MATH 100 V1A
October 1st – Practice problems
Solutions
1. Find the equation of the line tangent to y = (ax + b)3 at x = ab .
Solution: Let f (x) = (ax + b)3 . Then f
b
a
= 8b3 , and using the Chain Rule,
f 0 (x) = 3(ax + b)2 · a.
So, f 0 ab = 12b2 a. The equation of the line tangent to y = (ax + b)3 at x =
therefore given by
y − 8b3 = 12b2 a(x − a).
2. Let h(x) = (f (x))3 , and suppose f (0) =
tangent to y = h(x) at x = 0.
1
2
b
a
is
and f 0 (0) = 73 . Find the equation of the line
Solution: Using the Chain Rule, h0 (x) = 3 (f (x))2 · f 0 (x). So, the slope of the line
tangent to h at x = 0 is
2
7
1
7
2
0
0
h (0) = 3 (f (0)) · f (0) = 3
· = .
2
3
4
Now, h(0) = (f (0))3 = 81 , so the equation of the line tangent to h at x = 0 is given by
y−
7
1
= x.
8
4
3. (a) If S(h) (in units m3 ) is the amount of snowfall on Grouse Mountain as a function
of altitude h (in m), and h(t) is altitude as a function of time t (in s), what is an
equation for the amount of snowfall as a function of time?
Solution: The amount of snowfall as a function of time can be written as S(h(t)).
Note that this is not the same as S(t), which is the amount of snowfall at an
altitude of t metres.
(b) Fill in the blanks:
Solution: The answers are in red.
dS
is
dh
3
•
the rate of change of amount of snowfall with respect to altitude in units
m /m.
•
dS
dt
3
is the rate of change of amount of snowfall with respect to time in units
m /s.
•
dh
dt
is the rate of change of altitude with respect to time in units m/s.
4. Imagine a road on which the speed limit is specified at every single point. In other
words, there is a certain function L such that the speed limit x kilometres from the
beginning of the road is L(x). Two cars, A and B, are driving along this road; car A’s
position at time t is a(t), and car B’s position at time t is b(t).
(a) What equation expresses the fact that car A always travels at the speed limit?
(Hint: The answer is not a0 (t) = L(t)).
Solution: The equation that expresses the fact that car A always travels at the
speed limit is a0 (t) = L(a(t)). Since L is a function of distance, L(t) represents the
speed limit t kilometres from the beginning of the road.
(b) Suppose that A always goes at the speed limit, and that B’s position at time t
is A’s position at time t − 1. Show that B is also going at the speed limit at all
times.
Solution: Since A always goes the speed limit, we know that a0 (t) = L(a(t)) for
every t. If B’s position at time t is A’s position at time t − 1, then b(t) = a(t − 1).
From the Chain Rule, b0 (t) = a0 (t − 1), and a0 (t − 1) = L(a(t − 1)) since A always
travels the speed limit. So, b0 (t) = L(a(t − 1)) = L(b(t)), which means that B
always travels the speed limit too.
(c) Suppose B always stays a constant distance behind A. Under what conditions will
B still always travel the speed limit?
Solution: Note that we’re still assuming that A always travels at the speed limit. If
B is always a constant distance behind A, then b(t) = a(t) − D, for some constant
D > 0. So, b0 (t) = a0 (t) = L(a(t)) = L(b(t) + D). If B travels at the speed limit,
then b0 (t) = L(b(t)), so we want L(b(t)) = L(b(t) + D). That is, B will still travel
the speed limit if L is a periodic function with period D.
2