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 = maxmin 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 = (maxmin) = (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 = midline = (min yvalue) + 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.
 Fall '07
 FABBRI,MARC
 Unit Circle

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