### Problem Description: A baseball player hits the ball and then runs down the first base line at 22 ft/s. The first baseman fields the ball and then runs toward first base along the second base line at 18 ft/s. **Task:** Determine how fast the distance between the two players is decreasing at a moment when the hitter is 16 ft from first base and the first baseman is 12 ft from first base. (Use decimal notation. Give your answer to one decimal place.) **Graph/Diagram Explanation:** The diagram depicts a baseball diamond with players running: - Player running along the first base line (hitter). - First baseman running along the second base line toward first base. The labels indicate: - \( x \) for the hitter's distance from home base. - \( y \) for the first baseman's distance from first base. - \( d \) for the distance between the two players. The current positions: - The hitter is 16 ft from first base. - The first baseman is 12 ft from first base. ### Formula: Using the Pythagorean theorem since the paths form a right triangle: \[ d^2 = x^2 + y^2 \] Differentiate with respect to time \( t \): \[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] Rearrange to solve for \( \frac{dd}{dt} \) which represents the rate of change of the distance between the players: \[ \frac{dd}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{d} \] Substitute the known values: - \( x = 16 \) ft - \( y = 12 \) ft - \( \frac{dx}{dt} = 22 \) ft/s - \( \frac{dy}{dt} = -18 \) ft/s (negative because the baseman is moving toward the first base) Calculate \( d \): \[ d = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \text{ ft} \] Substitute into the differentiated equation: \[ \frac{dd}{dt} = \frac{(16 \times 22) + (12 \

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Chapter1: Units, Trigonometry. And Vectors
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### Problem Description:

A baseball player hits the ball and then runs down the first base line at 22 ft/s. The first baseman fields the ball and then runs toward first base along the second base line at 18 ft/s.

**Task:**

Determine how fast the distance between the two players is decreasing at a moment when the hitter is 16 ft from first base and the first baseman is 12 ft from first base.

(Use decimal notation. Give your answer to one decimal place.)

**Graph/Diagram Explanation:**

The diagram depicts a baseball diamond with players running:

- Player running along the first base line (hitter).
- First baseman running along the second base line toward first base.

The labels indicate:
- \( x \) for the hitter's distance from home base.
- \( y \) for the first baseman's distance from first base.
- \( d \) for the distance between the two players.

The current positions:
- The hitter is 16 ft from first base.
- The first baseman is 12 ft from first base.

### Formula:

Using the Pythagorean theorem since the paths form a right triangle:
\[ d^2 = x^2 + y^2 \]

Differentiate with respect to time \( t \):
\[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]

Rearrange to solve for \( \frac{dd}{dt} \) which represents the rate of change of the distance between the players:
\[ \frac{dd}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{d} \]

Substitute the known values:
- \( x = 16 \) ft
- \( y = 12 \) ft
- \( \frac{dx}{dt} = 22 \) ft/s
- \( \frac{dy}{dt} = -18 \) ft/s (negative because the baseman is moving toward the first base)

Calculate \( d \):
\[ d = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \text{ ft} \]

Substitute into the differentiated equation:
\[ \frac{dd}{dt} = \frac{(16 \times 22) + (12 \
Transcribed Image Text:### Problem Description: A baseball player hits the ball and then runs down the first base line at 22 ft/s. The first baseman fields the ball and then runs toward first base along the second base line at 18 ft/s. **Task:** Determine how fast the distance between the two players is decreasing at a moment when the hitter is 16 ft from first base and the first baseman is 12 ft from first base. (Use decimal notation. Give your answer to one decimal place.) **Graph/Diagram Explanation:** The diagram depicts a baseball diamond with players running: - Player running along the first base line (hitter). - First baseman running along the second base line toward first base. The labels indicate: - \( x \) for the hitter's distance from home base. - \( y \) for the first baseman's distance from first base. - \( d \) for the distance between the two players. The current positions: - The hitter is 16 ft from first base. - The first baseman is 12 ft from first base. ### Formula: Using the Pythagorean theorem since the paths form a right triangle: \[ d^2 = x^2 + y^2 \] Differentiate with respect to time \( t \): \[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] Rearrange to solve for \( \frac{dd}{dt} \) which represents the rate of change of the distance between the players: \[ \frac{dd}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{d} \] Substitute the known values: - \( x = 16 \) ft - \( y = 12 \) ft - \( \frac{dx}{dt} = 22 \) ft/s - \( \frac{dy}{dt} = -18 \) ft/s (negative because the baseman is moving toward the first base) Calculate \( d \): \[ d = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \text{ ft} \] Substitute into the differentiated equation: \[ \frac{dd}{dt} = \frac{(16 \times 22) + (12 \
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