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

The Brachistochrone Curve

Paige R MacDonald May 16, 2014

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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

• nly the force of gravity, neglecting friction.

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Solutions were found by Leibniz, L’Hopital, Newton, and Jacob and Johann Bernoulli.

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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”.

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Johann Bernoulli’s Solution

Johann Bernoulli’s solution begins by considering a problem in optics. x c − x a v1 v2 α1 α2 c b A P B

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Using time = distance velocity , the total time from A to B is T = √ a2 + x2 v1 +

• b2 + (c − x)2

v2 .

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Using time = distance velocity , the total time from A to B is T = √ a2 + x2 v1 +

• b2 + (c − x)2

v2 . Taking the derivative of each side leads to dT dx = x v1 √ a2 + x2 − c − x v2

• b2 + (c − x)2

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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 v1 √ a2 + x2 = c − x v2

• b2 + (c − x)2

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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 v1 √ a2 + x2 = c − x v2

• b2 + (c − x)2

and from the figure, sin α1 = x √ a2 + x2 and sin α2 = c − x

• b2 + (c − x)2

so sin α1 v1 = sin α2 v2 . .

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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 v1 √ a2 + x2 = c − x v2

• b2 + (c − x)2

and from the figure, sin α1 = x √ a2 + x2 and sin α2 = c − x

• b2 + (c − x)2

so sin α1 v1 = sin α2 v2 . . This is Snell’s Law of Refraction and demonstrates Fermat’s principle of least time.

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Now we divide the plane into more layers.

v1 v2 v3 v4 α1 α2 α2 α3α3 α4

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Applying Snell’s law to the divided layers, sin α1 v1 = sin α2 v2 = sin α3 v3 = sin α4 v4 . If we continue dividing the layers into smaller and smaller sections, the path approaches a smooth curve and sinα v = constant.

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Returning to the brachistochrone problem, we set up a coordinate system with an inverted y-axis.

x y A B y α β

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Due to the assumption that there is no friction, energy is conserved and Ki + Ui = Kf + Uf 1 2mv 2 = mgy.

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Due to the assumption that there is no friction, energy is conserved and Ki + Ui = Kf + Uf 1 2mv 2 = mgy. Solving for velocity, v =

• 2gy.

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Due to the assumption that there is no friction, energy is conserved and Ki + Ui = Kf + Uf 1 2mv 2 = mgy. Solving for velocity, v =

• 2gy.

From the figure, sin α = cos β = 1 sec β = 1 1 + tan2 β = 1

• 1 + (y ′)2

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Substituting v =

• 2gy

and sin α = 1

• 1 + (y ′)2

into sinα/v = constant and simplifying, y[1 + (y ′)2] = c.

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Substituting v =

• 2gy

and sin α = 1

• 1 + (y ′)2

into sinα/v = constant and simplifying, y[1 + (y ′)2] = c. This is the differential equation of the brachistochrone.

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Substituting v =

• 2gy

and sin α = 1

• 1 + (y ′)2

into sinα/v = constant and simplifying, y[1 + (y ′)2] = c. This is the differential equation of the brachistochrone. Using Leibniz notation y

• 1 +

dy dx 2 = c

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Substituting v =

• 2gy

and sin α = 1

• 1 + (y ′)2

into sinα/v = constant and simplifying, y[1 + (y ′)2] = c. This is the differential equation of the brachistochrone. Using Leibniz notation y

• 1 +

dy dx 2 = c Which simplifies to dy dx = c − y y 1/2

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Separating variables and solving for dx dx =

• y

C − y 1/2 dy

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Separating variables and solving for dx dx =

• y

C − y 1/2 dy Let

• y

C − y 1/2 = tan φ and solving for y y = 2c sin φ cos φ

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Separating variables and solving for dx dx =

• y

C − y 1/2 dy Let

• y

C − y 1/2 = tan φ and solving for y y = 2c sin φ cos φ Substituting back into the differential equation dx = 2c sin2 φdφ

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Integrating both sides and simplifying x = c 2(2φ − sin 2φ) + c1

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Using the initial condition, c1 = 0 and letting a = c/2 and θ = 2φ, we obtain x = a(θ − sinθ) and y = a(1 − cosθ).

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Using the initial condition, c1 = 0 and letting a = c/2 and θ = 2φ, we obtain x = a(θ − sinθ) and y = a(1 − cosθ). These are the parametric equations of the cycloid.

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The Cycloid

x y a (x, y) 2πa θ

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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.

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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

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In the calculus of variations, the quantity that varies is itself a function.

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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 I = x2

x1

f (x, y, y ′)dx

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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:

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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

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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.

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x y x1 x2 (x1, y1) (x2, y2) x y(x) ¯ y(x) = y(x) + αη(x) η(x) η(x) αη(x) η(x1) = η(x2) = 0, α a small parameter

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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).

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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). For any choice η(x), y(x) belongs to the family and corresponds to α = 0. Substituting ¯ y(x) into the integral I(α) = x2

x1

f (x, ¯ y, ¯ y ′)dx I(α) = x2

x1

f [x, y(x) + αη(x), y ′(x) + αη′(x)]dx.

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When α = 0, ¯ y(x) = y(x) and since y(x) minimizes the the integral, I(α) must have a minimum when α = 0, or when I ′(α) = 0.

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When α = 0, ¯ y(x) = y(x) and since y(x) minimizes the the integral, I(α) must have a minimum when α = 0, or when I ′(α) = 0. I ′(α) = x2

x1

∂ ∂αf (x, ¯ y, ¯ y ′)dx

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When α = 0, ¯ y(x) = y(x) and since y(x) minimizes the the integral, I(α) must have a minimum when α = 0, or when I ′(α) = 0. I ′(α) = x2

x1

∂ ∂αf (x, ¯ y, ¯ y ′)dx 0 = x2

x1

∂f ∂x ∂x ∂α + ∂f ∂¯ y ∂¯ y ∂α + ∂f ∂¯ y ′ ∂¯ y ′ ∂α

• dx

which integrates to give Euler’s equation d dx ∂f ∂y ′

• − ∂f

∂y = 0

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Special Case

In the case where x is missing from the function f , Euler’s equation can be integrated to a less general form y ′ ∂f ∂y ′ − f = c which is known as Beltrami’s identity.

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Euler’s Solution

Returning to the brachistochrone problem velocity = ds dt dt = ds v So the integral to be minimized is x2

x1

• 1 + (y ′)2

√2gy dx Since x does not appear explicitly in the function to be minimized, we can use Beltrami’s identity.

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y ′ ∂f ∂y ′ − f = c

• ∂

∂y ′

• 1 + (y ′)2

√2gy

• y ′ −
• 1 + (y ′)2

√2gy = c

• 1

√2gy 1 2[1 + (y ′)2]−1/22y ′

• y ′ −
• 1 + (y ′)2

√2gy = c

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Which simplifies to

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Which simplifies to y[1 + (y ′)2] = C

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The end.

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Bibliography

Till Tantau. Tikz and PGF Packages. David Arnold. Writing Scientific Papers in Latex. George Simmons Differential Equations with Applications and Historical Notes. Douglas S. Shafer. 2007 The Brachistochrone: Historical Gateway to the Calculus of Variations. Nils P. Johnson. 2009 The Brachistochrone Problem J J O’Connor and E F Robertson 2002 History Topic: The Brachistochrone Problem http://www.history.mcs.stand. ac.uk/PrintHT/Brachistochrone.html Unknown author. The Brachistochrone Problem http://www.math.utk.edu/~freire/teaching/m231f08/ m231f08brachistochrone.pdf

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