微分在实际情境中的应用

Once you can compute derivatives, you use them to describe the real world. The derivative $f'(x)$ is the instantaneous rate of change 瞬时变化率 of $f$ with respect to 关于 its input. Reading and reporting this rate correctly is a graded skill.

Units matter. The unit of $f'(x)$ is the unit of $f$ divided by the unit of $x$. If $C(t)$ is a number of acres and $t$ is in weeks, then $C'(t)$ is in acres per week. On the exam, “Using correct units, interpret the meaning of $g'(140)$” wants a full sentence: the value, the quantity, the rate word “per”, and the moment. For example: “$g'(140)=2.3$ means that at $x=140$, the quantity is increasing at about $2.3$ units per unit of $x$.”

Common mistakes. Reading $g'(140)$ as an amount (“the quantity is $2.3$”) instead of a rate; leaving out the units or the moment ($x=140$); and calling a negative derivative “increasing”.

For a particle 质点 moving on a line, three functions of time are linked by differentiation 求导:

• position 位置 $s(t)$;
• velocity 速度 $v(t)=s'(t)$ – signed; its sign gives direction;
• acceleration 加速度 $a(t)=v'(t)=s''(t)$.

Key readings (frequent exam parts):

• The particle is at rest 静止 when $v(t)=0$.
• It moves right/up when $v(t)>0$ and left/down when $v(t)<0$; it changes direction 改变方向 where $v$ changes sign.
• Speed 速率 is $|v(t)|$. Speed is increasing when $v$ and $a$ have the same sign (the particle is speeding up 加速), and decreasing when they have opposite signs.

Distinguish carefully between velocity (has direction) and speed (does not) – the exam tests this exact difference.

Common mistakes. Saying the particle speeds up because $a>0$: compare the signs of both $v$ and $a$. Saying it changes direction wherever $v=0$: $v$ must change sign. With $v(t)=(t-2)^2$ the particle stops for an instant at $t=2$, then keeps moving the same way.

Worked example. A particle moves with $s(t)=t^3-6t^2+9t$. Then $v(t)=3(t-1)(t-3)$, so it is at rest at $t=1$ and $t=3$ and changes direction at each. At $t=2$, $v=-3<0$ and $a(2)=6(2)-12=0$; just after, $a>0$ while $v<0$, so the particle is slowing down there.

The same derivative idea models any changing quantity: a draining tank, a spreading population, a cooling cup. Whenever a problem says "the rate at which...", it is describing a derivative. Read the units to know which quantity's rate you have, then interpret in context.

In a related rates 相关变化率 problem, several quantities change together over time, and you know some rates but want another. The engine is the chain rule 链式法则: differentiate a relationship with respect to time $t$. Every variable becomes a function of $t$, so each derivative picks up a “$\,/\,dt$” factor. The product rule 乘积法则 and quotient rule 商法则 may also be needed.

A reliable procedure – and a full-credit template on the exam:

1. Name the variables and write down the given rates and the unknown rate (e.g. “$\dfrac{dh}{dt}=-2$ cm/day, find $\dfrac{dV}{dt}$”).
2. Write an equation relating the quantities (often a geometric 几何 or volume 体积 formula).
3. Differentiate both sides with respect to $t$ (chain rule) – before substituting numbers.
4. Substitute the known values at the instant 时刻 of interest, and solve for the unknown rate.
5. State the answer with units and the correct sign (a decreasing quantity has a negative rate).

Substituting numbers too early is the classic error: differentiate the general relationship first, then plug in.

Worked example. A spherical balloon's volume grows at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$. From $V=\tfrac43\pi r^3$, differentiate first: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. At $r=5$, $100=4\pi(25)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt}=\dfrac{1}{\pi}\approx0.32\ \text{cm/s}$.

Worked example: sliding ladder. A $10$ m ladder leans against a wall. Its foot slides away from the wall at $0.5$ m/s. How fast is the top sliding down the wall when the foot is $6$ m from the wall?

