To solve this problem, we will calculate the rate at which the distance between the two ships is changing using the concept of related rates in calculus.

Step 1: Define variables

Let:

• $x(t)$: the position of Ship A relative to Ship B in the east-west direction.
• $y(t)$: the position of Ship B relative to Ship A in the north-south direction.
• $z(t)$: the distance between the ships.

Given:

• At noon, $x(0) = -180 \, \text{km}$ (Ship A is 180 km west of Ship B).
• Ship A is moving east at $\frac{dx}{dt} = 35 \, \text{km/h}$.
• Ship B is moving north at $\frac{dy}{dt} = 25 \, \text{km/h}$.
• The relationship between $x(t)$, $y(t)$, and $z(t)$ is: $z(t)^2 = x(t)^2 + y(t)^2$

Step 2: Position of the ships at 4:00 p.m.

At 4:00 p.m., 4 hours have passed since noon:

1. Position of Ship A: $x(4) = -180 + 35 \times 4 = -180 + 140 = -40 \, \text{km}.$

2. Position of Ship B: $y(4) = 25 \times 4 = 100 \, \text{km}.$

Step 3: Distance between the ships at 4:00 p.m.

Substitute $x(4)$ and $y(4)$ into the distance formula: $z(4) = \sqrt{x(4)^2 + y(4)^2} = \sqrt{(-40)^2 + 100^2} = \sqrt{1600 + 10000} = \sqrt{11600} = 107.703 \, \text{km}.$

Step 4: Differentiate the distance equation

Differentiate both sides of the equation $z(t)^2 = x(t)^2 + y(t)^2$ with respect to $t$: $2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}.$

Simplify: $z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}.$

Solve for $\frac{dz}{dt}$: $\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}.$

Step 5: Substitute known values at 4:00 p.m.

At 4:00 p.m., we have:

• $x(4) = -40$,
• $y(4) = 100$,
• $z(4) = 107.703$,
• $\frac{dx}{dt} = 35$,
• $\frac{dy}{dt} = 25$.

Substitute into the equation: $\frac{dz}{dt} = \frac{(-40)(35) + (100)(25)}{107.703}.$

Simplify the numerator: $(-40)(35) + (100)(25) = -1400 + 2500 = 1100.$

Substitute: $\frac{dz}{dt} = \frac{1100}{107.703} \approx 10.213 \, \text{km/h}.$

Final Answer:

The rate at which the distance between the ships is changing at 4:00 p.m. is approximately: $\boxed{10.213 \, \text{km/h}}$

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Would you like further explanation or clarification on any part of this solution? Here are related questions for practice:

1. How do related rates apply to other real-world motion problems?
2. How would the solution change if the ships had different speeds?
3. What happens if both ships are traveling in the same direction?
4. Can you derive a general formula for such problems without specific numbers?
5. How do we handle situations where the ships are initially at an angle to each other?

Tip: Always draw a diagram for related rates problems to visualize relationships between quantities!