# Work done along a path

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• Nov 27th 2012, 04:10 PM
richardsim10
Work done along a path
Let the vector field be F(x,y,z) = (z^2+e^(2z))i + (z^3+1)j + (2xz+2xe^(2z)+3yz^2)k calculate the work done along the segment from point (0,1,0) to (1,0,3) using 2 different methods. which method do you prefer, and why?

I am unsure as to which two methods to use. I know one of them is direct definition of work, that is W = Fd, but I have no idea about the other method.
• Nov 27th 2012, 04:21 PM
HallsofIvy
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?
• Nov 27th 2012, 04:31 PM
richardsim10
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
• Nov 27th 2012, 04:46 PM
jakncoke
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 )
• Nov 27th 2012, 05:15 PM
richardsim10
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?
• Nov 27th 2012, 05:37 PM
richardsim10
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?
• Nov 27th 2012, 05:39 PM
jakncoke
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
• Nov 27th 2012, 06:24 PM
richardsim10
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?
• Nov 27th 2012, 06:40 PM
jakncoke
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))
• Nov 27th 2012, 06:46 PM
richardsim10
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
• Nov 27th 2012, 06:48 PM
richardsim10
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.
• Nov 27th 2012, 06:52 PM
richardsim10
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
• Nov 27th 2012, 06:53 PM
jakncoke
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))
• Nov 27th 2012, 06:56 PM
richardsim10
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.
• Nov 27th 2012, 07:01 PM
richardsim10
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.