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