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Chapter 1: Problem 6

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?

Short Answer

Expert verified
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 by step solution

01

- 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 \(t=15\).
02

- 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.
03

- 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.
04

- 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.
05

- Define average velocity formula

The average velocity \(v_{avg}\) over a time interval \([t_1, t_2]\) is given by \(v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}\).
06

- 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.
07

- 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.
08

- 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.
09

- 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.
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).
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.
12

- Analyze velocity values

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties. a. \(y=f(x)\) such that - \(f(-2)=2\) and \(\lim _{x \rightarrow-2} f(x)=1\) \- \(f(-1)=3\) and \(\lim _{x \rightarrow-1} f(x)=3\) \- \(f(1)\) is not defined and \(\lim _{x \rightarrow 1} f(x)=0\) \- \(f(2)=1\) and \(\lim _{x \rightarrow 2} f(x)\) does not exist. b. \(y=g(x)\) such that \- \(\text { - } g(-2)=3, g(-1)=-1, g(1)=-2, \text { and } g(2)=3\) \- At \(x=-2,-1,1\) and \(2, g\) has a limit, and its limit equals the value of the function at that point. \- \(g(0)\) is not defined and \(\lim _{x \rightarrow 0} g(x)\) does not exist.

A bungee jumper's height \(h\) (in feet) at time \(t\) (in seconds) is given in part by the table: $$ \begin{array}{llllllllllll} t & 0.0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 3.5 & 4.0 & 4.5 & 5.0 \\ \hline h(t) & 200 & 184.2 & 159.9 & 131.9 & 104.7 & 81.8 & 65.5 & 56.8 & 55.5 & 60.4 & 69.8 \\ t & 5.5 & 6.0 & 6.5 & 7.0 & 7.5 & 8.0 & 8.5 & 9.0 & 9.5 & 10.0 \\ \hline h(t) & 81.6 & 93.7 & 104.4 & 112.6 & 117.7 & 119.4 & 118.2 & 114.8 & 110.0 & 104.7 \end{array} $$ a. Use the given data to estimate \(h^{\prime}(4.5), h^{\prime}(5),\) and \(h^{\prime}(5.5) .\) At which of these times is the bungee jumper rising most rapidly? b. Use the given data and your work in (a) to estimate \(h^{\prime \prime}(5)\). c. What physical property of the bungee jumper does the value of \(h^{\prime \prime}(5)\) measure? What are its units? d. Based on the data, on what approximate time intervals is the function \(y=h(t)\) concave down? What is happening to the velocity of the bungee jumper on these time intervals?

The temperature, \(H\), in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by \(H=f(t),\) where \(t\) is in minutes since the coffee was put on the counter. (a) Is \(f^{\prime}(t)\) positive or negative? [Choose: positive | negative] (b) What are the units of \(f^{\prime}(25) ?\) \(\square\) Suppose that \(\left|f^{\prime}(25)\right|=0.6\) and \(f(25)=65 .\) Fill in the blanks (including units where needed) and select the appropriate terms to complete the following statement about the temperature of the coffee in this case. At \(\square\) minutes after the coffee was put on the counter, its [Choose: derivative | temperature | change in temperature] is \(\square\) and will [Choose: increase | decrease] by about \(\square\) in the next 60 seconds.

A diver leaps from a 3 meter springboard. His feet leave the board at time \(t=0\), he reaches his maximum height of \(4.5 \mathrm{~m}\) at \(t=1.1\) seconds, and enters the water at \(t=2.45\). Once in the water, the diver coasts to the bottom of the pool (depth \(3.5 \mathrm{~m}\) ), touches bottom at \(t=7\), rests for one second, and then pushes off the bottom. From there he coasts to the surface, and takes his first breath at \(t=13\). a. Let \(s(t)\) denote the function that gives the height of the diver's feet (in meters) above the water at time \(t\). (Note that the "height" of the bottom of the pool is -3.5 meters.) Sketch a carefully labeled graph of \(s(t)\) on the provided axes in Figure 1.1.5. Include scale and units on the vertical axis. Be as detailed as possible. b. Based on your graph in (a), what is the average velocity of the diver between \(t=2.45\) and \(t=7 ?\) Is his average velocity the same on every time interval within [2.45,7]\(?\) c. Let the function \(v(t)\) represent the instantaneous vertical velocity of the diver at time \(t\) (i.e. the speed at which the height function \(s(t)\) is changing; note that velocity in the upward direction is positive, while the velocity of a falling object is negative). Based on your understanding of the diver's behavior, as well as your graph of the position function, sketch a carefully labeled graph of \(v(t)\) on the axes provided in Figure \(1.1 .6 .\) Include scale and units on the vertical axis. Write several sentences that explain how you constructed your graph, discussing when you expect \(v(t)\) to be zero, positive, negative, relatively large, and relatively small. d. Is there a connection between the two graphs that you can describe? What can you say about the velocity graph when the height function is increasing? decreasing? Make as many observations as you can.

Evaluate the limit $$\lim _{x \rightarrow-6} \frac{x^{2}-36}{x+6}$$ If the limit does not exist enter DNE. Limit =\(\square\)
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