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].(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Particle undergoes a displacement

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