Chapter 10 · Parametric Equations and Polar Coordinates

10.3 Polar Coordinates

Locate points by a distance and a direction, $(r,\theta)$ — and describe circles, cardioids, roses and spirals with remarkably simple equations.

A radar screen does not show aircraft on a square grid. It shows each echo as a distance from the antenna and a direction (a bearing). A garden sprinkler, a lighthouse beam, the pickup pattern of a microphone and the spiral of a nautilus shell are all described most naturally in the same way: how far out, and in which direction. Rectangular coordinates $(x,y)$ can describe these shapes, but the equations are often clumsy — a cardioid becomes the quartic $\left(x^2+y^2-x\right)^2=x^2+y^2$.

In this section we introduce polar coordinates $(r,\theta)$. We learn to move between polar and rectangular descriptions (with care over which quadrant an angle lies in), to sketch the classic polar curves — circles, cardioids, limaçons, rose curves, lemniscates and spirals — using symmetry, and finally to compute slopes of tangent lines with the derivative. In Section 10.4 we will use the same ideas to find areas and arc lengths.

The Polar Coordinate System

Fix a point $O$, called the pole (it plays the role of the origin), and a ray starting at $O$, called the polar axis (we always draw it along the positive $x$-axis). Any point $P$ can be reached by turning through an angle $\theta$ from the polar axis and then walking a distance $r$ along that direction.

A point $P$ is located by its distance $r$ from the pole and the angle $\theta$ measured anticlockwise from the polar axis.

📘 Definition 10.4 — Polar coordinates

The point $P$ with polar coordinates $(r,\theta)$ is found by rotating the polar axis through the angle $\theta$ (anticlockwise if $\theta\gt0$, clockwise if $\theta\lt0$) and then moving a directed distance $r$ along the rotated ray:

if $r\gt0$, $P$ lies on the ray at angle $\theta$, a distance $r$ from $O$;
if $r\lt0$, $P$ lies on the opposite ray (angle $\theta+\pi$), a distance $|r|$ from $O$;
if $r=0$, $P$ is the pole, whatever the value of $\theta$.

Angles are measured in radians.

⚠️ One point, infinitely many names

Unlike rectangular coordinates, polar coordinates are not unique. Adding a full turn does not move the point, and reversing the direction while also reversing the sign of $r$ does not move it either. For every integer $k$,

$$(r,\theta)=(r,\theta+2k\pi)=(-r,\theta+\pi+2k\pi),$$

and the pole is $(0,\theta)$ for every $\theta$. This flexibility is what lets a single equation $r=f(\theta)$ draw a whole closed curve — but it also means that two curves can meet at a point even though no single pair $(r,\theta)$ satisfies both equations. Keep this in mind in Section 10.4.

✏️ Example 1 — From polar to rectangular, and other names for a point

Find the rectangular coordinates of the points with polar coordinates (a) $\left(4,\tfrac{2\pi}{3}\right)$, (b) $\left(-3,\tfrac{\pi}{4}\right)$, (c) $\left(2,-\tfrac{\pi}{2}\right)$. (d) Give three other polar names for the point $\left(2,\tfrac{\pi}{3}\right)$, two of them with $r\lt0$.

💡 Show solution▶

Tool From the figure, $x=r\cos\theta$ and $y=r\sin\theta$. These formulas remain true when $r\lt0$, because moving $|r|$ along the opposite ray multiplies both coordinates by $-1$.

(a) $x=4\cos\tfrac{2\pi}{3}=4\left(-\tfrac12\right)=-2$, $\;y=4\sin\tfrac{2\pi}{3}=4\cdot\tfrac{\sqrt3}{2}=2\sqrt3$. The point is $\left(-2,\,2\sqrt3\right)$, in the second quadrant as expected for an angle of $120^\circ$.

(b) $x=-3\cos\tfrac{\pi}{4}=-\tfrac{3\sqrt2}{2}$, $\;y=-3\sin\tfrac{\pi}{4}=-\tfrac{3\sqrt2}{2}$. The angle points into the first quadrant, but $r\lt0$ sends us the other way: the point $\left(-\tfrac{3\sqrt2}{2},-\tfrac{3\sqrt2}{2}\right)\approx(-2.1213,-2.1213)$ is in the third quadrant.

(c) $x=2\cos\left(-\tfrac{\pi}{2}\right)=0$, $\;y=2\sin\left(-\tfrac{\pi}{2}\right)=-2$: the point $(0,-2)$ (a negative angle is measured clockwise).

(d) Add $2\pi$: $\left(2,\tfrac{7\pi}{3}\right)$. Reverse the ray and the sign of $r$: $\left(-2,\tfrac{\pi}{3}+\pi\right)=\left(-2,\tfrac{4\pi}{3}\right)$, and subtracting $2\pi$ from that angle, $\left(-2,-\tfrac{2\pi}{3}\right)$.

Check All four names give $x=1$, $y=\sqrt3$; for instance $-2\cos\tfrac{4\pi}{3}=-2\left(-\tfrac12\right)=1$ and $-2\sin\tfrac{4\pi}{3}=-2\left(-\tfrac{\sqrt3}{2}\right)=\sqrt3$. ✓

📐 Theorem 10.5 — Converting between polar and rectangular coordinates

If the point $P$ has polar coordinates $(r,\theta)$ and rectangular coordinates $(x,y)$, then

$$x=r\cos\theta,\qquad y=r\sin\theta,$$ $$r^2=x^2+y^2,\qquad \tan\theta=\frac{y}{x}\quad(x\ne0).$$

To convert $(x,y)\ne(0,0)$ to polar form with $r\gt0$, take $r=\sqrt{x^2+y^2}$ and choose $\theta$ so that $\cos\theta=\dfrac{x}{r}$ and $\sin\theta=\dfrac{y}{r}$ — that is, an angle in the same quadrant as $(x,y)$. If $x=0$, take $\theta=\tfrac{\pi}{2}$ when $y\gt0$ and $\theta=-\tfrac{\pi}{2}$ when $y\lt0$.

