Comparing Work on Two Specified Routes

Why Compare Work Along Different Routes?

When a force acts on an object moving between two points, the work done is determined by both the force field and the motion along the route. Two routes can have the same starting and ending points and still produce different work. To compare them, compute the work along each specified route separately, keeping track of its direction and parametrization.

This procedure is useful even before asking broader questions about whether a field has the same work along every possible route. Here the task is narrower: the routes are given, so evaluate each one and compare the results. A difference for these routes shows that the work values for this particular pair are different; equal values show only that this pair gives equal work.

Definition: Let $\mathbf{F}$ be a continuous force field and let $\mathbf{r}(t)$, $a\leq t\leq b$, be a continuously differentiable parametrization of an oriented path. The work along that path is $$ W=\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$ For a path made of finitely many smooth pieces, calculate the work on each piece and add the results.

The vector $\mathbf{r}'(t)$ gives the direction and rate of motion along the path. The dot product with the force extracts the component of the force in the direction of motion. Reversing the direction of travel reverses the sign of the work, so both routes must be oriented from the stated starting point to the stated ending point for a fair comparison.

How to Set Up a Route Comparison

For each route, confirm its endpoints, find its velocity by differentiating its parametrization, evaluate the force along the route, and then integrate the dot product. For a piecewise route, apply the same calculation on each segment and add the segment values.

Check the endpoints and orientation.
Verify that both parametrizations start and end at the specified points, in the stated direction.

Compute each velocity.
Differentiate the route parametrization to obtain $\mathbf{r}'(t)$.

Evaluate the force and dot product.
Substitute the route into $\mathbf{F}$, then calculate $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$.

Integrate and compare.
Find each work value and subtract or state their difference clearly.

Why the Parametric Formula Gives Work

Over a short parameter interval from $t$ to $t+\Delta t$, the displacement is approximately $\mathbf{r}'(t)\Delta t$. The force at the point is approximately $\mathbf{F}(\mathbf{r}(t))$. The work over that small displacement is therefore approximately the force dotted with the displacement:

\mathbf{F}(\mathbf{r}(t))\cdot\bigl(\mathbf{r}'(t)\Delta t\bigr) = \bigl(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\bigr)\Delta t.

Adding these approximations over a partition of the parameter interval gives a Riemann sum for the integral of the force–velocity dot product. As the partition is refined, continuity of the force along the smooth path ensures that the sum converges to the work integral:

W=\int_a^b\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.

This explains why the derivative of the parametrization is essential: it describes the small displacement per unit change in parameter. The same formula applies to each smooth segment of a piecewise route, and the total work is the sum because the total sequence of displacements is divided into those segments.

Worked Examples

Worked Example: A Diagonal Route Compared with Two Segments

Let $\mathbf{F}(x,y)=(-y,x)$. Compare the work from $(0,0)$ to $(1,1)$ along the diagonal with the work along the route that first moves horizontally to $(1,0)$, then vertically to $(1,1)$.

Diagonal route. Use $\mathbf{r}_1(t)=(t,t)$, $0\leq t\leq1$. Its endpoints are $\mathbf{r}_1(0)=(0,0)$ and $\mathbf{r}_1(1)=(1,1)$, and its velocity is $\mathbf{r}_1'(t)=(1,1)$. Along the route, $\mathbf{F}(\mathbf{r}_1(t))=(-t,t)$. Thus

W_1=\int_0^1(-t,t)\cdot(1,1)\,dt =\int_0^1(-t+t)\,dt =\int_0^1 0\,dt =0.

Two-segment route. On the horizontal segment, use $\mathbf{r}_2(t)=(t,0)$, $0\leq t\leq1$. Its force is $(0,t)$ and its velocity is $(1,0)$, so the work is

W_{2,\mathrm{horizontal}} =\int_0^1(0,t)\cdot(1,0)\,dt =\int_0^1 0\,dt =0.

On the vertical segment, use $\mathbf{r}_3(t)=(1,t)$, $0\leq t\leq1$. Its force is $(-t,1)$ and its velocity is $(0,1)$. Therefore

W_{2,\mathrm{vertical}} =\int_0^1(-t,1)\cdot(0,1)\,dt =\int_0^1 1\,dt =1.

The work on the two-segment route is $W_2=0+1=1$. Both routes have the required endpoints, but their work values are $0$ and $1$, respectively. The calculation on each route is verified by evaluating its force–velocity dot product before integration.

Worked Example: Comparing a Diagonal with an L-Shaped Route

Let $\mathbf{F}(x,y)=(xy,x^2)$. Compare the work from $(0,0)$ to $(1,1)$ along the diagonal and along the route that travels first along the $x$-axis and then vertically.

