Differentiate implicitly defined relations and solve related rates problems using the chain rule, including contexts involving volumes and surface areas of cones, spheres and cylinders

What this dot point is asking

QCAA wants you to differentiate relations that are not solved for $y$, and to link the rates of change of two quantities that vary together through time. Related rates problems are the classic application: a balloon inflating, a shadow lengthening, water draining from a cone. The chain rule connects the rates, and the assessable skill is setting up that connection correctly.

The answer

Implicit differentiation

When a relation such as $x^2 + y^2 = 25$ defines $y$ in terms of $x$ without being solved for $y$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Each $y$ term picks up a factor $\frac{dy}{dx}$ by the chain rule:

The chain rule for related rates

If two quantities depend on time $t$, their rates link through the chain rule. For $V$ a function of $r$,

A known rate (say $\frac{dV}{dt}$) and a geometric relationship between the variables determine the unknown rate.

The four-step method

Solve related rates problems systematically. First, write the equation relating the variables (often a volume or area formula). Second, differentiate both sides with respect to time, applying the chain rule to every variable. Third, substitute the known values and rates at the instant in question. Fourth, solve for the unknown rate and state units.

Geometric formulas

Common contexts use the volume of a sphere $V = \frac{4}{3}\pi r^3$, of a cone $V = \frac{1}{3}\pi r^2 h$, and of a cylinder $V = \pi r^2 h$, with surface areas as needed. Differentiating these with respect to time produces the rate relationships. For a cone draining water, you often eliminate one variable using the fixed ratio of radius to height before differentiating.

Differentiate, then substitute

The crucial order is to differentiate first and substitute the specific values only afterward. Substituting a numerical value (such as a fixed radius) before differentiating treats a varying quantity as constant and loses its rate term, a frequent and heavily penalised error.

Stating the answer

Report the rate with its sign and units; a negative rate means the quantity is decreasing. Include the instant the rate applies to, since related rates are usually evaluated at one moment.

Implicit derivatives of products and special functions

Implicit differentiation combines with the product rule whenever a term mixes $x$ and $y$. Differentiating $xy^2$ with respect to $x$ gives $y^2 + 2xy\dfrac{dy}{dx}$: the first piece treats $y^2$ as a constant times $x$, the second uses the chain rule on $y^2$. Inverse trigonometric functions appear too, with $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}$, $\dfrac{d}{dx}\arccos x = \dfrac{-1}{\sqrt{1 - x^2}}$ and $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}$. When the argument is a function of $x$, multiply by its derivative, so $\dfrac{d}{dx}\arccos(2x) = \dfrac{-2}{\sqrt{1 - 4x^2}}$.

Eliminating a variable before differentiating

In cone-draining problems the radius and height both vary, but they keep a fixed ratio set by the cone's shape, for example $r = \tfrac{1}{2}h$. Substituting this relation into $V = \tfrac{1}{3}\pi r^2 h$ before differentiating reduces the volume to a single variable, $V = \tfrac{1}{12}\pi h^3$, so the time derivative $\dfrac{dV}{dt} = \tfrac{1}{4}\pi h^2 \dfrac{dh}{dt}$ involves only one rate. Choosing which variable to keep is the key planning decision: keep the one whose rate the question asks about.

Tangents and normals from an implicit derivative

Implicit differentiation gives the gradient $\dfrac{dy}{dx}$ at any point of a curve even when the curve cannot be written as $y = f(x)$. Substitute the point's coordinates into the implicit derivative to get the numerical gradient $m$, then the tangent is $y - y_1 = m(x - x_1)$ and the normal has gradient $-\dfrac{1}{m}$. A common task is to find horizontal tangents (where the numerator of $\dfrac{dy}{dx}$ is zero) or vertical tangents (where the denominator is zero), which locate turning points and cusps of an implicitly defined curve.