In The Difference Quotient you met the slope of a secant line, $\frac{f(x+h) - f(x)}{h}$: an average rate of change over an interval. A speedometer shows something different: how fast you are going at one instant. Calculus gets the instant from averages by letting the interval shrink, using limits.
Key idea
A ball is thrown upward; its height after $t$ seconds is $s(t) = -16t^2 + 64t$ feet. Its average velocity over a short time interval starting at $t = 1$ is:
| interval | $[1, 2]$ | $[1, 1.1]$ | $[1, 1.01]$ | $[1, 1.001]$ |
|---|---|---|---|---|
| average velocity (ft/s) | 16 | 30.4 | 31.84 | 31.984 |
The averages close in on 32. Algebra explains it: $$s(1 + h) - s(1) = -16(1 + 2h + h^2) + 64 + 64h - 48 = 32h - 16h^2$$, so the average over $[1, 1 + h]$ is $32 - 16h$. As $h \to 0$ this tends to 32. The ball's velocity at $t = 1$ is 32 ft/s.
Why it works
Worked example: An average rate from a formula
- $f(4) = 16 - 12 = 4$ and $f(1) = 1 - 3 = -2$.
- Average rate $= \frac{f(4) - f(1)}{4 - 1} = \frac{4 - (-2)}{3} = 2$.
Answer: $2$: on average, $f$ rises 2 units per unit of $x$ on this interval.
Worked example: From averages to an instant
- $$f(2 + h) = (2 + h)^2 - 3(2 + h) = 4 + 4h + h^2 - 6 - 3h = -2 + h + h^2$$.
- $f(2) = -2$, so $f(2 + h) - f(2) = h + h^2$.
- Average rate $= \frac{h + h^2}{h} = 1 + h$ (for $h \ne 0$).
- As $h \to 0$ this tends to 1.
Answer: Average rate $1 + h$; instantaneous rate $1$.
Common mistake