Polar Coordinates

Learning Objectives

In this section, you will:

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.

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

An illustration of a boat on the polar grid.

Figure 1.

Plotting Points Using Polar Coordinates

When we think about plotting points in the plane, we usually think of $\,\left(x,y\right)\,$ 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 to polar coordinates, which are points labeled $\,\left(r,\theta \right)\,$ 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 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 $\,\left(2,\frac{\pi }{4}\right),$ we would move $\,\frac{\pi }{4}\,$ units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in (Figure).

Polar grid with point (2, pi/4) plotted.

Figure 2.

Plotting a Point on the Polar Grid

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

Show Solution

The angle $\,\frac{\pi }{2}\,$ is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the $\,\frac{\pi }{2}\,$ direction, as shown in (Figure).

Polar grid with point (3, pi/2) plotted.

Figure 3.

Try It

Plot the point $\,\left(2,\,\frac{\pi }{3}\right)\,$ in the polar grid.

Show Solution

Plotting a Point in the Polar Coordinate System with a Negative Component

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

Show Solution

We know that $\,\frac{\pi }{6}\,$ 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 $\,\left(2,\frac{\pi }{6}\right)\,$ by moving $\,\frac{\pi }{6}\,$ 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 $\,\frac{\pi }{6}\,$ in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.

See (Figure)(a). Compare this to the graph of the polar coordinate $\,\left(2,\frac{\pi }{6}\right)\,$ shown in (Figure)(b).

Two polar grids. Points (2, pi/6) and (-2, pi/6) are plotted. They are reflections across the origin in Q1 and Q3.

Figure 4.

Try It

Plot the points $\,\left(3,-\frac{\pi }{6}\right)$ and $\,\left(2,\frac{9\pi }{4}\right)\,$ on the same polar grid.

Show Solution

Converting from Polar Coordinates to Rectangular Coordinates

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

\begin{matrix} \begin{matrix} c o s θ = \frac{x}{r} \to x = r c o s θ \end{matrix} s i n θ = \frac{y}{r} \to y = r s i n θ \end{matrix}

$\begin{array}{l}\begin{array}{l}\\ \mathrm{cos}\,\theta =\frac{x}{r}\to x=r\mathrm{cos}\,\theta \end{array}\hfill \\ \mathrm{sin}\,\theta =\frac{y}{r}\to y=r\mathrm{sin}\,\theta \hfill \end{array}$

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

Comparison between polar coordinates and rectangular coordinates. There is a right triangle plotted on the x,y axis. The sides are a horizontal line on the x-axis of length x, a vertical line extending from thex-axis to some point in quadrant 1, and a hypotenuse r extending from the origin to that same point in quadrant 1. The vertices are at the origin (0,0), some point along the x-axis at (x,0), and that point in quadrant 1. This last point is (x,y) or (r, theta), depending which system of coordinates you use.

Figure 5.

Converting from Polar Coordinates to Rectangular Coordinates

To convert polar coordinates $\,\left(r,\,\theta \right)\,$ to rectangular coordinates $\,\left(x,\,y\right),$ let

c o s θ = \frac{x}{r} \to x = r c o s θ

$\mathrm{cos}\,\theta =\frac{x}{r}\to x=r\mathrm{cos}\,\theta$

s i n θ = \frac{y}{r} \to y = r s i n θ

$\mathrm{sin}\,\theta =\frac{y}{r}\to y=r\mathrm{sin}\,\theta$

How To

Given polar coordinates, convert to rectangular coordinates.

Given the polar coordinate $\,\left(r,\theta \right),$ write $\,x=r\mathrm{cos}\,\theta \,$ and $\,y=r\mathrm{sin}\,\theta .$
Evaluate $\,\mathrm{cos}\,\theta \,$ and $\,\mathrm{sin}\,\theta .$
Multiply $\,\mathrm{cos}\,\theta \,$ by $\,r\,$ to find the x-coordinate of the rectangular form.
Multiply $\,\mathrm{sin}\,\theta \,$ by $\,r\,$ to find the y-coordinate of the rectangular form.

Writing Polar Coordinates as Rectangular Coordinates

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

Show Solution

Use the equivalent relationships.

\begin{matrix} \begin{matrix} x = r c o s θ \end{matrix} x = 3 c o s \frac{π}{2} = 0 y = r s i n θ y = 3 s i n \frac{π}{2} = 3 \end{matrix}

$\begin{array}{l}\begin{array}{l}\\ x=r\mathrm{cos}\,\theta \end{array}\hfill \\ x=3\mathrm{cos}\,\frac{\pi }{2}=0\hfill \\ y=r\mathrm{sin}\,\theta \hfill \\ y=3\mathrm{sin}\,\frac{\pi }{2}=3\hfill \end{array}$

The rectangular coordinates are $\,\left(0,3\right).\,$ See (Figure).

Illustration of (3, pi/2) in polar coordinates and (0,3) in rectangular coordinates - they are the same point!

Figure 6.

Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates $\,\left(-2,0\right)\,$ as rectangular coordinates.

Show Solution

See (Figure). Writing the polar coordinates as rectangular, we have

\begin{matrix} x = r c o s θ x = - 2 c o s (0) = - 2 y = r s i n θ y = - 2 s i n (0) = 0 \end{matrix}

$\begin{array}{l}x=r\mathrm{cos}\,\theta \hfill \\ x=-2\mathrm{cos}\left(0\right)=-2\hfill \\ \hfill \\ y=r\mathrm{sin}\,\theta \hfill \\ y=-2\mathrm{sin}\left(0\right)=0\hfill \end{array}$

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

Illustration of (-2, 0) in polar coordinates and (-2,0) in rectangular coordinates - they are the same point!

Figure 7.

Try It

Write the polar coordinates $\,\left(-1,\frac{2\pi }{3}\right)\,$ as rectangular coordinates.

Show Solution

$\left(x,y\right)=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

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.

Converting from Rectangular Coordinates to Polar Coordinates

Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in (Figure).

\begin{matrix} c o s θ = \frac{x}{r} or x = r c o s θ s i n θ = \frac{y}{r} or y = r s i n θ r^{2} = x^{2} + y^{2} t a n θ = \frac{y}{x} \end{matrix}

$\begin{array}{l}\mathrm{cos}\,\theta =\frac{x}{r}\,\,\text{ or}\,\,\,x=r\mathrm{cos}\,\theta \hfill \\ \mathrm{sin}\,\theta =\frac{y}{r}\,\,\text{ or}\,\,\,y=r\mathrm{sin}\,\theta \hfill \\ \,\,\,\,\,\,{r}^{2}={x}^{2}+{y}^{2}\hfill \\ \mathrm{tan}\,\theta =\frac{y}{x}\,\hfill \end{array}$

Figure 8.

Writing Rectangular Coordinates as Polar Coordinates

Convert the rectangular coordinates $\,\left(3,3\right)\,$ to polar coordinates.

Show Solution

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

\begin{matrix} t a n θ = \frac{3}{3} t a n θ = 1 {t a n}^{- 1} (1) = \frac{π}{4} \end{matrix}

$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\mathrm{tan}\,\theta =\frac{3}{3}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\mathrm{tan}\,\theta =1\hfill \\ \,\,\,{\mathrm{tan}}^{-1}\left(1\right)=\frac{\pi }{4}\hfill \end{array}$

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 $\,\frac{\pi }{4}\,$ is in the first quadrant. Thus

\begin{matrix} \begin{matrix} r = \sqrt{3^{2} + 3^{2}} \end{matrix} r = \sqrt{9 + 9} r = \sqrt{18} = 3 \sqrt{2} \end{matrix}

$\begin{array}{l}\begin{array}{l}\\ r=\sqrt{{3}^{2}+{3}^{2}}\end{array}\hfill \\ r=\sqrt{9+9}\hfill \\ r=\sqrt{18}=3\sqrt{2}\hfill \end{array}$

So, $\,r=3\sqrt{2}\,\,$ and $\,\theta \text{=}\frac{\pi }{4},\,$ giving us the polar point $\,\left(3\sqrt{2},\frac{\pi }{4}\right).\,$ See (Figure).

