# Polar Coordinates, Equations, and Graphs

(Note that we talk about converting back and forth from Polar Complex Form to Rectangular Complex form here in the Trigonometry and the Complex Plane section. Also note that we discussed Parametric Equations here, which may seem similar to Polar Equations, since they both have applications in Trigonometry.)

# Plotting Points Using Polar Coordinates

So far, we’ve plotted points using rectangular (or Cartesian) coordinates, since the points since we are going back and forth $$x$$ units, and up and down $$y$$ units. In the Polar Coordinate System, we go around the origin or the pole a certain distance out, and a certain angle from the positive $$x$$axis:

The ordered pairs, called polar coordinates, are in the form $$\left( {r,\theta } \right)$$, with $$r>0$$ being the number of units from the origin or pole, like a radius of a circle, and $$\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$axis (polar axis), going counter-clockwise. If $$r<0$$, the point is $$r$$ units in the opposite direction (across the origin or pole) of the angle $$\theta$$. If $$\theta <0$$, you go clockwise with the angle, starting with the positive $$x$$axis, instead of counter-clockwise.

To plot a point, you typically circle around the positive $$x$$axis $$\theta$$ degrees first, and then go out from the origin or pole $$r$$ units. If $$r$$ is negative, you go in the oppositive direction (180°) $$r$$ units. If both $$r$$ and the angle  are negative, you have to make sure you go clockwise to get the angle, but in the opposite direction $$r$$ units.

Here’s a polar graph with some points on it; note that we typically count in increments of 15°, or $$\displaystyle \frac{\pi }{{12}}$$:

For a point $$\left( {r,\theta } \right)$$, do you see how you always go counter-clockwise (or clockwise, if you have a negative angle) until you reach the angle you want, and then out from the center $$r$$ units, if $$r$$ is positive? Again, if $$r$$ is negative, you go in the opposite direction from the angle $$r$$ units. If both $$r$$ and the angle $$\theta$$ are negative, you have to make sure you go clockwise to get the angle, but in the opposite direction $$r$$ units.

You may be asked to rename a point in several different ways, for example, between $$\left[ {-2\pi ,2\pi } \right)$$ or $$\left[ {-360{}^\circ ,360{}^\circ } \right)$$. For example, if we wanted to rename the point $$\left( {6,240{}^\circ } \right)$$ three other different ways between $$\left[ {-360{}^\circ ,360{}^\circ } \right)$$, by looking at the graph above, we’d get $$\left( {6,-120{}^\circ } \right)$$ (subtract 360°), $$\left( {-6,60{}^\circ } \right)$$ (make $$r$$ negative and subtract 180°), and $$\left( {-6,-300{}^\circ } \right)$$(subtract another 360°). (Remember that 240° and –120°, and 60° and –300° are co-terminal angles). To get these, if the first number ($$r$$) is negative, go in the opposite direction, and if the angle is negative, go clockwise instead of counterclockwise from the positive $$x$$axis. Note that $$\left( {r,\theta } \right)$$ and $$\left( {-r,-\theta } \right)$$ aren’t the same point!

# Polar-Rectangular Point Conversions

You will probably be asked to convert coordinates between polar form and rectangular form.

## Converting from Polar to Rectangular Coordinates

Let’s first convert from polar to rectangular form; to do this we use the following formulas, using Right Triangle Trigonometry:

This conversion is pretty straight-forward; examples are below.

$$\begin{array}{l}x\,\,=\,\,r\,\cos \,\theta \\y\,\,=\,\,r\,\sin \,\theta \end{array}$$

## Converting from Rectangular to Polar Coordinates

Converting from rectangular coordinates to polar coordinates can be a little trickier since we need to check the quadrant of the rectangular point to get the correct angle; the quadrants must match. Here are the formulas:

$$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\,\,\,\,\,\,\,\,\,\text{(this will be positive)}$$

$$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)\,\,\,\,\,\,\text{(check for correct quadrant)}$$

One word of caution: If $$x=0$$, we’ll get an error when trying to obtain $$\theta$$. In these cases, graph the point to get the angle; it will either be $$\displaystyle \frac{\pi }{2}\,\,(90{}^\circ )$$ or $$\displaystyle \frac{{3\pi }}{2}\,\,(270{}^\circ )$$, depending on where the point lies on the $$y$$-axis.

Again, we need to check quadrants when using the calculator to get $${{\tan }^{{-1}}}$$. We’ll have to add the following degrees or radians when the point is in the following quadrants (this is because the  $${{\tan }^{{-1}}}$$ function on the calculator only gives answers back in the interval $$\displaystyle \left( {-\frac{\pi }{2},\frac{\pi }{2}} \right)$$), as shown in the Inverse Trigonometric Functions section):

Note that there can be multiple “answers” when converting from rectangular to polar, since polar points can be represented in many different ways (co-terminal angles, positive or negative “$$r$$”, and so on). Thus, it is typically easier to convert from polar to rectangular.

## Examples

Here are some examples of conversions both way; note you may be asked to convert back to polar into degrees or radians. For converting back to polar, make sure answers are either between 0 and 360° for degrees or 0 to $$2\pi$$ for radians. (And again, note that when we convert back to polar coordinates, we may not always get the same representation of the polar point we started out with.)

