For x = t^3, y = t^2, find (dy)/(dx) in terms of t. [2 marks]

What identity eliminates θ from x = cosθ, y = sinθ? [1 mark]

CCEA CCEA 2020-style practice: A curve is given parametrically by x = t^2, y = 2t. Find (dy)/(dx) in terms of t, and the Cartesian equation of the curve.

Differentiate each: (dx)/(dt) = 2t and (dy)/(dt) = 2. Then (dy)/(dx) = (dy/dt)/(dx/dt) = (2)/(2t) = (1)/(t). For the Cartesian form, eliminate t: from y = 2t we get t = (y)/(2), so x = ((y)/(2))^2 = (y^2)/(4), that is y^2 = 4x. Markers reward both derivatives, the quotient for the gradient, eliminating t, and the Cartesian equation.

CCEA CCEA 2019-style practice: A curve has parametric equations x = 3cosθ, y = 2sinθ. Find the Cartesian equation, and the gradient at θ = (π)/(4).

From the equations, cosθ = (x)/(3) and sinθ = (y)/(2). Using sin^2θ + cos^2θ = 1: (x^2)/(9) + (y^2)/(4) = 1, an ellipse. For the gradient: (dx)/(dθ) = -3sinθ, (dy)/(dθ) = 2cosθ, so (dy)/(dx) = (2cosθ)/(-3sinθ). At θ = (π)/(4), sinθ = cosθ = (1)/(sqrt(2)), so (dy)/(dx) = (2)/(-3) = -(2)/(3). Markers reward using the Pythagorean identity to eliminate the parameter, the ellipse equation, both derivatives, and the gradient.

Northern IrelandMathsSyllabus dot point

How do parametric equations describe a curve through a parameter, and how do we convert and differentiate them?

Curves defined parametrically, converting between parametric and Cartesian form, and finding the gradient of a parametric curve using the chain rule.

A CCEA A-Level Mathematics answer on curves defined parametrically, converting between parametric and Cartesian form by eliminating the parameter, and finding the gradient of a parametric curve using dy by dx equals dy by dt divided by dx by dt.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

CCEA wants you to understand curves defined by parametric equations, convert between parametric and Cartesian form by eliminating the parameter, and find the gradient of a parametric curve using the chain rule. Parametric curves describe paths that a single Cartesian equation cannot, and the differentiation technique here is needed for motion and modelling.

The answer

Parametric curves

Converting to Cartesian form

To find the Cartesian equation, eliminate the parameter. If the parameter is easy to make the subject (for example $t = \frac{y}{2}$ ), substitute it into the other equation. For trigonometric parameters, use an identity such as $\sin^2\theta + \cos^2\theta = 1$ to eliminate $\theta$ , which often produces a circle or an ellipse. Sometimes the parameter appears in both equations in a way that needs a little manipulation first, such as squaring one equation or adding the two, before the identity can be applied; the goal is always a single equation relating $x$ and $y$ with no parameter left.

Parametric differentiation

Worked example: tangent to a parametric curve

Examples in context

Example 1. Projectile path. A projectile has $x = (u\cos\alpha)t$ and $y = (u\sin\alpha)t - \tfrac{1}{2}gt^2$ , parametric in time. Eliminating $t$ gives the parabolic Cartesian path, linking the parametric description to the trajectory seen in mechanics.

Example 2. An ellipse for a planetary orbit. Writing $x = a\cos\theta$ , $y = b\sin\theta$ traces an ellipse as $\theta$ runs from $0$ to $2\pi$ , a natural parametrisation of an orbit. The trigonometric parameter sweeps the curve smoothly, which a Cartesian equation cannot do directly, and it also records the direction of travel, something the Cartesian form loses entirely.

Try this

Q1. A curve has $x = t + 1$ , $y = t^2$ . Find the Cartesian equation. [2 marks]

Cue. $t = x - 1$ , so $y = (x - 1)^2$ .

Q2. For $x = t^3$ , $y = t^2$ , find $\frac{dy}{dx}$ in terms of $t$ . [2 marks]

Cue. $\frac{dy}{dx} = \frac{2t}{3t^2} = \frac{2}{3t}$ .

Q3. What identity eliminates $\theta$ from $x = \cos\theta$ , $y = \sin\theta$ ? [1 mark]

Cue. $\sin^2\theta + \cos^2\theta = 1$ , giving $x^2 + y^2 = 1$ .

Exam-style practice questions

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

CCEA 20206 marksA curve is given parametrically by

x = t^2

y = 2t

. Find

\frac{dy}{dx}

in terms of

t

, and the Cartesian equation of the curve.

Show worked answer →

Differentiate each: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$ .

Then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$ .

For the Cartesian form, eliminate $t$ : from $y = 2t$ we get $t = \frac{y}{2}$ , so $x = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4}$ , that is $y^2 = 4x$ .

Markers reward both derivatives, the quotient for the gradient, eliminating $t$ , and the Cartesian equation.

CCEA 20196 marksA curve has parametric equations

x = 3\cos\theta

y = 2\sin\theta

. Find the Cartesian equation, and the gradient at

\theta = \frac{\pi}{4}

Show worked answer →

From the equations, $\cos\theta = \frac{x}{3}$ and $\sin\theta = \frac{y}{2}$ . Using $\sin^2\theta + \cos^2\theta = 1$ :

$\frac{x^2}{9} + \frac{y^2}{4} = 1$ , an ellipse.

For the gradient: $\frac{dx}{d\theta} = -3\sin\theta$ , $\frac{dy}{d\theta} = 2\cos\theta$ , so $\frac{dy}{dx} = \frac{2\cos\theta}{-3\sin\theta}$ .

At $\theta = \frac{\pi}{4}$ , $\sin\theta = \cos\theta = \frac{1}{\sqrt{2}}$ , so $\frac{dy}{dx} = \frac{2}{-3} = -\frac{2}{3}$ .

Markers reward using the Pythagorean identity to eliminate the parameter, the ellipse equation, both derivatives, and the gradient.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)