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minimising t he curved surface area of a conical tent " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 279 8 "Question" }{TEXT -1 3 ": " }}{PARA 0 "" 0 "" {TEXT -1 45 "The volume enclosed inside a conical tent is " }{XPPEDIT 18 0 "27*Pi" "6#*&\"#F\"\"\"%#PiGF%" } {TEXT -1 92 " cubic metres. Find the radius of the base so that the ar ea of canvas required is a minimum." }}{PARA 0 "" 0 "" {TEXT -1 47 "No te: The floor of the tent is to be excluded. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 280 8 "Solution" }{TEXT -1 3 " : " }}{PARA 0 "" 0 "" {TEXT -1 50 "Let the radius of the base of the \+ conical tent be " }{TEXT 281 1 "r" }{TEXT -1 27 " metres, let the heig ht be " }{TEXT 282 1 "h" }{TEXT -1 45 " metres, let the slant height o f the cone be " }{TEXT 293 1 "R" }{TEXT -1 29 " metres, and let the ar ea be " }{TEXT 283 1 "A" }{TEXT -1 16 " square metres. 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Fj^m7$$!3*******z^p9a)F?$\"30+++gaW=>F-7$$!31+++.++]5F-$\"36+++YLl==F- 7$$!3$******4.\\VB\"F-Fi_m7$$!3%******fFu^S\"F-$\"3)******>8/1c\"F-7$$ !3')*****R8/1c\"F-Fc`m7$$!3%********oN*)p\"F-$\"3'******z-\\VB\"F-7$$! 3*)******\\Ll==F-$\"3/+++'*****\\5F-Fh[l-%%TEXTG6&7$$Fb\\lF^]m$\"\"*F^ ]mQ\"q6\"-%'COLOURG6&F[\\lF)F)F)-%%FONTG6$%'SYMBOLG\"#5-F`em6&7$$!#VF^ ]m$\"#;F^]mQ\"RFgemFhem-F\\fm6$%*HELVETICAGFeem-F`em6&7$$\"#XF^]m$F-F^ ]mFgfmFhemFhfm-%+AXESLABELSG6%Q!FgemFdgm-F\\fm6#%(DEFAULTG-%(SCALINGG6 #%.UNCONSTRAINEDG-%*AXESSTYLEG6#%%NONEG-%%VIEWG6$FggmFggm" 1 2 0 1 10 0 2 9 1 1 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curv e 3" "Curve 4" "Curve 5" "Curve 6" }}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 149 "The arc of a circle which forms the curved part of the b oundary of this sector has length equal to the circumference of the ba se of the cone, namely " }{XPPEDIT 18 0 "2*Pi*r" "6#*(\"\"#\"\"\"%#PiG F%%\"rGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 84 "This means \+ that the radian measure of the angle between the radii of this sector \+ is " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "theta=2*Pi*r/R " "6#/%&thetaG**\"\"#\"\"\"%#PiGF'%\"rGF'%\"RG!\"\"" }{TEXT -1 10 " ra dians. " }}{PARA 257 "" 0 "" {TEXT -1 9 "The area " }{TEXT 284 1 "A" } {TEXT -1 82 " of the sector (which is the area of the curved surface o f the cone) is given by " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "A = R^2*theta/2;" "6#/%\"AG*(%\"RG\"\"#%&thetaG\"\"\"F' !\"\"" }{XPPEDIT 18 0 "`` = Pi*r*R;" "6#/%!G*(%#PiG\"\"\"%\"rGF'%\"RGF '" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 58 "Alternatively, since areas of various sectors with radius " }{TEXT 295 1 "R" }{TEXT -1 147 " are in direct proportion with the curved boundary arc of the sec tor, the area of the sector that corresponds to the curved surface of \+ the cone is " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "A=2*P i*r/(2*Pi*R)" "6#/%\"AG**\"\"#\"\"\"%#PiGF'%\"rGF'*(F&F'F(F'%\"RGF'!