DOCSLIB.ORG

7.4 Arc Length and Surfaces of Revolution 467

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
7.4 Arc Length and Surfaces of Revolution 467 466 Chapter 7 Applications of Integration 7. 4 Arc Length and Surfaces of Revolution Find the arc length of a smooth curve. Find the area of a surface of revolution. Arc Length In this section, definite integrals are used to find the arc lengths of curves and the areas of surfaces of revolution. In either case, an arc (a segment of a curve) is approximated by straight line segments whose lengths are given by the familiar Distance Formula ϭ Ί͑ Ϫ ͒2 ϩ ͑ Ϫ ͒2 d x2 x1 y2 y1 . A rectifiable curve is one that has a finite arc length. You will see that a sufficient condition for the graph of a function f to be rectifiable between ͑a, f ͑a͒͒ and ͑b, f ͑b͒͒ is that fЈ be continuous on ͓a, b͔. Such a function is continuously differentiable on ͓a, b͔, and its graph on the interval ͓a, b͔ is a smooth curve. Consider a function y ϭ f ͑x͒ that is continuously differentiable on the interval ͓a, b͔. You can approximate the graph of f by n line segments whose endpoints are determined by the partition ϭ . ϭ a x0 < x1 < x2 < < xn b ⌬ ϭ Ϫ ⌬ ϭ Ϫ CHRISTIAN HUYGENS (1629–1695) as shown in Figure 7.37. By letting xi xi xiϪ1 and yi yi yiϪ1, you can The Dutch mathematician approximate the length of the graph by Christian Huygens, who invented n the pendulum clock, and James Ϸ Ί͑ Ϫ ͒2 ϩ ͑ Ϫ ͒2 s ͚ xi xiϪ1 yi yiϪ1 Gregory (1638–1675), a Scottish iϭ1 mathematician, both made early n ϭ Ί͑⌬ ͒2 ϩ ͑⌬ ͒2 contributions to the problem of ͚ xi yi finding the length of a rectifiable iϭ1 curve. n ⌬y 2 ϭ Ί͑⌬x ͒2 ϩ ΂ i΃ ͑⌬x ͒2 See LarsonCalculus.com to read ͚ i ⌬x i more of this biography. iϭ1 i n ⌬ 2 ϭ ϩ yi ͑⌬ ͒ ͚Ί1 ΂⌬ ΃ xi . iϭ1 xi This approximation appears to become better and better as ʈ⌬ʈ → 0 ͑n → ϱ͒. So, the y length of the graph is (x , y ) 2 2 Δ − (x , y ) y = y2 y1 n ⌬ 2 1 1 ϭ ϩ yi ͑⌬ ͒ (x , y ) s lim ͚Ί1 ΂ ΃ xi . 0 0 ʈ⌬ʈ→0 ϭ ⌬x (x , y ) i 1 i Δ − n n x = x2 x1 Ј͑ ͒ ͑ ͒ Because f x exists for each x in xiϪ1, xi , the Mean Value Theorem guarantees the ͑ ͒ existence of ci in xiϪ1, xi such that ͑ ͒ Ϫ ͑ ͒ ϭ Ј͑ ͒͑ Ϫ ͒ f xi f xiϪ1 f ci xi xiϪ1 x f ͑x ͒ Ϫ f ͑x ͒ a = x x x b = x i iϪ1 ϭ Ј ͑ ͒ 0 1 2 n Ϫ f ci xi xiϪ1 ⌬ yi y ϭ fЈ͑c ͒. y = f(x) ⌬ i s xi Because fЈ is continuous on ͓a, b͔, it follows that Ί1 ϩ ͓ fЈ͑x͔͒ 2 is also continuous (and s = length of therefore integrable) on ͓a, b͔, which implies that curve from n a to b Ί 2 s ϭ lim 1 ϩ ͓ fЈ͑c ͔͒ ͑⌬ x ͒ ʈ⌬ʈ→ ͚ i i 0 iϭ1 b ϭ ͵ Ί ϩ ͓ Ј͑ ͔͒2 x 1 f x dx ab a Figure 7.37 where s is called the arc length of f between a and b. Bettmann/Corbis Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 Arc Length and Surfaces of Revolution 467 Definition of Arc Length Let the function y ϭ f ͑x͒ represent a smooth curve on the interval ͓a, b͔. The arc length of f between a and b is b ϭ ͵ Ί ϩ ͓ Ј͑ ͔͒ 2 s 1 f x dx. a Similarly, for a smooth curve x ϭ g͑y͒, the arc length of g between c and d is d ϭ ͵ Ί ϩ ͓ Ј͑ ͔͒ 2 s 1 g y dy. c FOR FURTHER INFORMATION To see how arc length can be used to define trigonometric functions, see the article “Trigonometry Requires Calculus, Not Vice Versa” by Yves Nievergelt in UMAP Modules. Because the definition of arc length can be applied to a linear function, you can check to see that this new definition agrees with the standard Distance Formula for the length of a line segment. This is shown in Example 1. The Length of a Line Segment ͑ ͒ ͑ ͒ Find the arc length from x1, y1 to x2, y2 on the graph of y f ͑x͒ ϭ mx ϩ b x y ( 2, 2) as shown in Figure 7.38. Solution Because y − y 2 1 Ϫ y2 y1 (x1, y1) ϭ Ј͑ ͒ ϭ m f x Ϫ x2 x1 x − x 2 1 it follows that f(x) = mx + b x2 ϭ ͵ Ί ϩ ͓ Ј ͑ ͔͒2 Formula for arc length x s 1 f x dx x1 x 2 Ϫ 2 The formula for the arc length of the ϭ ϩ y2 y1 ͑ ͒ ͑ ͒ ͵ Ί1 ΂ ΃ dx graph of f from x1, y1 to x2, y2 is the x Ϫ x x1 2 1 same as the standard Distance Formula. x ͑ Ϫ ͒2 ϩ ͑ Ϫ ͒2 2 Figure 7.38 x2 x1 y2 y1 ϭ Ί ͑x͒ Integrate and simplify. ͑x Ϫ x ͒2 ΅ 2 1 x1 ͑ Ϫ ͒2 ϩ ͑ Ϫ ͒2 x x1 y2 y1 ϭ 2 ͑x Ϫ x ͒ Ί ͑ Ϫ ͒2 2 1 x2 x1 ϭ Ί͑ Ϫ ͒2 ϩ ͑ Ϫ ͒2 x2 x1 y2 y1 which is the formula for the distance between two points in the plane. TECHNOLOGY Definite integrals representing arc length often are very difficult to evaluate. In this section, a few examples are presented. In the next chapter, with more advanced integration techniques, you will be able to tackle more difficult arc length problems. In the meantime, remember that you can always use a numerical integration program to approximate an arc length. For instance, use the numerical integration feature of a graphing utility to approximate arc lengths in Examples 2 and 3. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 468 Chapter 7 Applications of Integration Finding Arc Length y Find the arc length of the graph of 3 1 y =+x 6 2x x3 1 y ϭ ϩ 2 6 2x ͓1 ͔ on the interval 2, 2 , as shown in Figure 7.39. 1 Solution Using dy 3x 2 1 1 1 x ϭ Ϫ ϭ 2 Ϫ 2 ΂x 2΃ 1 2 3 dx 6 2x 2 x The arc length of the graph of y on yields an arc length of ͓1 ͔ 2, 2 b 2 Figure 7.39 dy s ϭ ͵ Ί1 ϩ ΂ ΃ dx Formula for arc length a dx 2 1 1 2 ϭ ͵ ϩ 2 Ϫ Ί1 ΄ ΂x 2΃΅ dx 1͞2 2 x 2 1 1 ϭ ͵ 4 ϩ ϩ Ί ΂x 2 4΃ dx 1͞2 4 x 2 1 1 ϭ ͵ 2 ϩ Simplify. ΂x 2΃ dx 1͞2 2 x 1 x3 1 2 ϭ ΄ Ϫ ΅ Integrate. 2 3 x 1͞2 1 13 47 ϭ ΂ ϩ ΃ 2 6 24 33 ϭ . 16 Finding Arc Length y Find the arc length of the graph of ͑y Ϫ 1͒3 ϭ x2 on the interval ͓0, 8͔, as shown in (8, 5) 5 Figure 7.40. 4 Solution Begin by solving for x in terms of y: x ϭ ±͑y Ϫ 1͒3͞2. Choosing the 3 positive value of x produces 2 (y − 1)3 = x2 dx 3 1 (0, 1) ϭ ͑y Ϫ 1͒1͞2. x dy 2 123 456 7 8 The x-interval ͓0, 8͔ corresponds to the y-interval ͓1, 5͔, and the arc length is The arc length of the graph of y on ͓0, 8͔ d dx 2 s ϭ ͵ 1 ϩ dy Formula for arc length Figure 7.40 Ί ΂ ΃ c dy 5 3 2 ϭ ͵Ί1 ϩ ΄ ͑y Ϫ 1͒1͞2΅ dy 1 2 5 9 5 ϭ ͵Ί y Ϫ dy 1 4 4 1 5 ϭ ͵ Ί Ϫ Simplify. 9y 5 dy 2 1 1 ͑9y Ϫ 5͒3͞2 5 ϭ ΄ ΅ Integrate. 18 3͞2 1 1 ϭ ͑40 3͞2 Ϫ 4 3͞2͒ 27 Ϸ 9.073. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 Arc Length and Surfaces of Revolution 469 Finding Arc Length See LarsonCalculus.com for an interactive version of this type of example.
Load more

Recommended publications
  • Polar Coordinates, Arc Length and the Lemniscate Curve
    Ursinus College Digital Commons @ Ursinus College Transforming Instruction in Undergraduate Calculus Mathematics via Primary Historical Sources (TRIUMPHS) Summer 2018 Gaussian Guesswork: Polar Coordinates, Arc Length and the Lemniscate Curve Janet Heine Barnett Colorado State University-Pueblo, [email protected] Follow this and additional works at: https://digitalcommons.ursinus.edu/triumphs_calculus Click here to let us know how access to this document benefits ou.y Recommended Citation Barnett, Janet Heine, "Gaussian Guesswork: Polar Coordinates, Arc Length and the Lemniscate Curve" (2018). Calculus. 3. https://digitalcommons.ursinus.edu/triumphs_calculus/3 This Course Materials is brought to you for free and open access by the Transforming Instruction in Undergraduate Mathematics via Primary Historical Sources (TRIUMPHS) at Digital Commons @ Ursinus College. It has been accepted for inclusion in Calculus by an authorized administrator of Digital Commons @ Ursinus College. For more information, please contact [email protected]. Gaussian Guesswork: Polar Coordinates, Arc Length and the Lemniscate Curve Janet Heine Barnett∗ October 26, 2020 Just prior to his 19th birthday, the mathematical genius Carl Freidrich Gauss (1777{1855) began a \mathematical diary" in which he recorded his mathematical discoveries for nearly 20 years. Among these discoveries is the existence of a beautiful relationship between three particular numbers: • the ratio of the circumference of a circle to its diameter, or π; Z 1 dx • a specific value of a certain (elliptic1) integral, which Gauss denoted2 by $ = 2 p ; and 4 0 1 − x p p • a number called \the arithmetic-geometric mean" of 1 and 2, which he denoted as M( 2; 1). Like many of his discoveries, Gauss uncovered this particular relationship through a combination of the use of analogy and the examination of computational data, a practice that historian Adrian Rice called \Gaussian Guesswork" in his Math Horizons article subtitled \Why 1:19814023473559220744 ::: is such a beautiful number" [Rice, November 2009].
    [Show full text]
  • 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]
  • NOTES Reconsidering Leibniz's Analytic Solution of the Catenary
    NOTES Reconsidering Leibniz's Analytic Solution of the Catenary Problem: The Letter to Rudolph von Bodenhausen of August 1691 Mike Raugh January 23, 2017 Leibniz published his solution of the catenary problem as a classical ruler-and-compass con- struction in the June 1691 issue of Acta Eruditorum, without comment about the analysis used to derive it.1 However, in a private letter to Rudolph Christian von Bodenhausen later in the same year he explained his analysis.2 Here I take up Leibniz's argument at a crucial point in the letter to show that a simple observation leads easily and much more quickly to the solution than the path followed by Leibniz. The argument up to the crucial point affords a showcase in the techniques of Leibniz's calculus, so I take advantage of the opportunity to discuss it in the Appendix. Leibniz begins by deriving a differential equation for the catenary, which in our modern orientation of an x − y coordinate system would be written as, dy n Z p = (n = dx2 + dy2); (1) dx a where (x; z) represents cartesian coordinates for a point on the catenary, n is the arc length from that point to the lowest point, the fraction on the left is a ratio of differentials, and a is a constant representing unity used throughout the derivation to maintain homogeneity.3 The equation characterizes the catenary, but to solve it n must be eliminated. 1Leibniz, Gottfried Wilhelm, \De linea in quam flexile se pondere curvat" in Die Mathematischen Zeitschriftenartikel, Chap 15, pp 115{124, (German translation and comments by Hess und Babin), Georg Olms Verlag, 2011.
