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?
Best Answer
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.