Solutions to Exam 1 - Mathematics

1 Solutions to Exam 11. Letf(x, y) =x2+y2+xy x+y.(a) Show thatx= 1, y= 1 is the only critical point :To find the critical point, setfx= 0 andfy= 0 to obtain theequations 2x+y 1 = 0,2y+x+ 1 = 0 which has the solutionx= 1, y= 1.(b) Use the second derivative test to show thatx= 1, y= 1 is a local mini-mum (and thus an absolute minimum it is the only critical point) :Sincefxx= 2 andD=fxxfyy f2xy= 2 2 1 = 3>0, thesecond derivative test says thatfis a local minimum.(c) Letz=L(x, y) be the equation of the tangent plane ofz=f(x, y) atthe critical point. Without evaluating any integrals, explain why the followinginequality holds: Rf(x, y)dA RL(x, y)dAwhereRis any rectangle [a, b] [c, d].Solution:The tangent plane atx= 1, y= 1 is parallel to thexy-plane(sincefx= 0 =fyat that point) and has the equation of the formL(x, y) =kwherekis a constant equal tof(1, 1).

2 Since the local minimumx= 1, y= 1is the only critical point off, it is an absolute minimum and hencef(x, y) immediately implies that Rf(x, y)dA Rk dA= RL(x, y)dA2. Letg1(x, y, z) =x2+y3+z4 2 andg2(x, y, z) =xy y4+z.(a) Leth(u, v) = (u2+ 1, v2) andg(x, y, z) = (g1(x, y, z), g2(x, y, z)). Use thechain rule to compute the derivative ofh gat pointx= 0, y= 1, z= :Leth1(u, v) =u2+ 1 andh2(u, v) =v2. ThenDh=[ h1 u h1 v h2 u h2 v]=[2u00 2v]Additionally,Dg=[ g1 x g1 y g1 z g2 x g2 y g2 z]=[2x3y24z3y x 4y31]Atx= 0, y=z, z= 1, we haveu=g1(0,1,1) = 0,v=g2(0,1,1) = 0, andhence,D(h g)(0,1,1) =Dh( 2,0)Dg(0,1,1)=[0 00 0][0 0 01 4 1]=[0 0 00 0 0](b) Let the surfaceS1be given byg1(x, y, z) = 0 and the surfaceS2byg2(x, y, z) = 0.

3 Let the curveCbe the intersection ofS1andS2. Showthat the pointx= 0, y= 1, z= 1 isnota critical point of the functionf(x, y, z) = 3xy2+y2+z4restricted :Ifx= 0, y= 1, z= 1 is a critical point, then f(0,1,1) = 1 g1(0,1,1) + 2 g2(0,1,1)( )for some scalars 1, 2. This leads the following system of equations:3 = 22 = 3 1 4 24 = 4 1+ 2 The first two equations imply 1= 14/3 and 2= 3. But plugging this into thethird equation gives 4 = 4(14/3) + 3 = 65/3, a contradiction.(c) Give an example of a functionf(x, y, z) which, when restricted toC, has acritical point atx= 0, y= 1, z= :We need a function satisfying equation (*) above for some scalars 1and 1. For example, if we set 1= 2= 1, (*) gives f(0,1,1) = (0,3,4) + (1, 4,1) = (1, 1,5)Thusf(x, y, z) =x y+ 5zwill do.

4 (Note: There are many possible answers.)3. Letf(x, y, z) =yz+xz 6.(a) At the pointx= 1, y= 1, z= 1, find the unit vector that points in thedirection for whichfis increasing at the fastest :The functionfincreases at the fastest rate in the direction of f(1,1,1)| f(1,1,1)|=(1,1,2) 6.(b) Forx= (1,1,1), find a vectorvfor whichddtf(x+tv) = 17 att= 0. (Thereare many possible answers. Just find one.)Solution:If we letv= (a, b, c), then17 =ddt(f(x+tv)|t=0= f(1,1,1) v= (1,1,2) (a, b, c) =a+b+ , one possible answer isv= (17,0,0).(c) Supposec(t) is a flow line of fwithc(0) = (1,1,1). Calculate the ac-celeration of the curvec(t) att= :Since f= (z, z, x+y), if we setc(t) = (x(t), y(t), z(t)), thenc (t) = f(c(t)) impliesx (t) =z(t)y (t) =z(t)z (t) =x(t) +y(t)Thus,x (t) =z (t)=x(t) +y(t)y (t) =z (t)=x(t) +y(t)z (t) =x (t) +y (t) =2z(t)andx (0) =x(0) +y(0) = 2y (t) =x(0) +y(0) = 2z (t) =2z(0)= 24.)

5 For each of the four questions below, state whether the assertion istrueorfalse. If it is true,justifyand if it is false,give a counterexample.(a) Ifaandbare vectors, thea bis perpendicular :This is true. The vector triple product (a b) bwherea= (a1, a2, a3)andb= (b1, b2, b3) is given by the determinant a1a2a3b1b2b3b1b2b3 which is equal to zero since the last two rows coincide. Two vectors whose dotproduct is 0 are perpendicular.(b) IfFis a vector field, then Fis perpendicular :This is flse. IfF(x, y, z) =xi+zj+zk, then F= i. Onthe other hand, ( F) F= x.(c) IfF,GandHare vector fields so thatFis a gradient field andGis acurl of some vector field, then (divG)H= :This is true. SinceGis a curl of some vector field divG= a gradient field, curlF=~0.

6 So both the left hand side and the righthand side is equal to the zero vector.(d) Assume f(x, y, z)6=~0 for all (x, y, z). Ifc(t) is a flow line of f, then thefunctionf(c(t)) is an increasing function :This is true. Using the chain rule and the fact thatc (t) = f(c(t)),we haveddtf(c(t)) = f(c(t)) c (t) =c (t) c (t) =|c (t)|2>0and hencef(c(t)) is an increasing function.