Polar Coordinates

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 10.3 – Polar Coordinates

Expected Educational Results

• Objective 16–01: I can convert a point between polar coordinates and cartesian coordinates.

• Objective 16–02: I can convert equations between polar coordinates and cartesian coordinates.

• Objective 16–03: I can use technology to sketch a polar curve.

• Objective 16–04: I can find the derivative of a polar curve.

• Objective 16–05: I can use the derivative of a polar curve to find the equation of the tangent line to the curve, points where the curve has horizontal tangent lines, and points where the curve has vertical tangent lines.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Polar Coordinate System

Four Cartesian Points Represented in Polar Coordinates

Points in the Polar Coordinate System

Definition: Point in the Polar Coordinate System

A point in the polar coordinate system has the form $(r,\theta )$$\left(r,\theta\right)$, where $r$$r$ is the radial coordinate and $\theta$$\theta$ is the angular coordinate.

Definition: Radial Coordinate

The radial coordinate the distance the point is from the center of the polar coordinate system. The radial coordinate is represented by $r$$r$. $r$$r$ is the dependent variable in the polar coordinate system.

Definition: Angular Coordinate

The angular coordinate is the angle, in radians, of the point from the positive $x$$x$-axis in the counter-clockwise direction. The angular coordinate is represented by $\theta$$\theta$. $\theta$$\theta$ is the independent variable in the polar coordinate system.

Converting Between $2D$$2D$ Coordinate Systems

Convert from Cartesian Coordinates to Polar Coordinates

Consider the blue triangle in Figure 2. This blue triangle is formed with the cartesian points $(0,0)$$(0,0)$, $(-2,0)$$(-2,0)$, and $(-2,2\sqrt{3})$$\left(-2,2\sqrt{3}\right)$. To convert the cartesian point $(-2,2\sqrt{3})$$\left(-2,2\sqrt{3}\right)$, we need to find the values for the radial coordinate, $r$$r$, and angular coordinate, $\theta$$\theta$.

Finding the Radial Coordinate

The radial coordinate is the distance $(-2,2\sqrt{3})$$\left(-2,2\sqrt{3}\right)$ is from the origin $(0,0)$$(0,0)$ which the length along the hypotenuse of the blue right triangle.

What is the relationship among the cartesian coordinates, $x$$x$ and $y$$y$, and the radial coordinate, $r$$r$? This is the Pythagorean Theorem:

$r^2=x^2+y^2$$r^2=x^2+y^2$

So, $r^2=(-2{)}^2+(2\sqrt{3}{)}^2\Rightarrow r^2=4+12=16\Rightarrow r=\pm \sqrt{16}=\pm 4$$r^2=(-2)^2+(2\sqrt{3})^2\Rightarrow r^2=4+12=16\Rightarrow r=\pm\sqrt{16} =\pm 4$

Let's use $r=4$$r=4$. Note the point ${\displaystyle (-2,2\sqrt{3})}$$\displaystyle\left(-2,2\sqrt{3}\right)$ in Figure 2 above is on the circle with radius $4$$4$.

Finding the Angular Coordinate

What is the relationship among the cartesian coordinates, $x$$x$ and $y$$y$, and the angular coordinate, $\theta$$\theta$?

This is the trigonometric function, $\tan (\theta )$$\tan{(\theta)}$:

$\tan (\theta )={\displaystyle \frac{\text{length of the opposite side}}{\text{length of adjacent side}}}={\displaystyle \frac{y}{x}}$$\tan{(\theta)}=\dfrac{\text{length of the opposite side}}{\text{length of adjacent side}} = \dfrac{y}{x}$

So, $\tan (\theta )={\displaystyle \frac{y}{x}}\Rightarrow \tan (\theta )={\displaystyle \frac{2\sqrt{3}}{-2}}\Rightarrow \theta ={\tan }^{-1}\left({\displaystyle \frac{2\sqrt{3}}{-2}}\right)=-{\displaystyle \frac{\pi }{3}}$$\tan{(\theta)}=\dfrac{y}{x}\Rightarrow \tan{(\theta)}=\dfrac{2\sqrt{3}}{-2}\Rightarrow \theta = \tan^{-1}{\left(\dfrac{2\sqrt{3}}{-2}\right)}=-\dfrac{\pi}{3}$

The reference angle is ${\displaystyle {\theta }_b=\frac{\pi }{3}}$$\displaystyle \theta_b=\frac{\pi}{3}$; however, angles are always measured from the positive $x$$x$-axis. In Quadrant II, angles are ${\displaystyle \pi -{\theta }_{\text{reference}}}$$\displaystyle \pi-\theta_{\text{reference}}$. So, the angular coordinate is ${\displaystyle \theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}}$$\displaystyle \theta = \pi-\frac{\pi}{3}=\frac{2\pi}{3}$.

Therefore, the cartesian point ${\displaystyle (-2,2\sqrt{3})}$$\displaystyle\left(-2,2\sqrt{3}\right)$ is the same as the polar point ${\displaystyle (4,\frac{2\pi }{3})}$$\displaystyle\left(4,\frac{2\pi}{3}\right)$.

Formulas to Convert from Cartesian Coordinates to Polar Coordinates

• ${\displaystyle r^2=x^2+y^2}$$\displaystyle \color{red}r^2=x^2+y^2$

• ${\displaystyle \tan (\theta )=\frac{y}{x}}$$\displaystyle \color{red}\tan{(\theta)}=\frac{y}{x}$

Investigation 04

Convert the following to polar coordinates:

1. $(3,-3)$$(3,-3)$

2. $(1,-2)$$(1,-2)$

3. $(-1,\sqrt{3})$$(-1,\sqrt{3})$

4. $(-2,-2)$$(-2,-2)$

Convert from Polar Coordinates to Cartesian Coordinates

Consider the polar point ${\displaystyle (3,\frac{4\pi }{3})}$$\displaystyle\left(3,\frac{4\pi}{3}\right)$ shown in the Figure below:

Finding the $x$$x$-Coordinate

What is the relationship among the cartesian coordinate, $x$$x$ and the radial coordinate, $r$$r$, and angular coordinate $\theta$$\theta$? This is the definition of $\cos (\theta )$$\cos{(\theta)}$ on a right triangle:

$\cos (\theta )={\displaystyle \frac{\text{length of adjacent side }}{\text{length of hypotenuse}}}$$\cos{(\theta)}=\dfrac{\text{length of adjacent side }}{\text{length of hypotenuse}}$

So, ${\displaystyle \cos \left(\frac{4\pi }{3}\right)={\displaystyle \frac{x}{3}}\Rightarrow x=3\cos \left({\displaystyle \frac{4\pi }{3}}\right)\Rightarrow x=-{\displaystyle \frac{3}{2}}}$$\displaystyle\cos{\left(\frac{4\pi}{3}\right)}=\dfrac{x}{3}\Rightarrow x=3\cos{\left(\dfrac{4\pi}{3}\right)}\Rightarrow x=-\dfrac{3}{2}$

Finding the $y$$y$-Coordinate

What is the relationship among the cartesian coordinate, $y$$y$ and the radial coordinate, $r$$r$, and angular coordinate $\theta$$\theta$? This is the definition of $\sin (\theta )$$\sin{(\theta)}$ on a right triangle:

$\sin (\theta )={\displaystyle \frac{\text{length of opposite side }}{\text{length of hypotenuse}}}$$\sin{(\theta)}=\dfrac{\text{length of opposite side }}{\text{length of hypotenuse}}$

So, ${\displaystyle \sin \left(\frac{4\pi }{3}\right)=\frac{y}{3}\Rightarrow y=3\sin \left(\frac{4\pi }{3}\right)\Rightarrow y=-{\displaystyle \frac{3\sqrt{3}}{3}}}$$\displaystyle\sin{\left(\frac{4\pi}{3}\right)}=\frac{y}{3}\Rightarrow y=3\sin{\left(\frac{4\pi}{3}\right)}\Rightarrow y=-\dfrac{3\sqrt{3}}{3}$

Therefore, the polar point ${\displaystyle (3,\frac{4\pi }{3})}$$\displaystyle\left(3,\frac{4\pi}{3}\right)$ is the same as the cartesian point ${\displaystyle (-\frac{3}{2},-\frac{3\sqrt{3}}{3})}$$\displaystyle\left(-\frac{3}{2},-\frac{3\sqrt{3}}{3}\right)$.

Formulas to Convert from Polar Coordinates to Cartesian Coordinates

• ${\displaystyle x=r\cos (\theta )}$$\displaystyle \color{red}x=r\cos{(\theta)}$

• ${\displaystyle y=r\sin (\theta )}$$\displaystyle \color{red}y=r\sin{(\theta)}$

Investigation 05

Convert the following to cartesian (rectangular) coordinates:

1. ${\displaystyle (2,\frac{\pi }{6})}$$\displaystyle \left(2,\frac{\pi}{6}\right)$

2. ${\displaystyle (-3,\frac{5\pi }{3})}$$\displaystyle \left(-3,\frac{5\pi}{3}\right)$

3. $(1,30)$$(1,30)$

4. $(-3,\pi )$$(-3,\pi)$

5. $(2,0)$$(2,0)$

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Created: Wednesday, 20 October 2021 - 04:49 (EDT)