⚠️ The quadrant trap

The equation $\tan\theta=\dfrac yx$ has two solutions in every interval of length $2\pi$, pointing in opposite directions. A calculator's $\tan^{-1}$ always returns an angle in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, i.e. a direction in the first or fourth quadrant. So:

if $x\gt0$: $\theta=\tan^{-1}\left(\dfrac yx\right)$;
if $x\lt0$: $\theta=\tan^{-1}\left(\dfrac yx\right)+\pi$ (the calculator's direction must be reversed).

Always sketch the point and check that your angle points into the right quadrant.

✏️ Example 2 — From rectangular to polar, watching the quadrant

Find polar coordinates with $r\gt0$ and $0\le\theta\lt2\pi$ for the points (a) $(1,1)$, (b) $(-3,4)$, (c) $(0,3)$, (d) $\left(-\sqrt3,-1\right)$.

💡 Show solution▶

(a) $r=\sqrt{1+1}=\sqrt2$. The point is in quadrant I and $\tan\theta=1$, so $\theta=\tfrac{\pi}{4}$: $\left(\sqrt2,\tfrac{\pi}{4}\right)$.

(b) $r=\sqrt{9+16}=5$. Here $x\lt0$, so the point is in quadrant II. The calculator gives $\tan^{-1}\left(-\tfrac43\right)\approx-0.9273$, which points into quadrant IV — the wrong way. Add $\pi$: $\theta\approx-0.9273+\pi\approx2.2143$ (about $126.87^\circ$). Answer: $(5,\ 2.2143)$, exactly $\left(5,\ \pi-\tan^{-1}\tfrac43\right)$.

(d) $r=\sqrt{3+1}=2$. Now $\tan\theta=\dfrac{-1}{-\sqrt3}=\dfrac{1}{\sqrt3}$, and $\tan^{-1}\tfrac{1}{\sqrt3}=\tfrac{\pi}{6}$ — but the point is in quadrant III. So $\theta=\tfrac{\pi}{6}+\pi=\tfrac{7\pi}{6}$: $\left(2,\tfrac{7\pi}{6}\right)$.

Check (d): $2\cos\tfrac{7\pi}{6}=2\left(-\tfrac{\sqrt3}{2}\right)=-\sqrt3$ and $2\sin\tfrac{7\pi}{6}=2\left(-\tfrac12\right)=-1$. ✓ The tempting answer $\left(2,\tfrac{\pi}{6}\right)$ is the point $\left(\sqrt3,1\right)$ — the mirror image through the pole.

Polar Equations and Their Graphs

A polar equation is an equation in $r$ and $\theta$, most often of the form $r=f(\theta)$. Its graph is the set of all points that have at least one polar name $(r,\theta)$ satisfying the equation. Theorem 10.5 lets us translate an equation from one system to the other — useful both for recognising a polar curve and for finding a simple polar description of a familiar shape.

✏️ Example 3 — Converting equations

(a) Write $r=4\sin\theta$ in rectangular form and identify the curve. (b) Describe the graph of $\theta=\tfrac{\pi}{3}$. (c) Write the circle $x^2+y^2=6x$ in polar form. (d) Write the parabola $y=x^2$ in polar form.

💡 Show solution▶

(a) Multiply by $r$ to create the combinations $r^2$ and $r\sin\theta$: $r^2=4r\sin\theta$, so $x^2+y^2=4y$. Completing the square, $$x^2+(y-2)^2=4,$$ a circle with centre $(0,2)$ and radius $2$. (Multiplying by $r$ could add the pole, but the pole is already on the curve: $\theta=0$ gives $r=0$.)

(b) All points whose direction is $\tfrac{\pi}{3}$, with any $r$ — positive $r$ gives the ray into quadrant I and negative $r$ the opposite ray into quadrant III. Together they form the full line through the pole with slope $\tan\tfrac{\pi}{3}=\sqrt3$: $\;y=\sqrt3\,x$.

(c) Substitute $x^2+y^2=r^2$ and $x=r\cos\theta$: $r^2=6r\cos\theta$, so $r=0$ or $r=6\cos\theta$. The pole is already on $r=6\cos\theta$ (at $\theta=\tfrac{\pi}{2}$), so the circle is simply $$r=6\cos\theta.$$

(d) $r\sin\theta=r^2\cos^2\theta$. Dividing by $r$ (the pole is still included, at $\theta=0$): $$r=\frac{\sin\theta}{\cos^2\theta}=\tan\theta\sec\theta,\qquad -\tfrac{\pi}{2}\lt\theta\lt\tfrac{\pi}{2}.$$ Here the polar form is less pleasant than the rectangular one — polar coordinates are a tool for shapes organised around a centre.

