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Stitz-Zeager_College_Algebra_e-book

# Of course we could just as easily associate z with a

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Unformatted text preview: ify. This is the technique we employ below. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3)2 +y 2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.6 (r cos(θ) − 3)2 + (r sin(θ))2 = 9 r2 cos2 (θ) − 6r cos(θ) + 9 + r2 sin2 (θ) = 9 r2 cos2 (θ) + sin2 (θ) − 6r cos(θ) = 0 Subtract 9 from both sides. r2 − 6r cos(θ) = 0 Since cos2 (θ) + sin2 (θ) = 1 r(r − 6 cos(θ)) = 0 Factor. We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x − 3)2 + y 2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our ﬁnal answer.7 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r cos(θ) = −r sin(θ). Rearranging, we get r cos(θ) + r sin(θ) = 0 or r(cos(θ) + sin(θ)) = 0. This gives r = 0 or cos(θ) + sin(θ) = 0. Solving the latter equation for θ, we get θ = − π + πk for integers 4 k . As we did in the previous example...
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