Even after reading several answers to similar question I am having (A question about vector representation in polar coordinates., how to compute length in polar coordinates?), I feel still confused. Lets say, we have the following problem of finding the length of a vector $|\textbf{a}|$ in polar coordinates. If I define $a_{\theta} = a \sin(\theta)$ and $a_{r} = a \cos(\theta)$, $$|\textbf{a}|^2 =a_{\theta} ^2 + a_{r}^2 $$
Then I guess, it is not the formula in polar coord. but in local cartesian coord. defined at the point A. Following the answers: https://math.stackexchange.com/q/4325357 or https://math.stackexchange.com/q/1444388, I guess it is rather expressed in polar coord. as $$|\textbf{a}|^2 = r^2 a_{\theta} ^2 + a_{r}^2 $$ Then I don't know what is meant by $a_{\theta}$ or how to compute it. Please elaborate on that!
I was reading Mary L. Boas (Chapter 10, section 10)and it looks like my confusion originates from ignoring the fact that the contravariant and physical components of a vector are different in general. If we consider the line element in polar coordinates then the physical components are $dr$ and r $d\theta$ seen by,
$$d\textbf{s} = dr \textbf{e}_r + r d{\theta} \textbf{e}_{\theta}$$
But if I calculate the components using the definition of contravariant vectors ($ds^{\hat{i}} = \frac{\partial x^\hat{i}}{\partial x^i} ds^{i}$), I get $$ds^{\hat{1}}= \cos(\theta) dx + \sin(\theta) dy = dr$$ $$ds^{\hat{2}} = -\sin(\theta) dx + \cos(\theta) dy = d\theta$$
Therefore in my question $a_{\theta}$ stands for the contravariant component of $\textbf{a}$ not the physical component with which I would find the length. We have to multiply the covariant component with the scale factor to find the physical component. Taking this into account
$$|\textbf{a}|^2 = (a^{phys.}_{\theta})^2 + (a_r)^2$$
Now, it is just a Pythagorean theorem in a locally constructed cartesian coordinate.