At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 17 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.) knots

Step 1 ：This is a problem of related rates in calculus. We can model the situation with a right triangle, where the horizontal leg represents the path of ship A, the vertical leg represents the path of ship B, and the hypotenuse represents the distance between the two ships. We are given the rates at which the lengths of the two legs are changing (the speeds of the ships), and we are asked to find the rate at which the length of the hypotenuse is changing (the rate at which the distance between the ships is changing).

Step 2 ：We can use the Pythagorean theorem to relate the lengths of the sides of the triangle: \(a^2 + b^2 = c^2\), where a is the distance of ship A from the starting point, b is the distance of ship B from the starting point, and c is the distance between the ships. We can differentiate this equation with respect to time to find the rates at which these quantities are changing.

Step 3 ：At 3 PM, ship A has been sailing for 3 hours at 21 knots, so it is 63 nautical miles from its starting point. Similarly, ship B is 51 nautical miles from its starting point. We can substitute these values into the differentiated equation to find the rate at which the distance between the ships is changing.

Step 4 ：Given that a = 73, b = 51, da/dt = 21, db/dt = 17, and c = 89.05, we can calculate dc/dt = 26.95.

Step 5 ：Final Answer: The rate at which the distance between the ships is changing at 3 PM is approximately \(\boxed{26.95}\) knots.