Polar Coordinates: Graphs

The planets move through space in elliptical, periodic orbits about the sun. They are in constant motion, so fixing an exact position of any planet is valid only for a moment — we can fix only its instantaneous position. This is one application of polar coordinates, represented as $(r,\theta)$. We interpret $r$ as the distance from the center of the sun and $\theta$ as the planet’s angular bearing, or its direction from the center of the sun. In this section, we focus on the polar system and the graphs that are generated directly from polar coordinates.

Testing Polar Equations for Symmetry

Just as a rectangular equation such as $y=x^2$ describes the relationship between $x$ and $y$ on a Cartesian grid, a polar equation describes a relationship between $r$ and $\theta$ on a polar grid. Recall that the coordinate pair $(r,\theta)$ indicates that we move counterclockwise from the polar axis (positive $x$-axis) by an angle of $\theta$, and extend a ray from the pole (origin) $r$ units in the direction of $\theta$. All points that satisfy the polar equation are on the graph.

Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we can see how to apply the properties of symmetry to polar equations. Further, we use symmetry — in addition to plotting key points, zeros, and maximums of $r$ — to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line $\theta=\tfrac{\pi}{2}$ ($y$-axis). We replace $(r,\theta)$ with $(-r,-\theta)$ to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation $r=2\sin\theta$.

This equation exhibits symmetry with respect to the line $\theta=\tfrac{\pi}{2}$.

In the second test, we consider symmetry with respect to the polar axis ($x$-axis). We replace $(r,\theta)$ with $(r,-\theta)$ or $(-r,\pi-\theta)$ to determine equivalency between the tested equation and the original. For example, suppose we are given the equation $r=1-2\cos\theta$.

The graph of this equation exhibits symmetry with respect to the polar axis.

In the third test, we consider symmetry with respect to the pole (origin). We replace $(r,\theta)$ with $(-r,\theta)$ to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation $r=2\sin(3\theta)$.

The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line $\theta=\tfrac{\pi}{2}$, the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.

(a) A graph is symmetric with respect to the line $\theta=\tfrac{\pi}{2}$ ($y$-axis) if replacing $(r,\theta)$ with $(-r,-\theta)$ yields an equivalent equation.

(b) A graph is symmetric with respect to the polar axis ($x$-axis) if replacing $(r,\theta)$ with $(r,-\theta)$ or $(-r,\pi-\theta)$ yields an equivalent equation.

(c) A graph is symmetric with respect to the pole (origin) if replacing $(r,\theta)$ with $(-r,\theta)$ yields an equivalent equation.

Example. Test the equation $r=2\sin\theta$ for symmetry.

Solution. Test for each of the three types of symmetry.

Analysis. Using a graphing calculator, we can see that the equation $r=2\sin\theta$ is a circle centered at $(0,1)$ with radius $r=1$ and is indeed symmetric to the line $\theta=\tfrac{\pi}{2}$. We can also see that the graph is not symmetric with the polar axis or the pole. See below.

Graphing Polar Equations by Plotting Points

To graph in the rectangular coordinate system we construct a table of $x$ and $y$ values. To graph in the polar coordinate system we construct a table of $\theta$ and $r$ values. We enter values of $\theta$ into a polar equation and calculate $r$. However, using the properties of symmetry and finding key values of $\theta$ and $r$ means fewer calculations will be needed.

Finding Zeros and Maxima

To find the zeros of a polar equation, we solve for the values of $\theta$ that result in $r=0$. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for $x$. We use the same process for polar equations. Set $r=0$, and solve for $\theta$.

For many of the forms we will encounter, the maximum value of a polar equation is found by substituting into the equation those values of $\theta$ that result in the maximum value of the trigonometric functions. Consider $r=5\cos\theta$; the maximum distance between the curve and the pole is $5$ units. The maximum value of the cosine function is $1$ when $\theta=0$, so our polar equation is $5\cos\theta$, and the value $\theta=0$ yields the maximum $\lvert r\rvert$.

Similarly, the maximum value of the sine function is $1$ when $\theta=\tfrac{\pi}{2}$, and if our polar equation is $r=5\sin\theta$, the value $\theta=\tfrac{\pi}{2}$ yields the maximum $\lvert r\rvert$. We may find additional information by calculating values of $r$ when $\theta=0$. These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.

Example. Using the equation in the previous example, find the zeros and maximum $\lvert r\rvert$ and, if necessary, the polar axis intercepts of $r=2\sin\theta$.

Solution. To find the zeros, set $r$ equal to zero and solve for $\theta$.

Substitute any one of the $\theta$ values into the equation. We will use $0$.

The points $(0,0)$ and $(0,\pm n\pi)$ are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.

To find the maximum value of the equation, look at the maximum value of the trigonometric function $\sin\theta$, which occurs when $\theta=\tfrac{\pi}{2}\pm2k\pi$, resulting in $\sin\left(\tfrac{\pi}{2}\right)=1$. Substitute $\tfrac{\pi}{2}$ for $\theta$.

Analysis. The point $\left(2,\tfrac{\pi}{2}\right)$ is the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See below.

Investigating Circles

Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.

There are five classic polar curves: cardioids, limaçons, lemniscates, rose curves, and Archimedes’ spirals. We briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.

(a) $r=a\cos\theta$, $a>0$.

(b) $r=a\cos\theta$, $a<0$.

(c) $r=a\sin\theta$, $a>0$.

(d) $r=a\sin\theta$, $a<0$.

Example. Sketch the graph of $r=4\cos\theta$.

Solution. First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum $\lvert r\rvert$ for $r=4\cos\theta$. First, set $r=0$ and solve for $\theta$. Thus, a zero occurs at $\theta=\tfrac{\pi}{2}\pm k\pi$. A key point to plot is $\left(0,\tfrac{\pi}{2}\right)$.

To find the maximum value of $r$, note that the maximum value of the cosine function is $1$ when $\theta=0\pm2k\pi$. Substitute $\theta=0$ into the equation:

The maximum value of the equation is $4$. A key point to plot is $(4,0)$.

As $r=4\cos\theta$ is symmetric with respect to the polar axis, we only need to calculate $r$-values for $\theta$ over the interval $[0,\pi]$. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to the one below.

Investigating Cardioids

While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we discuss the cardioid on its own.

(a) $r=a+b\cos\theta$.

(b) $r=a-b\cos\theta$.

(c) $r=a+b\sin\theta$.

(d) $r=a-b\sin\theta$.

Example. Sketch the graph of $r=2+2\cos\theta$.

Solution. First, testing the equation for symmetry, we find that the graph of this equation is symmetric about the polar axis. Next, we find the zeros and maximums. Setting $r=0$, we have $\theta=\pi+2k\pi$. The zero of the equation is located at $(0,\pi)$. The graph passes through this point.

The maximum value of $r=2+2\cos\theta$ occurs when $\cos\theta$ is a maximum, which is when $\cos\theta=1$ or when $\theta=0$. Substitute $\theta=0$ into the equation, and solve for $r$.

The point $(4,0)$ is the maximum value on the graph.

We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval $[0,\pi]$. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, and then we plot the points and draw the graph.

Investigating Limaçons

The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled limaçons when $1<\tfrac{a}{b}<2$ and convex limaçons when $\tfrac{a}{b}\ge2$.

(a) $r=a+b\cos\theta$.

(b) $r=a-b\cos\theta$.

(c) $r=a+b\sin\theta$.

(d) $r=a-b\sin\theta$.

Example. Graph the equation $r=4-3\sin\theta$.

Solution. First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line $\theta=\tfrac{\pi}{2}$, yet it fails all three symmetry tests. A graphing calculator immediately illustrates the graph’s reflective quality.

Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting $r=0$ results in $\theta$ being undefined. What does this mean? How could $\theta$ be undefined? The angle $\theta$ is undefined for any value of $\sin\theta>1$. Therefore, $\theta$ is undefined because there is no value of $\theta$ for which $\sin\theta>1$. Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating $r$ when $\theta=0$.

So, there is at least one polar axis intercept at $(4,0)$.

Next, as the maximum value of the sine function is $1$ when $\theta=\tfrac{\pi}{2}$, we substitute $\theta=\tfrac{\pi}{2}$ into the equation and solve for $r$. Thus, $r=1$.

Make a table of the coordinates similar to the one below.

Analysis. This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving $\sin\theta$ is likely symmetric with respect to the line $\theta=\tfrac{\pi}{2}$, evaluating more points helps to verify that the graph is correct.

Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer (1471–1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal (1588–1651), rediscovered it.

(a) $r=a+b\cos\theta$, $a<b$.

(b) $r=a-b\cos\theta$, $a<b$.

(c) $r=a+b\sin\theta$, $a<b$.

(d) $r=a-b\sin\theta$, $a<b$.

Example. Sketch the graph of $r=2+5\cos\theta$.

Solution. Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when $r=0$, $\theta=1.98$. The maximum $\lvert r\rvert$ is found when $\cos\theta=1$ or when $\theta=0$. Thus, the maximum is found at the point $(7,0)$.

Even though we have found symmetry, the zero, and the maximum, plotting more points helps to define the shape, and then a pattern emerges. See the table below.

Investigating Lemniscates

The lemniscate is a polar curve resembling the infinity symbol $\infty$ or a figure eight. Centered at the pole, a lemniscate is symmetrical by definition.

