work (force vector moving particle along line)

[mcwang719]

Let r(t)=<x(t),y(t),z(t)> be the position vector function which corresponds to the 3D line in space which passes through the point P(-3,4,1) and is parallel to the vector v=<-2,3,6>. Suppose the force vector F=<2,1,3> moves the particle from its position at t=0 along the 3Dline to its final position at t=2. Find the number of work units accomplished by F.

so this is what i did. i found the parameterization which was r(t)=<-2t-3,3t+4,6t+1>. so then r(0)=<-3,4,1> and r(2)=<-10,10,13> so the displacement vector would be D=<-7,6,12>. so <2,1,3>dot<-7,6,12>= -14+6+36= 28j.
did i do this right??? thanks!!!!

 

[Unco]

Nice parameterising, Mc. The rest of your work is quite not-so-good, though. Indeed, the force is conservative, so, if we wanted to, we could find its potential function and evaluate it at the endpoints to get the work done by the force field on the particle. I think this is what you were thinking.

The potential function would just be U(x, y z) = 2x + y + 3z + some constant.

This is not a vector quantity, nor is work.


Alternatively, you could use your parameterisation to find x'(t), y'(t) and z'(t) and used the usual method for evaluating work line integrals.