Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: ircle, centered at the origin, with a radius of 2. The graph of r = 3 cos(θ) is also a circle - but this one is centered at the point 3 with rectangular coordinates 3 , 0 and has a radius of 2 . 2 y 2 −2 2 3 x −2 r = 2 and r = 3 cos(θ ) We have two intersection points to find, one in Quadrant I and one in Quadrant IV. Proceeding as above, we first determine if any of the intersection points P have a representation (r, θ) which satisfies both r = 2 and r = 3 cos(θ). Equating these two expressions for r, we get cos(θ) = 2 . To solve this equation, we need the arccosine function. We get 3 11 We are really using the technique of substitution to solve the system of equations r r = = 2 sin(θ) 2 − 2 sin(θ) 11.5 Graphs of Polar Equations 811 2 θ = arccos 2 + 2πk or θ = 2π − arccos 3 + 2πk for integers k . From these solutions, we get 3 2 2, arccos 3 as one representation for our answer in Quadrant I, and 2, 2π − arccos 2 3 as one representation for our answer in Quadrant IV. The reader is enc...
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