🧭 Problem-solving strategy — sketching a polar curve $r=f(\theta)$

Symmetry. Apply the tests of Theorem 10.6 below; symmetry can halve or quarter the work.
Where does the curve reach the pole? Solve $f(\theta)=0$. These directions are often tangent lines at the pole (Theorem 10.7).
Extremes. Find where $|r|$ is largest — the points farthest from the pole.
Table of values. Tabulate $r$ at convenient angles over one period ($\theta$ in steps of $\tfrac{\pi}{6}$ or $\tfrac{\pi}{12}$). Note where $r$ is negative: those points are drawn on the opposite ray.
Join the points in order of increasing $\theta$; it helps to sketch the ordinary graph of $r$ against $\theta$ first (Interactive 1 shows both side by side).

📐 Theorem 10.6 — Symmetry tests for polar curves

The graph of a polar equation is symmetric

about the polar axis (the $x$-axis) if replacing $(r,\theta)$ by $(r,-\theta)$, or by $(-r,\pi-\theta)$, gives an equivalent equation;
about the line $\theta=\tfrac{\pi}{2}$ (the $y$-axis) if replacing $(r,\theta)$ by $(r,\pi-\theta)$, or by $(-r,-\theta)$, gives an equivalent equation;
about the pole if replacing $(r,\theta)$ by $(-r,\theta)$, or by $(r,\theta+\pi)$, gives an equivalent equation.

Each pair names the same reflected point in two ways. The tests are sufficient but not necessary: because a point has many names, a curve can be symmetric even when one form of a test fails, so try both forms before concluding there is no symmetry.

For example, $r=1+\cos\theta$ is unchanged by $\theta\to-\theta$ (cosine is even), so it is symmetric about the polar axis; $r=1+\sin\theta$ is unchanged by $\theta\to\pi-\theta$, so it is symmetric about the $y$-axis. The table below collects the curves you will meet most often.

Circles$r=a$: centre the pole, radius $|a|$.
$r=2a\cos\theta$: centre $(a,0)$, radius $|a|$.
$r=2a\sin\theta$: centre $(0,a)$, radius $|a|$.

Lines$\theta=\alpha$: the line through the pole at angle $\alpha$.
$r=a\sec\theta$: the line $x=a$.
$r=a\csc\theta$: the line $y=a$.

Limaçons $r=a+b\cos\theta$ or $a+b\sin\theta$ ($a,b\gt0$)$a\lt b$: inner loop.
$a=b$: cardioid (heart shape, a cusp at the pole).
$b\lt a\lt2b$: dimpled.
$a\ge2b$: convex (oval).

Roses $r=a\cos n\theta$ or $a\sin n\theta$ ($n\ge2$ an integer)$n$ odd: $n$ petals, traced once for $0\le\theta\le\pi$.
$n$ even: $2n$ petals, traced once for $0\le\theta\le2\pi$.
Every petal has length $|a|$.

Lemniscates$r^2=a^2\cos2\theta$ or $r^2=a^2\sin2\theta$: a figure-eight through the pole, defined only where the right-hand side is $\ge0$.

Spirals$r=a\theta$ ($\theta\ge0$): Archimedean spiral, successive turns $2\pi a$ apart.
$r=ae^{b\theta}$: logarithmic (equiangular) spiral, turns grow geometrically.

✏️ Example 4 — Graphing a four-petal rose

Sketch the graph of $r=3\sin2\theta$. Identify its symmetries, the points farthest from the pole, and the order in which the petals are drawn as $\theta$ runs from $0$ to $2\pi$.

💡 Show solution▶

Symmetry Replace $(r,\theta)$ by $(-r,-\theta)$: $-r=3\sin(-2\theta)=-3\sin2\theta$, i.e. $r=3\sin2\theta$ ✓ — symmetric about the $y$-axis. Replace $\theta$ by $\theta+\pi$: $\sin(2\theta+2\pi)=\sin2\theta$ ✓ — symmetric about the pole. A curve symmetric about the $y$-axis and the pole is also symmetric about the $x$-axis. (Note that the simple test $\theta\to-\theta$ fails here, yet the curve is symmetric about the polar axis — the tests are not necessary conditions.)

Table On $0\le\theta\le\tfrac{\pi}{2}$:

$\theta$	$0$	$\tfrac{\pi}{12}$	$\tfrac{\pi}{6}$	$\tfrac{\pi}{4}$	$\tfrac{\pi}{3}$	$\tfrac{5\pi}{12}$	$\tfrac{\pi}{2}$
$r$	$0$	$1.5$	$2.598$	$3$	$2.598$	$1.5$	$0$

The curve leaves the pole along $\theta=0$, swings out to the tip $\left(3,\tfrac{\pi}{4}\right)$ and returns to the pole along $\theta=\tfrac{\pi}{2}$: one petal in quadrant I.

Negative $r$ For $\tfrac{\pi}{2}\lt\theta\lt\pi$, $\sin2\theta\lt0$, so $r\lt0$: the angle points into quadrant II but the points are drawn in quadrant IV. The petals appear in the order I, IV, III, II, with tips at distance $3$ in the directions $\tfrac{\pi}{4},\tfrac{7\pi}{4},\tfrac{5\pi}{4},\tfrac{3\pi}{4}$.

Pole $r=0$ when $\sin2\theta=0$, i.e. $\theta=0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2}$: the curve passes through the pole four times, tangent to the coordinate axes there (by Theorem 10.7 below, since $f'(\theta)=6\cos2\theta\ne0$ at these angles). Since $n=2$ is even, there are $2n=4$ petals, traced once for $0\le\theta\le2\pi$.

Check Select “four-petal rose” in Interactive 1: the side graph of $r$ against $\theta$ dips below zero on $\left(\tfrac{\pi}{2},\pi\right)$ and the arm turns red, pointing backwards into quadrant IV. ✓

As $\theta$ sweeps, the blue arm has length $r=f(\theta)$ and draws the curve on the left; the right panel is the ordinary graph of $r$ against $\theta$. When $r\lt0$ the arm turns red and points the opposite way (the dashed ray shows the direction $\theta$ itself). Watch which parts of the right-hand graph produce which loops or petals.

Curve:$\theta$:

The curve in the $xy$-plane

$r=f(\theta)$ as an ordinary graph

$\theta$—

$r=f(\theta)$—

$(x,y)$—

point drawn…—

Choose a family and move the parameters. For limaçons, Play slides $a$ from $0$ to $3$ so you can watch the inner loop shrink to a cusp (cardioid), then a dimple, then a convex oval. For roses, Play steps through $n=1,2,\dots,8$ and draws each rose: count the petals and compare odd with even $n$. Parts drawn with $r\lt0$ are red.

Family:ab

equation—

type—

symmetry—

one full trace—

Slopes and Tangent Lines

A polar curve $r=f(\theta)$ is really a parametric curve with parameter $\theta$: $$x=f(\theta)\cos\theta,\qquad y=f(\theta)\sin\theta.$$ So we can use the parametric slope formula $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ from Section 10.2. By the product rule, $$\frac{dx}{d\theta}=f'(\theta)\cos\theta-f(\theta)\sin\theta,\qquad \frac{dy}{d\theta}=f'(\theta)\sin\theta+f(\theta)\cos\theta.$$ Note: $dr/d\theta$ is not the slope of the curve — it measures how fast the distance from the pole changes, not the steepness in the $xy$-plane.

📐 Theorem 10.7 — Slope of a polar curve; tangents at the pole

Let $f$ be differentiable at $\theta$ and consider the curve $r=f(\theta)$. If $\dfrac{dx}{d\theta}\ne0$ at $\theta$, the slope of the tangent line at the point with parameter $\theta$ is

$$\frac{dy}{dx}=\frac{f'(\theta)\sin\theta+f(\theta)\cos\theta}{f'(\theta)\cos\theta-f(\theta)\sin\theta}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}.$$

Horizontal tangent where $\dfrac{dy}{d\theta}=0$ and $\dfrac{dx}{d\theta}\ne0$; vertical tangent where $\dfrac{dx}{d\theta}=0$ and $\dfrac{dy}{d\theta}\ne0$. If both vanish, examine the limit of $\dfrac{dy}{dx}$.
Tangents at the pole. If $f(\alpha)=0$ and $f'(\alpha)\ne0$, then the line $\theta=\alpha$ is tangent to the curve at the pole. (With $r=0$ the formula reduces to $\dfrac{f'(\alpha)\sin\alpha}{f'(\alpha)\cos\alpha}=\tan\alpha$.)

✏️ Example 5 — Slope and tangent line on a cardioid

Find the slope of the cardioid $r=1+\sin\theta$ at $\theta=\tfrac{\pi}{3}$, and the equation of the tangent line there in rectangular form.

💡 Show solution▶

Values $f\left(\tfrac{\pi}{3}\right)=1+\tfrac{\sqrt3}{2}$, $\;f'(\theta)=\cos\theta$, so $f'\left(\tfrac{\pi}{3}\right)=\tfrac12$; also $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$, $\cos\tfrac{\pi}{3}=\tfrac12$.

Numerator $\dfrac{dy}{d\theta}=\tfrac12\cdot\tfrac{\sqrt3}{2}+\left(1+\tfrac{\sqrt3}{2}\right)\tfrac12=\tfrac{\sqrt3}{4}+\tfrac12+\tfrac{\sqrt3}{4}=\tfrac{1+\sqrt3}{2}.$

Denominator $\dfrac{dx}{d\theta}=\tfrac12\cdot\tfrac12-\left(1+\tfrac{\sqrt3}{2}\right)\tfrac{\sqrt3}{2}=\tfrac14-\tfrac{\sqrt3}{2}-\tfrac34=-\tfrac{1+\sqrt3}{2}.$ So $$\frac{dy}{dx}=\frac{(1+\sqrt3)/2}{-(1+\sqrt3)/2}=-1.$$

Line The point is $x_0=r\cos\tfrac{\pi}{3}=\tfrac12+\tfrac{\sqrt3}{4}\approx0.9330$, $\;y_0=r\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}+\tfrac34\approx1.6160$. The tangent line is $y-y_0=-(x-x_0)$, i.e. $$y=-x+\tfrac54+\tfrac{3\sqrt3}{4}\approx-x+2.5490.$$

Check A numerical slope from two nearby points, $\theta=\tfrac{\pi}{3}\pm0.001$, gives $\dfrac{\Delta y}{\Delta x}\approx-1.0000$. ✓

✏️ Example 6 — Horizontal and vertical tangents, and the cusp at the pole

Find all points on the cardioid $r=1+\cos\theta$, $0\le\theta\lt2\pi$, where the tangent line is horizontal or vertical. What happens at the pole?

