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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Spring 2008 \hfil \pp
\hfil Mid-Term Exam Number Two\hfil  \pp
\hfil May 22, 2008 \hfil \pp
\hfil Answers \hfil \pp
\normalsize
\medskip

\begin{enumerate}

\item The plane is $-2x+7y-6z+5=0$.

\item (a) The point $(21,12,33)$ lies on both lines. 

(b) If $a=\frac{15}{2}$, the line will not intersect the plane.

\item (a) The point is the one given by $t=\frac{1}{4}$, i.e. $(\frac{1}{4},\frac{63}{16},-\ln 4)$.

(b) The line is $x=\frac{1}{4}+t, y=\frac{63}{16}-\frac{1}{2}t, z=-\ln 4+4t$.

\item The times are $\dsps t=\pm 9^{1/10}$.

\item The surface has two critical points, $(-\frac{1}{2}, 1)$, and $(\frac{1}{2}, -1)$.
They are both saddle points.

\item The curve has two points of maximum curvature:
\[\dsps \left( \left( \frac{1}{45} \right)^\frac{1}{4} , \left( \frac{1}{45} \right)^\frac{3}{4} \right)
\mbox{ and }
 \left( -\left( \frac{1}{45} \right)^\frac{1}{4} , -\left( \frac{1}{45} \right)^\frac{3}{4} \right)
\]
\end{enumerate}

\end{document}


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