# The Evolute of a Curve Examples 1

Recall from The Evolute of a Curve page that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function that traces the smooth curve $C$, then the evolute of $\vec{r}(t)$ is the vector-valued function $\vec{r_c}(t) = \vec{r}(t) + \rho (t) \hat{N}(t)$ where $\rho (t)$ is the radius of curvature, and $\hat{N}(t)$ is the unit normal vector. In essence, the evolute is a vector-valued function traced by the centers of the osculating circles of $\vec{r}(t)$.

We will now look at some more examples of finding the equation of the evolute of a function.

## Example 1

**Find the equation of the evolute of $\vec{r}(t) = (\cos t, 2 \sin t, 0)$.**

We first need to calculate the radius of curvature. First let's calculate the curvature with the formula $\kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$. First $\vec{r'}(t) = (-\sin t, 2 \cos t, 0)$ and $\vec{r''}(t) = (-\cos t, -2\sin t, 0)$. The cross product between these two vectors yields:

(1)Thus $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(2)^2} = 2$. We also have that $\| \vec{r'}(t) \| = \sqrt{\sin ^2 t + 4 \cos ^2 t} = \sqrt{1 + 3 \cos ^2 t}$ and so $\| \vec{r'}(t) \|^3 = \left ( \sqrt{1 + 3 \cos ^2 t} \right )^3$ . Therefore:

(2)Now let's calculate the unit normal vector $\hat{N}(t) = \hat{B}(t) \times \hat{T}(t)$.

The unit tangent vector is:

(3)The unit binormal vector is:

(4)Therefore the unit normal vector is:

(5)Therefore the equation of the evolute is:

(6)