Problem 8
Question
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?
Step-by-Step Solution
Verified Answer
a. Max rising at 5.5s (23.9 ft/s). b. \( h''(5) = 9.6 \text{ ft/s}^2 \). c. Measures acceleration, unit: \( \text{ft/s}^2 \). d. Concave down: 3.0-4.5s, velocity decreasing.
1Step 1: Understanding the Problem
You are given a table of heights at different times for a bungee jumper. The goal is to estimate the derivatives at specific times and infer the physical properties and behavior of the bungee jumper.
2Step 2: Estimating h'(4.5)
Use the central difference formula to estimate the derivative: \[ h'(4.5) = \frac{{h(5.0) - h(4.0)}}{{5.0 - 4.0}} = \frac{{69.8 - 55.5}}{{1.0}} = 14.3 \text{ feet/second} \]
3Step 3: Estimating h'(5)
Use the central difference formula: \[ h'(5) = \frac{{h(5.5) - h(4.5)}}{{5.5 - 4.5}} = \frac{{81.6 - 60.4}}{{1.0}} = 21.2 \text{ feet/second} \]
4Step 4: Estimating h'(5.5)
Again, use the central difference formula: \[ h'(5.5) = \frac{{h(6.0) - h(5.0)}}{{6.0 - 5.0}} = \frac{{93.7 - 69.8}}{{1.0}} = 23.9 \text{ feet/second} \]
5Step 5: Determining the Time of Maximum Rising
Compare the estimated derivatives: At 4.5s: 14.3, at 5.0s: 21.2, at 5.5s: 23.9. The bungee jumper is rising most rapidly at 5.5 seconds.
6Step 6: Estimating h''(5)
Use the change in the first derivatives: \[ h''(5) = \frac{{h'(5.5) - h'(4.5)}}{{5.5 - 4.5}} = \frac{{23.9 - 14.3}}{{1.0}} = 9.6 \text{ feet/second}^2 \]
7Step 7: Physical Property and Units
The value of \( h''(5) \) represents the acceleration of the bungee jumper. Its units are \( \text{feet/second}^2 \).
8Step 8: Determining Concavity and Velocity Behavior
Determine concavity from the data: The function \( h(t) \) is concave down where the second derivative is negative. This occurs approximately between 3.0 and 4.5 seconds. During these times, the velocity (first derivative) is decreasing.
Key Concepts
Central Difference FormulaFirst Derivative EstimationSecond Derivative InterpretationConcavity and Velocity
Central Difference Formula
To understand how the bungee jumper's height changes over time, we estimate the rate of change or the derivative of the height function, known as the first derivative. The central difference formula is an effective method for calculating this derivative using discrete data points.
It involves taking averages of slopes between neighboring points to approximate the derivative at a given point. The formula is:
\[ h'(t) = \frac{h(t + \triangle t) - h(t - \triangle t)}{2\triangle t} \]
In our exercise, we used a time step \triangle t = 0.5 seconds to compute the derivatives.
It involves taking averages of slopes between neighboring points to approximate the derivative at a given point. The formula is:
\[ h'(t) = \frac{h(t + \triangle t) - h(t - \triangle t)}{2\triangle t} \]
In our exercise, we used a time step \triangle t = 0.5 seconds to compute the derivatives.
First Derivative Estimation
Estimating the first derivative provides insights into the jumper's velocity or the rate of change in height. Here, we find the first derivative at specific times using the central difference formula.
For example, to estimate \( h'(4.5) \):
\[ h'(4.5) = \frac{h(5.0) - h(4.0)}{5.0 - 4.0} = \frac{69.8 - 55.5}{1.0} = 14.3 \text{ feet/second} \]
Similarly, we estimate:
For example, to estimate \( h'(4.5) \):
\[ h'(4.5) = \frac{h(5.0) - h(4.0)}{5.0 - 4.0} = \frac{69.8 - 55.5}{1.0} = 14.3 \text{ feet/second} \]
Similarly, we estimate:
- \( h'(5) \): 21.2 feet/second
- \( h'(5.5) \): 23.9 feet/second
Second Derivative Interpretation
The second derivative, denoted as \( h''(t) \), represents the acceleration of the bungee jumper. It tells us how the velocity is changing over time. To estimate \( h''(5) \), we calculated the change in the first derivatives.
This involves finding the difference between the first derivatives around our point of interest:
\[ h''(5) = \frac{h'(5.5) - h'(4.5)}{5.5 - 4.5} = \frac{23.9 - 14.3}{1.0} = 9.6 \text{ feet/second}^2 \]
This value indicates the bungee jumper's acceleration at 5 seconds, with units of feet per second squared.
This involves finding the difference between the first derivatives around our point of interest:
\[ h''(5) = \frac{h'(5.5) - h'(4.5)}{5.5 - 4.5} = \frac{23.9 - 14.3}{1.0} = 9.6 \text{ feet/second}^2 \]
This value indicates the bungee jumper's acceleration at 5 seconds, with units of feet per second squared.
Concavity and Velocity
Understanding concavity helps us determine the shape of the height function and how the velocity behaves. When the function \( h(t) \) is concave down (\( h''(t) < 0 \)), it means the jumper's velocity is decreasing.
To identify where the function is concave down, we examine the data and its second derivatives. In this exercise, the concavity shows up clearly between 3.0 and 4.5 seconds.
During these intervals, the bungee jumper's velocity decreases, suggesting a deceleration phase in the bungee jump.
Concavity intervals reflect important physical behaviors like the moments of rapid descent and slow-down, helping us grasp the dynamics of the jump.
To identify where the function is concave down, we examine the data and its second derivatives. In this exercise, the concavity shows up clearly between 3.0 and 4.5 seconds.
During these intervals, the bungee jumper's velocity decreases, suggesting a deceleration phase in the bungee jump.
Concavity intervals reflect important physical behaviors like the moments of rapid descent and slow-down, helping us grasp the dynamics of the jump.
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