We have that the curvature of a curve $\gamma (t)$ is given by $K(t)=\|\gamma ''(t)\|$ iff $\|\gamma '(t)\|=1$.
If $\|\gamma '(t)\| \neq 1$, then we find the arclength $s(t)=\int_0^t \|\gamma '(u)\|du=g(t)$, then we solve for $t=g^{-1}(s)$. Then we have that $\gamma (s)=\gamma (g^{-1}(s)) \Rightarrow \|\gamma '(s)\|=1$. So we find the curvature by the formula $K(s)=\|\gamma ''(s)\|$.
When the curvature of a regular curve $\gamma (t)$ is everywhere $>0$, then show that the curvature is a smooth function of $t$.
Could you give me some hints how we could show this?