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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 63 "Problem Se
t D 3-7 Solutions Taken Mainly from Your Problem Sets" }}{PARA 0 "" 0 
"" {TEXT -1 56 "(with some additional comments and revisions made by m
e)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "with(DEtools):with(plots):" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 79 "Number 3 (Similar to Calman and J
ohn C's, Melinda's and Rebekah's Problem Sets)" }}{PARA 0 "" 0 "" 
{TEXT -1 446 "In problem #3 we are asked to solve the given d.e. and g
raph that solution with different values of amplitude (a).   The intia
l conditions theta(0)=a, and theta'(0)=0.  We used the following value
s of a: a=0.1, 0.7, 1.5, 3.0 We looked at how the period varies with a
mplitude.  We were then asked to determine the period of oscillation f
or each value.  To do this we attempted to find an exact value but in \+
the end had to settle for an estimation." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "restart: with(plots): with(DEtools):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "eq1:=\{diff(x(t),t$2)+sin(x(t))=0,x
(0)=0.1,D(x)(0)=0\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G<%/,&-%
%diffG6$-%\"xG6#%\"tG-%\"$G6$F.\"\"#\"\"\"-%$sinG6#F+F3\"\"!/-F,6#F7$F
3!\"\"/--%\"DG6#F,F:F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "s
ol1:=dsolve(eq1,x(t),numeric);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%s
ol1GR6#%(rkf45_xG6'%\"iG%(rkf45_sG%)outpointG%#r1G%#r2G6#%aoCopyright~
(c)~1993~by~the~University~of~Waterloo.~All~rights~reserved.G6\"C&>8&-
%&evalfG6#9$@$52-%$absG6#,$F3!\"\"-F<6#,&&%,loc_controlG6#\"\"#\"\"\"F
3F?4-%'memberG6$&FD6#\"\"'<*$FG\"\"!$FFFQF?FG!\"#$FSFQ$F?FQFFC%>FD-%%c
opyG6#=F06#;FG\"#LE\\[lB\"#=FQ\"#8FQ\"#9FQFNFG\"#>FQFhnFQ\"#FFQ\"#CFQ
\"#6FQ\"#5FQ\"#GFQ\"#BFQ\"#<FQ\"#7FQ\"#@FQ\"\"%$FG!\")\"\")\"&++$\"#IF
Q\"\"&Fho\"\"$FQ\"\"($FG!\"*\"#:FQ\"#JFQ\"#EFQ\"#?$FGF?FFFQ\"#KFQ\"#;F
Q\"#HFQ\"#DFQ\"#AFQ\"\"*\"%+5FGFF>%'loc_y0G-FY6#=F06#;FGFFE\\[l#FFFQFG
Ffp>%'loc_y1G-FY6#=F0FcqE\\[l!@$0F;FQC$>&FD6#F^pF3@%1%'DigitsG-%'evalh
fG6#FdrC$>8%-%*traperrorG6#-Ffr6#-%=dsolve/numeric_solnall_rkf45G6,%&l
oc_FG-%$varG6#FD-Fes6#F_q-Fes6#Fgq-Fes6#%'loc_F1G-Fes6#%'loc_F2G-Fes6#
%'loc_F3G-Fes6#%'loc_F4G-Fes6#%'loc_F5G-Fes6#%)loc_workG@$/Fjr%*laster
rorGC%>8'-%+searchtextG6$.Ffr-%(convertG6$-%#opG6$FG7#Fjr%%nameG>8(-Fd
u6$.%)hardwareGFgu@%50FbuFQ0F`vFQ-Fas6,FcsFDF_qFgqF]tF`tFctFftFitF\\u-
%&ERRORG6#FjrFiv7$/%\"tGF7-%$seqG6$/&%$ordG6#,&8$FGFGFG&F_q6#Fiw/FiwFd
qF06%FDF_qFgqF0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "a:=odepl
ot(sol1,[t,x(t)],0..17,color=green):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 53 "eq2:=\{diff(x(t),t$2)+sin(x(t))=0,x(0)=0.7,D(x)(0)=0
\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "sol2:=dsolve(eq2,x(t
),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "e:=odeplot(s
ol2,[t,x(t)],0..17,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 53 "eq3:=\{diff(x(t),t$2)+sin(x(t))=0,x(0)=1.5,D(x)(0)=0\}:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "sol3:=dsolve(eq3,x(t),numeri
c):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "c:=odeplot(sol3,[t,x
(t)],0..17,color=magenta):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
53 "eq4:=\{diff(x(t),t$2)+sin(x(t))=0,x(0)=3.0,D(x)(0)=0\}:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "sol4:=dsolve(eq4,x(t),numeri
c):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "d:=odeplot(sol4,[t,x
(t)],0..17,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "d
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F27$Faw$\"17B)Qk^jK\"F[fl7$Ffw$\"1]'>+;O8\"=F[fl7$F[x$\"1qr^vH#4=#F[fl
7$F`x$\"1a+.dK9_CF[fl7$Fex$\"1.&y8g_lk#F[fl7$Fjx$\"1eAdrfS$y#F[fl7$F_y
$\"1ty?R:rxGF[fl7$Fdy$\"1!GOsBL.%HF[fl7$Fiy$\"1\"R.Q3h'yHF[fl7$F^z$\"1
&o!4\\9F(*HF[fl7$Fcz$\"1=3h!G)R)*HF[fl7$Fhz$\"1)=Y<1x@)HF[fl7$F][l$\"1
%oLp')fm%HF[fl-Fb[l6&Fd[lFe[lF(F(" 2 185 163 163 2 0 1 0 2 9 0 4 2 1 
45 45 10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 
0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 460 "The graph abov
e displays the period of oscillation varying with amplitude in a manne
r that would be expected.  We would expect that as the amplitude incre
