Introduction

Welcome to topic 3.14!

In this topic, you’ll learn how to construct graphs of polar functions.

Understanding Polar Coordinates and Polar Functions

Last topic, you learned about Polar coordinates, which describe point by an $\theta$ angle and distance $r$. 

A polar function is an equation of the form: $r=f\left(\theta \right)$,

Where the input $\theta$ (an angle) maps the output $r$ (a radius). Thus the graph of $r=f\left(\theta \right)$ consists of input-output pairs (like a rectangular equation). $\theta$ behaves like the “independent variable” (like $x$ in the cartesian plane) and $r$ behaves like the “dependent variable” (like $y$). 

The most basic form of this is demonstrated in Archimedean Spiral, $r=\theta$

As the angle $\theta$ grows, the radius $r$ grows with it.

Other simple forms of Polar Functions include:

$r=c$, where $c$ is a constant

This means that every point is a fixed distance from the polar. It is simply a circle.

$\theta =c$, where $c$ is a constant

This means all points lie along a fixed angle. This is shown through a line that crosses through the pole.

How to Graph a Polar Function

Graphing $r=f\left(\theta \right)$ is much like plotting a Cartesian function: we pick values of $\theta$, compute $r$, plot those points in polar form, and connect them.

The most reliable step-by-step method is simply by creating a table of values:

First, choose a domain (or in this case, your range for $\theta$). Usually, we’d choose one full cycle (one full cycle is typically $2\pi$, but as you’ll see later on, some functions repeat on an interval of $\pi$).

If the question asks for only a specific interval, then you should pick said interval.

Then, pick sample angles. It is best to pick angles that you should have memorized from the unit circle, like $0,\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2},\text{etc.}$

If $f\left(\theta \right)$ involves $\cos \theta$, pick angles where $\cos$ has simple values like $0,\frac{\pi }{3},\frac{\pi }{2},\text{...}$

If $f\left(\theta \right)$ involves $\sin \theta$, pick angles where $\sin$ has simple values like $0,\frac{\pi }{6},\frac{\pi }{4},\text{...}$

For $\tan \theta$, it is often easier to compute $\tan$ from known $\sin$ and $\cos$ values, since $\tan$ has no standard values except $0,\frac{\pi }{4},\text{...}$

After you pick your sample angles (a.k.a., your domain), you need to compute $r$ at each angle.

To do this, simply take your function $r=f\left(\theta \right)$ and beging plugging in your chosen values of $\theta$.

Now that you have all your coordinates, you can begin plotting them.

I went over how to plot polar points in the last unit, so I won’t go too deep here.

For a quick summary, though, you first draw a ray at angle $\theta$ from the origin, then measure out a distance $\left|r\right|$ along that ray.

If $r>0$, you go out from the pole.

If $r<0$, you go in the opposite direction.

Finally, connect your points with a smooth curve. This is easier said than done, so hold on before you start answering any problems; we must first go through some of the common types of polar graphs, because without them, you will not be able to correctly connect certain points.

Common Types of Polar Curves

Polar graphs include several “special” curves/graphs with distinctive shapes. The following listed below are some of the most common types of Polar Functions.

Circles (Polar Form)

Now, you already know how to graph a circle with center at $\left(0,0\right)$ with a radius $r$. You can convert that into polar and you’d technically have a circle graph.

However, a circle in polar form can also appear as $r=a\cos \theta$ OR $r=a\sin \theta$, where $a\ne 0$. These equations both give circles of DIAMETER $a$.

The difference between them is their orientation.

$r=a\cos \theta$ creates a circle centered on the positive $x$-axis. 

$r=a\sin \theta$ creates a circle centered on the positive $y$-axis.

Both types pass through the pole when $r=0$.

They are also both symmetric about the polar axis (note, both the “$x$” and “$y$” axis are called the polar axis. $r=a\sin \theta$ is symmetric about the “$y$-axis” while $r=a\cos \theta$ is symmetric about the “$x$-axis”).

