{VERSION 3 0 "APPLE_PPC_MAC" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Keith Seal , Sarah thompso n " }}{PARA 0 "" 0 "" {TEXT -1 8 "P 82 #17" }}{PARA 0 "" 0 "" {TEXT -1 6 "2-9-00" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 286 "In a lake, there is 7 million tons fish. The fish incr ease portional to the amount of fish in the lake. The rate of increas e is two fish per year proportional to the amount of fish. Also , the amount of fish remmoved comerically is a constant rate of 15 million \+ tons of fish per year." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Our DE is dp/dt = 2*p - 15,000,000" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "p:=unapply(rhs(dsolve(\{diff (p(t),t)=(2*p(t)-15000000),p(0)=7000000\},p(t))),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&\"(++](\" \"\"-%$expG6#,$9$\"\"#!'++]F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "Here we solved for time We got the same answer by hand as Maple d id. We just had to simply manipulate our equation to get the same ans wer maple did ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(p( t)=0,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#\"#:#\"\"\"\"\"# " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "Here we solve for the condtio n of how long it would take for the fish to deplenish completely." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#$\"+,^-a8!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "Maple answer of 1.35 years is the same answer we got when done by hand. It would take 1.35 years for lake to be empty of fish." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 " In the next part of the problem, we want to solve for the rate of incr ease to where the amount of fish stays the same as each year that goes by, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "fsolve((k*7000000-15000000)=7000000,k);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#$\"+Vr&G9$!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Maple is bei ng difficult but by hand we got the equation to be: 7 = 7k - 15 and \+ solving for k, we got 3.1428" }}{PARA 0 "" 0 "" {TEXT -1 112 "Actually , without using the unapply command, this problem was quite easy. Sin ce we used the original function " }}{PARA 0 "" 0 "" {TEXT -1 108 "dp /dt = kp - 15 and plugged in the known, t=1, dp/dt=7, p=7. Then we j ust solved for the unknown rate, k. " }}}}{MARK "13" 0 }{VIEWOPTS 1 1 0 1 1 1803 }