Envelope

Blue-aerial-shell
Field: Geometry
Image Created By: skylighter.com
Website: skylighter.com

Blue-aerial-shell

This is a beautiful blue-aerial-shell firework filling the sky. Each particle of the firework follows a parabolic trajectory, and together they sweep an area with the red curve as its boundary. This red boundary is then called the envelope of those parabolas. What's more, as we are going to see in the following sections, this envelope also turns out to be a parabola.

Basic Description

In geometry, an envelope of a family of curves is the boundary of these curves' "sweeping area". In most cases, the envelope is tangent to each member of the family at some point.

[A note on terminology: Click to show the definition of a family of curves.]

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[Click to hide the definition of a family of curves.]

A family of curves is a set of curves described by the same function, except for one or more variable parameters. For example, $y = -x + a$ with variable a is a family of straight lines. See image:

Figure 1-1: Ladder and Man

Here is a real world example of envelopes. Suppose there is a ladder leaning on a wall. The ladder starts to slide down because someone steps on it. What will be the shape of the area "swept" by the ladder before it hits the ground? A simulation of this process is shown below:

Figure 1-2: Demonstration of a sliding ladder

Figure 1-3: The complete astroid

Surprisingly, as shown in Figure 1-2, the area swept by a moving straight line does not necessarily have a straight boundary. In fact, its envelope is the first-quadrant portion of an astroidAstroid is the image of implicit function $x^{2/3} + y^{2/3} = a^{2/3}$. One may notice the Astroid is always tangent to the ladder at some point during the sliding process, as stated in the definition of an envelope.

If we slide the ladder in the other three quadrants, we will get a complete star-shaped envelope, as shown in Figure 1-3. In fact, the name astroid comes from the Greek word for "star".

For the math behind this envelope, please go to the More Mathematical Explanation section.

A gallery of beautiful envelopes

Envelopes of lines

As we have seen in the ladder example, a moving straight line can have a curve as its envelope. Here are more examples:

Figure 2-1

Figure 2-2 Gif animation of Parabola envelope

In Figure 2-1, line m, our sweeping line, is perpendicular to segment OA at its midpoint M. Point O is fixed in space.

If we slide point A along line l, line m will sweep out a Parabola, with point O as its focus and line l as its directrix.

[A note on terminology: Click to show the definition of a parabola's focus and directrix.]

[...]

[Click to hide the definition of a parabola's focus and directrix.]

A parabola can be defined as the set of points that are equidistant from a point and a straight line. This point is called the parabola's focus, and this straight line called the parabola's directrix. As shown in the following image, the green and orange segments have the same length by definition. For more about parabola, please go here.

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Figure 2-3

Figure 2-4 Gif animation of Ellipse envelope

Similar to what we did in Figure 2-1, our sweeping line is still the perpendicular bisector of segment AB. The only difference is that point A now slides on a circle, rather a straight line.

The result is an Ellipse with O and B as its foci, as shown in Figure 2-4.

[A note on terminology: Click to show the definition of an ellipse's foci.]

[...]

[Click to hide the definition of an ellipse's foci.]

An ellipse can be defined as the set of points that have a constant sum of distances to two other points. These two points are called the ellipse's foci. As shown in the following image, AF1 + AF2 is constant for all points A on the ellipse. For more about Ellipse and its foci, please go here.

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Figure 2-5

Figure 2-6 Gif animation of Hyperbola envelope

Similar to what we did in the previous example, our sweeping line is still the perpendicular bisector of segment AB. The only difference is that point B is outside the circle.

The result is a hyperbola with O and B as its foci, as shown in Figure 2-6.

[A note on terminology: Click to show the definition of a hyperbola's foci.]

[...]

[Click to hide the definition of a hyperbola's foci.]

An hyperbola can be defined as the set of points that have a constant difference of distances to two other points. These two points are called the hyperbola's foci. As shown in the following image, AF1 – AF2 is constant for all points A on the left half of hyperbola. For more about Hyperbola and its foci, please go here.

