Exam3_2 8 - e 2 x we see that 5 e 2 x-3 e-2 x 3 e 2 x + 4...

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 8 1. s ( t ) = 8 - 5 sin t - 6 cos t 2. s ( t ) = - 3 + 5 cos t - 6 sin t 3. s ( t ) = 7 - 5 cos t - 6 sin t correct 4. s ( t ) = 7 - 5 cos t + 6 sin t 5. s ( t ) = 8 + 5 sin t - 6 cos t 6. s ( t ) = - 4 - 5 sin t + 6 cos t Explanation: Since s 0 ( t ) = v ( t ) and d dt sin t = cos t, d dt cos t = - sin t, we see that s ( t ) = C - 5 cos t - 6 sin t with C an arbitrary constant. But then s (0) = 2 = C - 5 = 2 because cos 0 = 1 , sin 0 = 0 . Consequently, s ( t ) = 7 - 5 cos t - 6 sin t . keywords: antiderivative, application to me- chanics, velocity 014 (part 1 of 1) 10 points Deterrmine if lim x →-∞ µ 5 e 2 x - 3 e - 2 x 3 e 2 x + 4 e - 2 x exists, and if it does, Fnd its value. 1. limit = - 2 7 2. limit does not exist 3. limit = - 3 4 correct 4. limit = 2 7 5. limit = - 4 3 6. limit = 3 4 7. limit = 4 3 Explanation: Multiplying top and bottom by
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Unformatted text preview: e 2 x we see that 5 e 2 x-3 e-2 x 3 e 2 x + 4 e-2 x = 5 e 4 x-3 3 e 4 x + 4 . On the other hand, lim x - e ax = 0 for all a > 0. But then by properties of limits, lim x - 5 e 4 x-3 3 e 4 x + 4 =-3 4 . Consequently, the limit exists and limit =-3 4 . keywords: exponential function, limit as in-Fnity 015 (part 1 of 1) 10 points When f, g, F and G are functions such that lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , lim x 1 F ( x ) = 2 , lim x 1 G ( x ) = , which, if any, of A. lim x 1 f ( x ) g ( x ) ;...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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