8/11/2011 FINAL EXAM PRACTICE VI Maths 21a, O. Knill, Summer 2011Name:• 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 un s t ap l e the packet.• Please try to write nea t ly. Answers which are illegible for the grader ca n not be givencredit.• No not 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 h ave 180 minutes 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 qu ad r a ti c surface −x2+ y2+ z2= −5 is a one-sheeted hyperboloid.Solution:It is a two sheeted hyperboloid.2)T F|~u×~v| < |~u·~v| implies that the angle α between u and v satisfies |α | < π/4.Solution:Indeed, the condition means that |tan(α)| < 1 whi ch impli es that the angle is smallerthan π/4.3)T FR30R2π0r sin( θ) dθ dr is t h e area of a disc of radius 3.Solution:There i s a factor sin(θ) too much.4)T FIf a vector field~F (x, y, z) satisfies curl(F )(x, y, z) = 0 for all points (x, y, z)in sp a ce, then~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 cu r ve ~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 the curve ~r(t) = h3 + sin(t) , t, t2i is half of the curvatureof th 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 curvatu r e getssmaller if we scale, not larger.7)T FThe cur ve ~r(t) = hsin(t), t, cos(t)i for t ∈ [0, π] is a helix.Solution:True. Indeed, one can check that x(t)2+ z( t )2= 1.8)T FIf a functi on u(t, x) is a sol u t i on of the partial differential equation utx= 0,then it is constant.Solution:No, any function u(t, x) = at + bx is also a solution too and this solution is not constant.9)T FThe unit tangent vector~T of a curve is always perpendicular to the a ccel -eration vector.Solution:It is per pendicular 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 gradient of f at(x0, y0).Solution:It is parallel11)T FAt a critical point for which fxx> 0, the discriminant D determines whetherthe point is a local maximum or a local m i n i mum.Solution:No, it is still possib le that D = 0.312)T FIf a vector field~F (x, y) is a gradi ent field, then for any closed curve C theline integralRC~F ·~dr is zer o.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, thenRC~F ·~dr = 0.Solution:The gra d i ent field is perpendi cular 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 vector 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, wh i ch 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 par am e te r ize s 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-dimensi o n alcurl Qx− Pyof~F = hP, Qi is equal to 2.17)T FThe flux of the vector field~F (x, y, z) = h0, 0, zi through the bound ar y S ofa solid toru s E is equal to the volume the torus.4Solution:It is the volume of the sol i d toru s.18)T FIf~F is a vector field in space and S is the boundary of a cube then the fluxof~curl(~F ) thr ough S is 0.Solution:This i s true by Stokes theorem.19)T FIf div(~F )(x, y, z) = 0 for all (x, y, z) and S is a half sphere then the flux of~F throu g h S is zero.Solution:It would be consequence of the divergence theorem if the surface were closed, but it isnot.20)T FIf the cur 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. Its curl is zero but its divergence is not.Problem 2) (1 0 po ints)a) Mat ch the following contour surface maps with t h e functio n s 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) Mat ch the following parametrized surfaces with their definitions6I 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) (1 0 po ints)7No justifications are required in this p r oblem. The first p i ct u r e sh ows a gr ad i ent vectorfield~F (x, y) = ∇f(x, y).ABCThe critical points o f f(x, y) are call ed 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 l i n e integral of~Falong a strai ght line path from P to Q.statement True FalseRA→B~F~dr ≥ 0RA→C~F~dr ≥RA→B~F~drThe second picture again shows an oth er gradient vector field~F = ∇f(x, y) of a differentfunction f(x, y).ACBDWe want t o identify th e maximum of f(x, y)subject to th e constraint g(x, y) = x2+ y2=1. The soluti on s of the Lagrange equ at i on sin this case are labeled A, B, C, D. At whichpoint on the circle if f maximal?point maximumABCDRγ~F ·~dr denotes the line integral of~F alongthe circle γ : x2+ y2= 1, oriented counterclockwise.Rγ~F ·~r> 0 < 0 = 0Check if …
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