exam2_solnsw10

# exam2_solnsw10 - Math 116 Second Midterm Name EXAM...

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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 = - 0 . 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 = - 0 . 6 π q 19 . 6 V 25 π = - 0 . 6 . 784 πV . Using separation of variables, we have dV V 1 / 2 = - 0 . 6 . 784 πdt ≈ - 0 . 9416 dt 2 V 1 / 2 = - 0 . 6 . 784 πt + C V 1 / 2 = - 0 . 3 . 784 πt + C V = ( - 0 . 3 . 784 πt + C ) 2 ( - 0 . 4708 t + C ) 2 When t = 0, V = 500 π , so V = 500 π = C 2 , giving C = 500 π 39 . 6333 , so V = ( - 0 . 3 . 784 πt + 500 π ) 2 ( - 0 . 4708 t + 39 . 6333) 2 . When t = 10, V = 1219 . 7747 inches 3 , so h 15 . 5307 inches.
Math 116 / Exam 2 (March 23, 2010) page 3 2 . [10 points] Determine if each of the following integrals diverges or converges. If the integral converges, find the exact answer. If the integral diverges, write ”DIVERGES.” Show ALL

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