Polar coordinates

  • Thread starter putongren
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  • #1
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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.

33d65d19869f5b9f1637af2da465801c.png


a7c26879c9644e95490242bd842a87eb.png
 

Answers and Replies

  • #2
jgens
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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.
 
  • #3
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I'm new to vector calculus too. What does e1 and e2 mean?
 
  • #4
jgens
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They're unit vectors.
 
  • #5
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I'm new to vector calculus too. What does e1 and e2 mean?

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.
 
  • #6
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Here's a good link I found on deriving those equations: http://mathworld.wolfram.com/PolarCoordinates.html" [Broken]
 
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  • #7
jgens
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Which is pretty much exactly what I showed!
 

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