# Sunday Function

The vast majority of the functions we've talked about over all these Sundays have been ones that are expressible as a relationship between two numbers x and y. Sometimes the relationship is simple, sometimes it's fairly complicated. Mostly though, we just work according to the function as though it were a little machine. Take a number x, do to it what the function tells you to do, and the result is the number y. This is a function where the relationships between x and y are given in a very clear and explicit form.

But a function doesn't have to be given in that way. The only requirement for a function is that each x is associated with one and only one y. There's no requirement at all that the relationship between x and y be written in an explicit y = f(x) form. Maybe x and y are both written in terms of some other variable, and the relationship between x and y is thus only implicitly given. How about a sample?  To graph this implicit relationship, you would take the variable t to vary within some range and plot the (x,y) points you get out. For instance, let the scaling factor r = 1 for convenience, and let t = π We get x = π and y = 2. So plot the point (π,2) and continue repeating for each t within your range. Let's do this for all t between 0 and 2π. This is our Sunday Function, and it's a curve called a cycloid. If you were to attach a little glowing Christmas light to the outer edge of your car tire, this is the path the light would trace out over and over as you drove down the road.

And this is a function, by the way. Each x is associated with exactly one y. In general there's no guarantee this will be true for whatever weird implicit relationship we right down between x, y, and t.

While this "curve traced out by a point on the rim of a wheel" property is interesting, the cycloid also has some significantly more important properties in the world of physics. If you were to flip the cycloid over and focus on just one half of it, you're looking at the solution to an important problem in the history of the calculus of variations. Let's take that look before stating the problem: If you were to take a little marble and build a straight ramp such that the marble rolls down from one specific point to a lower one, you'd of course find that the marble takes a particular amount of time to roll down the ramp. You might want to speed things up. Maybe you could curve the ramp in such a way as to get the marble going faster - perhaps by making the first part of the ramp steeper. Any curve away from the straight line path will increase the distance the marble has to travel, but the extra slope at the top might more than compensate for that. Of all these infinite possibilities for the shape of the curved ramp, which is the fastest? It's the inverted cycloid. There's no shape that will get your marble down faster. (The end point won't always be the bottom of the cycloid, incidentally. Depending on how the start and end points are oriented with respect to each other, it may be (and usually is) the case that you'll end up having to use a greater or lesser portion of one complete cycloid.)

This general process of picking out a curve that minimizes a particular quantity (in this case time) turns out to be a surprisingly fundamental part of physics. From Fermat's principle in optics to the principle of least action in classical mechanics to Feynman's path integrals in quantum mechanics, you're never able to avoid these kinds of variational problems and very often their associated implicit functions. Nor indeed would you want to. These paths lead to some of the most elegant and astonishing descriptions of physics

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Notethat convincing oneself that if one assumes a minimum for the marble problem exists then it is easier to see that it needs to be the cycloid. Some of the difficulty is in establishing that there really is a minimum curve.

That should be "let t = pi We get x = pi...".

[Thanks! -Matt]

You should also note that a cycloid can be written in the form x=f(y) by solving for t and then substituting. It may not be too pretty, but it works.

By Nathan (not verified) on 20 Sep 2009 #permalink

Shouldn't it be y = r(1 - cost)?

By Fred (not verified) on 20 Sep 2009 #permalink

Geez, apparently I can't type. It should indeed be y = r(1 - cos(t)), I'll get it fixed. However that would be the correct way to write the inverted cycloid that appears next in the post.

One reason to optimize: the best answer may not be the obvious one. Given a constant volume of homogeneous isotropic mass, define a 3-D shape such that a point test mass resting on its surface experiences maximum gravitation. Hint: You can do almost 2.6% better than a solid sphere.

The cycloid also solves the isochrone problem ... the marble, when released from rest, will take the same time to reach the bottom no matter what the release point.

By kiwi (not verified) on 21 Sep 2009 #permalink