Unit 4 · Topic 4.1 · about 30 minutes · 7 parts · 13 scored checks, 1 of them multiple choice

AP Precalculus 4.1: Parametric Functions

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Not assessed on the AP Exam. Unit 4 is part of the course, but nothing in it is on the exam.

The graph of the parametric function f(t) = (t squared minus 1, t cubed minus 3t) for t from -2 to 2. Five points from a table of values are labeled with their t values: (3, -2) at t = -2, (0, 2) at t = -1, (-1, 0) at t = 0, (0, -2) at t = 1 and (3, 2) at t = 2. The path starts at (3, -2), marked with a square, climbs to the left through (0, 2), loops through (-1, 0), comes down through (0, -2), crosses itself at (2, 0) on the positive x-axis and ends at (3, 2), marked with a red dot. Arrows along the path point the way t increases.

Part 1 Think first A drone over a field

Guess before you read on. A wrong guess still helps the right answer stick.

Predict first

During a 4-second test flight, a drone's position $t$ seconds in is $x = (t - 2)^{2}$ meters east of a fence post and $y = t - 2$ meters north of it, for $0 \leq t \leq 4$ . Seen from above, what shape is its path?

In short. A parametric function $f (t) = (x (t), y (t))$ uses one independent variable, the parameter $t$ , to give both coordinates of a point in the plane.

To make a table of values, choose several values of $t$ in the domain and evaluate both $x (t)$ and $y (t)$ at each one.
To sketch the graph, join the points from the table in order of increasing $t$ , which is often not the order from left to right.
When the domain is restricted, as in $- 2 \leq t \leq 2$ , the graph has a start point at the least value of $t$ and an end point at the greatest.

Part 2 Learn it Four short ideas, each checked before the next

2a. Two equations, one input

Until now, $y$ has depended on $x$ . A parametric function works differently. Both $x$ and $y$ depend on a third variable, $t$ , called the parameter. You get two parametric equations, one for each coordinate:

x (t) = 4 - t^{2} y (t) = t^{2} + t

Here $x$ and $y$ are the names of two functions, and $t$ is the one independent variable they share. Put in one value of $t$ and you get one $x$ and one $y$ , which make an ordered pair. So the two equations act as a single function:

f (t) = (x (t), y (t)) = (4 - t^{2}, t^{2} + t)

Its outputs are points in $ℝ^{2}$ , the set of all ordered pairs of two real numbers. For example, $f (1) = (4 - 1, 1 + 1) = (3, 2)$ .

A point on its own does not show you $t$ . Looking at $(3, 2)$ on a graph, nothing tells you it came from $t = 1$ . You keep track of $t$ yourself.

Quick check · Fill the statement

The parametric function is $f (t) = (t^{2} - 1, 3 t)$ . Tap a blank, then tap an answer.

The single independent variable of $f$ is , called the parameter. In $f (t) = (x (t), y (t))$ , the letters $x$ and $y$ are names of two of it. The output $f (2)$ is the point .

2b. A table of values

To see what a parametric function draws, build a table. Choose several values of $t$ in the domain, and evaluate both $x (t)$ and $y (t)$ at each one. For $f (t) = (4 - t^{2}, t^{2} + t)$ with domain $- 2 \leq t \leq 2$ , the whole numbers give five rows.

A table of values for $f (t) = (4 - t^{2}, t^{2} + t)$
$t$	$x (t) = 4 - t^{2}$	$y (t) = t^{2} + t$	Point
$- 2$	$0$	$2$	$(0, 2)$
$- 1$	$3$	$0$	$(3, 0)$
$0$	$4$	$0$	$(4, 0)$
$1$	$3$	$2$	$(3, 2)$
$2$	$0$	$6$	$(0, 6)$

Be careful with negative values of $t$ . When $t = - 2$ , $t^{2}$ is 4, not $- 4$ , so $x (- 2) = 4 - 4 = 0$ and $y (- 2) = 4 - 2 = 2$ .

Two rows have $x = 3$ , and two have $y = 0$ . That is fine. Each of $x$ and $y$ is a function of $t$ , and nothing requires each $x$ to go with only one $y$ . Put both ends of the domain in your table. You will need them in a moment.

Quick check · Calculate

The parametric function is $f (t) = (3 - t^{2}, 2 t - 1)$ . What is the $x$ -coordinate of the point $f (- 2)$ ?

Your answer

2c. Join the points in order of t

To sketch the graph, plot the points from the table and connect them in order of increasing $t$ . Not left to right, and not bottom to top. The order is set by $t$ .

For $f (t) = (4 - t^{2}, t^{2} + t)$ that order is $(0, 2)$ , then $(3, 0)$ , $(4, 0)$ , $(3, 2)$ and $(0, 6)$ . The curve swings right, turns at $(4, 0)$ , and comes back left while it climbs. Small arrows along a sketch show the way $t$ increases.

Join the same five points from left to right instead, and you get $(0, 2)$ to $(0, 6)$ to $(3, 0)$ and so on. That zigzag is not the graph of $f$ .

More rows, such as $t = - 1.5$ or $t = 0.5$ , show more of where the curve bends. They never change the rule: every point goes between the points for the values of $t$ just below and just above its own.

Quick check · Put it in order

The parametric function $g (t) = (t^{2} - 2 t, t^{2} - 4)$ has domain $- 1 \leq t \leq 3$ , and its graph passes through each of these five points once. Tap them in the order a sketch joins them.

2d. Where the graph starts and stops

The domain of a parametric function is often restricted, like $- 2 \leq t \leq 2$ . Then the graph does not go on forever. It has a start point, at the least value of $t$ , and an end point, at the greatest.

For $f (t) = (4 - t^{2}, t^{2} + t)$ on $- 2 \leq t \leq 2$ , the start point is $f (- 2) = (0, 2)$ and the end point is $f (2) = (0, 6)$ . The start point is not the lowest point of the graph, and it is not $f (0)$ . It is wherever $t$ begins.

Change the domain and the ends move. On $- 1 \leq t \leq 1$ the same function starts at $(3, 0)$ , ends at $(3, 2)$ , and draws only the middle of the curve. In a sketch, mark both ends and stop the curve there.

Quick check · Multiple choice

The parametric function $h (t) = (1 - 2 t, t^{2} - 3)$ has domain $- 1 \leq t \leq 2$ . What are the start point and the end point of its graph?

Part 3 Watch it done From a formula to a sketch

Read each step, then do the last one yourself.

Build a table of values for $f (t) = (t^{2} - 1, t^{3} - 3 t)$ on the domain $- 2 \leq t \leq 2$ , using the whole numbers and $t = 0.5$ , and sketch its graph.

Step 1 · Choose values of t
The domain runs from $- 2$ to 2, so take the whole numbers $t = - 2, - 1, 0, 1, 2$ . That includes both ends. The extra row for $t = 0.5$ goes between the rows for $t = 0$ and $t = 1$ , where the curve bends.
Step 2 · Evaluate x(t)
$x (t) = t^{2} - 1$ gives 3, 0, $- 1$ , 0 and 3 at the whole numbers. The two negative values of $t$ give the same $x$ as their positive partners, because squaring removes the sign. The extra row has $x (0.5) = 0.25 - 1 = - 0.75$ .
Step 3 · Evaluate y(t)
$y (t) = t^{3} - 3 t$ . A cube keeps its sign: $y (- 2) = - 8 + 6 = - 2$ , $y (- 1) = - 1 + 3 = 2$ , $y (0) = 0$ , $y (1) = 1 - 3 = - 2$ and $y (2) = 8 - 6 = 2$ . So the whole-number rows give the points $(3, - 2)$ , $(0, 2)$ , $(- 1, 0)$ , $(0, - 2)$ and $(3, 2)$ .
Why this step?
When the points are joined, why does the sketch go from $(0, 2)$ to $(- 1, 0)$ , and not straight down to $(0, - 2)$ , the point directly below it?
$(0, 2)$ comes from $t = - 1$ . The next value of $t$ is 0, which gives $(- 1, 0)$ . The point $(0, - 2)$ comes from $t = 1$ , two steps later, so the path reaches it only after passing through $(- 1, 0)$ . Sharing an $x$ -coordinate says nothing about the order.
Your step · Calculate

Finish it. One cell is left in the table of $f (t) = (t^{2} - 1, t^{3} - 3 t)$ : the $y$ -coordinate in the row for $t = 0.5$ , where $x (0.5) = - 0.75$ . What is $y (0.5)$ ? Give a decimal.

