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int cal 4 - pokharel(yp624 HW04 Radin(57410 This print-out...

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pokharel (yp624) – HW04 – Radin – (57410) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points A Calculus student begins walking in a straight line from the RLM building towards the PCL Library. After t minutes his velocity v = v ( t ) is given (in multiples of 10 yards per minute) by the function whose graph is -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 2 4 6 8 10 - 2 2 4 6 - 2 t v How far is the student from the RLM build- ing at time t = 4? 1. dist = 85 yards 2. dist = 95 yards 3. dist = 65 yards 4. dist = 75 yards correct 5. dist = 115 yards Explanation: The student is walking towards the PCL Library whenever v ( t ) > 0 and towards RLM whenever v ( t ) < 0. The distance he has walked is given by 10 times the area between the graph of v and the t axis since 1 unit is equivalent to 10 yards. Now at t = 4 this area is 7 . 5 units, so the student has walked a distance of 75 yards. 002 (part 2 of 4) 10.0 points How far is the student from the RLM build- ing at time t = 8? 1. dist = 235 yards correct 2. dist = 255 yards 3. dist = 275 yards 4. dist = 225 yards 5. dist = 245 yards Explanation: The distance the student has walked is given by the area between the graph of v and the t axis. Now at t = 8 this area is 23 . 5 units, so the student has walked a distance of 235 yards. 003 (part 3 of 4) 10.0 points What is the total distance walked by the student from time t = 0 to t = 10? 1. dist = 300 yards 2. dist = 260 yards 3. dist = 270 yards correct 4. dist = 280 yards 5. dist = 250 yards Explanation: The distance the student has walked is given by the area between the graph of v and the t axis. Now at t = 10 this area is 27 units (area is always positive remember), so the student has walked a distance of 270 yards. 004 (part 4 of 4) 10.0 points How far is the student from the RLM build- ing when he turns back? 1. dist = 300 yards 2. dist = 280 yards
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pokharel (yp624) – HW04 – Radin – (57410) 2 3. dist = 230 yards 4. dist = 250 yards 5. dist = 260 yards correct Explanation: The student will turn back when his veloc- ity changes from postive to negative, i.e. , at t = 9. At this time he is 260 yards from the RLM building. keywords: distance, time, graph, velocity, area 005 10.0 points The graph of f is shown in the figure 2 4 6 8 2 4 6 If F is an anti-derivative of f and integraldisplay 8 3 f ( x ) dx = 69 4 , find the value of F (8) - F (2). 1. F (8) - F (2) = 93 4 2. F (8) - F (2) = 24 3. F (8) - F (2) = 45 2 4. F (8) - F (2) = 21 correct 5. F (8) - F (2) = 87 4 Explanation: We already know that the area under the graph on the interval 3 x 8 is equal to 69 4 , alternatively, by the Fundamental Theorem of Calculus we can say that F (8) - F (3) = 69 4 . On the other hand, integraldisplay 8 2 f ( x ) dx = integraldisplay 3 2 f ( x ) dx + integraldisplay 8 3 f ( x ) dx. Thus we need to find integraldisplay 3 2 f ( x ) dx = F (3) - F (2) . Now integraldisplay 3 2 f ( x ) dx = integraldisplay 3 2 3 2 x dx = 3 4 bracketleftBig x 2 bracketrightBig 3 2 = 15 4 .
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