Look at original problem statement and question in Description below.

Question

A 2000 gallon container initially holds 1000 gallons of a brine (salt water) solution containing 40 lbs of salt. Brine containing 0.02 lbs/gal of salt flows into the container at a rate of 50 gal/min. The solution is kept thoroughly mixed, and the mixture flows out at a rate of 25 gal/min. Set up (but DO NOT SOLVE) a differential equation that can be used to solve for the amount of salt in lbs, S(t), that is in the tank at any time t in minutes. Be sure to

include the initial condition and the time interval for which the equation is valid.

Answer 1

Suppose S(t) is the amount of salt in lbs at time t.

Then S(0) = 40.

Suppose V(t) is the volume in the container at time t

Then V(0) = 1000, and V(t) = 1000 + 25t (we have a net flow of 25 gallons per minute in)

Consider what happens in the time interval (t, t + Δt) where Δt is small

50Δt gallons of brine flows in, 0.02 lbs / gallon, so 0.02 x 50Δt lbs of salt flows in, ie Δt lbs

25Δt gallons flow out. The concentration of salt is approximately S(t)/ V(t) so 25Δt S(t)/V(t) lbs of salt flows out

So ΔS = Δt - 25Δt S(t)/V(t)

Or ΔS/Δt = 1 - 25 S(t)/ V(t) = 1 - 25 S(t) / (1000 + 25t)

As Δt > 0 the left tends to dS/dt

dS/dt = 1 - (25 S ./ (1000 + 25t))

Only valid until the tank is full which happens after 1000 / 25 = 40 minutes

So valid for t in the interval [0,40]

Mike