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by: Claudine Friesen


Claudine Friesen
GPA 3.56


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Class Notes
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This 15 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 E.So in Fall. Since its upload, it has received 23 views. For similar materials see /class/215394/math114-university-of-pennsylvania in Mathematics (M) at University of Pennsylvania.


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Date Created: 09/28/15
Math 114 Chapter 15 Section 6 Question 28 Shuvra Gupta October 26 2006 Please refer to the question mentioned above Let Vf denote the gradient of f We agreed Vf 2x ycosyi mosey So in particular at the point 10 Vf 2i j Solution I Let u aibj and since u is a unit vector we know xa2 b2 1 which in particular implies a2 b2 1 We also know Duf Vfu 2Z39j az39bj 2a b But we are given that Duf 1 which implies 2a b 1 So b 17 2a Using the fact that a2 b2 1 and substituting the expression we got for b we have a2172a2 1 7 512 7 4a 0 i a 0 g The corresponding values of b are 1 and 7 respectively So the two directions in which the directional derivative equals 1 are j and gi 7 gj I think this method above is the easiest but we can also do it by the other method Solution II Since u is a unit vector we can express u as cos 6i sin 6j where 6 is the angle u makes with the positive x axis Now 7 Duf Vfu 2239 jcos 6i sin 6j 2cos6 sin6 But we know Duf 1 so that implies 2cos6 sin6 1 So sin6 17 2cos6 Squaring both sides we get sin26 17 2cos 62 i sin26 14cos2674cos6 17cos26 14cos2674cos6 5cos2674cos6 0 7 cos 6 0 g The corresponding values of sin 6 are 1 and 7 respectively Because sin6 17 2cos6 So the two directions in which the directional derivative equals 1 are j and gi 7 gj PS One could also solve this problem using the formula Duf Vfu lVfKCOSCvL where 04 is the angle between Vf and u but this method is messier than the above two Hope this helps clarify the confusion in class 135 Equations of Lines and Planes MathllA39R mmquot winwax 135 Equations of Lines and Planes In order to find the equation of a line we need A a point on the line P0 x0 yo zo B a direction vector for the line v 1117 0 r r0 v vector equation of line L x yzgt x0 y0z0gttltabc equating components xx0at y y0bt zz0ct parametric equations of the line L eliminating the parameter I solve for t r0 ltx0 yo 10gt in each then equate the results x x I ltxyz 707 yoziz ZO 4 a b 0 P0P V 1 ab C symmetric equations of the line L Fm Math 1147 Rimmer m 135 Equations of Lines and Planes Find parametric equations of the line containing 513 and 3 24 Math 114 Rimmer 135 Equations of Lina and Plana Two lines in 3 space can interact in 3 ways A Parallel Lines their direction vectors are scalar multiples of each other B Intersecting Line there is a specific I and s so that the lines share the same point C Skew Lines their direction vectors are not parallel and there is no values of t and s that make the lines share the same point Determine whether the lines L1 and L are parallel skeW glm ha and Planes or intersecting If they intersect find the point of intersection L1 12 x 3 7 t x 8 2s y 5 313 y 76 7 4s z 71 7 413 z 5 s set the x39 s This should happen at the same time I so set these equal 1 1 4 3 t 825 5 Zs z S Check to make sure that thez 7t 5 2s values are e ual for this I and s 15 6s 11 4s q t 5 2s 6S44s 1 4t5s Setthey sz 2s4 14152 5 3 6 4 S2 33 check t S Now find the pt of intersection 3t 11 4s t 5 2 2 usingg 711 4s x 3 7 71 t 1 t 3 y 7 5 371 423 z 71 7 471 l Math 1147 Rimmer Determine Whether the lines L1 and l are parallel skew 135 Equations of Lines and Planes or intersecting If they intersect find the point of intersection 11 12 x 7 4 t x 3 2s y 78 7 2t y 71 s z 12t z 3 3S pal Math 11A Rimmer W 135 Equations of Lines and Planes Planes In order to find the equation of a plane we need A a point on the plane P0 x0 yo zo thisvecmiscaued B a vector that is orthogonal to the plane 11 1117 0 the normal View to the plane 11 r r0 0 n a r n r0 vector eguation of the plane nr r0O ax x0by y0czZUO scalar eguation of the plane r0 ltx0ay0azogt x rltxyzgt dX axobybyoczCZU0 4 axbycziax0by0cz00 P0PltXX0 yyo zzogt axbyczd0 n lt11 1 cgt linear eguation of the plane Determine the equation of the plane that contains the lines L1 and IT Math 1147 Rimmer N 135 Equations of Lines and Planes 11 12 x 3 7 t x 8 2S In order to nd the equat1on of a plane we need y 5 3t y 76 7 4s A a point on the plane Z 1 7 4t Z 5 S B a vector that is orthogonal