Polar Coordinates: Videos & Practice Problems

In polar coordinates, points are represented as (r, θ), where r is the distance from the pole and θ is the angle from the polar axis. To convert between polar and rectangular coordinates, use the equations: $x_{=}r\cos \left(\theta \right)$ and $y_{=}r\sin \left(\theta \right)$. Understanding these concepts is essential for graphing shapes like cardioids, limacons, roses, and lemniscates, which have specific equations and characteristics based on their parameters.

Intro to Polar Coordinates

In mathematics, the polar coordinate system offers a unique way to plot points using the parameters \( r \) and \( \theta \), as opposed to the traditional Cartesian coordinates of \( x \) and \( y \). In this system, \( r \) represents the distance from the pole (the origin, or point \( (0,0) \)), while \( \theta \) denotes the angle measured from the polar axis, which corresponds to the positive x-axis in Cartesian coordinates.

To understand polar coordinates, it is essential to visualize the system. The pole is located at \( r = 0 \), and as \( r \) increases, it indicates points further away from the pole. The angle \( \theta \) is measured counterclockwise from the polar axis, similar to how angles are measured on the unit circle. For example, an angle of \( \frac{\pi}{2} \) radians corresponds to the positive y-axis.

When plotting a point in polar coordinates, the ordered pair is written as \( (r, \theta) \). For instance, to plot the point \( (5, \frac{\pi}{6}) \), one would first locate the angle \( \frac{\pi}{6} \) on the polar coordinate system and then move outwards from the pole a distance of 5 units.

Negative angles in polar coordinates indicate a clockwise measurement from the polar axis. For example, to plot the point \( (5, -\frac{\pi}{3}) \), one would measure \( \frac{\pi}{3} \) radians clockwise from the polar axis and then count 5 units away from the pole to plot the point.

Additionally, when the radius \( r \) is negative, it signifies a movement in the opposite direction along the line defined by the angle \( \theta \). For example, to plot the point \( (-3, \frac{\pi}{6}) \), one would first locate the angle \( \frac{\pi}{6} \) and then count 3 units in the opposite direction from the pole.

Understanding the signs of both \( r \) and \( \theta \) is crucial when working with polar coordinates, as they dictate the direction and distance from the pole. This system is particularly useful for representing circular and rotational patterns, making it a valuable tool in various fields of mathematics and physics.

Intro to Polar Coordinates Example 1

In polar coordinates, each point is defined by a radius (r) and an angle (θ). When plotting points, it's essential to first identify the angle before determining the distance from the origin. Angles can be expressed in both radians and degrees, which is crucial for accurate plotting.

For example, consider point A, which has a radius of 5 and an angle of 60 degrees (or \(\frac{\pi}{3}\) radians). To plot this point, locate the angle of 60 degrees on the polar coordinate system. From this angle, count out 5 units along the line to reach point A.

Next, point B is defined by a radius of 3 and an angle of 90 degrees (or \(\frac{\pi}{2}\) radians). Again, start by locating the angle of 90 degrees, then move 3 units outward along that line to find point B.

Point C presents a different scenario with a radius of 0 and an angle of \(-\frac{5\pi}{3}\) radians. The negative angle indicates a clockwise measurement. However, since the radius is 0, point C remains at the origin, regardless of the angle.

Finally, point D has a radius of 2 and an angle of \(\frac{7\pi}{3}\) radians. This angle represents a full rotation (2π) plus an additional \(\frac{\pi}{3}\) radians. After completing the rotation, count out 2 units along the corresponding line to locate point D.

By understanding how to interpret the radius and angle, you can effectively plot points in polar coordinates, enhancing your grasp of this coordinate system.

Plot the point on the polar coordinate system.
$(5,210°)$

• A.
• B.
• C.
• D.

Answer: (A)

Plot the point on the polar coordinate system.
$(-3,-90°)$

• A.
• B.
• C.
• D.

Answer: (C)

Plot the point on the polar coordinate system.
$(6,-\frac{11\pi}{6})$

• A.
• B.
• C.
• D.

Answer: (A)

Plot the point on the polar coordinate system.
$(-2,\frac{2\pi}{3})$

• A.
• B.
• C.
• D.

Answer: (D)

Determining Different Coordinates for the Same Point

In the study of polar coordinates, it's essential to understand that a single point can be represented by multiple ordered pairs. This concept mirrors the idea of coterminal angles in the unit circle, where angles like \( \frac{\pi}{6} \) and \( \frac{13\pi}{6} \) correspond to the same position. To find these different representations, we can add or subtract multiples of \( 2\pi \) to the angle while keeping the radius \( r \) constant.

For example, if we start with the point represented by \( \left(4, \frac{\pi}{6}\right) \), we can find another ordered pair by adding \( 2\pi \) to the angle, resulting in \( \left(4, \frac{13\pi}{6}\right) \). This method can be extended indefinitely, allowing us to generate an infinite number of ordered pairs that correspond to the same point.

Additionally, we can alter the radius \( r \) by making it negative. When \( r \) is negative, we must adjust the angle \( \theta \) by adding \( \pi \) to ensure the point remains in the same location. For instance, if we take \( \left(-4, \frac{\pi}{6}\right) \), we need to convert it to \( \left(-4, \frac{7\pi}{6}\right) \) to find the correct position on the graph.

