Rates of Change in Applied Contexts Other Than Motion · 除运动以外的实际应用中的变化率
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| economics/ˌiːkəˈnɒmɪks/ | 经济学 | jīng jì xué |
Rates are everywhere, not just motion
- A derivative measures how fast something changes — and that "something" needn't be position.
- Money, populations, temperature, chemical concentration: all change at rates you can model with a derivative.
- The same $f'(a)$ = rate idea applies; only the units and story change.
- This lesson practices reading derivatives across many fields.
变化率无处不在,不只是运动
- 导数度量某物变化多快——而那个"某物"不必是位置。
- 金钱、人口、温度、化学浓度:都以你能用导数建模的速率变化。
- 同样的 $f'(a)$ = 变化率的想法适用;只有单位和故事在变。
- 这一课练习在许多领域里读导数。
Economics, biology, physics — same tool
- Economics 经济学: if $C(x)$ is cost, $C'(x)$ is marginal cost — the cost of one more unit ($\tfrac{\$}{\text{unit}}$).
- Biology: if $P(t)$ is a population, $P'(t)$ is the growth rate ($\tfrac{\text{organisms}}{\text{time}}$).
- Physics: if $Q(t)$ is charge, $Q'(t)$ is current ($\tfrac{\text{coulombs}}{\text{second}}$ = amperes).
- Whatever the field, the derivative's units are always output per input.
经济学、生物学、物理学——同一个工具
- 经济学:若 $C(x)$ 是成本,$C'(x)$ 是边际成本——多生产一件的成本($\tfrac{\$}{\text{item}}$)。
- 生物学:若 $P(t)$ 是人口,$P'(t)$ 是增长率($\tfrac{\text{organisms}}{\text{time}}$)。
- 物理学:若 $Q(t)$ 是电荷,$Q'(t)$ 是电流($\tfrac{\text{coulombs}}{\text{s}}$ = 安培)。
- 无论哪个领域,导数的单位永远是输出每输入。
$C(x)$ is cost in dollars for $x$ items. The units of the marginal cost $C'(x)$ are... · $C(x)$ 是生产 $x$ 件物品的成本（美元）。边际成本 $C'(x)$ 的单位是...
Output per input: dollars per item. · 输出比输入：美元/件。
In economics, the derivative of a cost function is called the ____ cost. · 在经济学中，成本函数的导数被称为____成本。
Marginal cost = cost of one more unit. · 边际成本 = 多生产一个单位的成本。
Match each quantity to the meaning of its derivative. · 将每个数量与其导数的含义匹配。
The derivative's units are always output-per-input for the field. · 导数的单位总是该领域的输出比输入。
A rate function describes behavior over time
- Often the derivative is itself a function of time, $f'(t)$, describing how fast the quantity changes at each moment.
- Where $f'(t)$ is large, the quantity changes fast; where $f'(t)\approx0$, it is nearly steady.
- A graph of $f'(t)$ tells the story of the change — peaks are moments of fastest change.
- Reading that story is a core exam skill.
变化率函数描述随时间的行为
- 导数本身常是时间的函数 $f'(t)$,描述每一刻该量变化多快。
- $f'(t)$ 大之处,该量变化得快;$f'(t)\approx0$ 之处,它几乎稳定。
- $f'(t)$ 的图讲述变化的故事——峰值是变化最快的时刻。
- 读懂那个故事是核心的考试技能。
A draining tank's volume · 排水罐的体积
y = ax² + c
This falling curve is volume $V(t)$; its (negative) slope is the drain rate $V'(t)$ — steeper means draining faster. · 这条下降曲线是体积 $V(t)$；其（负）斜率是排水速率 $V'(t)$ —— 越陡峭意味着排水越快。
Sign and size, in context
- Positive rate: the quantity is growing (population rising, tank filling).
- Negative rate: shrinking (cooling, draining, a declining balance).
- The magnitude compares speeds: a rate of $-50$ is faster change than $-5$.
- Always pair the number with a units-carrying sentence — that's what earns the marks.
情境中的符号与大小
- 正变化率:该量在增长(人口上升、水箱注满)。
- 负变化率:在缩减(冷却、排空、余额下降)。
- 大小比较快慢:$-50$ 的变化比 $-5$ 快。
- 永远把数字与一句带单位的话配对——那才能拿分。
For · 支持 $V(t)=100-4t^2$, the rate is $V'(t)=-8t$. Find $V'(3)$ (litres per minute). · 对于 $V(t)=100-4t^2$，速率是 $V'(t)=-8t$。求 $V'(3)$（升/分钟）。
$V'(3)=-8(3)=-24$; draining at $24$ L/min. · $V'(3)=-8(3)=-24$；以 $24$ L/min 的速度排水。
A negative rate of change means the modeled quantity is decreasing. · 负变化率意味着被建模的量正在减少。
Negative derivative → the quantity falls. · 负导数 → 该量下降。
Two balances change at rates $-5$ and $-50$ dollars/day. Which is changing faster? · 两个余额分别以 $-5$ 和 $-50$ 美元/天的速率变化。哪一个变化更快？
Magnitude sets speed: $|-50|>|-5|$, so $-50$ changes faster. · 绝对值决定快慢：$|-50|>|-5|$，所以 $-50$ 变化更快。
Keep units straight, especially in economics. Marginal cost $C'(x)$ has units of dollars per item, not dollars — it's the cost of the next item, not the total. Reporting a rate as if it were an amount (or dropping the units entirely) loses the meaning the question is testing.
保持单位清晰,尤其在经济学里。边际成本 $C'(x)$ 的单位是每件多少美元,而非美元——它是下一件的成本,不是总成本。把变化率当作数量来报告(或干脆丢掉单位)会失去题目所考的含义。
A tank drains so its volume is $V(t)=100-4t^2$ litres ($t$ in minutes).
- Rate of change: $V'(t)=-8t\ \tfrac{\text{L}}{\text{min}}$.
- At $t=3$: $V'(3)=-24\ \tfrac{\text{L}}{\text{min}}$.
- Interpretation: at $3$ minutes, the water is draining out at $24$ litres per minute (negative = decreasing).
一个水箱排水,体积为 $V(t)=100-4t^2$ 升($t$ 以分钟计)。
- 变化率:$V'(t)=-8t\ \tfrac{\text{L}}{\text{min}}$。
- 在 $t=3$:$V'(3)=-24\ \tfrac{\text{L}}{\text{min}}$。
- 解读:在第 $3$ 分钟,水以每分钟 $24$ 升排出(负 = 递减)。
A derivative models a rate of change in any field — economics (marginal cost), biology (growth rate), physics (current) — always with units of output per input. The sign says grow vs. shrink; the magnitude compares speeds. Interpret each rate in a full, units-carrying sentence.
导数在任何领域都建模变化率——经济学(边际成本)、生物学(增长率)、物理学(电流)——单位永远是输出每输入。符号说增长还是缩减;大小比较快慢。用一句完整、带单位的话解读每个变化率。