\documentclass[12pt]{report} 

\usepackage{palatino}
\usepackage{epsfig}

\pagestyle{empty} %%% this results in no pagenumbers (footer is empty}

\addtolength{\oddsidemargin}{-1.0in}
\addtolength{\evensidemargin}{-1.0in}
\addtolength{\textwidth}{1.5in}
\addtolength{\topmargin}{-0.5in}
\addtolength{\textheight}{1.0in}

\baselineskip=20pt

\newcommand{\dsps}{\displaystyle}
\newcommand{\pp}{\par \noindent}
\newcommand{\newp}{\vfil \eject}
%\newcommand{\newp}{\bigskip}

\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Winter 2006 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil January 31, 2006 \hfil \pp
\hfil Solutions \hfil \pp
\normalsize


\begin{enumerate}


\item It is necessary to find the first two derivatives of $f(x)$:
\[
f'(x) = \frac{1}{x \ln x}
\]
\[
f''(x) = - \frac{1+ \ln x}{(x \ln x)^2}
\]

Evaluating $f,f',$ and $f''$ at $x=e$ gives the coefficients of
$T_2(x)$ for $f(x)$:
\[
T_2(x) = \frac{1}{e}(x-e) -\frac{1}{e^2} (x-e)^2
\]


\item 

The first four non-zero terms of the Taylor series for $\sin x$ are
\[
x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}
\]
so the first four non-zero terms of the Taylor series for $\sin x^2$ are 
\[
x^2- \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!}
\]
Integrating this from 0 to 2 gives
\[
\left. \left( \frac{1}{3}x^3 - \frac{1}{7 \cdot 3!} x^7 + \frac{1}{11 \cdot 5!} x^{11} 
- \frac{1}{15 \cdot 7!} x^{15} \right) \right|_0^2
=
\frac{8}{3}-\frac{128}{42}+\frac{2048}{1320} -\frac{32768}{75600}
=
0.7371236.
\]



\item (Version 1) We use 
\[
\frac{1}{1-x} = 1+x+x^2+x^3+ ...
\]
so that
\[
\frac{1}{1-(-5x)} = 1 - 5x + (5x)^2 - (5x)^3  + ...
\]
and
\[
\frac{1}{3+x} = \left( \frac{1}{3} \right) \frac{1}{1-(-\frac{1}{3}x)}
= \frac{1}{3} \left( 1 - \frac{1}{3}x + 
\left( \frac{1}{3}x \right)^2 
- \left( \frac{1}{3}x \right)^3 + ... \right)
= \frac{1}{3} - \frac{1}{9} x + \frac{1}{27} x^2 -\frac{1}{81} x^3 + ...
\]
Adding these together, we have
\[
\frac{1}{1+5x} + \frac{1}{3+x} =
\frac{4}{3} - \frac{46}{9} x + \frac{676}{27} x^2 - \frac{10126}{81} x^3 + ...
\]

(Version 2) Using a similar technique,

\[
\frac{1}{1+7x} = 1 -7x + (7x)^2 - (7x)^3 + ...
\]
and
\[
\frac{1}{2+x} = \left( \frac{1}{2} \right)
\frac{1}{1 + \frac{x}{2}} 
=
\frac{1}{2} \left( 1 + \frac{x}{2} +\left( \frac{x}{2} \right)^2 + \left( \frac{x}{2} \right)^3
+ ...\right)
\]
Adding these together, we have
\[
\frac{1}{2+x} + \frac{1}{1+7x}
=
\frac{3}{2} - \frac{27}{4} x + \frac{393}{8} x^2 - \frac{5489}{16} x^3 + ...
\]


\item (Version 1) 
\[
\theta = \cos^{-1} \left( \frac{4}{\sqrt{26} \sqrt{11}} \right)
= 1.80958408790828020.
\]
(Version 2)
\[
\theta = \cos^{-1} \left( \frac{2}{\sqrt{89}\sqrt{69}} \right)
=
1.5452718055317992.
\]

\item Since the vectors are orthogonal,
\[
<x,3,2>\cdot<2,3,x> = 0
\]
so 
\[
2x+9+2x=0
\]
from which we conclude that $x=-\frac{9}{4}$.


\item Finding vectors parallel (or ``in") the plane, we get
\[
\vec{a} = <4,3,-3> \mbox{ and } 
\vec{b} = <1,-12,6>
\]
Their cross product is a normal vector to the plane:
\[
\vec{a} \times \vec{b} = <-18,-27,-51>
\]
so the equation of the plane is
\[
-18(x+3)-27(y-4) -51z = 0.
\]




\item (Version 1)
The direction vector of the line is orthogonal to the normal
vectors of both planes so is parallel to 
\[
<1,1,-1> \times <2,-3,4> = <1,-6,-5>
\]
This vector is parallel to the plane we're looking for.
To get another vector, we need another point in the plane.
We can get a point that's on the line, and hence on the plane,
by setting $x=0$ and solving.  We find $(0,17,14)$ is 
on the line, and on the plane.
The vector extending from this point to $(-2,7,3)$ is 
\[
<-2,-10,-11>
\]
and so the normal vector to the plane we seek is parallel to
\[
<-2,-10,-11> \times <1,-6,-5> 
=
<-16,-21,22>
\]
So, the equation of the plane is
\[
-16(x+2)-21(y-7)+22(z-3)=0
\]

(Version 2) The direction vector of the line is orthogonal to the normal
vectors of both planes so is parallel to 
\[
<1,1,-1> \times <2,-3,4> = <1,-6,-5>
\]
This vector is parallel to the plane we're looking for.
To get another vector, we need another point in the plane.
We can get a point that's on the line, and hence on the plane,
by setting $x=0$ and solving.  We find $(0,22,16)$ is 
on the line, and on the plane.
The vector extending from this point to $(-2,7,3)$ is 
\[
<-2,-15,-13>
\]
and so the normal vector to the plane we seek is parallel to
\[
<1,-6,-5> \times <-2,-15,-13>
=
<3,23,-27>
\]
So, the equation of the plane is
\[
3x+23(y-22)-27(x-16)=0.
\]




\end{enumerate}




\end{document}


% \epsfig{file=enclosure01.eps, 
% width=11cm,
% angle=0 }