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\underline{~~~~~~~~~~~~~Solution~~~~~~~~~~~~~~~~~~~~}\\
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\flushright Class (circle one) ~~~~~~~~~~~~~~ 9:40 ~~~~~~ 10:45\\
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%\flushright Discussion Instructor
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\begin{center}

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

{\large {\bf EXAM I}}

\vspace{2ex}
Friday, September 18, 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.~~ & 10 & \\
\hline
2.~~ & 15 & \\
\hline
3.~~ & 20 & \\
\hline
4.~~ & 30 &\\
\hline
5.~~ & 10 &\\
\hline
6.~~ & 15 &\\
\hline

&TOTAL&\\
\hline
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\begin{enumerate}

\item[(10 points) ~~1.] 

\begin{enumerate}
\item[(a)] Find the equation of the line containing the two points
 $P=(-1,2)$ and $Q=(1,1)$ in the form $y=mx+b$.


\vspace{10ex}

The slope of the line is given by: \\
\vspace{2ex}
$\ds \frac{rise}{run}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-2}{1-(-1)}=-\frac{1}{2}$.\\
\vspace{2ex}
We know the line passes through the point $Q=(1,1)$, so when $x=1$ we have $y=1$.\\ Using this fact to solve for $b$ in the equation $y=mx+b$ we get:

\begin{center}
$1=-\frac{1}{2}(1)+b$\\
$\frac{3}{2}=b$
\end{center}
\vspace{5ex}
Thus our final equation for the line is $\mathbf{{y=-\frac{1}{2}x+\frac{3}{2}}}$.

\vspace{10ex}



\item[(b)] Find the derivative $\ds \frac{dy}{dx}$ 
           of the expression for $y(x)$ you found in (a). 
          
\vspace{10ex}
The derivative of a line is a constant equal to the line's slope. In this case, the derivative will be $\mathbf{-\frac{1}{2}}$.


\end{enumerate}


\eject


\item[(15 points) ~~2.]  Find the following limits.


\begin{enumerate}
\item[(a)] $\ds \lim_{\Delta x \rightarrow 0} 
                (2 + 5\Delta x) $

\vspace{5ex}

$\ds \lim_{\Delta x \rightarrow 0} (2 + 5\Delta x) = 2 + 5 \lim_{\Delta x \rightarrow 0} \Delta x= 2+0 = \mathbf{2}$.

\vspace{10ex}


\item[(b)] $\ds \lim_{h \rightarrow 0} 
        \frac{4xh+5h^3}{h}$

\vspace{5ex}

$\ds \lim_{h \rightarrow 0} \frac{4xh+5h^{3}}{h} = \ds \lim_{h \rightarrow 0} (4x+5h^{2}) = 4x+5(0) = \mathbf{4x}$.

\vspace{10ex}

\item[(c)] $\ds \lim_{h \rightarrow 0}
       \frac{\sqrt{x+h}\;-\;\sqrt{x}}{h}$ 
%      \left({\frac{1}{x+h}}-{\frac{1}{x}}\right)
%       \left(\frac{1}{h}\right)$

\vspace{5ex}

$\ds \lim_{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$\\
\vspace{1ex}
$= \ds \lim_{h \rightarrow 0} \left(\frac{\sqrt{x+h}-\sqrt{x}}{h}\right)\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)$\\
\vspace{1ex}
$= \ds \lim_{h \rightarrow 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$\\
\vspace{1ex}
$= \ds \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})}$\\
\vspace{1ex}
$= \ds \lim_{h \rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}$\\
\vspace{1ex}
$= \ds \frac{1}{\sqrt{x}+\sqrt{x}}$\\
\vspace{1ex}
$= \ds \mathbf{\frac{1}{2\sqrt{x}}}$

\vfill

\end{enumerate}
\eject

\item[(20 points) ~~3. (a)]  Let $f(x) = x^2+5$.  Using the {\it
      definition} of the derivative, calculate $f'(x)$.
      Be sure to show all of your work.
      
\vspace{5ex}

$\ds f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$\\
\vspace{2ex}
$\ds f'(x)=\lim_{h \rightarrow 0} \frac{(x+h)^{2}+5-x^{2}-5}{h}$\\
\vspace{2ex}
$\ds f'(x)=\lim_{h \rightarrow 0} \frac{x^{2}+2xh+h^{2}+5-x^{2}-5}{h}$\\
\vspace{2ex}
$\ds f'(x)=\lim_{h \rightarrow 0} \frac{2xh+h^{2}}{h}$\\
\vspace{2ex}
$\ds f'(x)=\lim_{h \rightarrow 0} 2x+h$\\
\vspace{2ex}
$\ds f'(x)=\mathbf{2x}$

\vspace{5ex}

\item[(b)]  Using your result from (a), find the equation of
      the line tangent to the graph of $f(x) = x^2+5$ at 
      $x = 2$. 

\vspace{5ex}

The slope of the tangent line is given by $f'(x)=2x$. So, at $x=2$ we have $f'(2)=4$. So, the slope of the tangent line is 4.\\
\vspace{2ex}
We also require that the tangent line pass through the point $(2,f(2))$ with $f(2)=2^{2}+5=9$.\\
\vspace{2ex}
So, our line is given by $y=4x+b$, where $9=4(2)+b$ and therefore $b=1$.\\
\vspace{2ex}
So, our line equation is $\mathbf{y=4x+1}$.

