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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Amanda Smith" }}{PARA 0 "
" 0 "" {TEXT -1 12 "Katie Causby" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 8 "pp.96 #7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 352 "We had a parachutist with a ma
ss of 75kg drops from a helicopter 2000m above the ground. The forces \+
are the influence of gravity and air resistance. When the chute is clo
sed, the constant k1 = 30 kg/sec. When the chute is open, the constant
 k2 = 90 kg/sec. The chute does not open until after 20 m/sec. After h
ow many seconds will she reach the ground?" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "First we solved the differential e
quation for (m*dv/dt=m*g-k*v) where the initial v= 0." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "v1:=unapply(rhs(dsolve(\{diff(V(t),
t)=(m*g-k1*V(t))/m,V(0)=0\},V(t))),t);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#v1GR6#%\"tG6\"6$%)operatorG%&arrowGF(,&*&*&%\"mG\"\"\"%\"gGF0
\"\"\"%#k1G!\"\"F0*&*(-%$expG6#,$*&*&F3F09$F0F2F/F4!\"\"F0F/F2F1F2F2F3
F4F>F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "m:=75; g:=9.8
1; k1:=30;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\"#v" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"gG$\"$\")*!\"#" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#k1G\"#I" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "Now plugging
 in the other variables for mass(m), gravity(g), and constant(k1), we \+
solved for v after 20 seconds." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 14 "evalf(v1(20));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+ysn^C!\"
)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Now we want to integrate v(
t), solve for c with the initial position,x(t), at 0. Then solve the x
(t) for the 20 seconds." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "
x1:=unapply(int(v1(t),t)+c,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x
1GR6#%\"tG6\"6$%)operatorG%&arrowGF(,(9$$\"+++]_C!\")-%$expG6#,$F-$!++
+++S!#5$\"+++DJhF0%\"cG\"\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "c:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG\"\"!" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "x1(20);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$\"+\"o0_!\\!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
184 "After the shoot opens, solve the differential equation with V(0) \+
= initial velocity value v1(20). Then substitute in the second constan
t, k2, and integrate to get x2(t). Solve for c2. " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 83 "v2:=unapply(rhs(dsolve(\{diff(V(t),t)=(m*g-k
2*V(t))/m,V(0)=evalf(v1(20))\},V(t))),t);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%#v2GR6#%\"tG6\"6$%)operatorG%&arrowGF(,&*&\"\"\"F.%#k
2G!\"\"$\"+-+]dt!\"(*&*&-%$expG6#,$*&F/\"\"\"9$F;#!\"\"\"#vF;,&$!+,+vy
OF;F;F/$\"+R'QeA\"\"\"!F;F.F/F0$\"+++++?!#<F(F(F(" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 7 "k2:=90;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
#k2G\"#!*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "x2:=unapply(in
t(v2(t),t)+c2,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x2GR6#%\"tG6\"
6$%)operatorG%&arrowGF(,(9$$\"+-++v\")!\"*-%$expG6#,$F-$!+++++7F0$!+)R
9=O\"!\")%#c2G\"\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "c2:=solve(x2(0)=x1(20),c2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#
c2G$\"+@rQT]!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 231 "Since we wan
t to know when the parachutist will reach the ground, we subtracted th
e distance the parachutist fell when the parchute was closed from the \+
original height of 2000m. We then set x2 equal to this amount and solv
e for t. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "solve(x2(t)=20
00-x1(20),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+\"eu(H7!\"(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "The parachutist reached the ground
 in 122.98 seconds." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}}
{MARK "19 0 0" 44 }{VIEWOPTS 1 1 0 1 1 1803 }