MA1104 Multivariable Calculus Lecture Notes 1. Wong Yan Loi

Transcription

1 MA114 Multivrible lculus Lecture Notes 1 Wong Yn Loi eprtment of Mthemtics, Ntionl Universit of ingpore, ingpore This notes is written eclusivel for students tking the modules MA114, Multivrible lculus, t the Ntionl Universit of ingpore The contents follow closel the reference [1]

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3 ontents 1 Nottions 4 2 Vectors in R linders nd Qudric urfces 17 4 lindricl nd phericl oordintes 2 5 Vector Functions 23 6 Functions of severl vribles 27 7 Limits nd ontinuit 3 8 Prtil erivtives 34 9 Mimum nd Minimum Vlues 45 1 Lgrnge Multipliers Multiple Integrls urfce Are Triple Integrls 6 14 hnge of Vribles in Multiple Integrls Vector Fields 7 16 Line Integrls Line Integrls of Vector Fields The Fundmentl Theorem for Line Integrls Independence of Pth 76 2 Green s Theorem The url nd ivergence of Vector Field Prmetric urfces nd their Ares 9 23 Oriented urfces urfce Integrls of Vector Fields tokes Theorem The ivergence Theorem Further Eercises 15 Bibliogrph 17 3

4 1 Nottions The collection of ll rel numbers is denoted b R Thus R includes the integers emester I (213/14), 2, 1,, 1, 2, 3, the rtionl numbers, p/q, where p nd q re integers (q ), nd the irrtionl numbers, like 2, π, e, etc Members of R m be visulied s points on the rel-number line s shown in Figure Figure The Number Line 2 2 e 3 π 4 We write R to men is member of the set R In other words, is rel number Given two rel numbers nd b with < b, the closed intervl [, b] consists of ll such tht b, nd the open intervl (, b) consists of ll such tht < < b imilrl, we m form the hlf-open intervls [, b) nd (, b] The bsolute vlue of number R is written s nd is defined s = { if if < For emple, 2 = 2, 2 = 2 ome properties of re summried s follows: 1 = for ll R 2, for ll R 3 For fied r >, < r if nd onl if ( r, r) 4 2 =, R 5 (Tringle Inequlit) + + for ll, R A function f : A B is rule tht ssigns to ech A one specific member f() of B The fct tht the function f sends to f() is denoted smbolicll b f() For emple, f() = 2 /(1 ) ssigns the number 2 /(1 ) to ech 1 in R We cn specif function f b giving the rule for f() The set A is clled the domin of f nd B is the codomin of f The rnge of f is the subset of B consisting of ll the vlues of f Tht is, the rnge of f = {f() B A} Given f : A R, it mens tht f ssigns vlue f() in R to ech A uch function is clled rel-vlued function For rel-vlued function f : A R defined on subset A of R, the grph of f consists of ll the points (, f()) in the -plne c eprtment of Mthemtics 4

5 emester I (213/14) grph of f (, f()) A = domin of f Figure 2 The grph of f Eercise 11 Let r > Prove tht < r if nd onl if ( r +, + r) Eercise 12 Prove the tringle inequlit + + Eercise 13 Prove tht for n, R, 2 Vectors in R 3 21 The Eucliden 3-spce The Eucliden 3-spce denoted b R 3 is the set {(,, ),, R} To specif the loction of point in R 3 geometricll, we use right-hnded rectngulr coordinte sstem, in which three mutull perpendiculr coordinte es meet t the origin It is common to use the nd es to represent the horiontl coordinte plne nd the -is for the verticl height (,, ) (, ) Figure 3 A right-hnded coordinte sstem We usull denote point P with coordintes (,, ) b P (,, ) The distnce P 1 P 2 between two points P 1 ( 1, 1, 1 ) nd P 2 ( 2, 2, 2 ) is given b (1 2 ) 2 + ( 1 2 ) 2 + ( 1 2 ) 2 An eqution in,, describes surfce in R 3 c eprtment of Mthemtics 5

6 emester I (213/14) Emple 21 () = 3 is the eqution of horiontl plne t level 3 bove the -plne (b) = 2 is the eqution of verticl plne prllel to the -coordinte plne Ever point of this plne hs coordinte equl to 2 (c) imilrl = 2 is the eqution of verticl plne prllel to the -coordinte plne = 3 3 = = 2 () (b) Figure 4 (c) Emple 22 An eqution of sphere with centre O(, b, c) nd rdius r is ( ) 2 + ( b) 2 + ( c) 2 = r 2 r O P Figure 5 A sphere Eercise 23 how tht = is the eqution of sphere escribe its intersection with the plne = 1 olution Using the method of completing squre, the given eqution cn be written s ( + 2) 2 + ( 3) 2 + ( + 1) 2 = 8 Hence, it is the eqution of sphere centred t ( 2, 3, 1) with rdius 8 c eprtment of Mthemtics 6

7 emester I (213/14) ( + 2) 2 + ( 3) 2 = 4, = 1 is circle O Figure 6 A circle ling on sphere To find the intersection with the plne = 1, set = 1 in the bove eqution We obtin ( + 2) 2 + ( 3) 2 = 4 Therefore, it is circle ling on the horiontl plne = 1 with centre t ( 2, 3, 1) nd rdius 2 22 Vectors A 3-dimensionl vector is n ordered triple = 1, 2, 3 of rel numbers 1, 2, 3 re clled the components of A vector = 1, 2, 3 cn be represented b n rrow from n point P (,, ) to the point Q( + 1, + 2, + 3 ) in R 3 In this cse, we s tht the vector = 1, 2, 3 hs representtion P Q P (,, ) Q( + 1, + 2, + 3 ) Figure 7 Vector Insted of using n rrow on top of P Q or, we shll suppress the rrow but write PQ or in bold to denote the vector = P Q If P is the origin O, is clled the position vector of the point Q The position vectors of (1,, ), (, 1, ) nd (,, 1) re denoted b i, j nd k respectivel In other word, i = 1,,, j =, 1, nd k =,, 1 i, j nd k re clled the stndrd bsis vectors Therefore, if Q = (,, ), then q =,, = i + j + k uppose P = ( 1, 1, 1 ) nd Q = ( 2, 2, 2 ) The mgnitude of vector PQ is defined to be PQ = ( 2 1 ) 2 + ( 2 1 ) 2 + ( 2 1 ) 2 A vector PQ lso hs direction determined b the orienttion tht the rrow is pointing The ero vector,, is denoted b lerl = We s tht two vectors re equl if nd onl if the hve the sme direction nd the sme mgnitude This condition m be epressed lgebricll b sing tht if v 1 = 1, 1, 1 nd v 2 = 2, 2, 2, then v 1 = v 2 if nd onl if 1 = 2, 1 = 2 nd 1 = 2 If v 1 = 1, 1, 1 nd v 2 = 2, 2, 2, then define the sum v 1 + v 2 to be the vector 1 + 2, 1 + 2, If λ is n rel number nd v =,,, then define the sclr multiple λv to be the vector λ, λ, λ c eprtment of Mthemtics 7

8 emester I (213/14) Also v is defined to be ( 1)v lerl ( v) = v, nd v +v = Also u v = u+( 1)v Addition of vectors cn lso be described b the prllelogrm lw: The sum v 1 + v 2 is represented b the position vector which is the digonl of the prllelogrm determined b v 1 nd v 2 v 2 v 1 v 1 + v 2 O v 2 Figure 8 Vector Addition It is strightforwrd to check tht the set of ll position vectors in R 3 forms vector spce over R Eercise 24 Prove the tringle inequlit v 1 + v 2 v 1 + v 2 Proposition 25 Properties of vectors 1 + b = b (b + c) = ( + b) + c 3 + = 4 + = 5 α( + b) = αb + α 6 α = α 7 (α + β) = α + β 8 (αβ) = α(β) 9 1 = 1 α = α efinition 26 A unit vector is vector whose length is 1 1 For n nonero vector, = is unit vector tht hs the sme direction s ometimes, in order specif vector is of unit length, it is written s â Emple 27 Find the unit vector in the direction of the vector 2i j 2k olution 2i j 2k = ( ( 1) 2 + ( 2) 2) 1 2 (2i j 2k) is 1 3 = 9 = 3 Therefore the required unit vector 23 The ot Product efinition 28 Let = 1, 2, 3 nd b = b 1, b 2, b 3 The dot product or sclr product of nd b is the number b = 1 b b b 3 Emple 29 Let = 1, 2, 3 nd b = 1,, 1 Find b olution b = (1)( 1) + (2)() + (3)( 1) = 4 lerl, we hve i j = i k = j k = nd i i = j j = k k = 1 c eprtment of Mthemtics 8

9 emester I (213/14) Proposition 21 Properties of the ot Product 1 = 2 2 b = b 3 (b + c) = b + c 4 (α) b = α( b) = (αb) 5 = Proof Let s prove 1 Let = 1, 2, 3 Then = = 2 Theorem 211 If θ is the ngle between the vectors nd b, then b = b cos θ, θ π Proof Let OA = nd OB = b, where O is the origin nd θ = AOB -b Ḅ A b θ Figure 9 Angle between two vectors Appling cosine rule to OAB, we hve O b 2 = 2 + b 2 2 b cos θ As b 2 = ( b) ( b) = 2 2 b + b 2, it follows tht b = b cos θ or cos θ = b b Two vectors nd b re sid to be orthogonl or perpendiculr if the ngle between them is 9 In other words, nd b re orthogonl b = Emple 212 2i + 2j k is orthogonl to 5i 4j + 2k becuse (2i + 2j k) (5i 4j + 2k) = (2)(5) + (2)( 4) + ( 1)(2) = Let = 1, 2, 3 = The ngles α, β, γ in [, π] tht mkes with the,, es respectivel re clled the direction ngles of = 1, 2, 3 γ α β Figure 1 irection Angles c eprtment of Mthemtics 9

10 emester I (213/14) The cosines of these ngles, cos α, cos β, cos γ re clled the direction cosines of We m epress vector = 1, 2, 3 in terms of its mgnitude nd the direction cosines imilrl, Thus, cos α = i i = 1, 2, 3 1,, 1, 2, 3 1,, = 1 cos β = 2 nd cos γ = 3 = cos α, cos β, cos γ Net, we shll discuss the projection of vector long nother vector Let nd b be two vectors in R 3 Let s represent s PQ nd b s PR R b θ P b cos θ Q Then Figure 11 b cos θ = b Vector Projection = b efinition The sclr projection of b onto is b cos θ = ( ) b 2 The vector projection of b onto is b = b 2 Note tht the sclr projection is negtive if θ > 9 Moreover, in figure 11 R = PR P = b b Thus the distnce from R to the line P Q is given b 2 R = b b 2 Emple 214 Find the sclr nd vector projection of b = 1, 1, 2 onto = 2, 3, 1 olution = ( 2) = 14 Thus the sclr projection of b onto is b 1 14 (( 2)(1) + (3)(1) + (1)(2)) = 3/ 14 The vector projection of b onto is 3 14 = 3 14 = 3 7, 9 14, 3 14 Eercise 215 Find the ngle between two long digonls of unit cube = [75 ] Eercise 216 Prove the uch-chwr inequlit: b b etermine when equlit holds c eprtment of Mthemtics 1

11 24 The ross Product emester I (213/14) efinition 217 If = 1, 2, 3 nd b = b 1, b 2, b 3, then the cross product or vector product of nd b is b = 2 b 3 3 b 2, 3 b 1 1 b 3, 1 b 2 2 b 1 i j k = b 1 b 2 b 3 = 2 3 b 2 b 3 i 1 3 b 1 b 3 j b 1 b 2 k Emple 241 Let = 1, 3, 4 nd b = 2, 7, 5 Find b olution lerl, we hve i j k b = = i j k = 43i + 13j + k i j = k, j k = i, k i = j nd i i = j j = k k = Theorem 218 Let = 1, 2, 3, b = b 1, b 2, b 3 nd c = c 1, c 2, c 3 Then (b c) = b 1 b 2 b 3 c 1 c 2 c 3 Proof Eercise orollr 219 b c is perpendiculr to both b nd c Proof b (b c) = c (b c) = b 1 b 2 b 3 b 1 b 2 b 3 c 1 c 2 c 3 c 1 c 2 c 3 b 1 b 2 b 3 c 1 c 2 c 3 = = Theorem 22 If θ is the ngle between nd b, θ π, then b = b sin θ Proof First we need the following identit ( 2 b 3 3 b 2 ) 2 +( 3 b 1 1 b 3 ) 2 +( 1 b 2 2 b 1 ) 2 = ( )(b 2 1 +b 2 2 +b 2 3) ( 1 b b b 3 ) 2 which cn be esil verified b direct simplifiction of both sides Using this identit, we hve b 2 = ( 2 b 3 3 b 2 ) 2 + ( 3 b 1 1 b 3 ) 2 + ( 1 b 2 2 b 1 ) 2 = ( )(b2 1 + b2 2 + b2 3 ) ( 1b b b 3 ) 2 = 2 b 2 ( b) 2 = 2 b 2 2 b 2 cos 2 θ = 2 b 2 sin 2 θ c eprtment of Mthemtics 11

12 emester I (213/14) ince θ π, sin θ, we hve b = b sin θ It follows from this result tht b = b sin θ is the re of the prllelogrm determined b nd b b θ Figure 12 Are= b sin θ b is vector perpendiculr to the plne spnned b nd b with mgnitude b sin θ, where θ π is the ngle between nd b b θ Figure 13 b There re two possible choices of such vector It is the one determined b the right-hnd rule: b is directed so tht right-hnd rottion bout b through n ngle θ will crr to the direction of b To see this, first observe tht the cross product is independent of the choice of the coordinte sstem ee Eercise 229 To determine the direction of b, choose the -is long the direction of nd choose the -is so tht the vector b lies on the -plne nd let be the is perpendiculr to the -plne so tht,, form right-hnded coordinte sstem With this choice of coordinte sstem, = 1,, with 1 >, nd b = b 1, b 2, Thus, b = 1 b 2 k Therefore, the direction of b is long the -is nd it is long the positive or negtive direction of k ccording to whether b 2 is positive or negtive respectivel This is precisel the right-hnd rule described in the lst prgrph orollr 221 nd b re prllel if nd onl if b = Proposition 222 Properties of the ross Product 1 b = b 2 (α) b = α( b) = (αb) 3 (b + c) = b + c 4 ( + b) c = c + b c 5 (b c) = ( b) c 6 (b c) = ( c)b ( b)c Proof Eercise The reltion (b c) = ( b) c cn be proved b direct epnsion in component form Alterntivel, it cn be deduced b the propert of the determinnt: If two rows of determinnt re switched, the determinnt chnges sign Therefore, c eprtment of Mthemtics 12 b

