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Math Help - Related Rates

  1. #1
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    Related Rates

    As a circular griddle is being heated, its diameter changes at a rate of 0.01 cm/min. When the diameter is 30 cm, at what rate is the area of the griddle changing?

    I know my formula is V=PI*r^2
    and i know since i am given a diameter i would use (h/2) for my radius.

    but once i take my derivative of the area formula and plug in my diameter and my DD/DR as .01 cm/min, i am given .942...
    the answer is .471

    what am i doing wrong?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rhcp1231 View Post
    As a circular griddle is being heated, its diameter changes at a rate of 0.01 cm/min. When the diameter is 30 cm, at what rate is the area of the griddle changing?

    I know my formula is V=PI*r^2
    and i know since i am given a diameter i would use (h/2) for my radius.

    but once i take my derivative of the area formula and plug in my diameter and my DD/DR as .01 cm/min, i am given .942...
    the answer is .471

    what am i doing wrong?
    Your formula is: A = \pi r^2 = \frac {\pi D^2}4

    And so, by differentiating implicitly with respect to time, you get: \frac {dA}{dt} = \frac {\pi D}2 \cdot \frac {dD}{dt}

    Now, you want \frac {dA}{dt} when D = 30 and \frac {dD}{dt} = 0.01, just plug those in and compute.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Your formula is: A = \pi r^2 = \frac {\pi D^2}4

    And so, by differentiating implicitly with respect to time, you get: \frac {dA}{dt} = \frac {\pi D}2 \cdot \frac {dD}{dt}

    Now, you want \frac {dA}{dt} when D = 30 and \frac {dD}{dt} = 0.01, just plug those in and compute.
    it was so obvious lol thanks man
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