Envelopes and Astroids

By Jean Pierre Mutanguha
tags: Calculus

Envelopes kept cropping up in my doodles but I never gave them much attention. Up until the day I started drawing line segments of a constant length, say \(1\), from one side of a piece of paper to an adjacent side (See Fig 1). It appeared that these lines were approximating a circle. However a quick calculation shows that the distances could not match. The distance from the \(45^\circ\) line (red line) to the corner was longer than expected.

Fig 1. Each line segment (or hypotenuse) has unit length.

Claim. The lines in the square do not form the edge of a circle

Proof: Diagonal of the unit square has length \(\sqrt{2}\). So if the \(45^\circ\) line was tangent to the unit circle then the distance from the line to the corner would be \(\sqrt2 - 1 \approx 0.414\). But remember that the line segment has length \(1\). It also forms an isosceles triangle with the edges of the square. One can show that the altitude of a right isosceles triangle is half the length of the hypotenuse, \(0.5 \neq \sqrt2-1\). \(\Box\)

This immediately grabbed my attention. If it is not a circle then what is it? What was the defining attribute for the points on this "boundary" curve? Soon I had the insight that for any of these points, I could not draw a hypotenuse through it whose length was less than \(1\). This was their defining aspect. I was stuck at this for a while until I took to the internet to do some research.

Shortly after, I stumbled on the topic of envelopes in mathematics; Envelope is the standard name for what I had called boundary curve. Finally, in some (I lost the reference) Google book, I found a proof for the following theorem in multivariable calculus.

Theorem. Let \(f(x,y,\alpha)=0\) is a family of curves in the \(xy\)-plane. \(g(x,y)=0\) is their envelope if and only if it is a solution for \[ \begin{array}{rcl} f(x,y, \alpha)=0 \\ f_\alpha(x,y,\alpha)=0 \end{array} \]

In this theorem, \(f_\alpha(x,y,\alpha)\) stands for the partial derivative of \(f\) with respect to \(\alpha\). I'll reserve the proof of this theorem for another post and skip straight to how it helped me solve my problem.

Fig 2.

Solution 1: I had to characterize these lines in the implicit form \(f(x,y,\alpha)~=~0\). Using Fig 2, each line can be expressed as \[ y = -\cot(\alpha) x + \cos(\alpha) \] and implicitly \[ \cos(\alpha) x +\sin(\alpha)y- \sin(\alpha)\cos(\alpha) = 0\ \\ \ \\ \ (1) \] Implicitly differentiating this equation with respect to \(\alpha\) gives \[ -\sin(\alpha) x +\cos(\alpha)y- (\cos^2(\alpha)-\sin^2(\alpha)) = 0\ \\ \ \\ \ (2) \] The solution of these simultaneous equations is \[ \begin{aligned} x &= \sin^3(\alpha) \\ y &= \cos^3(\alpha) \end{aligned} \] Lastly, its non-parametric form becomes: (using the identity \(\sin^2(\alpha)+\cos^2(\alpha)=1\)) \[ x^{\frac{2}{3}} + y^{\frac{2}{3}}=1 \]

Notice that this solution never used my earlier insight. I was determined to find another solution that did take advantage of this insight. This time around the solution lay in differential equations. For each point \((x_0,y_0)\) on the envelope, let \(m(x_0,y_0)\) be the gradient of the line through \((x_0,y_0)\) with length \(1.\) By the insight, \(m(x_0,y_0)\) is also the gradient of the hypotenuse through \((x_0,y_0)\) with minimum length. Lastly, the unit line through \((x_0,y_0)\) is also tangent to the envelope at that \((x_0,y_0)\). Therefore, the curve satisfies the following differential equation \[ \frac{dy}{dx} = m(x,y) \]

Fig 3.

Solution 2: I need to find the gradient of the \(minimal\) hypotenuse for each point in the square. For any gradient \(m\), the length of the hypotenuse \(h\) through \((x_0,y_0)\) is determined by the following equation (see Fig 3): \[ (y_0-mx_0)^2+\left(x_0-\frac{y_0}{m}\right)^2=h^2 \] Since we are varying \(m\) to minimize \(h\), I'll implicitly differentiate with respect to \(m\) while treating \(x_0\) and \(y_0\) as constants. \[ -2x_0(y_0-mx_0)+2\frac{y_0}{m^2}\left(x_0-\frac{y_0}{m}\right) = 2h \frac{dh}{dm} \] If a minimum exists, the laws of calculus require that \(\frac{dh}{dm}=0\) at that point. \[ \begin{aligned} 0 &= x_0(y_0-mx_0)-\frac{y_0}{m^2}\left(x_0-\frac{y_0}{m}\right) \\ \implies 0&= x_0y_0m^3-x_0^2m^4-x_0y_0m+y_0^2 \\ &=(x_0m^3+y_0)(y_0-x_0m)\end{aligned} \] \[ m = - \left(\frac{y_0}{x_0}\right)^\frac{1}{3} \text{ or } \frac{y_0}{x_0} \] Since the hypotenuse slopes downwards, \(m\) must be negative. So the hypotenuse is minimum at \(m = - \left(\dfrac{y_0}{x_0}\right)^\frac{1}{3}\). We now have this separable first-order differential equation: \[ \frac{dy}{dx} = - \left(\frac{y}{x}\right)^\frac{1}{3} \] whose solution is \[ x^{\frac{2}{3}} + y^{\frac{2}{3}}=h^{\frac{2}{3}} \]

As expected, the equation of those points whose \(minimal\) hypotenuse \(h=1\) is \[ x^{\frac{2}{3}} + y^{\frac{2}{3}}=1 \]

The curve given by \(x^{\frac{2}{3}} + y^{\frac{2}{3}}=1\) is called an astroid. It also has another seemingly unrelated property. In Fig 4, a circle of radius \(0.25\) rolls in a circle of radius \(1\). The path traced out by the point on the smaller circle is the envelope of all line segments between the axes with unit length.

Fig 4. The envelope is an astroid