How to draw a polar curve without converting it to rectangular form. Establishes the Fundamental Graphing Principle for polar equations and the two shapes given by a constant coordinate, then develops the book's two-plane method: sketch r against theta on ordinary axes and read that sketch as instructions for the polar plane. Applies it to limacons, a four-petalled rose, and a lemniscate, with the negative-r intervals handled explicitly throughout.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.5 Graphs of Polar Equations, pp. 938-951
Objectives
Five outcomes, and the second is the one that makes every later graph possible.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 938-951 — the pages these objectives are drawn from
Warm-up
In rectangular coordinates, a point is on a graph when its coordinates satisfy the equation. That statement needs care here.
Discussion prompt
The point named (-2, pi) also has the name (2, 0). Is it on the graph of the equation r equals 2?
Hint: Does it have a name that satisfies the equation?
Answer:
Yes. The name 2 comma 0 satisfies r equals 2, so the point is on the graph. That the name negative 2 comma pi does not satisfy the equation is irrelevant.
So the principle has to read: a point is on the graph if some representation of it satisfies the equation, not if every one does. That is a genuinely weaker condition.
This is the price of the non-uniqueness from the last two lessons, and it will matter most when finding where two polar curves intersect — a problem the next lesson takes up.
Concept
Sketching a polar curve directly is hard. Sketching r as an ordinary function of theta on rectangular axes is easy, and that sketch says exactly how far out the curve is at each angle.
Fundamental Graphing Principle for Polar Equations — A point P is on the graph of a polar equation if and only if there is some representation of P whose coordinates satisfy the equation.
\[ r = f(\theta): \quad \text{sketch on } \theta r\text{-axes}, \; \text{then transfer} \]
The book is explicit that this is less precise than plotting points and markedly faster, and that working through each curve this way prepares you for calculus better than memorising families.
Figure (svg): The two-plane method: r plotted against theta on rectangular axes on the left, and the resulting polar curve on the right, with matching arrows
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 938-941
Section
Section 1
Concept
If only one of r and theta appears in the equation, the other is free. Each case gives one of the two shapes the polar system describes most naturally.
Theorem 11.8 — For constants a and alpha with a not zero: the graph of r equal to a is a circle centred at the origin with radius the absolute value of a, and the graph of theta equal to alpha is the whole line containing the terminal side of alpha.
\[ r = a \;\Longrightarrow\; \text{circle}, \qquad \theta = \alpha \;\Longrightarrow\; \text{line} \]
The line case is the one students get wrong. Restricting r to be positive would give only a ray, but nothing in the equation restricts r, so both halves of the line are included.
Figure (svg): The two simplest polar equations graphed side by side: a constant r giving a circle centred at the pole, and a constant theta giving a line through the pole
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 938-940
Picture it
Fixing the distance gives a circle; fixing the direction gives a line.
Figure (svg): The two simplest polar equations graphed side by side: a constant r giving a circle centred at the pole, and a constant theta giving a line through the pole
These are the two shapes for which polar form is unbeatable, and both are one symbol long.
Worked example
Example 11.5.1, part 2. The sign does not survive into the picture.
\[ \text{Graph } r = -3\sqrt{2}. \]
Note which variable is free
Why: Theta does not appear, so it takes every value.
Describe the points
Why: Every point named with first coordinate negative three root two.
Find their distance from the pole
Why: The distance is the absolute value.
\[ 3 \sqrt{2} \]
Name the shape
Why: All points at a fixed distance from the pole.
Figure (svg): The solution to Worked example a negative constant radius shown as a ladder of expressions, one row per legal move
\[ \text{a circle of radius } 3\sqrt{2} \approx 4.24, \text{ centred at the pole} \]
Verify: compare with the positive version
Why: The equation r equals positive three root two describes the same circle, since each point of one has a name satisfying the other. Different equations, same set of points — exactly what the Fundamental Graphing Principle allows.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 938-939
Matching
One constant, two possible shapes.
Match the pairs
Why: Both pairs collapse to a single shape. The two radii differ only in sign, and the absolute value is what sets the circle. The two angles differ by pi, and since a line contains the terminal sides of both an angle and its opposite, the graphs coincide.
Worked example
Example 11.5.1, part 4. An angle that names an axis.
\[ \text{Graph } \theta = -\tfrac{3\pi}{2}. \]
Note which variable is free
Why: R does not appear, so it takes every value.
