At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?


To solve the problem, we can use the principles of related rates in calculus. 

1. **Position of the Ships at Noon**:
   - Ship A is 170 km west of Ship B.
   - Let the position of Ship B be the origin (0, 0).
   - Position of Ship A: \( (-170, 0) \).

2. **Speeds**:
   - Ship A is sailing east at 35 km/h, so its x-coordinate increases.
   - Ship B is sailing north at 20 km/h, so its y-coordinate increases.

3. **Positions at 4:00 PM**:
   - Time from noon to 4 PM is 4 hours.
   - Position of Ship A at 4 PM:
     \[
     x_A = -170 + 35 \times 4 = -170 + 140 = -30 \text{ km}
     \]
   - Position of Ship B at 4 PM:
     \[
     y_B = 0 + 20 \times 4 = 80 \text{ km}
     \]

4. **Distance Between the Ships**:
   Let \( d \) be the distance between the two ships:
   \[
   d = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}
   \]
   In our case:
   \[
   d = \sqrt{(-30 - 0)^2 + (0 - 80)^2} = \sqrt{(-30)^2 + (-80)^2} = \sqrt{900 + 6400} = \sqrt{7300} \approx 85.44 \text{ km}
   \]

5. **Using Related Rates**:
   Let \( x = x_A + 170 \) (position of Ship A from Ship B) and \( y = y_B \). Then:
   \[
   x = -30 + 170 = 140 \text{ km} \quad (since \, x_A \, is \, west \, of \, B)
   \]
   The rate of change of \( x \) with respect to time \( t \) is \( \frac{dx}{dt} = 35 \, \text{km/h} \) (east) and the rate of change of \( y \) is \( \frac{dy}{dt} = 20 \, \text{km/h} \) (north).

6. **Differentiating the Distance Formula**:
   Using the formula for distance:
   \[
   d^2 = x^2 + y^2
   \]
   Differentiating with respect to time \( t \):
   \[
   2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
   \]
   Rearranging gives:
   \[
   \frac{dd}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{d}
   \]

7. **Substituting Values**:
   At 4 PM:
   - \( x = 140 \text{ km} \)
   - \( y = 80 \text{ km} \)
   - \( d \approx 85.44 \text{ km} \)
   - \( \frac{dx}{dt} = 35 \, \text{km/h} \)
   - \( \frac{dy}{dt} = 20 \, \text{km/h} \)

   Now calculate:
   \[
   \frac{dd}{dt} = \frac{(140)(35) + (80)(20)}{85.44}
   \]
   \[
   = \frac{4900 + 1600}{85.44} = \frac{6500}{85.44} \approx 76.06 \text{ km/h}
   \]

Therefore, the distance between the ships is increasing at approximately **76.06 km/h** at 4:00 PM.

Would you like more details on any part of this problem, or do you have other questions? Here are some related questions you might find interesting:

1. How would the distance change if Ship A was sailing west instead?
2. What would be the distance between the ships at 5 PM?
3. How would the problem change if Ship B was sailing south instead of north?
4. What is the total distance covered by each ship by 4 PM?
5. How can we visualize the movement of the ships over time?

**Tip**: Always diagram the situation for related rates problems; it can clarify how the distances and directions relate to each other.

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

Explore how to determine the rate of distance change between two ships sailing at different speeds. This problem involves related rates in calculus, featuring Ship A moving east at 35 km/h and Ship B moving north at 20 km/h. Learn to calculate that the distance between them is increasing at approximately 76.06 km/h at 4:00 PM.

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