Work as Tangential Force Accumulation

Work Depends on the Direction of Motion

A force can be large without doing much work along a particular path. If the force points directly across the direction of motion, its component along the path is zero. If it points partly along the motion, only that parallel part contributes to work.

This viewpoint gives a geometric interpretation of the work integral: work accumulates the signed tangential component of force as the object travels along the curve. The unit tangent identifies the direction of motion, while arc length measures the distance traveled.

The Unit Tangent and Tangential Force

Let a smooth path be parametrized by \(\mathbf{r}(t)\), where \(a\le t\le b\), and suppose the path is regular, meaning \(\mathbf{r}'(t)\ne\mathbf{0}\) throughout the interval. Its speed is \(\|\mathbf{r}'(t)\|\). The unit tangent points in the direction of motion and has length one.

\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}.

Let \(\mathbf{F}\) be a force field defined along the path. The signed tangential component of the force is the scalar projection \(\mathbf{F}\cdot\mathbf{T}\). A positive value means the force has a component along the direction of travel; a negative value means its tangential component points against that direction.

Definition: The tangential force along a regular parametrized path is the scalar function \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{T}(t)\). The work done by the force along the path is the integral of this tangential component with respect to arc length.

Since the arc-length element is \(ds=\|\mathbf{r}'(t)\|\,dt\), the work can be written in either of the following equivalent forms:

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

The regularity condition matters for the unit tangent: if \(\mathbf{r}'(t)=\mathbf{0}\), the displayed definition of \(\mathbf{T}\) is not available at that parameter value. For a regular path, the work integral also has the familiar component form from the parametrization of vector line integrals.

Why Only the Tangential Component Contributes

At each point on a regular path, any force vector can be split into a part parallel to the unit tangent and a part perpendicular to it. The parallel part is \((\mathbf{F}\cdot\mathbf{T})\mathbf{T}\). Subtracting it from the force leaves a vector whose dot product with \(\mathbf{T}\) is zero.

\mathbf{F} = (\mathbf{F}\cdot\mathbf{T})\mathbf{T} + \left[\mathbf{F}-(\mathbf{F}\cdot\mathbf{T})\mathbf{T}\right].

Indeed, because \(\mathbf{T}\cdot\mathbf{T}=1\), the second vector has dot product

\left[\mathbf{F}-(\mathbf{F}\cdot\mathbf{T})\mathbf{T}\right]\cdot\mathbf{T} = \mathbf{F}\cdot\mathbf{T}-(\mathbf{F}\cdot\mathbf{T})(\mathbf{T}\cdot\mathbf{T}) =0.

To see how this decomposition produces the work formula, note that for \(dt>0\), the infinitesimal displacement is \(d\mathbf{r}=\mathbf{r}'(t)\,dt=\mathbf{T}\,ds\). Therefore, the force dotted with that displacement is \(\mathbf{F}\cdot d\mathbf{r}=(\mathbf{F}\cdot\mathbf{T})\,ds\). The perpendicular part contributes zero because its dot product with \(\mathbf{T}\) is zero. Summing these infinitesimal contributions along the path gives the work integral.

Key result: For a regular path, work accumulates the signed tangential force over distance: \(W=\int_C(\mathbf{F}\cdot\mathbf{T})\,ds\). A perpendicular force component contributes no work along that path.

Worked Examples

Worked Example: Work Along a Parabolic Path

Let \(\mathbf{r}(t)=(t,t^2)\), \(0\le t\le1\), and let the force field be \(\mathbf{F}(x,y)=(0,x)\). Find the tangential force and the work along the path.

First, evaluate the force on the path and calculate its velocity and speed:

\mathbf{F}(\mathbf{r}(t))=(0,t),\qquad \mathbf{r}'(t)=(1,2t),\qquad \|\mathbf{r}'(t)\|=\sqrt{1+4t^2}.

The unit tangent is therefore \(\mathbf{T}(t)=(1,2t)/\sqrt{1+4t^2}\). The signed tangential force is

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{T}(t) = (0,t)\cdot\frac{(1,2t)}{\sqrt{1+4t^2}} = \frac{2t^2}{\sqrt{1+4t^2}}.

Multiplying by \(ds=\sqrt{1+4t^2}\,dt\) cancels the speed factor. Thus

W = \int_0^1 \frac{2t^2}{\sqrt{1+4t^2}}\sqrt{1+4t^2}\,dt = \int_0^1 2t^2\,dt = \left[\frac{2t^3}{3}\right]_0^1 = \frac{2}{3}.

For a check, the component-form integrand is \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=(0,t)\cdot(1,2t)=2t^2\), giving the same integral and value. Also, differentiating the antiderivative \(2t^3/3\) gives \(2t^2\), the integrand used above.

Worked Example: Separating Tangential and Perpendicular Force on a Circle

Consider the circular path \(\mathbf{r}(t)=(2\cos t,2\sin t)\), \(0\le t\le\pi/2\), in the force field \(\mathbf{F}(x,y)=(-y,x)+(x,y)\). Identify the force components and calculate the work.

Along the path, the velocity is \(\mathbf{r}'(t)=(-2\sin t,2\cos t)\), so the speed is \(2\), and the unit tangent is \(\mathbf{T}(t)=(-\sin t,\cos t)\). Evaluating the force gives

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

The first vector in this sum is \(2\mathbf{T}(t)\), so it is tangential. The second is perpendicular to the tangent: \((\cos t,\sin t)\cdot(-\sin t,\cos t)=0\). Including its factor of \(2\), the force decomposition is \(\mathbf{F}=2\mathbf{T}+2(\cos t,\sin t)\). Consequently, the signed tangential force is \(\mathbf{F}\cdot\mathbf{T}=2\).

The arc-length element is \(ds=2\,dt\). Therefore,

W = \int_0^{\pi/2}2(2\,dt) = \int_0^{\pi/2}4\,dt = 4\left[\ t\ \right]_0^{\pi/2} = 2\pi.

As a check using the parametrized work integrand, \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=4\): the tangential part contributes \(2\mathbf{T}\cdot2\mathbf{T}=4\), while the perpendicular part has dot product zero with \(\mathbf{r}'(t)=2\mathbf{T}(t)\). Thus the work is \(\int_0^{\pi/2}4\,dt=2\pi\), as before.

Worked Example: Work Along a Helix

Let \(\mathbf{r}(t)=(\cos t,\sin t,t)\), \(0\le t\le2\pi\), and let \(\mathbf{F}(x,y,z)=(-y,x,2)\). Find the tangential and perpendicular parts of the force along the path and then compute the work.

The velocity and its magnitude are

\mathbf{r}'(t)=(-\sin t,\cos t,1), \qquad \|\mathbf{r}'(t)\|=\sqrt{2}.

Thus \(\mathbf{T}(t)=(-\sin t,\cos t,1)/\sqrt{2}\). On the path, \(\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t,2)\), and its tangential component is

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{T}(t) = \frac{\sin^2 t+\cos^2 t+2}{\sqrt{2}} = \frac{3}{\sqrt{2}}.

The vector part of the force parallel to the path is \((\mathbf{F}\cdot\mathbf{T})\mathbf{T}=\frac{3}{2}(-\sin t,\cos t,1)\). Subtracting this from the full force leaves the perpendicular part:

\mathbf{F}_{\perp} = \mathbf{F}-(\mathbf{F}\cdot\mathbf{T})\mathbf{T} = \left(\frac{\sin t}{2},-\frac{\cos t}{2},\frac12\right).

Its dot product with the velocity is \(-\sin^2 t/2-\cos^2 t/2+1/2=0\), confirming that it is perpendicular to the path. Since \(ds=\sqrt{2}\,dt\), the work is

W = \int_0^{2\pi}\frac{3}{\sqrt{2}}\sqrt{2}\,dt = \int_0^{2\pi}3\,dt = \left[3t\right]_0^{2\pi} = 6\pi.

For a direct check, \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=\sin^2 t+\cos^2 t+2=3\), so the parametrized integral also equals \(\int_0^{2\pi}3\,dt=6\pi\).

A Common Mistake: Using the Force Magnitude

The magnitude \(\|\mathbf{F}\|\) is not generally the tangential force. It measures the full length of the force vector, including any part that points perpendicular to the path. Replacing \(\mathbf{F}\cdot\mathbf{T}\) with \(\|\mathbf{F}\|\) can therefore give an incorrect result.

For a concrete example, take the unit circle \(\mathbf{r}(t)=(\cos t,\sin t)\), \(0\le t\le2\pi\), with force \(\mathbf{F}(x,y)=(x,y)\). Along the circle, the force has magnitude \(1\), so using its magnitude as the tangential force would incorrectly give

\int_0^{2\pi}\|\mathbf{F}(\mathbf{r}(t))\|\,ds = \int_0^{2\pi}1\cdot1\,dt = 2\pi.

But the unit tangent is \(\mathbf{T}(t)=(-\sin t,\cos t)\), and the force on the circle is \((\cos t,\sin t)\). Their dot product is \(-\cos t\sin t+\sin t\cos t=0\). The force is entirely perpendicular to the motion, so the actual work is \(\int_0^{2\pi}0\,ds=0\). The correction is to project the force onto the unit tangent before accumulating it over distance.

Why This View Is Useful

The tangential interpretation helps distinguish two different questions: how large is the force, and how much of it acts along the path? The first is answered by \(\|\mathbf{F}\|\); the second by \(\mathbf{F}\cdot\mathbf{T}\). Work depends on the second quantity.

It also clarifies why the work integrand can be positive, negative, or zero at a point. A force aiding the motion has positive tangential component; one opposing the motion has negative tangential component; and one perpendicular to the motion has zero tangential component. Along an entire curve, these signed contributions are accumulated with respect to the distance traveled.

Takeaway: To interpret or calculate work along a regular path, first find the unit tangent, project the force onto it, and integrate that signed component with respect to arc length. The perpendicular component does not contribute to work along the path.

Check Your Understanding

Use the unit tangent and the tangential-force interpretation to answer these questions.

For a regular path \(\mathbf{r}(t)\), write the unit tangent in terms of \(\mathbf{r}'(t)\) and its magnitude.
Why does the vector \(\mathbf{F}-(\mathbf{F}\cdot\mathbf{T})\mathbf{T}\) have zero contribution to work along the path?
If the tangential component of force is negative over part of a path, what does its sign say about the force relative to the direction of travel?
For the unit-circle example with \(\mathbf{F}(x,y)=(x,y)\), why is the work zero even though the force magnitude is \(1\) everywhere on the curve?
For a regular parametrization, explain why \((\mathbf{F}\cdot\mathbf{T})\,ds\) equals \(\mathbf{F}\cdot\mathbf{r}'(t)\,dt\).