1. Coordinate Systems

c. Polar Coordinates - 2D

4. 2D Polar Plots

Before you can graph polar curves, you need to understand the graphs of the basic trig functions and how to shift them. These are covered in the review chapter on Trigonometry.

You can also review graphs of the basic trig functions and how to shift them by using 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

The rectangular graph of an equation is the set of all points whose rectangular coordinates \((x,y)\) satisfy the equation. For example, the graphs of the circle \(x^2+y^2=4\) and the cusp \(x^2=y^3\) are

Circle
eg_circle
Cusp
eg_cusp

Similarly, the polar graph of a polar equation is the set of all points for which some pair of polar coordinates \((r,\theta)\) satisfy the equation. The simplest examples make one coordinate constant. For example, the graphs of the circle \(r=2\) and the ray (or line, if \(r\) can be negative) \(\theta=\dfrac{\pi}{6}\) are

Circle 
Ray 

In the graph of \(\theta=\dfrac{\pi}{6}\), if we require \(r \ge 0\), then the graph is only the blue ray. If we allow \(r \lt 0\), then the graph is the whole red and blue line.

We want to be able to graph more complicated polar equations. So let's do it by examples:

We first graph the part of the curve with \(\theta \ge 0\). We start at \(r=\theta=0\) which is the origin. As \(r=\theta\) gets larger, we travel around counterclockwise and the radius also gets larger. So we trace out a spiral as shown in the first plot.

eg_spiral_pos_anim

As \(r=\theta\) gets negative, we travel around clockwise and since \(r\) is also negative, we measure \(|r|\) backwards, which traces out a spiral which is the mirror image through the \(y\)-axis of the first half as shown in the second plot.

eg_spiral_all_anim

If we restrict to the part of the curve, \(r=\theta\), with \(-\,\dfrac{3\pi}{2} \le \theta \le \dfrac{3\pi}{2}\), we get a heart shape.

Happy Valentine's Day!

rem_valentine

Plot the curve \(r=|\theta|\).

ex_spiral_abs_anim

Start with the graph of \(r=\theta\) for \(\theta \ge 0\).

eg_spiral_pos_anim

Then take its mirror image through \(\theta=0\) which is the positive \(x\)-axis.

ex_spiral_abs_anim

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

eg_r=costheta_rect/>

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

eg_r=1+costheta_rect

We first plot the polar points for these interesting points:

\(\theta=\) \(0\) \(\dfrac{\pi}{2}\) \(\pi\) \(\dfrac{3\pi}{2}\) \(2\pi\)
\(r=\) \(2\) \(1\) \(0\) \(1\) \(2\)
eg_r=1+costheta_pts

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:

eg_r=1+costheta_I

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

eg_r=1+costheta_II

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 from this that the second half of the cardioid is a mirror image of the first. This finishes the plot:

eg_r=1+costheta_IV
Cardioid:   \(r=1+\cos\theta\)

Because it is shaped like a heart!

Watch the animation of the cardioid being drawn:

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

Cardioid:   \(r=1+\sin\theta\)

ex_r=1+sintheta_polar
ex_r=1+sintheta_rect
Cardioid:   \(r=1+\sin\theta\)

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 first graphing \(r\) as a function of \(\theta\) as rectangular coordinates:

eg_r=1+2costheta_rect

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 our list of important points.

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

\(\theta=\) \(0\) \(\dfrac{\pi}{2}\) \(\dfrac{2\pi}{3}\) \(\pi\) \(\dfrac{4\pi}{3}\) \(\dfrac{3\pi}{2}\) \(2\pi\)
\(r=\) \(3\) \(1\) \(0\) \(-1\) \(0\) \(1\) \(3\)
eg_r=1+2costheta_pts

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 \(\theta \lt \theta \lt \pi\) with the curve in red when \(r\) is negative:

eg_r=1+2costheta_II

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

eg_r=1+2costheta_IV
Limaçon: \(r=1+2\cos\theta\)

Watch the animation of the limaçon being drawn.

The word limaçon is French for snail, but you can also remember that it looks like a lima bean.

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

Plot the limaçon \(r=1+2\sin\theta\)

Limaçon   \(r=1+2\sin\theta\)

ex_r=1+2sintheta_polar
ex_r=1+2sintheta_rect
Limaçon   \(r=1+2\sin\theta\)

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 \(r=\dfrac{1+\cos 2\theta}{2}\) makes it easier to plot. Here is a table of important values and a rectangular plot:

\(\theta=\) \(0\) \(\dfrac{\pi}{4}\) \(\dfrac{\pi}{2}\) \(\dfrac{3\pi}{4}\) \(\pi\) \(\dfrac{5\pi}{4}\) \(\dfrac{3\pi}{2}\) \(\dfrac{7\pi}{4}\) \(2 \pi\)
\(r=\) \(1\) \(0.5\) \(0\) \(0.5\) \(1\) \(0.5\) \(0\) \(0.5\) \(1\)
r=cos^2theta_rect

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

r=cos^2theta_polar
Infinity:   \(r=\cos^2\theta\)

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

Figure Eight:   \(r=\sin^2\theta\)

r=sin^2theta_polar
r=sin^2theta_rect
Figure Eight:   \(r=\sin^2\theta\)

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

Notice that \(r=0\) 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:

\(\theta=\) \(0\) \(\dfrac{\pi}{4}\) \(\dfrac{\pi}{2}\) \(\dfrac{3 \pi}{4}\) \(\pi\) \(\dfrac{5 \pi}{4}\) \(\dfrac{3\pi}{2}\) \(\dfrac{7 \pi}{4}\) \(2 \pi\)
\(r=\) \(1\) \(0\) \(-1\) \(0\) \(1\) \(0\) \(-1\) \(0\) \(1\)
r=cos2theta_rect

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

r=cos2theta_polar
\(4\)-Leaf Rose:   \(r=\cos(2\theta)\)

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 \(r=\cos{3 \theta}\)

3-Leaf Rose:   \(r=\cos(3\theta)\)

ex_r=cos3theta_polar

Here is a table of values and the rectangular plot:

\(\theta =\) \(0\) \(\dfrac{\pi }{6}\) \(\dfrac{\pi }{3}\) \(\dfrac{\pi }{2}\) \(\dfrac{2\pi }{3}\) \(\dfrac{5\pi }{6}\) \(\pi \)
\(r=\) \(1\) \(0\) \(-1\) \(0\) \(1\) \(0\) \(-1\)
ex_r=cos3theta_rect
3-Leaf Rose:   \(r=\cos(3\theta)\)

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, \(n\) with a positive radius and \(n\) with a negative radius.

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

ex_r=cos4theta_polar
8-Leaf Rose:   \(r=\cos(4\theta)\)

The function \(\cos(n \theta)\) is \(1\) when the argument is a multiple of \(2\pi\), i.e. \(n \theta=2k\pi\). On the interval \(0 \le \theta \le 2\pi\), this is at \(\theta=\dfrac{2k\pi}{n}\) for \(k=1,2,\cdots,n\). These are the angles at the tip of each "positive" leaf. Notice that these are the even multiples of \(\dfrac{\pi}{n}\).

Similarly, on the interval \(0 \le \theta \le 2\pi\), we have \(\cos(n \theta)=-1\) when \(\theta=\dfrac{(2k+1)\pi}{n}\) for \(k=1,2,\cdots,n\). These are the angles at the tip of each "negative" leaf. However, since the radius is negative, the leaf actually occurs at: \[ \theta=\dfrac{(2k+1)\pi}{n}+\pi =\dfrac{(2k+1+n)\pi}{n} \] Since \(n\) is even, these are the odd multiples of \(\dfrac{\pi}{n}\).

All together, there are \(2n\) leaves \(n\) positive and \(n\) negative. For further evidence, the polar plot of \(r=\cos 4\theta \) has \(8\) leaves:

ex_r=cos4theta_rect
8-Leaf Rose:   \(r=\cos(4\theta)\)

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. The \(n\) negative leaves coincide with the \(n\) positive leaves.

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

ex_r=cos5theta_polar
5-Leaf Rose:   \(r=\cos(5\theta)\)

On the interval \(0 \le \theta \le 2\pi\), the function \(\cos(n \theta)\) is \(1\) when \(\theta=\dfrac{2k\pi}{n}\) for \(k=1,2,\cdots,n\). These are the angles at the tip of each "positive" leaf. Notice that these are the even multiples of \(\dfrac{\pi}{n}\).

Similarly, on the interval \(0 \le \theta \le 2\pi\), we have \(\cos(n \theta)=-1\) when \(\theta=\dfrac{(2k+1)\pi}{n}\) for \(k=1,2,\cdots,n\). These are the angles at the tip of each "negative" leaf. However, since the radius is negative, the leaf actually occurs at: \[ \theta=\dfrac{(2k+1)\pi}{n}+\pi =\dfrac{(2k+1+n)\pi}{n} \] Since \(n\) is odd, these are the even multiples of \(\dfrac{\pi}{n}\) which are at the same angles as the positive leaves.

All together, there are only \(n\) leaves. The \(n\) negative leaves coincide with the \(n\) positive leaves. For further evidence,the polar plot of \(r=\cos 5\theta \) has \(5\) leaves. The \(5\) negative leaves overwrite the \(5\) positive leaves.

ex_r=cos5theta_rect
5-Leaf Rose:   \(r=\cos(5\theta)\)

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