Newton's laws in polar coordinates

[MikeN232]

I need explanation of these formulas for polar coordinate system where position of an object is characterized by 2 vectors: r - from the origin to the object, and Φ - perpendicular to r, in the direction of rotation.
https://drive.google.com/file/d/0ByKDaNybBn_eakJmS3dUVXVZUDA/view?usp=sharing

The 1st one is how we define a vector: product of magnitude by unit vector (which gives a direction), clear;
the 2nd we try to express change in unit vector, which can only change direction = angle of rotation, but why we multiply it then by unit vector Φ, and why ≈ ? Need explanation;
the 3rd we found derivative of position with respect to time;
the 4th, differentiated the 1st formula;
the 5th, substituted 3 into 4;
the 7th, differentiated it for the second time to find acceleration and here we need derivative of unit vector Φ;
8 found derivative of unit vector Φ; how they did it and why unit vector r is here? Need explanation.
9,10 substituted Φ, found acceleration

 

[fresh_42]

There seem to be something wrong with the upload of your pictures. Perhaps you should try to type them rather than to upload it.

 

[jtbell]

Specifically, we encourage people to use LaTeX for equations:

https://www.physicsforums.com/help/latexhelp/

 

[vanhees71]

$\newcommand{\dd}{\mathrm{d}}$
$\newcommand{\vv}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix}}$ $\newcommand{\vvv}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix}}$ $\newcommand{\bvec}[1]{\boldsymbol{#1}}$
The link works. Let's do the calculation in more detail. First you define your generalized coordinates by expressing the Cartesian components in terms of these generalized coordinates. For polar coordinates, it's
$$\vec{x}=\vv{r \cos \phi}{r \sin \phi}.$$
The coordinate basis (holonomous basis) is given by the tangents of the coordinate lines, i.e.,
$$\vec{b}_r=\frac{\partial \vec{x}}{\partial r}=\vv{\cos \phi}{\sin \phi}, \quad \vec{b}_{\phi} = \frac{\partial \vec{x}}{\partial \phi} = r \vv{-\sin \phi}{\cos \phi}.$$
Now you see that these basis vectors are everywhere perpendicular to each other, where the coordinates are well defined (i.e., for $\vec{x} \neq 0$):
$$\vec{b}_r \cdot \vec{b}_{\phi}=0.$$
Then it is usual and convenient to use a normalized basis. Then you have a Cartesian coordinate system in any point, but it depends on the point (i.e., it's a function of the generalized coordinates). Here you have
$$\vec{e}_r=\vec{b}_r=\vv{\cos \phi}{\sin \phi}, \quad \vec{b}_{\phi} = \frac{1}{r} \vec{b}_{\phi}=\vv{-\sin \phi}{\cos \phi}.$$
Now you describe the trajectory by making $r=r(t)$ and $\phi=\phi(t)$. Then you get
$$\vec{x}(t)=r(t) \vec{e}_r(t).$$
It is important to keep in mind that through the dependence of the basis vectors $\vec{e}_r$ and $\vec{e}_{\phi}$ on the generalized coordinates, these also become time dependent, i.e., you have to take also time derivative of those, and not only for the components. Using the product rule you get
$$\vec{v}=\dot{\vec{x}}=\dot{r} \vec{e}_r + r \dot{\vec{e}}_{r}.$$
Now we have (using the chain rule)
$$\dot{\vec{e}}_r=\frac{\mathrm{d}}{\mathrm{d} t} \vv{\cos \phi}{\sin \phi}=\dot \phi \vv{-\sin \phi}{\cos \phi}=\dot{\phi} \vec{e}_{\phi},$$
and thus
$$\vec{v}=\dot{r} \vec{e}_r + r \dot{\phi} \vec{e}_{\phi}.$$
Now, you should try to use this method to also calculate $\vec{a}=\dot{\vec{v}}$ yourself!