So far we have got all of the three Conic Section Curves as envelopes of straight lines. However, the sweeping curve for envelopes is in no way restricted to be a straight line. Circles, ellipses, and other curves can make sweeping curves for fantastic envelopes as well.

Envelope of circles

This section shows some interesting envelopes generated by moving a circle around.

Figure 3-1

Figure 3-2 Gif animation of Cardioid envelope

In Figure 3-1, we begin with a base circle O, which is fixed in space, then select two points A and B on the base circle. Our sweeping circle is centered at A, and passes through B.

If we fix point B and slide point A along the fixed circle, circle A will sweep out a Cardioid, as shown in Figure 3-2.

The name "Cardioid" comes from the Greek word for "heart-shaped". For more information about the Cardioid, please go to this page.

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Figure 3-3

Figure 3-4 Gif animation of Nephroid envelope

Similar to what we did in Figure 3-1, we still have a fixed base circle O, and a sweeping circle that has its center A sliding on the base circle. The only difference is that our sweeping circle is now tangent to a vertical line l, rather than passes through a fixed point.

The result is a Nephroid, which is the Greek word for "kidney-shaped". For more information about Nephroid please go here.

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Figure 3-5

Figure 3-6 Gif animation of Lemniscate envelope

In Figure 3-5, we begin with a hyperbola, with points F1 and F2 as its foci and A as its center. Our sweeping circle has its center O on the hyperbola, and passes through A.

If we slide O along the hyperbola, we will get a Lemniscate as the envelope of the sweeping circle.

The Lemniscate is an eight-shaped curve discovered by Jacob Bernoulli . For more information about Lemniscates please go here.

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Figure 3-7 a variation of "lemniscate"

Figure 3-8 Aha! I have ears like a lemniscate!

In Figure 3-5, if instead of having A as center of the hyperbola, we move it to an arbitrary position between the hyperbola's two halves, then we will get a variation of "lemniscate", which has a funny shape like a bunny's ears.

More complicated envelopes

The following envelopes have more complicated mechanisms than previous ones. But as a result they are even more interesting.

1. The Astroid again, but this time using ellipses

Recall that in Figure 1-1, we showed how to construct an astroid using a line segment sliding on coordinate axes. Actually there is another way to generate the same astroid: using a family of ellipses.

The animation below shows this process. Here the variable parameter is c, which determines the shape of the ellipse. As we can see, when c varies continuously from 0 to 1, the varying ellipse sweeps out an Astroid.

Figure 4-3
Astroid as envelope of ellipses

[Click to show the mechanism behind this]

First, let's loo [...]

[Click to conceal the mechanism behind this]

First, let's look at the family of ellipses ${x^2 \over c^2} + {y^2 \over (1-c)^2 } = 1$:

Figure 4-1

Figure 4-2

Here, because there is a variable parameter c in the equation of the ellipse, we get a family of curves out of this. For every different c value, this equation gives us a different ellipse.

For example, in Figure 4-1, since c < 1/2 , then (1 - c )² > c², so here the major axis is the y - axis, and the minor axis is the x - axis.

However, if we choose another c = 0.89, which is larger than 1/2, then we will have (1 - c )² < c², which makes y - axis the major axis.

To get the envelope of this family of ellipses, we can let c vary continuously from 0 to 1 and trace the area swept by these ellipses. As shown in the previous animation, this envelope turns out to be an astroid, exactly like the one we constructed before in the ladder example.

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2. A Deltoid as the envelope of Wallace-Simson lines

The Wallace-Simson line is related to an interesting theorem in geometry proposed by William Wallace in 1796. The theorem itself is not hard to prove, and with a little manipulation we can get one of the most beautiful envelopes out of it.

The following animation shows this process. Here M is the point that anchors the Wallace-Simson line. When M moves around the circle, that line sweeps out a Deltoid:

Figure 5-3
Deltoid as envelope of Wallace-Simson Line

[Click to show the mechanism behind this]

Here is a brief [...]

