I'm looking at page 16 of Fleisch's Student's Guide to Vectors and Tensors. The author is talking about the relationship between the unit vector in 2D rectangular vs polar coordinate systems. They give these equations:
\begin{align}\hat{r} &= \cos(\theta)\hat{i} + \sin(\theta)\hat{j}\\ \hat{\theta} &= -\sin(\theta)\hat{i} + \cos(\theta)\hat{j}\end{align}
I'm just not getting it. I understand how, in rectangular coordinates, $x = r \cos(\theta)$, but the unit vectors are just not computing.
To understand the formula:
1) draw a Cartesian coordinates with a unit circle (centred at origin with a radius of 1).
2) draw a vector pointing towards 2-oclock (so we have $\theta = 30$, this is arbitrary but I prefer this 30 degree angle as I tend to confuse the sin and cos on a 45 degree angle) with a length of 1, so this vector ($\vec{V}$) ends on the unit circle.
3) The key point to remember is that we are converting the unit vectors of the Cartesian system ($\hat{i}$ and $\hat{j}$) to those of the polar system ($\hat{r}$ and $\hat{\theta}$). Unit vector by definition is length 1.
4) To get to the radial unit vector $\hat{r}$: move $1\times cos(\theta)$ units along the x direction ($cos(\theta) \hat{i}$), then move $1\times sin(\theta)$ units along the y direction ($sin(\theta) \hat{j}$). That is the 1st equation: $\hat{r} = cos(\theta) \hat{i} + sin(\theta) \hat{j}$. Note that this how vector additions work. Draw it out on your paper and you will figure it out immediately.
5) Now to get the tangential unit vector ($\hat{\theta}$): that is, by definition, right-angle to $\hat{r}$, with a length of 1. So you will know that all you need to do is switch the positions of $cos(\theta)$ and $sin(\theta)$ in your 1st equation, and add a minus sign to one of them (so the dot product of these 2 resultant vectors is 0, equivalent to perpendicular). As we define anti-clock wise as the positive direction, the minus sign goes to the $\hat{i}$ direction. So here it is the 2nd equation: $\hat{\theta} = -sin(\theta) \hat{i} + cos(\theta） \hat{j}$.
6) If you dislike the step (5) way. Draw that $\hat{\theta}$ out on your unit circle: starting from origin, pointing towards 11-oclock (right angle to $\hat{r}$) and ending on the unit circle (length=1). To get to that end point by moving only along x- and y- directions, first move $-1\times sin(\theta)$ units on x-, then $1\times cos(\theta)$ units on y-. You get the same formula.
To reverse the conversion: from polar to Cartesian, you could simply do some pure algebra on the 2 equations we just derived, or use geometry to "move to" the target point you wish to derive.
The symbols on the left side of those equations don't make any sense. If you wanted to change to a new pair of coordinates $(\hat{u}, \hat{v})$ by rotating through an angle $\theta$, then you would have $$ \left\{\begin{align} \hat{u} &= (\cos \theta) \hat{\imath} + (\sin \theta)\hat{\jmath} \\ \hat{v} &= (-\sin \theta) \hat{\imath} + (\cos \theta)\hat{\jmath}. \end{align}\right. $$
I too got stuck there, and found the answer in a YouTube video at: http://www.youtube.com/watch?v=WwQTTJdAJP8. The difficulty I had was confusing the geometry of polar coordinates, where x = rcos(theta) etc, with the vector situation, where the radius vector is resolved into its components in the x and y directions, so, applying the formulae for expressing a vector in terms of its components to the case of the polar unit radial vector, we get the equation quoted: r^=cos(θ)i^+sin(θ)j^. Attached is an image from the video that gives the geometry of this equation.
The (endpoint of the) vector $(\cos\theta,\,\sin\theta)$ is on the unit circle, exactly at angle $\theta$ (if angles are measured from the $x$-axis, towards the $y$-axis). So, in polar form, we can say $r=1$ and $\theta=\theta$.
The other one, $(-\sin\theta,\,\cos\theta)$ is its rotated version, by $+90^\circ$. So, this has $\hat r=1$ and $\hat\theta=\theta+90^\circ$.
I think that the right way to understand this entire thing is through this way.
Note that we are using the notation $\hat{x},\hat{y}$ to denote vectors fields. For example, $\nabla f=\frac{\partial f}{\partial x}\hat{x}+\frac{\partial f}{\partial y}\hat{y}$ describe a vector on every single points in $\mathbb{R}^2$ with $\nabla f(x_0,y_0)=\frac{\partial f}{\partial x}(x_0,y_0)\hat{x}+\frac{\partial f}{\partial y}(x_0,y_0)\hat{y}$
For any given point $(x,y)=(r,\phi)\in \mathbb{R}^2$, we should have a it's own basis transformation formula, which is where the $\phi$ comes from in the equation. Note that $\hat{r}$ for the point $(x,y)$ is simply the unit vector in $\mathbb{R}^2$ pointing to point, and the vector $\hat{\phi}$ for the point $(x,y)$ is the unit vector orthogonal to $\hat{r}$ for the point $(x,y)$.
Using vector addition, we are done.
In our case here, $\hat{x},\hat{y}$ happens to be the same for every point in $\mathbb{R}$, while $\hat{r},\hat{\phi}$ is not necessarily the same for every point in $\mathbb{R}^2$
Caution: do not confuse $\hat{r},\hat{\phi}$ with the differential geometry object basis of tangent planes $\frac{\partial}{\partial r},\frac{\partial}{\partial \phi}$. $\hat{\phi}$ and $\frac{\partial}{\partial \phi}$ point in the same direction, but the former is unit vector while the latter is not.