To summarize, when given a point \( (r, \theta) \), we can find multiple ordered pairs that represent the same point by either:

• Keeping \( r \) the same and adding or subtracting multiples of \( 2\pi \) to \( \theta \).
• Changing \( r \) to a negative value, which requires adding \( \pi \) to \( \theta \) and then adding or subtracting multiples of \( 2\pi \).

This understanding allows for flexibility in working with polar coordinates, enabling the identification of various ordered pairs that all map back to the same point. As you practice, remember to apply these principles to explore the relationships between different representations in polar coordinates.

Plot the point $(3,\frac{\pi}{2})$ & find another set of coordinates, $(r,θ)$, for this point, where:
(A) $r≥0,2π≤θ≤4π$,
(B) $r≥0,-2π≤θ≤0$,
(C) $r≤0,0≤θ≤2π$.

• A. $(3,\frac{5\pi}{2}),(-3,-\frac{3\pi}{2}),(-3,\frac{3\pi}{2})$
• B. $(3,\frac{5\pi}{2}),(3,-\frac{3\pi}{2}),(-3,\frac{3\pi}{2})$
• C. $(-3,\frac{5\pi}{2}),(-3,-\frac{3\pi}{2}),(-3,\frac{\pi}{2})$
• D. $(3,\frac{5\pi}{2}),(3,-\frac{3\pi}{2}),(-3,\frac{\pi}{2})$

Answer: (B)

Determining Different Coordinates for the Same Point Example 2

In this exercise, we explore the concept of polar coordinates by plotting the point \((-2, \frac{5\pi}{3})\) and finding multiple equivalent sets of coordinates for the same point. In polar coordinates, a point is represented by an ordered pair \((r, \theta)\), where \(r\) is the radial distance from the origin and \(\theta\) is the angle measured from the positive x-axis.

To begin, we plot the angle \(\frac{5\pi}{3}\). This angle is located in the fourth quadrant, and since the radial distance \(r\) is negative, we move in the opposite direction along the line defined by the angle. Thus, we count 2 units in the direction opposite to \(\frac{5\pi}{3}\), placing our point at \((-2, \frac{5\pi}{3})\).

Next, we can generate additional sets of coordinates by manipulating the angle while keeping the radial distance the same. One method is to add multiples of \(2\pi\) to the angle \(\theta\). For instance, adding \(2\pi\) to \(\frac{5\pi}{3}\) gives us:

\[\frac{5\pi}{3} + 2\pi = \frac{5\pi}{3} + \frac{6\pi}{3} = \frac{11\pi}{3}\]

This results in the first new set of coordinates: \((-2, \frac{11\pi}{3})\).

Continuing this process, we can add another \(2\pi\) to \(\frac{11\pi}{3}\):

\[\frac{11\pi}{3} + 2\pi = \frac{11\pi}{3} + \frac{6\pi}{3} = \frac{17\pi}{3}\]

This gives us a second new set of coordinates: \((-2, \frac{17\pi}{3})\).

To find a third set of coordinates, we can change the radial distance \(r\) from negative to positive. When \(r\) is made positive, we must adjust the angle by adding \(\pi\) to the original angle to maintain the same point. Thus, we calculate:

\[\frac{5\pi}{3} + \pi = \frac{5\pi}{3} + \frac{3\pi}{3} = \frac{8\pi}{3}\]

This results in the third set of coordinates: \((2, \frac{8\pi}{3})\).

In summary, we have derived three sets of coordinates for the same point: \((-2, \frac{5\pi}{3})\), \((-2, \frac{11\pi}{3})\), and \((-2, \frac{17\pi}{3})\), as well as \((2, \frac{8\pi}{3})\). Each of these coordinates represents the same location in the polar coordinate system, illustrating the concept of periodicity and the flexibility of polar coordinates in representing points in a plane.

Plot the point $(5,-\frac{\pi}{3})$, then identify which of the following sets of coordinates is the same point.

• A. $(-5,-\frac{\pi}{3})$
• B. $(-5,\frac{\pi}{3})$
• C. $(-5,\frac{2\pi}{3})$
• D. $(-5,\frac{5\pi}{3})$

Answer: (C)

Plot the point $(-3,-\frac{\pi}{6})$, then identify which of the following sets of coordinates is the same point.

• A. $(-3,\frac{11\pi}{6})$
• B. $(-3,\frac{5\pi}{6})$
• C. $\left(3,\frac{11\pi}{6}\right)$
• D. $(3,\frac{\pi}{6})$

Answer: (A)

Convert Points from Polar to Rectangular

Understanding the conversion between polar and rectangular coordinates is essential in mathematics, particularly in trigonometry and geometry. Polar coordinates are expressed as \( (r, \theta) \), where \( r \) is the distance from the origin and \( \theta \) is the angle from the positive x-axis. To convert these coordinates into rectangular coordinates \( (x, y) \), we can utilize the relationships derived from right triangles.

