Rose (Mathematics)

The rose, or rhodonea curve, is a mathematical marvel, a sinusoid plotted in polar coordinates. Think of it as a shape born from the elegant dance of sine and cosine waves, devoid of the messy complications of a phase angle. It's the brainchild of Guido Grandi, an Italian mathematician who, between 1723 and 1728, apparently found himself with far too much time and a penchant for beautiful, intricate curves. He named them "rhodonea," which, if you care about such things, translates to "rose."

General Overview

Specification

The fundamental definition of a rose resides in its polar equation:

$r = a \cos(k\theta)$

Or, if you prefer the more mundane Cartesian coordinates, it can be expressed parametrically:

$x = r \cos(\theta) = a \cos(k\theta) \cos(\theta)$ $y = r \sin(\theta) = a \cos(k\theta) \sin(\theta)$

Naturally, the sine function can also be employed. This isn't some fundamental difference, mind you. It merely results in a rotation. As the identity $\sin(k\theta) = \cos(k\theta - \pi/2)$ shows, a sine-based rose is simply its cosine counterpart spun counter-clockwise by $\pi/(2k)$ radians, which is a mere quarter of the sinusoid's period.

So, what we're looking at are graphs of sinusoids, typically visualized in polar coordinates. The angular frequency, $k$, and the amplitude, $a$, dictate the radial coordinate $r$ based on the polar angle $\theta$. When $k$ is a rational number, these curves can, with some effort, be described algebraically, making them algebraic curves.

Petals

The defining characteristic of a rose, as its name suggests, is its petals. Each petal corresponds to a half-cycle of the sinusoid that defines it. A full cycle, with a period $T = 2\pi/k$, consists of a positive half-cycle (where $r \ge 0$) and a negative half-cycle (where $r \le 0$). Each half-cycle is $T/2 = \pi/k$ long.

The shape of each petal is identical, a testament to the inherent symmetry of the sinusoidal function. For $r = a \cos(k\theta)$, the primary petal is formed by the positive half-cycle bounded by $-T/4 \le \theta \le T/4$. This petal is symmetric about the polar axis. All other petals are simply rotations of this fundamental shape around the pole.

Now, plotting points in polar coordinates has its quirks. When $r$ is negative, the point isn't plotted at its polar angle $\theta$. Instead, you add $\pi$ radians to the angle and use the absolute value of $r$. This means positive and negative half-cycles can overlap, creating a more complex figure. Roses are always inscribed within the circle $r = a$.

The petal's form depends crucially on the period $T$. If $T \le 4\pi$ (or $|k| \ge 1/2$), the petal is a single, closed loop. This is because the total angular range for a polar plot is $2\pi$, and the half-cycle's angular width is less than or equal to $2\pi$. However, when $T > 4\pi$ (or $|k| < 1/2$), the plot begins to spiral. A half-cycle can take more than one full circuit around the pole before it closes. Consequently, when $4\pi < T \le 8\pi$ (or $1/4 \le |k| < 1/2$), each petal forms two loops. For $8\pi < T \le 12\pi$ (or $1/6 \le |k| < 1/4$), each petal has three loops, and so on. Roses like $k = 1/3, 1/5, 1/7$ exhibit single petals with multiple loops.

A critical distinction: if $k$ is a non-zero integer, the petals of a rose will not intersect each other. If $k$ is not an integer, intersection is inevitable.

Symmetry

Roses are inherently symmetrical, a reflection of the periodic and symmetrical nature of their underlying sinusoids.

• A rose defined by $r = a \cos(k\theta)$ is symmetric about the polar axis (the line $\theta = 0$). This is due to the trigonometric identity $\cos(k\theta) = \cos(-k\theta)$, which ensures that the curve plotted for a positive angle is identical to that plotted for the corresponding negative angle.
• Similarly, a rose defined by $r = a \sin(k\theta)$ exhibits symmetry about the line $\theta = \pi/2$. This stems from the identity $\sin(k\theta) = \sin(\pi - k\theta)$.
• Symmetry about the pole is not a universal property; it depends on specific values of $k$.
• Each individual petal possesses symmetry about the line passing through the pole and the petal's peak, mirroring the symmetry of the sinusoid's half-cycle. Roses with a finite number of petals are, by definition, rotationally symmetric, as each petal is identical in shape and rotated by a consistent angle around the pole.

