Work Across Piecewise-Defined Force Fields

Work When the Force Changes by Region

A force field may be defined by different formulas in different parts of space. A particle following a path through those regions experiences one force formula, then another. To calculate the total work, keep track of where along the path each formula applies.

The key step is to find the parameter values at which the path meets a boundary between regions. These values divide the path’s parameter interval into pieces. On each piece, use the force formula for the region containing that part of the path, and add the resulting work contributions.

Definition: Suppose a path is parametrized by $\mathbf{r}(t)$, for $a\le t\le b$, and the force at a point $\mathbf{x}$ is $\mathbf{F}(\mathbf{x})$. The work along the path is $$ W=\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt, $$ provided the integrand is integrable. If the force has different formulas in different regions, split this integral at the parameter values where the path crosses their interfaces.

Here $\mathbf{r}'(t)$ is the path’s velocity vector with respect to its parameter. The dot product measures the component of force in the direction of motion. A positive contribution adds work, while a negative contribution means the force acts partly against the motion.

How to Find the Right Intervals

Region boundaries are usually described by equations such as $x=1$ or $y=x$. Substitute the path coordinates into the boundary equation and solve for $t$. Sort the resulting parameter values, then test the path on each open interval to determine which region it occupies. The order of the regions along the path—not simply their labels—determines which formula to use.

Find the interfaces.
Write the equations defining the boundaries between force regions.

Find the crossing parameters.
Substitute $\mathbf{r}(t)$ into each boundary equation and solve within the path’s parameter interval.

Identify each region.
Use a test parameter value in each resulting interval to select the applicable force formula.

Integrate and add.
Evaluate the work integral on each interval separately, then sum the contributions.

For this procedure, it is enough that the path is piecewise smooth and that the force composed with the path is integrable on each piece. In particular, if a path crosses an interface only at finitely many parameter values and the force is bounded and piecewise continuous, the force’s assigned value at those isolated crossing points does not change the integral. A path that lies along an interface for an interval is different: the force specified on that interface must be used there.

Why Splitting the Integral Works

The reason for splitting is the additivity of definite integrals over adjacent intervals. It lets us calculate the total work as a sum of regional contributions without changing the work being measured.

Theorem: Let $a=t_0<t_1<\cdots<t_n=b$ divide the path at every relevant interface crossing. If the applicable force formula on each open interval $(t_{i-1},t_i)$ makes the work integrand integrable there, then $$ W=\sum_{i=1}^{n}\int_{t_{i-1}}^{t_i} \mathbf{F}_i(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt, $$ where $\mathbf{F}_i$ is the force formula applicable on that interval.

To justify the result, write $g(t)=\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$ wherever the force is defined along the path. The work is the integral of $g$ from $a$ to $b$. Additivity gives $\int_a^b g(t)\,dt=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}g(t)\,dt$. On each interval, the path lies in a region with a specified force formula, so $g(t)$ there is computed using $\mathbf{F}_i$. Substitution gives the displayed sum. If the formulas disagree at an isolated crossing, changing an integrand at finitely many points does not change its integral. This proves the result under the stated integrability conditions.

Worked Examples

Worked Example: Crossing a Vertical Interface

A path is $\mathbf{r}(t)=(t,2t)$, $0\le t\le2$. The force is $\mathbf{F}_1(x,y)=(y,x)$ when $x<1$, and $\mathbf{F}_2(x,y)=(2x-y,x+y)$ when $x\ge1$. Find the work.

The interface is $x=1$. Along the path, $x=t$, so $t=1$ is the crossing parameter. The path uses $\mathbf{F}_1$ for $0\le t<1$ and $\mathbf{F}_2$ for $1\le t\le2$. Its velocity is $\mathbf{r}'(t)=(1,2)$.

On the first interval, $\mathbf{F}_1(\mathbf{r}(t))=(2t,t)$. Thus the dot product is $ (2t,t)\cdot(1,2)=2t+2t=4t$, and the work there is $\int_0^1 4t\,dt=[2t^2]_0^1=2$.

On the second interval, $\mathbf{F}_2(\mathbf{r}(t))=(2t-2t,t+2t)=(0,3t)$. The dot product is $(0,3t)\cdot(1,2)=6t$, so the work there is $\int_1^2 6t\,dt=[3t^2]_1^2=12-3=9$.

The total work is $2+9=11$ in force-times-distance units. As a check, the two interval contributions add to the work integral over the full path, and the field used in each contribution matches the region containing that segment.

Worked Example: Passing Through Three Regions

Let $\mathbf{r}(t)=(t,0)$, $-1\le t\le2$. Define the force by $\mathbf{F}(x,y)=(x,1)$ for $x<0$, $\mathbf{F}(x,y)=(2,x)$ for $0\le x<1$, and $\mathbf{F}(x,y)=(x,-1)$ for $x\ge1$. Calculate the work.

The path has $x=t$, so it crosses the interfaces $x=0$ and $x=1$ at $t=0$ and $t=1$. Also, $\mathbf{r}'(t)=(1,0)$. Since the path is horizontal, the dot product with the velocity selects the first component of the force.

For $-1\le t<0$, the first component is $t$, giving $\int_{-1}^{0}t\,dt=[t^2/2]_{-1}^{0}=0-\tfrac12=-\tfrac12$.

For $0\le t<1$, the first component is $2$, giving $\int_0^1 2\,dt=[2t]_0^1=2$.

For $1\le t\le2$, the first component is $t$, giving $\int_1^2 t\,dt=[t^2/2]_1^2=2-\tfrac12=\tfrac32$.

Therefore, $-\tfrac12+2+\tfrac32=3$. The negative contribution in the first region is retained; dropping it would incorrectly increase the total. The sum $3$ is the work across all three intervals.

Worked Example: A Curved Path Crossing a Diagonal Boundary

Let $\mathbf{r}(t)=(t,t^2)$, $-1\le t\le1$. In the region $y>x$, let $\mathbf{F}_A(x,y)=(y,x)$. In the region $y\le x$, let $\mathbf{F}_B(x,y)=(x,y)$. Find the work.

The interface is $y=x$. Substituting the path gives $t^2=t$, or $t(t-1)=0$, so the crossings are $t=0$ and $t=1$. To identify the regions, compare $t^2$ and $t$: for $-1\le t<0$, $t^2>t$, while for $0<t<1$, $t^2<t$. Thus use $\mathbf{F}_A$ on $[-1,0)$ and $\mathbf{F}_B$ on $(0,1]$. The velocity is $(1,2t)$.

On the first interval, $\mathbf{F}_A(\mathbf{r}(t))=(t^2,t)$. Its dot product with the velocity is $t^2+2t^2=3t^2$. The work is $\int_{-1}^0 3t^2\,dt=[t^3]_{-1}^0=0-(-1)=1$.

On the second interval, $\mathbf{F}_B(\mathbf{r}(t))=(t,t^2)$. Its dot product with the velocity is $t+2t^3$. The work is $\int_0^1(t+2t^3)\,dt=[t^2/2+t^4/2]_0^1=\tfrac12+\tfrac12=1$.

The total is $1+1=2$. The path meets the interface at $t=0$ and $t=1$; these isolated values do not contribute separate work terms. The two integrals account for the complete path.

Boundary Values and Paths Along Interfaces

At a single crossing, it is possible for two piecewise formulas to assign different force values to the same boundary point. If there are only finitely many such parameter values and the integral is otherwise well-defined, the assignment at those points does not affect the work. What matters is which formula applies on the intervals on either side.

Do not apply that observation to a path that follows the interface over a whole interval. In that case, the path spends a nonzero amount of parameter time on the boundary, and the force there affects the integral. Use a formula explicitly specified for the boundary. If the force has not been defined there, the work along that portion has not been specified by the given data.

A Common Mistake

In the first example, one might substitute the whole path into $\mathbf{F}_1$, obtaining an integrand $4t$, and then calculate $\int_0^2 4t\,dt=[2t^2]_0^2=8$. That calculation is wrong because $\mathbf{F}_1$ applies only while $x<1$, or $t<1$. After the path crosses $x=1$, the force formula changes.

Correction: Find the crossing parameter before integrating. Use $\mathbf{F}_1$ on $0\le t<1$ and $\mathbf{F}_2$ on $1\le t\le2$; their contributions are $2$ and $9$, respectively, so the correct total is $11$.

Why This Method Matters

Piecewise force models describe situations where different laws apply in different locations. The calculation is not a new kind of work integral: it is the ordinary vector line integral, with careful attention to which force acts where. Correctly locating each crossing prevents a formula from being used outside its region and makes the contribution from each part of the route visible.

Before evaluating, check three things: the crossing parameters are in the correct order, each interval is paired with the right force formula, and every interval’s dot product uses the path velocity. Once those are secure, integrate each piece and add the signed results.

Check Your Understanding

For each question, identify the interval split or work expression that is appropriate before carrying out any calculation.

A path $\mathbf{r}(t)=(2t,1)$, $0\le t\le2$, crosses the interface $x=3$. At what parameter value does the crossing occur?
Why must a path crossing two interfaces usually be split into three parameter intervals?
If one regional contribution to work is negative, should it be replaced by its absolute value before adding the other contributions? Explain.
Why does changing a force value at one isolated crossing parameter not affect an otherwise integrable work integral?
How does the calculation change if a path follows an interface for an entire interval?