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If I[y]=\int_{x_0}^{x_1}F(x,y,y') \mathrm dx Where F=y^{-\frac{1}{2}}(1+(y')^2)^\frac{1}{2} Then I have shown the Euler-Lagrange equation implies that y(1+(y')^2)=2a For some $a\geq0$. If originally $x=y=0$ can anyone help me show that the parametric solution to this problem is the following? $x=a(t-\operatorname{sin}t)$ $y=a(1-\operatorname{cos}t)$

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    Thank you, I was being an idiot there!2012-01-18

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$x=a(t-\operatorname{sin}t)$ $y=a(1-\operatorname{cos}t)$ Implies $\dot{x}=a(1-\operatorname{cos}t)$ $\dot{y}=a\operatorname{sin}t$ Then y'=\frac{\operatorname{sin}t}{1-\operatorname{cos}t} Therefore y(1+(y')^2)=\frac{a(1-2\operatorname{cos}t+1)}{1-\operatorname{cos}t}=2a Therefore it parametrically describes the solution.