4.1 - Parametric Functions

Heads up: Unit 4 isn't assessed on the AP Precalculus Exam, but it's part of the course framework, and it sets you up for calculus and physics.

Overview

So far, every function has taken one input

x

and given one output

y

. A parametric function works differently: both

x

and

y

depend on a third variable,

t

, called the parameter. Think of

t

as time. As the clock runs, a point moves around the plane, and the parametric function tells you where it is at each moment. This topic covers what a parametric function is, how to build a table of values, and how to sketch its graph.

What a parametric function is

A parametric function in

\mathbb{R}^2

(the set of all ordered pairs of real numbers, which is just the coordinate plane) is a set of two parametric equations:

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

Both

x

and

y

are dependent variables. Each one depends on the same single independent variable, the parameter

t

Input: one value of $t$ .

Output: one point $(x, y)$ in the plane.

Graph: all the points $(x(t), y(t))$ for every $t$ in the domain. The parameter itself isn't on either axis.

Writing it as one function

The point at time

t_i

(x_i, y_i) = (x(t_i), y(t_i))

, so you can package the two equations into one function whose output is an ordered pair:

f(t) = \big(x(t),\, y(t)\big)

Here

x

and

y

are the names of two functions, not just coordinates. For example,

f(t) = (t^2 - 3, 2t)

means

x(t) = t^2 - 3

and

y(t) = 2t

. At

t = 1

f(1) = (1 - 3, 2) = (-2, 2)

Why use a parameter?

It tracks motion. A regular graph shows a path but not when the object was at each spot. A parametric function tells you where and when.

It draws curves that aren't functions of $x$ . Sideways parabolas, circles, and loops fail the vertical line test, so no single $y = f(x)$ can draw them. A parametric function can, because each output depends only on $t$ .

It splits motion into parts. You can study the horizontal part $x(t)$ and the vertical part $y(t)$ one at a time. That idea drives the rest of this unit.

Making a table of values

To build a numerical table for

f(t) = (x(t), y(t))

, evaluate both functions at several values

t_i

in the domain.

Choose $t$ -values spread across the domain. Include the endpoints if the domain is restricted.

Plug each $t$ into $x(t)$ to get the $x$ -coordinate.

Plug the same $t$ into $y(t)$ to get the $y$ -coordinate.

Each row gives one point $(x, y)$ , labeled with its $t$ .

Sketching the graph

Plot the points from the table and connect them in order of increasing

t

, not in order of increasing

x

. The order matters because the curve is the path the point actually travels.

Then add arrows along the curve to show which way the point moves as

t

increases. This direction is called the orientation of the curve, and it's part of the graph.

Restricted domains: Start and end points

The domain of a parametric function is often restricted to an interval like

a \le t \le b

. Then the graph has a start point

f(a) = (x(a), y(a))

and an end point

f(b) = (x(b), y(b))

. Draw both as filled dots, since the curve includes them, and don't extend the curve past them. With no restriction, the curve keeps going in both directions.

Example 1: Table and graph

Make a table of values and sketch the graph of

f(t) = (t^2 - 3, 2t)

for

-2 \le t \le 3

Step 1: Evaluate

x(t) = t^2 - 3

and

y(t) = 2t

at each integer from

t = -2

t = 3

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

Step 2: Plot the points and connect them in order from

t = -2

t = 3

. Add arrows for the direction of motion.

Answer: The graph is a sideways parabola that opens to the right. It starts at

(1, -4)

when

t = -2

, moves left to the vertex

(-3, 0)

t = 0

, then moves right and up to end at

(6, 6)

when

t = 3

This graph fails the vertical line test (the line

x = 1

hits it twice), but it's still a perfectly good parametric function: each

t

gives exactly one point.

Check: Since

y = 2t

, you get

t = \frac{y}{2}

. Substituting into

x(t)

gives

x = \frac{y^2}{4} - 3

, a parabola that opens right with vertex

(-3, 0)

. This move is called eliminating the parameter. It confirms the shape, but it throws away the timing and the direction.

Example 2: Solving for the parameter

For the same function,

f(t) = (t^2 - 3, 2t)

with

-2 \le t \le 3

: (a) At what value(s) of

t

is the point on the line

x = 1

? (b) Is the point

(6, -6)

on the graph?

