At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 19 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

To find the rate at which the distance between the ships is changing at 3 PM, we need to calculate the derivative of the distance function with respect to time.

Let's analyze the situation. At noon, ship A is 30 nautical miles due west of ship B. This means that at that moment, the horizontal distance between the ships is 30 nautical miles, and the vertical distance is 0 nautical miles.

Ship A is sailing west at a constant speed of 18 knots. Since the horizontal distance is decreasing, this gives us a rate of -18 knots for the horizontal distance.

Ship B is sailing north at a constant speed of 19 knots. Since the vertical distance is increasing, this gives us a rate of +19 knots for the vertical distance.

To find the rate of change of the distance between the ships at 3 PM, we can use the Pythagorean theorem. Let's call the horizontal distance x(t) and the vertical distance y(t) at time t.

Using the Pythagorean theorem, we have:
x(t)^2 + y(t)^2 = d(t)^2

We are interested in finding the rate of change of d(t) at 3 PM, so we differentiate the equation with respect to t:

2x(t) * x'(t) + 2y(t) * y'(t) = 2d(t) * d'(t)

Now, let's plug in the given rates:
x'(t) = -18 knots
y'(t) = 19 knots
x(t) = 30 nautical miles
y(t) = 0 (since the vertical distance remains constant)

Substituting these values, we have:
2 * 30 nautical miles * (-18 knots) + 2 * 0 * 19 knots = 2 * d(t) * d'(t)

Simplifying the equation:
-36 * 30 = 2 * d(t) * d'(t)

Solving for d'(t), we find:
d'(t) = -36 * 30 / 2 = -18 * 30 = -540 knots

Therefore, the rate at which the distance between the ships is changing at 3 PM is -540 knots.