Illustration of (3rad2, pi/4) in polar coordinates and (3,3) in rectangular coordinates - they are the same point!

Figure 9.

Analysis

There are other sets of polar coordinates that will be the same as our first solution. For example, the points $\,\left(-3\sqrt{2},\,\frac{5\pi }{4}\right)\,$ and $\,\left(3\sqrt{2},-\frac{7\pi }{4}\right)\,$ will coincide with the original solution of $\,\left(3\sqrt{2},\,\frac{\pi }{4}\right).\,$ The point $\,\left(-3\sqrt{2},\,\frac{5\pi }{4}\right)\,$ indicates a move further counterclockwise by $\,\pi ,\,$ which is directly opposite $\,\frac{\pi }{4}.\,$ The radius is expressed as $\,-3\sqrt{2}.\,$ However, the angle $\,\frac{5\pi }{4}\,$ 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 $\,\left(3\sqrt{2},\,\,\frac{\pi }{4}\right).\,$ The point $\,\left(3\sqrt{2},\,-\frac{7\pi }{4}\right)\,$ is a move further clockwise by $\,-\frac{7\pi }{4},\,$ from $\,\frac{\pi }{4}.\,$ The radius, $\,3\sqrt{2},\,$ 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.

How To

Given an equation in polar form, graph it using a graphing calculator.

Change the MODE to POL, representing polar form.
Press the Y= button to bring up a screen allowing the input of six equations: $\,{r}_{1},\,\,{r}_{2},\,\,.\,\,.\,\,.\,\,,\,\,{r}_{6}.$
Enter the polar equation, set equal to $\,r.$
Press GRAPH.

Writing a Cartesian Equation in Polar Form

Write the Cartesian equation $\,{x}^{2}+{y}^{2}=9\,$ in polar form.

Show 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 $\,\left(x,y\right)\,$ and $\,\left(r,\theta \right).\,$ Since $\,x=r\mathrm{cos}\,\theta \,$ and $\,y=r\mathrm{sin}\,\theta ,\,$ we can substitute and solve for $\,r.$

\begin{matrix} {(r c o s θ)}^{2} + {(r s i n θ)}^{2} = 9 r^{2} {c o s}^{2} θ + r^{2} {s i n}^{2} θ = 9 r^{2} ({c o s}^{2} θ + {s i n}^{2} θ) = 9 r^{2} (1) = 9 & {Substitute cos}^{2} θ + {s i n}^{2} θ = 1. r = \pm 3 \begin{matrix} \end{matrix} & Use the square root property . \end{matrix}

$\begin{array}{ll}\text{ }{\left(r\mathrm{cos}\,\theta \right)}^{2}+{\left(r\mathrm{sin}\,\theta \right)}^{2}=9\hfill & \hfill \\ \text{ }{r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta =9\hfill & \hfill \\ \text{ }{r}^{2}\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)=9\hfill & \hfill \\ \,\text{ }{r}^{2}\left(1\right)=9 \hfill & {\text{Substitute cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1.\hfill \\ \text{ }r=±3\begin{array}{cccc}& & & \end{array}\hfill & \text{Use the square root property}.\hfill \end{array}$

Thus, $\,{x}^{2}+{y}^{2}=9,r=3,$ and $\,r=-3\,$ should generate the same graph. See (Figure).

Plotting a circle of radius 3 with center at the origin in polar and rectangular coordinates. It is the same in both systems.

Figure 10. (a) Cartesian form $\,{x}^{2}+{y}^{2}=9\,$ (b) Polar form $\,r=3$

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

$\begin{array}{l}\begin{array}{l}\\ {x}^{2}+{y}^{2}=9\end{array}\hfill \\ \text{ }{y}^{2}=9-{x}^{2}\hfill \\ \text{ }y=±\sqrt{9-{x}^{2}}\hfill \end{array}$

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.

