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SectionPolar Coordinates

The coordinate system we are most familiar with is called the Cartesian coordinate system, a rectangular plane divided into four quadrants by horizontal and vertical axes. Cartesian coordinates identify a point on this plane by how far left or right it lies from the origin (the \(x\) coordinate) and how far above or below the origin it lies (the \(y\) coordinate).

xy grid

In earlier chapters, we found the Cartesian coordinates of a point on the unit circle at a given angle from the positive horizontal axis. Sometimes an angle, along with a point's distance from the origin provides a more useful way of describing the point's location than conventional Cartesian coordinates.

SubsectionPolar Coordinates

The polar coordinates of a point consist of an ordered pair, \((r,\theta)\text{,}\) where \(r\) is the distance from the point to the origin and \(\theta\) is the angle measured in standard position.

Notice that if we were to "grid" the plane for polar coordinates, it would look like the graph below, with circles at incremental radii and rays drawn at incremental angles.

polar grid
Example72

Plot the polar point \(\displaystyle \left(3,\frac{5\pi}{6}\right)\)

Solution

This point will be a distance of 3 from the origin at an angle of \(5\pi/6\text{.}\) Plotting this point we get the graph shown below.

polar grid with (r,theta)=(3, 5pi/6)
Example73

Plot the polar point \(\displaystyle \left(2,-\frac{\pi}{4}\right)\)

Solution

This point will be a distance of 2 from the origin at an angle of \(-\pi/4\text{.}\) Plotting this point we get the graph shown below.

polar grid with (r,theta)=(2, -pi/4)
Note74

In the previous example, we plotted the polar point \((2,-\pi/4)\text{.}\) Notice that the resulting point on the graph is the same as the polar point

\begin{equation*} \left(2,\frac{7\pi}{4}\right) \end{equation*}

Any Cartesian point can be represented by an infinite number of different polar coordinates by adding or subtracting full rotations to the \(\theta\) value of these points. For example, the same point could also be represented as

\begin{equation*} \left(2,\frac{15\pi}{4}\right) \hspace{.25in} \text{ or } \hspace{.25in} \left(2,-\frac{9\pi}{4}\right) \end{equation*}

The number of different ways to represent a polar point is infinite.

Plot the following points given in polar coordinates and label them on a polar grid.

\begin{equation*} A=\left(3,\frac{\pi}{6}\right) \hspace{.5in} B=\left(1,\frac{4\pi}{3}\right) \hspace{.5in} C=\left(4,\frac{3\pi}{4}\right) \end{equation*}

SubsectionConverting Between Polar and Cartesian Coordinates

To convert between polar coordinates and Cartesian coordinates, we can use the trigonometric relationships that we encountered in Chapter 1.

To convert between polar \((r,\theta)\) and Cartesian \((x,y)\) coordinates, we can use the following relationships

\begin{align*} \cos(\theta)\amp=\frac{x}{r} \hspace{.6in} x=r\cos(\theta) \\ \\ \sin(\theta)\amp=\frac{y}{r} \hspace{.6in} y=r\sin(\theta) \\ \\ \tan(\theta)\amp=\frac{y}{x} \hspace{.6in} x^2+y^2=r^2 \end{align*}
triangle with (x,y) coordinates and (r,theta) labeled

From these relationships and our knowledge of the unit circle, if \(r=1\) and \(\theta=\pi/3\text{,}\) the polar coordinates would be

\begin{equation*} (r,\theta)=\left(1,\frac{\pi}{3}\right) \end{equation*}

and the corresponding Cartesian coordinates would be

\begin{equation*} (x,y)=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) \end{equation*}

graph with (r,theta)=(1,\pi/3) labeled
Note76

Remembering your unit circle values will come in very handy as you convert between Cartesian and polar coordinates.

Example77

Find the Cartesian coordinates of a point with the polar coordinates

\begin{equation*} (r,\theta)=\left(5,\frac{2\pi}{3}\right) \end{equation*}
Solution

Using the above relationships, the \(x\) and \(y\) coordinates of the point are

\begin{align*} x \amp= r\cos(\theta) = 5\cos\left(\frac{2\pi}{3}\right)=5\left(-\frac{1}{2}\right)=-\frac{5}{2}\\ \\ y \amp= r\sin(\theta) = 5\sin\left(\frac{2\pi}{3}\right)=5\left(\frac{\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{2} \end{align*}

Thus, the Cartesian coordinates are

\begin{equation*} (x,y)=\left(-\frac{5}{2},\frac{5\sqrt{3}}{2}\right) \end{equation*}

Note that since \(2\pi/3\) is a common angle on the unit circle, we can find the exact Cartesian coordinates rather than using a calculator to find approximate values for \(x\) and \(y\text{.}\)

graph with (r,theta)=(5,2pi/3)
Example78

Find the Cartesian coordinates of a point with the polar coordinates

\begin{equation*} (r, \theta)=(2, \, 4.3) \end{equation*}
Solution

Using the above relationships, the \(x\) and \(y\) coordinates of the point are

\begin{equation*} x=r\cos(\theta)=2\cos(4.3) \hspace{.25in} \text{ and } \hspace{.25in} y=r\sin(\theta)=2\sin(4.3) \end{equation*}

