91影视

We are given,

$\mathbf{r}\left(t\right)=鉄�t,伪\sin 尾t,伪\cos 尾t鉄�,t=0$

Finding the center and radius of the osculating circle,

$$\begin{gathered} \mathbf{r}\left(t\right)=鉄�t,伪\sin 尾t,伪\cos 尾t鉄� \\ r\left(0\right)=鉄�0,伪\sin 0,伪\cos 0鉄� \\ =鉄�0,0,伪鉄� \\ {r}^{\text{'}}\left(t\right)=鉄�1,伪尾\cos 尾t,-伪尾\sin 尾t鉄� \\ ||r^{\text{'}}\left(t\right)||=\sqrt{1+{伪}^2{尾}^2\left({\cos }^2尾t+{\sin }^2尾t\right)} \\ =\sqrt{1+{伪}^2{尾}^2} \end{gathered}$$

The unit tangent vector is given by,

The unit normal vector is given by,

The next step is to determine the curvature k at $t=0$.

For this $||T^{\text{'}}\left(0\right)||$and $||r^{\text{'}}\left(0\right)||$are to be calculated.

since $||T^{\text{'}}\left(t\right)||=\frac{伪{尾}^2}{\sqrt{1+{伪}^2{尾}^2}}$,

so $||T^{\text{'}}\left(0\right)||=\frac{伪{尾}^2}{\sqrt{1+{伪}^2{尾}^2}}$

Since $||r^{\text{'}}\left(t\right)||=\sqrt{1+{伪}^2{尾}^2}$

so $||r^{\text{'}}\left(0\right)||=\sqrt{1+{伪}^2{尾}^2}$

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

role="math" localid="1649898338360" $$\begin{gathered} k=\frac{||T^{\text{'}}\left(0\right)||}{||r^{\text{'}}\left(0\right)||} \\ k=\frac{\frac{伪{尾}^2}{\sqrt{1+{伪}^2{尾}^2}}}{\sqrt{1+{伪}^2{尾}^2}} \\ =\frac{伪{尾}^2}{1+{伪}^2{尾}^2} \end{gathered}$$

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

$$\begin{gathered} 蚁=\frac{\frac{1}{伪{尾}^2}}{1+伪{尾}^2} \\ =\frac{1+{伪}^2{尾}^2}{伪{尾}^2} \end{gathered}$$

The center of the osculating circle is given by $r\left(0\right)+\frac{1}{k}N\left(0\right)$.