# Conservative Vector Fields and Associated Potential

• sandy.bridge
In summary, the conversation is about finding the conservative vector field's potential and the attempt at solving it. The solution involves a scalar function, but there was a mistake in taking the partial with respect to z. The conversation ends with the issue being resolved.
sandy.bridge

## Homework Statement

Having issues determining what I am doing wrong, so perhaps one of you can pin point it. I have the solution, and I am extremely close to the same result, however, I am nonetheless wrong.

Find the conservative vector fields potential.

$$\vec{F}(x, y, z)=[(2xy-z^2), 2yz+x^2), y^2-2zx)]$$

## The Attempt at a Solution

$$\vartheta=\int(2xy-z^2)dx=x^2y-z^2x+C(y, z)$$
then we have
$$2yz+x^2=x^2+∂C(y, z)/∂y$$
therefore,
$$C(y, z)=zy^2+C(z)$$
It's at the following step that I mess something up.
$$y^2-2zx=y^2+∂C(z)/∂z$$
In the solutions however, they have,
$$y^2-2zx=y^2-2zx+∂C(z)/∂z$$
However, wouldn't the "-2zx" term go with ∂C(z)/∂z?

Your scalar function ##\phi = x^2y - z^2x + c(y,z)## Then take the partial of this with respect to ##z## and equate with the ##z## component of ##\underline{F}##. Did you mistakenly take the partial with respect to z of something else? You can write $$\frac{\partial c(z)}{\partial z} = c'(z).$$

I'm betting that you're going to kick yourself for this !

sandy.bridge said:

## Homework Statement

Having issues determining what I am doing wrong, so perhaps one of you can pin point it. I have the solution, and I am extremely close to the same result, however, I am nonetheless wrong.

Find the conservative vector fields potential.

$$\vec{F}(x, y, z)=[(2xy-z^2), 2yz+x^2), y^2-2zx)]$$

## The Attempt at a Solution

$$\vartheta=\int(2xy-z^2)dx=x^2y-z^2x+C(y, z)$$
then we have
$$2yz+x^2=x^2+∂C(y, z)/∂y$$
therefore,
$$C(y, z)=zy^2+C(z)$$
It's at the following step that I mess something up.
So, at this point you have $\displaystyle \ \ \vartheta=x^2y-z^2x+C(y, z)=x^2y-z^2x+zy^2+C(z)\ .$

Now look at $\displaystyle \ \ ∂\vartheta/∂z\ .$
$$y^2-2zx=y^2+∂C(z)/∂z$$
In the solutions however, they have,
$$y^2-2zx=y^2-2zx+∂C(z)/∂z$$
However, wouldn't the "-2zx" term go with ∂C(z)/∂z?

Perfect! Not entirely sure how I didn't see that. Figured I should ask rather than stare at it for long. Thanks!

## 1. What is a conservative vector field?

A conservative vector field is a type of vector field in which the line integral of the vector field along any closed path is always zero. This means that the work done by the vector field on a particle moving along a closed path is independent of the path taken.

## 2. What is the significance of conservative vector fields?

Conservative vector fields are important in physics and engineering because they represent physical quantities, such as force or velocity, that are dependent only on the position and not on the path taken. This allows for easier analysis and calculation of these physical quantities.

## 3. How can you determine if a vector field is conservative?

A vector field is conservative if its curl is equal to zero. Another way to determine if a vector field is conservative is to check if it satisfies the condition of path independence, meaning the line integral along any closed path is zero.

## 4. What is the relationship between conservative vector fields and potential functions?

Conservative vector fields are associated with potential functions, which are scalar functions whose gradient is equal to the vector field. In other words, the potential function for a conservative vector field can be found by taking the line integral of the vector field along any path.

## 5. How are conservative vector fields used in practical applications?

Conservative vector fields are used in many practical applications, such as in fluid dynamics, electromagnetism, and mechanics. They allow for easier analysis and calculation of physical quantities, as well as the prediction and understanding of the behavior of systems.

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