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I am in the process of proving that if a space curve (in $R^3$) has infinite length and the curvature tends towards $0$ as the natural parameter $s$ tends to infinity, the curve must be unbounded - i.e. not contained in any sphere of finite radius. This seems correct intuitively, but I have no guarantee it is correct, unless I am missing something obvious. One way to prove my hunch, I have deduced, is to use a lemma that any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly $2π$, but it could be greater for all I know).

How might one go about proving such an upper bound exists, or if it exists? It might also be nice to know what the bound specifically is, too. I've thought it might be possible to pose this as a variational problem - maximizing length - and then reducing it into a simpler problem, but that appears to be hellishly complicated. Thoughts?

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    If your curvature goes to zero with increasing arclength, in the limit you have something that looks like a straight line, which is unbounded. I'll leave it up to other people to make this argument rigorous.2010-11-30
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    That's basically my thoughts. But it only locally looks like a line. Technically you could have a curve spiral around and never asymptotically approach being a line but still have curvature -> 0. And that doesn't even bother to touch on what torsion could do to the situation.2010-11-30
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    My gut feeling is that torsion can't make things any worse. Given a space curve $\gamma_1$ with curvature $\kappa(t)$ and torsion $\tau(t)$, suppose you construct another curve $\gamma_2$ whose torsion is identically zero, and whose curvature is $\lvert\tau(t)\rvert$. In other words, you're stuffing $\gamma_1$ into a plane while preserving the magnitude of its curvature. I conjecture that $\gamma_2$ is contained in a ball of no larger radius than any ball containing $\gamma_1$. If true, this would reduce the scope of the problem to purely plane curves.2010-11-30
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    That's false. Take helices, for instance, which have both constant curvature and constant torsion. Curves with constant curvature and zero torsion are simply circles. The hypothetical helix can be any size while the circle's size is invariant to extending the helix. (Although being able to reduce this problem to plane curves would be vastly helpful.)2010-11-30
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    Sorry, I think I wasn't very clear in my previous comment. I was looking at the contrapositive of the problem: for a given length and a bound on curvature, what's the curve that fits inside the smallest ball? If curves with nonzero torsion (e.g. helices) are always "bigger" than the corresponding plane curves (e.g. circles), then that's a *good* thing for your problem.2010-11-30
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    @chroma: By the way, if you are replying to a comment, starting your comment with "@username-you-are-responding-to" will notify the commenter about your response.2010-11-30
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    Anyway, on reading your problem more carefully, I realized I was thinking about a much more general problem. Surely there are simpler solutions in your case. Sorry for the bother.2010-11-30
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    To be clear, the bounty I set is for showing that "any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly 2π, but it could be greater for all I know)." It is not for the motivating problem of proving that the curve is unbounded, which already follows from my weaker answer.2011-01-31

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