91影视

We are given,

Finding the center and radius of the osculating circle,

The unit normal vector is given by,

Thus, the principal unit normal vector is,

$N\left(\frac{蟺}{6}\right)=鉄�0,-1,0鉄�$

The next step is to determine the curvature k at $t=\frac{蟺}{6}$.

For this $||T^{\text{'}}\left(\frac{蟺}{6}\right)||$and $||r^{\text{'}}\left(\frac{蟺}{6}\right)||$are to be calculated.

Since $||T^{\text{'}}\left(t\right)||=\frac{45}{\sqrt{226}}$so $||T^{\text{'}}\left(\frac{蟺}{6}\right)||=\frac{45}{\sqrt{226}}$

Since $||r^{\text{'}}\left(t\right)||=\sqrt{226}$so $||r^{\text{'}}\left(\frac{蟺}{6}\right)||=\sqrt{226}$

The curvature k of C at a point on the curve is given by,

$$\begin{gathered} k=\frac{||T^{\text{'}}\left(\frac{蟺}{6}\right)||}{||r^{\text{'}}\left(\frac{蟺}{6}\right)||} \\ =\frac{{\displaystyle \frac{{\displaystyle 45}}{{\displaystyle \sqrt{226}}}}}{{\displaystyle \sqrt{226}}} \\ =\frac{{\displaystyle 45}}{{\displaystyle 226}} \end{gathered}$$

The radius of the curvature $蚁$ of C is given by $蚁=\frac{1}{k}$.

$$\begin{gathered} 蚁=\frac{1}{k} \\ =\frac{1}{\frac{45}{226}} \\ =\frac{226}{45} \end{gathered}$$

The center of the osculating circle is given by $r\left(\frac{蟺}{6}\right)+\frac{1}{k}N\left(\frac{蟺}{6}\right)$.