Stitz-Zeager_College_Algebra_e-book

Our goal is to show that p k 1 is true or that z k1

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Unformatted text preview: below. θ 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π 2 r = 6 cos(θ) 6 √ 32 (r, θ) (6, 0) √π 3 2, 4 0 √ −3 2 3 0, π 2 √ 3π −3 2, 4 −6 √ −3 2 y (−6, π ) √ 5π −3 2, 4 0 √ 32 6 0, 3π √ 72 π 3 2, 4 3 −3 (6, 2π ) For a review of these concepts and this process, see Sections 1.5 and 1.7. 6 x 11.5 Graphs of Polar Equations 799 Despite having nine ordered pairs, we only get four distinct points on the graph. For this reason, we employ a slightly different strategy. We graph one cycle of r = 6 cos(θ) on the θr plane3 and use it to help graph the equation on the xy -plane. We see that as θ ranges from 0 to π , r ranges from 6 2 to 0. In the xy -plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y -axis (θ = π ). The 2 arrows drawn in the figure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the xy -plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but it is markedly faster. y r 6 θ runs from 0 to π 2 3 π 2 π 3π 2 2π x θ −3 −6 Next, we repeat the process as θ ranges from π to π . Here, the r values are all negative. This 2 means that...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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