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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257
"" 0 "" {TEXT -1 46 "Lesson 3: Applications of Integration 1: Work" }
}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "
When a force moves an object, we say the force does work. If the for
ce " }{XPPEDIT 18 0 "F;" "6#%\"FG" }{TEXT -1 53 " is constant, the wor
k done is given by the equation " }{XPPEDIT 18 0 "W = Fd;" "6#/%\"WG%#
FdG" }{TEXT -1 8 ", where " }{XPPEDIT 18 0 "d;" "6#%\"dG" }{TEXT -1
67 " is the distance moved. What happens if the force is not constant
?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Supp
ose, for definiteness, that a force " }{XPPEDIT 18 0 "F(x);" "6#-%\"FG
6#%\"xG" }{TEXT -1 22 " moves an object from " }{XPPEDIT 18 0 "a;" "6#
%\"aG" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 11 "
along the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 17 "-axis. Altho
ugh " }{XPPEDIT 18 0 "F;" "6#%\"FG" }{TEXT -1 12 " depends on " }
{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 29 ", we can divide the interva
l " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 43 " into smal
l subintervals, and suppose that " }{XPPEDIT 18 0 "F;" "6#%\"FG" }
{TEXT -1 40 " is almost constant on each subinterval." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Let's see why \+
the expression " }{XPPEDIT 18 0 "sum(F(a+k*(b-a)/n),k = 1 .. n);" "6#-
%$sumG6$-%\"FG6#,&%\"aG\"\"\"*(%\"kGF+,&%\"bGF+F*!\"\"F+%\"nGF0F+/F-;F
+F1" }{XPPEDIT 18 0 "(b-a)/n;" "6#*&,&%\"bG\"\"\"%\"aG!\"\"F&%\"nGF("
}{TEXT -1 77 " should be an approximation to the total work done in mo
ving the object from " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 4 " to \+
" }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 28 "When we divide the interval " }{XPPEDIT 18 0 "[a, b]
;" "6#7$%\"aG%\"bG" }{TEXT -1 6 " into " }{XPPEDIT 18 0 "n;" "6#%\"nG
" }{TEXT -1 49 " equal subintervals, each subinterval has length " }
{XPPEDIT 18 0 "(b-a)/n;" "6#*&,&%\"bG\"\"\"%\"aG!\"\"F&%\"nGF(" }
{TEXT -1 6 ". If " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 72 " is la
rge, each subinterval will be very short, and so the force on the " }
{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT -1 161 "-th subinterval can be app
roximated by its value at any point in the subinterval. We will choos
e the right-hand endpoint of the subinterval, which is the point " }
{XPPEDIT 18 0 "a+k*(b-a)/n;" "6#,&%\"aG\"\"\"*(%\"kGF%,&%\"bGF%F$!\"\"
F%%\"nGF*F%" }{TEXT -1 21 ". The force on the " }{XPPEDIT 18 0 "k;"
"6#%\"kG" }{TEXT -1 43 "-th subinterval is therefore approximately " }
{XPPEDIT 18 0 "F(a+k*(b-a)/n);" "6#-%\"FG6#,&%\"aG\"\"\"*(%\"kGF(,&%\"
bGF(F'!\"\"F(%\"nGF-F(" }{TEXT -1 107 ", and the work done in moving a
cross this subinterval, using the constant-force formula, is approxim
ately " }{XPPEDIT 18 0 "F(a+k*(b-a)/n);" "6#-%\"FG6#,&%\"aG\"\"\"*(%\"
kGF(,&%\"bGF(F'!\"\"F(%\"nGF-F(" }{XPPEDIT 18 0 "(b-a)/n;" "6#*&,&%\"b
G\"\"\"%\"aG!\"\"F&%\"nGF(" }{TEXT -1 38 ". The total work done in mo
ving from " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 4 " to " }
{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 23 " is given by adding up " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 101 " of these terms, one for e
ach subinterval, which gives the formula in the statement of the ques
tion." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 14 "Let's write a " }{TEXT 284 5 "Maple" }{TEXT -1 10 " funct
ion " }{TEXT 286 7 "worksum" }{TEXT -1 25 ", which takes a function "
}{XPPEDIT 18 0 "F;" "6#%\"FG" }{TEXT -1 14 ", an interval " }{XPPEDIT
18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 14 " and a number " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 17 ", and returns an " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 57 "-subinterval approximation \+
to the work done by the force " }{XPPEDIT 18 0 "F;" "6#%\"FG" }{TEXT
-1 26 " in moving an object from " }{XPPEDIT 18 0 "a;" "6#%\"aG" }
{TEXT -1 4 " to " }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 55 " (i.e. \+
translate the formula given in Question 1 into " }{TEXT 285 5 "Maple"
}{TEXT -1 9 " syntax)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "w
orksum := (F,a,b,n)-> sum(F(a + k*(b-a)/n)*(b-a)/n, k=1..n) ;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(worksumGR6&%\"FG%\"aG%\"bG%\"nG6\"6
$%)operatorG%&arrowGF+-%$sumG6$*&*&-9$6#,&9%\"\"\"*&*&%\"kGF9,&9&F9F8!
