Watching an elevator An observer is 20 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 20 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 the elevator rises at a rate of 5 m/s, what is the rate of change of the angle of elevation when the elevator is 10 m above the ground? When the elevator is 40 m above the ground? <IMAGE>

Verified step by step guidance

1. Define the variables: Let h be the height of the elevator above the ground, and θ be the angle of elevation from the observer to the elevator. The observer is at a height of 20 m and 20 m horizontally from the elevator shaft.
2. Use the tangent function to relate the angle of elevation to the heights: tan(θ) = (h - 20) / 20. This equation expresses the angle of elevation in terms of the height of the elevator.
3. Differentiate both sides of the equation with respect to time t to find the relationship between the rate of change of the angle of elevation (dθ/dt) and the rate of change of the height of the elevator (dh/dt). Use the chain rule for differentiation.
4. Substitute the known values into the differentiated equation: When the elevator is at 10 m and 40 m above the ground, calculate the corresponding height (h) and the rate of change of height (dh/dt = 5 m/s) to find dθ/dt.
5. Evaluate the expression for dθ/dt at both heights (10 m and 40 m) to determine the rate of change of the angle of elevation at those specific heights.