Question

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 A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by
cin()-2+sin(t/4) lb/gal. 40-18000, sin(t)- cos(t)-46324e( )

Answer 1

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)

Step-by-step 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)