Question

6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at a rate of 3 gallons per minute, and the well-stirred mixture flows out at a rate of 4 gallons per minute. Find the amount (in lb) of salt Q in the solution as a function of t in minutes.

Answer 1

The answer is below

Explanation:

Let Q represent the amount of salt in the tank at time t.

`(dQ)/(dt)= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= (4Q)/(100-t) \\\\(dQ)/(dt)= 6-(4Q)/(100-t) \\\\(dQ)/(dt)+ (4Q)/(100-t)=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {(4)/(100-t) } \, dt }=e^{-4\int\limits {(-1)/(100-t)}=e^(-4ln(100-t))=(100-t)^(-4)}\\\\Multiplying\ through \ by\ IF: \\\\(100-t)^(-4)(dQ)/(dt)+ (100-t)^(-4)(4Q)/(100-t)=6(100-t)^(-4)\\\\`

`Integrating:\\\\A(100-t)^(-4)=-2(100-t)^(-3)+c\\\\A=-2(100-t)+(c)/((100-t)^(-4)) \\\\at, t=0,A=0\\\\0=-2(100-0)+(c)/((100-0)^(-4))\\\\c=0.02\\\\A=-2(100-t)+(0.02)/((100-t)^(-4))`