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Stitz-Zeager_College_Algebra_e-book

# 6 2 3 5 6 8cis 5 8 cos 5 6 i sin 2 we use

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Unformatted text preview: terval [0, π ] results in retracting some portion of the curve.4 We present the ﬁnal graph below. y r 6 3 3 π π 2 3 θ 6 x −3 −3 −6 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy -plane Example 11.5.2. Graph the following polar equations. 1. r = 4 − 2 sin(θ) 2. r = 2 + 4 cos(θ) 4. r2 = 16 cos(2θ) 3. r = 5 sin(2θ) Solution. 1. We ﬁrst plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr axes. To help us visualize what is going on graphically, we divide up [0, 2π ] into the usual four subintervals 0, π , π , π , 2 2 π π π , 32 and 32 , 2π , and proceed as we did above. As θ ranges from 0 to π , r decreases from 2 4 to 2. This means that the curve in the xy -plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y -axis. y r θ runs from 0 to 6 π 2 x 4 2 π 2 4 π 3π 2 2π θ The graph of r = 6 cos(θ) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3. 11.5 Graphs of P...
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