# For the unit vector hattheta, geometrically show that hattheta = -sinthetahati + costhetahatj? Essentially, converting from cartesian to polar, how would I determine the unit vector for vectheta in terms of theta, hati, and hatj?

## I've been able to show that $\hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$: $\cos \theta = \frac{\hat{i}}{\hat{r}}$ $\sin \theta = \frac{\hat{j}}{\hat{r}}$ $\implies | | \hat{r} | | = \sqrt{\hat{r} \cdot \hat{r} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)}$ $= \sqrt{\hat{r} \cdot \hat{r} {\cos}^{2} \theta + \hat{r} \cdot \hat{r} {\sin}^{2} \theta}$ $= \sqrt{\hat{r} \cos \theta \cdot \hat{i} + \hat{r} \sin \theta \hat{j}}$ $\implies \hat{r} \cdot \hat{r} = | | \hat{r} | {|}^{2} = \hat{r} \cos \theta \cdot \hat{i} + \hat{r} \sin \theta \cdot \hat{j}$ $= \hat{r} \cdot \left(\cos \theta \hat{i} + \sin \theta \hat{j}\right)$ Thus, $\hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$. But how would I do it for $\hat{\theta}$? I'm probably just missing something really simple, like where the $\hat{\theta}$ vector points.

Feb 19, 2017

See below.

#### Explanation:

Considering $p = \left(r \cos \theta , r \sin \theta\right) = r \left(\cos \theta , \sin \theta\right) = r \hat{r}$ where $\hat{r} = \left(\cos \theta , \sin \theta\right)$

We have then

$\dot{p} = \dot{r} \hat{r} + r \dot{\hat{r}}$

but $\dot{\hat{r}} = \left(- \sin \theta , \cos \theta\right) \dot{\theta} = \dot{\theta} \hat{\theta}$

Here for convenience, we call $\left(- \sin \theta , \cos \theta\right) = \hat{\theta}$

$\hat{r} , \hat{\theta}$ form a basis of orthogonal unit vectors. They can be also called $\hat{n} , \hat{\tau}$ instead.

so we have $\dot{p} = \dot{r} \hat{r} + r \dot{\hat{r}} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$

deriving again

$\ddot{p} = \ddot{r} \hat{r} + \dot{r} \dot{\theta} \hat{\theta} + \dot{r} \dot{\theta} \hat{\theta} + r \ddot{\theta} \hat{\theta} + r \dot{\theta} \dot{\hat{\theta}}$

Here $\hat{\theta} = \left(- \sin \theta , \cos \theta\right)$ so

$\dot{\hat{\theta}} = - \left(\cos \theta , \sin \theta\right) \dot{\theta} = - \dot{\theta} \hat{r}$

and finally

$\ddot{p} = \left(\ddot{r} - r {\left(\dot{\theta}\right)}^{2}\right) \hat{r} + \left(2 \dot{r} \dot{\theta} + r \ddot{\theta}\right) \hat{\theta}$

Concluding $\hat{r} , \hat{\theta}$ are used for convenience. They have a strong geometric appeal. Also they obey all rules of vector and differential calculus.

Feb 20, 2017

#### Explanation:

As a matter of convenience, I use $\alpha$, instead of $\theta$.

The unit vector in the direction $\theta = \alpha$ is

$\cos \alpha \vec{i} + \sin \alpha \vec{j}$

$= < x , y > = < \cos \alpha , \sin \alpha >$, in Cartesian form.

The parallel position vector through the origin O ( r = 0 ) is

$\vec{O P}$, where $P \left(\cos \alpha , \sin \alpha\right)$, is the radius vector of

the unit circle r = 1 , in the direction $\theta = \alpha$.

P' is at $\left(\cos \left(\frac{\pi}{2} + \alpha\right) , \sin \left(\frac{\pi}{2} + \alpha\right)\right) = \left(- \sin \alpha , \cos \alpha\right)$.

$\vec{O P '} = < - \sin \alpha , \cos \alpha >$ is constructed as the radius

vector of the unit circle, in the direction $\theta = \frac{\pi}{2} + \alpha .$.

graph{(x^2+y^2-1)(y-x/sqrt3)(y+sqrt3x)=0 [-1, 1, -.05, 1]}

Here $\theta = \frac{\pi}{6}$. Any vector of length 1, in the direction

$\alpha = \theta + \frac{\pi}{2} = \frac{2}{3} \pi$ (shown as a radius ),

would represent $- \sin \left(\frac{\pi}{6}\right) \vec{i} + \cos \left(\frac{\pi}{6}\right) \vec{j}$.

In brief, $\vec{O P '}$ is $\vec{O P}$ turned through $\frac{\pi}{2}$, in the (

$\theta \uparrow$) anti-clockwise sense. OP is in ${Q}_{1}$and OP' is in ${Q}_{2}$.