An observer is 28 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 26 m horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that 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 6 m/s, what is the rate of change of the angle of elevation when the elevator is 15 m above the ground? When the elevator is 54 m above the ground?
when the elevator is x meters up, then we have the angle of elevation θ is
tanθ = (x-28)/26 for 0<= x <= (some max height)
sec^2θ dθ/dt = 1/26 dx/dt
so, when x=15,
tanθ = -13/26 = -1/2
sec^2θ = 1 + tan^2θ = 5/4
and we have
5/4 dθ/dt = 1/26 * 6
dθ/dt = 12/65 rad/s
When the elevator is 54 m above the ground?
c'mon dude - use x=54
you gotta do some of the work!
Post your work if you get stuck.
To find the rate of change of the angle of elevation, we need to differentiate the angle with respect to time. Let's denote the angle of elevation as θ and the time as t.
We are given that the elevator rises at a rate of 6 m/s. This means the rate of change of height with respect to time is dh/dt = 6 m/s.
We need to find dθ/dt, the rate of change of the angle of elevation with respect to time.
To begin, let's draw the figure and label the given values:
Atrium
________
| |
28m | |
observer --------------->| |
|______|
________ ________
| | | |
| | | |
| |----15m------Elevator------39m---->| |
| | | |
|________| |________|
From the figure, let's label the height of the elevator at any time t as h. We know that initially h = 0.
We are given the horizontal distance between the observer and the elevator is 26m. Therefore, we can use this information to find θ in terms of h:
tan(θ) = h / 26 (1)
Next, we need to differentiate equation (1) with respect to time t to find dθ/dt.
Using implicit differentiation:
sec^2(θ) * dθ/dt = (1/26) * dh/dt
dθ/dt = (1/26) * (dh/dt) * cos^2(θ)
To find dθ/dt, we need to find cos(θ). Since we have tan(θ) = h/26, we can use this to find cos(θ).
Using the trigonometric identity:
1 + tan^2(θ) = sec^2(θ)
Substituting tan(θ) = h/26:
1 + (h/26)^2 = sec^2(θ)
Taking the square root:
√(1 + (h/26)^2) = cos(θ)
Now we can substitute this value of cos(θ) into the expression for dθ/dt:
dθ/dt = (1/26) * (dh/dt) * √(1 + (h/26)^2)
Now we can find the rate of change of the angle of elevation at two specific heights: when the elevator is 15m above the ground (h = 15), and when it is 54m above the ground (h = 54).
When h = 15:
dθ/dt = (1/26) * (6 m/s) * √(1 + (15/26)^2)
When h = 54:
dθ/dt = (1/26) * (6 m/s) * √(1 + (54/26)^2)
Now we can calculate the numerical values of the rates of change of the angle of elevation at these two heights.