Average vs Instantaneous Rate of Change Explained

What Is The Difference Between Average And Instantaneous Rate Of Change?

The difference between average and instantaneous rate of change is the size of the window. The average rate of change compares two points (the full method is on the average rate of change page), so it describes how a function changed across an interval. The instantaneous rate of change shrinks that interval to a single point, so it describes how fast the function is changing at that exact input. Students on Brainly and in Briggs' calculus textbook are asked to "explain the difference" in almost these words, and the short answer is: interval versus point, secant versus tangent.

For a function $f$ on the interval $[a, b]$, the average rate of change is

$$\text{average rate} = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}$$

where $f(a)$ and $f(b)$ are the outputs at the two ends and $b - a \neq 0$ is the width of the interval. The instantaneous rate of change at $x = a$ is the limit of those averages as the second point slides into the first:

$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

provided this limit exists. Here $h$ is the width of a shrinking interval, and $f'(a)$ is the derivative of $f$ at $a$. Both rates have the same units: units of output per unit of input, such as metres per second or dollars per item.

How Does An Average Rate Become An Instantaneous Rate?

An average rate becomes an instantaneous rate by taking averages over smaller and smaller intervals and watching where they settle. Take $f(x) = x^2$ and fix the left end at $x = 3$. The average over $[3, 3 + h]$ is

$$\frac{(3 + h)^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h, \qquad h \neq 0$$

The averages close in on $6$, and

$$f'(3) = \lim_{h \to 0} (6 + h) = 6$$

You cannot just set $h = 0$ in the original fraction, because $\frac{0}{0}$ has no value; the algebra has to cancel the $h$ first. That expression $\frac{f(a + h) - f(a)}{h}$ is the difference quotient, and its limit is exactly the definition of the derivative. So the instantaneous rate is not a separate idea from the average rate. It is where the averages are heading, which is why the derivative as a rate of change reads every derivative as a speed of change.

How Do The Secant And Tangent Lines Show Each Rate?

Geometrically, the average rate of change is the slope of a secant line, a line through two points on the curve. The instantaneous rate of change is the slope of the tangent line, the line that touches the curve at one point and matches its direction there.

• Secant through $(a, f(a))$ and $(b, f(b))$. Its slope, rise over run, is $\frac{f(b) - f(a)}{b - a}$, the same number as the average rate, computed exactly as for the slope of a line.

• Tangent at $(a, f(a))$. Its slope is $f'(a)$. Once you know it, the tangent line equation is $y - f(a) = f'(a)(x - a)$.

• The link between them. Slide $b$ toward $a$ and the secant pivots about $(a, f(a))$. In the limit it becomes the tangent, which is the picture behind the table above.

Instantaneous Vs Average Rate Of Change: How Do They Compare Side By Side?

The two rates differ in what they need, what they mean and where they exist. Here is the whole comparison in one table.

Two rows deserve a second look. The units row is identical: both are rates, so both read "per". The "exists for" row is where they really differ, because an average always exists between two defined points, while an instantaneous rate can fail to exist at a corner such as the vertex of $y = |x|$, where the function is not differentiable.

When Are The Average And Instantaneous Rates Equal?

They are equal everywhere for a straight line, and at least once on any interval where the function is smooth enough. That second fact is the mean value theorem: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there is at least one $c$ in $(a, b)$ with

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

In words, somewhere inside the interval the tangent is parallel to the secant. Brainly threads ask this in several forms, such as "How do you find where the instantaneous rate of change is equal to the average rate of change?" and "Find the value of c where the instantaneous rate of change is the same as the average rate of change". The procedure is the same each time:

1. Compute the average rate $\frac{f(b) - f(a)}{b - a}$ on the given interval.

2. Differentiate to get $f'(x)$.

3. Solve $f'(x) = \text{average rate}$.

4. Keep only solutions that lie strictly inside $(a, b)$.

The theorem promises at least one such $c$; it does not promise exactly one, and Example 5 below finds two.

Examples Of Average Vs Instantaneous Rate Of Change

Example 1

For $f(x) = x^2$, find the average rate of change on $[1, 3]$ and the instantaneous rates at $x = 1$, $x = 2$ and $x = 3$.

$$\text{average} = \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4$$

$$f'(x) = 2x$$

$$f'(1) = 2, \qquad f'(2) = 4, \qquad f'(3) = 6$$

Final answer: the average rate is $4$. The instantaneous rate climbs from $2$ to $6$ across the interval and equals the average at the midpoint $x = 2$. (For any quadratic, the matching point is always the midpoint of the interval.)

Example 2

A stone dropped from rest falls $s(t) = 4.9t^2$ metres in $t$ seconds. How fast is it falling at $t = 2$?

