Minimizing Quantity Part 3: The Brachistochrone

Last post, I finished the post cryptically, mentioning that there is a certain path that an object falling will take less time to travel than any other. Now, one of the solutions, as I mentioned, is just a straight fall. That is only one of the solutions, but only if the endpoint is directly underneath the starting point. What if the endpoint is to the side, what curve would the fastest path follow?

brachsetup

Well, there is a classical method to solve this that involves Snell’s Law. This derivation was done by Johann Bernoulli as part of a challenge in the late 17th century. Why use Snell’s law? Because as I said on the first part of this series of posts, Snell’s law is the condition you get when an object travels in the path of least time. In this case, the setup is created such that the object passes through infinitely thin layers, with the object going slightly faster as it passes through each layer:

brachpath

Note, the picture was exaggerated to make a point. Like I said, the layer is infinitely small in order to simulate acceleration.

Now, first of all, it will be useful to know what the velocity is depending on the height. Knowing this is possible thanks to conservation of energy, so:

\frac{1}{2}mv^2=mgy

v=\sqrt{2gy}

Thanks to this, we can pinpoint at any layer and know the velocity. Now that we have that, we can begin the derivation by setting up Snell for the infinitely thin layer:

C=\frac{sin(\theta)}{v}=\frac{1}{v} \frac{dx}{ds}

Because the layer is infinitely thin, the component of sin of angle turns infinitesimally small, with ds signifying the infinitesimally small path the object took. Using the pythagorean theorem, ds can be expanded in terms of dy and dx, which we will see later. Second of all, the equation is equal to a constant because you can take any two layer and make the same kind of equation and make them equal to each other. You can read a bit more on Snell’s law on part two of my post if you need a refresher. Now, what is the constant equal to? Well, there is a simple way to find out by making the angle equal to 90 degrees, which happens at the bottom of the fall. Let’s call that fall distance H (note, not the same as the distance traveled, since we are looking for a curve here). Since sin(90)=1, and at the bottom the velocity is the highest, it is 1/maximum velocity. Finally, by inserting the velocity equation which I first mentioned, this gets us:

\frac{1}{\sqrt{mgH}}=\frac{1}{\sqrt{mgy}} \frac{dx}{\sqrt{dx^2+dy^2}}

Squaring both sides, eliminating m and g from both sides, and rearranging things gets us:

dx^2H=y(dx^2+dy^2)

All it takes is some more rearrangement to get us a very special differential equation:

dx^2(H-y)=dy^2y

(\frac{dy}{dx})^2=\frac{H-y}{y}

(\frac{dy}{dx})^2=\frac{H}{y}-1

You may be wondering why this differential equation is special. The reason is that the solution of this differential equation is two parametric equations that represents the cycloid. And since I said that the y direction goes positive in the downward direction, the cycloid is upside down, unlike the picture in the wikipedia link.

Let me show you that the cycloid is the solution to that differential equation. First of all, these two are the equations for the cycloid:

(1) x=r(t-sin(t))

(2) y=r(1-cos(t))

Where r=H/2. Also:

(3) dx=r(1-cos(t))dt

(4) dy=rsin(t)dt

Inserting these to the left hand side of the differential equation gets us:

(5) [\frac{sin(t)}{1-cos(t)}]^2

Next use the trigonometric identity sin^2+cos^2=1:

(6) [\frac{\sqrt{1-cos^2(t)}}{1-cos(t)}]^2

Now, according to the equation of cycloid for y (equation 2), you can rearrange it so that:

(7) \frac{y}{r}=1-cos(t)

(8) \sqrt{1-\frac{y}{r}}=cos(t)

Inserting (7) to the bottom of (6) and (8) to the cos at the top of (6) gets us:

(9) [\frac{\sqrt{1-(1-y/r)^2}}{y/r}]^2

(10) \frac{r^2}{y^2}-(\frac{r}{y}-1)^2

(11)  \frac{r^2}{y^2}-\frac{r^2}{y^2}+\frac{2r}{y}-1=\frac{2r}{y}-1

This is the exact same thing as the left hand side of the differential equation, which means the cycloid is the solution to the fastest path.

There are interesting things to notice about the solution. In the cycloid equations (1) and (2), by making t=0, x=0, which just leaves y=r. So the fastest way to make an object fall right beneath you is to just drop the object straight down. Also, if the location of where object lands is farther away in the x direction than the y direction, then the object will have to go down below the level in which it stops, and then go up again until it reaches the stopping point.

In the next post, you will see that there is a general equation that will allow us to find the path which minimizes quantities. This equation can be used to solve this problem and obtain Snell’s law.

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