Parametric equations

Parametric equations

A curve can be given by two equations that give $x$ and $y$ in terms of a third variable $t$, called the parameter:

$x = f(t), \qquad y = g(t)$

Each value of $t$ gives one point $(x, y)$. As $t$ changes, the point moves along the curve.

Points on the curve

To find where the curve crosses an axis, set $x = 0$ or $y = 0$, solve for $t$, and use that $t$ in the other equation.

On the $x$-axis $y = 0$, so $t^2 = 1$ and $t = \pm 1$. Then $x = \pm 2$. The curve crosses the $x$-axis at $(2, 0)$ and $(-2, 0)$.

Converting to Cartesian form

To remove the parameter, make $t$ the subject of one equation and substitute it into the other.

From $x = 2t$, $t = \dfrac{x}{2}$. Substitute:

$y = \left(\frac{x}{2}\right)^2 - 1 = \frac{x^2}{4} - 1$

This is a parabola.

When both equations involve trigonometric functions of the parameter, use an identity such as $\sin^2\theta + \cos^2\theta = 1$ instead.

$\cos\theta = \dfrac{x}{3}$ and $\sin\theta = \dfrac{y}{2}$. Then

$\frac{x^2}{9} + \frac{y^2}{4} = 1$

This is an ellipse.

Restrictions on the parameter

If the parameter has a restricted range, only part of the Cartesian curve is traced out. State the range of $x$ or $y$ that the parameter allows.

From $y = 2t$, $t = \dfrac{y}{2}$, so $x = \dfrac{y^2}{4}$, or $y^2 = 4x$. Since $t \geqslant 0$, $y \geqslant 0$ and $x \geqslant 0$. Only the upper half of the parabola is on the curve.

Converting from Cartesian form

A Cartesian curve can be given a parameter by choosing one of $x$ or $y$ to equal a simple expression in $t$.

With $x = t$, $y = t^2 + 3$.

Parametric equations in problems

Problems often ask for where two curves meet, or for a point on a curve with a given property. Substitute the parametric equations into the other curve's equation and solve for $t$.

Substitute: $2t = t^2 - 3$, so $t^2 - 2t - 3 = 0$, which gives $(t - 3)(t + 1) = 0$ and $t = 3$ or $t = -1$. The points are $(9, 6)$ and $(1, -2)$.