Your answer

Now join the six points in order of $t$ . Start at $(3, - 2)$ , where $t = - 2$ . Go to $(0, 2)$ , then $(- 1, 0)$ , then $(- 0.75, - 1.375)$ , then $(0, - 2)$ , and end at $(3, 2)$ , where $t = 2$ . The path climbs to the left, loops round through $(- 1, 0)$ , dips through the new point, and then heads right and up, crossing its own earlier stretch on the positive $x$ -axis.

The row for $t = 0.5$ never changes the order rule. It slots in by its value of $t$ and shows more of where the curve bends. The diagram on this page is this same graph, with the start point marked by a square and the end point by a red dot.

Part 4 Try it Move the ends of a graph

Warm up in the lab first. It graphs the parametric function with $x (t) = (t - 1)^{2} - 2$ and $y (t) = t$ . The top graph is the path in the plane, and the two below it are $x$ against $t$ and $y$ against $t$ .

Lab

Parametric Grapher

Drag Parameter t and watch the point move along the path in the order $t$ gives. The square is the start point and the filled dot is the end point. Then move Interval of t starts at and Interval of t ends at, and watch both ends of the graph move while the curve between them stays on the same track.

Open the full Parametric Grapher lab

Now a new function, on paper. Make your own table for $q (t) = (t^{2} - 2 t, 3 - t)$ with domain $0 \leq t \leq 4$ , using the whole numbers from 0 to 4. Then answer the two questions.

Try it · Calculate

The parametric function $q (t) = (t^{2} - 2 t, 3 - t)$ has domain $0 \leq t \leq 4$ . If the domain is stretched to $0 \leq t \leq 5$ , what is the $x$ -coordinate of the new end point of its graph?

Your answer

Try it · Decide and justify

For $q (t) = (t^{2} - 2 t, 3 - t)$ on $0 \leq t \leq 4$ , a table at $t = 0, 1, 2, 3, 4$ gives the points $(0, 3)$ , $(- 1, 2)$ , $(0, 1)$ , $(3, 0)$ and $(8, - 1)$ . Mia sketches the graph of $q$ by joining those points from left to right. Is her sketch the graph of $q$ ?

Which reason supports your answer?

Part 5 Explain it Put it in writing

Nothing you type leaves this page. Compare your answer with the model and tick off each point on the list.

Unit 4 is not on the AP Exam, so this is written practice for the course.

Written practice

The parametric function $g$ is given by $g (t) = (t^{2} - 2 t, t + 1)$ for $- 1 \leq t \leq 3$ . (i) Construct a table of values for $g$ at $t = - 1, 0, 1, 2$ and 3. (ii) Identify the start point and the end point of the graph of $g$ , and explain how you know which is which. (iii) Describe how to sketch the graph of $g$ from your table.

Your answer stays on this page. It is never sent or saved.

Model answer

(i) $t = - 1$ : $(3, 0)$ . $t = 0$ : $(0, 1)$ . $t = 1$ : $(- 1, 2)$ . $t = 2$ : $(0, 3)$ . $t = 3$ : $(3, 4)$ . (ii) The start point is $(3, 0)$ and the end point is $(3, 4)$ . The domain runs from $t = - 1$ to $t = 3$ , the start point comes from the least value of $t$ and the end point from the greatest. (iii) Plot the five points and join them in order of increasing $t$ : $(3, 0)$ to $(0, 1)$ to $(- 1, 2)$ to $(0, 3)$ to $(3, 4)$ . The path moves left, turns at $(- 1, 2)$ and comes back right, so joining the points from left to right would draw a different picture. The graph begins at $(3, 0)$ and stops at $(3, 4)$ , because the domain is restricted.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Gives the five points $(3, 0)$ , $(0, 1)$ , $(- 1, 2)$ , $(0, 3)$ and $(3, 4)$ , each with its value of $t$ .
Identifies $(3, 0)$ as the start point and $(3, 4)$ as the end point, because they come from the least and greatest values of $t$ in the domain.
Joins the points in order of increasing $t$ , not from left to right, so the path turns back at $(- 1, 2)$ .
Stops the sketch at the start and end points because the domain is restricted.

Part 6 Check yourself A robot vacuum on a grid

The two case questions use the same robot. Answer them from its formulas and table.

A robot vacuum cleans a room. Its position $t$ seconds after it starts is $v (t) = (x (t), y (t))$ , in meters from the room's west wall and south wall, where $x (t) = 5 - t$ and $y (t) = t^{2} - 4 t + 5$ , for $0 \leq t \leq 4$ . The table gives four of its positions.

