We are tasked with finding the rate at which the distance between two ships is changing at 4:00 p.m. The ships are moving with different velocities, and we need to apply related rates to solve the problem.

Step 1: Understanding the problem

• At noon, ship A is 170 km west of ship B. This means the initial horizontal distance between the two ships is 170 km.
• Ship A is sailing east at a speed of 40 km/h.
• Ship B is sailing north at a speed of 35 km/h.
• We are interested in the rate of change of the distance between the two ships at 4:00 p.m., which is 4 hours after noon.

Step 2: Establish the variables

• Let $x(t)$ be the horizontal distance between ship A and ship B at time $t$, where $t$ is the time in hours after noon.
• Let $y(t)$ be the vertical distance between ship A and ship B at time $t$.
• Let $z(t)$ be the straight-line distance between the two ships at time $t$.

We want to find $\frac{dz}{dt}$ at $t = 4$ hours.

Step 3: Express $x(t)$, $y(t)$, and $z(t)$

• At noon, $x(0) = 170$ km, and since ship A is sailing east at 40 km/h, the horizontal distance decreases over time. Thus, after $t$ hours: $x(t) = 170 - 40t$
• At noon, $y(0) = 0$ km, and since ship B is sailing north at 35 km/h, the vertical distance increases over time. Thus, after $t$ hours: $y(t) = 35t$
• The straight-line distance $z(t)$ between the ships is given by the Pythagorean theorem: $z(t) = \sqrt{x(t)^2 + y(t)^2}$

Step 4: Differentiate with respect to time

To find $\frac{dz}{dt}$, we differentiate $z(t)$ with respect to $t$: $\frac{dz}{dt} = \frac{1}{2} \cdot \frac{2x(t) \cdot \frac{dx}{dt} + 2y(t) \cdot \frac{dy}{dt}}{\sqrt{x(t)^2 + y(t)^2}}$ 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: Plug in the values at $t = 4$

At 4:00 p.m. (i.e., $t = 4$):

• $x(4) = 170 - 40 \times 4 = 170 - 160 = 10$ km.
• $y(4) = 35 \times 4 = 140$ km.
• $\frac{dx}{dt} = -40$ km/h (since ship A is moving east, reducing the distance).
• $\frac{dy}{dt} = 35$ km/h (since ship B is moving north, increasing the distance).

Now, we can calculate $z(4)$: $z(4) = \sqrt{x(4)^2 + y(4)^2} = \sqrt{10^2 + 140^2} = \sqrt{100 + 19600} = \sqrt{19700} \approx 140.356 \, \text{km}$

Next, substitute into the equation for $\frac{dz}{dt}$: $\frac{dz}{dt} = \frac{10 \times (-40) + 140 \times 35}{\sqrt{19700}} = \frac{-400 + 4900}{140.356} = \frac{4500}{140.356} \approx 32.059 \, \text{km/h}$

Thus, the distance between the ships is changing at approximately 32.059 km/h at 4:00 p.m.

Would you like more details on any part of the solution?

Here are 5 related questions:

1. How would the answer change if ship A were moving faster?
2. What if both ships started from the same location?
3. How is the rate of change of distance affected if ship B were also moving east?
4. What would happen if both ships moved in opposite directions?
5. Can you calculate the distance between the ships at 3:00 p.m.?

Tip: Related rates problems often rely on differentiating geometric relationships, like the Pythagorean theorem, with respect to time.