Envelope

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- In geometry, an '''envelope''' of a [[#Family|family of curves]] is the boundary of these curves' "sweeping area". In most cases, it's tangent to each member of the family at some point. + In geometry, an '''envelope''' of a [[#Family|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. Line 258: Line 258: |NumChars=0}} |NumChars=0}} -

+
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 [http://www.rac.es/ficheros/doc/00045.pdf here][http://www.rac.es/ficheros/doc/00045.pdf M. de Guzman, 2001, ''A simple proof of the Steiner theorem on the deltoid'']. This is a simplified version of Jakob Steiner's proof.. 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 [http://www.rac.es/ficheros/doc/00045.pdf here][http://www.rac.es/ficheros/doc/00045.pdf M. de Guzman, 2001, ''A simple proof of the Steiner theorem on the deltoid'']. This is a simplified version of Jakob Steiner's proof.. Line 299: Line 299: - In general, level sets are more powerful than graph 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. Some level sets, like the circle in [[#Figure6-1|Figure 6-1]], can not be rewritten as y = f(x) unless we use multiple functions. In a more extreme example, the level set$x^5 + y +\cos y = 1$ in [[#Figure6-2|Figure 6-2]] is not even possible to be reduced to y = f(x) form, because $\cos y$ is a [[#Transcendental|transcendental function]]. + In general, level sets are more powerful than graph 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, like the circle in [[#Figure6-1|Figure 6-1]], can not be rewritten as y = f(x) unless we use multiple functions. In a more extreme case, the level set$x^5 + y +\cos y = 1$ in [[#Figure6-2|Figure 6-2]] is not even possible to be reduced to y = f(x) form, because $\cos y$ is a [[#Transcendental|transcendental function]]. {{{!}}border="0" cellpadding=20 cellspacing=20 {{{!}}border="0" cellpadding=20 cellspacing=20 Line 330: Line 330: - ==Resolve the second problem: the boundary condition== + ==Resolving the second problem: the boundary condition==
Line 339: Line 339: - ::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
${\partial F(x,y,t) \over \partial t} = 0$ . + ::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:
${\partial F(x,y,t) \over \partial t} = 0$ . - To see why this is true, let's look more carefully at the [[#Beginning|ladder problem]]. The following two images are different phases of a ladder captured in the sliding process. + This condition is easy to prove using '''the implicit function theorem''', which states that, if we have a level set $F(x,y) = 0$, then ''y'' can be viewed as an ''implicit function'' of ''x''. If the value of ''y'' changes, the value of ''x'' also has to change, since the condition $F(x,y) = 0$ must always be satisfied. Sometimes we can derive an explicit function $y = f(x)$ out of it, sometimes we can't, as shown in the [[#Transcendental|transcendental function example]]. But the failure of derivation doesn't mean ''x'' and ''y'' are unrelated. They are still related through this "implicit function". - {{{!}}border="0" cellpadding=20 cellspacing=20 - {{!}}{{Anchor|Reference=Figure6-3|Link=[[Image:Ladderclose1.png|center|thumb|350px|Figure 6-3
Intersections lying on envelope]]}} - {{!}}} - In [[#Figure6-3|Figures 6-3 and 6-4]], different phases of a sliding ladder are indicated using different colors. One may immediately notice that if we connect the intersections of different phases, we will get a curve close to the astroid envelope. It is not exactly the envelope because the time intervals between different captured phases are finite. However, if we make the time intervals shorter, the curve connecting them will get closer and closer to the envelope. + This theorem can be generalized to functions of three or more variables. For example, look at a function $F(x,y,z) = 0$ of three variables. If we fix the value of one variable, say ''x'', then we will have ''y'' as an implicit function of ''z''. - This is not mere coincidence. Same thing happens in the [[#Ellipticenvelope|elliptic envelope of astroid]] (see [[#Figure6-5|Figure 6-5]]) . Actually it can be shown that if we choose two phases with '''infinitely small time interval''', then their intersection must lie on the envelope. The argument goes as following: + "Well", one may ask, "this is an interesting theorem, but how does it relate to envelopes?" The trick is to do the same thing to the level set that's going to sweep our envelope. In the level set $F(x,y,t) = C$, if we fix the value of x, then y will be an implicit function of t. More importantly, the maximum and minimum value of that implicit function must lie on the envelope, since the envelope should also be the boundary of this implicit function by definition. - {{Anchor|Reference=Figure6-5|Link=[[Image:Ellipsecloseed1.png|center|thumb|500px|Figure 6-5
Elliptic envelope revisited]]}} + For example, let's revisit the [[#Beginning|ladder problem and Astroid envelope]]. First, we can fix an ''x'' value by drawing a vertical line, as shown in [[#Figure6-3|Figure 6-3]]. Each phase of the ladder intersects this line at a different point, and the y-coordinate of the intersection is an implicit function of the ladder's position. What we want, however, is the highest and lowest among these intersections, because they must lie on the envelope. If the position of these two points can be determined, then we can locate two points like this for each fixed x value, and the envelope is just a collection of these highest and lowest points (see [[#Figure6-4|Figure 6-4]]). - In Figure 6-5, it's obvious that in one phase, the only segment that contributes to the envelope is between its intersections with the previous and next phase. For example, in the green ellipse, which has variable parameter t = 0.40, the segment above point A is covered by the orange envelope, thus not contributing to the envelope. Similarly the segment below point B does not contribute since it's covered by the blue ellipse. So the envelope is actually consisted of many "chunks" of segments that's not covered by other phases. + {{{!}}border="0" cellpadding=20 cellspacing=20 + {{!}}{{Anchor|Reference=Figure6-3|Link=[[Image:Astroid_x.png|center|thumb|350px|Figure 6-3
y as an implicit function of t]]}}{{!}}{{!}}{{Anchor|Reference=Figure6-4|Link=[[Image:Astroid_x_2.png|center|thumb|350px|Figure 6-4
Highest and lowest points lying on envelope]]}} + {{!}}} - Now consider what happens if we increase the number of captures, and shorten each time interval: + Now, the only problem remains is to determine the maximum and minimum value of this implicit function. This could be solved by using the '''chain rule''' in multivariable calculus. The chain rule is a formula 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[http://en.wikipedia.org/wiki/Chain_rule The Chain Rule], from Wikipedia. This is a more thorough introduction to the chain rule in multivariable calculus. . - :*Two neighboring phases will approximate each other. - :*Two intersections A and B will be closer. - :*Segment AB will be shorter and smoother. - + If we apply the chain rule to the level set $F(x,y,t) = C$ with variable x fixed, we will get: - Eventually, when the number of captures goes to infinity, points A and B will meet together, and segment AB will be "squeezed" into '''one point on the envelope''' (call this point P). Recall that P is the intersection of two neighboring phases: + - + - + - : $F(x,y,t) = C$ and $F(x,y,t + dt) = C$ + - + - + - which gives us: +

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

- : ${\partial F(x,y,t) \over \partial t} dt = 0$ + Since ${dx \over dt} = 0$ (x is fixed), and ${dt \over dt} = 1$, this expression can be reduced to: +

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

- : ${\partial F(x,y,t) \over \partial t} = 0$ + On the other hand, since $F(x,y,t) = C$ is a constant function, we have: - +

- This is the [[#Boundary|boundary condition]] we are trying to prove. The same argument goes for every other envelope. + :${dF(x,y,t) \over dt} = 0$ - +
- + So we can get: - One may also notice that, since the envelope only touches each curve at this single "squeezed" point, it is '''tangent''' to every curve in the family. In fact, envelope can also be defined as '''a curve that is tangent to every one of a family of curves'''. +

+ :${\partial F(x,y,t) \over \partial y}{dy \over dt} + {\partial F(x,y,t) \over \partial t} = 0$ +
+ in which ${dy \over dt}$ is the derivative of the implicit function y(t). +

+ For points on the envelope, 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|boundary condition]] we are trying to prove. Line 412: Line 420: {{EquationRef2|Eq. 1}}$F(x,y,t) = {x \over t} + {y \over \sqrt {a^2 - t^2}} = 1$ {{EquationRef2|Eq. 1}}$F(x,y,t) = {x \over t} + {y \over \sqrt {a^2 - t^2}} = 1$

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

:${\partial F(x,y,t) \over \partial t} = -{x \over t^2} + {yt \over (\sqrt {a^2 - t^2})^3} = 0$ :${\partial F(x,y,t) \over \partial t} = -{x \over t^2} + {yt \over (\sqrt {a^2 - t^2})^3} = 0$ Line 420: Line 428: {{EquationRef2|Eq. 2}}${x \over t^3} = {y \over (\sqrt {a^2 - t^2})^3}$ {{EquationRef2|Eq. 2}}${x \over t^3} = {y \over (\sqrt {a^2 - t^2})^3}$

- Substitute Eq.1 into Eq.2 twice, first substitute x, and then y, we get: + Divide Eq.1 by Eq.2, use appropriate sides of Eq.2, we can get: +

+ :${x \over t} / {x \over t^3} + {y \over \sqrt {a^2 - t^2}}/ {y \over (\sqrt {a^2 - t^2})^3} = 1/{x \over t^3} = 1/{y \over (\sqrt {a^2 - t^2})^3}$ +
+ Which gives us: +

+ :$t^2 + (a^2 - t^2) = {t^3 \over x} = {(\sqrt {a^2 - t^2})^3 \over y}$ +
+ Keep reducing: +

+ :$a^2 = {t^3 \over x} = {(\sqrt {a^2 - t^2})^3 \over y}$ +
+ And keep reducing:

- :${t \over x^{1/3}} = {\sqrt {a^2 - t^2} \over y^{1/3}} = a^{2/3}$ + :$t = x^{1/3}a^{2/3}$ $;$ $\sqrt {a^2 - t^2} = y^{1/3}a^{2/3}$

- Substitute back into Eq.1, we get: + Substituting back into Eq.1:

:${x^{2/3} \over a^{2/3}} + {y^{2/3} \over a^{2/3}} = 1$ :${x^{2/3} \over a^{2/3}} + {y^{2/3} \over a^{2/3}} = 1$ Line 430: Line 450: Which leads to: Which leads to:

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

- Other proofs are similar.[http://en.wikipedia.org/wiki/Envelope_(mathematics) Envelope], from Wikipedia. This page was particularly helpful for me in the More Mathematical Explanation section. It also has proof for some more envelopes. + Other proofs are similar.[http://en.wikipedia.org/wiki/Envelope_(mathematics) Envelope], from Wikipedia. This page was particularly helpful for me in the More Mathematical Explanation section. It also has proof for some more envelopes. |other=Calculus |other=Calculus Line 455: Line 475: - For more about the Envelope Theorem, please go here[http://www.economics.utoronto.ca/osborne/MathTutorial/MEEF.HTM Martin J. Osborne, ''Mathematical methods for economic theory: a tutorial by Martin J. Osborne'', 2011]. This is a brief introduction to Envelope Theorem in Microeconomics.. + For more about the Envelope Theorem, please go here[http://www.economics.utoronto.ca/osborne/MathTutorial/MEEF.HTM Martin J. Osborne, ''Mathematical methods for economic theory: a tutorial by Martin J. Osborne'', 2011]. This is a brief introduction to Envelope Theorem in Microeconomics.. Line 466: Line 486: - 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 [[#Figure7-3| Figure 7-3]]). For more about broadcasting, please go here [http://en.wikipedia.org/wiki/Amplitude_modulation Amplitude Modulation Broadcasting], from Wikipedia. This page is a more extensive introduction to AM broadcasting.. + 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 [[#Figure7-3| Figure 7-3]]). For more about broadcasting, please go here [http://en.wikipedia.org/wiki/Amplitude_modulation Amplitude Modulation Broadcasting], from Wikipedia. This page is a more extensive introduction to AM broadcasting.. Line 500: Line 520: :$y = x{\tan \theta} - x^2{{g \over 2v^2}(1 + \tan^2 \theta)}$, :$y = x{\tan \theta} - x^2{{g \over 2v^2}(1 + \tan^2 \theta)}$,

- in which $\theta$ is the angle of projection. This is a result from simple mechanics. For more about projectile trajectory, please go here[http://en.wikipedia.org/wiki/Trajectory Trajectory], from Wikipedia. This is the physics behind projectile motions.. + in which $\theta$ is the angle of projection. This is a result from simple mechanics. For more about projectile trajectory, please go here[http://en.wikipedia.org/wiki/Trajectory Trajectory], from Wikipedia. This is the physics behind projectile motions..

