Parametric Functions — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A drone keeps a flight log: every second it records how far east it is and how far north it is. Neither list alone shows the flight. Put the two numbers together for each second and plot them, and the path appears, with a direction of travel. That is a parametric function.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $g(t) = t^2 - 2t$ , what is $g(3)$ ?

$3$
$9$
$15$
$-3$

Show the answer

Substitute with parentheses: $g(3)$ $= (3)^2 - 2(3)$ $= 9 - 6$ $= 3$ .

Correct: $3$ :
$9$ :
$15$ :
$-3$ :

2. The point $(-3, 2)$ is how far from the $y$ -axis, and on which side?

3 units, to the left
3 units, to the right
2 units, above
2 units, below

Show the answer

The first coordinate, $-3$ , says 3 units left of the $y$ -axis.

Correct: 3 units, to the left:
3 units, to the right:
2 units, above:
2 units, below:

Part 4 · See it

See it first

The curve x(t) = t squared minus 2t, y(t) = 2t minus 1 for t from -1 to 3, a parabola opening to the right. Points are marked at t = -1 (3, -3), t = 0 (0, -1), t = 1 (-1, 1), t = 2 (0, 3) and t = 3 (3, 5). Arrows show the point moves left and up until t = 1, then right and up. — The curve $x(t) = t^2 - 2t$ , $y(t) = 2t - 1$ for $-1 \le t \le 3$ : each $t$ gives one point, and the arrows show the order the points come in. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Pick a value of the parameter $t$ $x(t)$ and $y(t)$ give one point $(x(t), y(t))$
Let $t$ run through an intervalthe points trace a curve in the plane
Plot the points in order of increasing $t$ the curve gets a direction of travel, shown with arrows
Two values of $t$ give the same $x$ but different $y$ the curve is not the graph of a function of $x$ , even though $x(t)$ and $y(t)$ are both functions of $t$

Part 6 · Key ideas

Key ideas

A parametric function $f(t) = (x(t), y(t))$ turns each input $t$ , the parameter, into a point of the plane.
To sketch one, make a table: values of $t$ , then $x(t)$ and $y(t)$ , then plot the points in order of increasing $t$ and add arrows.
The curve can go left, turn around, or cross itself. One $x$ with two $y$ -values means the curve is not the graph of a function of $x$ .
If $x(t)$ is easy to solve for $t$ , substitute into $y(t)$ to write $y$ as a function of $x$ .

Part 7 · Misconception

A common mistake

The wrong idea: The graph of $(x(t), y(t))$ is the graph of $y(t)$ against $t$ .

What actually happens: The curve plots $y$ against $x$ . The parameter $t$ is not on either axis: it only says which point comes when. The graph of $y(t)$ against $t$ is a different picture.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A drone flight log

Position of a drone at selected times: x(t) is meters east of the launch pad and y(t) is meters north of it
$t$ (seconds)	0	1	2	3	4
$x(t)$ (meters)	2	5	6	5	2
$y(t)$ (meters)	1	2	4	7	11

1. Where is the drone at $t = 3$ seconds?

$(5, 7)$
$(7, 5)$
$(5, 4)$
$(3, 5)$

Show the answer

Read both rows in the column $t = 3$ : $x(3) = 5$ and $y(3) = 7$ . The point is $(x(3), y(3)) = (5, 7)$ , 5 meters east and 7 meters north.

Correct: $(5, 7)$ : $(x(3), y(3)) = (5, 7)$ .
$(7, 5)$ : The coordinates are swapped; $x$ comes first.
$(5, 4)$ : $4$ is $y(2)$ , from the wrong column.
$(3, 5)$ : $3$ is the time, not a coordinate.

2. Between which two readings does the drone first move west?

Between $t = 2$ and $t = 3$
Between $t = 1$ and $t = 2$
Between $t = 3$ and $t = 4$
Between $t = 0$ and $t = 1$

Show the answer

West means $x$ decreases. $x$ goes $2, 5, 6, 5, 2$ : it first falls from $6$ to $5$ , between $t = 2$ and $t = 3$ .