Positions of the robot vacuum, in meters
$t$ (seconds)	$x (t)$	$y (t)$
$0$	$5$	$5$
$1$	$4$	$2$
$3$	$2$	$2$
$4$	$1$	$5$

Case question 1 · Calculate

The robot vacuum's position is $v (t) = (5 - t, t^{2} - 4 t + 5)$ for $0 \leq t \leq 4$ . What is the $y$ -coordinate of $v (2)$ ?

Your answer

Case question 2 · Select all

The robot vacuum's position is $v (t) = (5 - t, t^{2} - 4 t + 5)$ for $0 \leq t \leq 4$ . Which two statements about the graph of $v$ are true? Pick two.

Select 2 answers, then check.

The rest stand on their own and do not use the case.

Check yourself · Spot the error

A student finds the point of $f (t) = (t^{2} + t, 4 - 2 t)$ for $t = - 3$ . Tap the first line that goes wrong.

Check yourself · Sort it

Each card describes one letter in a parametric function $f (t) = (x (t), y (t))$ . Tap a card, then tap the letter it describes.

The parameter $t$

$x$ or $y$

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 3.15 Rates of Change in Polar Functions · Spot the error

A polar function $r = f (θ)$ has $f (1) = 2.5$ and $f (2.5) = - 0.5$ . A student finds the average rate of change of $r$ with respect to $θ$ from $θ = 1$ to $θ = 2.5$ . Tap the first line that goes wrong.

Work this one on paper. A graphing app plots $h (x) = \sin x + \sqrt{3} \cos x$ , and the curve looks like an ordinary sine wave, only taller and shifted. The sine sum identity says why. Expand $2 \sin (x + \frac{π}{3})$ :

2 (\sin x \cos \frac{π}{3} + \cos x \sin \frac{π}{3}) = 2 (\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x) = \sin x + \sqrt{3} \cos x

So $h (x) = 2 \sin (x + \frac{π}{3})$ for every $x$ . Use that form to read off what the first form hides.

From 3.12 Equivalent Representations of Trigonometric Functions · Calculate

The function $h (x) = \sin x + \sqrt{3} \cos x$ is equal to $2 \sin (x + \frac{π}{3})$ for every $x$ . What is the greatest value of $h (x)$ ?

Your answer

In three lines

A parametric function $f (t) = (x (t), y (t))$ gives both coordinates from one independent variable, the parameter $t$ .
A table of values evaluates $x (t)$ and $y (t)$ at several values of $t$ , and the sketch joins those points in order of increasing $t$ .
A restricted domain gives the graph a start point at the least value of $t$ and an end point at the greatest.

Vocabulary

Parametric function: A function $f (t) = (x (t), y (t))$ in $ℝ^{2}$ that gives the two dependent variables $x$ and $y$ from a single independent variable $t$ .
Parametric equations: The two equations of a parametric function, one for $x (t)$ and one for $y (t)$ .
Parameter: The single independent variable $t$ that both $x$ and $y$ depend on.
Numerical table of values: A table made by evaluating $x_{i}$ and $y_{i}$ at several values $t_{i}$ within the domain.
Start and end points: The points of the graph at the least and greatest values of $t$ , which exist when the domain of $f$ is restricted.

Course alignment, for teachers

AP Precalculus topic 4.1, Unit 4: Functions Involving Parameters, Vectors, and Matrices.

Skill 1.A: Solve equations and inequalities represented analytically, with and without technology.
Skill 2.B: Construct equivalent graphical, numerical, analytical, and verbal representations of functions that are useful in a given mathematical or applied context, with and without technology.

This lesson maps all 5 essential knowledge statements of the topic to the part that teaches each one: 5 are checked by a scored question. The map is in the lesson file.

Part 1 Think first A drone over a field

Part 2 Learn it Four short ideas, each checked before the next

2a. Two equations, one input

2b. A table of values

2c. Join the points in order of t

2d. Where the graph starts and stops

Part 3 Watch it done From a formula to a sketch

Part 4 Try it Move the ends of a graph

Parametric Grapher

Part 5 Explain it Put it in writing

Part 6 Check yourself A robot vacuum on a grid

Part 7 Keep it Bring back earlier ideas

Vocabulary

Labs that use this topic