
To calculate: To test for symmetry with respect to the line θ=π2 , the polar axis and the pole

Answer to Problem 21E
The polar equation is NOT symmetric with respect to the line θ=π2 and polar axis, while equation is symmetric with respect to pole
Explanation of Solution
Given information: Polar equation is r2=16sin2θ
Formula Used:
Test for Symmetry:
To test symmetry with respect to the line θ=π2 , replace (r,θ) by (r,π−θ)
To test symmetry with respect to the polar axis, replace (r,θ) by (r,−θ)
To test symmetry with respect to the pole, replace (r,θ) by (r,π+θ)
Calculation:
Polar equation is given as follows:
r2=16sin2θ
To test Symmetry respect with respect to the line θ=π2 :
Replacing (r,θ) by (r,π−θ) , polar equation is
r2=16sin2(π−θ)r2=16sin(2π−2θ)r2=16(−sin2θ)r2=−16sin2θ
Since above equation is NOT same as the given polar equation
Thus, polar equation is NOT symmetric with respect to the line θ=π2
To test Symmetry with respect to the polar axis
Replacing (r,θ) by (r,−θ) , polar equation is
r2=16sin2(−θ)r2=16sin(−2θ)r2=16(−sin2θ)r2=−16sin2θ
Since above equation is not same as the given polar equation
Thus, polar equation is NOT symmetric with respect to the polar axis
To test Symmetry with respect to the pole
Replacing (r,θ) by (r,π+θ) , polar equation is
r2=16sin2(π+θ)r2=16sin(2π+2θ)r2=16(sin2θ)r2=16sin2θ
Since above equation is same as the given polar equation
Thus, polar equation is with respect to the pole
Conclusion:
Hence, polar equation is NOT symmetric with respect to the line θ=π2 and polar axis, while equation is symmetric with respect to pole
Chapter 9 Solutions
Precalculus with Limits: A Graphing Approach
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