A tank contains 200 gallons of water in which 5 pounds of salt is dissolved. A b
ID: 2999955 • Letter: A
Question
A tank contains 200 gallons of water in which 5 pounds of salt is dissolved. A brine solution containing 3 pounds of salt per gallon is pumped into the tank at a rate of 6 gallons per minute. The solution is well mixed and pumped out at the same rate. Let A(t) represent the amount of salt in the tank at time t. Solve the initial value problem for A(t). How much salt is present in the tank after 20 minutes? Is my soln correct the weight seems on the high side?
dA/dt=Rin-Rout A(0)=5
Rin=(3lb/gal)(6gal/min)=18lb/min
Rout=(A/200(lb/gal))(6gal/min)=(3A/100)(lb/min)
dA/dt=18-3A/100
dA/dt+3A/100=18
i.f.=e^(3/100)t
d/dt[Ae^(3/100)t]=18e^(3/100)t
Ae^(3/100)t=600e^(3/100)t+C
A(t)=600+Ce^-(3/100)t
A(0)=5=600+Ce^-(3/100)(0)
yields C=-595
A(t)=600-595e^(-3/100)t
A(20)=600-595e^(-3/100)(20)
A(20)=273.46lbs
Explanation / Answer
correct
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