Unit 9 · Topic 9.6

Motion in the Plane

For a particle in the plane, velocity and acceleration are the first and second derivatives of the position vector, and speed is the magnitude of the velocity.

Practice 2: Connecting RepresentationsPractice 3: JustificationPractice 4: Communication and Notation

Part 1 · Hook

Why this matters

A soccer ball curls from the corner flag into the box. Where is it after one second, how fast is it going, how far has it flown, and how far is it from where it started? Each question has its own tool: the position vector, the speed, the integral of speed, and the integral of velocity. Motion in the plane puts all of Unit 9 together, and it is a regular free-response question on the BC exam.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. On a line, total distance traveled from t=0t = 0 to t=3t = 3 is

  1. ∫03|v(t)|dt\int_0^3 |v(t)|\,dt
  2. ∫03v(t)dt\int_0^3 v(t)\,dt
  3. |s(3)−s(0)||s(3) - s(0)|
  4. v(3)−v(0)v(3) - v(0)
Show the answer

Distance adds up the size of the velocity, so backtracking counts.

  • Correct: ∫03|v(t)|dt\int_0^3 |v(t)|\,dt:
  • ∫03v(t)dt\int_0^3 v(t)\,dt:
  • |s(3)−s(0)||s(3) - s(0)|:
  • v(3)−v(0)v(3) - v(0):

2. If v(t)=⟨2t,1⟩\mathbf{v}(t) = \left\langle 2t, 1 \right\rangle and r(0)=⟨3,0⟩\mathbf{r}(0) = \left\langle 3, 0 \right\rangle, then r(2)=\mathbf{r}(2) =

  1. ⟨7,2⟩\left\langle 7, 2 \right\rangle
  2. ⟨4,2⟩\left\langle 4, 2 \right\rangle
  3. ⟨7,1⟩\left\langle 7, 1 \right\rangle
  4. ⟨4,1⟩\left\langle 4, 1 \right\rangle
Show the answer

r(2)\mathbf{r}(2)=⟨3,0⟩= \left\langle 3, 0 \right\rangle+∫02⟨2t,1⟩dt+ \int_0^2 \left\langle 2t, 1 \right\rangle\,dt=⟨3+4,0+2⟩= \left\langle 3 + 4, 0 + 2 \right\rangle.

  • Correct: ⟨7,2⟩\left\langle 7, 2 \right\rangle:
  • ⟨4,2⟩\left\langle 4, 2 \right\rangle:
  • ⟨7,1⟩\left\langle 7, 1 \right\rangle:
  • ⟨4,1⟩\left\langle 4, 1 \right\rangle:

3. The length of x=3tx = 3t, y=4ty = 4t for 0≤t≤20 \le t \le 2 is

  1. 1010
  2. 1414
  3. 55
  4. 77
Show the answer

∫029+16dt\int_0^2 \sqrt{9 + 16}\,dt=10= 10.

  • Correct: 1010:
  • 1414:
  • 55:
  • 77:

Part 4 · See it

See it first

The path r(t) = 2t, 4t minus t squared for t from 0 to 4, an arch from (0, 0) up to (4, 4) and down to (8, 0), traced left to right. A dashed gray position vector goes from the origin to (2, 3), the point at t = 1. From that point a blue velocity vector 2, 2 is drawn; its length, the speed, is 2 root 2. A vermillion displacement vector 8, 0 runs from the start to the end, length 8, while the distance along the path, the integral of speed, is about 11.83.
For r(t)=⟨2t,4t−t2⟩\mathbf{r}(t) = \left\langle 2t, 4t - t^2 \right\rangle, 0≤t≤40 \le t \le 4: at t=1t = 1 the particle is at (2,3)(2, 3) with velocity ⟨2,2⟩\left\langle 2, 2 \right\rangle and speed 222\sqrt{2}. The displacement vector is ⟨8,0⟩\left\langle 8, 0 \right\rangle, of length 8, but the distance traveled along the arch is about 11.83. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Differentiate the position vectorvelocity v(t)=⟨x′(t),y′(t)⟩\mathbf{v}(t) = \left\langle x'(t), y'(t) \right\rangle, then acceleration a(t)=⟨x″(t),y″(t)⟩\mathbf{a}(t) = \left\langle x''(t), y''(t) \right\rangle
  2. Take the magnitude of the velocitythe speed (x′(t))2+(y′(t))2\sqrt{(x'(t))^2 + (y'(t))^2}, a number and never negative
  3. Integrate the velocity over [a,b][a, b]the displacement vector r(b)−r(a)\mathbf{r}(b) - \mathbf{r}(a); add the start to get the position
  4. Integrate the speed over [a,b][a, b]the total distance traveled along the path

Part 6 · Key ideas

Key ideas

  • Speed is the magnitude of the velocity, (x′(t))2+(y′(t))2\sqrt{(x'(t))^2 + (y'(t))^2}. It is not |dydx|\left|\frac{dy}{dx}\right|, which measures steepness, not how fast.
  • Position: x(b)x(b)=x(a)= x(a)+∫abx′(t)dt+ \int_a^b x'(t)\,dt, and the same for yy. Add the initial value, keep the dtdt.
  • Distance traveled is ∫ab(x′(t))2+(y′(t))2dt\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt; displacement is ∫abv(t)dt\int_a^b \mathbf{v}(t)\,dt, a vector.
  • "yy is decreasing" reads the sign of dydt\frac{dy}{dt}; "moving left" reads dxdt<0\frac{dx}{dt} < 0. A collision needs the same point at the same time.

Part 7 · Misconception

A common mistake

The wrong idea: At t=2t = 2, dxdt=3\frac{dx}{dt} = 3 and dydt=−4\frac{dy}{dt} = -4, so the speed is |dydx|=|−43|=43\left|\frac{dy}{dx}\right| = \left|\frac{-4}{3}\right| = \frac{4}{3}.

What actually happens: dydx\frac{dy}{dx} compares the two rates to give the slope of the path; it does not measure how fast the particle moves. Speed is the length of the velocity vector: 32+(−4)2=5\sqrt{3^2 + (-4)^2} = 5.

Part 8 · Check yourself

Check yourself

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

Graph

A path with the velocity vector drawn at one point

A particle moves along the path x(t)=4−t2x(t) = 4 - t^2, y(t)=t3−4ty(t) = t^3 - 4t for −2≤t≤2-2 \le t \le 2. The graph shows the path, the direction of motion, and the velocity vector drawn at the point where t=1t = 1.
The path x(t) = 4 minus t squared, y(t) = t cubed minus 4t for t from -2 to 2. It starts at the origin at t = -2, swings up and right through (3, 3), reaches (4, 0) at t = 0, swings down through (3, -3) at t = 1 and returns to the origin at t = 2: a closed loop traced clockwise. Arrows show the direction of increasing t. At the point where t = 1, (3, -3), a blue velocity vector is drawn pointing 2 units left and 1 unit down.
LevlPrep original diagram.

1. What is the velocity vector of the particle at t=1t = 1?

  1. ⟨3,−3⟩\left\langle 3, -3 \right\rangle
  2. ⟨−2,−1⟩\left\langle -2, -1 \right\rangle
  3. ⟨−2,6⟩\left\langle -2, 6 \right\rangle
  4. ⟨2,−1⟩\left\langle 2, -1 \right\rangle
Show the answer

v(t)=⟨−2t,3t2−4⟩\mathbf{v}(t) = \left\langle -2t, 3t^2 - 4 \right\rangle, so v(1)=⟨−2,−1⟩\mathbf{v}(1) = \left\langle -2, -1 \right\rangle: 2 units left and 1 down per unit of time, as the arrow in the graph shows.

  • ⟨3,−3⟩\left\langle 3, -3 \right\rangle: That is the position (x(1),y(1))=(3,−3)(x(1), y(1)) = (3, -3), where the vector starts.
  • Correct: ⟨−2,−1⟩\left\langle -2, -1 \right\rangle: Right: differentiate each component.
  • ⟨−2,6⟩\left\langle -2, 6 \right\rangle: That is the acceleration a(1)=⟨−2,6t⟩\mathbf{a}(1) = \left\langle -2, 6t \right\rangle at t=1t = 1.
  • ⟨2,−1⟩\left\langle 2, -1 \right\rangle: ddt(4−t2)=−2t\frac{d}{dt}(4 - t^2) = -2t, which is −2-2 at t=1t = 1: the particle moves left.

2. Find the speed of the particle at t=1t = 1.

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

Show the answer

Speed =‖v(1)‖= \left\| \mathbf{v}(1) \right\|=(−2)2+(−1)2= \sqrt{(-2)^2 + (-1)^2}=5= \sqrt{5}.

  • Answer: 5\sqrt{5} ≈ 2.236

3. Which statement about the motion from t=−2t = -2 to t=2t = 2 is correct?

  1. The displacement is ⟨0,0⟩\left\langle 0, 0 \right\rangle, so the distance traveled is 0
  2. The displacement is ⟨4,0⟩\left\langle 4, 0 \right\rangle, the farthest point from the start
  3. The displacement is ⟨3,−3⟩\left\langle 3, -3 \right\rangle, the position at t=1t = 1
  4. The displacement is ⟨0,0⟩\left\langle 0, 0 \right\rangle, but the distance traveled is positive
Show the answer

∫−22v(t)dt\int_{-2}^{2} \mathbf{v}(t)\,dt=r(2)−r(−2)= \mathbf{r}(2) - \mathbf{r}(-2)=⟨0,0⟩−⟨0,0⟩= \left\langle 0, 0 \right\rangle - \left\langle 0, 0 \right\rangle: the particle ends where it started. It still went around the loop, so the distance traveled, ∫−22(x′)2+(y′)2dt\int_{-2}^{2} \sqrt{(x')^2 + (y')^2}\,dt, is positive.

  • The displacement is ⟨0,0⟩\left\langle 0, 0 \right\rangle, so the distance traveled is 0: A zero displacement means the start and end coincide; the particle still traveled around the loop.
  • The displacement is ⟨4,0⟩\left\langle 4, 0 \right\rangle, the farthest point from the start: ⟨4,0⟩\left\langle 4, 0 \right\rangle is the position at t=0t = 0; the displacement compares the end with the start.
  • The displacement is ⟨3,−3⟩\left\langle 3, -3 \right\rangle, the position at t=1t = 1: That is a position along the way, not the change from start to end.
  • Correct: The displacement is ⟨0,0⟩\left\langle 0, 0 \right\rangle, but the distance traveled is positive: Right: net change zero, distance positive.

4. Which expression gives the total distance traveled by the particle from t=−2t = -2 to t=2t = 2?

  1. ∫−22(4−t2)2+(t3−4t)2dt\int_{-2}^{2} \sqrt{(4 - t^2)^2 + (t^3 - 4t)^2}\,dt
  2. ∫−224t2+(3t2−4)2dt\int_{-2}^{2} \sqrt{4t^2 + (3t^2 - 4)^2}\,dt
  3. (∫−22−2tdt)2+(∫−22(3t2−4)dt)2\sqrt{\left(\int_{-2}^{2} -2t\,dt\right)^2 + \left(\int_{-2}^{2} (3t^2 - 4)\,dt\right)^2}
  4. ∫−22|3t2−4−2t|dt\int_{-2}^{2} \left|\frac{3t^2 - 4}{-2t}\right|\,dt
Show the answer

Distance traveled is the integral of speed: ∫−22(−2t)2+(3t2−4)2dt\int_{-2}^{2} \sqrt{(-2t)^2 + (3t^2 - 4)^2}\,dt, and (−2t)2=4t2(-2t)^2 = 4t^2.

  • ∫−22(4−t2)2+(t3−4t)2dt\int_{-2}^{2} \sqrt{(4 - t^2)^2 + (t^3 - 4t)^2}\,dt: This squares the position components; speed squares the velocity components.
  • Correct: ∫−224t2+(3t2−4)2dt\int_{-2}^{2} \sqrt{4t^2 + (3t^2 - 4)^2}\,dt: Right: integrate the speed over the time interval.
  • (∫−22−2tdt)2+(∫−22(3t2−4)dt)2\sqrt{\left(\int_{-2}^{2} -2t\,dt\right)^2 + \left(\int_{-2}^{2} (3t^2 - 4)\,dt\right)^2}: This is the length of the displacement vector, which is 0 here because the path is closed.
  • ∫−22|3t2−4−2t|dt\int_{-2}^{2} \left|\frac{3t^2 - 4}{-2t}\right|\,dt: This integrates |dydx|\left|\frac{dy}{dx}\right|, the steepness of the path, not the speed.

5. A particle moves with x(t)=t2−4tx(t) = t^2 - 4t and y(t)=12t−t3y(t) = 12t - t^3 for 0<t<40 < t < 4. On which interval is the particle's yy-coordinate decreasing?

  1. 0<t<20 < t < 2
  2. 2<t<42 < t < 4
  3. 0<t<20 < t < 2 and 2<t<42 < t < 4
  4. Nowhere on 0<t<40 < t < 4
Show the answer

yy is decreasing where dydt=12−3t2<0\frac{dy}{dt} = 12 - 3t^2 < 0, that is t>2t > 2: the interval 2<t<42 < t < 4. The sign of dydx\frac{dy}{dx} does not decide this.

  • 0<t<20 < t < 2: On 0<t<20 < t < 2, xx is decreasing (x′=2t−4<0x' = 2t - 4 < 0) but y′=12−3t2>0y' = 12 - 3t^2 > 0: yy increases.
  • Correct: 2<t<42 < t < 4: Right: read the sign of dydt\frac{dy}{dt}.
  • 0<t<20 < t < 2 and 2<t<42 < t < 4: This reads dydx\frac{dy}{dx}=12−3t22t−4= \frac{12 - 3t^2}{2t - 4}, which is negative on both intervals. That says the path falls as xx increases, not that yy decreases in time.
  • Nowhere on 0<t<40 < t < 4: dydt=12−3t2\frac{dy}{dt} = 12 - 3t^2 is negative for 2<t<42 < t < 4.

6. A particle moves with r(t)=⟨3cos⁡t,3sin⁡t⟩\mathbf{r}(t) = \left\langle 3\cos t, 3\sin t \right\rangle. Find the total distance it travels from t=0t = 0 to t=2πt = 2\pi.

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

Show the answer

v(t)=⟨−3sin⁡t,3cos⁡t⟩\mathbf{v}(t) = \left\langle -3\sin t, 3\cos t \right\rangle, so the speed is 9(sin2⁡t+cos2⁡t)\sqrt{9(\sin^2 t + \cos^2 t)}=3= 3. Distance =∫02π3dt= \int_0^{2\pi} 3\,dt=6π= 6\pi, one trip around a circle of radius 3.

  • Answer: 6π6\,\pi ≈ 18.850

7. Particle A has position rA(t)=⟨2t,t2−1⟩\mathbf{r}_A(t) = \left\langle 2t, t^2 - 1 \right\rangle and particle B has position rB(t)=⟨t+2,3t−3⟩\mathbf{r}_B(t) = \left\langle t + 2, 3t - 3 \right\rangle for t≥0t \ge 0. Do they collide, and if so, when and where?

  1. Yes, at t=2t = 2, at the point (4,3)(4, 3)
  2. Yes, at (8,15)(8, 15), where their paths cross
  3. Yes, at t=1t = 1, where both have y=0y = 0
  4. No, they do not collide
Show the answer

A collision needs the same point at the same time. 2t=t+22t = t + 2 gives t=2t = 2, and then t2−1=3=3t−3t^2 - 1 = 3 = 3t - 3. Both are at (4,3)(4, 3) at t=2t = 2.

  • Correct: Yes, at t=2t = 2, at the point (4,3)(4, 3): Right: both coordinates match at the same time.
  • Yes, at (8,15)(8, 15), where their paths cross: The paths do cross at (8,15)(8, 15), but A is there at t=4t = 4 and B at t=6t = 6: different times, no collision.
  • Yes, at t=1t = 1, where both have y=0y = 0: At t=1t = 1 the yy-coordinates match but the xx-coordinates are 2 and 3.
  • No, they do not collide: They do collide: at t=2t = 2 both xx and yy agree.

Part 9 · Summary

Summary

For a particle in the plane, velocity and acceleration are the first and second derivatives of the position vector, and speed is the magnitude of the velocity. The integral of velocity is the displacement vector, so the position at a time is the start plus that integral; the integral of speed is the total distance traveled. Direction questions read the signs of dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}, and calculator answers carry their setup and three decimal places.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic