Average and Instantaneous Rates of… — AP® Calculus AB Notes

Unit 1 asked whether change can happen at an instant and built the tool to answer it: the limit. This page puts that tool to work. You will measure change over an interval, shrink the interval, and let a limit finish the job. Everything in Unit 2 grows out of the one formula you meet here.

Average rate of change: a change over a change

Suppose a delivery drone is 14 meters above the street at $t = 2$ seconds and 47 meters up at $t = 6$ seconds. It climbed $47 - 14 = 33$ meters in $6 - 2 = 4$ seconds, so on average it climbed $\frac{33}{4} = 8.25$ meters per second. That number is an average rate of change.

For any function $f$ on an interval $[a, b]$ :

$\text{average rate of change} = \frac{f(b) - f(a)}{b - a}.$

Three habits keep this right every time:

Subtract in the same order on top and bottom. Later value minus earlier value, over later time minus earlier time.
Divide by the change in input. The difference $f(b) - f(a)$ is how much $f$ changed; it is not yet a rate.
Name the units. The units are output units per input unit: meters per second, gallons per minute, dollars per item.

An average describes a whole interval. The drone might have climbed fast at first and slowly later; 8.25 meters per second hides that. It also does not tell you how high the drone was. A sentence that interprets an average rate names the quantity, its units and the interval: "Between $t = 2$ and $t = 6$ seconds, the drone's altitude increased by an average of 8.25 meters per second."

Worked example 1. The volume of water in a tank is $V(t)$ gallons after $t$ minutes, with $V(3) = 120$ and $V(8) = 95$ . Find the average rate of change of $V$ on $[3, 8]$ and interpret it.

Change in volume: $V(8) - V(3)$ $= 95 - 120$ $= -25$ gallons. Change in time: $8 - 3 = 5$ minutes.

Average rate: $\frac{-25}{5} = -5$ gallons per minute.

Meaning: between $t = 3$ and $t = 8$ minutes, the volume of water decreased by an average of 5 gallons per minute. The negative sign means the tank was losing water.

The picture: a secant line

Plot the two points $(a, f(a))$ and $(b, f(b))$ on the graph of $f$ and draw the line through them. That line is a secant line. Its slope is rise over run, $\frac{f(b) - f(a)}{b - a}$ , which is exactly the average rate of change. So every average rate you compute is the slope of a secant line, and every secant slope you read from a graph is an average rate.

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. — **Figure 1.** Secant lines from $P$ on the graph of $f(x) = \frac{x^2}{4} + 1$ , with their slopes. As $Q$ moves toward $P$ , the slopes close in on 0.5. LevlPrep original diagram.

Shrinking the interval: the difference quotient

To learn how fast $f$ changes at one input $x = a$ , look at shorter and shorter intervals that start at $a$ . Write the far end as $a + h$ , where $h$ is the width of the interval. The average rate on that interval is the difference quotient:

$\frac{f(a + h) - f(a)}{h}.$

The number $h$ can be positive (an interval to the right of $a$ ) or negative (an interval to the left). Look at the figure: for $f(x) = \frac{x^2}{4} + 1$ and $a = 1$ , the secant slopes for $h = 3, 2, 1, 0.4$ are $1.25, 1, 0.75, 0.6$ . As $h$ shrinks, the second point slides toward $P$ and the secant lines pivot toward one line. Their slopes close in on a single number.

The instantaneous rate of change is a limit

You cannot simply put $h = 0$ into the difference quotient: that gives $\frac{f(a) - f(a)}{0} = \frac{0}{0}$ , an indeterminate form. Instead, take the limit. The instantaneous rate of change of $f$ at $x = a$ is

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

when that limit exists. The method is the limit algebra of Unit 1: expand, simplify until the factor $h$ cancels, then let $h \to 0$ .

Worked example 2. Find the instantaneous rate of change of $f(x) = \frac{x^2}{4} + 1$ at $x = 1$ .

Values: $f(1) = \frac{1}{4} + 1 = \frac{5}{4}$ and $f(1 + h)$ $= \frac{(1 + h)^2}{4}$ $+ 1$ $= \frac{1 + 2h + h^2}{4}$ $+ 1$ .

Difference: $f(1 + h) - f(1)$ $= \frac{2h + h^2}{4}$ . Keep the middle term $2h$ when you square $1 + h$ ; dropping it is the most common slip.

Quotient, for $h \ne 0$ : $\frac{2h + h^2}{4h}$ $= \frac{h(2 + h)}{4h}$ $= \frac{2 + h}{4}$ .

Limit: $\lim_{h \to 0} \frac{2 + h}{4}$ $= \frac{2}{4}$ $= \frac{1}{2}$ . The rate of change of $f$ at $x = 1$ is $\frac{1}{2}$ , the number the secant slopes in the figure were closing in on.

Worked example 3. A cart rolls so that it has traveled $s(t) = 5t^2$ meters after $t$ seconds. How fast is it moving at $t = 2$ ?

Difference quotient: $\frac{s(2 + h) - s(2)}{h}$ $= \frac{5(4 + 4h + h^2) - 20}{h}$ $= \frac{20h + 5h^2}{h}$ $= 20$ $+ 5h$ .

For $h = 0.1$ the average is 20.5 meters per second; for $h = 0.01$ it is 20.05. The limit is $\lim_{h \to 0} (20 + 5h)$ $= 20$ , so at the instant $t = 2$ the cart is moving at 20 meters per second.

Average versus instantaneous

	Average rate of change	Instantaneous rate of change
Describes	an interval $[a, b]$	one input $x = a$
Formula	$\frac{f(b) - f(a)}{b - a}$	$\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$
On the graph	slope of a secant line	the limit of secant slopes
Needs	two values of $f$	a limit
Units	output per input	output per input

Both rates have the same units. The difference is what they describe: a stretch of time, or one moment.

Common mistakes

A value over a time is not a rate. The drone's altitude at $t = 5$ divided by 5 ignores where it started. A rate always compares two changes.
Substituting $h = 0$ too early. That gives $\frac{0}{0}$ , which says only that more algebra is needed.
Losing the middle term. $(a + h)^2 = a^2 + 2ah + h^2$ .
Treating an average as an instant. An average on $[2, 6]$ is not the rate at $t = 4$ .

The next topic gives the instantaneous rate its own name and notation, and turns it into a whole new function.