💡 Show solution▶

Derivatives With $f=1+\cos\theta$, $f'=-\sin\theta$: $$\frac{dy}{d\theta}=-\sin^2\theta+(1+\cos\theta)\cos\theta=2\cos^2\theta+\cos\theta-1=(2\cos\theta-1)(\cos\theta+1),$$ $$\frac{dx}{d\theta}=-\sin\theta\cos\theta-(1+\cos\theta)\sin\theta=-\sin\theta\,(1+2\cos\theta).$$

Horizontal $\dfrac{dy}{d\theta}=0$ when $\cos\theta=\tfrac12$ ($\theta=\tfrac{\pi}{3},\tfrac{5\pi}{3}$) or $\cos\theta=-1$ ($\theta=\pi$). At $\tfrac{\pi}{3}$ and $\tfrac{5\pi}{3}$, $\dfrac{dx}{d\theta}=\mp\tfrac{\sqrt3}{2}\cdot2\ne0$, so the tangents are horizontal at $r=\tfrac32$: the points $\left(\tfrac34,\tfrac{3\sqrt3}{4}\right)$ and $\left(\tfrac34,-\tfrac{3\sqrt3}{4}\right)$.

Vertical $\dfrac{dx}{d\theta}=0$ when $\sin\theta=0$ ($\theta=0,\pi$) or $\cos\theta=-\tfrac12$ ($\theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$). At $\theta=0$: $\dfrac{dy}{d\theta}=(1)(2)=2\ne0$ — vertical tangent at $(2,0)$. At $\tfrac{2\pi}{3},\tfrac{4\pi}{3}$: $\dfrac{dy}{d\theta}=(-2)\left(\tfrac12\right)=-1\ne0$, $r=\tfrac12$ — vertical tangents at $\left(-\tfrac14,\pm\tfrac{\sqrt3}{4}\right)$.

The pole At $\theta=\pi$ both derivatives vanish, and $f'(\pi)=0$, so the pole test does not apply. Put $\theta=\pi+h$: then $\cos\theta+1\approx\tfrac{h^2}{2}$, $\;2\cos\theta-1\approx-3$, $\;\sin\theta\approx-h$, $\;1+2\cos\theta\approx-1$, so $$\frac{dy}{dx}\approx\frac{(-3)\,h^2/2}{-h}=\frac{3h}{2}\to0.$$ The cardioid has a cusp at the pole with a horizontal tangent (along the polar axis).

Check Interactive 1 with the cardioid: the highest point occurs at $\theta=60^\circ$, where $y=\tfrac{3\sqrt3}{4}\approx1.2990$, and the curve flattens into the pole along the $x$-axis. ✓

⚠️ Common mistakes with polar slopes

Using $dr/d\theta$ as the slope. On the circle $r=2$, $dr/d\theta=0$ everywhere, yet the slope takes every value.
Forgetting the $f\sin\theta$ and $f\cos\theta$ terms. Both $x$ and $y$ are products, so each derivative has two terms.
Declaring a tangent “horizontal” when $dy/d\theta=dx/d\theta=0$. That is a $\tfrac00$ form — usually the pole or a cusp — and needs a limit, as in Example 6.
Missing tangents at the pole. Every root $\alpha$ of $f(\theta)=0$ with $f'(\alpha)\ne0$ gives a tangent line $\theta=\alpha$ at the pole; a rose passes through the pole between every pair of petals.

Exercises

Points. In 1–5 convert from polar to rectangular coordinates. In 6–9 find polar coordinates with $r\gt0$ and $0\le\theta\lt2\pi$ (give exact angles where possible, otherwise four decimal places).

1 $\left(3,\tfrac{\pi}{6}\right)$
2 $\left(-2,\tfrac{\pi}{4}\right)$
3 $(5,\pi)$
4 $\left(4,-\tfrac{2\pi}{3}\right)$
5 $\left(-1,\tfrac{7\pi}{6}\right)$
6 $(2,2)$
7 $\left(-1,\sqrt3\right)$
8 $\left(3,-3\sqrt3\right)$
9 $(-4,-3)$
10 Give three other polar names for $\left(3,\tfrac{\pi}{4}\right)$, at least two with $r\lt0$.

💡 Show solutions▶

1 $x=3\cos\tfrac{\pi}{6}=\tfrac{3\sqrt3}{2}$, $y=3\sin\tfrac{\pi}{6}=\tfrac32$: $\left(\tfrac{3\sqrt3}{2},\tfrac32\right)\approx(2.5981,1.5)$.

2 $x=-2\cdot\tfrac{\sqrt2}{2}=-\sqrt2$, $y=-\sqrt2$: $\left(-\sqrt2,-\sqrt2\right)$ — quadrant III because $r\lt0$.

3 $(5\cos\pi,5\sin\pi)=(-5,0)$.

4 $x=4\cos\left(-\tfrac{2\pi}{3}\right)=-2$, $y=4\sin\left(-\tfrac{2\pi}{3}\right)=-2\sqrt3$: $\left(-2,-2\sqrt3\right)$.

5 $x=-\cos\tfrac{7\pi}{6}=\tfrac{\sqrt3}{2}$, $y=-\sin\tfrac{7\pi}{6}=\tfrac12$: $\left(\tfrac{\sqrt3}{2},\tfrac12\right)$ — the angle points into quadrant III, the negative $r$ brings the point to quadrant I.

6 $r=\sqrt8=2\sqrt2$, quadrant I, $\tan\theta=1$: $\left(2\sqrt2,\tfrac{\pi}{4}\right)$.

7 $r=2$; quadrant II; $\tan^{-1}\left(-\sqrt3\right)=-\tfrac{\pi}{3}$, add $\pi$: $\left(2,\tfrac{2\pi}{3}\right)$.

8 $r=\sqrt{9+27}=6$; quadrant IV; $\tan^{-1}\left(-\sqrt3\right)=-\tfrac{\pi}{3}$, and to lie in $[0,2\pi)$ add $2\pi$: $\left(6,\tfrac{5\pi}{3}\right)$.

