# 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. Nov 4 '16 at 18:16

## 1 Answer

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

Do you know how to find the volume of a solid of revolution?

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

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

Otherwise, 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 $$dh/dt$$