Minimizing integral functions
In order to find a function that extremizes the value of integral functional: we can use the Euler equation: Second Euler equation (or Beltrami’s identity): To find a function that extremizes where i.e. the function does not explicitly depend on x.
Geodesics (shortest path between two points)
The length of the path is given by For a planar surface (cartesian coordinates), For a spherical surface (spherical coordinates), For a cylindrical surface (cylindrical coordinates)
Brachistochrone problem (shortest time between two points)
Finding the path which minimizes the time taken between two points Solving using the second Euler equation, The solution to this second order differential equation is the cycloid function
x = \frac{k^2}{2} (\theta - \sin \theta) + c \\ y = \frac{k^2}{2} (1 - \cos \theta) \end{aligned}$$ Substituting this into the equation for time, $$\begin{aligned} t = \frac{1}{\sqrt{ 2g }} \int_{0}^{\theta_{B}} \frac{\sqrt{ 1 + \cot^2 \frac{\theta}{2} }}{\frac{k^2}{2}(1-\cos \theta)} \frac{k^2}{2}(1-\cos \theta) \ d\theta \\ \boxed{t= \frac{k}{\sqrt{ 2g }} \theta_{B}} \rightarrow \boxed{\theta_{B} = \frac{\sqrt{ 2g }}{k} t = wt} \end{aligned}$$ where $k = \frac{1}{2gc^2}$ ### Tautochrone Problem Finding the path where a particle starting from rest takes the same amount to time to travel no matter where it starts $$t = \int_{A}^B \frac{\sqrt{ 1 + y'^2 }}{\sqrt{ 2g(y-y_{A}) }} \ dx$$ $$\begin{aligned} t = \frac{1}{\sqrt{ 2g }} \int_{0}^{\theta_{B}} \frac{\sqrt{ 1 + \cot^2 \frac{\theta}{2} }}{(1-\cos \theta - 1 + \cos \theta_{A})} (1-\cos \theta) \ d\theta \\ \boxed{t= \frac{\pi k}{\sqrt{2g }}} \end{aligned}$$