Partial derivative in spherical coordinates

by jonathanpun
Tags: derivative, implicit, partial, spherical
 P: 6 I am facing some problem about derivatives in spherical coordinates in spherical coordinates: x=r sinθ cos$\phi$ y=r sinθ sin$\phi$ z=r cosθ and r=$\sqrt{x^{2}+y^{2}+z^{2}}$ θ=tan$^{-1}$$\frac{\sqrt{x^{2}+y{2}}}{z}$ $\phi$=tan$^{-1}$$\frac{y}{x}$ $\frac{\partial x}{\partial r}$=sinθ cos$\phi$ then $\frac{\partial r}{\partial x}$=$\frac{1}{sinθ cos \phi }$ but if i calculate directly from r: $\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}$ substitute: =$\frac{r sinθ cos \phi }{r}$ = sinθ cos$\phi$ Why do the results are different? what i did wrong? From http://www.physicsforums.com/showthread.php?t=63886 not this case is the second case? but why the inverse still not true?
 Sci Advisor P: 6,076 ∂r/∂x is defined for constant x and y. ∂x/∂r is defined for constant θ and φ. There is no reason that they should be reciprocal.
Math
Emeritus
Thanks
PF Gold
P: 39,564
 Quote by mathman ∂r/∂x is defined for constant x and y.
You mean "for constant y and z" don't you?

 ∂x/∂r is defined for constant θ and φ. There is no reason that they should be reciprocal.