dx

7.

11.

OF INTEGRATION)

TECHNIQUES

number a >

+ x

2

dx

= ![x

0 we

2 J a

have)

+ x

2

+ a

2

1og(x

+

2 J a

2

+ x )].)))

after

XII)

CHAPTER

of

Applications

of these.

scription that

the

you

will

find

centuries, and leave objects, among which reasons

practical

by

of

the

of

There are

can be

mathematics

other

hand,

which we come into

contact

we simply state know it are those which of the past two journals reasons for studying these

(some people like them),

and

applied). the empirical world

The empirical world through

de-

all-encompassing

in describing

consists

structures.

mathematical

we

many

reasons

certain objects and

a definition,

mathematical

aesthetic

are

an

give

such

mathematics as

the

at that.

it

(some

Physics, on means

impossible to

Hence,instead

of study of described in

objects

and describing

discovering

is essentially

It

structures.

with

in

consists

Mathematics

Integration)

our

senses,

the

is

world

experi-

through

to makes a good physicist is the ability ments, measurements, etc. What and the ones mathematical structures choose, among many objects, I should which can be used to describethe empirical world. of course the above assertion in two the First, immediately qualify ways: description of physical situations structures can only be done by mathematical within the degree of accuracy provided by the experimental apparatus. should certain aesthetic criteria (simplicity, Second, the description satisfy After of all results of all experiments all, a complete listing elegance). a of the but is quite a distinct is world, performed description physical will from at stroke a which one thing giving single general principle account simultaneously for the results of all these experiments. For psychological reasons, it is impossible (for most people) to learn certain

notion,

we

two

Hence

in

this

without book,

seeing before

first

a

geometric

introducing

introduce one of its geometric or two, however, should not be confused. as shown on the following columns, page.))) frequently

These

tations. make

theories

mathematical

interpretation.

or physical

a mathematical physical

Thus

interpre-

we

might

as the

far

As

column

made up umn, other

the

needs

it than

satisfaction

aesthetic

pure

is used,

column motivate the first (because in a way that to understand something To provide applications for the second).

such

in

(granting

that

Mathematics)

Physics and

N urn bers)

Points

Derivative)

Slope

of

like

you

our the

however, brain is col-

first

first

column, the subject).)

geometry)

line)

curve

of a

Rate -df = Kf(x)

on a

we could

concerned,

column

second

The

entirely.

[XII])

To

purposes:

many

of our courseis

logical development

omit the second for

OF INTEGRATION)

APPLICATIONS

380)

change)

decay)

Exponential

dx)

Length

Integral)

Area Volume

Work)

to keep

it IS important

Nevertheless,

the

limit)

. 1 1m

mind

- f(x)

+ h)

f(x

in

the

that

as

derivative,

'

h

h-O)

are as a unique number between upper and lower sums, our a slope or an area, respectively. It is simply or geometric which interprets the mathematical notion in physical mind to the several such terms. Besides, we frequently interpretations assign notion same mathematical (viz. the integral being interpreted as an area, or as the work done by a force). which are about physics and the above remarks And by the way, and

the

integral,

not to be

confused with

mathematics

nor to

to physics

neither

belong

mathematics.

They

belong

to philosophy.)

Experience shows formula

Taylor's

of integration associated

given in with Taylor's

basic applications like polar

coordinates

about

the

or

physical ing is that

others,

for

that

one

in

cannot which

term, the

book,

as

lacking to cover well as to cover

formula and of

length

with As the

an

curve,

be omitted. deal

(work). except for doing concepts

which deals

a course

time is

estimate

volume

One

then

with

integration

the

all the

and

applications computations

The of its remainder. of revolution, area in has to make a choice

concepts (area of revolution) in the foreword, my already

geometric stated section

on

work,

if

time

is lacking,

feel-

it

is)))

[XII,

best to omit other the computations

XII,

from

of

plenty

to handle

time

formula.)

Taylor's

REVOLUTION)

OF

VOLUMES

\0371.

to have

in order

applications resulting

381)

OF REVOLUTION)

VOLUMES

\0371])

of revolutions. The main reason applications with volumes is that the integrals to be evaluated come out easier than in other appliwe derive cations. But ultimately, systematically the lengths, areas, and

We start our

the standard

all

of

volumes

be a

= .f(x) Y

Let

that

Assume

> 0

f(x)

figures\037

of

some interval a < x < b. revolve the curve y = f(x) we wish to compute.) volume

x on

If we

interval.

this

in

obtain a solid,whose

the x-axis, we

around

geometric

continuous function

f(Ci))

, \\

,

,

, I)

X.I)

Xi+ 1

,

, I I I)

of [a,

a partition

Take

b],

a =

Let

Ci

be

maximum

interval these

a minimum of f in that

cylinders

and f(d;) for

the

Xi

big

Xo


for

x-axis,

between

x = 1

and

x =

B for

x-axis,

between

x = 1

and

x =

B for

B>1.)

14. The

region

B>1.) the next problems, find the volving a number a > 0, and approaches O. If it does, state

= 1/fi,

region

by

bounded

by y

whether

find

determined

this volume

in-

bounds

by

approaches a

as

limit

l/fi,

the

between

x-axis,

x = a

and

x = 1 for

O

0

is)

o.)

figure.)

equal

to)

71t/6

A =

(1 +

t f

2 sin e)2 de

-1t/6)

1t/2

= 2

.t

(1 + 4 f

We

use

0 +

sin

4 sin 2

0) dO.

-1t/6)

the identity)

. SIn

2

l) =

1

- cos

20 .)

{7

2

The

integral

is then

easily evaluated, and

we

leave

this

to the

reader.)))

OF INTEGRATION)

APPLICATIONS

390)

XII,

\0372.

Find

the area

= 2(

1. ,

[XII, 93])

EXERCISES)

enclosed by

cos

1 +

the

curves:)

following

8))

2a cos 8)

3.

, =

5.

, = 1

7.

, = 2 + cos8)

+

sin

2.

2 ,2 = a sin

4.

, =

, =

8.

\0372.

Find

the areas of

the

< 8

n/6

< n/6

28)

sin

-n/6 < 8

2 cos 38,

< n/6)

EXERCISES)

SUPPLEMENTARY

XII,

-

cos 38,

= 1+

6. ,

8)

28 (a > 0))

bounded

regions,

following

the

by

curve

gIven

In

polar

coordinates.) 1.

, =

lOcos8)

3.

, =

J l

5.

, =

sin

7.

, = 1

9.

, =

Find

11.

y

-

2

cos

8)

8)

+ 2 sin

8)

cos38) area between

= 4

- x2 , 2

13. y

= x3 +

14.

y

= x

15.

y =

16.

y

, = 1

4.

, =

6.

, = 1

8.

, = 1 , =

10.

the

12. y=4-x

2.

,

x

2

- x2 ,

y =

0,

the

2

,

y=8-2x , y Y

= x

3

= -x,

x2 ,

Y =

x + 1,

= x3

and

y

= x

x =

between

- 2

between

1, between

+

between x = the

between

+ 6

between

2+

sin

28)

-

sin

8)

+

sin

28)

2+

given

curves,

following

- cos8)

cos8) in

x = 2

and

x=

-2

and x=2

x =

-1

and

0 and

coordinates.)

rectangular

x = 1

x = 2)

two

points

where

x =

0 and

the

the

value

two

of x

curves

> 0

intersect.)

where

the

two

curves intersect.)

XII,

\0373.

LENGTH

OF

CURVES)

be a differentiable function over some interval [a, b] (with assume that its derivative is continuous. We wish to deterf' mine the length of the curve described by the graph. The main idea is to approximate the curve by small line segments and add these up.)))

Let

y

a


assume

and

c\

As

our task

large.)

may

no/2,

Next,

numerically.

experimenting

by

result.)

3.1. Let c be

Theorem

Thus c

answer

general

becomes

and

small

10... 10 (10)... (10) < 1 . 2 .. . 20 (21)... (n)

approaches 0 as

our fraction

write)

We

n

becomes

10

n- 20

20

20!

large.)))

! (

2)

)

approaches

TA YLOR'S

442)

see that

the theorem we

From

0 as n

Sometimes a one,

expansIon. In the next is: such

that

< M

be continuous

one in the next

for all x

example occur with

connection

using

by

indefinite

Taylor's

Then)




a number

bn

and

an

479)

POSITIVE TERMS)

WITH

SERIES

\0372])

>

0

all

for

Assume

n.

IS

there

that

that)

such

> Cb n)

an

00

large, and

n sufficiently

all

for

that

L

b n does

not

converge.

Then

n=l)

00

an diverges.

L n=l)

Assume

Proof

the partial

an

> Cb n

n >

for

no. Since

L bn

we

diverges,

can make

sums) N

\"b

n

\037

=b

no)

+...+b

N

n=no)

as N becomes

large

arbitrarily

N

>

the

L Cbn

=

C L

n=no)

n=no)

Hence

N

N

an

L

But)

large.

arbitrarily

bn .)

n=no)

sums)

partial

N

L

an

= al

... + aN

+

n=l) 00

are

as N becomes

large

arbitrarily

and

large,

arbitrarily

hence

L

n=l)

as was

diverges,

2.

Example

to be

shown.)

whether

Determine

the series)

00

L n= 1)

n2

n3

+

1)

converges.

We

write) n n

Then

we

3

2

+

1

1

n + 1/n

2

-

n

see that) n2 n

3

+

1

>=

1

1

1 2n.)))

(

1 + l/n

3

.)

)

an

SERIES)

480)

Since

does not converge,

L 1/n

does not

con

00

we can

Indeed,

converges.

2

n

2n 4

+ 7

-

2

n

+ 7

n

n

+

2

(1 +

7/n

3)

2

1 +

1

)

- (1/n)3+

4 n (2

3)

+

n

write)

4

n

))

3/n

2 2

2

7/n

- (1/n)3+ 3/n4.)

2

1+ 2

bounded,

certainly

serieswith

1/n2

to

and in see that

7/n

- (1/n)3

is near

fact it

+

4) 3/n

Hence we can compareour L 1/n2 converges, and

t.

because

converges,

is bounded.)

factor

XIV,

Example 2

the factor)

n large,

the

serIes of

3. The series.) L 4n=l) 2n

is

the

that

follows

it

\0373])

either.)

verge

Example

For

[XIV,

EXERCISES)

\0372.

