Brachistochrone
Introduction
To find the shape of the curve which the time is shortest possible. . .
Theory
The time from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_a} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_b} is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t = \int_{P_a}^{P_b} \frac 1 v ds } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ds= \sqrt{1+y'{^2}}dx} is the Pythagorean distance measure and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} is determined from the the law of conservation of energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac12 mv^2 = mgy } . giving Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = \sqrt{2gy}} . Plugging these in, we get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t = \int_{P_a}^{P_b} \sqrt{\frac{1+y'^{2}}{2gy}}dx = \int_{P_a}^{P_b} f dx} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f=f(y,y')} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial f}{\partial y} - \frac{d}{d x}\frac{\partial f}{\partial y'} = 0} is satisfied.
No Friction
We get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial f}{\partial y'} = \frac{\partial }{\partial y'} \sqrt{\frac{1+y'^{2}}{2gy}} = \frac1{\sqrt{2gy}} \frac{\partial }{\partial y'} \sqrt{1+y'^{2}} = \frac{2y'}{\sqrt{2gy}} \frac1 {\sqrt{1+y'^{2}}} } . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} does not depend on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} , we may use the simplified E--L formula Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f-y' \frac{\partial f}{\partial y'}= \text{Constant}} .
Thus, we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f - y' \frac{\partial f}{\partial y'} = f' - \frac{2y'{^2}}{\sqrt{2gy}} \frac1 {\sqrt{1+y'^{2}}} = }
Friction
Rolling Ball
References
https://mathworld.wolfram.com/BrachistochroneProblem.html
https://physicscourses.colorado.edu/phys3210/phys3210_sp20/lecture/lec04-lagrangian-mechanics/
