Why Piecewise Motion Needs Boundary Checks
Particle-motion models often use different formulas on different time intervals. A change of formula may represent a change in the particle’s motion, but the formula change alone does not tell us what happens at the boundary time. The particle’s position might be continuous there, or the model might assign a jump in position. Even when position is continuous, the particle may have different left-hand and right-hand velocities, so its instantaneous velocity at that exact time may not exist.
The key is to keep three questions separate: What is the position at the boundary? What position do the formulas approach from each side? And do the one-sided rates of change agree?
Piecewise Position and Boundary Values
A piecewise position function assigns a position \(s(t)\) using different formulas on different parts of its time domain. The interval notation matters: for example, \(0\leq t\leq 2\) includes \(t=2\), whereas \(2<t\leq 5\) does not. At a boundary \(t=c\), exactly one rule might specify \(s(c)\), or two adjoining rules might both include it. If both include it, they must assign the same value for the function to be well-defined.
The left-hand limit of position at \(c\), written \(\lim_{t\to c^-}s(t)\), is the value approached as time approaches \(c\) from below. The right-hand limit, \(\lim_{t\to c^+}s(t)\), is the value approached from above. Position is continuous at \(c\) precisely when both one-sided limits and the position value agree:
If the position has a finite jump, the model does not describe a continuous path through that boundary. Whether that jump is physically realistic depends on the context, but it is mathematically important: the ordinary two-sided derivative, and hence ordinary instantaneous velocity, cannot exist at a discontinuity.
Velocity at an Interior Boundary
At an interior time \(c\), instantaneous velocity is the derivative of position, defined by a two-sided limit. The left-hand and right-hand difference quotients test the behavior separately:
If both limits exist as finite numbers and are equal, their common value is \(s'(c)\), the instantaneous velocity at \(c\). If they differ, the two-sided derivative does not exist, even if the position itself is continuous. If position is discontinuous, the derivative cannot exist there.
On an interval where a position formula is differentiable, its derivative gives velocity at interior points of that interval. But at a boundary, the derivative must be checked using the boundary position and values from both sides. At an endpoint of the entire time domain, only one side may be available; a one-sided rate can describe the model’s behavior there, but it is not a two-sided derivative.
Worked Example: Continuous Position with Different Boundary Slopes
Worked Example: Continuous Position with Different Boundary Slopes
Suppose, for \(0\leq t\leq 5\), a particle’s position in meters is
First check the boundary position. The first formula gives \(s(2)=2^2=4\). The second formula approaches \(4+3(2-2)=4\) as \(t\to 2^+\). The left-hand limit is also \(2^2=4\). Thus position is continuous at \(t=2\).
For \(t<2\), differentiating the first formula gives \(s'(t)=2t\), so the left-hand velocity approaches \(2(2)=4\) meters per second. For \(t>2\), differentiating the second formula gives \(s'(t)=3\) meters per second, so the right-hand velocity is \(3\) meters per second.
Check directly with difference quotients. For \(h<0\), \(s(2+h)=(2+h)^2\), and therefore
For \(h>0\), \(s(2+h)=4+3h\), so
The one-sided velocities are \(4\) and \(3\) meters per second, so \(s'(2)\) does not exist. The model gives a continuous position but a corner at \(t=2\); it does not assign one instantaneous velocity at that time. As a check, the formulas agree in position at the boundary, while their derivatives on the two open intervals give the distinct one-sided rates just calculated.
A Boundary Test and Why It Works
When the position formulas extend smoothly to a boundary from each side, the boundary test can be summarized precisely. Continuity is needed to ensure that both difference quotients measure change from the same boundary position. Once continuity holds, the left and right slopes determine whether a single derivative exists.
This follows from the definition of derivative, not from a special rule for piecewise functions. For nonzero \(h\), the difference quotient uses values to the left of \(c\) when \(h<0\), and values to the right when \(h>0\). A two-sided limit exists precisely when the limits from those two directions both exist and equal the same finite value. That common limit is the derivative by definition. If they disagree, the quotient approaches different values from the two sides and has no two-sided limit.
Evaluate \(s(c)\) from the rule that includes \(c\), then compare it with the left- and right-hand limits.
Differentiate each formula on its interval to obtain the rates approaching \(c\), or compute the one-sided difference quotients directly.
If position is continuous and the one-sided rates agree, that common rate is the instantaneous velocity. Otherwise, the two-sided velocity is undefined at \(c\).
Worked Example: A Position Jump at the Boundary
Worked Example: A Position Jump at the Boundary
Consider the position model, in meters, for \(0\leq t\leq 4\):
The first rule does not include \(t=2\); the second does. Thus \(s(2)=6-2=4\). From the left, the position approaches \(2+1=3\). From the right, the second formula approaches \(6-2=4\). Therefore, the position jumps from a left-hand limiting value of \(3\) meters to a value of \(4\) meters at \(t=2\); it is not continuous there.
