Rate of Change

If a quantity $y$ depends on $x$, then $\frac{dy}{dx}$ is the rate of change of $y$ with respect to $x$; when both depend on time, the chain rule connects the rates: $$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$$

The board routine: (1) write the geometric formula connecting the quantities ($A = \pi r^2$, $V = x^3$, $V = \frac{4}{3}\pi r^3$, $x^2 + y^2 = l^2$ for a ladder); (2) differentiate the whole relation with respect to $t$; (3) substitute the instant's values only after differentiating. Substituting first is the classic error — it freezes the variable and kills the rate. Always carry units: cm²/s for areas, cm³/s for volumes, and state whether the quantity is increasing (positive rate) or decreasing (negative rate).

Increasing and Decreasing Functions

On an interval, $f$ is increasing if larger inputs give larger outputs, and the derivative decides:

1. $f'(x) > 0$ on $(a, b)$ $\Rightarrow$ $f$ is increasing on $(a, b)$.
2. $f'(x) < 0$ on $(a, b)$ $\Rightarrow$ $f$ is decreasing on $(a, b)$.

The sign-table routine: compute $f'$, factor it completely, mark the roots of $f'$ on a number line, record the sign of each factor in each interval, and read off increasing/decreasing. For $f' = 6(x - 3)(x + 2)$: positive-negative-positive across $-2$ and $3$, so increasing on $(-\infty, -2)$ and $(3, \infty)$, decreasing on $(-2, 3)$. Presenting the factored derivative and the sign table is what earns the method marks.

Maxima and Minima

Critical point: $f'(c) = 0$ (or $f'$ undefined). Extreme values of differentiable functions live only at critical points — but not every critical point is an extremum ($f(x) = x^3$ at $0$ is the standing counterexample).

First derivative test: if $f'$ changes sign from positive to negative across $c$, then $c$ is a local maximum; negative to positive, a local minimum; no sign change, neither (a point of inflection).

Second derivative test — usually faster: at a critical point $c$, $$f''(c) < 0 \Rightarrow \text{local maximum}, \qquad f''(c) > 0 \Rightarrow \text{local minimum}$$ and if $f''(c) = 0$ the test is inconclusive — fall back to the first derivative test.

Absolute (global) extrema on a closed interval $[a, b]$: evaluate $f$ at every critical point inside the interval and at both endpoints; the largest value is the absolute maximum, the smallest the absolute minimum. Forgetting the endpoints is the standard mark-loser here.

Word problems — the five-step template: (1) name the variables and draw the figure; (2) write the target quantity as a function of one variable using the given constraint; (3) state the domain; (4) find critical points and classify with the second derivative test; (5) answer the actual question asked — the number, the dimensions, or the maximum value — with units. Standard results worth remembering: among rectangles of given perimeter the square has the largest area, and the open box cut from a square sheet of side $a$ has maximum volume when the cut square has side $\frac{a}{6}$.

Worked Examples — Rates of Change

Example 1 — Expanding circle

The radius of a circle is increasing at 3 cm/s. How fast is the area increasing when the radius is 10 cm?

Step 1 — relation and differentiate: $A = \pi r^2$, so $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$.

Step 2 — substitute the instant: $2\pi(10)(3) = 60\pi$.

Answer: the area grows at $60\pi$ cm²/s — differentiate first, substitute after.

Example 2 — Growing cube

The edge of a cube increases at 3 cm/s. Find the rate of increase of the volume when the edge is 10 cm.

Step 1 — relation: $V = x^3$, so $\frac{dV}{dt} = 3x^2\frac{dx}{dt}$.

Step 2 — substitute: $3(100)(3) = 900$.

Answer: 900 cm³/s.

Example 3 — Inflating balloon (rate reversed)

Air is pumped into a spherical balloon at 900 cm³/s. Find the rate at which the radius increases when the radius is 15 cm.

Step 1 — relation and differentiate: $V = \frac{4}{3}\pi r^3$ gives $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.

Step 2 — solve for the unknown rate: $900 = 4\pi(225)\frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{900}{900\pi}$.

Answer: $\dfrac{1}{\pi}$ cm/s — here the volume rate was given and the radius rate asked; the same equation serves both directions.

Example 4 — The sliding ladder

A 5 m ladder leans against a wall. Its foot is pulled away at 2 m/s. How fast is the top sliding down when the foot is 4 m from the wall?

Step 1 — relation: $x^2 + y^2 = 25$; differentiating, $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.

Step 2 — the instant: at $x = 4$, $y = \sqrt{25 - 16} = 3$, with $\frac{dx}{dt} = 2$: $$\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{4}{3}(2) = -\frac{8}{3}$$

Answer: the top slides down at $\dfrac{8}{3}$ m/s — the minus sign is the "downward" in disguise, and saying so earns the interpretation mark.

Worked Examples — Increasing and Decreasing

Example 5 — A linear warm-up

Show that $f(x) = 3x + 17$ is increasing on $\mathbb{R}$.

Step 1 — differentiate: $f'(x) = 3 > 0$ for every $x$.

Answer: increasing on all of $\mathbb{R}$ — one line, but the line must mention the sign of $f'$.

Example 6 — A quadratic split

Find the intervals on which $f(x) = x^2 - 4x + 6$ is (a) increasing, (b) decreasing.

Step 1 — differentiate and find the root: $f'(x) = 2x - 4 = 2(x - 2)$, zero at $x = 2$.

Step 2 — signs: negative for $x < 2$, positive for $x > 2$.

Answer: decreasing on $(-\infty, 2)$, increasing on $(2, \infty)$ — the vertex of the parabola is exactly the switch point.

Example 7 — The full sign-table pattern

Find the intervals on which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is increasing or decreasing.

Step 1 — differentiate and factor: $f'(x) = 6x^2 - 6x - 36 = 6(x - 3)(x + 2)$.

Step 2 — sign table across $-2$ and $3$: both factors negative-then-mixed-then-positive gives $f'$ positive on $(-\infty, -2)$, negative on $(-2, 3)$, positive on $(3, \infty)$.

Answer: increasing on $(-\infty, -2)$ and $(3, \infty)$; decreasing on $(-2, 3)$ — factored derivative plus sign table is the full-marks presentation.

Example 8 — A trigonometric interval

Show that $f(x) = \sin x$ is increasing on $\left(0, \dfrac{\pi}{2}\right)$ and decreasing on $\left(\dfrac{\pi}{2}, \pi\right)$.

Step 1 — differentiate: $f'(x) = \cos x$.

Step 2 — signs on each interval: $\cos x > 0$ on $\left(0, \frac{\pi}{2}\right)$ and $\cos x < 0$ on $\left(\frac{\pi}{2}, \pi\right)$.

Answer: as claimed — the graph rises to the crest at $\frac{\pi}{2}$ and falls after, exactly as the derivative's sign says.

Worked Examples — Maxima and Minima

Example 9 — Second derivative test

Find the local maximum and local minimum values of $f(x) = x^3 - 3x + 3$.

Step 1 — critical points: $f'(x) = 3x^2 - 3 = 0$ gives $x = \pm 1$.

Step 2 — classify: $f''(x) = 6x$; $f''(-1) = -6 < 0$ (local max), $f''(1) = 6 > 0$ (local min).

Step 3 — values: $f(-1) = 5$ and $f(1) = 1$.

Answer: local maximum value $5$ at $x = -1$; local minimum value $1$ at $x = 1$. (In general a function's local minimum value can even exceed one of its local maximum values elsewhere — "local" only compares against nearby points.)

Example 10 — Absolute extrema on a closed interval

Find the absolute maximum and minimum of $f(x) = x^3 - 12x$ on $[-3, 5]$.

Step 1 — critical points inside: $f'(x) = 3x^2 - 12 = 0$ gives $x = \pm 2$, both in the interval.

Step 2 — evaluate at critical points and endpoints: $f(-3) = 9$, $f(-2) = 16$, $f(2) = -16$, $f(5) = 65$.

Answer: absolute maximum $65$ at $x = 5$; absolute minimum $-16$ at $x = 2$ — the maximum sits at an endpoint, which is precisely why endpoints are non-negotiable in this checklist.

Example 11 — Splitting a number

Find two positive numbers $x$ and $y$ with $x + y = 60$ such that $xy^3$ is maximum.

Step 1 — one variable: $x = 60 - y$, so maximise $f(y) = (60 - y)y^3$ for $0 < y < 60$.

Step 2 — critical points: $f'(y) = 180y^2 - 4y^3 = 4y^2(45 - y) = 0$ gives $y = 45$ (the root $y = 0$ is out of domain).

Step 3 — classify: $f'$ changes from positive to negative across $45$ — a maximum.

Answer: $x = 15$, $y = 45$.

Example 12 — The open box

From a square tin sheet of side 18 cm, squares of side $x$ are cut from the corners and the flaps folded up to form an open box. Find $x$ for maximum volume, and the maximum volume.

Step 1 — the volume function: $V(x) = x(18 - 2x)^2$, for $0 < x < 9$.

Step 2 — critical points: $V'(x) = (18 - 2x)^2 + x \cdot 2(18 - 2x)(-2) = (18 - 2x)(18 - 6x)$, zero at $x = 9$ (boundary, rejected) and $x = 3$.

Step 3 — classify and evaluate: $V'$ changes positive to negative at $3$ — maximum; $V(3) = 3(12)^2 = 432$.

Answer: cut squares of side 3 cm for a maximum volume of 432 cm³ — matching the general rule $x = \frac{a}{6}$ with $a = 18$.

Example 13 — The square beats all rectangles

Show that among all rectangles with a given perimeter, the square has the largest area.

Step 1 — one variable: perimeter $P$ fixed; sides $x$ and $\frac{P}{2} - x$, so $A(x) = x\left(\frac{P}{2} - x\right)$.

Step 2 — critical point: $A'(x) = \frac{P}{2} - 2x = 0$ gives $x = \frac{P}{4}$.

Step 3 — classify: $A''(x) = -2 < 0$ — a maximum, and then both sides equal $\frac{P}{4}$.

Answer: the rectangle of maximum area is the square of side $\frac{P}{4}$. ∎ A one-page proof the board asks in both directions (given perimeter, and its twin: given area, the square minimises perimeter).