Thread: more Differential Equations

A pool of 60,000 gal water is contaminated with 5kg of dye. The filter can remove the dye and pumps at a rate of 200 gal/min find the initial value problem. let q(t) be the amount of dye in the pool at any time t.

My work:
dq/dt=-rq/V, where r is the rate of filtration and V is the volume of the pool
dq/q=-rdt/V
lnq=-rt/V+C
at t=o, q=5000g
C=ln5000
q=5000e^(3.33E-3t)

Book Answer:
dq/dt=-q/300, q(0)=5000g

The rate of filtration is being taken out of the volume of the pool. Why would the book choose to use the ratio -V/r?

The rate of filtration is being taken out of the volume of the pool. Why would the book choose to use the ratio -V/r?

The book isn't taking -V/r

The problem here is you are assuming the pool is being drained and no water is being added.

The book is assuming that pure water is being pumped in at the same rate the pool is being drained.

Therefore the volume is a constant 60,000gallons.

dq/dt = -q (r/V) = -q (200/60,000) = -q/300

nm, I just realized that 1/300=3.33E-3

By the way if you just drain it you would have

dq/dt = - q (200/(60,000-200t) = - q/(300-t)