Download Solutions - Purdue Math

MA16020 - Monday, February 16th
Directions: Work on the following problems. Feel free to work in a group with the people around you or to
move somewhere else in the room. Your work will not be graded, so the primary goal is to make sure you
understand how to solve each problem, not that you just get the right answer from someone you’re working with.
The problems are designed to be ordered by increasing difficulty - that is, the first problem is relatively easy and
the last problem is hard. Feel free to ask me questions.
1. The marginal cost of producing x crates of a product is
M (x) =
√
1 + x ln(1 + x).
Given that the base cost is $100, how much does it cost to produce x crates?
Solution: Marginal cost is the derivative of the cost function, so we have
√
dC
= 1 + x ln(1 + x), C(0) = 100.
dx
Separating variables,
dC =
√
1 + x ln(1 + x) dx.
We use integration by parts on the RHS to integrate. Let u = ln(1 + x) and dv =
1
du = 1+x
dx and v = 32 (1 + x)3/2 (we used a u-substitution here), so
2
2
C = (1 + x)3/2 ln(1 + x) −
3
3
(1 + x)3/2
2
2
dx = (1 + x)3/2 ln(1 + x) −
1+x
3
3
Z
Z
√
√
1 + x dx. Then
1 + x dx
2
4
= (1 + x)3/2 ln(1 + x) − (1 + x)3/2 + C1 .
3
9
Using the initial condition, 100 = − 94 + C1 , so C1 = 100 + 94 . The cost of producing x crates is
4
4
2
C(x) = (1 + x)3/2 ln(1 + x) − (1 + x)3/2 + 100 + .
3
9
9
2. A day’s temperature, in ◦ F, was modeled by the function
T (t) = 43 − t cos
π
t,
12
where t is given in hours and t = 0 corresponds to midnight. Find the average temperature for the day.
Solution: We need to find the average value of the function T on the interval [0, 24], which is
1
24
24 Z
0
π 1
43 − t cos t dt =
12
24
Z
24
0
1
43 dt −
24
Z
0
24
1
π
t cos t dt = 43 −
12
24
Z
24
t cos
0
π
t dt.
12
π
To compute the antiderivative of the integrand, let u = t and dv = cos 12
t dt. Then du = dt and v =
so
Z
t cos
π 12
π
12
t dt = t sin
t −
12
π
12
π
Using the Fundamental Theorem of Calculus,
Z
sin
π 144
π π
12
t dt = t sin
t + 2 cos
t + C.
12
π
12
π
12
12
π
π
sin 12
t,
24
Z
t cos
0
π
144 144
t dt = 2 − 2 = 0.
12
π
π
The average temperature is thus
43 − 0 = 43◦ F.
3. The rate (in grams per day) that a harmful chemical enters a water supply is given by
dC
= 1 + 2t2 e−t .
dt
Find the total amount of the chemical that enters the water supply during the first 3 days.
Solution: We need to compute
3
Z
2 −t
1 + 2t e
Z
dt =
0
3
3
Z
dt +
2 −t
2t e
0
3
Z
dt = 3 + 2
0
t2 e−t dt.
0
To compute the remaining integral, we first find the antiderivative of the integrand. First, let u = t2 and
dv = e−t dt. Then du = 2t dt and v = −e−t , so we have
Z
2 −t
t e
2 −t
dt = −t e
Z
+2
te−t dt.
We evaluate this second integral using IBP now. Let u = t and dv = e−t dt. Then du = dt and v = −e−t , so
we have
Z
2 −t
t e
2 −t
dt = −t e
Z
−t
−t
+ 2 −te + e dt = −t2 e−t − 2te−t − 2e−t + C.
Applying the Fundamental Theorem of Calculus,
Z
3+2
3
2 −t
t e
0
17
dt = 3 + 2 2 − 3
e
≈ 5.307 g.
4. A damping force affects the vibration of a spring so that the displacement of the spring is given by
y = e−4t (cos 2t + 5 sin 2t).
Find the average value of y on the interval [0, π].
Solution: We need to compute
1
π
Z
π
e−4t (cos 2t + 5 sin 2t) dt =
0
1
π
Z
π
e−4t cos 2t dt +
0
5
π
Z
π
e−4t sin 2t dt.
0
We take the antiderivative of the first integrand first. Using IBP once, with u = cos 2t and dv = e−4t dt (so
du = −2 sin 2t dt and v = − 41 e−4t ) gives
Z
e
−4t
1
1
cos 2t dt = − e−4t cos 2t −
4
2
Choosing u = sin 2t and dv = e−4t in the new integral gives us
Z
e−4t sin 2t dt.
Z
e
−4t
Z
1 −4t
1
−4t
− e sin 2t +
e cos 2t dt =
4
2
Z
1 −4t
1 −4t
1
− e cos 2t + e sin 2t −
e−4t cos 2t dt.
4
8
4
1
1
cos 2t dt = − e−4t cos 2t −
4
2
Adding the integral on the RHS over and multiplying through by 4/5 gives us
Z
1
1
e−4t cos 2t dt = − e−4t cos 2t + e−4t sin 2t + C.
5
10
A similar computation for the second integral gives
Z
1
1
e−4t sin 2t dt = − e−4t sin 2t − e−4t cos 2t + C.
5
10
Using this information and the FTC, our original integral is
1
π
Z
0
π
e−4t cos 2t dt +
5
π
Z
0
π
e−4t sin 2t dt =
1 − e−4π
1 − e−4π
+
≈ 0.223.
2π
5π