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Let $\alpha : I \rightarrow \mathbb{R^2}$ be a curve parametrized by arc length. Show that all normal lines of $\alpha$ are equidistant from a fixed point if and only if there exist numbers $a,b \in \mathbb{R}$ such that $k(s) = \pm \cfrac{1}{\sqrt{as+b}}$ $\forall s \in I$, where $k(s)$ denotes the curvature of $\alpha$ at the point $s$.

I'm kinda lost on this one and don't know where to start or how to attack the problem so any hints or ideas would be greatly appreciated.

Ernie060
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3 Answers3

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(1) If $c(t)$ has unit speed, then normal line is $$ c(t)+sn(t)$$ Assume that fixed point is origin. So the distance square is $$d(t):=|c(t) -c(t)\cdot nn |^2 = |c|^2-(c\cdot n)^2 $$ by considering critical point of $f(s)=|c(t)+sn |^2$ Since $d$ is independent of $t$, then $$ n'=-kc'\Rightarrow c\cdot c' (1+c\cdot c'') =0 $$ If $c\cdot c'=0$, then $|c|$ is constant so that $k$ is constant. If $c\cdot c'' =-1$, then $$ d= |c + \frac{n}{k}|^2 $$ Hence $\frac{1}{k^2} =|c|^2+ C_1$ And $(c\cdot c')'=0\Rightarrow |c|^2= C_2t+ C_3 $ This complete the proof.

(2) Assume that $k(s)= \pm \frac{1}{\sqrt{as+b}} \ \ast$ Note that by above a curve whose normal lines have equidistance from a fixed point satisfies $\ast$ Then the proof is completed by fundamental theorem of the local theory of curves.

HK Lee
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The envelope of the normals to the curves (C) we are looking after is thus a circle or an arc of a circle, centered on the given point. The envelope of the normals to a curve is called its evolute. As we are working the other way, the curves (C) we are looking for are the inverse of an evolute, i.e., an involute, i.e., we are looking for curves (C) that are...

"Circle involutes", which are classical curves (used in a certain number of devices, in particular gears):

Link

Their simplified intrinsic (or Cesaro) equation are known to be: $\rho^2=2 a s.$

http://mathworld.wolfram.com/CesaroEquation.html

This equation matches yours, taking $k=1/\rho$.

The presence of the additive constant $b$ account for the arbitrary origin of the arc length.

Glorfindel
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Jean Marie
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We can rewrite

$$ k(s) = \cfrac {d \varphi}{ds} = 1/R(s) = \pm \frac{1}{\sqrt{as+b}} \tag{1} $$

into standard form as

$$ R^2 = R_i^2 + 2 a s \tag{2} $$

which is the natural/intrinsic Cesàro equation of an involute formed by unwinding a taut tangent (always tangential normal lines of $\alpha$ ) arc length $s$ of radius of curvature $1/R$ on a circular spool radius $a$.It has period length of arc spool circumference and constant width $2 \pi a$. Center of spool is the fixed point.Arc length is proportional to rotation $ \varphi $.

InvoluteMW

For special case at start $s=0, R=R_i $.It satisfies a very simple relation

$ s^2 + a^2 = r^2 \tag{3}$

where $r$ is polar coordinate radius, and from which (1),(2) can be derived back.

Narasimham
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