What are mismatched units?

Keep units consistent so the final rate carries the correct units, such as cm^2\,s^-1 for an area rate.

Given A = π r^2, write dAdr. [1 mark]

The radius of a circle grows at 0.5\ cm s^-1. Find dAdt when r = 4\ cm. [3 marks]

A cube's side increases at 2\ cm s^-1. Find the rate of increase of its volume when the side is 5\ cm. [3 marks]

SEAB Original-style practice: The radius of a circle increases at a constant rate of 0.2\ cm s^-1. Find the rate of increase of the area when the radius is 5\ cm.

Area A = π r^2, so dAdr = 2π r. By the chain rule, dAdt = dAdr×drdt = 2π r × 0.2. At r = 5: dAdt = 2π(5)(0.2) = 2π ≈ 6.28\ cm^2\,s^-1. Markers reward dAdr, the chain-rule link, substituting r = 5 and the given drdt, and the value 2π.

SEAB Original-style practice: A spherical balloon is inflated so that its volume increases at 50\ cm^3\,s^-1. Find the rate of increase of the radius when the radius is 10\ cm. Take the volume as V = 43π r^3.

Differentiate: dVdr = 4π r^2. By the chain rule, dVdt = dVdr×drdt, so drdt = dV/dtdV/dr = 504π r^2. At r = 10: drdt = 504π(100) = 50400π = 18π ≈ 0.0398\ cm s^-1. Markers reward dVdr, the chain-rule rearrangement, substituting the values, and the final rate.

SingaporeAdditional MathematicsSyllabus dot point

How does the chain rule connect the rates at which two related quantities change with time?

Use the chain rule to relate connected rates of change, finding one rate from another for two quantities linked by an equation

A focused answer to the O-Level A-Maths outcome on connected rates of change. Using the chain rule to link the rates at which related quantities vary with time, and solving practical rate problems.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to link the rates at which two related quantities change over time. If two variables are connected by an equation, then their time-rates are connected by the chain rule. Given one rate, you can find the other, which is the basis of practical problems about expanding balloons, rising water levels and growing shadows.

The answer

The chain-rule link

If $y$ depends on $x$ , and both change with time $t$ , the chain rule connects their rates:

\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}.

The rate you want is the product of the rate that links the variables ( $\dfrac{dy}{dx}$ , from differentiating the connecting equation) and the rate you are given.

The standard recipe

Write the equation connecting the two quantities (often a formula for area or volume).
Differentiate it to get the linking rate, such as $\dfrac{dA}{dr}$ .
Write the chain-rule relation between the time-rates.
Substitute the given rate and the value of the variable at the instant asked, then solve for the unknown rate.

Rearranging for the unknown rate

The chain rule can be rearranged to isolate whichever rate you need. To find $\dfrac{dr}{dt}$ from $\dfrac{dV}{dt}$ :

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

Dividing by the linking rate gives the wanted rate.

Reading the question for the given rate

The phrase "increases at" or "decreases at" gives you a time-rate directly. A decrease is a negative rate. Match the units to confirm which rate is given and which is wanted.

Examples in context

Example 1. Water filling a tank. As water pours into a conical tank at a known volume rate, the chain rule links that rate to the rate at which the water level rises, letting an engineer predict how fast the depth changes at any height.

Example 2. A lengthening shadow. As a person walks away from a lamppost, the rate at which their shadow lengthens connects to their walking speed through the geometry of similar triangles, a classic connected-rates application.

Try this

Q1. Given $A = \pi r^2$ , write $\dfrac{dA}{dr}$ . [1 mark]

Cue. $2\pi r$ .

Q2. The radius of a circle grows at $0.5\ \text{cm s}^{-1}$ . Find $\dfrac{dA}{dt}$ when $r = 4\ \text{cm}$ . [3 marks]

Cue. $\dfrac{dA}{dt} = 2\pi r \times 0.5 = 2\pi(4)(0.5) = 4\pi\ \text{cm}^2\,\text{s}^{-1}$ .

Q3. A cube's side increases at $2\ \text{cm s}^{-1}$ . Find the rate of increase of its volume when the side is $5\ \text{cm}$ . [3 marks]

Cue. $V = x^3$ , $\dfrac{dV}{dx} = 3x^2$ , so $\dfrac{dV}{dt} = 3(25)(2) = 150\ \text{cm}^3\,\text{s}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksThe radius of a circle increases at a constant rate of

0.2\ \text{cm s}^{-1}

. Find the rate of increase of the area when the radius is

5\ \text{cm}

Show worked answer →

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

By the chain rule, $\dfrac{dA}{dt} = \dfrac{dA}{dr}\times\dfrac{dr}{dt} = 2\pi r \times 0.2$ .

At $r = 5$ : $\dfrac{dA}{dt} = 2\pi(5)(0.2) = 2\pi \approx 6.28\ \text{cm}^2\,\text{s}^{-1}$ .

Markers reward $\dfrac{dA}{dr}$ , the chain-rule link, substituting $r = 5$ and the given $\dfrac{dr}{dt}$ , and the value $2\pi$ .

Original6 marksA spherical balloon is inflated so that its volume increases at

50\ \text{cm}^3\,\text{s}^{-1}

. Find the rate of increase of the radius when the radius is

10\ \text{cm}

. Take the volume as

V = \dfrac{4}{3}\pi r^3

Show worked answer →

Differentiate: $\dfrac{dV}{dr} = 4\pi r^2$ .

By the chain rule, $\dfrac{dV}{dt} = \dfrac{dV}{dr}\times\dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{dV/dt}{dV/dr} = \dfrac{50}{4\pi r^2}$ .

At $r = 10$ : $\dfrac{dr}{dt} = \dfrac{50}{4\pi(100)} = \dfrac{50}{400\pi} = \dfrac{1}{8\pi} \approx 0.0398\ \text{cm s}^{-1}$ .

Markers reward $\dfrac{dV}{dr}$ , the chain-rule rearrangement, substituting the values, and the final rate.