# Applications of Differentiation

> Rates of change, motion in a straight line, and optimisation problems that find the best possible answer.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/applications-of-differentiation/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The cheapest can
A drink company wants a can that holds 375 mL. A tall skinny can uses lots of metal on the side. A short wide can uses lots on the top and bottom. Somewhere in between is the can that uses the **least** metal, and on millions of cans that saves a fortune.

Finding the best value, the cheapest, biggest, fastest or shortest, is called **optimisation**. The derivative does it: the best value sits at a stationary point, where the gradient is zero.

**In short:** Differentiation finds rates of change and best values in real problems. Velocity is the derivative of displacement and acceleration is the derivative of velocity, and to find a maximum or minimum you write one formula in one variable, set its derivative to zero and check the endpoints.

## Starting point: Years 9 and 10

### Formulas you need

| Shape | Perimeter or surface area | Area or volume |
|---|---|---|
| Rectangle | $P = 2l + 2w$ | $A = lw$ |
| Circle | $C = 2\pi r$ | $A = \pi r^2$ |
| Rectangular box | $SA = 2(lw + lh + wh)$ | $V = lwh$ |
| Cylinder | $SA = 2\pi r^2 + 2\pi rh$ | $V = \pi r^2 h$ |

### Speed, distance and time

$$\text{speed} = \frac{\text{distance}}{\text{time}}$$

On a distance-time graph, the gradient is the speed. On a velocity-time graph, the gradient is the acceleration.

### Best values with quadratics

A quadratic with $a < 0$ has a maximum at its vertex, $x = -\dfrac{b}{2a}$.

### Worked example: Biggest pen
A farmer has 60 m of fence to make a rectangular pen. Width $x$ means length $30 - x$. Area $A = x(30 - x) = 30x - x^2$. The vertex is at $x = 15$, so a 15 m by 15 m square gives the largest area, 225 m².

### Exam tip: Write one variable
Optimisation questions always have two unknowns and one fact linking them. Use the fact to write the thing you want in terms of **one** variable.

**Mind map**

```text
Applications
  Rates of change
    dy/dx is a rate
    Units: y per x
  Motion
    x: displacement
    v = dx/dt
    a = dv/dt
  Optimisation
    One variable formula
    Set derivative to 0
    Check it is max or min
    Check endpoints
```

### Year 11: Rates of change
Every derivative is a rate. If $V$ is the volume of water in a tank after $t$ minutes, then $\dfrac{dV}{dt}$ is how fast the volume changes, in litres per minute.

**Example.** A balloon's volume is $V = 4t^2 + t$ cm³ after $t$ seconds. The rate at $t = 3$ is $\dfrac{dV}{dt} = 8t + 1 = 25$ cm³ per second.

**Velocity.** If $x$ is the position of a particle at time $t$, its velocity is $v = \dfrac{dx}{dt}$. For $x = t^2 - 4t$, $v = 2t - 4$. At $t = 2$ the particle is at rest because $v = 0$.

### Year 12: Motion in a straight line
- **Displacement** $x$: position relative to the origin.
- **Velocity** $v = \dfrac{dx}{dt}$: rate of change of displacement. Its sign shows direction.
- **Acceleration** $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$: rate of change of velocity.

**Example.** $x = t^3 - 6t^2 + 9t$ metres.

$v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$, so it is at rest at $t = 1$ and $t = 3$.

$a = 6t - 12$, so acceleration is zero at $t = 2$.

**Speed** is the size of velocity. A particle is speeding up when $v$ and $a$ have the same sign.

Going backwards, integration turns acceleration into velocity and velocity into displacement.

**Which way does it move?** The sign of $v$ gives the direction: $v > 0$ means moving right (in the positive direction), $v < 0$ means moving left.

**Total distance** is not always the change in displacement, because the particle can turn back. Find when $v = 0$, then add the distance of each part. For $x = t^2 - 4t + 3$ on $0 \leq t \leq 5$: $v = 2t - 4 = 0$ at $t = 2$. The positions are $x(0) = 3$, $x(2) = -1$ and $x(5) = 8$. Distance $= 4 + 9 = 13$ m, even though the displacement only changed by 5.

### Year 12: Optimisation step by step
1. Draw a diagram and name the variables.
2. Write the quantity to optimise in terms of **one** variable.
3. Differentiate and set the derivative equal to zero.
4. Check it is a maximum or minimum (second derivative or table).
5. Check the endpoints of the allowed domain. Answer the question with units.

**The cheapest can.** $V = \pi r^2 h = 375$, so $h = \dfrac{375}{\pi r^2}$. The metal used is

$$S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \frac{750}{r}$$

$\dfrac{dS}{dr} = 4\pi r - \dfrac{750}{r^2} = 0$ gives $r^3 = \dfrac{750}{4\pi}$, so $r \approx 3.91$ cm.

$\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{1500}{r^3} > 0$, so this is a minimum. Then $h = \dfrac{375}{\pi(3.91)^2} \approx 7.81$ cm. The cheapest can has its height equal to its diameter.

**Related rates.** When two quantities change with time, link their rates with the chain rule. A cube's volume is $V = s^3$, so $\dfrac{dV}{dt} = \dfrac{dV}{ds} \times \dfrac{ds}{dt} = 3s^2\dfrac{ds}{dt}$. If $s = 4$ cm and the side shrinks at 0.1 cm/min, $\dfrac{dV}{dt} = 3(16)(-0.1) = -4.8$ cm³/min.

**Costs that depend on time.** For a trip of 400 km at $v$ km/h, the time is $\dfrac{400}{v}$ hours. If running costs $C(v)$ dollars per hour, the trip costs $\dfrac{400}{v} \times C(v)$. Minimise that.

**Shortest distance to a curve.** The distance from $(0, 3)$ to a point $(x, x^2)$ is $\sqrt{x^2 + (x^2 - 3)^2}$. Minimise the **square** of the distance instead, $D = x^2 + (x^2 - 3)^2$, because it is smallest at the same place and has no root to differentiate. $D' = 2x + 4x(x^2 - 3) = 2x(2x^2 - 5) = 0$ gives $x^2 = \frac{5}{2}$, and the shortest distance is $\sqrt{\frac{5}{2} + \frac{1}{4}} = \frac{\sqrt{11}}{2}$.

## Quick summary

- A derivative is a rate of change. Its units are "$y$ per $x$".
- $v = \dfrac{dx}{dt}$ and $a = \dfrac{dv}{dt}$. At rest means $v = 0$.
- For optimisation, write one formula in one variable, set the derivative to zero, and check the nature and the endpoints.
- A quadratic's best value is at its vertex, $x = -\dfrac{b}{2a}$.

## Quiz

Try these questions from the chapter quiz.

1. A rectangle has perimeter 40 cm and width $x$. Its length is:
   - A. $40 - x$
   - B. $20 - x$
   - C. $40 - 2x$
   - D. $\dfrac{40}{x}$
   - Answer: B. $20 - x$. $2l + 2x = 40$, so $l = 20 - x$.

2. What is the volume of a cylinder with radius 3 and height 10?
   - A. $30\pi$
   - B. $60\pi$
   - C. $90\pi$
   - D. $300\pi$
   - Answer: C. $90\pi$. $\pi r^2 h = \pi \times 9 \times 10$.

3. On a distance-time graph, the gradient represents:
   - A. speed
   - B. acceleration
   - C. distance
   - D. time
   - Answer: A. speed. Distance per time.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/applications-of-differentiation/#quiz