Lesson Objectives

- Learn about the polar coordinate system
- Learn how to convert between rectangular and polar coordinates
- Learn how to plot polar coordinates on a polar coordinate grid
- Learn how to find alternative forms for coordinates of a point

## How to Work with Polar Coordinates

Up to this point, we have used the rectangular coordinate system to graph points (ordered pairs) and equations. Recall that our rectangular coordinate system features two perpendicular axes, the x-axis, which is horizontal, and the y-axis which is vertical. When working with the rectangular coordinate system, each point in the plane (x, y) is specified by giving two numbers wrapped inside of parentheses and separated by a comma. This notation is referred to as an ordered pair and gives us the directed distances from the origin. As an example, suppose we wanted to plot the point (5, 6). This tells us from the origin, we want to move 5 units to the right and 6 units up. Let's consider another way to think about arriving at our point (5, 6). Let's begin by drawing a position vector with a terminal point at (5, 6). We can see from the image above, that our point (5, 6) can also be located by giving the directed angle (50.19°) from the positive x-axis to our position vector and the magnitude of our position vector ($\sqrt{61}$).

In other words, consider the following steps:

The ordered pair (r, θ) gives the polar coordinates of the point P. When r > 0, the point P will lie on the terminal side of θ.

When r < 0, the point P will lie on a ray pointing in the opposite direction of the terminal side of θ. The distance will be |r| from the pole.

In other words, if r is positive, we can think of this as walking forwards and if r is negative, we can think of this as walking backwards.

Example #1: Convert each pair of rectangular coordinates to polar coordinates and plot on the polar coordinate plane.

r > 0 and 0° ≤ θ < 360° $$\left(-\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}\right)$$ We need to find r and θ: $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^{2}+ \left(-\frac{3\sqrt{2}}{2}\right)^{2}}$$ $$r=\sqrt{\frac{9}{2}+ \frac{9}{2}}$$ $$r=\sqrt{\frac{18}{2}}$$ $$r=\sqrt{9}$$ $$r=3$$ To find θ, we can use our inverse tangent function.

Since both x and y coordinates are negative, we know our point lies in quadrant III. Let's find our reference angle. $$\text{tan}^{-1}(1)=45°$$ To find our angle in quadrant III, we add 180° to our reference angle of 45°: $$θ=180° + 45°=225°$$ Our coordinates in polar form: $$\left(3, 225°\right)$$ We can plot this point by finding the meeting point of our circle with radius 3 and the 225° angle. Example #2: Convert each pair of rectangular coordinates to polar coordinates and plot on the polar coordinate plane.

r > 0 and 0° ≤ θ < 360° $$(-2\sqrt{3}, 2)$$ We need to find r and θ: $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{(-2\sqrt{3})^{2}+ (2)^{2}}$$ $$r=\sqrt{12 + 4}$$ $$r=\sqrt{16}$$ $$r=4$$ To find θ, we can use our inverse tangent function.

Since x is negative and y is positive we know our point lies in quadrant II. Let's find our reference angle. $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ To find our angle in quadrant II, we subtract 180° minus our reference angle of 30°: $$θ=180° - 30°=150°$$ Our coordinates in polar form: $$\left(4, 150°\right)$$ We can plot this point by finding the meeting point of our circle with radius 4 and the 150° angle. In some cases, we will be given polar coordinates and be asked to find the rectangular coordinates of the point. Let's look at an example.

Example #3: Plot each point on the polar coordinate plane and find the rectangular coordinates of the point. $$(6, 75°)$$ First, let's plot this point on the polar coordinate plane. We can plot this point by finding the meeting point of our circle with radius 6 and the 75° angle. Now, let's convert our point over to rectangular coordinates. $$x=6 \cdot \text{cos}\hspace{.1em}75°$$ $$x=\frac{3\sqrt{6}- 3\sqrt{2}}{2}$$ $$y=6 \cdot \text{sin}\hspace{.1em}75°$$ $$y=\frac{3\sqrt{2}+ 3\sqrt{6}}{2}$$ Our rectangular coordinates are: $$\left(\frac{3\sqrt{6}- 3\sqrt{2}}{2}, \frac{3\sqrt{2}+ 3\sqrt{6}}{2}\right)$$ What happens when we are given a negative value for r? Let's look at an example.

