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-1 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 0 0 1 }1 1 0 0 0 0 2 0 2 0 2 2 -1 1 }{PSTYLE "" -1 212 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 0 0 1 }1 1 0 0 0 0 2 0 2 0 2 2 -1 1 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 205 "" 0 "" {TEXT 200 37 "Applications of the Definite Integral" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 4 "" 0 "" {TEXT 242 38 "Areas. Volumes. Arc Length. \+ Total Mass" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 369 "In this section we study applications of the definite integra l to calculating areas, volumes, arc length, the total mass given its \+ density, etc. As you know, very often the main difficulty in such appl ications is setting up an appropriate integral. Maple will not help yo u in this respect. But it may be helpful in many other ways, as illust rated by the examples below." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 203 11 "Example 1. " }{TEXT 241 62 " Consider the region R bounded by the graphs of the functions " }{XPPEDIT 18 0 "f(x) = exp(-x^2)*cos(x);" "6#/-%\"fG6#%\"xG*&-%$expG6#,$*$)F'\"\"#\" \"\"!\"\"F0-%$cosGF&F0" }{TEXT 241 6 " and " }{XPPEDIT 18 0 "g(x) = x ^2;" "6#/-%\"gG6#%\"xG*$)F'\"\"#\"\"\"" }{TEXT 241 2 ". " }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 34 "(a) Find the area of the region R." }}{PARA 0 "" 0 "" {TEXT 241 28 "(b) Find the perime ter of R." }}{PARA 0 "" 0 "" {TEXT 241 85 "(c) Find the volume of the \+ solid obtained by revolving the region R about the x axis." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 46 "The units on ea ch of the axes are centimeters." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 222 "As you know, the first step in problems similar to Exam ple 1 usually consists of plotting the region under consideration. It \+ wouldn't be easy in this case without Maple's help. Let's define and p lot the functions involved." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "f:=x->exp(-x^2)*cos(x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrow GF(*&-%$expG6#,$*$)F'\"\"#\"\"\"!\"\"F4-%$cosGF&F4F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "g:=x->x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*$)F'\"\"#\"\" \"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 75 "Now we plot the func tions, say, for x between -5 and 5, to start somewhere." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plot( [f(x),g(x)],x=-5..5,color=[red,blue]);" }}{PARA 13 "" 1 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT 241 191 "Now we see where the region R is. The picture wi ll be much clearer in a smaller range for x. Since units on both axes \+ have physical meaning and happen to be the same, let's use the command \"" }{TEXT 204 19 "scaling=constrained" }{TEXT 241 56 "\" that tells \+ Maple to use the same scale on both axes. " }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "plot([f(x),g(x)] ,x=-1..1,color=[red,blue],scaling=constrained, title=\"REGION R\");" } }{PARA 13 "" 1 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 196 "Obviously, to answer any of the questions (a)-(c), we have to find the points of intersect ion of the two graphs. The equations involved are much too difficult t o handle by hand, or by using the \"" }{TEXT 205 5 "solve" }{TEXT 241 69 "\" command that attempts to find exact solutions. We shall use the \"" }{TEXT 206 6 "fsolve" }{TEXT 241 143 "\" command that works numer ically, and allows us to specify the range in which we want a solution . Let's label the points of intersection by \"" }{TEXT 207 1 "a" } {TEXT 241 9 "\" and \"" }{TEXT 208 1 "b" }{TEXT 241 17 "\", respective ly." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "a:=fsolve(f(x)=g(x),x,-1..0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$!+9TI7p!