Why do we use $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$ as tangent vectors from Polar Coordinates?
The picture makes sense, but from differential manifold's point of view, given the manifold $M=\mathbb{R}^2$, we have two global charts given by $f:{M} \rightarrow \mathbb{R}^2$ where $$f=id =(x,y),$$ and $g:{M} \rightarrow \mathbb{R}^2$ where $$g(m_1, m_2)=(\sqrt{m_1^2 + m_2^2},\arctan(m_2/m_1)) = (r,\theta).$$
Let $\partial_{r_1}=(1,0), \partial_{r_2}=(0,1)$ be the canonical tangent vectors in $\mathbb{R}^2$, then the tangent vectors on $(M,f=(x,y))$ are $$\partial_x = df^{-1}(\partial_{r_1}) = (1,0)$$ $$\partial_y = df^{-1}(\partial_{r_2}) = (0,1)$$ The tangent vectors on $(M,g=(r,\theta))$ are $$\partial_r = dg^{-1}(\partial_{r_1}) = (\cos\theta,\sin\theta)$$ $$\partial_\theta = dg^{-1}(\partial_{r_2}) = (-r\sin\theta, r\sin\theta)$$
And most calculus books use $e_r = (\cos\theta,\sin\theta)$, $e_\theta = (-\sin\theta, \sin\theta)$, why can we just drop the factor $r$ in $e_\theta$? And the gradient, divergence, flux formula are given in this basis too...
Shouldn't it be more natural to keep the $r$. For example if we have a vector field $v$ in $(M,f)$, and we would like to do a change of coordinate to polar, the new vector field in $(M,g)$ is just given by $$\tilde v = d(g^{-1} \circ f) v = \tilde v_1 \partial_r + \tilde v_2 \partial_\theta$$ and the divergence in polar $$div(\tilde v) = \frac{\partial \tilde v_1}{\partial r} + \frac{\partial \tilde v_2}{\partial \theta}.$$
If we drop the $r$ in $e_\theta$, I believe we have this complicated formula $$div(\bar v) = \frac{1}{r}\frac{\partial r\bar v_1}{\partial r} + \frac{1}{r}\frac{\partial \bar v_2}{\partial \theta}.$$