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Polar Calculus and Applications · Tutorial 994 of 998

Arc Length of Rose Curves

Use the polar arc-length formula and petal symmetry to calculate the length of one petal and the total length of a rose curve.

Advanced 9 min read

What You'll Learn

  • Set bounds that trace exactly one petal of a cosine or sine rose.
  • Derive the one-petal length integral from the polar arc-length formula.
  • Express the resulting integral using a defined elliptic-integral notation.
  • Count petals correctly for odd and even values of the rose parameter.
  • Calculate one-petal and total lengths, including the special case of a circle.
  • Avoid common errors involving parameter changes and petal counts.

Rose-Curve Arc Length

A rose curve can have several congruent petals, but its length is not obtained by applying the polar arc-length formula over an arbitrary full revolution: that can trace parts of the curve more than once. The reliable approach is to identify the parameter interval for one petal, calculate its length, and then use the number of petals.

We consider \(r=a\cos(n\theta)\) and \(r=a\sin(n\theta)\), where \(a>0\) and \(n\in\mathbb{N}\). The parameter \(a\) sets the curve’s scale, while \(n\) affects both its shape and its number of petals.

Definition: A petal is one loop of the rose curve, bounded by consecutive parameter values at which \(r=0\). Its arc length is the length of that loop, traced once.

From Polar Speed to the Length of One Petal

For a polar curve \(r=r(\theta)\), the arc-length formula is

$$ L=\int_{\alpha}^{\beta}\sqrt{r(\theta)^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta, $$

provided the interval traces the curve segment in question once. For either rose, differentiating with respect to \(\theta\) gives a speed involving both \(r\) and its derivative:

$$ \begin{aligned} r&=a\cos(n\theta) &\Longrightarrow\quad \frac{dr}{d\theta}&=-an\sin(n\theta),\\ r&=a\sin(n\theta) &\Longrightarrow\quad \frac{dr}{d\theta}&=an\cos(n\theta). \end{aligned} $$

For the cosine rose, substituting into the speed gives \(a\sqrt{\cos^2(n\theta)+n^2\sin^2(n\theta)}\). For the sine rose it gives \(a\sqrt{\sin^2(n\theta)+n^2\cos^2(n\theta)}\). The one-petal integrals for these two curves are equal: after changing variables, their integrands differ only by a shift of the variable by \(\pi/2\), and each has the same symmetric half-period integral.

Key formula: For either rose, the length of one petal is $$ L_{\mathrm{petal}}=\frac{2a}{n}\int_0^{\pi/2}\sqrt{1+(n^2-1)\sin^2u}\,du. $$

Here is how the bounds and factor arise. For \(r=a\cos(n\theta)\), a petal centered on \(\theta=0\) is traced between the consecutive zeros \(\theta=-\pi/(2n)\) and \(\theta=\pi/(2n)\). Set \(u=n\theta\), so \(d\theta=du/n\). The bounds become \(-\pi/2\) and \(\pi/2\), and the speed becomes \(a\sqrt{\cos^2u+n^2\sin^2u}\). The integrand is even, and \(\cos^2u+n^2\sin^2u=1+(n^2-1)\sin^2u\). Thus the integral over the petal is twice the integral from \(0\) to \(\pi/2\), with the additional factor \(1/n\) from the change of variable.

For \(r=a\sin(n\theta)\), a petal is traced from \(\theta=0\) to \(\theta=\pi/n\). The same substitution gives \(u\) from \(0\) to \(\pi\). The resulting speed is \(a\sqrt{\sin^2u+n^2\cos^2u}\); its integral over \([0,\pi]\) is twice its integral over \([0,\pi/2]\). Replacing \(u\) by \(\pi/2-u\) in that half-interval integral gives the same value as for the cosine rose. This establishes the formula for both families.

For \(n>1\), this definite integral generally has no elementary antiderivative. We can name it precisely using the complete elliptic integral of the second kind, with the parameter convention

$$ E(m)=\int_0^{\pi/2}\sqrt{1-m\sin^2u}\,du. $$

In this convention, the argument \(m\) is a parameter, not its square root. Since \(1+(n^2-1)\sin^2u=1-(1-n^2)\sin^2u\), the one-petal length can be written \(L_{\mathrm{petal}}=(2a/n)E(1-n^2)\). This notation is simply a compact name for the stated integral.

Why the Integral Gives the Arc Length

The polar arc-length formula follows by viewing the curve in Cartesian coordinates, \(x=r\cos\theta\) and \(y=r\sin\theta\). Differentiating and adding the squares of the two coordinate derivatives gives

$$ \begin{aligned} \left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2 &=\left(\frac{dr}{d\theta}\cos\theta-r\sin\theta\right)^2 +\left(\frac{dr}{d\theta}\sin\theta+r\cos\theta\right)^2\\ &=\left(\frac{dr}{d\theta}\right)^2+r^2. \end{aligned} $$

The cross terms cancel, and \(\sin^2\theta+\cos^2\theta=1\). The length of a parametrized curve segment is the integral of its speed, so its speed is \(\sqrt{r^2+(dr/d\theta)^2}\). In particular, the squared terms ensure that a negative value of \(r\) does not make the speed negative.

Counting Petals and Finding Total Length

The cosine rose has \(n\) petals when \(n\) is odd and \(2n\) petals when \(n\) is even. The sine rose has the same count. For either curve, each petal has the same length: shifting the parameter by an appropriate multiple of \(\pi/n\) reproduces the same speed profile on the next petal.

