Exploration of Polar Coordinates
By Thuy Nguyen
In this exploration we would like to explore the following five parametric equations:
I. r = 2asin(kq)
II. r = 2acos(kq)
III. r = 2acos(kq) + b
IV. r = 2asin(kq) + b
V. r = c / [acos(kq) + bsin(kq)]
Our goal is to generalize the graphs that these equations will produce with different values of a, b, c, and k.
Note: Although all polar coordinates are of the form (r, q), we will use the x-axis and y-axis in describing our graphs.
I. r =
2asin(kq)
Fixing k = 1
red: a = 1
purple: a = 2
blue: a = -1/2
Now let k be an integer. We then have the following pattern:
- If
k is odd, there are k petals
- If
k is even, there are 2k petals
The following are graphs for k = 5 and k = 10, respectively,
with the same a values as above.
II. r =
2acos(kq)
Again, fixing k = 1
red: a = 1
purple: a = 2
blue: a = -1/2
For integers k, we have the same pattern as in the case
above, but the petals are in different positions.
III. r =
2acos(kq)
+ b
For the following graph, fix k = b = 1 and vary the a values.
Red: a = 0
Purple: a = 0.3
Blue: a = 0.5
Green: a = 0.9
Now vary a = 1, 2, 3, 4 in red, purple, blue, and green
respectively.
We see that as a increases, the function gets larger in size and the inside loop gets closer to the outside loop. For negative a values, function is reflected across the y-axis.
IV. r =
2asin(kq)
+ b
We begin by varying the values of a just as we did
previously and set b=k=1:
We see in the graphs above that we obtain the same graphs as the ones produced by varying a in part II, but this time the graphs are rotated 90 degrees counterclockwise. Now fix k = a = 1 and vary b = 1, 2, 3, 4: (negative b values are the same as positive b values)
We already know what the graph looks like when a=b=k=1 (in
red above), so weÕll start with k = 2:
Now letÕs look to see what happens when k = 6:
We see that there are 6 large and 6 small petals, a total of 12 petals. LetÕs now try an odd k value, say k = 11:
We see here that there are 11 large and 11 small petals, a
total 22 petals, but notice that this time the small petals are inside the
large petals.
Conclusion:
For any nonzero positive integer k, then:
For k = 0, we get a circle of radius b, and for k non-positive, reflect the graph over the x-axis.
V. r = c /
[acos(kq)
+ bsin(kq)]
Let us first fix b = k = 1 and vary the values of a and c:
In the following graph,
Red line: a = 1, c = 1
Purple line: a = 2, c = 1
Blue line: a = 3, c = 5
Green line: a = 3, c = -5
We see that the line always intersects the y-axis at y = c and it intersects the x-axis at c/a. Now we will use the same values of a and c as above, but weÕll now vary b as well.
In the following graph,
Red line: a = 1, c = 1, b = -1
Purple line: a = 2, c = 1, b = 1/2
Blue line: a = 3, c = 5, b = 2
Green line: a = 3, c = -5, b = 3
We see that the lines still intersect the x-axis at c/b, but now they intersect the y-axis at c/b.
The only value we have left to vary is k. LetÕs first examine what happens when k is an odd integer. WeÕll let a, b, and c be the same values as the function in red, so a = 1, c = 1, b = -1, and let k = 3:
And now for k = 4, an even integer:
Conclusion:
In the parametric equation r = c / (acos(kq) +
bsin(kq))
we have that for any nonzero integer k,
if k is even, then the parametric equation yields 2k non-intersecting curves,
and exactly one of these curves intersect the x-axis at x = c/a, exactly one
(different from the first) intersects the x-axis at x = -c/a, exactly one
intersects the y-axis at y = c/a, and exactly one intersects the y axis at y =
-c/a, where a and c any nonzero real numbers. If k is odd, the equation yields k non-intersecting curves
(for k = 1, we have a line), and exactly one of these curves will intersect the
x-axis at either positive or negative c/b, and exactly one curve (different
from the first) will intersect the y-axis at either positive or negative c/b,
where b and c are nonzero real numbers.
If one curve intersects an axis at a positive c/b, then the second curve
must intersect the other axis at negative c/b, and vice versa.