What the Dot Product Tells Us
An object moving through space can change its velocity in two ways: its direction can change, its speed can change, or both can happen at once. Acceleration records the change in velocity, but its components do not by themselves say whether the object is speeding up or slowing down.
The dot product of velocity and acceleration provides a direct test for the instantaneous rate of change of speed. The test works even when acceleration is not parallel to velocity, provided the velocity at the instant being studied is nonzero.
Definitions and the Sign Test
Let \(\mathbf{r}(t)\) be a position vector that is twice differentiable. Its velocity and acceleration are \(\mathbf{v}(t)=\mathbf{r}'(t)\) and \(\mathbf{a}(t)=\mathbf{v}'(t)\), respectively. The speed is the nonnegative scalar \(s(t)=\|\mathbf{v}(t)\|\). In three-dimensional Cartesian coordinates, for example, if \(\mathbf{v}=\langle v_1,v_2,v_3\rangle\) and \(\mathbf{a}=\langle a_1,a_2,a_3\rangle\), then \(\mathbf{v}\cdot\mathbf{a}=v_1a_1+v_2a_2+v_3a_3\).
When \(\mathbf{v}(t)\ne\mathbf{0}\), the speed is differentiable at that instant, and the sign test is:
| Value of \(\mathbf{v}\cdot\mathbf{a}\) | Instantaneous conclusion when \(\mathbf{v}\ne\mathbf{0}\) |
|---|---|
| Positive | \(s'(t)>0\): speed is increasing at that instant. |
| Negative | \(s'(t)<0\): speed is decreasing at that instant. |
| Zero | \(s'(t)=0\): speed has zero instantaneous rate of change. |
A zero value does not mean that acceleration is zero, or that the speed stays constant over an interval. It means only that the speed derivative is zero at the instant in question. Also, a positive or negative derivative at one instant is a local rate statement; to claim that speed increases or decreases throughout an interval, the derivative’s sign must be established across that interval.
Why the Test Works
The key is to differentiate the square of the speed. The Euclidean norm satisfies \(s(t)^2=\mathbf{v}(t)\cdot\mathbf{v}(t)\). Differentiating both sides and using the product rule for a dot product gives the required relationship.
Because the velocity is nonzero, its norm \(s(t)\) is positive. Dividing by \(2s(t)\) is therefore valid, and yields
The denominator is positive, so \(s'(t)\) and \(\mathbf{v}\cdot\mathbf{a}\) have the same sign. This proves the sign test. The derivation also explains why the condition \(\mathbf{v}\ne\mathbf{0}\) matters: if velocity is zero, the division by speed is not allowed.
Applying the Test Step by Step
For a motion given by a position function, first differentiate to find velocity and acceleration. Then evaluate both at the time of interest, check that velocity is nonzero, and calculate their dot product. The sign gives the instantaneous speed classification. Computing \(s'(t)\) separately can provide a useful check.
Differentiate position to obtain \(\mathbf{v}(t)\), then differentiate velocity to obtain \(\mathbf{a}(t)\).
Substitute that time into both vectors and verify that \(\mathbf{v}\ne\mathbf{0}\).
Use the component products and sum them, keeping their signs.
Positive means increasing speed, negative means decreasing speed, and zero means zero instantaneous speed derivative.
Worked Examples
Worked Example: Speed Increasing with Nonparallel Acceleration
Consider the position, in metres, \(\mathbf{r}(t)=\langle t,t^2\rangle\), with \(t\) measured in seconds. Determine whether speed is increasing or decreasing at \(t=1\) second.
Differentiate position and velocity: \[ \mathbf{v}(t)=\langle 1,2t\rangle,\qquad \mathbf{a}(t)=\langle 0,2\rangle. \] At \(t=1\), the values are \(\mathbf{v}(1)=\langle1,2\rangle\ \mathrm{m/s}\) and \(\mathbf{a}(1)=\langle0,2\rangle\ \mathrm{m/s^2}\). The velocity is nonzero. Their dot product is \[ \mathbf{v}(1)\cdot\mathbf{a}(1) =(1)(0)+(2)(2)=4\ \mathrm{m^2/s^3}. \] Since this is positive, speed is increasing at that instant. The vectors are not parallel: the first component of acceleration is zero, whereas the first component of velocity is not.
Check by differentiating speed directly: \(s(t)=\sqrt{1+4t^2}\), so \[ s'(t)=\frac{4t}{\sqrt{1+4t^2}}, \qquad s'(1)=\frac{4}{\sqrt{5}}\ \mathrm{m/s^2}>0. \] This agrees with the dot-product test.
Worked Example: Speed Decreasing with Nonparallel Acceleration
Let \(\mathbf{r}(t)=\langle2t,t^2\rangle\) metres, with \(t\) in seconds. Classify the speed at \(t=-1\) second.
Differentiation gives \(\mathbf{v}(t)=\langle2,2t\rangle\ \mathrm{m/s}\) and \(\mathbf{a}(t)=\langle0,2\rangle\ \mathrm{m/s^2}\). At \(t=-1\), \[ \mathbf{v}(-1)=\langle2,-2\rangle\ \mathrm{m/s}, \qquad \mathbf{a}(-1)=\langle0,2\rangle\ \mathrm{m/s^2}. \] The velocity is nonzero, and \[ \mathbf{v}(-1)\cdot\mathbf{a}(-1) =(2)(0)+(-2)(2)=-4\ \mathrm{m^2/s^3}. \] The dot product is negative, so speed is decreasing at this instant, even though the acceleration is not parallel to the velocity.
