The concept of the Average Rate of Change (ARC) is foundational in calculus, enabling us to understand how one quantity changes in relation to another over a given interval. This calculation divides the change in one variable by the change in another, offering a quantifiable measure of change between two points. Through calculus, the ARC concept evolves, especially as we approach intervals where the change in the independent variable tends toward zero, presenting challenges in traditional calculations.

Understanding Average Rate of Change

Definition: The ARC between two points on a function $y = f(x)$ is defined as:

$ARC = \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

where $\Delta y$ is the change in the function's value, and $\Delta x$ is the change in the independent variable's value.

Importance: This concept is crucial for understanding how the output of a function changes relative to changes in the input, providing a basis for more complex calculus concepts, including the derivative.

Transitioning to Calculus

• Limits and the ARC: As the interval $\Delta x$ approaches zero, we encounter the concept of a limit. Calculus allows us to explore what happens to the ARC as $x_2$ approaches $x_1$, moving towards an instantaneous rate of change.
  • Example of applying limits to ARC:

$\lim_{\Delta x \rightarrow 0} \frac{f(x_1 + \Delta x) - f(x_1)}{\Delta x}$

Calculating Average Rate of Change: Examples

Example 1: Linear Function

Consider the function $f(x) = 2x + 3$ over the interval [1, 4].

• Step 1: Identify $x_1$ and $x_2$, $x_1 = 1$, $x_2 = 4$.
• Step 2: Calculate $f(x_1)$ and $f(x_2)$, $f(1) = 5$, $f(4) = 11$.
• Step 3: Compute ARC,

$\begin{gathered} &&&&& ARC = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \\ &&&&& = \frac{11 - 5}{4 - 1} \\ &&&&& = \frac{6}{3} \\ &&&&& = 2 \end{gathered}$<p></p><ul><li><strong>Result</strong>: The ARC of$f(x)$over [1, 4] is 2.</li></ul><p></p><ul><li><strong>Graphical Representation</strong></li></ul><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/e7afb681-f43c-448e-92c5-da423c8af9f8-file.png" alt="Linear Function Graph" style="width: 500px; height: 386px" width="500" height="386"><p></p><h3><strong>Example 2: Quadratic Function</strong></h3><p>Consider$f(x) = x^2$over the interval [2, 5].</p><ul><li><strong>Step 1</strong>: Identify points,$x_1 = 2$,$x_2 = 5$.</li><li><strong>Step 2</strong>: Evaluate$f(x_1)$and$f(x_2)$,$f(2) = 4$,$f(5) = 25$.</li><li><strong>Step 3</strong>: Calculate ARC, </li></ul><p></p>$\begin{gathered} &&&&& ARC = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \\ &&&&& = \frac{25 - 4}{5 - 2} \\ &&&&& = \frac{21}{3} \\ &&&&& = 7 \end{gathered} $<p></p><ul><li><strong>Result</strong>: The ARC of$f(x)$over [2, 5] is 7.</li></ul><p></p><ul><li><strong>Graphical Representation:</strong></li></ul><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/585fefb1-8aea-41c6-a24c-0f74789a5f0e-file.png" alt="Quadratic Function Graph" style="width: 500px; height: 398px" width="500" height="398"><p><strong><br></strong></p><h3><strong>Example 3: Using Limits to Explore ARC's Approach to Zero</strong></h3><p>Consider$f(x) = x^2$as$\Delta x$approaches 0 from [3,$3 + \Delta x$].</p><p></p><ul><li><strong>Step 1</strong>: Express ARC as a limit, </li></ul><p></p>$\begin{aligned} && & \lim{\Delta x \rightarrow 0} \frac{f(3 + \Delta x) - f(3)}{\Delta x} \\ && = & \lim{\Delta x \rightarrow 0} \frac{(3 + \Delta x)^2 - 9}{\Delta x} \\ && = & \lim{\Delta x \rightarrow 0} \frac{9 + 6\Delta x + \Delta x^2 - 9}{\Delta x} \\ && = & \lim{\Delta x \rightarrow 0} \frac{6\Delta x + \Delta x^2}{\Delta x} \\ && = & \lim_{\Delta x \rightarrow 0} 6 + \Delta x \\ && = & 6 \end{aligned}$<ul><li><strong>Result</strong>: As$\Delta x$approaches 0, the Average Rate of Change of$f(x) = x^2$at$x = 3$approaches 6. This reflects the instantaneous rate of change, or derivative, of$f(x)$at$x = 3$.</li></ul><p></p><ul><li><strong>Graphical Representation:</strong></li></ul><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/059dfc4b-4797-4326-a37f-ad0162249e41-file.png" alt="Using Limits to Explore ARC's Approach to Zero" style="width: 500px; height: 396px" width="500" height="396"><p></p><h3><strong>Example 4: Real-World Application</strong></h3><p>Consider a vehicle's distance traveled, described by$f(t) = 4t^2 + 2t$where$f(t)$is distance in meters and (t) is time in seconds, over the interval [1, 4] seconds.</p><p></p><ul><li><strong>Step 1</strong>: Identify$t_1 = 1$,$t_2 = 4$.</li><li><strong>Step 2</strong>: Calculate$f(t_1)$and$f(t_2)$,$f(1) = 6$,$f(4) = 72$.</li><li><strong>Step 3</strong>: Compute ARC, </li></ul><p></p>$ \begin{aligned} && ARC = \frac{f(t_2) - f(t_1)}{t_2 - t_1} \\ && = \frac{72 - 6}{4 - 1} \\ && = \frac{66}{3} \\ && = 22 \end{aligned}$<p></p><ul><li><strong>Result</strong>: The vehicle's average speed over the interval [1, 4] seconds is 22 meters per second.</li></ul><p></p><h3><strong>Example 5: Changing Rates</strong></h3><p>For a balloon being inflated, the volume$V$in cubic centimeters is given by$V(t) = \frac{4}{3}\pi(6t)^3$where$t$is time in minutes. Calculate the ARC of volume change between 2 and 5 minutes.</p><ul><li><strong>Step 1</strong>: Identify$t_1 = 2$,$t_2 = 5$.</li><li><strong>Step 2</strong>: Calculate$V(t_1)$and$V(t_2)$, </li></ul><p></p>$\begin{aligned} &&&&& V(2) & = \frac{4}{3}\pi(6 \cdot 2)^3, \\ &&&&& V(5) & = \frac{4}{3}\pi(6 \cdot 5)^3. \end{aligned}$<p></p><ul><li><strong>Step 3</strong>: Compute ARC, </li></ul><p></p>$\begin{aligned} ARC & = \frac{V(t_2) - V(t_1)}{t_2 - t_1} \\ & = \frac{\frac{4}{3}\pi(6 \cdot 5)^3 - \frac{4}{3}\pi(6 \cdot 2)^3}{5 - 2} \\ & = \frac{4}{3}\pi \left( \frac{(6 \cdot 5)^3 - (6 \cdot 2)^3}{3} \right) \\ & = 468\pi \end{aligned}$<p></p><ul><li><strong>Result</strong>: The average rate of volume change between 2 and 5 minutes is$468\pi$cubic centimeters per minute, illustrating how volumes of three-dimensional shapes can change over time.</li></ul><h2 id="practice-questions"><strong>Practice Questions</strong></h2><h3><strong>Question 1</strong></h3><p>Given the function$f(x) = 3x^2 - 2x + 5$, calculate the average rate of change from$x = 1$to$x = 4$.</p><h3><strong>Question 2</strong></h3><p>A spherical balloon is being inflated so that its volume at any time$t$in seconds is given by$V(t) = \frac{4}{3}\pi t^3$. Calculate the average rate of change of the balloon's volume from$t = 2$seconds to$t = 5$seconds.</p><h3><strong>Question 3</strong></h3><p>For the function$g(x) = \ln(x)$, compute the average rate of change between$x = 1$and$x = e$, where$e$is the base of natural logarithms.</p><p></p><h2 id="solutions-to-practice-questions"><strong>Solutions to Practice Questions</strong></h2><h3><strong>Solution to Question 1</strong></h3><p><strong>1. Identify the function and points</strong>:$f(x) = 3x^2 - 2x + 5, x_1 = 1, x_2 = 4$.</p><p><strong>2. Calculate </strong>$f(x_1)$<strong> and </strong>$f(x_2)$: </p><p></p>$f(1) = 3(1)^2 - 2(1) + 5 = 6, \\ f(4) = 3(4)^2 - 2(4) + 5 = 41.$<p></p><p><strong>3. Compute the average rate of change</strong>: </p><p></p>$\begin{aligned} ARC &= \frac{f(x_2) - f(x_1)}{x_2 - x_1} \\ &= \frac{41 - 6}{4 - 1} \\ &= \frac{35}{3} \\ &= 11\frac{2}{3}. \end{aligned}$<p></p><p><strong>4. Result</strong>: The average rate of change from$x = 1$to$x = 4$is$11\frac{2}{3}$.</p><p></p><h3><strong>Solution to Question 2</strong></h3><p><strong>1. Given volume function</strong>:$V(t) = \frac{4}{3}\pi t^3$, with$t_1 = 2$and$t_2 = 5$.</p><p><strong>2. Find </strong>$V(t_1)$<strong> and </strong>$V(t_2)$: </p><p></p>$V(2) = \frac{4}{3}\pi (2)^3 = \frac{32}{3}\pi, \\ V(5) = \frac{4}{3}\pi (5)^3 = \frac{500}{3}\pi.$<p></p><p><strong>3. Calculate ARC</strong>: </p><p></p>$\begin{aligned} ARC &= \frac{V(t_2) - V(t_1)}{t_2 - t_1} \\ &= \frac{\frac{500}{3}\pi - \frac{32}{3}\pi}{5 - 2} \\ &= \frac{468\pi}{3} \\ &= 156\pi. \end{aligned}$<p></p><p><strong>4. Result</strong>: The average rate of change of the volume from$t = 2$to$t = 5$seconds is$156\pi$cubic units per second.</p><p></p><h3><strong>Solution to Question 3</strong></h3><p><strong>1. Identify the function and interval</strong>:$g(x) = \ln(x)$, from$x = 1$to$x = e$.</p><p><strong>2. Calculate </strong>$g(x_1)$<strong> and </strong>$g(x_2)$: </p><p></p>$g(1) = \ln(1) = 0, \\ g(e) = \ln(e) = 1.$<p></p><p><strong>3. Compute the ARC</strong>: </p><p></p>$\begin{aligned} ARC &= \frac{g(x_2) - g(x_1)}{x_2 - x_1} \\ &= \frac{1 - 0}{e - 1} \\ &= \frac{1}{e - 1}. \end{aligned}$<p></p><p><strong>4. Result</strong>: The average rate of change of$g(x)$between$x = 1$and$x = e$is$\frac{1}{e - 1}$.