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\input{tcilatex}
\begin{document}
\section{\protect\vspace{1pt}Newton's Law of Cooling}
First\_order linear differential equations can be used to solve a variety of
problems that involve temperature .For example:
\begin{enumerate}
\item a medical examiner can find the time of death in a homicide case,
\item a chemist can determine the time required for a plastic mixture to
cool to a hardening \ \ \ \ \ \ temperature
\item and an engineer can design the cooling and heating system of a
manufacturing facility although distinct;
\end{enumerate}
Each of these problems depends on a basic principle that is used to develop
the associated \ differential equation, we discuss this important law now.
\qquad
\begin{itemize}
\item \textsl{\ basic concept of Newton's law of cooling}
\end{itemize}
Newton's law of cooling states that the rate at which the temperature $T(t)$
changes in a cooling body is proportional to the differences between the
temperature of the body and the constant temperature $Ts$ of the surrounding
medium; this situation is represented as the first\_order initial \_value
problem
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $%
\begin{array}{c}
\ \frac{dT}{dt}=k(T-Ts); \\
\ \ \ \ \ T(0)=To%
\end{array}%
$ $\ \ \ \ \ \ \ \ \ \ \ \ \ $
Where $To$ is the initial temperature of the body and $k$ is the constant of
proportionality; we investigate problems involving Newton's law of cooling
in the following examples
In the ideal case , the temperature of the surroundings was assumed to be
constant;
However ,this does not have to be the case ;for example, consider the
problem of heating and cooling a building ;over the span of a twenty-four
hour day, the outside temperature varies ;the problem of determining the
temperature inside the building ,therefore ,becomes more complicated ;for
the meantime,lets's assume that the building has no heating or air
conditioning system ;hence the differential equation that should be solved
to find the temperature $u(t)$ at time $t$ inside the building is
\vspace{1pt} $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{%
du}{dt}=k(C(t)-u(t))$
Where $C(t)$ is a function that describes the outside temperature and $k>0$
is a constant that depends on the insulation of the building; according to
this equation ,if $C(t)>u(t)$,then $\frac{du}{dt}>0,$which implies that $u$
increase ;conversely ,if $C(t)