An observer is 31m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 22m 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 3/ms, what is the rate of change of the angle of elevation when the elevator is 20 m above the ground? When the elevator is 53m above the ground?

Question

An observer is 31m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 22m 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 3/ms, what is the rate of change of the angle of elevation when the elevator is 20 m above the ground? When the elevator is 53m above the ground?

please show work

Answer 1

see the related questions below. The first is the identical problem -- just change the numbers.

Answer 2

not enough detail for me to solve it in other one.

Answer 3

To find the rate of change of the angle of elevation, we need to determine how the angle is related to the height of the elevator.

Let's start by drawing a diagram:

O (Observer)
|\
| \
| \
| \ E (Elevator)
|θ \
|________\
D

In this diagram, O represents the observer, E represents the elevator, and D represents the ground floor of the hotel atrium.

From the given information, we know that the observer is 31m above the ground floor, and the elevator is 22m horizontally from the observer. We can use this information to determine the distance OD.

Using the Pythagorean theorem, we have:

OD² = OE² + DE²

OD² = (31m)² + (22m)²

OD² = 961m² + 484m²

OD² = 1445m²

OD ≈ 38.01m

Now, let's define some variables:

h represents the height of the elevator above the ground.
α represents the angle of elevation.

From the diagram, we can see that:

tan(α) = h / OD

To find the rate of change of the angle of elevation, we can differentiate both sides of this equation with respect to time t:

d(tan(α))/dt = d(h/OD)/dt

Now, let's find the derivative of each term separately:

d(tan(α))/dt = d(h/OD)/dt

Using the quotient rule, we have:

d(tan(α))/dt = (OD * d(h)/dt - h * d(OD)/dt) / OD²

Given that the elevator rises at a rate of 3 m/s, we can substitute d(h)/dt with 3 m/s:

d(tan(α))/dt = (OD * 3 - h * d(OD)/dt) / OD²

Next, let's determine the value of d(OD)/dt:

We know that OD can be expressed as a function of h:

OD = √(31² + (22 + h)²)

Differentiating both sides with respect to time t, we have:

d(OD)/dt = d(√(31² + (22 + h)²))/dt

Using the chain rule, we get:

d(OD)/dt = (1/2) * (31² + (22 + h)²)^(-1/2) * (0 + 2(22 + h) * d(h)/dt)

Simplifying this expression, we have:

d(OD)/dt = (22 + h) * d(h)/dt / √(31² + (22 + h)²)

Now we can substitute this result back into the previous equation:

d(tan(α))/dt = (OD * 3 - h * (22 + h) * d(h)/dt / √(31² + (22 + h)²)) / OD²

We are interested in finding the rate of change of the angle of elevation when the elevator is 20m and 53m above the ground. We can substitute these values for h to find the respective rates:

For h = 20m:

d(tan(α))/dt = (38.01m * 3 - 20m * (22 + 20) * d(h)/dt / √(31² + (22 + 20)²)) / (38.01m)²

For h = 53m:

d(tan(α))/dt = (38.01m * 3 - 53m * (22 + 53) * d(h)/dt / √(31² + (22 + 53)²)) / (38.01m)²

By calculating these expressions, you can find the respective rates of change of the angle of elevation at those heights.

Answer 4

of course there is...

when the elevator is x meters up,

tanθ = (x-31)/22

So,

sec^2θ dθ/dt = 1/22 dx/dt

when x=20, tanθ = -11/22, so sec^2θ = 5/4

5/4 dθ/dt = 1/22 (3)
dθ/dt = 55/6 m/s

Is one of the steps unclear?