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\flushright Name
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\\
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\flushright Student ID \#
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\\
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\flushright Class (circle one) ~~~~~~~~~~~~~~ 9:40 ~~~~~~ 10:45\\
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%\flushright Class Section
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%\flushright Discussion Instructor
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\begin{center}

Math 1210\\
Fall 2009\\
K. M. Golden
\vspace{5ex}

{\bf EXAM II}\\
Friday, October 23, 2009
\vspace {15ex}

\begin{tabular}{|l|l|l|l|l|l|l|}
%\hline
%\multicolumn{2}{|c|}{Dept. Use Only}\\
%\multicolumn{2}{|c|}{Exam Scores}\\
\hline
Problem &Points & Score~~~~~~\\
\hline
1.~~ & 30 & \\
\hline
2.~~ & 25 & \\
\hline
3.~~ & 20 &\\
\hline
4.~~ & 15 &\\
\hline
5.~~ & 10 &\\
%\hline
%6.~~ & 10 &\\
%\hline
%7.~~ & 10 &\\
  \hline

&TOTAL&\\
\hline
\end{tabular}
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\eject

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\begin{enumerate}
\item[(30 points) ~~1.]  Calculate the following limits, and be sure to
      show all of your work.  
  If a particular limit does not exist, state this clearly and tell why.


\begin{enumerate}
\item[(a)] $\ds \lim_{x \rightarrow 2} 
                \frac{x^2-x-2}{x-2} = \lim_{x \rightarrow 2} \frac{(x-2)(x+1)}{x-2} = \lim_{x \rightarrow 2} (x+1) = 3$

%\item[(a)] $\ds \lim_{x \rightarrow 0^+} 
%                x\sin{\left(\frac{1}{x}\right)}$
\vfill
\item[(b)] $\ds \lim_{x \rightarrow +\infty} 
    \frac{7 x^5+\pi x^4-10^{20}x^3-x+1}{(x+1)^2(x+2)^3} = \lim_{x \rightarrow +\infty} \frac{7x^{5}}{x^{5}} = 7$
\vfill
\item[(c)] $\ds \lim_{x \rightarrow 0} \frac{1-\cos{x}}{x} $ \\
           $\ds = \lim_{x \rightarrow 0} \frac{1-\cos{x}}{x} \cdot \frac{1+\cos{x}}{1+\cos{x}} = \lim_{x \rightarrow 0} \frac{1-\cos^2{x}}{x(1+\cos{x})} = \lim_{x \rightarrow 0} \frac{\sin^2{x}}{x(1+\cos{x})} $\\
           $\ds = \lim_{x \rightarrow 0} \frac{\sin{x}}{x} \cdot \lim_{x \rightarrow 0} \frac{\sin{x}}{1+\cos{x}} = 0$
%      \left({\frac{1}{x+h}}-{\frac{1}{x}}\right)
%       \left(\frac{1}{h}\right)$
\vfill

\item[(d)]
$\ds \lim_{x \rightarrow 0} f(x)$, ~~~ 
where $f (x) = \left \{
\begin{array}{ll}
\sin{x}, & x < 0 \\
\cos{x}, & x \geq 0 
\end{array}
\right.$ \\
\vspace{2ex}
$\ds \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} \cos{x} = \cos{0} = 1$

$\ds \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \sin{x} = \sin{0} = 0$

Since $\ds \lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)$, the limit does not exist.

\vfill
\item[(e)] $\ds \lim_{x \rightarrow 0}\frac{\sin{(\pi x)}}{3x} $ \\
           $\ds = \lim_{x \rightarrow 0}\frac{\sin{(\pi x)}}{3x} \cdot \frac{\pi x}{\pi x} = \lim_{x \rightarrow 0}\frac{\sin{(\pi x)}}{\pi x} \cdot \frac{\pi x}{3x} = \lim_{x \rightarrow 0}\frac{\sin{(\pi x)}}{\pi x} \cdot \lim_{x \rightarrow 0}\frac{\pi x}{3x} $
           $\ds = \frac{\pi}{3}$
\vfill
\item[(f)]
$\ds \lim_{x \rightarrow 3} f(x)$, ~~~ 
where $f (x) = \left \{
\begin{array}{ll}
1, & x \;\; \mbox{rational} \\
0, & x \;\; \mbox{irrational} 
\end{array}
\right.$ \\
\vspace{1ex}
The limit does not exist. Because we have 
$\ds \lim_{x \rightarrow 3} f(x) = 1$ where $x$ is rational and $\ds \lim_{t \rightarrow 3} f(t) = 0$ where $t$ is irrational.

