wn 1 and they are given by the formula wk n rcis 2 k

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Unformatted text preview: the curve pulls back to the origin. r y 5 π 4 −5 π 2 3π 4 π x θ 11.5 Graphs of Polar Equations 807 Even though we have finished with one complete cycle of r = 5 sin(θ), if we continue plotting beyond θ = π , we find that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(θ) which continues on from the first. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy -plane. y r 5 3 4 π1 5π 4 2 3π 2 3 7π 4 x 4 2π θ 1 2 −5 We have the final graph below. y r 5 5 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −5 θ −5 5 x −5 r = 5 sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy -plane 4. Graphing r2 = 16 cos(2θ) is complicated by the r2 , so we solve to get r = ± 16 cos(2θ) = ±4 cos(2θ). How do we sketch such a curve? First off, we sketch a fundamental period of r = cos(2θ) which we have dotted in the figure below. When cos(2θ) < 0, cos(2θ) is π undefined, so we don’t have any values on the interval π , 34 . On the intervals which remain, 4 cos(2θ) ranges from 0 to 1, inclusive. Hence,...
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