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# CALCULUSII MATH114

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This 288 page Class Notes was uploaded by Claudine Friesen on Monday September 28, 2015. The Class Notes belongs to MATH114 at University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/215403/math114-university-of-pennsylvania in Mathematics (M) at University of Pennsylvania.

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Date Created: 09/28/15

Math 114 Rimmer a9131 3D Cartesian Coord 131 Three Dimensional Coordinate Systems Cartesian plane ZZC plane x a Fg Math 114 Rimmer Emquot56131 3D Cartesian Coord Pomts have coordinates x M Example Z A456 B 3 3 1 A 4 5 6 C 2 20 C quot 5 2 20 2 I 39 1 E E i 39 V u Ir B 5 x i w Math 11A7 Rimmer Distance between p01nts P1 1 x1 yl zl M34343 Cartesian Coord P2 1 xz yz zz A456 diRP2i x2xi2y2y12zzzl2 32373 1 C 2 20 Find AB and AC i Which is larger AB 4 325 326 12 AB16449 x114 40677 Acy 4 225 226 02 AC 364936 121211 Math 1147 Rimmer Equat1on Of a Em quot 3 i313D Cartesian Coord x h2y k2z m2 r2 Center hkm Radius r Find the equation of the sphere with center at O 36 and radius 5 x2y32 z 62 3 Find the center and radius of the sphere that has the given equation 4x2 4y2 4z2 4x8y 3 O 4x2 xi4y22yi4zz3Li 4x 24y12 4z2 8 x2y12z22 Center l0 Radius J5 Math 114 004 Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania December 10 2009 Tong Zhu Math llArDDA Fall 2009 Vector Calculus Green39s Theorem Tong Zhu Math llArDDA Fall 2009 Vector Calculus Green39s Theorem It is about a line integral de l Qdy along a C 0 positively oriented Tong Zhu Math llArDDA Fall 2009 Vector Calculus Green39s Theorem It is about a line integral de l Qdy along a C 0 positively oriented 9 piecewise smooth Tong Zhu Math llArDDA Fall 2009 Vector Calculus Green39s Theorem It is about a line integral de l Qdy along a C 0 positively oriented 9 piecewise smooth 9 simple closed curve C Tong Zhu Math llArDDA Fall 2009 Vector Calculus Green39s Theorem It is about a line integral PdX Qdy along a C 0 positively oriented 9 piecewise smooth 9 simple closed curve C C Theurem Green39s Theurem Let C be a positively oriented piecewise smooth simple Closed Curve in the plane and D be the region bounded by C Theurem Green39s Theurem Let C be a positively oriented piecewise smooth simple Closed Curve in the plane and D b t e region bounded by C lfP and Q have Continuous partial derivatives on an open region that Contains D then Theurem Green39s Theurem Let C be a positively oriented piecewise smooth simple Closed Curve in the plane and D b t e region bounded by C lfP and Q have Continuous partial derivatives on an open region that Contains D then 60 6P CdeQdyD Eiiwx 0y Theurem Green39s Theurem Let C be a positively oriented piecewise smooth simple Closed Curve in the plane and D be t e region bounded by C lfP and Q have Continuous partial derivatives on an open region that Contains D then 60 6P CPdltQdyDEiadl1 Other notations for Closed Curve PdX Qdy C Il Example 1 Evaluate 35C xydx l X2y3dy where C is the triangle with vertices 00 10 and 12 and is positively oriented Tong Zhu Math llArDDA Fall 2009 Green39s Theorem does not only work for simple region like 39 but also works for regions with holes like Tong Zhu Math 114004 Fall 2009 Example 2 Evaluate 35C y3dx 7 X3dy where C is the union of the two circles of radius 1 and radius 2 centered at the origin with positive orientation Tong Zhu Math llAr An Application of Green39s Theorem Tong Zhu Math llArDDA Fall 2009 An Application of Green39s Theorem The area of a plane region D ADdA Tong Zhu Math llArDDA Fall 2009 An Application of Green39s Theorem The area of a plane region D A dAlxdyijllydxljlle7ydx D c c 2 c Tong Zhu Math llArDDA Fall 2009 Math 114 004 Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania October 29 2009 Tong Zhu Mad 114mm Fall 2mm The Chain Rule Tong Zhu Mad 114mm Fall zuug The Chain Rule If 2 fXy X gty ht Eu E 00 quot a Tony Zhu Mad 114mm Fall zuug The Chain Rule If 2 fXy X gty ht ii dti lxdt 8ydt7 Tong Zhu Mad 114mm Fall zuug The Chain Rule If 2 fXy X gty ht dti lxdt 8ydt78xdt 8ydt Tong Zhu Mad 114mm Fall zuug Example 1 Ifz 1X2 y2 X In t y cost find when t7r Mad 114 Example 1 Ifz 1X2 y2 X In t y cost find when t 7r Solution amp ampy dt 7 8x dt 8y dt Tong Zhu Mad 114mm Fall zuug Example 1 Ifz 1X2 y2 X In t y cost find when t7r Solution amp ampy dt 7 8x dt 8y dt 82 2X X ai2x1X2y2 1x2y2 Tong Zhu Mad 114mm Fall zuug Example 1 Ifz 1X2 y2 X In t y cost find when t7r Solution dz 7 82 dx 82 dy E EEEE 82 7 2X 7 X 8X 2V1XZy2 1x2y2 82 2y y 52x1X2y2 1x2y2 Tong Zhu Mad 114mm Fall zuug Example 1 Ifz 1X2 y2 X In t y cost find when t7r Solution dz 7 82 dx 82 dy E EEEE 827 2X 7 X EHW W W 1327 2y 7 y EHW W W dx 1 E Tong Zhu Mad 114mm Fall zuug Example 1 Ifz 1X2 y2 X In t y cost find when t7r Solution dz 7 82 dx 82 dy E EEEE 827 2X 7 X EHW w 827 2y 7 y E a i 7 V irp dx 1 dy E E7smt Tong Zhu Mad 114mm Fall zuug Example 1 Ifz 1X2 y2 X In t y cost find when t7r Solution amp ampy dt 7 8x dt 8y dt 82 7 2X 7 X 8X 2V1XZy2 1x2y2 82 7 2y 7 y 8y 2V1XZy2 1x2y2 dx 71 dy 7 t dt 7 t dt 7 539 So dz 7 Xt ysint Xtiysint df 1X2y27 1X2y2 1x2y2 Tong Zhu Mad 114mm Fall zuug When t7rxn7r Ton Zhu Mad 11 When t7rxn7rysin7r Mad 114 14 Fall zuug When t7rxn7rysin7r0 So dz Xtiysint n7r E7 1x2y2 7 39I139 1n7r2 Tong Zhu Mad 114mm Fall zuug A special case 2 fxy and y gx what is 7 Mad 114 A special case 2 fxy and y gx what is 7 amp gy dx 7 8X dx 8y dx Tong Zhu Mad 114mm Fall zuug A special case 2 fxy and y gx what is 7 amp amp 1 dx 7 8X dx 8y dx 7 8X ayg X Tong Zhu Mad 114mm Fall zuug Chain Rule If 2 fX y X g5 t y h5 t Eu E 00 quot a Tony Zhu Mad 114mm Fall zuug Chain Rule If 2 fX y X g5 t y h5 t g7 828x 828y as 55 5 Eu T a V3 Tong Zhu Mad 114mm Fall zuug Chain Rule If 2 fX y X g5 t y h5 t aziazax 823y aziazax 823y E E lw aaawm Tong Zhu Mad 114mm Fall zuug If 2 f39lt y lt g t y 75 1 Bzi zax Bz y 327323x323y 35 3x35 ayes ar axar 3y3t z E az a 3 inlermedmle variable X y nfz 3 3 6 3i a a E a s t s t Mth mu Fall 2m Example 2 Find an ifz excosy and X st y V52 t2 Mad 114 Example 2 Find and if z excosy and X st y V52 t2 Solution 5 Mad 114 Example 2 Find and ifz excosy and X st y v s2 t2 Solution 82 X 79 cosy x Mad 114 Example 2 Find and if z excosy and X st y V52 t2 Solution ziexcos 8 7 X7 y 8y7 Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y52t2 Solution 82 82 ix 77x X ecosy 8y esmy Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y V52 t2 Solution 7 x Q x 8X79 cosy ayi esmy Q 85 7 Tony Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y V52 t2 Solution 82 X 82 X 7 7 7 sm 8X 9 cosy 8y 9 y 85 7 Tony Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y V52 t2 Solution 8 87excosy 877exsiny Q t a 85 8t 5 Tony Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y52t2 Solution 8 7 7 X i 7 X 8X79 cosy ayi esmy 8X 8X 7 i 5 85 8t 81 85 7 Tony Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y52t2 Solution 8 if X if 7 X axie cosy 8y7 9 smy 8X 8X 7 i5 85 81 it 85W Tong Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y W Solution 82 8 eXcosy a iexsiny Q 1 8X s 85 8t 8y 5 8y EW a Tony Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y W Solution 82 8 eXcosy 579Xsiny 8X 8X EW irw Tong Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st Fm Solution 8 if X i i X axis cosy 8y 9 smy at 1x 85 8 it it 85W ef ng So gigg g i axcos t7exsin 75 85 7 8X85 8y85 7 y y 152t2 Tong Zhu Mad 114mm Fall zuug Example 2 Find and ifz excosy and X st y W Solution 2 82 8Xexcosy 57exsiny Q t 81 s 85 8t 8y 5 8y 1 EW irw 827 828x 823y 7 7 7 if if 7 excosy 7 eXsiny 85 8X 85 8y 85 152 2 82 7 828x 132 8y X X t a 7 5 EE 7 e cosys7 e smy7s2 t2 Tong Zhu Mad 114mm Fall zuug The idea of chain rule for functions With more than two variables is the same Ton Zhu Mad 11 um Fall 2mm The idea of chain rule for functions With more than two variables is the same Example 3 If W fxyz7 t and X Xu7 v y yu7 v z zuv and t tuv find 37quot and 35 Tony Zhu Mad 114mm Fall zuug The idea of chain rule for functions With more than tWo variables Is he same Example 3 If W fgtltyz t and x xu v y yuv z zuv and t tuv find 27 and 27 Solution w xyw awiawg away 3W32 3W3 E Eau ya a a w xyt 3W78W3xawayawazawat au axau 3y3u azau azau 3J l 3v 7 3x 3v 8y 8v 32 3v 3 3v I l Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 fx2 7 6 and 292 7 78 find when t 3 Tony Zhu Mad 114mm Fall 2mm Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 fx2 7 6 and 292 7 78 find when t 3 Solution e y dt iaxdt 8ydt 7 Tony Zhu Mad 114mm Fall 2mm Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 fx2 7 6 and 292 7 78 find when t 3 Solution dz 8f dx 8f dy 7 iiii iii dt ax dt 8y dt mx mg t WW t Tong Zhu Mad 114mm Fall 2mm Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 78 find when t 3 1 27 7 6 and 1 y27 7 Solution W g E ax dt 8y dt XX yg t WW t 15gt htg t fygt hth t Tong Zhu Mad 114mm Fall 2mm Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 78 find when t 3 1 27 7 6 and 1 y27 7 Solution W g dt ax dt 8y dt mx mg t WW t Mg mgf fygf7 hfhf When 1 3 since g3 2 and h3 7 X 2 and y 7 at t3 Tong Zhu Mad 114mm Fall 2mm Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 78 find when t 3 1 27 7 6 and 1 y27 7 Solution W g dt ax dt 8y dt mx mg t WW t Mg mgf fygf7 hfhf When 1 3 since g3 2 and h3 7 X 2 and y 7 at t 3 dz age hang3 M3 ham3 E t3 Tong Zhu Mad 114mm Fall 2mm Example 4 If 2 fxy where f is differentiable and X gt y ht also known g3 2 h3 7 g 3 5 h 3 74 78 find when t 3 1 27 7 6 and 1 y27 7 Solution wingr fyixiyih m 13gt7 htg t 1 ygt7 htht When 1 3 since g3 2 and h3 7 X 2 and y 7 at t 3 g H M3 hang3 M3 ham3 6x578x7462 Tong Zhu Mad 114mm Fall 2mm Implicit Differentiation Tong Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation FX7 y 0 defines y implicitly as a function of X how to find 7 Tong Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation FX7 y 0 defines y implicitly as a function of X how to find 7 8Fdx adeio Ea Ti7 Tong Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation FX7 y 0 defines y implicitly as a function of X dy how to find E7 8F dx 8F dy 7 0 aa 87 7 d Solve this for 8F 9 7 2 FF dx 3 Tony Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation FX7 y 0 defines y implicitly as a function of X how to find 7 8F dx 8F dy 7 0 aa 87 7 d Solve this for 8F 9 2 Q dx Fy Tong Zhu Mad 114mm Fall 2mm Example 5 Find dydx if Ty 1x2y Tong Zhu Mad 114mm Fall zuug Example 5 Find dydx if Ty 1x2y Solution Let FXyW717X2y0 Tong Zhu Mad 114mm Fall zuug Example 5 Find dydx if Ty 1x2y Solution Let FXyW717X2y0 So y FX 72M 72X Tong Zhu Mad 114mm Fall zuug Example 5 Find dydx if Ty 1x2y Solution Let FXyW717X2y0 So y X 2 Wi2xy Fy7ix F X Tong Zhu Mad 114mm Fall zuug Example 5 Find dydx if Ty 1x2y Solution Let FXyW717X2y0 So y X 7 7 7 7 2 lairX7 2xy airW X y52xym dx Fy zkixz 7 X72X2M Tong Zhu Mad 114mm Fall zuug Implicit Differentiation If an equation Fxyz 0 defines z implicitly as a function of Xy how to find 37 and 37 Tony Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation Fxyz 0 defines z implicitly as a function of Xy how to find 37 and Si Take partial derivatives on both sides of the equation i8 M 6 0 8X 8X 8y 8X 82 8X 7 Tony Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation Fxyz 0 defines z implicitly as a function of Xy how to find 37 and Si Take partial derivatives on both sides of the equation i8 wig 8X8X 8y8x azaxi EEE 8X 82 8X 7 Tony Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation Fxyz 0 defines z implicitly as a function of Xy how to find 37 and Si Take partial derivatives on both sides of the equation i8 M 6 0 8X 8X 8y 8X 82 8X 7 E ioj 8X azaxi Tong Zhu Mad 114mm Fall 2mm Implicit Differentiation If an equation Fxyz 0 defines z implicitly as a function of Xy how to find 37 and Si Take partial derivatives on both sides of the equation i8 M 6 0 8X 8X 8y 8X 82 8X 7 E ioj 8X azaxi a FX r327 Fy ax Fl 6 F2 Tong Zhu Mad 114mm Fall 2mm Example 6 Find 37 and 37 if X2 y2 z2 3xyz Mad 114 Example 6 Find 37 and 37 if X2 y2 z2 3xyz Solution Let FXy7 z X2 y2 z2 7 3xyz 0 Tony Zhu Mad 114mm Fall zuug Example 6 Find 37 and 37 if X2 y2 z2 3xyz Solution Let FXyz X2 y2 z2 7 3xyz 0 Then FX 2xi3yz Tong Zhu Mad 114mm Fall zuug Example 6 Find 37 and 37 if X2 y2 z2 3xyz Solution Let FXyz X2 y2 z2 7 3xyz 0 Then FX2xi3yz Fy2y73xz Tong Zhu Mad 114mm Fall zuug Example 6 Find 37 and 37 if X2 y2 z2 3xyz Solution Let FXyz X2 y2 z2 7 3xyz 0 Then FX2xi3yz Fy2y73xz Fz2zi3xy Tong Zhu Mad 114mm Fall zuug Example 6 Find 37 and 37 if X2 y2 z2 3xyz Solution Let FXyz X2 y2 z2 7 3xyz 0 Then FX2xi3yz Fy2y73xz Fz2zi3xy 827 if 2xi3yz 8X FZ 7 22 7 3xy Tong Zhu Mad 114mm Fall zuug Example 6 Find 37 and 37 if X2 y2 z2 3xyz Solution Let FXyz X2 y2 z2 7 3xyz 0 Then FX2xi3yz Fy2y73xz Fz2zi3xy 82 FX 2X 7 3yz 82 F 4 FZ 7 2273xy 8X FZ 7227 3xy 8y 7 Tony Zhu Mad 114mm Fall zuug Math 114 004 Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania October 8 2009 Tong Zhu Mad 114mm Fall 2mm Motion in Space Mad 114 14 Fall zuug Motion in Space An application of Vector Functions and its Derivatives Ton Zhu Mad 11 um Fall 2mm Motion in Space An application of Vector Functions and its Derivatives Recall t represents the motion ofa particle moving So it is the position function for this particle Tong Zhu Mad 114mm Fall zuug Motion in Space An application of Vector Functions and its Derivatives Recall t represents the motion ofa particle moving So it is the position function for this particle Now let39s look at its derivative a tAt7t 7 HERO m Tong Zhu Mad 114mm Fall zuug Motion in Space An application of Vector Functions and its Derivatives Recall t represents the motion ofa particle moving So it is the position function for this particle Now let39s look at its derivative a tAt7t s Hing m V This is also the velocity vector at time t Tong Zhu Mad 114mm Fall zuug Motion in Space An application of Vector Functions and its Derivatives Recall t represents the motion ofa particle moving So it is the position function for this particle Now let39s look at its derivative a tAt7t 7 Hing m 7t This is also the velocity vector at time t Why Isn t it the tangent vector Tong Zhu Mad 114mm Fall zuug Motion in Space An application of Vector Functions and its Derivatives Recall t represents the motion ofa particle moving So it is the position function for this particle Now let39s look at its derivative a tAt7t 7 HERO m W This is also the velocity vector at time t Why Isn t it the tangent vector What is the speed then Tong Zhu Mad 114mm Fall zuug Motion in Space An application of Vector Functions and its Derivatives Recall t represents the motion ofa particle moving So it is the position function for this particle Now let39s look at its derivative a tAt7t 7 HERO m W This is also the velocity vector at time t Why Isn t it the tangent vector What is the speed then l7tl Tong Zhu Mad 114mm Fall zuug How about the acceleration Ton Zhu Mad 11 14 Fall ZIIIIQ How about the acceleration Ton Zhu Mad 11 14 Fall ZIIIIQ How about the acceleration 3t v t r t Newton s Second Law of Motion Tong Zhu Mad 114mm Fall zuug How about the acceleration 3f V f WU Newton s Second Law of Motion given the mass of an object say In and the force acting on the object at time 1 say Ft the acceleration 51 can be found by n m3t Tong Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Tong Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is Tony Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is Tony Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is 7t 3tdt2t7 sin tf cos2tldt Tong Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is 7t 3tdt2t7 sin tf cos2tldt s s 1 s s tziicostj sin2tk C1 Tong Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is 7t 3tdt2t7 sin tf cos2tldt s s 1 s s tziicostj sin2tk C1 Tong Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is 7t 3tdt2t7 sin tf cos2tldt s s 1 s s tziicostj sin2tk C1 Tong Zhu Mad 114mm Fall 2mm Example 1 If the acceleration of an object is 31 2t7 sin til cos 2H and its initial velocity is 90 initial position is 0 j find the object39s velocity and position functions Solution The velocity function is 7t 3tdt2t7 sin tf cos2tldt s s 1 s s tziicostj sin2tk C1 s a 1 s 7t t2 1i7 cost 71j Esin2tk Tong Zhu Mad 114mm Fall 2mm The position function Ton Zhu Mad 11 14 Fall ZIIIIQ The position function Ft7tdtt2177cost71sin2tldt Tong Zhu Mad 114mm Fall zuug The position function 7 7 7 v 1 rt vtdt t2 1I7cost711 ism 214 dt 1 7 7 7 7 51 3 ti 7 sin 1 7 tj 7 icos2tk C2 Tong Zhu Mad 1144M Fall zuug The position function 7 7 7 v 1 rt vtdt t2 1I7cost711 ism 214 dt 1 7 7 7 7 51 3 ti 7 sin 1 7 tj 7 icos2tk C2 Tong Zhu Mad 1144M Fall zuug The position function 7 7 7 