Since \(\theta=4.3\) radians is not a common angle on the unit circle, we need to use our calculators to find approximate values for \(x\) and \(y\text{.}\) Using our calculators, we get

\begin{align*} x \amp= 2\cos(4.3) \approx -0.8016\\ \\ y \amp= 2\sin(4.3) \approx -1.8323 \end{align*}

Thus, the Cartesian coordinates are

\begin{equation*} (x,y)\approx(-0.8016, -1.8323) \end{equation*}
graph with (r,theta)=(2,4.3)
Example79

Find the polar coordinates of a point with the Cartesian coordinates

\begin{equation*} (x,y)=(3,-4) \end{equation*}
Solution

Let's start by plotting the point \((x,y)=(3,-4)\) on a graph and using the Pythagorean theorem to find the corresponding radius of the point. We get that

\begin{align*} r^2 \amp= x^2 + y^2 \\ \\ r^2 \amp= 3^2 + (-4)^2 \\ \\ r^2 \amp= 9 + 16 \\ \\ r^2 \amp= 25 \\ \\ r \amp= 5 \end{align*}
graph with (x,y)=(3,-4)

Now that we know the radius, we can find the angle using any of the three trigonometric relationships. Keep in mind that there may be more than one solution when solving for \(\theta\) and we will need to consider the quadrant that our \((x,y)\) point is in to decide which solution to use.

Using the cosine function, we get that

\begin{equation*} \cos(\theta) = \frac{x}{r} = \frac{3}{5} = 0.6 \end{equation*}

Since this cosine value does not correspond to a common angle on the unit circle, we must use the inverse cosine function to solve for \(\theta\text{,}\) which gives us

\begin{equation*} \theta = \cos^{-1}(0.6) \approx 0.927 \text{ radians} \end{equation*}

Since \(0.927\) is greater than 0 and less than \(\pi/2 \approx 1.571\text{,}\) we know that \(\theta=0.927\) lies in Quadrant I. Since the point \((x,y)=(3,-4)\) is located in Quadrant IV, we must find the other angle on the unit circle with \(\cos(\theta) = 0.6\)

unit circle with cos(\theta)=0.6 one angle shown and labeled

Recall from Section that we can use the symmetry of the unit circle to find this second angle. By symmetry, the magnitudes of the two angles shown below are equal. Thus, \(\theta=-0.927\) is another angle that satisfies \(\cos(\theta) = 0.6\)

unit circle with cos(\theta)=0.6 two angles shown and labeled

Therefore, the polar coordinates of the point are

\begin{equation*} (r,\theta)=(5, -0.927) \end{equation*}
graph with (x,y)=(3,-4)
Example80

Find the polar coordinates of a point with the Cartesian coordinates

\begin{equation*} (x,y)=\left(-4\sqrt{2},4\sqrt{2}\right) \end{equation*}
Solution

Let's start by plotting the point \((x,y)=(-4\sqrt{2},4\sqrt{2})\) on a graph and using the Pythagorean theorem to find the corresponding radius of the point. We get that

\begin{align*} r^2 \amp= x^2 + y^2 \\ \\ r^2 \amp= (-4\sqrt{2})^2 + (4\sqrt{2})^2 \\ \\ r^2 \amp= 32 + 32 \\ \\ r^2 \amp= 64 \\ \\ r \amp= 8 \end{align*}
graph with (x,y)=(-4sqrt{2},4sqrt{2})

Now that we know the radius, we can find the angle using any of the three trigonometric relationships. Keep in mind that there may be more than one solution when solving for \(\theta\) and we will need to consider the quadrant that our \((x,y)\) point is in to decide which solution to use.

Using the sine function, we get that

\begin{equation*} \sin(\theta) = \frac{y}{r} = \frac{4\sqrt{2}}{8}=\frac{\sqrt{2}}{2} \end{equation*}

Since \(y=\sqrt{2}/2\) corresponds to a common angle on the unit circle, we can find an exact angle. The two angles that have a sine value of \(\sqrt{2}\2\) on the unit circle are

\begin{equation*} \theta=\frac{\pi}{4} \hspace{.25in} \text{ and } \hspace{.25in} \theta=\frac{3\pi}{4} \end{equation*}

Since the point \((x,y)=(-4\sqrt{2},4\sqrt{2})\) is located in Quadrant II, we can determine that \(\theta=3\pi/4\text{.}\) Thus, the polar coordinates are

\begin{equation*} (r,\theta)=\left(8, \frac{3\pi}{4}\right) \end{equation*}
graph with (x,y)=(-4sqrt{2},4sqrt{2})

Convert the following

  1. Convert the polar coordinates \((r,\theta)=(2,\pi)\) to Cartesian coordinates.

  2. Convert the Cartesian coordinates \((x,y)=(0,-4)\) to polar coordinates.

SubsectionCurves Described Using Polar Equations

Just as a Cartesian equation like \(y=x^2\) describes a relationship between \(x\) and \(y\) values on a Cartesian grid, a polar equation can be written describing a relationship between \(r\) and \(\theta\) values on a polar grid.

