Question

A tank contains 40 gallons of a solution composed of water and bleach. At the start there are 4 gallons of bleach in the tank. A second solution containing half water and half bleach is added tothe tank at the rate of 4 gallons per minute. The solution is kept throughly mixed and drains from the tank at the rate of 4 gallons per minute.

(a) How much blcach is in the tank after t minutes?
(b) Will there be at least 14 gallons of bleach in the tank after 10 minutes?
(c) How much bleach will there be in the tank after a very long time has passed?

Answer 1

a) x* e∧ t/10 = 20 *e∧ t/10 + 2

b) x = 14,2 gall. In 10 min. we get 14,2 gallons of bleach in the tank

c) After a very long time we will find 20 gallons of bleach in the tank

Explanation:

The variation of bleach in water in the tank is Δ(x)t:

Δ(x)t = [Amount of x added to the tank - Amount of x drain out of the tank]*Δt

Now the amount of x added to the tank is:

Rate of adding* the concentration

Rate of adding = 4 gallons per minute

Concentration half and half ( water + bleach) = 0,5

Rate of draining = 4 gallons per minute

Concentration of draining: Unknown but can be expressed as:

x(t)/40. According to this

Δ(x)t = [4 * 0,5 - 4 * x(t)/40]*Δt

Dividing by Δt on both sides of the equation and tacking limits we get

dx/ dt = 2 - x / 10

dx/dt + x/10 =2

Multyiplying by e∧ t/10 on both sides

e∧ t/10* [ dx/dt + x/10 ] = 2 *e∧ t/10

To solve it, integrating the first member is

x* e∧ t/10 = 20 *e∧ t/10 + C

for initial condition t = 0

4 = 20 + C

C = -16

a) x* e∧ t/10 = 20 *e∧ t/10 - 16

b) in 10 minutes

x* e∧ t/10 = 20 *e∧ t/10 - 16

x *e = 20*e - 16

x = 20 - 16/e

x = 20 - 16/2,718

x = 20 - 5,88

x = 14,2 gallons.

c) After a very long time we will find 20 gallons of bleach in the tank