Sarah Hofmann

EMAT 6680 Summer 2006

Assignment 10 Problem 8

For problem 8 we are go evaluate the parametric equation (a cos(t),b sin(t)) from 0 < t < 2pi for different values of a and b.  We will begin with integer values where a = b = -3..2.

We can see that there are only three circles showing.  Why?  Well, when a = b = 0, we can see that (a cos(t),b sin(t)) = (0 cos(t), 0 sin(t)) = (0,0).  So there is no graph to show.  Using the odd and even properties of sine and cosine we can see that (-a cos(t), -b sin(t)) = (a(-cos(t)), b(-sin(t))) = (a cos(t), b sin(-t)) = (a cos(t), -b sin(t)).  But since an entire rotation of sin(t) or sin(-t) is traversed by t = 0..2Pi, we can see that (-a cos(t), -b sin(t)) = (a cos(t), b sin(-t)) trace the same curve, one clockwise, the other counter clockwise.  Thus we only need to consider positive integers in our current problem.  So we now consider (a cos(t), b sin(t)), for integers from a = b = 0..5.  They are represended by the following colors 0=red, 1=green, 2=yellow, 3=blue, 4=purple.

Again, the red circle or the case where a = b = 0 is not showing.  Further we can see each of these graphs is a circle of radius a = b.

We will now consider what happens when a < b. We will do this by first considering a = 0  and integers b = 1..5 again the colors will appear in the order red, green, yellow, blue, purple for the numbers 1 through 5.

As we may have anticipated, when a = 0 we do not see a graph, because the graphs are all points of the form (0, t).  Let us now consider the cases where a = 1 and integers b = 1..5.  Again the colors will be in order red, green, yellow, blue, and purple.

Here we can quite easily see that the number b represents where the graph will cross the vertical axis.  This may be because they are the points (cos(Pi/2),b sin(Pi/2)) = (0,b) and (cos(3Pi/3),b sin(3Pi/2)) = (0, -b) which doesn't depend on a.  All of the graphs cross the x axis at 1 and -1 because (cos(0),b sin(0)) = (1,0) and (cos(Pi), b sin(Pi)) = (-1, 0) does not depend on b.

Lets try the other direction.  Lets hold b = 1 and let integers a = 1..5.  Then we can see what happens in the other direction.

As we hoped, the value of a determines where the graph will cross the x-axis, for reasons similar to described above.  

This means that if we graph the equation with a = 2 and b = 5, the graph will be an ellipse with major axis on the y-axis of length 10 and minor axis on the x-axis of length 2.  Let's try it.

It worked!

We must now describe what changes for small numbers -3 < h < 3, when we change the equations to (a cos(t) + h sin(t), b sin (t) + h cos(t)).  Lets start simple with a = b = 1 and let h be all quarter values from -0.75 to 0.5.

It seems the negative values for h stretch the graph out on the line y = -x and the positive stretch the graph out on the line y = x.  Lets test this hypothesis by looking at all the halves between -2.5 and 0 and then all the graphs between 0 and 2.5.

In the first graph above the blue line represents when a = b = h = -1 and in the second graph the yellow line represents when a = b = h = 1.  That these should look linear makes sense because they are of the form (cos(t) - sin(t), sin(t) - cos (t)) = (x, -x) and (cos(t) + sin(t), sin(t) + cos(t)) = (x,x) respectively.  This confirms our thoughts about the values of h stretching about the line y = x and y = -x for positive and negative values of h respectively.

Finally we need to investigate the above for different values of a and b,  our hypothesis is that for any positive value of h, the equation (a cos(t) + h sin(t), b cos(t) + h sin(t)) will stretch around the line y = (b/a)x and negative values of h will stretch around the line y = -(b/a)x.  Lets explore this idea by fixing h and choosing different combinations of a and b.  We will graph h = 1.5 and the following pairs of (a,b) (1,1), (3,3) (1,2) (2,4) and (5,3), (2.5,1.5).  Then repeat the same pairs of (a,b) for h = -1.5.  Again the graphs will appear in the order, red, green, yellow, blue, purple, and teal.

Well that didn't quite come out how we had anticipated it would.  We can however say, for sure that the values of a and b differentiated the circles with respect to their slopes, however only clearly around a line in the case where a = b.