00

that the

1. Show

3

series L

I1n

converges.

n=1)

2. (a)

(b)

Show

00

L

n=1

following

L-

n=1

\037 \037

9.

series

test.)))

series L

series

(log n)2

L

00

7.

(log n)ln.]

converges.

n\0371

2

00

5.

4+ n

2

\"3 + \"

n

L

n= 1 n +

n)

+

8.

\037 \037

nl

Isin

n= 1 n

2)

1)

2

+

1)

nl

Icos

+

\0373.

n

n= 1 n

n

2

In

Estimate

[Hint:

converges. 3

for convergence:) 00

n + 5)

3

L (log n)ln

4.

n)

THE

We continue a

the

series

----.--{2 n)

n= 1 n

XIV,

the

1

00

6.

that

Show

Test the 3.

that

with

RATIO

TEST)

to consideronly series a geometric series, the

with simplest

> O.

terms test

To compare such

is given

by

the

ratio

THE RATIO

93])

[XIV,

481)

TEST)

00

Let

test.

Ratio

L

series

a

be

an

an >

with

0 for

all n.

that

Assume

n=1

is a

there

c

number

0
...> =

and


0, number ratio test. Let)

n anx

series)

L

any the

... +

we

when

00

converges

+

series

above

to prove

suffice

will

for 0

converges

< x < R.

that

for

We

use

xn b=n

n.),

Then)

bn + b n)

When particular tion.)))

n

is sufficiently is < t, so that

n!)

xn+1

1 (n

large, we

1)! X

+

it

can

n)

follows

apply

x




number

0 such

series)

the

that

00 ll

lalllr

L 11=1)

for all x such

T hen

converges.


s.)))

absolutely

of convergence if

0


< x < c.

1. Let)

n

x.

Then)

b.+ 1 =

bn

log(n + (n

+

1)

1) X.+l \037 \037 =

2

log

n

x

n

+ 1) 10 g n

log(n

n

(n +

2X' 1)))

)

power

SERIES)

492)

+ l)/log it follows that

Since

log(n

large,

nand if c

+

(nl(n

< c1
c then for all n sufficiently large, b n + llb n > 1, whence This is so for all c > 1, the series does not converge. and hence the series does not converge if x > 1. Hence 1 is the radius of hence

and

c
0 an

c


0, there Xo + h in

and

()

L as h

is

=

lim f(x)

L)

to S),)

respect

(with

X -+ XQ)

lim

(with respect to S'),)

= M)

f(x)

X -+ XQ)

then

Proof

L =

and Ix -

!vI. In particular,

Given xol




0, there

the

exists b 1 >

there

exists

0 such

that

whenever

()l we have)

If(x)

and

is unique.)

limit

()2 >

0 such

that

IJ(x)

-

LI
0 such

a number

is

there

533)

VECTORS)

LOCATED

\0372J)

vectors.)

B) B)

/Q) Q)

direction

Same

(a)

(b)

direction

Opposite

12

Figure

2. Let

Example

P = (3,7)

and

= (

Q

- 4, 2).

Let

A =

B=

and

(5, 1)

(-16, -14).

Then

Q

- P

-

= (-7,

carried

over

shall define

to what

to

and

located it

means

AB,

--+

-

B

and

--+

--+

Hence PQ is parallel --+ we even see that PQ In a similar manner,

5)

B

because

-

A

A

=

-15).)

(-21,

=

3(Q

- P).

AB have the same direction. made concerning any definition

vectors.

F or

for n-tuples

instance,

to be

in

the

perpendicular.)

B\037

o)

Figure

13)))

:/)

>

0,

n-tuples can be next

Q

Q-P

Since 3

section,

we

VECTORS)

534)

[XV, \037

\037

we can say

Then if

B

-

is

A

of such

XV, In

vectors

\0372.

= (1, -1),

2. P

= (1,4),

Q

Q = (-3,5), A

6.

= (2,3, -4),

=

= (

Q

= (1, -1),

P = (1,4),

Q

= (4,

8.P = 9.

(2,

3, -4),

these

Q-

the points

\0373.

It is

=

( -1,

(-

A =

Also draw A, P - Q, and

-

that

A

=

(a l ,

3-space,

let

6 on

vectors

-

28).)

a sheet of paper to ----+ Draw QP and BA.

----+

A-B.)

a2 )

and

B =

vectors always

we select of

think

may

(b l , b2 ).

We

cases

the

n

=

in

2 and

define

their scalar

A =

(a l ,

a

2

, a 3)

l bl and

+a

2

We

define their

b 2 .)

B = (b l ,

b 2 , b 3).

be)

A.

.B

and

3,

be)

scalar product to

A

5,

located

the

space. You

A.B=a

In

( -11,

a discussion

throughout

n-dimensional

In 2-space, let

and

B =

3, -1),

2,

-17).

PRODUCT)

n = 3 only.

In

parallel.)

1).

vectors of Exercises1, 2,

P, B

3, 8).) are

(3,1,1), B = (-3,9,

A =

5),

3,

and

PQ

5,

----+ AB

B = (9,6).

(5,7),

-4),

equivalent.)

5,10).

- 2, 3, -1), B = ( vectors

exercises.

understood

product to

= (

(-1, 5), B = (7,

3), A =

SCALAR

same

the

Q =

(3,1,1), B = (0,

A =

located

which

are

(1,8).

----+

(-2,3,

the located

Draw illustrate

XV,

=

B =

-1, 3, 5), A

----+ AB

and

PQ

B = (5, 2).

5),

(5,7),

-4),

Q = (-3,5), A

7. P = (1,-1,5),Q

( -1,

=

(-2,3,

case, determine

each

5. P

----+ vectors

located

which

3), A =

= (4,

3. P = (1,-1,5),Q

In

a picture

EXERCISES)

1. P

4. P

perpendicular

drawn

have

plane.)

case, determine

each

- P. In Fig. 13,we

to Q

perpendicular in the

PQ are

AB and

vectors

located

two

that

\0373])

n-space, covering B = (b l , ... ,b n ) be

B =

a 1b 1 +

both cases two vectors.

with

a2 b 2

+ a 3b3

one

notation,

We define their

to be)

a 1b 1 +

... + an

.)

b n.)))

let A = scalar

or

(a l , ... ,an)

dot product

[XV,

This

PRODUCT)

SCALAR

\0373])

is a

product

For

number.

if)

instance,

4, -

B = ( -1,

and)

A=(1,3,-2))

535)

3),)

then)

A. B

the

For

we do not give a geometric interpretation scalar to this We shall do this later. We derive first some important proper-

moment,

product.

ties. The SP 1.

A. B

have

We

are:)

ones

basic

2. If

SP

3. If x is

a

4. If

=

A

A.)

three

= A.B + A.C = (B

shall

is the

zero vector, then

prove these

now

Concerning

the

first,

we

a 1b 1 + because first

for

any

property.

For

SP 2, let

C).A.)

>

A.

A =

+

... +

0, and

.

B).)

otherwise)

O.)

properties.

have)

... + an

bn =

two numbers

a, b,

C = (cl' ...,cn).

Then)

B +

= x( A

A . (xB)

and)

A . A

We

+

then)

= x(A.B))

0

then)

vectors,

+ C)

number,

(xA).B

SP

= B.

C are

A, B,

A.(B SP

-1 + 12+ 6 = 17.)

=

C = (b1 +

b 1a 1

ab =

have

we

c l' .. . ,b

b n

n

+

an')

ba.

This proves

c n))

and)

A . (B

+ C)

= a 1(bi

+

= a 1b 1 + Reordering

the

terms

c 1)

a 1C 1

+

... + an(b

+ ...

+

an

n

+

bn +

cn)) an c n .)

yields)

a 1b 1 +

... + a

nb n

+

a 1c 1

+

... + an cn.)))

the

VECTORS)

536)

is

which

other

none

A. B +

than

. C.

A

property SP 3 as an exercise. for SP 4, we observe that Finally, eq ual to 0, then there is a term af i=

93])

[XV,

This

proves

wanted.

we

what

We leave

A.

Since every

> 0, it

IS

term

=

A

+

ai

one

if

and

0

follows

a i of coordinate af > 0 in the scalar

A

is not

product)

... + a;.) sum

the

that

>

IS

0,

as was

to be

shown.)

the

only

work

of the

much

In

use

of addition,

the four properties of these later. discussion

of

cise, verify

identities:)

and

do concerning

shall

we

which

properties

ordinary

scalar

the

vectors, we shall by

multiplication

We

product.

shall give

numbers,

a

formal

For the moment, observe that there are other can be added, familiar and which with which are subtracted, you objects on an functions instance the continuous for and multiplied numbers, by Exercise interval 6). [a, b] (cf. Insteadof writing A. A for the scalar product of a vector with itself, it 2 we A . (This is the only instance when also to write be convenient will As an exersuch a notation. Thus A 3 has no meaning.) ourselves allow the

following

(A + -

(A

A

dot

A. B

product

B being the

zero

A

B)2 =

A

For

2

2

2A.

-

2A. B

B2 ,)

+ B2 .)

be equal

to 0

either

without

A or

let)

instance,

(1, 2, 3))

B +

+

well

very

may

vector.

A =

B)2 =

B =

and)

(2, 1,

-1).)

Then)

A.B=O)

We

define

A.

say, orthogonal), this definition plane, if

perpendicularity. We Here we merely note E

be

the

three

B to be perpendicular (or as we shall also For the moment, it is not clear that in the notion of coincides with our intuitive geometric in section. it does the next shall that convince you

two vectors

1

=

A,

B = O.

an example.

(1, 0,

0),)

unit vectors,

as

E2 shown

Say =

3

in

R

(0, 1,

0),)

on

the

, let)

E3 =

diagram

(0,0, 1)) (Fig.

14).)))

[XV,

OF

NORM

THE

\0374])

537)

VECTOR)

A

z)

E3)

E 2) y)

x)

14)

Figure

we see

Then

vectors look

these the

is

i-th

dot

of

product

perpendicular to Ei the

dot

XV,

product)

\0373.

= 0,

if

(a 1, a 2 ,

a.I =

A.E.