Correct: Between $t = 2$ and $t = 3$ : $x$ drops from $6$ to $5$ here: the first move west.
Between $t = 1$ and $t = 2$ : $x$ rises from $5$ to $6$ here, so the drone moves east.
Between $t = 3$ and $t = 4$ : The drone does move west here, but it already did between $t = 2$ and $t = 3$ .
Between $t = 0$ and $t = 1$ : $x$ rises from $2$ to $5$ , so this move is east.

3. The drone is at $x = 5$ at two different times. What does that tell you?

At $x = 5$ the heights are $2$ and $7$ , so the path is not the graph of a function of $x$
The drone was at one and the same point at $t = 1$ and again at $t = 3$
The drone hovered in place from $t = 1$ to $t = 3$ before it moved west
$x(t)$ is not a function of $t$ , since two times share the output $5$

Show the answer

At $t = 1$ the drone is at $(5, 2)$ and at $t = 3$ at $(5, 7)$ : one $x$ -value, two $y$ -values. The path in the $xy$ -plane fails to be the graph of a function of $x$ , while $x(t)$ and $y(t)$ are still functions of $t$ .

Correct: At $x = 5$ the heights are $2$ and $7$ , so the path is not the graph of a function of $x$ : $(5, 2)$ and $(5, 7)$ share $x = 5$ with different $y$ .
The drone was at one and the same point at $t = 1$ and again at $t = 3$ : The $y$ -values differ ( $2$ and $7$ ), so they are different points.
The drone hovered in place from $t = 1$ to $t = 3$ before it moved west: The drone went east to $x = 6$ and back; it did not stay put.
$x(t)$ is not a function of $t$ , since two times share the output $5$ : Each $t$ has one $x(t)$ , so $x(t)$ is a function of $t$ ; repeated outputs are allowed.

4. A curve is given by $x(t) = 2t + 1$ and $y(t) = t^2$ . Write $y$ as a function of $x$ .

Type an expression for $y =$ .

Show the answer

Solve $x = 2t + 1$ for $t$ : $t = \frac{x - 1}{2}$ . Substitute into $y = t^2$ : $y = \left(\frac{x - 1}{2}\right)^2$ .

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

5. A curve is given by $x(t) = 3 - t$ and $y(t) = t^2 + 1$ . What is $y$ at the point where the curve crosses the $y$ -axis?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

On the $y$ -axis, $x = 0$ : $3 - t = 0$ , so $t = 3$ . Then $y(3) = (3)^2 + 1 = 10$ .

Answer: 10

6. Which function traces the same path as $f(t) = (t, t^2)$ , $-1 \le t \le 2$ , but in the opposite direction?

$(-t, t^2)$ for $-2 \le t \le 1$
$(t, -t^2)$ for $-1 \le t \le 2$
$(-t, t^2)$ for $-1 \le t \le 2$
$(t^2, t)$ for $-1 \le t \le 2$

Show the answer

For $(-t, t^2)$ , $y = t^2 = (-t)^2 = x^2$ . With $-2 \le t \le 1$ , $x = -t$ runs from $2$ down to $-1$ , so the path goes from $(2, 4)$ to $(-1, 1)$ : the same points, backward.

Correct: $(-t, t^2)$ for $-2 \le t \le 1$ : It starts at $(2, 4)$ and ends at $(-1, 1)$ , the reverse of $f$ .
$(t, -t^2)$ for $-1 \le t \le 2$ : $y = -x^2$ is the parabola reflected across the $x$ -axis, a different path.
$(-t, t^2)$ for $-1 \le t \le 2$ : With $-1 \le t \le 2$ , $x = -t$ runs from $1$ to $-2$ : a different piece of the parabola.
$(t^2, t)$ for $-1 \le t \le 2$ : Swapping the coordinates gives $x = y^2$ , a parabola opening right.

Part 9 · Summary

Summary

A parametric function

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

sends each value of the parameter

t

to a point. As

t

increases the points trace a curve with a direction. The curve is drawn in the

xy

-plane, and it need not be the graph of a function of

x

.

Part 10 · Up next

What comes next

Part 11 · Connections