See also: Related Rates Circle Problem.

Before we start, he important things to initially do are to express the volume or surface area as a function of

**one variable**: time. $$V = 16 + 410t$$ You already know that the rate the volume changes is $410 ft^3/sec$, so you don't have to differentiate that. So, we now write volume as a function of radius, to solve the first part, and we write radius as a function of time, by rewriting the volume equation.

So, here, we're just finding the radius at t = 0. $$V = \dfrac{4}{3}\pi{}r^3$$ $$16 = \dfrac{4}{3}\pi{}r^3 \to 3.8197 = r^3 \to 1.5632 = r$$

Now, we do it in general. $$\dfrac{4}{3}\pi{}r^3 = 16 + 410t \to r^3 = 3.8197 + 97.88t$$ $$r = (3.8196 + 97.88t)^{\frac{1}{3}}$$ In order to find the rate at which the radius changes, we must differentiate r(t). $$\dfrac{dr}{dt} = \dfrac{1}{3}(3.8197 + 97.88t)^{-\frac{2}{3}} * 97.88$$ At t = 0, $\dfrac{dr}{dt} = 13.352$. So, the radius changes at 13.352 feet/second at $t = 0$, or when $r = 1.5632$. Now, we want to find the surface area in terms of time.