Solve Brachistrone Problem: Find Path for Least Time Taken

[pardesi]

i have this doubt in teh famous brachistrone problem
The Problem
we have to find teh path $$y(x)$$ connecting two fixed points so that a body sliding along it under the influence of gravity only from rest should take the least possible?

The Proof

consider any length $$ds$$ along the path the time taken to cover it is $$\frac {ds}{v}$$ where $$v$$ is teh speed of teh body at that point then we have teh toatal tiem taken to coevr teh entire distance is $$T = \int_{1}^{2} \frac {ds}{v}$$
now we have by conservation of energy $$v = \sqrt {2gy}$$
thus $$T = \int_{1}^{2}\frac {ds}{\sqrt {2gy}}$$
clearly $$y$$ is a function of $$s$$ also we have by the calculus of variations a standard resukt that if
$$I = \int_{x_{1}}^{x_{2}} f(\dot{y},y,x)dx$$ is an extremum then we must have $$\frac {d (\frac {\partial f}{\partial \dot{y}})}{dx} = \frac {\partial f}{\partial y}$$
where $$y \equiv y(x)$$
here similarly we have $$f$$ as $$\frac {1}{\sqrt {2gy}}$$ and $$s$$ as $$y$$ and if we proceed so we get absurd results instead
if we put $$ds = \sqrt {1 + \dot{y}^{2}}dx$$ and then apply the result we get the right answer
my doubt is why does the result fail to hold in teh first case?

 

[cesiumfrog]

..teh speed of teh body at that point then we have teh toatal tiem taken to coevr teh..

Have you tried using less coffee?

 

[pardesi]

yeah sry for that i was at a library so had to hurrily type that
the corrected version "..the speed of the body at that point then we have the total time taken to cover the.."

 

[K.J.Healey]

s would be x in that case, not y.

 

[Nick R]

I have been trying to learn this sort of thing on my own lately - please say if I am making mistakes or have misconceptions.

The absurd result you get in the first case is y(s) = 0 right?

First time through I get

$$\frac {\partial{F}}{\partial{y}} = -\frac {{y}^{-3/2}}{2 \sqrt {2g}}$$

$$\frac {d}{ds} \frac {\partial{F}}{\partial{\dot{y}}} = \frac {d(0)}{ds} = 0$$

since $$\dot {y}$$ does not appear in $$F; F = \frac {1}{\sqrt {2gy}}$$

So it would appear that $$-\frac {{y}^{-3/2}}{2 \sqrt {2g}} = 0$$

Giving $$y = 0$$

Consider that F is simply $$F = \frac {1}{v}$$

And that $$\frac {1}{v} = \frac {dt}{ds} = \frac {dy}{ds}\frac{dt}{dy} = {\dot{y}} \frac {dt}{dy}$$

So we can write F in terms of $$\dot{y}$$

then $$\frac {\partial{F}}{\partial{\dot{y}}} = \frac{dt}{dy}$$

Clearly, the y-component velocity $$\frac{dy}{dt}$$ of the particle corresponds to a point along the curve it is falling.

Said differently, $$\frac{dy}{dt} = G(s)$$
so $$\frac {\partial{F}}{\partial{\dot{y}}} = \frac {1}{G(s)}$$

I am having difficulty thinking of what exactly $$G(s)$$ would be but it seems intuitive that

$$\frac {d}{ds} \frac {\partial{F}}{\partial{\dot{y}}} \neq 0$$

so

$$-\frac {{y}^{-3/2}}{2 \sqrt {2g}} \neq 0$$

and ultimately

$$y \neq 0$$

It seems that introduction of x and it's relation to y eliminates all the convolution here.

Is this correct?

 

[dy-e]

hi

the relation between ds, dx and dy :
$$ds = \sqrt{dy^2 + dx^2}$$
which you wrote in your first post is needed in deriving brachistrome because we need an independent variable to calculate. And, with that in mind, variable s is not a good one because it depends on x and y (which one is dependent of another).
Moreover, ds is an elementary distance which particle traveled, you can't say that $$ds = dy$$ because there's a motion in x-axis also!
In the Euler's type of equation
$$I = \int_{x_{1}}^{x_{2}} f(\dot{y},y,x)dx$$

$$\dot{y}(x), y(x)$$ must be a function of only x (and s is a function of x and y).
Since the variable s doesn't fit the Euler-Lagrange equation, you cannot assume that the solutions will fit to the connected condition:
$$\frac {d (\frac {\partial f}{\partial \dot{y}})}{dx} = \frac {\partial f}{\partial y}$$


I don't fully understand the meaning of function G(s) in your derivation but i think that the problem with it was much earlier.

 

[Nick R]

Let me try to be more thorough, so there is no confusion as to what I am doing.

We are minimizing an integral of the form

$$I = \int_{0}^{l} f(\dot{y}(s),y(s),s)ds$$

by calculus of variations,

$$\frac {d (\frac {\partial F}{\partial \dot{y}})}{ds} = \frac {\partial F}{\partial y}$$

Is true iff y(s) is such that it minimizes or maximizes $$I$$

Specifically we are minimizing

$$T = \int_{0}^{l} \frac {ds}{v}$$

which is the time taken by the particle in falling down the "ramp"

So $$F$$ in the context of the euler-lagrange equation is $$F = \frac{1}{v}$$

No forces besides gravity are acting on the particle, therefore

$$v = \sqrt {2gy}$$

it follows that, $$\frac {\partial{F}}{\partial{y}} = -\frac {{y}^{-3/2}}{2 \sqrt {2g}}$$

which is one side of the euler-lagrange equation.

In solving for the other side we must see that $$F = \frac {1}{v} = \frac {dt}{ds} = \frac {dy}{ds}\frac{dt}{dy} = {\dot{y}} \frac {dt}{dy}$$

so $$\frac {\partial{F}}{\partial{\dot{y}}} = \frac{dt}{dy}$$

well, $$\frac{dt}{dy}$$ is simply $$\frac{1}{V_y}$$

and $$V_y = \frac{ds}{dt}\frac{dy}{ds} = V \frac {d(y)}{ds} = \frac {1}{\sqrt {2gy}}\frac {d(y)}{ds}$$

Then the other side of the euler-lagrange equation is $$\frac {d(\frac {1}{\sqrt {2gy}}\frac {d(y)}{ds})}{ds}$$

And the solution to the brachistrone problem is

$$-\frac {{y}^{-3/2}}{2 \sqrt {2g}} = \frac {d(\frac {1}{\sqrt {2gy}}\frac {d(y)}{ds})}{ds}$$

The introduction of x is not needed to solve the problem (although it seems to be necessary to get a nice looking result...)

 

[Nick R]

I just spotted a mistake I made, this site doesn't appear to let me edit my posts so I'm forced to post another message...

The solution should be

$$-\frac {{y}^{-3/2}}{2 \sqrt {2g}} = \frac {d(\sqrt {2gy} \frac {1}{\frac {d(y)}{ds}})}{ds}$$

 

[dy-e]

hmm. Let's assume that's ok ;). But even if so, as i calculated, it seems that cycloid isn't the solution of this equation. I don't have time to write it now but have you checked it also? I checked the parametric formulas:
$$x = r(t - sin(t)), y = r(1- cos(t))$$