Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.