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 The curve is known as The Brachistochrone from the Greek
brachistos for“shortest” chronos for “time”.
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 a v1
α1
x P c − x
α2 v 2 b
B c
Paige R MacDonald The Brachistochrone Curve May 16, 2014 4 / 1 Taking the derivative of each side leads to dT x c − x = √ − 2 2 p 2 2 dx v1 a + x v2 b + (c − x)
Using distance time = , velocity the total time from A to B is √ a2 + x 2 pb2 + (c − x)2 T = + . v1 v2
Paige R MacDonald The Brachistochrone Curve May 16, 2014 5 / 1 Using distance time = , velocity the total time from A to B is √ a2 + x 2 pb2 + (c − x)2 T = + . v1 v2 Taking the derivative of each side leads to dT x c − x = √ − 2 2 p 2 2 dx v1 a + x v2 b + (c − x)
Paige R MacDonald The Brachistochrone Curve May 16, 2014 5 / 1 and from the figure, x c − x sin α1 = √ and sin α2 = a2 + x 2 pb2 + (c − x)2 so sin α sin α 1 = 2 . v1 v2 . This is Snell’s Law of Refraction and demonstrates Fermat’s principle of least time.
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 √ = 2 2 p 2 2 v1 a + x v2 b + (c − x)
Paige R MacDonald The Brachistochrone Curve May 16, 2014 6 / 1 This is Snell’s Law of Refraction and demonstrates Fermat’s principle of least time.
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 √ = 2 2 p 2 2 v1 a + x v2 b + (c − x) and from the figure, x c − x sin α1 = √ and sin α2 = a2 + x 2 pb2 + (c − x)2 so sin α sin α 1 = 2 . v1 v2 .
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 √ = 2 2 p 2 2 v1 a + x v2 b + (c − x) and from the figure, x c − x sin α1 = √ and sin α2 = a2 + x 2 pb2 + (c − x)2 so sin α sin α 1 = 2 . v1 v2 . 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 α1
α v2 2 α2
v3 α3 α3
v4 α4
Paige R MacDonald The Brachistochrone Curve May 16, 2014 7 / 1 Applying Snell’s law to the divided layers, sin α sin α sin α sin α 1 = 2 = 3 = 4 . v1 v2 v3 v4 If we continue dividing the layers into smaller and smaller sections, the path approaches a smooth curve and sinα = 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
β α B y
Paige R MacDonald The Brachistochrone Curve May 16, 2014 9 / 1 Solving for velocity, v = p2gy. From the figure, sin α = cos β 1 = sec β 1 = 1 + tan2 β 1 = p1 + (y 0)2
Due to the assumption that there is no friction, energy is conserved and Ki + Ui = Kf + Uf 1 mv 2 = mgy. 2
Paige R MacDonald The Brachistochrone Curve May 16, 2014 10 / 1 From the figure, sin α = cos β 1 = sec β 1 = 1 + tan2 β 1 = p1 + (y 0)2
Due to the assumption that there is no friction, energy is conserved and Ki + Ui = Kf + Uf 1 mv 2 = mgy. 2 Solving for velocity, v = p2gy.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 10 / 1 Due to the assumption that there is no friction, energy is conserved and Ki + Ui = Kf + Uf 1 mv 2 = mgy. 2 Solving for velocity, v = p2gy. From the figure, sin α = cos β 1 = sec β 1 = 1 + tan2 β 1 = p1 + (y 0)2
Paige R MacDonald The Brachistochrone Curve May 16, 2014 10 / 1 This is the differential equation of the brachistochrone. Using Leibniz notation " # dy 2 y 1 + = c dx
Which simplifies to dy c − y 1/2 = dx y
Substituting 1 v = p2gy and sin α = p1 + (y 0)2 into sinα/v = constant and simplifying,
y[1 + (y 0)2] = c.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1 Using Leibniz notation " # dy 2 y 1 + = c dx
Which simplifies to dy c − y 1/2 = dx y
Substituting 1 v = p2gy and sin α = p1 + (y 0)2 into sinα/v = constant and simplifying,
y[1 + (y 0)2] = c.
This is the differential equation of the brachistochrone.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1 Which simplifies to dy c − y 1/2 = dx y
Substituting 1 v = p2gy and sin α = p1 + (y 0)2 into sinα/v = constant and simplifying,
y[1 + (y 0)2] = c.
This is the differential equation of the brachistochrone. Using Leibniz notation " # dy 2 y 1 + = c dx
Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1 Substituting 1 v = p2gy and sin α = p1 + (y 0)2 into sinα/v = constant and simplifying,
y[1 + (y 0)2] = c.
This is the differential equation of the brachistochrone. Using Leibniz notation " # dy 2 y 1 + = c dx
Which simplifies to dy c − y 1/2 = dx y
Paige R MacDonald The Brachistochrone Curve May 16, 2014 11 / 1 Let y 1/2 = tan φ C − y and solving for y y = 2c sin φ cos φ Substituting back into the differential equation
dx = 2c sin2 φdφ
Separating variables and solving for dx
y 1/2 dx = dy C − y
Paige R MacDonald The Brachistochrone Curve May 16, 2014 12 / 1 Substituting back into the differential equation
dx = 2c sin2 φdφ
Separating variables and solving for dx
y 1/2 dx = dy C − y
Let y 1/2 = tan φ C − y and solving for y y = 2c sin φ cos φ
Paige R MacDonald The Brachistochrone Curve May 16, 2014 12 / 1 Separating variables and solving for dx
y 1/2 dx = dy C − y
Let y 1/2 = tan φ C − y and solving for y y = 2c sin φ cos φ Substituting back into the differential equation
dx = 2c sin2 φdφ
Paige R MacDonald The Brachistochrone Curve May 16, 2014 12 / 1 Integrating both sides and simplifying c x = (2φ − sin 2φ) + c 2 1
Paige R MacDonald The Brachistochrone Curve May 16, 2014 13 / 1 These are the parametric equations of the cycloid.
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).
Paige R MacDonald The Brachistochrone Curve May 16, 2014 14 / 1 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.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 14 / 1 The Cycloid
y
a θ x 2πa (x, y)
Paige R MacDonald The Brachistochrone Curve May 16, 2014 15 / 1 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
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 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
Z x2 I = f (x, y, y 0)dx x1
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
Z x2 I = f (x, y, y 0)dx x1
Paige R MacDonald The Brachistochrone Curve May 16, 2014 17 / 1 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.
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 Assuming a function y(x) exists that minimizes the integral, we consider a function η(x) that “disturbs” y(x) slightly.
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 , y ) 1 1 y¯(x) = y(x) + αη(x)
αη(x)
y(x) (x2, y2)
η(x) η(x) x x1 x x2
η(x1) = η(x2) = 0, α a small parameter
Paige R MacDonald The Brachistochrone Curve May 16, 2014 19 / 1 For any choice η(x), y(x) belongs to the family and corresponds to α = 0.
Substitutingy ¯(x) into the integral
Z x2 I (α) = f (x, y¯, y¯0)dx x1 Z x2 I (α) = f [x, y(x) + αη(x), y 0(x) + αη0(x)]dx. x1
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 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.
Substitutingy ¯(x) into the integral
Z x2 I (α) = f (x, y¯, y¯0)dx x1 Z x2 I (α) = f [x, y(x) + αη(x), y 0(x) + αη0(x)]dx. x1
Paige R MacDonald The Brachistochrone Curve May 16, 2014 20 / 1 Z x2 ∂ I 0(α) = f (x, y¯, y¯0)dx x1 ∂α Z x2 ∂f ∂x ∂f ∂y¯ ∂f ∂y¯0 0 = + + 0 dx x1 ∂x ∂α ∂y¯ ∂α ∂y¯ ∂α which integrates to give Euler’s equation
d ∂f ∂f − = 0 dx ∂y 0 ∂y
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 0(α) = 0.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 21 / 1 Z x2 ∂f ∂x ∂f ∂y¯ ∂f ∂y¯0 0 = + + 0 dx x1 ∂x ∂α ∂y¯ ∂α ∂y¯ ∂α which integrates to give Euler’s equation
d ∂f ∂f − = 0 dx ∂y 0 ∂y
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 0(α) = 0.
Z x2 ∂ I 0(α) = f (x, y¯, y¯0)dx x1 ∂α
Paige R MacDonald The Brachistochrone Curve May 16, 2014 21 / 1 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 0(α) = 0.
Z x2 ∂ I 0(α) = f (x, y¯, y¯0)dx x1 ∂α Z x2 ∂f ∂x ∂f ∂y¯ ∂f ∂y¯0 0 = + + 0 dx x1 ∂x ∂α ∂y¯ ∂α ∂y¯ ∂α which integrates to give Euler’s equation
d ∂f ∂f − = 0 dx ∂y 0 ∂y
Paige R MacDonald The Brachistochrone Curve May 16, 2014 21 / 1 Special Case
In the case where x is missing from the function f , Euler’s equation can be integrated to a less general form ∂f y 0 − f = c ∂y 0 which is known as Beltrami’s identity.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 22 / 1 Euler’s Solution
Returning to the brachistochrone problem ds velocity = dt ds dt = v So the integral to be minimized is
Z x2 p1 + (y 0)2 √ dx x1 2gy Since x does not appear explicitly in the function to be minimized, we can use Beltrami’s identity.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 23 / 1 ∂f y 0 − f = c ∂y 0 ! ∂ p1 + (y 0)2 p1 + (y 0)2 √ y 0 − √ = c ∂y 0 2gy 2gy
1 1 p1 + (y 0)2 √ [1 + (y 0)2]−1/22y 0 y 0 − √ = c 2gy 2 2gy
Paige R MacDonald The Brachistochrone Curve May 16, 2014 24 / 1 y[1 + (y 0)2] = C
Which simplifies to
Paige R MacDonald The Brachistochrone Curve May 16, 2014 25 / 1 Which simplifies to y[1 + (y 0)2] = C
Paige R MacDonald The Brachistochrone Curve May 16, 2014 25 / 1 The end.
Paige R MacDonald The Brachistochrone Curve May 16, 2014 26 / 1 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.pdfPaige R MacDonald The Brachistochrone Curve May 16, 2014 27 / 1 Frank Porter. Calculus of Variations. http://www.hep.caltech.edu/~fcp/math/ variationalCalculus/variationalCalculus.pdf