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Computing curvatures

by rsq_a
Tags: computing, curvatures
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rsq_a
#1
May3-12, 12:36 PM
P: 110
I'm having a really hard time working with non-standard expressions for the curvature. This deals with two expressions for the curvature, one for a 2D plane curve, and the other for an axi-symmetric surface in 3D.

The plane curve

Suppose we have a plane curve given by [itex]x = h(y)[/itex] for both x and y positive. In a paper I'm reading, the author writes:
The surface curvature can be expressed in terms of the slope angle, [itex]\beta[/itex] of the surface as per
[tex]
\kappa = -\frac{d( \cos\beta)}{dy}.
[/tex]

The geometrical relation,
[tex]
\frac{dh}{dy} = \frac{1}{\tan \beta}
[/tex]

expresses the dependence of h on [itex]\beta[/itex]
The second expression is more or less clear for me (modulo whether it should be negated or not). The first expression is not. How do you go from the standard definition:
[tex]
\kappa = \frac{h''}{(1+(h')^2)^{3/2}},
[/tex]

to this result?

Surface curvature

This one is from another paper. The author assumes that there is an axi-symmetric surface [itex]S(z,r) = 0[/itex], which is only a function of [itex]z[/itex] and [itex]r[/itex] in spherical coordinates. He states that the mean curvature is
[tex]
\kappa = (S_z^2 +S_r^2)^{-3/2} \left[ S_z^2 S_{rr} - 2S_z S_r S_{rz} + S_r^2 S_{zz} + r^{-1} S_r(S_r^2 + S_z^2)\right].
[/tex]

If we denote the downward angle of the slope at an arbitrary position on the drop surface by [tex]\delta[/tex], so that
[tex]
\cos\delta = \frac{S_z}{\sqrt{S_z^2+S_r^2}} \qquad
\sin\delta = \frac{S_r}{\sqrt{S_z^2+S_r^2}}
[/tex]

then the curvature can be written as
[tex]
\kappa = \frac{1}{r} \frac{d}{dr} \left(r \sin\delta\right)
[/tex]

along [itex]S = 0[/itex]

Again, I could really use some help in seeing how these expressions were derived. Or if it's not trivial, perhaps a source where the work is shown.
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