Update 03_euler_lagrange_equation.tex

[Kaiwen-S]

There is an error in the final result for the geodesics on a sphere. We should get $$\pm\dfrac{\sqrt{1-k^2}}{k}\cos(\phi-\phi_0)=\cot\theta$$ by the integral; the file right now is missing the square root sign on the numerator.

For reference, I will attach my calculation below.

We have $$\phi=\pm\int\dfrac{kd\theta}{\sin\theta\sqrt{\sin^2\theta-k^2}}=\pm\int\dfrac{k(-\csc^2\theta d\theta)}{-\sin\theta\csc^2\theta\sqrt{\sin^2\theta-k^2}}=\pm\int\dfrac{k(-\csc^2\theta d\theta)}{-\sqrt{1-k^2\csc^2\theta}}=\pm\int\dfrac{k(-\csc^2\theta d\theta)}{-\sqrt{1-k^2-k^2\cot^2\theta}}$$ Substitute $u=\cot\theta$, $du=-\csc^2\theta d\theta$ to obtain $$\phi=\pm\int\dfrac{-k du}{\sqrt{1-k^2-k^2u}}=\pm\int\dfrac{-du}{\sqrt{\frac{1-k^2}{k^2}-u^2}}=\arccos\left(\dfrac{k}{\sqrt{1-k^2}}u\right)+\phi_0$$ Rearranging, $$\pm\dfrac{\sqrt{1-k^2}}{k}\cos(\phi-\phi_0)=\cot\theta$$

[zeramorphic]

Thank you for this fix, this looks great!