Why does Taylor take the derivative or only $r \bf{\hat r}$ for the velocity/derivative of the position vector?

Question

Why does Taylor take the derivative or only $r \bf{\hat r}$ for the velocity/derivative of the position vector? If the position of an object in Polar coordinates is given by (r, theta), why is the 'position vector' here just $r \bf{\hat r}$ ? What of the angle the vector makes with the horizontal? Isn't that like saying the y-axis position is given because you somehow know the x position, in Cartesian coordiantes?

I am trying to derive the full equation for acceleration in Polar coordinates. If I took $\bf r = r \bf\hat r + \theta \bf{\hat{\theta}}$, I get a wrong answer; I get two extra terms in the r-direction for acceleration, at least.

I think I understand the difference between r, the distance radially away from the origin, and r, the position of the particle in the coordinate system (a vector), so what gives?

Page below (pg 27):

I looked in a class's lecture notes and it is the same, I do not understand:

Edit: It was this question, basically, and I like those answers, too.

Answer 1

The information of angle $\theta$ is entirely "encoded" in the position unit vector $\mathbf{\hat{r}}$, even though it isn't encoded in the radial variable $r$. You can see that if you convert to cartesians vectors: $$ \mathbf{\hat{r}} = \cos(\theta) \mathbf{\hat{i}} + \sin(\theta) \mathbf{\hat{j}} \Longrightarrow \mathbf{r} = r(\cos(\theta) \mathbf{\hat{i}} + \sin(\theta) \mathbf{\hat{j}}) = x\mathbf{\hat{i}} + y \mathbf{\hat{j}} $$ You can check using trig that these definitions agree to what you call polar coordinates, that is, $x = r \cos(\theta)$ and $y = r \sin (\theta)$.

Answer 2

Vector is a thing that has direction and length. $\hat{\boldsymbol{r}}$ is unit vector in the direction of $\boldsymbol{r}$ and encodes all the information about its direction, while r is its length. Thus all the information is encoded in $$\boldsymbol{r}=r\hat{\boldsymbol{r}}.$$

With forces, you could do the same. Take unit vector in the direction of the force and multiply it by its length $$\boldsymbol{F}=F\hat{\boldsymbol{F}},$$ but this is not always usefull. Usually you work in some coordinate basis with unit vectors $\hat{\boldsymbol{x}}, \hat{\boldsymbol{y}}$ in the case of cartesian coordinates, or $\hat{\boldsymbol{r}}, \hat{\boldsymbol{\theta}}$ in the case of polar coordinates. In that case, the direction of the force might be different than the direction of your basis vectors and thus the force will be linear combination of both vectors $$\boldsymbol{F}=F_r\hat{\boldsymbol{r}}+F_{\theta}\hat{\boldsymbol{\theta}}$$.

In general any vector can be written as a linear combination of all basis vectors. Thus you can also write $$\boldsymbol{r}=r_r\hat{\boldsymbol{r}}+r_\theta\hat{\boldsymbol{\theta}},$$ but the component $r_\theta$ is zero for the position vector, because the unit vector $\hat{\boldsymbol{\theta}}$ is always perpendicular to it, as can be seen from the picture you provided.

The same can happen to a force. For example, gravitational force by which some source at the origin of coordinate system acts on some other object is always in the direction of position vector $$\boldsymbol{F}_G=F_r\hat{\boldsymbol{r}}=-G\frac{mM}{r^2}\hat{\boldsymbol{r}}$$

Answer 3

$x=r\cos(\theta)$

$y=r\sin(\theta)$

Use the Chain Rule and Product Rule.

$\dot{x}= \dot{r} \cos{\theta} -r \sin(\theta) \dot{\theta}$

$\dot{y}= \dot{r} \sin{\theta} +r \cos(\theta) \dot{\theta}$

$\ddot{x}= \ddot{r} \cos{\theta} -\dot{r} \sin(\theta) \dot{\theta}-\dot{r} \sin(\theta) \dot{\theta}-r\cos(\theta)\dot{\theta}^2-r\sin(\theta)\ddot{\theta}$

$\ddot{y}= \ddot{r} \sin{\theta} +\dot{r} \cos(\theta) \dot{\theta}+\dot{r} \cos(\theta) \dot{\theta}-r\sin(\theta)\dot{\theta}^2+r\cos(\theta)\ddot{\theta}$

$\text{acceleration}=(\ddot{x},\ddot{y})=\ddot{x} \hat{i}+\ddot{y} \hat{j}=f(r,\theta) \hat{r} + g(r,\theta) \hat{\theta}$

Use the fact that basis vectors transform like covariant tensors of rank 1.

$\hat{r}= \frac{dx}{dr} \hat{i} + \frac{dy}{dr} \hat{j} \\$

$\hat{\theta}= \frac{dx}{d\theta} \hat{i}+\frac{dy}{d\theta} \hat{j} $

And

$$\hat{r} \cdot \text{acceleration} = f(r,\theta)$$

$$\hat{\theta} \cdot \text{acceleration} = g(r,\theta)$$.

Therefore,

$f(r,\theta)=\cos(\theta)[\ddot{r} \cos{\theta} -\dot{r} \sin(\theta) \dot{\theta}-\dot{r} \sin(\theta) \dot{\theta}-r\cos(\theta)\dot{\theta}^2 -r\sin(\theta)\ddot{\theta}]+\\ \sin(\theta)[\ddot{r} \sin{\theta} +\dot{r} \cos(\theta) \dot{\theta}+\dot{r} \cos(\theta) \dot{\theta}-r\sin(\theta)\dot{\theta}^2 +r\cos(\theta)\ddot{\theta}] =\ddot{r}-r\dot{\theta}^2$

(using $\cos^2 (\theta) + \sin^2 (\theta)=1)$