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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}

General Directions: This is an optional problem set.  You 
may submit solutions for some or all of the following problems.
Any points you earn will count as extra credit on the 
problem set component of your semester average.
You must show all work for credit on these problems. 

\begin{enumerate}
\item The metal making up a rod has density $\rho(x) = 4 + \sin(x)$
at location $x$.  The rod extends from $x = 0$ to $x=\frac{3\pi}{2}$ along
the $x$ axis.  Determine the total mass and the location of the
center of mass of the rod.

\soln
The total mass is 
\[
M = \int_0^{3\pi/2} 4 + \sin(x)\, dx = \left. 4x - \cos(x)\right|_0^{3\pi/2} = 6\pi + 1.
\]
The moment with respect to the $y$-axis is 
\[
M_y = \int_0^{3\pi/2} x(4 + \sin(x))\, dx.
\]
For this, we expand out and integrate $\int x \sin(x)\, dx$ by parts with $u = x$, $dv = \sin(x)\, dx$.
This gives
\begin{eqnarray*}
M_y &=& \int_0^{3\pi/2} x(4 + \sin(x))\, dx\\
&=& \int_0^{3\pi/2} 4x  + x\sin(x)\, dx\\
&=& \left. 2x^2 - x\cos(x) + \sin(x)\right|_0^{3\pi/2}\\
&=& \frac{9\pi^2}{2} - 1.
\end{eqnarray*}
The center of mass is located at 
\[
\overline{x} = \frac{M_y}{M} = \frac{\frac{9\pi^2}{2} - 1}{6\pi + 1} \doteq 2.187.
\]
(As might be expected from the form of the density function, this is located
to the left of the midpoint of the wire.)

\item A metal plate with constant density $5$ gm/cm${}^2$ has the 
shape of the region bounded by $y = \sqrt{x^2+1}$ and the $x$-axis,
for $0 \le x \le 4$.  Find the location of the center of mass
of the plate.  

\soln
The total mass is
\[
M = \int_0^4 5 \sqrt{x^2 + 1}\, dx.
\]
For this integral, we can either use the tangent substitution $x = \tan\theta$
or consult the table of integrals.  This is the form of \#21, with $u = x$ and $a = 1$.
The value is 
\[
M = \left.\frac{5x}{2}\sqrt{1+x^2} + \frac{5}{2}\ln(x + \sqrt{1 + x^2})\right|_0^4 = 10\sqrt{17} + 
\frac{5}{2}\ln(4 + \sqrt{17}).
\]
The moment with respect to the $y$-axis then given by an integral we can evaluate
with the substitution $u = 1 + x^2$, $du = 2x\,dx$:
\[
M_y = \int_0^4 5x\sqrt{1 + x^2}\, dx = \frac{5}{2}\int_{u=1}^{u=17} u^{1/2}\, du = 
\frac{5}{3}(17\sqrt{17} - 1).
\]
Finally, the moment with respect to the $x$-axis is 
\[
M_x = \int_0^4 5\cdot \frac{1}{2} (\sqrt{1+x^2})^2\, dx =\frac{5}{2} \int_0^4 1 + x^2\, dx = \frac{190}{3}.
\]
Hence the center of mass is the point with coordinates
\begin{eqnarray*}
\overline{x} &=& \frac{M_y}{M} \doteq 2.478,\\
\overline{y} &=& \frac{M_x}{M} \doteq 1.363.
\end{eqnarray*}

\item The probability of a certain type of transistor 
failing between $t = a$ and $t = b$ (months), for $0 < a < b$, 
is given by
\[
P(a \le t \le b) = c\int_a^b e^{-ct}\ dt.
\]
\begin{enumerate}
\item[(a)] If the probability of failure within the first six
months is $.10$, what is the value of $c$?

\soln
(Note:  The function 
\[
f(t) = c e^{-ct}, \quad \text{for} \quad t > 0
\]
and $f(t) = t$ for $t \le 0$ is the pdf.)  The given information says
\[
.1 = \int_0^6 c e^{-ct}\, dt = \left. -e^{-ct}\right|_0^6 = 1 - e^{-6c}.
\]
So $c = \frac{ln(.9)}{-6} \doteq .01756$.

\item[(b)] Given the value of $c$ in part (a), what is the
probability that the transistor lasts at least 6 months?

\soln
This is the same as the probability that the time to failure is
a number $\ge 6$ months.  That probability is computed by
\[
\int_6^\infty c e^{-ct}\, dt = 1 - \int_0^6 c e^{-ct}\, dt = 1 - .1 = .9.
\]

\item[(c)] Given the value of $c$ in part (a), what is
the \emph{mean} life of this type of transistor?

\soln
The mean life is
\begin{eqnarray*}
\overline{t} &=& \int_0^\infty t ce^{-ct}\, dt\\
&=& \lim_{b\to\infty} \int_0^b t ce^{-ct}\, dt\\
&=& \lim_{b\to\infty} \left. -t e^{-ct} - \frac{1}{c}e^{-ct}\right|_0^b \quad \text{(using parts: $u = t$, 
$dv = ce^{-ct}$)}\\
&=& \lim_{b\to\infty} \frac{-b}{e^{bc}} - \frac{1}{ce^{bc}} + \frac{1}{c}\\
&=& \frac{1}{c}\\
&\doteq& 56.95 \, \text{months}.
\end{eqnarray*}
 
\item[(d)] Given the value of $c$ in part (a), what is
the \emph{median} life of this type of transistor?

