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A magnitude is a function of the position on the plane, if it is known that in polar coordinates $r$ and $\theta$ the function verifies the equation on partial derivatives $$sin(\theta) \frac{\delta F}{\delta r}+\frac {cos(\theta)}{r} \frac {\delta F}{\delta \theta}=0$$

How do I find the same equation in cartesian coordintes?

How do I transform the partial derivatives from $r$ and $\theta$ to $x$ and $y$? My guess is that I could serve from $x=rcos(\theta)$ and $y=rsin(\theta)$ but I am not sure how to handle the partil derivatives

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By the chain rule we have \begin{eqnarray*} \frac{ \partial F}{\partial r } = \frac{ \partial x}{\partial r } \frac{ \partial F}{\partial x } +\frac{ \partial y}{\partial r } \frac{ \partial F}{\partial y } \\ \frac{ \partial F}{\partial \theta } = \frac{ \partial x}{\partial \theta } \frac{ \partial F}{\partial x } +\frac{ \partial y}{\partial \theta } \frac{ \partial F}{\partial y } \\ \end{eqnarray*} Now use $x=r \cos \theta$ and $y=r \sin \theta$ to calculate the $4$ derivatives $\frac{ \partial (x,y)}{\partial (r , \theta)}$ and substitute into the equation and we have \begin{eqnarray*} \sin \theta \left( \cos \theta \frac{ \partial F}{\partial x } +\sin \theta \frac{ \partial F}{\partial y } \right) + \cos \theta \left( -\sin \theta \frac{ \partial F}{\partial x } +\cos \theta \frac{ \partial F}{\partial y } \right)=0 \end{eqnarray*} So your equation simplifies to \begin{eqnarray*} \color{blue}{\frac{ \partial F}{\partial y } =0}. \end{eqnarray*}

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