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# 8-3. Polar Coordinates

Polar Coordinates
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 [link]). 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 class="small" id="Figure_08_03_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"></figure>

# Plotting Points Using Polar Coordinates

When we think about plotting points in the plane, we usually think of rectangular coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the radius or length of the directed line segment from the pole. The angle<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] measured in radians, indicates the direction of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]We move counterclockwise from the polar axis by an angle of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]and measure a directed line segment the length of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in the direction of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Even though we measure<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]first and then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] the polar point is written with the r-coordinate first. For example, to plot the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2, π 4 ),we would move<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4  units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in [link].

<figure class="small" id="Figure_08_03_002"></figure>
Plotting a Point on the Polar Grid

Plot the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3, π 2 ) on the polar grid.

The angle<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  direction, as shown in [link].

<figure class="small" id="Figure_08_03_003"></figure>

Plot the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,  π 3 ) in the polar grid.

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

Plot the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2,  π 6 ) on the polar grid.

We know that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 6  is located in the first quadrant. However,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>−2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]We can approach plotting a point with a negative<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in two ways:

1. Plot the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2, π 6 ) by moving<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 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;
2. Move<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 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 [link](a). Compare this to the graph of the polar coordinate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2, π 6 ) shown in [link](b).

<figure class="medium" id="Figure_08_03_005"></figure>

Plot the points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,− π 6 )and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2, 9π 4 ) 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x r →x=rcos θ sin θ= y r →y=rsin θ

Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in [link]. An easy way to remember the equations above is to think of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as the adjacent side over the hypotenuse and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as the opposite side over the hypotenuse.

<figure class="small" id="Figure_08_03_007"></figure>
Converting from Polar Coordinates to Rectangular Coordinates

To convert polar coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] r, θ ) to rectangular coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x, y ), let

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x r →x=rcos θ
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] y r →y=rsin θ

Given polar coordinates, convert to rectangular coordinates.

1. Given the polar coordinate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] r,θ ), write<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mi>r</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>r</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
2. Evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
3. Multiply<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]to find the x-coordinate of the rectangular form.
4. Multiply<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]to find the y-coordinate of the rectangular form.
Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3, π 2 ) as rectangular coordinates.

Use the equivalent relationships.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><mi>x</mi><mo>=</mo><mi>r</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x=3cos  π 2 =0 y=rsin θ y=3sin  π 2 =3

The rectangular coordinates are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,3 ). See [link].

<figure id="Figure_08_03_008"></figure>
Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2,0 ) as rectangular coordinates.

See [link]. Writing the polar coordinates as rectangular, we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>x</mi><mo>=</mo><mi>r</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x=−2cos( 0 )=−2 y=rsin θ y=−2sin( 0 )=0

The rectangular coordinates are also<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2,0 ).

<figure class="medium" id="Figure_08_03_009"></figure>

Write the polar coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1, 2π 3 ) as rectangular coordinates.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y )=( 1 2 ,− 3 2 )

# 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 [link].

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x r    or   x=rcos θ sin θ= y r    or   y=rsin θ        r 2 = x 2 + y 2 tan θ= y x
<figure class="small" id="Figure_08_03_010"></figure>
Writing Rectangular Coordinates as Polar Coordinates

Convert the rectangular coordinates<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,3 ) to polar coordinates.

We see that the original point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,3 ) is in the first quadrant. To find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]use the formula<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] y x . This gives

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3 3            tan θ=1     tan −1 (1)= π 4

To find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we substitute the values for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the formula<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 . We know that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]must be positive, as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4  is in the first quadrant. Thus

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><mi>r</mi><mo>=</mo><msqrt/></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3 2 + 3 2 r= 9+9 r= 18 =3 2

So,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>3</mn><msqrt/></mrow></annotation-xml></semantics>[/itex] 2   and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext>=</mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4 , giving us the polar point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 2 , π 4 ). See [link].

<figure id="Figure_08_03_011"></figure>
Analysis

There are other sets of polar coordinates that will be the same as our first solution. For example, the points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3 2 ,  5π 4 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 2 ,− 7π 4 ) will coincide with the original solution of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 2 ,  π 4 ). The point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3 2 ,  5π 4 ) indicates a move further counterclockwise by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>π</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]which is directly opposite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4 . The radius is expressed as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mn>3</mn><msqrt/></mrow></annotation-xml></semantics>[/itex] 2 . However, the angle<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] 5π 4  is located in the third quadrant and, as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 2 ,   π 4 ). The point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 2 , − 7π 4 ) is a move further clockwise by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 7π 4 , from<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4 . The radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><msqrt/></mrow></annotation-xml></semantics>[/itex] 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.

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:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] r 1 ,   r 2 ,  .  .  .  ,   r 6 .
3. Enter the polar equation, set equal to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
4. Press GRAPH.
Writing a Cartesian Equation in Polar Form

Write the Cartesian equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 =9 in polar form.

The goal is to eliminate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]from the equation and introduce<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Ideally, we would write the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]To obtain the polar form, we will use the relationships between<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] r,θ ). Since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mi>r</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>r</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we can substitute and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>   </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (rcos θ) 2 + (rsin θ) 2 =9     r 2 cos 2 θ+ r 2 sin 2 θ=9     r 2 ( cos 2 θ+ sin 2 θ)=9                             r 2 (1)=9  Substitute cos2 θ+ sin 2 θ=1.                                  r=±3 Use the square root property.

Thus,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 =9,r=3,and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mo>−</mo><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]should generate the same graph. See [link].

<figure class="medium" id="Figure_08_03_016"> <figcaption>(a) Cartesian form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 =9 (b) Polar form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>3</mn></mrow></annotation-xml></semantics>[/itex]</figcaption> </figure>

To graph a circle in rectangular form, we must first solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><msup><mi>x</mi></msup></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =9          y 2 =9− x 2           y=± 9− x 2

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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] Y 1 = 9− x 2  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] Y 2 =− 9− x 2 . PressGRAPH.

Rewriting a Cartesian Equation as a Polar Equation

Rewrite the Cartesian equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 =6y as a polar equation.

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:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] r 2 =6y Use  x 2 + y 2 = r 2 .                                   r 2 =6rsin θ Substitute y=rsin θ.         r 2 −6rsin θ=0Set equal to 0.       r(r−6sin θ)=0 Factor and solve.                                     r=0 We reject r=0, as it only represents one point, (0,0).                              or r=6sin θ

Therefore, the equations<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 =6y and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>6</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] should give us the same graph. See [link].

<figure id="Figure_08_03_012"> <figcaption>(a) Cartesian form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 =6y(b) polar form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>6</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]</figcaption> </figure>

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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a polar equation.

We will use the relationships<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mi>r</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>r</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>                        </mtext><mi>y</mi><mo>=</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>2</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]                  rsin θ=3rcos θ+2  rsin θ−3rcos θ=2 r(sin θ−3cos θ)=2 Isolate r.                          r= 2 sin θ−3cos θ Solve for r.

Rewrite the Cartesian equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] y 2 =3− x 2  in polar form.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>sec</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] to a rectangular equation, and draw its corresponding graph.

The conversion is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>           </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>sec</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]             r= 2 cos θ  rcos θ=2                 x=2

Notice that the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>sec</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]drawn on the polar grid is clearly the same as the vertical line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]drawn on the rectangular grid (see [link]). Just as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mi>c</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the standard form for a vertical line in rectangular form,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>c</mi><mi>sec</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the standard form for a vertical line in polar form.

<figure class="medium" id="Figure_08_03_013"> <figcaption>(a) Polar grid (b) Rectangular coordinate system</figcaption> </figure>

A similar discussion would demonstrate that the graph of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>csc</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] will be the horizontal line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]In fact,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>c</mi><mi>csc</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] is the standard form for a horizontal line in polar form, corresponding to the rectangular form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>c</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 3 1−2cos θ  as a Cartesian equation.

The goal is to eliminate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]and introduce<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]We clear the fraction, and then use substitution. In order to replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] we must use the expression<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 = r 2 .

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>                   </mtext><mi>r</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3 1−2cos θ r(1−2cos θ)=3       r( 1−2( x r ) )=3 Use cos θ= x r  to eliminate θ.             r−2x=3                    r=3+2xIsolate r.                   r 2 = (3+2x) 2 Square both sides.           x 2 + y 2 = (3+2x) 2 Use  x 2 + y 2 = r 2 .

The Cartesian equation is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 = ( 3+2x ) 2 . However, to graph it, especially using a graphing calculator or computer program, we want to isolate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>x</mi></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 + y 2 = ( 3+2x ) 2          y 2 = ( 3+2x ) 2 − x 2           y=± ( 3+2x ) 2 − x 2

When our entire equation has been changed from<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>θ</mi><mtext>  </mtext></mrow></annotation-xml></semantics>[/itex]to <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we can stop, unless asked to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or simplify. See [link].

<figure id="Figure_08_03_015"></figure>

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.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext>                             </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 = (3+2x) 2             x 2 + y 2 − (3+2x) 2 =0 x 2 + y 2 −(9+12x+4 x 2 )=0     x 2 + y 2 −9−12x−4 x2 =0              −3 x 2 −12x+ y 2 =9 Multiply through by −1.                   3 x 2 +12x− y 2 =−9        3( x 2 +4x+         )− y 2 =−9Organize terms to complete the square for x.            3( x 2 +4x+4)− y 2 =−9+12                   3 (x+2) 2 − y 2 =3                     (x+2) 2 − y 2 3 =1

Rewrite the polar equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] in Cartesian form.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =2y or, in the standard form for a circle,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 + ( y−1 ) 2 =1

Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2θ ) in Cartesian form.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>                 </mtext><mi>r</mi><mo>=</mo><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] Use the double angle identity for sine.                  r=2sin θcos θ Use cos θ= x r  and sin θ= y r .                 r=2( x r )( y r ) Simplify.                  r= 2xy r 2  Multiply both sides by  r 2 .                 r 3 =2xy ( x 2 + y 2 ) 3=2xy As  x 2 + y 2 = r 2 ,r= x 2 + y 2 .

This equation can also be written as

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 ) 3 2 =2xy or  x 2 + y 2 = ( 2xy ) 2 3

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

# Key Equations

 Conversion formulas cosθ=[/itex] x r →x=rcos θ sin θ= y r →y=rsin θ         r 2 = x 2 + y 2 tan θ= y x

# 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] r,θ ), θ>0, move in a counterclockwise direction from the polar axis by an angle of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and then extend a directed line segment from the pole the length of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] in the direction of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is negative, move in a clockwise direction, and extend a directed line segment the length of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] in the direction of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] See[link].
• If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is negative, extend the directed line segment in the opposite direction of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]See [link].
• To convert from polar coordinates to rectangular coordinates, use the formulas<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mi>r</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>r</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]See [link]and [link].
• To convert from rectangular coordinates to polar coordinates, use one or more of the formulas:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x r ,sin θ= y r ,tan θ= y x , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 . See [link].
• Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See [link], [link], and [link].
• 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 [link], [link], and [link].

# Section Exercises

## Verbal

How are polar coordinates different from rectangular coordinates?

For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.

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

Explain how polar coordinates are graphed.

Determine<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the point, then move<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]units from the pole to plot the point. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is negative, move<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]units from the pole in the opposite direction but along the same angle. The point is a distance of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]away from the origin at an angle of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]from the polar axis.

How are the points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3, π 2 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3, π 2 ) related?

Explain why the points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3, π 2 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,− π 2 ) are the same.

The point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3, π 2 ) has a positive angle but a negative radius and is plotted by moving to an angle of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  and then moving 3 units in the negative direction. This places the point 3 units down the negative y-axis. The point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,− π 2 ) has a negative angle and a positive radius and is plotted by first moving to an angle of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] π 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>></mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>0</mn><mo>≤</mo><mi>θ</mi><mo>≤</mo><mn>2</mn><mi>π</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Remember to consider the quadrant in which the given point is located when determining<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the point.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 7, 7π 6 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5,π )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −5,0 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 6,− π 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3, π 6 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] − 3 3 2 ,− 3 2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4, 7π 4 )

For the following exercises, convert the given Cartesian coordinates to polar coordinates with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mn>0</mn><mo>≤</mo><mi>θ</mi><mo><</mo><mn>2</mn><mi>π</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Remember to consider the quadrant in which the given point is located.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4,2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 5 , 0.464 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −4,6 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,−5 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 34 ,5.253 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −10,−13 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 8,8 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 8 2 , π 4 )

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>3</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>4</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>4</mn><mi>csc</mi><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>2</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mroot/></mrow></annotation-xml></semantics>[/itex] sinθ 2co s 4 θ 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =4y

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =3x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>3</mn><mi>cos</mi><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 − y 2 =x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 − y 2 =3y

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 3sinθ cos( 2θ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =9

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 =9y

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 9sinθ cos 2 θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>y</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 =9x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>9</mn><mi>x</mi><mi>y</mi><mo>=</mo><mn>1</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 1 9cosθsinθ

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.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>3</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>4</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =4x or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( x−2 ) 2 4 + y 2 4 =1;circle

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 4 sin θ+7cos θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 6 cos θ+3sin θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>3</mn><mi>y</mi><mo>+</mo><mi>x</mi><mo>=</mo><mn>6</mn><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]line

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>2</mn><mi>sec</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>3</mn><mi>csc</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>3</mn><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] line

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] rcos θ+2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>r</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 =4sec θ csc θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mi>y</mi><mo>=</mo><mn>4</mn><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]hyperbola

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>4</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>r</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 =4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =4; circle

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 4cos θ−3sin θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 3 cos θ−5sin θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>−</mo><mn>5</mn><mi>y</mi><mo>=</mo><mn>3</mn><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]line

## Graphical

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3, 3π 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5,π )

For the following exercises, plot the points.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2, π 3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1,− π 2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3.5, 7π 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −4, π 3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5, π 2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4, −5π 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3, 5π 6 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1.5, 7π 6 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2, π 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1, 3π 2 )

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>5</mn><mi>x</mi><mo>−</mo><mi>y</mi><mo>=</mo><mn>6</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 6 5cosθ−sinθ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mi>x</mi><mo>+</mo><mn>7</mn><mi>y</mi><mo>=</mo><mo>−</mo><mn>3</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + ( y−1 ) 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>2</mn><mi>sin</mi><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></annotation-xml></semantics>[/itex] x+2 ) 2 + ( y+3 ) 2 =13

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>2</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 cosθ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =5y

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =3x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>3</mn><mi>cos</mi><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>6</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mo>−</mo><mn>4</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + y 2 =16

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>θ</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2π 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mi>x</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mi>sec</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>−10</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 + ( y+5 ) 2 =25

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>3</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

## Technology

Use a graphing calculator to find the rectangular coordinates of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,− π 5 ). Round to the nearest thousandth.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1.618,−1.176 )

Use a graphing calculator to find the rectangular coordinates of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3, 3π 7 ). Round to the nearest thousandth.

Use a graphing calculator to find the polar coordinates of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −7,8 ) in degrees. Round to the nearest thousandth.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 10.630,131.186° )

Use a graphing calculator to find the polar coordinates of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,−4 ) in degrees. Round to the nearest hundredth.

Use a graphing calculator to find the polar coordinates of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2,0 ) in radians. Round to the nearest hundredth.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,3.14 )or( 2,π )

## Extensions

Describe the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>a</mi><mi>sec</mi><mtext> </mtext><mi>θ</mi><mo>;</mo><mi>a</mi><mo>></mo><mn>0.</mn></mrow></annotation-xml></semantics>[/itex]

Describe the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>a</mi><mi>sec</mi><mtext> </mtext><mi>θ</mi><mo>;</mo><mi>a</mi><mo><</mo><mn>0.</mn></mrow></annotation-xml></semantics>[/itex]

A vertical line with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]units left of the y-axis.

Describe the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>a</mi><mi>csc</mi><mtext> </mtext><mi>θ</mi><mo>;</mo><mi>a</mi><mo>></mo><mn>0.</mn></mrow></annotation-xml></semantics>[/itex]

Describe the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>=</mo><mi>a</mi><mi>csc</mi><mtext> </mtext><mi>θ</mi><mo>;</mo><mi>a</mi><mo><</mo><mn>0.</mn></mrow></annotation-xml></semantics>[/itex]

A horizontal line with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]units below the x-axis.

What polar equations will give an oblique line?

For the following exercise, graph the polar inequality.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo><</mo><mn>4</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>0</mn><mo>≤</mo><mi>θ</mi><mo>≤</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4 , r ≥ 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] π 4 , r ≥−3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>0</mn><mo>≤</mo><mi>θ</mi><mo>≤</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] π 3 , r < 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mo>−</mo><mi>π</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 6 <θ≤ π 3 ,−3<r < 2

## 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] r,θ ), where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]indicates the angle of rotation from the polar axis and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]represents the radius, or the distance of the point from the pole in the direction of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]
pole
the origin of the polar grid