Graphing the Ecliptic:

On this map, dark areas are where it is night, light areas are where it is day.
The red curve is a weird once because the Earth is a sphere that's axis is tilted.
I found the equation for the curve, graphed it then added it to the map.

Discussion:

Ecliptic: If the red line were drawn on a globe, it would just be a great circle dividing the Earth exactly in half.  (Assuming that the Earth is a sphere, and the suns rays are parallel.)  Since the Earth's axis is tilted, the entire south pole and Antarctica is in sun during our winter as you can clearly see on the map.
The angle that the earth is tilted relative to the sun is called the Ecliptic, it is also (phi) in my formula.  This angle changes with the seasons.  At the time the date was November 5, the ecliptic then is about 15 degrees, or around 1.3 radians (almost any globe with tell you this).  t in my formula is time in days.  It just shifts the red graph over the map as Earth spins.  In 1 day, it will go 360 degrees or 2(pi) radians.

Map Projection: I found the equation for a map with no projection, that means that the (x, y) coordinates on the map correspond to (longitude, Latitude) coordinates on the globe.  I originally thought this was a Mercator projection but it is not.  (To get equations for other map projections, simple plug in my equation for theta into the given maps equation for y(theta).

Derivation: Please don't hesitate to ask me to forward you this (I have it as a word document), it is really a very short and easy derivation.

Finishing: Once I had the shape of the curve (with the current date's ecliptic value), I shrunk my graph until the poles (y=-pi/2,pi/2) and the ends of the Earth (x=-pi, pi) on  lined up with the right spots on the Earth on the map.  Now that I had the size right, I shifted everything over to the right until it lined up (I could have played around with different values of 't' until they lined up, but this was was much easier.  I could have also found the exact time for midday at the equator, and calculated from there, but I didn't think of it then)

Tracking Satellites:

This is a map of the path of a Satellite.  Each point represents the point on earth directly below the satellite at a given time.
(Side point: If its in a circular orbit i.e. constant speed, Why do the points get further spaced apart near the poles?  hover mouse over picture for answer.)
Satellites travel in *great circular orbits, the kinds of circles the equation deals with.  In other words, if the Earth were not spinning (relative to the satellite) then we would get the same boring graph as before.  Adding motion to the earth will make things much more complicated though.

It ends up that I needed to derive an entirely new equation in parametric form (using the same diagram (not pictured here), because everything needs to be in terms of time (in minutes).
x(t)=arctan(cos(Phi)*tan(a*t)) ± (1/4)*t
y(t)=arcsin(sin(Phi)*sin(a*t))

Notice: . . . ± (1/4)*t  this takes into account the earth spinning 360 degrees in 24 hours or 1440 minutes.  it is really reduced from ± (360/1440)*t. + if the Earth is spinning away from the satellite, - if its traveling with the satellite. I forgot about this for quite a while and it drove me absolutely crazy, I had the Earth spinning the wrong way!
a*t is how fast the satellite is going around earth, like its angular speed, this depends on how far up it is.
 

Finding Phi:
This particular picture was taken from The Champ Sattelite's Home Page. It is used to map Earth's gravitational and magnetic fields.  I researched the satellite on the wonderful heavens-above.com (Champ satellite detail) and found the value of Phi is 87.2 degrees.

Finding 'a':
I could cheat and look it up on the website and calculate that a=3.920632233, or I could be more creative:
The site is updated every 5 mins so I think that each Dot is is about 5 mins apart.  I counted the number of dots it took to cross 60 degrees latitude near the equator (because it looks like a pretty good linear approximation there) to get 'a' approximately (2/3) or 2 degrees in 3 minutes.  That is pretty far off, it could be because the dots were more that 5 mins apart, but that is not the only thing.  I think I know the other reason, if you you can figure it out I'd be happy to talk to you about it.

I then tried using the map to approximate how far the Earth had spun while the satellite went 360 degrees, its full orbit.  The earth spun about 25 degrees, and I know the Earths angular speed so the time taken was around 100 mins. The satellite's angular velocity was then near 360/100=3.6, not a bad approximation.

plugging in Phi=87.2, a=3.92 we get:
x(t)=arctan(cos(87.2)*tan(3.92063*t))+(1/4)*t
y(t)=arcsin(sin(87.2)*sin(3.92063*t))

Dot Mode:
I tried graphing it in Dot Mode, t-step at 5 min, the dots were spaced the same as this graph, very exciting. Unfortunately I was using GraphCalc and it won't let you increase the dot size so I didn't post it here.  NJIT makes us use so many different math programs I can't use any of them efficiently.

I simply added the red line like I usually do.  The "Previous" black line, flows right into the "Predicted" blue line in continuation of the map, just like it should.

Note: Click Picture or Here, to get an unbroken version in new window.

Astronomy:

This is a map of where that eclipse we had a while back was visible.  They yellow lines were there originally.  I added the Red line on top of a yellow line.  It is such a nice model that it covers they yellow line completely.

  Since the moon's orbit is parallel to the ecliptic, it's rising and setting follows the same pattern as the sun.  Now the eclipse is only visible when the moon is visible, just like it is day only when the sun is visible.  So its the same exact curve!  The Earth turns a bit between different stages of the eclipse so the moon rises and sets in different places during that interval making it no longer visible or newly visible.  That's why you have a bunch of curves for different times.