Answer:
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
Step-by-step 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
