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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.
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.

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zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So do you find r'? Which would end up as 64π?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Careful, you're finding S'(r) here and yes it ends up being 64pi

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1You mean I might be learning something?! Finally!! Thank you everyone

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the instantaneous rate of change of the surface area\[ \frac{dS}{dr} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \frac{dS}{dr} = \frac{d\left(4\pi r^2\right)}{dr} \]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Then the product rule if I remember the name correctly.. Right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use the product rule, but the power rule is intersting.

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Sorry, I meant the power rule. Where you multiply the value of the exponent by the coefficient and then decrease the exponent by 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When you substituted in \(r=8\), you should have put in the \(\text {ft}\) as well.
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