# Conservative Vector Fields - Is this right?

1. Mar 7, 2008

### EngageEngage

Conservative Vector Fields -- Is this right?

1. The problem statement, all variables and given/known data

G = <(1 + x)e$$^{x+y}$$, xe$$^{x+y}$$+2z, -2y>
Evaluate $$\int$$$$_{C}$$G.dR
where C is the path given by:

x = (1 - t)e$$^{t}$$, y = t, z = 2t, 1=>t>=0
2. Relevant equations

3. The attempt at a solution

First, i noticed that there is a scalar potential associated with several terms (not all) in
the above vector field:

$$\varphi$$ = xe$$^{x + y}$$
$$\nabla$$$$\varphi$$ = <(1 + x)e$$^{x + y}$$, xe$$^{x + y}$$,0>

So, i then separated the initial vector field G into two parts:

G=<0, 2z, -2y> + <(1 + x)e$$^{x + y}$$, xe$$^{x + y}$$,0>

First, I compute some derivatives and then evaluate the first line integral:
$$\frac{dc}{dt}$$ =
{dx = -te$$^{t}$$dt
dy = 1dt
dz = 2dt}

so,

$$\int$$$$_{C}$$G$$_{1}$$ . $$\frac{dc}{dt}$$dt =
$$\int$$$$^{1}_{0}$$<0, 2(2t), -2(t)> . <-te$$^{t}$$, 1, 2>
=

$$\int$$0dt = 0

I then computed the second part of G, keeping in mind that this portion of the vector field is conservative. Therefore, i just made a path from point t = 0 to t = 1 with a straight line r, parametrized like so:

r = { x = 1-t, y = t, z = 2t}
$$\frac{dr}{dt}$$ = {dx = -1, dy = 1, dz = 2}

my second integral then becomes;

$$\int$$$$^{1}_{0}$$<(1 + 1 - t)e$$^{1 - t + t}$$, (1-t)e$$^{1-t + t}$$,0> . <-1, 1, 2> dt =

$$\int$$$$^{1}_{0}$$<(2-t)e, (1-t)e,0> . <-1, 1, 2> dt =

So, is my final answer simply this integral evaluated, or am I doing this completely wrong? Sorry if this is a bad question, my book just has no example problems at all, and integrating the theory can be confusing. Any feedback would be appreciated greatly. Thank you

2. Mar 7, 2008

### Dick

That looks like a reasonable approach to me. I didn't check all of the details, though.

3. Mar 7, 2008

### EngageEngage

The final answer that i got was -e, by computing this integral and adding the two portions (o & -e)

4. Mar 7, 2008

### EngageEngage

Great, thanks. I don't want anyone to check the details, I just wanted to see if what I was doing was reasonable, or if i was doing it completely wrong.

5. Mar 7, 2008

### Dick

I get -e as well.

6. Mar 7, 2008

### EngageEngage

Thanks a lot, I appreciate the help!