A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of . At what rate are the people moving apart 15 min after the woman starts walking?
The Correct Answer and Explanation is :
Let’s break this problem into parts and solve it step by step using related rates.
Step 1: Define variables
Let ( t ) represent the time in seconds since the woman started walking. The problem states that the woman starts walking 5 minutes after the man, which is equivalent to 300 seconds. So, the time ( t = 0 ) corresponds to 5 minutes after the man starts walking.
- Let ( x(t) ) represent the east-west distance between the two people, which is constant because the man is walking north, and the woman is walking south. Initially, ( x = 500 ) ft, and this value does not change.
- Let ( y_m(t) ) represent the position of the man along the north-south axis (measured in feet). Since the man starts walking north at 4 ft/s, his position at time ( t ) is given by ( y_m(t) = 4(t + 300) ), because the man has been walking for 300 seconds before the woman starts walking.
- Let ( y_w(t) ) represent the position of the woman along the north-south axis. Since she starts walking south at 5 ft/s, her position at time ( t ) is given by ( y_w(t) = 5t ).
Step 2: Use the distance formula
The distance ( D(t) ) between the two people at time ( t ) is given by the Pythagorean theorem, because the two people are moving along perpendicular paths (north-south and east-west):
[
D(t) = \sqrt{x(t)^2 + (y_m(t) – y_w(t))^2}
]
Since ( x(t) = 500 ) ft is constant, we have:
[
D(t) = \sqrt{500^2 + (y_m(t) – y_w(t))^2}
]
Substitute the expressions for ( y_m(t) ) and ( y_w(t) ):
[
D(t) = \sqrt{500^2 + \left( 4(t + 300) – 5t \right)^2}
]
Simplify the expression inside the parentheses:
[
D(t) = \sqrt{500^2 + \left( 4t + 1200 – 5t \right)^2}
]
[
D(t) = \sqrt{500^2 + (-t + 1200)^2}
]
Step 3: Differentiate with respect to time
To find the rate at which the distance between the two people is changing, differentiate ( D(t) ) with respect to time ( t ):
[
\frac{dD}{dt} = \frac{d}{dt} \left( \sqrt{500^2 + (-t + 1200)^2} \right)
]
Apply the chain rule:
[
\frac{dD}{dt} = \frac{1}{2} \left( 500^2 + (-t + 1200)^2 \right)^{-1/2} \cdot 2(-t + 1200)(-1)
]
[
\frac{dD}{dt} = \frac{(-t + 1200)}{\sqrt{500^2 + (-t + 1200)^2}}
]
Step 4: Calculate at ( t = 900 ) seconds
The problem asks for the rate at which the people are moving apart 15 minutes (900 seconds) after the woman starts walking. Plug ( t = 900 ) into the expression:
[
\frac{dD}{dt} = \frac{(-900 + 1200)}{\sqrt{500^2 + (-900 + 1200)^2}}
]
[
\frac{dD}{dt} = \frac{300}{\sqrt{500^2 + 300^2}}
]
Now calculate the square root:
[
\frac{dD}{dt} = \frac{300}{\sqrt{250000 + 90000}} = \frac{300}{\sqrt{340000}} \approx \frac{300}{583.095}
]
[
\frac{dD}{dt} \approx 0.514 \, \text{ft/s}
]
Final Answer:
The rate at which the two people are moving apart 15 minutes after the woman starts walking is approximately 0.514 ft/s.