Solve Height of Water in Cubical Tank

[KillerZ]

Homework Statement

Suppose water is leaking from a tank through a circular hole of area A_h at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to cA_{h}\sqrt{2gh}, where c (0 < c < 1) is a empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown. The radius of the hole is 2 in., and g = 32ft/s²

Homework Equations

cA_{h}\sqrt{2gh}

\frac{dV}{dt} = -cA_{h}\sqrt{2gh}

A_{h} = (\pi)r^{2} = (\pi)2^{2} = 4\pi

The Attempt at a Solution

I think have to find the volume and take the derivative of it with respect to time.

V = h^{3} not sure if this is the right volume

\frac{dV}{dt} = 3h^{2}\frac{dh}{dt}

3h^{2}\frac{dh}{dt} = -cA_{h}\sqrt{2gh}

\frac{dh}{dt} = \frac{-cA_{h}\sqrt{(2)(32)h}}{3h^{2}}

\frac{dh}{dt} = \frac{-cA_{h}8\sqrt{h}}{3h^{2}}

\frac{dh}{dt} = \frac{-c32\pi\sqrt{h}}{3h^{2}}

 

[tiny-tim]

Hi KillerZ!

V = h^{3} not sure if this is the right volume

No, V is proportional to h, isn't it?

 

[KillerZ]

Is it something like V = (a)(b)(h) then? So a and b both are 10 but h is changing.

 

[tiny-tim]

That's right!

 

[KillerZ]

so like this:

A_{h} = (\pi)r^{2} = (\pi)(\frac{2}{12})^{2} = \frac{4}{144}\pi I missed converting into ft in the first post.

(a) = (b) = 10

V = (a)(b)(h) = 100h

\frac{dV}{dt} = 100\frac{dh}{dt}

100\frac{dh}{dt} = -cA_{h}\sqrt{2gh}

\frac{dh}{dt} = \frac{-cA_{h}\sqrt{(2)(32)h}}{100}

\frac{dh}{dt} = \frac{-cA_{h}8\sqrt{h}}{100}

\frac{dh}{dt} = \frac{-c\frac{32}{144}\pi\sqrt{h}}{100}

\frac{dh}{dt} = \frac{-c\pi\sqrt{h}}{450}

 

[tiny-tim]

Looks good!