# Is the following curve $\gamma(s)$ closed?

Let $\gamma(s)$ be a curve (parametrized by arclength) whose image lies on the circular cylinder $x^2+y^2=1$ in $R^3$, given that curvature $\kappa(s)>0$ and that torsion $\tau(s)=0$ for all $s$.

Some friends and I were having difficulty interpreting this question, so any help will be quite helpful.

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## 2 Answers

You should be able to find a theorem saying that if the torsion is everywhere $0$, then the curve lies in a plane. So it's in the intersection of a plane with a right circular cone. All such intersections are circles, ellipses, or lines or pairs of parallel lines. The curvature is not positive if the curve is a straight line or a union of two of those. So it's a circle or an ellipse.

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Thanks, we seemed to have had some disagreement over interpreting that the image lied on the cylinder. – katari Nov 5 '12 at 1:09

Prove the following:

Proposition: Let $C = \gamma(s)$ be a plane curve such that $\kappa(s) \neq 0$, then $\tau(s) = 0$ for all $s \in I$.

Corollary: Let $C = \gamma(s)$ be a regular curve such that $\kappa(s) \neq 0$ and $\tau(s) = 0$ for all $s \in I$, then $C$ belongs to a fixed plane.

Then you can prove that your curve is the intersection of a plane with the cylinder, and therefore must be either a circle or an ellipse (by the reasoning of Michael's answer above).

And yes, the curve will be closed.

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