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 the water remaining. The tank intitially contains 300 liters and 23 liters leak out during the first day.

A. When will the tank be half empty? t= ? days

B. How much water will remain in the tank after 3 days? volume= ? Liters

Given that 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 the water remaining. The tank intitially contains 300 liters and 23 liters leak out during the first day

Let V represent the volume at time t.

Rate of change of V is [tex]\frac{dV}{dt} = k\sqrt{v}[/tex]

where k represents the constant of proportionality

We separate the variables and integrate

[tex]\frac{dV}{\sqrt{v} } =kdt\\2\sqrt{V} =kt +C\\[/tex]

Use the fact when t =0, V =300

[tex]2\sqrt{300} =C\\[/tex]

Now use the fact when t =1 V= 300-23 = 277

[tex]2\sqrt{277} =k+2\sqrt{300} \\k = 2(-0.67719) = -1.3544[/tex]

k is negative because there is leak

So equation is

[tex]2\sqrt{V} =-1.3544t+2\sqrt{300} \\V= (-0.6772t+\sqrt{300} )^2[/tex]

Using this

A) when half empty V = 150

t = [tex]\frac{2\sqrt{150} -2\sqrt{300}}{-1.3544} \\=3.75[/tex]

so it takes 3.75 days

B) When t =3

V = 233.7515