[Click to conceal the mechanism behind this]

Here is a brief introduction of the Wallace-Simson line:

Figure 5-1 Perpendicular Projections

Figure 5-2 The Wallace-Simson Line (orange)

The two figures above shows the construction process of Wallace-Simson line. In Figure 5-1, we start by drawing an arbitrary triangle and its circumscribed circle O. Then we select an arbitrary point M on the circumscribed circle, and make perpendicular projections of M onto the 3 sides of the triangle (extend line segment if not inside triangle), intersecting at P, Q, and R.

Wallace claimed that the three projections are on the same straight line (see the orange line in Figure 5-2). This line is called Wallace-Simson Line. A proof of this theorem can be found here.

Since M is an arbitrary point on circle O, we can move it along the circle. Points P, Q, and R are also going to move, since they are perpendicular projections of point M. So we will have a sweeping Wallace-Simson Line, and its envelope is a deltoid, as shown in the previous animation.

[Special note: Is this an envelope or not? (click to show)]

[...]

[Special note: Is this an envelope or not? (click to hide)]

One may be puzzled by the fact that, in Figure 5-3, the Wallace-Simson line actually sweeps across the whole plane. If envelope is defined as "the boundary of area swept by a family of curves", then in this case there should be no envelope at all! So where does this Deltoid come from? And why do people call it an envelope? To answer these questions, we have to look at the animation more carefully. A more thorough examination of the sweeping process reveals the fact that area inside the Deltoid is swept 3 times, while area outside is swept only once. In fact, as shown in Figure 5-4, this sweeping process can be divided into 3 parts, so that in each part the Wallace-Simson line sweeps out 1/3 of the whole Deltoid as strict envelope, without lines from other parts sticking out. The whole Deltoid can be viewed as these segments put together. Although it's not a single, perfect envelope, this doesn't affect its appearance. Figure 5-4 (a) Figure 5-4 (b) Figure 5-4 (c)

This envelope was firstly discovered and proved by Swiss mathematician Jakob Steiner. In 1856 he published a paper, giving a lengthy proof of why we get a Deltoid when moving the Wallace-Simson Line. A simplified version of this proof can be found here.

The Astroid, Cardioid, Nephroid, and Deltoid all belong to the Roulette Family, which means they can also be constructed by rolling one circle around another. For more information about Roulettes, please go to this page.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Calculus

[Click to view A More Mathematical Explanation]

At the beginning of this page, we gave the following definition:

• An envel [...]

[Click to hide A More Mathematical Explanation]

At the beginning of this page, we gave the following definition:

• An envelope of a family of curves is the boundary of their sweeping area.

However, if we want a more mathematical explanation of envelope, we have to redefine it in a more mathematical way, because some problems arise with the original definition when we dive into more math:

First, we need a more general way to represent curves. Before this section we have been describing curves as graphs of a single variable function y= f(x). However, not all curves can be written in this form. For example, a circle of radius a cannot be represented as y= f(x) unless we use two functions, $y = \sqrt{a^2 - x^2}$ and $y = - \sqrt{a^2 - x^2}$. Same thing happens for ellipses, hyperbolas, and other complicated curves. So we need a more general expression to include them in our discussion.

Second, "boundary of the sweeping area" is a rough description in everyday language. It's not something that we can use to derive mathematical formula and equations. So we have to be clear about we mean by "sweep", "boundary", and so on.

In the rest of this section, we are going to deal with these two problems one by one, and show how can we get a good mathematical explanation of envelopes using the new definition.

Resolving the first problem: the power of level sets

The first problem is easily resolved if we describe 2-D curves using level sets, rather than graphs of single variable functions y= f(x).

