# Envelope

(Difference between revisions)
 Revision as of 14:20, 11 June 2012 (edit)← Previous diff Revision as of 12:29, 20 June 2012 (edit) (undo)Next diff → Line 46: Line 46: =A gallery of beautiful envelopes= =A gallery of beautiful envelopes= - ==Envelope of lines== + ==Envelopes of lines==

- As we have seen the ladder example, a moving straight line can have a curve as its envelope. Here are more examples: + As we have seen in the ladder example, a moving straight line can have a curve as its envelope. Here are more examples: Line 171: Line 171: If we slide ''O'' along the hyperbola, we will get a ''Lemniscate'' as the envelope of the sweeping circle. 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 [http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli Lemniscate of Bernoulli], from Wikipedia. This is an introduction to Lemniscate and how it was discovered.. For more information about Lemniscate please go [http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli here]. + The Lemniscate is an eight-shaped curve discovered by Jacob Bernoulli [http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli Lemniscate of Bernoulli], from Wikipedia. This is an introduction to Lemniscate and how it was discovered.. For more information about Lemniscates please go [http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli here].

---- ---- Line 197: Line 197: {{HideShowThis|ShowMessage=Click to show the mechanism behind this|HideMessage=Click to conceal the mechanism behind this|HiddenText= {{HideShowThis|ShowMessage=Click to show the mechanism behind this|HideMessage=Click to conceal the mechanism behind this|HiddenText= -
First, let's look at the family of ellipses ${x^2 \over c^2} + {y^2 \over (1-c)^2 } = 1$: First, let's look at the family of ellipses ${x^2 \over c^2} + {y^2 \over (1-c)^2 } = 1$: {{{!}}border="0" cellpadding=20 cellspacing=20 {{{!}}border="0" cellpadding=20 cellspacing=20 Line 216: Line 215: - ===2. The Deltoid as the envelope of Wallace-Simson lines=== + ===2. A Deltoid as the envelope of Wallace-Simson lines===

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- Here is a brief introduction of the Wallace-Simson line, which will be our sweeping line: Here is a brief introduction of the Wallace-Simson line, which will be our sweeping line: Line 276: Line 273: - 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: + 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, 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. + First, we need a more general way to represent curves, because not all curves we have encountered so far can be expressed using a single, closed form function ''y= f''(''x''). For example, a circle of radius ''a'' cannot be represented in this way unless we use two functions, $y = \sqrt{a^2 - x^2}$ and $y = - \sqrt{a^2 - x^2}$. Line 297: Line 294: 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 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 [[#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]]. - - -
- {{HideShowThis|ShowMessage='''A note on terminology: Click to show the definition of''' transcendental funtions.|HideMessage='''Click to hide the definition of''' transcendental funtions.|HiddenText= - - {{{!}} border=1 cellpadding=0 cellspacing=0 - {{!}}
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 pi$(x)$ are polynomial functions of x with rational coefficients. +
- + 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 [[#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, since there is no closed-form, algebraic solution to this transcendental equation. For more about transcendental equations, please go to [http://en.wikipedia.org/wiki/Transcendental_equation this page]. - 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. +

- + -
+ - [[Image:Transcendental.png|center|thumb|350px|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. + -
+ - {{!}}} + - + - |NumChars=0}}
How to represent circles and lines]]}}{{!}}{{!}}{{Anchor|Reference=Figure6-2|Link=[[Image:Levelset4.png|center|thumb|350px|Figure 6-2
How to represent a complicated 2-D curve]]}} {{!}}{{Anchor|Reference=Figure6-1|Link=[[Image:Levelset3.png|center|thumb|350px|Figure 6-1
How to represent circles and lines]]}}{{!}}{{!}}{{Anchor|Reference=Figure6-2|Link=[[Image:Levelset4.png|center|thumb|350px|Figure 6-2
How to represent a complicated 2-D curve]]}} Line 341: Line 317: - 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. As shown in the [[#Transcendental|previous section]], sometimes we can derive an explicit function $y = f(x)$ from the level set, sometimes we can't. But the failure of derivation doesn't mean ''x'' and ''y'' are unrelated. They are still related through this "implicit function". + We will now 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 $F(x,y) = C$, then ''y'' can be viewed as an ''implicit function'' of ''x'', because their values are interrelated. 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 [[#Transcendental|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. The trick is to fix the value of one or more variables, with only two variables left to change. Then these two variables must be related to each other through the implicit function. For example, in the function $F(x,y,z) = 0$, if we fix the value of ''x'', then we will have ''y'' as an implicit function of ''z''. + 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''. - The implicit function theorem can tremendously simplify the derivation of an 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 [[#Figure6-3|Figure 6-3]], which is the ladder problem revisited, we can fix an ''x'' value by drawing a vertical line. Each phase of the ladder intersects this line at a different point, whose y-coordinate is an implicit function of the ladder's position. Now the problem is reduced to finding the highest and lowest point among these intersections, because they must lie on the envelope, as shown in [[#Figure6-4|Figure 6-4]]. + Now we are going to use this theorem 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 [[#Figure6-3|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 [[#Figure6-4|Figure 6-4]]. Line 355: Line 331: - Now, the only problem remains is to determine the maximum and minimum value of y as an implicit function of t. This could be done by using the '''chain rule'''. . The chain rule is a formula in Multivariable Calculus for computing the derivative of the composition of two or more functions. For example, if we have a function + 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

:$F(x(t),y(t))$ :$F(x(t),y(t))$ Line 366: Line 342: - If we apply the chain rule to the level set $F(x,y,t) = C$ with variable x fixed, we will get: + If we apply the chain rule to the level set $F(x,y,t) = C$ with variable ''x'' fixed, we will get:

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

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

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

## Revision as of 12:29, 20 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 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 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. 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 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.

 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 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.

 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 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.

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 equation produces 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 the y-axis, and the minor axis is the 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. 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.

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 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 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].

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

• An envelope of a [...]

• 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, because not all curves we have encountered so far can be expressed using a single, closed form function y= f(x). For example, a circle of radius a cannot be represented in this way unless we use two functions, $y = \sqrt{a^2 - x^2}$ and $y = - \sqrt{a^2 - x^2}$.

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$x^5 + y +\cos y = 1$ 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 6-1How to represent circles and lines Figure 6-2How 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 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 now 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 $F(x,y) = C$, then y can be viewed as an implicit function of x, because their values are interrelated. 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.

Now we are going to use this theorem 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 6-3Fix an x value by drawing a vertical line Figure 6-4Highest 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[4] .

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

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

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 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: that the envelope of a sliding ladder is an Astroid.

### Proof for the Astroid envelope

Figure 6-6

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:

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

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, using 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 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 answer these questions, 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.

## 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 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[7].

## Application in String Art

 Figure 7-4String arts use straight lines to reprensent curves Figure 7-5A 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.

# 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 $\theta$ is the angle of projection. For the physics behind this equation and 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        ${{\partial F(x,y,t)} \over {\partial t}}= {gx^2 \over v^2}t - x = 0$

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 domains[9]in projectile motion.

# 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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