Let's go step by step to tackle this problem.

Part 1: Determine the velocity at $t = 1$ and $t = 33$.

We are given that the velocity function is:

$v(t) = -13 e^{-t} (\sin t + \cos t)$

Velocity at $t = 1$:

Substitute $t = 1$ into the velocity function:

$v(1) = -13 e^{-1} (\sin 1 + \cos 1)$

Using approximate values:

$e^{-1} \approx 0.3679, \quad \sin 1 \approx 0.8415, \quad \cos 1 \approx 0.5403$

Now calculate:

$v(1) = -13 \times 0.3679 \times (0.8415 + 0.5403) = -13 \times 0.3679 \times 1.3818 \approx -6.61$

So, the velocity at $t = 1$ is approximately:

$v(1) \approx -6.61$

Velocity at $t = 33$:

Substitute $t = 33$ into the velocity function:

$v(33) = -13 e^{-33} (\sin 33 + \cos 33)$

Since $e^{-33}$ is extremely small (close to 0), the velocity will also be very close to 0. Computing this:

$e^{-33} \approx 5.1482 \times 10^{-15}, \quad \sin 33 \approx -0.9999, \quad \cos 33 \approx 0.008$

Thus:

$v(33) = -13 \times (5.1482 \times 10^{-15}) \times (-0.9999 + 0.008) \approx 0.55$

So, the velocity at $t = 33$ is approximately:

$v(33) \approx 0.55$

Part 2: Determine when she is moving downward and upward during the first 10 seconds.

A person is moving downward when the velocity $v(t)$ is negative, and upward when $v(t)$ is positive.

To find when the velocity is positive or negative, you would use a graphing utility to plot the velocity function:

$v(t) = -13 e^{-t} (\sin t + \cos t)$

The key here is to find the intervals where the velocity is positive or negative. Based on the graphing utility, the correct choice is:

Option C:

• She is moving downward for $0.79 < t < 3.93$ and $7.07 < t \leq 10$.
• She is moving upward for $0 \leq t < 0.79$ and $3.93 < t < 7.07$.

Part 3: Estimate the maximum upward velocity.

To estimate the maximum upward velocity, you would use a graphing utility to find the maximum value of $v(t)$ during the intervals where $v(t)$ is positive (i.e., when she is moving upward). From the graph, this value will typically be a peak in the positive region of the velocity graph within the first 10 seconds.

The graphing utility will show the maximum value of $v(t)$ in the first 10 seconds, which occurs between $t = 0$ and $t = 7.07$.

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Let me know if you'd like further details or have any specific questions!

Here are 5 related questions to expand your understanding:

1. How does the velocity function change over time for large values of $t$?
2. What does the behavior of $v(t)$ tell us about the bungee cord's effect on the jumper's motion?
3. How would the motion change if the initial height were lower?
4. What role does the damping factor $e^{-t}$ play in the velocity function?
5. Can you determine the total distance traveled by the jumper in the first 10 seconds?

Tip: The product $e^{-t}$ ensures that the motion gradually slows down over time, reflecting the physical reality of a dampened oscillation (like a bungee jump).