Brachistochrone: Difference between revisions

From wikiluntti
Line 32: Line 32:
\frac{2y'}{\sqrt{2gy}}
\frac{2y'}{\sqrt{2gy}}
\frac1 {\sqrt{1+y'^{2}}}
\frac1 {\sqrt{1+y'^{2}}}
</math>. Because the previous statement do not contain explicit <math>x</math>, the derivative is zero, thus
</math>.  
<math>
\frac{2y'}{\sqrt{2gy}}
\frac1 {\sqrt{1+y'^{2}}}
= C
</math>
giving
<math>
\frac{2y'}{\sqrt{2gy}}
\frac1 {\sqrt{1+y'^{2}}}
= C
</math>
 
Since <math>f</math> does not depend on <math>x</math>, we may use the simplified E--L formula <math>f-y' \frac{\partial f}{\partial y'}= \text{Constant}</math>.  
Since <math>f</math> does not depend on <math>x</math>, we may use the simplified E--L formula <math>f-y' \frac{\partial f}{\partial y'}= \text{Constant}</math>.  



Revision as of 18:56, 16 February 2021

Introduction

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

Theory

The time from to is where is the Pythagorean distance measure and is determined from the the law of conservation of energy . giving . 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}} .

Because 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 x}=0} ,

Friction

Rolling Ball

References

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

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