\" \"" }{TEXT -1 1 " " }{TEXT 296 1 "x" }{TEXT -1 1 " " }{XPPEDIT 18 0 "P i*R^2 = Pi*r*R;" "6#/*&%#PiG\"\"\"*$%\"RG\"\"#F&*(F%F&%\"rGF&F(F&" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 31 "Making use of the relat ion (i) " }{XPPEDIT 18 0 "R = sqrt(h^2+r^2);" "6#/%\"RG-%%sqrtG6#,&*$% \"hG\"\"#\"\"\"*$%\"rGF+F," }{TEXT -1 9 " we have " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "A=Pi*r*sqrt(h^2+r^2)" "6#/%\"AG*(%#Pi G\"\"\"%\"rGF'-%%sqrtG6#,&*$%\"hG\"\"#F'*$F(F/F'F'" }{TEXT -1 16 " -- ----- (ii). " }}{PARA 0 "" 0 "" {TEXT -1 32 "Since the volume of the c one is " }{XPPEDIT 18 0 "27*Pi" "6#*&\"#F\"\"\"%#PiGF%" }{TEXT -1 30 " we also have the constraint: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "Pi*r^2*h/3=27*Pi" "6#/**%#PiG\"\"\"*$%\"rG\"\"#F&%\"hGF &\"\"$!\"\"*&\"#FF&F%F&" }{TEXT -1 17 " ------- (iii). " }}{PARA 0 " " 0 "" {TEXT -1 26 "The equation (iii) gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "h=81/r^2" "6#/%\"hG*&\"#\")\"\"\"*$%\"r G\"\"#!\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 27 "Thus we c an substitute for " }{TEXT 297 1 "h" }{TEXT -1 19 " in (ii) to obtain \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "A=Pi*r*sqrt(81^2/ r^4+r^2)" "6#/%\"AG*(%#PiG\"\"\"%\"rGF'-%%sqrtG6#,&*&\"#\")\"\"#*$F(\" \"%!\"\"F'*$F(F/F'F'" }{TEXT -1 16 " ------- (iv). " }}{PARA 0 "" 0 " " {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "We can now plot a graph to show how the area " }{TEXT 290 1 "A" }{TEXT -1 20 " changes as we vary " }{TEXT 291 1 "r" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "g := r -> Pi*r*sqrt(81^2/(r^4)+r^2):\n'g(r)'=g(r);\np lot(g(r),r=0..8,A=0..400);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"gG6 #%\"rG*(%#PiG\"\"\"F'F*,&*&\"%hlF*F'!\"%F**$)F'\"\"#F*F*#F*F1" }} {PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7dp7 $$\"3/+++X&)G\\a!#?$\"3Uc)\\all(pY!#87$$\"3-+++4x&)*3\"!#>$\"3eG\\sFG) [L#F-7$$\"31+++klyM;F1$\"35Q0M^&)ec:F-7$$\"3-+++=arz@F1$\"3ctC'QTTu;\" F-7$$\"3)******>FWYs#F1$\"3X/Rh78`R$*!#97$$\"3%******f7t&pKF1$\"3YNUYh F%Hy(FC7$$\"3!*******z>]9QF1$\"3QwY7'z$4rmFC7$$\"3'******\\$3VfVF1$\"3 =eVlqq?PeFC7$$\"3$)*****zofV!\\F1$\"3)e!eq3&G')=&FC7$$\"3K+++S&)G\\aF1 $\"3iBRufcwpYFC7$$\"3m*****\\RQr9*QFC7$$\"3w*****\\5vS3(F1$\"3ym%y!Hu7#f$FC7$$\"3\")* *****fR+HwF1$\"3bq;6)*oaNLFC7$$\"3')*****\\\"G$R<)F1$\"3Hlhl/r<8JFC7$$ \"3!*******p;')=()F1$\"3MhLUNNg=HFC7$$\"3a******>0zj#*F1$\"3SNP&e(4#pu #FC7$$\"3e*****\\P>(3)*F1$\"3'or!zaUJ%f#FC7$$\"3'******H#[ON5!#=$\"3%* =:9*RrxX#FC7$$\"31+++3x&)*3\"Fbq$\"3@'[l,$G)[L#FC7$$\"3(******Rf]V9\"F bq$\"3e.;UKzpBAFC7$$\"3#*******yM%))>\"Fbq$\"3\"p$4ba2iA@FC7$$\"3/+++k jL`7Fbq$\"3'o_gk*GLI?FC7$$\"3%*******\\#HyI\"Fbq$\"3KeHq\"pNd%>FC7$$\" 37+++O@Ki8Fbq$\"3'pqKME1z'=FC7$$\"3&******4-:oT\"Fbq$\"3_p!Hcrjgz\"FC7 $$\"31+++1zIr9Fbq$\"3>02/QGaHz+e_\"Fbq$\"3#4[M1Xtxm\" FC7$$\"3(******Hc'yM;Fbq$\"3uMqba&)ec:FC7$$\"3)******RLsPu\"Fbq$\"3u3& *yqFbq$\"3xO= @4W:;8FC7$$\"3$******R9c#G?Fbq$\"3'fWiVn>YD\"FC7$$\"3/+++!3%3B@Fbq$\"3 3u_wg=e)>\"FC7$$\"32+++`*RFJ#Fbq$\"3'='Qa]BH+6FC7$$\"35+++EeR-DFbq$\"3 &QJy6^,p,\"FC7$$\"3;+++svq\")GFbq$\"3pjHdiS\\I))!#:7$$\"39+++<$>5E$Fbq $\"3x\\x?1&eL!yF_w7$$\"35+++Y]f(o$Fbq$\"3%yX:u#yn+pF_w7$$\"39+++w2<9TF bq$\"3'pZ%G9c=&='F_w7$$\"35+++0luSXFbq$\"3nam7Gk7/cF_w7$$\"31+++MAKn\\ Fbq$\"3eN9+$pmG7&F_w7$$\"32+++N*Gh#eFbq$\"3YD(*[CJtnVF_w7$$\"3)****** \\jN\\o'Fbq$\"3T154b&Hm!QF_w7$$\"3U+++=bQ%R)Fbq$\"3cV`_=**\\JIF_w7$$\" 3e*****\\Qk#z**Fbq$\"3%3qU2]p,b#F_w7$$\"35+++l9.i6!#<$\"35G6IUTF!>#F_w 7$$\"3#*******=\"\\F_w7$$\"3.+++&[A4]\"F`z$\"3%* HDk!)Q*op\"F_w7$$\"3!******p3Q\\n\"F`z$\"3_gtw[$G=_\"F_w7$$\"33+++C6@G =F`z$\"3M?&y`mceR\"F_w7$$\"3&)******f-w+?F`z$\"3i,:'y%Q1y7F_w7$$\"3%** *******y,u@F`z$\"3yPP\"**>\"))z6F_w7$$\"3)*******RP)4M#F`z$\"3!H2z%fqc +6F_w7$$\"3-+++>Zg#\\#F`z$\"3&)GX#z/)QR5F_w7$$\"3#******ps'*Gn#F`z$\"3 +Ywtq6M\"y*!#;7$$\"3@+++2xiDGF`z$\"31D+mhk`[$*F`]l7$$\"3))*****f9!H.IF `z$\"3SVN0I/GM*)F`]l7$$\"35+++2:bgJF`z$\"3YScv>8PT')F`]l7$$\"3\")***** f9#4LLF`z$\"3i$>TCofXR)F`]l7$$\"39+++O;R(\\$F`z$\"30]d+$[u$G#)F`]l7$$ \"37+++<4#)oOF`z$\"3a:2[>JRB\")F`]l7$$\"3*******>^Yi#QF`z$\"3sj<^.#yg3 )F`]l7$$\"3)******RG^g*RF`z$\"3D#Qt\"HBn1\")F`]l7$$\"3G+++>2VsTF`z$\"3 %=@fs4o>>)F`]l7$$\"3n*****f`pfK%F`z$\"3@\"RCYpjmJ)F`]l7$$\"3k*****Rm&z \"\\%F`z$\"3:*4&QzIF,&)F`]l7$$\"3!)*****4G5Jm%F`z$\"3.cd]\\uKV()F`]l7$ $\"39+++6#32$[F`z$\"30'e#\\s#Ru-*F`]l7$$\"3M+++Ey'G*\\F`z$\"32h@\"3.kR M*F`]l7$$\"3E+++J%=H<&F`z$\"30ru'4'\\8S(*F`]l7$$\"3V+++3>qM`F`z$\"3e3b y^nN85F_w7$$\"3$******4ISu]&F`z$\"3YE_qka,f5F_w7$$\"3o******ep'Rm&F`z$ \"3=eX&3cWM5\"F_w7$$\"3&******4%>4NeF`z$\"3cm'4T\\U^:\"F_w7$$\"3;+++8s 5'*fF`z$\"3%=>u&RLg17F_w7$$\"3g*****fc9W;'F`z$\"3;Vd%p&=;j7F_w7$$\"3x* ****zwl*GjF`z$\"3R%)y!)fF1@8F_w7$$\"3r*****fiN7]'F`z$\"39\\qp\\nJ%Q\"F _w7$$\"3M+++v>:nmF`z$\"33&=9sG'oZ9F_w7$$\"3\"******\\SDo$oF`z$\"3,KG\" )z`)[^\"F_w7$$\"3)******H&Q40qF`z$\"3wbB!e\\UQe\"F_w7$$\"3))*****43:(f rF`z$\"3!\\H*o-==\\;F_w7$$\"3/+++e%GpL(F`z$\"3Qd`OH^LE%*GV+2^(=F_w7$$ \"3M+++F_w7$$\"\")\"\"!$\"3e\"\\qM#\\iN?F_w-%'CO LOURG6&%$RGBG$\"#5!\"\"$F_glF_glFigl-%+AXESLABELSG6$Q\"r6\"Q\"AF^hl-%% VIEWG6$;FiglF]gl;Figl$\"$+%F_gl" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "It looks as though the minimum value of " }{TEXT 287 1 "A " }{TEXT -1 18 " is obtained when " }{TEXT 286 1 "r" }{TEXT -1 1 " " } {TEXT 285 1 "~" }{TEXT -1 6 " 3.9. " }}{PARA 0 "" 0 "" {TEXT -1 22 "To find this value of " }{TEXT 292 1 "r" }{TEXT -1 52 " more precisely w e use the fact that the derivative " }{XPPEDIT 18 0 "dA/dr=0" "6#/*&%# dAG\"\"\"%#drG!\"\"\"\"!" }{TEXT -1 38 " at the minimum point on the g raph of " }{XPPEDIT 18 0 "A = Pi*r*sqrt(6561/(r^4)+r^2);" "6#/%\"AG*(% #PiG\"\"\"%\"rGF'-%%sqrtG6#,&*&\"%hlF'*$F(\"\"%!\"\"F'*$F(\"\"#F'F'" } {TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "A = Pi*r*(6561*r^(-4)+r^2)^(1/2);" "6#/%\"AG*(%#PiG\"\"\"%\"rGF'),&*&\"%h lF')F(,$\"\"%!\"\"F'F'*$F(\"\"#F'*&F'F'F2F0F'" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 4 "Then" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "dA/dr = Pi*(6561*r^(-4)+r^2)^(1/2)+Pi*r*``(1/2)*(6561*r ^(-4)+r^2)^(-1/2)*(-26244*r^(-5)+2*r);" "6#/*&%#dAG\"\"\"%#drG!\"\",&* &%#PiGF&),&*&\"%hlF&)%\"rG,$\"\"%F(F&F&*$F1\"\"#F&*&F&F&F5F(F&F&*,F+F& F1F&-%!G6#*&F&F&F5F(F&),&*&F/F&)F1,$F3F(F&F&*$F1F5F&,$*&F&F&F5F(F(F&,& *&\"&Wi#F&)F1,$\"\"&F(F&F(*&F5F&F1F&F&F&F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=Pi*sqrt(6561/(r^4)+r^2)+Pi*(r^2-18122/r^4)/sqrt(6561/(r^4)+r^2) " "6#/%!G,&*&%#PiG\"\"\"-%%sqrtG6#,&*&\"%hlF(*$%\"rG\"\"%!\"\"F(*$F0\" \"#F(F(F(*(F'F(,&*$F0F4F(*&\"&A\"=F(*$F0F1F2F2F(-F*6#,&*&F.F(*$F0F1F2F (*$F0F4F(F2F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=Pi*(6561/(r^4)+r^2 +r^2-18122/(r^4))/sqrt(6561/(r^4)+r^2)" "6#/%!G*(%#PiG\"\"\",**&\"%hlF '*$%\"rG\"\"%!\"\"F'*$F,\"\"#F'*$F,F0F'*&\"&A\"=F'*$F,F-F.F.F'-%%sqrtG 6#,&*&F*F'*$F,F-F.F'*$F,F0F'F." }{XPPEDIT 18 0 "`` = Pi*(2*r^2-6561/(r ^4))/sqrt(6561/(r^4)+r^2);" "6#/%!G*&,$*&%#PiG\"\"\",&*&\"\"#F)*$)%\"r GF,F)F)F)*&\"%hlF)*$)F/\"\"%F)!\"\"F5F)F)F)-%%sqrtG6#,&*&F1F)*$%\"rGF4 F5F)*$F\"F\\p$!+S RUk&*Fgo-%&STYLEG6#%,PATCHNOGRIDGF0-%*THICKNESSG6#\"\"#-F$6'7$7$FAFeo7 $$F[qF4Fao7%7$$\"+C.*p!=F\\p$!*1wbV$F\\pF`q7$$\"+w'4I*=F\\p$!*%RUkYF\\ pFdpF0Fhp-%%TEXTG6&7$$FboF4$\"#vF`oQ\"r6\"F0-%%FONTG6$%*HELVETICAG\"#5 -F^r6&7$$!#:FcoFbrQ-~0~<~r~<~r~*FerF0Ffr-F^r6&7$$F4F4FbrQ$r~*FerF0Ffr- F^r6&7$$\"#:FcoFbrQ(r~>~r~*FerF0Ffr-F^r6&7$$\"\"$F4FbrQ\"3FerF0Ffr-F^r 6&7$Far$F`tFcoQ#dAFerF0Ffr-F^r6&7$FarFdsQ#drFerF0Ffr-F^r6&7$Far$\"#HF` oQ#__FerF0Ffr-F^r6&7$Fds$F[qFcoQ\"0FerF0Ffr-F^r6&7$F^s$\"\"%FcoQ\"_Fer F0-Fgr6$Fir\"#9-F^r6&7$FisFetQ\"+FerF0F\\v-%*AXESSTYLEG6#%%NONEG-%+AXE SLABELSG6$Q!FerFjv-%%VIEWG6$;$!#NFco$\"#DFco;$!#8Fco$\"#6Fco" 1 2 0 1 10 0 2 9 1 1 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "C urve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "C urve 10" "Curve 11" "Curve 12" "Curve 13" "Curve 14" "Curve 15" "Curve 16" "Curve 17" "Curve 18" "Curve 19" "Curve 20" "Curve 21" }}{TEXT -1 1 " " }}{PARA 257 "" 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 50 "We can use Maple to obtain the critical value of " }{TEXT 288 1 "r" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 227 "r := 'r':\ng := r -> Pi*r*sqrt(81^2/(r^4 )+r^2):\n'g(r)'=g(r);\ndA/dx=diff(rhs(%),r);\n``=simplify(rhs(%));\nrh s(%)=0;\nsolve(%):\nop(remove(has,[%],Complex(1))):\nr[min] := op(remo ve(_U->evalb(signum(_U)=-1),[%]));\nevalf(evalf[14](%));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/-%\"gG6#%\"rG*(%#PiG\"\"\"F'F*,&*&\"%hlF*F'!\"% F**$)F'\"\"#F*F*#F*F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%#dAG\"\" \"%#dxG!\"\",&*&%#PiGF&,&*&\"%hlF&%\"rG!\"%F&*$)F/\"\"#F&F&#F&F3F&*,F3 F(F+F&F/F&F,#F(F3,&*&\"&Wi#F&F/!\"&F(*&F3F&F/F&F&F&F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G**%#PiG\"\"\",&\"%hl!\"\"*&\"\"#F')%\"rG\"\"'F 'F'F'F.!\"%*&,&F)F'*$F-F'F'F'F.F0#F*F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/**%#PiG\"\"\",&\"%hl!\"\"*&\"\"#F&)%\"rG\"\"'F&F&F&F-!\"%*&,&F( F&*$F,F&F&F&F-F/#F)F+\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"rG6 #%$minG,$**\"\"$\"\"\"\"\"#!\"\"F*#F+F*F,#\"\"&\"\"'F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+!)[paQ!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 28 "The corresponding value of " }{TEXT 289 1 "A" }{TEXT -1 38 " can then be obtained by substitution." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "g(r[min]);\nevalf(evalf[14](%));\nr := 'r':" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*0\"\"*\"\"\"\"\"%!\"\"%#PiGF&\"\"$#F&F*\"\"##\"\"&\" \"'F/#F&F,*&)F*#F,F*F&)F,F3F&F0F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$ \"+&G&>&3)!\")" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "The area of the cone is a minimum when the radius of the base is approximately 3.854694880 metr es. " }}{PARA 0 "" 0 "" {TEXT -1 61 "The minimum area is approximately 80.85195285 square metres. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 71 "Example 2 - maximising the volume of a cylinder inscrib ed in a sphere " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {PARA 0 "" 0 "" {TEXT 263 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 78 "A right circular cylinder has to be designed to fit \+ inside a sphere of radius " }{XPPEDIT 18 0 "3*sqrt(2)" "6#*&\"\"$\"\" \"-%%sqrtG6#\"\"#F%" }{TEXT -1 82 " cm. so that both the top and botto m touch the sphere all around the circular rim." }}{PARA 0 "" 0 "" {TEXT -1 111 "Find the radius and height of the cylinder if it is to h ave maximum volume, and determine this maximum volume. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 264 8 "Solution" }{TEXT -1 3 ": " }}{PARA 0 "" 0 "" {TEXT -1 34 "Let the radius of the cylind er be " }{TEXT 265 1 "r" }{TEXT -1 24 " cm., let the height be " } {TEXT 266 1 "h" }{TEXT -1 38 " cm., let the radius of the sphere be " }{TEXT 304 1 "R" }{TEXT -1 27 " cm.,and let the volume be " }{TEXT 303 1 "V" }{TEXT -1 15 " cubic metres. 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" }}{PARA 0 "" 0 "" {TEXT -1 22 "To find this value of " }{TEXT 274 1 "r" }{TEXT -1 52 " more precisely we use the fact t hat the derivative " }{XPPEDIT 18 0 "dV/dr = 0;" "6#/*&%#dVG\"\"\"%#dr G!\"\"\"\"!" }{TEXT -1 39 " at the maximum point on the graph of " } {XPPEDIT 18 0 "V = 2*Pi*r^2*sqrt(18-r^2)" "6#/%\"VG**\"\"#\"\"\"%#PiGF '%\"rGF&-%%sqrtG6#,&\"#=F'*$F)F&!\"\"F'" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "V = 2*Pi*r^2*(18-r^2)^(1/2)" "6 #/%\"VG**\"\"#\"\"\"%#PiGF'%\"rGF&),&\"#=F'*$F)F&!\"\"*&F'F'F&F.F'" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dV/dr = 4*Pi*r*(18-r^2)^(1/2)+2*Pi*r ^2*``(1/2)*(18-r^2)^(-1/2)*(-2*r);" "6#/*&%#dVG\"\"\"%#drG!\"\",&**\" \"%F&%#PiGF&%\"rGF&),&\"#=F&*$F-\"\"#F(*&F&F&F2F(F&F&*.F2F&F,F&F-F2-%! G6#*&F&F&F2F(F&),&F0F&*$F-F2F(,$*&F&F&F2F(F(F&,$*&F2F&F-F&F(F&F&" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = 4*Pi*r*sqrt(18-r^2)-2*Pi*r^3/sqrt( 18-r^2);" "6#/%!G,&**\"\"%\"\"\"%#PiGF(%\"rGF(-%%sqrtG6#,&\"#=F(*$F*\" \"#!\"\"F(F(**F1F(F)F(F*\"\"$-F,6#,&F/F(*$F*F1F2F2F2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "`` = (4*Pi*r*(18-r^2)-2*Pi*r^3)/sqrt(18-r^2);" "6#/%!G* &,&**\"\"%\"\"\"%#PiGF)%\"rGF),&\"#=F)*$F+\"\"#!\"\"F)F)*(F/F)F*F)F+\" \"$F0F)-%%sqrtG6#,&F-F)*$F+F/F0F0" }{XPPEDIT 18 0 "`` = (72*Pi*r-6*Pi* r^3)/sqrt(18-r^2);" "6#/%!G*&,&*(\"#s\"\"\"%#PiGF)%\"rGF)F)*(\"\"'F)F* F)F+\"\"$!\"\"F)-%%sqrtG6#,&\"#=F)*$F+\"\"#F/F/" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 2 " \+ " }{XPPEDIT 18 0 "dV/dr=6*Pi*r*(12-r^2)/sqrt(18-r^2)" "6#/*&%#dVG\"\" \"%#drG!\"\"*,\"\"'F&%#PiGF&%\"rGF&,&\"#7F&*$F,\"\"#F(F&-%%sqrtG6#,&\" #=F&*$F,F0F(F(" }{TEXT -1 15 " ------- (v). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "Now " }{XPPEDIT 18 0 "dA/d r=0" "6#/*&%#dAG\"\"\"%#drG!\"\"\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "r*(12-r^2)=0" "6#/*&%\"rG\"\"\",&\"#7F&*$F%\"\"#!\"\"F&\"\"!" } {TEXT -1 14 ", which gives " }{XPPEDIT 18 0 "r=-2*sqrt(3),0" "6$/%\"rG ,$*&\"\"#\"\"\"-%%sqrtG6#\"\"$F(!\"\"\"\"!" }{TEXT -1 4 " or " } {XPPEDIT 18 0 "2*sqrt(3)" "6#*&\"\"#\"\"\"-%%sqrtG6#\"\"$F%" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 23 "The negative value for " } {TEXT 275 1 "r" }{TEXT -1 36 " is inapplicable to the problem and " } {XPPEDIT 18 0 "V=0" "6#/%\"VG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "r=0" "6#/%\"rG\"\"!" }{TEXT -1 19 ", so this value of " }{TEXT 276 1 "r" }{TEXT -1 45 " certainly does not give a maximum value for " } {TEXT 305 1 "V" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 10 "Consid er " }{XPPEDIT 18 0 "r^`*` = 2*sqrt(3);" "6#/)%\"rG%\"*G*&\"\"#\"\"\" -%%sqrtG6#\"\"$F)" }{TEXT -1 1 " " }{TEXT 306 1 "~" }{TEXT -1 14 " 3.4 64101615. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "By investigating the sign of the derivative we can verify that \+ this value of " }{TEXT 278 1 "r" }{TEXT -1 38 " gives a maximum value \+ for the volume " }{TEXT 277 1 "V" }{TEXT -1 2 ". " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{GLPLOT2D 395 122 122 {PLOTDATA 2 "6=-%'CURVESG6$7$7 $$!3++++++++N!#<$!3+++++++]7F*7$$\"3++++++++bF*F+-%'COLOURG6&%$RGBG\" \"!F4F4-F$6$7$7$F($\"3++++++++]!#=7$F.F9F0-F$6$7$7$F($\"\"\"F47$F.FAF0 -F$6$7$F'F@F0-F$6$7$7$$!3++++++++DF*F+7$FKFAF0-F$6$7$7$$!3++++++++]F;F +7$FRFAF0-F$6$7$7$F9F+7$F9FAF0-F$6$7$7$$\"3++++++++DF*F+7$FhnFAF0-F$6' 7$7$$!\"#F4$!\"\"F47$Fao$!\"$Fbo7%7$$!+w'4I>\"!\"*$!*1wbV$FjoFco7$$!+C .*p5\"Fjo$!*%RUkYFjo-%&STYLEG6#%,PATCHNOGRIDGF0-%*THICKNESSG6#\"\"#-F$ 6'7$7$FAFdo7$$FipF4$!#5Fbo7%7$$\"+w'4I*=Fjo$!+ggdN$)FaqF^q7$$\"+C.*p!= Fjo$!+SRUk&*FaqFbpF0Ffp-%%TEXTG6&7$$FeoF4$\"#vF`oQ\"r6\"F0-%%FONTG6$%* HELVETICAG\"#5-F^r6&7$$!#:FboFbrQ,0~<~r~<~r~*FerF0Ffr-F^r6&7$$F4F4FbrQ $r~*FerF0Ffr-F^r6&7$$\"#:FboFbrQ/r~*~<~r~<~~/18FerF0Ffr-F^r6&7$$\"\"$F 4FbrQ\"3FerF0Ffr-F^r6&7$Far$F`tFboQ#dVFerF0Ffr-F^r6&7$FarFdsQ#drFerF0F fr-F^r6&7$Far$\"#HF`oQ#__FerF0Ffr-F^r6&7$Fds$FipFboQ\"0FerF0Ffr-F^r6&7 $F_tFduFeuF0Ffr-F^r6&7$$\"$z\"F`o$\"\"(FboQ\"vFerF0-Fgr6$Fir\"\")-F^r6 &7$F^sFetQ\"+FerF0-Fgr6$Fir\"#9-F^r6&7$Fis$\"\"%FboQ\"_FerF0Fhv-F^r6&7 $$\"#XFboFduFgvF0Fhv-%*AXESSTYLEG6#%%NONEG-%+AXESLABELSG6$Q!FerF]x-%%V IEWG6$;$!#NFbo$\"#DFbo;$!#8Fbo$\"#6Fbo" 1 2 0 1 10 0 2 9 1 1 2 1.000000 46.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+ 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" "Curve 12" "Curve 13" "Curve 14" "Curve 15" "Curve 16" "Curve 17 " "Curve 18" "Curve 19" "Curve 20" "Curve 21" "Curve 22" "Curve 23" "C urve 24" }}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 59 "The volume of the cylinder is a maximum when its r adius is " }{XPPEDIT 18 0 "2*sqrt(3)" "6#*&\"\"#\"\"\"-%%sqrtG6#\"\"$F %" }{TEXT -1 1 " " }{TEXT 309 1 "~" }{TEXT -1 17 " 3.464101615 cm. " } }{PARA 0 "" 0 "" {TEXT -1 57 "The corresponding value of the height of the cylinder is " }{XPPEDIT 18 0 "2*sqrt(6);" "6#*&\"\"#\"\"\"-%%sqrt G6#\"\"'F%" }{TEXT -1 1 " " }{TEXT 307 1 "~" }{TEXT -1 18 " 4.89897948 6 cm., " }}{PARA 0 "" 0 "" {TEXT -1 42 "and the maximum volume of the \+ cylinder is " }{XPPEDIT 18 0 "24*Pi*sqrt(6)" "6#*(\"#C\"\"\"%#PiGF%-%% sqrtG6#\"\"'F%" }{TEXT -1 1 " " }{TEXT 308 1 "~" }{TEXT -1 20 " 184.68 71755 cu.cm. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "We can use Maple to obtain the critical value of " } {TEXT 270 1 "r" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 223 "r := 'r':\ng := r -> 2*Pi*r ^2*sqrt(18-r^2):\n'g(r)'=g(r);\ndA/dx=diff(rhs(%),r);\n``=simplify(rhs (%));\nrhs(%)=0;\nsolve(%);\nop(remove(has,[%],Complex(1))):\nr[min] : = op(remove(_U->evalb(signum(_U)<=0),[%]));\nevalf(evalf[14](%));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"gG6#%\"rG,$**\"\"#\"\"\"%#PiGF+)F 'F*F+,&\"#=F+*$F-F+!\"\"#F+F*F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*& %#dAG\"\"\"%#dxG!\"\",&**\"\"%F&%#PiGF&%\"rGF&,&\"#=F&*$)F-\"\"#F&F(#F &F2F&**F2F&F,F&F-\"\"$F.#F(F2F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%! G,$*,\"\"'\"\"\"%#PiGF(%\"rGF(,&\"#7!\"\"*$)F*\"\"#F(F(F(,&\"#=F(F.F-# F-F0F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*,\"\"'\"\"\"%#PiGF'%\"rG F',&\"#7!\"\"*$)F)\"\"#F'F'F',&\"#=F'F-F,#F,F/F,\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"!,$*&\"\"#\"\"\"\"\"$#F'F&F',$*&F&F'F(F)!