# Related rates of change are killing me!

1. Mar 16, 2006

### KingNothing

Hi. I am getting absolutely embarassed by these related rates problems. Here is one that I simply keep getting wrong:

The volume of an expanding sphere is increasing at a rate of 12 cubed cm per second. When the volume is $$36\pi$$, how fast is the surface area increasing?

$$V=\frac {4*pi*r^3}{3}$$$$S=4*pi*r^2$$
(how the heck do you use pi in latex? I know it's \pi, but that doesn't work right when Iput it in!)

$$\frac {dV}{dt}=4 \cdot \pi \cdot r^2 \cdot \frac {dr}{dt}$$
Since volume is $$36 \cdot \pi$$, $$r=3$$. Correct?

Last edited: Mar 16, 2006
2. Mar 16, 2006

### Staff: Mentor

Start by finding $dV/dt$ and $dS/dt$ in terms of r and $dr/dt$. (Hint: Chain rule)