(a) $r^2=a^2\cos2\theta$.

(b) $r^2=-a^2\cos2\theta$.

(c) $r^2=a^2\sin2\theta$.

(d) $r^2=-a^2\sin2\theta$.

Example. Sketch the graph of $r^2=4\cos2\theta$.

Solution. The equation exhibits symmetry with respect to the line $\theta=\tfrac{\pi}{2}$, the polar axis, and the pole.

Let’s find the zeros. It should be routine by now, but we approach this equation a little differently by making the substitution $u=2\theta$.

So, the point $\left(0,\tfrac{\pi}{4}\right)$ is a zero of the equation.

Now let’s find the maximum value. Since the maximum of $\cos u=1$ when $u=0$, the maximum $\cos2\theta=1$ when $2\theta=0$. Thus,

We have a maximum at $(2,0)$. Since this graph is symmetric with respect to the pole, the line $\theta=\tfrac{\pi}{2}$, and the polar axis, we only need to plot points in the first quadrant.

Analysis. Making a substitution such as $u=2\theta$ is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.

Some of the points on this graph may not show up using the Trace function on a graphing calculator, and the calculator table may show an error for these same points of $r$. This is because there are no real square roots for these values of $\theta$. In other words, the corresponding $r$-values of $\sqrt{4\cos(2\theta)}$ are complex numbers because there is a negative number under the radical.

Investigating Rose Curves

The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.

(a) $r=a\cos n\theta$, $n$ even.

(b) $r=a\sin n\theta$, $n$ odd.

Example. Sketch the graph of $r=2\cos4\theta$.

Solution. Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line $\theta=\tfrac{\pi}{2}$ and the pole.

Now we find the zeros. First make the substitution $u=4\theta$.

The zero is $\theta=\tfrac{\pi}{8}$. The point $\left(0,\tfrac{\pi}{8}\right)$ is on the curve.

Next, we find the maximum $\lvert r\rvert$. We know that the maximum value of $\cos u=1$ when $\theta=0$. Thus,

The point $(2,0)$ is on the curve. The graph of the rose curve has unique properties, which are revealed in the table below.

Analysis. When these curves are drawn, it is best to plot the points in order, as in the table. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.

Example. Sketch the graph of $r=2\sin(5\theta)$.

Solution. The graph of the equation shows symmetry with respect to the line $\theta=\tfrac{\pi}{2}$. Next, find the zeros and maximum. We want to make the substitution $u=5\theta$.

The maximum value is calculated at the angle where $\sin\theta$ is a maximum. Therefore,

Thus, the maximum value of the polar equation is $2$. This is the length of each petal. As the curve for $n$ odd yields the same number of petals as $n$, there are five petals on the graph. See below.

Create a table of values similar to the one below.

Investigating the Archimedes’ Spiral

The final polar equation we discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE–c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.

(a) $r=\theta$, $[0,2\pi]$.

(b) $r=\theta$, $[0,4\pi]$.

Example. Sketch the graph of $r=\theta$ over $[0,2\pi]$.

Solution. As $r$ is equal to $\theta$, the plot of the Archimedes’ spiral begins at the pole at the point $(0,0)$. While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.

Create a table such as the one below.

Analysis. The domain of this polar curve is $[0,2\pi]$. In general, however, the domain of this function is $(-\infty,\infty)$. Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex.

Summary of Curves

We have explored a number of seemingly complex polar curves in this section. Each was drawn and sketched above; the table below collects their formulas and conditions in one place.

Key concepts

• It is easier to graph polar equations if we can test the equations for symmetry with respect to the line $\theta=\tfrac{\pi}{2}$, the polar axis, or the pole.
• There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry.
• Polar equations may be graphed by making a table of values for $\theta$ and $r$.
• The maximum value of a polar equation is found by substituting the value of $\theta$ that leads to the maximum value of the trigonometric expression.
• The zeros of a polar equation are found by setting $r=0$ and solving for $\theta$.
• Some formulas that produce the graph of a circle in polar coordinates are given by $r=a\cos\theta$ and $r=a\sin\theta$.
• The formulas that produce the graphs of a cardioid are given by $r=a\pm b\cos\theta$ and $r=a\pm b\sin\theta$, for $a>0$, $b>0$, and $\tfrac{a}{b}=1$.
• The formulas that produce the graphs of a one-loop limaçon are given by $r=a\pm b\cos\theta$ and $r=a\pm b\sin\theta$ for $1<\tfrac{a}{b}<2$.
• The formulas that produce the graphs of an inner-loop limaçon are given by $r=a\pm b\cos\theta$ and $r=a\pm b\sin\theta$ for $a>0$, $b>0$, and $a<b$.
• The formulas that produce the graphs of a lemniscate are given by $r^2=a^2\cos2\theta$ and $r^2=a^2\sin2\theta$, where $a\ne0$.
• The formulas that produce the graphs of rose curves are given by $r=a\cos n\theta$ and $r=a\sin n\theta$, where $a\ne0$; if $n$ is even, there are $2n$ petals, and if $n$ is odd, there are $n$ petals.
• The formula that produces the graph of an Archimedes’ spiral is given by $r=\theta$, $\theta\ge0$.

