7.3 Polar Coordinates

1. Polar Coordinates

Title

Calculus II · Section 7.3

Naming a point by how far away it is and which way it lies

2. What this lesson gives you

Objectives

Every point so far has been named by a horizontal and a vertical distance. This lesson names the same points a second way, by a distance and a direction, and shows which curves become simple when you do.

• Locate a point from its distance r and angle θ, including a negative r
• Convert points between rectangular and polar coordinates, choosing the right quadrant
• Explain why every point has infinitely many polar names
• Convert equations in both directions and recognise lines, circles, limaçons, roses and spirals
• Test a polar equation for symmetry and use it to sketch the curve

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 642-661 — learning objectives 7.3.1 to 7.3.5

Rectangular coordinates describe a point by saying how far across and how far up it is. That is perfect for lines and parabolas, and awkward for anything that turns: circles about the origin, spirals, flowers of petals. Polar coordinates describe the same point by how far away it is and in which direction it lies.

You will learn to move points and equations back and forth between the two systems, and you will meet the one genuine difficulty of the section early: going from rectangular to polar, the inverse tangent gives an angle that is right only half the time, and you must decide the quadrant yourself.

The second half is about curves. A handful of polar equations produce a whole zoo of shapes, and three symmetry tests plus the rule for negative distances explain nearly everything strange about them.

3. Before anything new: how far and which way is (3, 4)?

Warm-up

Write your answer before revealing. You need nothing new here: the distance is the hypotenuse of a right triangle with legs 3 and 4, and the angle is the one whose tangent is the opposite leg over the adjacent leg.

The point of the exercise is the last sentence of the answer. The pair five and 0.927 locates the point exactly as well as the pair three and four. It is simply a different address for the same place, and it is the address this whole lesson is about.

Notice also that you used the inverse tangent without any trouble, because the point is in the first quadrant. Keep an eye on that. Later in the lesson the same calculation, done on a point in another quadrant, will give a wrong answer with total confidence.

4. Defining polar coordinates

Section

Part 1

5. How far, and which way

Concept

Join the origin to a point P. The segment has a length r, and it makes an angle θ with the positive x-axis. The legs of the right triangle are x and y.

\[ \cos\theta = \frac{x}{r} \;\Longrightarrow\; x = r\cos\theta \]

\[ \sin\theta = \frac{y}{r} \;\Longrightarrow\; y = r\sin\theta \]

polar coordinates — The ordered pair (r, θ): the radial coordinate r is the distance from the origin, and the angular coordinate θ is the angle from the positive x-axis.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 642 — Figure 7.27

Look at the triangle in the picture. The segment from the origin to P is the hypotenuse, of length r. The horizontal leg is x and the vertical leg is y, and the angle at the origin is theta.

Right-triangle trigonometry now does all the work. Cosine is adjacent over hypotenuse, so cos theta is x over r, and multiplying through gives x equals r cos theta. Sine is opposite over hypotenuse, which gives y equals r sin theta. These two formulas are the entire bridge from polar to rectangular.

The names matter a little: r is called the radial coordinate and theta the angular coordinate. Although the picture shows a first-quadrant point, the formulas hold everywhere, because cosine and sine carry the signs for you in the other quadrants.

6. Running the triangle backwards

Concept

To go from x and y back to r and θ, use the same triangle the other way round: Pythagoras for the hypotenuse, and a ratio of the legs for the angle.

\[ x^2 + y^2 = r^2\cos^2\theta + r^2\sin^2\theta \]

\[ x^2 + y^2 = r^2\left(\cos^2\theta + \sin^2\theta\right) = r^2 \]

\[ \frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \tan\theta \]

The first line gives r exactly. The second gives only the tangent of θ, and a tangent value belongs to two opposite directions. That gap is where most conversion mistakes come from.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 642-643 — Equations 7.7 and 7.8

Going the other way, you start with x and y and want r and theta. Square the two conversion formulas and add them. The factor r squared comes out, and what is left is cos squared plus sin squared, which is always one. So r squared is x squared plus y squared: Pythagoras again.

Dividing the second formula by the first cancels r and leaves the tangent of theta. That is a complete description of theta only up to a half turn, because tangent repeats every half turn. The points (3, 4) and its opposite point through the origin have the same tangent.

So the two directions are not equally easy. Polar to rectangular is plain substitution. Rectangular to polar always needs one extra decision: which of the two possible directions is the right one. You make that decision by looking at the signs of x and y.

7. Reading Theorem 7.4 piece by piece

Notation

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 643 — Theorem 7.4

Step through the notes one at a time. The first pair of formulas, equation 7.7, is the easy direction, and it works even for strange polar names such as a negative r or an angle bigger than a full turn. Whatever pair you are given, substitution gives the one correct rectangular point.

The second pair, equation 7.8, is the reverse direction. The distance formula is safe. The tangent formula is where you must slow down: it tells you the slope of the line through the origin and the point, not which end of that line the point is on.

The last note covers the case the formula cannot handle at all. On the vertical axis x is zero, the ratio is undefined, and you read the angle straight from a sketch: a quarter turn if the point is above the origin, three quarters of a turn if below.

8. The polar grid: circles and rays

Concept

pole and polar axis — The pole is the origin, where r is 0. The polar axis is the ray from the pole along the positive x-axis, where θ is 0.

Positive angles turn counterclockwise from the polar axis, negative angles clockwise. To plot a point, turn to the angle first, then walk the distance r along that ray.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 645-646 — Figure 7.28

Rectangular graph paper is a mesh of vertical and horizontal lines. Polar graph paper is a mesh of circles centred at the pole and rays leaving it. Each circle is a set of points at one fixed distance, and each ray is a set of points in one fixed direction.

The highlighted circle is every point with r equal to 2, and the highlighted ray is every point with theta equal to pi over 3. They meet at exactly one point, which is the point with polar coordinates two and pi over 3. That is what naming a point by polar coordinates means.

The order of operations when plotting is: turn first, then walk. Turn counterclockwise from the polar axis for a positive angle, clockwise for a negative one, and then walk out along the ray you are facing.

9. Example 7.10(a) and (b): two quadrants, two treatments

Worked example

Convert two rectangular points to polar coordinates.

\[ (1, 1) \quad\text{and}\quad (-3, 4) \]

\[ r = \sqrt{1^2 + 1^2} = \sqrt2 \]

\[ \tan\theta = \frac11 = 1 \;\Longrightarrow\; \theta = \frac{\pi}{4}, \quad (1,1) = \left(\sqrt2, \tfrac{\pi}{4}\right) \]

\[ r = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5 \]

\[ \tan\theta = \frac{4}{-3} = -\frac43 \]

\[ \theta = \pi - \arctan\frac43 \approx 3.1416 - 0.9273 = 2.214 \]

\[ 5\cos(2.2143) \approx -3.000, \quad 5\sin(2.2143) \approx 4.000 \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 643-644 — Example 7.10a and b

Part (a) is the easy case. The distance is root two, and since both coordinates are positive the point is in the first quadrant, which is exactly where the inverse tangent answers. So pi over 4 is correct as it stands.

