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Math Help - small changes in a function

  1. #1
    Gul
    Gul is offline
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    small changes in a function

    Volume of water in a dam is given by:

    V(h) = 750h^3 - 600h^2, where h is the depth of water at the dam wall

    If on a particular day the water drops by 0.08m from 22m, what will be the difference in volume. Okay, so apart from actually plugging and chugging into the equation, then subtracting, i can use the small changes formula. But I'd like to know why it works.

    dV/dh = DV/Dh

    dV = DV/Dh * dh

    dV = 2250(22)^2 - 600(22) * 0.08
    dV = 85,008

    Which is quite close to:
    ((750 * (22^3)) - (600 * (22^2))) - ((750 * (21.92^3)) - (600 * (21.92^2))) = 84 695.424

    How do I derive the:
    dV/dh = DV/Dh

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Gul View Post
    Volume of water in a dam is given by:

    V(h) = 750h^3 - 600h^2, where h is the depth of water at the dam wall

    If on a particular day the water drops by 0.08m from 22m, what will be the difference in volume. Okay, so apart from actually plugging and chugging into the equation, then subtracting, i can use the small changes formula. But I'd like to know why it works.

    dV/dh = DV/Dh

    dV = DV/Dh * dh

    dV = 2250(22)^2 - 600(22) * 0.08
    dV = 85,008

    Which is quite close to:
    ((750 * (22^3)) - (600 * (22^2))) - ((750 * (21.92^3)) - (600 * (21.92^2))) = 84 695.424

    How do I derive the:
    dV/dh = DV/Dh

    Thanks.
    Linear approximation of a function at a point (first two terms of Taylor series):

    V(h+\delta h)\approx V(h)+ \delta h ~V'(h)

    That's all there is to it.

    CB
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