DOCSLIB.ORG

PHYS 705: Classical Mechanics Calculus of Variations I 2

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
PHYS 705: Classical Mechanics Calculus of Variations I 2 1 PHYS 705: Classical Mechanics Calculus of Variations I 2 Calculus of Variations: The Problem Determine the function y(x) such that the following integral is min(max)imized. xB I F y( x ), y '( x ); x dx xA Comments: 1. Notation: • x is the independent variable (x =t in our mech prob) • y(x) is a function of x and y’(x) = dy/dx •F{ } is called a functional on y(x) 2. Intuition: Let say y(x) denote a route in the x-y plane and I is the amount of gas needed for the trip. The problem is to find the route which uses the least gas. 3 Calculus of Variations: The Problem Comments (cont): 3. To keep it simple, we will fix the end points (xA and xB). We assume that they are known. Method can be extended to situation where (xA and xB) can be varied as well. 4. The method does not explicitly give y(x). Instead, we will get a diff eq for y(x). 5. F is assumed to be at least twice differentiable, i.e., C 2 in all its arguments. 6. x is assumed to be unidirectional between xA and xB. If not, break it into different pieces. 4 Calculus of Variations Consider a family of functions y ( a , x ) parameterized by a. y desired path yB yxyx(,)a (0,) a () x y(0,x) a is a parameter a(x ) (xC ) is a 2 -smooth variation at x yA (xA ) ( x A ) 0 x xA xB Note: y(,a xA ) y A for all a, i.e., all sampled paths has the same end points, x and x y(,a xB ) y B A B y(0, x ) is the desired path 5 Calculus of Variations A necessary, but not sufficient, condition for the min(max)imization of I is: dI 0 da a 0 Taking the derivative of I wrt a, dI d xB F y( x ), y '( x ); x dx da d a xA xB d F y( x ), y '( x ); x dx (the end points xA and xB are fixed) da xA xB Fy Fy ' dx (x is the independent variable and ya y ' a does not vary with a) xA 6 Calculus of Variations Recall, we have yxyx(,)a (0,) a () x y So, (x ) a And, since we also have yxyx '(a , ) '(0, ) a '( x ) y' d So, a dx dIxB F F d This gives, dx da yy ' dx xA Now, to continue, we will integrate the second term by parts. 7 dI xB F F d dx da y y ' dx Calculus of Variations xA To integrate by parts, udv uv vdu (with the fixed limits x and x ) A B F dF Let, u du dx y' dx y ' d dv dx v dx xxB x So, BFd F B dF dx() x () x dx xA x B 0 y'' dx y dx y ' xAxA x A no variations @ end points udv uv vdu 8 Calculus of Variations xBFd x B dF So, we have, dx ( x ) dx y' dx dx y ' xA x A dI Putting this back into our expression for , da dIxB F dF (x ) ( x ) dx 0 da y dx y ' xA xB F d F (x ) dx 0 y dx y ' xA Since this has to be zero for any arbitrary C 2 variation ( x ) , F d F Euler-Lagrange Equation (1744) 0 y dx y ' (fundamental lamma of Calc. of Var.) 9 Calculus of Variations (Euler-Lagrange Equation) F d F 0 Euler-Lagrange Equation y dx y ' Comments: 1. This is the diff eq whose solution y(x) is the function we seek to min(max)imize with respect to F. dI 2. We only used a necessary condition for an extremum 0 da a 0 I might end up to be an inflection point. We can always check afterward. In Hamilton’s principle (later), the condition is only for a stationary value. 3. It is important to keep track of which variable is the independent variable. Here, it is x. 10 Euler-Lagrange Equation (Special Case) We will now derive an alternative form of the EL equation for the case when F does not explicitly depend on x, i.e., F( yy , ') (obviously, F depends on x implicitly thru y) Observe this: d F d F F dF yFy' ' y '' dx y' dx y '' y dx Fy Fy' F yx yx' x d F dF F F F y' F y' y'' y ' y '' dx y ' dx y ' y ' y y ' 11 Euler-Lagrange Equation (Special Case) Cancelling and grouping like-color terms, dF dFF F F yFy' ' y '' y '' dx y' dx y ' y y ' y ' Now, if y(x) satisfies the Euler-Lagrange Equation, the remaining term on the RHS will be zero as well. Thus, we have, d F F y' F 0 or y' F const dx y ' y ' This is a much easier equation to solve since it is 1st order only. We basically have done one integration already. 