Recognizing a Rose Curve
A rose curve is a polar graph described by a sine or cosine function of an integer multiple of the angle. Its repeated radial maxima create distinct, rounded petals that meet at the pole. The number and orientation of those petals depend on the multiplier \(n\) and on whether the equation uses cosine or sine.
The coefficient \(a\) sets the greatest distance of the graph from the pole. The integer \(n\) controls how rapidly the radial value oscillates as \(\theta\) changes. To sketch a rose, locate its zeros and its maximum positive or negative radial values. Negative radii must be interpreted using the polar-coordinate equivalence \((r,\theta)=(-r,\theta+\pi)\): a negative radius places the point in the direction opposite to the displayed angle.
Petal Counts and Directions
For the standard equations with \(a>0\), the number of distinct petals follows a parity rule. The direction formulas below give the angle of each petal’s tip, measured from the positive horizontal axis. Angles differing by \(2\pi\) represent the same direction.
| Equation | Parity of \(n\) | Number of petals | Petal directions |
|---|---|---|---|
| \(r=a\cos(n\theta)\) | Odd | \(n\) | \(\theta=2k\pi/n\), for \(k=0,\ldots,n-1\) |
| \(r=a\cos(n\theta)\) | Even | \(2n\) | \(\theta=k\pi/n\), for \(k=0,\ldots,2n-1\) |
| \(r=a\sin(n\theta)\) | Odd | \(n\) | \(\theta=(\pi/2+2k\pi)/n\), for \(k=0,\ldots,n-1\) |
| \(r=a\sin(n\theta)\) | Even | \(2n\) | \(\theta=(2k+1)\pi/(2n)\), for \(k=0,\ldots,2n-1\) |
These formulas follow by finding where the radial function has a positive maximum or a negative minimum. A positive maximum \(r=a\) gives a tip in the direction \(\theta\). A negative minimum \(r=-a\) gives a tip in direction \(\theta+\pi\), after converting to a positive radius. In particular, maxima of cosine occur when \(n\theta=2k\pi\), while maxima of sine occur when \(n\theta=\pi/2+2k\pi\). The negative extrema must also be included, with their directions shifted by \(\pi\).
Why the Parity Rule Holds
The zeros divide the angle parameter into intervals on which the curve runs from the pole out to a tip and back to the pole. In one full turn, each function has \(2n\) zeros, and the intervals between consecutive zeros give \(2n\) radial arches. Whether these arches represent \(2n\) distinct petals or include pairs tracing the same petal depends on the parity of \(n\).
For either sine or cosine, \(r(\theta+\pi)=(-1)^n r(\theta)\).
Then \(r(\theta+\pi)=-r(\theta)\). The points \((r(\theta),\theta)\) and \((r(\theta+\pi),\theta+\pi)\) are equivalent polar coordinates, so the arches pair up. Thus \(2n\) arches give \(n\) petals.
Then \(r(\theta+\pi)=r(\theta)\). The angle has changed by \(\pi\), so these arches lie in opposite directions rather than tracing the same petal. All \(2n\) arches give distinct petals.
To find the radial length, use \(|\sin(n\theta)|\leq 1\) and \(|\cos(n\theta)|\leq 1\). Therefore \(|r|\leq a\), and equality occurs at each radial maximum or minimum. The tip is distance \(a\) from the pole, so every petal has radial length \(a\).
Sketching a Cosine Rose with Odd \(n\)
Worked Example: \(r=3\cos(3\theta)\)
Here \(a=3\) and \(n=3\), which is odd. The petal count is \(n=3\), and each petal’s radial length is \(3\).
For positive tips, set \(\cos(3\theta)=1\): \(3\theta=2k\pi\), so \(\theta=2k\pi/3\). Taking \(k=0,1,2\) gives the directions \(0,\ 2\pi/3,\ 4\pi/3\).
Check how a negative extremum fits the same set of tips. At \(\theta=\pi/3\), \(r=3\cos(\pi)=-3\). Its equivalent positive-radius coordinate is \((3,\pi/3+\pi)=(3,4\pi/3)\), one of the three directions already listed.
