SUBJECT:ENGINEERING MATHEMATICS-I SUBJECT CODE :SMT1101 UNIT III FUNCTIONS OF SEVERAL VARIABLES. Jacobians

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1 SUBJECT:ENGINEERING MATHEMATICS-I SUBJECT CODE :SMT0 UNIT III FUNCTIONS OF SEVERAL VARIABLES Jacobians Changing ariable is something e come across er oten in Integration There are man reasons or changing ariables bt the main reason or changing ariables is to conert the integrand into something simpler and also to transorm the region into another region hich is eas to ork ith When e conert into a ne set o ariables it is not alas eas to ind the limits So beore e moe into changing ariables ith mltiple integrals e irst need to see ho the region ma change ith a change o ariables In order to change ariables in an integration e ill need the Jacobian o the transormation I n are n dierentiable nctions o n ariables n then the determinant n n n n n n is deined as the Jacobian o n ith respect to the n ariables n and is denoted b n n I and are nctions o and then J=

2 I and are nctions o and then J= Properties o the Jacobian Chain Rle or Jacobians: I and are nctions o independent ariables r and s and each o r and s are nctions o the ariables and then and are nctions o and Frther the jacobians satis the chain rle s r s r I and are nctions o and then and can be soled in terms o and Then I and are nctions o and and i are nctionall related or dependent then 0 Problems I = = + + = + + ind J Soltion J = Thereore I = + + = + = sho that 5 Soltion: = = =

3 = = = = 4 = 4 = c c c c c c I tan tan and sho that 0 Soltion: Let = tanθ and = tanφ tan tan tan tan tan = then And = θ + φ θ = sec θ + φ φ = sec θ + φ θ = φ = 0 sec sec 0 Ths

4 4 Sho that = = + + and = are nctionall dependent and also ind the relation Soltion: I are nctionall dependent then 0 0 To colmns areidentical Since = is the relation beteen the three ariables taking commonrom c and romc 5 I = e cos = e sin here = lr + sm = mr sl eri i r s r s 6 e e e cos sin e e sin cos r s l m m l l m = e lr+sm cosmr-sl = e lr+sm sinmr-sl r = le lr+sm cosmr-sl - m e lr+sm sinmr-sl s = me lr+sm cosmr-sl + l e lr+sm sinmr-sl r = le lr+sm sinmr-sl + m e lr+sm cosmr-sl s = me lr+sm sinmr-sl - l e lr+sm cosmr-sl r s r s r s e l m r s

5 Usel Links or this topic c/jacpol/jacpolhtml 4 B&rl=http%A%F%Ftcced%FVML%FMth6%Fdocments%FJacobian sppt&ei=dksvzbg867as_t4cabg&sg=afqjcnhqdmfptkpu6sc6wtkoeufa&bm=b dce 5 Talor's Series Statement : Let be a nction o to ariables hich possess continos partial deriaties at all points Then Another orm o Talor series : Maclarin's series : The Talor series epansion o abot the point 0 0 is called Maclarin's series

6 + Problems Find the Talor series epansion o pto second degree terms Soltion: Talor series epansion o pto second degree term is gien b Let Talor series epansion o pto second degree terms ie Epand in poers o and pto and inclding second degree term Soltion: Talor series epansion o abot the point ie in poers o and pto second degree term is gien b

7 Let Talor series epansion o in poers o and pto second degree term is gien b ie ie Epand as a Talor series in the neighborhood o pto second degree term Soltion: Talor series epansion o in the neighborhood o pto second degree term is gien b Let

8 Talor series epansion o term is gien b in the neighborhood o pto second degree ie ie 4 Using Talor series sho that Soltion: Talor series epansion o pto third degree term is gien b Let

9 Talor series epansion o pto third degree term is gien b ie ie 5 Epand in poers o pto orth degree terms Soltion: Talor series epansion o pto third degree term is gien b Let

10 Talor series epansion o in poers o pto orth degree terms ie Maima and Minima o nctions o to ariables The problem o determining the maimm or minimm o a nction is encontered in geometr mechanics phsics and other ields and as one o the motiating actors in the deelopment o the calcls in the seenteenth centr A nction o to ariables can be ritten in the orm = A critical point is a point a b sch that the to partial deriaties and are ero at the point a b A relatie maimm or a relatie minimm occrs at a critical point A critical point is a maimm i the ale o at that point is greater than its ale at all its sicientl close neighboring points

11 A critical point is a minimm i the ale o at that point is less than its ale at all its sicientl close neighboring points A critical point is a saddle point i the ale o at that point is greater than its ale at some neighboring point and i the ale o at that point is less than its ale at some other neighboring point Saddle point is a point hich is neither a maimm nor a minimm Working rle or identiing critical points o the nction = and to classi them Step : Find the partial deriaties and critical points ab at hich a maima or minima ma eist Soling 0 and 0 gies the Step : Find the ale o r s step and t at all the points ab got in Step : i I r < 0 and rt s > 0 the has a maimm point at ab and the corresponding maimm ale is ab ii I r > 0 and rt s > 0 the has a minimm point at ab and the corresponding minimm ale is ab iii I rt s < 0 the has neither a maimm nor a minimm point at ab and the point is called a saddle point i I rt s = 0 the rther inestigation is reqired to classi the point Problems Find the maima and minima o the nction i an or the nction = Soltion: = = + 4 Eqate and to ero = = 0 Soling eqations and e get the critical points 00 and /-4/ r = = t = = s = = 4

