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\begin{center}
Math 132: Calculus for Physical and Life Sciences 2\\ 
Extra Credit Problem Set 10\\ 
Due Wednesday, April 30, no later than 5:00pm
\end{center}
\vspace{10pt}

\noindent
\emph{Selected Answers}

\begin{enumerate}
\item[1.] 
Center of mass at
\[
\overline{x} = \frac{M_y}{M} = \frac{\frac{9\pi^2}{2} - 1}{6\pi + 1} \doteq 2.187.
\]


\item[3.] 
\begin{enumerate}
\item[(a)]  $c = \frac{ln(.9)}{-6} \doteq .01756$.

\item[(c)] $\overline{t} \doteq 56.95$ months.

 \end{enumerate}

\item[4.] 
\begin{enumerate}
\item[(a)] About $42\%$ of the students will take between $1.5$ and $2.5$ hours.
 
\item[(c)] Median time $ = 2.523$ hours
hours.

\end{enumerate}

\item[5.]
\begin{enumerate}
\item[(a)] 
\[
f(x) = \frac{1}{\sqrt{450 \pi}} e^{-(x-100)^2/450}.
\]
\item[(c)] About $0.3\%$ of the population.  
\end{enumerate}


\item[6.] 
The constant $\frac{1}{\sqrt{2\pi\sigma^2}}$ 
in the formula for the normal density is irrelevant for the location
of the critical number(s) and the inflection points, so we will ignore 
it and consider the following function $g(x)$ and its derivatives:
\begin{eqnarray*}
g(x) &=& e^{-(x-\mu)^2/(2\sigma^2)}\\
\Rightarrow g'(x) &=& -e^{-(x-\mu)^2/(2\sigma^2)}\cdot (x-\mu)/\sigma^2 \quad \text{(chain rule)}\\
\Rightarrow g''(x) &=& -e^{-(x-\mu)^2/(2\sigma^2)}
\left(\frac{1}{\sigma^2} - \frac{(x-\mu)^2}{\sigma^4}\right)\quad \text{(product rule)}.
\end{eqnarray*}
\end{enumerate}

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