Unit 4 · Topic 4.2

4.2 Parametric Functions Modeling Planar Motion

A parametric function can model a particle moving in a plane: (x(t), y(t)) is its position at time t. Its leftmost and rightmost points come from the extremes of x(t), its highest and lowest from the extremes of y(t), and its intercepts from the zeros of each function.

Key terms

planar motion
position
horizontal and vertical extrema
intercepts

Position at time t

In planar motion, a particle moves around a flat plane, and its position at time t is (x(t), y(t)). The graph shows the path; the parameter tells you when the particle is at each point.

Treat x(t) and y(t) as two separate functions of time. Everything you learned in unit 1 about where a function is largest, smallest or zero applies to each one.

Horizontal and vertical extremes

The particle is farthest right when x(t) is at its maximum, and farthest left when x(t) is at its minimum. These are the horizontal extrema of the motion.

The particle is highest when y(t) is at its maximum, and lowest when y(t) is at its minimum. These are the vertical extrema.

On a restricted time interval, check the endpoints too. The farthest-right point might be where the motion starts or stops.

To find them by hand, use what you know about each function's type: a quadratic x(t) has its extreme at its vertex, and a sinusoidal y(t) has its extremes at its peaks and valleys. With a calculator, graph x(t) and y(t) separately against t and use the maximum and minimum features.

Intercepts

The path meets the y-axis where x = 0. So the zeros of x(t) give the times when the particle is on the y-axis, and plugging those times into y(t) gives the y-intercepts.

Likewise, the zeros of y(t) give the times when the particle is on the x-axis, and x(t) at those times gives the x-intercepts.

It's easy to mix these up. A y-intercept needs x = 0, so it comes from x(t), not y(t).

A path can cross the same axis several times, so find every zero in the time interval, not just the first.

Modeling motion

In this course, parametric functions model planar motion: a thrown ball, a car on a track, a point on a wheel. For a thrown object near Earth's surface, the horizontal position often changes at a constant rate, x(t) = v·t, while the height follows a quadratic, y(t) = −4.9t² + (initial vertical speed)·t + (initial height), with distances in meters and time in seconds.

Always report a point as coordinates and say what time it happens, with units.

The model also answers “when” questions. To find when a thrown ball is 10 meters high, solve y(t) = 10. There are often two answers, one on the way up and one on the way down.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Extremes and intercepts
A particle moves with x(t) = t² − 4t and y(t) = 3 − t for 0 ≤ t ≤ 5. Find its leftmost, rightmost, highest and lowest points, and where it crosses each axis.
Show the solutionHide the solution
1. Step 1: x(t) = (t − 2)² − 4 has its minimum at t = 2: x = −4, y = 1. Leftmost point (−4, 1).
2. Step 2: Compare the endpoints for the maximum: x(0) = 0 and x(5) = 5. Rightmost point at t = 5: (5, −2).
3. Step 3: y(t) = 3 − t decreases the whole time, so the highest point is at t = 0, (0, 3), and the lowest is at t = 5, (5, −2).
4. Step 4: y-intercepts: x(t) = t(t − 4) = 0 at t = 0 and t = 4, giving (0, 3) and (0, −1).
5. Step 5: x-intercept: y(t) = 0 at t = 3, giving (x(3), 0) = (−3, 0).
Answer: Leftmost (−4, 1) at t = 2; rightmost (5, −2) at t = 5; highest (0, 3) at t = 0; lowest (5, −2) at t = 5; crosses the y-axis at (0, 3) and (0, −1), and the x-axis at (−3, 0).
Example 2Calculator allowed
A thrown ball
A ball is thrown from 1.5 meters above the ground. Its position after t seconds is x(t) = 20t, y(t) = −4.9t² + 15t + 1.5, in meters. Find its maximum height and how far it travels horizontally before it hits the ground.
Show the solutionHide the solution
1. Step 1: The height y(t) is a downward parabola with its vertex at t = 15/(2 · 4.9) ≈ 1.531 seconds.
2. Step 2: Maximum height: y(1.531) ≈ 12.980 meters.
3. Step 3: It lands when y(t) = 0. Using the quadratic formula (or a calculator) and taking the positive root: t ≈ 3.158 seconds.
4. Step 4: Horizontal distance: x(3.158) ≈ 20(3.158) ≈ 63.163 meters.
Answer: Maximum height ≈ 12.980 m (at t ≈ 1.531 s); it lands about 63.163 m away (at t ≈ 3.158 s).
Example 3
Trap: which function gives which intercept
For x(t) = t − 2, y(t) = t² − 1 with t ≥ 0, find the y-intercept and the x-intercept of the path.
Show the solutionHide the solution
1. Step 1: y-intercept: the path is on the y-axis when x(t) = 0, so t = 2. Then y(2) = 3. The y-intercept is (0, 3).
2. Step 2: x-intercept: the path is on the x-axis when y(t) = 0, so t = ±1. Only t = 1 is allowed. Then x(1) = −1. The x-intercept is (−1, 0).
3. Step 3: The trap is to set y(t) = 0 and call the result a y-intercept.
Answer: y-intercept (0, 3); x-intercept (−1, 0).

Common mistakes

Using zeros of y(t) to find y-intercepts. Zeros of x(t) give y-intercepts.
Forgetting to check the endpoints of the time interval when finding the farthest left, right, up or down.
Reporting a time when the question asks for a point, or the reverse.

On the exam

Unit 4 is not on the AP Precalculus Exam, so you'll see this topic on class tests rather than in May.
Class tests often ask for the highest point, the time a particle reaches an axis, or where it is at a given time. Give coordinates and times with units.

Connected topics

Videos

AP Precalculus – 4.2 Parametric Functions Modeling Planar Motion
The AlgebrosWatch on YouTube (opens in a new tab)
Parametric equations for projectile motion | Precalculus | Khan Academy
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Identifying vertical extremum of parametric planar motion functions | Khan Academy
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4.2-A Parametric Functions Modeling Planar Motion Video Lesson
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AP PreCalculus - 4.2& 4.3 Parametric Motion & Rate of Change
COACH MEIDENBAUERWatch on YouTube (opens in a new tab)