I)

the

i-th

with

A

If

our

to

(according and

only

(-1,

A. B

for

each

only the four given in the text

Using ties

And

i =1= j.

if

we observe

unit vector. We see that A is definition of perpendicularity with

is equal to

o.)

of the

n-tuples. (b) A = (-1, A = (-1, (d) A = (15, (f)

1)

above

properties for (A

3), B = (0,4) - 2, 3), B = (-1, 3, -4) -2,4), B = (n, 3, -1))

n-tuples.

scalar

of the

+ B)2

and

(A

verify

product,

in detail

the

identi-

- B)2.

4. Which of the following pairs of vectors are perpendicular? and (2,1,5) (b) (1, -1,1) and (2,3,1) (a) (1, -1,1) - n, 0)) and -1, (d) (n,2, 1) and (2, 2) (3, (c) (- 5, 2, 7) 5. Let

XV,

We

A

\0374.

define

that

EXERCISES)

A. A for each of the following 1) (a) A = (2, -1), B = (c) A = (2, -1, 5), B = (-1, 1, (e) A = (n, 3, -1), B = (2n, -3,7)

3.

= 0

a 3), then

i-th component

if its

1. Find

2. Find

E i.Ej

similarly

namely)

A,

of

and

A =

perpendicular.

component

the

E 1.E 2

that

be

a vector

THE

the

perpendicular to

NORM

norm

of a

OF

every

vector

X.

Show that A = O.)

A VECTOR)

vector A,

and

denote

IIAII=\037.)))

by

II A

II, the

number)

538)

VECTORS)

Since

A. A

times

called

the n =

When

can take

the

ma\037nitude of

A.)

> 0, we

2 and

A

=

in

=

picture (Fig.

the following

The

root.

square

norm is

2 J a

+

b

2 ,)

15).)

b)

J)

y a)

15)

Figure

Example

1. If A

= (1, 2), then)

When

n =

3 and

=

A

(aI'

2.

If A

= (-

=

J ai

3),

then

1,2,

II A II

If

n =

3, then the

J5.)

a2, a 3 ), then) IIAII

Example

J l + 4=

=

IIAII

=

J 1+

+

a\037 +

a\037.

4 + 9 = Jl4.)

picture looks like

Fig. 16,with

A =

A)

v w2+z2 =

v x2+y2+\037) z)

./

w', -----------\037

,

\" (x,

Figure

16)))

\"

y)

./ ,,\"

./

\"

./

\037])

also some-

then

(a, b),

IIAII

as

[XV,

(x, y, z).)

THE NORM OF

94])

[XV,

segment

between

0) and

(0,

Then again

of

norm

the

J n =

when

3, our

etry of the Pythagoras

In

components (x,y), t hen the length of the 2 (x, y) is equal to w = J x + y2, as indicated. be) A by the Pythagoras theorem would

first look at the two

If we

Thus

of

terms

w

2

A #

then

0,

# 0

Observe

for

that

any

to

is due

( because

1)2 =

(-

the

with

geom-

=

vector

+... + a;.)

J ai

some coordinate ai

because so

that)

see

we

# 0,

so

that

al

> 0,

/lAII # o. A we have)

=

II

-

All.)

that)

fact

the

is compatible

(a l' . . . ,an)

A =

IIAII

This

Z2.)

theorem.

coordinates,

II A II

y2 +

+

norm

of

definition

ai + ... + a; > 0,

and hence

2 J x

Z2 =

+

IIAII

If

539)

VECTOR)

A

- a 1)2 +

... +

(

1. Of course,this

- a = n )2 ai is

as

+

... + a;,)

it should

be

from

the

picture:)

A)

-A)

Figure

they

ing

A and

that

Recall

have

the same

17)

- A are said to have opposite direction. norm (magnitude, as is sometimes said when

However,

speak-

of vectors).

Let

A,

B

be

two

We define the

points.

between

distance

be)

IIA

-

BII

=

J (A

- B).(A -

B).)))

A and

B to

VECTORS)

540)

This

in

our geometric intuition with 18). It is the same -----+ thing located vector BA .)

coincides

definition

points

the

-----+ (Fig.

plane

located vector

or

AB

[XV,

the

when

the

as

A,

94])

Bare

of the

length

B)

=

Length

3.

located vector

Let -----+

A

(

- 1, 2) - A II.

II B

is

AB

=

In

the

vertical intuition

picture,

we

side has derived

see

from

liB

-

A II)

A

=

the

Then

of the

length

Thus)

(4, 2).

16+4=J20.) our

Thus

2.

-

has

side

horizontal

the

that

length

B

=J

=

B\\I

and B = (3,4). But

liB-Ail

-

18)

Figure

Example

IIA

4

length

and

the

reflect our geometric

definitions

Pythagoras.)

B)

A)

-3

-2

2

-1) 19)

Figure

a point X such that) points Let

P be

in

the

and

plane,

IIX

be called the X such that) will

open disc of

-

radius

IIX

-

3)

PII

let a




O.

The

set of

a)

a centered

PII
O. If a is a right end the is for considered only point quotient h < O. Then the usual rules for differentiation of functions are true in this 1 through 4 below, and the chain rule and thus Rules greater generality, of 92 remain true also. [An example of a statement which is not always true for curves defined over is given in Exercise 11 (b ).1 closed intervals curves. We consider the Newton quotient) Let us try to differentiate

is

taken

for those

the interval.

X(t

+ h)

- X(t)

h)

Its

numerator

is illustrated

in

Fig.

3.)

X(t))

Figure

3)))

OF VECTORS)

DIFFERENTIATION

568)

0, we

h approaches

As

see geometrically

\0371])

that)

- X(t)

+ h)

X(t

[XVI,

h)

should

a vector pointing

approach

the Newton

write

X(t +

h)

quotient -

X(t)

X

=

h

see

and

We

(

assume

- x 1(t)

h

a

each component is

that

coordinate.

,...,

xn(t

+

h)

-

We

can

Xn(t)

h)

Newton

for

quotient

xi(t) is differentiable.

each

that

curve.

of coordinates,)

+ h)

1 (t

of the

direction

the

in

terms

in

)

the corresponding

Then each quo-

tient)

- xi(t)

+ h)

Xi(t

h)

the

approaches tive

dX/dt

to

derivatives

dX

=

we could

also say

1

( dt the

limit

of the Newton

' .. .,

the

that

dX

is

1

( dt

dt

fact,

this

we

reason,

define

the deriva-

be)

dX

In

For

dxJdt.

,...,

dX n .) dt

)

vector)

dXn dt)

)

quotient)

+ h)

X(t

- X(t)

h)

as

h approaches

o. Indeed,

as

h

Xi(t

0, each

approaches

+

h)

component)

- xi(t)

h)

approaches

dxJdt.

Hence

the Newton quotient 1 ( d:r

,...,

n

d:r ).)))

approaches

the

vector)

[XVI,

569)

DERIVATIVE)

\0371])

=

4. If X (t)

Example

t,

(COS

sin t,

t)

then)

. -dX = ( -SIn t,

t, 1 ).

cos

dt)

dX / dt

denote

often

Physicists

could also write)

= (

X(t)

the

define

We

- sin

t,

cos

t, 1)

previous example, we

= X'(t).)

curve at

vector of the

velocity

in the

thus

X;

by

be the.

t to

time

vector

X'(t).) Example 5.

When

= (cos

X(t)

= (-

X'(t)

the velocity

vector at

t =

n

t, sin

and

for

velocity

1);)

=

(0,

-1,

1),)

we get) =

X'(n/4)

The

cos t,

t,

is)

X'(n)

t = n/4

sin

then)

t),

t,

vector

to the point X(t), nex t figure.)

then

1/J2,

(-I/J2,

1).)

is located at the origin, but when we visualize it as tangent to the

we

translate

curve, as in

it the

X(t)+X'(t))

Figure

We

passing

define through

the

line

tangent

X(t)

in

the

define a interpretations for X'(t):) Otherwise,

we don't

to

a curve

direction tangent

X'(t) is the velocity X'(t) is parallel to

4)

X at time

of X'(t), We

line.

at time a tangent

t

that

provided

have

to

therefore

t; vector at time

t.)))

be

the X'(t)

line =1=

o.

given two

OF VECTORS)

DIFFERENTIATION

570)

as

the

vector.

tangent

6.

Example

curve X(t) = (sin

Find a t, cos

t) at

G

down

t), so

= P

L(t)

+

this

+ X(t)(X(t) located vector

the

tangent

another

use

letter

L(t) =

)

X'(t)) each

Find

=

,

f

t

tJ3-

t.)

2

of the

X(t) =

(cost,

Figure

n

(

3)

of

the

}.)

2:

perpendicular

plane

sin

t,

t))

5)

be)

P=X

equation

=

cos

(

of

the

line as)

2'

2

=

G, f

the tangent

-J31 + -

the equation

point

D.)

already occupied.)In terms write

can

=

+

D

}('(1r/3)= N)

given

,

(f

a parametric

Then

L because X is

( t)

we get)

t = n13.)

the

line to the

is)

(x(t), y(t)), we

y(t)

7.

;

XG)

=

tA

X

Example

t =

at

that

))

(

coordinates

of

and)

f

)=G,-

and A = X'(nI3). line at the required point

tangent

Let

although

vector,

n13.

= X(nI3)

P

when

vector

equation

t =

-sin

X'

(We

tangent

located

write

parametric

X'(t) = (cos t,

We have

Let

to

However,

call X'(t) a the

\0371])

cumbersome.)

is

time

refer to

we should

speaking,

strictly

we sometimes

of language,

abuse

By

[XVI,

-n 3'

. SIn

n -

3'

n

-

3

')))

)

to the spira])

so

more

that

simply,)

then

must

\037

).)

the

at

given

P.

point

We have X'tt) =

\037 \037ln

L, co\037 L, 1), so)

n X,

( The

of the

equation

plane

!

J3 '

= _

3)

P

through

the

equation

of the desired

)

to N

perpendicular

is)

,)

plane is)

1

J3

= N .)

1

2'

2

(

X.N=P.N

so

to the plane

N perpendicular

vector

a

find

,

f

p=G, We

571)

DERIVATIVE)

91])

[XVI,

J3J3n - -+-+4 4

--x+-y+z= 2 2

3)

1t)

3.)

We

define If we

vector.

the speed of the curve denote the speed by

then

v(t),

=

v( t)

to be the

X(t)

/I X'

I)

thus)

V(t)2 =

We

we

( t) /I ,

I

and

norm of

by definition

can

also

omit the

t

the

from

V

Example

8. The

speed of

X'(t)2 =

2 =

the

norm of the velocity v(t)

=

and

notation, x

,.

t)2

on the circle)

moving

(cost,

X'(t) = ( -

J ( -sin

write)

x , =) X ,2

bug

X(t) = is the

X'(t). X'(t).)

sin

sin

t,

+

(cos

t))

cos t), 2

t)

=

and so 1.)))

is)

the have)

velocity

speed of the bug

The

9.

Example

X(t) = is

OF VECTORS)

DIFFERENTIATION

572)

of the velocity

norm

the

=

We define the

(

J ( -sin

=

v(t)

(cos t, =

X/(t)

on

moving

sin

- sin +

t)2

t,

spiral)

2

(cos

and so

t, 1),

t) +

is)

1

J2.)

derivative)

to be the

vector

acceleration

(t)

dX/

= X\"(t) ')

dt

also of course that X' is differentiable. We shall acceleration vector by X\"(t) as above. We shall now discuss acceleration. There are two possible for a scalar acceleration: First there is the rate of change of the speed, that is) dv

=

v

is the norm of

the

two

These

are

(t).)

acceleration

II X\"(t)

Warning.

that

vector,

is)

II.)

not

usually

definitions

I

dt

there

the

denote

provided

Second,

\0371])

t))

cos

t,

the

[XVI,

Almost

equal.

any

example

will

this.)

show

10. Let)

Example

X(t) =

(cost,

sin

t).)

Then:)

v(t) =

= (-

X\"(t)

say

when we

if and

Thus

one

which

acceleration,

but

one

fact

that

A

- sin

t))

so)

dv/dt

so)

IIX\"(t)11

= o. =

1.)

must

refer we must always to scalar acceleration, One could use the notation for scalar a(t) two which of the specify possibilitiesa(t) de-

the above two

cal interpretation.

1

need to

notes. The

cos t,

mean.

we

=

IIX/(t)II

bug

quantities are

moving

around

not

equal

a circle at

reflects uniform

the physispeed

has)))

[XVI,

dv/dt =

573)

DERIVATIVE)

\0371])

O. However, the acceleration vector is not 0, because the changing. Hence the norm of the acceleration

velocity

vector

is constantly

vector

is not equal to

O.

list the rules for differentiation. These will concern sums, and the chain rule which is postponed to the next section. The derivative of a curve is defined componentwise. Thus the rules for the derivative will be very similar to the rules for differentiating funcshall

We

products,

ti ons.)

Let X(t) and

Rule 1.

values

same

of

sum

two differentiable curves (definedfor X(t) + Y(t) is differentiable, and)

+

Y(t))

be

Y(t)

t). Then

the

d(X(t)

dX

2. Let

c be a

let

d(cX(t))

= c

dt

values

derivative

curves

differentiable

a

is

Y(t)] = X(t)

[X(t).

dt

is formally analogous

namely the the derivative duct.

.)

(defined

the

times

first

the

of

to

except

first,

. Y'(t)

differentiable

+

of

second

the

the

that

X'(t)

of a

derivative

the

derivative

the

whose

product

.

Y(t).)

of functions, times

product plus

the second

is now

a scalar

pro-

)

As an example of the proofs we lea ve the others to you as exercises.

Let for

shall

give

the third

one

in

detail,

simplicity)

X(t)

= (x 1(t),

x 2 (t)))

Y(t) = (y 1(t),

and)

Y2(t)).)

Then)

d

dt

X(t).

Y(t)

=

d dt

=

combining

+

[X1(t)Yl(t)

= x1(t)

by

for

function

is)

d

(This

dX dt

X(t) and Y(t) be two of t). Then X(t). Y(t)

Let

3.

Rule

Then cX(t) is

be differentiable.

X(t)

and)

differentiable,

same

dt)

dt

and

number,

dY

=-+-.

dt Rule

the

X(t).

dYl (t)

dX

+

dt

Y'(t) +

the appropriate

terms.)))

dt

X 2 (t)Y2(t)]

1

Yl(t) + x 2 (t)

X'(t) . Y(t),)

dY2 dt

+

dX 2

dt

Y2(t)

and

OF VECTORS)

DIFFERENTIATION

574)

The

and

for

proof

inserting...

3-space or n-space is obtained in the middle to take into account

Example 11.

3 or n, coordinates.)

2 by

replacing

by

the

other

in X(t) comes up frequently as the square of because it can be interpreted from the origin. Using the rule for the derivative of

The

= X(t).

X(t)2

square

for instance

applications,

of X(t)

distance

the

[XVI, 91])

a product,

formula)

the

find

we

d

. X'(t).)

= 2X(t)

X(t)2 dt

this formula

memorize

should

You

Suppose that sphere of constant

k.

radius

This means the square yields)

constant.

is

IIX(t)11

Taking

that

we

constant.

is also

is, X(t)2

X'(t)

=

distance from

the

vector

position

X(t)

the

origin,

-

\037 .\"\",.--..

---- ---

Curve

vector

X(t)

is a

we may X(t).)

Example

----.........,

i

on a

curve and also form

12. Let

curve

=

which

X(t)

remains

the

Then

X'(t).)

velocity

X'(t))

\037)

If

IIX(t)1I

2

= J

X(r) 1

then

X'(r).)

sphere)

f(t) the

is

f(t)X(t)

sin

= (et

for the same values of of the number f(t) by the

defined

a function,

product

X(t) = (cost, .f(t)X(t)

cos

= et ,

and f(t)

t, t) t

t, e sin

then

t, ett),)

and)

f(n)X(n)

0)

\"X(t)11 = k is constant.

i.e.

is perpendicular to

X(t)

//

X(t) . X'(t)

therefore)

moves along a

a bug

at constant

If

a

respect to t.

with

sides

both

Differentiate

and

0)

Suppose

Interpretation.

then

on

lies

X(t)

obtain)

2X(t).

t,

that

= k 2)

X(t)2

Then

loud.)

it out

repeating

by

= (e1t(

-1),

e

1t

(O),

e1t n )

=

(-e

1t

, 0,

e

1t n).)))

= (x(t), y(t),

If X(t)

then)

z(t)),

= (f(t)x(t),

f(t)X(t)

f(t)y(t),

4. If both are differentiable, Rule

and

f(t)

are defined over

X(t)

= f(t)X'(t)

f(t)X(t)

nurn

same

interval,

and

for Rule

3.)

a fixed vector, and let f be an ordinary differ= f'(t)A. Let F(t) = f(t)A. Then F'(t) = are fixed A where b and a, (a, b) (cos t)A

Let A be

13.

Example

entiable function For instance,

as

same

the

3.)

+ f'(t)X(t).)

dt

the

Rule

and)

is f(t)X(t),

so

then

d

The proof is just

f(t)z(t)).)

such differentiation analogous to

a rule for

have

We

575)

DERIVATIVE)

91])

[XVI,

if

variable.

one

of

=

F(t)

then)

bers'!

F(t) =

(a cos t,

b cos

t))

thus)

and

F'(t) if A,

Similarly,

=

(-a sin t,

B are fixed

-

t) =

b sin

(-sin

t)A.)

vectors, and) = (cos

G(t)

t)A + (sin

t)B,)

then)

= (-

G'(t)

XVI, Find

\0371.

cos

3. (cos

t,

5. (a) In the

(b)

In

t,

sin

sin

t))

the

(cos

t)B.)

and

vector.

position

3 and

Exercises

A, B

curves.)

following

t)

Exercises 3

direction 6. Let

+

EXERCISES)

the velocity of

1. (e',

sin t)A

from

be two

the

constant

4,

4,

show

position

2t, log(l

4. (cos

3t,

the

that

show

Is this

2. (sin

also the that

the

sin

in Exercises

acceleration

t)

3t)) vector

velocity

case

+ t),

is perpendicular

1 and

vector is

in

to

2? the

opposite

vector.

vectors.

What is

X=A+tB?)))

the

velocity

vector

of the curve)

576) 7.

Let the

X(t) be a

X'(t) at the point or also at the point

vector

velocity

at the point t to the curves of

8. (a)

the

Find

Exercises3 of a

equation

4

and

plane

point t =

the

at

\0371])

line

is said

the curve)

to

normal

(e',

(b) Samequestion

[XVI,

which is perpendicular to be normal to the curve to X(t) Find the equation of a line normal X(t). at the point n/3.

or

A plane

curve.

differentiable

VECTORS)

OF

DIFFERENTIATION

2 t ))

t,

1. t =

the point

at

o.

P be the point (1, 2, 3, 4) and Q the point (4, 3, 2, 1). Let A be the P and parallel to A. vector (1, 1, 1, 1). Let L be the line passing through between a point X on the line L, compute the distance Q and X (a) Given t). (as a function of the parameter that this X 0 on the line such (b) Show that there is precisely one point

9. Let

distance achieves a

(c)

10. Let P be the vector (1, -

exceptthat 11. Let

down

on

is a

minimum,

as

is

curve

an

on

defined

1, -

1,

2).

the

A be

Let

the preceding

in

problem,

J 146/15. open interval.

be a

Let Q

curve.

the

value of t such show that the

point X{to)' tance. ] If X(t) is the exists

point (1,

distance

minimum

formula

the

Q the

same questions

for

the distance

that

the distance

between

Q

between

Q

and

an arbitrary

the curve.

on to

1) and

Solve the

case the

this

is not

Write point

(c)

- 1, 3,

(1, 1).

2,

the

to

perpendicular

is 2J5.

this minimum line.

that

and

minimum,

Q is

be a differentiable

which

(b) If

-

0

point 3,

in

X(t)

point

(a)

X

that

Show

[Hint:

vector Q -

parametric value to

a unique

minimum

the

Investigate

the

that

of the

and X(t o) is to the curve, at

square of

of a straight line, show distance between Q and

representation such

is normal

X{to)

a

at

the dis-

the

there

that

is a

X{to)

minimum.)

12. Let

N

vector, c

a non-zero

be

of intersection X .N = plane

of

c.

Show

Prove the

that then

Poll


if x

and

2

x).)

and

if

< 0

only

if x

= 0

x


1,

an

with

nn

x


Also

2x.

are

points

integer n, so x =

n/4

+ nn/2

x = n/4. Min at Strictly increasing for

points:

3n/4

at

-)3 < x < O. -1 < x < O.

Down

for

Down

for

x


0,

point.)

L= so

if

2r

+ rO.)

P can

be expressed in

terms

of r only

by)

P(r)

= 2r

+

2Alr.)

2 P'(r) = 2 - 2Alr , and P'(r) = 0 if and only if r = A 1/2. So P has and one critical per) -+ 00 as r -+ 00 and as r -+ O. Hence P has a point, only for and that minimum, is at the critical point. This is a minimum minimum 2 all values of r > O. In part (a), we have 0 < n so r > 2A/n, and the data limits us to the interval)

We

have

J 2A/n


4Aln,

the critical

is at

minimum

the

us to

limits

which


JA,

J4Aln

In

point.

have

(b), we

part

interval)

r.)

the end

is at

minimum

the

A39)

TO EXERCISES)

ANSWERS

r =

point

J4Aln

Graph

of)

P(r) =

2r + 2Alr)

.)

J4A

ft)

It

38. x = 20.

The

are given

profits

by)

= SOx

P(x)

Then P'(x) = - 3x 2 + only if x = 20 or x = know how to the maximum is at

should So

39. 18

units,

40. (50 + 5)94)/3. Same 41. 30 units; $8900

42. 20.

Let

g(x) =

The x =

VII,

of

graph

gives

p.

1. Yes;

all

7. Yes;

for

y

for

-1

p.

224)

13. Yes;

\0372,

O. Let

do.

Also

graph

that

of a

cubic,

and P(O) is

is negative,

P(10)

P'(x) =

formula,

quadratic of P(x) is

lex) =

>

x(l000

-

38

as Problems

profits. Since p -

lOx)

is a parabola the maximum

3. Yes;

9. Yes;

1


0

first

the

by

and

formula,

g\"(y)

< 0 by

the second.)

229)

the

View

this

+

! or

j\"(x)g'(y).)

f'(X)2

as defined on

cosine

o < On

+ 2')

-1

f'(g(y\302\2732

VII,

3x\037

6. + 1

=

j\"(g(y\302\273g'(y)

J'(x)

is

of intervals.)

10.

-1

If

there

is the in-

g

J'(g(y\302\273)'

=

g\"(y)

-

1)

5. + 1

9. l4

If

1

1

--

11 \302\267 g (y)

1

+ J\"S)/2.

1

=

choice

4. + 1

-1

Hence

definition.

= (-1

Xl

then)

1

the

of

interval

present

case,

g'(2) =

(Alternative

A41)

TO EXERCISES)

ANSWERS

the cosine is

interval,

the

interval)

x


there

crease We

This

. -1

have:)

f'(x)

So

Then:)

O.

justifies

If

x

---+

If

x

---+ 0)

00

all the items

then then

in

log

where f'

0 for log x f(x)

the

and

point,

all

---+

\037 0)

graph.)))

x


Since

0,

that

follows

it

1 + 2

log x

>

EXERCISES)

graph

> 0

f'(x)

if

if and

0)

is strictly increasing

f

Exercise

From

Hence

f'ex)

x


the when

intervals

So

of

there

are two

regions

of

between

1/2)

1/2.)

is strictly

decreasing

x approaches

x log

0,

0 as x

approaches

means the curve

which

e

3 + 2

if

3+

210gx.)

words, log x = 0, or, in other inflection point occurs for of the graph. features

the

Thus

3/2.

2 =

all the indicated O. Then

1

x)

- + x)

(log X)2

x)(2 + log x).)

= 0

critical

increase

x =

0 or

log

x =

1

or

log x =

x =

1

or

x=

and

vary

2 + log x =

points

will

-

according

2

e- 2.)

= 1 and x = e - 2 . We decrease corresponding to at x

a table

make

intervals

the

0 1

and

approaches 1. x > 1. So x/log

then

x/log

denominator,

the

to

contributes

\037

fraction

one

decrease

critical

\037 00.

The

Proof:

and

x

numerator is

positive

for

x \037 00.

1 and

increase and

x

log x approaches 0,

denominator

The

x < 1 then x x/log x approaches 1, and the denominator for x < 1, so x/log x \037 - 00. the This already justifies graph If x

only

f(log

and x

o.

approaching

If

15.)

x > 0,

for

log x becomes large negative, a large negative number

is because

This

if

up

drawn.

log

as

Since)

bends

graph

behavior

.

,)

f'(x) = 0

approaches 1 (sincethe

and x

1

,)

by Exercise

is positive

e-

\037 00.

f(x)

)

denominator

the

xe-

\037 00

graph as

of the

features

the

out

x =

if

point. The

inflection e 1 .)

an

is

and only

-1)

x


log

2

-

x) =

= 0 if

f\"(x)

down

bends

and

0 so

< 0

f\"(x)

it

x

0

f\"(x) >

-1 (2 + log x

- +

x)

2/x >

have

we

too bad:)

is not

f\"(x) = (log

A53)

TO EXERCISES)

ANSWERS

are

point).

concerned,

Let

us

now

\037

- 00.

Pro\037f:

log x

as drawn

and look

Again

approaches 0

for

in

the

at the

so

far

the

numerator

but

is negative

as regions of

critical

point

regions

of

(there is

bending

up)))

TO EXERCISES)

ANSWERS

A54)

the

We write

down.

and

in the

derivative

first

-

=

f'(x)

-

1

1

x

log

form)

(log

. X)2)

Then)

-1

f\"(x) =

1 -1 - (-2)(1ogX)-3X x

X)2

(log

-1

= (I

1

3 og x )

-

(log

x)

X

-

2).

Therefore:)

= 0

f\"(x)

now analyze the 2 0, 1, and e

shall

We

between the points f\"(x) change sign. determined by minus sign in If x > e2

log x - 2 > makes f\"(x) If 1 < are

Note

that

signs

of

the

e 2.)

x =

of f\"(x) are the

sign

taken intervals, factors of

various

in

the points where the sign of f\"(x) (plus or minus) - 2, together log x, x, and log x

then

< 0

f\"(x)

log x

which

e2

then

and

f\"(x)

the

> 0

minus

the

and

x

and

negative.

positive,

2

will

be the

with

front.

0, both

x


X

] du

0)

f

x\"e- X

3/ 2

f

integral

definite

U

Then)

first.

f Then

15

x)

X'e- x dx

-

2

=

G

2

1/ 2

1

f

x m dx.

xr-

n(log

5/2 ]

Let x

+ 6).

[U

0

US/2

-

[

xe- P dx =

13.

3/ 2

3/2

1

=

o

1

-

du =

f

--[

2

11.

121

f

I/2

U)U

using dx

that

= n

B\"e

x\"-'efoa>

-

B

-+ 0,

X

dx.)))

we

find:)

1).)

e-

X

dx,

so)

Let

x

So xne-

=

In

last

This

dx.

1019 ; to

19

form)

1 .)

oo

_, -_ n., 10 -

n.

n! = n(n - 1)(n final integral is easily

2). . . 3

For way,

dx,)

0

is the

.2 . 1

B

dx =

e-X o

n

integers.

B

=

e-X dx

lim

f 0

B-+oo

=

first

the

of

product

namely)

evaluated,

oo

f

-x

e

I

This

the

get)

In

where

in

the evaluation of the integral to the next step. = 91 8 ; 18 = 81 7 ; and so on. Continuing in this

reduced

we have

instance, 110 = it takes n steps

be rewritten

can

equality

In = nl n Thus

A67)

TO EXERCISES)

ANSWERS

X

e-

0)

B-+oo

- [e

lim

_

lim

-

B-1] = 1.

B-+oo)

XI,

1. -f

3.

3

sin

3

sin

354)

p.

\0373,

sins x

x

3

- 2J2 cos8/2 x

15. arcsin (a)

an =

=

Co

0

19. (b)

XI,

1. log 2.

3.

.

sIn x

tan 2 x

- coseX

16u

arcsin

bx

x

sin

14.arcsin \037

Let x = au/b,

-;;.

dx

= (a/b)

) cos

(4In nn

= 0)

Co

2 - sin

b

j

7. nr 2

> 0)

b

x +

nn. -(2/n)cos nn, bn = 0 all n. 2 1)lnn , b n = 0 all n.

2

= 2(cos

x sin

p. 356)

Exercises, x

2

Write

-

-

Supplementary

11. k arcsin

15.

n, b n =

all

an

an and

all

\0373,

n12,

1 17.

2x)

a,

2

13. -log cosx

sin 8/2

2J2

arcsin(y

an =

= n 2 /3, Co

(b) (c) Co=

(b)

6. nab (if

M

16.!

J3

8n

5.

2. i cos

x + -ix

x cos

sin

-i

4. 3n

5

8. (a)

18.

-

x

x cos

4.

x

1/ 2

=

tan

Let x

- kx(l -

+ 1U3/2

2

- 1

x + 1

=

2u

2x

where

, dx

2

)J u

l

and

n

5. -

= 2du

- x2

that

note

4

13.

\037

2 =

16

- x 2)))

d tan xldx n

-

7 \302\267

tan

2

x + 1.

n

9 \302\267

4

14.

=

16

-arcsin

x

_! x

2 Jl - x

duo

TO EXERCISES)

ANSWERS

A68)

16. Let

x 2.

1 +

+

u

Then)

X3

Jl + x

2

I

1

=

dx

2x

X2

2.

I

+ x

Jl

1

=

dx 2

u

2.

rest of the exercises are done by dx = a cos 8 dO. We gi ve the answers, but

I

The

17.

- 1

log

---;;

19.

J

a

a+J [

2a 2x 2

x=a

32a

a+J

log

x = a sin

0 or

cos

]

1

-

arcsi

2 2

a

The

O.

x = a

-

is

principle

sin

3/ 2

Ul/2

-

2. [ 312

1/2]

sin 8 or x

19 in

choice of

.)

= a sin

8,

full.

to

whether

as

us do

Let

same.

the

let

usual,)

0 dO.)

= a cos

dx

(J,)

U

- x2

2

have a

We

.

]

x =

!x Ja

- X2

1

=

o ut Exerc ise

work

n(xla)

x

[

1 du U 1/2

letting

a2

18.

x

- x2

2

- X2

2

a

-

Then)

1 x

I

3

J

a2 -

x

a

I

but we show of sine, we used powers a pain,

It's

Thus

3

. sln 3 O(a cos

integration

Recall parts. We

by

analogy

II

I . 3 dO = \037 0 I Sin I sin O

with the

sin

2

to

that

.

a3 I

sln

try

I 3

0

I

---:-sin) 0

CSC

2

2

= -csc

0 dO.

0,)

let)

we

u =

1

sin 0

du =

- .

sin)

csc 2 0 dO,)

dv =

')

1 2

0

v =

cos 0 dO,

-cot

Then)

I =

_

cot 0 _ sin 0

COS2

I

0

sin 3 0

dO =

_

cos 0 sin

2

0

O.)

1-

_

I

sin

sin

3

so)

I =

0 - cos . 2 () sIn

- I

1

+ I

---:-- dO, sin) 0

whence)

I=

1

2[

-

cos 0 sin

2

(J

-

log(csc 0

+ cot 0)

.)))

]

2

0 dO

0)

dO. 0)

positive method here.

integrate

a similar

I

dO =

dO

and

-I

have)

we

tangent,

'

d cot 0

so

=

0)

do it.

to

how

a cos 8 dO

let)

I =

In

1

=

dx 2

You

the answers

leave

may

want the

terms

in

answer

= - ,

of

terms

in

of x,

cos

=

fJ

a)

a

1

csc()=-=-

20.

-

1 cot

2\"

x =

where

fJ

x

fJ

sin

cot

')

a sin

=

fJ

if you

But

J 1 cos

-

fJ

sin

fJ

2

sin

fJ

1 = a)

a2

=J

2 J a

x 2,

-

- x2 x)

cos fJ dfJ.

= a

dx

fJ,

done.

is usually

this

fJ,

then use:)

x

fJ

sin

A69)

TO EXERCISES)

ANSWERS

a)

21.

- x2

J l

1 +

-

J

log

23.

XI,

1.

at, dx

x =

x

-

2 J a

-x

\0374,

p.

= a dt, . x-

7 18

1)+

Exercise 19.

14.

to Exercise

reduce

and

(x +

log

use

_

and

same as

370)

. Don't 3) du = 2x dx.

2

is the

a)

-1 2(x

. The method )

arcsin

2

-i log(x -

2.

2

x

(

22. Let

+ X

t

7) fractions

partial

use the

here,

3. (a) ![log(x - 3) - log(x + 2)] (b) log(x + 1)4. -! ]0g( x + 1) + 2 log(x + 2) - \037]0 g( x + 3) 1 5. 2 log x - ]og(x + 1) 6. ]og(x + 1) +

u =

substitution

x2 -

+ 2)

log(x

x+l

7. -log(x

+ 1)+

8. log(x

1) + log(x

10. (a)

-

1

x

X2-\037

-1

1

12. 2 2x 14. t

1

+ 9

+

log

X

[ 2(X 1

x2 + 3

15 \302\267 C 1 --

- 1= 33 - 100 '

16. (a) Let x

+ 1)

x

18 x 2

+ 9

=

+

2)

(x -

1)(x2

bt,

dx

1)

+

+

3

8

arctan \037

+

x

x

]

1 54

arctan

x

1 13.

arctan\"3

x

8 x

2

+ 16

+

1

x arctan

4

32

arctan x. Factorization:

1

C 2 --

x+2) x +! arctan x 2 (x2 + 1 )

x

2X

1)2 -1

(x +

9.

2)

8 (x2 +

1)2

- 3

11.

3

+

4 (x2 +

-

+ 2)

log(x

2

+ x

+ 1))

and)

x

4

- 1 = (x +

1)(x- 1)(x2 +

11 C -- - 130 C -- - 110 e = ll - 100 ' 3 100' 4 100 ' 5 100 = bdt (b) Let x + a = bt, dx = bdt.)))

1).)

3

17. (a)

-!

(b)

!

-

1. 2J 3.

log

p.

y

- i

2

+ 1) -

-

arctan

J3

arctan

;;

J3 + 1))

4.

)

I + eX + 1) + log( J I + eX - 1) -log( J I + eX + 1) + log( J I + eX -

f(x)

= !(e X -

then

is large

>0

all

for

eX

then

positive,

f(x) the intermediate

Hence

the inverse

f -is

x, so

is small,

and e eX

X

1

f

(x)

1

as

x

as

x -+

-

2

Hence)

is small.

00.)

1 sinh

f (x) is large positive.

of f(x) consist for all numbers

values

J

all x. If x is large

for

\037 00,)

defined

= x

cosh x.)

increasing

00)

value theorem, the x = g(y) is

----;--= cosh

) =

positive, so -X positive, and e

function

g'(y) =

1)

is large

\037 00

-

X

e-

strictly

is large

-+

- log(1 + eX)

2. x

x. Then)

= sinh

eX)

f(x)

By

J3

log( J

But cosh x If x

1)]

2x + 1

1

X f'(x) = !(e +

negative,

+

2x + 1

1

+ x

- log(x2

1)

)

y3 2

-

377)

X

=

2

(b) ![log(x

+ x

log(x

1

+ log(x

1)

1

( x+1

-

2

x+x+

l + eX

arctan(e

5. Let

1)

! log

x2

19. -log(x -

\0375,

-

X

x +

arctan

18. (a) t log(x

XI,

TO EXERCISES)

ANSWERS

A70)

graph

f(x) =

J

y.

numbers.

all

We

1

=. x + 1

of

y2

+

1)

of

!(e.r -

e-.r)

= sinh x)))

have)

6. Let y

=

= !(e X

f(x)

X

cosh x.

) =

+ e-

= !-(e x

f'ex)

If x >

0

then

eX

strictly increasing,

x= We

exists.

= 1. As x \037 00, values of f(x) consist of all g is

- 111 =

=

g'(y)

X

and e- \037 0, so f(x) > 1 when x > 0. Hence

numbers

> 1.

let

eX

The

graph

and

there

at

the

u =

= u

of f(x) two

are

=

y

y =

Then

eX.

to

equation

is

f

J cosh

the

1 J y2-1')

x-l

of

graph

t(u + Iju).

\037 00.

have)

We

= 2

f(x) =

Finally,

Hence

O.

y)

eX \037 00

1

quadratic

x >

all

for

= x

sinh

f'(X)

> 0

all numbers

for

defined

x.

= sinh

= arccosh

g(y)

have f(O)

the inverse function Hence

- e- X)

1 so f'(x) function)

inverse

the

and

Then)

e- X


0 and

x = is the the

inverse functions

minus

sign. Then

for by

-

Indeed

J

y

- Jy 2

-

1))

y2

-

1))

suppose you can

algebra,

simple

2

J

taking)

log(y

x < 0.

y

+

log(y

-

1


0

for

u in

it

1,

terms of

7.

=

1. Since

y

>

1, it

-

y2

1,)

-

y

J

1)
1.

Area

=

r We

want

use

the

the area

to make

the

2 Jx

1

you

+ y

get into

cosh

t)

a perfect

into

dx = sinh t

and)

integral

We

dt.)

is of powers of et and e t which consIstIng limits of integration in the way explained in the t Let u = e and solve a quadratic equation for u.

J B2 +

1) + B J B2 + 1

= a cosh(xja).

Then)

dx

-d

2

dx

y 2

-1 = -1 cosh(x/a) = cosh(xja). a

1

=

-1 a)

= az,

dx

= adz,

= sinh(x/a),

a)

= a

Let x

square.

-

an

1 -dy = a sinh(x/a).-

14.

graph

dx.)

the square

under

expression

-

easy to evaluate. Change the last example of the section. You will fi nd the given answer .

12. 10g(B 13. Let

the

under

substitution)

x = Then

Hence

B is)

1 and

between

y=

J l + sinh

J l+

a)

2

(x/a)

(dyjdx)2.

and reduce to

the

worked-out

case.)))

XII,

p. 384)

1 \037

2

8

8. n[2(log2)2 2 log x x

\037

-

(b) \037

e\037

(

n

n

-

24

3B

15.

}

=R

1. f(x)

(log X)2

\037

)B2

c >

all

3. 12n 4.

-

J

a

f(X)2 dx -

n

comes

2)2 -

-

\037 1 13 \302\267 3

\037

B3 )

(

1

15. n

-

16.

XII,

limit

out

- 1)

- x 2.

2

a

O.

as a -+

to

- x 2 dx

=

2

6.

-5n 7. -n 8.4na 14 3 J3

Ra

2

2

is

volume

.)

n

9. _ (e 2

b

1

12.

-

2

- e-

n as

10

)

B -+ 00

\037

n(

-\037]

).

log B volume

The

O.

2n

3

14. n

--+

equ al

a2

{ 00

The

9(X)2 dx

J

11.

2]

--+

as

limit

- cosa

The

increases

without

increases

volume

without

bound.

bound.

+

log(j2

- 1) -Iog(csc

a-

cot

a)]

limit as a

\0372,

l. 6n

B

as

3

No

f

No

\037

\037), 2n

1).

nG {

'

410g2 +

No

-. a

log

17.

-2n

5.

2n

-

2n(1

p. 385)

3

10. n[2(log

(log X)2,

fa

f:a _32n 5

12.

3

1/2, n/(2c

g(x) = R

and

4nR

2.

=

u

2c))

algebra

easy

nr 2 h

h

fa after

by parts,

2

yes 4: 2

}

is

Volume

x.

Exercises,

x2

e12

r

=

is y

V = n

which

dx

ne

7.

2)

2: 2

(

14. For

2J a

+

(c)

4: 4

Supplementary

\0371,

yes

e\037B

24

6. n(e -

3

dx

1/2, n/(1 -

c
0, the

problem

3.

na

n

2

12

XII,

\0372,

3n

n

10.

4

XII,

\0373,

1. 287 (10

3. Je 4

5. J 1

+

6.

l7

8. We

9n

397

3/ 2

- I)

1 + 2

(31

2

3/2 work

+ t

2. -

fi

log

- 133 / 2 ) out

{j

8. n/3)

9n/2

p. 390

3n

5.

2

3n

6.

8

7. 2n

2

8.

2

3n

2

16.

10)

6

+

1 +

J 1+

e

2

J l +e

2

J e

- 1 1

+

fi 4

( 1 +

7. e

2;1 )

IOg( 41:

+ IOg

1

+

- v h2

)

IOg

'j

(

17

fi - 1

log

+ !

(

+4

4. 2J17 +

+ 1)

fi

- -1

e

Exercise 8

full.)

in

3/4 length

3J3

+

14. 34 15. sJ5

13. 34

2

10 3

12.

3

p.

e

4.

11.102

2

+

9n

3. n

2

7.

3n/2

Exercises,

Supplementary

1. 25n 2. 9.

6.

5. 3n/2

4.

A75)

TO EXERCISES)

ANSWERS

= f 0

2x _ 2 (1 x )

1 +

(1 _

J

3/4

J

-

J l

3/4

-_

2X2

X

2

dx )2

+ x

4

+ 4x 2 dx

2

1 -x)

fo

2

J (1 + X )2 dx 1 -x) 2

3/4

-_

dx)

4X2

=

f 0

2

1 +

fo

3/4 1 + x 2

_-

1

fo

- x2

3/4

-_

3/4 x 2

2 dx 2

1 -x

fo

\037

f

= -log(1 1 +

= log (

1 _

+

- x) + log(1 3/4 3/4 )

- 1 2

o-X1

dx

3/4

1

1

2 ( I-x

0

+ f

3/4

=2

dx

l+x )

_

dx

7

-

0

3/4

3

o)

4

+ x)

-1 = log

dX f

3/4)))

- 4)

1/4

-

9. \037

10. log(2 + )3))

n

(e

XII, *4,

407)

p.

2

3. J2(e

2. 21tr

2 log

+

82

J2(e

22. 1t

23.

1 (b) 3

2J5

9.

8

8. 5

7. 2)3

+ 2)

+ 10g(J5

12. J2(e 2

10. 4a

-

e)

1)

[

1 +

]

(e-

4

T

4 sin

18.

J5

If7

IS.

5312)

8 + 2ft

log

S.

i

- e-

= 2J2

s

31t

)

16.\"\"4

- J2 19.

20.

4)

2)

2)3)

p. 415)

*5,

l. 121ta2 j5

2. \037

6.

-

14. (8 3/2 -

- e 81 )

+

XII,

4. (a)

e)

2 \037

ft 17.J5 - 4

21. 8

-

+ 1

6. 4J2 13.

TO EXERCISES)

ANSWERS

A76)

-

(17ft

-

(lOJiO

1) 3.

2; (26)26

2J2) 4.

2

41t

a

2

2aR

S. 41t

1))

\037

XII,

\0376,

1. 5

418)

p.

in.-Ib

2.

6.

c

1- 1 [r1

2 x

]

106

8. (a) -90

and

4.

\037

10. 1500

Ib/in.;

\037

sin

3. c

180

10 80

Ib/in.;

yes;

r1

99c

200

E

CmM

dyne-cm

6

where c is

(b)

log 2 in.-pounds. If A is the pressure, then) P(x)

is the

9 CmM

Ax =

=

But

C =

20. 75 =

Q

1,500

in.-lb J

rIty

dx

=

C=

the

) of the

constant.)

dx = f Q

data.

'2

section

cross

2Q

P(x)A

initial

2-

dyne-cm 9. C \037 r ( 1

f Q from

0 f proportlona

constant

2Q

Force f

9

75 in),

when the volume is

2Q

Work

1t csc

.

the

area of

is the

the cylinder

of

length

1 2

in.-Ib

P(x). If a

1t cot 9

9

5.

7.

pound-miles

--;- [

This gives

-C X)

the

dx

=

then)

Clog

answer.)))

2.

cylinder,

TO EXERCISES)

ANSWERS

XII,

4

- 54

154

1. \037

3 15 _ 53)

(

XIII,

423)

p.

\0377,

3. 10flog 3)

10

2.

)

p. 434)

1, \037

1\302\267 (a ) f (k) ( x ) (b)

A77)

= (

1)k + 1( k - 1) '. (1 + xt l(k -

I)!

= (_I)k+

f(k)(O)

(c) Since

- 1)!/k!

(k

=

get froln

we

Ilk,

f(k)(O)

(

(b))

- l)k + 1(k -

k!)

I)!

(

_

k!)

l)k

+

1

k)

and) II

f(k)(o)

=

p.(x)

This

is

proves

gi ven

(

_

XIII,

f(x) =

1. 1 -

cos x,

-x

polynomial

Taylor

+1 Xk,)

Use this

f(II+4)(X).

function f(x)

l)k

the

and

formula

for

= cos x.)

to

PII(X)

446)

p.

x4

2

2!

=

f(II)(X)

PII(x) for the

\0373,

n-th

the

k

k\037O

For

.)

by) II

derive

It

= 10g(1+ x) then

that when f(x)

P.(x) =

2.

x

k!

k\037O

-

+

2. I f(n)(c)1

4! < 1 for

all

n

c so

numbers

all

and

follows

estimate

the

from

Theorem 2.1. 3. 1

-

5. /R4/ 6.

0.01

2

+

R 4 (0.1)

= 0.995+

R

4. /R3/
O)))

21.

22.

23.

24.

25.

26.

27.

29.

30.

31.

32.

33.

34.

36.

37.

38. 39.

40.

41.

v a2 +

f f

2 x V a2 + x 2

V a2

43.

x2

v' a2 +

x2

dx =

x2 x

f V a2

= V a2 +

dx

x+

f

x2

-

V a2

f x

2

v' a 2 -

V a2 -

-1

sin

-

x2

dx =

x

v' a2 f

-

x2

dx

-

-

x:l

- a 2)\"

(v'

=

2

2

x2

a2

xV

- a2

x(v'x 2

_a

2 )2-n

- n)a 2

(2

( V- 2

)\"dx=

2

)

+ c

+ c

dx j x 2V a 2 -

35.

-1 x

cosh

a

a 21+

a 2 )n-2

n-3

-

-

v'a 2

x2

x2

a 2-x)

+ C

C

(v' x 2 -

f

n\03721

= _

+ C

-

(n

2

+ C

- x 2 -;- C

V x2 -

Ix +

a 2 )ft :2+- 1

a)

+c

2

xcv'

-1 X-

1

1

a

-

2x2)

x2

+c

In

a

:2

a2)

x(v'x 2 -a

!

a

-

x

x

\037-

2

a -+- v' a 2 -

V a2 -

-

-: + C =

X \302\243 / V x2

dx =

n

-

x2

x2 (a

In

a+v'a2-x2 x l

a

= cosh -1

dx f

-

-1

sin

a 2

- a 2 dx =

x2

2

2

=_!In

dx

f V x2 -

(v'

a

x2

-

xVa2

\302\243 / V x2

1 X

-sin

= 81n. x2

+ c

xv' a 2 _

a

dx f v' a 2 -

\037

\037

x2 -

-

z::

dx =

dx

/

-1

sin \037

a

x2

v'a 2

28 .

I

x2

f

-

C

+

+ C

\037

v' a2

c

1

x2 +

\0374

f

dx 2

2)a

/

_ (V x 2

n \037 -1)

dx,

,

n\0372)

a 2 )\"-2

n+2 n\037-2

+C,

xn\037\037

x

2\302\243 / v x2 -

x2

f

a 2 dx =

- a2 x

Vx 2 f

x2

a 2x

x 2 dx =

+ c

x2

x

vi a2 +

x 2 dx =

+

V a2 +

a +

x2

-1

x V a 2 + x2 + 2

\037

a

\037

v' a2

f

-1

I

= _

+ c

\037

v' a 2 + x 2 + C x

a

In

a

-1

sinh \0374

a sinh

2

_!

-=

f x2Va2 +

f

+ c

\037

I \037I

\037 sinh

x2

+

-

x2

2

_

dx =

dx

f

1

a 2 + x2 _

-1 \037_

sinh

2

dx f x v' a 2 +

f

2 x(a +

dx =

-

sinh \037

2X\037V

f

f

x2 +

V a2 +

\037

vi

42.

x 2 dx =

-

x2

dx =

X

8

(2x

v. / x 2

-

\302\243 / 2 - a 2 )v x

2

a2 -

a sec

a2 -

-1

x I \037 I

a2

dx

==

cosh

-1 x a

2- Vx x)

a2

+

a

4

8

cosh

-1 x a

+ C

c

+c Con tinued

overleaf.)))

2

X

dx

44.

1

%2 _

V

==

1 xV x2 2

1

48.

x

1 I

dx

50 .

1 (V2ax -

51 .

- / 2ax - x2

I

x2

V2a;2V 2ax

1 55.

56. 58.

60.

ax dx

sin

f

. 2 ax

sin

1

.\"

sin ax

1

11

cos

61 .

1 62.

=

x

ax dx

==

cos

..

sin ax sin bx dx

1

.

sIn

==

==

cos ax cos bx dx =

1

ax cos ax dx

=

(a

sin (a

\"

sin ax cos ax

. ,,+1

dx =

n

n

-

sin

(n +

ax

l)a

)

+ c

+C x 2 ),,_2

dx,

dx

1

_ x 2 )\"-2)

(V2ax

-1 X-- - a + C

a

dx =

-

1

b)x

+

C)

+

C,)

sin

11-2

cos

_

b)x

b)

(a + b)x

sin (a

+ b)x

2(a +

a

+ c,

I

a

,)

a

2(a + b)

+

ax dx ==

x 2

+

C

sin 2ax 4 a)

+C

ax dx

cos(a 2(a

sin ax+

\037

ax dx

sin

1

1

2 COS

f

. \"-2

1

n)

n)

2( a _ b)

4a

.

+

b)

-

(

a

+C

59.

- b)x -

_ 2(a

a

a a)+c 2

- cos (a + b)x 2(a + b) sin

-

X

+ c

+

cos 2ax

-

1

1

ax

na

-1

+C

fcosax

ax cos ax

11-1. ax sin

2

57.

na ==

-a3 sin.

+

sin

x2

- 3)

x2

- sin-1

==

(V 2ax _

I

- 2)a2

(n

a

4a)

. ,,-1

1)

(n

- a

- sin 2ax + C

-sin

n+

+

+ C

cos ax \037

d x == 2

dx

x

a'\\J

x2

x 2 )2-\"

)

2

na +

(x

- x

/ 2a

_!

x 2 )\"

-v2ax-x

a

sin ax cos boC dx

1

(c)

64.

l x

- a

.

(a)

(b)

63.

-

xV2ax

x -:

- a

x

2' sin-a(

. -l x a sin

+

2a

x2

-

x2

V 2ax -

1

l

- 3a)V2ax6

a)(2x

-

2

- 2)a2

_2\037 ==asin-

-

dx

1

-

dx =

xdx

54.

a)(V2ax

-/ dx == v 2 ax

x

53.1

1

= (x+

dx

a

- z2 +

n+

(n

dx

47.

- a)(V2ax-

-

(x

Ix I

a

2az

x2)\"

V2tlX-X2

52.

V

2

--

xv

1

x-a

C

! cos- 1 \037 + c

==

c

+

- a2 +

z2

+ C

a 2x

(x x2 )\" dx =

-

(V 2ax

Ia I

a2

- x 2 dx =

2ax

V

49.

V

1 \037

V x2 - a 2

= -

x2

X + -2 V

a

!seca

a2

dx

46.

2

a2

dx

45.

2

- cosh-1 x-

a

==

2

2 \037b

2 2 -' r- b

2

2 \037 b

b)

n \037-1)

This table is

continued on

the endpapers

at the

back.)))

65. 66.

67.

cos ax

f

f

1

. = - In ISln axl + C

dx -;-sin ax

a)

-

sin ax f cos ax

f

n-I

. sin

-

dx =

ax

axl + C

Icos

a)

sin ax cos

n \037-1)

(n+ l)a

1 - -In

=

dx

\"'\"

68.

n+1 - cos ax + C,)

n. cos ax sin ax dx =

ax cos

+

a(m

\",+1

ax

-

1

m+

n)

n

+

n)

. n-2

n \037-m)

69.

70.

f

. n '\" sm ax cos ax

b+::max

J

b

f

1+

71.

72.

73. 74.

75.

76.

78.

so.)

= _!

d:

sin

ax

b + C cos ax

f

b + C cos ax

x sin ax

cLx

n

f

ax 2

tan

=

sin ax

:2

-

x

sin ax dx

=

cos ax dx

= x.

n

81.)

82.

84.

n X

f

ax

ftan

f

87. 88.

f

f

sin

-

=

ax

: tan

(

-1

-

-

C

b +

e

{b

[ 'J e

b cos ax

+

b

cot

=

ax dx

ax

see ax

dx =

cLx

=

tan

0-1ax

cos ax \037

n

+

In lsec ax

\037f

+

2

x

n-l

x

n-l.

cos ax

sin ax

b

+C,

2)

2

< c

2

+ c, b2

b

2

> e

sin ax

f

85.

axl + C

- cos ax

x cos ax

2

+

C'

=

- -1

l

a

cLx ==

< e

b

-ax +

cot

C

2)

cos ax +

:2

2

sin ax \037

+

C)

d.x

83.

n-2

f

1

2

dx)

+ C

tan

> c

l

c cos ax

79.

C)

\037f

- n

-

No. 87.))

use

b

C,

dx

tan ax dx, a(n _ 1) f 0-1 ax - cot - cot n-2 ax dx, a(n _ 1) f

\037

2]

77.

ax +

-n,

b2cosax

+ Ve2 -

+

axdx=\037tanax-x+C

tan

ax

tan

+ C

Icosaxl

:- - -a;)] +

ax d x,

+ C

I

b2

-

m =

(If

\037c2 C sIn ax

+

1

2

tan

n

f

-;

\037In

n 86.

cLx

tan

n

-; cos

\037-n)

\",-2

n

sm ax cos

86.))

+ C

2)

In

aV c2 _ a

.

ax

1

1 = ax

ax

2)

2

=

cos

X

!+

-

aV b2 _ e2

d.x

dx

( 4

( 4

=

J

!

tan

tan

a

ax

sin

dx

f

!

=

d:

a

b

l

b2

m

bsinax+

e+

In

+

.

J

m+n)

[ \037:

-1

aV c2 _

+

t.U,

-m, use No.

n =

(If

- 1

m

ax

n\037-I

-I

tan

=

dx.

f 1+

am ( +

aVb\037\037C2

+ C sin ax

f 1-

=

cLx

=

.

J

. 0+1 sin ax cos

ax ax cos\"'..J_

sin

f

n

f

2

f

cot

cLx '\"'

In Isin axl

\037

ax cLx =

- -1 cot a

ax

+ C - x+

C

\037 1

n

89.

cotax

f

\037 1

csc ax

cLx

=

- -1 In {J.

Icsc ax

+ cot axl

Con tlnued

+ C

overleaf.)))

90.

92. 93.

94.

95.

96.

97. 98.

99.

100.

101.

102.

104.

106. 107.

108. 109.

110. 111. 113. 115. 11.7

2

f

sec

see\"-2ax

dx --

-

\"

see ax

f

\"

csc ax cot ax

f

-1

oos-l ax l

I

x cos

f f

f f

f

x tan

f

e

a%

x

\"+1

n

o.

ea.

1 e \037

az

+

cos bx dx

=

bx

ax dx

x

x

=

ax

dx =

dx =

C

a \037b

f

b2

-

x

2

n

ax dx

sin h ax

dx

=

=

f

VI

+ 1f

VI

-a +

1

+

i

(In

x\"+l

n-I

dx

'

n \037-1)

a2x2)

dx

..

,

f

b

+ C

1 b

=

dx

\037-1)))

az

\037In o,

xne

dx =

b

+ C,

xne

o,

> 0,

b

\037f

\037

x

-

,,+1

1)2

+

+, C

112.

C

+ C

-

+

C

ax coshax na

f

-

n

cosh

116.

cosh2 ax

- 1 n)

x I\037ax

114.

f

x 2

\037 -1

n

.

f

f

sin h

= In Iin axl + C

ax dx

=

ax +

sinh \037

n- 2

ax, dx

dx =

n

2ax

sinh

4a -Jt ..,...

0

+

b \037 1

Xn-leo, dx

b \037 1

> 0,

cos bx)

n

+ C

ax)2 +

4a

f

b

b sin bx) + C

n \037 -1)

a-

dx,

(a cos bx +

(n

sinh 2ax

o.

,

a 2 x 2)

-

105. xn-lb

dx

-

1 + a 2x 2

f

1

b

In ax

cosh ax

sinh

\"+1

X\"+I

(a sin bx

,,+1

\037

sinh

.

ax

x

a n

n

1) +

a 2\037 b 2

In ax

+ C

103.

a 2 \037

n

x-lln

a n+ I

ax _

1

-

:::0: =

In

-I

sin

C

2 + a i)

In (1

1

+

-

(ax

dx

\"

-

C

+

::

ax dx =

1

- a2 x 2 +

V I

x\"+1 -I ax dx = tan ax -

-I

dx =

ea. sin

-

ax

1

n \037 1)

dx,

- a2 x 2 + C

V I \037

2\037

-

dx =

xnb

.

f

ax

l

ax

n\037O)

c,

+

x\"+1 -I axdx = cos ax+-

dx =

xe\"'

fsinh

f

ax dx =

n

fin f

l

cos-

csc\"-2

n \037 1)

ax

ax +

n +

\"

f

\"

\037

-I

n

f

dx = x

ax dx = x tan-

sin-

x\"

l

sin-

2

+ C)

n\037O)

C,

+

csc

_

-

;; _ 1 f

cot ax

\037

ax

na

dx =

dx = x

ax

ftan-

f

n

see

=

1f

1

-

2 cse ax dx =

f

n-2 see ax dx,

2

n

+

1)

na)

fsin

f

ax cot ax

_ a( n

ax dx

tan

n-

n

+

1)

\"-2

csc

91.

tan ax

an(

n csc ax dx =

f

+ C)

tan ax \037

see\" ax

f

=

ax dx

C x

2+

c

118.)

119.)

120.)

121.)

122.)

123.

125.

f f

f f f f

X

136.)

137.)

138.)

cosh ax dx

=

tanh

\037-:)a

f

dx =

_

-I

sin

=

In

2 csch

f

ax dx

\037I

I

-

=

f

seeh n ax

dx =

csch n ax

dx =

tanh

ax dx

n

csch ax coth ax

f

(n

-

sech1& ax

csch

e

az

az e

f

-e

=

n

e

-az 2

r(n) =

2 1f

n

ax

e

(n

ax dx \"\" x

-

coth \037

n y4

+ C ax +

C

1)

n \037 1)

2

ax dx \"\"

fsech

n --

-

see h

csc 1)

f

n\037O)

+ C,

n\037O)

]

+

C,

a

+

C,)

a

-h

+

a _ b]

-

I)!,

,t-2

2

e

[a + b

2

+ C,

- -a-b-h

bz

e

e

=

cosh bx dx

xn-1e-' dx =

lo

e

2 [ a+b az h

2\"

QO

-

az

.

sInh bx dx

n --

+

n

na)

f

coth

In Isinh axl

ax +

tanh \037

C

C

- l)a

-

f

ax dx \"\"

\037

dx,)

131.

na

dx =

ax

+ C

- l)a

=

coth

C

cschn-2 ax coth ax

-

f

11.-2 ax dx,

cothn-2

f

n\"'O)

dx)

126. tanh

f

sechn-2 ax tanh ax (n

n

sech ax

f

+

ax +

coth \037

f

ax I)a

tanh

\037

sInh ax

124.

(tanh ax) +

\037

ax dx

csch

n-l

cot h (n

sech ax

n-l.

C

+

d x,

ax

cosh ax dx

+ C

ax

tanh

-

x

ax +

tanh \037

(n

II. coth ax dx =

n-l

\037f

-

-

=

ax dx

x

n

-

ax

In (cosh ax)

= x

ax dx

n

h

tan

f

2

h

-t 1 n r-

ax d x ')

n-2

ax

dX,

n \037 1)

2) \037b

2

2) \037b

n > O.)

dx=-

1

7r

-,

2 \037a)

a >

0)

I, 3 .

r/2.

r/2

i

. sinh

-;

n-2

+ C

\037f

n

x

cosh

f

ax + C

n

-

ax

cosh

1

n

cosh

:2

n

-

n

sinh ax

a

\037

\037

1\"

141.)

-

-;

ax dx \"\"

tanh

139.)

140.)

=

X

f

135.)

sinh ax

X

sIuh ax dx

2

132.

-

a

\037

f

f

ax

\037 cosh

cosh ax dx \"\"

x

130.

134.

=

ax dx

ax sinh ax + na

coshn-l

ax d x =

x sinh

n

128.)

133.

n

II. .

127.

129.

cos h

. II. SIn x

dx =

n

i

cos

x

5 . . . (n

-

I)

.

2.4.6...n

d.x =

2 .

{

4

.

6

7r

_ ,

if

2

\302\267 . .

(n

- 1) ,

3.5.7\"'n)

if

n is an

n is an

even

odd

integer

integer

> 3)

> 2,)))