### The Evolute: Envelope of Normals

Every curve in the plane has several other curves associated with it. One of the most interesting and important of these is the evolute. Sin t (blue) and its evolute (red).

Suppose the curve ${\gamma}$ is specified in parametric form ${(x(t), y(t))}$ for ${t \in [0,1]}$. The centre of curvature ${\Gamma = (X, Y)}$ at a particular point ${P}$ corresponds to the centre of the osculating circle ${C}$, the circle that best approximates the curve ${\gamma}$ in the neighbourhood of the point ${P}$. The centre of curvature is a function of the parameter: ${\Gamma(t) = (X(t), Y(t))}$, so it traces out a curve. This is the evolute, the locus of the centre of curvature of the curve ${\gamma}$.

Since the tangents to the curve ${\gamma}$ and its osculating circle ${C}$ at point ${P}$ coincide, the centre of curvature is on the normal at ${P}$. Thus, as two neighbouring points on ${\gamma}$ coalesce, the point of intersection of their normals tends to the centre of curvature. So, the evolute ${\Gamma}$ may be described as the envelope of the normals (an envelop of a family of curves is a curve that is tangent to each of them at some point).

Evolutes were first considered by the Greek geometer Apollonius (c. 200 BC) in his famous treatise on conics. Their theory was developed in greater detail by Christiaan Huygens in his study of optics.

From its definition, the equation of the evolute is given by $\displaystyle \Gamma(t) = \gamma(t) + \rho(t) N(t) = \gamma(t) + \frac{1}{\kappa(t)} N(t)$

where ${\rho(t)}$ is the radius of curvature and ${\kappa(t) = 1/\rho(t)}$ is the curvature and ${N}$ is the unit normal to ${\gamma}$. Expanding in cartesian coordinates, we get $\displaystyle \rho = (x^{\prime 2}+y^{\prime 2}) / (x^{\prime}y^{\prime\prime}-x^{\prime\prime}y^{\prime}) \\ X = x - \rho y^\prime \\ Y = y + \rho x^\prime$

where primes are derivatives with respect to ${t}$.

Evolute of a Parabola

If we choose ${\gamma(t) = (x(t),y(t)) = (t, t^2)}$, the curve is a parabola. The slope is ${d y/d x = y^\prime/x^\prime = 2t}$, so the slope of the normal is ${-1/2t}$. The Figure below (Left panel) shows the parabola and a family of curves normal to it. We can see the “caustic curve” or envelope which is tangent to a normal at every point. This is the evolute. A zoomed view (Right panel) shows that it has a cusp at ${(0,0.5)}$. From the explicit expressions above,  we can easily get an equation for the evolute: $\displaystyle Y = \frac{1}{2} + 3X^{2/3}$

a semi-cubical parabola. The cusp corresponds to the vertex of the parabola, the point where ${d\rho/d t = 0}$.

The Four Vertex Theorem The Four Vertex Theorem states that any simple closed smooth plane curve has at least four vertices or points where the curvature is an extremum. The simplest example is an ellipse, with two points of maximum curvature (the pointy’ ends) and two minima (the flattest’ points). Since a vertex corresponds to a cusp of the evolute, an ellipse should have an evolute with four cusps. The Figure shows a family of normals to an ellipse. It is clear that their envelope has four cusps. If we write the ellipse in parametric form ${x=a\cos\phi, y=b\sin\phi}$, and substitute this in the expressions for the evolute, we get $\displaystyle \begin{array}{rcl} aX &=& (a^2-b^2) \cos^3\phi \nonumber \\ bY &=& (a^2-b^2) \sin^3\phi \nonumber \end{array}$

This is an astroid, a stretched form of a four-cusped hypocycloid. It is depicted in the Figure.

Another Example

With some toil, you can derive the equations for the evolute in polar coordinates ${(r,\phi)}$ where ${x = r\cos\phi}$ and ${y=r\sin\phi}$. As a simple application we take the limacon $\displaystyle r = \frac{1}{2} + \cos t \,, \qquad \phi = t$

Using the expressions for the evolute, we obtain the equations for a nephroid, an epicycle of two cusps (see Figure). The two cusps correspond to the maximum and minimum curvature of the limacon. Notice that it has just two vertices (one maximum and one minimum of curvature), which appears to conflict with the Four Vertex Theorem. But the limacon intersects itself and is not a simple curve, so the theorem does not apply.