Changing the Clock Without Changing the Curve
A parametrized curve describes positions in terms of a parameter. When that parameter represents time, changing the time variable changes how the motion is carried out. The object may move faster or slower, or traverse the same route in the opposite direction, even though the underlying geometric curve is unchanged.
To analyze such a change, let \(\mathbf r:I\to\mathbb R^n\) be a twice-differentiable position function on an interval \(I\). Replace its parameter \(u\) by an affine function of time, \(u=\phi(t)=ct+d\). The transformed position is \(\mathbf q(t)=\mathbf r(ct+d)\), defined on an interval \(J\) for which \(ct+d\in I\).
This distinction matters. Two motions may have the same geometric path while moving along it at different rates, traversing it in opposite directions, or making different numbers of passes.
The Chain Rule for Affine Time Changes
For \(\mathbf q(t)=\mathbf r(ct+d)\), the inner parameter has derivative \(du/dt=c\). Applying the chain rule to each component gives the velocity and acceleration of the transformed motion.
Here \(\mathbf r'(u)\) and \(\mathbf r''(u)\) are the original velocity and acceleration when the parameter is time. If speed is the magnitude of velocity, then the transformed speed is
The absolute value is essential: speed is nonnegative, so reversing the direction of motion does not make speed negative. For \(c\ne0\), the sign of \(c\) determines whether the new motion follows or reverses the original parameter order. The factor \(c^2\) in acceleration means that changing the sign of \(c\) does not change the acceleration vector at the corresponding original parameter value.
Why the Acceleration Factor Is Squared
The velocity formula follows by differentiating the composition once. Differentiating again requires the chain rule a second time: the factor \(c\) from the first derivative remains, and differentiating \(\mathbf r'(ct+d)\) contributes another factor \(c\).
Since \(c\) is constant, it can be taken outside the derivative in the second line. The sign of the acceleration factor is therefore always nonnegative, whether \(c\) is positive or negative.
Worked Example: Speeding Up a Parabolic Motion
Worked Example: Speeding Up a Parabolic Motion
Let \(\mathbf r(u)=(u,u^2)\) for \(0\le u\le2\). Define \(\mathbf q(t)=\mathbf r(3t)\) for \(0\le t\le2/3\). Find the new velocity, acceleration, and speed, and check that the geometric path is unchanged.
First, differentiate the original position:
The time change is \(u=3t\), so \(c=3\). The chain-rule formulas give
The speed is therefore
For a check by direct substitution, \(\mathbf q(t)=(3t,9t^2)\). Differentiating this expression gives \(\mathbf q'(t)=(3,18t)\) and \(\mathbf q''(t)=(0,18)\), as above. At \(t=1/3\), the original parameter is \(u=1\). The original velocity is \((1,2)\), with speed \(\sqrt5\); the new velocity is \((3,6)\), with speed \(\sqrt{45}=3\sqrt5\). Thus the speed has been multiplied by \(3\).
Finally, as \(t\) ranges from \(0\) to \(2/3\), \(u=3t\) ranges from \(0\) to \(2\). Hence the transformed position set is exactly \(\mathbf r([0,2])\): the same path is traversed three times as fast in the new time variable.
Slowing Down and Reversing Direction
If \(0<|c|<1\), the transformed speed at a corresponding point is smaller by the factor \(|c|\). If \(|c|>1\), it is larger by that factor. When \(c<0\), increasing \(t\) makes \(u=ct+d\) decrease, so the motion follows the original parameterization in reverse order. When \(c=0\), the parameter is fixed at \(d\): the position is constant, and both velocity and acceleration are zero.
For \(c\ne0\), the affine map \(t\mapsto ct+d\) is one-to-one. This means distinct times in its domain give distinct parameter values. It does not, by itself, mean that every value in \(I\) is reached. That depends on the chosen time interval \(J\).
Worked Example: Reversing a Quarter-Circle Motion
Let \(\mathbf r(u)=(\cos u,\sin u)\) for \(0\le u\le\pi/2\). Reverse its traversal using \(u=\pi/2-2t\) on \(0\le t\le\pi/4\). Find the transformed velocity, acceleration, and speed.
Write \(u=\pi/2-2t\), so \(c=-2\). The original derivatives are
Consequently,
At \(t=\pi/8\), \(u=\pi/4\), so the position is \((\sqrt2/2,\sqrt2/2)\). At that time the velocity and acceleration are
To check the signs directly, substituting the parameter gives \(\mathbf q(t)=(\sin 2t,\cos 2t)\). Its derivatives are \((2\cos 2t,-2\sin 2t)\) and \((-4\sin 2t,-4\cos 2t)\), which produce the same vectors at \(t=\pi/8\). As \(t\) increases from \(0\) to \(\pi/4\), \(u\) decreases from \(\pi/2\) to \(0\). The new motion therefore traces the same quarter-circle in reverse, at twice the original speed.
