#physics
$\text{require}physics$

Cartesian $$\vec{\pu{ pos }} = (\vec{x}, \vec{y}, \vec{z})$$$$\vec{\dd s} = (\vec{\dd x}, \vec{\dd y}, \vec{\dd z})$$$$\dd V = \dd x \cdot\dd y \cdot \dd z$$

Constant unit vectors of global frame

Polar $$\begin{align}\vec{\pu{ pos }} &= (\vec{r}, {\theta}, \vec{z}) \ \vec{r} &= |r|\hat{r} \ \hat{r} &= \hat{x}\cos \theta + \hat{y} \sin \theta \ \hat{\theta} &= -\hat{x}\sin \theta + \hat{y}\cos \theta\end{align}$$

We always require three independent variables for fully describing position.
Notice that $\hat{\theta }$ is also a vector, and is important

Trying to represent the magnitude of $(x,y)$ and the vector $(\hat{x},\hat{y})$ in terms of polar variables :

Varying unit vectors, observer frame.
Since the unit vectors aren't constant, they have finite time-derivatives.

$$\begin{align} \dot{\hat{r}} &= \dot{\theta}\hat{\theta} \ \dot{\hat{\theta}} &= -\dot{\theta}\hat{r} \end{align}$$

Velocity and acceleration (2D)

Cartesian : $$\begin{align} \vec{v} &= \frac{dx}{dt}\hat{x} + \frac{dy}{dt}\hat{y} \ \ \vec{a} &= \frac{d^2x}{dt^2}\hat{x} + \frac{d^2y}{dt^2}\hat{y}\end{align}$$

Polar : $$\begin{align}\vec{v} &= \frac{d(|r|\hat{r})}{dt}\\implies \vec{v} &= \frac{d|r|}{dt}\hat{r} + |r| \frac{d\hat{r}}{dt} \\pu{ Since, } \ \frac{d\hat{r}}{dt} &= \frac{d\theta}{dt} \cdot \hat{\theta} \ \therefore\ \ \vec{v} &= \frac{d|r|}{dt}\hat{r} + |r| \frac{d\theta}{dt}\hat{\theta} \end{align}$$

Velocity : $$\boxed{ \vec{v} = \dot{r}\hat{r} + |r|\dot{\theta}\hat{\theta} }$$

Acceleration : $$\boxed{\vec{a} = \underbrace{ (\ddot{r} - r\dot{\theta}^2) \hat{r}}{\text{Linear + Centripetal}} + \underbrace{(r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}}{\text{Tangential + Coriolis}}}$$

Infinitesimal Line Element in plane polar

Change in position from $(r,\theta )$ to $(r+\mathrm{\Delta }r,\theta +\mathrm{\Delta }\theta )$

We consider motion in two mutually perpendicular directions

1. Change in $r$, keeping $\theta$ constant
   Contribution : $|dr|\hat{r}$
2. Change in $\theta$, keeping $r$ constant
   Contribution : $|rd\theta |\hat{\theta }$
   Total line element $$\boxed{\vec{\dd l} = \dd r \cdot \hat{r} + r \dd \theta \cdot\hat{\theta}}$$

Area element $$\vec{A} = r \cdot\dd r \cdot\dd \theta \cdot\hat{n}$$

3D polar

Introduction of $z$ as independent variable.

Line element $$\boxed{\vec{\dd l} = \dd r \cdot \hat{r} + r \dd \theta \cdot\hat{\theta} + \dd z \cdot\hat{z} }$$

Area element $$\vec{A} = r \cdot\dd \theta \cdot \dd z \cdot \hat{r}$$

Volume element $$V = r\cdot\dd r\cdot\dd \theta \cdot \dd z$$

Spherical coordinate $$\begin{align} &(r, \theta, \phi) \ \ r &= \sqrt{ x^2 + y^2 + z^2 } \ \ z &= r\cos \theta \ \theta &= \tan^{-1} \left( \frac{\sqrt{ x^2 +y^2}}{z} \right) \ \phi &= \tan^{-1}\left( \frac{y}{x} \right)\end{align}$$

$\theta$ is from $z$ axis and $\varphi$ is from $x$ axis.

For unit vectors $$\begin{align} \hat{r} &= (\sin \theta \cos \phi) \cdot\hat{x} + (\sin \theta \sin \phi) \cdot \hat{y} + \cos \theta \cdot \hat{z} \ \hat{\theta} &= (\cos \theta \cos \phi)\cdot \hat{x} + (\cos \theta \sin \phi)\cdot \hat{y} - \sin \theta \cdot\hat{z} \ \hat{\phi} &= -\sin \phi \cdot \hat{x} + \cos \phi \cdot \hat{y} + 0\cdot \hat{z}\end{align}$$
In matrix form, which is easier to remember, $$\begin{align}\begin{bmatrix}\hat{r} \ \hat{\theta} \ \hat{\phi}\end{bmatrix} &= \begin{bmatrix}\sin \theta \cos \phi& \sin \theta \sin \phi & \cos \theta \ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \ -\sin \phi & \cos \phi & 0\end{bmatrix}\cdot\begin{bmatrix}\hat{x} \ \hat{y}\\hat{z}\end{bmatrix} \ \ \ \begin{bmatrix}\hat{x} \ \hat{y} \ \hat{z}\end{bmatrix} &= \begin{bmatrix}\sin \theta \cos \phi & \cos \theta \sin \phi & -\sin \phi \ \sin \theta \sin \phi & \cos \theta \sin \phi & \cos \phi \ \cos \theta & -\sin \theta & 0\end{bmatrix} \cdot \begin{bmatrix}\hat{r} \ \hat{\theta} \ \hat{\phi}\end{bmatrix} \end{align}$$
The respective derivatives are easy to find, and I am therefore omitting.

Line element $$\vec{\dd l} = \dd r\cdot \hat{r} + r \dd \theta \cdot \hat{\theta} + r\sin \theta \dd \phi \cdot \hat{\phi}$$

Area element $$\vec{\dd A} = (r^2\sin \theta\dd \theta \dd \phi)\cdot \hat{r}$$

Volume element $$\dd V = (r^2\sin \theta\ \dd r\ \dd \theta\ \dd \phi)\cdot \hat{r} $$

Conversion matrices

Between different coordinate systems

Cartesian <-> Cylindrical $$\begin{align} \begin{bmatrix} \hat e_\rho \ \hat e_\phi \ \hat e_z \end{bmatrix} &= \begin{bmatrix} \cos\phi & \sin\phi & 0 \ -\sin\phi & \cos\phi & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat e_x \ \hat e_y \ \hat e_z \end{bmatrix} \ \ \ \begin{bmatrix} \hat e_x \ \hat e_y \ \hat e_z \end{bmatrix} &= \begin{bmatrix} \cos\phi & -\sin\phi & 0 \ \sin\phi & cos\phi & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat e_\rho \ \hat e_\phi \ \hat e_z \end{bmatrix} \end{align}$$

Spherical <-> Cartesian $$\begin{align} \begin{bmatrix} \hat e_r \ \hat e_\theta \ \hat e_\phi \end{bmatrix} &= \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \ -\sin\phi & \cos\phi & 0 \end{bmatrix} \begin{bmatrix} \hat e_x \ \hat e_y \ \hat e_z \end{bmatrix} \ \ \ \ \begin{bmatrix} \hat e_x \ hat e_y \ \hat e_z \end{bmatrix} &= \begin{bmatrix} \sin\theta\cos\phi & \cos\theta\cos\phi & -\sin\phi \ \sin\theta\sin\phi & \cos\theta\sin\phi & \cos\phi \ \cos\theta & -\sin\theta & 0 \end{bmatrix} \begin{bmatrix} \hat e_r \ \hat e_\theta \ \hat e_\phi \end{bmatrix} \end{align}$$

Spherical <-> Cylindrical $$\begin{align} \begin{bmatrix}\hat e_r \ \hat e_\theta \ \hat e_\phi \end{bmatrix}&=\begin{bmatrix}\sin\theta & \cos\theta & 0 \\cos\theta & -\sin\theta & 0 \0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat e_\rho \ \hat e_z \ \hat e_\phi \end{bmatrix} \ \ \ \begin{bmatrix} \hat e_\rho \ \hat e_z \ \hat e_\phi \end{bmatrix} &= \begin{bmatrix} \sin\theta & \cos\theta & 0 \ \cos\theta & -\sin\theta & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat e_r \ \hat e_\theta \ \hat e_\phi \end{bmatrix} \end{align}$$