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.

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. (http://www.lmnoeng.com/Volume/InclinedCyl.htm) here's a discussion on that
when you do find the volume, just differentiate with respect to theta

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}