This question is a follow-up to a question posted here
After tweaking the parametrization a little bit, I decided to go with this: $$x=\frac{1}{2}\cos(t) , y=\frac{1}{2}+\frac{1}{2}\sin(t) , z=\frac{1}{2}+\frac{1}{2}\sin(t)$$ Without doing arc length parametrization( trying to do so I got $s=\frac{1}{2}\int_{0}^{t}\left | sin(\tau) \right |d\tau$, which I didn't know how to solve), I tried doing it the other way round. I found that : $$\frac{d\vec{r}}{dt}=(-\frac{1}{2}\sin(t), \frac{1}{2}\cos(t),\frac{1}{2}\cos(t))$$ The binormal vector is now given by $$\vec{B}=\frac{d\vec{r}}{dt}\times \frac{d^{2}\vec{r}}{dt}$$ and the normal vector is simply $$\vec{N}=\vec{B}\times \vec{T}$$ The equations for them become very nasty expressions, so I am very relcutant to type them out. There's two things that are unclear to me now:
- If I wanted to compute the curvature, $\kappa $, at some point which formula do I use? By definition we have that $\kappa =\left | \frac{d^{2}\vec{r}}{ds} \right | $ which is in terms of the natural parameter. How can I use my non-parametrized equations to obtain the curvature?
2.By computation I got that $$\vec{B}=\left ( 0,\frac{\cos^{2}t}{4}- \frac{\sin^{2}t}{4}, -\frac{\cos^{2}t}{4}+\frac{\sin^{2}t}{4}\right )$$ Can't I just write it as $\vec{B}=\left ( 0,1,-1\right )$?
Because your curve lies in the plane $y = z$ (and in fact is an ellipse whose major axis is $\sqrt{2}$ times the minor axis, since $y = x^{2} + y^{2}$ defines a right circular cylinder with axis parallel to the $z$-axis), its binormal vector is indeed constant. In your cross product formula, you need to normalize. So $B = \frac{1}{\sqrt{2}}(0, 1, -1)$ up to an overall sign. This should make your curvature expression a lot nicer. :)
(Separately, because your curve is an ellipse, it would be hopeless to seek an explicit arc length parametrization....)