13 emester I (213/14) (b c) = b 1 b 2 b 3 c 1 c 2 c 3 = c 1 c 2 c 3 b 1 b 2 b 3 = c 1 c 2 c b 1 b 2 b 3 = c ( b) = ( b) c In fct, (b c) = b 1 b 2 b 3 is the lgebric or sign volume of the prllelepiped c 1 c 2 c 3 determined b, b, c b c Figure 14 h = cos θ θ c b Volume = b c cos θ = (b c) Are of the bse prllelogrm = b c orollr 223 The vectors,b,c re coplnr (ie the ll lie on plne) if nd onl if (b c) = b c Figure 15,b,c re coplnr Emple 224 how tht the vectors = 1, 4, 7, b = 2, 1, 4, c =, 9, 18 re coplnr olution As (b c) = =, it follows from 222 tht, b nd c re coplnr 9 18 Emple 225 uppose rigid bod rottes with ngulr velocit w bout n is l through point O The ngulr velocit w is represented b vector long l If r is the position vector from O of point inside the rigid bod, then the velocit t this point is given b w r Eercise 226 how tht (b c) + b (c ) + c ( b) = Eercise 227 how tht ( b) (c d) = ( (c d)) b (b (c d)) = ( (b d)) c ( (b c)) d Eercise 228 uppose the vectors,b,c,d re coplnr how tht ( b) (c d) = Eercise 229 Let P = (p ij ) be n 3 3 orthogonl mtri (P t = P 1 ) with determinnt 1 Let {e 1, e 2, e 3 } be n orthonorml bsis of R 3 Let e j = p 1je 1 + p 2j e 2 + p 3j e 3, j = 1, 2, 3 uppose = 1 e e e 3 = 1 e e e 3 nd b = b 1e 1 + b 2 e 2 + b 3 e 3 = b 1 e 1 + b 2 e 2 + b 3 e 3 Prove tht e 1 e 3 e b 1 b 2 b 3 = e 1 e 3 e b 1 b 2 b 3 c eprtment of Mthemtics 13

14 emester I (213/14) Tht is if we denote the cross products with respect to the bses {e 1, e 2, e 3 } nd {e 1, e 2, e 3 } b nd respectivel, then b = b 25 Lines nd Plnes Let L be line pssing through point P (,, ) in the direction of the vector v =, b, c Then n point P on L hs position vector r = r + tv for some t R P v,, = r r P L Figure 16 Vector eqution of line Vector eqution of line: r = r + tv Write r =,, Then r = r + tv is equivlent to,, =,, + t, b, c Prmetric equtions of line: = + t, = + bt, = + ct Eliminting t, we obtin mmetric equtions of line: = b = c The numbers, b, c re clled the direction numbers of the stright line If, b or c is ero, we m still write the smmetric eqution of the line For emple, if =, we shll write the smmetric equtions s =, which is line ling on the plne = Emple 23 how tht the lines b =, c L 1 : = 1 + t, = 2 + 3t, = 4 t, L 2 : = 2s, = 3 + s, = 3 + 4s, re skew, ie the do not intersect nd re not prllel Hence the do not lie in the sme plne olution L 1 nd L 2 re not prllel becuse the corresponding vectors 1, 3, 1 nd 2, 1, 4 re not prllel The lines L 1 nd L 2 intersect if nd onl if the sstem 1 + t = 2s 2 + 3t = 3 + s 4 t = 3 + 4s hs (unique) solution in s nd t The first two equtions give t = 11/5, s = 8/5 But these vlues of t nd s do not stisf the lst eqution Thus, L 1 nd L 2 do not intersect L 2 L 1 Figure 17 kew Lines c eprtment of Mthemtics 14

15 emester I (213/14) onsider plne in R 3 pssing through point P (,, ) with norml vector n Let P (,, ) be point on the plne Let r nd r be the position vectors of P nd P respectivel n P r r r P r Figure 18 n (r r ) = Then vector eqution of the plne is given b n (r r ) = If n =, b, c, then the bove vector eqution cn be written s ( ) + b( ) + c( ) = In generl, liner eqution in,,, ie + b + c + d = is n eqution of plne in R 3 Emple 231 Find n eqution of the plne pssing through the points P (1, 3, 2), Q(3, 1, 6) nd R(5, 2, ) olution P Q= 3 1, 1 3, 6 2 = 2, 4, 4 P R= 4, 1, 2 Thus norml vector n to the plne is given b P Q P i j k R= = 12, 2, Therefore, n eqution of the plne is given b 1, 3, 2 12, 2, 14 = Tht is = 5 Eercise 232 () Find the ngle θ, ( θ 9 ) between the plnes + + = 1 nd = 1 (b) Find the smmetric equtions for the line of intersection of the plnes in () Proposition 233 The distnce from point P 1 ( 1, 1, 1 ) to the plne + b + c + d = is 1 + b 1 + c 1 + d 2 + b 2 + c 2 Proof Pick point P (,, ) on the plne Let b = P P 1 = 1, 1, 1 c eprtment of Mthemtics 15

16 emester I (213/14) P b N P 1 n Then Figure 19 istnce from point to plne NP 1 = projection of b long n = n b n = ( 1 )+b( 1 )+c( 1 ) 2 +b 2 +c 2 = ( 1+b 1 +c 1 ) ( +b +c ) 2 +b 2 +c 2 = 1+b 1 +c 1 +d 2 +b 2 +c 2 Emple 234 Find the distnce between the prllel plnes = 5 nd 5+ = 1 olution The plnes re prllel becuse their norml vectors 1, 2, 2 nd 5, 1, 1 re prllel Pick n point on the plne = 5 For emple, (1/2,, ) is point on = 5 Then the distnce between the two plnes is 5(1/2) + (1) + ( 1) ( 1) 2 = Emple 235 Find the distnce between the skew lines: 3 6 L 1 : = 1 + t, = 2 + 3t, = 4 t L 2 : = 2s, = 3 + s, = 3 + 4s olution As L 1 nd L 2 re skew, the re contined in two prllel plnes respectivel A norml to these two prllel plnes is given b i j k = 13, 6, Let s = in L 2 We get the point (, 3, 3) on L 2 Therefore, n eqution of the plne contining L 2 is, 3, ( 3) 13, 6, 5 = Tht is = Let t = in L 1 We get the point (1, 2, 4) on L 1 Thus, the distnce between L 1 nd L 2 is give b 13(1) 6( 2) 5(4) + 3 = ( 6) 2 + ( 5) 2 23 Eercise 236 Find the eqution of the stright line pssing through the point P (1, 5, 1) nd perpendiculr to the lines L 1 : = 5 + t, = 1 t, = 2t nd L 2 : = 11t, = 7t, = 2t [ 1 2 = 5 4 = +1 3 ] c eprtment of Mthemtics 16

17 3 linders nd Qudric urfces emester I (213/14) A clinder is surfce tht consists of ll lines (clled rulings) tht re prllel to given line nd pss through plne curve Emple 31 A prbolic clinder = 2 Figure 2 A prbolic clinder Emple 32 irculr clinders Figure = = 1 A qudric surfce is the grph of second degree eqution in,, : A 2 + B E + F + G + H + I + J = Using trnsltion nd rottion, the eqution cn be epressed in one of the following two stndrd forms: A 2 + B J = nd A 2 + B 2 + I = Emple 33 The grph of the eqution = 1 is n ellipsoid c eprtment of Mthemtics 17

18 emester I (213/14) (,, 2) (, 3, ) (1,, ) Figure = 1 The verticl trces (or sections) in = k re ellipses: = 1 k2, where 1 < k < 1 The verticl trces in = k re ellipses: = 1 k2 9, where 3 < k < 3 The horiontl trces in = k re lso ellipses: = 1 k2 4, where 2 < k < 2 Emple 34 The grph of the eqution = is n ellipticl prboloid Figure 23 = The horiontl trces in = k re ellipses: = k, where k > The verticl trces in = k re prbols: = 2 + 4k 2 imilrl, the verticl trces in = k re prbols: = k 2 Emple 35 ketch the surfce = 1 Figure = 1 The trces in = k re ellipses: = 1 + k2 4 c eprtment of Mthemtics 18

