Change in r-hat by change in theta

[Storm6436]

Homework Statement

The motion of a particle is described in standard planepolar coordinates. Derive
an expression for each of $\frac{d\hat{r}}{d\theta}$ and $\frac{d\hat{\theta}}{d\theta}$, each in terms of one or both of $\hat{r}$ and $\hat{\theta}$ as necessary. Derive both results through using each of two
methods:

i.First method: Start by expressing $\hat{r}$ and $\hat{\theta}$ in terms of Cartesian unit vectors. Then differentiate each expression, make a diagram and use geometry or trigonometry, etc. Clearly show all
necessary steps of the argument for each.

(Omitting second method because I'm pretty sure once I figure the first out, I'll be able to use the rotating frames argument on my own)

Homework Equations

$$y=r\sin(\theta)$$
$$\frac{dy}{d\theta}=r\cos(\theta)$$
$$x=r\cos(\theta)$$
$$\frac{dx}{d\theta}=-r\sin(\theta)$$
$$r\hat{r}=x\cos{\theta}\hat{x} + y\sin{\theta}\hat{y}$$

The Attempt at a Solution

So I chucked a $\frac{d}{d\theta}$ at the last equation listed, which spits out: $$ \frac{dr}{d\theta}\hat{r} +r \frac{d\hat{r}}{d\theta} = \left(\frac{dx}{d\theta}\cos{\theta}- x\sin{\theta}\right)\hat{x} + \left( \frac{dy}{d\theta}\sin{\theta}+ y\sin{\theta}\right)\hat{y}$$

I then figured if we're rolling only in $\theta$, then by the fact that $\hat{\theta}$ is orthogonal to $\hat{r}$, that makes $\frac{dr}{d\theta}\hat{r} = 0$

With that in mind, I subbed in the two equations for $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ which yielded:

$$r \frac{d\hat{r}}{d\theta} = \left(-r\sin{\theta}\cos{\theta}-r\sin{\theta}\cos{\theta}\right)\hat{x} + \left( r\sin{\theta}\cos{\theta}+r\sin{\theta}\cos{\theta}\right)\hat{y}$$

Which simplifies to:

$$r \frac{d\hat{r}}{d\theta} = -2r\sin{\theta}\cos{\theta}\hat{x} +2r\sin{\theta}\cos{\theta}\hat{y}$$

I guess where I'm stuck is trying to boil it down further into just $\hat{r}$ and $\hat{\theta}$. It's been way too long since I've done vector calc, so I'm not sure if I can just divide out all the r's or where else I should go from this step, provided I even did it right in the first place. I mean, it'd be nice if I could snarf the cos off the x and the sin off the y and be all voila, $\hat{r}$ but I'm pretty certain that's not how this works.

Similarly, I'm expecting this to end up something like $\frac{d\hat{r}}{d\theta}=d\theta$ and this doesn't look to me like it's heading that way at all.

 

[Chestermiller]

$$\hat{r}=\cos{\theta}\hat{i}+\sin{\theta}\hat{j}$$

 

[Storm6436]

You're kidding, right? *facepalm* Leave it to me to overcomplicate things and totally miss the obvious. Thanks, Chester. At least I managed to learn a decent amount of LaTeX last night while I was spinning my wheels.

 

[Storm6436]

For what it's worth, now that I've finished both parts of the above problem, it popped into my head where I went wrong above. For the use-case I was attempting, X and Y are fixed values, so $\frac{dx}{d\theta}=0$ and $\frac{dy}{d\theta}=0$ which plops out the answer I should have been getting if you correct the error I made above on the second term of $\hat{y}$, which should have been $\cos{\theta}$ not $\sin{\theta}$.