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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 28 "Area, Displacement, Distance" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 19 "Calculus
 II Project" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" 
{TEXT 285 10 "Objective:" }}{PARA 0 "" 0 "" {TEXT -1 92 "To  learn how
 to use integrals to compute areas,  displacements, and distances. You
 can use " }{TEXT 266 5 "Maple" }{TEXT -1 68 " to graph functions, to \+
solve equations, and to evaluate integrals. " }}}{PARA 4 "" 0 "" 
{TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" {TEXT 286 17 "Solved Example 1
:" }}{PARA 0 "" 0 "" {TEXT -1 40 "Compute the area enclosed by the cur
ve  " }{TEXT 259 3 " y " }{TEXT -1 2 "= " }{XPPEDIT 18 0 "1-x-x^2;" "6
#,(\"\"\"F$%\"xG!\"\"*$F%\"\"#F&" }{TEXT -1 11 "  and the  " }{TEXT 
260 1 "x" }{TEXT -1 6 "-axis." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 9 "So
lution:" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
87 "First let's plot the curve to get an idea of  the region whose are
a we need to compute." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 24 "plot(1-x-x^2,x=-2.5..1);" }}{PARA 13 "" 
1 "" {GLPLOT2D 256 158 158 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$!3++++++++D
!#<$!3+++++++]FF*7$$!3vmmTg*4PU#F*$!3y$H1=,g1X#F*7$$!3YLe*[SItN#F*$!3Q
?0&HBw'*>#F*7$$!3!om\"H_'zEG#F*$!34QZpUn%z#>F*7$$!3pm;/mS`2AF*$!31(y#p
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!4Kz(FP7$$!3OLL$eres6\"F*$\"31WyFxn\"**o)FP7$$!3\")*\\il=s</\"F*$\"3%o
(Gny*G[c*FP7$$!3E(*\\(of6*)p*FP$\"3gOv/)H-#H5F*7$$!3_JLe*[3*[*)FP$\"3y
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LvLKj'=\"F*7$$!3?m;H2chXnFP$\"3T(z::EG&>7F*7$$!3-**\\P%G$)Q2'FP$\"3u*R
:pun%Q7F*7$$!3=lm\"H(RR[`FP$\"3Y`jR;iy[7F*7$$!35+++DD*))f%FP$\"3I\"*\\
$z7\"R[7F*7$$!33)**\\7e^c'QFP$\"3v$z[m`KrB\"F*7$$!3?(*\\ilK?cJFP$\"3m!
3Bg8/g@\"F*7$$!3)3+](o=[oBFP$\"33D!\\K7^2=\"F*7$$!3!eLL$e\"z1m\"FP$\"3
V1T;j$*[Q6F*7$$!3'o&***\\(o[\\!*Fdo$\"3#*>*zua0B3\"F*7$$!3k#o;HdX9?#Fd
o$\"3F:\"o%>)H:-\"F*7$$\"3m\\++vBF&G&Fdo$\"3)f0A%e'QNW*FP7$$\"3xl;/^!p
HB\"FP$\"3)fnv3o4]h)FP7$$\"3?,](=s8$p>FP$\"3)>[3YimGk(FP7$$\"3Snm;H_A*
o#FP$\"3o!o9v`\"e(e'FP7$$\"3Q)*\\Pfe!HW$FP$\"3cJJ(\\1M<P&FP7$$\"3ARLL$
))*yoTFP$\"35@%Gv?HL4%FP7$$\"37NLeR666\\FP$\"3M7F)yf()pn#FP7$$\"3mr;/,
cGZcFP$\"3/p])H$4`j6FP7$$\"3i/++vMvBjFP$!3L#ouDw#RFKFdo7$$\"37pm\"z*>1
*4(FP$!3#*HJ%H7I(Q@FP7$$\"3![LLL=2Dz(FP$!3K(\\jN+C['QFP7$$\"3$3+vVQk=`
)FP$!31i;rr`86eFP7$$\"3*H+DccB&R#*FP$!3]z*)fPJSwxFP7$$\"\"\"\"\"!$!\"
\"Fgz-%'COLOURG6&%$RGBG$\"#5Fiz$FgzFgzF`[l-%+AXESLABELSG6$Q\"x6\"Q!6\"
-%%VIEWG6$;$!#DFizFez%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 13 "" 1 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 44 "To compute the area we have to find the \+
two " }{TEXT 261 1 "x" }{TEXT -1 23 "-intercepts. We'll use " }{TEXT 
267 5 "Maple" }{TEXT -1 25 " to solve the equation   " }{TEXT 262 4 "y
