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 1). 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.

Figure 1

Plotting Points Using Polar Coordinates

When we think about plotting points in the plane, we usually think of rectangular coordinates (x,y)(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 to polar coordinates, which are points labeled (r,θ)(𝑟,𝜃) 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 r𝑟 is the radius or length of the directed line segment from the pole. The angle θ,𝜃, measured in radians, indicates the direction of r.𝑟. We move counterclockwise from the polar axis by an angle of θ,𝜃, and measure a directed line segment the length of r𝑟 in the direction of θ.𝜃. Even though we measure θ𝜃 first and then r,𝑟, the polar point is written with the r-coordinate first. For example, to plot the point (2,π4),(2,𝜋4), we would move π4𝜋4 units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in Figure 2.

Figure 2

EXAMPLE 1

Plotting a Point on the Polar Grid

Plot the point (3,π2)(3,𝜋2) on the polar grid.

Solution

The angle π2𝜋2 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 π2𝜋2 direction, as shown in Figure 3.

Figure 3

TRY IT #1

Plot the point (2,π3)(2,𝜋3) in the polar grid.

EXAMPLE 2

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

Plot the point (−2,π6)(−2,𝜋6) on the polar grid.

TRY IT #2

Plot the points (3,−π6)(3,−𝜋6) and (2,9π4)(2,9𝜋4) on the same polar grid.

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,𝑥,𝑦,𝑟, and θ.𝜃.

cosθ=xr→x=rcosθsinθ=yr→y=rsinθcos𝜃=𝑥𝑟→𝑥=𝑟cos𝜃sin𝜃=𝑦𝑟→𝑦=𝑟sin𝜃

Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure 5. An easy way to remember the equations above is to think of cosθcos𝜃 as the adjacent side over the hypotenuse and sinθsin𝜃 as the opposite side over the hypotenuse.

Figure 5

CONVERTING FROM POLAR COORDINATES TO RECTANGULAR COORDINATES

To convert polar coordinates (r,θ)(𝑟,𝜃) to rectangular coordinates (x,y),(𝑥,𝑦), let

cosθ=xr→x=rcosθcos𝜃=𝑥𝑟→𝑥=𝑟cos𝜃

sinθ=yr→y=rsinθsin𝜃=𝑦𝑟→𝑦=𝑟sin𝜃

HOW TO

Given polar coordinates, convert to rectangular coordinates.

1. Given the polar coordinate (r,θ),(𝑟,𝜃), write x=rcosθ𝑥=𝑟cos𝜃 and y=rsinθ.𝑦=𝑟sin𝜃.
2. Evaluate cosθcos𝜃 and sinθ.sin𝜃.
3. Multiply cosθcos𝜃 by r𝑟 to find the x-coordinate of the rectangular form.
4. Multiply sinθsin𝜃 by r𝑟 to find the y-coordinate of the rectangular form.

EXAMPLE 3

Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates (3,π2)(3,𝜋2) as rectangular coordinates.

Solution

Use the equivalent relationships.

x=rcosθx=3cosπ2=0y=rsinθy=3sinπ2=3𝑥=𝑟cos𝜃𝑥=3cos𝜋2=0𝑦=𝑟sin𝜃𝑦=3sin𝜋2=3

The rectangular coordinates are (0,3).(0,3). See Figure 6.

Figure 6

EXAMPLE 4

Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates (−2,0)(−2,0) as rectangular coordinates.

Solution

See Figure 7. Writing the polar coordinates as rectangular, we have

x=rcosθx=−2cos(0)=−2y=rsinθy=−2sin(0)=0𝑥=𝑟cos𝜃𝑥=−2cos(0)=−2𝑦=𝑟sin𝜃𝑦=−2sin(0)=0

The rectangular coordinates are also (−2,0).(−2,0).

Figure 7

TRY IT #3

Write the polar coordinates (−1,2π3)(−1,2𝜋3) as rectangular coordinates.

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 8.

cosθ=xrx=rcosθsinθ=yr ory=rsinθr2=x2+y2tanθ=yxcos𝜃=𝑥𝑟𝑥=𝑟cos𝜃sin𝜃=𝑦𝑟 or𝑦=𝑟sin𝜃𝑟2=𝑥2+𝑦2tan𝜃=𝑦𝑥

Figure 8

EXAMPLE 5

Writing Rectangular Coordinates as Polar Coordinates

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

Solution

We see that the original point (3,3)(3,3) is in the first quadrant. To find θ,𝜃, use the formula tanθ=yx.tan𝜃=𝑦𝑥. This gives

tanθ=33tanθ=1θ=tan−1(1)θ=π4tan𝜃=33tan𝜃=1𝜃=tan−1(1)𝜃=𝜋4

To find r,𝑟, we substitute the values for x𝑥 and y𝑦 into the formula r=x2+y2−−−−−−√.𝑟=𝑥2+𝑦2. We know that r𝑟 must be positive, as π4𝜋4 is in the first quadrant. Thus

r=32+32−−−−−−√r=9+9−−−−√r=18−−√=32–√𝑟=32+32𝑟=9+9𝑟=18=32

So, r=32–√𝑟=32 and θ=π4,𝜃=𝜋4, giving us the polar point (32–√,π4).(32,𝜋4). See Figure 9.