On the open interval \(0<t<2\), the velocity is the derivative of \(t+1\), which is \(1\) meter per second. On \(2<t<4\), the velocity is the derivative of \(6-t\), which is \(-1\) meter per second. These interval velocities do not give an instantaneous velocity at \(t=2\).
The right-hand difference quotient at the boundary is \(-1\), since for \(h>0\), \(s(2+h)=4-h\), and \((s(2+h)-s(2))/h=((4-h)-4)/h=-1\). For \(h<0\), \(s(2+h)=3+h\), so the quotient is \(((3+h)-4)/h=(-1+h)/h=-1/h+1\), which has no finite limit as \(h\to0^-\). Hence the two-sided derivative does not exist. The position jump itself already rules out continuity and differentiability; the quotient confirms this directly.
This example is a mathematical model with a discontinuous position. A physical particle moving continuously through space would not make this instantaneous jump, so the discontinuity may indicate that the model or its interpretation needs attention.
When Velocity Is Given Instead
A piecewise velocity model gives the rate of change of position on each interval. The value of velocity at the boundary is determined by the rule that includes that time, if one is specified. The left- and right-hand limits of velocity describe the rates approaching the boundary; they need not equal the assigned value at the boundary. If position is obtained by accumulating this velocity from an initial position, then a finite jump in velocity does not itself force a jump in position. It can, however, prevent position from having a derivative at that time.
Worked Example: Matching Velocities at a Boundary
Worked Example: Matching Velocities at a Boundary
Let time be measured in seconds, and suppose velocity in meters per second is
Assume \(s(0)=1\) meter and that position changes according to this velocity. On the first interval, accumulating velocity gives \(s(t)=1+\frac{t^2}{2}\), because its derivative is \(t\) and it has the required initial value. At \(t=2\), this position is \(1+\frac{2^2}{2}=3\) meters.
On the second interval, start from that boundary position and accumulate the second velocity:
At the boundary, the first position formula gives \(1+\frac{2^2}{2}=3\); the second gives \(-\frac{2^2}{2}+4(2)-3=-2+8-3=3\). Thus position is continuous. Differentiating the first position formula gives \(t\), and differentiating the second gives \(4-t\); these reproduce the stated velocities on their respective intervals. The one-sided velocity limits at \(t=2\) are \(2\) and \(4-2=2\), and the assigned value is \(v(2)=4-2=2\).
Verify the boundary derivative using difference quotients. For \(h<0\), \(s(2+h)=1+(2+h)^2/2=3+2h+h^2/2\), so \((s(2+h)-s(2))/h=2+h/2\), which tends to \(2\). For \(h>0\), \(s(2+h)=-(2+h)^2/2+4(2+h)-3=3+2h-h^2/2\), so the quotient is \(2-h/2\), which also tends to \(2\). Therefore \(s'(2)=2\) meters per second, in agreement with \(v(2)\). The position formulas, their derivatives, and both one-sided quotients check consistently.
A Common Boundary Mistake
A frequent error is to differentiate only the formula whose interval includes the boundary and treat that derivative as the instantaneous velocity there. For the first worked example, the rule for \(t>2\) is \(4+3(t-2)\). A tempting but incorrect calculation is “its derivative is \(3\), so \(s'(2)=3\).” This uses only the right-hand piece and ignores the behavior from the left.
The correction is to compare both sides using the boundary value \(s(2)=4\). The left difference quotient tends to \(4\), while the right difference quotient equals \(3\). Since \(4\ne3\), the two-sided derivative does not exist. The value \(3\) is the right-hand velocity, not the instantaneous velocity at the boundary.
Why Boundary Analysis Matters
Interval-by-interval analysis prevents a piecewise formula from hiding important changes in the motion. A continuous position with unequal one-sided velocities can have a sharp corner. A discontinuous position has a jump in the model. Matching one-sided velocities, together with continuity and the stated differentiability on each side, allows a single instantaneous velocity at the boundary. These are different situations, even if the formulas look similar.
At the beginning or end of the full time domain, do not invent a missing side. For example, at the initial time \(t=0\), a model defined only for \(t\geq0\) can have a right-hand velocity, but it has no left-hand behavior in its domain. State that one-sided interpretation when it is relevant, and reserve the ordinary two-sided derivative for interior times where both sides are defined.
Check Your Understanding
For each question, pay attention to which interval includes the boundary and distinguish position values from one-sided behavior.
- A position function has left- and right-hand limits of \(5\) at \(t=3\), but its assigned value is \(s(3)=6\). Is it continuous there? Can it have an ordinary derivative there?
- At an interior boundary, position is continuous and the one-sided velocities are \(2\) and \(2\). What is the instantaneous velocity?
- A continuous position model has one-sided velocities \(-1\) and \(4\) at a boundary. Does the two-sided instantaneous velocity exist?
- Why is differentiating only the piece that includes \(t=c\) insufficient to establish \(s'(c)\)?
- At the initial endpoint of a time domain, what kind of rate can be defined if the model is given only for later times?
