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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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If you have the solution, you can just plug it in and check that it is true. Note that $y'= \frac{\dot y}{\dot x}$ –  Fabian Jan 18 '12 at 16:46
    
Thank you, I was being an idiot there! –  Freeman Jan 18 '12 at 21:40

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up vote 1 down vote accepted

$$x=a(t-\operatorname{sin}t)$$ $$y=a(1-\operatorname{cos}t)$$ 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.

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