Exam 2 solutions

Calculus I
Math 141 Fall 2009
Professor Ben Richert
Exam 2
Solutions
Problem 1 (20pts) Recall that a cylinder with radius r and height h has surface area A = 2πrh+2πr2 and volume V = πr2 h.
Do each of the following problems.
(a–10pts) The night before the exam you fall asleep on your desk (while studying) and dream that a cylindrical CokeTM can
sitting on your desk begins to grow! How fast is its volume increasing when the height is 8 cm if the height grows
at a rate of 1/3 cm per second and the radius is always exactly 1/2 the height?
Solution. First, a picture.
dh
1
Let r be the radius of the cylinder, h be the height, and V be the volume. Then
= cm per second. Now
dt
3
the chain rule says
dV
dV dh
=
,
dt
dh dt
and the formula for volume is
2
h
πh3
V = πr2 h = π
h=
2
4
2
dV
3πh
h
=
, and thus have
since r = . We compute that
2
dh
4
dV dh 3πh2 1
π82
dV =
=
=
= 16π
dt h=8
dh h=8 dt h=8
4 h=8 3
4
cubic cm per second.
(b–10pts) What are the dimensions of a cylindrical can with volume 100cm3 and smallest possible surface area?
Solution. First, a picture.
We want to minimize area given volume.
The objective function is: A = 2πrh + 2πr2 , 0 < r < ∞.
The constraint equations is: 100 = V = πr2 h.
100
Rearranging the latter gives h = 2 , so
πr
100
200
A = 2πr 2 + 2πr2 =
+ 2πr2 .
πr
r
We compute that
−200
+ 4πr
A′ =
r2
and
400
A′′ = 3 + 4π.
r
−200
+ 4πr = 0 for 0 < r < ∞, so
The critical points of A that are in the domain occur when
r2
200
4πr =
r2
200
r3 =
4π
r
r
3 50
3 200
=
.
r =
4π
π
r
50
, and
π
since, as we pointed out in class, any continuous function (which A is on (0, ∞)) having only one min on an open
interval takes its absolute value at that min. So the cylinder has smallest area when
r
3 50
r=
π
and
100
100
,
h= 2 =
50 (2/3)
πr
π π
whatever that is.
√
Problem 2 (10pts) Use a linear approximation (or differentials) to estimate 25.01. Give a decimal answer.
√
Solution. Let f (x) = x. Then we know (suspect?) that f (x) ≈ L25 (x) for x near 25, and thus f (25.01) ≈ L25 (25.01).
Recall that
L25 (x) = f (25) + f ′ (25)(x − 25).
1
Since f ′ (x) = √ , we have
2 x
√
1
1
L25 (x) = 25 + √ (x − 25) = 5 + (x − 25).
10
2 25
Since the second derivative is always positive for r > 0, it follows that we have a minimum at r =
3
Thus
f (25.01) ≈ L25 (25.01) = 5 +
1
1 1
1
1
(25.01 − 25) = 5 + (0.01) = 5 +
=5+
= 5 + .001 = 5.001.
10
10
10 100
1000
Problem 3 (10pts) Find all inflection points of the function f (x) = 3x5 − 10x3 + 12x + 14.
Solution. We know that a twice differentiable function f (x) has an inflection point at x = a if and only if f ′′ (a) = 0 and the
sign of f ′′ (x) changes from positive to negative or negative to positive at x = a. So computing derivatives:
f (x)
f ′ (x)
f ′′ (x)
= 3x5 − 10x3 + 12x + 14
= 15x4 − 30x2 + 12
= 60x3 − 60x = 60x(x2 − 1) = 60x(x − 1)(x + 1)
Thus the three possible inflection points of f (x) occur at x = −1, x = 0, and x = 1. We check the concavity on each
of (−∞, −1), (−1, 0), (0, −1), and (1, ∞) by computing f ′′ (x) on a point from each interval (this is sufficient because the
possible inflection points give the only places where the concavity, and hence the sign of f ′′ (x), could change). So
f ′′ (−2) = 60(−2)((−2)2 − 1) < 0
f (−1/2) = 60(−1/2)((−1/2)2 − 1) > 0
′′
f ′′ (1/2) = 60(1/2)((1/2)2 − 1) < 0
f ′′ (2) = 60(2)((2)2 − 1) > 0
and we see that f ′′ (x) does in fact change sign at each of x = −1, x = 0, and x = 1. Thus f (x) has three inflection points,
(−1, f (−1)) = (−1, 9), (0, f (0)) = (0, 14), and (1, f (1)) = (1, 19).
Problem 4 (10pts) Suppose that your father wants you to look over the data for the business he started 5 years ago. It
−t
+ 3 for 0 ≤ t ≤ 5 where t is in years.
turns out that his profits over the years have been given by the function P (t) = 2
t +1
During what time period(s) were his profits increasing?
Solution. We know that profits are increasing when the derivative is positive. To decide where the derivative is positive, it is
enough check the sign of P ′ (t) on the intervals delineated by the critical points (since the critical points are the places where
the derivative can change sign). Using the quotient rule, we have
P ′ (t) =
(t2 + 1)(−1) − (−t)(2t)
t2 − 1
= 2
.
2
2
(t + 1)
(t + 1)2
Since the denominator is never zero (as t2 + 1 > 0 for all t), the only critical points occur when P ′ (t) = 0, which in turn
happens if and only if t2 − 1 = 0, that is, if and only if t = 1 or t = −1. So the intervals we are interested in are (0, 1), and
(1, 5). We find that
(1/2)2 − 1
<0
P (1/2) =
((1/2)2 + 1)2
and
22 − 1
P (2) = 2
> 0,
(2 + 1)2
so profits are increasing during years 1 through 5.
Problem 5 (10pts) Consider the following graph, then 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. This is the only problem on the exam for which you need
show no work and for which no English is required.
B
E
D
lim f (x)
x→2
(A) 5
lim f (x)
x→−∞
(B) ∞
lim [f (x) − (x + 1)]
x→−∞
(C) (1, 2)
1
2
F
M > 0 such that if x > M , then |f (x) − 1| <
A
M > 0 such that f (x) > M whenever 0 < |x − 2| <
G
c satisfying the conclusion of the Mean Value Theorem for the interval [3, 5].
H
interval on which f (x) takes its absolute minimum.
C
interval on which f (x) does not take its absolute minimum.
(D) 0
1
2
(E) −∞
(F) 7
(G) 7/2
(H) (0, 2)

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