For now, let's leave Physics behind and focus on the curves themselves. In this trajectory, $\theta$ is the variable parameter. If we denote $\tan \theta$ by $t$, we can write out a family of curves (in level set form): For now, let's leave Physics behind and focus on the curves themselves. In this trajectory, $\theta$ is the variable parameter. If we denote $\tan \theta$ by $t$, we can write out a family of curves (in level set form): Line 517: Line 537: - 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 somewhere else, such as fountains or explosions. This analysis is also useful in the study of "safe domains" in projectile motion[http://arxiv.org/pdf/physics/0410034v1.pdf Jean-Marc Richard, ''Safe domain and elementary geometry'', 2008]. This is a study about safe domains in projectile motion.. + 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 somewhere else, such as fountains or explosions. This analysis is also useful in the study of "safe domains" in projectile motion[http://arxiv.org/pdf/physics/0410034v1.pdf Jean-Marc Richard, ''Safe domain and elementary geometry'', 2008]. This is a study about safe domains in projectile motion..

Revision as of 16:02, 7 June 2012

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 a parabolic area.

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 family of curves is a set of curves having 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 astroid. 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 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

Envelope of lines

As we have seen 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 parabola can be defined as the set of points that are equidistant to 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. For more about parabola, please go here.

 Figure 2-3 Figure 2-4Gif 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.

 An ellipse can be defined as the set of points that have a constant sum of distance to other two 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.

 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.

 An hyperbola can be defined as the set of points that have a constant difference of distance to other two 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 of envelopes are in no ways restricted to straight lines. Circle, Ellipse, and other curves can make 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.

 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.

 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.

 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 mechanism than previous ones. But as a result they are also more interesting.

1.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.

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, this function refers to a family of curves. For every different c value, we will get a different ellipse.

For example, in Figure 4-1, since $c < {1\over2}$, then $(1-c)^2 > c^2$, so here the major axis is y-axis, and the minor axis is x-axis.

However, if we choose another $c = 0.89$, which is larger than $1\over2$, then we will have $c^2 > (1-c)^2$, 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 following gif animation, the envelope turns out to be an astroid, exactly like the one we constructed in the ladder example:

Figure 4-3
Astroid as envelope of ellipses

2.Deltoid as 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.

Here is a brief introduction of the Wallace-Simson line, which will be our sweeping line:

 Figure 5-1Perpendicular Projections Figure 5-2The 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[2].

Since M is an arbitrary point on circle O, we canmove 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 following animation:

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

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[3].

Notice that 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

An envelope of [...]

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 two serious problems arise with the original definition when we dive into more math:

First, not all 2-D curves can be represented as a single function y = f(x) with a variable parameter t. For example, a circle $x^2 + y^2 = a^2$ cannot be expressed unless we use two functions, $y = \sqrt{a^2 - x^2}$ and $y = - \sqrt{a^2 - x^2}$. So we have to introduce new ways to describe curves.

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 graph 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, like 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$x^5 + y +\cos y = 1$ in Figure 6-2 is not even possible to be reduced to y = f(x) form, because $\cos y$ is a transcendental function.

 Figure 6-1How to represent circles and lines Figure 6-2How to represent a complicated 2-D curve
 In mathematics, functions are divided into algebraic functions and transcendental functions. An algebraic function is informally a function that satisfies a polynomial equation whose coefficients are themselves polynomials with rational coefficients. For example, an algebraic function in one variable x is a solution y for an equation: $p$n$(x)y^n + p$n-1$(x)y^n-1 + ... + p$0$(x) = 0$ where the coefficients $p$i$(x)$ are polynomial functions of x with rational coefficients. A function which is not algebraic is called a transcendental function. In other words, a transcendental function "transcends" algebra because it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction. Since transcendental functions are not algebraic, equations containing transcendental functions may not have well-expressed algebraic solutions. For example, we know that equation $x = \cos x$ must have a solution, because they intersect with each other in the following image. However, because $\cos x$ is a transcendental function, there is no way to express that solution in terms of addition, subtraction, multiplication, division, and root extraction of rational numbers. Probably the best way to solve it is to measure the x-coordinate of A directly. There is no better way to solve y = cos(x) than direct measurement. Similarly, we can not reduce the level set $x^5 + y + \cos y = 1$ into a simple function $y = f(x)$, because we can't solve y in terms of x algebraically.