19 emester I (213/14) The trces in = k re hperbols: = 1 k2 4 The trces in = k re hperbols: = 1 k2 1 < k < 1 Figure 25 k = 1 Verticl Trces in = k of = 1 Emple 36 Identit nd sketch the surfce = The eqution cn be rewritten in the stndrd form: hperboloid of 2 sheets long the direction of the -is k < 1 or k > 1 = 1 It is therefore Figure = Emple 37 lssif the qudric the surfce = (3, 1, ) Figure = B the method of completing squres, the eqution cn be written s 1 = ( 3) If we mke chnge of coordintes: = 3, = 1, = so tht the new origin is t (3, 1, ), then the eqution becomes = Therefore it is n elliptic prboloid with verte t (3, 1, ) Eercise 38 escribe the trces of the surfce = c eprtment of Mthemtics 19

20 emester I (213/14) Eercise 39 ketch the surfce = 4 Eercise 31 Find the eqution of the surfce obtined b rotting the curve = 2 bout the -is The grphs of the qudric surfces re summried in the following tble b c 2 = 1 The trces re ellipses When = b = c, it is sphere 2 c 2 = b 2 Horiontl trces re ellipses Verticl trces in = k nd = k re hperbols for k but pirs of lines for k = ellipsoid Elliptic Prboloid c = b 2 Horiontl trces re ellipses Verticl trces re prbols ouble cone b 2 2 c 2 = 1 Horiontl trces re ellipses Verticl trces re hperbols Hperboloid of one sheet Hperbolic Prboloid c = b 2 Horiontl trces re hperbols Verticl trces re prbols Here c < b c 2 = 1 Horiontl trces in = k re ellipses if k > c or k < c Verticl trces re hperbols The two minus signs indicte two sheets Hperboloid of Two sheets Tble 1 4 lindricl nd phericl oordintes 41 Polr oordintes Given point P with rtesin coordintes (, ), we m use its distnce r from the origin nd the ngle θ mesured in the counterclockwise sense mde with the -is to locte its position This gives the polr coordintes (r, θ) of P c eprtment of Mthemtics 2

21 emester I (213/14) r θ O (, ) Figure 28 Polr coordinte The reltions between rtesin nd polr coordintes re given b the following formuls: r 2 = 2 + 2, tn θ = = r cos θ, = r sin θ The convention is tht θ is positive if it is mesured in the counterclockwise sense, nd is negtive otherwise If r <, the rdius is mesured t the sme distnce r from the origin, but on opposite side of the origin Notice tht ( r, θ) represents the sme point s (r, θ + π) For emple, for the polr coordintes of the point Q below, we m write either ( 1, π 4 ) or (1, 5π 4 ) O 1 π 4 P (1, π 4 ) ( 1, π 4 ) = (1, 5π 4 ) Q Figure 29 Polr coordintes of P = (1, π 4 ) Eercise 41 Find the eqution in polr coordintes of the curve = 2 [Answer : r = 2 cos θ] 42 lindricl oordintes Given point P with rtesin coordintes (,, ) in 3-dimensionl spce, we m use the polr coordintes (r, θ) for the position of the foot of the perpendiculr from P onto the -plne Then the triple (r, θ, ) determines the position of P, it is clled the clindricl coordintes of P c eprtment of Mthemtics 21

22 emester I (213/14) P (r, θ, ) θ (r, θ, ) Figure 3 lindricl coordintes The reltions between rtesin nd clindricl coordintes re given b the following formuls: r 2 = 2 + 2, tn θ =, = = r cos θ, = r sin θ, = Emple 42 The surfce whose eqution in clindricl coordintes is = r is double cone with the origin s the verte O P (r, θ, ) Q Figure 31 A double cone Let P be point on this surfce with clindricl coordintes (r, θ, ) ince = r, the tringle OP Q in which OQP is right ngle is isosceles with OQ = r = = P Q Thus the cone opens up n ngle of 45 with the -is To convert the eqution to rtesin form, we cn squre both sides of = r, thus 2 = is the rtesin eqution of the double cone If we tke positive squre root on both sides, the grph of the resulting eqution = is the inverted cone on the upper hlf spce Eercise 43 Find the eqution of the ellipsoid = 1 in clindricl coordintes [Answer : 2 = 1 4r 2 ] 43 phericl oordintes Another coordinte sstem in 3-dimensionl spce is the sphericl coordinte sstem Given point P (,, ) with P the foot of the perpendiculr from P onto the -plne, let ρ be its distnce from the origin, θ the ngle tht OP mkes with the -is nd φ the ngle tht OP mkes with the -is Here θ is mesured in the counterclockwise sense from the -is with θ 2π, nd φ is mesured from the -is with φ π Note tht OP = ρ sin φ nd P P = ρ cos φ c eprtment of Mthemtics 22

23 emester I (213/14) φ θ ρ P (ρ, θ, φ) P (r, θ, ) Figure 32 phericl coordintes ρ, θ π The reltions between rtesin nd sphericl coordintes re given b the following formuls: ρ = , cos φ = ρ, φ π cos θ = ρ sin φ = ρ sin φ cos θ, = ρ sin φ sin θ, = ρ cos φ Emple 44 The point (, 2 3, 2) is in rtesin coordintes Find the sphericl coordintes of this point olution First, we hve ρ = 2 + (2 3) 2 + ( 2) 2 = 4 Net, cos φ = /ρ = 1/2 As φ π, we hve φ = 2π/3 Lstl, cos θ = /(ρ sin φ) = Thus, θ = π/2 or 3π/2 As = 2 3 >, θ 3π/2 Tht is θ = π/2 Therefore the sphericl coordintes of the point is (4, π/2, 2π/3) Emple 45 The surfce whose eqution in sphericl coordintes is ρ = R, where R is positive constnt, is sphere of rdius R centred t the origin Emple 46 Find the rtesin eqution of the surfce whose eqution in sphericl coordintes is ρ = sin θ sin φ olution = ρ 2 = ρ sin θ sin φ = ompleting squres, we hve 2 +( 1 2 )2 + 2 = 1 4 Therefore, the surfce is sphere with centre (, 1 2, ) nd rdius Vector Functions efinition 51 A vector function r(t) is function whose domin is set of rel numbers nd whose rnge is set of vectors In other word, f, g, h re clled the component functions of r r(t) = f(t), g(t), h(t) = f(t)i + g(t)j + h(t)k Emple 52 onsider the vector function r(t) = t 3, ln(3 t), t For ech of the component functions to be defined, we must hve 3 t > nd t Thus the domin of r is [, 3) The imge of r trces out curve in R 3 In generl if r(t) = f(t), g(t), h(t) is vector function, then = f(t), = g(t), = h(t) give the prmetric equtions of curve in R 3 c eprtment of Mthemtics 23

24 R t r r(t) R 3 emester I (213/14) Figure 33 A vector function Emple 53 The vector function r(t) = 1 + t, 2 + 5t, 1 + 6t defines curve which is stright line in R 3 Emple 54 ketch the curve whose vector eqution is r(t) = cos t, sin t, t olution The prmetric equtions of the curve re = cos t, = sin t, = t onsider point P (,, ) on this curve ince the, nd coordintes of P stisf the reltion = 1, it lies on the clinder = 1 Figure 34 A heli Moreover, P lies directl bove the point (,, ), which moves counterclockwise round the circle = 1 ince = t, the curve spirls upwrd round the clinder s t increses The curve is Heli Emple 55 Find the vector function tht represents the curve of intersection of the clinder = 1 nd the plne + = 2 olution ince lies on the clinder which projects onto the circle = 1 on the -plne, we cn write = cos t, = sin t with t 2π ince lso lies on the plne, its,, coordintes should stisf the eqution of the plne Thus, = 2 = 2 sin t onsequentl, the vector eqution of is r(t) = cos t, sin t, 2 sin t c eprtment of Mthemtics 24

