00010001010008main.ACT0002020012eActivity Save.EAC010000001b9fÉ slopeoftangent.EAC†Žline.ACT†)†,†2ˆ5†9ŒŒŠ" ~Å"[ˆ+Sˆ] of the TŒc ŽL†^ at a point Œ5\Œ= Authoríˆ^ˆlí’r[†ŒŒd ObjectiveˆX†ÒWe shall investigateˆ‚†aŠ relationship betweenˆ¢ ˆT‡‹sˆ¹secant ‰ s ŽAandŠÒ‹<ß’KŠ4–ß.ŽŠg‰| Example.ºG†ÇŠ‹ graph‘:¹*function below, find the slope ofˆ tangent Œline atˆ1 point Ž7B=(-0.1,0.79) byŽ3using various approaches.\Œp¨pΈŒ†Îˆ–ˆ ˆž `– – –<ˆÛvŽˆr† Ž A† 'LŽ'‰C˜ u˜x˜yŽ ˆ­ˆqˆ¶5ÆH  ˜ºÄH!ƒŠC@‡nÐ6–°†’В޲‹†”¾ ‰œ ÊÈ  † †  -(x+-1)^2+2” `†Œ0 Y™"iI%†H ˜††QÄ@–Q††g ÊÈ!Œ@†vÐ6–°†} •q!†@Y˜” ™–—RŠd†3 †©ÆH Š©B–Cˆ=  ™˜މ †G†G”ŸˆYˆŸ¶ˆŽ"ˆ¯ˆ. ‡.ȘÍ †3ÈÀ–ŽD †ã•Jšã6‚H…A‹/yG!xŠã“€ru ¢nˆ\ˆr¢ˆˆrˆŒ  ††† ކ1Ä@  † † ††9– Žˆˆ7†Ž2<ˆ?[†pMethod 1. Animation. ŽThe lengths BC, Œ4 slope ofˆ and Ot†2’ˆ Ž4tangent line atˆgpoi† B is capturedˆ8‡ below: ’±Part 1Žderiv†¶ve fromŒs†+‡Kleft:ŽI column 1=ˆ£xˆ¾CŠ·‡uŽ2=Œî†íŽ7Ž3=ŒõBˆ5<Œ¡-2Œª 8.01836† 14.Ž -1.967798305Œ 7.824042579Œ/ 4.0ž$ˆ7355966ŽS†66317652 4.0š#Šl0339491”l 441527443”lž$†£87119322’¢2533283H 3.9š#ˆØ8389†j2”؆Æ1670’l†6œ$ˆ£8†ì8‡, 6.883‡768”7œ$‰F 774588136’770095460¶3.8œ$‰}742386‡Œ7 6.5209016‘3.842386441 -1.710184746Œ 6.342882907Œ% †7œ$ˆ767798305ŽJ†716689745I3.7œ$ˆn6457813’% 5.992944273’n7œ$ˆ¥6†37966ޏ†7 821022315’¥7œ$ˆÜ5813†¨6Žn 5.6511–¶3.6œ#ˆ¤54917627”m4832†ñ“$3.6œ$ˆ¤516ˆÿ7”Û31743287•I6œ$ˆ¤48477‡q”Û 153624718¥ 3.584772881 -1.452571186Œ 4.99184204Œ$ †6œ#Š620369492Œ6 4.832083577”7œ$ˆm3†{6779% 4.67434801ކn4œ$ˆ¤35596610Žn4.5†±3”¢4œ!ˆ¢32376440”k36494012´3.4œ$‰29156271”Ù21326†ê8“3œ$ˆ¢2593†¡1”n06360691k†¥3œ$ˆ¥2†j593–“91†ë4509’¥3”$ -1.194957627ˆ 3.770336105’2œ$Š7†255932’762672002”2œ$Šn3055423”n 485114521’€2œ$ˆ¥098352†9Œ% 3.3455178’£1œ"ˆ£06615084”Ú20792798”ì1œ$ˆ£†ý‡153“07234314”£1œ$‰H00174745“ 2.93876124–¥œ$ -0.96954™€ 2.807180214’¦0Ž%Ž“ -0.9373440678 2.677597867Ž3.0Ž%Ž$ ˆ8 051423729’855001194†8Ž%3Œ8 †p8†2œo42442†91’9œ$ˆo 840738983”& 300819915’¹9%”§ 808537288”^17920882–p%”ß 7763355932Œ& 