v 1 rt vtdt t2 1I7cost711 ism 214 dt 1 7 7 7 7 51 3 ti 7 sin 1 7 tj 7 icos2tk C2 Tong Zhu Mad 1144M Fall zuug The position function ti7tdt12177cost71sin2tzdt 1 3 7 j 1 7 7 Et i t75Int7tJ7Zcos2tkC2 By 70 Hence the position function is 1 1 a 1 1 t gt3 ti 7 sin 1 7 t 71j 7 Zcos2t 71k Tong Zhu Mad 1147mm Fall 21mg Projectile Motion Mad 114 14 Fall zuug Projectile Motion A projectile is fired Ton Zhu Mad 11 um Fall 2mm Projectile Motion A projectile is fired an interesting website httpphet cooradoedusimsprojectie motionprojectie motion swf Tong Zlm Mall 1144quot Fall lellg Objective to find the parametric equations of the trajectory Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m Ton Zhu Mad 11 um Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 70 Tony Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 Tony Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m initial velocity 90 v the initial angle of elevation 00 the air resistance is negligible the gravity is the only force acting on the projectile Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector I 7mg Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector I 7mg Also by Newton39s second law of motion I m5 7mg Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector I 7mg Also by Newton39s second law of motion I m5 7mg a 5 7g Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector I 7mg Also by Newton39s second law of motion I m5 7mg i 5 gj 7t 3dt Tong Zhu Mad 114mm Fall 2mm So the velocity Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector I 7mg Also by Newton39s second law of motion I m5 7mg i 5 gj So the velocity m 3dt igtjl C1 Tong Zhu Mad 114mm Fall 2mm Objective to find the parametric equations of the trajectory Assume gt mass m gt initial velocity 90 v gt the initial angle of elevation 00 gt the air resistance is negligible gt the gravity is the only force acting on the projectile What is the force vector I 7mg Also by Newton39s second law of motion I m5 7mg i 5 gj So the velocity m 3dt igtjl C By 90 61 6 Tony Zhu Mad 114mm Fall 2mm So the velocity function is 7t igtj 70 Ton Zhu Mad 11 um Fall 2mm So the velocity function is i7t igtj39i v6 The position function Fm 7dt Tong Zhu Mad 114mm Fall 2mm So the velocity function is i7t igtj39i v6 The position function a 1 a a t i7dt7gtjiidt7 gt2jt C2 Tong Zhu Mad 114mm Fall 2mm So the velocity function is i7t igtj39i v6 The position function a a v a 1 2 a rt vdt igfj vodt figt j tvo C2 What is the initial position Tong Zhu Mad 114mm Fall 2mm So the velocity function is i7t igtj39i v6 The position function a a v a 1 2 a rt vdt igfj vodt figt j tvo C2 What is the initial position 0 6 Tony Zhu Mad 114mm Fall 2mm So the velocity function is i7t igtj39i v6 The position function a a v a 1 2 a rt vdt igfj vodt figt j tvo C2 What is the initial position 0 6 a 52 6 Tony Zhu Mad 114mm Fall 2mm So the velocity function is i7t igtj39i v6 The position function a a v a 1 2 a rt vdt igfj vodt figt j tvo C2 What is the initial position 0 6 a 52 6 l 1 v a t EEtZJ We Tong Zhu Mad 114mm Fall 2mm Now consider the components of 70 Ton Zhu Mad 11 14 Fall ZIIIIQ Now consider the components of 70 since the angle of elevation is 00 3 3 v v cos0ol v Sln001 Ton Zhu Mad 11 um Fall 2mm Now consider the components of v6 since the angle of elevation is 00 a v0 liBl cos007 liBl sin00 Plug it back to the position function we get i x T 1 v 7 7t llvBlCOSQolfl liBlsin00t 7 ngzlj Tong Zhu Mad 114mm Fall 2mm Now consider the components of v6 since the angle of elevation is 00 v liBl cos00i l liBl sin00j Plug it back to the position function we get i x T 1 v 7 7t illBlcos 39olt liBlsin00t 7 ngzlj So the parametric equations of the trajectory are a a 1 Xt lvol cos00t yt lvolsin00t 7 Egt2 Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations Ton Zhu Mad 11 um Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector by ltXyzgtxiyjzk Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector by ltXyzgtxiyjzk 1 For the velocity i7 Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector by ltXyzgtxiyjzk so 1 For the velocity i7 known i7t Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector byltxyzgtxiyjzk 1 For the velocity i7 known i7t so a Flt 7 Vlf Tlt rm vir vt Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector byltxyzgtxiyjzk 1 For the velocity i7 known i7t so s 7 Pa 7 7 w T rm ivmi W 2 How about the acceleration Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector byltxyzgtxiyjzk 1 For the velocity i7 known i7t so m4W0 T mi ivmi W 2 How about the acceleration an Jt Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector byltxyzgtxiyjzk 1 For the velocity i7 known i7t so s 7 Pa 7 7 w T mi ivmi W 2 How about the acceleration an Jt un t vtft Tong Zhu Mad 114mm Fall 2mm Tangential and Normal Components of Accelerations The 717 N frame gives us a way to analyze the motion in space by decomposing vectors this isjust like how we represent a vector byltxyzgtxiyjzk 1 For the velocity i7 known i7t so s 7 Pa 7 7 w T mi ivmi W 2 How about the acceleration an Jt un t vtft a s T IS In the frame What IS T t V Known Tong Zhu Mad 114mm Fall zuug Known Tong Zhu Mad 114mm Fall zuug Known Also known Tong Zhu Mad 114mm Fall zuug Known Also known a a So T t ravtNt We get an mmr MU t or 5 aTT 3va where 37 v and 3N nvz Tong Zhu Mad 114mm Fall zuug Ton Zhu Mad 11 Is there any other method to compute 37 and 3N if we are only given t Mad 114 Is there any other method to compute 37 and 3N if we are only given t 1 First since aN av and known Hr txr7 a wrw mr QJrerm WW 5 WW 0 WM