J.)=,dd +(I)  1Courier New1Symbol New1Courier New1Courier NewTimes New RomanTimes New Roman1Courier New1Courier New1Courier New1Courier NewP`NSP`NSP`NSP`NS?/ <```@@@?~~~~~ԱޱޱԱ޿޿ph (((Pdd{ d< p {2P7z?(D@1AVAp@5tkү@6A_p???XN@ > @Ia@ׁsF0*Dzv^)@ѽoΎ@<`5Β@033333??@^ߛOw?HzG???rhsH@2@@?@ @@@?@@?@?@?@xClʡE𘽃aG@8V"K@X@@02??@$$$??jP-?0|{b*@0Ր@ %c@@???dYL4Њ ~ s p:u Dzd T_h (c 0r Qs Nj 2e^t2te^tt^2-t+3!x = 2e^t; y = 2te^t; z = t^2-t+3P52??@$$$xU`?XdR???@@?@???@@l 8W %,L  \CV ] ,A\ 203(x,y,z) = (2,0,3)52??@$$$8>@hO.y=?? 1?1NE'@p? c+FE'@???pΓ?1NE'@ 4̣ RLd +\ [ 2e^p2pe^pp^2-p+3(x,y,z) = (2e^p,2pe^p,p^2-p+3)42??@$$$??Z~c?k xJ@ȡ0 ?r[(uw@???jg ?m a 4V O : pd % L 2e^p2pe^pp^2-p+302e^p+2e^p/sqrt(4e^(2p)+4(p+1)^2e^(2p)+(2p-1)^2)seg (2e^p,2pe^p,p^2-p+3) to (2e^p+2e^p/sqrt(4e^(2p)+4(p+1)^2e^(2p)+(2p-1)^2),2pe^p+2e^p*(p+1)/sqrt(4e^(2p)+4(p+1)^2*e^(2p)+(2p-1)^2),p^2-p+3+(2p-1)/sqrt(4e^(2p)+4(p+1)^2*e^(2p)+(2p-1)^2))82pe^p+2e^p*(p+1)/sqrt(4e^(2p)+4(p+1)^2*e^(2p)+(2p-1)^2)6p^2-p+3+(2p-1)/sqrt(4e^(2p)+4(p+1)^2*e^(2p)+(2p-1)^2)02??@$$$@?(33333pfffff?̬@@`fffff@???\ D T 8ܵʒ Dy t 2+2/3*t2/3*t3-1/3*t$x = 2+2/3*t; y = 2/3*t; z = 3-1/3*t42??@$$$?? I`?@,c@0P@@F?dg5@???jg ? X IJ ƵP  2e^p2pe^pp^2-p+32e^p+(-e^(2p)*(p+1)+2p^2-2p+.75 )/sqrt( (-e^(2p)*(p+1)+2p^2-2p+.75 )^2+(e^(2p)+p^3-2.25*p+1 )^2+(e^p*(-p^3-1.5*p^2+.5p+3.5) )^2 )seg (2e^p,2pe^p,p^2-p+3) to (2e^p+(-e^(2p)*(p+1)+2p^2-2p+.75 )/sqrt( (-e^(2p)*(p+1)+2p^2-2p+.75 )^2+(e^(2p)+p^3-2.25*p+1 )^2+(e^p*(-p^3-1.5*p^2+.5p+3.5) )^2 ),2pe^p+(e^(2p)+p^3-2.25*p+1 )/sqrt( (-e^(2p)*(p+1)+2p^2-2p+.75 )^2+(e^(2p)+p^3-2.25*p+1 )^2+(e^p*}2pe^p+(e^(2p)+p^3-2.25*p+1 )/sqrt( (-e^(2p)*(p+1)+2p^2-2p+.75 )^2+(e^(2p)+p^3-2.25*p+1 )^2+(e^p*(-p^3-1.5*p^2+.5p+3.5) )^2 )p^2-p+3+(e^p*(-p^3-1.5*p^2+.5p+3.5) )/sqrt( (-e^(2p)*(p+1)+2p^2-2p+.75 )^2+(e^(2p)+p^3-2.25*p+1 )^2+(e^p*(-p^3-1.5*p^2+.5p+3.5) )^2 )42??@$$$??pv ߒ?PܑH@좝?\@D mmM@???jg ?x{ Ho c 4] H | ~ 2e^p2pe^pp^2-p+3=2e^p+(-4p^2-2p+8)/sqrt((-4p^2-2p+8)^2 + (4p-6)^2 +16e^(2p) )seg (2e^p,2pe^p,p^2-p+3) to (2e^p+(-4p^2-2p+8)/sqrt((-4p^2-2p+8)^2 + (4p-6)^2 +16e^(2p) ),2pe^p+(4p-6)/sqrt((-4p^2-2p+8)^2 + (4p-6)^2 +16e^(2p) ),p^2-p+3+4e^p/sqrt((-4p^2-2p+8)^2 + (4p-6)^2 +16e^(2p) ))82pe^p+(4p-6)/sqrt((-4p^2-2p+8)^2 + (4p-6)^2 +16e^(2p) )8p^2-p+3+4e^p/sqrt((-4p^2-2p+8)^2 + (4p-6)^2 +16e^(2p) )xyzHXT@@K?@@@@@@@@@@@@@@@@?@@@@@@@@@@ @ @ @Project 2: Problem 4 Space Curvep/sqrt(4e^2p+4(p+1)^2e^2p+(E: <0@ڱ~l@ErT@Times New Romane Unit Binormal Vector is BluerveRdNOu$%@@М|]d@(18@Times New Romandg The Tangent line at (2, 0, 3) is Green*e^(2p)+(2p-1)^2),2pe^pŪ@`+11<@K@{2@Times New Romanh Unit Tangent Vector is Red(&Z d\hiN(tZ@RŜB@0p !@ǡI@Times New Romanh Unit Normal Vector is Purple&&e`N?z.=ƌ0ˑ@`@ = W@~@Times New Romanx = 2e^t; y = 2te^t; z = t^2-t+30OP`ߒ @ b7 @Dj (x,y,z) = (2,0,3)`ߒ 0-@0!t?w @j (x,y,z) = (2e^p,2pe^p,p^2-p+3)`ߒ @d@@}審0` 7 @k seg (2e^p,2pe^p,p^2-p+3) to (2e^p+2e^p/sqrt(4e^(2p)+4(p+1)^2 `ߒ kjH@nT˂ @ll x = 2+2/3*t; y = 2/3*t; z = 3-1/3*t*# `ߒ `Y@c-hFnR @,*#?$m seg (2e^p,2pe^p,p^2-p+3) to (2e^p+(-e^(2p)*(p+1)+2p^2-2p+.75`ߒ Bk?0ܴ0ʼ @OOxx5O_ m seg (2e^p,2pe^p,p^2-p+3) to (2e^p+(-4p^2-2p+8)/sqrt((-4p^2-2`ߒ ِǶ@J#/j 1Ģ @