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anonymous

  • one year ago

The surface area of a right circular cylinder of height 5 feet and radius r feet is given by S(r)=2πrh+2πr2. Find the instantaneous rate of change of the surface area with respect to the radius, r, when r = 6.

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  1. zepdrix
    • one year ago
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    \[\Large\rm S(r)=2\pi r h+2\pi r^2\]Hey Camila! :) Have you learned your derivative shortcuts at this point? Or do we have to use the limit definition for the derivative to find our instantaneous rate of change? -_-

  2. anonymous
    • one year ago
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    I know how to derive.

  3. zepdrix
    • one year ago
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    We're holding the height, h, constant at 5.\[\Large\rm S(r)=2\pi r \cdot 5+2\pi r^2\]Simplifies our function a little bit, which is nice.\[\Large\rm S(r)=10\pi r+2\pi r^2\]

  4. zepdrix
    • one year ago
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    So apply your power rule to each term, should be pretty straight forward :) Are you getting confused because of all that pi nonsense mixed in?

  5. anonymous
    • one year ago
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    Yeah that was the part that confused me.

  6. zepdrix
    • one year ago
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    \[\Large\rm (10\pi x)'=10\pi(x)'=10\pi (1)\]Normal power rule, ignoring the constant coefficients :)

  7. anonymous
    • one year ago
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    So I would get 10π+4πr and plug in 6 for r?

  8. zepdrix
    • one year ago
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    \[\Large\rm S'(r)=10 \pi+4\pi r\]Good good good.

  9. anonymous
    • one year ago
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    So when I plug that in it would be 34π right?

  10. zepdrix
    • one year ago
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    Yayyy good job! *\c:/* Camila <*c:/*

  11. anonymous
    • one year ago
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    Haha okay thank you so much!

  12. zepdrix
    • one year ago
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    np

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