\nopagenumbers
\def\pt#1{\hbox{#1) }}
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\input amssym.tex
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\headline={ESP Math 408D - AP\hfil Fall 1996}
\footline={\ifnum\pageno>1 \hfil Worksheet 3\hskip.3in Page \folio \fi}
\footnote{}{Mike Oehrtman}

\centerline
{\bf WORKSHEET 3} 

\bigskip 

\item{1.}  Compute the following indefinite integrals:
$$
\eqalign{
\pt a &\int x\cos x\,dx\cr
\pt d &\int \sqrt x\ln x\,dx\cr}\qquad
\eqalign{
\pt b &\int x^2e^x\,dx\cr
\pt e &\int x^3e^{x^2}\,dx\cr}\qquad
\eqalign{
\pt c &\int e^x\cos3x\,dx\cr
\pt f &\int\sin(\ln x)\,dx\cr}\hskip1.5in
$$

\medskip 
\item{2.}  Let $f$ and $g$ be differentiable functions.  Find
$$
\int f(x)g'(x)\,dx+\int g(x)f'(x)\,dx.
$$

\medskip
\item{3.} Compute
$$
\eqalign{
\pt a & \int_{-1}^0{2x\over(x^2+1)(x-1)^2}\,dx \cr
\pt b & \int {x^2\over x^2-1}\,dx\cr}\hskip3.4in
$$

\medskip
\item{4.}  Sociologists sometimes use the phrase {\it social
diffusion} to describe the way information spreads through a
population.  The information might be a rumor, a cultural fad, or news
about a technical innovation.  In a sufficiently large population, the
number of people $x$ who have the information is treated as a
differentiable function of time $t$.  The rate of diffusion, $dx/dt$,
is assumed to be proportional to the number of people who have the
information times the number of people who do not.  This leads to the
differential equation
$$
{dx\over dt}=kx(N-x),
$$
where $N$ is the number of people in the population.

\item{} Suppose $t$ is measured in days, $k=1/250$, and two people
start a rumor at time $t=0$ in a population of $N=1000$ people.

\medskip
\itemitem{a)}  Find $x$ as a function of $t$.

\medskip
\itemitem{b)}  When will half the population have heard the rumor?

\medskip
\itemitem{c)}  When will the rumor be spreading the fastest?

\eject
\item{5.}  Many chemical reactions are the result of interaction of
two molecules that undergo a change to produce a new product.  The
rate of the reaction typically depends on the concentrations of the
two kinds of molecules.  If $a$ is the amount of substance $A$, and
$b$ is the amount of substance $B$ at time $t=0$, and if $x$ is the
amount of product at time $t$, then the rate of formation of $x$ may
be given by the differential equation
$$
{dx\over dt}=k(a-x)(b-x),
$$
or
$$
{1\over(a-x)(b-x)}{dx\over dt}=k
$$
where $k$ is a constant specific to the reaction.  Integrate both
sides of this equation with respect to $t$ to obtain a relation
between $x$ and $t$

\medskip
\itemitem{a)} if $a=b$, and

\medskip
\itemitem{b)} if $a\neq b$.  

\medskip
\item{}  Assume in each case that $x=0$ when $t=0$.  What does this
assumption mean physically?

\medskip
\item{6.}  A catalyst for a chemical reaction is a substance that
controls the rate of the reaction without undergoing any permanent
change in itself.  An autocatalytic reaction is one whose product is a
catalyst for its own formation.  Such a reaction may proceed slowly at
first if the amount of catalyst present is small, and slowly again at
the end when most of the original substance is used up.  But in
between, when both the substance and its product are abundant, the
reaction may proceed at a faster rate.  In some cases it is reasonable
to assume that $\nu$ is proportional to $ax-x^2$ where
$$
\eqalign{
x&=\hbox{ the amount of product,}\cr
a&=\hbox{ the amount of substance at the beginning.}\cr}
$$
The equation that describes the autocatalytic reaction can be written
as 
$$
{dx\over dt}=kx(a-x).
$$
Solve this equation to find $x$ as a function of $t$.  Assume that
$x=x_0$ when $t=0$.





\bye