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Math Help - minimise/maximise distance between (0,4) and y=x^2/4

  1. #1
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    Smile minimise/maximise distance between (0,4) and y=x^2/4

    I am trying to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3.

    This is what i have so far

    y(distance)=√((x-0)^2+((x^2)/4)^2)
    =(x^2+(X^4)/16+16)^0.5
    then calculate y'
    then make y'=0
    then determine if max or min using 2nd derivative test

    The problem i am having is that the answers i get for x when y' = 0 are imaginary. i have computed this problem using matlab and get the same answers except that they are real and have isolated the problem to my simplification of y.

    matlab gives it as y=0.25(-16x^2+x^4+256)^0.5

    Could some one please explain how to simplify y=(x^2+(X^4)/16+16)^0.5 to y=0.25(-16x^2+x^4+256)^0.5 or suggest a better way to approach this problem

    Thanks in advance
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  2. #2
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    Quote Originally Posted by Poppy View Post
    I am trying to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3.

    This is what i have so far

    y(distance)=√((x-0)^2+((x^2)/4)^2)
    =(x^2+(X^4)/16+16)^0.5
    then calculate y'
    then make y'=0
    then determine if max or min using 2nd derivative test

    The problem i am having is that the answers i get for x when y' = 0 are imaginary. i have computed this problem using matlab and get the same answers except that they are real and have isolated the problem to my simplification of y.

    matlab gives it as y=0.25(-16x^2+x^4+256)^0.5

    Could some one please explain how to simplify y=(x^2+(X^4)/16+16)^0.5 to y=0.25(-16x^2+x^4+256)^0.5 or suggest a better way to approach this problem

    Thanks in advance
    Your original equation is wrong.

    Remember that the distance between two points (x_1, y_1) and (x_2, y_2) is given by

    d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    (this is easily proven by Pythagoras).


    In your case (x_1, y_1) = (0, 4) and (x_2, y_2) = \left(x, \frac{x^2}{4}\right).


    So the distance is

    d = \sqrt{(x - 0)^2 + \left(\frac{x^2}{4} - 4\right)^2}

     = \sqrt{x^2 + \left(\frac{x^2 - 16}{4}\right)^2}

     = \sqrt{x^2 + \frac{(x^2 - 16)^2}{16}}

     = \sqrt{\frac{16x^2 + (x^2 - 16)^2}{16}}

     = \frac{\sqrt{16x^2 + x^4 - 32x + 256}}{4}

     = \frac{\sqrt{x^4 + 16x^2 - 32x + 256}}{4}.


    Now you can find the minimum and maximum...
    Last edited by Prove It; December 9th 2009 at 03:16 PM.
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  3. #3
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    Thanks-just one error

    thanks for that it was very helpful- only one minor correction -when you expanded (x^2-16)^2 it should have been to x^4-32x+256 not x^2-32x+256 which makes the next equation ((16x^2^2+x^4-32x+256)/4)^1/2 not ((17x^2-32x+256)/4)^1/2

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  4. #4
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    Edited, thanks. It's well known that my arithmetic is shocking...

    However, to correct YOUR latest post, it actually becomes

    \frac{(x^4 + 16x^2 - 32x + 256)^{\frac{1}{2}}}{4}

    not

    \left(\frac{x^4 + 16x^2 - 32x + 256}{4}\right)^\frac{1}{2}.
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