Parametric Equations Overview – A Level Maths Year 2

Up to now every curve has been described by a single equation linking $x$ and $y$. Parametric equations describe a curve differently: both $x$ and $y$ are given in terms of a third variable, the parameter.

Imagine a dot moving across the page. At each moment $t$ it has an $x$-coordinate and a $y$-coordinate. Writing $x = \mathrm{f}(t)$ and $y = \mathrm{g}(t)$ describes not just the path, but the journey along it.

DfE content area C · Coordinate geometry (Year 2)

Understand and use the parametric equations of curves and conversion between Cartesian and parametric forms
Use parametric equations in modelling in a variety of contexts

Where it leads. Later this year you will find gradients of parametric curves on Implicit and Parametric Differentiation, and model moving objects in Projectiles.

Word	Meaning	Example
Parameter	The third variable that $x$ and $y$ both depend on.	$t$ in $x = t^2$, $y = 2t$
Parametric equations	The pair of equations $x = \mathrm{f}(t)$, $y = \mathrm{g}(t)$.	$x = 3\cos\theta$, $y = 3\sin\theta$
Cartesian equation	An equation linking $x$ and $y$ only.	$x^2 + y^2 = 9$
Eliminate	Remove the parameter to get the Cartesian equation.	$t = \tfrac{y}{2} \Rightarrow x = \tfrac{y^2}{4}$
Domain (of the curve)	The set of $x$ values the curve covers.	$x = \sin\theta$ gives $-1 \leqslant x \leqslant 1$
Range (of the curve)	The set of $y$ values the curve covers.	$y = t^2$ gives $y \geqslant 0$

The Cartesian equation can describe more than the curve. $x = \sin\theta$, $y = \cos 2\theta$ gives $y = 1 - 2x^2$, but only for $-1 \leqslant x \leqslant 1$. Always carry the restriction across.

1 — Plotting

Choose values of $t$, tabulate $x$ and $y$, plot and join in order of $t$

2 — Eliminating the parameter

Algebraic: make $t$ the subject of one equation and substitute
Trigonometric: isolate $\sin\theta$, $\cos\theta$ (or $\tan\theta$, $\sec\theta$) and use an identity

3 — Domain and range

Find the possible $x$ and $y$ values from the allowed values of the parameter

4 — Intersections

Axes: solve $x(t) = 0$ or $y(t) = 0$ · a line: substitute $x(t)$, $y(t)$ into its equation

Worked Example — Several techniques in one question

A curve has equations $x = t^2 - 1$, $y = 2t$. (a) Find a Cartesian equation. (b) Find where it crosses the $y$-axis. (c) Find where it meets the line $y = x - 2$.

①(a) $t = \dfrac{y}{2}$, so $x = \dfrac{y^2}{4} - 1$, i.e. $y^2 = 4(x + 1)$.

②(b) $x = 0$: $t^2 = 1$, $t = \pm 1$, giving $(0, 2)$ and $(0, -2)$.

③(c) $2t = (t^2 - 1) - 2 \Rightarrow t^2 - 2t - 3 = 0 \Rightarrow (t - 3)(t + 1) = 0$.

④$t = 3$: $(8, 6)$. $t = -1$: $(0, -2)$. Check: $6 = 8 - 2$ ✓, $-2 = 0 - 2$ ✓.

For (c) you could use the Cartesian equation instead, but substituting the parametric forms gives a simpler equation in $t$ — and the points follow at once.

CParametric and Cartesian FormsConverting to Cartesian form (including with trigonometric identities), sketching, intersections, domain and range.
CParametric ModellingThrown balls, big wheels, rolling wheels and boats: interpreting the parameter and restricting its values.

Leaving the parameter in. A Cartesian equation must contain only $x$ and $y$.

Forgetting to square the coefficient. From $x = 3\cos\theta$, $\cos^2\theta = \dfrac{x^2}{9}$, not $\dfrac{x^2}{3}$.

Range from end-points only. For $y = t^2$ with $-1 \leqslant t \leqslant 2$ the least value is $0$, at $t = 0$, not $1$.

Confusing paths crossing with objects meeting. Two moving objects collide only if they are at the same point for the same value of $t$.

Degrees and radians. In modelling with $\sin(\omega t)$, $\omega t$ is in radians unless you are told otherwise. Set your calculator accordingly.

Parametric

$x = \mathrm{f}(t)$, $y = \mathrm{g}(t)$.

