Math 234, Practice Test #2Show your work in all the problems.1. In what directions is the derivative off(x, y) =x2− y2x2+ y2at P = (1, 1) equal to zero ?2. Find an equation fo r the level surfa ce of the f unction through the givenpoint P .(a)f(x, y, z) = z − x2− y2, P = (3, −1, 1)(b)f(x, y, z) =Zyxdt√1 − t2+Zz√2dtt√t2− 1, P = (−1, 1/2, 1)3. Compute the limits of the following expressions if they exist. If youthink t hey don’t, consider different paths of approach to show that theydo not.(a)√x −√y + 1x − y −1, (x, y) → (4, 3) , x 6= y + 1(b)x2+ y2xy, (x, y) → (0, 0) , xy 6= 04. Compute all second order partial derivatives of the functionf(x, y) = x sin y + y sin x + xy5. Finddwdtif w = sin(xy + π), x = etand y = ln(t + 1). Then evaluate att = 0.1Solutions1. We first compute the gradient vector of f :∇f = 4xy2(x2+ y2)2, −4x2y(x2+ y2)2!Evaluating at (1, 1) yields∇f(1, 1) = (1, −1).The directions u in which the directional derivative (Duf)(1, 1) is zeroare the unit vectors ortho gonal to ∇f(1, 1) = (1, −1), i.e. u = (u1, u2)has to satisfy∇f(1, 1) • u = (1, −1) • (u1, u2) = u1− u2= 0and u21+ u22= 1. We getu =1√2(1, 1) and − u = −1√2(1, 1).2. (a) The level surfaces are given by the equationsc = z − x2− y2where c is a constant. In order to single out the one which passesthrough the po int P we need to insert the coor dina tes of P andcompute the constant c. Hencec = 1 − 32− (−1)2= −9so that the desired equation is −9 = z − x2− y2.(b) We compute the integrals firstZyxdt√1 −t2= sin−1(y) − sin−1(x)andZz√2dtt√t2− 1= sec−1|z| − sec−1(√2) = sec−1|z| −π42The latter follows from the fact that sec−1(√2) is the ’angle’ θwhere cos θ = 1/√2 which is π/4. We insert now x = −1, y = 1/2and z = 1:c = f(−1, 1/2, 1)= sin−1(1/2) − sin−1(−1) + sec−1|1| −π4=π6+π2+ 0 −π4=5π12The equation of the level surface is then5π12= sin−1(y) − sin−1(x) + sec−1|z| −π4or2π3= sin−1(y) − sin−1(x) + sec−1|z|3. (a) We use the following formula to simplify the given expression(√x−qy + 1)(√x+qy + 1) = (√x)2−(qy + 1)2= x−(y+1) = x−y−1Then√x −√y + 1x − y − 1=1√x +√y + 1and the limit can simply be computed by inserting (x, y) = (4, 3)which yields 1/4.(b) No further simplification as in problem (a) is possible here. Welook at the level curvesc =x2+ y2xyor x2+ y2− cxy = 0Solving this for y (quadratic equation) we gety =12(c ±√c2− 4) x = kx.3Note that all these curves pass through the origin, so there is nolimit for (x, y) → (0, 0). In order to confirm, we insert y = kx andwe getx2+ y2xy=x2+ k2x2kx2=1 + k2kTaking the limit x → 0 still yields (1 + k2)/k, i.e. the valuedepends on the direction in which we approach the origin.4. We havef(x, y) = x sin y + y sin x + xyThe first order derivatives are given byfx= sin y + y cos x + y , fy= x cos y + sin x + xThenfxx= −y sin x , fyy= −x sin y , fxy= fyx= cos y + cos x + 15. We compute∂w∂x= y cos(xy + π) ,∂w∂y= x cos(xy + π)andx′(t) = et, y′(t) =1t + 1.Thendwdt=∂w∂xx′(t) +∂w∂yy′(t)= etln(t + 1) cos(etln(t + 1) + π) +ett + 1cos(etln(t + 1) + π)=ln(t + 1) +1t + 1etcos(etln(t + 1) + π)Evaluating at t = 0 yieldsdwdtt=0= cos(π) =