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? station

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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
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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