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "b:=fsolve(f(x)=g(x),x,0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"+9TI7p!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 241 37 "Not surprisingly, a=-b. Observe that " }{XPPEDIT 18 0 "f(x);" "6#-%\"fG6#%\"xG" }{TEXT 241 5 " and " }{XPPEDIT 18 0 "g(x);" "6#-%\"gG6#%\"xG" }{TEXT 241 220 " are both eve n. Hence, region R is symmetric with respect to the y axis. With the p oints of intersection found, we are prepared to set up the definite in tegral that represents the area of the region. We label the area \"" } {TEXT 209 1 "A" }{TEXT 241 3 "\"." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "A:=Int(f(x)-g(x),x=a..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%$IntG6$,&*&-%$expG6#,$*$)%\" xG\"\"#\"\"\"!\"\"F2-%$cosG6#F0F2F2F.F3/F0;$!+9TI7p!#5$\"+9TI7pF;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 241 109 "Most of integrals in this worksh eet cannot be found using the Fundamental Theorem. Hence, we shall use the \"" }{TEXT 210 5 "evalf" }{TEXT 241 70 "\" command that, whenever applied, provides a numerical approximation." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(A);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+)[;$*)))!#5" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 46 "The a rea of our region is approximately .889 " }{XPPEDIT 18 0 "cm^2;" "6#* $)%#cmG\"\"#\"\"\"" }{TEXT 241 2 ". " }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 184 "To find the perimeter P of t he region we have to add the arc length of the two pieces. We use the \+ familiar formula for the arc length to set up the appropriate integral s.Recall that \"" }{TEXT 211 1 "D" }{TEXT 241 38 "\" stands for the de rivative function." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "P1:=Int(sqrt(1+D(f)(x)^2),x=a..b);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#P1G-%$IntG6$*$-%%sqrtG6#,&\"\"\"F-* $),&**\"\"#F-%\"xGF--%$expG6#,$*$)F3F2F-!\"\"F--%$cosG6#F3F-F:*&F4F--% $sinGF=F-F:F2F-F-F-/F3;$!+9TI7p!#5$\"+9TI7pFE" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "P2:=Int(sqrt(1+D(g)(x)^2),x=a..b);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#P2G-%$IntG6$*$-%%sqrtG6#,&\"\"\"F-*&\"\"%F-)% \"xG\"\"#F-F-F-/F1;$!+9TI7p!#5$\"+9TI7pF7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 84 "Now we can find the numerical value of the perimeter P b y evaluating both integrals." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "P:=evalf(P1)+evalf(P2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"+x4z?N!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 50 "The perimeter is approximately equal to 3.521 c m. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT 241 134 "To find the volume of revolution, we slice the s olid perpendicularly to the x axis. The approximate volume of each \"w asher\" slab is " }{XPPEDIT 18 0 "Pi*f(x)^2*Delta*x-Pi*g(x)^2*Delta*x; " "6#,&**%#PiG\"\"\"*$)-%\"fG6#%\"xG\"\"#F&F&%&DeltaGF&F,F&F&**F%F&*$) -%\"gGF+F-F&F&F.F&F,F&!\"\"" }{TEXT 241 69 ". The corresponding volume of the solid is then given by the integral" }}{PARA 0 "" 0 "" {TEXT 241 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "V:=Int(Pi*(f(x) ^2-g(x)^2),x=a..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG-%$IntG6$ *&%#PiG\"\"\",&*&)-%$expG6#,$*$)%\"xG\"\"#F*!\"\"F5F*)-%$cosG6#F4F5F*F **$)F4\"\"%F*F6F*/F4;$!+9TI7p!#5$\"+9TI7pFB" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(V);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+tM Q7F!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 34 "The volume is approximately 2.712 " } {XPPEDIT 18 0 "cm^3;" "6#*$)%#cmG\"\"$\"\"\"" }{TEXT 241 1 "." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 212 9 "Example 2" }{TEXT 241 59 ". Suppose that region R has va rying mass density given by " }{XPPEDIT 18 0 "m(x) = 25*x^2;" "6#/-%\" mG6#%\"xG*&\"#D\"\"\"*$)F'\"\"#F*F*" }{TEXT 241 11 " grams per " } {XPPEDIT 18 0 "cm^2;" "6#*$)%#cmG\"\"#\"\"\"" }{TEXT 241 161 " , so, a s you see, the density of the region increases as you move away from t he y axis and is constant along vertical lines. Find the total mass o f the region." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 83 " This problem is very easy. We slice the region into thi n vertical strips of width " }{XPPEDIT 18 0 "Delta*x;" "6#*&%&DeltaG\" \"\"%\"xGF%" }{TEXT 241 28 " corresponding to values of " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT 241 78 " in the interval [a,b]. The area of a vertical strip corresponding to a given " }{XPPEDIT 18 0 "x;" "6#%\"x G" }{TEXT 241 18 " is approximately " }{XPPEDIT 18 0 "(f(x)-g(x))*Delt a*x;" "6#*(,&-%\"fG6#%\"xG\"\"\"-%\"gGF'!\"\"F)%&DeltaGF)F(F)" }{TEXT 241 111 ", and the density is approximately constant, with value m(x) . Hence, the mass of this strip is approximately " }{XPPEDIT 18 0 "m( x)*(f(x)-g(x))*Delta*x;" "6#**-%\"mG6#%\"xG\"\"\",&-%\"fGF&F(-%\"gGF&! \"\"F(%&DeltaGF(F'F(" }{TEXT 241 50 " . We add all the masses and pass to the limit as " }{XPPEDIT 18 0 "Delta*x;" "6#*&%&DeltaG\"\"\"%\"xGF %" }{TEXT 241 66 " approaches 0. The resulting integral that gives the total mass is" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 34 "M:=Int(25*x^2*(f(x)-g(x)),x=a..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG-%$IntG6$,$*(\"#D\"\"\")%\"xG\"\"#F+,&*& -%$expG6#,$*$F,F+!\"\"F+-%$cosG6#F-F+F+F5F6F+F+/F-;$!+9TI7p!#5$\"+9TI7 pF>" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 33 "Its numerical value, in g rams, is" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+**G' )Q?!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 97 "The next problem, whe re the density is constant along horizontal lines, is much more challe nging." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 213 12 "Example 3. " }{TEXT 241 70 "Suppose now that \+ our same region R has varying mass density given by " }{XPPEDIT 18 0 "m(y) = 30*y^2;" "6#/-%\"mG6#%\"yG*&\"#I\"\"\"*$)F'\"\"#F*F*" }{TEXT 241 63 ", so that now the density increases as you move away from the " }{TEXT 214 6 "x axis" }{TEXT 241 23 " and is constant along " } {TEXT 215 10 "horizontal" }{TEXT 241 43 " lines. Find the total mass \+ of the region." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 170 "This time we have to slice the region into horizon tal strips as the density is constant along horizontal lines. The situ ation in the lower part of the region bounded by " }{XPPEDIT 18 0 "g( x);" "6#-%\"gG6#%\"xG" }{TEXT 241 71 " differs from the situation in t he upper part of the region bounded by " }{XPPEDIT 18 0 "f(x);" "6#-% \"fG6#%\"xG" }{TEXT 241 80 ". Shortly speaking, we have to divide the \+ region in two by the horizontal line " }{XPPEDIT 18 0 "y = f(a);" "6# /%\"yG-%\"fG6#%\"aG" }{TEXT 241 180 ". Recall that a and b are the x c oordinates of the intersections of the two graphs, which we found abov e. (Clearly, from the definition of a and b, and the symmetry of R, w e have " }{XPPEDIT 18 0 "f(a) = f(b);" "6#/-%\"fG6#%\"aG-F%6#%\"bG" } {TEXT 241 1 "=" }{XPPEDIT 18 0 "g(a) = g(b);" "6#/-%\"gG6#%\"aG-F%6#% \"bG" }{TEXT 241 10 ".) Define" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "c:=f(a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\"+=[*zx%!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "plot([f(x),g(x),c],x=-1..1,color=[red,blue,green]);" }}{PARA 13 "" 1 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 111 "Finding the total m ass of the lower region is easy. We slice the lower region into horiz ontal strips of width " }{XPPEDIT 18 0 "Delta*y;" "6#*&%&DeltaG\"\"\"% \"yGF%" }{TEXT 241 28 " corresponding to values of " }{XPPEDIT 18 0 "y ;" "6#%\"yG" }{TEXT 241 73 " from the interval [0,c]. The length of a \+ strip corresponding to a given " }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 20 " in the interval is " }{XPPEDIT 18 0 "2*sqrt(y);" "6#*&\"\"#\" \"\"-%%sqrtG6#%\"yGF%" }{TEXT 241 55 " and the density is approximatel y constant with value " }{XPPEDIT 18 0 "30*y^2;" "6#*&\"#I\"\"\"*$)% \"yG\"\"#F%F%" }{TEXT 241 41 ". Hence, the mass of each such strip is \+ " }{XPPEDIT 18 0 "30*y^2*2*sqrt(y)*Delta*y;" "6#*.