An Analysis of the Ellipse

8/13/2019 An Analysis of the Ellipse

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Created by: Aravind Narayan, (Kottoor, TC/76/2169, Anayara P.O, Trivandrum, Kerala, India: 695029)

AN ANALYSIS OF THE ELLIPSE, ITS GEOMETRY AND INFERENCE

OF THE RESULTS

PROLOGUE

This document is intended to present two geometric methods of deriving the Arc length of an EllipticalArc Segment. This Arc length of the Elliptical Arc Segment is derived using Geometric Intuition. The

Initial Two Chapters details the two methods of deriving the Arc length.

The next section sketches out an intuitive explanation of the current Mathematical knowledge and

explores the Analogues relationship between the various topics in mathematics. As such there is a short

description of the mathematical logic with which the thought processes of arriving at the geometric

methods are detailed.

Contents

SECTION A: This Section details the following primary chapters

1. Perimeter of the elliptical arc (arc length) a geometric method Transformation Method2. Closed form approximation to the elliptical integral of second kind - Transformation Method3. Perimeter of the elliptical arc (arc length) a geometric method -Chord Length Method4. Closed form approximation to the elliptical integral of second kind -Chord Length Method

SECTION B: This Section details the following secondary chapters

5. Review of Differential Equation6. The Analogy Table7. Geometric Intuition for the Elliptical Arc length8. The Question of Closed form solutions to Integrals


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Chapter -1

PERIMETER OF THE ELLIPTICAL ARC (ARC LENGTH) A GEOMETRIC METHOD

(TRANSFORMATION METHOD)

Abstract

There are well known formulas approximating the circumference of the Ellipse given in various times.

The most famous formulas (approximations) were presented before by Kepler (1609), Euler (1773), Muir

(1883), Lindner (1904), Ramanujan (1914), Hudson (1917), and Cantrell (2001). However there lacks an

Approximation to calculate the Arc length of a given Arc segment of an Ellipse. The Arc length of the

Elliptical Arc is presently given by the Incomplete Elliptical Integral of the Second Kind however a

closed form solution of the Elliptical Integral is not known. The current solution methods are numerical

approximation, based on series expansions of the Elliptical Integral is well developed. This paper

introduces a Geometric Method (procedure) to approximate the Arc length of any given Arc segment

within a quadrant Arc of the ellipse. An analytical procedure of the defined geometric method is detailed.

For a circular Arc the Arc length of the Circular Segment is given through Angular relations. Howeversuch formulae for the Arc length of the Ellipse, derived using known trigonometric Identities is not

known. The Paper translates the defined geometric procedure to analytical method using the known

trigonometric identities. This method thus gives a method to approximate the Incomplete Elliptical

Integral of the Second Kind using trigonometric identities.

The proposal is based on a geometric discovery and the underlying intuitive reasons for the

approximation are as yet not developed. Thus an analytical research into the geometric approximation

shall result in the development of the general method and a refinement of the proposed method.

Introduction

Let us define the Ellipse in the Canonical Position (Major Axis aligned with the x- axis, and the Center at

the origin). Let the Major Axis length be 2a and Minor Axis length be 2b. The Semi Major Axis

(Major Radius) is then a and the Semi Minor Axis (Minor Radius) is then b

The equation of the Ellipse in parametric form is written as

x(t) = cos()y(t) = sin()

The parameter t is called the eccentric anomaly and it is not the angle between the Radius and the X-

axis.

Intent

Our intent is to determine the Arc Length (Le) of the given Arc AB (Perimeter of the Elliptical Arc AB)

lying within a Quadrant Arc of the defined Ellipse.


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Conjure

In order to determine the Arc Length of the Elliptical Arc we transform the Elliptical Arc to a Circular

Arc such that the Circular Arc has the Same Arc length as that of the Elliptical Arc. Now to determine

this Circular Arc we first transform the given elliptical Arc to another Elliptical Arc. For this we define a

transformation on the Ellipse. The transformation acts on a point and transform it to another point on the

ellipse (i.e. its image points). On the given Arc we apply the transformation to identify the image points

of the End points of the original Arc. This Image points are the end points of the Transformed Arc. The

Transformation is defined as

The transformation is such as to covert the parametric value (Eccentric Anomaly) of the Original

Point to the Radial Angle (Angle that the Radius at the point makes with the X- axis) of its Image

Point

Once the transformed Elliptical Arc is defined the Equivalent Circular Arc can be determined

The Equivalent Circular Arc will have a Radius equal to the Radius of the Ellipse having a negative

Slope as that of the Chord of the transformed Elliptical Arc and will subtend an angle equal to the

angle subtended by the transformed Elliptical Arc

From the above conjure we can define a geometric procedure for determining the Arc length of the

Ellipse.

THE GEOMETRIC PROCEDURE FOR DETERMINING THE ARC LENGTH

Procedure

Identify the coordinates of the End points A and B Transform the end points of the ellipse to two new points such that the parametric values of the

end points of the Arc are the Radial Angle of the two new points. Let the two new points be Aiand Bi

Draw a Chord between the two new points obtained by the transformation .i.e. Chord AiBi Determine the Slope of this Chord AiBi Identify the Radius of the Ellipse Rc having the negative slope as that of the Chord AiBi Identify the Angle Subtended by the two new transformed points Ai and Bi at the Center of

the Ellipse.

Draw a Circular Arc with this Radius Rc Subtending an angle . Determine the Arc length of this circular Arc using the Arc length formulae for the Circular Arc This Arc length is equal to the Arc length of the Elliptical Arc

In the following Section an Analytical Method is detailed giving the Analytical application of the

Geometric Procedure.


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ANALYTICAL PROCEDURE AND FORMULA

Review

Given a point on the Ellipse, let the parametric value of the point be ti and the Radial angle (the angle

that the Radius of the Ellipse (at the point) makes with the x- axis), be t. Then the Relation between the

parametric value ti and the Radial Angle t is given as

tan()= .tan()Reference: Appendix B

Identify the coordinates of the End points

Let us consider an Arc formed by two points A and B lying in the first Quadrant. Let the Coordinates

of the point A be (x1,y1) and that of B be (x2,y2). Let the Point Ahave the parametric value t1 the

Point Bhave the parametric value t2 then the coordinates values of Aand Bare

x1= aCos(t1), y1= bSin(t1)

&

x2= aCos(t2), y2= bSin (t2)

Determine the Image points of the End points by applying the transformation

Given the Arc ABhaving end points Aand B, we identify the Image points Ai and Bi of Aand B

respectively. The Coordinate Values (x1,y1) & (x2,y2) forA & B, is transformed to,(x1i,y1i) & (x2i,y2i)

forAi & Bi respectively, andthe parametric value t1 for A is transformed to parameter t1i for Aiand the parametric value t2 for Bis transformed to parameter t2i for Bi

From the definition of transformation, the Transformation is mathematically given as

For point A: (x1,y1) toAi: (x1i,y1i)

tan()= .tan()t1i=[

.tan(1)]

For point B: (x2,y2) toBi: (x2i,y2i)

tan()= .tan()t2i=[ .tan(2)]


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Coordinates of Image Points Ai & Bi are given as

Ai: x1i= aCos(t1i), y1i= bSin(t1i)

Bi: x2i= aCos(t2i), y2i= bSin (t2i)

The Original ArcAB

(Fig 1)

Transformation

The End Points of the Original Arc is transformed to two new points, Ai&Bi shown below,

[Fig 2]

The image points of the Original Elliptical Arc gives two new points. This image points cuts a new Arc.

Thus we have a transformed Arc with End Points the Ai and Bi with Coordinates (x1i,y1i) and (x2i,y2i)

respectively.