\eject



\item[(30 points) ~~4.]  Find the following.

\begin{enumerate}
\item[(a)] $\ds \frac{d}{dx} \left( 3x^5+x^4-2x^2+6 \right)$

\vspace{5ex}
$\ds \frac{d}{dx} \left( 3x^5+x^4-2x^2+6 \right) = \mathbf{15x^4+4x^3-4x}$.\\
\vspace{5ex}


\item[(b)]  $\ds \int{(3x+2)(x-1)\;dx}$

\vspace{5ex}
$\ds \int{(3x+2)(x-1)dx} = \int{(3x^{2}-x-2)} \; dx$\\ 
\vspace{2ex}
$= \mathbf{x^{3}-\frac{x^{2}}{2}-2x+C}$.\\
\vspace{5ex}


\item[(c)]  $v(t)$, $\;$ where $\;$ $x(t)=-16t^2+4t+10\;\;$ and $\;$
$\ds v(t)= \frac{dx}{dt}$

%\item[(d)]   $\ds \frac{d^{12}}{dx^{12}}(33x^{11}+18x^{10}-7x^9+4.5x^8-76x^6-2x^5
  %                     -6x^3+x^2-5x+\pi)$

\vspace{5ex}
$\mathbf{v(t)=-32t+4}$.\\
\vspace{5ex}


\item[(d)] $\ds \int_0^2{ (x^2 + 1)  \;dx}$


\vspace{5ex}
$\ds \int_0^2{(x^{2}+1)dx}=\frac{x^{3}}{3}+x\vert_0^2=\frac{2^{3}}{3}+2-\frac{0^{3}}{3}-0=\frac{8}{3}+2=\mathbf{\frac{14}{3}}$.\\
\vspace{5ex}


\item[(e)] $x(t)$, where $\ds \frac{dx}{dt}=-32t+64$ 
           and $x(0)=50$.

\vspace{5ex}
$\frac{dx}{dt}=-32t+64 \rightarrow x(t)=-16t^{2}+64t+C$\\
\vspace{2ex}
Given $x(0)=50$ we get $x(0)=-16(0^{2})+64(0)+C=50$\\ 
$\rightarrow C=50$.\\
\vspace{2ex}
So, our final equation is $\mathbf{x(t)=-16t^{2}+64t+50}$.

\end{enumerate}
\eject

\item[(10 points) ~~5.] 
%On nearby planet $\Psi$,
%scientists have discovered how to vary their
%%planet's gravitational force with time so that
%the acceleration is $a(t)=-t$, which leads
%to the expression $x(t)=-t^3/6+v_0t+x_0$ for the position
%of a particle in one dimensional vertical motion, 
%where $x_0$ and $v_0$ are the initial position and velocity.
%Using this expression for $x(t)$, find how long it will 
%take for a ball, thrown upward from the ground on $\Psi$
%at $t=0$ with initial velocity 6$\;f/s$, to hit the ground. 
A pebble is dropped from a height of 32 feet. 
Its height above the ground at time $t$ 
is given by $x(t)=-16t^2+32.$ Find $t^*$, the 
time when the pebble hits the ground. 
%Is $t^*$ rational or irrational? 
Find the velocity of the pebble
$v(t^*)=x'(t^*)$ when it strikes the ground.

\vspace{5ex}

When the pebble hits the ground $x(t^*)=0$. Solving $x(t)=-16t^{2}+32$ for this value we get:\\
\vspace{2ex}
$0=-16(t^{*})^{2}+32 \rightarrow 16(t^{*})^{2}=32 \rightarrow (t^{*})^{2} = 2 \rightarrow \mathbf{t^*=\sqrt{2}}$.\\
\vspace{3ex}
To find the velocity we must calculate the derivative of position:\\
\vspace{2ex}
$\ds v(t) = \frac{d}{dt}(-16t^{2}+32) = -32t$. For $t^{*}=\sqrt{2}$ we get $v(t^{*}) = \mathbf{-32\sqrt{2}}$.\\
\vspace{2ex}

\item[(15 points) ~~6.]  
\begin{enumerate}
\item[(a)] Find the antiderivative of $3x^2-2x+2$ that has the 
         value 10 when $x=2$.

\vspace{1ex}

$\ds \int{(3x^2-2x+2)}\;dx=x^3-x^2+2x+C$.\\
\vspace{2ex}
We know $x^3-x^2+2x+C=10$ when $x=2$, so we can use this to solve for $C$.\\
\vspace{1ex}
$2^3-2^2+2\cdot2+C=10 \rightarrow C=2$.\\
\vspace{1ex}
So, the antiderivative is: $\mathbf{x^3-x^2+2x+2}$.
\vspace{2ex}

\item[(b)] Find the position $x(t)$ satisfying Newton's law $$\ds \frac{d^2 x}{dt^2}=-32$$
such that $x(0)=0$ and $x'(0)=v(0)=32$. Be sure to show all of your work.

\vspace{3ex}
$\ds \frac{d^2 x}{dt^2}=-32$
\vspace{1ex}
$\ds \rightarrow \frac{dx}{dt} = -32t + C_{1}$
\vspace{1ex}
$\ds \rightarrow x(t)=-16t^2 + C_{1}t + C_2.$
\vspace{2ex}
We're given $x(0)=0$, so 
$x(0)= -16\cdot0 + C_{1}\cdot0+C_{2}$, and thus $C_{2}=0$.\\
\vspace{1ex}
We're given $x'(0)=32$, so $32=-32\cdot0 + C_{1}$, and thus $C_{1}=32$.\\
\vspace{2ex}
Therefore, $\mathbf{x(t)=-16t^2+32t}$.

\end{enumerate}

%\eject

%\item[(10 points) ~~7.] 
%Use the Intermediate Value Theorem to show that
%$x^3+3x-2=0$ has a real solution between 0 and 1.

\end{enumerate}

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