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HARVARD MATH 21A - Final Exam Practice V

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8/5/2010 FINAL EXAM PRACTICE V 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.• Do not detach pages from this exam packet or unstaple the packet.• Please try to write neatly. Answers which are illegible for the grader can not be givencredit.• No n ot es, books, calculators, computers, or other electronic aids are allowed.• Problems 1-3 do not r equ i r e any justifications. For the rest of the p r ob l em s you have toshow your work. Even correct answers without derivation can not be given credit.• You have 180 m i nutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 1011 1012 1013 1014 10Total: 1501Problem 1) (20 points)1)T FThe q u adratic surface −x2+ y2+ z2= −5 i s a one-sheeted hyperboloid.Solution:It i s a two sheeted hyperboloid.2)T F|~u×~v| < |~u·~v| implies that the angl e α between u and v satisfies |α| < π/4.Solution:Indeed, the condition means tha t |tan(α)| < 1 whi ch impl i es that the angle is smallerthan π/4.3)T FR30R2π0r sin (θ) dθ dr is the area of a disc of radius 3.Solution:There is a factor sin(θ) too much.4)T FIf a vector field~F (x, y, z) satisfies cur l( F ) ( x , y, z) = 0 for all points (x, y, z)in space, t hen~F is a gradient field.Solution:True. We have derived this from Stokes theorem.5)T FThe acce ler a t io n of a parameterized curve ~r(t) = hx(t), y(t), z( t ) i is zero ifthe curve ~r(t) is a line.Solution:The acceleration can be parallel to the velocity. An acceleration para l lel to the velocityproduces a line.6)T FThe curvature of th e curve ~r(t) = h3 + sin(t), t, t2i is half of the curvatureof t h e curve ~s(t) = h6 + 2 sin(t), 2t, 2t2i.2Solution:If we scale a curve by a factor λ, the curvature gets scaled by 1/λ. The curvature getssmaller if we scale, not larger.7)T FThe c u r ve ~r(t) = hsin(t), t, cos(t)i for t ∈ [0, π] is a helix.Solution:True. Indeed, one can check t hat x( t )2+ z(t)2= 1.8)T FIf a funct i on u(t, x) is a solution of the partial differential equation utx= 0,then it i s constant.Solution:No, any function u(t, x) = at + bx i s also a solution too and this solution is not co n st a nt.9)T FThe unit tangent vector~T of a curve is always perpendicular to the accel-eration vector.Solution:It i s perpendicular to the normal vector~N.10)T FLet (x0, y0) b e the maximum of f(x, y) under the constraint g(x, y) = 1.Then t h e gradient of g at (x0, y0) is perpendicular to the gradi ent of f at(x0, y0).Solution:It i s parall el11)T FAt a cri t i cal point for which fxx> 0, the discriminant D determines whetherthe point is a local ma xi mum or a local minimum.Solution:No, it is still possible that D = 0.312)T FIf a vector field~F (x, y) is a gr a d i ent field, then for any closed curve C theline integralRC~F ·~dr is zero.Solution:A very basic fact.13)T FIf C is part of a level curve of a function f(x, y) and~F = hfx, fyi is thegradient field of f, the nRC~F ·~dr = 0.Solution:The g r ad i ent field is perpendicular to the level curves.14)T FThe gradient of the divergence of a vector field~F (x, y, z) = ∇f (x, y, z) isalways the zero vector field.Solution:The gradient of the divergence is not zero in general. Take F (x, y, z) = hx2, 0, 0i forexample.15)T FThe line integral of the vect o r field~F (x, y, z) = hx, y, zi along a line segmentfrom (0, 0, 0) to (1, 1, 1) is 1.Solution:By the fundamental theorem of line integrals, we can take the difference of the potentialf(x, y, z) = x2/2 + y2/2 + z2/2, which is 1/2 + 1/2 + 1/2 = 3/2.16)T FIf~F (x, y) = hx2− y, xi and C : ~r(t) = hqcos(t),qsin(t)i p a ra m et er i zes theboundary of the region R : x4+ y4≤ 1, thenRC~F ·~dr is twice the area ofR.Solution:This is a direct consequence of Green’s theorem and the fact that the two-dimension a lcurl Qx− Pyof~F = hP, Qi is equal to 2.17)T FThe flux of the vector field~F (x, y, z) = h0, 0, zi t hrough the boundary S ofa sol id torus E is equal to the volume the torus.4Solution:It i s the volume of the solid torus.18)T FIf~F is a vector field in space and S is the boundary of a cube then the fluxof~curl(~F ) th r o u gh S is 0.Solution:This is true by Stokes theorem.19)T FIf div(~F )(x, y, z) = 0 for all (x, y, z) and S is a hal f sphere then the flux o f~F through S is zero.Solution:It would be consequence of the divergence theorem i f the surface were closed, but it isnot.20)T FIf the cu r l of a vector field is zero everywhere, then its divergence is zeroeverywhere too.Solution:Take a gradient field F (x, y, z) = hx, y, zi. I ts curl is zero but its divergence is not.Problem 2) (10 points)a) M at ch the following contour surface maps with the functions f(x, y, z)5I IIIII IVEnter I,II,III,IV here Functionf(x, y, z) = −x2+ y2+ zf(x, y, z) = x2+ y2+ z2f(x, y, z) = −x2− y2+ zf(x, y, z) = −x2− y2+ z2b) Match the following parametrized surfaces with their definition s6I IIIII IVEnter I,II,III,IV here Function~r(u, v) = hu − v, u + 2v, 2u + 3v sin ( uv)i~r(u, v) = hcos(u) sin(v), 4 sin(u) sin(v), 3 cos(v)i~r(u, v) = h(v4− v2+ 1) cos(u), (v4− v2+ 1) sin(u), vi~r(u, v) = hu, v, sin(uv)iSolution:a) IV, I, III,IIb) II,I,III,IVProblem 3) (10 points)7No justifications are required in th is problem. The first picture shows a gr ad i ent vectorfield~F (x, y) = ∇f(x, y).ABCThe critical points of f( x , y) are called A, Band C. What can you say about the natureof these three critical points? Which one is alocal max, which a local min, which a saddle.point local max local min saddleABCRP →Q~F ·~dr denotes the line integral of~Falong a str ai ght line path from P to Q.statement True FalseRA→B~F~dr ≥ 0RA→C~F~dr ≥RA→B~F~drThe second picture again s h ows an other gradient vector field~F = ∇f(x, y) of a diff er entfunction f(x, y).ACBDWe want to identify the maximum of f(x, y)subject to the constraint g(x, y) = x2+ y2=1. The soluti o n s of the Lagrange equ at i on sin this case are labeled A, B, C, D. At whichpoint on the ci r cl e if f maximal?point maximumABCDRγ~F ·~dr denotes the line integral of~F alongthe circle γ : x2+ y2= 1, orient ed counterclockwise.Rγ~F ·~r > 0 < 0 = 0Check if true:8Solution:It …


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