1. Each edge of a cube is expanding at the rate of 1 centimeter (cm) per
second. How fast is the volume changing when the length of each edge is (a) 5
cm? (b) 10 cm? (c)

cm?
2. An airplane flies in level flight at constant velocity, eight miles above the ground. (In this exercise assume the earth is flat.) The flight path passes directly over a point P on the ground. The distance from the plane to P is decreasing at the rate of 4 miles per minute at the instant when this distance is 10 miles. Compute the velocity of the plane in miles per hour.
3. A baseball diamond is a 90-foot square. A ball is batted along the third-base line at a constant speed of 100 feet per second. How fast is its distance from first base changing when (a) it is halfway to third base? (b) it reaches third base?
4. A boat sails parallel to a straight beach at a constant speed of

miles per hour, staying

miles offshore. How fast is it approaching a lighthouse on the shoreline at
the instant it is exactly

miles from the lighthouse ?
5. A reservoir has the shape of a right-circular cone. The altitude is 10 feet, and the radius of the base is 4 ft. Water is poured into the reservoir at a constant rate of 5 cubic feet per minute. How fast is the water level rising when the depth of the water is 5 feet if (a) the vertex of the cone is up? (b) the vertex of the cone is down?
6. A water tank has the shape of a right-circular cone with its vertex down.
Its altitude is 10 feet and the radius of the base is 15 feet. Water leaks out
of the bottom at a constant rate of 1 cubic foot per second. Water is poured
into the tank at a constant rate of

cubic feet per second. Compute

so that the water level will be rising at the rate of 4 feet per second at the
instant when the water is 2 feet deep.
7. Water flows into a hemispherical tank of radius

feet (flat side up). At any instant, let

denote the depth of the water, measured from the bottom,

the radius of the surface of the water, and

the volume of the water in the tank. Compute

at the instant when

feet. If the water flows in at a constant rate of

cubic feet per second, compute

,
the rate at which

is changing, at the instant

when

feet.
8. A variable right triangle

in the

-plane
has its right angle at vertex

,
a fixed vertex

at the origin, and the third vertex

restricted to lie on the parabola

The point

starts at the point

at time

and moves upward along the

-axis
at a constant velocity of

cm/sec. How fast is the area of the triangle increasing when

sec?
9. The radius of a right-circular cylinder increases at a constant rate. Its
altitude is a linear function of the radius and increases three times as fast
as the radius. When the radius is 1 foot the altitude is 6 feet. When the
radius is 6 feet, the volume is increasing at a rate of 1 cubic foot per
second. When the radius is 36 feet, the volume is increasing at a rate of

cubic feet per second, where n is an integer. Compute

.
10. A particle is constrained to move along a parabola whose equation is

.
(a) At what point on the curve are the abscissa and the ordinate changing at
the same rate? (b) Find this rate if the motion is such that at time

we have

and

.
11. The equation

defines

as one or more functions of

.
(a) Assuming the derivative

exists, and without attempting to solve for

,
show that

satisfies the equation

.
(b) Assuming the second derivative

exists, show that

whenever

.
12. If

,
the equation

defines

as a function of

.
Without solving for

,
show that the derivative

has a fixed sign. (You may assume the existence of

.)
13. The equation

defines

implicitly as two functions of

if

.
Assuming the second derivative

exists, show that it satisfies the equation

.
14. The equation

defines

implicitly as a function of

.
Assuming the derivative

exists, show that it satisfies the equation

15. If

,
where

is a rational number, say

,
then

.
Assuming the existence of the derivative

,
derive the formula

using implicit differentiation and the corresponding formula for integer
exponents.
The Cauchy Mean Value Theorem is the basic tool needed to prove a theorem
which facilitates evaluation of limits of the form

when

Theorem. Suppose that

and and suppose also that

exists. Then

exists, and

PROOF. The hypothesis that

exists contains two implicit assumptions:
(1) there is an interval

such that

and

exist for all

in

except, perhaps, for

,
(2) in this interval

with, once again, the possible exception of

On the other hand,

and

are not even assumed to be defined at

.
If we define

(changing the previous values of

and

,
if necessary), then

and

are continuous at

.
If

,
then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to

and

on the interval

(and a similar statement holds for

).
First applying the Mean Value Theorem to

,
we see that

,
for if

there would be some

in

with

,
contradicting (2). Now applying the Cauchy Mean Value Theorem to

and

,
we see that there is a number

.
in

such that

or

Now

approaches

as

approaches

,
because

,
is in

;
since

exists, it follows that

Definition. The function

is uniformly continuous on an interval

if for every

there is some

such that, for all

and

in


Lemma. Let

and let

be continuous on the interval

Let

,
and suppose that statements (i) and (ii) hold:
(i) if

and

are in

and



,
then


,
(ii) if

and

are in

and



,
then

Then there is a



0 such that

Proof. Since

is continuous at

,
there is a


0 such that

It follows that
(iii) if

and

then

Choose

to be the minimum of

and

.
We claim that this

works. In fact, suppose that

and

are any two points in

with

.
If

and

are both in

then

by
(i); and if

and

are both in

then

by
(ii). The only other possibility is that

or

In either case, since


, we also have


and


So

by
(iii).
THEOREM If

is continuous on

,
then

is uniformly continuous on

PROOF: For

let's say that

is

good
on

if there is some

such that, for all

and

in

if


,
then

.
Then we're trying to prove that

is

-good
on

for all

Consider any particular

.
Let

Then

(since
a is in

),
and

is bounded above (by

),
so

has a least upper bound

.
We really should write

,
since

and

might depend on

.
But we won't since we intend to prove that

,
no matter what

is.
Suppose that we had

.
Since

is continuous at

,
there is some



such that, if


,
then

.
Consequently, if


and

,
then

So

is surely

-good
on the interval

.
On the other hand, since

is the least upper bound of

,
it is also clear that

is

-good
on

.
Then the above Lemma implies that

is

-good
on

,
so

is in A, contradicting the fact that

is an upper bound.
To complete the proof we just have to show that

is actually in

.
The argument for this is practically the same: Since

is continuous at

,
there is some

such that, if

,
then

So

is

-good
on

.
But

is also

-good
on

,
so the Lemma implies that

is

-good
on

.