Distance between evolute of $\alpha$ and $\alpha$ I'm trying to solve this exercise, can someone give me a hint?
Let $ \alpha : I \subset \mathbb{R} \rightarrow \mathbb{R} $ a arc length parametrization of a curve and $\kappa(s)$, the curvature at $\alpha(s)$ is positive and nondecreasing.
Let $\beta$ be $\alpha$ 's evolute.
For $x \in I$ show that
$$
 | \alpha(s) - \beta(x)| \leq \frac{1}{\kappa(x)},
$$ 
for all $s \in I, s \geq x$.

Geometrically it seems right, since $|\alpha(s) - \beta(s)| = \frac{1}{\kappa(s)}$, which means that the distance between $\alpha(s)$ and $\beta(s)$ is the radius $\frac{1}{\kappa(s)}$ of a circle.
And thus, $|\alpha(t) - \beta(t)|$ as $t$ increases, are decreasing radius of circles.
So $|\alpha(s) - \beta(x)|$ is a "diagonal" of a rectangle, with opposite sides of $\frac{1}{\kappa(x)}$ and $\frac{1}{\kappa(s)}$.
 A: Let $\beta(s)=\alpha(s)+\frac{1}{\kappa(s)}n(s)$ be the evolute curve and $t(s)=\alpha'(s)$ the tangent vector. Note that $$\beta'(s)=t(s)+\frac{-\kappa(s)^2t(s)-\kappa'(s)n(s)}{\kappa(s)^2}=\frac{-\kappa'(s)}{\kappa(s)^2}n(s).$$
Therefore, the arclength of the evolute curve between two point $\beta(t_1)$, $\beta(t_2)$ is given by $$\int_{t_1}^{t_2}\frac{\kappa'(s)}{\kappa(s)^2}ds=-\int_{t_1}^{t_2}\left(\frac{1}{\kappa(s)}\right)'ds=R_1-R_2,$$
where $R_1=1/\kappa(t_1)$ and $R_2=1/\kappa(t_2)$. Note that $R_1,R_2$ are the radius of curvature from the osculate circle, hence, as $R_1\ge R_2$, we conclude that the circle with radius $R_2$ is contained in the closed ball, which boundary is the circle with radius $R_1$.
Now, it remains to prove that if the curvature is positive nondeacreasing, and if $C_x$ is the osculate circle at the point $x$, then the curve must enter (or stay) in the the osculate circle $C_x$ for $s\ge x$. You can achieve it by using Taylor formula.
A: It's trivial from the geometric point of view!  
Imagine a circle C. It radius is $1/k(x)$. It center is $\beta (x)$. And our curve L touches C in point $\alpha (x)$. So because curvature of L is positive and nondecreasing L must lie inside circle C for all $s > x$. That is a distance from center of circle C to a next point $\alpha (s)$ of curve L which is equals to
$|\beta(x) - \alpha (s)|$ must be less than circle radius:
$$|\beta(x) - \alpha (s)| \le \frac{1}{k(x)}$$ 
Thus, in language of formulas you get your inequality! That's all. 
