Rose (mathematics)

A rose is the set of points in polar coordinates specified by the polar equation[2]

${\displaystyle r=a\cos(k\theta )}$

or in Cartesian coordinates using the parametric equations

${\displaystyle {\begin{aligned}x&=r\cos(\theta )=a\cos(k\theta )\cos(\theta )\\y&=r\sin(\theta )=a\cos(k\theta )\sin(\theta )\end{aligned}}}$

Roses can also be specified using the sine function.[3] Since

${\displaystyle \sin(k\theta )=\cos \left(k\theta -{\frac {\pi }{2}}\right)=\cos \left(k\left(\theta -{\frac {\pi }{2k}}\right)\right)}$.

Thus, the rose specified by r = a sin(kθ) is identical to that specified by r = a cos(kθ) rotated counter-clockwise by π/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 as polar coordinate (rather than Cartesian coordinate) graphs of sinusoids that have angular frequency of k and an amplitude of a that determine the radial coordinate r given the polar angle θ (though when k is a rational number, a rose curve can be expressed in Cartesian coordinates since those can be specified as algebraic curves[4]).

Artistic depiction of roses with different parameter settings

Roses are directly related to the properties of the sinusoids that specify them.

• Graphs of roses are composed of petals. 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 period T = 2π/k long and consists of a positive half-cycle, the continuous set of points where r ≥ 0 and is T/2 = π/k long, and a negative half-cycle is the other half where r ≤ 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 by r = 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 are rotations of this petal about the pole, including those for roses specified by the sine function with same values for a and k.[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 coordinate r is negative. The point is plotted by adding π 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 circle r = a.
  • When the period T of the sinusoid is less than or equal to 4π, the petal's shape is a single closed loop. A single loop is formed because the angle interval for a polar plot is 2π and the angular width of the half-cycle is less than or equal to 2π. When T > 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 when 4π < T ≤ 8π (or 1/4 ≤ |k| < 1/2), three loops when 8π < T ≤ 12π (or 1/6 ≤ |k| < 1/4), etc. Roses with only one petal with multiple loops are observed for k = 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 frequency k is a non-zero integer; otherwise, petals intersect one another.

All roses display one or more forms of symmetry due to the underlying symmetric and periodic properties of sinusoids.

• A rose specified as r = a cos(kθ) is symmetric about the polar axis (the line θ = 0) because of the identity a cos(kθ) = a cos(−kθ) that makes the roses specified by the two polar equations coincident.

• A rose specified as r = a sin(kθ) is symmetric about the vertical line θ = π/2 because of the identity a 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.

A rose with k = 1 is a circle that lies on the pole with a diameter that lies on the polar axis when r = 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

${\displaystyle \left(x-{\frac {a}{2}}\right)^{2}+y^{2}=\left({\frac {a}{2}}\right)^{2}}$

and

${\displaystyle x^{2}+\left(y-{\frac {a}{2}}\right)^{2}=\left({\frac {a}{2}}\right)^{2}}$

respectively.

A rose with k = 2 is called a quadrifolium because it has 4 petals. In Cartesian coordinates the cosine and sine specifications are

${\displaystyle \left(x^{2}+y^{2}\right)^{3}=a^{2}\left(x^{2}-y^{2}\right)^{2}}$

and

${\displaystyle \left(x^{2}+y^{2}\right)^{3}=4\left(axy\right)^{2}}$

respectively.

A rose with k = 3 is called a trifolium[9] because it has 3 petals. The curve is also called the Paquerette de Mélibée. In Cartesian Coordinates the cosine and sine specifications are

${\displaystyle \left(x^{2}+y^{2}\right)^{2}=a\left(x^{3}-3xy^{2}\right)}$

and

${\displaystyle \left(x^{2}+y^{2}\right)^{2}=-a\left(x^{3}-3xy^{2}\right)}$

respectively.[10] (See the trifolium being formed at the end of the next section.)

The total area of a rose with polar equation of the form r = a cos(kθ) or r = a sin(kθ), where k is a non-zero integer, is[11]

${\displaystyle {\begin{aligned}{\frac {1}{2}}\int _{0}^{2\pi }(a\cos(k\theta ))^{2}\,d\theta &={\frac {a^{2}}{2}}\left(\pi +{\frac {\sin(4k\pi )}{4k}}\right)={\frac {\pi a^{2}}{2}}&&\quad {\text{for even }}k\\[8px]{\frac {1}{2}}\int _{0}^{\pi }(a\cos(k\theta ))^{2}\,d\theta &={\frac {a^{2}}{2}}\left({\frac {\pi }{2}}+{\frac {\sin(2k\pi )}{4k}}\right)={\frac {\pi a^{2}}{4}}&&\quad {\text{for odd }}k\end{aligned}}}$

When k is even, there are 2k petals; and when k is odd, there are k petals, so the area of each petal is πa²/4k.

A rose with k = 1/2 is called the Dürer folium, named after the German painter and engraver Albrecht Dürer. The roses specified by r = a cos(θ/2) and r = a sin(θ/2) are coincident even though a cos(θ/2) ≠ a sin(θ/2). In Cartesian coordinates the rose is specified as[17]

${\displaystyle \left(x^{2}+y^{2}\right)\left(2\left(x^{2}+y^{2}\right)-a^{2}\right)^{2}=a^{4}x^{2}}$

The Dürer folium is also a trisectrix, a curve that can be used to trisect angles.

A rose with k = 1/3 is a limaçon trisectrix that has the property of trisectrix curves that can be used to trisect angles. The rose has a single petal with two loops. (See the animation below.)

Examples of roses r = cos(kθ) created using gears with different ratios.
The rays displayed are the polar axis and θ = π/2.
Graphing starts at θ = 2π when k is an integer, θ = 2dπ otherwise, and proceeds clockwise to θ = 0.

The circle, k = 1 (n = 1, d = 1). The rose is complete when θ = π is reached (half a revolution of the lighter gear).

The limaçon trisectrix, k = 1/3 (n = 1, d = 3), has one petal with two loops. The rose is complete when θ = 3π is reached (3/2 revolutions of the lighter gear).

The trifolium, k = 3 (n = 3, d = 1). The rose is complete when θ = π is reached (half a revolution of the lighter gear).

The 8 petals of the rose with k = 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 at θ = 10π (five revolutions of the lighter gear).