9 $r=5$; quadrant III; $\tan^{-1}\tfrac34\approx0.6435$, add $\pi$: $\theta=\pi+\tan^{-1}\tfrac34\approx3.7851$. Answer $(5,\ 3.7851)$.

10 For example $\left(3,\tfrac{9\pi}{4}\right)$, $\left(-3,\tfrac{5\pi}{4}\right)$, $\left(-3,-\tfrac{3\pi}{4}\right)$ (any $\left(3,\tfrac{\pi}{4}+2k\pi\right)$ or $\left(-3,\tfrac{5\pi}{4}+2k\pi\right)$ is correct).

Equations. In 1–7 convert to a rectangular equation and identify the curve where possible. In 8–12 convert to a polar equation, solved for $r$ (or $r^2$) where possible.

1 $r=5$
2 $\theta=\tfrac{3\pi}{4}$
3 $r=6\cos\theta$
4 $r=-4\sin\theta$
5 $r=\dfrac{2}{\cos\theta+\sin\theta}$
6 $r=2\csc\theta$
7 $r^2=\sin2\theta$
8 $x^2+y^2=4y$
9 $y=3x$
10 $x^2-y^2=4$
11 $y^2=8x$
12 $\left(x^2+y^2\right)^2=9\left(x^2-y^2\right)$

💡 Show solutions▶

1 $r^2=25$: $x^2+y^2=25$, the circle with centre the origin and radius $5$.

2 $\tan\theta=\tan\tfrac{3\pi}{4}=-1$, and negative $r$ gives the opposite ray: the whole line $y=-x$.

3 $r^2=6r\cos\theta$: $x^2+y^2=6x$, i.e. $(x-3)^2+y^2=9$ — circle, centre $(3,0)$, radius $3$.

4 $r^2=-4r\sin\theta$: $x^2+y^2=-4y$, i.e. $x^2+(y+2)^2=4$ — circle, centre $(0,-2)$, radius $2$.

5 $r\cos\theta+r\sin\theta=2$: $x+y=2$, a straight line.

6 $r\sin\theta=2$: the horizontal line $y=2$.

7 $r^2=2\sin\theta\cos\theta$; multiply by $r^2$: $r^4=2(r\sin\theta)(r\cos\theta)$, so $\left(x^2+y^2\right)^2=2xy$ — a lemniscate with loops in quadrants I and III.

8 $r^2=4r\sin\theta$: $r=4\sin\theta$ (the pole is included at $\theta=0$).

9 $\tan\theta=3$: $\theta=\tan^{-1}3\approx1.2490$ — with all real $r$ this is the whole line.

10 $r^2\left(\cos^2\theta-\sin^2\theta\right)=4$: $r^2\cos2\theta=4$, i.e. $r^2=4\sec2\theta$.

11 $r^2\sin^2\theta=8r\cos\theta$, so $r=\dfrac{8\cos\theta}{\sin^2\theta}=8\cot\theta\csc\theta$ (the vertex, the pole, occurs at $\theta=\tfrac{\pi}{2}$).

12 $r^4=9r^2\left(\cos^2\theta-\sin^2\theta\right)=9r^2\cos2\theta$, so $r^2=9\cos2\theta$ — a lemniscate (dividing by $r^2$ loses nothing, since the pole is on the curve at $\theta=\tfrac{\pi}{4}$).

III

Graphing. For 1–7, identify the curve, state its symmetries, find the maximum value of $|r|$ and the angles at which the curve passes through the pole, and sketch it. Then check your sketch with the plotter below.

1 $r=2+2\cos\theta$
2 $r=1+2\sin\theta$
3 $r=3+2\cos\theta$
4 $r=4\cos3\theta$
5 $r=2\sin4\theta$
6 $r^2=9\sin2\theta$
7 $r=\theta$, $\theta\ge0$ (where does the spiral cross the $x$-axis?)
8 Conceptual. Show that the test “replace $\theta$ by $-\theta$” fails for $r=\sin2\theta$, yet the curve is symmetric about the polar axis. Which form of the test succeeds?

Check your sketch: trace the curve and compare

Part:$\theta$:

$\theta$—

$r$—

$(x,y)$—

💡 Show solutions▶

1 Cardioid ($a=b=2$). Symmetric about the polar axis ($\cos$ is even). Maximum $r=4$ at $\theta=0$; passes through the pole once, at $\theta=\pi$ (a cusp pointing left).

2 Limaçon with an inner loop ($a=1\lt b=2$). Symmetric about $\theta=\tfrac{\pi}{2}$. Maximum $|r|=3$ at $\theta=\tfrac{\pi}{2}$, the point $(0,3)$. $r=0$ when $\sin\theta=-\tfrac12$: $\theta=\tfrac{7\pi}{6},\tfrac{11\pi}{6}$. For $\tfrac{7\pi}{6}\lt\theta\lt\tfrac{11\pi}{6}$, $r\lt0$ and the small inner loop is drawn, reaching $(0,1)$ at $\theta=\tfrac{3\pi}{2}$ (where $r=-1$).

3 Dimpled limaçon ($2\lt3\lt4$). Symmetric about the polar axis. $1\le r\le5$: maximum $5$ at $\theta=0$, minimum $1$ at $\theta=\pi$. Never passes through the pole.

