# 10.3 Polar coordinates

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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.

## Plotting points using polar coordinates

When we think about plotting points in the plane, we usually think of rectangular coordinates $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\left(r,\theta \right)\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is the radius or length of the directed line segment from the pole. The angle $\text{\hspace{0.17em}}\theta ,$ measured in radians, indicates the direction of $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ We move counterclockwise from the polar axis by an angle of $\text{\hspace{0.17em}}\theta ,$ and measure a directed line segment the length of $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ in the direction of $\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Even though we measure $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ first and then $\text{\hspace{0.17em}}r,$ the polar point is written with the r -coordinate first. For example, to plot the point $\text{\hspace{0.17em}}\left(2,\frac{\pi }{4}\right),$ we would move $\text{\hspace{0.17em}}\frac{\pi }{4}\text{\hspace{0.17em}}$ units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in [link] .

## Plotting a point on the polar grid

Plot the point $\text{\hspace{0.17em}}\left(3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ on the polar grid.

The angle $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ direction, as shown in [link] .

Plot the point $\text{\hspace{0.17em}}\left(2,\text{\hspace{0.17em}}\frac{\pi }{3}\right)\text{\hspace{0.17em}}$ in the polar grid .

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

Plot the point $\text{\hspace{0.17em}}\left(-2,\text{\hspace{0.17em}}\frac{\pi }{6}\right)\text{\hspace{0.17em}}$ on the polar grid.

We know that $\text{\hspace{0.17em}}\frac{\pi }{6}\text{\hspace{0.17em}}$ is located in the first quadrant. However, $\text{\hspace{0.17em}}r=-2.\text{\hspace{0.17em}}$ We can approach plotting a point with a negative $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ in two ways:

1. Plot the point $\text{\hspace{0.17em}}\left(2,\frac{\pi }{6}\right)\text{\hspace{0.17em}}$ by moving $\text{\hspace{0.17em}}\frac{\pi }{6}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\frac{\pi }{6}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\left(2,\frac{\pi }{6}\right)\text{\hspace{0.17em}}$ shown in [link] (b).

Plot the points $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{6}\right)$ and $\text{\hspace{0.17em}}\left(2,\frac{9\pi }{4}\right)\text{\hspace{0.17em}}$ on the same polar grid.

## 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 $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}r,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta .$

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how to solve x²=2x+8 factorization?
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
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
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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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