Math Core

An equation like $x^2 + y^2 = 1$ tells you where a path goes, but not when you are at each point or which way you travel. Parametric equations add a third variable, usually time, so you can describe motion: a ball in flight, a car on a track, a point on a spinning wheel.

Curves described by a parameter

Definition

Parametric equations

A pair of equations

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

describes a plane curve: as the parameter $t$ runs through an interval, the point $(f(t), g(t))$ traces the curve. The direction the point moves as $t$ increases is the curve's orientation.

Think of $t$ as a clock. At each instant, $f(t)$ tells you the horizontal position and $g(t)$ tells you the vertical position.

Graphing by making a table

Consider $x = t^2 - 2$ , $y = t + 1$ for $-2 \le t \le 2$ .

$t$	$-2$	$-1$	$0$	$1$	$2$
$x$	$2$	$-1$	$-2$	$-1$	$2$
$y$	$-1$	$0$	$1$	$2$	$3$

Plot the points $(x, y)$ in order of increasing $t$ and connect them. The curve is a parabola opening to the right, traced from bottom to top. When you sketch it, draw arrows along the curve to show the orientation.

Notice that the $t$ values never appear on the graph. They are hidden information about timing, which is exactly what the rectangular equation leaves out.

Eliminating the parameter

To find the rectangular equation of the path, eliminate $t$ . The most common method: solve one equation for $t$ and substitute into the other.

Worked example: Eliminating by substitution

Eliminate the parameter from $x = t^2 - 2$ , $y = t + 1$ .

Solution. From the second equation, $t = y - 1$ . Substitute into the first:

x = (y - 1)^2 - 2.

This is a parabola with vertex $(-2, 1)$ opening to the right, matching the table above.

When the equations involve sine and cosine, use the Pythagorean identity instead of solving for $t$ .

Circles and ellipses

For $x = h + a\cos t$ , $y = k + b\sin t$ , $0 \le t \le 2\pi$ :

\cos t = \dfrac{x - h}{a}, \quad \sin t = \dfrac{y - k}{b}, \quad \text{so} \quad \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1.

The curve is centered at $(h, k)$ . It is a circle of radius $a$ when $a = b$ , and an ellipse otherwise. It starts at $(h + a, k)$ and moves counterclockwise.

(cos t, 2 sin t) is an ellipse; (3 cos t, 3 sin t) is a circle of radius 3. Both start on the positive x-axis and go counterclockwise.Open in grapher →

Worked example: A shifted circle

Eliminate the parameter from $x = 1 + 2\cos t$ , $y = -3 + 2\sin t$ , and describe the curve.

Solution. $\cos t = \dfrac{x - 1}{2}$ and $\sin t = \dfrac{y + 3}{2}$ . Since $\cos^2 t + \sin^2 t = 1$ ,

\dfrac{(x - 1)^2}{4} + \dfrac{(y + 3)^2}{4} = 1 \quad\Longrightarrow\quad (x - 1)^2 + (y + 3)^2 = 4.

It is a circle with center $(1, -3)$ and radius 2.

Common mistake

Eliminating the parameter can lose restrictions on $x$ and $y$ . For $x = \sqrt{t}$ , $y = t$ , you get $y = x^2$ , but $x = \sqrt{t}$ is never negative. The curve is only the right half of the parabola ( $x \ge 0$ ). Always ask what values $x$ and $y$ can actually take.

Writing parametric equations

Going the other way, a single curve has many parametrizations. For the graph of $y = f(x)$ , the simplest is $x = t$ , $y = f(t)$ .

For the line segment from $P(x_1, y_1)$ to $Q(x_2, y_2)$ , start at $P$ and add a fraction $t$ of the trip:

x = x_1 + (x_2 - x_1)t, \qquad y = y_1 + (y_2 - y_1)t, \qquad 0 \le t \le 1.

At $t = 0$ you are at $P$ ; at $t = 1$ you are at $Q$ ; at $t = \tfrac{1}{2}$ you are at the midpoint.

Worked example: Parametrizing a segment

Parametrize the segment from $(-2, 5)$ to $(4, 2)$ , and find the point one-third of the way along.

Solution. The changes are $4 - (-2) = 6$ and $2 - 5 = -3$ , so

x = -2 + 6t, \qquad y = 5 - 3t, \qquad 0 \le t \le 1.

At $t = \tfrac{1}{3}$ : $x = -2 + 2 = 0$ and $y = 5 - 1 = 4$ . The point is $(0, 4)$ .

Projectile motion

Parametric equations shine for motion. If an object is launched from height $h$ feet with speed $v_0$ feet per second at angle $\alpha$ above horizontal, and air resistance is ignored, then after $t$ seconds

x = (v_0\cos\alpha)\,t, \qquad y = h + (v_0\sin\alpha)\,t - 16t^2.

The horizontal motion is steady; gravity only pulls on the vertical motion. (The $16$ is half of $32 \text{ ft/s}^2$ ; in meters use $4.9$ .)

Worked example: How far does it go?

A ball is kicked from the ground so that $x = 48t$ and $y = 36t - 16t^2$ (in feet). How long is it in the air, how far does it travel horizontally, and what is its maximum height?

Solution. It lands when $y = 0$ : $36t - 16t^2 = 4t(9 - 4t) = 0$ , so $t = \tfrac{9}{4} = 2.25$ seconds. The horizontal distance is $x = 48(2.25) = 108$ feet.

The height is greatest halfway through the flight, at $t = 1.125$ : $y = 36(1.125) - 16(1.125)^2 = 40.5 - 20.25 = 20.25$ feet.

The path of the ball from t = 0 to t = 2.25 secondsOpen in grapher →

Tip

To find when something happens, solve the equation for the coordinate that describes the event (landing means $y = 0$ , reaching a wall at $x = 30$ means $x = 30$ ), then plug that $t$ into the other equation.

Practice

Practice 1

A curve is given by $x = 2t + 1$ , $y = t - 3$ . Find the point on the curve when $t = 2$ .

Enter a point like (2, -3)

Practice 2

Eliminate the parameter from $x = 2t + 1$ , $y = t - 3$ . Write $y$ as a function of $x$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Describe the curve $x = 4\cos t$ , $y = 4\sin t$ , $0 \le t \le 2\pi$ .

Practice 4

The curve $x = 1 + 3\cos t$ , $y = -2 + 3\sin t$ is a circle. Find its center.

Enter a point like (2, -3)

Practice 5

Which describes the curve $x = \sqrt{t}$ , $y = t + 1$ for $t \ge 0$ ?

Practice 6

The segment from $(1, 2)$ to $(7, -1)$ is parametrized by $x = 1 + 6t$ , $y = 2 - 3t$ , $0 \le t \le 1$ . Find the point where $t = \tfrac{1}{3}$ .

Enter a point like (2, -3)

Practice 7

A projectile follows $x = 40t$ , $y = 64t - 16t^2$ , with distances in feet and $t$ in seconds. How far does it travel horizontally before it hits the ground?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

The curve $x = t^2 - 4$ , $y = t^3 - 4t$ passes through the origin twice, making a loop. Find both values of $t$ where the point is at $(0, 0)$ .

The curve for -π ≤ t ≤ π. The loop is traced for -2 ≤ t ≤ 2.Open in grapher →

Separate answers with commas, e.g. 2, -5