A metal sphere is dissolving in acid. It remains spherical and the rate at which its volume decreases is proportional to its surface area. Show that its radius is decreasing at a constant rate.
Here's a kick off
$\displaystyle \displaystyle \frac{dV}{dt}\propto 4\pi r^2 $
$\displaystyle \displaystyle \frac{dV}{dt}= 4k\pi r^2 $
$\displaystyle \displaystyle \frac{\frac{dV}{dr}}{\frac{dt}{dr}}= 4k\pi r^2 $
Now firstly find $\displaystyle \displaystyle \frac{dV}{dr} $ from $\displaystyle \displaystyle V = \frac{4}{3}\pi r^3$