Test the polar equation for symmetry with respect to the polar axis, the pole, and the line 0 = 1 (Select all that apply.) 2 r= 9 sec(O) O symmetric with respect to the polar axis symmetric with respect to the pole O symmetric with respect to the line 0 = 1 7 not symmetric with respect to any of these – [-12.12 Points] DETAILS SPRECALC7 8.2.015. Test the polar equation for symmetry with respect to the polar axis, the pole, and the line 0 = 1. (Select all that apply.) p2 = 16 cos(20) symmetric with respect to the polar axis O symmetric with respect to the pole O symmetric with respect to the line 0 = 1 7 O not symmetric with respect to any of these
The Correct Answer and Explanation is :
Let’s analyze the given polar equations for symmetry with respect to the polar axis, the pole, and the line ( \theta = 1 ).
Polar Equation 1: ( r = 9 \sec(\theta) )
Symmetry with respect to the polar axis:
- For symmetry with respect to the polar axis, we check if replacing ( \theta ) with ( -\theta ) results in the same equation.
- The equation becomes:
[
r = 9 \sec(-\theta)
]
Since ( \sec(-\theta) = \sec(\theta) ), the equation remains the same. - Therefore, this equation is symmetric with respect to the polar axis.
Symmetry with respect to the pole:
- For symmetry with respect to the pole, we check if replacing ( r ) with ( -r ) results in the same equation.
- The equation becomes:
[
-r = 9 \sec(\theta)
]
This equation is not the same as the original, so it does not have symmetry with respect to the pole.
Symmetry with respect to the line ( \theta = 1 ):
- To check for symmetry with respect to the line ( \theta = 1 ), we would typically replace ( \theta ) with ( 2 – \theta ) and see if the equation remains unchanged.
- Substituting into the equation:
[
r = 9 \sec(2 – \theta)
]
This does not simplify to the original equation, so it is not symmetric with respect to the line ( \theta = 1 ).
Polar Equation 2: ( r^2 = 16 \cos(2\theta) )
Symmetry with respect to the polar axis:
- For symmetry with respect to the polar axis, we replace ( \theta ) with ( -\theta ) and check if the equation stays the same:
[
r^2 = 16 \cos(-2\theta) = 16 \cos(2\theta)
]
Since the equation remains unchanged, this equation is symmetric with respect to the polar axis.
Symmetry with respect to the pole:
- For symmetry with respect to the pole, we replace ( r ) with ( -r ) and check if the equation remains the same:
[
(-r)^2 = 16 \cos(2\theta)
]
Since ( (-r)^2 = r^2 ), the equation remains unchanged. Therefore, this equation is symmetric with respect to the pole.
Symmetry with respect to the line ( \theta = 1 ):
- Substituting ( 2 – \theta ) for ( \theta ):
[
r^2 = 16 \cos(2(2 – \theta)) = 16 \cos(4 – 2\theta)
]
This does not simplify to the original equation, so the equation is not symmetric with respect to the line ( \theta = 1 ).
Conclusion
- For the first equation ( r = 9 \sec(\theta) ), it is symmetric with respect to the polar axis, but not symmetric with respect to the pole or the line ( \theta = 1 ).
- For the second equation ( r^2 = 16 \cos(2\theta) ), it is symmetric with respect to the polar axis and the pole, but not symmetric with respect to the line ( \theta = 1 ).