Given that $\beta$ is parametrised by arc length (with strictly positive curvature), show that the curve $\alpha (s) = \beta (s) - s \beta{'}(s)$ is a regular curve. To show that it's regular I have to show that $|| \alpha'(s) || > 0$ for all $s$. To try to do this I found the derivative of $\alpha$ to be $$\alpha'(s) = \beta'(s) - \beta'(s) - s\beta''(s) = -s\beta''(s) = -s\kappa(s)N(s)$$ where $\kappa$ and $N$ are the curvature and normal vector of $\beta$. My problem is now where to go from here. It is never stated whether or not the interval being worked with contains $0$ or not. If it does, clearly $\alpha'(0) = 0$ so that $\alpha$ is not regular.
Is my work up until this point correct, or is the question just wrong? The question was given to me by another student of the same course, so it may have been written down incorrectly.