Roses with Non-Zero Integer Values of $k$

When $k$ is a non-zero integer, the resulting curve is undeniably rose-shaped. The number of petals is $2k$ if $k$ is even, and $k$ if $k$ is odd. This is a specific instance of the more general case of rational $k$.

• The rose is contained within or inscribed by the circle $r = a$, which represents the maximum radial reach.
• Since polar plots are typically confined to angles from $0$ to $2\pi$, we observe $k$ full cycles of the sinusoid. No additional points are needed because the radial coordinate at $\theta = 0$ is the same as at $\theta = 2\pi$.
• If $k$ is even (and non-zero): The rose boasts $2k$ petals. Each peak of the sinusoid within the $2\pi$ range corresponds to a petal. Connecting the peaks of successive petals with line segments forms a regular polygon with $2k$ vertices, centered at the pole, with lines radiating through each vertex.
  • These roses are symmetric about the pole.
  • They are symmetric about every line passing through the pole and a petal's peak. The angle between successive peaks is $2\pi/(2k) = \pi/k$ radians. Consequently, these roses exhibit rotational symmetry of order $2k$.
  • Symmetry also extends to the lines that bisect the angles between successive peaks. These lines correspond to the half-cycle boundaries and the apothems of the inscribed polygon.
• If $k$ is odd: The rose has $k$ petals. Again, each peak corresponds to a point on the circle $r = a$. The coincidence of positive and negative half-cycles means that plotting only the positive (or only the negative) half-cycles is sufficient to draw the entire curve. Alternatively, any continuous interval of $\pi$ radians, such as $0$ to $\pi$, will trace the complete curve. Connecting successive peaks forms a regular polygon with $k$ vertices.
  • These roses possess rotational symmetry of order $k$, with the angle between successive peaks being $2\pi/k$ radians.
  • Crucially, the petals do not overlap.
  • These curves can be represented by algebraic curves of order $k+1$ when $k$ is odd, and $2(k+1)$ when $k$ is even.

The Circle

When $k=1$, the rose degenerates into a circle. Specifically, $r = a \cos(\theta)$ describes a circle with diameter $a$ lying on the polar axis, passing through the pole. This single "petal" is the entire curve. In Cartesian coordinates, this is represented by:

$(x - a/2)^2 + y^2 = (a/2)^2$

The sine equivalent, $r = a \sin(\theta)$, yields a circle of the same size, but centered on the line $\theta = \pi/2$.

The Quadrifolium

A rose with $k=2$ is known as a quadrifolium. It has $2k=4$ petals and forms a shape resembling a square with curved sides. Its Cartesian equation is:

$(x^2 + y^2)^3 = a^2(x^2 - y^2)^2$

The sine version is:

$(x^2 + y^2)^3 = 4(axy)^2$

The Trifolium

With $k=3$, we get the trifolium, characterized by $k=3$ petals that form an equilateral triangle. It's also sometimes called the Paquerette de Mélibée. The Cartesian equations are:

$(x^2 + y^2)^2 = a(x^3 - 3xy^2)$ for the cosine version.

$(x^2 + y^2)^2 = a(3x^2y - y^3)$ for the sine version.

The Octafolium

For $k=4$, the rose becomes an octafolium, possessing $2k=8$ petals and outlining an octagon. Its Cartesian representation is:

$(x^2 + y^2)^5 = a^2(x^4 - 6x^2y^2 + y^4)^2$

And the sine form:

$(x^2 + y^2)^5 = 16a^2(xy^3 - yx^3)^2$

The Pentafolium

When $k=5$, the rose is a pentafolium, with $k=5$ petals forming a regular pentagon. The Cartesian equations are:

$(x^2 + y^2)^3 = a(x^5 - 10x^3y^2 + 5xy^4)$

$(x^2 + y^2)^3 = a(5x^4y - 10x^2y^3 + y^5)$

The Dodecafolium

A rose with $k=6$ is termed a dodecafolium, featuring $2k=12$ petals that delineate a dodecagon. The Cartesian equations are:

