# Explainer: Average and Instantaneous Rates of Change

In this explainer, we will learn how to find the average rate of change of a function between two -values and use limits to find the instantaneous rate of change.

Rates of change are an important concept which is central to many areas of mathematics, physics, chemistry, and other scientific disciplines. In fact, it is not just limited to science; in everyday life, we often deal with concepts such as speed, acceleration and interest, which, at their core, are rates of change. Sometimes we are interested finding rates of change, or predicting the future based on the rate of change in the present.

When we say that a car is traveling at 60 miles per hour, what do we really mean? Can we only define such a concept if we travel for an hour? Of course, we could say we travel one mile a minute. However, if we are traveling for less than a minute at this speed, is it still reasonable to talk about traveling at a mile a minute? Certainly, if no time passes, we do not travel any distance. However, we are still happy talking about traveling at 60 miles per hour even if for an instance. When we say we are traveling at a particular speed at any given instance, we are talking about an instantaneous rate of change. In this explainer, we will look at how we can define this mathematically and how we can calculate it.

Let us first recap the definition of the average rate of change.

### Definition: Average Rate of Change

The average rate of change of a function over an interval is defined by

Often, we consider the average rate of change over the interval as a function of :

Notice that if , the average rate of change is over the interval .

We will look at one example where we consider the average rate of change from a given point

### Example 1: Average Rates of Change

For the function , list the average rates of change of over the interval , where , evaluated to 4 decimal places at most.

We begin by finding an expression for the average rate of change of . Recall that the average rate of change of a function over the interval is given by

Using the function we have been given, we can write this as

Expanding the parentheses, we have

Given that is not equal to zero, we can cancel the common factor of from the numerator and denominator, which yields

Therefore, the average rate of change at is

 𝐴(ℎ)=4+3ℎ. (1)

To find the average rate of change over the interval , where , we substitute . We summarize the results in the table below.

 𝑘 Interval 𝐴(ℎ) 1 2 3 4 [3,3.1] [3,3.01] [3,3.001] [3,3.0001] 4.3 4.03 4.003 4.0003

As we can see in the previous example, the value of is getting closer and closer to 4 as approaches 0. Mathematically, we formalize this by saying that the limit of as approaches zero is 4 and we write

The value we get by this process is referred to as the instantaneous rate of change of the function at .

We might think we can simply set equal to zero in equation (1) to get . However, we actually made the assumption that was nonzero in the process of deriving the expression for in equation (1). For this reason and to be mathematically precise, we talk about the limit. On first impression, this might seem a little pedantic. However, this is actually an extremely important concept which is central to calculus and, in particular, our understanding of instantaneous rates of change.

Recall that we think of the rates of change of a function as the change in the function for a given change in the input. However, if there is no change in input, there is no change in the output. Therefore, to precisely define instantaneous rate of change, we need to introduce the concept of a limit.

### Definition: Instantaneous Rate of Change

The instantaneous rate of change of a function at a point is defined as when this limit exists.

For many of the functions we are familiar with, such as polynomial, trigonometric, exponential, logarithmic, and rational functions, it is possible to find their instantaneous rate of change for values of in their domain. However, there are many functions where this is not possible; we will introduce examples of these functions when we discuss the derivative in more details.

We will now consider a graphical example.

### Example 2: Secants and Rates of Change

Consider the function .

1. Find the slopes of the secant lines passing through the points on the graph of where and for .
2. What does this suggest about the gradient of the tangent line at ?
1. We cannot find the slopes of the tangent line since the slopes of the secant lines are changing in the vicinity of .
2. Since the slopes of the secant lines are tending to zero, we cannot find the slopes of the tangent line.
3. Since the slopes of the secant lines are tending to zero, the slopes of the tangent at will be zero.
4. We cannot make any conclusion about the slope of a tangent line by considering the slopes of secants.

Part 1

When , . Therefore, all the secant lines will pass through the point . Recall that the slope of a line passing through and is defined as

Therefore, the slope of the secant line passing through and is simply given by

By substituting for , we can find the slopes of the secant lines. We summarize the details of this in the table below.

 𝑥 𝑓(𝑥) Slope of Secant 𝑓(𝑥)𝑥 2 1 12 14 2 14 132 1256 1 14 116 164

We can represent this visually by considering the graphs.

Part 2

As we can see, the slope of the secant lines is tending to zero as tends to zero. In the limit, the secant line becomes a tangent. Hence, the slope of the tangent at is zero. Therefore, the correct answer is (C).

The previous example demonstrates that the instantaneous rates of change at a point can be visualized as the limit of the slopes of secant lines between and as gets arbitrarily close to . In the limit, this process results in the slope of the tangent line to the graph at .

### Example 3: Taking the Limits of Average Rates of Change

Consider the average rate of change of the function over the interval with small values of .

1. Simplify the expression .
2. The average rate of change gets closer and closer to as becomes smaller and smaller. Simplify the expression for the difference between and .
3. For what values of is the difference exactly , , ? Give your answer as a fraction.

Part 1

Using the function we have been given, we can write

We can express the expression in the parentheses as a single fraction over a denominator of as follows:

Since is nonzero, we can cancel this factor from the numerator and denominator, which results in

Part 2

We define

Using our expression from part 1, we can rewrite this as

We can express this as a single fraction over a denominator of as follows:

Part 3

To find the value of for which is equal to the fraction , we can rearrange the formula for to make the subject. We start by setting equal to :

Multiplying through by , we get

Subtracting from both sides gives

Finally, we can divide by , which yields

Using this equation, we can find the values of for given values of . In particular, when .

Similarly, when and when .

The previous example demonstrates that, for a given value , we can find such that the difference is less than if . This idea is actually the core idea of what we mean to say, that as , although this will be covered in more detail as you study calculus further.

We will now look at a couple of examples where we find the instantaneous rate of change by taking the limit of the average rate of change.

### Example 4: Instantaneous Rates of Change

If the function , find .

Before we try to find the limit, we will first find an expression for the average rate of change:

Using the function we have been given, we have

To help simplify this expression to take the limit, we need to expand the parentheses . We can do this using the binomial theorem which states that where represents the binomial coefficients. Substituting in , , and , we have

Substituting this into our expression yields

The first term (the term) of this sum equals . We can take this term out of the sum as follows:

Then, we can cancel the common terms:

Since is nonzero, we can cancel this factor from the terms on the top and bottom, which yields

Finally, we notice that the only term which is independent of is the first term (the term) of this sum. For clarity, we can take this term out of the sum and we have

We can now take the limit as . Since all of the terms in the sum have at least one factor of , these terms will all tend to zero. Hence,

### Example 5: Instantaneous Rates of Change

Find the instantaneous rate of change of at .

Recall that the instantaneous rate of change of a function at a point is defined as when this limit exists. Using the function we have been given, we have

To evaluate this limit, we can multiply the numerator and denominator by the conjugate of the numerator , which yields

Since the expression in the numerator is the difference of squares, we can expand the parentheses as follows:

Finally, since is nonzero, we can cancel this common factor from the numerator and denominator, which gives us

Using the rules of finite limits, we can rewrite this as

### Key Points

1. To define an instantaneous rate of change, we introduce the concept of a limit. In particular, we take the average rate of change over smaller and smaller intervals. Formally, we define the instantaneous rate of change of a function at a point by when this limit exists.
2. We can also interpret the instantaneous rate of change of at as the slope of the tangent line to the graph of at .
3. We can take the limit of the slopes of secants of the curve passing through the graph of the function at and as .
4. To evaluate the limit and find the instantaneous rate of change, we often require some algebraic manipulation of the expression to eliminate from its denominator.
5. Although instantaneous rates exist for many functions, we are familiar with, there are, in fact, many functions for which the limit does not exist and, as a result, we are unable to define the notion of instantaneous rate of change for them. A number of examples of such functions will be introduced as you study calculus further.