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Unformatted text preview: Brachistochrone Challenge (Birth of Calculus of Variations) Simon Stevinus (15481620) Dutch  father of statics Galileo Galilei (15641642) Italian  father of dynamics Isaac Newton (16421727) Gottfried Wilhelm Leibnitz (16461716) John Bernoulli (16671748) In June, 1696, John Bernoulli challenged the following problem before the scholars of his time: Given are two points A and B (B is not directly below A) in a vertical plane. A movable particle m is to descend from A to B without friction following the law of gravity (natural fall). Determine the path of m to reach B from A in the shortest time. 1 Solution: Suppose the points A and B lie in the xy plane, the y axis directed vertically downward, and the x axis horizontal, with passage from A to B marked by an increase in x. Let the extremizing path have the equation ( ) y y x = . It is assumed that the initial speed v 1 of the particle to be given in the statement of the problem. Let the points A and B have the coordinates ( ) , x y 1 1 and ( ) , x y 2 2 , respectively, so that ( ) y x y = 1 1 and ( ) y x y = 2 2 . Since the speed along the curve is given by ( ) / v ds dt = , the total time of descent ( ) x x x x x x y dx ds I v v = = + = = 2 2 1 1 2 1 . where: ( ) ( ) dy ds dx dy dx y dx dx = + = + = + 2 2 2 2 2 1 1 Assuming frictionless descent, for simplicity, of the particle of mass m with a constant gravitational acceleration g, the velocity v of the natural fall of the particle can be computed in terms of coordinates by invoking the principle of conservation of energy, i.e., a decrease of potential energy is equal to an increase of kinetic energy. Hence, ( ) mv mv mg y y = 2 2 1 1 1 1 2 2 where: v gh g y y = = 2 2 and v g y y = 1 1 2 v y y g = 2 1 1 2 . It is clear that y y 1 is the vertical distance through which the particle must descend from rest to achieve the speed v 1 , and if v = 1 , then y y = 1 . Thus, the time of descent is 2 x x y I dx g y y + = 2 1 2 1 1 2 ....
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 Summer '10
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