Let $\Sigma \subset \mathbb{R}^2$ be a smooth, embedded, compact hypersurface. Then it can be locally written as a graph of a function. With that, I mean that for any $p \in \Sigma$ there is a neighborhood a ball $B_{r_x}(p) \subset \mathbb{R}^2$ of radius $r_p>0$ around $x$ such that the hypersurface is locally given as the graph of a function $u$, $$ \Sigma \cap B_{r_x}(x) = \{(z,u(z)): z \in [-r_p,r_p]\}. $$
What I want to know is whether it is possible to bound the radius $r_p$ from below by the curvature of the hypersurface. The fact that it is possible to find such a representation follows from the implicit function theorem but it doesn't yield a lower bound on $r_p$. Is there some quantitative version of the implicit function theorem?
My idea would be the following.
After rotating and translating, it is possible to arrange that in $p=(0,0)$ the gradient of $u'(0)=0$. If we assume that the second fundamental form of $\Sigma$ is bounded by a constant, $|A| \leq C$, I am able to bound, for any $\epsilon>0$, in a ball of radius $(4 \epsilon^{-1}C)^{-1}$ $$ |u'| \leq \epsilon, \quad |u''| \leq 2 C. $$ It seems to be possible to deduce from here, that for $Id$ being the identity map on $\mathbb{R}^2$, we get the following operator norm bound, $$ |(z,u(z))-Id| \leq \epsilon. $$ Hence, $(z,u(z))$ is invertible in $[-(4 \epsilon^{-1}C)^{-1},(4 \epsilon^{-1}C)^{-1}]$. Now I don't see the arguments but I think it should be possible to bound the radius $r_p$ from below by $(4 \epsilon^{-1}C)^{-1}$.