# 10.4 Polar coordinates: graphs  (Page 4/16)

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## Formulas for a cardioid

The formulas that produce the graphs of a cardioid    are given by $\text{\hspace{0.17em}}r=a±b\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a±b\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b>0,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\frac{a}{b}=1.\text{\hspace{0.17em}}$ The cardioid graph passes through the pole, as we can see in [link] .

Given the polar equation of a cardioid, sketch its graph.

1. Check equation for the three types of symmetry.
2. Find the zeros. Set $\text{\hspace{0.17em}}r=0.$
3. Find the maximum value of the equation according to the maximum value of the trigonometric expression.
4. Make a table of values for $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta .$
5. Plot the points and sketch the graph.

## Sketching the graph of a cardioid

Sketch the graph of $\text{\hspace{0.17em}}r=2+2\mathrm{cos}\text{\hspace{0.17em}}\theta .$

First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}$ we have $\text{\hspace{0.17em}}\theta =\pi +2k\pi .\text{\hspace{0.17em}}$ The zero of the equation is located at $\text{\hspace{0.17em}}\left(0,\pi \right).\text{\hspace{0.17em}}$ The graph passes through this point.

The maximum value of $\text{\hspace{0.17em}}r=2+2\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ occurs when $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is a maximum, which is when $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =1\text{\hspace{0.17em}}$ or when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$ Substitute $\text{\hspace{0.17em}}\theta =0\text{\hspace{0.17em}}$ into the equation, and solve for $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$

$\begin{array}{l}\begin{array}{l}\\ r=2+2\mathrm{cos}\left(0\right)\end{array}\hfill \\ r=2+2\left(1\right)=4\hfill \end{array}$

The point $\text{\hspace{0.17em}}\left(4,0\right)\text{\hspace{0.17em}}$ is the maximum value on the graph.

We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval $\text{\hspace{0.17em}}\left[0,\text{\hspace{0.17em}}\pi \right].\text{\hspace{0.17em}}$ The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in [link] , and then we plot the points and draw the graph. See [link] .

 $\theta$ $0$ $\frac{\pi }{4}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\pi$ $r$ 4 3.41 2 1 0

## Investigating limaçons

The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled limaçons when $\text{\hspace{0.17em}}1<\frac{a}{b}<2\text{\hspace{0.17em}}$ and convex limaçons when $\text{\hspace{0.17em}}\frac{a}{b}\ge 2.\text{\hspace{0.17em}}$

## Formulas for one-loop limaçons

The formulas that produce the graph of a dimpled one-loop limaçon are given by $\text{\hspace{0.17em}}r=a±b\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a±b\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ where All four graphs are shown in [link] .

Given a polar equation for a one-loop limaçon, sketch the graph.

1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted.
2. Find the zeros.
3. Find the maximum values according to the trigonometric expression.
4. Make a table.
5. Plot the points and sketch the graph.

## Sketching the graph of a one-loop limaçon

Graph the equation $\text{\hspace{0.17em}}r=4-3\mathrm{sin}\text{\hspace{0.17em}}\theta .$

First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},\text{\hspace{0.17em}}$ yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality.

Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting $\text{\hspace{0.17em}}r=0\text{\hspace{0.17em}}$ results in $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ being undefined. What does this mean? How could $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ be undefined? The angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is undefined for any value of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta >1.\text{\hspace{0.17em}}$ Therefore, $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is undefined because there is no value of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta >1.\text{\hspace{0.17em}}$ Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating $r$ when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$

$\begin{array}{l}r\left(0\right)=4-3\mathrm{sin}\left(0\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=4-3\cdot 0=4\hfill \end{array}$

So, there is at least one polar axis intercept at $\text{\hspace{0.17em}}\left(4,0\right).$

Next, as the maximum value of the sine function is 1 when $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},\text{\hspace{0.17em}}$ we will substitute $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}\text{\hspace{0.17em}}$ into the equation and solve for $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ Thus, $\text{\hspace{0.17em}}r=1.$

Make a table of the coordinates similar to [link] .

 $\theta$ $0$ $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi$ $r$ 4 2.5 1.4 1 1.4 2.5 4 5.5 6.6 7 6.6 5.5 4

The graph is shown in [link] .

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Manifoldee
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