# Find spherical polar components of $\frac{\partial \textbf{F}}{\partial \phi}$

by fazan   Last Updated September 21, 2018 18:20 PM

I have to find the partial derivatives in spherical form of $$\textbf{F}=r[\textbf{e}_{\theta}+\textbf{e}_{\phi}],$$ and I managed to find all the others except for the one over $$\phi$$.

I got this far: $$\frac{\partial \textbf{F}}{\partial \phi}=r[\cos(\theta)\textbf{e}_\phi+\cos(\phi)\textbf{x}-\sin(\phi)\textbf{y}]$$ , with the last two terms being the partial derivative of $$\textbf{e}_{\phi}$$ over $$\phi$$.

Since $$\textbf{e}_{\phi}$$ is an unit vector, its derivative has to be perpendicular to it, and as such must be of the form $$A\textbf{e}_{\theta}+B\textbf{e}_r$$, but if I try doing that I get that B must be both positive and negative. Where do I go wrong?

Edit: calculation of coefficients

$$[\cos(\phi),-\sin(\phi),0]=[A\sin(\theta)\cos(\phi)+B\cos(\theta)\cos(\phi), A\sin(\theta)\sin(\phi)+B\cos(\theta)\sin(\phi), A\cos(\theta)-B\sin(\theta)]$$

From the third components: $$A\cos(\theta)=B\sin(\theta)$$ we get $$A=\frac{\sin(\theta)}{\cos(\theta)}B$$.

Looking at the first components, we get $$B=\cos(\theta)$$, but looking at the second components we get $$B=-\cos(\theta)$$.

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solved, I misplaced a sign in $$\textbf{e}_\phi$$