An airplane flies at an altitude of 2 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation θ is changing when the angle is θ = 30°, θ = 60°, and θ = 70°.

cannot see figure.

To find the rate at which the angle of elevation θ is changing, we need to relate the given information to the change in θ with respect to time. We need to consider the vertical change (altitude) and horizontal change (distance) of the airplane.

Let's define some variables to make the explanation clearer:
- θ: angle of elevation
- h: altitude of the airplane
- d: horizontal distance between the observer and the airplane

We are given that the airplane flies at an altitude of 2 miles, so h = 2 miles. The speed of the plane is 600 miles per hour, which means that the horizontal distance d changes with time. Let's denote the rate of change of d with respect to time as dz/dt.

To relate θ, h, and d, we can use the tangent function:
tan(θ) = h / d

Now, to find the rate at which θ is changing with respect to time (dθ/dt), we can differentiate both sides of the equation with respect to time:
sec^2(θ) * dθ/dt = (dh/dt * d - h * dz/dt) / d^2

Since we know h and d are constant (h = 2 miles and d = 600 miles per hour), and dh/dt is 0 (since the altitude of the airplane is constant), the equation simplifies to:
sec^2(θ) * dθ/dt = -h * dz/dt / d^2

Now, let's substitute the given values into the equation to find the rates at which the angle of elevation is changing for different θ values.

1. θ = 30°:
- h = 2 miles
- d = 600 miles per hour
- dz/dt is not given

Substituting these values into the equation and solving for dθ/dt, we get:
sec^2(30°) * dθ/dt = -2 * dz/dt / (600^2)
dθ/dt = -2 * dz/dt / [(600 miles per hour)^2 * sec^2(30°)]

2. θ = 60°:
- h = 2 miles
- d = 600 miles per hour
- dz/dt is not given

Substituting these values into the equation and solving for dθ/dt, we get:
sec^2(60°) * dθ/dt = -2 * dz/dt / (600^2)
dθ/dt = -2 * dz/dt / [(600 miles per hour)^2 * sec^2(60°)]

3. θ = 70°:
- h = 2 miles
- d = 600 miles per hour
- dz/dt is not given

Substituting these values into the equation and solving for dθ/dt, we get:
sec^2(70°) * dθ/dt = -2 * dz/dt / (600^2)
dθ/dt = -2 * dz/dt / [(600 miles per hour)^2 * sec^2(70°)]

Note: To find the rates at which the angle of elevation is changing, you need to know the value of dz/dt, the rate at which the horizontal distance between the observer and the airplane is changing.