(x1i, y1i)(x2i, y2i)

t2t1

BiAi

(x2, y2)

(x1, y1)

A

B


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We can now determine the Chord AiBi

Determine the Slope of this ChordAiBi

In order to determine the slope let us construct the Right Triangle AiBiCishow in [Fig 3a] & [Fig 3b]. Thedifferential length of the x & y coordinates of the two points Ai and Bi is the sides of the triangle

given as

BiCi= xi= [x2i x1i] =[ cos()- cos()]AiCi= yi= [y2i y1i] = [ sin()- sin()]

Now the angle between side AiBiand BiCican be determined since we have the length BiCiand AiCi.

Therefore we can determine the Subtended Angle, [i.e. Angle AiBiCi= ]tan=

The Ellipse showing the construction of the triangle AiBiCiis shown in(Fig 3a)

(Fig 3a)

Ci

(x1i, y1i) (x2i, y2i)

BiAi


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A detailed drawing of the analysis of transformed arc is given in (Fig 3b)

(Fig 3b)

Determine the Radius of the Ellipse Rc having the negative slope as that of the Chord A iBi

With reference to the Conjure the Radius Rcwill have a Radial angle equal to (-). Since we havealready determined () we can determine the length of Radius Rc as

Rc= R(-) = [{()}{(()}](Formulae to determine the Length of the Radius of the Ellipse given the Radial Angle )

(x1i, y1i)

(x2i, y2i)

Bi

Ai

Ci


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The Radius Rc of the Circular Arc (Fig 4)

(Fig 4)

Determine the Angle Subtended by the two new transformed points A i and Bi at the Center

of the Ellipse

Let the Angle subtended by the end points of the transformed Arc Ai Bi at the center be , since the

Radial Angles of the points Aiand Bi are t1 and t2 respectively (Refer [Fig 2]), We can determine

as

=t1t2

Define the Circular Arc Lc with this Radius Rc Subtending an angle

(Fig 5)

The Arc Length of the Circular Arc is Lc is given as

Lc= Rc.

Rc

Rc

Rc

-

(xc, yc)

Lc


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Determine the Arc Length Leof the Elliptical Arc using the Arc length formulae for the Circular

Arc

Appendices

Appendix A:Ellipse in Canonical Position

Appendix B:Relation between the Radial Angle and Parametric Value (Eccentric Anomaly)

Appendix C:Sample Problem; Application of Method

Appendix D:Salient Features of the Method

Elliptical Arc Length (Le) Rc.


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Appendix A: Ellipse in Canonical Position

We define the Ellipse in the Canonical Position (Major Axis aligned with the x- axis and the Center at the

origin). Let the Major Axis Length be 2a and Minor Axis Length be 2b. The Semi Major Axis (Major

Radius) is then a and the Semi Minor Axis (Minor Radius) is then b. See [Fig 1]. Below

(Fig 1)


x(t) = a Cos(t )

y(t) = b Sin(t)


axis.


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Appendix B: Relation between the Radial Angle and Parametric Value (Eccentric Anomaly)

Consider a point on the Ellipse in canonical position, given by the coordinates (xi, yi). The Semi

major and Semi minor axis lengths is a and b respectively. Let the point be parameterized by

the parametric Value ti. We can now define a relationship between the Radial Angle i.e. the

Angle that the radius of the Ellipse (at the given point) makes with the x- axis and the parametricvalue ti. Let this radial angle be defined as t.

The relation between parametric value and Radial Angle (Fig 2)

(Fig 2)

The Coordinate values in terms of parametric value ti is

xi= a Cos (ti)

yi= b Sin (ti)

From (Fig 2), the Tangent of the Radial Angle is given as

tan() =

tan()= ()()tan()= .(()())

(xi, yi)

t

yi

i


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tan()= .tan()

.tan()= tan()

tan()= .tan()

Definition: - Radial Angle: is the angle that the Radius at a given point makes with the x- axis

ti= [ .tan()]


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Appendix C: Sample Problem; Application of Method

Problem Description

It is required to calculate the Arc Length (Le) of an Elliptical Arc given the Coordinates of the End points

of the Arc. Let the End points be designated A & B and the Coordinates be (x1, y1) and (x2, y2)

respectively. We are required to determine the Arc length.

The Semi Major Axis and Semi Minor Axis are a and b respectively. The Ellipse is aligned in the

Canonical Position i.e. center at origin and Major axis along the x- axis.

Solution

Step 1: Determine the parametric value of the end points of the given Arc

We can identify the parametric value for the End Points. Let the Parametric value be given as t1 for

point A and t2 for point B.

The parametric values are given as

x1= a Cos (t1)

Cos (t1)=

t1= 1 ------------------------------------------(1)x2= a Cos (t2)

Cos (t2)= t2= 2 ------------------------------------------(2)

Step 2: Determine the parametric value of the (Transformed) Image points of the given end points

As a next step we apply the transformation equations to the end points in order to determine the Image

points of the End points of Arc. The Image point for A is given as Ai and for B is given as Bi. The

parametric value t1i for Ai and the parametric value t2i for Bi can be determined

By applying the Relation between Radial Angle at a point and its parametric value

The transformation for Point A to Ai is given as :

tan()= .tan()t1i=[ .tan(1)] ---------------------------------------(3)


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And for Point B to Bi as

tan(

)=

.tan(

)

t2i=[ .tan(2)] --------------------------------------(4)The Coordinate Values of the image points are given as

Ai: x1i= aCos(t1i), y1i= bSin(t1i) ----------------------------------(5)

Bi: x2i= aCos(t2i), y2i= bSin (t2i) ----------------------------------(6)

Step 3: Determine the Radial angle -

of the Equivalent Circular radius

The Slope of the Chord AiBibe tan(), can be determined using the equationtan()=

tan()= ---------------------------------------(7)= tan 2121 ---------------------------------------(8)

The Radial angle for the equivalent Circular Arc is -Step 4: Determine the parametric value tc for the point of the ellipse at which the Radius of equivalent

Circular Arc will occur, from the Radial Angle - asForm 1

tan()=

.tan()= (-1).

.tan()tc= tan[(1). .tan()] -------------------------------(9)Refer: Appendix B


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Form 2

Substituting for tan() from Equation (7)in Equation (9)we gettc= tan

[(

1).

.

21

2

1

] -------------------------(10)

Form 3

Substituting for (x2i, x1i , y2i,y1i) from equation (5)and (6) in equation(10)

tc= tan[(1). .()()()()]

tc= tan[(1). .{()()}{()()}]

tc= tan[(1).

. .{

(

)

(

)}

{()()}]tc= tan

[(1). ()()()()] -----------------------------(11)Step 5: Determine the Coordinate Values (xc , yc) of point on the ellipse at which the Radius of the

Equivalent Circular Arc will be, using the parametric value tc.

xc= a Cos(tc) --------------------------------------(12)

yc= b Sin(tc) -------------------------------------(13)

Step 6: Determine the Length of the Radius of the Equivalent Circular Arc Rc as

Form 1: Using Coordinate Values (xc, yc)

Rc= + -----------------------------------(14)Substituting from equation (12)& (13)

Rc= cos()+ sin() ----------------------------------(15)


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Form 2: Using the Radial angle - the radius Rc can be determined asRc= R() = [{()}{(()}] --------------------(16)

Step 7: Determine the Subtended angle of the Equivalent Circular Arc

The subtended angle is the Angle subtended by the image points Ai & Bi at the center of the Ellipse

and is given as

=t1t2 ----------------------------------------(17)

Step 8: Determine Arc Length Leof the given Elliptical Arc as



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Appendix D: Salient Features of the Method

Suppose the major and minor axis lengths are made equal the Ellipse reduces to a circle. Now ifthe Arc length is determine using the Method the Result Match with Arc length for the Circular

Arc

When the Limits are applied to cover one Quadrant Arc and we Scale it four times we get theCircumference of the Ellipse. This Circumference matches with the Circumference obtained with

Euler Method

An analysis of the Output from the Formulae with a Numerical Approach gave the followingNoticeable features

The Error between the Numerical Output and the Result from the formulae is Equal to theError with the Euler Formulae

However when an Arc length smaller than the Quadrant Arc is considered the Error isalways less than the Error for the Complete Quadrant Arc. However the distribution of

error is not uniform.