\"#I\"\"\"*$)%\"yG \"\"#F%F%F)F%-%%sqrtG6#F(F%%&DeltaGF%F(F%" }{TEXT 241 140 ". The resu lting integral that gives the total mass of the lower region and thus \+ the numerical value of the mass, in grams, is as follows: " }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "M 1:=Int(30*y^2*2*sqrt(y),y=0..c); evalf(M1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#M1G-%$IntG6$,$*&\"#g\"\"\")%\"yG#\"\"&\"\"#F+F+/F-; \"\"!$\"+=[*zx%!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Xs`#H\"!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 92 "Observe that we were able to set up the integral beca use we could find the inverse function " }{XPPEDIT 18 0 "g^(-1);" "6#) %\"gG,$\"\"\"!\"\"" }{TEXT 241 2 "( " }{TEXT 216 2 "y " }{TEXT 241 4 " ) = " }{XPPEDIT 18 0 "sqrt(y);" "6#-%%sqrtG6#%\"yG" }{TEXT 241 6 " fo r " }{XPPEDIT 18 0 "g(x);" "6#-%\"gG6#%\"xG" }{TEXT 241 23 ". (More pr ecisely, for " }{XPPEDIT 18 0 "g(x);" "6#-%\"gG6#%\"xG" }{TEXT 241 27 " restricted to nonnegative " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT 241 174 ".) The value of the inverse multiplied by 2 gave us the lengt h of a horizontal strip. We cannot do the same for the upper region. W e can see from the graph that the inverse " }{XPPEDIT 18 0 "f^(-1);" "6#)%\"fG,$\"\"\"!\"\"" }{TEXT 241 2 "( " }{XPPEDIT 18 0 "y;" "6#%\"yG " }{TEXT 241 20 " ) for the function " }{XPPEDIT 18 0 "f(x);" "6#-%\"f G6#%\"xG" }{TEXT 241 28 " (restricted to nonnegative " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT 241 21 ") is defined for any " }{XPPEDIT 18 0 "y ;" "6#%\"yG" }{TEXT 241 40 " in the interval [c,1]. Indeed, for any " }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 76 " in this interval the cor responding horizontal line intersects the graph of " }{XPPEDIT 18 0 "f (x);" "6#-%\"fG6#%\"xG" }{TEXT 241 131 " over the positive half-line a t exactly one point. The problem is that we can't find the formula for the inverse as no formula for " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT 241 13 " in terms of " }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 19 " f or a solution to " }{XPPEDIT 18 0 "f(x) = y;" "6#/-%\"fG6#%\"xG%\"yG" }{TEXT 241 97 " can be found. The mass of the upper region is equal to the integral from c to 1 of the function " }{XPPEDIT 18 0 "30*y^2*2*f ^(-1);" "6#**\"#I\"\"\"*$)%\"yG\"\"#F%F%F)F%)%\"fG,$F%!\"\"F%" }{TEXT 241 1 "(" }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 42 "). We can't, h owever, find a formula for " }{XPPEDIT 18 0 "f^(-1);" "6#)%\"fG,$\"\" \"!\"\"" }{TEXT 241 1 "(" }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 97 "). Is there anything that we can do at this point? Yes, there is. We can't find the formula for " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT 241 13 " in terms of " }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 20 ", \+ but for any given " }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 29 " we c an find the solution to " }{XPPEDIT 18 0 "f(x) = y;" "6#/-%\"fG6#%\"xG %\"yG" }{TEXT 241 11 " using the " }{TEXT 217 6 "fsolve" }{TEXT 241 103 " command. Hence, we can set up left or right Riemann sums for the integral from c to 1 of the function " }{XPPEDIT 18 0 "30*y^2*2*f^(-1 );" "6#**\"#I\"\"\"*$)%\"yG\"\"#F%F%F)F%)%\"fG,$F%!