How Is the Derivative of Basis Vectors Computed in Polar Coordinates?

[Mathematicsresear]

Homework Statement

I am unsure as to how the partial derivative of the basis vector e_r with respect to theta is (1/r)e_theta in polar coordinates

Homework Equations

The Attempt at a Solution

differentiating gives me -sin(theta)e_x+cos(theta)e_y however I'm not sure how to get 1/r.

 

[vela]

You shouldn't get a factor of 1/r. Why do you think there is one?

 

[Orodruin]

You shouldn't get a factor of 1/r. Why do you think there is one?

Based on some of his other threads, he is not using normalised basis vectors but $\vec e_i = \partial_i \vec x$ (which I usually would denote $\vec E_i$ or similar to underline that they are not necessarily unit vectors - however, there are some places in the literature where $\vec e_i$ is used and instead $\hat i$ or similar is used for normalised basis vectors). With my preferred notation, where $\vec e_i$ are normalised and $\vec E_i = \partial_i \vec x$, you would have $\vec E_r = \vec e_r$ but $\vec E_\theta = r \vec e_\theta$ so indeed you would have $\partial_\theta \vec E_r = \vec E_\theta/r$.