From Force Sums to the Vector Line Integral

From Small Displacements to Work

When a force is constant and an object moves through a displacement, work is the dot product of the force and the displacement. Along a curved path, however, the force may change from point to point, and the direction of motion changes as well. The natural approach is to divide the path into small pieces and add the work done over each piece.

For a short displacement $\Delta\mathbf{r}$ near a point where the force is $\mathbf{f}$, the corresponding work is approximated by $\mathbf{f}\cdot\Delta\mathbf{r}$. The dot product accounts for direction: the part of the force perpendicular to the displacement contributes no work. Refining the pieces of the path leads to the vector line integral.

Definition: Let $\mathbf{r}:[a,b]\to\mathbb{R}^n$ be a continuously differentiable curve, and let $\mathbf{f}$ be a continuous force field on a region containing the curve. The work of $\mathbf{f}$ along the parametrized curve is $$ W=\int_a^b \mathbf{f}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$ This is also written $\int_C \mathbf{f}\cdot d\mathbf{r}$, where $C$ denotes the curve with the direction specified by $\mathbf{r}$.

The definition uses only continuity of the force field along the curve and continuous differentiability of the curve. It does not require the velocity $\mathbf{r}'(t)$ to be nonzero at every parameter value. If the curve is stationary over an interval, then $\mathbf{r}'(t)=\mathbf{0}$ there, so that interval contributes zero work.

The Discrete Force–Displacement Sum

Choose a partition $a=t_0<t_1<\cdots<t_m=b$. On the $i$-th subinterval, the curve moves from $\mathbf{r}(t_{i-1})$ to $\mathbf{r}(t_i)$. Its displacement is the vector difference between these endpoints. Choose a sample parameter $\tau_i\in[t_{i-1},t_i]$, and approximate the force over that small piece by its value at $\mathbf{r}(\tau_i)$. The resulting total approximate work is

W_P=\sum_{i=1}^m \mathbf{f}(\mathbf{r}(\tau_i))\cdot \bigl(\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\bigr).

Each summand has a direct interpretation: sampled force dotted with the actual endpoint-to-endpoint displacement for that piece. The piece of the path may be curved, but when the pieces are small, the chord between its endpoints approximates its local motion. The dot product adds the signed contribution in the direction of displacement.

To make the limiting argument precise, write $\Delta t_i=t_i-t_{i-1}$, and let the mesh of the partition be the largest of these subinterval lengths. Since $\mathbf{r}$ is continuously differentiable, the fundamental theorem of calculus gives

\mathbf{r}(t_i)-\mathbf{r}(t_{i-1}) =\int_{t_{i-1}}^{t_i}\mathbf{r}'(t)\,dt.

Thus the discrete sum is close to a Riemann sum for the function $g(t)=\mathbf{f}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$. The next derivation shows why its limit is exactly the integral in the definition.

Why the Limit Gives the Work Integral

Derivation: For a partition with sample points $\tau_i$, the $i$-th term of the force–displacement sum can be rewritten as $$ \mathbf{f}(\mathbf{r}(\tau_i))\cdot \bigl(\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\bigr) = \int_{t_{i-1}}^{t_i} \mathbf{f}(\mathbf{r}(\tau_i))\cdot\mathbf{r}'(t)\,dt. $$

The corresponding Riemann-sum term for $g$ is $g(\tau_i)\Delta t_i$. The difference between these two quantities is

\int_{t_{i-1}}^{t_i} \mathbf{f}(\mathbf{r}(\tau_i))\cdot \bigl(\mathbf{r}'(t)-\mathbf{r}'(\tau_i)\bigr)\,dt.

The curve and force field are continuous on the compact parameter interval, so $\mathbf{f}(\mathbf{r}(t))$ is bounded there; call a bound for its magnitude $M$. The derivative $\mathbf{r}'$ is uniformly continuous. If $\omega(\delta)$ is the largest change in $\mathbf{r}'$ between parameter values at most $\delta$ apart, then $\omega(\delta)\to0$ as $\delta\to0$. By the dot-product inequality, the magnitude of the difference above is at most $M\omega(\delta)\Delta t_i$ when the partition mesh is at most $\delta$. Summing over all subintervals bounds the total difference by

M\omega(\delta)\sum_{i=1}^m\Delta t_i =M\omega(\delta)(b-a),

which tends to zero as the mesh tends to zero. Meanwhile, $g$ is continuous, so its Riemann sums converge to $\int_a^b g(t)\,dt$. Therefore the force–displacement sums converge to that same integral:

\lim_{\lVert P\rVert\to0}W_P = \int_a^b\mathbf{f}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.

This explains both the formula and its assumptions. Continuity ensures that the force does not jump unpredictably along the curve, and continuous differentiability ensures that the curve has a well-defined, continuously varying velocity for the local displacement approximation.

Worked Examples

Worked Example: A Field Along a Parabolic Path

Let coordinates be measured in meters and force in newtons. Consider the force field $\mathbf{f}(x,y)=(x,y)$ and the curve $\mathbf{r}(t)=(t,t^2)$, for $0\le t\le1$. Find the work.

First evaluate the force along the curve:

\mathbf{f}(\mathbf{r}(t)) =\mathbf{f}(t,t^2) =(t,t^2).