\"\"F9F99'F?F9F9F=F9F9F@F?/F<;F9F@F+F+F+" }}}{SECT 0 {PARA 4 "" 0 ""
{TEXT -1 31 "Example 1: Riemann Sum approach" }}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 36 "The force felt by an object of mass " }{XPPEDIT 18 0 "m
;" "6#%\"mG" }{TEXT -1 32 " at the surface of the Earth is " }
{XPPEDIT 18 0 "-mg;" "6#,$%#mgG!\"\"" }{TEXT -1 8 ", where " }
{XPPEDIT 18 0 "g = 9.82;" "6#/%\"gG$\"$#)*!\"#" }{TEXT -1 172 " m/s^2 \+
is the 'accelerationn due to gravity'. Of course, the force felt by t
he object lessens as it moves away from the Earth. In fact, the corre
ct force law is given by " }{TEXT 290 28 "Newton's Law of Gravitation:
" }{TEXT -1 0 "" }}{PARA 258 "" 0 "" {XPPEDIT 18 0 "F = -GMm/(r^2);" "
6#/%\"FG,$*&%$GMmG\"\"\"*$%\"rG\"\"#!\"\"F," }{TEXT -1 2 " ." }}{PARA
0 "" 0 "" {TEXT -1 6 "Here, " }{XPPEDIT 18 0 "F;" "6#%\"FG" }{TEXT -1
34 " is the force felt by the object, " }{XPPEDIT 18 0 "m;" "6#%\"mG"
}{TEXT -1 14 " is its mass, " }{XPPEDIT 18 0 "M;" "6#%\"MG" }{TEXT -1
27 " is the mass of the Earth, " }{XPPEDIT 18 0 "r;" "6#%\"rG" }{TEXT
-1 65 " is the distance of the object from the centre of the Earth, an
d " }{XPPEDIT 18 0 "G;" "6#%\"GG" }{TEXT -1 41 " is a universal consta
nt. The values of " }{XPPEDIT 18 0 "M;" "6#%\"MG" }{TEXT -1 5 " and \+
" }{XPPEDIT 18 0 "G;" "6#%\"GG" }{TEXT -1 73 " are known, but we will \+
not need them, because of the following argument." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 287 "(a). The kilometre was o
riginally defined as 1/10000 of the distance from the North Pole to th
e Equator along the meridian which runs through Paris. Hence the circ
umference of the Earth is almost exactly 40000 kilometres. Find the r
adius of the Earth and assign it to the variable R. " }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "R := evalf(40000/(2*Pi));" }}{PARA
11 "" 1 "" {XPPMATH 20 "6#>%\"RG$\"+Ax>mj!\"'" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 34 "(b). At the surface of the Earth, " }{XPPEDIT 18 0 "r \+
= R;" "6#/%\"rG%\"RG" }{TEXT -1 11 ". Putting " }{XPPEDIT 18 0 "r = R
;" "6#/%\"rG%\"RG" }{TEXT -1 137 " in Newton's Law of Gravitation give
s one expression for the gravitational force at the surface of the Ear
th. Equate this expression to " }{XPPEDIT 18 0 "-mg;" "6#,$%#mgG!\"\"
" }{TEXT -1 42 ", and hence find the value of the product " }{XPPEDIT
18 0 "GM;" "6#%#GMG" }{TEXT -1 28 ". (Be careful with units: " }
{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 82 " is expressed in terms of m
etres/second^2, but other distances are in kilometres.)" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 294 9 "Solut
ion." }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 "It is best to sta
rt here with pencil and paper. Equating the two expressions for the f
orce at the surface of the Earth gives" }}{PARA 0 "" 0 "" {XPPEDIT 18
0 "-GMm/(R^2) = -mg;" "6#/,$*&%$GMmG\"\"\"*$%\"RG\"\"#!\"\"F+,$%#mgGF+
" }{TEXT -1 14 ". Cancelling " }{XPPEDIT 18 0 "-m;" "6#,$%\"mG!\"\""
}{TEXT -1 29 " and solving for the product " }{XPPEDIT 18 0 "GM;" "6#%
#GMG" }{TEXT -1 7 " gives " }{XPPEDIT 18 0 "GM = R^2*g;" "6#/%#GMG*&%
\"RG\"\"#%\"gG\"\"\"" }{TEXT -1 28 ". We will use the value of " }
{XPPEDIT 18 0 "R;" "6#%\"RG" }{TEXT -1 34 " computed above, and the va
lue of " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 73 " given in the que
stion, but to make the units consistent we will express " }{XPPEDIT
18 0 "g;" "6#%\"gG" }{TEXT -1 11 " in km/s^2:" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 13 "g := 0.00982;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"gG$\"$#)*!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "GM :=
R^2 * g;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#GMG$\"+#4'*)zR!\"%" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "(c) At a height of 42377 kilometr
es above the centre of the Earth, a satellite revolves in a " }{TEXT
291 13 "geostationary" }{TEXT -1 226 " orbit: it takes exactly 24 hour
s to revolve once around the Earth, and so it is always directly above
the same point on the Earth. Communications satellites, for example,
are always placed in geostationary orbits. Use your " }{TEXT 292 5 "
Maple" }{TEXT -1 377 " function from Question 2 to compute approximati
ons to the amount of work that must be done to raise a 250-kilogram sa
tellite from the surface of the Earth to a geostationary orbit. Use a
pproximations with 100 and 1000 subintervals. Your function may retur
n a negative value, although it clearly takes a positive amount of wor
k to raise a satellite into orbit. Explain this." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 296 10 "Solution.
" }{TEXT -1 192 " We have to raise the satellite from the surface of \+
the Earth to geostationary orbit. Since heights are being measured fr
om the centre of the Earth, we must raise the satellite from a height \+
" }{XPPEDIT 18 0 "R;" "6#%\"RG" }{TEXT -1 33 " to a height of 42377 ki
lometres." }}{PARA 0 "" 0 "" {TEXT -1 22 "We want the work done " }
{TEXT 298 7 "against" }{TEXT -1 124 " the force of gravity in moving b
etween these heights. Our function worksum will give us (approximatio
ns to) the work done " }{TEXT 297 2 "by" }{TEXT -1 68 " the force. Fi
rst, of course, we have to tell it what the force is:" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "f := r->-GM*250/r^2 ;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"rG6\"6$%)operatorG%&arrowGF(,$*&%#GM
G\"\"\"*$)9$\"\"#F/!\"\"!$]#F(F(F(" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 23 "worksum(f,R,42377,100);" }}{PARA 11 "" 1 "" {XPPMATH
20 "6#$!+K\\t&G\"!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "wo
rksum(f,R,42377,1000);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+M#)zB8!\"
&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Remember that this is the wo
rk done " }{TEXT 299 2 "by" }{TEXT -1 205 " the force, which is the ne
gative of the work that has to be done in lifting the satellite into o
rbit. (The gravitational force will do a positive amount of work when
the satellite crashes back to Earth.)" }}{PARA 0 "" 0 "" {TEXT -1 86
"We should probably only keep 3 significant figures in our answers, si
nce the value of " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 151 " was o
nly given to this accuracy. Let's say, then, that with 1000 subinterv
als we estimate the necessary amount of work to be 13200 Newton-kilome
tres," }}{PARA 0 "" 0 "" {TEXT -1 48 "or (in more usual units) 1320000
0 Newton-metres." }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "We've seen t
hat the expression " }{XPPEDIT 18 0 "sum(F(a+k*(b-a)/n),k = 1 .. n);"
"6#-%$sumG6$-%\"FG6#,&%\"aG\"\"\"*(%\"kGF+,&%\"bGF+F*!\"\"F+%\"nGF0F+/
F-;F+F1" }{XPPEDIT 18 0 "(b-a)/n;" "6#*&,&%\"bG\"\"\"%\"aG!\"\"F&%\"nG
F(" }{TEXT -1 132 " is an approximation to the work done. On physical
grounds, we would expect that this approximation would get better and
better as " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 120 " gets larger
