Note how … corresponds to

To find (and remember) $\mathbf{v}$ and $\mathbf{a}$ in polar coordinates, start with the complex number $r{e}^{i\theta }$ and take its derivatives:
$$\mathbf{R}=r{e}^{i\theta }\frac{d\mathbf{R}}{dt}=\frac{dr}{dt}{e}^{i\theta }+ir\frac{d\theta }{dt}{e}^{i\theta }$$
$$\frac{d^2\mathbf{R}}{d{t}^2}=\text{\_\_\_\_}+\text{\_\_\_\_}+\text{\_\_\_\_}+\text{\_\_\_\_}+\text{\_\_\_\_}.$$
Key idea: The coefficients of $e^{i\theta }$ and $i{e}^{i\theta }$ are the ${\mathbf{u}}_r$ and ${\mathbf{u}}_{\theta }$ components of $\mathbf{R}$, $\mathbf{v}$, $\mathbf{a}$:
$$\mathbf{R}=r{\mathbf{u}}_r+0{\mathbf{u}}_{\theta }\mathbf{v}=\frac{dr}{dt}{\mathbf{u}}_r+r\frac{d\theta }{dt}{\mathbf{u}}_{\theta }\mathbf{a}=\text{\_\_\_\_}.$$
(a) Fill in the five terms from the derivative of $d\mathbf{R}/dt$
(b) Convert $e^{i\theta }$ to ${\mathbf{u}}_r$ and $i{e}^{i\theta }$ to ${\mathbf{u}}_{\theta }$ to find $\mathbf{a}$
(c) Compare $\mathbf{R}$, $\mathbf{v}$, $\mathbf{a}$ with formulas (5–6)
(d) (for instructors only) Why does this method work?

Note how $e^{i\theta }=\cos \theta +i\sin \theta$ corresponds to ${\mathbf{u}}_r=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}$. This is one place where electrical engineers are allowed to write $j$ instead of $i$ for $\sqrt{-1}$.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §12.4 Polar Coordinates and Planetary Motion, #14