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

r=6sin⁡θ−cos⁡θr=\frac{6}{\sin\theta-\cos\theta}r=sinθ−cosθ6

6r=sin⁡θ−cos⁡θ6r=\sin\theta-\cos\theta6r=sinθ−cosθ

r=sin⁡θ−cos⁡θ6r=\frac{\sin\theta-\cos\theta}{6}r=6sinθ−cosθ

r=6(sin⁡θ−cos⁡θ)r=6\left(\sin\theta-\cos\theta\right)r=6(sinθ−cosθ)

Convert each equation to its polar form.3y−5x=23y-5x=2

r=23sin⁡θ−5cos⁡θr=\frac{2}{3\sin\theta-5\cos\theta}r=3sinθ−5cosθ2

r=2sin⁡θ−cos⁡θr=\frac{2}{\sin\theta-\cos\theta}r=sinθ−cosθ2

r=3sin⁡θ−5cos⁡θr=3\sin\theta-5\cos\thetar=3sinθ−5cosθ

r=35sin⁡θr=\frac35\sin\thetar=53sinθ

Convert each equation to its polar form.x2+y2=$$2yx^2$+y^2=2y$$

r=2sin⁡θr=2\sin\thetar=2sinθ

r=4sin⁡θr=4\sin\thetar=4sinθ

r=2cos⁡θr=2\cos\thetar=2cosθ

Convert each equation to its polar form.x2+(y−2)2=$$4x^2$+(y-2)^2=4$$

r=4sin⁡θr=4\sin\thetar=4sinθ

r2=4sin⁡θ$$r^2=4$$\sin\thetar2=4sinθ

r=2sin⁡θr=2\sin\thetar=2sinθ

r=cos⁡θ−sin⁡θr=\cos\theta-\sin\thetar=cosθ−sinθ

8. Polar Coordinates and Complex Numbers

Convert Equations Between Polar & Rectangular Forms

8. Polar Coordinates and Complex Numbers

Convert Equations Between Polar & Rectangular Forms: Videos & Practice Problems

Video Lessons Worksheet

Concept

Convert Equations from Rectangular to Polar

Video duration:

Convert Equations from Rectangular to Polar Video Summary

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 is straightforward and can be accomplished through algebraic manipulation.

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

$r \sin \theta = 5$.

To isolate $r$, divide both sides by $\sin \theta$:

$r = \frac{5}{\sin \theta}$.

Recognizing that $\frac{1}{\sin \theta} = \csc \theta$, we can express the final polar form as:

$r = 5 \csc \theta$.

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

$r \sin \theta = r \cos \theta + 1$.

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

$r \sin \theta - r \cos \theta = 1$.

Factoring out $r$ results in:

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

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

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

Lastly, for the equation $x^2 + y^2 = 25$, we can utilize the identity $x^2 + y^2 = r^2$. Thus, we can directly substitute:

$r^2 = 25$.

Taking the square root of both sides gives:

$r = 5$.

This indicates a circle of radius 5 in polar coordinates.

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

Study Smarter with Worksheets.

Follow along with each video using our printable worksheets

Example

Convert Equations from Rectangular to Polar Example 1

Video duration:

Convert Equations from Rectangular to Polar Example 1 Video Summary

In this discussion, we explore the conversion of equations from rectangular to polar form, utilizing the relationships between Cartesian coordinates and polar coordinates. The key relationships to remember are that $ 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 right side 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} $

This expression represents the polar form of the original equation.

In the second example, we are given the equation $ 4xy = 2 $ and tasked with solving for $ r^2 $. Again, we substitute $ x $ and $ y $ with their polar equivalents:

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

This simplifies to:

$ 4r^2 \cos \theta \sin \theta = 2 $

Dividing both sides by 2 yields:

$ 2r^2 \cos \theta \sin \theta = 1 $

Recognizing that $ \cos \theta \sin \theta $ can be expressed using the double angle identity, we rewrite it as:

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

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

$ 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) $

This final expression represents the polar form of the original equation. Understanding these conversions is crucial for solving problems in polar coordinates effectively.

