Let $\alpha(s)$ be a plane Frenet curve such that the angle $\theta$ between $\alpha(s)$ and its tangent vector $e_1(s)$ is constant. Show that either $\kappa = 0$ or
$$\kappa(s) = \frac{1}{\cot \theta + c}$$
where $c$ is a constant.
Hint: It holds that
$$\big< \alpha, e_1 \big> = |\alpha| \cdot \cos \theta$$
as well as
$$\big< \alpha, e_2 \big> = \pm |\alpha| \cdot \sin \theta$$
where $\dot{e}_1(s) = \kappa(s) e_2(s)$. Differentiate the first equation …
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I tried solving this by using the hint given but I always ended up with a term containing $| \alpha |$, which I cannot get rid of. I already considered that maybe $| \alpha|$ is the constant $c$ and I therefore do not even have to get rid of it. But then again I cannot assume $|\alpha|$ to be constant a-priori although It may be (as for e.g. $\theta = \pi/2$).
I humbly ask for your help.
Best Answer
This problem is just wrong. The only planar curves with constant curvature are circles, and only when the circle is centered at the origin will the angle between $\alpha$ and $e_1$ be constant (as you surmised, $\pi/2$). So $\cot\theta$ must be $0$ and $\kappa=1/c$.
I tried solving this the way I usually approach all such problems. I wrote the arclength parametrized curve $\alpha = \lambda e_1 + \mu e_2$ for some functions $\lambda$ and $\mu$, differentiated and applied the Frenet equations. The hypothesis in the problem tells us that $\lambda = \pm\cot\theta \mu$ (or $\mu=0$, which leads to $\kappa=0$). Assuming $\kappa\ne 0$, we get the equations \begin{alignat*}{3} \kappa\cot\theta&\mu + &\mu' &= 0 \\ -\kappa&\mu + \cot\theta&\mu' &= 1, \end{alignat*} and so $\mu = \dfrac{-1}{\kappa\csc^2\theta}=\dfrac{-\sin^2\theta}{\kappa}$ and $\mu' = \dfrac{\kappa\cot\theta}{\kappa\csc^2\theta} = \cos\theta\sin\theta$. This tells us that $\mu(s)=(\cos\theta\sin\theta)s + c$, and so $$\kappa(s) = \frac{-\sin^2\theta}{\mu(s)} = \frac{-\sin^2\theta}{(\cos\theta\sin\theta) s + c},$$ very different from the statement of the problem. By the way, since $\kappa>0$, we should probably have opted for the negative sign choice.