Practice a free-response question on parametric planar motion. Write your response, then reveal the model answer to see how to reason through each part.
Free Response Question · Unit 4 · Parametric Planar Motion
Note: Unit 4 is not assessed on the AP Precalculus Exam. This question is provided to build fluency for later courses.
An object moves in the coordinate plane so that at time t seconds (t ≥ 0) its position is given by the parametric equations x(t) = 2 + 3t and y(t) = 1 + t², where x and y are measured in meters. Selected positions are shown below.
t (seconds)
(x, y) (meters)
0
(2, 1)
1
(5, 2)
2
(8, 5)
3
(11, 10)
A
Find the object's position at t = 0 and at t = 2, and describe the general direction of its motion during this interval.
✓ Model answer
At t = 0 the position is (2, 1), and at t = 2 it is x = 2 + 3(2) = 8, y = 1 + 2² = 5, so (8, 5). Because x increases steadily and y increases as well, the object moves to the right and upward throughout the interval.
Why it earns credit: Correctly evaluates both positions with units and describes the direction (right and up) consistent with x and y both increasing.
B
Eliminate the parameter to express y as a function of x, and identify the type of curve the object follows.
✓ Model answer
Solve x = 2 + 3t for t: t = (x − 2)/3. Substitute into y = 1 + t²: y = 1 + ((x − 2)/3)² = 1 + (x − 2)²/9. This is a parabola opening upward with vertex at (2, 1).
Why it earns credit: Correctly solves for t, substitutes to get y in terms of x, AND identifies the curve as an upward-opening parabola. Note that the parametric form also records the direction of travel, which the rectangular equation alone does not.
C
Find the average rate of change of x and of y with respect to t over the interval from t = 0 to t = 2, and use them to describe how the object's motion changes compared with its motion near t = 0.
✓ Model answer
Average rate of change of x: (8 − 2)/(2 − 0) = 3 meters per second. Average rate of change of y: (5 − 1)/(2 − 0) = 2 meters per second. The horizontal position changes at a constant rate of 3 m/s because x is linear in t, but the vertical rate is not constant: since y = 1 + t², the object moves upward slowly near t = 0 and faster as t increases. So the path steepens over time — the object gains vertical speed while its horizontal speed stays the same.
Why it earns credit: Computes both average rates with units, recognizes that x changes at a constant rate while y speeds up (because y is quadratic in t), and connects that to the path steepening — a description grounded in the structure of the equations.
How to reason through parametric FRQs
Evaluate carefully and keep units. Substitute the given t-values and report positions as ordered pairs with units.
Eliminate the parameter cleanly. Solve the simpler equation for t, then substitute — and remember the rectangular form can lose direction and restrictions.
Separate horizontal and vertical behavior. Analyze how x changes with t and how y changes with t independently.
Tie conclusions to structure. A linear component changes at a constant rate; a quadratic component speeds up or slows down — say why.
Remember this unit is not on the AP Exam, but the reasoning habits transfer directly to calculus.