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Math Help - Yet More Optimization

  1. #1
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    Yet More Optimization

    Find the coordinates of the point on the parabola y = x^2 which is the closest to the point (3,0).

    I'm just wondering if I did this correctly:
    a point on the curve would be (x,x^2)

    So, i found the distance between  (3,0) and  (x,x^2).
    d = \sqrt{(0-x^2)^2+(3-x)^2}
    d = \sqrt{x^4 + 9 - 6x + x^2}

    Then, I took the derivative to minimize the distance.
    d' = \frac{1}{2} (x^4 + 9 - 6x + x^2)^{-\frac{1}{2}} * (4x^3 - 6 + 2x)

    Set it equal to zero to find where the minimum occurs.
    \frac{4x^3 - 6 + 2x}{2 \sqrt{x^4 + 9 - 6x + x^2}} = 0
    4x^3 - 6 + 2x = 0
    x = 1

    Plug it back in to get the y-value
    y(1) = 1^2 = 1

    The point on  y = x^2 which is closest to the point (3,0) is (1,1).

    Is this correct?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jacobpm64 View Post
    Find the coordinates of the point on the parabola y = x^2 which is the closest to the point (3,0).

    I'm just wondering if I did this correctly:
    a point on the curve would be (x,x^2)

    So, i found the distance between  (3,0) and  (x,x^2).
    d = \sqrt{(0-x^2)^2+(3-x)^2}
    d = \sqrt{x^4 + 9 - 6x + x^2}

    Then, I took the derivative to minimize the distance.
    d' = \frac{1}{2} (x^4 + 9 - 6x + x^2)^{-\frac{1}{2}} * (4x^3 - 6 + 2x)


    Set it equal to zero to find where the minimum occurs.
    \frac{4x^3 - 6 + 2x}{2 \sqrt{x^4 + 9 - 6x + x^2}} = 0
    4x^3 - 6 + 2x = 0
    x = 1

    Plug it back in to get the y-value
    y(1) = 1^2 = 1

    The point on  y = x^2 which is closest to the point (3,0) is (1,1).

    Is this correct?
    I didn't check the numbers, but the format is correct.

    -Dan
    Attached Thumbnails Attached Thumbnails Yet More Optimization-parabola.jpg  
    Last edited by topsquark; November 15th 2006 at 04:26 PM. Reason: Made a prettier picture.
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    I didn't check the numbers, but the format is correct.

    -Dan
    You do realize that you can insert points into the graph through the graph program?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    You do realize that you can insert points into the graph through the graph program?
    Thank you. I figured there had to be a way, but I didn't want to bother looking it up while answering a question.

    -Dan
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