The Error in comparison with the Numerical Result is more when the Arc considered isnear to the End point of the Major Axis

Note: Explanation: The Error with numerical method is more for Arcs nearer to the Major

Axis Quadrant point. This is because the Numerical method considers differential straight

lines. Since the Curvature varies more towards the end point of the Ellipse the numerical

Accuracy is bound to reduce when this numerical method is applied to Arcs nearer to the

Major Axis Quadrant point.

When an Arc nearer to the Minor Axis Quadrant point is considered then the NumericalError is bound to be less. The Formulae Result Matched closely with the Numerical

Result for Arc Segment in this Region.

In case an Arc length is to be determines which crosses a Quadrant point we split the Arc to two arcs with

the Quadrant point as the separation point. We can then sum the two arcs to obtain the Complete Arc

length.


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Chapter -2

CLOSED FORM APPROXIMATION TO THE ELLIPTICAL INTEGRAL OF SECOND KIND

(TRANSFORMATION METHOD)

Abstract

Elliptic Integrals are a class of Integrals which arose with the problem of giving the Arc length of the

Ellipse. Modern mathematics generalizes the Elliptic Integral to a general class of Integral functions. In

general elliptic Integrals cannot be expressed in terms of elementary functions.

The Elliptical Integral arises in mathematical modeling of many physical situations. Examples of physical

problems modeled by the class of elliptical Integrals include period of a simple pendulum, Deflection of

thin elastic bar etc.

The form of the Elliptical Integral which gives the Arc length of an Elliptical Arc segment is known as

the Incomplete Elliptical Integral of the Second Kind. Applying the limits so as to cover the Arc Segment

of one complete quadrant of the Ellipse gives the complete Elliptical Integral of the Second Kind.

Introduction

The Incomplete Elliptical Integral of the Second Kind, gives the Arc length of the Elliptical Arc. Hence a

geometric method to determine the Arc length shall give a method to obtain a closed form approximation

of the Elliptical Integral of the Second Kind. This paper presents a closed form approximation to the

Elliptical Integral of the Second kind using a newly discovered geometric intuition to approximate the

Arc Length of an Elliptical Arc segment within a Quadrant Arc of the Ellipse.

Closed Form Approximation to the Incomplete Elliptical Integral of Second Kind

A brief description of the geometric intuition as well as application of the intuition used to arrive at the

closed form solution to the Integral is detailed below

The Geometric Intuition

Given an Elliptical Arc segment within a quadrant Arc of an Ellipse we can define a Circular Arc having

the same Arc length as that of the Elliptical Arc. This circular Arc will have the following parameters.

The Angle of the Circular Arc will be equal to the difference in the Eccentric anomaly of the End points

of the given Elliptical Arc. The Radius of this Circular Arc will be the Radius of the Ellipse which has a

negative slope as that of the Chord of the Ellipse whose end points have the their Radial Angles equal tothe eccentric anomaly of the end points of the given Arc segments. The Radial angle at a point is the

angle that the Radius of the Ellipse at that point makes with the x-axis. The circumference of the Circular

Arc can then be determined using the standard formulae for the Arc length of the Circular Arc. This will

give the Arc length of the Elliptical Arc.


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Incomplete Elliptical Integral of Second Kind

The Elliptical Integral giving the Arc length of the Elliptical Arc segment in trigonometric form is given

as

Le=

[1

sin(

)

]

Where e is the Eccentricity of the Ellipse, the current known solution to the integral is through

numerical Integration. A method to approximate a closed form solution to this integral using the

geometric intuition is detailed below.

Approximation to the Incomplete Elliptical Integral of Second Kind

From the Geometric Method determining the Arc Length, we can approximate the incomplete

Elliptical Integral as well as the Complete Elliptical Integral of the Second Kind. In our Method

of determining the Arc Length we approximate the Elliptical Arc to an equivalent Circular Arc.

Let the Elliptical Arc length be represented Le and the Circular Arc length be Lc

LeLc

Le= (1 sin() () =Lc

Lc=() = () = ()( )Let

( ) = Then

() = (). (1 sin() () =().

Result

[1 sin()] R().


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Analyzing the geometric intuition the Radius, R ()is given as() = [{()}{(()}]

Where is given as = tan 2121Wherex1i,y1i,x2i,&y2iis given as

x1i= aCos(t1i), y1i= bSin(t1i)

x2i= aCos(t2i), y2i= bSin (t2i)

Wheret1i,&t2iis given as

t1i=[ .tan(1)]t2i=[ .tan(2)]

And is given as

=t1t2

Where t1& t2 are the limits of the incomplete, Elliptical Integral of Second kind

Conclusion and Future Prospects

Developing an analytical theory to explain the geometric result can be a precursor to developing a method

of solving the Elliptical Integrals.

Appendices

Appendix-A: Incomplete Elliptical Integral of Second Kind, derived as the Arc Length of Ellipse

Appendix-B: Complete Elliptical Integral of Second Kind from the Incomplete Elliptical Integral of

Second Kind


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The Arc Length of any Arc can be given by taking the Integral of the differential length. Let the Elliptical

arc length be represented as (Le). Let the Differential length of the elliptical arc be given as dl The

Elliptical Arc length is given by the integral

Le= Now dl =() +()

Le= () +() ---------------------------------------(1)Consider the ellipse with the Semi Major Axis and Semi Minor Axis, a and b respectively. The Ellipse

is aligned in the Canonical Position i.e. Center at Origin and Major axis along the x- axis.

The Equation of the Ellipse in Parametric form is

x (t)= a Cos (t)

y (t)= b Sin (t)

The Parameter t is the eccentric anomaly of the ellipse

Differentiating,x (t)with respect to t we get

= sin() = sin() ----------------------------------------(2)

Differentiating, y (t) with respect to t we get

= cos() = cos() ----------------------------------------(3)

Supposed we want to determine the Arc length between two points A & B on the Ellipse. Let the

Parametric value for the points A be t1 and the parametric value for the points B be t2.Now

substituting equation (2) & (3) in the Integral Equation (1) and applying the parametric values of the end

point of the Arc as the limits of the Integral will give the Integral equation for the Arc length of theElliptical Arc AB

Substituting Equation (2) and (3) in Equation (1)

Le= { sin() + cos() Le={ sin() + cos() } ------------------------------------(4)


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Consider the trigonometric identity

cos()= 1 sin()Substituting the Trigonometric identity in Equation (4) and Analyzing

Le= { sin() + (1 sin()2)} Le= { sin() + ()2)} Le= { +( sin() ()2)}

Le= { +( 2) ()2} Le= { (

2

) ()2} Le= [ {1}{()}] Le= (1 [{1}{()}]) Le= (1 [{1 }{()}]) Le=b (1 [{1}{()}]) -------------------------------(5)

Eccentricity of the ellipse commonly denoted e is given as

= ( ) =1

= ( )

= {1

}

Substituting in the eccentric anomaly e in Equation (5) we get

Le= [1 sin()] -----------------------------(6)This equation is known as the in complete Elliptical Integral of the Second Kind


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Second Kind

If we consider one Quadrant Arc of the Ellipse, the arc length is obtained by applying the limits

t2=0 and t1= we get the Complete Elliptical Integral of the Second Kind as

Le= (1 sin() ----------------------------(7)Once we have the Elliptical Arc length of the Quadrant Arc we can determine the Circumference as the

Quadrants are symmetric.


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Chapter -3

PERIMETER OF THE ELLIPTICAL ARC (ARC LENGTH) A GEOMETRIC METHOD

(METHOD OF CHORD LENGTH)

Abstract

There are well known formulas approximating the circumference of the Ellipse given in different periods

in history. However there lacks a formula to calculate the Arc length of a given Arc segment of an

Ellipse. The Arc length of the Elliptical Arc is presently given by the Incomplete Elliptical Integral of the

Second Kind, however a closed form solution of the Elliptical Integral is not known. The current solution

methods are numerical approximation methods, based on series expansions of the Elliptical Integral. This

paper introduces a Geometric Method (procedure) to approximate the Arc length of any given Arc

segment of the ellipse based on the Chord Length. An analytical procedure of the defined geometric

method is detailed.

For a circular Arc the Arc length of the Circular Segment is given through Angular relations. However a

similar formula for the Arc length of the Ellipse, using known trigonometric Identities is not known. ThePaper translates the defined geometric procedure to analytical, algebraic method using the known

trigonometric identities. This method thus gives a method to estimate the Incomplete Elliptical Integral of

the Second Kind using trigonometric identities.

The formulae for circular Arc can be considered to be based on the intuitive reasoning Equal Arcs Cuts

Equal Angles at the center for a Circle. This paper presents a similar intuitive logic for the Elliptical

Arc. The proposal is based on a geometric discovery based on intuitive reasoning and the algebraic theory

explaining the intuitive logic is as such not yet developed. An analytical research into the intuitive logic

behind the geometric approximation shall result in the development of the general theory and an

analytical, algebraic proof of the proposed method.

Introduction

Let us define the Ellipse in the Canonical Position (Major Axis aligned with the x- axis, and the Center at

the origin). Let the Major Axis length be 2a and Minor Axis length be 2b. The Semi Major Axis

(Major Radius) is then a and the Semi Minor Axis (Minor Radius) is then b.

The equation of the Ellipse in parametric form is given as

x(t) = cos()y(t) =

sin(

)

The parameter t is called the eccentric anomaly and it is not the angle between the Radius and the x-axis.

Intent

Our intent is to determine the Arc length (Le) of the given Elliptical Arc Segment AB (Perimeter of the

Elliptical Arc AB).


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The Conjure

In order to determine the Arc Length of the Elliptical Arc we transform the Elliptical Arc to a Circular

Arc such that the Circular Arc has the Same Arc length as that of the Elliptical Arc. The following

Conjure defines this Equivalent Circular Arc.

The Equivalent Circular Arc having the same Arc length as that of a given Elliptical Arc, will have aChord length equal to the Chord length of the given Elliptical Arc and it (Circular Arc) will subtend an

angle at the center whose value in radians is equal to the difference in the Eccentric anomalies of the

end points of the given Elliptical Arc

From the above conjure we can define a geometric procedure for determining the Arc length of the

Ellipse.

The Geometric Procedure for Determining the Arc Length

Procedure

Identify the coordinates of the End points A and B of the given Elliptical Arc Segment Draw a Chord between the two End points A & B of the Elliptical Arc segment .i.e. Chord

AB

Determine the length of this ChordAB Determine the eccentric anomalies (Parametric Values) of the End points A and B of the given

Elliptical Arc Segment

Determine the difference between the Eccentric Anomaly Values as angle in Radians. Construct a Circular Arc whose Chord length is equal to the Chord length AB of the given

Elliptical Arc and Subtending an angle equal to the difference between the Eccentric Anomaly

Values of the end points of the Elliptical Arc. Now for constructing this Circular Arc the length of

radius of the Circular Arc can be determined by the following procedure

o Construct an isosceles triangle whose base length is equal to the length of the Chord ofthe Elliptical arc, and the Angle opposite to the base is . The two equal angles are

then. (All angles in Radians)o Determine the length of the Equal Side of this Isosceles triangle, and Radius of the

Circular Arc is equal to this length of the equal side of the Isosceles triangle

The Arc length of this Circular Arc is equal to the Arc length of the Elliptical Arc.

In the following Section an Analytical Method is detailed giving the Analytical application of the

Geometric Procedure.


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Analytical Procedure and Formula

Identify the coordinates of the End points

Let us consider an Arc formed by two points A and B lying in the first Quadrant. Let the Coordinates

of the point A be (x1,y1) and that of B be (x2,y2). Let the Point Ahave the parametric value t1 and

Point Bhave the parametric value t2 , then the coordinates values of Aand Bare

x1= aCos(t1), y1= bSin(t1)

&

x2= aCos(t2), y2= bSin (t2)

The Elliptical ArcAB

(Fig 1)

Chord Length ABcan be determined by constructing the Right triangle ABC. The chord AB forms the

hypotenuse of the Right triangle ABC. Side AC is given by the differential x-coordinate length, and

Side BC is given by the differential y-coordinate lengths, of the End points A: (x1, y1) & B: (x2, y2)

of the given Elliptical Arc

AC = x=x2-x1

BC = y=y2-y1

h = AB = () +()

(x2, y2)

(x1, y1)A

BC


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Equivalent Circular Arc

The Equivalent Circular Arc will have a Chord length equal to AB. Now for a Circular Arc the Radius

will always be same. Hence a Chord of the Circular Arc always cut an Isosceles Triangle with the two

equal sides, equal to the Radius.

Determining the Subtended Angle

at the Center of the Circular Arc

Form our Conjure the Equivalent Circular Arc will cut an angle equal to the difference between the

parametric values of the End points of the Circular Arc. Let this Angle be .

Then is calculated as

= t1 t2

Determining the Radius of the Circular Arc

In order to determine the Radius of the Circular Arc we now define the isosceles triangle formed by the

two radiuses of the Circular Arc and the Chord of the Circular Arc.

From the Conjure, the Circular Arcs Chord length is the same as that of the Elliptical Arc. Hence we

know the base length h of the Isosceles Triangle as length of AB. Also we know the angle opposite to

the base of the Isosceles triangle as from the Conjure.

We can now algebraically determine the length of the Side of the Isosceles triangle denoted Rc which

will be equal to the Radius of the Circular Arc using the formulae

=

Where h is the Chord length and is the subtended angle opposite to the base.

For derivation;Reference: Appendix A

Determine the Arc length

We now have the Radius and Subtended angle of the Circular Arc. We can now define the Circular Arc

(Fig 2)

h

Rc

Rc


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Result

Once we have defined the Circular Arc whose Arc length is equal to the Elliptical Arc we can determine

the Elliptical Arc length Le using the Arc length formula for Circular Arc as

Elliptical Arc Length (Le) =Rc.Substituting for Rc, we get the Equation in Single form as

() = .

Substituting forh and we get the formulae in terms of the coordinates of the End points of the

Elliptical Arc and the parametric values of the End points of the Elliptical Arc as

() = ( ) +( ) .{ }

Conclusions

Thus we have a formula in standard trigonometric functions to determine the Arc length of given

Elliptical Arc segment.

Suppose the Major and Minor axis lengths are made equal the Ellipse reduces to a circle. In this case thederived formula reduces to the standard Arc length formula for the Circular Arc.

Appendices

Appendix A:Determine the side of an Isosceles Triangle given the Angle opposite to the Base and

Base length

Appendix B:Sample Problem; Application of Method


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Appendix A: Determine the Side of an Isosceles Triangle from the Angle opposite to the

Base and Base length

Consider the Isosceles triangle PQR. Let the Base length QR be given as h and the length of the equal

Sides PQ & PR be given as Rc. Let the angle RPQbe given as . Then the equal angles PQR&

QRP are given as, (Since sum of the angles of a triangle is equal to ).

From Law of Sines we get

( ) = ( ) =

()

From Equation (1) Consider the Identity

( ) = ()

Rc Rc

h

P

Q R


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Let us analyze

( ) = ()


= (The Angle A is in Radians)

Applying the identity to Equation (3) we get

( ) = ()

Substituting result (4) in Equation (2) we get

= ()

We know from the Double Angle Formulae

=Applying this to

=

= ()


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Substituting result (6) in Equation (5) we get

=

= ()

Result

=


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Appendix B: Sample Problem; Application of Method

Problem Description

It is required to calculate the Arc Length (Le) of an Elliptical Arc given the Coordinates of the End points

of the Arc. Let the End points be designated A & B and the Coordinates be (x1, y1) and (x2, y2)

respectively. We are required to determine the Arc length.

The Semi Major Axis and Semi Minor Axis are a and b respectively. The Ellipse is aligned in the

Canonical Position i.e. center at origin and Major axis along the x- axis.

Solution

Step 1: Determine the parametric value of the end points of the given Arc

Let the Parametric value (Eccentric Anomaly) be given as t1 for point A: (x1, y1) and t2 for point B:

(x2, y2).

The parametric values are then determined as

= cos()cos() =

= 1 ------------------------------------------(1)

= cos()cos() =

= 2 ------------------------------------------(2)Step 2: Determine the Subtended angle of the Equivalent Circular Arc

The subtended angle has a value in radians equal to the difference between the parametric values

(Eccentric Anomalies) of the End points of the Given Arc, i.e. t1 for Point A & t2 for Point B The

value of in Radians is determined as

=t1t2


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Step 3: Verify the y coordinate values of the end points A & B of the Elliptical Arc as

= sin = sin

Step 4: Determine Length of the Chord of the Equivalent Circular Arc h

The Chord of the Equivalent Circular Arc will have the same Arc length as that of the Chord of the

Elliptical Arc. Since they have the same length let us denote the Chord of the Elliptical Arc also as h.

The length h is then given as

= + = ( ) +( )

Step 5: Determine Radius of the Equivalent Circular Arc Rc as

=

Step 6: Determine Arc Length of the given Arc Elliptical Arc Leas



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Chapter -4

CLOSED FORM APPROXIMATION TO THE ELLIPTICAL INTEGRAL OF SECOND KIND

METHOD OF CHORD LENGTH

Abstract

Elliptic Integrals are a class of Integrals which arose with the problem of giving the Arc length of the

Ellipse. Modern mathematics generalizes the Elliptic Integral to a general class of Integral functions. In

general elliptic Integrals cannot be expressed in terms of elementary functions.

The Elliptical Integral arises in mathematical modeling of many physical situations. Examples of physical

problems modeled by the class of elliptical Integrals include period of a simple pendulum, deflection of

thin elastic bar etc.

The form of the Elliptical Integral which gives the Arc length of an Elliptical Arc segment is known as

the Incomplete Elliptical Integral of the Second Kind. Applying the limits so as to cover the Arc Segment

of one complete quadrant of the Ellipse gives the complete Elliptical Integral of the Second Kind.

Introduction

The Incomplete Elliptical Integral of the Second Kind, gives the Arc length of the Elliptical Arc. Hence a

geometric method and formula to determine the Arc length shall give a method to obtain a closed form

approximation of the Elliptical Integral of the Second Kind. This paper presents a closed form

approximation to the Elliptical Integral of the Second kind using a newly discovered geometric method

based on intuition to estimate the Arc length of an Elliptical Arc segment.

Closed Form Approximation to the Incomplete Elliptical Integral of Second Kind

A brief description of the geometric intuition as well as application of the intuition used to arrive at the

closed form solution to the Integral is detailed below

The Geometric Intuition

Given an Elliptical Arc segment of an Ellipse we can define a Circular Arc having the same Arc length as

that of the Elliptical Arc. This circular Arc will have the following parameters. The Angle of the Circular

Arc in Radians will be equal to the difference in the Eccentric anomaly values of the end points of the

given Elliptical Arc. The Circular Arc will have a chord length equal to the chord length of the Elliptical

Arc. Thus we now know the Chord length and the Angle subtended at the center of the Equivalent

Circular Arc. From these parameters we can derive the Arc length of the Circular Arc which will be equal

to the Arc length of the Elliptical Arc.


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Incomplete Elliptical Integral of Second Kind

The Elliptical Integral giving the Arc length of the Elliptical Arc segment in trigonometric form is given

as

Le=

[1

sin(

)

]

Where e is the Eccentricity of the Ellipse, the current known solution to the integral is through

numerical Integration. A method to estimate a closed form solution to this integral using the geometric

intuition is detailed below.

Estimating the Incomplete Elliptical Integral of Second Kind from the Geometric Intuition

From the Geometric method determining the Arc Length, we can estimate the incomplete

Elliptical Integral as well as the Complete Elliptical Integral of the Second Kind. In our method

of determining the Arc Length we transform the Elliptical Arc to an equivalent Circular Arc with

the same Arc length. Let the Elliptical Arc length be represented Le and the Circular Arc lengthbe Lc

LeLc

Le= (1 sin() () =LcLc=() = () = ()( )

Then

(1 sin() (t) =().( )

Result

Analyzing the geometric intuition the Radius, R (t)is given as

() = ( ) +( )

[1 sin()] R(t).( )


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Wherex1,y1,x2,&y2is given as

x1= a Cos(t1), y1= b Sin(t1)

&

x2= a Cos(t2), y2= b Sin (t2)

Where a & b are Constants (i.e. semi-major and semi-minor axis of ellipse), and t1& t2 are

the limits of the Incomplete, Elliptical Integral of Second kind.

Solution

The Closed form solution in Single form is given as

[1 sin()] = (

)

+ (

)

{1 2}

[1 sin()] = ( ) +( ) {1 2}

Conclusion and Future Prospects

Developing an analytical, algebraic theory to explain the geometric result based on the geometric intuition

can be a precursor to developing a method of solving the Elliptical Integrals. The Method can be further

refined to encompass the whole class of functions defined as the Elliptic function. The geometric fact

certifies an analogy

Appendices


Appendix-B: Complete Elliptical Integral of Second Kind from the Incomplete Elliptical Integral ofSecond Kind


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The Arc Length of any Arc can be given by taking the Integral of the differential length. Let the Elliptical

arc length be represented as (Le). Let the Differential length of the elliptical arc be given as dl The

Elliptical Arc length is given by the integral

Le= Now dl =() +()

Le= () +() ---------------------------------------(1)Consider the ellipse with the Semi Major Axis and Semi Minor Axis, a and b respectively. The Ellipse

is aligned in the Canonical Position i.e. Center at Origin and Major axis along the x- axis.

The Equation of the Ellipse in Parametric form is

x (t)= a Cos (t)

y (t)= b Sin (t)

The Parameter t is the eccentric anomaly of the ellipse

Differentiating,x (t)with respect to t we get

= sin() = sin() ----------------------------------------(2)

Differentiating, y (t) with respect to t we get

= cos() = cos() ----------------------------------------(3)

Supposed we want to determine the Arc length between two points A & B on the Ellipse. Let the

Parametric value for the points A be t1 and the parametric value for the points B be t2.Now

substituting equation (2) & (3) in the Integral Equation (1) and applying the parametric values of the end

point of the Arc as the limits of the Integral will give the Integral equation for the Arc length of theElliptical Arc AB

Substituting Equation (2) and (3) in Equation (1)

Le= { sin() + cos() Le={ sin() + cos() } ------------------------------------(4)


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cos()= 1 sin()Substituting the Trigonometric identity in Equation (4) and Analyzing

Le= { sin() + (1 sin()2)} Le= { sin() + ()2)} Le= { +( sin() ()2)}

Le= { +( 2) ()2} Le= { (

2

) ()2} Le= [ {1}{()}] Le= (1 [{1}{()}]) Le= (1 [{1 }{()}]) Le=b (1 [{1}{()}]) -------------------------------(5)

Eccentricity of the ellipse commonly denoted e is given as

= ( ) =1

= ( )

= {1

}

Substituting in the eccentric anomaly e in Equation (5) we get

Le= [1 sin()] -----------------------------(6)This equation is known as the in complete Elliptical Integral of the Second Kind


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Second Kind

If we consider one Quadrant Arc of the Ellipse, the arc length is obtained by applying the limits

t2=0 and t1= we get the Complete Elliptical Integral of the Second Kind as

Le= [1 sin()] ----------------------------(7)Once we have the Elliptical Arc length of the Quadrant Arc we can determine the Circumference as the

Quadrants are symmetric.


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Chapter -5

A REVIEW OF DIFFERENTIAL EQUATION

Analysis Preceding Differential Equations

We start our mathematics with algebra. Taking a geometric representation view of algebra we

have studied functions, the geometric representation of functions in 2D, 3D space or in higher

dimensions. We have seen that the functions can represent a curve in 2D or 3D and a Surface

in 3D. We also developed the vector functions as a equivalent form of Cartesian system to

easily represent a function or a surface or shape.

