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Shortest distance?

  1. Nov 27, 2007 #1
    shortest distance??

    1. The problem statement, all variables and given/known data

    find the shortest distance of y=x^2 from (4,0)

    2. Relevant equations



    3. The attempt at a solution

    if y=x^2, then
    y'=2x

    where x=4
    the slope is 8

    the solution is

    y=8(x-4)
    if that is the tangent line \, the normal line would be

    y=(1/8)(x-4)

    after that.. i'm stuck!!
     
  2. jcsd
  3. Nov 27, 2007 #2

    EnumaElish

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    One way is to write out the Euclidian distance from vector [x, x^2] to vector [4, 0], then minimize w/r/t x.
     
  4. Nov 27, 2007 #3

    HallsofIvy

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    EnumaElish, what rhey is trying to do is, I believe, more fundamental and a "nicer" method.

    The shortest distance from (4,0) to y= x2 will be along the line that is perpendicular to the graph. HOWEVER, rhey, x= 4 is for the point (4, 0), not any point on the graph! Let [itex](x_0, x_0^2)[/itex] be the point on y= x2 closest to (4, 0). The derivative there, and the slope of the tangent line, is 2x0 so the slope of the normal line is [itex]-1/(2x_0)[/itex]. The equation of the line through (4, 0), perpendicular to y= x2 at [itex](x_0,x_0^2)[/itex] is [itex]y= -1/(2x_0)(x- 4)[/itex]. If that is to pass through [itex](x_0, x_0^2)[/itex], you must have y= [itex]x_0^2= -1/(2x_0)(x_0- 4)[/itex]. Solve that equation for x0 and you are almost done!

    Hmm, that leads to a rather difficult equation for x0! Darn it, EnumaElish may be right!
     
    Last edited: Nov 27, 2007
  5. Nov 27, 2007 #4
    i really cant understand what are u trying to tell me.. honestly, i'm not that good in math! but i'm trying..
     
  6. Nov 27, 2007 #5
    1) Setup an equation for distance between point (0,4) and the function, [tex]f(x) = x^2[/tex]

    [tex]d = \sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2}[/tex]
    [tex]d = \sqrt{(x_2 - 4)^2 - (y_2)^2}[/tex]
    [tex]d = \sqrt{(x_2 - 4)^2 - (x_2)^4}[/tex]

    2) Take the derivative of d and set it equal to zero

    [tex]d' = 0 = \frac{4(x_2)^3 - 2(x_2 - 4)}{2\sqrt{(x_2 - 4)^2 - (x_2)^4}}[/tex]
    [tex]0 = 4(x_2)^3 - 2(x_2 - 4)[/tex]
    [tex]4(x_2)^3 = 2(x_2 - 4)[/tex]
    [tex]{x_2}^3 = x_2 - 4[/tex]

    As others have said, the problem is still difficult to solve from here...
     
    Last edited: Nov 27, 2007
  7. Nov 28, 2007 #6

    HallsofIvy

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    And, as EnumaElish said, it's far easier to minimize the square of the distance, not the distance itself! Oh, and it should be y- 4, not x- 4.
     
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