Torsion coefficient We went over the Frenet-Serret formulas today in class and the professor wrote
$$d\mathbf{\hat{B}}/dt=-\tau{}\mathbf{\hat{N}}.$$
He said that the coefficient is always negative (so that tau is always positive). 
I don't understand why this has to be and would quite like an explanation.
 A: In my experience with the Frenet-Serret apparatus it is the cuvrature which must always be positive; it is the torsion which may change sign along a regular curve.  To explain:
Let $\alpha(s) \subsetneq \Bbb R^3$ be a unit-speed curve; $s$ is the arc-length along $\alpha$; then as is well-known the unit tangent vector to $\alpha(s)$ is
$T(s) = \dot \alpha(s); \tag 1$
since
$T(s) \cdot T(s) = 1, \tag 2$
it follows in the usual manner that
$T(s) \cdot \dot T(s) = 0; \tag 3$
assuming that
$\dot T(s) \ne 0, \tag 4$
we may set
$\dot T(s) = \kappa(s) N(s), \tag 5$
where $N(s)$ is also a unit vector,
$\Vert N(s) \Vert^2 = N(s) \cdot N(s) = 1; \tag 6$
and by definition
$\kappa(s) > 0; \tag 7$
we have,
$\kappa(s) = \kappa(s) \Vert N(s) \Vert = \Vert \kappa(s)N(s) \Vert = \Vert \dot T(s) \Vert > 0; \tag 8$
$\kappa(s)$ is the magnitude of $\dot T(s)$; $N(s)$ is the direction.
We proceed to examine $\dot N(s)$; since
$T(s) \cdot N(s) = 0, \tag 8$
we have its derivative
$\dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag 9$
which yields, via (5), (6) and (8),
$\kappa(s) + T(s) \cdot \dot N(s) = \kappa(s) N(s) \cdot N(s) + T(s) \cdot \dot N(s)$
$= \dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag{10}$
and thus
$T(s) \cdot \dot N(s) = -\kappa(s); \tag{11}$
that is, the component of $\dot N(s)$ along $T(s)$ is $-\kappa(s)$.
Now
$\alpha(s) \subsetneq \Bbb R^3, \tag{12}$
and therefore we cannot assume that $\dot N(s)$ is collinear with $T(s)$; in general we must allow for a component of $\dot N(s)$ normal to both $N(s)$ and $T(s)$; we note however that (6) implies $\dot N(s)$ has no component along $N(s)$ itself, since it yields upon differentiation
$2N(s) \cdot \dot N(s) = 0 \Longrightarrow N(s) \cdot \dot N(s) = 0; \tag{13}$
bearing these facts in mind, we define the binormal vector
$B(s) = T(s) \times N(s), \tag{14}$
and noting that
$T(s) \cdot B(s) = N(s) \cdot B(s) = 0, \tag{15}$
and by virtue of (8) that
$B(s) \cdot B(s) = 1, \tag{16}$
we define the torsion $\tau(s)$ of $\alpha(s)$ as the component of $\dot N(s)$ along $B(s)$, so that
$\dot N(s) = -\kappa(s) T(s) + \tau (s) B(s), \tag{17}$
and we may thus also write
$\tau(s) = \tau(s) B(s) \cdot B(s) = 
\dot N(s) \cdot B(s); \tag{18}$
we now have, since (14) implies
$T(s) \times B(s) = -N(s), \tag{19}$
$\dot B(s) = \dot T(s) \times N(s) + T(s) \times \dot N(s)$
$= \kappa(s) N(s) \times N(s) + T(s) \times \dot N(s)$
$= \tau(s) T(s) \times B(s) = -\tau(s) N(s). \tag{20}$
In the above derivation of the Frenet-Serret equations (5), (17) and (20), we have nowhere found reason to stipulate
$\tau(s) > 0; \tag{21}$
indeed, it is well-known that 
$\tau(s) = 0, \; \dot B(s) = 0, \tag{22}$
when $\alpha(s)$ is a planar curve; see this question and my answer for a detailed explanation. 
We may in fact reverse the sign of the torsion of $\alpha(s)$ via the transformation
$\alpha(s) \to -\alpha(s); \tag{23}$
then
$T(s) = \dot \alpha(s) \to -T(s), \tag{24}$
and
$N(s) \to -N(s), \; \dot N(s) \to -\dot N(s); \tag{25}$
however,
$B(s) \to (-T(s)) \times (-N(s))$
$= T(s) \times N(s) = B(s); \tag{26}$
that is, $B(s)$ remains invariant under (23), hence also
$\dot B(s) \to \dot B(s); \tag{27}$
it now follows from 
$\dot B(s) = -\tau(s) N(s) \tag{28}$
that
$\tau(s) \to -\tau(s) \tag{29}$
under (23); therefore the sign of $\tau(s) \ne 0$ may always be reversed by taking the transformation (23); it follows that a curve of negative torsion exists if and only if a curve of positive torsion also exists; thus (21) is in fact false in general.
For more on this topic, the reader may consult the answer I gave to this question.
Torsion coefficient
