To solve this problem, we need to follow these steps:

1. Find the time $t$ when the speeds of the hare and the tortoise are the same for the first time after $t = 0$.

   The speed of the hare is given by $H(t) = -7t^2 + 43t$ and the speed of the tortoise is given by $T(t) = t$. Set $H(t) = T(t)$:

   $-7t^2 + 43t = t$

   Simplify this equation:

   $-7t^2 + 43t - t = 0$ $-7t^2 + 42t = 0$

   Factor out $t$:

   $t(-7t + 42) = 0$

   This gives two solutions:

   $t = 0 \quad \text{or} \quad t = 6$

   Since we are looking for the first time after $t = 0$, we choose $t = 6$.

2. Set up the integral to find the area between the curves from $t = 6$ to $t = 7$.

   The distance between the tortoise and the hare, or the area between the curves, is given by:

   $\text{Area} = \int_6^7 |H(t) - T(t)| \, dt$

   Since for this interval, the hare's speed $H(t)$ is greater than the tortoise's speed $T(t)$, we can write:

   $\text{Area} = \int_6^7 (H(t) - T(t)) \, dt$

   Substitute $H(t) = -7t^2 + 43t$ and $T(t) = t$:

   $\text{Area} = \int_6^7 (-7t^2 + 43t - t) \, dt$ $= \int_6^7 (-7t^2 + 42t) \, dt$

3. Evaluate the integral.

   Compute the integral:

   $\int_6^7 (-7t^2 + 42t) \, dt = \left[ -\frac{7t^3}{3} + 21t^2 \right]_6^7$

   Evaluate at $t = 7$ and $t = 6$:

   [ \left[ -\frac{7(7)^3}{3} + 21(7)^2 \right] - \left[ -\frac{7(6)^3}{3} + 21(6)^2 \right] ]

   Calculate each term:

   • At $t = 7$:

     $-\frac{7 \cdot 343}{3} + 21 \cdot 49 = -\frac{2401}{3} + 1029$ $= -800.33 + 1029 = 228.67$

   • At $t = 6$:

     $-\frac{7 \cdot 216}{3} + 21 \cdot 36 = -504 + 756 = 252$

   Subtract:

   $228.67 - 252 = -23.33$

   Since the area can't be negative, take the absolute value:

   $\text{Area} = 23.33 \text{ square miles}$

Therefore, the exact area between the curves from $t = 6$ to $t = 7$ is $\frac{70}{3}$ square miles. This value represents the distance between the tortoise and the hare over that time period.

Would you like further details or have any questions?

Here are 5 related questions:

1. How would the solution change if the speeds of the hare and tortoise were functions of $t^3$?
2. What is the significance of finding the area between two curves?
3. How can this problem be applied in real-world scenarios?
4. What happens if we extend the interval to $t = 8$?
5. How would the problem change if the hare's speed was $-5t^2 + 40t$ instead?

Tip: When solving problems involving areas between curves, always determine which curve is on top over the interval to ensure accurate calculation of the area.