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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Spring 2007 \hfil \pp
\hfil Mid-Term Exam Number One Solutions\hfil  \pp
\hfil April 19, 2007 \hfil \pp
\normalsize

\begin{enumerate}

\item \textit{Let $f(x) = e^x \sin x$.}

\begin{enumerate}
\item \textit{Find the second-order Taylor polynomial $T_2(x)$ for $f(x)$ based at $b=0$.}

\[
T_2(x) = x+x^2
\]

\item \textit{Give a bound on the error $|f(x)-T_2(x)|$ for $x$ in the interval $-0.1 \le x \le 0.1$.}

We may use the fact that 
\[
|f'''(x)| = |2e^x(\cos x - \sin x)| \le 4e^x \le 4e^{0.1} 
\]
on the interval $-0.1 \le x \le 0.1$, so the error is no more than
\[
\frac{4e^{0.1}}{3!}|0.1|^3 = 0.005894.... .
\]

\end{enumerate}



\item \textit{Find the first four non-zero terms of the Taylor series for}
\[
f(x) = xe^{x^2} - \frac{1}{4+x^2}
\]
\textit{based at $b=0$.}

Since 
\[e^x = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
\]
for all $x$, we have
\[
e^{x^2} = 1+ x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots
\]
and
\[
xe^{x^2} = x + x^3 + \frac{x^5}{2!} + \frac{x^7}{3!} + \cdots
\]

Also,
\[
\frac{1}{4+x^2} = \left( \frac{1}{4} \right) \frac{1}{1- \left( - \left( \frac{x}{2} \right)^2 \right)}
= \frac{1}{4}  \sum_{k=0}^{\infty} \left( -\left( \frac{x}{2} \right)^2 \right)^k
= 
\frac{1}{4} - \frac{x^2}{4 \cdot 4} + \frac{x^4}{4\cdot 2^4} - 
\frac{x^6}{4 \cdot 2^6} + \cdots
\]

Combining these results yields
\[
f(x) = -\frac{1}{4}+x + \frac{x^2}{4 \cdot 4} + x^3 + \cdots.
\]


\item \textit{Find the equation of the plane containing the line of intersection of the two planes}
\[
x+y+z+5=0 
\mbox{ and }
3x+2y-z+2=0
\]
\textit{and the point $(1,2,1)$.}

We find the direction vector of the line of intersection: it is the cross product
of the planes' normal vectors
\[
\langle 1,1,1 \rangle \times \langle 3,2,-1 \rangle
= \langle -3,4,-1 \rangle
\]
Then we need a point on the line (which is hence on the plane we seek).  Choosing
$x=0$, we can find the point
\[
\left( 0, -\frac{7}{3}, - \frac{8}{3} \right)
\]
on the line.  The vector extending from this point to $(1,2,1)$ (the other point we know is
on the plane) is
\[
\langle 1, \frac{13}{3}, \frac{11}{3} \rangle
\]
or, equivalently (since direction is all we need here)
\[
\langle 3, 13, 11 \rangle.
\]
The normal vector of the plane we seek is then
\[
\langle 3, 13, 11 \rangle \times \langle -3,4,-1 \rangle =
\langle -57, -30, 51 \rangle
\]
and so the equation of the plane is
\[
-57x-30y+51z + d = 0
\]
Since $(1,2,1)$ is on the line, we can find $d=66$, so the plane is
\[
-57x-30y+51z+66=0.
\]




\item \textit{Find the point of intersection of the two lines}
\[
x=4-t, y=6+2t, z=-1+3t 
\mbox{ and }
x=1+2t, y=14-8t, z=7-4t.
\]

Suppose $(a,b,c)$ is the point of intersection.
Then there is a $t$ value, say $t_1$ such that
\[
a=4-t_1, b=6+2t_1, c=-1+3t_1
\]
and a $t$ value, say $t_2$ such that
\[
a=1+2t_2, b = 14-8t_2, c=7-4t_2.
\]
Hence, we have
\[
4-t_1=1+2t_2
\]
and
\[
6+2t_1 = 14-8t_2
\]
This is a pair of equations in two unknowns which we can solve to get
$t_1= 2$ and $t_2=\frac{1}{2}$.  This then gives us the point of intersection
\[
(4-2,6+2\cdot 2, -1+3 \cdot 2) = (2, 10, 5).
\]


\item \textit{Let $S$ be the surface defined as the set of points $p$ 
(in three-dimensional space) such that the
distance from $p$ to the plane $y=5$ equals the distance from $p$
to the line }
\[
y=1, z=2.\]

\begin{enumerate}
\item \textit{Find an equation for $S$.}

The distance from a point $(x,y,z)$ to the plane $y=5$ is
\[
|y-5|
\]
and the distance from $(x,y,z)$ to the line $y=1, z=2$ is
\[
\sqrt{(y-1)^2+(z-2)^2}
\]
so an equation of the surface (i.e., an equation that is satisfied by 
all points on the surface) is
\[
|y-5| = \sqrt{(y-1)^2+(z-2)^2}.
\]
After squaring the equation
\[
(y-5)^2 = (y-1)^2 + (z-2)^2
\]
we can simplify it to
\[
-8y+24 = (z-2)^2
\]

\item \textit{Find the equation of the trace of $S$ in the plane $z=6$.  Describe the trace (i.e. 
what kind of curve is it?).}

If $z=6$ then the equation of the surface simplifies to
\[
-8y+24 = 4^2 = 16
\]
i.e.
\[
y = 1
\]
In the plane $z=6$, the set of points satisfying the equation $y=1$ is a \textbf{line}.


\end{enumerate}


\end{enumerate}

\end{document}


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