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# 8-4. Polar Coordinates: Graphs

Polar Coordinates: Graphs
In this section you will:
• Test polar equations for symmetry.
• Graph polar equations by plotting points.

The planets move through space in elliptical, periodic orbits about the sun, as shown in [link]. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneousposition. This is one application of polar coordinates, represented as <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] We interpret<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 the distance from the sun 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]as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.

<figure class="medium" id="Figure_08_04_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;"> <figcaption>Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA/JPL-Caltech)</figcaption> </figure>

# Testing Polar Equations for Symmetry

Just as a rectangular equation such as<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><msup/></mrow></annotation-xml></semantics>[/itex] x 2  describes the relationship between<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]on a Cartesian grid, a polar equationdescribes a relationship between<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]on a polar grid. Recall that the coordinate pair<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]indicates that we move counterclockwise from the polar axis (positive x-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 extend a ray from the pole (origin)<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 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]All points that satisfy the polar equation are on the graph.

Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums 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]to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  (y-axis). We replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><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><mo stretchy="false">(</mo><mi>−</mi><mi>r</mi><mo>,</mo><mo>−</mo><mi>θ</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]to determine if the new equation is equivalent to the original equation. For example, suppose we are given 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>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><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −r=2sin(−θ) Replace (r,θ) with (−r,−θ). −r=−2sin θ Identity: sin(−θ)=−sin θ.      r=2sin θ Multiply both sides by−1.

This equation exhibits symmetry with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 .

In the second test, we consider symmetry with respect to the polar axis (<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi></mrow></annotation-xml></semantics>[/itex]-axis). We replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] r,−θ ) or<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,π−θ ) to determine equivalency between the tested equation and the original. For example, suppose we are given 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>1</mn><mo>−</mo><mn>2</mn><mi>cos</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><mi>r</mi><mo>=</mo><mn>1</mn><mo>−</mo><mn>2</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] r=1−2cos(−θ) Replace (r,θ) with (r,−θ). r=1−2cos θ Even/Odd identity

The graph of this equation exhibits symmetry with respect to the polar axis.

In the third test, we consider symmetry with respect to the pole (origin). We replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −r,θ ) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given 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>sin</mi><mo stretchy="false">(</mo><mn>3</mn><mi>θ</mi><mo stretchy="false">).</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><mtr><mtd><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>r</mi><mo>=</mo><mn>2</mn><mi>sin</mi><mo stretchy="false">(</mo><mn>3</mn><mi>θ</mi><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −r=2sin(3θ)

The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.

Symmetry Tests

A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in [link].

<figure id="Figure_08_04_002"> <figcaption>(a) A graph is symmetric with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  (y-axis) if replacing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><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><mo stretchy="false">(</mo><mo>−</mo><mi>r</mi><mo>,</mo><mo>−</mo><mi>θ</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing<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 ) with<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 ) or<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,π−θ ) yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><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><mo stretchy="false">(</mo><mn>−</mn><mi>r</mi><mo>,</mo><mi>θ</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]yields an equivalent equation.</figcaption> </figure>

Given a polar equation, test for symmetry.

1. Substitute the appropriate combination of components for<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 ):<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 ) for<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  symmetry;<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 ) for polar axis symmetry; 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 ) for symmetry with respect to the pole.
2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.
Testing a Polar Equation for Symmetry

Test 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>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] for symmetry.

Test for each of the three types of symmetry.

 1) Replacing(r,θ)[/itex]with(r,θ)[/itex]yields the same result. Thus, the graph is symmetric with respect to the lineθ=[/itex] π 2 . r=2sin(θ)[/itex] −r=−2sin θ Even-odd identity     r=2sin θ Multiply by −1Passed 2) Replacingθ[/itex]withθ[/itex]does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. r=2sin(θ)[/itex] r=−2sin θ Even-odd identityr=−2sin θ≠2sin θ Failed 3) Replacingr[/itex]withr[/itex]changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. r=2sinθ[/itex]   r=−2sin θ≠2sin θ Failed
Analysis

Using a graphing calculator, we can see 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>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] is a circle centered at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mn>1</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with radius<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>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and is indeed symmetric to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 . We can also see that the graph is not symmetric with the polar axis or the pole. See [link].

