Your solution will involve L and n constant of proportionality, k. An aquarium p

Question

Your solution will involve L and n constant of proportionality, k. An aquarium pool has volume 2-10^6 liters. The pool initially contains pure fresh water. At t=0 minutes, water containing 10 grams/liter of salt is poured into the pool at a rate of 60 liters/minute. The salt water instantly mixes with the fresh water, and the excess mixture is drained out of the pool at the same rate (60 liters/minute). Write and solve a differential equation for S(t), the mass of salt in the pool at time t. What happens to S(t) as t rightarrow infinity?

Explanation / Answer

V = 2,106 litres

10g/litre * 60litres/minute = 600g of salt is flowing into the pool per minute

Mass of salt in pool at time t is S(t)
60 litres flow out every minute
Salt flowing out: (60/2,106) * S(t)

Initial condition: S(0) = 0

dS/dt = 600 - (60/2,106)S
dS/dt = 600 - 10S/351
dS/dt + (10/351)S = 600

This is a linear differential equation:
Integrating factor = e^(10/351)t.dt = 35.1 e^(10/351)t

35.1S e^(10/351)t = 35.1 * 600 e^(10/351)t
35.1S e^(10/351)t = 35.1² * 600 e^(10/351)t + k
S e^(10/351)t = 35.1 * 600 e^(10/351)t + c

S = 600*35.1 + c.e^(-10/351)t
S(0) = 0
0 = 21,060 + c
c = -21,060

S = 21,060 - 20160e^(-10/351)t

As t goes to infinity, the mass of salt in the pool will approach the asymptotic value of 21,060g.

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