Average and Instantaneous Rates of Change

How fast is something changing? Over a stretch of time, that’s an average rate of change: total change divided by time. But a car’s speedometer shows how fast you’re going right now, at one instant. Calculus starts exactly here: we use limits to turn average rates into an instantaneous rate of change, which is the big idea behind the derivative.

Key ideas

Average rate of change

The average rate of change of $f$ on the interval from $x = a$ to $x = b$ is

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

On a graph, this is the slope of the secant line through the points $(a, f(a))$ and $(b, f(b))$.

The units are “output units per input unit”. If $f(t)$ is litres of water and $t$ is in minutes, the average rate of change is in litres per minute.

The difference quotient

If the second point is a small step $h$ away from $a$, then $b = a + h$ and the average rate of change becomes the difference quotient:

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

It is still the slope of a secant line, just written with $h$ instead of $b$.

Instantaneous rate of change

Now slide the second point toward the first, so $h$ gets closer and closer to $0$. The secant lines approach the tangent line, and their slopes approach the instantaneous rate of change at $x = a$:

$$\text{instantaneous rate at } a = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

You can’t just substitute $h = 0$, because that gives $\tfrac{0}{0}$. Instead, simplify the quotient first (usually the $h$ cancels), then take the limit.

Secant lines through $P(1, 1)$ on $y = x^2$. As the second point moves toward $P$, the slopes $4, 3, 2.5, \dots$ approach the tangent slope $2$.

On the next page, this limit gets its official name: the derivative.

	Average rate of change	Instantaneous rate of change
Over	an interval $[a, b]$	a single point $x = a$
Graph	slope of a secant line	slope of the tangent line
Formula	$\dfrac{f(b) - f(a)}{b - a}$	$\displaystyle\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Worked examples

Example 1: Average rate from a formula

Find the average rate of change of $f(x) = x^2 - 3x$ on the interval $[1, 4]$.

Solution. Find the outputs at the endpoints:

$$f(4) = 16 - 12 = 4 \qquad f(1) = 1 - 3 = -2$$ $$\frac{f(4) - f(1)}{4 - 1} = \frac{4 - (-2)}{3} = \frac{6}{3} = 2$$

The secant line from $x = 1$ to $x = 4$ has slope $2$.

Example 2: Average rate from a table

Water drains from a tank. The volume $V(t)$, in litres, after $t$ minutes is shown below.

$t$ (min)	$0$	$2$	$5$	$9$
$V(t)$ (L)	$120$	$104$	$86$	$62$

Find the average rate of change of $V$ from $t = 2$ to $t = 9$, and explain what it means.

Solution.

$$\frac{V(9) - V(2)}{9 - 2} = \frac{62 - 104}{7} = \frac{-42}{7} = -6 \text{ L/min}$$

From $t = 2$ to $t = 9$ minutes, the volume decreases by an average of $6$ litres per minute. The negative sign means the volume is going down.

Example 3: Secant slopes closing in

For $f(x) = x^2$, estimate the instantaneous rate of change at $x = 1$ using secant slopes, then find it exactly.

Solution. The secant slope from $x = 1$ to $x = 1 + h$ is $\dfrac{(1 + h)^2 - 1}{h}$. Try smaller and smaller $h$, from both sides:

$h$	$2$	$1$	$0.5$	$0.1$	$0.01$	$-0.01$	$-0.1$
secant slope	$4$	$3$	$2.5$	$2.1$	$2.01$	$1.99$	$1.9$

The slopes are closing in on $2$. To get the exact value, simplify first and then take the limit:

$$\begin{aligned} \frac{(1 + h)^2 - 1}{h} &= \frac{1 + 2h + h^2 - 1}{h} \\ &= \frac{h(2 + h)}{h} \\ &= 2 + h && (h \ne 0) \end{aligned}$$ $$\lim_{h \to 0} (2 + h) = 2$$

The instantaneous rate of change at $x = 1$ is exactly $2$. These are the secant lines in the figure above.

Example 4: Instantaneous velocity

A ball is thrown upward. Its height in metres after $t$ seconds is $s(t) = 20t - 4.9t^2$. Find its instantaneous velocity at $t = 1$.

Solution. Velocity is the rate of change of height. Start with the difference quotient at $t = 1$. First, $s(1) = 20 - 4.9 = 15.1$, and

$$\begin{aligned} s(1 + h) &= 20(1 + h) - 4.9(1 + h)^2 \\ &= 20 + 20h - 4.9 - 9.8h - 4.9h^2 \\ &= 15.1 + 10.2h - 4.9h^2 \end{aligned}$$

So

$$\frac{s(1 + h) - s(1)}{h} = \frac{10.2h - 4.9h^2}{h} = 10.2 - 4.9h$$ $$\lim_{h \to 0} (10.2 - 4.9h) = 10.2$$

At $t = 1$ s, the ball is rising at $10.2$ m/s.

Common mistakes

Subtracting in a different order on the top and bottom. $\dfrac{f(b) - f(a)}{b - a}$ is right, and so is $\dfrac{f(a) - f(b)}{a - b}$. Mixing them, like $\dfrac{f(b) - f(a)}{a - b}$, flips the sign.

Substituting $h = 0$ too early. That always gives $\tfrac{0}{0}$, which tells you nothing. Expand and simplify until the $h$ in the denominator cancels, then let $h \to 0$.