<figure class="small" id="Figure_08_04_003"></figure>

Test the equation for symmetry:<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>2</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

The equation fails the symmetry test with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  and with respect to the pole. It passes the polar axis symmetry test.

# Graphing Polar Equations by Plotting Points

To graph in the rectangular coordinate system we construct a table of<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]values. To graph in the polar coordinate system we construct a table 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]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]values. We enter values 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] into a polar equation and calculate<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]However, using the properties of symmetry and finding key values 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]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]means fewer calculations will be needed.

## Finding Zeros and Maxima

To find the zeros of a polar equation, we solve for the values 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]that result in<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] Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for<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></mrow></annotation-xml></semantics>[/itex]We use the same process for polar equations. Set<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></mrow></annotation-xml></semantics>[/itex]and solve for<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]

For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values 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]into the equation that result in the maximum value of the trigonometric functions. Consider<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>5</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when<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><mn>0</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]so our polar equation is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>5</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and the value<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><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] will yield the maximum<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 |.

Similarly, the maximum value of the sine function is 1 when<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , and if our polar equation is<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>5</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the value<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  will yield the maximum<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 |. We may find additional information by calculating values 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]when<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><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.

Finding Zeros and Maximum Values for a Polar Equation

Using the equation in [link], find the zeros and maximum<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 | and, if necessary, the polar axis intercepts 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><mn>2</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

To find the zeros, set<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]equal to zero and solve for<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"><mrow><mn>2</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]    sin θ=0            θ= sin −1 0            θ=nπ where n is an integer

Substitute any one of the<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]values into the equation. We will use<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>0.</mn></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>r</mi><mo>=</mo><mn>2</mn><mi>sin</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] r=0

The points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mn>0</mn><mo stretchy="false">)</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><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mo>±</mo><mi>n</mi><mi>π</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.

To find the maximum value of the equation, look at the maximum value of the trigonometric function<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><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]which occurs when<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 ±2kπ resulting in<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] π 2 )=1. Substitute<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  for<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]

<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>r</mi><mo>=</mo><mn>2</mn><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] π 2 ) r=2(1) r=2
Analysis

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 2 ) will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See[link] and [link].

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>θ</mi></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>2</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"><mi>r</mi></annotation-xml></semantics>[/itex]
0 <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><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></annotation-xml></semantics>[/itex] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mn>0</mn></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>π</mi></mfrac></mrow></annotation-xml></semantics>[/itex] 6 <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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] π 6 )=1 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mn>1</mn></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>π</mi></mfrac></mrow></annotation-xml></semantics>[/itex] 3 <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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] π 3 )≈1.73 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>1.73</mn></mrow></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>π</mi></mfrac></mrow></annotation-xml></semantics>[/itex] 2 <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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] π 2 )=2 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mn>2</mn></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>π</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 <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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2π 3 )≈1.73 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>1.73</mn></mrow></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><mi>π</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 6 <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><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5π 6 )=1 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mn>1</mn></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>π</mi></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>2</mn><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] π )=0 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mn>0</mn></annotation-xml></semantics>[/itex]
<figure id="Figure_08_04_004"></figure>

Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values 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] r |: <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><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Tests will reveal symmetry about the polar axis. The zero is<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, π 2 ), and the maximum value is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>3</mn><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

## Investigating Circles

Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.

There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.

Formulas for the Equation of a Circle

Some of the formulas that produce the graph of a circle in polar coordinates are given by<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>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>r</mi><mo>=</mo><mi>a</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<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]is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is<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] | a | 2 , or one-half the diameter. 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><mi>a</mi><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>, </mo></mrow></annotation-xml></semantics>[/itex] the center is<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] a 2 ,0 ). 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><mi>a</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] the center is<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] a 2 ,π ). [link] shows the graphs of these four circles.

<figure id="Figure_08_04_005"></figure>
Sketching the Graph of a Polar Equation for a Circle

Sketch 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><mn>4</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum<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 | 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><mn>4</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]First, set<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></mrow></annotation-xml></semantics>[/itex]and solve for<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]. Thus, a zero occurs at<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 ±kπ. A key point to plot is<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,​​ π 2 ) .

To find the maximum value 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></mrow></annotation-xml></semantics>[/itex] note that the maximum value of the cosine function is 1 when<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><mn>0</mn><mo>±</mo><mn>2</mn><mi>k</mi><mi>π</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Substitute<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><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the equation:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><mi>r</mi><mo>=</mo><mn>4</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]      r=4cos(0)        r=4(1)=4

The maximum value of the equation is 4. A key point to plot is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>4</mn><mo>,</mo><mtext> </mtext><mn>0</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

As<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>4</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] is symmetric with respect to the polar axis, we only need to calculate r-values for<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]over the interval<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">[</mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>π</mi><mo stretchy="false">]</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to [link]. The graph is shown in [link].

 θ[/itex] 0 π[/itex] 6 π[/itex] 4 π[/itex] 3 π[/itex] 2 2π[/itex] 3 3π[/itex] 4 5π[/itex] 6 π[/itex] r[/itex] 4 3.46 2.83 2 0 −2 −2.83 −3.46 4
<figure class="small" id="Figure_08_04_006"></figure>

## Investigating Cardioids

While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own.

Formulas for a Cardioid

The formulas that produce the graphs of a cardioid are given by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>b</mi><mo>></mo><mn>0</mn><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><mfrac/></mrow></annotation-xml></semantics>[/itex] a b =1. The cardioid graph passes through the pole, as we can see in [link].

<figure id="Figure_08_04_007"></figure>

Given the polar equation of a cardioid, sketch its graph.

1. Check equation for the three types of symmetry.
2. Find the zeros. Set<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></mrow></annotation-xml></semantics>[/itex]
3. Find the maximum value of the equation according to the maximum value of the trigonometric expression.
4. Make a table of values for<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></mrow></annotation-xml></semantics>[/itex]
5. Plot the points and sketch the graph.
Sketching the Graph of a Cardioid

Sketch 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><mn>2</mn><mo>+</mo><mn>2</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting<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></mrow></annotation-xml></semantics>[/itex]we have<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><mi>π</mi><mo>+</mo><mn>2</mn><mi>k</mi><mi>π</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The zero of the equation is located at<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,π ). The graph passes through this point.

The maximum value 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><mn>2</mn><mo>+</mo><mn>2</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]occurs when<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] is a maximum, which is when<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><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or when<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><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Substitute<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><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the equation, 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><mtext> </mtext></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>r</mi><mo>=</mo><mn>2</mn><mo>+</mo><mn>2</mn><mi>cos</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] r=2+2(1)=4

The point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>4</mn><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the maximum value on the graph.

We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">[</mo><mn>0</mn><mo>,</mo><mtext> </mtext><mi>π</mi><mo stretchy="false">]</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in [link], and then we plot the points and draw the graph. See [link].

 θ[/itex] 0[/itex] π[/itex] 4 π[/itex] 2 2π[/itex] 3 π[/itex] r[/itex] 4 3.41 2 1 0
<figure class="small" id="Figure_08_04_008"></figure>

## Investigating Limaçons

The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled limaçons when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1</mn><mo><</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b <2 and convex limaçons when<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] a b ≥2.

Formulas for One-Loop Limaçons

The formulas that produce the graph of a dimpled one-loop limaçon are given by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mi>b</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mtext>and 1<</mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] a b <2. All four graphs are shown in [link].

<figure id="Figure_08_04_009"> <figcaption>Dimpled limaçons</figcaption> </figure>

Given a polar equation for a one-loop limaçon, sketch the graph.

1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted.
2. Find the zeros.
3. Find the maximum values according to the trigonometric expression.
4. Make a table.
5. Plot the points and sketch the graph.
Sketching the Graph of a One-Loop Limaçon

Graph 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>4</mn><mo>−</mo><mn>3</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality.

Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting<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]results in<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]being undefined. What does this mean? How could<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]be undefined? 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><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is undefined for any value of<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><mn>1.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Therefore,<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 undefined because there is no value 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]for which<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><mn>1.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi></mrow></annotation-xml></semantics>[/itex] when<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><mn>0.</mn><mtext> </mtext></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><mi>r</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mo>−</mo><mn>3</mn><mi>sin</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]        r=4−3⋅0=4

So, there is at least one polar axis intercept at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>4</mn><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Next, as the maximum value of the sine function is 1 when <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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , we will substitute<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2   into the equation 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><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Thus,<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>1.</mn></mrow></annotation-xml></semantics>[/itex]

Make a table of the coordinates similar to [link].

 θ[/itex] 0[/itex] π[/itex] 6 π[/itex] 3 π[/itex] 2 2π[/itex] 3 5π[/itex] 6 π[/itex] 7π[/itex] 6 4π[/itex] 3 3π[/itex] 2 5π[/itex] 3 11π[/itex] 6 2π[/itex] r[/itex] 4 2.5 1.4 1 1.4 2.5 4 5.5 6.6 7 6.6 5.5 4

The graph is shown in [link].

<figure class="small" id="Figure_08_04_010"> <figcaption>One-loop limaçon</figcaption> </figure>
Analysis

This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving<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] is likely symmetric with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , evaluating more points helps to verify that the graph is correct.

Sketch 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><mn>3</mn><mo>−</mo><mn>2</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it.

Formulas for Inner-Loop Limaçons

The formulas that generate the inner-loop limaçons are given by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>b</mi><mo>></mo><mn>0</mn><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><mtext> </mtext><mi>a</mi><mo><</mo><mi>b</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See [link]for the graphs.

<figure id="Figure_08_04_012"></figure>
Sketching the Graph of an Inner-Loop Limaçon

Sketch 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><mn>2</mn><mo>+</mo><mn>5</mn><mtext>cos</mtext><mtext> </mtext><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when<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><mi>θ</mi><mo>=</mo><mn>1.98.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] The maximum<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 | is found when<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><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or when<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><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Thus, the maximum is found at the point (7, 0).

Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.

 θ[/itex] 0[/itex] π[/itex] 6 π[/itex] 3 π[/itex] 2 2π[/itex] 3 5π[/itex] 6 π[/itex] 7π[/itex] 6 4π[/itex] 3 3π[/itex] 2 5π[/itex] 3 11π[/itex] 6 2π[/itex] r[/itex] 7 6.3 4.5 2 −0.5 −2.3 −3 −2.3 −0.5 2 4.5 6.3 7

As expected, the values begin to repeat after<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><mi>π</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The graph is shown in [link].

<figure class="small" id="Figure_08_04_013"> <figcaption>Inner-loop limaçon</figcaption> </figure>

## Investigating Lemniscates

The lemniscate is a polar curve resembling the infinity symbol<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>∞</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.

Formulas for Lemniscates

The formulas that generate the graph of a lemniscate are given by<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] r 2 = a 2 cos 2θ and<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] r 2 = a 2 sin 2θ where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>≠</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The formula<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] r 2 = a 2 sin 2θ is symmetric with respect to the pole. The formula<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] r 2 = a 2 cos 2θ is symmetric with respect to the pole, the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , and the polar axis. See [link] for the graphs.

<figure id="Figure_08_04_014"></figure>
Sketching the Graph of a Lemniscate

Sketch the graph of<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] r 2 =4cos 2θ.

The equation exhibits symmetry with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , the polar axis, and the pole.

Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>u</mi><mo>=</mo><mn>2</mn><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><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><mn>0</mn><mo>=</mo><mn>4</mn><mi>cos</mi><mtext> </mtext><mn>2</mn><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]                0=4cos u                0=cos u cos −1 0= π 2                u= π 2 Substitute 2θ back in for u.             2θ= π 2               θ= π 4

So, 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] 0, π 4 )is a zero of the equation.

