This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 116 — Second Midterm March 23, 2010 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this exam until you are told to do so. 2. This exam has 10 pages including this cover. There are 9 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 TI92 (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 10 2 10 3 6 4 12 5 10 6 12 7 12 8 14 9 14 Total 100 Math 116 / Exam 2 (March 23, 2010) page 2 1 . [10 points] There is a bucket, shaped like a cylinder, with a radius of 5 inches and a height of 20 inches. It has a circular hole in the bottom which has a radius of 1 inch. The bucket begins full of water, but it flows out the hole in the bottom. Let t be the number of seconds since the water began dripping from the bucket, and let V ( t ) denote the volume (in inches 3 ) of water remaining in the bucket at time t. Let h ( t ) be the depth of the water in the bucket at time t. a . [2 points] Write a formula for the volume of water in the bucket, V ( t ), as a function of the depth of the water in the bucket, h ( t ). Solution: V ( t ) = 25 πh ( t ) b . [8 points] The volume of water changes such that it satisfies the differential equation dV dt = . 6 π √ 19 . 6 h. Solve for the depth of the water at time t = 10 . Be sure to include units in your answer. Solution: We know that V = 25 πh, so h = V 25 π . Substituting this we get that dV dt = . 6 π q 19 . 6 V 25 π = . 6 √ . 784 πV . Using separation of variables, we have dV V 1 / 2 = . 6 √ . 784 πdt ≈  . 9416 dt 2 V 1 / 2 = . 6 √ . 784 πt + C V 1 / 2 = . 3 √ . 784 πt + C V = ( . 3 √ . 784 πt + C ) 2 ≈ ( . 4708 t + C ) 2 When t = 0, V = 500 π , so V = 500 π = C 2 , giving C = √ 500 π ≈ 39 . 6333 , so V = ( . 3 √ . 784 πt + √ 500 π ) 2 ≈ ( . 4708 t + 39 . 6333) 2 . When t = 10, V = 1219 . 7747 inches...
View
Full
Document
This note was uploaded on 01/28/2012 for the course MATH 116 taught by Professor Irena during the Fall '07 term at University of Michigan.
 Fall '07
 Irena
 Math, Calculus

Click to edit the document details