Unit 4 · Topic 4.2 · about 30 minutes · 7 parts · 12 scored checks, 1 of them multiple choice

AP Precalculus 4.2: Parametric Functions Modeling Planar Motion

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 path of a drone, the graph of f(t) = (t squared minus 4t, 6t minus t squared) for t from 0 to 6, in meters. It starts at the origin, swings up and to the left through (-3, 5) at t = 1 to its leftmost point (-4, 8) at t = 2, reaches its highest point (-3, 9) at t = 3, crosses the y-axis at (0, 8) at t = 4, passes (5, 5) at t = 5, and ends on the x-axis at (12, 0) at t = 6, its rightmost point. A dashed vertical segment marks the leftmost point and a dashed horizontal segment marks the highest point.

Part 1 Think first Farthest west, highest up

Commit to a guess first. Being wrong now still helps the answer stick.

Predict first

A survey drone lifts off from a dock. It climbs while drifting west, swings back east, and lands 16 meters east of where it took off. At the moment it is farthest west of its takeoff spot, is it also at its highest?

In short. A parametric function $f (t) = (x (t), y (t))$ can model a particle moving in the plane, and its graph shows where the particle is at each time $t$ .

The least and greatest values of $x (t)$ give the leftmost and rightmost points of the motion, and the least and greatest values of $y (t)$ give the lowest and highest points.
The real zeros of $x (t)$ give the $y$ -intercepts of the path, and the real zeros of $y (t)$ give the $x$ -intercepts.
When the domain is restricted, check the start and the end of the path as well, because an extreme can sit at either end.

Part 2 Learn it Three ideas, each checked before the next

2a. The graph shows where the particle is

Put the drone on a grid with the pad at the origin. At time $t$ seconds its position is $(x (t), y (t))$ , where $x (t)$ is how many meters east of the pad it is (negative means west) and $y (t)$ is its height in meters. For this flight, $x (t) = t^{2} - 4 t$ and $y (t) = 6 t - t^{2}$ , for $0 \leq t \leq 6$ .

Written as one function, $f (t) = (x (t), y (t))$ . Put in a time and you get a point. At $t = 1$ , $f (1) = (1 - 4, 6 - 1) = (- 3, 5)$ : one second in, the drone is 3 meters west of the pad and 5 meters up. Any particle moving in the plane can be modeled the same way, and the graph of $f$ , the path in the diagram, is every position it takes.

The drone's position
$t$ (seconds)	$x (t)$ (meters)	$y (t)$ (meters)
$0$	$0$	$0$
$1$	$- 3$	$5$
$2$	$- 4$	$8$
$3$	$- 3$	$9$
$4$	$0$	$8$
$5$	$5$	$5$
$6$	$12$	$0$

The graph leaves one thing out: time. A point on the path says where the drone was, not when. To know when, you go back to $t$ , which is why the dots in the diagram carry labels.

Quick check · Match

A particle's position at time $t$ is $f (t) = (t^{2} - 1, 3 t - t^{2})$ . Tap a time, then the particle's position at that time.

2b. Farthest left, right, down and up

How far west does the drone get? Only $x$ measures east and west, so the answer comes from $x (t)$ alone. The least value of $x (t)$ gives the leftmost point of the path and the greatest value gives the rightmost. In the same way, the least and greatest values of $y (t)$ give the lowest and highest points.

Here $x (t) = t (t - 4)$ is a quadratic with zeros at 0 and 4, and a quadratic turns halfway between its zeros, at $t = 2$ . Since $x (2) = - 4$ and $y (2) = 8$ , the leftmost point is $(- 4, 8)$ . The height $y (t) = t (6 - t)$ turns halfway between 0 and 6, at $t = 3$ , where $y (3) = 9$ and $x (3) = - 3$ . The highest point is $(- 3, 9)$ .

Each extreme comes from its own function, so they can happen at different moments, and here they do.

Check the ends too. Because $x (t)$ falls until $t = 2$ and rises after, its greatest value is at an end: $x (0) = 0$ or $x (6) = 12$ . The rightmost point, $(12, 0)$ , is where the flight ends. The lowest points are the two ends, $(0, 0)$ and $(12, 0)$ , where the height is 0.

Quick check · Calculate

A particle's position at time $t$ is $f (t) = (t^{2} - 6 t + 5, 2 t + 1)$ for $0 \leq t \leq 5$ . What is the least $x$ -coordinate the particle reaches, the farthest left its path goes?

Your answer

2c. Crossing the axes

Every point on the $y$ -axis has $x$ -coordinate 0. So the path meets the $y$ -axis exactly when $x (t) = 0$ , and the real zeros of $x (t)$ give the $y$ -intercepts. In the same way, the real zeros of $y (t)$ give the $x$ -intercepts. The letters cross over, and that is the part people get backward.

For the drone, $x (t) = t (t - 4)$ is 0 at $t = 0$ and $t = 4$ . Put those times into $y (t)$ : $y (0) = 0$ and $y (4) = 8$ . The $y$ -intercepts are $(0, 0)$ and $(0, 8)$ . Four seconds in, the drone is right over the pad, 8 meters up.

The height $y (t) = t (6 - t)$ is 0 at $t = 0$ and $t = 6$ , which gives the $x$ -intercepts $(0, 0)$ and $(12, 0)$ : takeoff and landing.

A zero of $x (t)$ is a time, not a point. To get the intercept you still need the other coordinate at that time. Only zeros inside the domain count, too. If $x (t)$ had a zero at $t = - 1$ , that would be a moment before the drone took off, and the path has no point there.

Quick check · Fill the statement

A particle moves along the path $f (t) = (t^{2} - 9, 2 t - 4)$ for $- 4 \leq t \leq 4$ . Tap a blank, then tap an answer.

The path meets the $y$ -axis where = 0, and it does so at . It meets the $x$ -axis where = 0, and it does so at .

Part 3 Watch it done A drone over a soccer field

Read each step, then do the last one yourself.

A camera drone films a soccer practice. Its position $t$ seconds after takeoff is $f (t) = (x (t), y (t))$ , with $x (t) = t^{2} - 5 t$ meters east of a corner flag (negative values are west) and $y (t) = 10 t - t^{2}$ meters above the ground, for $0 \leq t \leq 10$ . Find its highest point, the points where it is directly above the flag, and how far west of the flag it gets.

Step 1 · Highest point
Height is $y (t) = t (10 - t)$ , a quadratic with zeros 0 and 10. It turns halfway between them, at $t = 5$ , where $y (5) = 25$ . Then $x (5) = 25 - 25 = 0$ , so the highest point is $(0, 25)$ .
Step 2 · Directly above the flag
Directly above the flag means on the $y$ -axis, where $x = 0$ . Solve $x (t) = t (t - 5) = 0$ : $t = 0$ or $t = 5$ . Then $y (0) = 0$ and $y (5) = 25$ , so the $y$ -intercepts are $(0, 0)$ and $(0, 25)$ . The drone is over the flag at takeoff and again at the top of its climb.
Why this step?
Why do the zeros of $x (t)$ , and not the zeros of $y (t)$ , give the points on the $y$ -axis?
Every point on the $y$ -axis has $x$ -coordinate 0, so the drone is on that axis exactly when $x (t) = 0$ . The zeros of $y (t)$ are the times it is at ground level, which puts it on the $x$ -axis.
Step 3 · Farthest west
Farthest west means the least value of $x (t)$ . Since $x (t) = t (t - 5)$ has zeros 0 and 5, it turns halfway between them, at $t = 2.5$ . It opens upward, so that turning point is its least value, and the ends, $x (0) = 0$ and $x (10) = 50$ , cannot go lower.
Your step · Calculate

Finish it. A drone's distance east of a flag, in meters, is $x (t) = t^{2} - 5 t$ for $0 \leq t \leq 10$ , with negative values west of the flag. What is the least value of $x (t)$ ?

Your answer

So the drone gets 6.25 meters west of the flag at $t = 2.5$ , when it is $y (2.5) = 18.75$ meters up. Its farthest west point, $(- 6.25, 18.75)$ , is not its highest point, which comes later, right over the flag.

Part 4 Try it A train on an oval track

A toy train runs on an oval track on a classroom floor. Two taped lines cross at the origin: the $x$ -axis runs east, the $y$ -axis runs north, and distances are in feet. The train's position at time $t$ is $f (t) = (2 + 4 \cos t, 1 + 2 \sin t)$ for $0 \leq t \leq 2 π$ , which is one lap.

Warm up in the lab first. It shows a different oval, $(- 1 + 2.5 \cos t, 2 + \sin t)$ , not the train's track, with the graphs of $x$ and $y$ against $t$ under it and the readouts off. Move the Parameter t slider slowly through one lap. Each time the point is farthest left or right, look at what the $x$ graph is doing. Each time the point crosses the vertical axis, look at where the $x$ graph crosses 0.

Then work the two questions on paper, for the train's track, $x (t) = 2 + 4 \cos t$ and $y (t) = 1 + 2 \sin t$ .

Lab

Parametric Grapher

This is a warm-up oval, $(- 1 + 2.5 \cos t, 2 + \sin t)$ for $0 \leq t \leq 2 π$ , not the train's track. You can move $t$ and the interval but not the oval.

Open the full Parametric Grapher lab

Try it · Calculate

A toy train's position on a floor, in feet, is $f (t) = (2 + 4 \cos t, 1 + 2 \sin t)$ for $0 \leq t \leq 2 π$ , with $x$ measured east and $y$ measured north. What is the least value of $x (t)$ , the $x$ -coordinate of the train's farthest point west?

Your answer

Try it · Calculate

A toy train's position on a floor, in feet, is $f (t) = (2 + 4 \cos t, 1 + 2 \sin t)$ for $0 \leq t \leq 2 π$ , with $t$ in radians. Its path crosses the $y$ -axis at two points. What is the $y$ -coordinate of the crossing with the greater $y$ -coordinate? Give a decimal to three places.

Your answer

Part 5 Explain it Write it out

Nothing you type leaves this page. Check your answer against the model and each point on the list.

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

Written practice

A drone's position $t$ seconds after takeoff is $f (t) = (x (t), y (t))$ , with $x (t) = t^{2} - 2 t - 3$ meters east of a fence post (negative values are west) and $y (t) = 8 t - 2 t^{2}$ meters above the ground, for $0 \leq t \leq 4$ . (i) Identify the leftmost point of the drone's path, and explain how you found it. (ii) Identify the highest point of the path. (iii) Identify every point where the path meets the $y$ -axis, and explain why one zero of $x (t)$ gives no point of the path. (iv) Describe what the $x$ -intercepts of the path mean for the drone.

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

Model answer

(i) The leftmost point comes from the least value of $x (t)$ . $x (t) = t^{2} - 2 t - 3$ is a quadratic that opens upward with its vertex at $t = 1$ , where $x (1) = - 4$ . The ends give $x (0) = - 3$ and $x (4) = 5$ , both larger. Since $y (1) = 6$ , the leftmost point is $(- 4, 6)$ . (ii) The highest point comes from the greatest value of $y (t) = 2 t (4 - t)$ , which turns halfway between its zeros 0 and 4, at $t = 2$ . $y (2) = 8$ and $x (2) = - 3$ , so the highest point is $(- 3, 8)$ . (iii) The path meets the $y$ -axis where $x (t) = 0$ : $(t + 1) (t - 3) = 0$ , so $t = - 1$ or $t = 3$ . The time $t = - 1$ is before takeoff, outside $0 \leq t \leq 4$ , so it gives no point. $y (3) = 6$ , so the only $y$ -intercept is $(0, 6)$ : at $t = 3$ the drone is directly above the post, 6 meters up. (iv) The $x$ -intercepts come from the zeros of $y (t)$ , at $t = 0$ and $t = 4$ . They are $(- 3, 0)$ , where the drone takes off 3 meters west of the post, and $(5, 0)$ , where it lands 5 meters east of it.