For any point given in polar coordinates, the conversion formulas are:

\[x = r \cdot \cos(\theta)\]\[y = r \cdot \sin(\theta)\]

To illustrate this, consider the polar coordinate \( (5, \frac{\pi}{3}) \). Here, \( r = 5 \) and \( \theta = \frac{\pi}{3} \). By forming a right triangle, we can identify the hypotenuse as \( r \) and the angles as \( \theta \). Using the cosine and sine functions, we can find \( x \) and \( y \):

For \( x \):

\[x = 5 \cdot \cos\left(\frac{\pi}{3}\right) = 5 \cdot \frac{1}{2} = \frac{5}{2}\]

For \( y \):

\[y = 5 \cdot \sin\left(\frac{\pi}{3}\right) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}\]

Thus, the rectangular coordinates corresponding to the polar point \( (5, \frac{\pi}{3}) \) are \( \left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right) \).

Next, consider the polar coordinate \( (-3, \frac{\pi}{6}) \). The negative \( r \) value indicates that we will move in the opposite direction of the angle. First, we plot the angle \( \frac{\pi}{6} \) and then move three units in the opposite direction. The conversion to rectangular coordinates follows the same formulas:

For \( x \):

\[x = -3 \cdot \cos\left(\frac{\pi}{6}\right) = -3 \cdot \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2}\]

For \( y \):

\[y = -3 \cdot \sin\left(\frac{\pi}{6}\right) = -3 \cdot \frac{1}{2} = -\frac{3}{2}\]

Thus, the rectangular coordinates for \( (-3, \frac{\pi}{6}) \) are \( \left(-\frac{3\sqrt{3}}{2}, -\frac{3}{2}\right) \).

Lastly, when the polar coordinate is \( (0, -\frac{\pi}{6}) \), the \( r \) value of zero indicates that the point is at the origin. Therefore, both \( x \) and \( y \) will be zero:

\[x = 0 \cdot \cos\left(-\frac{\pi}{6}\right) = 0\]

\[y = 0 \cdot \sin\left(-\frac{\pi}{6}\right) = 0\]

Consequently, the rectangular coordinates are \( (0, 0) \), which corresponds to the origin in both coordinate systems.

In summary, converting between polar and rectangular coordinates involves applying trigonometric functions to the radius and angle, allowing for a seamless transition between these two representations of points in a plane.

Convert the point to rectangular coordinates.
$(-2,-\frac{\pi}{4})$

• A. $(\sqrt{2},-\sqrt2)$
• B. $(-1,1)$
• C. $(-\sqrt{2},\sqrt{2})$
• D. $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

Answer: (C)

Convert the point to rectangular coordinates.
$(4,\frac{\pi}{6})$

• A. $(2\sqrt{3},2)$
• B. $(4\sqrt{3},4)$
• C. $(2,2\sqrt{3})$
• D. $(2,\sqrt{3})$

Answer: (A)

Convert the point to rectangular coordinates.
$(-3,0)$

• A. $(-3,0)$
• B. $(0,-3)$
• C. $(0,0)$
• D. $(3,0)$

Answer: (A)

Convert the point to rectangular coordinates.
$(0,\frac{7\pi}{4})$

• A. $(0,\frac74)$
• B. $(-4,0)$
• C. $(0,0)$
• D. $(0,7)$

Answer: (C)

Convert Points from Rectangular to Polar

To convert points from rectangular coordinates to polar coordinates, we utilize the relationships between the two systems, primarily through the Pythagorean theorem and trigonometric functions. A point in rectangular coordinates is represented as (x, y), while in polar coordinates, it is represented as (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.

To find the polar coordinates, we start by calculating r using the formula:

\[ r = \sqrt{x^2 + y^2} \]

For example, for the point (3, 4), we calculate:

\[ r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Next, we determine θ using the tangent function, which relates the opposite side (y) to the adjacent side (x):

\[ \tan(θ) = \frac{y}{x} \]

For the point (3, 4), we find:

\[ θ = \tan^{-1}\left(\frac{4}{3}\right) \]

Calculating this gives approximately 53 degrees. Thus, the polar coordinates for the point (3, 4) are (5, 53°).

When dealing with points located on the axes or in different quadrants, we must be cautious. For instance, the point (-4, 0) lies on the negative x-axis. Here, we calculate r:

\[ r = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 \]

Since the point is on the negative x-axis, θ is π radians (or 180 degrees), giving us polar coordinates (4, π).

For a point like (-1, √3), we first plot the point in the second quadrant. We calculate r:

\[ r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Next, we find θ. The tangent function gives:

\[ θ = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) \]

This results in an angle in the fourth quadrant, so we adjust by adding π to find the correct angle in the second quadrant:

\[ θ = -\frac{\pi}{3} + \pi = \frac{2\pi}{3} \]

Thus, the polar coordinates for the point (-1, √3) are (2, 2π/3).

In summary, the process for converting from rectangular to polar coordinates involves plotting the point, calculating r using the Pythagorean theorem, and determining θ using the tangent function while considering the quadrant in which the point lies. This systematic approach ensures accurate conversion regardless of the point's location.

Convert Points from Rectangular to Polar Example 1

To convert the rectangular coordinates (3, -3) into polar coordinates, we follow a systematic approach. First, we plot the point on the rectangular coordinate system, which is located in the fourth quadrant.

