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This problem involves related rates, where we need to find how fast the distance between two ships is changing at a specific time.

### Setup
At noon, the initial positions of the ships are:
- Ship A is 10 nautical miles west of Ship B.
- Ship A moves west at 15 knots (15 nautical miles per hour).
- Ship B moves north at 23 knots (23 nautical miles per hour).

Let:
- \( x(t) \) represent the distance between Ship A and the vertical line through Ship B (i.e., how far west Ship A is),
- \( y(t) \) represent the distance between Ship B and the horizontal line through Ship A (i.e., how far north Ship B is),
- \( z(t) \) represent the distance between the two ships.

At \( t = 0 \) (noon):
- \( x(0) = 10 \) nautical miles,
- \( y(0) = 0 \) nautical miles,
- At 6 PM, \( t = 6 \) hours.

We need to find \( \frac{dz}{dt} \) (the rate at which the distance between the ships is changing) at \( t = 6 \) hours.

### Step 1: Define the relationship
The distance between the ships, \( z \), forms the hypotenuse of a right triangle with legs \( x(t) \) and \( y(t) \):
\[
z^2 = x^2 + y^2
\]

### Step 2: Take the derivative with respect to \( t \)
Differentiate both sides of the equation with respect to \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplify:
\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]

### Step 3: Evaluate at \( t = 6 \)
At \( t = 6 \):
- Ship A's position west of B: \( x(6) = 10 + 15 \times 6 = 100 \) nautical miles,
- Ship B's position north of A: \( y(6) = 23 \times 6 = 138 \) nautical miles.

Thus, the distance between the ships at \( t = 6 \):
\[
z(6) = \sqrt{100^2 + 138^2} = \sqrt{10000 + 19044} = \sqrt{29044} \approx 170.43 \text{ nautical miles.}
\]

Also:
- \( \frac{dx}{dt} = 15 \) knots (westward),
- \( \frac{dy}{dt} = 23 \) knots (northward).

Substitute these values into the differentiated equation:
\[
170.43 \cdot \frac{dz}{dt} = 100 \cdot (-15) + 138 \cdot 23
\]
\[
170.43 \cdot \frac{dz}{dt} = -1500 + 3174 = 1674
\]
\[
\frac{dz}{dt} = \frac{1674}{170.43} \approx 9.82 \text{ knots}.
\]

### Final Answer:
The distance between the ships is changing at approximately **9.82 knots** at 6 PM.

Would you like further explanation or details on any part of this? Here are some related questions to explore:

1. How would the result change if the ships were moving in opposite directions?
2. What if Ship A were moving east instead of west?
3. Can you generalize the formula for any given starting distances and speeds?
4. How would you calculate the distance at any given time \( t \)?
5. How does the Pythagorean theorem apply to this problem?

**Tip:** Always double-check units when working with rates, especially in problems involving speed and time!

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 23 knots. How fast (in knots) is the distance between the ships changing at 6 PM?

Solve a related rates problem where two ships are moving at different speeds and directions, with ship A moving west and ship B moving north. Learn how to calculate how fast the distance between them changes after 6 hours using differentiation and the Pythagorean theorem.

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