Three graphs have been constructed above. The red cardioid has the equation
r=1+cos(t). The green cardioid has the equation r=2+2cos(t). The blue cardioid
has the equation r=5+5cos(t). Notice that the graph is symmetric with the
Now, keeping the value of k equal to 1, let's see what happens when a<b:
Red Limacon: r=1-2cos(t)
Cyan Limacon: r=8+7cos(t)
Here, the green "rose" has the equation r=5cos(9t). Notice that there are 9 "leaves" of the green rose and that when b=5, the graph is located between values of x=-5 and x=5. The red rose has the equation r=2cos(9t). Again, there are k=9 leaves of the rose, but this time, the graph lies between x=-2 and x=2.
Red : r=5cos(5t/2)
Keeping the same values for a, b, and k, let's view the sine graph:
A variation of the k value changes the number of leaves on the rose when a and b are equal and k is an integer. A change in the values of a and b results in a change of the sizes of the leaves of the rose, how far the rose spans along the x and y axes (this is further illustrated below), and the location of the leaves on the rose with respect to the origin. When k is not an integer, let k=x/y. In these cases, as seen above, there are y sets of roses.
Green rose: r=2cos(4t)
Red rose: r=3+2cos(4t)
Blue rose: r=5+2cos(4t)
Whereas the green rose spanned to x and y values of -2 and 2, the red rose spans from x,y =-5 to 5 and the blue rose spans from x,y=-7 to 7. Notice that the value of a stretches the graph a units from the original equation.
Keeping the same values of a, b, and k for the sine graph shifts the green, red, and blue roses so that they are no longer symmetric to the y axis, but symmetric to the origin: