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Homework Help: Derivative of r^hat

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data

    I am following along my classical mechanics book by John Taylor on page 27 in anyone has it. He is saying:

    del r^hat ~ del phi * phi^hat

    then..

    del r^hat ~ D[phi] * del t * phi^hat

    Then...

    dr^hat/dt = D[phi] * phi^hat

    Sorry for the horrible formating but I don't know how to make this better.

    3. The attempt at a solution

    I have a few problems with this first why in the first equation is the phi needed? It would seem to me that del r^hat would just be equal to del phi^hat and the magnitude of phi would not be needed. Then how does the phi just turn into a derivative of phi in the second equation and now there is a del t? I understand the last line.
     
  2. jcsd
  3. Jan 24, 2010 #2
    1. click new reply
    2. click the capital sigma all the way to the right of the toolbar
    3. format your post using latex
     
  4. Jan 24, 2010 #3
    Those are capital deltas [tex]\Delta[/tex], not dels [tex]\nabla [/tex]. Two VERY different things.

    I don't like how Taylor develops the acceleration and velocity in polar coordinates using pictures and "approximate" arguments. Marion and Thornton do it this way too.

    Instead, consider this argument. (It will go strait to eqn. (1.43) on page 28.)

    We have [tex] \mathbf{r} = r \mathbf{e}_r[/tex], and

    [tex] \mathbf{e}_r = \cos \left(\phi\right) \hat{\mathbf{i}} + \sin \left(\phi\right) \hat{\mathbf{j}} [/tex]

    and

    [tex] \mathbf{e}_\phi = - \sin \left(\phi \right) \hat{\mathbf{i}} + \cos \left(\phi \right) \hat{\mathbf{j}}. [/tex]

    Now,

    [tex] \frac{d \mathbf{r}}{dt} = \frac{d}{dt} \left( r \mathbf{e}_r} \right) = \dot{r}\mathbf{e}_r + r \frac{d \mathbf{e}_r}{dt}. [/tex]

    But, by the chain rule,

    [tex] \frac{d\mathbf{e}_r}{dt} = \frac{\partial \mathbf{e}_r}{\partial r}\dot{r} + \frac{\partial \mathbf{e}_r}{\partial \phi} \dot{\phi}.[/tex]

    From above, then,

    [tex] \frac{\partial \mathbf{e}_r}{\partial r} = 0 [/tex]

    and

    [tex] \frac{\partial \mathbf{e}_r}{\partial \phi} = - \sin \left(\phi\right) \hat{\mathbf{i}}+ \cos \left(\phi\right) \hat{\mathbf{j}} = \mathbf{e}_\phi.[/tex]

    Perhaps you can finish from here....
     
  5. Jan 26, 2010 #4
    Not sure why you are using the chain rule in this step since [tex]
    {e}_r
    [/tex] doesn't have a time dependence.
     
  6. Jan 27, 2010 #5
    I did make a minor mistake: the chain rule for [tex]\mathbf{e}_r[/tex] is actually

    [tex]\frac{d\mathbf{e}_r}{dt} = \frac{d\mathbf{e}_r}{d\phi}\dot{\phi}.[/tex]

    We don't need that other term in there, and the derivative is an ordinary derivative, not a partial. They're basically the same in this case however.

    But, to answer your question: if [tex]\mathbf{e}_r[/tex] didn't depend on time, then we wouldn't be in this mess. It would be constant with respect to time like the standard unit vectors in cartesian coordinates.

    If
    [tex]\mathbf{e}_r = \cos \left(\phi\right) \hat{\mathbf{i}} + \sin \left(\phi\right) \hat{\mathbf{j}},[/tex]

    it's important to note that [tex]\phi = \phi\left(t\right).[/tex] So, you can look at [tex]\mathbf{e}_r[/tex] as being a vector valued function [tex]\mathbf{e}_r \left( \phi \left(t\right)\right)[/tex]; hence, the chain rule is needed.
     
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