Find the torsion and curvature of a curve

Question

Find the curvature and torsion of $\alpha(t)=(2t,t^2,\frac{t^3}{3})$ for $t\in\mathbb{R}$

I have already found the Frenet frame:

$T=\frac{\alpha'}{||\alpha'||}=(\frac{2}{t^2+2}, \frac{2t}{t^2+2}, \frac{t^2}{t^2+2})$

$B=\frac{\alpha' \times \alpha''}{||\alpha' \times \alpha''||}=(\frac{t^2}{t^2+2}, \frac{-2t}{t^2+2}, \frac{2}{t^2+2})$

$N=-(T \times B)=(\frac{-2t^3-4t}{(t^2+2)^2}, \frac{-t^4+4}{(t^2+2)^2}, \frac{-2t^3-4t}{(t^2+2)^2})$

Now $T'=(\frac{-4t}{(t^2+2)^2}, \frac{4-2t^2}{(t^2+2)^2}, \frac{4t+2t^3-2t^4}{(t^2+2)^2})$

Attempting $||T'||$ gives a crazy long equation that I am struggling to simplify.

So knowing $T'=kN$ I attempted $k=\frac{T'}{N}$

Obviously this results in a vector $k=(\frac{-4t}{-2t^3-4t}, \frac{4-2t^2}{-t^4+4}, \frac{4t+2t^3-2t^4}{-2t^3-4t})$

As k shouldn't be a vector, what do I need to do/where have I gone wrong?

Answer 1

I would suggest using the formulas $$\kappa = \frac{\|\alpha' \times \alpha''\|}{\|\alpha'\|^3} \qquad \text{and} \qquad \tau = \frac{\det(\alpha',\alpha'',\alpha''')}{\|\alpha' \times \alpha''\|^2}$$ where $\kappa$ is the curvature and $\tau$ is the torsion. First you have to check that $\alpha$ is a strongly regular curve in order to use the formulas, i.e. that $c' \times c'' \neq 0$ on $\mathbb{R}$.