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musicmar
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Homework Statement
Solve problem 23 under the assumption that the solution is pumped out at a faster rate of 10 gal/min. When is the tank empty?
23. A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.
The Attempt at a Solution
If I'm just looking for when the tank is empty, why do I need to care about salt? I solved it with a really simple D.E. using only the volume rates:
dV/dt = Rate in - Rate out
dV/dt = 5 gal/min - 10 gal/min ; V(0) = 500 gal
V = -5t + c
500 = 0 + c
c= 500
V = -5t + 500
0 = -5t + 500
t = 100 min.
I know this is the right answer for t, but the book also gives the equation for A(t) = 1000 - 10t - (1/10)(100-t)2
Is it really necessary to worry about the amount of salt, and if so, could someone help me?
Here is my work for solving for A(t):
dA/dt = 10 lb/min - (A(t)/500 gal) * (10 gal/min)
dA/dt = 5 - (1/50) A(t)
solve as linear; integrating factor = e(1/50)t
(A*e(1/50)t)' = 5*e(1/50)t
A*e(1/50)t = 250*e(1/50)t + c
c = 500
A = 250 + 500*e(1/50)t
Clearly, this is not the same answer. If someone could help me fix this, that would be great. Thank you.