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7.3 Polar coordinates  (Page 5/16)

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Suppose a curve is described in the polar coordinate system via the function r = f ( θ ) . Since we have conversion formulas from polar to rectangular coordinates given by

x = r cos θ y = r sin θ ,

it is possible to rewrite these formulas using the function

x = f ( θ ) cos θ y = f ( θ ) sin θ .

This step gives a parameterization of the curve in rectangular coordinates using θ as the parameter. For example, the spiral formula r = a + b θ from [link] becomes

x = ( a + b θ ) cos θ y = ( a + b θ ) sin θ .

Letting θ range from − ∞ to ∞ generates the entire spiral.

Symmetry in polar coordinates

When studying symmetry of functions in rectangular coordinates (i.e., in the form y = f ( x ) ) , we talk about symmetry with respect to the y -axis and symmetry with respect to the origin. In particular, if f ( − x ) = f ( x ) for all x in the domain of f , then f is an even function and its graph is symmetric with respect to the y -axis. If f ( − x ) = − f ( x ) for all x in the domain of f , then f is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.

Symmetry in polar curves and equations

Consider a curve generated by the function r = f ( θ ) in polar coordinates.

  1. The curve is symmetric about the polar axis if for every point ( r , θ ) on the graph, the point ( r , − θ ) is also on the graph. Similarly, the equation r = f ( θ ) is unchanged by replacing θ with − θ .
  2. The curve is symmetric about the pole if for every point ( r , θ ) on the graph, the point ( r , π + θ ) is also on the graph. Similarly, the equation r = f ( θ ) is unchanged when replacing r with − r , or θ with π + θ .
  3. The curve is symmetric about the vertical line θ = π 2 if for every point ( r , θ ) on the graph, the point ( r , π − θ ) is also on the graph. Similarly, the equation r = f ( θ ) is unchanged when θ is replaced by π − θ .

The following table shows examples of each type of symmetry.

This table has three rows and two columns. The first row reads “Symmetry with respect to the polar axis: For every point (r, θ) on the graph, there is also a point reflected directly across the horizontal (polar) axis” and it has a picture of a cardioid with equation r = 2 – 2 cosθ: this cardioid has points marked (r, θ) and (r, −θ), which are symmetric about the x axis, and the entire cardioid is symmetric about the x axis. The second row reads “Symmetry with respect to the pole: For every point (r, θ) on the graph, there is also a point on the graph that is reflected through the pole as well” and it has a picture of a skewed infinity symbol with equation r2 = 9 cos(2θ – π/2): this figure has points marked (r, θ) and (−r, θ), which are symmetric about the pole, and the entire figure is symmetric about the pole. The third row reads “Symmetry with respect to the vertical line θ = π/2: For every point (r, θ) on the graph, there is also a point reflected directly across the vertical axis” and there is a picture of a cardioid with equation r = 2 – 2 sinθ: this figure has points marked (r, θ) and (r, π − θ), which are symmetric about the vertical line θ = π/2, and the entire cardioid is symmetric about the vertical line θ = π/2.

Using symmetry to graph a polar equation

Find the symmetry of the rose defined by the equation r = 3 sin ( 2 θ ) and create a graph.

Suppose the point ( r , θ ) is on the graph of r = 3 sin ( 2 θ ) .

  1. To test for symmetry about the polar axis, first try replacing θ with − θ . This gives r = 3 sin ( 2 ( − θ ) ) = −3 sin ( 2 θ ) . Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing r with − r and θ with π − θ yields
    − r = 3 sin ( 2 ( π − θ ) ) − r = 3 sin ( 2 π − 2 θ ) − r = 3 sin ( −2 θ ) − r = −3 sin 2 θ .

    Multiplying both sides of this equation by −1 gives r = 3 sin 2 θ , which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.
  2. To test for symmetry with respect to the pole, first replace r with − r , which yields − r = 3 sin ( 2 θ ) . Multiplying both sides by −1 gives r = −3 sin ( 2 θ ) , which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing θ with θ + π gives
    r = 3 sin ( 2 ( θ + π ) ) = 3 sin ( 2 θ + 2 π ) = 3 ( sin 2 θ cos 2 π + cos 2 θ sin 2 π ) = 3 sin 2 θ .

    Since this agrees with the original equation, the graph is symmetric about the pole.
  3. To test for symmetry with respect to the vertical line θ = π 2 , first replace both r with − r and θ with − θ .
    − r = 3 sin ( 2 ( − θ ) ) − r = 3 sin ( −2 θ ) − r = −3 sin 2 θ .

    Multiplying both sides of this equation by −1 gives r = 3 sin 2 θ , which is the original equation. Therefore the graph is symmetric about the vertical line θ = π 2 .

This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of θ between 0 and π / 2 and then reflect the resulting graph.

θ r
0 0
π 6 3 3 2 ≈ 2.6
π 4 3
π 3 3 3 2 ≈ 2.6
π 2 0

This gives one petal of the rose, as shown in the following graph.

A single petal is graphed with equation r = 3 sin(2θ) for 0 ≤ θ ≤ π/2. It starts at the origin and reaches a maximum distance from the origin of 3.
The graph of the equation between θ = 0 and θ = π / 2 .

Reflecting this image into the other three quadrants gives the entire graph as shown.

A four-petaled rose is graphed with equation r = 3 sin(2θ). Each petal starts at the origin and reaches a maximum distance from the origin of 3.
The entire graph of the equation is called a four-petaled rose.
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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