Structural Equation for Frenet Frame of a Curve Show that the structural equation for the Frenet frame of a curve in $R^3$ can be put into the form
$t'(s) = w(s) \times t(s),$ $n'(s) = w(s) \times n(s),$ $b'(s) = w(s) \times b(s)$
for a certain vector $w(s)$. Find this vector. 
I am unsure of how to go about solving this problem.
 A: Given that $T(s)$, $N(s)$ and $B(s)$ satisfy the Frenet-Serret equations
$\dot T(s) = \kappa N(s), \tag 1$
$\dot N(s) = -\kappa T(s) + \tau B(s), \tag 2$
and
$\dot B(s) = -\tau N(s), \tag 3$
we seek a vector $W(s)$ such that 
$\dot T(s) = W(s) \times T(s), \tag 4$
$\dot N(s) = W(s) \times N(s), \tag 5$
and
$\dot B(s) = W(s) \times B(s) \tag 6$
hold; set
$W(s) = \alpha T(s) + \beta N(s) + \gamma B(s); \tag 7$
then since $T(s) \times T(s) = 0$, 
$\kappa N(s) = \dot T(s)  = W(s) \times T(s) = \beta N(s) \times T(s) + \gamma B(s) \times T(s) = -\beta B(s) + \gamma N(s), \tag 8$
where we have used the defining equation for $B(s)$,
$B(s) = T(s) \times N(s) = -N(s) \times T(s) \Longleftrightarrow B(s) \times T(s) = N(s) = -T(s) \times B(s); \tag 9$
from (8),
$\beta = 0, \; \gamma = \kappa; \tag{10}$
then
$W(s) = \alpha T(s) + \kappa B(s); \tag{11}$
next, using $B(s) \times B(s) = 0$ and (9),
$-\tau N(s) = \dot B(s) = W(s) \times B(s) = \alpha T(s) \times B(s) = -\alpha N(s), \tag{13}$
and so it follows that
$\alpha = \tau, \tag{14}$
whence
$W(s) = \tau T(s) + \kappa B(s); \tag{15}$
it is clear from what we have done that
$\dot T(s) = W(s) \times T(s), \tag{16}$
and
$\dot B(s) = W(s) \times B(s); \tag{17}$
we also see that, using $T(s) = -B(s) \times N(s)$,
$\dot N(s) = -\kappa T(s) + \tau B(s)$
$= \kappa B(s) \times N(s)   + \tau T(s) \times N(s) = (\kappa B(s) + \tau T(s)) \times N(s) = W(s) \times N(s). \tag{18}$
Question: How does $W(s)$ evolve?  Does it satisfy a nice, intuitively clear equation?
