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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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{\bf Question}

\begin{description}
\item[(a)] Find the constants a,b,c so that the force feild
defined by $$(x + 2y + az){\bf i} + (bx - 3y - z) {\bf j} + (4x +
cy + 2z) {\bf k}$$ is conservative.
\item[(b)] What is the potential associated with this force field
when $a,b,c$ are chosen to make it conservative.
\end{description}

\vspace{.25in}

{\bf Answer}

We require $\nabla \times {\bf F} = 0$

\begin{eqnarray*} 0 & = & \left| \begin{array}{ccc} {\bf i} & {\bf j}
& {\bf k} \\ \frac{\pl}{\pl x} & \frac{\pl}{\pl y} &
\frac{\pl}{\pl z} \\ (x + 2y + az) & (bx - 3y - z) & (4x + cy +
2z) \end{array} \right| \\ & = & {\bf i}(c +1) + {\bf j} (a - 4) +
{\bf k}(b - 2) \end{eqnarray*}

Therefore $a = 4, b = 2, c = -1.$

\begin{eqnarray*} {\bf F} & = & (x + 2y + 4z){\bf i} + (2x - 3y - z){\bf
j}+ (4x - 1y + 2z){\bf k} \\ & = & -\frac{\pl U}{\pl x}{\bf i} -
\frac{\pl U}{\pl y}{\bf j} - \frac{\pl U}{\pl z}{\bf k}
\end{eqnarray*}

$\begin{array}{rclr} \ds -\frac{\pl u}{\pl x} & = & -(x + 2y + 4z)
\\ \Rightarrow  U & = & -\frac{1}{2}x^2 - 2xy - 4xz + f_1(y,z)
&(1)
\\ \\\ds -\frac{\pl U}{\pl y} & = & -(2x - 3y -2)  \\ \Rightarrow  U & =
& -2xy + \frac{3}{2}y^2 + yz +f_2(x,z) & (2) \\ \\ \ds -\frac{\pl
U}{\pl z} & = & -(4x - y + 2z) \\ \Rightarrow  U & = & -4xz + yz +
\frac{1}{2} z^2 +f_3(x,z) & (3) \end{array}$

Comparing:

$\begin{array}{rcrcl} (1) {\rm \ and\ } (2) & \Rightarrow &
f_1(y,z) & = & \frac{3}{2}y^2 + yz + g(z) \\  & & f_2(x,z) & = &
-4xz -\frac{1}{2}x^2 g(z) \\  (2) {\rm \ and\ } (3) & \Rightarrow
& f_3(x,y) & = & -2xy + \frac{3}{2}y^2 - \frac{1}{2}x^2 , g(z) =
-\frac{1}{2}z^2 \end{array}$

Therefore $ U = -4x^2 + y^2 - 2xy - \frac{1}{2}x^2 +
\frac{3}{2}y^2 - \frac{1}{2} z^2$





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