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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] .

3 graphs side by side. (A) shows a ray extending into Q 1 and its symmetric version in Q 2. (B) shows a ray extending into Q 1 and its symmetric version in Q 4. (C) shows a ray extending into Q 1 and its symmetric version in Q 3. See caption for more information.
(a) A graph is symmetric with respect to the line θ = π 2 ( y -axis) if replacing ( r , θ ) with ( r , θ ) yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis ( x -axis) if replacing ( r , θ ) with ( r , θ ) or ( r , π− θ ) yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing ( r , θ ) with ( r , θ ) yields an equivalent equation.

Given a polar equation, test for symmetry.

  1. Substitute the appropriate combination of components for ( r , θ ) : ( r , θ ) for θ = π 2 symmetry; ( r , θ ) for polar axis symmetry; and ( 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 r = 2 sin θ for symmetry.

Test for each of the three types of symmetry.

1) Replacing ( r , θ ) with ( r , θ ) yields the same result. Thus, the graph is symmetric with respect to the line θ = π 2 . r = 2 sin ( θ ) r = −2 sin θ Even-odd identity r = 2 sin θ Multiply by −1 Passed
2) Replacing θ with θ 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 = 2 sin ( θ ) r = −2 sin θ Even-odd identity r = −2 sin θ 2 sin θ Failed
3) Replacing r with r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. r = 2 sin θ    r = −2 sin θ 2 sin θ Failed
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Test the equation for symmetry: r = 2 cos θ .

The equation fails the symmetry test with respect to the line θ = π 2 and with respect to the pole. It passes the polar axis symmetry test.

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Graphing polar equations by plotting points

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

Finding zeros and maxima

To find the zeros of a polar equation, we solve for the values of θ that result in r = 0. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for x . We use the same process for polar equations. Set r = 0 , and solve for θ .

For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of θ into the equation that result in the maximum value of the trigonometric functions. Consider r = 5 cos θ ; the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when θ = 0 , so our polar equation is 5 cos θ , and the value θ = 0 will yield the maximum | r | .

Similarly, the maximum value of the sine function is 1 when θ = π 2 , and if our polar equation is r = 5 sin θ , the value θ = π 2 will yield the maximum | r | . We may find additional information by calculating values of r when θ = 0. These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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