\end{enumerate}

\eject





\item[(25 points) ~~2.]  Find the following.  Be sure to
show all of your work.

\begin{enumerate}
\item[(a)] $\ds \frac{d}{dx}
           \left( x \sin{x} \right) = \sin{x} + x\cos{x}$ \\
           \vspace{3ex}
           Product Rule
\vfill
\item[(b)] $\ds \frac{d}{dx}
           \left(\frac{x}{1+x^2} \right) = \frac{1\cdot(1+x^{2}) - 2x \cdot x}{(1+x^{2})^2} = \frac{1-x^{2}}{(1+x^{2})^{2}}$ \\
           \vspace{3ex}
           Qutient Rule
\vfill
\item[(c)] $\ds \frac{d^2x}{dt^2}\;\;$ where $\;\;
          x(t) = \sin({ \sqrt{2}\;t})$ \\
          \vspace{1ex}
          $\ds \frac{dx}{dt} = \sqrt{2}\cos{(\sqrt{2}t)}$ \\
          \vspace{1ex}
          $\ds \frac{d^2x}{dt^2} = \sqrt{2} \left( \sqrt{2} \left(-\sin{(\sqrt{2}t)}\right)\right) = -2\sin{(\sqrt{2}t)}$
\vfill
\item[(d)] $\ds \frac{dm}{dv}\;\;$  % $\ds \frac{dm}{dv}\;,\;\;\;\;
            where $\;\; \ds 
       m(v)=m_0 \left( 1-\frac{v^2}{c^2}\right)^{-1/2}$
       
       $\ds \frac{dm}{dv} = -\frac{1}{2}m_{0} \left(1-\frac{v^{2}}{c^{2}}\right)^{-\frac{3}{2}} \cdot \left(-\frac{2v}{c^{2}}\right) = \frac{m_{0}v}{c^{2}}\left(1-\frac{v^{2}}{c^{2}}\right)^{-\frac{3}{2}}$ \\
       \vspace{3ex}
       Chain Rule
   %\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$
%\vfill
%\item[(e)] $\ds \lim_{x \rightarrow \infty} 
%             \frac{\sin{x}}{x}$
\vfill
\item[(e)] $\ds \frac{dy}{dx}\;\;$  % $\ds \frac{dm}{dv}\;,\;\;\;\;
            where $\; \ds x^2 + y^2 =1$ \\
            \vspace{1ex}
            By implicit differentiation, we have $\ds 2x + 2y\cdot\frac{dy}{dx} = 0$. Therefore, we get 
            $\ds \frac{dy}{dx} = -\frac{x}{y}$.
\end{enumerate}
\eject
\item[(20 points) ~~3.]  
Consider 
$f (x) = \left \{
\begin{array}{ll}
|x+1|, & x \leq 0 \\
5-(x-2)^2, & x > 0 
\end{array}
\right.,   \;\;$ 
defined on all of $\mathbb{R}$.

\begin{enumerate}
\item[(a)]  Sketch the graph of $f(x)$.
\begin{figure}[ht]
  \center
  \includegraphics[scale=0.4]{problem3a.png}
\end{figure}

