# The curvature of a tangent to a curve considered as a curve in its own right

Suppose we have a curve $$c:I \rightarrow \mathbb{R^3}$$ that is parametrised with respect to arc length. We then consider $$T,B,N$$ (tangent, binormal and normal) as curve in there own right each defined at a time $$t$$. The task is to express the curvature of the curve $$T$$ in terms of the curvature and torsion of the orginal curve $$c$$. I am struggling to do this. So far I have:

\begin{align*} \kappa_T &= \frac{\|T' \times T''\|}{\|T'\|^3} \\ &=\frac{\|T' \times T''\|}{\kappa^3}\\ &=\frac{\|c'' \times c'''\|}{\kappa^3}. \end{align*}

I need a expression for that top product, I am trying to link it to the alternative formula for torsion given by: $$\tau = \frac{\det(c'(t),c''(t),c'''(t)}{\|c' \times c''\|^2}.$$

Any help welcome :)

I think I may have it now:

We can rewrite the numerator of the torsion formula in the following way $$\tau = \frac{det(c'(t),c''(t),c'''(t)}{||c' \times c''||^2} = \frac{c'\cdot (c'' \times c''')}{||c' \times c''||^2}$$ Now we use that the denominator is actually $$\kappa^2$$, multiply up and take norms of both sides:

$$||\tau \kappa^2|| = ||c'||||c'' \times c'''||\ = ||c'' \times c'''||$$

And substituting this into what I had earlier, I get:

$$\kappa_T = \frac{||\tau||}{\kappa}$$

• I am afraid this answer is wrong. The norm of the inner product $c'\cdot (c''\times c''')$ is not equal to $\|c'\| \|c''\times c'''\|$. I have added the correct answer. Commented Jul 15, 2022 at 9:54

For clarity, denote the curve given by the tangent $$T$$ by $$\gamma$$. We will express $$\gamma'$$ and $$\gamma''$$ in term of the Frenet frame $$T$$, $$N$$, $$B$$ of $$c$$. We have \begin{align*} \gamma' &= \kappa N \\ \gamma'' &= -\kappa^2 T + \kappa' N + \kappa \tau B\\ \gamma' \times \gamma'' &= \kappa (-\kappa^2 N \times T + \kappa \tau N\times B) \\ &= \kappa^2 (\tau T + \kappa B). \end{align*}

So the curvature $$\kappa_\gamma$$ of the curve $$\gamma=T$$ becomes \begin{align*} \kappa_\gamma&= \frac{\|\gamma' \times \gamma''\|}{\|\gamma'\|^3} \\ &= \frac{\kappa^2 \sqrt{\tau^2 + \kappa^2}}{\kappa^3} \\ &= \sqrt{1 + \left(\frac{\tau}{\kappa}\right)^2}. \end{align*}