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Rates of change and related rates: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

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

8 min read

Quick answer

The derivative dydx\dfrac{dy}{dx} is the instantaneous rate of change of yy with respect to xx. When two quantities are linked, the chain rule connects their rates: dydt=dydx⋅dxdt\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}, which lets you find one rate from another.

Jump to a section
  1. What this dot point is asking
  2. Average versus instantaneous rate
  3. Related rates through the chain rule
  4. Choosing the right link
  5. Motion as a rate of change
  6. Exam-style questions

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 yy over an interval is the gradient of the chord, ΔyΔx\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.

Rate of change

If QQ depends on time tt, then dQdt\dfrac{dQ}{dt} is the instantaneous rate of change of QQ, with units of QQ per unit time. A positive rate means QQ is increasing; a negative rate means it is decreasing.

Instantaneous rate in context

A tank with volume V(t)=200−4t2V(t)=200-4t^{2} litres

Find the rate at which the volume is changing at t=3t=3 seconds.

Differentiate.

V′(t)=−8t.V'(t) = -8t.

Evaluate at t=3t=3.

V′(3)=−8×3=−24.V'(3) = -8\times 3 = -24.

The volume is decreasing at 2424 litres per second at t=3t=3. The negative sign shows the tank is emptying. Markers reward the derivative, the substitution, and the interpretation including units and direction.

When two variables are linked by an equation and both change with time, their rates are connected by the chain rule. If yy depends on xx and xx depends on tt, then

dydt=dydx⋅dxdt.\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 drdt=0.5\dfrac{dr}{dt}=0.5 m/s. Find the rate of increase of its area when r=4r=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=πr2.A = \pi r^{2}.

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

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

dAdt=dAdr⋅drdt=2πr⋅drdt.\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=4r=4 and drdt=0.5\dfrac{dr}{dt}=0.5 to find the instantaneous rate at the required moment.

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

Final answer: The area is increasing at approximately 12.5712.57 m2^{2} per second.

The key skill is identifying the equation that connects the quantities (here A=πr2A=\pi r^{2}) and the rate that is given (here drdt\dfrac{dr}{dt}). Differentiate the relationship with respect to tt 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 dVdt=−12\dfrac{dV}{dt}=-12 cm3^{3}/s. For this cone the volume in terms of the water depth hh is V=π3h3V=\dfrac{\pi}{3}h^{3}. Find the rate at which the depth is falling when h=2h=2 cm.

Differentiate with respect to tt using the chain rule:

dVdt=πh2dhdt.\frac{dV}{dt}=\pi h^{2}\frac{dh}{dt}.

Rearrange for the depth rate:
dhdt=1πh2dVdt.\frac{dh}{dt}=\frac{1}{\pi h^{2}}\frac{dV}{dt}.

Substitute h=2h=2 and dVdt=−12\dfrac{dV}{dt}=-12:
dhdt=−12π(4)=−3π≈−0.95 cm/s.\frac{dh}{dt}=\frac{-12}{\pi(4)}=\frac{-3}{\pi}\approx -0.95\text{ cm/s}.

The depth is falling at about 0.950.95 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)=dxdtv(t)=\dfrac{dx}{dt} is the rate of change of displacement and acceleration a(t)=dvdta(t)=\dfrac{dv}{dt} is the rate of change of velocity. The particle is momentarily at rest where v(t)=0v(t)=0 and its speed is greatest in magnitude where a(t)=0a(t)=0 between rest points. This link between derivatives and motion threads through both Unit 3 and the integration topics of Unit 4.

Common errors
Substituting values before differentiating
Differentiate the general relationship first; substituting r=4r=4 too early would treat rr as constant and give a derivative of zero.
Confusing average and instantaneous rate
The average rate uses two points; the instantaneous rate is the derivative at one point.
Omitting units or sign
A rate of change answer needs units and, in context, a statement of whether the quantity is increasing or decreasing.
Forgetting the chain-rule factor drdt\dfrac{dr}{dt}
In related rates, dAdt=dAdr⋅drdt\dfrac{dA}{dt}=\dfrac{dA}{dr}\cdot\dfrac{dr}{dt}; the given rate must be included.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2021-style6 marks
Calculator-assumed. A spherical balloon is inflated so that its volume increases at dVdt=30\dfrac{dV}{dt}=30 cm3^{3}/s. The volume is V=43πr3V=\dfrac{4}{3}\pi r^{3}. Find the rate at which the radius is increasing when r=5r=5 cm.
Show worked answer →

A related-rates problem requiring the chain rule.

Differentiate V=43πr3V=\dfrac{4}{3}\pi r^{3} with respect to tt: dVdt=4πr2drdt\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}.

Rearrange for drdt\dfrac{dr}{dt}: drdt=14πr2dVdt\dfrac{dr}{dt}=\dfrac{1}{4\pi r^{2}}\dfrac{dV}{dt}.

Substitute r=5r=5 and dVdt=30\dfrac{dV}{dt}=30: drdt=304π(25)=30100π≈0.0955\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 drdt\dfrac{dr}{dt}, and the correct evaluated rate with units.

WACE 2023-style4 marks
Calculator-free. The displacement of a particle is x(t)=t3−6t2+9tx(t)=t^{3}-6t^{2}+9t metres, with tt in seconds. (a) Find the velocity v(t)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)=dxdt=3t2−12t+9v(t)=\dfrac{dx}{dt}=3t^{2}-12t+9.

(b) At rest means v(t)=0v(t)=0: 3t2−12t+9=3(t2−4t+3)=3(t−1)(t−3)=03t^{2}-12t+9=3(t^{2}-4t+3)=3(t-1)(t-3)=0, so t=1t=1 s and t=3t=3 s.

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

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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