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Math Help - Optimizing Distance

  1. #1
    Junior Member
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    Oct 2005
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    Optimizing Distance

    Hi,
    Here is my question:
    Find the point P on the parabola y=(x^2) closest to the point (3,0).
    I solved it... but I am not sure what my domain restrictions should be...
    Thanks,

    D = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

    Let (x1, y1) = (3, 0) and (x2, y2) = (x, y). Then

    D = sqrt( (x - 3)^2 + (y - 0)^2 )
    D = sqrt( (x - 3)^2 + y^2 )

    *y = x^2

    D = sqrt( (x - 3)^2 + (x^2)^2 )
    D = sqrt( (x - 3)^2 + x^4 )

    We want to minimize this by making dD/dx = 0.
    We *could* differentiate this directly, but when we have to deal with derivatives with roots followed by the chain rule. I will instead square both sides, and then differentiate implicitly.

    D^2 = (x - 3)^2 + x^4

    Differentiate implicitly with respect to x,

    2D (dD/dx) = 2(x - 3) + 4x^3

    Make dD/dx = 0, to get

    0 = 2(x - 3) + 4x^3

    Solve for x.

    0 = 2x - 6 + 4x^3
    0 = 4x^3 + 2x - 6
    0 = 2x^3 + x - 3

    x=1... sub into original -> y=1. Therefore (1,1)
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  2. #2
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    Apr 2009
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    R = \sqrt{(x - 3)^2 + y^2} = \sqrt{(x - 3)^2 + (x^2)^2} = \sqrt{x^4 + x^2 - 6x + 9}

    R' = \frac{4x^3 + 2x - 6}{2\sqrt{x^4 + x^2 - 6x + 9}}

    4x^3 + 2x - 6 = 2x^3 + x - 3 = 0

    whoop, thunderstorm, gotta check out. =p I know I didn't help, but hey, at least you know it wasn't any more effort not to do it implicitly.

    Edit: Well, that brushed off quickly. Anyway, all your asking is what your "domain" restrictions should be? You already established a relationship between x and y when you substituted for x^2 and unlike a lot of optimization problems, this one doesn't require to be answers along a certain interval.
    Last edited by derfleurer; April 20th 2009 at 04:48 PM.
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  3. #3
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    Alright, so no specific domain... that makes it easier
    Thanks,
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