Question

A tank in the form of a right-circular cylinder of radius 2feet and height 10feet is standing on end. If the tank is initially full of water and water leaks from a circular hole of radius 12inch at its bottom, determine a differential equation for the height h of the water at time t>0. Ignore friction and contraction of water at the hole.

Short answer

The differential equation for the height h of the water at time t>0 is $\frac{dh}{dt}=-\frac{\sqrt{h}}{288}$.

Step-by-step solution

Determine the differential equation for the height

Let the volume of water in the tank at any moment be$V\left(t\right)=A_{w}h$.

Here,$A_w$(in feet) is the constant area of the upper surface of the water. So,

$\begin{array}{c}\frac{dV\left(t\right)}{dt}=A_w\frac{dh}{dt}\\ \frac{1}{A_w}\frac{dV\left(t\right)}{dt}=\frac{dh}{dt}\end{array}$

Using Torricelli’s law, then the equation becomes,

$\frac{dV\left(t\right)}{dt}=-A_h\sqrt{2gh}$

Obtain the equation from the above two equations.

$\frac{dh}{dt}=-\frac{A_h}{A_w}\sqrt{2gh}$

Determine the surface area of the hole and the area of the cylinder

Let the surface area of the hole be,

$\begin{array}{c}{A}_h=\pi r^2\\ =\pi {\left(\frac{0.5}{12}\right)}^2\\ =\frac{\pi }{576}f{t}^2\end{array}$

Let the area of the cross-section of the right-circular cylinder be,

$\begin{array}{c}{A}_w=\pi R^2\\ =\pi (2{)}^2\\ =4\pi f{t}^2\end{array}$

Determine the value of the differential equation for the height.

Substitute all the known values in the differential equation for the height.

$\begin{array}{c}\frac{dh}{dt}=-\frac{\frac{\pi }{576}}{4\pi }\sqrt{64h}\\ =-\frac{8}{2304}\sqrt{h}\\ \frac{dh}{dt}=-\frac{\sqrt{h}}{288}\end{array}$

Hence, the differential equation for the height of the water is $\frac{\mathrm{dh}}{\mathrm{dt}}=-\frac{\sqrt{\mathrm{h}}}{288}$.