To find how fast the distance between the ships is changing at 4 PM, we can use the concept of related rates in calculus. Let’s solve step by step:

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Step 1: Define variables

• Let $t$ be the time in hours after noon.
• Let $x(t)$ be the distance of Ship A from its starting point (eastward direction). Initially, $x(0) = 0$.
• Let $y(t)$ be the distance of Ship B from its starting point (northward direction). Initially, $y(0) = 0$.
• The distance between the ships at time $t$ is denoted by $z(t)$.

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Step 2: Express the positions of the ships

• Ship A travels east at 40 miles/hour:
  $x(t) = 260 - 40t$.
  At $t = 0$, Ship A is 260 miles west of Ship B.

• Ship B travels north at 60 miles/hour:
  $y(t) = 60t$.
  At $t = 0$, Ship B is at its initial position.

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Step 3: Distance between the ships

Using the Pythagorean theorem: $z(t) = \sqrt{x(t)^2 + y(t)^2}.$ Substituting $x(t)$ and $y(t)$: $z(t) = \sqrt{(260 - 40t)^2 + (60t)^2}.$

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Step 4: Differentiate with respect to $t$

We need $\frac{dz}{dt}$ (rate of change of distance between the ships). Differentiating $z(t)$: $\frac{dz}{dt} = \frac{1}{2\sqrt{(260 - 40t)^2 + (60t)^2}} \cdot 2 \left[ (260 - 40t)(-40) + (60t)(60) \right].$ Simplify: $\frac{dz}{dt} = \frac{(260 - 40t)(-40) + (60t)(60)}{\sqrt{(260 - 40t)^2 + (60t)^2}}.$

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Step 5: Evaluate at $t = 4$

At $t = 4$:

• $x(4) = 260 - 40(4) = 100$.
• $y(4) = 60(4) = 240$.
• $z(4) = \sqrt{100^2 + 240^2} = \sqrt{10000 + 57600} = \sqrt{67600} = 260$.

Now substitute into $\frac{dz}{dt}$: $\frac{dz}{dt} = \frac{(260 - 40(4))(-40) + (60(4))(60)}{\sqrt{(260 - 40(4))^2 + (60(4))^2}}.$ Simplify: $\frac{dz}{dt} = \frac{(100)(-40) + (240)(60)}{260}.$ $\frac{dz}{dt} = \frac{-4000 + 14400}{260}.$ $\frac{dz}{dt} = \frac{10400}{260} = 40 \, \text{miles/hour}.$

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Final Answer:

The distance between the ships is increasing at 40 miles/hour at 4 PM.

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Would you like further details or clarification on any step?

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Related Questions:

1. What would be the rate of change of distance at 2 PM?
2. How would the problem change if Ship A sailed west instead of east?
3. Can you derive a general formula for $\frac{dz}{dt}$ based on arbitrary speeds?
4. How does the distance between the ships evolve over time graphically?
5. What happens if both ships travel in perpendicular directions but at the same speed?

Tip:

Always double-check units and ensure consistent directions when solving related rates problems.