For the flow $A = \frac{c}{r}$ with $r=\sqrt{x^2+y^2+z^2}$ I wanted to calculate the velocity field with $\nabla A$
As a result I get $(-\frac{c}{r^2},0,0)$. So far so good. When I tried converting it to cartesian coordinates I'm stuck.
I would have inserted the given $r$ in my solution however wolfram alpha & the solution suggest the solution in cartesian coordinates is:
Converting with $x=\sqrt{x^2+y^2+z^2} \cos(0)$ does not really work. I'm sure I'm currently having a major blackout in seeing something elemental missing here.
You cannot simply convert your expression for polar coordinates to cartesian because the operator is different in the systems. But the cartesian calcuation is easy (using $c=1$): $$\nabla A = \left(\frac{\partial}{\partial x}A, \frac{\partial}{\partial y}A\right) = \left(\frac{\partial}{\partial x}\frac{1}{\sqrt{x^2+y^2}}, \frac{\partial}{\partial y}\frac{1}{\sqrt{x^2+y^2}}\right) = \left(\frac{-x}{(x^2+y^2)^{3/2}}, \frac{-y}{(x^2+y^2)^{3/2}}\right) $$