15. Polar Coordinates

b. Graphs of Polar Equations

2. Plotting Polar Equations

Polar equations often contain trigonometric functions. So you need to understand the graphs of the basic trig functions and how to shift and scale them. You can review by practicing with the following Maplets (requires Maple on the computer where this is executed):

Basic 6 Trigonometric Functions   Rate It

Shifting Trigonometric Functions   Rate It

Properties of Sine and Cosine Curves   Rate It

We have seen the graphs of a few simple polar equations. We want to be able to graph more complicated polar equations. So we'll do it by examples. The procedure is to make a table of points and/or make a rectangular plot of the polar equation and use them to draw the polar plot.

First think of \(r\) and \(\theta\) as rectangular coordinates. The graph of \(r=\cos\theta\) is

This plot shows the graph of r = cosine theta in rectangular coordinates, with theta on the horizontal axis from 0 to 2 pi and r on the vertical axis from -1 to 1. The curve starts at r = 1 when theta = 0, decreases to r = 0 at theta = pi over 2, reaches a minimum of r = -1 at theta = pi, and returns to r = 1 at theta = 2 pi, showing the cosine shape.

We lift this up by \(1\) so that the graph of \(r=1+\cos\theta\) is

This plot shows with the same plot as above but now the cosine curve is shifted upward by 1. The graph shows theta from 0 to 2 pi and r from 0 to 2. The curve reaches a maximum of 2 at theta =0 and 2 pi and a minimum of r = 0 at theta = pi.

We first plot the polar points for these interesting points:

\(\theta=\) \(0\) \(\dfrac{\pi}{2}\) \(\pi\) \(\dfrac{3\pi}{2}\) \(2\pi\)
\(r=\) \(2\) \(1\) \(0\) \(1\) \(2\)
This plot shows points where the x values are 2, 0, 0, 0, and 2, and the corresponding y values are 0, 1, 0, -1, and 0, respectively.

We now connect the dots in each quadrant. From the rectangular plot we see that as \(\theta\) increases from \(0\) to \(\dfrac{\pi}{2}\), the value of r decreases from \(2\) to \(1\). We add this to the plot:

This plot starts with the same points as above, but now they are connected smoothly to form part of the polar curve r = 1 plus cosine theta. As theta increases from 0 to pi over 2, the curve moves from x = 2 and y = 0 to x = 0 and y = 1, creating an arc in the first quadrant.

On the interval \(\dfrac{\pi}{2} \lt \theta \lt \pi\) we see that \(r\) decreases from \(1\) to \(0\). We add this section next:

This plot starts with the same curve as above, but now continues tracing the polar curve. As theta increases from pi over 2 to pi, the curve moves from x = 0 and y = 1 toward the origin, adding the another arc of the cardioid in the second quadrant.

Finally, we see that as \(\theta\) increases from \(\pi\) to \(\dfrac{3\pi}{2}\) to \(2\pi\), the value of \(r\) reverses and goes from \(0\) to \(1\) and finally back to \(2\). We conclude that the second half of the cardioid is a mirror image of the first. This finishes the plot:

This plot shows the same curve as above, but now includes the same traced section on the bottom half of the plane. The curve is symmetric about the x axis, forming a heart shaped polar curve with the cusp on the left.
\(r=1+\cos\theta\)

Because it is shaped like a heart!

Watch an animation of the cardioid being drawn below.

\(r=1+\cos\theta\)

Plot the cardioid \(r=1+\sin\theta\).

This plot shows the complete cardioid r = 1 + sine theta. The curve is a heart shape, is symmetric about the y axis with a cusp at the bottom at the origin.
\(r=1+\sin\theta\)
This plot shows the graph of r = 1 + sine theta with theta on the horizontal axis and r on the vertical axis. The curve starts at r = 1 when theta = 0, increases to a maximum of r = 2 at theta = pi over 2, decreases to a minimum of r = 0 at theta = 3 pi over 2, and then increases back to r = 1 at theta = 2 pi.

Frequently when plotting polar functions we allow \(r\) to be negative by measuring backward. In particular, if the direction is \(\theta\), then

We follow the same procedure as last time by graphing \(r\) and \(\theta\) as rectangular coordinates:

This plot shows the graph of r = 1 + 2 cosine theta with theta horizontal and r vertical. The curve starts at r = 3 when theta = 0, decreases to a minimum of r = -1 at theta = pi, and increases back to r = 3 at theta = 2 pi.

Notice that \(r=1+2\cos\theta=0\) when \(\cos\theta=-\,\dfrac{1}{2}\) or \(\theta=\dfrac{2\pi}{3}\) and \(\dfrac{4\pi}{3}\). We add these to the axes for our list of important points.

We plot the polar points at the important values of \(\theta\):

\(\theta=\) \(0\) \(\dfrac{\pi}{2}\) \(\dfrac{2\pi}{3}\) \(\pi\) \(\dfrac{4\pi}{3}\) \(\dfrac{3\pi}{2}\) \(2\pi\)
\(r=\) \(3\) \(1\) \(0\) \(-1\) \(0\) \(1\) \(3\)
This plot shows selected polar points for r = 1 + 2 cosine theta with x values 0, 0, 0, 1, and 3, and the corresponding y values 1, 0, -1, 0, and 0, respectively.

Notice that the origin occurs twice when \(\theta=\dfrac{2\pi}{3}\) and \(\dfrac{4\pi}{3}\). Also when \(\theta=\pi\), which is the negative \(x\)-axis, \(r=-1\) which we measure backwards along the positive \(x\)-axis. Be careful to check that you know how each of these points is plotted!

Next, we connect the dots by observing that \(r\) decreases from \(3\) to \(0\) on the interval \(0 \lt \theta \lt \dfrac{2\pi}{3}\). Also notice that \(r\) is negative on the interval \(\dfrac{2\pi}{3} \lt \theta \lt \dfrac{4\pi}{3}\), so we measure \(r\) opposite to the direction of \(\theta\) on this interval. The result is plotted for \(0 \lt \theta \lt \pi\):

This plot starts with the same points as above, but now they are connected to form half of the limacon r = 1 + 2 cosine theta. Starting from x = 3 and y = 0, the curve is traced counterclockwise as theta goes from 0 to 2 pi over 3 and r decreases from 3 to 0, forming an an arc in the upper half plane. Then theta continues counterclockwise from 2 pi over 3 to pi as r increases from 0 to 1 but drawn backwards, so it is in the fourth quadrant.

Finally, we note again that the second half is a mirror image of the first. Thus by symmetry our final plot is:

This plot shows the same curve as above, but now includes its reflection through the x axis. It forms a heart-shaped polar curve with a cups on the left but with an additional inner loop inside the heart.
\(r=1+2\cos\theta\)

The word limaçon is French for snail, but you can also remember that it looks like a lima bean. View an animation of this limaçon being plotted below.

\(r=1+\cos2\theta\)

Notice the radial line is blue when the radius is positive and it is red when the radius is negative.

Plot the limaçon \(r=1+2\sin\theta\)

This plot shows the complete limacon r = 1 + 2 sine theta. The curve is symmetric about the vertical axis and consists of a large heart shaped outer loop with the cusp at the bottom with a smaller inner loop above the x axis. The curve touches the x-axis at the origin and reaches a maximum height near y = 3.
\(r=1+2\sin\theta\)
This plot shows the graph of r = 1 + 2 sine theta on theta and r axes. The curve starts at r = 1 when theta = 0, increases to a maximum of r = 3 at theta = pi over 2, decreases to a minimum of r = -1 at theta = 3 pi over 2, and returns to r = 1 at theta = 2 pi.

When there is a multiple of \(\theta\) inside the sine or cosine, the horizontal scale of the rectangular plot stretches or shrinks, changing the period.

From the equation \(r=\cos^2\theta\), we find \(r=0\) when \(\theta=\dfrac{\pi}{2}\) or \(\dfrac{3\pi}{2}\) and \(r=1\) when \(\theta=0\) or \(\pi\). Also notice there are no negative values of \(r\). The equation \(r=\dfrac{1+\cos2\theta}{2}\) makes it easier to plot. Here is a table of important values and a rectangular plot:

\(\theta=\) \(0\) \(\dfrac{\pi}{4}\) \(\dfrac{\pi}{2}\) \(\dfrac{3\pi}{4}\) \(\pi\) \(\dfrac{5\pi}{4}\) \(\dfrac{3\pi}{2}\) \(\dfrac{7\pi}{4}\) \(2 \pi\)
\(r=\) \(1\) \(0.5\) \(0\) \(0.5\) \(1\) \(0.5\) \(0\) \(0.5\) \(1\)
This plot shows the graph of r = cosine squared theta with theta horizontal and r vertical. The curve oscillates between r = 0 and 1, reaching maximum values at theta = 0, pi, and 2 pi, and minimum values at theta = pi over 2 and 3 pi over 2.

Notice the plot repeats with a period of \(\pi\). Here is the polar plot:

This plot shows the complete polar curve r = cosine squared theta on x and y axes. The curve consists of two symmetric loops, on the left and right side of the origin, touching at the origin. The shape is symmetric about both x and y axes, forming a sideways figure eight pattern.
\(r=\cos^2\theta\)

Plot the curve \(r=\sin^2\theta\).

This plot shows the complete polar curve r = sine squared theta. The curve consists of two symmetric loops, one above and one below the x axis, touching at the origin, forming a figure eight pattern.
\(r=\sin^2\theta\)
This plot shows the graph of r = sine squared theta with theta horizontal and r vertical. The curve oscillates between r = 0 and 1, reaching maximum values at theta = pi over 2 and 3 pi over 2, and minimum values at theta = 0, pi and 2 pi.

Now let's combine a change of period with negative values of \(r\).

Notice that \(r=0\) when \(2\theta=\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}\) or \(\dfrac{7\pi}{2}\) or when \(\theta=\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}\) or \(\dfrac{7\pi}{4}\). Also \(r=1\) when \(\theta=0\) or \(\pi\) and \(r=-1\) when \(\theta=\dfrac{\pi}{2}\) or \(\dfrac{3\pi}{2}\). Also note that the plot repeats with a period of \(\pi\). Here is a table of important values and a rectangular plot:

\(\theta=\) \(0\) \(\dfrac{\pi}{4}\) \(\dfrac{\pi}{2}\) \(\dfrac{3 \pi}{4}\) \(\pi\) \(\dfrac{5 \pi}{4}\) \(\dfrac{3\pi}{2}\) \(\dfrac{7 \pi}{4}\) \(2 \pi\)
\(r=\) \(1\) \(0\) \(-1\) \(0\) \(1\) \(0\) \(-1\) \(0\) \(1\)
This plot shows the graph of r = cosine of 2 theta with theta horizontal and r vertical. The curve oscillates between r = 1 and -1, with maxima at theta = 0, pi and 2 pi and minima at theta = pi over 2 and 3 pi over 2.

Also notice that the plot has \(2\) positive bumps and \(2\) negative bumps in the interval \([0,2\pi]\). Here is the polar plot:

This plot shows the complete polar curve r = cosine of 2 theta. The curve forms four symmetric petals centered at the origin, with one petal on each side of each positive and negative axis. The shape is symmetric about both axes.
\(r=\cos(2\theta)\)

In the animation, the radial line is blue when \(r\) is positive and red when \(r\) is negative. They trace out different leaves.

Plot the 3-leaf rose \(r=\cos{3\theta}\)

This plot shows the complete polar curve r = cosine of 3 theta, which looks like a 3 leaf clover with one leaf (petal) along the positive x axis and one leaf in each of the second and third quadrants. The petals are equally spaced around the origin, and symmetric across the x axis.
\(r=\cos(3\theta)\)

Here is a table of values and the rectangular plot:

\(\theta=\) \(0\) \(\dfrac{\pi }{6}\) \(\dfrac{\pi }{3}\) \(\dfrac{\pi }{2}\) \(\dfrac{2\pi }{3}\) \(\dfrac{5\pi }{6}\) \(\pi\)
\(r=\) \(1\) \(0\) \(-1\) \(0\) \(1\) \(0\) \(-1\)
This plot shows the graph of r = cosine of 3 theta with theta horizontal and r vertical. The curve oscillates between r = 1 and r = -1, completing three full cycles over the interval from 0 to 2 pi.

There are \(3\) positive bumps and \(3\) negative bumps in the interval \([0,2\pi]\). Here is the polar plot:

In the animation, the radial line is blue when \(r\) is positive and red when \(r\) is negative. Notice that each leaf of the rose is traced out twice, once when \(r\) is positive and once when \(r\) is negative.

How many leaves are there on the rose \(r=\cos(n\theta)\) if \(n\) is even? Why?

The rose \(r=\cos(n\theta)\) with even \(n\) has \(2n\) leaves.

For example, the rose \(r=\cos(4\theta)\) has \(8\) leaves:

This plot shows the complete polar curve r = cosine of 4 theta. The curve forms eight symmetric petals centered at the origin, with petals evenly spaced around the circle with one patel along each axis and one petal along each 45 degree line. The shape is symmetric across both axes.
\(r=\cos(4\theta)\)

Looking at the above example, we see that when \(n\) is even the polar plot has \(2n\) leaves because \(r\) becomes negative at \(r=\dfrac{\pi}{2n}\) and traces a leaf in the opposite quadrant before becoming positive again at \(r=\dfrac{3\pi}{2n}\), and this repeats for \(n\) cycles. So \(r\) traces \(2n\) leaves from \(\theta=0\) to \(\theta=2\pi\). For further evidence, the polar plot of \(r=\cos(4\theta)\) is below, which has \(8\) leaves:

This plot shows the graph of r = cosine of 4 theta with theta horizontal and r vertical. The curve oscillates between r = 1 and r = -1, completing four full cycles over the interval from 0 to 2 pi.

How many leaves are there on the rose \(r=\cos(n\theta)\) if \(n\) is odd? Why is this not the same answer as in the even case?

The rose \(r=\cos(n\theta)\) with odd \(n\) has \(n\) leaves.

For example, the rose \(r=\cos(5\theta)\) has \(5\) leaves:

This plot shows the complete polar curve r = cosine of 5 theta. The curve forms five symmetric petals centered at the origin, evenly spaced around the circle with one along the positive x axis. The shape is symmetric across the x axis.
\(r=\cos(5\theta)\)

When \(n\) is odd the polar plot has \(n\) leaves because when \(r\) becomes negative it overwrites a piece of the graph in the opposite quadrant. For example, in the rectangular plot of \(r=\cos(5\theta)\) below, there are \(5\) positive bumps and \(5\) negative bumps. Each time \(r\) becomes negative, the graph retraces a leaf in the opposite quadrant. So the polar plot has only \(5\) leaves.

This plot shows the graph of r = cosine of 5 theta with theta horizontal and r vertical. The curve oscillates between r = 1 and r = -1, completing five full cycles over the interval from 0 to 2 pi.

You can review the graphs and equations of polar curves and practice identifying polar curves from their graphs by using the following Maplets (requires Maple on the computer where this is executed):

Basic 14 Polar Curves   Rate It

Identify a Polar Curve from its Graph   Rate It

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Supported in part by NSF Grant #1123255

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