Related rates of change: VCE Specialist Mathematics Unit 4

Syllabus dot point

“Related rates of change problems, the use of the chain rule to connect the rates of change of related variables, the setting up of a relating equation from the geometry or context, and the evaluation of an unknown rate at a given instant”

7 min readUpdated 2 Jun 2026

Quick answer

A related-rates problem connects two changing quantities by an equation, then uses the chain rule to relate their time derivatives. The standard procedure is: write an equation relating the variables (from geometry or the context), differentiate both sides with respect to time $t$ using the chain rule, substitute the known rate and the values at the instant in question, and solve for the unknown rate. The key chain-rule link is $\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}$ . Substitute numerical values only after differentiating, never before, or you lose the variable you need to differentiate.

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What this dot point is asking
The chain-rule link
A reliable procedure
Choosing the relating equation
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to solve related-rates problems: two or more quantities change with time, and given the rate of one you must find the rate of another. The engine is the chain rule, which links the rates through an equation relating the quantities. You set up that equation from the context, differentiate with respect to time, and substitute the instant of interest.

The chain-rule link

If two quantities $V$ and $r$ are related and both change with time, their rates are connected by the chain rule:

\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}.

This says the rate at which $V$ changes equals how fast $V$ changes with $r$ , times how fast $r$ changes with time. Given any two of the three rates, you can find the third.

A reliable procedure

Identify the variables and rates. Note which rate is given and which is wanted, with units.
Write a relating equation. Use the geometry or the context to connect the variables, for example $V = \frac{4}{3}\pi r^3$ for a sphere.
Differentiate with respect to time. Apply the chain rule to every variable, since each depends on $t$ .
Substitute the instant. Put in the known rate and the values of the variables at the moment asked about.
Solve and interpret. State the unknown rate with correct units and sign (positive for increasing, negative for decreasing).

The order matters: differentiate symbolically first, then substitute numbers, because substituting a numerical value for a variable before differentiating would freeze it and lose its rate.

Choosing the relating equation

The relating equation comes from the situation. Volume and radius of a sphere give $V = \frac{4}{3}\pi r^3$ . A right-angled triangle (such as a ladder against a wall) gives Pythagoras $x^2 + y^2 = L^2$ . A cone draining gives the cone volume formula, often with the radius expressed in terms of the height using similar triangles to reduce to one variable.

Worked example

Inflating a spherical balloon

Air is pumped into a spherical balloon so that its volume increases at $100\ \text{cm}^3/\text{s}$ . Find the rate at which the radius is increasing when the radius is $5\ \text{cm}$ .

Variables and rates. Given $\frac{dV}{dt} = 100\ \text{cm}^3/\text{s}$ ; wanted $\frac{dr}{dt}$ when $r = 5\ \text{cm}$ .

Relating equation. The volume of a sphere is

V = \frac{4}{3}\pi r^3.

Differentiate with respect to time. Using the chain rule,

\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.

Substitute the instant. Put $\frac{dV}{dt} = 100$ and $r = 5$ :

100 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}.

Solve.

\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318\ \text{cm/s}.

The radius increases at $\frac{1}{\pi}\ \text{cm/s}$ at that instant. The answer is positive, consistent with a growing balloon, and the units cm/s are correct for a rate of change of length.

Common exam traps

Substituting before differentiating: Putting $r = 5$ into the volume formula first gives a constant, whose derivative is zero. Differentiate symbolically, then substitute.
Forgetting the chain rule: $\frac{d}{dt}[r^3] = 3r^2\frac{dr}{dt}$ , not $3r^2$ . Every variable carries its own time derivative.
Not reducing to one variable: In cone problems, use similar triangles to write the radius in terms of the height before differentiating, so only one rate is unknown.
Wrong sign: A decreasing quantity has a negative rate. Watch for "draining" or "shrinking" wording.

Examples in context

Example 1. For a square of side $s$ and area $A = s^2$ , $\frac{dA}{dt} = 2s\frac{ds}{dt}$ .

Example 2. A ladder of length $L$ sliding down a wall satisfies $x^2 + y^2 = L^2$ , so $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ .

Try this

Q1. Write the related-rates equation linking the area and radius of a circle. [2 marks]

Cue. $A = \pi r^2$ , so $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$ .

Q2. A cube's edge grows at $2\ \text{cm/s}$ . Find $\frac{dV}{dt}$ when the edge is $4\ \text{cm}$ . [3 marks]

Cue. $V = s^3$ , $\frac{dV}{dt} = 3s^2\frac{ds}{dt} = 3(16)(2) = 96\ \text{cm}^3/\text{s}$ .

Q3. Why must you differentiate before substituting the instant? [1 mark]

Cue. Substituting first turns a variable into a constant, losing its rate of change.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

The radius of a circular oil slick increases at a constant rate of

0.5\ \text{m/s}

. Find the rate at which the area is increasing at the instant the radius is

20\ \text{m}

Show worked answer →

Relating equation: $A = \pi r^2$ . Differentiate with respect to time using the chain rule:

$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$ .

Substitute $\dfrac{dr}{dt} = 0.5$ and $r = 20$ :

$\dfrac{dA}{dt} = 2\pi (20)(0.5) = 20\pi \approx 62.8\ \text{m}^2/\text{s}$ .

Markers reward the area relation, differentiating before substituting, and the value $20\pi$ with correct units.

Practice5 marks

Water drains from an inverted right circular cone of height

12\ \text{cm}

and base radius

6\ \text{cm}

. When the water depth is

h

, the surface radius is

r

. (a) Use similar triangles to express

r

in terms of

h

, and hence write

V

in terms of

h

. (b) If the volume decreases at

8\ \text{cm}^3/\text{s}

, find the rate at which the depth is falling when

h = 4\ \text{cm}

Show worked answer →

(a) By similar triangles, $\dfrac{r}{h} = \dfrac{6}{12} = \dfrac{1}{2}$ , so $r = \dfrac{h}{2}$ . Then $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}$ .

(b) Differentiate: $\dfrac{dV}{dt} = \dfrac{\pi}{12}\cdot 3h^2\,\dfrac{dh}{dt} = \dfrac{\pi h^2}{4}\,\dfrac{dh}{dt}$ .

The volume decreases, so $\dfrac{dV}{dt} = -8$ . At $h = 4$ : $-8 = \dfrac{\pi (16)}{4}\,\dfrac{dh}{dt} = 4\pi\,\dfrac{dh}{dt}$ , so $\dfrac{dh}{dt} = \dfrac{-8}{4\pi} = -\dfrac{2}{\pi} \approx -0.637\ \text{cm/s}$ .

Markers reward the similar-triangles reduction to one variable, the differentiation, the negative rate for draining, and the value $-\frac{2}{\pi}$ .