On Surfaces of Revolution whose Mean Curvature is Constant

Similar documents
CONIC SECTIONS 1. Inscribe a parabola in the given rectangle, with its axis parallel to the side AB

UNIT I PLANE CURVES AND FREE HAND SKETCHING CONIC SECTIONS

Engineering Graphics, Class 5 Geometric Construction. Mohammad I. Kilani. Mechanical Engineering Department University of Jordan

RECTANGULAR EQUATIONS OF CONICS. A quick overview of the 4 conic sections in rectangular coordinates is presented below.

10.1 Curves defined by parametric equations

2.1 Partial Derivatives

Practice problems from old exams for math 233

Math 122: Final Exam Review Sheet


Lecture 4 : Monday April 6th

C.2 Equations and Graphs of Conic Sections

i + u 2 j be the unit vector that has its initial point at (a, b) and points in the desired direction. It determines a line in the xy-plane:

Test Yourself. 11. The angle in degrees between u and w. 12. A vector parallel to v, but of length 2.

Math Final Exam - 6/11/2015

1.6. QUADRIC SURFACES 53. Figure 1.18: Parabola y = 2x 2. Figure 1.19: Parabola x = 2y 2

This early Greek study was largely concerned with the geometric properties of conics.

(3,4) focus. y=1 directrix

Pre Calc. Conics.

MATH Review Exam II 03/06/11

You identified, analyzed, and graphed quadratic functions. (Lesson 1 5) Analyze and graph equations of parabolas. Write equations of parabolas.

INSTITUTE OF AERONAUTICAL ENGINEERING

14.4. Tangent Planes. Tangent Planes. Tangent Planes. Tangent Planes. Partial Derivatives. Tangent Planes and Linear Approximations

SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR

Hyperbolas Graphs, Equations, and Key Characteristics of Hyperbolas Forms of Hyperbolas p. 583

Math 148 Exam III Practice Problems

ENGINEERING CURVES (Week -2)

University of California, Berkeley Department of Mathematics 5 th November, 2012, 12:10-12:55 pm MATH 53 - Test #2

On the. Geometry. of Orbits

MATH 12 CLASS 9 NOTES, OCT Contents 1. Tangent planes 1 2. Definition of differentiability 3 3. Differentials 4

Lecture 3: Geometrical Optics 1. Spherical Waves. From Waves to Rays. Lenses. Chromatic Aberrations. Mirrors. Outline

Pre-Calc. Slide 1 / 160. Slide 2 / 160. Slide 3 / 160. Conics Table of Contents. Review of Midpoint and Distance Formulas

Name: ID: Section: Math 233 Exam 2. Page 1. This exam has 17 questions:

EXERCISES CHAPTER 11. z = f(x, y) = A x α 1. x y ; (3) z = x2 + 4x + 2y. Graph the domain of the function and isoquants for z = 1 and z = 2.

WESI 205 Workbook. 1 Review. 2 Graphing in 3D

As the Planimeter s Wheel Turns

47. Conservative Vector Fields

Chapter 4: The Ellipse

Mock final exam Math fall 2007

VectorPlot[{y^2-2x*y,3x*y-6*x^2},{x,-5,5},{y,-5,5}]

Solutions to the problems from Written assignment 2 Math 222 Winter 2015

VECTOR CALCULUS Julian.O 2016

The Geometric Definitions for Circles and Ellipses

Directional Derivative, Gradient and Level Set

Calculus II Fall 2014

Section 15.3 Partial Derivatives

6.00 Trigonometry Geometry/Circles Basics for the ACT. Name Period Date

M.V.S.R. ENGINEERING COLLEGE, NADERGUL HYDERABAD B.E. I/IV I - Internal Examinations (November 2014)

Engineering Graphics. Practical Book. Government Engineering College Bhuj (Kutch - Gujarat) Department of Mechanical Engineering

Pre-Calc Conics

JUNIOR CERTIFICATE 2009 MARKING SCHEME TECHNICAL GRAPHICS HIGHER LEVEL

Review guide for midterm 2 in Math 233 March 30, 2009

MATH 259 FINAL EXAM. Friday, May 8, Alexandra Oleksii Reshma Stephen William Klimova Mostovyi Ramadurai Russel Boney A C D G H B F E

Estimating Areas. is reminiscent of a Riemann Sum and, amazingly enough, will be called a Riemann Sum. Double Integrals

Definitions and claims functions of several variables

MAT01B1: Calculus with Polar coordinates

Unit 6 Task 2: The Focus is the Foci: ELLIPSES

Exam 2 Summary. 1. The domain of a function is the set of all possible inputes of the function and the range is the set of all outputs.

