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?

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.