Part (b) is the case to learn from. The distance is 5, and the tangent is minus four thirds. Your calculator will return about minus 0.927, which points into the fourth quadrant. But the point has a negative x and a positive y, so it lives in the second quadrant. The fix is to add a half turn, which gives about 2.214.

The book's printed working says the angle is minus the arctangent of four thirds, which would be minus 0.927; that line is a misprint, although the number the book finally reports, 2.21, is right. The check at the end settles it independently: converting 5 and 2.2143 back gives minus 3 and 4, the point you started with.

10. Why the inverse tangent sees only half the plane

Picture it

The two marked points are opposite each other through the origin. Their ratios y over x are equal, so their tangents are equal, and the inverse tangent, which only ever answers between minus a quarter turn and plus a quarter turn, returns the right-hand one.

Look at the two dots on the dashed line. One is the point you want; the other is its reflection through the origin. Divide y by x for either of them and you get the same number, minus four thirds, because flipping both signs leaves a quotient unchanged.

The inverse tangent has to give one answer for that number, and by convention it always gives an angle between minus a quarter turn and plus a quarter turn, which is the right-hand half of the plane. So for any point on the left of the vertical axis it points you to the wrong end of the line.

The rule that follows is short. If x is positive, trust the inverse tangent. If x is negative, add a half turn. If x is zero, read the answer off the axis. A sketch before you calculate makes this automatic.

11. Example 7.10(c) and (d): an axis point and a clockwise angle

Worked example

Convert two more rectangular points to polar coordinates.

\[ (0, 3) \quad\text{and}\quad \left(5\sqrt3, -5\right) \]

\[ r = \sqrt{0^2 + 3^2} = 3 \]

\[ \tan\theta = \frac{3}{0} \quad \text{undefined} \]

\[ (0, 3) = \left(3, \tfrac{\pi}{2}\right) \]

\[ r = \sqrt{\left(5\sqrt3\right)^2 + (-5)^2} = \sqrt{75 + 25} = 10 \]

\[ \tan\theta = \frac{-5}{5\sqrt3} = -\frac{1}{\sqrt3} \;\Longrightarrow\; \theta = -\frac{\pi}{6} \]

\[ 10\cos\left(-\tfrac{\pi}{6}\right) = 5\sqrt3, \quad 10\sin\left(-\tfrac{\pi}{6}\right) = -5 \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 644 — Example 7.10c and d

In part (c) the tangent formula breaks down entirely: you would have to divide by zero. That does not mean the angle is undefined. It means the formula cannot see it, and the picture can. The point sits straight up from the origin, a quarter turn from the polar axis.

Part (d) lands in the fourth quadrant, where x is positive and y negative. That is the half of the plane where the inverse tangent is right, so its answer, minus pi over 6, can be used directly. A negative angle simply means the turn is clockwise.

The picture shows all four points of the example with their rays. The length of each ray is r and its direction is theta. If you prefer angles between zero and a full turn, the answer to (d) can also be written with angle 11 pi over 6; both name the same point.

12. Trap: trusting the inverse tangent

Trap

13. Example 7.10(e) to (g): polar to rectangular

Worked example

Convert three polar points to rectangular coordinates. This direction is pure substitution.

\[ \left(3, \tfrac{\pi}{3}\right), \quad \left(2, \tfrac{3\pi}{2}\right), \quad \left(6, -\tfrac{5\pi}{6}\right) \]

\[ x = 3\cos\tfrac{\pi}{3} = \frac32, \quad y = 3\sin\tfrac{\pi}{3} = \frac{3\sqrt3}{2} \]

\[ x = 2\cos\tfrac{3\pi}{2} = 0, \quad y = 2\sin\tfrac{3\pi}{2} = -2 \]

\[ x = 6\cos\left(-\tfrac{5\pi}{6}\right) = 6\left(-\tfrac{\sqrt3}{2}\right) = -3\sqrt3 \]

\[ y = 6\sin\left(-\tfrac{5\pi}{6}\right) = 6\left(-\tfrac12\right) = -3 \]

\[ \sqrt{\left(-3\sqrt3\right)^2 + (-3)^2} = \sqrt{27 + 9} = 6 \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 644-645 — Example 7.10e to g

This is the comfortable direction. Every step is a substitution into x equals r cos theta and y equals r sin theta, and you need only the standard values of sine and cosine at multiples of pi over 6 and pi over 4.

Part (f) is worth a moment: three quarters of a turn faces straight down, so cosine is zero and sine is minus one, and the point is two units below the origin. Part (g) uses a clockwise angle. Turning 5 pi over 6 clockwise puts you in the third quadrant, where both cosine and sine are negative, which is why both coordinates come out negative.

The check is one you can use on every conversion: the distance of your answer from the origin must equal the r you started with. Here root of 27 plus 9 is 6.

14. Checkpoint 7.10: one conversion each way

Worked example

Convert the first point to polar coordinates and the second to rectangular coordinates.

\[ (-8, -8) \quad\text{and}\quad \left(4, \tfrac{2\pi}{3}\right) \]

\[ r = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt2 \]

\[ \tan\theta = 1, \quad \theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]

\[ x = 4\cos\tfrac{2\pi}{3} = 4\left(-\tfrac12\right) = -2 \]

\[ y = 4\sin\tfrac{2\pi}{3} = 4\cdot\tfrac{\sqrt3}{2} = 2\sqrt3 \]

\[ 8\sqrt2\cos\tfrac{5\pi}{4} = 8\sqrt2\left(-\tfrac{\sqrt2}{2}\right) = -8 \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 645 — Checkpoint 7.10

The first half of this checkpoint is the trap slide done properly. The distance is root 128, which simplifies to 8 root 2. The signs put the point in the third quadrant, so the angle is pi over 4 plus a half turn.

The second half is substitution: cosine of 2 pi over 3 is minus one half and sine is root 3 over 2, so the point is minus 2 across and 2 root 3 up, in the second quadrant, exactly where an angle of two thirds of a half turn should put it.

The final check converts the polar answer back. Cos of 5 pi over 4 is minus root 2 over 2, and multiplying by 8 root 2 gives minus 8. The same calculation with sine gives the other minus 8. You land where you began, so the angle was right.

15. Before reaching for the inverse tangent

Step zero

\[ \text{Convert } (-1, -1) \text{ to polar coordinates.} \]

Write your answer before revealing. This slide is here to build one habit: decide the quadrant before you calculate the angle.

Both coordinates are negative, so the point is in the third quadrant, and its angle must lie between a half turn and three quarters of a turn. Once you have written that range down, any answer outside it is caught immediately, including the inverse tangent's pi over 4.

The distance is root 2, and the reference angle, the acute angle to the nearest part of the x-axis, is pi over 4. Adding a half turn places it in the third quadrant: 5 pi over 4.

16. Complete the conversions

Comparison

Fill in each blank before checking. Two of the rows are on axes, where the tangent formula either divides by zero or gives zero, so read those angles from a mental picture: straight down is three quarters of a turn, and straight left is a half turn.

