I would do it like this:
We have
$c(t) = \int_0^t r(u)du, \tag{1}$
whence
$\dot c = \dfrac{dc}{dt} = r(t), \tag{2}$
and since $r(t)$ lies in the sphere of radius $R$, we have
$\Vert r(t) \Vert = R \tag{3}$
for all $t$. This shows that the tangent vector $\dot c$ has constant magnitude $R$:
$\Vert \dot c(t) \Vert = \Vert r(t) \Vert = R, \tag{4}$
holding for all $t$. But $\Vert \dot c \Vert$ is the rate of change of arc-length $s$ along $c(t)$ with respect to the parameter $t$:
$\dfrac{ds}{dt} = \Vert \dot c(t) \Vert = R, \tag{5}$
and (5) implies
$\dfrac{dt}{ds} = \dfrac{1}{R} \tag{6}$
along $c(t)$ as well. By (2) and (3) we see that the unit tangent field $\mathbf t$ to $c(t)$ is
$\mathbf t = \dfrac{\dot c}{R} = \dfrac{r(t)}{R}; \tag{7}$
the curvature $\kappa$ of $c(t)$ is thus given by the Frenet-Serret equation
$\dfrac{d\mathbf t}{ds} = \kappa \mathbf n, \tag{8}$
where $\Vert \mathbf n \Vert = 1$. From (6)-(8), using the chain rule for derivatives,
$\kappa \mathbf n = \dfrac{d\mathbf t}{ds} = \dfrac{dt}{ds} \dfrac{d\mathbf t}{dt} = \dfrac{1}{R}\dfrac{\dot r}{R} = \dfrac{\dot r}{R^2}. \tag{9}$
Since $r(t)$ is a unit speed curve, $\Vert \dot r \Vert = 1$, so taking the norm of each side of (9) yields
$\kappa = \dfrac{1}{R^2}, \tag{10}$
that is, the curvature of $c(t)$ is $R^{-2}$. QED!!!
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Suppose $(T,N,B)$ is a given Frenet-Serret frame for the curve and suppose $T\cdot u=\cos(\theta)$ for some constant vector $u$. It is sufficient to show $\tau/\kappa=\cot(\theta)$.
Differentiating $T\cdot u=\cos(\theta)$ yields $N\cdot u=0$. This implies $u=\cos(\theta)T+\sin(\theta)B$ because we can assume $u$ has unit length.
Differentiate this equation to obtain $0=\kappa cos(\theta)N-\tau \sin(\theta)N$ and so $\tau/\kappa=\cot(\theta)$.
To show the converse, first find a $\theta$ such that $\tau/\kappa=\cot(\theta)$ and work backwards through the proof above.
Best Answer
The shape of the trajectory does not depend on the speed, so it needn't be unit. (Think that tough you can drive at different speeds, the road remains unchanged. :)
The curvature and torsion formulas are established by computing the curvilinear abscissa, which "normalizes the speed away".