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

Finds the least value of $x (t)$ , $- 4$ at $t = 1$ , checks the ends, and identifies the leftmost point $(- 4, 6)$ .
Identifies the highest point $(- 3, 8)$ from the greatest value of $y (t)$ at $t = 2$ .
Solves $x (t) = 0$ , rejects $t = - 1$ as outside the domain, and identifies $(0, 6)$ .
Uses the zeros of $y (t)$ to give $(- 3, 0)$ and $(5, 0)$ and says they are where the drone takes off and lands.

Part 6 Check yourself A seabird's dive

The two case questions use the same seabird dive. Answer them from it.

A seabird dives for a fish next to a buoy. Its position $t$ seconds into the dive is $f (t) = (x (t), y (t))$ in meters, with $x (t) = 4 t - 8$ measured horizontally from the buoy and $y (t) = t^{2} - 6 t + 8$ measured up from the water's surface, for $0 \leq t \leq 5$ . A negative $y$ means the bird is under water.

The seabird's position at a few times
$t$ (seconds)	$x (t)$ (meters)	$y (t)$ (meters)
$0$	$- 8$	$8$
$1$	$- 4$	$3$
$2$	$0$	$0$
$5$	$12$	$3$

Case question 1 · Multiple choice

What is the lowest point of the seabird's path?

Case question 2 · Calculate

How far from the buoy, measured horizontally in meters, is the seabird when it comes back up through the surface of the water?

Your answer

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

Check yourself · Spot the error

A particle moves along the path $f (t) = (t^{2} - 4, t + 1)$ for $- 3 \leq t \leq 3$ . A student finds the $x$ -intercepts of the path. Tap the first line that goes wrong.

Check yourself · Sort it

A particle's position is $f (t) = (t^{3} - 4 t, (t^{2} - 1) (t - 3))$ for $- 3 \leq t \leq 3$ . Each card names a point where its path meets an axis. Tap a card, then the function whose zero gives the time the particle is at that point.

Use $x (t)$

Use $y (t)$

Part 7 Keep it From earlier lessons

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

From 4.1 Parametric Functions · 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.

From 3.5 Sinusoidal Functions · Calculate

The graph of a sinusoidal function k has a maximum at $x = 1.5$ , then its next minimum at $x = 5.5$ , then its next maximum at $x = 9.5$ . What is the period of k?

Your answer

In three lines

A parametric function $f (t) = (x (t), y (t))$ gives a particle's position at time $t$ , and its graph is the path.
The least and greatest values of $x (t)$ give the leftmost and rightmost points; those of $y (t)$ give the lowest and highest, ends included.
The real zeros of $x (t)$ give the $y$ -intercepts and the real zeros of $y (t)$ give the $x$ -intercepts.

Vocabulary

Parametric function: A function $f (t) = (x (t), y (t))$ that gives a point in the plane for each value of $t$ ; in planar motion, its graph shows the position of a particle at time $t$ .
Planar motion: The motion of a particle in the plane, modeled by a parametric function $f (t) = (x (t), y (t))$ .
Horizontal extrema: The leftmost and rightmost positions of a particle's motion, found from the minimum and maximum values of $x (t)$ .
Vertical extrema: The lowest and highest positions of a particle's motion, found from the minimum and maximum values of $y (t)$ .
Real zeros: The values of $t$ where $x (t)$ or $y (t)$ is 0; the real zeros of $x (t)$ correspond to $y$ -intercepts and those of $y (t)$ to $x$ -intercepts.

Course alignment, for teachers

AP Precalculus topic 4.2, Unit 4: Functions Involving Parameters, Vectors, and Matrices.

Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.
Skill 3.B: Apply numerical results in a given mathematical or applied context.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 are checked by a scored question. The map is in the lesson file.

Part 1 Think first Farthest west, highest up

Part 2 Learn it Three ideas, each checked before the next

2a. The graph shows where the particle is

2b. Farthest left, right, down and up

2c. Crossing the axes

Part 3 Watch it done A drone over a soccer field

Part 4 Try it A train on an oval track

Parametric Grapher

Part 5 Explain it Write it out

Part 6 Check yourself A seabird's dive

Part 7 Keep it From earlier lessons

Vocabulary

Labs that use this topic