The zeros satisfy \(\cos(3\theta)=0\), so \(3\theta=\pi/2+k\pi\), giving \(\theta=\pi/6+k\pi/3\). For example, on the interval from \(0\) to \(\pi/3\), the graph reaches the pole at \(\pi/6\); the positive-radius portion first extends to the tip at angle \(0\), and the negative-radius portion extends to the tip represented at \(4\pi/3\). These key values confirm the three-petal orientation and the tip distance.
Check: Substituting the tip direction \(\theta=0\) gives \(r=3\cos(0)=3\), so its distance from the pole is \(3\), as predicted.
Sketching an Even Sine Rose
Worked Example: \(r=2\sin(4\theta)\)
Here \(a=2\) and \(n=4\), which is even. The curve therefore has \(2n=8\) petals, each of radial length \(2\).
Positive tips occur when \(\sin(4\theta)=1\). Thus \(4\theta=\pi/2+2k\pi\), or \(\theta=\pi/8+k\pi/2\). For \(k=0,1,2,3\), these directions are \(\pi/8,\ 5\pi/8,\ 9\pi/8,\ 13\pi/8\).
Negative tips occur when \(\sin(4\theta)=-1\), so \(4\theta=3\pi/2+2k\pi\), or \(\theta=3\pi/8+k\pi/2\). Each negative radius points in direction \(\theta+\pi\). The combined set of positive-tip directions and converted negative-tip directions is \((2k+1)\pi/8\), for \(k=0,\ldots,7\). This gives eight equally spaced directions.
The zeros satisfy \(4\theta=k\pi\), so \(\theta=k\pi/4\). In particular, \(r=0\) at \(\theta=0\) and \(\theta=\pi/4\), while \(r=2\) at \(\theta=\pi/8\). These values show one arch running from the pole to a tip and back to the pole between those consecutive zeros.
Check: At \(\theta=\pi/8\), \(r=2\sin(\pi/2)=2\). At \(\theta=3\pi/8\), \(r=2\sin(3\pi/2)=-2\), which is represented by a positive radius in direction \(3\pi/8+\pi=11\pi/8\). Both tip distances are \(2\).
Sketching an Even Cosine Rose
Worked Example: \(r=\frac{3}{2}\cos(2\theta)\)
Here \(a=3/2\) and \(n=2\), so there are \(2n=4\) petals. The radial length of each is \(3/2\).
For even \(n\), the cosine tip directions are \(\theta=k\pi/n\). With \(n=2\) and \(k=0,1,2,3\), the directions are \(0,\ \pi/2,\ \pi,\ 3\pi/2\).
The zeros satisfy \(\cos(2\theta)=0\), so \(2\theta=\pi/2+k\pi\), giving \(\theta=\pi/4+k\pi/2\). These zeros divide the graph into radial arches. At \(\theta=0\), \(r=3/2\), and at \(\theta=\pi/2\), \(r=-3/2\). The latter is equivalent to a positive radius \(3/2\) at direction \(3\pi/2\).
Check: At each listed tip direction, the absolute value of \(\cos(2\theta)\) is \(1\), so \(|r|=(3/2)\cdot1=3/2\). The four directions and the tip distances agree with the count and length rules.
A Common Petal-Count Mistake
A frequent error is to assume that the number of petals is always \(n\), because \(n\) appears inside the sine or cosine. For example, the incorrect calculation “\(r=2\sin(4\theta)\) has \(4\) petals because \(n=4\)” overlooks the parity rule. Since \(4\) is even, the correct count is \(2n=8\).
Another source of confusion is treating a negative radial value as if it pointed in the angle’s stated direction. For instance, \(r=-2\) at \(\theta=\pi/3\) is not a tip in direction \(\pi/3\); it is the point \((2,4\pi/3)\). Convert negative-radius tips before listing directions or drawing the graph.
Check Your Understanding
Use the parity rule, tip directions, and radial length to answer these questions.
- How many petals does \(r=5\cos(5\theta)\) have, and what is the radial length of each?
- Find the directions of the petals of \(r=3\cos(2\theta)\).
- How many petals does \(r=4\sin(3\theta)\) have, and in which directions do its tips lie?
- For \(r=2\sin(6\theta)\), determine the petal count and write the general formula for its tip directions.
- At \(\theta=\pi/4\), the equation \(r=3\cos(2\theta)\) gives a negative radius. What positive-radius coordinate represents the same point?