Problem

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

$6r=$\sin\[\theta$-$\cos\]\theta$$

$r=$\frac{\sin\theta-\cos\theta}{6}$$

$r=6$\left$($\sin\]\theta$-$\cos\[\theta\]\right$)$

$r=$\frac{6}{\sin\theta-\cos\theta}$$

Problem

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

$r=$\frac{2}{3\sin\theta-5\cos\theta}$$

$r=$\frac{2}{\sin\theta-\cos\theta}$$

$r=3$\sin\[\theta$-5$\cos\]\theta$$

$r=$\frac$35$\sin\]\theta$$

Problem

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

$r=$\sqrt{2}$$

$r=2$\sin\]\theta$$

$r=4$\sin\]\theta$$

$r=2$\cos\]\theta$$

Problem

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

$r^2=4$\sin\]\theta$$

$r=4$\sin\]\theta$$

$r=2$\sin\]\theta$$

$r=$\cos\[\theta$-$\sin\]\theta$$

Concept

Convert Equations from Polar to Rectangular

Video duration:

Convert Equations from Polar to Rectangular Video Summary

Converting equations from polar to rectangular form involves a systematic approach, as there isn't a single operation that applies universally. In polar coordinates, points are represented by $ r $ (the radius) and $ \theta $ (the angle), while in rectangular coordinates, they are represented by $ x $ and $ y $. The key relationships to remember are:

1. $ x = r \cos(\theta) $

2. $ y = r \sin(\theta) $

3. $ r^2 = x^2 + y^2 $

To convert a polar equation to rectangular form, the first step is to manipulate the equation to include one of these terms. For example, consider the polar equation $ r = 4 $. Since it does not contain any of the necessary terms, we can square both sides to obtain:

$ r^2 = 16 $

Next, we replace $ r^2 $ with $ x^2 + y^2 $, leading to:

$ x^2 + y^2 = 16 $

This equation represents a circle with a radius of 4, centered at the origin, and is already in standard form.

In another example, for the polar equation $ r = \sec(\theta) $, we can rewrite the secant function as:

$ r = \frac{1}{\cos(\theta)} $

To eliminate the fraction, multiply both sides by $ \cos(\theta) $:

$ r \cos(\theta) = 1 $

Substituting $ r \cos(\theta) $ with $ x $ gives:

\( x = 1 \

This represents a vertical line at $ x = 1 $, which is also in standard form.

For the polar equation $ r = 6 \sin(\theta) $, we start by multiplying both sides by $ r $ to obtain:

\( r^2 = 6r \sin(\theta) \

Substituting gives:

\( x^2 + y^2 = 6y \

To rewrite this in standard form, we rearrange it to:

\( x^2 + y^2 - 6y = 0 \

Next, we complete the square for the $ y $ terms. The expression $ y^2 - 6y $ can be transformed into a perfect square by adding 9:

\( x^2 + (y - 3)^2 = 9 \

This represents a circle centered at $ (0, 3) $ with a radius of 3, now in standard form.

By following these steps and strategies, you can effectively convert polar equations into rectangular form and identify their geometric representations. Practice with various equations will enhance your understanding and proficiency in this conversion process.

Example

Convert Equations from Polar to Rectangular Example 2

Video duration:

Convert Equations from Polar to Rectangular Example 2 Video Summary

To convert the polar equation $ r = \frac{2}{1 + \cos \theta} $ into its rectangular form, we follow a systematic approach. The goal is to express the equation using $ x $ and $ y $ instead of $ r $ and $ \theta $. We start by eliminating the fraction by multiplying both sides by the denominator $ 1 + \cos \theta $, leading to:

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

This simplifies to:

$ 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 use the relationship $ r = \sqrt{x^2 + y^2} $ to substitute for $ r $. Thus, we rewrite the equation as:

$ \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, resulting in:

$ y^2 = 4 - 4x $

This equation can be rearranged to:

$ y^2 = -4x + 4 $

Recognizing the form of this equation, we identify it as a horizontal parabola. The standard form of a horizontal parabola is $ (y - k)^2 = 4p(x - h) $, where $ (h, k) $ is the vertex. In this case, the vertex is at $ (1, 0) $ and it opens to the left, confirming that the graph of the original polar equation is indeed a horizontal parabola.

Problem