What are substitute the known values?

Given (dV)/(dt) = 100 and r = 5:

QLDSpecialist MathematicsSyllabus dot point

Topic 2: Rates of change and differential equations

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

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on implicit differentiation and related rates. Covers differentiating relations not solved for y, the chain rule link between rates, the four-step related-rates method, and geometric volume contexts, with a verified worked example and the differentiate-then-substitute trap.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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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:

\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx} = 0 \ \Rightarrow\ \frac{dy}{dx} = -\frac{x}{y}.

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 $r$ ,

\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}.

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.

Worked example

Air is pumped into a spherical balloon at $100\ \text{cm}^3/\text{s}$ . How fast is the radius increasing when $r = 5\ \text{cm}$ ?

Relating equation. The volume of a sphere is $V = \dfrac{4}{3}\pi r^3$ .

Differentiate with respect to time (chain rule, since $r$ depends on $t$ ):

\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \frac{dr}{dt}.

Substitute the known values. Given $\frac{dV}{dt} = 100$ and $r = 5$ :

100 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}.

Solve for the unknown rate.

\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318\ \text{cm/s}.

Check. The rate is positive, consistent with the balloon expanding, and the units cm/s are correct for a rate of change of length. At a larger radius the same inflow would give a smaller $\frac{dr}{dt}$ , since $4\pi r^2$ grows, which makes physical sense.

Exam-style practice questions

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

2024 QCAA5 marksA drone travels vertically from point A at a constant speed of 8 m s^-1 over time t for t greater than or equal to 0 seconds. Observation of the drone is made from point B, which is 50 m horizontally from point A. When the drone is y metres above point A, it is z metres from point B. a) Determine an equation expressing z^2 in terms of y^2. b) State the value of dy/dt. c) Use your results to determine the rate at which z is increasing with respect to time when the drone is 20 metres above point A.

Show worked answer →

a) [1 mark] B is 50 m horizontally from A and the drone rises vertically, so by Pythagoras z^2 = 50^2 + y^2 = 2500 + y^2.

b) [1 mark] The drone rises at a constant 8 m s^-1, so dy/dt = 8.

c) [3 marks] Differentiate z^2 = 2500 + y^2 implicitly with respect to t:
2z (dz/dt) = 2y (dy/dt), so dz/dt = (y/z)(dy/dt).
When y = 20: z = sqrt(2500 + 400) = sqrt(2900) = 53.85 m.
dz/dt = (20 / 53.85) * 8 = 2.97 m s^-1.
So z is increasing at about 2.97 m s^-1 when the drone is 20 m above A.

2023 QCAA6 marksA curve modelled by the relation x*y^2 - y + arccos(2x) = 1, where -0.35 less than or equal to x less than or equal to 0.27 and 0 less than or equal to y less than or equal to 1, intersects the y-axis at point A. Determine the equation of the tangent to the curve at point A.

Show worked answer →

Find A first [point]: on the y-axis x = 0, so 0 - y + arccos(0) = 1, giving -y + pi/2 = 1, so y = pi/2 - 1. Hence A = (0, pi/2 - 1).

Differentiate implicitly [2 marks]: treat y as a function of x.
d/dx[xy^2] = y^2 + 2xy(dy/dx); d/dx[-y] = -(dy/dx); d/dx[arccos(2x)] = -2/sqrt(1 - 4x^2).
So y^2 + 2xy*(dy/dx) - (dy/dx) - 2/sqrt(1 - 4x^2) = 0.

Substitute A = (0, pi/2 - 1) [2 marks]: with x = 0 the 2xy term vanishes and sqrt(1 - 0) = 1:
(pi/2 - 1)^2 - (dy/dx) - 2 = 0, so dy/dx = (pi/2 - 1)^2 - 2 = -1.67 (to 2 d.p.).

Tangent line [2 marks]: y - (pi/2 - 1) = [(pi/2 - 1)^2 - 2](x - 0), i.e. y = [(pi/2 - 1)^2 - 2] x + (pi/2 - 1), approximately y = -1.67x + 0.57.