Question

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

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

Short answer

The center of the osculating circle is and radius is $\frac{226}{45}$.

Step-by-step solution

Step 1. Given Information   

We are given,

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

Step 2. Find the center and radius of the osculating circle 

Finding the center and radius of the osculating circle,

Step 3. Find the center and radius of the osculating circle 

The unit normal vector is given by,

Thus, the principal unit normal vector is,

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

Step 4. Find the center and radius of the osculating circle 

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 $\rho =\frac{1}{k}$.

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

Step 5. Find the center and radius of the osculating circle 

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