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Polar coordinates

  1. Aug 4, 2009 #1
    Dear All,

    How do you derive both equations below. Let r be the position vector (rcos(θ), rsin(θ)), with r and θ depending on time t.

    These equations can be found in wiki under polar coordinates.


  2. jcsd
  3. Aug 4, 2009 #2


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    This looks just like a staight-forward application of the chain and product rules . . .

    Proof: [tex]\mathbf{r} = [rcos(\theta), rsin(\theta)][/tex]

    [tex]\mathbf{\hat{r}} = [cos(\theta),sin(\theta)][/tex]

    [tex]\mathbf{\hat{\theta}} = [-sin(\theta),cos(\theta)][/tex]

    [tex]\mathbf{r} = rcos(\theta)\mathbf{e_1} + rsin(\theta)\mathbf{e_2}[/tex]

    [tex]\frac{d\mathbf{r}}{dt} = [\dot{r}cos(\theta) - rsin(\theta)\dot{\theta}]\mathbf{e_1} + [\dot{r}sin(\theta) + rcos(\theta)\dot{\theta}]\mathbf{e_2}[/tex]

    Expressing this in terms of our previously defined unit vectors we have that.

    [tex]\frac{d\mathbf{r}}{dt} = \dot{r}\mathbf{\hat{r}} + r\dot{\theta}\mathbf{\hat{\theta}}[/tex]

    As desired. A similar method could probably be used to get the second result. This is probably a bit sloppy but I'm just learning vector calculus.
  4. Aug 5, 2009 #3
    I'm new to vector calculus too. What does e1 and e2 mean?
  5. Aug 5, 2009 #4


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    They're unit vectors.
  6. Aug 5, 2009 #5
    The notation [tex]e_i[/tex] is often used for the i-th vector in the standard basis. So [tex]e_1[/tex] is the vector that points in the positive x direction, and [tex]e_2[/tex] to the positive y direction, etc.
  7. Aug 6, 2009 #6
    Here's a good link I found on deriving those equations: http://mathworld.wolfram.com/PolarCoordinates.html" [Broken]
    Last edited by a moderator: May 4, 2017
  8. Aug 6, 2009 #7


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    Which is pretty much exactly what I showed!
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