Sinusoidal Spiral

Q: What is a sinusoidal spiral?

It is the general name for the family of curves expressed in polar coordinates as rⁿ = aⁿ sin(nθ). Depending on the value of the exponent n, classical curves appear as special cases: the lemniscate (n=2), the circle (n=1), a straight line (n=-1), the rectangular hyperbola (n=-2), and others.

Q: What curve is the sinusoidal spiral when n=2?

r² = a² sin(2θ) coincides with Bernoulli's lemniscate. This figure-eight curve has the property that the product of the distances from two foci is constant.

Q: How is the radius of curvature of a sinusoidal spiral expressed?

The radius of curvature of the sinusoidal spiral rⁿ = aⁿ sin(nθ) is ρ = aⁿ / ((n+1)·rⁿ⁻¹). In particular, for n=1 (the circle) the radius of curvature is the constant a/2.

A family of classical curves from one polar equation

Intermediate (high-school to early-undergraduate level)

Goal of this page

Understand the polar equation $r^n = a^n \sin(n\theta)$ of the sinusoidal spiral, and learn the classical special-case curves that appear for different values of the exponent $n$.

1. Definition and polar equation

Definition of the sinusoidal spiral

For a constant $a > 0$ and a real number $n \neq 0$, the curve given in polar coordinates $(r, \theta)$ by

$$r^n = a^n \sin(n\theta) \tag{1}\label{eq:def}$$

is called the sinusoidal spiral. Replacing $\sin$ with $\cos$ gives $r^n = a^n \cos(n\theta)$, a curve of the same family rotated by $\pi/(2n)$ in $\theta$.

Because the Scottish mathematician Colin Maclaurin (1698–1746) studied these curves, they are also called Maclaurin's spirals.

Despite its name, for many values of $n$ the curve is a circle, a line, or a hyperbola, so in general it is not a spiral-shaped curve.

Figure 1: An example of a sinusoidal spiral. For a positive integer $n$ the curve has $n$ petals (here $n=3$). Varying the exponent $n$ produces classical curves—circles, parabolas, hyperbolas—one after another.

2. Special cases

Varying the exponent $n$ yields a variety of classical curves as special cases. For representative values of $n$, we see which curve the polar equation $\eqref{eq:def}$ reduces to, together with the derivation and a graph.

In the derivations below we repeatedly use the following relation between polar coordinates $(r,\theta)$ and Cartesian coordinates $(x,y)$ (Figure 2).

$$x = r\cos\theta,\quad y = r\sin\theta \tag{2}\label{eq:xy}$$

This is the basic identity that translates polar coordinates—which locate a point by its distance $r$ from the origin and the angle $\theta$ measured from the $x$-axis—into Cartesian coordinates.

Figure 2: Relationship between polar coordinates $(r,\theta)$ and Cartesian coordinates $(x,y)$. The point P lies at distance $r$ from the origin O and at angle $\theta$ from the $x$-axis; the right triangle OQP gives $x = r\cos\theta$ and $y = r\sin\theta$.

$n = -2$: Rectangular hyperbola

Setting $n=-2$ in $\eqref{eq:def}$ gives $r^{-2} = a^{-2}\sin(-2\theta)$. Since $\sin$ is odd, $\sin(-2\theta) = -\sin 2\theta$, so $r^{-2} = -a^{-2}\sin 2\theta$, i.e. $\dfrac{1}{r^2} = -\dfrac{\sin 2\theta}{a^2}$. Multiplying both sides by $a^2 r^2$ and rearranging moves the $r^{-2}$ on the left to the right, giving $r^2\sin 2\theta = -a^2$. Using the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ together with $\eqref{eq:xy}$,

$$r^2\sin 2\theta = r^2\cdot 2\sin\theta\cos\theta = 2(r\cos\theta)(r\sin\theta) = 2xy$$

so $2xy = -a^2$, i.e. $xy = -\dfrac{a^2}{2}$. This has the form "$xy =$ constant": a rectangular hyperbola (a hyperbola whose asymptotes meet at right angles) with the coordinate axes as asymptotes. The curve appears in the second and fourth quadrants.

Figure 3: $n=-2$ is the rectangular hyperbola $xy = -a^2/2$, with the coordinate axes as asymptotes.

$n = -1$: Straight line

Setting $n=-1$ in $\eqref{eq:def}$ gives $r^{-1} = a^{-1}\sin(-\theta) = -a^{-1}\sin\theta$. Hence $r\sin\theta = -a$, and by $\eqref{eq:xy}$, $y = -a$. This is a straight line at a fixed distance from the origin (the $\cos$ version gives the vertical line $x = a$). It is the special case of identically zero curvature.

Figure 4: $n=-1$ is the straight line $y = -a$.