4 Three-petal rose ($n=3$ odd), petals of length $4$, traced once on $[0,\pi]$. Symmetric about the polar axis. Tips in the directions $0,\tfrac{2\pi}{3},\tfrac{4\pi}{3}$ (the tip at $\theta=\tfrac{\pi}{3}$ has $r=-4$ and is drawn in direction $\tfrac{4\pi}{3}$). Through the pole at $\theta=\tfrac{\pi}{6},\tfrac{\pi}{2},\tfrac{5\pi}{6}$.

5 Eight-petal rose ($n=4$ even, so $2n=8$ petals) of length $2$, traced once on $[0,2\pi]$. Symmetric about both axes and the pole. Tips where $|\sin4\theta|=1$: directions $\tfrac{\pi}{8}+\tfrac{k\pi}{4}$. Through the pole at $\theta=\tfrac{k\pi}{4}$, $k=0,\dots,7$.

6 Lemniscate; real $r$ only when $\sin2\theta\ge0$, i.e. $0\le\theta\le\tfrac{\pi}{2}$ or $\pi\le\theta\le\tfrac{3\pi}{2}$, so the loops lie in quadrants I and III. Symmetric about the pole ($r\to-r$ leaves $r^2$ unchanged) and about the line $y=x$. Maximum $|r|=3$ at $\theta=\tfrac{\pi}{4}$; through the pole at $\theta=0,\tfrac{\pi}{2}$.

7 Archimedean spiral, starting at the pole when $\theta=0$; $|r|$ is unbounded. It crosses the positive $x$-axis at $\theta=2\pi,4\pi,\dots$ (distances $2\pi\approx6.28$, $4\pi\approx12.57$, …) and the negative $x$-axis at $\theta=\pi,3\pi,\dots$ (distances $\pi$, $3\pi$, …): successive turns are $2\pi$ apart. With $\theta\ge0$ only there is no symmetry.

8 $\theta\to-\theta$ gives $r=\sin(-2\theta)=-\sin2\theta$, a different equation. The other form, $(r,\theta)\to(-r,\pi-\theta)$, gives $-r=\sin(2\pi-2\theta)=-\sin2\theta$, i.e. $r=\sin2\theta$ ✓. So the rose is symmetric about the polar axis — a failed test proves nothing.

Slopes, tangents and applications.

1 Find the slope of $r=2\sin\theta$ at $\theta=\tfrac{\pi}{6}$, and check it using the rectangular equation of the circle.
2 Find the slope of the spiral $r=\theta$ at $\theta=\tfrac{\pi}{2}$.
3 Find the tangent line to $r=3\cos\theta$ at $\theta=\tfrac{\pi}{4}$ (in rectangular form).
4 Find all points on the cardioid $r=1+\sin\theta$, $0\le\theta\lt2\pi$, where the tangent line is horizontal. Describe the curve at the pole.
5 Find the tangent lines at the pole for the rose $r=\sin3\theta$.
6 Find the tangent lines at the pole for the limaçon $r=1+2\cos\theta$.
7 Applied — air-traffic radar. A radar at the pole records an aircraft at distance $12$ km, direction $150^\circ$ (measured anticlockwise from east), and one minute later at $13$ km, direction $140^\circ$. Find the rectangular positions, the distance flown and the average ground speed in km/h.
8 Challenge — the equiangular spiral. Let $\psi$ be the angle between the radius arm and the tangent line of $r=f(\theta)$. Show that $\tan\psi=\dfrac{r}{dr/d\theta}$, and deduce that for $r=e^{\theta}$ the angle $\psi$ is the same at every point. (This is why the nautilus shell and the flight path of a moth circling a lamp follow logarithmic spirals.)

💡 Show solutions▶

1 $f=1$, $f'=2\cos\tfrac{\pi}{6}=\sqrt3$. $\dfrac{dy}{dx}=\dfrac{\sqrt3\cdot\tfrac12+1\cdot\tfrac{\sqrt3}{2}}{\sqrt3\cdot\tfrac{\sqrt3}{2}-1\cdot\tfrac12}=\dfrac{\sqrt3}{1}=\sqrt3$. Check: the curve is $x^2+(y-1)^2=1$, the point is $\left(\tfrac{\sqrt3}{2},\tfrac12\right)$, and implicit differentiation gives $\dfrac{dy}{dx}=-\dfrac{x}{y-1}=-\dfrac{\sqrt3/2}{-1/2}=\sqrt3$. ✓

2 $f=\tfrac{\pi}{2}$, $f'=1$: $\dfrac{dy}{dx}=\dfrac{1\cdot1+\tfrac{\pi}{2}\cdot0}{1\cdot0-\tfrac{\pi}{2}\cdot1}=-\dfrac{2}{\pi}\approx-0.6366$.

3 $f=\tfrac{3}{\sqrt2}$, $f'=-\tfrac{3}{\sqrt2}$; numerator $-\tfrac{3}{\sqrt2}\cdot\tfrac{1}{\sqrt2}+\tfrac{3}{\sqrt2}\cdot\tfrac{1}{\sqrt2}=0$, denominator $-\tfrac32-\tfrac32=-3\ne0$. Slope $0$; the point is $\left(\tfrac32,\tfrac32\right)$, so the tangent line is $y=\tfrac32$ — the top of the circle with centre $\left(\tfrac32,0\right)$ and radius $\tfrac32$. ✓

4 $\dfrac{dy}{d\theta}=\cos\theta\sin\theta+(1+\sin\theta)\cos\theta=\cos\theta(1+2\sin\theta)$; $\dfrac{dx}{d\theta}=\cos^2\theta-\sin\theta-\sin^2\theta=(1-2\sin\theta)(1+\sin\theta)$. Horizontal: $\theta=\tfrac{\pi}{2}$ ($dx/d\theta=-2$) at $(0,2)$; $\sin\theta=-\tfrac12$, i.e. $\theta=\tfrac{7\pi}{6},\tfrac{11\pi}{6}$ ($dx/d\theta=1$, $r=\tfrac12$) at $\left(\mp\tfrac{\sqrt3}{4},-\tfrac14\right)$. At $\theta=\tfrac{3\pi}{2}$ both derivatives vanish: it is the pole, a cusp, and $\left|\tfrac{dy}{dx}\right|\to\infty$ there, so the tangent is vertical (the line $\theta=\tfrac{3\pi}{2}$).

5 $\sin3\theta=0$ for $\theta=0,\tfrac{\pi}{3},\tfrac{2\pi}{3}$ in $[0,\pi)$, and $f'=3\cos3\theta=\pm3\ne0$ there. Tangent lines at the pole: $\theta=0$, $\theta=\tfrac{\pi}{3}$, $\theta=\tfrac{2\pi}{3}$, i.e. $y=0$, $y=\sqrt3\,x$, $y=-\sqrt3\,x$.

6 $1+2\cos\theta=0$ at $\theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$; $f'=-2\sin\theta=\mp\sqrt3\ne0$. Tangents: $\theta=\tfrac{2\pi}{3}$ and $\theta=\tfrac{4\pi}{3}$, i.e. $y=-\sqrt3\,x$ and $y=\sqrt3\,x$ (the inner loop crosses itself at the pole).

7 $P_1=(12\cos150^\circ,12\sin150^\circ)=\left(-6\sqrt3,6\right)\approx(-10.392,6.000)$ km; $P_2=(13\cos140^\circ,13\sin140^\circ)\approx(-9.959,8.356)$ km. Distance $\sqrt{(0.434)^2+(2.356)^2}\approx2.396$ km (or by the law of cosines, $\sqrt{12^2+13^2-2\cdot12\cdot13\cos10^\circ}\approx2.396$ km). Speed $\approx2.396\times60\approx143.7$ km/h.

8 The tangent makes angle $\phi$ with the $x$-axis where $\tan\phi=\dfrac{dy}{dx}$, and $\psi=\phi-\theta$. Using $\tan(\phi-\theta)=\dfrac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$ with Theorem 10.7 and simplifying with $\sin^2\theta+\cos^2\theta=1$: numerator $\dfrac{f}{\cos\theta\,(f'\cos\theta-f\sin\theta)}$, denominator $\dfrac{f'}{\cos\theta\,(f'\cos\theta-f\sin\theta)}$, so $\tan\psi=\dfrac{f}{f'}=\dfrac{r}{dr/d\theta}$. For $r=e^\theta$, $dr/d\theta=e^\theta$, so $\tan\psi=1$ and $\psi=\tfrac{\pi}{4}$ at every point.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why do sound engineers choose cardioid, supercardioid or hypercardioid microphones, and where exactly is each one “deaf”?

Why it matters. A microphone's sensitivity in each direction is plotted as a polar curve — its polar pattern. Broadcasters, concert venues and hearing-aid designers worldwide choose patterns that pick up a speaker while rejecting noise from the audience or loudspeakers. The standard first-order patterns are the limaçons $r=a+(1-a)\cos\theta$, $0\le a\le1$.

What to do.

Use Interactive 2 (limaçon $r=a+b\cos\theta$ with $b=1-a$) to sketch the omnidirectional ($a=1$), cardioid ($a=0.5$), supercardioid ($a\approx0.37$), hypercardioid ($a=0.25$) and figure-eight ($a=0$) patterns.
For each, find the null angle where the microphone hears nothing: solve $r=0$, i.e. $\cos\theta=-\dfrac{a}{1-a}$.
Explain, using the limaçon classification, why a rear pickup lobe (an inner loop) appears exactly when $a\lt0.5$.
A stage monitor loudspeaker sits $125^\circ$ off-axis. Which pattern rejects it best? Justify with your null angles.

Tools: Interactive 2, calculator, optionally a free audio-measurement app and a phone loudspeaker.

Hand in: A one-page note with the five sketches, a table of null angles, and a recommendation for the stage set-up.

Going further. Measure your own phone's microphone: play a steady tone from 12 directions at equal distance, record the level, and plot the results on polar axes. Which limaçon fits best?

Instructor note: Null angles — cardioid $180^\circ$, supercardioid $\cos^{-1}(-0.37/0.63)\approx126.0^\circ$, hypercardioid $\cos^{-1}(-1/3)\approx109.5^\circ$, figure-eight $90^\circ$; the omnidirectional pattern has none. The supercardioid's null is almost exactly at $125^\circ$. Inner loop $\iff a\lt b=1-a\iff a\lt0.5$.

🧪 Self-check quiz

0/5 correct

Q1. The point with polar coordinates $\left(-2,\tfrac{\pi}{2}\right)$ has rectangular coordinates

Q2. Polar coordinates of $(-1,-1)$ with $r\gt0$ and $0\le\theta\lt2\pi$ are

Q3. How many petals does the rose $r=5\cos6\theta$ have?

Q4. The graph of $r=4\sin\theta$ is

Q5. For $r=f(\theta)$ with $f(\alpha)=0$ and $f'(\alpha)\ne0$, the tangent line at the pole is

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