12 Few Classic Problems (#1: a Straight Line) 1. What plane curve connecting two given points has the shortest length? y Arc length is given by: B I ds A where, 2 2 2 dy x x x ds dx dy or ds 1 dx A B dx Note: ds is written in So, we need to find y(x) that minimizes: Euclidian space. If it is B not (e.g. sphere), the result I1 y ' 2 dx (geodesic) will be different A NOT a straight line. 13 Few Classic Problems (#1: a Straight Line) To solve this problem, we apply the Euler-Lagrange equation, F d F 2 0 with F1 y ' y dx y ' FF 12 1/2 y ' 0 1'y 2' y 2 y y ' 2 1 y ' EL equation gives: d F F y ' 0 const dx y ' y ' 1 y ' 2 14 Few Classic Problems (#1: a Straight Line) Solving for y’ gives y’ being equals to another constant, y' const y mx b (it is a line and m and b to be determined by the endpoints) Alternatively, since F does not depend on x explicitly, we can use the second form for the EL equation, d F F y' F 0 y' F const dx y ' y ' y ' 2 y' 1' y const 1 y ' 2 15 Few Classic Problems (#1: a Straight Line) 2 Multiplying 1 y ' to the whole equation, we have, y'1'2 ycy 2 1' 2 (c being a constant) 1c 1 y ' 2 y' const and the same result, y mx b 16 Few Classic Problems (#2: Catenoid) 2. Minimum Surface of Revolution Problem (Soap Film bet 2 Wire Loops) Two points in an xy-plane are given. A curve in the plane connecting the two points is revolved about the x-axis. Find the curve that results in the minimum surface area of revolution. (Goldstein is not correct on this) y ds dx2 dy 2 1 2 ds da2 y ds 21' y y 2 dx y Area to be minimized is: x 2 x2 I da2 y 1 y ' 2 dx element of 1 x surf area da 1 (note: x is our indep var and it is unidirectional) 17 Few Classic Problems (#2: Catenoid) For the given integrant, the EL equation is: F d F 2 0 with Fyyx(,';) y 1 y ' y dx y ' With F not explicitly depends on x (the independent variable), we can use the alternative form for the EL equation instead: F y' Fc where c is an integration constant y ' F12 1/2 yy ' y1 y ' 2' y 2 y ' 2 1 y ' 18 Few Classic Problems (#2: Catenoid) Putting it into the alternative form of the EL equation: yy ' 2 y' yyc 1' 1 y ' 2 yy '2 y1 y ' 2 c 1 y ' 2 2 y 2 y c c21 y ' y 2 or y ' 1 2 1 y ' c 19 Few Classic Problems (#2: Catenoid) So the diff eq that we need to solve is now: dy 1 y2 c 2 dx c (both c and b are integration dy y x dx c ccosh1 b constants which will depend on 2 2 y c c the end points x1 and x2) 1 y This curve is called a x b 2 y ccosh Catenary and the surf. c of revolution is called a Catenoid. x 20 Few Classic Problems (#2: Catenoid) y A rendering of the catenoid using Mathematica Soap film between two loops 21 Few Classic Problems (#2: Catenoid) There are more interesting subtleties to this problem: 1. When one considers a family of catenaries that go through one fixed point (let say the left one : pt 1) … 2 2’ - All the caternaries will tangent on y an envelope curve (blue) which is a 1 parabola with its focus at pt #1 fixed x - Interesting observations: If we choose pt 2 at this location (other There will be no soln similar intersection pts), there will be 2 solns. here. 22 Few Classic Problems (#2: Catenoid) 2. What we have considered so far involve only twice-differentiable solutions. There is also a nondifferentiable solution that has physical relevance ! Smooth Caternary Solution Goldschmidt Solution y 1 y 1 2 2 x x Soap film bet two rings Keeps moving the two loops apart until the red surf area > blue surf area 23 Few Classic Problems (#3: Brachistochrone problem) 1 x Find the path that goes between pt 1 and pt 2 g v 2 with the least time under gravity.