Differentiate the curve:

\mathbf{r}'(t)=(1,2t).

The dot product is

\mathbf{f}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(t,t^2)\cdot(1,2t) =t+2t^3.

Integrate over the parameter interval:

W=\int_0^1(t+2t^3)\,dt =\left[\frac{t^2}{2}+\frac{t^4}{2}\right]_0^1 =\left(\frac12+\frac12\right)-0 =1\text{ J}.

Check: differentiating the antiderivative gives $t+2t^3$, the required integrand. Substituting the endpoints gives $1$ joule.

Worked Example: A Different Force Field and Curve

Let $\mathbf{f}(x,y)=(y,2x)$, and let $\mathbf{r}(t)=(t,t^3)$ for $0\le t\le1$. Take the coordinates in meters and the force values in newtons.

Substitute the curve into the force field, and differentiate the curve:

\mathbf{f}(\mathbf{r}(t)) =\mathbf{f}(t,t^3) =(t^3,2t), \qquad \mathbf{r}'(t)=(1,3t^2).

Then the work integrand is

\mathbf{f}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(t^3,2t)\cdot(1,3t^2) =t^3+6t^3 =7t^3.

Consequently,

W=\int_0^1 7t^3\,dt =\left[\frac{7t^4}{4}\right]_0^1 =\frac74\text{ J}.

Check: the derivative of $7t^4/4$ is $7t^3$, and evaluating at $1$ and $0$ gives $7/4$ joules.

Worked Example: Recovering the Integral from Finite Sums

Let $\mathbf{f}(x,y)=(0,x)$ and $\mathbf{r}(t)=(t,t^2)$ for $0\le t\le1$. Use a uniform partition of $n$ subintervals and sample the force at each left endpoint. Coordinates are in meters and force in newtons.

On the $k$-th subinterval, where $k=0,\ldots,n-1$, the left endpoint parameter is $k/n$. The force there is $(0,k/n)$, and the displacement to the next curve point is

\mathbf{r}\left(\frac{k+1}{n}\right)-\mathbf{r}\left(\frac{k}{n}\right) = \left(\frac1n,\frac{(k+1)^2-k^2}{n^2}\right) = \left(\frac1n,\frac{2k+1}{n^2}\right).

The work sum is therefore

W_n=\sum_{k=0}^{n-1} (0,k/n)\cdot\left(\frac1n,\frac{2k+1}{n^2}\right) =\frac{1}{n^3}\sum_{k=0}^{n-1}(2k^2+k).

Using $\sum_{k=0}^{n-1}k=n(n-1)/2$ and $\sum_{k=0}^{n-1}k^2=n(n-1)(2n-1)/6$, we obtain

W_n =\frac{1}{n^3} \left( \frac{n(n-1)(2n-1)}{3} +\frac{n(n-1)}{2} \right) =\frac{(n-1)(4n+1)}{6n^2}.

Taking the limit gives $\lim_{n\to\infty}W_n=2/3$ joules. Now check against the integral formula: along the curve, $\mathbf{f}(\mathbf{r}(t))=(0,t)$ and $\mathbf{r}'(t)=(1,2t)$, so their dot product is $2t^2$. Hence

\int_0^1 2t^2\,dt =\left[\frac{2t^3}{3}\right]_0^1 =\frac23\text{ J}.

The finite-sum limit agrees with the integral. Differentiating $2t^3/3$ also recovers the integrand $2t^2$.

A Common Mistake: Using Force Magnitude

Work is not generally the integral of force magnitude times distance. That calculation discards the force direction. For the first example, at $t=1/2$, the force is $(1/2,1/4)$ and the curve velocity is $(1,1)$. The correct instantaneous contribution per unit parameter is

(1/2,1/4)\cdot(1,1)=\frac34.

By contrast, multiplying magnitudes gives

\lVert(1/2,1/4)\rVert\,\lVert(1,1)\rVert =\frac{\sqrt{5}}{4}\sqrt{2} =\frac{\sqrt{10}}{4},

which is not $3/4$. The magnitude product would be appropriate only if the force and motion were aligned in the relevant direction; the dot product handles general directions correctly, including negative work when force points partly against the motion.

Why the Vector Line Integral Is Useful

The discrete-sum viewpoint connects a continuous calculation to the physical meaning of work. Each small contribution measures how much of the force acts along the local displacement, and the integral accumulates these signed contributions over the entire path. The same construction applies in any dimension: the force and velocity are vectors in $\mathbb{R}^n$, and their dot product is a scalar.

For practical calculation, the essential sequence is to evaluate the force field on the given curve, find the curve’s derivative, take their dot product, and integrate over the parameter interval. Each step has a distinct role: the field gives the force at each location, the derivative gives the local direction and rate of motion, and the dot product produces the work density in the parameter.

Check Your Understanding

Use the definition and derivation above to answer these questions.

In a force–displacement sum, what physical quantities are used in each summand?
Why does the dot product, rather than the product of magnitudes, measure work?
What continuity and differentiability assumptions were used to justify the limit?
For $\mathbf{r}(t)=(t,t^2)$, $0\le t\le1$, and $\mathbf{f}(x,y)=(y,0)$, what is the work integral’s integrand?
Why does a portion of the curve with $\mathbf{r}'(t)=\mathbf{0}$ contribute no work at that parameter value?