Let $s : (a,b) \rightarrow \mathbb{R^2} $ be a regular parameterized curve ( arc length parameterized)
with curvature $k(x) \neq 0 \forall x\in (a,b) $. We define the evolute of $s$
$e(x) :=s(x)+\frac{1}{k(x)}n(x),\quad x\in (a,b)$
$1)$ Show that the tangent at $x$ of the evolute of $s$ is the normal to $s$ at $x$.
$2)$ Now assume that $k'(x) < 0 \forall x \in(a,b) $ and show that: $$L( e_{[i,j]}) = \frac{1}{k(j)} - \frac{1}{k(i)}$$ for $ a < i \leq j < b $.
For $1)$: I know that we have to show: $\langle e'(x),s'(x)\rangle = 0 $.
$e'(x) = s'(x)-\frac{k'(x)}{k(x)^2}n(x)+\frac{1}{k(x)}n'(x). $ Now I read that I have to use: $\langle s''(x), n(x) \rangle = - \langle s'(x),n'(x) \rangle $ and $s''(x) = k(x)n(x) $ . I have already shown both equations but how can I use them to solve this exercise?
For $2)$ $L( e_{[i,j]}) = \int_{i}^{j} ||e'(x)|| dx = \int_{i}^{j} ||s'(x)-\frac{k'(x)}{k(x)^2}n(x)+\frac{1}{k(x)}n'(x)|| dx $ but what can I do now? We know that $k'(x) < 0 $ and maybe we can use $1)$ ?
1) $(n',s')=(n,s')' - (n,s'')=0-k \Rightarrow n'= -ks'$ so that $$ e'=-\frac{k'}{k^2} n,\ (e',s') =0$$
2) Assume that $k'|[a,b]<0$.
Then $$L(e) =\int\ |e'| =\int\ \frac{-k'}{k^2} = \frac{1}{k(a)} - \frac{1}{k(b)} $$