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# AnalyticGeometryandCalculusC MATH243

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This 16 page Class Notes was uploaded by Mustafa Wilderman on Saturday September 19, 2015. The Class Notes belongs to MATH243 at University of Delaware taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/207117/math243-university-of-delaware in Mathematics (M) at University of Delaware.

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Date Created: 09/19/15

Name ID Midterm 1 Math 243 Spring 2009 Duration 50 minutes Please close your books tum o your calculators and put them away You can use one page of notes To get full credit you should eccplam your answers 1 Consider the vectors 17 lt17 27 0gt and 17 lt17 737 2gt a 3 points Find two unit vectors that are perpendicular on both 17 and 17 Solution The cross products 17 gtlt 17 and 17 gtlt 17 are two vectors perpendicular on both 17 and 17 We have that Z 5 I def 7amp20s10a12 ugtlt17712 07173 2j12 k173 173 2 74707270k7372 477237513 The vector 17 gtlt 17 equals 717 gtlt 17 7417 277 57 To obtain two unit vectors perpendicular to both 17 and 17 we divide the vectors 17 gtlt 17 and 17 gtlt 17 by their length Thus7 the two unit vectors perpendicular to both 17 and 17 are lt4772775gt lt4772775gt lt 4 2 5 gt 42 722 75 7 WE M457 457 45 and lt747275gt i lt747275gt ilt 4 2 5 474 22 52 T 45 T 457 M457 45 b 2 points Find the angle between 17 and 17 and the length of the vector 17 17 If the angle is 07 it is enough to calculate any trigonometric function sin7 cos or tan of 0 Solution lf 6 is the angle between 17 and 177 then 1217 1760 cos0 f lullvl 122202 123222 i5 5 am 39 The vector 1717 equals lt1172 737 02gt lt27 717 2gt The length of this vector equals 2271222 3 2 Consider the line L that contains the points 1317 23 and Q27172 a 3 points Find the parametric equations of the line L Use these equations to determine the coordinates of the intersection point between L and the zy plane Solution A direction vector for the line L is the vector Pb lt2 7117 22 7 3gt lt17 717 71gt The pararnetric equations of L are the following z1t y27t 237t b 2 points There are exactly two points on the line L at distance 2 from P Find the coordinates of these two points Solution Let Luz be a point on the line L at distance 2xg from P Because Luz is on the line L7 it follows that there is a real number t such that z 1 t7y 2 7 122 3 7 t Because the distance between z7y72 and 17 23 is 2xg7 we deduce that m ltz71gt2lty72gt2lt273gt27lt1t71gt2lt27t72gt2lt37t73gt2 Mt 71t2 7t x3t2 This equation implies that t2 4 which means that t 72 or t 2 When It 727 we get that Luz 1 122 7 t7 7 t 7145 When It 27 we get that Luz 1t727t737t37071 3 a 3 points Find an equation of the plane that passes through the line of intersection of the planes 2 7 2 1 and y 22 3 and is perpendicular to the plane 2 y 7 22 1 Solution Let P denote the plane we are trying to nd Let P1 be the plane whose equation is z 7 2 1 and P2 be the plane whose equation of y 22 3 A normal vector to P1 is n 10 71 A normal vector to P2 is n 012 The intersection line between P1 and P2 has direction 71 gtlt n which equals 10 171 a gt7 OH 1 H H s E 1 0 One can nd a point on the intersection line between P1 and P2 by nding a solution of the system 2 7 2 1 and y 22 3 When 2 0 one nds z 1 and y 3 Thus 13 0 is on the intersection line between P1 and P2 Because P contains this line it means that P contains 130 Let Q be the plane whose equation is xy 7 22 1 A normal vector to Q is lt1 1 72 Because P and Q are perpendicular it follows that lt1 1 72 is contained in P The plane P contains the point 130 and the vectors lt1721 and 1172 This means that a normal vector for P is the following vector 239 lt1721gt gtlt lt1172gt 1 i2 1 