## zeesbrat3 one year ago 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.

1. zeesbrat3

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

2. ganeshie8

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

3. Plasmataco

yes, I think so

4. zeesbrat3

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

5. anonymous

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

6. anonymous

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

7. zeesbrat3

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

8. anonymous

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

9. zeesbrat3

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

10. anonymous

See what happens

11. zeesbrat3

8πr 8π(8) 64π

12. anonymous

Remember units

13. zeesbrat3

ft^3 right?

14. anonymous

Nope

15. anonymous

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