Determine the Average Rate of Change

Click For Summary

Discussion Overview

The discussion revolves around determining the average rate of change (AROC) of the function y = 2cos(x - $\pi$/3) + 1 over the interval from $\pi$/3 to $\pi$/2. Participants are exploring the calculations involved in finding the AROC and addressing discrepancies between their results and those presented in a textbook.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant attempts to calculate the AROC but reports an incorrect result of 1.4.
  • Another participant provides a detailed calculation of the AROC, arriving at approximately 1.39811405542801, which matches the first participant's calculation.
  • A third participant notes that the textbook provides a different value of -0.5157, suggesting that the textbook may be incorrect.
  • A later reply identifies an error in the previous calculations, correcting the AROC to approximately -0.511745261674736 after graphing the function and the secant line.

Areas of Agreement / Disagreement

Participants do not reach consensus, as there are competing calculations and interpretations of the average rate of change, with some results differing significantly from others.

Contextual Notes

Participants' calculations depend on the accuracy of their evaluations of the cosine function and the interpretation of the interval endpoints. There are unresolved discrepancies between participant results and the textbook value.

eleventhxhour
Messages
73
Reaction score
0
Determine the average rate if change of the function y = 2cos(x - $\pi$/3) + 1 for the interval $\pi$/3 $\le$ x $\le$ $\pi$/2

I tried finding the exact values of the two (0 and 0.5) and subbing them into the AROC equation but I keep getting the wrong answer (1.4)
 
Physics news on Phys.org
We find:

$$\frac{\Delta y}{\Delta x}=\frac{y\left(\dfrac{\pi}{2}\right)-y\left(\dfrac{\pi}{3}\right)}{\dfrac{\pi}{2}-\dfrac{\pi}{3}}=\frac{\left(2\cos\left(\dfrac{\pi}{6}\right)+1\right)-\left(2\cos\left(0\right)+1\right)}{\dfrac{\pi}{6}}=\frac{6\left(\sqrt{3}-1\right)}{\pi}\approx1.39811405542801$$
 
MarkFL said:
We find:

$$\frac{\Delta y}{\Delta x}=\frac{y\left(\dfrac{\pi}{2}\right)-y\left(\dfrac{\pi}{3}\right)}{\dfrac{\pi}{2}-\dfrac{\pi}{3}}=\frac{\left(2\cos\left(\dfrac{\pi}{6}\right)+1\right)-\left(2\cos\left(0\right)+1\right)}{\dfrac{\pi}{6}}=\frac{6\left(\sqrt{3}-1\right)}{\pi}\approx1.39811405542801$$

Okay, that's what I got. The textbook has it as -0.5157 so I guess it's just wrong?
 
We both made an error...it should be:

$$\frac{\Delta y}{\Delta x}=\frac{6\left(\sqrt{3}-2\right)}{\pi}\approx-0.511745261674736$$

I discovered my error when graphing the function and the resulting secant line:

View attachment 3547
 

Attachments

  • aroc.png
    aroc.png
    13.9 KB · Views: 111

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K