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\noindent
\vfil \noindent
\large
\hfil Math 126 C - Spring 2010 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil April 20, 2010 \hfil \pp
\hfil Answers \hfil \pp
\normalsize



\begin{enumerate}

\item \textit{Determine whether or not the line
\[
x = 4t-7, y= 5t-16, z=-2t+14
\]
and the line
\[
x =t+7, y=-3t-7, z=7t+22
\]
intersect.  If they do, give the point of intersection.
}

The lines intersect at the point (5,-1,8), corresponding to 
$t=3$ for the first line and $t=-2$ for the second line.


\item \textit{Let $P$ be the plane containing the points
$(1,5,2), (2,3,6)$ and $(7,4,1)$.  
Find the intersection of $P$ with the $y$-axis.}

The plane $P$ is given by
\[
6x+25y+11z=153.
\]
The $y$-axis consists of all points satisfying
\[
x=0, z=0
\]
so the intersection with the $y$-axis is the point $x=0, z=0$ and
\[
25y=153
\]
i.e., the point $\dsps (0,\frac{153}{25},0)$.

\item \textit{Consider the polar curve
\[
r = \sin \theta \tan \theta.
\]
}

\begin{enumerate}

\item \textit{Find an equivalent cartesian equation for this curve.}
\item \textit{The curve has a vertical asymptote.  What is the equation
of the asymptote?}
\end{enumerate}



(a) An equivalent cartesian equation is
\[
x(x^2+y^2)=y^2.
\]
(b) Rearranging, we have
\[
y^2 = \frac{-x^3}{x-1}
\]
We see that the right-hand side is unbounded as $x$ approaches 1; hence the curve
has a vertical asymptote at $x=1$.

\item \textit{Let $S$ be the surface in 3D consisting of all points which are twice
as far from the $z$-axis as they are from the $x$-axis.}

\begin{enumerate}
\item \textit{Give an example of a point on this surface, other than the origin.}

\item \textit{Give an equation for this surface.}

\item \textit{Describe this surface (if it is a quadric surface, categorizing it (i.e., 
ellipsoid, eliptic paraboloid, etc.) is sufficient).}
\end{enumerate}


(a) The point $(2,0,1)$ is such a point.
(b) From 
\[
\sqrt{x^2+y^2} = 2\sqrt{y^2+z^2}
\]
we can arrive at
\[
x^2-3y^2-4z^2=0
\]
(c) The surface is a quadric surface.
Setting $y=0$ or $z=0$, we see the traces are pairs of degenerate hyperbolas.
With $x$ set to a constant, we see traces which are ellipses.  We may conclude 
that the surface is a cone.

\item \textit{Let $P$ be the point in the first quadrant on the curve}
\[
x = \cos t, y = \csc t
\]
\textit{such that the tangent line to the curve at $P$ passes
through the origin.  Find the coordinates of $P$.}

\epsfig{file=graph01.eps, 
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By setting $\frac{dy}{dx}=\frac{y}{x}$ we find
\[
\frac{\cos t}{\sin^3 t}=\frac{1}{\cos t \sin t}
\]
which gives us
\[
\cos^2 t = \sin^2 t.
\]
This yields the solution
\[
t = \frac{\pi}{4}
\]
and so 
\[
P = \left( \frac{\sqrt{2}}{2},\frac{2}{\sqrt{2}} \right).
\]


\end{enumerate}


\end{document}


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