we have things to say. hopefully they are good.

before you start with this article, realize its a math one. i don’t think its inaccessible to most people, but i have a different idea of what inaccessible is.

This idea was first presented to me by Dr. Lowell Boone in a first year physics course at the College of Wooster. It has stuck with me since then, because I think its pretty nifty.

Imagine you are standing at point A with two ice cream cones on a hot summer day. Across a long, barren stretch of pavement, your buddy is patiently waiting for his cone. You want to get to him as quickly as possible to avoid having ice cream running down your hands. The fastest route you can take is a straight line. We could talk about path integrals to prove this, but I don’t think we need to do that for such a commonly accepted statement.

To calculate the time it takes you to get from point A to point B, we just multiply the distance times your velocity

Now let’s tweak the situation just slightly. Imagine you are going to carry the ice cream cones to your friend again, except some of the trip will happen on cement and some on sand. You still want to get to your friend as quickly as possible, but the ideal path isn’t a straight line anymore. Since you can walk faster on the pavement than on the sand, you will have to take that into account.

The two different turfs will effect your velocity. Instead of specifying your velocity on each surface, we start with your maximum velocity (which we’ll call ‘v’), and multiply it by an index (which we’ll call ‘n_i). For example, you walk at a maximum speed of 5 meters per second (so v=5m/s). On the cement, your index is 1 (so n_cement=1). Thus on the cement you will walk 5m/s (5*1). However, the index for sand is 0.5 (n_sand=0.5), so your velocity on sand is 2.5m/s (5*0.5).

Now back to the lesson at hand. To solve for the fastest path, we are going to break our route into two parts. We will walk a straight line on the pavement from point A to a point X on the boarder, and then a straight line from point X to our friend at point B. Our total time will be the time it takes us to walk the first path plus the time it takes us to walk the second path.

We can find the total time based on the speed indices and distances, since the time of each path is the velocity on the path multiplied by the distance traveled.

We will label the distance between Points A & B and the boarder as h_a and h_b. Now, the distance that we walk in both the cement and the sand is dependent on where point X is and the length of the boarder between A and B.

Using the distance formula, we can write d_a in terms of h_a and x and d_b in terms of h_b and l-x. Since the length of the boarder and the distance between the points and the boarder are defined by the problem, we now have the total time as a function of where we choose Point X to be, t[x].

We can plot t[x] for specific scenarios. For example, if your max velocity is 1m/s, the velocity indices are 1 and 0.5, Points A and B are 5 meters from the boarder, and the boarder is 20 meters long, than our graph of the total time versus the position of Point X looks like this

As you can see, there is a clear minimum on this plot, where the total time is the lowest. To find where Point X should be for the fastest path, we would want to find the x value of the minimum on this graph. If you have taken calculus, then you know that this means taking the derivative of t[x]. (The derivative of t[x], called t’[x], tells us the slope of t[x] for all values of x). We will then set t’[x]=0, meaning the slope of t[x] is a horizontal line, which only happens at minimums and maximums.

This equation represents the relationship between x and the other constants (n_a, h_a, l, etc.) for the fastest path. We can manipulate it to find some interesting properties of this route.

If we revisit the trip from A to X, we see that we have a triangle with sides h_a and x and a hypotenuse of d_a. Further, there is an angle between the hypotenuse and the boarder and its complimentary angle, which we’ll label φ_a and θ_a.

If we look at the left side of equation 7, we see a ratio between one of the sides and the hypotenuse, which indicates a trig relationship. In this case, we can replace the ratio with cos[φ_a] or sin[θ_a]. The same argument can be made for the right side of equation 7, leaving us with

We now have a relationship between the two velocity indices and the angles you will be walking at to deliver the ice cream if you want to take the fastest route to your friend. In the real world, you probably don’t care enough to do the calculations to figure out the fastest route. You probably don’t know the velocity indices, and you probably can’t do ridiculous mental math, and you already have your ice cream. However, this ice cream situation is analogous to another one, one that relates to the physics of light.

Instead of a person walking over cement and then sand, imagine light passing through water and then air. The light will pass through the water slower than through air. The light will bend, and it will bend just as predicted in equation 8. This means that light takes the fastest path when passing through different mediums.

That is pretty amazing, and we have noticed it. Sometimes, the world is just awesome.