\"\" " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"rG6#%$minG,$*&\"\"#\"\"\"\"\" $#F+F*F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+:;5kM!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "The corresponding values of " }{TEXT 310 1 "h" }{TEXT -1 5 " and " }{TEXT 271 1 "V" } {TEXT -1 38 " can then be obtained by substitution." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Eval(2*sqr t(18-r^2),r=2*sqrt(3));\n``=value(%);\n``=evalf(evalf[14](rhs(%)));" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#-%%EvalG6$,$*&\"\"#\"\"\",&\"#=F)*$)% \"rGF(F)!\"\"#F)F(F)/F.,$*&F(F)\"\"$F0F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G,$*&\"\"#\"\"\"\"\"'#F(F'F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G$\"+'[z*)*[!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "Eval(2*Pi*r^2*sqrt(18-r^2),r=2*sqrt (3));\n``=value(%);\n``=evalf(evalf[14](rhs(%)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%%EvalG6$,$**\"\"#\"\"\"%#PiGF))%\"rGF(F),&\"#=F)*$F+F )!\"\"#F)F(F)/F,,$*&F(F)\"\"$F1F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/ %!G,$*(\"#C\"\"\"%#PiGF(\"\"'#F(\"\"#F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G$\"+b<(o%=!\"(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 6 "Tasks " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q1 " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "_______________________________ ______ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 38 "____________________ _________________ " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 17 "Code for pi ctures" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 31 "pictures of cone and sector. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1532 " p1 := plot([cos(t),0.3*sin(t),t=-Pi..0],color=COLOR(RGB,.6,.4,.3)):\np 2 := plot([cos(t),0.3*sin(t),t=0..Pi],\n color=COLO R(RGB,.6,.4,.3),linestyle=2):\np3 := plot([[[1,0],[0,2]],[[-1,0],[0,2] ]],color=COLOR(RGB,.6,.4,.3)):\na := evalf(Pi/60):\nc1 := seq([cos(j*a ),0.3*sin(j*a)],j=-12..0):\nc2 := seq([cos(j*a),0.3*sin(j*a)],j=-27..- 12):\nc3 := seq([cos(j*a),0.3*sin(j*a)],j=-33..-27):\nc4 := seq([cos(j *a),0.3*sin(j*a)],j=-48..-33):\nc5 := seq([cos(j*a),0.3*sin(j*a)],j=-6 0..-48):\np4 := plots[polygonplot]([c1,[0,2]],\n style=patchnogrid, color=COLOR(RGB,1,.95,.78)):\np5 := plots[polygonplot]([c2,[0,2]],\n \+ style=patchnogrid,color=COLOR(RGB,1,.95,.81)):\np6 := plots[polygonp lot]([c3,[0,2]],\n style=patchnogrid,color=COLOR(RGB,1,.953,.83)): \np7 := plots[polygonplot]([c4,[0,2]],\n style=patchnogrid,color=CO LOR(RGB,1,.95,.81)):\np8 := plots[polygonplot]([c5,[0,2]],\n style= patchnogrid,color=COLOR(RGB,1,.95,.78)):\nc11 := seq([cos(2*j*a),0.3*s in(2*j*a)+2],j=0..60):\np9 := plot([[[cos(-40*a),0.3*sin(-40*a)],[0,2] ],\n [[1,0],[0,0],[0,2]]],color=COLOR(RGB,.6,.4,.3),linestyle=2): \np10 := plottools[arrow]([.6,0],[1,0],\n 0,.07,.15,arrow,col or=COLOR(RGB,.75,.35,.5)):\np11 := plot([[0,0],[.4,0]],color=COLOR(RGB ,.75,.35,.5)):\np12 := plot([[[0,0]]$3],style=point,\n symbol=[cro ss,diamond,circle],color=COLOR(RGB,.75,.35,.5)):\nt1 := plots[textplot ]([[.6,1.,`R`],[.5,.1,`r`],[.08,.9,`h`]],\n \+ font=[HELVETICA,9],color=black):\nplots[display]([p1,p2,p3,p4,p5,p6, p7,p8,p9,t1],axes=none);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 527 "a := evalf(Pi/60):\nc1 := seq([cos (j*a),sin(j*a)],j=-4..50):\np1 := plot([[0,0],c1,[0,0]],color=COLOR(RG B,.6,.4,.3)):\np2 := plots[polygonplot]([[0,0],c1,[0,0]],\n style=p atchnogrid,color=COLOR(RGB,1,.953,.78)):\nc2 := seq([.21*cos(2*j*a),.2 1*sin(2*j*a)],j=-2..25):\np3 := plot([c2],color=COLOR(RGB,.6,.4,.3)): \nt1 := plots[textplot]([.03,.09,`q`],font=[SYMBOL,10],color=black):\n t2 := plots[textplot]([[-.43,.16,`R`],[.45,-.18,`R`]],font=[HELVETICA, 9],color=black):\nplots[display]([p1,p2,p3,t1,t2],axes=none,scaling=un constrained);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 860 "rmin := 'rmin':\np1 := plot([[[-3.5,-1.25],[5.5 ,-1.25]],[[-3.5,.5],[5.5,.5]],\n [[-3.5,1],[5.5,1]],[[-3.5,-1.25], [-3.5,1]],\n [[-2.5,-1.25],[-2.5,1]],[[-.5,-1.25],[-.5,1]],\n \+ [[.5,-1.25],[.5,1]],[[2.5,-1.25],[2.5,1]]],color=black):\np2 := plotto ols[arrow]([-2,-.3],[-1,-1],\n 0,.15,.15,arrow,color=black,thic kness=2):\np3 := plottools[arrow]([1,-1],[2,-.3],\n 0,.15,.15,a rrow,color=black,thickness=2):\nt1 := plots[textplot]([[-3,.75,`r`],[- 1.5,.75,` 0 < r < r *`],\n [0,.75,`r *`],[1.5,.75,`r > r *`],[3,.75 ,`3`]],color=black,font=[HELVETICA,10]):\nt2 := plots[textplot]([[-3,. 3,`dA`],[-3,.0,`dr`],[-3,.29,`__`],[0,.2,`0`]],\n color=black,f ont=[HELVETICA,10]):\nt3 := plots[textplot]([[-1.5,.4,`_`],[1.5,.3,`+` ]],\n color=black,font=[HELVETICA,14]):\nplots[display]([p1,p2, p3,t1,t2,t3],axes=none,\n view=[-3.5..2.5,-1.3..1.1]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 32 "pictures for cylinder in sphe re " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "h := 'h':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 592 "a := evalf(sqrt(3)/2): b := .5:\np1 := plot([cos(t),sin(t),t=0..2 *Pi],color=black):\np2 := plot([[[a,-b],[a,b]],[[-a,-b],[-a,b]]],lines tyle=2,color=black):\np3 := plot([[a*cos(t),-b+.2*a*sin(t),t=0..2*Pi], \n [a*cos(t),b+.2*a*sin(t),t=0..2*Pi]],linestyle=2,color=black):\np 4 := plot([[[0,-b],[0,b],[a,b],[0,0]]],linestyle=2,color=black):\nt1 : = plots[textplot]([[.4,.58,`r`],[.43,.17,`R`],[-.1,.26,`h`],\n [-. 1,.25,`_`],[-.1,-.13,`h`],[-.1,-.14,`_`]],font=[HELVETICA,10]):\nt2 := plots[textplot]([[-.1,.13,`2`],[-.1,-.26,`2`]],font=[HELVETICA,9]):\n plots[display]([p1,p2,p3,p4,t1,t2],axes=none);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 946 "p1 := plot( [[[-3.5,-1.25],[5.5,-1.25]],[[-3.5,.5],[5.5,.5]],\n [[-3.5,1],[5.5 ,1]],[[-3.5,-1.25],[-3.5,1]],\n [[-2.5,-1.25],[-2.5,1]],[[-.5,-1.2 5],[-.5,1]],\n [[.5,-1.25],[.5,1]],[[2.5,-1.25],[2.5,1]]],color=bl ack):\np2 := plottools[arrow]([-2,-1],[-1,-.3],\n 0,.15,.15,arr ow,color=black,thickness=2):\np3 := plottools[arrow]([1,-.3],[2,-1],\n 0,.15,.15,arrow,color=black,thickness=2):\nt1 := plots[textplo t]([[-3,.75,`r`],[-1.5,.75,`0 < r < r *`],\n [0,.75,`r *`],[1.5,.75 ,`r * < r < /18`],[3,.75,`3`]],color=black,font=[HELVETICA,10]):\nt2 \+ := plots[textplot]([[-3,.3,`dV`],[-3,.0,`dr`],[-3,.29,`__`],[0,.2,`0`] ,[3,.2,`0`]],\n color=black,font=[HELVETICA,10]):\nt3 := plots[ textplot]([1.79,.7,`v`],color=black,font=[HELVETICA,8]):\nt4 := plots[ textplot]([[-1.5,.3,`+`],[1.5,.4,`_`],[4.5,.2,`+`]],\n color=bl ack,font=[HELVETICA,14]):\nplots[display]([p1,p2,p3,t1,t2,t3,t4],axes= none,\n view=[-3.5..2.5,-1.3..1.1]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "6 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }