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.

**Further Reading**

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