340
||||
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
tational attractions of the earth and the sun) balance each
other. These locations are called libration points. (A solar
research satellite has been placed at one of these libration
points.) If m1 is the mass of the sun, m 2 is the mass of the
earth, and r 苷 m 2兾共m1 m 2 兲, it turns out that the x-coordinate of L 1 is the unique root of the fifth-degree equation
p共 x兲 苷 x 5 共2 r兲 x 4 共1 2r兲 x 3 共1 r兲 x 2
2共1 r兲 x r 1 苷 0
and the x-coordinate of L 2 is the root of the equation
p共 x兲 2rx 2 苷 0
4.9
Using the value r ⬇ 3.04042 10 6, find the locations of the
libration points (a) L 1 and (b) L 2.
y
L¢
sun
earth
L∞
L¡
L™
x
L£
ANTIDERIVATIVES
A physicist who knows the velocity of a particle might wish to know its position at a given
time. An engineer who can measure the variable rate at which water is leaking from a tank
wants to know the amount leaked over a certain time period. A biologist who knows the
rate at which a bacteria population is increasing might want to deduce what the size of
the population will be at some future time. In each case, the problem is to find a function
F whose derivative is a known function f. If such a function F exists, it is called an antiderivative of f.
A function F is called an antiderivative of f on an interval I if
F共 x兲 苷 f 共 x兲 for all x in I .
DEFINITION
For instance, let f 共 x兲 苷 x 2. It isn’t difficult to discover an antiderivative of f if we keep
the Power Rule in mind. In fact, if F共 x兲 苷 13 x 3, then F共 x兲 苷 x 2 苷 f 共 x兲. But the function
G共 x兲 苷 13 x 3 100 also satisfies G共 x兲 苷 x 2. Therefore both F and G are antiderivatives
of f . Indeed, any function of the form H共 x兲 苷 13 x 3 C, where C is a constant, is an antiderivative of f . The question arises: Are there any others?
To answer this question, recall that in Section 4.2 we used the Mean Value Theorem to
prove that if two functions have identical derivatives on an interval, then they must differ
by a constant (Corollary 4.2.7). Thus if F and G are any two antiderivatives of f , then
y
y= 3 +3
˛
F共 x兲 苷 f 共 x兲 苷 G共 x兲
˛
y= 3 +2
so G共 x兲 F共 x兲 苷 C, where C is a constant. We can write this as G共 x兲 苷 F共 x兲 C, so we
have the following result.
˛
y= 3 +1
y= ˛
0
x
3
˛
y= 3 -1
˛
y= 3 -2
FIGURE 1
Members of the family of
antiderivatives of ƒ=≈
1 THEOREM If F is an antiderivative of f on an interval I , then the most general
antiderivative of f on I is
F共 x兲 C
where C is an arbitrary constant.
Going back to the function f 共 x兲 苷 x 2, we see that the general antiderivative of f is
C. By assigning specific values to the constant C, we obtain a family of functions
whose graphs are vertical translates of one another (see Figure 1). This makes sense
because each curve must have the same slope at any given value of x.
1 3
3x
SECTION 4.9 ANTIDERIVATIVES
||||
341
EXAMPLE 1 Find the most general antiderivative of each of the following functions.
(a) f 共 x兲 苷 sin x
(b) f 共 x兲 苷 1兾x
(c) f 共 x兲 苷 x n, n 苷 1
SOLUTION
(a) If F共 x兲 苷 cos x, then F共 x兲 苷 sin x, so an antiderivative of sin x is cos x. By
Theorem 1, the most general antiderivative is G共 x兲 苷 cos x C.
(b) Recall from Section 3.6 that
1
d
共ln x兲 苷
dx
x
So on the interval 共0, 兲 the general antiderivative of 1兾x is ln x C. We also learned
that
1
d
共ln ⱍ x ⱍ兲 苷
dx
x
for all x 苷 0. Theorem 1 then tells us that the general antiderivative of f 共 x兲 苷 1兾x is
ln ⱍ x ⱍ C on any interval that doesn’t contain 0. In particular, this is true on each of the
intervals 共, 0兲 and 共0, 兲. So the general antiderivative of f is
F共 x兲 苷
再lnln
x C1
共x兲 C2
if x 0
if x 0
(c) We use the Power Rule to discover an antiderivative of x n. In fact, if n 苷 1, then
d
dx
冉 1冊
x n1
n
苷
共n 1兲 x n
苷 xn
n1
Thus the general antiderivative of f 共 x兲 苷 x n is
F共 x兲 苷
x n1
C
n1
This is valid for n 0 since then f 共 x兲 苷 x n is defined on an interval. If n is negative
(but n 苷 1), it is valid on any interval that doesn’t contain 0.
M
As in Example 1, every differentiation formula, when read from right to left, gives rise
to an antidifferentiation formula. In Table 2 we list some particular antiderivatives. Each
formula in the table is true because the derivative of the function in the right column
appears in the left column. In particular, the first formula says that the antiderivative of a
constant times a function is the constant times the antiderivative of the function. The second formula says that the antiderivative of a sum is the sum of the antiderivatives. (We use
the notation F 苷 f , G 苷 t.)
2
TABLE OF
ANTIDIFFERENTIATION FORMULAS
To obtain the most general antiderivative from
the particular ones in Table 2, we have to add a
constant (or constants), as in Example 1.
N
Function
Particular antiderivative
cf 共 x兲
cF共 x兲
f 共 x兲 t共 x兲
F共 x兲 G共 x兲
x n1
n1
x n 共n 苷 1兲
1兾x
ex
cos x
ln ⱍ x ⱍ
ex
sin x
Function
Particular antiderivative
sin x
sec2x
sec x tan x
1
s1 x 2
1
1 x2
cos x
tan x
sec x
sin1x
tan1x
342
||||
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
EXAMPLE 2
Find all functions t such that
2 x 5 sx
t共 x兲 苷 4 sin x SOLUTION
x
We first rewrite the given function as follows:
t共 x兲 苷 4 sin x 2x 5
x
sx
x
苷 4 sin x 2 x 4 1
sx
Thus we want to find an antiderivative of
t共 x兲 苷 4 sin x 2 x 4 x1兾2
Using the formulas in Table 2 together with Theorem 1, we obtain
t共 x兲 苷 4共cos x兲 2
x5
5
x1兾2
1
2
C
苷 4 cos x 25 x 5 2 sx C
M
In applications of calculus it is very common to have a situation as in Example 2, where
it is required to find a function, given knowledge about its derivatives. An equation that
involves the derivatives of a function is called a differential equation. These will be
studied in some detail in Chapter 9, but for the present we can solve some elementary differential equations. The general solution of a differential equation involves an arbitrary
constant (or constants) as in Example 2. However, there may be some extra conditions
given that will determine the constants and therefore uniquely specify the solution.
Figure 2 shows the graphs of the function f in
Example 3 and its antiderivative f . Notice that
f 共 x兲 0, so f is always increasing. Also notice
that when f has a maximum or minimum, f
appears to have an inflection point. So the graph
serves as a check on our calculation.
N
Find f if f 共 x兲 苷 e x 20共1 x 2 兲1 and f 共0兲 苷 2.
The general antiderivative of
20
f 共 x兲 苷 e x 1 x2
EXAMPLE 3
SOLUTION
is
40
f 共 x兲 苷 e x 20 tan1 x C
To determine C we use the fact that f 共0兲 苷 2:
fª
f 共0兲 苷 e 0 20 tan1 0 C 苷 2
_2
3
f
Thus we have C 苷 2 1 苷 3, so the particular solution is
f 共 x兲 苷 e x 20 tan1 x 3
_25
FIGURE 2
Find f if f 共 x兲 苷 12 x 2 6 x 4, f 共0兲 苷 4, and f 共1兲 苷 1.
The general antiderivative of f 共 x兲 苷 12 x 2 6 x 4 is
V EXAMPLE 4
SOLUTION
x3
x2
4 x C 苷 4 x 3 3x 2 4 x C
3
2
Using the antidifferentiation rules once more, we find that
f 共 x兲 苷 12
f 共 x兲 苷 4
x4
4
3
x3
3
6
4
x2
2
Cx D 苷 x 4 x 3 2 x 2 Cx D
M
SECTION 4.9 ANTIDERIVATIVES
||||
343
To determine C and D we use the given conditions that f 共0兲 苷 4 and f 共1兲 苷 1. Since
f 共0兲 苷 0 D 苷 4, we have D 苷 4. Since
f 共1兲 苷 1 1 2 C 4 苷 1
we have C 苷 3. Therefore the required function is
f 共 x兲 苷 x 4 x 3 2 x 2 3x 4
If we are given the graph of a function f , it seems reasonable that we should be able to
sketch the graph of an antiderivative F. Suppose, for instance, that we are given that
F共0兲 苷 1. Then we have a place to start, the point 共0, 1兲, and the direction in which we
move our pencil is given at each stage by the derivative F共 x兲 苷 f 共 x兲. In the next example
we use the principles of this chapter to show how to graph F even when we don’t have a
formula for f . This would be the case, for instance, when f 共 x兲 is determined by experimental data.
y
y=ƒ
0
1
2
3
4
x
FIGURE 3
y
y=F(x)
2
1
0
1
FIGURE 4
M
x
V EXAMPLE 5 The graph of a function f is given in Figure 3. Make a rough sketch of
an antiderivative F, given that F共0兲 苷 2.
SOLUTION We are guided by the fact that the slope of y 苷 F共 x兲 is f 共 x兲. We start at the
point 共0, 2兲 and draw F as an initially decreasing function since f 共 x兲 is negative when
0 x 1. Notice that f 共1兲 苷 f 共3兲 苷 0, so F has horizontal tangents when x 苷 1 and
x 苷 3. For 1 x 3, f 共 x兲 is positive and so F is increasing. We see that F has a local
minimum when x 苷 1 and a local maximum when x 苷 3. For x 3, f 共 x兲 is negative
and so F is decreasing on 共3, 兲. Since f 共 x兲 l 0 as x l , the graph of F becomes flatter as x l . Also notice that F 共 x兲 苷 f 共 x兲 changes from positive to negative at x 苷 2
and from negative to positive at x 苷 4, so F has inflection points when x 苷 2 and x 苷 4.
We use this information to sketch the graph of the antiderivative in Figure 4.
M
RECTILINEAR MOTION
Antidifferentiation is particularly useful in analyzing the motion of an object moving in a
straight line. Recall that if the object has position function s 苷 f 共t兲, then the velocity function is v共t兲 苷 s共t兲. This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is a共t兲 苷 v共t兲, so the velocity function is
an antiderivative of the acceleration. If the acceleration and the initial values s共0兲 and v共0兲
are known, then the position function can be found by antidifferentiating twice.
V EXAMPLE 6
A particle moves in a straight line and has acceleration given by
a共t兲 苷 6 t 4. Its initial velocity is v共0兲 苷 6 cm兾s and its initial displacement is
s共0兲 苷 9 cm. Find its position function s共t兲.
SOLUTION Since v共t兲 苷 a共t兲 苷 6 t 4, antidifferentiation gives
v共t兲 苷
6
t2
2
4t C 苷 3t 2 4t C
Note that v 共0兲 苷 C. But we are given that v 共0兲 苷 6, so C 苷 6 and
v 共t兲 苷
3t 2 4t 6
344
||||
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
Since v 共t兲 苷 s共t兲, s is the antiderivative of v :
s共t兲 苷 3
t3
3
4
t2
2
6t D 苷 t 3 2t 2 6t D
This gives s共0兲 苷 D. We are given that s共0兲 苷 9, so D 苷 9 and the required position
function is
s共t兲 苷 t 3 2 t 2 6 t 9
M
An object near the surface of the earth is subject to a gravitational force that produces
a downward acceleration denoted by t. For motion close to the ground we may assume that
t is constant, its value being about 9.8 m兾s2 (or 32 ft兾s2 ).
A ball is thrown upward with a speed of 48 ft兾s from the edge of a cliff
432 ft above the ground. Find its height above the ground t seconds later. When does it
reach its maximum height? When does it hit the ground?
SOLUTION The motion is vertical and we choose the positive direction to be upward. At
time t the distance above the ground is s共t兲 and the velocity v 共t兲 is decreasing. Therefore,
the acceleration must be negative and we have
EXAMPLE 7
a共t兲 苷
dv
苷 32
dt
Taking antiderivatives, we have
v 共t兲 苷 32 t C
To determine C we use the given information that v 共0兲 苷 48. This gives 48 苷 0 C, so
v 共t兲 苷 32 t 48
The maximum height is reached when v 共t兲 苷 0, that is, after 1.5 s. Since s共t兲 苷 v 共t兲, we
antidifferentiate again and obtain
s共t兲 苷 16 t 2 48t D
Using the fact that s共0兲 苷 432, we have 432 苷 0 D and so
s共t兲 苷 16 t 2 48t 432
Figure 5 shows the position function of the
ball in Example 7. The graph corroborates the
conclusions we reached: The ball reaches its
maximum height after 1.5 s and hits the ground
after 6.9 s.
The expression for s共t兲 is valid until the ball hits the ground. This happens when s共t兲 苷 0,
that is, when
500
or, equivalently,
N
16 t 2 48t 432 苷 0
t 2 3t 27 苷 0
Using the quadratic formula to solve this equation, we get
t苷
0
FIGURE 5
8
3 3 s13
2
We reject the solution with the minus sign since it gives a negative value for t. Therefore
the ball hits the ground after 3(1 s13 )兾2 ⬇ 6.9 s.
M
SECTION 4.9 ANTIDERIVATIVES
4.9
13.
345
EXERCISES
1–20 Find the most general antiderivative of the function. (Check
your answer by differentiation.)
1
1. f 共 x兲 苷 x 3
2. f 共 x兲 苷 2 x 2 2 x 6
1
3
4
3. f 共 x兲 苷 2 4 x 2 5 x 3
4. f 共 x兲 苷 8 x 9 3 x 6 12 x 3
5. f 共 x兲 苷 共x 1兲共2 x 1兲
6. f 共 x兲 苷 x 共 2 x 兲 2
7. f 共 x兲 苷 5x 1兾4 7x 3兾4
8. f 共 x兲 苷 2 x 3x 1.7
6
4 3
x
x s3 x 4
9. f 共 x兲 苷 6 sx s
10. f 共 x兲 苷 s
5 4x 3 2x 6
10
11.
||||
f 共 x兲 苷
x9
u 4 3 su
f 共u兲 苷
u2
cos 5 sin f 共 x兲 苷 5e x 3 cosh x
x 5 x 3 2x
12. t共 x兲 苷
14.
15. t共 兲 苷
16.
17.
18.
19. f 共 x兲 苷
x
4
20.
x6
f 共 x兲 苷 3e x 7 sec 2 x
43.
44.
45.
46.
47.
48.
f 共 x兲 苷 2 cos x, f 共0兲 苷 1, f 共兾 2兲 苷 0
f 共 t 兲 苷 2 e t 3 sin t, f 共0兲 苷 0, f 共 兲 苷 0
f 共 x兲 苷 x 2, x 0, f 共1兲 苷 0, f 共2兲 苷 0
f 共 x 兲 苷 cos x, f 共 0 兲 苷 1, f 共0兲 苷 2, f 共0兲 苷 3
Given that the graph of f passes through the point 共1, 6兲
and that the slope of its tangent line at 共 x, f 共 x兲兲 is 2 x 1,
find f 共2兲.
Find a function f such that f 共 x兲 苷 x 3 and the line x y 苷 0
is tangent to the graph of f .
The graph of a function f is shown. Which graph is an
antiderivative of f and why?
49–50
f 共 t 兲 苷 sin t 2 sinh t
f 共 x兲 苷 2sx 6 cos x
2 x2
f 共 x兲 苷
1 x2
49.
y
50.
f
y
f
b
a
a
x
x
b
c
; 21–22 Find the antiderivative F of f that satisfies the given con-
dition. Check your answer by comparing the graphs of f and F.
21. f 共 x兲 苷 5x 4 2 x 5, F共0兲 苷 4
22. f 共 x兲 苷 4 3共1 x 2 兲1, F共1兲 苷 0
Find f .
f 共 x兲 苷 6 x 12 x 2
The graph of a function is shown in the figure. Make a rough
sketch of an antiderivative F, given that F 共0兲 苷 1.
51.
y
y=ƒ
23– 46
23.
24.
25. f 共 x兲 苷
2 2兾3
3x
26.
27. f 共t兲 苷
et
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
f 共 x兲 苷 2 x x
f 共 x兲 苷 6 x sin x
f 共t兲 苷 t st
3
6
f 共 x兲 苷 1 6 x, f 共0兲 苷 8
f 共 x兲 苷 8 x 3 12 x 3, f 共1兲 苷 6
f 共 x兲 苷 sx 共6 5 x兲, f 共1兲 苷 10
f 共 x兲 苷 2 x 3兾x 4, x 0, f 共1兲 苷 3
f 共 t 兲 苷 2 cos t sec 2 t, 兾 2 t 兾 2, f 共兾3兲 苷 4
f 共 x 兲 苷 共x 2 1兲兾x, f 共 1 兲 苷 21, f 共1兲 苷 0
f 共 x 兲 苷 x 1兾3, f 共 1 兲 苷 1, f 共1兲 苷 1
f 共 x兲 苷 4兾s1 x 2 , f ( 12 ) 苷 1
f 共 x兲 苷 24 x 2 2 x 10, f 共1兲 苷 5, f 共1兲 苷 3
f 共 x兲 苷 4 6 x 40 x 3, f 共0兲 苷 2, f 共0兲 苷 1
f 共 兲 苷 sin cos , f 共0兲 苷 3, f 共0兲 苷 4
f 共 t 兲 苷 3兾s t , f 共4兲 苷 20, f 共4兲 苷 7
f 共 x兲 苷 2 12 x, f 共0兲 苷 9, f 共2兲 苷 15
f 共 x兲 苷 20 x 3 12 x 2 4, f 共0兲 苷 8, f 共1兲 苷 5
c
0
x
1
The graph of the velocity function of a particle is shown in
the figure. Sketch the graph of the position function.
52.
√
0
53.
t
The graph of f is shown in the figure. Sketch the graph of f
if f is continuous and f 共0兲 苷 1.
y
2
y=fª(x)
1
0
_1
1
2
x
346
||||
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
; 54. (a) Use a graphing device to graph f 共 x兲 苷 2 x 3 sx .
E and I are positive constants that depend on the material
of the board and t 共 0兲 is the acceleration due to gravity.
(a) Find an expression for the shape of the curve.
(b) Use f 共L兲 to estimate the distance below the horizontal at
the end of the board.
(b) Starting with the graph in part (a), sketch a rough graph
of the antiderivative F that satisfies F共0兲 苷 1.
(c) Use the rules of this section to find an expression for F共 x兲.
(d) Graph F using the expression in part (c). Compare with
your sketch in part (b).
; 55–56 Draw a graph of f and use it to make a rough sketch of
the antiderivative that passes through the origin.
sin x
55. f 共 x兲 苷
, 2 x 2
1 x2
56. f 共 x兲 苷 sx 4 2 x 2 2 1, 1.5 x 1.5
A particle is moving with the given data. Find the position of the particle.
57. v共t兲 苷 sin t cos t, s共0兲 苷 0
58. v共t兲 苷 1.5 st , s共4兲 苷 10
59. a共t兲 苷 t 2, s共0兲 苷 1, v共0兲 苷 3
60. a共t兲 苷 cos t sin t, s共0兲 苷 0, v共0兲 苷 5
61. a共t兲 苷 10 sin t 3 cos t, s共0兲 苷 0, s共2兲 苷 12
62. a共t兲 苷 t 2 4 t 6, s共0兲 苷 0, s 共1兲 苷 20
y
0
69.
57–62
63.
64.
A stone is dropped from the upper observation deck (the
Space Deck) of the CN Tower, 450 m above the ground.
(a) Find the distance of the stone above ground level at time t.
(b) How long does it take the stone to reach the ground?
(c) With what velocity does it strike the ground?
(d) If the stone is thrown downward with a speed of 5 m 兾s,
how long does it take to reach the ground?
Show that for motion in a straight line with constant acceleration a, initial velocity v 0 , and initial displacement s 0 , the displacement after time t is
70.
71.
66.
67.
68.
An object is projected upward with initial velocity v 0 meters
per second from a point s0 meters above the ground. Show
that
关v共t兲兴 2 苷 v02 19.6 关s共t兲 s0 兴
Two balls are thrown upward from the edge of the cliff in
Example 7. The first is thrown with a speed of 48 ft兾s and the
other is thrown a second later with a speed of 24 ft兾s. Do the
balls ever pass each other?
A stone was dropped off a cliff and hit the ground with a
speed of 120 ft兾s. What is the height of the cliff?
If a diver of mass m stands at the end of a diving board with
length L and linear density , then the board takes on the
shape of a curve y 苷 f 共 x兲, where
EIy 苷 m t共L x兲 12 t共L x兲2
A company estimates that the marginal cost (in dollars per
item) of producing x items is 1.92 0.002 x. If the cost of
producing one item is $562, find the cost of producing 100
items.
The linear density of a rod of length 1 m is given by
共 x兲 苷 1兾s x , in grams per centimeter, where x is measured
in centimeters from one end of the rod. Find the mass of
the rod.
Since raindrops grow as they fall, their surface area increases
and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 m兾s and its
downward acceleration is
再90
if 0 t 10
if t 10
If the raindrop is initially 500 m above the ground, how long
does it take to fall?
a苷
72.
73.
s 苷 12 at 2 v 0 t s 0
65.
x
74.
75.
76.
0.9t
A car is traveling at 50 mi兾h when the brakes are fully
applied, producing a constant deceleration of 22 ft兾s2. What
is the distance traveled before the car comes to a stop?
What constant acceleration is required to increase the speed
of a car from 30 mi兾h to 50 mi兾h in 5 s?
A car braked with a constant deceleration of 16 ft兾s2, producing skid marks measuring 200 ft before coming to a stop.
How fast was the car traveling when the brakes were first
applied?
A car is traveling at 100 km兾 h when the driver sees an accident 80 m ahead and slams on the brakes. What constant
deceleration is required to stop the car in time to avoid a
pileup?
A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a共t兲 苷 60 t, at which time
the fuel is exhausted and it becomes a freely “falling” body.
Fourteen seconds later, the rocket’s parachute opens, and the
(downward) velocity slows linearly to 18 ft兾s in 5 s. The
rocket then “floats” to the ground at that rate.
(a) Determine the position function s and the velocity function v (for all times t). Sketch the graphs of s and v.