25 emester I (213/14) + = = 1 Figure 35 An ellipse The curve is n ellipse with centre (,, 2) nd it inclines t n ngle 45 to the horiontl plne Let r(t) = f(t), g(t), h(t) The limit of r(t) s t tends to is defined b: lim t r(t) = lim t f(t), lim g(t), lim h(t) t t Emple 56 Let r(t) = 1 + t 3, te t, sin t t Find lim r(t) t olution lim r(t) = lim 1 + t 3, lim te t sin t, lim = 1,, 1 t t t t t efinition 57 A vector function r(t) is continuous t t = if lim r(t) = r() t Tht is r(t) = f(t), g(t), h(t) is continuous t if nd onl if f(t), g(t), h(t) re continuous t 51 erivtive of vector function Given vector function r(t) Its derivtive is defined b: dr dt = r(t + h) r(t) r (t) = lim h h P PQ = r(t + h) r(t) Q P r (t) Q r(t+h) r(t) h r(t) r(t + h) O O Figure 36 erivtive of vector function If r (t) eists nd is nonero, we cll it tngent vector to the curve defined b r(t) t the point P ee figure 36 In this cse, T(t) = r (t)/ r (t) is clled the unit tngent vector Theorem 58 Let r(t) = f(t), g(t), h(t), where f, g, h re differentible functions of t Then r (t) = f (t), g (t), h (t) = f (t)i + g (t)i + h (t)k c eprtment of Mthemtics 25

26 emester I (213/14) Emple 59 Let r(t) = 1 + t 3, 2t, 1 Find the unit tngent vector to the curve defined b r(t) t the point where t = olution First we hve r (t) = 3t 2, 2, Thus, r () =, 2, = 2j Therefore, T() = 2j 2 = j Emple 51 Find prmetric equtions for the tngent line l to the heli with prmetric equtions = 2 cos t, = sin t, = t t t = π 2 olution l (, 1, π 2 ) Figure 37 The tngent to the heli The vector eqution of the heli is r(t) = 2 cos t, sin t, t Thus, r (t) = 2 sin t, cos t, 1 nd r ( π 2 ) = 2,, 1 is tngent vector to the heli t t = π 2 Therefore, the prmetric equtions of the tngent line l re given b: = + ( 2)t, = 1 + ()t, = π 2 + (1)t Tht is = 2t, = 1, = π 2 + t Given vector function r(t), we m compute successivel r (t), r (t), r (t) etc, provided the eist Theorem 511 Let u nd v be differentible vector functions of t, c sclr nd f relvlued function Then we hve the followings: d 1 dt (u(t) + v(t)) = u (t) + v (t) d 2 dt (cu(t)) = cu (t) d 3 dt (f(t)u(t)) = f (t)u(t) + f(t)v (t) d 4 dt (u(t) v(t)) = u (t) v(t) + u(t) v (t) d 5 dt (u(t) v(t)) = u (t) v(t) + u(t) v (t) 6 (hin Rule) d dt (u(f(t))) = f (t)u (f(t)) Eercise 512 uppose r(t) = c, where c is positive constnt how tht r(t) is orthogonl to r (t) for ll t Let r(t) = f(t)i + g(t)j + h(t)k be continuous vector function The definite integrl of r(t) from t = 1 to t = b is defined s: b ( b r(t)dt = ) ( b ) ( b f(t)dt i + g(t)dt j + Emple 513 Let r(t) = 2 cos ti + sin tj + 2tk Find π 2 r(t)dt ) h(t)dt k c eprtment of Mthemtics 26

27 emester I (213/14) olution π 2 r(t)dt = [2 sin t] π 2 i [cos t] π 2 j + [ t 2] π 2 k = 2i + j + π2 4 k 6 Functions of severl vribles 61 Functions of 2 vribles efinition 61 A function f of 2 vribles is rule tht ssigns to ech ordered pir of rel numbers (, ) in set unique rel number denoted b f(, ) Here is clled the domin of f The set of vlues tht f tkes on is clled the rnge of f Tht is Rnge of f = {f(, ) (, ) } We usull write = f(, ) to indicte tht is function of nd Moreover,, re clled the independent vribles nd is clled the dependent vrible R f(, ) (, ) f Figure 38 f : R Emple 62 Find the domin of f(, ) = ln( 2 ) olution The epression ln( 2 ) is defined onl when 2 > Tht is 2 > The curve 2 = seprtes the plne into two regions, one stisfing the inequlit 2 >, the other stisfing 2 < To find out which region is determined b the inequlit 2 > Pick n point in one of the regions nd test whether it stisfies the inequlit If it does, then b connectivit, tht whole region is the one stisfing 2 >, otherwise, it must be the other region For emple, pick the point (3, 2) ince 2 2 > 3, the region stisfing 2 > is the one contining (3, 2) Thus, domin of f is {(, ) R 2 2 > } 2 = (3, 2) Figure 39 omin of ln( 2 ) Emple 63 Find the domin nd rnge of g(, ) = olution The domin of g is {(, ) R } = {(, ) R } which is circulr disk of rdius 3 ince g(, ) = , the rnge of g lies in [,3] lerl ever number in [,3] cn be epressed s g(, ) for certin (, ) Therefore the rnge of g is the intervl [,3] c eprtment of Mthemtics 27

28 emester I (213/14) efinition 64 Let f be function of 2 vribles with domin The grph of f is the set of ll points (,, ) R 3 such tht = f(, ) f(, ) (,, f(, )) Figure 4 The grph of f In generl, the grph of f(, ) is surfce in R 3 Emple 65 The grph of f(, ) = is plne ee Figure 41 (2,, ) (, 3, ) (,, 6) Figure 41 = Emple 66 The grph of h(, ) = is n elliptic prboloid Figure 42 = = k, k > The domin of h is R 2 ince , the rnge of h is [, ) Ech horiontl trce is n ellipse with eqution given b = k, where k > 62 Level urves efinition 67 The level curves of function of 2 vribles re the curves in the -plne with eqution f(, ) = K, where K is constnt (K is in the rnge of f) c eprtment of Mthemtics 28

29 emester I (213/14) Figure 43 Level curves O 45 f(, ) = 2 = f(, ) Emple 68 ketch the level curves of f(, ) = for K = 6,, 6, 12 olution The level curves re = K which re stright lines Figure 44 Level curves of f(, ) = K = 12 K = 6 K = 6 Emple 69 ketch some level curves of h(, ) = olution If k <, then = K hs no solution in (, ) Therefore, there is no level curves for K < If K =, then = hs onl one solution (, ) Thus, the level curve consists of one single point t (, ) Figure 45 Level curves of f(, ) = = k, k > h(, ) = If K >, the, = K is n ellipse We m write this eqution in the stndrd form: 2 ( K 2 )2 + 2 ( K) 2 = 1 c eprtment of Mthemtics 29