Before we analyze functions we define spaces and meaning of functions. We have three spaces

we can define geometrically, the (One Dimensional) 1D Space, (Two Dimensional) 2D Space

& (Three Dimensional) 3D Space. The 1D Space is geometrically represented as a number line

(x-axis), the 2D Space is geometrically represented as an infinite Straight Plane (represented x-y plane using x & y axis). The 3D Space is geometrically represented as an infinite Straight

Volume (represented x-y-z Euclidean space using x, y & z axis). In each of the Spaces are

defined as a collection of points and by defining the space we mean that we can mathematically

define each point by defining the coordinate. A point in the 1D plane is defined by defining the

coordinate value in the number line. A point in the 2D space is defined by defining the (x, y)

coordinates on the x-y-axiss and any point in the 3D space is defined by defining the three

coordinates the (x, y, z) on the x-y-z- axis.

We explained previously that a geometric object can be defined on a space. However for any

given space the geometric object that can be defined on the space will start with objects of one

dimension less. This means if we have an n-dimensional space. We can define geometricobjects having dimensions [(n-1), (n-2) ----, 0] in this space. Thus in a 3D Space we can define

a 2D object (surface), a 1D object (Curve), a 0D object the point. Similarly in a 2D Space we

can define a 1D object (Curve), a 0D object the point etc.

Now that we have defined spaces as points we can see how we can attribute properties to these

spaces. We can add properties to these spaces through functions. By saying that we add

properties to this space we mean that we attribute properties to each point on the space. The

property added to a point in the space can be of two forms. The property assigns only a scalar

value to the point. The property assigned to the point adds a scalar value to a point in a

particular direction that is we assign a vector value to a point on the space. This is achieved by

defining a function on the space, for the former case we use a scalar function and for the latter

we use a vector function.

A physical application of the above mathematical concept is consider a flat rectangular metal

plate. We can define this rectangular Plate as a 2D space by defining the edges as the

coordinate axis. With the coordinates we can define any point on the plate geometrically by

defining the coordinate of the point in the reference axis. Now we have used the concept of

space to define each point on the plate uniquely. Now we attribute property to teach of the


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point. Let us consider the property of Temperature at each point on the plate. If one end of the

plate is heated the temperature will vary from point to point as the plate is not heated

uniformly. The temperature is a scalar value. Now a scalar function can be defined in the form

f(x, y) such that you plug in the values of the geometric coordinates of a point in the plate into

the function f and we get the temperature at that point as the output. Similarly we can define

another physical property at each point as a quantity defining the direction of maximum rate ofchange of temperature and the magnitude of this rate of change. This physical property can be

defined using a vector, where the assigned vector gives the direction and magnitude of

maximum rate of change of temperature. This is defined by a vector function of the form F(x,

y) which assigns a vector to any point defined by the geometric coordinates (x, y).

Defining a function on a space attributes values to each point in our geometric space. Such a

geometric space with a property attached to each point in the space is called a Field. If the

property is a scalar value then we have our space converted to a Scalar field and if the property

attached to each point is a vector then we have a vector field. When such a property is attached

to our space it is no more a simple geometric space but a Field. Now we go to visualizing a

Field in a geometric sense.

A scalar filed in particular dimensional space can be seen as a geometric object of the same

dimension in the next higher dimension. That is if we have a scalar field defined as function in

an n-dimensional space, then this can be visualized as an n-dimensional geometric object in

(n+1) dimensional space. Example a scalar field defined on a one dimensional number line

can be defined as 1D Curve on a 2D space. The curve is a geometric object. Similarly a scalar

field defined on a two dimensional x-y plane can be defined as 2D Surface (possibly Curved)

on a 3D space. Thus a 1variable function representing a curve in 2D space can be a scalar field

in a 1D space and similarly a 2variable function representing a surface in 3D space can be a

scalar field in a 2D space.

For a Vector function 2varaiable vector function is represented in 2D space itself and is

visualized by assigning arrows showing the vectors to each point in the 2D space. Similarly a

3varaiable vector function is represented in 3D space itself and is visualized by assigning

arrows showing the vectors to each point in the 3D space. Thus an n-variable vector function

assigns n-dimensional vectors to n- dimensional space.

We have seen the algebraic equations, the equations as functions and the meaning of functions

as scalar field or vector fields and also as geometric objects in space.

Next we went on to the manipulations of the functions. In this we studied Differentiation of the

functions. That is splitting a function into infinitely small parts. If our function represented acurve then our differentiation split our curve into infinitely small elements (for this we define

tangents to the curve or tangent planes if it is a surface). Thus if our curve was not straight the

infinitely small elements can be approximated to be straight. We then analyzed the Integration

that is adding together of these infinitely small elements. These methods helped us to analyze

and arrive at certain Qualities or factors of these functions that we studied such as the length of

the curve or, The Area of Surface, Volume of a closed region (bounded by a surface) etc.


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We developed two forms of representation of functions the analytical model that is using

algebraic functions and then the vector model where Vector functions where used to represent

points curves figures etc.

We even went on to study vector fields where we defined vectors functions and then

Differentiated and integrated them. The differentiation and integration allowed us to arrive at

quantities of the function which cannot be found by other means Such as area of a given

surface or Volume of a surface.

In vector cases also we can differentiate and integrate vector fields using Del Operators.

However in all the studies we had a function which we defined on a 2D or 3D space and then

study the properties of the functions.

Geometrically we define objects such as Points, Curves, Surfaces etc. then we use the tools to

study and arrive at the properties of theses geometric objects. In order to study the properties of

the geometric object we define the object as algebraic functions and use Calculus

(differentiation (Differentiation for 2D and Partial Differentiation for 3D) and integration(Simple integration, Double Integration, Triple Integration depending on the dimension and

application). Alternatively we also define the objects in vector form and apply vector calculus.

In the Vector Space form we first defined another form of geometric objects called as vectors.

We then went on to define another form of Space called the Vector Space. This was our normal

2D or 3D Space but with vectors assigned to each point in the space. The geometric

representation of the space is still in the same form as using the coordinate axes to define each

point in the space with its coordinate values. However unlike the scalar field instead of scalar

values for the space we assigned vector to each point defined by the coordinates. Thus our

normal space with points was converted to a Vector Field. We then analyzed the properties of

this defined (the Space) Field again by doing Differentiation and Integration of vector spaces.

The differentiation of vector spaces was different from the differentiation of the functions as a

vector was differentiated. This was achieved using the Del Operator and we defined two

forms of differentiation of the Vector space as the Curl and Divergence. The Integration of

these spaces was achieved using the double and Triple Integrals as usually however we had a

vector derivative as the differential for the integration. We also developed several theoremsto simply the Integration. The important theorems are the Greens Theorem, Stokes Theorem &

Gauss Divergence Theorem. (Stokes theorem applied to Differentiation as Curl and Divergence

applied for Differentiation as Divergence)

This is a note on the development in defining the geometric space and objects and studying theproperties of objects.

Differential Equations

With differential equation we have to do the opposite. We are given certain conditions instead

of functions. We need to define a space and a function in the space that will satisfy our

conditions. Thus if our Equation involves two variables only it is obvious that out function lies


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in the 2D space. But the important thing is that we should have sufficient conditions to define a

geometric object in the space that will satisfy our conditions.

In our above discussion we had seen that a geometric object is represented by functions if our

space can be defined geometrically that is 1D, 2D or 3D.Hence the given some conditions the

solutions of differential equations are functions that satisfy the conditions. The Differential

Equations give conditions that not only the function but also the differentials of the function

should satisfy. The solutions are functions that satisfy the conditions specified. Now since we

have seen that functions can be represented geometrically as geometric objects. The solutions

of our differential equation can also be represented and visualized as geometric objects if it is

within the realm of the 3D space.

