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Calculus AB · Unit 1 · 1.1
AP Calculus · Limits and Continuity
Can Change Occur at an Instant?
An average rate needs an interval. Shrink the interval to nothing and the limit answers the question.
key idea$\text{Instantaneous rate}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$
Four views of one idea
An average rate of change needs an interval — it is a change divided by a change. At a single instant both are zero, and $\tfrac00$ is meaningless. Calculus answers the question by taking averages over shrinking intervals and asking what they approach.
Graphical
Numerical| interval | $\Delta s$ | avg rate |
|---|
| $[1,3]$ | $40$ | $20$ |
| $[1,1.5]$ | $6.25$ | $12.5$ |
| $[1,1.05]$ | $0.5125$ | $10.25$ |
| $[1,1.001]$ | $0.010005$ | $10.005$ |
Analytical$s(t)=5t^{2}$, so the average rate on $[1,1+h]$ is
$\dfrac{5(1+h)^{2}-5}{h}=\dfrac{10h+5h^{2}}{h}=10+5h$ for $h\neq0$.
As $h\to0$ this approaches $\mathbf{10}$ m/s.
Verbal“Over shorter and shorter intervals beginning at $t=1$, the average speed settles on $10$ m/s. That number is the speed at the instant $t=1$.”
Translating between views
The same curve at three zoom levels. Each pair of dots defines a secant whose slope is an average rate. As the interval shrinks the arc between them straightens, and the secant slopes close in on a single number — the instantaneous rate. | Average rate of change | Instantaneous rate of change |
|---|
| needs | an interval $[a,b]$ | a single point $a$ |
| formula | $\dfrac{f(b)-f(a)}{b-a}$ | $\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ |
| on the graph | slope of the secant line | slope of the tangent line |
| at a single instant | undefined — gives $\tfrac00$ | well defined, via the limit |
| units | output units per input unit | the same units |
The right-hand column is the whole of differential calculus. Everything later is a way of computing that limit efficiently.
the core move
- An average rate divides a change by a change — both vanish at a single instant.
- Shrinking the interval is what makes the answer well defined.
- Secant slopes approach the tangent slope; that is the picture behind the limit.
- The instantaneous rate carries the same units as the average rate.
Worked examples
EX 1
Average rate from a formula
$[1,3]$For $s(t)=5t^{2}$ metres, find the average velocity on $[1,3]$.
1
Evaluate both endpoints
$s(1)=5$ and $s(3)=45$.
2
Apply the formula
$\dfrac{s(3)-s(1)}{3-1}=\dfrac{45-5}{2}$.
3
Compute
$=\dfrac{40}{2}=20$.
4
State with units
The average velocity on $[1,3]$ is $20$ m/s — metres divided by seconds.
EX 2
Shrink the interval
numerical → valueUsing the table above, estimate the instantaneous velocity at $t=1$.
1
Read the trend
Average rates $20$, $12.5$, $10.25$, $10.005$ as the interval shrinks.
2
Identify what they approach
The values are closing in on $10$.
3
State the estimate
The instantaneous velocity at $t=1$ is approximately $10$ m/s.
4
Note the status
This is an estimate from finitely many intervals. The algebra in the next example makes it exact.
EX 3
Make it exact
analytical → valueFor $s(t)=5t^{2}$, compute the instantaneous velocity at $t=1$ exactly.
1
Set up the average over $[1,1+h]$
$\dfrac{s(1+h)-s(1)}{h}=\dfrac{5(1+h)^{2}-5}{h}$.
2
Expand the numerator
$5\left(1+2h+h^{2}\right)-5=10h+5h^{2}$.
3
Cancel the $h$
$\dfrac{10h+5h^{2}}{h}=10+5h$ for $h\neq0$ — and a limit never uses $h=0$.
4
Take the limit
$\lim_{h\to0}(10+5h)=10$.
5
Answer
The velocity at $t=1$ is exactly $10$ m/s, confirming the numerical estimate.
EX 4
Why the naive question fails
verbal → reasoningA student argues: “At the instant $t=1$ no time passes and no distance is covered, so the speed must be $\tfrac00$, which is undefined. Instantaneous speed is meaningless.” Respond.
1
Grant what is correct
The observation is right: at a single instant $\Delta t=0$ and $\Delta s=0$, and $\tfrac00$ is indeed undefined.
2
Locate the flaw
The argument assumes instantaneous speed must be computed at the instant. Calculus never does that.
3
Give the replacement
It is defined as a limit of averages over intervals of positive length, as that length shrinks toward zero.
4
Point at the evidence
Those averages are $20$, $12.5$, $10.25$, $10.005$ — a clear, single destination even though the interval never actually reaches zero.
5
Conclude
The quantity is well defined precisely because the limit exists. The $\tfrac00$ form is what the limit is designed to sidestep.
Watch out
Average rate needs two points
“The average rate at $t=1$” is not a thing. If a question says average, look for the interval.
$h\neq0$ throughout the algebra
Cancelling $h$ from $\tfrac{10h+5h^{2}}{h}$ is legal because a limit never evaluates at $h=0$. Say so if asked to justify.
Secant and tangent are different lines
A secant cuts the curve at two points; the tangent touches at one. Average rate is a secant slope, instantaneous rate a tangent slope.
Carry the units
Metres per second, litres per minute, dollars per unit. A rate stated without units is an incomplete answer on the free-response section.