Next, we calculate the radial distance r using the formula:

r = \sqrt{x^2 + y^2}

Substituting the values, we have:

r = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}

Now, we determine the angle θ. Since the point is in the fourth quadrant, we use the inverse tangent function:

θ = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-3}{3}\right) = \tan^{-1}(-1)

This simplifies to:

θ = -\frac{\pi}{4}

Thus, the polar coordinates of the point are:

(r, θ) = (3\sqrt{2}, -\frac{\pi}{4})

It's important to note that polar coordinates can have multiple representations. For instance, the angle can also be expressed as 7π/4, which is equivalent to -π/4 in the context of the unit circle. Therefore, both representations, (3√2, -π/4) and (3√2, 7π/4), are valid and will correspond to the same rectangular coordinates (3, -3) when converted back.

Convert the point to polar coordinates.
$(0,5)$

• A. $(0,0)$
• B. $(5,0)$
• C. $(-5,\frac{\pi}{2})$
• D. $(5,\frac{\pi}{2})$

Answer: (D)

Convert the point to polar coordinates.
$(-2,2)$

• A. $(2\sqrt{2},\frac{3\pi}{4})$
• B. $(2\sqrt{2},-\frac{\pi}{4})$
• C. $(2\sqrt{2},\frac{\pi}{4})$
• D. $(-2\sqrt{2},\frac{3\pi}{4})$

Answer: (A)

Convert the point to polar coordinates.
$(1,1)$

• A. $(1,\frac{\pi}{4})$
• B. $(\sqrt{2},\frac{\pi}{4})$
• C. $(\sqrt{2},\frac{5\pi}{4})$
• D. $(\sqrt{2},-\frac{\pi}{4})$

Answer: (B)

Convert the point to polar coordinates.
$(-1,-\sqrt{3})$

• A. $(2,\frac{\pi}{3})$
• B. $(2,\frac{7\pi}{6})$
• C. $(2,\frac{4\pi}{3})$
• D. $(-2,\frac{4\pi}{3})$

Answer: (C)

Convert Equations from Rectangular to Polar

Converting equations from rectangular to polar form involves substituting the rectangular coordinates \(x\) and \(y\) with their polar equivalents, \(r \cos \theta\) and \(r \sin \theta\), respectively. This process allows us to express equations in terms of \(r\) and \(\theta\), which is essential for understanding polar coordinates.

For example, consider the equation \(y = 5\). By substituting \(y\) with \(r \sin \theta\), we rewrite the equation as:

\(r \sin \theta = 5\).

To isolate \(r\), we divide both sides by \(\sin \theta\), resulting in:

\(r = \frac{5}{\sin \theta}\).

Recognizing that \(\frac{1}{\sin \theta} = \csc \theta\), we can simplify this to:

\(r = 5 \csc \theta\).

Next, consider the equation \(y = x + 1\). Substituting both \(x\) and \(y\) gives:

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

To solve for \(r\), we rearrange the equation by moving \(r \cos \theta\) to the left side:

\(r \sin \theta - r \cos \theta = 1\).

Factoring out \(r\) yields:

\(r(\sin \theta - \cos \theta) = 1\).

Dividing both sides by \((\sin \theta - \cos \theta)\) gives:

\(r = \frac{1}{\sin \theta - \cos \theta}\).

Lastly, for the equation \(x^2 + y^2 = 25\), we can utilize the identity \(x^2 + y^2 = r^2\). Thus, we rewrite the equation as:

\(r^2 = 25\).

Taking the square root of both sides results in:

\(r = 5\).

This indicates a circle of radius 5 in polar coordinates, confirming the relationship between the rectangular and polar forms.

In summary, when converting equations from rectangular to polar form, remember the key substitutions: \(x = r \cos \theta\), \(y = r \sin \theta\), and \(x^2 + y^2 = r^2\). Mastering these conversions enhances your understanding of polar coordinates and their applications.

Convert Equations from Rectangular to Polar Example 1

In this exercise, we focus on converting equations from rectangular to polar form by substituting the Cartesian coordinates with their polar equivalents. The polar coordinates are defined as \( x = r \cos \theta \) and \( y = r \sin \theta \), where \( r \) is the radius and \( \theta \) is the angle.

For the first example, we start with the equation \( y = x^2 \). By substituting \( y \) with \( r \sin \theta \) and \( x \) with \( r \cos \theta \), we rewrite the equation as:

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

Squaring the cosine term gives us:

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

To isolate \( r \), we can divide both sides by \( r \) (assuming \( r \neq 0 \)), leading to:

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

Rearranging this, we find:

\( r = \frac{\sin \theta}{\cos^2 \theta} \)

Next, we move to the second example, where we need to solve for \( r^2 \) in the equation \( 4xy = 2 \). Substituting \( x \) and \( y \) gives us:

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

This simplifies to:

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

Dividing both sides by 2 results in:

\( 2r^2 \cos \theta \sin \theta = 1 \)

Recognizing that \( \cos \theta \sin \theta \) is part of the double angle identity, we can express it as:

\( r^2 \cdot \sin(2\theta) = 1 \)

To isolate \( r^2 \), we divide both sides by \( \sin(2\theta) \), yielding:

\( r^2 = \frac{1}{\sin(2\theta)} \)

Since \( \frac{1}{\sin(2\theta)} \) is equivalent to \( \csc(2\theta) \), we conclude that:

\( r^2 = \csc(2\theta) \)

Through these examples, we have successfully converted rectangular equations into their polar forms, demonstrating the process of substitution and the application of trigonometric identities.

Convert each equation to its polar form.
$y-x=6$

• A. $6r=\sin\theta-\cos\theta$
• B. $r=\frac{\sin\theta-\cos\theta}{6}$
• C. $r=6\left(\sin\theta-\cos\theta\right)$
• D. $r=\frac{6}{\sin\theta-\cos\theta}$

Answer: (D)

Convert each equation to its polar form.
$3y-5x=2$

• A. $r=\frac{2}{3\sin\theta-5\cos\theta}$
• B. $r=\frac{2}{\sin\theta-\cos\theta}$
• C. $r=3\sin\theta-5\cos\theta$
• D. $r=\frac35\sin\theta$

Answer: (A)

Convert each equation to its polar form.
$x^2+y^2=2y$

• A. $r=\sqrt{2}$
• B. $r=2\sin\theta$
• C. $r=4\sin\theta$
• D. $r=2\cos\theta$

Answer: (B)

Convert each equation to its polar form.
$x^2+(y-2)^2=4$

• A. $r^2=4\sin\theta$
• B. $r=4\sin\theta$
• C. $r=2\sin\theta$
• D. $r=\cos\theta-\sin\theta$

Answer: (B)

Convert Equations from Polar to Rectangular

Converting equations from polar to rectangular form involves a systematic approach, particularly when dealing with equations that include the variables \( r \) and \( \theta \). The key to this conversion lies in recognizing the relationships between polar and rectangular coordinates, specifically that \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( r^2 = x^2 + y^2 \).

To illustrate the conversion process, consider the polar equation \( r = 4 \). Since this equation does not initially contain any of the terms \( r \cos(\theta) \), \( r \sin(\theta) \), or \( r^2 \), we can manipulate it by squaring both sides, yielding \( r^2 = 16 \). This allows us to substitute \( r^2 \) with \( x^2 + y^2 \), resulting in the equation \( x^2 + y^2 = 16 \). This equation represents a circle with a radius of 4, confirming that it is already in standard form.

Next, consider the polar equation \( r = \sec(\theta) \). To convert this, we first rewrite the secant function as \( r = \frac{1}{\cos(\theta)} \). Multiplying both sides by \( \cos(\theta) \) eliminates the fraction, leading to \( r \cos(\theta) = 1 \). Substituting \( r \cos(\theta) \) with \( x \) gives us the equation \( x = 1 \), which is the equation of a vertical line.

For the polar equation \( r = 6 \sin(\theta) \), we start by multiplying both sides by \( r \) to obtain \( r^2 = 6r \sin(\theta) \). This results in \( r^2 \) and \( r \sin(\theta) \) terms, which we can replace with \( x^2 + y^2 \) and \( y \), respectively. Thus, we have \( x^2 + y^2 = 6y \). To rewrite this in standard form, we rearrange it to \( x^2 + y^2 - 6y = 0 \) and complete the square for the \( y \) terms. Adding 9 to both sides gives us \( x^2 + (y - 3)^2 = 9 \), which describes a circle centered at \( (0, 3) \) with a radius of 3.

Through these examples, we see that converting from polar to rectangular form requires careful manipulation and an understanding of the relationships between the coordinates. By applying these strategies, we can effectively identify the shapes of the graphs represented by the equations.

Convert Equations from Polar to Rectangular Example 2

To convert the polar equation \( r = \frac{2}{1 + \cos \theta} \) into its rectangular form, we start by eliminating the fraction. We do this by multiplying both sides by the denominator \( 1 + \cos \theta \), resulting in:

\( r(1 + \cos \theta) = 2 \)

Distributing \( r \) gives us:

\( r + r \cos \theta = 2 \)

Next, we recognize that \( r \cos \theta \) can be replaced with \( x \) (since \( r \cos \theta = x \)), and we also need to express \( r \) in terms of \( x \) and \( y \). We know from the relationship between polar and rectangular coordinates that:

\( r = \sqrt{x^2 + y^2} \)

Substituting these into our equation yields:

\( \sqrt{x^2 + y^2} + x = 2 \)

To eliminate the square root, we first isolate it:

\( \sqrt{x^2 + y^2} = 2 - x \)

Next, we square both sides to remove the square root:

\( x^2 + y^2 = (2 - x)^2 \)

Expanding the right side gives:

\( x^2 + y^2 = 4 - 4x + x^2 \)

We can simplify this by canceling \( x^2 \) from both sides:

\( y^2 = 4 - 4x \)

This equation can be rearranged to:

\( y^2 = -4(x - 1) \)

This is the standard form of a horizontal parabola that opens to the left, centered at the point (1, 0). The equation indicates that the vertex of the parabola is at (1, 0) and the focus is located at (0, 0), confirming the shape of the graph.

