The surface area, S, of a sphere of radius r feet is S = S(r) = 4πr2. Find the instantaneous rate of change of the surface area with respect to the radius r at r = 8.

Question

zeesbrat3 · Answer

So do you find r'? Which would end up as 64π?

ganeshie8 · Answer

Careful, you're finding S'(r) here and yes it ends up being 64pi

plasmataco · Answer

yes, I think so

zeesbrat3 · Answer

You mean I might be learning something?! Finally!! Thank you everyone

anonymous · Answer

the instantaneous rate of change of the surface area$$
\frac{dS}{dr}
$$

anonymous · Answer

$$
\frac{dS}{dr} = \frac{d\left(4\pi r^2\right)}{dr}
$$

zeesbrat3 · Answer

Then the product rule if I remember the name correctly.. Right?

anonymous · Answer

You can use the product rule, but the power rule is intersting.

zeesbrat3 · Answer

Sorry, I meant the power rule. Where you multiply the value of the exponent by the coefficient and then decrease the exponent by 1

anonymous · Answer

See what happens

zeesbrat3 · Answer

8πr
8π(8)
64π

anonymous · Answer

Remember units

zeesbrat3 · Answer

ft^3 right?

anonymous · Answer

Nope

anonymous · Answer

When you substituted in $r=8$, you should have put in the $\text {ft}$ as well.