1. ## another related rates

Here's another related rates problem if anyone would like to chew on it.

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.

2. Originally Posted by galactus
Here's another related rates problem if anyone would like to chew on it.

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.
ill give it a try.. is it..

$V = \pi r^2 h - \pi r^3 \tan \theta$, where $\theta$ is the angle with respect to the vertical.. so
$dV = -\pi r^3 \sec^2 \theta d\theta$

3. You're certainly on the right track, k.

Except, r is constant, so we differentiate wrt theta.

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:

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

4. Originally Posted by kalagota
ill give it a try.. is it..

$V = \pi r^2 h - \pi r^3 \tan \theta$, where $\theta$ is the angle with respect to the vertical.. so
$dV = -3\pi r^2 \sec^2 \theta d\theta$
ah yeah, a typo error on my scratch, i didnt notice it while typing here.. Üi've edited it.. Ü