Question

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

Short answer

The equation of the osculating circle is,

$(x+143{)}^2+{\left(y-\frac{241}{12}\right)}^2=\frac{(145{)}^3}{144}$

Step-by-step solution

Step 1. Given Information  

We are given,

Step 2. Finding the equation.

Calculate the graph of vector function $r\left(2\right)$, normal vector $N\left(2\right)$ and curvature k should be calculated,

The unit tangent vector is given by,

Step 3. Finding the equation.

Using the Quotient Rule for derivatives, we get

Normal vector is given by,

Step 4. Finding the equation.

Thus the principal unit normal vector to $r\left(t\right)$at $t=2$is,

The next step is to determine the curvature. For this $||T^{\text{'}}\left(2\right)||$and $||r^{\text{'}}\left(2\right)||$are to be calculated. Since $||T^{\text{'}}\left(t\right)||=\frac{6t}{1+9t^4}$, then

role="math" localid="1649827409088" $$\begin{gathered} ||T^{\text{'}}\left(2\right)||=\frac{6\left(2\right)}{1+9(2{)}^4} \\ =\frac{12}{145} \end{gathered}$$

since $||r^{\text{'}}\left(t\right)||=\sqrt{1+9t^4}$, then

role="math" localid="1649827433692" $$\begin{gathered} ||r^{\text{'}}\left(2\right)||=\sqrt{1+9(2{)}^4} \\ =\sqrt{145} \end{gathered}$$

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

$$\begin{gathered} k=\frac{||T^{\text{'}}\left(t_0\right)||}{||r^{\text{'}}\left(t_0\right)||} \\ =\frac{||T^{\text{'}}\left(2\right)||}{||r^{\text{'}}\left(2\right)||} \\ k=\frac{12}{145}·\frac{1}{\sqrt{145}} \\ =\frac{12}{(145{)}^{\frac{3}{2}}} \end{gathered}$$

The radius of curvature is,$\frac{1}{k}=\frac{(145{)}^{\frac{3}{2}}}{12}$.

Step 5. Finding the equation of the osculating circle.

The center of the osculating circle is given by,

The osculating circle will have centre $\left(-143,\frac{241}{12}\right)$and radius $\frac{(145{)}^{\frac{3}{2}}}{12}$.

So, the equation will be,

$$\begin{gathered} (x+143{)}^2+{\left(y-\frac{241}{12}\right)}^2={\left(\frac{(145{)}^{\frac{3}{2}}}{12}\right)}^2 \\ (x+143{)}^2+{\left(y-\frac{241}{12}\right)}^2=\frac{(145{)}^3}{144} \end{gathered}$$