Water drained from a spherical tank

Question

I found a very similar question but the question did not have an answer.

The problem

A spherical tank with a radius of $5 dm$ is filled with water at a pace of 3 litres/minute (which is $3dm^3/min$ but let's ignore units). How fast does the water surface rise in the tank at the moment when the depth is $2 dm$?

$$V(d) = \pi \frac{15d^2-d^3}{3}$$

My attempt at a solution

Let's summarize what we know from the problem text:

$$\frac{dV}{dt} = 3$$

and we want to know $\frac{dd}{dt}$ (where the d stands for depth)

So we set up the 'equation':

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

Now all we should have to do is find the inverse of $V(d)$ and derive that. Except that can't be done in a neat (as in orderly) way.

Then I thought that maybe we could set up something like

$$V(t) = \pi \frac{15d(t)^2-d(t)^3}{3}$$

and use implicit differentiation to get $d'(t)$, then you can use the fact that we know $V'(t)$ to be constant.

That didn't really work, I got this as my final result:

$$3 = 10d'(t) - d'(t)^2$$

and now I'm stuck. Maybe you're supposed to use some kind of knowledge of geometry to solve this one (so some other formula is needed), I'm not sure.

Answer 1

Alright, so I was almost there. Jens nudged me in the right direction. If I had to re-do this problem I'd give this as a hint to myself: Set up all the derivatives you have and the one you need and first try and find every equation you can with those, this will make it easier to find which one is viable.

Solution:

We have $$\frac{dV}{dt} = \frac{dd}{dt}\frac{dV}{dd}$$

We know that $$\frac{dV}{dt} = 3$$ at all times.

We can easily find $\frac{dV}{dd}$ (only solution shown):

$V'(d) = 10\pi d - \pi d^2$

and we were looking for $\frac{dd}{dt}$ when $d=2$ so $V'(2) = 16\pi$

which results in the following answer:

$\frac{dd}{dt} = 3/16\pi$ when $d=2$