 Polar Coordinate Convert from Polar to Rectangular Position Convert from Rectangular to Polar $$\displaystyle \left( {2,\frac{{4\pi }}{3}} \right)$$ \begin{align}x&=2\cos \frac{{4\pi }}{3}=2\left( {-\frac{1}{2}} \right)=-1\\y&=2\sin \frac{{4\pi }}{3}=2\left( {-\frac{{\sqrt{3}}}{2}} \right)=-\sqrt{3}\end{align} $$\left( {-1,-\sqrt{3}} \right)$$ $$r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{1+3}}=2$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{{-\sqrt{3}}}{{-1}}} \right)=\frac{{4\pi }}{3}\text{ (3rd quadrant)}$$ $$\displaystyle \left( {2,\frac{{4\pi }}{3}} \right)$$ $$\left( {4,\pi } \right)$$ \begin{align}x&=4\cos \pi =4\left( {-1} \right)=-4\\y&=4\sin \pi =4\left( 0 \right)=0\end{align} $$\left( {-4,0} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{16+0}}=4$$   $$\displaystyle \begin{array}{c}\theta ={{\tan }^{{-1}}}\left( {\frac{0}{{-4}}} \right)=0\text{ (between 2nd }\\\text{ and 3rd quadrants)}\\\left( {4,\pi } \right)\end{array}$$ $$\left( {-7,45{}^\circ } \right)$$ \displaystyle \begin{align}{c}x&=-7\cos \left( {45{}^\circ } \right)=-\frac{{7\sqrt{2}}}{2}\\y&=-7\sin \left( {45{}^\circ } \right)=-\frac{{7\sqrt{2}}}{2}\end{align} $$\displaystyle \left( {-\frac{{7\sqrt{2}}}{2},-\frac{{7\sqrt{2}}}{2}} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{\frac{{49\left( 2 \right)}}{4}+\frac{{49\left( 2 \right)}}{4}}}=7$$   $$\theta ={{\tan }^{{-1}}}\left( 1 \right)=225{}^\circ \text{ (3rd quadrant)}$$ $$\left( {7,225{}^\circ } \right),\text{ which is the same as }\left( {-7,45{}^\circ } \right)$$ $$\left( {5,150{}^\circ } \right)$$ \displaystyle \begin{align}x&=5\cos \left( {150{}^\circ } \right)=5\left( {-\frac{{\sqrt{3}}}{2}} \right)=-\frac{{5\sqrt{3}}}{2}\\y&=5\sin \left( {150{}^\circ } \right)=5\left( {\frac{1}{2}} \right)=\frac{5}{2}\end{align} $$\displaystyle \left( {-\frac{{5\sqrt{3}}}{2},\frac{5}{2}} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{\frac{{25\left( 3 \right)}}{4}+\frac{{25}}{4}}}=5$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{{\frac{5}{2}}}{{-\frac{{5\sqrt{3}}}{2}}}} \right)=\text{150}{}^\circ \text{ (2nd quadrant)}$$ $$\displaystyle \left( {5,150{}^\circ } \right)$$ $$\displaystyle \left( {3,-\frac{{\pi }}{2}} \right)$$ \displaystyle \begin{align}x&=3\cos \left( {-\frac{\pi }{2}} \right)=3\left( 0 \right)=0\\y&=3\sin \left( {-\frac{\pi }{2}} \right)=3\left( {-1} \right)=-3\end{align} $$\displaystyle \left( {0,-3} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{0+9}}=3$$   $$\displaystyle \begin{array}{c}\theta ={{\tan }^{{-1}}}\left( {\frac{-3}{0}} \right)=\frac{3\pi }{2}\text{ (we want where tan is }\\\text{ undefined, between 3rd and 4th quadrants)}\end{array}$$ $$\displaystyle \left( {3,\frac{3\pi }{2}} \right),\text{ which is the same as }\left( {3,-\frac{{\pi }}{2}} \right)$$

Here are some problems going from Rectangular to Polar without special angles (angles found on the Unit Circle). Note that we may have to add $$\pi$$ or $$2\pi$$ to our answer, depending on which Quadrant the point lies in:

.

 Rectangular Point Convert from Rectangular to Polar $$\left( {-1,5} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{1+25}}=\sqrt{{26}}$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{5}{{-1}}} \right)\approx -1.373+\pi \approx 1.768\text{ (2nd quadrant)}$$ $$\displaystyle \left( {\sqrt{{26}},1.768} \right)$$ $$\left( {4,3} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{16+9}}=5$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{3}{4}} \right)\approx .644\text{ (1st quadrant)}$$ $$\displaystyle \left( {5,.644} \right)$$ $$\left( {3,-1.5} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{9+11.25}}=\sqrt{{11.25}}$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{{-1.5}}{3}} \right)\approx -.464+2\pi \approx 1.768\text{ (4th quadrant)}$$ $$\displaystyle \left( {\sqrt{{11.25}},5.812} \right)$$ $$\left( {-3,-5} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{9+25}}=\sqrt{{34}}$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{{-5}}{{-3}}} \right)\approx 1.030+\pi \approx 1.768\text{ (3rd quadrant)}$$ $$\displaystyle \left( {\sqrt{{34}},4.172} \right)$$

### Angle Conversions in Calculator

Note that you can also use “2nd APPS (ANGLE)” on your graphing calculator to do these conversions, but you won’t get the answers with the roots in them (you’ll get decimals that aren’t “exact”). You have to solve for the $$x$$ and $$y$$, or $$r$$ and $$\theta$$ separately, using the “,” above the 7 for the comma. Make sure you have your calculator either in DEGREES or RADIANS (in MODE), depending on what you’re working with. Here are some examples:

# Drawing Polar Graphs

I find that drawing polar graphs is a combination of part memorizing and part knowing how to create polar t-charts. Below are tables of some of the more common polar graphs, including t-charts in both degrees and radians.

Note that you can also put these in your graphing calculator, using radians or degrees. Here is an example with radians: mode: RADIAN, POLAR (instead of FUNCTION). Then, in window, setθ = [0, 2π], θstep = π/24 or .1, X = [–10, 10], Y = [–6, 6]. Then using “Y=” to put in the equation, using for theta. You can also use 2nd window (tblset) to set up the start and delta points in a table, and 2nd graph (table) to see the points.

You can also just set the mode to POLAR, put in the graph, and use zoom, ZStandard (option 6) and then zoom, ZTrig (option 7) to set the window (also try zoom, ZoomFit (option 0) if it doesn’t look right). By putting in smaller values of θstep, the graph is drawn more slowly and more accurately; to redraw graph, you can turn the graph off and back on by going to “=” and un-highlighting and highlighting it back again before hitting “graph”. Here is the result:

### Polar Graphs

Type of Polar Function T-Chart Graph

Circle

$$r=\text{p}$$ (radius)

$$r=5$$

Note: This makes sense since $$r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}$$, and the equation of a circle is $${{x}^{2}}+{{y}^{2}}={{5}^{2}}$$.

Note that $$r=-5$$ would produce the same graph.

 $$r$$ θ      ° 5 0       0 5 $$\displaystyle \frac{\pi }{2}$$    90 5 $$\pi$$    180 5 $$\displaystyle \frac{{3\pi }}{2}$$   270 5 $$2\pi$$   360

Shifted Circle

$$r=b\cos \theta ,\,\,r=b\sin \theta$$

$$r=4\cos \theta$$

Circle symmetric with $$x$$-axis with diameter 4.

If negative (such as $$r=-4\cos \theta$$), reflect over $$y$$-axis, so it’s on left-hand side.

 $$r$$ θ      ° 4 0      0 $$2\sqrt{2}\approx 2.8$$ $$\displaystyle \frac{\pi }{4}$$    45 0 $$\displaystyle \frac{\pi }{2}$$    90 $$-2\sqrt{2}\approx -2.8$$ $$\displaystyle \frac{{3\pi }}{4}$$ 135 –4 $$\pi$$   180

Shifted Circle

$$r=b\cos \theta ,\,\,r=b\sin \theta$$

$$r=-6\sin \theta$$

Circle symmetric with $$y$$-axis with diameter 6.

