The length element in cylindrical coordinates

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Homework Statement



Show that in cylindrical coordinates

[itex]x = \rho cos \theta[/itex]
[itex]y = \rho sin \theta[/itex]
[itex]z = z[/itex]

the length element ds is given by

[itex]ds^{2} = dx^{2} + dy^{2} + dz^{2} = d \rho^{2} + \rho^{2} d \theta ^{2} + dz^{2}[/itex]

Homework Equations



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The Attempt at a Solution



Notice [itex]\rho = \sqrt{x^{2} + y^{2}}[/itex]

I have found expressions for ∂x/∂θ, ∂x/∂ρ, ∂y/∂θ, ∂y/∂ρ, ∂ρ/∂x, ∂ρ/∂y, ∂θ/∂x, ∂θ/∂y, and of course it is trivial that dz = dz. Given all of these, how do i start using chain rule and total derivatives to find ds2? I guess because i have 8 partials but end up only ρ, θ, and z, i don't really know which ones to start with.

Any help is greatly appreciated.
 

Answers and Replies

  • #2
pasmith
Homework Helper
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599

Homework Statement



Show that in cylindrical coordinates

[itex]x = \rho cos \theta[/itex]
[itex]y = \rho sin \theta[/itex]
[itex]z = z[/itex]

the length element ds is given by

[itex]ds^{2} = dx^{2} + dy^{2} + dz^{2} = d \rho^{2} + \rho^{2} d \theta ^{2} + dz^{2}[/itex]

Homework Equations



--

The Attempt at a Solution



Notice [itex]\rho = \sqrt{x^{2} + y^{2}}[/itex]

I have found expressions for ∂x/∂θ, ∂x/∂ρ, ∂y/∂θ, ∂y/∂ρ, ∂ρ/∂x, ∂ρ/∂y, ∂θ/∂x, ∂θ/∂y, and of course it is trivial that dz = dz. Given all of these, how do i start using chain rule and total derivatives to find ds2? I guess because i have 8 partials but end up only ρ, θ, and z, i don't really know which ones to start with.

Any help is greatly appreciated.
Start from
[tex]
dx = \frac{\partial x}{\partial \rho}d\rho + \frac{\partial x}{\partial \theta}d\theta
[/tex]
Squaring both sides gives [itex]dx^2[/itex].
 
  • #3
265
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yeah i got it. thanks.
 

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