Curvature

Let $C$ be a graph of a $C^1$ vector-valued function $\boldsymbol{r}$, parametrized by arc length $s$, with $T'(s)$ existing. The curvature of $C$ at $s$ is

Since $T(s) = \boldsymbol{r}'(s)$ for an arc-length parametrization, if $\boldsymbol{r}''(s)$ exists,

If $\boldsymbol{r}(t)$ is twice differentiable and $\boldsymbol{r}'(t) \ne \vec{0}$, the curvature of $C$ is

Equivalently,

Examples:

• The curvature of a straight line is $0$.
• The curvature of a circle with radius $a > 0$ is $\frac{1}{a}$.

Osculating Circle

If a curve $C$ in two-dimensional space has non-zero curvature $\kappa$ at a point $P$, the circle of radius

that shares a common tangent with $C$ at $P$ and is centered on the concave side of the curve at $P$ is the osculating circle, or circle of curvature, at $P$.

Center of Curvature

The center of curvature of $C$ at a point $P = \boldsymbol{r}(s)$ with $\kappa(s) \ne 0$ is the center of the osculating circle at $P$:

Since $N(s)$ points toward the concave side of $C$, the center of curvature lies at distance $\rho$ from $P$ on that side.

Worked Examples

Helix

Find the curvature and radius of curvature of $\boldsymbol{r}(t) = \cos t\,\boldsymbol{i} + \sin t\,\boldsymbol{j} + t\,\boldsymbol{k}$.

Differentiate twice.

Compute the cross product.

Compute the norms.

Apply the formula.

The radius of curvature is $\rho = 2$ at every point.

Parabola

Find the curvature of $y = x^2$, and the osculating circle at the origin.

Parametrize as $\boldsymbol{r}(t) = t\,\boldsymbol{i} + t^2\,\boldsymbol{j}$ and differentiate twice.

Compute the cross product.

Apply the formula.

At $t = 0$, $\kappa = 2$ and $\rho = \frac{1}{2}$.

The tangent at the origin is $\boldsymbol{i}$, so $N(0) = \boldsymbol{j}$ points toward the concave side. The center of curvature is

The osculating circle is $x^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{4}$.