--- title: Polar Coordinates module: Parametric Equations and Polar Coordinates moduleNumber: 8 lessonNumber: 2 order: 802 summary: > Polar coordinates locate a point by a distance from the pole and an angle from the polar axis, giving circles, spirals, and flower-shaped curves short equations. Conversion between the two systems is right-triangle trigonometry, and treating a polar curve as a parametric curve in the angle yields the tangent, area, and arc-length formulas. topics: [Parametric Equations and Polar Coordinates] sources: - book: Stewart ref: "Ch. 10 — Parametric Equations and Polar Coordinates; §10.3 Polar Coordinates, §10.4 Areas and Lengths in Polar Coordinates" --- Cartesian coordinates measure a point by two perpendicular distances. Many curves that are awkward in that grid become simple when a point is instead located by how far it is from a center and in what direction. That is the polar system. > **Definition (Polar coordinates).** Fix a point $O$ called the **pole** and a > ray from $O$ called the **polar axis**. A point $P$ has polar coordinates > $(r, \theta)$ where $r$ is the distance $|OP|$ and $\theta$ is the angle from > the polar axis to the line $OP$, measured positive counterclockwise. The pole > itself is $(0, \theta)$ for any $\theta$. $$ % caption: A point $P$ at polar coordinates $(r, \theta)$: the radius $r$ is its % distance from the pole $O$, and $\theta$ is the angle its radial line makes % with the polar axis. \begin{tikzpicture}[>=stealth, scale=0.95, font=\small] \definecolor{acc}{HTML}{4A6FA5} \foreach \rad in {1,2,3} \draw[black] (0,0) circle (\rad); \foreach \a in {0,30,...,330} \draw[black] (\a:0.34) -- (\a:3); \draw[->, black, thick] (0,0) -- (3.4,0) node[right, font=\footnotesize] {polar axis}; \coordinate (P) at (50:2.5); \draw[acc, very thick] (0,0) -- (P) node[midway, above left, font=\footnotesize] {$r$}; \fill[acc] (P) circle (2.6pt) node[above right, font=\footnotesize] {$P$}; \draw[acc, ->] (0.9,0) arc (0:50:0.9); \fill[black] (0,0) circle (1.6pt) node[below left, font=\footnotesize] {$O$}; \end{tikzpicture} $$ **Non-uniqueness.** Unlike Cartesian coordinates, a polar point has many names. A full turn adds nothing, and a negative radius points the opposite way: $$ (r, \theta) = (r, \theta + 2\pi n) = (-r, \theta + (2n+1)\pi), \qquad n \in \mathbb{Z}. $$ The convention for $r < 0$ is that $(-r, \theta)$ lies at distance $|r|$ from $O$ on the ray opposite to $\theta$. So $(2, \pi/3)$, $(2, 7\pi/3)$, and $(-2, 4\pi/3)$ all name the same point. ## Converting between systems Placing the pole at the origin and the polar axis along the positive $x$-axis lines up the two systems. Right-triangle trigonometry reads off the relations in both directions. - **Polar to Cartesian:** $x = r\cos\theta$, $y = r\sin\theta$. - **Cartesian to polar:** $r^2 = x^2 + y^2$, $\tan\theta = \dfrac{y}{x}$. The first pair is unambiguous. The second is not: $\tan\theta$ repeats every $\pi$, so a value of $\tan\theta$ leaves two candidate angles, and the quadrant of $(x, y)$ decides which one. > **Worked example.