Convert each equation to its rectangular form.
$r=-4\cos\theta$

• A. $r=-4x$
• B. $x^2+y^2=4$
• C. $(x+2)^2+y^2=2$
• D. $(x+2)^2+y^2=4$

Answer: (D)

Convert each equation to its rectangular form.
$r=\frac{2}{1-\sin\theta}$

• A. $y^2=4-4x$
• B. $x^2+y^2=2y$
• C. $y=\frac14x^2-1$
• D. $x^2-1=y$

Answer: (C)

Introduction to Common Polar Equations

When exploring polar equations, four primary shapes frequently arise: the cardioid, limacon, rose, and lemniscate. Understanding the equations and characteristics of these shapes is essential for classification and graphing.

The cardioid and limacon share similar equations, expressed as r = a \pm b \cos(\theta) or r = a \pm b \sin(\theta). For a cardioid, both a and b must be greater than zero, with a equal to b. This equality results in the heart-like shape of the cardioid. In contrast, for the limacon, while a and b are still positive, a can be either greater than or less than b. If a is greater than b, the limacon has a dimple; if a is less than b, it features an inner loop.

The rose curve, resembling a flower, is defined by the equations r = a \cos(n\theta) or r = a \sin(n\theta). Here, a must be non-zero, and n is an integer greater than or equal to two, indicating the number of petals on the rose.

Lastly, the lemniscate, which resembles an infinity symbol, is represented by the equations r^2 = \pm a^2 \cos(2\theta) or r^2 = \pm a^2 \sin(2\theta). In this case, a cannot be zero, and the presence of r squared is a distinctive feature that helps identify lemniscate equations.

To classify these shapes, consider the specific characteristics of the equations. For example, the equation r = 1 + \cos(\theta) indicates addition, suggesting it is either a cardioid or limacon. Since both a and b are equal to one, this equation represents a cardioid. Conversely, the equation r = 4 \sin(2\theta) features a sine function with a coefficient of two in the argument, which aligns with the rose curve format. Thus, this equation describes a rose with two petals.

By mastering the classification of these polar equations, one can effectively predict the corresponding graph shapes, enhancing both understanding and application in polar coordinates.

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.
$r^2=-25\cos2\theta$

• 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

When graphing polar equations, one of the key shapes to understand is the cardioid, which is characterized by its specific equation format. The general form of a cardioid is given by:

r = a ± b \cos(\theta) \quad \text{or} \quad r = a ± b \sin(\theta)

In these equations, it is important to note that the values of a and b are equal, which is a defining feature of the cardioid shape. To successfully graph a cardioid, two main steps are essential: determining the symmetry of the graph and plotting points at the quadrantal angles.

To identify the symmetry, check whether the equation includes cosine or sine. For example, in the equation r = 1 + \cos(\theta), the presence of cosine indicates that the graph will be symmetric about the polar axis.

Next, we plot points at the quadrantal angles, which are 0, π/2, π, and 3π/2. For each angle, we substitute the value of θ into the equation to find the corresponding r value:

• For θ = 0: r = 1 + \cos(0) = 1 + 1 = 2 (Point: (2, 0))
• For θ = π/2: r = 1 + \cos(\pi/2) = 1 + 0 = 1 (Point: (1, π/2))
• For θ = π: r = 1 + \cos(π) = 1 - 1 = 0 (Point: (0, π))
• For θ = 3π/2: Using symmetry, we reflect the point from θ = π/2 to get (1, 3π/2).

After plotting these points, the final step is to connect them with a smooth, continuous curve. The resulting graph will exhibit the characteristic "bump" of a cardioid, extending outward from the pole. While additional points can be calculated for greater accuracy, this method provides a solid foundation for graphing cardioids effectively.

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 \( (2, 0) \).

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 \( \theta = \pi \):

By symmetry, we can reflect the point found at \( \theta = \frac{\pi}{2} \) to \( \theta = \pi \). However, we can also calculate:

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

This gives us the point \( (2, \pi) \).

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

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

This indicates that the point is at the pole, \( (0, \frac{3\pi}{2}) \).

With these points plotted: \( (2, 0) \), \( (4, \frac{\pi}{2}) \), \( (2, \pi) \), and \( (0, \frac{3\pi}{2}) \), we can now connect them with a smooth curve. The shape of the cardioid will have its furthest point from the pole at the top, creating a rounded dimple at the top and tapering down towards the pole.

In summary, the graph of \( r = 2 + 2 \sin \theta \) is a cardioid that opens upwards, with its maximum radius at \( \theta = \frac{\pi}{2} \) and a minimum radius at the pole when \( \theta = \frac{3\pi}{2} \). This understanding of the graph's symmetry and key points allows for a clear representation of the cardioid's shape.

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

• A.
• B.
• C.
• D.

Answer: (B)

Limacons

To graph a limacon, we start by identifying its equation, which is typically in the form r = a ± b cos(θ) or r = a ± b sin(θ). The values of a and b determine the characteristics of the limacon, specifically whether it has a dimple or an inner loop. If a is greater than b, the limacon will have a dimple; if a is less than b, it will have an inner loop. For example, in the equation r = 3 - 2 sin(θ), since a = 3 and b = 2, we can conclude that this limacon will have a dimple.

Next, we analyze the symmetry of the graph. Because the equation contains a sine function, the graph will be symmetric about the line θ = π/2. This symmetry helps in plotting points at the quadrantal angles: 0, π/2, π, and 3π/2.

Calculating the points:

• For θ = 0: r = 3 - 2 sin(0) = 3, so the point is (3, 0).
• For θ = π/2: r = 3 - 2 sin(π/2) = 1, so the point is (1, π/2).
• For θ = π: By symmetry, we reflect the point from θ = π/2, resulting in (3, π).
• For θ = 3π/2: r = 3 - 2 sin(3π/2) = 5, so the point is (5, 3π/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 its characteristics. This equation represents a limacon because it includes an addition and the coefficients \( a \) and \( b \) are not equal. Here, \( a = 1 \) and \( b = 2 \), indicating that the limacon will have an inner loop since \( a < b \).

Next, we determine the symmetry of the graph. Since the equation contains a cosine function, it will exhibit symmetry about the polar axis. This is an important aspect to consider as we plot points.

We begin plotting points using quadrantal angles:

• For \( \theta = 0 \):

  Calculating \( r \): \( r = 1 + 2 \cos(0) = 1 + 2(1) = 3 \). Thus, the point is \( (3, 0) \).

• For \( \theta = \frac{\pi}{2} \):

  Calculating \( r \): \( r = 1 + 2 \cos\left(\frac{\pi}{2}\right) = 1 + 2(0) = 1 \). Thus, the point is \( (1, \frac{\pi}{2}) \).

• Using symmetry, we reflect this point to \( \theta = \frac{3\pi}{2} \) with \( r = 1 \), giving us another point at \( (1, \frac{3\pi}{2}) \).
• For \( \theta = \pi \):

  Calculating \( r \): \( r = 1 + 2 \cos(\pi) = 1 + 2(-1) = -1 \). A negative \( r \) means we plot in the opposite direction, resulting in the point \( (1, \pi) \) or \( (-1, 0) \) on the polar axis.

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 \( r = 1 + 2 \cos(\theta) \) is a limacon with an inner loop, showcasing both symmetry and distinct features that arise from the relationship between \( a \) and \( b \).

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 two. The value of n determines the number of petals in the rose. Specifically, if n is even, the number of petals is given by $\mathrm{2n}$, while if n is odd, the number of petals is simply n.

For example, consider the equation $r=4cos(2\theta )$. Here, a is 4 and n is 2, which is even. Thus, the rose will have $\mathrm{2n}=4$ petals.

Next, we determine the length of each petal, which is equal to the value of a. Since a is 4, each petal will extend to a radius of 4 units. The angle θ for the first petal is determined by the cosine function, which starts at θ=0. Therefore, the first petal is located at the point (4, 0).

To find the positions of the remaining petals, we calculate the spacing between them. The spacing is determined by dividing $\mathrm{2\pi }$ by the number of petals. In this case, $\mathrm{2\pi }/\; 4\; =\pi /2$. Starting from the angle of the first petal (0), we add $\pi /2$ to find the angles for the subsequent petals:

• Second petal: $\theta =0\; +\pi /2\; =\pi /2$ (4, $\pi /2$)
• Third petal: $\theta =\pi /2\; +\pi /2\; =\pi$ (4, $\pi$)
• Fourth petal: $\theta =\pi +\pi /2\; =\; 3\pi /2$ (4, 3$\pi /2$)

After plotting all four petals at their respective angles, we connect them with a smooth curve, resulting in a complete rose graph. This process can be repeated for other rose equations by following the same steps, ensuring a clear understanding of how to graph these beautiful curves.

Roses Example 3

To graph the polar equation \( r = 2 \sin(3\theta) \), we start by identifying its structure, which is characteristic of a rose curve. The general form for a rose curve is \( r = a \sin(n\theta) \), where \( a \) represents the length of the petals and \( n \) indicates the number of petals. In this case, \( n = 3 \), which is an odd number, indicating that the graph will have exactly three petals.

Next, we determine the value of \( a \), which is 2 in this equation. This means each petal will extend to a radius of 2 units from the origin. To find the angles at which the petals are located, we calculate the angle for the first petal using the formula \( \theta = \frac{\pi}{2n} \). Substituting \( n = 3 \), we find:

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

This angle corresponds to the first petal, which we plot at \( r = 2 \) and \( \theta = \frac{\pi}{6} \). To find the positions of the remaining petals, we need to determine the spacing between them. The petals are spaced apart by:

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

Starting from the first petal's angle \( \frac{\pi}{6} \), we add the spacing to find the angles for the second and third petals:

For the second petal:

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

For the third petal:

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

Now we can plot the second petal at \( r = 2 \) and \( \theta = \frac{5\pi}{6} \), and the third petal at \( r = 2 \) and \( \theta = \frac{3\pi}{2} \). All petals have the same length, so we plot them accordingly. Finally, we connect these points with smooth curves, ensuring that the graph passes through the origin, as is characteristic of rose curves.

The resulting graph of \( r = 2 \sin(3\theta) \) will display three symmetrical petals, each extending out to a radius of 2, creating a visually appealing rose shape.

