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

r = 6 sin ⁡ θ − cos ⁡ θ r=\frac{6}{\sin\theta-\cos\theta} r = s i n θ − c o s θ 6

6 r = sin ⁡ θ − cos ⁡ θ 6r=\sin\theta-\cos\theta 6 r = sin θ − cos θ

r = sin ⁡ θ − cos ⁡ θ 6 r=\frac{\sin\theta-\cos\theta}{6} r = 6 s i n θ − c o s θ

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

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

r = 2 3 sin ⁡ θ − 5 cos ⁡ θ r=\frac{2}{3\sin\theta-5\cos\theta} r = 3 s i n θ − 5 c o s θ 2

r = 2 sin ⁡ θ − cos ⁡ θ r=\frac{2}{\sin\theta-\cos\theta} r = s i n θ − c o s θ 2

r = 3 sin ⁡ θ − 5 cos ⁡ θ r=3\sin\theta-5\cos\theta r = 3 sin θ − 5 cos θ

r = 3 5 sin ⁡ θ r=\frac35\sin\theta r = 5 3 sin θ

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

r = 2 sin ⁡ θ r=2\sin\theta r = 2 sin θ

r = 2 r=\sqrt{2} r = 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

r = 4 sin ⁡ θ r=4\sin\theta r = 4 sin θ

r = 2 cos ⁡ θ r=2\cos\theta r = 2 cos θ

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

r = 4 sin ⁡ θ r=4\sin\theta r = 4 sin θ

r 2 = 4 sin ⁡ θ r^2=4\sin\theta r 2 = 4 sin θ

r = 2 sin ⁡ θ r=2\sin\theta r = 2 sin θ

r = cos ⁡ θ − sin ⁡ θ r=\cos\theta-\sin\theta r = cos θ − sin θ

Convert each equation to its rectangular form. r = − 4 cos ⁡ θ r=-4\cos\theta r = − 4 cos θ

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

x 2 + y 2 = 4 x^2+y^2=4 x 2 + y 2 = 4

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

Convert each equation to its rectangular form. r = 2 1 − sin ⁡ θ r=\frac{2}{1-\sin\theta} r = 1 − s i n θ 2

y = 1 4 x 2 − 1 y=\frac14x^2-1 y = 4 1 x 2 − 1

y 2 = 4 − 4 x y^2=4-4x y 2 = 4 − 4 x

x 2 + y 2 = 2 y x^2+y^2=2y x 2 + y 2 = 2 y

x 2 − 1 = y x^2-1=y x 2 − 1 = y

15. Polar Equations

Convert Equations Between Polar and Rectangular Forms

15. Polar Equations

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

Video Lessons Practice Worksheet

Topic summary

To convert equations from rectangular to polar form, replace x and y with $r^{2}$ and $r \sin θ$ respectively. For example, the equation $y = 5$ becomes $r = 5 \csc^{(θ)}$ . Understanding these conversions is essential for graphing and analyzing polar equations effectively.

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 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:

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

This can be further simplified using the cosecant function, yielding:

$r = 5 \csc \theta$.

Next, let's convert the equation $y = x + 1$. Substituting both $x$ and $y$ gives us:

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

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 $r$ results in:

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

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

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

Lastly, consider the equation $x^2 + y^2 = 25$. Recognizing that $x^2 + y^2$ is equivalent to $r^2$, we can directly substitute:

$r^2 = 25$.

Taking the square root of both sides leads to:

$r = 5$.

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.

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, a crucial skill in understanding the relationship between Cartesian coordinates and polar coordinates. The polar coordinates are defined using the relationships $ 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} \

In the second example, we convert 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 solve for $ 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:

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

Through these examples, we see how to effectively convert rectangular equations into polar form, utilizing trigonometric identities and algebraic manipulation to achieve the desired results.

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.
$3y-5x=2$

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

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

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

$r=\frac35\sin\theta$

Problem

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

$r=\sqrt{2}$

$r=2\sin\theta$

$r=4\sin\theta$

$r=2\cos\theta$

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 using $ r $ (the radius) and $ \theta $ (the angle), while rectangular coordinates use $ 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 $.

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

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

Replacing $ r^2 $ with $ x^2 + y^2 $ and $ r \sin(\theta) $ with $ y $ results in:

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

To rewrite this in standard form, we rearrange the equation:

\( 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.

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 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 $

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 know that:

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

Substituting these into the equation gives us:

$ \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 results in:

$ 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 = -4x + 4 $

Recognizing the form of this equation, we see that it represents a horizontal parabola. The standard form of a horizontal parabola is given by $ (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.

In summary, the polar equation $ r = \frac{2}{1 + \cos \theta} $ converts to the rectangular form $ y^2 = 4 - 4x $, which describes a horizontal parabola.

Problem

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

$r=-4x$

$x^2+y^2=4$

$(x+2)^2+y^2=2$

$(x+2)^2+y^2=4$

Problem

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

$y^2=4-4x$

$x^2+y^2=2y$

$y=\frac14x^2-1$

$x^2-1=y$

Do you want more practice?

More sets

Convert Equations Between Polar and Rectangular Forms

Go over this topic definitions with flashcards

More sets

Convert Equations Between Polar and Rectangular Forms definitions
15. Polar Equations
15 Terms

Here’s what students ask on this topic:

To convert the equation $y = 5$ from rectangular to polar form, replace $y$ with $r \sin θ$ . This gives $r \sin θ = 5$ . Solve for $r$ by dividing both sides by $\sin θ$ , resulting in $r = \frac{5}{\sin}$ . Recognize that $\frac{1}{\sin}$ is the cosecant function, so the polar form is $r = 5 \csc θ$ .

To convert the polar equation $r = 4$ to rectangular form, start by squaring both sides to get $r^{2} = 16$ . Recognize that $r^{2}$ is equivalent to $x^{2} + y^{2}$ in rectangular coordinates. Thus, replace $r^{2}$ with $x^{2} + y^{2}$ to obtain $x^{2} + y^{2} = 16$ . This equation represents a circle with a radius of 4 centered at the origin.

To convert $r = 6 \sin θ$ to rectangular form, multiply both sides by $r$ to get $r^{2} = 6 r \sin θ$ . Replace $r^{2}$ with $x^{2} + y^{2}$ and $r \sin θ$ with $y$ , resulting in $x^{2} + y^{2} = 6 y$ . Rearrange to $x^{2} + y^{2} - 6 y = 0$ and complete the square for $y$ to get $x^{2} + (y - 3)^{2} = 9$ , representing a circle centered at $(0, 3)$ with radius 3.

To convert $y = x + 1$ to polar form, replace $y$ with $r \sin θ$ and $x$ with $r \cos θ$ . This gives $r \sin θ = r \cos θ + 1$ . Rearrange to $r (\sin θ - \cos θ) = 1$ and solve for $r$ by dividing both sides by $\sin θ - \cos θ$ , resulting in $r = \frac{1}{\sin}$ .