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Position of a Particle

  1. Jan 18, 2006 #1
    Can someone please check my work on this?

    Problem: Suppose that a particle following the given path c(t) flies off on a tangent at t = t_0. Compute the position of the particle at the given time t_1.
    Given: [itex]\vec{c}(t)=(e^t, e^{-t}, \cos{t}), t_0=1, t_1=2[/itex]

    Here's how I did it:
    The tangent to the position vector at any time is the velocity vector, so:
    [tex]\vec{v}(t)=(e^t, -e^{-t}, -\sin{t})[/tex]
    [tex]\vec{v}(t_0)=(e, -\frac{1}{e}, -\sin{1}) = \vec{c}_0 (t_0)[/tex]
    where I define [itex]\vec{c}_0 (t)[/itex] as the position along the tangent at time t.
    Then, the position at t_1 is:
    [tex]\vec{c}(t_0)+\vec{c}_0 (t_0) = (2e,0, \cos{1}-\sin{1})[/tex]

    Does this result make sense?

    Edit: Actually, I think the proper final equation should be [itex]\vec{c}(t_0)+(t_1-t_0)\vec{c}_0 (t_0) = (2e,0, \cos{1}-\sin{1})[/itex] but in this case the answer comes out the same.
    Last edited: Jan 18, 2006
  2. jcsd
  3. Jan 18, 2006 #2


    Staff: Mentor

    I got the same answer, but your notation (using c and c0 instead of c and v) confused me for a moment.

  4. Jan 19, 2006 #3
    Thanks, Dale. Sorry about the notation, it made more sense in my head :tongue2:
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