solve: A man starts walking north at 5 ft/s from a point P. Five minutes later a woman starts walking south at 6 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 minutes after the woman starts walking? (Round your answer to two decimal places.)

This is a related rates problem. We are asked to find the rate at which the distance between two people is changing as they walk in opposite directions from specific points.

### Step-by-step process:

1. **Define Variables:**
   - Let \( x(t) \) be the distance the man has walked north from point \( P \) at time \( t \).
   - Let \( y(t) \) be the distance the woman has walked south from her starting point 500 feet east of \( P \).
   - Let \( z(t) \) be the distance between the two people at any time \( t \).

2. **Relations Between Variables:**
   Since the man walks directly north and the woman walks south from a point 500 feet east of the man's starting point, their relative positions form a right triangle with:
   - One leg of the triangle as the horizontal distance between them (which remains 500 feet),
   - The other leg as the vertical distance between them (which is the sum of the distances they've each walked),
   - The hypotenuse as \( z(t) \), the distance between the two people.

   Using the Pythagorean Theorem:
   \[
   z(t)^2 = x(t)^2 + 500^2
   \]

3. **Distances:**
   - The man is walking north at 5 ft/s. So, after \( t \) seconds, his distance from point \( P \) is:
     \[
     x(t) = 5t \quad \text{(in feet)}
     \]
   - The woman starts walking after 5 minutes (300 seconds), so after \( t \) seconds, she has walked south for \( t - 300 \) seconds (assuming \( t \geq 300 \)). Therefore:
     \[
     y(t) = 6(t - 300) \quad \text{for } t \geq 300.
     \]
   - The total vertical distance between them is \( x(t) + y(t) \), but since they are walking in opposite directions, the vertical leg is the sum of their distances:
     \[
     \text{Vertical distance} = x(t) + y(t) = 5t + 6(t - 300)
     \]

4. **Differentiate with Respect to Time \( t \):**
   We need to find the rate of change of \( z(t) \) (i.e., \( \frac{dz}{dt} \)) when \( t = 15 \) minutes after the woman starts walking, which is 900 seconds total from the start (600 seconds since she started walking).

   \[
   \frac{d}{dt}(z(t)^2) = \frac{d}{dt}(x(t)^2 + 500^2)
   \]
   Using the chain rule:
   \[
   2z(t) \frac{dz}{dt} = 2x(t) \frac{dx}{dt}
   \]

Let’s calculate it further.The rate at which the two people are moving apart 15 minutes after the woman starts walking is approximately **10.99 ft/s**.

Would you like further details on how this was calculated or have any questions?

Here are 5 related questions for further exploration:
1. How does the distance between them change if they were walking at the same speed?
2. How would the answer change if the woman started walking 10 minutes after the man?
3. What would the rate be if both people started walking at the same time?
4. How would the rate change if the man was walking diagonally rather than strictly north?
5. How does the related rates method apply to other scenarios involving distances and speeds?

**Tip:** When solving related rates problems, always define all variables clearly and express the relationships using appropriate geometry (such as the Pythagorean Theorem).

A man starts walking north at 5 ft/s from a point P. Five minutes later a woman starts walking south at 6 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 minutes after the woman starts walking? (Round your answer to two decimal places.)

Solve a related rates problem involving a man walking north at 5 ft/s and a woman walking south at 6 ft/s from a point 500 feet away. Using the Pythagorean Theorem and differentiation, learn how to calculate the rate at which they are moving apart 15 minutes after the woman starts walking. Suitable for Grades 10-12.

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