Screen shot 2010-12-13 at 8.49.05 AM.png

Final Project!!!!

By:

Alex Moore

Its been a long and difficult road, but we are here. The Final project! For the first part of our project we journey on one more exploration. Have you ever considered the function x³+y³ = a, for some real number a? Of course we have. What’s the fun exploring this little guppy? Lets investigate a function that is much more interesting!

Consider the function x³+y³ = 3xy. To get a feel for the function let us graph it in the x,y-plane.

Screen shot 2010-12-13 at 8.59.03 AM.png

Not the nicest looking function, but there is one great quality it has: this graph contains the origin! Why is this good? This opens up a way for us to parametrize this curve. Recall that parametrized curves are curves whose points are defined by (x(t), y(t)) where x(t) and y(t) are continuous functions. In order to parametrize any curve we need some variable to measure. In this case, we can let our parameter t be the slope of the line passing through the origin, y = tx. For each value of the line y = tx intersects the line at exactly 1 point (excluding the origin), call point (x(t), y(t)).

Screen shot 2010-12-13 at 9.09.17 AM.png

The green point represents (x(n), y(n)) for t = n. Since we are using y = tx, we substitute and solve for x(t):

Screen shot 2010-12-13 at 9.16.49 AM.png

Not that when we divided both sides by x² in the 5^th line we can do so with no fear since x=0 implies t=0, so we have the x=0 case covered and so for the purposes for above we have x is nonzero. Now that we have x(t) we substitute into our original equation to solve for y(t):

Screen shot 2010-12-13 at 9.25.53 AM.png

A minor detail here (that really isn’t so minor) is that our parametric equation has a singularity. Since (-1)³ = -1, our denominator is equal to 0 at t = -1. Since t corresponds to slope, let us graph y = (-1)x.

Screen shot 2010-12-13 at 9.30.00 AM.png

This portion of the graph hints at why we have problems at t = -1. It appears that the line y = -x intersects the graph only at the origin. What will happen when t= -1 when we graph our parametric equations? Let us see. The graph on the left has t ranging from -0.9 to 20 and the graph on the right has t ranging from -15 to -1.1.

Screen shot 2010-12-13 at 9.41.22 AM.png

As we can see we get the whole graph between the two values of t. Let us now attempt to include t=-1.

Screen shot 2010-12-13 at 9.47.15 AM.png

Where did this extra red line come from?!? This is the program’s attempt to add in the asymptote to the “tails” of the function. Why did it do that? Let us remind ourselves that we understand the concept of limits of a function, and lets compute the limits of x(t) and y(t) and t approaches -1.

Screen shot 2010-12-13 at 9.55.21 AM.png

Clearly this limit does not exist. As we approach -1 from the right, (x, y) limit to (-inf, +inf) and as we approach -1 from the left we approach (+inf, -inf). This red line in our graph is not only an attempt to show as an asymptote but it is also the program trying to connect the limits at infinity. As we know from assignment 10, parametrizations are not unique. Does there exists a parametrization for this function without a singularity anywhere? The fact that the graph is self-intersecting certainly lowers our hopes but perhaps there is!

Let us move on to the next part of the final project! We wish to find the area of the shaded region below.

Screen shot 2010-12-13 at 11.58.22 AM.png

This region is a square with a quarter circle arc from each corner producing a region bordered by four smaller arcs. First we fiddle around a little to help set up a strategy. It appears as though this region is the union of a square and four partial arc sectors, as depicted below.

Screen shot 2010-12-13 at 12.05.31 PM.png

Clearly the 4 smaller regions are portions of sectors and the area of those pieces is easily calculated. We focus for the time being on the larger portion of the region. Label the 4 points of intersection A, B, C, D.

Screen shot 2010-12-13 at 12.46.21 PM.png

Now we use a little bit of our trigonometry knowledge. We know the point B has horizontal distance (½)s and the distance from the bottom left vertex to B is s by construction. Therefore, by the Pythagorean Theorem we can calculate the vertical distance from B to the bottom side.

Screen shot 2010-12-13 at 12.57.47 PM.png

Therefore, the distance from B to the top side is the same as the distance from D to the bottom side is

Screen shot 2010-12-13 at 1.04.06 PM.png

Hence, the distance from B to D is

Screen shot 2010-12-13 at 1.04.17 PM.png

Consider the new picture, EFGH the large square with midpoints I, J, K, L.

Screen shot 2010-12-13 at 1.22.55 PM.png

By definition of cosine,

Screen shot 2010-12-13 at 1.27.55 PM.png

and by the symmetry of the problem we conclude angle LEC has measure 30 degrees. That means that B and C subtend the arc from F to H into three equal parts and therefore angle LEF is trisected. Hence, the perpendicular bisector of BC is the angle bisector of angle LED! By symmetry we also know that perpendicular bisector of AB is the angle bisector of angle LHG!

Screen shot 2010-12-13 at 1.35.53 PM.png

Therefore, since we have right angles at angle BNP and BMP, we conclude that BNPM is a square! By the symmetry of the problem can conclude that the area of BMPN is one-fourth the area of ABCD. Since d(P,B) = ½ d(B,D), we have:

Screen shot 2010-12-13 at 1.46.14 PM.png

Now we find the area of each of the sector portions.

Screen shot 2010-12-13 at 1.50.39 PM.png

Since angle CEB has measure 30 degrees, which is pi/6 radians, the area of the sector CEB is (½)s²(pi/6) = s²(pi/12). The area of triangle EBC is (½)(s)(s*sin(pi/6))=(1/4)s² and so the area of the four partial sectors and the total shaded area is:

Screen shot 2010-12-13 at 2.05.44 PM.png

Screen shot 2010-12-13 at 2.04.31 PM.png

Therefore, the answer to our problem is that the shaded region has area equal to where s is side length of the given square!