Section 5.1 Polar Coordinates

Supplemental Videos.

The main topics of this section are also presented in the following videos:

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.

Subsection 5.1.1 Polar 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

Example 5.2.

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)

Example 5.3.

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)

Note 5.4.

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.

Subsection 5.1.2 Converting 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

Note 5.5.

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

Example 5.6.

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)

Example 5.7.

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)

Example 5.8.

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 Inverse Trigonometric Functions 1.6 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)

Example 5.9.

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})

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})

Example 5.10.

Convert the following

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

Solution.

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

\begin{equation*} x=r\cos(\theta)=2\cos(\pi)=-2 \hspace{.25in} \text{ and } \hspace{.25in} y=r\sin(\theta)=2\sin(\pi)=0 \end{equation*}

Therefore the Cartesian coordinates are \((x,y)=(2,0)\text{.}\)
Let’s start by plotting the point \((x,y)=(0,-4)\) on a graph.

graph with (x,y)=(0,-4)

Just by looking at the graph, we can see that the distance from the origin is \(4\) and the angle is \(\frac{3\pi}{2}\text{.}\) Thus the polar coordinates are \((r,\theta)=\left(4,\frac{3\pi}{2}\right)\text{.}\)

Subsection 5.1.3 Curves 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.

Example 5.11.

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

Example 5.12.

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

Subsection 5.1.4 Describing 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.

Example 5.13.

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

Example 5.14.

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

Example 5.15.

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*}

Example 5.16.

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*}

Exercises 5.1.5 Exercises

1. Plotting Polar Points.

For each of the following coordinates, choose the corresponding point on the polar coordinate axes above. Then convert the point from polar coordinates to Cartesian coordinates.

The Cartesian coordinates of the point are \((x,y)=\) ( , ).

What are the Cartesian coordinates of the general point \((r,\theta)\text{?}\)

The \(x\) coordinate is...

\(\displaystyle r\sin(\theta)\)
\(\displaystyle \sin(\theta)\)
\(\displaystyle \frac{\cos(\theta)}{r}\)
\(\displaystyle \cos(\theta)\)
\(\displaystyle r\cos(\theta)\)
None of the above

The \(y\) coordinate is...

\(\displaystyle r\cos(\theta)\)
\(\displaystyle \frac{\cos(\theta)}{r}\)
\(\displaystyle \sin(\theta)\)
\(\displaystyle r\sin(\theta)\)
\(\displaystyle \cos(\theta)\)
None of the above

2. Plotting Polar Points.

For each of the following coordinates, choose the corresponding point on the polar coordinate axes above. Then convert the point from polar coordinates to Cartesian coordinates.

The Cartesian coordinates of the point are \((x,y)=\) ( , ).

The angles are \(\pi\) radians apart.
The points plotted are the same.
The points are opposites of each other.
The points plotted are reflections of each other.
The radii for each point are the same.
The angles of both points differ by \(2\pi\) and hence represent the same angle.
The angles are negatives of each other, so they represent opposite angles.
None of the above

(d) Select all points that are equivalent to the point with polar coordinates (2,\(\frac{7\pi}{4}\)).

(-2,\(\frac{7\pi}{4}\)).
(2,\(-\frac{3\pi}{4}\)).
(2,\(\frac{5\pi}{3}\)).
(2,\(\frac{15\pi}{4}\)).
(2,\(-\frac{\pi}{4}\)).
(2,\(\frac{7\pi}{4}\)).
None of the above

3. Polar Regions.

Fill in the blanks to give polar inequalities describing each of the regions below. The two angles that you enter should be between 0 and \(2\pi\text{.}\)

The region below is described by the polar inequalities

\(\leq\theta\leq\)

\(\leq r \leq\)

Hint.

How much of the circle is shaded when starting at the angle \(0\text{?}\) What are the radii of the inner and outer circles?

4. Polar Regions.

Fill in the blanks to give polar inequalities describing each of the regions below. The two angles that you enter should be between 0 and \(2\pi\text{.}\) Note that all angles involved in this problem are multiples of \(\frac{\pi}{4}\text{.}\)

The region below is described by the polar inequalities

\(\leq\theta\leq\)

\(\leq r \leq\)

Hint.

How much of the circle is shaded when starting at the angle \(0\text{?}\) What are the radii of the inner and outer circles?

5. Polar Regions.

Fill in the blanks to give polar inequalities describing each of the regions below. The angles that you give should be between \(0\) and \(2\pi\text{.}\) Note that all angles involved in this problem are multiples of \(\frac{\pi}{4}\text{.}\)

The region below is described by the polar inequalities

\(\leq\theta\leq\)

\(\leq r \leq\)

Hint.

How much of the circle is shaded when starting at the angle \(0\text{?}\) What are the radii of the inner and outer circles?

6. Writing Polar Inequalities.

For each of the following descriptions, write a set of polar inequalities. The units for the angles should be given in radians. For each problem, we recommend sketching a picture on a separate piece of paper. The angles you enter should be between \(0\) and \(2\pi\text{.}\)

(a) A flat washer with an internal diameter hole of 4 millimeters and an external diameter of 6.4 mm

\(\leq\theta\leq\)

mm \(\leq r \leq\) mm

(b) A pizza has a diameter of 40 centimeters, and has been cut into 12 equal slices. Describe the region taken up by 3 slices of pizza.

\(0\) \(\leq\theta\leq\)

cm \(\leq r \leq\) cm

(c) A pizza has a diameter of 46 centimeters, and has been cut into 8 equal slices. Furthermore, the crust at the edge of the pizza takes up the last 2 centimeters of each slice. Describe the region covered by toppings of 3 slices of pizza.

\(0 \leq\theta\leq\)

cm \(\leq r \leq\) cm

7. Converting Cartesian to Polar.

Consider the point \((x,y)=\)(1, \(\sqrt{3}\)) plotted in the graph below.

(a) Find the value of \(r\) in the equivalent \((r,\theta)\) polar coordinates.

\(r=\)

(b) Now that we have the values of \(x\text{,}\) \(y\text{,}\) and \(r\text{,}\) which of the following equations could help us solve for \(\theta\text{?}\)

\(\displaystyle \tan(\theta)=\frac{x}{y}\)
\(\displaystyle \cos(\theta)=\frac{x}{r}\)
\(\displaystyle r\tan(\theta)=\frac{x}{y}\)
\(\displaystyle r\tan(\theta)=\frac{y}{x}\)
\(\displaystyle \cos(\theta)=\frac{y}{r}\)
\(\displaystyle \sin(\theta)=\frac{x}{r}\)
\(\displaystyle \tan(\theta)=\frac{y}{x}\)
\(\displaystyle \sin(\theta)=\frac{y}{r}\)
None of the above

\(\theta=\)

(d) Which of the following are also possible values of \(\theta\text{?}\)

\(\displaystyle -\frac{5\pi}{3}\)
\(\displaystyle \frac{7\pi}{3}\)
\(\displaystyle \frac{4\pi}{3}\)
\(\displaystyle \frac{5\pi}{3}\)
\(\displaystyle \frac{2\pi}{3}\)
None of the above

(e) Find all possible values of \(\theta\text{.}\)

\(\theta=\) + \(\cdot k\)

where \(k\) is any integer and the base angle is between 0 and \(2\pi\text{.}\)

Hint.

a) Note that the triangle above is a right triangle.

b) Recall that \(x = r\cos(\theta)\) and \(y = r \sin(\theta)\text{.}\) Is there another trigonometric function we can use to solve for \(\theta\text{?}\)

8. Converting Polar Coordinates.

Consider the following diagram:

(a) Suppose \((x,y)=(4,3).\) What is the value of \(r\text{?}\)

\(r=\)

(b) Give a possible value for the angle \(A\) between 0 and \(2\pi\text{.}\) Note that there are infinitely many possibilities for \(A\) outside this interval.

\(A=\)

(c) Suppose we were to consider one of the other three points in the diagram. Would the value of \(r\) change? What about the value of \(A\text{?}\) Use symmetry to find the polar coordinates of the other three points. Each angle \(A\) should be given in radians between 0 and \(2\pi\text{.}\)

Point Q has polar coordinates

\((r,A)=\) (, )

Point R has polar coordinates

\((r,A)=\) (, )

Point S has polar coordinates

\((r,A)=\) (, )

(d) Use your work above to convert the Cartesian coordinates \((-5,12)\) to polar coordinates. [Hint: take note of what quadrant the point is in.]

\((r,A)=\) (, )

(e) In general, if \((x,y)\) is an arbitrary point in the first quadrant, how can you find an equivalent polar coordinate \((r,A)\text{?}\)

\(\displaystyle r^2=x^2-y^2\)
\(A=\arccos(\frac{y}{r})\) (using calculator)
\(A=\pi-\arcsin(\frac{y}{r})\) (using calculator)
\(A=\arcsin(\frac{y}{r})\) (using calculator)
\(A=-\arcsin(\frac{y}{r})\) (using calculator)
\(A=2\pi-\arcsin(\frac{y}{r})\) (using calculator)
\(\displaystyle r=\sqrt{x^2+y^2}\)
None of the above

What if \((x,y)\) is in the second quadrant?

\(A=-\arcsin(\frac{y}{r})\) (using calculator)
\(A=2\pi-\arcsin(\frac{y}{r})\) (using calculator)
\(\displaystyle r^2=x^2-y^2\)
\(A=\arcsin(\frac{y}{r})\) (using calculator)
\(A=\pi+\arcsin(\frac{y}{r})\) (using calculator)
\(A=\pi-\arcsin(\frac{y}{r})\) (using calculator)
\(\displaystyle r=\sqrt{x^2+y^2}\)
None of the above

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