Appendix I to Ch. 3 The Eikonal Equation in Optics

Review Sheet for Math 230, Midterm exam 2. Fall 2006

INTEGRATION OVER NON-RECTANGULAR REGIONS. Contents 1. A slightly more general form of Fubini s Theorem

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 6 - Tues 17th Oct 2017 Functions of Several Variables and Partial Derivatives

WJEC LEVEL 2 CERTIFICATE 9550/01 ADDITIONAL MATHEMATICS

Pre-Calc. Midpoint and Distance Formula. Slide 1 / 160 Slide 2 / 160. Slide 4 / 160. Slide 3 / 160. Slide 5 / 160. Slide 6 / 160.

RECOMMENDATION ITU-R S.1257

Lecture 19. Vector fields. Dan Nichols MATH 233, Spring 2018 University of Massachusetts. April 10, 2018.

REVIEW SHEET FOR MIDTERM 2: ADVANCED

Calculus IV Math 2443 Review for Exam 2 on Mon Oct 24, 2016 Exam 2 will cover This is only a sample. Try all the homework problems.

Lecture 2: Geometrical Optics. Geometrical Approximation. Lenses. Mirrors. Optical Systems. Images and Pupils. Aberrations.

The Compensating Polar Planimeter

Chapter 1. Trigonometry Week 6 pp

Algebra II B Review 3

On a Geographic Projection of the Surface of the Sphere

a) 2, 4, 8, 14, 22, b) 1, 5, 6, 10, 11, c) 3, 9, 21, 39, 63, d) 3, 0, 6, 15, 27, e) 3, 8, 13, 18, 23,

B.E. 1 st Year Engineering Graphics ( )

BVRIT HYDERABAD College of Engineering for Women Department of Basic Sciences and Humanities

Conic and Quadric Surface Lab page 4. NORTHEASTERN UNIVERSITY Department of Mathematics Fall 03 Conic Sections and Quadratic Surface Lab

Appendix. Springer International Publishing Switzerland 2016 A.Y. Brailov, Engineering Graphics, DOI /

MATH 8 FALL 2010 CLASS 27, 11/19/ Directional derivatives Recall that the definitions of partial derivatives of f(x, y) involved limits

Technical Drawing Paper 1 - Higher Level (Plane and Solid Geometry)

Experiment O11e Optical Polarisation

Lecture 2: Geometrical Optics. Geometrical Approximation. Lenses. Mirrors. Optical Systems. Images and Pupils. Aberrations.

Waves & Oscillations

FUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DIFFERENTIATION

ENGINEERING GRAPHICS (Engineering Drawing is the language of Engineers)

Exam 1 Study Guide. Math 223 Section 12 Fall Student s Name

Solution of a geometric problem.

Practice problems from old exams for math 233

Engineering Graphics UNIVERSITY OF TEXAS RIO GRANDE VALLEY JAZMIN LEY HISTORY OF ENGINEERING GRAPHICS GEOMETRIC CONSTRUCTION & SOLID MODELING

Discussion 8 Solution Thursday, February 10th. Consider the function f(x, y) := y 2 x 2.

The Ellipse. PF 1 + PF 2 = constant. Minor Axis. Major Axis. Focus 1 Focus 2. Point 3.4.2

CHAPTER 10 Conics, Parametric Equations, and Polar Coordinates

MATH Exam 2 Solutions November 16, 2015

Introduction to CATIA V5

Section 3: Functions of several variables.

LOCKWOOD'S "CURVES" ON A GRAPHICS CALCULATOR

TIME SCHEDULE. Module Topic Periods 1 Importance of Engineering Graphics Drawing Instruments Drawing Standards Lettering and Numbering

DIFFERENTIAL EQUATIONS. A principal model of physical phenomena.

33. Riemann Summation over Rectangular Regions

MAT187H1F Lec0101 Burbulla

DWG 002. Blueprint Reading. Geometric Terminology Orthographic Projection. Instructor Guide

Transcription:

On Surfaces of Revolution whose Mean Curvature is Constant Ch. Delaunay May 4, 2002 When one seeks a surface of given area enclosing a maximal volume, one finds that the equation this surface must satisfy is the second order partial differential equation () 2 u x 2 ( + ( u y ) ) 2 ( )( ) u u 2 2 u x y x y + 2 u y 2 + ( ) + u 2 3/2 =0; a (constant mean curvature) 2a ( + r( + q 2 ) 2pqs + t( + p 2 )+ a ( + p2 + q 2 ) 3/2 =0 ( ) ) 2 u x the question is therefore reduced to integrating this equation, which has not yet been done in general. In the particular case where a =, equation () becomes that of the minimal surface which was integrated by Monge; but the complicated form of the integral he gave, from which one derives no benefit, suggests that if ever one could completely integrate equation (), the integral would be of no use. If one adds to equation () the condition that the surface be one of revolution, the difficulty entirely disappears, and not only can one find the general equation of the surface, one can also give a very simple geometric definition of its meridian curve. This is what I propose to show in this paper. I first observe that equation () expresses that the sum of the principal curvatures is constant and equals. Now one knows that at each point a

of a surface the sum of the curvatures of two normal sections which are perpendicular to each other, is equal to the sum of principal curvatures. One can therefore define the mean curvature of a surface at a point as half the sum of the principal curvatures at this point; in this way the surface represented by equation () is that where the mean curvature is constant and equal to, and it s this surface that we propose to find in the particular case where 2a it is one of the revolution. The radii of principal curvature at a point of a surface of revolution are, as one knows, the radius of curvature of the meridian curve at this point, and the portion of the normal to the surface included between the point and the axis; it follows therefore that if one references the meridian curve of the desired surface to a pair of rectangular coordinate axes, of which one, the x-axis, is the axis of the surface, that curve will be determined by the condition (2) ρ + N = a, where ρ represents its radius of curvature and N the portion of its normal (see text). Let x and y be the coordinates of an arbitrary point on the curve, x and y the coordinates of the center of the osculating circle at that point, or equivalently the coordinates of the corresponding point on the evolute, and s the arclength of the evolute starting from a fixed origin up to the point (x,y ). Now, according to the properties of evolutes, ρ = b s b is a constant which depends on the origin of the arc s. Therefore the portion of the tangent to evolute at the point (x,y ) included between the point and the x-axis is equal to y ds ; adding ρ to y, we get the value of N dy which is N = y ds dy + b s ; by means of these values, equation (2) becomes b s + = y ds + b s dy a, an equation which will help us determine the evolute of the desired curve. One obtains first, by integration, y 2 = α(b s )(2a b + s ), 2

where α is an arbitrary constant. Upon resolving for s and differentiating, one finds dy = α a 2 y 2 α ; ds y and as one has ( ) dx dy 2 =, ds ds one deduces dx y 2 ( + α) a = 2 α 2. ds y Upon inspection of the values of dy and dx, one sees that the constant α ds ds can take all the possible positive values, and that, in this case, the values of y aα must be included between the limits +α and α a; but that if α is negative, it must be included between 0 and, and, in this case, y can take all the values greater than aα +α, which indicates that the evolute has infinite branches. Let ϕ be the angle that the tangent to the desired meridian curve at the point (x, y) makes with the x-axis; one will have dx =sinϕ, dy = cos ϕ, ds ds dx = dy tan ϕ. It follows (that) (3) y = aα +α sin 2 ϕ, and x = dy tan ϕ = y tan ϕ + (as well) or really, replacing y by its value, y dϕ cos 2 ϕ ; (4) x = β aα tan ϕ ϕ +α sin 2 ϕ + aα dϕ 0 cos 2 ϕ +α sin 2 ϕ, where β is a new arbitrary constant. These two equations (3) and (4) represent the evolute. One can eliminate the angle ϕ, and one obtains thereby the equation of the curve; one can also express the integral which enters into the value of x by means of elliptic functions; but it is preferable, for that which follows, to leave (to) these equations (in) the form which we have come (one comes) to give (of them). 3

