Question

A 2100​-gal tank can support no more than 160 guppies. Eight guppies are introduced into the tank. Assume that the rate of growth of the population is
dP/dt=0.0015(160−P)P​, where time t is in weeks.

a. Find a formula for the guppy population in terms of t.
b. How long will it take for the guppy population to be​ 100? 125?

Answer 1

The rate of growth equation is separable:
\[\frac{\mathrm{d}P}{\mathrm{d}t}=\frac{3}{2000}(160-P)P\iff\frac{\mathrm{d}P}{P(160-P)}=\frac{3}{2000}\,\mathrm{d}t\]Integrate both sides and use the initial condition \(P(0)=8\) (the initial amount of guppies) to find the particular solution \(P(t)\).

Part (b) is a matter of finding \(t\) such that \(P(t)=100\) and \(P(t)=125\).