A fish tank initially contains 10 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 4 liters per minute. The solution is mixed well and drained at 4 liters per minute.

a. Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c.

dx/dt =

b. Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of c and t.

x (t) =

c. In 25 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incoming brine?

a) dx/dt = 4*c - (4*x / 10) = 4*c - 0.4*x

b) x = 10*c - 10*c*e^ (-0.4*t)

c) c = 2.500 g / Liters

Step-by-step explanation:

Given:

- Initial volume of pure water V_o = 10 Liters

- Inflow of brine Q_in = 4 Liters / min

- Outflow of mixture Q_out = 4 liters / min

Find:

a) Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c.

b) Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of c and t.

c) In 25 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incoming brine?

Solution:

- We will define a constant (c) as concentration of salt as amount of salt in grams per liter.

- dx / dt is the amount of salt in grams per unit time t.

- We will construct a net flow balance:

dx / dt = Q_in*c - (amount of salt / V_o) * Q_out

- Denoting the amount of salt in grams to be x. then we have:

dx/dt = 4*c - (4*x / 10) = 4*c - 0.4*x

- Once we have formulated the ODE, we will solve it to get the amount of salt x in grams at any time t, as follows:

dx/dt = 4*c - 0.4*x

- Separate variables:

dx / (4*c - 0.4*x) = dt

- Integrate both sides:

- (10 / 4) * Ln | 4*c - 0.4*x | = t + K

- Evaluate constant K, when t = 0, x = 0.

- (10 / 4) * Ln|4c| = K

- Input constant K back in solution:

Ln |4*c - 0.4*x| = - 0.4*t + Ln|4c|

4*c - 0.4*x = 4c*e^ (-0.4*t)

x = 10*c - 10*c*e^ (-0.4*t)

- Evaluate using the derived expression t = 25 mins and x = 25, compute c?

x = 10*c - 10*c*e^ (-0.4*t)

25/10 = c (1 - e^ (-0.4*25))

c = 2.5 / (1 - e^ (-0.4*25))

c = 2.500 g / Liters