# How to get the gradient potential in polar coordinate

In polar coordinate,

$$\nabla U = \frac{\partial U}{\partial r}\hat{\mathbf{r}} + \frac{1}{r}\frac{\partial U}{\partial \theta}\hat{\mathbf{\theta}} .$$

Can anyone show me how to get this result?

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Assuming that you're working in two dimensions, with Cartesian coordinates $x,y$, the result is obtained by starting with the expression for the gradient in the $x,y$ system:

\begin{align} \nabla U &= \vec{e}_1 \frac{\partial U}{\partial x} + \vec{e}_2\frac{\partial U}{\partial y}, \end{align} where $\vec{e}_i$ are the unit vectors in $x (i=1)$ and $y (i=2)$ directions. Then use the transformation between Cartesion and (plane-)polar coordinates: \begin{align} x &= r \cos\theta,\\ y &= r \sin\theta, \end{align} and the chain rule. For example: $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}$.

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