 = " }{TEXT 268 1 "0" }{TEXT 269 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "xint:=[solve(1-x-x^2,x)]
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%xintG7$,&#!\"\"\"\"#\"\"\"*&#F
*F)F**$-%%sqrtG6#\"\"&F*F*F(,&F'F**&#F*F)F*F.F*F*" }}}{PARA 11 "" 1 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Since  " }{TEXT 263 1 "y
" }{TEXT -1 21 "  is positive for all" }{TEXT 256 2 " x" }{TEXT -1 75 
" between the two intercepts, the area in question is the definite int
egral." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "area:=Int(1-x-x^2,x=xint[2]..xint[1]);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%%areaG-%$IntG6$,(\"\"\"F)%\"xG!\"\"*$)F*\"\"#F
)F+/F*;,&#F+F.F)*&#F)F.F)-%%sqrtG6#\"\"&F)F),&F2F)*&#F)F.F)*$F5F)F)F+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "area:=int(1-x-x^2,x=xin
t[2]..xint[1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%areaG,,*$-%%sqrt
G6#\"\"&\"\"\"!\"\"*&#F+\"\"#F+*$),&#F,F/F+*&#F+F/F+F&F+F,F/F+F+F,*&#F
+F/F+),&F3F+*&F7F+F'F+F+F/F+F+*&#F+\"\"$F+*$)F2F=F+F+F,*&#F+F=F+)F9F=F
+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "simplify(%); " }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#\"\"&\"
\"\"#!\"&\"\"'" }}}{PARA 12 "" 1 "" {TEXT -1 0 "" }{TEXT 264 20 "     \+
  The area is  " }{TEXT -1 0 "" }{XPPEDIT 18 0 "5/6*sqrt(5);" "6#*(\"
\"&\"\"\"\"\"'!\"\"-%%sqrtG6#F$F%" }{TEXT -1 2 " ." }}}{PARA 5 "" 0 "
" {TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 17 "Solved Example \+
2." }}{PARA 0 "" 0 "" {TEXT -1 70 "A particle is moving along a straig
ht line. The velocity function is  " }{XPPEDIT 18 0 "v(t) = t^4-5*t^3-
t+15;" "6#/-%\"vG6#%\"tG,**$F'\"\"%\"\"\"*&\"\"&F+*$F'\"\"$F+!\"\"F'F0
\"#:F+" }{TEXT -1 25 "   in meters per second. " }}{PARA 0 "" 0 "" 
{TEXT -1 50 "a) Find the displacement in the time period from  " }
{TEXT 270 2 "t " }{TEXT -1 9 "= 1  to  " }{TEXT 271 1 "t" }{TEXT -1 7 
" = 5.2." }}{PARA 0 "" 0 "" {TEXT -1 56 "b) Find the distance traveled
 in the time period  from  " }{TEXT 272 1 "t" }{TEXT -1 10 " = 1  to  \+
" }{TEXT 273 1 "t" }{TEXT -1 7 " = 5.2." }}}{PARA 11 "" 1 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" {TEXT 
-1 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 21 "v:=t->t^4-5*t^3-t+15;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"vGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,**$)9$\"\"%\"
\"\"F1*&\"\"&F1)F/\"\"$F1!\"\"F/F6\"#:F1F(F(F(" }}}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 2 "a)" }{TEXT 274 1 " " }
{TEXT -1 33 "The displacement is the integral:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Int(v(t),t=1
..5.2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,**$)%\"tG\"\"%\"
\"\"F+*&\"\"&F+)F)\"\"$F+!\"\"F)F0\"#:F+/F);F+$\"#_F0" }}}{PARA 11 "" 
1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "int(v(t
),t=1..5.2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+g$R^-\"!\"(" }}}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
265 42 "       The displacement is  -102.5 meters." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 2 "b)" }{TEXT -1 46 "  The d
istance traveled by the particle is    " }{XPPEDIT 18 0 "int(abs(v(t))
,t = 1 .. 5.2);" "6#-%$intG6$-%$absG6#-%\"vG6#%\"tG/F,;\"\"\"-%&FloatG
6$\"#_!\"\"" }{TEXT -1 61 ". To evaluate this integral we have to look
 at the graph of  " }{TEXT 275 1 "v" }{TEXT -1 1 "(" }{TEXT 276 1 "t" 
}{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 28 "plot(v(t),t=0..7,y=-80..50);" }}{PARA 13 "" 1 
"" {GLPLOT2D 350 212 212 {PLOTDATA 2 "6%-%'CURVESG6$7W7$$\"\"!F)$\"#:F
)7$$\"3gmmm\"z+e_\"!#=$\"3qC202,-$[\"!#;7$$\"3sLL3->R`GF/$\"3E(HX'RI^g
9F27$$\"3mmm;apSYVF/$\"3(>18u=]!>9F27$$\"3Onm;z'=$\\eF/$\"3K\\DM6t9`8F
27$$\"3!RL$3Ft3XtF/$\"3_F\"zEi?vD\"F27$$\"3tmmTNj&=t)F/$\"3fv)['[L$z8
\"F27$$\"33+](=`xn,\"!#<$\"3'ob=RgJhz*FO7$$\"3#omT&y/Gl6FO$\"3u8$Q%4n+
nxFO7$$\"3++]PurI88FO$\"3%[=W\"HCuN`FO7$$\"3aLL$e#3dl9FO$\"3fhBfKAT3CF
O7$$\"3ymm\"Ht%o*f\"FO$!3gpE7;nW\">&F/7$$\"3K++]F_m]<FO$!3u6hcNb$\\=%F
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FO$!3!p_d\")yyD]$F27$$\"3P+](oibk\"HFO$!3!of&olt>gRF27$$\"3a+]i!o<-1$F
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<(zmU\"\\@IaF27$$\"3?+]7VLA&y$FO$!3_6o(3$fymaF27$$\"3'pm;a?@.$RFO$!3'4
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yt+\"F27$$\"3Mm;a)3rf&\\FO$\"3W!Q<]LHXo%FO7$$\"3*4++vW0d5&FO$\"3Y-QW=S
L'R#F27$$\"3;L$3-\"QfY_FO$\"3d>S_?$*oOXF27$$\"3C+]PWF'QR&FO$\"30q>(R37
99(F27$$\"3[LL$e/Xy`&FO$\"3a:l@%fg!35!#:7$$\"3m**\\(=<\"e)o&FO$\"3+_SB
*zp1O\"Fgw7$$\"3%ymmm(zvLeFO$\"3aSX7c[*pu\"Fgw7$$\"3-nm\"zAAA)fFO$\"3Y
gKf^\\(H>#Fgw7$$\"3LM$3-7d%HhFO$\"3#HQNJ0y'*o#Fgw7$$\"3#4++]p]ZE'FO$\"
3WC_OEo.(>$Fgw7$$\"3$QL$e*R7)>kFO$\"3Wq(4$=!GC%QFgw7$$\"3S+]7=p:*['FO$
\"3gcs?`LFaTFgw7$$\"3'pmmmV,&elFO$\"3#)\\NLZl0\"[%Fgw7$$\"3,M3xcrVKmFO
$\"3WyFev\\SY[Fgw7$$\"3<+](o(GP1nFO$\"3H$>qe:L(H_Fgw7$$\"3#3++]zQrx'FO
$\"3R\\s29C&Rh&Fgw7$$\"3g+]78Z!z%oFO$\"3%*)GvhT\"f:gFgw7$$\"3I+DccB&R#
pFO$\"3IJV,*\\mrY'Fgw7$$\"\"(F)$\"$%pF)-%'COLOURG6&%$RGBG$\"#5!\"\"F(F
(-%+AXESLABELSG6$Q\"t6\"Q\"yFi\\l-%%VIEWG6$;F(Fj[l;$!#!)F)$\"#]F)" 1 
2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}
{PARA 13 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "The dist
ance is    " }{XPPEDIT 18 0 "A[1]-A[2]+A[3];" "6#,(&%\"AG6#\"\"\"F'&F%
6#\"\"#!\"\"&F%6#\"\"$F'" }{TEXT -1 23 ".  We have to find the " }
{TEXT 277 1 "x" }{TEXT -1 28 "-intercepts of the function." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "solv
e(v(t),t);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6&-%'RootOfG6$,**$)%#_ZG\"
\"%\"\"\"F+*&\"\"&F+)F)\"\"$F+!\"\"F)F0\"#:F+/%&indexGF+-F$6$F&/F3\"\"
#-F$6$F&/F3F/-F$6$F&/F3F*" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 
11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "Perhaps  " }
{TEXT 278 6 "fsolve" }{TEXT -1 19 "  will work better." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "fsolve(
v(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+*RGpd\"!\"*$\"+)Gl]\"
\\F%" }}}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+*RGpd\"!\"*$\"+)Gl]\"\\F%
" }}{PARA 0 "" 0 "" {TEXT -1 38 "It does. Let's name the two solutions
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "xinter:=[fsolve(v(t),t)];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%'xinterG7$$\"+*RGpd\"!\"*$\"+)Gl]\"\\F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "distance:=Int(v(t),t=1..xinter[1])-
Int(v(t),t=xinter[1]..xinter[2])+Int(v(t),t=xinter[2]..5.2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%)distanceG,(-%$IntG6$,**$)%\"tG\"\"%\"\"\"
F.*&\"\"&F.)F,\"\"$F.!\"\"F,F3\"#:F./F,;F.$\"+*RGpd\"!\"*F.-F'6$F)/F,;
F7$\"+)Gl]\"\\F9F3-F'6$F)/F,;F>$\"#_F3F." }}}{PARA 11 "" 1 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "distance:=int(v(t)
,t=1..xinter[1])-int(v(t),t=xinter[1]..xinter[2])+int(v(t),t=xinter[2]
..5.2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)distanceG$\"+W*G3>\"!\"(
" }}}}{PARA 257 "" 1 "" {TEXT -1 79 "_________________________________
______________________________________________" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 10 "ASSIGNMENT" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 258 1 " " 
}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Problem 1." }}{PARA 0 "" 0 "" 
{TEXT -1 39 " Find the area enclosed by the curve   " }{TEXT 279 1 "y
" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x^2/sqrt(x+5)-2;" "6#,&*&%\"xG\"\"
#-%%sqrtG6#,&F%\"\"\"\"\"&F+!\"\"F+F&F-" }{TEXT -1 12 "   and the  " }
{TEXT 280 1 "x" }{TEXT -1 6 "-axis." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 
"Problem 2." }}{PARA 0 "" 0 "" {TEXT -1 73 "A particle is moving along
 a straight line. The velocity function is     " }{XPPEDIT 18 0 "v(t) \+
= 4*t^5-25*t^4-10*t^2+300*t;" "6#/-%\"vG6#%\"tG,**&\"\"%\"\"\"*$F'\"\"
&F+F+*&\"#DF+*$F'F*F+!\"\"*&\"#5F+*$F'\"\"#F+F1*&\"$+$F+F'F+F+" }
{TEXT -1 27 "     in meters per second. " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "a) Find the displacement in the ti
me period from  " }{TEXT 281 1 "t" }{TEXT -1 10 " = 1  to  " }{TEXT 
282 1 "t" }{TEXT -1 5 " = 5." }}{PARA 0 "" 0 "" {TEXT -1 56 "b) Find t
he distance traveled in the time period  from  " }{TEXT 283 1 "t" }
{TEXT -1 11 " = 1   to  " }{TEXT 284 1 "t" }{TEXT -1 5 " = 5." }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 60 "_______________________________________________
_____________" }}{PARA 0 "" 0 "" {TEXT -1 177 "MSIP Grant #P120A80089-
98: \"Three Urban Calculus Reform Programs: Adopting the Best,\" 1998-
2001; MSEIP Grant #P120AA010031:  \"Four Colleges: Calculus + Enhancem
ents\", 2001-2004 " }}}{MARK "21 1" 1 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 
1 }{PAGENUMBERS 0 1 2 33 1 1 }