The tempting route is distance divided by time:

$$\frac{s(2)}{2} = \frac{4.9 \times 4}{2} = 9.8\ \text{m/s}$$

Check it against the physics. The stone starts at $0\ \text{m/s}$ and speeds up the whole way, so its speed at the end of the 2 seconds must be higher than its average speed over them. And $9.8\ \text{m/s}$ is exactly that average:

$$\frac{s(2) - s(0)}{2 - 0} = \frac{19.6 - 0}{2} = 9.8\ \text{m/s}$$

Dividing the position by the time answered "average from $t = 0$ to $t = 2$", not "how fast at $t = 2$". The instantaneous rate needs the derivative:

$$s'(t) = 9.8t$$

$$s'(2) = 9.8 \times 2 = 19.6\ \text{m/s}$$

Short intervals agree: the average over $[2, 2.1]$ is $20.09\ \text{m/s}$ and over $[2, 2.01]$ is $19.649\ \text{m/s}$, closing in on $19.6$.

Final answer: $19.6\ \text{m/s}$, twice the average speed over the first 2 seconds.

Example 3

For $f(x) = \sqrt{x}$ on $[0, 4]$, find the average rate of change and the point where the instantaneous rate equals it.

$$\text{average} = \frac{\sqrt{4} - \sqrt{0}}{4 - 0} = \frac{2}{4} = \frac{1}{2}$$

$$f'(x) = \frac{1}{2\sqrt{x}}$$

$$\frac{1}{2\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 1 \implies x = 1$$

Final answer: the average rate is $\frac{1}{2}$, and the tangent at $x = 1$ is parallel to the secant. Note that $f'(0)$ does not exist (the tangent there is vertical), yet the mean value theorem still applies, because it only needs differentiability inside $(0, 4)$.

Example 4

A greenhouse temperature is logged each hour: $12^\circ\text{C}$ at $t = 0$, $15$ at $t = 1$, $19$ at $t = 2$, $24$ at $t = 3$ and $26$ at $t = 4$. Find the average rate over the 4 hours and estimate the instantaneous rate at $t = 2$.

With data and no formula, you can only estimate the instantaneous rate, by averaging over the narrowest interval centred on $t = 2$:

$$\text{average over } [0, 4] = \frac{26 - 12}{4 - 0} = \frac{14}{4} = 3.5^\circ\text{C per hour}$$

$$\text{estimate at } t = 2 \approx \frac{T(3) - T(1)}{3 - 1} = \frac{24 - 15}{2} = 4.5^\circ\text{C per hour}$$

Final answer: $3.5^\circ\text{C}$ per hour on average, about $4.5^\circ\text{C}$ per hour at $t = 2$. The one-sided estimates, $5$ (forward) and $4$ (backward), bracket the centred one.

Example 5

For $f(x) = x^3 - 3x$ on $[-2, 2]$, find every $c$ where the instantaneous rate equals the average rate.

$$\text{average} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{2 - (-2)}{4} = 1$$

$$f'(x) = 3x^2 - 3$$

$$3x^2 - 3 = 1 \implies x^2 = \frac{4}{3} \implies x = \pm\frac{2}{\sqrt{3}}$$

Final answer: $c = \pm\frac{2\sqrt{3}}{3} \approx \pm 1.1547$. Both lie inside $(-2, 2)$, so both count.

Example 6

A workshop's cost to make $q$ chairs is $C(q) = 1000 + 5q + 0.01q^2$ dollars. Compare the average rate of change of cost from $q = 100$ to $q = 200$ with the instantaneous rate (marginal cost) at $q = 100$, $150$ and $200$.

$$C(100) = 1000 + 500 + 100 = 1600$$

$$C(200) = 1000 + 1000 + 400 = 2400$$

$$\text{average} = \frac{2400 - 1600}{200 - 100} = 8\ \text{dollars per chair}$$

$$C'(q) = 5 + 0.02q$$

$$C'(100) = 7, \qquad C'(150) = 8, \qquad C'(200) = 9$$

Final answer: the extra 100 chairs cost $8$ dollars each on average, while the marginal cost rises from $7$ to $9$ dollars per chair across the range, matching the average at $q = 150$.

Why Do We Need Two Kinds Of Rate?

"How can anything move during an instant?" Zeno of Elea's arrow paradox, written around 2,500 years ago, asked exactly that: at any single instant an arrow occupies one position, so in that instant it seems to be at rest. The Stanford Encyclopedia of Philosophy notes that speed at a single instant cannot be computed directly as distance over time, because both are zero. The instantaneous rate of change is calculus's answer.

• The average rate is easy but blurry. Two measurements and one division give it, which is why it suits trip logs, data tables and reports.

• The instantaneous rate is sharp but needs a limit. It is defined as the value the averages approach, which sidesteps $\frac{0}{0}$ instead of dividing by zero.

• Real decisions need the sharp one. A speed limit, a drug dose rate or a production decision depends on the rate now, not the rate over the last hour.

The averages are how you measure; the instantaneous rate is what you mean. That pairing runs through all of differential calculus, from velocity and acceleration to optimisation.

Where Do You Use Each Rate In Real Life?

The two rates usually appear in pairs: the average as the summary, the instantaneous as the live reading.

• Driving. Average speed is total distance over total time; the speedometer shows instantaneous speed. Average-speed enforcement cameras on some motorways use the first, a handheld radar gun the second.