Figure 9

Analysis

There are other sets of polar coordinates that will be the same as our first solution. For example, the points (−32–√,5π4)(−32,5𝜋4) and (32–√,−7π4)(32,−7𝜋4) will coincide with the original solution of (32–√,π4).(32,𝜋4). The point (−32–√,5π4)(−32,5𝜋4) indicates a move further counterclockwise by π,𝜋, which is directly opposite π4.𝜋4. The radius is expressed as −32–√.−32. However, the angle 5π45𝜋4 is located in the third quadrant and, as r𝑟 is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as (32–√,π4).(32,𝜋4). The point (32–√,−7π4)(32,−7𝜋4) is a move further clockwise by −7π4,−7𝜋4, from π4.𝜋4. The radius, 32–√,32, 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.

1. Change the MODE to POL, representing polar form.
2. Press the Y= button to bring up a screen allowing the input of six equations: r1,r2,…,r6.𝑟1,𝑟2,…,𝑟6.
3. Enter the polar equation, set equal to r.𝑟.
4. Press GRAPH.

EXAMPLE 6

Writing a Cartesian Equation in Polar Form

Write the Cartesian equation x2+y2=9𝑥2+𝑦2=9 in polar form.

Solution

The goal is to eliminate x𝑥 and y𝑦 from the equation and introduce r𝑟 and θ.𝜃. Ideally, we would write the equation r𝑟 as a function of θ.𝜃. To obtain the polar form, we will use the relationships between (x,y)(𝑥,𝑦) and (r,θ).(𝑟,𝜃). Since x=rcosθ𝑥=𝑟cos𝜃 and y=rsinθ,𝑦=𝑟sin𝜃, we can substitute and solve for r.𝑟.

   (rcosθ)2+(rsinθ)2=9   r2cos2θ+r2sin2θ=9r2(cos2θ+sin2θ)=9r2(1)=9                                 r=±3Substitute cos2θ+sin2θ=1.Use the square root property.   (𝑟cos𝜃)2+(𝑟sin𝜃)2=9   𝑟2cos2𝜃+𝑟2sin2𝜃=9𝑟2(cos2𝜃+sin2𝜃)=9𝑟2(1)=9Substitute cos2𝜃+sin2𝜃=1.                                 𝑟=±3Use the square root property.

Thus, x2+y2=9,r=3,𝑥2+𝑦2=9,𝑟=3, and r=−3𝑟=−3 should generate the same graph. See Figure 10.

Figure 10 (a) Cartesian form x2+y2=9𝑥2+𝑦2=9 (b) Polar form r=3𝑟=3

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

x2+y2=9        y2=9−x2          y=±9−x2−−−−−√𝑥2+𝑦2=9        𝑦2=9−𝑥2          𝑦=±9−𝑥2

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 Y1=9−x2−−−−−√𝑌1=9−𝑥2 and Y2=−9−x2−−−−−√.𝑌2=−9−𝑥2. Press GRAPH.

EXAMPLE 7

Rewriting a Cartesian Equation as a Polar Equation

Rewrite the Cartesian equation x2+y2=6y𝑥2+𝑦2=6𝑦 as a polar equation.

EXAMPLE 8

Rewriting a Cartesian Equation in Polar Form

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

Solution

We will use the relationships x=rcosθ𝑥=𝑟cos𝜃 and y=rsinθ.𝑦=𝑟sin𝜃.

                        y=3x+2                 rsinθ=3rcosθ+2rsinθ−3rcosθ=2r(sinθ−3cosθ)=2r=2sinθ−3cosθIsolate r.Solve for r.                        𝑦=3𝑥+2                 𝑟sin𝜃=3𝑟cos𝜃+2𝑟sin𝜃−3𝑟cos𝜃=2𝑟(sin𝜃−3cos𝜃)=2Isolate 𝑟.𝑟=2sin𝜃−3cos𝜃Solve for 𝑟.

TRY IT #4

Rewrite the Cartesian equation y2=3−x2𝑦2=3−𝑥2 in polar form.

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 9

Graphing a Polar Equation by Converting to a Rectangular Equation

Covert the polar equation r=2secθ𝑟=2sec𝜃 to a rectangular equation, and draw its corresponding graph.