** Convert $(2, \pi/3)$ to Cartesian coordinates and $(1, -1)$ > to polar coordinates. > > _Polar to Cartesian._ With $r = 2$, $\theta = \pi/3$, > > $$ > x = 2\cos\tfrac{\pi}{3} = 2\cdot\tfrac12 = 1, \qquad > y = 2\sin\tfrac{\pi}{3} = 2\cdot\tfrac{\sqrt3}{2} = \sqrt3, > $$ > > so the point is $(1, \sqrt3)$. > > _Cartesian to polar._ With $x = 1$, $y = -1$, $r = \sqrt{1 + 1} = \sqrt2$ and > $\tan\theta = -1$. The point $(1, -1)$ lies in the fourth quadrant, so choose > $\theta = -\pi/4$ rather than $3\pi/4$, giving $(\sqrt2, -\pi/4)$. (Taking > $r = -\sqrt2$ with $\theta = 3\pi/4$ names the same point.) ## Polar curves The graph of $r = f(\theta)$ is the set of points with at least one polar representation satisfying the equation. A few equations that would be clumsy in Cartesian form are immediate here. - $r = a$ is the circle of radius $|a|$ centered at $O$. - $\theta = \theta_0$ is the line through $O$ at angle $\theta_0$. - $r = 2a\cos\theta$ is a circle of radius $a$ through the pole; substituting $r\cos\theta = x$ and $r^2 = x^2 + y^2$ gives $(x-a)^2 + y^2 = a^2$. > **Worked example.** Convert $r = 2\sin\theta$ to Cartesian form, and the line > $x + y = 1$ to polar form. > > Multiply $r = 2\sin\theta$ by $r$ to introduce $r^2$ and $r\sin\theta$: > $r^2 = 2r\sin\theta$, so $x^2 + y^2 = 2y$, i.e. > > $$ > x^2 + (y - 1)^2 = 1, > $$ > > a circle of radius $1$ centered at $(0, 1)$. For the line, substitute > $x = r\cos\theta$, $y = r\sin\theta$: from $r\cos\theta + r\sin\theta = 1$, > > $$ > r = \frac{1}{\cos\theta + \sin\theta}. > $$ More elaborate curves are read off a Cartesian plot of $r$ against $\theta$: as $\theta$ advances, watch $r$ grow and shrink, and swing a radial arm to that length. Three families recur often enough to name. > **Definition (Cardioid, rose, limaçon).** > - A **cardioid** $r = a(1 \pm \sin\theta)$ or $a(1 \pm \cos\theta)$ is a > heart-shaped curve passing through the pole with a single cusp. > - A **rose** $r = a\cos n\theta$ or $a\sin n\theta$ has $n$ petals if $n$ is > odd and $2n$ petals if $n$ is even. > - A **limaçon** $r = b + a\cos\theta$ has an inner loop when $|a| > |b|$, a > cusp (cardioid) when $|a| = |b|$, and a dimple or convex oval otherwise. $$ % caption: Three polar families: the cardioid $r = 1+\sin\theta$ (one cusp), the % four-petal rose $r = \cos 2\theta$, and the limaçon $r = 1+2\cos\theta$ with an % inner loop. \begin{tikzpicture}[>=stealth, scale=0.85, font=\small] \definecolor{acc}{HTML}{4A6FA5} % cardioid r = 1 + sin t \begin{scope} \draw[black] (0,0) circle (1); \draw[acc, very thick] plot[domain=0:6.2832, samples=160] ({(1+sin(deg(\x)))*cos(deg(\x))},{(1+sin(deg(\x)))*sin(deg(\x))}); \node[font=\footnotesize, anchor=north] at (0,-1.4) {cardioid}; \end{scope} % four-leaved rose r = cos 2t \begin{scope}[xshift=4.2cm] \draw[black] (0,0) circle (1); \draw[acc, very