Studying the family of curves $\beta(s,r) = \alpha(s) + r\,{\bf N}(s)$ I'm reviewing some stuff on plane curves, just because, and I would like to confirm some things. The whole exercise is:

Let $\alpha(s) = (x(s),y(s))$ be a unit-speed parametrized curve, ${\bf N}(s)$ its normal vector and $\kappa(s)$ its curvature. Consider the family of curves $\beta(s,r) = \alpha(s) +r\,{\bf N}(s)$, $-\epsilon \leq r \leq \epsilon$. Check that $\beta(s,r_0)$ and $\beta(s_0, r)$ are regular curves for $\epsilon > 0$ small enough, check that they're orthogonal and that the curvature $\overline{\kappa}$ of $\beta(s,r_0)$ is $\frac{\kappa}{1+r_0\kappa}$.

A direct computation gives: $$\left\|\frac{\partial \beta}{\partial s}(s,r_0)\right\| = |1-r_0\kappa(s)|, \quad \left\|\frac{\partial \beta}{\partial r}(s_0, r)\right\| = 1,$$ so that $\beta(s_0, r)$ is always regular. My first problem is with the first curve. If $|\kappa(s)|$ attained a maximum, say, $|\kappa^\ast|$, then I can take $0 < \epsilon < 1/|\kappa^\ast|$ and from this $|1-r_0\kappa(s)| > 0$ for all $s$.

The exercise was written sloppily and I should assume that $\alpha$ is defined in a closed interval, so that I can obtain this maximum $|\kappa^\ast|$? Or there's a trick around it?

The second part is too easy, no problems. 
For the final part, I think that the formula given is wrong. Because $$\frac{\partial \beta}{\partial s}(s,r_0) = (1-r_0\kappa(s))\,{\bf T}(s), \quad \frac{\partial^2\beta}{\partial s^2}(s,r_0)= -r_0\kappa'(s)\,{\bf T}(s) + (1-r_0\kappa(s))\kappa(s)\,{\bf N}(s)$$will give: $$\det\left(\frac{\partial \beta}{\partial s}(s,r_0),\frac{\partial^2\beta}{\partial s^2}(s,r_0)\right) = (1-r_0\kappa(s))^2\kappa(s)$$ and so: $$\overline{\kappa}(s) = \frac{\kappa(s)}{|1-r_0\kappa(s)|}.$$

Did I miss anything? In this question, for example, they take $\alpha (s)- r\,{\bf N}(s)$ in the beginning, which is coherent with my work above. But I think I have too many absolute values there, too. Is there a way to get rid of them?

 A: Since your curve is parametrized by the arc length the following relations hold: ${\bf N}'(s)= -k(s) {\bf T}(s)$ and $\alpha'(s)={\bf T}(s)$. So for your final part the relation $\frac{\partial \beta}{\partial s}(s,r_0) = (1-r_0\kappa(s))\,{\bf T}(s)$ is correct as well as the second derivative of $\beta$. We can then conclude that the curvature  of $\beta(s)$ is $\overline{\kappa}(s) = \frac{\kappa(s)}{1-r_0\kappa(s)}$. The curvature can effectively be computed by using the cross product  $\beta'(s) \times  \beta''(s)$ and by taking the component in the $z$ direction. Since ${\bf T}(s)$ is a unit vector its module is $1$. When reversing the sign in your original curve, i.e. when considering a parallel curve at a distance $r$ as $\beta(s)=\alpha (s)- r\,{\bf N}(s)$, the corresponding geodesic curvature can be obtained from the previous one by replacing $r$ by $-r$.
Concerning the regularity condition of $\beta(s)$, this is satisfied if $\frac{1}{r}$ is not between the extremal values of the geodesic curvature.
This means that if we define $k_{min}=\min \{\overline {k(s)} \}$ and $k_{max}=\max \{ \overline{k(s)} \}$ then $\beta(s)$ is regular if and only if $\frac 1 r \not \in [ k_{min},k_{max} ]$.
