Sarah Hofmann

EMAT 6680 Summer 2006

Assignment 11

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 r = 2*a*sin(k*theta) for varrying a and k.  We begin by fixing a = 1 and varrying k through the non-zero integers -3..3from least to greatest.

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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.

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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 r = 2*a*sin(k*theta)  by fixing k = 5 and varrying a through the non-zero integers from -3..3, beginning with the smallest integer and working up.

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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.

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Success!  The petal has a radius of 44, as we suspected.

We must now investigate the equation r = 2*a*cos(k*theta) for various values of k and a.  We will begin as above by fixing a = 1and varrying k from -3..3

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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.

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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 r = 2*a*sin(k*theta)+b 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.

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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

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Indeed, it 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.

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Success!  Our hypothesis seems to hold.  

We are now to repeat our investigation with the equation r = 2*a*cos(k*theta)+b 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.

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Since our results are as we expected, we now check for values of a = 5, k = 3, and integers b = 8..12

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Our results are as we hypothesized.  You can click here for an animation showing a similar phenomenon as seen for the sine equation.

We conclude our exploration with an investigation of the equation r = c/(a*cos(k*theta)+b*sin(k*theta)) .  We will proceed by fixing three of the variables (a,b,c, and k) and varying the fourth. 

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.









We can immediately see that changing k = 4 to k = -4 and k = 5 to k = -5 flips the graph about the x-axis.  This is consistant with our knowledge of the fact that sin starts increasing and cosine starts decreasing when moving positively away from 0, however both sine and cosine are decreasing when moving negatively away from zero, so the picture would be shifted counter-clockwise by a positive k value and clockwise by a negative k value.

In evaluating the above pictures, it also seems that k has similar effects on this equation that it had on our previous equations.  It seems that when k is an odd integer, there is a k point star in the middle, but when k is an even integer there is a 2k point star in the middle.  Let's explore this phenomenon and see what happens when k is not an integer via the animation below.



We can observe a few different things when watching the animation.  First, our assumption about the number of points of the star as k varries is correct for non-zero and non-unit values of k.  Second, we can see that when k is not between -1 and 1, as k varries, the graph seems to spin and create new stars on the odd integers that double immediately after the odd integers.  Finally we notice that there is a fixed point of (1,0) for all of the graphs.  We also notice that the stars completely fall apart when the value of k is between -1 and 1.  Probably because the functions 1/(sin(x)) and 1/(cos(x)) vary quickly between -1 and 1. 

We continue our exploration by fixing b = c  = 1, k = 5, and varying a through the animation below.



We can see that a affects the size of the star and how the star is twisted.  As the value of a moves away from 0, the star gets smaller.  So the star is biggest when a = 0.  We also notice that as the value of a increases, the star turns clockwise. 

We continue with our exploration by holding a = c = 1, k = 5, and varying b.



We can see in this animation, that like a, varying b affects the size and the orientation of the star.  As b moves away from 0, the star gets smaller and smaller.  We also notice that when b = 0 the graph is oriented with a point on the positive x-axis, but as b moves bigger positively or negatively, the graph moves so a point is centered either on the positive y-axis or the negative y-axis respectively.  We note again that the graph moves in a clockwise direction as the value of b increases.

We conclude our exploration by fixing a = b = 1, k = 5 and varying c.



We can immediately see that varying c has different effects than a and b.  For instance, the difference between c and -c is a precise rotation about the origin of 180 degrees.  We also notice that when c = 0, the graph is empty.  Which makes sense, since 0/n = 0 for any n.  We also notice that the value of c mainly affects the size of the star.  As c moves away from 0, the star gets bigger and bigger, regardless of whether c is moving positively away from 0 or negatively away from 0.

This concludes our exploration.