Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. The tank initially contains 300 liters and 19 liters leak out during the first day. When will the tank be half empty? t= 9.100 How much water will remain in the tank after 3 days? volume =
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.
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