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Amy Ferguson Moncure

Calculus BC · Unit 9

Motion in the Plane

Use the signs of the velocity components for direction. Integrate speed for distance and velocity for displacement.

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Key Ideas Easy Medium Hard

\(\mathrm{distance}=\int_a^b |\mathbf{v}(t)|\,dt\)

What to Remember

Direction comes from the components

Moving right when \(x'(t)>0\) and left when \(x'(t)<0\). Moving up when \(y'(t)>0\).

At rest means the velocity vector is zero

One component being zero is not enough. The object can still be moving vertically or horizontally.

Displacement versus distance

Displacement is \(\left\langle \int x', \int y'\right\rangle\). Distance is \(\int |\mathbf{v}|\,dt\).

Position needs a starting point

Add the integral of velocity to the initial position vector.

Level 1

Easy

Read one component to decide left, right, up, or down.

Example
\(x=t^2-4t\) and \(y=t^2\). When is the particle moving left?

\(x'(t)=2t-4=2(t-2)\)

\(x'(t)<0\) when \(t<2\).

Answer: for \(t<2\)

1
For that same motion, when is the particle moving up?
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\(y'(t)=2t\)

For \(t>0\), \(y'>0\).

At \(t=0\) the vertical velocity is 0.

Answer: for \(t>0\)

2
Is the particle ever at rest if \(x'=2(t-2)\) and \(y'=2t\)?
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\(y'=0\) at \(t=0\), but then \(x'=-4\).

\(x'=0\) at \(t=2\), but then \(y'=4\).

Never at rest.

Answer: no

3
\(\mathbf{v}(t)=\langle 3, -2\rangle\). Describe the direction of motion.
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The horizontal component is positive: moving right.

The vertical component is negative: moving down.

Answer: right and down

4
Speed of \(\langle 3, -2\rangle\) is constant. Find it.
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\(\sqrt{9+4}\)

\(\sqrt{13}\)

Answer: \(\sqrt{13}\)

5
The particle starts at \(\langle 1, 5\rangle\) with velocity \(\langle 3, -2\rangle\). Find its position at \(t=2\).
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\(\langle 1+3\cdot 2,\ 5+(-2)\cdot 2\rangle\)

\(\langle 7, 1\rangle\)

Answer: \(\langle 7, 1\rangle\)

Level 2

Medium

Total distance on a line in the plane, or a stop.

Example
\(\mathbf{v}(t)=\langle t-1,\ 2(t-1)\rangle\) and the particle starts at the origin. Find the distance traveled from \(t=0\) to \(t=2\).

Speed \(\sqrt{5}\,|t-1|\).

\(\sqrt{5}\left(\int_0^1 (1-t)\,dt+\int_1^2 (t-1)\,dt\right)\)

Each integral equals \(1/2\).

\(\sqrt{5}\)

Answer: \(\sqrt{5}\)

1
Find the position at \(t=2\) for that same velocity, starting at the origin.
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Displacement is \(\langle 0, 0\rangle\).

The particle is back at the origin.

Answer: \(\langle 0, 0\rangle\)

2
\(x=t^3-3t\) and \(y=t^2-1\). Find the velocity when \(t=1\), and say whether the particle is at rest.
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\(x'=3t^2-3\) and \(y'=2t\).

At \(t=1\): \(\langle 0, 2\rangle\).

The vertical component is not zero, so it is not at rest.

Answer: \(\langle 0, 2\rangle\), not at rest

3
For \(x=\cos t\) and \(y=\sin t\), the particle moves on the unit circle. Find the velocity at \(t=\pi/2\).
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\(\langle -\sin t, \cos t\rangle\)

At \(\pi/2\): \(\langle -1, 0\rangle\).

It is moving left, with speed 1.

Answer: \(\langle -1, 0\rangle\)

4
Total distance along that unit-circle path from \(t=0\) to \(t=2\pi\).
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Speed is 1.

\(2\pi\)

Answer: \(2\pi\)

5
A particle has \(x'(t)<0\) and \(y'(t)=0\). Describe the motion.
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No vertical change.

The horizontal component is negative.

It moves straight left.

Answer: straight left

Level 3

Hard

Split the speed integral, and keep displacement separate.

Example
\(\mathbf{v}(t)=\langle 4\cos t,\ 3\cos t\rangle\) for \(0\le t\le \pi\), starting at \(\langle 0, 0\rangle\). Find the distance traveled and the final position.

Speed \(\sqrt{16\cos^2 t+9\cos^2 t}=5|\cos t|\).

On \([0,\pi/2]\), \(\cos t\ge 0\). On \([\pi/2,\pi]\), \(\cos t\le 0\).

\(5\int_0^{\pi/2}\cos t\,dt+5\int_{\pi/2}^{\pi}(-\cos t)\,dt\)

Each integral of the cosine piece equals 1.

\(10\)

Position: \(\langle 4\sin t,\ 3\sin t\rangle\).

At \(t=\pi\): \(\langle 0, 0\rangle\).

Answer: distance 10, back at the origin

1
At what time in \([0,\pi]\) is that particle at rest?
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\(\cos t=0\) makes both components zero.

\(t=\pi/2\)

Answer: \(t=\pi/2\)

2
Position at the rest time \(t=\pi/2\) for \(\mathbf{r}(t)=\langle 4\sin t,\ 3\sin t\rangle\).
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\(\langle 4, 3\rangle\)

Answer: \(\langle 4, 3\rangle\)

3
\(x'(t)=6t\) and \(y'(t)=2\), with \(\mathbf{r}(0)=\langle -3, 1\rangle\). Find when the particle crosses the y-axis, and its height then.
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\(x(t)=3t^2-3\)

\(3t^2-3=0\) at \(t=1\) for \(t\ge 0\).

\(y(t)=2t+1\)

\(y(1)=3\)

Answer: at \(t=1\), height 3

4
Explain the difference between the answers “displacement \(\langle 0,0\rangle\)” and “distance \(\sqrt{5}\)” for \(\mathbf{v}(t)=\langle t-1, 2(t-1)\rangle\) on \([0,2]\).
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Displacement is the net change in the position vector. The particle ends where it started.

Distance adds the speed along the whole trip, including the turnaround at \(t=1\).

Answer: net change can be zero while the path has length

5
A particle moves right when \(t<1\) and left when \(t>1\), and its vertical velocity is always positive. Can it be at rest at \(t=1\)?
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At rest requires \(y'(1)=0\) as well.

The vertical velocity is always positive, so it is never zero.

It is not at rest at \(t=1\). It is moving straight up at that instant.

Answer: no

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