Parametric equations of curves, converting between parametric and Cartesian forms, and differentiating parametrically to find gradients and tangents.

What this dot point is asking

WJEC wants you to describe a curve with parametric equations (both coordinates given in terms of a third variable, the parameter), to convert between parametric and Cartesian form by eliminating the parameter, and to differentiate parametrically to find the gradient $\dfrac{dy}{dx}$ and hence tangents and normals. Parametric methods describe curves, like ellipses, that are awkward to write as $y = f(x)$.

The answer

Parametric form

In parametric form, both coordinates are functions of a parameter, usually $t$ (or an angle): $x = f(t)$ and $y = g(t)$. As $t$ runs through its values, the point $(x, y)$ traces the curve. This is the natural description for motion (where $t$ is time) and for curves like circles and ellipses.

Converting to Cartesian form

To eliminate the parameter, either make $t$ the subject of the simpler equation and substitute, or use an identity. For trig parametrisations, the Pythagorean identity is the standard tool.

Parametric differentiation

The gradient comes from the chain rule rearranged:

Examples in context

Example 1. Projectile path

A projectile has $x = 20t$ and $y = 15t - 5t^2$ (metres, with $t$ in seconds). Eliminating $t = \dfrac{x}{20}$ gives $y = 15\left(\dfrac{x}{20}\right) - 5\left(\dfrac{x}{20}\right)^2 = 0.75x - 0.0125x^2$, the parabolic trajectory. The parameter is time, and elimination reveals the shape of the path.

Example 2. An ellipse

The curve $x = 4\cos t$, $y = 3\sin t$ has Cartesian equation $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$, an ellipse with semi-axes $4$ and $3$. Its gradient is $\dfrac{dy}{dx} = \dfrac{3\cos t}{-4\sin t}$, which is zero at the top and bottom (where $\cos t = 0$). Parametric form makes both the shape and the gradient accessible.

Example 3. A point of zero gradient

For $x = t^3 - 3t$, $y = t^2$, the gradient is $\dfrac{dy}{dx} = \dfrac{2t}{3t^2 - 3} = \dfrac{2t}{3(t^2 - 1)}$. This is zero when $t = 0$, giving the point $(0, 0)$, and undefined (a vertical tangent) when $t = \pm 1$. Reading the special points off the parameter is far easier than from a Cartesian equation, which is why parametric form is preferred for motion and curve-sketching problems.

Try this

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

• Cue. $t = \dfrac{x}{2}$, so $y = \left(\dfrac{x}{2}\right)^2 = \dfrac{x^2}{4}$.

Q2. For $x = t + 1$, $y = t^2$, find $\dfrac{dy}{dx}$ in terms of $t$. [2 marks]

• Cue. $\dfrac{dx}{dt} = 1$, $\dfrac{dy}{dt} = 2t$, so $\dfrac{dy}{dx} = 2t$.

Q3. A curve has $x = \sin t$, $y = \cos t$. State its Cartesian equation. [2 marks]

• Cue. $x^2 + y^2 = \sin^2 t + \cos^2 t = 1$, the unit circle.