MHB Determine the Average Rate of Change

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SUMMARY

The average rate of change (AROC) of the function y = 2cos(x - π/3) + 1 over the interval π/3 ≤ x ≤ π/2 is calculated using the formula Δy/Δx. The correct computation yields an AROC of approximately -0.5117, contrasting with the erroneous value of 1.4 initially reported. The error was identified through graphing the function and analyzing the secant line, confirming that both the textbook and the initial calculation contained mistakes.

PREREQUISITES
  • Understanding of trigonometric functions, specifically cosine.
  • Familiarity with the concept of average rate of change in calculus.
  • Ability to perform basic calculus operations, including limits and derivatives.
  • Experience with graphing functions and interpreting secant lines.
NEXT STEPS
  • Study the properties of trigonometric functions, focusing on transformations.
  • Learn how to derive the average rate of change for different types of functions.
  • Explore graphing techniques for visualizing functions and their secant lines.
  • Investigate common errors in calculus calculations and how to avoid them.
USEFUL FOR

Students and educators in calculus, mathematicians analyzing trigonometric functions, and anyone seeking to improve their understanding of average rates of change in mathematical contexts.

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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)
 
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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
 

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