For a check, \(s(t)=\sqrt{4+4t^2}\ \mathrm{m/s}\), and \[ s'(t)=\frac{4t}{\sqrt{4+4t^2}}\ \mathrm{m/s^2}. \] Thus \(s'(-1)=-4/\sqrt{8}=-\sqrt{2}\ \mathrm{m/s^2}<0\), matching the test.
Worked Example: Zero Dot Product but Nonzero Acceleration
Consider \(\mathbf{r}(t)=\langle t,\tfrac12t^2\rangle\) metres, with \(t\) in seconds. What does the dot product say at \(t=0\)?
Here \(\mathbf{v}(t)=\langle1,t\rangle\ \mathrm{m/s}\) and \(\mathbf{a}(t)=\langle0,1\rangle\ \mathrm{m/s^2}\). At zero, \(\mathbf{v}(0)=\langle1,0\rangle\ \mathrm{m/s}\) and \(\mathbf{a}(0)=\langle0,1\rangle\ \mathrm{m/s^2}\). Therefore \[ \mathbf{v}(0)\cdot\mathbf{a}(0)=(1)(0)+(0)(1)=0. \] Since velocity is nonzero, the test applies: \(s'(0)=0\). This says the speed has zero instantaneous rate of change, not that acceleration vanishes.
Indeed, \(s(t)=\sqrt{1+t^2}\ \mathrm{m/s}\), so \[ s'(t)=\frac{t}{\sqrt{1+t^2}}\ \mathrm{m/s^2}, \qquad s'(0)=0. \] The speed is not constant: for example, \(s(0)=1\ \mathrm{m/s}\), while \(s(1)=\sqrt{2}\ \mathrm{m/s}\). The zero derivative describes the rate at one instant only.
When Velocity Is Zero
If \(\mathbf{v}(t_0)=\mathbf{0}\), then \(\mathbf{v}(t_0)\cdot\mathbf{a}(t_0)=0\) automatically, regardless of the acceleration. The sign test cannot classify the speed derivative by dividing by \(\|\mathbf{v}(t_0)\|\), because that denominator is zero. The object is instantaneously at rest, but the speed might begin to rise immediately; the dot product at that instant alone does not determine the behavior.
Worked Example: Why Zero Velocity Needs Separate Analysis
Take \(\mathbf{r}(t)=\langle t^2,t^3\rangle\) metres, with \(t\) in seconds. Then \[ \mathbf{v}(t)=\langle2t,3t^2\rangle\ \mathrm{m/s}, \qquad \mathbf{a}(t)=\langle2,6t\rangle\ \mathrm{m/s^2}. \] At \(t=0\), \(\mathbf{v}(0)=\langle0,0\rangle\), so the object is at rest, and \[ \mathbf{v}(0)\cdot\mathbf{a}(0)=(0)(2)+(0)(0)=0. \] That zero does not give the sign of the speed derivative through the instant.
The speed is \[ s(t)=\sqrt{4t^2+9t^4}=|t|\sqrt{4+9t^2}\ \mathrm{m/s}. \] For \(t>0\), the quotient \(s(t)/t\) tends to \(2\ \mathrm{m/s^2}\) as \(t\) approaches zero from the right. For \(t<0\), the quotient \(s(t)/t\) tends to \(-2\ \mathrm{m/s^2}\) from the left. The two-sided derivative at zero does not exist. This confirms that the zero dot product at a zero-velocity instant is inconclusive.
Common Pitfall: Confusing Acceleration with Speed Change
A common error is to assume that any nonzero acceleration makes speed increase. For example, in the decreasing-speed example, acceleration is \(\langle0,2\rangle\ \mathrm{m/s^2}\), which is nonzero, but the dot product at \(t=-1\) is \(-4\ \mathrm{m^2/s^3}\). Speed is decreasing. The direction of acceleration by itself is not the classification; its dot product with velocity is.
Another error is to treat a zero dot product as proof that the motion has constant speed. The third example shows why this fails: the speed derivative is zero at \(t=0\), yet speed takes different values at nearby times. To establish constant speed over an interval, one must show that speed has zero derivative throughout the interval (and that the relevant differentiability conditions hold), not merely at one instant.
Check Your Understanding
For each question, distinguish what the dot product establishes at the specified instant from what would require information over a time interval.
- If \(\mathbf{v}=\langle2,-1\rangle\) and \(\mathbf{a}=\langle1,-3\rangle\), what is \(\mathbf{v}\cdot\mathbf{a}\), and what does its sign imply when velocity is nonzero?
- Can acceleration be nonzero when the speed derivative is zero? Give the condition on \(\mathbf{v}\) and \(\mathbf{a}\) for a nonzero velocity.
- Why can the sign test not be applied by dividing by speed when \(\mathbf{v}=\mathbf{0}\)?
- If \(\mathbf{v}\cdot\mathbf{a}=0\) at one instant with \(\mathbf{v}\ne\mathbf{0}\), does that prove speed is constant? Explain.
- What does a negative dot product imply about the instantaneous speed derivative when velocity is nonzero?