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Implicit differentiation and related rates in QCE Specialist Mathematics Unit 4

Syllabus dot point

“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”

QCESpecialist MathematicsUnit 4: Further calculus, and…8 min read

Quick answer

Differentiate an unsolved relation implicitly, attaching dydx\frac{dy}{dx} to each yy term by the chain rule. For related rates, connect quantities through dVdt=dVdr⋅drdt\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}: write the relating equation, differentiate with respect to time, then substitute known values and solve for the unknown rate. Use volume formulas for spheres, cones and cylinders. Always differentiate before substituting numbers, and state the sign and units of the final rate.

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  1. What this dot point is asking
  2. The answer
  3. Exam-style questions

What this dot point is asking

QCAA wants you to differentiate relations that are not solved for yy, 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 x2+y2=25x^2 + y^2 = 25 defines yy in terms of xx without being solved for yy, differentiate both sides with respect to xx, treating yy as a function of xx. Each yy term picks up a factor dydx\frac{dy}{dx} by the chain rule:

ddx(x2+y2)=2x+2ydydx=0 ⇒ dydx=−xy.\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 tt, their rates link through the chain rule. For VV a function of rr,

dVdt=dVdr⋅drdt.\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}.

A known rate (say dVdt\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=43πr3V = \frac{4}{3}\pi r^3, of a cone V=13πr2hV = \frac{1}{3}\pi r^2 h, and of a cylinder V=πr2hV = \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 xx and yy. Differentiating xy2xy^2 with respect to xx gives y2+2xydydxy^2 + 2xy\dfrac{dy}{dx}: the first piece treats y2y^2 as a constant times xx, the second uses the chain rule on y2y^2. Inverse trigonometric functions appear too, with ddxarcsin⁡x=11−x2\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}, ddxarccos⁡x=−11−x2\dfrac{d}{dx}\arccos x = \dfrac{-1}{\sqrt{1 - x^2}} and ddxarctan⁡x=11+x2\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}. When the argument is a function of xx, multiply by its derivative, so ddxarccos⁡(2x)=−21−4x2\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=12hr = \tfrac{1}{2}h. Substituting this relation into V=13πr2hV = \tfrac{1}{3}\pi r^2 h before differentiating reduces the volume to a single variable, V=112πh3V = \tfrac{1}{12}\pi h^3, so the time derivative dVdt=14πh2dhdt\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 dydx\dfrac{dy}{dx} at any point of a curve even when the curve cannot be written as y=f(x)y = f(x). Substitute the point's coordinates into the implicit derivative to get the numerical gradient mm, then the tangent is y−y1=m(x−x1)y - y_1 = m(x - x_1) and the normal has gradient −1m-\dfrac{1}{m}. A common task is to find horizontal tangents (where the numerator of dydx\dfrac{dy}{dx} is zero) or vertical tangents (where the denominator is zero), which locate turning points and cusps of an implicitly defined curve.

Worked example

Air is pumped into a spherical balloon at 100 cm3/s100\ \text{cm}^3/\text{s}. How fast is the radius increasing when r=5 cmr = 5\ \text{cm}?

Relating equation. The volume of a sphere is V=43πr3V = \dfrac{4}{3}\pi r^3.

Differentiate with respect to time (chain rule, since rr depends on tt):

dVdt=ddt(43πr3)=4πr2drdt.\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 dVdt=100\frac{dV}{dt} = 100 and r=5r = 5:

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

Solve for the unknown rate.

drdt=100100π=1π≈0.318 cm/s.\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 drdt\frac{dr}{dt}, since 4πr24\pi r^2 grows, which makes physical sense.

Common trap

The most penalised error is substituting the given radius into the volume formula before differentiating, which turns the varying radius into a constant and destroys the drdt\frac{dr}{dt} term. Differentiate the general relation first, then substitute r=5r = 5. A second slip is forgetting the chain-rule factor: differentiating r3r^3 with respect to tt gives 3r2drdt3r^2\frac{dr}{dt}, not 3r23r^2. Every time-dependent variable contributes its own rate factor.

Exam-style questions

Questions in the style of QCAA 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.

Practice5 marks
Paper 2 (complex familiar). A drone rises vertically from AA at a constant 88 m s−1^{-1}. Point BB is 5050 m horizontally from AA. When the drone is yy m above AA it is zz m from BB. (a) Express z2z^2 in terms of yy. (b) State dydt\dfrac{dy}{dt}. (c) Determine the rate at which zz increases when y=20y = 20 m.
Show worked answer →

(a) By Pythagoras, z2=502+y2=2500+y2.z^2 = 50^2 + y^2 = 2500 + y^2.

(b) The drone rises at a constant rate, so dydt=8\dfrac{dy}{dt} = 8 m s−1^{-1}.

(c) Differentiate implicitly in tt: 2zdzdt=2ydydt2z\dfrac{dz}{dt} = 2y\dfrac{dy}{dt}, so dzdt=yzdydt\dfrac{dz}{dt} = \dfrac{y}{z}\dfrac{dy}{dt}. When y=20y = 20, z=2900≈53.85z = \sqrt{2900} \approx 53.85, so dzdt=2053.85×8≈2.97\dfrac{dz}{dt} = \dfrac{20}{53.85}\times 8 \approx 2.97 m s−1^{-1}.

Markers reward the Pythagorean relation, the stated rate, implicit differentiation, and the evaluated rate with units.

Practice6 marks
Paper 2 (complex unfamiliar). A curve has relation xy2−y+arccos⁡(2x)=1xy^2 - y + \arccos(2x) = 1 and meets the yy-axis at AA. Determine the equation of the tangent to the curve at AA.
Show worked answer →

At AA, x=0x = 0: 0−y+arccos⁡0=10 - y + \arccos 0 = 1, so −y+π2=1-y + \tfrac{\pi}{2} = 1, giving y=π2−1y = \tfrac{\pi}{2} - 1. Thus A=(0, π2−1).A = (0,\ \tfrac{\pi}{2} - 1).

Differentiate implicitly: ddx(xy2)=y2+2xydydx\dfrac{d}{dx}(xy^2) = y^2 + 2xy\dfrac{dy}{dx}; ddx(−y)=−dydx\dfrac{d}{dx}(-y) = -\dfrac{dy}{dx}; ddxarccos⁡(2x)=−21−4x2\dfrac{d}{dx}\arccos(2x) = \dfrac{-2}{\sqrt{1 - 4x^2}}. So y2+2xydydx−dydx−21−4x2=0.y^2 + 2xy\dfrac{dy}{dx} - \dfrac{dy}{dx} - \dfrac{2}{\sqrt{1 - 4x^2}} = 0.

At AA the 2xy2xy term vanishes and 1−0=1\sqrt{1 - 0} = 1: (π2−1)2−dydx−2=0(\tfrac{\pi}{2} - 1)^2 - \dfrac{dy}{dx} - 2 = 0, so dydx=(π2−1)2−2≈−1.67.\dfrac{dy}{dx} = (\tfrac{\pi}{2} - 1)^2 - 2 \approx -1.67.

Tangent: y=−1.67x+(π2−1)≈−1.67x+0.57.y = -1.67x + (\tfrac{\pi}{2} - 1) \approx -1.67x + 0.57.

Markers reward locating AA, the implicit derivative of each term (including the inverse cosine), substitution, and the tangent equation.

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