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T-7
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Homework Statement
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).
Homework 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]
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,