Question

Your swimming pool containing 60,000 gal of water has been contaminated by 6 kg of a non toxic dye that leaves a swimmer's skin an unattractive green. The pool filtering system can take water from the pool, remove the dye, and return the water to the pool at a flow rate of 200 gal/min. (a) What is the initial value problem for the filtering process; let q(t) be the amount of dye in the pool at any time t. What is the initial amount of dye in the pool?

Answer 1

You can treat this is like a mixing problem where pure water (not containing the dye) is pumped into a tank that hold 60,000 gal of water. The rates at which the amount of dye is flowing in and out of the pool are, respectively,

${q_{\rm in}}'=0$

${q_{\rm out}}'=\left(\dfrac q{60,000}\dfrac{\rm kg}{\rm gal}\right)\left(200\dfrac{\rm gal}{\rm min}\right)=\dfrac q{300}\dfrac{\rm kg}{\rm min}$

Then the net rate at which the amount of dye in the pool changes is $q(t)$ such that

$q'=-\dfrac q{300}$

Initially, the pool contains $q(0)=6\,\mathrm{kg}$ of dye.

The ODE is separable as

$\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{300}$

$\implies\ln|q|=-\dfrac t{300}+C$

$\implies q=e^{C-t/300}$

$\implies q(t)=Ce^{-t/300}$

Given that $q(0)=6$, you get $C=6$, so the amount of dye in the pool at any time $t$ is governed by

$\boxed{q(t)=6e^{-t/300}}$