91影视

We are given,

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

Calculate the graph of vector function $r\left(0\right)$, normal vector $N\left(0\right)$ and curvature k should be calculated,

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

The unit tangent is given by.

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

Thus the principal unit normal vector is,

$N\left(0\right)=鉄�0,-1鉄�$

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 $||T^{\text{'}}\left(t\right)||=尾$

so $||T^{\text{'}}\left(0\right)||=尾$

since $||r^{\text{'}}\left(t\right)||=伪尾$

so $||r^{\text{'}}\left(0\right)||=伪尾$

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

$$\begin{gathered} k=\frac{||T^{\text{'}}\left(0\right)||}{||r^{\text{'}}\left(0\right)||} \\ k=\frac{尾}{伪尾} \\ =\frac{1}{伪} \end{gathered}$$

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

$$\begin{gathered} 蚁=\frac{1}{\frac{1}{伪}} \\ =伪 \end{gathered}$$

The center of the osculating circle is given by

$$\begin{gathered} r\left(0\right)+\frac{1}{k}N\left(0\right)=鉄�0,伪鉄�+伪鉄�0,-1鉄� \\ =鉄�0,伪-伪鉄� \\ =鉄�0,0鉄� \end{gathered}$$

So the osculating circle is,

$$\begin{gathered} (x-0{)}^2+(y-0{)}^2={伪}^2 \\ {x}^2+y^2={伪}^2 \end{gathered}$$