How can both equations for polar coordinates be derived?

putongren
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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.

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

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

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

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


\mathbf{r} = rcos(\theta)\mathbf{e_1} + rsin(\theta)\mathbf{e_2}

\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}

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

\frac{d\mathbf{r}}{dt} = \dot{r}\mathbf{\hat{r}} + r\dot{\theta}\mathbf{\hat{\theta}}

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.
 
I'm new to vector calculus too. What does e1 and e2 mean?
 
They're unit vectors.
 
putongren said:
I'm new to vector calculus too. What does e1 and e2 mean?

The notation e_i is often used for the i-th vector in the standard basis. So e_1 is the vector that points in the positive x direction, and e_2 to the positive y direction, etc.
 
Here's a good link I found on deriving those equations: http://mathworld.wolfram.com/PolarCoordinates.html"
 
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Which is pretty much exactly what I showed!
 
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