In Multivariable Calculus, the level set of a two-variable function F(x,y) at height C is defined as the set of points (x,y) that satisfy the condition F(x,y) = C. For example, instead of writing y = 2x - 1 for a line, we could write 2x - y = 1, in which F(x,y) = 2x - y and C = 1.

In general, level sets are more powerful than graphs of single functions when we need to describe 2-D curves, since all single variable functions y = f(x) can be written in the level set form F(x,y) = f(x) - y = 0 , but the converse is not true. For example, the circle in Figure 6-1, can not be rewritten as y = f(x) unless we use multiple functions. In a more extreme case, the level set x⁵ + y + cos y = 1 in Figure 6-2 is not even possible to be reduced to y = f(x) form, since there is no closed-form, algebraic solution to this transcendental equation. For more about transcendental equations, please go to this page.

Figure 6-1 How to represent circles and lines

Figure 6-2 How to represent a complicated 2-D curve

Because of these advantages of level sets, in the rest of this section we will use F(x,y) = C, rather than y = f(x), to describe a family of curves. At least for the purpose of envelopes, the method of level sets is sufficient to describe all 2-D curves that we care about.

Resolving the second problem: the boundary condition

The next question is, given a family of level set curves F(x,y,t) = C with variable parameter t, how can we find its boundary and express it in mathematical language?

The answer is given by the boundary condition, which states that:

For a family of level set curves F(x,y,t) = C with variable parameter t, it's envelope, or boundary of sweeping area, must satisfy the condition:
${\partial F(x,y,t) \over \partial t} = 0$ .

We will prove this condition using the implicit function theorem, which is an important theorem in calculus. The implicit function theorem states that, if we have a level set F(x,y) = C that satisfies some mild conditions, then y can be viewed as an implicit function of x, because their values are interrelated with each other. If the value of y changes, the value of x also has to change, since the condition F(x,y) = C must always be satisfied. As shown in the previous section, sometimes we can derive an explicit function y = f(x) from the level set, sometimes we can't. But the failure of deriving this explicit expression doesn't mean that x and y are unrelated. They are still related through this "implicit function".

This theorem can be generalized to functions of three or more variables. For example, in the level set F(x,y,t) = C , x can be viewed as an implicit function of y and t, y can be viewed as an implicit function of x and t, and so on. Moreover, if we fix one of the variables in this level set, then it's reduced to the 2-variable case. Say, if we fix the value x in the lever set F(x,y,t) = C , then y is an implicit function of t.

With the implicit function theorem in hand, we are now equipped to prove the boundary condition, and find the envelope. Rather than looking at the whole family F(x,y,t) = C , we can fix the value of x, and focus on y as an implicit function of t. For example, in Figure 6-3, which is the ladder problem revisited, we can fix an x value by drawing a vertical line, so that each phase of the ladder intersects this line at a different point. The height, or y-coordinate of this point is an implicit function of the ladder's position, which is in turn determined by the variable parameter t. Now the problem is reduced to finding the highest and lowest ones among all these intersections, because they must lie on the envelope, as shown in Figure 6-4.

Figure 6-3 Fix an x value by drawing a vertical line

Figure 6-4 Highest and lowest points lying on envelope

The maximum and minimum y values can be determined using the chain rule, which is a formula in calculus for computing the derivative of the composition of two or more functions. For example, if we have a function

$F(x(t),y(t))$

in which $x(t)$ and $y(t)$are differentiable functions of t. Then the chain rule claims that:

${dF \over dt} = {\partial F \over \partial x}{dx \over dt} + {\partial F \over \partial y}{dy \over dt}$

Same for function of three or more variables .

