Polar Coordinates

Over $12$ kilometers from port, a sailboat encounters rough weather and is blown off course by a $16$-knot wind. How can the sailor indicate his location to the Coast Guard? In this section, we investigate a method of representing location that is different from a standard coordinate grid.

Plotting Points Using Polar Coordinates

When we think about plotting points in the plane, we usually think of rectangular coordinates $(x,y)$ in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce polar coordinates, which are points labeled $(r,\theta)$ and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.

The polar grid is scaled as the unit circle with the positive $x$-axis now viewed as the polar axis and the origin as the pole. The first coordinate $r$ is the radius or length of the directed line segment from the pole. The angle $\theta$, measured in radians, indicates the direction of $r$. We move counterclockwise from the polar axis by an angle of $\theta$, and measure a directed line segment the length of $r$ in the direction of $\theta$. Even though we measure $\theta$ first and then $r$, the polar point is written with the $r$-coordinate first. For example, to plot the point $\left(2,\tfrac{\pi}{4}\right)$, we would move $\tfrac{\pi}{4}$ units in the counterclockwise direction and then a length of $2$ from the pole. This point is plotted on the grid below.

Example. Plot the point $\left(3,\tfrac{\pi}{2}\right)$ on the polar grid.

Solution. The angle $\tfrac{\pi}{2}$ is found by sweeping in a counterclockwise direction $90^\circ$ from the polar axis. The point is located at a length of $3$ units from the pole in the $\tfrac{\pi}{2}$ direction, as shown below.

Example. Plot the point $\left(-2,\tfrac{\pi}{6}\right)$ on the polar grid.

Solution. We know that $\tfrac{\pi}{6}$ is located in the first quadrant. However, $r=-2$. We can approach plotting a point with a negative $r$ in two ways:

• Plot the point $\left(2,\tfrac{\pi}{6}\right)$ by moving $\tfrac{\pi}{6}$ in the counterclockwise direction and extending a directed line segment $2$ units into the first quadrant. Then retrace the directed line segment back through the pole, and continue $2$ units into the third quadrant;
• Move $\tfrac{\pi}{6}$ in the counterclockwise direction, and draw the directed line segment from the pole $2$ units in the negative direction, into the third quadrant.

The two constructions land on the same point, shown below.

Compare this to the graph of the polar coordinate $\left(2,\tfrac{\pi}{6}\right)$, shown below.

Converting from Polar Coordinates to Rectangular Coordinates

When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables $x$, $y$, $r$, and $\theta$.

Dropping a perpendicular from the point in the plane to the $x$-axis forms a right triangle, as illustrated below. An easy way to remember the equations above is to think of $\cos\theta$ as the adjacent side over the hypotenuse and $\sin\theta$ as the opposite side over the hypotenuse.

Example. Write the polar coordinates $\left(3,\tfrac{\pi}{2}\right)$ as rectangular coordinates.

Solution. Use the equivalent relationships.

The rectangular coordinates are $(0,3)$. See below.

Example. Write the polar coordinates $(-2,0)$ as rectangular coordinates.

Solution. Writing the polar coordinates as rectangular, we have

The rectangular coordinates are also $(-2,0)$.

Converting from Rectangular Coordinates to Polar Coordinates

To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.

Example. Convert the rectangular coordinates $(3,3)$ to polar coordinates.

Solution. We see that the original point $(3,3)$ is in the first quadrant. To find $\theta$, use the formula $\tan\theta=\tfrac{y}{x}$. This gives

To find $r$, we substitute the values for $x$ and $y$ into the formula $r=\sqrt{x^2+y^2}$. We know that $r$ must be positive, as $\tfrac{\pi}{4}$ is in the first quadrant. Thus

So, $r=3\sqrt2$ and $\theta=\tfrac{\pi}{4}$, giving us the polar point $\left(3\sqrt2,\tfrac{\pi}{4}\right)$. See below.

Analysis. There are other sets of polar coordinates that will be the same as our first solution. For example, the points $\left(-3\sqrt2,\tfrac{5\pi}{4}\right)$ and $\left(3\sqrt2,-\tfrac{7\pi}{4}\right)$ will coincide with the original solution of $\left(3\sqrt2,\tfrac{\pi}{4}\right)$. The point $\left(-3\sqrt2,\tfrac{5\pi}{4}\right)$ indicates a move further counterclockwise by $\pi$, which is directly opposite $\tfrac{\pi}{4}$. The radius is expressed as $-3\sqrt2$. However, the angle $\tfrac{5\pi}{4}$ is located in the third quadrant and, as $r$ is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as $\left(3\sqrt2,\tfrac{\pi}{4}\right)$. The point $\left(3\sqrt2,-\tfrac{7\pi}{4}\right)$ is a move further clockwise by $-\tfrac{7\pi}{4}$, from $\tfrac{\pi}{4}$. The radius, $3\sqrt2$, is the same.

Transforming Equations between Polar and Rectangular Forms

We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.

Example. Write the Cartesian equation $x^2+y^2=9$ in polar form.

Solution. The goal is to eliminate $x$ and $y$ from the equation and introduce $r$ and $\theta$. Ideally, we would write the equation $r$ as a function of $\theta$. To obtain the polar form, we will use the relationships between $(x,y)$ and $(r,\theta)$. Since $x=r\cos\theta$ and $y=r\sin\theta$, we can substitute and solve for $r$.

Thus, $x^2+y^2=9$, $r=3$, and $r=-3$ should generate the same graph.

To graph a circle in rectangular form, we must first solve for $y$.

Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form $Y_1=\sqrt{9-x^2}$ and $Y_2=-\sqrt{9-x^2}$. Press GRAPH.

Example. Rewrite the Cartesian equation $x^2+y^2=6y$ as a polar equation.

Solution. This equation appears similar to the previous example, but it requires different steps to convert the equation. We can still follow the same procedures we have already learned and make the following substitutions.

Therefore, the equations $x^2+y^2=6y$ and $r=6\sin\theta$ should give us the same graph.

Example. Rewrite the Cartesian equation $y=3x+2$ as a polar equation.

Solution. We will use the relationships $x=r\cos\theta$ and $y=r\sin\theta$.

Identify and Graph Polar Equations by Converting to Rectangular Equations

We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.

Example. Convert the polar equation $r=2\sec\theta$ to a rectangular equation, and draw its corresponding graph.

Solution. The conversion is

Notice that the equation $r=2\sec\theta$ drawn on the polar grid is clearly the same as the vertical line $x=2$ drawn on the rectangular grid, below. Just as $x=c$ is the standard form for a vertical line in rectangular form, $r=c\sec\theta$ is the standard form for a vertical line in polar form.

A similar discussion would demonstrate that the graph of the function $r=2\csc\theta$ will be the horizontal line $y=2$. In fact, $r=c\csc\theta$ is the standard form for a horizontal line in polar form, corresponding to the rectangular form $y=c$.

Example. Rewrite the polar equation $r=\tfrac{3}{1-2\cos\theta}$ as a Cartesian equation.

Solution. The goal is to eliminate $\theta$ and $r$, and introduce $x$ and $y$. We clear the fraction, and then use substitution. In order to replace $r$ with $x$ and $y$, we must use the expression $x^2+y^2=r^2$.

The Cartesian equation is $x^2+y^2=(3+2x)^2$. However, to graph it, especially using a graphing calculator or computer program, we want to isolate $y$.

When our entire equation has been changed from $r$ and $\theta$ to $x$ and $y$, we can stop, unless asked to solve for $y$ or simplify.

Analysis. In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola’s standard form. To do this, we can start with the initial equation.

The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we investigate further in a later chapter.

Example. Rewrite the polar equation $r=\sin(2\theta)$ in Cartesian form.

Solution.

