### The volume of a sphere is increasing at a rate of 410 ft3/sec. At the instant when the volume is 16 cubic feet, calculate the length of the radius, the rate at which the radius changes, and the rate at which the surface area changes.

Before we start, he important things to initially do are to express the volume or surface area as a function of **one variable**: time.

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.

David Witten