Identify the terminal side
Why: Negative three halves of pi is coterminal with pi over 2.
Include the negative r values
Why: Those give the opposite half.
Name the shape
Why: Both halves plus the pole.
Figure (svg): The solution to Worked example a constant angle shown as a ladder of expressions, one row per legal move
\[ \theta = -\tfrac{3\pi}{2} \;\Longrightarrow\; \text{the whole } y\text{-axis} \]
Verify: check the three parts
Why: Positive r gives the upper half, negative r the lower half, and r equal to zero the origin. Together those are every point of the y-axis and nothing else.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 939-940
Trap
\[ \theta = \tfrac{5\pi}{4} \;\Longrightarrow\; \text{a ray from the pole into Quadrant III} \]
Draw the terminal side of the angle
Why: That is what the angle picks out, so it should be the graph.
But the terminal side is a ray, and the equation places no restriction on r. Negative values of r are permitted and they produce the opposite half.
Let r run over every real number. Positive r gives the third-quadrant half, negative r the first-quadrant half, and zero gives the pole.
So the graph is the full line containing the terminal side, which is what Theorem 11.8 says.
The habit worth building: when a variable is absent from an equation, it is free, and free means every value including the negative ones. This same reading is what made theta equals negative pi over 4 the whole line y equals negative x in the last lesson.
Prediction
You are asked to graph theta equal to pi over 6 with the restriction r at least 0.
Predict first
How does the graph differ from the unrestricted one?
Correct: It is a ray rather than a line.
Why: Without the restriction, negative r values produce the half of the line pointing into quadrant three, so the graph is the whole line. Restricting r to be non-negative deletes that half, leaving only the terminal side itself — a ray from the pole into quadrant one. This is exactly why the restriction has to be stated when it is intended.
Sorting
Theorem 11.8 covers only the constant cases.
Sort into buckets
Sort each equation.
Socratic
The rectangular principle says a point is on the graph when its coordinates satisfy the equation.
Discussion prompt
Why must the polar version say some representation rather than the representation?
Hint: How many representations does a point have?
Answer:
Because a point has infinitely many representations, and there is no reason for all of them to satisfy a given equation. The point named 2 comma 0 is also named negative 2 comma pi, and only the first satisfies r equals 2.
If the principle demanded that every representation satisfy the equation, almost no point would be on almost any graph — the definition would be useless.
The consequence to carry forward: two different equations can have exactly the same graph, as r equals 3 and r equals negative 3 do. Sets of ordered pairs and sets of points in the plane are no longer the same thing, and it is the points that a graph is made of.
Section
Section 2
Concept
Plot r as an ordinary function of theta on rectangular axes. That sketch tells you, for each angle, how far out the curve is and whether it is moving in or out — which is exactly the information needed to draw it in the polar plane.
Plotting a table of points is the alternative and it is worse: for r equal to 6 cosine theta, nine ordered pairs gave only four distinct points, because different angles kept naming the same place.
Figure (svg): The two-plane method: r plotted against theta on rectangular axes on the left, and the resulting polar curve on the right, with matching arrows
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 940-942
Picture it
The arrows on the left become radial distances on the right.
Figure (svg): The two-plane method: r plotted against theta on rectangular axes on the left, and the resulting polar curve on the right, with matching arrows
The left-hand picture looks exactly like an ordinary cosine graph, and for good reason: at that stage the relationship between r and theta has not yet been interpreted as coordinates.
Worked example
The book's first full application of the method.
\[ \text{Graph } r = 6\cos(\theta). \]
Sketch r against theta
Why: One cycle of an ordinary cosine, amplitude 6.
Read the first quarter
Why: As theta goes 0 to pi over 2, r falls from 6 to 0.
Read the second quarter
Why: As theta goes pi over 2 to pi, r falls from 0 to -6.
Notice the retracing
Why: Beyond theta equal to pi the curve repeats itself.
\[ \text{complete on } [0, \pi] \]
Figure (svg): The solution to Worked example a circle through the pole shown as a ladder of expressions, one row per legal move
\[ r = 6\cos(\theta) \;\Longrightarrow\; \text{circle of radius } 3 \text{ centred at } (3, 0) \]
Verify: check against the conversion
Why: Multiplying by r gives r squared equal to 6 r cosine theta, so x squared plus y squared equal to 6x, which completes the square to x minus 3 squared plus y squared equal to 9. That is the circle described, confirming the sketch.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 940-942
Prediction
On the theta-r plane, r decreases from 4 to 0 as theta runs from 0 to pi over 2.
Predict first
What does the polar curve do over that interval?
Correct: Spirals in from the polar axis to the pole.
Why: The angle sweeps counter-clockwise from the polar axis to the positive y-axis while the distance shrinks from 4 to 0. So the curve starts 4 units out on the polar axis and arrives at the pole just as the angle reaches pi over 2. Reading a falling r as moving inward is the single translation the whole method rests on.
Worked example
The same curve, attempted by plotting.
\[ \text{Tabulate } r = 6\cos(\theta) \text{ at the eight standard angles and count the distinct points.} \]
Evaluate at the first four angles
Why: Six, three root two, zero, negative three root two.
Evaluate at the rest
Why: Negative six, negative three root two, zero, three root two.
Plot them
Why: Nine ordered pairs in all, counting 2 pi.
Count distinct locations
Why: Many pairs name the same place.
Figure (svg): The solution to Worked example why the table of points fails shown as a ladder of expressions, one row per legal move
\[ 9 \text{ ordered pairs} \;\longrightarrow\; 4 \text{ distinct points} \]
Verify: identify a collision
Why: The pair 3 root 2 comma pi over 4 and the pair negative 3 root 2 comma 5 pi over 4 name the same point, since the r values are opposite and the angles differ by pi. Four such collisions account for the shortfall, and no amount of extra table rows fixes the underlying problem.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 940-941
Error analysis
A student graphs a polar curve from a table, gets a poor picture, and doubles the number of rows.
Annotate
On: \( 8 \text{ rows} \;\longrightarrow\; 16 \text{ rows} \;\longrightarrow\; \text{still a poor picture} \)
The failure is not one of effort. A table answers where, and a polar curve needs where and in what order and going which way. The theta-r sketch answers all three at once.
Faded example
For r equal to 6 cosine theta, describe the interval from pi over 2 to pi.
Fill in the blanks
Over that interval r falls from 0 to -6, so the curve is drawn in Quadrant IV rather than the quadrant the angle points to.
Why: The cosine is negative on that interval, reaching negative 6 at theta equal to pi. Since the angle is sweeping through quadrant two and r is negative, the curve appears in the diagonally opposite quadrant, which is four.
Sorting
Translate each feature of the theta-r sketch.
Sort into buckets
Sort each feature by what it does on the polar plane.
Edge cases
The method says to keep going until the curve retraces itself.
Discussion prompt
How do you recognise retracing, and what happens if you stop too early or too late?
Hint: What would you see on the theta-r sketch?
Answer:
Retracing shows up as the polar curve arriving back at a point it has already drawn, moving in the same direction. On the theta-r sketch it often corresponds to the function repeating, but not always, which is the trap.
Stopping too early leaves an incomplete curve — the standard error with roses, where half a turn draws only half the petals. Stopping too late costs nothing but time, since the extra angles simply redraw what is there.
So when unsure, go further rather than less. The asymmetry of the two errors is worth exploiting: one produces a wrong answer and the other only wasted effort.
Section
Section 3
Concept
Where the theta-r sketch dips below the horizontal axis, r is negative. The angle still sweeps where it sweeps, but the curve is drawn half a turn away, in the diagonally opposite quadrant.
The book notes that the curve hugs the line theta equals the zero angle as it approaches the pole. That is the tangents at the pole result from calculus, quoted here for the drawing rather than proved.
Figure (svg): The portion of a polar curve where r is negative, shown being drawn in the quadrant opposite the one the angle points to
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 941-945
Picture it
One translation, applied wherever the sketch dips.
Figure (svg): The portion of a polar curve where r is negative, shown being drawn in the quadrant opposite the one the angle points to
Getting this backwards produces a curve that looks plausible and is wrong, which is why it is worth marking the negative intervals on the sketch before transferring anything.
Worked example
Example 11.5.2, part 2. The angles where the curve meets the pole matter first.
\[ \text{Graph } r = 2 + 4\cos(\theta). \]
Find where r is zero
Why: Set the expression to zero and solve for theta.
\[ \cos = -\frac{1}{2} \]
Solve on a full turn
Why: The two solutions.
\[ \theta = 2 \pi / 3, 4 \pi / 3 \]
Split the interval at those angles
Why: Together with the quadrant boundaries, six subintervals.
Transfer each, watching the sign
Why: The negative interval draws the inner loop.
