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A rescue team can travel 50 miles per ho ur on the road but only 10 miles per hour in the desert. The rescue truck is 60 miles from the point P on the road which is exactly 15 mi les from the crash site. It can leave the road at the point P but pe rhaps it can arrive faster by leaving the road sooner. What should th e rescue team do?" }{TEXT 263 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {OLE 1 33809 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :fyyyyyAB;F;N;V;^;f;;JAjA>:[Z:F=F=N=V=^=f =n=v=>>AFANAVA^AfAn A;jYJyKyky;:::::::::::::::::::::::::::::::::::::::::_lqvGcMJ:::::::JEf:yyyxIN::;`:Z@[::JbIk?@xl=X]Aj;J:@:<:=j[vGUMrvC?MoJ: :::::::JCNZ;^:vYxI>:<::::::j`J:j:vCSmlF@[KaFFcmnnHEM:>:::::::oJ;@j:j;> Z:vYxY:B::::::^:^^qK:<:=:jRj^^HEmpnCfGEM:>::::::n=?R:yyyyyy:>:<::::::? :EJ:>:F:;JyKy;vY:::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::F:wyyAbR<:Tn=JbDNsPqs\\`A:::Zn IalHh:J:::::::::::=:;::::::::::F:D:<:::::::::::::::::::::::::::::::::: ::::::::::::::::::JJXA?TMorW>Z:JUL:K;=a:FJF:K:;JlEAWh=>uHJ:<::>j:@j:B: ;F:>J>:J;NZBr:Ar:B:yI:;ryyI:;J:ryB:<: yQ:yyy?J;B:yyy;ny=R:?j:vy:>b:xYgWZUY\\iyiOu@KjyyyyyY:m>>kyKsM_:vF:SewcWZ;suqlZxkvyejaPoB:Z:^iSvxy`fyK\\]ygF`RvxyxfRoA;J:yQ:Z:jq> :WfBpywdEJswn?Nry?e>vd>J:Z:Nyu>BWN:>Z:JHJRMd:`rbhE=a:^gbhE;J:>j:@j:>Z:>j:J:K:BZB: j:ky=JZBJ;F:yyy>r:Ar:B:yI:;ryyI:;J:ryB:<:yQ:wY;N:b:xYyi?SIZ[GUYx?uMxyPfe`vsQfBMMl=rM`N<=:SB:yKPfv [=^_;x[AwvQfB>uMP:yi?[UZ:;jyyY:t\\C::;:KZ:N:>:^:;J:v;LZD@IL:F:;B:[@;B:YT:[hRuLZD@oY ZevYJZDpN`QyXOExNExnMQJXDRPAkeLJOuK;uKAtJA:::::::::::::::::::::::::::: :::::::::<:Wb=Wb=::::::::::::::::::::::::::::::::::::Z::t::::::::::::: ::::::::::::::::::::::::B:ZH:::::::::::::::::::::::::::::::::::::<:b=: :::::::::::::::::::::::::::::::::::Z::t::::::::::::::::::::::::::::::: ::::::B:ZH:::::::::::::::::::::::::::::::::::::<:b=::::::::::::::::::: :::::::::::::::::Z::t:::::::::::::::::::::::::::::::::::::B:ZH:::::::: :::::::::::::::::::::::::::::<:b=::::::::::::::::::::::::::::::::::::Z ::t:::::::::::::::::::::::::::::::::::::B:ZH:::::::::::::::::::::::::: :::::::::::<:b=::J:n[:n[Hb=;:>:;JAn;b=W::::::::::::::::::::::::::::::< J:ZH:::;b=<:W:;B:WB:n;WB:b=t::::::::::::::::::::::::::::::B:;B:tZH:::; b=<:W:B:tZHB::b=::J:tZ:JA:ZHB:;b=n;WZ:b=B:b=t:::::::: ::::::::::::::::::::::B:;B:b=::B:B:<:t:WJ:<:WJAZHn;::::::::::::::::::: ::::::::::B:;:t:::>ZHJA::B:b=:::::::::::::::::::::::::::::::b=::::::::::::::::::::::::::: :::::::::Z::t:::::::::::::::::::::::::::::::::::::B:ZH:::::::::::::::: :::::::::::::::::::::<:b=::::::::::::::::::::::::::::::::::::Z::t::::: ::::::::::::::::::::::::::::::::B:ZH:::::::::::::::::::::::::::::::::: :::<:b=::J:t:WB:;:::b=;Z:B:tJAt::::::::::::::::::::::::::::Z::t:::>ZH: J:n;>:W:;JA:WJA::::::::::::::::::::::::::::Z:>:b=::J:t::WJ:tJAt:;b=W:;JAJ:n;>:W:n; :::::::::::::::::::::::::::Z:>Z:ZH:::;b=;JAZHJA:::::::::::ZHJA :::::::::::ZHJA::::::::::::ZH:n;: :::::::::ZH:n;::::::::::ZH:n;:::::::::::ZHJA:<:t::::::::::::::::::::::::::::: :::Z:>:b=::;b=;B:<::::::::::::::::::::::::::::::::t:W::ZHZ:::: :::::::::::::::::::::::::::::WZH:Z:>:b=::;b=WJA:<:t::::::::::::::::::: ::>:n;::::::::;B:WZH:Z:B:b=::;b=n;Z:ZH:::::::::::::::::::::;:W:::::::: :::ZHJA:<:t:::::::::::::::::::::>:n;::::::::::B:;:t::J:B::n;t: ::::::::::::::::::::>ZHJA::::::::::Z:>:b=:::::::::::::::::::::::::>ZHJ A::::::::::Z:>:b=:::::::::::::::::::::::::>ZHJA::::::::::Z:>:b=::::::: ::::::::::::::::::>ZHJA::::::::::Z:>:b=:::::::::::::::::::::::::>ZHJA: :::::::::Z:>:b=::::::::::J:::WJ:tZ::;b=Wb=W::::::::::J:JA::::::::::Z:> :b=::::::::::B:W::n;::::::::::B:;:t::::::::::::<:>:Wb=W::::::::::J:JA::::::::::Z:>:b=::::::::::B:>ZHB:WJAZH>Z:n[:J :tJ:n[Hn;::::::::::;:W:::::::::::Z:B:WJ: n[Hn;::::::::::;:W::::::::::::b=::::::::::J:::W::;b=n;::::::::::;:W:<:tJ:<:n;;b=<:>:W b=W:::::::b =:::::::::::::::::::::::::>ZHJAZ:ZHZ:ZH>:<:>:Wb=W::::::ZHJA>:WB:>ZH>:Wb=W::::::ZHJAB::;B::Wb=W::::::ZHJA::::::::B:;B:b=:::::::::::::::::::::::::>ZHJAZ:ZH>Z:B:WZH: J:t:W:::::::n;::::::::Z:>:b=:::::::::::::::::::::::::>ZHJA:;:W:::::::::B: ;:t::::::::::::::::::::::::::;b=n;J:>:WJ:n;;J:tJ:n[Hb=Z:n;;b=;:WJ:>:tZH>:W:n;::::Z:>Z:b=t: :::::::::::::::::::::::::;b=n;J::;:W:;JAJ:n;W:::::::JA:;b=J:n[:>:;b=;ZHb=;B::>ZH>ZHJAt:;JAJ:n;JA::::::WJ:<:B:WJA:t:::::Z::t::::::::::::::::::::: ::::B:WJ:Z:ZH:::::::::::::::::::::::::: n;:::::::Z:>:b=::::::::::::::::::::::::Z:n;ZH::::::::::Z:>:b=::::::::: ::::::::::::::::;:n;::::::::::B:ZH:::::::::::::::::::::::::::::::::::: :<:b=::::::::::::::::::::::::::::::::::::Z::t::::::::::::::::::::::::: ::::::::::::B:ZH:::::::::::::::::::::::::::::::::::::<:b=::::::::::::: :::::::::::::::::::::::Z::t:::::::::::::::::::::::::::::::::::::B:ZH:: :::::::::::::::::::::::::::::::::::<:b=::::::::::::::::::::::::::::::: :::::Z::t:::::::::::::::::::::::::::::::::::::B:ZH:::::::::::::::::::: :::::::::::::::::<:b=::::::::::::::::::::::::::::::::::::Z::t::::::::: ::::::::::::::::::::::::::::B:ZH:::::::::::::::::::::::::::::::::::::< :b=::::::::::::::::::::::::::::::::::::Z::t::::::::::::::::::::::::::: ::::::::::B:ZH:::::::::::::::::::::::::::::::::::::<:b=::::::::::::::: :::::::::::::::::::::Z:J::::::::::::::::::::::::::::::::::::::j:vCSmlB :;:::::::::N;?B:yyyyyy:>:;B::::::V:;JD>:;::::::::::::vYxI:;Z:::::::::: ::::::::::yay=J:B:::::::::::::::::::jysy:>:<::::::::::6:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 251 "Let x denote the distance from the initial position of the rescue truck t o the point at which the truck leaves the road and turns into the dese rt. It will then travel a distance of x miles along the road and (b y the Pythagorean theorem) a distance " }{XPPEDIT 18 0 "sqrt((60-x)^2+ 15^2);" "6#-%%sqrtG6#,&*$,&\"#g\"\"\"%\"xG!\"\"\"\"#F**$\"#:F-F*" } {TEXT -1 96 " miles through the desert. The travel time along the roa d is then (by time = distance/speed) " }{XPPEDIT 18 0 "x/50;" "6#*&% \"xG\"\"\"\"#]!\"\"" }{TEXT -1 53 " hours. Similarly, the time throug h the desert is " }{XPPEDIT 18 0 "sqrt((60-x)^2+15^2)/10;" "6#*&-%%s qrtG6#,&*$,&\"#g\"\"\"%\"xG!\"\"\"\"#F+*$\"#:F.F+F+\"#5F-" }{TEXT -1 158 " hours. Thus we want to determine x, the distance that the truc k travels along the road before it turns to the desert, for which that the total travel time " }{XPPEDIT 18 0 "T(x) = x/50+sqrt((60-x)^2+15^ 2)/10;" "6#/-%\"TG6#%\"xG,&*&F'\"\"\"\"#]!\"\"F**&-%%sqrtG6#,&*$,&\"#g F*F'F,\"\"#F**$\"#:F5F*F*\"#5F,F*" }{TEXT -1 273 " is minimal. In oth er words, we are looking for the global minimum of the function T(x) i n the interval [0,60]. Indeed, the only values of x which make sense p hysically are those in between 0 and 60. Here's the solution using Map le. We begin by defining the function T(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "T:= x->x/50+sqrt(( 60-x)^2+15^2)/10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"TGf*6#%\"xG6 \"6$%)operatorG%&arrowGF(,&9$#\"\"\"\"#]-%%sqrtG6#,&*$),&\"#gF/F-!\"\" \"\"#F/F/\"$D#F/#F/\"#5F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 " Let's plot the function T(x) in the interval [0,60] to see where the g lobal minimum might be." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "plot(T(x),x=0..60,labels=[\"x-miles before turning to desert\", \"T-total travel time\"]);" }}{PARA 13 " " 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 242 "We see that the global minimum is somewhere between 50 and 60. That is, the truck should leave the road sometime after having traveled fifty miles along the road. Let's plot T(x) on the smaller interval to have a better idea when it might be. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "plot(T(x),x=50..60,labels=[\"x-miles before turning t o desert\",\"T-total travel time\"]);" }}{PARA 13 "" 1 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 "Now we see more clearly that the global minimum is \+ between 56 and 58 miles. To find the exact value we have to find the c orresponding zero of the derivative D(T)(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(T);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#f*6#%\"xG6\"6$%)operatorG%&arrowGF&,&#\"\"\"\"# ]F,*&,&!$?\"F,9$\"\"#F,-%%sqrtG6#,&*$),&\"#gF,F1!\"\"F2F,F,\"$D#F,F;#F ,\"#?F&F&F&" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "bestx:=solve(D(T)(x)=0,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&bestxG,&\"#g\"\"\"*$-%%sqrtG6#\"\"'F'#!\"&\"\"%" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf(bestx);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#$\"+#y8Qp&!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 101 " We see that the tr uck should leave the road after having traveled about 56.94 miles alo ng the road." }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT 265 10 "Example 2." }{TEXT -1 196 " If the truck could go, say, 15 mph in the desert, should it leave the road before having traveled 56.94 miles, or after? More generally, if the truck can travel r mph through the desert, where " }{TEXT 266 1 " " }{TEXT -1 44 "r is a giv en positive speed, 0 " 0 "" {MPLTEXT 1 0 33 "S:=x-> x/50+ sqrt((60-x)^2+225)/r;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"SGf*6#%\" xG6\"6$%)operatorG%&arrowGF(,&9$#\"\"\"\"#]*&-%%sqrtG6#,&*$),&\"#gF/F- !\"\"\"\"#F/F/\"$D#F/F/%\"rGF:F/F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Let's find the derivative D(S) with respect to x." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(S);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#f*6#%\"xG6\"6$%)operator G%&arrowGF&,&#\"\"\"\"#]F,*&,&!$?\"F,9$\"\"#F,*&-%%sqrtG6#,&*$),&\"#gF ,F1!\"\"F2F,F,\"$D#F,F,%\"rGF,F<#F,F2F&F&F&" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 201 "Next, we find zeros of the derivative. They are possib le candidates for the global minimum of S(x) in [0,60], that is, possi ble candidates for the optimal, the best, x corresponding to a given s peed r." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "bestxforr:=solve (D(S)(x)=0,x); " }{TEXT -1 1 "\000" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%*bestxforrG,&*&-%%sqrtG6#,$*&\"\"\"F,,&*$)%\"rG\"\"#F,F,!%+DF,!\"\"F 3F,F0F,!#:\"#gF," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 359 "We found a zero of the derivative D(S), but we don't know if it is a global minimum of the function S(x ) in [0,60]. For all we know, it could be only a local minimum, or max imum, or neither. Since S(x) contains a parameter r, we cannot draw th e graph of S(x) to determine its behavior in [0,60]. Let's calculate \+ and simplify the second derivative D(D(S)). " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "simplify(D( D(S))(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"\"F%*&),(\"%DQF% %\"xG!$?\"*$)F*\"\"#F%F%#\"\"$F.F%%\"rGF%!\"\"\"$D#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 " We see that the second derivative S''(x) is always positive. (It is ea sy to check that the quadratic function " }{XPPEDIT 18 0 "3825-120*x+x ^2;" "6#,(\"%DQ\"\"\"*&\"$?\"F%%\"xGF%!\"\"*$F(\"\"#F%" }{TEXT -1 683 " is always positive.) That means that the derivative S'(x) is always \+ increasing. Hence, bestxforr is the only zero of S'(x). Left of bestxf orr, S'(x) is negative, to the right of bestxforr, S'(x) is positive. \+ (Can you see why?) Therefore, S(x) is decreasing left of bestxforr, an d increasing right of bestxforr. Does it mean that x = bestxforr is th e solution to our optimization problem for any given r and represents \+ the optimal distance after which the truck should turn into the desert ? Yes, as long as x = bestxforr falls within the interval [0,60]. Othe rwise, the global minimum over [0,60] is at x = 60 or at x=0. Let's p lot values of bestxforr for speeds r between 0 and 50." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "plot(b estxforr,r=0..50,labels=[\"r-speed through desert\",\"bestxforr\"]);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1212 " We see that bestxfo rr falls in the interval [0,60] for all values of r between 0 and some where shortly before 50. For those values of r, bestxforr is, indeed, our optimal solution. Hence, as evident from the graph, for speed thr ough the desert r close to zero, the optimal x is near 60; that is, th e truck should go almost to the point P before turning to the desert. \+ With increasing values of r, the truck should turn to the desert soone r. Observe that bestxforr is always strictly smaller that 60, except \+ for r=0, which is not under consideration. Hence, surprisingly, the tr uck should never travel right up to the point P. We see that for value s of r near 50 mph the critical point bestxforr is outside the interva l [0,60]. More precisely, bestxforr is negative, hence, it is to the \+ left of the interval [0,60]. In that case, bestxforr does not represen t the optimal solution. Since S'(x) is positive to the right of the cr itical point bestxforr, the function S(x) is increasing throughout [0, 60]. Thus, the global minimum of S(x) in [0,60] is at x =0, which mean s that the truck should leave the road immiediately and travel entirel y through the desert. Let's find the values of r for which that happen s." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "solve(bestxforr=0,r); evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#\"#<\"\"\" #\"$+#F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"++Dr][!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 256 "What is our conclusion then? For all speeds r between about 48 .51 and 50, bestxforr is negative and x=0 is the global minimum of S( x) in the interval [0,60]. In that case, the truck should leave the ro ad immediately and travel entirely through the desert." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 17 "Homework Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 256 11 "Problem 1. " }{TEXT -1 290 " A man in a rowboat finds himself at a point P which is 1 mile from the nearest p oint A on a straight shoreline. His goal is to reach a point Q which \+ is 1 mile directly inland from a point B on the shore 2 miles from A. \+ He decides to row directly to a point on the shore between A and B, " }{TEXT 260 1 "x" }{TEXT -1 159 " miles from A, and then walk directly to Q. His rowing speed is 2 mph. His walking speed is 5 mph. Using \+ similar methods as in Example 1, find the value of " }{TEXT 259 1 "x" }{TEXT -1 56 " which results in the shortest travel time from P to Q. \+ " }}{PARA 0 "" 0 "" {METAFILE 238 207 207 1 "iXmGh`B:VoCngB>w=Z:^aOG:M Z=Nv[F:=jGAl:>:E:Qb:B:E:Sb:YPqWm[jZhJJ^SYPq?:;ja;;JNJqka?:=JFJ::::jyyy yyI::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::>Z:vYxI::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::>:ry:: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 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::::::::::::::::::::::::::::::::krwyay=::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::jD`sysy::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :f`?wYxI:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::;r=vY::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::VhdvYxI:::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::OB:U[:vXHJDJ:>Jj:>J>:;NZ=VZ=JZmWZ=JMeVrwyuAX;GbCO:=R:=jy=Z:>ZymyyY:>Z:vy:vy:J:xyywy yyYC:cc:VZ:J:3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 267 11 "Proble m 2. " }{TEXT -1 121 " At noon ship A was 100 miles due east of ship B . Ship A is sailing west at 11 mph and ship B is sailing south at r mp h. " }}{PARA 0 "" 0 "" {TEXT -1 120 " (a) Assume that r = 9.5 mph . At what time will the ships be nearest to one another and what will \+ this distance be?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 583 " (b) This time you know only that r is a given posit ive speed. Let t =besttforr be the time, in hours after noon, at which the ships are nearest to one another for a given speed r. Find the fo rmula for besttforr in terms of r. Plot besttforr as a function of r, \+ for r, say, between 0 and 100 mph. Comment on the behavior of the func tion as r gets close to 0, and as r gets larger and larger. Explain, i n practical terms, what you see. Find a formula for the minimal distan ce between the ships as a function of r. Plot the function and interpr et its behavior in practical terms. " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 268 28 "Problem 3. (Extra credit). " } {TEXT -1 166 "Look again at the man in the rowboat as described in Pro blem 1. Suppose that this time we are not given his walking speed and \+ his rowing speed. We know only that the " }{TEXT 269 6 "ratio " } {TEXT -1 65 "of his rowing speed to his walking speed is r, where 0 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0 0" 8 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }