# The Radius of Curvature 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 Radius of Curvature at $P$ is $\rho (t) = \frac{1}{\kappa (t)}$ provided that $\kappa (t) \neq 0$. |

*For functions $y = f(x)$ with curvature $\kappa (x)$, the radius of curvature is $\rho (x) = \frac{1}{\kappa (x)}$. Notice that in both of these cases that $\kappa \neq 0$. We saw from The Curvature of Straight Lines and Circles page that $\kappa = 0$ only when the curve $C$ is a straight line.*

Simply put, the radius of curvature is just the reciprocal of the curvature at a given point.

## Example 1

**Let $\vec{r}(t) = (t, t^2, t^3)$. Find the radius of curvature at $t = 0$.**

We must first find the curvature of $\vec{r}(t)$. To do so, we will use the formula $\kappa (t) = \frac{ \| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$.

We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = (1, 2t, 3t^2)$, and once again to get $\vec{r''}(t) = (0, 2, 6t)$. Therefore $\vec{r'}(t) \times \vec{r''}(t)$ is given by the following formula:

(1)Thus $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{36t^4 + 36t^2 + 4}$. We also have that $\| \vec{r'}(t) \|^3 = \left (\sqrt{1 + 4t^2 + 9t^4}\right )^3$. Therefore the curvature is given by:

(2)So the radius of curvature is:

(3)Plugging in $t = 1$ and we have that $\rho (1) = \frac{1}{2}$.