thick] plot[domain=0:6.2832, samples=200] ({cos(deg(2*\x))*cos(deg(\x))},{cos(deg(2*\x))*sin(deg(\x))}); \node[font=\footnotesize, anchor=north] at (0,-1.4) {rose}; \end{scope} % limaçon r = 1 + 2 cos t \begin{scope}[xshift=8.8cm] \draw[black] (0,0) circle (1); \draw[acc, very thick] plot[domain=0:6.2832, samples=200] ({(1+2*cos(deg(\x)))*cos(deg(\x))},{(1+2*cos(deg(\x)))*sin(deg(\x))}); \node[font=\footnotesize, anchor=north] at (0,-1.6) {limacon}; \end{scope} \end{tikzpicture} $$ Three more curves round out the standard catalog: the odd rose keeps its petal count, the lemniscate is a figure-eight through the pole, and the Archimedean spiral winds outward at a rate set by the angle. $$ % caption: More polar families: the three-petal rose $r = \cos 3\theta$ (odd $n$ % gives $n$ petals), the lemniscate $r^2 = \cos 2\theta$ (a figure-eight through % the pole), and the Archimedean spiral $r = \theta$. \begin{tikzpicture}[>=stealth, scale=0.8, font=\small] \definecolor{acc}{HTML}{4A6FA5} \begin{scope} \draw[black] (0,0) circle (1); \draw[acc, very thick] plot[domain=0:360, samples=220] ({cos(3*\x)*cos(\x)},{cos(3*\x)*sin(\x)}); \node[font=\footnotesize, anchor=north] at (0,-1.4) {rose $n=3$}; \end{scope} \begin{scope}[xshift=4.2cm] \draw[acc, very thick] plot[domain=-45:45, samples=90] ({sqrt(cos(2*\x))*cos(\x)},{sqrt(cos(2*\x))*sin(\x)}); \draw[acc, very thick] plot[domain=135:225, samples=90] ({sqrt(cos(2*\x))*cos(\x)},{sqrt(cos(2*\x))*sin(\x)}); \node[font=\footnotesize, anchor=north] at (0,-1.4) {lemniscate}; \end{scope} \begin{scope}[xshift=8.6cm] \draw[acc, very thick] plot[domain=0:720, samples=220] ({0.0028*\x*cos(\x)},{0.0028*\x*sin(\x)}); \node[font=\footnotesize, anchor=north] at (0,-1.4) {spiral}; \end{scope} \end{tikzpicture} $$ **Sketching the cardioid $r = 1 + \sin\theta$.** As $\theta$ goes $0 \to \pi/2$, $r$ rises $1 \to 2$; as $\theta$ goes $\pi/2 \to \pi$, $r$ falls $2 \to 1$; as $\theta$ goes $\pi \to 3\pi/2$, $r$ falls $1 \to 0$, pinning the cusp at the pole; and as $\theta$ goes $3\pi/2 \to 2\pi$, $r$ rises $0 \to 1$. The trace closes into a heart. ### Symmetry Recognizing symmetry halves the plotting work. Each rule is an invariance of the equation under a substitution. | Substitution leaving the equation unchanged | Symmetry | | --- | --- | | $\theta \to -\theta$ | about the polar axis | | $\theta \to \pi - \theta$ | about the vertical line $\theta = \pi/2$ | | $r \to -r$, or $\theta \to \theta + \pi$ | about the pole | Since $\cos(-\theta) = \cos\theta$, the circle $r = 2\cos\theta$ and the rose $r = \cos 2\theta$ are symmetric about the polar axis; the cardioid $r = 1 + \sin\theta$, unchanged under $\theta \to \pi - \theta$, is symmetric about $\theta = \pi/2$. ## Tangents to polar curves Treat $\theta$ as a parameter. Writing $x = r\cos\theta = f(\theta)\cos\theta$ and $y = f(\theta)\sin\theta$ and applying the parametric slope rule with the product rule gives the slope in terms of $r$ and $\d r/\d\theta$.