A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom.When friction and contraction of the water at the hole are taken into account,the height h of water in the tank is described by 

\f[\frac{dh}{dt}=-c\frac{A_{h}}{A_{w}}\sqrt{2gh}\f]

where 0 <c <1. How long will it take the tank take to empty if c =0.6? Given that the tank is 10 feet high and has radius 2 feet and the circular hole has radius 0.5 inch and g=32 ft/s^2.

We begin by combining all the constants as follows:

\f[A_{h}=pi*(0.5)^{2}=0.785 ft^{2}\f]

\f[A_{w}=pi*(2)^{2}=12.566 ft^{2}\f]

therefore,

\f[c\frac{A_{h}}{A_{w}}\sqrt{2g}=0.6*\frac{0.785}{12.566}*\sqrt{2*32}=0.3\f]

The DE then becomes;

\f[\frac{dh}{dt}=-0.3\sqrt{h}\f]

separating the variables,

\f[ \frac{dh}{\sqrt{h}}=-0.3dt\f]

integrating,

\f[ \int \frac{dh}{\sqrt{h}}=-0.3\int dt\f]

\f[2\sqrt{h}=-0.3t+k\f]

we can rewrite the above equation as follows

Where L is the constant of integration.

but h(0)=H ; substituting this into the above equation we have

Thus the equation becomes

\f[\sqrt{h}=-0.15t+\sqrt{H}\f]

For the tank to empty , h=0 .therefore,

\f[0=-0.15t+\sqrt{H}\f]

solving for t ;

\f[t=\frac{\sqrt{H}}{0.15}=\frac{\sqrt{10}}{0.15}=21.08 seconds\f]