ased, as long as other variables remained the same,  the period would \+
increase.  Here we see that when a=0.1, the period is about t=6.31 s. \+
 For a=0.7 we see that the period is about t=6.49 s.  When a=1.5, we s
ee that the period is about t=7.32 s.  When the a=3.0 we see that the \+
period is about t=16.15 s." }}{PARA 0 "" 0 "" {TEXT -1 836 "We attempt
ed to do determine the periods exactly in a number of ways allthough a
ll were unsuccessful.  We first attempted to use the maximize command \+
where a function, here the derivitive of our d.e. set to 0, and a rang
e of t values are defined.  The command is meant to locate the maximum
 points in the function over the given interval.  Next we tryed the ev
alf function over the necessary intervals using the derivative of the \+
given d.e.  However maple couldn't make heads or tales of the second p
artial term.  This meant it couldn't solve for t when the derivative e
qualled 0.  We attempted taking the derivitive of our dsolved d.e. but
 couldn't make it work.  We also attempted unapplying and simplifying \+
the dsolved d.e. and then taking the derivative.  This again did not w
ork.  Thus we were forced to simply estimate the period." }}{PARA 0 "
" 0 "" {TEXT -1 153 "Notice that the above won't work because numerica
l methods just give us a bunch of points - they do not give us a smoot
h function that we can work with. " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT 
-1 34 "Deeper Exploration of the Accuracy" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Let's look at the linear pl
ots:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "leq1:=\{diff(x(t),t
$2)+x(t)=0,x(0)=0.1,D(x)(0)=0\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
%leq1G<%/,&-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F.\"\"#\"\"\"F+F3\"\"!/-F,6#
F4$F3!\"\"/--%\"DG6#F,F7F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "lsol1:=dsolve(leq1,x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "la:=odeplot(lsol1,[t,x(t)],0..17,color=black):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "leq2:=\{diff(x(t),t$2)+x(t)=
0,x(0)=0.7,D(x)(0)=0\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "
lsol2:=dsolve(leq2,x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "le:=odeplot(lsol2,[t,x(t)],0..17,color=black):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "leq3:=\{diff(x(t),t$2)+x(t)=
0,x(0)=1.5,D(x)(0)=0\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "
lsol3:=dsolve(leq3,x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "lc:=odeplot(lsol3,[t,x(t)],0..17,color=black):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "leq4:=\{diff(x(t),t$2)+x(t)=
0,x(0)=3.0,D(x)(0)=0\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "
lsol4:=dsolve(leq4,x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "ld:=odeplot(lsol4,[t,x(t)],0..17,color=black):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Let's look at a plot of .1 and .7 \+
as y(0)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "display(\{a,la,e,le\});
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0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 138 "The above graph repres
ents 3.0 with red as non-linear and black as linear.  We can see that \+
the right phase shift has increased even more." }}}{SECT 0 {PARA 5 "" 
0 "" {TEXT -1 0 "" }{TEXT 270 0 "" }{TEXT 271 44 "Finding the Phase Sh
ift Using Graphical Info" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "For a
 delta in a given range, found on the graph as the region around the" 
}}{PARA 0 "" 0 "" {TEXT -1 87 "point where the equation peaks, we can \+
use fsolve to find the phase shift of each graph" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 108 "For an amplitude of .1, we look between 5 and 8. \+
 We can see that delta is 6.28, which is approximately 2Pi." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "fsolve(.1*cos(6.28-delta)=.1
,delta=5..8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"$G'!\"#" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 222 "For am amplitude of .7 and a delt
a between 6 and 10, I find detla to equal 6.53.  This is larger than 2
Pi, but smaller than 25Pi/12, which is only just over 2Pi.  The accura
cy is about 96% found by dividing 2Pi by 25Pi/12:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 58 "fsolve(.7*cos(6.53-delta)=.7,delta=6..10);ev
alf(25*Pi/12);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"$`'!\"#" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+&p%)\\a'!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 204 "For an amplitude of 1.5 and a delta between 6 and 10, de
lta is found to be 7.41.  Quite larger than 2Pi but smaller than 5Pi/2
.  The accuracy has decreased and is about 80% found by dividing 2Pi b
y 5Pi/2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "fsolve(1.5*cos(
7.41-delta)=1.5,delta=6..10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"$T
(!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 201 "And even Pi shift would
 be 6Pi.  The Pi shift is larger than 5Pi but not quite 6Pi.  The accu
racy will be off largely.  To find the approximate accuracy, divide 2P
i by 6Pi, which is about 33% accurate." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 69 "fsolve(3*cos(16.19-delta)=3,delta=16..18);evalf(5*Pi)
;evalf(11*Pi/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"%>;!\"#" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Fjzq:!\")" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+gf(ys\"!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
114 "Overall the above numbers indicate that the larger the Amplitude \+
the less accurate the period (2Pi--6.28) becomes." }}}{SECT 0 {PARA 5 
"" 0 "" {TEXT -1 0 "" }{TEXT 272 0 "" }{TEXT 273 1 " " }{TEXT 274 48 "
Using Elliptic Inegrals to Find the Exact Period" }}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 199 "Using a non-elementary integral called an elliptic \+
integral, we could find the exact period using the following formula, \+
where alpha is our initial angle:  The proof is beyond the scope of th
e class." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "4*Int(1/(sqrt(1
-sin(alpha/2)^2*sin(x)^2)),x=0..Pi/2);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$-%$IntG6$*&\"\"\"F(*$-%%sqrtG6#,&\"\"\"F.*&)-%$sinG6#,$%&alpha
G#F.\"\"#F7F()-F26#%\"xGF7F(!\"\"F(!\"\"/F;;\"\"!,$%#PiGF6\"\"%" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "evalf(4*Int(1/(sqrt(1-sin(1/
2)^2*sin(x)^2)),x=0..Pi/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+kc
(**p'!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "evalf(4*Int(1/
(sqrt(1-sin(.7/2)^2*sin(x)^2)),x=0..Pi/2));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+;(*=\"['!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 58 "evalf(4*Int(1/(sqrt(1-sin(1.5/2)^2*sin(x)^2)),x=0..Pi/2));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+S\\'3I(!\"*" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 56 "evalf(4*Int(1/(sqrt(1-sin(3/2)^2*sin(x)^2)),x=
0..Pi/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+PRb:;!\")" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Our graphical estimations are good
!" }}}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 70 "Number 4 (Similar to Eri
c's and Gabriel, Gabe and Chris' Problem Sets)" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 211 "We are asked to d
efine a Maple function of v that will numerically solve the de E\"+sin
(E)=0 using the initial conditions E(0)=0, E'(0)=v.  I set up two ivps
 for the linear equation and the non-linear equation.  " }}{PARA 0 "" 
0 "" {TEXT -1 99 "We are then asked to plot both the linear and nonlin
ear equations for v=1, 1.99,2,2.01 from t=0..40" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 27 "eq2:=diff(x(t),t$2)+x(t)=0;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$eq2G/,&-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\"
\"F*F2\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "linivp:=v->(
\{eq2,x(0)=0,D(x)(0)=v\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'liniv
pGR6#%\"vG6\"6$%)operatorG%&arrowGF(<%/-%\"xG6#\"\"!F1/--%\"DG6#F/F09$
%$eq2GF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "nonlinivp:=
v->(\{eq1,x(0)=0,D(x)(0)=v\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*n
onlinivpGR6#%\"vG6\"6$%)operatorG%&arrowGF(<%/-%\"xG6#\"\"!F1/--%\"DG6
#F/F09$%$eq1GF(F(F(" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 3 "v=1" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "e:=odeplot(dsolve(linivp(1.0
