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Math Help - Rate of Change Problem

  1. #1
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    Rate of Change Problem

    Hi all, first time here. Not sure whether this is the right category...

    I am working with water levels and a truncated cone.
    I have been told that V = (pi/3)(((x+2)^3)/4)-2000).
    I am then told to find dV/dx, which = (pi(x^2+40x+400)/4).

    This is where I am stuck, because it then asks "Hence find the rate of change of height (x) in terms of x.
    Would I use Chain Rule, or is it simply dV/dx inverse?
    I'm not sure at all.

    Thanks for the help.

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  2. #2
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    Sorry, but I disagree with your derivative. If

     V = \frac{\pi}{3}(\frac{(x+2)^3}{4} - 2000) , then

     \frac{dV}{dx} = \pi \frac{x^2 + 4x + 4}{4} /
    Last edited by Math Major; July 30th 2010 at 09:38 PM.
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  3. #3
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    Whoops, meant to write (x+20)^3, not (x+2)^3.
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  4. #4
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    Quote Originally Posted by Etherlite View Post
    Hi all, first time here. Not sure whether this is the right category...

    I am working with water levels and a truncated cone.
    I have been told that V = (pi/3)(((x+20)^3)/4)-2000).
    I am then told to find dV/dx, which = (pi(x^2+40x+400)/4).

    This is where I am stuck, because it then asks "Hence find the rate of change of height (x) in terms of x.
    Would I use Chain Rule, or is it simply dV/dx inverse?
    I'm not sure at all.
    any information given about how the volume is changing w/r to time?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    any information given about how the volume is changing w/r to time?
    Don't worry all, I realised on a previous page that they gave me the rate at which water is dripping into the bucket.
    Apply chain rule, and the answer's there.

    Thanks all.
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