$n = -1/2$: Parabola

Setting $n=-1/2$ in $\eqref{eq:def}$ gives $r^{-1/2} = a^{-1/2}\sin(-\theta/2)$. Squaring both sides and rearranging gives $r = \dfrac{a}{\sin^2(\theta/2)} = \dfrac{2a}{1-\cos\theta}$. This is the polar equation of a conic of eccentricity $e = 1$ with focus at the origin—that is, a parabola.

Figure 5: $n=-1/2$ is the parabola $r = 2a/(1-\cos\theta)$ (focus at the origin).

$n = 1/2$: Cardioid

Setting $n=1/2$ in $\eqref{eq:def}$ gives $r^{1/2} = a^{1/2}\sin(\theta/2)$. Squaring gives $r = a\sin^2(\theta/2) = \dfrac{a}{2}(1-\cos\theta)$, a heart-shaped curve with a single cusp: the cardioid. It is also the locus traced by a point on a circle as it rolls around a fixed circle of the same radius.

Figure 6: $n=1/2$ is the cardioid $r = \frac{a}{2}(1-\cos\theta)$, with its cusp at the origin.

$n = 1$: Circle

Setting $n=1$ in $\eqref{eq:def}$ gives $r = a\sin\theta$. Multiplying both sides by $r$ gives $r^2 = a r\sin\theta$. With $r^2 = x^2+y^2$ and $\eqref{eq:xy}$ ($y = r\sin\theta$) we get $x^2 + y^2 = a y$, i.e. $x^2 + \left(y - \dfrac{a}{2}\right)^2 = \left(\dfrac{a}{2}\right)^2$. This is a circle through the origin with center $\left(0, \dfrac{a}{2}\right)$ and radius $\dfrac{a}{2}$.

Figure 7: $n=1$ is the circle through the origin $x^2 + (y-a/2)^2 = (a/2)^2$.

$n = 2$: Bernoulli's lemniscate

Setting $n=2$ in $\eqref{eq:def}$ gives $r^2 = a^2\sin 2\theta$. With $r^2 = x^2 + y^2$ and $r^2\sin 2\theta = 2xy$ (from $\eqref{eq:xy}$) we get $(x^2 + y^2)^2 = 2a^2 xy$. This is the figure-eight lemniscate of Bernoulli, the locus of points the product of whose distances to two fixed points (foci) is constant.

Figure 8: $n=2$ is Bernoulli's lemniscate $(x^2+y^2)^2 = 2a^2xy$.

Thus, for a positive integer $n$ the curve has $n$ petals (Figure 1, $n=3$, is an example), while for a fractional $n = p/q$ (in lowest terms) the shape becomes more intricate. Replacing $\sin$ with $\cos$ gives the same curve rotated about the origin.

Gallery

The following table summarizes how the shape changes as $n = p/4$ with the integer $p$ running from $-8$ to $8$ (the case $p=0$ is excluded, since $r=0$ is degenerate). When $p$ is a multiple of $4$ (so $n$ is an integer) we recover the classical curves above; otherwise petal-like curves ($n>0$) or unbounded curves ($n<0$) appear.

$n = -2$

$n = -\tfrac{7}{4}$

$n = -\tfrac{3}{2}$

$n = -\tfrac{5}{4}$

$n = -1$

$n = -\tfrac{3}{4}$

$n = -\tfrac{1}{2}$

$n = -\tfrac{1}{4}$

$n = \tfrac{1}{4}$

$n = \tfrac{1}{2}$

$n = \tfrac{3}{4}$

$n = 1$

$n = \tfrac{5}{4}$

$n = \tfrac{3}{2}$

$n = \tfrac{7}{4}$

$n = 2$

Figure 9: The sinusoidal spiral $r^n = a^n\sin(n\theta)$ for $n = p/4$ ($-8 \le p \le 8$, $p \neq 0$). The $n$ in each cell is shown as a reduced fraction. All cells use the same scale; the larger $|n|$ is (especially for $n<0$), the farther the unbounded branches extend.

3. Curvature

Curvature of the sinusoidal spiral

The radius of curvature $\rho$ of the sinusoidal spiral $r^n = a^n \sin(n\theta)$ is

$$\rho = \frac{a^n}{(n+1)\,r^{n-1}} \tag{3}\label{eq:rho}$$

(valid for $n \neq -1$). This radius-of-curvature formula is a classical result for the sinusoidal (Maclaurin) spiral.

$n=1$ (circle): a check

$r = a\sin\theta$ is a circle of radius $a/2$. By $\eqref{eq:rho}$, $\rho = a/(2 \cdot r^0) = a/2$ (constant curvature), matching the radius of curvature $a/2$ of the circle.

$n=2$ (lemniscate): radius of curvature

By $\eqref{eq:rho}$, $\rho = a^2/(3r)$. At the maximum $r = a$, $\rho = a/3$; the radius of curvature at the vertex is $a/3$.

Frequently asked questions

Q1. What is a sinusoidal spiral?