0),x(t),numeric,maxfun=8000),[t,x(t)],0..40,color=RED):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "f:=odeplot(dsolve(nonlinivp(1.00),x
(t),numeric,maxfun=8000),[t,x(t)],0..40,color=BLUE):" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 128 "before examining the graph, we must remember t
hat the solution plot for the linear equation is in red and the non-li
near in blue" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "display(\{e
,f\});" }}{PARA 13 "" 1 "" {INLPLOT "6$-%'CURVESG6$7T7$\"\"!F(7$$\"+1`
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[l6&F_[lF(F(F`[l" 2 355 190 190 2 0 1 0 2 9 0 4 2 1 45 45 10030 0 
10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 325 32 0 0 0 
0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "for initial velocity v = \+
1.00" }}{PARA 0 "" 0 "" {TEXT -1 170 "it appears that the non-linear r
epresentation has both a larger amplitude and a longer period.  the tw
o start in phase, but by t = 40 are almost perfectly out of phase.  " 
}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 6 "v=1.99" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 83 "g:=odeplot(dsolve(linivp(1.99),x(t),numeric,maxf
un=8000),[t,x(t)],0..40,color=RED):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 87 "h:=odeplot(dsolve(nonlinivp(1.99),x(t),numeric,maxfun
=8000),[t,x(t)],0..40,color=BLUE):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "display(\{g,h\});" }}{PARA 13 "" 1 "" {INLPLOT "6$-%'
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}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 341 "the trend we saw in the last g
raph is greatly accentuated with a larger initial velocity.  as the in
itial velocity gets bigger, i would trust our linear approximation les
s and less, because it is based on the assumption that the displacemen
t is small for the pendulum's motion; one would expect larger displace
ments with an initial velocity." }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 
5 "v=2.0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "i:=odeplot(dsol
ve(linivp(2),x(t),numeric,maxfun=8000),[t,x(t)],0..40,color=RED):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "j:=odeplot(dsolve(nonlinivp(
2),x(t),numeric,maxfun=8000),[t,x(t)],0..40,color=BLUE):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "display(\{i,j\});" }}{PARA 13 "" 1 
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\"+Ln$=f$Feo$\"1J=;_c@a&*F/7$$\"+'QpMn$Feo$\"1C&Hfjqrr*F/7$$\"+R?5bPFe
o$\"12TmCR935!#97$$\"+#pMn$QFeo$\"1pxG8g*e3\"F\\z7$$\"+XtO=RFeo$\"1M3t
y([PA\"F\\z7$$\"+)*******RFeo$\"1^col=`y8F\\z-%'COLOURG6&%$RGBGF(F($\"
*++++\"Feo-F$6$7TF'7$F*$\"1@?Cq&psX\"F/7$F1$\"1fx?w\\<'*>F/7$F7$\"1gjB
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$Fio$!1XHltXU:=F/7$F^p$!1vaNSEG>jFa\\l7$Fcp$\"1fr!Rih!)\\*Fa\\l7$Fhp$
\"1MON!RxH$>F/7$F]q$\"1X3(>q&*zp\"F/7$Fbq$\"1Rw'33a%HRFa\\l7$Fgq$!1\\
\")Rtrtf6F/7$F[r$!1Pvmd!f:)>F/7$F`r$!15d*\\47Yb\"F/7$Fer$!1D>pZ<cz9Fa
\\l7$Fjr$\"1#*f$R,T>N\"F/7$F_s$\"1S(fCe^)**>F/7$Fds$\"1:#HUWluQ\"F/7$F
is$!1_$y5.l%H**!#<7$F^t$!1N?:](zM_\"F/7$Fct$!113=a`d()>F/7$Fht$!1#>n+n
5\"*>\"F/7$F]u$\"1Omk>rF]MFa\\l7$Fbu$\"1r\"QGKJ<n\"F/7$Fgu$\"1W%z'=!=
\\%>F/7$F\\v$\"1HQ16(oU#**Fa\\l7$Fav$!1%f+Qao[&eFa\\l7$Ffv$!17,$4jHWz
\"F/7$F[w$!1I;P:;`s=F/7$F`w$!1')GtZJt0xFa\\l7$Few$\"10nml`'*p\")Fa\\l7
$Fjw$\"14]Un\"*p*)=F/7$F_x$\"1q?f2E_r<F/7$Fdx$\"1-=9e9Tp`Fa\\l7$Fix$!1
'*Ge9!=g.\"F/7$F^y$!12v*zo$3c>F/7$Fcy$!14/io\\VV;F/7$Fhy$!1pk76b,^HFa
\\l7$F^z$\"1Nx\"HV.#R7F/7$Fcz$\"1,]dq%oD*>F/7$Fhz$\"1`8z#eE-\\\"F/-F][
l6&F_[lF`[lF(F(" 2 210 165 165 2 0 1 0 2 9 0 4 2 1 45 45 10030 0 
10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 262 292 0 0 0 
0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 321 "It appears that the init
ial velocity v = 2 is a critical velocity to send the pendulum into a \+
circular or orbital motion rather than the standard oscillation.  sinc
e this is maximum displacement, i would say our linear approximation i
s out the window.  by rotating the axis, we could still obtain the per
iod if we wish.  " }}{PARA 0 "" 0 "" {TEXT -1 403 "Notice that the pen
dulum rotates to 180 degrees (ie the \"top\" of the swing) and then st
ays there for a bit, debating what to do next.  Then, it keeps rotatin
g in the same direction until it gets to 180+2Pi degrees, and then it \+
pauses again before rotating in the same direction. Notice that the no
n-linear de recognizes the difference between an angle of 0,6.28~2Pi,4
Pi,... while the linear de does not." }}}{SECT 0 {PARA 4 "" 0 "" 
{TEXT -1 6 "v=2.01" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "k:=od
eplot(dsolve(linivp(2.01),x(t),numeric,maxfun=8000),[t,x(t)],0..40,col
or=RED):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "l:=odeplot(dsol
ve(nonlinivp(2.01),x(t),numeric,maxfun=8000),[t,x(t)],0..40,color=BLUE
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "display(\{k,l\});" }}
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Z/bjH,DFf^l7$F]w$\"1l0+eQl_EFf^l7$Fbw$\"1U-l`^4ZFFf^l7$Fgw$\"1Mq2$o\"z
$z#Ff^l7$F\\x$\"1xe-.Yb<GFf^l7$Fax$\"1Tv<]rRMGFf^l7$Ffx$\"1tQ@^V7cGFf^
l7$F[y$\"1!>ryTpv*GFf^l7$F`y$\"1JC-!efB)HFf^l7$Fey$\"1M-qq,jEJFf^l7$Fj
y$\"1e6f!ff'yKFf^l7$F_z$\"1gK*pP/UP$Ff^l7$Fdz$\"1-C=Bf_@MFf^l7$Fiz$\"1
84$o!)RbW$Ff^l-F^[l6&F`[lF(F(Fa[l" 2 247 96 96 2 0 1 0 2 9 0 4 2 1 45 
45 10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 0 
0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 309 "In graph 4, we c
an tell that the frequency has increased with a larger initial velocit
y.  The non-linear pendulum just keeps swinging around and around and \+
around all in the same direction, and so the angle keeps increasing.  \+
The linear de pendulum does not recognize the difference in angle that
 is occuring." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 
0 "" {TEXT -1 54 "Number 5 (Similar to Katie C and Amanda's Problem Se
t)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }{TEXT 258 0 "" }{TEXT 259 75 "These are the non-linear equations
 using the different values for velocity." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 48 "eq5:=diff(x(t),t$2)+.5*diff(x(t),t)+sin(x(t))=0:" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "h1:=dsolve(\{eq5,x(0)=0,D(
x)(0)=4\},x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 
"q:=odeplot(h1,[t,x(t)],0..20,color=red):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 48 "eq17:=diff(x(t),t$2)+1*diff(x(t),t)+sin(x(t))=0:" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "h2:=dsolve(\{eq17,x(0)=0,D
(x)(0)=4\},x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
47 "plot5:=odeplot(h2,[t,x(t)],0..20,color=yellow):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 48 "eq18:=diff(x(t),t$2)+2*diff(x(t),t)+sin(x
(t))=0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "h3:=dsolve(\{eq1
8,x(0)=0,D(x)(0)=4\},x(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "plot6:=odeplot(h3,[t,x(t)],0..20,color=blue):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 0 "" }{TEXT 261 51 "Th
is is the solution for the non-linear equations. " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 25 "display(\{q,plot5,plot6\});" }}{PARA 13 "" 
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F/-Fb[l6&Fd[lFe[lF(F(" 2 203 96 96 2 0 1 0 2 9 1 4 2 1 45 45 10030 0 
10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 233 103 0 0 0 
0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 0 "" }{TEXT 
265 535 "In the case b=0.5 (red graph), the pendulum rotates 6.28~2Pi=
 360 degrees and then oscillates around the equilibrium position (the \+
de solution recognizes that this is 2Pi and not 0 as the angle).  The \+
angle hovers around this equilibrium position, so we see the graph hov
ering around 2PI.  In b=1 (yellow graph), the pendulm rotates almost t
o 180 degrees, it drops the reverse direction to a negative angle, and
 oscillates constantly at about x=0.  In b=2 (blue graph), the pendulu
m rotates about 90 degrees and then levels off at x=0." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 262 
0 "" }{TEXT 263 68 "These are the linear equations for the different v
alues of velocity." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "eq6:=
diff(x(t),t$2)+.5*diff(x(t),t)+x(t)=0:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "h4:=dsolve(\{eq6,x(0)=0,D(x)(0)=4\},x(t),numeric):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "r:=odeplot(h4,[t,x(t)],0..