The other two rows need the quadrant rule. Minus 2 and 2 is in the second quadrant with a reference angle of pi over 4, giving 3 pi over 4. One and minus root 3 is in the fourth quadrant with reference angle pi over 3, so within one turn its angle is 5 pi over 3. Its distance is 2, since one plus three is four.

17. One point, many names

Section

Part 2

18. A full turn changes nothing

Concept

Turning through a whole extra revolution brings you back to facing the same way, so adding any multiple of a full turn to the angle names the same point.

\[ \left(2, \tfrac{\pi}{3}\right) \text{ and } \left(2, \tfrac{7\pi}{3}\right) \]

\[ 2\cos\tfrac{7\pi}{3} = 2\cos\left(\tfrac{\pi}{3} + 2\pi\right) = 2\cos\tfrac{\pi}{3} = 1 \]

\[ 2\sin\tfrac{7\pi}{3} = 2\sin\tfrac{\pi}{3} = \sqrt3 \quad\Longrightarrow\quad \text{both are } \left(1, \sqrt3\right) \]

Rectangular coordinates give each point exactly one name. Polar coordinates give each point infinitely many, and that is not a defect: it is what lets a single equation trace a curve more than once.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 645 — the non-uniqueness of polar coordinates

In rectangular coordinates each point has exactly one name. In polar coordinates it has infinitely many. Adding a full turn to the angle leaves you facing the same way, so you walk to the same point. That is why 7 pi over 3 and pi over 3 name the same place, and so do minus 5 pi over 3 and 13 pi over 3.

The algebra on the slide confirms it: cosine and sine repeat every full turn, so the rectangular coordinates come out identical.

This is not an inconvenience to be apologised for. When an equation such as r equals 4 sin theta is traced over more than one period, it passes through the same points again. Understanding that one point has many names is what makes those repeated traces unsurprising.

19. A negative r walks backwards

Concept

The book allows r to be negative. The rule: turn to the angle, then walk that far in the opposite direction, through the pole.

\[ x = -2\cos\tfrac{4\pi}{3} = -2\left(-\tfrac12\right) = 1 \]

\[ y = -2\sin\tfrac{4\pi}{3} = -2\left(-\tfrac{\sqrt3}{2}\right) = \sqrt3 \]

\[ \left(-r, \theta\right) = \left(r, \theta + \pi\right) \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 645-646 — negative r and the plotting rule

A distance cannot really be negative, but polar equations produce negative values of r all the time, so the book gives them a meaning. Face the direction theta, then walk backwards: through the pole and out the other side.

The picture shows it for minus 2 at 4 pi over 3. The dashed orange ray points into the third quadrant. Walking two units backwards along it lands you in the first quadrant, on the ray pi over 3. The conversion formulas agree without any special treatment: minus 2 times cos of 4 pi over 3 is 1, and minus 2 times its sine is root 3.

The general rule is on the last line. A negative r at angle theta is the positive r at angle theta plus a half turn. You will use this identity every time a curve dips below zero.

20. Example 7.11: plotting three points

Worked example

Plot these three points on the polar plane, then confirm each by converting to rectangular coordinates.

\[ \left(2, \tfrac{\pi}{4}\right), \quad \left(-3, \tfrac{2\pi}{3}\right), \quad \left(4, \tfrac{5\pi}{4}\right) \]

\[ \left(2\cos\tfrac{\pi}{4}, 2\sin\tfrac{\pi}{4}\right) = \left(\sqrt2, \sqrt2\right) \approx (1.414, 1.414) \]

\[ \left(-3, \tfrac{2\pi}{3}\right) = \left(3, \tfrac{5\pi}{3}\right) \]

\[ x = -3\left(-\tfrac12\right) = 1.5, \quad y = -3\cdot\tfrac{\sqrt3}{2} \approx -2.598 \]

\[ \left(4\cos\tfrac{5\pi}{4}, 4\sin\tfrac{5\pi}{4}\right) = \left(-2\sqrt2, -2\sqrt2\right) \approx (-2.828, -2.828) \]

\[ (+,+), \quad (+,-), \quad (-,-) \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 646 — Example 7.11 and Figure 7.29

Plot each point by turning first and walking second. The first point is easy: an eighth of a turn, two units out. The third is similar: five eighths of a turn puts you in the third quadrant, and four units out.

The middle point is the one to slow down on. You face 2 pi over 3, in the second quadrant, and then walk three units backwards, which carries you through the pole into the fourth quadrant. The hollow dot in the picture marks where you would have landed if you had walked forwards.

The check compares signs with quadrants. It is cheap, and it catches the most common slip with negative distances: plotting them as if they were positive.

21. Checkpoint 7.11: where does (−3, −7π/2) land?

Prediction

\[ \left(4, \tfrac{5\pi}{3}\right) \quad\text{and}\quad \left(-3, -\tfrac{7\pi}{2}\right) \]

\[ -\tfrac{7\pi}{2} + 4\pi = \tfrac{\pi}{2}, \quad x = -3\cos\tfrac{\pi}{2} = 0, \quad y = -3\sin\tfrac{\pi}{2} = -3 \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 646 — Checkpoint 7.11

Commit to an answer first. The angle looks alarming, but a few full turns can always be added or removed without changing the direction. Adding two full turns to minus 7 pi over 2 gives pi over 2, so after all that clockwise turning you are facing straight up.

The negative distance then sends you three units straight down, to the point zero and minus three on the negative y-axis. The picture shows the direction you face as a dashed ray and the point where you actually end up.

The first point of the checkpoint, 4 at 5 pi over 3, is a routine fourth-quadrant point. If you picked the positive y-axis for the second one, your turning was right and only the walking was wrong, which is the more common of the two errors.

22. Every name of one point

Fill the middle

\[ \left(3, \tfrac{\pi}{6}\right) = \left(3, \tfrac{13\pi}{6}\right) = \left(3, -\tfrac{11\pi}{6}\right) = \left(-3, \tfrac{7\pi}{6}\right) = \left(-3, -\tfrac{5\pi}{6}\right) \]

The formula line lists five names for a single point, and every one of them comes from two moves. Adding or subtracting a full turn never changes the point. Flipping the sign of r and moving the angle by a half turn never changes it either.

Fill in the blanks with words. Once you can say these two rules in a sentence, you can generate any polar name you need, and you can check whether two given pairs name the same point by asking whether one can be turned into the other using only these moves.

23. Match each polar name to its point

Matching

Work each one out by turning and walking, not by formula. For the first, face an eighth of a turn and walk backwards two units: third quadrant, at minus root 2 in both coordinates.

The second turns a quarter turn clockwise and walks forward two, straight down. The third faces left, a half turn, and walks backwards one unit, which takes it to the right, onto the positive x-axis. The fourth is an eighth of a turn plus a full turn, which is just the first-quadrant point at distance two.

If you mixed up the first and last, look again at the sign of r: the two pairs face the same direction, but one walks forwards and the other backwards.

24. Break the claim: different pairs, different points

Counterexample

\[ \text{Claim: if } r_1 = r_2 \text{ and } \theta_1 - \theta_2 \text{ is not a multiple of } 2\pi, \text{ the points differ.} \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 645 — the pole

Try to find a counterexample before revealing. The claim sounds reasonable: if the distances agree and the angles are genuinely different, the points ought to be different.

