Wemazi

Topic 9.6 · BC

Interpret motion in the plane

Builds on calculating distance traveled. 8 minutes.

A drone that flies past the origin

A drone flies along a straight line with position \(\langle 2t - 4,\ 3\rangle\) and velocity \(\langle 2,\ 0\rangle\). The dashed segment joins the origin \(O\) to the drone, and its length is the distance from the origin.

Drag the slider. The x-part of the velocity is 2, so the drone always moves to the right. The y-part is 0, so it never moves up or down.

The speed is \(\sqrt{2^2 + 0^2} = 2\) at every time, but the distance from the origin falls from 5 to 3 and then rises to 5.

The direction, the speed and the distance from the origin are separate quantities, so each needs its own test.

Three questions, three signs

Drag the point. The first graph shows the two parts of the velocity. A positive part means to the right or up, and a negative part means to the left or down.

The other graphs show \(S'\), the derivative of the speed \(S\), and \(D'\), the derivative of the distance \(D\) from the origin. Each graph is green where it is positive and red where it is negative.

The speed rises where \(S'\) is positive, and the distance rises where \(D'\) is positive. The two signs are independent.

Notation

  1. The signs of the two parts of the velocity give the direction of motion.

    \[\begin{aligned} \cy{x'} > 0&:\ \text{right}, \qquad x' < 0:\ \text{left} \\ \cy{y'} > 0&:\ \text{up}, \qquad y' < 0:\ \text{down} \end{aligned}\]

    The particle reverses its horizontal direction where \(x'\) changes sign, and its vertical direction where \(y'\) changes sign.

  2. The chain rule gives \(S'\), where \(x''\) and \(y''\) are the parts of the acceleration. The denominator \(S\) is positive, so the speed rises where the numerator is positive.

    \[\begin{aligned} S &= \sqrt{x'^2 + y'^2} \\ \dfrac{dS}{dt} &= \dfrac{\cy{x'x'' + y'y''}}{S} \end{aligned}\]

  3. The same steps with \(D = \sqrt{x^2 + y^2}\) give \(D'\). If the numerator is positive, the particle moves away from the origin. If it is negative, the particle moves toward it.

    \[\dfrac{dD}{dt} = \dfrac{\cy{xx' + yy'}}{D}\]

When it fails

If the path goes through \(O\), then \(D = 0\) at that time, and the formula for \(D'\) divides by 0.

On this line, \(D = |2t - 4|\) has a sharp corner at \(t = 2\), so \(D'\) does not exist there. It is negative before \(t = 2\) and positive after.

Example 1: A circle that does not surround the origin

A particle has position \(\vec r(t) = \langle 3 + 2\cos t,\ 2\sin t\rangle\). Describe the direction of motion, the speed and the distance from the origin.

  1. Differentiate to find the velocity.

    \[x' = -2\sin t,\qquad y' = 2\cos t\]

  2. The x-part is negative for \(0 < t < \pi\) and positive for \(\pi < t < 2\pi\), so the particle moves to the left and then to the right. The y-part is positive for \(0 < t < \pi/2\) and \(3\pi/2 < t < 2\pi\), so the particle moves up there and down between.

  3. The speed is the same at every time, because \(\sin^2 t + \cos^2 t = 1\).

    \[S = \sqrt{4\sin^2 t + 4\cos^2 t} = 2\]

  4. The numerator of \(D'\) is negative for \(0 < t < \pi\), so the distance falls, and it is positive for \(\pi < t < 2\pi\), so the distance rises.

    \[\begin{aligned} xx' + yy' &= (3 + 2\cos t)(-2\sin t) \\ &\quad + (2\sin t)(2\cos t) \\ &= \cy{-6\sin t} \end{aligned}\]

Example 2: Speeding up toward the origin

A particle has position \(\vec r(t) = (4 - t^2)\langle 3,\ 4\rangle\) for \(0 \le t < 2\). At \(t = 1\), is the speed increasing? Is the particle moving toward or away from the origin?

  1. Find the position and the velocity at \(t = 1\). Both parts of the velocity are negative, so the particle moves left and down.

    \[\begin{aligned} \vec r(1) &= \langle 9,\ 12\rangle \\ \vec v(t) &= \langle -6t,\ -8t\rangle \\ \vec v(1) &= \langle -6,\ -8\rangle \end{aligned}\]

  2. The speed is \(10t\). Its derivative is 10, which is positive, so the speed is increasing.

    \[\begin{aligned} S &= \sqrt{(-6t)^2 + (-8t)^2} = 10t \\ S' &= 10 \end{aligned}\]

  3. A tempting wrong move: The speed is increasing, so the particle moves away from the origin.

    The dashed segment from \(O\) is 20 long at \(t = 0\) and 15 long at \(t = 1\), so the distance falls while the speed rises.

  4. The distance has its own test. The numerator of \(D'\) is negative, so the particle moves toward the origin.

    \[\begin{aligned} xx' + yy' &= (9)(-6) + (12)(-8) \\ &= \cr{-150} \\ D' &= \dfrac{-150}{15} = -10 \end{aligned}\]

Practice

A particle has position \(\vec r(t) = \langle 2t,\ t^2 - 3\rangle\) for \(t > 0\). Does it ever reverse its horizontal or vertical direction? Does the speed increase? When does the particle move toward the origin?

Find the velocity and the acceleration. Then use the three tests. The graphs of \(S\) and \(D\) show whether your answers are correct.

Solution

  1. Differentiate twice.

    \[\vec v = \langle 2,\ 2t\rangle,\qquad \vec a = \langle 0,\ 2\rangle\]

  2. The x-part is 2, so the particle always moves to the right. The y-part \(2t\) is positive, so it always moves up. Neither direction reverses.

    \[\dfrac{dy}{dx} = \dfrac{2t}{2} = t\]

  3. The numerator of \(S'\) is positive for \(t > 0\), so the speed increases.

    \[\begin{aligned} x'x'' + y'y'' &= (2)(0) + (2t)(2) \\ &= 4t \end{aligned}\]

  4. The numerator of \(D'\) is negative for \(0 < t < 1\), so the particle moves toward the origin there. After \(t = 1\), it moves away.

    \[\begin{aligned} xx' + yy' &= (2t)(2) + (t^2 - 3)(2t) \\ &= \cy{2t(t^2 - 1)} \end{aligned}\]

Questions

The figure shows a particle at moment A or B. The dashed arrow is the position from the origin \(O\), the orange arrow is the velocity, and the chalk arrow is the acceleration. The top row of numbers gives \(x'x''\) and \(y'y''\), and the bottom row gives \(xx'\) and \(yy'\).

  1. Which statement is true at moment A?

    • The particle moves left and up, the speed is increasing, and the distance from \(O\) is not changing at that instant.
    • The particle moves left and up, the speed is increasing, and the distance from \(O\) is increasing.
    • The particle moves right and up, the speed is increasing, and the distance from \(O\) is not changing.
    Answer

    The particle moves left and up, the speed is increasing, and the distance from \(O\) is not changing at that instant.

  2. Which statement is true at moment B?

    • The particle moves right and up, the speed does not change at that instant, and the distance from \(O\) is increasing.
    • The particle moves right and up, the speed is decreasing, and the distance from \(O\) is increasing.
    • The particle moves right and up, the speed does not change at that instant, and the distance from \(O\) is not changing.
    Answer

    The particle moves right and up, the speed does not change at that instant, and the distance from \(O\) is increasing.

  3. A particle has position \(\langle t,\ 6 - t\rangle\). At \(t = 2\), is it moving toward the origin or away from it?

    • Toward, because \(xx' + yy' = -2\)
    • Away, because the x-coordinate increases
    • Neither, because the speed is constant
    Answer

    Toward, because \(xx' + yy' = -2\)

Summary

The signs of \(x'\) and \(y'\) give the direction of motion, and the sign of \(x'x'' + y'y''\) tells whether the speed rises.

The sign of \(xx' + yy'\) tells whether the distance from the origin rises, which is a separate question from the speed.

You use these tests to describe the motion at a moment and to find when a particle is nearest to the origin.

This is one of five skills in topic 9.6 of Which & Why: AP Calculus.