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Computing curvatures 
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#1
May312, 12:36 PM

P: 110

I'm having a really hard time working with nonstandard expressions for the curvature. This deals with two expressions for the curvature, one for a 2D plane curve, and the other for an axisymmetric 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: [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 axisymmetric 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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