Math 171
Exam III Solutions
Fall 2002
1
1. (a) cos(tan−1 5) = √
26
5π
(b) arccos(cos( 4 )) = 3π
4
2. (a) The points that we use are (0, 5000) and (3, 15000) and the formula is y = Ao ekt with
Ao = 5000
15000 = 5000e3k
3 = e3k
ln(3) = 3k
k = ln(3)
3 ≈ 0.3662040962
Answer y = 5000e0.3662040962t
(b) plug in a 1 for t and you get y = 7211.24785. Answer: approximately 7221
3. (a) y ′ =
5
1 + (5x)2
(b) y = (x2 + 2)x
ln y = x ln(x2 + 2)
y′
2x2
2
y = ln(x + 2) + x2 +2
2
′
x
y = (x + 2)
"
2x2
ln(x + 2) + 2
x +2
2
4. a(t) = 2et + cos t, t + 3 sin t
#
The velocity function is the antiderivative of the acceleration function.
v(t) =
*
t2
− 3 cos(t) + K
2e + sin(t) + J,
2
t
+
Since v(0) = 5i − 2j we can compute J and K.
02
2
2e0 + sin(0) + J = 5
2+J =5
J =3
Thus v(t) =
*
− 3 cos(0) + K = −2
0 − 3 + K = −2
K=1
+
t2
− 3 cos(t) + 1
2et + sin(t) + 3,
2
5. (12 points) Compute the exact values of these limits.
(a) lim
n→∞
lim
n→∞
n
X
3 i
i=1
n n
+ 4 = lim
n→∞
n
n
12
3 X
12 X
+
=
lim
1
i
+
n→∞ n2
n2
n
n i=1
i=1
n X
3i
i=1
"
3 n(n + 1) 12
3n2 + 3n
+
n
=
lim
+ 12 = 1.5 + 12 = 13.5
n→∞
n2
2
n
2n2
4
(b) lim (2 − x) x−1 = lim eln(2−x)
x→1+
x→1+
Now compute this limit.
4 ln(2 − x) LH
lim
= lim
x−1
x→1+
x→1+
4
−4
2−x
1
Answer: lim (2 − x) x−1 = e−4
x→1+
4
x−1
lim
= ex→1+
= −4
4 ln(2−x)
x−1
#
6. The domain of the function is all real numbers greater than zero.
f ′ (x) =
x∗
1
x
− ln(x)
1 − ln(x)
=
x2
x2
f ′ (x) = 0
1−ln(x)
=0
x2
1 − ln(x) = 0
1 = ln x
x = e1
This is the only critical value of the function.
Decreasing on (e1 , ∞)
7. (10 points) For the function f (x) = 36 − x2 and the partition P = {0, 1, 4, 5}.
(a) P (0) ∗ 1 + P (1) ∗ 3 + P (4) ∗ 1 = 36 + 105 + 20 = 161
(b) Since the function is a parabola and is increasing on the interval (−∞, 6) and decreasing
on the interval (6, ∞), a left sum on the interval will always be an underestimate.
8. The possible inflection values are x = −3, 2, and 9.
f ′′
+
−5 −3
−
0
+
+
2
5
9
10
Concave up: (−∞, −3), (2, 9), (9, ∞)
concave down: (−3, 2)
inflection values at x = −3 and x = 2
9. Remember this is the graph of f ′ (x).
(a) There is a relative max at the critical value of x = −3. The critical value of x = 3 is neither
a max nor a min.
(b) (−3, −2.1)
(c) (−5, −3) and (0, 3)
10. The variables are labeled on the picture. The formula for the area of the poster is Ap = xy and
the formula for the area of the printed area is A = (y − 2)(x − 3). Since the Area of the poster
is fixed at 180 in2 , we know that xy = 180. Solving for y and substituting into the printed area
formula gives.
540
180
− 2) = 186 − 2x −
Now this
A = (x − 3)(
y−2
x
x
formula is only good for the interval (3, 90) The smallest value for x would be 3 (this would give no printed
area). The largest value that x can be is 90 (this
would be that y is at its smallest value of 2).
x
x−3
Printed Area
A′ = −2 + 540
2
x
2 = 540
x2
x2 = √
270
√
x = 270 ≈ 16.43167673 Note: − 270 is not used
since it is not in the interval.
y
√
Since A′′ = −1080
and A′′ ( 270) < 0, we have that this critical value will be a local max. If we
x3
look at a sign chart, we will be able to make the same conclusion.
√
Dimensions: 270 in by √180
in or 16.4317 in by 10.9544 in
270