Curvature of a regular curve is a smooth function of parameter if it does not vanish

Question

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?

Answer 1

If a vector, $v$, is smooth and does not vanish, then $$ \left|v\right|=\sqrt{v\cdot v}\tag{1} $$ is also smooth since $\sqrt{x}$ is smooth away from $0$.

The curvature is the length of the vector $$ \frac{\gamma'\times\gamma''}{\left|\gamma'\right|^3}\tag{2} $$ which is smooth because $\gamma$ is smooth and $\gamma'\ne0$.

Therefore, if it does not vanish, the absolute value of the vector in $(2)$ is smooth. That is $$ \kappa=\frac{\left|\gamma'\times\gamma''\right|}{\left|\gamma'\right|^3}\tag{3} $$ is smooth. Thus, if neither $\gamma'$ nor $\kappa$ vanish, $\kappa$ is smooth.

Answer 2

Hint: Show first that if the curve is smooth, then the arc length parametrization is also smooth. Hence, assume the smooth curve satisfies $\|\dot{\gamma}\|=1.$ The point is that the map $\ddot{\gamma}\mapsto\|\ddot{\gamma}\|$ is smooth whenever $\ddot{\gamma}$ does not vanish.