Week 10 Cheatsheet — Applications of Differentiation

How this week breaks down

Three techniques, all applications of $f^{\prime }(x)$. Skim this once, then revise from the in-depth note.

Topic	What you do
Critical points	Solve $f^{\prime }(x)=0$ (and find where $f^{\prime }$ is undefined). Classify with $f^{\prime \prime }$ or sign-change.
Optimization	Translate the word problem into a single-variable function. Critical-point that function.
Related rates	Differentiate both sides of a geometric equation w.r.t. time. Plug in the instant.

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1 — Finding critical points

Definition. A critical point of $f(x)$ is an $x$ where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined.

Classification table

Test	Verdict
$f^{\prime \prime }(c)>0$	local min at $c$
$f^{\prime \prime }(c)<0$	local max at $c$
$f^{\prime \prime }(c)=0$	inconclusive — use first-derivative sign change

First-derivative (sign change) test

Inspect $f^{\prime }$ just to the left and just to the right of $c$:

• $-,+$  →  min
• $+,-$  →  max
• same sign on both sides  →  neither

Worked snippets

Function	Critical-point work	Classification
$f(x)=x^3-6x+1$	$f^{\prime }(x)=3x^2-6=0\Rightarrow x=\pm \sqrt{2}$	$f^{\prime \prime }(x)=6x$, so min at $(\sqrt{2},1-4\sqrt{2})$ and max at $(-\sqrt{2},1+4\sqrt{2})$
$f(x)=x\ln (x)$, $x>0$	$f^{\prime }(x)=1+\ln (x)=0\Rightarrow x=\frac{1}{e}$	$f^{\prime \prime }(1/e)=e>0$, so min at $(1/e,-1/e)$
$f(x)=3x^5-5x^3$	Critical points at $x=-1,0,1$	Min at $(1,-2)$, max at $(-1,2)$, and $x=0$ is neither after a sign check on $f^{\prime }$

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2 — Optimization

The five-step recipe

1. Define variables and a sketch.
2. Write the constraint linking them (e.g. $xy=192$).
3. Write the objective to minimise/maximise (e.g. $S=x+3y$).
4. Use the constraint to reduce to one variable.
5. Critical-point that function. Don’t forget endpoints if the domain is bounded.

Standard setups

Problem	Constraint	Objective
Two positive numbers, product known	$xy=k$	linear combination $S=ax+by$
Rectangle with a wall	$2x+y=P$	area $A=xy$
Open-top box, square base	$s^2+4sh=S$	volume $V=s^{2}h$
Open-top tank, fixed width	$4Lh=V$	\text{cost}=(\text{base }\/\text{m}^2)\cdot\text{area}+(\text{side }$/\text{m}^2)\cdot\text{area}$
Open cylinder	$\pi r^{2}h=V$	$\text{cost}=(\text{base})\pi r^2+(\text{side})2\pi rh$
Pole-rope / shoreline pipe	distance variables	$L=\sqrt{x^2+a^2}+\sqrt{(d-x{)}^2+b^2}$

Quick exemplars

Problem	Model	Result
Rectangle, $120\text{m}$ fence, one side is wall	$A=x(120-2x)$	Max at $x=30\text{m}$; area $1800{\text{m}}^2$
Open-top box, surface $108{\text{cm}}^2$	$V=s^{2}h$ with $s^2+4sh=108$	$s=6\text{cm}$, $h=3\text{cm}$, $V=108{\text{cm}}^3$
Open-top tank, $36{\text{m}}^3$, width $4\text{m}$	$4Lh=36$	$3\times 4\times 3\text{m}$ at \330$
Open cylinder, $24\pi {\text{m}}^3$	$\pi r^{2}h=24\pi$	$r=2\text{m}$, $h=6\text{m}$, cost \5.65$

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3 — Related rates

The recipe

1. Write the geometric / physical relationship without time (e.g. $A=\pi r^2$).
2. Differentiate both sides with respect to $t$, attaching $dQ/dt$ for every time-varying quantity.
3. Plug in the values at the instant of interest.
4. Solve for the unknown rate. Sign matters.

Standard geometric derivatives

Quantity	Relationship	$dQ/dt$
Circle area	$A=\pi r^2$	$2\pi r\dot{r}$
Square area	$A=s^2$	$2s\dot{s}$
Cube volume	$V=s^3$	$3s^2\dot{s}$
Sphere volume	$V=\frac{4}{3}\pi r^3$	$4\pi r^2\dot{r}$
Cone volume ($r$ fixed)	$V=\frac{1}{3}\pi r^{2}h$	$\frac{1}{3}\pi r^2\dot{h}$

(Using $\dot{r}=dr/dt$ shorthand.)

Worked snippets

Problem	Substitute	Rate
Artery	$\dot{r}=0.02\text{mm/month}$, $r=3\text{mm}$	$\dot{A}=2\pi r\dot{r}=0.12\pi {\text{mm}}^2/\text{month}$
Dissolving cube	$\dot{s}=-0.5\text{mm/min}$, $s=8.2\text{mm}$	$\dot{V}=3s^2\dot{s}=-100.86{\text{mm}}^3/\text{min}$

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Common mistakes (all three sub-topics)

1. Treating $f^{\prime \prime }(c)=0$ as conclusive. It is inconclusive — fall back to the first-derivative test.
2. Differentiating after substituting numeric values in related rates. Numbers are constant; their derivative is zero. Differentiate symbolically first, then plug in.
3. Forgetting the constraint in optimisation — you can’t differentiate a two-variable expression with single-variable calculus.
4. Skipping endpoints on a closed-interval optimisation.
5. Dropping the sign or the units in the final answer.

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Key formulas

$$f^{\prime }(c)=0\text{(critical point)}{f}^{\prime \prime }(c)>0\Rightarrow \text{min}{f}^{\prime \prime }(c)<0\Rightarrow \text{max}$$ $$\text{Optimise:}\text{objective}(x)\stackrel{d/dx}{\to }0\stackrel{\text{classify}}{\to }\text{answer}$$ $$\text{Related rate:}Q=g(P)\stackrel{d/dt}{\to }\dot{Q}=g^{\prime }(P)\dot{P}$$

For the why and many more worked examples, see the in-depth note.