1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Particle undergoes a displacement

  1. Oct 31, 2006 #1
    Lets say we have [tex] \vec{r} = x\vec{i} + y\vec{j} + z\vec{k} [/tex]. If the particle undergoes a displacement [tex] \Delta \vec{r} [/tex] in time [tex] \Delta t [/tex] then we know that [tex] \Delta r \doteq \frac{d\vec{r}}{dt} \Delta t = \vec{v}\Delta t [/tex].

    How is [tex] \vec{v} \Delta t [/tex] tangent to the particles trajectory, when we take [tex] \Delta t \rightarrow 0 [/tex]? Wouldn't the expression become 0? I can see how [tex] \Delta \vec{r} \rightarrow 0 [/tex] as [tex] \Delta t \rightarrow 0 [/tex].

    Also lets say we have the following:

    [tex] \vect{r} = A(e^{\alpha t}\vec{i} + e^{-\alpha t}\vec{j}) [/tex]

    I know that [tex] \vec{v} = A(\alpha e^{\alpha t}\vec{i} + e^{-\alpha t}\vec{j}) [/tex]

    As [tex] t\rightarrow \infty, e^{\alpha t} \rightarrow \infty [/tex] and [tex] e^{-\alpha t} \rightarrow 0 [/tex]. From this how do we come to the conclusion that [tex] \vec{r} \rightarrow Ae^{\alpha t}\vec{i} [/tex]? I thought that [tex] \vec{r} \rightarrow \infty [/tex].

    Similarily, [tex] \vec{v} \rightarrow \alpha Ae^{\alpha t}\vec{i} [/tex] when it seems like it should approach [tex] \infty [/tex]. Perhaps it has to do something with the x-components?

    [tex] \alpha [/tex] is a constant.

    Thanks for your help
    Last edited: Oct 31, 2006
  2. jcsd
  3. Oct 31, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    The expression [tex]v\Delta t[/tex] would approach zero. It's the vector [tex]v=\lim_{t\to 0}\Delta r/Delta t[/tex] that is tangential to the particle's trajectory.

    I`m not sure what you want to do here. There's an error in the expression for v (not sure if that's a typo). Then (assuming alpha>0) you take the limit as t-> infinity (what for?).

    It's true that r-> infinity if t-> infinity, but If you want to approximate the trajectory for large t, you can drop the second term since it is vanishingly small in comparison with the first term.
  4. Oct 31, 2006 #3
    They wanted me to sketch the trajectory. But [tex] \vec{r} = Ae^{\alpha t }\vec{i}} + Ae^{-\alpha t}\vec{j} [/tex]?
    Then the second term approaches 0 and the first term approaches [tex] \infty [/tex]. So then doesnt [tex] \vec{r} \rightarrow \infty [/tex]?
    Last edited: Oct 31, 2006
  5. Oct 31, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

    Yes, but that doesn't help in sketching the trajectory. You want the asymptotic behaviour of r, not the limit as t-> infinity.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Particle undergoes a displacement