Business Calc Homework w answers_Part_59

06 2005 005 0 0585 the overall rate then is 12 004875

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Unformatted text preview: k → x dm yields . For k dm 2: /2 ∑mky . Taking dm ∑mk My (b) y M [(4 0 (4 dA as mk and letting (2 sin 2x)] dx 4 sin 2x) dx 0 /2 4x dA → 0, k → yields 2 sin 2x) /2 2 cos 2x y dm 2 4 0 . dm 2. It appears that the areas for k 2 4. 3 will continue to be /k 34. By symmetry, x 0. For y, use horizontal strips: 3. Ak [(2k k sin kx) k sin kx] dx 0 /k y dm y dA (2k y dA 2k sin kx) dx 0 y dm dA dA If we make the substitution u kx, then du k dx and the 4 y(2 y) dy u-limits become 0 to . Thus, 0 4 2 /k y dy 2 5 2k sin kx) dx 0 /k 4 0 0 2 3/2 4 y 3 0 0 2 y 5/2 2 (2k Ak 0 (2 (2 4. 2 12 5 2 sin kx)k dx 2 sin u) du. 4 5. Because the amplitudes of the sine curves are k, the kth butterfly stands 2k units tall. The vertical edges alone have lengths (2k) that increase without bound, so the perimeters are tending to infinity. 35. By symmetry, y 0. For x, use vertical strips: Quick Review 7.2 x dm x dA x dA 1. x dm dA sin x dx 0 x(2x) dx [1 1] 2 0 1 2 cos x 0 dA 2. 1 2x e 2 e2x dx 0 1 0 12 (e 2 1) 3.1...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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