If we apply the chain rule to the level set $F(x,y,t) = C$ with variable x fixed, we will get:

${dF(x,y,t) \over dt} = {\partial F(x,y,t) \over \partial y}{dy \over dt} + {\partial F(x,y,t) \over \partial t}{dt \over dt}$

The expression ${dx \over dt}$ didn't appear on the right side because variable x is fixed. Moreover, since ${dt \over dt} = 1$, this expression can be further reduced to:

${dF(x,y,t) \over dt} = {\partial F(x,y,t) \over \partial y}{dy \over dt} + {\partial F(x,y,t) \over \partial t}$

On the other hand, since $F(x,y,t) = C$ is a constant function, we have:

${dF(x,y,t) \over dt} = 0$

So we can get:

${\partial F(x,y,t) \over \partial y}{dy \over dt} + {\partial F(x,y,t) \over \partial t} = 0$

in which ${dy \over dt}$ is the derivative of the implicit function y(t).

For most envelopes, y is at its maximum or minimum as discussed before, so ${dy \over dt} = 0$. And the previous equation is reduced to:

• ${\partial F(x,y,t) \over \partial t} = 0$

which is the boundary condition we are trying to prove.

Conclusion and Application

Now we have the family of level set curves:

• $F(x,y,t) = C$

and the boundary condition:

• ${\partial F(x,y,t) \over \partial t} = 0$

Since every point on the envelope must satisfy both equations, we can combine them to solve for a 2-D envelope curve. However, the calculation involved is rather long and complicated, so here I will only prove a simple case: that the envelope of a sliding ladder is an Astroid.

Proof for the Astroid envelope

Figure 6-6
Model of ladder on wall

As shown in Figure 6-6, the length of the ladder is a. For simplicity we will only consider the envelope in the first quadrant.

Choose the x-coordinate of point A as the variable parameter t. So the y-coordinate of point B is $\sqrt{a^2 - t^2}$

Thus the equation of line AB, our sweeping curve, is:

Eq. 1        $F(x,y,t) = {x \over t} + {y \over \sqrt {a^2 - t^2}} = 1$

Differentiate Eq. 1 with regard to $t$ to get the boundary condition:

${\partial F(x,y,t) \over \partial t} = -{x \over t^2} + {yt \over (\sqrt {a^2 - t^2})^3} = 0$

Move the x term to the right side to get a neater equation:

Eq. 2        ${x \over t^3} = {y \over (\sqrt {a^2 - t^2})^3} = k$

in which we introduced a new variable k to represent their values for convenience. As we have discussed before, Eq. 1 and Eq. 2 already define a 2-D envelope curve. The following steps are just some manipulations that help us to eliminate t and get a direct relationship between x and y.

Substituting Eq. 2 into Eq. 1, we can get:

$kt^2 + k(a^2-t^2) = 1$

which gives us:

$k = {1 \over a^2}$

Substituting back into Eq. 2:

${x \over t^3} = {y \over (\sqrt {a^2 - t^2})^3} = {1 \over a^2}$

We can derive the values of t and $\sqrt {a^2 - t^2}$ from the equation above:

$t = x^{1/3}a^{2/3}$ , and $\sqrt {a^2 - t^2} = y^{1/3}a^{2/3}$

Substituting these values back into Eq.1:

${x^{2/3} \over a^{2/3}} + {y^{2/3} \over a^{2/3}} = 1$

Which leads to:

$x^{2/3} + y^{2/3} = a^{2/3}$ , finally, the equation of an Astroid.

Other proofs are similar.

Why It's Interesting

Although the envelope concept looks like pure math, it does have some interesting applications in various areas, such as Microeconomics, Applied Physics, and String Art.

Application in Microeconomics: the Envelope Theorem

Figure 7-1
The Envelope Theorem

Economists often deal with maximization or minimization problems: to maximize benefit, minimize cost, maximize social revenue, and so on. However, the issue is that there are so many variable parameters in economics. How many men should I hire? How much land should I buy or rent? Should I invest more money to buy new machines, or should I just make with old ones? Because of all these variable parameters, economists often end up doing maximization or minimization of a family of curves, rather than a single curve (see Figure 7-1)

To analyze all these curves at once, economists introduced the Envelope Theorem, which allows them to find the envelope of a family of curves first, and then determine the maximum or minimum value on the envelope. Since no points go beyond the envelope, this point must be the absolute extremum among the whole family of curves.

