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Let's work through this problem step by step.

We have a baseball diamond, which is a square with sides of length 90 feet. The batter is running toward first base at a speed of 23 ft/s.

For part (a), we need to determine at what rate his distance from second base is changing when he is halfway to first base. 

### Step-by-Step Solution:

1. **Set up the problem:**
   - The baseball diamond is a square, and the distance from home plate to first base is 90 feet.
   - The batter's speed toward first base is given as \( \frac{dx}{dt} = 23 \, \text{ft/s} \), where \(x\) represents his distance to first base.
   - At halfway, \(x = 45\) feet.

2. **Distance from second base:**
   - The distance from second base to the batter can be found using the Pythagorean theorem. Since second base is directly across the diagonal from first base, the distance from second base to the batter is the hypotenuse of a right triangle where one leg is 90 feet (the distance from second to first base), and the other leg is \(x\), the distance from the batter to first base.

   Let \( d \) be the distance from the batter to second base. The Pythagorean theorem gives:
   \[
   d^2 = 90^2 + x^2
   \]
   Taking the derivative of both sides with respect to time \(t\), we get:
   \[
   2d \frac{dd}{dt} = 2x \frac{dx}{dt}
   \]
   Simplifying:
   \[
   d \frac{dd}{dt} = x \frac{dx}{dt}
   \]
   Solving for \( \frac{dd}{dt} \):
   \[
   \frac{dd}{dt} = \frac{x \frac{dx}{dt}}{d}
   \]

3. **Substitute values:**
   - At the moment the batter is halfway to first base, \(x = 45 \, \text{ft}\).
   - Using the Pythagorean theorem again to find \(d\) when \(x = 45\):
   \[
   d = \sqrt{90^2 + 45^2} = \sqrt{8100 + 2025} = \sqrt{10125} \approx 100.62 \, \text{ft}
   \]
   - Now, substitute \(x = 45 \, \text{ft}\), \( \frac{dx}{dt} = 23 \, \text{ft/s} \), and \(d \approx 100.62 \, \text{ft}\) into the equation for \( \frac{dd}{dt} \):
   \[
   \frac{dd}{dt} = \frac{45 \times 23}{100.62} \approx \frac{1035}{100.62} \approx 10.29 \, \text{ft/s}
   \]

Thus, the rate at which his distance from second base is changing is approximately **10.29 ft/s**.

### Part (b) follows a similar approach:
1. The distance from third base forms another right triangle where one leg is 90 feet (the distance from third base to first base), and the other leg is \(x = 45 \, \text{ft}\).

2. Using the Pythagorean theorem for this triangle:
   \[
   d_{\text{third base}}^2 = 90^2 + x^2
   \]
   Similarly, differentiate and solve using the same steps as in part (a).

Would you like to go through part (b) in detail? Let me know if you'd like further clarification or help with it! Here are some related questions:

1. How would the rate of change in distance from home plate differ if the batter was running faster?
2. What happens to the rates of change if the baseball diamond had different dimensions?
3. How does calculus help in analyzing the changing rates of motion in baseball?
4. Could the distances from first, second, and third base all be changing at the same rate at any point?
5. What if the batter were running directly toward second base? How would the rates of change compare?

**Tip:** Related rates problems often involve the Pythagorean theorem when distances change in triangular setups!

A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 23 ft/s. At what rate is his distance from second base changing when he is halfway to first base? At what rate is his distance from third base changing at the same moment?

Solve a related rates problem involving a baseball diamond. Calculate the rate at which a batter's distance from second base and third base is changing as he runs towards first base at 23 ft/s. This problem applies concepts of the Pythagorean theorem and differentiation, suitable for high school calculus students.

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