91Ó°ÊÓ

We are given,

Calculate the graph of vector function $r\left(2\right)$, normal vector $N\left(2\right)$ and curvature k should be calculated,

The unit tangent vector is given by,

Using the Quotient Rule for derivatives, we get

Normal vector is given by,

Thus the principal unit normal vector to $r\left(t\right)$at $t=2$is,

The next step is to determine the curvature. For this $||T^{\text{'}}\left(2\right)||$and $||r^{\text{'}}\left(2\right)||$are to be calculated. Since $||T^{\text{'}}\left(t\right)||=\frac{6t}{1+9t^4}$, then

role="math" localid="1649827409088" $$\begin{gathered} ||T^{\text{'}}\left(2\right)||=\frac{6\left(2\right)}{1+9(2{)}^4} \\ =\frac{12}{145} \end{gathered}$$

since $||r^{\text{'}}\left(t\right)||=\sqrt{1+9t^4}$, then

role="math" localid="1649827433692" $$\begin{gathered} ||r^{\text{'}}\left(2\right)||=\sqrt{1+9(2{)}^4} \\ =\sqrt{145} \end{gathered}$$

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

$$\begin{gathered} k=\frac{||T^{\text{'}}\left(t_0\right)||}{||r^{\text{'}}\left(t_0\right)||} \\ =\frac{||T^{\text{'}}\left(2\right)||}{||r^{\text{'}}\left(2\right)||} \\ k=\frac{12}{145}·\frac{1}{\sqrt{145}} \\ =\frac{12}{(145{)}^{\frac{3}{2}}} \end{gathered}$$

The radius of curvature is,$\frac{1}{k}=\frac{(145{)}^{\frac{3}{2}}}{12}$.

The center of the osculating circle is given by,

The osculating circle will have centre $\left(-143,\frac{241}{12}\right)$and radius $\frac{(145{)}^{\frac{3}{2}}}{12}$.

So, the equation will be,

$$\begin{gathered} (x+143{)}^2+{\left(y-\frac{241}{12}\right)}^2={\left(\frac{(145{)}^{\frac{3}{2}}}{12}\right)}^2 \\ (x+143{)}^2+{\left(y-\frac{241}{12}\right)}^2=\frac{(145{)}^3}{144} \end{gathered}$$