Since it’s negative, reflect over the $$x$$-axis, so it’s on the “bottom”.

(Positive would be above the $$x$$-axis).

 $$r$$ θ      ° 0 0       0 $$-3\sqrt{2}\approx -4.2$$ $$\displaystyle \frac{\pi }{4}$$    45 –6 $$\displaystyle \frac{\pi }{2}$$    90 $$-3\sqrt{2}\approx -4.2$$ $$\displaystyle \frac{{3\pi }}{4}$$ 135 0 $$\pi$$   180

Here are some polar equations that result in lines:

Type of Polar Function T-Chart Graph

Line

$$\theta =\text{angle}$$

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

Note: There is no “$$r$$” in the equation; just draw line at $$\displaystyle \frac{\pi }{4}$$. These graphs always go through the pole (center).

Note that $$\displaystyle \theta =\frac{{5\pi }}{4}$$ and $$\displaystyle \theta =-\frac{{3\pi }}{4}$$ would produce the same graph.

 $$r$$ θ        ° n/a $$\displaystyle \frac{\pi }{4}$$     45 n/a $$\displaystyle \frac{\pi }{4}$$     45 n/a $$\displaystyle \frac{\pi }{4}$$      45

n/a = not applicable; can be anything

Vertical Line

$$r=a\sec \theta$$

$$r=3\sec \theta$$

I remember that this line is vertical since it’s the same as $$r\cos \theta =3$$. This is the same as $$x=3$$ in rectangular form, which is a vertical line.

If negative (such as $$r=-3\sec \theta$$), reflect over the $$y$$-axis, so it’s on the left-hand side.

 $$r$$ θ        ° 3 0        0 $$\approx 4.2$$ $$\displaystyle \frac{\pi }{4}$$      45 und $$\displaystyle \frac{\pi }{2}$$      90 $$\approx -4.2$$ $$\displaystyle \frac{{3\pi }}{4}$$    135 –3 $$\pi$$      180

und = undefined

Horizontal Line

$$r=a\csc \theta$$

$$r=-4\csc \theta$$

I remember that this line is horizontal since it’s the same as $$r\sin \theta =-4$$. This is the same as is $$y=-4$$ in rectangular form, which is a horizontal line.

Since it’s negative, reflect over the $$x$$-axis, so it’s on the “bottom”. (Positive would be above the $$x$$-axis).

 $$r$$ θ        ° und 0         0 $$\approx -5.7$$ $$\displaystyle \frac{\pi }{4}$$      45 –4 $$\displaystyle \frac{\pi }{2}$$      90 $$\approx -5.7$$ $$\displaystyle \frac{{3\pi }}{4}$$    135 und $$\pi$$     180

Here are graphs that we call Cardioids and Limacons. They are in the form $$r=a+b\cos \theta$$ or $$r=a+b\sin \theta$$, and $$b$$ can be positive or negative. Note that, unlike their rectangular equivalents, $$r=a+b\cos \theta$$ and $$r=-a+b\cos \theta$$ (same with sin) are the same polar graph! Try it!

First, the Cardioids (Hearts); note that these and the LimeconLoops” touch the pole (origin), while the LimeconBeans” do not:

Type of Polar Function T-Chart Graph

Cardioid or Heart

$$\begin{array}{l}r=a+b\cos \theta ,\,\,\,a=b\\r=a+b\sin \theta ,\,\,\,a=b\end{array}$$

$$r=3+3\cos \theta$$

Note: I remember that when $$a=b$$, things are in harmony, like a heart.

The heart goes out $$3+3=6$$ units on the (positive) $$x$$-axis and hits 3 and –3 on the $$y$$-axis.

If cos is negative (such as $$r=3-3\cos \theta$$), reflect over the $$y$$-axis, so it’s on the left-hand side.

 $$r$$ θ        ° 6 0         0 3 $$\displaystyle \frac{\pi }{2}$$       90 0 $$\pi$$       180 3 $$\displaystyle \frac{{3\pi }}{2}$$    270

Note that $$r=-3+3\cos \theta$$ would make same graph.

Cardioid or Heart

$$\begin{array}{l}r=a+b\cos \theta ,\,\,\,a=b\\r=a+b\sin \theta ,\,\,\,a=b\end{array}$$

$$r=4-4\sin \theta$$

Note: I remember that when $$a=b$$, things are in harmony, like a heart.

The heart goes out $$4+4=8$$ units on the (negative) $$y$$-axis and hits 4 and –4 on the $$x$$-axis.

Since sin is negative, it’s reflected over the $$x$$-axis, so it’s on the “bottom”. (Positive would be above the $$x$$-axis).

 $$r$$ θ        ° 4 0         0 0 $$\displaystyle \frac{\pi }{2}$$       90 4 $$\pi$$       180 8 $$\displaystyle \frac{{3\pi }}{2}$$    270

Note that $$r=-4-4\sin \theta$$ would make same graph.

Here are the Limacons:

Type of Polar Function T-Chart Graph

Limacon (Loop)

$$\begin{array}{l}r=a+b\cos \theta ,\,\,\,a<b\\r=a+b\sin \theta ,\,\,\,a<b\end{array}$$

$$r=3-5\cos \theta$$

Note: I remember that when $$a$$ is less than $$b$$ ($$a<b$$), it’s a loop (the l’s match).

The limacon goes out $$3+5=8$$ units on the (negative) $$x$$-axis and hits 3 and –3 on the $$y$$-axis. It also hits $$5-3=2$$ in the loop.

Since cos is negative, it’s reflected over $$y$$-axis, so it’s on the left-hand side. (Positive would be on the right-hand side).

 $$r$$ θ         ° –2 0         0 3 $$\displaystyle \frac{\pi }{2}$$       90 8 $$\pi$$       180 3 $$\displaystyle \frac{{3\pi }}{2}$$    270

Note that $$r=-3-5\cos \theta$$ would make same graph.

Limacon (Bean)

$$\begin{array}{l}r=a+b\cos \theta ,\,\,\,a>b\\r=a+b\sin \theta ,\,\,\,a>b\end{array}$$

$$r=4+3\sin \theta$$

Note: I remember that when $$b$$ is the smallest, it’s a “bean”.

The bean goes up $$4+3=7$$ units on the (positive) $$y$$-axis and hits 4 and –4 on the $$x$$. It also hits $$4-3=1$$ on the negative $$y$$-axis.

If sin is negative (such as $$r=4-3\sin \theta$$), it’s reflected over $$x$$-axis, so it’s on the bottom.

 $$r$$ θ         ° 4 0         0 7 $$\displaystyle \frac{\pi }{2}$$       90 4 $$\pi$$       180 1 $$\displaystyle \frac{{3\pi }}{2}$$    270

Note that $$r=-4+3\sin \theta$$ would make same graph.