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }{TEXT 266 0 "" }{TEXT 267 369 "In the case b=0.5 (red graph),
 the pendulum rotates a little less than 180 degrees,  then oscillates
 constantly below and above the x-axis.  In b=1 (yellow graph), the pe
ndulm rotates between 90 and 180 degrees, it drops to a negative angle
, and oscillates constantly at about x=0.  In b=2 (blue graph), the pe
ndulum rotates about 90 degrees and then levels off at x=0." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 268 0 "" }{TEXT 269 134 "T
he linear solutions do not rotate as far around as the non-linear solu
tions and the linear solutions oscillate more below the x-axis." }}}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 56 "Number 6 (Similar to David's and \+
Rebekah's Problem Sets)" }}{PARA 0 "" 0 "" {TEXT -1 168 "In this probl
em, we look at the effect of a periodic external foce on the pendulum.
  The book has chosen a damping coefficient that is more typical of ai
r resistance.  " }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 4 "D.E." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "restart:with(plots):with(plottools)
:with(DEtools):" }}{PARA 7 "" 1 "" {TEXT -1 37 "Warning, new definitio
n for translate" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "The omega is o
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "eq1:=diff(y(t),t$2)+.05*diff(y(t),t
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$6$7TF'7$F*$\"1&=f`&\\%)*e\"F/7$F1$\"1Qa'\\_#y\\8F/7$F6$!19lzRY!>p$F/7
$F;$!16N!*[hXhZF/7$F@$\"1\\EhI^%R0$F/7$FE$\"1U,#4?9!Q#)F/7$FJ$\"1>_ted
l\\uFgn7$FO$!1$zMxu+'o#*F/7$FT$!1AFkgVZveF/7$FZ$\"1&QuXsskk'F/7$Fin$\"
17\">*)QWU/*F/7$F^o$!1DI&f![M!*=F/7$Fco$!1X&Qp.S-z)F/7$Fho$!1A@(R&[<9>
F/7$F]p$\"1l#y\"HQQKjF/7$Fbp$\"1Cx.oRz/KF/7$Fgp$!18rQpfa6QF/7$F\\q$!1'
HGB&R70DF/7$Faq$\"11\"pH6_)REF/7$Ffq$\"1k\\h(z7O;\"F/7$F[r$!1`9H^<kqIF
/7$F`r$!1rAPzzI>ZFgn7$Fer$\"1EZTB/L*[%F/7$Fjr$\"15uqud\">B\"F/7$F_s$!1
6x*)oG*\\w&F/7$Fds$!1S$G3arYX$F/7$Fis$\"1?!37UZ_s&F/7$F^t$\"1ol[zs#o>'
F/7$Fct$!1/$HFEF;$QF/7$Fht$!1ph<B@\"\\,)F/7$F]u$\"1b1yB0aBqFgn7$Fbu$\"
1BNt:\"**G.)F/7$Fgu$\"1Z\\n)*zH8AF/7$F\\v$!1;;4%R7ta'F/7$Fav$!1R&f%\\#
>_#QF/7$Ffv$\"1J/u\\iIJYF/7$F[w$\"1l%QE$G`\"4%F/7$F`w$!1Rf%o.>wE$F/7$F
ew$!1b3A#>\"f&p$F/7$Fjw$\"1o@_yA^*z#F/7$F_x$\"1PTE<!)*pW$F/7$Fdx$!1OK>
*pZ7%HF/7$Fix$!1/LmcmBhQF/7$F^y$\"1'R?y:e.1$F/7$Fcy$\"1WF<x&y8'\\F/7$F
hy$!1k#z!=#48_#F/7$F]z$!1DiG*eK9G'F/7$Fbz$\"1_2Lj4AF5F/7$Fgz$\"1)R=F@p
R6(F/-F\\[l6&F^[l$\")viobFV$\")!\\DP\"FV$\")%yg>%FV" 2 195 131 131 2 
0 1 0 2 9 0 4 2 1 45 45 10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 
0 0 0 0 1 1 0 0 0 276 260 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 88 "The frequency that moves the pendulum farthest away from the eq
uilibrium point is 0.8.  " }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 36 "Li
near functions with varying omega." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
75 "This equation represents the linear function with the omega variab
e of 0.6." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "eq2a:=diff(y(t),t$2)+.
