Average and Instantaneous Rates of Change

Part 1 · Hook

Why this matters

A weather balloon climbs 600 meters in its first 2 minutes. That tells you its average climb, 300 meters per minute, but not how fast it was rising at the moment a gust hit it. Calculus begins by turning an average over a stretch of time into a rate at a single instant, and the tool that does it is a limit.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\lim_{h \to 0} (6 + h)$ ?

$6$
$0$
$6 + h$
It does not exist

Show the answer

The expression is continuous in $h$ , so substitute $h = 0$ .

Correct: $6$ :
$0$ :
$6 + h$ :
It does not exist:

2. For $h \ne 0$ , $\frac{h^2 + 5h}{h}$ simplifies to

$h + 5$
$h^2 + 5$
$5$
$h + 5h$

Show the answer

Factor $h$ from the numerator and cancel it: $\frac{h(h + 5)}{h} = h + 5$ .

Correct: $h + 5$ :
$h^2 + 5$ :
$5$ :
$h + 5h$ :

3. A quantity changes from 20 to 50 while time goes from 3 hours to 9 hours. Its average rate of change is

5 per hour
30 per hour
10 per hour
6 per hour

Show the answer

Change in quantity over change in time: $\frac{50 - 20}{9 - 3} = 5$ per hour.

Correct: 5 per hour:
30 per hour:
10 per hour:
6 per hour:

Part 4 · See it

See it first

The parabola f(x) = x squared over 4, plus 1, with the point P at x = 1. Four secant lines join P to points Q at x = 4, 3, 2 and 1.4, with slopes 1.25, 1, 0.75 and 0.6; a table adds h = 0.01 with slope 0.5025. As Q moves toward P, the secant slopes approach 0.5, the instantaneous rate of change of f at x = 1. — Secant lines from $P$ on the graph of $f(x) = \frac{x^2}{4} + 1$ . As the second point $Q$ moves toward $P$ , the secant slopes close in on 0.5, the rate of change of $f$ at $P$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Two points on a graph, at $x = a$ and $x = b$ give a secant line whose slope is the average rate of change $\frac{f(b) - f(a)}{b - a}$
The second point is written $a + h$ the average rate becomes the difference quotient $\frac{f(a + h) - f(a)}{h}$
$h$ shrinks toward 0the interval shrinks to the single instant $x = a$ and the secant lines pivot toward one limiting line
The limit $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ existsits value is the instantaneous rate of change of $f$ at $x = a$

Part 6 · Key ideas

Key ideas

The average rate of change of $f$ on $[a, b]$ is $\frac{f(b) - f(a)}{b - a}$ : the change in output over the change in input, with units of output per input.
A secant line joins two points of a graph. Its slope is the average rate of change between them.
The difference quotient $\frac{f(a + h) - f(a)}{h}$ is the average rate on the interval from $a$ to $a + h$ ; $h$ can be negative.
The instantaneous rate of change at $x = a$ is $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ , when that limit exists. Simplify the quotient first; substituting $h = 0$ at the start gives $\frac{0}{0}$ .

Part 7 · Misconception

A common mistake

The wrong idea: To find how fast something is changing at $t = 5$ , divide its value by 5.

What actually happens: A rate of change compares two changes: a change in the quantity over a change in time. $A(5) \div 5$ is a value over a time, not a change over a change. Use $\frac{A(b) - A(a)}{b - a}$ for an average rate, and a limit of such quotients for the rate at one instant.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A drone climbing after takeoff

A delivery drone takes off from a rooftop. Its altitude above the street is measured at a few times.

Altitude $A(t)$ of a drone, in meters, at selected times $t$ , in seconds
$t$ (seconds)	0	2	5	6	9
$A(t)$ (meters)	6	14	41	47	62

1. Find the average rate of change of the drone's altitude from $t = 2$ to $t = 6$ seconds. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\frac{A(6) - A(2)}{6 - 2}$ $= \frac{47 - 14}{4}$ $= \frac{33}{4}$ $= 8.25$ meters per second.

Answer: $\frac{33}{4}$ ≈ 8.250 meters per second

2. Which sentence correctly interprets $\frac{A(6) - A(2)}{6 - 2}$ $= 8.25$ ?

Between $t = 2$ and $t = 6$ seconds, the altitude rose 8.25 meters per second on average.
At $t = 4$ seconds, the drone was rising at exactly 8.25 meters per second.
The drone climbed a total of 8.25 meters between $t = 2$ and $t = 6$ seconds.
Between $t = 2$ and $t = 6$ seconds, the drone was 8.25 meters above the street.

Show the answer

An average rate describes a whole interval: the total climb, 33 meters, spread over 4 seconds.

Correct: Between $t = 2$ and $t = 6$ seconds, the altitude rose 8.25 meters per second on average.: It names the interval, the units and the word "average".
At $t = 4$ seconds, the drone was rising at exactly 8.25 meters per second.: An average over an interval says nothing exact about one instant, even the midpoint; the climb may have been uneven.
The drone climbed a total of 8.25 meters between $t = 2$ and $t = 6$ seconds.: The total climb was $47 - 14 = 33$ meters; 8.25 is that climb per second.
Between $t = 2$ and $t = 6$ seconds, the drone was 8.25 meters above the street.: 8.25 is measured in meters per second, a rate, not an altitude in meters.

3. Over which interval of the table was the average rate of change of $A$ greatest?

$[2, 5]$
$[6, 9]$
$[0, 2]$
$[5, 6]$

Show the answer

Average rates: $[0, 2]$ : $\frac{8}{2} = 4$ ; $[2, 5]$ : $\frac{27}{3} = 9$ ; $[5, 6]$ : $\frac{6}{1} = 6$ ; $[6, 9]$ : $\frac{15}{3} = 5$ meters per second. The greatest is 9, on $[2, 5]$ .

Correct: $[2, 5]$ : The climb of 27 meters over 3 seconds gives 9 meters per second, the largest.
$[6, 9]$ : The altitude is highest here, but the climb is only 15 meters in 3 seconds, 5 meters per second.
$[0, 2]$ : The climb is 8 meters in 2 seconds, 4 meters per second.
$[5, 6]$ : The climb is 6 meters in 1 second, 6 meters per second: a short interval is not a fast one.

4. A student says the drone was rising at $\frac{A(5)}{5} = \frac{41}{5} = 8.2$ meters per second at $t = 5$ . What is wrong with this reasoning?

It divides an altitude by a time instead of dividing a change in altitude by a change in time.
It should divide by 6 seconds, the next time in the table.
Nothing; an altitude divided by its time is the rate at that time.
It should use $A(9)$ , the last altitude, because more data gives a better rate.

Show the answer

A rate of change compares two changes. The drone started 6 meters up, so even the average from $t = 0$ is $\frac{41 - 6}{5} = 7$ meters per second, not 8.2.

Correct: It divides an altitude by a time instead of dividing a change in altitude by a change in time.: Value over time ignores where the altitude started; rates use differences.
It should divide by 6 seconds, the next time in the table.: Dividing by a different time still divides a value, not a change.
Nothing; an altitude divided by its time is the rate at that time.: $A(5) \div 5$ is not a rate of change; it would be the average from $t = 0$ only if the drone started at altitude 0.
It should use $A(9)$ , the last altitude, because more data gives a better rate.: Data far from $t = 5$ describes other moments; a rate at $t = 5$ needs times close to 5.

5. Let $g(x) = x^2 + 3x$ . Simplify $\frac{g(2 + h) - g(2)}{h}$ for $h \ne 0$ . Type an expression in $h$ .

Type an expression for $\frac{g(2 + h) - g(2)}{h} =$ .

Show the answer

$g(2 + h)$ $= (2 + h)^2$ $+ 3(2 + h)$ $= 10$ $+ 7h$ $+ h^2$ and $g(2) = 10$ , so the quotient is $\frac{7h + h^2}{h} = 7 + h$ .

Answer: $\frac{g(2 + h) - g(2)}{h} =$ $7 + h$

6. For $g(x) = x^2 + 3x$ , find the instantaneous rate of change of $g$ at $x = 2$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\lim_{h \to 0} \frac{g(2 + h) - g(2)}{h}$ $= \lim_{h \to 0} (7 + h)$ $= 7$ .

Answer: 7

7. What is the average rate of change of $k(x) = \frac{1}{x}$ on $[1, 4]$ ?

$-\frac{1}{4}$
$-\frac{3}{4}$
$\frac{1}{4}$
$-4$

Show the answer

$\frac{k(4) - k(1)}{4 - 1}$ $= \frac{\frac{1}{4} - 1}{3}$ $= \frac{-\frac{3}{4}}{3}$ $= -\frac{1}{4}$ .

Correct: $-\frac{1}{4}$ : The change $-\frac{3}{4}$ divided by 3.
$-\frac{3}{4}$ : That is the change in $k$ ; divide by the change in $x$ , 3.
$\frac{1}{4}$ : The sign: $k$ falls from 1 to $\frac{1}{4}$ , so the rate is negative.
$-4$ : That is $\frac{4 - 1}{\frac{1}{4} - 1}$ $= \frac{3}{-\frac{3}{4}}$ , run over rise, upside down.

Part 9 · Summary

Summary

An average rate of change is a change in output over a change in input, the slope of a secant line. Writing the second input as

a + h

gives the difference quotient. Letting

h \to 0

shrinks the interval to one instant, and the limit of the difference quotient, when it exists, is the instantaneous rate of change.

Part 10 · Up next

What comes next

Part 11 · Connections