# Related Rates of Change

> When one quantity changes, others linked to it change too. Use the chain rule to find how fast, for ripples, balloons, tanks and ladders.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/related-rates/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The ripple in the pond
Drop a stone into a still pond. A circle of ripples spreads out at a steady 3 cm every second. The radius grows at a fixed rate, but the area inside the circle grows faster and faster. Early on it gains a little each second. Later it gains a lot.

Radius and area are linked by $A = \pi r^2$. Because they are linked, their rates of change are linked too. The chain rule is the tool that connects them.

**In short:** Related rates link how fast two connected quantities change, using the chain rule: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$. Write one formula linking the two quantities, differentiate it, then substitute the values you know.

## Starting point: what you already know

- **A rate of change** is a derivative. $\dfrac{dr}{dt}$ means "how fast $r$ changes each second".
- **The chain rule:** $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$. Think of the $dx$ parts "cancelling".
- **Formulas for shapes:** circle area $\pi r^2$, sphere volume $\dfrac{4}{3}\pi r^3$, sphere surface area $4\pi r^2$, cone volume $\dfrac{1}{3}\pi r^2 h$, cube volume $x^3$.
- **Similar triangles:** matching sides are in the same ratio.
- **Signs:** a quantity that is getting smaller has a negative rate.

## The method

### Key idea: Related rates in four steps
1. Write down the rate you know and the rate you want, with units.
2. Write a formula that links the two quantities.
3. Differentiate the formula with respect to the quantity in it.
4. Use the chain rule to join the rates, then substitute the moment you are asked about.

### Worked example: The ripple
$\dfrac{dr}{dt} = 3$ cm/s. Find $\dfrac{dA}{dt}$ when $r = 10$ cm.

$A = \pi r^2$, so $\dfrac{dA}{dr} = 2\pi r$.

$$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 2\pi r \times 3 = 6\pi r$$

At $r = 10$: $\dfrac{dA}{dt} = 60\pi \approx 188$ cm² per second.

## Turning the chain the other way

Often you know how fast the "big" quantity changes and want the small one. Flip the derivative.

### Worked example: Blowing up a balloon
Air goes into a round balloon at 100 cm³/s. How fast is the radius growing when $r = 5$ cm?

$V = \dfrac{4}{3}\pi r^3$, so $\dfrac{dV}{dr} = 4\pi r^2$, which is $100\pi$ when $r = 5$.

$$\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt} = \frac{1}{100\pi} \times 100 = \frac{1}{\pi} \approx 0.32 \text{ cm/s}$$

### Exam tip: Write the chain so the units cancel
$\dfrac{\text{cm}}{\text{s}} = \dfrac{\text{cm}}{\text{cm}^3} \times \dfrac{\text{cm}^3}{\text{s}}$. If the units do not cancel to what you want, the chain is wrong.

### Watch out: Substitute last
Differentiate first, while $r$ is still a letter. If you put $r = 5$ in too early, the formula becomes a constant and its derivative is zero.

**Mind map**

```text
Related rates
  Idea
    Linked quantities
    Linked rates
  Tool
    Chain rule
    dy/dt = dy/dx × dx/dt
  Steps
    Know and want
    Linking formula
    Differentiate
    Substitute last
  Classic problems
    Ripples
    Balloons
    Filling a cone
    Sliding ladder
```

### Year 11: Filling a cone-shaped tank
Water pours at 2 m³/min into a tank shaped like an upside-down cone. Its radius is always half its depth, so $r = \dfrac{h}{2}$. How fast does the depth rise when $h = 4$ m?

Use similar triangles to remove $r$: $V = \dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}$.

$\dfrac{dV}{dh} = \dfrac{\pi h^2}{4} = 4\pi$ at $h = 4$.

$\dfrac{dh}{dt} = \dfrac{1}{4\pi} \times 2 = \dfrac{1}{2\pi} \approx 0.16$ m/min. The water rises more slowly as the tank gets wider.

### Year 12: The sliding ladder
A 5 m ladder leans on a wall. Its foot slides away at 0.5 m/s. How fast does the top slide down when the foot is 3 m from the wall?

