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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 25 "A Distribution of Weights" }}
{PARA 257 "" 0 "" {TEXT -1 9 "Jim Herod" }}{PARA 258 "" 0 "" {TEXT -1
12 "P O Box 1038" }}{PARA 259 "" 0 "" {TEXT -1 25 "Grove Hill, Alabama
36451" }}{PARA 260 "" 0 "" {TEXT -1 21 "herod@math.gatech.edu" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 438 " Most of us have experiences with balancin
g a series of weights on a line segment and know that the balance poin
t may not occur at the geometric center of the line segment. The balan
ce point is located at a point that may be computed as a combination o
f sums and products of weights and distances from the balance point. I
n particular, a light weight far away from the balance point off-sets \+
a heavy weight close to the balance point." }}{PARA 0 "" 0 "" {TEXT
-1 18 " Locating the " }{TEXT 256 14 "center of mass" }{TEXT -1
84 " for a collection of points is an important idea, for most objects
do not appear as " }{TEXT 257 6 "points" }{TEXT -1 15 " but appear as
" }{TEXT 258 6 "solids" }{TEXT -1 218 ". For the purposes of the scie
nces, however, we often like to consider the solid as being concentrat
ed at a single point. This center of mass is often computed as an inte
gral. We illustrate this idea in this worksheet." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 16 "Weighted Points.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 179 " \+
The balance point or center of mass of three point-forces located at \+
2, -2, and -3 with weights 3, 1, and 2, respectively, is found as a \+
special case of the general formula" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 24 " " }{XPPEDIT 18
0 "Sum(m[i]*x[i],i=1..n)/Sum(m[i],i=1..n)" "6#*&-%$SumG6$*&&%\"mG6#%\"
iG\"\"\"&%\"xG6#F+F,/F+;F,%\"nGF,-F%6$&F)6#F+/F+;F,F2!\"\"" }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "where the " }
{XPPEDIT 18 0 "m[i]" "6#&%\"mG6#%\"iG" }{TEXT -1 35 "'s are weights (o
r masses) and the " }{XPPEDIT 18 0 "x[i]" "6#&%\"xG6#%\"iG" }{TEXT -1
41 "'s are distances from a reference point. " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 " We find the center
of mass for this example." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "(3*2+1*(-2)+2*(-3))/(3+1+2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 46 "Point masses from the p
erspective of integrals" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 328 " There is a classical function that can be used
to describe this idea of point masses in a satisfying way. The functi
on is called the Dirac delta function. Indeed, the Dirac is not a func
tion at all in the sense that one typically describes functions. The D
irac function is defined by its integration property: Suppose that " }
{TEXT 259 1 "a" }{TEXT -1 122 " is a number, F is any function smooth \+
enough that it can be integrated on an interval containing a. Then the
integral of " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 64 " Dirac( x - a
) F(x) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
23 "has an easy evaluation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
15 "readlib(Dirac):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "int(
Dirac(x-a)*F(x),x=-infinity..infinity);" }}}{PARA 0 "" 0 "" {TEXT -1
143 " With this idea, the distance that the three masses in the pr
evious problems are from the 0 reference point can be obtained as an i
ntegral." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "int(Dirac(t-2)*
t,t=-infinity..infinity);\nint(Dirac(t+2)*t,t=-infinity..infinity);\ni
nt(Dirac(t+3)*t,t=-infinity..infinity);" }}}{PARA 0 "" 0 "" {TEXT -1
75 "In a similar manner, the sum of the weights can be computed as an \+
integral." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "int((3*Dirac(t-
2)+1*Dirac(t+2)+2*Dirac(t+3)),t=-infinity..infinity);" }}}{PARA 0 ""
0 "" {TEXT -1 109 "This means that the center of mass for these three \+
point-masses can be computed as the quotient of integrals." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "int(3*Dirac(t-2)*t+1*Dirac(t+2)*t+
2*Dirac(t+3)*t,t=-4..4)/\n int(3*Dirac(t-2)+1*Dirac(t+2)+2*Dirac(t+
3),t=-4..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Fat Wei
ghts." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 346
" Having the realization of center of mass as quotients of integra
ls, we see that there is another way to think of the three point-masse
s. We could conceive of them as fat-points -- as, rectangles with widt
h one, with height the weight of the point-mass, and with center the p
oint where the mass was concentrated. We illustrate with a picture." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "FW:=t->piecewise(t >-3.5 an
d t <-2.5,2,t > -2.5 and t <-1.5,1,\n t > 1.5 and t < 2.5,3,0);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "plot(FW(x),x=-4..4);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 133 " If the generalization carries over, the x coordinate for \+
center of mass for these rectangles should be a quotient of integrals.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 " \+
" }{XPPEDIT 18 0 "int(FW(x)*x,x=-4..4)/int(FW(x),x=-4..4)" "6
#*&-%$intG6$*&-%#FWG6#%\"xG\"\"\"F+F,/F+;,$\"\"%!\"\"F0F,-F%6$-F)6#F+/
F+;,$F0F1F0F1" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 54 "The bottom of this quotient is the sum of
the weights." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "int(FW(t),t
=-4..4);" }}}{PARA 0 "" 0 "" {TEXT -1 35 "The center of mass is the qu
otient." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "int(FW(x)*x,x=-4.
.4)/int(FW(x),x=-4..4);\n int(FW(x),x=-4..4);" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 32 "Continuously dist
ributed weights" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 107 " Perhaps it is clear how the center of mass for a co
ntinuously distributed weight w(x) should be found." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 66 "Int(x*w(x),x=-infinity..infinity)/Int(w(x),
x=-infinity..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT
-1 8 "Example." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 140 "The usual bell-shaped curve familiar in statistical dist
ributions has center of mass under the peak. Just look at the curve to
believe this." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 24 "plot(exp(-x^2),x=-5..5);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 " We ask where the ce
nter of mass for the right half of the curve should be. Will it be mor
e or less than 1, ... than 1/2?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 60 "int(exp(-x^2)*x,x=0..infinity)/int(exp(-x^2),x=0..infinity);"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Exercises fo
r the reader." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 ""
{TEXT -1 11 "Exercise 1." }}{PARA 0 "" 0 "" {TEXT -1 13 "The graph of \+
" }{XPPEDIT 18 0 "1+sin(x)/x" "6#,&\"\"\"F$*&-%$sinG6#%\"xGF$F)!\"\"F$
" }{TEXT -1 242 " , x > 0, seems symmetric about the line y = 1. Over
the interval [ 0 , 20], how close is the x coordinate of the center o
f mass of the area under the curve of this function to the center of m
ass of the rectangle with height 1 and length 20?" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "f:=x->1+sin(
x)/x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x),x=0..20)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "int(f(x)*x,x=0..20)/in
t(f(x),x=0..20);\nevalf(%);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 11 "Exercis
e 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "F
ind the x coordinate for center of mass of the triangle as can be draw
n below" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "top:=x->piecewis
e(x > 1 and x < 2,2*x-1, x > 2 and x <= 4, -x/2+4);\nbottom:=x->piecew
ise(x > 1 and x <= 4, (x+2)/3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 32 "plot(\{top(x),bottom(x)\},x=1..4);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1
1 1 }{PAGENUMBERS 0 1 2 33 1 1 }