What is the instantaneous rate of change of the package after 2.5 days of travel?
Speed of iPhone During Travel
|
Days of Travel |
Speed (MPH) |
|
.5 |
23.00
|
|
1 |
0.00
|
|
1.5 |
137.00
|
|
2 |
540.00
|
|
2.5 |
52.00
|
|
3 |
0.00
|
|
3.5 |
600.00
|
|
4 |
50.00
|
|
4.5 |
35.00
|
Average Rate of Change
(2.5, 52) to (.5, 23)
(23 – 52) / (.5 – 2.5) = 14.5
(23 – 52) / (.5 – 2.5) = 14.5
(2.5, 52) to (4, 50)
(50 – 52) / (4 – 2.5) = -1.33
(50 – 52) / (4 – 2.5) = -1.33
(2.5, 52) to (.5, 23)
(137 – 52) / (1.5 – 2.5) = -85
I used the slope formula to find the average rate of change between the points on my graph. I used t=2.5 as a point in each formula. The average rate of change between these lines can become quite large depending on the point used. The numbers in my graph vary greatly and so do the average rates of change. Depending on which days I choose to compare, the ARC can be either positive or negative because the speed go up and down on a day to day and hour by hour basis. As the speeds get closer together, the ARC goes down, and the farther apart the speeds get, the higher the ARC goes.
Instantaneous Rate of Change (IRC)
The target line being at the points t=2.5 and t=3.5 tells me that those are the points I should use for my instantaneous rate of change formula. I plug those points and their values into the slope formula to find the instantaneous rate of change.
(2.5, 52) to (3.5, 600)
(600 – 52) / (3.5 – 2.5) = 548
(600 – 52) / (3.5 – 2.5) = 548
My instantaneous rate of change at the slope of t=2.5 is 548. What this means in the context of my graph is that at the point t=2.5, the slope of the line at t=2.5 has a rate of change of 548.
I know this works because at that point of my graph, the calculations around it are getting closer and closer together, until they reach t=2.5. In this example, the calculations between t=3 and t=2.5, and the calculations between t=2 and t=2.5 each contribute to finding the instantaneous rate of change at the slope of t=2.5.
Hi Max, I liked how you explained the steps and method of how you went about on you blog post. Although I hate UPS for losing my packages, your blog was a fun read! good job!
ReplyDeleteI thought the information and the calculations on the blog were explained well and they were organized and easy to follow. You chose an interesting topic also
ReplyDeletei like this, max! what an interesting and creative topic! your calculations look correct and your table is clear and well done. i couldn't view your graph, unfortunately.
ReplyDeleteyour explanation of the IRC is correct, you just need to include the units and explain what that IRC value means in terms of units, which according to what you have here, it means that the after 2.5 days the speed of the package is increasing at a rate of 548 mph per day? this value seems to large, especially considering your secant values were much smaller. so i wonder a bit about the location of your tangent line.
other than that, a good post.
prof little