Question

For the amusement of the guests, some hotels have elevators on the outside of the building. One such hotel is 400 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 20 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. 4 (a) Find a formula for h(t), the elevator's height above the ground as it descends from the top of the hotel. h(t) = (b) Using your answer to part (a), express θ as a function of time t. θ(t) = Find the rate of change of θ with respect to t. dθ dt = (c) The rate of change of θ is a measure of how fast the elevator appears to you to be moving. At what time does the elevator appear to be moving fastest? time = seconds At what height does the elevator appear to be moving fastest?

Answer 1

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Answer:

  a.  h(t) = -20t +400

  b.  θ(t) = arctan(2 -2/15t); dθ/dt = -30/(1125 -120t +4t^2)

  c.  15 seconds; 100 ft

Step-by-step explanation:

a. The initial height of the elevator is 400 ft. The rate of change of height is -20 ft/s, so the height equation can be ...

  h(t) = -20t +400

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b. The tangent of the angle above the line of sight is "opposite"/"adjacent":

  tan(θ) = (h(t) -100)/(150) = -2/15t +2

  θ(t) = arctan(2 -2/15t) . . . . radians

The derivative of the angle function is ...

  dθ/dt = 1/(1+(2 -2/15t)^2)(-2/15)

  dθ/dt = -30/(1125 -120t +4t^2)

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c. The value of dθ/dt will have a peak where the denominator has a minimum, at t = -(-120)/2(4)) = 15. (The quadratic vertex coordinate is t=-b/(2a).)

The elevator appears to be moving fastest at t=15 seconds.

The height at that time is ...

  h(15) = 400 -20(15) = 100

The elevator appears to be moving fastest when it is at eye level, 100 ft above the ground.