I'm trying to strengthen my intuition about covariant derivatives by solving some examples. The example I chose is very trivial but I'm still getting things wrong.
I considered a vector field in polar coordinates: $$\vec{W}(r,\theta) = r\vec{e_r} + \theta \vec{e_\theta}$$
which basically takes the vector pointing towards the point and attaches it to it. I also considered a simple path: $$\gamma(t) = t\vec{e_r} + c\vec{e_\theta}$$
which is a line passing through the origin making a constant angle. Intuitively, I know that $\frac{d\vec{W}}{dt} = \vec{e_r}$ because the vector field grows in this direction along this path. However, my calculation shows something different:
$$\frac{d\vec{W}}{dt} = \frac{dr}{dt}\frac{\partial \vec{W}}{\partial r} + \frac{d\theta}{dt}\frac{\partial \vec{W}}{\partial \theta} = \frac{dr}{dt}\frac{\partial \vec{W}}{\partial r} = \frac{\partial \vec{W}}{\partial r} = \frac{\partial}{\partial r}(r\vec{e_r} + \theta \vec{e_\theta}) = \vec{e_r} + \theta\frac{\partial \vec{e_\theta}}{\partial r} = \vec{e_r} + \frac{c}{r} \vec{e_\theta} = \vec{e_r} + \frac{c}{t} \vec{e_\theta}$$
Now the first term I got correctly. However, why is there a dependence on $\vec{e_\theta}$? I'm sure this result is wrong but I can't find the wrong step. Thank you very much.
With the help of the commentator Graham Kemp, I found the solution. The problem was that I used two different conventions. There is a convention that $\vec{e_\theta}$ scales with r, and a convention that keeps it constant with respect to $r$.
Since I am following the convention that $\vec{e_\theta} = -r\sin\theta \vec{e_x} + r\cos\theta \vec{e_y}$, the vector field I should be using is: $\vec{W}(r,\theta) = r \vec{e_r} + \frac{\theta}{r}\vec{e_\theta}$ so there won't be a change in the angle.
This makes the cancellation happens as:
$$\frac{\partial}{\partial r} (\frac{\theta}{r}\vec{e_\theta}) = -\frac{\theta}{r^2}\vec{e_\theta} + \frac{\theta}{r^2}\vec{e_\theta} = 0$$