140Bquiz2csols

# 140Bquiz2csols - 3 and t π 3 = 5 π 3 or equivalently t =...

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1 MATH 140B QUIZ#2 Solutions Date: Sept 10, 2007 Instructor: Dr. M.Fabbri 1. i) False . The angle 7 π/ 6 is in quadrant 3, where sin is always negative, so this is impossible. ii) True . The angle 4 π/ 3 is in quadrant 3 and intersects the unit circle at ( - 1 / 2 , - 3 / 2) , so sin(4 π/ 3) = - 3 / 2 , and cos(4 π/ 3) = - 1 / 2 . Therefore, tan(4 π/ 3) = - 3 / 2 - 1 / 2 = 3 . iii) True . The angle 2 π/ 3 is in quadrant 2 and intersects the unit circle at ( - 1 / 2 , 3 / 2) , so cos(2 π/ 3) = - 1 / 2 . Therefore, sec(2 π/ 3) = - 2 . Therefore, D) is the answer. 2. i) False . The maximum y -value is - 3 + 4 = 1 . ii) False . The frequency is b 2 π = π/ 2 2 π = 1 4 . iii) False . The mid-line is k = - 3 . Therefore, E) is the answer. 3. We need to solve sin θ = - 3 / 2 where θ = t + π/ 3 . The only values of θ between 0 and 2 π satisfying sin θ = - 3 / 2 are θ = 4 π/ 3 and θ = 5 π/ 3 . Since θ = t + π/ 3 here, we have t + π/ 3 = 4 π/
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Unformatted text preview: 3 and t + π/ 3 = 5 π/ 3 , or equivalently, t = π and t = 4 π/ 3 . Finally, to get all solutions, we take into account that sine has period 2 π. Therefore, C) is the answer. 4. i) True . a = max-min 2 = 2-(-6) 2 = 8 2 = 4 . ii) True . By inspection of the graph, 2 π b = period = 6 π, and so b = 2 π 6 π = 1 3 . iii) False . k = max + min 2 = 2 + (-6) 2 =-4 2 =-2 . Therefore, B) is the answer. 5. i) True 2 a = (max-min) = (volume inhaled) = (tidal volume) = 3400 , so a = 1700 . ii) False . b 2 π = frequency = 1 2 breath per second , so b = π. iii) False . k = mid-line = (min y-value) + a = 1100 + 1700 = 2800 . Therefore, A) is the answer....
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## This note was uploaded on 03/30/2008 for the course MATH 140B taught by Professor Fabbri,marc during the Fall '07 term at Penn State.

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