Speeding Up and Slowing Down Workshop

What the Dot Product Tells Us About Speed

An object moving along a space curve has a velocity vector \(\mathbf{v}(t)\) and an acceleration vector \(\mathbf{a}(t)\). Velocity describes the direction and rate of motion; speed is the nonnegative scalar quantity \(\|\mathbf{v}(t)\|\). To decide whether an object is speeding up or slowing down, we need the rate at which that scalar changes.

The key calculation is the dot product \(\mathbf{v}(t)\cdot\mathbf{a}(t)\). Its sign tells us whether the speed is increasing or decreasing whenever the velocity is nonzero. This gives a practical method even when the motion is not along a straight line: we do not need to find the direction of the path or split acceleration into components.

Definition: An object is speeding up on an interval if its speed \(\|\mathbf{v}(t)\|\) is increasing there. It is slowing down on an interval if its speed is decreasing there. These descriptions concern speed, not whether individual components of velocity are increasing or decreasing.

The Speed-Change Rule

Assume the velocity is differentiable and \(\mathbf{a}(t)=\mathbf{v}'(t)\). At any time when \(\mathbf{v}(t)\ne\mathbf{0}\), differentiate the speed. The resulting formula gives the sign test.

Theorem: Wherever \(\mathbf{v}(t)\ne\mathbf{0}\), the speed \(s(t)=\|\mathbf{v}(t)\|\) satisfies

s'(t)=\frac{\mathbf{v}(t)\cdot\mathbf{a}(t)}{\|\mathbf{v}(t)\|}.

The denominator is positive, so \(s'(t)\) has the same sign as \(\mathbf{v}(t)\cdot\mathbf{a}(t)\). A positive dot product means speed is increasing; a negative dot product means speed is decreasing. If the dot product is zero, the instantaneous rate of change of speed is zero, provided the velocity is nonzero.

Why the Rule Works

The proof follows by differentiating the square of the speed. This avoids differentiating a square root until the velocity is known to be nonzero.

Write speed squared.
By the definition of the Euclidean norm, \(s(t)^2=\mathbf{v}(t)\cdot\mathbf{v}(t)\).

Differentiate both sides.
The product rule for the dot product gives \(2s(t)s'(t)=2\mathbf{v}(t)\cdot\mathbf{v}'(t)\).

Use acceleration and divide only when allowed.
Since \(\mathbf{v}'(t)=\mathbf{a}(t)\), if \(\mathbf{v}(t)\ne\mathbf{0}\), then \(s(t)>0\). Dividing by \(2s(t)\) gives the stated formula.

For completeness, the key differentiation can be written as

\frac{d}{dt}\bigl(\mathbf{v}\cdot\mathbf{v}\bigr) =\mathbf{v}'\cdot\mathbf{v}+\mathbf{v}\cdot\mathbf{v}' =2\mathbf{v}\cdot\mathbf{a}.

At a time when \(\mathbf{v}(t)=\mathbf{0}\), division by the speed is not valid. The dot product is then automatically zero, but that fact alone does not tell us whether the object is speeding up or slowing down on either side of that instant. Check the sign on the neighboring intervals, and, when useful, examine the speed directly.

A Reliable Interval-Classification Method

To classify motion on a specified time interval, find the scalar function \(q(t)=\mathbf{v}(t)\cdot\mathbf{a}(t)\). Its zeros and any points where the velocity is zero are candidates for dividing the time interval into pieces. Test the sign of \(q\) on each resulting open interval. The sign is constant on each such interval when \(q\) is continuous and has no zero there.

Compute the dot product.
Multiply corresponding components of velocity and acceleration, then add.

Find critical times.
Solve \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\) and note any times when \(\mathbf{v}(t)=\mathbf{0}\).

Make sign intervals.
Use the critical times to partition the time range, then test the dot product in each open interval.

Translate signs into motion.
A positive sign means speeding up; a negative sign means slowing down. Treat a zero as a boundary or an instant to inspect, not automatically as an entire interval of constant speed.

Worked Examples

Worked Example: A Single Change from Slowing Down to Speeding Up

Let \(\mathbf{v}(t)=\langle t,2\rangle\) and \(\mathbf{a}(t)=\langle 1,0\rangle\), for \(-3<t<3\). These are consistent because differentiating the velocity gives the stated acceleration.

First compute the dot product:

\mathbf{v}(t)\cdot\mathbf{a}(t) =\langle t,2\rangle\cdot\langle 1,0\rangle =t(1)+2(0)=t.

The dot product is negative for \(-3<t<0\) and positive for \(0<t<3\). Therefore the object slows down on \((-3,0)\) and speeds up on \((0,3)\). At \(t=0\), the velocity is \(\langle 0,2\rangle\), which is nonzero, and the dot product is zero. This is an instant when the rate of change of speed is zero, not an interval of constant speed.

Check by calculating and differentiating the speed:

s(t)=\|\mathbf{v}(t)\|=\sqrt{t^2+4}, \qquad s'(t)=\frac{2t}{2\sqrt{t^2+4}} =\frac{t}{\sqrt{t^2+4}}.

The denominator is positive, so \(s'(t)\) has the sign of \(t\), exactly as the dot-product test predicts. At \(t=0\), the speed is \(2\) and its derivative is \(0\).

Worked Example: Multiple Changes and Times at Rest

Let \(\mathbf{v}(t)=\langle t^2-1,0\rangle\) and \(\mathbf{a}(t)=\langle 2t,0\rangle\), for \(-2<t<2\). Since the derivative of \(t^2-1\) is \(2t\), these functions are consistent.

