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.) \[ -23.601 \mathrm{~km} / \mathrm{h} \]

Step 1 ：Let's denote the distance of ship A from the starting point (170 km west of ship B) as \(x\), the distance of ship B from the starting point as \(y\), and the distance between the two ships as \(z\).

Step 2 ：At 4:00 p.m., ship A has been sailing for 4 hours and ship B has been sailing for 4 hours as well. So, \(x = 170 - 30 \times 4 = 50\) km and \(y = 20 \times 4 = 80\) km.

Step 3 ：We can use the Pythagorean theorem to find \(z\): \(z = \sqrt{x^2 + y^2} = \sqrt{50^2 + 80^2}\).

Step 4 ：Then we differentiate the equation \(z^2 = x^2 + y^2\) with respect to time \(t\) to find \(\frac{dz}{dt}\), the rate of change of the distance between the ships.

Step 5 ：The derivative of \(z^2\) with respect to time is \(2z \frac{dz}{dt}\), the derivative of \(x^2\) with respect to time is \(2x \frac{dx}{dt}\), and the derivative of \(y^2\) with respect to time is \(2y \frac{dy}{dt}\). So, the differentiated equation is \(2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}\).

Step 6 ：We can solve this equation for \(\frac{dz}{dt}\) to find the rate of change of the distance between the ships.

Step 7 ：Final Answer: The rate at which the distance between the ships is changing at 4:00 p.m. is approximately \(\boxed{1.060 \mathrm{~km/h}}\).