"1) a) Let V bean inner product space and let v ?V Then?<v,v> is called the length of the v or norm of v itisdenoted by ? ?v? ? If it satisfies the following properties 1) ? ?v? ?>=0 and ? ?v? ?= 0 ? v = 02)|| ?v|| = | ?| || v || 3)|| u + v|| <= ||u || + || v || n b) Let x , x , ..., x be vectors in R , and1 2 mlet a , a , ..., a be real numbers. 1 2 m F issaid to be linear if there exist y such thaty = a x + a x + ... + a x1 1 2 2 m m Y is called a linear combination of the vectors; thescalars a are called the coefficients of the lineari combination. If all the a are zero the lineari combination is the trivial one. A set of vectors x , x ,1 2 n ..., x in R are linearly independent if the only linearm combination of them that adds up to zero, is the trivialone. Otherwise they are called linearly dependent. If aset of vectors x , x , ..., x is linearly dependent, there1 2 m is a nontrivial set of coefficientsa , a , ..., a such that 1 2 m a x + a x + ... + a x = 01 1 2 2 m m Note that in the above equation, the a's are scalars andthe x's are vectors.Linear Transformations (special type of function fromn m RtoR )n m T is a linear transformation from RtoRif the w 's iare linear functions in the x 's.i w = a x +a x + a x + ... + a x 1 11 112 2 13 3 1n n w = a x +a x + a x + ... + a x 2 21 122 2 23 3 2n n ..w = a x + a x + a x + ... + a x m m1 1 m2 2 m3 3 mn n c) Thefunction f is continuous at the point a in itsdomain if 1) lim f( x ) exists and 2) lim f( x ) = f( a ) x->a d) Let F (t0++?t) and F ( t ) be neighboring pointsThen F(t0+ ?t)- F( t0) / ?t= F’ ( t0 ) + ? where ? ?0 as?t ?0ThereforeF(t0+ ?t)=F( t0)+ (F’ ( t0 ) + ?) ?tWhen ?t approaches 0 , F ( t0 + ?t) approaches f ( t0) So F is differentiable at t=t0e)Sufficient condition for the function to bedifferentiable is Lim= Lim=limx?x0- x?x0+x ?x0if f has a derivative at the point a, then, by definitionof the usual derivative for numeric functions: f)Gradient of a scalar-valued function n F:R -> R. The gradient is a column vectorwhose components are the same as for theJacobian derivative, which is a row vector. T T grad F = (F') = (DF) g) Directional derivative of a scalar-valuedn function f:R -> R along the direction u. ?f/?u = grad f · u From this we know how find the directions ofmaximum increase and decrease for a scalar- valued function at a point.h)Thedirectional derivative of F along theline with direction cosines l,m,n is derivedbelow?F/ ?S= l( ?F/ ?x)+m( ?F/ ?y)+n( ?F/ ?z)i)If u , v, w are the functions of a singlevariable t , then F is ultimately a function of tand then the total derivative of F with respectto t as dF/dt= ?F/ ?u(du/dt) + ?F/ ?v(dv/dt)+ ?F/ ?w( dw/dt)j) Derivative (Jacobian) matrix, F'. This is anm×n matrix. It is also denoted by DF. DV=J ( x , y , x/ r , ?, ? ) dr d ?d?is the jacobian matrixin spherical co –ordiantes 2)Solution :Let V bean inner product space and let v ?V Then ?