Tutorials › Calculus III › Polygonal Routes Through Multiple Force Regions

Home  /  Tutorials  /  Work and Vector Line Integrals

Work and Vector Line Integrals · Tutorial 768 of 999

Polygonal Routes Through Multiple Force Regions

Learn a reliable segment-by-segment method for finding total work when a polygonal route crosses regions with different force formulas.

Intermediate 10 min read

What You'll Learn

  • Parametrize a directed line segment between two vertices
  • Find where a segment crosses a force-region boundary
  • Evaluate work separately on each interval with a fixed force formula
  • Keep track of how route direction affects the work contribution
  • Add segment contributions and check the result

Why Polygonal Routes Need Care

A polygonal route consists of directed straight segments joined at vertices. If the force is constant, work along a segment can be found from the force and the displacement. But when the force varies with position, or follows different formulas in different regions, each part of the route must be evaluated using the force that applies there.

The important details are the direction of travel, the points where the route crosses a region boundary, and the force formula on each resulting piece. A vertex is not necessarily the only place where a force formula changes: a single straight segment may pass through several regions.

Key idea: Parametrize each directed segment, divide it wherever the force formula changes, compute the work on every resulting piece, and add all contributions in route order.

Work Along a Directed Segment

Let a segment start at \(P\) and end at \(Q\), where \(P\) and \(Q\) are points in the plane or in space. Its standard parametrization is

$$ \mathbf{r}(t)=P+t(Q-P),\qquad 0\le t\le 1. $$

Here \(Q-P\) is the displacement vector, and \(\mathbf{r}'(t)=Q-P\). For a force field \(\mathbf{F}\), the work along this directed segment is the line integral

$$ W_{P\to Q}=\int_0^1 \mathbf{F}(\mathbf{r}(t))\cdot (Q-P)\,dt. $$

This formula uses the force evaluated at the current position and its component in the direction of motion. The dot product matters: force perpendicular to the motion contributes no work. Reversing the endpoints reverses the direction of travel and, for the same force along the reversed segment, changes the sign of the work.

Definition: For a polygonal route made of directed segments \(C_1,\ldots,C_n\), its total work is the sum of the work integrals on those segments. If a segment crosses a boundary between force regions, its integral is split at each crossing and evaluated using the applicable force formula on each interval.

Why Splitting at a Boundary Works

Suppose a parametrized segment uses one force formula for \(0\le t\le a\) and another for \(a\le t\le1\). The work integral is an ordinary definite integral of the corresponding piecewise-defined integrand. By additivity of definite integrals, its value is the sum of the integrals on those parameter intervals.

$$ \int_0^1 \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^a \mathbf{F}_1(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt + \int_a^1 \mathbf{F}_2(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$

The split point \(a\) is found by substituting the segment parametrization into the equation of the boundary. This is the reason the procedure is systematic: each integral uses a single force formula throughout its interval. A boundary point itself does not affect the value of the integral, since changing the integrand at one parameter value does not change a definite integral.

1
List the directed segments.
Write each segment from its starting vertex to its ending vertex, preserving the route order.
2
Parametrize each segment.
Use \(\mathbf{r}(t)=P+t(Q-P)\) for \(0\le t\le1\), and record the constant velocity \(Q-P\).
3
Locate formula changes.
Substitute \(\mathbf{r}(t)\) into the region-boundary equation and solve for the crossing parameter values.
4
Integrate and add.
On every interval use its applicable force formula, take the dot product with the segment velocity, integrate, and sum all pieces.

Worked Example: Crossing a Vertical Boundary

Worked Example: Crossing a Vertical Boundary

A particle travels from \((0,0)\) to \((2,0)\), then to \((2,1)\). In joules, with position measured in metres, the force field is \(\mathbf{F}(x,y)=(1,y)\) when \(x\le1\), and \(\mathbf{F}(x,y)=(x,1+y)\) when \(x>1\). Find the total work.

For the first segment, \(P=(0,0)\), \(Q=(2,0)\), so \(\mathbf{r}_1(t)=(2t,0)\), \(\mathbf{r}_1'(t)=(2,0)\). The boundary \(x=1\) is reached when \(2t=1\), or \(t=\tfrac12\).

For \(0\le t\le\tfrac12\), the force is \((1,0)\), so the integrand is \((1,0)\cdot(2,0)=2\). For \(\tfrac12<t\le1\), the force is \((2t,1)\), so the integrand is \((2t,1)\cdot(2,0)=4t\). Thus

$$ W_1=\int_0^{1/2}2\,dt+\int_{1/2}^1 4t\,dt =1+\left[2t^2\right]_{1/2}^1 =1+\left(2-\frac12\right) =\frac52\text{ J}. $$

For the second segment, \(P=(2,0)\), \(Q=(2,1)\), so \(\mathbf{r}_2(t)=(2,t)\), \(\mathbf{r}_2'(t)=(0,1)\). It lies entirely in \(x>1\). The force is \((2,1+t)\), and its dot product with the velocity is \(1+t\). Therefore

$$ W_2=\int_0^1(1+t)\,dt =\left[t+\frac{t^2}{2}\right]_0^1 =\frac32\text{ J}. $$

The total is \(W_1+W_2=\tfrac52+\tfrac32=4\text{ J}\). To check the substitutions, the first segment’s dot products are \(2\) and \(4t\), as obtained directly from the two force formulas and its velocity \((2,0)\); the second segment’s dot product is \(1+t\). The antiderivative differences give the stated contributions, which sum to \(4\text{ J}\).

