MATH SOLVE

6 months ago

Q:
# 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) 300-25t (b) Using your answer to part (a), express θ as a function of time e. Find the rate of change of θ with respect to t. de dt to you to be moving. At what time does the elevator appear to be moving fastest? The rate of change of θ is a measure of how fast the elevator appears time tc) seconds At what height does the elevator appear to be moving fastest?

Accepted Solution

A:

Answer:(a) h(t) = 300 - 30t(b) θ = [tex]tan^{-1}[/tex]((200 - 30t)150) [tex]-\frac{3}{35-3t}[/tex](c) t = [tex]6\frac{2}{3} sec[/tex]h = 100m above groundStep-by-step explanation:(a) change of distance of the elevator is governed by equation of speedspeed = distance/timedistance = speed x time= 30 x t = 30tsince the elevator starts at t = 0 on top of the hotel at 300ft, and the distance between the elevator keep decreasing to the ground, the equation becomesh(t) = 300 - 30t(b)θ = angle between the line of horizon of the observer to the line of sight to elevatorsince the observer is 100 feet above the ground, the vertical distance between observer and the top of the hotel is 200ft.hence, the opposite side of the angle θ is 300 - 30t - 100 = 200 - 30ttherefore, tan θ = (200 - 30t)/150and θ = [tex]tan^{-1}[/tex]((200 - 30t)150)Differentiate θ with respect to time, t.Use chain rule: [tex]\frac{de}{dt}=\frac{de}{du}. \frac{du}{dt}[/tex]let u = (200 - 30t)/150u' = -1/5θ = [tex]tan^{-1}[/tex]uθ' = [tex]\frac{1}{1+u^{2} }[/tex][tex]\frac{de}{dt} = \frac{15}{35-3t}.(-\frac{1}{5})[/tex]= [tex]-\frac{3}{35-3t}[/tex](c) The elevator will seem to be moving the fastest at θ = 0 since that is the fastest change of θ as the elevator move downwardtan θ = (200 - 30t)/150tan (0) = 0(200 - 30t) = 0200 = 30tt = 200/30 = [tex]6\frac{2}{3} sec[/tex]height = 100m above ground