Tong Zhu Mad 114mm Fall zuug Is there any other method to compute 37 and 3N if we are only given t 1 First since aN av and known a r tf rw 3N r qx MUMWNQ rmr mm 0 3 W0 3 V U 2 For 37 consider the dot product of 7 and 3 W5 V7iV71HV2N vv Trltav37i vv Tong Zhu Mad 114mm Fall zuug Is there any other method to compute 37 and 3N if we are only given t 1 First since aN av and known a irqx r ti 3N ir qx r ti t 2 W r ti W 3 MW rt 3 2 For 37 consider the dot product of i7 and 3 i75 v7iv71rvzlil vv Trltav37i vv What does this tell us Tong Zhu Mad 114mm Fall 2mm Is there any other method to compute 37 and 3N if we are only given t 1 First since aN av and known 7 t 7 t 7 t 7 t a 7t 7 t K H w i WM H w NW N y i ir f3 ir f3 MW 2 For 37 consider the dot product of i7 and a What does this tell us We can get i7 5 r r 7 a 37 V W Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk Solution Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk a Solution r t 7 sin t7 cos tj E r7 t Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk a Solution r t 7 sin t7 cos tj E r7 t 7 cos 1 77 sin ti Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk Solution r7 t 7 sin t7 cos tj E r7 t 7 cos 1 77 sin ti and Wm Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk Solution r7 t 7 sin t7 cos tj E r7 t 7 cos 1 77 sin ti and lr t So the tangential component of 3 is 2 2 ar a 0 W Tong Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk Solution r7 t 7 sin t7 cos tj E r7 t 7 cos 1 77 sin ti and lr tl So the tangential component of 3 is 7 7 3T r or 0 W And t t 7 I E r t x r t isint cost 1 icost isint 0 Tony Zhu Mad 114mm Fall 2mm Example 2 Find the tangential and normal components of the acceleration vector of the helix t cos ti i sin tj l tk Solution r7 t 7 sin t7 cos tj E r7 t 7 cos 1 77 sin ti and lr tl So the tangential component of 3 is 2 2 3T s 0 W And t t 7 I E r t x r t isin t cost 1 sin tii cos tj icost isint 0 Hence the normal component of 3 is Tony Zhu Mad 114mm Fall 2mm Math 114 004 Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania November 12 2009 Tong Zhu Mad 114mm Fall 2mm Double Integrals in Polar Coordinates Ton Zhu Mad 11 um Fall 2mm Double Integrals in Polar Coordinates Example 1 Evaluate ffD xydA where D is the disk centered at the origin with radius 3 Tony Zlm Mad 114mm Fall 2mm Double Integrals in Polar Coordinates Example 1 Evaluate ffD xydA where D is the disk centered at the origin with radius 3 Tony Zlm Mad 114mm Fall 2mm Polar Coordinate System Tong Zhu Mad 114mm Fall zuug Polar Coordinate System POW 0r Pr9 Tong Zhu Math 1144quot Fall lellg Polar Coordinate System POW 0r Pr9 Xrcos yrsin0 Tong Zhu Math 1144quot Fall lellg How can we use polar coordinates to evaluate ll Ton Zhu Mad 11 14 Fall ZIIIIQ How can we use polar coordinates to evaluate fx ydA t V D This is in fact a special case of change of variables in multiple integrals Which Will be shown in later section Mad 114mm Fall zuug How can we use polar coordinates to evaluate fx ydA t V D This is in fact a special case of change of variables in multiple integrals Which Will be shown in later section Geometrically what is dA in the polar coordinates Tong Zhu Mad 114mm Fall 2mm How can we use polar coordinates to evaluate fx ydA t V D This is in fact a special case of change of variables in multiple integrals Which Will be shown in later section Geometrically what is dA in the polar coordinates Tong Zhu Mad 114mm Fall 2mm How can we use polar coordinates to evaluate fx ydA t V D This is in fact a special case of change of variables in multiple integrals Which Will be shown in later section Geometrically what is dA in the polar coordinates dA rdrd0 Tong Zhu Mad 114mm Fall 2mm How can we use polar coordinates to evaluate fx ydA t V D This is in fact a special case of change of variables in multiple integrals Which Will be shown in later section Geometrically what is dA in the polar coordinates dArdrd0 fxydA frcos0rsin0rdrd0 D D Tong Zhu Mad 114mm Fall 2mm Revisit Example 1 Evaluate ffD xydA where D is the disk centered at the origin with radius 3 Tony Zhu Mad 114mm Fall 2mm Example 2 Find the area of the region that lies inside r 3 i 2sin0 and outside r 2 Tony Zhu Mad 114mm Fall 2mm Example 3 Find the volume of the solid under the cone 2 xXZ y2 and above the disk X2 ly2 4 Tony Zhu Mad 114mm Fall 2mm 32 Math 114 Rimmer CONIC SECTIONS They are called conic sections or conics because they result from intersecting a cone with a plane A fitiquot Math 114 Rimmer P A R A B 0 LA 115 Conic Sections A parabola is the set of points in a plane that are equidistant from a fixed point F the focus and a fixed line the directrix axis l parabola focus F i directrix vertex nnnnnnnnnnnnnnnnnnn an Math 114 Rimmer gas 115 Conic Sections An ellipse is the set of points in a plane the sum of whose distances from two fixed points F1 and F2 is a constant These two fixed points are called the foci mmmmmmmmmmmmm on HYPERBOLA A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points F1 and F2 the foci is a constant yll Math 114 Rimmer wg rwi su 115 Conic Sections Standard Equation y aX h2 k VGITGXZ hk ELLIPSE 11216 hk 00 Math 114 Rimmer gag2a 115 Conic Sections Standard Equation Hyperbola X W y k2 1 I 7 Q L y IA 7a 0 a 0 7139 0 0 c 0 rm here hk 00 1re hk 00 Math 114 004 Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania November 3 2009 Tong Zhu Mad 114mm Fall 2mm Maximum and Minimum Values Ton Zhu Mad 11 um Fall 2mm Maximum and Minimum Values absolute maximum local nbsalme V minimum maximum Recall For a one variable function y fx how