There are many types of curves that can be drawn using trigonometric functions. In this section, we focus on drawing circles and rays.

Example82

Sketch a graph of the polar equation \(r=2\text{.}\)

Solution

Recall that when a variable does not show up in the equation, it does not matter what value that variable has. The output for the equation will remain the same.

For example, the Cartesian equation \(y=2\) describes all the points where \(y=2\text{,}\) no matter what the \(x\) values are, which produces a horizontal line.

Likewise, this polar equation describes all points at a distance of 2 from the origin, no matter the value of angle \(\theta\text{.}\) Therefore, this equation produces a circle of radius 2 on a polar graph.

polar graph with r=2

Note that we graphed the polar equation \(r=2\) on a polar grid. However, polar graphs are often graphed on a Cartesian coordinate system, as shown below. We will use both types of graphs in future examples.

xy graph with r=2
Example83

Sketch a graph of the polar equation

\begin{equation*} \theta=\frac{\pi}{3} \end{equation*}
Solution

Like the previous example, this polar equation involves only one variable, \(\theta\text{.}\) Therefore, \(\theta=\pi/3\) describes all points at an angle of \(\pi/3\) from the origin, regardless of the value of the radius, \(r\text{.}\) This equation produces a ray through the origin which makes an angle of \(\pi/3\) with the positive horizontal axis.

polar graph with theta=pi/3

SubsectionDescribing Regions with Polar Inequalities

In addition to graphing polar equations, we can also write polar inequalities to describe regions in the plane. Sometimes, this is not easily done with Cartesian coordinates.

Example84

Shade the region described by the polar inequalities

\begin{align*} 0 \amp\leq r \lt \infty \\ \\ \frac{\pi}{6} \amp\leq \theta \leq \frac{\pi}{3} \end{align*}
Solution

In the shaded region below, \(\theta\) is restricted to be between \(\pi/6\) and \(\pi/3\) (including those angles). Within that range for \(\theta\text{,}\) all positive \(r\) values are allowed. Thus, the shaded region shown below is represented by the polar inequalities

\begin{align*} 0 \amp\leq r \lt \infty \\ \\ \frac{\pi}{6} \amp\leq \theta \leq \frac{\pi}{3} \end{align*}
polar graph with pi/6 leq theta leq pi/3
Example85

Shade the region described by the polar inequalities

\begin{align*} 1 \amp\lt r \lt 2.5 \\ \\ 0 \amp\leq \theta \leq 2\pi \end{align*}
Solution

In the shaded region below, \(r\) is restricted to be between \(1\) and \(2.5\text{.}\) Notice that we do not want to include the points corresponding to \(r=1\) and \(r=2.5\) so we use dashed lines instead of solid ones. Any value of \(\theta\) between \(0\) and \(2\pi\) is allowed. Thus, the washer shaped region shaded below is represented by the polar inequalities

\begin{align*} 1 \amp\lt r \lt 2.5 \\ \\ 0 \amp\leq \theta \leq 2\pi \end{align*}
polar graph with 1 lt r lt 2
Example86

Write a set of polar inequalities to describe the shaded region shown below.

polar graph with 0 lt r lt 2 and pi/2 leq theta leq 4pi/3
Solution

The shaded region begins at the origin and has an outer radius of 2. Therefore,

\begin{equation*} 0 \leq r \leq 2 \end{equation*}

The shaded region stretches from the angle \(\pi/2\) to the angle \(4\pi/3\text{,}\) so

\begin{equation*} \frac{\pi}{2} \leq \theta \leq \frac{4\pi}{3} \end{equation*}

Putting together these bounds for \(r\) and \(\theta\) gives us the polar inequalities

\begin{align*} 0 \amp\leq r \leq 2 \\ \\ \frac{\pi}{2} \amp\leq \theta \leq \frac{4\pi}{3} \end{align*}
Example87

Write a set of polar inequalities to describe the shaded region shown below.

polar graph with 2 lt r lt 3 and -pi/2 leq theta leq 5pi/6
Solution

The shaded region has an inner radius of 2 and an outer radius of 3. Since there are dotted lines at these two radii, we need to use strict inequalities. Therefore,

\begin{equation*} 2 \lt r \lt 3 \end{equation*}

The shaded region stretches from the angle \(3\pi/2\) to the angle \(5\pi/6\text{.}\) However, \(3\pi/2\) is greater than \(5\pi/6\text{,}\) so in order to describe the shaded region, \(\theta\) cannot be greater than \(3\pi/2\) and less than \(5\pi/6\text{.}\) Instead, we can use the angle \(-\pi/2\) as our lower bound. This gives us the inequality

\begin{equation*} -\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6} \end{equation*}

Notice that \(\theta\) can equal \(-\pi/2\) and \(5\pi/6\text{,}\) as represented by the solid lines. Putting together the bounds for \(r\) and \(\theta\) gives us the polar inequalities

\begin{align*} 2 \amp\lt r \lt 3 \\ \\ -\frac{\pi}{2} \amp\leq \theta \leq \frac{5\pi}{6} \end{align*}