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To determine the value of \( t \) that minimizes the distance \( d(t) = \sqrt{13 - 104t + 260t^2} \), we can simplify the problem by focusing on minimizing the expression inside the square root, as the square root function is monotonic (increasing).

Let:
\[
f(t) = 13 - 104t + 260t^2
\]
We need to find the value of \( t \) that minimizes \( f(t) \), as this will also minimize \( d(t) \).

### Steps to Solve

1. **Find the derivative of \( f(t) \):**
   \[
   f'(t) = -104 + 520t
   \]

2. **Set the derivative equal to zero to find the critical points:**
   \[
   -104 + 520t = 0
   \]
   \[
   520t = 104
   \]
   \[
   t = \frac{104}{520} = \frac{1}{5} = 0.2
   \]

3. **Verify that this is a minimum by checking the second derivative or by analyzing the behavior of \( f(t) \):**

   The second derivative of \( f(t) \) is:
   \[
   f''(t) = 520
   \]
   Since \( f''(t) = 520 \) is positive, this indicates that \( f(t) \) is concave up, confirming that \( t = 0.2 \) is a minimum.

### Conclusion
The value of \( t \) that minimizes the distance between the two boats is:
\[
t = 0.2 \text{ hours after noon, or } 12:12 \text{ PM.}
\]

Would you like further details on this solution?

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### Related Questions:

1. How can we confirm the minimum distance without using derivatives?
2. What would the distance \( d(t) \) be at \( t = 0.2 \) hours?
3. How does changing coefficients in \( f(t) \) affect the time \( t \) of minimum distance?
4. What if the function had a cubic term—how would that change the approach?
5. How do we interpret the physical meaning of a minimum distance in terms of the boat paths?

### Tip:
For any minimum or maximum problem, start by finding the derivative and set it to zero. This gives critical points, which can then be analyzed further.

What is the value of t such that the distance between the two boats is a minimum?

Solve for the time after noon that minimizes the distance between two boats using calculus techniques. This problem involves quadratic functions, optimization through derivatives, and the second derivative test. Suitable for Grades 11-12.

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