to find maxmin values Ton Zhu Mad 11 14 Fall ZIIIIQ Recall For a one variable function y fx how to find maxmin values Find all the points satisfying y fX 0 Tony Zhu Mad 114mm Fall 2mm Recall For a one variable function y fx how to find maxmin values Find all the points satisfying y fX 0 How to classify max or min points Tong Zhu Mad 114mm Fall 2mm Recall For a one variable function y fx how to find maxmin values Find all the points satisfying y fX 0 How to classify max or min points 0 If f x O and f x gt 0 then f has a local minimum at X 9 If f x O and f x lt 0 then f has a local maximum at X Tong Zhu Mad 114mm Fall 2mm Recall For a onevariable function y fX how to find maxmin values Find all the points satisfying y f X 0 How to classify max or min points 0 If fX O and f X gt 0 then f has a local minimum at X 9 If f X O and f X lt 0 then f has a local maximum at X Tong Zhu Math 1144quot Fall 2mm If fxy has a local maximum or local minimum at a7 b and the first order derivatives exists here then 1 a7 b O and sam0 Tong Zhu Mad 114mm Fall zuug If fxy has a local maximum or local minimum at a7 b and the first order derivatives exists here then 1 a7 b O and sam0 If 1 a7 b O and fya7 b 0 the point a7 b is called a critical point Tong Zhu Mad 114mm Fall 2mm If fX7 y has a local maximum or local minimum at a7 b and the rst order derivatives exists here then fxa7 b O and Ma b 0 If fxa7 b O and fya7 b O the point a7 b is called a critical point 39 I ii ll I i all W N Mo t 1 WWquot M Tong Zhu Math 1144quot Fall 2mm Another type of critical point Ie fx 0 and fy 0 but not local max or local m 5 Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Tong Zhu Mad 114mm Fall 2mm Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Let D 0a b ma bma b e we b Tong Zhu Mad 114mm Fall 2mm Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Let D 00quot mlavbfwavbifxya7b12 l l i l Tong Zhu Mad 114mm Fall 2mm Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Let DDabfmab ya7b ya7b2 fxx fxy fyxfyy a If D gt O and 15037 b gt 0 then fa7 b is a local minimum Tong Zhu Mad 114mm Fall 2mm Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Let D 00quot mlavbfwavbifxya7b12 l l i l a If D gt O and 15037 b gt 0 then fa7 b is a local minimum b If D gt O and 15037 b lt 0 then fa7 b is a local maximum Tong Zhu Mad 114mm Fall 2mm Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Let D 037 b 6037 algaav b 7 6937 bl2 7 may of a If D gt O and 15037 b gt 0 then fa7 b is a local minimum b If D gt O and 15037 b lt 0 then fa7 b is a local maximum c If D lt 0 then fa7 b is a saddle point not a local maximum or minimum Tong Zhu Mad 114mm Fall 2mm Second Derivatives Test If 37 b is a critical point of fxy ie 1 a7 b O and 1amp37 b 0 and the second derivatives of f are continuous on a disk with center 37 b Let D 00quot mlavbfwavbifxya7b12 l l i l a If D gt O and 15037 b gt 0 then fa7 b is a local minimum b If D gt O and 15037 b lt 0 then fa7 b is a local maximum c If D lt 0 then fa7 b is a saddle point not a local maximum or minimum d If D 0 the test gives no information Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y0 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 yx3 y3x Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 3X gtX9XgtXOOI 10I 71 Tony Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fX7y X4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 3 i3iixx gtX9XgtXOOI 10I 71 Three critical points 00 11 and 7171 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 3 y X gtX9XgtXOOI 10 1 y3 X Three critical points 00 11 and 7171 Then check the second order derivatives XX 12x2 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 3 y X gtX9XgtXOOI 10 1 y3 X Three critical points 00 11 and 7171 Then check the second order derivatives XX 12x2 fxy 74 Tony Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 3 i3iixx gtX9XgtXOOI 10I 71 Three critical points 00 11 and 7171 Then check the second order derivatives XX 12x2 fxy 74 fyy 12y2 Tong Zhu Mad 114mm Fall 2mm Example 1 Find and classify the critical points of the function fxyx4y474xy2 Solution First find the critical points amp4X374y07fy4y374x0 3 i3iixx gtX9XgtXOOI 10I 71 Three critical points 00 11 and 7171 Then check the second order derivatives XX 12x2 fxy 74 fyy 12y2 Tong Zhu Mad 114mm Fall 2mm The second derivatives test D fxxfyy 7 ny12 144X2y2 716 Tony Zhu Mad 114mm Fall zuug The second derivatives test D fxxfyy 7 ny12 144X2y2 716 At 11 and 7171 D 144716 gt 0 and fxx gt 0 Tony Zhu Mad 114mm Fall zuug The second derivatives test D fxxfyy 7 ny12 144X2y2 716 At 11 and 7171 D 144716 gt 0 and XX gt 0 So 11 and 7171 are local minima Tong Zhu Mad 11441114 Fall 21mg The second derivatives test D fxxfyy 7 ny12 144X2y2 716 At 11 and 7171 D 144716 gt 0 and XX gt 0 So 11 and 7171 are local minima At 00 D 716 lt 0 Tony Zhu Mad 11441114 Fall 21mg The second derivatives test D fxxfyy 7 602 144X2y2 716 At 11 and 7171 D 144716 gt 0 and XX gt 0 So 11 and 7171 are local minima At 00 D 716 lt 0 so 0 is a saddle point Tong Zhu Mad 11441114 Fall 21mg The second derivatives test D fxxfyy 7 602 144X2y2 716 At 11 and 7171 D 144716 gt 0 and XX gt 0 So 11 and 7171 are local minima At 00 D 716 lt 0 so 0 is a saddle point Tong Zhu Mad 11441114 Fall 21mg Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Tony Zhu Mad 114mm Fall zuug Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d lt22Y12Z12 Tong Zhu Mad 114mm Fall 2mm Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d x722y712z12 For a point Xyz on the plane X y 7 z 1 there is z X i y 71 Tony Zhu Mad 114mm