05*diff(y(t),t)+(y(t))=.3*cos(.6*t):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "ivpL.6:=\{eq2a,y(0)=0,D(y)(0)=0\}:" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 75 "This equation represents the linear function wi
th the omega variabe of 0.8." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "eq2
b:=diff(y(t),t$2)+.05*diff(y(t),t)+y(t)=.3*cos(.8*t):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "ivpL.8:=\{eq2b,y(0)=0,D(y)(0)=0\}:
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "This equation represents the \+
linear function with the omega variabe of 1." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 55 "eq2c:=diff(y(t),t$2)+.05*diff(y(t),t)+y(t)=.3*cos(1*t
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "ivpL1:=\{eq2c,y(0)=0,
D(y)(0)=0\}:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "This equation rep
resents the linear function with the omega variabe of 1.2." }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 57 "eq2d:=diff(y(t),t$2)+.05*diff(y(t),t)+y(t)=
.3*cos(1.2*t):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "ivpL1.2:=
\{eq2d,y(0)=0,D(y)(0)=0\}:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "Thi
s will plot the function with the omega of .6 in coral." }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 86 "inL.6:=odeplot(dsolve(ivpL.6,y(t),numeric,maxf
un=4000000),[t,y(t)],0..60,color=coral):" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 59 "This will plot the function with the omega of .8 in viole
t." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "inL.8:=odeplot(dsolve(ivpL.8,
y(t),numeric,maxfun=4000000),[t,y(t)],0..60,color=violet):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 58 "This will plot the function with the omeg
a of 1 in orange." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "inL1:=odeplot(
dsolve(ivpL1,y(t),numeric,maxfun=4000000),[t,y(t)],0..60,color=orange)
:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "This will plot the function \+
with the omega of 1.2 in navy." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "i
nL1.2:=odeplot(dsolve(ivpL1.2,y(t),numeric,maxfun=4000000),[t,y(t)],0.
.60,color=navy):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "display
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1 0 2 9 0 4 2 1 45 45 10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 0 
0 0 0 1 1 0 0 0 185 281 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 84 "The frequency that moves the pendulum farthest away from the eq
uilibrium point is 1." }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 20 "Freque
ncies together" }}{PARA 0 "" 0 "" {TEXT -1 154 "By comparing each freq
uency graph together, I can tell that the graph where omega is .8 beha
ves vary differently between the linear and non-linear graphs." }}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 11 "Omega of .6" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 22 "display(inNL.6,inL.6);" }}{PARA 13 "" 1 "" 
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{PARA 0 "" 0 "" {TEXT -1 442 "As in example 10, to determine which ome
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 Notice that b,m and k can be solved for by looking at the de setup fr
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7v@TLCF47$Fct$!1JqM2u+&=&F47$Fht$!1'\\!Qc_toNF47$F]u$\"1[f1$f(e.WF47$F
bu$\"1s_jml&H`%F47$Fgu$!1xx^0$R5T$F47$F\\v$!1<5_^o%)z_F47$Fav$\"1Kh$Q#
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ew$!19GBdZb4KF-7$Fjw$\"1U.HFiD?fF47$F`x$\"1&4eggqsh\"F47$Fex$!1MB\"GLM
hc&F47$Fjx$!126nei/OGF47$F_y$\"1o$3$fg9X\\F47$Fdy$\"1uk?e'[)=RF47$Fiy$
!1-_9zW1(3%F47$F^z$!1%*[j@Kw8[F47$Fcz$\"1Fg%e,II.$F47$Fhz$\"1\\F3$e$)y
Z&F47$F][l$!1+kd#)zdL=F4-Fb[l6&Fd[l$\")+++!)Fg[l$\")Vyg>Fg[lFeel" 2 
206 200 200 2 0 1 0 2 9 0 4 2 1 45 45 10030 0 10056 10074 0 0 0 20530 
0 12020 0 0 0 0 0 0 0 1 1 0 0 0 318 27 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 318 "As can be seen, the non-linear function resonates
 and stays within -2 and 2.  So lets consider the linear function.  Th
e linear function is resonating back and forth between 6 and -6.  It i
s not showing signs of going beyond either of its boundary points, thu
s we can assume that the amplitude will not go to infinity." }}{PARA 
0 "" 0 "" {TEXT -1 133 "Note that we need to increase maxfun in order \+
to get it to plot out this far.  Otherwise, the numerical plot will ju
st stop plotting." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 
17 "Extra Exploration" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "od
eplot(dsolve(ivp.8,y(t),numeric,maxfun=4000000),[t,y(t)],0..1000,color
=violet);" }}{PARA 13 "" 1 "" {INLPLOT "6$-%'CURVESG6#7T7$\"\"!F(7$$\"
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=fzFD$\"1<<#)[rC7LF@7$$\"+>`Ej\")FD$\"12N6ebzEHF@7$$\"+_pMn$)FD$\"1!z)
RF*oBL$F@7$$\"+&eG9d)FD$\"1IE$**\\m^/$F@7$$\"+=-^v()FD$\"1@x97m&\\4$F@
7$$\"+^=fz*)FD$\"1fmT>j!eI$F@7$$\"+%[tO=*FD$\"1zv>s\"yx#HF@7$$\"+<^v(Q
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aaj5$F@7$$\"+-+++5!\"'$\"1P6n%H$)))H$F@-%'COLOURG6&%$RGBG$\")#R!)4$F,$
