Math Core

A speedometer shows how fast you're going right now, but every measurement of speed you can actually make uses two moments: distance traveled divided by time elapsed. Calculus closes that gap. In this lesson you'll see how the average rate of change over an interval turns into an instantaneous rate of change at a single point, using the limits you learned in Unit 1.

Average rate of change

Suppose a quantity $y = f(x)$ changes as $x$ goes from $a$ to $b$ . The average rate of change compares the change in output to the change in input.

Definition

Average rate of change

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

\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}

Geometrically, it is the slope of the secant line through the points $\big(a, f(a)\big)$ and $\big(b, f(b)\big)$ .

The units of an average rate are always "output units per input unit." If $s(t)$ is position in meters and $t$ is time in seconds, the average rate of change of $s$ is an average velocity in meters per second.

Worked example: Average rate from a formula

Find the average rate of change of $f(x) = x^3 - 2x$ on $[1, 3]$ .

$f(3) = 27 - 6 = 21$ and $f(1) = 1 - 2 = -1$ , so

\frac{f(3) - f(1)}{3 - 1} = \frac{21 - (-1)}{2} = \frac{22}{2} = 11

On average, $f$ increases 11 units for every 1 unit increase in $x$ on this interval.

From secant lines to a tangent line

An average rate hides everything that happens inside the interval. To learn how fast $f$ is changing at exactly $x = a$ , make the interval smaller and smaller. Let the second point be $a + h$ , where $h$ is a small nonzero number. The average rate on the interval from $a$ to $a + h$ is the difference quotient

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

As $h \to 0$ , the second point slides toward the first, and the secant line pivots toward a single line that just touches the curve at $x = a$ : the tangent line.

The curve y = x², the secant line through x = 1 and x = 3 (dashed, slope 4), and the tangent line at x = 1 (slope 2).Open in grapher →

In the graph, the secant line from $x = 1$ to $x = 3$ has slope $\dfrac{9 - 1}{3 - 1} = 4$ . Bring the right-hand point closer to $x = 1$ and the slope drops toward 2, the slope of the tangent line.

Instantaneous rate of change

The instantaneous rate of change of $f$ at $x = a$ is the limit of average rates of change as the interval shrinks to zero width:

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

provided this limit exists. It equals the slope of the tangent line to the graph of $f$ at $x = a$ .

You can't just plug in $h = 0$ : that gives $\dfrac{0}{0}$ . This is exactly the kind of indeterminate limit you simplified algebraically in Unit 1.

Worked example: Watching the average rates settle down

A ball's position is $s(t) = 4t^2$ meters after $t$ seconds. Estimate its velocity at $t = 1$ using shorter and shorter intervals.

interval	$\Delta s$	$\Delta t$	average velocity (m/s)
$[1, 1.1]$	$4.84 - 4 = 0.84$	$0.1$	$8.4$
$[1, 1.01]$	$4.0804 - 4 = 0.0804$	$0.01$	$8.04$
$[1, 1.001]$	$4.008004 - 4 = 0.008004$	$0.001$	$8.004$

The average velocities approach 8, so the instantaneous velocity at $t = 1$ appears to be $8$ m/s.

To confirm it exactly, simplify the difference quotient:

\begin{aligned} \frac{s(1 + h) - s(1)}{h} &= \frac{4(1 + 2h + h^2) - 4}{h} \\ &= \frac{8h + 4h^2}{h} = 8 + 4h \quad (h \ne 0) \end{aligned}

As $h \to 0$ , $8 + 4h \to 8$ . The velocity at $t = 1$ is exactly $8$ m/s.

Estimating from a table

On the AP exam you'll often get a function only as a table of values. You can't take a limit of a table, but you can estimate an instantaneous rate with the average rate over the smallest interval that contains the point. When the point sits between two data values, use the two values that bracket it.

Worked example: Rate of change from data

The temperature $T(t)$ of a cup of coffee, in degrees Fahrenheit, is measured $t$ minutes after it is poured.

$t$ (minutes)	0	2	5	9
$T(t)$ (°F)	180	168	153	137

Estimate the rate at which the temperature is changing at $t = 4$ .

The data values closest to $t = 4$ on either side are $t = 2$ and $t = 5$ :

\frac{T(5) - T(2)}{5 - 2} = \frac{153 - 168}{3} = -5

At $t = 4$ , the temperature is decreasing at about $5$ degrees Fahrenheit per minute. The negative sign means the temperature is going down; the units are °F per minute.

Common mistake

Don't confuse the average rate of change with the average of the function's values. The average rate is $\dfrac{f(b) - f(a)}{b - a}$ , a slope. Adding $f(a)$ and $f(b)$ and dividing by 2 answers a completely different question.

Tip

Always attach units and a direction word when you interpret a rate: "decreasing at 5 °F per minute at $t = 4$ " earns credit on an AP free-response question; " $-5$ " by itself usually doesn't.

Practice

Practice 1

Find the average rate of change of $f(x) = x^2 + 3x$ on the interval $[0, 2]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the average rate of change of $g(x) = \sqrt{x}$ on $[4, 9]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A particle moves along a line so that its position at time $t$ is $s(t) = t^3 - t$ . What is the average velocity of the particle on the interval $1 \le t \le 3$ ?

Practice 4

Water drains from a tank. The volume $V(t)$ , in liters, is recorded at selected times $t$ , in minutes.

$t$ (minutes)	0	2	5	9
$V(t)$ (liters)	40	34	25	13

Use the data to estimate the rate of change of $V$ at $t = 3.5$ , in liters per minute.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Let $f(x) = x^2 - 4x$ . Find the instantaneous rate of change of $f$ at $x = 5$ by evaluating $\displaystyle \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find the instantaneous rate of change of $f(x) = \dfrac{1}{x}$ at $x = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Which expression gives the instantaneous rate of change of a function $f$ at $x = 3$ ?