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galactus · Nov 10, 2007, 05:46 AM

Few calc problems are posted on this site. For us mathophiles, here's a related rates problem if anyone would like to have a go:

A cylindrical beaker has radius r and height h. Water is pouring out as the beaker is tilted. Find a formula that relates the rate of change of the volume of water in the beaker and the rate of change of the angle the beaker makes with the vertical.

jehan60188 · Nov 10, 2007, 10:49 AM

this isn't a very difficult question as far as related rates go, but finding the volume of the liquid is pretty complicated
Inclined cylinder volume calculation for tanks and pipes. here's a discussion on that
when you do find the volume, just differentiate with respect to theta

galactus · Nov 10, 2007, 03:07 PM

Thanks for that link. That's interesting.

You're correct, it's not too bad. But here's my solution to the problem:

If we let, say, d = the depth of the water at the axis of the cylinder, then

$d=h-rtan{\theta}$

Then we have:

$V={\pi}r^{2}d$

$V={\pi}r^{2}(h-rtan{\theta})$

$V={\pi}r^{2}h-{\pi}r^{3}tan{\theta}$

r and h are constant, so differentiating wrt theta:

$\frac{dV}{dt}=-{\pi}r^{3}sec^{2}{\theta}\frac{d{\theta}}{dt}$