SCSA WACE 2021-style practice: Calculator-assumed. A spherical balloon is inflated so that its volume increases at dVdt=30 cm^3/s. The volume is V=43π r^3. Find the rate at which the radius is increasing when r=5 cm.

A related-rates problem requiring the chain rule. Differentiate V=43π r^3 with respect to t: dVdt=4π r^2drdt. Rearrange for drdt: drdt=14π r^2dVdt. Substitute r=5 and dVdt=30: drdt=304π(25)=30100π≈ 0.0955 cm/s. Markers reward differentiating before substituting, the chain-rule factor drdt, and the correct evaluated rate with units.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How does the derivative describe how fast a quantity changes, and how do we link the rates of related quantities?

Interpret the derivative as an instantaneous rate of change and use the chain rule to relate the rates of change of connected quantities

WACE Year 12 Mathematics Methods Unit 3 rates of change: the derivative as instantaneous rate, average versus instantaneous rate, related rates through the chain rule, with worked SCSA-style examples in context.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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What this dot point is asking
Average versus instantaneous rate
Related rates through the chain rule
Choosing the right link
Motion as a rate of change

What this dot point is asking

SCSA Unit 3 frames the derivative as a rate of change and then connects rates of related quantities. This dot point is examined in both sections, with related-rates problems appearing most often in the calculator-assumed section.

Average versus instantaneous rate

The average rate of change of $y$ over an interval is the gradient of the chord, $\dfrac{\Delta y}{\Delta x}$ . The instantaneous rate is the gradient of the tangent, the limit of the chord gradient, which is the derivative. When a question asks for the rate at a specific instant, evaluate the derivative at that point.

When two variables are linked by an equation and both change with time, their rates are connected by the chain rule. If $y$ depends on $x$ and $x$ depends on $t$ , then

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

This lets you transfer a known rate into an unknown one.

A related-rates problem

A circular ripple has radius increasing at $\dfrac{dr}{dt}=0.5$ m/s. Find the rate of increase of its area when $r=4$ m.

Step 1: Write the geometric relationship

Because the ripple is circular, the area depends on the radius through the standard formula. This gives us the link between the two quantities whose rates we want to connect.

A = \pi r^{2}.

Step 2: Differentiate with respect to time using the chain rule

Neither $A$ nor $r$ is fixed; both change with $t$ . Differentiating the geometric relation with respect to $t$ via the chain rule introduces both rates into a single equation. Numerical values must not be substituted until after differentiating.

\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt} = 2\pi r\cdot\frac{dr}{dt}.

Step 3: Substitute the given values and evaluate

Now that the general rate equation is established, we substitute $r=4$ and $\dfrac{dr}{dt}=0.5$ to find the instantaneous rate at the required moment.

\frac{dA}{dt} = 2\pi(4)(0.5) = 4\pi \approx 12.57.

Final answer: The area is increasing at approximately $12.57$ m $^{2}$ per second.

Choosing the right link

The key skill is identifying the equation that connects the quantities (here $A=\pi r^{2}$ ) and the rate that is given (here $\dfrac{dr}{dt}$ ). Differentiate the relationship with respect to $t$ first, and only substitute the specific values at the very end, so that variable quantities are not frozen too early.

A related-rates problem with a volume link

A draining cone

Water drains from an inverted cone so its volume falls at $\dfrac{dV}{dt}=-12$ cm $^{3}$ /s. For this cone the volume in terms of the water depth $h$ is $V=\dfrac{\pi}{3}h^{3}$ . Find the rate at which the depth is falling when $h=2$ cm.

Differentiate with respect to $t$ using the chain rule:

\frac{dV}{dt}=\pi h^{2}\frac{dh}{dt}.

Rearrange for the depth rate:

\frac{dh}{dt}=\frac{1}{\pi h^{2}}\frac{dV}{dt}.

Substitute

h=2

and

\dfrac{dV}{dt}=-12

\frac{dh}{dt}=\frac{-12}{\pi(4)}=\frac{-3}{\pi}\approx -0.95\text{ cm/s}.

The depth is falling at about

0.95

cm/s. The negative sign confirms the water level is dropping. Markers reward differentiating before substituting and the correct sign and units.

Motion as a rate of change

The single most common rate-of-change context is straight-line motion, where velocity $v(t)=\dfrac{dx}{dt}$ is the rate of change of displacement and acceleration $a(t)=\dfrac{dv}{dt}$ is the rate of change of velocity. The particle is momentarily at rest where $v(t)=0$ and its speed is greatest in magnitude where $a(t)=0$ between rest points. This link between derivatives and motion threads through both Unit 3 and the integration topics of Unit 4.

Exam-style practice questions

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

WACE 20216 marksCalculator-assumed. A spherical balloon is inflated so that its volume increases at

\dfrac{dV}{dt}=30

^{3}

/s. The volume is

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

. Find the rate at which the radius is increasing when

r=5

cm.

Show worked answer →

A related-rates problem requiring the chain rule.

Differentiate $V=\dfrac{4}{3}\pi r^{3}$ with respect to $t$ : $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$ .

Rearrange for $\dfrac{dr}{dt}$ : $\dfrac{dr}{dt}=\dfrac{1}{4\pi r^{2}}\dfrac{dV}{dt}$ .

Substitute $r=5$ and $\dfrac{dV}{dt}=30$ : $\dfrac{dr}{dt}=\dfrac{30}{4\pi(25)}=\dfrac{30}{100\pi}\approx 0.0955$ cm/s.

Markers reward differentiating before substituting, the chain-rule factor $\dfrac{dr}{dt}$ , and the correct evaluated rate with units.

WACE 20234 marksCalculator-free. The displacement of a particle is

x(t)=t^{3}-6t^{2}+9t

metres, with

t

in seconds. (a) Find the velocity

v(t)

. (b) Find the times at which the particle is at rest.

Show worked answer →

A rate-of-change application to motion.

(a) Velocity is the derivative of displacement: $v(t)=\dfrac{dx}{dt}=3t^{2}-12t+9$ .

(b) At rest means $v(t)=0$ : $3t^{2}-12t+9=3(t^{2}-4t+3)=3(t-1)(t-3)=0$ , so $t=1$ s and $t=3$ s.

Markers reward differentiating displacement and factorising to find both rest times.