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Up to similarity, these curves can all be expressed by a polar equation of the form
or, alternatively, as a pair of Cartesian parametric equations of the form
If k is an integer, the curve will be rose-shaped with
- 2k petals if k is even, and
- k petals if k is odd.
When k is even, the entire graph of the rose will be traced out exactly once when the value of theta, θ, changes from 0 to 2. When k is odd, this will happen on the interval between 0 and . (More generally, this will happen on any interval of length 2 for k even, and for k odd.)
If k is a half-integer (e.g. 1/2, 3/2, 5/2), the curve will be rose-shaped with 4k petals. Example: n=7, d=2, k= n/d = 3.5, as θ changes from 0 to 4.
If k can be expressed as n ± 1/6, where n is a nonzero integer, the curve will be rose-shaped with 12k petals.
If k can be expressed as n/3, where n is an integer not divisible by 3, the curve will be rose-shaped with n petals if n is odd and 2n petals if n is even.
If k is rational, then the curve is closed and has finite length. If k is irrational, then it is not closed and has infinite length. Furthermore, the graph of the rose in this case forms a dense set (i.e., it comes arbitrarily close to every point in the unit disk).
for all , the curves given by the polar equations
are identical except for a rotation of /2k radians.
A rose whose polar equation is of the form
where k is a positive integer, has area
if k is even, and
if k is odd.
The same applies to roses with polar equations of the form
since the graphs of these are just rigid rotations of the roses defined using the cosine.
How the parameter k affects shapes
In the form k = n, for integer n, the shape will appear similar to a flower. If n is odd, half of these will overlap, forming a flower with n petals. However, if n is even, the petals will not overlap, forming a flower with 2n petals.
When d is a prime number, then n/d is a least common form and the petals will stretch around to overlap other petals. The number of petals each one overlaps is equal to the how far through the sequence of primes this prime is + 1, i.e. 2 is 2, 3 is 3, 5 is 4, 7 is 5, etc.
In the form k = 1/d when d is even, it will appear as a series of d/2 loops that meet at 2 small loops at the center touching (0, 0) from the vertical and is symmetrical about the x-axis. If d is odd, then it will have d/2 loops that meet at a small loop at the center from either the left (when in the form d = 4n − 1) or the right (d = 4n + 1).
If d is not prime and n is not 1, then it will appear as a series of interlocking loops.
If k is an irrational number (e.g. , , etc.) then the curve will have infinitely many petals, and it will be dense in the unit disc.
Adding an offset parameter c, so the polar equation becomes
alters the shape as illustrated at right. In the case where the parameter k is an odd integer, the two overlapping halves of the curve separate as the offset changes from zero.
k <- 4 t <- seq(0, 4*pi, length.out=500) x <- cos(k*t)*cos(t) y <- cos(k*t)*sin(t) plot(x,y, type="l", col="blue")
MATLAB and OCTAVE
function rose(del_theta, k, amplitude) % inputs: % del_theta = del_theta is the discrete step size for discretizing the continuous range of angles from 0 to 2*pi % k = petal coefficient % if k is odd then k is the number of petals % if k is even then k is half the number of petals % amplitude = length of each petal % outputs: % a 2D plot from calling this function illustrates an example of trigonometry and 2D Cartesian plotting theta = 0:del_theta:2*pi; x = amplitude*cos(k*theta).*cos(theta); y = amplitude*cos(k*theta).*sin(theta); plot(x,y)
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