Defining Average and Instantaneous Rates of Change at a Point

A ball is thrown upward from a 20-meter ledge, and its height after $t$ seconds is $s(t) = 20 + 15t - 4.9t^2$ meters. Asking how fast it is rising over the first two seconds has an easy answer: divide the change in height by the elapsed time. Asking how fast it is rising at exactly $t = 1$ is harder, because at a single instant no time passes and nothing changes, so the division becomes $\frac{0}{0}$. Calculus answers the second question by taking a limit of the first.

Average rate of change: the difference quotient

The average rate of change of $f$ over an interval from $a$ to $b$ is the change in output divided by the change in input:

Geometrically, it is the slope of the secant line, the line through the two points $(a, f(a))$ and $(b, f(b))$ on the graph. Its units are output units per input unit: for height in meters and time in seconds, meters per second.

For the ball, over $[0, 2]$:

This says the ball gained height at 5.2 meters per second on average over those two seconds. It does not say the ball was moving at that speed at any particular moment. The ball rose faster at first and slower later.

When the interval starts at a fixed point $a$, two ways of naming the other endpoint are standard, and they give the same quotient:

• Call the other endpoint $x$. The average rate over the interval between $a$ and $x$ is $\dfrac{f(x) - f(a)}{x - a}$.
• Call the length of the step $h$, so the other endpoint is $a + h$. The average rate is $\dfrac{f(a + h) - f(a)}{h}$.

Either expression is called a difference quotient. Substituting $x = a + h$ turns the first into the second. The step $h$ may be negative, which puts the second point to the left of $a$; the quotient is still the slope of the secant line between the two points. It is never allowed to be $0$, since then there would be no interval and the quotient would be $\frac{0}{0}$.

From average to instantaneous: the limit

To get the rate at exactly $t = 1$, compute the average rate over the interval from $1$ to $1 + h$ for a general $h$, and then watch what happens as the interval shrinks. Expand first:

Since $s(1) = 30.1$,

Dividing by $h$ is legal because $h \neq 0$. Now the quotient is a simple expression in $h$, and you can see where it goes:

From both sides, the average rates close in on $5.2$, and the formula $5.2 - 4.9h$ shows exactly why: the error is $4.9h$, which goes to $0$ with $h$. So at the instant $t = 1$, the ball is rising at 5.2 meters per second.

Look at what the limit did. Plugging $h = 0$ into the original quotient gives $\frac{0}{0}$, which carries no information. The algebra cancels the factor of $h$ that causes the $\frac{0}{0}$ (the same move as the limits of topic 1.6), and then direct substitution into $5.2 - 4.9h$ finishes the job.

The figure shows the same thing geometrically. As the second point $Q$ slides along the curve toward $P$, the secant line through $P$ and $Q$ rotates, and its slope approaches the slope of the tangent line at $P$, the line that touches the curve at $P$ and matches its direction there. Topic 2.2 writes equations for tangent lines.

"Provided the limit exists" is a real condition. For $f(x) = |x|$ at $a = 0$, the difference quotient is $\frac{|h|}{h}$, which equals $1$ for every $h > 0$ and $-1$ for every $h < 0$. The one-sided limits disagree, so there is no instantaneous rate of change at $0$, which matches the sharp corner in the graph. Topic 2.4 sorts out when derivatives fail to exist.

Reading a limit as a derivative

Because the definition has a fixed shape, a limit in that shape can be recognized as a derivative without evaluating it. In

the pieces match $\frac{f(a + h) - f(a)}{h}$ with $f(x) = x^2$ and $a = 4$, because $(4 + h)^2$ is $f(4 + h)$ and $16$ is $f(4)$. So the limit is $f'(4)$, the instantaneous rate of change of $x^2$ at $x = 4$, which is also the slope of the parabola $y = x^2$ at the point $(4, 16)$. Expanding gives $\frac{8h + h^2}{h} = 8 + h \to 8$.

The test is that the subtracted term must be the same function evaluated at the base point. In $\lim\limits_{x \to 4}\frac{x^2 - 16}{x - 4}$, the term $16$ is $f(4)$ for $f(x) = x^2$ and the denominator is $x - 4$, so this is the same derivative in the other form, and factoring $x^2 - 16 = (x - 4)(x + 4)$ gives $8$ again.

Worked examples

Where students lose points

On the exam

Average rates of change appear constantly in free-response contexts, usually as "find the average rate of change of $f$ over the interval" followed by "using correct units, interpret." Multiple-choice questions often present a limit in the shape of the definition and ask for its value; when the function is one whose derivative you know, recognize it rather than expanding. Questions that say "use the definition of the derivative" require the limit, written with $\lim$ on every line until $h$ (or $x - a$) is gone.