- #1
tandoorichicken
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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.
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.
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