Question

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?

Answer 1

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

Answer 2

When the elevator is 54 m above the​ ground?

Answer 3

c'mon dude - use x=54

you gotta do some of the work!
Post your work if you get stuck.

Answer 4

[AI-generated answer — Explain Bot]

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.