Originally Posted by

**galactus** Here is a decent one. I like problems like this (it's not that bad):

**The conical tank in the drawing loses water out of the hole in the bottom.**

The cross-sectional area of the hole is 1/4 ft^2. Determine a DE representing

the height of the water h at ant time. Ignore friction and contraction of the

water stream at the hole.

Hint: you can use

$\displaystyle \frac{dh}{dt}=\frac{-A_{0}}{A_{w}}\sqrt{2gh}$ to

represent the height at any time, but use the given data.

Torricelli's law tells us that the speed of the water which flows out of the tank is $\displaystyle \sqrt{2gh}$. (the hole is small enough so that this is correct) As there's no "contraction of the water stream at the hole" (conservation of volumetric flow rate)

$\displaystyle -\frac{1}{4}\sqrt{2gh}=A_{\omega}\frac{\mathrm{d}h} {\mathrm{d}t}$

Substituting $\displaystyle A_{\omega}=\pi\left(\frac{8}{20}h\right)^2=\pi\fra c{4}{25}h^2$ gives us

$\displaystyle

\frac{1}{4}\sqrt{2gh}=-\pi\frac{4}{25}h^2\frac{\mathrm{d}h}{\mathrm{d}t}$

which can be rewritten

$\displaystyle \mathrm{d}t=-\pi\frac{16}{25\sqrt{2g}}h^{\frac{3}{2}}\mathrm{d} h$

Integrating :

$\displaystyle t=\pi\frac{16}{25\sqrt{2g}}\frac{2}{5}\left(8^{\fr ac{5}{2}}-h^{\frac{5}{2}}\right) \Longrightarrow h=\left(8^{\frac{5}{2}}-\frac{125\sqrt{2g}}{32\pi}t\right)^{\frac{2}{5}}$

Using the graph, one can see that the function $\displaystyle h(t)$ is linear for $\displaystyle t<60\,\mathrm{ seconds}$ which could be interesting to make an hourglass. (it'd be a large one

)