Unformatted text preview: st quadrant. For the point of intersection
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
for which πk = π + 2πk which means while the origin is on both graphs, the point is never
reached simultaneously. In any case, we have determined the three points of intersection to
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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