Parametric Equations

A third variable that places each point

A curve is usually given as an equation between x and y. A parametric curve gives x and y separately, each in terms of a third variable called the parameter, often written t. For example, $x=t^2$x = t² and y = 2t.

Each value of t gives one value of x and one value of y, and so one point. At t = 2 the point is $\left(2^2,2\times 2\right)=\left(4,4\right)$(2², 2 × 2) = (4, 4). As t runs through its values, the points run along a path, and that path is the curve.

It helps to think of t as time. Then the two equations say where a moving point is at each moment: $x=t^2$x = t² tells you how far across it is, and y = 2t how far up.

Start with a table

Substitute some values of t and work out both coordinates. At t = 0 the point is (0, 0). At t = 1 it is (1, 2), at t = 2 it is (4, 4), and at t = 3 it is (9, 6). The same value of t goes into both equations every time.

Negative values of t count too. At t = −1 the point is (1, −2), because ${\left(-1\right)}^2=1$(−1)² = 1 and 2 × (−1) = −2. At t = −2 it is (4, −4), and at t = −3 it is (9, −6).

Plot the seven points in order of t and the path appears: it comes in from the lower right, passes through the origin at t = 0, and leaves to the upper right. The curve has a direction, the direction in which t increases.

The path of $x=t^2$x = t², y = 2t, with the points for t = −3 to 3. The positive values of t trace the gold upper half and the negative values the plain lower half. The points at t = 2 and t = −2 share x = 4.

Eliminate the parameter

To get the Cartesian equation, the ordinary equation between x and y, remove t. Solve the simpler equation for t, then substitute into the other.

Here y = 2t is the simpler one, so $t=\frac{y}{2}$t = y/2. Put that into $x=t^2$x = t²: $x={\left(\frac{y}{2}\right)}^2=\frac{y^2}{4}$x = (y/2)² = y²/4. Check a point from the table: at (9, 6), $\frac{y^2}{4}=\frac{36}{4}=9$y²/4 = 36/4 = 9, which is x.

The Cartesian equation keeps the shape and loses the timing. $x=\frac{y^2}{4}$x = y²/4 says which points are on the curve, but not which value of t reaches each one or in which direction the curve is traced.

Two more curves

Take x = t + 1 and $y=t^2$y = t². At t = 3 the point is $\left(3+1,3^2\right)=\left(4,9\right)$(3 + 1, 3²) = (4, 9). To eliminate t, solve the first equation: t = x − 1. Then $y={\left(x-1\right)}^2$y = (x − 1)², a parabola with its lowest point at (1, 0), where t = 0.

Take x = 3 cos t and y = 3 sin t. Here neither equation is easy to solve for t, but an identity removes it: ${\cos }^{2}t+{\sin }^{2}t=1$cos²t + sin²t = 1, so ${\left(\frac{x}{3}\right)}^2+{\left(\frac{y}{3}\right)}^2=1$(x/3)² + (y/3)² = 1, which is $x^2+y^2=9$x² + y² = 9. The curve is the circle of radius 3 about the origin.

At t = 0 the point is (3, 0), and at $t=\frac{\pi }{2}$t = π/2 it is (0, 3), so the point goes round the circle counterclockwise. At $t=\frac{\pi }{4}$t = π/4 it is $\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)$(3/√2, 3/√2), about (2.121, 2.121), and ${\left(\frac{3}{\sqrt{2}}\right)}^2+{\left(\frac{3}{\sqrt{2}}\right)}^2=\frac{9}{2}+\frac{9}{2}=9$(3/√2)² + (3/√2)² = 9/2 + 9/2 = 9.

The circle x = 3 cos t, y = 3 sin t, with the points for t = 0, $\frac{\pi }{4}$π/4, $\frac{\pi }{2}$π/2 and $\pi$π. As t increases the point goes round counterclockwise, and every point is 3 from the origin.

The usual mistakes

Forgetting to square. For $x=t^2$x = t², y = 2t at t = 2, x is $2^2=4$2² = 4, not 2. The point is (4, 4), not (2, 4).

Putting t into one equation only. Both coordinates come from the same value of t. For x = t + 1, $y=t^2$y = t² at t = 3, the point is (4, 9); (3, 9) leaves out the + 1, and (4, 6) doubles where it should square.

Swapping the roles of x and y. Eliminating t from x = t, $y=t^2$y = t² gives $y=x^2$y = x², because y is the one that carries the square.

Reading one value of t as the whole curve. A single t gives a single point; the curve is every point as t runs through its values.