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Math Help - related rates of change

  1. #1
    Newbie
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    Oct 2009
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    related rates of change

    Hi

    I am new to this forum so I hope i post this question at the right level.

    A spherical vacuole (something in a cell) is decreasing in size so that its volume decreases at a constant rate of 100 cubic micrometres per seconds. At what rate is the radius decreasing at the instant when the radius of 5 micrometre.

    This question needs the chain rule and is based on related rates of change.

    So, the [change in the radius wrt time] is equal to the [change in the radius wrt to the volume] * [change in the volume wrt time]

    dr /dt = dr/dv * dv/dt

    The change in volume wrt time is constant at 100, so can substitute dv/dt for 100 (or -100 because its a decrease?)

    dr /dt = dr/dv * 100

    Now to find how the radiu changes wrt volume

    v = 4/3 * pi * r^3

    r = (3/4 * 1/pi * v) ^1/3

    when the radius is 5, the volume is 523.6

    to find dr/dv, use teh chain rule

    r = u^1/3 where u = (3/4 * 1/pi * v)

    dr/du = 1/3u^-2/3
    du/dv = (3/4 * 1/pi )
    = 1/3 (3/4 * 1/pi * v)^-2/3 * (3/4 * 1/pi )
    = 1/4 * pi * (3/4 * 1/pi * v)^-2/3

    so that's how r changes wrt to the volume. I'm sure that is wrong!

    If i then substitute the volume for then the radius is 5

    = 1/4 * pi * (3/4 * 1/pi * 523.6 )^-2/3

    then multiply * 100 i should have the answer.

    But the answer is 1/pi!!!

    I have no idea how they get that. All of the previous examples in the book i am using have been based on the approach i've just used.

    many thanks for any help
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  2. #2
    MHF Contributor
    Joined
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    Welcome to MHF!

    You're making this way harder than it needs to be.

    V=\frac{4}{3} \pi r^3

    \frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}

    You are given dV/dt and r, so use that info to solve for dr/dt.
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