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

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?

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.

• At which point do we use the fact that the curvature is $>0$ everywhere? Oct 5, 2015 at 20:33
• We use the fact that $\kappa\gt0$ everywhere when we say that it does not vanish. As defined in $(3)$, $\kappa\ge0$. In two dimensions, there is the idea of left and right curvature, and if $\kappa\ne0$ and is smooth, then the Intermediate Value Theorem says we need either $\kappa\gt0$ always or $\kappa\lt0$ always.
– robjohn
Oct 5, 2015 at 20:45
• I see... When we know that $\gamma '$ and $\gamma ''$ are smooth, why is $\gamma ' \times \gamma ''$ also smooth? Oct 8, 2015 at 15:32
• Because the product of smooth functions is smooth: $(fg)'=f'g+fg'$.
– robjohn
Oct 8, 2015 at 15:35
• When we hadn't the restriction $\kappa >0$ to find an example at which $\kappa$ isn't smooth, do we have to find a curve $\gamma$ such that $\gamma ' \times \gamma ''=0$ ? Oct 8, 2015 at 17:06

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.

• That the curve is smooth means that $\gamma '(t) \neq 0$ and so $||\gamma '(t)||\neq 0$, right? Then we have that $s(t)=\int_0^t ||\gamma (u)||du \neq 0$. Can we show in this way that the arc length parametrization is also smooth? Oct 1, 2015 at 17:50
• @MaryStar Yes, this is the right way. Oct 1, 2015 at 17:53
• So we have to show that $\gamma '(s)\neq 0$, or not? How can we get this result? Oct 1, 2015 at 19:17
• Do we have to apply the chain rule? $$\frac{d\gamma}{ds}=\frac{d\gamma}{dt}\frac{dt}{ds}$$ Is this correct? Oct 3, 2015 at 3:20
• @MaryStar Actually, if $\gamma(s)$ is an arc length parametrization, then $\|\dot{\gamma}\|=1$, and in particular, $\dot{\gamma}\neq0$. Oct 4, 2015 at 11:01