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The Brachistochrone Curve Introduction Johann Bernoulli's . Euler's Solution Home Page Title Page The Brachistochrone Curve JJ II Paige MacDonald J I May 16, 2014 Page1 of 21 Go Back Full Screen Close Quit Introduction Johann Bernoulli's . Euler's Solution Home Page Abstract Title Page This papers explores the brachistochrone problem, which is to find the path of quickest descent of a particle moving between two fixed points in a vertical plane. A brief history of the problem is given, followed by different approaches to solving the problem. JJ II J I Page1 of 21 Go Back Full Screen Close Quit 1. Introduction The brachistochrone problem seeks to find the curve between two points, A and B, in a vertical plane and not in the same vertical line, along which a particle will slide in Introduction the shortest amount of time under the force of gravity and neglecting friction. Johann Bernoulli's . Johann Bernoulli posed the problem in 1696. Solutions were found by Gottfried Wil- Euler's Solution helm von Leibniz, Isaac Newton, Guillaume de l'Hopital, Jacob Bernoulli, and Johann Bernoulli himself. All of their answers agreed, although each used different methods of Home Page derivation. In 1744, Euler published a work generalizing the work done by the Bernoulli brothers and came up with what is now known as the Euler-Lagrange differential equation (which Title Page Lagrange later independently derived) in order to minimize the value of a definite integral over a family of functions, which led to the calculus of variations. Lagrange provided an analytic method to solve the brachistochrone problem and JJ II other problems of its type, and introduced partial derivatives to the equation. J I 2. Johann Bernoulli's Solution A creative solution like Johann Bernoulli's uses an indirect approach, but results in Page1 of 21 mathematics that are much easier to understand, so this will be a good starting point. Consider a ray of light that travels from A to P with velocity v1 and then from P to Go Back B with velocity v2 as in Figure1. Using the equation distance time = velocity Full Screen and solving for time leads to the following (in terms of the notation in Figure1): p a2 + x2 pb2 + (c − x)2 Close T = + v1 v2 Quit Introduction A Johann Bernoulli's . v1 Euler's Solution a α1 Home Page x P c − x Title Page α 2 JJ II v2 b J I Page2 of 21 B c Go Back Figure 1: A ray of light travels from A to P with velocity v1 and then from P to B with velocity v2. Full Screen Close Quit Taking the derivative of each side gives: p ! dT d a2 + x2 pb2 + (c − x)2 = + Introduction dx dx v1 v2 Johann Bernoulli's . 1 2 2 − 1 1 2 2 − 1 (a + x ) 2 (2x) (b + (c − x) ) 2 (2)(c − x)(−1) Euler's Solution = 2 + 2 v1 v2 x c − x = p − Home Page 2 2 p 2 2 v1 a + x v2 b + (c − x) In order to minimize T , set dT=dx = 0. So, Title Page x c − x 0 = p − 2 2 p 2 2 v1 a + x v2 b + (c − x) (1) JJ II x c − x p = 2 2 p 2 2 v1 a + x v2 b + (c − x) J I Referring back to Figure1, it can be seen that x sin α1 = p (2) Page3 of 21 a2 + x2 and c − x Go Back sin α2 = (3) pb2 + (c − x)2 Substituting equations (2) and (3) into equation (1) results in the following: Full Screen sin α sin α 1 = 2 (4) v1 v2 Close Quit Introduction Johann Bernoulli's . Euler's Solution v Home Page 1 α1 Title Page α2 v2 α2 JJ II α3 v3 α3 J I v Page4 of 21 4 α4 Figure 2 Go Back Full Screen Close Quit Equation (4) is known as Snell's law of refraction and demonstrates Fermat's prin- ciple of least time, which says that light travels from one point to another along the path requiring the least time. If the layers are divided as in Figure2, the velocity Introduction of light in each layer is constant, but decreases from one layer to the layer below it. Applying Snell's law gives Johann Bernoulli's . Euler's Solution sin α sin α sin α sin α 1 = 2 = 3 = 4 v1 v2 v3 v4 As the layers are divided into smaller and smaller sections, the path approaches a Home Page smooth curve and the velocity decreases continuously so that sin α = c (5) Title Page v where c is a constant. Returning to the Brachistochrone problem, assume that the particle moves along a JJ II curve from A to B in the least possible time as in Figure3. Due to the assumption that there is no friction, the total energy of the particle at point A is that same as the total energy at B, which is the kinetic energy plus the potential energy. Since the particle J I starts from rest, Ki + Ui = Kf + Uf Page5 of 21 1 0 + mgy = mv2 + 0 2 1 Go Back mv2 = mgy: 2 Solving for velocity results in v = p2gy: (6) Full Screen From Figure3, 1 1 1 Close sin α = cos β = = p = p : (7) sec β 1 + tan2 β 1 + (y0)2 Quit x A Introduction Johann Bernoulli's . Euler's Solution y Home Page Title Page β JJ II α B J I y Figure 3 Page6 of 21 Substituting equations (6) and (7) into equation (5) results in: Go Back Full Screen Close Quit 1 p = c 2gyp1 + (y0)2 p 1 p2gy 1 + (y0)2 = Introduction c Johann Bernoulli's . p p 1 y 1 + (y0)2 = p Euler's Solution c 2g 1 y[1 + (y0)2] = 2c2g Home Page Noting that the right hand side of the equation will be replaced by some other arbitrary constant, the differential equation of the brachistochrone is Title Page y[1 + (y0)2] = C (8) JJ II Using Leibniz notation, the differential equation of the brachistochrone becomes: " 2# dy J I y 1 + = C dx dy 2 C Page7 of 21 1 + = dx y 2 dy C Go Back = − 1 dx y dy 2 C − y = Full Screen dx y dy C − y 1=2 = Close dx y Quit Separating variables and solving for dx results in y 1=2 dx = dy (9) Introduction C − y Johann Bernoulli's . Introducing a new variable and setting Euler's Solution y 1=2 = tan φ, (10) Home Page C − y and then solving for y gives Title Page y = tan2 φ C − y y = (C − y) tan2 φ JJ II y = C tan2 φ − y tan2 φ y(1 + tan2 φ) = C tan2 φ J I y sec2 φ = C tan2 φ Page8 of 21 y = C sin2 φ. Taking the derivative of each side with respect to φ leads to Go Back dy = 2C sin φ cos φdφ. (11) Next, substitute equations (10) and (11) into equation (9) to obtain Full Screen dx = (tan φ)(2C sin φ cos φdφ) Close dx = 2C sin2 φdφ Quit Integrating both sides results in Z Z dx = 2c sin2 φdφ Introduction Z 1 − cos 2φ Johann Bernoulli's . x = 2C dφ 2 Euler's Solution Z Z x = C 1dφ − C cos 2φdφ Home Page 1 x = Cφ − c sin 2φ + c 2 1 C Title Page x = (2φ − sin 2φ) + c 2 1 Using the initial condition (0; 0) in equation (10) and solving for φ gives φ = 0 when JJ II the particle is at point A. Therefore, c1 is 0 and C x = (2φ − sin 2φ) (12) 2 J I and y = C sin2 φ Page9 of 21 1 − cos 2φ y = C 2 (13) C Go Back y = (1 − cos 2φ): 2 Letting a = C=2 and θ = 2φ, equations (12) and (13) become Full Screen x = a(θ − sin θ) (14) and Close y = a(1 − cos θ): (15) Quit Introduction Johann Bernoulli's . Euler's Solution y Home Page Title Page JJ II a θ J I x 2πa (x; y) Page 10 of 21 Figure 4 Go Back Full Screen Close Quit These are the parametric equations of the curve traced out by a point on the circum- ference of a circle of radius a rolling along the x axis (the cycloid) with the condition that the curve passes through B for some value of a. The result is shown in Figure4. Introduction Johann Bernoulli's . 3. Euler's Solution Euler's Solution Euler's approach to solving this problem is more analytic and involves minimizing a function of a function, called a functional. Problems of this type belong to The Calculus Home Page of Variations, which Euler's work on this problem has been credited with leading to. This is similar to optimization problems in elementary calculus, except in calculus the Title Page quantity that varies when finding extrema of f(x) is a simple variable, x. In calculus of variations, the quantity that varies is itself a function of the form f(x; y; y0). The functional assigns a numerical value (time, in the case of the brachistochrone problem) JJ II to each function in the family of candidate functions. For the brachistochrone problem, referring to Figure5, suppose two points A and B are in a vertical plane. An infinite number of curves can be drawn connecting the J I points, but which of these curves minimizes the time for a frictionless particle to move from point A to point B under gravity? The problem, then, is to find a function that minimizes the family of functions that are differentiable and whose derivatives and Page 11 of 21 second derivatives are continuous and pass through the points A and B-these are the candidate, or admissable, functions.
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