# Trouble Understanding Line Integrals of Vector Fields Formula

In my textbook, it explains that "wok done by a variable force $$f(x)$$ in moving a particle from $$a$$ to $$b$$ is $$W = \int_a^b{f(x)}\ dx$$." Makes sense to me. However, we've reached the section where we are to calculate work on a curve.

Now suppose that $$\vec{F} =\ $$ is a continuous force field on $$\mathbb{R}^3$$ ... We wish to compute the work done by this force in moving a particle along a smooth curve $$C$$. We divide $$C$$ into subarcs with length $$\Delta s_i$$. ... If $$\Delta s_i$$ is small, ... then the particle moves from approximately in the direction of $$\vec{T}(t_i)$$, the tangent vector [at that point]. Thus the work done by the force $$\vec{F}$$ in moving the particle [across $$\Delta s$$] is approximately $$\vec{F}(x_i, y_i,z_i) \cdot [\Delta s_i \vec{T}(t_i)]$$.

Eventually it explains the integral to be $$W = \int_C \vec{F}\cdot\vec{T}\ ds$$.

I suppose that makes some sense, but why isn't the formula just $$W = \int_C \vec{F}(x,y,z)\ ds$$? Why is the tangent required if $$ds$$ is infinitely small? Is it better to consider the tangent to be the distance in $$W = \vec{F} \cdot \vec{D}$$? and if that's the case, what does that make $$ds$$?

## 1 Answer

Suppose your path were a segment of the $$x-$$ axis but your force field were in the $$y-$$ direction. Then moving the particle along the path doesn't require (or generate) any work. You need the dot product with the tangent to the curve because that's the only component of the force that actually affects the amount of work performed.

• Your answer made me realize a major misunderstanding. Thank you! – npengra317 Feb 14 '19 at 0:51