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Unformatted text preview: syed (sms3768) Quest HW 7 seckin (56425) 1 This print-out should have 17 questions. Multiple-choice 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 mid-day the storm will be at a distance of s ( t ) = 8 + 2 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 =- 8 mph 2. storm speed = 7 mph 3. storm speed = 8 mph 4. storm speed = 6 mph correct 5. storm speed =- 7 mph 6. storm speed =- 6 mph Explanation: The storm will reach Austin when s ( t ) = (4- t )(2 + t ) = 0 , i.e. , when t = 4. Now the velocity of the storm is given by s ( t ) = 2- 2 t, the negative sign indicating that the storm is moving towards Austin. Thus at t = 4, s parenleftBig 4 parenrightBig = 2- 8 =- 6 . Consequently, since speed is always non- negative, storm speed = 6 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 9 feet away from the base of the wall? 1. speed = 8 ft/sec 2. speed = 7 ft/sec 3. speed = 31 4 ft/sec 4. speed = 15 2 ft/sec correct 5. speed = 29 4 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 . syed (sms3768) Quest HW 7 seckin (56425) 2 In this case dy dt =- x y dx dt . But again by Pythagoras, if x = 9, then y = 12. Thus, if the foot of the ladder is moving away from the wall at a speed of dx dt = 10 ft/sec , and x = 9, then the velocity of the top of the ladder is given by dy dt =- 3 4 dx dt . Consequently, the speed at which the top of the ladder is falling is speed = vextendsingle vextendsingle vextendsingle dy dt vextendsingle vextendsingle vextendsingle = 15 2 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. away from RLM at 40 yds/min correct 2. away from RLM at 30 yds/min 3. away from RLM at 20 yds/min 4. towards RLM at 20 yds/min 5. towards RLM at 40 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- 6 4- 2 = 40 yds/min ....
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This note was uploaded on 04/29/2010 for the course MATH 408K taught by Professor Seckin during the Spring '10 term at University of Texas-Tyler.
- Spring '10