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-1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Author" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 54 "FAMILY OF SOLUTION CURVES OF A SECOND ORDER EQUATION I" }}{PARA 19 "" 0 "" {TEXT -1 12 "by Dr. \+ Fadyn" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }{TEXT 256 582 "We're going to look at an example of the family of s olution curves to a second-order linear homogeneous differential equat ion in this project. The problem we will look at is actually quite si mple. However, we will do a very careful analysis of properties share d by different members of this family, and this will take a bit of wor k. Toward the end of the project we will apply the knowledge gained f rom this analysis to predicting the motion of a mass-spring system for which the differential equation is a model. So let's get started by looking at our differential equation: " }{XPPEDIT 257 0 "diff(y(x),` $`(x,2))+6*diff(y(x),x)+8*y(x) = 0;" "6#/,(-%%diffG6$-%\"yG6#%\"xG-%\" $G6$F+\"\"#\"\"\"*&\"\"'F0-F&6$-F)6#F+F+F0F0*&\"\")F0-F)6#F+F0F0\"\"! " }{TEXT -1 1 " " }{TEXT 258 69 ". We will write this in Maple and na me the equation \"eq\" as follows:" }}{PARA 0 "" 0 "" {TEXT 259 0 "" } {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "eq:=diff(y( x),x$2)+6*diff(y(x),x)+8*y(x)=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 63 "The characteristic equation of this differential equation is: " }{XPPEDIT 261 0 "m^2+6* m+8 = 0;" "6#/,(*$%\"mG\"\"#\"\"\"*&\"\"'F(F&F(F(\"\")F(\"\"!" }{TEXT -1 1 " " }{TEXT 262 116 ". The roots of this equation are m = -2 and \+ m = -4, so we may write the solution of the differential equation as: \+ " }{XPPEDIT 263 0 "y(x) = C1*exp(-2*x)+C2*exp(-4*x);" "6#/-%\"yG6#%\" xG,&*&%#C1G\"\"\"-%$expG6#,$*&\"\"#F+F'F+!\"\"F+F+*&%#C2GF+-F-6#,$*&\" \"%F+F'F+F2F+F+" }{TEXT -1 2 " " }{TEXT 264 70 ". Maple's \"dsolve \" command can also solve the differential equation:" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sol:=dsol ve(eq,y(x));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 265 106 "We have named the solution \"sol\". T he remainder of this worksheet demands that _C1 be the coefficient of \+ " }{XPPEDIT 267 0 "exp(-2*x);" "6#-%$expG6#,$*&\"\"#\"\"\"%\"xGF)!\"\" " }{TEXT -1 0 "" }{TEXT 266 31 " and _C2 be the coefficient of " } {XPPEDIT 269 0 "exp(-4*x);" "6#-%$expG6#,$*&\"\"%\"\"\"%\"xGF)!\"\"" } {TEXT -1 0 "" }{TEXT 268 95 " . So , since there is no guarantee that \+ Maple has make this choice, we will do so by defining:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol := y(x) = _C1*exp(-2*x)+_C2*exp(-4*x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 270 498 "Observe that _ C1 and _C2 are arbitrary constants, and that for every different choi ce of _C1 and _C2 we a obtain different solution curve. The totality \+ of all these solution curves is the \"family\" of solution curves for \+ the differential equation. Let's look at some of these curves. To ge t a batch of them, we'll use Maple's sequence command (\"seq\") to rep lace both _C1 and _C2 by -1, 0, 1, and 2. We will generate a list of sixteen members of the family and name the result \"fam\" as follows: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "fam:=[seq(seq(subs(_C1=i,_C2=j,rhs(sol)),i=-1..2),j=- 1..2)];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 271 48 "Next let's plot this list of solution fun ctions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "plot(fam,x=-2..2,-8..8, title=`Some Members Of The Fa mily`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 272 342 "To obtain a closer view, click on the gr aph and then click on the \"magnifying glass\" icons at the top of the screen. You can also access individual members of the list by using \+ the notation fam[i] where i is an integer from 1 to 16. For exampl e, to access and plot the fourth member of the list fam, we give the f ollowing Maple commands:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "fam[4];" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "plot(fam[4],x=-1..2,-8..8);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 1 " " }{TEXT 273 13 "Question 1: " }{TEXT 258 334 "Which members \+ of this list have an x-intercept? Which members have a maximum? Whic h have a minimum? Which have a point of inflection? Which are strict ly increasing? Which are strictly decreasing? When you reference a m ember of the family in your answer, refer to it by the notation fam[i] where i is an integer between 1 and 16." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 274 68 "Among all members \+ of the family of solutions we have the functions " }{XPPEDIT 275 0 "y [1](x) = exp(-2*x);" "6#/-&%\"yG6#\"\"\"6#%\"xG-%$expG6#,$*&\"\"#\"\" \"F*F1!\"\"" }{TEXT -1 1 " " }{TEXT 276 5 "and " }{XPPEDIT 277 0 "y[2 ](x) = exp(-4*x);" "6#/-&%\"yG6#\"\"#6#%\"xG-%$expG6#,$*&\"\"%\"\"\"F* F1!\"\"" }{TEXT -1 2 " " }{TEXT 278 150 ". These two functions form \+ a \"fundamental set of solutions\" for the differential equation. The ir graphs are particularly simple and have no extrema:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "plot([ exp(-2*x),exp(-4*x)],x=-2..2,-2..12,color=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 279 233 "Next we will attempt to classify the members of this family w hich do have extrema. At the outset, we will suppose that neither _C1 \+ nor _C2 is equal to zero. We may do so because if one of them is zero , the functions have the form " }{XPPEDIT 281 0 "_C1*exp(-2*x);" "6# *&%$_C1G\"\"\"-%$expG6#,$*&\"\"#F%%\"xGF%!\"\"F%" }{TEXT -1 0 "" } {TEXT 280 5 " or " }{XPPEDIT 283 0 "_C2*exp(-4*x);" "6#*&%$_C2G\"\"\" -%$expG6#,$*&\"\"%F%%\"xGF%!\"\"F%" }{TEXT -1 0 "" }{TEXT 282 179 " , and these have no extrema. Let's use Maple to find the derivative of our solution (sol). We name the result \"dsol\", and then find the c ritical number and name it \"xextreme\":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "dsol:=diff(sol,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "xextreme:=solve(rhs(dsol) =0,x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 284 144 "Next we' ll find the value of y at xextreme (using Maples substitute (\"sub\") \+ command) , name the result \"yextreme\" and then simplify this result: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "yextreme:=subs(x=xextreme,rhs(sol));" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "yextreme :=simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 285 437 "Our next task is to determine when the point ( xextreme , yextreme ) is actually a maximum or a minimum point. To begin, no tice from the form of xextreme that _C1 and _C2 must have opposite sig ns. This must be true in order for the natural logarithm function to \+ exist. Remember that the domain of ln(x) is x>0. So assuming that _ C1 and _C2 have different signs, let's use the second derivative test \+ for relative extrema as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ddsol:=diff(sol,x$2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "subs(x=xextreme,rhs(ddsol)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 286 133 "As this \+ set of commands shows, when the critical number xextreme is substitu ted into the second derivative, we obtain the quantity " }{XPPEDIT 287 0 "2*_C1^2/_C2;" "6#*(\"\"#\"\"\"*$%$_C1G\"\"#F%%$_C2G!\"\"" } {TEXT -1 0 "" }{TEXT 288 6 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 289 0 "" }{TEXT 290 13 "Question \+ 2: " }{TEXT 291 548 "Determine appropriate conditions on _C1 and _C2 \+ so that y(x) has (a) a maximum value at xextreme and (b) a minimum val ue at xextreme. I don't just want examples here. You need to give me a general criterion involving _C1 and _C2. If you can't get Maple to do this it is not a problem. Just write out the solution steps by ha nd. In addition answer the following: (a) at a minimum value, is yext reme positive or negative? Can yextreme be equal to zero? (b) at a \+ maximum value, is yextreme positive or negative? Can yextreme be equa l to zero?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 292 209 "Let's look at some examples with which you can partially check your answers to question 2. This time we'll use a di fferent method of obtaining a list of family members. Consider the fo llowing Maple commands:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "f:=c->subs(\{_C1=c[1],_C2=c[2]\},rh s(sol));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "cvals:=[[-1,1], [2,-1],[-2,1],[1,-2],[1/2,-3],[2,-1/2],[-2,1/2]];" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 19 "fam2:=map(f,cvals);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 293 85 "The fi rst command above makes f into a function of c and will replace _C1 an d _C2 by " }{XPPEDIT 294 0 "c[1];" "6#&%\"cG6#\"\"\"" }{TEXT -1 1 " " }{TEXT 295 4 "and " }{TEXT -1 0 "" }{XPPEDIT 296 0 "c[2];" "6#&%\"cG6# \"\"#" }{TEXT 297 106 " in the right hand side of sol. The next assig nment just defines a list of constant values of the form [ " } {XPPEDIT 298 0 "c[1];" "6#&%\"cG6#\"\"\"" }{TEXT -1 1 " " }{TEXT 299 2 ", " }{XPPEDIT 300 0 "c[2];" "6#&%\"cG6#\"\"#" }{TEXT -1 1 " " } {TEXT 301 151 "] . The third command (\"map\") applies the function f to the list cvals and generates our family, \"fam2\". Let's look at \+ a plot of the members of fam2:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "plot(fam2,x=-1..1,-4..4,colo r=[red,green,blue,black,magenta,grey,cyan]);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 302 223 "You sh ould notice that every member of fam2 has an exremum. Make sure these results agree with your answers to question 2. It is also quite easy to use Maple to find the coordinates of the extremum of every curve i n fam2:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "xt:=c->subs(\{_C1=c[1],_C2=c[2]\},xextreme);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "map(xt,cvals);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "yxtreme:=c->subs(\{_C1=c[1],_C2=c[2]\},yextreme); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "map(yxtreme,cvals);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 303 113 "That thi rd command (\"evalf(%)\") above just evaluates the previous result to \+ floating point (i.e. decimal) form. " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 304 0 "" }{TEXT 305 13 "Quest ion 3: " }{TEXT 306 341 "Generate a list of at least seven solutions \+ to the differential equation whose graphs have no extrema. Call your \+ list \"fam3\". Then plot your list of functions to visually verify th at none of them have any extrema. Do your results confirm the conclus ions you made in question 2. If not, you may need to re-think your an swer to question 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 307 400 "Next we will be looking for conditions on _C1 and _C2 which will guarantee that the curve has an extremum (either a max or \+ a min) and the x-coordinate of the extremum is positive. We are headi ng toward using this differential equation to model a mass-spring syst em where y represents displacement and x is time. In the case that th e curve has an extremum with a positive x-coordinate we must have " } {XPPEDIT 308 0 "xextreme = -ln(-1*_C1/(2*_C2))/2;" "6#/%)xextremeG,$*& -%#lnG6#,$*(\"\"\"\"\"\"%$_C1GF-*&\"\"#F-%$_C2GF-!\"\"F2F-\"\"#F2F2" } {TEXT -1 0 "" }{TEXT 309 25 " > 0. This means that " }{XPPEDIT 256 0 "-1*_C1/(2*_C2)" "6#,$*(\"\"\"\"\"\"%$_C1GF&*&\"\"#F&%$_C2GF&!\"\"F+ " }{TEXT -1 2 " " }{TEXT 310 44 "< 1. You will continue this derivat ion in :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 311 0 "" }{TEXT 312 13 "Question 4: " }{TEXT 313 61 "For \+ xextreme to be positive, explain exactly why we must have" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }{TEXT 314 0 "" }{XPPEDIT 256 0 "-1*_C1/(2*_C2)" "6#,$*(\"\"\"\"\"\"%$_C1GF&*&\"\"#F&%$_C2GF&!\"\"F+" }{TEXT -1 2 " " }{TEXT 315 488 "< 1 . Next, find conditions on _C1 and _C2 which wil l guarantee that the corresponding solution curve has a maximum with a positive x-coordinate. Finally, find conditions on _C1 and _C2 which will guarantee that the corresponding solution curve has a minimum wi th a positive x-coordinate. I don't just want examples here. You nee d to give me a general criteria involving _C1 and _C2. If you can't g et Maple to do this it is not a problem. Just write out the solution \+ steps by hand." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 316 248 "The values of _C1 and _C1 which yield a \+ solution curve having an extremum with a positive x-coordinate can be \+ the illustrated graphically as a portion of the _C1-_C2 coordinate pl ane. Execute the following Maple commands to see this illustration:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 286 "with(plots):p[1]:=inequal(\{C1<-2*C2,C2<0,C1>0\},C1=-4..4,C2= -4..4,optionsexcluded=(color=grey),optionsfeasible=(color=orange),opti onsopen=(color=blue,thickness=2),labels=[\"C1\",\"C2\"]): pc1:=textplo t([2,-2.5,`max with xc>0`],color=yellow):pline1:=textplot([-1,1,` C1 = -2C2`],color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 280 "p [2]:=inequal(\{C1>-2*C2,C2>0,C1<0\},C1=-4..4,C2=-4..4,optionsexcluded= (color=grey),optionsfeasible=(color=orange),optionsopen=(color=blue,th ickness=2),labels=[\"C1\",\"C2\"]): pc2:=textplot([-2,2.5,`min. with x c>0`],color=yellow):pline2:=textplot([-0.9,0.9,` C1 = -2C2`],color=yel low):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "display(p[1],pc1,p line1);display(p[2],pc2,pline2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 317 13 "Another view:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "with(plots):P:=[p[1],p[2]]: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "display(P,insequence=true);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 318 340 "Click \+ on the graph above and then use the \"buttons\" near the top of the sc reen which look like they operate a CD-player to play the animation. \+ Try running it in \"continuous cycle mode\" for a while. Well, I hope that these pictures have provided you with the information you need t o determine whether your answer to question 4 is correct. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 319 138 " Of course in p ractice, the constants _C1 and _C2 are determined by initial condition s. Let's suppose that we add the initial conditions " }{XPPEDIT 320 0 "y(0) = alpha;" "6#/-%\"yG6#\"\"!%&alphaG" }{TEXT -1 1 " " }{TEXT 321 2 ", " }{XPPEDIT 322 0 "diff(y(0),x) = beta;" "6#/-%%diffG6$-%\"yG 6#\"\"!%\"xG%%betaG" }{TEXT -1 0 "" }{TEXT -1 2 " " }{TEXT 323 68 "to our problem. Let's use Maple to solve our initial value problem:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "deq:=diff(y(x),x$2)+6*diff(y(x),x)+8*y(x)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "ics:=y(0)=alpha,D(y)(0)=beta;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "dsolve(\{deq,ics\},y(x));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 324 56 "We now turn our attention from the _C1-_C2 plane to the \+ " }{XPPEDIT 325 0 "alpha;" "6#%&alphaG" }{TEXT 326 1 "-" }{XPPEDIT 327 0 "beta;" "6#%%betaG" }{TEXT -1 2 " " }{TEXT 328 504 "plane. We \+ are interested in identifying which regions of this plane correspond t o initial conditions which yield solution curves having extrema, whic h regions yield solution curves with maximums, which with minimums, wh ich yield solution curves with extrema having a positive x-coordinate, which yield solution curves with a negative x-coordinate, and which y ield solution curves with no extrema. The diagrams produced by the fo llowing Maple commands provide a comprehensive answer to these questio ns:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 286 "p1:=inequal(\{1/2*beta+2*alpha>0,-1/2*beta-alpha<0,1/2*beta+2*a lpha<-2*(-1/2*beta-alpha)\},alpha=-4..4,beta=-4..4,optionsexcluded=(co lor=grey),optionsfeasible=(color=orange),optionsopen=(color=blue,thick ness=2),labels=[\"alpha\",\"beta\"]):t2:=textplot([1.4,2,`max. with xc >0`],color=yellow):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 238 "p2: =inequal(\{1/2*beta+2*alpha<0,-1/2*beta-alpha>0,1/2*beta+2*alpha>-2*(- 