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You Try It: A rectangular box is to be placed in the ¬rst quadrant {(x, y) : x ≥
0, y ≥ 0} in such a way that one side lies on the positive x-axis and one side lies
on the positive y-axis. The box is to lie below the line y = ’2x + 5. Give the
dimensions of such a box having greatest possible area.


3.3 Related Rates
If a tree is growing in a forest, then both its height and its radius will be increasing.
These two growths will depend in turn on (i) the amount of sunlight that hits the
tree, (ii) the amount of nutrients in the soil, (iii) the proximity of other trees. We
may wish to study the relationship among these various parameters. For example,
if we know that the amount of sunlight and nutrients are increasing at a certain rate
then we may wish to know how that affects the rate of change of the radius. This
consideration gives rise to related rates problems.
EXAMPLE 3.11
A toy balloon is in the shape of a sphere. It is being in¬‚ated at the rate
of 20 in.3 /min. At the moment that the sphere has volume 64 cubic inches,
what is the rate of change of the radius?
SOLUTION
We know that volume and radius of a sphere are related by the formula
4π 3
V= (—)
r.
3
The free variable in this problem is time, so we differentiate equation (—) with
respect to time t. It is important that we keep the chain rule in mind as we do
so.1 The result is

dV dr
= · 3r 2 · . (——)
3
dt dt
1 The point is that we are not differentiating with respect to r.
92 CHAPTER 3 Applications of the Derivative

Now we are given that dV /dt = 20. Our question is posed at the moment that

V = 64. But, according to (—), this means that r = 3 48/π. Substituting these
values into equation (——) yields
2
4π dr
20 = ·3 ·
3
48/π .
3 dt
Solving for dr/dt yields
5
dr
= 2/3 1/3 .
48 · π
dt
EXAMPLE 3.12
A 13-foot ladder leans against a wall (Fig. 3.15).The foot of the ladder begins
to slide away from the wall at the rate of 1 foot per minute. When the foot is
5 feet from the wall, at what rate is the top of the ladder falling?




Fig. 3.15


SOLUTION
Let h(t) be the height of the ladder at time t and b(t) be the distance of the base
of the ladder to the wall at time t. Then the Pythagorean theorem tells us that
h(t)2 + b(t)2 = 132 .
We may differentiate both sides of this equation with respect to the variable t
(which is time in minutes) to obtain
2 · h(t) · h (t) + 2 · b(t) · b (t) = 0.
Solving for h (t) yields
b(t) · b (t)
h (t) = ’ .
h(t)
93
CHAPTER 3 Applications of the Derivative

At the instant the problem is posed, b(t) = 5, h(t) = 12 (by the Pythagorean
theorem), and b (t) = 1. Substituting these values into the equation yields
5·1 5
h (t) = ’ = ’ ft/min.
12 12
Observe that the answer is negative, which is appropriate since the top of the
ladder is falling.

You Try It: Suppose that a square sheet of aluminum is placed in the hot sun.
It begins to expand very slowly so that its diagonal is increasing at the rate of
1 millimeter per minute. At the moment that the diagonal is 100 millimeters, at
what rate is the area increasing?
EXAMPLE 3.13
A sponge is in the shape of a right circular cone (Fig. 3.16). As it soaks up
water, it grows in size. At a certain moment, the height equals 6 inches, and
is increasing at the rate of 0.3 inches per second. At that same moment, the
radius is 4 inches, and is increasing at the rate of 0.2 inches per second.
How is the volume changing at that time?




Fig. 3.16


SOLUTION
We know that the volume V of a right circular cone is related to the height
h and the radius r by the formula
1
V = π r 2 h.
3
Differentiating both sides with respect to the variable t (for time in seconds)
yields
1
dV dr dh
= π 2r h + r 2 .
3
dt dt dt
94 CHAPTER 3 Applications of the Derivative

Substituting the values for r, dr/dt, h, and dh/dt into the right-hand side yields
1 1 24π
dV
= π 2 · 4 · (0.2) · 6 + 42 · (0.3) = π [9.6 + 4.8] = .
3 3 5
dt
You Try It: In the heat of the sun, a sheet of aluminum in the shape of an equilateral
triangle is expanding so that its side length increases by 1 millimeter per hour. When
the side length is 100 millimeters, how is the area increasing?



3.4 Falling Bodies
It is known that, near the surface of the earth, a body falls with acceleration (due
to gravity) of about 32 ft/sec2 . If we let h(t) be the height of the object at time t
(measured in seconds), then our information is that
h (t) = ’32.
Observe the minus sign to indicate that height is decreasing.
Now we will do some organized guessing. What could h be? It is some function
whose derivative is the constant ’32. Our experience indicates that polynomials
decrease in degree when we differentiate them. That is, the degree goes from 5 to 4,
or from 3 to 2. Since, h is a polynomial of degree 0, we therefore determine
that h will be a polynomial of degree 1. A moment™s thought then suggests that
h (t) = ’32t. This works! If h (t) = ’32t then h (t) = [h (t)] = ’32. In fact
we can do a bit better. Since constants differentiate to zero, our best guess of what
the velocity should be is h (t) = ’32t + v0 , where v0 is an undetermined constant.
Now let us guess what form h(t) should have. We can learn from our experience
in the last paragraph. The “antiderivative” of ’32t (a polynomial of degree 1)
should be a polynomial of degree 2. After a little ¬ddling, we guess ’16t 2 . And this
works. The antiderivative of v0 (a polynomial of degree 0) should be a polynomial
of degree 1. After a little ¬ddling, we guess v0 t. And this works. Taking all this
information together, we ¬nd that the “antiderivative” of h (t) = ’32t + v0 is
h(t) = ’16t 2 + v0 t + h0 . (†)
Notice that we have once again thrown in an additive constant h0 . This does no
harm:
h (t) = [’16t 2 ] + [v0 t] + [h0 ] = ’32t + v0 ,
just as we wish. And, to repeat what we have already con¬rmed,
h (t) = [h (t)] = [’32t] + [v0 ] = ’32.
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CHAPTER 3 Applications of the Derivative

We now have a general formula (namely (†)) for the position of a falling body at
time t. [Recall that we were ¬rst introduced to this formula in Section 2.6.] See
Fig. 3.17.




Fig. 3.17

Before doing some examples, we observe that a falling body will have initial
velocity 0. Thus
0 = h (0) = ’32 · 0 + v0 .
Hence, for a falling body, v0 = 0. In some problems we may give the body an
initial push, and then v0 will not be zero.
EXAMPLE 3.14
Suppose that a falling body hits the ground with velocity ’100 ft/sec. What
was the initial height of the body?

SOLUTION
With notation as developed above, we know that velocity is given by
h (t) = ’32t + 0.
We have taken v0 to be 0 because the body is a falling body; it had no initial
push. If T is the time at which the body hits the ground, then we know that
’100 = h (T ) = ’32 · T .
As a result, T = 25/8 sec.
96 CHAPTER 3 Applications of the Derivative

When the body hits the ground, its height is 0. Thus we know that
0 = h(T ) = h(25/8) = ’16 · (25/8)2 + h0 .
We may solve for h0 to obtain
625
h0 = .
4
Thus all the information about our falling body is given by
625
h(t) = ’16t 2 + .
4
At time t = 0 we have
625
h(0) = .
4
Thus the initial height of the falling body is 625/4 ft = 156.25 ft.
Notice that, in the process of solving the last example, and in the discussion
preceding it, we learned that h0 represents the initial height of the falling body and
v0 represents the initial velocity of the falling body. This information will be useful
in the other examples that we examine.
EXAMPLE 3.15
A body is thrown straight down with an initial velocity of 10 feet per second.
It strikes the ground in 12 seconds. What was the initial height?

SOLUTION
We know that v0 = ’10 and that h(12) = 0. This is the information that we
must exploit in solving the problem. Now
h(t) = ’16t 2 ’ 10t + h0 .
Thus
0 = h(12) = ’16 · 122 ’ 10 · 12 + h0 .
We may solve for h0 to obtain
h0 = 2424 ft.
The initial height is 2424 feet.

You Try It: A body is thrown straight up with initial velocity 5 feet per second
from a height of 40 feet. After how many seconds will it hit the ground? What will
be its maximum height?
97
CHAPTER 3 Applications of the Derivative

EXAMPLE 3.16
A body is launched straight up from height 100 feet with some initial velocity.
It hits the ground after 10 seconds. What was that initial velocity?

SOLUTION
We are given that h0 = 100. Thus

h(t) = ’16t 2 + v0 t + 100.
Our job is to ¬nd v0 . We also know that
0 = h(10) = ’16 · 102 + v0 · 10 + 100.
We solve this equation to ¬nd that v0 = 150 ft /sec.

You Try It: On a certain planet, bodies fall with an acceleration due to gravity
of 10 ft/sec2 . A certain body is thrown down with an initial velocity of 5 feet per
second, and hits the surface 12 seconds later. From what height was it launched?




Exercises
1. Sketch the graph of f (x) = x/[x 2 + 3], indicating all local maxima and
minima together with concavity properties.
2. What is the right circular cylinder of greatest volume that can be inscribed
upright in a right circular cone of radius 3 and height 6?
3. An air mattress (in the shape of a rectangular parallelepiped) is being in¬‚ated
in such a way that, at a given moment, its length is increasing by 1 inch
per minute, its width is decreasing by 0.5 inches per minute, and its height
is increasing by 0.3 inches per minute. At that moment its dimensions are
= 100 , w = 60 , and h = 15 . How is its volume changing at that time?
4. A certain body is thrown straight down at an initial velocity of 15 ft /sec. It
strikes the ground in 5 seconds. What is its initial height?
5. Because of viral infection, the shape of a certain cone-shaped cell is
changing. The height is increasing at the rate of 3 microns per minute.
For metabolic reasons, the volume remains constantly equal to 20 cubic
microns. At the moment that the radius is 5 microns, what is the rate of
change of the radius of the cell?
6. A silo is to hold 10,000 cubic feet of grain. The silo will be cylindrical
in shape and have a ¬‚at top. The ¬‚oor of the silo will be the earth. What
dimensions of the silo will use the least material for construction?
7. Sketch the graph of the function g(x) = x·sin x. Show maxima and minima.
98 CHAPTER 3 Applications of the Derivative

8. A body is launched straight down at a velocity of 5 ft /sec from height
400 feet. How long will it take this body to reach the ground?
9. Sketch the graph of the function h(x) = x/(x 2 ’ 1). Exhibit maxima,
minima, and concavity.
10. A punctured balloon, in the shape of a sphere, is losing air at the rate of
2 in.3 /sec. At the moment that the balloon has volume 36π cubic inches,
how is the radius changing?
11. A ten-pound stone and a twenty-pound stone are each dropped from height
100 feet at the same moment. Which will strike the ground ¬rst?
12. A man wants to determine how far below the surface of the earth is the
water in a well. How can he use the theory of falling bodies to do so?
13. A rectangle is to be placed in the ¬rst quadrant, with one side on the x-axis
and one side on the y-axis, so that the rectangle lies below the line 3x +5y =
15. What dimensions of the rectangle will give greatest area?
14. A rectangular box with square base is to be constructed to hold 100 cubic
inches. The material for the base and the top costs 10 cents per square


Y
inch and the material for the sides costs 20 cents per square inch. What
FL
dimensions will give the most economical box?
15. Sketch the graph of the function f (x) = [x 2 ’1]/[x 2 +1]. Exhibit maxima,
minima, and concavity.
AM

16. On the planet Zork, the acceleration due to gravity of a falling body near the
surface of the planet is 20 ft /sec. A body is dropped from height 100 feet.
How long will it take that body to hit the surface of Zork?
TE
CHAPTER 4



The Integral

4.0 Introduction
Many processes, both in mathematics and in nature, involve addition. You are famil-
iar with the discrete process of addition, in which you add ¬nitely many numbers
to obtain a sum or aggregate. But there are important instances in which we wish
to add in¬nitely many terms. One important example is in the calculation of area”
especially the area of an unusual (non-rectilinear) shape. A standard strategy is to
approximate the desired area by the sum of small, thin rectangular regions (whose
areas are easy to calculate). A second example is the calculation of work, in which
we think of the work performed over an interval or curve as the aggregate of small
increments of work performed over very short intervals. We need a mathematical
formalism for making such summation processes natural and comfortable. Thus we
will develop the concept of the integral.



4.1 Antiderivatives and Indefinite
Integrals
4.1.1 THE CONCEPT OF ANTIDERIVATIVE
Let f be a given function. We have already seen in the theory of falling bodies
(Section 3.4) that it can be useful to ¬nd a function F such that F = f . We call

99
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100 CHAPTER 4 The Integral

such a function F an antiderivative of f . In fact we often want to ¬nd the most
general function F , or a family of functions, whose derivative equals f . We can
sometimes achieve this goal by a process of organized guessing.
Suppose that f (x) = cos x. If we want to guess an antiderivative, then we
are certainly not going to try a polynomial. For if we differentiate a polynomial
then we get another polynomial. So that will not do the job. For similar reasons
we are not going to guess a logarithm or an exponential. In fact the way that we
get a trigonometric function through differentiation is by differentiating another
trigonometric function. What trigonometric function, when differentiated, gives
cos x? There are only six functions to try, and a moment™s thought reveals that
F (x) = sin x does the trick. In fact an even better answer is F (x) = sin x + C. The
constant differentiates to 0, so F (x) = f (x) = cos x. We have seen in our study
of falling bodies that the additive constant gives us a certain amount of ¬‚exibility
in solving problems.
Now suppose that f (x) = x 2 . We have already noted that the way to get a
polynomial through differentiation is to differentiate another polynomial. Since
differentiation reduces the degree of the polynomial by 1, it is natural to guess that
the F we seek is a polynomial of degree 3. What about F (x) = x 3 ? We calculate
that F (x) = 3x 2 . That does not quite work. We seek x 2 for our derivative, but
we got 3x 2 . This result suggests adjusting our guess. We instead try F (x) = x 3 /3.
Then, indeed, F (x) = 3x 2 /3 = x 2 , as desired. We will write F (x) = x 3 /3 + C
for our antiderivative.
More generally, suppose that f (x) = ax 3 + bx 2 + cx + d. Using the reasoning
in the last paragraph, we may ¬nd fairly easily that F (x) = ax 4 /4 + bx 3 /3 +
cx 2 /2 + dx + e. Notice that, once again, we have thrown in an additive constant.

You Try It: Find a family of antiderivatives for the function f (x) = sin 2x ’
x 4 + ex .

4.1.2 THE INDEFINITE INTEGRAL
In practice, it is useful to have a compact notation for the antiderivative. What we
do, instead of saying that “the antiderivative of f (x) is F (x) + C,” is to write

f (x) dx = F (x) + C.

So, for example,

cos x dx = sin x + C

and
x4 x2
x + x dx = + +C
3
4 2
101
CHAPTER 4 The Integral

and
e2x
dx = + C.
2x
e
2
The symbol is called an integral sign (the symbol is in fact an elongated “S”)
and the symbol “dx” plays a traditional role to remind us what the variable is. We
call an expression like
f (x) dx

an inde¬nite integral. The name comes from the fact that later on we will have a
notion of “de¬nite integral” that speci¬es what value C will take”so it is more
de¬nite in the answer that it gives.
EXAMPLE 4.1
Calculate
sin(3x + 1) dx.

SOLUTION
We know that we must guess a trigonometric function. Running through
the choices, cosine seems like the best candidate. The derivative of cos x is
’ sin x. So we immediately see that ’ cos x is a better guess”its derivative
is sin x. But then we adjust our guess to F (x) = ’ cos(3x + 1) to take into
account the form of the argument. This almost works: we may calculate that
F (x) = 3 sin(3x + 1). We determine that we must adjust by a factor of 1/3.
Now we can record our ¬nal answer as
1
sin(3x + 1) dx = ’ cos(3x + 1) + C.
3
We invite the reader to verify that the derivative of the answer on the right-hand
side gives sin(3x + 1).
EXAMPLE 4.2
Calculate
x
dx.
x2 + 3

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