Position of a Particle

1. Jan 18, 2006

tandoorichicken

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: $\vec{c}(t)=(e^t, e^{-t}, \cos{t}), t_0=1, t_1=2$

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

Does this result make sense?

Edit: Actually, I think the proper final equation should be $\vec{c}(t_0)+(t_1-t_0)\vec{c}_0 (t_0) = (2e,0, \cos{1}-\sin{1})$ but in this case the answer comes out the same.

Last edited: Jan 18, 2006
2. Jan 18, 2006

Staff: Mentor

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

-Dale

3. Jan 19, 2006