# Conservative Vector Fields - Is this right?

• EngageEngage
In summary, this conservative vector field has a potential associated with it, and two line integrals can be used to find the value of the potential.
EngageEngage
Conservative Vector Fields -- Is this right?

## Homework Statement

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

## 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

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

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

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.

I get -e as well.

Thanks a lot, I appreciate the help!

## 1. What is a conservative vector field?

A conservative vector field is a type of vector field in which the line integral of the field over any closed path is equal to zero. This means that the work done by the field is independent of the path taken.

## 2. How do you determine if a vector field is conservative?

A vector field is conservative if it satisfies the condition of being path-independent, meaning that the line integral of the field over any closed path is equal to zero. This can be determined using the gradient theorem or by checking if the field satisfies the curl-free condition (curl = 0).

## 3. What are some real-world applications of conservative vector fields?

Conservative vector fields have many applications in physics, such as in electrostatics and gravitational fields. They can also be used in fluid mechanics to model the flow of a fluid without energy loss.

## 4. Can a vector field be both conservative and non-conservative?

No, a vector field cannot be both conservative and non-conservative. This is because the definition of a conservative vector field is that the line integral over any closed path is equal to zero, while a non-conservative field will have a non-zero line integral over some closed paths.

## 5. How can conservative vector fields be visualized?

Conservative vector fields can be visualized using vector field plots or streamlines. In a vector field plot, the vectors represent the direction and magnitude of the field at different points, while in a streamline plot, the lines represent the path of a particle moving in the field. These visualizations can help in understanding the behavior of conservative vector fields.

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