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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

the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
how can are find the domain and range of a relations
austin Reply
A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
more than 6000
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
Thanks po.
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
What is domain
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply
i want to sure my answer of the exercise
meena Reply
what is the diameter of(x-2)²+(y-3)²=25
Den Reply
how to solve the Identity ?
Barcenas Reply
what type of identity
Confunction Identity
how to solve the sums
hello guys
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
Shakeena Reply
by how many trees did forest "A" have a greater number?
Practice Key Terms 9

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