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Velocity Vector in Polar Coordinates (Kleppner p.30)

  1. Jun 6, 2013 #1
    In polar coordinates we have [itex] \vec{r} = r \hat{r} [/itex] [itex]\Rightarrow \vec{v} = \frac{d}{dt}({r \hat{r}}) = \dot{r}\hat{r} + r \frac{d \hat{r}}{dt} [/itex].

    In the book Introduction to Mechanics, K & K says the right term is the component of velocity directed radially outward. (Surely a typo, as the left term is the velocity component associated with the direction [itex] \hat{r} [/itex].) Then he goes on to say it's a good guess that the other term is the component in the tangential [itex] \left( \hat{\theta} \right) [/itex] direction. He proves this is so in 3 ways; namely by proving [itex] \frac{d\hat{r}}{dt} [/itex] is in the [itex] \hat{\theta} [/itex] direction). The first two ways I understand - It's the third one I'm stuck on.

    He starts by drawing two position vectors [itex] \vec{r} [/itex] and [itex] \vec{r} + \Delta \vec{r} [/itex] at the respective times [itex] t [/itex] and [itex] t + \Delta t [/itex], along with their respective unit vectors [itex] \hat{r}_{1} [/itex] , [itex] \hat{r}_{2} [/itex] , [itex] \hat{\theta}_{1} [/itex] , and [itex] \hat{\theta}_{2} [/itex]. From the geometry, we see that [itex] \Delta \hat{r} = \hat{\theta}_{1}sin\Delta\theta - \hat{r}_{1}(1-cos\Delta\theta) [/itex], where [itex] \Delta\theta [/itex] is the angle between the two position vectors).

    From this we see that [itex] \frac{\Delta\hat{r}}{\Delta t} = \hat{\theta}_{1}\frac{sin\Delta\theta}{\Delta t} - \hat{r}_{1}\frac{(1-cos\Delta\theta)}{\Delta t} = \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right)[/itex]. Almost there, just need to take the limit of this quantity as [itex] \Delta t [/itex] tends to 0. So we need to evaluate [itex] \frac{d \hat{r}}{dt} = \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\hat{r}}{\Delta t}[/itex].

    He make the following argument which concludes the proof:

    "In the limit [itex] \Delta t \to 0 [/itex], [itex]\Delta\theta [/itex] approaches zero, but [itex]\Delta\theta/\Delta t[/itex] approaches the limit [itex]d\theta/dt[/itex]. Therefore, [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]. The term in [itex] \hat{r} [/itex] entirely vanishes in the limit and we are left with [itex] \frac{d \hat{r}}{dt}= \dot{\theta} \hat{\theta} [/itex]."

    I understand that [itex] \Delta\theta/\Delta t[/itex] approaches [itex]d\theta/dt[/itex] as [itex] \Delta t \to 0 [/itex], but I'm lost after that. How does one come to the conclusion that [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]? Then, how does this lead us to the conclusion that [itex] \hat{r} [/itex] entirely vanishes in the limit?

    I've been trying my hardest to work through this text, but I tend to get snagged for quite some time on explanations like the above. Usually I can fill in the missing steps myself. I feel as though I cannot thoroughly penetrate this textbook as I have others.

    Thanks in advance for any guidance.
  2. jcsd
  3. Jun 6, 2013 #2


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    Maybe you can make use of the product law for limits if you think of ##\frac{\Delta \theta}{\Delta t}## and ##(\Delta \theta)^n## as functions of ##\Delta t##.
  4. Jun 6, 2013 #3
    Well, since [itex] \displaystyle \lim_{x \to a} f(x)g(x)= \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x) [/itex] then [itex]\Rightarrow \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot \lim_{\Delta t \to 0} (\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot 0 = 0[/itex].

    How does this help though?
  5. Jun 6, 2013 #4


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    I hope I'm not misunderstanding your original question. I am assuming you want to show that ##\displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n## equals zero. If so, isn't that what you have essentially shown?
  6. Jun 6, 2013 #5
    The goal is to prove that [itex] \frac{d \hat{r}}{dt} = \dot{\theta}\hat{\theta} [/itex].

    I do now understand why the previous limit holds. However, why is it relevant to proving that [itex] \frac{d \hat{r}}{dt} = \dot{\theta}\hat{\theta} [/itex]?

    Since a limit of products is identical to the product of the separated limits, any limit containing a factor of [itex] (\Delta\theta)^n [/itex] will result in 0. Is this why [itex] \hat{r} [/itex] vanishes? We can factor out a [itex] (\theta)^2 [/itex] from the second term of [itex] \frac{\Delta\hat{r}}{\Delta t} [/itex]. And, since [itex] n=2,>0 [/itex], then the limit is 0. Right?

    But this reasoning suggests that we can factor out a [itex] (\theta) [/itex] from the first term of [itex] \frac{\Delta\hat{r}}{\Delta t} [/itex]. And since [itex] n=1,>0 [/itex] then the limit is 0 for this one as well. Which is certainly not correct, since [itex] \hat{\theta} [/itex] is not supposed to vanish.

    EDIT: Hold on! I figured it out, I just need to type out all the details.
    Last edited: Jun 6, 2013
  7. Jun 6, 2013 #6


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    OK. Good.
  8. Jun 6, 2013 #7
    I think I got it:


    \frac{d\hat{r}}{dt} &= \displaystyle \lim_{\Delta t \to 0} \left( \frac{\Delta\hat{r}}{\Delta t} \right) \\

    &= \lim_{\Delta t \to 0} \left [ \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right) \right] \\

    & = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left( \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} - \frac{1}{6} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^3}{\Delta t} + \cdots \right) - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left( \frac{1}{2} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^2}{\Delta t} - \frac{1}{24} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^4}{\Delta t} + \cdots \right) \\

    & = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left[ \dot{\theta} - \frac{1}{6} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^2 \right) + \cdots \right] - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left[ \frac{1}{2} \lim_{\Delta t \to 0} \left ( \frac{\Delta\theta}{\Delta t} \cdot \Delta\theta \right) - \frac{1}{24} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^3 \right) + \cdots \right] \\

    & = \dot{\theta} \cdot \lim_{\Delta t \to 0} \hat{\theta}_{1} - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot 0 \ \left( \text{since} \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n \ \text{for} \ n>0 \right) \\

    &= \dot{\theta} \hat{\theta}


  9. Jun 7, 2013 #8


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    Yes, that looks OK. In going from the second to third line you don't really need to take the limits of ##\hat{\theta}_1## and ##\hat{r}_1## since they are fixed vectors that don't depend on ##\Delta t##. But, what you did is fine.
  10. Jun 7, 2013 #9
    I meant it implicitly in going from the 5th to the 6th line. Thanks TSny!
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