The length element in cylindrical coordinates

In summary, the length element ds in cylindrical coordinates can be expressed as ds^2 = dx^2 + dy^2 + dz^2 = d\rho^2 + \rho^2d\theta^2 + dz^2. Using the chain rule and total derivatives, we can find the length element by starting with the expression dx = \frac{\partial x}{\partial \rho}d\rho + \frac{\partial x}{\partial \theta}d\theta and squaring both sides.
  • #1
stripes
266
0

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.
 
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  • #2
stripes said:

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
yeah i got it. thanks.
 

FAQ: The length element in cylindrical coordinates

1. What is the formula for the length element in cylindrical coordinates?

The length element in cylindrical coordinates is given by the formula dl = √(dr2 + r22 + dz2), where dr and are the differentials of the radial and angular coordinates, and dz is the differential of the height coordinate.

2. How is the length element derived in cylindrical coordinates?

The length element in cylindrical coordinates is derived using the Pythagorean theorem. By considering a small displacement in each of the three coordinates, we can construct a right triangle with sides dr, rdθ, and dz. The length element is then the hypotenuse of this triangle, which can be found using the Pythagorean theorem.

3. What is the significance of the length element in cylindrical coordinates?

The length element in cylindrical coordinates is used to calculate the length of a curve in three-dimensional space. It is an essential element in many physical and mathematical applications, such as in calculating surface area, volume, and line integrals.

4. How is the length element different from the arc length in cylindrical coordinates?

The length element is a differential quantity, while the arc length is the integral of the length element over a curve. In other words, the length element gives the length of an infinitesimal segment of a curve, while the arc length gives the length of the entire curve.

5. Can the length element be generalized to other coordinate systems?

Yes, the length element can be generalized to other coordinate systems, such as spherical and polar coordinates. The formula for the length element may differ, but the concept remains the same - it is a differential quantity used to calculate the length of a curve in three-dimensional space.

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