Why an Intersection Is Not Always Enough
When two polar curves intersect, the intersection angles are natural places to divide the angle range into intervals. On each interval, we often want to know which curve is farther from the pole. That comparison matters when interpreting the geometry of the curves and setting up a radial comparison.
A sample angle can answer the question on an entire interval, provided the curves’ distances from the pole do not become equal anywhere inside that interval. For signed polar radii, however, equal distances can occur even when the curves do not intersect. It is therefore important to compare absolute radii and to check whether the comparison can change between the intersections.
Distance from the Pole and “Outer”
A polar curve given by \(r=f(\theta)\) assigns a signed radius to each angle. A positive radius places the point in the direction \(\theta\); a negative radius places it in the opposite direction. In either case, the distance from the pole is nonnegative.
When both radii are nonnegative, the points lie on the same ray from the pole. In that case, the curve with the larger radius is also the outer radial boundary along that ray. If either radius is negative, comparing absolute radii still tells us which point is farther from the pole, but the two points might lie on opposite rays. Thus “farther from the pole” and “outer boundary along a particular ray” are not automatically the same claim.
Using a Sample Angle
Suppose the curves meet at the ends of an open angular interval and neither curve has an equal-distance point inside it. Since the radius functions are continuous, their absolute values are continuous as well. The difference between those absolute values cannot switch from positive to negative without passing through zero. Consequently, one sample angle determines the distance ordering throughout the interval.
If \(D\) is positive at a sample angle and never equals zero on the interval, then it is positive throughout that interval. If it is negative at the sample angle and never equals zero, then it is negative throughout. A zero of \(D\) marks a place where the two curves are equally far from the pole.
Worked Example: Comparing Two Positive Radii
Worked Example: Comparing Two Positive Radii
Compare \(r_1=2+\cos\theta\) and \(r_2=2-\cos\theta\) over one revolution. Find their intersections and determine which curve is farther from the pole between them.
First solve the equality of the radii:
On \(0\leq\theta\leq2\pi\), the intersection angles are \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\). Both radii are positive throughout this range, because \(-1\leq\cos\theta\leq1\) gives \(1\leq r_1,r_2\leq3\). Thus comparing the radii directly also compares their distances from the pole.
Test \(\theta=\frac{\pi}{4}\) in the first interval:
Since \(\frac{\sqrt{2}}{2}>0\), the first value is larger, so \(r_1\) is farther from the pole on \((0,\frac{\pi}{2})\). Test \(\theta=\frac{5\pi}{4}\) in the next interval:
Here \(r_2\) is larger, so it is farther from the pole on \((\frac{\pi}{2},\frac{3\pi}{2})\). The ordering switches at each intersection, where the radii are equal. At the end of the revolution, the comparison on \((\frac{3\pi}{2},2\pi)\) matches the first interval: for example, \(\cos(\frac{7\pi}{4})=\frac{\sqrt{2}}{2}\), so \(r_1>r_2\). This checks the ordering over all three intervals.
Why the Sample-Angle Test Works
Here is the precise continuity argument. It also identifies the condition a sample-angle test requires.
Set \(D(\theta)=|r_1(\theta)|-|r_2(\theta)|\). If the radius functions are continuous, then \(D\) is continuous.
The ordering can change only where \(D(\theta)=0\), which is where \(|r_1(\theta)|=|r_2(\theta)|\). This holds exactly when \(r_1=r_2\) or \(r_1=-r_2\).
If \(D\) were positive at one angle and negative at another in an interval, continuity would force \(D\) to equal zero somewhere between them.
If all zeros of \(D\) have been excluded from the interval, its sign is constant there. One sample angle therefore establishes the distance ordering on the whole interval.
The key is to check zeros of the distance difference, not merely the curve intersections. In particular, a solution of \(r_1=-r_2\) can split an interval even though it does not represent a shared point of the two curves at that angle.
Worked Example: Equal Distance Without an Intersection
Worked Example: Equal Distance Without an Intersection
Compare \(r_1=2\cos\theta\) and \(r_2=1\) on the interval between their intersections \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\). Determine which is farther from the pole on each part of that interval.
The intersections occur when \(2\cos\theta=1\), or \(\cos\theta=\frac12\). On the stated interval, the solutions are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\). To find all possible changes in distance ordering, solve \(|2\cos\theta|=1\):
The first equation gives the intersections. The second gives \(\cos\theta=-\frac12\), with solutions \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\) inside the interval. At those extra angles, the absolute radii are equal, but the signed radii have opposite signs.
