Solving Related Rates Problems · 求解相关变化率问题
A repeatable recipe for related rates
- Related-rates word problems feel hard until you follow a fixed procedure.
- Same five steps every time: picture, equation, differentiate, substitute, solve.
- The one rule people break: substitute numbers only after differentiating.
- Master the recipe and every related-rates problem becomes routine.
相关变化率的可复用套路
- 相关变化率应用题在你遵循固定步骤前会显得很难。
- 每次都是同样的五步:画图、方程、求导、代入、求解。
- 人们最常破的一条规则:只在求导之后才代入数字。
- 掌握这个套路,每道相关变化率题都变成例行公事。
The five steps
- 1. Draw & label the changing quantities; note what's given and wanted.
- 2. Relate the variables with an equation (geometry/formula).
- 3. Differentiate both sides with respect to $t$ (chain rule → rate factors).
- 4. Substitute the known values at that instant.
- 5. Solve for the unknown rate; report with units and sign.
五个步骤
- 1. 画图并标注变化的量;记下已知与待求。
- 2. 用方程联系变量(几何/公式)。
- 3. 关于 $t$ 求导两边(链式法则 → 变化率因子)。
- 4. 代入那一瞬间的已知值。
- 5. 求解未知变化率;带单位和符号报告。
Volume grows fastest when big · 当半径较大时体积增长最快
y = a·x³ (V vs r)
Because · 因为 $\tfrac{dV}{dt}=4\pi r^2\tfrac{dr}{dt}$, a fixed radius rate makes the volume rate soar as $r$ grows. · 因为 $\tfrac{dV}{dt}=4\pi r^2\tfrac{dr}{dt}$，固定的半径变化率会使体积变化率随 $r$ 增大而飙升。
Put the related-rates steps in order. · 将相关变化率的步骤按顺序排列。
Picture, relate, differentiate, substitute-and-solve. · 作图、建立关系、求导、代入并求解。
Differentiate first, plug in last
- A number that is changing must stay a variable while you differentiate.
- If you set $r=5$ before differentiating, its rate $\tfrac{dr}{dt}$ vanishes — and the whole method collapses.
- Substitute the instant's values only at step 4, after the derivative is taken.
- The only things you may fix early are true constants (like a ladder's fixed length).
先求导,最后才代入
- 一个正在变化的数,在你求导时必须保持为变量。
- 若你在求导之前就设 $r=5$,它的变化率 $\tfrac{dr}{dt}$ 就消失了——整个方法崩溃。
- 只在第 4 步、求完导之后,才代入那一瞬间的值。
- 唯一可以提前固定的是真正的常数(如梯子固定的长度)。
With $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$, $r=5$, and $\dfrac{dr}{dt}=2$, find $\dfrac{dV}{dt}$ as a multiple of $\pi$ (enter the number before $\pi$). · 给定 $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$、$r=5$ 和 $\dfrac{dr}{dt}=2$，求 $\dfrac{dV}{dt}$ 作为 $\pi$ 的倍数（输入 $\pi$ 前面的数字）。
$4\pi(25)(2)=200\pi$.
When should you substitute the instant's numeric values? · 何时应代入瞬时数值？
Differentiate first; substitute numbers only afterward, or the rate terms vanish. · 先求导；仅在之后代入数值，否则变化率项会消失。
Setting $r=5$ before differentiating is a valid shortcut. · 求导前设定 $r=5$ 是一个有效的捷径。
It kills $\tfrac{dr}{dt}$ and breaks the method. · 这会消除 $\tfrac{dr}{dt}$ 并破坏该方法。
Check units and reasonableness
- The answer is a rate — attach units of output per input (e.g. $\tfrac{\text{cm}^3}{\text{s}}$).
- Check the sign: growing quantity → positive; shrinking → negative.
- Ask if the size is reasonable for the situation.
- A tidy conclusion sentence ("the volume is increasing at ... per second") earns the interpretation marks.
检查单位与合理性
- 答案是一个变化率——附上输出每输入的单位(如 $\tfrac{\text{cm}^3}{\text{s}}$)。
- 检查符号:增长的量 → 正;缩减的 → 负。
- 问一问对这个情形,大小是否合理。
- 一句整洁的结论("体积以每秒……增长")能拿到解读分。
A related-rates answer is a rate, so it must be reported with its ____. · 相关变化率的答案是一个变化率，因此必须附带其 ____ 报告。
E.g. cm³ per second. · 例如：立方厘米每秒。
A related-rates answer comes out $\dfrac{dV}{dt}=-30\,\tfrac{\text{cm}^3}{\text{s}}$. This means the volume is... · 相关变化率的答案为 $\dfrac{dV}{dt}=-30\,\tfrac{\text{cm}^3}{\text{s}}$。这意味着体积...
A negative rate means the volume is shrinking at $30$ cm³/s. · 负速率意味着体积正在以 $30$ cm³/s 的速度缩小。
The number-one related-rates mistake: plugging in the instant's values before differentiating. Keep every changing quantity symbolic through the differentiation step; substitute only afterward. Substituting early kills the rate terms and gives a wrong (often zero) answer.
相关变化率的头号错误:在求导前就代入那一瞬间的值。在求导这一步之前,让每个变化的量保持符号形式;之后再代入。过早代入会杀掉变化率项,给出错误(常为零)的答案。
A spherical balloon's radius grows at $\tfrac{dr}{dt}=2\,\tfrac{\text{cm}}{\text{s}}$. How fast is the volume growing when $r=5$ cm?
- Relate: $V=\tfrac43\pi r^3$. Differentiate: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
- Now substitute $r=5$, $\tfrac{dr}{dt}=2$: $\dfrac{dV}{dt}=4\pi(25)(2)=200\pi$.
- The volume is increasing at $200\pi\approx628\ \tfrac{\text{cm}^3}{\text{s}}$.
一个球形气球的半径以 $\tfrac{dr}{dt}=2\,\tfrac{\text{cm}}{\text{s}}$ 增大。当 $r=5$ cm 时,体积增长多快?
- 联系:$V=\tfrac43\pi r^3$。求导:$\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$。
- 现在代入 $r=5$、$\tfrac{dr}{dt}=2$:$\dfrac{dV}{dt}=4\pi(25)(2)=200\pi$。
- 体积以 $200\pi\approx628\ \tfrac{\text{cm}^3}{\text{s}}$ 增长。
Solve a related rates problem in five steps: draw/label, write a relating equation, differentiate with respect to $t$, substitute the instant's values only after differentiating, and solve for the unknown rate. Report it with correct units and sign, and check the result is reasonable.
用五步解相关变化率题:画图/标注、写联系方程、关于 $t$ 求导、只在求导后代入那一瞬间的值、求解未知变化率。带正确的单位和符号报告,并检查结果是否合理。