- The goal here is to develop a method for finding the envelope for a family of ellipses. The method is almost identical to what we did for families of lines. Here we write an equation for an ellipse as an implicit equation \[ A x^2 + B y^2 = C. \] with positive constants $A,$ $B$ and $C.$
-
We first establish a little piece of theory about implicit equations of ellipses. Let the green ellipse be given by the equation $Ax^2+By^2= C$ and the red ellipse by $A_1x^2+B_1y^2= C_1.$ Assume that these two ellipses intersect at the point $(x_0,y_0).$ Subtracting the equations for the two ellipses we get the equation of the orange curve
\[
(A_1-A) x^2+(B_1 - B) y^2 - (C_1 - C)= 0.
\]
The orange curve also goes through the point $(x_0,y_0).$ This is proved by a simple verification. Below is a picture that illustrates this fact.
- Now assume that we are given a family of ellipses \[ A(s) x^2 + B(s) y^2 = C(s). \] Here $A(s)$, $B(s)$ and $C(s)$ are differentiable positive functions defined on a common domain $D.$ In the picture below the ellipses in this family are represented by light gray lines. The picture below indicates that all the ellipses in the given family are tangent to some curve. This curve is called the envelope of the given family. Our goal is to determine the equation for the envelope.
- We use the following strategy to determine the equation of the envelope. We chose $s_0 \in D$ and focus on the ellipse \[ A(s_0) x^2 + B(s_0) y^2 = C(s_0). \] This is the green ellipse in the picture below. The goal is to find the point on the green ellipse which is also on the envelope. That is the green point in the picture below.
- To find the green point we explore the ellipses in the family which are nearby the green ellipse. That is we choose small $\Delta s$ and consider the ellipse \[ A(s_0 + \Delta s) x^2 + B(s_0+ \Delta s) y^2 = C(s_0+ \Delta s). \] This is the red ellipse in the picture below. The red ellipse and the green ellipse intersect at the orange point. Intuitively it is clear that as $\Delta s \to 0$ the orange point approaches the green point.
-
Now comes the brilliant idea: Replace the red ellipse with the orange curve
\[
\bigl( A(s_0+\Delta s) - A(s_0) \bigr) x^2 + \bigl( B(s_0+\Delta s) - B(s_0) \bigr) y^2 = \bigl( C(s_0+\Delta s) - C(s_0) \bigr).
\]
The orange curve also goes through the orange point. But, more importantly, the orange curve can be written as follows:
\[
\frac{ A(s_0+\Delta s) - A(s_0) }{\Delta s} x^2 + \frac{ B(s_0+\Delta s) - B(s_0) }{\Delta s} y^2 = \frac{ C(s_0+\Delta s) - C(s_0) }{\Delta s}.
\]
Taking the limit as $\Delta s \to 0$ in the last equation we get
\[
A'(s_0) x^2 + B'(s_0) y^2 = C'(s_0).
\]
This is the equation of the purple curve at the end of the animation below. This also means that the orange curves have a limiting curve as $\Delta s \to 0.$ Further this means that the limit of orange points in the animation below is the intersection of the purple curve and the green ellipse.
Place the cursor over the image to start the animation.
- Thus, to determine the green point on the green ellipse we intersect the green ellipse and the purple curve, that is the curves given implicitly by \[ A(s_0) x^2 + B(s_0) y^2 = C(s_0) \] and \[ A'(s_0) x^2 + B'(s_0) y^2 = C'(s_0). \] By doing this for every $s$ we get all the points on the envelope.
- Consider the family of ellipses given by \[ \frac{x^2}{\cos s} + \frac{y^2}{\sin s} = 1, \] where $s \in (0,\pi/2)$. Determine the envelope of this family of ellipses.
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For a given $s \in (0,\pi/2)$ the point on the envelope is on the intersection of the curves
\[
\frac{x^2}{\cos s} + \frac{y^2}{\sin s} = 1 \quad \text{and} \quad \frac{x^2}{(\cos s)^2} (\sin s) - \frac{y^2}{(\sin s)^2} (\cos s) = 0.
\]
Solving we get
\[
x^2 = (\cos s)^{3} \quad \text{and} \quad y^2 = (\sin s)^{3},
\]
and further
\[
x = \operatorname{sign}(\cos s) (\cos s)^{3/2} \quad \text{and} \quad y = \operatorname{sign}(\sin s) (\sin s)^{3}, \quad s \in [0,2\pi).
