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

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Math 116 — Second Midterm March 2011 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this exam until you are told to do so. 2. This exam has 12 pages including this cover. There are 8 problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck. 3. Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions, so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation) for each problem, so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3 00 × 5 00 note card. 7. If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers , and remove all headphones. Problem Points Score 1 14 2 14 3 14 4 12 5 10 6 14 7 7 8 15 Total 100
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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.
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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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