Problem 6 - Active Calculus

Question

A bungee jumper dives from a tower at time \(t=0\). Her height \(h\) (measured in feet) at time \(t\) (in seconds) is given by the graph in Figure 1.1.4. In this problem, you may base your answers on estimates from the graph or use the fact that the jumper's height function is given by \(s(t)=100 \cos (0.75 t) \cdot e^{-0.2 t}+100\). a. What is the change in vertical position of the bungee jumper between \(t=0\) and \(t=15 ?\) b. Estimate the jumper's average velocity on each of the following time intervals: [0,15] , \([0,2],[1,6],\) and \([8,10] .\) Include units on your answers. c. On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why? d. Estimate the jumper's instantaneous velocity at \(t=5\). Show your work and explain your reasoning, and include units on your answer. e. Among the average and instantaneous velocities you computed in earlier questions, which are positive and which are negative? What does negative velocity indicate?

Step-by-Step Solution

Answer

a. 95.6 feet, b. -6.37, -10.86, -11.52, 2.37 feet/second, c. [8,10], d. Approx. -15 feet/second, e. Velocities in [0,2] and [1,6] are negative, indicating downward motion.

Step 1 - Define the change in vertical position

The vertical position change is the difference in height at two specific time points: initial time (t=0) and final time (t=15). Use the height function given: \(s(t) = 100 \, \text{cos}(0.75t) \, e^{-0.2t} + 100\). Evaluate this function at \(t=0\) and \(t=15\).

Step 2 - Evaluate height at \(t=0\)

Substitute \(t=0\) into the height function: \(s(0) = 100 \, \text{cos}(0) \, e^{0} + 100 = 100 \, (1) \, (1) + 100 = 200\) feet.

Step 3 - Evaluate height at \(t=15\)

Substitute \(t=15\) into the height function: \(s(15) = 100 \, \text{cos}(11.25) \, e^{-3} + 100\). Use a calculator to evaluate this expression: \(s(15) \approx 100 \, (0.885) \, (0.0498) + 100 \approx 104.4\) feet.

Step 4 - Calculate the change in vertical position

The change in vertical position between \(t=0\) and \(t=15\) is \(s(0) - s(15) = 200 - 104.4 = 95.6\) feet.

Step 5 - Define average velocity formula

The average velocity \(v_{avg}\) over a time interval \([t_1, t_2]\) is given by \(v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}\).

Step 6 - Compute average velocity on [0,15]

Use \(t_1 = 0\) and \(t_2 = 15\): \(v_{avg, 0-15} = \frac{s(15) - s(0)}{15 - 0} = \frac{104.4 - 200}{15} \approx -6.37\) feet/second.

Step 7 - Compute average velocity on [0,2]

Evaluate \(s(2)\) using the height function: \(s(2) \approx 100 \, \text{cos}(1.5) \, e^{-0.4} + 100 \approx 178.29\). Use \(t_1 = 0\) and \(t_2 = 2\): \(v_{avg, 0-2} = \frac{s(2) - s(0)}{2 - 0} = \frac{178.29 - 200}{2} \approx -10.86\) feet/second.

Step 8 - Compute average velocity on [1,6]

Evaluate \(s(1)\) and \(s(6)\). Using the height function: \(s(1) \approx 188.14\), \(s(6) \approx 130.53\). Use \(t_1 = 1\) and \(t_2 = 6\): \(v_{avg, 1-6} = \frac{s(6) - s(1)}{6 - 1} = \frac{130.53 - 188.14}{5} \approx -11.52\) feet/second.

Step 9 - Compute average velocity on [8,10]

Evaluate \(s(8)\) and \(s(10)\). Using the height function: \(s(8) \approx 136.04\), \(s(10) \approx 140.78\). Use \(t_1 = 8\) and \(t_2 = 10\): \(v_{avg, 8-10} = \frac{s(10) - s(8)}{10 - 8} = \frac{140.78 - 136.04}{2} \approx 2.37\) feet/second.

