Does the line integral definition of Work involve distance or displacement?

My textbook reports the following definition of Work:

where ds is the infinitesimal displacement.

I know that an infinitesimal displacement is usually denoted by dr and I also know that the magnitude of dr is given by ds (infinitesimal distance) Now, if we are talking about displacement (in Work definition), why should we use ds instead of dr?

I ask this because my textbook always refers to infinitesimal displacement as dr. I have always associated 's' to distance, so I see ds as an infinitesimal "distance vector", but I am quite sure that distance is only a scalar quantity, not a vector.

• Of course it does... – user66452 Mar 17 '15 at 19:04

1 Answer

Its just a matter of what you use to call as displacement and as distance .

I have seen the usage of:

• dx
• ds
• dr

as the displacement too.

Wikipedia says :

The work done by a constant force of magnitude F on a point that moves a displacement (not distance) s in the direction of the force is the product,
W = Fs.

Note the usage of s as dispacement .
All in all , its the displacement that is used in calculating work and one may refer to it in many ways.(probably your textbook used different notations in different chapters)

And Distance is a scalar quantity.