Why Speed Can Change Even When Motion Has a Direction
For motion along a space curve, velocity records both the direction of motion and how quickly position changes. Speed records only the magnitude of velocity. Since velocity can change direction, speed need not change whenever velocity changes. The useful question is: how can we calculate the instantaneous rate at which speed changes?
The key is to differentiate the square of the speed. Squared speed is the dot product of velocity with itself, so its derivative connects speed to both velocity and acceleration. This gives a compact formula for deciding whether motion is speeding up or slowing down.
The speed \(s(t)\) is nonnegative. Its derivative, when it exists, is a signed rate: a positive value means speed is increasing, while a negative value means speed is decreasing.
Deriving the Rate of Change of Speed
Suppose \(\mathbf{v}(t)\ne\mathbf{0}\) at the time under consideration. Then \(s(t)=|\mathbf{v}(t)|>0\), and the speed is differentiable there. The squared magnitude satisfies \(s(t)^2=\mathbf{v}(t)\cdot\mathbf{v}(t)\). Differentiate both sides with respect to \(t\).
Since \(s=|\mathbf{v}|\), the definition of magnitude gives \(s^2=\mathbf{v}\cdot\mathbf{v}\).
The chain rule gives \(\frac{d}{dt}(s^2)=2ss'\). The product rule for a dot product gives \(\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})=\mathbf{a}\cdot\mathbf{v}+\mathbf{v}\cdot\mathbf{a}\).
The dot product is symmetric, so the right-hand side is \(2\mathbf{v}\cdot\mathbf{a}\). Because \(s>0\), division by \(2s\) is valid.
This derivation also explains why the dot product appears: differentiating the squared magnitude produces two equal terms, each involving velocity and acceleration. The factor of \(2\) then cancels the factor of \(2\) from differentiating \(s^2\).
Reading the Sign
At a time when \(\mathbf{v}\ne\mathbf{0}\), the denominator \(|\mathbf{v}|\) is positive. Therefore the sign of \(s'\) is the sign of \(\mathbf{v}\cdot\mathbf{a}\). This gives a direct test using velocity and acceleration.
A zero rate at one instant does not by itself mean speed is constant over a time interval. It says only that the derivative of speed at that instant is zero. To conclude that speed remains constant, one would need information over an interval, not just at a single time.
Worked Examples
Worked Example: Finding the Speed Rate from Position
Let \(\mathbf{r}(t)=(t,t^2,2t)\). Find the rate of change of speed at \(t=1\), and determine whether the motion is speeding up or slowing down at that time.
First differentiate position to find velocity, then differentiate velocity to find acceleration:
At \(t=1\), these are \(\mathbf{v}(1)=(1,2,2)\) and \(\mathbf{a}(1)=(0,2,0)\). The speed is
The dot product is
Since the velocity is nonzero, the formula applies:
The rate is positive, so the object is speeding up at \(t=1\). As a check, the speed for general \(t\) is \(s(t)=\sqrt{1+4t^2+4}=\sqrt{5+4t^2}\). Direct differentiation gives \(s'(t)=4t/\sqrt{5+4t^2}\), and \(s'(1)=4/\sqrt{9}=4/3\), as found above.
Worked Example: Speeding Up on One Side of a Time
Let \(\mathbf{r}(t)=(2t,t^2,0)\). Determine the rate of change of speed at \(t=-1\) and \(t=1\), and classify the motion at each time.
Differentiate to obtain
At \(t=-1\), velocity is \((2,-2,0)\), acceleration is \((0,2,0)\), and speed is \(\sqrt{2^2+(-2)^2}=2\sqrt{2}\). Thus
The negative rate means speed is decreasing at \(t=-1\). At \(t=1\), velocity is \((2,2,0)\), acceleration remains \((0,2,0)\), and speed is again \(2\sqrt{2}\). Therefore
The positive rate means speed is increasing at \(t=1\). To check both results, the speed at any \(t\) is \(s(t)=\sqrt{4+4t^2}=2\sqrt{1+t^2}\). Differentiating gives \(s'(t)=2t/\sqrt{1+t^2}\). Substitution gives \(s'(-1)=-\sqrt{2}\) and \(s'(1)=\sqrt{2}\), matching the dot-product calculation.
Worked Example: Zero Velocity and a Limitation of the Formula
Let \(\mathbf{r}(t)=(t^2,0,0)\). Find the speed and examine its behavior at \(t=0\).
The velocity and acceleration are
The speed is \(s(t)=|\mathbf{v}(t)|=|2t|=2|t|\). At \(t=0\), velocity is zero, so the rate formula cannot be used: its numerator and denominator are both zero. To check whether the speed itself has a derivative there, use the difference quotient. Since \(s(0)=0\), for \(h>0\) the quotient is \((2|h|-0)/h=2\), while for \(h<0\) it is \((2|h|-0)/h=-2\). The one-sided limits disagree, so \(s'(0)\) does not exist.
For \(t\ne0\), the formula does apply. The dot product is \(\mathbf{v}\cdot\mathbf{a}=4t\), and the speed is \(2|t|\). Hence it gives \(s'(t)=4t/(2|t|)\), which is \(2\) for \(t>0\) and \(-2\) for \(t<0\). These values agree with direct differentiation on each side of zero. The example confirms that the nonzero-velocity condition matters.
A Common Mistake: Dividing When Speed Is Zero
The formula was obtained by dividing by \(2s\), which is valid only when \(s>0\). A tempting but invalid calculation is to substitute a time with zero velocity into the formula and treat the resulting \(0/0\) as a rate. In the preceding example, at \(t=0\) that substitution gives \((\mathbf{v}\cdot\mathbf{a})/|\mathbf{v}|=0/0\), which is undefined. It does not establish that speed has a derivative or tell us its value.
At zero velocity, return to the definition of speed and, when needed, use the definition of derivative. Depending on the motion, speed may fail to be differentiable there, as in the example, or it may have a derivative. The formula itself makes no claim at such a time.
When the Formula Is Useful
Computing speed from velocity components and differentiating the resulting square root can be cumbersome. The dot-product formula often shortens the work: find velocity and acceleration, compute their dot product, and divide by speed. It also separates the direction of the speed change from its magnitude. A positive or negative result gives the direction of change; the numerical value tells how many units of speed change per unit of time.
The formula concerns the derivative of speed, not the magnitude of acceleration. Acceleration can be nonzero while the instantaneous rate of change of speed is zero. For example, if velocity and acceleration have dot product zero while velocity is nonzero, the formula gives \(s'=0\). The velocity is changing, but speed has zero instantaneous derivative at that time.
Check Your Understanding
Use the definition of speed and the rate formula where its condition is satisfied.
- If \(\mathbf{v}=(3,0,0)\) and \(\mathbf{a}=(2,1,0)\) at an instant, what is the rate of change of speed? Is the object speeding up or slowing down?
- For \(\mathbf{v}=(1,-2,0)\) and \(\mathbf{a}=(2,-1,0)\), calculate the speed rate and state its sign.
- Why is it necessary to assume \(\mathbf{v}\ne\mathbf{0}\) when deriving the formula?
- If the dot product of nonzero velocity and acceleration is zero at one instant, what can you conclude about the speed rate at that instant? What can you not conclude about a whole interval?
- For \(\mathbf{r}(t)=(t,t^2,0)\), find \(\mathbf{v}(t)\), \(\mathbf{a}(t)\), and the speed rate at \(t=1\).