Defining Average and Instantaneous Rates of Change at a Point
Introduced| English | Meaning |
|---|---|
| average rate of change/ˈævrɪdʒ reɪt ɒv tʃeɪndʒ/ | The change in a function's output divided by the change in its input across an interval. |
| instantaneous rate of change/ˌɪnstənˈteɪnɪəs reɪt ɒv tʃeɪndʒ/ | The rate of change at a particular input value, given by the derivative where it exists. |
| difference quotient/ˈdɪfrəns ˈkwəʊʃənt/ | The change in a function's output divided by the corresponding change in its input over an interval. |
| secant line/ˈsiːkənt laɪn/ | A straight line passing through two distinct points on a curve. |
| tangent line/ˈtændʒənt laɪn/ | A straight line through a point on a curve with the same gradient as the curve at that point. |
From "over an interval" to "at an instant"
- Unit 1 built the idea of a limit. Now we use it to measure change.
- Over an interval, the average rate of change 平均变化率 is total change ÷ interval width.
- At a single instant, the instantaneous rate of change 瞬时变化率 is what a speedometer shows.
- The bridge between them is a limit — and that bridge is the whole point of Unit 2.
Average rate = slope of a secant
- The average rate of change of $f$ from $x=a$ to $x=b$ is the difference quotient 差商:
-
$$\frac{f(b)-f(a)}{b-a}$$
- Geometrically this is the slope of the secant line 割线 joining $\big(a,f(a)\big)$ and $\big(b,f(b)\big)$.
- It is an average: it smooths over everything that happened between $a$ and $b$.
Find the average rate of change of $f(x)=x^2$ on $[1,3]$.
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$\dfrac{9-1}{3-1}=\dfrac{8}{2}=4$.
The average rate of change over an interval equals the slope of the...
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Average rate = rise over run between the two endpoints = secant slope.
Shrink the interval
- To get the rate at $x=a$, slide $b$ toward $a$ and watch the secant slope.
- As the interval shrinks, the secant pivots toward the tangent line 切线 at $a$.
- The limiting slope is the instantaneous rate of change at $a$.
- No time actually passes at an instant — the limit is what makes "rate at a point" meaningful.
Slide the secant into a tangent
y = x²
Move the point along $y=x^2$ — the tangent slope you read off is the instantaneous rate, the limit of the secant slopes.
The instantaneous rate of change at a point equals the slope of the ____ line there.
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It is the limit of secant slopes as the interval shrinks — the tangent slope.
Reading the rate in context
- The sign tells direction: positive rate → increasing, negative → decreasing.
- The size tells speed: a bigger magnitude means faster change.
- Units matter: for position in metres over time in seconds, the rate is in $\tfrac{\text{m}}{\text{s}}$ — a velocity.
- So average rate = secant slope (over an interval); instantaneous rate = tangent slope (at a point).
If the average rate of change of $f$ on $[0,4]$ is $0$, then $f$ must be constant on $[0,4]$.
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It only means $f(0)=f(4)$; $f$ could rise then fall (e.g. go out and come back).
For $f(x)=x^2$, average rates on $[3,3.1]$ and $[3,3.01]$ are $6.1$ and $6.01$. Estimate the instantaneous rate at $x=3$.
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The averages close in on $6$.
A position function has instantaneous rate $-5\,\tfrac{\text{m}}{\text{s}}$ at time $t$. Select all correct readings.
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Negative rate = moving in the negative direction (so position decreases); its magnitude is the speed; and it is the tangent slope.
Average and instantaneous rates are not interchangeable. The average rate over $[a,b]$ can be zero while the instantaneous rate is large at every interior point (imagine going out and coming back). Only in the limit of a shrinking interval does the average rate become the instantaneous rate.
For $f(x)=x^2$, find the average rate on $[2,4]$ and estimate the instantaneous rate at $x=2$.
- Average on $[2,4]$: $\dfrac{f(4)-f(2)}{4-2}=\dfrac{16-4}{2}=6$.
- Shrink toward $2$: on $[2,2.1]$, $\dfrac{4.41-4}{0.1}=4.1$; on $[2,2.01]$, $4.01$.
- The average rates close in on $4$ — the instantaneous rate at $x=2$ is $4$.
The average rate of change over $[a,b]$ is the difference quotient $\frac{f(b)-f(a)}{b-a}$ — the slope of a secant line. Shrinking the interval, the secant approaches the tangent line, whose slope is the instantaneous rate of change at that point.
Write from a secant slope of 8.1
- f(x) = x squared.
- The secant runs from 4 to 4.1.
What is the difference quotient, and what is the instantaneous rate at 4?
A short secant is close to the derivative 8.
The slope of x squared from 4 to 4.1 is what?
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(16.81 - 16) / 0.1 = 8.1.