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12.2 Polar Coordinates

In Rectangular / Cartesian coordinates, a point is located as \((x, y)\).

Cartesian coordinate system showing points (2, 3) in the first quadrant and (-2, -1) in the third quadrant.

Polar Coordinates

In Polar, a point is located as \((r, \theta)\).

  • \(r\): displacement from origin to the point
  • \(\theta\): angle formed by a line through origin and the point with the positive x-axis

For example, in polar: \((2, \pi/3)\)

\(r = 2\), \(\theta = \pi/3\)

Polar plot of the point (2, pi/3) showing a distance of 2 and an angle of pi/3 from the positive x-axis.
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Negative Values in Polar Coordinates

Both \(r\) and \(\theta\) can be negative.

For example, \((-2, -\pi/3)\)

When \(r < 0\), go away from the quadrant that \(\theta\) puts us in.

Graph showing point (-2, -pi/3). The angle -pi/3 is in quadrant 4, but r=-2 moves the point to quadrant 2.

Equivalent Representations

Notice \((-2, -\pi/3)\) can be expressed as \((2, 2\pi/3)\) also.

Graph showing the point (2, 2pi/3) in the second quadrant, equivalent to the previous example.

Also \((-2, 5\pi/3)\)

Graph showing the point (-2, 5pi/3). The angle 5pi/3 is in quadrant 4, r=-2 reflects it to quadrant 2.
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Polar Coordinates: Multiple Representations

Point: \( (-1, -7\pi/6) \)

Find two other ways to express location in polar:

Polar graph showing point (-1, -7pi/6) located in the fourth quadrant via a dashed line extension.
  • \( (1, -\pi/6) \)
  • \( (-1, +5\pi/6) \) (positive)
  • \( (1, 11\pi/6) \)

Conversion: Polar → Cartesian

\( (r, \theta) \rightarrow (x, y) \)

Coordinate system diagram showing a point (r, theta) and its projection onto x and y axes.
A right-angled triangle with hypotenuse r, base x, height y, and angle theta.
\( \cos \theta = \frac{x}{r} \rightarrow \)
\( x = r \cos \theta \)
\( \sin \theta = \frac{y}{r} \rightarrow \)
\( y = r \sin \theta \)

Also, \( x^2 + y^2 = r^2 \)

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Example: Polar to Cartesian Conversion

In polar: \( (-2, \pi/3) \)

In Cartesian?

Graph of angle pi/3 with a dashed line extending through the origin to point (-2, pi/3) in Quadrant III.

In QIII, we need \( x < 0, y < 0 \)

\( x = r \cos \theta = (-2) \cos(\pi/3) = (-2)(\frac{1}{2}) = -1 \)

\( y = r \sin \theta = (-2) \sin(\pi/3) = (-2)(\frac{\sqrt{3}}{2}) = -\sqrt{3} \)

Check: signs are right

\( x^2 + y^2 = r^2 \)?

\( (-1)^2 + (-\sqrt{3})^2 = (-2)^2 \)?

\( 1 + 3 = 4 \) yes

In Cartesian, this point is \( (-1, -\sqrt{3}) \)

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Transforming Equations

Equations can be transformed too.

Example: Polar to Cartesian

Consider the polar equation: \( r = 3 \) (polar)

Polar is good with circles and circle-like shapes.

\( r = 3 \rightarrow \) all points with \( r = 3 \) and \( \theta = \text{real number} \)

A circle centered at the origin on a Cartesian plane with a radius of 3 units.

In Cartesian:

\[ r = 3 \]\[ r^2 = 9 \]

From \( x^2 + y^2 = r^2 \):

\[ x^2 + y^2 = 9 \]
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Example: Converting a Complex Polar Equation

\[ r = \frac{1}{2 \cos \theta + 4 \sin \theta} \]

Each point \( (r, \theta) \) maps to \( \left( \frac{1}{2 \cos \theta + 4 \sin \theta}, \theta \right) \).

Go to Cartesian:

\( x = r \cos \theta \)
\( y = r \sin \theta \)

\[ r (2 \cos \theta + 4 \sin \theta) = 1 \]\[ 2r \cos \theta + 4r \sin \theta = 1 \]

Substituting \( x \) and \( y \):

\[ 2x + 4y = 1 \]

This is a line (good in Cartesian, not so good in Polar).

Each point \( (x, y) \) can be written as:

\[ (x, \frac{1}{4} - \frac{1}{2}x) \]
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Conversion: Cartesian → Polar

Converting from Cartesian coordinates \( (x, y) \) to Polar coordinates \( (r, \theta) \).

We know:

\[ \begin{cases} x = r \cos \theta \\ y = r \sin \theta \end{cases} \implies \frac{y}{x} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta \]\[ x^2 + y^2 = r^2 \]

So, \( \tan \theta = \frac{y}{x} \) or:

\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]

\[ r = \sqrt{x^2 + y^2} \]

or

\[ r = -\sqrt{x^2 + y^2} \]

check quadrants!

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Example

Cartesian: \( (-1, -\sqrt{3}) \)

quadrant check: Q III

Let's start w/ \( r \):

\[ r^2 = x^2 + y^2 \]\[ r^2 = (-1)^2 + (-\sqrt{3})^2 = 4 \]\[ \text{so, } r = 2 \quad \text{or} \quad r = -2 \]

Then \( \theta \):

\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]\[ = \tan^{-1} \left( \frac{-\sqrt{3}}{-1} \right) = \tan^{-1} \left( \frac{-\sqrt{3}/2}{-1/2} \right) \]\[ = \frac{\pi}{3} \quad \text{or} \quad \frac{4\pi}{3} \]

now pair \( r = 2, -2 \) and \( \theta = \frac{\pi}{3}, \frac{4\pi}{3} \) correctly \( \rightarrow \) Q III

\[ (2, 4\pi/3), \quad (-2, \pi/3) \]

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Example: Converting Cartesian to Polar Coordinates

Convert the equation y = 1/x to polar coordinates.

Basic Relationships:

  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( r^2 = x^2 + y^2 \)

Starting with the Cartesian equation:

\[ y = \frac{1}{x} \]

Substitute the polar relationships for \( x \) and \( y \):

\[ r \sin \theta = \frac{1}{r \cos \theta} \]

Multiply both sides by \( r \cos \theta \):

\[ (r \sin \theta)(r \cos \theta) = 1 \]
\[ r^2 \sin \theta \cos \theta = 1 \]

or

\[ r^2 = \frac{1}{\sin \theta \cos \theta} = \frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} = \csc \theta \cdot \sec \theta \]
\[ r^2 = \csc \theta \sec \theta \]