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We have to prove that a curve has constant curvature $\kappa = 1/r$ if and only if it is in a circular arc of radius $r$.

I am confused because doesn't a helix also have a constant curvature given by $\frac{a}{a^2 + b^2}$ where $a$ is the radius of the circle and $b$ is the rate of ascension? I feel like an additional assumption here is needed (such as that the curve is planar, thus torsion $\tau = 0$).

Indeed, using the assumption $\tau = 0$ and Frenet-Serret I found a differential equation involving the Normal vector $N$ with a trigonometric solution. I wasn't sure what to do from here, however.

Edit: The question definitely asks for curves (doesn't specify plane curve) so I'll ask the TA tomorrow. From now assume that it wants only planar curves, so $\tau = 0$. Can somebody help me with that solution?

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  • $\begingroup$ You are correct. Perhaps the problem statement refers to a plane curve in the first place? $\endgroup$ – Travis Willse Nov 18 '14 at 6:37
  • $\begingroup$ Maybe the question is about plane curves? Otherwise, you're right, the helix has also constant curvature. $\endgroup$ – Taladris Nov 18 '14 at 6:38
  • $\begingroup$ The question definitely asks for curves in general (not specifying plane curves), so I'll ask the TA tomorrow and for now I'll assume it only wants plane curves. Can somebody help me with that solution? $\endgroup$ – MCT Nov 18 '14 at 6:44
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Let this curve be given in a parametrized form as $\vec{r}(t)=(x(t),y(t))=x(t)i+y(t)j$. By definition the curvature of this curve is given by $$\kappa=\frac{||\vec{r}'(t)\times\vec{r}''(t)||}{||\vec{r}'(t)||^3}$$ Plugging in $\vec{r}'(t)=x'(t)i+y'(t)j$ and $\vec{r}''(t)=x''(t)i+y''(t)j$ yields $$\kappa=\frac{||\vec{r}'(t)\times\vec{r}''(t)||}{||\vec{r}'(t)||^3}=\frac{||(x'(t)i+y'(t)j)\times(x''(t)i+y''(t)j)||}{||x'(t)i+y'(t)j||^3}=\frac{||x'(t)y''(t)k-x''(t)y'(t)k||}{||x'(t)i+y'(t)j||^3}$$ The last equality is equivalent to $$\kappa=\frac{|x'(t)y''(t)-x''(t)y'(t)|}{((x'(t))^2+(y'(t))^2)^{3/2}}\Leftrightarrow \kappa^2=\frac{(x'(t)y''(t)-x''(t)y'(t))^2}{((x'(t))^2+(y'(t))^2)^{3}}$$ Now let the parametrized curve be a circle of radius $r$ then $$x^2(t)+y^2(t)=r^2\Rightarrow x'(t)x(t)+y'(t)y(t)=0\Rightarrow (x'(t))^2+(y'(t))^2=-(x''(t)x(t)+y''(t)y(t))$$ Therefore $$\kappa^2=\frac{(x'(t)y''(t)-x''(t)y'(t))^2}{-(x''(t)x(t)+y''(t)y(t))^3}$$ having in mind that $$x'(t)x(t)+y'(t)y(t)=0\Rightarrow \frac{x'(t)}{y'(t)}=-\frac{y(t)}{x(t)}$$ so $$\kappa^2=\frac{(x'(t)y''(t)-x''(t)y'(t))^2}{-(x''(t)x(t)+y''(t)y(t))^3}=-(\frac{y'(t)}{x(t)})^2\cdot\frac{(x''(t)x(t)+y''(t)y(t))^2}{(x''(t)x(t)+y''(t)y(t))^3}$$ $$\Rightarrow\kappa^2=-(\frac{y'(t)}{x(t)})^2\cdot\frac{1}{x''(t)x(t)+y''(t)y(t)}$$ $$\Rightarrow \kappa^2=(\frac{y'(t)}{x(t)})^2\cdot\frac{1}{(x'(t))^2+(y'(t))^2}=\frac{1}{x^2(t)}\cdot\frac{1}{(\frac{x'(t)}{y'(t)})^2+1}$$ $$\Rightarrow \kappa^2=\frac{1}{x^2(t)}\cdot\frac{1}{(\frac{x(t)}{y(t)})^2+1}=\frac{1}{x^2(t)}\cdot\frac{x^2(t)}{x^2(t)+y^2(t)}=\frac{1}{x^2(t)+y^2(t)}=\frac{1}{r^2}$$ $$\Rightarrow \kappa=\frac{1}{|r|}=constant$$ For the other direction of the statement one can assume that $\kappa$ is constant then it follows
$$\kappa^2=\frac{(x'(t)y''(t)-x''(t)y'(t))^2}{((x'(t))^2+(y'(t))^2)^{3}}$$ is constant. Denote by $z(t)=\frac{y'(t)}{x'(t)}$ then $$\kappa^2=\frac{1}{(x'(t))^2}\cdot\frac{(z'(t))^2}{(z^2(t)+1)^3}$$ Now we have two possibilities $$\kappa=\frac{1}{x'(t)}\cdot\frac{z'(t)}{(z^2(t)+1)^{3/2}}$$ or $$\kappa=-\frac{1}{x'(t)}\cdot\frac{z'(t)}{(z^2(t)+1)^{3/2}}$$ Lets deal with the first case (the other follows the same technique) $$\kappa=\frac{1}{x'(t)}\cdot\frac{z'(t)}{(z^2(t)+1)^{3/2}}\Rightarrow \kappa\int x'(t)\,dt=\int \frac{z'(t)}{(z^2(t)+1)^{3/2}}\,dt $$ $$\Rightarrow \kappa x(t)=\frac{z(t)}{\sqrt{z^2(t)+1}}+c$$ $$(\kappa x(t)-c)^2=\frac{z^2(t)}{z^2(t)+1}\Rightarrow z^2(t)=\frac{(\kappa x(t)-c)^2}{1-(\kappa x(t)-c)^2}\Rightarrow z(t)=\frac{|\kappa x(t)-c|}{\sqrt{1-(\kappa x(t)-c)^2}}$$ Therefore $$y'(t)=\frac{|\kappa x(t)-c|}{\sqrt{1-(\kappa x(t)-c)^2}}\cdot x'(t)\Rightarrow \int y'(t)\,dt=\int \frac{|\kappa x(t)-c|}{\sqrt{1-(\kappa x(t)-c)^2}}\cdot x'(t)\,dt $$ assume now $\kappa x(t)-c\geq 0$ then $$y(t)=\frac{1}{\kappa}\int \frac{u}{\sqrt{1-u^2}}\cdot \,du $$ where $u=\kappa x(t)-c$. After integration $$y(t)=-\frac{1}{\kappa}\sqrt{1-u^2}+c_1\Rightarrow (y(t)-c_1)^2+\frac{u^2}{\kappa^2}=\frac{1}{\kappa^2}$$ Substitute back $u=\kappa x(t)-c$ to get $$(y(t)-c_1)^2+(x(t)-\frac{c}{\kappa})^2=\frac{1}{\kappa^2}$$ This is the equation of a circle.

