# 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?

• 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 $b$. Sep 6 '19 at 19:46
• @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 Sep 7 '19 at 0:01

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.