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Homework Help: Setting up a Lagrangian

  1. Mar 8, 2008 #1
    [SOLVED] Setting up a Lagrangian

    1. The problem statement, all variables and given/known data
    Two points A and B are located in a uniform gravitational field. A is higher than B,
    and B is not directly below A. A friction free track is to be built from A to B in the
    plane in which they are both located. Down this track, friction free toy cars will be
    rolled, starting from rest at A. The cars stay in contact with the track at all times. In
    this problem we will figure out the shape of the track that will get the cars from A to
    B in the minimum possible time. According to wikipedia this problem was answered
    incorrectly by Galileo in 1638 before being solved in 1696 by Newton, Johann and
    Jacob Bernoulli, Leibniz and l'Hopital.
    Setting up the problem, a point on the track has coordinates (x([tex]\tau[/tex] ); y([tex]\tau[/tex] )), where x
    is the horizontal component of the displacement of the car from A, y is the vertical
    component of the downwards displacement from A, and [tex]\tau[/tex] is the time.

    Using ordinary arguments from classical mechanics, show that when
    the car has dropped through height y, the magnitude of its velocity is [tex]\sqrt{2gy}[/tex], where g is the acceleration due to gravity.

    Show that
    [tex]L=\frac{(\frac{dx}{d\tau})^2 + (\frac{dy}{d\tau})^2}{y^2}[/tex]
    is a suitable Lagrangian to minimize the total travel time of the car.

    2. Relevant equations

    3. The attempt at a solution
    I've solved the first part, equating KE and PE and rearranging to find v, exactly as the question asks.

    My main problem is with the second part, solving the Lagrangian.

    As I understand it, I need to find an expression for the time taken to get from A to B, and then differentiate this with respect to [tex]\tau[/tex].

    So from this starting point I've said that the time taken (squared) is equal to the distance moved divided by the velocity (obviously), which has given me
    [tex]\frac{x^2 + y^2}{2gy}[/tex], using the value for the velocity worked out in the previous equation.

    After this I've differentiated everything with respect to [tex]\tau[/tex] leaving me with:
    [tex]\frac{(2x\frac{dx}{d\tau} + 2y\frac{dy}{d\tau})2gy - (x^2 + y^2)2g\frac{dy}{d\tau}}{4g^2y^2}[/tex]

    From here I've gotten stuck. I can't see how this will expand and rearrange to give the equation in the form above. I know the constants can be omitted (minimising with constants in is the same as minimising without the constants), but other than that I can't see it.

    I differentiated using the quotient and product rules, which I'm pretty sure I've gotten right. The only other thing I can think of is that the velocity I used isn't quite right, as it seems to only take into acount the velocity in one direction. Or is that a case of me overthinking things again? Any help would be appreciated.
    Last edited: Mar 8, 2008
  2. jcsd
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