For more about the Envelope Theorem, please go here.

Application in Physics: Envelope of Waves

Figure 7-2 A typical beating wave with high frequency and slowly changing amplitude

Figure 7-3 An audio signal may be carried by an AM or FM radio wave

In physics, if we combine two waves of almost the same wavelength and frequency, we will get a beating wave (see Figure 7-2). For such a wave, physicists usually care more about its envelope, rather than the wave itself, since the envelope is what people will actually hear, or see. For example, the two branches of a tuning fork are almost, but not exactly, identical. So if a tuning fork starts to vibrate, its two branches will produce two slightly different sound waves. The superposition of these two waves is a beating sound wave with varying amplitude. This is why people can hear "beats" when they strike a tuning fork.

A similar mechanism is used in AM (Amplitude Modulation) broadcasting. Different waves are superposed with each other to form a sinusoidal carrier wave with changing amplitude, which can be used to carry audio signals (see Figure 7-3). For more about broadcasting, please go here.

Application in String Art

Figure 7-4 String arts use straight lines to reprensent curves

Figure 7-5 A 3-D String Art Product

String Art is a material representation of envelopes, in which people arrange colored straight strings to form complicated geometric figures.

For more about String Art please go here.

How the Main Image Relates

As pointed out in the main image, the envelope of all particles' trajectories in an exploding firework is a parabola. Here comes the explanation:

Figure 8-1
Simulation of the blue-aerial-shell firework

Figure 8-1 shows a simulation of the exploding process. Blue parabolas are trajectories of particles, and the red parabola is their envelope.

For the envelope to be parabolic, we have to make several assumptions:

• The firework is composed of many particles, each projected from the origin at the same time, with same velocity v.

• Particles are subject to constant gravity, with gravitational acceleration g.

• Air friction can be neglected. In fact, this turns out to be an contestable assumption. Most firework particles are relatively small and light, so they could be significantly deflected by air friction. However, the case with air friction is way too complicated for this page. Besides, air friction can be neglected, at least for some fireworks with big and heavy particles such as blue-aerial-shell. So we can still accept this assumption and see what happens.

With the assumptions above, we can write out the trajectory of one particular particle using simple mechanics:

$y = x{\tan \theta} - x^2{{g \over 2v^2}(1 + \tan^2 \theta)}$,

in which θ is the angle of projection. For the physics behind this equation and more about projectile trajectory, please go here.

For now, let's leave Physics behind and focus on the curves themselves. In this trajectory, θ is the variable parameter. If we denote tanθ by t , we can write out a family of curves (in level set form):

Eq 1        $F(x,y,t) = y - tx + {{g \over 2v^2}(1 + t^2)x^2} = 0$

Differentiate to get the boundary condition (see the More Mathematical Explanation section):

Eq 2        ${{\partial F(x,y,t)} \over {\partial t}}= {gx^2 \over v^2}t - x = 0$

Substitute Eq 2 into Eq 1 to eliminate t. After doing some algebra we can get:

$y = { v^2 \over 2g } - {gx^2 \over 2v^2}$,

which gives us the parabolic envelope in Figure 8-1.

As we have discussed before, this is not true for all fireworks. Because of air friction, most fireworks have an envelope more like a sphere. Nonetheless, this parabolic pattern can be seen elsewhere, such as in fountains or explosions. This analysis is also useful in the study of safe domainsin projectile motion.

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Related Links

Additional Resources

1.http://jwilson.coe.uga.edu/Texts.Folder/Envel/envelopes.html. I used this page as a starting point.

2.http://poncelet.math.nthu.edu.tw/disk3/summer01/work/861/02/ex2.html. Here are some animations for more cool envelopes.

3.http://www.dynamicgeometry.com/. This is a very helpful geometric software called Geometer's Sketchpad. I used this software to create most of my pictures.

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