At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast (in km/hr) is the distance between the ships changing at 4:00 p.m.? (Round your answer to three decimal places.)

To solve this problem, we can use the Pythagorean theorem and the concept of related rates.

Step 1: Set up the problem Let's denote the distance of ship A from the point directly north of ship B as x (km), and the distance of ship B from that point as y (km). The distance between the two ships is z (km).

At noon, ship A is 170 km west of ship B, so x = 170 km. Ship A is sailing east at 35 km/h, so dx/dt = -35 km/h (we take it as negative because the distance x is decreasing as ship A moves east). Ship B is sailing north at 25 km/h, so dy/dt = 25 km/h.

We want to find dz/dt, the rate at which the distance between the ships is changing at 4:00 p.m.

Step 2: Apply the Pythagorean theorem According to the Pythagorean theorem, we have x^2 + y^2 = z^2.

Step 3: Differentiate both sides with respect to time t By differentiating both sides with respect to time t, we get 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).

Step 4: Solve for dz/dt We can rearrange the equation to solve for dz/dt: dz/dt = (x(dx/dt) + y(dy/dt)) / z.

Step 5: Plug in the values at 4:00 p.m. At 4:00 p.m., 4 hours have passed since noon. In this time, ship A has moved 35 km/h * 4 h = 140 km east, so x = 170 km - 140 km = 30 km. Ship B has moved 25 km/h * 4 h = 100 km north, so y = 100 km.

The distance between the ships at 4:00 p.m. is z = sqrt(x^2 + y^2) = sqrt((30 km)^2 + (100 km)^2) = sqrt(10000 km^2) = 100 km.

Now we can plug these values into the equation for dz/dt: dz/dt = (30 km * -35 km/h + 100 km * 25 km/h) / 100 km = (-1050 + 2500) / 100 km/h = 14.5 km/h.

So, the distance between the ships is changing at a rate of 14.5 km/h at 4:00 p.m.