Unit 4 · Topic 4.2

Parametric Functions Modeling Planar Motion

A particle moving in the plane has position (x(t), y(t)) at time t .

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

Part 1 · Hook

Why this matters

A robot vacuum crosses a living room. Its app stores two numbers every second: how far it is along the east wall and how far along the north wall. From those two lists alone you can tell when it was heading left, when it was climbing toward the far wall, and when it doubled back.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. On which interval is g(t)=4t−t2g(t) = 4t - t^2 increasing?

  1. t<2t < 2
  2. t>2t > 2
  3. 0<t<40 < t < 4
  4. t<4t < 4
Show the answer

g(t)=−(t−2)2+4g(t) = -(t - 2)^2 + 4 is a parabola opening down with vertex at t=2t = 2, so it increases to the left of t=2t = 2.

  • Correct: t<2t < 2:
  • t>2t > 2:
  • 0<t<40 < t < 4:
  • t<4t < 4:

2. A curve is x(t)=t+1x(t) = t + 1, y(t)=3ty(t) = 3t. Where is the point at t=2t = 2?

  1. (3,6)(3, 6)
  2. (2,6)(2, 6)
  3. (6,3)(6, 3)
  4. (3,2)(3, 2)
Show the answer

x(2)=2+1=3x(2) = 2 + 1 = 3 and y(2)=3(2)=6y(2) = 3(2) = 6, so the point is (3,6)(3, 6).

  • Correct: (3,6)(3, 6):
  • (2,6)(2, 6):
  • (6,3)(6, 3):
  • (3,2)(3, 2):

Part 4 · See it

See it first

The path of a particle with x(t) = t cubed minus 6t squared plus 9t and y(t) = 4t minus t squared for t from 0 to 4. It starts at (0, 0) at t = 0, moves right and up to (4, 3) at t = 1, turns and moves left through (2, 4) at t = 2, the highest point, to (0, 3) at t = 3, then turns again and moves right and down to (4, 0) at t = 4, crossing its earlier path near (2, 1).
A particle with x(t)=t3−6t2+9tx(t) = t^3 - 6t^2 + 9t and y(t)=4t−t2y(t) = 4t - t^2 for 0≤t≤40 \le t \le 4. It turns around in xx at t=1t = 1 and t=3t = 3 and is highest at t=2t = 2. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. x(t)x(t) is increasing on an interval of timethe particle moves right during that time
  2. y(t)y(t) is decreasing on an interval of timethe particle moves down during that time
  3. x(t)x(t) changes from increasing to decreasingthe particle turns from moving right to moving left; that point is farthest right nearby
  4. One coordinate turns around while the other keeps changingthe particle is still moving, so it is not at rest

Part 6 · Key ideas

Key ideas

  • In planar motion the position at time tt is (x(t),y(t))(x(t), y(t)). Read the horizontal motion from x(t)x(t) and the vertical motion from y(t)y(t), one at a time.
  • x(t)x(t) increasing: moving right. x(t)x(t) decreasing: moving left. y(t)y(t) increasing: moving up. y(t)y(t) decreasing: moving down.
  • Where x(t)x(t) has a maximum the particle is farthest right; where y(t)y(t) has a maximum it is highest.
  • A model is only good for the times the context allows, such as from launch until the ball lands.

Part 7 · Misconception

A common mistake

The wrong idea: When x(t)x(t) stops increasing and starts decreasing, the particle stops moving.

What actually happens: Only the horizontal motion turns around. If y(t)y(t) is still changing, the particle is moving straight up or down at that moment, so it is not at rest.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Graphs of x(t)x(t) and y(t)y(t) for a moving particle

A particle moves in the plane with position (x(t),y(t))(x(t), y(t)) for 0≤t≤40 \le t \le 4, where x(t)=t3−6t2+9tx(t) = t^3 - 6t^2 + 9t and y(t)=4t−t2y(t) = 4t - t^2. The graph shows each coordinate against time.

12341234tpositionx(t)y(t)

1. On which interval is the particle moving to the left?

  1. 1<t<31 < t < 3
  2. 0<t<10 < t < 1
  3. 2<t<42 < t < 4
  4. 0<t<20 < t < 2
Show the answer

The particle moves left where x(t)x(t) is decreasing. The graph of x(t)x(t) falls from 44 at t=1t = 1 to 00 at t=3t = 3.

  • Correct: 1<t<31 < t < 3: x(t)x(t) decreases on (1,3)(1, 3): left.
  • 0<t<10 < t < 1: x(t)x(t) increases on (0,1)(0, 1), so the particle moves right.
  • 2<t<42 < t < 4: y(t)y(t) decreases on (2,4)(2, 4); that is when the particle moves down.
  • 0<t<20 < t < 2: y(t)y(t) increases on (0,2)(0, 2); that is when it moves up.

2. Where is the particle when it is highest?

  1. (2,4)(2, 4)
  2. (4,2)(4, 2)
  3. (4,3)(4, 3)
  4. (0,3)(0, 3)
Show the answer

The particle is highest where y(t)y(t) is largest: y(2)=4y(2) = 4. Its position then is (x(2),y(2))(x(2), y(2))=(8−24+18,4)= (8 - 24 + 18, 4)=(2,4)= (2, 4).

  • Correct: (2,4)(2, 4): At t=2t = 2: x(2)=2x(2) = 2, y(2)=4y(2) = 4.
  • (4,2)(4, 2): The coordinates are swapped; xx comes first.
  • (4,3)(4, 3): (4,3)(4, 3) is the point at t=1t = 1, where the particle is farthest right, not highest.
  • (0,3)(0, 3): (0,3)(0, 3) is the point at t=3t = 3, after the particle has started down.

3. On which interval is the particle moving left and down at the same time?

  1. 2<t<32 < t < 3
  2. 1<t<21 < t < 2
  3. 3<t<43 < t < 4
  4. 1<t<31 < t < 3
Show the answer

Left needs x(t)x(t) decreasing, on (1,3)(1, 3). Down needs y(t)y(t) decreasing, on (2,4)(2, 4). Both hold on the overlap, (2,3)(2, 3).

  • Correct: 2<t<32 < t < 3: The overlap of (1,3)(1, 3) and (2,4)(2, 4) is (2,3)(2, 3).
  • 1<t<21 < t < 2: On (1,2)(1, 2), y(t)y(t) is still increasing, so the particle moves left and up.
  • 3<t<43 < t < 4: On (3,4)(3, 4), x(t)x(t) increases again: right and down.
  • 1<t<31 < t < 3: This is the whole time it moves left, but for 1<t<21 < t < 2 it is moving up.

4. At t=1t = 1, x(t)x(t) changes from increasing to decreasing. Is the particle at rest at t=1t = 1?

  1. No: y(t)y(t) is still increasing then, so the particle is moving up
  2. Yes: x(t)x(t) stops changing, so the particle stops
  3. Yes: the particle is at its highest point then
  4. No: x(t)x(t) is still increasing at t=1t = 1
Show the answer

Only the horizontal motion turns around at t=1t = 1. y(t)=4t−t2y(t) = 4t - t^2 increases until t=2t = 2, so at t=1t = 1 the particle is still rising. It is not at rest.

  • Correct: No: y(t)y(t) is still increasing then, so the particle is moving up: The vertical motion continues: y(t)y(t) increases on (0,2)(0, 2).
  • Yes: x(t)x(t) stops changing, so the particle stops: A stop needs both coordinates to stop changing; y(t)y(t) keeps increasing.
  • Yes: the particle is at its highest point then: The highest point is at t=2t = 2, where y(2)=4y(2) = 4.
  • No: x(t)x(t) is still increasing at t=1t = 1: x(t)x(t) reaches its largest value at t=1t = 1; it is turning around, not still increasing.

5. How does the graph of x(t)x(t) against tt differ from the path of the particle?

  1. The graph of x(t)x(t) shows when the particle moves right or left; the path shows where it goes
  2. They are the same picture drawn with different labels on the axes
  3. The graph of x(t)x(t) shows the path, and the path shows the time
  4. The graph of x(t)x(t) shows where the particle is highest
Show the answer

The graph of x(t)x(t) puts time across and horizontal position up, so its rises and falls show right and left motion. The path plots yy against xx and leaves time out.

  • Correct: The graph of x(t)x(t) shows when the particle moves right or left; the path shows where it goes: Against tt: direction in xx. In the plane: the path.
  • They are the same picture drawn with different labels on the axes: The axes differ (tt and xx versus xx and yy), so the shapes usually differ too.
  • The graph of x(t)x(t) shows the path, and the path shows the time: The path is yy against xx; time is not on either of its axes.
  • The graph of x(t)x(t) shows where the particle is highest: Height is y(t)y(t); the graph of x(t)x(t) says nothing about it.

6. A particle moves with x(t)=2t−3x(t) = 2t - 3 and y(t)=t2−4ty(t) = t^2 - 4t for 0≤t≤50 \le t \le 5. What is its xx-coordinate when it is at its lowest point?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

y(t)y(t)=t2−4t= t^2 - 4t=(t−2)2−4= (t - 2)^2 - 4 is lowest at t=2t = 2. Then x(2)=2(2)−3=1x(2) = 2(2) - 3 = 1, so the lowest point is (1,−4)(1, -4).

  • Answer: 1

Part 9 · Summary

Summary

A particle moving in the plane has position (x(t),y(t))(x(t), y(t)) at time tt. Where x(t)x(t) increases it moves right, where y(t)y(t) increases it moves up, and the turning points of each coordinate give the farthest right, left, highest and lowest positions. Read each coordinate separately, then put the two together.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic