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{\bf Question}

Find the work done in moving a particle in the force field $${\bf
F} = 3x^2 {\bf i} + (2xz - y){\bf j} + z{\bf k}$$
\begin{description}
\item[(a)] along the straight line from (0,0,0) to (2,1,3),
\item[(b)] along the space curve $x = 2t^2, \, y = t, \, z = 4t^2 - t,$
from $t = 0 $ to $ t = 1.$
\item[(c)] Is the work done independent of the path?  Explain.
\end{description}


\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
Straight line $x = 2t$, $y = t$, $z = 3t$   $0 \leq t \leq 1$

\begin{eqnarray*} {\rm Work\ done} & = & \int {\bf F} \cdot d{\bf r} \\ & = &
\int_0^1 {\bf F} \cdot \frac{d {\bf r}}{dt} dt \\ & = & \int_0^1
[3(2t)^2 {\bf i} + (2.2t.3t - t){\bf j} + 3t{\bf k}] \cdot [2{\bf
i} + {\bf j} + 3{\bf k}] dt \\ & = & \int_0^1 [24t^2 + 12t^2 - t +
9t] dt \\ & = & \left[8t^3 + 4t^3 \frac{1}{2}t^2 +
\frac{9}{2}t^2\right]_0^1
\\ & = & 8 + 4 + 4 \\ & = & 16 \end{eqnarray*}



\item[(b)]
\begin{eqnarray*} {\rm Work\ done} & = & \int {\bf F} \cdot d{\bf r} \\ & = &
\int_0^1 {\bf F} \cdot \frac{d {\bf r}}{dt} dt \\ & = & \int_0^1
\left[3((2t)^2)^2 {\bf i} + (2.2t^2(4t^2 - t).(- t)){\bf j} +
 (4t^2 - t){\bf k}\right]\\ & &\hspace{.3in} \cdot   [4t{\bf i} + {\bf j} +
(8t-1){\bf k}] dt
\\ & = & \int_0^1 \left[48t^5 + 16 t^4 - 4t^3 - t + (4t^t - t)(8t -1)\right] dt \\
& = & \left[8t^6 + \frac{16}{5}t^5 - t^4 - \frac{1}{2}t^2 +
\frac{1}{2}(4t^2 - t)^2\right]_0^1 \\ & = & 8 + \frac{16}{5} - 1 -
\frac{1}{2} + \frac{9}{2} \\ & = & 14\frac{1}{5} \end{eqnarray*}


\item[(c)]
No.  The force is not conservative.
\end{description}



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