Re: Work done along a path

I hope that by "Fd" you mean the **integral** of the **dot** product of the vector force, F, and vector, d, from (0, 1, 0) to (1, 0, 3).

As for the "other method", Is it possible that this vector field is "conservative"? That is, that there exist some function, $\displaystyle \phi(x,y,z)$ such that $\displaystyle \nabla \phi= F$? If that is true then the $\displaystyle \phi$ is the "potential function" and the work done is $\displaystyle \phi(1,0,3)- \phi(0,1,0)$. How can you check that? How can you find $\displaystyle \phi$ from F?

Re: Work done along a path

Yes sorry, by Fd I mean the integral. And what you're suggesting, is it Green-riemann theory? Sorry I'm not familiar with English terms; I attend a French University in Quebec. If it is Green-Riemann do you need to check t

Re: Work done along a path

Well what he means is that work done along a curve c is equal $\displaystyle W = \int\limits_c F ds $ . And If there exists some function G, such that $\displaystyle \nabla G = F $ then

$\displaystyle W =\int\limits_c F ds = G(b) - G(a)$ where a and b are the starting an ending points of your curve, (This is called the path independence theorem or something like that )

Re: Work done along a path

Ok, that I understood. So basically, I need to prove that F is conservative, then find its potential function, and subtract the evaluation of it of point say b to point a. Something like that?

Re: Work done along a path

But then, would the definition of work that I mentionned, which is the dot product of F and d, the distance traveled between the two points, still be valid? And if so, what would be the upper and lower bounds?

Re: Work done along a path

Well if you find a potential function then that is equivalent to saying F is conservative. If you find the potential function, then you do not even need to integrate anything, so there would be no use for finding bounds.

The second method would be

to evaluate the intergral along the straight line segment from the vector A to the vector B, that you had.

Here i'll be more clear

First you have to show F is conservative, so find its potential function G(Scalar Field $\displaystyle G: \mathbb{R}^3 \to \mathbb{R} $)

Method 1:

W = G(b) - G(a)

Method 2:

The equation for a line from point a to point b can be defined at t $\displaystyle \in [0, 1] $ $\displaystyle c(t) = (b-a)t + a $

Then evauluate $\displaystyle \int_0^1 F(c(t)) \dot c'(t) dt $ = W

Re: Work done along a path

Ok, well I've done both methods, and they give me different answers. For method 1, I've found 9 + e^6 and for method 2 I've found 39 + 2e^6. Now I don't know if this is normal or not, but the following question asks which method do I prefer, and why? Is it related to the fact that they do not equal eachother?

Re: Work done along a path

Nono, i've computed both and they gave me the same answer, you must have had an error. The answer is $\displaystyle e^6 + 8 $. By asking which method is easier, obviously method number 1 is way easier, since you dont need to do integration.

List what you got for G, the potential function.

Also give me what you got for F(c(t))

Re: Work done along a path

For G I got G = xz^2+xe^(2z)+yz^3. For F(c(t)) I got (9t^2+e^(6t))i + (81t^3+1)j + (6t^2+2te^(6t)+27t^2)k

Re: Work done along a path

Ok I saw my mistake for F(c(t)). My c(t) is wrong to begin with. But I still cant see my mistake for G.

Re: Work done along a path

My F(c(t)) is now (9t^2+e^(6t))i + (27t^3+1)j + (6t^2+2te^(6t)+27t^2-27t^3)k

Re: Work done along a path

Your G is a bit wrong. i

It has to be G = $\displaystyle xz^2 + xe^{2z} + yz^{3} + y $ (you forgot the y).

Your F(c(t)) is also wrong.

Did you get c(t) = $\displaystyle \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix} t + \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $ ?

then do F(t, -t + 1, 3t) = F(c(t))

Re: Work done along a path

Yeah I got the c(t) right now. and I corrected my G, which now gives me the right answer. That's the constant h(x,y) that I didn't catch.

Re: Work done along a path

Ok, I've corrected everything and now it gives me the same answer for both methods. Thanks a lot, really a life saver.