[^param] $$ \frac{\d y}{\d x} = \frac{\dfrac{\d r}{\d\theta}\sin\theta + r\cos\theta} {\dfrac{\d r}{\d\theta}\cos\theta - r\sin\theta}. $$ Horizontal tangents occur where the numerator vanishes (and the denominator does not); vertical tangents where the denominator vanishes. At the pole, where $r = 0$, the formula collapses to $\d y/\d x = \tan\theta$, so the tangent line at the pole is $\theta = \theta_0$ for each $\theta_0$ with $f(\theta_0) = 0$. For the rose $r = \cos 2\theta$, which passes through the pole at $\theta = \pi/4$ and $3\pi/4$, those two angles are the tangent lines there. > **Worked example.** For the cardioid $r = 1 + \sin\theta$, find the slope of the > tangent at $\theta = \pi/3$ and the angles of the horizontal and vertical > tangents. > > With $\d r/\d\theta = \cos\theta$, the slope formula gives > > $$ > \frac{\d y}{\d x} > = \frac{\cos\theta\,\sin\theta + (1+\sin\theta)\cos\theta} > {\cos\theta\,\cos\theta - (1+\sin\theta)\sin\theta} > = \frac{\cos\theta\,(1 + 2\sin\theta)}{(1+\sin\theta)(1 - 2\sin\theta)}. > $$ > > At $\theta = \pi/3$, $\sin\theta = \tfrac{\sqrt3}{2}$ and $\cos\theta = \tfrac12$; > the numerator is $\tfrac12(1 + \sqrt3)$ and the denominator is > $(1 + \tfrac{\sqrt3}{2})(1 - \sqrt3)$, whose ratio is $-1$. > > A horizontal tangent needs a vanishing numerator: $\cos\theta = 0$ or > $\sin\theta = -\tfrac12$, giving $\theta = \tfrac{\pi}{2},\, \tfrac{7\pi}{6},\, \tfrac{11\pi}{6}$. > A vertical tangent needs a vanishing denominator: $\sin\theta = \tfrac12$, giving > $\theta = \tfrac{\pi}{6},\, \tfrac{5\pi}{6}$. At $\theta = 3\pi/2$ both vanish, and > a limit check shows the cusp at the pole points straight down. ## Area in polar coordinates Rectangular area sums thin rectangles. Polar area sums thin circular sectors. The area of a sector of radius $r$ and central angle $\theta$ is $\tfrac12 r^2\theta$, so a wedge of angular width $\d\theta$ at radius $r = f(\theta)$ contributes $\tfrac12 [f(\theta)]^2\,\d\theta$. $$ % caption: A polar region swept by a rotating ray is partitioned into thin % sectors; the shaded wedge of width $\Delta\theta$ at radius $f(\theta_i^{*})$ % has area about $\tfrac12 [f(\theta_i^{*})]^2\,\Delta\theta$. \begin{tikzpicture}[>=stealth, scale=1.0, font=\small] \definecolor{acc}{HTML}{4A6FA5} \fill[acc!12] (0,0) -- (40:2.35) arc (40:55:2.35) -- cycle; \draw[acc, very thick] plot[domain=0.2:2.0, samples=90] ({(1.6+0.9*sin(deg(2*\x)))*cos(deg(\x))},{(1.6+0.9*sin(deg(2*\x)))*sin(deg(\x))}); \foreach \a in {10,25,40,55,70,85,100} \draw[black] (\a:0.5) -- (\a:2.9); \draw[acc] (0,0) -- (40:2.35); \draw[acc] (0,0) -- (55:2.35); \fill[black] (0,0) circle (1.4pt) node[below left, font=\footnotesize] {$O$}; \node[font=\footnotesize] at (47.5:1.35) {$f$}; \end{tikzpicture} $$ Summing over sectors and passing to the limit is a Riemann sum for $\tfrac12[f(\theta)]^2$. > **Theorem (Polar area).