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 distinct petals. To graph a lemniscate, one can identify the locations of these petals, which can be achieved by determining the first petal's position and then reflecting it over the pole (the origin in rectangular coordinates).

The general equation for a lemniscate is expressed as:

\[ r^2 = \pm a^2 \cos(2\theta) \quad \text{or} \quad r^2 = a^2 \sin(2\theta) \]

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

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

To identify the parameters of the lemniscate, we first recognize that the constant 4 represents \( a^2 \). To find \( a \), we take the square root of 4, yielding \( a = 2 \). This indicates that the radius \( r \) for the first petal will be 2.

Next, we analyze the sine function in the equation. Since we have a positive coefficient for the sine function, we can determine the angle \( \theta \) for the first petal. The angle corresponding to the first petal is:

\[ \theta = \frac{\pi}{4} \]

With \( r = 2 \) and \( \theta = \frac{\pi}{4} \), we can plot the first petal at the coordinates derived from these values. To find the second petal, we reflect the first petal over the pole, resulting in a symmetrical point located directly opposite the first petal.

Finally, to complete the graph of the lemniscate, we connect the two petals with a smooth, continuous curve, forming the characteristic infinity shape. For greater accuracy, 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.

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. The presence of \( r^2 \) indicates that we are dealing with polar coordinates, which is typical for lemniscates.

The first step in graphing this lemniscate is to identify the value of \( a \). Since the equation is in the form \( r^2 = a^2 \cos(2\theta) \), we can find \( a \) by taking the square root of 25, which gives us \( a = 5 \).

Next, we determine the angle \( \theta \) where the first petal will be located. Given that we have a positive \( a^2 \) and a cosine function, the first petal occurs at \( \theta = 0 \). Thus, we plot the first petal at the point \( (5, 0) \) on the polar axis.

To complete the graph, we reflect this petal over the pole, which gives us the second petal. This reflection will place the second petal at \( (-5, \pi) \) in polar coordinates. Finally, we connect these points with a smooth curve, resulting in the characteristic shape of the lemniscate, resembling an infinity symbol or a propeller.

In summary, the graph of \( r^2 = 25 \cos(2\theta) \) consists of two petals, each extending outwards from the origin, and is symmetric about the polar axis. This graphing process highlights the relationship between the polar equation and its geometric representation.

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

• A.
• B.
• C.
• D.

Answer: (B)

Here’s what students ask on this topic:

How do you convert polar coordinates to rectangular coordinates?

To convert polar coordinates (r, θ) to rectangular coordinates (x, y), use the following equations:

$x=r\cos \left(\theta \right)$

and

$y=r\sin \left(\theta \right)$

Here, r is the distance from the pole, and θ is the angle measured counterclockwise from the polar axis. For example, if the polar coordinates are (5, π/3), calculate:

$x=5\cos \left(\frac{\pi }{3}\right)$

and

$y=5\sin \left(\frac{\pi }{3}\right)$

After solving, x = 5/2 and y = 5√3/2. The rectangular coordinates are (5/2, 5√3/2).

What are coterminal angles in polar coordinates?

Coterminal angles are angles that share the same terminal side when measured from the polar axis. In polar coordinates, a point can have multiple representations due to coterminal angles. To find coterminal angles, add or subtract multiples of 2π:

$\theta \text{'}=\theta \pm 2\pi n$

where n is an integer. For example, the angle π/6 has coterminal angles like 13π/6 (π/6 + 2π) and -11π/6 (π/6 - 2π). These angles map to the same position on the polar coordinate system. Additionally, you can change the sign of r and add π to θ to find alternate representations of the same point.

How do you graph a cardioid in polar coordinates?

To graph a cardioid, start with its equation: r = a ± b cos(θ) or r = a ± b sin(θ), where a = b. Follow these steps:

1. Determine symmetry: If the equation contains cos(θ), the graph is symmetric about the polar axis. If it contains sin(θ), it is symmetric about the line θ = π/2.

2. Plot points at quadrantal angles (0, π/2, π, 3π/2). For each angle, calculate r using the equation.

3. Connect the points with a smooth curve, ensuring the cardioid’s characteristic heart-like shape. For example, r = 1 + cos(θ) has points at (2, 0), (1, π/2), (0, π), and (1, 3π/2). The graph is symmetric about the polar axis.

What is the difference between a cardioid and a limacon?

Both cardioids and limacons have equations of the form r = a ± b cos(θ) or r = a ± b sin(θ). The difference lies in the relationship between a and b:

- Cardioid: a = b. The graph resembles a heart shape.

- Limacon: a ≠ b. If a > b, the graph has a dimple. If a < b, the graph has an inner loop.

For example, r = 2 + 2 cos(θ) is a cardioid, while r = 3 - 2 sin(θ) is a limacon with a dimple because a > b.

How do you convert rectangular equations to polar equations?

To convert rectangular equations to polar equations, replace x and y with their polar equivalents:

$x=r\cos \left(\theta \right)$

and

$y=r\sin \left(\theta \right)$

Also, use:

$x{2}^{}+y{2}^{}=r^2$

For example, to convert x² + y² = 25, replace x² + y² with r² to get r² = 25. Solving for r gives r = 5, which represents a circle of radius 5 in polar form.