Respuesta :
Answer:
The number A(t) of pounds of salt in the tank at time t = [tex]\frac{3t(100 - t)}{25}[/tex]
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 = [tex]\frac{d}{dt} A(t)[/tex]
= (Rate in flow )( Concentration In) - (Rate out flow )( Concentration out)
= (4) (3) - (8) ([tex]\frac{A(t)}{400 - 4t)}[/tex]
= 12 - [tex]\frac{2A(t)}{100 - t)}[/tex]
⇒[tex]\frac{d}{dt} A(t)[/tex] + [tex]\frac{2A(t)}{100 - t)}[/tex] = 12
Now,
Integrating Factor, I.F = [tex]e^{\int {\frac{2}{100 - t} } \, dt }[/tex]
= [tex]e^{-2ln(100 - t)}[/tex]
= (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) = [tex]12(100 - t)+ C(100 - t)^{2}[/tex]
Now,
We have A(0) = 0
⇒C = -12/100
∴ we get
A(t) = 12(100 - t) - [tex]\frac{12}{100}[/tex] (100 - t)²
= 12(100 - t) [ 1 - [tex]\frac{100 - t}{100}[/tex] ]
= 12(100 - t) ([tex]\frac{t}{100}[/tex])
= [tex]\frac{3t(100 - t)}{25}[/tex]
⇒A(t) = [tex]\frac{3t(100 - t)}{25}[/tex]
∴ we get
The number A(t) of pounds of salt in the tank at time t = [tex]\frac{3t(100 - t)}{25}[/tex]