** If $f$ is positive and continuous, the area bounded > by $r = f(\theta)$ and the rays $\theta = a$, $\theta = b$ (with > $0 < b - a \le 2\pi$) is > $$ > A = \int_a^b \tfrac12\,[f(\theta)]^2\,\d\theta = \int_a^b \tfrac12 r^2\,\d\theta. > $$ The right petal of the rose $r = \cos 2\theta$ is swept as $\theta$ runs from $-\pi/4$ to $\pi/4$, the angular window where $\cos 2\theta \ge 0$. $$ % caption: The right petal of $r = \cos 2\theta$ is traced as $\theta$ runs from % $-\pi/4$ to $\pi/4$; those rays, where $r$ returns to $0$, bound the area % integral. \begin{tikzpicture}[>=stealth, scale=1.5, font=\small] \definecolor{acc}{HTML}{4A6FA5} \fill[acc!12] plot[domain=-45:45, samples=80] ({cos(2*\x)*cos(\x)},{cos(2*\x)*sin(\x)}) -- (0,0); \draw[acc, very thick] plot[domain=-45:45, samples=80] ({cos(2*\x)*cos(\x)},{cos(2*\x)*sin(\x)}); \draw[black, dashed] (0,0) -- (45:1.2) node[anchor=south west, font=\footnotesize] {upper ray}; \draw[black, dashed] (0,0) -- (-45:1.2) node[anchor=north west, font=\footnotesize] {lower ray}; \fill[black] (0,0) circle (0.6pt); \end{tikzpicture} $$ > **Worked example.** Find the area enclosed by one petal of the four-leaved rose > $r = \cos 2\theta$. > > The right petal is traced for $-\pi/4 \le \theta \le \pi/4$, so > > $$ > A = \int_{-\pi/4}^{\pi/4} \tfrac12 \cos^2 2\theta\,\d\theta > = \int_0^{\pi/4} \cos^2 2\theta\,\d\theta > = \int_0^{\pi/4} \tfrac12(1 + \cos 4\theta)\,\d\theta = \frac{\pi}{8}. > $$ **Region between two polar curves.** For $f(\theta) \ge g(\theta) \ge 0$, the area inside $r = f$ and outside $r = g$ subtracts one sector integral from the other: $$ A = \int_a^b \tfrac12\bigl([f(\theta)]^2 - [g(\theta)]^2\bigr)\,\d\theta. $$ $$ % caption: The region inside the circle $r = 3\sin\theta$ and outside the cardioid % $r = 1+\sin\theta$ (shaded); the curves meet at $\theta = \pi/6$ and $5\pi/6$, % the limits of the area integral. \begin{tikzpicture}[>=stealth, scale=0.85, font=\small] \definecolor{acc}{HTML}{4A6FA5} \fill[acc!14] plot[domain=30:150, samples=90] ({3*sin(\x)*cos(\x)},{3*sin(\x)*sin(\x)}) -- plot[domain=150:30, samples=90] ({(1+sin(\x))*cos(\x)},{(1+sin(\x))*sin(\x)}) -- cycle; \draw[acc, very thick] plot[domain=0:180, samples=150] ({3*sin(\x)*cos(\x)},{3*sin(\x)*sin(\x)}); \draw[black, very thick] plot[domain=0:360, samples=200] ({(1+sin(\x))*cos(\x)},{(1+sin(\x))*sin(\x)}); \node[acc, font=\footnotesize, anchor=south] at (0,3.08) {circle}; \node[black, font=\footnotesize, anchor=north] at (0,-1.15) {cardioid}; \fill[black] (1.30,0.75) circle (1.6pt); \fill[black] (-1.30,0.75) circle (1.6pt); \fill[black] (0,0) circle (1.4pt); \end{tikzpicture} $$ > **Worked example.** Find the area inside the circle $r = 3\sin\theta$ and outside > the cardioid $r = 1 + \sin\theta$. > > The curves meet where $3\sin\theta = 1 + \sin\theta$, i.e. $\sin\theta = \tfrac12$, > at $\theta = \pi/6$ and $5\pi/6$. Subtracting the cardioid's sector integral from > the circle's and using symmetry about $\theta = \pi/2$, > > $$ > A = \int_{\pi/6}^{5\pi/6} \tfrac12\bigl[(3\sin\theta)^2 - (1+\sin\theta)^2\bigr]\,\d\theta > = \int_{\pi/6}^{\pi/2}\bigl(3 - 4\cos 2\theta - 2\sin\theta\bigr)\,\d\theta = \pi. > $$ **A caution on intersections.