Description of unit vectors in Cartesian vs. Polar Coordinates I'm trying to understand the following description in Classical Mechanics by John Taylor:

He goes on to write:

The vectors $\hat{\mathbf{x}}, \hat{\mathbf{y}}$ are the same at all
  points in the plane, whereas the new vectors $\hat{\mathbf{r}}, \hat{\mathbf{\phi}}$ change their directions as the position vector
  $\mathbf{r}$ moves around.

In image a) above, the unit vector $\hat{\mathbf{x}}$ is placed at the tip of $\mathbf{r}$, so that leads me to believe that it will move if $\mathbf{r}$ changes. Are we supposed imagine $\hat{\mathbf{x}}$ placed at the origin? Can someone explain this a little bit more? Are we supposed to image both sets of unit vectors (whether it be $\hat{\mathbf{x}}, \hat{\mathbf{y}}$ or $\hat{\mathbf{r}}, \hat{\mathbf{\phi}}$) placed at the origin, and it's just their direction that matters?
 A: I believe you got it right but were mislead, partly by the author, for two reasons.

the unit vector $\hat x$ is placed at the tip of $r$

Since a vector is not characterized with an origin but only with a direction and a magnitude, the point from which it is drawn could be arbitrary. What can be meant by movement though is a change in direction as you quoted.
As a side note, a difference is to be made between what he calls unit vectors and position vector. The first only holds a direction while the later is a difference of coordinates.
Then when he writes :

The vectors $\hat x, \hat y$ are the same at all points in the plane, whereas the new vectors $\hat r$ and $\hat\phi$ change their directions as the position vector  r moves around.

the underlying assumption is that we consider motion in the frame $(\hat x,\hat y)$, where, trivially,  $\hat x$ and $\hat y$ do not change direction.
If we were to consider the $(\hat r,\hat\phi)$ frame, then $\hat x$ and $\hat y$ would be changing direction as $\phi$ changes.
A: This is something that may make some people puzzled at first, so I'll try to give a complete answer to it.
Vectors in $\mathbb{R}^n$
First of all, let us consider the space $\mathbb{R}^n$. Mathematicaly this is just the vector space of all $n$-tuples $(x^1,\dots, x^n)$ with the addition and scalar multiplication given by
$$(x^1,\dots, x^n)+(y^1,\dots, y^n)=(x^1+y^1,\dots, x^n+y^n)\\ \lambda(x^1,\dots, x^n)=(\lambda x^1,\dots, \lambda x^n)$$
The cases you should think about are $n=2,3$. The first is the plane and the second is the three-dimensional euclidean space. The important thing is that we would like to think of the elements of $\mathbb{R}^n$ as points in one $n$-dimensional space. 
It turns out that this definition describes something that is more convenient than just points. It describes points with respect to one fixed location which we call the origin. The elements of $\mathbb{R}^n$ are more correctly thought as arrows fixed at this origin, and ending at the location of the point. If you think like this in the case $n=2$ you will see that the above rules for adding and scaling these elements is the same as the pictorial ones people learn in basic physics texts.
Among the elements of $\mathbb{R}^n$ there is a set of $n$ distinguished ones $\{e_1,\dots e_n\}$ which are defined as: $$e_i = (0,\dots, 1,\dots,0),$$ in other words they have all but the $i$-th entry equal to zero. They form a basis in the linear algebra sense, called canonical basis. They are the arrows pointing along $n$ mutual orthogonal axes which meet at the origin and their importance is that any vector $x\in \mathbb{R}^n$ is such that
$$x = \sum x^i e_i.$$
So: these vectors lie in the origin, they describe positions with respect to this origin and although we have a canonical basis, we have infinitely many other basis we could use. These basis are also located at the same origin as are all elements of $\mathbb{R}^n$
The Tangent Spaces and Vector Fields
It turns out though, that specially in Physics, we need to describe mathematicaly vector fields. They should be assignments of vectors like those we discussed above to each point of space. Why this is useful? To describe quantities which like location, need a direction to be specified.
Now, as I said, vectors in $\mathbb{R}^n$ as defined above are fixed on the chosen origin! So how can we proceed?
Essentialy the point is that how we interpret this origin is arbitrary and up to us. If I hand you the vector $x\in \mathbb{R}^n$ and not tell you where in reality the origin is, you just know one relative position, to one origin which you can put wherever you want.
So here is the idea: given $\mathbb{R}^n$ describing positions with respect to some observer at some origin, what I want a vector field to do is to associate to each $x\in \mathbb{R}^n$ another arrow. So we define the following space:
$$T_x \mathbb{R}^n = \{(x,v) : v\in \mathbb{R}^n\}$$
This is the set of all pairs $(x,v)$ with $x$ fixed on the first entry, and letting the second entry vary among all elements of $\mathbb{R}^n$.
What this means? Means the following: start with some $\mathbb{R}^n$ with its origin. Pick $x$ representing a point with respect to this chosen origin. Now copy $\mathbb{R}^n$ there. Thus $T_x\mathbb{R}^n$ is like you pick another $\mathbb{R}^n$ (another set of arrows pointing in all directions of all sizes) and place it at $x$.
In better terms, $T_x\mathbb{R}^n$ is the space of all possible directions one could follow if at the point $x$ with respect to the origin we choose. It is called the tangent space of $\mathbb{R}^n$ at $x$.
A vector field is then a function $V(x)$ such that $V(x)\in T_x\mathbb{R}^n$ for each $x$.
Moving Frames
In the begining of our discussion I said that you could have infinitely many basis on $\mathbb{R}^n$, but they were all fixed at the origin.
Now you have a copy of $\mathbb{R}^n$ on each point. Thus you have infinitely many basis on each point.
The basis like $\hat{r},\hat{\theta},\hat{\phi}$ are assignments of one such basis to each point of $\mathbb{R}^n$. In other words, each of these objects are vector fields! 
So what we are considering are sets of vector fields $\{V_1,\dots,V_n\}$ such that at each $x\in \mathbb{R}^n$ we have $\{V_1(x),\dots, V_n(x)\}$ one basis picked on $T_x\mathbb{R}^n$.
This set of vector fields can be called a moving frame, because you can think of it as a set of axes associated to each point.
One possible way to build a such a moving frame is to pick one basis on the origin and place the same basis at each point. Pick the canonical one $\{e_i\}$ and define $V_i(x)=(x,e_i)$. There you have it.
Now your doubt abou constancy is: you think of the basis as changing because of the base point is changing. And you have a point, after all $V_i(a)=(a,e_i)$ and $V_i(b)=(b,e_i)$ are basis of different vector spaces $T_a \mathbb{R}^n$ and $T_b \mathbb{R}^n$. But make no mistake, the vectors of the basis are actually the same, they just have been carried over to a different point, and it is in this sense that they are constant.
Now, although this construction is nice to be precise and actually understand what is going on, and although in more general spaces than $\mathbb{R}^n$ this construction is mandatory, in $\mathbb{R}^n$ all tangent spaces are naturally equivalent to $\mathbb{R}^n$. That because they all differ just by translation.
So what one usually does is: leave the base point to interpretation in $\mathbb{R}^n$. So you talk about a vector $v\in T_a\mathbb{R}^n$ just as one element of $\mathbb{R}^n$, but you keep in mind when interpreting things that it was located at $a$.
This is common with vector fields. The electric field for instance, one writes as an ordinary $\mathbb{R}^3$ vector and leaves to interpretation that $E(x)$ lies in $T_x\mathbb{R}^3$. The same for the moving frames and hence to the vectors $\hat{r},\hat{\theta},\hat{\phi}$.
