4.2: Helicopter Bungee Jump
This activity is intended to supplement Calculus, Chapter 3, Lesson 4.
ID: 11760
Time Required: 40 minutes
Activity Overview
In this activity, students will observe a simulation of a record breaking bungee jump, consider a mathematical model of the height as a function of time, and take the derivative to determine points of interest like the minimum height, maximum velocity, acceleration, and maximum jerk. Students will algebraically, numerically, graphically and verbally investigate higher order derivatives.
Topic: Higher order derivatives
- Interpret the derivative in context of velocity, speed, and acceleration
- Corresponding characteristics of the graph of \begin{align*}f, f', f''\end{align*}
Teacher Preparation and Notes
- Students will need to know that the derivative of position is velocity, and the derivative of velocity is acceleration. For the questions at the end, students are expected to be familiar with the concept of the definite integral. The Product Rule using exponential and trigonometric functions is also utilized.
- The data for the helicopter bungee jump problem is from Bungee Consultants International (http://www.bungeeconsultants.com/stuntengineering-stuntcredits.htm).
Associated Materials
- Student Worksheet: Helicopter Bungee Jump http://www.ck12.org/flexr/chapter/9728, scroll down to the second activity.
Part 1 – Bungee Jump
This activity begins by describing the record breaking bungee jump. Students enter a parametric graph of the situation in their graphing calculators. If students have trouble entering the parametric equations, then direct them to change the calculator mode to parametric (press MODE and change Graph to 2:PARAMETRIC).
On the student worksheet, students use the Product Rule to take the first and second derivatives. The derivatives can be verified on the HOME screen, but it is such a long function that it wouldn’t fit on the screen well. To begin, some students may find it helpful for organizational purposes to use CAS to solve for \begin{align*}y'(t) = 0, y''(t) = 0, y'''(t) = 0, y^{IV}(t) = 0\end{align*}.
The questions are designed to develop student’s proficiency with CAS functionality and deepen understanding of the connection between \begin{align*}y, y', y'', y'''\end{align*}.
When students are asked to enter the velocity, acceleration, and jerk functions into \begin{align*}y2, y3\end{align*}, and \begin{align*}y4\end{align*}, they can use the HOME screen to find the derivatives of the position function and to store the functions, as shown to the right. Students can arrow up to an answer and press ENTER to copy an answer into the entry line.
For Question 6, many compare and contrast discoveries can be made. Question 7 points out the limitations of a mathematical model. When the jumper is in freefall for first 4 seconds or so, the only acceleration should be \begin{align*}\text{gravity} = 32 \ ft/s^2\end{align*}.
Solutions
- It is expected that students will say \begin{align*}``v = 0\text{''}\end{align*} or \begin{align*}``y = \frac{max}{min}\text{''}\end{align*}. Solving when the first derivative equals zero for a position-time graph gives the time when the velocity is zero. This will give the maximum or minimum position.
- \begin{align*}y''(t) =\end{align*} acceleration
- \begin{align*}y''(t) = 0\end{align*} when \begin{align*}t =5.5 \ s\end{align*}, 21.2 s, and 36.9 s
- \begin{align*}y^{\text{IV}}(t) = 0\end{align*} for \begin{align*}t > 15.7 \ s\end{align*} when \begin{align*}t =16.6 \ s\end{align*}
- \begin{align*}y''(t) = 0\end{align*} when \begin{align*}t = 5.5 \ s\end{align*}. The speed is 619.98 ft/s. (Speed is scalar and should not be negative.)
- When the height \begin{align*}y\end{align*} is a minimum, the velocity graph is zero. The velocity in ft/s graph is identical to the velocity in mph except for the scale. The \begin{align*}y, v, a\end{align*}, and \begin{align*}j\end{align*} graphs all have a similar damped shape.
- Maximum acceleration is about \begin{align*}2.5 \ g \left (\frac{80 \ ft/s^2}{32 \ ft/s^2} = 2.5\right )\end{align*}
- Point of infection of \begin{align*}y(x)\end{align*} is (21.218, 4684.477).
Part 2 – Graphically Examine Another Situation
To test for understanding of calculus concepts, questions relating a graph of the derivative to the original function are effective. Students do not need to know the Power Rule for integration to do the first question. They do need to understand the concept of the integral and its notation. If students know how to find the area of a triangle, they can find this integral.
The only question in this series that uses the function is Question 12—using calculus to find when \begin{align*}v\end{align*} is a maximum.
For further thought and discussion
- Verify that \begin{align*}v(x)\end{align*} is continuous.
- Prove that \begin{align*}v(x)\end{align*} is differentiable.
Students will relate concavity with the graphs in Question 13. Remind students that they are finding when the graph of \begin{align*}s\end{align*} is concave up, not \begin{align*}v\end{align*} as shown in the graph on the student worksheet. Explaining answers “using calculus” is important to stress. Often, students justify that a function is increasing by saying “it is going up.” They should avoid the use of “it.” For example, “the derivative is positive, so the function is increasing.”
Question 14 asks when is the function s decreasing and to explain. This occurs when \begin{align*}s'< 0\end{align*}. Since \begin{align*}s' = v\end{align*} and \begin{align*}v\end{align*} is negative between 0 and 1, that is the solution.
For further thought and discussion
- When is \begin{align*}s\end{align*} increasing?
- When is \begin{align*}s\end{align*} concave up?
- When is \begin{align*}s'\end{align*} concave down?
Solutions
- \begin{align*}s(1) = \frac{1}{2}(1)(-2) = -1\end{align*}
- \begin{align*}s'(1) = v(1) = 0\end{align*}
- \begin{align*}s''(1) = v'(1) = \text{slope} = 2\end{align*}
- Max \begin{align*}v\end{align*} occurs when \begin{align*}v' = 0\end{align*}. Max \begin{align*}v\end{align*} occurs when \begin{align*}v' = -2x + 10 = 0\end{align*}. So \begin{align*}x = 5\end{align*}.
- \begin{align*}s'' > 0\end{align*} when \begin{align*}v' > 0\end{align*}, i.e., when the slope of the graph is positive or when \begin{align*}0 < x < 5\end{align*}.
- \begin{align*}s' = v < 0\end{align*}, when \begin{align*}0 < x < 1\end{align*}
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Date Created:
Feb 23, 2012Last Modified:
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