\]
This parametric equation can be written as an implicit equation $|x|^{4/3}+|y|^{4/3} = 1.$
- Consider the family of ellipses given by \[ \frac{x^2}{(\cos s)^2} + \frac{y^2}{(\sin s)^2} = 1, \] where $s \in (0,\pi/2)$. Determine the envelope of this family of ellipses.
Elliptic Integrals
Definition.
The function $E:(-\infty,1] \to \mathbb R$ defined for every $m \in (-\infty,1]$ by
\[
E(m) = \int_0^{\pi/2} \sqrt{1-m \, (\sin \theta)^2\ } d\theta
\]
is called the complete elliptic integral of the second kind. In Mathematica this function is EllipticE[].
Place the cursor over the image to start the animation.
- Problem 1. Evaluate $E(0)$ and $E(1).$
- Problem 2. Prove that the function $E$ is decreasing.
- Problem 3. Determine the range of the function $E.$
- Problem 4. For all $m \in (-\infty,1]$ we have \[ \int_0^{\pi/2} \sqrt{1-m \, (\cos \theta)^2\ } d\theta = E(m). \]
- Problem 5. For all $s \in (0, +\infty)$ we have \[ E\left(1- \frac{1}{s^2}\right) = \frac{1}{s} E\bigl(1-s^2\bigr). \]
- Problem 6. Let $a, b \in (0, +\infty).$ Use the complete elliptic integral of the second kind to express the arc length of the ellipse given by the implicit equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] as a function of $a$ and $b,$ Denote this function by $L(a,b).$ Prove that $L(a,b) = L(b,a).$
- Problem 7. Let $a, b \in (0, +\infty).$ Set $s = b/a.$ Use the complete elliptic integral of the second kind to express the arc length of the ellipse given by the implicit equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] as a function of $a$ and $s,$
- Problem 8. Let $a, b \in (0, +\infty).$ Set $s = b/a.$ Use the complete elliptic integral of the second kind to express the arc length of the ellipse given by the implicit equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] as a function of $b$ and $s,$
- Problem 9. Use Problems 7 and 8 to discover functions $\alpha(s)$ and $\beta(s)$ defined for all $s\in(0,+\infty)$ and such that for each $s\in(0,+\infty)$ the ellipse \[ \frac{x^2}{\alpha(s)^2} + \frac{y^2}{\beta(s)^2} = 1 \] has the arc length $2\pi.$
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The functions from Problem 9 enable us to create the animation below. The pictures below also include envelope of the family of ellipses. To calculate the envelope, we need to introduce the complete elliptic integral of the first kind.
Place the cursor over the image to start the animation.
Definition.
The function $K:(-\infty,1) \to \mathbb R$ defined for every $m \in (-\infty,1)$ by
\[
K(m) = \int_0^{\pi/2} \frac{1}{\sqrt{1-m \, (\sin \theta)^2\ }} d\theta
\]
is called the complete elliptic integral of the first kind. In Mathematica this function is EllipticK[].
Place the cursor over the image to start the animation.
- Problem 1. Evaluate $K(0).$
- Problem 2. Prove that the function $K$ is increasing.
- Problem 3. Determine the range of the function $K.$
- Problem 4. For all $m \in (-\infty,1)$ we have \[ \int_0^{\pi/2} \frac{1}{\sqrt{1-m \, (\cos \theta)^2\ }} d\theta = K(m). \]
- Problem 5. For all $s \in (0, +\infty)$ we have \[ K\left(1- \frac{1}{s^2}\right) = s K\bigl(1-s^2\bigr). \]
- Problem 6. Prove that the derivative of the complete elliptic integral of the second kind is given by the following formula \[ \frac{d}{dm} E(m) = \frac{E(m) - K(m)}{2m} \] for all $m \in (-\infty,1).$
- Problem 7. Find a parametric equation for the envelope of the family of all ellipses whose arc length is $2\pi.$ It is sufficient to find the parametrization of the part of the envelope in the first quadrant. Hint. In Problem 9 in "Complete Elliptic Integral of the second kind" you found functions $\alpha(s)$ and $\beta(s)$ so that \[ \frac{x^2}{\alpha(s)^2} + \frac{y^2}{\beta(s)^2} = 1 \] is a parametrization of the family of all the ellipses whose arc length is $2\pi.$ It turns out that this parametrization can be rewritten in an equivalent form as \[ A(s) x^2 + B(s) y^2 = C(s) \] with exceptionally simple functions $A(s)$ and $B(s).$ Use a parametrization like this to solve the problem.