Roses specified by the sinusoidr = cos(kθ) for various rational numbered values of the angular frequencyk =n/d. Roses specified byr = sin(kθ) are rotations of these roses by one-quarter period of the sinusoid in a counter-clockwise direction about the pole (origin). For proper mathematical analysis,k must be expressed in irreducible form.
A rose is the set of points in polar coordinates specified by thepolar equation[2]
or in Cartesian coordinates using the parametric equations
Roses can also be specified using the sine function.[3] Since
.
Thus, the rose specified byr =a sin(kθ) is identical to that specified byr =a cos(kθ) rotated counter-clockwise byπ/2k radians, which is one-quarter the period of either sinusoid.
Since they are specified using the cosine or sine function, roses are usually expressed aspolar coordinate (rather thanCartesian coordinate) graphs of sinusoids that haveangular frequency ofk and anamplitude ofa that determine the radial coordinater given the polar angleθ (though whenk is arational number, a rose curve can be expressed in Cartesian coordinates since those can be specified asalgebraic curves[4]).
Graphs of roses are composed ofpetals. A petal is the shape formed by the graph of a half-cycle of the sinusoid that specifies the rose. (A cycle is a portion of a sinusoid that is one periodT =2π/k long and consists of a positive half-cycle, the continuous set of points wherer ≥ 0 and isT/2 =π/k long, and a negative half-cycle is the other half wherer ≤ 0.)
The shape of each petal is same because the graphs of half-cycles have the same shape. The shape is given by the positive half-cycle with crest at(a,0) specified byr =a cos(kθ) (that is bounded by the angle interval−T/4 ≤θ ≤T/4). The petal is symmetric about the polar axis. All other petals arerotations of this petal about the pole, including those for roses specified by the sine function with same values fora andk.[5]
Consistent with the rules for plotting points in polar coordinates, a point in a negative half-cycle cannot be plotted at its polar angle because its radial coordinater is negative. The point is plotted by addingπ radians to the polar angle with a radial coordinate|r|. Thus, positive and negative half-cycles can be coincident in the graph of a rose. In addition, roses are inscribed in the circler =a.
When the periodT of the sinusoid is less than or equal to4π, the petal's shape is a single closed loop. A single loop is formed because the angle interval for a polar plot is2π and the angular width of the half-cycle is less than or equal to2π. WhenT > 4π (or|k| <1/2) the plot of a half-cycle can be seen as spiraling out from the pole in more than one circuit around the pole until plotting reaches the inscribed circle where it spirals back to the pole, intersecting itself and forming one or more loops along the way. Consequently, each petal forms two loops when4π <T ≤ 8π (or1/4 ≤ |k| <1/2), three loops when8π <T ≤ 12π (or1/6 ≤ |k| <1/4), etc. Roses with only one petal with multiple loops are observed fork =1/3,1/5,1/7, etc. (See the figure in the introduction section.)
A rose's petals will not intersect each other when the angular frequencyk is a non-zero integer; otherwise, petals intersect one another.
All roses display one or more forms ofsymmetry due to the underlying symmetric and periodic properties of sinusoids.
A rose specified asr =a cos(kθ) is symmetric about the polar axis (the lineθ = 0) because of theidentitya cos(kθ) =a cos(−kθ) that makes the roses specified by the two polar equations coincident.
A rose specified asr =a sin(kθ) is symmetric about the vertical lineθ =π/2 because of the identitya sin(kθ) =a sin(π −kθ) that makes the roses specified by the two polar equations coincident.
Only certain roses are symmetric about the pole.
Individual petals are symmetric about the line through the pole and the petal's peak, which reflects the symmetry of the half-cycle of the underlying sinusoid. Roses composed of a finite number of petals are, by definition,rotationally symmetric since each petal is the same shape with successive petals rotated about the same angle about the pole.
The roser = cos(4θ). Sincek = 4 is an even number, the rose has2k = 8 petals. Line segments connecting successive peaks lie on the circler = 1 and will form anoctagon. Since one peak is at(1,0) the octagon makes sketching the graph relatively easy after the half-cycle boundaries (corresponding to apothems) are drawn.The rose specified byr = cos(7θ). Sincek = 7 is an odd number, the rose hask = 7 petals. Line segments connecting successive peaks lie on the circler = 1 and will form aheptagon. The rose is inscribed in the circler = 1.
Whenk is a non-zero integer, the curve will be rose-shaped with2kpetals ifk is even, andk petals whenk is odd.[6] The properties of these roses are a special case of roses with angular frequenciesk that are rational numbers discussed in the next section of this article.
The rose is inscribed in the circler =a, corresponding to the radial coordinate of all of its peaks.
Because a polar coordinate plot is limited to polar angles between 0 and2π, there are2π/T =k cycles displayed in the graph. No additional points need be plotted because the radial coordinate atθ = 0 is the same value atθ = 2π (which are crests for two different positive half-cycles for roses specified by the cosine function).
Whenk is even (and non-zero), the rose is composed of2k petals, one for each peak in the2π interval of polar angles displayed. Each peak corresponds to a point lying on the circler =a. Line segments connecting successive peaks will form a regularpolygon with an even number of vertices that has its center at the pole and a radius through each peak, and likewise:
The roses are symmetric about the pole.
The roses are symmetric about each line through the pole and a peak (through the "middle" a petal) with the polar angle between the peaks of successive petals being2π/2k =π/k radians. Thus, these roses have rotational symmetry of order2k.
The roses are symmetric about each line that bisects the angle between successive peaks, which corresponds to half-cycle boundaries and theapothem of the corresponding polygon.
