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

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

We are given,

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

Finding the center and radius of the osculating circle,

At $t=\pi ,$

$||r^{\text{'}}\left(t\right)||=\sqrt{4{\pi }^2+5}$

The curvature at $t=\pi$is,

$$\begin{gathered} \frac{||r^{\text{'}}\left(t\right)\times r^{\text{'}\text{'}}\left(t\right)||}{{||r^{\text{'}}\left(t\right)||}^3}=\frac{\sqrt{16{\pi }^2+68{\pi }^2+80}}{{\left(\sqrt{4{\pi }^2+5}\right)}^3} \\ =\frac{2\sqrt{4{\pi }^2+17{\pi }^2+20}}{{\left(4{\pi }^2+5\right)}^{\frac{3}{2}}} \end{gathered}$$

Thus radius of the osculating circle is

$\frac{1}{k}=\frac{{\left(4{\pi }^2+5\right)}^{\frac{3}{2}}}{2\sqrt{4{\pi }^4+17{\pi }^2+20}}$

The center is given by,

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{1}{\sqrt{4t^2+5}}⟨1,2(tcsot+\sin t),2(-t\sin t+\cos t)⟩ \\ \text{At}t=\pi , \\ T\left(t\right)=T\left(\pi \right) \\ =\frac{1}{\sqrt{4{\pi }^2+5}}⟨1,-2\pi ,-2⟩ \end{gathered}$$

Using the Quotient rule for derivatives,

The binomial vector is,

The centre of the osculating circle is given by,

Hence. the center is,

role="math" localid="1649843878643"