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Work and Vector Line Integrals · Tutorial 751 of 999

Parametrizing a Vector Line Integral

Learn to evaluate work along a parametrized curve by combining the force along the path with the curve’s velocity.

Intermediate 9 min read

What You'll Learn

  • Identify the path, its direction of travel, and its parameter interval.
  • Evaluate a vector field at the points specified by a curve parametrization.
  • Differentiate the parametrization to obtain the velocity vector.
  • Set up the vector line integral as a dot product integrated over the parameter.
  • Check a calculation by verifying the force, velocity, and antiderivative.
  • Recognize and correct the error of integrating the force without the velocity factor.

From a Path to an Integral

A force can change as an object moves, so finding the total work along a curved path requires accounting for the force at each position and for the direction in which the object is moving. A parametrization makes those positions and directions available as functions of a single variable.

Let a curve be described by \(\mathbf{r}(t)\) for \(a\leq t\leq b\). At parameter value \(t\), the object is at \(\mathbf{r}(t)\), and its velocity with respect to the parameter is \(\mathbf{r}'(t)\). If \(\mathbf{F}\) is the force field, then the force acting at that point is \(\mathbf{F}(\mathbf{r}(t))\). The work integral is formed by taking the dot product of these two vectors and integrating in \(t\).

Definition: Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be continuously differentiable, and let \(\mathbf{F}\) be a continuous vector field defined at every point of the curve. The vector line integral of \(\mathbf{F}\) along the parametrized curve is
$$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$
When \(\mathbf{F}\) represents force, this integral gives the work done by that force along the specified traversal.

The notation \(\int_C\) refers to the curve being traversed, while the right-hand side is an ordinary one-variable integral. The parameter interval and the parametrization specify how the curve is followed. In particular, \(\mathbf{r}'(t)\) is a vector, not merely the speed: it records the direction of motion as well as the rate of change of position.

Why the Dot Product Appears

The work done by a nearly constant force over a small displacement is approximated by the force dotted with that displacement. This is the local calculation behind the vector line integral. As the displacement intervals become small, the sum of these approximations approaches the integral in the definition.

1
Divide the parameter interval.
Choose points \(a=t_0<t_1<\cdots<t_m=b\). Over a short interval, the displacement is approximately \(\mathbf{r}'(\tau_i)\Delta t_i\), where \(\Delta t_i=t_i-t_{i-1}\) and \(\tau_i\) lies in that interval.
2
Approximate work on each piece.
The force at a representative point is approximately \(\mathbf{F}(\mathbf{r}(\tau_i))\). Its work over the small displacement is therefore approximately \(\mathbf{F}(\mathbf{r}(\tau_i))\cdot\mathbf{r}'(\tau_i)\Delta t_i\).
3
Take the limit of the sums.
Adding these terms gives Riemann sums for the continuous function \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\). As the largest interval width tends to zero, the sums converge to its integral over \([a,b]\).

Thus the parameter derivative appears because it converts a small change in parameter into a small displacement along the curve. The force is evaluated at the actual position, and the dot product measures the force’s contribution in the direction of that displacement.

A Reliable Setup

For a particular problem, keep the geometric path, its parametrization, and the force field distinct until each has been used in the right way. The substitution \(\mathbf{F}(\mathbf{r}(t))\) means that every position input of the field is replaced by the corresponding coordinate of the curve. Then differentiate the curve, take the dot product, and integrate across the full parameter interval.

1
Record the path and interval.
Write \(\mathbf{r}(t)\) and the values \(a\leq t\leq b\). Make sure they describe the requested traversal.
2
Evaluate the field on the curve.
Substitute the coordinates of \(\mathbf{r}(t)\) into \(\mathbf{F}\), obtaining \(\mathbf{F}(\mathbf{r}(t))\).
3
Differentiate the curve.
Compute \(\mathbf{r}'(t)\), the velocity vector for this parametrization.
4
Integrate the dot product.
Form \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\) and integrate it from \(a\) to \(b\).

Worked Examples

Worked Example: A Polynomial Curve in the Plane

Find the work integral for \(\mathbf{F}(x,y)=(y,x)\) along \(\mathbf{r}(t)=(t,t^2)\), \(0\leq t\leq1\).

First substitute the curve coordinates into the force field. Since \(x=t\) and \(y=t^2\),

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

Next differentiate the parametrization:

$$ \mathbf{r}'(t)=\frac{d}{dt}(t,t^2)=(1,2t). $$

The dot product is

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

Therefore, the work is

$$ \int_C\mathbf{F}\cdot d\mathbf{r} =\int_0^1 3t^2\,dt =\left[t^3\right]_0^1 =1-0 =1. $$

Check: differentiating the antiderivative \(t^3\) gives \(3t^2\), the integrand. The force on the curve and the velocity have also been evaluated separately before taking their dot product.

Worked Example: A Quarter-Circle Path

Evaluate the work of \(\mathbf{F}(x,y)=(x+y,x-y)\) along \(\mathbf{r}(t)=(2\cos t,2\sin t)\), \(0\leq t\leq\frac{\pi}{2}\).

The parametrization traces the quarter-circle from \((2,0)\) to \((0,2)\). Substitution into the field gives

$$ \mathbf{F}(\mathbf{r}(t)) =\mathbf{F}(2\cos t,2\sin t) =(2\cos t+2\sin t,\,2\cos t-2\sin t). $$

The velocity vector is

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

Taking the dot product and collecting terms gives

$$ \begin{aligned} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) &=(2\cos t+2\sin t)(-2\sin t) +(2\cos t-2\sin t)(2\cos t)\\ &=-4\sin t\cos t-4\sin^2t +4\cos^2t-4\sin t\cos t\\ &=4(\cos^2t-\sin^2t)-8\sin t\cos t\\ &=4\cos(2t)-4\sin(2t). \end{aligned} $$

Integrate over the stated interval:

$$ \begin{aligned} \int_C\mathbf{F}\cdot d\mathbf{r} &=\int_0^{\pi/2}\bigl(4\cos(2t)-4\sin(2t)\bigr)\,dt\\ &=\left[2\sin(2t)+2\cos(2t)\right]_0^{\pi/2}\\ &=(0-2)-(0+2)\\ &=-4. \end{aligned} $$

Check: differentiating \(2\sin(2t)+2\cos(2t)\) gives \(4\cos(2t)-4\sin(2t)\), as required. The negative result is consistent with the computed dot product: its accumulated value over the interval is negative.

Worked Example: A Curve in Three Dimensions

Evaluate the vector line integral for \(\mathbf{F}(x,y,z)=(z,x,y)\) along \(\mathbf{r}(t)=(t,t^2,t^3)\), \(0\leq t\leq1\).

Substituting \(x=t\), \(y=t^2\), and \(z=t^3\) into the field gives

$$ \mathbf{F}(\mathbf{r}(t)) =\mathbf{F}(t,t^2,t^3) =(t^3,t,t^2). $$

Differentiate each coordinate of the curve:

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

The resulting dot product is

$$ \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(t^3,t,t^2)\cdot(1,2t,3t^2) =t^3+2t^2+3t^4. $$

Integrating each term yields

$$ \begin{aligned} \int_C\mathbf{F}\cdot d\mathbf{r} &=\int_0^1(t^3+2t^2+3t^4)\,dt\\ &=\left[\frac{t^4}{4}+\frac{2t^3}{3}+\frac{3t^5}{5}\right]_0^1\\ &=\frac14+\frac23+\frac35\\ &=\frac{15}{60}+\frac{40}{60}+\frac{36}{60}\\ &=\frac{91}{60}. \end{aligned} $$

Check: the derivative of the displayed antiderivative is \(t^3+2t^2+3t^4\), and its value at \(t=0\) is zero. The endpoint evaluation therefore gives the stated result.

A Common Error: Omitting the Velocity

A frequent mistake is to substitute the curve into the force field and then integrate that force vector as though it were already the work integrand. Work depends on displacement as well as force. The curve derivative supplies the displacement direction and rate needed for the dot product.

Common mistake: For \(\mathbf{F}(x,y)=(y,x)\) and \(\mathbf{r}(t)=(t,t^2)\), one might incorrectly integrate \(t^2+t\), the sum of the entries of \(\mathbf{F}(\mathbf{r}(t))\), obtaining
$$ \int_0^1(t^2+t)\,dt=\frac13+\frac12=\frac56. $$
This is not the work integral. The force must be dotted with \(\mathbf{r}'(t)=(1,2t)\), giving \(3t^2\), and the correct integral is \(\int_0^1 3t^2\,dt=1\).

Another way to detect the error is to inspect the quantities being combined. A force vector alone does not describe the displacement made during a parameter interval. The dot product with \(\mathbf{r}'(t)\) accounts for how the path moves at each parameter value; omitting it changes the mathematical quantity being calculated.

Checks Before Accepting a Result

Before finishing a vector line integral, verify that the force field has been evaluated at the curve, not at unrelated variables; that the derivative is taken with respect to the same parameter used in the interval; and that the dot product is integrated over the entire interval. These checks apply whether the path lies in the plane or in three-dimensional space.

For a physical force measured in newtons and a position measured in metres, the integral has units of newton-metres, the units of work. This provides a dimensional check in an applied problem. In a purely mathematical exercise, the same setup still follows: field along the curve, curve derivative, dot product, parameter integral.

Check Your Understanding

Use the parametrized-integral definition to answer the following questions.

  1. For a curve \(\mathbf{r}(t)\), what does \(\mathbf{F}(\mathbf{r}(t))\) represent?
  2. Why is \(\mathbf{r}'(t)\) needed in the work integral?
  3. Set up, but do not evaluate, the integral for \(\mathbf{F}(x,y)=(x,y)\) along \(\mathbf{r}(t)=(\cos t,\sin t)\), \(0\leq t\leq\frac{\pi}{2}\).
  4. What is the specific error in integrating the entries of \(\mathbf{F}(\mathbf{r}(t))\) without forming a dot product?
  5. If \(\mathbf{r}(t)=(t,t^2)\), compute \(\mathbf{r}'(t)\) and state how it enters the vector line integral.

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