1. Variables: $x$ = distance from wall to foot, $y$ = height of the top. Given $\dfrac{dx}{dt}=0.5$ m/s; find $\dfrac{dy}{dt}$ when $x=6$.
2. Equation (Pythagoras' theorem 勾股定理): $x^2+y^2=100$.
3. Differentiate with respect to $t$: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$.
4. At this instant $y=\sqrt{100-36}=8$, so $2(6)(0.5)+2(8)\dfrac{dy}{dt}=0$, which gives $\dfrac{dy}{dt}=-\dfrac{6}{16}=-0.375$ m/s.
5. The top is sliding down at $0.375$ m/s (the negative sign means $y$ is decreasing).

Note that $y=8$ came from the equation at that instant; it is not given in the question.

Common mistakes. Treating a constant as a variable or a variable as a constant: the ladder's length $10$ never changes (its derivative is $0$), but $x$ and $y$ both do. Dropping the sign: report $\dfrac{dy}{dt}=-0.375$ m/s, or say the top slides down at $0.375$ m/s, never “down at $-0.375$ m/s”.

Near a point of tangency 切点, a smooth curve 光滑曲线 looks like its tangent line 切线 – this is local linearity 局部线性. So the tangent line gives a linear approximation 线性近似 (linearization 线性化) of the function near that point:

Use it to estimate $f$ at an $x$ close to $a$.

Over- or underestimate? The answer follows from concavity 凹凸性. If the graph is concave up 上凹 near $a$ (it curves above its tangent), the tangent-line value is an underestimate 低估. If it is concave down 下凹, the tangent line lies above the curve, giving an overestimate 高估. Exam parts test this reasoning, so justify with the sign of $f''$.

Worked example. Estimate $\sqrt{4.1}$. Take $f(x)=\sqrt{x}$ and $a=4$: $f(4)=2$, and $f'(x)=\dfrac{1}{2\sqrt{x}}$ gives $f'(4)=\tfrac14$. Then

Here $f''(x)=-\dfrac{1}{4x^{3/2}}<0$, so the graph is concave down and $2.025$ is an overestimate (the true value is $2.0248\ldots$).

Common mistake. Using the slope at the new point, $f'(4.1)$, instead of the known point, $f'(4)$. The whole idea is that $f(a)$ and $f'(a)$ are easy to find exactly.

When direct substitution 直接代入 in a quotient of limits gives the indeterminate form 未定式 $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, you may use L'Hospital's Rule 洛必达法则:

provided the numerator 分子 and denominator 分母 are differentiable 可导 near the limiting input (excluding the input itself), the denominator derivative is nonzero there, and the derivative-ratio limit exists, finitely or infinitely. Use the corresponding one-sided 单侧 conditions when needed. Differentiate the top and bottom separately (this is not the quotient rule), then try the limit again. First confirm the form really is $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ – applying the rule to any other form is a mistake.

Worked example. $\displaystyle\lim_{x\to0}\frac{\sin x}{x}$ gives $\tfrac00$, so differentiate top and bottom: $\displaystyle\lim_{x\to0}\frac{\cos x}{1}=1$. And $\displaystyle\lim_{x\to0}\frac{e^{2x}-1}{x}$ is also $\tfrac00$; it becomes $\displaystyle\lim_{x\to0}\frac{2e^{2x}}{1}=2$.

Worked example. Evaluate $\displaystyle\lim_{x\to \infty}\frac{x^2}{e^{x}}$. Both top and bottom grow without bound: the form is $\tfrac{\infty}{\infty}$. Apply the rule, check the new form, and apply it again:

This proves the Unit 1 claim that an exponential 指数函数 beats any polynomial 多项式. Check the form before every use: once it stops being $\tfrac00$ or $\tfrac{\infty}{\infty}$, stop and substitute.

• In motion problems: velocity is the derivative of position, acceleration the derivative of velocity; speed increases when velocity and acceleration share a sign.
• For related rates, differentiate the relating equation with respect to time, then substitute the given values last.
• Use the tangent line for a linear approximation near a known point; it is accurate only close by.
• Read the sign of a rate: $\tfrac{dy}{dt}>0$ means $y$ is increasing at that instant, $\tfrac{dy}{dt}<0$ means it is decreasing. A falling height or a shrinking volume has a negative rate.
• Always state units and interpret the answer in context.