We will spend the day reviewing some of the problems from the December
midterm exam. In particular, we will take a closer look at problems 5 and 8.
If you wish to discuss another problem, please send me an email, or attend my
office hours.
We will begin with problem 5. Here is the problem statement:
Problem 0.1. A ball is floating on the surface of the sea. The ball’s height h
(in m) above the sea floor at time t (in s) is given by
h(t) = 5 + 0.1 sin(2t).
Answer the questions below. Evaluate all function values and provide justification for your claims. Include units.
• What is the range of possible values for the ball’s height?
• What is the (vertical) velocity of the ball at t =
• Is the ball moving up or down at t =
π
6?
• Is the ball speeding up or down at t =
π
6?
π
6?
Explain.
Explain.
• Calculate the initial position of the ball, and then determine the earliest
time at which the ball is back in its original position.
We will go over the answers to each part of the problem. First, we will
discuss part a.
What is the range of possible values for the ball’s height? We want to know
the values that are taken on by h(t). h(t) is a sum of two functions: the constant
function 5 and the trigonometric function 0.1 sin(2t).
We know that sin(t) (and hence sin(2t)) takes on every value between −1 and
1 . Therefore, 0.1 sin(2t) takes on every value between −0.1 and 0.1. Adding
this to the constant 5, we get that h(t) takes on every value from 4.9 to 5.1.
This is the range of h(t).
Now we will discuss part b. The vertical velocity of an object is the rate
of change of its height with respect to time; that is, the vertical velocity of an
object is the derivative of its height with respect to time.
We compute this derivative- the derivative of 5 is zero because 5 is a constant
function. What is the derivative of 0.1 sin(2t)? We need to use the chain rulethe inside is 2t and the outside is 0.1 sin(x). So the derivative is 0.1 cos(2t) ∗ 2 =
0.2 cos(2t).
Now, we plug in π6 - we have 0.2 cos π3 . What is cos π3 ? It is 12 (you should
be comfortable with trigonometric functions by now). So the vertical velocity
is 0.1.
Note that velocity is signed - if the velocity is positive, the ball is moving
upwards, and if the velocity is negative, the ball is moving downwards. Because
the velocity that we computed was positive, the ball is moving upward.
To answer part d, we need to know if the ball is speeding up or slowing
down- that is, if it’s accelerating or decelerating. Note that the ball is moving
1
upwards, so a positive acceleration would correspond to the ball speeding up,
and a negative acceleration would correspond to the ball slowing down.
The acceleration is the rate of change of velocity, so it can be computed by
taking the second derivative of h with respect to time. The first derivative was
0.2 cos(2t).
We differentiate this using the chain rule again: the derivative of the “outside”
0.2 cos(x) is −0.2 sin(x), and the derivative of the “inside”
2t is 2. So the
√
π
3
second derivative is −0.4 sin(2t). At t = 6 , this is −0.4 ∗ 2 , which is negative.
Therefore, the ball is slowing down.
For the last part, we must compute the initial position. To do this, we can
simply plug t = 0 into the function h(t) to get 5. We need to know when h(t)
will again be equal to 5, which happens when 0.1 sin(2t) = 0. Solving sin(2t) = 0
gives that 2t must be a multiple of π. Therefore, t is a multiple of π2 , and the
smallest positive multiple of π2 is π2 .
We will now discuss problem 8 on the December exam.
Problem 0.2. If a spherical tank of radius 2 meters is filled up to height h with
water, then the volume of water V in the tank is given by the following function:
h
.
V (h) = πh2 2 −
3
• What is the domain of the function V (h) in this context?
• As water is poured into the tank, at what rate is the volume of water in
the tank changing with respect to the height of the water when the water
is 1 meter deep? What are the units of this quantity?
• Over time, both the volume of water and the height of water in the tank
change. Suppose you observe that the height of the water is increasing at
a rate of 0.1 meters/min when the water in the tank is 1 meter deep. How
fast (in cubic meters per minute) is the volume increasing at that instant?
There are two kinds of restrictions that can be placed on the domain of
a function: restrictions arising from a function not being defined everywhere,
and restrictions arising from the problem statement. The function V (h) is
a polynomial, and therefore it is defined everywhere. Therefore, we need to
consider the constraints imposed by the statement of the problem.
The height of the water is certainly nonnegative, so we have h ≥ 0. Furthermore, the tank has radius 2, so the maximum height of the water is 4. Therefore,
the domain of V (h) is the set 0 ≤ h ≤ 4.
The second question is asking how the volume of the water in the tan k
changes with respect to the height. When you want to find the rate of change of
2
a function, it makes sense to differentiate it. We can differentiate the function
to get, by the product rule,
h
h2
2πh 2 −
+π .
3
3
The units here are cubic meters per meter, which simplifies to square meters.
Now, if we wanted to find the rate of change of the volume with respect to
time instead, we need to take into account the dependence of h on the time t.
We have that the volume is given by V (h(t)), where V is the function defined
above and h(t) is some function of time. We can obtain the derivative of the
volume with respect to time by applying the chain rule to this expression:
d
V (h(t)) = V ′ (h(t))h′ (t)
dt
We use the expression for V ′ (h) given above with the value of h given in the
question (1 meter), and the value for h′ (t) given in the question (0.1 meters per
minute), To get the value
5
1
2π
+π
0.1
3
3
which comes out to
1.1π
3 .
3