\item[(b)]  Sketch the graph of the derivative of 
      $f(x)$. 
\begin{figure}[ht]
  \center
  \includegraphics[scale=0.4]{problem3b.png}
\end{figure}
\item[(c)]  Where is $f(x)$ continuous? \\
\vspace{3ex}
$f(x)$ is continuous everywhere. \\
\vspace{3ex}
\item[(d)]  Where is $f(x)$ differentiable? \\
\vspace{3ex}
$f(x)$ is differentiable everywhere except $x=-1$ and $x=0$. \\
\vspace{3ex}
\item[(e)]  Use the derivative to find the maximum value of $f(x)$ on the interval $[0,\infty)$. 
           Where does the maximum occur? Justify your results. \\
           \vspace{3ex}
           On $[0,\infty)$, $f(x) = 5-(x-2)^2$. The maximum is achieved at the places where the derivative is 0 or it is the boundary. We have $f'(x) = -2x+4$. It equals $0$ if $x=2$. Moreover we have $f(2) = 5$ and $f(0) = 1$. Thus the maximum is achieved at $x=2$ and the maximum value is $5$.
\end{enumerate}

\eject

%\item[(10 points) ~~3.] Use implicit differentiation
%to find $\ds \frac{dy}{dx}$ for any point $(x,y)$ on the
%circle $x^2 + y^2 = 1$. Use your result to 
%find all points on the circle
%for which the tangent line is vertical, and
%then all points for which the tangent line
%is horizontal. 








%\item[(10 points) ~~4.] A mass connected to a spring moves 
%along the $x-$axis so that its $x-$coordinate at time $t$ is
%\begin{eqnarray*}
%x(t) = \sin{2t} + \sqrt{3}\cos{2t}.
%\end{eqnarray*}
%Find its velocity $\ds v(t)=\frac{dx}{dt}\;$.

%\vspace{20ex}




\item[(15 points) ~~4.]  
A primordial lightning bolt ignites a flame in the 
top of a 100 foot tall Cretaceous tree.
A 20 foot tall {\it Tyrannosaurus rex}
runs away from the flame
at a speed of 10 $f/s$. How fast is the tip of
his shadow moving when he is 150 feet away from the 
base of the tree? Be sure to show all your work. \\
\begin{figure}[ht]
  \center
  \includegraphics[scale=0.4]{problem4.png}
\end{figure}
The triangle made by the top of the tree, the base of the tree, and the tip of the shadow is similar to the triangle made by the top of the $Tyrannosaurus$, the base of the $Tyrannosaurus$, and the tip of the shadow. This implies:
\[
  \frac{T}{x+y} = \frac{R}{y}
\]
Thus, we have $\ds \frac{100}{x+y} = \frac{20}{y}$ and $\ds y = \frac{1}{4}x$. \\
Since $\ds \frac{dx}{dt}=10 f/s$, the speed of the tip of the shadow is \\
$\ds \frac{d(x+y)}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = \frac{dx}{dt} + \frac{1}{4}\frac{dx}{dt} = \frac{5}{4} \cdot 10 = 12.5 f/s$

%\eject

%\vspace{4in}
%\item[(10 points) ~~6.] Use the differential to approximate
%the increase in volume of a spherical bubble as its radius
%increases from 3 to 3.025 inches.

\vspace{10ex}


\item[(10 points) ~~5.]  Use linear approximation 
                 (the differential) to estimate $\sqrt{4.25}\;$. \\
                 \vspace{3ex}
                 The question is not completely clear on the point at which we should construct the tangent line, although the obvious choice is $x=4$. So, the solution uses $x=4$. However, if you set everything up correctly but used a different point, we gave you full credit. \\
                 \vspace{3ex}
                 The derivative of the function $\ds f(x) = \sqrt{x}$ is $\ds f'(x) = \frac{1}{2\sqrt{x}}$, and so $\ds f'(4) = \frac{1}{4}$. The tangent line to $f(x)$ at point $x=4$ is $\ds L(x) = \frac{1}{4}x+1$, and so $\ds f(4.25) \approx \frac{1}{4}(4.25) + 1 = \frac{33}{16}$.

%\vspace{40ex}
%\item[(10 points) ~~6.] Most of the gases in earth's atmosphere are contained
%in the first 20 miles of altitude above the surface of the earth. 
%The radius of the earth is about 4,000 miles. Use the differential
%to estimate the volume (in cubic miles) 
%of this main part of our atmsophere.

%estimate the amount of paint in cubic centimeters needed
%           to apply a coat of paint 0.04 cm thick to a hemispherical
%           dome with a diameter of 40 meters.

\end{enumerate}

\end{document}