Rewriting a Cartesian Equation as a Polar Equation

Rewrite the $\,{x}^{2}+{y}^{2}=6y\,$ as a polar equation.

Show 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:

$\begin{array}{ll}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{r}^{2}=6y\hfill & \text{Use }{x}^{2}+{y}^{2}={r}^{2}.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{r}^{2}=6r\mathrm{sin}\,\theta \hfill & \text{Substitute}\,y=r\mathrm{sin}\,\theta .\hfill \\ \text{ }{r}^{2}-6r\mathrm{sin}\,\theta =0\hfill & \text{Set equal to 0}.\hfill \\ \text{ }r\left(r-6\mathrm{sin}\,\theta \right)=0\hfill & \text{Factor and solve}.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r=0\hfill & \text{We reject }r=0,\,\text{as it only represents one point, }\left(0,0\right).\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{or}\,r=6\mathrm{sin}\,\theta \begin{array}{cccc}& & & \end{array}\hfill & \hfill \end{array}$

Therefore, the equations $\,{x}^{2}+{y}^{2}=6y\,$ and $\,r=6\mathrm{sin}\,\theta \,$ should give us the same graph. See (Figure).

Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are circles.

Figure 11. (a) Cartesian form $\,{x}^{2}+{y}^{2}=6y$ (b) polar form $\,r=6\mathrm{sin}\,\theta$

The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical.

Rewriting a Cartesian Equation in Polar Form

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

Show Solution

We will use the relationships $\,x=r\mathrm{cos}\,\theta \,$ and $\,y=r\mathrm{sin}\,\theta .$

$\begin{array}{ll}\text{ }y=3x+2\hfill & \hfill \\ \text{ }r\mathrm{sin}\,\theta =3r\mathrm{cos}\,\theta +2\hfill & \hfill \\ \,r\mathrm{sin}\,\theta -3r\mathrm{cos}\,\theta =2\hfill & \hfill \\ r\left(\mathrm{sin}\,\theta -3\mathrm{cos}\,\theta \right)=2\hfill & \text{Isolate }r.\hfill \\ \text{ }\,\,\,r=\frac{2}{\mathrm{sin}\,\theta -3\mathrm{cos}\,\theta }\begin{array}{cccc}& & & \end{array}\hfill & \text{Solve for }r.\hfill \end{array}$

Try It

Rewrite the Cartesian equation $\,{y}^{2}=3-{x}^{2}\,$ in polar form.

Show Solution

$r=\sqrt{3}$

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.

Graphing a Polar Equation by Converting to a Rectangular Equation

Covert the polar equation $\,r=2\mathrm{sec}\,\theta \,$ to a rectangular equation, and draw its corresponding graph.

Show Solution

The conversion is

$\begin{array}{l}\text{ }r=2\mathrm{sec}\,\theta \hfill \\ \,\,\,\,\text{ }r=\frac{2}{\mathrm{cos}\,\theta }\hfill \\ r\mathrm{cos}\,\theta =2\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\hfill \end{array}$

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

Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are lines.

Figure 12. (a) Polar grid (b) Rectangular coordinate system

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

Rewriting a Polar Equation in Cartesian Form

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

Show 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 $\,{x}^{2}+{y}^{2}={r}^{2}.$

$\begin{array}{llll}\text{ }r=\frac{3}{1-2\mathrm{cos}\,\theta }\hfill & \hfill & \hfill & \hfill \\ r\left(1-2\mathrm{cos}\,\theta \right)=3\hfill & \hfill & \hfill & \hfill \\ \,\,\,\,\,\,r\left(1-2\left(\frac{x}{r}\right)\right)=3\hfill & \hfill & \hfill & \text{Use }\mathrm{cos}\,\theta =\frac{x}{r}\text{ to eliminate }\theta .\hfill \\ \text{ }\,\,\,r-2x=3\hfill & \hfill & \hfill & \hfill \\ \text{ }r=3+2x\hfill & \hfill & \hfill & \text{Isolate }r.\hfill \\ \text{ }{r}^{2}={\left(3+2x\right)}^{2}\hfill & \hfill & \hfill & \text{Square both sides}.\hfill \\ \text{ }\,{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}\hfill & \hfill & \hfill & \text{Use }{x}^{2}+{y}^{2}={r}^{2}.\hfill \end{array}$

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

$\begin{array}{l}{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}\hfill \\ \text{ }{y}^{2}={\left(3+2x\right)}^{2}-{x}^{2}\hfill \\ \text{ }y=±\sqrt{{\left(3+2x\right)}^{2}-{x}^{2}}\hfill \end{array}$

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. See (Figure).

Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are hyperbolas.

Figure 13.

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

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.

$\begin{array}{ll}\,\text{ }{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}\hfill & \hfill \\ \,\,\text{ }{x}^{2}+{y}^{2}-{\left(3+2x\right)}^{2}=0\hfill & \hfill \\ {x}^{2}+{y}^{2}-\left(9+12x+4{x}^{2}\right)=0\hfill & \hfill \\ \,\text{ }{x}^{2}+{y}^{2}-9-12x-4{x}^{2}=0\hfill & \hfill \\ \,\text{ }-3{x}^{2}-12x+{y}^{2}=9\hfill & \text{Multiply through by }-1.\hfill \\ \,\,\text{ }\,3{x}^{2}+12x-{y}^{2}=-9\hfill & \hfill \\ \,\,\text{ }\,3\left({x}^{2}+4x+\,\,\,\,\,\,\,\,\,\right)-{y}^{2}=-9\hfill & \text{Organize terms to complete the square for}\,x.\hfill \\ \,\,\text{ }\,3\left({x}^{2}+4x+4\right)-{y}^{2}=-9+12\hfill & \hfill \\ \text{ }3{\left(x+2\right)}^{2}-{y}^{2}=3\hfill & \hfill \\ \,\text{ }{\left(x+2\right)}^{2}-\frac{{y}^{2}}{3}=1\hfill & \hfill \end{array}$

Try It

Rewrite the polar equation $\,r=2\mathrm{sin}\,\theta \,$ in Cartesian form.

Show Solution

${x}^{2}+{y}^{2}=2y\,$ or, in the standard form for a circle, $\,{x}^{2}+{\left(y-1\right)}^{2}=1$

Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation $\,r=\mathrm{sin}\left(2\theta \right)\,$ in Cartesian form.

Show Solution

$\begin{array}{ll}\text{ }r=\mathrm{sin}\left(2\theta \right)\hfill & \text{Use the double angle identity for sine}.\hfill \\ \text{ }r=2\mathrm{sin}\,\theta \mathrm{cos}\,\theta \begin{array}{cccc}& & & \end{array}\hfill & \text{Use }\mathrm{cos}\,\theta =\frac{x}{r}\text{ and }\mathrm{sin}\,\theta =\frac{y}{r}.\hfill \\ \text{ }r=2\left(\frac{x}{r}\right)\left(\frac{y}{r}\right)\hfill & \text{Simplify}.\hfill \\ \text{ }r=\frac{2xy}{{r}^{2}}\hfill & \text{ Multiply both sides by }{r}^{2}.\hfill \\ \text{ }{r}^{3}=2xy\hfill & \hfill \\ {\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{3}=2xy\hfill & \text{As}\,{x}^{2}+{y}^{2}={r}^{2},r=\sqrt{{x}^{2}+{y}^{2}}.\hfill \end{array}$

This equation can also be written as

${\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}=2xy\,\text{or}\,{x}^{2}+{y}^{2}={\left(2xy\right)}^{\frac{2}{3}}$

Access these online resources for additional instruction and practice with polar coordinates.

