The tortoise versus the hare:

The speed of the hare is given by the function
H
(
t
)
=
−
7
t
2
+
43
t
, and the speed of the tortoise is given by
T
(
t
)
=
t
. For both equations,
t
is time measured in hours and the speed is measured in miles per hour. Find the distance between the tortoise and the hare, or the area between the curves, from the first time
t
>
0
when they are traveling at the same speed until one hour after. Enter an exact answer.

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.

Explore how to calculate the exact distance between a tortoise and a hare using integral calculus. This problem involves finding the area between two curves represented by their respective speed functions. Learn step-by-step with a detailed solution.

Math Problem Statement

The speed of the hare is given by the function H ( t )

− 7 t 2 + 43 t , and the speed of the tortoise is given by T ( t )

Solution

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