Apply implicit differentiation to find $\frac{dy}{dx}$ from equations relating $x$ and $y$ that cannot be expressed in the form $y = f(x)$, and apply differentiation to related rates problems

What this dot point is asking

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. The dot point is heavily examined in PSMT (Problem-Solving and Modelling Task / IA2 in some configurations, IA3 setup) and in the EA Paper 2 short response.

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.

The PSMT is the place where related rates is most heavily examined. Strong PSMT 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.