This equation can also be written as

Key equations

Key concepts

• The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.
• To plot a point in the form $(r,\theta)$, $\theta>0$, move in a counterclockwise direction from the polar axis by an angle of $\theta$, and then extend a directed line segment from the pole the length of $r$ in the direction of $\theta$. If $\theta$ is negative, move in a clockwise direction, and extend a directed line segment the length of $r$ in the direction of $\theta$.
• If $r$ is negative, extend the directed line segment in the opposite direction of $\theta$.
• To convert from polar coordinates to rectangular coordinates, use the formulas $x=r\cos\theta$ and $y=r\sin\theta$.
• To convert from rectangular coordinates to polar coordinates, use one or more of the formulas $\cos\theta=\tfrac{x}{r}$, $\sin\theta=\tfrac{y}{r}$, $\tan\theta=\tfrac{y}{x}$, and $r=\sqrt{x^2+y^2}$.
• Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations.
• Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane.

Practice

Plot points using polar coordinates

Convert from polar coordinates to rectangular coordinates

Convert from rectangular coordinates to polar coordinates

Transform equations between polar and rectangular forms

Identify and graph polar equations by converting to rectangular equations

This section is adapted from Precalculus 2e, Section 8.3: Polar Coordinates 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 sailboat illustration opening the section (Figure 1), an ornamental scene-setter with no mathematics beyond its printed compass labels, and reworded the two sentences that pointed at it into a self-contained opener. Recreated every instructional figure as an accessible spec-first SVG: the polar-grid recipe used throughout (concentric circles at each integer radius plus the two diagonal grid lines, matching the source’s own grid) for every point-plotting figure; both panels of the negative-$r$ construction (the retrace-through-the-pole sweep and its direct equivalent); the generic $x,y,r,\theta$ right triangle used to introduce each conversion direction; a polar-grid-plus-rectangular-grid pair for every polar/rectangular point-equivalence example; the circle, line, and hyperbola equation-graph pairs, with the hyperbola’s polar branches, the horizontal-circle’s polar trace, and the vertical line’s polar trace each sampled or drawn from the exact solved equation (never freehand) — the hyperbola’s dashed asymptote lines use its own solved slope $\pm\sqrt3$. The source prints each polar panel’s equation directly on the curve; the $r=2\sec\theta$ and $r=6\sin\theta$ panels keep that label, but the hyperbola’s polar panel — the densest figure on the page, six rings, two diagonals, and two long curve branches — has no readable gap left for its 19-character label at any position the figure-overlap gate accepts, so that one label is omitted; the equation is still stated in the adjacent prose and the figure’s ariaLabel. Every retained “Try It” became a real fillin or multiplechoice component. The two “plot the point” Try Its (following Examples 1 and 2) became graph-mode multiple choice, since a polar answer cannot be graded by the interactive graphplot component (its snap lattice is rectangular, not polar) — this leaves the section with two graph-mode multiple-choice questions and no graphplot, which the corpus’s usual “one recognition multiple choice per section” convention does not cleanly cover, the same way intermediate algebra 3.4 could not convert its shading questions; flagged for the parent’s adjudication. The remaining three Try Its (rectangular-coordinate, polar-form, and Cartesian-form conversions) became fillin components with the answerForm their printed subject needs: exact-radical on the “rewrite in polar form” Try It, since its answer $r=\sqrt3$ has no decimal shape to fall back on, and circle-standard-form on the “rewrite in Cartesian form” Try It, since the source itself offers two equally correct forms and the standard-form one is the shape it prints last. Adapted ten selected end-of-section exercises into a closing Practice block, one group per objective: two “find the polar coordinates of the point” Graphical exercises (transcribed as unlabeled figures, since the printed figures carry no coordinate labels either) pinned to their representative with $r>0$, $0\le\theta<2\pi$ and the radians form; two Algebraic polar-to-rectangular and two rectangular-to-polar conversions, the latter pair also carrying radians since their answers name an angle; two plain equation transformations; and two “convert to Cartesian form and identify the conic” exercises, each split into its two natural asks — a fillin for the equation and a multiplechoice for the categorical conic name, since a conic name is never a gradable number. Every Practice item and Try It was independently re-derived (including by running the trigonometry and equation algebra in Node) rather than read off the source key.