Answer:
The number A(t) of pounds of salt in the tank at time t =
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 =
= (Rate in flow )( Concentration In) - (Rate out flow )( Concentration out)
= (4) (3) - (8) (
= 12 -
⇒ +
= 12
Now,
Integrating Factor, I.F =
=
= (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) =
Now,
We have A(0) = 0
⇒C = -12/100
∴ we get
A(t) = 12(100 - t) - (100 - t)²
= 12(100 - t) [ 1 - ]
= 12(100 - t) ()
=
⇒A(t) =
∴ we get
The number A(t) of pounds of salt in the tank at time t =