Example #4: Plot each point in the polar coordinate system and give the rectangular coordinates. $$(-4, 120°)$$ Working with negative values for r can be tricky. First, let's plot (4, 120°): To work with an r-value of -4, we would want to travel in the opposite direction from the pole 4 units: Notice how this corresponds to the polar coordinates (4, 300°).

Now, let's convert out point over to rectangular coordinates. $$x=-4 \hspace{.1em}\text{cos}\hspace{.1em}120°$$ $$x=-4 \cdot -\frac{1}{2}=2$$ $$y=-4 \hspace{.1em}\text{sin}\hspace{.1em}120°$$ $$y=-4 \cdot \frac{\sqrt{3}}{2}=-2\sqrt{3}$$ Our rectangular coordinates are: $$(2, -2\sqrt{3})$$

(-4, 120°) and (4, 300°) locate the same point.

In general:

(r, θ) and (r, θ + 360°) locate the same point.

(r, θ) and (r, θ - 360°) locate the same point.

(r, θ) and (-r, θ + 180°) locate the same point.

(r, θ) and (-r, θ - 180°) locate the same point.

(r, θ) and (-r, θ + 180° + 360°) locate the same point.

(r, θ) and (-r, θ + 180° - 360°) locate the same point.

(r, θ) and (-r, θ - 180° + 360°) locate the same point.

(r, θ) and (-r, θ - 180° - 360°) locate the same point.

We can summarize this into a nice statement: $$(r, θ)=(r, θ \pm 360n°)$$ $$(r, θ)=(-r, θ \pm 180°(2n + 1))$$ Where n is any integer.

Let's look at an example.

Example #5: Find all pairs of polar coordinates that describe the same point. $$(3, 240°)$$ Following our format from above: $$(3, 240° \pm 360n°)$$ and $$(-3, 240 ° \pm 180°(2n + 1))$$ Let's see if we can simplify more. Let's begin with the positive case: $$240 ° + 180° \cdot 2n + 180°$$ $$240° + 360n° + 180°$$ $$420° + 360n°$$ 420° is coterminal with 60°: $$60° + 360n°$$ Let's think about the negative case: $$240° - 180° \cdot 2n - 180°$$ $$240° - 360n° - 180°$$ $$60° - 360n°$$ We can write this as: $$(-3, 60° \pm 360n°)$$ Since n can be any integer, we can write our full answer as: $$(3, 240° + 360n°)$$ $$\text{and}$$ $$(-3, 60° + 360n°)$$

In other words, consider the following steps:

- Face the positive x-axis
- Rotate 50.19° counterclockwise
- Move forward by $\sqrt{61}$ units
- We will arrive at our point (5, 6)

### Polar Coordinate System

Let's now consider the polar coordinate system, which is based on a fixed point O called the pole (or origin), and an initial ray, called the polar axis. The polar axis is generally drawn in the direction of the positive x-axis. To form our polar coordinate system in the plane, we will fix a point (O), again called the pole (or origin), and construct from O an initial ray in the direction of the positive x-axis, which will be known as the polar axis. Once this is done, each point P in the plane can be assigned polar coordinates (r, θ), where:- r is the directed distance from O to P (pole to the point)
- θ is the directed angle, measured counterclockwise from the polar axis to ray OP

### Converting Between Rectangular and Polar Coordinates

Let's now return to our rectangular coordinate plane and place the pole (O) at the origin. This will allow our polar axis to coincide with the positive x-axis. If a point has rectangular coordinates (x, y) and polar coordinates (r, θ): $$x=r \hspace{.1em}\text{cos}\hspace{.1em}θ$$ $$y=r \hspace{.1em}\text{sin}\hspace{.1em}θ$$ $$r=\sqrt{x^{2}+ y^{2}}$$ $$\text{tan}\hspace{.1em}θ=\frac{y}{x}, x ≠ 0$$ Note: When using arctan (the inverse tangent function), we can only obtain angles in quadrants I or IV (rotating clockwise). Keep this in mind while solving problems.The ordered pair (r, θ) gives the polar coordinates of the point P. When r > 0, the point P will lie on the terminal side of θ.