\")t8V=F,Fb[l" 2 197 141 141 2 0 1 0 2 6 0 4 2 1 45 45 10030 0 10056 
10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 171 143 0 0 0 0 0 0 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Notice that we get some interes
ting behavior!" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 65 "Number 7 (Simila
r to Parts of  Melinda's and Eric's Problem Sets)" }}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 19 "Problem 7 Statement" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 101 "This problem considers a cable attached to posts at both
 ends, going thru the points (0,0) and (1,0)." }}{PARA 0 "" 0 "" 
{TEXT -1 102 "The height of the cable at any given point between is re
presented by the function y''=c(sqrt(1+y'^2))." }}{PARA 0 "" 0 "" 
{TEXT -1 324 "y(0)=0 and y(1)=0 are called boundary value conditions. \+
 The term c is a constant representing the length of the cable.  We wi
ll use c=1.  For some value of s, for with y(0)=0 and y'(0)=s, the con
dition y(0)=1 is satisfied.  The problem wants us to find a workable v
alue of s and determine that amount of \"dip\" in the cable." }}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 62 "Finding s so that the solution wi
th D(y)(0)=s goes thru (1,0)." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 
"Notice that s must be negative since otherwise the cable will not dip
 down (s is the slope of the tangent line at the point (0,0) ).  So, w
e will guess and check various negative values of s." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 6 "c7:=1:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "de7:=diff(y(t),t$2)-(c7*(sqrt(1+(diff(y(t),t))^2)))=0
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "y7:=s->dsolve(\{de7,y(
0)=0,D(y)(0)=s\},y(t),numeric):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 50 "p1:=odeplot(y7(-.51),[t,y(t)],0..1.02, color=red):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "p3:=odeplot(y7(-.52),[t,y(t)],0..1.
02, color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "p4:=ode
plot(y7(-.53),[t,y(t)],0..1.02, color=magenta):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 18 "display(p1,p3,p4);" }}{PARA 13 "" 1 "" 
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45 45 10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 
0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "Notice that the
 blue curve with D(y)(0)=.52 goes thru 0,1." }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 60 "We can zoom in to find the value of D(y)(0) to any accu
racy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "zoom1:=odeplot(y7(
-.520),[t,y(t)],0.96..1.02, color=red):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 58 "zoom2:=odeplot(y7(-.521),[t,y(t)],0.96..1.02, color=b
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"" {MPLTEXT 1 0 27 "display(zoom1,zoom2,zoom3);" }}{PARA 13 "" 1 "" 
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,8*Fip7$F^[l$\"18FQD#yRz*Fip-Fc[l6&Fe[lFf[lFi[lFf[l" 2 169 144 144 2 
0 1 0 2 9 0 4 2 1 45 45 10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 
0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
56 "So, we see that D(y)(0)=-.522 (blue curve) goes thru 0,1" }}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 23 "Finding the maximum dip" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "To find the maximum dip, the valu
e for s=D(y)(0) will be put into the original equation and we'll dsolv
e it. " }}{PARA 0 "" 0 "" {TEXT -1 61 "Notice that I am not using a nu
merical approximation anymore." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 77 "realsolution:=unapply(rhs(evalf(dsolve(\{de7,y(0)=0,D(y)(0)=-.
522\},y(t)))),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-realsolutionGR
6#%\"tG6\"6$%)operatorG%&arrowGF(,&-%%coshG6#,&9$\"\"\"$!+9:-3]!#5F2F2
$!+DV/G6!\"*F2F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 158 "Since th
e maximum deip will be at an inflection point, the point where the der
ivative of the realsolution is equal to zero will be the location of t
hat dip.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "realsolutiond
erivative:=diff(realsolution(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%7realsolutionderivativeG-%%sinhG6#,&%\"tG\"\"\"$!+9:-3]!#5F*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "Now we want to solve realsolution
derivative=0 for the t value.  We can estimate that the t value occurs
 at around .5, so we will use fsolve to do this" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 48 "a7:=fsolve(realsolutionderivative=0,t=0.4..0.6
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#a7G$\"+9:-3]!#5" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 95 "To obtain the actual depth of the dip, th
is value for t must be put back into the realsolution." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "realsolution(a7);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#$!*DV/G\"!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
103 "Notice that this value does agree with the approximate value of -
.13 that can be observed on the graph." }}}}}{MARK "4 0 0" 70 }
{VIEWOPTS 1 1 0 1 1 1803 }