Computing the geodesic curvature of the upper parallel of the torus

Question

How would this be done? In a previous exercise we were asked to determine whether the minimum, maximum and upper parallels of the torus were geodesics, asymptotic curves, or lines of curvature. The hint at the back says use the fact that the absolute value of the geodesic curvature is the absolute value of the projection onto the tangent plane of the usual curvature. But I don't really know what this means. I'm new to geodesics and I'm trying to get a better visual feel for them. I know it can be done by brute force (parametrizing the torus and setting u=pi/2 etc..) but I feel theres a much nicer way. Any help is appreciated.

Answer 1

By "upper parallel", I am assuming you mean the topmost circle on the torus (where a horizontal plane placed on top would make contact).

Hint: $$k_g = k \ \mathbf{N} \cdot (\mathbf{n} \times \mathbf{T})$$

Where $k_g$ is the geodesic curvature, $\mathbf{N}$ and $\mathbf{T}$ the normal and tangent vectors to the curve respectively, and $\mathbf{n}$ the surface normal. You don't have to do any horrendous calculations to find $\mathbf{n}$; it should be intuitive. :)

The upper parallel is just a circle; what is its radius? From there, what is its curvature? $\mathbf{N}$? $\mathbf{T}$?

If you get stuck on this part, remember you can parametrize the circle as $\langle r\cos(t), r\sin(t), z \rangle$, where $r$ is its radius and $z$ the constant height.

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As an aside, can you see why this circle would be an asymptotic curve?