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Work and Vector Line Integrals · Tutorial 773 of 999

Choosing Efficient Parametrizations for Work

Choose parametrizations that simplify work calculations while preserving the path’s endpoints and orientation.

Intermediate 10 min read

What You'll Learn

  • Parametrize a line segment directly from its initial point to its terminal point.
  • Choose angle intervals that encode a circle’s starting point, sweep, and orientation.
  • Adapt circle parametrizations to ellipses with different centers and semiaxes.
  • Set helix parameters to match both angular motion and vertical change.
  • Reparametrize an oriented curve and account for the parameter’s effect on velocity.
  • Detect errors caused by reversing direction or omitting a parameter scale factor.

Why Parametrization Choices Matter

A path can often be described by more than one parametrization. A good choice makes the endpoints, direction, and motion easy to see, and can keep the work integral simple. A poor choice can obscure the orientation or introduce needless algebra.

The goal is not merely to produce coordinates that lie on the desired geometric curve. A parametrization also specifies how the curve is traversed. For work, that direction matters: traversing a path in reverse generally changes the sign of the work.

For a piecewise \(C^1\) curve \(\mathbf r:[a,b]\to\mathbb R^n\) and a vector field \(\mathbf F\) defined along it, the work is

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

This formula makes the role of a convenient parametrization clear: it determines both the field values along the path and the velocity vector \(\mathbf r'(t)\).

Definition: An oriented parametrized curve consists of a parametrization together with its direction of travel, from the point at the initial parameter value to the point at the terminal parameter value. A change of parameter preserves orientation when it traverses the same path in the same direction.

A Reliable Choice for Line Segments

Let \(A\) be the initial point and \(B\) the terminal point. The direct parametrization is the affine interpolation from \(A\) to \(B\):

$$ \mathbf r(t)=A+t(B-A),\qquad 0\le t\le 1. $$

At \(t=0\), the position is \(A\); at \(t=1\), it is \(B\). Its velocity \(B-A\) is constant, so the parametrization has no unnecessary changes in direction or speed. If the path is instead required to go from \(B\) to \(A\), swap the endpoints. The segment as a geometric set is unchanged, but its orientation is reversed.

Worked Example: Work Along an Oriented Line Segment

Find the work of \(\mathbf F(x,y)=(y,x)\) along the segment from \(A=(1,-1)\) to \(B=(3,2)\).

First find the displacement:

$$ B-A=(3-1,\;2-(-1))=(2,3). $$

Use the direct parametrization:

$$ \mathbf r(t)=(1,-1)+t(2,3)=(1+2t,\,-1+3t),\qquad 0\le t\le1. $$

The endpoints and velocity are

$$ \mathbf r(0)=(1,-1)=A,\qquad \mathbf r(1)=(3,2)=B,\qquad \mathbf r'(t)=(2,3). $$

Along this path, the field is

$$ \mathbf F(\mathbf r(t))=(-1+3t,\;1+2t). $$

Therefore, its dot product with the velocity is

$$ \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =2(-1+3t)+3(1+2t) =-2+6t+3+6t =1+12t. $$

Integrate over the chosen interval:

$$ W=\int_0^1(1+12t)\,dt =\left[t+6t^2\right]_0^1 =(1+6)-0 =7. $$

As a check, differentiating the antiderivative \(t+6t^2\) gives \(1+12t\), the integrand above. The parametrization also checks the requested direction: it starts at \(A\) and ends at \(B\).

Choosing an Angle Interval for a Circle

A circle of radius \(R\) centered at \((h,k)\) can be parametrized by

$$ \mathbf r(t)=(h+R\cos t,\;k+R\sin t). $$

The angle \(t\) is measured from the positive \(x\)-direction relative to the center. Increasing \(t\) traces the circle counterclockwise. The interval specifies the portion of the circle and the starting point: for example, an interval from \(\alpha\) to \(\beta\), with \(\beta>\alpha\), begins at angle \(\alpha\) and sweeps counterclockwise through angle \(\beta-\alpha\). A full counterclockwise circuit uses an interval of length \(2\pi\).

Worked Example: One Counterclockwise Circuit

Calculate the work of \(\mathbf F(x,y)=(-y,x)\) around the circle of radius \(2\) centered at the origin, starting at \((2,0)\) and moving counterclockwise once.

The starting point has angle \(0\), and one full counterclockwise circuit increases the angle to \(2\pi\). Choose

$$ \mathbf r(t)=(2\cos t,\;2\sin t),\qquad 0\le t\le2\pi. $$

The endpoints and velocity are

$$ \mathbf r(0)=(2,0),\qquad \mathbf r(2\pi)=(2,0),\qquad \mathbf r'(t)=(-2\sin t,\;2\cos t). $$

Substitute the position into the field:

$$ \mathbf F(\mathbf r(t))=(-2\sin t,\;2\cos t). $$

Thus the integrand is

$$ \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =(-2\sin t)^2+(2\cos t)^2 =4\sin^2t+4\cos^2t =4. $$

The work is

$$ W=\int_0^{2\pi}4\,dt =4(2\pi-0) =8\pi. $$

Checking the direction, the initial velocity is \(\mathbf r'(0)=(0,2)\), which points upward from the rightmost point, as required for counterclockwise travel. The integral evaluates to \(8\pi\) because the integrand is the constant \(4\) over an interval of length \(2\pi\).

Ellipses: Scale the Coordinates Separately

An axis-aligned ellipse centered at \((h,k)\), with horizontal semiaxis \(a>0\) and vertical semiaxis \(b>0\), has the parametrization

$$ \mathbf r(t)=(h+a\cos t,\;k+b\sin t). $$

Like the circle parametrization, increasing \(t\) moves counterclockwise. The factors \(a\) and \(b\) stretch the horizontal and vertical coordinates by different amounts. The same angle-interval choices still control the starting point and sweep. In particular, \(0\le t\le2\pi\) starts at the rightmost point and makes one counterclockwise circuit.

Worked Example: Work Around an Ellipse

Find the work of \(\mathbf F(x,y)=(-y,x)\) around the ellipse centered at \((1,-2)\), with semiaxes \(3\) and \(2\), traversed counterclockwise once from its rightmost point.

The prescribed starting point is \((1+3,-2)=(4,-2)\). Choose the full-turn parametrization

$$ \mathbf r(t)=(1+3\cos t,\;-2+2\sin t),\qquad 0\le t\le2\pi. $$

It has the required endpoints and velocity:

$$ \mathbf r(0)=(4,-2),\qquad \mathbf r(2\pi)=(4,-2),\qquad \mathbf r'(t)=(-3\sin t,\;2\cos t). $$

The field along the ellipse is

$$ \mathbf F(\mathbf r(t)) =(2-2\sin t,\;1+3\cos t). $$

Taking the dot product gives

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

Integrating term by term,

$$ \begin{aligned} W &=\int_0^{2\pi}(6-6\sin t+2\cos t)\,dt\\ &=\left[6t+6\cos t+2\sin t\right]_0^{2\pi}\\ &=(12\pi+6+0)-(0+6+0)\\ &=12\pi. \end{aligned} $$

As a check, the derivative of \(6t+6\cos t+2\sin t\) is \(6-6\sin t+2\cos t\), the computed integrand. The initial velocity is \((0,2)\), pointing upward from the rightmost point, so the ellipse is traversed counterclockwise.

Helices: Coordinate the Turn and the Rise

A circular helix of radius \(R\), centered around the vertical axis, can be written as

$$ \mathbf r(t)=(R\cos t,\;R\sin t,\;ct+d), $$

where \(c\) is the vertical change per unit increase in angle and \(d\) sets the initial height. The angle interval determines how much the helix turns; the third coordinate determines its height at each parameter value. To hit a specified terminal height, choose \(c\) and the interval together rather than choosing the angular motion alone.

Worked Example: Work Along a Rising Half-Turn

Find the work of \(\mathbf F(x,y,z)=(-y,x,1)\) along the helix of radius \(2\) that starts at \((2,0,0)\), rises to height \(\pi\), and makes a counterclockwise half-turn.

A half-turn has angular interval \(0\le t\le\pi\). The height must increase from \(0\) to \(\pi\), so use \(z=t\):

$$ \mathbf r(t)=(2\cos t,\;2\sin t,\;t),\qquad 0\le t\le\pi. $$

The endpoints and velocity are

$$ \mathbf r(0)=(2,0,0),\qquad \mathbf r(\pi)=(-2,0,\pi),\qquad \mathbf r'(t)=(-2\sin t,\;2\cos t,\;1). $$

The field evaluated along the helix is

$$ \mathbf F(\mathbf r(t))=(-2\sin t,\;2\cos t,\;1). $$

Its dot product with the velocity is

$$ \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =4\sin^2t+4\cos^2t+1 =5. $$

Therefore,

$$ W=\int_0^\pi5\,dt =5(\pi-0) =5\pi. $$

The endpoint check confirms both the half-turn and the required rise. The integrand is constant \(5\), and its integral over an interval of length \(\pi\) is \(5\pi\), which also checks the calculation directly.

When a Parameter Interval Changes

The same oriented curve can be traversed on a different parameter interval. If \(\mathbf r:[a,b]\to\mathbb R^n\) is piecewise \(C^1\), let \(\phi:[\alpha,\beta]\to[a,b]\) be \(C^1\) and strictly increasing, with \(\phi(\alpha)=a\) and \(\phi(\beta)=b\). Then \(\mathbf q(s)=\mathbf r(\phi(s))\) traverses the same path in the same direction. The work is unchanged.

Theorem: Under these conditions, if \(\mathbf F\) is continuous along the curve, then $$ \int_\alpha^\beta \mathbf F(\mathbf q(s))\cdot\mathbf q'(s)\,ds = \int_a^b \mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt. $$ The same conclusion holds for piecewise \(C^1\), strictly increasing changes of parameter, with the integrals taken piece by piece.

Here is the justification in the \(C^1\) case. The chain rule gives \(\mathbf q'(s)=\mathbf r'(\phi(s))\phi'(s)\). Substitute this into the work integral, then use the change of variable \(t=\phi(s)\):

$$ \begin{aligned} \int_\alpha^\beta \mathbf F(\mathbf q(s))\cdot\mathbf q'(s)\,ds &=\int_\alpha^\beta \mathbf F(\mathbf r(\phi(s)))\cdot \mathbf r'(\phi(s))\,\phi'(s)\,ds\\ &=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt. \end{aligned} $$

Strict increase preserves the order in which points are visited; the regularity condition makes the chain rule and change of variables applicable. If \(\phi\) is strictly decreasing and exchanges the endpoints, the same calculation gives the negative of the original work. Orientation is therefore part of the data, not a cosmetic feature of the parametrization.

A Common Mistake: Reversing Direction Without Changing the Sign

For the circle in the earlier example, replacing \((2\cos t,2\sin t)\) by \((2\cos t,-2\sin t)\) on \(0\le t\le2\pi\) changes the direction to clockwise. It is incorrect to keep the counterclockwise answer \(8\pi\) merely because the same circle is covered.

For the clockwise parametrization,

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

For \(\mathbf F(x,y)=(-y,x)\), the field along this path is \((2\sin t,2\cos t)\), so

$$ \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =(2\sin t)(-2\sin t)+(2\cos t)(-2\cos t) =-4. $$

Consequently, the clockwise work is \(\int_0^{2\pi}-4\,dt=-8\pi\), not \(8\pi\). The correction is to check the direction encoded by the parameter before using a work result.

A Practical Selection Checklist

Before integrating, identify the path information that the parametrization must preserve:

  • Endpoints: Substitute the initial and terminal parameter values into \(\mathbf r\) and compare with the requested points.
  • Direction: Check whether the parameter increases along the required direction. For circles and ellipses, increasing angle gives counterclockwise motion in the standard parametrization.
  • Extent: Choose an interval covering exactly the requested arc, number of turns, or segment—not more or less.
  • Velocity: Differentiate the full parametrization, including any scale factor from a changed parameter interval.
  • Work integrand: Compute \(\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\) using that velocity, then integrate over the same interval.

For example, replacing \(t\in[0,2\pi]\) by \(s\in[0,1]\) for one full circle requires the angle \(2\pi s\), not \(s\). The derivative then includes the factor \(2\pi\). Omitting this scale changes the velocity and produces an incorrect work integral even if the intended geometric path is recognizable.

Check Your Understanding

For each question, consider both the geometric path and its orientation.

  1. Write a direct parametrization of the segment from \((2,1)\) to \((-1,4)\) on the interval \([0,1]\). What is its constant velocity?
  2. What parameter interval and direction give one clockwise circuit of a circle starting at its rightmost point?
  3. Give a counterclockwise parametrization of an ellipse centered at \((0,3)\) with horizontal semiaxis \(4\) and vertical semiaxis \(1\), starting at its rightmost point.
  4. A helix makes two counterclockwise turns while rising from height \(1\) to height \(5\). If its angle parameter runs from \(0\) to \(4\pi\), what linear formula for its height gives the required endpoints?
  5. State the regularity and monotonicity conditions on a change of parameter that justify the work-invariance result above.

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