Work Along Helices

Work Along a Helical Path

A helical trajectory combines circular motion in the horizontal plane with steady motion in the vertical direction. Its position changes in all three coordinates, so calculating work requires tracking how each force component acts along the path.

The essential procedure is to describe the helix with a parameter, evaluate the force field at the resulting position, and take the dot product with the velocity. That dot product measures the force component acting in the direction of motion at each instant.

Key idea: Work along a helix is the integral of the force field evaluated on the helix, dotted with the helix’s velocity vector.

Parametrizing a Helix

A standard helix of radius $a>0$, with its axis along the $z$-axis, can be parametrized by

\mathbf{r}(t)=\langle a\cos t,\ a\sin t,\ bt\rangle, \qquad \alpha\le t\le\beta.

Here $a$ is the distance from the helix to the $z$-axis, and $b$ is the vertical change per unit increase in $t$. One complete revolution corresponds to an increase of $2\pi$ in $t$, so the vertical rise per revolution is $2\pi b$. The parameter interval determines which part of the helix is traversed and in which direction.

Differentiating each coordinate gives the velocity vector:

\mathbf{r}'(t)=\langle -a\sin t,\ a\cos t,\ b\rangle.

The first two velocity components describe the changing circular position, while the third records the vertical motion. In general, none of these components can be discarded when computing work.

Work as a One-Variable Integral

Let $\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle$ be a force field that is continuous along the helix. For a smooth parametrized path $\mathbf{r}(t)$, the work from $t=\alpha$ to $t=\beta$ is defined by integrating the force along the path dotted with the velocity.

Definition: The work done by $\mathbf{F}$ along the parametrized curve $\mathbf{r}(t)$, $\alpha\le t\le\beta$, is $$ W=\int_\alpha^\beta \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$ For the helix above, this becomes $$ W=\int_\alpha^\beta \left[ -aP(a\cos t,a\sin t,bt)\sin t +aQ(a\cos t,a\sin t,bt)\cos t +bR(a\cos t,a\sin t,bt) \right]dt. $$

The notation $\mathbf{F}(\mathbf{r}(t))$ means that the coordinates of the path are substituted into every component of the field. The dot product then pairs the resulting force components with the corresponding velocity components.

Why the Parametrized Formula Works

Over a small parameter change $\Delta t$, the displacement along the path is approximately $\mathbf{r}'(t)\Delta t$. The work over that small part is therefore approximately the force at the current position dotted with this displacement. Adding these contributions and taking the limit produces the work integral.

Approximate the displacement.
For a small interval of width $\Delta t$, differentiability gives $\mathbf{r}(t+\Delta t)-\mathbf{r}(t)\approx\mathbf{r}'(t)\Delta t$.

Approximate the work on that interval.
The force is approximately $\mathbf{F}(\mathbf{r}(t))$, so its work contribution is approximately $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\Delta t$.

Take the limit of the sums.
Summing these contributions over a partition and taking the limit gives the Riemann integral $\int_\alpha^\beta\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt$.

For the helix, substituting its velocity into the dot product gives the three terms in the displayed formula: the $x$-force paired with $-a\sin t$, the $y$-force paired with $a\cos t$, and the $z$-force paired with $b$.

Worked Examples

Worked Example: A Rotational Field with a Vertical Component

Find the work done by $\mathbf{F}(x,y,z)=\langle -y,x,2z\rangle$ along one turn of the helix $\mathbf{r}(t)=\langle 2\cos t,2\sin t,3t\rangle$, for $0\le t\le2\pi$.

First, differentiate the path:

\mathbf{r}'(t)=\langle -2\sin t,2\cos t,3\rangle.

Now evaluate the field on the path. Since $x=2\cos t$, $y=2\sin t$, and $z=3t$,

\mathbf{F}(\mathbf{r}(t)) =\langle -2\sin t,2\cos t,6t\rangle.

The dot product is

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(-2\sin t)(-2\sin t)+(2\cos t)(2\cos t)+(6t)(3) =4\sin^2t+4\cos^2t+18t =4+18t.

Therefore,

W=\int_0^{2\pi}(4+18t)\,dt =\left[4t+9t^2\right]_0^{2\pi} =8\pi+36\pi^2.

As a check, differentiating the antiderivative $4t+9t^2$ gives $4+18t$, the integrand. Substituting the endpoints gives $4(2\pi)+9(2\pi)^2=8\pi+36\pi^2$, as calculated.

Worked Example: Coordinate Dependence Can Cancel in the Dot Product

Find the work done by $\mathbf{F}(x,y,z)=\langle xz,yz,x^2+y^2\rangle$ along $\mathbf{r}(t)=\langle\cos t,\sin t,t\rangle$, for $0\le t\le2\pi$.

The velocity is

\mathbf{r}'(t)=\langle-\sin t,\cos t,1\rangle.

Substituting the path coordinates into the field gives

\mathbf{F}(\mathbf{r}(t)) =\langle t\cos t,t\sin t,\cos^2t+\sin^2t\rangle =\langle t\cos t,t\sin t,1\rangle.

Taking the dot product,

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =-t\cos t\sin t+t\sin t\cos t+1 =1.

The first two terms cancel exactly, so

W=\int_0^{2\pi}1\,dt =\left[t\right]_0^{2\pi} =2\pi.

The check is direct: the derivative of the antiderivative $t$ is $1$, and its endpoint difference is $2\pi-0=2\pi$. The field components vary along the path even though their total tangential contribution simplifies to a constant.

Worked Example: A Partial Turn with Polynomial Dependence

Find the work done by $\mathbf{F}(x,y,z)=\langle y,z,x\rangle$ along $\mathbf{r}(t)=\langle2\cos t,2\sin t,t\rangle$, for $0\le t\le\pi$.

The path velocity and the force evaluated on the path are

\mathbf{r}'(t)=\langle-2\sin t,2\cos t,1\rangle, \qquad \mathbf{F}(\mathbf{r}(t))=\langle2\sin t,t,2\cos t\rangle.

Thus the integrand is

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

An antiderivative is obtained term by term:

\int -4\sin^2t\,dt=-2t+\sin(2t), \qquad \int 2t\cos t\,dt=2t\sin t+2\cos t, \qquad \int 2\cos t\,dt=2\sin t.

For the middle integral, differentiating $2t\sin t+2\cos t$ gives $2\sin t+2t\cos t-2\sin t=2t\cos t$. Combining the antiderivatives and evaluating,

\begin{aligned} W &=\left[-2t+\sin(2t)+2t\sin t+2\cos t+2\sin t\right]_0^\pi\\ &=(-2\pi+0+0-2+0)-(0+0+0+2+0)\\ &=-2\pi-4. \end{aligned}

For a full check, differentiating the combined antiderivative gives $-2+2\cos(2t)+2t\cos t+2\cos t$. Since $-2+2\cos(2t)=-4\sin^2t$, this is the original integrand. The endpoint substitution yields the stated work.

A Common Pitfall: Omitting the Vertical Contribution

In the first example, it would be incorrect to keep only the contributions from the $x$- and $y$-components of the force. Those terms give $4\sin^2t+4\cos^2t=4$, whose integral over one turn is $8\pi$. But the field also has vertical component $2z=6t$, and the path has vertical velocity $3$. Their dot-product contribution is $(6t)(3)=18t$, with integral $36\pi^2$. The correct total is $8\pi+36\pi^2$, not $8\pi$.

Practical check: Write the force and velocity as three-component vectors before taking their dot product. This makes it harder to overlook a coordinate, especially the vertical one.

Using the Formula Reliably

For a helix and a continuous force field, the calculation follows a consistent sequence:

Record the parameter interval and write the position vector $\mathbf{r}(t)$.
Differentiate every coordinate to obtain $\mathbf{r}'(t)$.
Substitute all three coordinates into every component of $\mathbf{F}$.
Compute $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$ explicitly.
Integrate over the stated parameter interval and check the antiderivative and endpoint values.

The parameter bounds are part of the path description: changing them can change the portion traversed. The velocity vector also reflects the direction of traversal, so its signs must match the chosen parametrization. The final work has the units of force multiplied by distance when the field represents physical force.

Check Your Understanding

For each question, identify the path, the field values along it, and the velocity before evaluating or interpreting the work.

For $\mathbf{r}(t)=\langle3\cos t,3\sin t,2t\rangle$, what is $\mathbf{r}'(t)$, and what is the vertical rise over one complete turn?
For $\mathbf{F}(x,y,z)=\langle z,0,y\rangle$ and $\mathbf{r}(t)=\langle\cos t,\sin t,t\rangle$, write the work integrand.
Why must the vertical force component be paired with $b$ in the standard helix formula?
For the field and path in question 2, simplify the dot product before integrating.
What is the error in calculating work using only the first two components of the force and velocity?

What You'll Learn