\"\"F%" }{TEXT 241 1 "(" }{XPPEDIT 18 0 "y;" "6#%\"yG" }{TEXT 241 494 "). Indeed, Riemann sums require values of the integrand at a finite number of points onl y. And having Maple to perform calculations, we can set up Riemann sum s for a large number of divisions, which will give us a good approxima tion of the desired integral. This is what we do below. Actually, we a re calculating the left Riemann sum for fifty divisions. (It should be noted that one could find the total mass of the region using so calle d multiple integrals, which we shall learn in MTH 243.) " }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 159 "In physical term s, we shall divide the interval [c,1] into fifty subintervals, calcula te the length of each horizontal strip by \"fsolving\" the correspondi ng " }{XPPEDIT 18 0 "f(x) = y;" "6#/-%\"fG6#%\"xG%\"yG" }{TEXT 241 227 " equation, calculate the mass of each strip and add those masses \+ up. Since the number of subintervals is large, we shall obtain a reaso nably good approximation of the mass of the top region. Let's try to d o it \"by hand\" using" }{TEXT 218 6 " lists" }{TEXT 241 62 ", which w e introduced in the previous worksheet and Maple's \"" }{TEXT 219 3 "s um" }{TEXT 241 9 "\" and \"" }{TEXT 220 3 "seq" }{TEXT 241 114 "\" com mands. Let's take the division of the interval [c,1] into 50 subinterv als. Denote the width of each strip by" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "h:=(1-c)/50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hG$\"+O5SW5!#6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 76 "Take the list of the corresponding lower endpoints of subintervals in [c,1]." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "PL:=[seq(c+(i-1)*h,i=1..50)];" }}{PARA 206 "" 1 "" {XPPMATH 20 "6#>%#PLG7T$\"+=[*zx%!#5$\"+A\\V#)[F($\"+D](o)\\F($\"+H^J\"4&F($\"+K_ v&>&F($\"+O`>+`F($\"+Saj/aF($\"+Vb24bF($\"+Zc^8cF($\"+]d&zr&F($\"+aeRA eF($\"+ef$o#fF($\"+hgFJgF($\"+lhrNhF($\"+oi:SiF($\"+sjfWjF($\"+wk.\\kF ($\"+zlZ`lF($\"+$o;zl'F($\"+'ycBw'F($\"+!*ozmoF($\"+%*pBrpF($\"+(4xc2( F($\"+,s6!=(F($\"+/tb%G(F($\"+3u**)Q(F($\"+7vV$\\(F($\"+:w(yf(F($\"+>x J-xF($\"+Ayv1yF($\"+Ez>6zF($\"+I!Qc,)F($\"+L\"y+7)F($\"+P#=XA)F($\"+S$ e*G$)F($\"+W%)RL%)F($\"+[&Qy`)F($\"+^'yAk)F($\"+b(=nu)F($\"+e)e6&))F($ \"+i*)fb*)F($\"+m!R+1*F($\"+p\"zW;*F($\"+t#>*o#*F($\"+w$fLP*F($\"+![*z x%*F($\"+%eRAe*F($\"+(ozmo*F($\"+\"z>6z*F($\"+%*)fb*)*F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 61 "Let's find the corresponding length of each horizontal strip." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "LL:=[seq(2*fsolve(f(x)=PL[i],x,0..1),i=1..50)];" }}{PARA 207 "" 1 "" {XPPMATH 20 "6#>%#LLG7T$\"+B3Y#Q\"!\"*$\"+c$HFO\"F($\"+/I4V8F( $\"+a4aB8F($\"++E1/8F($\"+.vk%G\"F($\"+W`Gl7F($\"+%)e'fC\"F($\"+?*ymA \"F($\"+SUT27F($\"+%eh\")=\"F($\"+&f5*o6F($\"+w2l\\6F($\"+O9PI6F($\"+X ;166F($\"+o,r\"4\"F($\"+8aIs5F($\"+g`$G0\"F($\"+)[(GL5F($\"+#p[O,\"F($ \"+C>0R**!#5$\"+!GC/u*FQ$\"+%R\\/a*FQ$\"+-B'*Q$*FQ$\"+%4'yN\"*FQ$\"+]+ tI*)FQ$\"+#[(eB()FQ$\"+[J89&)FQ$\"+u,7-$)FQ$\"+qjF(3)FQ$\"+5(*HpyFQ$\" +OG&yk(FQ$\"+7ibAuFQ$\"+[&zH>(FQ$\"+[4jepFQ$\"+%*G%*=nFQ$\"+))RDtkFQ$ \"+mYy?iFQ$\"+mUggfFQ$\"+oae\"p&FQ$\"+G%RBT&FQ$\"+)y?67&FQ$\"+CLo:[FQ$ \"+k20$\\%FQ$\"+?F8\\TFQ$\"+UX.yPFQ$\"+oupqLFQ$\"+A\"[=\"HFQ$\"+]/lrBF Q$\"+Z#4Hn\"FQ" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 74 "Now we shall m ake a list of masses of each of the fifty horizontal strips." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 " ML:=[seq(30*PL[i]^2*LL[i]*h,i=1..50)];" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#>%#MLG7T$\"+aEb)))*!#6$\"+]-#y,\"!#5$\"+8)Hl/\"F+$\"+CT%\\2\"F+$ \"+gC-.6F+$\"+KJsI6F+$\"+.M+e6F+$\"+'R>[=\"F+$\"+'*f767F+$\"+Sn(oB\"F+ $\"+2P-i7F+$\"+\"QV:F+$\"+\\?IW:F+$\"+CN:V:F+$\"+1reR:F+$\"+?ZUL:F+$\"+ \"*zYC:F+$\"+^U\\7:F+$\"+M9D(\\\"F+$\"+m9Xy9F+$\"+f:wb9F+$\"+-DzG9F+$ \"+$f#3(R\"F+$\"+r\\2g8F+$\"+7Z3<8F+$\"+f#\\sE\"F+$\"+J/X47F+$\"+%*[=U 6F+$\"+ANLj5F+$\"+h66(p*F($\"+ozlg&)F($\"+#psO7(F($\"+\"G]E8&F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 117 "Finally, let's sum up the masses of strips and obtain a n approximate value of the mass of the upper region, in grams." }} {PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "M2:=sum(ML[j],j=1..50);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#M2 G$\"+b)pl_'!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 241 294 "Of course, one can easily progra m a procedure that would calculate the approximate mass of the upper r egion for any number of divisions n. Could you do it yourself after se eing a few simple examples of procedures in the previous worksheet? If you are curious, such a procedure is defined below." }}{PARA 0 "" 0 " " {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 144 "AM2:= p roc(n) local h,L,i,k; global f,c; h:=(1-c)/n; L:=[seq(fsolve(f(x)=c+(i -1)*h,x,0..1),i=1..n)]; sum(30*(c+(k-1)*h)^2*2*L[k]*h,k=1..n); end:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 54 "Let's check our procedure for n=50 to see if it works." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "AM2(50);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+b)pl_'!\"*" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 162 "Our procedure works. We can calculate now the mass of t he upper region for, say, 70 divisions of the interval [c,1] and obtai n a better approximation, and so on." }}{PARA 0 "" 0 "" {TEXT 241 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "AM2(70);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Ng&o^'!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 168 "What else c ould we do to calculate the mass of the upper region? Instead of slici ng into horizontal strips, we could slice the region into vertical str ips whose length " }{XPPEDIT 18 0 "f(x)-c;" "6#,&-%\"fG6#%\"xG\"\"\"% \"cG!\"\"" }{TEXT 241 63 " is easy to find and compensate for the fac t that the density " }{XPPEDIT 18 0 "30*y^2;" "6#*&\"#I\"\"\"*$)%\"yG \"\"#F%F%" }{TEXT 241 401 " is not nearly constant along each strip by dividing each vertical strip horizontally into small rectangles. On e ach rectangle the density would be nearly constant, so we could calcul ate the mass of each rectangle, add them up, and so on. This sort of p rocedure would lead us directly to the notion of the Riemann integral \+ over an area and multiple integrals. We shall study this notion next s emester." }}{PARA 0 "" 0 "" {TEXT 241 1 " " }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 4 "" 0 "" {TEXT 242 17 "Parametric Curves" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 221 11 "Example 4. " }{TEXT 241 386 " Satellites in orbit around a central mass have elliptical orbits. For instance, \+ the orbits of the planets around the sun are elliptical (though nearly circular) while the orbits of artificial satellites around the earth \+ are often very elliptical. This example concerns a motion along an ell iptical path. As you know, motion is best described using parametric r epresentations of curves." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 110 "Consider a point moving in the plane so that it s x and y coordinates are functions of time t given as follows:" }} {PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "x:=2*cos(t); y:=3*sin(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"xG,$*&\"\"#\"\"\"-%$cosG6#%\"tGF(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG,$*&\"\"$\"\"\"-%$sinG6#%\"tGF(F(" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 61 "The point \+ (x,y) will describe an ellipse as t goes from 0 to " }{XPPEDIT 18 0 "2 *Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT 241 200 ". We plot the elliptica l path below. Note the syntax appropriate for parametric plots, especi ally the location of square brackets. In order to see clearly the elli ptical shape of the path, we use the " }{TEXT 222 21 "\"scaling=constr ained" }{TEXT 241 10 "\" option." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot([x,y,t=0..2*Pi],scaling =constrained);" }}{PARA 13 "" 1 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 80 "Supp ose that your position at time t is given by the above parametric equa tions." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 46 "(a) What is your position and speed at t=3.5?" }}{PARA 0 "" 0 "" {TEXT 241 70 "(b) How far along the ellipse do you move between time \+ t=0 and t=3.5?" }}{PARA 0 "" 0 "" {TEXT 241 126 "(c) Given that you st arted at (2,0) when t=0, at what time will you have traveled exactly 4 units of length along the ellipse?" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 241 123 "To answer (a), observe that the position at 3.5 is ( x(3.5), y(3,5)), and by the familiar formula, the speed at time t is \+ " }{XPPEDIT 18 0 "v(t) = sqrt((dx/dt)^2+(dy/dt)^2);" "6#/-%\"vG6#%\"tG -%%sqrtG6#,&*$)*&%#dxG\"\"\"%#dtG!\"\"\"\"#F0F0*$)*&%#dyGF0F1F2F3F0F0" }{TEXT 241 19 ". Hence, we obtain" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "evalf(subs(t=3.5,x)); eva lf(subs(t=3.5,y));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+vL\"H(=!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+$o\\B0\"!\"*" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 88 "Our position at t=3.5 is, approximately, (-1.873, -1.052). (We had to use the command \"" }{TEXT 223 5 "evalf" }{TEXT 241 12 "\" before \"" } {TEXT 224 4 "subs" }{TEXT 241 78 "\" in order to obtain a numerical va lue. Try to see what happens if you use \"" }{TEXT 225 4 "subs" }{TEXT 241 13 "\" without \"" }{TEXT 226 5 "evalf" }{TEXT 241 50 "\".) Let' s set up the expression for the speed v:" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "v:=sqrt((diff(x,t))^ 2+(diff(y,t))^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"vG*$-%%sqrtG6 #,&*&\"\"%\"\"\")-%$sinG6#%\"tG\"\"#F,F,*&\"\"*F,)-%$cosGF0F2F,F,F," } }}{EXCHG {PARA 0 "" 0 "" {TEXT 241 26 "The speed at t=3.5 is then" }} {PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "evalf(subs(t=3.5,v));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+_Uk &*G!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 189 "To answer (b) we need the formula that \+ says that the arc length of a parametric curve is the integral of the \+ speed. We use it below to calculate the distance traveled between t=0 \+ and t=3.5." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "AL:=Int(v,t=0..3.5); evalf(AL);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%#ALG-%$IntG6$*$-%%sqrtG6#,&*&\"\"%\"\"\")-%$sinG6#% \"tG\"\"#F/F/*&\"\"*F/)-%$cosGF3F5F/F/F//F4;\"\"!$\"#N!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+5cM&**)!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 158 "We have tr aveled almost 9 units of length. To answer (c) we have to set up an eq uation and solve it numerically. The length traveled up to time T is g iven by:" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "ALT:=Int(v,t=0..T);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ALTG-%$IntG6$*$-%%sqrtG6#,&*&\"\"%\"\"\")-%$sinG6#%\"tG\"\"#F/F/ *&\"\"*F/)-%$cosGF3F5F/F/F//F4;\"\"!%\"TG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 235 "We want ALT to be equal to 4, so we have to solve numer ically for T the equation ALT=4. Since ALT is strictly increasing ther e is only one solution to this equation and we do not have to worry ab out specifying the range for a solution." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "fsolve(ALT=4,T);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+(Q:we\"!\"*" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 241 86 "At t \+ equal approximately to 1.588 units of time, we have traveled 4 units o f distance." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 227 319 "IMPORTANT! Since we defined x and y in terms of t \+ in the example above we must restore x and y to be simply letters, oth erwise we will have difficulties with plotting functions of x or y and with many other commands. You may have to do this in other situation s as well, for example, after working on Problem 3 below" }{TEXT 228 1 "." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 241 53 "This is how you restore x and y to be simply letters:" }}{PARA 0 "" 0 "" {TEXT 241 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "x:=' x'; y:='y';" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xGF$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yGF$" }}}{EXCHG {PARA 209 "" 0 "" {TEXT 200 95 "IMPORTANT! Re-execute the last command line otherwise you may get some strange error messages!" }}}{EXCHG {PARA 4 "" 0 "" {TEXT 242 0 " " }}{PARA 4 "" 0 "" {TEXT 242 17 "Homework Problems" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 229 11 "Problem 1. " }{TEXT 241 49 " Consider the region G bounded by the graphs of " }{XPPEDIT 18 0 "y = 2*cos(x);" "6#/%\"yG*&\"\"#\"\"\"-%$cosG6#%\"xGF'" }{TEXT 241 6 " and " }{XPPEDIT 18 0 "y = x^4/2;" "6#/%\"yG*&)%\"xG\"\"%\"\" \"\"\"#!\"\"" }{TEXT 241 33 " . Units on both axes are inches." }} {PARA 0 "" 0 "" {TEXT 241 20 "(a) Plot the region." }}{PARA 0 "" 0 "" {TEXT 241 94 "(b) Express in terms of definite integrals and find the \+ numerical value of the perimeter of G." }}{PARA 0 "" 0 "" {TEXT 241 108 "(c) Express in terms of definite integrals (or one integral) and \+ find the numerical value of the area of G." }}{PARA 0 "" 0 "" {TEXT 241 147 "(d) Express in terms of definite integrals and find the numer ical value of the volume of the solid obtained by revolving region G a bout the x axis." }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT 241 0 "" }}{PARA 0 "" 0 "" {TEXT 230 12 "Problem 2. " } {TEXT 241 4 "(a) " }{TEXT 231 1 " " }{TEXT 241 77 "Suppose that region G defined in Problem 1 has varying mass density given by " }{XPPEDIT 18 0 "m(x) = 10*(x^2+1);" "6#/-%\"mG6#%\"xG*&\"#5\"\"\",&*$)F'\"\"#F*F *F*F*F*" }{TEXT 241 125 " grams per inch square. Express in terms of d efinite integrals and find the numerical value for the total mass of t he region." }}{PARA 0 "" 0 "" {TEXT 241 63 " (b) Su ppose that region G has mass density " }{XPPEDIT 18 0 "p(y) = 11*y^2;" "6#/-%\"pG6#%\"yG*&\"#6\"\"\"*$)F'\"\"#F*F*" }{TEXT 241 117 " grams p er inch square. Express in terms of definite integrals and find the nu merical value for the total mass of G. " }}{PARA 0 "" 0 "" {TEXT 232 7 "HINT: " }{TEXT 233 4 "Both" }{TEXT 234 144 " functions that bound \+ the region G have inverses (when restricted to appropriate intervals) . Formulas for the inverses can be easily found for " }{TEXT 235 4 "bo th" }{TEXT 236 145 " functions. You do not have to use lists, Riemann \+ sums etc. as in Example 3 of this worksheet. Maple's syntax for the i nverse cosine function is" }{TEXT 241 3 " \"" }{TEXT 237 7 "arccos(" } {TEXT 238 8 "variable" }{TEXT 239 1 ")" }{TEXT 241 3 "\"." }}{PARA 0 " " 0 "" {TEXT 241 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 240 11 "Problem \+ 3. " }{TEXT 241 91 " Suppose you are traveling along the path on the x y plane given by the parametric equations" }}{PARA 0 "" 0 "" {TEXT 241 0 "" }}{PARA 210 "" 0 "" {XPPEDIT 18 0 "x = 3*sin(t);" "6#/%\"xG*& \"\"$\"\"\"-%$sinG6#%\"tGF'" }{TEXT 200 7 " , " }{XPPEDIT 18 0 "y \+ = 2*cos(t);" "6#/%\"yG*&\"\"#\"\"\"-%$cosG6#%\"tGF'" }{TEXT 200 3 " , " }}{PARA 0 "" 0 "" {TEXT 241 17 "where t is time. " }}{PARA 0 "" 0 "" {TEXT 241 29 " (a) Plot the path." }}{PARA 0 "" 0 "" {TEXT 241 56 " (b) What is your position and speed at t=4.1?" }} {PARA 0 "" 0 "" {TEXT 241 94 " (c) During a single orbit, h ow many times will your speed be the same as at t=4.1?" }}{PARA 0 "" 0 "" {TEXT 241 54 " (d) What is the length of the whole orbi t?" }}{PARA 0 "" 0 "" {TEXT 241 134 " (e) What was your ini tial position? After what time have you traveled exactly 3 units of le ngth from the initial position?" }}{PARA 0 "" 0 "" {TEXT 241 43 " \+ (See the note before Problem 1.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 211 "" 0 "" {TEXT 200 76 "MTH 142 Ma ple Worksheets written by B.Kaskosz and L. Pakula, Copyright 1999." }} }{EXCHG {PARA 212 "" 0 "" {TEXT 200 26 "Last modified August 1999." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }