Osculating circles: Find the center and radius of the osculating circle to the given vector function at the specified value of t.

$\mathbf{r}\left(t\right)=⟨t,5\sin 3t,5\cos 3t⟩,t=\frac{\pi }{6}$

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,

$N\left(\frac{\pi }{6}\right)=⟨0,-1,0⟩$

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

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

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

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

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

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

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

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

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