The purpose of a first course in calculus is to teach the student the basic notions of derivative and integral, and the
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Data
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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
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in 1978,
1973, 1968,1964by
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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,)))