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Can we solve for x(t) and y(t) with respect to time?

  1. Jan 17, 2017 #1
    1. The problem statement, all variables and given/known data
    I would like to ask a question I was discussing the other day with a friend of mine. Suppose you have a point mass m, sliding on the friction-free curve [itex]y = e^{-x}[/itex] starting from position [itex]x(0) = 0[/itex] and [itex]y(0) = 1[/itex] with zero initial velocity. Can we find in closed-form the exact position [itex](x,y)[/itex] with respect to time?

    2. The attempt at a solution
    Going down the usual path, from conservation of energy I found that [itex]v = \sqrt{2g(1-y)}[/itex]. Assuming [itex]φ[/itex] is the angle between [itex]\vec{v}[/itex] and the downward vertical direction (the direction of negative ordinates), and using that [itex]\dot{x} = v \sin(φ)[/itex], [itex]\dot{y} = v \cos(φ)[/itex], [itex]φ = \tan^{-1}(e^{x})[/itex], I reached

    [tex]\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-e^{x}}{e^{2x}+1}}[/tex]
    [tex]\dot{y} = -y\sqrt{2g}\sqrt{\frac{1-y}{1+y^2}}[/tex]

    From these two, I need to solve only one. Choosing the first, separating [itex]dx[/itex] and [itex]dy[/itex] and integrating, yields

    [tex]\int_0^{x(t)}\sqrt{\frac{e^{2x} + 1}{e^{2x} - e^{x} }}dx = \sqrt{2g}t[/tex]

    However, plugging the integral in the LHS in Mathematica doesn't compute anything, most likely meaning that this integral is not solvable in closed form, let alone invertible in closed-form.

    Choosing now the second equation and doing the same yields:
    [tex]\int_1^{y(t)}\sqrt{\frac{1+y^2}{1 - y}}\frac{dy}{y} = - \sqrt{2g}t[/tex]

    Again, trying to compute the integral in the LHS, gives an assortment of complex elliptic integrals that are gain not invertible in some closed-form. Any thoughts about what can be done?
     
    Last edited: Jan 17, 2017
  2. jcsd
  3. Jan 17, 2017 #2

    BvU

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    You reached $$
    \dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-1}{e^{2x}+1}}$$
    but I reach something else ( from ##\ \ e^{2x}(1-e^{-x})\ \ ## under the square root in the numerator ).
     
  4. Jan 17, 2017 #3
    Thank you for replying BvU! You are absolutely correct ##\dot{x}## is not the one I wrote but
    $$\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-e^{x}}{e^{2x}+1}} = \sqrt{2g}\sqrt{\frac{1-e^{-x}}{1+e^{-2x}}}$$
    I also noted that ##\dot{y}## needs a minus in order for ##\dot{y} = -e^{-x}\dot{x} = -y \dot{x}## to hold. I will correct the initial post! However, after this correction, things get even more challenging, since the integral now is not analytically computable!
     
  5. Jan 17, 2017 #4

    BvU

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    :welcome:

    Did someone promise you there would be an elegant looking analytical solution ?
    If you want that, you might look at a cycloid instead of an exponential :rolleyes:
     
  6. Jan 17, 2017 #5
    Haha! :smile:None indeed BvU! As I said, it is just an inquiry I had with a friend of mine the other day. It surprised me that the only method I could find involved integration and then inversion of the result so I thought that I might be going down the wrong path. At best, there should be a way to acquire the solution in some integral form and then let numerical integration take over!

    Thank you for your warm welcoming message! :smile:
     
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