Load more

Recommended publications
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • Surface Area of a Sphere in This Example We Will Complete the Calculation of the Area of a Surface of Rotation
    Surface Area of a Sphere In this example we will complete the calculation of the area of a surface of rotation. If we’re going to go to the effort to complete the integral, the answer should be a nice one; one we can remember. It turns out that calculating the surface area of a sphere gives us just such an answer. We’ll think of our sphere as a surface of revolution formed by revolving a half circle of radius a about the x-axis. We’ll be integrating with respect to x, and we’ll let the bounds on our integral be x1 and x2 with −a ≤ x1 ≤ x2 ≤ a as sketched in Figure 1. x1 x2 Figure 1: Part of the surface of a sphere. Remember that in an earlier example we computed the length of an infinites­ imal segment of a circular arc of radius 1: r 1 ds = dx 1 − x2 In this example we let the radius equal a so that we can see how the surface area depends on the radius. Hence: p y = a2 − x2 −x y0 = p a2 − x2 r x2 ds = 1 + dx a2 − x2 r a2 − x2 + x2 = dx a2 − x2 r a2 = dx: 2 2 a − x The formula for the surface area indicated in Figure 1 is: Z x2 Area = 2πy ds x1 1 ds y z }| { x r Z 2 z }| { a2 p 2 2 = 2π a − x dx 2 2 x1 a − x Z x2 a p 2 2 = 2π a − x p dx 2 2 x1 a − x Z x2 = 2πa dx x1 = 2πa(x2 − x1): Special Cases When possible, we should test our results by plugging in values to see if our answer is reasonable.
    [Show full text]
  • Villarceau-Section’ to Surfaces of Revolution with a Generating Conic
    Journal for Geometry and Graphics Volume 6 (2002), No. 2, 121{132. Extension of the `Villarceau-Section' to Surfaces of Revolution with a Generating Conic Anton Hirsch Fachbereich Bauingenieurwesen, FG Stahlbau, Darstellungstechnik I/II UniversitÄat Gesamthochschule Kassel Kurt-Wolters-Str. 3, D-34109 Kassel, Germany email: [email protected] Abstract. When a surface of revolution with a conic as meridian is intersected with a double tangential plane, then the curve of intersection splits into two con- gruent conics. This decomposition is valid whether the surface of revolution inter- sects the axis of rotation or not. It holds even for imaginary surfaces of revolution. We present these curves of intersection in di®erent cases and we also visualize imaginary curves. The arguments are based on geometrical reasoning. But we also give in special cases an analytical treatment. Keywords: Villarceau-section, ring torus, surface of revolution with a generating conic, double tangential plane MSC 2000: 51N05 1. Introduction Due to Y. Villarceau the following statement it is valid (compare e.g. [1], p. 412, [3], p. 204, or [4]): The curve of intersection between a ring torus ª and any double tangential plane ¿ splits into two congruent circles. We assume that r is the radius of the meridian circles k of ª and that their centers are in the distance d, d > r, from the axis a of rotation. We generalize and replace k by a conic which may also intersect the axis a. Under these conditions it is still true that the intersection with a double tangential plane ¿ is reducible.
    [Show full text]
  • Calculus 2 Tutor Worksheet 12 Surface Area of Revolution In
    Calculus 2 Tutor Worksheet 12 Surface Area of Revolution in Parametric Equations Worksheet for Calculus 2 Tutor, Section 12: Surface Area of Revolution in Parametric Equations 1. For the function given by the parametric equation x = t; y = 1: (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1; (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0; (c) What is the Cartesian equation for this function? (d) What is the surface of rotation in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape. c 2018 MathTutorDVD.com 1 2. For the function given by the parametric equation x = t; y = 2t + 1: (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1; (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0; (c) What is the Cartesian equation for this function? (d) What is the surface of rotation in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape. 3. For the function given by the parametric equation x = 2t + 1; y = t: (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1; c 2018 MathTutorDVD.com 2 (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0; (c) What is the Cartesian equation for this function? (d) What is this function in geometric terms? Compare the results of the above ques- tions to the geometric formula for the surface area of this shape.
