The complete question is;
A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine the amount of salt in the tank at any time
t, if the concentration in the inflow is variable and given by
c(t) = 2 + sin(t/4) lb/gal.
Answer:
dA/dt = (8 + 4sin(t/4)) - (A_t/100)
Explanation:
Rate is given as;
dA/dt = R_in - R_out
R_in = (concentration of salt inflow) x (input rate of brine)
So, R_in = (2 + sin(t/4)) x 4 = (8 + 4sin(t/4))
The solution is being pumped out at the same rate, thus it is accumulating at the same rate.
After t minutes, there will be 400 + (0 x t) gallons left = 400 gallons left
Thus,
R_out =(concentration of salt outflow) x (output rate of brine)
R_out = (A_t/400) x 4 = A_t/100
Thus,
A_t = (100/t)(8t - 16cos(t/4) - 60)
Since we want to find the amount of salt, A(t), let's integrate;
Thus, A = 8t - 16cos(t/4) - (A_t/100)t
A = 60 from the question. Thus,
60 = 8t - 16cos(t/4) - (A_t/100)t
Let's try to make A_t the subject;
(A_t/100)t = 8t - 16cos(t/4) - 60
A_t = (100/t)(8t - 16cos(t/4) - 60)