At noon, ship A is $170 \mathrm{~km}$ west of ship B. Ship A is sailing east at $30 \mathrm{~km} / \mathrm{h}$ and ship B is sailing north at $20 \mathrm{~km} / \mathrm{h}$. How fast (in $\mathrm{km} / \mathrm{hr}$ ) is the distance between the ships changing at 4:00 p.m.? (Round your answer to three decimal places.)
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Step 1 ：Let's denote the distance of ship A from the initial point of ship B as x (in km), the distance of ship B from its initial point as y (in km), and the distance between the two ships as z (in km).

Step 2 ：At noon, x = 170 km, \(\frac{dx}{dt}\) = -30 km/hr (since ship A is moving towards ship B), and \(\frac{dy}{dt}\) = 20 km/hr.

Step 3 ：We want to find \(\frac{dz}{dt}\) at 4:00 p.m., which is 4 hours after noon.

Step 4 ：By the Pythagorean theorem, we have \(z^2 = x^2 + y^2\). Differentiating both sides with respect to time t, we get \(2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}\).

Step 5 ：We can solve this equation for \(\frac{dz}{dt}\), but first we need to find the values of x, y, and z at 4:00 p.m.

Step 6 ：At 4:00 p.m., x = 170 km - 30 km/hr * 4 hr = 50 km, and y = 20 km/hr * 4 hr = 80 km. Then we can find z using the Pythagorean theorem.

Step 7 ：After finding x, y, and z, we can substitute these values and the given rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into the equation for \(\frac{dz}{dt}\).

Step 8 ：\(\frac{dx}{dt}\) = -30, \(\frac{dy}{dt}\) = 20, x = 50, y = 80, z = \(\sqrt{50^2 + 80^2}\) = 94.339 km

Step 9 ：Substituting these values into the equation for \(\frac{dz}{dt}\), we get \(\frac{dz}{dt} = \frac{50*(-30) + 80*20}{2*94.339}\) = -23.601 km/hr

Step 10 ：Final Answer: The rate at which the distance between the ships is changing at 4:00 p.m. is approximately \(\boxed{-23.601}\) km/hr. The negative sign indicates that the distance between the ships is decreasing.