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In this section, you will:
  • Plot points using polar coordinates.
  • Convert from polar coordinates to rectangular coordinates.
  • Convert from rectangular coordinates to polar coordinates.
  • Transform equations between polar and rectangular forms.
  • Identify and graph polar equations by converting to rectangular equations.

Over 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind (see [link] ). How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid.

An illustration of a boat on the polar grid.

Plotting points using polar coordinates

When we think about plotting points in the plane, we usually think of rectangular coordinates ( x , y ) in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates    , which are points labeled ( r , θ ) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole    , or the origin of the coordinate plane.

The polar grid is scaled as the unit circle with the positive x- axis now viewed as the polar axis    and the origin as the pole. The first coordinate r is the radius or length of the directed line segment from the pole. The angle θ , measured in radians, indicates the direction of r . We move counterclockwise from the polar axis by an angle of θ , and measure a directed line segment the length of r in the direction of θ . Even though we measure θ first and then r , the polar point is written with the r -coordinate first. For example, to plot the point ( 2 , π 4 ) , we would move π 4 units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in [link] .

Polar grid with point (2, pi/4) plotted.

Plotting a point on the polar grid

Plot the point ( 3 , π 2 ) on the polar grid.

The angle π 2 is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the π 2 direction, as shown in [link] .

Polar grid with point (3, pi/2) plotted.
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Plot the point ( 2 , π 3 ) in the polar grid .

Polar grid with point (2, pi/3) plotted.

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Plotting a point in the polar coordinate system with a negative component

Plot the point ( 2 , π 6 ) on the polar grid.

We know that π 6 is located in the first quadrant. However, r = −2. We can approach plotting a point with a negative r in two ways:

  1. Plot the point ( 2 , π 6 ) by moving π 6 in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;
  2. Move π 6 in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.

See [link] (a). Compare this to the graph of the polar coordinate ( 2 , π 6 ) shown in [link] (b).

Two polar grids. Points (2, pi/6) and (-2, pi/6) are plotted. They are reflections across the origin in Q1 and Q3.
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Plot the points ( 3 , π 6 ) and ( 2 , 9 π 4 ) on the same polar grid.

Points (2, 9pi/4) and (3, -pi/6) are plotted in the polar grid.
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Converting from polar coordinates to rectangular coordinates

When given a set of polar coordinates    , we may need to convert them to rectangular coordinates . To do so, we can recall the relationships that exist among the variables x , y , r , and θ .

Questions & Answers

sebd me some questions about anything ill solve for yall
Manifoldee Reply
how to solve x²=2x+8 factorization?
Kristof Reply
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
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Prashant Reply
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Amit
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Dorbor
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Biswajit
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Gaurav
Find the possible value of 8.5 using moivre's theorem
Reuben Reply
which of these functions is not uniformly cintinuous on (0, 1)? sinx
Pooja Reply
which of these functions is not uniformly continuous on 0,1
Basant Reply
solve this equation by completing the square 3x-4x-7=0
Jamiz Reply
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
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Modress
-x=7
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siame
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deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
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solve x
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you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
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how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
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What is Indices
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If one side only of a triangle is given is it possible to solve for the unkown two sides?
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Practice Key Terms 3

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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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