Graphs of Roses produce “petals” and are in the form $$r=a\cos \left( {b\theta } \right)$$ or $$r=a\sin \left( {b\theta } \right)$$. Note that since we have the starting point for these graphs, and the distance between the petals, the t-chart isn’t that helpful. The t-charts do tell us the order in which the petals are drawn; they are drawn in loops like figure eights. (In the t-charts, I made the $$\Delta$$ angle the same as the distance between petals).

Type of Polar Function T-Chart Graph

Rose (“$$\boldsymbol {b}$$” is even)

$$r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)$$

$$r=7\cos \left( {4\theta } \right)$$

Since $$b$$ (4) is even, we have $$2b$$ petals, or 8 petals. The length of each petal is $$a$$ (7).

With positive cos, they start at the positive $$\boldsymbol { x}$$-axis and they are $$\displaystyle \frac{{360}}{8}$$, or 45° apart, going counterclockwise (I find it easier to do these divisions using degrees instead of radians). For negative cos, we have the same graph when $$b$$ is even!

 $$r$$ θ      ° 7 0      0 –7 $$\displaystyle \frac{\pi }{4}$$    45 7 $$\displaystyle \frac{\pi }{2}$$    90 –7 $$\displaystyle \frac{{3\pi }}{4}$$   135 7 $$\pi$$    180 –7 $$\displaystyle \frac{{5\pi }}{4}$$    225 7 $$\displaystyle \frac{{3\pi }}{2}$$   270 –7 $$\displaystyle \frac{{7\pi }}{4}$$   315

Rose (“$$\boldsymbol {b}$$” is odd)

$$r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)$$

$$r=4\cos \left( {3\theta } \right)$$

Since $$b$$ (3) is odd, we have $$b$$ petals, or 3 petals (we don’t multiply by 2). The length of each petal is $$a$$ (4).

With positive cos, they start at the positive $$\boldsymbol { x}$$-axis and are $$\displaystyle \frac{{360}}{3}$$, or 120° apart, going counterclockwise.

 $$r$$ θ      ° 4 0      0 4 $$\displaystyle \frac{{2\pi }}{3}$$   120 4 $$\displaystyle \frac{{4\pi }}{3}$$   240

Rose (“$$\boldsymbol {b}$$” is odd)

$$r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)$$

$$r=-6\cos \left( {5\theta } \right)$$

Since $$b$$ (5) is odd, we have $$b$$ petals, or 5 petals (we don’t multiply by 2). The length of each petal is $$a$$ (6).

With negative cos, they start at the negative positive $$\boldsymbol { x}$$-axis (reflect over $$\boldsymbol { y}$$-axis) and are $$\displaystyle \frac{{360}}{5}$$, or 72° apart, going counterclockwise.

 $$r$$ θ      ° –6 0      0 –6 $$\displaystyle \frac{{2\pi }}{5}$$   72 –6 $$\displaystyle \frac{{4\pi }}{5}$$   144 –6 $$\displaystyle \frac{{6\pi }}{5}$$  216 –6 $$\displaystyle \frac{{8\pi }}{5}$$   288

($$r$$’s can also be positive, if you start at $$\pi$$)

And here are some sine Rose graphs:

Type of Polar Function T-Chart Graph

Rose (“$$\boldsymbol {b}$$” is even)

$$r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)$$

$$r=8\sin \left( {4\theta } \right)$$

Since $$b$$ (4) is even, we have $$2b$$ petals, or 8 petals. The length of each petal is $$a$$ (8).

With positive sin, they start at $$\displaystyle \frac{{90}}{b}=\frac{{90}}{4}=22.5{}^\circ$$ up from the positive $$\boldsymbol { x}$$-axis (memorize this) and they are $$\displaystyle \frac{{360}}{8}$$, or 45° apart, going counterclockwise. For negative sin, we have the same graph when $$b$$ is even!

 $$r$$ θ       ° 8 $$\displaystyle \frac{\pi }{8}$$    22.5 –8 $$\displaystyle \frac{3\pi }{8}$$    67.5 8 $$\displaystyle \frac{{5\pi }}{8}$$  112.5 –8 $$\displaystyle \frac{{7\pi }}{8}$$   157.5 8 $$\displaystyle \frac{{9\pi }}{8}$$   202.5 –8 $$\displaystyle \frac{{11\pi }}{8}$$  247.5 8 $$\displaystyle \frac{{13\pi }}{8}$$  292.5 –8 $$\displaystyle \frac{{15\pi }}{8}$$  337.5

Rose (“$$\boldsymbol {b}$$” is odd)

$$r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)$$

$$r=5\sin \left( {3\theta } \right)$$

Since $$b$$ (3) is odd, we have $$b$$ petals, or 3 petals (we don’t multiply by 2). The length of each petal is $$a$$ (5).

With positive sin, they start at $$\displaystyle \frac{{90}}{3}=\frac{{90}}{3}=30{}^\circ$$ up from the positive $$\boldsymbol { x}$$-axis (reflect over $$x$$-axis) and are $$\displaystyle \frac{{360}}{3}$$, or 120° apart, going counterclockwise.

 $$r$$ θ      ° 5 $$\displaystyle \frac{{\pi }}{6}$$    30 5 $$\displaystyle \frac{{5\pi }}{6}$$   150 5 $$\displaystyle \frac{{3\pi }}{2}$$  270

Rose (“$$\boldsymbol {b}$$” is odd)

$$r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)$$

$$r=-6\sin \left( {5\theta } \right)$$

Since $$b$$ (5) is odd, we have $$b$$ petals, or 5 petals (we don’t multiply by 2). The length of each petal is $$a$$ (6).

With negative sin, they start at $$\displaystyle \frac{{90}}{b}=\frac{{90}}{5}=18{}^\circ$$ down from the positive $$\boldsymbol { x}$$-axis (reflect over the $$x$$-axis) and are $$\displaystyle \frac{{360}}{5}$$, or 72° apart, going counterclockwise.

 $$r$$ θ      ° 6 $$\displaystyle \frac{{-\pi }}{10}$$   –18 6 $$\displaystyle \frac{{3\pi }}{10}$$   54 6 $$\displaystyle \frac{{7\pi }}{10}$$   126 6 $$\displaystyle \frac{{11\pi }}{10}$$  198 6 $$\displaystyle \frac{{3\pi }}{2}$$  270

Note: For a rose graph, you may be asked to name the order that petals are drawn. One way to do this is to use the angle measurements $$\displaystyle 0,\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4},\,\frac{{7\pi }}{4}$$, solve for $$r$$, and observe the order of the petals. You can also use the graphing calculator as shown above, but make the θstep smaller to slow down the drawing of the graph.

Note: You may also see a combination of a rose and a limacon in the form $$r=a+b\cos \left( {k\theta } \right),\,\,r=a+b\sin \left( {k\theta } \right),\,\,k>1$$. In these cases, you may see graphs that don’t meet at the origin; try these on your calculator!

Here are a couple more polar graphs (Spirals and Lemniscates) that you might see:

Type of Polar Function T-Chart Graph

Spiral

$$r=a\theta$$

$$r=2\theta$$

Note that in the t-chart, the $$r$$ values are in radians; they can also be in degrees.

The smaller the $$r$$, the tighter the spiral.

 $$r$$ θ         ° 0 0         0 3.14 $$\displaystyle \frac{\pi }{2}$$       90 6.28 $$\pi$$       180 9.42 $$\displaystyle \frac{{3\pi }}{2}$$    270

Lemniscate  (Figure “8”)

$${{r}^{2}}={{a}^{2}}\cos \left( {2\theta } \right),\,\,{{r}^{2}}={{a}^{2}}\sin \left( {2\theta } \right)$$

$$r=\sqrt{{{{4}^{2}}\cos \left( {2\theta } \right)}}=\sqrt{{16\cos \left( {2\theta } \right)}}=4\sqrt{{\cos \left( {2\theta } \right)}}$$

With cos, graph is horizontal across the $$\boldsymbol {x}$$-axis.

The graph has 2 petals and the length of each petal is $$a$$ (4).

 $$r$$ θ         ° 4 0         0 und $$\displaystyle \frac{\pi }{2}$$       90 4 $$\pi$$       180 und $$\displaystyle \frac{{3\pi }}{2}$$    270

und = undefined

Lemniscate  (Figure “8”)

$${{r}^{2}}={{a}^{2}}\cos \left( {2\theta } \right),\,\,{{r}^{2}}={{a}^{2}}\sin \left( {2\theta } \right)$$

$$r=\sqrt{{{{7}^{2}}\sin \left( {2\theta } \right)}}=\sqrt{{49\sin \left( {2\theta } \right)}}=7\sqrt{{\sin \left( {2\theta } \right)}}$$

With sin, graph is along $$\displaystyle \frac{\pi }{4}$$.

The graph has 2 petals and the length of each petal is $$a$$ (7).

 $$r$$ θ         ° 0 0         0 7 $$\displaystyle \frac{\pi }{4}$$      45 7 $$\displaystyle \frac{5\pi }{4}$$    180 0 $$\displaystyle \frac{{3\pi }}{2}$$    270

You might be asked to obtain the equation of a polar function from a graph:

Graph Get Polar Equation from Graph
. We see that the graph is a Rose pattern with an odd number of petals, so we don’t have to divide the number of petals by 2 to get $$b$$ in the formula $$r=a\cos \left( {b\theta } \right)$$ or $$r=a\sin \left( {b\theta } \right)$$.

The petals are 4 units long ($$a$$) and don’t start on the $$x$$-axis, so we have either $$r=4\sin \left( {5\theta } \right)$$ or $$r=-4\sin \left( {5\theta } \right)$$.

With 5 petals and with sin, the first petal should start at $$\displaystyle \frac{{90}}{b}=\frac{{90}}{5}=18{}^\circ$$ up from the positive $$x$$-axis; since they start 18 degrees below (reflected) over the $$x$$-axis, the graph is $$r=-4\sin \left( {5\theta } \right)$$.

The graph is a Bean (Limacon – no loop) and not a Cardioid or Heart, since it doesn’t go through the origin.

The graph is horizontal and is reflected over the $$\boldsymbol {y}$$-axis, so we have $$r=a-b\cos \theta ,\,\,\,a>b$$.

Since the graph hits 5 ($$a$$) on the $$\boldsymbol {y}$$-axis and goes out to 9 to the left, we have $$r=5-4\cos \theta$$, since $$b=9-a$$. Also, since $$a-b \,\,(5-4=1)$$ is where it hits the $$x$$-axis, it looks good!

 $$\Leftarrow \text{ A Line at }r=4\csc \left( \theta \right)$$ $$\Leftarrow \text{ A Line at }\theta =60{}^\circ$$

Here are more types of questions you may get asked when studying polar graphs:

 Polar Equation Question Solution For the rose polar graph $$5\sin \left( {10\theta } \right)$$:   Find the length of each petal, number of petals, spacing between each petal, and the tip of the 1st petal in Quadrant I. The length of each petal in the rose polar graph is $$a$$, so this length is 5. Since $$b$$ (10) is even, we need to double it to get the number of petals, so we get 20. The spacing between each petal is $$\displaystyle \frac{{360{}^\circ }}{{20}}=18{}^\circ$$, since there are 360° in a circle. The tip of the 1st petal in Quadrant I is at $$\displaystyle \frac{{90{}^\circ }}{b}=\frac{{90{}^\circ }}{{10}}=9{}^\circ$$. Graph to check the answer! Write a polar equation for a spiral that goes through the polar point $$\left( {6,60{}^\circ } \right)$$. Since the equation of a polar spiral graph is $$r=a\theta$$ and we know a point on the graph is $$\left( {6,60{}^\circ } \right)$$, we can solve for $$a$$: $$\displaystyle r=a\theta ;\,\,6=a60;\,\,a=\frac{1}{{10}}$$. Thus, the polar equation is $$\displaystyle r=\frac{1}{{10}}\theta$$ or $$\displaystyle r=\frac{\theta }{{10}}$$ (in degrees). Can a rose have 14 petals? Explain why or why not. No. Since, for a rose, if the number of pedals is even, there are twice that number of petals. If the number of petals is odd, you have exactly that number of petals. Since twice 7 is 14, but 7 is odd, this can’t happen. Graph the rectangular and polar equation $$r=-6\cos \left( {5\theta } \right)$$. We learned from the Transformations of Trig Functions section that a rectangular graph in the form $$y=a\cos b\left( {x-c} \right)+d$$ has an amplitude of $$\left| a \right|$$ (flipped over the $$x$$-axis if $$a$$ is negative), has a period of $$\displaystyle \frac{{2\pi }}{b}$$, a horizontal shift of $$c$$, and a vertical shift of $$d$$. In our case, the amplitude is 6, it’s flipped over the $$x$$-axis, and the period is $$\displaystyle \frac{{2\pi }}{b}=\frac{{2\pi }}{5}$$.   We already saw that the polar graph is a rose with an odd number of petals (5), each 6 units long, and starts at $$\displaystyle -\frac{{90}}{b}=-\frac{{90}}{5}=-18{}^\circ$$. You can see how different these graphs look:

# Converting Equations from Rectangular to Polar

To convert Rectangular Equations (Cartesian Coordinates) to Polar Equations, we want to get rid of the $$x$$’s and $$y$$’s and only have $$r$$’s and/or $$\theta$$’s in the answer. We do this with the following equations:

$$\begin{array}{l}x=r\cos \,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\\y=r\,\sin \,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\end{array}$$

Here are some examples; note that we want to solve for $$r$$ if we can; in the case of quadratics or higher degrees, this may involve moving everything to one side and factoring.