Expanding $f(a + h)$ wrong. $(1 + h)^2$ is $1 + 2h + h^2$, not $1 + h^2$. Put $a + h$ in brackets everywhere $x$ appears, then expand carefully.

Leaving out units or meaning. On the AP exam, a rate in context needs units and a sentence: “the volume is decreasing at an average rate of $6$ litres per minute from $t = 2$ to $t = 9$”.

Averaging the outputs instead of finding the rate. The average rate of change is the change in output divided by the change in input. It is not $\tfrac{f(a) + f(b)}{2}$.

Practice

1. (Warm-up) Find the average rate of change of $g(x) = 2x^2 + 1$ on $[0, 3]$.

Solution

$g(3) = 19$ and $g(0) = 1$, so

$$\frac{g(3) - g(0)}{3 - 0} = \frac{19 - 1}{3} = 6$$

2. (Warm-up) The temperature outside was $4$ °C at 6:00 a.m., $10$ °C at 9:00 a.m., and $13$ °C at noon. Find the average rate of change of the temperature from 6:00 a.m. to noon, with units.

Solution

From 6:00 a.m. to noon is $6$ hours:

$$\frac{13 - 4}{6} = \frac{9}{6} = 1.5 \text{ °C per hour}$$

On average, the temperature rose $1.5$ °C per hour.

3. (Warm-up) Simplify the difference quotient $\dfrac{f(a + h) - f(a)}{h}$ for $f(x) = 5x - 2$. Why does your answer make sense?

Solution

$$\frac{5(a + h) - 2 - (5a - 2)}{h} = \frac{5a + 5h - 2 - 5a + 2}{h} = \frac{5h}{h} = 5$$

The graph is a line with slope $5$, so every secant line (and the tangent line) has slope $5$, no matter which points you pick.

4. (Core) For $f(x) = x^3$, find the average rate of change on $[1, 1.1]$ and on $[1, 1.01]$. Use your answers to guess the instantaneous rate of change at $x = 1$.

Solution

$$\frac{1.1^3 - 1}{0.1} = \frac{1.331 - 1}{0.1} = 3.31$$$$\frac{1.01^3 - 1}{0.01} = \frac{1.030301 - 1}{0.01} = 3.0301$$

The slopes are approaching $3$, so the instantaneous rate at $x = 1$ appears to be $3$.

5. (Core) Use the limit of the difference quotient to find the instantaneous rate of change of $f(x) = x^2 + 4x$ at $x = 2$.

Solution

$f(2) = 12$ and $f(2 + h) = (2 + h)^2 + 4(2 + h) = 4 + 4h + h^2 + 8 + 4h = 12 + 8h + h^2$.

$$\lim_{h \to 0} \frac{12 + 8h + h^2 - 12}{h} = \lim_{h \to 0} \frac{h(8 + h)}{h} = \lim_{h \to 0} (8 + h) = 8$$

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

Solution

Combine the fractions in the numerator over a common denominator:

$$\begin{aligned} \frac{\frac{1}{2 + h} - \frac{1}{2}}{h} &= \frac{\frac{2 - (2 + h)}{2(2 + h)}}{h} \\ &= \frac{-h}{2h(2 + h)} \\ &= \frac{-1}{2(2 + h)} \end{aligned}$$$$\lim_{h \to 0} \frac{-1}{2(2 + h)} = -\frac{1}{4}$$

7. (Core) A bacteria population is modelled by $P(t) = 200 + 30t^2$, where $t$ is in hours.

• (a) Find the average rate of change of $P$ from $t = 2$ to $t = 5$.
• (b) Find the instantaneous rate of change at $t = 2$.

Solution

(a) $P(5) = 200 + 750 = 950$ and $P(2) = 200 + 120 = 320$:

$$\frac{950 - 320}{5 - 2} = \frac{630}{3} = 210 \text{ bacteria per hour}$$

(b) $P(2 + h) - P(2) = 30(2 + h)^2 - 30(4) = 30(4h + h^2)$, so

$$\lim_{h \to 0} \frac{30(4h + h^2)}{h} = \lim_{h \to 0} 30(4 + h) = 120$$

At $t = 2$ hours, the population is growing at $120$ bacteria per hour.

8. (Challenge) Find the instantaneous rate of change of $f(x) = \sqrt{x}$ at $x = 4$. (Hint: multiply the top and bottom by the conjugate.)

Solution

$$\begin{aligned} \frac{\sqrt{4 + h} - 2}{h} &= \frac{\sqrt{4 + h} - 2}{h} \cdot \frac{\sqrt{4 + h} + 2}{\sqrt{4 + h} + 2} \\ &= \frac{(4 + h) - 4}{h\left(\sqrt{4 + h} + 2\right)} \\ &= \frac{1}{\sqrt{4 + h} + 2} \end{aligned}$$$$\lim_{h \to 0} \frac{1}{\sqrt{4 + h} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$$

9. (Challenge) Let $f(x) = x^2$.

• (a) Show that the average rate of change of $f$ on $[a, b]$ is $a + b$.
• (b) Find $b \gt 1$ so that the average rate of change on $[1, b]$ is $7$.

Solution

(a) Factor the difference of squares:

$$\frac{b^2 - a^2}{b - a} = \frac{(b - a)(b + a)}{b - a} = a + b$$

(b) $1 + b = 7$, so $b = 6$. Check: $\dfrac{36 - 1}{6 - 1} = \dfrac{35}{5} = 7$.