Graphing Other Common Polar Equations: Videos & Practice Problems

Polar equations often represent four key shapes: cardioids, limacons, roses, and lemniscates. Cardioids and limacons share similar equations, with the distinction that cardioids have equal parameters (a = b). Roses, defined by $r_{=}a\cos (n\theta )$ or $r_{=}a\sin (n\theta )$, have petals determined by n. Lemniscates, resembling an infinity symbol, are characterized by $r_{²}=\pm a²cos(2\theta )$ or $r_{²}=\pm a²sin(2\theta )$.

Introduction to Common Polar Equations

When exploring polar equations, four primary shapes frequently arise: the cardioid, limacon, rose, and lemniscate. Each of these shapes is characterized by specific equations that include variables such as r, θ, and trigonometric functions.

The cardioid resembles a heart shape and is defined by the equations r = a ± b cos(θ) or r = a ± b sin(θ), where both a and b are positive and equal (i.e., a = b). In contrast, the limacon shares the same equation form but allows for a to be either greater than or less than b. If a > b, the limacon has a dimple; if a < b, it features an inner loop. Notably, these equations are the only ones in this context that involve addition or subtraction.

The rose curve, which resembles a flower, is represented by the equations r = a cos(nθ) or r = a sin(nθ). Here, a must be non-zero, and n is an integer greater than or equal to 2, determining the number of petals on the rose. For example, if n = 2, the rose will have two petals.

Lastly, the lemniscate, which looks like an infinity symbol, is defined by the equations r² = ±a² cos(2θ) or r² = ±a² sin(2θ). In this case, a cannot be zero, and the presence of r² is a distinctive feature that helps identify this shape.

To classify these equations, consider the following examples: For the equation r = 1 + cos(θ), the addition indicates it could be a cardioid or limacon. Since both a and b are equal to 1, this equation represents a cardioid. In another example, r = 4 sin(2θ), the presence of 2θ suggests a lemniscate, but since it lacks the r² term, it is classified as a rose with n = 2.

Understanding these classifications allows for accurate graphing of polar equations, enhancing comprehension of their geometric representations.

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.
$r=4\sin2\theta$

• A. Cardioid
• B. Limacon
• C. Rose
• D. Lemniscate

Answer: (C)

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.
$r=3+2\cos\theta$

• A. Cardioid
• B. Limacon
• C. Rose
• D. Lemniscate

Answer: (B)

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.
$r^2=-25\cos 2\theta$

• A. Cardioid
• B. Limacon
• C. Rose
• D. Lemniscate

Answer: (D)

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.
$r=1-\sin\theta$

• A. Cardioid
• B. Limacon
• C. Rose
• D. Lemniscate

Answer: (A)

Cardioids

In polar coordinates, a cardioid is a specific type of curve defined by the equation $r^{=}a+bcos\theta$ or $r^{=}a+bsin\theta$, where the constants $a$ and $b$ are equal. The orientation and position of the cardioid can vary based on the specific equation used, but its overall shape remains consistent.

To graph a cardioid, the first step is to determine its symmetry. If the equation includes $cos$, the graph will be symmetric about the polar axis. Conversely, if it includes $sin$, the symmetry will be about the vertical line $\theta =\pi /2$.

Next, plot points at the quadrantal angles: $0,\pi /2,\pi ,\; and3\pi /2.\; For\; example,\; using\; the\; equation$ r^{=}1+cos\theta $,\; we\; can\; calculate:$

• For $\theta =0$: $r=1+1=2$, plot at $(2,0)$.
• For $\theta =\pi /2$: $r=1+0=1$, plot at $(1,\pi /2)$.
• For $\theta =\pi$: $r=1+(-1)=0$, plot at the pole $(0,\pi )$.
• For $\theta =3\pi /2$: Reflect the point from $\pi$ across the polar axis, resulting in $(1,3\pi /2)$.

Finally, connect these points with a smooth curve to complete the cardioid shape, which features a distinctive "bump" at the top and extends outward. While additional points can be calculated for greater accuracy, this method provides a clear representation of the cardioid's structure.

Cardioids Example 1

To graph the polar equation \( r = 2 + 2 \sin \theta \), we recognize that this represents a cardioid. The presence of the sine function indicates that the graph will be symmetric about the line \( \theta = \frac{\pi}{2} \). Understanding this symmetry is crucial for accurately plotting the graph.

We begin by evaluating the function at key quadrantal angles:

1. For \( \theta = 0 \):

\[ r = 2 + 2 \sin(0) = 2 + 2 \cdot 0 = 2 \]

This gives us the point \( (2, 0) \) in polar coordinates.

