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I understand that $\theta $ basis vector have variable length proportional to the radius. Text

Some change in $\theta$ trace longer curve the further it away from the center. So the same $\Delta\theta$ make bigger impact on the function change the larger the current $r$ is.

What I don't understand intuitively is why this behaviour shouldn't propagate to the gradient as well.

The correct formula for gradient in polar coordinates $$\nabla f = \frac{\partial f}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial f}{\partial \theta}\hat{\theta}$$ explicitly compensates for this behaviour.

But why $$\nabla f = \frac{\partial f}{\partial r}\hat{r} + \frac{\partial f}{\partial \theta}\hat{\theta}$$ wouldn't be the "direction and rate of fastest increase" in polar coordinates?

I've seen the derivations, and I understand why they are correct, but I have the problem that incorrect formula still seems more natural for me intuitively as "direction and rate of fastest increase" in polar coordinates.

simd
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    I think you should try explaining in more detail why the last equation is more intuitive to you. That will help you get better answers. – Nicholas Todoroff Mar 30 '23 at 06:23
  • @NicholasTodoroff I've only worked with gradient in cartesian coordinates before and probably became too accustomed to think about it as just "a vector of all partial derivatives". Before learning about polar coordinates this simplistic approach made perfect intuitive and visual sense to me as "direction and rate of fastest increase". Now it seems to me that I don't understand what the gradient really is. At least generally. Maybe this is the problem. – simd Mar 30 '23 at 11:24

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Make an explicit example. Consider the function $f(r, \theta)=\theta$. It is clear that the gradient $\nabla f$ must have the direction $e_\theta$ everywhere. But what would its magnitude be, as the correct formula says $1/r$ while yours predicts $1$?

Just integrate it. You must have the fundamental formula $$ f(p_2)-f(p_1)=\int_\gamma \nabla f\cdot \vec{ds}, $$ where $\gamma$ is a line that connects the two points $p_2$ and $p_1$. Suppose these points have angular coordinates $\theta_2$ and $\theta_1$ respectively, so the left--hand side of the latter equation reads $\theta_2-\theta_1$. Suppose these points are joined by an arc of circumference $\gamma$ of radius $r$, so the line element reads $\vec{ds}= e_\theta\, rd\theta$. Then, plugging in your formula for $\nabla f$, the right--hand integral reads $$ \int_{\theta_1}^{\theta_2} r\,dr=r(\theta_2-\theta_1), $$ which is off by a factor of $r$ from the correct one. This $r$ is precisely the factor you missed in your formula.

Giuseppe Negro
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