Let $a(s)$ be an arc paramterized plane curve. The goal of this problem is to show that that the unique evolute for $a(s)$ that is in the same plane as $a(s)$ is given by:
$b(s)= a(s) + \frac{N}{\kappa}$
where $N$ is the unit normal vector for $a(s)$ and $\kappa$ is the curvature.
Since $b(s)$ is the evolute of $a(s)$, $a(s)$ is the involute of $b(s)$. I showed earlier that this means that $a(s)$ is of the form:
$a(s)= \frac{b'(s)}{|b'(s)|}(c-s)+b(s)$
Also:
$a'(s) \cdot b'(s)=0$.
where $b'(s)$ is given by:
$b'(s)=a'(s) + \frac{N'}{\kappa}-\frac{N\kappa'}{\kappa^2}$,
Anyway, I would appreciate some insight into this. How do i show uniqueness?
So:
$a'(s) \cdot b'(s)=a'(s)\cdot a'(s)+a'(s) \cdot \frac{N'}{\kappa}- a'(s) \cdot \frac{N\kappa'}{\kappa^2}$
= $1+a'(s) \cdot \frac{(-\kappa a'(s))}{\kappa}+0$
= $1-1=0$
so $b(s)$ appears be an evolute of $a(s)$... but how do we show uniqueness? and what gaurentees that $b(s)$ is in the same plane as $a(s)$?
Let $b$ be a evolute lying in the plane of $a$. Since $a' \cdot b' = T \cdot b' = 0$, by the fact that $T$ and $N$ are perpendicular and $b$ lies in $span\{T, N\}$, $b'$ and $N$ are parallel.
Say $b - a = \mu N$ for some scalar function $\mu(s)$, differentiate both side we obtain $$b' - a' = b' - T = \mu' N + \mu (-\kappa T)$$
Again, since $b'$ and $N$ are parallel and $\{T, N\}$ is linearly independent, we know $\mu = 1 / \kappa$.
Hence, any $b$ should suffice that $b = a + \frac{N}{\kappa}$.