Let this image be in the x,y(x) plane. I am trying to compute the minimum time taken to get from A to B. I have determined that, using Euler-Lagrange methods,$$t=\sqrt{\frac{R}{g}}\int_{x_A}^{x_B}\frac{\sqrt{1+y'^2}}{\sqrt{R^2-x^2-y^2}}dx,$$ where $R$ is the radius of the circle and $g$ is acceleration due to gravity. The Euler-Lagrange is given by $$\frac{\partial F}{\partial x}-\frac{d}{dy}\Big(\frac{\partial F}{\partial x'}\Big)=0.$$ I have computed that $$\frac{\partial F}{\partial x}=\frac{x\sqrt{1+y'^2}}{(R^2-y^2-x^2)^{3/2}},$$ but I am struggling to see how to compute the rest. Is $\frac{\partial F}{\partial x'}=0$? This would make sense, except I know that the solution is as follows: $$x(\theta)=R\Big[(1-b)\cos{\theta}+b\cos{\frac{1-b}{b}\theta}\Big],$$ $$y(\theta)=R\Big[(1-b)\sin{\theta}-b\sin{\frac{1-b}{b}\theta} \Big],$$ with $b \in [0,1]$. I am struggling to see how this would be obtained. The book says that the calculation is tedious, but can someone please show me how to do this?