Fall 2mm Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d Xi22y712z12 For a point Xyz on the plane X y 7 z 1 there is z X i y 71 So the distance function becomes d x722y712xy2 Tong Zhu Mad 114mm Fall 2mm Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d Xi22y712z12 For a point Xyz on the plane X y 7 z 1 there is z X i y 71 So the distance function becomes d x722y712xy2 Consider the function fX7y d2 Xi2 y712xy2 Tong Zhu Mad 114mm Fall 2mm Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d Xi22y712z12 For a point Xyz on the plane X y 7 z 1 there is z X l y 71 So the distance function becomes d x722y712xy2 Consider the function fX7y d2 Xi22y712xy2 Check the first order partial derivatives fx2X22Xy4X2y740 Tong Zhu Mad 114mm Fall 2mm Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d Xi22y712z12 For a point Xyz on the plane X y 7 z 1 there is z X l y 71 So the distance function becomes d x722y712xy2 Consider the function fX7y d2 Xi22y712xy2 Check the first order partial derivatives fx2X22Xy4X2y740 fy2y712xy 2x4y720 Tong Zhu Mad 114mm Fall 2mm Example 2 Find the shortest distance from the point 2171 to the planexyiz 1 Solution The distance from any point Xyz to 21771 is d Xi22y712z12 For a point Xyz on the plane X y 7 z 1 there is z X l y 71 So the distance function becomes d x722y712xy2 Consider the function fX7y d2 Xi22y712xy2 Check the first order partial derivatives fx2X22Xy4X2y740 fy2y712xy 2x4y720 Tong Zhu Mad 114mm Fall 2mm X1y0 There is only one critical point 17 0 Ton Zhu Mad 11 14 Fall ZIIIIQ There is only one critical point 17 0 Check the second order partial derivatives 6a476lty27fyy4 Tong Zhu Mad 114mm Fall 2mm There is only one critical point 17 0 Check the second order partial derivatives ax4ampy2gy4 0ampxfyy7ay21674712gt0 Tong Zhu Mad 114mm Fall 2mm There is only one critical point 17 0 Check the second order partial derivatives ax4ampy2gy4 0ampxfyy7ay21674712gt0 By the Second Derivatives Test 10 is a local minimum So the shortest dista nce is dmin Tong Zhu Mad 114mm Fall 2mm Absolute Maximum and Absolute Minimum Tong Zhu Mad 114mm Fall 2mm Absolute Maximum and Absolute Minimum 0 Absolute Maximum at 37 b if for all X7 y in the domain of f fxy fab Tong Zhu Mad 114mm Fall 2mm Absolute Maximum and Absolute Minimum 0 Absolute Maximum at 37 b if for all X7 y in the domain of f fxy fab 0 Absolute Minimum at 37 b if for all X7y in the domain of f fxy 2 fa7 b Tong Zhu Mad 114mm Fall 2mm Absolute Maximum and Absolute Minimum 0 Absolute Maximum at 37 b if for all X7 y in the domain of f fxy fab 0 Absolute Minimum at 37 b if for all X7y in the domain of f fxy 2 fa7 b zx24y279inthedomain 71 X 37 y Nico Tong Zhu Mad 114mm Fall 2mm Absolute Maximum and Absolute Minimum 0 Absolute Maximum at 37 b if for all X7 y in the domain of f fXy g fab 0 Absolute Minimum at 37 b if for all X7y in the domain of f fXy 2 fab zX24y279inthedomain ilgxg3 iggyg Mia nbsalule maximum mum mmimum Tong Zhu Mall 1144quot Fall lellg Extreme Value Theorem for Functions of Two Variables If f is continuous on a closed bounded set D in R2 then f attains an absolute maximum value fx1y1 and an absolute minimum value fx2y2 at some point thl and X27y2 in D Tong Zhu Mad 114mm Fall zuug Extreme Value Theorem for Functions of Two Variables If f is continuous on a closed bounded set D in R2 then f attains an absolute maximum value fx1y1 and an absolute minimum value fx2y2 at some point thl and X27y2 in D What is a closed bounded set Tong Zhu Mad 114mm Fall zuug Extreme Value Theorem for Functions of Two Variables If f is continuous on a closed bounded set D in R2 then f attains an absolute maximum value fX1y1 and an absolute minimum value fX2y2 at some point X17y1 and X27y2 in D What is a closed bounded set Tong Zhu Math 1144quot Fall lellg Extreme Value Theorem for Functions of Two Variables If f is continuous on a closed bounded set D in R2 then f attains an absolute maximum value fX1y1 and an absolute minimum value fX2y2 at some point X17y1 and X27y2 in D What is a closed bounded set dew Tong Zhu Math 1144quot Fall lellg Extreme Value Theorem for Functions of Two Variables If f is continuous on a closed bounded set D in R2 then f attains an absolute maximum value fX1y1 and an absolute minimum value fX2y2 at some point X17y1 and X27y2 in D What is a closed bounded set closed Xes Tong Zhu Math 1144quot Fall lellg Math 114 quot4 Fall zuug closed7 Math 114 Hill Fall zuug closed7 No Math 114 Hill Fall zuug Definition Closed Set A closed set in R2 is one that contains all its boundary points Tong Zhu Mad 114mm Fall 2mm Definition Closed Set A closed set in R2 is one that contains all its boundary points what is a boundary point Tong Zhu Mad 114mm Fall 2mm Definition Closed Set A closed set in R2 is one that contains all its boundary points what is a boundary point Definition Bounded Set A bounded set in R2 is one that is contained within some disk Or it is finite in extent Tong Zhu Mad 114mm Fall 2mm How to find the extreme values of a two variable function on a closed bounded domain D7 Ton Zhu Mad 11 14 Fall ZIIIIQ How to find the extreme values of a two variable function on a closed bounded domain D7 0 Find the critical points of fx7 y in D ie the points satisfying 1Q f 0 Tony Zhu Mad 114mm Fall 2mm How to find the extreme values of a two variable function on a closed bounded domain D7 0 Find the critical points of fx7 y in D ie the points satisfying 1Q f 0 9 Find the extreme values f on the boundary of D Tong Zhu Mad 114mm Fall 2mm How to find the extreme values of a two variable function on a