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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.

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  1. zeesbrat3
    • one year ago
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    So do you find r'? Which would end up as 64π?

  2. ganeshie8
    • one year ago
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    Careful, you're finding S'(r) here and yes it ends up being 64pi

  3. Plasmataco
    • one year ago
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    yes, I think so

  4. zeesbrat3
    • one year ago
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    You mean I might be learning something?! Finally!! Thank you everyone

  5. anonymous
    • one year ago
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    the instantaneous rate of change of the surface area\[ \frac{dS}{dr} \]

  6. anonymous
    • one year ago
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    \[ \frac{dS}{dr} = \frac{d\left(4\pi r^2\right)}{dr} \]

  7. zeesbrat3
    • one year ago
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    Then the product rule if I remember the name correctly.. Right?

  8. anonymous
    • one year ago
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    You can use the product rule, but the power rule is intersting.

  9. zeesbrat3
    • one year ago
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    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
    • one year ago
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    See what happens

  11. zeesbrat3
    • one year ago
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    8πr 8π(8) 64π

  12. anonymous
    • one year ago
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    Remember units

  13. zeesbrat3
    • one year ago
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    ft^3 right?

  14. anonymous
    • one year ago
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    Nope

  15. anonymous
    • one year ago
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    When you substituted in \(r=8\), you should have put in the \(\text {ft}\) as well.

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