Solution

The conversion is

rrrcosθx====2secθ2cosθ22𝑟=2sec𝜃𝑟=2cos𝜃𝑟cos𝜃=2𝑥=2

Notice that the equation r=2secθ𝑟=2sec𝜃 drawn on the polar grid is clearly the same as the vertical line x=2𝑥=2 drawn on the rectangular grid (see Figure 12). Just as x=c𝑥=𝑐 is the standard form for a vertical line in rectangular form, r=csecθ𝑟=𝑐sec𝜃 is the standard form for a vertical line in polar form.

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

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

EXAMPLE 10

Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation r=31−2cosθ𝑟=31−2cos𝜃 as a Cartesian equation.

Solution

The goal is to eliminate θ𝜃 and r,𝑟, and introduce x𝑥 and y.𝑦. We clear the fraction, and then use substitution. In order to replace r𝑟 with x𝑥 and y,𝑦, we must use the expression x2+y2=r2.𝑥2+𝑦2=𝑟2.

rr(1−2(xr))r−2xr−2xrr2x2+y2=======31−2cosθ3333+2x(3+2x)2(3+2x)2Usecosθ=xrtoeliminateθIsolaterSquarebothsidesUsex2+y2=r2𝑟=31-2cos𝜃𝑟(1-2(𝑥𝑟))=3𝑟-2𝑥=3Usecos𝜃=𝑥𝑟𝑡𝑜eliminate𝜃𝑟-2𝑥=3𝑟=3+2𝑥Isolate𝑟𝑟2=(3+2𝑥)2Squarebothsides𝑥2+𝑦2=(3+2𝑥)2Use𝑥2+𝑦2=𝑟2

The Cartesian equation is x2+y2=(3+2x)2.𝑥2+𝑦2=(3+2𝑥)2. However, to graph it, especially using a graphing calculator or computer program, we want to isolate y.𝑦.

x2+y2=(3+2x)2        y2=(3+2x)2−x2          y=±(3+2x)2−x2−−−−−−−−−−−√𝑥2+𝑦2=(3+2𝑥)2        𝑦2=(3+2𝑥)2−𝑥2          𝑦=±(3+2𝑥)2−𝑥2

When our entire equation has been changed from r𝑟 and θ𝜃 to x𝑥 and y,𝑦, we can stop, unless asked to solve for y𝑦 or simplify. See Figure 13.

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.

x2+y2x2+y2−(3+2x)2×2+y2−(9+12x+4×2)x2+y2−9−12x−4×2−3×2−12x+y23x2+12x−y23(x2+4x+)−y23(x2+4x+4)−y23(x+2)2−y2(x+2)2−y23==========(3+2x)20009−9−9−9+1231Multiplythroughby−1Organizetermstocompletethesquareforx𝑥2+𝑦2=(3+2𝑥)2𝑥2+𝑦2-(3+2𝑥)2=0𝑥2+𝑦2-(9+12𝑥+4𝑥2)=0𝑥2+𝑦2-9-12𝑥-4𝑥2=0-3𝑥2-12𝑥+𝑦2=9Multiplythroughby-13𝑥2+12𝑥-𝑦2=-93(𝑥2+4𝑥+)-𝑦2=-93(𝑥2+4𝑥+4)-𝑦2=-9+12𝑂𝑟𝑔𝑎𝑛𝑖𝑧𝑒termstocompletethesquarefor𝑥3(𝑥+2)2-𝑦2=3(𝑥+2)2-𝑦23=1

TRY IT #5

Rewrite the polar equation r=2sinθ𝑟=2sin𝜃 in Cartesian form.

EXAMPLE 11

Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation r=sin(2θ)𝑟=sin(2𝜃) in Cartesian form.

Solution

                 r=sin(2θ)                 r=2sinθcosθ                 r=2(xr)(yr)                 r=2xyr2               r3=2xy(x2+y2−−−−−−√)3=2xyUse the double angle identity for sine.Use cosθ=xr and sinθ=yr.Simplify. Multiply both sides by r2.Asx2+y2=r2,r=x2+y2−−−−−−√.                 𝑟=sin(2𝜃)Use the double angle identity for sine.                 𝑟=2sin𝜃cos𝜃Use cos𝜃=𝑥𝑟 and sin𝜃=𝑦𝑟.                 𝑟=2(𝑥𝑟)(𝑦𝑟)Simplify.                 𝑟=2𝑥𝑦𝑟2 Multiply both sides by 𝑟2.               𝑟3=2𝑥𝑦(𝑥2+𝑦2)3=2𝑥𝑦As𝑥2+𝑦2=𝑟2,𝑟=𝑥2+𝑦2.

This equation can also be written as

(x2+y2)32=2xyorx2+y2=(2xy)23(𝑥2+𝑦2)32=2𝑥𝑦or𝑥2+𝑦2=(2𝑥𝑦)23