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{SECT 0 {PARA 260 "" 0 "" {TEXT -1 19 "A Model for Cooling" }}{PARA
256 "" 0 "" {TEXT -1 9 "Jim Herod" }}{PARA 257 "" 0 "" {TEXT -1 12 "P \+
O Box 1038" }}{PARA 258 "" 0 "" {TEXT -1 25 "Grove Hill, Alabama 36451
" }}{PARA 259 "" 0 "" {TEXT 256 19 "herod@math.gatech.e" }{TEXT 258 2
"du" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 261 "" 0 "" {TEXT -1 11 "The Problem" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 462 " We model the cooling of \+
a pitcher of water due to radiation of heat to the surrounding medium.
The model is based on the following episode: between seven and eight \+
o'clock on the morning of March 9, 1996, the temperature on my porch w
as about 15 degrees Fahrenheit. A pitcher of hot water was allowed to \+
cool in this environment. Every three minutes, the falling temperature
of the pitcher of water was measured. A plot of this data can be made
as follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 101 "minute:=[seq(3*n,n=0..20)];\ntemp:=[112,104,99,94
,88,85,82,78,75,72,69,65,63,60,58,57,55,53,51,50,48];" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot([seq([minute[i],temp[i]],i=1..
21)],style=POINT,color=BLACK);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 341 "Using the simple model for coolin
g called Newton's Model for Cooling, it is supposed that the rate of c
ooling is proportioned to the difference between the temperature of th
e object and the temperature of the surrounding medium. Thus, the temp
erature of the pitcher of water is modeled as though its temperature, \+
T(t), satisfied the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 59 " T '(t) = k \+
( T(t) - 15)" }}{PARA 0 "" 0 "" {TEXT -1 52 " \+
T(0) = 112." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 41 "Solutions for this equation are standard." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
49 "dsolve(\{diff(T(t),t)=k*(T(t)-15),T(0)=112\},T(t));" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "We take the soluti
on to have the form " }{XPPEDIT 18 0 "T(t) = 15+97*exp(k*t)" "6#/-%\"T
G6#%\"tG,&\"#:\"\"\"*&\"#(*F*-%$expG6#*&%\"kGF*F'F*F*F*" }{TEXT -1
114 ". Of course, the problem is we don't know k. The number k should \+
be negative, since the temperature is decreasing." }}{PARA 0 "" 0 ""
{TEXT -1 86 " The search for k is made by choosing k to minimize t
he following function of K. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Sum((Temp[ii]-(15+97*exp(K*M
inute[ii])))^2,ii=1..21);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA
0 "" 0 "" {TEXT -1 145 " This function of K is a standard measure o
f how much the approximating function differs from the data. It is usu
al to call this function the " }{TEXT 257 15 "error function." }{TEXT
-1 43 " We call it ERR and seek the minimum value." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "ERR:=K->sum(
(temp[ii]-(15+97*exp(K*minute[ii])))^2,ii=1..21);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 203 " One way to find the
minimum value of this function is to draw a graph. Draw this graph, f
ind K to give the minimum error, and compare the graph of the data wit
h the resulting approximating function." }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(ERR(K),K=-0.1..0);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "fsolve(diff(ERR(k),k)=0
,k,k=-0.1..0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "k:=%;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "with(plots):\nJJ:=plot([seq
([minute[i],temp[i]],i=1..21)],style=POINT,color=BLACK):\nKK:=plot(97*
exp(k*t)+15,t=0..60):\ndisplay(\{JJ,KK\});" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 463 "This model is onl
y an approximation for what will happen in such a large body of water.
The extent to which this model and this solution is appropriate may b
e judged from the above two graphs. Alternate models are described by \+
considering the problem as a boundary value problem. Such notions are \+
described with ideas that come later in undergraduate mathematics. The
construction of an appropriate model is a problem you can return to w
hen you have better tools." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8
"restart:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 24 "Exercise for the student" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 178 "During Saturday afternoon, July 20, \+
1997, the temperature on my porch was considerably warmer than the mor
ning more than a year ago. I repeated the experiment. Here is the data
. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99
"minute:=[seq(3*n,n=0..20)];\ntemp:=[38,38,39,40,41,42,43,44,45,46,46,
48,50,50,50,52,52,53,54,55,55];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 65 "plot([seq([minute[i],temp[i]],i=1..21)],style=POINT,color=BLAC
K);" }}}{PARA 0 "" 0 "" {TEXT -1 106 "1. Set up and solve the associat
ed differential equation to find the nature of the approximating funct
ion." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "dsolve(\{diff(T(t),t
)=k*(T(t)-87),T(0)=38\},T(t));" }}}{PARA 0 "" 0 "" {TEXT -1 33 "2. Use
the data to approximate k." }}{PARA 0 "" 0 "" {TEXT -1 43 "3. Graph t
he data and your fit to the data." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "0 0" 0 }
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