Normal unit vector

Question

I am looking at the following exercise:

I have done the following about the second part, about the signed curvature of $\iota$ :

The signed curvature of $\gamma$ is different from the signed curvature of $\iota$, right?

So, let $\kappa_s$ be the signed curvature of $\gamma$ and $\kappa_{s, \iota}$ the signed curvature of $\iota$.

We have $$\iota '(s)=\gamma '(s)-\gamma '(s)+(l-s)\gamma ''(s) \Rightarrow \iota '(s)=(l-s)\gamma '' (s) \tag{1}$$

We define as the unit tangent vector $\textbf{t}$, the signed unit normal vector $\textbf{n}_{s, \iota}$ and the signed curvature $\kappa_{s, \iota}$ of $\iota$ the corresponding quantities of the unit speed reparametrization of $\tilde{\iota}(a)$, where $a$ is the arc length of $\iota$. So, $$\textbf{t}=\frac{\iota ' (s)}{\|\iota ' (s)\|}=\frac{\iota '(s)}{a'(s)}$$ Therefore, $$\iota '(s)=a'(s)\textbf{t}$$

We have that if $\iota$ is a unit-speed plane curve, then $$\textbf{n}_{s, \iota } ' =-\kappa_{s, \iota} \iota'$$

Generalizing this formula for a regular curve, not necessarily unit-speed, we have $$\textbf{n}_{s,\iota } ' =-\kappa_{s,\iota}a'(s)\textbf{t} \tag{2}$$

$$(1) \Rightarrow a'(s)\textbf{t}=(l-s)\gamma ''(s) \Rightarrow a'(s) \textbf{t}=(l-s)\kappa_s \textbf{n}_s$$

$$(2) \Rightarrow \textbf{n}_{s, \iota } '=-\kappa_{s, \iota} (l-s)\kappa_s \textbf{n}_s$$

Which is the relation between the normal unit vector of the curve and normal unit vector of the involute of the curve?

Answer 1

Yummy. A bounty. :)

We have: $$\iota=\gamma+(l-s)\gamma'$$ $$\dot\iota=(l-s)\gamma''$$ $$\ddot\iota=-\gamma'' + (l-s)\gamma'''$$

Since it is given that $\kappa_s > 0$, it follows that the curve bends continuously in the counter clock direction, meaning $n_s=n$ and $\kappa_s = \kappa$. Since the string is wound around the curve, we can conclude that $\iota$ is also counter clock wise, meaning $n_{\iota,s} = n_\iota$ and $\kappa_{\iota,s} = \kappa_\iota$.

With the general formula for $\kappa_\iota$ we get: $$\kappa_\iota = \frac{\|\dot\iota \times \ddot\iota\|}{\|\dot\iota\|^3} = \frac{\|(l-s)\gamma'' \times (-\gamma'' + (l-s)\gamma''')\|}{\|(l-s)\gamma''||^3} = \frac{\|\gamma'' \times \gamma'''\|}{(l-s)\kappa^3} = \frac{\|\kappa n \times (\kappa'n + \kappa n')\|}{(l-s)\kappa^3} = \frac{\|n \times n'\|}{(l-s)\kappa} = \frac{\|n \times -\kappa t\|}{(l-s)\kappa} = \frac{1}{l-s} $$

Btw, the unit vectors are perpendicular.

Answer 2

It looks like you closed the bounty early. Is that legal? Well, I'll show you what I did for this anyway.

For convenience I'll use $g$ in place of $\gamma$ and $h$ for the involute of $g.$ I'll often suppress the arc-length parameter $s.$ Some observations:

i)$h' = (l-s)g''$

ii) $|g''|' = (g''\cdot g''')/|g''|$

iii) $g''' = -|g''|^2g' + cg''.$

iii) deserves some explanation. We know $g',g''$ are orthogonal, so $g'\cdot g'' \equiv 0.$ Differentiate that to get $|g''|^2 + g'\cdot g''' \equiv 0.$ Now $g'$ is a unit vector. So resolving $g'''$ as an orthogonal sum with respect to $g',g''$ gives iii), where $c$ is a function of $s$ that we don't need too much information about.

Denote the unit tangent vector of $h$ by $T.$ By i) we have $T= g''/|g''|.$ Using ii) and iii) we get

$$\tag 1 \frac{dT}{ds} = \frac{|g''|g''' - g''|g''|'}{|g''|^2} = -|g''|g'.$$

Let $s_1$ denote arc length along the curve $h.$ Then $d s_1/ ds = (l-s)|g''(s)|.$ Using this and $(1),$ we get

$$\frac{dT}{ds_1} = \frac{dT/ds}{ds_1/ds } = \frac{-|g''|g'}{(l-s)|g''(s)|} = \frac{-1}{l-s}.$$

So the curvature along the involute $h$ is $1/(l-s).$ How do we know this is the signed curvature? That's easy: $g''$ is a positive multiple of $ig'$ (thinking of $g',g''$ as complex numbers), and the same relation therefore holds for $dT/d s_1$ and $T$ because the first is a negative mulitple of $g'$ and the second is a positive multiple of $g''.$