In a spherical coordinate system my unit vectors are:
$\vec{e_r}=\begin{pmatrix}\sin\theta\cdot \cos\phi \\ \sin\theta \cdot \sin\phi \\ \cos\theta \end{pmatrix}$; $\vec{e_\theta}=\begin{pmatrix}\cos\theta\cdot \sin\phi \\ \cos\theta \cdot \sin\phi \\ -\sin\theta \end{pmatrix}$; $\vec{e_\phi}=\begin{pmatrix}-\sin\phi\\ \cos\phi\\ 0\end{pmatrix}$
My position vector is:
$\vec{r}=r \cdot \vec{e_r}$
If $\theta, \phi, r$ are functions of time.
$\Rightarrow$ $\phi(t), \theta(t), r(t)$
I want to find the first and second derivative $\frac{d}{dt}\vec{r}=\vec{v}$ and $\frac{d}{dt}\vec{v}=\vec{a}$
For the first derivative I get:
$\frac{d}{dt}\vec{r}=\frac{d}{dt}(r \cdot \vec{e_r})=r'(t)\cdot \vec{e_r}+r(t)\begin{pmatrix}\cos\theta(t)\cos\phi(t)\cdot \theta'(t)-\sin\theta(t)\sin\phi(t)\cdot \phi'(t)\\\cos\theta(t)\sin\phi(t)\cdot \theta'(t)+\sin\theta(t)\cos\phi(t)\cdot \phi'(t)\\ -\sin\theta(t)\cdot \theta'(t)\end{pmatrix}$
$\vec{v}=r'(t)\vec{e_r}+r(t)\cdot \vec{e_\theta}\cdot \theta'(t)+\phi'(t)\cdot \sin\theta(t)\cdot \vec{e_\phi}\cdot r(t)$
This was already a relatively messy calculation. I was wondering if there is faster and more elegant method for finding $\frac{d}{dt}\vec{v}?$
Thanks in advance
EDIT: So I brute forced it but I am coming up with a really ugly answer and I am not sure if it's correct.
$\vec{a}=\frac{d}{dz}\vec{v}$
$\vec{v}=r'(t)\vec{e_r}+r(t)\cdot \vec{e_\theta}\cdot \theta'(t)+\phi'(t)\cdot \sin\theta(t)\cdot \vec{e_\phi}\cdot r(t)$
$\Leftrightarrow \frac{d}{dt}\vec{v}=\frac{d}{dt}(r'(t)\vec{e_r}+r(t)\cdot \vec{e_\theta}\cdot \theta'(t)+\phi'(t)\cdot \sin\theta(t)\cdot \vec{e_\phi}\cdot r(t))$
$\Leftrightarrow \frac{d}{dt}\vec{v}=\color{red}{\frac{d}{dt}[r'(t)\vec{e_r}]}+\color{blue}{\frac{d}{dt}[r(t)\cdot \vec{e_\theta}\cdot \theta'(t)]}+\color{green}{\frac{d}{dt}[\phi'(t)\cdot \sin\theta(t)\cdot \vec{e_\phi}\cdot r(t)]} $
I tried to color code the terms so it is easier to see. I tried to differentiate each one of those terms and then summarize them.
$\color{red}{\frac{d}{dt}[r'(t)\vec{e_r}]}=r''(t)\vec{e_r}+r'(t)[\vec{e_\theta}\theta'(t)+\phi'(t)\sin\theta(t)\vec{e_\phi}r(t)]$
$\color{blue}{\frac{d}{dt}[r(t)\cdot \vec{e_\theta}\cdot \theta'(t)]}=r'(t)\vec{e_\theta}\theta'(t)+r(t)\vec{e_\theta}\theta''(t)-\theta'(t)\vec{e_r}+\cos\theta(t)\vec{e_\phi}$
$\color{green}{\frac{d}{dt}[\phi'(t)\cdot \sin\theta(t)\cdot \vec{e_\phi}\cdot r(t)]}=\phi''(t)\sin\theta(t)\vec{e_\phi}r(t)+\phi'(t)\cos\theta(t)\theta'(t)\vec{e_\phi}r(t)+\phi'(t)\sin\theta(t)\vec{e_\phi}r'(t)+\phi'(t)sin\theta(t)r(t)\begin{pmatrix}-\cos\phi(t)\phi'(t)\\ -\sin\phi(t)\phi'(t)\\ 0\end{pmatrix}$
So basically this monster is supposed to be the second derivative. I am not really sure if I have done it correctly. Maybe someone with a little more experience can tell me if I did something wrong or if there is a simpler solution to the problem. Thanks.