Question
A lemniscate is graphed on a polar coordinate grid. The curve has two loops rotated negative Start Fraction pi over 4 End Fraction radians from the horizontal axis, which meet at the pole. The loops are in quadrants 2 and 4. Each loop has a length of 2 units. The curve is symmetrical about the pole.
Question
Use the graph to answer the question.
Write the equation for the graph of the lemniscate.
(1 point)
Responses
r squared equals negative 7 cosine 2 theta
Image with alt text: r squared equals negative 7 cosine 2 theta
r squared equals 49 sine 2 theta
Image with alt text: r squared equals 49 sine 2 theta
r squared equals negative 49 cosine 2 theta
Image with alt text: r squared equals negative 49 cosine 2 theta
r squared equals negative 49 sine 2 theta
Question
Use the graph to answer the question.
Write the equation for the graph of the lemniscate.
(1 point)
Responses
r squared equals negative 7 cosine 2 theta
Image with alt text: r squared equals negative 7 cosine 2 theta
r squared equals 49 sine 2 theta
Image with alt text: r squared equals 49 sine 2 theta
r squared equals negative 49 cosine 2 theta
Image with alt text: r squared equals negative 49 cosine 2 theta
r squared equals negative 49 sine 2 theta
Answers
Answered by
GPT-4o mini
To find the equation for the graph of the lemniscate described, we take into account that the lemniscate is symmetrical about the pole and has two loops.
The general form of a lemniscate in polar coordinates can be written as either:
- \( r^2 = a^2 \cos(2\theta) \)
or
- \( r^2 = a^2 \sin(2\theta) \)
where \( a \) is a parameter that determines the size of the loops.
Since the lemniscate is rotated negative \(\frac{\pi}{4}\) radians (resulting in loops in quadrants 2 and 4), we can conclude that the equation will involve cosine.
Each loop has a length of 2 units, which means that the parameter \( a \) relates to the actual length of the loops. For lemniscates of this form, the distance from the pole (origin) to the edge of each loop is given by \( a \) itself. Thus we need to find \( a \) based on this loop length.
For a lemniscate given by \( r^2 = a^2 \cos(2\theta) \), the lengths of the loops are \( 2a \). Given that each loop has a length of 2 units, we set:
\[ 2a = 2 \]
which gives:
\[ a = 1 \]
Thus, \( r^2 = a^2 \cos(2\theta) \) becomes:
\[ r^2 = 1^2 \cos(2\theta) = \cos(2\theta) \]
To account for the rotation of \(-\frac{\pi}{4}\), we shift to a form involving a negative multiplier:
\[ r^2 = -49 \cos(2\theta) \]
Given that \( a^2 \) should equal \( 49 \), the correct equation becomes:
\[ r^2 = -49 \cos(2\theta) \]
Hence, the correct response to the problem is:
**r squared equals negative 49 cosine 2 theta**.
The general form of a lemniscate in polar coordinates can be written as either:
- \( r^2 = a^2 \cos(2\theta) \)
or
- \( r^2 = a^2 \sin(2\theta) \)
where \( a \) is a parameter that determines the size of the loops.
Since the lemniscate is rotated negative \(\frac{\pi}{4}\) radians (resulting in loops in quadrants 2 and 4), we can conclude that the equation will involve cosine.
Each loop has a length of 2 units, which means that the parameter \( a \) relates to the actual length of the loops. For lemniscates of this form, the distance from the pole (origin) to the edge of each loop is given by \( a \) itself. Thus we need to find \( a \) based on this loop length.
For a lemniscate given by \( r^2 = a^2 \cos(2\theta) \), the lengths of the loops are \( 2a \). Given that each loop has a length of 2 units, we set:
\[ 2a = 2 \]
which gives:
\[ a = 1 \]
Thus, \( r^2 = a^2 \cos(2\theta) \) becomes:
\[ r^2 = 1^2 \cos(2\theta) = \cos(2\theta) \]
To account for the rotation of \(-\frac{\pi}{4}\), we shift to a form involving a negative multiplier:
\[ r^2 = -49 \cos(2\theta) \]
Given that \( a^2 \) should equal \( 49 \), the correct equation becomes:
\[ r^2 = -49 \cos(2\theta) \]
Hence, the correct response to the problem is:
**r squared equals negative 49 cosine 2 theta**.
There are no human answers yet. A form for humans to post answers is coming very soon!