12 Critical Point r rt s Classiication Saddle point /-4/ 0 4 Minimm point The point / -4/ is a minimm point o the nction and the minimm ale F/-4/ = -4/7 Find the maima and minima o the nction = a - Soltion: = a = a Eqate and to ero a = 0 a = 0 Soling eqations and e get the critical points 00 a0 0a and a/ a/ r = = - t = = - s = = a - - Critical Point r rt s Classiication a Saddle point a0 0 -a Saddle point 0a -a -a Saddle point a/ a/ -a/ a / Maimm or minimm point a/ a/ is the onl point hich cold be either be a maimm or a minimm r depends on the ale o a r = -a/ < 0 i a is positie r = -a/ > 0 i a is negatie Hence has a maimm at a/ a/ i a is positie and has a minimm at a/ a/ i a is negatie a The ale is a/ a/ = 7 Eamine the nction = or etreme ales Soltion: = = 6 0 Eqate and to ero

13 = = 0 Soling eqations and e get the critical points and 60 r = = 6-0 t = = 6-0 s = = 6 Critical Point r rt s Classiication Saddle point Saddle point Maimm point Minimm point 4 0 is a maimm point and the maimm ale is 40 = 6 0 is a minimm point and the minimm ale is 60 = 08 4 Find the etreme ales o the nction Soltion: = -0-8 = 8-0 Eqate and to ero -0-8 = = 0 Since 00 satisies both and 0 0 is a critical point To get the other points rerite eqation From e get Sbstitte this in Soling 4 e get = - - Sbstitte these ales in e get the critical points r = = -0 t = = -0 s = = 4-8

14 Critical Point r rt s Classiication Maimm point Saddle point Saddle point 8-70 Saddle point Saddle point 00 is a maimm point and the maimm ale is 00 = 0 5 Sho that = a/ = a/ makes the nction = a 4 a maimm Soltion: a a 4 4 Pt = a/ and =a/ in both eqations and Since both and are ero at a/a/ it is a critical point 6a a 6a r at a/ a/ = t at a/ a/ = s at a/ a/ = 4 a hich is negatie or an ale o a 9 4 a 8 4 a 8 a rt s at a/ a/ = 44 hich is positie or an ale o a Since r is negatie and rt s is positie the point a/ a/ is a maimm point

15 Usel Links or this topic Maimiingpd 5 %0Fnctions%0o%0To%0Variablespd Constrained Maima and Minima Sometimes e ma reqire to ind the etreme ales o a nction o three or more ariables sa hich are not independent bt are connected b a relation sa constrained etreme ales In sch sitations e se The etreme ales o a nction in sch a sitation is called to eliminate one o the ariables sa rom the gien nction ths conerting the nction o three ariables as a nction o onl to ariables Then e ind the nconstrained maima and minima o the conerted nction When this procedre cannot be sed e se Lagrange's method Lagrange's Mltiplier Method Sometimes e ma reqire to ind the maimm and minimm ales o a nction here sbject to the constraint We deine a nction here λ is the Lagrange's mltiplier independent o The neccessar condition or a maimm or minimm are 4 Soling the or eqations and 4 e get the ales o λ hich gie the etreme ales o Problems

16 Proe that the stationar ales o here occr at Soltion: Let and = implies implies implies From and e get ie ie = = = = consider = consider = consider = Ths is stationar at this point Find three positie constants sch that their sm is a constant and their prodct is maimm Soltion: Let the three positie constants be sch that Let and We hae to maimie Let = sbject to the constraint

17 implies implies implies 4 From and 4 e get Consider Consider Thereore Sbstitting in e get Thereore Hence the three nmbers are Split 4 into three parts sch that contined prodct o the irst sqare o the second and cbe o the third ma be minimm Soltion: Let the three parts be sch that Let We hae to minimie Let = implies sbject to the constraint implies implies 4 From and 4 e get Consider Consider Sbstitting in e get Thereore Hence the three parts o 4 are

18 4 Find the shortest distance o the point - rom the plane + = + 4 Soltion: Let the oot o the perpendiclar rom the point - to the plane be be Shortest distance rom - to the point distance on the plane is the perpendiclar We hae to ind the minimm distance eqialentl sbject to the constraint Let and Let = implies implies implies 4 From and 4 e get Consider Consider Sbstitting this in e get + = 4 ie and Shortest distance rom - to the point on the plane is gien b = 5 Find the points on the srace nearest to the origin Soltion: Let the point on the srace hich is nearest to the origin be Distance rom this point to the origin is Bt Using in e get No

19 r = t = To ind the point at hich maimm and minimm occrs e eqate Soling and 4 e get Sbstitting or in e get Thereore stationar points are 0 0 and At the stationar point 0 0 the nction has a minimm at 0 0 At the stationar point the nction has a minimm at 0 0- Hence the points on the srace nearest to the origin are 0 0 and 0 0-4