7 11 a 172 1 i2 quot7 3Z3j 313 721 a It follows that the equation of P is 32 7 1 3y 7 3 32 7 0 0 which is the same as 2 y 2 4 b 2 points Find pararnetric equations for the line through the point 012 that is parallel to the plane 2 y 2 2 and perpendicular to the line 2 1ty 1 7 t 2 2t Solution Let L denote the line we are trying to nd Because L is parallel to 2y2 2 it means that L is perpendicular to 111 Thus L is perpendicular to 111 and lt1 71 2 This implies that a direction vector for L is 111 gtlt lt1 712 This equals 5 5 3 lt17171gtXlt177L2gt71 1 121 17j11k1 1 712 12 171 1712 3575721 The line L passes through 012 and has direction lt371 72 Its pararnetric equations are 23t y17t 2272t 4 Use Maple to draw the following surfaces 21 2 points z2y272216775 x 5776 y 6777 z 7 b 2 points the points Px7y7z whose distance from 17273 equals 4 and 73 S x 5772 y 6771 z 7 c 1 point the points Qx7y7z that are equidistant from the points 07273 and 27172 when 713953177131 1771 Z L MM2i72 25me WUWWW of f MWx w PW Wmquot Vfrltfx7lt 7y 31gt my vfmca 7907 quot 7 my 0 6 gt quota 9 ltquotquot23921gt 9 r I uf lmI a c lt35 25 04 339 32gt 5 lt 0 0 1p lt0 0 53 T L of vfm is 31w gw quota 27 I39X7L7 fry w 4 0 r i Y X A1 Em0 quot2104 gt739lt Igtz xquot o mvmx gt 95er 39 KOMU lt542 wiltBM Cdv f OHO examof Mm m 4 W MW Wm X MM nkf t39X Ci Q 9y2O1 a f M as w W 7 76 l Hmf 2 o n jim 0514 W Myst an MaiW vsHQ 0am frmq QL wys x mamaw n1 I 07g WC um un t1 W W x XV WWW M a 71 Waf i 7 fr f 44 44 0Mu 7 W Af w39 a944 oMWwV 7c WfXDyflPzi fwd LariKym nt n o M m h 3 M 39 m M W m 5 quot7 7 W 1quot K107 W172 39 5 a WEmma c rswmaf k fwzfm Diva7 W39enfl WQM Myth Wt ms 0 mama wazfm 151 W W fC5n ltr f xldmat mx 3 k W2Wy3 gt gvdn x 2 1 0 ampx LEMg 31 JX Wm 2 33 x 2422 5 Lamp 3apxj ig2xgAB 2 sf lf hj na ampgtlt L 4Samp 4amp3 43J Z 77 3 9431 xmya n C af f m aw610 amp x 7ampwa 39 f Gig 3 2 Shh C 3 6 C2 O Id cm zxzwztz 139 gB C39cgf a 915 27gt 4 Wo t 19 I ogt9 og39tsx I 39 6 Cz V6 IIOgt1ttlt 37 f0 921er 39ltt39 o H 652 at max L W at O 8 fa lt2quot ngI JXoamp39 ffc cv o39xongwycg WM g gf 7U M Lgtuzw 5 z a W Joztw waamp f l LD ageVOGUQH 603805 MCWH lttf w fquotgt 994370 l LowwRH QianZH Macao o M2V Q mtwrwo yeast0 44 0H mo 39 e oro MP2 998 5535 cog oo13 WG 19 0599 f5 f6 051 Ef gy 32 C max af m 7AM 3 o eleven 24 U 2Igtt0 20 A Wm 1317M Q0002 yx C I 2 60 3 2 o 9 16 0quot 000gt Jjlt 2f I lt f Lt vi 763 225 g27t chcgblinazo t Liz quotZ 05163 97 CL Hfquot fr396 239Igt tlt 3 2 039gt 3 0 2 t Igtlt2t 2m ticHi 2 639 oz M g 2 vst39 jftl xgg f 1206 a quot 39 2 2 4 143227 1 WM fcq xv X jg gif Xo 1 gig zf yam zt ao n t wgb o 2 H0Ltf720 220056H 396 05 915wa ab r K g gg M M z to f o mWWQWIZMN choawMC WOW aw Wt magkgzw oagj Migk rwgtmwj h osjfz Wen WI W m 0f 6 Es 0176 15 276 pgts j 1 we lama WW t2t3 ttt3j Wk 239ij ffj r l Mm Hay Wk TM f MPf2lttf 4tltrzt5o r C39 0 39 y w ts z a O it 5 quotJ5 476 2 571 A ammimmm mfm w 1 Ma A an a 4 in gcw szM f amwgth c arvs Wk C 6W 2 at 96 JW WIS amp wc f 93078 o gt lt I zx a ff 908 2731 zEg m jf lamww 0quot 4M4 14046 mm CamM a J46 mffadw f Mm as am a w qu 9M0 hOff Anyi MM MM WW2 mM wg WW MI Jana man gymxx in gm 79 Wm F lt 0 g hgtd gtltol 0 III A 92 uh a o W We 7 26ft 22345 gas3 KWQ 23 in W W 24 o 22 fax W was Mm 327 Xo 222 760 BLxo 30 sz 0 w fogZD 0 HFTW lt20 7Q gt905 0 17i 1th W I gj37g Name ID Midterm 2 Math 243 Spring 2009 Duration 50 minutes Please close your books turn o your phones and calculators and put them away You can use one page of handwritten notes To get full credit you should eccplaz39n your answers 1 Consider the curve rt lt3 sint4t3cos tgt a 3 points Repararnetrize the curve rt with respect to arc length measured from the point where t 0 in the direction of increasing t Solution We have that t t sst ruldu x3cosu24273sinu2du 0 0 t t q9cos2usin2u16du V916du 0t 0 5du5t 0 Thus7 s 5t which means that t Thus7 the repararnetrized curve is 4 rs lt3sin Es cos b 2 points What is the curvature of rt at the point 047r 73 7 Solution At 047r 73 the value oft is 7139 We have that rt lt3 cos t 4 73 sin tgt and r t lt73 sin t 0 73cos tgt At t 7T7 we have that r 7r lt734 0gt and r 7r lt003gt It follows that 25 a a WXr 73 4 0 12j9 003 msw Alsm Wm W5 Thus7 the curvature at t 7T equals wmxwmMMM 7 WWW 7 W mwwiw73 W M 012 2 Let ay 2 y2 74 75 a 3 points Find and sketch the domain of f Find the range of f Solution The domain is formed by all points Ly such that 2 y2 2 4 These points form lie outside the circle x2 y2 4 The range is 75 00 b 2 points Draw a contour map of the function showing four labeled level curves Solution The level curve at height k is formed by the points Ly such that k ay x2 y 7 475 This means that k5 x2 y 7 4 which implies x2y2 4k52 This is a circle of center 0 0 and radius M4 k 5 When k 75 the level curve is a circle of center 0 0 and radius 2 When k 73 the level curve is a circle of center 0 0 and radius 4 4 2W When k 07 the level curve is a circle of center 0 0 and radius 4 2 When k 17 the level curve is a circle of center 0 0 and radius 4 36 2m 3 a 3 points The temperature at a point Ly is Ty measured in degrees Celsius A bug crawls so that its position after t seconds is given by z 41 ty 2 it where z and y are measured in centimeters The temperature function satis es Tm23 4 and Ty23 3 How fast is the temperature rising on the bugs path after 3 seconds 7 Solution We apply chain rule The function T depends on x and y which depend on t Thus7 T is a function of t Tm mm ylttgtgtz lttgt Tyltzlttgtylttgtgtzlttgt We have that x t 1t N and y t 2 it At t 37 we have that z 3 2V1 i and y 3 Also7 at t 37 we have that 23 41 2 and y3 2 3 3 Thus7 T3 Tm3ay33 Ty3ay3y3 m2 3 Ty2 3 l l 47 372 41L 3 b 2 points lf 2 5x2 y2 and Ly changes from 12 to 11 21 compare the values of A2 and dz Solution First7 10x and 37 2y We have that A2 5112 212 4 512 22 5121 4414 9 606 4414 9 1046 i 9 146 and 32 32 dz 7 12dz 67y12dy 1001 401 14 4 5 points A ball is thrown with angle of elevation 60quot What is the initial speed if the maximum height of the ball is 30 m 7 Solution Let 110 denote the initial speed The position vector of the ball is rt lt00 cos 60t10 sin 60t7 NR gt Thus7 the height of the ball is yt vo sin 60t 7 When the height is maximum7 we have that y t 0 This means that the time when the ball is at maximum height satis es the equation 9t 0 yt 10 sin 60t 7 110 sin 60 7 gt 110 Sin 60 Thus7 at the maximum height we have that t 9 Because the maximum height is 30 m7 it follows that 110 sin 60 113 sin2 60 30 y 7 7 9 29 This implies 609 sin2 6039 2 007 i N i 7 Taking g N 10 m per second squared and because sin 60 7 77 we get 600 v0 m T 800 20 Z Name ID Midterm 3 Math 243 Spring 2009 Duration 50 minutes Please close your books turn o your phones and calculators and put them away You can use one page of handwritten notes To get full credit you should eccplaz39n your answers 1 5 points Calculate the double integral idA R2y2 where R 12 gtlt 01 Solution We have that 1 2 11 2 2 dA dxdy MW 12 11 0 1 y 0 2 11n4 y 1n1 12 dy 0 To calculate f ln1 112dy7 we use integration by parts Let u ln1 yz and 1 y We have that 2y 112d d7d1127 d nyy uv uv vu yny yly2y 22 2 yln1y27idyyln1y27lt27 dy 1y2 1 1 1 2 i2 2 d yn 11 21 1y2 y yln1 yz 7 2y 2 arctany C To calculate fln4 yz dy7 we use the previous result and the substitution y 2x This implies 4 y2 4 2x2 41 x2 and dy 2dz We have that ln4 12 dy ln41 222 dz 2 ln4 ln1 22 dz 2 ln4z 2 x ln1 x2 7 2x 2 arctan C 2 ln4z 2zln1 x2 7 4x 4 arctanz C 2 yln4yln 1 7 2y4arctan C y ln4 y ln4 12 7 y ln4 7 2y 4arctan C y ln4 yz 7 2y 4 arctan C

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