In: Math
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.
(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) =
tan−1(2−t5)
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 height is the
elevator when it appears to be moving fastest?
h =