Two boats leave the same dock at the same time. One of the boats travels north at 45 mph, while the other travels west at 30 mph. How fast is their distance from each other changing after 2 hours have passed? Hint: Consider the distance each boat will travel after this time has passed.
Two boats leave the same dock at the same time. One of the boats travels north at 45 mph, while the other travels west at 30 mph. How fast is their distance from each other changing after 2 hours have passed? Hint: Consider the distance each boat will travel after this time has passed.
Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
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Two boats leave the same dock at the same time. One of the boats travels north at 45 mph, while the other travels west at 30 mph. How fast is their distance from each other changing after 2 hours have passed?
Hint: Consider the distance each boat will travel after this time has passed.
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![### Problem Statement
**Two boats leave the same dock at the same time. One of the boats travels north at 45 mph, while the other travels west at 30 mph. How fast is their distance from each other changing after 2 hours have passed?**
**Hint:**
Consider the distance each boat will travel after this time has passed.
### Explanation and Solution:
1. **Distances Traveled:**
- The boat traveling north at 45 mph will have traveled:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 45 \, \text{mph} \times 2 \, \text{hours} = 90 \, \text{miles}
\]
- The boat traveling west at 30 mph will have traveled:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 30 \, \text{mph} \times 2 \, \text{hours} = 60 \, \text{miles}
\]
2. **Using Pythagorean Theorem:**
To find the distance between the two boats after 2 hours, we can use the Pythagorean theorem since the paths form a right triangle.
\[
d = \sqrt{(90)^2 + (60)^2} = \sqrt{8100 + 3600} = \sqrt{11700} \approx 108.17 \, \text{miles}
\]
3. **Rate of Change of Distance:**
Let \( x \) be the position of boat traveling north and \( y \) be the position of boat traveling west. The distance between the boats at any time \( t \) can be represented as \( s \). Using related rates:
\[
s^2 = x^2 + y^2
\]
Differentiating with respect to time \( t \):
\[
2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplify:
\[
s \frac{ds}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]
Plugging in the values:
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Transcribed Image Text:### Problem Statement
**Two boats leave the same dock at the same time. One of the boats travels north at 45 mph, while the other travels west at 30 mph. How fast is their distance from each other changing after 2 hours have passed?**
**Hint:**
Consider the distance each boat will travel after this time has passed.
### Explanation and Solution:
1. **Distances Traveled:**
- The boat traveling north at 45 mph will have traveled:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 45 \, \text{mph} \times 2 \, \text{hours} = 90 \, \text{miles}
\]
- The boat traveling west at 30 mph will have traveled:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 30 \, \text{mph} \times 2 \, \text{hours} = 60 \, \text{miles}
\]
2. **Using Pythagorean Theorem:**
To find the distance between the two boats after 2 hours, we can use the Pythagorean theorem since the paths form a right triangle.
\[
d = \sqrt{(90)^2 + (60)^2} = \sqrt{8100 + 3600} = \sqrt{11700} \approx 108.17 \, \text{miles}
\]
3. **Rate of Change of Distance:**
Let \( x \) be the position of boat traveling north and \( y \) be the position of boat traveling west. The distance between the boats at any time \( t \) can be represented as \( s \). Using related rates:
\[
s^2 = x^2 + y^2
\]
Differentiating with respect to time \( t \):
\[
2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplify:
\[
s \frac{ds}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]
Plugging in the values:
\
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