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

This condition is easy to prove using the implicit function theorem, which states that, if we have a level set $F(x,y) = 0$, then y can be viewed as an implicit function of x. If the value of y changes, the value of x also has to change, since the condition $F(x,y) = 0$ must always be satisfied. Sometimes we can derive an explicit function $y = f(x)$ out of it, sometimes we can't, as shown in the transcendental function example. But the failure of derivation doesn't mean 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, look at a function $F(x,y,z) = 0$ of three variables. If we fix the value of one variable, say x, then we will have y as an implicit function of z.

"Well", one may ask, "this is an interesting theorem, but how does it relate to envelopes?" The trick is to do the same thing to the level set that's going to sweep our envelope. In the level set $F(x,y,t) = C$, if we fix the value of x, then y will be an implicit function of t. More importantly, the maximum and minimum value of that implicit function must lie on the envelope, since the envelope should also be the boundary of this implicit function by definition.

For example, let's revisit the ladder problem and Astroid envelope. First, we can fix an x value by drawing a vertical line, as shown in Figure 6-3. Each phase of the ladder intersects this line at a different point, and the y-coordinate of the intersection is an implicit function of the ladder's position. What we want, however, is the highest and lowest among these intersections, because they must lie on the envelope. If the position of these two points can be determined, then we can locate two points like this for each fixed x value, and the envelope is just a collection of these highest and lowest points (see Figure 6-4).

 Figure 6-3y as an implicit function of t Figure 6-4Highest and lowest points lying on envelope

Now, the only problem remains is to determine the maximum and minimum value of this implicit function. This could be solved by using the chain rule in multivariable calculus. The chain rule is a formula 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[4] .

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

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

Since ${dx \over dt} = 0$ (x is fixed), and ${dt \over dt} = 1$, this expression can be 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 points on the envelope, 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: the envelope of a sliding ladder is an Astroid.

Proof for the Astroid envelope

Figure 6-6

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

Choose x-coordinate of point A as the variable parameter t. So 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:

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

From which we can get:

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

Divide Eq.1 by Eq.2, use appropriate sides of Eq.2, we can get:

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

Which gives us:

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

Keep reducing:

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

And keep reducing:

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

Substituting back into Eq.1:

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

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

Other proofs are similar.[5]

Why It's Interesting

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

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 problem 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 do 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)

So here comes the Envelope Theorem. It allows economists 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 extrumum among the whole family of curves

Application in Physics: Envelope of Waves

 Figure 7-2A typical beating wave with high frequency and slowly changing amplitude Figure 7-3An audio signal may be carried by an AM or FM radio wave

In physics, if we combine two wave 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 [7].

Application in String Arts

 Figure 7-4String arts use straight lines to reprensent curves Figure 7-5A 3-D String Art Product

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

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 consisted of many shinning 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 arguable 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:

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

in which $\theta$ is the angle of projection. This is a result from simple mechanics. For more about projectile trajectory, please go here[8].

For now, let's leave Physics behind and focus on the curves themselves. In this trajectory, $\theta$ is the variable parameter. If we denote $\tan \theta$ 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        $F(x,y,t) = {gx^2 \over v^2}t - x = 0$

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 a parabolic envelope of the particles' trajectories.

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 somewhere else, such as fountains or explosions. This analysis is also useful in the study of "safe domains" in projectile motion[9].

Teaching Materials

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.

References

1. Lemniscate of Bernoulli, from Wikipedia. This is an introduction to Lemniscate and how it was discovered.
2. Simson Line, from Wikipedia. This is a simple proof of the existence of Wallace-Simson line.
3. M. de Guzman, 2001, A simple proof of the Steiner theorem on the deltoid. This is a simplified version of Jakob Steiner's proof.
4. The Chain Rule, from Wikipedia. This is a more thorough introduction to the chain rule in multivariable calculus.
5. Envelope, from Wikipedia. This page was particularly helpful for me in the More Mathematical Explanation section. It also has proof for some more envelopes.
6. Martin J. Osborne, Mathematical methods for economic theory: a tutorial by Martin J. Osborne, 2011. This is a brief introduction to Envelope Theorem in Microeconomics.
8. Trajectory, from Wikipedia. This is the physics behind projectile motions.
9. Jean-Marc Richard, Safe domain and elementary geometry, 2008. This is a study about safe domains in projectile motion.

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