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Brachistrone problem

  1. Aug 7, 2008 #1
    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?
     
  2. jcsd
  3. Aug 7, 2008 #2
    Have you tried using less coffee? :wink:
     
  4. Aug 7, 2008 #3
    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.."
     
  5. Aug 7, 2008 #4
    s would be x in that case, not y.
     
  6. Aug 8, 2008 #5
    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

    [tex]\frac {\partial{F}}{\partial{y}} = -\frac {{y}^{-3/2}}{2 \sqrt {2g}}[/tex]

    [tex]\frac {d}{ds} \frac {\partial{F}}{\partial{\dot{y}}} = \frac {d(0)}{ds} = 0[/tex]

    since [tex]\dot {y}[/tex] does not appear in [tex]F; F = \frac {1}{\sqrt {2gy}}[/tex]

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

    Giving [tex] y = 0 [/tex]

    Consider that F is simply [tex]F = \frac {1}{v}[/tex]

    And that [tex] \frac {1}{v} = \frac {dt}{ds} = \frac {dy}{ds}\frac{dt}{dy} = {\dot{y}} \frac {dt}{dy} [/tex]

    So we can write F in terms of [tex] \dot{y} [/tex]

    then [tex]\frac {\partial{F}}{\partial{\dot{y}}} = \frac{dt}{dy}[/tex]

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

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

    I am having difficulty thinking of what exactly [tex]G(s)[/tex] would be but it seems intuitive that

    [tex]\frac {d}{ds} \frac {\partial{F}}{\partial{\dot{y}}} \neq 0[/tex]

    so

    [tex]-\frac {{y}^{-3/2}}{2 \sqrt {2g}} \neq 0[/tex]

    and ultimately

    [tex] y \neq 0 [/tex]

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

    Is this correct?
     
    Last edited: Aug 8, 2008
  7. Aug 9, 2008 #6
    hi

    the relation between ds, dx and dy :
    [tex]ds = \sqrt{dy^2 + dx^2}[/tex]
    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 [tex]ds = dy[/tex] because there's a motion in x axis also!
    In the Euler's type of equation
    [tex]
    I = \int_{x_{1}}^{x_{2}} f(\dot{y},y,x)dx
    [/tex]

    [tex]\dot{y}(x), y(x)[/tex] 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:
    [tex]
    \frac {d (\frac {\partial f}{\partial \dot{y}})}{dx} = \frac {\partial f}{\partial y}
    [/tex]


    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.
     
  8. Aug 9, 2008 #7
    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

    [tex]I = \int_{0}^{l} f(\dot{y}(s),y(s),s)ds[/tex]

    by calculus of variations,

    [tex]\frac {d (\frac {\partial F}{\partial \dot{y}})}{ds} = \frac {\partial F}{\partial y}[/tex]

    Is true iff y(s) is such that it minimizes or maximizes [tex]I[/tex]

    Specifically we are minimizing

    [tex]T = \int_{0}^{l} \frac {ds}{v}[/tex]

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

    So [tex]F[/tex] in the context of the euler-lagrange equation is [tex]F = \frac{1}{v}[/tex]

    No forces besides gravity are acting on the particle, therefore

    [tex]v = \sqrt {2gy}[/tex]

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

    which is one side of the euler-lagrange equation.

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

    so [tex]\frac {\partial{F}}{\partial{\dot{y}}} = \frac{dt}{dy}[/tex]

    well, [tex]\frac{dt}{dy}[/tex] is simply [tex]\frac{1}{V_y}[/tex]

    and [tex] V_y = \frac{ds}{dt}\frac{dy}{ds} = V \frac {d(y)}{ds} = \frac {1}{\sqrt {2gy}}\frac {d(y)}{ds}[/tex]

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

    And the solution to the brachistrone problem is

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

    The introduction of x is not needed to solve the problem (although it seems to be necessary to get a nice looking result...)
     
    Last edited: Aug 9, 2008
  9. Aug 10, 2008 #8
    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

    [tex]-\frac {{y}^{-3/2}}{2 \sqrt {2g}} = \frac {d(\sqrt {2gy} \frac {1}{\frac {d(y)}{ds}})}{ds} [/tex]
     
  10. Aug 10, 2008 #9
    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:
    [tex] x = r(t - sin(t)), y = r(1- cos(t))
    [/tex]
     
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