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For a plane curve $$\gamma:\quad s\mapsto{\bf z}(s)=\bigl(x(s),y(s)\bigr)$$ parametrized by arc length one has $$\dot{\bf z}(s)=\bigl(\cos\theta(s),\sin\theta(s)\bigr)\ ,\tag{1}$$ where $\theta(s)$ denotes the argument of $\dot{\bf z}(s)$. The curvature $\kappa(s)$ is then given by $$\kappa(s):=\dot\theta(s)\ .\tag{2}$$ That a circular arc of radius $r>0$ has $$\kappa(s):=\dot\theta(s)\equiv{1\over r}$$ follows from elementary geometric considerations.

Conversely: Assume that an $r>0$ is given, and that $$\kappa(s)\equiv{1\over r}\qquad(-a<s<a)\ .\tag{3}$$ Since these data do not determine the exact location of $\gamma\subset{\mathbb R}^2$ we may assume that $${\bf z}(0)=(r,0),\quad\theta(0)={\pi\over2}\ .\tag{4}$$ Using $(2)$ and $(3)$ we then obtain $$\theta(s)={s\over r}+{\pi\over2}\ .$$ Plugging this into $(1)$ we get $$\dot x(s)=\cos\theta(s)=-\sin{s\over r},\qquad \dot y(s)=\sin\theta(s)=\cos{s\over r}\ .$$ On account of the initial conditions $(4)$ this leads to $$x(s)=r\cos{s\over r},\quad y(r)=r\sin{s\over r}\qquad(-a<s<a)\ ,$$ which is obviously an arc of radius $r$ centered at $(0,0)$.

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