Skip to main content
\( \newcommand{\lt}{ < } \newcommand{\gt}{ > } \newcommand{\amp}{ & } \)

Section5.3Rates of Change

Subsection5.3.1Rate of Change Definitions

Consider the point–slope equation of a line \begin{equation*} y-b = m(x-a), \end{equation*} which is the equation of a line with slope \(m\) and passing through a point \((a,b)\). If we solve for \(y\), then we can think of \(y\) as a function of \(x\) \begin{equation*} y = f(x) = m(x-a)+b. \end{equation*} In particular, we have \(f(a) = b\), and the original equation of the line is showing that the change in the output of the function \begin{equation*} \Delta f = f(x) - f(a) \end{equation*} is proportional to the change in the input \begin{equation*} \Delta x = x-a. \end{equation*} The slope is interpreted as the proportionality constant, the ratio of the changes, or more commonly the rate of change.

Only in the case of a linear function do we actually find that the ratio of changes \(\Delta f / \Delta x\) (ratio of the change in output to the change in input) is a constant. What do we do for other functions? There are two key concepts we can consider. We can look at the average rate of change over a particular interval or we can consider the instantaneous rate of change at a particular point.

Definition5.3.1

The average rate of change of a function \(f(x)\) over an interval \([a,b]\) is defined as the ratio \begin{equation*} \left. \frac{\Delta f}{\Delta x} \right|_{[a,b]} = \frac{f(b)-f(a)}{b-a}. \end{equation*}

This definition is really saying that you compute the slope of the line that joins the two points \((a,f(a))\) and \((b,f(b))\).

Example5.3.2

Find the average rate of change of \(f(x)=x^3-4x\) over the interval \([1,2]\).

Solution

A rate of change for a physical quantity has units corresponding to the ratio it describes. One of the most common examples involves calculating the average rate of change of position with respect to time as an average velocity. That is, the average velocity over an interval is the ratio of the change in position divided by the change in time.

Example5.3.3

If the position (height, in feet) of an object above the ground is defined by the function of time (in seconds) \begin{equation*} h(t) = 5+64t-16t^2, \end{equation*} then what is the average velocity over the interval \([1,4]\)?

Solution

Subsection5.3.2Instantaneous Rate of Change

The instantaneous rate of change of a function is the rate of change at a particular point, which corresponds to the slope of a tangent line. A tangent line is defined in terms of a single point such that the line is the line that best approximates the function near that point. Because only one point is involved, we can't use the ordinary notion of slope. Instead we consider what happens when the second point is sequentially chosen to be closer and closer to the point in question. This is the meaning behind the definition.

Definition5.3.4

The instantaneous rate of change of a function \(f(x)\) at \(x=a\) is defined as the limit of the average rate of change when the width of the interval is made arbitrarily small: \begin{equation*} \left. \frac{df}{dx} \right|_{a} = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \end{equation*} or \begin{equation*} \left. \frac{df}{dx} \right|_{a} = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \end{equation*}

The limit that appears in this statement will be made more formal later. For now, we are going to think of it as saying that we want to consider a sequence of intervals that include \(a\) and have a width that decreases to zero. We will start by considering an example.

Example5.3.5

Find the instantaneous rate of change of \(f(x) = x^3-4x\) at 1.

Solution

We can also visualize the calculation graphically. The figure below is a dynamic plot. The function \(f(x)=x^3-4x\) from the example is shown and the point at \((1,f(1))\) is labeled \(A\). There is a slider for the value of \(h\), the displacement of the second point \(B\) from \(x=1\). The value for \(h\) is allowed to be negative to allow for points on the left of 1. The average rate of change on each interval corresponds to the slope of the chord or secant line joining the two points. As \(h \to 0\), this slope approaches the value of the instantaneous rate of change.

In the previous example, the limit of the average rate of change was an easily recognized value because it was an integer. When the limit is not an integer, you need to look for a decimal value that is approximated. This often works better if we consider a sequence of intervals on both sides of the point of interest.

A slope does not depend on which point we think of as first or second since \begin{equation*} \frac{y_2-y_1}{x_2-x_1} = \frac{y_1-y_2}{x_1-x_2}. \end{equation*} So let us define the average rate of change between a point \(a\) and a second point a displacement \(h\) away from \(a\) instead of using the interval notation (where we have to worry about the ordering of the points). \begin{equation*}\left. \frac{\Delta f}{\Delta x} \right|_{a; h} = \frac{f(a+h)-f(a)}{h}.\end{equation*}

Example5.3.7

Find the instantaneous rate of change of \(f(x) = 3^x\) at 1.

Solution

A dynamic graph of this calculation is illustrated below.

Subsection5.3.3Using Algebra for Rates of Change

In the previous sections, we computed the instantaneous rate of change by looking for a limiting value in the average rate of change for a sequence of intervals that had decreasing width. It is important to realize that this is the fundamental definition of the instantaneous rate of change. In many cases involving basic algebraic functions, like polynomials, it is possible to determine the limiting value of this process algebraically. This is because we can find a formula for the average rate of change for any value of the spacing \(h\) and determine what will happen when \(h \to 0\).

The process for using this strategy is to consider the same calculations that we used to form the tables, but to determine the formula for each step instead of numerical values for the particular values of \(h\).

  • Identify the function \(f(x)\) and the point of interest \(a\).
  • Use composition to find the formula for \(f(a+h)\) and expand.
  • Compute the change in the function \(\Delta f = f(a+h) - f(a)\) and simplify.
  • Compute the average rate of change, \begin{equation*}\frac{\Delta f}{\Delta x} = \frac{f(a+h)-f(a)}{h},\end{equation*} and use factoring in order to find a formula that does not divide by \(h\). (You can do this step all at once after some practice.)
  • The limit or instantaneous rate of change is the value of this final formula when \(h \to 0\).
  • If it is not possible to rewrite without dividing by \(h\), then you can use the formula itself in a table to see what happens when \(h \to 0\) by testing a sequence of values for \(h\) that approach 0.

Can you see that these steps correspond to the columns in our most recent table-based computations?

Example5.3.10

Find the instantaneous rate of change for \(f(x)=x^2+3x\) at \(2\).

Solution
Example5.3.11

Find the instantaneous rate of change for \(f(x)=\frac{1}{3x+1}\) at \(1\).

Solution

The rate of change often has a physical interpretation. For example, if we know the position (e.g., height) as a function of time, then the rate of change corresponds to the velocity of the object. In economics, if we know how an equation relating the revenue that corresponds to the number of items being sold, then the rate of change of revenue with respect to the number of items is called the marginal revenue and corresponds to the amount of revenue change per extra item sold.

The units of a rate of change are determined by the units in the ratio. Since velocity is the rate of change of position with respect to time, the units of velocity are the units of length divided by the units of time, such as kilometers per hour or meters per second. Marginal revenue is the rate of change of revenue with respect to items sold, so the units would be a monetary unit per item, such as dollars per item.