Fresh water is poured at a rate of 2 gal/min into a tank A, which initially cont

Question

Fresh water is poured at a rate of 2 gal/min into a tank A, which initially contains 100 gallons of a salt solution with concentration 0.5 lb/gal. The stirred mixture flows out of tank A at the same rate and into a second tank B that initially contained 100 gal of fresh water. The mixture in tank B is also stirred and flows out at the same rate.

a) Determine an IVP satisfied by the amount of salt in tank A.

b) Find the amount of salt in tank A at any time.

c) Find the IVP satisfied by the amount of salt in tank B.

d) Determine the amount of salt in tank B as a function of time.

Explanation / Answer

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A tank originally contains 100 gallons of fresh water. Then water containing .5 lbs of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. How much salt is in the tank at the end of 10 minutes?

ds/dt = (0.5) (2) - s(2)/100 ....... change in salt = salt in - salt out
ds/dt = 1 - s/50
d/dt [ e^(t/50) s ] = e^(t/50)
....... if you don't get that, then separate variables and integrate both sides
....... ? 50 / (50 - s) ds = ? dt
e^(t/50) s = 50 e^(t/50) + c
s = 50 + c e^(-t/50)
0 = 50 + c ........ s[0] = 0 (initially fresh water)
c = -50
s = 50 (1 - e^(-t/50))
s(10) = 50 (1 - e^(-10/50)) ? 9.063462346 lbs

Answer: ? 9.06 lbs

PS. That's an elementary differential equations problem.
It's odd you're doing that in Calculus II.

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