A plane flies at a speed $560 \mathrm{~km} / \mathrm{hr}$ at a constant height of $11 \mathrm{~km}$. How rapidly is the angle of elevation to the plane changing when the plane is directly above a point $115 \mathrm{~km}$ away from the observer? The angle of elevation is changing at radians/hr (enter a positive value). Round your answer to 3 decimal places.

Step 1 ：Given that the plane flies at a speed of \(560 \mathrm{~km} / \mathrm{hr}\) at a constant height of \(11 \mathrm{~km}\). We are asked to find how rapidly the angle of elevation to the plane is changing when the plane is directly above a point \(115 \mathrm{~km}\) away from the observer.

Step 2 ：We can use the Pythagorean theorem to relate the distance of the plane from the observer (d), the height of the plane (h), and the horizontal distance of the plane from the observer (x). We have: \[d^2 = h^2 + x^2\]

Step 3 ：Differentiating both sides with respect to time (t), we get: \[2d \frac{dd}{dt} = 2h \frac{dh}{dt} + 2x \frac{dx}{dt}\]

Step 4 ：Since the height of the plane is constant, \(\frac{dh}{dt} = 0\). We can solve for \(\frac{dd}{dt}\) to find the rate at which the distance of the plane from the observer is changing.

Step 5 ：The angle of elevation (θ) can be found using the relation: \[\tan(θ) = \frac{h}{x}\]

Step 6 ：Differentiating both sides with respect to time, we get: \[\sec^2(θ) \frac{dθ}{dt} = \frac{h \frac{dx}{dt} - x \frac{dh}{dt}}{x^2}\]

Step 7 ：We can solve for \(\frac{dθ}{dt}\) to find the rate at which the angle of elevation is changing.

Step 8 ：Substituting the given values: h = 11, x = 115, dx/dt = 560, we find d = 115.52488909321661, theta = 0.09536204690240899, dd/dt = 557.4556314703394, and dtheta/dt = 0.462

Step 9 ：Final Answer: The rate at which the angle of elevation is changing when the plane is directly above a point 115 km away from the observer is \(\boxed{0.462}\) radians/hr.