Key Equations

Conversion formulas

$\begin{array}{ll}\hfill & \mathrm{cos}\,\theta =\frac{x}{r}\to x=r\mathrm{cos}\,\theta \hfill \\ \hfill & \mathrm{sin}\,\theta =\frac{y}{r}\to y=r\mathrm{sin}\,\theta \hfill \\ \hfill & \,\,\,\,\,\,\,{r}^{2}={x}^{2}+{y}^{2}\hfill \\ \hfill & \mathrm{tan}\,\theta =\frac{y}{x}\hfill \end{array}$

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 $\,\left(r,\theta \right),\,\theta >0,\,$ 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 $\,\theta .$ See (Figure).
If $\,r\,$ is negative, extend the directed line segment in the opposite direction of $\,\theta .\,$ See (Figure).
To convert from polar coordinates to rectangular coordinates, use the formulas $\,x=r\mathrm{cos}\,\theta \,$ and $\,y=r\mathrm{sin}\,\theta .\,$ See (Figure) and (Figure).
To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: $\,\mathrm{cos}\,\theta =\frac{x}{r},\mathrm{sin}\,\theta =\frac{y}{r},\mathrm{tan}\,\theta =\frac{y}{x},\,$ and $\,r=\sqrt{{x}^{2}+{y}^{2}}.\,$ See (Figure).
Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See (Figure), (Figure), and (Figure).
Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane. See (Figure), (Figure), and (Figure).

Section Exercises

Verbal

How are polar coordinates different from rectangular coordinates?

Show Solution

How are the polar axes different from the x– and y-axes of the Cartesian plane?

Explain how polar coordinates are graphed.

Show Solution

Determine $\,\theta \,$ for the point, then move $\,r\,$ units from the pole to plot the point. If $\,r\,$ is negative, move $\,r\,$ units from the pole in the opposite direction but along the same angle. The point is a distance of $\,r\,$ away from the origin at an angle of $\,\theta \,$ from the polar axis.

How are the points $\,\left(3,\frac{\pi }{2}\right)\,$ and $\,\left(-3,\frac{\pi }{2}\right)\,$ related?

Explain why the points $\,\left(-3,\frac{\pi }{2}\right)\,$ and $\,\left(3,-\frac{\pi }{2}\right)\,$ are the same.

Show Solution

The point $\,\left(-3,\frac{\pi }{2}\right)\,$ has a positive angle but a negative radius and is plotted by moving to an angle of $\,\frac{\pi }{2}\,$ and then moving 3 units in the negative direction. This places the point 3 units down the negative y-axis. The point $\,\left(3,-\frac{\pi }{2}\right)\,$ has a negative angle and a positive radius and is plotted by first moving to an angle of $\,-\frac{\pi }{2}\,$ and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y-axis.

Algebraic

For the following exercises, convert the given polar coordinates to Cartesian coordinates with $\,r>0\,$ and $\,0\le \theta \le 2\pi .\,$ Remember to consider the quadrant in which the given point is located when determining $\,\theta \,$ for the point.

$\left(7,\frac{7\pi }{6}\right)$

$\left(5,\pi \right)$

Show Solution

$\left(-5,0\right)$

$\left(6,-\frac{\pi }{4}\right)$

$\left(-3,\frac{\pi }{6}\right)$

Show Solution

$\left(-\frac{3\sqrt{3}}{2},-\frac{3}{2}\right)$

$\left(4,\frac{7\pi }{4}\right)$

For the following exercises, convert the given Cartesian coordinates to polar coordinates with $\,r>0,\,\,0\le \theta <2\pi .\,$ Remember to consider the quadrant in which the given point is located.

$\left(4,2\right)$

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$\left(2\sqrt{5}, 0.464\right)$

$\left(-4,6\right)$

$\left(3,-5\right)$

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$\left(\sqrt{34},5.253\right)$

$\left(-10,-13\right)$

$\left(8,8\right)$

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$\left(8\sqrt{2},\frac{\pi }{4}\right)$

For the following exercises, convert the given Cartesian equation to a polar equation.

