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Math Help - closest point on graph

  1. #1
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    closest point on graph

    Find the Point(s), P(x,y) on the graph of y=4-x^2 so that the distance between two points P and Q(0,2) is a minimum.

    i think what you do is

    d=squareroot[(x-0)+(y-2)]

    d^2=[(x-0)+(y-2)]

    then you plug in either x or y and differentiate?

    the answer is supposed to be sqareroot(3/2), -squareroot(3/2), and 5/2
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  2. #2
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    Quote Originally Posted by jeph View Post
    Find the Point(s), P(x,y) on the graph of y=4-x^2 so that the distance between two points P and Q(0,2) is a minimum.

    i think what you do is

    d=squareroot[(x-0)+(y-2)]

    d^2=[(x-0)+(y-2)]

    then you plug in either x or y and differentiate?

    the answer is supposed to be sqareroot(3/2), -squareroot(3/2), and 5/2
    Correct, you need to minimize,
    x^2+(y-2)^2
    Thus,
    x^2+(4-x^2-2)^2
    x^2+(2-x^2)^2
    x^2+4-4x^2+x^4
    x^4-3x^2+4
    Call this a function,
    f(x)=x^4-3x^2+4
    Thus, how do you minimize a differenciable function?
    You make its derivatives zero,
    f'(x)=4x^3-6x
    Thus,
    2x(2x^2-3)=0
    Thus,
    x=0
    2x^2-3\to x=\pm \sqrt{3/2}
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  3. #3
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    ok I understand that now, but how did he get the 5/2?
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  4. #4
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    Quote Originally Posted by jeph View Post
    ok I understand that now, but how did he get the 5/2?
    I do not know.
    I did not get it when I did the problem.
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  5. #5
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    Quote Originally Posted by jeph View Post
    ok I understand that now, but how did he get the 5/2?
    Hello, Jeph,

    you were looking for a point. Therefore you need two numbers: the x-value, which TPH has already calculated for you and the y-value.
    Plug in the x-value into the equation of your function:

    y=4- \left( \sqrt{\frac{3}{2}} \right)^2 = \frac{5}{2}

    EB
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  6. #6
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    oh! thanks =D
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