An observer is 20m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 18m 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 11 m above the ground? When the elevator is 38m above the ground?

Question

An observer is 20m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 18m 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 11 m above the ground? When the elevator is 38m above the ground?

please show work and don't tell me to look at the other problems please there is not enough information from the others for me to figure out mine.

Answer 1

see the related questions below - just change the numbers

Answer 2

The two solutions below have exactly the same problem, and each step is shown. Where do you get stuck?

Answer 3

To solve this problem, we can use the concept of trigonometry and related rates. Let's start by labeling the given information:

- The observer is 20m above the ground floor.
- The elevator shaft is 18m horizontally from the observer.
- The elevator rises at a rate of 3 m/s.

Let's introduce some variables:
- Let h(t) represent the height of the elevator at time t.
- Let θ(t) represent the angle of elevation at time t.

We need to find the rate of change of θ(t) with respect to time t when the elevator is at two different heights: 11m and 38m above the ground.

To determine the value of θ(t), we can use the tangent function. Tangent is defined as the ratio of the opposite side to the adjacent side in a right triangle. In this case, the opposite side of the triangle is the vertical distance (h(t) - 20) and the adjacent side is the horizontal distance 18.

Using the tangent function, we have:

tan(θ(t)) = (h(t) - 20) / 18

To find the rate of change of θ(t) with respect to time t, we will differentiate both sides of the above equation with respect to t using the chain rule:

sec^2(θ(t)) * dθ/dt = ((dh/dt) / 18)

Now, we can rearrange the equation to solve for dθ/dt:

dθ/dt = ((dh/dt) / 18) * sec^2(θ(t))

Given that the elevator rises at a rate of 3 m/s (dh/dt = 3) and using the information that the elevator is at two different heights, we can substitute these values into the equation to find the rate of change of θ(t):

a) When the elevator is 11m above the ground:
dθ/dt = ((3) / 18) * sec^2(θ(t))

We need θ(t) in order to substitute it into the equation. To find θ(t), we can use the tangent function again:

tan(θ(t)) = (11 - 20) / 18
tan(θ(t)) = -9/18
θ(t) = atan(-9/18)

Now we can substitute θ(t) into the equation and solve for dθ/dt:

dθ/dt = ((3) / 18) * sec^2(atan(-9/18))

b) When the elevator is 38m above the ground:
dθ/dt = ((3) / 18) * sec^2(θ(t))

Again, we need θ(t) to substitute it into the equation:

tan(θ(t)) = (38 - 20) / 18
tan(θ(t)) = 18/18
θ(t) = atan(1)

Now we can substitute θ(t) into the equation and solve for dθ/dt:

dθ/dt = ((3) / 18) * sec^2(atan(1))

To get the actual values for dθ/dt, you would need to calculate the sec^2 values for the angles θ(t):

a) When the elevator is 11m above the ground:
dθ/dt = ((3) / 18) * sec^2(atan(-9/18))

b) When the elevator is 38m above the ground:
dθ/dt = ((3) / 18) * sec^2(atan(1))

You can use a scientific calculator, trigonometric tables, or software to calculate the sec^2 values and then compute the final values of dθ/dt.