Question

A plane flies at a speed 540 km/h at a constant height of 10 km. How rapidly is the angle of elevation to the plane changing when the plane is directly above a point 130 km away from the observer?

Answer 1

To find the rate at which the angle of elevation is changing, we can use trigonometry and related rates. By setting up a right triangle and using the tangent function, we can find an equation that relates the rate of change of the angle of elevation to the rate of change of the distance between the observer and the point directly below the plane. Solving this equation will give us the answer.

Step-by-step explanation:

To find how rapidly the angle of elevation is changing, we can use trigonometry and the concept of related rates. Let's call the angle of elevation theta and the distance between the observer and the point directly below the plane x. We want to find d(theta)/dt, the rate of change of theta with respect to time.

Since the height of the plane is constant at 10 km, we can set up a right triangle with the side opposite theta as 10 km and the side adjacent to theta as x km. Then, we can use the tangent function: tan(theta) = 10/x. Differentiating both sides with respect to time, we get sec^2(theta) * d(theta)/dt = -10/x^2 * dx/dt.

Given that x = 130 km, dx/dt = -540 km/h (since the plane is flying away from the observer), and we can use the equation we derived earlier to solve for d(theta)/dt.

d(theta)/dt = -10/(130^2) * (-540) / (sec^2(theta)) = 0.047 rad/h.