Now let’s find the maximum value. Since the maximum 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>u</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>u</mi><mo>=</mo><mn>0</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the maximum<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><mn>2</mn><mi>θ</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>θ</mi><mo>=</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><mtext> </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] r 2 =4cos(0)     r 2 =4(1)=4           r=± 4  =2

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to [link].

 θ[/itex] 0 π[/itex] 6 π[/itex] 4 π[/itex] 3 π[/itex] 2 r[/itex] 2 2[/itex] 0 2[/itex] 0

Plot the points on the graph, such as the one shown in [link].

<figure class="small" id="Figure_08_04_015"> <figcaption>Lemniscate</figcaption> </figure>
Analysis

Making a substitution such as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>u</mi><mo>=</mo><mn>2</mn><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.

Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points 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]This is because there are no real square roots for these values 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]In other words, the corresponding r-values of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msqrt/></mrow></annotation-xml></semantics>[/itex] 4cos(2θ)   are complex numbers because there is a negative number under the radical.

## Investigating Rose Curves

The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.

Rose Curves

The formulas that generate the graph of a rose curve are given by<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>cos</mi><mtext> </mtext><mi>n</mi><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><mi>a</mi><mi>sin</mi><mtext> </mtext><mi>n</mi><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>≠</mo><mn>0.</mn><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>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is even, the curve has<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]petals. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is odd, the curve has<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]petals. See [link].

<figure class="medium" id="Figure_08_04_016"></figure>
Sketching the Graph of a Rose Curve (n Even)

Sketch 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><mn>2</mn><mi>cos</mi><mtext> </mtext><mn>4</mn><mi>θ</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  and the pole.

Now we will find the zeros. First make the substitution<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>u</mi><mo>=</mo><mn>4</mn><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><mtr><mtd><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><mn>0</mn><mo>=</mo><mn>2</mn><mi>cos</mi><mtext> </mtext><mn>4</mn><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]                            0=cos 4θ                         0=cos u cos −1 0=u                  u= π 2               4θ= π 2                   θ= π8

The zero is<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 8 . 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] 0, π 8 ) is on the curve.

Next, we find the maximum<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 |. We know that the maximum value 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>u</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]when<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><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>r</mi><mo>=</mo><mn>2</mn><mi>cos</mi><mo stretchy="false">(</mo><mn>4</mn><mo>⋅</mo><mn>0</mn><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mtd></mtr></mtable></annotation-xml></semantics>[/itex] r=2cos(0) r=2(1)=2

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] 2,0 ) is on the curve.

The graph of the rose curve has unique properties, which are revealed in [link].

 θ[/itex] 0 π[/itex] 8 π[/itex] 4 3π[/itex] 8 π[/itex] 2 5π[/itex] 8 3π[/itex] 4 r[/itex] 2 0 −2 0 2 0 −2

As<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]when<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 8 , it makes sense to divide values in the table by<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 8  units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at<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></mrow></annotation-xml></semantics>[/itex]each petal extends out 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><mo>=</mo><mn>2</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and then turns back to zero<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]times for a total of eight petals. See the graph in[link].

<figure class="small" id="Figure_08_04_017"> <figcaption>Rose curve,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]even</figcaption> </figure>
Analysis

When these curves are drawn, it is best to plot the points in order, as in the [link]. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.

Sketch 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><mn>4</mn><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2θ ).

The graph is a rose curve,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]even

Sketching the Graph of a Rose Curve (n Odd)

Sketch 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><mn>2</mn><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5θ ).

The graph of the equation shows symmetry with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 . Next, find the zeros and maximum. We will want to make the substitution<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>u</mi><mo>=</mo><mn>5</mn><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><mtr><mtd><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><mn>0</mn><mo>=</mo><mn>2</mn><mi>sin</mi><mo stretchy="false">(</mo><mn>5</mn><mi>θ</mi><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]                       0=sin u sin −1 0=0               u=0            5θ=0               θ=0

The maximum value is calculated at the angle where<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] is a maximum. Therefore,

<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><mn>2</mn><mi>sin</mi><mrow><mo>(</mo></mrow></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 5⋅ π 2 ) r=2(1)=2

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]odd yields the same number of petals as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]there will be five petals on the graph. See [link].

Create a table of values similar to [link].

 θ[/itex] 0 π[/itex] 6 π[/itex] 3 π[/itex] 2 2π[/itex] 3 5π[/itex] 6 π[/itex] r[/itex] 0 1 −1.73 2 −1.73 1 0
<figure class="small" id="Figure_08_04_019"> <figcaption>Rose curve,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]odd</figcaption> </figure>

Sketch the graph of<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><mo stretchy="false">(</mo><mn>3</mn><mi>θ</mi><mo stretchy="false">).</mo></mrow></annotation-xml></semantics>[/itex]

Rose curve,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]odd

## Investigating the Archimedes’ Spiral

The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.

Archimedes’ Spiral

The formula that generates the graph of the Archimedes’ spiral is given by<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>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] for<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><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]As<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]increases,<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] increases at a constant rate in an ever-widening, never-ending, spiraling path. See [link].

<figure class="medium" id="Figure_08_04_020"></figure>

Given an Archimedes’ spiral over<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] 0,2π ],sketch the graph.

1. Make a table of values for<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]over the given domain.
2. Plot the points and sketch the graph.
Sketching the Graph of an Archimedes’ Spiral

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

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 equal to<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]the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.

Create a table such as [link].

 θ[/itex] π[/itex] 4 π[/itex] 2 π[/itex] 3π[/itex] 2 7π[/itex] 4 2π[/itex] r[/itex] 0.785 1.57 3.14 4.71 5.50 6.28

Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in [link].

<figure class="small" id="Figure_08_04_021"> <figcaption>Archimedes’ spiral</figcaption> </figure>
Analysis

The domain of this polar curve is<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] 0,2π ]. In general, however, the domain of this function is<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] −∞,∞ ). Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex.

Sketch 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><mo>−</mo><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]over the interval<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] 0,4π ].

# Summary of Curves

We have explored a number of seemingly complex polar curves in this section. [link] and [link] summarize the graphs and equations for each of these curves.

<figure id="Figure_08_04_023"></figure> <figure id="Figure_08_04_024"></figure>

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

# Key Concepts

• It is easier to graph polar equations if we can test the equations for symmetry with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 , the polar axis, or the pole.
• There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See [link].
• Polar equations may be graphed by making a table of values for<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]
• The maximum value of a polar equation is found by substituting the value<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]that leads to the maximum value of the trigonometric expression.
• The zeros of a polar equation are found by setting<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 solving for<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].
• Some formulas that produce the graph of a circle in polar coordinates are given by<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>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>r</mi><mo>=</mo><mi>a</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]See [link].
• The formulas that produce the graphs of a cardioid are given by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>b</mi><mo>></mo><mn>0</mn><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><mfrac/></mrow></annotation-xml></semantics>[/itex] a b =1. See [link].
• The formulas that produce the graphs of a one-loop limaçon are given by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1</mn><mo><</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b <2. See[link].
• The formulas that produce the graphs of an inner-loop limaçon are given by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>b</mi><mo>></mo><mn>0</mn><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>a</mi><mo><</mo><mi>b</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] See [link].
• The formulas that produce the graphs of a lemniscates are given by<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] r 2 = a 2 cos 2θ and<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] r 2 = a 2 sin 2θ, where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>≠</mo><mn>0.</mn></mrow></annotation-xml></semantics>[/itex]See [link].
• The formulas that produce the graphs of rose curves are given by<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>cos</mi><mtext> </mtext><mi>n</mi><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><mi>a</mi><mi>sin</mi><mtext> </mtext><mi>n</mi><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>≠</mo><mn>0</mn><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>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is even, there are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]petals, and if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is odd, there are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]petals. See [link] and [link].
• The formula that produces the graph of an Archimedes’ spiral is given by<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>θ</mi><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>θ</mi><mo>≥</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]See [link].