The pole breaks it. At r equal to zero you never leave the origin, so it does not matter which way you face. Every angle names the same point. The claim is true everywhere except there.

This is more than a curiosity. It is why a polar curve can pass through the pole several times at different angles, as the roses do at every gap between petals, and why the zeros of r mark where a curve touches the pole.

25. Polar curves and their equations

Section

Part 3

26. A curve is a rule r = f(θ)

Concept

A rectangular graph is a rule giving y for each x. A polar graph is a rule giving r for each angle: the curve is every point whose distance matches the rule in its own direction.

\[ r = f(\theta) \]

1. Make a table with two columns, θ and r
2. Choose angles, usually multiples of π/6 and π/4 over one period
3. Compute r for each angle
4. Plot each (r, θ), remembering that a negative r is plotted backwards
5. Connect the points in order of increasing θ and look for a pattern

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 647 — Problem-Solving Strategy: Plotting a Curve in Polar Coordinates

In rectangular coordinates, a function assigns a height to each horizontal position, and its graph is all the points that obey that rule. A polar function assigns a distance to each direction, and its graph is all the points whose distance in their own direction is exactly what the rule says.

The book's strategy is the same one you used for graphing in precalculus: make a table, plot the points, join them in order. Two things are different. You should choose angles at the special values where sine and cosine are easy, and you must plot any negative r backwards.

Periodicity saves work. Many of these functions repeat every full turn, or even every half turn, so a table over one period is all you ever need.

27. Example 7.12: r = 4 sin θ, point by point

Worked example

Graph the curve by tabulating r over one full period.

\[ r = 4\sin\theta, \quad 0 \le \theta \le 2\pi \]

\[ \theta = \tfrac{\pi}{2}: \quad r = 4 \;\Longrightarrow\; (x, y) = (0, 4) \]

\[ \left(-2, \tfrac{7\pi}{6}\right) = \left(2, \tfrac{\pi}{6}\right) = \left(\sqrt3, 1\right) \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 647-649 — Example 7.12 and Figure 7.30

The table is built from the special values of sine. On the first half turn sine rises from zero to one and falls back, so r rises from zero to four and falls back. Plotted in order, those points climb away from the pole, reach the top at (0, 4), and come back.

On the second half turn every value of r is negative. Look at the check step: at 7 pi over 6 the radius is minus two, and walking backwards from that direction lands you at 2 at pi over 6, a point already on the curve.

The picture shows only the first half turn, and already the curve is a closed circle. So what does the second half turn add? Nothing new, as the next slide shows.

28. Why the circle is traced twice

Picture it

On the left, r is graphed against θ in ordinary rectangular axes: a positive hump, then a negative one. On the right, the negative hump is plotted backwards, and every one of its points lands on the circle the positive hump already drew.

The left panel is a new way to look at a polar function: graph r against theta in ordinary axes, exactly as you would graph y against x. You see a blue hump above the axis and an orange hump below.

The right panel is the polar picture of the same function. The blue hump draws the circle. The orange hump, where r is negative, is plotted backwards, and every one of its points lands on the circle the blue hump has already drawn. That is why the orange dashes sit exactly on top of the blue.

Keep this pair of pictures in mind. The graph of r against theta is the easiest way to see where r is positive, where it is negative, and where it is zero, and those three facts decide the shape of every polar curve in this lesson.

29. Example 7.12: the same circle in x and y

Worked example

Rewrite the equation in rectangular coordinates and identify the curve.

\[ r = 4\sin\theta \]

\[ r^2 = 4r\sin\theta \]

\[ x^2 + y^2 = 4y \]

\[ x^2 + y^2 - 4y = 0 \]

\[ x^2 + \left(y^2 - 4y + 4\right) = 4 \]

\[ x^2 + (y - 2)^2 = 4 \]

\[ \left(\sqrt3\right)^2 + (1 - 2)^2 = 3 + 1 = 4 \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 649 — Example 7.12

The table suggested a circle; converting to rectangular coordinates proves it. The move that makes the conversion work is multiplying both sides by r. On its own, sin theta has no rectangular name, but r sin theta is y and r squared is x squared plus y squared.

After substituting, what remains is completing the square, a precalculus skill. Half of minus four is minus two, its square is four, and adding four to both sides gives a circle of radius two centred at (0, 2).

Did multiplying by r add anything? It could only add the pole, and the pole is already on the curve, at theta equal to zero. The check with the table point confirms the equation with numbers.

30. Checkpoint 7.12: the cardioid r = 4 + 4 cos θ

Worked example

Graph the curve from a table. Cosine is even, so the values for negative angles mirror the positive ones.

\[ r = 4 + 4\cos\theta \]

\[ r(0) = 8 \Rightarrow (8, 0), \qquad r(\pi) = 0 \Rightarrow \text{the pole} \]

\[ r(2\pi - \theta) = 4 + 4\cos(2\pi - \theta) = 4 + 4\cos\theta = r(\theta) \]

\[ r\left(\tfrac{\pi}{2}\right) = 4 + 0 = 4 \;\Longrightarrow\; (0, 4) \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 649 — Checkpoint 7.12

Cosine is largest at theta equal to zero and smallest at a half turn, so this curve reaches out to 8 on the right and shrinks to 0 on the left. The table fills in the values between.

Because cosine is even, the lower half of the curve is a mirror image of the upper half. That is the symmetry idea from the end of the lesson, used informally: once you have the values from zero to a half turn, the rest comes for free.

The shape is a heart, which is what cardioid means. The pinch at the pole is a cusp, where r touches zero without going negative. Check the crossing of the vertical axis by putting theta equal to pi over 2: cosine is zero, so r is 4.

31. Example 7.13(a) and (b): a constant angle, a constant distance

Worked example

Rewrite each equation in rectangular coordinates and identify the graph.

\[ \text{(a) } \theta = \frac{\pi}{3} \qquad \text{(b) } r = 3 \]

\[ \tan\theta = \tan\frac{\pi}{3} = \sqrt3 \]

\[ \frac{y}{x} = \sqrt3 \;\Longrightarrow\; y = \sqrt3\,x \]

\[ r^2 = 9 \;\Longrightarrow\; x^2 + y^2 = 9 \]

\[ \left(-3, \tfrac{\pi}{3}\right) = \left(3, \tfrac{4\pi}{3}\right) \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 650 — Example 7.13a and b

The simplest polar equations fix one coordinate. Fixing theta at pi over 3 gives every point in that direction, and taking the tangent turns the equation into y equals root 3 times x. The picture shows why the answer is a whole line and not just a ray: negative values of r supply the half in the third quadrant.

Fixing r at 3 gives every point at distance three, a circle. Squaring produces x squared plus y squared equals 9.

The book adds a warning here, and it deserves attention. Squaring can introduce new solutions: the squared equation also allows r equal to minus 3. In this case nothing new is added, because minus 3 at any angle is plus 3 half a turn away, a point already on the circle. Always ask that question when you square.

32. Before converting r = 6 cos θ − 8 sin θ

Step zero

\[ r = 6\cos\theta - 8\sin\theta \]

Write down your first move before revealing. Cosine and sine on their own have no rectangular names, but r cos theta and r sin theta do. So multiply both sides by r.

Multiplying an equation by something that can be zero can add solutions, so it is fair to ask whether this is safe. The only point it could add is the pole. And the pole is already on this curve: set the right-hand side to zero and you get tangent theta equals three quarters, an angle at which r really is zero.

That short argument is worth writing down on an exam. It is the difference between a conversion that happens to be right and one you have shown to be right.

33. Example 7.13(c): a circle through the pole

Worked example

\[ r = 6\cos\theta - 8\sin\theta \]

\[ r^2 = 6r\cos\theta - 8r\sin\theta \]

\[ x^2 + y^2 = 6x - 8y \]

\[ \left(x^2 - 6x\right) + \left(y^2 + 8y\right) = 0 \]

\[ \left(x^2 - 6x + 9\right) + \left(y^2 + 8y + 16\right) = 25 \]

\[ (x - 3)^2 + (y + 4)^2 = 25 \]

\[ (6-3)^2 + (0+4)^2 = 25, \quad (0-3)^2 + (-8+4)^2 = 25 \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 650 — Example 7.13c

After multiplying by r and substituting, you have x squared plus y squared equals 6x minus 8y. Bring everything to one side, group the x terms and the y terms, and complete both squares: add nine for the x terms and sixteen for the y terms.

The result is a circle with centre (3, minus 4) and radius 5. The picture shows a nice consequence: the distance from the pole to the centre is also 5, by the three-four-five triangle, so the circle passes through the pole.

The check uses two polar points. At theta equal to zero, r is 6, giving (6, 0). At a quarter turn, r is minus 8, so you face up and walk backwards eight units, reaching (0, minus 8). Both lie on the circle.

34. Checkpoint 7.13: r = sec θ tan θ

Worked example

\[ r = \sec\theta\tan\theta \]

\[ r = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\cos^2\theta} \]

\[ r\cos^2\theta = \sin\theta \]

\[ r^2\cos^2\theta = r\sin\theta \]

\[ (r\cos\theta)^2 = r\sin\theta \;\Longrightarrow\; x^2 = y \]

\[ r = \sqrt2 \cdot 1 = \sqrt2, \quad \left(\sqrt2\cos\tfrac{\pi}{4}, \sqrt2\sin\tfrac{\pi}{4}\right) = (1, 1), \quad 1^2 = 1 \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 650 — Checkpoint 7.13

Start by writing everything in sines and cosines: secant times tangent is sine over cosine squared. Clearing the denominator gives r cos squared theta equals sin theta.

Now multiply by r once more, so the left side is r cos theta, squared, and the right side is r sin theta. Those are x squared and y. The curve is the parabola y equals x squared, a curve whose rectangular equation is far simpler than its polar one.

The picture draws it straight from the polar equation. As theta approaches a right angle from either side, secant blows up, r grows without bound, and the arms of the parabola climb. The check at pi over 4 gives the point (1, 1), which satisfies y equals x squared.

35. Trap: replacing sin θ by y

Trap

36. Rectangular to polar: the circle x² + y² = 4x

Worked example

Go the other way: convert a rectangular equation to polar form, and make sure no point is lost.

\[ x^2 + y^2 = 4x \]

\[ r^2 = 4r\cos\theta \]

\[ r\left(r - 4\cos\theta\right) = 0 \]

\[ r = 0 \quad\text{or}\quad r = 4\cos\theta \]

\[ \theta = \tfrac{\pi}{2}: \quad r = 4\cos\tfrac{\pi}{2} = 0 \]

\[ 1^2 + \left(\sqrt3\right)^2 = 4 = 4(1) \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 661 — Exercises 158 to 164

This is the reverse direction, rectangular to polar. The substitution itself is immediate: the left side is r squared and the right is 4r cos theta.

The step to learn is the next one. Dividing both sides by r would assume r is not zero, and could lose the pole. Factoring is honest: either r is zero, or r equals 4 cos theta. Then ask whether the second equation already contains the pole. It does, at theta equal to pi over 2, so nothing was lost and the answer is simply r equals 4 cos theta.

The picture marks exactly that: the pole is on the circle, reached at a quarter turn. The table point at pi over 3 confirms the equation numerically.

37. Exercises 158 to 161: rectangular to polar

Comparison

Substitute x equals r cos θ and y equals r sin θ, then solve for r where you can.

\[ x^2 - y^2 = r^2\left(\cos^2\theta - \sin^2\theta\right) = r^2\cos 2\theta \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 661 — Exercises 158 to 161

Fill in the blanks before checking. The method is the same every time: replace x by r cos theta and y by r sin theta, simplify, and solve for r if you can.

The circle is immediate. The vertical line becomes r cos theta equals 8, and dividing by cosine gives 8 sec theta. The hyperbola uses a double-angle identity, shown on the formula line: cos squared minus sin squared is cos 2 theta. The oblique line leaves a combination of cosine and sine in the denominator.

Notice how the results feel. Circles about the origin get simpler; lines and hyperbolas get messier. That is the point of having two systems.

38. Every polar curve is a parametric curve

Concept

The conversion formulas turn any polar equation into a parametrization with the angle as the parameter, which connects this section to Section 7.1.

\[ r = f(\theta) \quad\Longrightarrow\quad x = f(\theta)\cos\theta, \quad y = f(\theta)\sin\theta \]

\[ r = a + b\theta: \quad x = (a + b\theta)\cos\theta, \quad y = (a + b\theta)\sin\theta \]

Letting the angle run over all real numbers traces the entire spiral. Everything you learn about parametric curves, including slopes and arc length in the next section, now applies to polar curves.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 655 — polar curves as parametric curves

In Section 7.1 you described curves by giving x and y as functions of a parameter. A polar curve is already one of those. Substitute r equals f of theta into the conversion formulas and you get x and y as functions of theta, with theta as the parameter.

The spiral on the slide is an example: the distance grows steadily with the angle, and the parametrization follows at once. Letting theta run over all real numbers traces the whole spiral, inward for negative angles and outward for positive ones.

This observation pays off in the next section. Slopes of tangent lines, areas and arc lengths of polar curves are all computed by treating them as parametric curves with parameter theta.

39. Which system gives the simpler equation?

Sorting

Sort each curve before checking. The question to ask is whether the curve is naturally described by distance and turning, or by horizontal and vertical positions.

Circles about the origin, spirals and roses are the polar family. A rose's rectangular equation is a sixth-degree polynomial equation, and the spiral cannot be written with x or y alone at all, as you will see in Example 7.14.

Horizontal lines, parabolas and general lines belong to rectangular coordinates. Each of them can be written in polar form, but the result involves cosecants, secants or fractions of trig functions. Neither system is better; each is better for some curves.

40. A catalogue of polar curves

Section

Part 4

41. Circles and lines

Concept

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 650-651 — Figure 7.31

This is the first part of the book's catalogue, the curves you should recognise on sight. A constant r is a circle about the pole. A multiple of cos theta is a circle through the pole with its centre on the x-axis; a multiple of sin theta has its centre on the y-axis. You proved the sine case in Example 7.12.

A constant angle is a line through the pole, and a secant is a vertical line: r sec theta equals a is the same as r cos theta equals a, which says x equals a. Similarly a cosecant gives a horizontal line.

The table gives the rectangular form of each, so if you ever meet one in disguise you can convert it and check.

42. Limaçons: r = a + b cos θ

Concept

A limaçon adds a constant to a multiple of cosine or sine. Its shape depends only on how a compares with b. When a is smaller than b, the radius goes negative over part of each turn, and those backwards points make an inner loop.

\[ r = a + b\cos\theta \quad\text{or}\quad r = a + b\sin\theta \]

cardioid — The limaçon with a equal to b (or a equal to minus b): the radius just touches zero once per turn, which makes a cusp at the pole.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 651-652 — Figures 7.31 and 7.32

A limaçon adds a constant to a multiple of cosine or sine. The four curves in the picture all have b equal to 2; only a changes. Watch what happens as a grows from 1 to 4.

At a equal to 1 the radius swings between 3 and minus 1. The stretch where it is negative is drawn backwards, and becomes the small inner loop. At a equal to 2 the radius just touches zero once, and the inner loop shrinks to a cusp: that is the cardioid. At a equal to 3 there is only a dent, called a dimple, and by a equal to 4 the curve is smooth and convex.

So the whole family is controlled by one number, the ratio of a to b. The next slide lets you move it.

43. Push a down to b: when does the inner loop appear?

Edge cases

\[ \min_{\theta}(a + 2\cos\theta) = a - 2 < 0 \iff a < 2 \]

The slider draws r against theta in ordinary axes, the same view as the left panel of the circle picture. Start at a equal to 3 and drag it down slowly. Watch the lowest point of the curve.

The minimum of a plus 2 cos theta happens at a half turn, where cosine is minus one, and it equals a minus 2. While a is above 2 the whole graph stays above the axis, so the polar curve never reaches the pole. At exactly 2 the graph just touches the axis, and the polar curve has a cusp. Below 2 part of the graph dips under the axis, and that negative stretch is drawn backwards as the inner loop.

This is the general lesson of the slide: the sign of r against theta tells you the shape of the polar curve.

44. Roses: r = a sin kθ

Concept

\[ r = a\sin(k\theta) \quad\text{or}\quad r = a\cos(k\theta), \qquad k \text{ a whole number} \]

Every petal has length a, the largest value of r. The surprise is the odd case: three petals for k equal to 3, not six.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 652-653 — Figures 7.32 and 7.33

A rose is a sine or cosine of a whole-number multiple of theta. Every petal has the same length, the amplitude a, and the petals are spaced evenly around the pole.

The petal count is the famous puzzle. For even k you get twice k petals: 4 for k equal to 2, 8 for k equal to 4. For odd k you get exactly k: 3 for k equal to 3, 5 for k equal to 5. The book leaves the reason as an exploration, and the next two slides explore it.

The last column of the table is a practical point: an odd rose is completely traced by the time theta reaches a half turn, so there is no need to tabulate further.

45. Where the negative arcs go

Picture it

Each hump of the graph of r against θ becomes one petal-shaped arc. The orange humps have negative r, so they are drawn half a turn away from where they point. With k even they fall into empty gaps and add new petals; with k odd they land exactly on petals already drawn.

The top row graphs r against theta for k equal to 2 and k equal to 3. Both have the same number of positive humps as negative ones: two and two, then three and three. Every hump draws one petal-shaped arc in the polar picture.

The difference is where the negative arcs land. A negative hump is drawn half a turn away from the direction it points. For k equal to 2 that places it in the empty gap between two blue petals, so it adds a new orange petal. For k equal to 3 half a turn lands it exactly on top of a blue petal, so it adds nothing new: the dashed orange retraces the blue.

That is the whole explanation of the even and odd rule, and you can now derive it rather than memorise it.

46. Slide k and count the humps

Tweak it

\[ \text{humps over } [0, 2\pi]: 2k \quad\Longrightarrow\quad \text{petals} = \begin{cases} 2k & k \text{ even} \\ k & k \text{ odd} \end{cases} \]

Again the slider shows r against theta. Move k one step at a time and count the humps over one full turn. You will always find twice k of them, half above the axis and half below.

For even k, every hump is a separate petal, so twice k petals. For odd k, each hump below the axis retraces one above it, so only k distinct petals. Try k equal to 1: two humps, but the polar curve r equals 3 sin theta is a single circle traced twice, exactly as in Example 7.12. That is the odd rule with one petal.

47. Trap: doubling an odd rose

Trap

48. Rational k closes, irrational k never does

Intuition

When k is a fraction, the radius repeats after a whole number of turns, so the curve eventually returns to where it started and closes. When k is irrational, it never repeats: the curve keeps adding new loops forever and, in the limit, fills the whole disc of radius three.

\[ k = \tfrac37: \quad \sin\left(\tfrac37(\theta + 14\pi)\right) = \sin\left(\tfrac{3\theta}{7} + 6\pi\right) = \sin\tfrac{3\theta}{7} \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 653 — Figure 7.34 and Exercise 182

So far k has been a whole number. The book goes further. When k is a fraction such as three sevenths, sine of k theta repeats after theta has increased by 14 pi, seven full turns, as the formula line shows. After that the curve retraces itself, so it is closed. The left panel shows the complete closed curve.

When k is irrational, like pi, no whole number of turns ever makes the sine repeat. The curve never returns to its starting point, and every new loop is slightly shifted from the others. The right panel shows only forty radians of it, and it is already filling the disc.

The book calls this a space-filling curve: traced forever, it passes arbitrarily close to every point of the disc of radius three.

49. Example 7.14: describing the spiral

Worked example

A point moves counterclockwise around the origin and its distance from the origin is a constant multiple k of the angle. Find the curve's equation.

\[ d(P, O) = k\theta \]

\[ \sqrt{x^2 + y^2} = k\arctan\frac{y}{x} \]

\[ y = x\tan\left(\frac{\sqrt{x^2 + y^2}}{k}\right) \]

\[ \text{no explicit } y = g(x) \]

\[ r = k\theta, \qquad \text{or with an offset } r = a + k\theta \]

\[ r(\theta + 2\pi) - r(\theta) = 2\pi k, \quad 4\pi - 2\pi = 2\pi \approx 6.283 \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 654 — Example 7.14, the Archimedean spiral

This example is the chapter opener, the shell of the nautilus. The description is simple: as the point turns, its distance from the centre grows in proportion to the angle.

Try to write that in rectangular coordinates and you get a square root on one side and an inverse tangent on the other. Solving for y leaves y on both sides, wrapped in a tangent, and no algebra will untangle it. The rectangular description is hopeless.

In polar coordinates the condition is the equation: r equals k theta. The check shows its signature property. Each full turn adds the same distance, two pi k, so the coils are evenly spaced, which you can see in the picture where the crossings of the positive x-axis are 2 pi apart.

50. The nautilus and the logarithmic spiral

Real world

\[ r = 1.2\left(1.25^{\theta}\right) \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 654-655 — Figure 7.36, the logarithmic spiral

Write your answer before revealing. The real nautilus is not Archimedean. Its shell follows a logarithmic spiral, where the distance is a constant times a power of a fixed base.

Increasing theta by a full turn multiplies r by 1.25 raised to the power 2 pi, about 4.06, no matter where you start. The picture marks three points one turn apart on the polar axis, and each is about four times farther out than the last.

Multiplying every distance by the same factor is exactly what scaling a picture does, so each coil is a scaled copy of the one inside it. That is why a nautilus can keep growing and keep its shape. An Archimedean spiral adds a fixed amount per turn instead, so it looks different at different sizes.

51. Symmetry

Section

Part 5

52. Three reflections of a point

Concept

A curve is symmetric about a line or a point when reflecting every point of the curve gives another point of the curve. In polar coordinates each of the three reflections has a simple formula.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 655-656 — symmetry in polar coordinates

Symmetry of a graph means that reflecting it leaves it unchanged. Three reflections matter for polar curves: in the polar axis, in the vertical line through the pole, and through the pole itself.

The picture shows where each sends a point. Reflecting in the polar axis keeps the distance and negates the angle. Reflecting through the pole keeps the distance and adds a half turn. Reflecting in the vertical line keeps the distance and replaces the angle by a half turn minus it.

The table gives the rectangular version of each move, which is the easiest way to remember them: flip y, flip both, or flip x.

53. Reading Theorem 7.5 piece by piece

Notation

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 655-656 — Theorem 7.5

Each test asks whether the equation survives the substitution that performs one reflection. If replacing theta by minus theta gives back the same equation, then every reflected point is also on the curve.

The practical shortcuts are in the notes. Equations involving only cosines of theta pass the polar-axis test, because cosine is even. Equations of the form a plus b sin theta pass the vertical-line test, because the sine of a half turn minus theta equals the sine of theta.

The last note is the one that trips people up. Passing a test proves symmetry, but failing one particular substitution does not disprove it, because each reflected point has other polar names. The next slide explains why.

54. Every reflection has two polar names

Concept

A negative radius plus half a turn is the same point, so each reflected point can be written a second way. Theorem 7.5 lets you use either name.

\[ (r, -\theta) = (-r, \pi - \theta) \qquad \text{polar axis} \]

\[ (r, \pi + \theta) = (-r, \theta) \qquad \text{pole} \]

\[ (r, \pi - \theta) = (-r, -\theta) \qquad \text{vertical line} \]

Each identity is the rule for a negative radius applied once: flip the sign of r and move the angle by half a turn. When one form of a test fails, the other form may still succeed, and Example 7.15 needs exactly that.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 657 — the alternative tests used in Example 7.15

Recall the rule for a negative radius: flip the sign of r and move the angle by a half turn, and you have the same point. Apply that to each reflected point and you get a second name for it.

So the polar-axis reflection can be tested either by replacing theta with minus theta, or by replacing r with minus r and theta with a half turn minus theta. The pole can be tested by adding a half turn to theta, or by replacing r with minus r. The vertical line can be tested by replacing theta with a half turn minus theta, or by replacing r and theta with their negatives.

When one form fails, try the other before concluding. Example 7.15 is a curve where one form fails and the other succeeds.

55. Example 7.15: testing r = 3 sin 2θ for symmetry

Worked example

\[ r = 3\sin 2\theta \]

\[ 3\sin(-2\theta) = -3\sin 2\theta \ne 3\sin 2\theta \]

\[ -r = 3\sin(2\pi - 2\theta) = -3\sin 2\theta \;\Longrightarrow\; r = 3\sin 2\theta \]

\[ 3\sin(2\theta + 2\pi) = 3\sin 2\theta \]

\[ -r = 3\sin(-2\theta) = -3\sin 2\theta \;\Longrightarrow\; r = 3\sin 2\theta \]

\[ \left(2.598, -\tfrac{\pi}{6}\right) = \left(-2.598, \tfrac{5\pi}{6}\right), \quad 3\sin\tfrac{5\pi}{3} \approx -2.598 \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 657 — Example 7.15

The first substitution fails: replacing theta by minus theta turns 3 sin 2 theta into its negative. A careless reader would stop there and say the rose is not symmetric about the polar axis. The second form, with minus r and a half turn minus theta, gives back the original equation, so it is symmetric after all.

The pole test is easy because sin 2 theta repeats every half turn. The vertical-line test succeeds with the form that negates both r and theta: the two sign changes cancel.

The check makes the conclusion concrete. Take the point at pi over 6, reflect it in the polar axis, and rename the result with a negative radius at 5 pi over 6. Evaluating the equation there gives exactly that negative radius, so the reflected point really is on the rose.

56. One petal, then reflect

Picture it

With all three symmetries known, a quarter turn of table is enough. The five values draw one petal in the first quadrant; reflecting it in the two axes gives the whole four-petal rose.

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, pp. 657-659 — Figures 7.37 and 7.38

Knowing all three symmetries, you only need the curve on a quarter turn. The table covers theta from zero to pi over 2, and r rises from zero to three at pi over 4 and falls back to zero. Plotted, those five points make one petal in the first quadrant, shown on the left.

Reflecting that petal in the vertical line and the polar axis, as on the right, produces the other three petals. You get the whole four-petal rose from five table values.

This is how symmetry pays off in practice: it cuts the table down and tells you in advance what the finished picture must look like.

57. Checkpoint 7.14: the three-petal rose r = 2 cos 3θ

Worked example

Determine the symmetry of the graph and create a graph.

\[ r = 2\cos 3\theta \]

\[ 2\cos(-3\theta) = 2\cos 3\theta \;\checkmark \]

\[ 2\cos(3\theta + 3\pi) = -2\cos 3\theta, \qquad -r = 2\cos 3\theta \]

\[ 2\cos(3\pi - 3\theta) = -2\cos 3\theta, \qquad -r = 2\cos(-3\theta) \]

\[ \text{tips at } \theta = 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3} \]

\[ |r| = 2 \iff \cos 3\theta = \pm 1 \iff \theta = \tfrac{n\pi}{3}, \quad \text{tips at } 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3} \text{ only} \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 659 — Checkpoint 7.14

The polar-axis test passes in one line because cosine is even. For the pole and the vertical line, both forms of each test give the negative of the original equation, so neither symmetry can be proved, and the check shows that the pole symmetry really does fail.

The graph is an odd rose: three petals of length two, one pointing along the polar axis and the others a third of a turn either side. The picture draws the polar axis dashed, and the curve is its own mirror image in that line.

The check is the useful skill here. To show a symmetry fails, find one point whose reflection is missing. The tip at (2, 0) reflected through the pole would be at (minus 2, 0). But the only directions where the distance is 2 are the three tips, and none of them points left.

58. Find the error: one failed substitution

Error analysis

Read the argument and find the flaw before revealing the notes. The substitution is done correctly, so the error is not in the algebra.

The error is in the logic. Theorem 7.5 says that if the equation survives the substitution, the curve is symmetric. It does not say that if the equation fails the substitution, the curve is not symmetric. The same reflected point has another name, and under that name the equation may well survive.

For this rose the other name works, and the picture of the full rose settles it visually. The general habit: never conclude no symmetry from a single failed test. Either try the other form or find a specific point whose reflection is missing.

59. Which symmetry does each curve have?

Sorting

Sort each curve before checking. Look first at whether the equation involves cosine or sine, and then ask whether r appears only squared.

The cosine-only equations are unchanged when theta is negated, so they are symmetric about the polar axis. The cardioid and the vertical line x equals 2 have nothing more: neither looks the same after a half-turn rotation. The sine limaçons are the mirror case, symmetric about the vertical line only.

The last two have all three symmetries. The four-petal cosine rose is like the sine rose of Example 7.15. For r squared equals 9 cos theta, replacing r by minus r changes nothing, which gives the pole, and cosine gives the polar axis. Any two of the three symmetries force the third, because two reflections combine into the third.

60. Exercise 175: the lemniscate r² = 4 cos 2θ

Worked example

Sketch the curve and identify its symmetry.

\[ r^2 = 4\cos 2\theta \]

\[ \cos 2\theta \ge 0 \iff -\tfrac{\pi}{4} \le \theta \le \tfrac{\pi}{4} \text{ (or the opposite range)} \]

\[ (-r)^2 = r^2, \qquad \cos(-2\theta) = \cos 2\theta \]

\[ r = \pm 2\sqrt{\cos 2\theta} \]

\[ (x^2 + y^2)^2 = 4(x^2 - y^2): \quad (1.5 + 0.5)^2 = 4 = 4(1.5 - 0.5) \;\checkmark \]

OpenStax Calculus Volume 2, §7.3 Polar Coordinates §7.3, p. 661 — Exercise 175

This curve is new: the equation gives r squared, not r. Since a square cannot be negative, the curve exists only where cos 2 theta is at least zero, within an eighth of a turn of the polar axis or of the opposite direction.

Symmetry does most of the work. Only r squared appears, so negating r leaves the equation unchanged, and cosine is even. All three symmetries hold. Tabulate the positive square root on a quarter of the range and reflect.

The result is a figure eight on its side, called a lemniscate. The picture colours the two square roots differently: the positive root draws the right lobe, and the negative root, plotted backwards, draws the left. The check converts to rectangular form by multiplying through by r squared and tests the table point at pi over 6.

61. Putting it together

Section

Part 6

62. Pattern: sketching r = f(θ)

Pattern

1. Symmetry first. Test θ → −θ, θ → π − θ and θ → θ + π, trying both names of each reflection. Each symmetry you find cuts the table.
2. Zeros of r. Solve f(θ) = 0: these are the angles at which the curve passes through the pole.
3. Largest |r|. Find where the curve reaches farthest from the pole.
4. Table and plot. Compute r at multiples of π/6 and π/4; plot negative r backwards.
5. Name it if you can. Multiply by r and convert: a circle or a line is often hiding in the equation.

This is the order to work in whenever you meet a new polar equation. The picture applies every step to r equals 1 plus 2 cos theta.

Symmetry first, because it halves or quarters the work: this limaçon involves only cosine, so it is symmetric about the polar axis. Then the zeros of r: here cosine equals minus one half, at 2 pi over 3 and 4 pi over 3, which are the dashed rays where the curve passes through the pole. Then the farthest point: r is 3 at theta equal to zero.

Between the zeros, r is negative, and those points are drawn backwards to make the inner loop. With symmetry, zeros, extremes and a handful of table values, you can sketch almost any curve in this section, and converting to rectangular form will often name it for you.

63. Put the conversion to polar in order

Ranking

Drag the steps into order before checking. The main decision is where the quadrant goes, and the answer is first. Deciding it before computing the angle means you cannot be misled by the inverse tangent.

Computing the reference angle from the absolute value of y over x, and then placing it in the right quadrant, is the most reliable version of the calculation. The final check is two multiplications, and it catches every quadrant error.

64. Check: converting a point

Check

Sketch the point first: negative x, positive y, second quadrant. That rules out one of the options before any computation.

The distance is 2, and the reference angle is pi over 6, so the second-quadrant angle is 5 pi over 6. Two of the wrong options name the opposite point through the origin, the classic inverse tangent mistake made two different ways.

65. Check: counting petals

Check

Identify k before anything else: here it is 6. The number in front of the sine is the length of each petal, not the number of them.

Six is even, so every negative hump of r becomes a new petal, and the count is twice six, twelve. If you are ever unsure, remember the explanation, not just the rule: for even k, half a turn moves a negative hump into a gap.

66. Check: a rectangular form

Check

Rewrite everything in sines and cosines first. Cotangent times cosecant is cosine over sine squared.

Clearing the denominator and multiplying by r gives r squared sine squared equals r cosine, which is y squared equals x: a parabola opening to the right. It is Checkpoint 7.13 with the roles of x and y swapped, which is exactly why option B is tempting.

67. Explain the negative arcs

Explain it to yourself

Write your explanation before revealing. The aim is to give the reason for the petal rule in your own words, using the one idea of this lesson that makes the rule work.

A good answer mentions three things: over a full turn there are twice k humps of r; a negative hump is plotted half a turn away from where it points; and whether half a turn lands in a gap or on an existing petal depends on whether k is even or odd.

68. Exit ticket

Exit ticket

\[ r = 1 - 2\sin\theta \]

This combines the whole second half of the lesson in one curve. Work through the sketching pattern: symmetry, zeros, extremes, then name the shape.

The equation involves only sine, and sine of a half turn minus theta equals sine of theta, so the curve is symmetric about the vertical line. The radius is zero where sine is one half, at pi over 6 and 5 pi over 6. The farthest point is at three quarters of a turn, where sine is minus one and r is 3, which is the point three units straight down.

Because the constant is smaller than the coefficient of sine, the radius is negative between the two zeros, so this is a limaçon with an inner loop, opening downward.

69. Recap

Recap

Next, Section 7.4 uses the parametric form of a polar curve to find areas enclosed by these curves and the lengths of their arcs.

Stewart, Calculus: Early Transcendentals 8e, §10.3 Polar Coordinates §10.3, pp. 658-668 — the same material in Stewart

Two conversion formulas go from polar to rectangular with no thought at all. The reverse direction needs Pythagoras for the distance and a decision about the quadrant for the angle, because the inverse tangent only ever answers on the right-hand half of the plane.

Every point has infinitely many polar names: add full turns, or flip the sign of r and add a half turn. The pole has every angle. Those facts explain retraced circles, odd roses and inner loops.

Converting equations usually starts by multiplying by r. The catalogue of circles, lines, limaçons, roses and spirals, together with the three symmetry tests, lets you sketch most polar curves quickly. In the next section these curves are treated as parametric curves, and you will find the areas they enclose and the lengths of their arcs.