# 10.3 Polar coordinates  (Page 5/8)

 Page 5 / 8

How are the polar axes different from the x - and y -axes of the Cartesian plane?

Explain how polar coordinates are graphed.

Determine $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for the point, then move $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ units from the pole to plot the point. If $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is negative, move $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ units from the pole in the opposite direction but along the same angle. The point is a distance of $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ away from the origin at an angle of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ from the polar axis.

How are the points $\text{\hspace{0.17em}}\left(3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ related?

Explain why the points $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ are the same.

The point $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ has a positive angle but a negative radius and is plotted by moving to an angle of $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ and then moving 3 units in the negative direction. This places the point 3 units down the negative y -axis. The point $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ has a negative angle and a positive radius and is plotted by first moving to an angle of $\text{\hspace{0.17em}}-\frac{\pi }{2}\text{\hspace{0.17em}}$ and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y -axis.

## Algebraic

For the following exercises, convert the given polar coordinates to Cartesian coordinates with $\text{\hspace{0.17em}}r>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0\le \theta \le 2\pi .\text{\hspace{0.17em}}$ Remember to consider the quadrant in which the given point is located when determining $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for the point.

$\left(7,\frac{7\pi }{6}\right)$

$\left(5,\pi \right)$

$\left(-5,0\right)$

$\left(6,-\frac{\pi }{4}\right)$

$\left(-3,\frac{\pi }{6}\right)$

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

$\left(4,\frac{7\pi }{4}\right)$

For the following exercises, convert the given Cartesian coordinates to polar coordinates with $\text{\hspace{0.17em}}r>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \theta <2\pi .\text{\hspace{0.17em}}$ Remember to consider the quadrant in which the given point is located.

$\left(4,2\right)$

$\left(-4,6\right)$

$\left(3,-5\right)$

$\left(\sqrt{34},5.253\right)$

$\left(-10,-13\right)$

$\left(8,8\right)$

$\left(8\sqrt{2},\frac{\pi }{4}\right)$

For the following exercises, convert the given Cartesian equation to a polar equation.

$x=3$

$y=4$

$r=4\mathrm{csc}\theta$

$y=4{x}^{2}$

$y=2{x}^{4}$

$r=\sqrt[3]{\frac{sin\theta }{2co{s}^{4}\theta }}$

${x}^{2}+{y}^{2}=4y$

${x}^{2}+{y}^{2}=3x$

$r=3\mathrm{cos}\theta$

${x}^{2}-{y}^{2}=x$

${x}^{2}-{y}^{2}=3y$

$r=\frac{3\mathrm{sin}\theta }{\mathrm{cos}\left(2\theta \right)}$

${x}^{2}+{y}^{2}=9$

${x}^{2}=9y$

$r=\frac{9\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta }$

${y}^{2}=9x$

$9xy=1$

$r=\sqrt{\frac{1}{9\mathrm{cos}\theta \mathrm{sin}\theta }}$

For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented.

$r=3\mathrm{sin}\text{\hspace{0.17em}}\theta$

$r=4\mathrm{cos}\text{\hspace{0.17em}}\theta$

${x}^{2}+{y}^{2}=4x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{{\left(x-2\right)}^{2}}{4}+\frac{{y}^{2}}{4}=1;$ circle

$r=\frac{4}{\mathrm{sin}\text{\hspace{0.17em}}\theta +7\mathrm{cos}\text{\hspace{0.17em}}\theta }$

$r=\frac{6}{\mathrm{cos}\text{\hspace{0.17em}}\theta +3\mathrm{sin}\text{\hspace{0.17em}}\theta }$

$3y+x=6;\text{\hspace{0.17em}}$ line

$r=2\mathrm{sec}\text{\hspace{0.17em}}\theta$

$r=3\mathrm{csc}\text{\hspace{0.17em}}\theta$

$y=3;\text{\hspace{0.17em}}$ line

$r=\sqrt{r\mathrm{cos}\text{\hspace{0.17em}}\theta +2}$

${r}^{2}=4\mathrm{sec}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}\theta$