The dot product factors as

\mathbf{v}(t)\cdot\mathbf{a}(t) =(t^2-1)(2t) =2t(t-1)(t+1).

The critical times are \(t=-1,0,1\). Testing one value in each interval gives:

Interval	Test value and dot product	Classification
\((-2,-1)\)	\(t=-\frac32:\;2(-\frac32)(\frac54)=-\frac{15}{4}<0\)	Slowing down
\((-1,0)\)	\(t=-\frac12:\;2(-\frac12)(-\frac34)=\frac34>0\)	Speeding up
\((0,1)\)	\(t=\frac12:\;2(\frac12)(-\frac34)=-\frac34<0\)	Slowing down
\((1,2)\)	\(t=\frac32:\;2(\frac32)(\frac54)=\frac{15}{4}>0\)	Speeding up

At \(t=-1\) and \(t=1\), the velocity is zero, so the object is momentarily at rest. At \(t=0\), the velocity is \(\langle-1,0\rangle\), which is nonzero; there the dot product changes from positive to negative.

Directly, the speed is \(s(t)=|t^2-1|\). On \((-2,-1)\), this is \(t^2-1\), so \(s'(t)=2t<0\). On \((-1,0)\), it is \(1-t^2\), so \(s'(t)=-2t>0\). On \((0,1)\), \(s'(t)=-2t<0\), and on \((1,2)\), \(s'(t)=2t>0\). These derivatives verify all four classifications, including the intervals adjoining the times at rest.

Worked Example: Three Sign Changes in Two Dimensions

Let \(\mathbf{v}(t)=\langle t^2-1,t\rangle\) and \(\mathbf{a}(t)=\langle 2t,1\rangle\), for \(-1<t<1\). Differentiating \(\mathbf{v}\) confirms that \(\mathbf{v}'(t)=\mathbf{a}(t)\).

The dot product is

\begin{aligned} \mathbf{v}(t)\cdot\mathbf{a}(t) &=(t^2-1)(2t)+t(1)\\ &=2t^3-2t+t\\ &=t(2t^2-1). \end{aligned}

Its zeros are \(t=-1/\sqrt{2}\), \(t=0\), and \(t=1/\sqrt{2}\). Test the four intervals:

Interval	Test value and dot product	Classification
\((-1,-1/\sqrt{2})\)	\(t=-\frac34:\;(-\frac34)(\frac18)=-\frac{3}{32}<0\)	Slowing down
\((-1/\sqrt{2},0)\)	\(t=-\frac12:\;(-\frac12)(-\frac12)=\frac14>0\)	Speeding up
\((0,1/\sqrt{2})\)	\(t=\frac12:\;(\frac12)(-\frac12)=-\frac14<0\)	Slowing down
\((1/\sqrt{2},1)\)	\(t=\frac34:\;(\frac34)(\frac18)=\frac{3}{32}>0\)	Speeding up

The velocity is nonzero at each of the three critical times: at \(t=0\) it is \(\langle-1,0\rangle\), and at either root of \(2t^2-1\) its first component is \(-1/2\). To check the result, compute

s(t)=\sqrt{(t^2-1)^2+t^2} =\sqrt{t^4-t^2+1}, \qquad s'(t)=\frac{4t^3-2t}{2\sqrt{t^4-t^2+1}} =\frac{t(2t^2-1)}{\sqrt{t^4-t^2+1}}.

The denominator is positive for every \(t\), so the derivative has the same signs as the dot product. This confirms the alternating slowing-down and speeding-up intervals.

Common Pitfall: Looking at Acceleration Alone

A nonzero acceleration does not by itself mean an object is speeding up. For example, in the first worked example, acceleration is \(\langle1,0\rangle\) throughout, so its magnitude is always \(1\). But at \(t=-1\),

\mathbf{v}(-1)=\langle-1,2\rangle, \qquad \mathbf{v}(-1)\cdot\mathbf{a}(-1) =\langle-1,2\rangle\cdot\langle1,0\rangle=-1<0.

The object is slowing down at that time, despite having nonzero acceleration. The relevant question is not whether acceleration exists or how large it is, but whether its component in the direction of velocity is positive or negative. The dot product measures precisely that relationship.

Takeaway: On intervals where \(\mathbf{v}(t)\ne\mathbf{0}\), classify motion using the sign of \(\mathbf{v}(t)\cdot\mathbf{a}(t)\). Positive means speeding up; negative means slowing down. At a zero of the dot product or a time when velocity vanishes, inspect the neighboring intervals rather than assigning a classification from that single value alone.

Check Your Understanding

For each question, use the dot product and explain how its sign supports your conclusion.

If \(\mathbf{v}(t)\cdot\mathbf{a}(t)>0\) and \(\mathbf{v}(t)\ne\mathbf{0}\), what is happening to speed?
For \(\mathbf{v}(t)=\langle t,1\rangle\) and \(\mathbf{a}(t)=\langle1,0\rangle\), identify the speeding-up and slowing-down intervals on \((-2,2)\).
Why is it invalid to use \(s'(t)=(\mathbf{v}\cdot\mathbf{a})/\|\mathbf{v}\|\) at a time when \(\mathbf{v}=\mathbf{0}\)?
Can a zero dot product at one instant establish that speed is constant on an interval? Explain.
For \(\mathbf{v}(t)=\langle t^2-1,0\rangle\), identify the times at rest and the intervals where speed increases or decreases on \((-2,2)\).