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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT 256 38 "Problem 23 in Section 2.3 Boyce's book" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "First need to l
oad library of routines where DEplot is contained." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }{TEXT -1 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
257 19 "BALL THROWN UPWARD " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 84 "There are two ODE modelling this phenomena one \+
corresponds to the upward motion and " }}{PARA 0 "" 0 "" {TEXT -1 35 "
another one to the downward motion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "foeq:=diff(v(t),t)=-9.8-(v
(t))^2/(1325*.15);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 44 "Obtain the direction field for this equation" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 64 "DEplot(foeq, v(t),t=0..1,v=-1..20,arrows=THIN,title=`Figure 1`);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "To obtain the a numerical app
rox to our IVP, we need to incorporate the initial condition." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 73 "DEplot(foeq, v(t),t=0..3,v=-1..20,\{[0,20]\},arrows=THIN,title=`
Figure 1`);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 50 "The colored  line is the solution to our problem. " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT 258 19 "ANALYTICAL SOLUTION" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 107 "The equation to be sol
ved is a separable equation and can be easily computed by hand. Here, \+
we use MAPLE to" }}{PARA 0 "" 0 "" {TEXT -1 114 "obtain its solution. \+
After separation of variable this is the expression in terms of the va
riable \"v\" that results" }}{PARA 0 "" 0 "" {TEXT -1 12 "in the left.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "Int1:=1/(v^2+a^2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "int(Int1,v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "dsolve(foeq,v(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 
"evalf(dsolve(foeq,v(t)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 154 "This expresion represents the \"general \+
solution\". To obtain a particular solution, we need to include an ini
tial value inside the above command as follows" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "evalf(dsolv
e(\{foeq,v(0)=20\},v(t)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 133 "We can obtain a function reperesenting t
he above particular solution, and later plot that function performing \+
the following procedure" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 47 "First separate the rhs of the above expression." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 48 "tempv:= rhs(evalf(dsolve(\{foeq,v(0)=20\},v(t))));" }{TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "Then, use that rhs to define a fun
ction v(t) using the command unapply." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "v1:=unapply(tempv,t);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
105 "To obtain the value of  \"t\" in which the ball reach its maximum
 height. We set v1(t)=0 and solve for \"t\"." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "tmax:=fsolve(v1(t
)=0,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 113 "The position function is obtained by integration of v1(t
) with respect to t and we add the value 30 corresponding" }}{PARA 0 "
" 0 "" {TEXT -1 60 "to the distance from the ground to the roof of the
 building." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 16 "Type of integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "v11:=a*tan(b*t+c);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "int(v11,t);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 31 "Position:=int(v1(s),s=0..t)+30;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Positionf:=unapply(Position,t);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Maximum Height:" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Positionf
(tmax);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 53 "Now, we can make a graphic of this position function." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 66 "plot(Positionf(t), t=0..tmax, Positionf=25..50, title=`Figura 1`
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "GOING DOWN THE EQUATION CHA
NGE BECAUSE THE AIR RESISTANCE CHANGE SIGN." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "foeq:=diff(v(t),t
)=-9.8+(v(t))^2/(1325*.15);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 44 "Obtain the direction field for this equa
tion" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "DEplot(foeq, v(t),t=0..1,v=-20..0,arrows=THIN,title=`
Figure 1`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "To obtain the a
 numerical approx to our IVP, we need to incorporate the initial condi
tion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 72 "DEplot(foeq, v(t),t=0..5,v=-50..0,\{[0,0]\},arrows=TH
IN,title=`Figure 1`);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 50 "The colored  line is the solution to our problem. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT 259 19 "ANALYTICAL SOLUTION" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "The eq
uation to be solved is a separable equation and can be easily computed
 by hand. Here, we use MAPLE to" }}{PARA 0 "" 0 "" {TEXT -1 114 "obtai
n its solution. After separation of variable this is the expression in
 terms of the variable \"v\" that results" }}{PARA 0 "" 0 "" {TEXT -1 
12 "in the left." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 18 "Int2:=1/(v^2-a^2);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 12 "int(Int2,v);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "dsolve(foeq,v(t));" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(dsolve(foeq,v(t)));
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
154 "This expresion represents the \"general solution\". To obtain a p
articular solution, we need to include an initial value inside the abo
ve command as follows" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "evalf(dsolve(\{foeq,v(0)=0\},v(t)))
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
131 "This expresion represent a particular solution for the given init
ial value. Unfortunately, it is not a function v(t) yet. In fact, " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 133 "We can obtain a function reperesenting
 the above particular solution, and later plot that function performin
g the following procedure" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 47 "First separate the rhs of the above expression." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 47 "tempv:= rhs(evalf(dsolve(\{foeq,v(0)=0\},v(t))));" }{TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "Th
en, use that rhs to define a function v(t) using the command unapply.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "v1:=unapply(tempv,t);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Now, we can make a graphi
c of this function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 49 "plot(v1(t), t=0..5, v1=-50..0, title=`Fig
ura 1`);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
110 "To find the time from th highest point where the velocity is zero
 to the ground. We proceed by trial an error." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Position2:=i
nt(v1(t),t=0..3.277723);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Therefore t=3.277723 is that \+
time. The total time in the way up plus the way down is " }}{PARA 0 "
" 0 "" {TEXT -1 58 "                                                  \+
        " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "TotalTime:= 3.2
77723+tmax;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "
65" 0 }{VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }