After setting the calculator in Polar mode, we enter the equation [latex]r=2\theta[/latex] and enter the window settings

[latex]\theta\text{min}=0~~~~~~~~~~\theta\text{max}=6\pi~~~~\theta\text{step}=0.1\\ \text{Xmin}=-25~~~~\text{Xmax}=25~~~~\text{Xstep}=5\\ \text{Ymin}=-25~~~~\text{Ymax}=25~~~~~~\text{Yscl}=5[/latex]

We then press Zoom 5 to set a square window. The graph is shown at right.

Studying a table of values can help us understand the shape of the graph.

As [latex]\theta[/latex] increases, [latex]r[/latex] increases also, at a constant rate. We wind our way around the pole, steadily increasing our distance from the pole as we go. We spiral outward, tracing the graph shown above.

We’ll graph the equation in stages in order to see how the graph is traced out. Begin with the window settings

[latex]\theta\text{min}=0~~~~~~~~~~~\theta\text{max}=\dfrac{\pi}{3}~~~~~~\theta\text{step}=0.5\\ \text{Xmin}=-4.5~~~~\text{Xmax}=4.5~~~~~~\text{Xstep}=1\\ \text{Ymin}=-3~~~~~~~\text{Ymax}=3~~~~~~~~~~~~\text{Yscl}=1[/latex]

Watch as your calculator produces the graph shown in figure (a).

Observe that as [latex]\theta[/latex] increases from [latex]0[/latex] to [latex]\dfrac{\pi}{3}\text{,}[/latex] [latex]r[/latex] first increases, reaching its maximum value of 2 at [latex]\theta = \dfrac{\pi}{6}[/latex] and then decreases back to 0. You can verify the values in the table below, which shows the points at multiples of [latex]\dfrac{\pi}{18}\text{.}[/latex] These points create the first loop of the graph.

Next, change [latex]\theta\text{max}[/latex] to [latex]\dfrac{2\pi}{3}\text{,}[/latex] and graph again. This time the calculator traces out two loops, as shown in figure (b). For [latex]\theta[/latex] between [latex]\dfrac{\pi}{3}[/latex] and [latex]\dfrac{2\pi}{3}\text{,}[/latex] [latex]r[/latex] is negative, so the second loop lies in the third and fourth quadrants.

Finally, change [latex]\theta\text{max}[/latex] to [latex]\pi\text{,}[/latex] and graph again. For [latex]\theta[/latex] between [latex]\dfrac{2\pi}{3}[/latex] and [latex]\pi\text{,}[/latex] the graph traces out a third loop, as shown in figure (c). For [latex]\theta[/latex] between [latex]\pi[/latex] and [latex]2\pi\text{,}[/latex] the entire graph is traced a second time. The finished graph is a rose with 3 petals of length 2.

Referring to the Catalog of Polar Graphs, we see that the graph of this equation is a rose, with petal length [latex]a=3[/latex] and four petals, because [latex]2n=4\text{.}[/latex] If we can locate the tips of the petals, we can use them as guide points to sketch the graph.
Now, the points at the tips of the petals have [latex]r=3\text{,}[/latex] so we substitute [latex]r=3[/latex] into the equation of the rose to find the values of [latex]\theta[/latex] at those points.

[latex]3 =3\sin 2\theta \qquad \text{Divide both sides by 3.}\\ 1 = \sin 2\theta \qquad \text{Apply the arcsine function to both sides.}\\ \dfrac{\pi}{2} = 2\theta \qquad \text{Divide both sides by 2.}\\ \dfrac{\pi}{4} = \theta[/latex]

Thus, one of the petal tips is located at [latex]\theta = \dfrac{\pi}{4}\text{.}[/latex] Because the 4 petals are evenly spaced around the pole, the angle between the petals is [latex]\dfrac{2\pi}{4} = \dfrac{\pi}{2}\text{,}[/latex] and the other petal tips occur at

[latex]\theta = \dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3\pi}{4}\\ \theta = \dfrac{\pi}{4} + 2\left(\dfrac{\pi}{2}\right) = \dfrac{5\pi}{4}\\ \theta = \dfrac{\pi}{4} + 3\left(\dfrac{\pi}{2}\right) = \dfrac{7\pi}{4}[/latex]

We plot the tips of the petals as guide points, at [latex]\left(3, \dfrac{\pi}{4}\right)\text{,}[/latex][latex]~ \left(3, \dfrac{3\pi}{4}\right)\text{,}[/latex] [latex]~ \left(3, \dfrac{5\pi}{4}\right)[/latex]and [latex](3, \dfrac{7\pi}{4})\text{.}[/latex] Now we can sketch a rose with 4 petals of length 3, as shown at right.

This is the equation for a limaçon, with [latex]a=3[/latex] and [latex]b=2\text{.}[/latex] Because [latex]a \gt b\text{,}[/latex] the limaçon will have a dent, like a lima bean, rather than a loop. As guide points, we locate the points at the four quadrantal angles.

We plot the guide points and connect them in order with a smooth curve, as shown below. Note that this limaçon involving cosine is symmetric about the [latex]x[/latex]-axis; limaçons involving sine are symmetric about the [latex]y[/latex]-axis.

1. The graph is a circle centered at [latex](-2,0)[/latex] and with radius 2, so we choose the equation [latex]r=2a\cos \theta\text{,}[/latex] with [latex]a=-2\text{.}[/latex] Thus, [latex]r=-4\cos \theta\text{.}[/latex]
2. The graph is a cardioid with its axis of symmetry on the [latex]x[/latex]-axis, and the “bottom” of the heart points in the positive[latex]x[/latex]direction, so its equation has the form [latex]r=a+a\cos \theta\text{.}[/latex] At [latex]\theta=0,~ r=2\text{,}[/latex] so we can solve for [latex]a\text{:}[/latex][latex]2 = a + a\cos 0\\ 2 = a + a(1) =2a[/latex]Thus, [latex]a=1[/latex] and we choose the equation [latex]r=1+\cos \theta\text{.}[/latex]

We equate the two expressions for [latex]r[/latex] and solve for [latex]\theta\text{.}[/latex]

[latex]r=2+2\sin \theta = r=2+2\cos \theta \\ \sin \theta = \cos \theta \qquad \text{Divide both sides by} \qquad \cos\theta. \\ \tan \theta = 1 \qquad \text{Solve for} \qquad \theta.\\ \theta = \dfrac{\pi}{4},~\dfrac{5\pi}{4}[/latex]

We evaluate either expression for [latex]t[/latex] to find the other coordinate of the intersection point.

[latex]\begin{gather*} \text{when}~~~~\theta = \dfrac{\pi}{4},~~~~r = 2 + 2\sin \dfrac{\pi}{4} = 2 + \left(\dfrac{1}{\sqrt{2}}\right) = 2 + \sqrt{2}\\ \text{when}~~~~\theta = \dfrac{5\pi}{4},~~~~r = 2 + 2\sin \dfrac{5\pi}{4} = 2 + \left(\dfrac{-1}{\sqrt{2}}\right) = 2 - \sqrt{2} \end{gather*} Thus, two of the intersection points are [latex]\left(2 + \sqrt{2}, \dfrac{\pi}{4}\right)[/latex] and [latex]\left(2 - \sqrt{2}, \dfrac{5\pi}{4}\right)\text{,}[/latex] as shown. However, you can see in the figure that the graphs also appear to intersect at the pole. To verify that the pole indeed lies on both graphs, we can solve for [latex]\theta[/latex] in each equation when [latex]r=0\text{.}[/latex]

However, you can see in the figure that the graphs also appear to intersect at the pole. To verify that the pole indeed lies on both graphs, we can solve for [latex]\theta[/latex] in each equation when [latex]r={0}\text{.}[/latex]

[latex]{0} =2+2\sin \theta \qquad \qquad {0} =2+2\cos \theta \qquad \qquad\\ -1 =\sin \theta -1 = \cos \theta \\ \theta = \dfrac{3\pi}{2} \theta =\pi[/latex]

Both points, [latex]\left(0,\dfrac{3\pi}{2}\right)[/latex] and [latex](0,\pi)\text{,}[/latex] represent the pole. Thus, the graphs intersect at three points: [latex]\left(2 + \sqrt{2}, \dfrac{\pi}{4}\right)\text{,}[/latex] [latex]\left(2 - \sqrt{2}, \dfrac{5\pi}{4}\right)\text{,}[/latex] and the pole.