A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 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.)


To solve this problem, we'll use the concept of related rates and the Pythagorean Theorem, since the movement of the man and the woman forms a right triangle.

### Step 1: Set up the coordinate system
- Assume point \( P \) is at the origin \((0, 0)\).
- The man starts walking north at a rate of \( 3 \) ft/s from \( P \). 
- The woman starts walking south at a rate of \( 4 \) ft/s from a point that is \( 500 \) ft directly east of \( P \). 

### Step 2: Express the positions as functions of time
Let:
- \( x(t) \) be the horizontal distance between the man and the woman, which remains \( 500 \) ft at all times because neither of them changes their horizontal position.
- \( y_m(t) \) be the vertical distance of the man from \( P \). Since he starts at \( P \) and walks at \( 3 \) ft/s, and given that the woman starts 5 minutes later (300 seconds):
  \[
  y_m(t) = 3t \quad \text{(for \( t \) in seconds)}
  \]
- \( y_w(t) \) be the vertical distance of the woman from \( P \). Since she starts walking 5 minutes (300 seconds) after the man:
  \[
  y_w(t) = 4(t - 300) \quad \text{for \( t \geq 300 \) seconds}
  \]

### Step 3: Calculate the vertical separation between them
The vertical separation between the man and the woman, \( y(t) \), is:
\[
y(t) = y_m(t) + y_w(t) = 3t + 4(t - 300)
\]
Simplify:
\[
y(t) = 3t + 4t - 1200 = 7t - 1200
\]

### Step 4: Apply the Pythagorean Theorem
The distance \( D(t) \) between the man and the woman is:
\[
D(t) = \sqrt{x(t)^2 + y(t)^2} = \sqrt{500^2 + (7t - 1200)^2}
\]

### Step 5: Differentiate to find the rate of change of \( D(t) \)
Differentiate both sides with respect to time \( t \):
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{500^2 + (7t - 1200)^2}} \cdot 2(7t - 1200) \cdot 7
\]
\[
\frac{dD}{dt} = \frac{7(7t - 1200)}{\sqrt{500^2 + (7t - 1200)^2}}
\]

### Step 6: Evaluate \( \frac{dD}{dt} \) at \( t = 1200 \) seconds
Since the woman starts walking 5 minutes after the man, 15 minutes after she starts walking corresponds to \( 15 + 5 = 20 \) minutes total from when the man started walking. This is \( t = 1200 \) seconds.

Substituting \( t = 1200 \) into the equation:
\[
y(1200) = 7(1200) - 1200 = 7200 - 1200 = 6000
\]
\[
D(1200) = \sqrt{500^2 + 6000^2} = \sqrt{250000 + 36000000}
\]
\[
D(1200) = \sqrt{36250000}
\]

Now, substitute \( t = 1200 \) into \( \frac{dD}{dt} \):
\[
\frac{dD}{dt}\Bigg|_{t=1200} = \frac{7(6000)}{\sqrt{36250000}}
\]
\[
\frac{dD}{dt}\Bigg|_{t=1200} = \frac{42000}{6010.41} \approx 6.99 \text{ ft/s}
\]

### Final Answer
The rate at which the people are moving apart 15 minutes after the woman starts walking is approximately **6.99 ft/s**.

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Do you want more details or have any questions?

Here are 5 related questions:
1. What would be the rate if the man and woman started walking simultaneously?
2. How would the solution change if the woman started walking west instead of south?
3. Can we find the time when the rate of separation is maximum?
4. How do we calculate the distance between the man and woman at any given time?
5. What if the man walked at a different speed, say 5 ft/s? How would the answer change?

**Tip:** For problems involving movement and distance over time, set up a coordinate system and apply the Pythagorean Theorem for clarity.

A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 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.)

Explore how to solve a related rates problem using the Pythagorean Theorem. A man walks north at 3 ft/s, while a woman walks south at 4 ft/s from a point 500 ft east. Learn how to calculate the rate at which they move apart 15 minutes after the woman starts walking. Suitable for high school students in Grades 10-12.

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