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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?

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    $\begingroup$ That formula you used for $t$ is not correct. It implies $|\tau| = |\kappa|$ by taking the norm of the last step, which is certainly not true. Instead take the second derivative of $b$. $\endgroup$ Sep 6 '19 at 19:46
  • $\begingroup$ @NinadMunshi why is not correct? By the second Frenet formula, $$n'=-\kappa t-\tau b\implies n'+\tau b=-\kappa \tau\implies t=\frac{-n'-\tau b}{\kappa}$$ So, i just did a mistake on the signal $\endgroup$ Sep 7 '19 at 0:01
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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.

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