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.