Also note that when we also get $$r=0$$ (the pole) for the answers, this is one point only, and in these cases, the pole is included in the other part of the answers. Thus, we can typically discard $$r=0$$.

 Rectangular Equation Convert to Polar Equation $$x=-3$$ $$\displaystyle \begin{array}{c}r\cos \theta =-3\\r=\frac{{-3}}{{\cos \theta }};\,\,\,\,\,\,\,\,\,\,\underline{{r=-3\sec \theta }},\,\,\,\cos \theta \ne 0\end{array}$$ $$y={{x}^{4}}$$ $$\begin{array}{c}r\sin \theta ={{\left( {r\cos \theta } \right)}^{4}}\\r\sin \theta ={{r}^{4}}{{\cos }^{4}}\theta \\r\sin \theta -{{r}^{4}}{{\cos }^{4}}\theta =0\\r\left( {\sin \theta -{{r}^{3}}{{{\cos }}^{4}}\theta } \right)=0\end{array}$$              $$\require {cancel} \displaystyle \begin{array}{c}\xcancel{{r=0}}\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\sin \theta -{{r}^{3}}{{\cos }^{4}}\theta =0\\\,r=\sqrt[3]{{\frac{{\sin \theta }}{{{{{\cos }}^{4}}\theta }}}};\,\,\,\,r=\sqrt[3]{{\tan \theta {{{\sec }}^{3}}\theta }}\\\underline{{r=\sec \theta \,\sqrt[3]{{\tan \theta }}}};\,\,\,\,\cos \theta \ne 0\end{array}$$ $${{x}^{2}}+{{y}^{2}}=-3x$$ $$\displaystyle \begin{array}{c}\text{Note that }{{x}^{2}}+{{y}^{2}}={{r}^{2}}:\\{{r}^{2}}=-3\left( {r\cos \theta } \right)\\{{r}^{2}}+3r\cos \theta =0\\r\left( {r+3\cos \theta } \right)=0\end{array}$$              $$\xcancel{{r=0\,}}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\underline{{r=-3\cos \theta }}$$ $$2x+y=3$$ $$\begin{array}{c}2\left( {r\cos \theta } \right)+r\sin \theta =3\\r\left( {2\cos \theta +\sin \theta } \right)=3\end{array}$$                $$\displaystyle \underline{{r=\frac{3}{{2\cos \theta +\sin \theta }}}}$$ $${{x}^{2}}+{{y}^{2}}=49$$ $$\begin{array}{c}\text{Use }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\,\,\text{identity:}\\{{\left( {r\cos \theta } \right)}^{2}}+{{\left( {r\sin \theta } \right)}^{2}}=49\\{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =49\end{array}$$                    $$\begin{array}{c}{{r}^{2}}\left( {{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta } \right)=49\\{{r}^{2}}\left( 1 \right)=49\\\underline{{r=\pm 7}}\end{array}$$ $$y=-x$$ $$\begin{array}{c}r\sin \theta =-r\cos \theta \\r\sin \theta +r\cos \theta =0\\r\left( {\sin \theta +\cos \theta } \right)=0\end{array}$$            $$\displaystyle \begin{array}{c}\xcancel{{r=0}}\,\,\,\,\,\text{or}\,\,\,\,\sin \theta =-\cos \theta \\\,\tan \theta =-1\\\,\underline{{\theta =\frac{{3\pi }}{4}}}\,\,\,\,\,\text{(or}\,\,\frac{{7\pi }}{4}\text{)}\end{array}$$

# Converting Equations from Polar to Rectangular

To convert Polar Equations to Rectangular Equations (Cartesian Coordinates), we want to get rid of the $$r$$’s and $$\theta$$’s and only have $$x$$’s and/or $$y$$’s in the answer. We do this with the following equations, depending on what we have in the polar equation:

$$\displaystyle r=\sqrt{{{{x}^{2}}+\,{{y}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta \,\,=\,\,\frac{x}{r}=\frac{x}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}$$

$$\displaystyle \theta \,\,=\,\,{{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \theta \,\,=\,\,\frac{y}{r}\,\,=\,\,\frac{y}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}$$

Here are some examples. Note that sometimes we may be asked to Complete the Square to get the equation in a circle (or other conic) form; we learned how to do this in the Factoring Quadratics and Completing the Square section here.

 Polar Equation Convert to Rectangular Equation; Identify the Graph $$r=4\sin \theta$$ $$\displaystyle \begin{array}{c}\sqrt{{{{x}^{2}}+{{y}^{2}}}}=4\left( {\frac{y}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}} \right)\\{{\left( {\sqrt{{{{x}^{2}}+{{y}^{2}}}}} \right)}^{2}}=4y\\{{x}^{2}}+{{y}^{2}}=4y\end{array}$$           $$\begin{array}{c}\text{Complete the square to get circle}\\\text{Center: }\left( {0,2} \right),\,\,\text{radius}:2\\{{x}^{2}}+{{y}^{2}}=4y\\{{x}^{2}}+{{y}^{2}}-4y=0\\{{x}^{2}}+\left( {{{y}^{2}}-4y+4} \right)=0+4\\\underline{{{{x}^{2}}+{{{\left( {y-2} \right)}}^{2}}=4}}\end{array}$$ $$\theta =45{}^\circ$$ $$\displaystyle {{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)=45{}^\circ ;\,\,\,\,\,\frac{y}{x}=1;\,\,\,\,\,\,\,\,\,\,\,\,\underline{{y=x}}$$    line $$\displaystyle \theta =-\frac{\pi }{6}$$ $$\displaystyle {{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)=-\frac{\pi }{6};\,\,\,\,\,\frac{y}{x}=-\frac{1}{{\sqrt{3}}};\,\,\,\,\,\,\,\,\,\,\,\,\underline{{y=-\frac{x}{{\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}}}=-\frac{{x\sqrt{3}}}{3}$$    line $$r=5$$ $$\sqrt{{{{x}^{2}}+{{y}^{2}}}}=5;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{{{x}^{2}}+{{y}^{2}}=25}}$$    circle $$\displaystyle r=\frac{4}{{2+\cos \theta }}$$ I’m going to simplify before substituting for r, but you don’t have to. Also, it may be up to your teacher as to how far you have to simplify after substituting in the rectangular variables. I didn’t complete the square for the ellipse, but you can see that it has positive $${{x}^{2}}$$ and $${{y}^{2}}$$ that don’t have the same coefficients. $$\displaystyle \begin{array}{c}r=\frac{4}{{2+\frac{x}{r}}};\,\,\,\,\,\,\,\,\,r\left( {2+\frac{x}{r}} \right)=4;\,\,\,\,\,\,\,\,\,\,2r+x=4\\2\sqrt{{{{x}^{2}}+{{y}^{2}}}}+x=4;\,\,\,\,\,\,\,2\sqrt{{{{x}^{2}}+{{y}^{2}}}}=4-x\\{{\left( {2\sqrt{{{{x}^{2}}+{{y}^{2}}}}} \right)}^{2}}={{\left( {4-x} \right)}^{2}};\,\,\,\,\,\,\,\,\,\,4\left( {{{x}^{2}}+{{y}^{2}}} \right)=16-8x+{{x}^{2}}\\4{{x}^{2}}+4{{y}^{2}}=16-8x+{{x}^{2}};\,\,\,\,\,\,\,\underline{{3{{x}^{2}}+4{{y}^{2}}+8x=16}}\end{array}$$    ellipse $${{r}^{2}}\sin \left( {2\theta } \right)=4$$ $$\require {cancel} \begin{array}{c}{{r}^{2}}\cdot 2\sin \theta \cos \theta =4\\\cancel{{{{r}^{2}}}}\cdot 2\left( {\frac{y}{{\cancel{r}}}} \right)\left( {\frac{x}{{\cancel{r}}}} \right)=4\\2xy=4\\\underline{{y=\frac{2}{x}}}\end{array}$$   inverse        $$\displaystyle \begin{array}{c}\text{Note that we used the identity}\\\sin \left( \theta \right)=2\sin \theta \cos \theta .\\\\\text{Note also that we left }r\,\text{in the equation and}\\\text{we could simplify before substituting}\text{.}\end{array}$$ $$r=2\sin \theta +3\cos \theta$$ $$\displaystyle \begin{array}{c}r=2\left( {\frac{y}{r}} \right)+3\left( {\frac{x}{r}} \right)\\{{r}^{2}}=2y+3x\\{{x}^{2}}+{{y}^{2}}=2y+3x\end{array}$$          $$\begin{array}{c}\text{Complete the square to get circle}\\\text{Center: }\left( {\frac{3}{2},1} \right),\,\,\text{radius}:\frac{{\sqrt{{13}}}}{2}\text{:}\\{{x}^{2}}-3x+{{y}^{2}}-2y=0\\\left( {{{x}^{2}}-3x+\frac{9}{4}} \right)+\left( {{{y}^{2}}-2y+1} \right)=0+\frac{9}{4}+1\\\underline{{{{{\left( {x-\frac{3}{2}} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}=\frac{{13}}{4}}}\end{array}$$ $$\displaystyle r=\frac{1}{2}\cdot \frac{{\sin \theta }}{{\cos \theta }}\cdot \frac{1}{{\cos \theta }}=\frac{1}{2}\cdot \frac{{\left( {\frac{y}{{\cancel{r}}}} \right)}}{{\left( {\frac{x}{{\cancel{r}}}} \right)}}\cdot \frac{1}{{\left( {\frac{x}{r}} \right)}}=\frac{1}{2}\cdot \frac{y}{{\frac{{{{x}^{2}}}}{r}}}=\frac{1}{2}\cdot \frac{y}{1}\cdot \frac{r}{{{{x}^{2}}}}$$ $$\displaystyle r=\frac{{\tan \theta \sec \theta }}{2}$$ $$\displaystyle r=\frac{1}{2}\cdot \frac{{\sin \theta }}{{\cos \theta }}\cdot \frac{1}{{\cos \theta }}=\frac{1}{2}\cdot \frac{{\left( {\frac{y}{{\cancel{r}}}} \right)}}{{\left( {\frac{x}{{\cancel{r}}}} \right)}}\cdot \frac{1}{{\left( {\frac{x}{r}} \right)}}=\frac{1}{2}\cdot \frac{y}{{\frac{{{{x}^{2}}}}{r}}}=\frac{1}{2}\cdot \frac{y}{1}\cdot \frac{r}{{{{x}^{2}}}}$$ $$\displaystyle \cancel{r}=\frac{{\cancel{r}y}}{{2{{x}^{2}}}};\,\,\,\,1=\frac{y}{{2{{x}^{2}}}};\,\,\,\,y=2{{x}^{2}}$$    Parabola

# Polar Graph Points of Intersection

To find the intersection points for sets of polar curves, it’s helpful to draw the curves and also to solve algebraically. To solve algebraically, we just set the $$r$$’s together and solve for $$\theta$$. Note also that after we solve for one variable (like $$\theta$$), we have to plug it back in either equation to get the other coordinate (like $$r$$).

We also have to be careful since there are “phantom” or “elusive” points, typically at the pole. The reason these points are “phantom” is because, although we don’t necessarily get them algebraically, we can see them on a graph. This is because, with an “$$r$$” of 0, the $$\theta$$ could really be anything, since we aren’t going out any distance. We will also see phantom points when one of the equations is “$$r=$$ constant”, since another way to write this is “$$r=$$ the negative of that constant”. Note that with “phantom” points, both equations do not have to work; I know, it’s weird. To get all these elusive points, you put in the $$r$$ value in both curves to see what additional points you get.

Find the intersection points for the following sets of polar curves (algebraically) and also draw a sketch. Find the intersections when $$\theta$$ is between 0 and $$\boldsymbol {2\pi}$$, and $$r$$ is positive.

 Polar Equations Intersecting Points Graph $$\begin{array}{l}r=-\sin \theta \text{ }\\r=\cos \theta \end{array}$$ $$\displaystyle \begin{array}{c}-\sin \theta =\cos \theta ;\,\,\,\,\,\,\tan \theta =-1\\\theta =\frac{{3\pi }}{4}\,\,\left( {r=\cos \left( {\frac{{3\pi }}{4}} \right)=-\frac{{\sqrt{2}}}{2}} \right)\text{ (duplicate)},\,\,\,\,\frac{{7\pi }}{4}\,\,\left( {r=\frac{{\sqrt{2}}}{2}} \right)\\\underline{{\left( {\frac{{\sqrt{2}}}{2},\frac{{7\pi }}{4}} \right),\,\,\left( {0,0} \right)\text{ (”phantom” point)}}}\end{array}$$ $$\begin{array}{l}r=\cos \theta \\r=\cos 2\theta \end{array}$$ $$\displaystyle \begin{array}{c}\cos \theta =\cos 2\theta \\\cos \theta =2{{\cos }^{2}}\theta -1\,\,\text{(identity)}\\2{{\cos }^{2}}\theta -\cos \theta -1=0;\,\,\,\,\,\,\left( {2\cos +1} \right)\left( {\cos \theta -1} \right)=0\\\cos \theta =-\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\cos \theta =1\\\theta =\frac{{2\pi }}{3}\,\,\left( {r=\cos \left( {\frac{{2\pi }}{3}} \right)=-\frac{1}{2}} \right),\,\,\frac{{4\pi }}{3}\,\,\left( {r=-\frac{1}{2}} \right),\,\,\,\theta =0\,\,\left( {r=1} \right)\\\text{which is}\\\underline{{\left( {\frac{1}{2},\frac{{5\pi }}{3}} \right),\,\,\left( {\frac{1}{2},\frac{\pi }{3}} \right),\,\,\,\left( {1,0} \right),\,\,\left( {0,0} \right)\text{ (”phantom” point) }}}\end{array}$$ $$\begin{array}{c}r=3\\r=-6\sin \theta \end{array}$$ $$\displaystyle \begin{array}{c}-6\sin \theta =3;\,\,\,\,\,\,\sin \theta =-\frac{1}{2}\\\theta =\frac{{7\pi }}{6}\,\,\left( {r=3} \right),\,\,\,\,\frac{{11\pi }}{6}\,\,\left( {r=3} \right)\\\underline{{\left( {3,\frac{{7\pi }}{6}} \right),\,\,\left( {3,\frac{{11\pi }}{6}} \right)}}\end{array}$$   No “phantom” points $$\begin{array}{l}r=\sin 2\theta \\r=\cos \theta \end{array}$$ $$\displaystyle \begin{array}{c}\sin 2\theta =\cos \theta \\2\sin \theta \cos \theta \,\,\text{(identity)}=\cos \theta \\2\sin \cos \theta -\cos \theta =0;\,\,\,\cos \theta \left( {2\sin \theta -1} \right)=0\\\cos \theta =0\,\,\,\,\,\,\,\,\,\,\sin \theta =\frac{1}{2}\\\theta =\frac{\pi }{2}\,\,\left( {r=\cos \left( {\frac{\pi }{2}} \right)=0} \right),\,\,\,\,\frac{{3\pi }}{2}\,\,\left( {r=0} \right)\,\,\,\text{(duplicate)},\,\,\,\\\theta =\frac{\pi }{6}\,\,\left( {r=\frac{{\sqrt{3}}}{2}} \right)\text{ },\,\,\,\,\frac{{5\pi }}{6}\,\,\left( {r=-\frac{{\sqrt{3}}}{2}} \right)\\\underline{{\left( {0,\frac{\pi }{2}} \right)\text{ (which is the same }\left( {0,0} \right)),\,\,\,\left( {\frac{{\sqrt{3}}}{2},\frac{\pi }{6}} \right),\,\,\left( {\frac{{\sqrt{3}}}{2},\frac{{11\pi }}{6}} \right)}}\end{array}$$

It also might be good to know the sequence in which the polar graphs are drawn; in other words, from 0 to $$2\pi$$, which parts of the graphs are drawn before the other graphs. (Check it out on a graphing calculator, where you can see it!) You can use a t-chart, or set the polar equation to 0 if the graph crosses the pole, and test points in between. Here are some examples:

Problem and Solution T-Chart Graph
The inner loop for $$2+4\cos \theta$$ is formed between what two values of $$\theta$$?

Solution:

Find out the two cases when $$r=0$$, since that’s before and after the graph draws its inner loop:

$$\displaystyle 0=2+4\cos \theta ;\,\,\,\,\,\cos \theta =-\frac{1}{2}$$

$$\displaystyle \theta =\frac{{2\pi }}{3};\,\,\,\,\,\theta =\frac{{4\pi }}{3}$$

Since the end of the inner loop is at $$\left( {-2,\pi } \right)$$ (same as $$\displaystyle \left( {2,0{}^\circ } \right)$$), and this is between $$\displaystyle \frac{{2\pi }}{3}$$ and $$\displaystyle \frac{{4\pi }}{3}$$, the inner loop is formed when

$$\displaystyle \frac{{2\pi }}{3}<\theta <\frac{{4\pi }}{3}$$

 r θ        ° 6 0         0 2 $$\displaystyle \frac{\pi }{2}$$       90 0 $$\displaystyle \frac{{2\pi }}{3}$$    120 –2 $$\pi$$      180 0 $$\displaystyle \frac{{4\pi }}{3}$$    240 2 $$\displaystyle \frac{{3\pi }}{2}$$    270
The leftmost petal for $$\cos 2\theta$$ is drawn between what two values of $$\theta$$? In what order are the petals drawn?

Solution:

With this problem, we can create the following t-chart and see the sequence of petals being drawn. To determine which values to use in the t-chart, set $$\cos 2\theta$$ to 0 to see what $$\theta$$ values are between each petal:

$$\displaystyle \begin{array}{c}0=\cos 2\theta \\0=2{{\cos }^{2}}\theta -1\,\,\,\text{(identity)}\,\,\,\,\,\end{array}$$

$$\displaystyle \text{cos}\theta \,\,\text{=}\,\,\pm \sqrt{{\frac{1}{2}}}=\pm \frac{{\sqrt{2}}}{2}$$

$$\displaystyle \theta =\frac{\pi }{4};\,\,\,\,\,\theta =\frac{{3\pi }}{4};\,\,\,\,\,\theta =\frac{{5\pi }}{4};\,\,\,\,\,\theta =\frac{{7\pi }}{4}$$

The leftmost (3rd) petal is formed when

$$\displaystyle \frac{{3\pi }}{4}<\theta <\frac{{5\pi }}{4}$$

 r θ        ° 1 0         0 0 $$\displaystyle \frac{\pi }{4}$$        45 –1 $$\displaystyle \frac{\pi }{2}$$        90 0 $$\displaystyle \frac{{3\pi }}{4}$$     135 1 $$\displaystyle \pi$$        180 0 $$\displaystyle \frac{{5\pi }}{4}$$     225 –1 $$\displaystyle \frac{{3\pi }}{2}$$     270 0 $$\displaystyle \frac{{7\pi }}{4}$$     315

Learn these rules, and practice, practice, practice!

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