Calculus of variation help

by matpo39
Tags: calculus, variation
 P: 120 use chain rule... $$dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \phi}d\phi+\frac{\partial x}{\partial \theta}d\theta$$ same with $dy$
 P: 120 Calculus of variation help Ok.. first of all, you are in 3 dimentions so the lenght of a curve is given by $$L=\int_{t_{1}}^{t_{2}}{\sqrt{{dx}^2+{dy}^2+{dz}^2}}$$ now, given the spherical coordinates $$x=r\cos\phi\sin\theta$$ $$y=r\sin\phi\sin\theta$$ $$z=r\cos\theta$$ you have that $${dx}^2+{dy}^2+{dz}^2={dr}^2+r^2\sin^2\theta{d\phi}^2+r^2{d\theta}^2$$ but you are over a sphere right? so $dr=0$ and $r=R$ using all this we have $$L=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{\sin^2\theta{d\phi}^2+{d\theta} ^2}}=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{\sin^2\theta{\left(\frac{d\p hi}{d\theta}\right)}^2{d\theta}^2+{d\theta}^2}}$$ so $$L=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{1+\sin^2\theta{\left(\frac{d\ph i}{d\theta}\right)}^2}d\theta}$$