AS & A Level · Practice questions · Parametric and implicit differentiation

Every curve has a gradient.

Six original Cambridge-style questions on parametric and implicit differentiation, at Paper 2 and Paper 3 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Differentiate everything, then rearrange.

Parametric: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\); find \(t\) at the point first.
Implicit: differentiate every term with respect to \(x\); each \(y\) term gains \(\dfrac{dy}{dx}\).
Products such as \(xy\) need the product rule.
Collect the \(\dfrac{dy}{dx}\) terms on one side and factorise.
Stationary points: numerator zero, then substitute back into the curve.

Recall

[2 marks]

A curve has parametric equations \(x = 4t\), \(y = \dfrac{2}{t}\). Find \(\dfrac{dy}{dx}\) in terms of \(t\).

\(\dfrac{dx}{dt} = 4\), \(\dfrac{dy}{dt} = -\dfrac{2}{t^2}\). ✓
\(\dfrac{dy}{dx} = -\dfrac{2}{t^2} \div 4 = -\dfrac{1}{2t^2}\). ✓

Calculation

[4 marks]

A curve has parametric equations \(x = \theta - \sin\theta\), \(y = 1 - \cos\theta\). Find the equation of the tangent at the point where \(\theta = \dfrac{\pi}{2}\).

\(\dfrac{dx}{d\theta} = 1 - \cos\theta\), \(\dfrac{dy}{d\theta} = \sin\theta\), so \(\dfrac{dy}{dx} = \dfrac{\sin\theta}{1 - \cos\theta}\). ✓
At \(\theta = \dfrac{\pi}{2}\): gradient \(= \dfrac{1}{1 - 0} = 1\); point \(\left(\dfrac{\pi}{2} - 1, 1\right)\). ✓
\(y - 1 = x - \dfrac{\pi}{2} + 1\). ✓
\(y = x + 2 - \dfrac{\pi}{2}\). ✓

Calculation

[4 marks]

The curve \(y^2 + xy = 6\) passes through the point \((1, 2)\). Find the equation of the normal at this point, in the form \(ax + by = c\).

\(2y\dfrac{dy}{dx} + x\dfrac{dy}{dx} + y = 0\), so \(\dfrac{dy}{dx} = -\dfrac{y}{2y + x}\). ✓
At \((1, 2)\): \(\dfrac{dy}{dx} = -\dfrac{2}{5}\), so the normal gradient is \(\dfrac{5}{2}\). ✓
\(y - 2 = \dfrac{5}{2}(x - 1)\), so \(2y - 4 = 5x - 5\). ✓
\(5x - 2y = 1\). ✓

Calculation

[3 marks]

The curve \(xe^{y} + y = 2\) passes through \((2, 0)\). Find the gradient of the curve at this point.

\(e^{y} + xe^{y}\dfrac{dy}{dx} + \dfrac{dy}{dx} = 0\) (product rule on \(xe^{y}\)). ✓
\(\dfrac{dy}{dx} = -\dfrac{e^{y}}{xe^{y} + 1}\). ✓
At \((2, 0)\): \(-\dfrac{1}{2 + 1} = -\dfrac{1}{3}\). ✓

Analysis

[5 marks]

Find the coordinates of the points on the curve \(x^2 + 2xy + 3y^2 = 18\) at which the tangent is parallel to the \(x\)-axis.

\(2x + 2x\dfrac{dy}{dx} + 2y + 6y\dfrac{dy}{dx} = 0\), so \(\dfrac{dy}{dx} = -\dfrac{x + y}{x + 3y}\). ✓
Horizontal tangent: \(x + y = 0\), so \(y = -x\). ✓
Substitute: \(x^2 - 2x^2 + 3x^2 = 18\), so \(2x^2 = 18\) and \(x = \pm 3\). ✓
Points \((3, -3)\) and \((-3, 3)\) (the denominator \(x + 3y = \mp 6 \ne 0\)). ✓

Analysis

[7 marks]

A curve has parametric equations \(x = t^2 + 2t\), \(y = t^3 - 12t\).

(a) Find \(\dfrac{dy}{dx}\) in terms of \(t\).

(b) Find the coordinates of the stationary points of the curve.

(a) \(\dfrac{dy}{dx} = \dfrac{3t^2 - 12}{2t + 2}\). ✓
(b) \(3t^2 - 12 = 0\), so \(t = \pm 2\) (the denominator is not zero there). ✓
\(t = 2\): \((8, -16)\). \(t = -2\): \((0, 16)\). ✓
(c) At \(t = 1\): point \((3, -11)\), gradient \(\dfrac{3 - 12}{4} = -\dfrac{9}{4}\). ✓
\(y + 11 = -\dfrac{9}{4}(x - 3)\), so \(4y + 44 = -9x + 27\). ✓
\(9x + 4y + 17 = 0\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.