If a curve has velocity and acceleration of constant magnitude, then its curvature is constant. If a curve has velocity and acceleration of constant magnitude, then its curvature is constant.
I think it's too easy a problem, but I'm not sure my solution is correct:
Let $\alpha (t) = (x (t), y (t)) $ be the curve, having constant velocity and acceleration implies that $ x' (t) = c_1 $, $ y' (t) = c_2 $, $ x'' (t) = c_3 $, $ y'' (t) = c_4 $.
Using the following equation for curvature in which the parametrization does not matter and assuming that the curve is regular:
$$
\kappa=\dfrac{x'y''-x''y'}{(x'^2+y'^2)^{\frac{3}{2}}}=\dfrac{c_1c_4-c_3c_2}{(c_1^2+c_2^2)^{\frac{3}{2}}}.
$$
So with this we conclude that the curvature is constant since the denominator is different from 0.
Is the proof correct?
 A: Hint 1. Show that the angle between $\alpha''$ and $J(\alpha')= (-y',x')$ is constant. Subhint: For this, show that the angle between $\alpha''$ and $\alpha'$ is constant.
Hint 2. Note that
$$
\kappa = \frac{\alpha'' \cdot J\alpha'}{\|\alpha'\|^3}.
$$
A: Here is a solution which works for curves
$\alpha(t) \in \Bbb R^n \tag 1$
for any $n \in \Bbb N$, $n \ge 2$:
We have the velocity of $\alpha(t)$:
$\alpha'(t) = \dfrac{d\alpha(t)}{dt} = \dfrac{ds}{dt}\dfrac{d\alpha(t)}{ds} = \dfrac{ds}{dt}T(t), \tag 2$
$T$ being the unit tangent vector, $\vert T \vert = 1$, whence the acceleration,
$\alpha''(t) = \dfrac{d}{dt} \left (\dfrac{ds}{dt}T(t) \right ) = \dfrac{d^2s}{dt^2} T(t) + \dfrac{ds}{dt} \dfrac{dT}{dt}$
$= \dfrac{d^2s}{dt^2} T(t) + \dfrac{ds}{dt} \dfrac{ds}{dt} \dfrac{dT}{ds} = \dfrac{d^2s}{dt^2} T(t) + \left (\dfrac{ds}{dt} \right )^2 \dfrac{dT}{ds}; \tag 3$
the first Frenet-Serret equation (which extends to any number of dimensions $n$) reads
$\dfrac{dT}{ds} = \kappa N, \tag 4$
and thus (3) becomes
$\alpha''(t) = \dfrac{d^2s}{dt^2} T(t) + \left( \dfrac{ds}{dt} \right )^2 \kappa N; \tag 5$
now by definition
$\dfrac{ds}{dt} = \vert \alpha'(t) \vert, \tag 6$
and we are given that $\vert \alpha'(t) \vert$ is constant, hence so is $ds/dt$, hence
$\dfrac{d^2s}{dt^2} = \dfrac{d}{dt}\dfrac{ds}{dt} = 0, \tag 7$
and thus (5) yields
$\alpha''(t) = \left( \dfrac{ds}{dt} \right )^2 \kappa N; \tag 8$
taking the norm:
$\vert \alpha''(t) \vert = \left( \dfrac{ds}{dt} \right )^2 \kappa \tag 9$
by virtue of the fact that
$\vert N \vert = 1; \tag{10}$
being given the hypothesis that $\vert \alpha''(t) \vert$ is constant, and having seen that so is $ds/dt$, the conclusion that $\kappa$ is constant immediately follows.
