I take a curve $\vec\gamma:[a,b]\longrightarrow \Delta$, where $\Delta \subseteq \mathbb{R}^2$ is the disk of radius $r>0$. If the curve has length $L>0$ does exist an upper bound (in terms of $r$ and $L$) for the curvature of the curve?

Any reference is very appreciated.

EDIT: The answer of John Hughes has solved the problem.

Now, what happen if the curve is closed?


1 Answer 1


No. Consider the curve $$ \gamma(t) = (s \cos t, s \sin t), 0 \le t \le \frac{L}{2\pi s} $$ where $s < r$. its curvature is $\frac{1}{s}$, which is clearly unbounded.

If you don't like that the path intersects itself, just make $s$ a very slowly increasing function of $t$ with mean $S$. Then the curvature will be approximately $\frac{1}{S}$.

  • $\begingroup$ Thanks. Good example. What happen if I take a closed curve? $\endgroup$ Jun 27, 2016 at 18:12
  • $\begingroup$ Equally bad: You wind around the origin, as above, at radius near $S$, for $L/2$ units of distance, make a hairpin turn, and follow your tracks back again, at a slightly smaller radius, and then make another hairpin turn to complete the path with length approximately $L$; small adjustments can thne make it have length $L$. The curvature along most of the path is about $1/S$, except at the hairpin turns, where it's much higher. Hence: no upper bound. $\endgroup$ Jun 27, 2016 at 19:01
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    $\begingroup$ By the way...adding conditions after you've gotten an answer really isn't a good idea, as it tends to make the answers appear inadequate. Ask a new question instead. See this: meta.stackexchange.com/questions/43478/… $\endgroup$ Jun 27, 2016 at 19:04

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