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Final Project!!!!

By:

Alex Moore

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            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.

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       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)). 

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

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Not that when we divided both sides by x2 in the 5th 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):

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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.

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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.

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

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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.

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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. 

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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.

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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.

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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.

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

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Hence, the distance from B to D is

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

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By definition of cosine,

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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!

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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:

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

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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:

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Therefore, the answer to our problem is that the shaded region has area equal to Screen shot 2010-12-13 at 2.07.47 PM.png where s is side length of the given square!