Calculus I, Sections 1-5, Exam 2, Version C
Wednesday, March 30, 2011
SOLUTIONS
√
1. (4 pts) Use an appropriate linear approximation to approximate 3 1003. Your answer
should be a decimal number. Show your work.
√
√
1
Solution: Let f (x) = 3 x = x 3 and a = 1000. Then compute f (a) = 3 1000 = 10,
2
1
2
1
f 0 (x) = 13 x− 3 and f 0 (a) = 13 · 1000− 3 = 13 (1000 3 )−2 = 13 · 10−2 = 31 · 100
. Then the linear
approximation
L(x) = f (a) + f 0 (a)(x − a) = 10 + 31 ·
1
(x
100
− 1000),
and so
L(1003) = 10 + 13 ·
1
(1003
1001
− 1000) = 10 + 31 ·
1
(3)
100
= 10 +
1
100
= 10 + 0.01 = 10.01
2. (6 pts) Compute the following limits. Each answer should be a real number, +∞, −∞,
or “does not exist.” Show your work.
ln x
.
x→∞ 10 + 1
x
Solution: As x → ∞, ln x → ∞, while 10 +
+∞.
(a) lim
1
x
→ 10 + 0 = 10. Thus the limit is
(b) lim cot t.
t→0
t
and so if t = 0, we get the quotient 10 . So we expect the
Solution: cot t = cos
sin t
limit to be infinite, and we must determine the signs of each one-sided limit. As
t → 0+ , both sin t and cos t are positive, and so limt→0+ cot t = +∞. As t → 0− ,
both sin t < 0, while cos t > 0, and so the limit limt→0+ cot t = −∞. Since the
two one-sided limits disagree, limt→0 cot t does not exist.
3. (9 pts) Compute the following derivatives. Show your work.
(a)
d
sin−1 (2x).
dx
Solution: Use the Chain Rule to compute the derivative
1
p
(b)
1−
(2x)2
·2= √
2
.
1 − 4x2
d t
(e ln t).
dt
Solution: Use the Product Rule
d t
1
et
(e ln t) = et + et ln t = + et ln t.
dt
t
t
1
(c) (f −1 )0 (0), where f (x) = 2x − sin x.
Solution: Since f (0) = 2 · 0 − sin 0 = 0, we also have f −1 (0) = f −1 (f (0)) = 0.
Also, f 0 (x) = 2 − cos x. So compute
(f −1 )0 (0) =
1
f 0 (f −1 (0))
=
1
1
=
= 1.
f 0 (0)
2 − cos 0
4. (8 pts) Consider a rectangular box with a square base, which will consist of 4 vertical
sides, the base, and the top. If the surface area of the box is constrained to be 150 in2 ,
what are the dimensions of the box with maximum volume? Show your work.
Solution: The volume of a box of dimensions `, w, h is V = `wh, and the surface area
is A = 2`w + 2`h + 2wh. For a square base, we have ` = w, and so V = w2 h and
A = 2w2 + 4wh.
The constraint is 150 = A = 2w2 + 4wh. Solving for h gives
h=
75
w
150 − 2w2
=
− .
4w
2w
2
The objective function
2
V =w h=w
2
w
75
−
2w
2
=
w3
75
·w−
.
2
2
The interval of w follows from the conditions
√ h > 0 and w > 0. The formula for
√ h in
terms of w shows that h > 0 when w < 5 3. So the interval for w is 0 < w < 5 3.
To maximize V , compute
75 3 2
dV
=
− w
dw
2
2
75
3 2
and the critical points are when 2 − 2 w = 0, w2 = 25, w = 5 (we need only consider
√
positive w). We can also compute for the endpoints of the interval V (0) = V (5 3) = 0.
· 5 − 12 · 53 = 125 is a global maximum. Thus w = ` = 5 in, and
Thus V (5) = 75
2
75
5
h = 10 − 2 = 5 in also. The volume is 125 in3 .
2
3
5. (8 pts) Consider the function g(x) = x + ln(x − 1).
(a) What is the domain of g(x)?
Solution: The domain of ln x is x > 0. So the domain of ln(x − 1) is x − 1 > 0,
or x > 1. This is the domain of g(x).
(b) Find all the intervals on which g is increasing. Also find all intervals on which g
is decreasing. Find all critical points of g. Show your work.
1
1
and so g 0 (x) = 0 if 1 + x−1
, (x − 1) + 1 = 0,
Solution: Compute g 0 (x) = 1 + x−1
x = 0. This is outside the domain of g(x), and so there are no critical points. On
1
= 2 > 0. So g 0 (x) > 0 on the entire
the interval (1, ∞), check g 0 (2) = 1 + 2−1
domain (1, ∞), and g(x) is increasing there.
(c) Find all intervals on which g is concave up. Find all intervals on which g is concave
down. Show your work.
1
Solution: Compute g 00 (x) = − (x−1)
2 . This is always negative. So g(x) is concave
down on its domain (0, ∞).
4