Problem: Show that if $\kappa(s) = 0$ for all $s$, then the curve $\mathbf{r} = \mathbf{r}(s)$ is a straight line. (Here, $\kappa$ represents curvature.)
Attempt at solution: If $\kappa(s) = 0$ for all $s$, then \begin{align*} \frac{d\hat{T}}{ds} = \kappa \hat{N} = 0, \end{align*} where $\hat{N}$ is the unit principal normal. This means, however, that $\hat{T}(s)$ is constant.
So can I now just conclude that $\mathbf{r}$ is a straight line?
Best Answer
You have argued with the normal vector ${\bf n}$, but this vector is only defined in points where $\kappa\ne0$. This brings us into a vicious circle. Argue as follows instead:
If the curve $\gamma:\>s\mapsto{\bf r}(s)$ is parametrized with respect to arc length then by definition $\kappa(s):=|\ddot{\bf r}(s)|$. The assumption $\kappa(s)\equiv0$ then implies $\ddot{\bf r}(s)\equiv{\bf 0}$, and this leads to ${\bf r}(s)={\bf a}+s {\bf u}$, where ${\bf u}$ is a unit vector.