Limits and the motion of objects ii
- 2. In part one of Limits and the Motion of Objects you plotted data that you were given and used limits to approximate the velocity of the object at a specific point by finding the slope of the graph. In part II of the project you will use your calculator to find a function that approximates the data. You will then use that function and the analytic capabilities of your calculator to find the velocity of the object at a specific point by finding the slope of the function
- 3. So far, if you connected the plotted points, you should have a graph that looks something like this,
- 4. But the plotted points you were given were incomplete. Adding additional points gives you the graph,
- 5. Clearly this is a fifth degree or quintic function. This is unfortunate because although your calculator has quadratic, cubic and quartic regression capabilities, it does not have quintic regression capabilities.
- 6. A quintic function is of the form, 𝑓 𝑥 = 𝑎5 𝑥5 + 𝑎4 𝑥4 + 𝑎3 𝑥3 + 𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 To find this particular function we need to find, 𝑎5, 𝑎4, 𝑎3, 𝑎2, 𝑎1, and 𝑎0 Six unknowns will require six equations and six equation will require six points on the line. But which six points should be chosen?
- 7. The point you want to chose would be the critical points. In calculus we will rigorously define critical points, but for now we will define them as the points of extrema and the points of inflection. For the graph drawn we have extrema at (-1.4, 161), (0.9, 0.0), (2.4, 34.1), and (4.0, -32.7). For the graph drawn we have points of inflections at (0.0, 45.5), (1.6, 15.7), and (3.2, 4.1). This give us 7 points, one more then we need so solve the system.
- 8. I will chose to disregard the maximum at 𝑥 = 2.4 mainly because the two adjacent points of inflection guarantee the maximum will exist in the solution to the system of equations. So the system becomes, 𝑎5 −1.4 5 + 𝑎4 −1.4 4 + 𝑎3 −1.4 3 + 𝑎2 −1.4 2 + 𝑎1 −1.4 + 𝑎0 = 161 𝑎5 0.0 5 + 𝑎4 0.0 4 + 𝑎3 0.0 3 + 𝑎2 0.0 2 + 𝑎1 0.0 + 𝑎0 = 45.5 𝑎5 0.9 5 + 𝑎4 0.9 4 + 𝑎3 0.9 3 + 𝑎2 0.9 2 + 𝑎1 0.9 + 𝑎0 = 0.0 𝑎5 1.6 5 + 𝑎4 1.6 4 + 𝑎3 1.6 3 + 𝑎2 1.6 2 + 𝑎1 1.6 + 𝑎0 = 15.7 𝑎5 3.2 5 + 𝑎4 3.2 4 + 𝑎3 3.2 3 + 𝑎2 3.2 2 + 𝑎1 3.2 + 𝑎0 = 4.1 𝑎5 4.0 5 + 𝑎4 4.0 4 + 𝑎3 4.0 3 + 𝑎2 4.0 2 + 𝑎1 4.0 + 𝑎0 = −32.7