<v,v>is called the length of the v or norm of v itis denotedby ? ?v? ?a) First we have to prove thatc ? ? ? ?is also a norm in Vfrom the properties of norm we know that 1)? ?v? ?>=0 and? ?v? ?= 0 ? v = 02)|| ?v|| = | ?| || v || 3)|| u + v|| <= ||u || + || v ||Now consider c |||| 1 It is given that || || 1is a norm Therefore, c || || > 0and c || || = 0 ? v = 0C || ?v|| = | c ? | || v ||And c|| u + v || ^ 2 < c(u + v ) , c ( u + v )>Cc- < u+v,u+v>< = | c | ^ 2 || u || + || v ||< = || cu|| + || cv ||Hence c || || is also a norm in v b)It is given that || ||1 and || || 2 be norms in vNow we have to prove that || || + || || is also a norm in VNow consider|||| 1+|||| 2 It is given that || || 1,|||| 2arenorms in VTherefore, || ||1, |||| 2> 0and|| || 1+|||| 2> 0And|| || 1+|||| 2= 0 ? v = 0 || ?v||+ || ?v|| =| ? | | ?| || v ||And|| u + v ||+ || w, z|| <= || u + w || + || v + z ||Hence |||| 1+|||| 2is also a norm in V 3) a) Solution :Thefunction f is continuous at the point a in its domain if 3) lim f( x ) exists and 4) lim f( x ) = f( a ) x->aFor the given function limit does not exist at ( 0 , 0 )That is Limlim x ?o+ ?x->o-Hence the given function is not continuous at the point ( 0 , 0 )b) Now we have to find the directionalderivative Let P ( x , y ) and Q ( x+ ?x, y + ?y)be 2 neighboring pts onPQ If l , m , n be the direction cosines of any line PQ andthe directional derivative at P is Lim{f ( Q) – f (P)}/ PQ=Lim [ l fx(x’,y’)+m fy(x’,y’)]Q->P ?s->0= l ?f/ ?x+m ?f/ ?ywhere the unit vector a = ( l , m ,n ) and u = ( fx’,fy’) andfs=a.u c)Now we have to prove thatf ( x ) = x^2 y/( x^4+y^2)f’(x)= 1/( x^4+y^2)^2[x^4+y^2(2xy)-x^2y(4x^3)]? f = 1/(/( x^4+y^2)^2[x^4+y^2(2xy)-x^2y(4x^3)+y^2]Hence4) Solution :a) Prove thatfis differentiable everywhereIff is differentiable at each point then it is said to bedifferentiable everywhere Let F( t0 )=tan-1(x^2+y^2) and F(t0+ ?t)= [tan-1( x+ ?x)^2, y + ?y)^2]Then F(t0+ ?t)- F( t0) / ?t= F’ ( t0 ) + ? where ? ?0 as ?t?0ThereforeF(t0+ ?t)=F( t0)+ (F’ ( t0 ) + ?) ?tWhen ?t approaches 0 , F ( t0 + ?t) approaches f ( t0) So F is differentiable at t=t0 b) Here we have to find the equation of the tangentplane to Z = f ( x , y ) at ( 4/5,3/5)Solution :The partial derivative for At( 4/5 , 3/5) are :(4/5,3/5) = {1/( x^2 + y^2 ) * ( 2x) }at (4/5,3/5) ={ 1/ ( 4/5)^2 +(3/5)^2}*2(4/5)= { 1/(16/25)+(9/25)}*(8/5)=(8/5)/(16+9)/25=(8/5)/25/25=8/5(4/5,3/5) = {1/( x^2 + y^2 ) * ( 2y) }at (4/5,3/5) ={ 1/ ( 4/5)^2 +(3/5)^2}*2(3/5)= { 1/(16/25)+(9/25)}*(6/5)=(6/5)/(16+9)/25=(6/5)/25/25=6/5Thus (grad f) at (4/5 ,3/5) is (8/5)i + ((6/5)j------( 1 ) Now the equation of the tangent plane is (8/5)i + ((6/5)j { ( x-(4/5) i+(y-(3/5)j)}=08/5( x-(4/5) )+6/5(y-(3/5)) = 0(8/5)*(5x-4)/5+(6/5)*(5y-3)/5=0(40x-32)/25+(30y-18)/25=040x+30y-50= 0 dividing through out by 10 we get 4x+3y-5=0 or 4x+3y= 5 is the required equation .c) In which direction is f( x,y) decreasing mostrapidly at the point ( 4/5, 3/5 ) Solution :The greatest decrease at ( 4/5, 3/5) isin thedirection of the negativegradient Nowgrad f =From ( 1) ofpart( b ) we knowthat (grad f) at (4/5 ,3/5) is (8/5)i + ((6/5)jHence (-grad f) at (4/5 ,3/5) is (-8/5)i + (-6/5)j5)a)Solution :Here we haveto prove thatF is a conservativevector field We know that in a conservative field F , curl F = 0 Curl F = ? X F = 0Here F = F1 + F2 And curl F = ? X F=-i ( ?F2 / ?z) +j ( ?F1/ ?z)+ k {( ?F2 / ?x) – ( ?F1/ ?y)} Therefore n.curlF = k.k({( ?F2 / ?x) – ( ?F1/ ?y)}Since the unit normal vectoron the X-Y plane is kWe have n.curl F ={( ?F2 / ?x) – ( ?F1/ ?y)}Here F = F1i+F2jThereforeF1 =x / ( x^2 + y ^ 2 ) andF2 = y / ( x^2 + y ^2 ) Hence ( ?F1/ ?y) = 1 / ( x^2 + y ^ 2 ) ^2( - x ( 2y )) = - 2xy/ ( x^2 + y ^ 2 )^2and ( ?F2/ ?x) = 1 / ( x^2 + y ^ 2 ) ^2( - y( 2x )) = - 2xy/ ( x^2 + y ^ 2 )^2Therefore ? {( ?F2 / ?x) – ( ?F1/ ?y)}becomes-2xy/(x^2+y^2)^2-(-2xy/(x^2+ y^2)^2 ) -2xy/(x^2+y^2)^2+2xy/(x^2+ y^2)^2 )2xy-2xy / ( x^2+y^2)^2 = 0Hence n. curl F = 0 ? curl F = 0 Hence the givenfunction F is a conservative vectorfieldHence the proof.b)Solution : we have to find a function f such that F = ?fr2 r2F = ? fresults? F.dr = ? df = f ( r2 – r1 )r1r1Hence f ( r ) = ? F.dr = ? x /( x ^ 2 + y ^ 2) dx +? y/ ( x ^ 2 + y ^ 2 ) dy]Multiplying and dividing by 2 we get½ { 2x/( x^2 + y^2} +1/2{ 2y(x^2+y^2)}½ { log ( x^2+ y^2 )log ?( x^2 + y^2 )Hencefx, y ) = log ?( x^2 + y^2 )c)Solution :We have to evaluate ? F.dr where C is the linesegment from ( 0,1 ) to ( 2,0)From Stokes theorem We have ? F.dr = ? ? n .( ? X F )dSF.dr = (i F1 +jF2 ).( idx+jdy) = F1 dx +F2 dyTherefore ? F1 dx +F2 dy = ? ? {( ?F2 / ?x) – ( ?F1/ ?y)}dS = ? ? {( ?F2 / ?x) – ( ?F1/ ?y)}dx dyHere ThereforeF1 =x / ( x^2 + y ^ 2 ) andF2 = y / ( x^2 + y ^2 ) Hence ( ?F1/ ?y) = 1 / ( x^2 + y ^ 2 ) ^2( - x ( 2y )) = - 2xy/ ( x^2 + y ^ 2 )^2and ( ?F2/ ?x) = 1 / ( x^2 + y ^ 2 ) ^2( - y( 2x )) = - 2xy/ ( x^2 + y ^ 2 )^2Therefore ? ? {( ?F2 / ?x) – ( ?F1/ ?y)}dx dy becomes? ?-2xy/(x^2+y^2)^2dxdy- ? ?