    [Show full text]
  • The Arc Length of a Random Lemniscate
    The arc length of a random lemniscate Erik Lundberg, Florida Atlantic University joint work with Koushik Ramachandran [email protected] SEAM 2017 Random lemniscates Topic of my talk last year at SEAM: Random rational lemniscates on the sphere. Topic of my talk today: Random polynomial lemniscates in the plane. Polynomial lemniscates: three guises I Complex Analysis: pre-image of unit circle under polynomial map jp(z)j = 1 I Potential Theory: level set of logarithmic potential log jp(z)j = 0 I Algebraic Geometry: special real-algebraic curve p(x + iy)p(x + iy) − 1 = 0 Polynomial lemniscates: three guises I Complex Analysis: pre-image of unit circle under polynomial map jp(z)j = 1 I Potential Theory: level set of logarithmic potential log jp(z)j = 0 I Algebraic Geometry: special real-algebraic curve p(x + iy)p(x + iy) − 1 = 0 Polynomial lemniscates: three guises I Complex Analysis: pre-image of unit circle under polynomial map jp(z)j = 1 I Potential Theory: level set of logarithmic potential log jp(z)j = 0 I Algebraic Geometry: special real-algebraic curve p(x + iy)p(x + iy) − 1 = 0 Prevalence of lemniscates I Approximation theory: Hilbert's lemniscate theorem I Conformal mapping: Bell representation of n-connected domains I Holomorphic dynamics: Mandelbrot and Julia lemniscates I Numerical analysis: Arnoldi lemniscates I 2-D shapes: proper lemniscates and conformal welding I Harmonic mapping: critical sets of complex harmonic polynomials I Gravitational lensing: critical sets of lensing maps Prevalence of lemniscates I Approximation theory:
    [Show full text]
  • Discrete and Smooth Curves Differential Geometry for Computer Science (Spring 2013), Stanford University Due Monday, April 15, in Class
    Homework 1: Discrete and Smooth Curves Differential Geometry for Computer Science (Spring 2013), Stanford University Due Monday, April 15, in class This is the first homework assignment for CS 468. We will have assignments approximately every two weeks. Check the course website for assignment materials and the late policy. Although you may discuss problems with your peers in CS 468, your homework is expected to be your own work. Problem 1 (20 points). Consider the parametrized curve g(s) := a cos(s/L), a sin(s/L), bs/L for s 2 R, where the numbers a, b, L 2 R satisfy a2 + b2 = L2. (a) Show that the parameter s is the arc-length and find an expression for the length of g for s 2 [0, L]. (b) Determine the geodesic curvature vector, geodesic curvature, torsion, normal, and binormal of g. (c) Determine the osculating plane of g. (d) Show that the lines containing the normal N(s) and passing through g(s) meet the z-axis under a constant angle equal to p/2. (e) Show that the tangent lines to g make a constant angle with the z-axis. Problem 2 (10 points). Let g : I ! R3 be a curve parametrized by arc length. Show that the torsion of g is given by ... hg˙ (s) × g¨(s), g(s)i ( ) = − t s 2 jkg(s)j where × is the vector cross product. Problem 3 (20 points). Let g : I ! R3 be a curve and let g˜ : I ! R3 be the reparametrization of g defined by g˜ := g ◦ f where f : I ! I is a diffeomorphism.
    [Show full text]
  • Superellipse (Lamé Curve)
    6 Superellipse (Lamé curve) 6.1 Equations of a superellipse A superellipse (horizontally long) is expressed as follows. Implicit Equation x n y n + = 10<b a , n > 0 (1.1) a b Explicit Equation 1 n x n y = b 1 - 0<b a , n > 0 (1.1') a When a =3, b=2 , the superellipses for n = 9.9 , 2, 1 and 2/3 are drawn as follows. Fig1 These are called an ellipse when n= 2, are called a diamond when n = 1, and are called an asteroid when n = 2/3. These are known well. parametric representation of an ellipse In order to ask for the area and the arc length of a super-ellipse, it is necessary to calculus the equations. However, it is difficult for (1.1) and (1.1'). Then, in order to make these easy, parametric representation of (1.1) is often used. As far as the 1st quadrant, (1.1) is as follows . xn yn 0xa + = 1 a n b n 0yb Let us compare this with the following trigonometric equation. sin 2 + cos 2 = 1 Then, we obtain xn yn = sin 2 , = cos 2 a n b n i.e. - 1 - 2 2 x = a sin n ,y = bcos n The variable and the domain are changed as follows. x: 0a : 0/2 By this representation, the area and the arc length of a superellipse can be calculated clockwise from the y-axis. cf. If we represent this as usual as follows, 2 2 x = a cos n ,y = bsin n the area and the arc length of a superellipse can be calculated counterclockwise from the x-axis.
    [Show full text]
  • Arc Length and Curvature
    Arc Length and Curvature For a ‘smooth’ parametric curve: xftygt= (), ==() and zht() for atb≤≤ OR for a ‘smooth’ vector function: rijk()tftgtht= ()++ () () for atb≤≤ the arc length of the curve for the t values atb≤ ≤ is defined as the integral b 222 ('())('())('())f tgthtdt++ ∫ a b For vector functions, the integrand can be simplified this way: r'(tdt ) ∫a Ex) Find the arc length along the curve rjk()tt= (cos())33+ (sin()) t for 0/2≤ t ≤π . Unit Tangent and Unit Normal Vectors r'(t ) Unit Tangent Vector Æ T()t = r'(t ) T'(t ) Unit Normal Vector Æ N()t = (can be a lengthy calculation) T'(t ) The unit normal vector is designed to be perpendicular to the unit tangent vector AND it points ‘into the turn’ along a vector curve’s path. 2 Ex) For the vector function rij()tt=++− (2 3) (5 t ), find T()t and N()t . (this blank page was NOT a mistake ... we’ll need it) Ex) Graph the vector function rij()tt= (2++− 3) (5 t2 ) and its unit tangent and unit normal vectors at t = 1. Curvature Curvature, denoted by the Greek letter ‘ κ ’ (kappa) measures the rate at which the ‘tilt’ of the unit tangent vector is changing with respect to the arc length. The derivation of the curvature formula is a rather lengthy one, so I will leave it out. Curvature for a vector function r()t rr'(tt )× ''( ) Æ κ= r'(t ) 3 Curvature for a scalar function yfx= () fx''( ) Æ κ= (This numerator represents absolute value, not vector magnitude.) (1+ (fx '( ))23/2 ) YOU WILL BE GIVEN THESE FORMULAS ON THE TEST!! NO NEED TO MEMORIZE!! One visualization of curvature comes from using osculating circles.
    [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]
  • 11.1 Circumference and Arc Length
    11.1 Circumference and Arc Length EEssentialssential QQuestionuestion How can you fi nd the length of a circular arc? Finding the Length of a Circular Arc Work with a partner. Find the length of each red circular arc. a. entire circle b. one-fourth of a circle y y 5 5 C 3 3 1 A 1 A B −5 −3 −1 1 3 5 x −5 −3 −1 1 3 5 x −3 −3 −5 −5 c. one-third of a circle d. fi ve-eighths of a circle y y C 4 4 2 2 A B A B −4 −2 2 4 x −4 −2 2 4 x − − 2 C 2 −4 −4 Using Arc Length Work with a partner. The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. LOOKING FOR The diameter of the tire is 25 inches. Is the REGULARITY front tire still in contact with the IN REPEATED painted box? Explain. REASONING To be profi cient in math, you need to notice if 3 ft calculations are repeated and look both for general methods and for shortcuts. CCommunicateommunicate YourYour AnswerAnswer 3. How can you fi nd the length of a circular arc? 4. A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution? Section 11.1 Circumference and Arc Length 593 hhs_geo_pe_1101.indds_geo_pe_1101.indd 593593 11/19/15/19/15 33:10:10 PMPM 11.1 Lesson WWhathat YYouou WWillill LLearnearn Use the formula for circumference.
    [Show full text]