Implicit differentiation and related rates (Specialist Mathematics content): Notes for QCE Maths Methods

What this dot point is asking

Syllabus scope. In the QCAA 2025 syllabuses, implicit differentiation and related rates of change sit in Specialist Mathematics (Unit 4, Topic 3: Rates of change and differential equations), not in Mathematical Methods. Methods students use the chain rule for rates of change in context, so this page is extension for Methods and core content in our QCE Specialist Mathematics notes.

QCAA wants you to differentiate equations relating $x$ and $y$ that cannot be solved explicitly for $y$ (implicit differentiation), and to apply differentiation to related-rates word problems where multiple time-dependent quantities are linked by a geometric or algebraic constraint. In Specialist Mathematics it can be assessed in internal assessment and in the external assessment.

Implicit differentiation

Most functions in Unit 3 are given explicitly: $y = f(x)$. Many real curves (circles, ellipses, products of $x$ and $y$) are given implicitly: $F(x, y) = 0$ or similar.

Implicit differentiation is the technique for finding $dy/dx$ without first solving for $y$.

The key idea

Treat $y$ as a function of $x$: $y = y(x)$. Differentiate both sides of the equation with respect to $x$, applying the chain rule whenever $y$ appears.

In particular:

• $\frac{d}{dx}(y) = \frac{dy}{dx}$.
• $\frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx}$ (chain rule).
• $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$.
• $\frac{d}{dx}(x y) = y + x \frac{dy}{dx}$ (product rule).

After differentiating, isolate $\frac{dy}{dx}$ algebraically.

Worked example. Circle equation

$x^2 + y^2 = 25$.

Differentiate both sides. $2x + 2y \frac{dy}{dx} = 0$.

Solve. $\frac{dy}{dx} = -\frac{x}{y}$.

At point $(3, 4)$: $\frac{dy}{dx} = -3/4$. The tangent line is perpendicular to the radius at $(3, 4)$.

Worked example. Product term

$x^2 + xy + y^3 = 7$.

Differentiate. $2x + (y + x \frac{dy}{dx}) + 3 y^2 \frac{dy}{dx} = 0$.

Collect. $2x + y + (x + 3 y^2) \frac{dy}{dx} = 0$.

Solve. $\frac{dy}{dx} = -\frac{2x + y}{x + 3 y^2}$.

Related rates

Related-rates problems link two or more time-dependent quantities by a geometric or algebraic equation, and ask for one rate given the others.

The four-step procedure

Step 1. Identify the variables and relate them. Write an equation linking the time-dependent quantities. Often a geometric formula (volume of a sphere, area of a circle, Pythagoras).

Step 2. Differentiate both sides with respect to time $t$. Use the chain rule on each variable: $\frac{d}{dt}(r^3) = 3 r^2 \frac{dr}{dt}$.

Step 3. Substitute the known values at the specific moment of interest. NOTE: differentiation comes before substitution. Substituting before differentiating treats the variable as constant.

Step 4. Solve for the unknown rate and state units.

Standard contexts

Sphere inflation

$V = \frac{4}{3} \pi r^3$. $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.

Expanding circle (area)

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

Cone water tank (with similar triangles)

Tank: height $H$, top radius $R$. Water depth $h$, surface radius $r = h R / H$. So $V = \frac{1}{3} \pi r^2 h = \frac{\pi R^2}{3 H^2} h^3$. $\frac{dV}{dt} = \frac{\pi R^2}{H^2} h^2 \frac{dh}{dt}$.

Sliding ladder

$x^2 + y^2 = L^2$. $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$. $\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$.

Shadow length from a walker and a fixed light source

Similar-triangle setup; differentiate the linear constraint.

Worked example. Inflating balloon

$\frac{dV}{dt} = 50$ cm$^3$/s. Find $\frac{dr}{dt}$ when $r = 10$ cm.

$V = \frac{4}{3} \pi r^3$, so $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.

Substitute. $50 = 4 \pi (10)^2 \frac{dr}{dt} = 400 \pi \frac{dr}{dt}$.

Solve. $\frac{dr}{dt} = \frac{50}{400 \pi} = \frac{1}{8 \pi}$ cm/s $\approx 0.0398$ cm/s.

Worked example. Cone tank

A conical tank with height 4 m and top radius 2 m fills with water at 0.5 m$^3$/min. Find the rate at which water depth rises when $h = 1$ m.

Geometry. $r = h/2$ (similar triangles: $\frac{r}{h} = \frac{R}{H} = \frac{2}{4}$).

Volume. $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (h/2)^2 h = \frac{\pi h^3}{12}$.

Differentiate. $\frac{dV}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$.

Substitute. $0.5 = \frac{\pi (1)^2}{4} \frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{2}{\pi}$ m/min $\approx 0.637$ m/min.

Decreasing rates

If a quantity is decreasing, its rate is negative. "Water draining at 5 L/min" gives $\frac{dV}{dt} = -5$. The negative sign carries through the calculation.

PSMT applications

The QCE Mathematical Methods PSMT (problem-solving and modelling task, IA3) often involves a real-world related-rates scenario embedded in a multi-step modelling problem. The investigation may ask you to:

• Set up a mathematical model relating two or more variables (often using a real geometric or physical context).
• Use related rates to find a rate of change at a specific moment.
• Solve, interpret, and evaluate the result in the original real-world context.
• Discuss model limitations and refinements.

In a problem-solving and modelling task, strong responses identify the variables and relationships clearly, apply the four-step procedure with attention to units, and discuss the model's validity at the boundary cases.