closed bounded domain D7 0 Find the critical points of fx7 y in D ie the points satisfying 1Q fy 0 9 Find the extreme values f on the boundary of D 9 The largest value from Step 1 and Step 2 is the absolute maximum and the smallest value from Step 1 and Step 2 is the absolute minimum Tong Zhu Mad 114mm Fall 2mm Example 3 Find the absolute maximum and minimum values of X7 X2 y2 X2y 4 on the set D X7Ylle S 17 lYl 1 Tony Zhu Mad 114mm Fall 2mm Example 3 Find the absolute maximum and minimum values of X7 X2 y2 X2y 4 on the set D X7Ylle S 17 lYl 1 Solution First find the critical points of 1 Tony Zhu Mad 114mm Fall 2mm Example 3 Find the absolute maximum and minimum values of fxy X2 ly2 x2y l 4 on the set D X7Yllxl S 17 lYl S 1 Solution First find the critical points of f 2x2xy0 Tong Zhu Mad 114mm Fall 2mm Example 3 Find the absolute maximum and minimum values of X7 X2 y2 X2y 4 on the set D X7Ylle S 17 lYl 1 Solution First find the critical points of f amp2x2xy0 fy2yxz0 Tong Zhu Mad 114mm Fall 2mm Example 3 Find the absolute maximum and minimum values of X7 X2 y2 X2y 4 on the set D X7Ylle S 17 lYl 1 Solution First find the critical points of f amp2x2xy0 fy2yxz0 Tong Zhu Mad 114mm Fall 2mm Example 3 Find the absolute maximum and minimum values of X7 X2 y2 X2y 4 on the set D X7Ylle S 17 lYl 1 Solution First find the critical points of f 1Q 2X l 2xy O fy 2y i X2 O X Oy O X i iy 1 Three critical points 00 771 and 7amp771 Note 771 and 7amp771 are not in D So only one critical point in D 00 f07 O 4 Tony Zhu Mad 114mm Fall 2mm Secondly find the extreme values on the boundary of D Ton Zhu Mad 11 um Fall 2mm Secondly find the extreme values on the boundary of D Where is D7 Ton Zhu Mad 11 um Fall 2mm Secondly find the extreme values on the boundary of D Where is D Secondly find the extreme values on the boundary of D Where is D What are the equations of the four boundaries Secondly find the extreme values on the boundary of D Where is D What are the equations of the four boundaries L1y77171 X 139 Math 114 quot4 Fall zuug L1y7171 X 1 L2X171 y 1 L3y171 X 1 L4X7171 y 1 Tong Zhu Math 1144quot Fall lellg On L1 y 7171 x 1plugy 71 into X7 X2 y2 X2y 4 we get Tong Zhu Mad 114mm Fall zuug On L1 y71 ilgxg 1 plugy71 into X7 X2y2X2y4 we get x71x217x245 Tong Zhu Mad 114mm Fall zuug On L1 y71 ilgxg 1 plugy71 into X710 X2y2X2y4 we get x71x217x245 On L2 X 1 71 y 1 plugx 1 into fxy Tong Zhu Mad 114mm Fall zuug On L1 y71 ilgxg 1 plugy71 into X710 X2y2X2y4 we get x71x217x245 On L2 X 1 71 y 1 plugx 1 into fxy 1 f17Y1y24y2y5 f 2y10y Tong Zhu Mad 114mm Fall zuug On L1 y71 ilgxg 1 plugy71 into X710 X2y2X2y4 we get x71x217x245 On L2 X 1 ilgyg 1 plugx 1 into fxy 1 f17Y1y24y2y5 f 2y10y The value at 17 f17 Tong Zhu Mad 114mm Fall zuug On L1 y71 ilgxg 1 plugy71 into flxd X2y2X2y4 we get x71x217x245 On L2 X 1 ilgyg 1 plugx 1 into fxy 1 f17Y1y24y2y5 f 2y10y The value at 17 f17 We also need to check the boundary of L2 Tong Zhu Mad 114mm Fall zuug On L1 y 1 1 3x3 1 plugy 1 into fXy X2y2x2y4 we get fx 1X21 X245 On L2 X 1 1 gyg 1 plugx 1 into fXy 1 f1y 1y2y4y2y5gt f2y1 0gty E The value at 1 f1 Q We also need to check the 5 4 boundary of L2 f1 1 f1l 7 Why as 71 r 1 Tong Zhu Math 114 004 Fall 2009 On L3y171 X 1 Ton Zhu Mad 11 14 Fall ZIIIIQ On L3y171 X 1 fx1X21X242X25f 4x0x0 Tong Zhu Mad 114mm Fall zuug On L3y171 X 1 fQJJx21x2425gtF4x0 x0 amm4 Tong Zhu Mad 114mm Fall zuug On L3y171 X 1 fx1X21X242X25f 4x0x0 00 4 f711 7 f1 1 7 Tony Zhu Mad 114mm Fall zuug On L3 y171 X 1 fx1X21X242X25f 4x0x0 f00 4 f711 7 f1l7 On L4 X7171 y 1 f71y1y2y4 Same as on L1 7171 5 711 7 On D the absolute maximum is fi11 7 and the absolute minimum is f00 4 Tony Zhu Mad 114mm Fall 2mm Example 4 Find the absolute maximum and minimum VaUes of Xiy 2x3 y4 on the set D X7y xz yz S 1 Solution 1 1 First find the critical points of f in D 26x207 x0 fy4y307 y0 Only one critical point 00 and f00 0 2 On the boundary of D X7ylx2 y2 1 So y2 17 X2 fxy fx 2X317X22X42X372X2171 X 1 7w 4x3 6x2 7 4x 2x2x2 3x 7 2 2x2x 71x 2 0 gtX O or g or 7 2 X72 isnotin D Whenx0yilwhenxyi nib Tong Zhu Mad 1147mm Fall 2mm 1 V3 13 i aii i 0 l l f2 2 16 Note we also need to check the boundary X 71 and X 1 of fXX42X372X217 ilgxgl VVhy 72 Tom Zhu Math 1144quot Fall lellg When Xl1 y0 f1 0 2 f7l 0 72 So absolute max f10 2 and absolute min f717 0 72 Tony Zhu Mad 114mm Fall zuug Solution 2 Use polar coordinates X cos 0 y sin 0 0 g 0 g 27139 fxy 1 09 2cos30 l 17 C05202 cos402cos307 2cos201 f0 74cos30sin0 7 6cos20sin0 4cos0sin0 72sin0cos03cos0 7 2 l 2cos20 72 sin0cos02 cos0 71cos0 2 f 00sin000r cos000r cos0 1 ii X7y i17 0 or X7 y 07 i1 or X7 y E 2 The rest is the same as Solution 1 Tony Zhu Mad 114mm Fall zuug W Math 114 Rimmer Z f x 156 Directional Derivatives and Gradients fx x0y0 is the rate of change of z in the direction parallel to the x axis fy x0y0 is the rate of change of z in the direction parallel to the y axis z4 x2 2y2 slope fx 11 What about the rate of change of z in other directions 39 Math 114 Rimmer Want to nd the rate of change of z at x0 yo zo w 156 Directional Derivatives 39 39 39 39 d G d39 t 1n the d1rect10n of an arb1trary un1t vector u 19 5 quot Va 399quot 5 Surface S with equation z f x y Z0 f 960 yo The vertical plane that passes through P P 960 yo Z0 in the direction of u intersects the S in a curve C The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u Let Q x y z be another point on C maysz Project P and Q onto the xy plane to get P and Q W hu hahbgt hax x0 and hi y yo xx0ha andy y0hb Z fxy fx0hay0hb 7272009

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