Unit Vector polar in terms of cartesian

[aaj92]

Homework Statement

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.

Homework Equations

x = r cos(theta)
y = r sin(theta)
r = sqrt(x^2 + y^2)
theta = arctan(y/x)

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}.

 

[Simon Bridge]

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)

 

[elfmotat]

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). x^μ represents the Cartesian coordinates x and y (i.e. x¹=x, x²=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 e₁=(1,0) and e₂=(0,1) times the appropriate values, you'll get basis vectors for r and θ.

 

[kusiobache]

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.