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8.4 Polar coordinates: graphs  (Page 2/16)

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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.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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