In the algebraic case we represented the function as s geometric object in the corresponding

geometric space. Then the coordinate values of the points of the geometric object gave the

coordinates of the algebraic function. In fact the algebraic function defines the conditions

satisfied by the geometric object in the space and correspondingly we define the geometric

object. However with the differential equation case we have seen that the differential equationspecifies the condition for the entire space and not just a geometric object. Hence the

conditions of the differential equation define the space itself which means every point in the

space and the coordinates of the space. The conditions defined by the differential equation to

each point in the space is geometrically represented as a vector field by assigning vectors to

every point in the geometric space with each vector satisfying the condition specified by the

differential equation at that point. (The condition specified by the differential equation i.e. the

derivative is the slope of the vector and hence the direction of the vector. The representation

vectors only represent the direction and is hence vector field is also called as slope field or

direction field)

We should also note that the solution depends in the conditions defined by the differentialequation. If we have sufficient conditions and the form in which we have the conditions we can

define solution of differential equations. The solutions depend on the conditions specified. For

a general differential equation we normally have a number of solutions that is a number of

functions satisfying the specified conditions. This means we can say that we can have more

than one geometric object that can be defined in a space that will satisfy the conditions

specified in the differential equation. Hence we have to have specific conditions with which we

can define a particular solution. That is if we are given only a differential equation we may

have a number of functions that can satisfy the condition of the differential equation and hence

we can plot a number of geometric objects that will satisfy the conditions specified in the

Differential Equation. However in order to identify a particular function or the geometric

object as the solution we should have additional conditions. Sometimes this is given as initial

value problem that is the additional conditions are initial values. Thus gives the start point of

the solution hence we have the particular function which is the solution we are looking for. In

the geometric sense the condition gives the start point of the geometric object. In another case

this gives the boundary conditions for the solution. This will allow us to fit a function that will

satisfy the conditions of our differential equation.


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Just as for algebraic function given a coordinate x the corresponding y value satisfying the

function can be determined by identifying the point on the geometric curve representing the

algebraic function having the given x coordinate. The y coordinates of this point is our

corresponding y value satisfying the function. Similarly if we plot the direction filed of a

given differential equation which is the geometric plot of our space satisfying the differential

equation we have a vector field. If we are given conditions then we can define the point on thespace satisfying the conditions. We start with the vectors satisfying the differential equation.

We take an infinitesimally small differential length. At the end point of this vector we have

another vector defined by the differential equation and at the end point of the second one

another new vector and so on. By connecting the vectors starting from the defined point we can

get a curve which is satisfying the differential equation subject to the condition. By making the

differential length infinitesimally small we can get a smooth curve which satisfy the condition

of the differential equation and is the solution of the differential equation for the given

conditions. This is analogous to identifying the y value satisfying a algebraic function for a

given x value by plotting the algebraic function as a curve and indentifying the point on the

curve have the x coordinate equal to the given x value. The corresponding y coordinate

value is our solution satisfying the algebraic function for the given x coordinate. The same is

done with differential equation. However the only difference is that the initial condition

represents a point on the space and the solution represents a curve instead of a value as was the

case with the algebraic equation.

Since the solution of our differential equations were curves (not values (y-coordinate value) as

was the case with algebraic equation) the solution of the differential when written algebraically

will be functions. (Since functions represent the curves geometrically)

Thus a differential equation gives us certain conditions and we have to fit functions satisfying it. These

functions form the solutions of the differential equation. A number of physical situations are modeled as

differential equations. The solutions are functions satisfying the differential equations and hence for 2D

and 3D the solutions can be plotted as geometric objects.

Analyzing a Differential Equation in the Geometric frame

We can analyze a differential equation by representing the differential equation geometrically.

We did this for a normal equation by representing the equation as function in the x-y Cartesian

plane. However for the differential equation we do not represent the equation as a function

Instead the differential Equation is represented as a Vector field called the direction filed. The

solutions to the differential equation are then curves on this space. This is analogues to our

algebraic equation in that if an algebraic equation is represented as a curve (function) given anx value the corresponding y value can be found by plotting the point on the curve

corresponding to x value and then finding the y coordinate of that point on the curve. For a

solution for the differential equation also we will do a same procedure only difference is that

given a point on the vector field representing the differential equation we first identify the vector

at the point and connecting the vectors we will plot the solution curve for the point. In the

algebraic case our solution was a Value the y- coordinate corresponding to the given x-


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coordinate. However in the vector case our solution is a curve (a function) corresponding to the

given initial conditions.

Representation of Differential Equation as Vector Field

We are familiar with representing a function as a curve in the x-y plane. An Algebraic equation

can be converted to a function form and represented as a curve geometrically. Now we will

explain how a differential equation is represented as a Vector field.

A geometric interpretation of the derivative is as the slope of a tangent. Now if we have a curve

representing a function the derivative of that function will give the slope of the tangent to the

curve. By substituting the point into the derivative function we can get the slope of the curve at

that point. Thus using the derivative we can define the tangent to the point on the curve with the

slope.

If we represent the curve as a vector function with the vector equation giving the position vector

for a unique point on the curve. The tangential vector can be given as a derivative of the vector

function. Thus the tangent vector at a point can be obtained by taking the derivative of the

position vector of the particular point.

Now as we discussed before the algebraic equation gives the condition for a geometric object

(curve) in a space i.e. defines the curve in the space e.g., a 1D curve in the 2D space.

Analogously a differential equation defines not a geometric object in the space but the entire

space itself. Hence we can define the tangent vectors from the derivative at each point in the

space.

We have also seen that the tangent vector is defined by the first derivative only. Hence if wehave a first order differential equation of the form

y = f(y, t)

This will not be a problem as we can define a y-t plane and plugging in value of y & t for

points on the plane we can get the corresponding slope values as y and with the slope values

plot the direction vectors at each point on the y-t space.

But what if our differential equation is of a higher order that is say a second order

equation of the form

y=f(y,y, t )

In this case y cannot be thought as the slope as it is only the first order equation which

form the slope of a differential equation. However in this case we can do some algebraic

substitutions. With these substitutions we can get the second order differential equation as a

system of two fist order equation with two unknown variables to be solved. Given sufficient

conditions we can define these two first order equations. Here since the solution to differential


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equations is actually functions the unknowns of the system of differential equations are also

functions.

Now this system of differential equation can be written in matrix form. With this we get

matrix equations. In our analysis of Linear Algebra we know that a vector can be written as a

column matrix and operated by using Matrices. Writing this System of Differential Equation inMatrix form will give us just this that is a Vector Equation with a Matrix as the Vector operator.

Thus our Second order differential equation is converted to a first order matrix

differential equation with a 2x2 matrix A as the operator. Thus if we input a vector into theequation the Operator A acts on to give the derivative vector .

As with our preceding discussion the derivative vector

represents a tangent vector.

Hence if we define a Space whose coordinates are the entries of vector which is theunknown functions of our system of equation that is the solution of our second order differentialequation, we can define a vector space where the vector this space is given by the derivative

vectors that is the tangent vectors. Thus we get a tangent vector space. This is the geometricrepresentation of the Second order differential equation.

We can generalize this to any order differential equation with the dimensions of the vectors in the Vector

equation of the corresponding differential equation being equal to order of the differential equation. So if

we have a 3rd

Order Differential equation we will have a system of 3 First Order Differential

Equations and the vector in the phase plot will be three dimensional vectors. Thus the Geometricplot will be three dimensional vector fields.

Thus for an nth order Differential Equation the Vector field will be an n-dimensional

vector fieldcurve or surface.

General Solution & Numerical Solution

We have seen that if given sufficient conditions to identify a point in the vector field then we

can identify a curve satisfying the differential equation by connecting vectors to get the

solution for the given conditions.

In some cases we will be able to identify an algebraic solution to the differential equation in the

form of a function. We then form the general solution to the differential equation in the form of

algebraic function. However there will still be some unknown terms in the general solution.

