# Vector Field Conceptual Question

Given that: $$F = \langle yz-2xy^2, axz-2x^2y+z, xy+y \rangle$$ in which $a$ is some constant.

Now, for what $a$ would make the vector field of $F$ conservative? Why is there only one, or are there many? How can we find an $f$ with $\nabla f=F$? Also for what $a$ would $F$ be the curl of another vector field?

For to find for which $a$ the vector field is conservative, do we have to go through the process of finding partial derivative, or is there a much shorter approach. I don't know how to approach my other questions either.

• "long process of finding partial derivative". It's really so long for you? You want to find some magic method? – Michael Galuza Aug 3 '15 at 17:03
• @MichaelGaluza Wrong choice of words. Sorry – gauss's son Aug 3 '15 at 17:04
• a necessary condition: the Jacobian of $F$ is symmetric. – user251257 Aug 3 '15 at 17:04
• or just integrate – user251257 Aug 3 '15 at 17:04
• @user251257 Can you show what you mean? – gauss's son Aug 3 '15 at 17:05

## 2 Answers

$\newcommand{\i}{\hat{\mathbf{i}}} \newcommand{\j}{\hat{\mathbf{j}}} \newcommand{\k}{\hat{\mathbf{k}}} \renewcommand{\p}{\partial\,} \renewcommand{\b}{\begin} \renewcommand{\e}{\end} \renewcommand{\R}{\mathbb R}$

## 1. Determine if $\boldsymbol F$ is conservative

HINT: Recall for any scalar vector field $f$ is the curl of its gradient is zero: $\nabla \times \big( \nabla f \big) = 0$.

Since $F$ is conservative iff $\ \exists\; f : \mathbb R^3 \to \mathbb R$ such that $F = \nabla f$, we can easily check if such $f$ exists by taking curl of $F$ and comparing it to zero

If you are not convinced in the accuracy of the last statement, I propose to address something like this website.

Indeed,

• $\Rightarrow$ If $F$ is conservative, then $\exists$ a scalar field $f$ s.t. $F = \nabla f$. Taking curl of both sides of the last equality, we get $\nabla \times F = \nabla \times \big(\nabla f\big) = 0$. Thus we established that $$\boxed{ \ F \ \text{ is conservative } \implies \nabla \times F = 0 \ }$$
• $\Leftarrow$ If $\nabla \times F = 0$, then $F$ is conservative, because a vector field is conservative if and only if it is irrotational, which equivalent to $F$ having zero curl in 3D space. Therefore we can write $$\boxed{ \ \nabla \times F = 0 \implies \exists \, f \ \text{ s.t. } \ F = \nabla f \iff F \ \text{ is conservative } \ }$$

### 1.1 Solution

Now, let us use these facts in order to determine value of parameter $a$ for which your vector field $$F = \begin{bmatrix} F_x \\ F_y \\ F_z \end{bmatrix} = \begin{bmatrix} yz-2xy^2 \\ axz-2x^2y+z \\ xy+y \end{bmatrix}$$ is conservative. In other words, we need to find values of $a$ such that the curl of $F$ will be zero.

Denote $\i, \j, \k$ orthonormal basis vectors of $\mathbb R^3$. Then the curl of $F$ can be computed as

$$\nabla \times F = \begin{bmatrix} \i & \j & \k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ F_x & F_y & F_z \end{bmatrix} = \left(\dfrac{\partial F_z}{\partial y}-\dfrac{\partial F_y}{\partial z}\right)\i + \left(\dfrac{\partial F_x}{\partial z}-\dfrac{\partial F_z}{\partial x}\right)\j + \left(\dfrac{\partial F_y}{\partial x}-\dfrac{\partial F_x}{\partial y}\right)\k.$$ Let us compute partial derivatives of $F_x, F_y, F_z$ components of $F$: \begin{aligned} \dfrac{\partial F_x}{\partial y} & = z - 4 xy, & \dfrac{\partial F_x}{\partial z} & = y \\ \dfrac{\partial F_y}{\partial x} & = a z - 4 x y & \dfrac{\partial F_y}{\partial z} & = a x + 1 \\ \dfrac{\partial F_z}{\partial x} & = y & \dfrac{\partial F_z}{\partial y} & = x + 1 \end{aligned}

Substituting it into the $\nabla \times F$ formula, we get $$\nabla \times F = \Big(\left(x+1\right) - \left(ax + 1\right) \Big)\i + \Big( y - y \Big)\j + \Big(\left( az - 4 xy \right) - \left(z - 4 xy\right) \Big)\k = \begin{bmatrix} \left(1-a\right) x \\ 0 \\ \left( a - 1 \right) z \end{bmatrix}$$

If $F$ is a conservation field, then $\nabla \times F = 0$ for any $(x,y,z) \in \mathbb R^3$, so we write

$$\nabla \times F = \begin{bmatrix} \left(1-a\right) x \\ 0 \\ \left( a - 1 \right) z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \ \forall \, x,y,z \in \mathbb R \implies \boxed{ \ a - 1 = 0 \ }$$

Therefore we finally conclude that $F$ is conservation field if and only if $a = 1$.

## 2. Find potential function for $\boldsymbol F$

In the section above we established that $F$ is conservative for $a=1$, i.e. that there exists a potential function $f:\Bbb R^3 \to \Bbb R$ such that $$\nabla f = \left[\begin{matrix} \frac{\p f}{\p x} \\ \frac{\p f}{\p x} \\ \frac{\p f}{\p x} \end{matrix}\right] = F\vert_{a=1} = \left.\left[\begin{smallmatrix} F_x \\ F_y \\ F_z \end{smallmatrix}\right]\right|_{a=1} = \begin{bmatrix} yz-2xy^2 \\ xz-2x^2y+z \\ xy+y \end{bmatrix}$$

Thus, one can find potential function $f$ by integrating components of $F$ and matching results.

### 2.1 Solution

First, integrate $F_x = \dfrac{\p f}{\p x}$ with respect to variable $x$. \begin{aligned} \frac{\p f}{\p x} &= yz - 2xy^2 &&\implies& f(x,y,z) &= yz - x^2 y^2 + h_(y,z), \end{aligned} where $h$ is unknown functions of $y$ and $z$.

Second, differentiate obtained expression w.r.t. $y$ and compare to $F_y$:

\begin{aligned} \frac{\p f}{\p y} = F_y & \implies \frac{\p }{\p y} \Big(yz - x^2 y^2 + h(y,z)\Big) =z - 2x^2 y + \frac{\p }{\p y} h(y,z) = axz - 2 x^2y + z \\ & \implies \frac{\p }{\p y} h(y,z) = axz \implies h(y,z) = axyz + g(z) \\ & \implies f(x,y,z) = yz - x^2 y^2 + axyz + g(z) \end{aligned} where $g$ is unknown functions of $z$.

Third, we differentiate obtained expression for $f$ w.r.t. $z$ and match it to $F_z$:

\begin{aligned} \frac{\p f}{\p z} = F_z & \implies \frac{\p }{\p z} \big(yz - x^2 y^2 + xyz + g(z)\big) = y + xy + \frac{\p }{\p z} g(z) = xy + y \\ & \implies \frac{\p }{\p z} g(z) = 0 \implies g(z) = c \ \ - \operatorname{const} \\ & \implies f(x,y,z) = yz - x^2 y^2 + xyz + c \end{aligned} where $c\in \R$ is an arbitrary constant.

Thus we have obtained the potential of $F$:

$$\boxed{ \ \ f = (1-x) yz - x^2 y^2 + c \ \ }$$

## 3. Determine if $\boldsymbol F$ is a curl of another vector field

HINT: Recall that the divergence of a curl of any vector field is zero: $\forall \ A:\R^3 \to \R^3 \quad \nabla \cdot \big( \nabla \times A\big) \equiv 0$.

Therefore in order to establish whether $F$ is a curl of another vector field it is sufficient to check whether $\nabla \cdot F = 0$ for all $(x,y,z) \in \R^3$.

### 2.1 Solution

All we need to do is to compute divergence of $F$, and find out if there values of $a$ for which $\nabla \cdot F \equiv 0$.

$$\nabla\cdot F = \dfrac{\p F_x}{\p x} + \dfrac{\p F_y}{\p y} + \dfrac{\p F_z}{\p z} = \dfrac{\p }{\p x} \Big(yz-2xy^2 \Big) + \dfrac{\p }{\p y} \Big( axz-2x^2y+z \Big) + \dfrac{\p }{\p z} \Big( xy+y \Big) = -2y^2 - 2 x^2 + 0 \not \equiv 0$$

Therefore we conclude that $F$ is not a curl of any other vector field.

• For the question, "is there any value of a such that F is the curl of some other vector field" is it for any value of a? – gauss's son Aug 3 '15 at 19:22
• Can you help me with this? – gauss's son Aug 5 '15 at 1:16
• @gauss'sson The easiest way to answer this question is to use the fact that the divergence of a curl of any vector field is zero: $\ \forall \ A:\mathbb R^3 \to \mathbb R^3 \ \ \nabla \cdot \big( \nabla \times A \big) \equiv 0$. Can you pick it up from here, or do you need me to write it out? – Vlad Aug 5 '15 at 4:18

In order to determine a so that $\vec{f}$ is conservative, you must find:

$$\oint \vec{F} \cdot d\vec{l} = 0$$

$$\int _{x_1} ^{x_2} F_x dx + \int _{y_1} ^{y_2} F_{y} dy + \int _{z_1} ^ {z_2} F_{z} dz = 0$$

For the second question:

$$\frac{\partial f}{\partial x} = F_x$$

$$\frac{\partial f}{\partial y} = F_y$$

$$\frac{\partial f}{\partial z} = F_z$$

Now you must solve the set of partial differential equation.

• Can you please expand on this? – gauss's son Aug 3 '15 at 17:17
• Is it clear now? – Gabs Aug 3 '15 at 17:48
• The third one isn't – gauss's son Aug 3 '15 at 17:51
• Are you having a conceptual problem or you can't find the differential equation solution (you know what to do but for some reason, you can't solve it)? – Gabs Aug 3 '15 at 17:53
• conceptual problem – gauss's son Aug 3 '15 at 17:53