$x=3$

$y=4$

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$r=4\mathrm{csc}\theta$

$y=4{x}^{2}$

$y=2{x}^{4}$

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$r=\sqrt[3]{\frac{sin\theta }{2co{s}^{4}\theta }}$

${x}^{2}+{y}^{2}=4y$

${x}^{2}+{y}^{2}=3x$

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$r=3\mathrm{cos}\theta$

${x}^{2}-{y}^{2}=x$

${x}^{2}-{y}^{2}=3y$

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$r=\frac{3\mathrm{sin}\theta }{\mathrm{cos}\left(2\theta \right)}$

${x}^{2}+{y}^{2}=9$

${x}^{2}=9y$

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$r=\frac{9\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta }$

${y}^{2}=9x$

$9xy=1$

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$r=\sqrt{\frac{1}{9\mathrm{cos}\theta \mathrm{sin}\theta }}$

For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented.

$r=3\mathrm{sin}\,\theta$

$r=4\mathrm{cos}\,\theta$

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${x}^{2}+{y}^{2}=4x\,$ or $\,\frac{{\left(x-2\right)}^{2}}{4}+\frac{{y}^{2}}{4}=1;$ circle

$r=\frac{4}{\mathrm{sin}\,\theta +7\mathrm{cos}\,\theta }$

$r=\frac{6}{\mathrm{cos}\,\theta +3\mathrm{sin}\,\theta }$

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$3y+x=6;\,$ line

$r=2\mathrm{sec}\,\theta$

$r=3\mathrm{csc}\,\theta$

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$y=3;\,$ line

$r=\sqrt{r\mathrm{cos}\,\theta +2}$

${r}^{2}=4\mathrm{sec}\,\theta \,\mathrm{csc}\,\theta$

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$xy=4;\,$ hyperbola

$r=4$

${r}^{2}=4$

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${x}^{2}+{y}^{2}=4;\,$ circle

$r=\frac{1}{4\mathrm{cos}\,\theta -3\mathrm{sin}\,\theta }$

$r=\frac{3}{\mathrm{cos}\,\theta -5\mathrm{sin}\,\theta }$

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$x-5y=3;\,$ line

Graphical

For the following exercises, find the polar coordinates of the point.

Polar coordinate system with a point located on the third concentric circle and pi/2.

Polar coordinate system with a point located on the third concentric circle and midway between pi/2 and pi in the second quadrant.

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$\left(3,\frac{3\pi }{4}\right)$

Polar coordinate system with a point located midway between the first and second concentric circles and a third of the way between pi and 3pi/2 (closer to pi).

Polar coordinate system with a point located on the fifth concentric circle and pi.

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$\left(5,\pi \right)$

Polar coordinate system with a point located on the fourth concentric circle and a third of the way between 3pi/2 and 2pi (closer to 3pi/2).

For the following exercises, plot the points.

$\left(-2,\frac{\pi }{3}\right)$

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$\left(-1,-\frac{\pi }{2}\right)$

$\left(3.5,\frac{7\pi }{4}\right)$

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$\left(-4,\frac{\pi }{3}\right)$

$\left(5,\frac{\pi }{2}\right)$

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$\left(4,\frac{-5\pi }{4}\right)$

$\left(3,\frac{5\pi }{6}\right)$

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$\left(-1.5,\frac{7\pi }{6}\right)$

$\left(-2,\frac{\pi }{4}\right)$

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$\left(1,\frac{3\pi }{2}\right)$

For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis.

$5x-y=6$

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$r=\frac{6}{5\mathrm{cos}\theta -\mathrm{sin}\theta }$

Plot of given line in the polar coordinate grid

$2x+7y=-3$

${x}^{2}+{\left(y-1\right)}^{2}=1$

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$r=2\mathrm{sin}\theta$

Plot of given circle in the polar coordinate grid

${\left(x+2\right)}^{2}+{\left(y+3\right)}^{2}=13$

$x=2$

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$r=\frac{2}{\mathrm{cos}\theta }$

Plot of given circle in the polar coordinate grid

${x}^{2}+{y}^{2}=5y$

${x}^{2}+{y}^{2}=3x$

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$r=3\mathrm{cos}\theta$

Plot of given circle in the polar coordinate grid.

For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane.

$r=6$

$r=-4$

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${x}^{2}+{y}^{2}=16$

Plot of circle with radius 4 centered at the origin in the rectangular coordinates grid.

$\theta =-\frac{2\pi }{3}$

$\theta =\frac{\pi }{4}$

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$y=x$

Plot of line y=x in the rectangular coordinates grid.

$r=\mathrm{sec}\,\theta$

$r=-10\mathrm{sin}\,\theta$

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${x}^{2}+{\left(y+5\right)}^{2}=25$ Plot of circle with radius 5 centered at (0,-5).

$r=3\mathrm{cos}\,\theta$

Technology

Use a graphing calculator to find the rectangular coordinates of $\,\left(2,-\frac{\pi }{5}\right).\,$ Round to the nearest thousandth.

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$\left(1.618,-1.176\right)$

Use a graphing calculator to find the rectangular coordinates of $\,\left(-3,\frac{3\pi }{7}\right).\,$ Round to the nearest thousandth.

Use a graphing calculator to find the polar coordinates of $\,\left(-7,8\right)\,$ in degrees. Round to the nearest thousandth.

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$\left(10.630,131.186°\right)$

Use a graphing calculator to find the polar coordinates of $\,\left(3,-4\right)\,$ in degrees. Round to the nearest hundredth.

Use a graphing calculator to find the polar coordinates of $\,\left(-2,0\right)\,$ in radians. Round to the nearest hundredth.

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$\,\left(2,3.14\right)or\left(2,\pi \right)\,$

Extensions

Describe the graph of $\,r=a\mathrm{sec}\,\theta ;a>0.$

Describe the graph of $\,r=a\mathrm{sec}\,\theta ;a<0.$

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A vertical line with $\,a\,$ units left of the y-axis.

Describe the graph of $\,r=a\mathrm{csc}\,\theta ;a>0.$

Describe the graph of $\,r=a\mathrm{csc}\,\theta ;a<0.$

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A horizontal line with $\,a\,$ units below the x-axis.

What polar equations will give an oblique line?

For the following exercise, graph the polar inequality.

$r<4$

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$0\le \theta \le \frac{\pi }{4}$

$\theta =\frac{\pi }{4},\,r\,\ge \,2$

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$\theta =\frac{\pi }{4},\,r\,\ge -3$

$0\le \theta \le \frac{\pi }{3},\,r\,<\,2$

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[latex]\frac{-\pi }{6}<\theta \le \frac{\pi }{3},-3

Glossary

polar axis: on the polar grid, the equivalent of the positive x-axis on the rectangular grid

polar coordinates: on the polar grid, the coordinates of a point labeled $\,\left(r,\theta \right),\,$ where $\,\theta \,$ indicates the angle of rotation from the polar axis and $\,r\,$ represents the radius, or the distance of the point from the pole in the direction of $\,\theta$

pole: the origin of the polar grid

Further Applications of Trigonometry

Learning Objectives

Plotting Points Using Polar Coordinates

Plotting a Point on the Polar Grid

Try It

Plotting a Point in the Polar Coordinate System with a Negative Component

Try It

Converting from Polar Coordinates to Rectangular Coordinates

Converting from Polar Coordinates to Rectangular Coordinates

How To

Writing Polar Coordinates as Rectangular Coordinates

Writing Polar Coordinates as Rectangular Coordinates

Try It

Converting from Rectangular Coordinates to Polar Coordinates

Converting from Rectangular Coordinates to Polar Coordinates

Writing Rectangular Coordinates as Polar Coordinates

Analysis

Transforming Equations between Polar and Rectangular Forms

How To

Writing a Cartesian Equation in Polar Form

Rewriting a Cartesian Equation as a Polar Equation

Rewriting a Cartesian Equation in Polar Form

Try It

Identify and Graph Polar Equations by Converting to Rectangular Equations

Graphing a Polar Equation by Converting to a Rectangular Equation

Rewriting a Polar Equation in Cartesian Form

Analysis

Try It

Rewriting a Polar Equation in Cartesian Form

Key Equations

Key Concepts

Section Exercises

Verbal

Algebraic

Graphical

Technology

Extensions

Glossary

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