# Curve in a disk

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?

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 $\frac{1}{S}$.
• 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. Jun 27, 2016 at 19:01