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The brachistochrone

  1. Mar 10, 2008 #1

    T-7

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

    I am after an expression for the time T it takes to slide down from height zero to -h of the brachistochrone. (Starts from x=y=0, slides along a distance -x, descends a height -h).

    2. Relevant equations

    I have deduced that

    [tex] T = \frac{1}{(2g)^{1/2}} \int _{0}^{-h} \frac{(x^{\prime}^{2}+1)^{1/2}}{(-y)^{1/2}} dy
    [/tex]

    Along the way, I have shown (using Euler-Lagrange Eqns) that

    [tex]
    \frac{x^{\prime}}{(-y)^{1/2}(1+x^{\prime 2})^{1/2}} = const = \frac{1}{\eta^{1/2}}
    [/tex]

    and

    [tex]
    x^{\prime} = - \frac{y^{1/2}}{(\eta-y)^{1/2}} =>
    x = - \int \frac{y^{1/2}}{(\eta-y)^{1/2}} dy
    [/tex]

    and, after a horrid bit of integration:

    [tex]
    x(\theta) = -\frac{\eta}{2}(\theta - sin\theta),
    y(\theta) = -\frac{\eta}{2}(1 - cos\theta)
    [/tex]

    3. The attempt at a solution

    The question suggests I combine the result

    [tex]
    \frac{x^{\prime}}{(-y)^{1/2}(1+x^{\prime 2})^{1/2}} = const = \frac{1}{\eta^{1/2}}
    [/tex]

    with [tex]\eta = h[/tex]

    with the equation I derived for T

    [tex] T = \frac{1}{(2g)^{1/2}} \int _{0}^{-h} \frac{(x^{\prime}^{2}+1)^{1/2}}{(-y)^{1/2}} dy
    [/tex]

    (I presume that I have got this equation for T correct?)

    On doing so, I simplified the integral to

    [tex] T = \frac{1}{(2gh)^{1/2}} \int _{0}^{-h} \frac{(x^{\prime}^{2}+1)}{x^{\prime}} dy
    [/tex]

    which, using [tex]x^{\prime} = - \frac{y^{1/2}}{(\eta-y)^{1/2}}[/tex] I boiled down to

    [tex] T = \frac{h}{(2gh)^{1/2}} \int _{0}^{-h} \frac{1}{y^{1/2}(h-y)^{1/2}} dy
    [/tex]

    I am at a loss to know how to solve this integral (assuming I've got it right so far, and need to solve it!).

    Could anyone suggest the next move to make? This is frustrating!

    Cheers,
     
  2. jcsd
  3. Mar 10, 2008 #2
    T-7

    I did not check your calculations, but assumed they were correct up to the end. The integral you have can be recast in the form of dx/sqrt(ax**2 + bx + c) by bringing the sqrt(y) into the right hand radical sign. Integrals of this form are listed in Gradshteyn and Ryzhik's "Table of Integrals and Series". All that you need to do is set the trinomial coefficients equal to those in your last equation with c=0.

    Jeff
     
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