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Let's break down this problem step by step to help you find the correct answer.

### Given Information:
- The baseball diamond is a square with sides of 90 feet.
- The player runs from first base to second base at 14 ft/s.
- We need to determine the rate at which the player's distance from home plate is increasing when the player is halfway between first and second base.

### Setting up the Problem:
Let’s set up a coordinate system:
- Home plate is at \((0, 0)\),
- First base is at \((90, 0)\),
- Second base is at \((90, 90)\),
- The player runs along the line from first base to second base.

Denote:
- \( x \) as the player's position along the path from first to second base. 
  - Since the player is halfway, \( x = \frac{90}{2} = 45 \text{ feet} \).
- \( y \) is the distance of the player from home plate (the quantity we are interested in).
- The rate of change of \( x \) is given by the speed of the player: \( \frac{dx}{dt} = 14 \, \text{ft/s} \).

### Step 1: Relationship Between \( y \) and \( x \)
We need to express \( y \), the distance from the player to home plate, in terms of \( x \). 

From the right triangle formed by the player's position, home plate, and the line between first and second base:
- One leg is 90 feet (the distance from home to first base),
- The other leg is \( x \) (the player’s distance along the line from first to second).

Using the Pythagorean theorem:
\[
y^2 = 90^2 + x^2
\]
\[
y^2 = 8100 + x^2
\]

### Step 2: Differentiate with Respect to Time
Now, we differentiate both sides with respect to time \( t \):
\[
2y \frac{dy}{dt} = 2x \frac{dx}{dt}
\]
Simplifying:
\[
y \frac{dy}{dt} = x \frac{dx}{dt}
\]

### Step 3: Solving for \( \frac{dy}{dt} \)
We want to solve for \( \frac{dy}{dt} \), which is the rate at which the player’s distance from home plate is increasing.

\[
\frac{dy}{dt} = \frac{x \frac{dx}{dt}}{y}
\]

### Step 4: Find \( y \) When \( x = 45 \)
At the halfway point, \( x = 45 \) feet. Using the equation for \( y \):
\[
y^2 = 8100 + 45^2
\]
\[
y^2 = 8100 + 2025 = 10125
\]
\[
y = \sqrt{10125} \approx 100.62 \text{ feet}
\]

### Step 5: Calculate \( \frac{dy}{dt} \)
Now, substitute the known values into the equation for \( \frac{dy}{dt} \):
\[
\frac{dy}{dt} = \frac{45 \times 14}{100.62} \approx 6.26 \, \text{ft/s}
\]

Thus, your answer of **6.26 ft/s** seems mathematically correct based on this setup. However, it appears to have been marked incorrect.

### Potential Issue:
The system might be expecting a more precise or slightly different answer based on rounding. The key steps are correct, so ensure that the problem doesn't expect further decimal places or different rounding rules.

Would you like any further clarification, or do you have any questions?

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Here are 5 related questions to deepen understanding:
1. How would the rate change if the player were 75% of the way to second base?
2. What would the rate of change be if the player ran slower, at 10 ft/s?
3. Can this problem be extended to three-dimensional motion?
4. What would happen if the diamond were a different size?
5. How does the Pythagorean theorem help in determining the rate of change?

**Tip**: Always keep track of significant figures in problems where precision is key.

At what rate is the player's distance from home base increasing if he is halfway between first to second base?

Learn how to calculate the rate of change of a player's distance from home plate using related rates and the Pythagorean theorem. This step-by-step solution explores the application of differentiation and geometry for a problem involving a baseball diamond.

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