Osculating circles: Find the equation of the osculating circle to the given function at the specified value of t.

$\mathbf{r}\left(t\right)=⟨3\sin 2t,3\cos 2t⟩,t=\frac{\pi }{3}$

We are given,

$\mathbf{r}\left(t\right)=⟨3\sin 2t,3\cos 2t⟩,t=\frac{\pi }{3}$

First calculate the graph of vector function $r\left(t\right)$, normal vector $N\left(t\right)$ and the curvature k at $t=\frac{\pi }{3}$.

role="math" localid="1649836096067"

Calculating unit tangent vector at t,

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨6\cos 2t,-6\sin 2t⟩}{6} \\ =⟨\cos 2t,-\sin 2t⟩ \\ {T}^{\text{'}}\left(t\right)=⟨-2\sin 2t,-2\cos 2t⟩ \\ ||T^{\text{'}}\left(t\right)||=\sqrt{(-2\sin 2t{)}^2+(-2\cos 2t{)}^2} \\ =\sqrt{4\left({\sin }^{2}2t+{\cos }^{2}2t\right)} \\ =2 \end{gathered}$$

The unit normal vector is given by,

The next step is to determine the curvature k at role="math" $t=\frac{\pi }{3}$.

For this $||T^{\text{'}}\left(\frac{\pi }{3}\right)||$and $||r^{\text{'}}\left(\frac{\pi }{3}\right)||$are to be calculated.

Since $\left|T^{\text{'}}\left(t\right)=2\right|$,

so $||T^{\text{'}}\left(\frac{\pi }{3}\right)||=2$

Since $||r^{\text{'}}\left(t\right)||=6$,

so $||r^{\text{'}}\left(\frac{\pi }{3}\right)||=6$

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

$$\begin{gathered} k=\frac{||T^{\text{'}}\left(\frac{\pi }{3}\right)||}{||r^{\text{'}}\left(\frac{\pi }{3}\right)||} \\ k=\frac{{\displaystyle 2}}{{\displaystyle 6}} \\ k=\frac{{\displaystyle 1}}{{\displaystyle 3}} \end{gathered}$$

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

$$\begin{gathered} \rho =\frac{1}{\frac{1}{3}} \\ \rho =3 \end{gathered}$$

The center of the osculating circle at $t=\frac{\pi }{3}$is given by

So the osculating circle is,

$$\begin{gathered} (x-0{)}^2+(y-0{)}^2=3^2 \\ {x}^2+y^2=9 \end{gathered}$$