# Determine whether the vector field F (x, y, z) = (4 x y + z^2) i + (2 x^2 + 6 y z) j + 2 x z k is...

## Question:

Determine whether the vector field {eq}\displaystyle F (x,\ y,\ z) = (4 x y + z^2)\mathbf i + (2 x^2 + 6 y z) \mathbf j + 2 x z\ \mathbf k {/eq} is conservative. If it is, find a potential function for the vector field.

## Conservative Vector Field:

The conservative vector field also known as path-independent vector field.

If {eq}\vec{F} {/eq} is a vector field defined on all of {eq}R^3 {/eq} whose component functions have continuous partial derivatives and {eq}curl ( \vec{F} ) =0 {/eq}, then {eq}\vec{F} {/eq} is conservative field.

The curl of the vector function {eq}\vec{F}(x,y,z)=f_1 i + f_2 j +f_3 k {/eq} is deined by

{eq}curl(\vec{F})=\begin{vmatrix} i &j &k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial }{\partial z} \\ f_1 &f_2 &f_3 \end{vmatrix} {/eq}.

#### Potential Function for any Field:

Let {eq}f(x,y,z) {/eq} be any given scalar function, then gradient of {eq}f {/eq} is given by,

{eq}grad(f) = \nabla f = \dfrac{{\partial f}}{{\partial x}}\mathbf{i} + \dfrac{{\partial f}}{{\partial y}}\mathbf{j}+\dfrac{{\partial f}}{{\partial z}}\mathbf{k} =\left\langle\dfrac{{\partial f}}{{\partial x}}, \dfrac{{\partial f}}{{\partial y}}, \dfrac{{\partial f}}{{\partial z}} \right\rangle. {/eq}

Whenever the {eq}grad(f) {/eq} is given in the question. We have to use the concept of partial integration and partial derivatives to get involved in the problems and solve them accordingly.

That is if:

{eq}grad(f)=\langle F_{1}, F_{2},F_{3}\rangle=\left\langle \dfrac{{\partial f}}{{\partial x}}, \dfrac{{\partial f}}{{\partial y}},\dfrac{{\partial f}}{{\partial z}}\right\rangle {/eq} ,then we can compare it as:

{eq}F_{1}=\dfrac{{\partial f}}{{\partial x}} ,F_{2}=\dfrac{{\partial f}}{{\partial y}},F_{3}=\dfrac{{\partial f}}{{\partial z}}. {/eq}

Hence then we can solve these equations accordingly.

## Answer and Explanation: 1

The curl of the vector field is given as:

{eq}\begin{array}{*{20}{l}} {\begin{array}{*{20}{l}} {Curl\left( {\bf{F}} \right) = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right|} \end{array}}\\ { = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {4xy + {z^2}}&{2{x^2} + 6yz}&{2zx} \end{array}} \right|}\\ { = \left\langle {\left( {\frac{\partial }{{\partial y}}\left( {2zx} \right) - \frac{\partial }{{\partial z}}\left( {2{x^2} + 6yz} \right)} \right),\left( {\frac{\partial }{{\partial z}}\left( {4xy + {z^2}} \right) - \frac{\partial }{{\partial x}}\left( {2zx} \right)} \right),\left( {\frac{\partial }{{\partial x}}\left( {2{x^2} + 6yz} \right) - \frac{\partial }{{\partial y}}\left( {4xy + {z^2}} \right)} \right)} \right\rangle }\\ { = \left\langle {0 - 6y,2z - 2z,4x - 4x} \right\rangle }\\ { = \left\langle { - 6y,0,0} \right\rangle }\\ { \ne \vec 0} \end{array} {/eq}

Hence the given vector field is not conservative.

So we can't calculate the Potential function in this case.

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Chapter 5 / Lesson 8Learn how to tell if a force is conservative and what exactly is being conserved. Then look at a couple of specific examples of forces to see how they are conservative.