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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "File: depart\\math\\maple
\\misc\\sfps5-1.mws Date: 17-apr-1998 By: bob jantzen" }}}{EXCHG
{PARA 18 "" 0 "" {TEXT -1 66 "Stewart, Focus on Problem Solving, Chapt
er 5, page 444, problem #1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 106 "Divide a 14 inch diameter pizza into (equal area
) thirds by 2 parallel cuts perpendicular to any diameter." }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 "This pizza probl
em is really about inverting a function defined by an integral with a \+
variable limit of integration." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 316 "In this case it is an area function of a
linear dimension variable, and we would like to determine values of t
he linear dimension that correspond to values of the area function. T
he mathematical model is a circle of radius 7 centered at the origin i
n the x-y plane. Find the area from the left diameter endpoint at " }
{XPPEDIT 18 0 "[-7,0]" "6#7$,$\"\"(!\"\"\"\"!" }{TEXT -1 33 " to the v
ertical line segment at " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 18 " \+
as a function of " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 90 ". Then f
ind the fractional area function by dividing this area function by the
total area " }{XPPEDIT 18 0 "49*Pi" "6#*&\"#\\\"\"\"%#PiGF%" }{TEXT
-1 29 " and determine the values of " }{XPPEDIT 18 0 "x" "6#%\"xG" }
{TEXT -1 189 " that correspond to equal area vertical sliced divisions
of the circle. In other words, given an equally spaced division of th
e possible fractional area values, determine the corresponding " }
{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 115 " values. For this specific \+
problem we choose thirds, but once we have set this up we can divide i
t however we like." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 19 "We plot the circle:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 19 "f:=x->sqrt(49-x^2);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 68 "plot(\{f(x),-f(x)\},x=-7..7,color=red,scaling=constra
ined); circle:=%:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 18 "the area function:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 48 "Area:=x->int(2*f(t),t=-7..x); Area(-7); Area(7
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "the fractional area functio
n:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "area:=x->int(2*f(t),t
=-7..x)/49/Pi; \narea(x);\narea(-7); area(0); area(7);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot(area(x),x=-7..7); area_plt:=%:
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 202 "Clearly this is a nice invertible function since it is m
onotonically increasing, but looking at the formula, clearly cannot be
inverted exactly. We can only do it numerically. For division into t
hirds:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot(\{1/3,2/3\},
x=-7..7,color=green): tickmarks:=%:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display(\{area_plt,t
ickmarks\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "X1:=fsolve(
area(x)=1/3,x=-7..0);\nX2:=fsolve(area(x)=2/3,x=0..7);\n" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 156 "plot([[X1,f(X1)],[X1,-f(X1)]],x=-7
..7,color=blue): X1_plt:=%:\nplot([[X2,f(X2)],[X2,-f(X2)]],x=-7..7,col
or=blue): X2_plt:=%:\ndisplay(\{circle,X1_plt,X2_plt\});" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
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