Math 222
Quiz 7
March 26, 2014
Name:
Solve each of the following problems. You must show your work to receive credit. All
work will be assessed unless it is crossed out. You may not use any notes, calculators,
electronic devices, etc.
Problem 1 (5 points): Let f (x) = sin(2x).
(a) Find T20 f (x).
Solution: We compute
f (x) = sin(2x)
f 0 (x) = 2 cos(2x)
f 00 (x) = −4 sin(2x)
Therefore we have f (0) = f 00 (0) = 0 and f 0 (0) = 2. We conclude that
T20 f (x) = 2x
π/4
(b) Find T2
f (x).
Solution: Using the derivatives we computed in part (a) we find f (π/4) = 1, f 0 (π/4) =
0, and f 00 (π/4) = −4. Therefore we obtain
π 2
π/4
T2 f (x) = 1 − 2 x −
4
Problem 2 (5 points): A 1000 gallon vat is full of a 25% solution of acid. Starting
at time t = 0 a 40% solution of acid is pumped into the vat at 20 gallons per minute.
The solution is kept well mixed and drawn off at 20 gallons per minute so as to maintain
the total value of 1000 gallons. Find the concentration of acid in the tank at time t.
Solution: Let C(t) be the concentration of acid in the tank and Q(t) be the quantity
of acid in the tank. Since there are 1000 gallons of acid in the tank at all times, Q and
C are related by C(t) = Q(t)/1000. We will formulate and solve a differential equation
for Q and then convert our answer to an expression for C using the above relation.
Since a 40% acid solution is flowing in at 20 gallons per minute, the rate in of acid is
40% of 20, or 8 gallons per minute.
Solution of concentration C is flowing out of the tank at 20 gallons per minute, so the
rate out of acid is 20C, or 20Q/1000 = Q/50.
Therefore, the differential equation that describes the mixing process is
Q
dQ
=8−
dt
50
The tank initially contains 1000 gallons of a 25% acid solution, so there is initially 25%
of 1000, or 250 gallons of acid in the tank. So the initial condition is Q(0) = 250.
The equation is both linear and separable, so we can write in standard form and find
an integrating factor, or we can separate variables. Separating variables
dQ
= dt
8 − Q/50
Integrating both sides yields
−50 ln |8 − Q/50| = t + C
or equivalently
ln |8 − Q/50| = −50t + C
Exponentiating both sides yields
8 − Q/50 = Ce−50t
We obtain the general solution by solving for Q
Q = 400 + Ce−50t
As usual, we use C to represent an arbitrary constant in each step, even though its value
may change, and we eliminate the absolute values when we exponentiate by allowing
both positive and negative values of C. Using our initial condition Q(0) = 250 yields
250 = 400 + C, so C = −150. So we obtain
Q(t) = 400 − 150e−50t
Finally, we want the concentration C(t), so we divide Q(t) by 1000 to obtain
C(t) = 0.4 − 0.15e−50t