Figure (svg): The solution to Worked example a limacon with an inner loop shown as a ladder of expressions, one row per legal move
\[ r = 0 \text{ at } \theta = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}; \text{ the interval between them gives the inner loop} \]
Verify: check the extreme values
Why: The maximum of r is 6, at theta equal to 0, and the minimum is negative 2, at theta equal to pi. The outer curve reaches 6 units out along the polar axis, and the inner loop reaches 2 units out — in the direction opposite pi, which is along the positive x-axis. That matches the picture.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 944-947
Faded example
For r equal to 3 plus 5 cosine theta, find where the curve meets the pole.
Fill in the blanks
3 + 5\cos\theta = 0 \;\Longrightarrow\; \cos\theta = -3/5 \;\Longrightarrow\; \theta = \arccos(2.21) \approx ___ \text___
Why: The cosine equals negative three fifths, whose arccosine is about 2.21 radians. The second solution on a full turn is 2 pi minus that, about 4.07. Since the coefficient 5 exceeds the constant 3, solutions exist and the limacon has an inner loop.
Worked example
Example 11.5.2, part 1. The same family, a different comparison.
\[ \text{Graph } r = 4 - 2\sin(\theta). \]
Ask whether r is ever zero
Why: Set the expression to zero.
\[ \sin = 2 \]
Note it has no solution
Why: The sine never exceeds 1.
Conclude the curve misses the pole
Why: R stays positive throughout.
\[ r\text{ between } 2\text{ and } 6 \]
Transfer the four quarters
Why: R runs 4 down to 2, back to 4, up to 6, back to 4.
Figure (svg): The solution to Worked example a limacon without one shown as a ladder of expressions, one row per legal move
\[ 2 \le r \le 6 \text{ throughout} \;\Longrightarrow\; \text{no inner loop, and the pole is never reached} \]
Verify: compare the coefficients
Why: The constant is 4 and the coefficient of the sine is 2. Since 4 exceeds 2, the expression can never reach zero, so the curve cannot reach the pole. The comparison predicted the shape before any sketching.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 942-944
Trap
\[ \theta \in \left[\tfrac{\pi}{2}, \pi\right], \; r < 0 \;\Longrightarrow\; \text{draw in Quadrant II} \]
Draw the arc in the quadrant the angle sweeps through
Why: The angle is in quadrant two, so the curve should be too.
But r is negative there, which is a half turn. The curve belongs in quadrant four, and drawing it in quadrant two produces a shape that is a reflection of the truth.
Negative r sends the arc to the diagonally opposite quadrant. Angle in quadrant two with negative r means the curve is drawn in quadrant four.
A useful habit: on the theta-r sketch, shade the intervals below the axis before transferring anything. Then the flip is impossible to forget.
The check afterwards: count how many times the curve crosses each axis and compare against the zeros of r. They must agree, and they will not if a negative arc was flipped the wrong way.
Prediction
A limacon has the form r equal to a plus b cosine theta with a and b positive.
Predict first
When does it have an inner loop?
Correct: When b is greater than a.
Why: An inner loop requires r to become negative, which requires the expression to reach below zero. The cosine bottoms out at negative 1, so the minimum of r is a minus b. That is negative exactly when b exceeds a. If a exceeds b, r stays positive and the curve never even reaches the pole; if they are equal, r touches zero once and the curve is a cardioid with a cusp rather than a loop.
Sorting
Compare the constant against the coefficient.
Sort into buckets
Sort each limacon.
Explain it to yourself
Two intervals of the theta-r sketch lie below the axis for a limacon with a loop.
Discussion prompt
Explain how those two intervals produce a single closed loop rather than two separate arcs.
Hint: What happens at the two angles where r is zero?
Answer:
There is really one interval, not two — the one strictly between the two zeros of r. The curve enters the pole at the first zero, is drawn in the opposite quadrants throughout that interval, and returns to the pole at the second zero.
Because it starts and ends at the pole, the arc closes on itself. That is precisely what makes it a loop rather than an open arc.
And it sits inside the rest of the curve because the absolute value of r there is smaller than elsewhere — the largest the loop gets is the absolute value of a minus b, which is less than the outer curve's maximum of a plus b. The two zeros of r are what create the loop, and their separation is what sets its size.
Section
Section 4
Concept
For a rose the function repeats after a fraction of a turn, but the curve does not. The interval required to complete a polar graph has to be discovered, not inferred from the period.
