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Problem: Suppose that a particle following the given pathc(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.

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# Position of a Particle

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