\soln
The cumulative distribution is given by $F(t) = 1 - e^{-ct}$ if $t > 0$ and $F(t) = 0$ if $t \le 0$.
Hence 
\[
\frac{1}{2} = F(t) = 1 - e^{-ct} \Rightarrow e^{-ct} = \frac{1}{2} \Rightarrow t = \frac{\ln(1/2)}{-c} = 
39.37\, \text{months}.
\]
\end{enumerate}

\item Suppose that $t$ is the time (in hours) it takes for 
calculus students to complete their final exam.  Assume
that all students finish within 3 hours and that the probability
density function for the time $t$ is 
\[
f(t) = \begin{cases}
{\displaystyle \frac{4x^3}{81}} & \text{ if } 0 < x < 3\\
0 & \text{otherwise}\\
\end{cases}
\]
\begin{enumerate}
\item[(a)] What proportion of the students take between $1.5$ and $2.5$
hours to complete the exam?

\soln
This is the same as the probability that the time is between $1.5$ and $2.5$:
\[
P(1.5 \le x \le 2.5) = \int_{1.5}^{2.5} \frac{4x^3}{81}\, dx = \left.\frac{x^4}{81}\right|_{1.5}^{2.5} 
\doteq .4198.
\]
About $42\%$ of the students will take between $1.5$ and $2.5$ hours.
 
\item[(b)] What is the \emph{mean time} for students to complete the exam?

\soln
The mean time is 
\begin{eqnarray}
\overline{x} &=& \int_0^3 x\cdot \frac{4x^3}{81}\, dx\\
&=& \int_0^3 \frac{4x^4}{81}\, dx\\
&=& \left.\frac{4x^5}{405}\right|_0^3\\
&=& \frac{12}{5} = 2.4\, \text{hours}.
\end{eqnarray}
\item[(c)] What is the \emph{median time} for students to complete the exam?

\soln
The cdf is the antiderivative of $\frac{4x^3}{81}$ which takes the value
$0$ at $0$, so $F(x) = \frac{x^4}{81}$.  We have 
$\frac{1}{2} = \frac{x^4}{81}$ when $x = \left(\frac{81}{2}\right)^{1/4} \doteq 2.523$
hours.

\item[(d)] At what time should the (evil) professor set the end of the
exam period if he wants to make sure that only $2/3$ of the students
have completed the exam when the papers are collected?

\soln
We need to solve $F(x) = \frac{2}{3}$.  This gives $x = (54)^{1/4} \doteq 2.71$
hours (about 2 hours and 43 minutes).  

\end{enumerate}

\item The distribution of scores on IQ exams is often modeled by 
a \emph{normal} distribution with mean $\mu = 100$ and standard
deviation $\sigma = 15$.
\begin{enumerate}
\item[(a)] Give the formula for the normal pdf that fits this description.

\soln
The appropriate normal pdf is
\[
f(x) = \frac{1}{\sqrt{450 \pi}} e^{-(x-100)^2/450}.
\]
\item Estimate the fraction of the population with IQ scores
between 115 and 120 by applying a Midpoint Riemann sum approximation
for the appropriate integral.  Use $n = 5$ subintervals in the Riemann sum.

\soln
This is the same as the probability that a randomly chosen individual
has IQ in this interval:
\[
P(115 \le x \le 120) = \int_{115}^{120} f(x)\, dx \doteq 0.06743971593
\]
(about $6.7\%$ of the population have IQ scores in this range).  
\item[(b)] Estimate the fraction of the population with IQ scores between
140 and 150 by the same method as in part (b).

\soln
\[
P(140 \le x \le 150) = \int_{140}^{150} f(x)\, dx \doteq 0.00338
\]
(about $0.3\%$ of the population).  
\end{enumerate}


\item Let $\mu$ and $\sigma^2 > 0$ be any two real constants.
\begin{enumerate}
\item[(a)]
Show that the normal pdf 
\[
f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}
\]
has exactly one critical number at $x = \mu$, and that $f(x)$ has a 
local maximum at $x = \mu$.

\soln
See below.

\item[(b)] Show that $f(x)$ has inflection points at $x = \mu \pm \sigma$.

\soln 
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*}
We have $g'(x) = 0$ if and only if $x = \mu$ which shows the first part of (a). 
By the second derivative test, 
\[
g''(\mu) = -\frac{1}{\sigma^2} < 0,
\]
which implies that $g(x)$ and $f(x)$ have maxima at $x = \mu$.   
We have $g''(x) = 0$ if and only if
\begin{eqnarray*}
0 &=& \frac{1}{\sigma^2} - \frac{(x-\mu)^2}{\sigma^4}\\
   \Rightarrow (x-\mu)^2 &=& \sigma^2\\
   \Rightarrow x -\mu &=& \pm \sigma\\
   \Rightarrow x &=& \mu \pm \sigma,
\end{eqnarray*}
which is what we wanted to show.

\item[(c)] Give qualitative sketches of $y = f(x)$ with $\mu = 4$ and $\sigma^2 = 4$,
and the cumulative distribution function for this normal distribution.

\soln
The graph $y = f(x)$ has a maximum at $x = 4$, inflections at $x = 4 \pm 2 = 2,6$.
The cumulative distribution $F(x)$ is increasing for all $x$ and
has a graph that is an ``S''-shaped curve with
$\ds \lim_{x\to -\infty} F(x) = 0$ and $\ds \lim_{x\to\infty} F(x) = 1$.  It 
has maximum positive slope at $x \ 4$.
\end{enumerate}


\end{enumerate}

\end{document}
