EMAT 6680 Summer 2006
For assignment 11, we have chosen to write up problem number 2 because in our opinion it produced the most interesting and beautiful graphs.
To begin we need to investigate the equation for varrying a and k. We begin by fixing a = 1 and varrying k through the non-zero integers -3..3from least to greatest.
We can conclude that the difference between a positive and a negative value for k is a rotation about the x-axis. This is consistant with sin(t) being an odd function. It also appears that the value of k dictates how many "leaves" the function will have. If k is odd, the function will have k leaves. If k is even, the function will have 2k leaves. The leaves all seem to start at the origin, tangent to the x-axis in the first quadrant. Let's test our hypothesis with a few large values of k, say 5 and 16.
Indeed, the above graphs have 5 and 2(16) = 32 leaves and they are tangent to the x-axis as believed.
Let us continue our investigation of the equation by fixing k = 5 and varrying a through the non-zero integers from -3..3, beginning with the smallest integer and working up.
We can draw two conclusions from the above graphs. We first conclude that the difference between a and -a is a rotation about the x-axis. This is consistant with our knowledge of flipping regular functions about the x-axis by multiplying by a negative. Second, we must be areful to pay attention to the scale of the above graphs. It seems that the radius of the leaves is equal to |2a|. Lets check our hypothesis with a larger number for a, say a = 22.
Success! The petal has a radius of 44, as we suspected.
We must now investigate the equation for various values of k and a. We will begin as above by fixing a = 1and varrying k from -3..3
Here we had some expected and unexpected results. It seems that again, if k is even, the graph has 2k leaves and if k is odd the graph has k leaves. Since cosine is an even function, we expected no difference between the graphs for positive values of k and negative values of k, which did occur. However here, the graphs seem to be "centered" about the positive x-axis instead of tangent to it. This makes sense, since sin(0) = 0 and cos(0) = 1.
We now explore various non-zero values of a from -3..3 as we did with the sine version, we will fix h = 5 again.
The results for varrying a seem to be the same as the sine version of the equation. A negative value of a rotates the graph about the y-axis and the length of a leaf is |2a|.
Let us continue our investigations by adding another variable to the function. We will now consider the function for various values of a, k and b. We already know how a and k affect the graph. So lets fix a = 2, k = 3, and vary through the integers 1..5.
That was definately unexpected! Something interesting seems to be happening around b = 4. Lets examine it further by looking at values close to 4, say b = 3.25, b=3.5, b=3.75, b=4, b=4.25, b=4.5, and b=4.75
seems that the "double" graph gets smaller and smaller as b approaches
4 from the left until the graph is normal at 4, then the graph begins
breaking apart at the origin as the value of b moves away from 4 to the
right. We can see this and the fact that the double graph appears
at b = 0 from the animation below.
The next question to answer is, why 4? Why not some other number? We hypothesis that this is because 4 = 2*2=2a. Let's check our hypothesis with a different value of a, say a = 5, k = 3, and b = 8..12.
Success! Our hypothesis seems to hold.
We are now to repeat our investigation with the equation and we expect our results to be exactly the same. We will verify this by checking the equation for a = 2, k = 3, and b = 3.25..4.75 ranging over the quarter values.
Since our results are as we expected, we now check for values of a = 5, k = 3, and integers b = 8..12
Our results are as we hypothesized. You can click here for an animation showing a similar phenomenon as seen for the sine equation.
our exploration with an investigation of the equation
. We will
proceed by fixing three of the variables (a,b,c, and k) and varying the
We will begin by fixing, a = b = c = 1, and exploring
k. Learning from our previous experience, we will begin by
looking at k = 4, -4, 5, and -5 to see if we can recognize any vertical
or horizontal flips by changing k from positive to negative.