8/5/2010 FINAL EXAM PRACTICE III 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.• Justify your answers. Answers without derivation can not be given credit.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly. Answers which are illegible for the gr ader can not be given credit.• No n ot es, books, calculators, computers, or other electronic aids can be allowed.• 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 1015 10Total: 1601Problem 1) (20 points)1)T FFor any two nonzero vectors ~v, ~w the vector ~v − ~w is perpendicular to ~v × ~w.Solution:Indeed, both ~v and ~w are perpendicular. So also their difference.2)T FThe c ro ss product satisfies the law (~u ×~v) × ~w = ~u × (~v × ~w).Solution:Take ~v = ~w, then the right hand side is the zero vector while the left hand side is notzero in gen er al (for example if u = i, v = j).3)T FIf the curvature of a smooth curve ~r(t) in space is defined and zer o for allt, then th e curve is part of a line.Solution:One can see that with t h e formula κ(t) = |r′(t) × r′′(t)|/|r′(t)|3which shows th at theacceleration r′′(t) is in the velocity direction at all times. One can also see it intuitivelyor with the defi nition κ(t) = |T′(t)|/|r′(t)|. If curve is not part of a line, then T has tochange which means that κ is not zero somewhere.4)T FThe curve ~r(t) = (1 − t)A + tB, t ∈ [0, 1] connects the point A with thepoint B.Solution:The curve is a parameterizat io n o f a line and for t = 0, one has ~r(0) = A and for t = 1one has ~r(1) = B.5)T FFor every c, the function u(x, t) = ( 2 cos(ct) + 3 sin(ct)) sin(x) is a solutio nto t he wave equation utt= c2uxx.Solution:Just differentiate.6)T FThe a r c length of ~r(t) = (t, sin(t)), t ∈ [0, 2π] isR2π0q1 + cos2(t)dt.2Solution:The s peed at time t is |~r′(t)| =q1 + cos2(t).7)T FLet (x0, y0) b e the maximum of f(x, y) under the cons tr a int g(x, y) = 1.Then fxx(x0, y0) < 0.Solution:While this would be true for g(x, y) = f(y), where the constraint is a straight line parallelto t he y axes, it is false in general.8)T FThe function f(x, y, z) = x2− y2− z2decreases in the direction(2, −2, −2)/√12 at the point (1, 1, 1).Solution:It increases in that direction.9)T F~F is a vector field for which |~F (x, y, z)| ≤ 1. For every curve C : ~r(t) witht ∈ [0, 1], the line integralRC~F ·~dr is ≤ the arc length of C.Solution:|~F ·~r′| ≤ |~F ||~r′| ≤ |~r′|.10)T FLet~Fbe a vector field and C is a curve which is a flow line, thenRC~F·~dr> 0.Solution:The vector field points in the same direct i on than the velocity vector so that the dotproduct is positive at each point.11)T FThe divergence of the gradient of any f(x, y, z) is al ways zero.Solution:div(grad( f )) = ∆f is the Laplacian of f.312)T FFor every function f, one has div(cu r l (g r ad ( f ))) = 0.Solution:Both because div(curl(F ) = 0 and curl( gr a d ( f ) = 0.13)T FIf for two vector fields~F a n d~G one has curl(~F ) = curl(~G), then~F =~G + (a, b, c), wher e a, b, c are con s ta nts.Solution:One can also have~F =~G + grad(f) which are vectorfields with the same curl.14)T FFor every vector field~Fthe identity grad(div(~F)) =~0holds.Solution:F = (x2, y2, z2) h as div(F ) = (2x, 2y, 2z) which has a nonzero gradient.15)T FIf a nonem p ty quadric surface g(x, y, z) = ax2+ by2+ cz2= 5 can becontained inside a finite box, then a, b, c ≥ 0.Solution:If one or two of the constants a, b, c are negative, we have a hyperboloid which all can notbe contained into a finite space. If all three are negative, then the surface is empty.16)T FIf~F is a vector field in space then the flux of~F throu gh any closed surfaceS is 0.Solution:While it is true that the flux of curl(F ) vanishes through every closed surface, th is is nottrue for~F itself. Take for example F = (x, y, z).17)T FIf div(~F )(x, y, z) = 0 for all (x, y, z), then curl(~F ) = (0, 0, 0) for all ( x, y, z).Solution:Take (−y, x, 0) for example.418)T FThe flux of the vector field~F (x, y, z) = (y + z, y, −z) through the boundaryof a solid region E is equal to the volume of E.Solution:By the divergence theorem, the flux through the boundary isR R REdiv(F ) dV butdiv(F ) = 0. So the flux is zero.19)T FIf in spherical coordinates the equa t io n φ = α (with a constant α) definesa p la n e, then α = π/2.Solution:Otherwise, it is would be a cone (or for α = 0 or α = π a h al f line).20)T FFor every function f(x, y, z), there exists a vector field~F such that div(~F ) =f.Solution:In o r d er to solve Px+ Qy+ Rz= f just take F = (0, 0,Rz0f(x, y, w) dw).5Problem 2) (10 points)-2 -1 1 2-2-112-1 -0.5 0.5 1-1-0.50.51I II-1 -0.5 0.5 1-1-0.50.51-1 -0.5 0.5 1-1-0.50.51III IVFor the sign of the curl or divergence, where either + (positive), − (negative) or 0 for zero.The vector fields are considered on the square [−1/2, 1/2]x[−1/2, 1/2] in this problem.6Enter I,II,III,IV here Vector field curl sign divergence signF (x, y) = (x, y2)F (x, y) = (1 − y, x)F (x, y) = (y − x, −y)F (x, y) = (−x, y3)Solution:Enter I,II,III,IV here Vector field curl sign divergence signIIF (x, y) = (x, y2) 0 +IF (x, y) = (1 − y, x) + 0IVF (x, y) = (y − x, −y) - -IIIF (x, y) = (−x, y3) 0 -Problem 3) (10 points)Mark with a cross in th e column below ”conservative” if a vector fields is conservative(that is if curl(~F )(x, y, z) = (0, 0, 0) for all points (x, y, z)). Similarly, mark th e fieldswhich are incompressible (t h at is if div (~F )(x, y, z) = 0 for all (x, y, z)). No just i fi cat i onsare needed.Vectorfield conservative incompressiblecurl(~F ) =~0 div(~F ) = 0~F (x, y, z) = (−5, 5, 3)~F (x, y, z) = (x, y, z)~F (x, y, z) = (−y, x, z)~F (x, y, z) = (x2+ y2, xyz, x − y + z)~F (x, y, z) = (x − 2yz, y − 2zx, z − 2xy)7Solution:Vectorfield conservative incompressiblecurl(~F ) =~0 div(~F ) = 0~F (x, y, z) = (−5, 5, 3)X X~F (x, y, z) = (x, y, z)X~F (x, y, z) …
View Full Document
Unlocking...