Geometric interpretation of $\beta_r(s) = \alpha(s) + r\mathbf{n}(s)$ Let $\alpha(s)$ be a smooth curve parameterised by arc-length and for fixed $r > 0$ define $\beta_r(s) = \alpha(s) + r\mathbf{n}(s)$, where $\mathbf{n}(s)$ is the unit normal vector to $\alpha$ at $s$.  What is the geometric interpretation of the result:
$$\left(\frac{d}{ds}\beta_r(s)\right)\cdot\mathbf{t}(s) = 0 \iff r = \frac{1}{\kappa(s)}$$
This question is from a course I am studying and isn't homework.  I have proved the result, but I am struggling to interpret it.  I can see that $\beta_r(s)$ would be the equation of the normal line at $s$ if $r$ was a parameter.  But $r$ being fixed has me somewhat confused, particularly since the result involves $r$ taking a value that is a function of the parameter $s$.  As far as I can make out, the result says that the direction of change of the normal (?) is perpendicular to the tangent vector when $r$ is one on the curvature.
 A: The parameter $r$ is fixed for $s$ fixed. When we change $s$ around, we also change $r$. If you plug $1/k(s)$ into the equation, you get $$\beta(s)=\alpha(s)+\frac{1}{k(s)}\mathbf n(s).$$ To begin thinking about the geometric interpretation, think about the term $1/k(s)$ and about circles. Intuitively, when a circumference is big, it has low curvature; when it is tiny, it has a very high curvature. It is an exercise to you to show that the curvature of a circle of radius $R$ (that's why they used $r$ in first place) is constant and equals $1/k=1/k(s)$.
Draw a part of $\alpha$, mark the point $\alpha(s)$ and draw the vector $\mathbf n(s)$. Draw a circle of radius $R=\left|\frac{\mathbf n}{k(s)}\right|$ and center $\beta(s)=\alpha(s)+\frac{1}{k(s)}\mathbf n$. When we study the curve $\beta$, we are studying the movement of the center of this circle for various inputs $s$.
At this point, you can interpret the term $\beta '$ and the tangent $\mathbf t$ geometrically. The former shows a tendency of the center of moving in the $\mathbf n$ direction; and $\beta \cdot \mathbf t=0$ is a consequence of the normal vector and the tangent being perpendicular.
The curve $\beta$ has a very special name: it is called an evolute of $\alpha$. As far as I know, the property $\beta'\cdot \mathbf t=0$ is not geometrically trivial, I don't know how to show it using only geometry. Some good using of Frenet equations can show it, though.
