The Brachistochrone Curve Paige R MacDonald May 16, 2014 Paige R - PowerPoint PPT Presentation

The Brachistochrone Curve Paige R MacDonald May 16, 2014 Paige R MacDonald The Brachistochrone Curve May 16, 2014 1 / 1

The Problem In 1696, Johann Bernoulli posed the following problem to the scientific community: Find the curve of quickest descent between two points in a vertical plane, and not in the same vertical line, for a particle to slide under only the force of gravity, neglecting friction. Paige R MacDonald The Brachistochrone Curve May 16, 2014 2 / 1

Solutions were found by Leibniz, L’Hopital, Newton, and Jacob and Johann Bernoulli. Paige R MacDonald The Brachistochrone Curve May 16, 2014 3 / 1

Solutions were found by Leibniz, L’Hopital, Newton, and Jacob and Johann Bernoulli. The curve is known as The Brachistochrone from the Greek brachistos for“shortest” chronos for “time”. Paige R MacDonald The Brachistochrone Curve May 16, 2014 3 / 1

Johann Bernoulli’s Solution Johann Bernoulli’s solution begins by considering a problem in optics. A v 1 a α 1 x c − x P α 2 v 2 b B c Paige R MacDonald The Brachistochrone Curve May 16, 2014 4 / 1

Using time = distance velocity , the total time from A to B is √ a 2 + x 2 b 2 + ( c − x ) 2 � T = + . v 1 v 2 Paige R MacDonald The Brachistochrone Curve May 16, 2014 5 / 1

Using time = distance velocity , the total time from A to B is √ a 2 + x 2 b 2 + ( c − x ) 2 � T = + . v 1 v 2 Taking the derivative of each side leads to dT x c − x dx = √ a 2 + x 2 − b 2 + ( c − x ) 2 � v 1 v 2 Paige R MacDonald The Brachistochrone Curve May 16, 2014 5 / 1

Assuming the ray of light is able to select the path from A to B that minimizes time T , then dT / dx = 0 and x c − x a 2 + x 2 = √ b 2 + ( c − x ) 2 � v 1 v 2 Paige R MacDonald The Brachistochrone Curve May 16, 2014 6 / 1

Assuming the ray of light is able to select the path from A to B that minimizes time T , then dT / dx = 0 and x c − x a 2 + x 2 = √ b 2 + ( c − x ) 2 � v 1 v 2 and from the figure, x c − x sin α 1 = sin α 2 = and √ a 2 + x 2 b 2 + ( c − x ) 2 � so sin α 1 = sin α 2 . v 1 v 2 . Paige R MacDonald The Brachistochrone Curve May 16, 2014 6 / 1

Assuming the ray of light is able to select the path from A to B that minimizes time T , then dT / dx = 0 and x c − x a 2 + x 2 = √ b 2 + ( c − x ) 2 � v 1 v 2 and from the figure, x c − x sin α 1 = sin α 2 = and √ a 2 + x 2 b 2 + ( c − x ) 2 � so sin α 1 = sin α 2 . v 1 v 2 . This is Snell’s Law of Refraction and demonstrates Fermat’s principle of least time . Paige R MacDonald The Brachistochrone Curve May 16, 2014 6 / 1

Now we divide the plane into more layers. v 1 α 1 α 2 α 2 v 2 α 3 α 3 v 3 v 4 α 4 Paige R MacDonald The Brachistochrone Curve May 16, 2014 7 / 1

Applying Snell’s law to the divided layers, sin α 1 = sin α 2 = sin α 3 = sin α 4 . v 1 v 2 v 3 v 4 If we continue dividing the layers into smaller and smaller sections, the path approaches a smooth curve and sin α = constant . v Paige R MacDonald The Brachistochrone Curve May 16, 2014 8 / 1

Returning to the brachistochrone problem, we set up a coordinate system with an inverted y -axis. x A y α β B y Paige R MacDonald The Brachistochrone Curve May 16, 2014 9 / 1

Due to the assumption that there is no friction, energy is conserved and K i + U i = K f + U f 1 2 mv 2 = mgy . Paige R MacDonald The Brachistochrone Curve May 16, 2014 10 / 1

Due to the assumption that there is no friction, energy is conserved and K i + U i = K f + U f 1 2 mv 2 = mgy . Solving for velocity, � v = 2 gy . Paige R MacDonald The Brachistochrone Curve May 16, 2014 10 / 1

Due to the assumption that there is no friction, energy is conserved and K i + U i = K f + U f 1 2 mv 2 = mgy . Solving for velocity, � v = 2 gy . From the figure, sin α = cos β 1 = sec β 1 = 1 + tan 2 β 1 = � 1 + ( y ′ ) 2 Paige R MacDonald The Brachistochrone Curve May 16, 2014 10 / 1

Substituting 1 � v = 2 gy sin α = and � 1 + ( y ′ ) 2 into sin α/ v = constant and simplifying, y [1 + ( y ′ ) 2 ] = c . Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1

Substituting 1 � v = 2 gy sin α = and � 1 + ( y ′ ) 2 into sin α/ v = constant and simplifying, y [1 + ( y ′ ) 2 ] = c . This is the differential equation of the brachistochrone. Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1

Substituting 1 � v = 2 gy sin α = and � 1 + ( y ′ ) 2 into sin α/ v = constant and simplifying, y [1 + ( y ′ ) 2 ] = c . This is the differential equation of the brachistochrone. Using Leibniz notation � � 2 � � dy y 1 + = c dx Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1

Substituting 1 � v = 2 gy sin α = and � 1 + ( y ′ ) 2 into sin α/ v = constant and simplifying, y [1 + ( y ′ ) 2 ] = c . This is the differential equation of the brachistochrone. Using Leibniz notation � � 2 � � dy y 1 + = c dx Which simplifies to � 1 / 2 dy � c − y dx = y Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1

Separating variables and solving for dx � 1 / 2 � y dx = dy C − y Paige R MacDonald The Brachistochrone Curve May 16, 2014 12 / 1

Separating variables and solving for dx � 1 / 2 � y dx = dy C − y Let � 1 / 2 � y = tan φ C − y and solving for y y = 2 c sin φ cos φ Paige R MacDonald The Brachistochrone Curve May 16, 2014 12 / 1

Separating variables and solving for dx � 1 / 2 � y dx = dy C − y Let � 1 / 2 � y = tan φ C − y and solving for y y = 2 c sin φ cos φ Substituting back into the differential equation dx = 2 c sin 2 φ d φ Paige R MacDonald The Brachistochrone Curve May 16, 2014 12 / 1

Integrating both sides and simplifying x = c 2(2 φ − sin 2 φ ) + c 1 Paige R MacDonald The Brachistochrone Curve May 16, 2014 13 / 1

Using the initial condition, c 1 = 0 and letting a = c / 2 and θ = 2 φ , we obtain x = a ( θ − sin θ ) and y = a (1 − cos θ ) . Paige R MacDonald The Brachistochrone Curve May 16, 2014 14 / 1

Using the initial condition, c 1 = 0 and letting a = c / 2 and θ = 2 φ , we obtain x = a ( θ − sin θ ) and y = a (1 − cos θ ) . These are the parametric equations of the cycloid. Paige R MacDonald The Brachistochrone Curve May 16, 2014 14 / 1

The Cycloid y a θ x 2 π a ( x , y ) Paige R MacDonald The Brachistochrone Curve May 16, 2014 15 / 1

The Calculus of Variations Euler’s work on the Brachistochrone problem has been crediting with leading to the Calculus of Variations , which is concerned with finding extrema of functionals . Paige R MacDonald The Brachistochrone Curve May 16, 2014 16 / 1

The Calculus of Variations Euler’s work on the Brachistochrone problem has been crediting with leading to the Calculus of Variations , which is concerned with finding extrema of functionals . This can be related to optimazation problems in elementary calculus, where the goal is to find a minimum of maximum of a function f ( x ), where a single variable, x is the quantity that varies. f : R − → R Paige R MacDonald The Brachistochrone Curve May 16, 2014 16 / 1

In the calculus of variations, the quantity that varies is itself a function. Paige R MacDonald The Brachistochrone Curve May 16, 2014 17 / 1

In the calculus of variations, the quantity that varies is itself a function. A functional , then, is a function that takes a function as its input, and returns a real number. I : f − → R So the integral to minimize is � x 2 I = f ( x , y , y ′ ) dx x 1 Paige R MacDonald The Brachistochrone Curve May 16, 2014 17 / 1

Admissable Functions In order to find the function that yields the smallest value for I , we must select from the family of functions only those satisfying certain conditions: Paige R MacDonald The Brachistochrone Curve May 16, 2014 18 / 1

Admissable Functions In order to find the function that yields the smallest value for I , we must select from the family of functions only those satisfying certain conditions: Passes through points A and B Of class C 2 Paige R MacDonald The Brachistochrone Curve May 16, 2014 18 / 1

Admissable Functions In order to find the function that yields the smallest value for I , we must select from the family of functions only those satisfying certain conditions: Passes through points A and B Of class C 2 Assuming a function y ( x ) exists that minimizes the integral, we consider a function η ( x ) that “disturbs” y ( x ) slightly. Paige R MacDonald The Brachistochrone Curve May 16, 2014 18 / 1

y ( x 1 , y 1 ) y ( x ) = y ( x ) + αη ( x ) ¯ αη ( x ) ( x 2 , y 2 ) y ( x ) η ( x ) η ( x ) x x 1 x x 2 η ( x 1 ) = η ( x 2 ) = 0, α a small parameter Paige R MacDonald The Brachistochrone Curve May 16, 2014 19 / 1

For any value of the function η ( x ), ¯ y ( x ) represents a one-parameter family of admissable functions whose vertical deviation from the minimizing curve y ( x ) is αη ( x ). Paige R MacDonald The Brachistochrone Curve May 16, 2014 20 / 1