In Write Up Eleven I have been exploring Polar Equations r = a + b cos( kt). To reiterate the conjectures found I found the following:
1. When a = b, the radius is a + b, and k = the number of petals.
2. When a < b, the radius remains a + b, however 2k = the number of petals. Furthermore, when k is odd there are 2k petals with k large petals and k small petals. The small petals are looped inside the larger ones. This looping inside seems to be related to negative values.
3. When a > b, the radius remains a + b. There are k petals but no looping inside or crossing over other petals occurs. The further apart a and b are the more stretched the petals appear to be. With the vertical shift "a" being greater than the amplitude "b" when both are positive there would be no negative values.
To extend my study I looked at what the results would be if a were positive and b negative. Of course this would still allow "a" to be greater than "b". The following graph will demonstrate what I think is a generalization. Note I used 2 + 1 cos (5 ø) and compared 2 - 1 cos(5ø) to find the change that occurs when b is negative. It appears to still have the same open petal design when b < o with no looping over or inside. However when I measured the two graphs to compare you will note there is a rotation of of pi/ k radians in the counterclockwise direction. See the graph below.
Note the value of ø is approximately pi / 5 where the black line is placed. This is to demonstrate the rotation of the red graph to form the blue graph. Find below several other graphs to support my conjecture that is b < 0 the graph if a rotation by pi / k units CCW.
Note in the next image there is a rotation that occurs, but something else is different.
Note in this graph I allow both a and b to be negative. Also important to note is that k is odd. I will show an additional graph to support my conjecture that if a and b are negative there will be 2k petals with k petals looped inside of k larger petals. The radius will be the absolute value of a + b and if k is even there will be 2k petals but no looping inside just as we saw with the other case of a < b. See graphs below to make your own conjectures.
Additional explorations are suggested so that students might make their own conjectures. To recap I found that allowing b to be negative when a > b produces a rotation by pi/k units CCW with k petals or loops and no looping inside. When a> b and both a and b are negative there is still a rotation by pi/k units CCW, but the value of k affects the petals. If k is even there are 2k petals with no looping inside but alternating small and large petal occur. When k is odd there is also 2k petals, however the smaller petals are looping inside the larger ones. The one constant that I have found is that the radius remains the absolute value of a + b.
I also explored some changes in the k value of the equation r = a + b cos (kø). The following will demonstrate my findings. I first compared 2 + 3 cos(ø) with 2 + 3 cos (.5ø) and found the following:
Note the blue curve seems to be a partial curve of the first. In the subsequent graphs I tried to detect a pattern. See below.
Note the graphs have 2.5, 3.5 and 4.5 petals respectively with a< b. The graphs appear to be a partial curve. This would be consistent with a Cartesian graph in which k controls the period or the amount of repetitions in the curve that one would see. Note again the radius is a + b and the number of petals correspond to k.