Rates of Change
Introduced| English | Meaning |
|---|---|
| rate of change/reɪt ɒv tʃeɪndʒ/ | The change in one quantity divided by the corresponding change in another. |
| 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. |
| interval/ˈɪntəvl/ | A continuous set of real numbers between specified bounds, which may include or exclude its endpoints. |
| slope/sləʊp/ | The change in a line's vertical coordinate divided by the corresponding change in its horizontal coordinate. |
| secant line/ˈsiːkənt laɪn/ | A straight line passing through two distinct points on a curve. |
| constant/ˈkɒnstənt/ | A quantity whose value remains fixed within a given problem, expression or program. |
How fast, exactly?
- A car's speedometer answers a question the odometer cannot: not how far, but how fast right now.
- "How fast is the output changing?" is the central question of this whole course.
- The answer is a number called a rate of change 变化率.
- This lesson measures it two ways: over a stretch, and at a single instant.
Average rate of change
- Over an interval 区间 from $x = a$ to $x = b$, compare how much the output changed to how much the input changed.
- The average rate of change 平均变化率 is $\dfrac{f(b) - f(a)}{b - a}$ — change in output over change in input.
- It is a ratio, so it always carries units: metres per second, dollars per year, °C per minute.
- Example in words: if temperature rose 12°C over 4 hours, the average rate was 3°C per hour.
The average rate of change of $f$ over the interval $[a, b]$ equals…
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It is the change in output divided by the change in input — a ratio, so it carries units of output per unit of input.
It is the slope of a secant line
- Draw the straight line joining the two endpoints $(a, f(a))$ and $(b, f(b))$: that is the secant line 割线.
- Its slope 斜率 — rise over run — is exactly the average rate of change.
- Steeper secant → faster average change; a downhill secant → a negative rate.
A runner's distance is $f(t)$ metres. If $f(2) = 10$ and $f(5) = 40$, what is the average speed over $[2, 5]$?
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Average speed $= \dfrac{40 - 10}{5 - 2} = \dfrac{30}{3} = 10$ m/s — the secant slope over that interval.
On a graph, the average rate of change over $[a, b]$ is the slope of the…
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The secant line joins $(a, f(a))$ and $(b, f(b))$; its slope is exactly the average rate of change.
The rate at a single point
- A secant needs two points. To get the rate at just one point, shrink the interval.
- Take average rates over smaller and smaller intervals around the point.
- As the interval closes toward zero, the secant tips over into a tangent — its slope is the rate at that exact point.
- This "rate at an instant" is the idea that grows into calculus next year.
Shrink the interval until the secant becomes a tangent
f(x) = 0.5x²
Drag the point to read the slope at a single spot — this is the limit of the average rate of change as the interval shrinks to zero.
To estimate the rate of change at a single point, average rates are taken over very ____ intervals around it.
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As the interval shrinks toward zero, the secant slope approaches the slope at that exact point.
Sign, size, and units
- A positive rate means the output is rising; a negative rate means it is falling; zero means no change.
- A bigger number (ignoring sign) means faster change.
- For a straight line the rate is the same everywhere — it is constant 常数.
- Always state the units: a rate of change with no units is only half an answer.
A negative average rate of change means the output increased over the interval.
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False. A negative ratio means the output ended lower than it started, so the function decreased on average over the interval.
Select all true statements about average rate of change.
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It can be positive, negative, or zero. Everything else is true: it is a slope, it carries units, and it approximates a point rate over small intervals.
Average rate of change is about the two endpoints only — it ignores everything in between. A hiker who climbs then descends back to the same height has an average rate of zero, even though they were moving the whole time.
A car travels 150 km between 1:00 and 3:00.
- Average speed $= \dfrac{150 \text{ km}}{2 \text{ h}} = 75$ km/h.
- That is the secant slope of the distance–time graph over those two hours.
- The car's actual speed changed constantly — 75 km/h is just the average.
The average rate of change over $[a, b]$ is $\dfrac{f(b) - f(a)}{b - a}$, the slope of the secant line through the endpoints. Shrink the interval toward zero and it becomes the rate at a single point. Always report the units.
Write from an average rate of 5
- f of 1 is 3 and f of 4 is 18.
- Find the average rate of change.
A runner's distance is f of t. f of 1 is 3 and f of 4 is 18. What is the average rate of change?
15 divided by 3 is 5.
What is the average rate of change?
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Output change over input change is 5.