3.4 Parametric equations and Cartesian conversion — Learn

Prerequisites: plotting coordinates; substituting into a formula; rearranging to change the subject; the identity $\cos^2 t + \sin^2 t = 1$ from GCSE.

By the end of this section you can: read the points of a curve from parametric equations; convert a parametric curve to Cartesian form by eliminating the parameter; convert a simple Cartesian equation to a parametric form; state the part of a curve that a restricted parameter covers.

1. Two equations, one parameter

So far every curve has been written as one equation linking $x$ and $y$ , such as $y = x^2$ . A curve can also be described by giving $x$ and $y$ separately in terms of a third quantity. That third quantity is the parameter, and the two equations are parametric equations.

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

Each value of $t$ gives one point $(x, y)$ . As $t$ varies over its range, the point traces the curve.

Example 1. Sketch the curve $x = 2t$ , $y = t^2$ .

Substitute a few values of $t$ and write down the points.

$t$	$x = 2t$	$y = t^2$	point
$-3$	$-6$	$9$	$(-6, 9)$
$-2$	$-4$	$4$	$(-4, 4)$
$-1$	$-2$	$1$	$(-2, 1)$
$0$	$0$	$0$	$(0, 0)$
$1$	$2$	$1$	$(2, 1)$
$2$	$4$	$4$	$(4, 4)$
$3$	$6$	$9$	$(6, 9)$

The points lie on a curve that opens upwards. Notice that $y$ is never negative, because $t^2 \ge 0$ . The curve is a parabola.

2. Eliminating the parameter

To find the Cartesian equation, eliminate $t$ : combine the two equations into one that links $x$ and $y$ only. The usual move is to make $t$ the subject of the simpler equation, then substitute it into the other.

Example 2. Find the Cartesian equation of $x = 2t$ , $y = t^2$ .

The first equation is simpler. Divide by $2$ :

t = \frac{x}{2}.

Substitute into the second equation:

y = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}.

The curve is the parabola $y = \frac{x^2}{4}$ . Check one point: when $t = 2$ , the parametric equations give $(4, 4)$ , and the Cartesian equation gives $y = \frac{16}{4} = 4$ . The point fits.

Example 3. Find the Cartesian equation of $x = 2 + t$ , $y = 5 - 3t$ .

Make $t$ the subject of the first equation:

t = x - 2.

Substitute into the second equation:

y = 5 - 3(x - 2) = 5 - 3x + 6 = 11 - 3x.

The curve is a straight line. The parameter traces it at a steady rate, so $t$ carries the timing as well as the position.

3. Eliminating with an identity

When the parameter appears in a sine and a cosine, the identity $\cos^2 t + \sin^2 t = 1$ removes it. Square both equations and add them.

Example 4. Find the Cartesian equation of $x = 5\cos t$ , $y = 5\sin t$ .

Make each trigonometric function the subject:

\cos t = \frac{x}{5}, \qquad \sin t = \frac{y}{5}.

Square both and add, using the identity:

\left(\frac{x}{5}\right)^2 + \left(\frac{y}{5}\right)^2 = \cos^2 t + \sin^2 t = 1.

Multiply by $25$ :

x^2 + y^2 = 25.

The curve is the circle of radius $5$ centred on the origin.

Example 5. Find the Cartesian equation of $x = 6 + 5\cos t$ , $y = 8 + 5\sin t$ .

Rearrange first, so the trigonometric functions stand alone:

\cos t = \frac{x - 6}{5}, \qquad \sin t = \frac{y - 8}{5}.

Square and add:

\left(\frac{x - 6}{5}\right)^2 + \left(\frac{y - 8}{5}\right)^2 = 1.

Multiply by $25$ :

(x - 6)^2 + (y - 8)^2 = 25.

This is the rim of the lake in the chapter problem, and the path of the drone. The centre is $(6, 8)$ and the radius is $5$ .

4. When the parameter is restricted

The range of $t$ decides which part of the curve is drawn. The Cartesian equation often describes a larger curve, so the range must be carried across.

Example 6. A curve is given by $x = t + 1$ , $y = t^2$ , for $0 \le t \le 3$ . Find its Cartesian equation and the part of the curve it covers.

Make $t$ the subject: $t = x - 1$ . Substitute:

y = (x - 1)^2.

The equation $y = (x - 1)^2$ describes a whole parabola. The parameter only runs from $0$ to $3$ , so $x = t + 1$ runs from $1$ to $4$ . The curve is the part of the parabola with $1 \le x \le 4$ .

The same care applies in reverse. When you choose parametric equations for a Cartesian curve, say which part you mean. For the circle $x^2 + y^2 = 16$ , the pair $x = 4\cos t$ , $y = 4\sin t$ with $0 \le t < 2\pi$ covers the whole circle exactly once.

5. Try it: the parametric explorer

Predict first. Three curves are offered: a parabola, a line, and a circle. For the parabola, predict what happens to the point when $t$ is negative. Write your prediction down before you move the slider.

Test. Choose a curve, then move the slider for $t$ . Watch the point move and read the coordinates. The panel shows the Cartesian equation that the curve is equivalent to.

Explain. For the circle, describe in words what happens to the point as $t$ goes from $0$ to about $6.3$ . Then write one sentence saying why the parameter has to be eliminated to get back to the equation of the circle.

Watch out. Two traps come with parameters.

The Cartesian form can cover more than the curve. Eliminating $t$ can produce an equation with extra points. The pair $x = t^2$ , $y = t^3$ traces only the right-hand branch, because $x = t^2 \ge 0$ , yet $y^2 = x^3$ is satisfied by extra points. Always state the range of $t$ or the range of $x$ .

A parametric curve need not be a function. For $x = t^2$ , $y = t^3$ , one value of $x$ can come from two values of $t$ . A parametric description handles curves that a single $y = f(x)$ cannot.

Re-check: pick one value of $t$ , work out the point from the parametric equations, and confirm that it satisfies the Cartesian equation.

You should now be able to

read points from a parametric curve and sketch it;
eliminate the parameter by substitution;
eliminate the parameter with $\cos^2 t + \sin^2 t = 1$ ;
convert a simple Cartesian equation into parametric form;
state the part of a curve covered by a restricted parameter;
explain why eliminating the parameter can change the curve.