I'm TA'ing multivariable this semester, and I just noticed that we always tend to normalize all our basis vectors when using polar coordinates. This is in stark contrast what I'm used to in differential geometry, as we'd prefer that our coordinate basis to transform by the law \begin{align*} \frac{\partial}{\partial x}&=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}\\ &=\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}, \end{align*} and similarly with $\frac{\partial}{\partial y}$. This $\frac{1}{r}$ factor makes up for the fact that if we travel in the angular direction, we cover more ground the further we are from the origin. So for example, the gradient in these "geometric" polar coordinates would take on the form $$\nabla f |_{(r,\theta)}=(\frac{\partial}{\partial r},\frac{1}{r^2}\frac{\partial}{\partial \theta})$$ which agrees with the usual way of defining gradients locally by $\nabla f=g^{ij}(\partial_if)\partial_j$. This in opposition to the more common $\frac{1}{r}$ factor which comes using the normalized polar coordinate system. So why are we normalizing these coordinates? If you insist on working in an orthonormal frame, why not call it a polar frame instead of polar coordinates to avoid bad practices in the future?
Edit: Let me put in an explicit computation in with the "geometric" (which I learned is called holonomic) basis. Consider $$f(x,y)=\frac{x}{x^2+y^2},$$ so that in polar coordinates, $$f(r,\theta)=\frac{\cos\theta}{r}.$$ One sees: \begin{align*} \nabla f&=\frac{\partial f}{\partial x}\bigg\vert_{(r,\theta)}\frac{\partial}{\partial x}+\frac{\partial f}{\partial y}\bigg\vert_{(r,\theta)}\frac{\partial}{\partial y}\\ &=\frac{\sin^2\theta-\cos^2\theta}{r^2}\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)-\frac{2\cos\theta\sin\theta}{r^2}\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)\\ &=\frac{\sin^2\theta\cos\theta-\cos^3\theta-2\cos\theta\sin^2\theta}{r^2}\frac{\partial}{\partial r}+\frac{-\sin^3\theta+\cos^2\theta\sin\theta-2\cos^2\theta\sin\theta}{r^3}\frac{\partial}{\partial \theta}\\ &=-\frac{\cos\theta}{r^2}\frac{\partial}{\partial r}-\frac{\sin\theta}{r^3}\frac{\partial}{\partial \theta}\\ &=\frac{\partial f}{\partial r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{\partial f}{\partial \theta}\frac{\partial}{\partial \theta}. \end{align*}