A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/minute. The well mixed solution is then pumped out at a rate of 10 gal/minute. Find the number A (t) of pounds of salt in the tank at time t. When is the tank empty?

A) A (t) = 10 (100 - t) + c (100 - t) ²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) * (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 * 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10) gal/min = - 5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) * (output rate of brine)

R_out = [A (t) / (500 - 5t) ]lb/gal * 10 gal/min

R_out = 10A (t) / (500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A (t) / (500 - 5t)

This simplifies to;

dA/dt = 10 - 2A (t) / (100 - t)

Rearranging, we have;

dA/dt + 2A (t) / (100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^ (∫2 / (100 - t)) = e^ (In (100 - t) ^ (-2)) = 1 / (100 - t) ²

Now, let's multiply the differential equation by the integrating factor 1 / (100 - t) ².

We have;

So, we;

(1 / (100 - t) ²) (dA/dt) + 2A (t) / (100 - t) ³ = 10 / (100 - t) ²

Integrating this, we now have;

A (t) / (100 - t) ² = ∫10 / (100 - t) ²

This gives;

A (t) / (100 - t) ² = (10 / (100 - t)) + c

Multiplying through by (100 - t) ², we have;

A (t) = 10 (100 - t) + c (100 - t) ²

B) At initial condition, A (0) = 0.

So, 0 = 10 (100 - 0) + c (100 - 0) ²

1000 + 10000c = 0

10000c = - 1000

c = - 1000/10000

c = - 0.1

Thus;

A (t) = 10 (100 - t) + - 0.1 (100 - t) ²

A (t) = 1000 - 10t - 0.1 (10000 - 200t + t²)

A (t) = 1000 - 10t - 1000 + 20t - 0.1t²

A (t) = 10t - 0.1t²

Tank will be empty when A (t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes