
To calculate: To identify and graph the polar equation. Also to identify any symmetry and zeros of r

Answer to Problem 36E
Type of polar equation is limacon with inner loop
Polar equation is symmetric with respect to polar axis
Zeros of polar equations are θ=π6,11π6
Explanation of Solution
Given information: Polar equation is r=√3−2cosθ
Formula Used:
Cardioid: Heart shape curve and general form is r=a+acosθ
Limacons: General form of cardioid and general form is r=a+bcosθ
When ab<1 , polar equation represents limacon with inner loop
When ab=1 , polar equation represents Cardiod
When 1<ab<2 , polar equation represents Dimpled limacon
When ab=2 , polar equation represents convex limacon
Rose Curves: Sinusoidal curve that have a flower shape and general equation is r=acos(kθ)
Archimedian Spirals: Curve that extends indefinitely outward from a pole and general form is r=a+bθ
Lemniscate: Eight shaped curve and general form is r2=a2cos(2θ)
Test for Symmetry:
1. To test symmetry with respect to the line θ=π2 , replace (r,θ) by (r,π−θ)
2. To test symmetry with respect to the polar axis, replace (r,θ) by (r,−θ)
3. To test symmetry with respect to the pole, replace (r,θ) by (r,π+θ)
Calculation:
Polar equation is given as follows:
r=√3−2cosθ
Type of Polar Equation:
Polar equation is of form r=a+bcosθ such that ab<1
Thus, graph of polar equation is a limacon with inner loop
Test Symmetric of Polar Equation:
To test Symmetry respect with respect to the line θ=π2 :
Replacing (r,θ) by (r,π−θ) , polar equation is
r=√3−2cos(π−θ)r=√3+2cosθ
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
r=√3−2cos(−θ)r=√3−2cosθ
Since above equation is same as the given polar equation
Thus, polar equation is 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
r=√3−2cos(π+θ)r=√3+2cosθ
Since above equation is NOT same as the given polar equation
Thus, polar equation is NOT symmetric with respect to the pole
Calculating Zeros of Polar Equation:
In order to calculate zeros of polar equation, substitute r=0
Thus,
√3−2cosθ=0cosθ=√32θ=cos−1√32θ=π6,11π6
Graph of Polar Equation:
Graph of polar equation is as follows:
Conclusion:
Hence, type of polar equation is limacon with inner loop
Polar equation is symmetric with respect to polar axis
Zeros of polar equations are θ=π6,11π6
Chapter 9 Solutions
Precalculus with Limits: A Graphing Approach
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