Find the tangent, normal and binormal vectors at the point $(1,1,1)$

Question

I'm having some trouble with the following question:

Let $\alpha:[0,2] \to \mathbb R^3$ with $\alpha(t)=(t,t^2,t)$. Find the tangent, normal and binormal vectors at the point $(1,1,1)$.

I first tried to reparametrize this curve by arc length. I got:

$$s(t) = \int _{0}^{t} \sqrt{2+4\tau^2} d\tau =\frac{1}{2} \left(t\sqrt{2 + 4 t^2} + \sinh^{-1}(\sqrt{2}t)\right)$$

And I don't think it's possible to explicitly find an inverse $t(s)$ for this function, so I don't think that it's possible to explicitly reparametrize this curve by arc length. How can we find the tangent, normal, and binormal vectors then?

Answer 1

Unit tangent vector is $$T(t)=\frac{\alpha'(t)}{|\alpha'(t)|}=\frac{(1,2t,1)}{|(1,2t,1)|}=\frac1{\sqrt{4t^2+2}}(1,2t,1)$$ At $(1,1,1)$ you have $t=1$, so $T(1)=\frac1{\sqrt 6}(1,2,1)$.

The normal unit vector is $$N(t)=\frac{T'(t)}{|T'(t)|}$$ The binormal vector $$B(t)=T(t)\times N(t)$$ Now all you need to do is calculate $T'(t)$ and plug the result in the above equations.