We have determined the values of the coordinates x,y of an arbitrary point of the evolute as a function of the auxiliary variable ϕ, itiseasyfrom this to deduce the values of the coordinates x, y, at the corresponding point of the desired curve, as a function of the same variable; as a result, one has y = y + ρ dy ds, x = x + ρ dx ds. Therefore one has also, by that which proceeded, ρ = b s = a a 2 y 2 α ; and if one replaces dy dx by cos ϕ,,bysinϕ,x and y by their values ds ds (pulled) from equations (3) and (4), one finds the following equations for representing the desired meridian curve: (5) { y = a cos ϕ + a +α sin 2 ϕ, x = β + a sin ϕ a tan ϕ +α sin 2 ϕ + ϕ 0 aαdϕ cos 2 ϕ +α sin. 2 ϕ The question is therefore now completely resolved from the analytic point of view; but we can proceed further along to interpret geometrically the result we have reached. The integral which enters into the value of x is the same form as the one which is present in the investigation of the arclength of an ellipse or a hyperbola; one is therefore naturally led to the idea that the curve represented by the equations (5) can well be a kind of cycloid generated by a point in the plane of an ellipse or hyperboloid which is rolling upon a straight (line). As it was desired to verify this premise, I have recognized that indeed the meridian curve one finds is that which is generated by the focus of an ellipse or a hyperbola which rolls upon the x-axis. To demonstrate this, it suffices to seek directly the equation of this curve, and from doing this it is seen that it is identical with the equations (5). Imagine therefore a hyperbola, for example, which rolls upon the x-axis, and consider this curve in an arbitrary position. Let s be the arclength of the curve (hyperbola) included between the vertex and the point of contact with the x-axis, r the radius vector which joins the point of contact to the 4

focus above the origin of the arc s, andϕthe angle that the radius vector makes with the normal to the curve at the point of contact: one will have for the coordinates of the focus. x = β + s r sin ϕ y = r cos ϕ. Now if x and y are the coordinates of the point of contact taken relative to the axes of the hyperbola, and if one represents by a transverse semi-axis, and by e the eccentricity of that curve, one will have y = a(e2 ) e tan ϕ, x = a e e cos ϕ 2 sin 2 ϕ, r = ex a; and in consequence r cos ϕ = a cos ϕ + a e 2 sin 2 ϕ, r sin ϕ = a sin ϕ + a tan ϕ e 2 sin 2 ϕ. On the other hand, one has ds = dx 2 + dy2 = a(e 2 )dϕ cos 2 ϕ e 2 sin 2 ϕ, from whence ϕ a(e 2 )dϕ s = 0 cos 2 ϕ e 2 sin 2 ϕ. One has therefore at last the following equations for representing the curve generated by the focus of the hyperbola as it rolls upon the x-axis: y = a cos ϕ + a e 2 sin 2 ϕ, ϕ x = β + a sin ϕ a tan ϕ e 2 sin 2 a(e 2 )dϕ ϕ + 0 cos 2 ϕ e 2 sin 2 ϕ. These equations evidently coincide with the equations (5) if one puts e = +α, 5

which can always be done when α is positive. Thus, in this case, the equations (5) represent the curve described by the focus of a hyperbola rolling upon the x-axis; a is the transverse semi-axis of this hyperbola, and +α is the eccentricity. A similar calculation shows that, in the case that α is negative, the equations (5) represent the curve described by the focus of an ellipse rolling upon the x-axis; a is then the semi-major axis of the ellipse, and +α is its eccentricity. One can therefore conclude (from) of everything that preceded, the following theorem: For finding the meridian curve of a surface of revolution of which the mean curvature is constant and equal to, it must be done by rolling upon 2a the axis of the surface an ellipse or a hyperbola such that the major axis or the transverse axis is equal to 2a, and the focus describes the desired curve. If the mean curvature of a surface of revolution of which one seeks the meridian curve is zero, one has 2a = ; (and) then the meridian curve will be generated by the focus of a parabola rolling upon the axis of the surface. But one knows that this curve is a catenoid: one recovers therefore in this way an already known theorem, that if one rolls a parabola upon a straight line, the focus of this parabola describes a catenoid. 6