· Part 1:
A. An example of a real world application of rate of change would be to examine the change in sea level of a scuba diver over a given period of time.
B. When scuba diving it is critical to track your change in depth as you descend further into the ocean over a certain period of time as to not move to quickly. If you do you will build up to much pressure in your brain and body which could be fatal. For this reason, this application has real world consequences. If I were to complete this experiment a question I would ask would be, how long does it take (in seconds) to descend 150 feet below sea level?
C.
Time in Seconds
|
0
|
5
|
10
|
15
|
20
|
25
|
30
|
Distance Below Sea Level in Feet
|
0
|
10
| 19 |
34
|
58
|
75
|
83
|
D.
Slope Calculations for three different secant lines using x=5.
=(52-35)/(20-15)=3.4 feet per second
=(85-52)/(25-20)=6.6 feet per second
=(85-35)/(25-15)=5.0 feet per second
I found through my calculations that by using the x value equal to 5 to calculate the average rate of change as the y values got bigger the output value of feet per second also increased. In terms of my experiment, this finding is important because it demonstrates that as the scuba diver descends further below sea level he will be moving at a faster rate.
F. The dotted linear line represents a tangent line that passes through Point P at t=15 which was used to calculate the slope in the previous problem.
G. The second point on this graph represented as Q will be used to calculate the IRC for PQ, P as defined in part F. Point P=(15, 30) and Point Q=(27, 60) where the x values represent time in seconds and the y values represent distance below sea level in feet.
Slope calculation= (60-30)/(27-15)=2.5
The slope equal to 2.5 is significant because it demonstrates the instantaneous rate of change to be equal to 2.5 for this given example. This can also be seen represented graphically.
H. In my experiment, I was attempting to identify the relationship for a scuba diver between descending further below sea level as it relates to time. I am able to know that the IRC was determined in Part G using point PQ. The tangent line passed through point P (15, 30) and Point Q (27, 60) by using these points it was possible to determine the instantaneous rate of change for this experiment. By using these points it was found that the line would increase at a rate of 2.5.
Excellent explanations and really cool application, I had trouble thinking through applications that were not just basic 'above ground' velocity. The only thing i would change, and i may be interpreting it incorrectly, is i would have made your secant lines straight. excellent work though!
ReplyDeleteMichael, first of all, excellent application topic! You did a great job explaining your experiment in detail and your visual aids are clear and we'll organized. Most of your calculations look great. However, there was a bit of confusion with some of the secant line approximarions. I guess I was confused as to whether you trying to find the IRC at x = 5, x = 10, or x = 15. If you were going g for x = 5 then your decant calculations should have been 34-10/15-5, 58-10/20-5, and 75-10/25-5, from the values in your table, resulting in 1.4, 3.2, and 3.25. From the values you got, you were still able to deduce that the rates of change are decreasing, but your tangent line value would have been closer to 1.4, if x = 5 was your target value. Also, if x =5 was your target value, then your tangent line should be going through the point x = 5. Let me know if you don't get the explanation I've given.
ReplyDeleteFor the values you got, your explanation does get the general idea of IRC across.
Professor little