# Sanity Check: Related Rates for Sphere

• January 12th 2013, 09:17 AM
Biff
Sanity Check: Related Rates for Sphere
Hello,

Working on the following problem:

Express the rate at which the volume of the sphere changes with respect to the surface area of the sphere (as a function of r).

My Work:
$S = 4\pi r^2 \Rightarrow r = \left( \frac{S}{4\pi}\right)^{\frac{1}{2}}$

$V = \frac{4}{3} \pi r^3 = \frac{4}{3}\pi \cdot \left( \frac{S}{4\pi}\right)^{\frac{3}{2}}$

$\frac{d}{dS} \left( V = \frac{4}{3} \pi \cdot \left( \frac{S}{4\pi}\right)^{\frac{3}{2}\right)$

$\frac{dV}{dS} = \frac{1}{2} \left(\frac{S}{4\pi}\right)^{\frac{1}{3}}$

$\text{Substitute } S:$
$\frac{dV}{dS} = \frac{1}{2} \left( \frac{4\pi r^2}{4\pi}\right)^{\frac{1}{3}} = \frac{1}{2} (r^2)^{\frac{1}{3}} = \frac{1}{2} \cdot r^{\frac{2}{3}}$

My brain is fried after doing a lot of calculus problems so I wanted a sanity check: does it look okay?
• January 12th 2013, 09:31 AM
MarkFL
Re: Sanity Check: Related Rates for Sphere
You have made a slight error. You should have:

$\frac{dV}{dS}=\frac{1}{2}\left(\frac{S}{4\pi} \right)^{\frac{1}{2}}$

You could also use the chain rule as follows:

$\frac{dV}{dS}=\frac{dV}{dr}\cdot\frac{dr}{dS}=(4 \pi r^2)\left(\frac{1}{8\pi r} \right)=\frac{r}{2}$