https://file.aiquickdraw.com/mathbot/file/172979661577906yf9mot.jpg

This problem involves related rates and the geometry of a square baseball diamond. Let’s break it down:

- The baseball diamond is a square with sides of 80 feet.
- The player is running from second base to third base with a speed of 20 ft/s.
- At the moment of interest, the player is 25 feet away from third base.

We are asked to find the rate at which the player's distance from home plate is changing at that instant.

### Step-by-Step Solution:

1. **Label the problem:**
   - Let \( x \) be the distance from the player to third base.
   - Let \( y \) be the distance from the player to home plate.
   - The player's speed is given as \( \frac{dx}{dt} = -20 \, \text{ft/s} \) (negative because the distance to third base is decreasing).

2. **Use the Pythagorean theorem:**
   Since the player is moving along the path from second to third base, and the diamond is a square, the player's distance from home plate (y) forms the hypotenuse of a right triangle. The sides are:
   - One leg is \( x \) (distance from third base),
   - The other leg is 80 feet (from third base to home plate along the base line).

   Thus, the Pythagorean theorem gives us the relationship:
   \[
   y^2 = x^2 + 80^2
   \]
   \[
   y^2 = x^2 + 6400
   \]

3. **Differentiate implicitly with respect to time \( t \):**
   \[
   2y \frac{dy}{dt} = 2x \frac{dx}{dt}
   \]
   Simplifying:
   \[
   y \frac{dy}{dt} = x \frac{dx}{dt}
   \]

4. **Substitute known values:**
   At the moment of interest, \( x = 25 \) feet, \( \frac{dx}{dt} = -20 \, \text{ft/s} \).
   
   First, find \( y \) when \( x = 25 \):
   \[
   y^2 = 25^2 + 6400 = 625 + 6400 = 7025
   \]
   So,
   \[
   y = \sqrt{7025} \approx 83.83 \, \text{ft}
   \]

5. **Solve for \( \frac{dy}{dt} \):**
   Now, using the relation \( y \frac{dy}{dt} = x \frac{dx}{dt} \):
   \[
   83.83 \frac{dy}{dt} = 25 \times (-20)
   \]
   \[
   83.83 \frac{dy}{dt} = -500
   \]
   \[
   \frac{dy}{dt} = \frac{-500}{83.83} \approx -5.96 \, \text{ft/s}
   \]

### Final Answer:
The player's distance from home plate is decreasing at approximately \( 5.96 \, \text{ft/s} \) at that instant.

Would you like a more detailed explanation on any part? 

Here are 5 related questions to practice:
1. How would the solution change if the player was running at a different speed?
2. What if the distance from third base was 30 ft? How would that affect \( \frac{dy}{dt} \)?
3. Can you set up a similar related rates problem for a different point on the baseball diamond?
4. How does the concept of implicit differentiation apply here?
5. What happens to the rate of change as the player approaches third base?

**Tip:** Always label each variable clearly in related rates problems. Understanding the geometry can help set up the correct relationships.

A baseball diamond is a square whose sides are 80 ft long. Suppose that a player running from second base to third base has a speed of 20 ft/s at the instant when he is 25 ft from third base. At what rate is the player’s distance from home plate changing at that instant?

Learn how to solve a related rates problem involving a baseball diamond, where a player runs from second to third base. The problem uses the Pythagorean theorem and implicit differentiation to calculate how the player's distance from home plate changes.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation