91Ó°ÊÓ

We are given,

$\mathbf{r}\left(t\right)=⟨t,2t\sin t,2t\cos tâŸ\copyright ,t=\mathrm{Ï€}$

Finding the center and radius of the osculating circle,

At $t=\mathrm{Ï€},$

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

The curvature at $t=\mathrm{Ï€}$is,

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

Thus radius of the osculating circle is

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

The center is given by,

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

Using the Quotient rule for derivatives,

The binomial vector is,

The centre of the osculating circle is given by,

Hence. the center is,

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