For the amusement of the guests, some hotels have elevators on the outside of the building. One such hotel is 300 feet high. You are standing by a window 100 feet above the ground and 150 feet away from the hotel, and the elevator descends at a constant speed of 30 ft/sec, starting at time t = 0, where t is time in seconds. Let θ be the angle between the line of your horizon and your line of sight to the elevator. (See Figure 4.101.)
- (a) Find a formula for ℎ(t), the elevator’s height above the ground as it descends from the top of the hotel.
- (b) Using your answer to part (a), express θ as a function of time t and find the rate of change of θ with respect to t.
- (c) The rate of change of θ is a measure of how fast the elevator appears to you to be moving. At what height is the elevator when it appears to be moving fastest?
Figure 4.101
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Chapter 4 Solutions
Calculus: Single And Multivariable
Additional Math Textbook Solutions
Calculus: Early Transcendentals (2nd Edition)
University Calculus: Early Transcendentals (4th Edition)
Single Variable Calculus: Early Transcendentals (2nd Edition) - Standalone book
University Calculus: Early Transcendentals (3rd Edition)
Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)
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