Worked Example: A Segment Crossing a Horizontal Boundary

Worked Example: A Segment Crossing a Horizontal Boundary

A route goes from \((0,0)\) to \((2,2)\), then to \((3,2)\). Let \(\mathbf{F}(x,y)=(y,x)\) when \(y\le1\), and \(\mathbf{F}(x,y)=(2,x+1)\) when \(y>1\). Find the total work.

On the diagonal segment, \(\mathbf{r}_1(t)=(2t,2t)\) and \(\mathbf{r}_1'(t)=(2,2)\). It crosses \(y=1\) when \(2t=1\), so the split is at \(t=\tfrac12\). Below the boundary, the force is \((2t,2t)\) and the dot product is \(8t\). Above the boundary, the force is \((2,2t+1)\) and the dot product is \(4+4t\). Hence

$$ \begin{aligned} W_1 &=\int_0^{1/2}8t\,dt+\int_{1/2}^1(4+4t)\,dt\\ &=\left[4t^2\right]_0^{1/2}+\left[4t+2t^2\right]_{1/2}^1\\ &=1+\left(6-\frac52\right) =\frac92\text{ J}. \end{aligned} $$

On the horizontal segment, \(\mathbf{r}_2(t)=(2+t,2)\), \(\mathbf{r}_2'(t)=(1,0)\), for \(0\le t\le1\). This lies wholly above \(y=1\). The force is \((2,3+t)\), so the dot product is \(2\), giving

$$ W_2=\int_0^1 2\,dt=2\text{ J}. $$

The total work is \(\tfrac92+2=\tfrac{13}{2}\text{ J}\). For a check, the diagonal dot products \(8t\) and \(4+4t\) follow from the segment velocity \((2,2)\); the horizontal dot product is \(2\) because its velocity is \((1,0)\). Evaluating the displayed antiderivatives at their interval endpoints reproduces each contribution and their sum.

Worked Example: Direction and a Sloping Boundary

Worked Example: Direction and a Sloping Boundary

A route travels from \((0,0)\) to \((3,0)\), then to \((0,1)\). The force is \(\mathbf{F}(x,y)=(x,1)\) when \(x+y<2\), and \(\mathbf{F}(x,y)=(1,y)\) when \(x+y\ge2\). Find the total work.

On the first segment, \(\mathbf{r}_1(t)=(3t,0)\), with velocity \((3,0)\). It reaches \(x+y=2\) when \(3t=2\), so \(t=\tfrac23\). Before that point the force is \((3t,1)\), and the dot product is \(9t\). After it the force is \((1,0)\), and the dot product is \(3\). Thus

$$ W_1=\int_0^{2/3}9t\,dt+\int_{2/3}^1 3\,dt =\left[\frac92t^2\right]_0^{2/3}+\left[3t\right]_{2/3}^1 =2+1=3\text{ J}. $$

On the second segment, \(\mathbf{r}_2(t)=(3-3t,t)\), with velocity \((-3,1)\). Along it, \(x+y=3-2t\), which equals \(2\) at \(t=\tfrac12\). The first half is in the region \(x+y\ge2\): the force is \((1,t)\), and the dot product is \(-3+t\). The second half is in \(x+y<2\): the force is \((3-3t,1)\), and the dot product is \(-3(3-3t)+1=-8+9t\). Therefore

$$ \begin{aligned} W_2 &=\int_0^{1/2}(-3+t)\,dt+\int_{1/2}^1(-8+9t)\,dt\\ &=\left[-3t+\frac{t^2}{2}\right]_0^{1/2} +\left[-8t+\frac92t^2\right]_{1/2}^1\\ &=-\frac{11}{8}-\frac58=-2\text{ J}. \end{aligned} $$

The negative contribution on the second segment is consistent with the force having a component opposite the direction of travel. The route’s total work is \(3+(-2)=1\text{ J}\). As a check, the velocity \((-3,1)\) gives dot products \(-3+t\) and \(-8+9t\) on the two correct intervals; evaluating their antiderivatives yields \(-2\text{ J}\), and adding the first segment gives \(1\text{ J}\).

A Common Mistake: Using One Formula for a Whole Segment

A frequent error is to identify the force formula at the segment’s starting point and use it all the way to the endpoint. In the first worked example, the route starts the horizontal segment in \(x\le1\), but crosses into \(x>1\) halfway through. If we incorrectly use \((1,y)\) throughout that segment, then \(y=0\) and the computed work would be

$$ \int_0^1 (1,0)\cdot(2,0)\,dt=2\text{ J}. $$

This is wrong: after \(t=\tfrac12\), the force is \((x,1+y)\), not \((1,y)\). The correct horizontal-segment work is \(\tfrac52\text{ J}\), found by splitting at \(2t=1\). The correction is not to choose a more convenient single formula, but to locate every region crossing and integrate each part with its own force law.

Why the Method Matters

The total work is sensitive to both the path’s direction and the force encountered along it. Even when the route consists of simple straight pieces, a force boundary can make a single segment behave like several distinct integration intervals. Careful parametrization keeps these issues visible: the velocity records the direction, and the parameter values identify precisely where the force law changes.

Before integrating, check that each segment starts and ends at the intended vertices, that every boundary crossing has been included, and that the force formula matches the region on each interval. After integrating, add the contributions with their signs. Negative work should not be discarded: it records that the force opposes some part of the motion.

Check Your Understanding

Use directed segment parametrizations and split at every force-region boundary.

  1. For a segment from \(P\) to \(Q\), what are its standard parametrization and velocity?
  2. How do you find the parameter value where a segment crosses a boundary \(x+y=3\)?
  3. Why must a segment crossing a force boundary be split into separate integrals?
  4. What does a negative work contribution indicate about the force and direction of motion?
  5. For a route with three directed segments, how is the total work obtained from the segment contributions?

Previous

Next

Back to Tutorials