$xy=4;\text{\hspace{0.17em}}$ hyperbola

$r=4$

${r}^{2}=4$

${x}^{2}+{y}^{2}=4;\text{\hspace{0.17em}}$ circle

$r=\frac{1}{4\mathrm{cos}\text{\hspace{0.17em}}\theta -3\mathrm{sin}\text{\hspace{0.17em}}\theta }$

$r=\frac{3}{\mathrm{cos}\text{\hspace{0.17em}}\theta -5\mathrm{sin}\text{\hspace{0.17em}}\theta }$

$x-5y=3;\text{\hspace{0.17em}}$ line

## Graphical

For the following exercises, find the polar coordinates of the point.

$\left(3,\frac{3\pi }{4}\right)$

$\left(5,\pi \right)$

For the following exercises, plot the points.

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

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

$\left(3.5,\frac{7\pi }{4}\right)$

$\left(-4,\frac{\pi }{3}\right)$

$\left(5,\frac{\pi }{2}\right)$

$\left(4,\frac{-5\pi }{4}\right)$

$\left(3,\frac{5\pi }{6}\right)$

$\left(-1.5,\frac{7\pi }{6}\right)$

$\left(-2,\frac{\pi }{4}\right)$

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

For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis.

$5x-y=6$

$r=\frac{6}{5\mathrm{cos}\theta -\mathrm{sin}\theta }$

$2x+7y=-3$

${x}^{2}+{\left(y-1\right)}^{2}=1$

$r=2\mathrm{sin}\theta$

${\left(x+2\right)}^{2}+{\left(y+3\right)}^{2}=13$

$x=2$

$r=\frac{2}{\mathrm{cos}\theta }$

${x}^{2}+{y}^{2}=5y$

${x}^{2}+{y}^{2}=3x$

$r=3\mathrm{cos}\theta$

For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane.

$r=6$

$r=-4$

${x}^{2}+{y}^{2}=16$

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

$\theta =\frac{\pi }{4}$

$y=x$

$r=\mathrm{sec}\text{\hspace{0.17em}}\theta$

$r=-10\mathrm{sin}\text{\hspace{0.17em}}\theta$

${x}^{2}+{\left(y+5\right)}^{2}=25$

$r=3\mathrm{cos}\text{\hspace{0.17em}}\theta$

## Technology

Use a graphing calculator to find the rectangular coordinates of $\text{\hspace{0.17em}}\left(2,-\frac{\pi }{5}\right).\text{\hspace{0.17em}}$ Round to the nearest thousandth.

$\left(1.618,-1.176\right)$

Use a graphing calculator to find the rectangular coordinates of $\text{\hspace{0.17em}}\left(-3,\frac{3\pi }{7}\right).\text{\hspace{0.17em}}$ Round to the nearest thousandth.

Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(-7,8\right)\text{\hspace{0.17em}}$ in degrees. Round to the nearest thousandth.

$\left(10.630,131.186°\right)$

Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(3,-4\right)\text{\hspace{0.17em}}$ in degrees. Round to the nearest hundredth.

Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(-2,0\right)\text{\hspace{0.17em}}$ in radians. Round to the nearest hundredth.

$\text{\hspace{0.17em}}\left(2,3.14\right)or\left(2,\pi \right)\text{\hspace{0.17em}}$

## Extensions

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{sec}\text{\hspace{0.17em}}\theta ;a>0.$

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{sec}\text{\hspace{0.17em}}\theta ;a<0.$

A vertical line with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units left of the y -axis.

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{csc}\text{\hspace{0.17em}}\theta ;a>0.$

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{csc}\text{\hspace{0.17em}}\theta ;a<0.$

A horizontal line with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units below the x -axis.

What polar equations will give an oblique line?

For the following exercise, graph the polar inequality.

$r<4$

$0\le \theta \le \frac{\pi }{4}$

$\theta =\frac{\pi }{4},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}2$

$\theta =\frac{\pi }{4},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\ge -3$

$0\le \theta \le \frac{\pi }{3},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}<\text{\hspace{0.17em}}2$

$\frac{-\pi }{6}<\theta \le \frac{\pi }{3},-3

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