** Solving two polar equations can miss crossing points, because a single geometric point has many $(r, \theta)$ names and the pole has no unique name at all. The circle $r = 3\sin\theta$ and cardioid $r = 1 + \sin\theta$ meet at the pole, yet the pole solves the first at $\theta = 0$ and the second at $\theta = 3\pi/2$, so no common $\theta$ reveals it. Always sketch both curves to find every intersection. > **Worked example.** Find all points of intersection of $r = \cos 2\theta$ and > $r = \tfrac12$. > > Setting $\cos 2\theta = \tfrac12$ gives $2\theta = \pm\tfrac{\pi}{3} + 2\pi k$, so > in $[0, 2\pi)$, > > $$ > \theta = \tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ \tfrac{7\pi}{6},\ \tfrac{11\pi}{6}, > $$ > > four points at $r = \tfrac12$. A sketch shows four more, missed because the > circle also satisfies $r = -\tfrac12$. Solving $\cos 2\theta = -\tfrac12$ gives > $\theta = \tfrac{\pi}{3},\ \tfrac{2\pi}{3},\ \tfrac{4\pi}{3},\ \tfrac{5\pi}{3}$, the > remaining four. Eight intersections in all, half invisible to the first equation. ## Arc length in polar coordinates The same parametric substitution converts the arc-length integral. With $x = r\cos\theta$, $y = r\sin\theta$, the sum of squared derivatives simplifies, using $\cos^2\theta + \sin^2\theta = 1$, to a clean expression in $r$ and $\d r/\d\theta$: $$ \left(\frac{\d x}{\d\theta}\right)^2 + \left(\frac{\d y}{\d\theta}\right)^2 = r^2 + \left(\frac{\d r}{\d\theta}\right)^2. $$ > **Theorem (Polar arc length).** If $f$ has a continuous derivative, the length > of $r = f(\theta)$ for $a \le \theta \le b$ (traced once) is > $$ > L = \int_a^b \sqrt{\,r^2 + \left(\frac{\d r}{\d\theta}\right)^2\,}\;\d\theta. > $$ **Length of the cardioid $r = 1 + \sin\theta$.** Here $\d r/\d\theta = \cos\theta$, so the integrand is $\sqrt{(1+\sin\theta)^2 + \cos^2\theta} = \sqrt{2 + 2\sin\theta}$, and over the full turn $0 \le \theta \le 2\pi$ the integral evaluates to $$ L = \int_0^{2\pi} \sqrt{2 + 2\sin\theta}\;\d\theta = 8. $$ ## Polar formulas from parametric ones Every polar formula above is a polar curve read as a parametric curve in $\theta$, with $x = r\cos\theta$ and $y = r\sin\theta$ substituted in. The slope, area, and length integrals are the parametric ones after that substitution. | Quantity | Parametric-in-$t$ form | Polar-in-$\theta$ form | | --- | --- | --- | | Slope | $\dfrac{\d y/\d t}{\d x/\d t}$ | $\dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$ | | Area | $\displaystyle\int g(t)\,f'(t)\,\d t$ | $\displaystyle\int \tfrac12 r^2\,\d\theta$ | | Arc length | $\displaystyle\int \sqrt{(x')^2 + (y')^2}\,\d t$ | $\displaystyle\int \sqrt{r^2 + (r')^2}\,\d\theta$ | [^param]: Stewart, §10.3–§10.4 — the polar slope, area, and arc-length formulas each come from writing $x = r\cos\theta$, $y = r\sin\theta$ and applying the [parametric formulas](/calculus/parametric-and-polar/parametric-curves-and-their-calculus) with $\theta$ as the parameter.