Construct complete signed-radius curves

Try

At angle θ=π, the rule r=1+2cosθ gives r=−1. Where is the point? The negative radius reverses the ray: x=r cosθ=1 and y=r sinθ=0. It is (1,0), not (−1,0). Its physical distance from the origin is 1, even though its signed radius is −1.

This lesson constructs complete noncircle curves, extending the earlier coordinate and polar-circle work. Always convert the pair together: (x,y)=(r cosθ,r sinθ). Angle alone does not determine the point’s half-plane when r can be negative. The rule and a specified angular domain describe a trace; a small point table illustrates it but does not prove every branch.

Angle θ	Signed r=1+2cosθ	Cartesian point
0	3	(3,0)
π/2	1	(0,1)
2π/3	0	(0,0)
π	−1	(1,0)
4π/3	0	(0,0)
3π/2	1	(0,−1)
2π	3	(3,0)

Predict

Predict which angular interval creates the small inner loop. Solve r=0 before drawing: cosθ=−1/2 gives θ=2π/3 and 4π/3 in 0≤θ≤2π. Radius is negative strictly between those zeros and positive on the remaining open portions. The negative interval starts and ends at the origin and passes through (1,0), so a complete construction must include it instead of deleting it as invalid.

At θ=5π/6, the radius is 1−√3. Since sinθ=1/2 is positive while r is negative, the Cartesian point is below the horizontal axis. Predict whether θ=7π/6 puts the corresponding inner point above or below it. The rule is symmetric under θ↦2π−θ: radius stays the same, x stays the same and y changes sign. Thus the two points are reflections across the horizontal axis.

Experiment

Use the full 0-to-2π trace of the limacon first. The initial active angle π gives signed radius −1 and Cartesian point (1,0). Reveal the trace, then choose angles on either side of π. Compare the angle ray, opposite-ray point, signed-radius value and origin distance. The graph panels are synchronized representations of the same rule, not separate measurements.

Trace signed polar coordinates

Predict the Cartesian point before revealing it. A polar point uses x=r cos θ and y=r sin θ. The signed radius r may be negative; distance from the origin is ∣r∣. The graph window must span at least 0.00001π radians so its labels remain distinguishable; this is a display limit, not a mathematical restriction. Angles below are entered as decimal multiples of π radians, so 1 means π, not one radian.

Curve familyLimaçon: r=1+2 cos θRose: r=2 cos(2θ)

Window start (θ/π, −2 to 2)Window end (θ/π, −2 to 2)Active angle (θ/π, inside window)

Reveal polar trace

Reset polar workbench

All controls are manual; nothing animates automatically. Editing a trace input clears the trace; editing an interpolation input clears that estimate. Reset restores every input and hides calculations. This temporary workbench does not save work, mark completion or award mastery. Retain your predictions, coordinate evidence and assumptions in the lesson response.

Restrict the angle window to [2π/3,4π/3] to isolate the inner loop; the tool’s angle controls use multiples of π, so enter approximately 0.666666666666667 and 1.333333333333333. Floating-point near-zero readouts do not decide exact endpoint signs: the algebra above establishes both zeros. Restrict further to [π,4π/3] to select the upper half of that loop. Although these input angles lie in the lower angular half-plane, their negative radii put the points above the axis.

Return to the full window and choose the rose r=2cos(2θ). Set Active angle to 0.5 and predict where the point at θ=π/2 goes. The radius is −2, so the point is (0,−2). Build the table of all four tips before accepting the complete picture below.

Understand

A full loop argument can use physical rays. On the negative-radius limacon interval, put θ=φ+π with −π/3≤φ≤π/3, allowing equivalent angle representatives. Then r=1−2cosφ≤0. Reversing the ray gives nonnegative physical radius ρinner=−r=2cosφ−1 at physical angle φ. It starts at zero when φ=−π/3, reaches 1 at φ=0 and returns to zero at φ=π/3. The outer curve on that same ray has radius ρouter=1+2cosφ, exactly 2 greater. Thus the traced negative-radius portion lies inside the corresponding outer-ray positions, not on a new curve to the left.

Within the nonzero part of this interval, distinct physical angles select distinct rays, so there is no unaccounted repeated arc. Continuity joins the two zero-radius endpoints at the origin. On the remaining parameter intervals the radius is positive; those portions connect the origin through the upper outer curve to (3,0), and through the lower outer curve back again. This uses signs, continuity and angular ranges to justify the complete trace; finite sampling alone is not that proof.

For the rose, zeros satisfy cos(2θ)=0, giving θ=π/4+kπ/2. Representative inputs for its four distinct tips are 0,π/2,π,3π/2; the endpoint 2π repeats the tip at 0. At the four representative inputs, their signed radii are 2,−2,2,−2 and Cartesian tips are (2,0),(0,−2),(−2,0),(0,2). Between consecutive zeros the radius has one sign. Each such interval traces from the origin to one tip and back, with a continuously turning physical ray. The right petal is split by the chosen domain boundary: [0,π/4] and [7π/4,2π] together give it. A contiguous alternative is [−π/4,π/4] if that extended angular domain is explicitly allowed. Do not claim that [0,π/2] traces one whole petal; it combines halves of two different petals.

Apply

For the limacon, find all θ in [0,2π] that place the point above or on the horizontal axis. Write the product y=(1+2cosθ)sinθ and make a sign chart using every zero of either factor. This tests whether your restriction follows the point rather than merely its nominal angle.

Check the complete angular restriction

The sign boundaries are 0,2π/3,π,4π/3,2π. The result is [0,2π/3] union [π,4π/3] union the single endpoint 2π. The second interval supplies the upper inner loop, while the first supplies the upper outer curve. The last endpoint repeats (3,0). On (2π/3,π), a positive sine and negative radius place the point below the axis. Every boundary point satisfies y=0, and no omitted open interval does.

Now construct the upper petal of r=2cos(2θ). Use [5π/4,7π/4], with the tip at 3π/2 where r=−2 and the point is (0,2). Explain how the same negative sign that placed the earlier π/2 tip below the axis now places this one above. Include both origin endpoints in your construction.

Continue

The workbench offers these two fixed formulas, not a general polar equation editor. Scaled or rotated curves in the written practice are manual constructions using exact coordinates and sign arguments.

Keep four records together: the radial rule, exact zero/sign intervals, Cartesian landmarks on equal scales, and the angular restriction for the requested portion. State when two input angles repeat a point and when a proposed window gives only part of a petal. Use polar coordinates for a coordinate refresher if needed.

Next, separate signed radius, distance and estimates. Moving toward more negative signed radius can mean moving farther from the origin, and an interpolation table does not specify a unique curve between its observations.