Determine whether speed increases

A particle that slows down and speeds up

A particle has velocity \(\langle 2t,\ t^2 - 4\rangle\) and acceleration \(\langle 2,\ 2t\rangle\). The orange arrow shows the velocity, and the chalk arrow with two heads shows the acceleration. The figure draws both arrows shorter than they are.

The first graph shows the speed \(S\), the length of the orange arrow, and an orange tangent at the current time. The second graph shows \(S'\), the slope of that tangent, which is green where the speed increases and red where it decreases.

Set \(t = 1\). The tangent slopes down, so the speed decreases, even though the acceleration points up and to the right.

Set \(t = 2\). The tangent slopes up, so the speed increases. Between these times, the speed is lowest at \(t = \sqrt2\), where the tangent is flat.

The sign of the derivative of the speed

Drag the point. The speed \(S\) is a function of time, so it increases where \(S'\) is positive and decreases where \(S'\) is negative.

The speed is \(\sqrt{x'^2 + y'^2}\), and the chain rule gives \(S'\) as a fraction with \(S\) in the denominator.

While the particle moves, \(S\) is positive, so \(S'\) has the same sign as the numerator of that fraction at every time.

Notation

1. The speed is the square root of the sum of the squares of the two parts of the velocity.

   \[S = \sqrt{x'^2 + y'^2}\]

2. Differentiate with the chain rule. The derivative of the sum under the root is \(2x'x'' + 2y'y''\), and its factor 2 cancels the 2 that the square root puts in the denominator.

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

3. The speed \(S\) is positive when the particle moves, so \(S'\) has the sign of the numerator.

   A positive numerator means the speed increases, and a negative one means it decreases. If it is zero, the tangent of the speed graph is flat at that instant.

4. The numerator adds the products of matching parts of the velocity and the acceleration, so it is the dot product \(\cy{\vec v \cdot \vec a}\). A product such as \(x'x''\) is positive when that part of the velocity grows in size. If one part grows and the other shrinks, the sign of the sum decides.

When it fails

In the motion that stops, the velocity is \(\langle 2t - 4,\ 3t - 6\rangle\) and the acceleration is \(\langle 2,\ 3\rangle\). At \(t = 2\), the velocity is zero, so \(S = 0\) and the formula divides by 0. The speed is \(\sqrt{4(t-2)^2 + 9(t-2)^2} = \sqrt{13}\,|t - 2|\), which has a sharp corner at \(t = 2\), so \(S'\) does not exist there.

Test the times on both sides instead. The numerator is \(13(t - 2)\), which is negative before \(t = 2\), so the speed falls, and positive after, so the speed rises.

Example 1: Two times

A particle has velocity \(\vec v(t) = \langle t^2,\ 6 - 2t\rangle\). Is its speed increasing or decreasing at \(t = 1\)? At \(t = 3\)?

1. Differentiate the velocity to find the acceleration.

   \[\vec a(t) = \langle 2t,\ -2\rangle\]

2. At t = 1, the velocity is ⟨1, 4⟩ and the acceleration is ⟨2, −2⟩. The numerator is negative, so the speed is decreasing.

   \[\begin{aligned} x'x'' + y'y'' &= (1)(2) + (4)(-2) \\ &= \cr{-6} \end{aligned}\]

3. At t = 3, the velocity is ⟨9, 0⟩ and the acceleration is ⟨6, −2⟩. The numerator is positive, so the speed is increasing.

   \[\begin{aligned} x'x'' + y'y'' &= (9)(6) + (0)(-2) \\ &= \cg{54} \end{aligned}\]

Example 2: Positive parts that slow the particle

A particle has velocity \(\vec v(t) = \langle t + 2,\ t^2 - 5\rangle\). Is its speed increasing or decreasing at \(t = 1\)?

1. Find the acceleration and evaluate both vectors at t = 1.

   \[\begin{aligned} \vec a(t) &= \langle 1,\ 2t\rangle \\ \vec v(1) &= \langle 3,\ -4\rangle \\ \vec a(1) &= \langle 1,\ 2\rangle \end{aligned}\]

2. A tempting wrong move: Both parts of the acceleration are positive, so the speed is increasing.

   The y-part of the velocity is \(-4\), and the y-part of the acceleration is 2. They have opposite signs, so \(y'y'' = -8\), which is larger in size than \(x'x'' = 3\). The tangent of the speed graph slopes down.

3. Add the two products.

   \[\begin{aligned} x'x'' + y'y'' &= (3)(1) + (-4)(2) \\ &= \cr{-5} \end{aligned}\]

4. The numerator is negative, so the speed is decreasing.

   As a check, the speed is 5 and \(S' = -5/5 = -1\).

Practice

A particle has velocity \(\vec v(t) = \langle -2\sin t,\ 3\cos t\rangle\) for \(0 < t < \pi\). When is the speed increasing? When is it decreasing?

Find the sign of \(x'x'' + y'y''\) on each part of the interval. The tangent of the speed graph shows the sign you should find.

Questions

The figure shows a particle at moment A or moment B, with the velocity as the orange arrow and the acceleration as the chalk arrow. The numbers under the figure are the products \(x'x''\) and \(y'y''\). At A, the velocity is \(\langle 4,\ 0\rangle\). At B, it is \(\langle 3,\ 2\rangle\). At both, the acceleration is \(\langle -2,\ 3\rangle\).

1. At moment A, is the speed increasing, decreasing, or not changing?

   • Decreasing
   • Increasing
   • Not changing
   Answer

   Decreasing

2. At moment B, what happens to the speed at that instant?

   • It does not change.
   • It decreases.
   • It increases.
   Answer

   It does not change.

3. A particle has velocity \(\langle t,\ 4 - t^2\rangle\). What happens to the speed at \(t = 1\)?

   • It decreases, because \(x'x'' + y'y'' = -5\).
   • It increases, because the x-part of the acceleration is positive.
   • It increases, because \(x'x'' + y'y'' = 7\).
   Answer

   It decreases, because \(x'x'' + y'y'' = -5\).

Summary

A particle speeds up when \(\vec v \cdot \vec a = x'x'' + y'y''\) is positive and slows down when it is negative.

This works because the chain rule gives \(S'\) as this sum divided by the speed, which is positive while the particle moves.

If the velocity is zero at a time, \(S'\) does not exist there, so you test the sign at times on both sides.

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