Consider the vector \mathbf F(x,y,z) = (2x+4y)\mathbf i +(4z+4x) \mathbf j + (4y + 2x) \mathbf k . a. Find a function f such that \mathbf F = \nabla f and f(0,0,0) = 0 . b. Suppose C is...

Consider the vector   \mathbf F(x,y,z) = (2x+4y)\mathbf i +(4z+4x) \mathbf j + (4y + 2x) \mathbf k . a. Find a function   f  such that   \mathbf F = \nabla f  and  f(0,0,0) = 0 . b. Suppose   C  is...

Exploring the Properties of a Vector Field

In this article, we will delve into the properties of a vector field and investigate its relationship with a potential function. Specifically, we will consider the vector field:

\mathbf F(x,y,z) = (2x+4y)\mathbf i +(4z+4x)\mathbf j + (4y + 2x)\mathbf k

Part a: Finding a Potential Function

Our first task is to find a function f such that the given vector field \mathbf F is the gradient of f, that is, \mathbf F = \nabla f. To do this, we need to find the partial derivatives of f with respect to x, y, and z, and then equate them to the corresponding components of the vector field.

Taking the partial derivatives, we have:

\frac{\partial f}{\partial x} = 2x + 4
\frac{\partial f}{\partial y} = 4y
\frac{\partial f}{\partial z} = 4z

Equating these to the components of the vector field, we get:

2x + 4y = 2x + 4
4z + 4x = 4z + 4x
4y + 2x = 4y + 2x

Integrating these equations, we obtain the potential function f(x,y,z):

f(x,y,z) = x^2 + 2xy + 2z^2 + 2xz + C

where C is an arbitrary constant. Since the problem statement requires that f(0,0,0) = 0, we set C = 0, and the final potential function is:

f(x,y,z) = x^2 + 2xy + 2z^2 + 2xz

Part b: Investigating the Vector Field

Now, suppose that C is a closed curve in the three-dimensional space. We can evaluate the line integral of the vector field \mathbf F along the curve C using the fundamental theorem of calculus for line integrals:

\int_C \mathbf F \cdot d\mathbf r = \int_C \nabla f \cdot d\mathbf r = f(B) - f(A)

where A and B are the starting and ending points of the curve C, respectively.

This means that the line integral of the vector field \mathbf F along any closed curve C is equal to the difference in the values of the potential function f evaluated at the starting and ending points of the curve. If the curve C is closed, then A = B, and the line integral will be zero.

This property of the vector field \mathbf F being the gradient of a potential function f is known as path independence. It means that the line integral of \mathbf F along any path connecting two points depends only on the endpoints and not on the specific path taken.

In conclusion, we have found a potential function f such that the given vector field \mathbf F is the gradient of f, and we have discussed the implications of this property for line integrals along closed curves.

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