2006a_x2a_sols

2006a_x2a_sols - Spring 2006 Math 152 Exam 2A Solutions c...

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Spring 2006 Math 152 Exam 2A: Solutions Mon, 27/Mar c ± 2006, Art Belmonte 1. (c) The arc length of the curve r ( t ) = [ x ( t ), y ( t ) ] = ± t 2 , t 2 + t ² , 0 t 1 , is represented by the following integral. L = ³ b a ´ ´ r 0 ( t ) ´ ´ dt = ³ 1 0 k [2 t , 2 t + 1] k = ³ 1 0 µ 4 t 2 + 4 t 2 + 4 t + 1 = ³ 1 0 µ 1 + 4 t + 8 t 2 2. (c) The differential equation dy dx = xy 2 + x 2 y = ( y + x ) is not separable. Other choices are separable, as follows. (a) Rewrite = sin x cos y as sec ydy = sin xdx . (b) Rewrite = + x 2 y = y x + x 2 · as 1 y = x + x 2 · . (d) Rewrite = e x + y = e x e y as e - y = e x . (e) Repeated factoring gives = 1 + x + y + = ( 1 + x ) + y ( 1 + x ) = ( 1 + x )( 1 + y ), whence 1 1 + y = ( 1 + x ) . 3. (d) The linear differential equation y 0 + ( 2sin2 x ) y = cos 4 x is already in standard linear form (SLF). Accordingly, an integrating factor is μ = exp ) = e - cos 2 x . 4. (b) Let y = y ( t ) be the amount of salt in the tank at time t . The classical balance law gives = rate in - rate out = ¹ 0 . 1 kg L × 10 L min º - ¹ y kg 100 L × 10 L min º = 1 - y 10 kg min . Since the tank initially contains pure water, we have y ( 0 ) = 0 kg of salt in the tank at the start. Therefore, = 1 - y 10 , y ( 0 ) = 0. 5. (a) The integral ³ 1 x 1 + x 4 dx converges by comparison to ³ 1 1 x 3 . First note that the integrand x 1 + x 4 is positive on [1 , ) .Wethenhave ³ 1 x
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This note was uploaded on 03/25/2008 for the course MATH 152 taught by Professor Teitler during the Spring '08 term at Texas A&M.

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2006a_x2a_sols - Spring 2006 Math 152 Exam 2A Solutions c...

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