7/22/2010 SECOND HOURLY PRACTICE II Maths 21a, O.Knill, Summer 2010Name:• Start by printing your name in the above box.• Try to answer each question on the same page as the question is asked. If needed, usethe back or the next empty page for work. If you need additional paper, write your nameon it.• Do not detach pages from this exam packet or unstaple the packet.• Provide details to all computations except for problems 1-3.• Please write neatly. Answers which are illegible for the grader can not be given credit.• No notes, books, calculators, computers, or other electronic aids can be allowed.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 10Total: 1101Problem 1) True/False questions (20 points)Mark for each of the 20 questions the correct letter. No justifications are needed.1)T FThe function f(x, y) = x + y has no critical points at all.Solution:Take f(x, y) = x + y for example.2)T FThe gradient of f at a point (x0, y0, z0) is orthogonal to the level surface off which contains (x0, y0, z0).Solution:It is a basic and important fact that ∇f is perpendicular to the level surface.3)T FIf (2, 3) is a local maximum for t he function f with discriminant D > 0,then fxx(2, 3) < 0.Solution:We can not have fxx(2, 3) > 0. If fxx(2, 3) = 0, then D ≤ 0.4)T FIf f satisfies the partial differential equation fx+ fy= 0 everywhere, thenthe discriminant D is zer o at every critical point.Solution:Because fxy= fxxand fyx= fyy, we have D = fxxfyy− f2xy= 0.5)T FIf f (x, y) has a saddle point, t hen −f(x, y) has a saddle point.Solution:The same critical point is also a saddle point, if we turn it upside down.6)T FThe value of the function f(x, y) = xy at (x, y) = (3.1, 5.2) can be estimatedas 15 + 0.5 + 0.6.2Solution:Use the formula for L(x, y).7)T FThe chain rule tells thatddtf(~r(t)) = ∇f(~r(t)) ·~r′(t)Solution:Unlike in the practice exam, this is now correct.8)T FThe gradient of f at a point (x0, y0) is parallel to the level curve of f whichcontains (x0, y0).Solution:It is a basic and important fact that ∇f is perpendicular to the level surface.9)T FIf (1, 1) is a critical point of f(x, y) and is not a critical point of g(x, y) then(1, 1) can not be a critical point of f under the constraint g.Solution:It not only can, it always is. ∇f = λ∇g is solved with λ = 0.10)T FIf an airship always moves in the direction opposite to the gradient of thepressure, then the pressure momentarily decreases.Solution:Use the chain rule.11)T FAt points, where the velocity of a curve is zero, the curvature is zero.Solution:The curvature is not defined at such points.12)T FIf D is the discriminant at a critical point and D < 0 then fxy≤ 0.3Solution:Take f(x, y) = −xy.13)T FIf (0, 0) is a critical point of f(x, y) and fxx(0, 0) > 0 and fyy(0, 0) > 0, then(0, 0) can not be a local minimum.Solution:Take x2+ y2− 100xy for example14)T FThe tangent plane to the graph f(x, y) = z at a point (x0, y0) is parallel toh0, 0, 1i if (x0, y0) is a critical point.Solution:The vector (2, 3, 1) is perp endicular to the surface as is −fx, −fy, 1.15)T FThe directional derivative D~vf is a vector perpendicular to ~v.Solution:The directional derivative is a scalar, not a vector.16)T FThe integralR10R10x2+ y2dxdy is th e volume of the solid bounded by the 5planes x = 0, x = 1, y = 0, y = 1, z = 0 and the paraboloid z = x2+ y2.Solution:In generalR RRf(x, y) dydx is the volume under the graph of f17)T FThe directional derivative Dvf(1, 1) is zero if v is a unit vector tangent tothe level curve of f which goes through (1, 1).Solution:The level curve is perp endicular to the gradient.18)T FIf (a, b) is a maximum of f(x, y) under the constraint g(x, y) = 0, thenthe Lagrange multiplier λ there has the same sign as the discriminant D =fxxfyy− f2xyat (a, b).4Solution:False, by changing g to −g, we can change the Lagrange multiplier, but the discriminantstays the same.19)T FThe surface area of a surface is independent of the parametrizationSolution:Yes, this is a basic property like arc length for curves.20)T FThe double integralR10R10xy dxdy is in polar coordinates the integralR10R10r2cos(θ) sin(θ) r drdθ.Solution:We also have to transform the parameter range.Problem 2) (10 points)Match the regions with the corresponding double integrals5a0.00.20.40.60.81.00.20.40.60.81.0b0.00.20.40.60.81.00.20.40.60.81.0c0.00.20.40.60.81.00.20.40.60.81.0d0.00.20.40.60.81.00.20.40.60.81.0e0.00.20.40.60.81.00.20.40.60.81.0f0.00.20.40.60.81.00.20.40.60.81.0Enter a,b,c,d,e or f Integral of Function f (x, y)R10Rxx/2f(x, y) dydxR10Ry0f(x, y) dxdyR10Rx/20f(x, y) dydxR10R1y/2f(x, y) dxdyR10Rx0f(x, y) dydxR10R11−xf(x, y) dydx6Solution:a,c,d,b,f,e.Problem 3) (10 points)a) (4 points) Find all the critical points of the function f (x, y) = xy in the interior of theelliptic domainx2+14y2< 1 .and decide for each point whether it is a maximum, a minimum or a saddle point.b) (4 points) Find th e extrema of f on the boundaryx2+14y2= 1 .of the same domain.c) (2 points) What is the global maximum and minimum of f on x2+14y2≤ 1.Solution:a) ∇f(x, y) = hy, xi = h0, 0i implies x = y = 0. The only critical point in the interior is(0, 0) . The discriminant is D = −12= −1. The point is a saddle point .b) With g(x, y) = x2+ y2/4, we have the Lagrange equationsy = λ2xx = λ2y/4x2+ y2/4 = 1Dividing the first equation by the second to get y/x = 4x/y which means y2= 4x2ory = ±2x. The third equation gives 2x2= 1 or x = ±1/√2. The third equation givesy = 2√1 − x2= ±√2. The critical points are{(1/√2),√2}, {(−1/√2),√2}, {(1/√2), −√2}, {−(1/√2), −√2} .The value of the function at these points are 1, −1, −1, 1. The first and last are maxima,the second and third are minima.c) There is no global maximum, nor a global minimum in the interior of the disc becausethere is no local maximum in the interior of the disc. The global maxima as well as theminima are on the boundary.7Problem 4) (10 points)Oliver rides his bike along streets in the Massachusetts. Since the streets can be quitebumpy, he tries to avoid critical points which are maxima (bumps) or minima (potholes)but aims to drive over saddle points (mountain passes). Assume the street is the graph ofthe functionf(x, y) =x44+y44−x22+y22.List all critical points and classify them as local maxima, local minima and saddle points.Solution:∇f(x, y) = x3− x, y3+ y = (0, 0) so that the critical points are
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