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Pre-Calc Homework Solutions 293

Pre-Calc Homework Solutions 293 - Section 7.2(12 0[3.6 2(12...

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32. } (1 2 2 (1 2 2) 0) } [3.6 1 2(4.0) 1 2(3.1) 1 2(2.8) 1 2(2.8) 1 2(3.2) 1 2(3.3) 1 2(3.1) 1 2(3.2) 1 2(3.4) 1 2(3.4) 1 2(3.9) 1 4.0] 5 40 thousandths or 0.040 33. (a) x w 5 } M M y } 5 . Taking dm 5 d dA as m k and letting dA 0, k yields . (b) y w 5 } M M y } 5 . Taking dm 5 d dA as m k and letting dA 0, k yields . 34. By symmetry, x w 5 0. For y w , use horizontal strips: y w 5 5 5 5 5 5 } 1 5 2 } 35. By symmetry, y 5 0. For x , use vertical strips: x 5 5 5 5 5 5 } 4 3 } Section 7.2 Areas in the Plane (pp. 374–382) Exploration 1 A Family of Butterflies 1. For k 5 1: E p 0 [(2 2 sin x ) 2 sin x ] dx 5 E p 0 (2 2 2 sin x ) dx 5 2 x 1 2 cos x 4 5 2 p 2 4 For k 5 2: E p /2 0 [(4 2 2 sin 2 x ) 2 (2 sin 2 x )] dx 5 E p /2 0 (4 2 4 sin 2 x ) dx 5 4 x 1 2 cos 2 x 4 5 2 p 2 4 2. It appears that the areas for k \$ 3 will continue to be 2 p 2 4. 3. A k 5 E p / k 0 [(2 k 2 k sin kx ) 2 k sin kx ] dx 5 E p / k 0 (2 k 2 2 k sin kx ) dx If we make the substitution u 5 kx , then du 5 k dx and the u -limits become 0 to p . Thus, A k 5 E p / k 0 (2 k 2 2 k sin kx ) dx 5 E p / k 0 (2 2 2 sin kx ) k dx 5 E p 0 (2 2 2 sin u ) du . 4. 2 p 2 4 5. Because the amplitudes of the sine curves are k , the k th butterfly stands 2 k units tall. The vertical edges alone have lengths (2 k ) that increase without bound, so the perimeters
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