
#1
Jan2312, 09:09 PM

P: 25

1. The problem statement, all variables and given/known data
Prove that the unit vector r{hat} of twodimensional polar coordinates is equal to r{hat}= x{hat}cosθ + y{hat}sinθ and find the corresponding expression for θ{hat}. all I need is the last part... I'm just not sure what θ{hat} is? How do I go about doing this? Nothing in my book even hints at how to do this. 2. Relevant equations x = r cos(theta) y = r sin(theta) r = sqrt(x^2 + y^2) theta = arctan(y/x) 3. The attempt at a solution I really just need help getting started... I've been staring at this for over an hour which I know is sad but r{hat} is significantly easier than theta{hat}. 



#2
Jan2312, 10:06 PM

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P: 11,009

How did you do the first bit?
What would be the analogous method for the second bit? You are not asked to prove it, just write it down. note: [itex]\text{\hat{a}} \rightarrow \hat{a}[/itex] ... rather than a{hat}. (welcome to PF) 



#3
Jan2312, 10:57 PM

P: 260

I'm not sure if this will help you, but the general form of the transformation of a basis vector is:
[tex]\vec{e}_{\bar{\nu}}=\sum_{\mu=1}^n \frac{ \partial x^\mu }{ \partial x^{\bar{\nu}}}\vec{e_\mu}[/tex] where n is the number of dimensions (in this case two). x^{μ} represents the Cartesian coordinates x and y (i.e. x^{1}=x, x^{2}=y). x^{ν} (with a bar over it  this distinguishes between Cartesian and polar coordinates) represents the polar coordinates r and θ. What you need to do is differentiate the Cartesian coordinates x and y with respect to r and θ (i.e. dx/dr, dx/dθ, dy/dr, and dy/dθ). When you sum the Cartesian basis vectors e_{1}=(1,0) and e_{2}=(0,1) times the appropriate values, you'll get basis vectors for r and θ. 



#4
Jan2412, 12:10 AM

P: 29

Unit Vector polar in terms of cartesian
Could someone give me a hint on the first part of this? Because I can derive it  that is just simple trigonometry  but I can't figure out how to concretely prove that [itex]\hat{r}= \hat{x}cosθ + \hat{y}sinθ[/itex]
Edit: I'm thinking illustrate that [itex]\vec{r} = r\hat{r}[/itex] in polar and then showing that in Cartesian [itex]\vec{r} = \hat{x}cos\phi +\hat{y}sin\phi[/itex] Edit2: Nope, I'm confused again.. I think elfmotat is correct, but I don't quite understand his explanation. Edit3: nevermind  I got it. 


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