Math Problem Statement

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.)

Solution

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 PP is at the origin (0,0)(0, 0).
  • The man starts walking north at a rate of 33 ft/s from PP.
  • The woman starts walking south at a rate of 44 ft/s from a point that is 500500 ft directly east of PP.

Step 2: Express the positions as functions of time

Let:

  • x(t)x(t) be the horizontal distance between the man and the woman, which remains 500500 ft at all times because neither of them changes their horizontal position.
  • ym(t)y_m(t) be the vertical distance of the man from PP. Since he starts at PP and walks at 33 ft/s, and given that the woman starts 5 minutes later (300 seconds): y_m(t) = 3t \quad \text{(for t in seconds)}
  • yw(t)y_w(t) be the vertical distance of the woman from PP. 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)y(t), is: y(t)=ym(t)+yw(t)=3t+4(t300)y(t) = y_m(t) + y_w(t) = 3t + 4(t - 300) Simplify: y(t)=3t+4t1200=7t1200y(t) = 3t + 4t - 1200 = 7t - 1200

Step 4: Apply the Pythagorean Theorem

The distance D(t)D(t) between the man and the woman is: D(t)=x(t)2+y(t)2=5002+(7t1200)2D(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)D(t)

Differentiate both sides with respect to time tt: dDdt=125002+(7t1200)22(7t1200)7\frac{dD}{dt} = \frac{1}{2\sqrt{500^2 + (7t - 1200)^2}} \cdot 2(7t - 1200) \cdot 7 dDdt=7(7t1200)5002+(7t1200)2\frac{dD}{dt} = \frac{7(7t - 1200)}{\sqrt{500^2 + (7t - 1200)^2}}

Step 6: Evaluate dDdt\frac{dD}{dt} at t=1200t = 1200 seconds

Since the woman starts walking 5 minutes after the man, 15 minutes after she starts walking corresponds to 15+5=2015 + 5 = 20 minutes total from when the man started walking. This is t=1200t = 1200 seconds.

Substituting t=1200t = 1200 into the equation: y(1200)=7(1200)1200=72001200=6000y(1200) = 7(1200) - 1200 = 7200 - 1200 = 6000 D(1200)=5002+60002=250000+36000000D(1200) = \sqrt{500^2 + 6000^2} = \sqrt{250000 + 36000000} D(1200)=36250000D(1200) = \sqrt{36250000}

Now, substitute t=1200t = 1200 into dDdt\frac{dD}{dt}: dDdtt=1200=7(6000)36250000\frac{dD}{dt}\Bigg|_{t=1200} = \frac{7(6000)}{\sqrt{36250000}} dDdtt=1200=420006010.416.99 ft/s\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.


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.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Pythagorean Theorem
Differentiation

Formulas

Pythagorean theorem: D(t) = √(x(t)^2 + y(t)^2)
Differentiation of distance: dD/dt = [7(7t - 1200)] / √(500^2 + (7t - 1200)^2)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12