The book states this explicitly and it is one of the most useful cautions in the section: the temptation to graph over one period of the function and stop is strong, and it is wrong about as often as it is right.
Figure (svg): A four-petalled rose traced by five times the sine of twice theta, with the interval of theta needed to complete it marked as a full turn
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 947-950
Picture it
Two petals come from the positive half of the function and two from the negative half.
Figure (svg): A four-petalled rose traced by five times the sine of twice theta, with the interval of theta needed to complete it marked as a full turn
Graphing only over a period of the function would have produced two petals and a confident wrong answer.
Worked example
Example 11.5.2, part 3. Following the book's four subintervals.
\[ \text{Graph } r = 5\sin(2\theta). \]
Split the first half turn at the quarter points
Why: Four subintervals of width pi over 4.
\[ [0, \frac{\pi}{4}], [\frac{\pi}{4}, \frac{\pi}{2}],... \]
First quarter: r rises 0 to 5
Why: The curve sweeps out to 5 units along the line at pi over 4.
Second quarter: r falls 5 to 0
Why: The curve returns to the pole hugging the y-axis.
Third and fourth: r goes negative then back
Why: Those draw a petal in Quadrant IV, opposite the angle.
Figure (svg): The solution to Worked example a four-petalled rose shown as a ladder of expressions, one row per legal move
\[ r = 5\sin(2\theta): \text{ four petals, needing } 0 \le \theta \le 2\pi \]
Verify: count the petals against the zeros
Why: The function is zero at 0, pi over 2, pi, 3 pi over 2 and 2 pi, so the curve visits the pole five times over a full turn, bounding four petals. Four zeros between visits, four petals — consistent.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 947-949
Prediction
Consider the rose r equal to a cosine of three theta.
Predict first
How many petals does it have?
Correct: Three.
Why: With an odd multiplier the curve completes in half a turn, and the second half turn retraces it — the negative arcs land exactly on the petals already drawn. With an even multiplier the negative arcs fall in the empty gaps and double the count. So an odd n gives n petals and an even n gives 2n, which is why sine of two theta gave four. The value of a sets only the petal length.
Worked example
Two curves where the relationship between period and required interval runs opposite ways.
\[ \text{Compare the interval needed for } r = 6\cos(\theta) \text{ and } r = 5\sin(2\theta). \]
Period of the first
Why: The cosine has period 2 pi.
\[ \text{period } 2 \pi \]
Interval needed for the first
Why: The circle is complete after half a turn.
Period of the second
Why: The sine of twice theta has period pi.
Interval needed for the second
Why: All four petals need a full turn.
\[ \text{needs } 2 \pi \]
Figure (svg): The solution to Worked example comparing the two intervals shown as a ladder of expressions, one row per legal move
\[ 6\cos\theta: \; \text{period } 2\pi, \text{ needs } \pi. \qquad 5\sin 2\theta: \; \text{period } \pi, \text{ needs } 2\pi. \]
Verify: explain each
Why: The circle finishes early because the second half turn retraces it with negative r values naming the same points. The rose finishes late because the negative half of the function draws two genuinely new petals. Both effects come from negative r, pulling in opposite directions.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 950-950
Error analysis
A student graphs the rose over one period of the function.
Annotate
On: \( r = 5\sin(2\theta), \; \text{period } \pi \;\Longrightarrow\; \text{graph over } [0, \pi] \;\Longrightarrow\; \text{two petals} \)
The period governs the function; the graph is governed by where the points land, and negative r values relocate them. The only reliable stopping rule is the appearance of retracing.
Fill the middle
Find the angles at which r equal to 5 sine of two theta meets the pole, over a full turn.
Fill in the blanks
5\sin(2\theta) = 0 \;\Longrightarrow\; 2\theta = \pi k \;\Longrightarrow\; \theta = pi k / 2 \text4 k = 0, 1, 2, 3, 4 \;\Longrightarrow\; ___ \text___
Why: The sine is zero when its argument is a multiple of pi, so theta is a multiple of pi over 2. Over a full turn that gives five visits to the pole at 0, pi over 2, pi, 3 pi over 2 and 2 pi, and the four intervals between them are the four petals.
Sorting
Compare the period against what the curve actually needs.
Sort into buckets
Sort each curve.
Counterexample
A student proposes: a polar curve is complete once theta has run through one period of the function.
