SOLUTIONS - UCSD Mathematics

1 SOLUTIONSP roblem the critical points of the functionf(x,y) = 2x3 3x2y 12x2 3y2and determine their type local min/local max/saddle point. Are there any global min/max?Solution:Partial derivativesfx= 6x2 6xy 24x,fy= 3x2 find the critical points, we solvefx= 0 = x2 xy 4x= 0 = x(x y 4) = 0 = x= 0orx y 4 = 0fy= 0 = x2+ 2y= 0we findy= 0from the second equation. In the second case, we solve the systembelow by substitutionx y 4 = 0,x2+ 2y= 0 = x2+ 2x 8 = 0= x= 2orx= 4 = y= 2ory= three critical points are(0,0),(2, 2),( 4, 8).To find the nature of the critical points, we apply the second derivative test. We haveA=fxx= 12x 6y 24, B=fxy= 6x, C=fyy= the point(0,0)we havefxx= 24,fxy= 0,fyy= 6 = AC B2= ( 24)( 6) 0>0 = (0,0)is local , we find(2, 2)is a saddle pointsinceAC B2= (12)( 6) ( 12)2=<0and( 4, 8)is saddlesinceAC B2= ( 24)( 6) (24)2< function has no global min sincelimy ,x=0f(x,y) = and similarly there is no global maximum sincelimx ,y=0f(x,y) =.

2 1 Problem the global max and min of the functionf(x,y) =x2 2x+ 2y2 2y+ 2xyover the compact region 1 x 1,0 y :We look for the critical points in the interior: f= (2x 2 + 2y,4y 2 + 2x) = (0,0) = 2x 2 + 2y= 4y 2 + 2x= 0 = y= 0,x= , the point(1,0)is not in the interior so we discard it for check the boundary. There are four lines to be considered: the linex= 1:f( 1,y) = 3 + 2y2 critical points of this function ofyare found by setting the derivative to zero: y(3 + 2y2 4y) = 0 = 4y 4 = 0 = y= 1withf( 1,1) = 1. the linex= 1:f(1,y) = 2y2 the derivative and setting it to0we find the critical pointy= 0. The corre-sponding point(1,0)is one of the corners, and we will consider it separately below.

3 The liney= 0:f(x,0) =x2 the derivative and setting it to0we find2x 2 = 0 = x= 1. This gives thecorner(1,0)as before. the liney= 2:f(x,2) =x2+ 2x+ 4with critical pointx= 1which is again a , we check the four corners( 1,0),(1,0),( 1,2),(1,2).The values of the functionfaref( 1,0) = 3,f(1,0) = 1,f( 1,2) = 3,f(1,2) = the boxed values we select the lowest and the highest to find the global min and global conclude thatglobal minimum occurs at(1,0)global maximum occurs at(1,2).Problem Lagrange multipliers, optimize the functionf(x,y) =x2+ (y+ 1)2subject to the constraint2x2+ (y 1)2 :We check for the critical points in the interiorfx= 2x,fy= 2(y+ 1) = (0, 1)is a critical second derivative testfxx= 2,fyy= 2,fxy= 0shows this a local minimum withf(0, 1) = check the boundaryg(x,y) = 2x2+ (y 1)2= 18via Lagrange multipliers.

4 We compute f= (2x,2(y+ 1)) = g= (4x,2(y 1)).Therefore2x= 4x = x= 0or =122(y+ 1) = 2 (y 1).In the first casex= 0we getg(0,y) = (y 1)2= 18 = y= 1 + 3 2,1 3 2with valuesf(0,1 + 3 2) = (2 + 3 2)2,f(0,1 3 2) = (2 3 2) the second case =12we obtain from the second equation2(y+ 1) =y 1 = y= ,g(x,y) = 18 = x= ( 1, 3), the function takes the valuef( 1, 3) = ( 1)2+ ( 3 + 1)2= comparing all boxed values, it is clear the(0, 1)is the minimum, while(0,1 + 3 2)is the functionw=ex2ywherex=u v, y= the chain rule, compute the derivatives w u, w :We have w x= 2xyexp(x2y) = 2u v1uv2exp(u2v 1uv2)=2v3/2exp(uv) w y=x2exp(x2y) =u2vexp(uv) x u= v, x v=u2 v y u= 1u2v2, y v= w u= w x x u+ w y y u=2v3/2exp(uv) v u2vexp(uv) 1u2v2==2vexp(uv) 1vexp(uv)=1vexp(uv).

5 Similarly, w u= w x x u+ w y y u= uv2exp(uv).Problem 5.(i)For what value of the parametera, will the planesax+ 3y 4z= 2, x ay+ 2z= 5be perpendicular?(ii)Find a vector parallel to the line of intersection of the planesx y+ 2z= 2,3x y+ 2z= 1.(iii)Find the plane through the origin parallel toz= 4x 3y+ 8.(iv)Find the angle between the vectorsv= (1, 1,2),w= (1,3,0).(v)A plane has equationz= 5x 2y+ what values ofais the vector(a,1,12)normal to the plane?Solution:(i) The normal vectors to the two planes aren1= (a,3, 4), n2= (1, a,2).The planes are perpendicular ifn1,n2are perpendicular. We compute the dot productn1 n2= 0 = a 1 + 3 ( a) + ( 4) 2 = 0 = 2a 8 = 0 = a= 4.(ii) The vectors normal to the two planes aren1= (1, 1,2), n2= (3, 1,2).

6 The line of intersection will be perpendicular to bothn1,n2. But so is the cross the line of intersection will be parallel to the cross productn1 n2= (1, 1,2) (3, 1,2) = (0,4,2).(iii) The second plane must have the same normal vector hence the same coefficients forx,y, it passes through the origin, the equation isz= 4x 3y.(iv) We compute the angle using the dot productcos =v w||v|| ||w||= 2 6 10= 1 15.(v) The plane has the equation5x 2y z= 7 = 52x+y+12z=72hence a normal vector is( 52,1,12).Comparing with the vector we are given, we see thata= 6.(i)Compute the second degree Taylor polynomial of the functionf(x,y) =ex2 yaround(1,1).(ii)Compute the second degree Taylor polynomial of the functionf(x) = sin(x2)aroundx=.

7 (iii)The second degree Taylor polynomial of a certain functionf(x,y)around(0,1)equals1 4x2 2(y 1)2+ 3x(y 1).Can the point(0,1)be a local minimum forf? How about a local maximum?Solution:(i) After computing all derivatives and substituting, we find the answer1 + 2(x 1) (y 1) + 3(x 1)2+12(y 1)2 2(x 1)(y 1).(ii) We havef( ) = first derivative isfx= 2xcosx2= fx( ) = 2 cos = 2 .The second derivative isfxx= 2 cosx2 2xsinx2= fxx( ) = Taylor polynomial is 2 (x ) (x )2= x2+ .(iii) From the Taylor polynomial we findfx(0,1) =fy(0,1) = 0so(0,1)is a critical point. We can find the second derivatives12fxx(0,1) = 4,12fyy(0,1) = 2, fxy(0,1) = the second derivative testAC B2= ( 8)( 4) 32>0, A= 8<0 = (0,1)is a local 7.

8 (i)The temperatureT(x,y)in a long thin plane at the point(x,y)satisfies Laplace s equationTxx+Tyy= the functionT(x,y) = ln(x2+y2)satisfy Laplace s equation?(ii)For the functionf(x,y) = sin(x2+y2) ln(x4y4+ 1) tan(xy)is it true thatfxyxyy=fyyxyx?Solution:(i) We computeTx=2xx2+y2= Txx=2y2 2x2(x2+y2)2Ty=2yx2+y2= Tyy=2x2 2y2(x2+y2) ,Txx+Tyy=2y2 2x2(x2+y2)2+2x2 2y2(x2+y2)2= 0.(ii) The two derivatives are equal as the order in which derivatives are computed is the functionf(x,y) =x2y4.(i)Carefully draw the level curve passing through(1, 1). On this graph, draw the gradient of thefunction at(1, 1).(ii)Compute the directional derivative offat(1, 1)in the directionu=(45,35).Use this calculationto estimatef((1, 1) +.)

9 01u).(iii)Find the unit directionvof steepest descent for the functionfat(1, 1).(iv)Find the two unit directionswfor which the derivativefw= :(i) The level isf(1,1) = level curve isf(x,y) =f(1,1) = 1 = x2=y4= x= level curve is a union of two parabolas through the origin. The gradient f=(2xy4, 4x2y5)= f(1, 1) = (2,4)is normal to the parabolas.(ii) We computefu= f u= (2,4) (45,35)= the approximation, we havef(1, 1) = 1andf((1, 1) +.01u) f(1, 1) +.01fu= 1 +.01 4 = (iii) The direction of steepest decrease is opposite to the gradient. We need to divide by thelength to get a unit vector:v= f|| f||= (2,4) 22+ 42=( 1 5, 2 5).(iv) Writew= (w1,w2).We havefw= f w= (2,4) w= 2w1+ 4w2= 0 = w1= unit lengthw21+w22= 1 = ( 2w2)2+w22= 1 = w2= 1 ( 2 5,1 5).

10 Problem the functionf(x,y) = ln(e2xy3).(i)Write down the tangent plane to the graph offat(2,1).(ii)Find the approximate value of the number ln( ( )3).Solution:(i) Using the chain rule, we computefx=122e2xe2x ln(e2xy3)=1 ln(e2xy3)= fx(2,1) =1 lne4=1 4= ,fy=123y2e2xy3e2x ln(e2xy3)=32y1 ln(e2xy3)= fy(2,1) =123 lne4=32 1 4= computef(2,1) = lne4= 4 = tangent plane isz 2 =12(x 2) +34(y 1) = z=12x+34y+14.(ii) The number we are approximating isf( , ) 12 +34 +14= thatz=e3x+2y, y= ln(3u w), x=u+ z v, z :By the chain rule z v= z x x v= 3e3x+2y 2 = 6e3xe2y= 6e3u+6ve2 ln(3u w)= 6e3u+6v(3u w) , z w= z y y w= 2e3x+2y 13u w= 2e3u+6ve2 ln(3u w) 13u w= 2e3u+6v(3u w)2 13u w= 2e3u+6v(3u w).