Use the three resulting subintervals. At \(\theta=\frac{\pi}{2}\), which lies in \((\frac{\pi}{3},\frac{2\pi}{3})\), the radii are \(r_1=0\) and \(r_2=1\). Thus \(r_2\) is farther. At \(\theta=\pi\), which lies in \((\frac{2\pi}{3},\frac{4\pi}{3})\), the radii are \(r_1=-2\) and \(r_2=1\), so \(|r_1|=2>|r_2|=1\); \(r_1\) is farther. At \(\theta=\frac{3\pi}{2}\), in \((\frac{4\pi}{3},\frac{5\pi}{3})\), the radii are \(r_1=0\) and \(r_2=1\), so \(r_2\) is farther again.
At \(\theta=\frac{2\pi}{3}\), \(r_1=-1\) and \(r_2=1\), and at \(\theta=\frac{4\pi}{3}\), the same pair of radii occurs. Each pair has equal absolute values. These are not intersections at those angles: the negative radius and the positive radius point in opposite directions. The check confirms that the distance ordering changes at these equal-distance angles, not just at the two intersections.
Worked Example: A Negative Radius and a False Comparison
Worked Example: A Negative Radius and a False Comparison
Compare \(r_1=1+2\cos\theta\) and \(r_2=1\) on \((\frac{\pi}{2},\frac{3\pi}{2})\). The curves meet at the endpoints because \(\cos\theta=0\) there. Determine which is farther from the pole inside the interval.
At \(\theta=\frac{2\pi}{3}\), \(\cos\theta=-\frac12\), so
At \(\theta=\pi\), \(\cos\theta=-1\), so
At \(\theta=\frac{4\pi}{3}\), the cosine is again \(-\frac12\), giving \(|r_1|=0<1=|r_2|\). To check whether the ordering changes, solve the equal-distance equation:
This requires \(1+2\cos\theta=1\) or \(1+2\cos\theta=-1\). The first gives \(\cos\theta=0\), the endpoint intersections; the second gives \(\cos\theta=-1\), or \(\theta=\pi\) in this interval. Thus the constant-radius curve \(r_2=1\) is farther throughout the open interval except at \(\theta=\pi\), where the two distances tie.
A common error is to say that \(r_2\) is farther at \(\theta=\pi\) because \(1>-1\). That compares signed radii, not distances. The correction is to compare \(|1|=1\) with \(|-1|=1\): the points are equally far from the pole.
A Reliable Comparison Procedure
For continuous polar curves, use the following procedure when deciding which is farther from the pole.
- Write the two radius functions as \(r_1(\theta)\) and \(r_2(\theta)\), and identify the angular interval of interest.
- Find the points in the interval where \(r_1=r_2\) and where \(r_1=-r_2\). These together identify possible equal-distance angles.
- Divide the interval at every such angle, including any endpoint needed to mark the boundaries.
- Choose one convenient sample angle in each resulting open subinterval.
- Evaluate both absolute radii at that angle and compare them. State any angles of equality separately.
- If the question concerns an outer boundary along a ray, also check that the compared points lie on that same ray; absolute-radius comparison alone establishes distance from the pole, not ray alignment.
When both radii are known to be nonnegative on an interval, the extra equation \(r_1=-r_2\) can only hold when both radii are zero. The comparison then often reduces to checking the intersections \(r_1=r_2\) and testing one angle between consecutive intersections. When signed radii occur, the additional equal-distance check is essential.
Why the Distinction Matters
A curve’s signed radius records direction as well as distance. Discarding the sign is appropriate for the question “Which point is farther from the pole?” but it does not preserve the point’s direction. Conversely, comparing signed radii can give the wrong answer to a distance question, as the last example shows.
The practical lesson is to state exactly what is being compared. For distance, use absolute radii. For a radial boundary on a specified ray, first ensure the points lie on that ray and then compare their nonnegative radial distances. Once the relevant equalities have been found, continuity justifies using a sample angle on each interval where the ordering cannot change.
Check Your Understanding
For each question, distinguish intersections from other angles at which the curves are equally far from the pole.
- Why is \(|r|\), rather than \(r\), the distance from the pole?
- For continuous radius functions, what must happen if the sign of \(|r_1(\theta)|-|r_2(\theta)|\) changes inside an interval?
- What two equations describe all possible equal-distance angles?
- How can \(r_1=-r_2\) give equal distances without giving an intersection at that angle?
- If both radii are positive, when does “farther from the pole” also identify the outer boundary along the same ray?