# Deriving a relationship between curvature, torsion and the curvature of tangent vector

Let $\alpha(s)$ be a regular and biregular curve in $\mathbb{R}^3$ parametrized in arc length with its curvature $k(s)$ and torsion $\tau(s)$. Then we can think of it's tangent vector $T(s)$ as another curve which range is contained on the unit sphere, let's call it $\beta(s)=\alpha'(s)$ with its curvature, say $k_1(s)$. I have to prove the following relation:

$$k_1^2=\frac {k^2+\tau^2}{k^2}$$

I've tried different approaches and I've derived some relations (maybe true as long as my calculation were right) using Frenet-Serres formula but I still cannot conclude.

Since the following discussion addresses multiple curves $$\gamma$$ each with its own tangent, normal, and binormal vectors, I will distinguish 'twixt the Frenet-Serret apparatus of different curves via subscripting: thus the unit tangent vector to $$\gamma$$ will be denoted by $$T_\gamma$$ and so forth.

We are considering the curve $$\beta(s)$$ given as the tangent vector $$T_\alpha(s)$$ to the unit speed curve $$\alpha(s)$$, so that

$$\beta(s) = T_\alpha(s) = \alpha'(s) \tag 1$$

with

$$\langle \beta(s), \beta(s) \rangle = \langle T_\alpha(s), T_\alpha(s) \rangle = 1, \tag 2$$

and we wish to compute the curvature $$\kappa_1(s)$$ of $$\beta(s)$$. One way of addressing this is to express the Frenet-Serret apparatus of $$\beta(s)$$ in terms of $$\alpha(s)$$ and its Frenet frame $$T_\alpha$$, $$N_\alpha$$, $$B_\alpha$$. We recall that $$s$$ is the arc-length along $$\alpha(s)$$; however, $$s$$ is not in general the arc-length along $$\beta(s) = T_\alpha(s)$$; that is, $$\beta(s)$$ is not in general a unit-speed curve. We thus donote the arc-length along $$\beta$$ by $$\sigma$$; then $$T_\beta$$ is given by

$$T_\beta(\sigma) = \beta'(\sigma) = \dfrac{dT_\alpha(s)}{ds}\dfrac{ds}{d\sigma} = \kappa(s)N_\alpha(s)\dfrac{ds}{d\sigma}, \tag 3$$

and since

$$\Vert T_\beta(\sigma) \Vert = \Vert N_\alpha(s) \Vert = 1, \tag 4$$

it follows from (3) that

$$\left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert = \left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert \Vert N_\alpha(s) \Vert = \left \Vert \kappa(s) \dfrac{ds}{d\sigma} N_\alpha(s) \right \Vert = \Vert T_\beta(\sigma) \Vert = 1;\tag 5$$

we conclude from (5) that

$$\kappa^2(s) \left ( \dfrac{ds}{d\sigma} \right )^2 = \left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert^2 = 1, \tag 6$$

a formula which we shall shortly deploy. Since $$\kappa(s) > 0$$, we may also write

$$\kappa(s) \left \vert \dfrac{ds}{d\sigma} \right \vert = \left \vert \kappa(s) \dfrac{ds}{d\sigma} \right \vert = 1, \tag 7$$

and from this we see that

$$\kappa(s) \dfrac{ds}{d\sigma} = \pm 1; \tag 8$$

by reversing the direction of traverse of $$\beta(\sigma)$$ with respect to $$s$$ if necessary we may also assume that $$ds / d\sigma > 0$$, so that in fact

$$\kappa(s) \dfrac{ds}{d\sigma} = 1, \tag 9$$

which, though not further necessary here, is worth noting in it's own right.

We compute the curvature $$\kappa_1(\sigma)$$ of $$\beta(\sigma)$$ via the Frenet-Serret equations; as applied to $$\beta(\sigma)$$, we find via (3) that

$$\kappa_1(\sigma) N_\beta(\sigma) = T_\beta'(\sigma) = \dfrac{d}{d\sigma} \left ( \kappa(s)N_\alpha(s)\dfrac{ds}{d\sigma} \right ); \tag{10}$$

now,

$$\dfrac{d}{d\sigma} \left ( \kappa(s)N_\alpha(s)\dfrac{ds}{d\sigma} \right ) = \dfrac{d}{d\sigma} \left ( \left (\kappa(s)\dfrac{ds}{d\sigma}\right )N_\alpha(s)\right )$$ $$= \dfrac{d}{d\sigma} \left (\kappa(s)\dfrac{ds}{d\sigma}\right ) N_\alpha(s) + \left ( \kappa(s)\dfrac{ds}{d\sigma}\right )\dfrac{d}{d\sigma} N_\alpha(s); \tag{11}$$

from (8), we see that $$(\kappa(s)ds / d\sigma)$$ must be constant, whence

$$\dfrac{d}{d\sigma} \left (\kappa(s)\dfrac{ds}{d\sigma}\right ) = 0; \tag{12}$$

combining (8), (10), (11) and (12):

$$\kappa_1(\sigma) N_\beta(\sigma) =\pm \dfrac{d}{d\sigma} N_\alpha(s); \tag{13}$$

we proceed:

$$\dfrac{dN_\alpha(s)}{d\sigma} = \dfrac{dN_\alpha(s)}{ds} \dfrac{ds}{d\sigma} = \dfrac{ds}{d\sigma} (-\kappa(s) T_\alpha(s) + \tau(s) B_\alpha(s)), \tag{14}$$

by virtue of the Frenet-Serret equations for $$\alpha(s)$$; from (13) and(14),

$$\kappa_1^2(\sigma) = (\kappa_1(\sigma) N_\beta(\sigma)) \cdot (\kappa_1(\sigma) N_\beta(\sigma)) = \left (\dfrac{d}{d\sigma} N_\alpha(s) \right) \cdot \left (\dfrac{d}{d\sigma} N_\alpha(s) \right)$$ $$= \left (\dfrac{ds}{d\sigma} \right )^2 (-\kappa(s) T_\alpha(s) + \tau(s) B_\alpha(s)) \cdot (-\kappa(s) T_\alpha(s) + \tau(s) B_\alpha(s))$$ $$= \left (\dfrac{ds}{d\sigma} \right )^2 (\kappa^2(s) + \tau^2(s)), \tag{15}$$

where we have used $$\Vert B_\alpha(s) \Vert = 1$$, $$T_\alpha(s) \cdot B_\alpha(s) = 0$$ as well as (2) in this derivation. Finally, via (6),

$$\kappa_1^2(\sigma) = \dfrac{\kappa^2(s) + \tau^2(s)}{\kappa^2(s)}, \tag{16}$$

the desired result.