At noon, ship A is 170 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 35 km/h. How fast (in km/hr) is the distance between the ships changing at 4:00 p.m.? (Round your answer to three decimal places.)

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.

Calculate how fast the distance between two ships is changing at 4:00 p.m. as they move at different speeds. This problem applies the Pythagorean theorem and differentiation to solve related rates. Ideal for students in Grades 10-12.

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