jkwiens.com: Answer: Brachistochrone Problem

Answer: Brachistochrone Problem

Saturday, 10 of November , 2007 at 5:13 pm

Question: A particle starts from rest and slides down a smooth curve under gravity. Find the shape of the curve that will minimize the time taken between two given points.

Answer: First, we need to find the distance dl of two neighboring points on the curve f(x).

$dl = \sqrt{dx^{2} + dy^{2}} = \sqrt{dx^{2} (1 + \frac{dy^{2}}{dx^{2}})} = \sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx$

We can then use the relationship $dt = \frac{dl}{v}$ to determine the time required to travel the distance dl. If we integrate dt over the entire path of f(x), we will obtain the function we want to minimize.

$t = \oint \frac{dl}{v} = \int \frac{\sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx}{v}$
In order to obtain the value v, we can use the fact that energy is conserved. Therefore the following relation will hold:
$\frac{1}{2}mv^{2} = mgy \rightarrow v = \sqrt{2gy}$
We can now insert this relationship into our integral to get the following:
$t = \int \frac{\sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx}{\sqrt{2gy}}$
In order to minimize t, we will need to minimize the following function
$\Psi(x,y) = \frac{\sqrt{1 + \frac{dy^{2}}{dx^{2}}}}{\sqrt{2gy}}$
This function, according to Calculus of Variations, will be an extrema when the Euler-Lagrange differential equation is satisfied.
$\frac{\partial\Psi}{\partial y} - \frac{\partial}{\partial x}\partial\frac{\Psi}{\partial \dot{y}} = 0$
If we insert $\Psi(x,y)$ into the Euler-Lagrange differential equation, we obtain the following differential equation (with a little bit of algebra).
$2\ddot{y}y + \dot{y}^{2} + 1 = 0$
With a little more manipulation, we can reduce this differential equation into something more manageable.
$2\ddot{y}y + \dot{y}^{2} + 1 = 0$
$2\ddot{y}y + \dot{y}^{3} + \dot{y} = 0$
$\frac{\partial}{\partial x}(y\dot{y}^{2}) + \frac{\partial y}{\partial x} = 0$
$\frac{\partial}{\partial x}(y\dot{y}^{2} + y) = 0$
If we integrate this differential equation, we get the following:
$y\dot{y}^{2} + y = C$
$\dot{y} = \sqrt{\frac{C - y}{y}}$
This equation is solved by the parametric equations:
$x = \frac{C}{2}(\theta - \sin{\theta})$
$y = \frac{C}{2}(1 - \cos{\theta})$
which is a cycloid.

Category: Answers, Calculus, Classical Mechanics

No Comments

No comments yet.

Leave a comment