Exam2W11FSol

# Exam2W11FSol - Math 116 Second Midterm March 2011 Name EXAM...

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Math 116 / Exam 2 (March 2011) page 2 1 . [14 points] Indicate if each of the following statements are true or false by circling the correct answer. Justify your answers. a . [2 points] The function z ( t ) = sin( at ) + at is a solution to the differential equation z 00 + a 2 z = a 3 t . True False Solution: z 0 = a cos( at ) + a z 00 = - a 2 sin( at ) z 00 + a 2 z = a 2 sin( at ) + a 2 (sin( at ) + at ) = a 3 t. b . [3 points] The motion of a particle is given by the parametric curve x = x ( t ), y = y ( t ) for 0 t 3 shown below. The arrows indicate the direction of the motion of the particle along the path. If the curve passes only twice through the origin, x (1) = x (2) = 0 and y (1) = y (2) = 0 then d dt dy dt dx dt ! dx dt > 0 for t = 1. x y True False Solution: The first time that the curve passes through the origin at t = 1, the curve has negative concavity. c . [3 points] Euler’s method yields an overestimate for the solutions to the differential equation dy dx = 4 x 3 + 2 x + 1. True False Solution: False, y 00 = 12 x 2 + 2 > 0 then Euler method is an underestimate since y is concave up.
Math 116 / Exam 2 (March 2011) page 3 d . [3 points] The graph of x = x ( t ) and y = y ( t ) for 0 t 2 is given below. If y 0 (1) = 0, then it must be the case that ( x (1) , y (1)) = (0 , 0). - 1.0 - 0.5 0.5 1.0 x - 0.4 - 0.2 0.2 0.4 0.6 0.8 1.0 y True False Solution: False. The particle can stop at any point at time t = 1 in the parabola. Then y 0 (1) = 0 and x 0 (1) = 0 without necessarily be the case that ( x (1) , y (1)) = (0 , 0).

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