Midterm Exam 2
Spring 2013
Math 0120
100 points total
Solutions
1. (a) [10 points] Find f 0 (2) if f (x) =
Solution:
√
√
e.
e is a constant. Hence, f 0 (x) = 0 for any x and, in particular, f 0 (2) = 0.
(b) [10 points] Find f 0 (1) if f (x) = ex ln(x2 ).
Solution:
1
f (x) = ex · 2 ln x = 2ex ln x. Product rule: f 0 (x) = 2ex ln x + 2ex · . Then
x
f 0 (1) = 2e · 0 + 2e · 1 = 2e.
2. [15 points] For the function f (x) = x3 − 9x2 + 24x make the sign diagram that contains both
the first and second derivatives and sketch its graph. On the graph mark all important points.
Solution:
f 0 (x) = 3x2 − 18x + 24 = 3(x2 − 6x + 8) = 3(x − 2)(x − 4).
CNs: f 0 (x) is defined everywhere, f 0 (x) = 0 when x = 2 and x = 4 (CNs).
f 0 (x) < 0 when 2 < x < 4, f 0 (x) > 0 when x < 2 or x > 4.
Concavity: f 00 (x) = 6x − 18 = 6(x − 3).
IP is x = 3. f 00 (x) < 0 when x < 3, f 00 (x) > 0 when x > 3.
Using this information make the sign diagram and sketch the graph of the function.
3. [15 points] Maximum Sustainable Yield:
Marine ecologists estimate the reproduction
curve for swordfish in the Georges Bank fishing grounds to be f (p) = −0.01p2 + 5p, where
p and f (p) are in hundreds. Find the population that gives maximum sustainable yield and
the size of the yield. Support every step in your solution. [Sustainable yield is reproduction
function minus population].
Solution:
Sustainable yield is Y (p) = f (p) − p = −0.01p2 + 5p − p = −0.01p2 + 4p.
We maximize the yield by finding its derivative and CNs. Y 0 (p) = −0.02p+4 = −0.02(p−200).
CNs: Y 0 (p) is defined everywhere, Y 0 (p) = 0 when p = 200 which is the only CNs of Y (p).
Y 00 (p) = −0.02 < 0. Hence the yield has a relative maximum at p = 200 hundreds or
1
p = 20, 000. This maximum is absolute b/c the graph of the yield is a parabola opened
down.
1
· 200 · 200 + 4 · 200 = −2 · 200 + 4 · 200 = 400 hundreds or Y (20, 000) = 40, 000.
Y (200) = −
100
Answer: Maximum sustainable yield is 40, 000 when the population is 20, 000.
4. [10 points] Use implicit differentiation to find y 0 (−1) if y 2 + x3 = 8 and y = 3.
Solution:
We differentiate the equation with respect to x:
d 2
[y + x3 = 8], 2yy 0 + 3x2 = 0.
dx
When x = −1, y = 3 we have 2 · 3 · y 0 + 3 · (−1)2 = 0, 2y 0 = −1, y 0 (−1) = −1/2.
5. Find the value of $2000 deposited in a bank at 12% interest for 6 years compounded
(a) [5 points] quarterly,
(b) [5 points] monthly,
(c) [5 points] continuously.
Write the result as a numeric formula and simplify it. Do not calculate.
6m
r 6m
0.12
Solution: r = 12% = 0.12, t = 6, V (6) = P 1 +
= 2000 1 +
m
m
(a) m = 4, V (6) = 2000 · (1.03)24 .
(b) m = 12, V (6) = 2000 · (1.01)72 .
(c) V (6) = 2000e0.12·6 = 2000e0.72 .
6. [15 points] A spherical snowball is melting such that the radius shrinks at a constant rate
of 2 cm per minute. How fast is the volume of the snowball shrinking when its diameter is 10
cm? [The volume of a sphere is V = 34 πr3 ].
4
dV
4
dr
We differentiate the equation V = πr3 with respect to t:
= · 3πr2 · ,
3
dt
3
dt
dV
dr
dr
= 4πr2 · . It is given that r = 10/2 = 5 cm and
= 2 cm/min.
dt
dt
dt
dV
Then
= 4π · 52 · 2 = 200π cm3 /min.
dt
Solution:
7. [10 points] A bank account grows at 8% compounded quarterly. How long will it take to
double the amount? [Use natural logarithm to find the number of years t. Leave ln in your
2
answer, simplify the result].
Solution:
V (t) = P
Let P be the present value. Then the value after t years will be
4t
0.08
1+
= P · 1.024t .
4
We need to find t when V (t) = 2P . Then P · 1.024t = 2P , 1.024t = 2.
To find t we apply ln to both sides: ln (1.024t ) = ln 2. Then 4 t ln 1.02 = ln 2 and
t=
ln 2
years.
4 ln 1.02
bonus problem. [15 points extra] Find y 00 (2) if x3 + y 3 = 7. [y = −1 when x = 2].
Solution:
Implicit differentiation of the equation with respect to x to gives
x2
d 3
3
2
2 0
0
[x + y = 8] ⇒ 3x + 3y y = 0 ⇒ y = − 2
dx
y
When x = 2, y = −1 we have y 0 (2) = −4
Differentiate the equation one more time:
d
[3x2 + 3y 2 y 0 = 0] ⇒ 6x + 6y(y 0 )2 + 3y 2 y 00 = 0.
dx
When x = 2, y = −1 we have
6 · 2 + 6(−1)(−4)2 + 3(−1)2 y 00 = 0, 6(2 − 16) + 3y 00 = 0 or 3y 00 = 6 · 14,
Finally, y 00 (2) = 28.
3