91影视

We are given,

$\mathbf{r}\left(t\right)=鉄�3\sin 2t,3\cos 2t鉄�,t=\frac{蟺}{3}$

Finding the binormal vector,

$$\begin{gathered} r\left(t\right)=鉄�3\sin 2t,3\cos 2t鉄� \\ {r}^{\text{'}}\left(t\right)=鉄�6\cos 2t,3\cos 2t鉄� \\ ||r^{\text{'}}\left(t\right)||=\sqrt{36{\cos }^{2}2t+36{\sin }^{2}2t} \\ =\sqrt{36\left({\cos }^{2}2t+{\sin }^{2}2t\right)} \\ =6 \\ T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{鉄�6\cos 2t,-6\sin 2t鉄�}{6} \\ =鉄�\cos 2t,-\sin 2t鉄� \end{gathered}$$

At $t=\frac{蟺}{3}$,

role="math" localid="1649681944639"

So, the unit tangent vector to r(t) at $t=\frac{蟺}{3}$is .

$$\begin{gathered} T^{\text{'}}\left(t\right)=鉄�-2\sin 2t,-2\cos 2t鉄� \\ ||T^{\text{'}}\left(t\right)||=\sqrt{(-2\sin 2t{)}^2+(-2\cos 2t{)}^2} \\ =\sqrt{4\left({\sin }^{2}2t+{\cos }^{2}2t\right)} \\ =2 \\ N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||} \\ =\frac{鉄�-2\sin 2t,-2\cos 2t鉄�}{2} \\ =鉄�-\sin 2t,-\cos 2t鉄� \end{gathered}$$

At $t=\frac{蟺}{3}$,

Thus, the principal unit vector is.

The principal unit vector is .

The equation of the osculating at $r\left(t=\frac{蟺}{3}\right)$is defined by,

Hence, the equation is $z=0$.