Pre-Calc Homework Solutions 293

# Dm 2 cos 2x 4 3 will continue to be 2 it appears that

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Unformatted text preview: e 2. It appears that the areas for k 2 4. /k 34. By symmetry, x y dm y 0. For y, use horizontal strips: dA dA y dA 3. Ak [(2k 0 /k k sin kx) k sin kx] dx (2k 0 2k sin kx) dx kx, then du k dx and the y dm 4 dA If we make the substitution u u-limits become 0 to . Thus, /k y(2 0 4 y) dy y dy 4 0 2 0 Ak (2k 0 /k 2k sin kx) dx 2 sin kx)k dx 2 sin u) du. 2 y 5/2 2 12 5 2 5 (2 0 2 3/2 4 y 3 0 (2 0 4. 2 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 witho...
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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