# Generating a sine wave with phase – answers

In a previous post I posed several mathematical problems... I'd like to go back and give some answers to them.

To reiterate, we take a sine wave period and wrap it around a cylinder, giving us this: The problems are to:

1. Prove that the curve is planar.
2. Prove that the curve is an ellipse.
3. Make a Keplerian observation.
Johannes Kepler was a mathematician/astronomer who made several observations relating to planetary orbits.

Let's define the curve. Since we're going to be making analogies to planetary orbit, let's define the curve
parametrically. The parametric point is the analog of a planet.

Peter piper picked a peck of pickled peppers. -- Mother Goose

Penny Pingleton, you know you are punished.
From now on you're wearing a giant P on your blouse EVERY DAY to school
so that the whole world knows that Penny Pingleton is permanently, positively, punished.

-- Prudence Pingleton, Hairspray

OK, now that we have that out of the way...
I want the parametric point to sweep its way around the unit cylinder with constant angular velocity,
as viewed from above. So if we project the path of the point into the unit cylinder with z = 0, then
r(t) = 1 for all t and θ(t) is linear in t. In fact let's have
θ(t) be t, so at time t = 2π
our parametric point has made a complete cycle.

Now we'll add the z component in. The curve is a wrapping of sin(t), so z = sin(t); we aren't affecting
the vertical component of the paper by wrapping it horizontally around a cylinder.

We're actually, in a sense, done. We have the formalized equations:

• r(t) = 1
• θ(t) = t
• z(t) = sin(t)

But let's convert to Cartesian
coordinates, since we'll be dealing with planes. The relevant formulae are:

• x2 + y2 = r2
• tan(θ) = y/x

A little intuition, checked by back-substitution, gives us these for x and y:

• x(t) = cos(t)
• y(t) = sin(t)

Putting these together with our formula for z, we have:

• x(t) = cos(t)
• y(t) = sin(t)
• z(t) = sin(t)

Note that, no matter what t is, y(t) = z(t)... that is, the parametric point travels entirely in
the plane given by the equation y = z. This proves that the parametric path is planar.
QED #1.

To tackle #2 we have to prove that this planar curve is, in fact, an ellipse. One way to do that would
be to come up with x' and y' coordinates in the plane of the curve, transform the parametric equations into
the new coordinate system, and verify that a(x')2 + b(y')2 = c2
for some a, b, and c... but when I came up
with the problem I had a much simpler proof in mind. There's also a third way of doing this, where you
note that the curve is the result of an affine transformation of the unit circle... and an affine transformation
of a circle is demonstratably an ellipse... but I hadn't thought of that.

Another astronomer/mathematician, Appollonius of Perga, made a study of "conic sections"... that is, the intersection of a plane with
a double-headed cone. He came up with an exhaustive list of the kinds of curves that resulted.

Basic conic sections:

For complicated reasons (to wit, that gravity diminishes as the square of the distance between two objects,)
all heavenly bodies trace either an ellipse or a hyperbola as they go around (or past) another heavenly body.

Parabolae are theoretically possible but would smack heavily of "design"; as yet none have been observed.

For completeness, there are other "designed" ways to intersect a double-headed cone with a plane in a coincidental fashion:

Degenerate conic sections:

Since our curve is the intersection of a plane and a cylinder, we're done! QED #2. Well, not quite.

Appolonius' results hold for cones. We have a cylinder. Those aren't quite the same thing. Right?

They aren't, quite... but a cylinder can be viewed as a degenerate form of cone. To wit, a cylinder is a cone whose
apex is infinitely far off.
I'll get back to a "proof" of this in a minute, but if you allow this, then we really are done.

Instead of the seven intersections of a double-headed cone and a plane, there are
only four ways a cylinder and a plane can intersect. Two of these are new.

Cylindrical conics:

Now for the proof that cylindrical intersection #1 really is an ellipse. Place two double-headed cones
coaxially with the cylinder. Arrange them so that:

• The apexes of both cones are in the "up" direction, from anywhere on the curve in question.
• The "top" cone intersects the highest point on the curve in question.
• The "bottom" cone intersects the lowest point on the curve in question.

Diagram: Notice that the entire curve lies in between the two cones. Consider the intersection of the plane that
determines the curve with each of the cones. We know both of these are ellipses; we know that the intersection
of the plane with the "upper" cone lies wholly outside our unknown curve; and we know that the intersection of
the plane with the "inner" cone lies wholly inside our unknown curve.

Diagram: The final stage of the proof lies in pushing the apexes of both cones to infinity, which "squeezes" our unknown curve
in between two known ellipses. Since the horizontal distance between the cones goes to zero as 23/2 / tan(θ), our
unknown curve cannot help but be elliptical-ish... to any degree of precision... and it is, thus, an ellipse. QED #2.

For the final problem, we must make a "Keplerian observation." The observation in question is that our parametric point,
like the planets, sweeps out equal areas in equal times. What makes this interesting is that the parametric point does not move in the same way
that a planet does... so it's a little odd that such a similar result should hold... but it does.

Let's talk a little about Kepler's second law. A planet moves around the sun in an elliptical orbit. Fine.
The sun lies at one of the foci of the ellipse. An imaginary line between the planet and the sun will sweep out area at a constant rate.
That is to say, when the planet is far from the sun, that line will be long... and it will move correspondingly slowly.
When the planet is near to the sun, that line will be short... and it will move correspondingly quickly. Conversely, our parametric point sweeps around the origin at constant angular velocity. So this is trivially
"equal areas in equal times". Right?

Not quite. It's true that our parametric point appears to have constant angular velocity, if you project
its path into the x-y plane
... that is, if you view it from directly above, from a point on the z-axis.

But that's a silly way to look at things. To get an idea of the point's actual motion, it's far more natural
to view it from an axis orthogonal to the plane of motion. So let's put an observer (Aranea) directly above the point, and
another observer (Nellie) in the more natural location. From Aranea's point of view, the motion is a simple constant-speed trip around a circle. But Nellie sees the parametric
point's true velocity. The parametric point's velocity has two components... a constant "clockwise" component in the
x-y plane, and a variable orthogonal component going "up" or "down". This is maximized when the curve crosses the x-y plane,
and minimized when the curve is topping or bottoming out. So Nellie sees this: It seems hard to believe, given that the parametric point is speeding up and slowing down according to different rules...
and the notion of "center" is different... but the fact is, the parametric point also sweeps out equal areas in equal times!

To see this most easily, look at Aranea's point of view again. Have Aranea draw a grid on her picture of the system.
She can count the area swept out in any given time by counting the number of squares.

Now consider what those squares look like to Nellie. They're rectangles. But they're all of equal area.
Aranea and Nellie can label the squares according to some naming scheme, and will agree that the same set of squares/rectangles were
swept out... but this means, since the rectangles are of equal area, that even in Nellie's point of view, equal areas are
swept out in equal times! QED #3.

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