30 emester I (213/14) Thus, lrger K gives rise to n ellipse with longer mjor nd minor es Eercise 61 ketch the level curves of p(, ) = for K = 4, 1,, 1, 4 The grph of p is shown in figure 46 Figure 46 p(, ) = 63 Functions of three or more vribles Let f : R 3 R be function of three vribles We cn describe f b emining the level surfces of f These re surfces in R 3 given b the equtions f(,, ) = K, where K R Emple 611 Let f(,, ) = The level surfces of f re concentric spheres with equtions of the form = K for K > If K =, then the level surfce reduces to point t the origin of R 3 For K <, there is no level surfce for f K = 2 K = 1 f 2 R = 2 Figure 47 Level surfces of f(,, ) = Eercise 612 ketch the level surfces of q(,, ) = for K = 4, 1,, 1, 4 7 Limits nd ontinuit efinition 71 Let f be function of two vribles whose domin includes points rbitrril close to (, b) We s tht the limit of f(, ) s (, ) pproches (, b) is L nd we write f(, ) = L if for n positive number ɛ, there is corresponding positive number δ lim (,) (,b) such tht (, ) nd < ( ) 2 + ( b) 2 < δ = f(, ) L < ɛ c eprtment of Mthemtics 3

31 L + ɛ L L ɛ f(, ) emester I (213/14) (, ) δ = rdius of the disc centred t (, b) Figure 48 lim f(, ) = L (,) (,b) Note tht f is not required to be defined t (, b) The ide is tht s (, ) pproches (, b), f(, ) pproches L In other words, f(, ) cn be mde s close to the number L s we wish b requiring (, ) sufficientl close to (, b) This is the mening of the bove definition b δ f ( ) L ɛ L L + ɛ R Figure 49 lim f(, ) = L (,) (,b) The impliction in definition 71 ss tht ll points (, ) which re inside the disc centred t (, b) with rdius δ re mpped b f into the intervl (L ɛ, L + ɛ) ee Figure 49 L (, b) Figure 5 f(, ) pproches L long different pths It cn be proved from the definition tht if lim f(, ) = L eists, then (,) (,b) (i) its vlue L is unique, nd (ii) L is independent of the choice of n pth pproching (, b) c eprtment of Mthemtics 31

32 emester I (213/14) Proposition 72 Let = α(t), = β(t) be the prmetric equtions of pth in R 2 such tht (α(t), β(t)) lies in the domin of f(, ) for ll t in certin open intervl contining t o nd lim α(t) = nd lim β(t) = b uppose f(, ) = L Then lim f(α(t), β(t)) = L t t o t to t t o lim (,) (,b) Proof Let ɛ n positive number ince lim f(, ) = L, there eists positive number δ (,) (,b) such tht f(, ) L < ɛ whenever < ( ) 2 + ( b) 2 < δ Now becuse lim α(t) = nd lim β(t) = b, there eists positive η such tht α(t) < δ/2 t to t to nd β(t) b < δ/2 for ll t stisfing < t t o < η Thus for ll t stisfing < t t o < η, we hve < (α(t) ) 2 + (β(t) b) 2 < δ 2 /4 + δ 2 /4 < δ so tht f(α(t), β(t)) L < ɛ This shows tht lim f(α(t), β(t)) = L t to Eercise 73 Prove tht if lim f(, ) eists, then there its vlue is unique Tht is (,) (,b) there is onl one number L stisfing the definition 71 Emple 74 how tht olution Let f(, ) = 2 2 lim (, ) (, ) long = lim (,) (,) Net let s pproch (, ) long the -is lim (, ) (, ) long = does not eist First let s pproch (, ) long the -is f(, ) = lim f(, ) = lim = lim 1 = f(, ) = lim f(, ) = lim = lim 1 = 1 ince f hs two different limits long 2 different pths, the given limit does not eist Emple 75 how tht lim (,) (,) 2 does not eist + 2 olution Let f(, ) = 2 First let s pproch (, ) long the -is + 2 lim (, ) (, ) long = Net let s pproch (, ) long the -is lim (, ) (, ) long = f(, ) = lim f(, ) = lim = f(, ) = lim f(, ) = lim = At this point, we cnnot conclude nthing s the limit m eist or m not eist Now let s pproch (, ) long the pth = lim (, ) (, ) long = 2 f(, ) = lim f(, ) = lim = 1 2 c eprtment of Mthemtics 32

33 emester I (213/14) ince f hs two different limits long 2 different pths, the given limit does not eist Emple 76 Let f(, ) = (, ) (, ) long = m 2 2 how tht + 4 lim f(, ) does not eist (,) (,) olution Let s pproch (, ) long the line = m, where m is n rel number (m) 2 lim f(, ) = lim 2 + (m) 2 = lim 3 m 2 2 (1 + m 2 2 ) = lim m m 2 2 = Thus, the limit s (, ) pproches to the origin long n stright line is ero However, we still cnnot conclude nthing s the limit m eist or m not eist Now let s pproch (, ) long the curve 2 = lim (, ) (, ) long 2 = 2 2 f(, ) = lim = 1 2 ince f hs two different limits long 2 different pths, the given limit does not eist 2 = Figure 51 Approch (, ) long 2 = Emple 77 Prove tht lim (,) (,) = olution Let ɛ positive number We wish to find positive number δ such tht < < δ = < ɛ In order to obtin the δ tht enbles the bove impliction to hold We begin b estimting the epression Note tht = Thus, if we choose δ = ɛ/3, then < < δ = < 3δ = ɛ B the definition of limit, we hve lim (,) (,) = Remrk 78 We remrk tht the usul limit theorems hold for limits of functions of two vribles For emple lim (f(, ) + g(, )) = lim (,) (,b) (,) (,b) f(, ) + lim (,) (,b) g(, ) c eprtment of Mthemtics 33

34 efinition 79 A function f of two vribles is sid to be continuous t (, b) if lim f(, ) = f(, b) (,) (,b) f is sid to be continuous on R 2 if f is continuous t ech point (, b) in emester I (213/14) Emple 71 Ever polnomil in, is continuous on R 2 Ech rtionl function is continuous in its domin For instnce, the rtionl function f(, ) = is continuous on = {(, ) R 2 + } Eercise 711 Let f(, ) = 32 2 Where is f continuous? + 2 { 3 2 if (, ) (, ) Eercise 712 Let f(, ) = how tht f is continuous on R 2 if (, ) = (, ) Remrk 713 One m compute limits using polr coordintes This is especill convenient for limits t the origin nd for those epressions tht re independent of θ More precisel, one cn prove tht lim (,) (,) f(, ) = lim f(r cos θ, r sin θ) r + Emple 714 Find lim (,) (,) (2 + 2 ) ln( ) olution We shll chnge to polr coordintes lim (,) (,) (2 + 2 ) ln( ) = lim r r2 ln (r 2 ) + 2 ln r = lim r + = lim r + r 2 2(1/r) ( 2)(1/r 3 ) using L Hôpitl s rule = lim r + r2 = Remrk 715 For functions of three or more vribles, there re similr definition of limits nd continuit ee section 151 of [1] More precisel, for functions of three vribles, these re stted s follows: efinition 716 such tht lim f(,, ) = L if for n ɛ >, there is corresponding δ > (,,) (,b,c) (,, ) nd < ( ) 2 + ( b) 2 + ( c) 2 < δ = f(,, ) L < ɛ efinition 717 A function f is clled continuous t (, b, c) if lim f(,, ) = f(, b, c) (,,) (,b,c) 8 Prtil erivtives efinition 81 Let f be function of two vribles The prtil derivtive of f with respect to t (, b) is f( + h, b) f(, b) f (, b) = lim h h The prtil derivtive of f with respect to t (, b) is f(, b + h) f(, b) f (, b) = lim h h c eprtment of Mthemtics 34

