Calculus for the Life Sciences I
Math 161 Winter 2012
Professor Ben Richert
Exam 2
Solutions
Problem 1 (10pts). Find all inflection points of the function f (x) = x(6 − x)2 .
Solution. We begin by computing derivatives
f (x)
= x(36 − 12x + x2 ) = x3 − 12x2 + 36x
f 0 (x)
=
3x2 − 24x + 36
f 00 (x)
=
6x − 24
because solutions to f 00 (x) = 0 give the possible points of inflection. In this case
0 = f 00 (x) = 6x − 24
gives x = 4 as the only possibility. Now to determine if there is an inflection point at x = 4 we consider the
concavity on either side. Note that
f 00 (0) = 6(0) − 24 = −24 < 0
while
f 00 (5) = 6(5) − 24 = 6 > 0
so that the concavity does change (from concave down to concave up) at the point x = 4. The point
(4, f (4)) = (4, 16) is thus the only point of inflection.
Problem 2 (10pts). Do the following two problems:
(a–5pts) Find all horizontal asymptotes of the function f (x) =
2x2 + x + 1
.
x2 − x + 1
Solution. Horizontal asymptotes occur when lim f (x) = L or lim f (x) = L. In this case,
x→∞
2
2x + x + 1
=
x→∞ x2 − x + 1
lim
1
2
lim x1
x→∞ 2
x
2
2x + x + 1
x2 − x + 1
x→−∞
2
2 xx2 + xx2 + x12
= lim x2
x
1
x→∞
x2 − x2 + x2
limx→∞ x12
2+0+0
1 = 1−0+0
limx→∞ x2
limx→∞ 2 + limx→∞ x1 +
limx→∞ 1 − limx→∞ x1 +
Thus there is a horizontal asymptote at y = 2. Similarly,
=
2+
2x2 + x + 1
= lim
2
x→−∞ x − x + 1
x→−∞ 1 −
lim
So there is exactly one horizontal asymptote at y = 2.
1
x
1
x
+
+
1
x2
1
x2
2+
x→∞ 1 −
= lim
1
x
1
x
+
+
1
x2
1
x2
= 2.
= 2.
(b–5pts) Find all vertical asymptotes of the function f (x) =
x2 + 1
.
x−1
Solution. The only place a vertical asymptote can occur is for a such that lim+ f (x) = ±∞ or
x→a
lim− f (x) = ±∞, and this can only occur where the function is not defined. Thus we need only
x→a
consider x = 1 since this is the only point not in the domain of
x2 + 1
. In this case,
x−1
x2 + 1
=∞
x−1
and “constant divided by small is big.” Thus there is a
lim+ f (x) =
x→1
2
since the x + 1 → 2 while x − 1 → 0+
vertical asymptote at x = 1.
Problem 3 (15pts). You need to design a rectangular shipping container with square base and no lid. The
base is to be made of plywood costing $2 per square foot while the sides are made of cardboard costing $(4/5)
per square foot. What are the dimensions of the cheapest container which has volume 10 cubic feet?
Solution. We want to minimize cost given cost of materials, shape of the box, and the required volume.
We let x be the length of the base and y be the height. Then the area of the base is x2 so that its cost is
2x2 . The area of a single side is xy, so the total area from the sides is 4xy, and the corresponding cost is
(4/5)4xy. Thus the optimization equation for cost is
C = 2x2 + (16/5)xy.
The constraint equation is 10 = x2 y. Note that x, y > 0 is required as x, y are measurements (hence positive)
and nonzero (else their product cannot be 10).
Solving the constraint equation for y yields
10
y = 2,
x
and substituting this into the optimization equations we obtain
10
32
C = 2x2 + (16/5)x 2 = 2x2 + .
x
x
Now optimizing C in the usual way, we compute the derivative, set it equation to zero, and solve for x to
find the critical points.
C 0 = 4x −
32
x2
=
⇐⇒ 4x =
⇐⇒ x3
=
⇐⇒ x =
Note that
0
32
x2
8
2
64
>0
x3
for x > 0, so f (x) has a relative min at x = 2 and is concave up for x > 0 (so that this min is the absolute
10
min). Thus the dimensions that minimize cost are x = 4 and y = 2 = 10/4 feet.
x
C 00 = 4 +
Problem 4 (10pts). The concentration of caffeine (in grams per milliliter) in the bloodstream of a certain
mathematician is given by
C(t) = 10 − ln(t2 + 1)
where t counts the number of hours since he finished his morning pot of coffee. Use differentials to estimate
the amount by which the concentration of caffeine in his bloodstream changes between t = 1 and t = 1.5
hours.
Solution. For the function C(t), we have
dC = C 0 (t) dt = −
t2
2t
dt
+1
f 0 (x)
d
ln(f (x)) =
). We want to estimate ∆C, the amount by which
dx
f (x)
the concentration of caffeine in the mathematicians bloodstream changes over the time interval ∆t = 1/2.
We are instructed to use differentials and since ∆t = dt = 1/2 is small, we suspect that ∆C ≈ dC (so that
the estimate is not a total affront to our sensibilities). Thus at t = 1 and with dt = 1/2, we have
1
1
2(1)
d =− .
∆C ≈ dC = − 2
1 +1
2
2
by definition, (we used the rule that
So the concentration in his bloodstream is estimated to have gone down by 1/2 a gram per milliliter.
Problem 5 (12pts). Those industrious gold minors are still working hard trying to make their fortunes,
beginning at 8am each morning as usual. The function f (t) = te−t/4 computes the number of grams of gold
they have found (on a given day) after working for t hours (0 ≤ t ≤ 4).
(a–6pts) At what rate are the minors finding gold (in grams per hour) at t = 2?
Solution. To find the rate that the minors are finding gold, we must compute the derivative (which
always calculates the rate of change): hence
t −t/4 2 −2/4
d −t/4 −t/4
−2/4
te
= e
− e
= e
− e
dt
4
4
t=2
t=2
d f (x)
grams per hour. Note the use of the product rule and the fact that
e
= ef (x) f 0 (x).
dx
(b–6pts) Is the rate at which they are finding gold (in grams per hour) increasing or decreasing at t = 2?
Solution. We already observed that the rate at which they are finding gold is computed by f 0 (t).
Therefore, to determine if the rate at which they are finding gold is increasing or decreasing, we
need to decide if f 0 (x) is increasing or decreasing. We do this in the usual way, by considering the
d
sign of f 0 (t). That is, we compute f 00 (t) and see if it is positive or negative at t = 2. In this case
dt
1
1
t
1 1
t
1
t
f 00 (t) = − e−t/4 − e−t/4 + e−t/4 = e−t/4 − − +
= e−t/4 − +
.
4
4
16
4 4 16
2 16
At t = 2 we find
1
2
6
00
−2/4
−2/4
f (2) = e
− +
=e
−
<0
2 16
16
(note that e−1/2 > 0). We determine that the rate at which the minors are finding gold (in grams
per hour) at time t = 2 is decreasing.
Problem 6. (7pts) Match each of the entries in the left column with one entry from the right column. Use
each entry from the right column exactly once. You do not need to show your work for this problem.
(A) 3
D
d
ln (x + 1)2 (x + 2)3
dx
C
log3 1
G
d x
e + log2 (x)
dx
(D)
d x
e log2 (x)
dx
2
3
+
x+1 x+2
(E)
5
2(x + 1) + 3(x + 2)
F
E
B
A
d
ln 2(x + 1) + 3(x + 2)
dx
d ln(x)
e
dx
log2 8
(B) 1
(C) 0
(F) ex log2 (x) +
(G) ex +
1
x ln 2
ex
x ln 2