Finding the work done ( calculus 3 )

Question

Find the work done by the force field F=<z,x,y> on moving a particle from the point (3,0,0) to (0,pi/2,3) along the helix x=3cost, y=t, z=3sint

Answer 1

Work is the line integral ∫_C F·dr (where F and dr are both vectors). C is the curve of your trajectory--for this problem it's the given helix.

 

Line integrals move through space, which means that our x, y, and z-coordinates are all simultaneously changing. It would be much nicer to deal with a single coordinate instead of the three independent ones. That's where the parameter t comes in--we have x(t), y(t), and z(t) already given to us as a description/parameterization of the helix. The functions for x/y/z are termed parametric equations. 

 

With a little inspection, we can identify the starting and stopping values for the parameter. There should be one particular value of t that we plug in to the parametric equations to yield the initial point (3,0,0), and a second value of t that likewise yields the terminal point (0, pi/2, 3). Verify that these values are t=0 and t=pi/2. These will become the limits of integration for the line integral over dt.

 

What remains is to convert the integrand from F·dr to [something]dt. We can recast F simply by substituting in the appropriate parametric equations: 

 

F = <z,x,y> = <3sin(t), 3cos(t), t>.

 

The dr  represents a small displacement in space. From this, we can rely on a calc1-based understanding of differentials to arrive at:

 

dr = <dx, dy, dz> = <(dx/dt)dt, (dy/dt)dt, (dz/dt)dt> = <dx/dt, dy/dt, dz/dt>dt = r' dt

 

r' = dr/dt = <-3sin(t), 1, 3cos(t)>

 

 

Finally, evaluate the line integral:

 

w = ∫_CF·dr = ∫₀^π/2 <3sin(t), 3cos(t), t>·<-3sin(t), 1, 3cos(t)> dt = ∫₀^pi/2 [-9sin²(t)+3cos(t)-3t cos(t)] dt