-2xy/(x^2+ y^2)^2 dxdyover the same limits from ( 0 , 1 ) to ( 2 , 0 ) Hence both the terms gets cancelled And value becomes 0. 6)Solution : The parametric equation of the x^2 + y^2 = 1 arex = cos ? and y = sin ? and z=0Therefore r=(cos ?,sin ? ,0)and dr/d ? =(sin ?,2 cos ?,0)F=( -ydx+xdy+zdz)Now? F.dr = ? F.(dr/d ?)d?F =x-y+z=cos ?-sin ? ? F.(dr/d ?)d? becomes?(cos ?sin ?)-2cos ?sin ?d?? ( - cos ?sin ?d?)-1/2( cos 2 ?/2)= - cos ?= -{cos(2 ?)-cos 0}= 1-1= 0--------------------------------( 1 )b) By using stokes theorem we have to prove that ? F.dr = ? ? n. ? X F dSTheparametric equations of the plane isx= u , y =-v and z = v-u ThenR= u –v +v-u = 0 Hence? ? n. ? X F dS= 0------------------( 2)Hence ( 1 ) = ( 2 )7) Solution: We have toverify divergence theoremthat is ? ? f.n dS = ? ? divf dvdiv F=i ( ?F / ?x) + j ( ?F / ?y) + k ( ?F / ?z)= i.i+j.j+k.2z? ? ? div. F dV= ? ? ? ( 1+1+2z) dzdydx = ? ? ? ( 2+2z) dzdydx = 3-------------( 1 ) = ? ? f.n ds =2 ? ? ? ( x+y+z) dxdydz= 2 ( 3 / 2 ) = 3 -----------( 2 )therefore (1) = (2)8) Soluition:Here x= r cos ? andy = rsin ?dx = cos ? dr- r sin d ?, dy = r cos d ?+ sin ?drThen Becomes rdr d ?= ?and henceI = ? ?9)a)Solution :Let B bethe closed unit ball in bethe spherical coordinate transformationSolution:We have to compute the jacobian matrix Here x = r sin ? cos ?, y = r sin ?sin ?,and z = rcos ?Hence x^2 = r^2 sin^2? cos ^2 ?,Y^2 = r^2 sin^2? sin ^2 ?,And z^2 = r^2 cos^2 ?dx = -r sin ?sin ? d ?+ r cos ?cos ?d?+ sin ?cos ?drdy=rsin ?cos ? d ?+ r cos ?sin ?d ?+ sin ?sin ?drdz=-r sin ?d?+cos ?drTherefore d s ^2 = dx^2 +dy^2=dz^2 = dr^2 + r^2d?^2 + r^2 sin^2 ?d ?^2 =h1dr^2+h2d ?^2+h3d ?^2Therefore h1= 1 , h2 = r and h3 = r sin ?Thevolume elementin the spherical polar co- ordinates is DV= h1h2h3 dudvdw= r^2 sin^2 ?drd ?d?= J ( x , y , x/ r , ?, ? ) dr d ?d ?is the jacobian matrixb) We have to evaluate the improper integral Here x = r sin ? cos ?, y = r sin ?sin ?,and z = rcos ?Hence x^2 = r^2 sin^2? cos ^2 ?,Y^2 = r^2 sin^2? sin ^2 ?,And z^2 = r^2 cos^2 ?Hence 1/ ( x^2 + y^2 +z^2 ) = 11/{r^2 sin^2 ? cos ^2 ?)+ r^2 sin^2 ? sin ^2 ?+ r^2cos^2 ?} 1/r^2 sin^2 ?( sin^2 ?+ cos^2 ?)+ r^2cos^2 ?1/{ r^2 sin^2 ?+r^2cos^2 ?}1/r^2( sin^2 ?+cos^ ?}1/r^2-------------( 1)and dx = -r sin ?sin ? d ?+ r cos ?cos ?d ?+ sin ?cos ?drdy=rsin ?cos ? d ?+ r cos ?sin ?d ?+ sin ?sin ?drdz=-r sin ?d?+cos ?drTherefore d s ^2 = dx^2 +dy^2=dz^2 = dr^2 + r^2d?^2 + r^2 sin^2 ?d ?^2= h1dr^2+h2d ?^2+h3d ?^2Therefore h1= 1 , h2 = r and h3 = r sin ?------( 2 )= ? ? ? (1/r^2) ( 1*r*r sin ?) dr d ? d ?where the integralranges from ( 0 , ?), ( 0 , 2 ? )and ( 0 , ? )Hence the given integral becomes ? ? ? (sin?)dr d ? d ?=- [cos( 2 ?)- cos 0]=1-1= 0 "