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!!!
Well, we have
$(\alpha - c) \cdot (\alpha - c) = r^2, \tag 1$
since $\alpha$ lies on the sphere of radius $r$ centered at $c$; this is just what equation (1) affirms. If we differentiate (1) with respect to $s$, the arc-length along $\alpha$, we obtain
$2\alpha' \cdot (\alpha - c) = \alpha' \cdot (\alpha - c) + (\alpha - c) \cdot \alpha' = 0, \tag 2$
whence
$\alpha' \cdot (\alpha - c) = 0; \tag 3$
since $\alpha$ is a unit-speed curve, we have
$\alpha' = T, \; \Vert T \Vert = 1, \tag 4$
the unit tangent vector to $\alpha$; thus (3) becomes
$T \cdot (\alpha - c) = 0, \tag 5$
which is nearly self-evident, since $T$ is tangent to the sphere (1), hence normal to the radial vector $\alpha - c$; in any event, we may differentiate (5) once again to obtain
$T' \cdot (\alpha - c) + T \cdot T = T' \cdot (\alpha - c) + T \cdot \alpha' = 0, \tag 6$
or
$T' \cdot (\alpha -c) + \Vert T \Vert^2 = 0; \tag 7$
we now use the Frenet-Serret equation
$T' = \kappa N, \tag 8$
where $N$ is the normal vector to $\alpha$, and (4) to re-write (7) as
$\kappa N \cdot (\alpha - c) + 1 = 0, \tag 9$
which since $\kappa > 0$ implies
$N \cdot (\alpha - c) = -\dfrac{1}{\kappa}; \tag {10}$
we may now differentiate (9) with respect to $s$ to find
$\kappa' N \cdot (\alpha - c) + \kappa N' \cdot (\alpha - c) + kN \cdot \alpha'= 0; \tag{11}$
we have
$N \cdot \alpha' = N \cdot T = 0, \tag{12}$
and also the Frenet-Serret equation
$N' = -\kappa T + \tau B, \tag{13}$
where $\tau$ is the torsion and $B = T \times N$ the binormal vector of $\alpha$; when (10), (12) and (13) are substituted into (11) we obtain
$-\dfrac{\kappa'}{\kappa} + \kappa (-\kappa T + \tau B) \cdot (\alpha - c) = 0, \tag{14}$
or
$-\dfrac{\kappa'}{\kappa} -\kappa^2 T \cdot (\alpha - c) + \kappa \tau B \cdot (\alpha - c) = 0, \tag{15}$
which by virtue of (5) reduces to
$-\dfrac{\kappa'}{\kappa} + \kappa \tau B \cdot (\alpha - c) = 0, \tag{16}$
and since $\kappa > 0 \ne \tau$ we have
$-\dfrac{\kappa'}{\kappa^2 \tau} + B \cdot (\alpha - c) = 0, \tag{17}$
whence
$B \cdot (\alpha - c) = \dfrac{\kappa'}{\kappa^2 \tau} = -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau}; \tag {18}$
(5), (10) and (18) express the components of the radial vector $\alpha - c$ in the orthonormal frame $T$, $N$, $B$, whence
$\alpha - c = -\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B, \tag{19}$
the requisite result.
We may find $r$ in terms of $\kappa$ and $\tau$ by inserting (19) into (1):
$r^2 = (\alpha - c) \cdot (\alpha - c)$
$= \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ) \cdot \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ); \tag{20}$
since
$\Vert N \Vert = \Vert B \Vert = 1, \; N \cdot B = 0, \tag{21}$
(20) reduces to
$r^2 = \dfrac{1}{\kappa^2} + \left ( \left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} \right )^2 = \dfrac{1}{\kappa^2} + \left ( -\dfrac{\kappa'}{\kappa^2} \right )^2 \dfrac{1}{\tau^2} = \dfrac{1}{\kappa^2} + \dfrac{(\kappa')^2}{\kappa^4 \tau^2} = \dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}, \tag{22}$
whence
$r = \sqrt{\dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}} = \dfrac{1}{\kappa^2 \tau} \sqrt{(\kappa')^2 + \kappa^2 \tau^2}, \tag{23}$
as the desired formula for $r$.
Best Answer
Suppose $ \tau $ is non-zero. Then $ B \cdot \alpha = 0 $, so that $ \alpha = -\frac{1}{\kappa} N $. Differentiating this once gives: $$ T = -\frac{1}{\kappa} \frac{dN}{dt} = -\frac{1}{\kappa} ( -\kappa T + \tau B ) = T -\frac{\tau}{\kappa} B $$ Which means $ \tau = 0 $!