Discussion prompt
Give a counterexample in each direction, and state what the correct stopping rule is.
Hint: One curve finishes early and one finishes late.
Answer:
Finishing early: r equals 6 cosine theta has period 2 pi, but the circle is complete after theta reaches pi. The second half turn produces negative r values that name points already drawn.
Finishing late: r equals 5 sine of two theta has period pi, but two of its four petals are drawn only after theta passes pi. The period is half of what the curve needs.
So the period is neither an upper nor a lower bound, and the rule is simply: continue until the curve begins to retrace, then stop. The book makes exactly this point, adding that there is in general no relation between the two intervals. Where a general rule does not exist, a check has to take its place, and that is the honest state of affairs here.
Section
Section 5
Concept
An equation giving r squared has to be solved for r, producing a plus-or-minus pair. Where the right side is negative there is no real r at all, and the curve simply does not exist for those angles.
lemniscate — The figure-eight curve given by r squared equal to a squared cosine of twice theta. It exists only on the intervals where the cosine of twice theta is non-negative, and it is empty elsewhere.
\[ r^2 = 16\cos(2\theta) \;\Longrightarrow\; r = \pm 4\sqrt{\cos(2\theta)} \]
For the lemniscate, the lines at pi over 4 and 3 pi over 4 are the zeros, and they serve as the guides the curve follows through the pole.
Figure (svg): The lemniscate given by r squared equal to sixteen cosine of twice theta, with the intervals where the right side is negative shown as gaps in the curve
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 949-950
Picture it
Between the dashed lines, the equation has no real solutions.
Figure (svg): The lemniscate given by r squared equal to sixteen cosine of twice theta, with the intervals where the right side is negative shown as gaps in the curve
The empty sectors are where the cosine of twice theta is negative, so the square root is undefined. There is nothing to draw there, and drawing something is the error.
Worked example
Example 11.5.2, part 4. The full analysis.
\[ \text{Graph } r^2 = 16\cos(2\theta). \]
Solve for r
Why: Take the square root of both sides, keeping both signs.
Find where the radicand is non-negative
Why: The cosine of twice theta must be at least zero.
Translate to intervals for theta
Why: Divide by 2.
\[ \theta\text{ in } [-\frac{\pi}{4}, \frac{\pi}{4}]\text{ and its opposite} \]
Note the zeros and transfer
Why: R is zero at pi over 4 and 3 pi over 4.
Figure (svg): The solution to Worked example the lemniscate shown as a ladder of expressions, one row per legal move
\[ r = \pm 4\sqrt{\cos(2\theta)}, \text{ defined only where } \cos(2\theta) \ge 0 \]
Verify: check the extreme point
Why: At theta equal to zero the cosine of twice theta is 1, so r is plus or minus 4. Those are the two tips of the figure-eight, four units out along the positive and negative x-axes. The widest points are on the axis the curve lies along, as the picture shows.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 949-950
Faded example
For r squared equal to 25 cosine of twice theta, find the first interval where the curve exists.
Fill in the blanks
\cos(2\theta) \ge 0 \;\Longrightarrow\; -\tfrac-pi/4pi/4 \le 2\theta \le \tfrac______ \;\Longrightarrow\; ___ \le \theta \le ___
Why: The cosine is non-negative between negative pi over 2 and pi over 2, and dividing by the 2 inside halves the interval to negative pi over 4 through pi over 4. That is one lobe of the lemniscate, centred on the polar axis.
Worked example
Finding where a curve fails to exist, before drawing anything.
\[ \text{For } r^2 = 9\sin(2\theta), \text{ find the angles contributing no points.} \]
Set the condition
Why: The right side must be non-negative for r to be real.
Solve the inequality
Why: The sine is non-negative on the first half of its period.
Divide by two
Why: The angles where the curve exists.
\[ \theta\text{ in } [0, \frac{\pi}{2}] \]
Add the second branch
Why: The next interval where the sine is non-negative.
\[ \theta\text{ in } [\pi, 3 \pi / 2] \]
Figure (svg): The solution to Worked example reading the gaps directly shown as a ladder of expressions, one row per legal move
\[ \text{empty for } \theta \in \left(\tfrac{\pi}{2}, \pi\right) \text{ and } \theta \in \left(\tfrac{3\pi}{2}, 2\pi\right) \]
Verify: check the total
Why: The curve exists on two intervals each of length pi over 2, so on half the full turn. A lemniscate has two lobes, one per interval, which matches. The lobes here lie along the line at pi over 4 rather than along an axis, because the sine peaks there.