Let the foot be $x$ m out and the top $y$ m up, so $x^2 + y^2 = 25$ and $y = \sqrt{25 - x^2}$.

$\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{25 - x^2}} = -\dfrac{3}{4}$ when $x = 3$.

$\dfrac{dy}{dt} = -\dfrac{3}{4} \times 0.5 = -0.375$ m/s. The top slides down at 0.375 m/s. The negative sign says $y$ is shrinking.

**Changing angles.** A balloon rises at 4 m/s, 50 m from an observer. With $\tan\theta = \frac{h}{50}$, differentiate with respect to $t$: $\sec^2\theta\dfrac{d\theta}{dt} = \dfrac{1}{50}\dfrac{dh}{dt}$. When $h = 50$, $\theta = \frac{\pi}{4}$ and $\sec^2\theta = 2$, so $\dfrac{d\theta}{dt} = \dfrac{4/50}{2} = 0.04$ rad/s.

**Areas with an angle.** For a triangle with two sides of length $a$ and the angle $\theta$ between them, $A = \frac{1}{2}a^2\sin\theta$, so $\dfrac{dA}{dt} = \frac{1}{2}a^2\cos\theta\dfrac{d\theta}{dt}$.

**Shadows.** Use similar triangles to link the shadow $s$ to the distance $x$ from the lamp. A 2 m person and a 5 m lamp give $\dfrac{s}{2} = \dfrac{x + s}{5}$, so $s = \frac{2}{3}x$ and $\dfrac{ds}{dt} = \frac{2}{3}\dfrac{dx}{dt}$, whatever the distance.

**Distance from an observer.** A plane at height 3 km flying level: $D^2 = x^2 + 9$, so $D\dfrac{dD}{dt} = x\dfrac{dx}{dt}$.

**Comparing two moments.** In a cone filling at a steady rate, $V = kh^3$, so $\dfrac{dh}{dt} = \dfrac{1}{3kh^2}\dfrac{dV}{dt}$: the depth rate is proportional to $\dfrac{1}{h^2}$. Doubling the depth divides the rate by 4.

**Three rates in a chain.** From a volume rate to a surface area rate for a sphere: $\dfrac{dS}{dt} = \dfrac{dS}{dr} \times \dfrac{dr}{dV} \times \dfrac{dV}{dt} = \dfrac{8\pi r}{4\pi r^2}\dfrac{dV}{dt} = \dfrac{2}{r}\dfrac{dV}{dt}$.

## Quick summary

- Linked quantities have linked rates.
- Chain rule: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$, and $\dfrac{dx}{dy} = 1 \div \dfrac{dy}{dx}$.
- Find a formula with only the two quantities you need. Use similar triangles or Pythagoras to remove a third.
- Differentiate first, substitute last. Check the units and the sign.

## Quiz

Try these questions from the chapter quiz.

1. The radius of a circle grows at 2 cm/s. Which expression is $\dfrac{dA}{dt}$?
   - A. $2\pi r$
   - B. $4\pi r$
   - C. $\pi r^2$
   - D. $2r$
   - Answer: B. $4\pi r$. $\dfrac{dA}{dr} \times \dfrac{dr}{dt} = 2\pi r \times 2$.

2. With the same circle, find $\dfrac{dA}{dt}$ when $r = 5$.
   - A. $10\pi$ cm²/s
   - B. $20\pi$ cm²/s
   - C. $25\pi$ cm²/s
   - D. $50\pi$ cm²/s
   - Answer: B. $20\pi$ cm²/s. $4\pi \times 5 = 20\pi$.

3. A cube's side grows at 0.1 cm/s. How fast does the volume grow when the side is 10 cm?
   - A. 10 cm³/s
   - B. 30 cm³/s
   - C. 100 cm³/s
   - D. 3 cm³/s
   - Answer: B. 30 cm³/s. $\dfrac{dV}{dx} = 3x^2 = 300$, times 0.1.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/related-rates/#quiz