The graph above represents the function $r=5\cos \theta$. Let’s see why this graph looks like this.

Filling out a table, we see that:

$\begin{array}{ccc}\mathbf{\theta }& \cos (\mathbf{\theta })& 5\cos \mathbf{\theta }\\ 0& 1& 5\\ \frac{\pi }{4}& \frac{\sqrt{2}}{2}& \frac{5\sqrt{2}}{2}\\ \frac{\pi }{2}& 0& 0\\ \frac{3\pi }{4}& -\frac{\sqrt{2}}{2}& -\frac{5\sqrt{2}}{2}\\ \pi & -1& -5\\ \frac{5\pi }{4}& -\frac{\sqrt{2}}{2}& -\frac{5\sqrt{2}}{2}\\ \frac{3\pi }{2}& 0& 0\\ \frac{7\pi }{4}& \frac{\sqrt{2}}{2}& \frac{5\sqrt{2}}{2}\\ 2\pi & 1& 5\end{array}$

Note, only $0\le \theta \le \pi$ is needed because for $\theta >\pi$ the points repeat (since $\cos (\theta +\pi )=-\cos \theta$, the corresponding $r$ is negative and sends you back to previous points). Thus, this graph can be modeled by finding points only up to $\pi$.

Additionally, notice that our value $a=5$ does indeed represent the diameter.

In general, these circles “lie” on the polar axis. As for which polar axis, $\sin \theta$ lies on the “$y$-axis,” while $\cos \theta$ lies on the “$x$-axis.”

If $a>0$, the graph will be either above or to the right of the pole/origin on the polar axis.

If $a<0$, it will be below or to the left of the pole/origin on the polar axis.

Rose Curves

The above is a graph of $r=5\sin (2\theta )$. It is an example of a rose.

Rose curves have the form: $r=a\cos (n\theta )$ or $r=a\sin (n\theta )$

Where $a\ne 0$ and $n$ is any integer with the condition that $|n|\ne 1$, since if it’s equal to $1$, it just becomes a circle.

The graph produces a flower-like shape with “petals.”

The two key parameters are $a$, which controls the petal length, and $n$, which controls the number (and as a result, the spacing) of petals.

There are some general guidelines you can follow to simplify the process:

Number of petals:

If $n$ is odd, the rose has $|n|$ petals.

If $n$ is even, it has $2|n|$ petals.

If you want a cool little experiment, try putting values of $n$ that approach $2$ or $3$ or whatever value; you will see them converge into the specified number of petals.

Orientation (cos vs sin):

A cosine-rose is symmetric about the polar (horizontal) axis, whereas a sine-rose is symmetric about the polar (vertical) axis.

For $r=a\cos (n\theta )$,

When $n$ is odd, $a\cos (n\theta )$ has one petal on the $x$-axis (to the right assuming $a$ is positive).

$(r=3\cos (3\theta ))$

When $n$ is even, $a\cos (n\theta )$ has a petal on both the $x$-axis and the $y$-axis.

$(r=3\cos (2\theta ))$

For $r=a\sin (n\theta )$,

When $n$ is odd, one petal lies on the $y$-axis.

$(r=3\sin (3\theta ))$

When $n$ is even, none of the petals will lie on either axis.

$(r=3\sin (2\theta ))$

(Note: Transformations can alter these shapes; they will be discussed later)

Petal Endpoints

All petals reach out a distance $|a|$ from the pole. In fact, the rose curve is inscribed in the circle $r=|a|$: each petal tip lies on that circle.

This makes sense, since the maximum output of any sine or cosine is $1$, so the maximum distance a rose function could go is $a\cdot 1=a$.

$(r=3\sin (3\theta )\text{ and }r=3)$

Periodic Behavior:

When $n$ is odd, the graph will repeat every $\pi$. This can also be interpreted as: you only need an interval of $(k,,k+\pi )$, where $k\in \mathbb{R}$, to graph the entire function, since after that, the function just begins drawing itself again.

However, when $n$ is even, the graph takes an interval of $(k,k+2\pi )$, where $k\in \mathbb{R}$, to completely draw the function.

The two main takeaways for this are:

When graphing for when $n$ is odd, you only need to find points up to $P+\pi$, where $P$ is the first point you choose (typically, we choose $P=0$ because it’s the simplest and it is useful for other types of problems).

When finding an interval, you will now be using $\pi$ rather than $2\pi$, if $n$ is odd (this will be elaborated later on in the article).

Now, knowing this, you can draw the actual graph. Simply plot the major points (If needed, plotting points that are multiples of $\frac{\pi }{4}$ can be helpful), then connect them.

Technically, if you’re familiar with the symmetry, you can draw half the petals and just flip them over, though this is not recommended.

However, if you need to find the domain of a specific petal, you must understand how it’s drawn.

Animations courtesy of National Curve Bank

The rose curve is not like a bouncy ball; it does not repeatedly bounce off of the pole while moving one direction drawing petals. Instead, it follows a certain tangent line, as shown in the graph above.

How the Curve is Traced (as θ increases)

As $\theta$ increases from $0$ to $2\pi$ (or $\pi$ when $n$ is odd), the rose is usually drawn petal by petal (except for one exception which we will talk about).

Each petal is traced when $r$ goes from $0$ up to $\pm a$ and back to $0$ over a $\theta$-interval of length $\frac{\pi }{n}$ or $\frac{2\pi }{n}$ (depending on whether the graph repeats over an interval of $\pi$ or $2\pi$).

We typically start a curve when $\theta =0$.

Let’s say we have a rose function of cosine:

$r=3\cos (3\theta )$

At $\theta =0$, we have $r=3\cos (0)=3$.

So the curve begins at the point $(r,\theta )=(3,0)$, which is on the positive $x$-axis.

Now, THIS IS THE EXCEPTION (When you are drawing the first petal of a cosine rose function)!

As $\theta$ increases, the function begins to go back towards the pole:  

However, instead of completing the current petal, it begins drawing the next:

Then, it will complete the next petal

And finally, it will loop around to finish the first petal

If you’ll notice, the lines/function are always being drawn in a counter-clockwise manner.

If you were to draw this by hand (in the correct order), you will see that your hand is continuously going counter-clockwise.

This will be true for all rose function, meaning you can graph any rose function (in correct order) by

A) finding the major points via a table, and

B) drawing a line through them in a counter-clockwise manner.

However, the main purpose of this is to find a specific interval of the function.

For simplicity, we’ll use a sine function this time.

Animations courtesy of National Curve Bank

As you can see, each petal is drawn one after another.

Unlike cosine, sine will always finish a petal before drawing the next.

To start, let’s find the interval of any specific petal.

Because the amount of petals in this function is even, we know that n is even. Specifically, we know that $n=4$, since there are $8$ petals and when $n$ is even, the number of petals is $2n$.

And for scenarios in which $n$ is even, the “period” of the function is $2\pi$.

Each petal is the same in terms of shape; the only difference is what way they face. So, intuitively, it makes sense that each petal is made over equal-length intervals.

Thus, to find the interval of each petal in this graph, we can simply divide $2\pi$ by the number of petals:

$\frac{2\pi }{\text{number of petals}}=\frac{2\pi }{8}=\frac{\pi }{4}$

According to our answer, every time $\theta$ increases by $\frac{\pi }{4}$ assuming we start from $\theta =0$, a new petal will be created.

Is it true? Let's see:

As we can see, every time we add a $\frac{\pi }{4}$ to our upper bound for the interval of $\theta$, one new petal is created, confirming that the interval-length for each petal is $\frac{\pi }{4}$.

Intuition can also guide you from here; the interval-length of half a petal is half of the petals’ interval-length. In other words, from the pole to the tip of a petal, there is an interval-length $\frac{\pi }{8}$ (Remember, this is for the function $\sin (4\theta )$).

Limaçons

Limaçons (coming from the French word, “snail”) are polar curves in the form

$r=a\pm b\cos \theta$

Or

$r=a\pm b\sin \theta$

Depending on the constants $a$ and $b$, they take on several shapes.

When $a=b$,

The Limaçon will be in the shape of a heart, and is thus called a heart shaped Limaçon. It is also called a Cardioid.

When $a<b$,

The curve has an inner loop, and is thus called an inner-loop Limaçon.

When $1<\frac{a}{b}<2$, 

This Limaçon is called a Dimpled Limacon. 

And when $\frac{a}{b}\ge 2$, 

(Note: Convex Limaçons are NOT to be confused with circles. As you will learn later, Convex Limaçons are not centered at the pole)

Graphing most Limaçons are not difficult; usually, you can simply find key points and graph from there (although some difficulty may arise with the inner loop, which will be elaborated upon later).

In the next section, you will read about the significance of these keypoints.

Key Points of Limaçons

Generally, for any Limacon, you must find the points at which the function crosses both polar axes. 

However, a useful fact for any limaçon is that the maximum radius is |a|+|b| and the minimum (if positive) is $|a|-|b|$.

For $1<|a/b|<2$ (dimpled case), the “dimple” point is at $r=|a|-|b|$ (will be elaborated upon later)

Cardioids/Heart shaped Limaçons

When graphing Cardioids, the points you absolutely need to find are the points at which $\theta =0$, $\frac{\pi }{2}$, $\pi$, $\frac{3\pi }{2}$ (note, you do not need $2\pi$ because that is when the function begins repeating itself).

To find the points, you simply plug in each value of $\theta$ and you will receive its corresponding $r$. After, you can simply graph it to get a general shape of your function.

As for other points you can graph to make your graph more accurate (which are HEAVILY recommended but are not completely necessary to understand the general shape), you can simply follow the same rule for roses:

If your function uses cosine (if the function looks like it’s shifted right or left), then use angles that are multiples of $\frac{\pi }{3}$ for values that come out cleanly.

If your function uses sine (if the function looks like it’s shifted up or down), then use angles that are multiples of $\frac{\pi }{6}$ for the same reason.

Once you have your points, you can simply connect the dots.

If you are asked for a specific interval, you can simply graph the points that have a in that interval (of course, if the interval starts with a value of $\theta$ that is not shared by one of your points, you will need to find the coordinate for the value of $\theta$ asked)

Inner-loop Limaçons

After some practice, knowing how to connect the dots for an Inner-loop Limaçon becomes easy (it’s really the only way you can get better at it, a.k.a., by going through more examples).

However, the hard part is finding specific intervals, which, if you’re given an inner-loop Limaçon, is most likely what’s being asked.

First, find where $r=0$ (the pole intersections).

There should be $2$ solutions. If there are not, you most likely do not have an inner-loop Limaçon.

Use these solutions as boundaries. If asked for a theta-specific interval, it typically (note, not always) starts at one of these solutions and ends at the other. Occasionally, however, it may start at one of the intersections with the polar axis. 

Typically, when asked for an interval, it will have bounds that lie at the farthest point from the pole (a multiple of $\frac{\pi }{2}$ or depending on whether your function is sine or cosine), one of two (or both angles) that lead to $r=0$ (aka the two intersections at the pole, or the farther point on the inner-loop from the pole.

There will be practice problems provided at the bottom of this article.

Convex Limaçons

Convex Limaçons are NOT the same circle polar functions you learned earlier.

The difference lies in the fact that they are not centered at $(0,0)$, or the pole.

Other than being able to tell the difference, there is not much to them. You simply plot your points and connect them in a circular fashion. 

Note: Convex Limaçons do not share the behavior of circles, that being, they take $2\pi$ to complete a revolution instead of $\pi$. This is true for all Limaçons.

Dimpled Limaçons

Dimpled Limaçons are actually a bit hard to graph, especially since you can’t exactly see the dimple sometimes.

First, determine whether the function uses cosine or sine.

If it is cosine, the dimple will be on either the left or right.

If it’s sine, then it will be on the top or bottom.

Second, you must find the minimum and maximum distance from the pole. The maximum and minimum r-values are crucial for seeing the dimple’s extent. 

Now, you do not need to find every major point on the graph to find the min and max distance.

Instead, you just need to know that,

$a+b$ is the maximum distance from the pole

And 

$a-b$ is the closest positive distance to the pole.

This point is the dimple.

After plotting the other major points previously mentioned that apply for any polar function, simply connect the points smoothly and you’ll be done.

For finding a specific interval, it’s essentially the same as a circle, except that now instead of the function repeating every $\pi$, it repeats every $2\pi$.

Finding Polar Equations Given a Graph

To find the specific equation $r=a\pm \cos (\theta )$ or $r=a\pm \sin (\theta )$ that is represented by a given graph, you must identity the type of symmetry (which trigonometric function is being used) and measure the radial extrema along the axis of symmetry.

First, we must establish the type of function. Symmetry about the horizontal polar axis implies the use of $\cos (\theta )$, while symmetry about the vertical polar axis implies the use of $\sin (\theta )$.

Next, the maximum and minimum radial distances along this axis must be measured.

For any Limaçon (not just the dimpled variety), the maximum distance from the pole to the farthest edge of the curve is $a+b$.

However, for the closest point to the curve, the method varies depending on the type. 

For Convex, Dimpled, or Cardoids, the positive distance between the closest point to the pole and the pole is given by $a-b$.

However, for Inner-Loop shapes, $a-b$ represents the distance to the tip of the inner loop, but the algebraic value $r$ is negative to form the loop.

We define $R_{closest}$ as the signed radius $a-b$. If the inner loop extends $2$ units, then $R_{closest}=-2$.

As for why this is true, simply creating a table of points and graphing their respective radius will show you why.

Once you have these two measurements, you can solve for a and b using this simple system:

$a=\frac{R_{max}+R_{closest}}{2}$

$b=\frac{R_{max}-R_{closest}}{2}$

Lemniscates

Lemniscates (Latin for “ribbon”) are figure-eight-shaped curves centered at the pole. They are fundamentally different from Limaçons and Roses because their equations involve $r^2$, which imposes special restrictions on where the curve can exist.

Note: Lemniscates rarely appear on the AP exam! So far, a lemniscate has not appeared on an AP exam, however, there may be one multiple-choice question on a future exam featuring a lemniscate.

The two standard forms for lemniscates are the following:

$r^2=a^2\cdot \cos (2\theta )$

$r^2=a^2\cdot \sin (2\theta )$

They look like this:

$r^2=25\cos (2\theta )$

In both equations,

$|a|$ represents the maximum distance from the pole to the tip of either loops.

Once again, orientation will be based on the trigonometric function (sine or cosine). 

If the function uses cosine, then the loops are aligned along the $x-$axis (the horizontal polar axis) or the $y-$axis (the vertical polar axis)

If the function uses sine, the loops are aligned diagonally along the lines $\theta =\frac{\pi }{4}$ or $\theta =\frac{5\pi }{4}$.

Graphing Lemniscates

The Domain Restriction

Since the equation for a lemniscate contains the term $r^2$, the right side of the equation (with the trig functions) must be positive for $r$ to be a real number. If it were negative, $r$ would be undefined. This requirement creates angular “gaps” in which the curve does not exist. To demonstrate this concept, let’s try graphing the function $r^2=25\cos (2\theta )$.

(Note: $r=\sqrt{(25\cos (2\theta ))}$ is the same as  $r^2=25\cos (2\theta )$ graphically. Since $r=-\sqrt{(25\cos (2\theta ))}$ is also the same graphically as the two polar functions above, there is no need to graph the square root variants of this lemniscate curve.)

The first interval is  $(0,\frac{\pi }{4})$. Intuitively, the next portion of the graph should be drawn when $\theta$ reaches $\frac{\pi }{2}$. However,

nothing changes. Let's see what is going on in the background.

As we can see, when $\theta$ goes from $0$ to $\frac{\pi }{4}$, the graph of $r=\sqrt{25\cos (2\theta )}$ is drawn. Then, when theta goes from $\frac{\pi }{4}$ to $\frac{\pi }{2}$, the function of $r=\sqrt{-25\cos \left(2\theta \right)}$ is drawn. This is because from $\frac{\pi }{4}$ to $\frac{\pi }{2}$, $\cos (2\theta )$ is negative, so if we add a negative sign inside the square root, we can see what’s happening between the “gaps.”

Let’s see what happens as we continue.

Thus, to trace one loop of a lemniscate, you must find the angles ($\theta$) where the curve begins and ends at the pole $r=0$. 

For $r^2=a^2\cos (2\theta )$, the curve is only traced where $\cos (2\theta )\ge 0$. One full loop is traced from $\theta =-\frac{\pi }{4}$ to $\theta =\frac{\pi }{4}$.

For $r^2=a^2\sin (2\theta )$, the curve is traced only where $\sin (2\theta )\ge 0$. One full loop is traced from $\theta =0$ to $\theta =\frac{\pi }{2}$.

Plotting the Two Loops

Fortunately, actually plotting the lemniscate is nowhere near as hard as graphing something like the inner-loop. When you solve the equation for $r$, you get $r=\pm \sqrt{a^{2}f(2\theta )}$. For every valid angle $\theta$, there are two corresponding radii: a positive $(r,\theta )$ and a negative $(-r,\theta )$ Plotting both of these points is what automatically generates the two loops of the lemniscate shape. 

Technically, if you’re careful, you can find one side of the lemniscate and simply “copy it over” whatever polar axis it’s symmetric to; this distance between the farthest point from the pole will be the same, only its direction will be reversed. As for connecting the points, there’s not much to it. Unlike the rose polar function or the dimpled limaçon polar function, lemniscates don’t have any special areas where they differ from the natural curve. If need be, you can find points in between the angles theta that correspond to the max distance from the pole, assuming the value theta is a valid angle.

Finding the Lemniscate Equation Given a Graph

To derive the equation from a graph, you only need to perform two steps:

First, identify the function type. Once again, if the loops lie flat on the $x-$axis or the $y-$axis, use the cosine function for lemniscates (Note: When it is on the $y-$axis, it means the inside of the square root has an added negative sign: $r^2=-a^2\cos (2\theta )$

If the loops lie diagonally, use $r^2=a^2\sin (2\theta )$. If the loops are in quadrants 1 and 3, then the normal equation $r^2=a^2\sin (2\theta )$ works. However, if they are in quadrants 2 and 4, then the function has a negative added in the square root: $r^2=a^2\sin (2\theta )$.

Then, measure the length $|a|$. Measure the distance from the pole to the tip of any loop. This value is $|a|$. Remember to square this distance to find $a^2$ for the final equation.

Practice Problems

For Questions 3-12, graph the functions given. It is suggested that you construct a table of values first before graphing each function.

For Questions 18-20, name and graph the following polar functions:

For Questions 21-23, convert the following polar functions into rectangular form:

For Questions 25-26, convert the following rectangular functions into polar form. Use trigonometric identities to simplify the final answer.

For Questions 27-28, write the equation that corresponds to the graph shown. Additionally, name the type of graph shown.

Question 27

Question 28

For Questions 29-31, provide the interval for the dashed portion of the graph. The domain of the function is $\left[0,2\pi \right]$. 

Question 29

Question 30

Question 31

Solutions to Practice Problems