When r < 0, the point P will lie on a ray pointing in the opposite direction of the terminal side of θ. The distance will be |r| from the pole.

In other words, if r is positive, we can think of this as walking forwards and if r is negative, we can think of this as walking backwards.

### Creating a Polar Coordinate Plane

In order to work with polar coordinates, we often create a polar grid or plane. We start with a pole, which will be the origin. Then create the polar axis which is drawn in the direction of the positive x-axis. From here, we would make a series of circles with the center at the pole where the radius is 1, then 2, then 3, so on and so forth. Lastly, we can create a series of angles marked at intervals of 15°: Image Source: Wikimedia Commons### Plotting Points with Polar Coordinates

Let's look at a few examples where we are asked to plot our polar coordinates and determine the rectangular coordinates.Example #1: Convert each pair of rectangular coordinates to polar coordinates and plot on the polar coordinate plane.

r > 0 and 0° ≤ θ < 360° $$\left(-\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}\right)$$ We need to find r and θ: $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^{2}+ \left(-\frac{3\sqrt{2}}{2}\right)^{2}}$$ $$r=\sqrt{\frac{9}{2}+ \frac{9}{2}}$$ $$r=\sqrt{\frac{18}{2}}$$ $$r=\sqrt{9}$$ $$r=3$$ To find θ, we can use our inverse tangent function.

Since both x and y coordinates are negative, we know our point lies in quadrant III. Let's find our reference angle. $$\text{tan}^{-1}(1)=45°$$ To find our angle in quadrant III, we add 180° to our reference angle of 45°: $$θ=180° + 45°=225°$$ Our coordinates in polar form: $$\left(3, 225°\right)$$ We can plot this point by finding the meeting point of our circle with radius 3 and the 225° angle. Example #2: Convert each pair of rectangular coordinates to polar coordinates and plot on the polar coordinate plane.

r > 0 and 0° ≤ θ < 360° $$(-2\sqrt{3}, 2)$$ We need to find r and θ: $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{(-2\sqrt{3})^{2}+ (2)^{2}}$$ $$r=\sqrt{12 + 4}$$ $$r=\sqrt{16}$$ $$r=4$$ To find θ, we can use our inverse tangent function.

Since x is negative and y is positive we know our point lies in quadrant II. Let's find our reference angle. $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ To find our angle in quadrant II, we subtract 180° minus our reference angle of 30°: $$θ=180° - 30°=150°$$ Our coordinates in polar form: $$\left(4, 150°\right)$$ We can plot this point by finding the meeting point of our circle with radius 4 and the 150° angle. In some cases, we will be given polar coordinates and be asked to find the rectangular coordinates of the point. Let's look at an example.

Example #3: Plot each point on the polar coordinate plane and find the rectangular coordinates of the point. $$(6, 75°)$$ First, let's plot this point on the polar coordinate plane. We can plot this point by finding the meeting point of our circle with radius 6 and the 75° angle. Now, let's convert our point over to rectangular coordinates. $$x=6 \cdot \text{cos}\hspace{.1em}75°$$ $$x=\frac{3\sqrt{6}- 3\sqrt{2}}{2}$$ $$y=6 \cdot \text{sin}\hspace{.1em}75°$$ $$y=\frac{3\sqrt{2}+ 3\sqrt{6}}{2}$$ Our rectangular coordinates are: $$\left(\frac{3\sqrt{6}- 3\sqrt{2}}{2}, \frac{3\sqrt{2}+ 3\sqrt{6}}{2}\right)$$ What happens when we are given a negative value for r? Let's look at an example.

Example #4: Plot each point in the polar coordinate system and give the rectangular coordinates. $$(-4, 120°)$$ Working with negative values for r can be tricky. First, let's plot (4, 120°): To work with an r-value of -4, we would want to travel in the opposite direction from the pole 4 units: Notice how this corresponds to the polar coordinates (4, 300°).

Now, let's convert out point over to rectangular coordinates. $$x=-4 \hspace{.1em}\text{cos}\hspace{.1em}120°$$ $$x=-4 \cdot -\frac{1}{2}=2$$ $$y=-4 \hspace{.1em}\text{sin}\hspace{.1em}120°$$ $$y=-4 \cdot \frac{\sqrt{3}}{2}=-2\sqrt{3}$$ Our rectangular coordinates are: $$(2, -2\sqrt{3})$$

### Alternative Forms for Coordinates of a Point

A given point in the rectangular coordinate plane can have an infinite number of pairs of polar coordinates. We saw in our last example:(-4, 120°) and (4, 300°) locate the same point.

In general:

(r, θ) and (r, θ + 360°) locate the same point.

(r, θ) and (r, θ - 360°) locate the same point.

(r, θ) and (-r, θ + 180°) locate the same point.

(r, θ) and (-r, θ - 180°) locate the same point.

(r, θ) and (-r, θ + 180° + 360°) locate the same point.

(r, θ) and (-r, θ + 180° - 360°) locate the same point.

(r, θ) and (-r, θ - 180° + 360°) locate the same point.

(r, θ) and (-r, θ - 180° - 360°) locate the same point.

We can summarize this into a nice statement: $$(r, θ)=(r, θ \pm 360n°)$$ $$(r, θ)=(-r, θ \pm 180°(2n + 1))$$ Where n is any integer.

Let's look at an example.

Example #5: Find all pairs of polar coordinates that describe the same point. $$(3, 240°)$$ Following our format from above: $$(3, 240° \pm 360n°)$$ and $$(-3, 240 ° \pm 180°(2n + 1))$$ Let's see if we can simplify more. Let's begin with the positive case: $$240 ° + 180° \cdot 2n + 180°$$ $$240° + 360n° + 180°$$ $$420° + 360n°$$ 420° is coterminal with 60°: $$60° + 360n°$$ Let's think about the negative case: $$240° - 180° \cdot 2n - 180°$$ $$240° - 360n° - 180°$$ $$60° - 360n°$$ We can write this as: $$(-3, 60° \pm 360n°)$$ Since n can be any integer, we can write our full answer as: $$(3, 240° + 360n°)$$ $$\text{and}$$ $$(-3, 60° + 360n°)$$

#### Skills Check:

Example #1

Convert to rectangular coordinates. $$(3, 315°)$$

Please choose the best answer.

A

$$\left(0, -4\right)$$

B

$$\left(-1, -3\right)$$

C

$$\left(-2\sqrt{2}, -2\sqrt{2}\right)$$

D

$$\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

E

$$\left(\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}\right)$$

Example #2

Convert to polar coordinates $$r > 0$$ $$0 ≤ θ < 360°$$ $$(-2\sqrt{2}, 2\sqrt{2})$$

Please choose the best answer.

A

$$(4, 135°)$$

B

$$(3, 0°)$$

C

$$(3, 150°)$$

D

$$(4\sqrt{2}, 135°)$$

E

$$(3\sqrt{5}, 150°)$$

Example #3

Find the alternative polar coordinates that describe the same point. $$(4, 150°)$$

Please choose the best answer.

A

$$(4, 330°)$$

B

$$(-4, -30°)$$

C

$$(-4, 510°)$$

D

$$(4, -30°)$$

E

$$(8, 300°)$$

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