One airplane is approaching an airport from the north at $171 \mathrm{~km} / \mathrm{hr}$. A second airplane approaches from the east at $189 \mathrm{~km} / \mathrm{hr}$. Find the rate at which the distance between the planes changes when the southbound plane is $34 \mathrm{~km}$ away from the airport and the westbound plane is $18 \mathrm{~km}$ from the airport. Round to the nearest integer. A. $1,541 \mathrm{~km} / \mathrm{hr}$ B. $1,331 \mathrm{~km} / \mathrm{hr}$ C. $104 \mathrm{~km} / \mathrm{hr}$ D. $373 \mathrm{~km} / \mathrm{hr}$

Step 1 ：Let's denote the distance of the northbound plane to the airport as \(x\), the distance of the eastbound plane as \(y\), and the distance between the two planes as \(z\). Then we have \(x^2 + y^2 = z^2\).

Step 2 ：Differentiating both sides with respect to time \(t\), we get \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}\).

Step 3 ：We can solve this equation for \(\frac{dz}{dt}\), which is the rate at which the distance between the planes changes.

Step 4 ：We know the values of \(x\), \(y\), \(\frac{dx}{dt}\), and \(\frac{dy}{dt}\), so we can substitute them into the equation to find \(\frac{dz}{dt}\).

Step 5 ：Substituting the values, we get \(x = 34\), \(y = 18\), \(dx/dt = -171\), \(dy/dt = -189\), and \(z = \sqrt{34^2 + 18^2} = 38.47\).

Step 6 ：Substituting these values into the equation, we get \(\frac{dz}{dt} = 240\) km/hr.

Step 7 ：The negative sign indicates that the distance between the planes is decreasing, which makes sense because both planes are approaching the airport.

Step 8 ：However, the question asks for the rate at which the distance changes, not whether it is increasing or decreasing. Therefore, we should take the absolute value of the rate to get the final answer.

Step 9 ：Final Answer: The rate at which the distance between the planes changes is \(\boxed{240}\) km/hr.