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 𝑥-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 𝑓(𝑥) over an interval [𝑥,𝑥] is defined by 𝑓(𝑥)𝑓(𝑥)𝑥𝑥.

Often, we consider the average rate of change over the interval [𝑥,𝑥+] as a function of : 𝐴()=𝑓(𝑥+)𝑓(𝑥).

Notice that if <0, the average rate of change is over the interval [𝑥||,𝑥].

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 𝑓(𝑥)=3𝑥14𝑥+7, list the average rates of change of 𝑓 over the interval 3,3+110, where 𝑘=1,2,3,4, evaluated to 4 decimal places at most.

Answer

We begin by finding an expression for the average rate of change of 𝑓. Recall that the average rate of change 𝐴() of a function 𝑓 over the interval [𝑥||,𝑥] is given by 𝐴()=𝑓(𝑥+)𝑓(𝑥).

Using the function we have been given, we can write this as 𝐴()=3(𝑥+)14(𝑥+)+73𝑥14𝑥+7.

Expanding the parentheses, we have 𝐴()=3𝑥+2𝑥+14𝑥14+73𝑥+14𝑥7=6𝑥+314.

Given that is not equal to zero, we can cancel the common factor of from the numerator and denominator, which yields 𝐴()=6𝑥14+3.

Therefore, the average rate of change at 𝑥=3 is

𝐴()=4+3.(1)

To find the average rate of change over the interval 3,3+110, where 𝑘=1,2,3,4, we substitute =110. We summarize the results in the table below.

𝑘1234
Interval[3,3.1][3,3.01][3,3.001][3,3.0001]
𝐴()4.34.034.0034.0003

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

The value we get by this process is referred to as the instantaneous rate of change of the function at 𝑥=3.

We might think we can simply set equal to zero in equation (1) to get 𝐴()=4. However, we actually made the assumption that was nonzero in the process of deriving the expression for 𝐴() 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 𝑓 at a point 𝑥 is defined as lim𝑓(𝑥+)𝑓(𝑥) 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 𝑥 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 𝑓(𝑥)=14𝑥.

  1. Find the slopes of the secant lines passing through the points on the graph of 𝑓 where 𝑥=0 and 𝑥=2 for 𝑛=0,1,2,3.
  2. What does this suggest about the gradient of the tangent line at 𝑥=0?
    1. We cannot find the slopes of the tangent line since the slopes of the secant lines are changing in the vicinity of 𝑥=0.
    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 𝑥=0 will be zero.
    4. We cannot make any conclusion about the slope of a tangent line by considering the slopes of secants.

Answer

Part 1

When 𝑥=0, 𝑓(𝑥)=0. Therefore, all the secant lines will pass through the point (0,0). Recall that the slope 𝑚 of a line passing through (𝑥,𝑦) and (𝑥,𝑦) is defined as 𝑚=𝑦𝑦𝑥𝑥.

Therefore, the slope of the secant line passing through (0,0) and (𝑥,𝑓(𝑥)) is simply given by 𝑚=𝑓(𝑥)𝑥=14𝑥.

By substituting 𝑥=2 for 𝑛=0,1,2,3, we can find the slopes of the secant lines. We summarize the details of this in the table below.

𝑥211214
𝑓(𝑥)2141321256
Slope of Secant 𝑓(𝑥)𝑥114116164

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 𝑥 tends to zero. In the limit, the secant line becomes a tangent. Hence, the slope of the tangent at 𝑥=0 is zero. Therefore, the correct answer is (C).

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

Example 3: Taking the Limits of Average Rates of Change

Consider the average rate of change of the function 𝑓(𝑥)=1𝑥 over the interval [3,3+] with small values of .

  1. Simplify the expression 𝑓(3+)𝑓(3).
  2. The average rate of change gets closer and closer to 19 as becomes smaller and smaller. Simplify the expression for the difference 𝛿() between 𝑓(3+)𝑓(3) and 19.
  3. For what values of is the difference 𝛿() exactly 110, 1100, 110? Give your answer as a fraction.

Answer

Part 1

Using the function 𝑓(𝑥)=1𝑥 we have been given, we can write 𝑓(3+)𝑓(3)=113+13.

We can express the expression in the parentheses as a single fraction over a denominator of 3(3+) as follows: 𝑓(3+)𝑓(3)=13(3+)3(3+)=19+3.

Since is nonzero, we can cancel this factor from the numerator and denominator, which results in 𝑓(3+)𝑓(3)=19+3.

Part 2

We define 𝛿()=𝑓(3+)𝑓(3)19.

Using our expression from part 1, we can rewrite this as 𝛿()=19+3+19.

We can express this as a single fraction over a denominator of 9(3+) as follows: 𝛿()=3(1)+(3+)9(3+)=27+9.

Part 3

To find the value of for which 𝛿() is equal to the fraction 1𝑑, we can rearrange the formula for 𝛿() to make the subject. We start by setting 𝛿() equal to 1𝑑: 1𝑑=𝛿()=27+9.

Multiplying through by 𝑑(27+9), we get 27+9=𝑑.

Subtracting 9 from both sides gives 27=(𝑑9).

Finally, we can divide by 𝑑9, which yields =27𝑑9.

Using this equation, we can find the values of for given values of 𝑑. In particular, 𝛿()=110 when =27.

Similarly, 𝛿()=1100 when =2791 and 𝛿()=110 when =279,991.

The previous example demonstrates that, for a given value 1𝑑, we can find such that the difference 𝛿()=𝑓(3+)𝑓(3)19 is less than 1𝑑 if 0<<. This idea is actually the core idea of what we mean to say, that 𝛿()0 as 0, 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 𝑓(𝑥)=3𝑥5, find lim𝑓(𝑥+)𝑓(𝑥).

Answer

Before we try to find the limit, we will first find an expression for the average rate of change: 𝐴()=𝑓(𝑥+)𝑓(𝑥).

Using the function we have been given, we have 𝐴()=3(𝑥+)5(3𝑥5)=3(𝑥+)+3𝑥.

To help simplify this expression to take the limit, we need to expand the parentheses (𝑥+). We can do this using the binomial theorem which states that (𝑎+𝑏)=𝑛𝑘𝑎𝑏, where 𝑛𝑘 represents the binomial coefficients. Substituting in 𝑎=𝑥, 𝑏=, and 𝑛=9, we have (𝑥+)=9𝑘𝑥.

Substituting this into our expression yields 𝐴()=39𝑘𝑥+3𝑥.

The first term (the 𝑘=0 term) of this sum equals 3𝑥. We can take this term out of the sum as follows: 𝐴()=3𝑥39𝑘𝑥+3𝑥.

Then, we can cancel the common terms: 𝐴()=39𝑘𝑥.

Since is nonzero, we can cancel this factor from the terms on the top and bottom, which yields 𝐴()=39𝑘𝑥.

Finally, we notice that the only term which is independent of is the first term (the 𝑘=1 term) of this sum. For clarity, we can take this term out of the sum and we have 𝐴()=391𝑥39𝑘𝑥=27𝑥39𝑘𝑥.

We can now take the limit as 0. Since all of the terms in the sum have at least one factor of , these terms will all tend to zero. Hence, lim𝑓(𝑥+)𝑓(𝑥)=27𝑥.

Example 5: Instantaneous Rates of Change

Find the instantaneous rate of change of 𝑓(𝑥)=𝑥 at 𝑥=𝑥>0.

Answer

Recall that the instantaneous rate of change of a function 𝑓 at a point 𝑥 is defined as lim𝑓(𝑥+)𝑓(𝑥) when this limit exists. Using the function we have been given, we have lim𝑥+𝑥.

To evaluate this limit, we can multiply the numerator and denominator by the conjugate of the numerator 𝑥++𝑥, which yields lim𝑥+𝑥𝑥++𝑥𝑥++𝑥.

Since the expression in the numerator is the difference of squares, we can expand the parentheses as follows: limlim𝑥+𝑥𝑥++𝑥=𝑥++𝑥.

Finally, since is nonzero, we can cancel this common factor from the numerator and denominator, which gives us lim1𝑥++𝑥.

Using the rules of finite limits, we can rewrite this as limlim1𝑥++𝑥=1𝑥++𝑥=12𝑥.

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 𝑓 at a point 𝑥 by lim𝑓(𝑥+)𝑓(𝑥) when this limit exists.
  2. We can also interpret the instantaneous rate of change of 𝑓 at 𝑥 as the slope of the tangent line to the graph of 𝑓 at 𝑥=𝑥.
  3. We can take the limit of the slopes of secants of the curve passing through the graph of the function at 𝑥 and 𝑥+ as 0.
  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.

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