This problem can be solved using the concept of related rates in calculus. Here are the steps:

1. First, we need to understand the problem and the relationship between the variables. The airplane is flying at a constant altitude of 5 miles and its speed is 700 miles per hour. We need to find the rate at which the angle of elevation (θ) is changing when θ = 30°.

2. The relationship between the variables can be expressed using the tangent function in trigonometry. The tangent of the angle of elevation (θ) is equal to the opposite side (the altitude of the plane, which is constant at 5 miles) over the adjacent side (the horizontal distance of the plane from the observer, which we'll call x). So, tan(θ) = 5/x.

3. Differentiate both sides of the equation with respect to time (t). Using the chain rule, the derivative of tan(θ) is sec²(θ) times dθ/dt (the rate of change of θ), and the derivative of 5/x is -5/x² times dx/dt (the rate of change of x, which is the speed of the plane). So, sec²(θ) * dθ/dt = -5/x² * dx/dt.

4. We know that dx/dt = 700 miles per hour (the speed of the plane), and we want to find dθ/dt when θ = 30°. We also know that sec(30°) = 2/√3, so sec²(30°) = 4/3. Substitute these values into the equation: 4/3 * dθ/dt = -5/x² * 700.

5. Solve for dθ/dt. First, multiply both sides of the equation by 3/4 to isolate dθ/dt on one side: dθ/dt = -15/4x² * 700.

6. Finally, we need to find the value of x when θ = 30°. From the original equation tan(θ) = 5/x, we can solve for x: x = 5/tan(30°) = 5√3. Substitute this value into the equation for dθ/dt: dθ/dt = -15/4*(5√3)² * 700 = -15/60 * 700 = -175 radians per hour.

So, the rate at which the angle of elevation is changing when θ = 30° is -175 radians per hour. This means the angle is decreasing at this rate, as the negative sign indicates.

Question

This problem can be solved using the concept of related rates in calculus. Here are the steps:

1. First, we need to understand the problem and the relationship between the variables. The airplane is flying at a constant altitude of 5 miles and its speed is 700 miles per hour. We need to find the rate at which the angle of elevation (θ) is changing when θ = 30°.

2. The relationship between the variables can be expressed using the tangent function in trigonometry. The tangent of the angle of elevation (θ) is equal to the opposite side (the altitude of the plane, which is constant at 5 miles) over the adjacent side (the horizontal distance of the plane from the observer, which we'll call x). So, tan(θ) = 5/x.

3. Differentiate both sides of the equation with respect to time (t). Using the chain rule, the derivative of tan(θ) is sec²(θ) times dθ/dt (the rate of change of θ), and the derivative of 5/x is -5/x² times dx/dt (the rate of change of x, which is the speed of the plane). So, sec²(θ) * dθ/dt = -5/x² * dx/dt.

4. We know that dx/dt = 700 miles per hour (the speed of the plane), and we want to find dθ/dt when θ = 30°. We also know that sec(30°) = 2/√3, so sec²(30°) = 4/3. Substitute these values into the equation: 4/3 * dθ/dt = -5/x² * 700.

5. Solve for dθ/dt. First, multiply both sides of the equation by 3/4 to isolate dθ/dt on one side: dθ/dt = -15/4x² * 700.

6. Finally, we need to find the value of x when θ = 30°. From the original equation tan(θ) = 5/x, we can solve for x: x = 5/tan(30°) = 5√3. Substitute this value into the equation for dθ/dt: dθ/dt = -15/4*(5√3)² * 700 = -15/60 * 700 = -175 radians per hour.

So, the rate at which the angle of elevation is changing when θ = 30° is -175 radians per hour. This means the angle is decreasing at this rate, as the negative sign indicates.

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