Trap
\[ r^2 = 16\cos(2\theta) \;\Longrightarrow\; r = 4\sqrt{\cos(2\theta)} \quad \text{(one branch)} \]
Take the square root
Why: The square root symbol denotes the non-negative root, so one branch seems right.
But the equation is r squared equals something, and both signs satisfy it. Discarding the negative branch here would matter for any curve where the two branches differ.
\[ r = \pm 4\sqrt{\cos(2\theta)} \]
Keep both signs and graph both
Why: Then check whether one branch retraces the other.
For this particular lemniscate the two branches happen to coincide as sets of points, since a negative r at angle theta is a positive r at theta plus pi and the curve is symmetric. But that has to be checked, not assumed — the book notes it as a remark rather than building it into the method.
Prediction
You are graphing r squared equal to some expression, and at theta equal to pi over 3 the expression is negative.
Predict first
What should appear on the graph at that angle?
Correct: Nothing at all.
Why: For r squared to equal a negative number, r would have to be imaginary, and the polar plane contains no such point. So the curve simply has no point at that angle. This is genuinely different from r equal to zero, which does put a point on the graph — at the pole. A gap and a visit to the pole look nothing alike and must not be confused.
Matching
Three different things can happen at a given angle.
Match the pairs
Why: A positive right side gives two real roots, opposite in sign, which are two points diametrically opposite one another. Zero gives the single root zero, the pole. A negative right side gives no real root at all. The size of the positive value sets how far out the two points are.
Real world
The radiation pattern of a half-wave dipole antenna is a figure-eight in the plane containing the antenna: strong broadside, and with deep nulls off the ends.
Discussion prompt
Explain why the polar form of that pattern is the useful one, and what the nulls correspond to mathematically.
Hint: What is being plotted against what?
Answer:
The pattern plots field strength against direction, which is a radius against an angle — polar form is again the literal shape of the data rather than a convenient choice.
The nulls are the angles where the radius reaches zero, which are exactly the zeros of the function, the angles at which the curve passes through the pole. On a lemniscate those are the lines at pi over 4 and 3 pi over 4; on the dipole they are the two directions along the antenna's axis.
This is why a dipole is oriented the way it is: the nulls are pointed at directions you do not want to serve, and the broadside lobes at those you do. Reading a null off a polar plot is the same act as finding a zero of r, and antenna engineers do it constantly.
It also explains why the plots are drawn on polar grids in every datasheet. A rectangular plot of the same data would hide the geometry — the very thing the engineer is trying to see.
Comparison
Fill the blanks from memory. Each row is one translation from the left picture to the right one.
Comparison matrix
| On the theta-r sketch | On the polar plane | Watch for |
|---|---|---|
| r increasing | the curve moves away from the pole | the direction of sweep |
| r crossing zero | the curve passes through the pole | it hugs the line at that angle |
| r below the axis | draw in the opposite quadrant | the most common error |
| the sketch repeating | possible retracing | possible, not certain |
The third row is where marks are lost and the fourth is where whole petals are lost. Marking the negative intervals on the sketch before transferring handles the first of those.
Pattern
Five moves, and they work on a curve you have never seen before.
If the equation gives r squared, solve for r with both signs and find where the right side is negative. Those angles contribute no points at all.
OpenStax Algebra and Trigonometry 2e, §10.4 Polar Coordinates: Graphs §10.4
Check
The constant cases.
Check your understanding
What is the graph of theta equal to 7 pi over 6?
Answer: B
Why: With theta fixed, r is free and may take any real value including the negatives. Positive r gives the third-quadrant half of the line and negative r gives the first-quadrant half, so the graph is the entire line containing that terminal side.
Check
Limacons.
Check your understanding
Does the curve r equal to 3 minus 4 sine theta have an inner loop?
Answer: A
Why: The value of r reaches zero when the sine of theta equals three quarters, which has solutions since three quarters is between negative 1 and 1. Beyond those angles r goes negative, producing the loop. The comparison is between the size of the coefficient and the size of the constant, and 4 exceeds 3.
Check
How far theta must run.
Check your understanding
How many petals does r equal to 2 sine of 4 theta have?
Answer: B
Why: The multiplier 4 is even, and for an even multiplier the negative arcs fall in the gaps between the positive ones rather than on top of them, doubling the count. So there are eight petals, and theta must run a full turn to draw them all.
Real world
A loudspeaker's directivity is measured by rotating it and recording the sound pressure at each angle. The result is plotted on a polar grid. A design under test produces a plot with two large lobes forward, two smaller ones behind, and narrow directions where the level collapses.
Discussion prompt
Explain what the acoustician is reading off that plot, and which features of this lesson correspond to which acoustic properties.
Hint: What are the lobes, and what are the collapses?
Answer:
The plot is a polar curve, r equal to level as a function of angle. The lobes are the intervals where r is large, and they are the directions the speaker actually serves. The narrow collapses are the zeros of r, the angles where the curve passes through the pole.
The four-lobe structure is exactly a rose. The number of lobes and their angular width are determined by the same arithmetic as a rose's petal count — driver spacing plays the role the multiplier plays in the equation, and doubling it doubles the lobes.
The rear lobes are what the negative-r intervals produce. In the acoustic case they are real radiated sound, arriving half a turn from where the design intended, and they are usually unwanted — the same relocation that this lesson keeps flagging as the commonest drawing error.
What the acoustician does with the plot is read off the null angles and the lobe widths, and neither is legible in any other representation. The polar graph is the measurement, not an illustration of it — which is a fair summary of why this section exists.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
On the theta-r sketch, r is negative while theta sweeps through quadrant one. Where is the curve drawn?
Correct: Quadrant III.
\[ (-r, \theta) = (r, \theta + \pi) \;\Longrightarrow\; \text{diagonally opposite} \]
Why: A negative r is a half turn, so the curve appears diagonally opposite the quadrant the angle points to. Quadrant one and quadrant three are diagonally opposite, so that is where the arc is drawn. Quadrants two and four would be reflections rather than half turns, and drawing in quadrant one ignores the sign entirely.
Explain it
They graphed r equal to 5 sine of two theta by plotting twenty points from a table and got a shape with two petals and some stray dots.
Discussion prompt
In no more than five sentences, explain why the table let them down and what to do instead.
Hint: What does a table not tell you?
Answer:
A table tells you where the curve goes but not in what order or in which direction, and for a polar curve those matter as much as the locations. It also gives no warning about where the curve passes through the pole, which is where the petals start and end.
Instead, sketch r against theta on ordinary axes — that is just an ordinary sine graph and takes ten seconds. Then read it in pieces: where it rises the curve moves out, where it falls the curve moves in, and where it dips below the axis the arc is drawn in the opposite quadrant.
The two missing petals were exactly the intervals where the sketch was below the axis, which is why they landed somewhere unexpected.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The constant cases are fixed by Theorem 11.8 and by remembering that a constant angle gives a line, not a ray. Transferring is fixed by working one subinterval at a time and saying out loud whether r is rising or falling. The negative arcs are fixed by shading those intervals on the sketch before transferring. The interval is fixed by continuing until retracing appears rather than trusting the period. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page write Theorem 11.8 and draw its two graphs, a circle and a line, noting beside the line why it is not a ray. Underneath, draw a pair of pictures side by side for the curve r equal to 2 plus 4 cosine theta: on the left the sketch of r against theta over a full turn with the two zeros marked and the negative interval shaded, and on the right the polar curve with its inner loop, labelling which arc came from which subinterval. In the bottom left, write the comparison that predicts whether a limacon has a loop, a cusp, or neither, with one example of each. In the bottom right, write the petal-count rule for roses and note the two curves from this lesson whose required interval was shorter and longer than the function's period. Finally, circle the one translation from the left picture to the right that you are most likely to get backwards.
Almost everyone circles the negative-r rule, and rightly. Every other translation preserves the quadrant; that one moves the arc diagonally across the plane, and nothing in the sketch reminds you unless you shade it first.
Recap
Five things, and the third is where the marks are.
| If the question says | Your first move is |
|---|---|
| Only r appears, or only theta | Apply Theorem 11.8; no work needed |
| Graph this polar equation | Sketch r against theta on ordinary axes |
| Where does it meet the pole? | Set r to zero and solve for theta |
| Does the limacon have a loop? | Compare the coefficient against the constant |
| The equation gives r squared | Solve with both signs, then find where it is negative |
Two curves can now be drawn individually. The next lesson asks where two of them meet, and the non-uniqueness of polar names makes that a genuinely harder question than solving the two equations simultaneously.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 938-951 — everything on these slides traces back here
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