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Variational mechanics

  1. Feb 10, 2006 #1
    for a geodesic right cylinder with radius R. Find the curve taht minimizes the distance between two points r1 and r2. where r = (R,phi, z) in cylindrical polar coordinates. Express your answer as z = z(phi)

    pardon the sloppy math
    not the most fun to get this type of question on a 50 minute test! Anyway,

    x = r cos t
    y = r sin t
    z = z

    [tex] \mbox{distance} = \int_{1}^{2} \sqrt{r^2 + \left( \frac{\partial z}{\partial \theta}\right)^2} [/tex]
    ist hat the distance between two points on a cylinder?
    Or would this distane be represented by something of a helix? I mean a cylinder could be considered as helix... right?
    Last edited: Feb 10, 2006
  2. jcsd
  3. Feb 10, 2006 #2


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    If your integration limits are angles [itex]\phi_1~and ~\phi_2[/itex] and you integrate w. r. to [itex]\phi[/itex], then you are good.
  4. Feb 20, 2006 #3
    ok so do i plug this into the euler lagrange equation?

    so [tex] f(z,\theta) = \sqrt{r^2 + \left( \frac{\partial z}{\partial \theta}\right)^2} [/tex]

    Euler Lagrange equation is
    [tex] \frac{\partial f}{\partial \theta} - \frac{d}{d\theta} \frac{\partial f}{\partial z} = 0 [/tex]
    is this the right way to go?
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