Contravariant components $a^k$ of the acceleration $\boldsymbol{a}$ of a particle Well I was studying the uses of Christoffel in physics and I got the following Idea :
Let us consider a particle moving on a trajectory defined by spherical coordinates $r, \theta, \varphi$, the acceleration $\boldsymbol{a}$ is a vector of course, if we wanted to determinate its contravariant components $a^k$. let the trajectory be defined by: $r=c,\ \theta=\omega t \ , \varphi= \pi/4 \ , t \ \text{is time}$.
Well to do this we need to determinate the Christoffel symbols, for this trajectory:
$$\Gamma_{1 \ 3}^3=1/r \ ; \ \Gamma_{2 \ 3}^3=\text{cotan} \ \theta \ ; \ \Gamma_{3 \ 3}^3=0$$
My questions :


*

*How can we identify the contravariant components of this particle ? 

*Can we just treat this particle as a point $M$ and determinate the contravariant components of $\frac{\partial^2 \boldsymbol{OM}}{\partial t^2}$ ? 


Thanks in advance !
 A: I tried to solve this problem and here's my attempts :

Quick recall :
  $$\begin{align}
\\ \mathbf{e_1} &= \partial_1 \mathbf{M}=\sin  \theta \cos  \varphi \mathbf{i}+\sin \theta \sin \varphi \mathbf{j}+ \cos \theta \mathbf{k}
\\ \mathbf{e_2}&= \partial_2 \mathbf{M}=r\cos\theta\cos \varphi\mathbf{i}+r\cos\theta \sin\varphi\mathbf{j}-r\sin \theta \mathbf{k} 
\\ \mathbf{e_3}&=\partial_3 \mathbf{M}=-r\sin\theta\sin \varphi \mathbf{i}+r\sin\theta\cos\varphi \mathbf{j} \end{align}$$

Well I had the same idea as yours which is let consider this particle to be a point let's call it $\mathbf{M}$ and we're going to determinate the components of $\mathbf{a}$ to do this we need the to determinate the Christoffel symbols values as @Gunter said :
Differentials of natural base vectors :
$\mathbf{i, j, k}$ are constants in their modules and directions, the differential of $\mathbf{e_1}$ is :
$$d\mathbf{e_1}=(\cos\theta\cos \varphi \mathbf{i}+\cos\theta\sin\varphi \mathbf{j})d\theta+(-\sin\theta \sin\varphi\mathbf{i}+\sin\theta\cos\varphi)d\varphi $$
We notice that the terms inside the brackets are respectively $\mathbf{e_2}/r$ and $\mathbf{e_3}/r$, thus:
$$d\mathbf{e_1}=\frac{d\theta}r \mathbf{e_2}+\frac{d\varphi}r \mathbf{e_3}$$
With differentiation also we get:
$$\begin{align}
d\mathbf{e_2}&=(-rd\theta)\mathbf{e_1}+(dr/r)\mathbf{e_2}+(\text{cotan}\ \theta d\varphi)\mathbf{e_3} \\
d\mathbf{e_3}&=(-r\sin^2 \theta d\varphi)\mathbf{e_1}+(-\sin\theta \cos \theta d\varphi)\mathbf{e_2}+((dr/r)+\text{cotan } \ \theta d\theta)\mathbf{e_3}
\end{align}$$
I'm still far from the answer but we concluded a really interesting :
$$d\mathbf{e_i}=a_i^k \mathbf{e_k} \qquad(1)$$
Where $a_i^k$ the contravariant components of $d\mathbf{e_i}$:
Let's start solving the problem :
Note the spherical coordinates : $u^1=r, \ u^2=\theta, \ u^3=\varphi$
So :   $du^1=dr, \ du^2=d\theta, \ du^3=d\varphi$ and the components can be written as :
$$a_i^j=\Gamma_{ki}^i du^k \qquad (2)$$
In $(1)$ we notice that the components $a_i^k$ are linear combination for example :
$$a_1^2=d\theta/r , \ a_3^3=(dr/r)+\text{cotan} \ \theta d\theta$$
In $(2)$ we have the quantities $\Gamma_{ki}^j$ are functions of $r,\theta, \varphi$ and we obtain simply by identifying $a_i^j$ for example :
 $$a_3^3=(dr/r)+\text{cotan} \ \theta d\theta=\Gamma_{1 \ 3}^3du^1+\Gamma_{2 \ 3}^3du^2+\Gamma_{3 \ 3}^3 du^3$$
Thus :
$$\Gamma_{1 \ 3}^3=1/r, \ \Gamma_{2 \ 3}^3=\text{cotan} \ \theta,\ \Gamma_{3 \ 3}^3=0$$ 
And here you go you have the identities and for the $\mathbf{a}$ components are :
$$\fbox{$a^1=\frac{d^2 u^1}{dt^2}+\Gamma_{ik}^1 \frac{du^i}{dt} \frac{du^k}{dt}, \ a^2=0, \ a^3=0$}$$
Hope it helps go trough these final results and you will solve your problem, but don't hurry, first of all determinate the values of Christoffel symbols at the trajectory with the system you gave. Good luck !  