35 emester I (213/14) There re different nottions for the prtil derivtive of function If = f(, ), we write f (, ) = f = f = f(, ) =, f (, ) = f = f = f(, ) = In other words, in order to find f, we m simpl regrd s constnt nd differentite f(, ) with respect to imilrl, to find f, one cn simpl regrd s constnt nd differentite f(, ) with respect to Tht is f (, b) = d d f(, b) = nd f (, b) = d d f(, ) =b Emple 82 Let f(, ) = Then f = nd f = Thus for emple, f (1, 1) = 5 nd f (1, 1) = 1 Geometricll, f (, b) mesures the rte of chnge of f in the direction of i t the point (, b) If we consider the line = b on the -plne prllel to the -is nd pssing through the point (, b), the imge of this line under f is curve 1 on the surfce = f(, ) Then f (, b) is just the grdient of the tngent line to 1 t (, b) imilrl, f (, b) is just the derivtive t (, b) of the curve 2 trced out s the imge of the line = under f f(, b) the tngent line to the curve 1 hs grdient f (, b) 1 = f(, ) (, b) the line = b f(, b) 2 the tngent line to the curve 2 hs grdient f (, b) (, b) Emple 83 Find nd Figure 52 Prtil derivtives if is defined implicitl s function of nd b = 1 olution Tke prtil derivtive with respect to on both sides: olving for, we hve ( + ) = = imilrl, = For functions of more thn two vribles, s such w = f(,, ), we cn similrl define f, f, f, f, f, f, or w, w, w c eprtment of Mthemtics 35

36 Eercise 84 Let f(,, ) = e ln Find f, f nd f how tht f = f emester I (213/14) As in the cse of function of one vrible, we m lso define higher order prtil derivtives of function of severl vribles Let f be function of nd Then f nd f re lso functions of nd Thus we m consider (f ), (f ), (f ) nd (f ) For convenience, we shll simpl denote them b f, f, f nd f respectivel These re the second order prtil derivtives of f There re other nottions for the higher order prtil derivtives uppose = f(, ) Then we lso write: f = ( ) f = 2 f 2 = 2 2 f = ( ) f = 2 f = 2 f = ( ) f = 2 f = 2 f = ( ) f = 2 f 2 = 2 2 Emple 85 Let f(, ) = Find f, f, f nd f olution First f = nd f = Thus, f = , f = 6 2 4, f = (f ) = 6 2 nd f = (f ) = 6 2 Theorem 86 (lirut s Theorem) Let f be defined on n open disk contining the point (, b) If f nd f re continuous t (, b), then f (, b) = f (, b) Proof We ssume f nd f re defined in smll open disk centered t (, b) Let (, ) be point in Fi nd consider the function [f(, ) f(, )] [f(, b) f(, b)] in Appl Men Vlue Theorem with respect to We hve [f(, ) f(, )] [f(, b) f(, b)] = [f (, ζ 1 ) f (, ζ 1 )]( b) for some ζ 1 between nd b Net, we ppl men vlue theorem to f (, ζ 1 ) with respect to We get [f(, ) f(, )] [f(, b) f(, b)] = f (ζ 2, ζ 1 )( )( b), for some ζ 2 between nd Now we cn rewrite the epression [f(, ) f(, )] [f(, b) f(, b)] s [f(, ) f(, b)] [f(, ) f(, b)] Appling men vlue theorem first with respect to nd then with respect to, we hve [f(, ) f(, )] [f(, b) f(, b)] = [f(, ) f(, b)] [f(, ) f(, b)] = f (ζ 3, ζ 4 )( b)( ), where ζ 3 is between nd nd ζ 4 is between nd b Thus we hve f (ζ 2, ζ 1 ) = f (ζ 3, ζ 4 ) ince f nd f re continuous t (, b), so b tking limit s (, ) tends to (, b), we obtin f (, b) = f (, b) Emple 87 Let On cn show tht f(, ) = f (, ) = { if (, ) (, ) if (, ) = (, ) { ( ) ( ) 2 if (, ) (, ) if (, ) = (, ), c eprtment of Mthemtics 36

37 emester I (213/14) nd f (, ) = But f (, ) = 1, while f (, ) = 1 { ( ) ( ) 2 if (, ) (, ) if (, ) = (, ) Eercise 88 Let f(,, ) = sin(3 + ) Find f nd f how tht f = f 81 Tngent Plne Let f be function of two vribles The grph of f is surfce in R 3 with eqution = f(, ) Let P (,, ) be point on this surfce Thus, = f(, ) Assuming tngent plne to the surfce eists, we shll find its eqution f(, ) (, ) P 1 2 = f(, ) Figure 53 The tngent plne Recll tht the eqution of plne pssing through P (,, ) is of the form A( ) + B( ) + ( ) = Assuming the plne is not verticl, we hve is not ero Thus we m write the eqution of the plne s = ( ) + b( ) The tngent line to 1 t P is obtined b tking = in the bove eqution Tht is = ( ) ince f (, ) is the grdient of the tngent line 1 t P, we hve = f (, ) imilrl, b = f (, ) onsequentl, the eqution of the tngent plne to the surfce = f(, ) t P is = + f (, )( ) + f (, )( ) Emple 89 Find the eqution of the tngent plne to the elliptic prboloid = t the point (1, 1, 3) olution Let f(, ) = Then f (, ) = 4 nd f (, ) = 2 so tht f (1, 1) = 4 nd f (1, 1) = 2 Hence, the eqution of the tngent plne t (1, 1, 3) is given b = 3 + 4( 1) + 2( 1) Tht is = Liner Approimtion ince the tngent plne to the surfce = f(, ) t P is ver close to the surfce t lest when it is ner P, we m use the function defining the tngent plne s liner pproimtion to f Recll tht the eqution of the tngent plne to the grph of f(, ) t P (, b, f(, b)) is = f(, b) + f (, b)( ) + f (, b)( b) c eprtment of Mthemtics 37

38 emester I (213/14) efinition 81 The liner function L whose grph is this tngent plne is given b L(, ) = f(, b) + f (, b)( ) + f (, b)( b) L is clled the linerition of f t (, b) The pproimtion f(, ) L(, ) = f(, b) + f (, b)( ) + f (, b)( b) is clled the liner pproimtion or tngent plne pproimtion of f t (, b) Emple 811 Let f(, ) = e Find the linerition of f t (1, ) Use it to pproimte f(11, 1) olution First we hve f (, ) = e + e nd f (, ) = 2 e Thus f (1, ) = 1 nd f (1, ) = 1 Then, L(, ) = f(1, )+f (1, )( 1)+f (1, )( ) = + The corresponding liner pproimtion is e + Therefore, f(11, 1) 11 + ( 1) = 1 The ctul vlue of f(11, 1) is round up to 5 deciml plces 83 The differentil Let = f(, ) As in the cse of functions of one vrible, we tke the differentils d nd d to be independent vribles f(, b) (, b) d = f(, ) d d ( + d, b + d) Figure 54 The differentil efinition 812 The differentil d, or the totl differentil, is defined to be d = f (, )d + f (, )d onsider the differentil of f t the point (, b) The tngent plne pproimtion of f t (, b) implies tht for smll chnge d of nd smll chnge d of b, the ctul chnge of is pproimtel equl to d In other words, d = f (, b)d + f (, b)d Emple 813 Let f(, ) = Find d If chnges from 2 to 25 nd chnges from 3 to 296, compre the vlues of nd d olution d = d + d = (2 + 3)d + (3 2)d At the point (2, 3), d = ((2)(2) + 3(3))d + ((3)(2) (2)(3))d Tht is d = 13d Now we tke d = 25 2 = 5 nd d = = 4 Thus, d = 13(5) = 65 For the ctul chnge in, we hve = f(25, 296) f(2, 3) = 6449 efinition 814 Let = f(, ) f is sid to be differentible t (, b) if = f (, b) + f (, b) + ɛ 1 + ɛ 2, c eprtment of Mthemtics 38

