Question

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

$\mathbf{r}\left(t\right)=⟨\alpha \sin \beta t,\alpha \cos \beta t⟩,t=0$

Short answer

The equation of the osculating circle is $x^2+y^2={\alpha }^2$.

Step-by-step solution

Step 1. Given Information  

We are given,

$\mathbf{r}\left(t\right)=⟨\alpha \sin \beta t,\alpha \cos \beta t⟩,t=0$

Step 2. Finding the equation.

Calculate the graph of vector function $r\left(0\right)$, normal vector $N\left(0\right)$ and curvature k should be calculated,

$$\begin{gathered} r\left(t\right)=⟨\alpha \sin \beta t,\alpha \cos \beta t⟩ \\ r\left(0\right)=⟨\alpha \sin 0,\alpha \cos 0⟩ \\ =⟨0,\alpha ⟩ \\ {r}^{\text{'}}\left(t\right)=⟨\alpha \beta \cos \beta t,-\alpha \beta \sin \beta t⟩ \\ ||r^{\text{'}}\left(t\right)||=\sqrt{(\alpha \beta \cos \beta t{)}^2+(-\alpha \beta \sin \beta t{)}^2} \\ =\sqrt{{\alpha }^2{\beta }^2} \\ =\alpha \beta \end{gathered}$$

The unit tangent is given by.

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ T\left(t\right)=\frac{⟨\alpha \beta \cos \beta t,-\alpha \beta \sin \beta t⟩}{\alpha \beta } \\ =⟨\cos \beta t,-\sin \beta t⟩ \\ {T}^{\text{'}}\left(t\right)=⟨-\beta \sin \beta t,-\beta \cos \beta t⟩ \\ ||T^{\text{'}}\left(t\right)||=\sqrt{(-\beta \sin \beta t{)}^2+(-\beta \cos \beta t{)}^2} \\ =\sqrt{{\beta }^2\left({\sin }^2\beta t+{\cos }^2\beta t\right)} \\ =\beta \\ N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||} \\ =\frac{⟨-\beta \sin \beta t,-\beta \cos \beta t⟩}{\beta } \\ =⟨-\sin \beta t,-\cos \beta t⟩ \\ N\left(0\right)=⟨-\sin 0,-\cos 0⟩ \\ =⟨0,-1⟩ \end{gathered}$$

Step 3. Finding the equation.

Thus the principal unit normal vector is,

$N\left(0\right)=⟨0,-1⟩$

The next step is to determine the curvature k at $t=0$.

For this $||T^{\text{'}}\left(0\right)||$and $||r^{\text{'}}\left(0\right)||$are to be calculated.

since $||T^{\text{'}}\left(t\right)||=\beta$

so $||T^{\text{'}}\left(0\right)||=\beta$

since $||r^{\text{'}}\left(t\right)||=\alpha \beta$

so $||r^{\text{'}}\left(0\right)||=\alpha \beta$

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

$$\begin{gathered} k=\frac{||T^{\text{'}}\left(0\right)||}{||r^{\text{'}}\left(0\right)||} \\ k=\frac{\beta }{\alpha \beta } \\ =\frac{1}{\alpha } \end{gathered}$$

Step 4. Finding the equation.

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

$$\begin{gathered} \rho =\frac{1}{\frac{1}{\alpha }} \\ =\alpha \end{gathered}$$

The center of the osculating circle is given by

$$\begin{gathered} r\left(0\right)+\frac{1}{k}N\left(0\right)=⟨0,\alpha ⟩+\alpha ⟨0,-1⟩ \\ =⟨0,\alpha -\alpha ⟩ \\ =⟨0,0⟩ \end{gathered}$$

So the osculating circle is,

$$\begin{gathered} (x-0{)}^2+(y-0{)}^2={\alpha }^2 \\ {x}^2+y^2={\alpha }^2 \end{gathered}$$