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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Spring 2007 \hfil \pp
\hfil Mid-Term Exam Number Two\hfil  \pp
\hfil Answers \hfil \pp
\hfil May 10, 2007 \hfil \pp
\normalsize

\begin{enumerate}

\item 
\begin{enumerate}
\item 
\[
T(1) = \langle \frac{4}{\sqrt{257}},\frac{15}{\sqrt{257}},\frac{4}{\sqrt{257}} \rangle
\]
\item One solutions is
\[
x = 4+4t, y = 5+15t, z=2+4t
\]
\end{enumerate}

\item No.  Basically you can show that if there was an intersection, it would
occur when $\sin \theta=1$.  However, there is no point on the curve where
$\sin \theta = 1$, so there is no intersection.

\item 
\begin{enumerate}
\item
\[
a_T = \frac{2}{\sqrt{6}}
\]
\item 
\[
t = \pm 2^{-1/6}
\]
\end{enumerate}

\item You can show that 
\[
\kappa = \frac{10}{(8t^2-4t+13)^{3/2}}
\]
This is maximal when $8t^2-4t+13$ is minimal, i.e., at $t=1/4$.

\item 
\begin{enumerate}
\item The region above the line $y=-x$ (the line itself is not included).

\item $\dsps f_{xy} (x,y) = e^y + \frac{1}{(x+y)^2}$.

\end{enumerate}


\end{enumerate}

\end{document}


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