Partial derivative in spherical coordinatesby jonathanpun Tags: derivative, implicit, partial, spherical 

#1
Oct1211, 11:27 AM

P: 6

I am facing some problem about derivatives in spherical coordinates
in spherical coordinates: x=r sinθ cos[itex]\phi[/itex] y=r sinθ sin[itex]\phi[/itex] z=r cosθ and r=[itex]\sqrt{x^{2}+y^{2}+z^{2}}[/itex] θ=tan[itex]^{1}[/itex][itex]\frac{\sqrt{x^{2}+y{2}}}{z}[/itex] [itex]\phi[/itex]=tan[itex]^{1}[/itex][itex]\frac{y}{x}[/itex] [itex]\frac{\partial x}{\partial r}[/itex]=sinθ cos[itex]\phi[/itex] then [itex]\frac{\partial r}{\partial x}[/itex]=[itex]\frac{1}{sinθ cos \phi }[/itex] but if i calculate directly from r: [itex]\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}[/itex] substitute: =[itex]\frac{r sinθ cos \phi }{r}[/itex] = sinθ cos[itex]\phi[/itex] 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? 



#2
Oct1211, 03:26 PM

Sci Advisor
P: 5,938

∂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. 



#3
Oct1211, 03:35 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890





#4
Oct1311, 06:20 PM

Sci Advisor
P: 5,938

Partial derivative in spherical coordinates 


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