Unit Vector polar in terms of cartesian

1. aaj92

25
1. The problem statement, all variables and given/known data

Prove that the unit vector r{hat} of two-dimensional 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. Simon Bridge

14,382
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:
$\text{\hat{a}} \rightarrow \hat{a}$ ... rather than a{hat}.

(welcome to PF)

Last edited: Jan 24, 2012
3. elfmotat

261
I'm not sure if this will help you, but the general form of the transformation of a basis vector is:

$$\vec{e}_{\bar{\nu}}=\sum_{\mu=1}^n \frac{ \partial x^\mu }{ \partial x^{\bar{\nu}}}\vec{e_\mu}$$

where n is the number of dimensions (in this case two). xμ represents the Cartesian coordinates x and y (i.e. x1=x, x2=y). xν (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 e1=(1,0) and e2=(0,1) times the appropriate values, you'll get basis vectors for r and θ.

4. kusiobache

29
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 $\hat{r}= \hat{x}cosθ + \hat{y}sinθ$

Edit: I'm thinking illustrate that $\vec{r} = r\hat{r}$ in polar and then showing that in Cartesian $\vec{r} = \hat{x}cos\phi +\hat{y}sin\phi$

Edit2: Nope, I'm confused again.. I think elfmotat is correct, but I don't quite understand his explanation.

Edit3: nevermind - I got it.

Last edited: Jan 24, 2012