Lesson Explainer: Average and Instantaneous Rates of Change Mathematics • Higher Education

In this explainer, we will learn how to find the average rate of change of a function between two ๐‘ฅ-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 ๐‘“(๐‘ฅ) over an interval [๐‘ฅ,๐‘ฅ]๏Šง๏Šจ is defined by ๐‘“(๐‘ฅ)โˆ’๐‘“(๐‘ฅ)๐‘ฅโˆ’๐‘ฅ.๏Šจ๏Šง๏Šจ๏Šง

Often, we consider the average rate of change over the interval [๐‘ฅ,๐‘ฅ+โ„Ž]๏Šง๏Šง as a function of โ„Ž: ๐ด(โ„Ž)=๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž.๏Šง๏Šง

Notice that if โ„Ž<0, the average rate of change is over the interval [๐‘ฅโˆ’|โ„Ž|,๐‘ฅ]๏Šง๏Šง.

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 ๐‘“(๐‘ฅ)=3๐‘ฅโˆ’14๐‘ฅ+7๏Šจ, list the average rates of change of ๐‘“ over the interval ๏”3,3+110๏ ๏‡, where ๐‘˜=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 ๐‘“. Recall that the average rate of change ๐ด(โ„Ž) of a function ๐‘“ over the interval [๐‘ฅโˆ’|โ„Ž|,๐‘ฅ]๏Šง๏Šง is given by ๐ด(โ„Ž)=๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž.๏Šง๏Šง

Using the function we have been given, we can write this as ๐ด(โ„Ž)=3(๐‘ฅ+โ„Ž)โˆ’14(๐‘ฅ+โ„Ž)+7โˆ’๏€น3๐‘ฅโˆ’14๐‘ฅ+7๏…โ„Ž.๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šง

Expanding the parentheses, we have ๐ด(โ„Ž)=3๏€น๐‘ฅ+2๐‘ฅโ„Ž+โ„Ž๏…โˆ’14๐‘ฅโˆ’14โ„Ž+7โˆ’3๐‘ฅ+14๐‘ฅโˆ’7โ„Ž=6๐‘ฅโ„Ž+3โ„Žโˆ’14โ„Žโ„Ž.๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šง๏Šง๏Šจ

Given that โ„Ž is not equal to zero, we can cancel the common factor of โ„Ž from the numerator and denominator, which yields ๐ด(โ„Ž)=6๐‘ฅโˆ’14+3โ„Ž.๏Šง

Therefore, the average rate of change at ๐‘ฅ=3๏Šง is

๐ด(โ„Ž)=4+3โ„Ž.(1)

To find the average rate of change over the interval ๏”3,3+110๏ ๏‡, where ๐‘˜=1,2,3,4, we substitute โ„Ž=110๏‡. We summarize the results in the table below.

๐‘˜1234
Interval[3,3.1][3,3.01][3,3.001][3,3.0001]
๐ด(โ„Ž)4.34.034.0034.0003

As we can see in the previous example, the value of ๐ด(โ„Ž) is getting closer and closer to 4 as โ„Ž approaches 0. Mathematically, we formalize this by saying that the limit of ๐ด(โ„Ž) as โ„Ž approaches zero is 4 and we write lim๏‚โ†’๏Šฆ๐ด(โ„Ž)=4.

The value we get by this process is referred to as the instantaneous rate of change of the function at ๐‘ฅ=3.

We might think we can simply set โ„Ž equal to zero in equation (1) to get ๐ด(โ„Ž)=4. However, we actually made the assumption that โ„Ž was nonzero in the process of deriving the expression for ๐ด(โ„Ž) 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 ๐‘“ at a point ๐‘ฅ๏Šง is defined as lim๏‚โ†’๏Šฆ๏Šง๏Šง๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž 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 ๐‘ฅ 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 ๐‘“(๐‘ฅ)=14๐‘ฅ๏Šฉ.

  1. Find the slopes of the secant lines passing through the points on the graph of ๐‘“ where ๐‘ฅ=0 and ๐‘ฅ=2๏Šง๏Šฑ๏Š for ๐‘›=0,1,2,3.
  2. What does this suggest about the gradient of the tangent line at ๐‘ฅ=0?
    1. We cannot find the slopes of the tangent line since the slopes of the secant lines are changing in the vicinity of ๐‘ฅ=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 ๐‘ฅ=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 ๐‘ฅ=0, ๐‘“(๐‘ฅ)=0. Therefore, all the secant lines will pass through the point (0,0). Recall that the slope ๐‘š of a line passing through (๐‘ฅ,๐‘ฆ)๏Šง๏Šง and (๐‘ฅ,๐‘ฆ)๏Šจ๏Šจ is defined as ๐‘š=๐‘ฆโˆ’๐‘ฆ๐‘ฅโˆ’๐‘ฅ.๏Šจ๏Šง๏Šจ๏Šง

Therefore, the slope of the secant line passing through (0,0) and (๐‘ฅ,๐‘“(๐‘ฅ)) is simply given by ๐‘š=๐‘“(๐‘ฅ)๐‘ฅ=14๐‘ฅ.๏Šจ

By substituting ๐‘ฅ=2๏Šง๏Šฑ๏Š for ๐‘›=0,1,2,3, we can find the slopes of the secant lines. We summarize the details of this in the table below.

๐‘ฅ211214
๐‘“(๐‘ฅ)2141321256
Slope of Secant ๐‘“(๐‘ฅ)๐‘ฅ114116164

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 ๐‘ฅ tends to zero. In the limit, the secant line becomes a tangent. Hence, the slope of the tangent at ๐‘ฅ=0 is zero. Therefore, the correct answer is (C).

The previous example demonstrates that the instantaneous rates of change at a point ๐‘ฅ๏Šง can be visualized as the limit of the slopes of secant lines between ๐‘ฅ๏Šง and ๐‘ฅ๏Šจ as ๐‘ฅ๏Šจ gets arbitrarily close to ๐‘ฅ๏Šง. In the limit, this process results in the slope of the tangent line to the graph at ๐‘ฅ๏Šง.

Example 3: Taking the Limits of Average Rates of Change

Consider the average rate of change of the function ๐‘“(๐‘ฅ)=1๐‘ฅ over the interval [3,3+โ„Ž] with small values of โ„Ž.

  1. Simplify the expression ๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Ž.
  2. The average rate of change gets closer and closer to โˆ’19 as โ„Ž becomes smaller and smaller. Simplify the expression for the difference ๐›ฟ(โ„Ž) between ๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Ž and โˆ’19.
  3. For what values of โ„Ž is the difference ๐›ฟ(โ„Ž) exactly 110, 1100, 110๏Šช? Give your answer as a fraction.

Answer

Part 1

Using the function ๐‘“(๐‘ฅ)=1๐‘ฅ we have been given, we can write ๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Ž=1โ„Ž๏€ผ13+โ„Žโˆ’13๏ˆ.

We can express the expression in the parentheses as a single fraction over a denominator of 3(3+โ„Ž) as follows: ๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Ž=1โ„Ž๏€ฝ3โˆ’(3+โ„Ž)3(3+โ„Ž)๏‰=1โ„Ž๏€ฝโˆ’โ„Ž9+3โ„Ž๏‰.

Since โ„Ž is nonzero, we can cancel this factor from the numerator and denominator, which results in ๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Ž=โˆ’19+3โ„Ž.

Part 2

We define ๐›ฟ(โ„Ž)=๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Žโˆ’๏€ผโˆ’19๏ˆ.

Using our expression from part 1, we can rewrite this as ๐›ฟ(โ„Ž)=โˆ’19+3โ„Ž+19.

We can express this as a single fraction over a denominator of 9(3+โ„Ž) as follows: ๐›ฟ(โ„Ž)=3(โˆ’1)+(3+โ„Ž)9(3+โ„Ž)=โ„Ž27+9โ„Ž.

Part 3

To find the value of โ„Ž for which ๐›ฟ(โ„Ž) is equal to the fraction 1๐‘‘, we can rearrange the formula for ๐›ฟ(โ„Ž) to make โ„Ž the subject. We start by setting ๐›ฟ(โ„Ž) equal to 1๐‘‘: 1๐‘‘=๐›ฟ(โ„Ž)=โ„Ž27+9โ„Ž.

Multiplying through by ๐‘‘(27+9โ„Ž), we get 27+9โ„Ž=๐‘‘โ„Ž.

Subtracting 9โ„Ž from both sides gives 27=(๐‘‘โˆ’9)โ„Ž.

Finally, we can divide by ๐‘‘โˆ’9, which yields โ„Ž=27๐‘‘โˆ’9.

Using this equation, we can find the values of โ„Ž for given values of ๐‘‘. In particular, ๐›ฟ(โ„Ž)=110 when โ„Ž=27.

Similarly, ๐›ฟ(โ„Ž)=1100 when โ„Ž=2791 and ๐›ฟ(โ„Ž)=110๏Šช when โ„Ž=279991.

The previous example demonstrates that, for a given value 1๐‘‘, we can find โ„Ž๏Šง such that the difference ๐›ฟ(โ„Ž)=๐‘“(3+โ„Ž)โˆ’๐‘“(3)โ„Žโˆ’๏€ผโˆ’19๏ˆ is less than 1๐‘‘ if 0<โ„Ž<โ„Ž๏Šง. This idea is actually the core idea of what we mean to say, that ๐›ฟ(โ„Ž)โ†’0 as โ„Žโ†’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 ๐‘“(๐‘ฅ)=โˆ’3๐‘ฅโˆ’5๏Šฏ, find lim๏‚โ†’๏Šฆ๏Šง๏Šง๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž.

Answer

Before we try to find the limit, we will first find an expression for the average rate of change: ๐ด(โ„Ž)=๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž.๏Šง๏Šง

Using the function we have been given, we have ๐ด(โ„Ž)=๏€บโˆ’3(๐‘ฅ+โ„Ž)โˆ’5๏†โˆ’(โˆ’3๐‘ฅโˆ’5)โ„Ž=โˆ’3(๐‘ฅ+โ„Ž)+3๐‘ฅโ„Ž.๏Šง๏Šฏ๏Šฏ๏Šง๏Šง๏Šฏ๏Šฏ๏Šง

To help simplify this expression to take the limit, we need to expand the parentheses (๐‘ฅ+โ„Ž)๏Šง๏Šฏ. We can do this using the binomial theorem which states that (๐‘Ž+๐‘)=๏„š๏€ฟ๐‘›๐‘˜๏‹๐‘Ž๐‘,๏Š๏Š๏‡๏Šฒ๏Šฆ๏Š๏Šฑ๏‡๏‡ where ๏€ฟ๐‘›๐‘˜๏‹ represents the binomial coefficients. Substituting in ๐‘Ž=๐‘ฅ๏Šง, ๐‘=โ„Ž, and ๐‘›=9, we have (๐‘ฅ+โ„Ž)=๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž.๏Šง๏Šฏ๏Šฏ๏‡๏Šฒ๏Šฆ๏Šฏ๏Šฑ๏‡๏Šง๏‡

Substituting this into our expression yields ๐ด(โ„Ž)=โˆ’3๏€ฟ๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž๏‹+3๐‘ฅโ„Ž.๏Šฏ๏‡๏Šฒ๏Šฆ๏Šฏ๏Šฑ๏‡๏Šง๏‡๏Šฏ๏Šง

The first term (the ๐‘˜=0 term) of this sum equals โˆ’3๐‘ฅ๏Šฏ๏Šง. We can take this term out of the sum as follows: ๐ด(โ„Ž)=โˆ’3๐‘ฅโˆ’3๏€ฟ๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž๏‹+3๐‘ฅโ„Ž.๏Šฏ๏Šง๏Šฏ๏‡๏Šฒ๏Šง๏Šฏ๏Šฑ๏‡๏Šง๏‡๏Šฏ๏Šง

Then, we can cancel the common terms: ๐ด(โ„Ž)=โˆ’3๏€ฟ๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž๏‹โ„Ž.๏Šฏ๏‡๏Šฒ๏Šง๏Šฏ๏Šฑ๏‡๏Šง๏‡

Since โ„Ž is nonzero, we can cancel this factor from the terms on the top and bottom, which yields ๐ด(โ„Ž)=โˆ’3๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž.๏Šฏ๏‡๏Šฒ๏Šง๏Šฏ๏Šฑ๏‡๏Šง๏‡๏Šฑ๏Šง

Finally, we notice that the only term which is independent of โ„Ž is the first term (the ๐‘˜=1 term) of this sum. For clarity, we can take this term out of the sum and we have ๐ด(โ„Ž)=โˆ’3๏€ฟ91๏‹๐‘ฅโˆ’3๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž=โˆ’27๐‘ฅโˆ’3๏„š๏€ฟ9๐‘˜๏‹๐‘ฅโ„Ž.๏Šฎ๏Šง๏Šฏ๏‡๏Šฒ๏Šจ๏Šฏ๏Šฑ๏‡๏Šง๏‡๏Šฑ๏Šง๏Šฎ๏Šง๏Šฏ๏‡๏Šฒ๏Šจ๏Šฏ๏Šฑ๏‡๏Šง๏‡๏Šฑ๏Šง

We can now take the limit as โ„Žโ†’0. Since all of the terms in the sum have at least one factor of โ„Ž, these terms will all tend to zero. Hence, lim๏‚โ†’๏Šฆ๏Šง๏Šง๏Šฎ๏Šง๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž=โˆ’27๐‘ฅ.

Example 5: Instantaneous Rates of Change

Find the instantaneous rate of change of ๐‘“(๐‘ฅ)=โˆš๐‘ฅ at ๐‘ฅ=๐‘ฅ>0๏Šง.

Answer

Recall that the instantaneous rate of change of a function ๐‘“ at a point ๐‘ฅ๏Šง is defined as lim๏‚โ†’๏Šฆ๏Šง๏Šง๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž when this limit exists. Using the function we have been given, we have lim๏‚โ†’๏Šฆ๏Šง๏Šงโˆš๐‘ฅ+โ„Žโˆ’โˆš๐‘ฅโ„Ž.

To evaluate this limit, we can multiply the numerator and denominator by the conjugate of the numerator โˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ๏Šง๏Šง, which yields lim๏‚โ†’๏Šฆ๏Šง๏Šง๏Šง๏Šง๏Šง๏Šง๏€บโˆš๐‘ฅ+โ„Žโˆ’โˆš๐‘ฅ๏†๏€บโˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ๏†โ„Ž๏€บโˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ๏†.

Since the expression in the numerator is the difference of squares, we can expand the parentheses as follows: limlim๏‚โ†’๏Šฆ๏Šง๏Šง๏Šง๏Šง๏‚โ†’๏Šฆ๏Šง๏Šง๐‘ฅ+โ„Žโˆ’๐‘ฅโ„Ž๏€บโˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ๏†=โ„Žโ„Ž๏€บโˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ๏†.

Finally, since โ„Ž is nonzero, we can cancel this common factor from the numerator and denominator, which gives us lim๏‚โ†’๏Šฆ๏Šง๏Šง1โˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ.

Using the rules of finite limits, we can rewrite this as limlim๏‚โ†’๏Šฆ๏Šง๏Šง๏‚โ†’๏Šฆ๏Šง๏Šง๏Šง1๏€บโˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ๏†=1โˆš๐‘ฅ+โ„Ž+โˆš๐‘ฅ=12โˆš๐‘ฅ.

Key Points

  • 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 ๐‘“ at a point ๐‘ฅ๏Šง by lim๏‚โ†’๏Šฆ๏Šง๏Šง๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)โ„Ž when this limit exists.
  • We can also interpret the instantaneous rate of change of ๐‘“ at ๐‘ฅ๏Šง as the slope of the tangent line to the graph of ๐‘“ at ๐‘ฅ=๐‘ฅ๏Šง.
  • We can take the limit of the slopes of secants of the curve passing through the graph of the function at ๐‘ฅ๏Šง and ๐‘ฅ+โ„Ž๏Šง as โ„Žโ†’0.
  • To evaluate the limit and find the instantaneous rate of change, we often require some algebraic manipulation of the expression to eliminate โ„Ž from its denominator.
  • 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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