$(x^2 + y^2)^7 = a^2(x^6 - 15x^4y^2 + 15x^2y^4 - y^6)^2$

$(x^2 + y^2)^7 = 4a^2(3x^5y - 10x^3y^3 + 3xy^5)^2$

Total and Petal Areas

The total area enclosed by a rose defined by $r = a \cos(k\theta)$ or $r = a \sin(k\theta)$, where $k$ is a non-zero integer, can be calculated. The integration yields different results depending on the parity of $k$:

For even $k$: Area $= \frac{1}{2} \int_{0}^{2\pi} (a \cos(k\theta))^2 \, d\theta = \frac{\pi a^2}{2}$

For odd $k$: Area $= \frac{1}{2} \int_{0}^{\pi} (a \cos(k\theta))^2 \, d\theta = \frac{\pi a^2}{4}$

Since there are $2k$ petals for even $k$ and $k$ petals for odd $k$, the area of a single petal is consistently $\pi a^2 / (4k)$. This offers a rather unique way to settle a game of He loves me... he loves me not – just calculate the area.

Roses with Rational Number Values for $k$

When $k$ is a rational number, expressed in its irreducible fraction form $k = n/d$ (where $n$ and $d$ are non-zero integers), the number of petals is determined by the denominator of $1/2 - 1/(2k)$, which simplifies to $(n-d)/(2n)$. More directly, the number of petals is $n$ if both $n$ and $d$ are odd, and $2n$ otherwise.

• If both $n$ and $d$ are odd: The positive and negative half-cycles of the sinusoid coincide. The entire curve is traced within a polar angle interval of $d\pi$.
• If $n$ is even and $d$ is odd, or vice versa: The complete rose is graphed within a polar angle interval of $2d\pi$. These roses exhibit symmetry about the pole, regardless of whether the sine or cosine function is used.
• If $n$ is odd and $d$ is even: Roses defined by $r = a \cos(k\theta)$ and $r = a \sin(k\theta)$ become identical. For these coincident pairs, the sine rose's crest aligns with the cosine rose's peak on the polar axis at either $\theta = d\pi/2$ or $\theta = 3d\pi/2$. This explains why roses with non-zero integer $k$ are never coincident between their sine and cosine forms.
• The rose is always inscribed within the circle $r = a$.

The Dürer Folium

A rose with $k = 1/2$ is known as the Dürer folium, named in honor of Albrecht Dürer. Intriguingly, both $r = a \cos(\theta/2)$ and $r = a \sin(\theta/2)$ describe the same curve, despite their differing equations. In Cartesian coordinates, it's given by:

$(x^2 + y^2)(2(x^2 + y^2) - a^2)^2 = a^4x^2$

This curve also holds the distinction of being a trisectrix, meaning it can be used to divide an angle into three equal parts.

The Limaçon Trisectrix

With $k = 1/3$, the rose becomes a limaçon trisectrix. Like the Dürer folium, it possesses the property of trisecting angles. This particular rose has a single petal with two loops, a characteristic that can be visually striking.

Roses with Irrational Number Values for $k$

When $k$ is an irrational number, the rose curve takes on an infinite number of petals. It never truly completes, never closes. Consider $r = a \cos(\pi\theta)$. Its period is $T=2$, giving a petal in the interval $-1/2 \le \theta \le 1/2$ with a peak on the polar axis. However, no other angle in the domain will plot at the coordinates $(a,0)$. These roses, defined by irrational frequencies, form a dense set within the disk $r \le a$, meaning they approach every point in the disk arbitrarily closely.

Rotations Required to Close the Curve

The number of rotations, or the total angular range, needed for a rhodonea curve to form a complete, closed figure is dependent on the ratio $k = n/d$.

• If $k$ is an integer: The curve closes after $\pi$ radians if $k$ is odd, and after $2\pi$ radians if $k$ is even.
• If $k$ is a rational number: The total rotation required is $d\pi / \text{gcd}(n,d)$ if $nd$ is odd, and $2d\pi / \text{gcd}(n,d)$ otherwise. This formula precisely dictates the angular sweep needed for the rose curve to repeat its pattern.