Applications of differentiation (VCE Maths Methods Unit 2)

What this dot point is asking

VCAA wants you to apply differentiation to function analysis: finding stationary points and classifying them, deriving tangent and normal equations, and solving optimisation problems by constructing a single-variable function and finding its extremum.

Stationary points

A point on $y = f(x)$ where $f'(x) = 0$. Three types:

• Local maximum: $f'$ changes from positive to negative.
• Local minimum: $f'$ changes from negative to positive.
• Stationary point of inflection: $f'$ does not change sign.

First-derivative sign test

Pick a point just to the left and just to the right of the stationary point; compute the sign of $f'$ at each.

Second-derivative test

If $f'(a) = 0$ and:

• $f''(a) > 0$: local minimum (concave up).
• $f''(a) < 0$: local maximum (concave down).
• $f''(a) = 0$: test is inconclusive; use the first-derivative sign test instead.

Tangent and normal lines

Tangent at $x = a$: $y - f(a) = f'(a)(x - a)$.

Normal at $x = a$: $y - f(a) = -\dfrac{1}{f'(a)}(x - a)$ (perpendicular to the tangent; provided $f'(a) \neq 0$).

If $f'(a) = 0$ the tangent is horizontal and the normal is vertical ($x = a$).

Optimisation procedure

1. Identify the quantity to optimise (area, volume, cost, distance).
2. Express it as a function of one variable using the constraints.
3. Differentiate; set $f'(x) = 0$; solve.
4. Verify with a sign test or the second-derivative test.
5. Check the domain boundaries (the global extremum might be at an endpoint).
6. State the answer in words with units.

Worked example (stationary points)

For $f(x) = x^3 - 6x^2 + 9x + 2$:

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$.

Stationary points: $x = 1, x = 3$.

$y$-coordinates: $f(1) = 1 - 6 + 9 + 2 = 6$; $f(3) = 27 - 54 + 27 + 2 = 2$.

Second-derivative test: $f''(x) = 6x - 12$. $f''(1) = -6 < 0$, so local max at $(1, 6)$. $f''(3) = 6 > 0$, so local min at $(3, 2)$.

Common traps

Forgetting to find $y$-coordinates

A stationary point is a point; both $x$ and $y$ are needed.

Skipping the classification

$f'(x) = 0$ identifies candidates; the test tells you which type.

Treating local extrema as global

Always check the domain endpoints.

Domain restriction in optimisation

A rectangle cannot have negative or zero side lengths. Enforce the practical domain.

In one sentence

Differentiation locates stationary points by setting $f'(x) = 0$, classifies them by the first-derivative sign test or the sign of $f''(x)$, gives tangent ($m = f'(a)$) and normal ($m = -1/f'(a)$) line equations, and solves optimisation problems by modelling the quantity to optimise as a single-variable function and finding its extremum (with domain endpoints checked).

Examples in context

Example 1. Minimising fencing. A farmer encloses a rectangular paddock of area $200 \text{ m}^2$ against a straight river (no fence needed along the river). If $x$ is the side perpendicular to the river, the depth is $\frac{200}{x}$ and fencing used is $L = 2x + \frac{200}{x}$. Then $L'(x) = 2 - \frac{200}{x^2} = 0$ gives $x^2 = 100$, so $x = 10$ m. Since $L''(x) = \frac{400}{x^3} > 0$, this is a minimum. Minimum fencing $L = 20 + 20 = 40$ m.

Example 2. Maximum profit. A company's profit (thousands of dollars) from producing $x$ hundred units is $P(x) = -x^3 + 12x^2 - 36x + 50$. Then $P'(x) = -3x^2 + 24x - 36 = -3(x - 2)(x - 6)$. Stationary points at $x = 2$ and $x = 6$. As $P''(x) = -6x + 24$, $P''(6) = -12 < 0$, so $x = 6$ is the local maximum: $P(6) = -216 + 432 - 216 + 50 = \$50{,}000$.

Try this

Q1. Find and classify the stationary points of $f(x) = x^3 - 3x^2$. [3 marks]

• Cue. $f'(x) = 3x^2 - 6x = 3x(x - 2)$; $x = 0$ (local max, $f''(0) = -6$) and $x = 2$ (local min, $f''(2) = 6$).

Q2. Find the equation of the tangent to $y = x^2 - 4x$ at $x = 3$. [3 marks]

• Cue. $f(3) = -3$, $f'(x) = 2x - 4$, $f'(3) = 2$; tangent $y + 3 = 2(x - 3)$, so $y = 2x - 9$.

Q3. A rectangle has perimeter $40$ cm. (a) Express its area $A$ in terms of one side $x$. (b) Find the dimensions giving maximum area. [2+2 marks]

• Cue. (a) $A = x(20 - x) = 20x - x^2$. (b) $A'(x) = 20 - 2x = 0 \Rightarrow x = 10$; a $10 \times 10$ square, area $100 \text{ cm}^2$.