Question

A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per gallon is added to the tank at 6 gal/min, and the resulting solution leaves at the same rate. Find the quantity Q(t) of salt in the tank at time t > 0.

Answer 1

Salt flows into the tank at a rate of

(1/2 lb/gal) * (6 gal/min) = 3 lb/min

and flows out at a rate of

(Q(t)/60 lb/gal) * (6 gal/min) = 6Q(t) lb/min

The net rate of change of the amount of salt in the tank at time $t$ is then governed by

$\dfrac{\mathrm dQ}{\mathrm dt}=3-6Q$

Solve for $Q$:

$\dfrac{\mathrm dQ}{\mathrm dt}+6Q=3$

$e^{6t}\dfrac{\mathrm dQ}{\mathrm dt}+6e^{6t}Q=3e^{6t}$

$\dfrac{\mathrm d}{\mathrm dt}[e^{6t}Q]=3e^{6t}$

$e^{6t}Q=\dfrac{e^{6t}}2+C$

$\implies Q=\dfrac12+Ce^{-6t}$

The tank starts with 10 lb of salt, so that Q(0) = 10. This gives us

$10=\dfrac12+C\implies C=\dfrac{19}2$

so that the amount of salt in the tank at time $t$ is given by

$\boxed{Q(t)=\dfrac{1+19e^{-6t}}2}$