1/2*beta-alpha)\},alpha=-4..4,beta=-4..4,optionsexcluded=(color=grey), optionsfeasible=(color=orange),optionsopen=(color=blue,thickness=2),la bels=[\"alpha\",\"beta\"]): : \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "t1:=textplot([2,2,`max. with xc>0`],color=red):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "t2:=textplot([-1.7,-2,`min. \+ with xc>0`],color=yellow):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "t3:=textplot([2.5,-1.5,`max with xc<0`],color=brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "t4:=textplot([-2.5,1.5,`min with xc <0`],color=brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "t5:=t extplot([[-1.5,3.8,`No`],[-1.2,3.2,`Ext.`]],color=brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "t6:=textplot([[1.5,-3.7,`Ext.`],[1. 2,-3.2,`No`]],color=brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "t11:=textplot([2,2,`max. with xc>0`],color=yellow):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "t22:=textplot([-2,-2,`min. with xc> 0`],color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "q1:=\{p1 ,t11,t22,t3,t4,t5,t6\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 " q2:=\{p2,t1,t2,t3,t4,t5,t6\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "display(q1);display(q2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 329 0 "" }{TEXT 330 13 "Question 5: " }{TEXT 331 163 "In the two figures above, what are the equations of those two blue lines? Remember that when you answer this question your answer \+ should be equations in terms of " }{XPPEDIT 332 0 "alpha;" "6#%&alphaG " }{TEXT -1 1 " " }{TEXT 333 2 " a" }{TEXT -1 0 "" }{TEXT 334 4 "nd \+ " }{XPPEDIT 335 0 "beta;" "6#%%betaG" }{TEXT -1 1 " " }{TEXT 336 86 ". There are six regions determined by the two figures. Choose an alph a-beta pair in " }{TEXT 337 4 "each" }{TEXT 338 294 " of these six reg ions, solve the corresponding initial-value problem and plot the solut ion. In each case comment on how your graph supports the conclusions \+ made in the two diagrams. Finally, determine what happens on the boun daries which separate the six regions. Pick an alpha-beta pair on " } {TEXT 339 4 "each" }{TEXT 340 112 " of the six (straight line) boundar ies, solve the corresponding initial-value problem and plot the soluti on. In " }{TEXT 341 4 "each" }{TEXT 342 205 " of these six cases, make a conjecture about the properties of solution curves corresponding to alpha-beta pairs chosen on that boundary. If you can back up your co njecture with proof, so much the better." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 343 207 "Next we consider \+ conditions on _C1 and _C2 which will guarantee that the solution curve has a positive x-intercept . First, we may assume that neither _C1 n or _C2 is equal to zero, since graphs of the form " }{XPPEDIT 344 0 "_ C1*exp(-2*x);" "6#*&%$_C1G\"\"\"-%$expG6#,$*&\"\"#F%%\"xGF%!\"\"F%" } {TEXT -1 1 " " }{TEXT 345 4 "and " }{XPPEDIT 346 0 "_C2*exp(-4*x);" "6 #*&%$_C2G\"\"\"-%$expG6#,$*&\"\"%F%%\"xGF%!\"\"F%" }{TEXT -1 1 " " } {TEXT 347 70 " have no x-intercept whatever. Consider the following M aple commands:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 42 "sol := y(x) = _C1*exp(-2*x)+_C2*exp(-4*x);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "xint:=solve(rhs(sol)=0,x); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 348 0 "" }{TEXT 349 13 "Question 6: " }{TEXT 350 507 "Det ermine general conditions on _C1 and _C2 which will guarantee that the solution curve has a positive x-intercept. Just as in question 4, I \+ don't just want examples here. You need to give me a general criterio n involving _C1 and _C2. Next sketch graphs in the _C1-_C2 plane show ing the regions which will yield a solution curve that has a positive \+ x-intercept. If you can't get Maple to do these things it is not a pr oblem. Just write out the solution steps by hand and/or sketch your g raphs by hand." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 351 101 "Here are some examples with which you can partially che ck the accuracy of your answers to question 6:" }}{PARA 0 "" 0 "" {TEXT 352 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol := y( x) = _C1*exp(-2*x)+_C2*exp(-4*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "f:=c->subs(\{_C1=c[1],_C2=c[2]\},rhs(sol));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "cvals:=[[-1,2],[-2,1],[-2,-4 ],[2,-3],[1,-1/2],[2,1]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "fam4:=map(f,cvals);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 " plot(fam4,x=-3..3,-5..5,color=[red,green,blue,black,magenta,grey]);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }{TEXT 353 0 "" }{TEXT 354 13 "Question 7: " }{TEXT 355 429 "For ea ch of the members of fam4 above, determine the regions in the _C1-_C2 \+ plane in which each ordered pair lies with relation to your answer to \+ question6. Do these curves support the answer you gave to question 6? Be specific in your answer...determine the region where each _C1-_C2 pair comes from and state specifically what behavior is expected in t his region, and what behavior is actually observed in the solution gra ph." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 356 244 "Now, suppose that in addition to having a positive x-in tercept, we require conditions on _C1 and _C2 which would also guarant ee that the solution curve has an extremum with a positive x-coordinat e. Consider the following set of Maple commands:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 285 "with(plots ):im[1]:=inequal(\{C1<-2*C2,C1<-C2,C2<0,C1>0\},C1=-4..4,C2=-4..4,optio nsexcluded=(color=grey),optionsfeasible=(color=orange),optionsopen=(co lor=blue,thickness=2),labels=[\"C1\",\"C2\"]): pimax1:=textplot([[1.3 \n,-2.5,`max with xc>0`],[1.5,-3.2,`intercept with xc>0`]],color=yello w):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 284 "with(plots):im[2]:= inequal(\{C1>-2*C2,C1>-C2,C2>0,C1<0\},C1=-4..4,C2=-4..4,optionsexclude d=(color=grey),optionsfeasible=(color=orange),optionsopen=(color=blue, thickness=2),labels=[\"C1\",\"C2\"]): pimin2:=textplot([[-1.3,2.8,`min with xc>0`],[-1.5,3.5,`intercept with xc>0`]],color=yellow):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "display(im[1],pimax1);display(im[2] ,pimin2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 357 21 "Again, some examples:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol := y(x) = _C1*exp(-2*x)+ _C2*exp(-4*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "f:=c->sub s(\{_C1=c[1],_C2=c[2]\},rhs(sol));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "cvals:=[[-2,3],[-2,1.5],[-2,0.5],[-2,-1],[2,-3],[2,-1 .5],[2,-0.5],[2,1]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "fam 5:=map(f,cvals);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot(fam5,x=-3..3,-5..5,color=[red, green,blue,black,magenta,grey,cyan,maroon]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 358 0 "" }{TEXT 359 13 "Question \+ 8: " }{TEXT 360 783 "In the plots in the _C1-_C2 plane, what are the \+ equations of those two blue lines? Note that the _C1-_C2 plane is div ided into eight regions by the two blue lines and the coordinate axes. Two of these regions have been labeled in the diagrams. Each of the \+ eight graphs in the family fam5 has _C1-_C2 values from a different on e of these eight regions. Using the graphs as examples, and perhaps s upplying some more of your own, classify the remaining six regions wit h respect to maxes, mins, positive x-intercepts or negative x-intercep ts. Show a completed diagram with all eight regions completely classi fied. Don't worry about doing the diagram in Maple. You can print ou t the (incomplete) diagrams above and just fill in the descriptions of the remaining six regions by hand." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 361 53 "Now we will convert to a lpha and beta. Recall that: " }{XPPEDIT 363 0 "_C1 = beta/2+2*alpha; " "6#/%$_C1G,&*&%%betaG\"\"\"\"\"#!\"\"F(*&\"\"#F(%&alphaGF(F(" } {TEXT -1 0 "" }{TEXT 362 4 " , " }{XPPEDIT 364 0 "_C2 = -beta/2-alpha ;" "6#/%$_C2G,&*&%%betaG\"\"\"\"\"#!\"\"F*%&alphaGF*" }{TEXT -1 0 "" } {TEXT 365 101 " . So let's say we want to find the alpha-beta region s where the x-intercept is positive. We have:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 305 "with(plots ):ii[2]:=inequal(\{1/2*beta+2*alpha>-(-1/2*beta-alpha),-1/2*beta-alpha >0,1/2*beta+2*alpha<0\},alpha=-4..4,beta=-4..4,optionsexcluded=(color= grey),optionsfeasible=(color=orange),optionsopen=(color=blue,thickness =2),labels=[\"alpha\",\"beta\"]): pib:=textplot([0.2,-3,`intercept wit h x>0`],color=black):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "display(ii [2],pib);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 307 "with(plots):ii[1]:=inequal(\{1/2*b eta+2*alpha<-(-1/2*beta-alpha),-1/2*beta-alpha<0,1/2*beta+2*alpha>0\}, alpha=-4..4,beta=-4..4,optionsexcluded=(color=grey),optionsfeasible=(c olor=orange),optionsopen=(color=blue,thickness=2),labels=[\"alpha\",\" beta\"]): pia:=textplot([-0.4,3.2,`intercept with x>0`],color=black): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "display(ii[1],pia);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 366 245 "Now let's tran slate the information about _C1 and _C2 to information concerning alph a and beta for which the corresponding solution curves have the TWO p roperties of a positive x-intercept and an extremum at a positive x-co ordinate. We obtain:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 378 "with(plots):im[1]:=inequal(\{1/2*b eta+2*alpha<-2*(-1/2*beta-alpha),1/2*beta+2*alpha<-(-1/2*beta-alpha),- 1/2*beta-alpha<0,1/2*beta+2*alpha>0\},alpha=-4..4,beta=-4..4,optionsex cluded=(color=grey),optionsfeasible=(color=orange),optionsopen=(color= blue,thickness=2),labels=[\"alpha\",\"beta\"]): pimax1a:=textplot([[-. 2\n,2.8,`max with xc>0`],[-.5,3.5,`intercept with xc>0`]],color=black) :" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "display(im[1],pimax1a);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 377 "with(plots):im[2]:=inequal( \{1/2*beta+2*alpha>-2*(-1/2*beta-alpha),1/2*beta+2*alpha>-(-1/2*beta-a lpha),-1/2*beta-alpha>0,1/2*beta+2*alpha<0\},alpha=-4..4,beta=-4..4,op tionsexcluded=(color=grey),optionsfeasible=(color=orange),optionsopen= (color=blue,thickness=2),labels=[\"alpha\",\"beta\"]): pimax1b:=textpl ot([[.3\n,-2.5,`min with xc>0`],[0,-3.2,`intercept with xc>0`]],color= black):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "display(im[2],pimax1b); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 367 162 "As \+ you can see, imposing the extra conditions about the extrema have not \+ changed our regions at all. The same regions are colored red in both \+ sets of diagrams. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 368 0 "" }{TEXT 369 13 "Question 9: " }{TEXT 370 628 "In this question you will complete the classification of the \+ eight regions defined by the two blue lines and the coordinate axes in the diagrams above with respect to extrema and intercepts. First of \+ all, what are the equations of those two blue lines in the diagrams? \+ Next, pick at least one point in each of the eight regions and use the resulting graphs to help you classify the remaining six regions. Onc e again, you can just print out one of the diagrams above and fill in \+ the remaining information by hand. The following output could help yo u answer this question, or you can just generate a set of graphs on yo ur own:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "solab := y(x) = (-1/2*beta-alpha)*exp(-4*x)+(1/2*beta +2*alpha)*exp(-2*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "g:= (alpha,beta)->(-1/2*beta-alpha)*exp(-4*x)+(1/2*beta+2*alpha)*exp(-2*x) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "cvals:=[(-1/2,3),(-1/2 ,1.2),(-1/2,0.25),(-1/2,-1),(1/2,-3),(1/2,-1.2),(1/2,-0.25),(1/2,1)]; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "L:='L':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "for i from 1 to 16 by 2 do j:=i-iqu o(i,2):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 " L[j]:=plot(g(op(i,cval s),op(i+1,cvals)),x=-2..2,-1..1)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 " od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "for j from 1 to 8 do L[j] od;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "display(L[3],L [4]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "display(L[7],L[8]) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 371 178 "We \+ should now have a complete picture of the behavior of the members of t he family of solutions with respect to extrema and x-intercepts as the y relate to the initial conditions " }{XPPEDIT 372 0 "y(0) = alpha;" " 6#/-%\"yG6#\"\"!%&alphaG" }{TEXT -1 0 "" }{TEXT 373 13 " , y ' (0) = \+ " }{XPPEDIT 374 0 "beta;" "6#%%betaG" }{TEXT -1 1 " " }{TEXT 375 113 " . Now we'll apply this information to the motion of a mass suspended \+ on a spring. The differential equation: " }{XPPEDIT 376 0 "diff(y(x ),`$`(x,2))+6*diff(y(x),x)+8*y(x) = 0;" "6#/,(-%%diffG6$-%\"yG6#%\"xG- %\"$G6$F+\"\"#\"\"\"*&\"\"'F0-F&6$-F)6#F+F+F0F0*&\"\")F0-F)6#F+F0F0\" \"!" }{TEXT -1 2 " " }{TEXT 377 243 "can be viewed as a model of a ma ss-spring system which has a mass of one slug suspended on a spring wi th spring constant 8 lb/ft with the system subjected to a damping forc e of 6 lb of resistance for each foot per second of velocity. Here we " }{TEXT 378 45 "take the positive direction to be downward. " } {TEXT 379 5 "Then " }{XPPEDIT 380 0 "y(0) = alpha;" "6#/-%\"yG6#\"\"!% &alphaG" }{TEXT -1 2 " " }{TEXT 381 65 "gives the initial displacemen t of the mass (below equilibrium if " }{XPPEDIT 382 0 "alpha;" "6#%&al phaG" }{TEXT -1 1 " " }{TEXT 383 37 "is positive and above equilibrium if " }{XPPEDIT 384 0 "alpha;" "6#%&alphaG" }{TEXT -1 1 " " }{TEXT 385 31 "is negative). Also, y ' (0) = " }{XPPEDIT 386 0 "beta;" "6#%% betaG" }{TEXT -1 2 " " }{TEXT 387 40 "gives the initial velocity (dow nward if " }{XPPEDIT 388 0 "beta;" "6#%%betaG" }{TEXT -1 0 "" }{TEXT 389 27 " is positive and upward if " }{XPPEDIT 390 0 "beta;" "6#%%beta G" }{TEXT -1 1 " " }{TEXT 391 15 "is negative). " }{TEXT -1 0 "" } {TEXT 392 69 "Let's look at alpha-beta space and the by now familiar e ight regions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "pmass:=plot(\{-4*alpha,-2*alpha\},alpha=-4..4,-4 ..4,labels=[\"alpha\",\"beta\"],color=[red,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "tmass:=textplot([[-0.5,3.5,`I`],[-1.2,3.3, `II`],[-2,1.5,`III`],[-2,-2,`IV`],[0.4,-3,`V`],[1,-2.7,`VI`],[2,-1.5,` VII`],[2,2,`VIII`]],color=orange):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "display(pmass,tmass);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 393 787 "Your next question will ask you to explain the motion of the mass if the initial conditions are chosen in each of th e regions I-VIII in alpha-beta space. For example, if the initial con ditions are chosen in region I, then the mass starts at alpha units ab ove equilibrium (since alpha is negative in region I and a negative in itial position is above equilibrium) with a downward velocity (since b eta is positive in region I and positive velocity is downward), passes through equilibrium, reaches a maximum distance below equilibrium and then approaches equilibrium asymptotically fairly rapidly at first an d then more and more slowly (due to the point of inflection) without e ver passing through equilibrium again. Well now that I've done region I for you as an example, it's your turn:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 394 0 "" }{TEXT 395 13 "Qu estion 10: " }{TEXT 396 293 " For all of the other regions II-VIII in \+ the diagram above, give a full explanation of the motion of the mass i f the initial conditions are chosen in the corresponding region. Fina lly, explain the motion of the mass if the initial conditions are chos en on the boundaries of these regions for " }{TEXT 397 5 "each " } {TEXT 398 202 "of the eight boundary lines. Observe that some of thes e curves have inflection points. Take these into account in your expl anations. By the way, what happens if both alpha and beta are equal t o zero?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }