We are given that a cylindrical tank with volume 300L is leaking from a hole in the bottom at a rate that is proportional to the remaining volume. We are given that after one day, 19L has leaked out.
Let t be the time in days, let V(t) be the volume in liters at time t. Then V(0)=300 and V(1)=281. The rate of change of the volume, `(dV)/(dt) ` , is proportional to the remaining volume so `(dV)/(dt)=ksqrt(V) ` where k is the constant of proportionality. This differential equation is separable so we rewrite as: `(dV)/sqrt(V)=kdt ` . Integrating both sides we get:
`2V^(1/2)=kt+C_1 " or " V^(1/2)=k/2 t + C ` where C is the constant of integration.
V(0)=300 so `C=sqrt(300)=10sqrt(3) `
V(1)=281 so `sqrt(281)=k/2(1)+10sqrt(3) ==> k~~-1.115 `
Thus `V^(1/2) =-.557t+10sqrt(3)`
(a) The time when the volume will be 1/2 the original volume:
`sqrt(150)=-.557t+10sqrt(3) ==> t~~9.1 `
(b) How much remains after three days? t=3 so
`sqrt(V)=-.557(3)+10sqrt(3) ==> V~~244.91"L" `
A graph of the model with time as the independent variable and volume the dependent variable.