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I was trying to study for an exam in differential geometry and got stuck on the following problem : Determine all the planar curves $\alpha(s)$ parametrized by arc length, such that the angle between $\alpha$ and $\alpha'$ is a constant $0<\theta<\pi$. My attempt : The curve is planar so the torsion is zero. What I had in mind is trying to find the curvature using the Frenet equation and then sole for such a curve in the xy plane and use the fact that any such curve is obtained by translation/rotation. Any hints? Thanks!

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Hint: We have:$$\cos \measuredangle(\alpha(s),\alpha'(s)) = \frac{\langle \alpha(s),\alpha'(s)\rangle}{\|\alpha(s)\|},$$and if the angle is constant, so is its cosine above. Try to work with this, differentiating this (multiplying both sides by $\|\alpha(s)\|$ before, it will be easier I think) and using $\tau \equiv 0$ as needed.

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  • $\begingroup$ Thanks for your reply! I tried this, but it seems that I did it wrong. Differentiating the inner product yield $<\alpha,\alpha''>+<\alpha',\alpha'>=0+1=1$. Differentiating $\Vert \alpha \Vert\cos \theta$ yields $\frac{<\alpha,\alpha'>}{\Vert \alpha \Vert } \cos\theta=\cos ^{2} \theta$ $\endgroup$
    – mich95
    Apr 11 '15 at 0:12
  • $\begingroup$ That cosine is constant, it will vanish. What I meant to do is: $$\langle \alpha, {\bf T}\rangle = c \sqrt{\langle \alpha,\alpha\rangle} \implies 1+ \kappa \langle \alpha, {\bf T}\rangle = c \frac{\langle \alpha, {\bf T}\rangle}{\sqrt{\langle \alpha, \alpha \rangle}} = c^2,$$ etc. You might have to differentiate more times to get something. $\endgroup$
    – Ivo Terek
    Apr 11 '15 at 0:30

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