This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: roberts (eer474) Quest HW 7 seckin (56425) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A storm system is headed towards Austin. The Weather Service Office forecasts that t hours after midday the storm will be at a distance of s ( t ) = 4 + 3 t t 2 miles from Austin. At what speed (in miles per hour) will the storm be moving when it reaches Austin? 1. storm speed = 5 mph correct 2. storm speed = 7 mph 3. storm speed = 6 mph 4. storm speed = 6 mph 5. storm speed = 7 mph 6. storm speed = 5 mph Explanation: The storm will reach Austin when s ( t ) = (4 t )(1 + t ) = 0 , i.e. , when t = 4. Now the velocity of the storm is given by s ( t ) = 3 2 t, the negative sign indicating that the storm is moving towards Austin. Thus at t = 4, s parenleftBig 4 parenrightBig = 3 8 = 5 . Consequently, since speed is always non negative, storm speed = 5 mph . 002 10.0 points A 15 foot ladder is leaning against a wall. If the foot of the ladder is sliding away from the wall at a rate of 10 ft/sec, at what speed is the top of the ladder falling when the foot of the ladder is 12 feet away from the base of the wall? 1. speed = 37 3 ft/sec 2. speed = 13 ft/sec 3. speed = 12 ft/sec 4. speed = 40 3 ft/sec correct 5. speed = 38 3 ft/sec Explanation: Let y be the height of the ladder when the foot of the ladder is x feet from the base of the wall as shown in figure x ft. 15 ft. We have to express dy/dt in terms of x, y and dx/dt . But by Pythagoras theorem, x 2 + y 2 = 225 , so by implicit differentiation, 2 x dx dt + 2 y dy dt = 0 . roberts (eer474) Quest HW 7 seckin (56425) 2 In this case dy dt = x y dx dt . But again by Pythagoras, if x = 12, then y = 9. Thus, if the foot of the ladder is moving away from the wall at a speed of dx dt = 10 ft/sec , and x = 12, then the velocity of the top of the ladder is given by dy dt = 4 3 dx dt . Consequently, the speed at which the top of the ladder is falling is speed = vextendsingle vextendsingle vextendsingle dy dt vextendsingle vextendsingle vextendsingle = 40 3 ft/sec . keywords: speed, ladder, related rates 003 (part 1 of 3) 10.0 points A Calculus student leaves the RLM build ing and walks in a straight line to the PCL Li brary. His distance (in multiples of 40 yards) from RLM after t minutes is given by the graph 2 4 6 8 10 2 4 6 8 10 t distance (i) What is his speed after 3 minutes, and in what direction is he heading at that time? 1. towards RLM at 40 yds/min 2. away from RLM at 60 yds/min 3. away from RLM at 80 yds/min correct 4. away from RLM at 40 yds/min 5. towards RLM at 80 yds/min Explanation: The graph is linear and has positive slope on [2 , 4], so the speed of the student at time t = 3 coincides with the slope of the line on [2 , 4]. Hence speed = 40 8 4 4 2 = 80 yds/min ....
View
Full
Document
This note was uploaded on 11/08/2010 for the course MATH 408K taught by Professor Gualdani during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Gualdani
 Differential Calculus

Click to edit the document details