AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/4/4-3)
Unit 4 · Topic 4.3
4.3 Parametric Functions and Rates of Change
You can read a particle's direction of motion from x(t) and y(t) separately: increasing x means moving right and increasing y means moving up. Average rates of change of x and y over a time interval combine to give the slope of the secant line between two points on the path.
Key terms
- direction of motion
- average rate of change
- secant line slope
- parametrization
Direction of motion
Look at x(t) and y(t) one at a time as t increases.
Where x(t) switches from increasing to decreasing, or the reverse, the particle changes horizontal direction. That's a rightmost or leftmost point (topic 4.2). Where y(t) switches, the particle reaches a highest or lowest point.
| x(t) is… | y(t) is… | The particle moves… |
|---|---|---|
| Increasing | Increasing | Right and up |
| Increasing | Decreasing | Right and down |
| Decreasing | Increasing | Left and up |
| Decreasing | Decreasing | Left and down |
Same point, different directions
A path can pass through the same point more than once, at different times. The particle may be heading a different way each time. So direction belongs to a time t, not just to a point on the curve.
The same curve can also be traced by different parametric functions. They might go in opposite directions, start in different places, or move at different speeds. For example, (t, 2t) for 0 ≤ t ≤ 1 and (1 − t, 2 − 2t) for 0 ≤ t ≤ 1 trace the same segment from opposite ends.
A figure-eight path, for example, passes through its center twice, heading a different way each time.
Average rates of change
Over a time interval [t₁, t₂], the average rate of change of x is (x(t₂) − x(t₁))/(t₂ − t₁), and the average rate of change of y is (y(t₂) − y(t₁))/(t₂ − t₁). Each one is measured per unit of time.
The slope of the secant line between the two points on the path is the change in y over the change in x. Since both rates share the same denominator, the slope equals (average rate of change of y)/(average rate of change of x).
This only works when the average rate of change of x isn't zero. If x(t₂) = x(t₁), the secant line is vertical and has no slope.
Watch the units. If x and y are in meters and t is in seconds, each average rate is in meters per second, but the secant slope is in meters per meter. The time units cancel because the slope describes the shape of the path, not the timing.
Why this matters
These ideas let you describe motion without eliminating the parameter. In calculus, shrinking the interval turns the secant slope into the slope of the curve at a point, a key BC topic.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Direction on intervals
A particle moves with x(t) = t² − 2t and y(t) = t³ − 3t. Describe its direction on 0 < t < 1 and on t > 1.
Show the solutionHide the solution
- Step 1: x(t) = (t − 1)² − 1 decreases for t < 1 and increases for t > 1.
- Step 2: y(t) = t³ − 3t has turning points at t = −1 and t = 1: it decreases on −1 < t < 1 and increases for t > 1.
- Step 3: On 0 < t < 1: x decreasing and y decreasing, so the particle moves left and down.
- Step 4: For t > 1: both increasing, so it moves right and up.
Answer: Left and down on 0 < t < 1; right and up for t > 1.
- Example 2
Secant slope from average rates
For x(t) = t² + 1 and y(t) = 3t − t², find the average rates of change of x and y on [1, 3], and the slope of the secant line between the corresponding points.
Show the solutionHide the solution
- Step 1: x(1) = 2 and x(3) = 10, so the average rate of change of x is (10 − 2)/(3 − 1) = 4 units per unit of time.
- Step 2: y(1) = 2 and y(3) = 0, so the average rate of change of y is (0 − 2)/2 = −1.
- Step 3: Secant slope = (−1)/4 = −1/4. Check directly: from (2, 2) to (10, 0), the slope is −2/8 = −1/4.
Answer: Rates 4 and −1; secant slope −1/4.
- Example 3
Trap: same path, different motion
P(t) = (t, 2t) and Q(t) = (2 − 2t, 4 − 4t), both for 0 ≤ t ≤ 1. Do they describe the same motion?
Show the solutionHide the solution
- Step 1: P starts at (0, 0) and ends at (1, 2). Q starts at (2, 4) and ends at (0, 0).
- Step 2: Both lie on the line y = 2x, but Q covers the segment from (0, 0) to (2, 4), which is twice as long, in the opposite direction.
- Step 3: So they trace different segments, in opposite directions, at different speeds. A matching equation y = 2x doesn't mean matching motion.
Answer: No. They lie on the same line, but P moves from (0, 0) to (1, 2) while Q moves from (2, 4) back to (0, 0).
Common mistakes
- Dividing the average rate of change of x by that of y. The secant slope is y's rate over x's rate.
- Assuming the particle always moves the same way through a given point.
- Assuming two parametrizations of the same curve describe the same motion.
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.
- Expect questions that give x(t) and y(t), or their graphs, and ask when the particle moves left, right, up or down, or ask for a secant slope from two times.
Connected topics
Videos
Check yourself
3 questions on 4.3 Parametric Functions and Rates of Change. Pick an answer to see if you got it, and why.
A particle moves with position x(t) = t² − 6t and y(t) = 9 − t² for 0 ≤ t ≤ 5. During which interval is the particle moving to the right and down?
A particle moves in the xy-plane so that its position at time t is (x(t), y(t)), where x(t) = t² − 4t and y(t) = t³ − 3t, for 0 ≤ t ≤ 3.
Which of the following describes the motion of the particle for 0 < t < 1?
What is the slope of the line through the particle's positions at t = 0 and t = 3?
0 of 3 answered