Deriving unit vector $\hat{\theta}$

Question

Consider the polar coordinates $$x=r\cos\theta\\y=r\sin\theta$$

Show that the unit vector $\hat{\theta}=(-\sin\theta, \cos\theta)$.

My attempt:

$$\vec{\theta} = \theta \hat{\theta}$$ I think that $$\hat{\theta}=\left(\frac1r \frac{\partial x}{\partial \theta}, \frac1r \frac{\partial y}{\partial \theta}\right)$$

I think that the factor $\frac1r$ must be added for normalization.

But this is not rigorous. Please give me a hint.

Answer 1

HINT:

The unit vector $\hat{\theta}$ is given by

$$ \hat{\theta} = \frac{\frac{d \mathbf{r}}{d\theta}}{|\frac{d \mathbf{r}}{d\theta}|},$$

where $\mathbf{r}$ is the radius vector. So, in Cartesian coordinates it is given by

$$ \mathbf{r} = \begin{bmatrix} x \\ y \end{bmatrix}, $$

in polar coordinates it is given by

$$ \mathbf{r} = \begin{bmatrix} r \cos \theta \\ r\sin \theta \end{bmatrix}. $$

Can you take it from here?

Answer 2

I think the easiest way to do it is to rely on the geometrical intuition, you know $$ \hat r=\frac{1}{r}(x,y)=(\cos\theta,\sin\theta), $$ and you know $\hat r \cdot \hat \theta=0$ then, $$ \hat \theta =\lambda(-\sin\theta,\cos\theta) $$ for some constant $\lambda\in\mathbb R$.

Moreover, you know $|\hat \theta|=1$, hence $\lambda =\pm 1$. Lastly, you remember that theta increases counterclockwise, hence $\lambda$ should be 1. Hence,

$$ \hat \theta =(-\sin\theta,\cos\theta) $$