A man starts walking north at 4 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 min after the woman starts walking? (Round your answer to two decimal places.)
 ft/s

To solve this related rates problem, we will use the Pythagorean theorem to express the distance between the man and the woman at any given time, and then differentiate it to find the rate at which they are moving apart. 

---

### Step 1: Variables and relationships
- **The man's motion:** 
  - Speed: \(4 \, \text{ft/s}\) 
  - Time: \(15 + 5 = 20 \, \text{min} = 1200 \, \text{seconds}\)  
    (The man has been walking for 20 minutes since he started earlier.)

- **The woman's motion:** 
  - Speed: \(6 \, \text{ft/s}\) 
  - Time: \(15 \, \text{min} = 900 \, \text{seconds}\)  
    (The woman has been walking for 15 minutes.)

- **Initial distance:**  
  The woman starts 500 ft to the east of point \(P\). The man's position is along the **north-south line**, and the woman’s position is along the **southward line**, starting east of \(P\). 

---

### Step 2: Coordinates and Distances
- Let \(x(t)\) be the **east-west distance** between the two people, which is a constant 500 ft (because the man walks only north, and the woman walks only south).
- Let \(y(t)\) be the **north-south distance** between the two people. 

- **Man’s northward position after 1200 s:**  
  \(4 \times 1200 = 4800 \, \text{ft}\)

- **Woman’s southward position after 900 s:**  
  \(6 \times 900 = 5400 \, \text{ft}\)

Thus, the total north-south distance between them is:
\[
y(t) = 4800 + 5400 = 10200 \, \text{ft}
\]

---

### Step 3: Pythagorean theorem to relate the total distance \(D\)
The total distance between the two people at time \(t\) is:
\[
D(t) = \sqrt{x(t)^2 + y(t)^2}
\]
\[
D(t) = \sqrt{500^2 + 10200^2}
\]
\[
D(t) = \sqrt{250000 + 104040000} = \sqrt{104290000} \approx 10212.25 \, \text{ft}
\]

---

### Step 4: Differentiate to find the rate of change of distance \(D(t)\)
We now differentiate the equation \(D(t) = \sqrt{x(t)^2 + y(t)^2}\) with respect to time \(t\):
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{x(t)^2 + y(t)^2}} \cdot 2 \left( x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt} \right)
\]

Since \(x(t) = 500 \, \text{ft}\) (constant), \(\frac{dx}{dt} = 0\). Thus, the equation simplifies to:
\[
\frac{dD}{dt} = \frac{1}{\sqrt{500^2 + 10200^2}} \cdot (10200 \cdot \frac{dy}{dt})
\]

Since both are walking in opposite directions along the north-south axis:
\[
\frac{dy}{dt} = 4 + 6 = 10 \, \text{ft/s}
\]

Thus:
\[
\frac{dD}{dt} = \frac{10200 \cdot 10}{\sqrt{104290000}}
\]
\[
\frac{dD}{dt} = \frac{102000}{10212.25} \approx 9.99 \, \text{ft/s}
\]

---

### Final Answer
The rate at which the two people are moving apart 15 minutes after the woman starts walking is:
\[
\boxed{9.99 \, \text{ft/s}}
\]

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Do you have any questions or want further details on any step?

### 5 Related Questions
1. How do you derive the Pythagorean theorem in related rates problems?
2. Why is the woman’s initial eastward distance kept constant during the problem?
3. How do you handle related rates problems where multiple variables change?
4. What happens if one person walks in a different direction, like northeast?
5. How do units affect calculations in related rates problems?

**Tip:** When dealing with related rates, always identify which quantities are constant and which ones are changing with respect to time.

A man starts walking north at 4 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 min after the woman starts walking? (Round your answer to two decimal places.)

Solve a related rates problem where a man walks north at 4 ft/s and a woman walks south at 6 ft/s from 500 ft east of his starting point. Learn how to apply the Pythagorean theorem and differentiation to find the rate at which they are moving apart 15 minutes after the woman starts walking. Includes detailed steps and explanations.

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