For the diagonal, take $\mathbf{r}_1(t)=(t,t)$, $0\leq t\leq1$. Its velocity is $(1,1)$, and the force along it is $(t^2,t^2)$. The work is

W_1=\int_0^1(t^2,t^2)\cdot(1,1)\,dt =\int_0^1 2t^2\,dt =\left[\frac{2t^3}{3}\right]_0^1 =\frac{2}{3}.

For the first segment of the L-shaped route, set $\mathbf{r}_2(t)=(t,0)$. The velocity is $(1,0)$, and the force is $(0,t^2)$. Hence

W_{2,\mathrm{horizontal}} =\int_0^1(0,t^2)\cdot(1,0)\,dt =\int_0^1 0\,dt =0.

For the second segment, set $\mathbf{r}_3(t)=(1,t)$. The velocity is $(0,1)$, and the force is $(t,1)$. Thus

W_{2,\mathrm{vertical}} =\int_0^1(t,1)\cdot(0,1)\,dt =\int_0^1 1\,dt =1.

The L-shaped route has total work $W_2=0+1=1$, while the diagonal has work $W_1=\frac{2}{3}$. Their difference, taking L-shaped work minus diagonal work, is

W_2-W_1=1-\frac{2}{3}=\frac{1}{3}.

Both parametrizations begin at $(0,0)$ and end at $(1,1)$. Adding the two segment integrals gives the full work on the L-shaped route, as required.

Worked Example: A Curved Route Compared with a Chord

Let $\mathbf{F}(x,y)=(y,0)$, and compare the quarter-circle arc from $(1,0)$ to $(0,1)$ with the straight chord joining the same points.

Parametrize the arc by $\mathbf{r}_1(t)=(\cos t,\sin t)$, $0\leq t\leq\frac{\pi}{2}$. It has the stated endpoints, velocity $\mathbf{r}_1'(t)=(-\sin t,\cos t)$, and force $\mathbf{F}(\mathbf{r}_1(t))=(\sin t,0)$. Therefore

W_1 =\int_0^{\pi/2}(\sin t,0)\cdot(-\sin t,\cos t)\,dt =-\int_0^{\pi/2}\sin^2t\,dt =-\frac{\pi}{4}.

For the chord, use $\mathbf{r}_2(t)=(1-t,t)$, $0\leq t\leq1$. It also starts at $(1,0)$ and ends at $(0,1)$. Its velocity is $(-1,1)$, and the force along it is $(t,0)$. Consequently,

W_2 =\int_0^1(t,0)\cdot(-1,1)\,dt =\int_0^1(-t)\,dt =-\left[\frac{t^2}{2}\right]_0^1 =-\frac{1}{2}.

Thus the arc work minus the chord work is $-\frac{\pi}{4}-(-\frac{1}{2})=\frac{1}{2}-\frac{\pi}{4}$. The routes share endpoints and orientation, but their computed work values differ. The trigonometric integral is exact; no rounding is needed.

A Common Error: Using Position Instead of Velocity

In the diagonal calculation for $\mathbf{F}(x,y)=(xy,x^2)$, the parametrization is $\mathbf{r}(t)=(t,t)$. A common mistake is to dot the force with the position $(t,t)$, rather than with the velocity $\mathbf{r}'(t)=(1,1)$. That incorrect calculation gives

(t^2,t^2)\cdot(t,t)=2t^3, \qquad \int_0^1 2t^3\,dt=\frac{1}{2}.

This is not the work integral for the given parametrization. The displacement rate is $(1,1)$, not the position $(t,t)$. The correct integrand and result are

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(t^2,t^2)\cdot(1,1) =2t^2, \qquad \int_0^1 2t^2\,dt=\frac{2}{3}.

Always differentiate the parametrization before forming the dot product. Also check that each route is traversed in the intended direction: swapping its endpoints reverses the sign of its work.

Interpreting the Comparison

A route comparison answers a specific question about two specified motions. If the computed values differ, the work along those two routes is different. If they agree, the result establishes equality only for the pair examined. Neither outcome, by itself, compares the work along every possible route between the endpoints.

When reporting a comparison, state the route orientation, each work value, and the convention used for the difference. For example, “route B minus route A” removes ambiguity about the sign. These habits make the calculation easy to check and prevent a correct integral from being attached to the wrong direction of travel.

Check Your Understanding

For each question, make sure the route’s velocity is included in the work integrand.

What vector is dotted with the force field in the parametrized work formula, and what does it represent?
For a route consisting of three smooth segments, how is its total work calculated?
Why do two routes with the same endpoints need not have the same work?
For $\mathbf{F}(x,y)=(y,0)$ and $\mathbf{r}(t)=(t,t^2)$, $0\leq t\leq1$, write the work integrand, without evaluating the integral.
If a route is traversed in reverse, what happens to the sign of its work?