91Ó°ÊÓ

We are given an function,

$\mathbf{r}\left(t\right)=⟨t,\mathrm{Î\pm }\sin \mathrm{β}t,\mathrm{Î\pm }\cos \mathrm{β}tâŸ\copyright ,t=0$

Principal unit normal vector:

If r(t) is a twice differentiable vector function, then the principal unit vector to r(t) is,

$N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||}$

Where $T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||}$

T(t) should be calculated first and then N(t) is to be evaluated.

We start by comparing $r^{\text{'}}\left(t\right)$first:

$$\begin{gathered} r^{\text{'}}\left(t\right)=⟨1,\mathrm{Î\pm }\mathrm{β}\cos \mathrm{β}t,-\mathrm{Î\pm }\mathrm{β}\sin \mathrm{β}tâŸ\copyright \\ \left|r^{\text{'}}\left(t\right)\right|=\sqrt{1+(\mathrm{Î\pm }\mathrm{β}\cos \mathrm{β}t{)}^2+1-(\mathrm{Î\pm }\mathrm{β}\sin \mathrm{β}t{)}^2} \\ =\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2} \end{gathered}$$

The unit tangent vector to r(t) is

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨1,\mathrm{Î\pm }\mathrm{β}\cos \mathrm{β}t,-\mathrm{Î\pm }\mathrm{β}\sin \mathrm{β}tâŸ\copyright }{\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2}} \end{gathered}$$

Dividing $T^{\text{'}}\left(t\right)$by its magnitde,

The principal unit normal vector.

At $t=0,$

role="math" localid="1649675362152" $$\begin{gathered} N\left(t\right)=N\left(0\right) \\ =⟨0,-\sin 0,-\cos 0âŸ\copyright \\ =⟨0,0,-1âŸ\copyright \end{gathered}$$

Thus, the principle unit normal vector of r(t) is $⟨0,0,-1âŸ\copyright$.