At noon, ship A is 180 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?

To solve this problem, we need to use the Pythagorean theorem and the concept of related rates in calculus.

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 180 km west of ship B, so x = 180 km. Ship A is sailing east at 40 km/h and ship B is sailing north at 15 km/h. So, dx/dt = -40 km/h (since x is decreasing as ship A moves east) and dy/dt = 15 km/h.

Step 2: Apply the Pythagorean theorem According to the Pythagorean theorem, we have z² = x² + y². Differentiating both sides with respect to time t gives 2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt).

Step 3: Find dz/dt at 4:00 PM At 4:00 PM, 4 hours have passed since noon. In this time, ship A has moved 40 km/h * 4 h = 160 km east, and ship B has moved 15 km/h * 4 h = 60 km north. So, x = 180 km - 160 km = 20 km and y = 60 km.

Substitute x, y, dx/dt, and dy/dt into the differentiated equation: 2z(dz/dt) = 220 km(-40 km/h) + 260 km(15 km/h) To find z, use the Pythagorean theorem: z = sqrt(x² + y²) = sqrt((20 km)² + (60 km)²) = 64 km.

Substitute z into the equation and solve for dz/dt: 264 km(dz/dt) = 220 km(-40 km/h) + 260 km(15 km/h) dz/dt = [220 km(-40 km/h) + 260 km(15 km/h)] / (2*64 km) dz/dt = -25 km/h

So, the distance between the ships is decreasing at a rate of 25 km/h at 4:00 PM.