to the plane We have two points in the plane from 113551 and from 218765 We have two vectors in the plane from l1lt7L 34 and from L2 241 nltabc Fill Math 1147 R Immer 135 Equations of Lines and Planes Determine the equation of the plane that passes through 123321 and 4522 1 Math 1147R139mmer Two distinct planes in 3 space either are 13 5E 1 quot 5 quot5 quotd 3quot parallel or intersect in a line Math 114 r Rimmer 135 Equations of Lines and Planes Find the line of intersection of the two planes x 2y z 0 2x 3 y Zz O W Math 114 r Rimmer 135 Equations of Lines and Planes If two planes intersect then you can determine the angle between them 4 between planes 4 between their normal vectors n n COS 6 M 1 lnzl Find the between the planes x 2y z O 2x 3 y Zz O Math 1147lemer Distance between a p01nt and a plane 135mm 0 Lines and Planes lnbl lax1by1cz1dl D b compnl lnl b D Math 114 Rimmer 132 Vectors Quantities such as area volume mass and time can be characterized by a single real variable Other quantities such as displacement velocity and force involve both magnitude and direction To represent these quantities we use a vector represented by a directed line segment arrow B mm 31 The magnitude of a vector pm is represented by Vor a V 2 AB Any other vector u that has the same magnitude and direction as v A lnlLla point is called an equivalent or equal vector 3 u v Math 114 Rimmer 132 Vectors We can multiply a vector by a real number C This is called scalar multiplication cv has a magnitude that is 0 times the magnitude of v cv has has the same direction as v if c gt 0 cv has has the opposite direction as V if c lt 0 3v V 2v 2 V 792009 Math m mm can add a vector v to another vector u animus Thls ls called vector addldon vu Connect the termlnal polnt ofthe flnt vector to the lnltlal polnt ofthe econd vector U I When connected dud way dre um I dre vector from dre mltlal pomt of dre flnt vector to me terrnmal pomt ofthe econd vector V a v Vector subtractlon vru lsjust Vector u addrtron ln drsgursev in V u ans can allbe summed up uslng the parallelogram determlned by v and u v vu v Math m Rimmu39 131mm So far we have studled vectors geometrrcallv We now want to look at vectors algebrarcallv The magnitude ofv is found by 112 12 1 1 the vector is called a unit vector Standard unit Vectors ilt10 14015 Now v can be written as v 792009 Now for 3 dimensions we have iltlv0v0gt v a1ia2ja3k j 01 0 k lt0 0 lgt Vltaigta2gta3 511512 and a3 are called the components of v More specifically a1 the i component of v a2 thej component of v a3 the k component of v Math IN 7 Rimmu Scalar Multiplication Wivn39zvmm V 01 a2 03gt scaled by a factor c cv ltca1ca2ca3 multiply each component by c Vector Addition v a1a2aj added to ultb1b2b3 V a1 b17a2 vaa3 b3 add componentwise pun it Mm J WM xmz 792009 792009 Math 11A Rimmer way 132 Vectols Vector from a pointA to another point B In 3dimensions V ltb1 a1b2 azb3 a3gt Scale the vector v by the reciprocal of V V l uzmz M lt0102703gt Flt y y gt M M M 1 Math 11A Rimmer 0 Find the component form and magnitude of the vector v With the ayuwluvmm initial point 320 and terminal point 415 0 Find 3v 0 Find a unit vector in the direction of v vlt4 3l 25 0gt lt1 15 ori j5k v12 1252 11 25 JE3J 3vlt313 135gt3 315gt or 3i 3j15k ult1 1 5 gtZlt 43 56gt NEE3 9 9 9 Math 114 Rimmer as 3 132 Vectors A 200 lb traffic light supported by two cables hangs in equilibrium As shown in figure b let the weight of the light be represented by w and the forces in the two cables by F1 and F2 As shown in figure c the forces can be arranged to form a triangle Equilibrium implies that the sum of the forces is 0 Find F1 and F2 and find the magnitudes ofF1 and F2 1 F1 ltHF1HCOS200 Flusin2 gt FZHSin15 I F1Hsin20 HF2Hsin15 200gtlt00gt F2 lt HF2Hcos15 W lt0 200gt F1 F2 w ltHF1Hc0s20 HF2Hc0s15 F7 HFzHCOSlly FZ cos 15 W7 cos20 w Zgsm20quotllellsm 15quot200 F cos 15 V V 1 HFZHCOSE20Ds1n20 s1n15 200 W c mew 2F2z32766 lbs HFle33681 lbs insln20 sml5 cos20 Math 114 Rimmer If we were interested in having these forces in Newtons we can just convert may 132 Vectors at the end using the coversion factor 1 lb force x 4448 Newtons Where does this conversion factor come from l 39 lNewtonzlkglm2 sec sec 1 lb force s1 slug So we need to convert slugs into kg and ft into meters sec2 1 lb force s1 14 5 9 kg 7033 m 24448221604 kg m 3 1 lb force 4448 Newtons 792009 Math 114 004 Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania Septem ber 15 2009 En Tong Zhu Mad 114mm Fall 2mm


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