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Math Help - Applications of DIfferentiation - Cylinder

  1. #1
    Member classicstrings's Avatar
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    Applications of DIfferentiation - Cylinder

    I keep getting the wrong answer for this question, help appreciated.

    A cylinder is designed so that it is always as high as its base diameter. It is expanding.

    What is the rate of change of its surface area when radius is 1cm?

    What is the rate of change of its volume for the same radius?
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  2. #2
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    Quote Originally Posted by classicstrings

    A cylinder is designed so that it is always as high as its base diameter. It is expanding.

    What is the rate of change of its surface area when radius is 1cm?

    What is the rate of change of its volume for the same radius?
    When the radius is 1 then the diamter is 2 thus the height is 2.
    You have,
    V=\pi r^2 h
    Thus,
    \frac{dV}{dt}=2rh\pi \frac{dr}{dt}+ \pi r^2\frac{dh}{dt}
    There is not enough infromation, i.e. what are the rates of change in radius and height. No wonder you cannot solve it.
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  3. #3
    Member Glaysher's Avatar
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    But h=2r

    So V=2 \pi r^3

    and \frac{dV}{dr} = 6 \pi r^2

    and \frac{dV}{dt}=\frac{dV}{dr} \frac{dr}{dt}

    Only need rate of change of r
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  4. #4
    Member classicstrings's Avatar
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    Thanks! ---> worked all the answers out
    Last edited by classicstrings; September 7th 2006 at 09:28 AM.
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    Quote Originally Posted by Glaysher
    But h=2r

    So V=2 \pi r^3

    and \frac{dV}{dr} = 6 \pi r^2

    and \frac{dV}{dt}=\frac{dV}{dr} \frac{dr}{dt}

    Only need rate of change of r
    You are right, I missed that.
    ---
    To add to Glaysher's post.
    Then,
    \frac{dh}{dt}=2\frac{dr}{dt}
    So you elimanate a variable.
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