Work Along Graph Curves

Work Along a Graph Curve

A curve is sometimes described by a graph instead of by a ready-made parametrization. In the plane, it may be given as \(y=f(x)\). In space, its coordinates may be given as functions of one parameter. To calculate work, the essential step is to turn that description into a parametrization and include the curve’s tangent in the integral.

The direction of travel matters. Traversing the same curve in reverse changes the sign of the work, so the parameter interval or integration limits must reflect the stated direction.

Working assumptions: The curve is traversed over a finite closed interval and has a piecewise \(C^1\) parametrization: it is continuously differentiable on each of finitely many subintervals, with appropriate one-sided values at their endpoints. The force field is continuous along the curve. These assumptions ensure that the work integrand is piecewise continuous and its definite integral exists.

Planar Curves Given by \(y=f(x)\)

Let a planar curve be the graph \(y=f(x)\) for \(a\le x\le b\), with \(f\) piecewise \(C^1\). If the curve is traversed with \(x\) increasing, a parametrization is \(\mathbf r(x)=\langle x,f(x)\rangle\). Its tangent vector is \(\mathbf r'(x)=\langle 1,f'(x)\rangle\).

Write the force field as \(\mathbf F(x,y)=\langle P(x,y),Q(x,y)\rangle\). Work is the integral of the force dotted with the displacement along the curve. Substituting the parametrization and its derivative gives the graph formula.

W=\int_a^b \mathbf F(\mathbf r(x))\cdot\mathbf r'(x)\,dx =\int_a^b \left[P(x,f(x))+Q(x,f(x))f'(x)\right]\,dx.

The term involving \(f'(x)\) accounts for the vertical change \(dy=f'(x)\,dx\). Equivalently, the work integral can be written as \(\int P\,dx+Q\,dy\) along the oriented curve. If travel goes from \(x=b\) to \(x=a\), use the limits \(b\) to \(a\); do not silently keep the increasing-\(x\) orientation.

Worked Example: A Parabolic Arc Traversed from Left to Right

Find the work of \(\mathbf F(x,y)=\langle y,x\rangle\) along \(y=x^2\), from \(x=0\) to \(x=1\).

Parametrize by \(x\): \(\mathbf r(x)=\langle x,x^2\rangle\), for \(0\le x\le1\). Then \(\mathbf r'(x)=\langle1,2x\rangle\), and the force along the curve is \(\mathbf F(\mathbf r(x))=\langle x^2,x\rangle\). Thus

\mathbf F(\mathbf r(x))\cdot\mathbf r'(x) =\langle x^2,x\rangle\cdot\langle1,2x\rangle =x^2+2x^2=3x^2.

Integrate over the stated direction:

W=\int_0^1 3x^2\,dx =\left[x^3\right]_0^1 =1^3-0^3=1.

As a check, the graph formula gives \(P(x,x^2)=x^2\), \(Q(x,x^2)=x\), and \(f'(x)=2x\). Its integrand is \(x^2+x(2x)=3x^2\), the same integrand obtained from the dot product. The work is \(1\).

Space Curves Given by Component Functions

For a space curve with coordinates depending on \(t\), write \(\mathbf r(t)=\langle x(t),y(t),z(t)\rangle\), where the coordinate functions are piecewise \(C^1\) on the parameter interval \([\alpha,\beta]\). For a continuous force field \(\mathbf F=\langle P,Q,R\rangle\) along the curve, the work is

W=\int_\alpha^\beta \mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt =\int_\alpha^\beta \left[ P(\mathbf r(t))x'(t)+Q(\mathbf r(t))y'(t)+R(\mathbf r(t))z'(t) \right]dt.

Each coordinate contributes its own rate of change. For example, when a curve is given as \(y=f(x)\) and \(z=g(x)\), and is traversed with \(x\) increasing, use \(\mathbf r(x)=\langle x,f(x),g(x)\rangle\). The tangent is \(\langle1,f'(x),g'(x)\rangle\), so the work integrand includes the change in both \(y\) and \(z\).

W=\int_a^b \left[ P(x,f(x),g(x)) +Q(x,f(x),g(x))f'(x) +R(x,f(x),g(x))g'(x) \right]dx.

This formula applies when the curve is traversed in the direction of increasing \(x\). If \(x\) decreases along the stated route, reverse the bounds or use a parameter that increases in the direction of travel. In either case, the tangent must point in the direction in which the object moves.

Worked Example: A Space Curve with Three Changing Coordinates

Find the work of \(\mathbf F(x,y,z)=\langle y,z,x\rangle\) along \(\mathbf r(t)=\langle t,t^2,2t\rangle\), for \(0\le t\le1\).

The tangent is \(\mathbf r'(t)=\langle1,2t,2\rangle\). Substituting the curve into the force field gives \(\mathbf F(\mathbf r(t))=\langle t^2,2t,t\rangle\). Therefore,

\mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =\langle t^2,2t,t\rangle\cdot\langle1,2t,2\rangle =t^2+4t^2+2t =5t^2+2t.

Now integrate over the parameter interval:

W=\int_0^1(5t^2+2t)\,dt =\left[\frac{5}{3}t^3+t^2\right]_0^1 =\left(\frac{5}{3}+1\right)-0 =\frac{8}{3}.

As a check on the endpoint evaluation, at \(t=1\) the antiderivative is \(5/3+1=8/3\), and at \(t=0\) it is \(0\). Their difference is \(8/3\), as reported.

Why the Tangent Appears

The work formula follows from approximating a short piece of motion by its displacement. Over a small parameter change \(\Delta t\), the displacement is approximately \(\mathbf r'(t)\Delta t\). The force is approximately \(\mathbf F(\mathbf r(t))\), so the work on that short piece is approximately \(\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\Delta t\). Adding these contributions and taking the limit produces the integral. The following argument makes the parameter dependence explicit.

Start with a parametrization.
On each smooth part of the path, write its position as \(\mathbf r(t)\). Its velocity, or tangent with respect to \(t\), is \(\mathbf r'(t)\).

Compute work per unit parameter.
A short displacement is approximately \(\mathbf r'(t)\Delta t\). The corresponding work is approximately \(\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\Delta t\).

Accumulate along the parameter interval.
Under the stated piecewise \(C^1\) and continuity assumptions, the integrand is piecewise continuous. Its integral sums the limiting contributions over the full route.

There is also a direct check that changing a regular parameter does not alter the work if the direction is preserved. Suppose \(t=\phi(u)\) is a differentiable, increasing change of parameter. The chain rule gives \(\frac{d}{du}\mathbf r(\phi(u))=\mathbf r'(\phi(u))\phi'(u)\). Substitution into the integral yields

\int \mathbf F(\mathbf r(\phi(u)))\cdot \frac{d}{du}\mathbf r(\phi(u))\,du = \int \mathbf F(\mathbf r(\phi(u)))\cdot \mathbf r'(\phi(u))\phi'(u)\,du.

With \(t=\phi(u)\), so \(dt=\phi'(u)\,du\), this is exactly the original work integral in the variable \(t\). If the change of parameter is decreasing, it reverses the direction and therefore changes the sign instead.

Direction and a Common Error

The limits in a graph formula encode orientation. For a graph traversed from \(x=a\) to \(x=b\), the integral runs from \(a\) to \(b\). Traversing the same graph back from \(b\) to \(a\) changes the limits and negates the integral. This is a consequence of the signed orientation of a definite integral, not a change in the force field.

Worked Example: Traversing a Parabola from Right to Left

Find the work of \(\mathbf F(x,y)=\langle y,2x\rangle\) along \(y=1-x^2\), starting at \(x=1\) and ending at \(x=-1\).

Here \(f'(x)=-2x\). The graph formula has integrand

$$ P(x,f(x))+Q(x,f(x))f'(x) =(1-x^2)+(2x)(-2x) =1-5x^2. $$

The route is from \(x=1\) to \(x=-1\), so

W=\int_1^{-1}(1-5x^2)\,dx =\left[x-\frac{5}{3}x^3\right]_1^{-1} =\left(-1+\frac{5}{3}\right)-\left(1-\frac{5}{3}\right) =\frac{2}{3}-\left(-\frac{2}{3}\right) =\frac{4}{3}.

As a direction check, reversing the route gives \(\int_{-1}^{1}(1-5x^2)\,dx=-4/3\), the negative of the work in the stated direction.

A frequent mistake is to include the force component in the direction of graph height but omit the rate \(f'(x)\). In the first example, that mistake would use only \(P(x,x^2)=x^2\) and report

\int_0^1 x^2\,dx=\frac{1}{3}.

That is not the work: it ignores the vertical displacement. The correct integrand includes \(Q(x,f(x))f'(x)=x(2x)=2x^2\), giving \(\int_0^1(x^2+2x^2)\,dx=1\). Always compute the complete dot product of the force with the curve’s tangent.

Practical Checks

Before evaluating a work integral along a graph, confirm that the force is evaluated at points on the curve and that every coordinate derivative is included. Then check that the parameter interval follows the actual travel direction. These checks catch common setup errors before integration begins.

For a planar graph, use the tangent \(\langle1,f'(x)\rangle\) when \(x\) increases.
For a space graph, include the derivatives of every component in the tangent.
For travel in the opposite direction, reverse the limits or choose a parameter that runs with the motion.
Under piecewise \(C^1\) regularity and continuity of the force along the curve, the work integrand is piecewise continuous, so its definite integral exists.

Check Your Understanding

Use the force–tangent dot product and the stated travel direction in each problem.

For \(y=x^3\) from \(x=0\) to \(x=1\), what tangent vector should be used?
Write the work integrand for \(\mathbf F(x,y)=\langle x,y\rangle\) along \(y=2x\), traversed with increasing \(x\).
How should the limits change if that same graph is traversed from \(x=3\) to \(x=0\)?
For \(\mathbf r(t)=\langle t, t^2, \sin t\rangle\), what is \(\mathbf r'(t)\)?
Why is integrating only the horizontal force component generally insufficient for a non-horizontal graph?

What You'll Learn