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)

\begin{align} \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 1 & 2t & 3t^2\\ 0 & 2 & 6t \end{vmatrix} = (12t^2 - 6t^2, -(6t), 2) = (6t^2, -6t, 2) \end{align}

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)

\begin{align} \kappa (t) = \frac{ \sqrt{36t^4 + 36t^2 + 4}}{\left (\sqrt{1 + 4t^2 + 9t^4}\right )^3} \end{align}

So the radius of curvature is:

(3)

\begin{align} \rho (t) = \frac{\left (\sqrt{1 + 4t^2 + 9t^4}\right )^3}{\sqrt{36t^4 + 36t^2 + 4}} \end{align}

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