Unformatted text preview: the curve pulls back to the origin.
y 5 π
4 −5 π
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.
4 2 3π
2 3 7π
4 x 4 2π θ 1 2
−5 We have the ﬁnal graph below.
5 5 π
4 π 5π
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
undeﬁned, so we don’t have any values on the interval π , 34 . On the intervals which remain,
cos(2θ) ranges from 0 to 1, inclusive. Hence,...
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