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This results in n roots spaced equally around the

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Unformatted text preview: nitely many representations. As a result, if a point P is on the graph of two different polar equations, it is entirely possible that the representation P (r, θ) which satisfies one of the equations does not satisfy the other equation. Here, more than ever, we need to rely on the Geometry as much as the Algebra to find our solutions. Example 11.5.3. Find the points of intersection of the graphs of the following polar equations. 1. r = 2 sin(θ) and r = 2 − 2 sin(θ) 3. r = 3 and r = 6 cos(2θ) 2. r = 2 and r = 3 cos(θ) 4. r = 3 sin θ 2 and r = 3 cos θ 2 Solution. 1. Following the procedure in Example 11.5.2, we graph r = 2 sin(θ) and find it to be a circle centered at the point with rectangular coordinates (0, 1) with a radius of 1. The graph of r = 2 − 2 sin(θ) is a special kind of lima¸on called a ‘cardioid.’10 c y 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ ) It appears as if there are three intersection points: one in the first quadrant, one in the second quadrant, and the origin. Our next task...
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