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Math Help - Radius of Speherical Ballon increasing at time

  1. #1
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    Radius of Speherical Ballon increasing at time


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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by CAIR View Post

    This is a related rate problem, which means that you need to differentiate a standard equation for the volume of a sphere with respect to time. Do you know how to do that?
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  3. #3
    Gul
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    Hello. This is a related rates problem.

    You are given the following:


    DV/dt = 40cm^3, when r = 10cm

    You want: dr/dt:


    Dv/dt = DV/dr . dr/dt

    -> 40 = 4. pi. r^2 . (dr/dt)

    -> 40 = 4 . pi . (100) . (dr/dt)

    -> dr/dt = 40 / ((400) . (pi))

    -> dr/dt = 1 / ((10) . (pi) )

    -> dr/dt = 0.031830988

    To test:

    DV/dt = DV/dr . (dr/dt)

    -> DV/dt = 400 pi . (0.031830988)
    -DV/dt = 40


    Therefore dr/dt does, in fact = 0.031830988



    Best of luck.
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