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In exercise 13 in section 87 we showed you how to

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Unformatted text preview: st quadrant. For the point of intersection π π in the second quadrant, we try θ = 56 . Both equations give us the point 1, 56 , so this is our answer here. What about the origin? We know from Section 11.4 that the pole may be represented as (0, θ) for any angle θ. On the graph of r = 2 sin(θ), we start at the origin when θ = 0 and return to it at θ = π , and as the reader can verify, we are at the origin exactly when θ = πk for integers k . On the curve r = 2 − 2 sin(θ), however, we reach the origin when θ = π , and more generally, when θ = π + 2πk for integers k . There is no integer value of k 2 2 for which πk = π + 2πk which means while the origin is on both graphs, the point is never 2 reached simultaneously. In any case, we have determined the three points of intersection to π be 1, π , 1, 56 and the origin. 6 2. As before, we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. The graph of r = 2 is a c...
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