## Oct 5 Related Rates Sphere Problem

MathJax TeX Test Page The volume of a sphere is increasing at a rate of $410 \text{ ft}^3$/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. $$V = 16 + 410t$$ 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.
MathJax TeX Test Page $$SA = 4\pi{}r^2$$ (for reference) $$r^2 = \left((3.8196 + 97.88t)^{\frac{1}{3}}\right)^2 = (3.8196 + 97.88t)^{\frac{2}{3}}$$ $$SA = 4\pi{}r^2 = 4\pi{}(3.8196 + 97.88t)^{\frac{2}{3}}$$ $$\dfrac{dS}{dt} = 4\pi{}(\frac{2}{3})(97.88)(3.8196 + 97.88t)^{-\frac{1}{3}}$$ At t = 0, the surface area changes by $524.569 ft^2/sec$.