| This is the 1st part of my write-up of Assignment #11 |
Brian R. Lawler
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| EMAT 6680 |
12/14/00
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The Problem
| Investigate varying a, b, c, and k. |
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The graph of 
with a = 1 and k = 1. Notice this defines a radius 1 circle
centered at (1,pi/2). |
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| The graph of
with a = 1 and k = 2. This is a 4 petal flower centered at
the origin with petals that appear to be 2 in length along their line of
symmetry. |
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| The graph of
with a = 1 and k = +/- 3. This is a 3 petal flower centered
at the origin with petals that appear to be 2 in length along their line
of symmetry. Notice the negative inverts the original flower. |
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| Continuing to change the values for k will just continue to add petals to the flowers observed. When k is an even number, there are always 2k as many petals. The first petal counterclockwise from the positive x-axis is at (2pi / 4k) radians and the next petals are (2pi / 2k) radians counterclockwise from there. When k is an odd number, there are k petals. One petal will be on the negative y axis, with the rest (2pi / k) radians counterclockwise from there. A negative value for k only has an observable effect on odd values of k, as seen in the final example above. This flips the image of the graph with positive k over the horizontal axis. | ||
Next to look at the graphs of the form  .
First consider when a = 1 and k = 1. Notice this defines a
radius 1 circle centered at (1, 0). As you may expect, there is a clockwise
rotations of pi/2 compared with the same function that replaces the cosine
with sine (see the green circle above). |
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| The graph of
with a = 1 and k = 2. This is a 4 petal flower centered at
the origin with petals that appear to be 2 in length along their line of
symmetry. Again, there is a clockwise rotations of pi/2 compared with the
same function that replaces the cosine with sine (see the yellow flower
above). |
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| The graph of
with a = 1 and k = +/- 3. This is a 3 petal flower centered
at the origin with petals that appear to be 2 in length along their line
of symmetry. Again, there is a clockwise rotations of pi/2 compared with
the same function that replaces the cosine with sine. However, notice that
both graphs do not appear. In fact, both are present and the same inversion
about the x-axis occurs as seen in the sine function above. However,
because of the pi/2 rotation, this inversion maps the flower right back
to itself. |
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| Continuing to change the values for k will just continue to add petals to the flowers. This behavior observed in the cosine functions summarily acts the same as the sine examples above. When k is an even number, there are always 2k as many petals. The first petal counterclockwise from the positive x-axis is at 0 radians of counterclockwise rotation and the next petals are (2pi / 2k) radians counterclockwise from there. When k is an odd number, there are k petals. One petal will be on the positive x axis, with the rest (2pi / k) radians counterclockwise from there. A negative value for k only has no observable effect on values of k, as seen in the final example above. | ||
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The a value adjusts that radius of these graphs. In fact, as written, each petal has a length of 2a. Notice that the circles that are generated when k = 1 have also have a diameter of 2a. At right is an animation of varying values for a. Here you can observe the effects of changing a. Click on the equation below to investigate further.
(an aside: I was curious to make sure my independent investigation of the two variables was complete. In other words, I wanted to make sure that I didn't miss something that might happen as both variables moved together. To check this, I ran a few more examples. I did not find anything behaving differently than described above.) |
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click to continue to Part II.
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Comments? Questions? e-mail me at blawler@coe.uga.edu |
| Last revised: December 28, 2000 |
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