An observer is 23m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 15m horizontally from the observer. The angle of elevation of the elevator is the angle of the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of 4/ms, what is the rate of change of the angle of elevation when the elevator is 18 m above the ground? When the elevator is 38m above the ground?

Question

Go Away Steve · Accepted Answer

Why are you on every bloody question Steve?

Steve · Answer

when the elevator is at height x,

tanθ = (x-23)/15
so,
sec^2θ dθ/dt = 1/15 dx/dt
(1+tan^2θ) dθ/dt = 1/15 dx/dt
(1 + ((x-23)/15)^2) dθ/dt = 1/15 dx/dt

Now just plug and chug. You have x, and dx/dt=4 ...