Contextual Applications of Differentiation

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.

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

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.

Worked example (estimating a rate from a table). Water is pumped into a tank. The volume $W(t)$, in litres, is measured at times $t$ in hours.

$t$ (hours)	$0$	$4$	$9$	$15$
$W(t)$ (litres)	$200$	$260$	$320$	$350$

Estimate $W'(6)$. Use the two measurements that bracket $t=6$:

Interpretation: at $t=6$ hours, the volume of water in the tank is increasing at about $12$ litres per hour. A full answer names the quantity, the time, increasing or decreasing, the size and the units.

A rate of a rate. "Is the rate at which water enters increasing at $t=6$?" asks about the sign of $W''(6)$, not $W'(6)$. Here the rate is $15$ L/h on $[0,4]$, $12$ L/h on $[4,9]$ and $5$ L/h on $[9,15]$, so the data suggest the rate is decreasing.

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).

Worked example. Air is pumped into a spherical balloon so its volume grows at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$. How fast is the radius growing when $r=5\ \text{cm}$? Start from $V=\tfrac43\pi r^3$ and differentiate with respect to $t$ first:

Substituting $r=5$ too early is the classic error – differentiate the general relationship, then plug in.

Worked example: conical tank. A tank is a cone with its vertex down, $12$ ft deep with a top radius of $3$ ft. Water drains out at $2$ cubic feet per minute. How fast is the depth falling when the water is $4$ ft deep?

The volume of the water is $V=\tfrac13\pi r^2h$, which has two changing variables. The similar triangles give $\dfrac rh=\dfrac{3}{12}$, so $r=\dfrac h4$. Substitute before differentiating:

Now substitute the instant: $\dfrac{dV}{dt}=-2$ (draining) and $h=4$:

The depth is falling at about $0.637$ feet per minute. Substituting $h=4$ before differentiating would turn $h$ into a constant and give $\dfrac{dV}{dt}=0$, which is impossible.

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 under? The tangent line lies below a concave-up curve and above a concave-down curve. So if $f''>0$ near $a$, the linear approximation is an underestimate; if $f''<0$, an overestimate.

Worked example. $f(2)=5$, $f'(2)=-1$ and $f''(x)>0$ for all $x$. Estimate $f(2.1)$: $L(2.1)=5+(-1)(0.1)=4.9$. Because $f$ is concave up, the estimate $4.9$ is an underestimate of $f(2.1)$.

Worked example. Estimate $\sqrt{4.1}$. Take $f(x)=\sqrt{x}$ and $a=4$: $f(4)=2$ and $f'(x)=\dfrac{1}{2\sqrt{x}}$ so $f'(4)=\dfrac14$. Then $L(4.1)=2+\tfrac14(4.1-4)=2.025$ (the true value is $2.0248\ldots$). Because $\sqrt{x}$ is concave down 下凹, the tangent lies above the curve, so this is a slight overestimate 高估 – justify over/under with the sign of $f''$.

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. Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$. Substituting gives $\tfrac00$, so differentiate top and bottom: $\displaystyle\lim_{x\to 0}\frac{\cos x}{1}=\cos 0=1$ – confirming the famous limit from Unit 1.

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.

Worked example (from given values). $f$ is differentiable with $f(2)=3$ and $f'(2)=8$. Find $\displaystyle\lim_{x\to2}\frac{f(x)-3}{x^2-4}$.

First show the form: $f$ is continuous, so $\displaystyle\lim_{x\to2}\big(f(x)-3\big)=3-3=0$, and $\displaystyle\lim_{x\to2}(x^2-4)=0$. The form is $\tfrac00$, so

(The last step needs $f'$ continuous at $2$, which exam questions provide; then the derivative-ratio limit exists.) Writing "$=\tfrac00$" as if it were a number, or skipping the check of both limits, loses the point for L'Hospital's Rule.

• 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.
• A tangent-line estimate is an underestimate where the curve is concave up, an overestimate where it is concave down.

Practice.

1. $P(t)$ is the number of people in a queue, $t$ in minutes. Interpret $P'(10)=-4$.
2. A particle has $s(t)=t^3-9t^2+24t$. When is it at rest, and is it speeding up or slowing down at $t=3$?
3. A circle's radius grows at $0.5$ cm/s. How fast is its area growing when $r=6$ cm?
4. A $1.8$ m person walks away from a $6$ m lamppost at $1.4$ m/s. How fast is the length of their shadow growing?
5. Use linearization to estimate $\sqrt[3]{8.3}$, and say whether it is too big or too small.
6. Evaluate $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}$.

Answers.

1. At $t=10$ minutes, the number of people in the queue is decreasing at $4$ people per minute.
2. $v=3t^2-18t+24=3(t-2)(t-4)$: at rest at $t=2$ and $t=4$. At $t=3$, $v=-3<0$ and $a=6t-18=0$, so it is neither speeding up nor slowing down at that instant.
3. $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(6)(0.5)=6\pi\approx18.8$ cm²/s.
4. Similar triangles: $\dfrac{s}{1.8}=\dfrac{x+s}{6}$, so $4.2s=1.8x$ and $s=\tfrac37x$. Then $\dfrac{ds}{dt}=\tfrac37(1.4)=0.6$ m/s.
5. $f(x)=x^{1/3}$, $a=8$: $f'(8)=\tfrac1{12}$, so $\sqrt[3]{8.3}\approx2+\tfrac{0.3}{12}=2.025$. $f''<0$ for $x>0$, so this is too big.
6. $\tfrac00$: $\displaystyle\lim_{x\to0}\frac{\sin x}{2x}$, still $\tfrac00$: $\displaystyle\lim_{x\to0}\frac{\cos x}{2}=\frac12$.