Practice

Test polar equations for symmetry

Graph polar equations by plotting points

This section is adapted from Precalculus 2e, Section 8.4: Polar Coordinates: Graphs by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the decorative solar-system illustration opening the section (Figure 1), a credit photograph with no mathematics beyond its printed planet labels, and reworded the sentence that pointed at it into a self-contained opener. Recreated every instructional figure as an accessible spec-first SVG: the three-panel symmetry-test diagram (a ray into Quadrant I and its mirror image under each of the three substitutions, with the shared angle marked on both rays); the polar-grid-with-marked-points recipe for every worked example (Examples 1–3 and 5–8’s circle, one-loop limaçon, inner-loop limaçon, lemniscate, and both rose curves) and for Example 1’s own rectangular-grid circle; and, for every one of the seven “Formulas for …” boxes, all four (or two, for the rose-curve and spiral boxes, which the source itself shows as two panels) orientation panels the source prints, each sampled from its own printed formula with a representative $a,b$ (or $n$) chosen to differ from the worked example immediately below it so the two figures are not identical. Every curve — circles, cardioids, limaçons, the lemniscate, both rose curves, and the spiral, in every figure on the page — is a polylines trace sampled directly from its exact polar formula with node, never a smoothed approximation; a lemniscate branch is sampled only over the $\theta$-interval where the radicand is non-negative, tracing both the $+\sqrt{\ }$ and $-\sqrt{\ }$ branches to draw both lobes. The one figure not recreated is the two-panel “Summary of Curves” recap (Figures 20–21): every shape and formula it collects was already drawn in full, with its own points and table, earlier on this page, so the summary is represented as a reference table of formulas and conditions instead of eight more panels of curves already shown. Every retained “Try It” became a real interactive component. The first (test $r=-2\cos\theta$ for symmetry) and the rose-curve Try It after Example 8 (sketch $r=4\sin(2\theta)$) became multiplechoice questions on which symmetries or curve family the equation exhibits, since a set of symmetries or a curve name is categorical, never a number. The Try It after Example 2, which bundles a symmetry test with a zero and a maximum for $r=3\cos\theta$, was split into three components — a multiplechoice for the symmetry, and two fillins (the zero, asked as the $\theta$-value alone to avoid the pole’s any-angle ambiguity at $r=0$; the maximum, asked as the ordered pair the source itself prints) — since the source bundles three separate results into one prompt. The Try It after Example 5 (sketch $r=3-2\cos\theta$) became a fillin for the maximum $\lvert r\rvert$ plus a multiplechoice naming the curve family, both independently derived and checked against the source’s printed answer image (a one-loop limaçon extending left); the Try It after Example 9 (sketch $r=3\cos(3\theta)$) became a graph-mode multiplechoice recognition question, since the source’s own answer is a rendered graph. The Try It after Example 10 (sketch $r=-\theta$ over $[0,4\pi]$) became a fillin asking for $r$ at $\theta=4\pi$, presented beside a recreation of the source’s answer figure. Adapted ten selected end-of-section exercises into a closing Practice block, one group per objective: three symmetry-test exercises (Graphical #7, #9, #11, whose printed answers are “polar axis only,” “all three,” and “$\theta=\tfrac{\pi}{2}$ only” respectively — the polar-axis and pole passes for #9 use the source’s own alternate substitutions $(-r,\pi-\theta)$ and the equivalent $(r,\theta+\pi)$, since the primary substitutions alone do not reduce to the original equation) as multiplechoice questions with the symmetry set as the option text; six curve-identification exercises (Graphical #17, #19, #27, #33, #37, #41, spanning all seven named families) as multiplechoice questions naming the shape; and Graphical #23 ($r=7+4\sin\theta$, printed answer “one-loop/dimpled limaçon”) as a graph-mode multiplechoice — the section’s one recognition-by-graph question, per the corpus convention, with distractors varying the sign (opens downward), the axis (opens right), and the family ($a<b$, an inner-loop limaçon) rather than only the marked point. Every Practice item, Try It, and worked example was independently re-derived — including by running every symmetry substitution, zero, and maximum in Node — rather than read off the source key; the polar-grid points plotted on every recreated figure were computed the same way, never estimated from the printed image.