• Economics. Average cost per extra unit over a production range versus marginal cost, the instantaneous rate, at one output level (Example 6).

• Medicine. Average change in a patient's blood-drug concentration over a dosing interval versus the rate of change at the moment of a reading.

• Population and finance. Average growth per year over a decade versus the growth rate at a given year, the quantity a model like $P'(t) = kP(t)$ describes.

• Sport. A sprinter's average speed over 100 m versus top speed at the fastest instant of the race.

Where Do Average And Instantaneous Rates Get Mixed Up?

Most mistakes on this topic come from answering one rate's question with the other rate's formula, and the result often looks perfectly reasonable.

Mistake 1: Dividing The Output By The Input

Where it slips in: "How fast is it moving at $t = 2$?" when the position formula is given.

Don't do this: Computing $\frac{s(2)}{2}$. That is the average rate from $t = 0$ to $t = 2$, not the rate at $t = 2$ (Example 2 gives $9.8$ instead of $19.6$).

The correct way: Differentiate and substitute: $s'(2)$. Use $\frac{f(b) - f(a)}{b - a}$ only when the question names an interval.

The first instinct with "speed equals distance over time" is to divide whatever two numbers are on the page. The rusher does this in a second and never notices that the answer is an average.

Mistake 2: Mixing The Order Of Subtraction

Where it slips in: Average rate on an interval with negative endpoints, such as $[-2, 2]$.

Don't do this: $\frac{f(2) - f(-2)}{-2 - 2}$, which flips the sign of the answer.

The correct way: Keep the same order on top and bottom: $\frac{f(b) - f(a)}{b - a}$, with $b$ the right end both times.

Mistake 3: Plugging In Zero Too Early

Where it slips in: Finding an instantaneous rate from the limit definition.

Don't do this: Substituting $h = 0$ into $\frac{f(a + h) - f(a)}{h}$ straight away, which gives $\frac{0}{0}$ and leads some students to answer "undefined".

The correct way: Simplify first so the $h$ in the denominator cancels (as $\frac{6h + h^2}{h} = 6 + h$ did above), then take the limit.

The memorizer knows the limit formula by heart but treats "$h \to 0$" as "put in $0$". Students who first build the shrinking-interval table by hand stop making this mistake, because they have watched the averages settle without ever reaching $h = 0$.

Mistake 4: Dropping Or Mixing Units

Where it slips in: Word problems with time in minutes and hours, or distance in feet and metres.

Don't do this: Reporting "the rate is 8" with no units, or combining a rate in metres per second with a distance in kilometres.

The correct way: Write the rate as output units per input unit, such as $8$ dollars per chair, and convert everything to one system before dividing.

Practice Problems On Average Vs Instantaneous Rate Of Change

1. For $f(x) = x^3$, find the average rate of change on $[0, 2]$ and the instantaneous rate at $x = 2$.

2. For $f(x) = \frac{1}{x}$, find the average rate of change on $[1, 4]$ and the instantaneous rate at $x = 2$.

3. For $f(x) = x^2 + 2x$ on $[0, 4]$, find the value of $c$ where the instantaneous rate equals the average rate.

4. A ball's height is $h(t) = -16t^2 + 64t$ feet. Find its average velocity on $[0, 2]$ and its instantaneous velocity at $t = 1$ and $t = 3$.

5. For $f(x) = e^{x}$ on $[0, 1]$, find the average rate of change to 4 decimal places and the $x$ where $f'(x)$ equals it.

Answers:

• Answer to Question 1: average $4$; instantaneous $12$.

• Answer to Question 2: average $-\frac{1}{4}$; instantaneous $-\frac{1}{4}$ (here $x = 2$ is exactly the mean value theorem point).

• Answer to Question 3: $c = 2$.

• Answer to Question 4: average $32$ ft/s; $32$ ft/s at $t = 1$ (rising) and $-32$ ft/s at $t = 3$ (falling).

• Answer to Question 5: average $e - 1 \approx 1.7183$; $x = \ln(e - 1) \approx 0.5413$.

Key Takeaways

• The average vs instantaneous rate of change comes down to interval versus point: $\frac{f(b) - f(a)}{b - a}$ against $f'(a)$.

• The average rate is the slope of a secant line; the instantaneous rate is the slope of the tangent line.

• The instantaneous rate is the limit of average rates as the interval shrinks to zero width.

• The mean value theorem guarantees the two rates match at least once inside an interval where $f$ is continuous and differentiable.

• Dividing an output by its input gives an average, not the rate at an instant.

Where To Go From Here

Work through practice problems 1 to 5, computing both rates every time; if problem 3 or 5 trips you up, return to the section on when the two rates are equal. To practise rates of change with a live Bhanzu trainer, book a free demo class: https://bhanzu.com/book-demo-lp?utm_source=blog&utm_medium=article&utm_campaign=average-vs-instantaneous-rate-of-change#book-demo

Read More

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• Derivative of a function

• Power rule

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