2. For \( \theta = \frac{\pi}{2} \):

\[ r = 2 + 2 \sin\left(\frac{\pi}{2}\right) = 2 + 2 \cdot 1 = 4 \]

This results in the point \( (4, \frac{\pi}{2}) \), which is at the top of the graph.

3. For \( \theta = \pi \):

Using the symmetry about \( \theta = \frac{\pi}{2} \), we can reflect the point found at \( \theta = 0 \) to get the point at \( (2, \pi) \).

4. For \( \theta = \frac{3\pi}{2} \):

\[ r = 2 + 2 \sin\left(\frac{3\pi}{2}\right) = 2 + 2 \cdot (-1) = 0 \]

This gives us the point at the pole, \( (0, \frac{3\pi}{2}) \).

With these points plotted, we can now connect them with a smooth, continuous curve. The general shape of the cardioid will have its furthest point from the pole at the top, creating a rounded bottom and a dimpled heart shape characteristic of cardioids. The orientation of the cardioid is determined by the sine function, which in this case opens upwards.

For a more precise graph, additional points can be plotted between these quadrantal angles. However, the key features of the cardioid are captured with the points calculated, illustrating the overall shape and positioning of the graph for \( r = 2 + 2 \sin \theta \).

Graph $r=2-2\cos\theta$

• A.
• B.
• C.
• D.

Answer: (B)

Limacons

To graph a limacon, we start with the general equation of the form \( r = a \pm b \cos \theta \) or \( r = a \pm b \sin \theta \). The values of \( a \) and \( b \) determine the characteristics of the graph, specifically whether it has a dimple or an inner loop. If \( a > b \), the limacon will have a dimple; if \( a < b \), it will have an inner loop. For example, in the equation \( r = 3 - 2 \sin \theta \), we see that \( a = 3 \) and \( b = 2 \), indicating that this limacon will have a dimple.

Next, we analyze the symmetry of the graph. Since the equation contains a sine function, the graph will be symmetric about the line \( \theta = \frac{\pi}{2} \). This symmetry helps in plotting points at the quadrantal angles: \( 0 \), \( \frac{\pi}{2} \), \( \pi \), and \( \frac{3\pi}{2} \).

Calculating the points involves substituting these angles into the equation:

• For \( \theta = 0 \): \[ r = 3 - 2 \sin(0) = 3 \] This gives the point \( (3, 0) \).
• For \( \theta = \frac{\pi}{2} \): \[ r = 3 - 2 \sin\left(\frac{\pi}{2}\right) = 1 \] This gives the point \( (1, \frac{\pi}{2}) \).
• For \( \theta = \pi \): \[ r = 3 - 2 \sin(\pi) = 3 \] By symmetry, this point is also \( (3, \pi) \).
• For \( \theta = \frac{3\pi}{2} \): \[ r = 3 - 2 \sin\left(\frac{3\pi}{2}\right) = 5 \] This gives the point \( (5, \frac{3\pi}{2}) \).

After plotting these points, we connect them with a smooth curve, ensuring to reflect the dimple at the top of the graph. The resulting shape will clearly illustrate the characteristics of a limacon with a dimple. For greater accuracy, additional points can be plotted as needed.

Limacons Example 2

To graph the polar equation \( r = 1 + 2 \cos(\theta) \), we first identify that this represents a limacon, specifically because the coefficients \( a = 1 \) and \( b = 2 \) are not equal, indicating the presence of an inner loop rather than a dimple.

In the first step, we analyze the values of \( a \) and \( b \). Since \( a < b \), we confirm that the limacon will indeed have an inner loop. This characteristic also implies that the graph will intersect the pole (origin) at least once, which we can plot as a point at the origin.

Next, we determine the symmetry of the graph. The presence of the cosine function indicates that the graph will be symmetric about the polar axis. This symmetry will help us in plotting points efficiently.

For the third step, we plot points using quadrantal angles. Starting with \( \theta = 0 \):

• When \( \theta = 0 \), \( r = 1 + 2 \cos(0) = 1 + 2(1) = 3 \). We plot the point at \( (3, 0) \).
• For \( \theta = \frac{\pi}{2} \), \( r = 1 + 2 \cos\left(\frac{\pi}{2}\right) = 1 + 2(0) = 1 \). This point is plotted at \( (1, \frac{\pi}{2}) \).
• Using symmetry, we reflect this point to \( \theta = \frac{3\pi}{2} \), giving us another point at \( (1, \frac{3\pi}{2}) \).
• Finally, for \( \theta = \pi \), \( r = 1 + 2 \cos(\pi) = 1 + 2(-1) = -1 \). A negative \( r \) value means we plot this point in the opposite direction, resulting in a point at \( (1, \pi) \) or \( (-1, 0) \).