Once we have the general solution we will go forward for a particular solution. Given an initial

condition we will substitute into the general solution to identify the unknown constant. This

general solution then transforms into a particular solution identifying a single function.


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The geometric meaning of the general solution is that if we plot the vector field of the

Differential equation that is the direction filed we have a general pattern for the solution curves

that is the curves formed by connecting vectors follow a general pattern. The Algebraic

function forming the general solution defines this pattern. Substituting values for the constant

terms in the general solution defines different curve. However since the function is the same

the patterns of the curves remains same. Giving the initial conditions defines a constant andthus defines one of these curves. Thus by finding a general solution we define a general

function which will give a particular shaped curve. This curve can be translated to cover the

entire space. Thus we have infinite number of curve of specific pattern spanning the entire

space. Giving initial conditions defines a particular curve of the infinitely possible individual

curves in the space.

In many cases however we do not have algebraic methods developed to define a general

solution that is a general shape of the curve. It is perfectly possible as we know that if we make

arbitrary hand drawing of a curve in 2D space it is not necessary that we have function

representing the shape. Thus if we write a differential equation and we plot a curve by

connecting vectors then it is not necessary that we have a algebraic function representing thefigure we define. Thus it is perfectly normal not to have a general solution for a given

differential equation. In fact it more generally the rule and most of the differential equations do

not have a solution defined as an algebraic function.

However with the geometric method of plotting the direction field (vector field) of the

differential equation we can identify the solution for any given condition. First we plot the

direction field of the differential equation. We define the point in the direction field satisfying

the given initial condition. We then plot our solution curve from this initial point by connecting

vectors. Thus we can define the curve which is the solution to our differential equation for the

given conditions. This solution is geometric one and there may not be algebraic function

representation of our geometric curve. However we have the solution defined for the given

differential equation for the given condition. In this procedure we have to note that the

accuracy of our solution depends on how small we consider the direction vectors i.e. the

differential length of the direction vectors. Reducing the differential length increases the

accuracy, smaller the length more accurate. However when we plot geometrically we have a

limitation on how small we can get. Hence there is always an error associated with this

approach.

We have seen the geometric method of arriving at a solution. However it is tedious to

geometrically plot a differential equation. Also if our initial condition is define away from the

origin we may need to have an extremely large plot which will become impossible. In order to

avoid this and generate a geometric solution without having to generate a complete field plot

we have developed numerical method

Given a differential equation the geometric method identifies the point corresponding to the initial

conditions for the solution. It then defines a vector at the point satisfying the differential equation. This

vector is made infinitesimally small and then the next vector is defined thus by connecting lengths we

define the solution. The numerical method follows an iterative method to define the vector. As we


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increase the number of iterations the smaller becomes the differential length and hence the accuracy

increase. However with the increase in iterations the computational time also correspondingly increases.

This represents a geometric meaning of the numerical solution of the differential equations also.

The classical numerical methods for first Ordinary (Single Variable) Differential Equations are the

Eulers Method, Midpoint Method & Runge Kutta Method etc.

However the solution curve obtained by the above methods is geometric one and we may not be able to

represent it algebraically. If we draw a smooth geometric curve in a plane it is not necessary that we are

able to write it as algebraic equation. In fact there are very few curves which can be written as an

algebraic equation. In this case also we will not get an algebraic solution to the curve. However we can

still use some curve fitting methods such as (Poissons distribution etc.) once we numerically plot our

geometric curve to obtain an algebraic representation of our geometric plot.

Partial Differential Equations

Partial differential equations just generalize this concept to higher dimensions. In partial differential

equations we can have more than one variable. If it is a two variable equation then we have our solution

in the 3D geometric space. The differential equation defines a 3D vector field. The solution is then a two

variable function that is a surface. Geometrically the solution of our differential equation forms a surface.

If we go for a graphical solution we can plot a surface. Now this is possible only with a computer

software program.

This is a more tedious process than the ODEs (Ordinary Differential Equations) where we plotted a

solution curve on a 2D direction field. Here our solution is a surface instead of a curve. Hence we will

define small surface elements and then connect them together to form our solution surface. This is

analogues to forming differential length vectors and connecting them together to form our solution curvefor ODEs.

If we go for a numerical approach we will have to define small surface elements instead of differential

lengths. These surface elements then connected together from the solution. This approach for partial

differential equations of two variables whose solution is defined as surface is called Finite Element

Analysis.

Now analogously for a three variable differential equation the solution is in the form of a geometric

volume. The geometric and corresponding Numerical approach is called as the Finite Volume Method.


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Chapter -6

THE ANALOGY TABLE


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Chapter -7

GEOMETRIC INTUITION FOR THE ELLIPTICAL ARC LENGTH

Geometric Intuition

The logic behind the geometric Intuition is based on the Transformation of triangles. For a circle theTangent to the circle at any point is always a right triangle. However for an Ellipse this is not the Case.

For an ellipse the tangential angle is always aligned at angle to the Radius. The angle between the tangent

and the Radius of the Ellipse is unique for each point on the Ellipse. Also for an ellipse the angle that the

tangent makes with the Radius keeps changing from point to point.

Thus if we consider a differential length on the Arc of an Ellipse we have an Acute angle triangle

whereas for a Circle it is always a right triangle. The acute angle triangle for the ellipse is unique for each

point of the Ellipse. For a circle the Differential Triangle is congruent for any point on the circle.

Analogy between the Circle and Ellipse

The Circle

Consider a circle with the center at the origin.

The Circular Equation in polar form is given as

x(t)= cosy(t)= sin

Where is the angle that the Radius makes with the x Axis

Given two points given as representing 1 & 2 then the Arc length (L) is given as

L= ( )That is radius subtended between the parametric values

The Ellipse

We define the Ellipse in the Canonical Position i.e. Major Axis aligned with the x- axis and the Center at

the origin.


x(t) = a Cos(t )

y(t) = b Sin(t)


axis.


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Analogy

Now for the circle we defined the Circular Arc length as the product between the constant r and the

difference between the parametric values. Here we can define this constant as the radius of the Circle.

Proceeding in analogously we should be able to define the Elliptical Arc length in a similar fashion as

both the Ellipse and Circle belong to the same family and circle is only a Special Case of Ellipse.

For the circle the parametric value is the radial angle itself however for an ellipse the parametric value is

the eccentric anomaly. Hence when considering the Ellipse we should consider the eccentric anomaly not

the Radial Angle.

Circle

For a circle given the Chord length C and the angle subtended by the two radius of the circle the

Radius R in terms of Chord Length C and the angle can be given as

R = 2sin2

An analogous intuitive thought leading to underlying geometric equations will yield the geometric

formulae for an Ellipse in a similar fashion relating the Chord defined with the parametric Values of the

End point and The Radial Angle.

The Geometric intuition begins with the thought process of Equal Arcs of Circles cutting equal angles.

However for ellipse equal angles in different locations cuts unequal Arcs on the same ellipse. However

they are related trough the Major and minor axis lengths which are constants ( They are analogous to the

radius of the Ellipse). The position of the points are related trough their parametric values. Assembling

these factor logically we can arrive at our Conjure.

The Conjure

Given an Elliptical Arc, the Arc length of the Elliptical Arc will be equal to that of a Circular Arc whose

Radius and Subtended Angle is given by the following conjure

The Equivalent Circular Arc having the same Arc length as that of a given Elliptical Arc, will have a

Chord length equal to the Chord length of the given Elliptical Arc and it (Circular Arc) will subtend an

angle at the center whose value in radians is equal to the difference in the Eccentric anomalies of theend points of the given Elliptical Arc

Aravind Narayan

30-April -2012

Using the above conjure we can define geometric methods for the determining the Arc length and

similarly we can define the Algebraic formulas also.


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Intuition

If we consider the relationship between the coordinate value of an ellipse and its Eccentric anomaly wecan identify that this relationship is analogous to the relationship between the Coordinate values and

Radial angle of a circle

Hence

The Relationship between a Circular Arc and The subtended angle at the center is analogous to the

Documents

An Analysis of the Ellipse