Write- up # 11

for

Jenni McIntire and Teresa Davis

Problem 1 in assignment 11

Investigate for a = b, a < b and a > b.

We first look at integer values for k when

In all the case above, a = b (they are both 1). We notive that the integer valu of k gives us the number of curves (or leaves) of each graph. Since a = 1 and b = 1, the length of each curve is 2 (1 + 1).

In these graphs, a < b. We kept k at the same value 5, so each graph has 5 curves. But having different values of a and b gave us each double curves. One curve has length a + b and the other has b - a.

Finally, we look at the case where a > b. We notice that we still have the same number of leaves, but each leaf is not totally section off. The main length is still a + b, but indention is a - b.

Second, we look at values of k = 1/2 , 3/2 , 5/2 , ... when .

When a = b in these graphs, we get a curve with an inward loop. The value of k still gives us the number of leaves, but it is the numerator that tells us this.

The same thing also happens when a < b. We get a double figure, but the second leaf is between the first (coming from the demoninator).

Similarly, when a > b, we get curves that are not fully closed.

Finally we look at values of k = 1/3 , 2/3 , 5/3, ... when .

We see that this case gives us 3 loops ( the denominator of k) for each leaf ( the numerator of k)

If a < b, we have the leaves doubled.

If a > b, the curves ar not fully closed.

Return