9. Line Integrals

e. Applications of Vector Line Integrals

1. Work

In a Physics course you learn that when a constant force, \(F\), moves an object a distance, \(D\), in a straight line then the work done is \(W=FD\) provided the force is in the direction of the motion. In the chapter on the Dot Product, you learned that when a constart force vector, \(\vec F\), moves an object by a displacement vector, \(\vec D\), the work done is \(W=\vec F\cdot\vec D\) even if they are not parallel. We now want to look at the case where the force may not be constant and the path may not be straight. If the object moves a small distance \(ds\) in the direction \(\hat{T}\) then the displacement is \(d\vec D=\hat{T}\,ds\). Assuming the force is essentially constant over this small distance, the small amount of work done is \(dW=\vec F\cdot d\vec D=\vec F\cdot\hat{T}\,ds\) and the total work is \[ W=\int_A^B \vec F\cdot\hat{T}\,ds=\int_A^B \vec F\cdot d\vec s \]

Notice that we used the theoretical version (3) of the line integral to derive the formula for work, but in the example below we will use version (1) to compute it.

Find the work done by the force \(\vec F=\langle y^2,z,x^2\rangle\) on an object which moves along the twisted cubic \(\vec r(t)=(t,t^2,t^3)\) from \((0,0,0)\) to \((1,1,1)\).

The restriction of the force to the curve is \(\vec F(\vec r(t)) =\left\langle (t^2)^2,(t^3),(t)^2\right\rangle =\langle t^4,t^3,t^2\rangle\) and the velocity is \(\vec v=\langle 1,2t,3t^2\rangle\). So the work is \[\begin{aligned} W&=\int_{(0,0,0)}^{(1,1,1)} \vec F\cdot d\vec s =\int_0^1 \vec F\cdot\vec v\,dt \\ &=\int_0^1 (t^4+2t^4+3t^4)\,dt =\left[\dfrac{6t^5}{5}\right]_0^1 =\dfrac{6}{5} \end{aligned}\]

You can practice finding the work done by a force to move an object along a curve using the following Maplet (requires Maple on the computer where this is executed):

Work Along a Curve   Rate It