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Ds/dt and dr/dt

  1. Jul 3, 2014 #1
    This is one very basic question. But I just need to confirm if I understood it right.

    Suppose a particle moves along a curve and crosses Δs path in Δt time. Then we can say the velocity of the particle is [itex]\vec{v}[/itex] = ds/dt [itex]\hat{u}[/itex]
    Where [itex]\hat{u}[/itex] is tangent to the curve.

    Also if the same particle, as it crosses Δs, goes through a displacement Δ[itex]\vec{r}[/itex] in the same time interval Δt we say [itex]\vec{v}[/itex] = d[itex]\vec{r}[/itex]/dt

    Is the V's calculated above are same (ie equal)?

    I know the question is silly, but at present this forum is the only place for me to get help.
     
  2. jcsd
  3. Jul 3, 2014 #2

    verty

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    Homework Helper

    The are the same. I think I know what you are thinking, that perhaps the second one is slightly smaller because the path may be curved, but any difference between ##\Delta{s}## and ##|\Delta\vec{r}|## goes to 0 in the limit, and any angle between ##\Delta\vec{r}## and the path goes to 0 in the limit as well. So in the limit they are perfectly the same, ##d\vec{r} \over dt## is exactly the vector rate of change of position which is velocity.
     
  4. Jul 3, 2014 #3
    Thanks. :smile:
     
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