Question

A large tank is filled to capacity with 400 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min. The well-mixed solution is pumped out at a rate of 8 gals/min. Find the number A(t) of pounds of salt in the tank at time t.

Answer 1

Answer:

The number A(t) of pounds of salt in the tank at time t = $\frac{3t(100 - t)}{25}$

Step-by-step explanation:

Given - A large tank is filled to capacity with 400 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min. The well-mixed solution is pumped out at a rate of 8 gals/min.

To find - Find the number A(t) of pounds of salt in the tank at time t.

Proof -

Given that,

Capacity of water = 400 gallons

rate in flow = 4 gal/min

rate out flow = 8 gal/min

Concentration in = 3 lbs/gal

Concentration out = (Amount of salt at time t )/ (Solution in tank at time t)

Now,

Let

A(t) = Amount of salt in the tank at time t

Now,

Initially tank has 400 gallons of water, So

A(0) = 0 ................(1)

Now,

Rate of change in the amount of salt = $\frac{d}{dt} A(t)$

                                                             = (Rate in flow )( Concentration In) - (Rate out flow )( Concentration out)

                                                             = (4) (3) - (8) ($\frac{A(t)}{400 - 4t)}$

                                                             = 12 - $\frac{2A(t)}{100 - t)}$

⇒$\frac{d}{dt} A(t)$ + $\frac{2A(t)}{100 - t)}$ = 12

Now,

Integrating Factor, I.F = $e^{\int {\frac{2}{100 - t} } \, dt }$

                                    = $e^{-2ln(100 - t)}$

                                    = (100 - t)⁻²

⇒I.F = (100 - t)⁻²

The solution becomes

A (I.F) = ∫ (I.F)(12) dt + C

⇒A (100 - t)⁻² = ∫12(100 - t)⁻² dt + C

⇒A (100 - t)⁻² = -12(100 - t)⁻¹ (-1) + C

⇒A(t) = $12(100 - t)+ C(100 - t)^{2}$

Now,

We have A(0) = 0

⇒C = -12/100

∴ we get

A(t) = 12(100 - t) - $\frac{12}{100}$ (100 - t)²

     = 12(100 - t) [ 1 - $\frac{100 - t}{100}$ ]

     = 12(100 - t) ($\frac{t}{100}$)

     = $\frac{3t(100 - t)}{25}$

⇒A(t) = $\frac{3t(100 - t)}{25}$

∴ we get

The number A(t) of pounds of salt in the tank at time t = $\frac{3t(100 - t)}{25}$