After plotting these points, we connect them with a smooth curve. The inner loop of the limacon is formed by the point closest to the pole, while the outer loop extends outward, creating the characteristic shape of a limacon with an inner loop.

In summary, the graph of the equation \( r = 1 + 2 \cos(\theta) \) reveals a limacon with an inner loop, showcasing the unique features of polar equations and their graphical representations.

Graph $r=1+2\sin\theta$

• A.
• B.
• C.
• D.

Answer: (A)

Roses

To graph a rose curve, we start with the general equation of the form $r=acos(n\theta )$ or $r=asin(n\theta )$, where $a$ is a non-zero constant and $n$ is an integer greater than or equal to 2. The number of petals in the rose is determined by the value of $n$. If $n$ is even, the number of petals is $2n$. If $n$ is odd, the number of petals is simply $n$.

For example, consider the equation $r=4cos(2\theta )$. Here, $n$ is 2, which is even, indicating that the rose will have $22=4$ petals. The constant $a$ is 4, meaning each petal will extend to a radius of 4 units.

To find the angles for the petals, we start with the first petal. Since the equation uses cosine, the first petal is located at $\theta =0$. The petals are spaced evenly around the circle, which can be calculated by dividing $2\pi$ by the number of petals. Thus, the spacing between petals is

Starting from $\theta =0$ for the first petal, we can find the angles for the remaining petals by adding

• Second petal:
• Third petal:
• Fourth petal:

With all four petals plotted at a radius of 4, the final step is to connect these points with a smooth curve, resulting in the characteristic shape of a rose. This process can be repeated for other rose equations by adjusting the values of $a$ and $n$ to explore different configurations and petal counts.

Roses Example 3

To graph the polar equation \( r = 2 \sin(3\theta) \), we recognize that it represents a rose curve of the form \( r = a \sin(n\theta) \). Here, \( a = 2 \) and \( n = 3 \). Since \( n \) is an odd number, the graph will have \( n \) petals, which means there will be 3 petals in this case.

To begin plotting the first petal, we calculate the angle where it will be located. The angle for the first petal can be determined using the formula \( \theta = \frac{\pi}{2n} \). Substituting \( n = 3 \), we find:

\( \theta = \frac{\pi}{2 \times 3} = \frac{\pi}{6} \)

Next, we evaluate the radius at this angle, which is given by the equation. Since \( r = 2 \), the first petal will be plotted at the coordinates corresponding to \( \left(2, \frac{\pi}{6}\right) \).

To find the positions of the other petals, we need to determine the spacing between them. The petals are spaced apart by an angle of:

\( \frac{2\pi}{n} = \frac{2\pi}{3} \)

Starting from the first petal's angle \( \frac{\pi}{6} \), we add \( \frac{2\pi}{3} \) to find the angle for the second petal:

\( \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{6} + \frac{4\pi}{6} = \frac{5\pi}{6} \)

At this angle, the radius remains \( r = 2 \), so the second petal is plotted at \( \left(2, \frac{5\pi}{6}\right) \).

For the third petal, we again add \( \frac{2\pi}{3} \) to the angle of the second petal:

\( \frac{5\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6} + \frac{4\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \)

Thus, the third petal is plotted at \( \left(2, \frac{3\pi}{2}\right) \).

Now that we have the angles and corresponding radii for all three petals, we can connect these points with smooth curves. The petals will extend from the origin, creating a rose-like shape. The final graph of \( r = 2 \sin(3\theta) \) will exhibit three symmetrical petals, each with a length of 2, radiating from the center.

Graph $r=3\cos4\theta$

• A.
• B.
• C.
• D.

Answer: (D)

Lemniscates

The lemniscate is a fascinating mathematical shape resembling an infinity symbol or a propeller with two petals. To graph a lemniscate, it is essential to identify the locations of these petals. The general equation for a lemniscate can be expressed in polar coordinates as:

\( r^2 = \pm a^2 \cos(2\theta) \) or \( r^2 = a^2 \sin(2\theta) \)

In this case, we are working with the specific equation:

\( r^2 = 4 \sin(2\theta) \)

This equation indicates that it is indeed a lemniscate since it features \( r^2 \). To begin graphing, we first need to determine the value of \( a \). The constant 4 in the equation represents \( a^2 \), so we find \( a \) by taking the square root:

\( a = \sqrt{4} = 2 \)

Next, we need to find the angle \( \theta \) for the first petal. Since we are using the sine function in our equation, we can deduce that the angle corresponding to the first petal is:

\( \theta = \frac{\pi}{4} \)

At this angle, the radius \( r \) for the first petal is 2. Thus, we plot the first petal at the coordinates corresponding to \( r = 2 \) and \( \theta = \frac{\pi}{4} \). To find the second petal, we reflect this point over the pole (the origin in rectangular coordinates), which gives us the coordinates for the second petal.

Finally, we connect these two points with a smooth curve, resulting in the characteristic shape of the lemniscate. The graph will resemble an infinity symbol, with both petals extending symmetrically from the pole. For more precision, additional points can be calculated by substituting various values of \( \theta \) into the original equation.

Understanding how to graph a lemniscate not only enhances your skills in polar coordinates but also deepens your appreciation for the beauty of mathematical shapes. Practice with different equations to become more familiar with this intriguing concept!

Lemniscates Example 4

To graph the equation \( r^2 = 25 \cos(2\theta) \), we recognize that this represents a lemniscate, a figure-eight shaped curve that is defined in polar coordinates. The presence of \( r^2 \) indicates that we are dealing with a polar equation, and the cosine function suggests the orientation of the petals.

First, we identify the value of \( a \) by taking the square root of the constant multiplied by the cosine function. In this case, since \( 25 \) is the coefficient, we find:

\( a = \sqrt{25} = 5 \).

Next, we determine the angle \( \theta \) for the first petal. Given that we have a positive coefficient and a cosine function, the first petal is located at \( \theta = 0 \). Thus, we plot the first petal at:

\( (r, \theta) = (5, 0) \), which corresponds to the point \( (5, 0) \) on the polar axis.

To complete the graph, we reflect this petal over the pole (the origin), which gives us the second petal. This reflection results in another petal located at:

\( (r, \theta) = (5, \pi) \), or \( (-5, 0) \) in Cartesian coordinates.

Finally, we connect these two points with a smooth curve, creating the characteristic shape of the lemniscate, which resembles an infinity symbol or a propeller. The graph will have two symmetrical petals, each extending outwards from the origin.

In summary, the key steps to graphing the lemniscate defined by \( r^2 = 25 \cos(2\theta) \) involve determining the value of \( a \), identifying the angles for the petals, and reflecting them appropriately to complete the shape.

Graph $r^2=9\sin2\theta$

• A.
• B.
• C.
• D.

Answer: (B)

Here’s what students ask on this topic:

What are the common shapes of polar equations and their corresponding forms?

Polar equations often represent four key shapes: cardioids, limacons, roses, and lemniscates. Cardioids have the form $r=a+bcos\theta$ or $r=a+bsin\theta$ with $a=b$. Limacons have the same form but with $a>b$ or $a<b$. Roses are defined by $r=acosn\theta$ or $r=asinn\theta$, where $n$ determines the number of petals. Lemniscates are characterized by $r{2}^{}=\pm a^{2}cos2\theta$ or $r{2}^{}=\pm a^{2}sin2\theta$.

How do you determine the number of petals in a rose polar equation?

The number of petals in a rose polar equation is determined by the integer $n$ in the equation $r=acosn\theta$ or $r=asinn\theta$. If $n$ is even, the rose will have $2n$ petals. If $n$ is odd, the rose will have $n$ petals. For example, if $n$ is 3, the rose will have 3 petals, but if $n$ is 4, the rose will have 8 petals.

What is the difference between a cardioid and a limacon in polar equations?

Both cardioids and limacons have equations of the form $r=a+bcos\theta$ or $r=a+bsin\theta$. The key difference is in the relationship between $a$ and $b$. For a cardioid, $a=b$, resulting in a heart-shaped graph. For a limacon, $a$ and $b$ are not equal. If $a>b$, the limacon has a dimple. If $a<b$, it has an inner loop.

How do you graph a lemniscate polar equation?

To graph a lemniscate, start with the equation $r{2}^{}=\pm a^{2}cos2\theta$ or $r{2}^{}=\pm a^{2}sin2\theta$. Identify $a$ by taking the square root of the constant term. Plot the first petal at $r=a$ and $\theta =\mathrm{\pi }/4$ for sine or $\theta =0$ for cosine. Reflect this petal over the pole to get the second petal. Connect the points with a smooth curve to form the infinity symbol shape.

How do you determine the symmetry of a polar graph?

To determine the symmetry of a polar graph, examine the equation. If the equation contains $cos\theta$, the graph is symmetric about the polar axis (horizontal axis). If it contains $sin\theta$, it is symmetric about the line $\theta =\mathrm{\pi }/2$ (vertical axis). For equations involving $r{2}^{}$, such as lemniscates, the graph is symmetric about the origin.