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Work and Vector Line Integrals · Tutorial 754 of 999

Constant Force and Endpoint Displacement

Learn why constant-force work depends only on the initial and final positions, and verify the result on straight and curved paths.

Intermediate 9 min read

What You'll Learn

  • Define work for a constant force along a smooth parametrized path
  • Derive the endpoint-displacement formula from the vector line integral
  • Compute work along a straight path and check it using displacement
  • Verify the same formula on curved paths in two and three dimensions
  • Distinguish work from force magnitude multiplied by path length

Introduction

A force can act while an object follows a route that bends, loops, or changes direction. To calculate the work, we account for how much of the force points along the object's motion at each instant. For a constant force, this accumulation has a particularly simple result: the work is the force dotted with the net displacement.

This does not mean that the object must travel in a straight line. The route may be curved. The result says that, for a fixed constant force and fixed endpoints, all smooth routes give the same work. We will derive that fact from the vector line integral and verify it with explicit calculations.

Work Along a Smooth Path

Let an object's path be described by a continuously differentiable position function \(\mathbf{r}:[a,b]\to\mathbb{R}^n\). Its velocity with respect to the parameter \(t\) is \(\mathbf{r}'(t)\). Suppose a constant force \(\mathbf{F}\in\mathbb{R}^n\) acts at every point along the path. The work is the vector line integral of the force along that path.

Definition: For a continuously differentiable path \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) and a constant force \(\mathbf{F}\), the work is $$ W=\int_a^b \mathbf{F}\cdot\mathbf{r}'(t)\,dt. $$ The initial position is \(\mathbf{r}(a)\), the final position is \(\mathbf{r}(b)\), and the net displacement is \(\mathbf{r}(b)-\mathbf{r}(a)\).

The dot product \(\mathbf{F}\cdot\mathbf{r}'(t)\) measures the component of force in the direction of motion, multiplied by the speed associated with the parameter. If force is measured in newtons and position in metres, work is measured in newton-metres, or joules.

The Endpoint-Displacement Formula

Theorem: If \(\mathbf{F}\) is constant and \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) is continuously differentiable, then $$ W=\mathbf{F}\cdot\bigl(\mathbf{r}(b)-\mathbf{r}(a)\bigr). $$ Thus the work depends on the force and the two endpoints, not on the particular smooth path between them.

The condition that the force is constant matters: it allows \(\mathbf{F}\) to be taken outside the accumulation as a fixed vector. The condition on the path ensures that its derivative exists and is continuous, so the integral and the fundamental theorem of calculus apply.

Why the Formula Works

Write the force and path in components as \(\mathbf{F}=\langle F_1,\ldots,F_n\rangle\) and \(\mathbf{r}(t)=\langle x_1(t),\ldots,x_n(t)\rangle\). The derivative of the path is \(\langle x_1'(t),\ldots,x_n'(t)\rangle\). By the definition of the dot product, the work integral separates into a sum of ordinary integrals.

1
Expand the integrand.
Because the force is fixed, \(\mathbf{F}\cdot\mathbf{r}'(t)=\sum_{i=1}^n F_i x_i'(t)\).
2
Integrate each component.
Each \(F_i\) is a constant, so \(\int_a^b F_i x_i'(t)\,dt=F_i\int_a^b x_i'(t)\,dt\).
3
Apply the fundamental theorem of calculus.
For every component, \(\int_a^b x_i'(t)\,dt=x_i(b)-x_i(a)\).

Combining those component calculations gives the result, since the list of component changes is exactly the net displacement:

$$ \begin{aligned} W &=\int_a^b \sum_{i=1}^n F_i x_i'(t)\,dt\\ &=\sum_{i=1}^n F_i\int_a^b x_i'(t)\,dt\\ &=\sum_{i=1}^n F_i\bigl(x_i(b)-x_i(a)\bigr)\\ &=\mathbf{F}\cdot\bigl(\mathbf{r}(b)-\mathbf{r}(a)\bigr). \end{aligned} $$

This is a complete derivation: the last equality is the component definition of the dot product. Any bends in the path affect the instantaneous velocity, but the component integrals add up to the total change in position.

Worked Example: Work Along a Straight Path

Worked Example: Work Along a Straight Path

A constant force \(\mathbf{F}=\langle 4,-1\rangle\) newtons acts along the path \(\mathbf{r}(t)=\langle 1+2t,3-t\rangle\) metres, where \(0\le t\le2\). Find the work from the line integral, then check it using endpoint displacement.

First differentiate the path:

$$ \mathbf{r}'(t)=\langle 2,-1\rangle. $$

The force dotted with the velocity is constant:

$$ \mathbf{F}\cdot\mathbf{r}'(t) =\langle 4,-1\rangle\cdot\langle 2,-1\rangle =4(2)+(-1)(-1) =8+1 =9. $$

Therefore the work integral is

$$ W=\int_0^2 9\,dt =9(2-0) =18\text{ J}. $$

Now evaluate the endpoints:

$$ \mathbf{r}(0)=\langle 1,3\rangle,\qquad \mathbf{r}(2)=\langle 5,1\rangle. $$

The displacement is \(\langle 5-1,1-3\rangle=\langle 4,-2\rangle\) metres. Its dot product with the force is

$$ \mathbf{F}\cdot\bigl(\mathbf{r}(2)-\mathbf{r}(0)\bigr) =\langle 4,-1\rangle\cdot\langle 4,-2\rangle =4(4)+(-1)(-2) =16+2 =18\text{ J}. $$

The two calculations agree, verifying the work value.

Worked Example: A Curved Quarter-Circle Route

Worked Example: A Curved Quarter-Circle Route

Let a constant force \(\mathbf{F}=\langle 3,4\rangle\) newtons act along the quarter-circle path \(\mathbf{r}(t)=\langle 2\cos t,2\sin t\rangle\) metres, for \(0\le t\le\frac{\pi}{2}\). Calculate the work directly and by the endpoint formula.

Differentiate the path:

$$ \mathbf{r}'(t)=\langle -2\sin t,2\cos t\rangle. $$

Then form the dot product with the force:

$$ \mathbf{F}\cdot\mathbf{r}'(t) =3(-2\sin t)+4(2\cos t) =-6\sin t+8\cos t. $$

Integrate over the stated parameter interval:

$$ \begin{aligned} W &=\int_0^{\pi/2}(-6\sin t+8\cos t)\,dt\\ &=\bigl[6\cos t+8\sin t\bigr]_0^{\pi/2}\\ &=(6\cdot 0+8\cdot 1)-(6\cdot 1+8\cdot 0)\\ &=8-6\\ &=2\text{ J}. \end{aligned} $$

For the independent endpoint check, \(\mathbf{r}(0)=\langle 2,0\rangle\) and \(\mathbf{r}(\frac{\pi}{2})=\langle 0,2\rangle\). Thus the displacement is \(\langle -2,2\rangle\) metres, and

$$ \mathbf{F}\cdot\bigl(\mathbf{r}(\tfrac{\pi}{2})-\mathbf{r}(0)\bigr) =\langle 3,4\rangle\cdot\langle -2,2\rangle =3(-2)+4(2) =-6+8 =2\text{ J}. $$

The direct integral and the endpoint calculation agree, even though the route is curved.

Worked Example: A Curved Path in Three Dimensions

Worked Example: A Curved Path in Three Dimensions

Consider the constant force \(\mathbf{F}=\langle 1,-2,3\rangle\) newtons and the path \(\mathbf{r}(t)=\langle t,t(1-t),2t\rangle\) metres for \(0\le t\le1\). This path curves in its second coordinate. Compute its work and check the result from the endpoints.

Differentiate each coordinate:

$$ \mathbf{r}'(t)=\langle 1,1-2t,2\rangle. $$

The force–velocity dot product is

$$ \begin{aligned} \mathbf{F}\cdot\mathbf{r}'(t) &=1(1)+(-2)(1-2t)+3(2)\\ &=1-2+4t+6\\ &=5+4t. \end{aligned} $$

Therefore,

$$ W=\int_0^1(5+4t)\,dt =\bigl[5t+2t^2\bigr]_0^1 =(5+2)-0 =7\text{ J}. $$

The endpoints are \(\mathbf{r}(0)=\langle 0,0,0\rangle\) and \(\mathbf{r}(1)=\langle 1,0,2\rangle\), so the displacement is \(\langle 1,0,2\rangle\) metres. Substitution into the endpoint formula gives

$$ \mathbf{F}\cdot\bigl(\mathbf{r}(1)-\mathbf{r}(0)\bigr) =\langle 1,-2,3\rangle\cdot\langle 1,0,2\rangle =1(1)+(-2)(0)+3(2) =1+0+6 =7\text{ J}. $$

This matches the integral, including the contribution from all three force and displacement components.

A Common Mistake: Using Path Length Instead

A frequent error is to multiply the magnitude of the force by the total distance travelled and call the result the work. That calculation ignores the direction of motion. Work accumulates the force component along the motion, not the full force magnitude at every point.

For the quarter-circle example, the force magnitude is \(\sqrt{3^2+4^2}=5\) newtons. The path has radius \(2\) metres and angle \(\frac{\pi}{2}\), so its length is \(2\cdot\frac{\pi}{2}=\pi\) metres. The incorrect calculation would be

$$ \text{incorrect work}=|\mathbf{F}|\,(\text{path length})=5\pi\text{ J}, $$

which is not the actual work, \(2\) joules. The correction is to use the dot product with the displacement, or equivalently to integrate the tangential component of force along the path. In general, force magnitude times distance is valid only when the force points along the motion throughout the route.

Why Endpoint Displacement Is Useful

For a constant force, the theorem can save substantial work: a complicated smooth route need not be parameterized or integrated if its endpoints are known. It also gives a clear geometric interpretation. A force component perpendicular to the net displacement contributes zero to the final dot product, while a component aligned with the displacement contributes positively and one opposed to it contributes negatively.

The path still describes how the object moves, and it is needed when calculating the line integral directly. The endpoint formula is a special simplification justified by the force being constant; it should not be assumed for a force that changes with position. For the situation treated here, the derivation proves that the integral along any continuously differentiable route equals the same force–displacement dot product.

Check Your Understanding

Use the endpoint formula where it applies, and distinguish displacement from distance travelled.

  1. A constant force \(\langle 2,5\rangle\) newtons acts while an object moves from \(\langle 1,0\rangle\) to \(\langle 4,-2\rangle\) metres. What is the work?
  2. In the theorem's proof, where is the assumption that the force is constant used?
  3. Why does the curved quarter-circle example have work \(2\) joules rather than \(5\pi\) joules?
  4. If two continuously differentiable routes have the same initial and final points and the same constant force acts along both, what can be concluded about their work?
  5. What regularity condition on the parametrized path was assumed in the theorem?

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