Geodesic curvature of a curve on a surface

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Let M⊂R3M \subset \mathbb {R}^{3} be the torus obtained by rotating the circle {(0,y,z):(y−2)2+z2=1}\{ (0, y, z) : (y-2)^2 + z^2 = 1\} around the zz-axis, and let c(t)=(2cos⁡t,2sin⁡t,1)c(t) = (2\cos t, 2 \sin t, 1) ("top circle"). Is this curve a geodesic on MM? Explain without long computations.

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Let M2⊂R3M^2 \subset \mathbb {R}^{3} be the cylinder x2+y2=1x^2+y^2=1. Suppose the curve α(s)⊂M2\alpha (s) \subset M^2 is parameterized by arclength.

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(a)

If k(s)>0k(s) > 0 and τ(s)=0\tau (s) = 0, show that α(s)\alpha (s) is a closed curve. (Here k,τk, \tau are curvature and torsion of α\alpha in R3\mathbb {R}^{3}.)

(b)

If kg(s)=1k_g(s) = 1, show that α(s)\alpha (s) is a closed curve (kgk_g is the geodesic curvature in MM).

References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.