91Ó°ÊÓ

We are given,

Finding the binormal vector,

$$\begin{gathered} \mathbf{r}\left(t\right)=⟨t,\mathrm{Î\pm }\sin \mathrm{β}t,\mathrm{Î\pm }\cos \mathrm{β}tâŸ\copyright ,t=0 \\ \mathbf{r}\left(t\right)=⟨t,\mathrm{Î\pm }\sin \mathrm{β}t,\mathrm{Î\pm }\cos \mathrm{β}tâŸ\copyright ,t=0 \\ ||r^{\text{'}}\left(t\right)||=\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2\left({\cos }^2\mathrm{β}t+{\sin }^2\mathrm{β}t\right)} \\ =\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2} \\ T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{1}{\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2}}⟨1,\mathrm{Î\pm }\mathrm{β}\cos \mathrm{β}t,-\mathrm{Î\pm }\mathrm{β}\sin \mathrm{β}tâŸ\copyright \end{gathered}$$

At $t=0$,

$$\begin{gathered} T\left(t\right)=T\left(0\right) \\ =\frac{1}{\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2}}⟨1,\mathrm{Î\pm }\mathrm{β},0âŸ\copyright \end{gathered}$$

Thus, the unit tangent vector is $\frac{1}{\sqrt{1+{\mathrm{Î\pm }}^2{\mathrm{β}}^2}}⟨1,\mathrm{Î\pm }\mathrm{β},0âŸ\copyright$.

By using the Quotient Rule for derivatives,

Thus, the principal unit normal vector to r(t) is $⟨0,0,-1âŸ\copyright$,

The binormal vector is given by,

The equation of the osculating at $r(t=0)$is defined by,

Hence, the equation is $-\mathrm{Î\pm }\mathrm{β}x+y=0$.