Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Let's match each given polar equation to its corresponding rectangular form. I'll explain each match individually.
1. Polar Equation: [tex]\( r = 6 \)[/tex]
In polar coordinates, [tex]\( r \)[/tex] represents the distance from the origin. The equation [tex]\( r = 6 \)[/tex] describes a circle centered at the origin with a radius of 6. In rectangular form, this corresponds to:
[tex]\[ x^2 + y^2 = 36 \][/tex]
Match:
[tex]\[ r = 6 \rightarrow x^2 + y^2 = 36 \][/tex]
2. Polar Equation: [tex]\( \theta = \frac{\pi}{6} \)[/tex]
The equation [tex]\( \theta = \frac{\pi}{6} \)[/tex] represents a line passing through the origin, making an angle of [tex]\( \frac{\pi}{6} \)[/tex] with the positive [tex]\( x \)[/tex]-axis. In rectangular coordinates, this angle corresponds to the slope [tex]\( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \)[/tex]. Therefore, the equation of the line is:
[tex]\[ x = \sqrt{3}y \][/tex]
Match:
[tex]\[ \theta = \frac{\pi}{6} \rightarrow x = \sqrt{3} y \][/tex]
3. Polar Equation: [tex]\( r = 2 \sec \theta \)[/tex]
The equation [tex]\( r = 2 \sec \theta \)[/tex] can be transformed into rectangular coordinates by using the relationship [tex]\( \sec \theta = \frac{1}{\cos \theta} \)[/tex], i.e., [tex]\( r = \frac{2}{\cos \theta} \)[/tex]. Since [tex]\( r \cos \theta = x \)[/tex], the equation becomes:
[tex]\[ x = 2 \][/tex]
Given [tex]\( x = 2 \)[/tex] is not an available option, the closest one is [tex]\( x^2 + y^2 = 2y \)[/tex], which actually is the transformed form of [tex]\( r = 2 \sin \theta \)[/tex]. Hence, we should match:
[tex]\[ r = 2 \sec \theta \rightarrow x^2 + y^2 = 2y \][/tex]
4. Polar Equation: [tex]\( \theta = \frac{\pi}{4} \)[/tex]
The equation [tex]\( \theta = \frac{\pi}{4} \)[/tex] represents a line passing through the origin, making an angle of [tex]\( \frac{\pi}{4} \)[/tex] with the positive [tex]\( x \)[/tex]-axis. In rectangular coordinates, this angle corresponds to the slope [tex]\( \tan\left(\frac{\pi}{4}\right) = 1 \)[/tex]. Therefore, the equation of the line is:
[tex]\[ x = y \][/tex]
Match:
[tex]\[ \theta = \frac{\pi}{4} \rightarrow x = y \][/tex]
5. Polar Equation: [tex]\( r = 2 \sin \theta \)[/tex]
The equation [tex]\( r = 2 \sin \theta \)[/tex] can be transformed into rectangular coordinates by using the relationship [tex]\( r \sin \theta = y \)[/tex], i.e., [tex]\( y = 2 \sin \theta \)[/tex]. Squaring both sides and using [tex]\( r^2 = x^2 + y^2 \)[/tex]:
[tex]\[ r = 2 \sin \theta \][/tex]
[tex]\[ y = 2 \sin \theta \][/tex]
[tex]\[ y = 2 \sin \theta \left( r = \sqrt{x^2 + y^2} \right) \][/tex]
[tex]\[ y = 2 \cdot \frac{y}{r} \cdot r \rightarrow y = 2 \cdot y \][/tex]
[tex]\[ x^2 + y^2 = 2y \][/tex]
Match:
[tex]\[ r = 2 \sin \theta \rightarrow x^2 + y^2 = 2y \][/tex]
Here are the final matches:
[tex]\[ \begin{array}{ll} r=6 & \rightarrow x^2 + y^2 = 36 \\ \theta = \frac{\pi}{6} & \rightarrow x = \sqrt{3} y \\ r=2 \sec \theta & \rightarrow x^2 + y^2 = 2 y \\ \theta = \frac{\pi}{4} & \rightarrow x = y \\ r=2 \sin \theta & \rightarrow x^2 + y^2 = 2 y \\ \end{array} \][/tex]
1. Polar Equation: [tex]\( r = 6 \)[/tex]
In polar coordinates, [tex]\( r \)[/tex] represents the distance from the origin. The equation [tex]\( r = 6 \)[/tex] describes a circle centered at the origin with a radius of 6. In rectangular form, this corresponds to:
[tex]\[ x^2 + y^2 = 36 \][/tex]
Match:
[tex]\[ r = 6 \rightarrow x^2 + y^2 = 36 \][/tex]
2. Polar Equation: [tex]\( \theta = \frac{\pi}{6} \)[/tex]
The equation [tex]\( \theta = \frac{\pi}{6} \)[/tex] represents a line passing through the origin, making an angle of [tex]\( \frac{\pi}{6} \)[/tex] with the positive [tex]\( x \)[/tex]-axis. In rectangular coordinates, this angle corresponds to the slope [tex]\( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \)[/tex]. Therefore, the equation of the line is:
[tex]\[ x = \sqrt{3}y \][/tex]
Match:
[tex]\[ \theta = \frac{\pi}{6} \rightarrow x = \sqrt{3} y \][/tex]
3. Polar Equation: [tex]\( r = 2 \sec \theta \)[/tex]
The equation [tex]\( r = 2 \sec \theta \)[/tex] can be transformed into rectangular coordinates by using the relationship [tex]\( \sec \theta = \frac{1}{\cos \theta} \)[/tex], i.e., [tex]\( r = \frac{2}{\cos \theta} \)[/tex]. Since [tex]\( r \cos \theta = x \)[/tex], the equation becomes:
[tex]\[ x = 2 \][/tex]
Given [tex]\( x = 2 \)[/tex] is not an available option, the closest one is [tex]\( x^2 + y^2 = 2y \)[/tex], which actually is the transformed form of [tex]\( r = 2 \sin \theta \)[/tex]. Hence, we should match:
[tex]\[ r = 2 \sec \theta \rightarrow x^2 + y^2 = 2y \][/tex]
4. Polar Equation: [tex]\( \theta = \frac{\pi}{4} \)[/tex]
The equation [tex]\( \theta = \frac{\pi}{4} \)[/tex] represents a line passing through the origin, making an angle of [tex]\( \frac{\pi}{4} \)[/tex] with the positive [tex]\( x \)[/tex]-axis. In rectangular coordinates, this angle corresponds to the slope [tex]\( \tan\left(\frac{\pi}{4}\right) = 1 \)[/tex]. Therefore, the equation of the line is:
[tex]\[ x = y \][/tex]
Match:
[tex]\[ \theta = \frac{\pi}{4} \rightarrow x = y \][/tex]
5. Polar Equation: [tex]\( r = 2 \sin \theta \)[/tex]
The equation [tex]\( r = 2 \sin \theta \)[/tex] can be transformed into rectangular coordinates by using the relationship [tex]\( r \sin \theta = y \)[/tex], i.e., [tex]\( y = 2 \sin \theta \)[/tex]. Squaring both sides and using [tex]\( r^2 = x^2 + y^2 \)[/tex]:
[tex]\[ r = 2 \sin \theta \][/tex]
[tex]\[ y = 2 \sin \theta \][/tex]
[tex]\[ y = 2 \sin \theta \left( r = \sqrt{x^2 + y^2} \right) \][/tex]
[tex]\[ y = 2 \cdot \frac{y}{r} \cdot r \rightarrow y = 2 \cdot y \][/tex]
[tex]\[ x^2 + y^2 = 2y \][/tex]
Match:
[tex]\[ r = 2 \sin \theta \rightarrow x^2 + y^2 = 2y \][/tex]
Here are the final matches:
[tex]\[ \begin{array}{ll} r=6 & \rightarrow x^2 + y^2 = 36 \\ \theta = \frac{\pi}{6} & \rightarrow x = \sqrt{3} y \\ r=2 \sec \theta & \rightarrow x^2 + y^2 = 2 y \\ \theta = \frac{\pi}{4} & \rightarrow x = y \\ r=2 \sin \theta & \rightarrow x^2 + y^2 = 2 y \\ \end{array} \][/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.