3
$\begingroup$

I am looking at the following exercise:

enter image description here

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?

$\endgroup$
3
+150
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ How did you get the last equality of the first line at the calculation of $\kappa_{\iota}$? Isn't it as follows? $$\frac{||(l-s)\gamma'' \times (-\gamma'' +(l-s)\gamma''')||}{||(l-s)\gamma''||^3}=\frac{||(l-s)\gamma'' \times (-1)\gamma''+(l-s)\gamma'' \times (l-s)\gamma''')||}{(l-s)||\gamma''||^3}\\ =\frac{||-(l-s)\gamma'' \times \gamma'' +(l-s)^2\gamma'' \times \gamma''')||}{(l-s)||\gamma''||^3}=\frac{||(l-s)^2\gamma'' \times \gamma''')||}{(l-s)||\gamma''||^3}\\ =\frac{(l-s)^2||\gamma'' \times \gamma''')||}{(l-s)||\gamma''||^3}=\frac{(l-s)||\gamma'' \times \gamma''')||}{||\gamma''||^3}$$ $\endgroup$ – Mary Star Nov 10 '15 at 18:57
  • 2
    $\begingroup$ Almost. In your first line we've lost a power of 3 for $(l-s)$. :eek: Oh, and let's assume that $l\ge s$. Furthermore, $||\gamma''|| = \kappa$. $\endgroup$ – Klaas van Aarsen Nov 10 '15 at 20:09
  • 2
    $\begingroup$ I see... Now I don't understand the last two equalities... We use the Frenet-Serret equalities, right? It as follows, right? $$\frac{|| n \times n'||}{(l-s)\kappa}=\frac{|| n \times (-\kappa t+\tau b)||}{(l-s)\kappa}=\frac{||(-\kappa n \times t+\tau n \times b)||}{(l-s)\kappa}$$ How do we continue? $\endgroup$ – Mary Star Nov 10 '15 at 21:52
  • 2
    $\begingroup$ Since we're in $\mathbb R^2$, $\tau=0$. And $||n\times t||$ is always 1. $\endgroup$ – Klaas van Aarsen Nov 10 '15 at 23:38
  • 2
    $\begingroup$ Is $n$ the signed unit normal vector or the unit normal vector? $\endgroup$ – Mary Star Nov 11 '15 at 0:17
0
$\begingroup$

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 $g''.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.