IDNLearn.com is your reliable source for expert answers and community insights. Our experts are ready to provide prompt and detailed answers to any questions you may have.
Sagot :
Let's solve the problem step-by-step:
### Step 1: Populate the Table of Values
We need to fill in the table with values of [tex]\( r \)[/tex] for different values of [tex]\( \theta \)[/tex] using the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex].
Using the given values of [tex]\(\theta\)[/tex]:
1. For [tex]\(\theta = 0\)[/tex]:
- [tex]\(\sin(0) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
2. For [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
3. For [tex]\(\theta = \frac{\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
4. For [tex]\(\theta = \frac{\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
5. For [tex]\(\theta = \frac{\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]
- [tex]\( r = \frac{2}{1} = 2 \)[/tex]
6. For [tex]\(\theta = \frac{2\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
7. For [tex]\(\theta = \frac{3\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
8. For [tex]\(\theta = \frac{5\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
9. For [tex]\(\theta = \pi\)[/tex]:
- [tex]\(\sin(\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
10. For [tex]\(\theta = \frac{3\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{2}\right) = -1 \)[/tex]
- [tex]\( r = \frac{2}{-1} = -2 \)[/tex]
11. For [tex]\(\theta = 2\pi\)[/tex]:
- [tex]\(\sin(2\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
Now, we populate the table:
[tex]\[ \begin{tabular}{|c|c|} \hline $\theta$ & $r=2 / \sin \theta$ \\ \hline 0 & \(\infty\) \\ \hline $\pi / 6$ & 4 \\ \hline $\pi / 4$ & 2.828 \\ \hline $\pi / 3$ & 2.309 \\ \hline $\pi / 2$ & 2 \\ \hline $2\pi / 3$ & 2.309 \\ \hline $3\pi / 4$ & 2.828 \\ \hline $5\pi / 6$ & 4 \\ \hline $\pi$ & \(\infty\) \\ \hline $3\pi / 2$ & -2 \\ \hline $2\pi$ & \(\infty\) \\ \hline \end{tabular} \][/tex]
### Step 2: Type of Graph
By observing the values, we notice that as [tex]\(\theta\)[/tex] approaches 0, [tex]\(\pi\)[/tex], or [tex]\(2\pi\)[/tex], [tex]\( r \)[/tex] tends to infinity.
When graphing these points in polar coordinates, you will see that for most [tex]\(\theta\)[/tex] values, the radius [tex]\(r\)[/tex] is positive and finite, and yet for [tex]\(\theta = 0\)[/tex], [tex]\(\pi\)[/tex], and [tex]\(2\pi\)[/tex], it becomes infinite. As [tex]\(\theta\)[/tex] moves from 0 to [tex]\(2\pi\)[/tex], the values stabilize and recreate a straight horizontal line.
### Step 3: Converting Polar Equation to Rectangular Equation
To justify the type of graph, we convert the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] to its rectangular form.
Given:
[tex]\[ r = \frac{2}{\sin \theta} \][/tex]
Multiplying both sides by [tex]\(\sin \theta\)[/tex]:
[tex]\[ r \sin \theta = 2 \][/tex]
Using the conversions for polar to rectangular coordinates:
[tex]\[ x = r \cos \theta \][/tex]
[tex]\[ y = r \sin \theta \][/tex]
We substitute [tex]\( r \sin \theta \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 2 \][/tex]
This is a linear equation of a horizontal line at [tex]\( y = 2 \)[/tex] in the rectangular coordinate system. Therefore, the graph of the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] corresponds to the line [tex]\(y = 2\)[/tex] in the rectangular coordinate system.
### Step 1: Populate the Table of Values
We need to fill in the table with values of [tex]\( r \)[/tex] for different values of [tex]\( \theta \)[/tex] using the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex].
Using the given values of [tex]\(\theta\)[/tex]:
1. For [tex]\(\theta = 0\)[/tex]:
- [tex]\(\sin(0) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
2. For [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
3. For [tex]\(\theta = \frac{\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
4. For [tex]\(\theta = \frac{\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
5. For [tex]\(\theta = \frac{\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]
- [tex]\( r = \frac{2}{1} = 2 \)[/tex]
6. For [tex]\(\theta = \frac{2\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
7. For [tex]\(\theta = \frac{3\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
8. For [tex]\(\theta = \frac{5\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
9. For [tex]\(\theta = \pi\)[/tex]:
- [tex]\(\sin(\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
10. For [tex]\(\theta = \frac{3\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{2}\right) = -1 \)[/tex]
- [tex]\( r = \frac{2}{-1} = -2 \)[/tex]
11. For [tex]\(\theta = 2\pi\)[/tex]:
- [tex]\(\sin(2\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
Now, we populate the table:
[tex]\[ \begin{tabular}{|c|c|} \hline $\theta$ & $r=2 / \sin \theta$ \\ \hline 0 & \(\infty\) \\ \hline $\pi / 6$ & 4 \\ \hline $\pi / 4$ & 2.828 \\ \hline $\pi / 3$ & 2.309 \\ \hline $\pi / 2$ & 2 \\ \hline $2\pi / 3$ & 2.309 \\ \hline $3\pi / 4$ & 2.828 \\ \hline $5\pi / 6$ & 4 \\ \hline $\pi$ & \(\infty\) \\ \hline $3\pi / 2$ & -2 \\ \hline $2\pi$ & \(\infty\) \\ \hline \end{tabular} \][/tex]
### Step 2: Type of Graph
By observing the values, we notice that as [tex]\(\theta\)[/tex] approaches 0, [tex]\(\pi\)[/tex], or [tex]\(2\pi\)[/tex], [tex]\( r \)[/tex] tends to infinity.
When graphing these points in polar coordinates, you will see that for most [tex]\(\theta\)[/tex] values, the radius [tex]\(r\)[/tex] is positive and finite, and yet for [tex]\(\theta = 0\)[/tex], [tex]\(\pi\)[/tex], and [tex]\(2\pi\)[/tex], it becomes infinite. As [tex]\(\theta\)[/tex] moves from 0 to [tex]\(2\pi\)[/tex], the values stabilize and recreate a straight horizontal line.
### Step 3: Converting Polar Equation to Rectangular Equation
To justify the type of graph, we convert the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] to its rectangular form.
Given:
[tex]\[ r = \frac{2}{\sin \theta} \][/tex]
Multiplying both sides by [tex]\(\sin \theta\)[/tex]:
[tex]\[ r \sin \theta = 2 \][/tex]
Using the conversions for polar to rectangular coordinates:
[tex]\[ x = r \cos \theta \][/tex]
[tex]\[ y = r \sin \theta \][/tex]
We substitute [tex]\( r \sin \theta \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 2 \][/tex]
This is a linear equation of a horizontal line at [tex]\( y = 2 \)[/tex] in the rectangular coordinate system. Therefore, the graph of the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] corresponds to the line [tex]\(y = 2\)[/tex] in the rectangular coordinate system.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
