The Osculating Circle at a Point on a Curve

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$, and let $P$ be the point on $C$ at $t$. The Osculating Circle at $P$ is the circle that best approximates $C$ at $P$. The osculating circle at $P$: 1) Contains the point $P$. 2) Has radius $\rho (t) = \frac{1}{\kappa (t)}$. 3) Has curvature $\kappa (t)$. 4) Shares the same tangent at $P$.

If $y = f(x)$ generates a curve $C$, then the osculating circle at $P$ is also defined to be the circle that best approximates $C$ at $P$ by having radius $\rho (x) = \frac{1}{\kappa (x)}$, curvature $\kappa (x)$, and shares the same tangent at $P$.

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$, and let $P$ be the point on $C$ at $t$. Then Center of Curvature at $P$ is the center of the osculating circle at $P$.

From the definition of an osculating circle, we can calculate the center of curvature which we will denote by $\vec{r_c}(t)$, by the following formula:

(1)

\begin{align} \mathrm{Center \: of \: Curvature} = \vec{r_c}(t) = \vec{r}(t) + \rho (t) \hat{N}(t) \end{align}

This can easily be seen as $\vec{r}(t)$ takes us to the point $P$, and then we travel $\rho (t)$ (the radius of the osculating circle at $P$) in the direction of the unit normal vector $\hat{N}(t)$.

Example 1

Let $\vec{r}(t) = \left ( t, \frac{1}{2}t^2 + 1, 0 \right )$. Find the equation of the osculating circle at $t = 1$.

We will first simplify this problem down. We should first notice that the graph of $\vec{r}(t)$ represents a plane curve - namely one that lies on the $xy$-plane. Since $x = t$, then $y = \frac{1}{2}t^2 + 1$ implies that $y = \frac{1}{2}x^2 + 1$. So $\vec{r}(t)$ represents the graph of the single variable function $f(x) = \frac{1}{2}x^2 + 1$.

We will first find the curvature of this function by using the formula $\kappa (x) = \frac{ \mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$ and then take the reciprocal to get the radius of curvature, $\rho (x) = \frac{1}{\kappa (x)}$. To use this formula, we must first calculate $f'(x)$ and $f''(x)$. We have that $f'(x) = x$, and $f''(x) = 1$, and so:

(2)

\begin{align} \kappa (x) = \frac{1}{(1 + x^2)^{3/2}} \\ \rho (x) = (1 + x^2)^{3/2} \end{align}

Since $x = t$, the given value $t = 1$ corresponds to $x = 1$. Plugging this into the formula above and we have that $\rho (1) = 2^{3/2}$.

We now need to find the center of curvature, i.e, the center of the osculating circle at $t = 1$. To do so, we need to find the unit normal vector $\hat{N}(1)$. To calculate the unit normal vector, we must first find the unit tangent vector $\hat{T}(1)$ and the unit binormal vector $\hat{B}(1)$ and then take the cross product $\hat{N}(1) = \hat{B}(1) \times \hat{T}(1)$.

First let's calculate the unit tangent vector with the formula $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. We note that $\vec{r'}(t) = (1, t, 0)$, and $\| \vec{r'}(t) \| = \sqrt{1 + t^2}$. Therefore $\hat{T}(t) = \left ( \frac{1}{\sqrt{1 + t^2}}, \frac{t}{\sqrt{1+t^2}}, 0 \right )$. Plugging in $t = 1$ and we have that $\hat{T}(1) = \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right )$.

Now let's calculate the unit binormal vector with the formula $\hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|}$. We note that $\vec{r'}(t) = (1, t, 0)$ and $\vec{r''}(t) = (0, 1, 0)$. Therefore $\vec{r'}(t) \times \vec{r''}(t) = (0, 0, 1)$. Clearly $\| \vec{r'}(t) \times \vec{r''}(t) \| = 1$, so $\hat{B}(t) = (0, 0, 1)$. Plugging in $t = 1$ and we get that $\hat{B}(1) = (0, 0, 1)$.

Therefore the unit normal vector can be obtained as $\hat{N}(1) = \hat{B}(1) \times \hat{T}(1)$:

(3)

\begin{align} \quad \quad \hat{N}(1) = (0,0,1) \times \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right ) = \left ( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right ) \end{align}

Therefore the center of curvature (the center of the circle) at $t = 1$ is:

(4)

\begin{align} \quad \vec{r_c}(1) = \vec{r}(1) + \rho (1) \hat{N}(1) \\ \quad \vec{r_c}(1) = \left (1, \frac{3}{2}, 0 \right) + 2^{3/2} \left ( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right ) \\ \quad \vec{r_c}(1) = \left (1, \frac{3}{2}, 0 \right ) + (-2, 2, 0)\\ \quad \vec{r_c}(1) = \left (-1, \frac{7}{2}, 0 \right ) \end{align}

In $\mathbb{R}^2$, the center of the osculating circle at $t = 1$ is thus $\left ( -1, \frac{7}{2} \right )$. We calculated the radius of the circle above ($\rho (1) = 2^{3/2}$) and so the equation of the osculating circle is:

(5)

\begin{align} (x + 1)^2 + \left ( y - \frac{7}{2} \right )^2 = (2^{3/2})^2 \\ (x + 1)^2 + \left ( y - \frac{7}{2} \right )^2 = 8 \end{align}

In $\mathbb{R}^3$, we can represent this circle with the set of parametric equations $\left\{\begin{matrix}x = 2^{3/2} \cos t - 1\\ y = 2^{3/2} \sin t + \frac{7}{2} \\ z = 0\\ \end{matrix}\right.$.