Stitz-Zeager_College_Algebra_e-book

# The true course of the boat is given by b r finding

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Unformatted text preview: impliﬁes 3 3 3 √ to x = 1 − 2 3. Similarly, y = −x sin(θ) + y cos(θ) = (−2) sin π + (−4) cos π which 3 √ √ √ √3 gives y = − 3 − 2 = −2 − 3. Hence P (x , y ) = 1 − 2 3, −2 − 3 . To check our answer √ √ algebraically, we use the formulas in Theorem 11.9 to convert P (x , y ) = 1 − 2 3, −2 − 3 back into x and y coordinates. We get x = x cos(θ) − y sin(θ) √ √ = (1 − 2 3) cos π − (−2 − 3) sin 3 √ √ 3 = 1 − 3 − − 3− 2 2 π 3 =2 Similarly, using y = x sin(θ) + y cos(θ), we obtain y = −4 as required. To check our answer graphically, we √ sketch in the x -axis and y -axis to see if the new coordinates P (x , y ) = √ 1 − 2 3, −2 − 3 ≈ (−2.46, −3.73) seem reasonable. Our graph is below. y x y π 3 π 3 x P (x, y ) = (2, −4) P (x , y ) ≈ (−2.46, −3.73) √ 2. To convert the equation 21x2 +10xy 3+31y 2 = √ to an equation in the variables x √nd y , 144 a y3 x3 π π x π π we substitute x = x cos 3...
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