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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Spring 2009 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil April 21, 2009 \hfil \pp
\hfil Solutions \hfil \pp
\normalsize

\begin{enumerate}

\item 

\begin{enumerate}

\item {\textit{Find the equation of the plane $P$ containing the point (1,2,3) which is parallel
to the plane containing the points (0,3,4), (3,2,1), and (5,4,2).}}

We first find two vectors extending between two pairs of points given.  

Two such vectors are $\langle 3,-1,-3 \rangle$ and $\langle 5, 1, -2 \rangle$.

Taking their cross product we have the vector $\langle 5, -9, 8 \rangle$.

This is the normal vector to the plane which is parallel to the plane $P$, and hence
is the normal vector for the plane $P$.

Since $P$ contains the point $(1,2,3)$, an equation for $P$ is
\[
5(x-1)-9(y-2)+8(z-3)=0
\]
which can be, optionally, simplified to
\[
5x-9y+8x=11.
\]

\item {\textit{Give an example of a line contained in plane $P$.}}

There are very many reasonable approaches to this problem.  One is to take the vector
$\langle 3,-1,3 \rangle$, known to be parallel to $P$, as the direction vector for the line.
Then, taking the point $(1,2,3)$, known to be in $P$, we have the line
\[
x=1+3t, y=2-t, z=3+3t
\]


\end{enumerate}


\item {\textit{Thoroughly describe the surface defined as the set of points which are twice as far from the $z$-axis
as they are from the $xy$-plane.  }}

The distance from the point $(x,y,z)$ to the $xy$-plane is $|z|$.

The distance from the point $(x,y,z)$ to the $z$-axis is $\sqrt{x^2+y^2}$.

Thus, the surface defined is the set of points satisfying the equation
\[
\sqrt{x^2+y^2} = 2|z|.
\]
Squaring this equation yields
\[
x^2+y^2=4z^2
\]
or
\[
\frac{x^2}{4}+\frac{y^2}{4}-z^2=0.
\]
This we recognize as the equation of a cone.  This is a cone with apex
at the origin and axis the $z$-axis. Traces parallel to the $xy$-plane are circles, while
traces parallel to the $xz$-plane or the $yz$-plane are hyperbolas, except
for those passing through the origin (such traces are pairs of lines through
the origin).


\item {\textit{The curve defined by the polar equation $r=\sin^2 \theta$ is shown in the figure below.}}

\epsfig{file=126mt1-1.ps, 
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\begin{enumerate}
\item {\textit{Find the slope of the tangent line to the curve at the point 
where $\dsps \theta=\frac{\pi}{4}$.}}

We have $x=r \cos \theta = \cos \theta \sin^2 \theta$ and
$y=r \sin \theta = \sin^3 \theta$.  From this we are able to conclude, after
simplification, that
\[
\frac{dy}{dx} = \frac{3 \sin \theta \cos \theta}{2 \cos^2 \theta - \sin^2 \theta}.
\]
Taking $\theta = \frac{\pi}{4}$, we find 
\[
\frac{dy}{dx} = 3.
\]


\item {\textit{What is the maximum $x$-coordinate for a point on this curve?}}
We see, assisted by the figure, that where the maximal $x$-coordinate occurs, 
\[
\frac{dx}{d\theta} = 0.
\]
Thus we need to solve 
\[
\sin \theta (2 \cos^2 \theta - \sin^2 \theta) = 0.
\]
Since $\sin \theta=0$ results in $r=0$ and $x=0$, we need only concern ourselves with the other
factor.  Since
\[
2 \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - (1- \cos^2 \theta) = 3 \cos^2 \theta - 1
\]
we may conclude that 
\[
\cos \theta = \frac{\pm 1}{\sqrt{3}}.
\]
Since $x=\cos \theta (1- \cos^2 \theta)$,
we conclude that the maximum $x$-coordinate is 
\[
\frac{1}{\sqrt{3}} \left( 1- \frac{1}{3} \right) = \frac{2}{3 \sqrt{3}} \approx 0.384900179... .
\]


\end{enumerate}



\item {\textit{Where does the line which passes through the points $(0,5,-3)$ and $(1,2,8)$
intersect the plane $x-3y+4z=11$?}}

We may begin by finding parametric equations for the line.  This will do:
\[
x=t, y=5-3t, z=-3+11t.
\]
Then, we seek a solution to
\[
t - 3(5-3t)+4(-3+11t) = 11.
\]
The solution is $t=\frac{19}{27}$.
Hence the point is $\dsps \left( \frac{19}{27}, \frac{26}{9}, \frac{128}{27} \right)$.


\item {\textit{Consider the curve with the vector equation}}
\[
\vec{r}(t)= \langle t^2, 2t^2-t, 3t-t^2 \rangle
\]
{\textit{Is there a point on this curve where the tangent line is parallel to the
vector $\langle 20, 38, -14 \rangle$? If so, find the point.  If not, explain why.}}

The tangent vector is 
\[
\vec{r}(t) = \langle 2t, 4t-1, 3-2t \rangle .
\]

If this vector is parallel to $\langle 20, 38, -14 \rangle$ for some $t$, then there exists
a scalar $k$ such that
\[
2t = 20k \mbox{ and } 4t-1 = 38k.
\]
Solving this pair of equations simultaneously yields $k=1/2$ and $t=5$.

Checking the $z$-components, we find that $3-2(5) = -7 = \frac{1}{2}(-14)$, so the 
direction vector at $t=5$ is, indeed, $1/2$ times the vector $\langle 20, 38, -14 \rangle$,
and so the direction vector is parallel to $\langle 20,38,-14 \rangle$.

The sought point on the line is thus $\langle 25, 45, -10 \rangle$.




\end{enumerate}

\end{document}


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