Step 10 - Identify interval of maximum average velocity

The greatest average velocity is on the interval [8,10] as it has the highest positive average velocity (2.37 feet/second).

Step 11 - Estimate instantaneous velocity at \(t=5\)

Instantaneous velocity can be estimated by the derivative: \(v(5) = s'(5)\). Using numerical differentiation or the tangent slope on a graph at \(t=5\), approximate: \(v(5) \approx -15\) feet/second.

Step 12 - Analyze velocity values

Positive velocities occur on [8,10]; negative velocities on [0,2] and [1,6]. Negative velocity indicates downward motion.

Key Concepts

Average Velocity

Average velocity represents the overall rate of change of position over a specific time interval. To calculate average velocity, we use the formula:

• \( v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \)

Here, \(s(t_1)\) and \(s(t_2)\) represent the positions at times \(t_1\) and \(t_2\) respectively. For the bungee jumper problem, this involves substituting specific values of time into the height function \(s(t)\), calculating the position at these times, and using these positions in the formula. For instance, to find the average velocity between \(t=0\) and \(t=15\), we substitute these into the function: \[ s(0) = 200 \text{ feet} \] and \[ s(15) \approx 104.4 \text{ feet} \]. The average velocity over this interval is then \( v_{\text{avg}} = \frac{104.4 - 200}{15 - 0} \approx -6.37 \text{ feet/second} \). Calculating average velocities over different intervals helps to understand how the jumper's speed changes over time.

Instantaneous Velocity

Instantaneous velocity is the rate of change of position at a specific moment in time. This concept can be understood as the velocity of the bungee jumper at a particular second. It is essentially the derivative of the position function with respect to time, \(s'(t)\). For the bungee jumper’s height function, to find the instantaneous velocity at, say \(t=5\), we need to calculate:

• \( v(5) = s'(5) \) by differentiating the height function \(s(t)\).

This can be estimated using numerical methods like drawing a tangent line to the graph at \(t=5\) and finding its slope, potentially yielding an approximation such as \(v(5) \approx -15 \text{ feet/second}\). This negative value indicates that the jumper is moving downward at that instant.

Vertical Position Change

The vertical position change refers to the difference in height from one time to another. For the bungee jumper problem, this is computed between \(t=0\) and \(t=15\). To determine this change, evaluate the height function at these times:

• \( s(0) = 200 \text{ feet}\), and
• \( s(15) \approx 104.4 \text{ feet}\).

The vertical position change is then the difference: \[ s(0) - s(15) = 200 - 104.4 = 95.6 \text{ feet}\]. This calculation shows that the jumper has fallen 95.6 feet between the time intervals \(t=0\) and \(t=15\). Understanding the vertical position change is crucial for analyzing the dynamics of the jumper's fall.

Velocity Calculation

Calculating both average and instantaneous velocities helps in understanding how the motion of the jumper changes over time. For average velocity, we apply the formula over different intervals as illustrated:

• \(v_{\text{avg, 0-2}} = \frac{s(2) - s(0)}{2 - 0} \approx -10.86 \text{ feet/second}\), indicating the fall speed over 2 seconds.
• \(v_{\text{avg, 1-6}} = \frac{s(6) - s(1)}{6 - 1} \approx -11.52 \text{ feet/second}\), depicting a quicker fall between \(t=1\) and \(t=6\).
• \(v_{\text{avg, 8-10}} = \frac{s(10) - s(8)}{10 - 8} \approx 2.37 \text{ feet/second}\), showing an upward motion.

Calculating the instantaneous velocity at specific points, like \(t=5\), where \(v(5) \approx -15 \text{ feet/second}\), gives insight into the speed at precise moments. Negative velocities indicate downward motion, while positive velocities point to upward movement.
The combination of these calculations provides a comprehensive understanding of the jumper's velocity profile throughout the fall.