Let's break down the problem and solve it step by step using related rates.

Step 1: Define variables

• Let $x(t)$ represent the east-west distance between ship A and the starting point (directly west of ship B at noon), so $x(t)$ measures how far ship A has moved east from its original position.
• Let $y(t)$ represent the north-south distance between ship B and the starting point (directly north of ship A at noon), so $y(t)$ measures how far ship B has moved north.
• Let $z(t)$ represent the distance between the two ships at time $t$.

At noon, ship A is 150 km west of ship B. So at $t = 0$, we have:

• $x(0) = 0$ km (since ship A starts at the origin for our reference frame).
• $y(0) = 0$ km (ship B starts directly north of ship A).

Step 2: Derivatives of positions

Ship A is moving east at 35 km/h, so: $\frac{dx}{dt} = 35 \, \text{km/h}$ Ship B is moving north at 20 km/h, so: $\frac{dy}{dt} = 20 \, \text{km/h}$

Step 3: Distance between ships

The distance $z(t)$ between the ships can be found using the Pythagorean theorem: $z(t) = \sqrt{x(t)^2 + y(t)^2}$

We need to find how fast the distance between the ships is changing at 4:00 PM (i.e., at $t = 4 \, \text{hours}$).

Step 4: Differentiate $z(t)$

To find the rate of change of $z(t)$, differentiate both sides of the equation with respect to time $t$:

$\frac{d}{dt}\left(z(t)\right) = \frac{d}{dt}\left(\sqrt{x(t)^2 + y(t)^2}\right)$

Using the chain rule: $\frac{dz}{dt} = \frac{1}{2} \cdot \frac{1}{\sqrt{x(t)^2 + y(t)^2}} \cdot 2\left(x(t)\frac{dx}{dt} + y(t)\frac{dy}{dt}\right)$

Simplifying: $\frac{dz}{dt} = \frac{x(t) \cdot \frac{dx}{dt} + y(t) \cdot \frac{dy}{dt}}{\sqrt{x(t)^2 + y(t)^2}}$

Step 5: Calculate the values at $t = 4$

At $t = 4$ hours:

• Ship A's position, $x(t)$, is: $x(4) = 35 \, \text{km/h} \times 4 \, \text{h} = 140 \, \text{km}$
• Ship B's position, $y(t)$, is: $y(4) = 20 \, \text{km/h} \times 4 \, \text{h} = 80 \, \text{km}$
• The distance between the ships, $z(t)$, is: $z(4) = \sqrt{x(4)^2 + y(4)^2} = \sqrt{140^2 + 80^2} = \sqrt{19600 + 6400} = \sqrt{26000} \approx 161.24 \, \text{km}$

Now, we can calculate the rate of change of the distance: $\frac{dz}{dt} = \frac{140 \times 35 + 80 \times 20}{\sqrt{140^2 + 80^2}} = \frac{4900 + 1600}{161.24} = \frac{6500}{161.24} \approx 40.3 \, \text{km/h}$

Final Answer:

The distance between the ships is changing at a rate of approximately 40.3 km/h at 4:00 PM.

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Would you like a more detailed explanation or clarification on any step? Here are a few related questions you might find interesting:

1. How would the rate of change of distance change if both ships were traveling in the same direction?
2. How can we apply related rates to other real-world problems involving motion?
3. What if the speeds of the ships were doubled, how would the rate of change of distance change?
4. How does the angle between the paths of the two ships affect the rate of change of the distance between them?
5. How could we set up a similar problem if the ships started at different locations?

Tip: When working with related rates problems, always keep track of your variables and make sure your initial conditions match the problem setup.