. Mathematically, on the other hand, we recognise that our approximat
ion is a Riemann sum for the integral " }{XPPEDIT 18 0 "int(F(x),x = a
.. b);" "6#-%$intG6$-%\"FG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 70 ", and \+
we know that the Riemann sums will converge to this integral as " }
{XPPEDIT 18 0 "proc (n) options operator, arrow; infinity end;" "6#R6#
%\"nG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 114 ". I
t seems reasonable to couclude, therefore, that the integral gives the
exact amount of work done by the force." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 37 "Example 2: Exact Integrat
ion Approach" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 301 "Use an appropriat
e integral to compute the exact amount of work required to lift the sa
tellite. Compare with the Riemann sum approximations. (For example, \+
you could ask: How good are the approximations? How many subinterval
s are necessary for the approximation to be within 1% of the true answ
er?)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}{TEXT 301 9 "Solution." }{TEXT -1 16 " (The function " }{XPPEDIT 18
0 "f;" "6#%\"fG" }{TEXT -1 42 " should still be defined from Question \+
3.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "exact_work := int(f,
R..42377);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+exact_workG$!+$\\5\"
G8!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "As in Question 1, this \+
is the work done " }{TEXT 302 2 "by" }{TEXT -1 165 " the force, which \+
is the negative of what we want. Rounding to 3 significant figures, l
et's say we have to do 13300000 Newton-metres of work to raise the sat
ellite." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
174 "Our approximations in Question 1 were pretty good. In particular
, a 1% error would mean (in Newton-kilometres) an error of no more tha
n 132 N-km, or an approximation between" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 35 "exact_work - 132; exact_work + 132;" }}{PARA 11 "" 1
"" {XPPMATH 20 "6#$!+$\\58M\"!\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$
!+$\\5\\J\"!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 223 "Our 1000-inte
rval approximation certainly meets this requirement, but the 100-inter
val one does not. How many intervals do we need? (Some trial and err
or was necessary to get the number of intervals in the next commands.)
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "worksum(f,R,42377,350);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+AQ$eJ\"!\"&" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 23 "worksum(f,R,42377,300);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$!+\"*3!QJ\"!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
143 "Somewhere between 300 and 350 intervals would be sufficient---you
can experiment further if you want to narrow the number down more acc
urately." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
467 "Question: In this question, it was easy to see how accurate our \+
approximations were, because we could compare them with the exact answ
er. Of course, this also makes the comparison unnecessary: if we have
the exact answer, we don't need to worry about approximations! Riema
nn sum approximations are most useful when we can't work out the exact
answer. In that situation, how do you think we could have confidence
that our approximations were sufficiently accurate?" }}}}}{MARK "0 1 \+
0" 46 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }