At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 7 PM

To find the rate at which the distance between the two ships is changing at 7 PM, we can break down the problem into several steps.

### Step 1: Define the Positions of the Ships
Let:
- \( A \) be the position of Ship A.
- \( B \) be the position of Ship B.
- \( x \) be the horizontal distance (east-west direction) between the ships.
- \( y \) be the vertical distance (north-south direction) between the ships.
  
We are given:
- At noon, \( x = 40 \) nautical miles (Ship A is 40 nautical miles due west of Ship B).
- Ship A is sailing west at 17 knots, meaning \( \frac{dx}{dt} = -17 \).
- Ship B is sailing north at 16 knots, meaning \( \frac{dy}{dt} = 16 \).

### Step 2: Determine \( x \) and \( y \) at 7 PM
Since the ships start moving at noon, 7 PM is 7 hours later.

#### For Ship A:
- \( x = 40 + (17 \text{ knots} \times 7 \text{ hours}) \)
- \( x = 40 + 119 = 159 \) nautical miles.

#### For Ship B:
- \( y = 16 \text{ knots} \times 7 \text{ hours} \)
- \( y = 112 \) nautical miles.

### Step 3: Use the Distance Formula
Let \( z \) be the distance between the ships. Using the Pythagorean theorem:

\[
z = \sqrt{x^2 + y^2}
\]

### Step 4: Differentiate \( z \) with Respect to \( t \)
To find \( \frac{dz}{dt} \), the rate of change of the distance between the ships, we differentiate both sides of the equation \( z = \sqrt{x^2 + y^2} \) with respect to \( t \):

\[
\frac{dz}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x\frac{dx}{dt} + 2y\frac{dy}{dt}
\]

Simplifying:

\[
\frac{dz}{dt} = \frac{x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}}{\sqrt{x^2 + y^2}}
\]

### Step 5: Substitute Known Values
Now we substitute \( x = 159 \), \( y = 112 \), \( \frac{dx}{dt} = -17 \), and \( \frac{dy}{dt} = 16 \):

1. First, calculate \( z \):
   \[
   z = \sqrt{159^2 + 112^2}
   \]
   \[
   z = \sqrt{25281 + 12544} = \sqrt{37825} \approx 194.5 \text{ nautical miles}
   \]

2. Then, substitute into \( \frac{dz}{dt} \):
   \[
   \frac{dz}{dt} = \frac{159(-17) + 112(16)}{194.5}
   \]
   \[
   = \frac{-2703 + 1792}{194.5} 
   \]
   \[
   = \frac{-911}{194.5} 
   \]
   \[
   \approx -4.68 \text{ knots}
   \]

### Answer:
The distance between the two ships is decreasing at a rate of approximately \( 4.68 \) knots at 7 PM.

Would you like further details or have any questions?

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Here are some related questions that may interest you:

1. How would the solution change if one of the ships changed course?
2. What if the ships were initially positioned differently, such as one being north or south instead of due west?
3. How would you calculate the distance if only the time period is different (e.g., 5 PM)?
4. Can we solve this problem using parametric equations instead?
5. What if both ships were moving in the same direction at different speeds?

**Tip:** When analyzing related rates problems, always start by defining what you know and finding an equation that links the variables you need to differentiate.

At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 7 PM?

Explore how to calculate the rate at which the distance between two ships changes over time using related rates. Given ship A moving west at 17 knots and ship B moving north at 16 knots, find the change in distance at 7 PM. Ideal for students in Grades 11-12 learning about calculus, related rates, and the Pythagorean theorem.

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