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

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}$

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)}$

Converting Equations

Converting Cartesian Equations into Polar Equations

Converting equations between the two coordinate systems is more important than converting points.

Example 01: Convert the cartesian equation $y=3+x^2$$y=3+x^2$ into a polar equation.

Solution:

Use the formulas: $x=r\cos (\theta )$$x=r\cos{(\theta)}$ and $y=r\sin (\theta )$$y=r\sin{(\theta)}$

Substitute the $x$$x$- and $y$$y$-variables:

$r\sin (\theta )=3+(r\cos (\theta ){)}^2$$r\sin{(\theta)}=3+(r\cos{(\theta)})^2$

This is an acceptable form for a polar equation. If you want you can solve for the dependent variable, $r$$r$; you will need to complete the square to solve for $r$$r$.

Investigation 06

Convert the following to polar coordinates:

NOTE: You may need to use some trigonometric identities.

1. $x=3$$x=3$

2. $y=-2$$y=-2$

3. $y=2x$$y=2x$

4. $xy=1$$xy=1$

5. $y=4x^2$$y=4x^2$

6. $x^2-y^2=1$$x^2-y^2=1$

Converting Polar Equations into Cartesian Equations

Example 02: Convert the polar equation $1=3\cos (2\theta )$$1=3\cos{(2\theta)}$ into a cartesian equation.

Solution: Use the formulas: $x^2+y^2=r^2$$x^2+y^2=r^2$, $\tan (\theta )={\displaystyle \frac{y}{x}}$$\tan{(\theta)}=\dfrac{y}{x}$, $x=r\cos (\theta )$$x=r\cos{(\theta)}$, and $y=r\sin (\theta )$$y=r\sin{(\theta)}$.

There is no conversion formula for $\cos (2\theta )$$\cos{(2\theta)}$. From trigonometry, we know $\cos (2x)={\cos }^2(x)-{\sin }^2(x)$$\cos{(2x)}=\cos^2{(x)}-\sin^2{(x)}$. From the conversion formulas, we know $\cos (x)={\displaystyle \frac{x}{r}}$$\cos{(x)}=\dfrac{x}{r}$ and $\sin (x)={\displaystyle \frac{y}{r}}$$\sin{(x)}=\dfrac{y}{r}$.

So, $1=3\cos (2\theta )\Rightarrow 1=3({\cos }^2(x)-{\sin }^2(x))\Rightarrow 1=3({\left({\displaystyle \frac{x}{r}}\right)}^2-{\left({\displaystyle \frac{y}{r}}\right)}^2)$$1=3\cos{(2\theta)}\Rightarrow 1=3\left(\cos^2{(x)}-\sin^2{(x)}\right)\Rightarrow 1=3\left(\left(\dfrac{x}{r}\right)^2-\left(\dfrac{y}{r}\right)^2\right)$

We are not done because a cartesian equation can only have $x$$x$- and $y$$y$-variables. We know $r^2=x^2+y^2$$r^2=x^2+y^2$. So,

$1=3({\left({\displaystyle \frac{x}{r}}\right)}^2-{\left({\displaystyle \frac{y}{r}}\right)}^2)\Rightarrow 1=3({\displaystyle \frac{x^2}{r^2}}-{\displaystyle \frac{y^2}{r^2}})\Rightarrow 1=3({\displaystyle \frac{x^2}{x^2+y^2}}-{\displaystyle \frac{y^2}{x^2+y^2}})$$1=3\left(\left(\dfrac{x}{r}\right)^2-\left(\dfrac{y}{r}\right)^2\right)\Rightarrow 1=3\left(\dfrac{x^2}{r^2}-\dfrac{y^2}{r^2}\right)\Rightarrow 1=3\left(\dfrac{x^2}{x^2+y^2}-\dfrac{y^2}{x^2+y^2}\right)$

This is an acceptable form for a cartesian equation. If you want you can solve for the dependent variable, $y$$y$; you will need to write $y$$y$ as a piecewise-defined function to solve for $y$$y$.

Investigation 07

Convert the following to cartesian coordinates:

NOTE: You may need to use some trigonometric identities.

1. $r=3$$r=3$

2. $\theta =\frac{\pi }{3}$$\theta=\frac{\pi}{3}$

3. $r=\sin (2\theta )$$r=\sin{(2\theta)}$

4. $r={\cos }^2(\theta )$$r=\cos^2{(\theta)}$

5. $r^2=\sin (\theta )\cos (\theta )$$r^2=\sin{(\theta)}\cos{(\theta)}$

6. $r=\frac{1}{\sin (\theta )}$$r=\frac{1}{\sin{(\theta)}}$

Homework

At this time, you should be able to complete the following assignments:

• Section 10.3: # 1, 3, 5, 7, 9, 11, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26.

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Created: Wednesday, 20 October 2021 - 05:10 (EDT)