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Partial derivative in spherical coordinates

by jonathanpun
Tags: derivative, implicit, partial, spherical
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jonathanpun
#1
Oct12-11, 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?
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mathman
#2
Oct12-11, 03:26 PM
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.
HallsofIvy
#3
Oct12-11, 03:35 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,564
Quote Quote by mathman View Post
∂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.

mathman
#4
Oct13-11, 06:20 PM
Sci Advisor
P: 6,076
Partial derivative in spherical coordinates

Quote Quote by HallsofIvy View Post
You mean "for constant y and z" don't you?
Correct - my typo.


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