Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Problem:** Train A leaves the station and travels east at 45 mph, while Train B leaves the same station and travels south at 60 mph. How fast are they separating one hour later?
**Diagram Explanation:**
The diagram illustrates a right triangle where:
- The horizontal line \( x \) represents the eastward distance traveled by Train A.
- The vertical line \( y \) represents the southward distance traveled by Train B.
- The hypotenuse \( z \) represents the distance between the two trains.
**Calculations:**
After one hour:
- Train A has traveled \( 45 \) miles east.
- Train B has traveled \( 60 \) miles south.
The distance between the trains, \( z \), can be calculated using the Pythagorean theorem:
\[
z = \sqrt{x^2 + y^2} = \sqrt{45^2 + 60^2} = \sqrt{2025 + 3600} = \sqrt{5625} = 75 \, \text{miles}
\]
**Rate of Separation:**
To find how fast they are separating, we calculate the derivative of \( z \) with respect to time. Using the Pythagorean relation:
\[
z^2 = x^2 + y^2
\]
Differentiating both sides with respect to time \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Substitute the known values:
- \( x = 45 \) miles, \( \frac{dx}{dt} = 45 \) mph
- \( y = 60 \) miles, \( \frac{dy}{dt} = 60 \) mph
- \( z = 75 \) miles
\[
2(75) \frac{dz}{dt} = 2(45)(45) + 2(60)(60)
\]
\[
150 \frac{dz}{dt} = 4050 + 7200
\]
\[
150 \frac{dz}{dt} = 11250
\]
\[
\frac{dz}{dt} = \frac{11250}{150} = 75 \, \text{mph}
\]
Therefore, the trains are separating at a rate of 75 mph after one](https://content.bartleby.com/qna-images/question/deff5ce5-c161-409f-a0ee-bdeaef8b0965/473610d2-575c-4630-a3c7-fe08b9ea50bb/53xtzyc_processed.jpeg)
Transcribed Image Text:**Problem:** Train A leaves the station and travels east at 45 mph, while Train B leaves the same station and travels south at 60 mph. How fast are they separating one hour later?
**Diagram Explanation:**
The diagram illustrates a right triangle where:
- The horizontal line \( x \) represents the eastward distance traveled by Train A.
- The vertical line \( y \) represents the southward distance traveled by Train B.
- The hypotenuse \( z \) represents the distance between the two trains.
**Calculations:**
After one hour:
- Train A has traveled \( 45 \) miles east.
- Train B has traveled \( 60 \) miles south.
The distance between the trains, \( z \), can be calculated using the Pythagorean theorem:
\[
z = \sqrt{x^2 + y^2} = \sqrt{45^2 + 60^2} = \sqrt{2025 + 3600} = \sqrt{5625} = 75 \, \text{miles}
\]
**Rate of Separation:**
To find how fast they are separating, we calculate the derivative of \( z \) with respect to time. Using the Pythagorean relation:
\[
z^2 = x^2 + y^2
\]
Differentiating both sides with respect to time \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Substitute the known values:
- \( x = 45 \) miles, \( \frac{dx}{dt} = 45 \) mph
- \( y = 60 \) miles, \( \frac{dy}{dt} = 60 \) mph
- \( z = 75 \) miles
\[
2(75) \frac{dz}{dt} = 2(45)(45) + 2(60)(60)
\]
\[
150 \frac{dz}{dt} = 4050 + 7200
\]
\[
150 \frac{dz}{dt} = 11250
\]
\[
\frac{dz}{dt} = \frac{11250}{150} = 75 \, \text{mph}
\]
Therefore, the trains are separating at a rate of 75 mph after one
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