Working out the The Unit Tangent $T$ and the Principal Normal $N$ of a curve $\vec{C}(t)$ = $(\frac{t^3}{3}, 2t-1,t^2+2)$

22 Views Asked by At

I am able to work this out. But in the end I get really complicated numbers to work with. This creates an issue when wanting to work out the Principal Normal N. Maybe I'm making a mistake.

$\vec{r} = \frac{t^3}{3}i + (2t-1)j + (t^2+2)k$

$\frac{dr}{dt} = t^2 +2j+2tk$

$\frac{ds}{dt} = |\frac{dr}{dt}| = \sqrt{t^4+4+2t^2}$

$T = \frac{dr}{ds} = \frac{dr}{dt}/\frac{ds}{dt} = \frac{t^2i+2j+2tk}{\sqrt{t^4+4+2t^2}}$

Any hints on how I could simplify this? Because when I want to work out the Principal Normal N, I have to do $\frac{dT}{dt}$ and it would be quite problematic for me to do that considering I have a square root terms in the denominator.