Consider the graphs of these equations, as k, a, and b vary:
r = (2a) sin (kq)
r = (2a) sin (kq) + b
As the title of this writeup implies, roses abound. But interesting variations occur for
Varying values of k.
Of course, when k = 0, r = (2a) sin (kq) = 0 (a degenerate circle), and r = (2a) sin (kq) + b = b, a circle with radius b. Below are the images of the two graphs for k = 1, 2, 3, and 4, with a = 1 and b = 1.
If you have done work with polar graphs before, you know that r = 2asin (kq) produces roses with petals of length 2a (length 2 in this case, since a = 1.) Furthermore, the rose will have k petals if k is odd and 2k petals if k is even. That is, focus on solely the purple graphs above: when k = 1, the graph is a circlea rose with 1 petal. When k = 2, the graph has 4 petals; it has 3 petals when k = 3, and 8 petals when k = 4.
Now focus on the blue graphs of r = 2sin (kq) + 1. For k = 1, the graph is a limacon with a loop, since b < 2a. For k > 1, the addition of b < 2a makes the petal length of these roses change, so that for k even or odd, the number of petals is 2k.
Next consider b = 2a (in this case b = 2; a = 1) as k varies. The graph of r = 2sin (kq) + 2 appears to always have k petals, for k odd or even. Below are the images of k = 1, 2, 3, and 4 with a = 1, b = 2. Notice that for k = 1, the graph of r = 2sin (kq) + 2 is a cardioid, and the petal length appears to be twice as long as the petal length of the roses produced by r = (2a) sin (kq). This relationship makes sense since the maximum value of r will be 2 + 2 = 4 (when kq equals multiples of p/ 2.) This case will be discussed further in the Varying values of b section.
Values of b > 2a will be discussed below in the Varying values of b section.
Download this Graphing Calculator File to see an animation of the graphs as k varies. Note:
Consider the graphs of r = (2a) sin (kq) and r = (2a) cos (kq) for k = 1, 2, 3, and 4, with a = 1 and b = 0.
Clearly, they are rotations of each other, and roses produced with cosine appear to always have a petal along the positive xaxis (or perhaps more correctly, positive polar axis.) The table below summarizes the patterns in the counterclockwise (ccw) angle of rotation in order to transform r = (2a) cos (kq) into r = (2a) sin (kq).

(in degrees, then radians) 


90 = 360/1 ÷ 4 OR p/ 2 = (2p/1)(1/4) 


45 = 360/4 ÷ 2 OR p/ 4 = (2p/4)(1/2) 


30 = 360/3 ÷ 4 OR p/ 6 = (2p/3)(1/4) 


22.5 = 360/8 ÷ 2 OR p/ 8 = (2p/8)(1/2) 


18 = 360/5 ÷ 4 OR p/10 = (2p/5)(1/4) 


15 = 360/12 ÷ 2 OR p/ 12 = (2p/12)(1/2) 


12 6/7 = 360/7 ÷ 4 OR p/14 = (2p/7)(1/4) 


11.25 = 360/16 ÷ 2 OR p/16 = (2p/16)(1/2) 


10 = 360/9 ÷ 4 OR p/18 = (2p/9)(1/4) 


9 = 360/20 ÷ 2 OR p/20 = (2p/20)(1/2) 

It appears that the angle of rotation can be determined by taking 2p (or 360), dividing by the number of petals, and then multiplying by 1/4 if k is odd, by 1/2 if k is even. (Alternately you could halve the number of petals when computing for k even and then you would multiply by 1/4, just as for k odd.) Certainly an interesting pattern! The same rotation holds when b is nonzero.
As stated previously, varying a causes the petal length to change, since when kq equals multiples of p/ 2, r = (2a) sin (kq) will be a maximum. Thus, greater absolute values of a will produce greater maximum distances from the pole and so greater petal length.
Below are graphs of both r and r for a = 3, 1, and 2, with k = 3, b = 1.
Notice that for the graphs of r = (2a) sin (3q), the petal length is 2*3 = 6, 2*1 = 2, and 2*2 = 4. (Even when a is negative, the petal length is positive since it is a distance.) Let's call 2a the "standard" petal length, when b = 0.
For the graphs of r = (2a) sin (3q) + 1, the longest petals are 1 more than the standard petal length. That is, they have length 2*a + 1 (i.e., 7, 3, and 5, respectively.) The shorter petals have lengths 2*a  1 (i.e., 5, 1, and 3, respectively) because of they way the equation operates when sine values reach their minimum (when 3q = 3p/ 2, i.e., q = p/ 2). For example, r = (2*3) sin (3p/ 2) + 1 = 6 + 1 = 5. However, when graphed, the point (5, p/ 2) appears identical to (5, 3p/ 2). Thus the shorter petal lengths are 1 less than the standard petal length.
Notice also that for negative values of a, the locations of the petals change, which also makes sense: when b < 2a, negative values of a produce negative values of r. Negative values of r locate the coordinates of the graph at angles that are p + q, in opposite quadrants. Thus the graph looks like it been reflected over the xaxis (or pole), even though technically, 180 degree rotation has occurred.
Now look at the graph for a = 0.3.
Clearly something unusual is going on here because this rose is degenerate! Notice that for this graph, b = 1 is greater than 2a = 0.6. This observation will be explored further in the last section of this writeup.
Download this Graphing Calculator File to see an animation of the graphs as a varies. Note:
Obviously, varying b only affects r = (2a) sin (kq) + b! However, I will provide the graph of r = (2a) sin (kq) in the images below for reference and contrast. So, observe the graphs of r and r for b = 4, 2, and 1 with a = 1 and k = 2.
As hinted at the end of the last section, for b > 2a, the rose is degenerate and the petals are "one big petal," so to speak (or a peanutlike shape.) For b = 2a, the graph appears to be on the cusp of degeneracy. And, as seen previously, for b < 2a, the rose has petals of two different lengths.
One way to examine this phenomenon is to consider the origin, or pole. The graphs of r = (2a) sin (kq) + b for b > 2a do not pass through the pole because for r to be zero, (2a) sin (kq) must be equal and opposite to b for some q. But (2a) sin (kq) will equal 2a at most (when kq = p/ 2 or q = p/ 2k.) Thus if b > 2a, r = (2a) sin (kq) + b can never equal zero.
Furthermore, if b = 2a, then the graph will intersect the pole precisely when q = p/ 2k, or in the case of the graph above, when q = p/ 4.
If b > 2a, then the graph will intersect the pole when kq = arcsin(b/2a.) In the case of the graph above, pole intersections occur when 2q = arcsin(1/2), or = p/12. Of course, the different lengths of petals produced are discussed above in the Varying values of a section.
Finally, notice also that the graph alters its orientation for b negative and b positive. Again, this rotation has to do with the negative values of r produced (and the subsequent graphing of points at angles of p + q.) For example, when b = 4, r values for q in the first quadrant will be negative. Thus, those values will appear on the graph in the third quadrant.
Furthermore, the values of r for positive values of sine in the first quadrant will be reduced by negative b values, wheras the values of r for positive values of sine in the first quadrant will be increased by positive b values. Thus the graph for negative values of b will have maximum r values in the quadrants where kq produces negative sine values (the second and fourth quadrants in the case of the above graphs since k = 2.) On the other hand, the maximum values of r for positive values of b will occur in the quadrants where kq produces positive sine values.
Download this Graphing Calculator File to see an animation of the graphs as b varies. Note: