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 $\frac{1}{S}$.