The parity distinction comes from how negative values of \(r\) map to points in the plane. The polar coordinates \((r,\theta)\) and \((-r,\theta+\pi)\) represent the same point. When \(n\) is odd, shifting \(\theta\) by \(\pi\) changes the sign of \(\cos(n\theta)\) and \(\sin(n\theta)\); negative-radius portions therefore trace petal directions already represented by positive-radius portions. When \(n\) is even, that sign change does not occur, and the opposite directions give additional petals. Thus the total length is the number of petals multiplied by the length of one petal.

Theorem: For \(a>0\) and \(n\in\mathbb{N}\), the one-petal length of either rose is $$ \frac{2a}{n}E(1-n^2). $$ The total length is $$ \begin{cases} 2aE(1-n^2), & n\text{ odd},\\ 4aE(1-n^2), & n\text{ even}. \end{cases} $$

Worked Example: A Cosine Rose with One Petal

Find the length of \(r=5\cos\theta\). Here \(a=5\) and \(n=1\), so the curve has one petal. Its one-petal length is

$$ \begin{aligned} L_{\mathrm{petal}} &=\frac{2(5)}{1}\int_0^{\pi/2}\sqrt{1+(1^2-1)\sin^2u}\,du\\ &=10\int_0^{\pi/2}1\,du\\ &=10\left(\frac{\pi}{2}\right)=5\pi. \end{aligned} $$

The total length is also \(5\pi\), since there is one petal. As a check, \(r=5\cos\theta\) describes a circle of radius \(5/2\), whose circumference is \(2\pi(5/2)=5\pi\). This agrees with the arc-length calculation.

Worked Example: One Petal and Total Length for an Even Rose

Find the one-petal and total lengths of \(r=3\cos(2\theta)\). Here \(a=3\), \(n=2\), and an even \(n\) gives \(2n=4\) petals. For one petal,

$$ \begin{aligned} L_{\mathrm{petal}} &=\frac{2(3)}{2}\int_0^{\pi/2}\sqrt{1+(2^2-1)\sin^2u}\,du\\ &=3\int_0^{\pi/2}\sqrt{1+3\sin^2u}\,du\\ &=3E(-3). \end{aligned} $$

There are four equal petals, so

$$ L_{\mathrm{total}}=4L_{\mathrm{petal}}=4\bigl(3E(-3)\bigr)=12E(-3). $$

By the definition of \(E\), this total is \(12\int_0^{\pi/2}\sqrt{1+3\sin^2u}\,du\). The factor \(3\) in the one-petal result comes from \(2a/n\); multiplying it by the four petals gives the displayed total. This verifies that the integral and petal count have both been applied.

Worked Example: A Sine Rose with Odd \(n\)

Find the one-petal and total lengths of \(r=2\sin(3\theta)\). Here \(a=2\), \(n=3\), and an odd \(n\) gives three petals. One petal is traced over \(0\leq\theta\leq\pi/3\). Substituting \(u=3\theta\) changes these bounds to \(0\leq u\leq\pi\). Therefore, directly from the sine-rose speed,

$$ \begin{aligned} L_{\mathrm{petal}} &=\frac{2}{3}\int_0^\pi\sqrt{\sin^2u+9\cos^2u}\,du\\ &=\frac{4}{3}\int_0^{\pi/2}\sqrt{\sin^2u+9\cos^2u}\,du\\ &=\frac{4}{3}\int_0^{\pi/2}\sqrt{1+8\sin^2v}\,dv\\ &=\frac{4}{3}E(-8). \end{aligned} $$

In the second line, symmetry about \(\pi/2\) halves the interval and contributes a factor of two. In the third line, \(v=\pi/2-u\), so \(\sin^2u=\cos^2v\) and \(\cos^2u=\sin^2v\). The total length is

$$ L_{\mathrm{total}}=3\left(\frac{4}{3}E(-8)\right)=4E(-8). $$

This also checks the bounds: the chosen interval runs between consecutive zeros of \(r\), so it traces one petal, and multiplying by three counts the three petals once each.

Common Errors and a Reliable Setup

A frequent error is to substitute \(u=n\theta\) in the integrand but forget to change the differential. For example, for \(r=3\cos(2\theta)\), a mistaken calculation might write

$$ L_{\mathrm{petal}}\overset{\text{wrong}}{=}6\int_0^{\pi/2}\sqrt{1+3\sin^2u}\,du. $$

This has omitted \(d\theta=du/2\), so it is twice the correct one-petal length. The correction is

$$ L_{\mathrm{petal}} =3\int_0^{\pi/2}\sqrt{1+3\sin^2u}\,du =3E(-3), $$

as in the worked example. Another error is to assume every rose has \(n\) petals. That fails when \(n\) is even: \(r=a\cos(2\theta)\), for instance, has four petals, not two. Determine the parity-based petal count before multiplying.

A dependable calculation keeps three decisions visible: use consecutive zeros to set one-petal bounds, retain the factor \(1/n\) when changing variables, and multiply by \(n\) or \(2n\) according to whether \(n\) is odd or even. These steps matter because the same polar expression can trace a loop more than once over a poorly chosen interval.

Check Your Understanding

Use the one-petal formula and the parity-based petal count to answer these questions.

  1. What parameter interval traces a petal centered at \(\theta=0\) for \(r=a\cos(n\theta)\)?
  2. Write the one-petal length of \(r=4\sin(2\theta)\) using \(E(m)\), and state the number of petals.
  3. Why does the substitution \(u=n\theta\) introduce a factor of \(1/n\) in the length integral?
  4. How many petals does \(r=a\cos(5\theta)\) have, and what is its total length in terms of \(E\)?
  5. Why does a negative radius not cause the arc-length speed to be negative?

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