Whenk is odd, the rose is composed of thek petals, one for each crest (or trough) in the2π interval of polar angles displayed. Each peak corresponds to a point lying on the circler =a. These rose's positive and negative half-cycles are coincident, which means that in graphing them, only the positive half-cycles or only the negative half-cycles need to plotted in order to form the full curve. (Equivalently, a complete curve will be graphed by plotting any continuous interval of polar angles that isπ radians long such asθ = 0 toθ =π.[7]) Line segments connecting successive peaks will form a regular polygon with an odd number of vertices, and likewise:
The roses are symmetric about each line through the pole and a peak (through the middle of a petal) with the polar angle between the peaks of successive petals being2π/k radians. Thus, these roses have rotational symmetry of orderk.
The rose’s petals do not overlap.
The roses can be specified by algebraic curves of orderk + 1 whenk is odd, and2(k + 1) whenk is even.[8]
A rose withk = 1 is acircle that lies on the pole with a diameter that lies on the polar axis whenr =a cos(θ). The circle is the curve's single petal. (See the circle being formed at the end of the next section.) In Cartesian coordinates, the equivalent cosine and sine specifications are
A rose withk = 2 is called aquadrifolium because it has2k = 4 petals and will form asquare. In Cartesian coordinates the cosine and sine specifications are
A rose withk = 3 is called a trifolium[9] because it hask = 3 petals and will form anequilateral triangle. The curve is also called the Paquerette de Mélibée. In Cartesian Coordinates the cosine and sine specifications are
and
respectively.[10] (See the trifolium being formed at the end of the next section.)
A rose withk = 4 is called anoctafolium because it has2k = 8 petals and will form anoctagon. In Cartesian Coordinates the cosine and sine specifications are
A rose withk = 5 is called apentafolium because it hask = 5 petals and will form aregular pentagon. In Cartesian Coordinates the cosine and sine specifications are
A rose withk = 6 is called adodecafolium because it has2k = 12 petals and will form adodecagon. In Cartesian Coordinates the cosine and sine specifications are
The totalarea of a rose with polar equation of the formr =a cos(kθ) orr =a sin(kθ), wherek is a non-zero integer, is[11]
Whenk is even, there are2k petals; and whenk is odd, there arek petals, so the area of each petal isπa2/4k.
As a consequence, if someone wanted to play the popular gameHe loves me... he loves me not on a rose like above, instead of counting the petals they could calculate the area of the rose to determine the result of the game.
In general, whenk is a rational number in theirreducible fraction formk =n/d, wheren andd are non-zero integers, the number of petals is the denominator of the expression1/2 −1/2k =n −d/2n.[12] This means that the number of petals isn if bothn andd are odd, and2n otherwise.[13]
In the case when bothn andd are odd, the positive and negative half-cycles of the sinusoid are coincident. The graph of these roses are completed in any continuous interval of polar angles that isdπ long.[14]
Whenn is even andd is odd, or visa versa, the rose will be completely graphed in a continuous polar angle interval2dπ long.[15] Furthermore, the roses are symmetric about the pole for both cosine and sine specifications.[16]
In addition, whenn is odd andd is even, roses specified by the cosine and sine polar equations with the same values ofa andk are coincident. For such a pair of roses, the rose with the sine function specification is coincident with the crest of the rose with the cosine specification at on the polar axis either atθ =dπ/2 or atθ =3dπ/2. (This means that rosesr =a cos(kθ) andr =a sin(kθ) with non-zero integer values ofk are never coincident.)
The rose is inscribed in the circler =a, corresponding to the radial coordinate of all of its peaks.
A rose withk =1/2 is called the Dürer folium, named after the German painter and engraverAlbrecht Dürer. The roses specified byr =a cos(θ/2) andr =a sin(θ/2) are coincident even thougha cos(θ/2) ≠a sin(θ/2). In Cartesian coordinates the rose is specified as[17]
The Dürer folium is also atrisectrix, a curve that can be used to trisect angles.
A rose withk =1/3 is alimaçon trisectrix that has the property oftrisectrix curves that can be used to trisect angles. The rose has a single petal with two loops. (See the animation below.)
Examples of rosesr = cos(kθ) created using gears with different ratios. The rays displayed are the polar axis andθ =π/2. Graphing starts atθ = 2π whenk is an integer,θ = 2dπ otherwise, and proceeds clockwise toθ = 0.
Thecircle,k = 1 (n = 1,d = 1). The rose is complete whenθ =π is reached (half a revolution of the lighter gear).
Thelimaçon trisectrix,k =1/3 (n = 1,d = 3), has one petal with two loops. The rose is complete whenθ = 3π is reached (3/2 revolutions of the lighter gear).
The trifolium,k = 3 (n = 3,d = 1). The rose is complete whenθ =π is reached (half a revolution of the lighter gear).
The 8 petals of the rose withk =4/5 (n = 4,d = 5) is each, a single loop that intersect other petals. The rose is symmetric about the pole. The rose is complete atθ = 10π (five revolutions of the lighter gear).
A rose curve specified with anirrational number fork has an infinite number of petals[18] and will never complete. For example, the sinusoidr =a cos(πθ) has a periodT = 2, so, it has a petal in the polar angle interval−1/2 ≤θ ≤1/2 with a crest on the polar axis; however there is no other polar angle in the domain of the polar equation that will plot at the coordinates(a,0). Overall, roses specified by sinusoids with angular frequencies that are irrational constants form adense set (that is, they come arbitrarily close to specifying every point in the diskr ≤a).
