When a Motion Formula Changes
A parametrization can describe different rules of motion on different time intervals. For example, an object might move according to one pair of coordinate formulas before a time \(t=c\), then follow another pair afterward. At the transition, it is important to check more than the formulas on either side: the position must fit together, and the rates of change from each side may or may not agree.
Write the position of an object at time \(t\) as the vector \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). When \(x(t)\) and \(y(t)\) are piecewise-defined, analyze them separately for position and continuity, then combine their rates of change into velocity vectors. This lets you distinguish a continuous motion with an abrupt change in velocity from a formula that describes a jump in position.
Position and Continuity at a Transition
Suppose the formulas change at \(t=c\). The left-hand position is the limit of \(\mathbf{r}(t)\) as \(t\) approaches \(c\) from below, and the right-hand position is the limit as \(t\) approaches \(c\) from above. The position is continuous at \(c\) precisely when both limits exist and equal the value \(\mathbf{r}(c)\).
Because \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), this vector equality holds exactly when it holds for each coordinate. In practice, check the \(x\)-coordinates and \(y\)-coordinates separately. A mismatch in either coordinate means the position vector is not continuous.
The value assigned at \(t=c\) matters. Even if the formulas on either side approach the same position, assigning a different value at the transition would make the position discontinuous there. A discontinuity means this piecewise rule does not describe an object traveling continuously through that time: the position changes abruptly in the mathematical model.
One-Sided Velocity
On an interval where the coordinate formulas are differentiable, velocity is the vector of coordinate derivatives:
At a formula change, calculate the velocity from each side. If position is continuous at \(c\), the one-sided velocities are given by the corresponding one-sided difference quotients:
Here \(\mathbf{v}_-(c)\) is the velocity approached from earlier times, and \(\mathbf{v}_+(c)\) is the velocity approached from later times. In routine calculations, differentiate the formula on each interval and evaluate the resulting one-sided values at \(c\), provided the relevant limits exist. The speed from either side is the magnitude of that velocity vector.
This condition concerns the whole vector: both coordinate rates must agree from the two sides. If the position is continuous but the one-sided velocities differ, there is no ordinary two-sided velocity at the transition, even though velocity exists on each adjacent interval.
Why Matching One-Sided Velocities Is the Test
The derivative at \(c\), if it exists, is the limit of the difference quotient as \(h\) approaches zero from either direction. Restricting that limit to \(h<0\) gives the left-hand derivative; restricting it to \(h>0\) gives the right-hand derivative. A two-sided limit exists exactly when these one-sided limits exist and are equal. Applying this reasoning to each coordinate proves the stated vector test.
For a piecewise formula, differentiating each branch gives the rates on its interval. If the position pieces meet continuously, the branch difference quotients at the join use the same position \(\mathbf{r}(c)\); their limits are therefore precisely the one-sided velocities. This is why the position check comes first. If the position jumps, those branch rates do not supply a single ordinary derivative of the position at the transition.
Use the branch that includes \(t=c\) to evaluate \(\mathbf{r}(c)\).
Evaluate the left- and right-hand limits of each coordinate and compare them with \(\mathbf{r}(c)\).
Differentiate the formulas on the left and right intervals, then take their values or limits at \(c\).
If they agree, there is a single velocity at \(c\); if not, velocity changes abruptly and the two-sided derivative does not exist.
Worked Examples
Worked Example: Continuous Position but Different Velocities
Let the motion be defined for \(t\) near \(2\) by
First evaluate the position at the transition using the \(t\le 2\) branch:
Now calculate the left-hand limits:
Calculate the right-hand limits from the other formulas:
Both limits equal \(\mathbf{r}(2)\), so position is continuous. Differentiate each branch. On the left, \(x'(t)=2t\) and \(y'(t)=3\), giving
On the right, \(x'(t)=4\) and \(y'(t)=2(t-2)\), giving
The velocities differ, so there is no two-sided velocity at \(t=2\). The one-sided speeds are \(\sqrt{4^2+3^2}=5\) and \(\sqrt{4^2+0^2}=4\). As a check, both position limits were \(\langle4,6\rangle\), but the coordinate rates \(3\) and \(0\) for \(y\) disagree at the join.
Worked Example: A Discontinuity in Position
Consider the motion near \(t=1\):
The assigned position at \(t=1\) comes from the \(t\ge 1\) branch:
From the left, \(x(t)\) approaches \(1+1=2\), while \(y(t)\) approaches \(1^2=1\). Thus
From the right, the position limit is \(\langle2(1)+1,1^2\rangle=\langle3,1\rangle\). Since the left limit disagrees with the assigned position, the position is discontinuous. The branch derivatives are \(\langle1,2t\rangle\) on the left and \(\langle2,2t\rangle\) on the right, with limiting values \(\langle1,2\rangle\) and \(\langle2,2\rangle\). These describe rates on the adjacent intervals, but they are not one-sided derivatives of a continuous position at \(1\). There is no ordinary velocity at the transition. Checking the left position limit against \(\mathbf{r}(1)\) confirms the discontinuity directly.
Worked Example: Different Formulas with a Matching Velocity
Let
At \(t=1\), the position from the included branch is \(\mathbf{r}(1)=\langle1,1\rangle\). The left position limit is \(\langle1^2,1^3\rangle=\langle1,1\rangle\). The right position limit is \(\langle2(1)-1,1+3(0)+0^2\rangle=\langle1,1\rangle\). Therefore position is continuous.
Differentiate the left branches: \(x'(t)=2t\) and \(y'(t)=3t^2\). Hence
On the right, \(x'(t)=2\). For the \(y\)-coordinate, differentiating each term gives \(y'(t)=3+2(t-1)\). Thus
The formulas change, but position and one-sided velocity both match. The ordinary velocity exists and is \(\mathbf{r}'(1)=\langle2,3\rangle\); the speed there is \(\sqrt{2^2+3^2}=\sqrt{13}\). As a check, both coordinate derivatives agree from the two sides, and their common vector has magnitude \(\sqrt{13}\).
Common Pitfall: Checking Only the Derivatives
A tempting but incorrect shortcut is to differentiate the two branches, see finite one-sided rates, and conclude that velocity exists at the transition. For the discontinuous example above, one might write:
The conclusion is wrong. These are rates on the adjacent intervals, not a velocity at the transition. The left position approaches \(\langle2,1\rangle\), while the assigned position is \(\langle3,1\rangle\). Since the position is discontinuous, it cannot have an ordinary derivative there. The correction is to check position continuity first, then compare one-sided velocities only when the pieces meet continuously.
Interpreting the Transition
There are three distinct outcomes to recognize:
| Position at \(c\) | One-sided velocities | Conclusion |
|---|---|---|
| Continuous | Equal | A single velocity exists at \(c\). |
| Continuous | Different | Position passes through \(c\), but the two-sided velocity does not exist. |
| Discontinuous | Any result from branch derivatives | The position jumps; there is no ordinary velocity at \(c\). |
At a transition that is an endpoint of the stated time interval, only the available one-sided behavior can be examined; there is no two-sided derivative within that interval. At an interior transition, continuity and matching one-sided velocities provide the test for an ordinary velocity. Keep position and velocity distinct: a continuous position does not guarantee continuous velocity, and matching branch derivatives do not repair a jump in position.
Check Your Understanding
For each question, check position before drawing a conclusion about velocity.
- What must be true of the left-hand position limit, the assigned position, and the right-hand position limit for continuity at a transition \(t=c\)?
- A continuous position has one-sided velocities \(\langle3,4\rangle\) and \(\langle3,0\rangle\). Does an ordinary velocity exist at the transition? What are the one-sided speeds?
- If the left-hand position limit differs from \(\mathbf{r}(c)\), can the ordinary velocity exist there, even if both branch derivatives are finite?
- For the motion in Worked Example 3, state the position, velocity, and speed at \(t=1\).
- Why must the position-continuity check come before comparing branch derivatives?
