Determining a curvature by knowing the binormal vector I need to prove that knowing the binormal vector $\vec{b}(s)$ of a regular curve $\alpha$ parametrized by arc length is sufficient to know $|\tau(s)|$ and $\kappa(s)$, respectively the module of torsion and the curvature of $\alpha$.
By the third Frenet Formula, $|\vec{b}'|=|\tau|\, ||\vec{n}||=|\tau|$. No problems here to determine $|\tau|$.
I'm working now in a way to determine $\kappa$. What I did was that:
$$\vec{b}=\vec{t}\wedge\vec{n}=\frac{1}{\kappa}(\vec{n}+\tau \vec{b})\wedge \vec{n}=\frac{1}{\kappa}(\vec{n}\wedge\vec{n})+\frac{\tau}{\kappa}(\vec{b}\wedge \vec{n})=\frac{\tau}{\kappa}(\vec{b}\wedge\vec{n}) $$
Am I on the right way? What can I do?
 A: We are given $B(s)$ for the arc-length parametrized curve $\alpha(s)$; we wish to find
$\vert \tau(s) \vert$ and $\kappa(s)$.
As pointed out by our OP Mateus Rocha in the text of the question itself, $\vert \tau(s) \vert$ is easily had; we simply exploit the Frenet-Serret equation
$\dot B(s) = -\tau(s) N(s) \tag 1$
to write
$\vert \tau(s) \vert = \vert \tau(s) \vert \vert N(s) \vert = \vert \tau(s) N(s) \vert = \vert \dot B \vert, \tag 2$
or
$\vert \tau(s) \vert = \vert \dot B(s) \vert, \tag 3$
giving us $\vert \tau(s) \vert$; we used 
$\vert N(s) \vert =1  \tag 4$
in deriving (2); next, we differentiate (1):
$\ddot B(s) = -\dot \tau(s) N(s) - \tau(s) \dot N(s)$
$= -\dot \tau(s) N(s) - \tau(s) (-\kappa(s) T(s) + \tau(s) B(s)) = -\dot \tau(s) N(s)  + \tau(s) \kappa T(s) - \tau^2(s) B(s), \tag 5$
where we have exploited the Frenet-Serret formula
$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s); \tag 6$
from (1),
$\dot \tau(s) \dot B(s) = -\tau(s) \dot \tau N(s); \tag 7$
from (5),
$\tau(s) \ddot B(s) = -\tau(s) \dot \tau(s) N(s) + \tau^2(s) \kappa(s) T(s) - \tau^3(s) B(s); \tag 8$
thus, subtracting (7) from (8),
$\tau(s) \ddot B(s) - \dot \tau(s) \dot B(s) = \tau^2(s) \kappa T(s) - \tau^3(s) B(s), \tag 9$
whence,
$\tau^2(s) \kappa(s) T(s) = \dot \tau(s) \dot B(s) - \tau(s) \ddot B(s) + \tau^3(s) B(s); \tag{10}$
this yields
$\tau^2(s) \kappa(s) = \tau^2(s) \kappa(s) \vert T(s) \vert = \vert \tau^2(s) \kappa(s) T(s) \vert = \vert \dot \tau(s) \dot B(s) - \tau(s) \ddot B(s) + \tau^3(s) B(s) \vert, \tag{11}$
since
$\vert T(s) \vert = 1; \tag{12}$
a minor re-arrangement results in
$\kappa(s) = \dfrac{\vert \dot \tau(s) \dot B(s) - \tau(s) \ddot B(s) + \tau^3(s) B(s) \vert}{\tau^2(s)}. \tag{13}$
Nota Bene:  We observe that (13) is invariant under reversal of the sign of $\tau(s)$; thus this formula is consistent with (3), which only gives us $\vert \tau(s) \vert$. Indeed, the sign of $\tau$ cannot be determined from (1)-(3); the derivation of the formula (13) for $\kappa(s)$, is independent of the sign of $\tau(s)$.  End of Note.
