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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 instantaneous position. This is one application of polar coordinates    , represented as ( r , θ ) . We interpret r as the distance from the sun and θ 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.

Illustration of the solar system with the sun at the center and orbits of the planets Mercury, Venus, Earth, and Mars shown.
Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA/JPL-Caltech)

Testing polar equations for symmetry

Just as a rectangular equation such as y = x 2 describes the relationship between x and y on a Cartesian grid, a polar equation describes a relationship between r and θ on a polar grid. Recall that the coordinate pair ( r , θ ) indicates that we move counterclockwise from the polar axis (positive x -axis) by an angle of θ , and extend a ray from the pole (origin) r units in the direction of θ . 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 r ) to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line θ = π 2 ( y -axis). We replace ( r , θ ) with ( r , θ ) to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation r = 2 sin θ ;

r = 2 sin θ r = 2 sin ( θ ) Replace ( r , θ ) with  ( r , θ ) . r = −2 sin θ Identity:  sin ( θ ) = sin θ . r = 2 sin θ Multiply both sides by −1.

This equation exhibits symmetry with respect to the line θ = π 2 .

In the second test, we consider symmetry with respect to the polar axis ( x -axis). We replace ( r , θ ) with ( r , θ ) or ( r , π θ ) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation r = 1 2 cos θ .

r = 1 2 cos θ r = 1 2 cos ( θ ) Replace  ( r , θ ) with ( r , θ ) . r = 1 2 cos 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 ( r , θ ) with ( r , θ ) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation r = 2 sin ( 3 θ ).

r = 2 sin ( 3 θ ) r = 2 sin ( 3 θ )

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 θ = 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.

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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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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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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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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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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