Brachistochrone - Maple Help

Brachistochrone${}$

Main Concept

Consider the following problem, posed by Johann Bernoulli in 1696 to the most renowned mathematicians of his time:

Given two points A and B, of which A is the higher, what curve minimizes the time for a bead starting from rest at A to descend to B under the force of gravity alone?

Bernoulli had already discovered the answer (a cycloid curve), but wished to test his contemporaries. He received correct solutions from Jacob Bernoulli, Gottfried Leibniz, Guillaume de l'Hôpital, and Isaac Newton, with each arriving at that answer by vastly different methods. Today, the problem is regarded as an excellent example of the calculus of variations, and is presented as follows.

Variational Solution

Let $A$ be the origin and $B$ some lower point not directly below $A$. Let $\left(x,y\right)$ be the horizontal and vertical distances of the bead from $A$, so that $y>0$ as the bead falls ($g>0$ as well). The problem stated is to minimize the time $t$ that it takes a bead to go from $A$ to $B$ under gravity alone. The travel time can be expressed as the integral of the inverse of the speed $v$ with respect to arclength $s$ along the bead's path:

To express the speed in terms of arclength, notice that the potential energy of the bead is  where the convention $V\left(0\right)=0$ is used,  and the kinetic energy is $T=\frac{1}{2}{\mathrm{mv}}^{2}$ the bead starts from rest,  as well. Using conservation of energy you get:

$0=$$\frac{1}{2}{\mathrm{mv}}^{2}$ -

Rearranging terms gives  Changing variables using:

,

where $\mathrm{ds}$ is the increment of length on the curve, the integral for minimization becomes:

where  It remains to find the function $y\left(x\right)$ which minimizes the time.

This problem can be approached using the calculus of variations. In particular, the function is varied until an extreme value for the integral is found. In most cases the Euler-Lagrange equation would be applied next:

$\frac{\partial F}{\partial y}-\frac{ⅆ}{ⅆx}\left(\frac{\partial F}{\partial y'}\right)=0.$

However, $F$ does not explicitly depend on $x$, and thus the simpler Beltrami identity is used instead:

$F-y'\frac{\partial F}{\partial y'}=C,$

where $C$ is an integration constant. Applying $F,$ you get:

Squaring both sides yields:

This differential equation is separable and can be integrated to give:

,

where $a=\frac{1}{2}{C}^{2}$, using $y\left(0\right)=0$ and $x\left(0\right)=0$ as initial conditions. Using the following definition:

$\mathrm{θ}=\frac{\mathrm{π}}{2}-\mathrm{arcsin}\left(1-\frac{y}{a}\right)$,

the solution can be rewritten parametrically as:

This is a parameterization of a cycloid! To fully solve the problem, $a$ and ${\mathrm{θ}}_{B}$ must be determined by satisfying the boundary conditions

,

which in general must be solved numerically.

The actual time can be then expressed in terms of $\mathrm{θ}$ as follows:

 $t$ $=$ $=$ $=$ .

 Derivation of the Beltrami Identity You can simplify the Euler-Lagrange equation for the case that $F$ does not explicitly depend on, that is, $\frac{\partial F}{\partial x}=0$. Consider the total derivative of $F$ with respect to $x:$ $\frac{ⅆF}{ⅆx}=\frac{\partial F}{\partial y}y'+\frac{\partial F}{\partial y'}y''+\frac{\partial F}{\partial x}.$   The last term must be zero since $F$ does not explicitly depend on $x$ by assumption. Solving for $\frac{\partial F}{\partial y}y'$ gives:   Multiplying the Euler-Lagrange equation by $y'$ and substituting the previous equation into it gives: $\frac{ⅆF}{ⅆx}-\frac{\partial F}{\partial y'}y''-y'\frac{ⅆ}{ⅆx}\left(\frac{\partial F}{\partial y'}\right)=0,$   where left hand side can be written as: $\frac{ⅆ}{ⅆx}\left(F-y'\frac{\partial F}{\partial y'}\right)=0.$   Integrating this gives the Beltrami identity: $F-y'\frac{\partial F}{\partial y'}=C,$ where $C$ is the integration constant.

Click or drag the endpoint and press "Play" to release the beads.

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