Brachistochrone

Introduction

To find the shape of the curve which the time is shortest possible. . .

Theory

The time from $P_{a}$ to $P_{b}$ is $t=\int _{P_{a}}^{P_{b}}{\frac {1}{v}}ds$ where $ds={\sqrt {1+y'{^{2}}}}dx$ is the Pythagorean distance measure and $v$ is determined from the the law of conservation of energy ${\frac {1}{2}}mv^{2}=mgy$. giving $v={\sqrt {2gy}}$. Plugging these in, we get $t=\int _{P_{a}}^{P_{b}}{\sqrt {\frac {1+y'^{2}}{2gy}}}dx=\int _{P_{a}}^{P_{b}}fdx$, where $f$ is the function subject to variational consideration.

According to the Euler--Lagrange differential equation the stationary value is to be found, if E-L equation ${\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}=0$ is satisfied.

No Friction

We get ${\frac {\partial f}{\partial y'}}={\frac {\partial }{\partial y'}}{\sqrt {\frac {1+y'^{2}}{2gy}}}={\frac {1}{\sqrt {2gy}}}{\frac {\partial }{\partial y'}}{\sqrt {1+y'^{2}}}={\frac {2y'}{\sqrt {2gy}}}{\frac {1}{\sqrt {1+y'^{2}}}}$. Because the previous statement do not contain explicit $x$, the derivative is zero, thus ${\frac {2y'}{\sqrt {2gy}}}{\frac {1}{\sqrt {1+y'^{2}}}}=C$ giving ${\frac {2y'}{\sqrt {2gy}}}{\frac {1}{\sqrt {1+y'^{2}}}}=C$

Because ${\frac {\partial f}{\partial x}}=0$,

References

https://mathworld.wolfram.com/BrachistochroneProblem.html

https://physicscourses.colorado.edu/phys3210/phys3210_sp20/lecture/lec04-lagrangian-mechanics/