Same Path Does Not Require the Same Parameter Range
If the transformed time interval \(J\) maps onto the original interval \(I\), then the two position sets are certainly equal: every original parameter value is reached, so every original position is reached. But this condition is sufficient, not necessary. If the original parameterization repeats positions, the transformed motion can cover a smaller parameter interval and still visit exactly the same set of locations.
Worked Example: Equal Circle Paths With Different Parameter Ranges
Let \(\mathbf r(u)=(\cos u,\sin u)\) for \(0\le u\le4\pi\), and let \(\mathbf q(t)=\mathbf r(2t)\) for \(0\le t\le\pi\). Compare their position sets and speeds.
The original parameter interval is \([0,4\pi]\), whereas \(u=2t\) on the new time interval takes values only in \([0,2\pi]\). The map does not cover the original interval. Nevertheless, \(\mathbf r([0,4\pi])\) is the unit circle, because the cosine-sine pair completes two revolutions. The positions \(\mathbf q(t)=(\cos 2t,\sin 2t)\) for \(0\le t\le\pi\) also complete one full revolution and give the unit circle. Therefore \(\mathbf r(I)=\mathbf q(J)\), even though \(\phi(J)\ne I\).
For the original motion, \(\mathbf r'(u)=(-\sin u,\cos u)\), so its speed is \(\sqrt{\sin^2u+\cos^2u}=1\). For the new motion, \(c=2\), and therefore \(\mathbf q'(t)=2\mathbf r'(2t)=(-2\sin2t,2\cos2t)\); its speed is \(2\). Direct differentiation of \((\cos2t,\sin2t)\) gives this same velocity. The new motion has the same geometric path but makes one revolution instead of two, at twice the speed.
Arc Length and the Time Interval
A change of speed does not change the length of a fixed traversal when the transformed interval corresponds to the same portion of the original motion. For a continuously differentiable curve, length over a parameter interval is the integral of speed. The chain-rule speed formula shows how this works for an affine time change.
If \(c>0\) and \(\phi\) maps the endpoints of \(J\) from \(u_0\) to \(u_1\), substitution \(u=ct+d\) gives the original length \(\int_{u_0}^{u_1}\|\mathbf r'(u)\|\,du\). If \(c<0\), the parameter decreases as time increases; reversing the integration bounds cancels the negative differential, and the length over the corresponding original interval is again unchanged. This statement concerns the matching parameter intervals. If a parametrization repeats a path, the total lengths may differ when one motion makes fewer passes, even though the position sets are equal.
A Common Sign and Scaling Mistake
A frequent error is to multiply acceleration by \(c\) instead of \(c^2\), or to treat a negative speed factor as a negative speed. Take \(\mathbf r(u)=(u,u^2)\) and reverse and rescale with \(u=2-2t\). The original acceleration is \(\mathbf r''(u)=(0,2)\), and \(c=-2\). The correct transformed acceleration is
The incorrect rule \(c\mathbf r''(u)\) would give \((0,-4)\), which has both the wrong magnitude and direction. Check by substitution: \(\mathbf q(t)=(2-2t,(2-2t)^2)=(2-2t,4-8t+4t^2)\), so \(\mathbf q'(t)=(-2,-8+8t)\) and \(\mathbf q''(t)=(0,8)\). Likewise, speed is \(\|\mathbf q'\|=|c|\|\mathbf r'\|\), not \(c\|\mathbf r'\|\); speed cannot be negative.
What to Track When Time Changes
When comparing two parametrized motions, distinguish three questions: which locations are visited, in what order they are visited, and how quickly they are visited. For an affine time change, the position set is controlled by the image interval \(\phi(J)\) together with any repeated positions in \(\mathbf r\). The sign of \(c\) controls the order of parameter values, and \(|c|\) controls the speed scale. Acceleration has the squared factor \(c^2\), evaluated at the corresponding original parameter value.
Identify \(u=ct+d\) and the time interval \(J\), checking that \(\phi(J)\subseteq I\).
Use \(\mathbf q'=c\mathbf r'(\phi(t))\) and \(\mathbf q''=c^2\mathbf r''(\phi(t))\).
Speed is multiplied by \(|c|\); a negative \(c\) reverses the parameter order.
Check whether \(\mathbf r(I)=\mathbf r(\phi(J))\); interval coverage is sufficient, but repeated positions can make it unnecessary.
Check Your Understanding
Use the chain-rule formulas and the distinction between a position set and a traversal to answer these questions.
- If \(\mathbf q(t)=\mathbf r(4t-1)\), what are \(\mathbf q'(t)\) and \(\mathbf q''(t)\) in terms of derivatives of \(\mathbf r\)?
- If \(c=-3\), by what factor does speed change at a corresponding parameter value, and does the motion follow or reverse the original parameter order?
- Why is a one-to-one affine parameter map not enough to show that the entire original parameter interval is reached?
- Give the condition for two parametrizations to have the same geometric path, and explain why mapping onto the original parameter interval is sufficient but not necessary.
- For \(\mathbf r(u)=(\cos u,\sin u)\), explain how a shorter parameter interval can still produce the full unit circle.