    [Show full text]
  • Arxiv:1509.00145V1 [Hep-Lat] 1 Sep 2015 I.Tectnr Solution Catenary the III
    Center Vortices, Area Law and the Catenary Solution Roman H¨ollwiesera1,2, † and Derar Altarawneh1,3, ‡ 1Department of Physics, New Mexico State University, PO Box 30001, Las Cruces, NM 88003-8001, USA 2Institute of Atomic and Subatomic Physics, Nuclear Physics Dept. Vienna University of Technology, Operngasse 9, 1040 Vienna, Austria 3Department of Applied Physics, Tafila Technical University, Tafila , 66110 , Jordan (Dated: June 28, 2018) We present meson-meson (Wilson loop) correlators in Z(2) center vortex models for the infrared sector of Yang-Mills theory, i.e., a hypercubic lattice model of random vortex surfaces and a con- tinuous 2+1 dimensional model of random vortex lines. In particular we calculate quadratic and circular Wilson loop correlators in the two models respectively and observe that their expectation values follow the area law and show string breaking behavior. Further we calculate the catenary solution for the two cases and try to find indications for minimal surface behavior or string surface tension leading to string constriction. PACS numbers: 11.15.Ha, 12.38.Gc Keywords: Center Vortices, Lattice Gauge Field Theory CONTENTS I. Introduction 2 II. Random center vortex models 2 A. Random vortex world-line model 2 B. Random vortex world-surface model 3 III. The Catenary Solution 4 A. Circular Wilson loops 5 B. Quadratic Wilson loops 6 IV. Measurements & Results 6 A. Quadratic Wilson loop correlators in the 4D vortex surface model 6 B. Circular Wilson loop correlators in the 3D vortex line model 7 V. Conclusions 11 A. Beltrami Identity and Catenary Solution 11 B. Catenary vs. Goldschmidt solution 11 arXiv:1509.00145v1 [hep-lat] 1 Sep 2015 Acknowledgments 12 References 12 a Funded by an Erwin Schr¨odinger Fellowship of the Austrian Science Fund under Contract No.
    [Show full text]
  • Area of a Surface of Revolution
    AREA OF A SURFACE OF REVOLUTION A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 7.2 and 7.3. We want to define the area of a surface of revolution in such a way that it corresponds to our intuition. If the surface area is A, we can imagine that painting the surface would require the same amount of paint as does a flat region with area A. Let’s start with some simple surfaces. The lateral surface area of a circular cylinder with radius r and height h is taken to be A ෇ 2␲rh because we can imagine cutting the cylin- cut h der and unrolling it (as in Figure 1) to obtain a rectangle with dimensions 2␲r and h. r Likewise, we can take a circular cone with base radius r and slant height l, cut it along the dashed line in Figure 2, and flatten it to form a sector of a circle with radius l and cen- tral angle ␪ ෇ 2␲r͞l. We know that, in general, the area of a sector of a circle with radius ␪ 1 2␪ l and angle is 2 l (see Exercise 67 in Section 6.2) and so in this case it is h 2␲r A ෇ 1 l 2␪ ෇ 1 l 2ͩ ͪ ෇ ␲rl 2πr 2 2 l FIGURE 1 Therefore, we define the lateral surface area of a cone to be A ෇ ␲rl.
    [Show full text]
  • Lengths of Curves and Surfaces of Revolution 14.4   
    ® Lengths of Curves and Surfaces of Revolution 14.4 Introduction Integration can be used to find the length of a curve and the area of the surface generated when a curve is rotated around an axis. In this Section we state and use formulae for doing this. Prerequisites • be able to calculate definite integrals Before starting this Section you should ... • find the length of curves Learning Outcomes • find the area of the surface generated when a On completion you should be able to ... curve is rotated about an axis HELM (2008): 27 Section 14.4: Lengths of Curves and Surfaces of Revolution 1. The length of a curve To find the length of a curve in the xy plane we first divide the curve into a large number of pieces. We measure (or, at least, approximate) the length of each piece and then by an obvious summation process obtain an estimate for the length of the curve. Theoretically, we allow the number of pieces to increase without bound, implying that the length of each piece will tend to zero. In this limit the summation process becomes an integration process. y δy δx y(x) a b x Figure 11 Figure 11 shows the portion of the curve y(x) between x = a and x = b. A small piece of this curve has been selected and can be considered as the hypotenuse of a triangle with base δx and height δy. (Here δx and δy are intended to be ‘small’ so that the curved segment can be regarded as a straight segment.) s 2 p δy Using Pythagoras’ theorem, the length of the hypotenuse is: δx2 + δy2 = 1 + δx δx By summing all such contributions between x = a and x = b, and letting δx → 0 we obtain an expression for the total length of the curve: x=b s 2 X δy lim 1 + δx δx→0 δx x=a But we already know how to write such an expression in terms of an integral.
    [Show full text]
  • Chapter 4 Surfaces
    Chapter 4 Surfaces In this chapter we turn to surfaces in general. We discuss the following topics. Describing surfaces with equations and parametric descriptions. ² Some constructions of surfaces: surfaces of revolution and ruled surfaces. ² Tangent planes to surfaces. ² Surface area of surfaces. ² Curvature of surfaces. ² Figure 4.1: In this design for the 1972 Olympic Games in Munich, Germany, Frei Otto illustrates the role of surfaces in modern architecture. (Photo: Wim Huisman) 91 92 Surfaces 4.1 Describing general surfaces 4.1.1 After degree 2 surfaces, we could proceed to degree 3, but instead we turn to surfaces in general. We highlight some important aspects of these surfaces and ways to construct them. 4.1.2 Surfaces given as graphs of functions Suppose f : R2 R is a di®erentiable function, then the graph of f consisting of all points of the form!(x; y; f(x; y)) is a surface in 3{space with a smooth appearance due to the di®erentiability of f. Since we are not tied to degree 1 or 2 expressions for f, the resulting surfaces may have all sorts of fascinating shapes. The description as a graph can 1 1 0.5 0.5 0 0 2 2 -0.5 -0.5 -1 -1 0 -6 0 -2 -4 -2 0 -2 0 -2 2 2 Figure 4.2: Two `sinusoidal' graphs: the graphs of sin(xy) and sin(x). be seen as a special case of a parametric description: x = u y = v z = f(u; v); where u and v are the parameters.
    [Show full text]
  • Dictionary of Mathematical Terms
    DICTIONARY OF MATHEMATICAL TERMS DR. ASHOT DJRBASHIAN in cooperation with PROF. PETER STATHIS 2 i INTRODUCTION changes in how students work with the books and treat them. More and more students opt for electronic books and "carry" them in their laptops, tablets, and even cellphones. This is This dictionary is specifically designed for a trend that seemingly cannot be reversed. two-year college students. It contains all the This is why we have decided to make this an important mathematical terms the students electronic and not a print book and post it would encounter in any mathematics classes on Mathematics Division site for anybody to they take. From the most basic mathemat- download. ical terms of Arithmetic to Algebra, Calcu- lus, Linear Algebra and Differential Equa- Here is how we envision the use of this dictio- tions. In addition we also included most of nary. As a student studies at home or in the the terms from Statistics, Business Calculus, class he or she may encounter a term which Finite Mathematics and some other less com- exact meaning is forgotten. Then instead of monly taught classes. In a few occasions we trying to find explanation in another book went beyond the standard material to satisfy (which may or may not be saved) or going curiosity of some students. Examples are the to internet sources they just open this dictio- article "Cantor set" and articles on solutions nary for necessary clarification and explana- of cubic and quartic equations. tion. The organization of the material is strictly Why is this a better option in most of the alphabetical, not by the topic.
    [Show full text]
  • A Brief Guide to Calculus Ii
    JIMMYBROOMFIELD ABRIEF GUIDETO CALCULUSII THEUNIVERSITYOFMINNESOTA Copyright © 2015 Jimmy Broomfield http://moodle.umn.edu License: Creative Commons CC BY-NC 4.0 http://creativecommons.org/licenses/by-nc/4.0/ First printing, March 2015 Contents Integration Techniques 5 Basic Integrals 5 Integration By Parts 6 Trigonometric Identities 9 Trigonometric Substitution 12 Partial Fraction 16 Improper Integrals 19 Applications of Integration 23 Arc Length 23 Area of a Surface of Revolution 25 Application to Physics and Engineering 27 Fluid Force and Fluid Pressure 28 Moments and Centers of Mass 30 Centroid of a Lamina 31 Differential Equations 35 Introduction 35 Slope Fields and Euler’s Method 38 Separable Equations 38 Models for Population Growth 38 Linear Equations 41 Predator-Prey Systems 42 Glossary of Terms 42 Integration Techniques Throughout this guide, we will present results without proof. If you would like a proof of the techniques and theorems that we use, please see "Calculus Early Transcendentals", volume I, 7E by James Stewart. To begin this guide, we will review basic integration formula that every calc II student should know, and then move to new techniques. Basic Integrals The integrals below are essential formulas the should be memorized. If you struggle with a few of them, please practice until you have committed them to memory. Also note that the constants have been left out of the table below for convenience. Z Z xn+1 2 xn dx = , n 6= −1 sec x dx = tan x n + 1 Z Z 1 csc2 x dx = − cot x dx = ln jxj x Z Z x x dx = x ex dx = ex sec tan sec Z Z 1 csc x cot x dx = − csc x ax dx = ax ln a Z Z 1 1 x dx = · tan−1 sin x dx = − cos x a2 + x2 a a Z Z 1 x cos x dx = sin x p dx = sin−1 a2 − x2 a 6 Along with these integral formulas, you should also be comfort- able with the technique of u-substitution.
    [Show full text]
  • 6: Parametrized Surfaces Implicit to Parametric: find the Center (A, B, C) and the Radius R Possibly by Completing the Square
    Math S21a: Multivariable calculus Oliver Knill, Summer 2011 2 Spheres: Parametric: ~r(u, v)= a, b, c + ρ cos(u) sin(v), ρ sin(u) sin(v), ρ cos(v) . h i h i Implicit: (x a)2 +(y b)2 +(z c)2 = ρ2. − − − Parametric to implicit: reading off the radius. 6: Parametrized surfaces Implicit to parametric: find the center (a, b, c) and the radius r possibly by completing the square. There is a second, fundamentally different way to describe a surface, the parametrization. A parametrization of a surface is a vector-valued function ~r(u, v)= x(u, v),y(u, v), z(u, v) , h i where x(u, v),y(u, v), z(u, v) are three functions of two variables. Because two parameters u and v are involved, the map ~r is also called uv-map. 3 Graphs: Parametric: ~r(u, v)= u,v,f(u, v) h i A parametrized surface is the image of the uv-map. Implicit: z f(x, y) = 0. Parametric to Implicit: think about z = f(x, y) − The domain of the uv-map is called the parameter do- Implicit to Parametric: use x and y as the parameterizations. main. If we keep the first parameter u constant, then v ~r(u, v) is a curve on the surface. Similarly, if v is constant, then u ~r(u, v) traces7→ a curve the surface. These curves are called grid curves. 7→ A computer draws surfaces using grid curves. The world of parametric surfaces is intriguing and complex. You can explore this world with the help of the computer algebra system Mathematica.
    [Show full text]
  • Calculus II, Section 8.2, #4 Area of a Surface of Revolution (A) Set up An
    Calculus II, Section 8.2, #4 Area of a Surface of Revolution (a) Set up an integral for the area of the surface obtained by rotating the curve about (i) the x-axis and (ii) the y-axis.1 x = ln (2y + 1) , 0 ≤ y ≤ 1 (b) Use the numerical integration capability of a calculator to evaluate the surface areas correct to four decimal places. In any calculus textbook, there are a lot of formulas associated with surface area. In the eighth edition of Calculus, Early Transcendentals by James Stewart, there are five formulas set out for finding the surface area of a solid of revolution. If we try to intuitively understand the surface area concept, we can spare ourselves the need to memorize these formulas. Much like the computation of the lateral area of a cylinder (circumference·length), the area of a surface of revolution is found by circumference · arc length where we use the formula for arc length and integrate over the length of the interval of integration. Thus we have x=b dy 2 S = 2π · radius · 1+ dx s dx Zx=a or y=d dx 2 S = 2π · radius · 1+ dy s dy Zy=c For this problem, we will be finding surface areas associated with revolving the graph of x = ln (2y + 1) , 0 ≤ y ≤ 1 about the x and y axes. Let’s do some preliminary work with the function. The graph of x = ln (2y + 1) , 0 ≤ y ≤ 1 is shown y at right. A simple computation confirms when y = 0, x = ln (2 · 0+1) = 0 and when y = 1, 1 x = ln (2 · 1 + 1) = ln (3).
    [Show full text]