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# Week 1 MATH 210

Humboldt

GPA 3.7

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This 6 page Class Notes was uploaded by Merissa Notetaker on Tuesday January 26, 2016. The Class Notes belongs to MATH 210 at Humboldt State University taught by Peter Goetz in Summer 2015. Since its upload, it has received 41 views. For similar materials see Calculus 3 in Math at Humboldt State University.

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Date Created: 01/26/16

Calculus 3 EX . Show that the equation ×2+y2+ -22-6×22=11 . 4y a sphere . Find center and radius if so . we around gives want move = ( X -a)2 + which is ' . / (X-3) + ( yt2)2t 1Z -1)2 . 25 a Sphere W/ center (3 , -2,1 ) radius 5 a square yes , c+2-2-22-+1 thy2t446×+9e+! ×2+bx+(b㱺 ?_ (b㱺2 3- Various Sub spaces of space (112) Coordinate Planes Z - : Z=O EX . Describe in words the set of points Xy plane equation 1123 8 and draw - . 1h that sahsty y< , ! XZ plane : equation y .o Z / y - plane : x.-o .8 plane in1123, line in1122 - yz equation y , × To , . R3={ (× ,y,z)/ × y ,z ER } ' ' parallel to xz plane 11 Containing ( 1,8 ,3) ⾨ Y × 8 YL is the set of all points in 1123 which he to the left of the .-8 . but not plane y on y¥ " pen half space shot lspace including boundary plane Cut in1/2 12.2 Vectors A vector is a that has a direction and an arrow representation quamty magnitude , using 11£11 Y Point Magnitude = length of arrow , 20 , µ,+,a| 1/22 i,to " . terminaa :}oint 1 X } 't'P j=u in the picture in general , T=h if and if 11Tltllhll and > have same direction only •X Describing Vector Algebra Addition to Tail Rule Tip Parallelogram Rule too % %t . :%t - T ! Math Fun Fact ! 1122 Friday Monty Hall Problem 3 game show doors class ywghoorIlln:e :#}ss , sw÷n switch doorsandor.on.si# .. what do you do ? ggg stay or switch . com 10 times switch time - 50% win loose Stay , play } every / Stats stay 66 .lu win for switch we a door W Chance we car. other doors has 213 chance : door dosh't chancek 1/3 got opening change of unpicked doors 12.2 vectors continued Zero vector 8 • 0+6 ¥ =L Scalar Multiplication r= some real number to make a new vector RT Case 2 : r< 0 : : easel r >o case 3 to , 08=8 rg " g. " oteillrvlrlrlnvn °€¥n9matghn,+ude=ryj magnitude = lryjy , ,, Subtraction of Vectors . A his F ::Oiit ... . * :# of Algebraic representation Vectors Z a 7 (a ,b,C) 8 in 3 . . µ space 0 to coordinate ze B we place tail at organ (0 ,0,01 y now it is defined as D= < a .b , (> X < Ex . Draw Vector I, -2,3 > , then move tail of Vector to ( Xiyiz ) Z ' ' ' 4-2/2-+3 > ( .pj= q<×+ / , - a( X -y, Z ) \ y x Vector From Coordinates - Given two points P= (X , ,, Z ,) Z. Q=( Xz , ,, Z .) , we get vector PQ , y y ,yz .-2 : 8 Xz - X,,,=( - y,, Z . -Z ,) o/Q=LXz yz X., ) Magnitude of Vector < a ,b, c) Unit Vector Coordinate Unit Vector Z ( a ,b,c) > W = < a, b ,c > is to 1 F< 1,010 > - magnitude equal I .-< o 1,0 ) . wy 11811 = ' 11611 : 1 , (o.O,o, NETC E=< ,) 0,0 , × y Vector yi. tt Unitizing in FE unit of I , making Vector in direction ÷ Proof : 11¥ 011 - " 811=1 th , ¥ Algebra wl coordmatized Vector li - L > it a ,b , c> J=< die ,f 8=( , ,bct+ef ) vii. ( ra > ,rb,rc Properties : Axioms of a vector Space . ' 1) uitv . Ttu 5) r( Itv )= rust rv litlw .it )= ( Prtw )tT 6) rts )X . ruitsu 2) ( 3) 8+8=8 7) ( vs )u>=r( sci ) 4) Ji ( ii)J=8 8) 1. ui=u 1125 Ex . > force P : Find resultant on F=t÷+F , ' = - > 11¥11 10lb F ,= 10 cos 45%+10 f- 552,552 ) F. 21 E s|n4S°j= o°j=( r F , 11FIH = 12lb Ez = 12 os 30 IZSM 60.3 6) .j⊥)g0 45 . ; ( .it } , P E , t¥z=( -552+653,552+6 )=E HE 11=5*6+1552.165-513.5 lb a- ' tan 'f¥2tt6ut)=76° 12.3 Dot Product . helps compute angles as = = La ,.dz , . ..an ) £ ( bi, bz , ... bn ) a%b=a . : scalar not a vector ,b tazbzt . . Ex . E. . ( 1,5 ) Tse ( - rz 3> Ex . as =P 6=25 . 3k 3 , .ij - space a=< 1,1 ,o> b- =< 0,2, 3> as .£=f| . - Feil ( 5.3 ) .' Bit 15 £ . } . 0.12+0=2 Properties r .tanbn real # any - 11 Ela ,, ... an> ai .ci . a ,2+ . .. tan = 11212 5) 8.2=0 as .B= 2) E.ci 3) as .( bit ) = oiob + as .E 4) ( rail . E=r( oiob ) Ex . Expression Does it make sense ? Result ( ciobloc no , cant dot scalar 3. Vector )E E ( oiib yes , you can scale vector Vector 11811 ( 5 ic ) scalar yes 8.5+2 no lloill (.tc ) no , cant dot scalar Is vector Angles Computi,g Pr00I : 0 < 0<180 ' Ther0= let 8 } } be nonzero vectors Law of cosine . Then T°T= 1151111811 COSO su .v )o ftp.ig#)hgC2=a2+b2-2abcosc • fyc- • } .ae O=co5' RH ' Huili11118?1 cos O ?_ ii. J.us -J - = iou us.T - 8 .b+ 8.8 utility 110412-26.81+11842 = . ) lid tlyvtlnylily flifolvlullllvllcoso - R.J. HRHHJIICOSO Ex . as =jEti , is = 35+25 - K Find btwn 83.5 angle ' as. B=6 . z -I O= cos ' ( ¥5 ) -~7l° Perpendicular or Vectors Orthogonal h[⊥#E°J=o EX . Law 1126 Parallelogram . .in "aaino÷€¥#÷:II:IIIH¥H¥I¥I¥¥¥on ,#¥ ... =( - ( Jot ) JD ) tcuoultcuyvtvyutiillt.FI )+( = 2( iou )i 2( To f) EX . let P= ( l, -3 , -2) Q= ( 2,0 , -4) , R=( 6 , -2 , -5) a do , ) they make a triangle ? parallel "e '? ' :n:aiamYiY I j T is a scalar PTR ¥ no scale relationship , is not multi of it a scale multiple of FQ . ie . f=rP Yes . its a triangle . p.a.ro#3Ffzett.3IYareimypara right angle IF • = 0 acute ( 900 positive dot b) Is DPQR a ? right triangle FQ . Frist : : 790 ' dot FR 3+6=14 to obtuse negative 2 L (POT , PT ) Tangle btw PQ and positive = cos-' ( PQ.P÷ ) PQTNFNI p : PQ .RP = -4.2 i ° - 5 = : ( , ) ( , -1,3 ) 20-2+3=21 acute 8 : do self , not your right angle Direction angles and Direction Cosines Ex . I=< 2 , 1,2> Find BZ . cos B cos ( B - - 2= < ( 8 ,T ) cosy ) )=<2#(o,i>= ÷ p=< ( 11<2,1 , " " -7 FEI .c<iiiflhrgE)sttoncdairsinison ;s}{ } u 7 8 ( ii , ' ' D= cos ( zt ) 470.5 : x i I Project roggb = projection of B onto 8 = > ra .¥÷ compaob = llprogoibllillbllcosotllbll = gijbprogab ,§j,Tan , , = unit vector in direction of a ) ,fj÷( Iai÷h¥i=iE÷÷:÷a

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