2.0595841”ð8$”§74413–ƒ 1.9†34327ŽI†§8%•N7‡8322034’882628320•`8%Œ -0.6797305085 1.712601018Ž2.7%Œ$ ˆ8 475288136’8 600893597”8Ž%4–815327118Ž& 1.49115771”oŽ$9”o 583125423] 1.38338998Ž]†§6Ž%–o5509†5†ªŒ8 1.277586Ž\†Ý6Ž#˜n18722033Ž‚†¥1737447–7Ž$Žm †¥4865–6‡K07185982”Ü5œ$ˆ¤ 454318644 0.971927993•5&¨ -0.4221169492 0.8739451291Œ 2.5&Œ% †9 389915254”9 7779068626’94&–9 577135593’r68380†89–9&–r25511864ŽM†« 5916456575’«4&”«293310†æ&0.5014†â5Ž„ 2.3Œ%7”â 261108474ŽÏ ‡.1†I52‡3Ž$Ž€ˆ§228‡779ŽK‡R 326717196Ž^ˆoŒ&8”¦ˆ05084އŠ2‡ž42941ާ‡Š2Ž&ˆñ -0.1645033898† † 596765176Œ 2.2Œ&9Œ$Š832301694 0.0790116285ŽL †9Ž'5Œ^*ˆs001Œ*2.416700428î-4Ž„†2†%0000Ž„ ˆ  Part 2. \Œ¯derivative from the rightΈÐ†çÎ†í† Šã† ††`– – –˜<vއ?r† Ž A† ŽcLŽ'‰SC˜ u˜x˜yŽ ˆ­ˆq‡£5ÆH  Œ‡µÄH # C@†Ð6–°† ˆ †„” Ž †9ÊÈ  / -(x+-1)^2+2˜EP†Q ™] Y˜U‘RT#rŠ ††ŠÄ@–Q††  ÈÀ–©sF#D64†Á Y™# Q‘† ™S”QddІ3†XÄH Š©AšìëY™  ™˜މ¢ŸˆYˆŸ ‡FŠ"ˆ¯ˆ.†·.Ș.†3”ÉBœBà†ÈÀ  † @†Ð6–°† 6‚H…A†% Y™yG!x† ™“€ruІ3†B.È@ ŒB†††¢ŽŽzˆ"ˆ@ˆˆ†’1Ä 2ˆP9Ž ˆ§œ"–"ˆ!†ÖÊÈ– y=-(x+-1)^2+2Ž3Ž,ˆ4ˆDˆ^ˆ„†ŽË<ˆ%Ø[Šãcolumn 1=the x of C ŽŽ 2=length BC ŒŽ3=slopeˆ<ŒW2Œ 2.11047388Œp0.1Œz 1.964408475 2.083299182Š 0.1355915254Œ 1.928816949Œ%ˆ75†88171”7 711830508’78932†G2I†n3539034”n 2067745763’n8ˆ389I†¥14310467’¥ 242366101ކ¥82204237”I9†²1003”¥2779†l2¸†Ü7864†µ4”I97700865•3‡4‹’¥7†é5ˆà•%604‡57 0.34914†à•7‡X†ð9”¶94498›3†Œ32203•€6796˜¶1.930495201 0.4203237288Œ 1.644084746’ 916794934”7 559152542Œ% 1.6†5932”90370839”l 915067797’l 572901695’~89106913Ž$†£ 527098305޶†£53731ˆ8’I 878716103’Ú56268Š8’Ú501718†à’ì8664940I†¥5982813†à’%46612711–n†ù5405Žþ‡H6338‰F1”\4305ˆHŽ“‡Z84185354l†¥6694‰p6•3‡jœ 829156692 0.7050559322Œ 1.3†5254” 816034309”7 406474576”723761017’I 802363898”n762ˆ31’n28†^949”€78†99543’¥81183†©85’¥2†¢7796”€77292177–7†¢22033ޏ†Ê21698644”€75693738‡883013‰’“1813†¤1Ž€ 1.7399792ŽÈ†¤9186‡L84Žî†$1458š~7219‡m04“†ç1†Å10•m1102†ð6Ž%‡H7027‡ƒ6‘0.9897881356 1.074620339’ 682375973”$25379661”63902†J4’H66066333–$6097118–l03437†8’H6†j72†•”96562712Œ¢ †µ 678457627’µ61303862”I132154†¹”7322Š”È58699820”€1677ˆ\Žì†n8†èˆ€1•55939359”·‰1–Ɇ¥8610Ší 1.530170—Y2389™50.8254Y49927764•2745¡ 0.7898881356ˆ 1.466667693’ 310111864Œ$ˆ7 542966102”73229587ŽI 1.34570339”6 187050847”$96120$†~ 381294915’l6831†¡5˜‘5810207Ž‘ˆµ168†¡41’£64752†[˜\18204238”µ52477†Ê’¤61193ˆ¥Ž‡27639249–ì8806†¦0.57633‡L•52326‡9•Y523‰31Žþ†¤54†Ä7‡'•~‡g90030•†ú252‡q¢ 0.5051559322 1.139161254’ 594844068Œ$ 0.46956–†7 089390793’I63043ˆ]”733972881”%03756390”I 666027119’n3983†4†¨’†657090”¸70†·8†–’¦36278†*05’¹ 927648238Ž‚ 1.73721017’$3†~Š61’J8†ð16†óއ'7†É†r9”\2‡806779”K 809242503ŽÍ†o8083•p‡L256015‡p”7468076†ã“–843984†„0.2204237288 0.6821946073Œ 1.879576271’%18483†=34”8153870†)’8 915167797Œ% 0.14924067”o 5463695589’o950759†_’\11364†N25’§ 475127552Ž‚†§98635084Žo 0.078ˆ‘†º”» 401647291Žó 2.0ˆò‡Žòˆ9424661†+Ž—‡325†Ñ791á†95753389>6‡+4‰+64î-ŽO†å 247920777/ˆt931254¿ -0.02871694’b‡Ž‡@650641 †®1'-0.06430847458ˆ † 8509439906Œ 2.1Ž(5Œ&*ˆ<9†#000001Œ*2.416517994î-4Œ†1999Œa  ŒrBy inspection, when the Ž)distance BC gets closerIto 0,ˆ3 slope of†*Ž^–1 †)’&’_†/†o †c gent line atŽž †Š poi†B.˜»‰.Method 2.Numerically,R‰]NDefˆOf(í¸)=-† -1)^2+2RŽ$done[†Z‰€ We let x=‡1ŽgŠR7 Define Q(h)=f(-0.1+h)-Œ )ŒhRŒ%doneRˆ/Œ)dŒIF(x)=listToMat(Q(ma† L†(x,1))) ¨EŒ{ F(Ž0.001Œ ˆ ˆ’Œ’*†¿’ Š Š’ Žd)Šãšƒ2.199Œ Š Ž ŠŽŽ†62’¹ˆ †62”dž6ˆ01—J[‡š!To find the slope ofˆ  tangent line atŠ)point Œ!(-0.1,0.79), we doR†9ŒAf(x)ŒLxŒT|x=ˆ5R†)ˆg2.2[ŽsOrˆHcan use˜ following:Ž~ŠlŒ«diff(ˆo,x,1,ˆ—)¢cŽ×à*ˆúabov‡ yntax meansŽÞtaking derivati†&‡ f ‘with respect to x,Љ<or†7 1‰6Œñ or6‹)‹M’Ôs:ˆÍde‡˜e f1(a)=œÚa’× doneR† † f1(a)RŒ-2î’Ž aŽ-Œ01ˆ=Œ=†=-0.1Œ@ŒQ2.2–`[ŽY  †o Exercise 1.Redo the problem by 9 changingŒ! oint B to†P†Ûo†8rŒs†R˜xœp2.Ž3 ReconstructŠvcurve©†|usˆpŒZ function,ʈ©seca†‰line andˆ½Ž‘ animaˆ2.eAct020011implicit diff.EAC010000001d9cÉø^„œ­[Implicit Differentiations[ -step 1. 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