# Section Exercises

## Verbal

Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane.

Symmetry with respect to the polar axis is similar to symmetry about the<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi></mrow></annotation-xml></semantics>[/itex]-axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line<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><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2  is similar to symmetry about the<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi></mrow></annotation-xml></semantics>[/itex]-axis.

Which of the three types of symmetries for polar graphs correspond to the symmetries with respect to the x-axis, y-axis, and origin?

What are the steps to follow when graphing polar equations?

Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at <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><mn>0</mn><mo>,</mo><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] π 2 ,  π  and  3π 2 , and sketch the graph.

Describe the shapes of the graphs of cardioids, limaçons, and lemniscates.

What part of the equation determines the shape of the graph of a polar equation?

The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.

## Graphical

For the following exercises, test the equation for symmetry.

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

symmetric with respect to the polar axis

<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><mo>+</mo><mn>2</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>3</mn><mi>sin</mi><mtext> </mtext><mn>2</mn><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

symmetric with respect to the polar axis, symmetric with respect to the line <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 2 , symmetric with respect to the pole

<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><mi>r</mi><mo>=</mo><mn>2</mn><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

no symmetry

<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><mfrac/></mrow></annotation-xml></semantics>[/itex] θ 2

<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 θ

no symmetry

<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><msqrt/></mrow></annotation-xml></semantics>[/itex] 1− cos 2 θ

<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] 5sin 2θ

symmetric with respect to the pole

For the following exercises, graph the polar equation. Identify the name of the shape.

<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]

<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>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

circle

<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><mo>+</mo><mn>2</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><mi>r</mi><mo>=</mo><mn>2</mn><mo>−</mo><mn>2</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

cardioid

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

cardioid

<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><mo>+</mo><mn>2</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>7</mn><mo>+</mo><mn>4</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

one-loop/dimpled limaçon

<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><mo>+</mo><mn>3</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><mi>r</mi><mo>=</mo><mn>5</mn><mo>+</mo><mn>4</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

one-loop/dimpled limaçon

<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><mo>+</mo><mn>9</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><mi>r</mi><mo>=</mo><mn>1</mn><mo>+</mo><mn>3</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

inner loop/two-loop limaçon

<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><mo>+</mo><mn>5</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>5</mn><mo>+</mo><mn>7</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

inner loop/two-loop limaçon

<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><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><mi>r</mi><mo>=</mo><mn>5</mn><mo>+</mo><mn>6</mn><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

inner loop/two-loop limaçon

<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 =36cos( 2θ )

<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 =10cos( 2θ )

lemniscate

<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 =4sin( 2θ )

<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 =10sin( 2θ )

lemniscate

<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><mtext>sin</mtext><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo></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><mtext>cos</mtext><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

rose curve

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>5</mn><mtext>sin</mtext><mo stretchy="false">(</mo><mn>3</mn><mi>θ</mi><mo stretchy="false">)</mo></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><mtext>sin</mtext><mo stretchy="false">(</mo><mn>4</mn><mi>θ</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

rose curve

<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><mtext>sin</mtext><mo stretchy="false">(</mo><mn>5</mn><mi>θ</mi><mo stretchy="false">)</mo></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>−</mi><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

Archimedes’ spiral

<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>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><mo>−</mo><mn>3</mn><mi>θ</mi></mrow></annotation-xml></semantics>[/itex]

Archimedes’ spiral

## Technology

For the following exercises, use a graphing calculator to sketch the graph of the polar equation.

<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 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] 1 θ

<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><mtext> </mtext><mi>θ</mi><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] a cissoid

<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><msqrt/></mrow></annotation-xml></semantics>[/itex] 1− sin 2 θ, a hippopede

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mn>5</mn><mo>+</mo><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4θ )

<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><mo>−</mo><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2θ )

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

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

For the following exercises, use a graphing utility to graph each pair of polar equations on a domain 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] 0,4π ]  and then explain the differences shown in the graphs.

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

They are both spirals, but not quite the same.

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

Both graphs are curves with 2 loops. The equation with a coefficient 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] has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>4</mn><mi>π</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]to get a better picture.

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

On a graphing utility, graph<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] 16 5 θ ) on<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] 0,4π ],[ 0,8π ],[ 0,12π ], 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] 0,16π ]. Describe the effect of increasing the width of the domain.

When the width of the domain is increased, more petals of the flower are visible.

On a graphing utility, graph and sketch<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><mtext> </mtext><mi>θ</mi><mo>+</mo><msup/></mrow></annotation-xml></semantics>[/itex] ( sin( 5 2 θ ) ) 3  on<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] 0,4π ].

On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs.

<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><msub><mi>r</mi></msub></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 =3sin(3θ) r 2 =2sin(3θ) r 3 =sin(3θ)

The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.

On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs.

<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><msub><mi>r</mi></msub></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 =3+3cos θ r 2 =2+2cos θ r 3 =1+cos θ

On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs.

<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><msub><mi>r</mi></msub></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 =3θ r 2 =2θ r 3 =θ

The graphs are spirals. The smaller the coefficient, the tighter the spiral.

## Extensions

For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 =3+2sin θ,  r 2 =2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 =6−4cos θ,  r 2 =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 ),( 4, 5π 3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 =1+sin θ,  r 2 =3sin θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 =1+cos θ,  r 2 =3cos θ

<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 2 , π 3 ),( 3 2 , 5π 3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 =cos( 2θ ),  r 2 =sin( 2θ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 = sin 2 ( 2θ ),  r 2 =1−cos( 4θ )

<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] 0, π 2 ), ( 0,π ), ( 0, 3π 2 ), ( 0,2π )

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 2 =sin θ, r 2 2 =cos θ

<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 4 2 , π 4 ), ( 8 4 2 , 5π 4 )  and at<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><mfrac/></mrow></annotation-xml></semantics>[/itex] 3π 4 ,   7π 4    since<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 squared

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>r</mi></msub></mrow></annotation-xml></semantics>[/itex] 1 =1+cos θ,  r 2 =1−sin θ

## Glossary

Archimedes’ spiral
a polar curve given by<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>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]When multiplied by a constant, the equation appears as<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>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]As<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>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the curve continues to widen in a spiral path over the domain.
cardioid
a member of the limaçon family of curves, named for its resemblance to a heart; its equation is given as<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<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] a b =1
convex limaҫon
a type of one-loop limaçon represented by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]such 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] a b ≥2
dimpled limaҫon
a type of one-loop limaçon represented by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1</mn><mo><</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b <2
inner-loop limaçon
a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi> </mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo><</mo><mi>b</mi></mrow></annotation-xml></semantics>[/itex]
lemniscate
a polar curve resembling a figure 8 and given by the 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] r 2 = a 2 cos 2θ and<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] r 2 = a 2 sin 2θ,  a≠0
one-loop limaҫon
a polar curve represented by<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><mo>±</mo><mi>b</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>r</mi><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] such that <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>a</mi><mo>></mo><mn>0</mn><mo>,</mo><mi>b</mi><mo>></mo><mn>0</mn><mo>,</mo></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><mfrac/></mrow></annotation-xml></semantics>[/itex] a b >1; may be dimpled or convex; does not pass through the pole
polar equation
an equation describing a curve on the polar grid.
rose curve
a polar equation resembling a flower, given by the equations<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>cos</mi><mtext> </mtext><mi>n</mi><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><mi>a</mi><mi>sin</mi><mtext> </mtext><mi>n</mi><mi>θ</mi><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]when <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] is even there are <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]petals, and the curve is highly symmetrical; when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is odd there are <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>n</mi></mrow></annotation-xml></semantics>[/itex] petals.