The Evolute of a Curve

# The Evolute of a Curve

Recall from The Osculating Circle at a Point on a Curve page that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function that traces out the smooth curve $C$, then the center of curvature $\vec{r_c}(t)$ can be obtained with the following formula:

(1)
\begin{align} \vec{r_c} (t) = \vec{r}(t) + \rho (t) \hat{N}(t) \end{align}

We note that $\rho (t)$ is the radius of curvature at $t$, and $\hat{N}(t)$ is the unit normal vector at $t$.

 Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$. The Evolute of $\vec{r}(t)$ is the curve generated by the vector-valued function $\vec{r_c}(t) = \vec{r}(t) + \rho (t) \hat{N} (t)$.

Essentially, the evolute of a vector-valued function is a curve traced out by the centers of the osculating circles of $C$ as $t$ varies.

We will now look at an example of finding the evolutes of a vector-valued function.

## Example 1

Determine the equation of the evolute of the function $\vec{r}(t) = (\cos t, \sin t, t)$.

We will first find the radius of curvature. To do so, let's first find the curvature $\kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$. We have that $\vec{r'}(t) = (-\sin t, \cos t, 1)$, and $\vec{r''}(t) = (-\cos t, -\sin t, 0)$. Therefore the cross product $\vec{r'}(t) \times \vec{r''}(t)$ is:

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

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

(3)
\begin{align} \quad \kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3} = \frac{\sqrt{2}}{(\sqrt{2})^3} = \frac{1}{(\sqrt{2})^2} = \frac{1}{2} \end{align}

The radius of curvature $\rho (t) = \frac{1}{\kappa (t)}$, and so:

(4)
\begin{align} \rho (t) = \frac{1}{\kappa (t)} = 2 \end{align}

Now we need to calculate the unit normal vector. Recall that $\hat{N}(t) = \hat{B}(t) \times \hat{T}(t)$.

We first calculate the unit tangent vector $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. We calculated earlier that $\vec{r'}(t) = (-\sin t, \cos t, 1)$ and that $\| \vec{r'}(t) \| = \sqrt{2}$. Therefore:

(5)
\begin{align} \quad \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \left ( -\frac{\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right ) \end{align}

We will now calculate the unit binormal vector $\hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|}$. We've already calculated the cross product $\vec{r'}(t) \times \vec{r''}(t) = (\sin t, -\cos t, 1 )$ and the norm of the cross product $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{2}$ above, and so:

(6)
\begin{align} \quad \hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|} = \left ( \frac{\sin t}{\sqrt{2}}, -\frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right ) \end{align}

Now we can finally find the unit normal vector $\hat{N}(t)$:

(7)
\begin{align} \quad \quad \hat{N}(t) = \hat{B}(t) \times \hat{T}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\sin t}{\sqrt{2}} & -\frac{\cos t}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ -\frac{\sin t}{\sqrt{2}} & \frac{\cos t}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \end{vmatrix} = \left (-\cos t, -\sin t, 0 \right ) \end{align}

Therefore the equation of the evolute of $\vec{r}(t)$ is given by:

(8)
\begin{align} \quad \vec{r_c}(t) = \vec{r}(t) + \rho (t) \hat{N}(t) \\ \quad \vec{r_c}(t) = (\cos t, \sin t, 1) + 2 (-\cos t, -\sin t, 0) \\ \quad \vec{r_c}(t) = (-\cos t, -\sin t, 1) \end{align}

So the evolute is a circle. The following graph represents the helix $\vec{r}(t)$ (in green), and the evolute $\vec{r_c}(t)$ (in yellow).