# Related rates - Water drained from a spherical tank

Problem

Water is being drained from a spherical tank with radius 5 meters.

When the depth of the water is 2 meters, it is drained at a rate of $0.5 \mathrm m^3 / \mathrm{min}$.

How fast does the depth decrease at this moment?

My thoughts

I immediately think I need a function for the height of the body of water, with respect to the volume, but I can't find such an expression.

I also start thinking I should express everything with respect to the time $t$, but even then I get stuck trying to find a function for the depth of the water.

Am I over-complicating it? Or maybe under-complicating it?

Any help appreciated!

• You can write a formula for volume as a function of the height of the body of water. Your formula doesn't need to have time in it, but you differentiate with respect to time. – T.J. Gaffney Nov 4 '16 at 18:16

You are right about that you need volume in terms of depth, but time variable isn't needed.

Do you know how you can find volume of revolution?

If so, try to justify why the volume $V$ of water at depth $h$ is given by

$$V(h)= \pi\int_{-5}^{h} 25-x^2\,dx$$

Otherwie I don't know of any method that might easily compute the volume.

Knowing this, recall that

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

by chain rule, and you are given $dV/dt$ so you should be able to compute $dh/dt$