(a) Set

x(t) = 1

and solve:

t^2 - 3 = 1 \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = \pm 2

Both values are in the domain, and

f(-2) = (1, -4)

and

f(2) = (1, 4)

Answer: The point is on the line

x = 1

t = -2

, where it's at

(1, -4)

, and at

t = 2

, where it's at

(1, 4)

(b) A point is on the graph only if one value of

t

gives both coordinates. Solve each coordinate separately:

x(t) = 6: \;\; t^2 - 3 = 6 \;\Rightarrow\; t = \pm 3 \qquad\qquad y(t) = -6: \;\; 2t = -6 \;\Rightarrow\; t = -3

The only value that works for both is

t = -3

, but

-3

is outside the domain

-2 \le t \le 3

Answer: No. The point

(6, -6)

would be on the graph if the domain included

t = -3

, but with this restricted domain it isn't.

Example 3: A curve that crosses itself

Make a table of values and sketch the graph of

f(t) = (t^2 - 4, \, t^3 - 4t)

for

-2.5 \le t \le 2.5

. Find any point the curve passes through more than once.

Step 1: Build the table. For example, at

t = -1

x(-1) = (-1)^2 - 4 = -3, \qquad y(-1) = (-1)^3 - 4(-1) = -1 + 4 = 3

$t$	$x(t) = t^2 - 4$	$y(t) = t^3 - 4t$	Point
$-2.5$	$2.25$	$-5.625$	$(2.25, -5.625)$
$-2$	$0$	$0$	$(0, 0)$
$-1$	$-3$	$3$	$(-3, 3)$
$0$	$-4$	$0$	$(-4, 0)$
$1$	$-3$	$-3$	$(-3, -3)$
$2$	$0$	$0$	$(0, 0)$
$2.5$	$2.25$	$5.625$	$(2.25, 5.625)$

Step 2: Connect the points in order of

t

. The curve starts at

(2.25, -5.625)

, moves up and left to the origin at

t = -2

, loops counterclockwise through

(-3, 3)

(-4, 0)

, and

(-3, -3)

, returns to the origin at

t = 2

, and then heads up and to the right to end at

(2.25, 5.625)

Step 3: Confirm the repeated point with algebra.

x(t) = 0

when

t^2 = 4

, so

t = \pm 2

. Then

y(2) = 8 - 8 = 0

and

y(-2) = -8 + 8 = 0

, so both times give

(0, 0)

Answer: The curve passes through the origin twice, at

t = -2

and at

t = 2

. A graph of

y = f(x)

can never do this, and it's why the table needs a

t

column: the point

(0, 0)

alone doesn't tell you which pass you're on.

On a graphing calculator

The CED expects you to graph parametric functions with technology, including setting a good window and the right parameter limits. On a TI-84:

Press MODE and change FUNC to PAR.

In Y=, enter $x(t)$ as X1T and $y(t)$ as Y1T. The X,T,θ,n key types $t$ .

In WINDOW, set Tmin and Tmax to the domain, and set Tstep small (like 0.05) so the curve looks smooth. Then set Xmin, Xmax, Ymin, and Ymax so the whole path fits.

Press GRAPH. TRACE moves along the curve in order of $t$ and shows $t$ , $x$ , and $y$ together, and TABLE lists the values by $t$ .

Watch out: the default window uses

0 \le t \le 2\pi

, which can cut off or extend your curve. Always set Tmin and Tmax yourself.

Common mistakes

Connecting points in order of $x$ . Always connect them in order of increasing $t$ . Otherwise loops and sideways curves come out wrong.

Leaving off the arrows. The direction of motion is part of the graph. Without arrows, nobody can tell where the curve starts.

Ignoring the domain. Draw the start and end points as filled dots and stop there. A point is on the graph only if its $t$ is in the domain.

Putting $t$ on an axis. The axes are $x$ and $y$ . The parameter shows up only in the table or as labels on points.

Matching coordinates from different times. For $(a, b)$ to be on the graph, the same $t$ has to give $x(t) = a$ and $y(t) = b$ .

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