Chapter 2, Problem 1RE

Explain the difference between average and instantaneous rates of change, and discuss how they are calculated.

To determine

The difference between the instantaneous and average rates of change.

Answer to Problem 1RE

The difference between the instantaneous and average rates of change is that the instantaneous rate of change is calculated at a point whereas the average rate of change is calculated over an interval.

Explanation of Solution

The instantaneous rate of change is calculated at a point.

Consider a function, $y=g\left(x\right)$ , and a point, $x_0$ .

Let instantaneous rate of change be denoted as $r_{ins}$ .

The instantaneous rate of change of $y$ with respect to $x$ at $x_0$ is determined by the formula shown below.

$r_{ins}=\underset{x\to x_0}{\lim }\frac{g\left(x\right)-g\left(x_0\right)}{x-x_0}$

The average rate of change is calculated over an interval.

Consider a function, $y=g\left(x\right)$ , and an interval, $\left[x_0,x_1\right]$ .

Let the average rate of change be denoted as $r_{avg}$ .

The average rate of change of $y$ with respect to $x$ over the interval, $\left[x_0,x_1\right]$ , is calculated by the formula given below.

$r_{avg}=\frac{g\left(x_1\right)-g\left(x_0\right)}{x_1-x_0}$

Example: Consider a function, $g\left(x\right)=4x^2$ , over an interval, $\left[0,5\right]$ , which represents the population of amoeba at a time of $x$ minutes.

The aim is to find the instantaneous rate of change in the population of amoeba at $x=1$ minutes as well as the average rate of change in the same over an interval, $\left[0,5\right]$ .

Use the formulae mentioned above for the instantaneous and average rates of changes.

$\begin{array}{c}{r}_{ins}=\underset{x\to 1}{\lim }\frac{g\left(x\right)-g\left(1\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{4x^2-4}{x-1}\\ =\underset{x\to 1}{\lim }\frac{4\left(x^2-1\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{4\bcancel{\left(x-1\right)}\left(x+1\right)}{\bcancel{\left(x-1\right)}}\end{array}$

Further,

$\begin{array}{c}{r}_{ins}=\underset{x\to 1}{\lim }4\left(x+1\right)\\ =4\left(1+1\right)=8\end{array}$

Now,

$\begin{array}{c}{r}_{avg}=\frac{g\left(5\right)-g\left(0\right)}{5-0}\\ =\frac{4{\left(5\right)}^2-0}{5}\\ =4\left(5\right)=20\end{array}$

Therefore, the instantaneous rate of change in the population of amoeba at $x=1$ minutes is 8 while the average rate of change over an interval, $\left[0,5\right]$ , is 20.

Geometrically, the instantaneous rate of change of y with respect to $x$ at $x_0$ is represented by a slope of a tangent line at point $\left(x_0,g\left(x_0\right)\right)$ .

The average rate of change of y with respect to $x$ over the interval, $\left[x_0,x_1\right]$ , is represented by a slope of the secant line through two points, $\left(x_0,g\left(x_0\right)\right)$ and $\left(x_1,g\left(x_1\right)\right)$ .

Therefore, the instantaneous rate of change is calculated at a point whereas the average rate of change is calculated over an interval.