Practice Problems 11 for 1110
December 10, 2016
Exercise 1. An article in the Wall Street Journal’s "Heard on the Street" column (Money and Investment August 1, 2001)
reported that investors often look at the "change in the rate of change" to help them "get into the market before any big rallies."
Your stock broker alerts you that the rate of change in a stock’s price is increasing. As a result you
(a) can conclude the stock’s price is decreasing
(b) can conclude the stock’s price is increasing
(c) cannot determine whether the stock’s price is increasing or decreasing.
Solution. (c) If f (t) is the stock’s price as a function of time, then the rate of change of the stock’s price is the first derivative
f 0 (t). The stock is increasing at the instant t if f 0 (t) is positive, and the stock is decreasing at the instant t if f 0 (t) is negative.
But the only information that we are given is that f 0 (t) itself increasing, which doesn’t tell us whether the value of f 0 (t) is
positive or negative.
Exercise 2. Imagine that you are skydiving. The graph of your speed as a function of time from the time you jumped out of
the plane to the time you achieved terminal velocity is
(a) increasing and concave up
(b) decreasing and concave up
(c) increasing and concave down
(d) decreasing and concave down.
Solution. (c) In free fall an object accelerates at the constant rate of 9.8 meters per second2 , so in a vacuum the object’s
velocity would be increasing the entire way down until it hit the ground. However, in the Earth’s atmosphere the air creates
air drag, which produces a counter force that becomes greater as the velocity increases, until finally the drag force is equal and
opposite the force of gravity. The velocity upon which the drag force and gravity force have equal magnitudes is called terminal
velocity. So the graph is always increasing because the skydiver’s speed is always increasing and it is concave down because the
air drag slows this increase as time progresses.
Exercise 3. Water is being poured into a "Dixie cup" (a standard cup that is smaller at the bottom than at the top). The
height of the water in the cup is a function of the volume of water in the cup. What does the graph of the function look like?
(a) Increasing and concave up.
(b) Increasing and concave down.
(c) A straight line with positive slope.
Solution. (b) The function h(v) is increasing because if there is more water in the cup then the height is greater. The function
is concave down because it takes more water to raise the height, say one millimeter, at the top of the cup rather than near the
bottom of the cup. More mathematically, the instantaneous rate of change of the height is smaller when there is more water in
the cup.
Exercise 4. On a toll road a driver takes a time stamped toll-card from the starting booth and drives directly to the end of the
toll section. After paying the required toll, the driver is surprised to receive a speeding ticket along with the toll receipt. Which
of the following best describes the situation?
(a) The booth attendant does not have enough information to prove that the driver was speeding.
(b) The booth attendant can prove that the driver was speeding during his trip.
(c) The driver will get a ticket for a lower speed than his actual maximum speed.
(d) Both (b) and (c).
Solution. (d) As a concrete example, suppose that the tolls are 80 miles apart and the speed limit on this entire stretch of road
is 70 miles per hour. Suppose that the driver reaches the second toll exactly 1 hour after leaving the first toll. If the driver was
going the maximum legal speed of 70 miles per hour for the entire hour, then the distance travelled is exactly 70 miles and hence
the driver would not have reached the second toll in this time. More mathematically, let f (t) is the position function of the car
with f (0) = 0 being the first toll and f (tfinish ) = d being the position of the second toll at the finish time. Then the mean value
theorem says that there is an instant in time s ∈ [0, tfinish ] such that
f 0 (s) =
d
f (tfinish ) − f (0)
=
,
tfinish − 0
tfinish
so if the distance between the tolls divided by the time it took to get there is greater than the speed limit, then the driver had
to be going faster than the speed limit at some point in time. This last sentence may have been intuitive for some people, but
notice that the mean value theorem allows us to formalize our intuition into formal mathematics.
d
is the speed at which the driver would have to be
To see that statement (c) also holds, note that the speed f 0 (s) = tfinish
traveling the entire time in order to cover the distance in tfinish amount of time. But the driver needs time to speed up when
d
at some point. To
leaving the first toll, so to make up for this smaller speed the driver would have to go even faster than tfinish
d
help your reasoning draw a graph of the constant function tfinish together with the graph of the velocity of the driver.
Exercise 5. Two racers start a race at the same moment and finish in a tie. Which of the following must be true?
(a) At some point during the race the two racers were not tied.
(b) The racers’ speeds at the end of the race must have been exactly the same.
(c) The racers must have had the same speed at exactly the same time at some point in the race.
(d) The racers had to have the same speed at some moment, but no necessarily at exactly the same time.
Solution. (c) The trick to this question is to let f1 (t) be the position function of the first racer with respect to time and f2 (t) be
the position function of the second racer with respect to time. Let’s suppose that they start the race at time t = 0 and end the
race at time t = 1 (it really doesn’t matter what number we choose for the finish time). Then f1 (0) = f2 (0) and f1 (1) = f2 (1)
because they start at the same time and finish in a tie. Now define a new function g(t) = f1 (t) − f2 (t), so we know that g(0) = 0
and g(1) = 0. Then by the Mean Value Theorem (or even Rolle’s Theorem) we know there is an instant in time s ∈ [0, 1] such
that
g(1) − g(0)
= 0.
g 0 (s) =
1−0
But then 0 = g 0 (s) = f10 (s) − f20 (s) implies that f10 (s) = f20 (s), so s is an instant of time when the velocities of the two runners
is exactly the same. Note that this is a question where your initial intuition could deceive you, in which case it is important to
define the functions f1 (t), f2 (t), and g(t), so that the mathematical formalism can do the hard work for you.
Exercise 6. True or False. If f 00 (a) = 0, then f has an inflection point at a.
Solution. False, consider f (x) = x4 . Then f 0 (x) = 4x3 and f 00 (x) = 12x2 , which satisfies f 00 (0) = 0 and yet f (x) = x4 does not
have an inflection point at x = 0.
Exercise 7. True or False. For f (x) =| x | on the interval [− 12 , 2], can you find a point c in (− 12 , 2) such that
f 0 (c) =
Solution. False,
f 0 (c) = 3/5.
f (2)−f (− 12 )
2−(− 21 )
=
3/2
5/2
f (2) − f (− 21 )
2 − (− 12 )
= 3/5, but f (x) only has points with slope 1 or −1, so there is no point c such that