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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Michelle and Keith" }}
{PARA 0 "" 0 "" {TEXT -1 10 "problem 10" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 289 "Our problem involved a 2kg mass tha
t was dropped from a height of 30 meters above some water.  We were gi
ven a k value for air and water.  The k value for air was 10 kg/s, and
 the kvalue for water was 100 kg/s.  We were supposed to find the dist
ance the mass had fallen at time=60 seconds." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 8 "restart;" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 78 "Our differential equation for while the mass was in the air was
 dv/dt=9.81-5v." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "eq:=unap
ply(rhs(dsolve(\{diff(v(t),t)=(9.8-5*v(t)),v(0)=0\},v(t))),t);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqGR6#%\"tG6\"6$%)operatorG%&arrowG
F(,&$\"++++g>!\"*\"\"\"-%$expG6#,$9$!\"&$!++++g>F/F(F(F(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 221 "We got the same answer by doing this by \+
hand using first order linear and seperable (either one).  We then too
k the integral of the velocity function to get the position function, \+
and solved for the constant using x(0)=0." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 29 "x:=unapply(int(eq(t),t)+c,t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"xGR6#%\"tG6\"6$%)operatorG%&arrowGF(,(9$$\"++++g>!
\"*-%$expG6#,$F-$!\"&\"\"!$\"++++?R!#5%\"cG\"\"\"F(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "c:=solve(x(0)=0,c);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"cG$!++++?R!#5" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 155 "We solved for the time when the object had fallen thirty
 meters because this is when it hit the water, which is when the diffe
rential equation will change." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 17 "solve(x(t)=30,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+XAh]:!
\")$!+=0*=\"))!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "sol1:=
15.50612245;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%%sol1G$\"+XAh]:!\")" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 151 "Here we solved for the velocity of the object at the ins
tant before it hits the water.  This, in turn became the intitial cond
ition for our second d.e." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "sol2:=eq(sol1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol2G$\"++++
g>!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "The new d.e. was dv/dt
=2.45-50v, with the initial condition of v(15.5)=1.96.  We also achiev
ed the same solution by doing this d.e. by hand using the separable me
thod" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "eq2:=unapply(rhs(ds
olve(\{diff(v(t),t)=(2.45-50*v(t)),v(sol1)=sol2\},v(t))),t);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%$eq2GR6#%\"tG6\"6$%)operatorG%&arrowGF(,&$
\"+++++\\!#6\"\"\"-%$expG6#,$9$!#]$\"+4!Qs#)*\"$F$F(F(F(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 137 "Just as before, we took the integral of \+
the velocity function to get the position function, and used the initi
al condition of x(15.5)=30." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
32 "x2:=unapply(int(eq2(t),t)+c1,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%#x2GR6#%\"tG6\"6$%)operatorG%&arrowGF(,(9$$\"+++++\\!#6-%$expG6#,$
F-$!#]\"\"!$!+-wWl>\"$E$%#c1G\"\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 26 "c1:=solve(x2(sol1)=30,c1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#c1G$\"++?%y#H!\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "finalsoln:=x2(60);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%*finalsolnG$\"++?%=A$!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "
Here we solved for when time is equal to 60 seconds and found that the
 total distance the object had fallen was 32.2 meters from the origina
l height." }}}}{MARK "17 0 0" 149 }{VIEWOPTS 1 1 0 1 1 1803 }