39 where lim ɛ 1 = nd lim ɛ 2 = (, ) (,) (, ) (,) Emple 815 Prove tht f(, ) = is differentible t (, b) emester I (213/14) olution First f (, ) = nd f (, ) = At the point (, b), we hve f (, b) = b nd f (, b) = = f( +, b + ) f(, b) = ( + )(b + ) b = b + + = f (, b) + f (, b) + Here ɛ 1 = nd ɛ 2 = lerl, lim ɛ 1 = nd (, ) (,) = Thus, f(, ) = is differentible t (, b) lim ɛ 2 = (, ) (,) lim (, ) (,) Note tht from the definition of differentibilit, if f(, ) is differentible t (, b), then f (, b) nd f (, b) eist However the converse is not necessril true { Eercise 816 Let f(, ) = if (, ) (, ) if (, ) = (, ) how tht f (, ) nd f (, ) eist but f is not differentible t (, ) Eercise 817 Prove tht if f(, ) is differentible t (, b), then f(, ) is continuous t (, b) Theorem 818 uppose f (, ) nd f (, ) eist in n open disk contining (, b) nd re continuous t (, b) Then f is differentible t (, b) Proof Write = f( +, b + ) f(, b) = [f( +, b + ) f( +, b)] + [f( +, b) f(, b)] B ssumption, f nd f eist ner (, b) Thus when nd re sufficientl smll, we m ppl the Men Vlue Theorem to ech of the bove differences Then, = f ( +, b + c 1 ) + f ( + c 2, b), where c 1, c 2 (, 1) Moreover, b the ssumption tht both f nd f re continuous t (, b), we hve f ( +, b + c 1 ) = f (, b) + ɛ 2, f ( + c 2, b) = f (, b) + ɛ 1, where lim ɛ 2 = nd lim ɛ 1 = Thus, = f (, b) + f (, ) + (, ) (,) (, ) (,) ɛ 1 + ɛ 2, with both the limits of ɛ 1 nd ɛ 2 tend to s nd tend to Therefore f is differentible t (, b) Tht f nd f eist in n open disk contining the point nd re continuous t tht point re onl sufficient conditions for the differentibilit of f t the point The re b no mens necessr ee the following eercise { ( ) sin if (, ) (, ) Eercise 819 Let f(, ) = how tht f is differentible t (, ) but f nd f re not continuous t (, if (, ) = (, ) ) c eprtment of Mthemtics 39

40 emester I (213/14) Theorem 82 (The chin rule, cse 1) uppose = f(, ) is differentible function of nd, where = g(t), = h(t) re both differentible functions of t Then is differentible function of t nd d dt = f d dt + f d dt Proof We give proof of the chin rule under the dditionl ssumption tht the prtil derivtives of f re continuous Note tht = f((t), (t)) is function of t Let s check the definition of differentibilit of f((t), (t)) t point t = t enote ((t ), (t )) s (, ) B definition, We m write f((t), (t)) f((t ), (t )) t t d dt (t f((t), (t)) f((t ), (t )) ) = lim t t t t = f((t), (t)) f((t ), (t)) t t + f((t ), (t)) f((t ), (t )) t t B men vlue theorem pplied to f s function of, we cn ssert tht for some c between nd, f(, ) f(, ) = f (c, )( ) Thus f((t), (t)) f((t ), (t )) = f t t (c, (t))(t) (t ) + f t t ((t ), d) (t) (t ), t t where c nd d lie between (t), (t ) nd (t), (t ) respectivel B tking limit s t t, nd using the continuit of the prtil derivtives f, f, nd the fct tht c nd d converge to (t ) nd (t ) respectivel, we obtin the required formul Emple 821 Let = , where = sin 2t, = cos t Find d dt olution d dt = d dt + d dt = ( )(2 cos 2t) + ( )( sin t) Theorem 822 (The chin rule, cse 2) uppose = f(, ) is differentible function of nd, where = g(s, t), = h(s, t) re both differentible functions of s nd t Then is differentible function of s nd t nd s = f s + f s, t = f t + f t Emple 823 Let = e sin, where = st 2, = s 2 t Find s olution s = s + t = t + nd t s = (e sin )t 2 + (e cos )(2st) t = (e sin )(2st) + (e cos )(s 2 ) c eprtment of Mthemtics 4

41 emester I (213/14) Eercise 824 uppose = f(, ) hs continuous 2nd order prtil derivtives nd = r 2 + s 2, = 2rs Find r nd 2 r 2 84 Implicit ifferentition uppose F (, ) = defines implicitl s function of Tht is = f() Then F (, f()) = Now we use the chin rule(cse 1) to differentite F with respect to Thus d F d + F d d = Therefore, d d = F F Emple 825 Find d d if = 6 olution Let F (, ) = The given eqution is simpl F (, ) = Therefore, d d = F F = Net, suppose is given implicitl s function of nd b n eqution F (,, ) = In other words, one m solve locll in terms of nd in the eqution F (,,, ) = to obtin = f(, ) Then F (,, f(, )) = We wish to find nd in terms of F, F nd F To do so, we use the chin rule to differentite the eqution F (,, ) = keeping in mind tht is regrded s function of nd We thus obtin: F + F + F = Note tht = 1 nd = Thus F + F = Hence, imilrl, = F F = F F Eercise 826 Three nts A, B nd crwl long the positive, nd es respectivel A nd B re crwling t constnt speed of 1 cm/s, is crwling t constnt speed of 3 cm/s nd the re ll trveling w from the origin Find the rte of chnge of the re of tringle AB when A is 2 cm w from the origin while B nd re 1 cm w from the origin [The re of the tringle A(,, )B(,, )(,, ) is given b ] [Answer : 4 cm 2 /s] 85 irectionl erivtives nd the Grdient Vector efinition 827 Let f be function of nd The directionl derivtive of f t (, ) in the direction of unit vector u =, b is if this limit eists u f(, ) = lim h f( + h, + hb) f(, ) h c eprtment of Mthemtics 41

42 emester I (213/14) P (, ) P (,, ) = f(, ) h û =, b Figure 52 irectionl erivtive Note tht i f(, ) = f (, ) nd j f(, ) = f (, ), where i nd j re the stndrd bsis vectors in R 2 Theorem 828 Let f be differentible function of nd Then f hs directionl derivtive in the direction of n unit vector u =, b nd u f(, ) = f (, ) + f (, )b = f (, ), f (, ) u Proof onsider g(h) = f( + h, + hb) lerl g () = u f(, ) B the chin rule, g (h) = f d dh + f d dh At h =, we hve g () = f (, ) + f (, )b Hence, u f(, ) = f (, ) + f (, )b Emple 829 Let f(, ) = Find u f(1, 2), where u is the unit vector mking n ngle of π 6 with the positive -is olution û = cos π 6, sin π 6 (1, 2) Figure 53 irectionl erivtive of f First, f = 3 2 3, f = Thus f (1, 2) = 3 nd f (1, 2) = 13 Therefore, u f(1, 2) = 3, 13 cos π 6, sin π 6 = (13 3 3)/2 efinition 83 Let f be differentible function of nd The grdient of f is the vector function f(, ) = f (, ), f (, ) = f i + f j Thus we hve the following formul for the directionl derivtive in terms of the grdient of f c eprtment of Mthemtics 42