Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To find the intersection points of the curves defined by the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex], we need to check if the given candidate points satisfy both equations simultaneously.
### Candidate Points:
1. [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
2. [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
3. [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
4. [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
5. [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
We will check each point one by one.
#### Check 1: [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 2: [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 3: [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(\frac{1}{2}\right) = 5 + 2 = 7 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(\frac{1}{2}\right) = -3 \)[/tex].
Since [tex]\(3 \neq 7\)[/tex] and [tex]\(3 \neq -3\)[/tex], [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 4: [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex] is an intersection point.
#### Check 5: [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex] is an intersection point.
### Conclusion:
The intersection points of the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
Thus, the correct points are:
[tex]\[ \left(3, \frac{7\pi}{6}\right), \left(3, \frac{11\pi}{6}\right) \][/tex]
### Candidate Points:
1. [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
2. [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
3. [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
4. [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
5. [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
We will check each point one by one.
#### Check 1: [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 2: [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 3: [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(\frac{1}{2}\right) = 5 + 2 = 7 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(\frac{1}{2}\right) = -3 \)[/tex].
Since [tex]\(3 \neq 7\)[/tex] and [tex]\(3 \neq -3\)[/tex], [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 4: [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex] is an intersection point.
#### Check 5: [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex] is an intersection point.
### Conclusion:
The intersection points of the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
Thus, the correct points are:
[tex]\[ \left(3, \frac{7\pi}{6}\right), \left(3, \frac{11\pi}{6}\right) \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
