Find the rate of change of radius of cylinder with respect of time Question:
A great amount of oil is spilled in the ocean. After a brief turbolence, the oil spreads forming a disk of radius $r$ and thickness $h$. The radius of the disk is increasing and the thickness is decreasing. Knowing that $h=\frac {c}{\sqrt t}$, prove that the $\frac {dr}{dt}  $ is inversely proportional to $t^{3/4} $.
Attempted solution:
We know that the volume of oil spilled ($V$) remains unchanged. Therefore:
$$\frac {dV}{dt} = \pi(r^2\frac{dh}{dt}+2rh\frac{dr}{dt}) = 0$$
$$\frac {dh}{dt} = \frac {-1}{2} \frac{c}{t^{3/2}}$$
Symplifying and substing:
$$ \frac {-1}{2} \frac{c}{t^{3/2}}r^2+2r\frac {c}{\sqrt t}\frac{dr}{dt} = 0 $$
Isolating the request:
$$ \frac{dr}{dt} = \frac {1}{2} \frac{c}{t^{3/2}}r^2 * \frac {t^{1/2}}{c *2r}$$
$$ \frac{dr}{dt} = \frac {1}{4} \frac{r}{t} $$
I think that the $r$ at the top is the problem, but I don't know how to go on. I've also tried with implicit differentiation and I get the same result.
Ps. Sorry for my bad English, which is obviously not my first language!

Everything you've done is fine. But you haven't used $V_0 = \pi r^2 h$ to get $r$ as a function of $t$.