Eliminate

$t$ = …, then substitute.

Pythagorean

$\sin^2\theta + \cos^2\theta = 1$.

Also useful

$1 + \tan^2\theta = \sec^2\theta$; $\cos 2\theta = 1 - 2\sin^2\theta$.

Circle

$x = a + r\cos\theta$, $y = b + r\sin\theta$.

Ellipse

$x = p\cos\theta$, $y = q\sin\theta$.

Restrictions

Domain and range come from the parameter.

Models

Interpret $t$; restrict it to sensible values.

Ten questions covering the whole topic. Work each one through before opening the solution.

Question 1

Find a Cartesian equation for $x = 3t - 1$, $y = t + 2$.

▶ Show solution

$t = y - 2$, so $x = 3(y - 2) - 1 = 3y - 7$.

Question 2

For $x = t^2$, $y = t^3$, find the points where $t = 0, 1, 2$ and a Cartesian equation.

▶ Show solution

$(0, 0)$, $(1, 1)$, $(4, 8)$.

$x^3 = t^6 = y^2$, so $y^2 = x^3$.

Question 3

Describe the curve $x = 2\cos\theta$, $y = 2\sin\theta$, $0 \leqslant \theta < 2\pi$.

▶ Show solution

$x^2 + y^2 = 4$: a circle, centre the origin, radius $2$, traced once anticlockwise starting from $(2, 0)$.

Question 4

Find $y$ in terms of $x$ for $x = t - 1$, $y = \dfrac{1}{t}$, $t > 0$, and state the domain.

▶ Show solution

$t = x + 1$, so $y = \dfrac{1}{x + 1}$.

$t > 0$ means $x > -1$.

Question 5

Find a Cartesian equation for $x = 4\cos\theta$, $y = 3\sin\theta$.

▶ Show solution

$\cos\theta = \tfrac{x}{4}$, $\sin\theta = \tfrac{y}{3}$, so $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ (an ellipse).

Question 6

Find where the curve $x = t^2$, $y = 4t$ meets the line $y = x$.

▶ Show solution

$4t = t^2 \Rightarrow t(t - 4) = 0 \Rightarrow t = 0$ or $t = 4$.

Points $(0, 0)$ and $(16, 16)$.

Question 7

A curve has $x = 3 + 2t$, $y = t^2$, $-1 \leqslant t \leqslant 2$. Find the domain and range.

▶ Show solution

$x$ increases with $t$: domain $1 \leqslant x \leqslant 7$.

$y = t^2$: least $0$ at $t = 0$, greatest $4$ at $t = 2$. Range $0 \leqslant y \leqslant 4$.

Question 8

Find a Cartesian equation for $x = 2\sin\theta$, $y = \cos 2\theta$, stating the domain.

▶ Show solution

$\cos 2\theta = 1 - 2\sin^2\theta$ and $\sin\theta = \tfrac{x}{2}$.

$y = 1 - 2\left(\tfrac{x}{2}\right)^2 = 1 - \dfrac{x^2}{2}$, for $-2 \leqslant x \leqslant 2$.

Question 9

Find a Cartesian equation for $x = \sec\theta$, $y = 2\tan\theta$.

▶ Show solution

$\sec^2\theta - \tan^2\theta = 1$, with $\sec\theta = x$ and $\tan\theta = \tfrac{y}{2}$.

$x^2 - \dfrac{y^2}{4} = 1$.

Question 10

A ball is kicked from ground level. After $t$ seconds it is at $x = 8t$, $y = 10t - 5t^2$ (metres).

(a) Find when and where it lands. (b) Find its greatest height. (c) Find the Cartesian equation of its path, with its domain. (d) Give one limitation of the model.

▶ Show solution

(a) $5t(2 - t) = 0 \Rightarrow t = 2$ s; then $x = 16$ m.

(b) $y = -5(t - 1)^2 + 5$: greatest height $5$ m at $t = 1$ s.

(c) $t = \tfrac{x}{8}$: $y = \dfrac{10x}{8} - \dfrac{5x^2}{64} = \dfrac{5x}{4} - \dfrac{5x^2}{64}$, for $0 \leqslant x \leqslant 16$.

(Check: $x = 16$ gives $20 - 20 = 0$ ✓; $x = 8$ gives $10 - 5 = 5$ ✓.)

(d) It ignores air resistance (or treats the ball as a particle, or ignores spin).